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How to Use the VAR.P Function in Excel: A Step-by-Step Guide
By Spencer Lanoue
July 11, 2025
Contents
Excel is a powerhouse when it comes to managing data and performing statistical analyses. One of the many functions that can help you make sense of your data is the VAR.P function. If you're working with a complete dataset and need to calculate the variance, this function will be your best friend. In this article, we'll walk through how to use the VAR.P function in Excel, breaking down its purpose and providing step-by-step instructions.
We’ll cover everything from understanding what variance is, to applying the function in your spreadsheets. By the time you reach the end, you'll be ready to put this function to work in your own projects. Let's get started on this journey of mastering the VAR.P function.
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Understanding Variance and Its Importance
Before jumping into how the VAR.P function works, it's helpful to understand what variance is and why it matters. Variance is a statistical measurement that tells you how much a set of numbers is spread out. In simpler terms, it measures how far each number in the set is from the mean (average) and therefore from every other number in the set.
Why is variance important? Well, it gives you insight into the variability within your dataset. For example, if you're looking at the test scores of students in a class, a high variance indicates that the scores vary widely, while a low variance suggests that the scores are clustered closely around the mean. This can be crucial when making decisions based on data, whether it's for business, education, or any other field.
The VAR.P function in Excel is designed to calculate the variance based on the entire population of a dataset. This is different from VAR.S, which calculates variance based on a sample. So, if you're dealing with complete data, VAR.P is the way to go.
Getting Acquainted with the VAR.P Function
Let’s look at the syntax of the VAR.P function. The function is straightforward and easy to use, which is one of the reasons it's so popular among Excel users.
=VAR.P(number1, [number2], ...)
The function takes a series of numbers as arguments:
number1: This is a required argument. It is the first number, cell reference, or range for which you want to calculate the variance.
number2, ... : These are optional additional numbers, cell references, or ranges. You can provide up to 254 numbers or ranges.
Now, you might wonder what happens if you mistakenly include text or logical values. No worries! The VAR.P function will ignore text or logical values in cell references. However, if you directly input logical values or text representations of numbers, Excel will throw an error. That's a handy reminder to always double-check your data inputs!
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Setting Up Your Data for VAR.P
Preparing your data is an essential step before diving into calculations. A clean and organized dataset ensures that your VAR.P function runs smoothly and provides accurate results.
Here are some tips to get your data ready:
Check for Errors: Ensure that your data range doesn't contain any errors like #N/A or #DIV/0! These can disrupt your calculations.
Remove Text: Remember, the VAR.P function only works with numbers. If your data range includes text, remove or relocate it.
Handle Blanks: Blank cells in your data range are ignored by the VAR.P function, but it's a good practice to know why they are blank.
Use Consistent Units: If your dataset includes measurements, ensure they're in the same unit (e.g., all in meters or all in kilometers).
Organizing your data might seem tedious, but it saves time and prevents errors in the long run. Think of it as setting the stage for a smooth performance!
Applying the VAR.P Function Step-by-Step
Now that your data is ready, it's time to apply the VAR.P function. Follow these steps to calculate variance in Excel:
Step 1: Select the Cell
Choose the cell where you want the variance result to appear. This is where the magic happens, so pick a spot that’s easily accessible for future reference.
Step 2: Input the Function
Type =VAR.P( into the selected cell. You'll notice Excel suggesting the function as you type, which is a neat little feature to ensure you’re on the right track.
Step 3: Enter Your Data Range
Next, select the range of cells that contain the data you want to analyze. You can do this by clicking and dragging your mouse over the cells or typing the range directly into the formula. For example, if your data is in cells A1 through A10, you’d type =VAR.P(A1:A10).
Step 4: Close the Parenthesis and Press Enter
Once your data range is in place, close the parenthesis and hit Enter. Excel will immediately calculate the variance for you.
And there you have it! You've just calculated the variance for your complete dataset using the VAR.P function. Pretty straightforward, right?
Practical Example: Calculating Variance for a Dataset
Let’s walk through a practical example to solidify this concept. Suppose you have a dataset representing the daily sales (in units) of a small retail store over ten days:
Day 1: 200
Day 2: 180
Day 3: 210
Day 4: 190
Day 5: 220
Day 6: 230
Day 7: 250
Day 8: 240
Day 9: 260
Day 10: 270
To find the variance of these sales figures, you would:
Enter the sales data into cells A1 through A10.
Select a cell for the variance result, say B1.
Type =VAR.P(A1:A10) into cell B1 and press Enter.
Excel will give you the variance of the store’s daily sales over these ten days. This value tells you how much the sales figures vary from the average daily sales, providing insights into sales consistency.
Common Errors and Troubleshooting
Even the best of us run into hiccups when working with Excel. Here are some common errors you might encounter while using the VAR.P function and how to troubleshoot them:
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#VALUE! Error
This error occurs if any of the cells in the range contain text or logical values. Ensure that your input range consists only of numeric data.
#DIV/0! Error
If your dataset is empty, Excel will return this error. Double-check that you’ve selected the correct range and that it contains data.
#NAME? Error
This happens if you misspell the function's name. Verify that you’ve typed VAR.P correctly.
Remember, errors are just part of the learning process. With a little patience and practice, you'll quickly learn how to avoid these common pitfalls.
VAR.P vs. VAR.S: Knowing the Difference
It’s crucial to differentiate between VAR.P and VAR.S to ensure you're using the right tool for your analysis. While both functions calculate variance, they serve different purposes:
VAR.P: Use this when your dataset represents the entire population. It gives a true measure of variance across the complete data.
VAR.S: This function is designed for sample data. If you’re working with a subset of a larger dataset, VAR.S provides an estimate of the variance.
Choosing the correct function depends on your dataset. If you're unsure, consider whether your data represents the whole group you're interested in or just a part of it.
Advanced Tips for Using VAR.P Like a Pro
Once you’ve mastered the basics, there are a few advanced tips and tricks to help you make the most of the VAR.P function:
Combine with Other Functions
Excel allows you to combine functions for more complex calculations. You can nest VAR.P within other functions like AVERAGE or SUM to perform multiple calculations simultaneously.
Use Conditional Formatting
Highlight outliers in your dataset by applying conditional formatting based on variance results. This visual cue can help you quickly identify areas that need attention.
Create Charts for Better Visualization
Use variance results to create charts that visually represent data spread. Charts like histograms can offer insights that numbers alone might not convey.
These tips can enhance your data analysis skills and make your spreadsheets more insightful and visually appealing.
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Exploring Real-World Applications
Understanding how to use the VAR.P function is great, but seeing how it applies to real-world scenarios can truly bring its value to life. Here are a few examples:
In Business
Businesses can use VAR.P to analyze sales data and determine consistency. A high variance might indicate fluctuations in sales, prompting a closer look at sales strategies or market conditions.
In Education
Teachers can apply the VAR.P function to student test scores to evaluate performance consistency. This can help identify if certain teaching methods are more effective than others.
In Research
Researchers use variance to study data variability, which is crucial in experiments and studies. VAR.P provides a mathematical foundation for analyzing data trends and patterns.
These examples highlight just a few of the ways VAR.P can be applied across various fields, showcasing its versatility and importance in data analysis.
Final Thoughts
In this article, we covered the use of the VAR.P function in Excel, from understanding its purpose to applying it in practical scenarios. The VAR.P function is an essential tool for calculating variance in a complete dataset, offering insights into data variability and consistency.
If you're looking to take your data analysis to the next level, consider using Bricks. This tool integrates spreadsheets, docs, and presentations, all powered by AI. Whether it's writing formulas, cleaning data, or creating visuals like dashboards and reports, Bricks can do it all in seconds. You won't need to be a spreadsheet expert to harness the full power of your data with Bricks by your side.
Spencer Lanoue
Spencer’s spent 10+ years building products and growing startups like Buffer, UserTesting, and Bump Health. He lives in spreadsheets—crunching data, building dashboards, and creating visuals to drive decisions. At Bricks, he’s focused on figuring out how to use AI to automate the busy work of spreadsheets, letting AI handle the heavy lifting for you.
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11001 | https://es.slideshare.net/slideshow/problemas-resueltos-tf-refrigeracion/13647934 | Cambiar idioma
Problemas resueltos tf refrigeracion
Este documento presenta 15 problemas resueltos relacionados con ciclos frigoríficos de compresión mecánica simples y múltiples. Los problemas cubren cálculos para ciclos estándar, compresión doble directa con inyección parcial y total, y compresión doble en cascada. Se calculan propiedades como potencia de compresión, calor de condensación, caudal y eficiencia energética para cada caso.
Colección de Problemas
Resueltos de Tecnología
Frigorífica
Versión 2.1 (septiembre de 2003)
Compresor Compresor
de baja de alta
1 2 3 4
Condensador
Evaporador
6
5
8 7
p (kPa)
7 5 4
3
6 2
8 1
h (kJ/kg)
Autor: Juan Francisco Coronel Toro
Profesor asociado del Grupo de Termotecnia
Dpto. de Ingeniería Energética y mecánica de Fluidos
Universidad de Sevilla
Este documento está basado en versiones anteriores desarrolladas por:
□ D. Ramón Velázquez Vila
□ D. José Guerra Macho
□ D. Servando Álvarez Domínguez
□ D. José Luis Molina Félix
□ D. David Velázquez Alonso
□ D. Luis Pérez-Lombard
□ D. Juan F. Coronel Toro
Todos ellos pertenecientes al Grupo de Termotecnia.
Parte de la información ha sido tomada de las siguientes referencias:
DEPARTMENT OF MECHANICAL ENGINEERING, TECHNICAL UNIVERSITY OF
DENMARK, COOLPACK, A collection of simulations tools for refrigeration, Versión
1.46 (2000).
STOECKER, W.F. Industrial Refrigeration Handbook. 1st ed. McGraw Hill (1998)
KLEIN, S.A. y ALVARADO, F.L., Engineering Equation Solver Software (EES),
Academia Versión 6.271 (20-07-2001).
Problemas Resueltos - Tecnología Frigorífica
Índice
Índice ......................................................................................................... 3
Ciclo simple de compresión mecánica:
Problema 1.................................................................................................. 4
Ciclos múltiples de compresión mecánica:
Problema 2.................................................................................................. 6
Problema 3................................................................................................ 12
Problema 4................................................................................................ 15
Problema 5................................................................................................ 19
Ciclo simple de compresión mecánica (compresores alternativos):
Problema 6................................................................................................ 23
Problema 7................................................................................................ 26
Problema 8................................................................................................ 28
Problema 9................................................................................................ 31
Ciclo simple de compresión mecánica (evaporadores y condensadores):
Problema 10 .............................................................................................. 33
Problema 11 .............................................................................................. 36
Problema 12 .............................................................................................. 39
Problema 13 .............................................................................................. 42
Problema 14 .............................................................................................. 44
Problemas combinados:
Problema 15 .............................................................................................. 47
3
Problemas Resueltos - Tecnología Frigorífica
Problema 1
Ciclo simple de compresión mecánica
Una máquina frigorífica utiliza el ciclo estándar de compresión de vapor. Produce 50 kW
de refrigeración utilizando como refrigerante R-22, si su temperatura de condensación es
40°C y la de evaporación -10°C, calcular:
a. Efecto frigorífico
b. Caudal de refrigerante
c. Potencia de compresión
d. Coeficiente de eficiencia energética
e. Relación de compresión
f. Caudal volumétrico de refrigerante manejado por el compresor
g. Temperatura de descarga del compresor
h. Coeficiente de eficiencia energética del ciclo inverso de Carnot con las mismas
temperaturas de evaporación y condensación
Solución:
Compresor
p (kPa)
1 2
2 Evaporador
3
Condensador
4 1
4 3
Válvula de
h (kJ/kg) expansión
Si trasladamos las temperaturas de evaporación (-10°C) y condensación (40°C)
sobre el diagrama P-h del R-22, obtenemos los siguientes valores:
Presiones: p cond = 1533.52 kPa p evap = 354.3 kPa
Entalpías: h1 = 401.56 kJ / kg h2 = 438.56 kJ / kg h3 = h4 = 249.67 kJ / kg
a. Efecto frigorífico: h1 − h4 = 151.89 kJ / kg
b. Caudal de refrigerante:
qf 50 kW
qf = mR (h1 − h4 );
& &
mR = = = 0.3292 kg / s
(h1 − h4 ) 151.89 kJ / kg
c. Potencia de compresión:
Pc = mR (h2 − h1 ) = 0.3292 kg / s(438.56 − 401.56 kJ / kg) = 12.18 kW
&
4
Problemas Resueltos - Tecnología Frigorífica
qf 50 kW
d. Coeficiente de eficiencia energética: COP = = = 4.105
Pc 12.18 kW
p cond 1533.52 kPa
e. Relación de compresión: rc = = = 4.328
p evap 354.3 kPa
f. Caudal volumétrico de refrigerante manejado por el compresor: Este siempre se
toma a la entrada al compresor y necesitamos el volumen específico en el punto 1:
v1 = 0.06535 m³ / kg
& &
V = m v = 0.3292 kg / s 0.06535 m³ / kg = 0.0215 m³ / s = 77.448 m³ / h
R R 1
g. Temperatura de descarga del compresor: Si miramos en el diagrama p-h la
isoterma que pasa por el punto 2 es aproximadamente t 2 = 64.3°C .
h. Coeficiente de eficiencia energética del ciclo inverso de Carnot con las mismas
temperaturas de condensación y evaporación.
1 1
COPCarnot = = = 5.263
Tcond 273.15 + 40 K
−1 −1
Tevap 273.15 − 10 K
5
Problemas Resueltos - Tecnología Frigorífica
Problema 2
Ciclos múltiples de compresión mecánica
Una aplicación de producción de frío demanda una potencia frigorífica de 100.000 frig/h,
su temperatura de evaporación debe ser -30°C y su temperatura de condensación 40°C. Si
se pretende usar en todos los casos R-22, calcular el trabajo de compresión, el calor de
condensación y el coeficiente de eficiencia energética en los siguientes casos:
a. Ciclo estándar de compresión mecánica simple.
b. Compresión doble directa con enfriador intermedio, inyección parcial. (Eficiencia del
enfriador intermedio 0.8)
c. Compresión doble directa con enfriador intermedio, inyección total.
d. Compresión doble en cascada.
Solución:
Comencemos por calcular el coeficiente de eficiencia energética de del ciclo teórico
de Carnot:
1 1
COPCarnot = = = 3.474
Tcond 273.15 + 40 K
−1 −1
Tevap 273.15 − 30 K
Este es el límite máximo para las eficiencias de todos los ciclos que vamos a
estudiar a continuación.
La potencia frigorífica en todos los ciclos debe ser:
qf = 100000 frig / h = 100000 kcal / h = 116.28 kW
a. Ciclo estándar de compresión mecánica simple.
Compresor
p (kPa)
1 2
2 Evaporador
3
Condensador
4 1
4 3
Válvula de
h (kJ/kg) expansión
Si trasladamos las temperaturas de evaporación y condensación sobre el diagrama
P-h del R-22, obtenemos los siguientes valores
Presiones: p cond = 1533.5 kPa p evap = 163.5 kPa
6
Problemas Resueltos - Tecnología Frigorífica
p cond 1533.5 kPa
Relación de compresión: rc = = = 9.38
p evap 163.5 kPa
Entalpías: h1 = 393.147 kJ / kg h2 = 451.021 kJ / kg h3 = h4 = 249.674 kJ / kg
Calculemos el caudal de refrigerante a partir del balance en el evaporador:
qf 116.28 kW
qf = mR (h1 − h4 ); mR =
& & = = 0.8105 kg / s
(h1 − h4 ) 393.147 − 249.674 kJ / kg
Trabajo de compresión:
Pc = mR (h2 − h1 ) = 0.8105 kg / s(451.021 − 393.147 kJ / kg) = 46.907 kW
&
Calor de condensación:
qc = mR (h2 − h3 ) = 0.8105 kg / s(451.021 − 249.674 kJ / kg) = 163.192 kW
&
qf 116.28 kW
Coeficiente de eficiencia energética: COP = = = 2.479
Pc 46.907 kW
b. Compresión doble directa con enfriador intermedio, inyección parcial.
Compresor Compresor
de baja de alta
1 2 3 4
Evaporador &
mRA
&
mRB
Condensador
6
5
8 7
p (kPa)
7 5 4
3
6 2
8 1
h (kJ/kg)
El primer paso es determinar la presión intermedia a la que trabaja el enfriador
intermedio, para así poder dibujar el ciclo:
7
Problemas Resueltos - Tecnología Frigorífica
pint = p cond p evap = 1533.5 · 163.5 = 500.73 kPa a esta presión le corresponde una
temperatura intermedia de t int = 0.19°C .
Al utilizar la media geométrica se consigue que la relación de compresión en el
compresor de alta y baja sean la misma.
p p
rc = cond = int = 3.06
pint p evap
Las entalpías de los puntos que hasta el momento podemos localizar sobre el
diagrama P-h son las siguientes:
h1 = 393.147 kJ / kg h2 = 420.11 kJ / kg h3 = 405.44 kJ / kg
h4 = 433.33 kJ / kg h5 = h6 = 249.67 kJ / kg
Balance de energía sobre el evaporador:
qf = mRB (h1 − h8 )
&
Balance de energía sobre el enfriador intermedio:
mRBh6 + mRBh2 + (mRA − mRB )h6 = mRAh3 + mRBh7
& & & & & &
Simplificando y sabiendo que las entalpías de los puntos 7 y 8 son iguales:
& & & &
mRAh6 + mRBh2 = mRAh3 + mRBh8
& &
Tenemos por tanto 2 ecuaciones con 3 incógnitas ( mRA , mRB , h8 ). Es necesario
plantear una nueva ecuación. La eficiencia del enfriador intermedio se plantea
como:
t − t7
ε = 0.8 = 5 ; t 7 = t 5 − ε(t 5 − t 6 ) = 40°C − 0.8(40 − 0.19°C ) = 8.152°C
t5 − t 6
Con esta temperatura del punto 7 y la presión de condensación obtenemos la
entalpía de este punto:
h7 = 209.66 kJ / kg = h8
Del balance del evaporador podemos ahora despejar el caudal de refrigerante sobre
el evaporador:
& qf 116.28 kW
mRB = = = 0.6337 kg / s
(h1 − h8 ) 393.147 − 209.66 kJ / kg
Y volviendo al balance sobre el enfriador intermedio obtenemos el caudal de
refrigerante sobre el condensador:
& &
mRA = mRB 8
(h − h2 ) = 0.8561 kg / s
(h6 − h3 )
Trabajo de compresión: Pc = mRA (h4 − h3 ) + mRB (h2 − h1 ) = 40.963 kW
& &
Calor de condensación: Q c = mRA (h4 − h5 ) = 157.231 kW
& &
qf 116.28 kW
Coeficiente de eficiencia energética: COP = = = 2.839
Pc 40.963 kW
8
Problemas Resueltos - Tecnología Frigorífica
c. Compresión doble directa con enfriador intermedio, inyección total.
Compresor Compresor
de baja de alta
1 2 3 4
Evaporador
&
mRB &
mRA
Condensador
6 5
8 7
P (kPa)
5 4
7 6 3
2
8 1
h (kJ/kg)
La presión intermedia es la misma que en el caso anterior.
pint = p cond p evap = 1533.5 · 163.5 = 500.73 kPa
Las entalpías de los puntos sobre el diagrama P-h son las siguientes:
h1 = 393.147 kJ / kg h2 = 420.11 kJ / kg h3 = 405.44 kJ / kg
h4 = 433.33 kJ / kg h5 = h6 = 249.67 kJ / kg h7 = h8 = 200.23 kJ / kg
Balance de energía sobre el evaporador:
qf 116.28 kW
qf = mRB (h1 − h8 )
& &
mRB = = = 0.6027 kg / s
(h1 − h8 ) 393.147 − 200.23 kJ / kg
Balance de energía sobre el enfriador intermedio:
& & & &
mRAh6 + mRBh2 = mRAh3 + mRBh7
despejando:
9
Problemas Resueltos - Tecnología Frigorífica
& & (h − h2 ) = 0.8508 kg / s
mRA = mRB 7
(h6 − h3 )
Trabajo de compresión: Pc = mRA (h4 − h3 ) + mRB (h2 − h1 ) = 39.978 kW
& &
Calor de condensación: qc = mRA (h4 − h5 ) = 156.258 kW
&
qf 116.28 kW
Coeficiente de eficiencia energética: COP = = = 2.909
Pc 39.978 kW
d. Compresión doble en cascada.
Compresor Compresor
de baja de alta
Evaporador & &
mRA
mRB
Intercambiador Condensador
intermedio
p (kPa)
7 6
3
2
8 5
4 1
h (kJ/kg)
Para el caso de refrigeración en cascada la temperatura de evaporación del ciclo
superior debe ser inferior a la temperatura de condensación del ciclo inferior, a
esta diferencia de temperaturas se la llama solape. Si utilizamos la temperatura
intermedia de los casos anteriores y un solape de 5°C, podremos suponer que:
t 3 = +2.7°C y t 5 = t 8 = −2.3°C
Las entalpías de los puntos sobre el diagrama P-h son las siguientes:
h1 = 393.147 kJ / kg h2 = 422.22 kJ / kg h3 = h4 = 203.18 kJ / kg
h5 = 404.52 kJ / kg h6 = 434.56 kJ / kg h7 = h8 = 249.67 kJ / kg
Balance de energía sobre el evaporador:
10
Problemas Resueltos - Tecnología Frigorífica
qf 116.28 kW
qf = mRB (h1 − h4 )
& &
mRB = = = 0.6121 kg / s
(h1 − h4 ) 393.147 − 203.18 kJ / kg
Balance de energía sobre el intercambiador intermedio:
& & & &
mRAh8 + mRBh2 = mRBh3 + mRAh5
despejando:
& & (h − h2 ) = 0.8658 kg / s
mRA = mRB 3
(h8 − h5 )
Trabajo de compresión: Pc = mRA (h6 − h5 ) + mRB (h2 − h1 ) = 43.804 kW
& &
Calor de condensación: qc = mRA (h6 − h7 ) = 160.078 kW
&
qf 116.28 kW
Coeficiente de eficiencia energética: COP = = = 2.655
Pc 43.804 kW
Resumen de resultados:
Caso qf (kW) qc (kW) Pc (kW) COP
Compresión mecánica simple 116.28 163.192 46.907 2.479
Con enfriador intermedio, inyección parcial 116.28 157.231 40.963 2.839
Con enfriador intermedio, inyección total 116.28 156.258 39.978 2.909
En cascada 116.28 160.078 43.804 2.655
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Problemas Resueltos - Tecnología Frigorífica
Problema 3
Ciclo de compresión mecánica múltiple
Un sistema de refrigeración utiliza R-22 con una capacidad frigorífica de 180 kW a una
temperatura de evaporación de -30°C y una presión de condensación de 1500 kPa.
Calcular:
a) Potencia absorbida por un sistema de compresión mecánica simple estándar.
b) Potencia absorbida por el ciclo múltiple de la figura, donde el enfriador intermedio
funciona a una presión de 600 kPa.
1 2
3
4
6 5
8 7
Solución:
a. Ciclo de compresión mecánica simple
Compresor
p (kPa)
1 2
2 Evaporador
3
Condensador
4 1
4 3
Válvula de
h (kJ/kg) expansión
Si trasladamos las temperaturas de evaporación y la presión de condensación sobre
el diagrama P-h del R-22, obtenemos los siguientes valores
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Problemas Resueltos - Tecnología Frigorífica
Presiones: p cond = 1500 kPa p evap = 163.5 kPa
p cond 1500 kPa
Relación de compresión: rc = = = 9.1743
p evap 163.5 kPa
Temperaturas: t cond = 39.1°C t evap = −30°C
Entalpías: h1 = 393.15 kJ / kg h2 = 450.38 kJ / kg h3 = h4 = 248.48 kJ / kg
Calculemos el caudal de refrigerante a partir del balance en el evaporador:
qf 180 kW
qf = mR (h1 − h4 ); mR =
& & = = 1.2442 kg / s
(h1 − h4 ) 393.15 − 248.48 kJ / kg
Trabajo de compresión:
Pc = mR (h2 − h1 ) = 1.2442 kg / s(450.38 − 393.15 kJ / kg) = 71.206 kW
&
qf 180 kW
Coeficiente de eficiencia energética: COP = = = 2.528
Pc 71.206 kW
b. Ciclo de la figura del problema.
1 2
3
&
mR1 4
&
mR 2
6 5
8 7
p (kPa)
5 4 2
7 6 3
8 1
h (kJ/kg)
La presión intermedia es pint = 600 kPa; t int = 5.88°C .
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Problemas Resueltos - Tecnología Frigorífica
Las entalpías de los puntos sobre el diagrama P-h son las siguientes:
h1 = 393.15 kJ / kg h2 = 450.38 kJ / kg h3 = 407.46 kJ / kg
h4 = 430.09 kJ / kg h5 = h6 = 248.48 kJ / kg h7 = h8 = 206.95 kJ / kg
Balance de energía sobre el evaporador:
qf 180 kW
qf = mR1 (h1 − h8 )
& &
mR1 = = = 0.9667 kg / s
(h1 − h8 ) 393.15 − 206.95 kJ / kg
Balance de energía sobre el enfriador intermedio:
(mR1 + mR2 )h6 = mR1h7 + mR2h3
& & & &
despejando:
& & (h − h6 ) = 0.2525 kg / s
mR 2 = mR1 7
(h6 − h3 )
Trabajo de compresión: Pc = mR1 (h2 − h1 ) + mR 2 (h4 − h3 ) = 61.038 kW
& &
qf 180 kW
Coeficiente de eficiencia energética: COP = = = 2.949
Pc 61.038 kW
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Problemas Resueltos - Tecnología Frigorífica
Problema 4
Ciclos múltiples de compresión mecánica
En un sistema de amoniaco con dos evaporadores y un compresor el evaporador de baja
temperatura suministra 180 kW de refrigeración con una temperatura de evaporación de
-30°C y el otro evaporador suministra 200 kW a 4°C. La temperatura de condensación
puede considerarse igual a 40°C
1. Calcular la potencia de compresión requerida y la eficiencia energética del ciclo.
2. Se sustituye el ciclo anterior por un ciclo con dos evaporadores y dos compresores
(ver figura), si se pretende suministrar la misma potencia frigorífica en ambos
evaporadores con las mismas temperaturas de evaporación y la misma temperatura
de condensación. Calcular la potencia de compresión requerida y la eficiencia
energética del ciclo.
3 4
1 2
Evaporador
de alta
Evaporador
de baja
Condensador
6 5
8 7
Nota: Suponer que no existen pérdidas de presión en los elementos del ciclo y que no
existe recalentamientos, ni subenfriamientos.
Solución:
1) A continuación se muestra un esquema de un sistema de refrigeración con dos
evaporadores y un compresor.
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Problemas Resueltos - Tecnología Frigorífica
5 8 1 2
7
Evaporador
Evaporador
de alta
de baja
Condensador
4 6 &
mRA
&
mRB
3
El diagrama p-h de este ciclo es el siguiente:
p (kPa)
3 2
6 7
4 5 1 8
Conocidas las temperaturas de evaporación de cada uno de los evaporadores y la
temperatura de condensación conocemos las entalpías de los siguientes puntos:
h3 = h4 = h6 = 386.43 kJ / kg h5 = 1422.46 kJ / kg
h7 = h8 = 1464.80 kJ / kg
Realizando balances de energía en ambos evaporadores podemos obtener los
caudales que circula por cada uno de ellos:
qfB 180 kW
qfB = mRB (h5 − h4 );
& &
mRB = = = 0.1737 kg / s
(h5 − h4 ) (1422.46 − 386.43 kJ / kg)
qfA 200 kW
qfA = mRA (h7 − h6 );
& &
mRA = = = 0.1855 kg / s
(h7 − h6 ) (1464.8 − 386.43 kJ / kg)
Por tanto el caudal total que debe mover el compresor y su entalpía se obtienen de
un sencillo balance de masa y energía en la mezcla de las dos corrientes:
& & &
mR = mRA + mRB = 0.1855 + 0.1737 kg / s = 0.3592 kg / s
& &
mRAh8 + mRBh5 0.1855·1464.8 + 0.1737·1422.46
& & &
mRh1 = mRAh8 + mRBh5 ; h1 = = = 1444.3 kJ / kg
&R
m 0.3592
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Problemas Resueltos - Tecnología Frigorífica
Si buscamos en el diagrama el punto 1 y seguimos su línea de entropía constante
hasta la presión de descarga tendremos: h2 = 1862.5 kJ / kg
El trabajo de compresión será:
Pc = mR (h2 − h1 ) = 0.3592 kg / s(1862.5 − 1444.3 kJ / kg) = 150.217 kW
&
El coeficiente de eficiencia energética del ciclo:
q + qfB 200 + 180 kw
COP = fA = = 2.5297
Pc 150.217 kw
2) Para el segundo sistema, el diagrama P-h será el que se muestra a
continuación:
3 4
1 2 &
mRC
Evaporador
de alta
&
mRB &
mRA
Evaporador
de baja
Condensador
6 5
8 7
5 4
P (kPa)
7 6 3
2
8 1
h (kJ/kg)
Conocidas las temperaturas de evaporación de cada uno de los evaporadores y la
temperatura de condensación conocemos las entalpías de los siguientes puntos:
h5 = h6 = 386.43 kJ / kg h3 = 1464.8 kJ / kg
h7 = h8 = 218.3 kJ / kg h1 = 1422.46 kJ / kg
h2 = 1617.04 kJ / kg h4 = 1627.0 kJ / kg
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Problemas Resueltos - Tecnología Frigorífica
Realizando un balance de energía en cada uno de los evaporadores obtenemos
qfB 180 kW
qfB = mRB (h1 − h8 );
& mRB =
& = = 0.1495 kg / s
(h1 − h8 ) (1422.46 − 218.3 kJ / kg)
qfA 200 kW
qfA = mRA (h3 − h6 );
& &
mRA = = = 0.1855 kg / s
(h3 − h6 ) (1464.8 − 386.43 kJ / kg)
Realizando un balance de energía sobre el enfriador intermedio:
mRBh2 + (mRC − mRA )h6 = mRBh7 + (mRC − mRA )h3
& & & & & &
Despejando el caudal de refrigerante por el compresor de alta:
& m h − mRBh2 + mRA (h6 − h3 ) 0.1495(218.3 − 1617.04) + 0.1855(386.43 − 1464.8)
& & &
mRC = RB 7 =
(h6 − h3 ) (386.43 − 1464.8)
&
mRC = 0.3794 kg / s
El trabajo de compresión será:
Pc = mRB (h2 − h1 ) + mRC (h4 − h3 ) = 0.1495(1617.04 − 1422.46 ) + 0.3794(1627.0 − 1464.8)
& &
Pc = 90.628 kW
El coeficiente de eficiencia energética del ciclo:
q + qfB 200 + 180 kw
COP = fA = = 4.193
Pc 90.628 kw
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Problemas Resueltos - Tecnología Frigorífica
Problema 5
Ciclos múltiples de compresión mecánica
Un ciclo estándar de compresión mecánica simple utiliza R-22 como refrigerante. La
capacidad frigorífica del evaporador es 180 kW a una temperatura de -30°C. La presión de
condensación del refrigerante es 1533,52 kPa. Más tarde el ciclo es revisado para funcionar
con los mismos parámetros pero siguiendo los esquemas (A) y (B), en ambos casos la
presión del deposito intermedio es 497.59 kPa.
Calcular la potencia de compresión necesaria y el COP para el ciclo simple y para las dos
configuraciones de ciclo compresión múltiple propuestas:
4 4
5
Condensador Condensador
3 3
5 2 6 2
6 7
8
1 1
7 9
Evaporador Evaporador
(A) (B)
Nota: Suponer que no existen pérdidas de presión en los elementos del ciclo, que no existe
recalentamientos, ni subenfriamientos en los evaporadores y condensadores y que los
compresores son ideales.
Solución:
Ciclo de compresión mecánica simple
Si trasladamos las temperaturas de evaporación y la presión de condensación sobre
el diagrama P-h del R-22, obtenemos los siguientes valores
t evap = −30°C p evap = 163.48 kPa
p cond = 1533.52 kPa t cond = 40°C
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Problemas Resueltos - Tecnología Frigorífica
Entalpías: h1 = 393.15 kJ / kg h3 = h4 = 249.67 kJ / kg
Compresor
p (kPa) 1 2
Evaporador
3 2
Condensador
4 1
4 3
Válvula de
h (kJ/kg) expansión
Sí el compresor es ideal la entalpía del punto 2 será: s1 = s2 = 1.8034 kJ / kg·K
h2 = 451 kJ / kg
Calculemos el caudal de refrigerante a partir del balance en el evaporador:
qf 180 kW
qf = mR (h1 − h4 ); mR =
& & = = 1.2545 kg / s
(h1 − h4 ) 393.15 − 249.67 kJ / kg
Trabajo de compresión:
Pc = mR (h2 − h1 ) = 1.2545 kg / s(451 − 393.15 kJ / kg) = 72.573 kW
&
qf 180 kW
Coeficiente de eficiencia energética: COP = = = 2.480
Pc 72.573 kW
Ciclo (A):
4
Condensador
3
p (kPa)
5 4
6 3 2
5 2
6
7 1
1
7
Evaporador h (kJ/kg)
(A)
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Problemas Resueltos - Tecnología Frigorífica
La presión de intermedia a la opera el depósito es pint = 497.59 kPa que
corresponde con una temperatura de cambio de fase de t int = 0°C
Las entalpías de los nuevos puntos, suponiendo los procesos de compresión
isentrópicos son las siguientes h2 = 419.95 kJ / kg h3 = 405.37 kJ / kg
h4 = 433.43 kJ / kg
Realizando un balance de energía sobre el evaporador obtenemos el caudal que
circula por el mismo, y por el compresor de baja presión. Como el salto de entalpía
es el mismo que en el caso del ciclo simple y demandamos la misma potencia
frigorífica, el caudal debe ser el mismo qf = mR ,B (h1 − h8 ); mR ,B = 1.2545 kg / s .
& &
Realizando un balance de energía en el depósito intermedio obtenemos el caudal
que circula por el compresor de alta:
mR ,B h2 + (mR ,A − mR ,B ) h6 = mR ,A h3 ;
& & & &
& & h − h6 419.95 − 249.67 kJ / kg
mR ,A = mR ,B 2 = 1.2545 kg / s = 1.372 kg / s
h3 − h6 405.37 − 249.67 kJ / kg
Por tanto, el trabajo de compresión será:
Pc = mR ,B (h2 − h1 ) + mR ,A (h4 − h3 )
& &
Pc = 1.2545 kg / s(419.95 − 393.15 kJ / kg) + 1.372 kg / s(433.43 − 405.37 kJ / kg) = 72.119 kW
qf 180 kW
Coeficiente de eficiencia energética: COP = = = 2.496
Pc 72.119 kW
Ciclo (B):
4
5
Condensador
3
p (kPa)
6 4 5 2
8 7 3
6 2
7
8 9 1
1
9
Evaporador
h (kJ/kg)
(B)
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Problemas Resueltos - Tecnología Frigorífica
Al igual que el caso (A), la presión de intermedia a la opera el depósito es
pint = 497.59 kPa que corresponde con una temperatura de cambio de fase de
t int = 0°C
Las entalpías de los puntos, suponiendo los procesos de compresión isentrópicos
son las siguientes:
h1 = 393.15 kJ / kg h2 = 451 kJ / kg h3 = 405.37 kJ / kg h4 = 433.43 kJ / kg
h6 = h7 = 249.67 kJ / kg h8 = h9 = 200 kJ / kg
Realizando un balance de energía sobre el evaporador obtenemos el caudal que
circula por el mismo, y por el compresor de baja presión (le llamaremos así aunque
realmente no sea de baja):
qf 180 kW
qf = mR ,B (h1 − h9 );
& mR ,B =
& = = 0.932 kg / s
(h1 − h9 ) (393.15 − 200 kJ / kg)
Realizando un balance de energía en el depósito intermedio obtenemos el caudal
que circula por el compresor de alta:
(m
& R ,A + mR ,B ) h7 = mR ,A h3 + mR ,B h8 ;
& & &
& & h − h8 249.67 − 200 kJ / kg
mR ,A = mR ,B 7 = 0.932 kg / s = 0.297 kg / s
h3 − h7 405.37 − 249.67 kJ / kg
Por tanto, el trabajo de compresión será:
Pc = mR ,B (h2 − h1 ) + mR ,A (h4 − h3 )
& &
Pc = 0.932 kg / s(451 − 393.15 kJ / kg) + 0.297 kg / s(433.43 − 405.37 kJ / kg) = 62.25 kW
qf 180 kW
Coeficiente de eficiencia energética: COP = = = 2.892
Pc 62.25 kW
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Problemas Resueltos - Tecnología Frigorífica
Problema 6
Ciclo simple de compresión mecánica (compresores alternativos)
Se necesita evacuar 150.000 kcal/h de cierta cámara frigorífica, para lo que se decide
instalar un sistema de producción de frío por compresión mecánica. La temperatura de la
cámara no puede superar los –3°C y el la diferencia de temperaturas a la entrada del
evaporador se estima en 7°C. Se dispone de un gran caudal de agua de pozo a 15°C que
piensa utilizarse como agente condensante. El fluido frigorígeno empleado es R-134a.
Para el funcionamiento de dicha instalación se adquirió un compresor alternativo de 2.250
cm³ de cilindrada, el cual aspira vapor con un recalentamiento en la tubería de aspiración de
10°C. Este compresor gira a 850 r.p.m. y su rendimiento volumétrico es de 0,8 para una
relación de compresión de 3,3.
Calcular:
El grado de subenfriamiento del fluido condensado para que pueda funcionar la
instalación con este compresor y si es posible su realización.
Nota: Considerar un salto máximo admisible en el agua de pozo de 5°C y un salto mínimo
de temperaturas en el condensador (entre fluido refrigerante y agua de pozo) de 5°C.
Solución:
a. Cálculo del grado de subenfriamiento
La temperatura del aire en la cámara debe ser inferior a –3°C, por lo tanto
Compresor
1 2 &
mw t w,ent
Evaporador
&
ma t a,ent
t a,sal
Condensador
t w,sal
4 3
Válvula de expansión
podemos suponer que esta es la temperatura de entrada del aire al evaporador.
t a,ent = −3°C . Luego si el salto a la entrada al evaporador debe ser de 7°C la
temperatura de evaporación será:
∆t = t a,ent − t evap = 7°C; t evap = t a,ent − ∆t = −3 − 7 = −10°C
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Problemas Resueltos - Tecnología Frigorífica
En cuanto al condensador la temperatura de entrada al condensador del agua de
pozo es de 15°C, el salto máximo en esta agua es de 5°C luego la temperatura del
agua a la salida del condensador será, t w,sal = 15 + 5 = 20°C y la temperatura de
condensación 5°C por encima de la temperatura más alta alcanzada en el
condensador:
∆t = t cond − t w,sal = 5°C; t cond = t w,sal + ∆t = 20 + 5 = 25°C
Con estos datos intentaremos dibujar el ciclo sobre un diagrama p-h de R-134a,
aunque desconocemos las entalpías de los puntos 3 y 4, ya que estas dependen del
grado de subenfriamiento que es nuestra incógnita. El punto 1 (entrada al
compresor) se encuentra a la presión de evaporación y sobre la isoterma de 0°C (-
10°C + 10°C).
Los valores de las entalpías de los diferentes puntos son:
h1 = 400.05 kJ / kg h5 = 391.32 kJ / kg
h2 = 425.98 kJ / kg h3 = h4 = ? kJ / kg
v1 = 0.10397 m³ / kg
p (kPa)
3 25°C
2
-10°C
1
4 5
h (kJ/kg)
La potencia frigorífica que debe suministrar este sistema es de 150000 kcal/h =
174.42 kW.
Con los datos físicos del compresor podemos calcular el desplazamiento volumétrico
de este:
&
Vt = Vcilindro ω = 2250 cm³ 850 rev / min = 0.0319 m³ / s
Para este caso la relación de presiones es 3.31 podemos decir que rendimiento
volumétrico del compresor va a ser aproximadamente 0.8.
&
VR1
ηvol = ; & &
VR1 = ηvol Vt = 0.8 0.0319 m³ / s = 0.0255 m³ / s
V&
t
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Problemas Resueltos - Tecnología Frigorífica
Siendo el volumen específico sobre el punto 1 v1 = 0.10397 m³ / kg , tendremos un
caudal másico:
V& 0.0255 m³ / s
&
mR = R1 = = 0.245 kg / s
v1 0.10397 m³ / kg
La potencia frigorífica sobre el evaporador es:
qf = mR (h5 − h4 ) ;
&
qf 174.42 kW
despejando: h4 = h5 − = 391.32 kJ / kg − = −320.6 imposible sería
&R
m 0.245 kg / s
necesario subenfriar el líquido a menos cientos de grados.
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Problemas Resueltos - Tecnología Frigorífica
Problema 7
Ciclo simple de compresión mecánica (compresores alternativos)
Los datos de catálogo de un compresor son los siguientes:
Refrigerante: R-22
Número de cilindros: 6
Velocidad de giro: 1740 r.p.m.
Diámetro del cilindro: 67 mm
Carrera: 57 mm
Porcentaje de espacio muerto: 4.8 %
Para las siguientes condiciones de operación:
Temperatura de evaporación: 5°C
Temperatura de condensación: 50°C
Subenfriamiento del líquido: 3°C
Recalentamiento del vapor: 8°C
La potencia frigorífica que indica el catálogo es 96.4 kW y la potencia absorbida 28.9 kW.
Calcular:
El rendimiento volumétrico teórico, el rendimiento volumétrico real y el rendimiento
isentrópico o de la compresión.
Solución:
La siguiente figura muestra el diagrama P-h del problema indicado con
anterioridad.
P (kPa)
3 50°C
2
5°C
1
4 5
h (kJ/kg)
Rendimiento volumétrico teórico:
v
ηvol,t = 1 − C suc − 1
v
des
26
Problemas Resueltos - Tecnología Frigorífica
El factor de espacio muerto o factor de huelgo C=0.048, y los volúmenes
específicos en la succión y la descarga:
v suc = v1 = 43.2 l / kg v des = v 2 = 14.13 l / kg
43.2 l / kg
ηvol,t = 1 − 0.048
14.13 l / kg − 1 = 0.9012
Desplazamiento volumétrico del compresor:
2
& πDc
Vt = Nc ω Lc
4
Donde:
Número de cilindros: Nc = 6
Velocidad de giro: ω = 1740 r.p.m. = 29 rev / s
Diámetro de cilindros: Dc = 0.067 m
Carrera: L c = 0.057 m
π(0.067 m)
2
&
Vt = 6 · 29 rev / s 0.057 m = 0.035 m³ / s = 35 l / s
4
Los valores de las entalpías de los diferentes puntos son:
h1 = 413.1 kJ / kg h2 = 445.5 kJ / kg h3 = h4 = 259.1 kJ / kg
Balance de energía sobre el evaporador:
qf 96.4 kW
qf = mR (h1 − h4 )
& &
mR = = = 0.6260 kg / s
(h1 − h4 ) 413.1 − 259.1 kJ / kg
El caudal volumétrico de refrigerante a la entrada al compresor (punto 1) será el
siguiente:
& &
V1 = mR v1 = 0.6260 kg / s 43.2 l / kg = 27.042 l / s
Por lo tanto el rendimiento volumétrico real será:
V& 27.04 l / s
ηvol,r = 1 = = 0.773
&
Vt 34.97 l / s
El trabajo de compresión teórico o isentrópica podemos calcularlo como:
Pc,s = mR (h2 − h1 ) = 0.6260 kg / s(444.5 − 413.1 kJ / kg) = 19.656 kW
&
Y por lo tanto el rendimiento de compresión o rendimiento isentrópico valdrá:
Pc,s 19.656 kW
ηs = = = 0.6801
Pc,r 28.9 kW
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Problemas Resueltos - Tecnología Frigorífica
Problema 8
Ciclo simple de compresión mecánica (compresores alternativos)
Los datos de catálogo del compresor SP4L220E son los siguientes:
Refrigerante: R-134a
Desplazamiento volumétrico: 86.1 m³/h
Para las siguientes condiciones de operación:
Temperatura de evaporación: -10°C
Temperatura de condensación: 50°C
Subenfriamiento del líquido: 5°C
Recalentamiento del vapor: 10°C
La potencia frigorífica que indica el catálogo es 23.7 kW y la potencia absorbida 10.0 kW.
Calcular:
La potencia frigorífica, el trabajo de compresión y el coeficiente de eficiencia energética, si
pretendemos utilizar este compresor en un ciclo con las mismas temperaturas de
evaporación y compresión pero sin subenfriamiento del líquido ni recalentamiento del
vapor.
Solución:
La siguiente figura muestra el diagrama P-h (R-134a) del problema con
subenfriamiento y recalentamiento. El punto “2s” es el punto de salida de un
proceso de compresión isentrópico
p (kPa)
3 50°C 2s
2
-10°C
1
4
h (kJ/kg)
Los valores de las entalpías de los puntos que pueden obtenerse son:
h1 = 400.049 kJ / kg h2s = 441.196 kJ / kg h3 = h4 = 263.712 kJ / kg
Balance de energía sobre el evaporador:
qf 23.7 kW
qf = mR (h1 − h4 )
& &
mR = = = 0.1738 kg / s
(h1 − h4 ) 400.049 − 263.712 kJ / kg
28
Problemas Resueltos - Tecnología Frigorífica
El trabajo de compresión puede obtenerse como: Pc = mR (h2 − h1 )
&
Despejando de aquí la entalpía del punto “2” tendremos:
Pc 10.0 kW
h2 = h1 + = 400.049 kJ / kg + = 457.586 kJ / kg
&
mR 0.1738 kg / s
El ciclo sobre el cual queremos instalar nuestro compresor es el siguiente:
3’ 2s’
4’ 1’
h (kJ/kg)
Los valores de las entalpías de los puntos son:
h1' = 391.321 kJ / kg h2s' = 430.328 kJ / kg h3' = h4' = 271.418 kJ / kg
Al mantenerse la presión de succión y presión de descarga entre las cuales trabaja
el compresor, tenemos que la relación de presiones es la misma que en el caso
anterior y puede considerarse una buena hipótesis suponer que el rendimiento
volumétrico y de la compresión del compresor se mantienen.
& &
V1 mR v1 0.1738 kg / s 0.104 m³ / kg
Rendimiento volumétrico: ηvol = = = = 0.7558
&
Vt &
Vt 86.1 m³ / h · 1 / 3600
Luego el caudal de refrigerante en el segundo caso será:
&
Vη 0.0239 m³ / s 0.7558
m′ = t vol =
&R = 0.1825 kg / s
v1' 0.09899 m³ / kg
La potencia frigorífica:
q′f = m′ (h1' − h4' ) = 0.1825 kg / s(391.321 − 271.418 kJ / kg) = 21.880 kW
&R
Del otro parámetro que podemos considerar constante, el rendimiento isentrópico o
rendimiento de la compresión, podemos obtener al trabajo absorbido por el
compresor en la segunda situación:
P m (h − h1 ) 0.1738 kg / s(441.196 − 400.049 kJ / kg)
&
ηc = c,s = R 2s = = 0.7151
Pc,r Pc,r 10.0 kW
′
Pc,s m′ (h2s′ − h1′ ) 0.1825 kg / s(430.328 − 391.321 kJ / kg)
&R
′
Pc,r = = = = 9.955 kW
ηc ηc 0.7151
Los valores del coeficiente de eficiencia energética, vale en ambos casos:
q 23.7 kW q′ 21.880 kW
COP = f = = 2.37 COP′ = f = = 2.198
Pc 10.0 kW ′
Pc 9.955 kW
29
Problemas Resueltos - Tecnología Frigorífica
30
Problemas Resueltos - Tecnología Frigorífica
Problema 9
Ciclo simple de compresión mecánica (compresores alternativos)
Un compresor hermético alternativo de 4 cilindros para R-22, tiene una velocidad de giro
de 29 rev/s. El diámetro de los cilindros es 87 mm y la carrera 70 mm. El rendimiento
volumétrico ha sido obtenido experimentalmente en función de la relación de compresión
(rc):
2
ηvol = 0.0016 rc − 0.0734 rc + 1.0117
Sí suponemos que la temperatura de condensación es constante e igual a 40°C, calcular la
potencia frigorífica para las siguientes temperaturas de evaporación: -20°C, -10°C, 0°C,
10°C
Nota: Suponer ciclo estándar sin sobrecalentamiento ni subenfriamiento
Solución:
El desplazamiento volumétrico del compresor será
& πD2
Vt = Nc ω c
Lc
4
Donde:
Número de cilindros: Nc = 4
Velocidad de giro: ω = 29 rev / s
Diámetro de cilindros: Dc = 0.087 m
Carrera: L c = 0.070 m
π(0.087 m)
2
&
Vt = 4 · 29 rev / s 0.070 m = 0.0483 m³ / s = 173.775 m³ / h
4
Comencemos por el primer caso t evap = −20°C
Dibujando el ciclo sobre un diagrama P-h obtenemos:
p (kPa)
3 2
4 1
h (kJ/kg)
Presiones: p cond = 1534 kPa p evap = 245.4 kPa
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![Problemas Resueltos - Tecnología Frigorífica
p cond 1534 kPa
rc = = = 6.252
p evap 245.4 kPa
Luego el rendimiento volumétrico valdrá:
ηvol = 0.0016 (6.252 ) − 0.0734 (6.252 ) + 1.0117 = 0.6153
2
Podemos obtener por tanto el caudal de refrigerante para este caso:
&
m v η V & 0.6145 · 0.0483 m³ / s
&
ηvol = R 1 ; mR = vol t = = 0.3207 kg / s
&
V v 0.09263 m³ / kg
t 1
El incremento de entalpía en el evaporador vale:
∆hevap = h1 − h4 = 396.9 − 249.8 kJ / kg = 147.1 kJ / kg
Por tanto la potencia frigorífica será:
&
qf = mR ∆hevap = 0.3207 kg / s 147.1 kJ / kg = 47.18 kW
Realizando los mismos cálculos para las otras tres temperaturas de evaporación
obtenemos los siguientes resultados:
t evap (°C) p evap (kPa) ηvol &
mR (kg / s) ∆hevap (kJ / kg) qf (kW)
-20 245.4 0.6153 0.3207 147.1 47.18
-10 354.9 0.7243 0.536 151.3 81.09
0 498.1 0.8008 0.821 155.2 127.4
10 681.2 0.8545 1.189 158.7 188.8
200
180
160
qf [kW]
140
120
100
80
60
40
-20 -15 -10 -5 0 5 10
tevap [°C]
32
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Problemas Resueltos - Tecnología Frigorífica
Problema 10
Ciclo simple de compresión mecánica (evaporadores y condensadores)
Se dispone de una máquina para enfriamiento de agua condensada por aire que realiza un
ciclo simple de compresión mecánica, sin recalentamiento del vapor ni subenfriamiento del
líquido, utilizando R-22. Según los datos del fabricante sí a dicha maquina se le suministra
un caudal de agua a enfriar de 0.19 kg/s a una temperatura de entrada de 20°C, siendo la
temperatura del aire a la entrada al condensador 25°C y su caudal, forzado por un
ventilador, 5500 m³/h. Entonces, la potencia frigorífica desarrollada por la máquina en las
condiciones anteriores es 8 kW y la potencia absorbida por el compresor 1.5 kW, el U·A
del evaporador es 883 W/K, y las características de los compresores alternativos son las
siguientes:
Nº de compresores: 2
Diámetro: 5 cm
Carrera: 5 cm
Rendimiento volumétrico: 0.822
Velocidad de giro: 750 rev/min
Calcular:
Temperatura de salida del agua, temperatura de salida del aire, temperatura de evaporación
del refrigerante, temperatura de condensación del refrigerante.
Solución:
La siguiente figura muestra una enfriadora de agua condensada por aire.
Compresor
&
mw t w,ent 1 2
Evaporador
t w,sal
Condensador
4 3
Válvula de expansión
La potencia frigorífica evacuada por el evaporador de la máquina es 8 kW, luego:
8000
8 kW = qf = mw cp (t w,ent − t w,sal )
& t w,sal = 20°C − °C = 9.927°C
4180·0.19
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Problemas Resueltos - Tecnología Frigorífica
Si estudiamos el evaporador como un intercambiador con cambio de fase,
tendremos:
tw,ent
T (°C)
UA 883
t w,ent − t w,sal −
&
mw c p −
ε= =1−e =1−e 0.19·4180
= 0.671
t w.ent − t evap tw,sal
tevap
Despejando de la expresión anterior obtenemos la temperatura de evaporación del
t − t w,sal
refrigerante: t evap = t w,ent − w,ent = 4.988 °C ≈ 5°C
ε
A través de los datos del compresor puedo calcular el caudal de refrigerante:
&
VR ,1 π 0.052
& & πD2 1
ηvol =
VR ,1 = ηvol Vt = ηvol Nc
; c
L c ω = 0.822 2 0.05·750 = 2.0175·10 −3 m³ / s
V& 4 4 60
t
Para calcular el caudal másico de refrigerante será necesario conocer el volumen
específico a la entrada del compresor del R-22 como vapor saturado.
v1 = 0.04036 m3 / kg .
&
VR ,1
&
mR = = 0.05 kg / s
v1
Si obtenemos la potencia frigorífica a través de los datos del refrigerante, podremos
despejar la entalpía del punto 4 (entrada al evaporador) que es igual a la del punto
3 (salida del condensador) por ser el proceso de expansión isentálpico.
8 kW = qf = mR (h1 − h4 )
&
p (kPa)
Como la temperatura de
evaporación es 5°C, podemos 3 2
calcular la entalpía del punto 1:
h1 = 407.15 kJ / kg
qf
h4 = h1 − = 247.15 kJ / kg 4 1
&
mR
h (kJ/kg)
Si interpolamos en la curva de
líquido saturado obtendremos la temperatura de condensación asociada al punto 3:
t cond − 35 247.15 − 243.10
= ; t cond = 38.082 °C
40 − 35 249.67 − 243.10
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Problemas Resueltos - Tecnología Frigorífica
Para calcular la temperatura de salida del aire en el condensador será necesario
realizar un balance de energía sobre este:
qcond = mR (h2 − h3 ) = macp (t a,sal − t a,ent )
& &
En la ecuación anterior no conocemos ni la entalpía del punto 2, ni la temperatura
de salida del aire, pero podemos calcular la potencia evacuada en el condensador
indirectamente, sumando la potencia frigorífica y el trabajo de compresión.
qc = qf + Pc = Vaρ acp (t a,sal − t a,ent )
&
qf + Pc 8000 + 1500
t a,sal = t a,ent + = 25 + = 30.182°C
& ρ c
Va a p 5500 / 3600·1.2·1000
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Problemas Resueltos - Tecnología Frigorífica
Problema 11
Ciclo simple de compresión mecánica (evaporadores y condensadores)
Una máquina frigorífica basada en un ciclo estándar de compresión mecánica desarrolla
una potencia frigorífica de 5 kW. El fabricante suministra el coeficiente de eficiencia
energética (COP) de dicha máquina como una función de la temperatura de condensación
del refrigerante en °C, COP = 2.5 − 0.01 (t cond − 30) .
El condensador de dicha máquina frigorífica es un intercambiador de carcasa y tubo, un
paso por carcasa y dos por tubos (el refrigerante circula por la carcasa con un coeficiente de
película de 10000 W/m²K). El intercambiador dispone de 50 tubos en forma de U, tienen
una longitud total 3 m, un diámetro interior de 20 mm y un espesor 1 mm. Están fabricados
en acero inoxidable de k = 15.1 W/mK, pueden considerarse despreciables las resistencias
de ensuciamiento.
Se dispone de un caudal de 0.2 kg/s de agua a 20°C para evacuar el calor de condensación.
Determinar:
1. Temperatura de condensación del refrigerante y calor total intercambiado en el
condensador
2. Caudal de agua necesario para conseguir una temperatura de condensación del
refrigerante de 30°C.
Nota: Suponer que el coeficiente de película interior en los tubos es 113 W/m²K,
independiente de la velocidad del fluido por encontrarse este en régimen laminar
Solución:
Apartado 1:
La siguiente figura muestra un esquema del condensador enfriadora de agua
condensada por aire.
Refrigerante, tcond
&
mw t w,ent
t w,sal
Si expresamos el COP en función de los datos del problema (Potencia frigorífica) y
calor de condensación tendremos:
36
Problemas Resueltos - Tecnología Frigorífica
qf qf 1 1
COP = = qc = qf 1 + = qf 1 +
Pc qc − qf COP
2.5 − 0.01(t cond − 30)
La ecuación anterior contiene las dos incógnitas del apartado 1, necesitamos por
tanto otra ecuación que nos permita cerrar el problema. Realizando un balance de
energía sobre el condensador tendremos:
refrigerante
T (°C)
tcond
tsal
tent agua
Área (m²)
−
UA
& wcp (t w,sal − t w,ent ) = mwcp 1 − e m c
qc = m & & w p (t − t w,ent )
cond
Luego si igualo esta ecuación con la anterior tendré una sola ecuación con una sola
incógnita, la temperatura de condensación.
1 −
UA
= mwcp 1 − e m c (t cond − t w,ent )
&
qf 1 + & w p
2.5 − 0.01(t cond − 30)
Para poder resolver esta ecuación necesito conocer UA.
Cálculo de UA:
1 1
UA = = = 1046.9 W / K
1 ln(Dext / Dint ) 1 9.646E − 6 + 6.6972E − 6 + 9.389E − 4
+ +
hext A ext 2πkNtL t hint A int
Donde:
A ext = Nt πDextL t = 10.367 m²
A int = Nt πDintL t = 9.425 m²
Despejando de la ecuación inicial y resolviendo queda una ecuación cuadrática de la
que la única solución válida es:
t cond = 28.7°C
y el calor evacuado en el condensador:
1
qc = qf 1 + = 6.989 kW
2.5 − 0.01(t cond − 30)
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Problemas Resueltos - Tecnología Frigorífica
Apartado 2:
Sí la temperatura de condensación del refrigerante es de 30°C, podemos calcular el
calor de condensación:
1 1
qc = qf 1 + = qf 1 + = 7 kW
2.5 − 0.01(t cond − 30)
2.5
Utilizando la otra expresión del calor de condensación tendríamos que:
−
UA
& w c p 1 − e m c
qc = m
& w p (t − t w,ent )
cond
Una ecuación fuertemente no lineal que debe de resolverse de forma iterativa:
&
1ª Iteración, supongo mw = 0.2 kg / s , despejo &
mw = 0.2343 kg / s
& w = 0.2343 kg / s , despejo
2ª Iteración, supongo m &
mw = 0.2549 kg / s
.... &
mw = 0.2884 kg / s
38
Problemas Resueltos - Tecnología Frigorífica
Problema 12
Evaporadores – Válvulas de expansión
Una máquina frigorífica de amoniaco es utilizada para enfriar una corriente de 31.6 l/min
de agua a 15°C, el agua sale del evaporador a 10.43°C en condiciones de evaporador limpio
(máquina recién instalada).
Se supone que con el paso del tiempo aparecerá una resistencia de ensuciamiento en el lado
del agua de aproximadamente Rensu=0.001 m² K/W, el área exterior del evaporador, la que
está en contacto directo con el agua, es 10 m².
Calcular el cambio que se produce en la potencia de compresión y en coeficiente de
eficiencia energética, suponiendo:
1. La temperatura de condensación permanece constante e igual a 40°C.
2. El evaporador es inundado y la válvula de expansión mantiene la temperatura de
salida del agua (10.43°C).
3. El rendimiento de la compresión es 0.7
4. El coeficiente de intercambio global del evaporador limpio es UAlimpio= 800 W/K
Solución:
El caudal de agua a la entrada, suponiendo una densidad del agua de 1 kg/l, es:
&
mw = 31.6 l / min = 0.5267 kg / s .
La potencia frigorífica suministrada será:
qf = mw cp (t w,ent − t w,sal ) = 0.5267 kg / s 4.18 kJ / kg(15 − 10.43°C ) = 10.061 kW
&
Esta potencia frigorífica va a ser la misma en el caso sucio, puesto que la válvula de
expansión va a mantener la misma temperatura de salida del agua.
Si planteamos la ecuación de transferencia en el intercambiador:
−UA lim pio
t w,ent − t w,sal &
mw c p
ε= =1−e
t w,ent − t evap
tw,ent
T (°C)
tw,sal
tevap
Despejando de la ecuación anterior la temperatura de evaporación:
39
Problemas Resueltos - Tecnología Frigorífica
t w,ent − t w,sal 15°C − 10.43°C
t evap = t w,ent − −UA lim pio
= 15°C − −800 W / K
≈ 0°C
0.5267 kg / s 4180 J / kg
1−e
&
mw cp
1−e
Con esta temperatura de evaporación ya podemos dibujar el ciclo estándar de
compresión sobre un diagrama P-h del amoniaco (R-717):
P (kPa)
3 40°C 2s
0°C
4 1
h (kJ/kg)
Las entalpías de los puntos son:
h1 = 1460.66 kJ / kg h2s = 1646 kJ / kg h3 = h4 = 386.43 kJ / kg
Realizando un balance de energía en el lado del refrigerante del evaporador:
qf 10.061 kW
qf = mR (h1 − h4 ); mR =
& & = = 0.0094 kg / s
h1 − h4 1460 .66 − 386 .43 kJ / kg
De la definición de rendimiento de la compresión podemos obtener el trabajo real
de compresión:
P P m (h − h1 ) 0.0094 kg / s(1646 − 1460.66 kJ / kg)
&
ηc = c,s ; Pc = c,s = R 2s = = 2.480 kW
Pc ηc ηc 0.7
El coeficiente de eficiencia energética valdrá para el caso limpio:
qf 10.061 kW
COPlim pio = = = 4.057
Pc,lim pio 2.480 kW
Para el caso del intercambiador sucio, el primer paso es calcular el valor del nuevo
1 1
UA: UA sucio = = = 740.741 W / K
1 R 1 0.001 m²K / W
+ ensu +
UA lim pio A ext 800 W / K 10 m²
En este caso la temperatura de evaporación cambiará puesto que la válvula de
expansión mantiene la potencia frigorífica:
t w,ent − t w,sal 15°C − 10.43°C
t evap = t w,ent − −UA sucio
= 15°C − −740.741 W / K
≈ −1°C
&
mw cp 0.5267 kg / s 4180 J / kg
1−e 1−e
Las entalpías de los puntos son:
40
Problemas Resueltos - Tecnología Frigorífica
h1 = 1459.59 kJ / kg h2s = 1650.52 kJ / kg h3 = h4 = 386.43 kJ / kg
Realizando un balance de energía en el lado del refrigerante del evaporador:
Pf 10.061 kW
Pf = mR (h1 − h4 ); mR =
& & = = 0.0094 kg / s
h1 − h4 1459 .59 − 386.43 kJ / kg
De la definición de rendimiento de la compresión podemos obtener el trabajo real
de compresión:
P P m (h − h1 ) 0.0094 kg / s(1650.52 − 1459.59 kJ / kg)
&
ηc = c,s ; Pc = c,s = R 2s = = 2.557 kW
Pc ηc ηc 0.7
El coeficiente de eficiencia energética valdrá para el caso sucio:
qf 10.061 kW
COPsucio = = = 3.935
Pc,sucio 2.557 kW
41
Problemas Resueltos - Tecnología Frigorífica
Problema 13
Evaporadores - Psicrometría
Una cámara frigorífica para almacenamiento se mantiene a una temperatura de 10°C y una
humedad relativa del 80%. El caudal de aire sobre el evaporador es 30000 m³/h y la
temperatura del aire medida a la salida del evaporador es de 5°C. En estas condiciones la
instalación desarrolla una potencia frigorífica de 100 kW.
Calcular la cantidad de agua de condensado que se forma en el evaporador en una hora.
Solución:
El aire de entrada al evaporador se encuentra en las condiciones medias de la
cámara frigorífica, t ent = 10°C φ ent = 80% .
&
El caudal de aire a la entrada al evaporador es Vair = 30000 m³ / h = 8.333 m³ / s
Realizando un balance de energía sobre el caudal de aire del evaporador tendremos
que: qf = mair (ha,ent − ha,sal ) .
&
Si suponemos que el caudal volumétrico de aire ha sido medido a la entrada al
evaporador, podemos decir que su densidad a 10°C es aproximadamente 1.247
& &
kg/m³, y por lo tanto: mair = Vair ρ air = 8.333 m³ / s 1.247 kg / m³ = 10.391 kg / s .
Podemos discutir en este punto si este caudal es de aire seco o aire húmedo, pero
la diferencia entre ambos será tan pequeña que puede considerarse que ambos
valen lo mismo y son iguales al valor anterior.
Si colocamos sobre un diagrama psicrométrico del aire a presión atmosférica el
punto de entrada podremos leer en el eje de entalpías cual es la entalpía del aire a
la entrada: ha,ent = 26 kJ / kg a.s.
Por tanto podemos despejar del balance de energía anterior la entalpía a la salida
del evaporador:
q 100 kW
ha,sal = ha,ent − f = 26 kJ / kg a.s. − = 16.376 kJ / kg a.s.
& air
m 10.391 kg / s
Ahora podemos colocar el punto de salida del aire sobre el diagrama psicrométrico
en el punto de intersección entre la línea de entalpía igual a la anterior y de
temperatura seca igual a 5°C.
Si miramos en el eje de humedades absolutas obtenemos:
wa,ent = 0.006 kg H2O / kg a.s. wa,sal = 0.0045 kg H2O / kg a.s.
42
Problemas Resueltos - Tecnología Frigorífica
h (kJ/kg a.s.)
w (kg H2O/kg a.s.)
φ=80%
ha,ent
ha,sal
wa,ent
w a,sal
t a,sal t a,ent t (°C)
Por lo tanto el caudal de agua condensada será igual a la cantidad de agua perdida
por el aire en su paso por el evaporador:
magua = mair (w a,ent − w a,sal ) = 10.391 kg / s(0.006 − 0.0045 kg H2O / kg a.s.) = 0.0156 kg H2O / s
& &
En una hora la cantidad de condensado sería:
0.0156 kg / s · 3600 s = 56.16 kg H2O / h
43
Problemas Resueltos - Tecnología Frigorífica
Problema 14
Ciclo simple de compresión mecánica (evaporadores y condensadores)
Se dispone de una máquina para enfriamiento de agua condensada por aire que realiza un
ciclo simple de compresión mecánica, sin recalentamiento del vapor ni subenfriamiento del
líquido, utilizando R-22. Según los datos del fabricante si a dicha maquina se le suministra
un caudal de agua a enfriar de 0.20 kg/s a una temperatura de entrada de 20°C, la
temperatura de evaporación del refrigerante es 6°C y la potencia consumida por el
compresor 1.5 kW, con una temperatura de condensación de 40°C. El desplazamiento
volumétrico del compresor vale 9 m³/h y el rendimiento volumétrico 0.8
Calcular:
1. Potencia frigorífica, potencia evacuada por el condensador y U·A del evaporador.
2. Si suponemos que el caudal de agua desciende a 0.18 kg/s, y que la máquina
funciona con una válvula de expansión automática (mantiene la temperatura de
evaporación constante), calcular la nueva potencia frigorífica, potencia de
compresión y potencia evacuada por el condensador.
Nota: suponer que el U·A del evaporador es proporcional al caudal de agua elevado a 0.8 y
que el rendimiento volumétrico e isentrópico del compresor en el segundo apartado son los
mismos que los del primer apartado.
Solución:
1) La siguiente figura muestra una enfriadora de agua.
Compresor
&
mw t w,ent 1 2
Condensador
Evaporador
t w,sal
4 3
Válvula de expansión
Conocidas las temperaturas de evaporación y condensación del refrigerante
podemos localizar sobre un diagrama P-h alguno de los puntos del ciclo con las
siguientes entalpías:
h1 = 407.5 kJ / kg h4 = h3 = 249.67 kJ / kg h2s = 430.6 kJ / kg
44
Problemas Resueltos - Tecnología Frigorífica
P (kPa)
3 40°C 2s
2
6°C
1
4
h (kJ/kg)
A través de los datos del compresor puedo calcular el caudal de refrigerante:
&
VR ,1 mR v1
& η V &
& 0.8·9 / 3600 m³ / s
ηvol = = ; mR = vol t = = 0.0511 kg / s
&
Vt &
Vt v1 0.03915 m³ / kg
Realizando un balance sobre el evaporador del lado del refrigerante:
qf = mR (h1 − h4 ) = 0.0511 kg / s(407.5 − 249.67 kJ / kg) = 8.065 kW
&
La potencia evacuada por el condensador puede obtenerse de un balance de
energía sobre toda la máquina:
qc = qf + Pc = 8.065 + 1.5 kW = 9.565 kW
Si realizamos un balance sobre el evaporador pero desde el lado del agua, podemos
obtener la temperatura de salida del agua:
tw,ent
T (°C)
tw,sal
tevap
qf 8.065
qf = mw cp (t w,ent − t w,sal )
& t w,sal = t w,ent − = 20°C − °C = 10.353 °C
&
mw c p 4.18·0.20
Y planteando la ecuación de transferencia sobre el evaporador:
t w,ent − t w,sal 20 − 10.353 °C
qf = UA ∆t lm = UA = UA = UA 8.258 °C
t w,ent − t evap 20 − 6
ln ln
t w,sal − t evap 10.353 − 6
45
Problemas Resueltos - Tecnología Frigorífica
qf 8.065 kW
UA = = = 0.9766 kW / °C
∆t lm 8.258 °C
&
2) Para este apartado suponemos que el nuevo caudal de agua es mw = 0.18 kg / s ,
las temperaturas de entrada del agua y de evaporación son la misma ya que el
sistema esta controlado por una válvula de expansión automática que mantiene la
presión, y por tanto la temperatura, de evaporación constante.
Si el rendimiento volumétrico y el desplazamiento volumétrico no cambian, con la
misma temperatura de evaporación tendremos el mismo caudal del refrigerante:
&
VR ,1 mR v1
& η V &
& 0.8·9 / 3600 m³ / s
ηvol = = ; mR = vol t = = 0.0511 kg / s
&
Vt &
Vt v1 0.03915 m³ / kg
También se nos indica que el UA del evaporador es proporcional al caudal de agua
elevado a 0.8, por tanto, el ratio entre el nuevo UA y el antiguo es igual a la
relación de caudales elevada a 0.8:
0.8 0.8
UA nuevo mw,nuevo
& 0.18
=
m
= = 0.9192; UA nuevo = UA·0.9192 = 0.8977 kW / °C
UA &w 0.2
Podemos ahora obtener la potencia frigorífica a través de la expresión:
−
UA
0.8977
& w cp (t w,ent − t w,sal ) = mw cp 1 − e m c (t w,ent − t evap ) = 0.18·4.181 − e 0.18·4.18 (20 − 6 ) = 7.339 kW
& −
qf = m & w p
Con esta potencia frigorífica y realizando un balance del lado del refrigerante
obtenemos la entalpía de entrada al evaporador:
q 7.339 kW
qf = mR (h1 − h4 ); h4 = h1 − f = 407.5 kJ / kg −
& = 263.9 kJ / kg
&
mR 0.0511 kg / s
Esta entalpía corresponde a una temperatura de condensación de:
52°C − 50°C t cond − 50°C
= ; t cond = 50.464°C
266.05 − 263.25 kJ / kg 263.9 − 263.25 kJ / kg
Para obtener el trabajo de compresión debemos suponer que el rendimiento
isentrópico permanece constante desde el apartado 1:
Pc,s mR (h2s − h1 ) 0.0511 kg / s(430.6 − 407.5 kJ / kg)
&
ηc = = = = 0.7869
Pc Pc 1.5 kW
Para el caso nuevo:
h2s = 437.0 kJ / kg
Pc,s m (h − h1 ) 0.0511 kg / s(437.0 − 407.5 kJ / kg)
&
Pc = = R 2s = = 1.916 kW
ηc ηc 0.7869
Y la potencia evacuada en el condensador será:
qc = qf + Pc = 7.339 + 1.916 kW = 9.255 kW
46
Problemas Resueltos - Tecnología Frigorífica
Problema 15
Problema combinado de compresión múltiple y torre de refrigeración
La instalación frigorífica de la figura utiliza amoniaco como refrigerante, consta de dos
evaporadores que mantienen diferentes temperaturas de conservación en sendas cámaras
frigoríficas. Se conocen los siguientes datos:
Evaporador baja: Potencia frigorífica: qf ,B = 30 kW
Presión de evaporación del refrigerante: p evap,B = 119.46 kPa
Evaporador alta: Potencia frigorífica: qf ,A = 15 kW
Presión de evaporación del refrigerante: p evap,A = 190.11 kPa
Condensador: Presión refrigerante salida del condensador: p cond = 1554.89 kPa
Depósito intermedio:Presión del refrigerante: pint = 429.41 kPa
Torre de Refrig.: Temperatura del agua a la entrada a la torre: 35°C
Temperatura seca del aire exterior: t ext = 35°C
Temperatura de bulbo húmedo del aire exterior: t bh,ext = 25°C
Humedad relativa del aire a la salida de la torre: φ a,sal = 90%
Caudal de aire seco de entrada en torre: &
Va = 15700 m³ / h
Cercanía de la torre: 5°C
Compresores (ambos): Rendimiento isentrópico: 0.8
Se pide:
A. Dibujar un esquema del diagrama p-h del refrigerante con todos los puntos de la
figura colocados en él.
B. Calcular la potencia consumida por cada uno de los compresores y el COP de la
instalación.
C. Caudal de agua de la bomba del circuito de condensación.
D. Caudal de agua de reposición (evaporado) en la torre.
47
Problemas Resueltos - Tecnología Frigorífica
Compresor Torre de
de baja Compresor refrigeración
1 2 de alta
3 4
9 12
Evaporador
de baja 11
Evaporador Condensador
de alta
8 10
6 5
7
Deposito Bomba del circuito
intermedio de condensación
Nota: Suponer que no existen pérdidas de presión en los elementos del ciclo y que no
existe recalentamientos, ni subenfriamientos.
Solución:
Las temperaturas asociadas a las presiones de cambio de fase del amoniaco
mostradas en el enunciado son las siguientes:
p evap,B = 119.46 kPa → t evap,B = −30°C
p evap,A = 190.11 kPa → t evap,A = −20°C
pint = 429.41 kPa → t evap,B = 0°C
p cond = 1554.89 kPa → t cond = 40°C
A. Diagrama p-h de la instalación frigorífica
48
Problemas Resueltos - Tecnología Frigorífica
p (kPa)
5 40°C 4s 4
7 6 0°C 3 2s
2
10 -20°C 11
-30°C 9
12
8 1
h (kJ/kg)
B. Mirando en las tablas de líquido / vapor saturado del amoniaco (R-717)
conseguimos las entalpías de los siguientes puntos:
h7 = h8 = h10 = 200 kJ / kg h3 = 1460.66 kJ / kg h11 = h12 = 1436.51 kJ / kg
h5 = h6 = 386.43 kJ / kg h9 = 1422.46 kJ / kg
Calculemos los caudales de refrigerante a partir de los balances en los
evaporadores:
qf ,B 30 kW
qf ,B = mR ,B (h11 − h10 );
& &
mRB = = = 0.0243 kg / s
(h11 − h10 ) 1436.51 − 200 kJ / kg
qf ,A 15 kW
qf ,A = mR ,A (h9 − h8 );
& &
mRA = = = 0.0123 kg / s
(h9 − h8 ) 1422.46 − 200 kJ / kg
Mezcla de las corrientes 9 y 12, para obtener la corriente 1:
& &
mR ,Ah12 + mR ,Bh9
h1 = = 1427.2 kJ / kg
& &
mR ,A + mR ,B
Si seguimos la isentrópica que parte del punto 1 hasta la presión del deposito
intermedio obtenemos: h2s ≈ 1610 kJ / kg . El rendimiento isentrópico de la
compresión es 0.8 y por tanto:
h2s − h1 h2s − h1
ηs = 0.8 = ; h2 = h1 + = 1655.7 kJ / kg
h2 − h1 ηs
Realicemos ahora un balance de energía en el deposito para calcular el caudal de
refrigerante que circula por el compresor de alta y el condensador:
49
Problemas Resueltos - Tecnología Frigorífica
(m
& RA + mRB ) h2 + mRCond h6 = (mRA + mRB ) h7 + mRCond h3
& & & & &
h − h7
mRCond = (mRA + mRB ) 2
& & & = 0.0496 kg / s
h3 − h6
Si seguimos la isentrópica que parte del punto 3 hasta la presión de condensación
obtenemos: h4s ≈ 1647 kJ / kg . El rendimiento isentrópico de la compresión es 0.8 y
por tanto:
h4s − h3 h4s − h3
ηs = 0.8 = ; h4 = h3 + = 1693.6 kJ / kg
h4 − h3 ηs
Podemos ya calcular las potencias de compresión de ambos compresores:
Pc,B = (mRA + mRB )(h2 − h1 ) = 8.36 kW
& &
Pc,A = mRCond (h4 − h3 ) = 11.55 kW
&
qf ,B + qf ,A
Y el COP de la instalación: COP = = 2.26 .
PC,A + PC,B
C. Realizando un balance de energía sobre el condensador:
qcond = mw cp (t w,ent − t w,sal )
&
La temperatura de agua a la entrada al condensador es la de salida de la torre y
viceversa. Luego:
t w,sal = 35°C, cercanía = 5°C = t w,ent − t bh,ext ; t w,ent = t bh,ext + 5°C = 30°C
La potencia de condensación puede ser calculada de dos formas:
qcond = mRCond (h4 − h5 ) = qf ,A + qf ,B + PC,A + PC,B = 64.91 kW
&
Despejando del balance de energía en el condensador tenemos:
& qcond 64.91 kW
mw =
cp (t w,ent − t w,sal ) = 4.175 kJ / kg·K (35 − 30°C) = 3.11 kg / s
D. Las condiciones del aire a la entrada a la torre son las siguientes:
t ext = 35°C; t bh,ext = 25°C . Si miramos en el diagrama psicrométrico del aire
húmedo: ha,ext = ha,ent = 76 kJ / kg a.s.; w a,ent = 0.016 kg agua / kg a.s.
El caudal másico de aire seco a la entrada a la torre es:
& & 15700 m³ / h
mas = Vas ρ = 1.146 kg / m³ = 5 kg / s
3600 s / h
Todo el calor cedido por el condensador será absorbido por el aire exterior luego:
50
Problemas Resueltos - Tecnología Frigorífica
qcond
qcond = mas (ha,sal − ha,ent );
& ha,sal = ha,ent + = 89 kJ / kg a.s.
&
mas
Con esta entalpía y la humedad relativa del 90% podemos colocar sobre el
diagrama psicrométrico el punto de salida del aire:
t a,sal = 29°C; wa,sal = 0.023 kg agua / kg a.s.
La diferencias de humedades absolutas entre el aire a la salida y a la entrada nos
permite calcular la cantidad de agua evaporada en la torre que es a su vez igual al
caudal de agua que es necesario reponer:
mw,rep = mas (w a,sal − wa,ent ) = 0.035 kg agua / s = 126 kg agua / h
& &
ha,sal φ=90%
ha,ent w (kg H2O/kg a.s.)
w a,sal
h (kJ/kg a.s.)
wa,ent
t (°C) t bh,ent t a,ent
51
Colección de Problemas
Resueltos de Tecnología
Frigorífica
Versión 2.1 (septiembre de 2003)
Compresor Compresor
de baja de alta
1 2 3 4
Condensador
Evaporador
6
5
8 7
p (kPa)
7 5 4
3
6 2
8 1
h (kJ/kg)
Autor: Juan Francisco Coronel Toro
Profesor asociado del Grupo de Termotecnia
Dpto. de Ingeniería Energética y mecánica de Fluidos
Universidad de Sevilla
Este documento está basado en versiones anteriores desarrolladas por:
□ D. Ramón Velázquez Vila
□ D. José Guerra Macho
□ D. Servando Álvarez Domínguez
□ D. José Luis Molina Félix
□ D. David Velázquez Alonso
□ D. Luis Pérez-Lombard
□ D. Juan F. Coronel Toro
Todos ellos pertenecientes al Grupo de Termotecnia.
Parte de la información ha sido tomada de las siguientes referencias:
DEPARTMENT OF MECHANICAL ENGINEERING, TECHNICAL UNIVERSITY OF
DENMARK, COOLPACK, A collection of simulations tools for refrigeration, Versión
1.46 (2000).
STOECKER, W.F. Industrial Refrigeration Handbook. 1st ed. McGraw Hill (1998)
KLEIN, S.A. y ALVARADO, F.L., Engineering Equation Solver Software (EES),
Academia Versión 6.271 (20-07-2001).
Problemas Resueltos - Tecnología Frigorífica
Índice
Índice ......................................................................................................... 3
Ciclo simple de compresión mecánica:
Problema 1.................................................................................................. 4
Ciclos múltiples de compresión mecánica:
Problema 2.................................................................................................. 6
Problema 3................................................................................................ 12
Problema 4................................................................................................ 15
Problema 5................................................................................................ 19
Ciclo simple de compresión mecánica (compresores alternativos):
Problema 6................................................................................................ 23
Problema 7................................................................................................ 26
Problema 8................................................................................................ 28
Problema 9................................................................................................ 31
Ciclo simple de compresión mecánica (evaporadores y condensadores):
Problema 10 .............................................................................................. 33
Problema 11 .............................................................................................. 36
Problema 12 .............................................................................................. 39
Problema 13 .............................................................................................. 42
Problema 14 .............................................................................................. 44
Problemas combinados:
Problema 15 .............................................................................................. 47
3
Problemas Resueltos - Tecnología Frigorífica
Problema 1
Ciclo simple de compresión mecánica
Una máquina frigorífica utiliza el ciclo estándar de compresión de vapor. Produce 50 kW
de refrigeración utilizando como refrigerante R-22, si su temperatura de condensación es
40°C y la de evaporación -10°C, calcular:
a. Efecto frigorífico
b. Caudal de refrigerante
c. Potencia de compresión
d. Coeficiente de eficiencia energética
e. Relación de compresión
f. Caudal volumétrico de refrigerante manejado por el compresor
g. Temperatura de descarga del compresor
h. Coeficiente de eficiencia energética del ciclo inverso de Carnot con las mismas
temperaturas de evaporación y condensación
Solución:
Compresor
p (kPa)
1 2
2 Evaporador
3
Condensador
4 1
4 3
Válvula de
h (kJ/kg) expansión
Si trasladamos las temperaturas de evaporación (-10°C) y condensación (40°C)
sobre el diagrama P-h del R-22, obtenemos los siguientes valores:
Presiones: p cond = 1533.52 kPa p evap = 354.3 kPa
Entalpías: h1 = 401.56 kJ / kg h2 = 438.56 kJ / kg h3 = h4 = 249.67 kJ / kg
a. Efecto frigorífico: h1 − h4 = 151.89 kJ / kg
b. Caudal de refrigerante:
qf 50 kW
qf = mR (h1 − h4 );
& &
mR = = = 0.3292 kg / s
(h1 − h4 ) 151.89 kJ / kg
c. Potencia de compresión:
Pc = mR (h2 − h1 ) = 0.3292 kg / s(438.56 − 401.56 kJ / kg) = 12.18 kW
&
4
Problemas Resueltos - Tecnología Frigorífica
qf 50 kW
d. Coeficiente de eficiencia energética: COP = = = 4.105
Pc 12.18 kW
p cond 1533.52 kPa
e. Relación de compresión: rc = = = 4.328
p evap 354.3 kPa
f. Caudal volumétrico de refrigerante manejado por el compresor: Este siempre se
toma a la entrada al compresor y necesitamos el volumen específico en el punto 1:
v1 = 0.06535 m³ / kg
& &
V = m v = 0.3292 kg / s 0.06535 m³ / kg = 0.0215 m³ / s = 77.448 m³ / h
R R 1
g. Temperatura de descarga del compresor: Si miramos en el diagrama p-h la
isoterma que pasa por el punto 2 es aproximadamente t 2 = 64.3°C .
h. Coeficiente de eficiencia energética del ciclo inverso de Carnot con las mismas
temperaturas de condensación y evaporación.
1 1
COPCarnot = = = 5.263
Tcond 273.15 + 40 K
−1 −1
Tevap 273.15 − 10 K
5
Problemas Resueltos - Tecnología Frigorífica
Problema 2
Ciclos múltiples de compresión mecánica
Una aplicación de producción de frío demanda una potencia frigorífica de 100.000 frig/h,
su temperatura de evaporación debe ser -30°C y su temperatura de condensación 40°C. Si
se pretende usar en todos los casos R-22, calcular el trabajo de compresión, el calor de
condensación y el coeficiente de eficiencia energética en los siguientes casos:
a. Ciclo estándar de compresión mecánica simple.
b. Compresión doble directa con enfriador intermedio, inyección parcial. (Eficiencia del
enfriador intermedio 0.8)
c. Compresión doble directa con enfriador intermedio, inyección total.
d. Compresión doble en cascada.
Solución:
Comencemos por calcular el coeficiente de eficiencia energética de del ciclo teórico
de Carnot:
1 1
COPCarnot = = = 3.474
Tcond 273.15 + 40 K
−1 −1
Tevap 273.15 − 30 K
Este es el límite máximo para las eficiencias de todos los ciclos que vamos a
estudiar a continuación.
La potencia frigorífica en todos los ciclos debe ser:
qf = 100000 frig / h = 100000 kcal / h = 116.28 kW
a. Ciclo estándar de compresión mecánica simple.
Compresor
p (kPa)
1 2
2 Evaporador
3
Condensador
4 1
4 3
Válvula de
h (kJ/kg) expansión
Si trasladamos las temperaturas de evaporación y condensación sobre el diagrama
P-h del R-22, obtenemos los siguientes valores
Presiones: p cond = 1533.5 kPa p evap = 163.5 kPa
6
Problemas Resueltos - Tecnología Frigorífica
p cond 1533.5 kPa
Relación de compresión: rc = = = 9.38
p evap 163.5 kPa
Entalpías: h1 = 393.147 kJ / kg h2 = 451.021 kJ / kg h3 = h4 = 249.674 kJ / kg
Calculemos el caudal de refrigerante a partir del balance en el evaporador:
qf 116.28 kW
qf = mR (h1 − h4 ); mR =
& & = = 0.8105 kg / s
(h1 − h4 ) 393.147 − 249.674 kJ / kg
Trabajo de compresión:
Pc = mR (h2 − h1 ) = 0.8105 kg / s(451.021 − 393.147 kJ / kg) = 46.907 kW
&
Calor de condensación:
qc = mR (h2 − h3 ) = 0.8105 kg / s(451.021 − 249.674 kJ / kg) = 163.192 kW
&
qf 116.28 kW
Coeficiente de eficiencia energética: COP = = = 2.479
Pc 46.907 kW
b. Compresión doble directa con enfriador intermedio, inyección parcial.
Compresor Compresor
de baja de alta
1 2 3 4
Evaporador &
mRA
&
mRB
Condensador
6
5
8 7
p (kPa)
7 5 4
3
6 2
8 1
h (kJ/kg)
El primer paso es determinar la presión intermedia a la que trabaja el enfriador
intermedio, para así poder dibujar el ciclo:
7
Problemas Resueltos - Tecnología Frigorífica
pint = p cond p evap = 1533.5 · 163.5 = 500.73 kPa a esta presión le corresponde una
temperatura intermedia de t int = 0.19°C .
Al utilizar la media geométrica se consigue que la relación de compresión en el
compresor de alta y baja sean la misma.
p p
rc = cond = int = 3.06
pint p evap
Las entalpías de los puntos que hasta el momento podemos localizar sobre el
diagrama P-h son las siguientes:
h1 = 393.147 kJ / kg h2 = 420.11 kJ / kg h3 = 405.44 kJ / kg
h4 = 433.33 kJ / kg h5 = h6 = 249.67 kJ / kg
Balance de energía sobre el evaporador:
qf = mRB (h1 − h8 )
&
Balance de energía sobre el enfriador intermedio:
mRBh6 + mRBh2 + (mRA − mRB )h6 = mRAh3 + mRBh7
& & & & & &
Simplificando y sabiendo que las entalpías de los puntos 7 y 8 son iguales:
& & & &
mRAh6 + mRBh2 = mRAh3 + mRBh8
& &
Tenemos por tanto 2 ecuaciones con 3 incógnitas ( mRA , mRB , h8 ). Es necesario
plantear una nueva ecuación. La eficiencia del enfriador intermedio se plantea
como:
t − t7
ε = 0.8 = 5 ; t 7 = t 5 − ε(t 5 − t 6 ) = 40°C − 0.8(40 − 0.19°C ) = 8.152°C
t5 − t 6
Con esta temperatura del punto 7 y la presión de condensación obtenemos la
entalpía de este punto:
h7 = 209.66 kJ / kg = h8
Del balance del evaporador podemos ahora despejar el caudal de refrigerante sobre
el evaporador:
& qf 116.28 kW
mRB = = = 0.6337 kg / s
(h1 − h8 ) 393.147 − 209.66 kJ / kg
Y volviendo al balance sobre el enfriador intermedio obtenemos el caudal de
refrigerante sobre el condensador:
& &
mRA = mRB 8
(h − h2 ) = 0.8561 kg / s
(h6 − h3 )
Trabajo de compresión: Pc = mRA (h4 − h3 ) + mRB (h2 − h1 ) = 40.963 kW
& &
Calor de condensación: Q c = mRA (h4 − h5 ) = 157.231 kW
& &
qf 116.28 kW
Coeficiente de eficiencia energética: COP = = = 2.839
Pc 40.963 kW
8
Problemas Resueltos - Tecnología Frigorífica
c. Compresión doble directa con enfriador intermedio, inyección total.
Compresor Compresor
de baja de alta
1 2 3 4
Evaporador
&
mRB &
mRA
Condensador
6 5
8 7
P (kPa)
5 4
7 6 3
2
8 1
h (kJ/kg)
La presión intermedia es la misma que en el caso anterior.
pint = p cond p evap = 1533.5 · 163.5 = 500.73 kPa
Las entalpías de los puntos sobre el diagrama P-h son las siguientes:
h1 = 393.147 kJ / kg h2 = 420.11 kJ / kg h3 = 405.44 kJ / kg
h4 = 433.33 kJ / kg h5 = h6 = 249.67 kJ / kg h7 = h8 = 200.23 kJ / kg
Balance de energía sobre el evaporador:
qf 116.28 kW
qf = mRB (h1 − h8 )
& &
mRB = = = 0.6027 kg / s
(h1 − h8 ) 393.147 − 200.23 kJ / kg
Balance de energía sobre el enfriador intermedio:
& & & &
mRAh6 + mRBh2 = mRAh3 + mRBh7
despejando:
9
Problemas Resueltos - Tecnología Frigorífica
& & (h − h2 ) = 0.8508 kg / s
mRA = mRB 7
(h6 − h3 )
Trabajo de compresión: Pc = mRA (h4 − h3 ) + mRB (h2 − h1 ) = 39.978 kW
& &
Calor de condensación: qc = mRA (h4 − h5 ) = 156.258 kW
&
qf 116.28 kW
Coeficiente de eficiencia energética: COP = = = 2.909
Pc 39.978 kW
d. Compresión doble en cascada.
Compresor Compresor
de baja de alta
Evaporador & &
mRA
mRB
Intercambiador Condensador
intermedio
p (kPa)
7 6
3
2
8 5
4 1
h (kJ/kg)
Para el caso de refrigeración en cascada la temperatura de evaporación del ciclo
superior debe ser inferior a la temperatura de condensación del ciclo inferior, a
esta diferencia de temperaturas se la llama solape. Si utilizamos la temperatura
intermedia de los casos anteriores y un solape de 5°C, podremos suponer que:
t 3 = +2.7°C y t 5 = t 8 = −2.3°C
Las entalpías de los puntos sobre el diagrama P-h son las siguientes:
h1 = 393.147 kJ / kg h2 = 422.22 kJ / kg h3 = h4 = 203.18 kJ / kg
h5 = 404.52 kJ / kg h6 = 434.56 kJ / kg h7 = h8 = 249.67 kJ / kg
Balance de energía sobre el evaporador:
10
Problemas Resueltos - Tecnología Frigorífica
qf 116.28 kW
qf = mRB (h1 − h4 )
& &
mRB = = = 0.6121 kg / s
(h1 − h4 ) 393.147 − 203.18 kJ / kg
Balance de energía sobre el intercambiador intermedio:
& & & &
mRAh8 + mRBh2 = mRBh3 + mRAh5
despejando:
& & (h − h2 ) = 0.8658 kg / s
mRA = mRB 3
(h8 − h5 )
Trabajo de compresión: Pc = mRA (h6 − h5 ) + mRB (h2 − h1 ) = 43.804 kW
& &
Calor de condensación: qc = mRA (h6 − h7 ) = 160.078 kW
&
qf 116.28 kW
Coeficiente de eficiencia energética: COP = = = 2.655
Pc 43.804 kW
Resumen de resultados:
Caso qf (kW) qc (kW) Pc (kW) COP
Compresión mecánica simple 116.28 163.192 46.907 2.479
Con enfriador intermedio, inyección parcial 116.28 157.231 40.963 2.839
Con enfriador intermedio, inyección total 116.28 156.258 39.978 2.909
En cascada 116.28 160.078 43.804 2.655
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Problemas Resueltos - Tecnología Frigorífica
Problema 3
Ciclo de compresión mecánica múltiple
Un sistema de refrigeración utiliza R-22 con una capacidad frigorífica de 180 kW a una
temperatura de evaporación de -30°C y una presión de condensación de 1500 kPa.
Calcular:
a) Potencia absorbida por un sistema de compresión mecánica simple estándar.
b) Potencia absorbida por el ciclo múltiple de la figura, donde el enfriador intermedio
funciona a una presión de 600 kPa.
1 2
3
4
6 5
8 7
Solución:
a. Ciclo de compresión mecánica simple
Compresor
p (kPa)
1 2
2 Evaporador
3
Condensador
4 1
4 3
Válvula de
h (kJ/kg) expansión
Si trasladamos las temperaturas de evaporación y la presión de condensación sobre
el diagrama P-h del R-22, obtenemos los siguientes valores
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Problemas Resueltos - Tecnología Frigorífica
Presiones: p cond = 1500 kPa p evap = 163.5 kPa
p cond 1500 kPa
Relación de compresión: rc = = = 9.1743
p evap 163.5 kPa
Temperaturas: t cond = 39.1°C t evap = −30°C
Entalpías: h1 = 393.15 kJ / kg h2 = 450.38 kJ / kg h3 = h4 = 248.48 kJ / kg
Calculemos el caudal de refrigerante a partir del balance en el evaporador:
qf 180 kW
qf = mR (h1 − h4 ); mR =
& & = = 1.2442 kg / s
(h1 − h4 ) 393.15 − 248.48 kJ / kg
Trabajo de compresión:
Pc = mR (h2 − h1 ) = 1.2442 kg / s(450.38 − 393.15 kJ / kg) = 71.206 kW
&
qf 180 kW
Coeficiente de eficiencia energética: COP = = = 2.528
Pc 71.206 kW
b. Ciclo de la figura del problema.
1 2
3
&
mR1 4
&
mR 2
6 5
8 7
p (kPa)
5 4 2
7 6 3
8 1
h (kJ/kg)
La presión intermedia es pint = 600 kPa; t int = 5.88°C .
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Problemas Resueltos - Tecnología Frigorífica
Las entalpías de los puntos sobre el diagrama P-h son las siguientes:
h1 = 393.15 kJ / kg h2 = 450.38 kJ / kg h3 = 407.46 kJ / kg
h4 = 430.09 kJ / kg h5 = h6 = 248.48 kJ / kg h7 = h8 = 206.95 kJ / kg
Balance de energía sobre el evaporador:
qf 180 kW
qf = mR1 (h1 − h8 )
& &
mR1 = = = 0.9667 kg / s
(h1 − h8 ) 393.15 − 206.95 kJ / kg
Balance de energía sobre el enfriador intermedio:
(mR1 + mR2 )h6 = mR1h7 + mR2h3
& & & &
despejando:
& & (h − h6 ) = 0.2525 kg / s
mR 2 = mR1 7
(h6 − h3 )
Trabajo de compresión: Pc = mR1 (h2 − h1 ) + mR 2 (h4 − h3 ) = 61.038 kW
& &
qf 180 kW
Coeficiente de eficiencia energética: COP = = = 2.949
Pc 61.038 kW
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Problemas Resueltos - Tecnología Frigorífica
Problema 4
Ciclos múltiples de compresión mecánica
En un sistema de amoniaco con dos evaporadores y un compresor el evaporador de baja
temperatura suministra 180 kW de refrigeración con una temperatura de evaporación de
-30°C y el otro evaporador suministra 200 kW a 4°C. La temperatura de condensación
puede considerarse igual a 40°C
1. Calcular la potencia de compresión requerida y la eficiencia energética del ciclo.
2. Se sustituye el ciclo anterior por un ciclo con dos evaporadores y dos compresores
(ver figura), si se pretende suministrar la misma potencia frigorífica en ambos
evaporadores con las mismas temperaturas de evaporación y la misma temperatura
de condensación. Calcular la potencia de compresión requerida y la eficiencia
energética del ciclo.
3 4
1 2
Evaporador
de alta
Evaporador
de baja
Condensador
6 5
8 7
Nota: Suponer que no existen pérdidas de presión en los elementos del ciclo y que no
existe recalentamientos, ni subenfriamientos.
Solución:
1) A continuación se muestra un esquema de un sistema de refrigeración con dos
evaporadores y un compresor.
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Problemas Resueltos - Tecnología Frigorífica
5 8 1 2
7
Evaporador
Evaporador
de alta
de baja
Condensador
4 6 &
mRA
&
mRB
3
El diagrama p-h de este ciclo es el siguiente:
p (kPa)
3 2
6 7
4 5 1 8
Conocidas las temperaturas de evaporación de cada uno de los evaporadores y la
temperatura de condensación conocemos las entalpías de los siguientes puntos:
h3 = h4 = h6 = 386.43 kJ / kg h5 = 1422.46 kJ / kg
h7 = h8 = 1464.80 kJ / kg
Realizando balances de energía en ambos evaporadores podemos obtener los
caudales que circula por cada uno de ellos:
qfB 180 kW
qfB = mRB (h5 − h4 );
& &
mRB = = = 0.1737 kg / s
(h5 − h4 ) (1422.46 − 386.43 kJ / kg)
qfA 200 kW
qfA = mRA (h7 − h6 );
& &
mRA = = = 0.1855 kg / s
(h7 − h6 ) (1464.8 − 386.43 kJ / kg)
Por tanto el caudal total que debe mover el compresor y su entalpía se obtienen de
un sencillo balance de masa y energía en la mezcla de las dos corrientes:
& & &
mR = mRA + mRB = 0.1855 + 0.1737 kg / s = 0.3592 kg / s
& &
mRAh8 + mRBh5 0.1855·1464.8 + 0.1737·1422.46
& & &
mRh1 = mRAh8 + mRBh5 ; h1 = = = 1444.3 kJ / kg
&R
m 0.3592
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Problemas Resueltos - Tecnología Frigorífica
Si buscamos en el diagrama el punto 1 y seguimos su línea de entropía constante
hasta la presión de descarga tendremos: h2 = 1862.5 kJ / kg
El trabajo de compresión será:
Pc = mR (h2 − h1 ) = 0.3592 kg / s(1862.5 − 1444.3 kJ / kg) = 150.217 kW
&
El coeficiente de eficiencia energética del ciclo:
q + qfB 200 + 180 kw
COP = fA = = 2.5297
Pc 150.217 kw
2) Para el segundo sistema, el diagrama P-h será el que se muestra a
continuación:
3 4
1 2 &
mRC
Evaporador
de alta
&
mRB &
mRA
Evaporador
de baja
Condensador
6 5
8 7
5 4
P (kPa)
7 6 3
2
8 1
h (kJ/kg)
Conocidas las temperaturas de evaporación de cada uno de los evaporadores y la
temperatura de condensación conocemos las entalpías de los siguientes puntos:
h5 = h6 = 386.43 kJ / kg h3 = 1464.8 kJ / kg
h7 = h8 = 218.3 kJ / kg h1 = 1422.46 kJ / kg
h2 = 1617.04 kJ / kg h4 = 1627.0 kJ / kg
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Problemas Resueltos - Tecnología Frigorífica
Realizando un balance de energía en cada uno de los evaporadores obtenemos
qfB 180 kW
qfB = mRB (h1 − h8 );
& mRB =
& = = 0.1495 kg / s
(h1 − h8 ) (1422.46 − 218.3 kJ / kg)
qfA 200 kW
qfA = mRA (h3 − h6 );
& &
mRA = = = 0.1855 kg / s
(h3 − h6 ) (1464.8 − 386.43 kJ / kg)
Realizando un balance de energía sobre el enfriador intermedio:
mRBh2 + (mRC − mRA )h6 = mRBh7 + (mRC − mRA )h3
& & & & & &
Despejando el caudal de refrigerante por el compresor de alta:
& m h − mRBh2 + mRA (h6 − h3 ) 0.1495(218.3 − 1617.04) + 0.1855(386.43 − 1464.8)
& & &
mRC = RB 7 =
(h6 − h3 ) (386.43 − 1464.8)
&
mRC = 0.3794 kg / s
El trabajo de compresión será:
Pc = mRB (h2 − h1 ) + mRC (h4 − h3 ) = 0.1495(1617.04 − 1422.46 ) + 0.3794(1627.0 − 1464.8)
& &
Pc = 90.628 kW
El coeficiente de eficiencia energética del ciclo:
q + qfB 200 + 180 kw
COP = fA = = 4.193
Pc 90.628 kw
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Problemas Resueltos - Tecnología Frigorífica
Problema 5
Ciclos múltiples de compresión mecánica
Un ciclo estándar de compresión mecánica simple utiliza R-22 como refrigerante. La
capacidad frigorífica del evaporador es 180 kW a una temperatura de -30°C. La presión de
condensación del refrigerante es 1533,52 kPa. Más tarde el ciclo es revisado para funcionar
con los mismos parámetros pero siguiendo los esquemas (A) y (B), en ambos casos la
presión del deposito intermedio es 497.59 kPa.
Calcular la potencia de compresión necesaria y el COP para el ciclo simple y para las dos
configuraciones de ciclo compresión múltiple propuestas:
4 4
5
Condensador Condensador
3 3
5 2 6 2
6 7
8
1 1
7 9
Evaporador Evaporador
(A) (B)
Nota: Suponer que no existen pérdidas de presión en los elementos del ciclo, que no existe
recalentamientos, ni subenfriamientos en los evaporadores y condensadores y que los
compresores son ideales.
Solución:
Ciclo de compresión mecánica simple
Si trasladamos las temperaturas de evaporación y la presión de condensación sobre
el diagrama P-h del R-22, obtenemos los siguientes valores
t evap = −30°C p evap = 163.48 kPa
p cond = 1533.52 kPa t cond = 40°C
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Problemas Resueltos - Tecnología Frigorífica
Entalpías: h1 = 393.15 kJ / kg h3 = h4 = 249.67 kJ / kg
Compresor
p (kPa) 1 2
Evaporador
3 2
Condensador
4 1
4 3
Válvula de
h (kJ/kg) expansión
Sí el compresor es ideal la entalpía del punto 2 será: s1 = s2 = 1.8034 kJ / kg·K
h2 = 451 kJ / kg
Calculemos el caudal de refrigerante a partir del balance en el evaporador:
qf 180 kW
qf = mR (h1 − h4 ); mR =
& & = = 1.2545 kg / s
(h1 − h4 ) 393.15 − 249.67 kJ / kg
Trabajo de compresión:
Pc = mR (h2 − h1 ) = 1.2545 kg / s(451 − 393.15 kJ / kg) = 72.573 kW
&
qf 180 kW
Coeficiente de eficiencia energética: COP = = = 2.480
Pc 72.573 kW
Ciclo (A):
4
Condensador
3
p (kPa)
5 4
6 3 2
5 2
6
7 1
1
7
Evaporador h (kJ/kg)
(A)
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Problemas Resueltos - Tecnología Frigorífica
La presión de intermedia a la opera el depósito es pint = 497.59 kPa que
corresponde con una temperatura de cambio de fase de t int = 0°C
Las entalpías de los nuevos puntos, suponiendo los procesos de compresión
isentrópicos son las siguientes h2 = 419.95 kJ / kg h3 = 405.37 kJ / kg
h4 = 433.43 kJ / kg
Realizando un balance de energía sobre el evaporador obtenemos el caudal que
circula por el mismo, y por el compresor de baja presión. Como el salto de entalpía
es el mismo que en el caso del ciclo simple y demandamos la misma potencia
frigorífica, el caudal debe ser el mismo qf = mR ,B (h1 − h8 ); mR ,B = 1.2545 kg / s .
& &
Realizando un balance de energía en el depósito intermedio obtenemos el caudal
que circula por el compresor de alta:
mR ,B h2 + (mR ,A − mR ,B ) h6 = mR ,A h3 ;
& & & &
& & h − h6 419.95 − 249.67 kJ / kg
mR ,A = mR ,B 2 = 1.2545 kg / s = 1.372 kg / s
h3 − h6 405.37 − 249.67 kJ / kg
Por tanto, el trabajo de compresión será:
Pc = mR ,B (h2 − h1 ) + mR ,A (h4 − h3 )
& &
Pc = 1.2545 kg / s(419.95 − 393.15 kJ / kg) + 1.372 kg / s(433.43 − 405.37 kJ / kg) = 72.119 kW
qf 180 kW
Coeficiente de eficiencia energética: COP = = = 2.496
Pc 72.119 kW
Ciclo (B):
4
5
Condensador
3
p (kPa)
6 4 5 2
8 7 3
6 2
7
8 9 1
1
9
Evaporador
h (kJ/kg)
(B)
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Problemas Resueltos - Tecnología Frigorífica
Al igual que el caso (A), la presión de intermedia a la opera el depósito es
pint = 497.59 kPa que corresponde con una temperatura de cambio de fase de
t int = 0°C
Las entalpías de los puntos, suponiendo los procesos de compresión isentrópicos
son las siguientes:
h1 = 393.15 kJ / kg h2 = 451 kJ / kg h3 = 405.37 kJ / kg h4 = 433.43 kJ / kg
h6 = h7 = 249.67 kJ / kg h8 = h9 = 200 kJ / kg
Realizando un balance de energía sobre el evaporador obtenemos el caudal que
circula por el mismo, y por el compresor de baja presión (le llamaremos así aunque
realmente no sea de baja):
qf 180 kW
qf = mR ,B (h1 − h9 );
& mR ,B =
& = = 0.932 kg / s
(h1 − h9 ) (393.15 − 200 kJ / kg)
Realizando un balance de energía en el depósito intermedio obtenemos el caudal
que circula por el compresor de alta:
(m
& R ,A + mR ,B ) h7 = mR ,A h3 + mR ,B h8 ;
& & &
& & h − h8 249.67 − 200 kJ / kg
mR ,A = mR ,B 7 = 0.932 kg / s = 0.297 kg / s
h3 − h7 405.37 − 249.67 kJ / kg
Por tanto, el trabajo de compresión será:
Pc = mR ,B (h2 − h1 ) + mR ,A (h4 − h3 )
& &
Pc = 0.932 kg / s(451 − 393.15 kJ / kg) + 0.297 kg / s(433.43 − 405.37 kJ / kg) = 62.25 kW
qf 180 kW
Coeficiente de eficiencia energética: COP = = = 2.892
Pc 62.25 kW
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Problemas Resueltos - Tecnología Frigorífica
Problema 6
Ciclo simple de compresión mecánica (compresores alternativos)
Se necesita evacuar 150.000 kcal/h de cierta cámara frigorífica, para lo que se decide
instalar un sistema de producción de frío por compresión mecánica. La temperatura de la
cámara no puede superar los –3°C y el la diferencia de temperaturas a la entrada del
evaporador se estima en 7°C. Se dispone de un gran caudal de agua de pozo a 15°C que
piensa utilizarse como agente condensante. El fluido frigorígeno empleado es R-134a.
Para el funcionamiento de dicha instalación se adquirió un compresor alternativo de 2.250
cm³ de cilindrada, el cual aspira vapor con un recalentamiento en la tubería de aspiración de
10°C. Este compresor gira a 850 r.p.m. y su rendimiento volumétrico es de 0,8 para una
relación de compresión de 3,3.
Calcular:
El grado de subenfriamiento del fluido condensado para que pueda funcionar la
instalación con este compresor y si es posible su realización.
Nota: Considerar un salto máximo admisible en el agua de pozo de 5°C y un salto mínimo
de temperaturas en el condensador (entre fluido refrigerante y agua de pozo) de 5°C.
Solución:
a. Cálculo del grado de subenfriamiento
La temperatura del aire en la cámara debe ser inferior a –3°C, por lo tanto
Compresor
1 2 &
mw t w,ent
Evaporador
&
ma t a,ent
t a,sal
Condensador
t w,sal
4 3
Válvula de expansión
podemos suponer que esta es la temperatura de entrada del aire al evaporador.
t a,ent = −3°C . Luego si el salto a la entrada al evaporador debe ser de 7°C la
temperatura de evaporación será:
∆t = t a,ent − t evap = 7°C; t evap = t a,ent − ∆t = −3 − 7 = −10°C
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Problemas Resueltos - Tecnología Frigorífica
En cuanto al condensador la temperatura de entrada al condensador del agua de
pozo es de 15°C, el salto máximo en esta agua es de 5°C luego la temperatura del
agua a la salida del condensador será, t w,sal = 15 + 5 = 20°C y la temperatura de
condensación 5°C por encima de la temperatura más alta alcanzada en el
condensador:
∆t = t cond − t w,sal = 5°C; t cond = t w,sal + ∆t = 20 + 5 = 25°C
Con estos datos intentaremos dibujar el ciclo sobre un diagrama p-h de R-134a,
aunque desconocemos las entalpías de los puntos 3 y 4, ya que estas dependen del
grado de subenfriamiento que es nuestra incógnita. El punto 1 (entrada al
compresor) se encuentra a la presión de evaporación y sobre la isoterma de 0°C (-
10°C + 10°C).
Los valores de las entalpías de los diferentes puntos son:
h1 = 400.05 kJ / kg h5 = 391.32 kJ / kg
h2 = 425.98 kJ / kg h3 = h4 = ? kJ / kg
v1 = 0.10397 m³ / kg
p (kPa)
3 25°C
2
-10°C
1
4 5
h (kJ/kg)
La potencia frigorífica que debe suministrar este sistema es de 150000 kcal/h =
174.42 kW.
Con los datos físicos del compresor podemos calcular el desplazamiento volumétrico
de este:
&
Vt = Vcilindro ω = 2250 cm³ 850 rev / min = 0.0319 m³ / s
Para este caso la relación de presiones es 3.31 podemos decir que rendimiento
volumétrico del compresor va a ser aproximadamente 0.8.
&
VR1
ηvol = ; & &
VR1 = ηvol Vt = 0.8 0.0319 m³ / s = 0.0255 m³ / s
V&
t
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Problemas Resueltos - Tecnología Frigorífica
Siendo el volumen específico sobre el punto 1 v1 = 0.10397 m³ / kg , tendremos un
caudal másico:
V& 0.0255 m³ / s
&
mR = R1 = = 0.245 kg / s
v1 0.10397 m³ / kg
La potencia frigorífica sobre el evaporador es:
qf = mR (h5 − h4 ) ;
&
qf 174.42 kW
despejando: h4 = h5 − = 391.32 kJ / kg − = −320.6 imposible sería
&R
m 0.245 kg / s
necesario subenfriar el líquido a menos cientos de grados.
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Problemas Resueltos - Tecnología Frigorífica
Problema 7
Ciclo simple de compresión mecánica (compresores alternativos)
Los datos de catálogo de un compresor son los siguientes:
Refrigerante: R-22
Número de cilindros: 6
Velocidad de giro: 1740 r.p.m.
Diámetro del cilindro: 67 mm
Carrera: 57 mm
Porcentaje de espacio muerto: 4.8 %
Para las siguientes condiciones de operación:
Temperatura de evaporación: 5°C
Temperatura de condensación: 50°C
Subenfriamiento del líquido: 3°C
Recalentamiento del vapor: 8°C
La potencia frigorífica que indica el catálogo es 96.4 kW y la potencia absorbida 28.9 kW.
Calcular:
El rendimiento volumétrico teórico, el rendimiento volumétrico real y el rendimiento
isentrópico o de la compresión.
Solución:
La siguiente figura muestra el diagrama P-h del problema indicado con
anterioridad.
P (kPa)
3 50°C
2
5°C
1
4 5
h (kJ/kg)
Rendimiento volumétrico teórico:
v
ηvol,t = 1 − C suc − 1
v
des
26
Problemas Resueltos - Tecnología Frigorífica
El factor de espacio muerto o factor de huelgo C=0.048, y los volúmenes
específicos en la succión y la descarga:
v suc = v1 = 43.2 l / kg v des = v 2 = 14.13 l / kg
43.2 l / kg
ηvol,t = 1 − 0.048
14.13 l / kg − 1 = 0.9012
Desplazamiento volumétrico del compresor:
2
& πDc
Vt = Nc ω Lc
4
Donde:
Número de cilindros: Nc = 6
Velocidad de giro: ω = 1740 r.p.m. = 29 rev / s
Diámetro de cilindros: Dc = 0.067 m
Carrera: L c = 0.057 m
π(0.067 m)
2
&
Vt = 6 · 29 rev / s 0.057 m = 0.035 m³ / s = 35 l / s
4
Los valores de las entalpías de los diferentes puntos son:
h1 = 413.1 kJ / kg h2 = 445.5 kJ / kg h3 = h4 = 259.1 kJ / kg
Balance de energía sobre el evaporador:
qf 96.4 kW
qf = mR (h1 − h4 )
& &
mR = = = 0.6260 kg / s
(h1 − h4 ) 413.1 − 259.1 kJ / kg
El caudal volumétrico de refrigerante a la entrada al compresor (punto 1) será el
siguiente:
& &
V1 = mR v1 = 0.6260 kg / s 43.2 l / kg = 27.042 l / s
Por lo tanto el rendimiento volumétrico real será:
V& 27.04 l / s
ηvol,r = 1 = = 0.773
&
Vt 34.97 l / s
El trabajo de compresión teórico o isentrópica podemos calcularlo como:
Pc,s = mR (h2 − h1 ) = 0.6260 kg / s(444.5 − 413.1 kJ / kg) = 19.656 kW
&
Y por lo tanto el rendimiento de compresión o rendimiento isentrópico valdrá:
Pc,s 19.656 kW
ηs = = = 0.6801
Pc,r 28.9 kW
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Problemas Resueltos - Tecnología Frigorífica
Problema 8
Ciclo simple de compresión mecánica (compresores alternativos)
Los datos de catálogo del compresor SP4L220E son los siguientes:
Refrigerante: R-134a
Desplazamiento volumétrico: 86.1 m³/h
Para las siguientes condiciones de operación:
Temperatura de evaporación: -10°C
Temperatura de condensación: 50°C
Subenfriamiento del líquido: 5°C
Recalentamiento del vapor: 10°C
La potencia frigorífica que indica el catálogo es 23.7 kW y la potencia absorbida 10.0 kW.
Calcular:
La potencia frigorífica, el trabajo de compresión y el coeficiente de eficiencia energética, si
pretendemos utilizar este compresor en un ciclo con las mismas temperaturas de
evaporación y compresión pero sin subenfriamiento del líquido ni recalentamiento del
vapor.
Solución:
La siguiente figura muestra el diagrama P-h (R-134a) del problema con
subenfriamiento y recalentamiento. El punto “2s” es el punto de salida de un
proceso de compresión isentrópico
p (kPa)
3 50°C 2s
2
-10°C
1
4
h (kJ/kg)
Los valores de las entalpías de los puntos que pueden obtenerse son:
h1 = 400.049 kJ / kg h2s = 441.196 kJ / kg h3 = h4 = 263.712 kJ / kg
Balance de energía sobre el evaporador:
qf 23.7 kW
qf = mR (h1 − h4 )
& &
mR = = = 0.1738 kg / s
(h1 − h4 ) 400.049 − 263.712 kJ / kg
28
Problemas Resueltos - Tecnología Frigorífica
El trabajo de compresión puede obtenerse como: Pc = mR (h2 − h1 )
&
Despejando de aquí la entalpía del punto “2” tendremos:
Pc 10.0 kW
h2 = h1 + = 400.049 kJ / kg + = 457.586 kJ / kg
&
mR 0.1738 kg / s
El ciclo sobre el cual queremos instalar nuestro compresor es el siguiente:
3’ 2s’
4’ 1’
h (kJ/kg)
Los valores de las entalpías de los puntos son:
h1' = 391.321 kJ / kg h2s' = 430.328 kJ / kg h3' = h4' = 271.418 kJ / kg
Al mantenerse la presión de succión y presión de descarga entre las cuales trabaja
el compresor, tenemos que la relación de presiones es la misma que en el caso
anterior y puede considerarse una buena hipótesis suponer que el rendimiento
volumétrico y de la compresión del compresor se mantienen.
& &
V1 mR v1 0.1738 kg / s 0.104 m³ / kg
Rendimiento volumétrico: ηvol = = = = 0.7558
&
Vt &
Vt 86.1 m³ / h · 1 / 3600
Luego el caudal de refrigerante en el segundo caso será:
&
Vη 0.0239 m³ / s 0.7558
m′ = t vol =
&R = 0.1825 kg / s
v1' 0.09899 m³ / kg
La potencia frigorífica:
q′f = m′ (h1' − h4' ) = 0.1825 kg / s(391.321 − 271.418 kJ / kg) = 21.880 kW
&R
Del otro parámetro que podemos considerar constante, el rendimiento isentrópico o
rendimiento de la compresión, podemos obtener al trabajo absorbido por el
compresor en la segunda situación:
P m (h − h1 ) 0.1738 kg / s(441.196 − 400.049 kJ / kg)
&
ηc = c,s = R 2s = = 0.7151
Pc,r Pc,r 10.0 kW
′
Pc,s m′ (h2s′ − h1′ ) 0.1825 kg / s(430.328 − 391.321 kJ / kg)
&R
′
Pc,r = = = = 9.955 kW
ηc ηc 0.7151
Los valores del coeficiente de eficiencia energética, vale en ambos casos:
q 23.7 kW q′ 21.880 kW
COP = f = = 2.37 COP′ = f = = 2.198
Pc 10.0 kW ′
Pc 9.955 kW
29
Problemas Resueltos - Tecnología Frigorífica
30
Problemas Resueltos - Tecnología Frigorífica
Problema 9
Ciclo simple de compresión mecánica (compresores alternativos)
Un compresor hermético alternativo de 4 cilindros para R-22, tiene una velocidad de giro
de 29 rev/s. El diámetro de los cilindros es 87 mm y la carrera 70 mm. El rendimiento
volumétrico ha sido obtenido experimentalmente en función de la relación de compresión
(rc):
2
ηvol = 0.0016 rc − 0.0734 rc + 1.0117
Sí suponemos que la temperatura de condensación es constante e igual a 40°C, calcular la
potencia frigorífica para las siguientes temperaturas de evaporación: -20°C, -10°C, 0°C,
10°C
Nota: Suponer ciclo estándar sin sobrecalentamiento ni subenfriamiento
Solución:
El desplazamiento volumétrico del compresor será
& πD2
Vt = Nc ω c
Lc
4
Donde:
Número de cilindros: Nc = 4
Velocidad de giro: ω = 29 rev / s
Diámetro de cilindros: Dc = 0.087 m
Carrera: L c = 0.070 m
π(0.087 m)
2
&
Vt = 4 · 29 rev / s 0.070 m = 0.0483 m³ / s = 173.775 m³ / h
4
Comencemos por el primer caso t evap = −20°C
Dibujando el ciclo sobre un diagrama P-h obtenemos:
p (kPa)
3 2
4 1
h (kJ/kg)
Presiones: p cond = 1534 kPa p evap = 245.4 kPa
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![Problemas Resueltos - Tecnología Frigorífica
p cond 1534 kPa
rc = = = 6.252
p evap 245.4 kPa
Luego el rendimiento volumétrico valdrá:
ηvol = 0.0016 (6.252 ) − 0.0734 (6.252 ) + 1.0117 = 0.6153
2
Podemos obtener por tanto el caudal de refrigerante para este caso:
&
m v η V & 0.6145 · 0.0483 m³ / s
&
ηvol = R 1 ; mR = vol t = = 0.3207 kg / s
&
V v 0.09263 m³ / kg
t 1
El incremento de entalpía en el evaporador vale:
∆hevap = h1 − h4 = 396.9 − 249.8 kJ / kg = 147.1 kJ / kg
Por tanto la potencia frigorífica será:
&
qf = mR ∆hevap = 0.3207 kg / s 147.1 kJ / kg = 47.18 kW
Realizando los mismos cálculos para las otras tres temperaturas de evaporación
obtenemos los siguientes resultados:
t evap (°C) p evap (kPa) ηvol &
mR (kg / s) ∆hevap (kJ / kg) qf (kW)
-20 245.4 0.6153 0.3207 147.1 47.18
-10 354.9 0.7243 0.536 151.3 81.09
0 498.1 0.8008 0.821 155.2 127.4
10 681.2 0.8545 1.189 158.7 188.8
200
180
160
qf [kW]
140
120
100
80
60
40
-20 -15 -10 -5 0 5 10
tevap [°C]
32
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Problemas Resueltos - Tecnología Frigorífica
Problema 10
Ciclo simple de compresión mecánica (evaporadores y condensadores)
Se dispone de una máquina para enfriamiento de agua condensada por aire que realiza un
ciclo simple de compresión mecánica, sin recalentamiento del vapor ni subenfriamiento del
líquido, utilizando R-22. Según los datos del fabricante sí a dicha maquina se le suministra
un caudal de agua a enfriar de 0.19 kg/s a una temperatura de entrada de 20°C, siendo la
temperatura del aire a la entrada al condensador 25°C y su caudal, forzado por un
ventilador, 5500 m³/h. Entonces, la potencia frigorífica desarrollada por la máquina en las
condiciones anteriores es 8 kW y la potencia absorbida por el compresor 1.5 kW, el U·A
del evaporador es 883 W/K, y las características de los compresores alternativos son las
siguientes:
Nº de compresores: 2
Diámetro: 5 cm
Carrera: 5 cm
Rendimiento volumétrico: 0.822
Velocidad de giro: 750 rev/min
Calcular:
Temperatura de salida del agua, temperatura de salida del aire, temperatura de evaporación
del refrigerante, temperatura de condensación del refrigerante.
Solución:
La siguiente figura muestra una enfriadora de agua condensada por aire.
Compresor
&
mw t w,ent 1 2
Evaporador
t w,sal
Condensador
4 3
Válvula de expansión
La potencia frigorífica evacuada por el evaporador de la máquina es 8 kW, luego:
8000
8 kW = qf = mw cp (t w,ent − t w,sal )
& t w,sal = 20°C − °C = 9.927°C
4180·0.19
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Problemas Resueltos - Tecnología Frigorífica
Si estudiamos el evaporador como un intercambiador con cambio de fase,
tendremos:
tw,ent
T (°C)
UA 883
t w,ent − t w,sal −
&
mw c p −
ε= =1−e =1−e 0.19·4180
= 0.671
t w.ent − t evap tw,sal
tevap
Despejando de la expresión anterior obtenemos la temperatura de evaporación del
t − t w,sal
refrigerante: t evap = t w,ent − w,ent = 4.988 °C ≈ 5°C
ε
A través de los datos del compresor puedo calcular el caudal de refrigerante:
&
VR ,1 π 0.052
& & πD2 1
ηvol =
VR ,1 = ηvol Vt = ηvol Nc
; c
L c ω = 0.822 2 0.05·750 = 2.0175·10 −3 m³ / s
V& 4 4 60
t
Para calcular el caudal másico de refrigerante será necesario conocer el volumen
específico a la entrada del compresor del R-22 como vapor saturado.
v1 = 0.04036 m3 / kg .
&
VR ,1
&
mR = = 0.05 kg / s
v1
Si obtenemos la potencia frigorífica a través de los datos del refrigerante, podremos
despejar la entalpía del punto 4 (entrada al evaporador) que es igual a la del punto
3 (salida del condensador) por ser el proceso de expansión isentálpico.
8 kW = qf = mR (h1 − h4 )
&
p (kPa)
Como la temperatura de
evaporación es 5°C, podemos 3 2
calcular la entalpía del punto 1:
h1 = 407.15 kJ / kg
qf
h4 = h1 − = 247.15 kJ / kg 4 1
&
mR
h (kJ/kg)
Si interpolamos en la curva de
líquido saturado obtendremos la temperatura de condensación asociada al punto 3:
t cond − 35 247.15 − 243.10
= ; t cond = 38.082 °C
40 − 35 249.67 − 243.10
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Problemas Resueltos - Tecnología Frigorífica
Para calcular la temperatura de salida del aire en el condensador será necesario
realizar un balance de energía sobre este:
qcond = mR (h2 − h3 ) = macp (t a,sal − t a,ent )
& &
En la ecuación anterior no conocemos ni la entalpía del punto 2, ni la temperatura
de salida del aire, pero podemos calcular la potencia evacuada en el condensador
indirectamente, sumando la potencia frigorífica y el trabajo de compresión.
qc = qf + Pc = Vaρ acp (t a,sal − t a,ent )
&
qf + Pc 8000 + 1500
t a,sal = t a,ent + = 25 + = 30.182°C
& ρ c
Va a p 5500 / 3600·1.2·1000
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Problemas Resueltos - Tecnología Frigorífica
Problema 11
Ciclo simple de compresión mecánica (evaporadores y condensadores)
Una máquina frigorífica basada en un ciclo estándar de compresión mecánica desarrolla
una potencia frigorífica de 5 kW. El fabricante suministra el coeficiente de eficiencia
energética (COP) de dicha máquina como una función de la temperatura de condensación
del refrigerante en °C, COP = 2.5 − 0.01 (t cond − 30) .
El condensador de dicha máquina frigorífica es un intercambiador de carcasa y tubo, un
paso por carcasa y dos por tubos (el refrigerante circula por la carcasa con un coeficiente de
película de 10000 W/m²K). El intercambiador dispone de 50 tubos en forma de U, tienen
una longitud total 3 m, un diámetro interior de 20 mm y un espesor 1 mm. Están fabricados
en acero inoxidable de k = 15.1 W/mK, pueden considerarse despreciables las resistencias
de ensuciamiento.
Se dispone de un caudal de 0.2 kg/s de agua a 20°C para evacuar el calor de condensación.
Determinar:
1. Temperatura de condensación del refrigerante y calor total intercambiado en el
condensador
2. Caudal de agua necesario para conseguir una temperatura de condensación del
refrigerante de 30°C.
Nota: Suponer que el coeficiente de película interior en los tubos es 113 W/m²K,
independiente de la velocidad del fluido por encontrarse este en régimen laminar
Solución:
Apartado 1:
La siguiente figura muestra un esquema del condensador enfriadora de agua
condensada por aire.
Refrigerante, tcond
&
mw t w,ent
t w,sal
Si expresamos el COP en función de los datos del problema (Potencia frigorífica) y
calor de condensación tendremos:
36
Problemas Resueltos - Tecnología Frigorífica
qf qf 1 1
COP = = qc = qf 1 + = qf 1 +
Pc qc − qf COP
2.5 − 0.01(t cond − 30)
La ecuación anterior contiene las dos incógnitas del apartado 1, necesitamos por
tanto otra ecuación que nos permita cerrar el problema. Realizando un balance de
energía sobre el condensador tendremos:
refrigerante
T (°C)
tcond
tsal
tent agua
Área (m²)
−
UA
& wcp (t w,sal − t w,ent ) = mwcp 1 − e m c
qc = m & & w p (t − t w,ent )
cond
Luego si igualo esta ecuación con la anterior tendré una sola ecuación con una sola
incógnita, la temperatura de condensación.
1 −
UA
= mwcp 1 − e m c (t cond − t w,ent )
&
qf 1 + & w p
2.5 − 0.01(t cond − 30)
Para poder resolver esta ecuación necesito conocer UA.
Cálculo de UA:
1 1
UA = = = 1046.9 W / K
1 ln(Dext / Dint ) 1 9.646E − 6 + 6.6972E − 6 + 9.389E − 4
+ +
hext A ext 2πkNtL t hint A int
Donde:
A ext = Nt πDextL t = 10.367 m²
A int = Nt πDintL t = 9.425 m²
Despejando de la ecuación inicial y resolviendo queda una ecuación cuadrática de la
que la única solución válida es:
t cond = 28.7°C
y el calor evacuado en el condensador:
1
qc = qf 1 + = 6.989 kW
2.5 − 0.01(t cond − 30)
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Problemas Resueltos - Tecnología Frigorífica
Apartado 2:
Sí la temperatura de condensación del refrigerante es de 30°C, podemos calcular el
calor de condensación:
1 1
qc = qf 1 + = qf 1 + = 7 kW
2.5 − 0.01(t cond − 30)
2.5
Utilizando la otra expresión del calor de condensación tendríamos que:
−
UA
& w c p 1 − e m c
qc = m
& w p (t − t w,ent )
cond
Una ecuación fuertemente no lineal que debe de resolverse de forma iterativa:
&
1ª Iteración, supongo mw = 0.2 kg / s , despejo &
mw = 0.2343 kg / s
& w = 0.2343 kg / s , despejo
2ª Iteración, supongo m &
mw = 0.2549 kg / s
.... &
mw = 0.2884 kg / s
38
Problemas Resueltos - Tecnología Frigorífica
Problema 12
Evaporadores – Válvulas de expansión
Una máquina frigorífica de amoniaco es utilizada para enfriar una corriente de 31.6 l/min
de agua a 15°C, el agua sale del evaporador a 10.43°C en condiciones de evaporador limpio
(máquina recién instalada).
Se supone que con el paso del tiempo aparecerá una resistencia de ensuciamiento en el lado
del agua de aproximadamente Rensu=0.001 m² K/W, el área exterior del evaporador, la que
está en contacto directo con el agua, es 10 m².
Calcular el cambio que se produce en la potencia de compresión y en coeficiente de
eficiencia energética, suponiendo:
1. La temperatura de condensación permanece constante e igual a 40°C.
2. El evaporador es inundado y la válvula de expansión mantiene la temperatura de
salida del agua (10.43°C).
3. El rendimiento de la compresión es 0.7
4. El coeficiente de intercambio global del evaporador limpio es UAlimpio= 800 W/K
Solución:
El caudal de agua a la entrada, suponiendo una densidad del agua de 1 kg/l, es:
&
mw = 31.6 l / min = 0.5267 kg / s .
La potencia frigorífica suministrada será:
qf = mw cp (t w,ent − t w,sal ) = 0.5267 kg / s 4.18 kJ / kg(15 − 10.43°C ) = 10.061 kW
&
Esta potencia frigorífica va a ser la misma en el caso sucio, puesto que la válvula de
expansión va a mantener la misma temperatura de salida del agua.
Si planteamos la ecuación de transferencia en el intercambiador:
−UA lim pio
t w,ent − t w,sal &
mw c p
ε= =1−e
t w,ent − t evap
tw,ent
T (°C)
tw,sal
tevap
Despejando de la ecuación anterior la temperatura de evaporación:
39
Problemas Resueltos - Tecnología Frigorífica
t w,ent − t w,sal 15°C − 10.43°C
t evap = t w,ent − −UA lim pio
= 15°C − −800 W / K
≈ 0°C
0.5267 kg / s 4180 J / kg
1−e
&
mw cp
1−e
Con esta temperatura de evaporación ya podemos dibujar el ciclo estándar de
compresión sobre un diagrama P-h del amoniaco (R-717):
P (kPa)
3 40°C 2s
0°C
4 1
h (kJ/kg)
Las entalpías de los puntos son:
h1 = 1460.66 kJ / kg h2s = 1646 kJ / kg h3 = h4 = 386.43 kJ / kg
Realizando un balance de energía en el lado del refrigerante del evaporador:
qf 10.061 kW
qf = mR (h1 − h4 ); mR =
& & = = 0.0094 kg / s
h1 − h4 1460 .66 − 386 .43 kJ / kg
De la definición de rendimiento de la compresión podemos obtener el trabajo real
de compresión:
P P m (h − h1 ) 0.0094 kg / s(1646 − 1460.66 kJ / kg)
&
ηc = c,s ; Pc = c,s = R 2s = = 2.480 kW
Pc ηc ηc 0.7
El coeficiente de eficiencia energética valdrá para el caso limpio:
qf 10.061 kW
COPlim pio = = = 4.057
Pc,lim pio 2.480 kW
Para el caso del intercambiador sucio, el primer paso es calcular el valor del nuevo
1 1
UA: UA sucio = = = 740.741 W / K
1 R 1 0.001 m²K / W
+ ensu +
UA lim pio A ext 800 W / K 10 m²
En este caso la temperatura de evaporación cambiará puesto que la válvula de
expansión mantiene la potencia frigorífica:
t w,ent − t w,sal 15°C − 10.43°C
t evap = t w,ent − −UA sucio
= 15°C − −740.741 W / K
≈ −1°C
&
mw cp 0.5267 kg / s 4180 J / kg
1−e 1−e
Las entalpías de los puntos son:
40
Problemas Resueltos - Tecnología Frigorífica
h1 = 1459.59 kJ / kg h2s = 1650.52 kJ / kg h3 = h4 = 386.43 kJ / kg
Realizando un balance de energía en el lado del refrigerante del evaporador:
Pf 10.061 kW
Pf = mR (h1 − h4 ); mR =
& & = = 0.0094 kg / s
h1 − h4 1459 .59 − 386.43 kJ / kg
De la definición de rendimiento de la compresión podemos obtener el trabajo real
de compresión:
P P m (h − h1 ) 0.0094 kg / s(1650.52 − 1459.59 kJ / kg)
&
ηc = c,s ; Pc = c,s = R 2s = = 2.557 kW
Pc ηc ηc 0.7
El coeficiente de eficiencia energética valdrá para el caso sucio:
qf 10.061 kW
COPsucio = = = 3.935
Pc,sucio 2.557 kW
41
Problemas Resueltos - Tecnología Frigorífica
Problema 13
Evaporadores - Psicrometría
Una cámara frigorífica para almacenamiento se mantiene a una temperatura de 10°C y una
humedad relativa del 80%. El caudal de aire sobre el evaporador es 30000 m³/h y la
temperatura del aire medida a la salida del evaporador es de 5°C. En estas condiciones la
instalación desarrolla una potencia frigorífica de 100 kW.
Calcular la cantidad de agua de condensado que se forma en el evaporador en una hora.
Solución:
El aire de entrada al evaporador se encuentra en las condiciones medias de la
cámara frigorífica, t ent = 10°C φ ent = 80% .
&
El caudal de aire a la entrada al evaporador es Vair = 30000 m³ / h = 8.333 m³ / s
Realizando un balance de energía sobre el caudal de aire del evaporador tendremos
que: qf = mair (ha,ent − ha,sal ) .
&
Si suponemos que el caudal volumétrico de aire ha sido medido a la entrada al
evaporador, podemos decir que su densidad a 10°C es aproximadamente 1.247
& &
kg/m³, y por lo tanto: mair = Vair ρ air = 8.333 m³ / s 1.247 kg / m³ = 10.391 kg / s .
Podemos discutir en este punto si este caudal es de aire seco o aire húmedo, pero
la diferencia entre ambos será tan pequeña que puede considerarse que ambos
valen lo mismo y son iguales al valor anterior.
Si colocamos sobre un diagrama psicrométrico del aire a presión atmosférica el
punto de entrada podremos leer en el eje de entalpías cual es la entalpía del aire a
la entrada: ha,ent = 26 kJ / kg a.s.
Por tanto podemos despejar del balance de energía anterior la entalpía a la salida
del evaporador:
q 100 kW
ha,sal = ha,ent − f = 26 kJ / kg a.s. − = 16.376 kJ / kg a.s.
& air
m 10.391 kg / s
Ahora podemos colocar el punto de salida del aire sobre el diagrama psicrométrico
en el punto de intersección entre la línea de entalpía igual a la anterior y de
temperatura seca igual a 5°C.
Si miramos en el eje de humedades absolutas obtenemos:
wa,ent = 0.006 kg H2O / kg a.s. wa,sal = 0.0045 kg H2O / kg a.s.
42
Problemas Resueltos - Tecnología Frigorífica
h (kJ/kg a.s.)
w (kg H2O/kg a.s.)
φ=80%
ha,ent
ha,sal
wa,ent
w a,sal
t a,sal t a,ent t (°C)
Por lo tanto el caudal de agua condensada será igual a la cantidad de agua perdida
por el aire en su paso por el evaporador:
magua = mair (w a,ent − w a,sal ) = 10.391 kg / s(0.006 − 0.0045 kg H2O / kg a.s.) = 0.0156 kg H2O / s
& &
En una hora la cantidad de condensado sería:
0.0156 kg / s · 3600 s = 56.16 kg H2O / h
43
Problemas Resueltos - Tecnología Frigorífica
Problema 14
Ciclo simple de compresión mecánica (evaporadores y condensadores)
Se dispone de una máquina para enfriamiento de agua condensada por aire que realiza un
ciclo simple de compresión mecánica, sin recalentamiento del vapor ni subenfriamiento del
líquido, utilizando R-22. Según los datos del fabricante si a dicha maquina se le suministra
un caudal de agua a enfriar de 0.20 kg/s a una temperatura de entrada de 20°C, la
temperatura de evaporación del refrigerante es 6°C y la potencia consumida por el
compresor 1.5 kW, con una temperatura de condensación de 40°C. El desplazamiento
volumétrico del compresor vale 9 m³/h y el rendimiento volumétrico 0.8
Calcular:
1. Potencia frigorífica, potencia evacuada por el condensador y U·A del evaporador.
2. Si suponemos que el caudal de agua desciende a 0.18 kg/s, y que la máquina
funciona con una válvula de expansión automática (mantiene la temperatura de
evaporación constante), calcular la nueva potencia frigorífica, potencia de
compresión y potencia evacuada por el condensador.
Nota: suponer que el U·A del evaporador es proporcional al caudal de agua elevado a 0.8 y
que el rendimiento volumétrico e isentrópico del compresor en el segundo apartado son los
mismos que los del primer apartado.
Solución:
1) La siguiente figura muestra una enfriadora de agua.
Compresor
&
mw t w,ent 1 2
Condensador
Evaporador
t w,sal
4 3
Válvula de expansión
Conocidas las temperaturas de evaporación y condensación del refrigerante
podemos localizar sobre un diagrama P-h alguno de los puntos del ciclo con las
siguientes entalpías:
h1 = 407.5 kJ / kg h4 = h3 = 249.67 kJ / kg h2s = 430.6 kJ / kg
44
Problemas Resueltos - Tecnología Frigorífica
P (kPa)
3 40°C 2s
2
6°C
1
4
h (kJ/kg)
A través de los datos del compresor puedo calcular el caudal de refrigerante:
&
VR ,1 mR v1
& η V &
& 0.8·9 / 3600 m³ / s
ηvol = = ; mR = vol t = = 0.0511 kg / s
&
Vt &
Vt v1 0.03915 m³ / kg
Realizando un balance sobre el evaporador del lado del refrigerante:
qf = mR (h1 − h4 ) = 0.0511 kg / s(407.5 − 249.67 kJ / kg) = 8.065 kW
&
La potencia evacuada por el condensador puede obtenerse de un balance de
energía sobre toda la máquina:
qc = qf + Pc = 8.065 + 1.5 kW = 9.565 kW
Si realizamos un balance sobre el evaporador pero desde el lado del agua, podemos
obtener la temperatura de salida del agua:
tw,ent
T (°C)
tw,sal
tevap
qf 8.065
qf = mw cp (t w,ent − t w,sal )
& t w,sal = t w,ent − = 20°C − °C = 10.353 °C
&
mw c p 4.18·0.20
Y planteando la ecuación de transferencia sobre el evaporador:
t w,ent − t w,sal 20 − 10.353 °C
qf = UA ∆t lm = UA = UA = UA 8.258 °C
t w,ent − t evap 20 − 6
ln ln
t w,sal − t evap 10.353 − 6
45
Problemas Resueltos - Tecnología Frigorífica
qf 8.065 kW
UA = = = 0.9766 kW / °C
∆t lm 8.258 °C
&
2) Para este apartado suponemos que el nuevo caudal de agua es mw = 0.18 kg / s ,
las temperaturas de entrada del agua y de evaporación son la misma ya que el
sistema esta controlado por una válvula de expansión automática que mantiene la
presión, y por tanto la temperatura, de evaporación constante.
Si el rendimiento volumétrico y el desplazamiento volumétrico no cambian, con la
misma temperatura de evaporación tendremos el mismo caudal del refrigerante:
&
VR ,1 mR v1
& η V &
& 0.8·9 / 3600 m³ / s
ηvol = = ; mR = vol t = = 0.0511 kg / s
&
Vt &
Vt v1 0.03915 m³ / kg
También se nos indica que el UA del evaporador es proporcional al caudal de agua
elevado a 0.8, por tanto, el ratio entre el nuevo UA y el antiguo es igual a la
relación de caudales elevada a 0.8:
0.8 0.8
UA nuevo mw,nuevo
& 0.18
=
m
= = 0.9192; UA nuevo = UA·0.9192 = 0.8977 kW / °C
UA &w 0.2
Podemos ahora obtener la potencia frigorífica a través de la expresión:
−
UA
0.8977
& w cp (t w,ent − t w,sal ) = mw cp 1 − e m c (t w,ent − t evap ) = 0.18·4.181 − e 0.18·4.18 (20 − 6 ) = 7.339 kW
& −
qf = m & w p
Con esta potencia frigorífica y realizando un balance del lado del refrigerante
obtenemos la entalpía de entrada al evaporador:
q 7.339 kW
qf = mR (h1 − h4 ); h4 = h1 − f = 407.5 kJ / kg −
& = 263.9 kJ / kg
&
mR 0.0511 kg / s
Esta entalpía corresponde a una temperatura de condensación de:
52°C − 50°C t cond − 50°C
= ; t cond = 50.464°C
266.05 − 263.25 kJ / kg 263.9 − 263.25 kJ / kg
Para obtener el trabajo de compresión debemos suponer que el rendimiento
isentrópico permanece constante desde el apartado 1:
Pc,s mR (h2s − h1 ) 0.0511 kg / s(430.6 − 407.5 kJ / kg)
&
ηc = = = = 0.7869
Pc Pc 1.5 kW
Para el caso nuevo:
h2s = 437.0 kJ / kg
Pc,s m (h − h1 ) 0.0511 kg / s(437.0 − 407.5 kJ / kg)
&
Pc = = R 2s = = 1.916 kW
ηc ηc 0.7869
Y la potencia evacuada en el condensador será:
qc = qf + Pc = 7.339 + 1.916 kW = 9.255 kW
46
Problemas Resueltos - Tecnología Frigorífica
Problema 15
Problema combinado de compresión múltiple y torre de refrigeración
La instalación frigorífica de la figura utiliza amoniaco como refrigerante, consta de dos
evaporadores que mantienen diferentes temperaturas de conservación en sendas cámaras
frigoríficas. Se conocen los siguientes datos:
Evaporador baja: Potencia frigorífica: qf ,B = 30 kW
Presión de evaporación del refrigerante: p evap,B = 119.46 kPa
Evaporador alta: Potencia frigorífica: qf ,A = 15 kW
Presión de evaporación del refrigerante: p evap,A = 190.11 kPa
Condensador: Presión refrigerante salida del condensador: p cond = 1554.89 kPa
Depósito intermedio:Presión del refrigerante: pint = 429.41 kPa
Torre de Refrig.: Temperatura del agua a la entrada a la torre: 35°C
Temperatura seca del aire exterior: t ext = 35°C
Temperatura de bulbo húmedo del aire exterior: t bh,ext = 25°C
Humedad relativa del aire a la salida de la torre: φ a,sal = 90%
Caudal de aire seco de entrada en torre: &
Va = 15700 m³ / h
Cercanía de la torre: 5°C
Compresores (ambos): Rendimiento isentrópico: 0.8
Se pide:
A. Dibujar un esquema del diagrama p-h del refrigerante con todos los puntos de la
figura colocados en él.
B. Calcular la potencia consumida por cada uno de los compresores y el COP de la
instalación.
C. Caudal de agua de la bomba del circuito de condensación.
D. Caudal de agua de reposición (evaporado) en la torre.
47
Problemas Resueltos - Tecnología Frigorífica
Compresor Torre de
de baja Compresor refrigeración
1 2 de alta
3 4
9 12
Evaporador
de baja 11
Evaporador Condensador
de alta
8 10
6 5
7
Deposito Bomba del circuito
intermedio de condensación
Nota: Suponer que no existen pérdidas de presión en los elementos del ciclo y que no
existe recalentamientos, ni subenfriamientos.
Solución:
Las temperaturas asociadas a las presiones de cambio de fase del amoniaco
mostradas en el enunciado son las siguientes:
p evap,B = 119.46 kPa → t evap,B = −30°C
p evap,A = 190.11 kPa → t evap,A = −20°C
pint = 429.41 kPa → t evap,B = 0°C
p cond = 1554.89 kPa → t cond = 40°C
A. Diagrama p-h de la instalación frigorífica
48
Problemas Resueltos - Tecnología Frigorífica
p (kPa)
5 40°C 4s 4
7 6 0°C 3 2s
2
10 -20°C 11
-30°C 9
12
8 1
h (kJ/kg)
B. Mirando en las tablas de líquido / vapor saturado del amoniaco (R-717)
conseguimos las entalpías de los siguientes puntos:
h7 = h8 = h10 = 200 kJ / kg h3 = 1460.66 kJ / kg h11 = h12 = 1436.51 kJ / kg
h5 = h6 = 386.43 kJ / kg h9 = 1422.46 kJ / kg
Calculemos los caudales de refrigerante a partir de los balances en los
evaporadores:
qf ,B 30 kW
qf ,B = mR ,B (h11 − h10 );
& &
mRB = = = 0.0243 kg / s
(h11 − h10 ) 1436.51 − 200 kJ / kg
qf ,A 15 kW
qf ,A = mR ,A (h9 − h8 );
& &
mRA = = = 0.0123 kg / s
(h9 − h8 ) 1422.46 − 200 kJ / kg
Mezcla de las corrientes 9 y 12, para obtener la corriente 1:
& &
mR ,Ah12 + mR ,Bh9
h1 = = 1427.2 kJ / kg
& &
mR ,A + mR ,B
Si seguimos la isentrópica que parte del punto 1 hasta la presión del deposito
intermedio obtenemos: h2s ≈ 1610 kJ / kg . El rendimiento isentrópico de la
compresión es 0.8 y por tanto:
h2s − h1 h2s − h1
ηs = 0.8 = ; h2 = h1 + = 1655.7 kJ / kg
h2 − h1 ηs
Realicemos ahora un balance de energía en el deposito para calcular el caudal de
refrigerante que circula por el compresor de alta y el condensador:
49
Problemas Resueltos - Tecnología Frigorífica
(m
& RA + mRB ) h2 + mRCond h6 = (mRA + mRB ) h7 + mRCond h3
& & & & &
h − h7
mRCond = (mRA + mRB ) 2
& & & = 0.0496 kg / s
h3 − h6
Si seguimos la isentrópica que parte del punto 3 hasta la presión de condensación
obtenemos: h4s ≈ 1647 kJ / kg . El rendimiento isentrópico de la compresión es 0.8 y
por tanto:
h4s − h3 h4s − h3
ηs = 0.8 = ; h4 = h3 + = 1693.6 kJ / kg
h4 − h3 ηs
Podemos ya calcular las potencias de compresión de ambos compresores:
Pc,B = (mRA + mRB )(h2 − h1 ) = 8.36 kW
& &
Pc,A = mRCond (h4 − h3 ) = 11.55 kW
&
qf ,B + qf ,A
Y el COP de la instalación: COP = = 2.26 .
PC,A + PC,B
C. Realizando un balance de energía sobre el condensador:
qcond = mw cp (t w,ent − t w,sal )
&
La temperatura de agua a la entrada al condensador es la de salida de la torre y
viceversa. Luego:
t w,sal = 35°C, cercanía = 5°C = t w,ent − t bh,ext ; t w,ent = t bh,ext + 5°C = 30°C
La potencia de condensación puede ser calculada de dos formas:
qcond = mRCond (h4 − h5 ) = qf ,A + qf ,B + PC,A + PC,B = 64.91 kW
&
Despejando del balance de energía en el condensador tenemos:
& qcond 64.91 kW
mw =
cp (t w,ent − t w,sal ) = 4.175 kJ / kg·K (35 − 30°C) = 3.11 kg / s
D. Las condiciones del aire a la entrada a la torre son las siguientes:
t ext = 35°C; t bh,ext = 25°C . Si miramos en el diagrama psicrométrico del aire
húmedo: ha,ext = ha,ent = 76 kJ / kg a.s.; w a,ent = 0.016 kg agua / kg a.s.
El caudal másico de aire seco a la entrada a la torre es:
& & 15700 m³ / h
mas = Vas ρ = 1.146 kg / m³ = 5 kg / s
3600 s / h
Todo el calor cedido por el condensador será absorbido por el aire exterior luego:
50
Problemas Resueltos - Tecnología Frigorífica
qcond
qcond = mas (ha,sal − ha,ent );
& ha,sal = ha,ent + = 89 kJ / kg a.s.
&
mas
Con esta entalpía y la humedad relativa del 90% podemos colocar sobre el
diagrama psicrométrico el punto de salida del aire:
t a,sal = 29°C; wa,sal = 0.023 kg agua / kg a.s.
La diferencias de humedades absolutas entre el aire a la salida y a la entrada nos
permite calcular la cantidad de agua evaporada en la torre que es a su vez igual al
caudal de agua que es necesario reponer:
mw,rep = mas (w a,sal − wa,ent ) = 0.035 kg agua / s = 126 kg agua / h
& &
ha,sal φ=90%
ha,ent w (kg H2O/kg a.s.)
w a,sal
h (kJ/kg a.s.)
wa,ent
t (°C) t bh,ent t a,ent
51
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Nonnegative Solutions to Linear Diophantine Equations in 2 Variables with Constraints
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$\begingroup$
I suppose this is more of an algorithms question, but given a linear Diophantine equation in 2 variables, how do I find a solution pair $(x, y)$ such that $x \geq 0, y \geq 0$, and $x + y$ is minimum of all possible nonnegative solutions?
I realize variations of this question have been asked before, but they either deal with the general case of the $n$ variable equations, or just wish to know if a solution exists, etc.
I want to know if there's an algorithmic way of computing such minimum nonnegative solutions?
EDIT: In fact, all I really want is the sum $x + y$ of these two nonnegative $x$ & $y$ if they exist; and the sum to be minimum.
Here's what I got:
$x = x_1 - \frac{rb}{d}$, and $y = y_1 + \frac{ra}{d}$, where $r \in \mathbb{Z}, \;\text{and}\; d = gcd(a, b)$.
This gives us the general solution. We want $x$, and $y$ to be nonnegative, so:
$$r \leq \frac{x_1d}{b},\;\; r \geq \frac{-y_1d}{a}$$
$$\lceil \frac{-y_1d}{a} \rceil \leq r \leq \lfloor \frac{x_1d}{b} \rfloor.$$
Also,
$$x + y = x_1 - rb/d + y_1 + ra/d = \frac{r}{d}(a - b) + (x_1 + y_1).$$
Combining the two,
$$\lceil \frac{-y_1d}{a} \rceil \frac{(a - b)}{d} + (x_1 + y_1) \leq x + y \leq \dots.$$
So the minimum sum is equal to $\lceil \frac{-y_1d}{a} \rceil \frac{(a - b)}{d} + (x_1 + y_1)$?
Unfortunately, it's not giving the correct answer on all tests.
EDIT 2: Big mistake on my part!!! I didn't check if $(a - b)$ is negative when multiplying!! It's correct now, apologies.
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edited Dec 18, 2017 at 13:59
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Suppose your linear diophantine equation is $ax+by=c$
Case 1: $a, b$ are both positive and are relatively prime. If a particular solution of the equation is given by $(x_0,y_0)$, then the general solution is $x=x_0-bt, y=y_0+at$.
This means that you can go from a solution to another solution either by decreasing the $x$-value by $b$ and increasing the $y$-value by $a$ (good for your optimization if $ab$).
So you should take the variable with the smaller coefficient to be as small as possible.
Other cases: If $a,b$ are positive and not relatively prime, divide the equation through by $\gcd(a,b)$ (assuming $c$ is divisible by this value--if not the equation has no solutions) and proceed as in case 1.
If $a$ and $b$ are of opposite signs, the solutions lie on a positive sloped line. Take the lowest point in the first quadrant (or on a nonnegative axis) that lands on the integer grid.
If $a$ and $b$ are both negative and $c$ is positive, there are no nonnegative integer solutions. If $a$ and $b$ are both negative and $c$ is negative, multiply the equation by $-1$ and proceed as in earlier cases.
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edited Dec 18, 2017 at 14:41
answered Dec 18, 2017 at 14:27
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11003 | https://vinequai.com/inversion | Introduction to inversion
Inversion is a simple geometric operation exchanging the inside and the outside of a circle or sphere.
Here is the definition for a sphere. Note that the same definition applies to circles. Given a circle or sphere of radius (r), inversion sends a point (P) to a point (P') where (OP \cdot OP' = r^2) and the two points lie on the same line going through the center of the inverting sphere. Note that points on the boundary are fixed. In its most basic definition, the inversion map is not defined at the center of the sphere. The definition can be extended to exchange the center with a point at infinity.
Inversion can also be given a more geometric definition which is shown in the visualization above. If the given point is in the interior of the sphere, we take a perpendicular plane to the point and intersect it with the sphere. The inversion maps to the unique point on the axial line that intersects the tangent plane there. If the point is on the outside of the sphere, we just reverse the process by first finding a tangent and then a perpendicular plane at the tangent point to arrive at a point on the axial line in the interior of the sphere.
From the visualization, if you select the sphere checkbox, we see a neat property of inversion: it maps spheres to spheres. It also takes circles to circles.
Applying inversion to the image of a point gives you back the original point. So you can think of the red point as being the inverted image of the blue point or vice versa.
Inversion may seem like a classical curiosity from acquity, but it is actually relevant for doing modern mathematics. For instance, inversions relate different models of hyperbolic geometry to each other and also represent isometries of hyperbolic space in the disk model.
Relationship to stereographic projection
Inversions also have a neat relationship with stereographic projection, which we looked at in another Vinequai post. Stereographic projection of a sphere from its north pole can be viewed as the restriction of an inversion of a containing sphere whose radius is twice as large. This is shown in the visualization above. You can view the red point as the inversion of the blue point with respect to the yellow sphere, or you can view it as the stereographic projection of the blue point with respect to the blue sphere.
What this means is that facts that we prove about inversion automatically apply to stereographic projection. For instance as we will show below, inversion maps circles to circles, which means that stereographic projection also maps circles to circles. We also saw this property of stereographic projection in the stereographic projection Vinequai post
Characterisation via orthogonal circles and spheres
Inversion also has an equivalent definition with respect to orthogonal spheres. It is the unique map that exchanges the inside of the sphere with the outside of the sphere and maps spheres that intersect orthogonally to themselves. This is a bit of a mouthful, so the visualization above can help show what I mean.
To see how this definition is connected to the original one. All orthogonal spheres passing through a given point in the inverting sphere share a unique point in common at the exterior of this sphere. This point at the exterior is the image of the given point.
The fact that the two definitions are the same boils down to the tangent secant theorem
This characterisation by orthogonal spheres is a useful one, as we can use it to prove interesting properties about inversion. For instance inversion preserves angles. For any point and any two direction verctors, we can find two orthogonal spheres that meet at that point along each of the direction vectors. By symmetry across the plane containing the centers of all orthogonal spheres, we see that the two orthogonal spheres meet at the same angle at the image point.
To see the fact that we can achieve any desired direction through the given point with an orthogonal sphere, you can use the visualization above. For shallow angles, we place the center of the orthogonal sphere near the midpoint of the red and blue points. For a steeper angle, we place the center farther away.
The orthogonal sphere characterisation can also be used to show that inversion takes circles to circles and spheres to spheres. Roughly speaking, since inversion preserves orthgonal spheres and preserves angles, it must take tangent spheres to tangent spheres.
For spheres that are not tangent, we can gradually blow them up or shrink them down until they are tangent and then apply the same reasoning.
All spheres passing through a point in the interior and orthogonal to the inverting sphere pass through a point on the exterior. This fact can be seen in the visualization above. These points are the inversion pair. Each orthogonal sphere (purple) intersects the inverting sphere (gold) in a circle at 90 degrees. The intrsection is indicated in green.
References
Material for this math demonstration comes from Chapters 1.2 and 2.2 of Bill Thurston's Three-Dimensional Geometry and Topology. This book is a treasure trove of beautiful geometric ideas and will be a rich source of inspiration for future Vinequai posts.
Have suggestions?
Send any ideas for visualizations that you would like to see to [email protected]. |
11004 | https://www.drdavidnaylor.net/uploads/5/2/9/6/52962947/chapter_2_part_4_hydrostatic_forced_on_curved_surfaces_d2l__2020_.pdf | MEC516/BME516: Fluid Mechanics I Chapter 2: Fluid Statics © David Naylor, 2019 Department of Mechanical & Industrial Engineering Hydrostatic Forces on Curved Surfaces • Hydrostatic Forces on Curved Surfaces • Theory (use of FBDs) • Two solved examples • Useful for the engineering design of: - Liquid containment structures (e.g. storage tanks, dams and levees) - Ships, submarine vehicles 2 Hydrostatic Forces on Curved Surfaces • Analysis is for liquids: - Pressure increases linearly with depth - Gauge pressure distribution for incompressible fluids: 𝑝𝑝= 𝛾𝛾ℎ - Recall that pressure acts normal to bounding surface • Goals of the analysis: 1. Calculate the resultant forces on the surface: 𝐹𝐹 𝐻𝐻, 𝐹𝐹 𝑉𝑉 2. Locate the line of action of each force 3 Hydrostatic Forces on Curved Surfaces • While possible, integrating the local pressure distribution would be laborious • Use a Free Body Diagram approach: I.
Isolate a section of fluid adjacent to the surface. Draw a Free Body Diagram II.
Decompose hydrostatic force into horizontal and vertical components III. Apply static equilibrium: ∑𝐹𝐹= 0 4 Free Body Diagram Hydrostatic Forces on Curved Surfaces • Example: Consider the liquid above the curved gate (AB) in the diagram Vertical Component: ∑𝐹𝐹 𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣= 0 𝐹𝐹 𝑣𝑣= 𝑊𝑊 1 + 𝑊𝑊 2 = γꓯ1 + γꓯ2 • Vertical component of the pressure force is equal to weight of the fluid above the surface • Note: FBD shows forces on the liquid • The hydrostatic force of the liquid on the gate 𝐹𝐹 𝑉𝑉 5 γ Hydrostatic Forces on Curved Surfaces Horizontal Component: ∑𝐹𝐹 ℎ𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜= 0 𝐹𝐹 𝐻𝐻= 𝐹𝐹 𝐴𝐴𝐴𝐴= γℎ𝐶𝐶𝐶𝐶𝐴𝐴𝐴𝐴𝐴𝐴 • Horizontal component of the pressure force equals the force on a projection of the curved surface into the vertical plane, AC • 𝐹𝐹 𝐻𝐻acts at the centre of pressure of the vertical face (AC) • Plane gate methods to calculate 𝐹𝐹 𝐻𝐻, 𝑦𝑦𝐶𝐶𝐶𝐶 6 Area, AAC Example Problem (Midterm 2015) 7 Water (ρ=998 kg/m3) is contained behind a semi-circular gate, AB with a radius of R=2.0m. The gate is hinged at point A. Neglect the weight of the gate.
Calculate: (a) The horizontal and vertical components of the hydrostatic force on gate (AB) per unit depth (into the page). Clearly indicate the directions of the forces.
(b) The horizontal force (FB) at point B needed to hold the gate in place (per unit depth) Figure 2.13 Example Problem (Midterm 2015) 8 Solution (a) I recommend drawing the hydrostatic pressure distribution on the gate • The direction of the forces on the gate: 𝐹𝐹 𝐻𝐻 𝐹𝐹 𝑣𝑣 𝑝𝑝 Example Problem (Midterm 2015) 9 (a) Free body diagram of the water adjacent to the gate • These are the forces on the water (opposite of forces on gate) • Static equilibrium in vertical direction 𝐹𝐹 𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣= 0 𝐹𝐹 𝑉𝑉= 𝑊𝑊= γꓯ 𝐹𝐹 𝑉𝑉= γꓯ= γ 𝜋𝜋𝑅𝑅2 𝑑𝑑 2 Example Problem (Midterm 2015) 10 𝐹𝐹 𝑉𝑉= γꓯ= γ 𝜋𝜋𝑅𝑅2 𝑑𝑑 2 Thus, the vertical force on the gate is: 𝐹𝐹 𝑉𝑉= 9790 𝑁𝑁 𝑚𝑚3 𝜋𝜋(2𝑚𝑚)2 1𝑚𝑚 2 = 61.51 𝑘𝑘𝑘𝑘↓ • 𝐹𝐹 𝑉𝑉acts in line with the weight at distance: 4𝑅𝑅 3𝜋𝜋= 4 (2.0𝑚𝑚) 3𝜋𝜋 = 0.8488 𝑚𝑚 4𝑅𝑅 3𝜋𝜋= 0.8488 𝑚𝑚 Example Problem (Midterm 2015) 11 • Static equilibrium in horizontal direction 𝐹𝐹 ℎ𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜= 0 𝐹𝐹 𝐻𝐻= 𝐹𝐹 𝐴𝐴𝐴𝐴 • Thus, 𝐹𝐹 𝐻𝐻is equal to the force on vertical plane surface AB 𝐹𝐹 𝐻𝐻= 𝐹𝐹 𝐴𝐴𝐴𝐴= γ ℎ𝐶𝐶𝐶𝐶𝐴𝐴𝐴𝐴𝐴𝐴 𝐹𝐹 𝐻𝐻= 9790 𝑁𝑁 𝑚𝑚3 2.0 𝑚𝑚4.0 𝑚𝑚2 𝐹𝐹 𝐻𝐻= 78.32 𝑘𝑘𝑘𝑘← Example Problem (Midterm 2015) 12 • Force 𝐹𝐹 𝐻𝐻acts in line with 𝐹𝐹 𝐴𝐴𝐴𝐴 • 𝐹𝐹 𝐴𝐴𝐴𝐴acts below the centroid of surface AB: 𝑦𝑦𝐶𝐶𝐶𝐶= −𝐼𝐼𝑥𝑥𝑥𝑥sin 𝜃𝜃 ℎ𝐶𝐶𝐶𝐶𝐴𝐴𝐴𝐴𝐴𝐴 𝐼𝐼𝑥𝑥𝑥𝑥= 𝑑𝑑(2𝑅𝑅)3 12 = 1𝑚𝑚(4𝑚𝑚)3 12 = 5.333 𝑚𝑚4 𝑦𝑦𝐶𝐶𝐶𝐶= −5.333 𝑚𝑚4 sin(90𝑜𝑜) 2.0𝑚𝑚(4.0 𝑚𝑚2) = −0.6667 𝑚𝑚 = 0.6667 m Example Problem (Midterm 2015) 13 (b) Force (FB) at point B needed to hold the gate in place • Free body diagram for the gate ∑𝑀𝑀 𝐴𝐴= 0 𝐹𝐹 𝐵𝐵2𝑅𝑅−𝐹𝐹 𝐻𝐻1.333𝑚𝑚−𝐹𝐹 𝑉𝑉0.8488𝑚𝑚= 0 𝐹𝐹 𝐵𝐵= 78.32 𝑘𝑘𝑘𝑘(1.333𝑚𝑚)+61.51𝑘𝑘𝑘𝑘(0.8488𝑚𝑚) 4.0 𝑚𝑚 𝐹𝐹 𝐵𝐵= 39.2 𝑘𝑘𝑘𝑘→ + Ans.
Example Problem 14 A liquid with specific weight (γ) is contained in a tank shown in the sketch. The tank has unit depth (into the page).
(a) Draw the hydrostatic pressure distribution on curved surface A-B (b) Derive expressions for the vertical and horizontal hydrostatic forces on curved surface A-B. Clearly indicate the directions of the forces Example Problem 15 Solution (a) hydrostatic pressure distribution on surface A-B 𝑝𝑝𝐵𝐵= 𝛾𝛾(ℎ−𝑅𝑅) 𝑝𝑝𝐴𝐴= 𝛾𝛾𝛾 Sketch must have: • 𝑝𝑝𝐴𝐴> 𝑝𝑝𝐵𝐵 • Arrow must be perpendicular to AB • Directions of the hydrostatic forces on gate AB: ℎ−𝑅𝑅 Example Problem 16 (b) Expressions for the vertical and horizontal hydrostatic forces • Free body diagram of the forces on the water (under the gate) 𝐹𝐹 𝐻𝐻 𝐹𝐹 𝐵𝐵𝐵𝐵 𝐹𝐹 𝑉𝑉 𝐹𝐹 𝐴𝐴𝐴𝐴 𝑊𝑊 Example Problem 17 • Static equilibrium: ∑𝐹𝐹 ℎ𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜= 0 𝐹𝐹 𝐻𝐻= 𝐹𝐹 𝐵𝐵𝐵𝐵= γℎ𝐶𝐶𝐶𝐶𝐴𝐴𝐵𝐵𝐵𝐵 • For surface BC: ℎ𝐶𝐶𝐶𝐶= ℎ− 𝑅𝑅 2 , 𝐴𝐴𝐵𝐵𝐵𝐵= 𝑅𝑅(1) • Horizontal force on surface AB is: 𝐹𝐹 𝐻𝐻= γ ℎ−𝑅𝑅 2 𝑅𝑅 ← 𝐹𝐹 𝐻𝐻 𝐹𝐹 𝑉𝑉 𝐹𝐹 𝐴𝐴𝐴𝐴 𝐹𝐹 𝐵𝐵𝐵𝐵 𝑊𝑊 ℎ−𝑅𝑅/2 CG Ans.
Example Problem 18 • Static equilibrium: ∑𝐹𝐹 𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣= 0 𝐹𝐹 𝑉𝑉= 𝐹𝐹 𝐴𝐴𝐴𝐴−𝑊𝑊 • For surface AC: 𝐹𝐹 𝐴𝐴𝐴𝐴= γℎ𝑅𝑅(1) • Weight of the water: 𝑊𝑊= γꓯ= γ 𝜋𝜋𝑅𝑅2 4 (1) • Vertical Force on surface AB is: 𝐹𝐹 𝑉𝑉= γ ℎ𝑅𝑅−γ 𝜋𝜋𝑅𝑅2 4 ↑ 𝐹𝐹 𝐻𝐻 𝐹𝐹 𝑉𝑉 𝐹𝐹 𝐴𝐴𝐴𝐴 𝐹𝐹 𝐵𝐵𝐵𝐵 𝑊𝑊 𝑝𝑝𝐴𝐴𝐴𝐴= γℎ Ans.
Pressure Distribution on a Curved Surface Which sketch of the pressure distribution is correct?
Answer: Distribution 2 What is wrong with the others?
19 Example Problem 20 The curved gate (AB) shown in the sketch has a radius of R=6m.
(a) Calculate the total hydrostatic force on the curved gate AB per meter of depth (into the page) (b) Find the line of action of the resultant force Watch the Video Solution END NOTES Presentation by Dr. David Naylor Department of Mechanical and Industrial Engineering Ryerson University, Toronto, Ontario Canada © David Naylor 2020. Please do not share these notes.
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11006 | https://trackingtime.co/best-practices/timeline-google-docs.html | CONTENT HUBS
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How to Create a Timeline Using Google Docs & Other Free Tools (+9 Free Templates)
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Everything is easier to understand when it’s visual—especially when it comes to schedules, deadlines, and timelines. In project management, a timeline is one of the most powerful tools for bringing visuals into the mix. It helps you organize tasks, track progress, and get a clear overview of your project’s status.
In this article, you’ll learn how to create your own timeline, plus we’ve gathered 9 ready-made templates for Google Docs, Google Sheets, and Microsoft Office—as well as options from other websites to make the process even simpler for you.
How to Create a Timeline
What Does a Timeline Include?
A timeline is a comprehensive tool that helps organize projects by incorporating several key components:
| Component | Description |
--- |
| ✅ Project Start and End Dates | Establish the overall duration of the project, providing a clear timeframe within which tasks must be completed. |
| ✅ List of Tasks | A detailed list of all activities to be completed during the project, ensuring that nothing is overlooked. |
| ✅ Task Start and End Dates | Each task is assigned specific start and end dates, helping team members know when to start and finish. |
| ✅ Task Durations | The time allocated for each task, helping project managers plan the timeline effectively. |
| ✅ Milestones | Significant events or achievements marking important stages of progress, often acting as checkpoints. |
| ✅ Dependencies | Identifying task relationships, particularly which tasks rely on others to be completed first. |
Different Types of Timelines
Timelines can take various formats, such as Gantt charts, flowcharts, or simple lists, depending on the project’s complexity and team preferences.
Gantt Chart
A Gantt chart is a timeline commonly used in project management to visually represent tasks and their durations with bars. It’s especially helpful for complex projects, as it clearly shows the start and end dates for each task.
This visual format makes it easier to track task dependencies and overlaps.
Roadmap Timeline
Used to outline strategic plans and milestones over time, roadmap timelines are often applied in business development or product launches.
✔️ They offer a high-level view of key objectives and milestones, helping teams stay aligned with long-term goals.
Chronological Timeline
This type of timeline displays events in the order they occurred, making it ideal for historical or biographical contexts.
✔️ It’s straightforward and focuses on the sequence of events rather than their duration.
Interactive Timeline
Dynamic and online, interactive timelines are especially useful for educational or informational purposes, offering an engaging way to present complex sequences of events.
✔️ They allow users to explore detailed information about each event.
How to Make a Timeline on Google Docs
Creating a timeline in Google Docs is simple and effective. Just follow these four steps:
1. Open a New Document
2. Select “Drawing”
3. Add Shapes and Lines
4. Finalize the Timeline
Free, Ready-to-Use Timeline Templates
With so many ready-to-use templates available, creating a timeline is much easier than starting from scratch. Instead of making one yourself in Google Docs, we recommend using one of the templates below.
Here are 9 timeline template examples from various industries and types, each with its own unique strengths and uses. You can access or download them below:
Marketing Campaign Timeline Template
🔘 Main Use: Planning and executing marketing campaigns effectively.
🔘 Strengths: Helps organize phases like planning, execution, and evaluation, ensuring all marketing activities are well-aligned.
Check out Miro’s Marketing Campaign Timeline Template for a detailed example.
Annual Timeline Template
🔘 Main Use: This template is designed to track and manage projects over a 12-month period, providing a comprehensive overview of tasks and milestones throughout the year.
🔘 Strengths: It’s particularly effective for monitoring progress, setting realistic deadlines, and keeping all stakeholders updated on the project’s status.
Monthly Business Operations Timeline Template by JustFreeSlide
🔘 Main Use: Managing ongoing business activities.
🔘 Strengths: Helps track recurring tasks and ensures that all business operations run smoothly throughout the month.
Event Planning Timeline Template by Smartsheet
🔘 Main Use: This template helps align team members and track progress throughout event planning. It presents the event schedule, milestones, and goals to stakeholders.
🔘 Strengths: The template includes slides for 30-, 60-, and 90-day timelines, highlighting key activities and allowing color-coding for tasks.
Construction Project Timeline Template
🔘 Main Use: This template helps schedule deliverables for a construction project in chronological order, ensuring tasks and milestones are managed efficiently.
🔘 Strengths: It features editable tasks (e.g., demo prep, framing, electrical) that can be sequenced, and it visually tracks due dates, durations, and statuses.
Roadmap Timeline for Product Launch
🔘 Main Use: Outlining strategic milestones for launching new products.
🔘 Strengths: Aligns marketing, development, and sales teams with a clear roadmap of milestones leading up to the launch.
Manufacturing Timeline Template by SweetProcess
🔘 Main Use: Designed to streamline production planning, this template helps you plan and track all stages, from product conception to distribution.
🔘 Strengths: It outlines clear milestones for each production stage, ensuring smooth transitions. Additionally, it keeps all team members informed of their roles and deadlines.
Risk Management Timeline Template
🔘 Main Use: This template helps organizations prepare for, manage, and recover from crises by outlining key actions in a structured format.
🔘 Strengths: The Crisis Management Plan Timeline Template ensures a coordinated response to unexpected events by visually organizing phases such as preparation, immediate response, communication, and recovery.
Generic Weekly Timeline Template by OfficeTimeline
🔘 Main Use: General project management for any type of project.
🔘 Strengths: Offers a simple structure for organizing tasks each week, making it versatile for various types of projects.
Tips on Creating a Timeline
Creating an effective timeline requires careful planning and attention to detail. Here are some tips to help you develop a useful timeline:
FAQs
Why does every team need a timeline?
What are some common methods for creating a timeline?
What are the key elements of a timeline?
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How do I export and share my timeline?
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11007 | https://micro.magnet.fsu.edu/cells/microfilaments/microfilaments.html | | |
| | | | |
| | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- | | | | Galleria | | License Info | | Image Use | | Custom Photos | | Partners | | Site Info | | Contact Us | | Publications | | Home | | | | The Galleries: | | Photo Gallery | | Silicon Zoo | | Pharmaceuticals | | Chip Shots | | Phytochemicals | | DNA Gallery | | Microscapes | | Vitamins | | Amino Acids | | Birthstones | | Religion Collection | | Pesticides | | BeerShots | | Cocktail Collection | | Screen Savers | | Win Wallpaper | | Mac Wallpaper | | Movie Gallery | | | | | Microfilaments Common to all eukaryotic cells, these filaments are primarily structural in function and are an important component of the cytoskeleton, along with microtubules and often the intermediate filaments. Microfilaments range from 5 to 9 nanometers in diameter and are designed to bear large amounts of tension. In association with myosin, microfilaments help to generate the forces used in cellular contraction and basic cell movements. The filaments also enable a dividing cell to pinch off into two cells and are involved in amoeboid movements of certain types of cells. Microfilaments are solid rods made of a protein known as actin. When it is first produced by the cell, actin appears in a globular form (G-actin; see Figure 1). In microfilaments, however, which are also often referred to as actin filaments, long polymerized chains of the molecules are intertwined in a helix, creating a filamentous form of the protein (F-actin). All of the subunits that compose a microfilament are connected in such a way that they have the same orientation. Due to this fact, each microfilament exhibits polarity, the two ends of the filament being distinctly different. This polarity affects the growth rate of microfilaments, one end (termed the plus end) typically assembling and disassembling faster than the other (the minus end). Unlike microtubules, which typically extend out from the centrosome of a cell, microfilaments are typically nucleated at the plasma membrane. Therefore, the periphery (edges) of a cell generally contains the highest concentration of microfilaments. A number of external factors and a group of special proteins influence microfilament characteristics, however, and enable them to make rapid changes if needed, even if the filaments must be completely disassembled in one region of the cell and reassembled somewhere else. When found directly beneath the plasma membrane, microfilaments are considered part of the cell cortex, which regulates the shape and movement of the cell's surface. Consequently, microfilaments play a key role in development of various cell surface projections (as illustrated in Figure 2), including filopodia, lamellipodia, and stereocilia. Illustrated in Figure 2 is a fluorescence digital image of an Indian Muntjac deer skin fibroblast cell stained with fluorescent probes targeting the nucleus (blue) and the actin cytoskeletal network (green). Individually, microfilaments are relatively flexible. In the cells of living organisms, however, the actin filaments are usually organized into larger, much stronger structures by various accessory proteins. The exact structural form that a group of microfilaments assumes depends on their primary function and the particular proteins that bind them together. For instance, in the core of surface protrusions called microspikes, microfilaments are organized into tight parallel bundles by the bundling protein fimbrin. Bundles of the filaments are less tightly packed together, however, when they are bound by alpha-actinin or are associated with fibroblast stress fibers (the parallel green fibers in Figure 2). Notably, the microfilament connections created by some cross-linking proteins result in a web-like network or gel form rather than filament bundles. Over the course of evolutionary history of the cell, actin has remained relatively unchanged. This, along with the fact that all eukaryotic cells heavily depend upon the integrity of their actin filaments in order to be able to survive the many stresses they are faced with in their environment, makes actin an excellent target for organisms seeking to injure cells. Accordingly, many plants, which are unable to physically avoid predators that might want to eat them or harm them in some other way, produce toxins that affect cellular actin and microfilaments as a defensive mechanism. The death cap mushroom, for example, produces a substance called phalloidin that binds to and stabilizes actin filaments, which can be fatal to cells. BACK TO ANIMAL CELL STRUCTURE BACK TO PLANT CELL STRUCTURE Questions or comments? Send us an email.© 1995-2025 by Michael W. Davidson and The Florida State University. All Rights Reserved. No images, graphics, software, scripts, or applets may be reproduced or used in any manner without permission from the copyright holders. Use of this website means you agree to all of the Legal Terms and Conditions set forth by the owners.This website is maintained by ourGraphics & Web Programming Teamin collaboration with Optical Microscopy at theNational High Magnetic Field Laboratory.Last modification: Friday, Nov 13, 2015 at 02:18 PMAccess Count Since October 1, 2000: 492901Microscopes provided by: | |
11008 | https://artofproblemsolving.com/wiki/index.php/2008_AIME_II_Problems/Problem_15?srsltid=AfmBOorENreszkj86_Zgw9rCAvZmzGb3ikC_s1ByaCMfY5iuxajNzYvB | Art of Problem Solving
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2008 AIME II Problems/Problem 15
Contents
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1 Problem
2 Solution
2.1 Solution 1
2.2 Solution 2
2.3 Solution 3 (Big Bash)
3 Video Solution
4 See also
Problem
Find the largest integer satisfying the following conditions:
(i) can be expressed as the difference of two consecutive cubes;(ii) is a perfect square.
Solution
Solution 1
Write , or equivalently, .
Since and are both odd and their difference is , they are relatively prime. But since their product is three times a square, one of them must be a square and the other three times a square. We cannot have be three times a square, for then would be a square congruent to modulo , which is impossible.
Thus is a square, say . But is also a square, say . Then . Since and have the same parity and their product is even, they are both even. To maximize , it suffices to maximize and check that this yields an integral value for . This occurs when and , that is, when and . This yields and , so the answer is .
Solution 2
Suppose that the consecutive cubes are and . We can use completing the square and the first condition to get: where and are non-negative integers. Now this is a Pell equation, with solutions in the form . However, is even and is odd. It is easy to see that the parity of and switch each time (by induction). Hence all solutions to the first condition are in the form: where . So we can (with very little effort) obtain the following: . It is an AIME problem so it is implicit that , so . It is easy to see that is strictly increasing by induction. Checking in the second condition works (we know is odd so we don't need to find ). So we're done.
Solution 3 (Big Bash)
Let us generate numbers to for the second condition, for squares. We know for to be integer, the squares must be odd. So we generate . cannot exceed since it is AIME problem. Now take the first criterion, let be the smaller consecutive cube. We then get:
Now we know either or must be factor of , hence or. Only satisfy this criterion. Testing each of the numbers in the condition yields as the largest that fits both, thus answer .
Video Solution
2008 AIME II #15
MathProblemSolvingSkills.com
See also
2008 AIME II (Problems • Answer Key • Resources)
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11009 | https://www.sciencedirect.com/science/article/pii/S1936879819319405 | Effect of Oxygen Therapy on Cardiovascular Outcomes in Relation to Baseline Oxygen Saturation - ScienceDirect
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JACC: Cardiovascular Interventions
Volume 13, Issue 4, 24 February 2020, Pages 502-513
Coronary
Effect of Oxygen Therapy on Cardiovascular Outcomes in Relation to Baseline Oxygen Saturation
Author links open overlay panelStefan K.James MD, PhD a b, David Erlinge MD, PhD c, Johan Herlitz MD, PhD d, Joakim Alfredsson MD, PhD e, Sasha Koul MD, PhD c, Ole Fröbert MD, PhD f, Thomas Kellerth MD f, Annica Ravn-Fischer MD, PhD g, Patrik Alström MD h, Ollie Östlund PhD b, Tomas Jernberg MD, PhD i, Bertil Lindahl MD, PhD a b, Robin Hofmann MD, PhD h
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Routine Supplementary Oxygen for Myocardial Infarction
JACC: Cardiovascular Interventions, Volume 13, Issue 4, 24 February 2020, Pages 514-516
Joseph Fisher, Duminda N. Wijeysundera
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Abstract
Objectives
The aim of this study was to determine the effect of supplemental oxygen in patients with myocardial infarction (MI) on the composite of all-cause death, rehospitalization with MI, or heart failure related to baseline oxygen saturation. A secondary objective was to investigate outcomes in patients developing hypoxemia.
Background
In the DETO2X-AMI (Determination of the Role of Oxygen in Suspected Acute Myocardial Infarction) trial, 6,629 normoxemic patients with suspected MI were randomized to oxygen at 6 l/min for 6 to 12 h or ambient air.
Methods
The study population of 5,010 patients with confirmed MI was divided by baseline oxygen saturation into a low-normal (90% to 94%) and a high-normal (95% to 100%) cohort. Outcomes are reported within 1 year. To increase power, all follow-up time (between 1 and 4 years) was included post hoc, and interaction analyses were performed with oxygen saturation as a continuous covariate.
Results
The composite endpoint of all-cause death, rehospitalization with MI, or heart failure occurred significantly more often in patients in the low-normal cohort (17.3%) compared with those in the high-normal cohort (9.5%) (p<0.001), and most often in patients developing hypoxemia (23.6%). Oxygen therapy compared with ambient air was not associated with improved outcomes regardless of baseline oxygen saturation (interaction p values: composite endpoint, p=0.79; all-cause death, p=0.33; rehospitalization with MI, p=0.86; hospitalization for heart failure, p=0.35).
Conclusions
Irrespective of oxygen saturation at baseline, we found no clinically relevant beneficial effect of routine oxygen therapy in normoxemic patients with MI regarding cardiovascular outcomes. Low-normal baseline oxygen saturation or development of hypoxemia was identified as an independent marker of poor prognosis. (An Efficacy and Outcome Study of Supplemental Oxygen Treatment in Patients With Suspected Myocardial Infarction; NCT01787110)
Central Illustration
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Key Words
cardiovascular outcomes
myocardial infarction
oxygen therapy
randomized clinical trial
reactive oxygen species
Abbreviations and Acronyms
CI
confidence interval
HF
heart failure
IQR
interquartile range
MI
myocardial infarction
PCI
percutaneous coronary intervention
STEMI
ST-segment elevation myocardial infarction
After decades of routine oxygen therapy in patients with acute myocardial infarction (MI) without proof of improved cardiovascular outcomes (1), guidelines have recently been changed on the basis of results from the DETO2X-AMI (Determination of the Role of Oxygen in Suspected Acute Myocardial Infarction) trial, demonstrating no evidence that routine oxygen therapy was associated with reduced mortality in normoxemic patients (defined as oxygen saturation between 90% and 100%) with suspected MI 2, 3. Currently, initiation of oxygen treatment is recommended in hypoxemic patients (defined as oxygen saturation<90%) only (3). Some experts call for a change in clinical practice 4, 5, whereas others remain cautious, requesting more evidence (6). Controversy remains concerning the clinical risk/benefit ratio in subgroups of patients in the low-normal range (defined as oxygen saturation between 90% and 94%) compared with the high-normal range (defined as oxygen saturation between 95% and 100%) at baseline (7).
In the low-normal range, the ideal cutoff to initiate supplemental oxygen remains uncertain. Prophylactic oxygen therapy in this subgroup may prevent the development of hypoxemia and alleviate the mismatch of oxygen supply and demand in the ischemic myocardium. Consequently, infarct size may be reduced, minimizing complications such as heart failure (HF) and arrhythmias (1).
On the contrary, in the high-normal range, supplemental oxygen leads to hyperoxemia (higher than normal oxygen tension in the blood), which may cause detrimental effects due to vasoconstriction of the vasculature and generation of reactive oxygen species, potentially leading to toxic myocardial contractile dysfunction, increased oxidative stress, reperfusion injury, and ultimately myocardial cell death (8).
Thus, the aim of this subgroup analysis in patients with confirmed MI from the DETOX-AMI trial was to determine the effect of supplemental oxygen on the composite of all-cause death, rehospitalization with MI, or HF in relation to baseline oxygen saturation. A secondary objective was to investigate outcomes in patients developing hypoxemia during the intervention period.
Methods
Study design
The DETO2X-AMI trial was a nationwide, multicenter, open-label, registry-based randomized clinical trial(9) comparing routine supplemental oxygen therapy versus ambient air in patients with suspected MI (2). The trial was based on the SWEDEHEART (Swedish Web System for Enhancement and Development of Evidence-Based Care in Heart Disease Evaluated According to Recommended Therapies) (10) registry for patient recruitment, trial procedures, and follow-up.
The study design (11), methods, and first results have been described in detail previously 2, 12, 13. The trial was approved by the ethical review authority (Gothenburg DNR 287-12) and the medical products agency of Sweden (EudraCT 2013-002882-20). The academic research organization (trial administration, data management, and statistical analyses) was supplied by the Uppsala Clinical Research Center at Uppsala University, which also provides the infrastructure and service of the SWEDEHEART registry.
Patient population
Patients were evaluated for enrollment at first medical contact with the ambulance services, emergency departments, coronary care units, or catheterization laboratories of participating hospitals. Eligible patients≥30 years of age with oxygen saturation≥90% on pulse oximetry had to have typical symptoms suggestive of MI (defined as chest pain or dyspnea) for<6 h and either ischemic changes on electrocardiography (14) or elevated cardiac troponin on admission (above the locally defined decision limit for MI). To allow complete follow-up through the Swedish National Population Registry (15), only Swedish residents with unique personal identification numbers were enrolled.
Patients were excluded if they had ongoing oxygen therapy or cardiac arrest prior to enrollment. The overall study population included patients with suspected MI (n=6,629). One-fourth of these patients received discharge diagnoses other than acute MI(2).
In the present study, we assessed the subgroup with confirmed MI (n=5,010), both non–ST-segment elevation MI and ST-segment elevation MI (STEMI), which implies superior data completeness, including coronary angiography and intervention, and a more defined pathophysiological risk/benefit ratio.
Study procedures
After oral informed consent was obtained, eligible patients were randomly assigned to either oxygen therapy at 6 l/min for 6 to 12 h delivered by open face mask or ambient air (unrestricted 1:1 ratio). Randomization was performed online following a computer-generated list and a customized randomization module within the SWEDEHEART database. The allocated treatment was initiated directly after randomization.
Oral consent was confirmed by signature within 24 h. All patients were treated according to standard of care. Oxygen saturation was documented at the beginning and the end of the randomized treatment period. Hypoxemia developed commonly because of circulatory or respiratory failure. If supplemental oxygen outside the protocol was provided, a protocol violation including the cause was entered into the study database.
Endpoints and follow-up
The primary endpoint of the main trial was all-cause mortality within 1 year in the intention-to-treat population with suspected MI (2). In this subgroup analysis, the primary objective was to study the effect of supplemental oxygen on the composite of all-cause death, rehospitalization with MI, or HF within 1 year stratified according to baseline oxygen saturation (90% to 94% vs. 95% to 100%), mirroring guidelines and a clinical trial at the time of planning 16, 17. Interaction analysis of oxygen therapy versus ambient air is also presented using all available follow-up time (median 2.1 years; interquartile range [IQR]: 1.0 to 3.7 years) with saturation as a continuous covariate.
As secondary objectives, individual components of the composite and myocardial injury measured by high-sensitivity cardiac troponin T were analyzed. The same outcomes were also investigated in patients who developed hypoxemia during the intervention period and are presented according to randomized group.
Mortality data were obtained from the Swedish population registry, which includes the vital status of all Swedish residents with 99.5% completeness within 1 month (15). Data on rehospitalization with MI were obtained from the SWEDEHEART registry and defined according to International Classification of Diseases codes I21 and I22. Data on hospitalization for HF were obtained from the Swedish inpatient registry (18), including all International Classification of Diseases codes from all admissions in Sweden and defined as code I50.
Data on myocardial injury by highest measured level of high-sensitivity cardiac troponin T during the hospitalization period were obtained from the SWEDEHEART registry. The end of follow-up was December 30, 2016, 365 days after randomization of the last patient. No central adjudication or study-specific patient follow-up was performed.
Statistical analysis
The sample-size calculations for the overall trial have been described in detail previously 2, 11. The subgroup analysis of 90% to 94% versus 95% to 100% in the intention-to-treat population with suspected MI (n=6,629) was pre-specified and presented in the supplementary appendix to the main publication (2). Here we present data from the subgroup with confirmed MI (n=5,010) using a Cox proportional hazards model adjusting for sex and age, with treatment, subgroup indicator, and treatment-subgroup interaction term. Because we had access to hospitalizations for HF (13) that were unavailable for the primary publication, the composite of death or hospitalization with MI or HF was chosen as the primary outcome for this analysis to include clinically relevant outcomes and increase statistical power. To further increase the power to detect an interactive effect, we decided: 1) to include all follow-up time for all patients in the statistical analyses; and 2) to avoid the original dichotomization of baseline saturation and perform the interaction analyses with saturation as a continuous covariate (19). Descriptively, we present data up to 1 year only, in line with the original trial design, and for the 2 original saturation categories.
The results were analyzed according to the intention-to-treat principle. Time to event within 365 days is presented using Kaplan-Meier curves, and the number and percentage of patients with events are presented in tables, divided into 90% to 94% and 95% to 100% oxygen saturation at baseline (Central Illustration, Table 3). Time to event for all follow-up is presented in the Online Appendix (Online Table 1, Online Figure S1). Number and percentage of patients with composite events, by subtype, are also presented graphically in the 11 classes from 90% to 100%. For this plot, a patient who died within 1 year is presented as dead even if death was preceded by another event type, and surviving patients with hospitalizations for both MI and HF are presented as MI. Statistical analysis was performed using a Cox proportional hazards model, with age in years as a linear covariate on the log-hazard scale, sex and randomized treatment as factors, baseline saturation in percentage modeled as a restricted cubic spline with knots at the 5th, 35the, 65th, and 95th percentile, and all treatment-saturation interaction terms (Figure 1). The same model was used for the primary composite and the individual event types, and in patients developing hypoxemia, which are also presented descriptively by treatment arm (Tables 1 and 2). Results are presented graphically as estimated hazard ratio (oxygen vs. ambient air) by baseline saturation with pointwise 95% confidence bands (Online Figure S2) and the interaction p values (Table 3).
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Central Illustration. Kaplan-Meier Curves for the Main Composite Endpoint at 365 Days for Patients With Confirmed Myocardial Infarction Assigned to Oxygen or Ambient Air
Kaplan-Meier curves are shown for the cumulative probability of the main composite endpoint of all-cause death, rehospitalization with myocardial infarction (MI), or hospitalization for heart failure within 365 days after randomization for patients with confirmed MI assigned to oxygen or ambient air. (Top) Stratified according to baseline oxygen saturation (90% to 94% vs. 95% to 100%). (Bottom) Stratified according to patients who developed hypoxemia (oxygen saturation<90%) versus those who remained normoxemic (oxygen saturation 90% to 100%).
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Figure 1. Post Hoc Analysis of the Relationship Between the Main Composite Endpoint at 365 Days for Patients With Confirmed Myocardial Infarction and Oxygen Level at Baseline
Proportion of patients with the main composite endpoint of all-cause death (black), rehospitalization with myocardial infarction (MI) (red), and hospitalization for heart failure (HF) (blue) within 365 days after randomization for patients with confirmed MI and oxygen level at baseline by treatment and baseline oxygen saturation.
Table 1. Baseline Characteristics, Clinical Presentation, and Final Diagnoses in Patients With Confirmed Myocardial Infarction, Stratified by Baseline Saturation, and the Subgroup Developing Hypoxemia After Randomization
| Empty Cell | DETO2X-AMI | By Baseline Saturation | Developed Hypoxemia Requiring Oxygen Outside the Protocol |
---
| Total (N = 5,010, 100%) | 90%–94% (n = 836, 16.7%) | 95%–100% (n=4,174,83.3%) | Oxygen (n = 57, 2.3%) | Ambient Air (n = 231, 9.1%) |
| Demographics | | | | | |
| Age, yrs | 68.0 (60.0–76.0) | 71.0∗ (64.0–80.0) | 68.0 (59.0–75.0) | 69.0 (62.0–77.0) | 72.0† (63.0–80.0) |
| Male | 3,622 (72.3) | 574 (68.7)∗ | 3,048 (73.0) | 37 (64.9) | 142 (61.5)† |
| Risk factors | | | | | |
| Body mass index, kg/m 2 | 27.1 ± 4.3 | 27.6 ± 4.5 | 27.0 ± 4.3 | 27.5 ± 4.5 | 26.8 ± 4.3 |
| Current smoking | 1,199 (23.9) | 195 (23.3) | 1,004 (24.1) | 10 (17.5) | 59 (25.5) |
| Hypertension | 2,358 (47.1) | 436 (52.2)∗ | 1,922 (46.0) | 31 (54.4) | 120 (51.9) |
| Diabetes | 932 (18.6) | 186 (22.2)∗ | 746 (17.9) | 12 (21.1) | 47 (20.3) |
| Previous CV disease | | | | | |
| MI | 895 (17.9) | 186 (22.2)∗ | 709 (17.0) | 12 (21.1) | 44 (19.0) |
| PCI | 694 (13.9) | 131 (15.7)∗ | 563 (13.5) | 7 (12.3) | 33 (14.3) |
| CABG | 262 (5.2) | 50 (6.0) | 212 (5.1) | 4 (7.0) | 13 (5.6) |
| Heart failure (EF≤50%) | 262 (5.2) | 55 (6.6)∗ | 207 (5.0) | 5 (8.8) | 21 (9.1) |
| Stroke | 235 (4.7) | 59 (7.1)∗ | 176 (4.2) | 1 (1.8) | 18 (7.8) |
| PAD | 211 (4.2) | 47 (5.6)∗ | 164 (3.9) | 2 (3.5) | 9 (3.9) |
| COPD | 117 (2.3) | 29 (3.5)∗ | 88 (2.1) | 1 (1.8) | 6 (2.6) |
| Cause of admission | | | | | |
| Chest pain | 4,807 (95.9) | 798 (95.5) | 4,009 (96.0) | 53 (93.0) | 215 (93.1) |
| Dyspnea | 73 (1.5) | 22 (2.6)∗ | 51 (1.2) | 0 | 9 (3.9)† |
| Medication on admission | | | | | |
| Aspirin | 1,269 (25.3) | 249 (29.8)∗ | 1,020 (24.4) | 16 (28.1) | 64 (27.7) |
| P2Y 12 receptor inhibitors | 221 (4.4) | 53 (6.3)∗ | 168 (4.0) | 1 (1.8) | 15 (6.5) |
| Beta-blockers | 1,416 (28.3) | 296 (35.4)∗ | 1,120 (26.8) | 15 (26.3) | 76 (32.9) |
| Statins | 1,208 (24.1) | 230 (27.5)∗ | 978 (23.4) | 11 (19.3) | 53 (22.9) |
| ACE inhibitor or AT II blocker | 1,735 (34.6) | 302 (36.1)∗ | 1,433 (34.3) | 22 (38.6) | 88 (38.1) |
| Calcium blockers | 916 (18.3) | 195 (23.3)∗ | 721 (17.3) | 15 (26.3) | 51 (22.1) |
| Diuretic agents | 698 (13.9) | 166 (19.9)∗ | 532 (12.7) | 12 (21.1) | 51 (22.1) |
| Presentation | | | | | |
| Ambulance to ED | 1,533 (30.6) | 274 (32.8)∗ | 1,259 (30.2) | 9 (15.8)† | 51 (22.1)† |
| Ambulance to CCU or catheterization laboratory | 1,983 (39.6) | 380 (45.5)∗ | 1,603 (38.4) | 41 (71.9)† | 137 (59.3)† |
| Systolic blood pressure, mm Hg | 149.5 ± 28.0 | 147.3 ± 28.9 | 149.9 ± 27.8 | 141.9 ± 26.1† | 138.0 ± 29.8† |
| Heart rate, beats/min | 77.4 ± 18.3 | 80.7 ± 20.5 | 76.8 ± 17.8 | 84.7 ± 19.3† | 77.5 ± 19.1 |
| Rales | 254 (5.1) | 64 (7.7)∗ | 190 (4.6) | 7 (12.3)† | 30 (13.0)† |
| Cardiogenic shock | 27 (0.5) | 7 (0.8) | 20 (0.5) | 3 (5.3)† | 7 (3.0)† |
| Oxygen saturation, % | 97.0 (95.0–98.0) | 93.0∗ (92.0–94.0) | 97.0 (96.0–99.0) | 97.0 (95.0–98.0) | 97.0 (95.0–98.0) |
Values are n median (interquartile range), n (%), or mean ± SD.
ACE=angiotensin-converting enzyme, AT=angiotensin; CABG=coronary artery bypass grafting; CCU=cardiac care unit; COPD=chronic obstructive pulmonary disease; CV=cardiovascular; DETO2X-AMI=Determination of the Role of Oxygen in Suspected Acute Myocardial Infarction; ED=emergency department; EF=ejection fraction; MI=myocardial infarction; PAD=peripheral artery disease; PCI=percutaneous coronary intervention.
∗
p<0.05 for the comparison between the low-normal saturation (90% to 94%) group and the high-normal saturation (95% to 100%) group.
†
p<0.05 for the comparison between patients who developed hypoxemia and those who remained normoxemic per randomized group.
Table 2. In-Hospital Procedural Data, Medication, Procedures, and Complications in Patients With Confirmed Myocardial Infarction Stratified by Baseline Saturation, and the Subgroup Developing Hypoxemia After Randomization
| Empty Cell | DETO2X-AMI | By Baseline Saturation | Developed Hypoxemia Requiring Oxygen Outside the Protocol |
---
| Total (N = 5,010, 100%) | 90%–94% (n = 836, 16.7%) | 95%–100% (n = 4,174, 83.3%) | Oxygen (n = 57, 2.3%) | Ambient Air (n = 231, 9.1%) |
| Trial procedural data | | | | | |
| Duration of oxygen therapy, hours per protocol | 11.68 (6.0–12.0) | 10.63∗ (6.0–12.0) | 11.75 (6.0–12.0) | 1.2† (0.6–2.5) | 1.2† (0.6–2.3) |
| Hypoxemia | 288 (5.7) | 95 (11.4)∗ | 193 (4.6) | 57 (100) | 231 (100) |
| Oxygen saturation at end of treatment period, % | 98.0 (96.0–99.0) | 97.0∗ (95.0–98.0) | 98.0 (96.0–99.0) | 81.0 (74.0–88.0) | 86.0 (85.0–88.0) |
| Hospital stay, days | 3.0 (3.0–4.0) | 4.0† (3.0–5.0) | 3.0 (3.0–4.0) | 5.0† (3.0–9.0) | 4.0† (3.0–7.0) |
| Laboratory measures | | | | | |
| eGFR, ml/min | 78.9 ± 20.5 | 75.6 ± 21.3† | 79.5 ± 20.3 | 76.0 ± 21.1 | 72.0 ± 23.4† |
| Hb, g/l | 140.2 ± 16.54 | 140.2 ± 15.98 | 140.2 ± 16.65 | 138.3 ± 22.05 | 134.0† ± 18.69 |
| CRP, mg/l | 4.00 (1.4–9.0) | 5.00∗ (2.0–11.0) | 3.60 (1.3–8.0) | 4.1 (1.65–7.5) | 5.0† (2.5–14.0) |
| hs-troponin T, ng/l | 964 (233–2,923) | 1,365.0∗ (328–3,809) | 921.0 (223–2,770) | 4,235† (1,025–8,870) | 2,569† (1,040–8,638) |
| Procedures | | | | | |
| Coronary angiography | 4,765 (95.1) | 776 (92.8)∗ | 3,989 (95.6) | 56 (98.2) | 223 (96.5) |
| Single-vessel disease | 2,195 (43.8) | 350 (41.9)∗ | 1,845 (44.2) | 19 (33.3)† | 96 (41.6)† |
| Multivessel disease | 2,328 (46.5) | 394 (47.1)∗ | 1,934 (46.3) | 36 (63.2)† | 120 (51.9)† |
| PCI | 4,228 (84.4) | 704 (84.2) | 3,524 (84.4) | 54 (94.7)† | 214 (92.6)† |
| CABG | 212 (4.2) | 27 (3.2) | 201 (4.8) | 4 (7.0)† | 2 (0.9) |
| Echocardiography, EF | | | | | |
| >40% | 3,638 (72.6) | 533 (63.8)∗ | 3,105 (74.4) | 23 (40.4)† | 119 (51.5)† |
| <40% | 741 (14.8) | 173 (20.7)∗ | 568 (13.6) | 29 (50.9)† | 80 (34.6)† |
| CPAP | 79 (1.6) | 25 (3.0)∗ | 54 (1.3) | 19 (33.3)† | 23 (10.0)† |
| In-hospital medication | | | | | |
| IV diuretic agents | 515 (10.3) | 145 (17.3)∗ | 370 (8.9) | 35 (61.4)† | 98 (42.4)† |
| IV inotropes | 108 (2.2) | 23 (2.8)∗ | 85 (2.0) | 38 (16.5)† | 15 (26.3)† |
| IV nitroglycerin | 408 (8.1) | 82 (9.8)∗ | 326 (7.8) | 14 (24.6)† | 34 (14.7)† |
| PCI procedural data | | | | | |
| Heparin | 3,110 (62.1) | 488 (58.4) | 2,622 (62.8) | 39 (68.4) | 146 (63.2) |
| Bivalirudin | 1,738 (34.7) | 307 (36.7) | 1,431 (34.3) | 24 (42.1) | 112 (48.5) |
| Abciximab | 107 (2.1) | 18 (2.2) | 89 (2.1) | 7 (12.3)† | 9 (3.9)† |
| Total stent length, mm | 26.0 (18.0–40.0) | 24.0 (18.0–38.0) | 26.0 (18.0–40.0) | 26.5 (16.0–55.0) | 24.0 (18.0–40.0) |
| Complete revascularization | 2,653 (53.0) | 427 (51.1) | 2,226 (53.3) | 122 (52.8) | 25 (43.9)† |
| Complications | | | | | |
| New-onset atrial fibrillation | 166 (3.3) | 36 (4.3) | 130 (3.1) | 5 (8.8)† | 20 (8.7)† |
| AV block grade II or III | 94 (1.9) | 13 (1.6) | 81 (1.9) | 2 (3.5) | 11 (4.8)† |
| Cardiogenic shock | 64 (1.3) | 12 (1.4) | 52 (1.2) | 23 (10.0)† | 12 (21.1)† |
| Cardiac arrest | 130 (2.6) | 30 (3.6)∗ | 100 (2.4) | 17 (29.8)† | 27 (11.7)† |
| Death | 89 (1.8) | 28 (3.3)∗ | 61 (1.5) | 6 (10.5)† | 18 (7.8)† |
| Discharge MI type | | | | | |
| STEMI/LBBB | 2,952 (58.9) | 552 (66.0)∗ | 2,400 (57.5) | 46 (80.7)† | 201 (87.0)† |
| NSTEMI | 2,058 (41.1) | 284 (34.0)∗ | 1,774 (42.5) | 11 (19.3)† | 30 (13.0)† |
Values are median (interquartile range), n (%), or mean ± SD.
AV=atrioventricular; CPAP=continuous positive airway pressure; CRP=C-reactive protein; eGFR=estimated glomerular filtration rate; Hb=hemoglobin; hs=high-sensitivity; IV=intravenous; LBBB=left bundle branch block; NSTEMI=non–ST-segment elevation myocardial infarction; STEMI=ST-segment elevation myocardial infarction; other abbreviations as in Table 1.
∗
p<0.05 for the comparison between the low-normal saturation (90% to 94%) group and the high-normal saturation (95% to 100%) group.
†
p<0.05 for the comparison between patients who developed hypoxemia and those who remained normoxemic per randomized group.
Table 3. Endpoints in Patients With Confirmed Myocardial Infarction Stratified by Baseline Saturation and the Subgroup Developing Hypoxemia Within 365 After Randomization
| Empty Cell | DETO2X-AMI (Total N=5,010) | By Baseline Saturation | By Baseline Saturation and Treatment |
---
| Oxygen | Ambient Air | p Value, Interaction Oxygen Versus Ambient Air |
| Oxygen (n=2,485,49.6%) | Ambient Air (n=2,525,50.4%) | HR (95%CI) | 90%–94% (n = 836, 16.7%) | 95%–100% (n=4,174,83.3%) | p Value, Interaction by Oxygen Saturation at Baseline∗ | 90%–94% (n = 399, 8.0%) | 95%–100% (n = 2,086, 41.6%) | 90%–94% (n = 327, 6.5%) | 95%–100% (n = 2,088, 41.7) |
| 365 days | | | | | | | | | | | |
| Composite of all-cause death, rehospitalization with MI, or HF | 264 (10.6) | 277 (11.0) | 0.95 (0.81–1.13), p=0.58 | 145 (17.3) | 396 (9.5) | <0.01 | 72 (18.0) | 192 (9.2) | 73 (16.7) | 204 (9.8) | 0.79 |
| All-cause death | 122 (4.9) | 133 (5.3) | 0.91 (0.71–1.17), p=0.47 | 72 (8.6) | 183 (4.4) | 0.13 | 33 (8.3) | 89 (4.3) | 39 (8.9) | 94 (4.5) | 0.33 |
| Rehospitalization with MI | 91 (3.6) | 94 (3.8) | 1.04 (0.78–1.39), p=0.77 | 29 (3.5) | 156 (3.6) | 0.48 | 16 (4.0) | 78 (3.7) | 13 (3.0) | 78 (3.7) | 0.86 |
| Rehospitalization for HF | 78 (3.1) | 85 (3.4) | 0.91 (0.67–1.24), p=0.57 | 56 (6.7) | 107 (2.6) | <0.01 | 28 (7.0) | 50 (2.4) | 28 (6.4) | 57 (2.7) | 0.35 |
Values are n (%) unless otherwise indicated.
HF=heart failure; other abbreviations as in Table 1.
∗
Interaction analysis adjusted for age, sex, hypertension, diabetes, previous MI, previous percutaneous coronary intervention, renal function; and ST-segment elevation myocardial infarction.
The highest measured level of high-sensitivity troponin T concentration during hospitalization was analyzed for observed cases using a linear model for the log-transformed concentration with the same covariates as the Cox model for events. Results are presented graphically as geometric mean ratio by baseline saturation with 95% pointwise confidence band and the interaction p value (Figure 2).
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Figure 2. Highest Measured Level of High-Sensitivity Troponin T Concentration During Hospitalization by Treatment Group and Stratified per Baseline Oxygen Saturation
Results are presented graphically as geometric mean ratio by baseline saturation with 95% pointwise confidence band (p=0.78).
The relationship between baseline saturation and future events was analyzed using a Cox proportional hazards model with baseline saturation modeled as a spline as described previously, adjusted for sex, hypertension, diabetes, STEMI, previous MI, percutaneous coronary intervention (PCI) or HF, age, and glomerular filtration rate as linear covariates. The results are presented graphically as estimated event rate at 1 year by saturation, for a “typical” patient with median age and estimated glomerular filtration rate, and the most common value for other cofactors, with 95% pointwise confidence bands and p value for an effect of saturation (Online Figure S3).
All analyses were conducted using SAS version 9.4 (SAS Institute, Cary, North Carolina).
Results
Study population
Of the 6,629 patients with suspected MI who were enrolled in the main trial, 5,010 (76%) received a primary discharge diagnosis of MI—2,952 (59%) STEMI and 2,058 (41%) non–ST-segment elevation MI—and were included in this analysis; 2,485 were allocated to oxygen, which was initiated immediately after randomization, and 2,525 were allocated to ambient air. Stratified by baseline oxygen saturation, 836 (17%) had low-normal oxygen saturation of 90% to 94%, and 4,174 (83%) had a high-normal oxygen saturation of 95% to 100%. Overall, among the patients assigned to oxygen, 57 (2%) developed hypoxemia, compared with 231 of the patients (9%) assigned to ambient air (study flowchart, Online Figure S4).
When comparing patients according to baseline oxygen saturation, patients in the low-normal cohort were older, more often female, more often had comorbid conditions, and more often had previous cardiovascular disease, and consequently, they had more intensive medical therapy at baseline. Other risk factors, such as body mass index and smoking, were equally distributed. At presentation, patients in the low-normal cohort were more often admitted by ambulance and more frequently had clinical signs of HF with pulmonary rales and dyspnea as the chief symptoms (Table 1).
Irrespective of randomized treatment, patients who developed hypoxemia had similar baseline characteristics compared with those who remained normoxemic, except for hypoxemic patients allocated to ambient air, who were significantly older. Clinically, patients who developed hypoxemia arrived more often by ambulance—in the great majority of cases directly to the cardiac care unit or catheterization laboratory—and presented more often with signs of acute HF with dyspnea, higher heart rates, lower blood pressure, pulmonary rales, and cardiogenic shock at arrival compared with patients who remained normoxemic (Table 1).
In-hospital procedures and complications
At the time of randomization, the median oxygen saturation was 93% (IQR: 92.0% to 94.0%) in the low-normal cohort and 97.0% (IQR: 95.0% to 98.0%) in the high-normal cohort. The median duration of oxygen therapy was shorter in the low-normal cohort, most likely because of a greater proportion of patients developing hypoxemia and receiving oxygen outside the protocol (11.4% vs. 4.6%, low-normal vs. high-normal, respectively). At the end of the intervention period, the median oxygen saturation was 97.0% (IQR: 95.0% to 98.0%) in the low-normal cohort and 98.0% (IQR: 96.0% to 99.0%) in the high-normal cohort.
Patients in the low-normal cohort presented more commonly with STEMI, had longer hospital stays and larger myocardial injuries measured by high-sensitivity cardiac troponin T, more frequently had multivessel coronary artery disease, and more often developed acute HF with reduced ejection fraction treated with medication as well as continuous positive airway pressure ventilation. Furthermore, complications such as cardiac arrest and death occurred more frequently in the low-normal cohort.
Irrespective of randomized treatment, patients who developed hypoxemia compared with those who remained normoxemic, similar to the low-normal cohort, most often had STEMI with underlying multivessel coronary disease leading to a larger infarct size and acute HF (Table 2).
Clinical endpoints
Patients stratified according to oxygen saturation at baseline per randomized group
In the cohort with low-normal oxygen saturation, the incidence of the composite of all-cause death, rehospitalization with MI, or rehospitalization for HF within 1 year was 18.0% (72 of 399) among patients allocated to oxygen treatment compared with 16.7% (73 of 437) among those allocated to ambient air (Central Illustration, Table 3). The rates of death from any cause were 8.3% and 8.9% in the 2 groups (oxygen and ambient air), respectively; the rates of rehospitalization with MI were 4.0% and 3.0%, respectively; and the rates of rehospitalization for HF were 7.0% and 6.4%, respectively (Table 3).
In the cohort with high-normal oxygen saturation, the incidence of the composite endpoint was 9.2% (192 of 2,086) among patients allocated to oxygen treatment compared with 9.8% (204 of 2,088) among patients allocated to ambient air (Central Illustration, Table 3). The rates of death from any cause were 4.3% and 4.5% in the 2 randomized groups (oxygen and ambient air), respectively; the rates of rehospitalization with MI were 3.7% and 3.7%, respectively; and the rates of rehospitalization for HF were 2.4% and 2.7%, respectively (Table 3).
Event rates for the composite endpoint differed significantly (p<0.01) with regard to baseline saturation with higher rates for low saturation, also after adjusting for additional relevant cofactors (hypertension, diabetes, glomerular filtration rate, STEMI, previous MI, PCI or HF) (p<0.01), numerically, driven mainly by rehospitalization for HF (Table 3).
There was no significant interaction between the randomized groups and oxygen saturation at baseline regarding the composite endpoint (p=0.79) or the individual components of all-cause death (p=0.33), rehospitalization with MI (p=0.86), or rehospitalization for HF (p=0.35) (Table 3).
Data for the highest measured level of high-sensitivity cardiac troponin T showed no significant difference between the low-normal cohort and the high-normal cohort, irrespective of randomized treatment (p=0.78) (Figure 2, Table 2).
Patients who developed hypoxemia
Irrespective of randomized treatment, patients who developed hypoxemia had significantly higher event rates of the main composite endpoint compared with those who remained normoxemic throughout the trial. In the oxygen group, the incidence of the composite endpoint was 22.8% (13 of 57) among patients developing hypoxemia compared with 10.3% (251 of 2,428) among patients who remained normoxemic. In the ambient-air group, the incidence of the composite endpoint within 1 year was 23.8% (55 of 231) among patients developing hypoxemia compared with 9.7% (222 of 2,294) among those who remained normoxemic (Central Illustration).
Discussion
In this pre-specified subgroup analysis of the DETO2X-AMI trial, a registry-based randomized clinical trial of supplemental oxygen or ambient air in normoxemic patients with acute MI, we found no significant treatment effect of oxygen therapy on the rate of the composite endpoint of all-cause death, rehospitalization with MI, or HF, irrespective of oxygen saturation at baseline. However, patients in the low-normal cohort (oxygen saturation 90% to 94%) had significantly higher event rates compared with those in the high-normal cohort (95% to 100%). Irrespective of baseline saturation or randomized treatment, patients who developed hypoxemia of any cause throughout the intervention period had a markedly increased infarct size and a doubling of event rates compared with patients who remained normoxemic.
Effect of supplemental oxygen in patients with high-normal oxygen saturation at baseline
For decades, liberal oxygen therapy has been a cornerstone of MI therapy irrespective of oxygen saturation at baseline (1), despite experimental studies demonstrating negative effects of excessive oxygen therapy on hemodynamic parameters caused by hyperoxemia-mediated direct vasoconstriction and an increased production of reactive oxygen species1, 8, 20. This is likely of particular importance in patients with MI with coronary atherosclerosis, in whom macrovascular hemodynamic effects of flow-limiting stenoses(21) are accompanied by endothelial and microvascular dysfunction (22). Results from a smaller clinical trial that reported increased early myocardial injury associated with oxygen therapy (17) and data from a recent meta-analysis in critically ill patients (including patients with MI) indicating increased mortality (23) have further fueled the oxygen risk/benefit debate. However, in previous analyses from the DETO2X-AMI trial, we could not demonstrate any clinically relevant adverse effects of oxygen therapy in the short, medium, or long term in patients with suspected MI 2, 13 or PCI-treated patients with STEMI (12).
In the present study, even in patients with confirmed MI in the subgroup with high-normal oxygen saturation at baseline—overall the clear majority of trial participants (83%) theoretically at highest risk for adverse events caused by inadvertent hyperoxemia—results remained unchanged. Thus, no clinically relevant harm was detected in our trial when oxygen was given in a moderate dose for a relatively short time. Nevertheless, dose-dependent toxicity may exist, and high-flow oxygen should probably be avoided (24).
Effect of supplemental oxygen in patients with low-normal oxygen saturation at baseline
Supplemental oxygen has been used routinely in the treatment of patients with MI with the rationale that oxygen therapy increases oxygen delivery to the ischemic myocardium and thereby reduces infarct size and subsequent complications (1). In the present trial, patients in the low-normal cohort had markedly increased event rates compared with those in the high-normal cohort, with the largest proportional difference in the early post-MI phase, possibly reflecting increased infarct size. As expected, the proportion of patients who developed hypoxemia was smaller in the oxygen-treated patients. Of note, however, the overall risk remained unchanged irrespective of oxygen treatment. It seems that lower oxygen saturation at baseline is an independent risk marker reflecting a higher prior cardiovascular disease burden with acute-on-chronic myocardial injury predicting worse prognosis.
Patients developing hypoxemia during the intervention period
By far the worst prognosis was found in patients who developed hypoxemia during the intervention period. Here, hypoxemia likely reflects a more severe and sudden onset of MI. Our data support this hypothesis, as these patients presented predominately with STEMI and were most often transported by ambulance directly to the catheterization laboratory and treated with PCI. The infarct size measured by cardiac troponin T was increased 2- to 4-fold compared with patients who remained normoxemic. Hypoxemia developed typically early on, within the first 2 h after randomization, and despite aggressive medical and procedural treatment, the prognosis remained poor. In this trial, supplemental oxygen did not significantly ameliorate outcomes.
Previous studies and other related clinical scenarios
To our knowledge, no other clinical trial has previously addressed the effect of oxygen therapy on cardiovascular outcomes in relation to baseline oxygen saturation.
However, controversy remains in many other related clinical scenarios—for example after cardiac arrest 25, 26, in critically ill patients (23), in patients with cardiogenic shock(27), and in those with HF (28)—for which high-flow oxygen is still common practice despite a lack of evidence.
Study limitations
General and conceptual limitations to the main study have previously been described in detail (2). The results and conclusions of the present study are drawn from a pre-specified subgroup analysis without formal power calculation. In particular, in the cohort with low-normal oxygen saturation as well the subgroup of patients who developed hypoxemia during the intervention, statistical power to detect differences in outcomes is limited, as cases and events were few, which reduces the discriminatory power of our analysis.
Overall, we could only measure the amount of oxygen supplied but not the amount of oxygen that was actually received by the patient. The physiologic effect of an oxygen mask at 6 l/min depends on how the mask sits on the face and how the patient is breathing, thus providing a flow rate between 0 l/min and a maximum 6 l/min of oxygen for inhalation. We used pulse oximetry to measure peripheral blood oxygenation, which reflects current clinical practice in the acute setting and is considered adequate within the normoxemic range (29). However, exhaled oxygen or arterial blood gas analysis would have been preferable being the more precise methods. Pulse oximetry, in general, tends to overestimate true arterial oxygen tension (30), and above-normal levels cannot be displayed at all. Consequently, this poses diagnostic difficulties at both ends of the spectrum: toward hypoxemia, we may have misclassified patients as being normoxemic and thus withheld guideline-recommended therapy; toward above-normal values, it is likely that patients with high-normal oxygen saturation at baseline enrolled in our trial had unintentional intermittent above-normal arterial oxygen tension.
Conclusions
Consistent with previous results from DETO2X-AMI, we could not demonstrate any clinically relevant beneficial effect of routine oxygen therapy in normoxemic patients with MI with respect to all-cause death, rehospitalization with MI, or HF at 1 year, irrespective of baseline oxygen saturation. Thus, our results corroborate current guidelines (3). Oxygen saturation at baseline in the low-normal range was identified as an important independent marker of poor prognosis that was not affected by oxygen treatment.
Perspectives
WHAT IS KNOWN? Routine oxygen therapy is not associated with reduced mortality in normoxemic patients with suspected MI.
WHAT IS NEW? Irrespective of oxygen saturation at baseline, no beneficial effect of routine oxygen therapy in normoxemic patients with MI with respect to all-cause death, cardiovascular death, rehospitalization with MI, or HF at 1 year could be demonstrated. Low-normal baseline oxygen saturation, however, was identified as a powerful independent predictor of poor prognosis.
WHAT IS NEXT? Uncertainty remains regarding oxygen treatment in a number of patient groups susceptible to hypo- or hyperoxemia, such as after cardiac arrest, critical illness, and HF. Future randomized clinical trials should address the clinical effect of oxygen therapy in specific groups.
Acknowledgments
This study was designed and conducted, and the paper written, by the authors, who vouch for the data, all analyses, and the fidelity of this report to the trial protocol and statistical analysis plan (Online Appendix). The authors thank the staffs at all centers participating in the DETO2X-AMI collaboration for their professionalism and commitment to this study. The authors are grateful for the assistance from personnel at Uppsala Clinical Research Center, Uppsala University, and the Swedish Research Council and the Swedish Heart-Lung Foundation for the funding of this trial.
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2025 ACC/AHA/ACEP/NAEMSP/SCAI Guideline for the Management of Patients With Acute Coronary Syndromes: A Report of the American College of Cardiology/American Heart Association Joint Committee on Clinical Practice Guidelines
2025, Journal of the American College of Cardiology Show abstract The “2025 ACC/AHA/ACEP/NAEMSP/SCAI Guideline for the Management of Patients With Acute Coronary Syndromes” incorporates new evidence since the “2013 ACCF/AHA Guideline for the Management of ST-Elevation Myocardial Infarction” and the corresponding “2014 AHA/ACC Guideline for the Management of Patients With Non–ST-Elevation Acute Coronary Syndromes” and the “2015 ACC/AHA/SCAI Focused Update on Primary Percutaneous Coronary Intervention for Patients With ST-Elevation Myocardial Infarction.” The “2025 ACC/AHA/ACEP/NAEMSP/SCAI Guideline for the Management of Patients With Acute Coronary Syndromes” and the “2021 ACC/AHA/SCAI Guideline for Coronary Artery Revascularization” retire and replace, respectively, the “2016 ACC/AHA Guideline Focused Update on Duration of Dual Antiplatelet Therapy in Patients With Coronary Artery Disease.” A comprehensive literature search was conducted from July 2023 to April 2024. Clinical studies, systematic reviews and meta-analyses, and other evidence conducted on human participants were identified that were published in English from MEDLINE (through PubMed), EMBASE, the Cochrane Library, Agency for Healthcare Research and Quality, and other selected databases relevant to this guideline. Many recommendations from previously published guidelines have been updated with new evidence, and new recommendations have been created when supported by published data.
### 2025 ACC/AHA/ACEP/NAEMSP/SCAI Guideline for the Management of Patients With Acute Coronary Syndromes: A Report of the American College of Cardiology/American Heart Association Joint Committee on Clinical Practice Guidelines
2025, Circulation
### Interplay of hypoxia-inducible factors and oxygen therapy in cardiovascular medicine
2023, Nature Reviews Cardiology
### German S3 Guideline - Oxygen Therapy in the Acute Care of Adult Patients
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### German S3 Guideline: Oxygen Therapy in the Acute Care of Adult Patients
2022, Respiration
### High flow oxygen and risk of mortality in patients with a suspected acute coronary syndrome: Pragmatic, cluster randomised, crossover trial
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This work was supported by the Swedish Research Council (grant VR20130307), the Swedish Heart-Lung Foundation (grant HLF20130262, HLF20160688), and the Swedish Foundation for Strategic Research (grant SSF KF10-0024). Drs. Hofmann and Östlund have received grants from the Swedish Research Council and the Swedish Heart-Lung Foundation. Dr. Hofmann was supported by the Stockholm County Council (clinical postdoctoral appointment). The funding agencies had no access to the study data and no role in trial design, implementation, or reporting. The authors have reported that they have no relationships relevant to the contents of this paper to disclose.
© 2020 by the American College of Cardiology Foundation. Published by Elsevier.
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11010 | https://dspace.mit.edu/bitstream/handle/1721.1/122634/JCMNS66.pdf?sequence=1&isAllowed=y | MIT Open Access Articles Models for the Phase Diagram of Palladium Hydride Including O-site and T-site Occupation The MIT Faculty has made this article openly available. Please share how this access benefits you. Your story matters.
Citation: Hagelstein, Peter L. "Models for the Phase Diagram of Palladium Hydride Including O-site and T-site Occupation." Journal of Condensed Matter Nuclear Science 20 (2016): 54-80 © 2016 ISCMNS As Published: www.iscmns.org/CMNS/JCMNS-Vol20.pdf Publisher: International Society of Condensed Matter Nuclear Scientists (ISCMNS) Persistent URL: Version: Final published version: final published article, as it appeared in a journal, conference proceedings, or other formally published context Terms of Use: Article is made available in accordance with the publisher's policy and may be subject to US copyright law. Please refer to the publisher's site for terms of use.
J. Condensed Matter Nucl. Sci. 20 (2016) 54–80 Research Article Models for the Phase Diagram of Palladium Hydride Including O-site and T-site Occupation Peter L. Hagelstein∗ Massachusetts Institute of Technology, Cambridge, MA, USA Abstract Early statistical mechanics models for palladium hydride allowed for a good description of the phase diagram based on a simple parameterization of the O-site energy. In this work we study generalizations of these models to include higher-order dependence on loading, temperature-dependent O-site energies, and also to include T-site occupation. Experimental data sets for 10 isotherms were assembled, and augmented with additional extrapolated points for the low-pressure α-phase region as well as the high pressure β-phase region. Loading-dependent O-site energies are optimized by minimizing the mean square error in the chemical potential between the model and data set. The resulting models give a good match to the phase diagram. If the O-site energy is allowed to be temperature dependent then the fit is better, but the resulting optimum is a mathematical optimum not so closely connected with the physical system. Models were studied in which O-site and T-site occupation occurs. When optimized these models are able to provide a good match to the phase diagram. When the O-site to T-site excitation energy is fixed according to estimates developed in earlier studies, the resulting temperature-dependent O-site energies are physically plausible. When the excitation energy are optimized together with the O-site energy, the resulting optimum is a mathematical one much less connected to the physical system.
An earlier analysis of solubility in the α-phase led to a strong argument that T-site occupation occurs in palladium hydride and in palladium deuteriude; the present study supports this conclusion based on an independent data set.
c ⃝2016 ISCMNS. All rights reserved. ISSN 2227-3123 Keywords: Mean field model, Palladium hydride, Phase diagram, Statistical mechanics, T-site occupation 1. Introduction Palladium hydride [1–5] is perhaps the most studied of the metal hydrides [6–14], in both experiment and theory.
Our interest in the problem is motivated by issues associated with the development of models for excess heat in the Fleischmann-Pons experiment [15–17]. In a realistic simulation models are needed for the loading and electrochemical reactions; for hydrogen and deuterium occupation of bulk sites; for vacancy creation and the occupation of vacancy sites. Fundamental to all of these problems are basic statistical mechanics models for hydrogen and deuterium in palladium, which is the focus of the study described in this work. Similar issues arise in the case of modeling nickel systems, which we hope to address in subsequent research.
∗E-mail: plh@mit.edu c ⃝2016 ISCMNS. All rights reserved. ISSN 2227-3123 P.L. Hagelstein / Journal of Condensed Matter Nuclear Science 20 (2016) 54–80 55 The first basic statistical mechanics models of this type were described long ago in papers now considered to be classics; by Fowler and Smithells ; and by Lacher . Since the chemical potential of hydrogen in gas was known theoretically, it was possible to construct relevant models for hydrogen in palladium since the chemical potentials are equal in equilibrium (the formulations used in the early papers seem more complicated, but are equivalent to matching the solid and gas phase chemical potentials). From a comparison of the model isotherms with data, it was possible to estimate the O-site energy of interstitial hydrogen in Pd, resulting in a statistical mechanics model for part of the phase diagram. In what follows we study generalizations of these models to include higher-order polynomial dependence on loading, temperature-dependent O-site energies, and T-site occupation, to develop models that can describe more of the phase diagram.
To proceed, a set of isotherms are first assembled, and then a set of Lacher models with polynomial dependence of the O-site energy on the loading are optimized. This classic model is found to provide a good match to the data set at elevated temperature, but noticeable deviations occur near room temperature.
In the literature one can find more sophisticated models for interstitial hydrogen in metals termed “lattice gas” models [20–26]. In this case a model Hamiltonian is used with the O-site energy and interactions between the hydrogen atoms specified; statistical mechanics allows for the calculation of the phase diagram or other observables. Model parameters are selected which give the best fit with experiment. The interaction between interstitial hydrogen atoms in the lattice gas model gives rise to a splitting of the energies of the different configurations which cannot be described in terms of a simple loading-dependent O-site energy (independent of temperature). The average O-site energy in this case increases with temperature as more energetic configurations contribute in a statistical mechanics calculation.
Lattice gas models are both interesting and desirable; however, the development and optimization of such models involves significant computation, and research is ongoing to develop models which are better able to match experi-ment. The focus here is focused on simpler mean field models (alternatively. empirical models) that can be used for applications; our interest in this study is not to determine a better lattice gas Hamiltonian. Nevertheless, we know from these studies, and also from density functional calculations, a substantial splitting of the configuration energies occurs [27,28], and this impacts the empirical models under discussion. In what follows models are considered in which the O-site energy is allowed to depend on temperature, to account approximately for this configurational splitting. The mean-field models when optimized are able to provide better matches to the data set than more basic Lacher models.
Unfortunately, the optimized solutions have O-site energies which do not increase with temperature for all loading, and hence are not consistent with a physical model for configurational splitting.
Interstitial hydrogen is known to occupy octahedral sites [29,30], which is consistent with the models discussed so far. In high pressure experiments it is thought that the H/Pd loading has exceeded unity [31,32], conditions under which the occupation of additional sites beyond the O-sites is required. Similar claims have been made for the deuterium loading in electrochemical Fleischmann–Pons experiments . At present the development of over-unity loading of H or D in bulk Pd in any experiment is not universally accepted. There is also not general agreement as to where excess interstitial H or D might go once the O-sites are fully occupied. We have recently studied new models in which both O-site and T-site occupation occurs [34,35], and compared them with experiment in different regimes. Models that include T-site occupation can account for the observed experimental solubility data in the α-phase over a wide range of temperatures , conditions where models with only O-site occupation have trouble. Estimates for the O-site to T-site excitation energy for α-phase PdHx and PdDx near 100 meV resulted from this study. The new models were used to compare with experiment in the β-phase and higher, which resulted in a parameterization of the loading-dependent O-site to T-site excitation energy near room temperature.
As indicated above the primary motivation for this work is to develop simple mean-field models that can be used for applications. However, another motivation was to see whether models developed for O-site to T-site occupation were compatible with the isotherms of the phase diagram. A related issue is whether T-site occupation is necessary to understand the phase diagram. We know from previous work that T-site occupation is required to model loading 56 P.L. Hagelstein / Journal of Condensed Matter Nuclear Science 20 (2016) 54–80 in the α-phase; however, whether T-site occupation is needed for the rest of the phase diagram has not previously been addressed. In the end, our results suggest that it may not be possible to develop an accurate model that is also acceptable on physical grounds without T-site occupation.
2. Isotherms and Phase Diagram The first task is to assemble isotherms for a reference phase diagram. Since palladium hydride is perhaps the best studied of the metal hydrides, there are many data sets with which to work. We made use of a subset of isotherms presented by Manchester et al. , which recommended itself since a sizeable collection of individual pressure-composition-temperature (PCT) data points were plotted. The online web digitization program of Ankit Rohagti allowed for the conversion of the published data points into digital form.
It was found necessary to extend the isotherms to lower loading in the α-phase, and also to higher loading in the β-phase. Reliable isotherms for the α-phase could be constructed based on an empirical fit to literature data as described in the following subsection. For high concentration a (less reliable) generalization of an empirical fit due to Baranowski et al. was used. This allowed for a “complete” set of isotherms at ten different temperatures between 20◦C and 433◦C which could be used for fitting statistical mechanics model parameters.
Extrapolated α-phase data points at low pressure are needed to provide an “anchor” for the model at low loading.
There is a differential relation between the O-site and T-site energies, and the chemical potential; consequently, a boundary condition is required in order to determine a unique solution for the chemical potential as a function of loading. The extrapolated low pressure points serve to fix the boundary condition in the optimizations. Good agreement using this approach is found with the results obtained previously based on a different data set (the data set of Clewley et al. ).
The biggest issue with the extrapolated points at high loading is that far fewer experimental studies have been published, so that the development of an accurate extrapolation is problematic. Another concern is that if T-site occupation is significant, then there is reason to be concerned whether the extrapolation used includes it (and there is no reason to believe it does). Consequently, it would not be expected that a reliable O-site to T-site excitation energy can be determined at high loading from the isotherms, since the impact of T-site occupation is greatest at high temperature where there is no data at high loading.
2.1. Empirical fit for the α-phase Empirical fits for the α-phase have been studied previously [18,37–43]; for example Simons and Flanagan give the empirical formula (in our notation) ln P(atm) = 13.04 −2327 T −11110 T θ + 2 ln θ 1 −θ.
(1) It was found that this formula is useful over a restricted temperature range. A new empirical formula was developed based on ln P(atm) = a−1 T + a0 + a1T + b−1 T + b0 + b1T θ + 2 ln θ 1 −θ.
(2) The fitting parameters were optimized to give a−1 = −2478.35 K, a0 = 14.1429, a1 = −0.00204315 K−1, P.L. Hagelstein / Journal of Condensed Matter Nuclear Science 20 (2016) 54–80 57 θ 0.001 0.01 P (atm) 10-4 10-3 10-2 10-1 100 273 K 363 K 413 K 513 K 807 K 348 K Figure 1.
Solubility data in the α-phase region of PdHx (red triangles); empirical fit (blue lines).
b−1 = −24147.1 K, b0 = 57.5628, b1 = −0.0534775 K−1.
(3) The resulting empirical fit is compared with data used for parameter optimization in Fig. 1. For the data sets at 273, 348 and 363 K we used the tabulated data of Simons and Flanagan ; the isotherms at 413 and at 513 K are from Oates et al. ; and the solubility data at 807 K is from Kleppa and Phutella .
2.2. Empirical fit at high loading At high loading there are empirical formulas in the literature, such as the empirical fit of Baranowski et al. given by ln fH2 = −(100.4 −90.1θ)(kJ/mol) RT + 106.4J/(mol K) R + 2 ln θ 1 −θ.
(4) Motivated by this, we decided to construct an empirical fit of the form ln fH2 = (a0 + b0θ) T + a1 + b1θ + T a2 + b2θ + 2 ln θ 1 −θ + 2PVH kBT .
(5) This fit includes the next order in temperature, and adds a P dV term. The fitting parameters are: 58 P.L. Hagelstein / Journal of Condensed Matter Nuclear Science 20 (2016) 54–80 θ 0.55 0.65 0.75 0.60 0.70 0.80 ln f (atm) -2 0 2 4 6 8 Figure 2.
Data points and fit for the empirical model at high pressure.
a0 = −18215.5 K, b0 = 17469.8 K, a1 = 35.3611, b1 = −22.1764, a2 = −0.0102061 K−1, b2 = 0.0038497 K−1 (6) with fugacity in atmospheres. This empirical fit is compared against the high loading data points from our data set in Fig. 2.
2.3. Isotherms The resulting isotherms are shown in Fig. 3. We selected data sets from Manchester et al. for 20, 70, 120, 160, 200, 243, 270, 300, 340, and 433◦C; and we extended the data set to both lower and higher H/Pd loadings using the extrapolations described in this section.
3. Model with Loading-dependent O-site Energy The first realistic model for the phase diagram of PdH was put forth in Lacher’s classic paper , which assumed that the O-site energy is quadratic in the loading and independent of temperature. Models of this kind have been used to predict phase diagrams previously [19,45,46]. In this section O-site energy models that are polynomial in loading are optimized against the reference isotherm data set to determine fitting parameters.
P.L. Hagelstein / Journal of Condensed Matter Nuclear Science 20 (2016) 54–80 59 θ 10-5 10-4 10-3 10-2 10-1 P (atm) 10-7 10-6 10-5 10-4 10-3 10-2 10-1 100 101 102 103 104 105 θ 0.2 0.4 0.6 0.8 1.0 10-7 10-6 10-5 10-4 10-3 10-2 10-1 100 101 102 103 104 105 Figure 3.
Data points used in this study for the isotherms of the phase diagram.
3.1. Model A model for O-site occupation can be derived by assuming that hydrogen is in equilibrium in the gas phase and solid phase, so that the chemical potential in the two phases match µs H = µg H = 1 2µg H2.
(7) The chemical potential in the gas phase is accurately known. For the solid phase and only O-site occupation we may write µH = EO + θ∂EO ∂θ −kBT ln 1 −θ θ −kBT ln zO + PVH, (8) where EO is the O-site energy which depends on the loading θ, and may also depend on temperature in some of the models under consideration.
3.2. O-site partition function For the modeling in this work a non-SHO O-site partition function is used as discussed in . For O-site occupation the partition function is taken to be 60 P.L. Hagelstein / Journal of Condensed Matter Nuclear Science 20 (2016) 54–80 zO = 4 n exp −En kBT 3 (9) with En = ℏωOns.
(10) The excitation energy and scaling parameter for O-site occupation are ℏωO = 69.0 meV, sO = 1.2.
(11) 3.3. Optimization For the optimization in this work, a measure of the error I is defined according to I = 1 N j µH(data) −µH(model) 2 j , (12) where the summation is over all of the data points of the isotherms in the phase diagram. The hydrogen in gas chemical potential [34,35] is determined at the loading of a given data point, and then compared with the chemical potential for the hydrogen in solid determined from the model under consideration. The model parameters are optimized from a minimization of the resulting mean square error I.
3.4. O-site energy results We optimized models with polynomial O-site energy of the form EO(θ) = a0 + a1θ + a2θ2 + a3θ3 + a4θ4 + a5θ5 + a6θ6 (13) for polynomial O-site energy models models with third- to sixth-order dependence on the loading θ. The errors associated with the different optimizations are given in Table 1. One sees that the errors are smaller for higher-order models.
Table 1.
Error associated with the opti-mization of the different models assuming θ-dependent O-site energies considered in this section.
Order I 3 4.22 × 10−4 4 1.63 × 10−4 5 1.37 × 10−4 6 1.34 × 10−4 P.L. Hagelstein / Journal of Condensed Matter Nuclear Science 20 (2016) 54–80 61 θ 0.0 0.2 0.4 0.6 0.8 1.0 EO(θ)+ ED/2 (meV) -160 -140 -120 -100 -80 -60 Figure 4.
O-site energy as a function of θ with only O-site occupation; third-order polynomial (dark blue line); fourth-order polynomial (blue line); fifth-order polynomial (cyan line); and sixth-order polynomial (red line).
Results for the O-site energy are shown in Fig. 4. There are minor differences in the different models, with a general convergence toward a curve close to the sixth-order polynomial model result. Fitting parameters for the sixth-order optimized O-site energy are given by a0 = −73.244, a1 = −211.382, a2 = 67.137, a3 = 93.793 a4 = −212.541, a5 = 404.335, a6 = −211.373 (14) with individual aj coefficients in meV.
3.5. Comparison of predicted phase diagram with experiment The phase diagram for the sixt-order polynomial fit for EO is shown in Fig. 5. In general this model does pretty well; however, the predicted isotherms lie noticeably above the experimental data points at the lower temperatures.
4. O-site Energy Dependent on Loading and Temperature In a more sophisticated lattice gas treatment the different configurations (corresponding to different arrangements of hydrogen atoms in different lattice sites) have different energies. Consequently, the use of a loading-dependent O-site energy itself constitutes a significant approximation (in the literature this is referred to as a mean-field approximation).
Here we make use of a mean field approximation where the (mean) O-site energy is allowed to depend both on loading and on temperature. As before the fitting parameters of the O-site energy are determined by optimization against the reference phase diagram.
62 P.L. Hagelstein / Journal of Condensed Matter Nuclear Science 20 (2016) 54–80 θ 10-5 10-4 10-3 10-2 10-1 P (atm) 10-8 10-7 10-6 10-5 10-4 10-3 10-2 10-1 100 101 102 103 104 105 θ 0.2 0.4 0.6 0.8 1.0 10-8 10-7 10-6 10-5 10-4 10-3 10-2 10-1 100 101 102 103 104 105 Figure 5.
Isotherms and data points for the temperature-independent sixth-order polynomial fit of EO.
Some intuition about how this model might work seems useful here. If there is a substantial splitting in the different configuration energies, we would expect this to be captured in the empirical model with model O-site energies lower at low temperature, and model O-site energies higher at high temperature. This will provide a check on the results, to make sure that the intended physics is captured in the optimization results.
4.1. Model and optimization We have optimized models with a polynomial dependence on loading as before, but now the model is expanded to assume in addition a linear dependence on the temperature according to EO(θ, T ) = a0 + a1θ + a2θ2 + a3θ3 + a4θ4 + a5θ5 + a6θ6 + T b0 + b1θ + b2θ2 + b3θ3 + b4θ4 + b5θ5 + b6θ6 .
(15) We have optimized the model parameters by minimizing the associated error I as described in the previous section.
The additional degrees of freedom have allowed for the error I to be significantly reduced, as can be seen by comparing the results of Table 1 with those of Table 2.
P.L. Hagelstein / Journal of Condensed Matter Nuclear Science 20 (2016) 54–80 63 Table 2.
Errors associated with the opti-mization of the different models with O-site occupation considered in this section.
a,b order I 3,3 1.67 × 10−4 4,4 4.40 × 10−5 5,5 2.06 × 10−5 6,6 2.01 × 10−5 4.2. O-site energies The optimized O-site energies that result for the model with sixth-order polynomial dependence on the loading is shown in Fig. 6. The associated expansion coefficients are a0 = −57.269, a1 = −269.674, a2 = 42.263, a3 = −10.436, a4 = 376.561, a5 = −222.779, a6 = −7.575, b0 = −0.0282103, b1 = 0.1068191, b2 = 0.3125901, b3 = −1.0341667, b4 = 1.0675563, b5 = −0.4938008, b6 = 0.0728168.
(16) The aj coefficients are in meV, and the bj coefficients are in meV/K.
We see a modest spread in the O-site energies, with the O-site energy increasing with temperature at high loading (as would be expected for a physical system). At low loading, the O-site energy appears to decrease with temperature, θ 0.1 0.3 0.5 0.7 0.9 0.0 0.2 0.4 0.6 0.8 1.0 EO(θ,T)+Ed/2 (meV) -180 -160 -140 -120 -100 -80 -60 Figure 6.
O-site energies as a function of loading for the different temperatures of the isotherms (blue lines); extrapolation to T = 0 (black line).
64 P.L. Hagelstein / Journal of Condensed Matter Nuclear Science 20 (2016) 54–80 θ 10-5 10-4 10-3 10-2 10-1 P (atm) 10-8 10-7 10-6 10-5 10-4 10-3 10-2 10-1 100 101 102 103 104 105 θ 0.2 0.4 0.6 0.8 1.0 10-8 10-7 10-6 10-5 10-4 10-3 10-2 10-1 100 101 102 103 104 105 Figure 7.
Isotherms and data points for the temperature-dependent sixth-order polynomial fit of EO.
an effect not consistent with our picture above involving a spread in O-site energies. Consequently, we recognize that this optimized model corresponds to a mathematical optimum, but one that is not particularly closely connected to a physical picture at low loading.
4.3. Model phase diagram Given the reduced error associated with the optimization of the models, the fit to the isotherms is expected to be improved. This can be seen in the associated phase diagram for the lowest error sixth-order polynomial model in Fig. 7. The agreement is now better for the lower temperature isotherms.
5. Inclusion of T-site Occupation with a Model Excitation Energy As discussed briefly in the Introduction the issue of T-site occupation in palladium hydride is controversial; based on the results of modern density functional calculations T-site occupation should be expected, but there is not at present an unambiguous experimental demonstration of T-site occupation. In previous work we made estimates for the O-site to T-site excitation in the α-phase and in the β-phase . In this section we make use of these earlier estimates for the excitation energy to construct a statistical mechanics model that can be fit to experimental data. The O-site to T-site excitation energy in this case is assumed to depend on loading, but not on temperature.
P.L. Hagelstein / Journal of Condensed Matter Nuclear Science 20 (2016) 54–80 65 5.1. Model for O-site and T-site occupation A model for the chemical potential in terms of the O-site and T-site loading θO and θT, in terms of the O-site and T-site energies EO and ET, was developed previously . The chemical potential for interstitial hydrogen in O-sites and in T-sites is given by µH = EO + θO ∂EO ∂θ + θT ∂ET ∂θ −kBT ln 1 −θO θO −kBT ln zO, (17) µH = ET + θT ∂ET ∂θ + θO ∂EO ∂θ −kBT ln 2 −θT θT −kBT ln zT.
(18) Subtracting these two relations leads to ET −EO = kBT ln 2 −θT θT θO 1 −θO + ln zT zO .
(19) This we solve to determine θT given θO, and then make use of Eq. (17) for the chemical potential of H in the solid phase.
5.2. T-site partition function In this analysis we have made use of a non-SHO T-site partition function of the form zT = 4 n exp −En kBT 3 (20) with En = ℏωTns.
(21) The excitation energy and scaling parameter for O-site occupation are ℏωT = 53.8 meV, sT = 1.2.
(22) 5.3. O-site to T-site excitation energy For T-site occupation we make use of a model for the O-site to T-site energy parameterized by ∆E(θ) = α0 + α1θ 1 + β1θ + 1 meV (23) with fitting parameters α0 = 100.676 meV, α1 = 824.259 meV K−1, β1 = 3.1108 K−1.
(24) 66 P.L. Hagelstein / Journal of Condensed Matter Nuclear Science 20 (2016) 54–80 θ 0.0 0.2 0.4 0.6 0.8 1.0 ∆E(θ) (meV) 0 50 100 150 200 250 300 Figure 8.
Empirical model for the O-site to T-site excitation energy as a function of loading θ.
This empirical model is shown in Fig. 8. Estimates for the excitation energy in the α-phase near θ = 0 and in the β-phase near θ = 1 were developed previously from the analyses reported in [34,35]. For the intermediate loading regime the neutron diffraction measurements of Pitt and Gray can be interpreted in terms of an excitation energy for palladium deuteride. Based on our earlier analysis we would expect that the O-site excitation energy is probably similar for palladium hydride near room temperature and above.
5.4. Models based on a temperature-independent O-site energies We have optimized a set of models with O-site and T-site occupation assuming a temperature-independent O-site energy model. The errors are given in Table 3.
The fitting coefficients (in meV) for the sixth-order polynomial solutions are a0 = −62.763, a1 = −249.626, a2 = 173.686, a3 = −91.330, a4 = 32.832, a6 = −110.942.
(25) Table 3.
Errors associated with the opti-mization of models with O-site and T-site oc-cupation considered in this subsection.
Order I 3 | ∆E model 4.22 × 10−4 4 | ∆E model 4.19 × 10−4 5 | ∆E model 4.09 × 10−4 6 | ∆E model 4.05 × 10−4 P.L. Hagelstein / Journal of Condensed Matter Nuclear Science 20 (2016) 54–80 67 θ 0.1 0.3 0.5 0.7 0.9 0.0 0.2 0.4 0.6 0.8 1.0 EO(θ,T)+Ed/2 (meV) -180 -160 -140 -120 -100 -80 -60 Figure 9.
O-site energies as a function of loading for the different temperatures of the isotherms (blue lines); extrapolation to T = 0 (black line).
In this case lower errors would have been obtained with no T-site occupation, as we can see by comparing with Table 1.
According to this there is no benefit (in terms of error reduction) to augmenting a model based on O-site occupation with a temperature-independent EO(θ) so that it includes T-site occupation based on the ∆E(θ) model given above.
5.5. Temperature-dependent O-site energy Next we consider models based on an O-site energy that depends on both loading and on temperature, combined with a T-site model based on the O-site to T-site excitation energy given above. The intention here is to account for the configurational splitting discussed above, and by doing so reduce the associated optimization error. Results for the errors are given in Table 4.
Results for the different O-site energies for the sixth-order version of the model are shown in Fig. 9. In contrast to the previous unphysical results for the temperature-dependent O-site energies in the last section (Fig. 6), these O-site energies are increasing with increasing temperature. Also, the increase in the O-site energy at θ = 0 is now comparable to what was found in our earlier analysis of the α-phase solubility in Ref. .
The fitting coefficients for the 6th-order polynomial model are Table 4.
Errors associated with the opti-mization of the different models with O-site and T-site occupation considered in this sec-tion.
a,b Order I 3,3 | ∆E model 8.07 × 10−5 4,4 | ∆E model 2.99 × 10−5 5,5 | ∆E model 2.65 × 10−5 6,6 | ∆E model 1.93 × 10−5 68 P.L. Hagelstein / Journal of Condensed Matter Nuclear Science 20 (2016) 54–80 θ 10-5 10-4 10-3 10-2 10-1 P (atm) 10-8 10-7 10-6 10-5 10-4 10-3 10-2 10-1 100 101 102 103 104 105 θ 0.2 0.4 0.6 0.8 1.0 10-8 10-7 10-6 10-5 10-4 10-3 10-2 10-1 100 101 102 103 104 105 Figure 10.
Isotherms and data points for the model with temperature-dependent 6th-order polynomial fit of EO and an empirical ∆E model.
a0 = −77.093, a1 = −92.584, a2 = −496.658, a3 = 361.628, a4 = 940.828, a5 = −1130.451, a6 = 325.675, b0 = 0.0330887, b1 = −0.3739481, b2 = 2.0569084, b3 = −3.9515725, b4 = 3.7471947, b5 = −1.9623168, b6 = 0.5147468, (26) where the aj coefficients are in meV, and the bj coefficients are in meV/K.
5.6. Phase diagram The phase diagram that results is shown in Fig. 10. Although there are minor issues, in general this appears to be a pretty solid match between model and data. We have in this model succeeded in obtaining a good match with the phase diagram with a model that is physically plausible.
P.L. Hagelstein / Journal of Condensed Matter Nuclear Science 20 (2016) 54–80 69 6. Unconstrained Models with Both O-site and T-site Occupation In the previous section we made use of a fixed model for O-site to T-site excitation energy based on earlier analyses with different data sets. In this section we allow for the loading-dependent excitation energy itself to be determined directly as a result of the optimization. The motivation for this is to see whether the phase diagram data set itself can be used to provide an independent estimate for the O-site to T-site excitation energy.
6.1. Models with temperature-dependent O-site energy We discussed above that there is a spread in the energies of the different configurations for partially loaded palladium hydride, and that a crude way to account for this might be to work with a temperature-dependent O-site energy. The O-site energy in this case is parameterized using fifth-order polynomials in loading according to EO(θ, T ) = a0 + a1θ + a2θ2 + a3θ3 + a4θ4 + a5θ5 + T b0 + b1θ + b2θ2 + b3θ3 + b4θ4 + b5θ5 .
(27) The O-site to T-site excitation energy is similarly parameterized according to ∆E(θ) = d0 + d1θ + d2θ2 + d3θ3 + d4θ4 + d5θ5.
(28) We have optimized a set of models of this kind, with the results of optimization listed in Table 5. We see that the associated errors are lower than what we found for all other models considered in this work.
θ 0.1 0.3 0.5 0.7 0.9 0.0 0.2 0.4 0.6 0.8 1.0 ∆E (meV) 0 100 200 300 Figure 11.
O-site to T-site excitation energy as a function of θ, for models with temperature-dependent O-site energy EO and unconstrained temperature-independent O-site to T-site excitation energy; ∆E linear in θ (dark blue line); second-order model (dark green line); third-order model (blue line); fourth-order model (red line); fifth-order model (pink line).
70 P.L. Hagelstein / Journal of Condensed Matter Nuclear Science 20 (2016) 54–80 The fitting coefficients for the lowest error fifth-order polynomial solution are: a0 = −65.426, a1 = −5.627, a2 = −1099.972, a3 = 1503.386, a4 = −266.445, a5 = −226.248, b0 = −0.007048, b1 = −0.364993, b2 = 2.052026, b3 = −2.207565, b4 = −0.033973, b5 = 0.596698, d0 = 178.523, d1 = 615.281, d2 = −2641.792, d3 = 3242.712, d4 = −1066.321, d5 = −17.063, (29) where the aj and dj coefficients are in meV, and the bj coefficients are in meV/K.
6.2. O-site to T-site excitation energy These results support the notion that T-site occupation occurs in palladium hydride, since we obtain better agreement with the phase diagram when T-site occupation is included. The optimized O-site to T-site excitation energies (shown in Fig. 11) are between about 100 and 300 meV, which are in the general range of what we estimated in earlier studies [34,35]. We note that the fifth-order polynomial curve with the lowest associated error lies well above the empirical model of Fig. 8 at θ = 0, where we have some confidence in the excitation energy based on our analysis of the α-phase experimental data. It also lies well above the empirical model of Fig. 8 where we have some confidence in the analysis of experimental results at high loading.
6.3. O-site energy Next consider the spread of O-site energies for the lowest error model, shown in Fig. 12. We recall that in the physical system there is a spread in the energies of the different configuration, which we seek to model with a temperature-dependent O-site energy. At high loading the O-site energies are seen to increase with temperature, consistent with the splitting effect under consideration. However, at low loading the situation is reversed. We recognize this solution as a mathematical optimum, and not as an acceptable physical solution.
Table 5.
Errors associated with the un-constrained optimization of different models with O-site and T-site occupation.
a,b | d order I 5,5 | 0 2.17 × 10−5 5,5 | 1 2.03 × 10−5 5,5 | 2 1.34 × 10−5 5,5 | 3 9.28 × 10−6 5,5 | 4 8.22 × 10−6 5,5 | 5 8.08 × 10−6 P.L. Hagelstein / Journal of Condensed Matter Nuclear Science 20 (2016) 54–80 71 θ 0.1 0.3 0.5 0.7 0.9 0.0 0.2 0.4 0.6 0.8 1.0 EO(θ,T)+Ed/2 (meV) -200 -180 -160 -140 -120 -100 -80 -60 Figure 12.
O-site energies as a function of loading for the different temperatures of the isotherms (blue lines); extrapolation to T = 0 (black line).
Above an H/Pd loading of 0.26 the curves are ordered with increasing excitation energy corresponding to increasing temperature.
6.4. Phase diagram The phase diagram for the model with the lowest error is shown in Fig. 13. The fit is very good, as expected.
7. Constrained Models with Both O-site and T-site Occupation The low errors found in the optimization of the previous section provide motivation to consider a constrained opti-mization, where the O-site to T-site excitation energy is again determined based on optimization, but now we constrain the excitation energy to values estimated in previous work at θ = 0 and at θ = 1. It is possible that an independent estimate for the excitation energy at intermediate loading can be determined from this kind of optimization.
7.1. Model In this case we use an excitation energy of the form ∆E(θ) = d0 + d1θ + d2θ2 + d3θ3 + d4θ4 + d5θ5 with d0 = 101.676 meV.
(30) For the O-site energy, we made use of fifth-order polynomials in θ and assumed a linear temperature dependence as before.
The optimization is carried out on the modified error J given by 72 P.L. Hagelstein / Journal of Condensed Matter Nuclear Science 20 (2016) 54–80 θ 10-5 10-4 10-3 10-2 10-1 P (atm) 10-8 10-7 10-6 10-5 10-4 10-3 10-2 10-1 100 101 102 103 104 105 θ 0.2 0.4 0.6 0.8 1.0 10-8 10-7 10-6 10-5 10-4 10-3 10-2 10-1 100 101 102 103 104 105 Figure 13.
Isotherms and data points for the model with temperature-dependent fifth-order polynomial fit of EO and optimized fifth-order model for ∆E(θ).
J = 1 N j µH(data) −µH(model) 2 j + γ 226 meV − 5 i=0 di 2 .
(31) We choose low values for γ in the vicinity of unity for the initial optimizations, and then later on increase γ to 100.
This leads to values of ∆E at θ = 1 close to 226 meV.
7.2. Results Constrained optimizations have been carried out on a set of models with the results given in Table 6. The errors I (see Eq. (12)) that result are larger than what we obtained in the unconstrained optimizations of the previous section (see Table 5). There is a minor penalty associated with imposing constraints on the model parameters.
The fitting coefficients for the lowest error model with fifth-order polynomial fits are a0 = −77.889, a1 = −6.933, a2 = −1102.321, a3 = 1616.798, a4 = −453.666, a5 = −149.614, P.L. Hagelstein / Journal of Condensed Matter Nuclear Science 20 (2016) 54–80 73 θ 0.1 0.3 0.5 0.7 0.9 0.0 0.2 0.4 0.6 0.8 1.0 ∆E (meV) 0 50 100 150 200 250 Figure 14.
O-site to T-site excitation energy as a function of θ, for models with temperature-dependent O-site energy EO and temperature-independent O-site to T-site excitation energy; ∆E linear in θ (dark blue line); second-order model (dark green line); third-order model (blue line); fourth-order model (red line); fifth-order model (pink line).
b0 = 0.034584, b1 = −0.454285, b2 = 2.306099, b3 = −2.428471, b4 = −0.179903, b5 = 0.805586, d1 = 413.703, d2 = −951.745, d3 = −612.564, d4 = 2793.066, d5 = −1517.832, (32) where the aj and dj are in meV, and the bj are in meV/K.
7.3. Excitation energy The excitation energy for the different versions of the model are shown in Fig. 14. The models converge slowly with order (and have not finished converging as a function of the polynomial order), with the O-site to T-site excitation Table 6.
Errors associated with the con-strained optimization of the different models with O-site and T-site occupation considered in this section.
a,b | d Order I 5,5 | 1 2.71 × 10−5 5,5 | 2 2.47 × 10−5 5,5 | 3 2.16 × 10−5 5,5 | 4 1.70 × 10−5 5,5 | 5 1.50 × 10−5 74 P.L. Hagelstein / Journal of Condensed Matter Nuclear Science 20 (2016) 54–80 θ 0.1 0.3 0.5 0.7 0.9 0.0 0.2 0.4 0.6 0.8 1.0 EO(θ,T)+Ed/2 (meV) -200 -180 -160 -140 -120 -100 -80 -60 Figure 15.
O-site energies as a function of loading for the different temperatures of the isotherms (blue lines); extrapolation to T = 0 (black line).
energy for the fifth-order model (in pink) first increasing with loading, then decreasing, and then increasing again. This reversal leads to low values for the excitation energy going down to about 117 meV at θ = 0.56. This reversal moves the excitation energy sufficiently low to be in disagreement with the neutron diffraction data for palladium deuteride.
7.4. O-site energy The O-site energies for the lowest error version of the model are shown in Fig. 15. Constraining the excitation energy at θ = 0 results in O-site energies which seem generally reasonable. We might also expect the largest change with temperature to occur generally in the vicinity of θ = 1/2, since this is where the splitting is probably largest. We see in Fig. 15 that the largest increase in temperature occurs at a loading closer to θ = 0.66, and that there is a decrease for larger θ. Such a behavior might be explained if there is an additional thermal effect, such as an increase in lattice expansion with temperature at high loading.
The spread in O-site energy for these models has gotten sufficiently large in this case that we would be concerned about the resulting model being acceptable on physical grounds. If the excitation energy were larger in the general vicinity of θ = 1/2, then the spread would be reduced. This suggests that an excitation energy more consistent with the neutron diffraction experiments in this regime would result in a more physical and better model.
There is a spread near θ = 0 which is worth some discussion. We note that we found a similar increase in the O-site energy with temperature in our previous analysis of solubility for α-phase PdHx and PdDx . There is agreement in the magnitude of this effect in the different models (of this work, and of the previous one), which are derived from different data sets; this is encouraging, and suggests consistency both in the experimental data and in the analysis. The origin of the effect lies in the anomalously large shift of the Fermi level which at T = 0 is part way down a sudden P.L. Hagelstein / Journal of Condensed Matter Nuclear Science 20 (2016) 54–80 75 θ 10-5 10-4 10-3 10-2 10-1 P (atm) 10-8 10-7 10-6 10-5 10-4 10-3 10-2 10-1 100 101 102 103 104 105 θ 0.2 0.4 0.6 0.8 1.0 10-8 10-7 10-6 10-5 10-4 10-3 10-2 10-1 100 101 102 103 104 105 Figure 16.
Isotherms and data points for the model with temperature-dependent fifth-order polynomial fit of EO and optimized constrained fifth-order model for ∆E(θ).
drop in the density of states (see Fig. 2 of Ref. ). We would not expect to see such a large shift due to the Fermi level at higher loading. This expectation is consistent with the rapid fall off of the spread with θ that can be seen in Fig. 15.
7.5. Phase diagram The phase diagram that results from the lowest error constrained optimization is shown in Fig. 16. This phase diagram is very similar to the one we obtained with the unconstrained optimization. It looks very good in general.
8. Discussion and Conclusions In this study we optimized a set of empirical models for the O-site energy, and in many cases also the O-site to T-site excitation energy, by minimizing the error between the model chemical potential and the isotherm data set. The general approach seems to be quite successful in giving models that match the isotherms of the data set, and also sheds light on the physical system. These are some of the simplest empirical models that can produce relatively high quality phase diagrams.
The optimization of a Lacher type of model with a temperature-independent O-site energy that is higher-order in θ is conceptually straightforward, and it produces better results than we had expected initially. We have not found such models studied systematically previously in the literature. The generalization to temperature-dependent O-site energies results in a phase diagram that is probably good enough to work with for some applications. The only problem 76 P.L. Hagelstein / Journal of Condensed Matter Nuclear Science 20 (2016) 54–80 with it is that it does not correspond to a plausible physical model since the O-site energies decrease with temperature at low loading.
The issue of T-site occupation remains somewhat controversial at this time, primarily due to the lack of completely unambiguous experimental evidence showing explicitly significant T-site occupation. There is substantial theoretical support in density functional calculations and in quantum chemistry calculations for sufficiently low T-site energies such that one would expect them to have noticeable occupation and impact the phase diagram. The results from neutron diffraction experiments in palladium deuteride (showing T-site occupation) seem clear enough; however, the significance of these experimental results do not appear at this time to be widely appreciated. Consequently, our use of models with both O-site and T-site occupation in this work is likely to be acceptable to the theorists, and objectionable to the experimentalists.
Unconstrained optimization assuming both O-site and T-site occupation leads to models with very low mean square errors, but which are not physically acceptable at low loading since the O-site energy decreases with temperature.
Constrained optimization with the excitation energy fixed at θ = 0 fixes this problem, and results in a good match with low error. The biggest headaches in this case are that the excitation energy seems a bit low relative to the PdD neutron diffraction experiments, and the spread in the O-site energies in the mid-range of loading is larger than what we would hope for.
When we make use of a model with fixed ∆E(θ) in Section 5, these problems are improved; the resulting phase diagram looks very good. In this case the only headache is that the error I is a bit larger, due mostly to the imposed boundary condition at θ = 1. This draws attention to the extrapolation that we used for high loading, based on a generalization of the approach of Baranowski et al. . The issue here is that in neither case does the extrapolation take into account possible T-site occupation. Additional experimental PCT data at high loading and at high temperature will be needed to resolve this issue.
In general these results strongly support the hypothesis that T-site occupation occurs and impacts the phase diagram.
One could argue that the results are not site specific, as any other site with a similar excitation energy would work equally well. However, at this point for us there are no other obvious competitive candidates.
It seems useful to consider the arguments in support of T-site occupation briefly here. We have made use of a temperature-dependent O-site energy to account for configurational splitting that would be associated with a more general lattice gas model. The optimization of the associated parameters leads to models in the absence of T-site occu-pation with the wrong temperature dependence at low loading (as discussed in Section 4). Including T-site occupation corrects this, in the case that the O-site to T-site excitation energy is close to the empirical model discussed in Section 5. One interpretation is that this means that we would not expect a lattice gas model based on O-site occupation to give as good of a match to the data set as an equivalent lattice gas model which includes T-site occupation. In our earlier analysis of α-phase solubility, where the O-site to T-site excitation energy is lowest, and where we have access to a larger spread in temperature, the analysis is unambiguous . With T-site occupation a good fit to the data can be obtained; without T-site occupation there is no hope for a physically acceptable model. Consequently, this earlier analysis weighs more heavily in support of T-site occupation, and the present study is supportive based on an analy-sis of different data over a larger range of loading. Our earlier analysis of PdH and PdD at high loading near room temperature also requires T-site occupation in order to make sense of published data. Here our basic conclusion is that the available PCT data for palladium hydride over the entire range of available temperature and loading independently provides support for T-site occupation.
The error I from the optimizations is the variance of the chemical potential. We can relate the associated standard deviation to the standard deviation of the fugacity approximately according to ∆µH kBT = 1 2 ∆f f .
(33) P.L. Hagelstein / Journal of Condensed Matter Nuclear Science 20 (2016) 54–80 77 The lowest errors from the different models correspond to values of ∆µH near 3–4 meV. At low pressure the fugacity is very nearly equal to the pressure; at high pressure the fugacity can be much larger. In this case it may be that optimizations based on ln P instead of µH would be preferred, as this would tend to reduce the weight associated with the (poorly known) high pressure points. This might be of interest in a future study.
Yet another issue that might be addressed in future work involves the issue of the statistical weight of the O-site states in the presence of splitting. Including temperature dependence of the O-site states as in the empirical models studied in this paper implies a splitting in the O-site states, where lower energy states have more occupation at low temperature, and with where higher energy states have more occupation at elevated temperature. In this kind of picture, we would expect a reduction in the associated entropy of the O-site states which would occur if the splitting were modeled explicitly. This has not been addressed in the models considered above. It may be possible to develop even better empirical models by including this effect.
We note that there are other approaches to modeling the phase diagram; for example, as in the recent analysis of Joubert and Thiébaut . In this work the parameterization is a physical chemistry model in terms of the Gibbs energy rather than a physics type of parameterization in terms of site energies as we have used.
The optimizations reported here lead to estimates for the O-site and T-site energies, which are not so readily available otherwise from experiment. These energies are of interest as many groups at present work with density functional codes and quantum chemistry codes which predict them.
Appendix: No Abrupt Change Near a Loading of Unity In the course of the review process the reviewer raised issues pointing out the absence of an abrupt change in the experimental data near a loading of unity, which underscores the absence of experimental support for T-site occupation at high loading.
In one communication the reviewer wrote: The upper phase limit for the beta phase cannot be obtained simply by measuring the H/Pd (D/Pd) ratio because a second hydrogen-rich phase can form, thereby creating a two phase mixture beyond the limit of the beta phase that would have an overall composition above the fcc limit of the beta phase. This behavior is seen in the other hydride systems as well as in the behavior of fcc compounds containing C and N. Instead, the upper limit has to be based on seeing an abrupt change in slope of some measured value. In fact, this change in slope is detected in the data near D/Pd=0.98 using the resistivity and the temperature effect on the resistivity combined with the overall composition.
In another communication the reviewer wrote: A phase boundary occurs at the composition at which another crystal structure forms with a greater concentration of the second atom. For example, the phase limit of beta-VH occurs at H/V=0.8 where another phase having the crystal composition of VH2 forms (page 15, Fukai). If only the overall composition were measured, the composition would be seen to increase beyond 0.8. This increase alone would not reveal that the increase resulted from formation of the VH2 phase rather than H being added to the beta phase. This limitation applies to PdH when the composition is noted to increase above H/Pd=1. In the absence of being able to detect the additional crystal structure, the boundary has to be identified by a change in behavior, such as a change in slope seen in the resistivity or chemical activity. Changes in slope of both are seen near D/Pd=1. Graphs showing the result of pressure reported by Baranowski et al. and the effect of temperature on the resistivity by Tripodi et al. are attached.
In other words, no evidence supports the ability of the beta phase to accommodate D or H above the normal filling of the O-sites. Extra D or H in the T-sites or in metal atom vacancies is not supported by the location of the upper phase boundary, at least to an amount of occupancy that the accuracy of the measurements permit.
78 P.L. Hagelstein / Journal of Condensed Matter Nuclear Science 20 (2016) 54–80 A person might argue that even though no extra H were present, the H in the lattice might be distributed between the O and T sites while retaining a H/Pd ratio below unity at the upper boundary. Why this novel distribution might form would provide a challenge for the paper to explain.
These comments of the reviewer are interesting, and worth some additional discussion. The issues under discussion are relevant to the analysis presented in an earlier work , and somewhat less so to the analysis of this paper since we included phase diagram data only below θ = 1.
When we began the earlier analysis we were expecting to see a sudden change in the chemical potential at the θ = 1 boundary, consistent with intuition that suggested the O-site to T-site excitation energy is large near θ = 1. A very large number of models were analyzed with an excitation energy between 350 meV and 600 meV, and the results showed an abrupt change where below θ = 1 only O-site occupation occurred, and above θ = 1 the O-sites were completely filled and only incremental T-site occupation occurred with increased loading. Predictions from these early models would correspond will to the expectation that one would expect a sharp boundary between the β-phase for θ < 1 and a T-site phase for θ > 1.
It took a long time to move past this initial perspective and investigate models with lower O-site to T-site excitation energies. Such models behave qualitatively differently in that there is no sharp boundary in the chemical potential, or in any readily observable model parameter, at θ = 1. The 226 meV excitation energy estimated in is sufficiently low that a smooth transition occurs with partial occupation of O-sites and T-sites below and above θ = 1.
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P.L. Hagelstein / Journal of Condensed Matter Nuclear Science 20 (2016) 54–80 79 J.R. Lacher, A theoretical formula for the solubility of hydrogen in palladium, Proc. Roy. Soc. London, Series A, Mathematical and Physical Sciences 161 (1937) 525–545.
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Minimizing the median, integer programming
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Suppose we want to
mini median(ai)
ai are real numbers
Does someone know how to pose this as an integer programming problem or point me in the direction of a resource?
optimization
integer-programming
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asked Aug 29, 2017 at 18:40
VogtsterVogtster
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If ai are real numbers, the median of them is a constant, what is there to minimize? Or are you minimizing over different ai, in which case, what are the possible ranges of ai? If ai∈R is unrestricted, the problem is unbounded below, by driving all ai→−∞.
– gt6989b
Commented
Aug 29, 2017 at 18:43
Sort of you can think I'm doing minximedian(ai(xi)). The Things I'm taking medians of are functions of the xi
– Vogtster
Commented
Aug 29, 2017 at 18:53
@Vogster, The median is minimized when each of its arguments are separately minimized, so if your problem is you are given n one-variable functions f1,…,fn and asked to find a1,…,an such that the median of the fi(ai) is minimized, the answer is to find the minimum of each fi separately. Or is each fi a function of all n variables?
– user326210
Commented
Aug 29, 2017 at 19:11
To take a median, you need a set, a way of comparing members of the set (a "less than" relation), and a way of "counting" the members of the set so you can distinguish the "lower half" of the set from the "upper half". The question and comments do not make it clear that you have any of those things. It's also not clear how minxi would be different from mini (if each i in whatever set i comes from has a corresponding xi, then trying each i means you try each xi).
– David K
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Aug 29, 2017 at 19:26
@user326210 its a function of all n variables.
– Vogtster
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Aug 29, 2017 at 20:57
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EDIT: Removing a simpler model which was incorrect.
The way median
```
y = median(a)
```
is implemented in the optimization modelling toolbox YALMIP is roughly by (writing in MATLAB pseudo code)
```
y = s(length(a)/2); s = sort(a);
```
Hence, to model median we need to model sort. This can be done by introducing a binary matrix Z with
```
s = Za, sum(Z) = 1, sum(Z') = 1, diff(s) >= 0
```
and we're down to model binary times continuous, which is done using standard big-M methods.
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edited Aug 29, 2017 at 19:46
answered Aug 29, 2017 at 19:12
Johan LöfbergJohan Löfberg
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Johan, could you be a bit more specific about the big-M methods that can be used here? When I do a search, all that comes up in this context is MIQCP methods. Is this what you are referring to, or are there simpler methods?
– Alex
Commented
Mar 20, 2019 at 23:41
In the final paragraph here yalmip.github.io/command/binmodel. Note that I am not saying that this model is the best way to represent median using an integer programming, it is just one simple
– Johan Löfberg
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Mar 21, 2019 at 7:38
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Given bounded variables a1,…,aN subject to some
constraints, you can minimize their median with N additional binary
variables z1,…,zN and one additional continuous variable
y if N is odd (or if you are a bit sloppy about the definition
of ``median''). You minimize y subject to the constraints y≥ai−Mizi,i=1,…,N
and ∑Ni=1zi=⌊N2⌋,
where the Mi are suitably large constants. The constraint force
y to be greater than or equal to ⌈N2⌉
of the ai; the objective will result in those being the ⌈N2⌉
smallest of them and in y being no bigger than the largest of that
set, making y the median.
If N is even and a(1),…,a(N) is the order statistic
of the a variables, the median is technically (a(⌊N2⌋)+a(⌈N2⌉))/2.
If you can live with using a(⌊N2⌋)
as the ``median'', the above should work. Otherwise, you need a
second set of binary variables (w1,…,wN) and a second
continuous variable (replace y with y1 and y2). You
minimize (y1+y2)/2 subject to the constraints
y1y2∑i=1Nzi∑i=1Nwi≥ai−Mizi∀i≥ai−Miwi∀i=N2−1=N2
where the Mi are as above. The constraints force y1 to
be at least as large as N2+1 of the ai and y2
to be at least as large as N2 of them. The minimum objective
value will occur when y1=a(N2+1) and y2=a(N2).
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answered Aug 31, 2017 at 14:57
prubinprubin
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11012 | https://www.ck12.org/flexi/precalculus/solving-trigonometric-equations/solve-the-equation-on-the-interval-0-2p)-lessmathgreater-beginarrayc-cos-x-1-cos-x-xfracpitext-fractext-pitext-endarray-less-by-mathgreater/ | Flexi answers - Solve the equation on the interval [0, 2π): @$\begin{align} \begin{array}{c} \cos x-1=-\cos x \ x=\frac{\pi}{\text{[?]}}, \frac{\text{[?]} \pi}{\text{[?]}} \end{array} \end{align}@$ | CK-12 Foundation
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Question-lessmathgreater-beginarrayc-cos-x-1-cos-x-xfracpitext-fractext-pitext-endarray-less-by-mathgreater/ "Solve the equation on the interval [0, 2π):
@$\begin{align}
\begin{array}{c}
\cos x-1=-\cos x \
x=\frac{\pi}{\text{[?]}}, \frac{\text{[?]} \pi}{\text{[?]}}
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Solve the equation on the interval [0, 2π): cosx−1=−cosx x=π[?],[?]π[?]
Flexi Says:
To solve the equation cosx−1=−cosx on the interval [0,2 π), we first combine like terms:
2 cosx=1
Then, we solve for x:
x=cos−1(1 2)
The cosine function has a period of 2 π, and it is positive in the first and fourth quadrants. Therefore, the solutions to the equation on the interval [0,2 π) are:
x=π 3,5 π 3
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How can you solve for the general solution of trigonometric equations?What are all the exact solutions of Unexpected text node: '\cos{x} = -\frac{\sqrt{3}}{2}' on the interval Unexpected text node: '\pi \le x \le 2\pi'? Give your answer in radians.Unexpected text node: '(\tan \theta+2)(2 \tan \theta+1)=5 \tan \theta+\sec ^{2} \theta' .
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11013 | https://homepages.gac.edu/~jsiehler/NoThree/noThree.html | The No-Three-In-Line Problem
A Puzzle
Mark six more points (making twelve in all) without allowing three along any straight line.
The figure to the right shows a 6×6 grid with six of the points marked as clues. You can mark (or unmark) additional grid points by clicking on them. The problem:
Mark six more points, making a total of 12.
You can't have three marked points along any straight line.
That includes not just horizontal and vertical lines, but lines of any slope.
Give it a try! You can reason the solution out logically from the given clues.
A Family of Problems
"The No-Three-In-Line Problem" is really a family of problems centered around the general question, "How many markers can be placed on an n×n grid with no three in a line?" Some basic observations:
You can't have more than 2 in any row, so the answer can't be more than 2n.
To get 2n markers on an n×n grid, you need 2 in each row (and 2 in each column).
For small values of n (that is, small grid sizes), it's not hard to find solutions that place 2n markers with no three in a line, even working by hand. We'll look at some of these small grids as puzzles in the next few pages.
For larger values of n, it is not known whether 2n markers can be placed, or whether the maximum number is smaller than 2n. This is one of the outstanding questions.
The 4×4 Solutions
Find all four ways to solve the 4×4 grid. The solutions display a variety of symmetry types.
In all of the following pages, when I refer to a “solution” on the n×n grid, I mean an arrangement of 2n markers with no three in line.
There are essentially four different solutions on the 4×4 grid. Each of the diagrams to the right gives enough clues to determine one of the four solutions uniquely.
You can get other solutions by rotating or reflecting these four. If you take all of those different orientations into account then there are 11 distinct solutions in all.
About the Numbers
It seems fair to assume that many mathematicians and puzzle enthusiasts have worked out the different solutions for small grids over the years. Beyond a certain point, enumerating the possibilities by hand becomes impractical, and computer searches have taken over.
The number of distinct solutions for each grid size up to 18×18 is known. These numbers are available in the Online Encyclopedia of Integer Sequences (Sequence A000769).
Although it is dated, the most comprehensive single site for information about this problem is still Achim Flammenkamp's No-Three-In-Line page. Among other things, you can find a table listing the number of solutions for different grid sizes, and you can download a file containing all the solutions known to Flammenkamp. That includes all of the solutions on grids up to 16×16, and certain solutions with larger sizes. Information about the complete set of solutions for n=17 and n=18 doesn't appear to have been published, other than the entry in the OEIS. If further information was once available on the web, it doesn't appear to be any longer.
The 5x5 Solutions, Part 1
Find three solutions on the 5×5 grid (with generous clues).
On the 5×5 grid, there are five essentially different solutions. Clues for three of the solutions are shown above, and they should be easy to solve. The other two are on the next page. They should be a little more difficult.
Counting slopes
The 5×5 grid is the smallest where lines other than horizontal, vertical, and diagonal (slope ±1) are a factor. As the grid size increases, the number of different slopes to consider increases as well. There's a formula for the number of different slopes determined by pairs of points in the n×n grid, namely, [\sum_{q=1}^n 4\varphi(q),] where (\varphi) denotes Euler's totient function. This sum increases whenever n increases, with the largest jumps occurring at prime values of n. The ever-increasing number of different slopes to consider is the essential reason why it is conjectured that placing 2n markers is impossible once n is sufficiently large. There are (conjecturally) just too many constraints to satisfy.
The 5x5 Solutions, Part 2
Find the remaining two solutions on the 5×5 grid (with slightly tougher clues).
Beyond the 5×5 grid, the number of solutions increases rapidly, although the growth is quirky. There are fewer solutions on the 9×9 grid than on the 8×8, for example, and the number of solutions for the 10×10 grid is nearly identical to the number for the 11×11. This seeming “quirkiness” is at least partially related to the prominent role of the Euler totient function in this problem (see above). Quirks aside, the number of solutions does grow rapidly. In fact, throughout the range where the number of solutions is currently known, the number of solutions appears to grow exponentially. If the outstanding conjecture is correct, however, this growth eventually must halt and, for sufficiently large grids, the number of solutions plummets to zero. Once that happens, the question is, "If we can't place 2n markers on the n×n grid when n is large, then how many can we place?"
An Oldie: Dudeney's Problem
The first instance of the no-three-in-line problem in print was Problem 317 ("A PUZZLE WITH PAWNS") in H. E. Dudeney's Amusements in Mathematics. His problem reads:
Place two pawns in the middle of the chessboard, one at Q4 and the other at K5. Now, place the remaining fourteen pawns (sixteen in all) so that no three shall be in a straight line in any possible direction.
Dudeney's clue doesn't quite determine the solution uniquely, so the version here adds one additional marker as a clue. Even with the third marker, the puzzle is still challenging. If you like, you can make it a bit easier.
Dudeney's problem was mentioned (along with a miscellany of other puzzles) by Martin Gardner in a 1977 article. The article was later anthologized, with an addendum summarizing progress on the generalized no-three-in-line problem, in Penrose Tiles to Trapdoor Ciphers (1989). Gardner also introduced a variant of the problem, asking for the minimum number of markers which can be placed on the n×n grid so that no further markers can be placed without producing three in a line.
Large Voids
The presence of large, regular areas with no markers is a notable feature of some solutions.
The 8x8 solution above is uniquely determined by the indicated square “voids” where no markers are to be placed. You can click here to simplify the puzzle by revealing a few markers.
The 9x9 solution below is uniquely determined by the 2×6 void in its upper left corner. You can click here to simplify by revealing another large void.
Special Symmetry 1
Place 16 markers with mirror symmetry.
It is very rare for a solution to have a horizontal or vertical line as its sole line of reflective symmmetry. Only three such solutions are known:
A 4x4, which you can see above (if you solved it)
The 8x8 to be solved here
A 26x26 solution
A square grid with odd dimensions cannot have any solution of this symmetry type. The reasoning is as follows:
On an odd grid, the central horizontal line is one of the grid lines.
That center line needs to have two markers on it.
Consider the vertical line through one of those markers.
If the remaining markers on that vertical line occur in mirrored pairs...
...then there's an odd number of markers in that column.
The same reasoning applies to mirror symmetry across a vertical line.
It does not apply to symmetry across diagonal lines, however. There are solutions with a single line of diagonal symmetry on both odd and even grids, and they are much less rare. The solution to Dudeney's problem (above) is an example on the 8×8 grid, and two of the 5×5 solutions have this symmetry as well.
Special Symmetry 2
Place 20 markers with full symmetry: horizontal, vertical and diagonal lines of reflection, and 90 degree rotation. This is much easier than it looks.
Solutions with all the symmetries of the square grid (horizontal, vertical, and diagonal lines of reflection, and 90 degree rotations) are also extremely rare. Again, only three such solutions are known:
The (rather silly) 2×2 solution
A 4×4, which you can see above
This remarkable 10×10
Any grid with a solution of this symmetry type must have even dimensions. The reasoning is the same as on the previous page.
It is conjectured that there are no solutions with full symmetry on any grid larger than 10×10. This has been verified up to n=82 by computer, but there's not yet any proof that there are no larger ones.
Knowing that the solution has full symmetry makes this rather too easy as a puzzle. Despite the intimidating look of the blank 10×10 grid, it gets filled in very quickly! Still, I think this must be counted as one of the most remarkable sporadic objects in geometry and combinatorics, particularly if it really is the largest of its kind.
An Impossible Symmetry Type?
Another unresolved question concerns a particular symmetry type. There are no known solutions which have horizontal and vertical lines of symmetry without having 4-fold rotational symmetry. It has been conjectured that there are no such solutions, but I don't know of any feasible strategy that has been proposed for making progress on proving this.
Unequal Frequencies
Not all grid points are equally likely to appear in solutions to the problem. In general, points near the corners and the center of the grid appear in fewer solutions than points which are on, or slightly inside, the inscribed circle touching the edges of the grid.
The images above illustrate this phenomenon for a few of the largest grid sizes where all solutions are known. Points are colored darker when they appear in more solutions.
This correlates roughly inversely with the number of collinear triples in the grid which contain the point. For example, the corner points occur in a large number of collinear triples, but few solutions. Points near the centers of the edges occur in relatively few collinear triples, and occur in correspondingly more solutions.
Among the known solutions, it's not hard to find individual examples where the markers cluster around the circle, and avoid the corners and center. There are some examples on the following pages.
Very Circular Solutions 1
Just for fun, here are a few selected solutions on medium-sized grids which show a tendency to cluster near the circle.
Very Circular Solutions 2
...and here are a few of the larger known solutions showing a high degree of circularity (the largest of these is a 36×36 grid).
Things You Could Do
In the following, n refers to the size of the grid, and "solution" means a configuration of 2n markers on the n×n grid with no three in line. None of these would be trivial, and some of them are almost certain to be ferociously hard, but at least some of them are feasible and potentially rewarding.
Find some new solutions which aren't already known. What is and isn't known:
All the solutions up to n=16 are already known, and available from Flammenkamp's site. Someone probably knows the solutions for n=17 and n=18, but they aren't published or accessible as far as I know.
For n=19 up to n=46, some (but not all) solutions are known. Those which are known have special symmetries. Any asymmetric solution with n>18 would be a new discovery.
There is only a single solution known for each of n=48, 50, and 52. These all have four-fold rotational symmetry. There are probably many others to be discovered.
There are no solutions known for n=47, 49, or 51. These odd sizes don't permit as many symmetries, making it harder to search for solutions.
No solutions are known for any n>52.
Find a general construction which can be used to place a large number of markers on the n×n grid with no three in line, even if it achieves less than 2n. A construction which achieves more than (3n/2) would be new. Existing constructions (see the references below) use the idea of identifying the n×n grid with ordered pairs of integers modulo n, and using pairs that satisfy some quadratic equation(s).
Prove or disprove: there are no solutions larger than 10x10 with full symmetry.
Prove or disprove: Every solution that has horizontal and vertical lines of symmetry has full symmetry (including 4-fold rotation).
Find some (large) value of n for which you can prove that there is no solution (i.e., the maximum number of markers with no three in line is less than 2n)
Prove that, for all sufficiently large n, there are no solutions. Or defy expectations and prove that there are solutions of arbitrarily large sizes.
Find the largest value of n for which a solution exists (if there is such a largest value).
References, Links, and Reading
Most of these references have a further list of references of their own. I've tried to select a few of the most useful starting points.
No Three in Line Problem (Achim Flammenkamp)
Although it hasn't been updated much in many years, Flammenkamp's page is still the most thorough clearinghouse for information solutions to the problem. There's not much theory there.
2. Sequence A00769 in the Online Encyclopedia of Integer Sequences
Gives the number of inequivalent solutions on the n×n grid (up to n=18, currently). Also has a list of references relevant to this problem and some of its close relatives.
3. Random no-three-in-line sets (David Eppstein)
Includes some more current discussion of the asymptotic problem (when n is large, how many markers should we expect to be able to place). Also, why it may be optimistic to expect that we will find the best (largest possible) solutions on large grids by algebraic constructions of the type found in Hall, Jackson, et al.
4. The No-Three-in-Line Problem (Richard K. Guy and Patrick Kelly, 1968)
This is the source for a number of the conjectures mentioned above, and includes an early discussion of the asymptotic problem. A much shorter version of this was published in the Canadian Mathematical Bulletin; the version I have linked includes more detail and interesting discussion.
5. Some advances in the no-three-in-line problem (Hall, Jackson, Sudbury and Wild, 1975)
Gives algebraic constructions for placing markers with no three in line. A simple construction places n markers on the n×n grid, and a slightly more complicated one achieves approximately (3/2)n markers when n is large.
6. Back From the Klondike and Other Problems (from Gardner, Penrose Tiles to Trapdoor Ciphers, 1989)
The book has been reissued a few times, but I think this chapter is unchanged since its appearence in 1989. |
11014 | https://openstax.org/books/prealgebra-2e/pages/1-introduction | Skip to ContentGo to accessibility pageKeyboard shortcuts menu
Prealgebra 2e
Introduction
Prealgebra 2eIntroduction
Search for key terms or text.
Figure 1.1 Purchasing pounds of fruit at a fruit market requires a basic understanding of numbers. (credit: Dr. Karl-Heinz Hochhaus, Wikimedia Commons)
Chapter Outline
1.1 Introduction to Whole Numbers
1.2 Add Whole Numbers
1.3 Subtract Whole Numbers
1.4 Multiply Whole Numbers
1.5 Divide Whole Numbers
Even though counting is first taught at a young age, mastering mathematics, which is the study of numbers, requires constant attention. If it has been a while since you have studied math, it can be helpful to review basic topics. In this chapter, we will focus on numbers used for counting as well as four arithmetic operations—addition, subtraction, multiplication, and division. We will also discuss some vocabulary that we will use throughout this book.
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11015 | https://www.youtube.com/watch?v=NG9TRBfZDJQ | Movement Of Floating Cylinder In A Liquid Vessel | Example Of SHM | Waves And Oscillations.
ENGINEERING TUTORIAL
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In this video, we are going to discuss another SHM example : Movement Of Floating Cylinder In A Liquid Vessel.
Check this playlist for more videos on this subject:
Waves And Oscillations :
Check out the other videos in the playlists below (updated regularly):
Sensors, Transducers and Instrumentation :
Computer Networking and Data Communication :
Communication Systems :
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Analog Electronics :
Digital Electronics :
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Electrical Machines :
Basic Physics :
Electricity And Magnetism :
Internet Of Things :
Neural Networks :
Fluid Mechanics And Hydraulic Machines :
Optoelectronics Devices And Systems :
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2 comments
Transcript:
Introduction hello friends welcome to engineering tutorial so we'll continue our discussion related to waves and oscillations and so far we are discussing about the examples of simple harmonic motion so we started with spring mass arrangement then simple pendulum and then the previous video we discussed about the flow of rise and fall of liquid in a youtube in this video we are going to discuss another example which is the up and down movement of a wooden cylinder which is uh floating on a liquid kept in a vessel so simple harmonic motion we know that it is the oscillatory to and through movement of an object in between two extreme points about a mean or equilibrium position so this up and down movement of uh cylinder in order to understand Oscillatory Motion this let us say we have a vessel containing a liquid of density rho in that ah vessel we have a wooden cylinder okay of mass m okay in that we have a wooden cylinder of mass m which is submerged such that a portion h of that cylinder is inside the liquid this is the initial equilibrium position now let us say this wooden cylinder is pushed further by an additional distance y okay by an additional distance y let us say this much the additional distance y now here when this wooden cylinder is pushed downwards there will be a restoring force acting on this wooden cylinder which will try to again push it upwards towards the equilibrium position further pushing it upward in this way by the same margin upward by a margin plus y such that this will be the final extreme position then again it will come down to this equilibrium position it will again go to this position again to this position so this will continue this oscillatory motion okay this oscillatory motion is simple harmonic motion so how this happens we'll try to understand it Floatation now according to the principle of floatation when an object floats on a liquid of a certain density the weight of the liquid which is displaced by the submerged portion okay the portion which is inside the liquid that is equal to the weight of the body so in this equilibrium position the weight of the cylinder okay the weight of this cylinder is equal to the weight of the liquid displaced by this this much portion okay the portion h which is inside so that is given by mg the mass of the cylinder into g that is the weight which is equal to rho which is the volume of the liquid multiplied with cross sectional area a of the cylinder cross sectional area a and h product of a and h which is the depth or the portion of the cylinder the length of the cylinder which is submerged okay that this is equal to the mass of the cylinder okay mass of the cylinder is this much row a h okay so here volume of the submerged portion is equal to a cross sectional area of the cylinder and h the portion which is inside the liquid and when we multiply rho with it okay rho h this becomes the mass of the cylinder okay this is the cylinder of the submerged portion volume of the submerged portion so this is the mass of the cylinder okay row h into and the weight of the cylinder is just to multiply g this is the weight of the cylinder so clear about this the according to the principle of floatation the weight of the object which is submerged is equal to the weight of the liquid displaced by the submerged portion so here this is the submerged portion this much portion which is inside the liquid so cross sectional area of the cylinder is a okay volume of the submerged portion is product of area and submerged length immersed length a h then density of this liquid is rho we know mass is equal to volume into density so mass is equal to volume into density rho a h okay weight is equal to mass into acceleration into gravity so weight of the cylinder is equal to this rho h into g weight of the cylinder is equal to rho a h g this now the restoring force which acts on this liquid that depends on the extra length to which it is pushed down the extra length to which it is pushed down from the equilibrium position is y h plus y this is the extra portion plus y so the restoring force which acts on this cylinder which tries to push it upwards okay which tries to push it upwards that is equal to the weight of the extra liquid which is displaced okay the weight of the extra liquid which is displaced okay weight of the extra liquid which is displaced okay and that is also in the negative direction the opposite direction upwards okay so the extra length to which it is submerged is y so the volume of that extra portion is given by cross sectional area a of the cylinder multiplied with y that is the volume of the extra portion the volume of the extra portion then the mass of the extra portion that is equal to rho a y that is the mass of the extra portion then the weight of the extra liquid which is displaced because of this pushing down by a factor y that is given by rho a y multiplied with g this and that is equal to the restoring force in the negative direction rho a y g so you get the point the restoring force which acts in the opposite direction of this extra displacement y the factor by which it is pushed down further because of that the extra volume which is created is cross sectional area of the cylinder a which is fixed multiplied with the extra length this length y okay this much portion let's say the extra length which is this this much portion which is extra now because of that the volume x which is created is a y the mass of this extra liquid which is displaced is given by rho a y weight is equal to simply multiplied with g so it becomes rho a y g the restoring force is in the opposite direction of the displacement that's why the negative sign so here this means the restoring force which is in the opposite direction that means SHM Example this will execute simple harmonic motion okay this is the equilibrium position it is pushed downwards and again it will move to this extreme other extreme position then it will come to equilibrium position and this will go on as an example of shm so this is the restoring force now the acceleration of the block the acceleration of this block is equal to restoring force by mass of the block okay the mass of the cylinder or this object okay the acceleration so restoring force we all know that is equal to minus rho a y g mass of the cylinder which we calculated here that is equal to ah rho a h mass of the cylinder is rho a h this so here rho a rho a will get cancelled out so acceleration is equal to minus g by h into y this is the acceleration of the cylinder so again if you notice here the acceleration is directly proportional to the displacement the extra displacement but in the opposite direction so that's why this means that this is the basis of shm motion simple harmonic motion the acceleration is directly proportional to the displacement instantaneous displacement but in the opposite direction now in order to find out the time period okay if you want to find out the time period so for that we know time period is equal to 2 pi root over of displacement by acceleration of the object in this case the cylinder so here if i have to calculate displacement by acceleration which is y by a this acceleration of the cylinder that will be equal to minus h by g okay here a comes to this side in the denominator and this goes g by h goes to the left hand side becomes h by g minus h by g so if we only consider the magnitude then acceleration displacement by acceleration that will be equal to just h by g this so here the time period of this floating cylinder okay let us call it fc time period of this floating cylinder is equal to 2 pi root over of displacement by acceleration which is equal to h by g okay time period of the floating cylinder is two pi root over of h by g and frequency of this floating cylinder is simply the reciprocal of the time period which is equal to 1 by 2 pi root over of g by h this so always understand from the start the cylinder of mass m cross sectional area a is submerged to a portion h length h in a liquid of density rho the weight of the cylinder is equal to the weight of the liquid displaced or the portion which it is submerged that is equal to rho a h volume of the liquid displaced cross sectional area into h which is submerged a h then we multiplied the density of the liquid to calculate the mass and that is equal to rho h that finished then it is pushed further by length y extra length y because of this the weight of the extra liquid which is displaced that produces a restoring force in the opposite direction that is equal to the weight of the extra liquid displaced the weight of the extra liquid displaced is given by the volume extra volume a y multiplied with again the density rho a y then it is again multiplied with the acceleration due to gravity g and that gave us the restoring force negative sign because of the opposite direction now in order to find out the acceleration of the block we have to do this restoring force by mass of the cylinder which is minus rho a y g by rho h rho rho i got cancelled and we got this so now we calculated displacement by acceleration ratio to calculate the time period and we got this so this oscillatory motion goes on and on and on equilibrium position push downward then again it is pushed upward because of the restoring force again it will go to this position equilibrium again downward again upward so this is a oscillatory 2 1 4 movement about this equilibrium position okay so here we have discussed the floating cylinder example of simple harmonic motion so i hope you like this video and please subscribe my channel engineering tutorial for more such videos related to engineering science and technology have a great day thank you very much you |
11016 | https://www.oed.com/dictionary/prosecution_n | Oxford English Dictionary
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11017 | https://math.stackexchange.com/questions/1879509/what-is-the-total-area-belonging-to-only-one-of-four-unit-circles | geometry - What is the total area belonging to only one of four unit circles? - Mathematics Stack Exchange
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What is the total area belonging to only one of four unit circles?
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Asked 9 years, 1 month ago
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Please forgive the crudeness of this diagram.
(I took an image from some psychobabble website and tried to delete the larger circle that's not relevant to my question).
Let's say these are four unit circles joined together such that each circle shares some area with two other circles.
Obviously the total area not shared with other circles is four times the area of this (again please forgive the crude diagram)
Or I could calculate the total are of a single "petal" and multiply that by 4 4. But I have truly forgotten all the calculus and trigonometry I was taught more than half a century ago.
Am I on the right track? Is there a better way than either of the two ideas I've had so far?
P.S. Not sure if osculating circle tag applies.
geometry
circles
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asked Aug 2, 2016 at 22:24
Mr. BrooksMr. Brooks
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Do all circles intersect / touch at the center of the big (erased) circle?Arthur –Arthur 2016-08-02 22:39:16 +00:00 Commented Aug 2, 2016 at 22:39
I wonder what search terms you used, I sure can't find that psychobabble website. I've found plenty of websites with images you could have copied straight into here with no editing, e.g., myy.haaga-helia.fi/~woljo/illustrator/flower-logo just before Step 4.David R. –David R. 2016-08-03 22:39:05 +00:00 Commented Aug 3, 2016 at 22:39
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For one "petal" only, consider this:
area of petal=area of square=(2–√r)2=2 r 2 area of petal=area of square=(2 r)2=2 r 2
For the entire figure:
total area=4 r 2+4⋅1 2 π r 2=4 r 2+2 π r 2 total area=4 r 2+4⋅1 2 π r 2=4 r 2+2 π r 2
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edited Aug 3, 2016 at 14:33
answered Aug 3, 2016 at 5:37
BlueBlue
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Nice one - you could explicitly mention the mirrored/rotated out half-petals to make it a bit more obvious though Tobias Kienzler –Tobias Kienzler 2016-08-03 05:52:12 +00:00 Commented Aug 3, 2016 at 5:52
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Another approach:
Consider the square enclosed in red. It has side length r r, and therefore area r 2 r 2. Notice that a quarter portion of two different circles overlap in this region to create a single petal, which has area p p.
So we have r 2=2(1 4 π r 2)−p r 2=2(1 4 π r 2)−p, which rearranges to p=(π 2−1)r 2 p=(π 2−1)r 2
Then the unshared area belonging to a single circle has area π r 2−2 p=2 r 2 π r 2−2 p=2 r 2
The area of the entire diagram is 4 π r 2−4 p=(2 π+4)r 2 4 π r 2−4 p=(2 π+4)r 2
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edited Aug 3, 2016 at 1:42
answered Aug 3, 2016 at 0:26
Marcus AndrewsMarcus Andrews
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If the circle has radius r r, the area of the "black diagram" is the area of the circle minus the area of two "petals". The area of two "petals" can be obtained as the area of a circle minus the area of the inscribed square. Therefore the area of the black diagram is just the area of the inscribed square 4(r 2/2))=2 r 2.4(r 2/2))=2 r 2.
P.S. My "petal" is the intersection of two circles.
P.P.S. The diagram is composed by two full circles and two black diagrams so its total area is 2 π r 2+2⋅2 r 2=(2 π+4)r 2.2 π r 2+2⋅2 r 2=(2 π+4)r 2.
P.P.P.S. The picture says it all:
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edited Aug 3, 2016 at 9:09
answered Aug 2, 2016 at 22:39
Robert ZRobert Z
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1 This is not correct, as the other answers have shown Ross Millikan –Ross Millikan 2016-08-03 04:01:09 +00:00 Commented Aug 3, 2016 at 4:01
2 @ Ross Millikan I calculated the area of the BLACK region as requested in the question and not the total area of the diagram.Robert Z –Robert Z 2016-08-03 07:11:18 +00:00 Commented Aug 3, 2016 at 7:11
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Draw two lines from the center of a circle, one to the center of the diagram, one to the center of the diagram, the other to a tip of the petal.
This sector has an angle of 90 degrees. The area of the sector = 1 4 1 4 the area of the cirlce or π 4 r 2 π 4 r 2 Now join the endpoints, creating an isosceles right triangle. The area of the triangle is 1 2 r 2.1 2 r 2.
1 2 1 2 petal = (π 4−1 2)r 2(π 4−1 2)r 2
There are 8 half petals.
The total area then is 4 π r 2−8(π 4−1 2)r 2=(2 π+4)r 2 4 π r 2−8(π 4−1 2)r 2=(2 π+4)r 2
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answered Aug 2, 2016 at 23:49
Doug MDoug M
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Let's say the circles each have radius 1 1. Rotating a bit for convenience, their centres could be at [1,0][1,0], [0,1][0,1], [−1,0][−1,0], [0,−1][0,−1]. The area of your black region is the area of a circle minus twice the area common to two adjacent circles. See e.g. here
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answered Aug 2, 2016 at 22:41
Robert IsraelRobert Israel
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11018 | https://radiopaedia.org/articles/ct-cerebral-venography-protocol?lang=us | Published Time: 2024-06-28T08:16:35.383Z
CT cerebral venography (protocol) | Radiology Reference Article | Radiopaedia.org
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CT cerebral venography (protocol)
Last revised by Arlene Campos on 28 Jun 2024
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Citation, DOI, disclosures and article data
Citation:
Ibrahim D, Campos A, Murphy A, et al. CT cerebral venography (protocol). Reference article, Radiopaedia.org (Accessed on 14 Jul 2025)
DOI:
Permalink:
rID:
42089
Article created:
5 Jan 2016, Dalia Ibrahim
Disclosures:
At the time the article was created Dalia Ibrahim had no recorded disclosures.
View Dalia Ibrahim's current disclosures
Last revised:
28 Jun 2024, Arlene Campos
Disclosures:
At the time the article was last revised Arlene Campos had no financial relationships to ineligible companies to disclose.
View Arlene Campos's current disclosures
Revisions:
16 times, by 10 contributors - see full revision history and disclosures
Systems:
Vascular, Central Nervous System
Sections:
Anatomy
Tags:
cases
Synonyms:
CT cerebral venogram
CT dural sinus venography
CTV head
Head CTV
More Cases Needed: This article has been tagged with "cases" because it needs some more cases to illustrate it. Read more...
CT cerebral venography(also known as aCTV head or CT venogram)is a contrast-enhanced examination with an acquisition delay providing an accurate detailed depiction of the cerebral venous system.
NB: This article is intended to outline some general principles of protocol design. The specifics will vary depending on CT hardware and software, radiologists' and referrers' preference, institutional protocols, patient factors (e.g. allergy) and time constraints.
On this page:
Article:
Indications
Technique
Data acquisition and analysis
Findings
Variant anatomy
Practical points
Related articles
References
Images:
Cases and figures
Indications
A CT venogram is obtained in a number of clinical scenarios where anatomy and patency of the cerebral veins is required. It is an alternative to MR venography.Indications include the diagnosis of cerebral venous thrombosisand preoperative anatomy particularly for posterior fossa surgery where the sigmoid sinuses may be compressed (e.g. retrosigmoid craniotomies).
Purpose
The purpose of this exam is to visualize the cerebral veins and venous sinuses filled with contrast opacified blood, allowing their anatomy and patency to be assessed.
Contraindications
IV iodinated contrast contraindications, such as chronic renal failureand allergy.
Technique
patient position
supine with their arms by their side
scout
CT to the vertex
scan extent
CT to the vertex
scan direction
caudocranial
contrast injection considerations
injection
75-100 ml of non-ionic iodinated contrast
scan delay
45 seconds (see practical points)
respiration phase
suspended
Data acquisition and analysis
images are analyzed on a dedicated workstation
proper evaluation of the dural sinuses indicates proper inspection of the axial thin-section contrast-enhanced source images of a helical CT scan
two-dimensional (2D) & three-dimensional (3D) multiplanar images, as well as rendering techniques such as maximum intensity projection (MIP), surface shaded displays (SSD) and volume rendering (VR) in a sagittal, coronal, and oblique planes
an essential step in CT venography is the removal of bone from the images, by graded subtraction
Findings
Venous sinus abnormalities
dural venous sinus thrombosis
thrombosis recanalization, the sinus shows an irregular appearance with multiple intrasinus channels and dural collateral vessels
sinus stenosis or occlusion secondary to tumor invasion (e.g. meningioma)
vascular malformations (e.g.AVM, dural arteriovenous fistula or developmental venous anomaly)
idiopathic intracranial hypertension: bilateral stenoses of the transverse sinuses, without definitive evidence of current or prior thrombosis
Parenchymal abnormalities
cerebral parenchymal abnormalities such as hemorrhagic infarction
Variant anatomy
sinus hypoplasia and aplasia
arachnoid granulations
sinus duplication or fenestration
variant anatomy of the sinuses (e.g. occipital sinus or persistent falcine sinus)
Practical points
Caution must be taken to achieve a well-timed acquisition in CT cerebral venography, particularly in cases of intracranial hypertension which can cause delayed filling of the venous sinus. A premature acquisition may create a false impression of thrombosis, which is simply due to contrast not yet reaching the venous sinuses - if the cerebral veins are not opacified on a CT cerebral venogram, premature acquisition should be suspected.
Quiz questions
Question 2124
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A patient without prior medical history was involved in a motor vehicle collision and arrives obtunded. The patient undergoes noncontrast CT of the head and cervical spine. At the acquisition scanner, you identify scattered convexal subarachnoid hemorrhage and fractures of the right petrous temporal bone, characterized as otic-capsule sparing, coursing to the sigmoid plate, and not clearly involving the petrous carotid canal on that side. Based on this information, which of the following imaging tests is the most appropriate next step?
CT angiography: arterial phase only of the head and neck
CT angiography: arterial phase of the head and neck plus venous phase of the head
CT of the head and neck with contrast (brain parenchymal/soft tissue phase)
MRI brain without contrast, MR angiography of the head without contrast, MR angiography of the neck with contrast plus axial T1 precontrast with fat saturation at the skull base, and MRI of the cervical spine without contrast
point-of-care ultrasound of the carotid arteries and jugular veins
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References
Leach J, Fortuna R, Jones B, Gaskill-Shipley M. Imaging of Cerebral Venous Thrombosis: Current Techniques, Spectrum of Findings, and Diagnostic Pitfalls. Radiographics. 2006;26 Suppl 1(suppl_1):S19-41; discussion S42-3. doi:10.1148/rg.26si055174 - Pubmed
Seo H, Choi DS, Shin HS et-al. Bone subtraction 3D CT venography for the evaluation of cerebral veins and venous sinuses: imaging techniques, normal variations, and pathologic findings. AJR Am J Roentgenol. 2014;202 (2): W169-75. doi:10.2214/AJR.13.10985 - Pubmed citation
Casey SO, Alberico RA, Patel M et-al. Cerebral CT venography. Radiology. 1996;198 (1): 163-70. doi:10.1148/radiology.198.1.8539371 - Pubmed citation
Majoie C, van Straten M, Venema H, den Heeten G. Multisection CT Venography of the Dural Sinuses and Cerebral Veins by Using Matched Mask Bone Elimination. AJNR Am J Neuroradiol. 2004;25(5):787-91. PMC7974470 - Pubmed
Rodallec MH, Krainik A, Feydy A et-al. Cerebral venous thrombosis and multidetector CT angiography: tips and tricks. Radiographics. 2006;26 Suppl 1 (suppl_1): S5-18. doi:10.1148/rg.26si065505 - Pubmed citation
Khandelwal N, Agarwal A, Kochhar R et-al. Comparison of CT venography with MR venography in cerebral sinovenous thrombosis. AJR Am J Roentgenol. 2006;187 (6): 1637-43. doi:10.2214/AJR.05.1249 - Pubmed citation
Incoming Links
Articles:
Deep cerebral vein thrombosis
Empty delta sign (dural venous sinus thrombosis)
Cranial vault
Dural venous sinus thrombosis
Computed tomography curriculum
Transverse sinus stenosis
Intracranial haemorrhage
MR cerebral venography
Superior ophthalmic vein thrombosis
Heparin-induced thrombocytopenia
Cerebral venous thrombosis
CT head (protocol)
Cases:
Cerebral venous thrombosis (CVT)
Dural venous sinus thrombosis - empty delta sign
Deep cerebral veins thrombosis and thalamic venous infarction
Cerebral venous sinus thrombosis
Sigmoid sinus thrombosis and subdural haematomas (infant)
Normal head CT angiogram
Multiple choice questions:
Question 2124
Related articles: Computed tomography
computed tomographyin practice
computed tomography overview
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11019 | https://brainly.com/question/42126006 | [FREE] 2) Three men and three women line up at a check-out counter at a store. a) In how many ways can they line - brainly.com
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Three men and three women line up at a check-out counter at a store.
a) In how many ways can they line up?
b) In how many ways can they line up if the first person in the line is a woman, and then the line alternates by gender?
c) Find the probability that the first person in line is a woman, then a man, and then alternating by gender.
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There are 720 ways they can line up in total. If the line alternates by gender and the first person is a woman, there are 24 possible ways they can line up. The probability that the first person is a woman, then a man, and then alternating by gender is 1/30 or approximately 0.0333.
Explanation
a) To determine the number of ways the three men and three women can line up at the check-out counter, we can use the concept of permutations. There are 6 people in total, and each one can occupy a different position in the line, so the total number of ways they can line up is 6! (6 factorial) which is equal to 720.
b) If the first person in line is a woman and the line alternates by gender, we can consider the two genders as separate groups. The group of women has 3 members and the group of men has 3 members. The first position in line must be occupied by a woman, so we have 3 choices. For the subsequent positions, we alternate between the two groups, so there are 2 choices for each position. Therefore, the total number of ways they can line up is 3 2 2 2 = 24.
c) To find the probability that the first person in line is a woman, then a man, and then alternating by gender, we can divide the number of ways they can line up with this condition by the total number of possible ways they can line up. From part b, we know that there are 24 ways they can line up with the given condition. From part a, we know that there are 720 total ways they can line up. So, the probability is 24/720, which simplifies to 1/30 or approximately 0.0333.
Learn more about Permutations and probability here:
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There are 720 ways for the three men and three women to line up. If the first person is a woman and they alternate genders, there are 36 arrangements. The probability of this specific arrangement is 0.05 or 5%.
Explanation
To solve the problem, let's break it down into the three parts given:
a) In how many ways can they line up?
To determine the number of ways the three men and three women can line up at the check-out counter, we can use the concept of permutations. There are 6 people in total (3 men + 3 women), and they can occupy different positions in the line. Therefore, the total number of ways they can line up is calculated using factorial:
6!=6×5×4×3×2×1=720
So, there are 720 unique ways for them to line up.
b) In how many ways can they line up if the first person in line is a woman, and then the line alternates by gender?
If the first person in line is a woman, we can start by selecting one of the 3 women to occupy the first position, which gives us 3 choices. The lineup must follow the pattern: Woman-Man-Woman-Man-Woman-Man. The number of ways to arrange them can be calculated as follows:
For the first position (woman): 3 choices
For the second position (man): 3 choices
For the third position (woman): 2 choices
For the fourth position (man): 2 choices
For the fifth position (woman): 1 choice
For the sixth position (man): 1 choice
The total number of arrangements is:
3×3×2×2×1×1=3×3×2×2=36
So, there are 36 ways for them to line up under these conditions.
c) Find the probability that the first person in line is a woman, then a man, and then alternating by gender.
To find this probability, we will take the number of favorable outcomes (ways that meet the given condition) and divide by the total outcomes (ways they can line up without restrictions). From part b, we found there are 36 favorable arrangements. From part a, we found there to be 720 total arrangements. Therefore, the probability is calculated as:
Probability=720 36=20 1=0.05
Thus, the probability that the first person is a woman, the second is a man, and they alternate genders is 0.05 or 5%.
Examples & Evidence
For example, if we had four students, two boys and two girls, and we wanted to find similar arrangements, we would use the same principles of permutations and combinations to figure out the total arrangements and specific orderings. This helps to illustrate the concepts in different contexts.
The calculations rely on the principles of factorials and basic permutation rules, which are standard within combinatorics in mathematics.
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Community Answer 5.0 3 men and 3 woman line up at a checkout counter in the store. find the probability that the first person in line is a woman in the line alternates by gender.
Community Answer Men and women line up at a checkout counter in a store. in how many ways can they line up if the first person in line is a manman?, and the people in line alternate manman?, womanwoman?, manman?, womanwoman?, and so? on?
Community Answer 3 Seven men and seven women line up at a checkout counter in a store. In how many ways can they line up if the first person in line is a man , and the people in line alternate man, woman, man, woman ,and so on
Community Answer 4.5 6 four men and four women line up at a checkout counter in a store. In how many different ways can they line up if the first person in line is a man, and the people in line alternate man, woman, man, woman, and so on?
Community Answer and four women line up at a checkout counter in a storeIn how many ways can they line up if the first person in line is a woman, and the people in line woman, man, woman man, and so on?
Community Answer Three men and three women line up at a checkout counter. find the probability that the first person is a women and then they alternate by gender?
Community Answer 4.6 12 Jonathan and his sister Jennifer have a combined age of 48. If Jonathan is twice as old as his sister, how old is Jennifer
Community Answer 11 What is the present value of a cash inflow of 1250 four years from now if the required rate of return is 8% (Rounded to 2 decimal places)?
Community Answer 13 Where can you find your state-specific Lottery information to sell Lottery tickets and redeem winning Lottery tickets? (Select all that apply.) 1. Barcode and Quick Reference Guide 2. Lottery Terminal Handbook 3. Lottery vending machine 4. OneWalmart using Handheld/BYOD
Community Answer 4.1 17 How many positive integers between 100 and 999 inclusive are divisible by three or four?
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Telescoping series
In mathematics, a telescoping series is a series whose partial sums eventually only have a finite number of terms after cancellation. This is often done by using a form of for some expression .
Contents
1 Example 1
2 Solution 1
3 Example 2
4 Solution 2
5 Problems
5.1 Introductory
5.2 Intermediate
5.3 Olympiad
6 See Also
Example 1
Derive the formula for the sum of the first counting numbers.
Solution 1
We wish to write for some expression . This expression is as .
We then telescope the expression:
.
(Notice how the sum telescopes— contains a positive and a negative of every value of from to , so those terms cancel. We are then left with , the only terms which did not cancel.)
Example 2
Find a general formula for , where .
Solution 2
We wish to write for some expression . This can be easily achieved with as by simple computation.
We then telescope the expression:
.
Problems
Introductory
When simplified the product becomes:
(Source)
The sum can be expressed as , where and are positive integers. What is ? (Source)
Which of the following is equivalent to (Hint: difference of squares!)
(Source)
Intermediate
Let denote the value of the sum can be expressed as , where and are positive integers and is not divisible by the square of any prime. Determine . (Source)
Olympiad
Find the value of , where is the Riemann zeta function
See Also
Algebra
Summation
Retrieved from "
Category:
Algebra
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Multiplying Polynomials
A polynomial looks like this:
example of a polynomial
this one has 3 terms
To multiply two polynomials:
multiply each term in one polynomial by each term in the other polynomial
add those answers together, and simplify if needed
Let us look at the simplest cases first.
1 term × 1 term (monomial times monomial)
To multiply one term by another term, first multiply the constants, then multiply each variable together and combine the result, like this (press play):
(Note: I used "·" to mean multiply. In Algebra we don't like to use "×" because it looks too much like the letter "x")
For more about multiplying terms, read Multiply and Divide Variables with Exponents
1 term × 2 terms (monomial times binomial)
Multiply the single term by each of the two terms, like this:
2 term × 1 terms (binomial times monomial)
Multiply each of the two terms by the single term, like this:
(I did that one a bit faster by multiplying in my head before writing it down)
2 terms × 2 terms (binomial times binomial)
Each of the two terms in the first binomial ...
... is multiplied by ...
... each of the two terms in the second binomial
That is 4 different multiplications ... Why?
Matching up Partners
Two friends (Alice and Betty) challenge
two other friends (Charles and David) to
individual tennis matches.
How many matches does that make?
Alice plays Charles, and then Alice plays David
Then Betty plays Charles and then Betty plays David
They could play in any order, so long as each of the first two friends
gets to play each of the second two friends.
It is the same when we multiply binomials!
Instead of Alice and Betty, let's just use a and b, and Charles and David can be c and d:
We can multiply them in any order so long as each of the first two terms gets multiplied by each of the second two terms.
But there is a handy way to help us remember to multiply each term called "FOIL".
It stands for "Firsts, Outers, Inners, Lasts":
Firsts: ac
Outers: ad
Inners: bc
Lasts: bd
So you multiply the "Firsts" (the first terms of both polynomials), then the "Outers", etc.
Let us try this on a more complicated example:
2 terms × 3 terms (binomial times trinomial)
"FOIL" won't work here, because there are more terms now. But just remember:
Multiply each term in the first polynomial by each term in the second polynomial
Like Terms
And always remember to add Like Terms:
Example: (x + 2y)(3x − 4y + 5)
(x + 2y)(3x − 4y + 5)
= 3x 2− 4xy + 5x + 6xy − 8y 2 + 10y
= 3x 2+ 2xy + 5x − 8y 2 + 10y
Note: −4xy and 6xy are added because they are Like Terms.
Also note: 6yx means the same thing as 6xy
Long Multiplication
You may also like to read about Polynomial Long Multiplication
Mathopolis:Q1)Q2)Q3)Q4)Q5)Q6)Q7)Q8)Q9)
PolynomialsIntroduction to AlgebraAlgebra - Basic DefinitionsAlgebra Index
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11022 | https://selectividad.intergranada.com/ESO/fq4/Clase/Prob_MRU.pdf | © Septiembre 2016 1 Problemas del Movimiento Departamento de Física y Química © Raúl González Medina Problemas M.R.U. 1.- Un corredor pedestre corre 200 m en 21,6 segundos. Calcular su velocidad en m/s, km/h y m/min. Sol: 9,26 m/s; 33,336 Km/h; 555,6 m/min 2.- La velocidad de un avión es de 970 Km/h; la de otro, es de 300 m/s. ¿Cuál es el más lento? Sol: El primero 3.- Expresa 72 Km/h en m/s; Km/min y cm/s. Sol: 20 m/s; 1,2 Km/min; 2000 cm/s 4.- Hallar la velocidad de un móvil que recorre 60 Km en 90 minutos. Expresar el resultado en Km/h, m/min y m/s. Sol: 40 Km/h = 11,11 m/s 5.- Qué distancia recorrerá un móvil durante 45 minutos si marcha con una velocidad de: a) 25 Km/h; b) 25 m/s. Sol: a) 18,75 Km b) 67500 m 6.- Que velocidad posee un atleta que para recorrer 5 Km emplea un tiempo de: a) 30 minutos; b) 1200 segundos. Expresar ambas velocidades en m/s y Km/h. Sol: a) 10 Km/h = 2,778 m/s b) 15 km/h = 4,1667 m/s 7.- Un coche inicia un viaje de 495 Km. a las ocho y media de la mañana con una velocidad media de 90 Km/h. ¿A qué hora llegará a su destino? Solución: a las dos de la tarde. 8.- Un móvil recorre 98 km en 2 h, calcular: a) Su velocidad. b) ¿Cuántos kilómetros recorrerá en 3 h? Solución: a) 13,6 m/s; b) 147 km 9.- ¿Cuánto tarda en llegar la luz del sol a la Tierra?, si la velocidad de la luz es de 300.000 km/s y el sol se encuentra a 150.000.000 km de distancia. Solución: 8,3 min 10.- Una persona sale de su casa y recorre en línea recta los 200m que la separan de la panadería a una velocidad constante de 1,4 m/s. Permanece en la tienda 2 min y regresa a su casa a una velocidad de 1,8 m/s. a) Calcula su velocidad media. b) ¿cuál ha sido su desplazamiento? c) ¿Qué espacio ha recorrido? d) Realiza una gráfica v-t. Solución: v=1,07 m/s; b) Δr=0; c) X=400m 11.- Dos trenes se cruzan perpendicularmente y hacen un recorrido durante cuatro horas, siendo la distancia que los separa al cabo de ese tiempo, de 100 km. Si la velocidad de uno es de 20 km/h, calcular la velocidad del otro tren. Solución: v = 15 km/h 12.- Dos vehículos cuyas velocidades son 10 Km/h y 12 Km/h respectivamente se cruzan perpendicularmente en su camino. Al cabo de seis horas de recorrido, ¿cuál es la distancia que los separa? Solución: 93,72 km. 13.- Dos automóviles que marchan en el mismo sentido, se encuentran a una distancia de 126 Km. Si el más lento va a 42 Km/h, calcular la velocidad del más rápido, sabiendo que le alcanza en seis horas. Solución: v = 63 km/h 14.- Un deportista sale de su casa en bici a las seis de la mañana. Al llegar a un cierto lugar, se le estropea la bici y ha de volver andando. Calcular a qué distancia ocurrió el percance sabiendo que las velocidades de desplazamiento han sido de 30 Km/h en bici y 6 Km/h andando y que llegó a su casa a la una del mediodía. Solución: 35 km 15.- Un deportista recorre una distancia de 1.000 km, parte en moto y parte en bici. Sabiendo que las velocidades han sido de 120 Km/h en la moto y 20 Km/h en bici, y que el tiempo empleado ha sido de 15 horas calcular los recorridos hechos en moto y en bici. Solución: la motocicleta 840 km y la bici 160 km. 16.- La velocidad de un remolcador respecto del agua de un río es de 12 Km/h. La velocidad de la corriente es de 1.25 m/s. Calcular el tiempo que durará el viaje de ida y vuelta entre dos ciudades situadas a 33 Km. de distancia en la misma orilla del río. Solución: t1 = 2 horas; t2=4,4 horas 17.- Un observador se halla a 510 m. de una pared. Desde igual distancia del observador y de la pared, se hace un disparo ¿al cabo de cuántos segundos percibirá el observador: a) el sonido directo? b) el eco? Velocidad del sonido 340 m/s. Solución: el sonido directo a 0,75 s, y el del eco a 2,25 s. 18.- Un ladrón roba una bicicleta y huye con ella a 20 km/h. Un ciclista que lo ve, sale detrás del mismo, tres minutos más tarde a 22 Km/h. ¿Al cabo de cuánto tiempo lo alcanzará? Solución: 30 minutos. 19.- Calcular la longitud de un tren cuya velocidad es de 72 Km/h y que ha pasado por un puente de 720 m de largo, si desde que penetró la máquina hasta que salió el último vagón han pasado ¾ de minuto. Solución: 180 metros. 20.- Una canoa invierte 20 minutos para bajar cierto trayecto de un río y 36 minutos para hacer el mismo recorrido en sentido contrario. Calcular las velocidades de la canoa en los dos casos si la longitud del recorrido ha sido 10,8 Km. Solución: 5 m/s y 9 m/s 21.- Se cruzan dos trenes en sentido contrario, uno de 50 m de longitud con velocidad de 60 Km/h y otro de 175 m de longitud y velocidad desconocida. Si tardan en cruzarse 6 segundos. Calcular la velocidad con que se mueve el segundo tren. Solución: V = 75 km/h 22.- Dos ciclistas pasan por una carretera rectilínea con velocidad constante. Cuando van en el mismo sentido, el primero adelanta al segundo 150 m/min.; cuando van en sentidos contrarios, el uno se acerca a otro 350 m. cada veinte segundos. Hallar la velocidad de cada ciclista. Solución: V1= 10 m/s V2= 7,5 m/s 23.- Dos coches salen a su encuentro, uno de Bilbao y otro de Madrid. Sabiendo que la distancia entre ambas capitales es de 443 Km. y que sus velocidades respectivas son 78 Km/h y 62 Km/h y que el coche de Bilbao salió hora y media más tarde, calcular: a) Tiempo que tardan en encontrarse b) ¿A qué distancia de Bilbao lo hacen? Solución: tardan en encontrarse 2,5 horas; a 195 km de Bilbao. 24.- Un barquero quiere cruzar un río de 100 m de anchura; para ello rema perpendicularmente a la corriente imprimiendo a la barca una velocidad de 2 m/s respecto del agua. La velocidad de la corriente es de 0,5 m/s. Calcula: a) El tiempo que tarda en atravesar el río. b) La velocidad de la barca respecto a la orilla del río. c) En qué punto de la orilla opuesta desembarcará. d) Qué longitud ha recorrido la barca al llegar a la orilla opuesta. Sol: a) 50s; b) 2,06m/s; c) 25 m más debajo de la perpendicular; d) 103 m 25.- Un móvil recorre dos km hacia el Norte, después 1 km hacia el Este, a continuación 4 km hacia el Sur, luego toma la dirección Oeste durante 3 km y por último sube un km más hacia el Norte. Calcula: a) Los desplazamientos parciales. b) El desplazamiento total. c) El espacio total recorrido. d) ¿A qué distancia del punto de partida se encuentra al final?. Solución: a) 2j, 1i, -4j, -3i, 1j; b) -2i – j; c) 11 km; d) 5 km 26.- Un cazador y su perro emprenden el camino hacia un refugio situado a 9 km de distancia. El cazador camina a 4 km/h y el perro a 8 km/h. El perro, que obviamente llega antes al refugio, da la vuelta y regresa hacia su amo. ¿Dónde se encuentran por primera vez?. A continuación, repite constantemente el viaje de ir al refugio y volver a buscar al amo, hasta que por fin llegan ambos definitivamente al final del trayecto. Calcula la distancia total que el perro ha recorrido. Sol: A 3 km del refugio; 18 km. |
11023 | https://anylearn.ai/guide/angle-of-inclination | Angle Of Inclination, a Course on AnyLearn
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Angle Of Inclination
1.Definition and Properties of Angle Of Inclination
2.Slope of a Line
3.Trigonometric Ratios
4.Coordinate Geometry
5.Applications in Real World Contexts
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The angle of inclination refers to the angle formed between a reference plane and a line or surface, often measured from the horizontal or vertical axis. It's commonly used in geometry, physics, and engineering to describe the steepness or slope of an object, such as a road, roof, or celestial orbit.
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Definition and Properties of Angle Of Inclination
Understanding the concept of inclination reveals the foundational relationship between angles and their geometric properties.
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Slope of a Line
The slope serves as a critical indicator of the steepness and direction of a line in a coordinate plane.
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Trigonometric Ratios
Trigonometric ratios provide essential tools for relating angles to side lengths in right triangles.
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Coordinate geometry bridges algebra and geometry, allowing for the analysis of geometric shapes using algebraic equations.
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11024 | https://brightchamps.com/en-us/math/data/variance-of-binomial-distribution | Table Of Contents
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Last updated on September 12, 2025
A Beginner’s Guide to Variance of a Binomial Distribution
The binomial distribution is used to measure how much the probabilities differ from the expected value (mean). This value shows the difference between the sampled observations and the expected value. In this topic, we are going to learn more about the variance of binomial distribution.
What is Binomial Distribution?
First, we need to understand what the binomial distribution is before learning about the variance of a binomial distribution. A binomial distribution is a discrete probability distribution that has only two outcomes. The two outcomes are typically expressed as 1 for success and 0 for failure in a given number of trials.
How to Calculate the Variance of Binomial Distribution?
The variance of a binomial distribution measures how spread out the probability values are around the mean. Variance defines how much the values differ from the mean value in a data set. When we calculate the variance of a binomial distribution, we have to follow certain steps. They are:
Step 1: The first step is to identify the total number of trials (n) and the probability of success in a single trial (p). These two parameters are crucial for defining a binomial distribution.
Here, n is the total number of trials. Each trial is independent, and its outcome does not change the outcome of other trials. Also, p represents the probability of success and the value is between 0 and 1.
Step 2: Use the variance formula.
The symbol for variance is σ2, and it represents the square of the standard deviation. When we find the values of n and p, we can use the formula:
Variance (σ2) = np (1 - p)
Derivation of Variance of Binomial Distribution
The binomial distribution represents the probability of getting a specific number of successes in independent trials. The possible outcomes of each trial are success and failure. In every trail the probability of success remains the same. Let X be the number of successes in the n trials. Then the variance of X can be calculated as:
σ2 = E (X2) - (E (X))2
Now, let’s find the E(X), then the mean of X, which is np, here n is the number of trials and p is the probability of success.
Then, we have to identify the E (X2). This refers to the squared values of X. Also, we need to find the expected value of X2. X2 has a distribution where each outcome is squared because X follows a binomial distribution.
(E (X^2) = \displaystyle\sum_{k=0}^{n}k^2). P (X = k)
Next, using the probability mass function (PMF) of the binomial distribution, we can find the probability of getting k successes in n trials.
P (X = k) = (\binom{n}{k}) pk (1 - p) n - k
We can substitute this formula into an equation for E (X2) and then analyze the sum. Finally, add the values of E (X) and E (X2) into the formula of variance. Then simplify it to get the variance of the binomial distribution as (σ2) = np (1 - p)
Real-life applications of the Variance of Binomial Distribution
The variance of a binomial distribution measures how much the number of successes deviates from the expected mean. In the fields of medical research, finance, sports analysis, and manufacturing, the role of the variance of the binomial distribution is vital.
Common Mistakes and How to Avoid Them on Variance of Binomial Distribution
The variance of the binomial distribution tells us how much our actual results differ from the expected value on average. However, some mistakes can lead to incorrect calculations and interpretations. By understanding the common mistakes of the variance of the binomial distribution, students can improve their statistical skills and practical knowledge.
The variance of the binomial distribution tells us how much our actual results differ from the expected value on average. However, some mistakes can lead to incorrect calculations and interpretations. By understanding the common mistakes of the variance of the binomial distribution, students can improve their statistical skills and practical knowledge.
Mistake 1
Using the incorrect formula
Using the incorrect formula
Students should ensure that they are using the correct formula for calculating the variance. The correct formula is:
Variance (σ2) = np (1 - p)
Where p is the probability of success and n is the number of trails. If students use the wrong formula, it can lead to errors and wrong conclusions.
Students should ensure that they are using the correct formula for calculating the variance. The correct formula is:
Variance (σ2) = np (1 - p)
Where p is the probability of success and n is the number of trails. If students use the wrong formula, it can lead to errors and wrong conclusions.
Mistake 2
Confusing the values of success and failure
Confusing the values of success and failure
Carefully find the values of the probability of success (p) and the probability of failure (q). Sometimes, kids mistakenly interpret the values of p and q. Before applying the formula, make sure the probability of failure is identified correctly as q = 1 - p.
Carefully find the values of the probability of success (p) and the probability of failure (q). Sometimes, kids mistakenly interpret the values of p and q. Before applying the formula, make sure the probability of failure is identified correctly as q = 1 - p.
Mistake 3
Ignoring the conditions of a binomial distribution
Ignoring the conditions of a binomial distribution
Remember that while applying the formula for the variance of the binomial distribution, the situation should meet the required conditions. They are trials of a binomial distribution are independent, with only two possible outcomes. If we use the formula for situations that do not meet the conditions, it will affect the final results.
Remember that while applying the formula for the variance of the binomial distribution, the situation should meet the required conditions. They are trials of a binomial distribution are independent, with only two possible outcomes. If we use the formula for situations that do not meet the conditions, it will affect the final results.
Mistake 4
Thinking the value of variance is a whole number
Thinking the value of variance is a whole number
Understand the fact that the value of variance can be a fraction or decimal. Kids expect the value of variance as a whole number and end up with incorrect conclusions. There is no rule that the variance will be a whole number. For example, if n = 9, p = 0.5, then the formula is:
Variance (σ2) = np (1 - p)
σ2 = 9 (0.5) (1 - 0.5)
σ2 = 9 (0.5) (0.5)
σ2 = 9 × 0.25
σ2 = 2.25
Here, the variance of the binomial distribution is 2.25, which is a decimal.
Understand the fact that the value of variance can be a fraction or decimal. Kids expect the value of variance as a whole number and end up with incorrect conclusions. There is no rule that the variance will be a whole number. For example, if n = 9, p = 0.5, then the formula is:
Variance (σ2) = np (1 - p)
σ2 = 9 (0.5) (1 - 0.5)
σ2 = 9 (0.5) (0.5)
σ2 = 9 × 0.25
σ2 = 2.25
Here, the variance of the binomial distribution is 2.25, which is a decimal.
Mistake 5
Forgetting the relationship between variance, n, and p
Forgetting the relationship between variance, n, and p
Before using the formula for the variance of a binomial distribution, kids should understand that variance depends on n and p. The value of variance increases with n. For example, if n = 12, p = 0.4, the variance will be 12 (0.4) (1 - 0.4) = 12 × 0.4 × 0.6
σ2 = 12 × 0.24 = 2.88
Next, if n = 30, p = 0.4, the variance will be 30 (0.4) (1 - 0.4) = 30 × 0.4 × 0.6
σ2 = 30 × 0.24 = 7.2
The value of variance increases with the value of n.
Before using the formula for the variance of a binomial distribution, kids should understand that variance depends on n and p. The value of variance increases with n. For example, if n = 12, p = 0.4, the variance will be 12 (0.4) (1 - 0.4) = 12 × 0.4 × 0.6
σ2 = 12 × 0.24 = 2.88
Next, if n = 30, p = 0.4, the variance will be 30 (0.4) (1 - 0.4) = 30 × 0.4 × 0.6
σ2 = 30 × 0.24 = 7.2
The value of variance increases with the value of n.
Tips and Tricks of Variance of Binomial Distribution
The variance of the binomial distribution helps us to understand how much the results fluctuate around the mean. Understanding the concepts of variance of binomial distribution is useful in working with statistics, probability, and risk assessment. Here are some of the tricks and tips that help us to effectively work with the fundamental concept.
Solved examples of Variance of Binomial Distribution
Problem 1
Find the variance of the binomial distribution having 15 trials and a probability of success of 0.6.
3.6
3.6
Explanation
We can use the formula for the variance of a binomial distribution:
Variance (σ2) = np (1 - p)
Here, n is the number of trials = 15
p is the probability of success = 0.6
Hence, the prob failure = 1 - p = 1 - 0.6 = 0.4
Now, we can substitute the values to the formula: (σ2) = np (1 - p)
15 × 0.6 × 0.4
15 × 0.24 = 3.6
The variance is 3.6.
It means that the number of successes will fluctuate around the mean with a variance of 3.6
We can use the formula for the variance of a binomial distribution:
Variance (σ2) = np (1 - p)
Here, n is the number of trials = 15
p is the probability of success = 0.6
Hence, the prob failure = 1 - p = 1 - 0.6 = 0.4
Now, we can substitute the values to the formula: (σ2) = np (1 - p)
15 × 0.6 × 0.4
15 × 0.24 = 3.6
The variance is 3.6.
It means that the number of successes will fluctuate around the mean with a variance of 3.6
Problem 2
A factory produces 10 bulbs daily. The probability of a defective bulb is 0.2. Find the variance of the defective bulbs per day.
1.6
1.6
Explanation
To find the variance of the defective bulbs per day, we can apply the binomial variance formula. Here,
n = 10
p = 0.2
1 - p = 1 - 0.2 = 0.8
The variance formula is:
Variance (σ2) = np (1 - p)
σ2 = 10 × 0.2 × 0.8 = 1.6
Hence, the variance of defective bulbs per day is 1.6
To find the variance of the defective bulbs per day, we can apply the binomial variance formula. Here,
n = 10
p = 0.2
1 - p = 1 - 0.2 = 0.8
The variance formula is:
Variance (σ2) = np (1 - p)
σ2 = 10 × 0.2 × 0.8 = 1.6
Hence, the variance of defective bulbs per day is 1.6
Problem 3
Felix takes 20 quizzes. The probability of passing each quiz is 0.8. Find the variance of the number of quizzes passed.
3.2
3.2
Explanation
Variance (σ2) = np (1 - p) is the formula for the variance of the binomial distribution.
Here, n = 20
p = 0.8
1 - p = 1 - 0.8 = 0.2
Now, we can substitute the values.
σ2 = 20 × 0.8 × 0.2
σ2 = 20 × 0.16 = 3.2
The number of quizzes Felix passes fluctuates around the mean with a variance of 3.2
Variance (σ2) = np (1 - p) is the formula for the variance of the binomial distribution.
Here, n = 20
p = 0.8
1 - p = 1 - 0.8 = 0.2
Now, we can substitute the values.
σ2 = 20 × 0.8 × 0.2
σ2 = 20 × 0.16 = 3.2
The number of quizzes Felix passes fluctuates around the mean with a variance of 3.2
Problem 4
A basketball player takes 30 free throws. The probability of making a basket is 0.6. Find the variance of successful shots.
7.2
7.2
Explanation
To find the answer, we can use the formula, Variance (σ2) = np (1 - p)
Where, n = 30
p = 0.6
1 - p = 1 - 0.6 = 0.4
So the formula will be:
σ2 = 30 × 0.6 × 0.4
σ2 = 30 × 0.24 = 7.2
The variance of successful shots is 7.2
To find the answer, we can use the formula, Variance (σ2) = np (1 - p)
Where, n = 30
p = 0.6
1 - p = 1 - 0.6 = 0.4
So the formula will be:
σ2 = 30 × 0.6 × 0.4
σ2 = 30 × 0.24 = 7.2
The variance of successful shots is 7.2
Problem 5
In a shop, 40 customers visit daily. The probability that a customer makes a purchase is 0.6. Find the variance of the number of customers making a purchase.
9.6
9.6
Explanation
To find the variance of the number of customers making a purchase, we can use the formula.
Here, n = 40
p = 0.6
1 - p = 1 - 0.6 = 0.4
The binomial variance formula is:
Variance (σ2) = np (1 - p)
σ2 = 40 × 0.6 × 0.4
σ2 = 40 × 0.24 = 9.6
The variance of the number of customers making a purchase is 9.6
To find the variance of the number of customers making a purchase, we can use the formula.
Here, n = 40
p = 0.6
1 - p = 1 - 0.6 = 0.4
The binomial variance formula is:
Variance (σ2) = np (1 - p)
σ2 = 40 × 0.6 × 0.4
σ2 = 40 × 0.24 = 9.6
The variance of the number of customers making a purchase is 9.6
FAQs on Variance of Binomial Distribution
1.Define the variance of the binomial distribution.
The variance of the binomial distribution explains how much the values differ from the mean. A binomial distribution involves n independent trials, and the probability of success is denoted as p. Also, the probability of failure is q (1 - p).
The variance of the binomial distribution explains how much the values differ from the mean. A binomial distribution involves n independent trials, and the probability of success is denoted as p. Also, the probability of failure is q (1 - p).
2.Explain the formula of variance of the binomial distribution.
To find the value of n and p, we can use the formula:
Variance (σ2) = np (1 - p)
Here n is the number of trials, then p is the probability of success, and 1 - p is the probability of failure. The symbol for variance is σ2 and is the square of the standard deviation.
To find the value of n and p, we can use the formula:
Variance (σ2) = np (1 - p)
Here n is the number of trials, then p is the probability of success, and 1 - p is the probability of failure. The symbol for variance is σ2 and is the square of the standard deviation.
3. What is variance in a binomial distribution?
Variance tells us how much the number of successes fluctuates from the mean. A higher variance means the result is far different from the mean, and a low variance means the results stay more consistent.
Variance tells us how much the number of successes fluctuates from the mean. A higher variance means the result is far different from the mean, and a low variance means the results stay more consistent.
4.Is it possible for the variable to be negative?
No, variance cannot be negative. It measures how much values differ from the mean, so the value of variance will always be zero or positive.
No, variance cannot be negative. It measures how much values differ from the mean, so the value of variance will always be zero or positive.
5.How are n and variance related?
As n increases, the variance also increases, provided that p remains constant. Here, n is the number of trials and p is the probability of success. Also, if n doubles, it results in the doubling of variance, only if p remains constant.
As n increases, the variance also increases, provided that p remains constant. Here, n is the number of trials and p is the probability of success. Also, if n doubles, it results in the doubling of variance, only if p remains constant.
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Enzymes that catalyse electron transfer in oxidation-reduction reactions. Oxidoreductases are classified into several groups according to their respective donors or acceptors.
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11026 | https://ec.europa.eu/programmes/erasmus-plus/project-result-content/0c2dbd0a-9ddd-45cd-950c-0edbbe848894/50_Chess_and_Mathematics_Exercises_CHAMPS_Final.pdf | 50 Chess and Mathematics Exercises for Schools A (chess) game-based approach to problem solving IO1 nnn ii iii INTRODUCTION This book provides material for the much-sought-after link between chess and mathematics for the classroom. We have fully tried out all these exercises and found that most children are enthusiastic – often more so even than their teachers! Chess is a classic board game that children enjoy at all levels. We use the chessboard and the chess pieces to convey mathematical insights consistent with the syllabus for primary school (i.e. children from age 6-11) mathematics in most countries. Only a basic knowledge of chess is required – how the pieces move. It is not necessary to be a chess player to use this book. The main emphasis is on problem solving. The 50 exercises are categorised by age and by the most natural grouping – individuals, pairs, quads (two pairs), groups or the whole class. The relevant topic in the mathematics syllabus is also displayed. Most exercises have preliminary questions or extensions. Solving the exercises should normally take no more than a lesson period. The materials required are minimal. Chess sets can be used but the exercises work just as well using 8x8 grids and coloured counters or markers. Printouts of board positions are useful for several exercises. Gradually work your way through the book at the pace of the children. Some children will far exceed the stated age range. Problem-solving requires persistence. The children are asked to complete a task or conduct an investigation or play a game. Hints are often provided to overcome intellectual hurdles. A solution method is provided for each problem but there is always more than one solution method – and children should be encouraged to think for themselves. The teacher can develop their own preliminary and extension exercises to suit the ability of the children. The later exercises require a higher level of abstraction and solutions are prone to error hence requiring more teacher intervention. The mathematical games are simple and easy to play. They illustrate some fundamental concepts such as parity and symmetry. The point of these games is not to win but to understand the underlying concepts. Children are fascinated to discover that many of these games can be won by recognising that there is an underlying pattern. This is part of the realisation that mathematics provides an underlying pattern to scientific laws. From a didactic perspective, children are delighted to learn “tricks” how to win a game or solve a puzzle. These exercises are drawn from traditional sources and supplemented by some original exercises from the authors. Many of the exercises will be new to recreational mathematicians. What marks them as distinctive is that they are developed within a constrained domain of the chessboard to create a sense of familiarity as well as accessibility. The primary school reference syllabus is from Singapore because this is regarded as being close to the problem-solving methodology promoted by this project. The Singapore “method” has gained widespread international attention following its success in the OECD Pisa rankings. The primary school syllabus is consistent with those in Europe as validated by the teaching experts on the project. In addition to mathematics, many of these exercises incorporate some principles of Game Theory which although not normally taught in primary school is within the capability of most children. iv Perhaps the greatest value in working through these exercises comes from having to structure the problems. A well-structured problem is a delight to solve. Teachers are encouraged to develop and expand these exercises and to share their classroom experiences. This project was funded through ErasmusPlus from the European Union. The partners on the CHAMPS project were the Slovak Chess Federation (co-ordinator), Chess in Schools and Communities (UK), Ludus (Portugal), University of Girona (Spain) and Veľká Ida Primary School, Slovakia. The following individuals also contributed to the project: Carlos Santos, Carme Saurina Canals, Josep Serra, Mark Szavin, Stefan Löffler, Alessandro Dominici, Malcolm Pein, Chris Fegan, Zdenek Gregor, Eva Repkova, Vladimir Szucs, Viera Kebluskova and Niki Vrbova. The project manager was Stefan Marsina. The chessboard diagrams were generated using the Logiq Board from LearningChess.net (Hungary). John Foley (@ChessScholar) Rita Atkins Carlos Santos Viera Haraštová March 2019 ©Copyright the authors. This work is licensed under the Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License. See Educators may use, reuse, adapt and share this material on a non-commercial basis provided that attribution is given to the authors. v THE 50 EXERCISES 1. EACH SQUARE HAS A NAME INDIVIDUAL 1 2. IDENTIFY THE SQUARE COLOUR WHOLE CLASS 2 3. PIECE CATEGORISATION GROUPS 3 4. CROSS THE LINE GAME PAIRS 4 5. DIAGONAL OPPOSITION GAME PAIRS 5 6. CHESS ARITHMETIC INDIVIDUAL 6 7. FOUR ROOKS PUZZLE PAIRS 7 8. THE FALLEN PIECES QUADS 8 9. CORNER ATTACK PUZZLE QUADS 9 10. PIECE POWER CONTOURS QUADS 10 11. AVOID THREE IN A LINE INDIVIDUAL 11 12. COUNTERS ON A LINE PAIRS 12 13. WYTHOFF’S GAME [Q] PAIRS 13 14. ROOK CORNER PUZZLE PAIRS 14 15. SHORTEST ROOK TOUR INDIVIDUAL 15 16. RACE TO THE CORNER [ROOK] PAIRS 16 17. CHOMP PAIRS 17 18. ARMY POWER PUZZLE PAIRS 18 19. MAXIMUM NUMBER OF KNIGHTS PAIRS 19 20. MYSTERY COMBINATION PIECE PAIRS 20 21. POSITIONAL LOGIC PAIRS 21 22. COUNTERS ON A LINE PAIRS 22 23. MINIMUM MOVES MAP PAIRS 23 24. HOW MANY ROUTES? [PAWN] PAIRS 24 25. THE LIGHT PAWN PAIRS 25 26. WYTHOFF’S GAME [QQ] PAIRS 26 27. THE DOMINO TILING PUZZLE PAIRS 27 28. THE TROMINO TILING PUZZLE PAIRS 28 29. TWO QUEENS PUZZLE QUADS 29 30. EQUIDISTANT CHECKMATE PAIRS 30 31. SUBTRACTION GAME PAIRS 31 32. NIM PAIRS 32 33. GOLDEN COIN GAME PAIRS 33 34. NORTHCOTT’S GAME PAIRS 34 35. SLIDING ROOKS GAME PAIRS 35 36. HOW MANY ROUTES? [ROOK] PAIRS 36 37. ROOK TOUR ON A MUTILATED BOARD PAIRS 37 38. BISHOP ZIGZAG PUZZLE PAIRS 38 39. FIVE QUEENS PUZZLE QUADS 39 40. DIVIDING THE BOARD INDIVIDUAL 40 41. TILING THE BOARD INDIVIDUAL 41 42. TWELVE KNIGHTS PROBLEM PAIRS 42 43. EIGHT QUEENS PUZZLE PAIRS 43 44. HOW MANY SQUARES ON A BOARD? PAIRS 44 45. TOURNAMENT SCORING PUZZLE INDIVIDUAL 45 46. KINGS RANDOM JUMP PAIRS 46 47. KING’S RANDOM WALK PAIRS 47 48. THE INVENTION OF CHESS GROUP 48 49. HOW MANY RECTANGLES ON A BOARD? GROUP 49 50. LEAPER PROBLEM PAIRS 50 age 6 7 8 9 10 11 PAGE vi vii MATHEMATICAL TOPICS 1. CO-ORDINATES, POSITIONS, MOVEMENTS, PENCIL 2. PARITY, VISUALISATION, PENCIL 3. PATTERNS, SORTING, ORDER, SEQUENCES, PENCIL 4. PARITY, SYMMETRY 5. PARITY, SYMMETRY 6. ARITHMETIC, SYMBOLS, EQUATIONS, PENCIL 7. GEOMETRY, SPATIAL NOTIONS, ENUMERATION, INTERSECTION 8. ENUMERATION, PIGEONHOLE PRINCIPLE, MAXIMUM/MINIMUM 9. ARITHMETIC, TRIAL AND ERROR, INPUT/OUTPUT 10. ENUMERATION, SYMMETRY 11. GEOMETRY, STRAIGHT LINES, SLOPE, RULER 12. GEOMETRY, STRAIGHT LINES, SLOPE, RULER, PENCIL 13. WORKING BACKWARDS FROM THE TARGET 14. PARITY, PENCIL 15. ENUMERATION 16. SYMMETRY 17. SYMMETRY, ELIMINATION 18. SHAPES, SYMMETRY 19. ENUMERATION, VENN DIAGRAMS, TRIAL AND ERROR 20. ENUMERATION, PARITY 21. DISJOINT PROPERTIES, UNION, SYMMETRY 22. LOGIC 23. STRAIGHT LINES, SLOPES, TRIAL AND ERROR 24. ENUMERATION, SPATIAL NOTIONS 25. ENUMERATION, SPATIAL NOTIONS 26. LOGIC, INFORMATION, TREE DIAGRAMS 27. SYMMETRY, WORKING BACKWARDS FROM THE TARGET 28. SHAPES, PARITY 29. TRIAL AND ERROR, SYMMETRY, MULTIPLES, DIVIDE TO CONQUER 30. ENUMERATION, TRIAL AND ERROR 31. GEOMETRY, DISTANCE MEASURES, PEER LEARNING 32. SYMMETRY, MULTIPLES 33. SYMMETRY, MULTIPLES, POWERS 34. SYMMETRY, MULTIPLES, TREE DIAGRAMS, WORKING BACKWARDS 35. SYMMETRY, REPRESENTATION 36. SYMMETRY, PARITY, WORKING BACKWARDS 37. ENUMERATION, PASCAL’S TRIANGLE 38. ENUMERATION, PASCAL’S TRIANGLE 39. ENUMERATION, PASCAL’S TRIANGLE 40. DECOMPOSITION OF SHAPES, TRIAL AND ERROR 41. DECOMPOSITION OF SHAPES, ENUMERATION, TRIAL AND ERROR, AREA MEASURES, SQUARE NUMBERS 42. ENUMERATION, TRIAL AND ERROR 43. ENUMERATION, TRIAL AND ERROR 44. ENUMERATION, SHAPES, ARITHMETIC, ORGANISING INFORMATION IN TABLES 45. LOGIC, ORGANISING INFORMATION IN TABLES 46. ENUMERATION, TRIAL AND ERROR, SYMMETRY 47. ENUMERATION, PROPORTION, ORGANISING INFORMATION IN TABLES 48. EXPONENTIAL GROWTH, GEOMETRIC SEQUENCE 49. ENUMERATION, SHAPES, ORGANISING INFORMATION IN TABLES 50. ENUMERATION, SYMMETRY, ANGLES viii 1 1. Each Square has a Name Individual Age 6+ Co-ordinates, positions, movements, pencil This exercise is fundamental to understanding location on the board. Hands-on exercise Explain each square is named according to its co-ordinates: column (file) letter and row (rank) number. Example b3 in diagram: Walk up the street (b) to the house number 3. For small children, draw a line from the square to its co-ordinates. Hand out printouts of a chessboard. Use pencils. Task: Write the name of each square on the chessboard. Games (a) Teacher calls out names of squares and ask class to put a piece on them (not easy at first). (b) Trace the route (e.g. a1-a5) (c) Make a movement (e.g. c8-h3) and ask class for the co-ordinates (d) Place a rook on a1 and ask class how it can get to f5 (e) Print a maze on the chessboard and ask for the co-ordinates for the escape route (f) Using Velcro strips, get children to place pieces on the squares blindfold using only touch. Figure 1: Empty Board b3 Naming the squares with a blindfold 2 2. Identify the Square Colour Whole Class Age 6+ Parity, Visualisation, Pencil Hand out a sheet with only the first row shaded. Explain alternating shading. Task: complete the chessboard pattern. Whole class exercise Teacher faces away from an empty chess display board and asks the children to call out a square. Teacher identifies the colour of the square e.g. c4 is white. Repeat with the children Class Discussion: What methods can you use to arrive at the answer? Possible methods: (a) Snake Pattern: a1 = black, b1 = white, c1 = black, …h1 = white, h2 = black, g2 = white etc. (b) Diagonals: The long diagonals (Black a1-h8; White a8-h1) travel through the board. Find the nearest square to the one you need and adjust colour accordingly. (c) Parity addition (age 11+) Relabel the rows from a, b, c … to 1, 2, 3 and add the co-ordinates of the required square. If the sum is even, then the square is black. Extension Visualisation (age 8+) Pupils close their eyes and raise their right hand for white or left hand for black square called out. Photo: Children answering c4 3 3. Piece Categorisation Groups Age 6+ Patterns, sorting, order, sequences, pencil Hands-on exercise Show the class that the pawns are all the same size and smaller than the pieces Show that the physical pieces get shorter the closer they are to the corner ♚=♛ > ♝> ♞> ♜ Tasks: find (a) 5 ways to classify chess pieces e.g. according to colour, whether they slide or jump etc. (b) 5 ways to order chess pieces e.g. according to their exchange value, size of their base etc. Using piece tokens makes sorting quicker. Extension Teacher shows children groups of pieces and asks class to explain the grouping principle Solution Method_ A combination of approaches is required to generate ideas • Brainstorming involves calling out ideas in a group • Compare and contrast two piece types • Discuss how a third piece type differs from another two types • Analyse the pieces alone, in their starting position, or during play. Answer Classification ¨ Colour of piece ¨ Can/cannot move from starting position ¨ Colour of starting square occupied ¨ Long-range v short-range ¨ Heavy v light ¨ Pawns v pieces ¨ Major v minor ¨ Sliders/Jumpers ¨ Whether they can stand upside down Ordering ¨ Height ¨ Diameter of base ¨ Exchange value ¨ Number of pieces of each type ¨ Fastest to other side of board from unblocked starting square ¨ Alphabetical (language dependent) ¨ Number of piece-type that can fit onto one playing square Solutions offered by children: 4 4. Cross the Line Game Pairs Age 6+ Parity, symmetry Place the two kings on their starting squares. The kings move as in chess. White goes first. Who wins – the first player or the second player? – in each of these positions. Game Strategy This game is related to the concept of Opposition in chess. A direct opposition occurs on a file when two kings face each other with only one square between them. The distant opposition occurs on a file when two kings face each other with an odd number of squares (>1) between them. Thus, this game is closely related to the concepts of parity and symmetry. Answers 4(a) If White goes first, 1. ♔e2 is the winning move. If we observe the blue mirror line, Black has to be the first to choose one side, leaving the way open for White to choose the other. If Black tries to stay on the mirror line, the fact that there is an odd number of squares between the kings proves to be the determining factor for White victory. 4(b) The best move is the same as in 4(a) 4(c) Black wins by keeping the distant opposition with 1..♚b8 Figure 4(a) The first player to cross the middle line is the winner Figure 4(b) White wins by reaching the 8th rank. Figure 4(b) Black to play and prevent White from reaching the 8th rank 4(a) and 4(b) The best move for White 5 5. The Diagonal Opposition Game Pairs Age 6+ Parity, symmetry Place the two kings on diagonally opposite corners. The kings move as in chess. White goes first. The first player to occupy the starting square of the opposing king or the marked square on the other side wins. Who wins - the first player or the second player? Game Strategy Opposition along a diagonal (instead of a rank or file) is called diagonal opposition. This concept is the main idea behind this game. Answer 1.Kb2 is the winning move. If we observe the blue mirror line, as in the previous game, Black has to be the first to choose one side, leaving the way open for White to choose the other. Note that, at some point, the diagonal opposition may turn into direct opposition. Consequently, the pupil can use previously learned knowledge in a new situation - something important in mathematics. 6 6. Chess Arithmetic Individuals Age 7+ Arithmetic, symbols, equations, pencil Explain the conventional points value for each chess piece. Explain: ♕ = ♖ + ♗ + ♙ because 9 = 5 + 3 + 1 Tasks: (a) ♞ +♜+ ♟ = ? (b) ♕ = ♗+♗+ ? (c) Can you find four pieces that add together to the same value as a queen? What difference would it make if the queen was worth 10? (d) These are the captured pieces during a game, which side is leading in material, black or white? 1♞ ♝ ♛ ♖ ♙ 2♕ ♞ ♞ ♜ 3♟♟♟♟♜♝♕♘♘♖♖ Answers (a) 9 (b) 3 (c) Not possible because the pieces all have odd numbers, assuming the king is not worth 0. It is possible to find four pieces adding to 10 e.g. ♗+♗+♗+♙ (d) 1 White 2 White 3 Black ♕ = 9 ♖ = 5 ♗ = 3 ♘ = 3 ♙ = 1 7 7. Four rooks puzzle Pairs Age 7+ Geometry, spatial notions, enumeration, intersection Tasks: (a) Place four rooks on black squares so that all the white squares are attacked. (b) Find another position to achieve the same task. Advise the children to check their solutions. Solution Method_ The chessboard has 32 white squares. Each rook covers 8 white squares (4x8 = 32). Note that if a rook stands on a white square it would only cover 6 other white squares. Standing on a black square a rook can cover 8 white squares. Therefore, we are seeking four black squares on which to place the rooks. If the rooks stand on black squares in adjacent lines e.g. on a1 and b2, then the coverage of the rooks would intersect on the white squares b1 and a2 which means that instead of covering 16 white squares between them, they would only cover 14 white squares, which is insufficient. By inspection we find that the rooks should be an even number of squares away from each other given that their intersections need to be restricted to black squares. Hint: Start by finding solutions on the long black diagonal. Then find the other solutions. Answer 8 8. The Fallen Pieces Quads Age 7+ Enumeration, pigeonhole principle, maximum/minimum After chess club there are many chess pieces on the floor. You pick them up and count them. How many chess sets did they come? (a) There are 17 black pawns ♟ ♟ ♟ ♟ ♟ ♟ ♟ ♟ ♟ ♟ ♟ ♟ ♟ ♟ ♟ ♟ ♟ (b) In addition to the 17 black pawns you also find 3 white knights. ♟ ♟ ♟ ♟ ♟ ♟ ♟ ♟ ♟ ♟ ♟ ♟ ♟ ♟ ♟ ♟ ♟ ♘ ♘ ♘ (c) The classroom has only just enough chess sets for 20 children to play in pairs. Investigate the maximum solutions for each scenario above. Solution Method_ Hands-on investigation using counters. Explain meaning of maximum and minimum and ask to find these for each question. Note that we are not talking about probable an outcome is, only if it is possible. Answer The answer should be a range: one or more pieces could have come from each set. It is important to be aware that not all answers are single numbers.- (a) Between 3 and 17 At least 3 sets are required because a set contains only 8 pawns. If one pawn belongs to each set, then we require the maximum number of sets. (b) Between 3 and 20 We do not need any more than 3 sets because they already include 3 white knights. 20 pieces are missing which could have come from 20 sets at worst. The colour is irrelevant. (c) Between 3 and 10 The minimum number does not change. The maximum number is restricted by the requirement that the number of boards is 10 (i.e. 20÷2). 9 9. Corner attack puzzle Quads Age 7+ Arithmetic, trial and error, input/output Each corner of an otherwise empty board is occupied by an unknown chess piece. These pieces may be under attack from one or more unknown pieces in the other corners. Ignore colours. Each of the corner squares contains a number. This is the total number of attacks on that square from the pieces in the other corner squares. Each piece attacks at least one other piece. Hands-on tasks: Deduce what the four corner pieces are from the information given below: (d) Is there any other answer to 9(c)? Extension Solution Method_ • Establish that the only relevant pieces are the long-distance pieces ♗♖♕ • Establish that the number of squares attacked by each piece is ♗=1 ♖=2 ♕=3 • Explain that no. of attacks output equals no. of attacks input i.e. the sum of the piece attack values = sum of the corner numbers Note that there is only one way to obtain totals of 4 and 12, being minimum and maximum respectively. Answers (a)♗♗♗♗ (b)♕♕♕♕ (c)♖♖♖♖ (d)♗♗♕♕ (e)♗♖♕♖ (f) ♕♗♗♖ Task 9(a) Task 9(b) Task 9(c) Task 9(e) Task 9(f) 10 10. Piece power contours Quads Age 7+ Enumeration, symmetry The power of a piece varies according to its position on the board. Its power is the number of squares the piece attacks. Hands-on Task: (using coloured pencils on a printed board or coloured counters on a board) Colour-code each square according to the number of squares attacked from that square. The resulting pattern is the “power contour” for that piece. Find the contours for: • Rook • Bishop • Queen Which piece is depicted by these power contours? Answers Rook 14 uniform distribution Bishop See diagram Queen = Bishop + Rook Add 14 to each square of the bishop contour 10(a) Knight 10(b) King Figure 10(a) Figure 10(b) Bishop Power Contour 11 11. Avoid Three in a Line Game Pairs Age 8+ Geometry, straight lines, slope, ruler Starting with an empty chessboard, two players take turns to place same-colour counters on the board. A player loses the game by making a straight line of three counters. There is no obligation to notify your opponent that you have reached a losing condition. Use a ruler placed through the centre of the counters to check for straight lines. The game is easy enough but there may be confusion about how to find out if a third counter is on the same line as the first two. Playing method Check the board after each move. A counter may be on a straight line of three even if it is some distance away. Explain that there are different directions for straight lines. The straight lines used in the sliding chess moves of the rook are the easiest to recognise: + Rook (Orthogonal) • vertical (up, down) • horizontal (left, right) The straight lines used in the sliding chess moves of the bishop are also easy to recognise: x Bishop (Diagonal) • slope bottom left to top right (slope = 1) • slope top left to bottom right (slope = -1) The slope between any two counters is the ratio between the vertical distance and the horizontal distance. If the slope between two counters is the same as the slope from them to a third counter, then the counters are on the same straight line. Lines pointing up to the right have a positive slope and those pointing down to the right have a negative slope. The knight moves in a 2:1 L-shape. Straight lines also radiate from the knight but these are harder to recognise. The slope is important when considering the knight. In the diagram on the right, the knights all appear on the same line and their slope is ½ because the vertical distance (1) is divided by the horizontal distance (2). In the diagram on the left, there are two counters (a4, d5) with a vertical (up) distance of 1 and a horizontal (across) distance of 3 (like a long reach knight). Hence their ratio is 1/3. By inspection, the square g6 is also on a 1/3 slope from d5. If a4-d5 is a 1/3 slope and d5-g6 is a 1/3 slope, then a line runs through the three squares a4-d5-g6. The player to moves should select another square than g6. Placing a counter on g6 loses The slope of the knight move is 1/2 12 12. Counters on a Line Pairs Age 8+ Geometry, straight lines, slope, ruler, pencil Hand out printouts of a chessboard with 11 counters in the pattern shown. Tasks (a) How many sets of three counters on a straight line can you find? (b) Put a twelfth counter on the board and create four more sets of three in a line. It is easier to complete this exercise by first playing the game in Exercise 11. Solution Method (a) The children can find the vertical and diagonal lines easily (a7-c6-e5; e5-e4-e3; g7-e5-a1). It is more difficult for them to find lines whose slope is other than diagonal. Place a ruler between the centre of any two counters. Ensure the ruler goes through the centres. Find if there is a third counter along the straight edge. Test for all pairs of counters. Alternatively, find the slope of the line between any two points. The slope is the ratio between the vertical and horizontal distance from one counter to the next e.g. two up and three across. Then from the second point find if there is another point which has the same slope. If so, then all three points are on the same line. Test using a ruler (more details on Exercise 11). (b) Move systematically counter-by-counter and, using a ruler, check if you can find some lines. Children tend lack a systematic approach. Hint: start at a1 and move to the nearest counter. From c1, the lines c1-c6-c8; c1-e3-h6; c1-b4-a7; c1-e4-g7 radiate. 13 13. Wythoff’s Game [Q] Pairs Age 8+ Working backwards from the target Place a queen on h5 on an empty chessboard. This queen can only move West, South West or South. The players take turns moving the queen. The first player to move the queen to a1 is the winner. Who wins – the first player or the second player? Solution Method_ First play the game to get a feeling of possible best play. Identify the ‘safe squares’ where you would like to move to in order to win. To do this, work backwards from a1. Use counters to mark the safe squares. There are three safe squares: b3, c2 and f4 (marked green). Answer The second player wins with best play. On their first move, the first player cannot avoid allowing the second player to either reach a safe square or go directly to a1. 14 14. Rook Corner Puzzle Pairs Age 8+ Parity, pencil Use pencil and paper with many prints of a chessboard with a rook on a1. Task: The rook must visit all the squares on the board and finish on h8. Condition: The rook is not allowed to visit the same square twice. Solution Method_ Having tried a few different routes the children will conclude that there is no solution. They may need some convincing to grasp that failing to find a solution does not mean that there is none to be found. Proving that something is impossible is a giant leap in children’s mathematical understanding. Hints and leading questions to guide towards the solution: • The rook should proceed in one-step moves on its journey. Use two colours: e.g. red for odd (1st, 3rd, 5th etc.) and blue for even moves as shown in the diagram. • How many squares does the rook visit on its way to h8? • What is the colour of the 1st, 2nd, 3rd, 4th etc. squares? • What is the colour of the 63rd square? • What is the colour of the h8 target square? Answer Every odd visited square is a light square, so the 63rd last square should also be light, but it is in fact a dark square so the rook can never complete the mission. Extension Question: On what size boards would the rook tour work? Answer: The task is only possible for odd sized boards (7x7, 5x5 3x3 etc.) 15 15. Shortest Rook Tour Individual Age 8+ Enumeration Use pencil and paper with many prints of a chessboard with a rook on a1. Alternatively, use plastic sheets over a board template. Show a rook on a1 and a marker on h8. Task: The rook must visit all the squares on the board and return to a1. Condition: The rook is not allowed to visit the same square twice. Suggested activities: • Find as many ways as you can to complete the rook's tour. • Draw each route on paper. • Find the length of each tour that you have drawn. Note that the squares passed through by the rook en route from one stopping place to another are 'visited' during that move. It is important to emphasise that the rook returns to its starting point, a1. Let children explore the various tours and state their lengths. Exploration_ • What is the smallest number of moves in which the tour can be completed? • What is the longest tour (the largest number of moves) that you can find Answer It can be proved that the shortest tour consists of 16 moves. Some interesting properties of the rook’s tour, however, can be investigated. For discussion: • If a tour starts with a vertical move, it must end with a horizontal move. • Vertical and horizontal moves alternate. • There are equal number of horizontal and vertical moves. • The length of each tour is even. A tour with the minimum 16 moves A tour with 28 moves The longest tour with 56 moves 56 moves - wrong answer! 16 16. Race to the corner [rook] Pairs Age 8+ Symmetry Use a chessboard, a rook, and counters to mark the moves. Place a rook on h8 and a marker on a1. This rook can only move South or West. Players take turns to move the rook. The first player to reach a1 wins the game. This is a game that children like to play. Who wins – the first player or the second player? Playing method_ The second player has the winning strategy. He /she must move the rook to the a1-h8 long diagonal. Here is how to guide children towards the solution: • First children should explore the game freely. Ask them to look for a way to win. Ask if they prefer to be the first or second player in this game. • Play as second player with correct strategy against a child or a group of children. • If the winning strategy is still not found, then tell them to think about the importance of the a1-h8 long diagonal. 17 17. Chomp Pairs Age 6+ Shapes, symmetry Using a printout of a board with a “poison” marker on h8. See Figure 18(a). Each player takes turns to fold (“chomp”) the paper chessboard horizontally or vertically along any marked line. See Figure 18(b). There are no pieces in this game. The section further from the marker is discarded. The loser is the one left holding the poison. In the original game they played with a large bar of chocolate! Who wins – the first or second player? Figure 18(a) Avoid holding h8 Figure 18(b) Chomping horizontally Playing method_ In order to win, a player should use a geometrical strategy. Whatever the first player does, the second player should fold the paper so as to leave a square-shaped board. See Figure 18(c). In other words, the winning strategy is to chomp along the diagonal to h8. The opponent will have no other choice but to fold into an oblong. Successively, in each move, the strategy of leaving square-shaped shapes is repeated until the poisoned square is achieved. Figure 18(c) The next player can only make an oblong 18 18. Army Power Puzzle Pairs Age 8+ Enumeration, Venn diagrams, Trial and Error Place one of each of the six piece types of the same colour on their starting squares. See Figure 19(a). Q1: How many squares does White attack? Squares attacked more than once count only once. Q2: Identify the squares which are attacked once, twice, thrice. Q3: Reposition the pieces to find the minimum number of squares that are being attacked. Q4: (advanced) Reposition the pieces to find the maximum number of squares under attack. Answers Q1: 25, see Figure 19(b) Q2: See Figure 19(c) Q3: The minimum number of attacks is 5, achieved various ways. See Figure 19(d). Figure 19(d) Minimum number of attacks Q4: The maximum number of squares attacked is 51. 6 squares are occupied by pieces. 7 squares are unattacked. Extension Q5 Give the locations unoccupied squares and ask the class to figure out the position of the pieces. Figure 19(a) Army Power Puzzle Figure 19(b) 25 squares covered Figure 19(c) Maximum number of attacks Figure 19(c) Multiple attacks 1 attack 2 attacks 3 attacks 19 19. Maximum number of knights Pairs Age 8+ Enumeration, parity What is the maximum number of knights that can be placed on a chessboard such that no knight attacks another? Use counters instead of knights. Solution Method_ Knights that do not attack each other are called independent knights. Notice that a knight changes colour on each move (from a light square to a dark one and vice versa). Knights placed on the same colours will be independent. Answer The maximum number is 32. 20 20. Mystery combination piece Pairs Age 8+ Disjoint properties, union, symmetry A combination piece combines powers from different chess pieces. A mystery combination piece is attacking the crossed squares (Fig 20): (a) Mark where the mystery piece is it located (b) Identify the pieces it combines Review Exercise 10 for piece patterns. Solution Method_ Detect the symmetry around d2. (a) Find the longest lines and join them together with a ruler and pencil. These are the ranges of the long-distance pieces. See where they intersect. d8-d1, a5-e1, h6-c1, a2-h2 (b) Notice that a queen on d2 is consistent with the pattern of crosses. Subtract the queen’s pattern and see what is left. This is consistent with the moves of a knight. Answers (a) d2 (b) queen + knight Figure 20: Mystery Combination Piece 21 21. Positional logic Pairs Age 8+ Logic Starting pieces ♚♛♜♝♞♟ This is a hands-on activity. The children must set up the correct position on the chessboard. Task: Set up a position with the above pieces obeying the following rules: 1. The pieces stand next to each other on the same line 2. The king is next to the queen which is on c4 3. The rook is between the bishop and a pawn 4. The bishop is the furthest piece from the queen 5. The knight stands on a higher rank than the king Solution Method_ This can be solved using guesswork, trial and error. Alternatively, a more structured approach can be deployed. The key is to find a good starting point. Make use of the information such that no guessing is necessary. For example: (a) From (5) we learn that the pieces are arranged vertically along the c file (b) From (2) we know that the position of the queen is fixed at c4 (c) From (3), there is a 3-piece segment bishop+rook+pawn There are only four squares above the c4 queen and three below. To satisfy (4), the bishop must be on c8 the furthest point from the queen on the c file (d) From (5) the knight is above the king which, since c4 is occupied, implies that it must be at c5 or above (e) By elimination, the king sits on c3. Extension Creative activity: Ask children to devise a positional logic problem. The ‘best problem’ award goes to the one with a unique solution! 21. Solution 22 22. Counters on a Line Pairs Age 8+ Straight lines, slopes, trial and error Children recognise horizontal, vertical or diagonal lines. They find it difficult to identify lines with other slopes especially when the counters are spaced apart. Task 1: Hand out printouts of these two positions and ask the class to find the spoiling lines where three counters are in a row. (a) (b) Task 2: Place 16 counters on the chessboard so that no three are in a line. Solution Method_ Solving this requires trial and error whilst checking for all possible lines. Here are some hints: • Start by putting the counters in groups (e.g. of four) which simplifies the search task. • Place two counters in each column and each row whilst checking they do not touch on the diagonals. Then check for the other slopes with a ruler through the centre of the counters. • If you get near a solution, try adjusting the adjacent counters systematically. Task 1 Task 2 One solution: Four groups of four 23 23. Minimum Moves Map Pairs Age 9+ Enumeration, spatial notions Task Use the coloured counters to build a colour-coded map of the board showing the minimum number of moves it takes a piece to reach every other square from d4 for (a) a king; (b) a knight. Solution Method_ Starting at d4, mark all the squares reachable in one move. Then from each of these repeat the process using different coloured counters. Repeat until there are no more squares. Answer 24 24. How Many Routes? [pawn] Pairs Age 9+ Enumeration, spatial notions The pawn can make any move one square forwards whether straight or diagonal. Note the pawn can also make a move by capturing an imaginary piece – hence diagonal moves must be counted as well. Task: (Using printouts of these positions) How many different routes are there for the pawn: From To (a) a2 Other side (b) d5 Other side (c) d4 d8 Solution Method_ For each possible position of the pawn count the number of ways the pawn can reach that square and write that number in the square. Start from the original position. Some leading questions: • How do you proceed to the next rank? • What do you do with the numbers when you move up a rank? • Is the pawn equally likely to occupy every possible square? Answer 21(a) 35 ways to reach back rank 21(b) 27 ways to reach the back rank 21(c) 19 ways to get to d8 25 25. The Light Pawn Pairs Age 9+ Logic, information, tree diagrams There are eight white pawns that look the same but one of the pawns is slightly lighter than the others. The difference is too small to judge by hand. You have an imaginary weighing balance in which you can place one or more pawns either side. Use this to find the light pawn. Question: How many weighings do you need to find the lighter pawn? Solution Method_ In this thought experiment, the key idea is that if you know that one item is a different weight, then if you weigh two items, you can derive conclusions about the third item. Consider eight pawns with labels A, B, C, D, E, F, G, H. Let your first weighing be ABC v DEF. If it is balanced, then you know that the lighter pawn is either G or H. In that case, execute the final single weighing G v H. If it is unbalanced, with ABC heavier than DEF, then you know that the lighter pawn is either D, E or F. In that case, execute the final single weighing D vs E (if it is balanced, the lighter pawn is F; if D is heavier than E, the lighter pawn is E; if E is heavier than D, the lighter pawn is D). Vice versa if the first weighing is unbalanced, with DEF heavier than ABC. A pedagogical way to proceed is to organize a tree on the whiteboard: The pupils should understand that the eight possibilities (eight pawns) match the eight terminals of the tree that describes the strategy. Answer Only two weighings are required. Extension What happens if there are nine pawns, one of them slightly lighter than the others? The strategy still works, with one more branch (for the balanced ABC vs DEF). There are three possible results (leftright). There are three ways to walk through a one-weighing tree and nine ways (3x3) to walk through a two-weighings tree, and so on. Note that 10 pawns cannot be solved with only two weighings – three weighings are required. 1st weighing 2nd weighing 26 26. Wythoff’s Game [QQ] Pairs Age 9+ Symmetry, working backwards from the target Place two queens of the same colour on h5 and g8 on an otherwise empty chessboard. Stipulate that these queens can only move West, South West or South. The players take turns moving either queen. Provide markers to place on the board. The first player to move the queen to a1 is the winner. Who wins – the first player or the second player? o See also Exercise 13 Wythoff’s Game with one queen. o First children should explore the game freely. Ask them to look for a way to win. Ask if they prefer to be the first or second player in this game. o Play as first player with correct strategy against a child or a group of children. o If the winning strategy is still not found, then tell them to think about the importance of the a1-h8 long diagonal Solution Method_ Set up a symmetric position around the a1-h8 main diagonal. Thereafter the first player copies the moves of the second player – a strategy known as Tweedledum-Tweedledee. Answer The first player can win by moving a queen to e8. Extension Let’s play with a single king starting from h8. Like before, the first player to move the king to a1 wins. Hint: Work backwards from the solution. Identify the squares where you do not want to land the king (marked in red). Winning strategy: avoid the red squares. The first player can win with best play: King to g7, then copy the opponent’s move. Wythoff’s king game: full pattern Wythoff’s king game: The final moves 27 27. The domino tiling puzzle Pairs Age 9+ Shapes, parity This exercise benefits from access to 32 dominoes of size 2x1 that cover exactly two squares of the board. These can be card cut-outs. Q1: Can the dominoes cover all 64 squares of the chess board? Q2: Can the dominoes cover all 62 squares of a board from which two opposite corners are removed? Q3: Can the dominoes cover all 62 squares of a board from which two different coloured squares are removed? Answers Q1 Yes, depicted in Figure 28(a). There are many ways to cover the board. Q2 No. The second question has an ingenious chromatic solution. A domino piece always covers one white square and one black square. So, a group of dominoes covers an equal number of squares of each color. The mutilated board (Figure 28(b)) has 30 white squares and 32 black squares, so the answer to the given question is negative. Extension Q3: Yes. Suppose that you remove two squares of different colors. Yes. There is an ingenious geometric solution. The line in Figure 28(c) goes through all squares. Figure 28(a) Covering a chessboard with dominoes Figure 28(b) Multilated chessboard Figure 28(c) Continuous line Figure 28(d) Continuous line in two segments Remove one black and one white square. The removal divides the line into two segments (See Figure 28(d)). One connects the yellow sides and the other connects the blue sides). Now, all that must be done is to cover each segment with dominoes. This is possible because there is a balanced number of black and white squares in each segment. This method works for any pair of removed opposite coloured squares. 28 28. The tromino tiling puzzle Individuals Age 9+ Trial and Error, symmetry, multiples, divide to conquer Using printouts of empty chessboard prints with pencil and 22 trominoes. A tromino is a shape of size 3x1. Each tromino covers exactly three squares of the board. Can you tile the chessboard with 3x1 trominoes if a corner square has been removed? This exercise is easy to explain and can be done alone. Reserve for a “special day”. (a) Can the trominoes cover all 64 squares of the chess board? (b) Remove one corner of the chessboard. Can the trominoes cover all 63 squares of the mutilated board? (c) Is there any other square, apart from the corner square, which you can remove to make the trominoes fit? Introductory questions • Can the trominoes cover all 64 squares of the chessboard? No, since 64 is not a multiple of 3. • How many trominoes do you need to cover 63 squares? 63 / 3 = 21: 21 trominoes will cover 63 squares. • Does it matter which corner of the chessboard is removed? No, it does not matter due to symmetry. Getting into the problem • Try to cover the chessboard that has a corner removed with trominoes. Make several attempts with different tile arrangements. [It is not possible to complete the task.] • Remove a different square from the chessboard (cross it out on your sheet). Perhaps try to remove a square from the long diagonal. Removing b7 does not seem to work as shown in Figure 29(a). Answer If c3, c6, f3 or f6 is removed, the tiling can be done! Solution Method_ With Trial and Error children will find one of the four symmetrical solutions - Figure 29(b). The proof is beyond primary school level. Figure 29(b) Figure 29(a) 29 Figure 30(b) minimum (i) Figure 30(b) minimum (ii) 29. Two queens puzzle Quads Age 9+ Enumeration, trial and error Tasks: Place two queens on a chessboard such that they attack the: (a) maximum number of squares (b) minimum number of squares Only count unoccupied squares. A second attack on a square is not counted. Ignore colours. Solution Method_ (a) We know from its power contour that the queen is strongest at the centre of the board. Start by placing a queen in one of the central squares (e.g. d4). Then systematically try placing the second queen as close as possible. The maximum number of unoccupied squares attacked 42 arises where the queens are (i) orthogonally adjacent in the centre, or (ii) are a knight’s distance away on the roomy side of the first queen (e.g. d4 & e6). (b) We know from the power contour that the queen is weakest on the edge of the board. We can also restrict movements by placing two queens on the same line. As a first attempt, placing the queens on the corners (opposite or lateral) gives 32. However, there is a lower number of 31 which arises with (i) a1 & a3/a5/a7 or (ii) b1 & d1/f1/h1 (plus symmetries). Figure 30(c) is a two queens joint contour map for the number of unoccupied squares under attack given one queen on a1 – which incorporates the solution in Figure 30(b)(i). Answer Figure 30(a) maximum (i) Figure 30(a) maximum (ii) Figure 30(c) Joint contour with Qa1 30 30. Equidistant checkmate Pairs Age 9+ Geometry, distance measures, peer learning This is a hands-on exercise with pieces and a chessboard best done in pairs or small groups. As it is necessary to know about checkmate to play chess, at least one member of the group or pair must know the rules of chess. After this investigation they will all know what checkmate is! Introduction Question: One piece gives check but what is the role of the second piece in checkmate? Answer: The second piece guards the escape squares i.e. holds the king ‘in prison’. Solution Method_ Set up positions with two white pieces that are the same distance away from the a1 square. Now place the white king on c3 and the black king on a1 i. Find two white pieces to deliver checkmate ii. Ensure that these two white pieces are the same distance away from the black king Answers There are a few different solutions with legal checkmate. In Figures 31(a) and 31(b) the queen can swap with the other piece and still deliver checkmate. Note Figure 31(c) is a false solution because the white queen can be captured. Nor is Figure 31(d) a solution because the position is illegal – the black king was already in check in the previous move. Figure 31(a) Figure 31(b) Figure 31(c) X Figure 31(d) X Figure 31(e) In Figure 31(e) the pieces are 5 units (a unit being the length of the side of a square) away from the black king; more precisely the centre point of a1. The centre points of a1, e1 and e4 form a right-angled triangle. The two shorter sides are of 3 and 4 unit length and, using Pythagoras’ theorem, the length of the hypotenuse must be 5 units. [Age 10+] 31 31. The Subtraction Game Pairs Age 9+ Symmetry, multiples Place 10 counters on three rows. Players take turns to remove 1, 2 or 3 counters from the end of any row. The player to remove the last counter is the winner. This game is related to the 3 times table (or the 4 times table if the maximum number of counters to be removed is 4, etc.) Preliminary question Find a winning move in a row of size 8 The winning move is the removal of one counter. In general, the winning play is to leave 4, 7, (10) … counters i.e. a multiple of 3 plus 1, which is one more than the maximum number of counters that can be removed. In the end, the opponent must leave 1, 2 or 3 counters after which the first player removes the last counter. Game strategy_ The best strategy in last-move-wins games is to create independent pairs of identical components. Thereafter, for each pair, mimic the move of your opponent - the “Tweedledum and Tweedledee” strategy (see also Exercise 26). Winning Plan The winning move is to remove the third row. In Figure 32(b) the pairs of identical components are bracketed together. From this position, copy anything the opponent does. Extension Play the position in 32(c). The winning move is to remove one counter from the first rank. For a solution method, see next Exercise 33 Nim. 1 7 6 5 4 3 2 8 Figure 32(a) Subtraction Game Figure 32(c) Pairs of identical components } } Figure 32(c) Extension exercise 32 32. Nim Pairs Age 10+ Symmetry, multiples, powers Place ten counters on three ranks: 5,3,2 Players take turns to remove one or more counters from the end of any rank. The person to remove the last counter is the winner. Who wins, the first player or the second player? Game strategy A winning plan is to try to obtain symmetry and then copy your opponent’s moves. Symmetry arises when there are pairs of rows with the same number of counters in each row. If your opponent removes one or more counters, you remove the same number of counters in the other row: Tweedledum and Tweedledee. Winning Plan The winning move for the first player is to remove four counters from the first row to leave {1,3,2}. Whatever your opponent does, you can create symmetry on your next turn. For example: • Your opponent removes row 1 leaving {0, 3, 2}. On your move, remove one counter from the second row to leave {0, 2, 2}. This is a symmetrical and you can win as described above. • Your opponent removes the third row leaving {0, 3, 5}. On your move, remove two counters from the first row to leave {0, 3, 3}. This is a symmetrical and you can win as described above. [Age 11+] More generally, the winning plan is to achieve symmetry regarding the number of “powers of 2” (1,2,4,8). For example, a row of 7 comprises three powers of 2: 1, 2, and 4. In the starting position, the groups are {5,3,2} decompose as follows: 5 = 4 + 0 + 1 3 = 2 + 1 2 = 2 + 0 This gives us There are two 1’s, two 2’s and one 4. By removing the one 4 we are left with these paired powers of 2: two 1’s and two 2’s Hence the winning play on the first move is to remove four counters from the first row. Powers of 2 1 2 4 No. of these 2 2 1 33 33. Golden Coin Game Pairs Age 10+ Symmetry, multiples, tree diagrams, working backwards Place ten red counters on three rows: 5,3,2. In addition to these counters, append one golden counter to the longest row. See Figure 34(a). Players take turns removing any number of counters from anywhere within any row. The golden counter must be the last one to be removed and it cannot be removed along with a red counter. The player to remove the golden counter is the winner. Who wins? This game benefits from having played the Subtraction Game (Exercise 32) and Nim (Exercise 33). Game strategy The conventional strategy when playing subtraction games is to be the person to remove the last counter. The twist in this game is that the person who removes the last red counter is the loser not the winner - this is the misère version of the subtraction game. The best strategy is to follow the conventional winning strategy as if the game were played just with red counters. However, just before the end, give the other side no choice but to remove the last red counter. Winning Plan To simplify matters, we ignore the golden counter and play to lose the game with red counters. Notice that the arrangement of red counters is the same as in Exercise 33. Hence, to take control of the game, the first player should remove 4 red counters from the first row. The play then follows the game tree in Figure 34(b) which sets out the responses from Player 1 to moves by Player 2. Player 1 wins by ultimately leaving one unit (Box 10) which is reached by various routes. Player 2 must pick up the last red counter after which Player 1 picks up the Golden Counter. Player 1 reaches a winning position by leaving either two pairs (Box 7) or three units (Box 8). Figure 34(a) Golden Coin Game 1 3 2 4 5 6 7 8 10 After Player 1 After Player 2 After Player 1 9 After Player 2 Figure 34(b) Game Tree for Golden Coin Game 34 34. Northcott’s Game Pairs Age 10+ Symmetry, representation Player 1 has counters on a4, a5 and a6 and Player 2 has counters on g4, f5 and e6. Moving alternately, each player moves one of their coloured counters along the rank any number of moves towards their opponent. No captures or jumping are permitted. The last person to move is the winner. Who wins? Solution Method_ This exercise is equivalent to Nim (Exercise 32) but with spaces representing counters. The horizontal distance between each of the differently coloured counters is {2,3,5}. Each move reduces the distance, equivalent to removing a counter in Nim. The playing strategy is therefore the same as in Nim. Northcott's Game 35 35. Sliding Rooks Game Pairs Age 10+ Symmetry, parity, working backwards Place eight white rooks on the left side of the board and eight black rooks on the right-hand side of the board (Figure 36). These rooks can only slide sideways, not up and down. The white rooks can only move rightwards, and the black rooks can only move leftwards. Capturing or jumping over pieces is not permitted. The last person to move is the winner. Who wins? Playing Method_ The rows are independent of each other because the rooks can only move horizontally. Work backwards and consider a single row and then take the other rows into account. The first player can win the first row by moving their rook next to their opponent’s rook. Similarly, the second player can win the second row. If there are only two rows, the second player is the last player to move and hence wins the game. The second player will win following this symmetrical strategy because there is an even number of rows. The first player must try and “lose a move” to obtain symmetry. One trick is for the first player is tempt the second player to unwittingly break symmetry. For example, on the first move, the first player could leave one square between their rook and their opponent’s (e.g. ♖a1-f1) tempting the second player to close the gap. However, the second player’s correct response would be to mirror the move on another row e.g. ♜h8-c8. As long as the second player is alert, he or she should win. Figure 36(a) Sliding Rooks Game Figure 36(b) White tries 1.♖f1 but Black responds 1..♜c8 36 36. How Many Routes? [rook] Pairs Age 10+ Enumeration, Pascal’s triangle Resources: Printouts of these positions with a rook on a1 and the target square marked in colour. Task The rook can only move North or East, alternating direction on every move. Calculate how many different routes are there for the rook to reach: Figure 37(a) Figure 37(b) Figure 37(c) Introductory questions • Why do you think that the rook can only move North or East? So that the rook gets closer to the target square on every move. Otherwise there could be countless ways to get there. • Why is it important to alternate direction on every move? We are looking for different routes, but subsequent moves in the same direction would be part of the same route. How many ways can the rook get to a5 from a1? (Figure 37(a) As it alternates direction on every move, there is only one route, the single move from a1 to a5. Solution Method_ • Start from the original position. For each possible position of the rook, count the number of ways it can reach that square and write that number in the square. • Solving trick: look for the neighbouring squares in the direction where the rook could have come from. For example, for the rook to reach c2 it could have come through b2 or c1, so the sum of the numbers in b2 and c1 gives the total number of routes to c2; see Figure 37(b). • Continue numbering each square this until you get to the target square. Answers There are 20 routes to d4 as shown on Figure 37(e). There are 330 ways to get to e8. Figure 37(d) Figure 37(e) 37 37. Rook’s Tour on a Mutilated Board Pairs Age 10+ Symmetry, Elimination Remove the opposite corners of a chessboard. Place a rook on any square. Can the rook visit all the squares of the mutilated chessboard without landing twice on the same square? This is an advanced version of Exercise 14, the Rook Corner Puzzle. Solution method Use pencil and paper and chessboard printouts. Children should have an unguided investigation first. Let them ask for clarification about the problem. Expect questions like these: • Does it matter where the rook starts? No, the rook can start from any square. • Does the rook have to go back to the starting square? No, the rook can finish on any square provided it has visited every square once and only once. The children will not be able to find a solution. Now you can guide them towards the proof with the following hints and questions: • How many dark and how many light squares does the rook have to visit? 30 dark and 32 light squares. • The rook should proceed in one-step moves on its journey. Use two colours, for example red for odd (1st, 3rd, 5th etc) and blue for even (2nd, 4th, 6th etc.) moves as shown in the diagram. • How many one-step moves would the rook have to do to complete the tour? 61 moves. • How many odd and how many even one-step moves would the rook have to do to complete the tour? 31 odd and 30 even moves. • Suppose the rook starts on a light square, such as a2. What can you say about the colour of the target square (where the rook lands) on odd and on even one-step moves? The target square of odd moves is always a dark square, while the target square of even moves is always a light square. • Suppose the rook starts on a light square, such as a2. Count the number of dark target squares it needs to complete the tour. Can you explain why the rook cannot complete the tour? The target square of odd moves is always a dark square and there are 31 odd moves altogether. Hence there are 31 dark target squares. The rook cannot complete its tour as it would need to visit only 30 dark squares. • Suppose the rook starts on a dark square, such as b2. Count the number of light target squares it needs to complete the tour. Can you explain why the rook cannot complete the tour? The target square of odd moves is always a light square and there are 31 odd moves altogether. Hence there are 31 light target squares. The rook cannot complete its tour as it would need to visit 32 light squares. Answer By a process of elimination the answer is no, the rook cannot complete a tour of the board. 38 38. Bishop Zigzag Puzzle Pairs Age 10+ Enumeration, Pascal’s triangle The diagram shows the bishop crossing the board diagonally. On each move, it moves up a rank and alternates direction from North-East to North-West and vice versa. Use pencil and chessboard prints. Questions: (a) How many routes from b1 to g8? (b) How many routes from b1 to the other side of the board? (c) What is the total number of white-square paths starting on the first rank and ending on the eighth rank? This problem has a similar solution method to that of Exercise 37. Children who understood the enumeration of the rook’s routes will succeed with this task about the bishop. Introductory questions • Why do you think that the bishop must move up a rank on each move? So that it would get closer to the target square on each move. • Why is it important to alternate direction on every move? We are looking for different routes, and just like in Exercise 37, subsequent moves in the same direction are part of the same route. Solution Method_ Use a chessboard to find ways for the bishop to get to g8 from b1. Write down your routes. Here are two examples: 1st route: b1→d3→c4→g8 2nd route: b1→e4→d5→ g8 How many different routes can you find? There are altogether seven. For each possible position of the bishop count the number of ways it can reach that square and write that number in the square. Start from the original position. Answers (a) There are 7 routes from b1 to g8. (b) There are 14+28+20+7=69 routes to the other side; see Figure 38(a). (c) There are 296 different routes. Add all numbers in the 8th rank of Figures 38(a), 38(b), 38(c) and 38(d). Figure 38(a) Figure 38(b) Figure 38(c) Figure 38(d) 39 39. Five queens domination puzzle Quads Age 10+ Enumeration, Pascal’s triangle Task Rearrange five queens such that they attack every square on the board at least once. The attacks must cover all the occupied squares. Pieces that attack every unoccupied square are said to dominate the chessboard. A minimum of five queens are required to dominate the chessboard. Introductory question How many squares can a queen attack? Note that the queen does not attack the square on which she stands. 21, 23, 25 or 27 depending on her position. See the queen’s contour in question 40. Solution Method_ Place five queens on the chessboard. Mark every square that is under attack by one or more queens with a counter. How many free (unoccupied and not attacked) squares can you count? Try to reduce the number of free squares by moving a queen to a different square. Can you reduce the number of free squares to zero? If so, then you have solved the task. Answers There are exactly 4860 different ways to accomplish the task. Let’s look at some interesting arrangements: Figure 42(a) All queens are placed on a long diagonal Figure 42(b) All queens are placed on a diagonal that is not the long diagonal Figure 42(c) All queens are placed along a single file of the chessboard Figure 42(d) None of the queens attack one another Note that in in (a) - (c) the attacks cover all the occupied squares as well. Figure 42(a) Figure 42(b) Figure 42(c) Figure 42(d) 40 40. Dividing the board Individual Age 10+ Decomposition of shapes, trial and error Hand out printouts of a chessboard marked up with crosses as in Figure 40(a). Task Divide the chessboard into eight smaller squares so that each one contains at least one cross. Not all the drawn squares need to be the same size. Solution Method_ Try different size squares from 2x2. There are several potential squares which contain one cross. Start from a corner of the board and then systematically move down and across. Answer Figure 40(b) shows the solution. Figure 40(a) Divide the board Figure 40(b) Board divided into 8 squares 41 41. Tiling the board Quads Age 10+ Decomposition of shapes, enumeration, Trial and Error, area measures, square numbers Display a chessboard tiled into 16 squares. Hand out printouts of an empty chessboard. Task: Show two other ways to partition a chessboard into 16 squares. Introductory questions 1. A single square of the chessboard has area of 1 unit. What is the area of the chessboard? 64 units. 2. Squares are drawn on the chessboard. List the area of these squares. What name is given to these numbers? 1, 4, 9, 16, 25, 36, 49 and 64. These are square numbers. Getting into the problem 1. Tile the chessboard with squares. Try to use many different sizes of squares. Count the number of squares that you used. 2. Try to use 16 squares to tile the chessboard. Find as many different solutions as you can. Solution Method_ Rearranging the same set of squares does not constitute a new answer. The task can be completed in eight different ways. The teacher may give hints to guide the children towards each solution. 1. All squares are the same size: Figure 41(a) 2. There is a 7x7 square: Figure 41(b) 3. Use a 5x5 square and three 3x3 squares: Figure 41(c) 4. Use three 4x4 squares: Figure 41(d) 5. There are two 4x4 and six 2x2 squares: Figure 41(e) 6. There is a 4x4 square and four squares of unit area: Figure 41(f) 7. Use a 4x4 square and the same number of 3x3 and 2x2 squares: Figure 41(g) 8. There are eight 2x2 squares in this solution: Figure 41(h) 42 42. Twelve knights problem Pairs Age 10+ Enumeration, Trial and Error Task Place twelve Knights on a chessboard so that every square is either attacked or occupied. If you do not have enough knights, use counters to represent them. Introductory question How many squares can a knight attack? 2, 3, 4, 6 or 8 depending on its position. See the knight’s contour in Exercise 10. Solution Method_ 1. Put twelve knights on the chessboard. Mark every square that is under attack (by one or more knights) with a counter. How many free (unoccupied and not attacked) squares can you count? a) Try to reduce the number of free squares by moving a knight to a different square. b) Can you further reduce the number of free squares? 2. Place three knights on b6, c6 and c5. Mark every square that is under attack by one or more knights with a counter. Now one quarter of the board is almost fully covered with knights and counters. Try to use this knight arrangement to cover the rest of the board. Answers If every square is either occupied or attacked by the knights, then we say that the knights dominate the chessboard. The two solutions are shown in Figure 39(a) and Figure 39(b). Figure 39(a) Figure 39(b) Note if we reflect either solution around the horizontal or vertical central axis, or along the main diagonals, then we get the other solution. Also, the diagrams have a rotational symmetry of order four (i.e. they look the same when rotated a quarter turn). 43 43. Eight queens puzzle Groups Age 10+ Enumeration, Trial and Error Task: Rearrange eight queens such that no queen attacks any other. You may find it helpful to mark the attacked squares with counters. This is the oldest and most famous chess & maths problem. Pieces that do not attack one another are called independent. Solution Method_ • Put two queens on the chessboard that do not attack each other e.g. a knight’s move apart • Add another queen such that no queen attacks any other • Add another queen. Check that still no queens are under attack. Increase the number of queens one by one. Can you get to eight queens? There is no formula for doing this. Computer scientists use a method called “backtracking”: every time you find a conflict, backtrack to the last placement and instead try the next square. In a simplified version of this investigation one, two or three queens are missing from the solution and the task is to place the remaining queens on the correct squares. Best done on paper with prepared diagrams. Answers Altogether there are 92 positions which satisfy the requirements. If we exclude all rotations and reflections, then there are only 12. Some of these are shown below: Figure 43(a) Four in a row solution Figure 43(b) Another four in a row Figure 43(c) The only symmetrical solution Figure 43(d) 44 44. How many Squares on a Chessboard? Individuals Age 10+ Enumeration, shapes, arithmetic, organising information in tables Task 1: How many squares of all sizes can you find in this grid? Size of square No. of squares 1 by 1 4 2 by 2 1 __ Total 5 Task 2: How many squares of all sizes can you find in this grid? Size of square No. of squares 1 by 1 9 2 by 2 3 by 3 1 _ Total 14 Solution Method Start from a small grid and step by step expand the size of the square. Task 3: How many squares of all sizes can you find in a 4x4 grid? Task 4: How many squares of all sizes can you find on a chessboard? Complete the grid. Number of squares Grid size 1x1 2x2 3x3 4x4 5x5 6x6 7x7 8x8 Total 1x1 1 1 2x2 4 1 5 3x3 9 4 1 14 4x4 16 9 4 1 30 5x5 25 16 9 4 1 55 6x6 36 25 16 9 4 1 91 7x7 49 36 25 16 9 4 1 140 8x8 64 49 36 25 16 9 4 1 204 At some point, the pattern may be noticed. The total number of geometrical squares on a chessboard is 12 + 22 + 32 + 42 + 52 + 62 + 72 + 82 Answers 1) 5 2) Missing number is 4 3) 30 4) 204 45 45. Tournament Scoring Puzzle Quads Age 10+ Logic, organising information in tables The scoring system for a school tournament is: Win = 3 points Draw = 2 points Loss = 1 point Four children (Albert, Bridget, Cecilia, and Dirk) play each other once in an all-play-all tournament. You are told that: 1) Bridget was the winner 2) Dirk came last 3) Cecilia scored a win, a draw and a loss 4) All the players scored a different number of points. How many points did Albert score? Introductory Question (the relevance of this will become apparent later) Distribute 10 counters into three squares subject to minimum=2 and maximum=4 per square. 4,4,2; 4,3,3 excluding permutations Solution Method_ There are intricate solutions going through all the game outcomes. We can take a shortcut by examining the implications of condition (4) that each player scored a different number of points. Establish that in a 4-player all-play-all there are 6 games (see diagram). Show that each game results in 4 points giving a tournament total of 24 points. (24 is an invariant.) We want to distribute 24 points across 4 players with a minimum of 3 and a maximum of 9. The combinations with unique point counts are: Winner Second Third Last Total 9 8 4 3 24 8 7 6 3 24 8 7 5 4 24 Only one of these combinations has 6 points. This gives the final points table. Played Points Results Bridget 3 8 WWD 3+3+2 Albert 3 7 WWL 3+3+1 Cecilia 3 6 WDL 3+2+1 Dirk 3 3 LLL 1+1+1 The corresponding all-play-all table is Bridget Albert Cecilia Albert Albert = Bridget Cecilia Bridget win Cecilia = Albert Dirk Bridget win Albert win Cecilia win Answer Albert scored 7 points. B D A C 1 6 3 5 2 4 46 46. Kings random jump Pairs Age 10+ Enumeration, Trial and Error, Symmetry Each square of a chessboard is occupied by a king. Each king moves randomly to an adjacent square. Assume each square can accommodate more than one king. What is the greatest number of empty squares that could remain? Hint: use counters that have different-coloured sides. This ensures that all jumps are ‘recorded’: if a king jumps, the counter is turned over and put on the arrival square. Older children can also work on paper with a set of empty chessboard prints. Introductory questions • What is the greatest number of kings that can end up on the same square? A square can be surrounded by a maximum of eight squares, but the resident king must jump away from this square, so the answer is 8. • Is it possible that all squares are occupied after the jumps? Yes, it is, for example if neighbouring kings on the same file swap places. • Is it possible to have an isolated king after the jumps? An isolated king has no neighbours on adjacent squares. No, it is not possible. If there were an isolated king, then the king that was originally on that square would have had nowhere to jump to. Solution Method_ 1. Place the counters on every square of the chessboard with the same colour up. Make every king jump to an adjacent square by moving each counter to an adjacent square and turning it over at the same time. When all kings jumped count the number of free squares on the board. 2. Repeat 1. and try to increase the number of unoccupied squares. Answer Having explored the problem children may come to the correct conclusion that at most 52 squares can remain empty, so all the kings can gather on 12 squares. See 46(a) and 46(b). Figure 46(a) Figure 46(b) Figure 46(c) Figure 46(d) Extension Ask the class to collect as many possible answers with 52 unoccupied squares as they can. Discuss how some of these are related to each other by symmetry. For example, Figure 46(c) is a reflection of Figure 46(b) around the vertical bisector of the board, and Figure 46(d) is a rotation by 90 degrees clockwise of Figure 46(b). Some solutions have their own symmetry: Figure 46(a) has a rotational symmetry of order 4 around the centre, while Figure 46(b) has a line symmetry with the horizontal bisector of the board as the mirror line. The colour of the squares is ignored. 47 47. King’s random walk Pairs Age 11+ Enumeration, Proportion, Organising information in tables A king moves randomly starting at a8. What is the probability that the king returns to a8 after: (a) Two moves? (b) Three moves? Introductory questions • How many squares can the king move to from a8? 3: a7, b7 and b8 • How many squares can the king move to from a7 (and b8 by symmetry)? 5: a6, b6, b7, b8 and a8 • How many squares can the king move to from b7? 8: a8, a7, a6, b6, c6, c7, c8 and b8 Solution Method_ (a) Complete this table to account for all possibilities: First move to No. of possible moves from there No. Of moves when king returns to a8 a7 5 1 b7 8 1 b8 5 1 TOTAL: 18 3 The king returns to a8 three times out of 18 i.e. 1 out of 6 times on average (16.7%). (b) Collect information about his majesty’s walk in a larger table: First 2 moves No. of possible moves from there No. of moves when king returns to a8 a7, a8 3 0 a7, b8 5 1 a7, b7 8 1 a7, b6 8 0 a7, a6 5 0 b7, a8 3 0 b7, b8 5 1 b7, c8 5 0 b7, c7 8 0 b7, c6 8 0 b7, b6 8 0 b7, a6 5 0 b7, a7 5 1 b8, a8 3 0 b8, a7 5 1 b8, b7 8 1 b8, c7 8 0 b8, c8 5 0 TOTAL: 105 6 The king returns to a8 six times out of 105 i.e. 2 out of 35 times on average (5.7%). 48 48. The Invention of Chess Groups Age 11+ Exponential Growth, Geometric Sequence There is a legend about the invention of Chess. The modern version is this. When the inventor showed the Persian king his new game, the Shah was very impressed and offered him a choice of two rewards. Either the inventor could have €1 million for every square on the chessboard i.e. €64 million for the whole board, or he could have 1 cent for the first square, 2 cents for the second square, 4 cents for the third square, doubling each time, all the way up to sixty-four squares. Which option would you choose? Small scale trial To understand the inventor’s dilemma, use a reduced chessboard (4x4). Fill all the squares of the board with rice grains. Children can «feel» the very rapid growth through this simple procedure. The growth rate gets faster the more rice grains we get – the illustration of exponential growth. With a standard 8x8 board, we get the following which is impossible to handle manually. Intermediate question Which is greater: a) The total of the grains of rice in squares 1-8 b) The number of grains of rice in square 9 Answer: b) Repeat question for comparisons {1-16 v 17} and {1-32 v 33} with the same answer. Solution Method_ This exercise involves doing a lot of long calculations. A calculator or spreadsheet should be used to save time and achieve accuracy. The formula for the number of grains of rice for n squares is 2n – 1. Multiply 2 by itself according to the number of squares required and subtract 1. To convert cents to euros, divide by 100. Answer The doubling option gives a much higher figure. Although starting slowly, once half the board is covered, the situation has changed dramatically. By square 33, the doubling total has reached nearly €43 million overtaking the first option of €33 million. Filling the complete board would cost more than all the money in the world. 49 49. How many rectangles on a chessboard? Pairs Age 11+ Enumeration, Shapes, Organising information in tables This is a challenging task for primary students, therefore best done as a teacher-led activity. First present this table without the numbers in the second column: Rectangle type No. of this rectangle on the chessboard 1x1 8x8 = 64 1x2 7x8x2 = 112 1x3 6x8x2 = 96 1x4 5x8x2 = 80 1x5 4x8x2 = 64 1x6 3x8x2 = 48 1x7 2x8x2 = 32 1x8 1x8x2 = 16 2x2 7x7 = 49 2x3 6x7x2 = 84 2x4 5x7x2 = 70 2x5 4x7x2 = 56 2x6 3x7x2 = 42 2x7 2x7x2 = 28 2x8 1x7x2 = 14 3x3 6x6 = 36 3x4 5x6x2 = 60 3x5 4x6x2 = 48 3x6 3x6x2 = 36 3x7 2x6x2 = 24 3x8 1x6x2 = 12 4x4 5x5 = 25 4x5 4x5x2 = 40 4x6 3x5x2 = 30 4x7 2x5x2 = 20 4x8 1x5x2 = 10 5x5 4x4 = 16 5x6 3x4x2 = 24 5x7 2x4x2 = 16 5x8 1x4x2 = 8 6x6 3x3 = 9 6x7 2x3x2 = 12 6x8 1x3x2 = 6 7x7 2x2 = 4 7x8 1x2x2 = 4 8x8 1x1 = 1 Total: 1296 Ask pairs of children to cut out one or more of the rectangle types from paper. Their task is to place this on the chessboard as many different ways as possible and record their result. Encourage them to find a systematic way to do the counting. Beware of pitfalls such as counting axb but not counting bxa rectangles. Collect all answers and correct if necessary. The total number of rectangles that can be found on a chessboard is 1296. Advanced Solution Method: A rectangle is bounded by two vertical and two horizontal lines as shown on Figure 49(a). Every rectangle is uniquely determined by a vertical and a horizontal pair of lines. There are as many horizontal pairs as vertical ones, so it is enough to count the total number of vertical pairs that can be drawn. There are 9 ways to choose the first vertical line and 8 ways to choose the second. We have double-counted each pair, so the number of vertical pairs is 9x8/2 = 36. There are 36 horizontal pairs as well. For each vertical pair we can choose any of the 36 horizontal pairs, hence there are 36x36 choices, giving a total of 1296 rectangles. Solution Method_ Figure 49(a) 50 50. Leaper problem Pairs Age 11+ Enumeration, symmetry, angles Leapers are pieces which move m squares in one direction and then n squares at a right angle. The only leaper used in chess is the knight which is a (2,1) leaper. Tasks (a) Produce a Power Contour for each of these leapers below : (Fig. 50) (2,1) Knight (2,2) Alfil (3,1) Camel (3,2) Zebra [optional] (3,3) Tripper [optional] (4,1) Giraffe [optional] Hint: To build a piece contour, take a look at some examples of piece power contours (Exercise 10). (b) Identify the leapers defined by the contour maps below Solution Method_ Visually compare the diagrams drawn in (a) with the contours given in (b). Solution (a) Alfil (2,2) (b) Camel (3,1) m n Fig. 50(a) Fig. 50(b) Fig. 50 Leapers |
11027 | https://proofwiki.org/wiki/Mood_of_Categorical_Syllogism/Examples/Socrates | Mood of Categorical Syllogism/Examples/Socrates - ProofWiki
Mood of Categorical Syllogism/Examples/Socrates
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Example of Mood of Categorical Syllogism
Consider the Socrates syllogism:
All men are mortal. Socrates is a man. Therefore, Socrates is mortal.
This has the moodA I I A I I.
Sources
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Examples of Moods of Categorical Syllogisms
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Skills - 8. Rounding, Limits of Accuracy and Bounds
Using Essential GCSE Maths
When the answer to a calculation is a decimal with a lot of digits, a decision has to be made about the accuracy to which the answer is stated, i.e. how many figures to write down.
Truncation, which is sometimes used by digital instruments such as calculators, is where all the decimal places after a cut-off point are simply discarded. For example, if 1.418263748 is truncated after the fourth decimal place, the digits 63748... are discarded to leave 1.4182.
Rounding is where a figure is approximated by the nearest value with the desired accuracy. Rounding may be to a set number of significant figures (sf) or decimal places (dp). The key principle is to look at one more digit than the desired accuracy requires. If the value of this digit is 5 or more, then when the number is rounded the final digit is rounded up by 1.
When rounding decimals where all the digits before the decimal point are zero, the first significant figure is the first non-zero digit after the decimal point.
When calculating a value from an expression or formula, it is important to round only the final answer. Rounding too early leads to a loss of accuracy, because subsequent steps in the calculation are using approximated values. Always state the number of significant figures to which you give an answer.
In real-world applications the precision of measurements is limited. The final answer to a calculation cannot be given to more significant figures than the values from which it is calculated. In general it is appropriate to find the value which is stated to the smallest number of significant figures, and give your answer to the same number of significant figures.
Stating a quantity to a set number of significant figures or decimal places means that the quantity lies within a range of values which would all round to the same answer. This range is called an error interval and can be expressed as an inequality between upper and lower bounds (limits).
In Example 4 the lower bound is 2.475mm, and the upper bound is 2.485mm. The inequality signs for the two bounds are different; p can be equal to the lower bound, but cannot be equal to the upper bound.
When dealing with a quantity that has to be an integer, there are two ways of writing the upper limit of the inequality.
When more than one variable is used in a calculation, upper and lower bounds for the final answer are found by considering the upper and lower bounds of the individual variables from which the answer is calculated. |
11029 | https://www.wyzant.com/resources/answers/801771/what-is-the-relationship-between-trapezoids-and-triangles | WYZANT TUTORING
Tim D.
What is the relationship between trapezoids and triangles?
What is the relationship between trapezoids and triangles?
1 Expert Answer
Kathy G.
answered • 12/05/20
Experienced Math tutor with 40 years experience
If you divide the trapezoid into 2 triangles both triangles have the same height but different bases.
The area of one triangle will be 1/2 b1 h. and the area of the second triangle will be 1/2 b2 h
When you add these 2 together and factor out the 1/2 and the h you get A = 1/2 h (b1 + b2)
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Chapter 1 The Living World
The Living World Class 11 Biology Chapter 1 CBSE Notes - 2025-26
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Biology Notes for Chapter 1 The Living World Class 11 - FREE PDF Download
The Living World Class 11 Notes are prepared to make learning easier for students. These notes simplify important topics like classification, taxonomy, and biodiversity by breaking them down into simple, easy-to-understand points. Each concept is explained clearly, with summaries and key facts that help students quickly learn and remember essential information. The notes also include examples and diagrams that further enhance understanding. Ideal for quick revision and thorough exam preparation, Class 11 Biology Notes ensure that students can confidently grasp the core ideas and perform well in their exams.
Download the FREE PDF for Class 11 Biology Chapter 1 Notes, prepared by experts at Vedantu and updated according to the latest CBSE Class 11 Biology Syllabus, to make study sessions more productive and efficient.
Competitive Exams after 12th Science
Access Revision Notes For Class 11 Biology Chapter 1 The Living World
Life is a unique process that is made from the aggregation of molecules. These molecules undergo various chemical reactions to perform their specific functions which are called metabolism. This results in the production and utilization of energy. The metabolism will result in the growth, development, reproduction, adaptations, etc of the living organisms through the production of various biomolecules.
All the living organisms that live in various habitats are found to share a somewhat similar genetic material that may be either terrestrial, aquatic, in mountains, deserts, oceans, forests, etc.
Living organisms contain certain important characteristics that include growth and development, body organization, homeostasis, reproduction, adaptation, and energy utilization.
1.2 Diversity in the Living World:
The Earth is the main area where living organisms live. The world consists of millions of living organisms that sometimes we cannot even see with our naked eyes. These organisms are found to be living in various habitats that include forests, oceans, deserts, lakes, mountains, and even hot water springs.
There are different types of plants, animals, insects in the world. This is very important and their variability is necessary for survival. The number of species identified and studied is 1.7-1.8 million. They all together make a natural diversity of life in the world which is generally called biological diversity or biodiversity.
Taxonomy: Taxonomy is the study of the classification, characterization, nomenclature, and identification of organisms and it is a branch of science. Systematics is another branch of science that includes the study of the classification, nomenclature, identification, and evolutionary history of an organism. Thus, the taxonomic characteristics of an organism along with its evolutionary history come under systematics. In 1813, A.P de Candolle was the first to introduce the term taxonomy while systematics was introduced as the time of human civilization.
The term Systematics is derived from the Latin word ‘systema’ which means the systematic arrangement of organisms. Linnaeus published his book Systema Naturae where the classification of plants, animals were based on taxonomy.
Neo-systematics is the branch of systematics that deals with the species to be the product of evolution. In 1940, Julia Huxley was the one who developed this concept. It involves the known characteristics of an organism and also the known evidence from different fields of biology.
Identification: It is the method of pacing the organisms in their exact place based on their classification. The identification of organisms can be done with the help of taxonomic keys.
Classification: The classification is the process of grouping various living organisms based on the common features that they share. A single group consists of those organisms that have similar common features. To make classification easier various groups are forms in which different organisms are placed depending upon their characteristics.
Characterization: The studying and understanding of characters of organisms and categorizing them like external and internal structure (morphology and anatomy), the structure of the cell (cytology), developmental process (embryology), and ecological information (ecology) of the organism.
Nomenclature (naming): The naming of living organisms is called nomenclature. There are two types of names, one is vernacular (common names) and the other is the scientific name. Local names are used in local languages or common language and are easy for the local peoples but these names are not used by biologists because:
For many species a single local name is often used.
The local names sometimes lead to incorrect meanings about the organism.
In different regions of the country or world, the different local names are used for one organism. Scientific names: The names are given according to certain rules and are followed by the biologist all over the world. To make it common around the world various international codes have been established.
These codes are:
• ICBN-International Code of Botanical Nomenclature
• ICZN-International Code of Zoological Nomenclature
• ICVN-International Code of Viral Nomenclature
ICBN/ICNB-International Code for Bacteriological Nomenclature or Nomenclature of Bacteria.
Binomial Nomenclature:
Carl Linnaeus is the one credited for the introduction of the binomial nomenclature of the plants and animals with his work in the book Species Plantarum in 1753. Binomial nomenclature is the biological system of naming the organisms in which the name is composed of two terms, where, the first term indicates the genus, and the second term indicates the species of the organism. E.g., Mangifera indica Linn. Mangifera is the genus name and indica is the species name. Linn indicates that this species was first described by Linnaeus Who can give scientific names: Identification, nomenclature, and classification of organisms are all involved in this branch of biology.
Rules:
A scientific name generally has two components (words) in Latin or is derived from Latin irrespective of their origin.
The First word of the biological name denotes the genus name whereas the second one denotes species.
When applying the binomial nomenclature system, the name of the species is written in italics or underlined separately when handwritten.
The generic name must start with a capital letter while a specific name should start with a small letter.
The name of the author is printed in Roman or an abbreviated form at the end after the species name.
Only one correct name must be assigned to each taxonomic group.
The scientific name selected should be such that it would be easy to pronounce, and short.
Eg: Mangifera indica- Mangifera is the genus name and indica is the species name.
1.3 Taxonomic Categories:
In 1956 the term Taxon was introduced and in 1964, Mayr defined taxon to be the various categories based on different characters of the organisms that consist of a taxonomic group of any rank.
Taxonomic Hierarchy:
Various organisms in different categories depending upon their common characters to make classification easier. These groups together are called taxonomic hierarchies. The taxonomic hierarchy includes. Kingdom, division of the kingdom, phylum, class, order, family, genus, and species. Species are the lowest while the kingdom is the highest rank within the hierarchy. It is also called the Linnaean hierarchy as it was first proposed by Carolus Linnaeus, the Father of Systematic Botany. The hierarchy includes seven obligate categories.
They are as follows-
Kingdom - Animalia
↑
Phylum - Chordata
(Division in case of plants)
↑
Class - Mammalia
↑
Order - Primata
↑
Family - Hominidae
↑
Genus - Homo
↑
Species – sapiens
Species: It is the lowest category of the taxonomic hierarchy. There are around 8.7 million species observed on earth till now while their rest are left undiscovered. It refers to a group of organisms that are similar in shape, form, generative options. Species may be more divided into subspecies. It was first defined by Ernst Mayr in 1964 that the species are the interbreeding populations that are reproductively isolated from other such groups. The term species was first introduced by the biologist John Ray.
E.g.: sapiens.
Genus: A category that is placed above species as they consist of a group of related species. Genus are of various types based on the number of species present like monotypic (one genus present), and polytypic (several species present). For e.g., the genus Panthera constitutes both lion and tiger.
Family: This taxonomic category consists of related genera having similar characteristics. For e.g., the families Canidae, Felidae, Ursidae, etc come under one order Carnivora.
Order or Cohort: This taxonomic category is more specific than the class as it consists of one or more similar families. The class Mammalia consists of around twenty-six orders that include primates, Carnivora, etc.
Class: It was the most general taxonomic category before the introduction of phyla. In the animal kingdom, there are around 108 classes that include Pisces, reptilia, aves, etc. The categories used in classification now are different from those of the Linnaeus taxonomy.
Phylum: This category is more specific than the kingdom. In the animal kingdom, there are around thirty-five phyla that include phylum Arthropoda, Chordata, etc.
Kingdom: The highest level of classification is the kingdom which is further divided into various subgroups. The total kingdoms of the living organisms are five in number that includes Monera, Protista, Fungi, Plantae, and Animalia.
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| | | |
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| Generic Name | Specific Epithet | Common Name |
| Mangifera | indica | Mango |
| Solanum | tuberosum | Potato |
| Solanum | nigrum | Nightshade |
| Panthera | leo | Lion |
| Panthera | tigris | Tiger |
| Homo | sapiens | Man |
| | | | | | | |
--- --- ---
| Common Name | Biological Name | Genus | Family | Order | Class | Phylum/Division |
| Man | Homo sapiens | Homo | Hominidae | Primata | Mammalia | Chordata |
| Housefly | Musca domestica | Musca | Muscidae | Diptera | Insecta | Arthropoda |
| Mango | Mangifera indica | Mangifera | Anacardiaceae | Sapindales | Dicotyledonae | Angiospermae |
| Wheat | Triticum aestivum | Triticum | Poaceae | Poales | Monocotyledonae | Angiospermae |
5 Important Topics of Biology Class 11 Chapter 1 You Shouldn’t Miss!
| | | |
---
| S.No. | Important Topic | Brief Description |
| 1 | Diversity in the Living World | Exploring the vast variety of living organisms on Earth and their classification. |
| 2 | Taxonomy | The science of classification, including taxonomy categories and binomial nomenclature. |
| 3 | Systematics | Study of the relationships among organisms and their evolutionary history. |
| 4 | Nomenclature | Rules and guidelines for naming organisms scientifically using binomial nomenclature. |
| 5 | Hierarchy of Biological Classification | Understanding the different levels of biological classification from domain to species. |
Importance of Living World Class 11 Notes
Understanding the basics of living organisms, their classification, and taxonomy is crucial for advanced biology topics.
Class 11 Biology Chapter 1 The Living World Notes provide a clear introduction to key concepts that will be built upon in subsequent chapters.
The notes break down complex concepts such as biodiversity, systematics, and nomenclature into easy-to-understand points.
With summaries and key facts, Class 11 Biology Ch 1 Notes are ideal for quick revision before exams.
The notes include clear explanations, examples, and diagrams that help in a better understanding of the living world, making it easier for students to connect theoretical concepts with practical examples.
By covering crucial topics and providing important details, The Living World Class 11 Short Notes helps students prepare thoroughly for exams.
Tips for Learning the Class 11 Biology Chapter 1 The Living World
Focus on grasping fundamental ideas such as the characteristics of living organisms, biodiversity, and the basics of taxonomy. Ensure you understand terms like taxonomy, systematics, and nomenclature.
Utilise diagrams, flowcharts, and tables to visualise the classification of organisms and the hierarchy of biological classification. Visual aids can help reinforce your understanding and memory.
Write concise summaries of each section of the chapter. Highlight key points, definitions, and examples. Summaries help in quick revision and improve learning.
Solve practice questions and past exam papers related to the chapter. This will help you apply your knowledge and improve your problem-solving skills.
Connect theoretical concepts with real-life examples of organisms and their classification. This makes the learning process more engaging and relevant.
Regularly review your notes and summaries to keep the information fresh in your mind. Consistent revision helps in better retention and recall during exams.
Conclusion
The Living World Class 11 Notes cover essential concepts of biology. The chapter explains what defines living organisms, their classification into different categories, and the system of naming them. It also discusses biodiversity and the various levels of biological classification. Key topics include taxonomy, systematics, and nomenclature. These notes break down complex ideas into simple, easy-to-understand points, making it easier for students to learn and remember important information.
Related Study Materials for Class 11 Biology Chapter 1 The Living World
Students can also download additional study materials provided by Vedantu for Biology Class 11, Chapter 1–
| | |
--- |
| S.No | Study Material Links for The Living World Class 11 |
| 1 | Class 11 The Living World Important Questions |
| 2 | Class 11 The Living World NCERT Solutions |
| 3 | Class 11 The Living World NCERT Exemplar |
Chapter-wise Class 11 Biology Notes PDF Download
| | |
--- |
| S. No | Chapter wise Class 11 Biology Revision Notes |
| 1 | Chapter 2: Biological Classification Notes |
| 2 | Chapter 3: Plant Kingdom Notes |
| 3 | Chapter 4: Animal Kingdom Notes |
| 4 | Chapter 5: Morphology of Flowering Plants Notes |
| 5 | Chapter 6: Anatomy of Flowering Plants Notes |
| 6 | Chapter 7: Structural Organisation in Animals Notes |
| 7 | Chapter 8: Cell the Unit of Life Notes |
| 8 | Chapter 9: Biomolecules Notes |
| 9 | Chapter 10: Cell Cycle and Cell Division Notes |
| 10 | Chapter 11: Photosynthesis in Higher Plants |
| 11 | Chapter 12: Respiration in Plants Notes |
| 12 | Chapter 13: Plant Growth and Development Notes |
| 13 | Chapter 14: Breathing and Exchange of Gases Notes |
| 14 | Chapter 15: Body Fluids and Circulation Notes |
| 15 | Chapter 16: Excretory Products and Their Elimination Notes |
| 16 | Chapter 17: Locomotion and Movement Notes |
| 17 | Chapter 18: Neural Control and Coordination Notes |
| 18 | Chapter 19: Chemical Coordination and Integration Notes |
Related Study Materials Links for Class 11 Biology
| | |
--- |
| S. No | Related Study Materials Links for Class 11 Biology |
| 1. | CBSE Class 11 Biology NCERT Solutions |
| 2. | CBSE Class 11 Biology NCERT Important Questions |
| 3. | CBSE Class 11 Biology NCERT Exemplar |
| 4. | CBSE Class 11 Biology NCERT Books |
| 5. | CBSE Class 11 Biology Sample Papers |
| 6. | CBSE Class 11 Biology NCERT Solutions in Hindi |
FAQs on The Living World Class 11 Biology Chapter 1 CBSE Notes - 2025-26
What are the core concepts covered in the revision notes for The Living World, Class 11?
These revision notes provide a comprehensive summary of Chapter 1, focusing on key areas such as the defining characteristics of living organisms, the concept of biodiversity, the necessity of classification, the principles of taxonomy and systematics, the rules of binomial nomenclature, and an overview of various taxonomical aids used for study.
What are the most important topics to focus on when revising Chapter 1, The Living World?
For a quick and effective revision, you should focus on:
The universal rules of binomial nomenclature.
The sequence and definition of each rank in the taxonomic hierarchy (Species, Genus, Family, Order, Class, Phylum, Kingdom).
The distinct functions of key taxonomical aids like Herbaria, Museums, and Zoological Parks.
The difference between defining and non-defining characteristics of life.
How do these revision notes explain the concept of binomial nomenclature?
The notes simplify binomial nomenclature as a two-part scientific naming system for organisms, established by Carolus Linnaeus. It consists of the Generic name (starting with a capital letter) and the specific epithet (starting with a small letter). The summary highlights that these names are typically in Latin and are italicised when printed to denote their origin.
Why are metabolism and consciousness considered defining features of life, while growth is not?
Metabolism (the sum total of all chemical reactions in the body) and consciousness (the ability to sense and respond to stimuli) are considered defining features because they occur exclusively in all living organisms. In contrast, growth, defined as an increase in mass, can also be observed in non-living objects like mountains or crystals through external accumulation. Therefore, it is not a unique or defining property of life.
How do the revision notes for Chapter 1 clarify the difference between taxonomy and systematics?
The notes clarify that taxonomy is the process of identifying, naming, and classifying organisms into different taxa. Systematics is a broader field that includes taxonomy but also incorporates the study of evolutionary relationships (phylogeny) between organisms. In essence, systematics aims to understand the diversity of life in the context of its evolutionary history.
What are the key taxonomic aids discussed in these notes, and what is their primary function in a quick revision?
The notes cover the essential taxonomic aids for quick recall:
Herbarium: A collection of dried, pressed plant specimens for reference.
Botanical Garden: A collection of living plants for identification.
Museum: A collection of preserved plant and animal specimens for study.
Zoological Park: A place where wild animals are kept in protected environments, similar to their natural habitats, to study their behaviour.
How does understanding the taxonomical hierarchy help in organising the vast biodiversity of the living world?
The taxonomical hierarchy acts as a systematic framework, similar to a multi-level filing system. By arranging organisms into progressively broader categories from species to kingdom based on shared characteristics, it simplifies the study of millions of life forms. This organisation not only makes identification easier but also helps reveal evolutionary relationships between different groups.
How can a student effectively use these revision notes to align with the latest CBSE/NCERT syllabus for 2025-26?
These revision notes are structured to mirror the CBSE/NCERT curriculum for 2025-26. To use them effectively, a student should first read the NCERT chapter thoroughly and then use these notes to quickly reinforce key definitions, recap complex concepts, and understand the logical flow between topics. They are designed to be a final-touch revision tool before exams.
Beyond memorisation, how do the concepts in 'The Living World' form the foundation for other Biology chapters?
This chapter is foundational because the principles of classification and nomenclature are not just for this unit; they are the language of biology. Understanding the taxonomic hierarchy is essential for navigating complex chapters like 'Biological Classification', 'Plant Kingdom', and 'Animal Kingdom'. The systematic approach learned here is applied throughout biology to understand the relationships and diversity of all life forms.
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11031 | https://ocw.mit.edu/courses/8-01sc-classical-mechanics-fall-2016/mit8_01scs22_chapter23.pdf | Chapter 23 Simple Harmonic Motion 23.1 Introduction: Periodic Motion............................................................................. 1 23.1.1 Simple Harmonic Motion: Quantitative ...................................................... 1 23.2 Simple Harmonic Motion: Analytic .................................................................... 3 23.2.1 General Solution of Simple Harmonic Oscillator Equation ...................... 6 Example 23.1: Phase and Amplitude ...................................................................... 7 Example 23.2: Block-Spring System ....................................................................... 9 23.3 Energy and the Simple Harmonic Oscillator ................................................... 10 23.3.1 Simple Pendulum: Force Approach ........................................................... 13 23.3.2 Simple Pendulum: Energy Approach ........................................................ 16 23.4 Worked Examples............................................................................................... 18 Example 23.3: Rolling Without Slipping Oscillating Cylinder........................... 18 Example 23.4: U-Tube ............................................................................................ 19 23.5 Damped Oscillatory Motion............................................................................... 21 23.5.1 Energy in the Underdamped Oscillator..................................................... 24 23.6 Forced Damped Oscillator ................................................................................. 26 23.6.1 Resonance ..................................................................................................... 27 23.6.2 Mechanical Energy ...................................................................................... 30 Example 23.5: Time-Averaged Mechanical Energy ............................................ 30 23.6.3 The Time-averaged Power .......................................................................... 34 23.6.4 Quality Factor .............................................................................................. 35 23.7 Small Oscillations................................................................................................ 36 Example 23.6: Quartic Potential ........................................................................... 39 Example 23.7: Lennard-Jones 6-12 Potential....................................................... 41 Appendix 23A: Solution to Simple Harmonic Oscillator Equation ....................... 42 Appendix 23B: Complex Numbers............................................................................ 45 Appendix 23C: Solution to the Underdamped Simple Harmonic Oscillator........ 48 Appendix 23D: Solution to the Forced Damped Oscillator Equation.................... 50 Chapter 23 Simple Harmonic Motion …Indeed it is not in the nature of a simple pendulum to provide equal and reliable measurements of time, since the wide lateral excursions often made may be observed to be slower than more narrow ones; however, we have been led in a different direction by geometry, from which we have found a means of suspending the pendulum, with which we were previously unacquainted, and by giving close attention to a line with a certain curvature, the time of the swing can be chosen equal to some calculated value and is seen clearly in practice to be in wonderful agreement with that ratio. As we have checked the lapses of time measured by these clocks after making repeated land and sea trials, the effects of motion are seen to have been avoided, so sure and reliable are the measurements; now it can be seen that both astronomical studies and the art of navigation will be greatly helped by them…1 Christian Huygens 23.1 Introduction: Periodic Motion There are two basic ways to measure time: by duration or periodic motion. Early clocks measured duration by calibrating the burning of incense or wax, or the flow of water or sand from a container. Our calendar consists of years determined by the motion of the sun; months determined by the motion of the moon; days by the rotation of the earth; hours by the motion of cyclic motion of gear trains; and seconds by the oscillations of springs or pendulums. In modern times a second is defined by a specific number of vibrations of radiation, corresponding to the transition between the two hyperfine levels of the ground state of the cesium 133 atom. Sundials calibrate the motion of the sun through the sky, including seasonal corrections. A clock escapement is a device that can transform continuous movement into discrete movements of a gear train. The early escapements used oscillatory motion to stop and start the turning of a weight-driven rotating drum. Soon, complicated escapements were regulated by pendulums, the theory of which was first developed by the physicist Christian Huygens in the mid 17th century. The accuracy of clocks was increased and the size reduced by the discovery of the oscillatory properties of springs by Robert Hooke. By the middle of the 18th century, the technology of timekeeping advanced to the point that William Harrison developed timekeeping devices that were accurate to one second in a century. 23.1.1 Simple Harmonic Motion: Quantitative 1 Christian Huygens, The Pendulum Clock or The Motion of Pendulums Adapted to Clocks By Geometrical Demonstrations, tr. Ian Bruce, p. 1. 23-1 One of the most important examples of periodic motion is simple harmonic motion (SHM), in which some physical quantity varies sinusoidally. Suppose a function of time has the form of a sine wave function, y(t) = Asin(2πt / T ) (23.1.1) where A > 0 is the amplitude (maximum value). The function y(t) varies between A and − A , because a sine function varies between +1 and −1. A plot of y( ) t vs. time is shown in Figure 23.1. Figure 23.1 Sinusoidal function of time The sine function is periodic in time. This means that the value of the function at time t will be exactly the same at a later time t′ = t + T , where T is the period. That the sine function satisfies the periodic condition can be seen from ⎡ 2π ⎤ ⎡ 2π ⎤ ⎡ 2π ⎤ y(t + T ) = Asin (t + T ) = Asin t + 2π = Asin t = y( ) t . (23.1.2) ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎣ T ⎦ ⎣ T ⎦ ⎣ T ⎦ The frequency, f , is defined to be f ≡ 1/T . (23.1.3) ⎡−1 The SI unit of frequency is inverse seconds, s ⎦ ⎤ , or hertz [Hz]. The angular frequency ⎣ of oscillation is defined to be ω0 ≡ 2π / T = 2π f , (23.1.4) and is measured in radians per second. (The angular frequency of oscillation is denoted by ω0 to distinguish from the angular speed ω = dθ / dt .) One oscillation per second, 1 Hz , corresponds to an angular frequency of 2π rad ⋅s−1 . (Unfortunately, the same 23-2 symbol ω is used for angular speed in circular motion. For uniform circular motion the angular speed is equal to the angular frequency but for non-uniform motion the angular speed is not constant. The angular frequency for simple harmonic motion is a constant by definition.) We therefore have several different mathematical representations for sinusoidal motion y(t) = Asin(2πt / T ) = Asin(2π f t) = Asin(ω0t) . (23.1.5) 23.2 Simple Harmonic Motion: Analytic Our first example of a system that demonstrates simple harmonic motion is a spring-object system on a frictionless surface, shown in Figure 23.2 Figure 23.2 Spring-object system The object is attached to one end of a spring. The other end of the spring is attached to a wall at the left in Figure 23.2. Assume that the object undergoes one-dimensional motion. The spring has a spring constant k and equilibrium length leq . Choose the origin at the equilibrium position and choose the positive x -direction to the right in the Figure 23.2. In the figure, x > 0 corresponds to an extended spring, and x < 0 to a compressed spring. Define x( ) t to be the position of the object with respect to the equilibrium position. The force acting on the spring is a linear restoring force, Fx = −k x (Figure 23.3). The initial conditions are as follows. The spring is initially stretched a distance l0 and given some initial speed v0 to the right away from the equilibrium position. The initial position of the stretched spring from the equilibrium position (our choice of origin) is x0 = (l0 − leq ) > 0 and its initial x -component of the velocity is vx,0 = v0 > 0. 23-3 Figure 23.3 Free-body force diagram for spring-object system Newton’s Second law in the x -direction becomes d 2 x −k x = m 2 . (23.2.1) dt This equation of motion, Eq. (23.2.1), is called the simple harmonic oscillator equation (SHO). Because the spring force depends on the distance x , the acceleration is not constant. Eq. (23.2.1) is a second order linear differential equation, in which the second derivative of the dependent variable is proportional to the negative of the dependent variable, d 2 x k 2 = − x . (23.2.2) dt m In this case, the constant of proportionality is k / m , Eq. (23.2.2) can be solved from energy considerations or other advanced techniques but instead we shall first guess the solution and then verify that the guess satisfies the SHO differential equation (see Appendix 22.3.A for a derivation of the solution). We are looking for a position function x(t) such that the second time derivative position function is proportional to the negative of the position function. Since the sine and cosine functions both satisfy this property, we make a preliminary ansatz (educated guess) that our position function is given by x(t) = Acos((2π / T )t) = Acos(ω0 t) , (23.2.3) where ω0 is the angular frequency (as of yet, undetermined). We shall now find the condition that the angular frequency ω0 must satisfy in order to insure that the function in Eq. (23.2.3) solves the simple harmonic oscillator equation, Eq. (23.2.1). The first and second derivatives of the position function are given by 23-4 ⎜ ⎟ dx = −ω0 Asin(ω0t) dt (23.2.4) d 2 x = −ω0 2 Acos(ω0 t) = −ω0 2 x. dt2 Substitute the second derivative, the second expression in Eq. (23.2.4), and the position function, Equation (23.2.3), into the SHO Equation (23.2.1), yielding −ω0 2 Acos(ω0 t) = − k Acos(ω0 t) . (23.2.5) m Eq. (23.2.5) is valid for all times provided that k ω0 = . (23.2.6) m The period of oscillation is then 2π m T = = 2π . (23.2.7) ω0 k One possible solution for the position of the block is ⎛ ⎞ k x(t) = Acos (23.2.8) t ⎜ ⎝ ⎟ ⎠ , m and therefore by differentiation, the x -component of the velocity of the block is ⎛ ⎞ k k v x (t) = − Asin (23.2.9) t ⎜ ⎝ ⎟ ⎠ . m m Note that at t = 0 , the position of the object is x0 ≡ x(t = 0) = A since cos(0) = 1 and the velocity is v ≡ v (t = 0) = 0 since sin(0) = 0 . The solution in (23.2.8) describes an x,0 x object that is released from rest at an initial position A = x0 but does not satisfy the initial velocity condition, v (t = 0) = v ≠ 0 . We can try a sine function as another possible x x,0 solution, ⎛ k ⎞ x( ) t = Bsin t . (23.2.10) ⎜ ⎟ m ⎝ ⎠ This function also satisfies the simple harmonic oscillator equation because 23-5 d 2 x k ⎛ k ⎞ 2 = − Bsin⎜ t⎟ = −ω0 x , (23.2.11) dt2 m ⎝ m ⎠ where ω0 = k / m . The x -component of the velocity associated with Eq. (23.2.10) is dx k ⎛ ⎞ k v x (t) = Bcos . (23.2.12) t ⎜ ⎝ ⎟ ⎠ = dt m m The proposed solution in Eq. (23.2.10) has initial conditions x0 ≡ x(t = 0) = 0 and v ≡ v (t = 0) = ( k / m)B , thus B = v / k / m . This solution describes an object that x,0 x x,0 is initially at the equilibrium position but has an initial non-zero x -component of the velocity, vx,0 ≠ 0 . 23.2.1 General Solution of Simple Harmonic Oscillator Equation Suppose x1( ) t and x2( ) t are both solutions of the simple harmonic oscillator equation, d 2 k x1(t) = − x1(t) dt2 m (23.2.13) d 2 k x2(t) = − x2(t). dt2 m Then the sum x( ) t = x1( ) t + x2( ) t of the two solutions is also a solution. To see this, consider d 2 x(t) d 2 d 2 x1(t) d 2 x2(t) = (x1(t) + x2(t)) = + . (23.2.14) dt2 dt2 dt2 dt2 Using the fact that x1( ) t and x2( ) t both solve the simple harmonic oscillator equation (23.2.13), we see that d 2 k k k x( ) t = − x ( ) t + − x ( ) t = − (x ( ) t + x ( ) t ) 2 1 2 1 2 dt m m m (23.2.15) k = − x( ).
t m Thus the linear combination x( ) t = x1( ) t + x2( ) t is also a solution of the SHO equation, Eq. (23.2.1). Therefore the sum of the sine and cosine solutions is the general solution, x(t) = C cos(ω0 t) + Dsin(ω0 t) , (23.2.16) 23-6 where the constant coefficients C and D depend on a given set of initial conditions x ≡ x(t = 0) and v ≡ v (t = 0) where and v are constants. For this general 0 x,0 x x0 x,0 solution, the x -component of the velocity of the object at time t is then obtained by differentiating the position function, dx vx (t) = = −ω0C sin(ω0 t) + ω0 D cos(ω0 t) . (23.2.17) dt To find the constants C and D , substitute t = 0 into the Eqs. (23.2.16) and (23.2.17). Because cos(0) = 1 and sin(0) = 0 , the initial position at time t = 0 is x0 ≡ x(t = 0) = C . (23.2.18) The x -component of the velocity at time t = 0 is v = v (t = 0) = −ω0C sin(0) + ω0 Dcos(0) = ω0 D . (23.2.19) x,0 x Thus v C = x and D = x,0 . (23.2.20) 0 ω0 The position of the object-spring system is then given by ⎛ ⎞ ⎛ ⎞ k k v x,0 k / m x(t) = x0 cos sin (23.2.21) t + t ⎜ ⎝ ⎟ ⎠ ⎜ ⎝ ⎟ ⎠ m m and the x -component of the velocity of the object-spring system is ⎛ ⎛ ⎞ ⎞ k k k v x (t) = − x0 sin (23.2.22) t⎟ ⎠ + v x,0 cos ⎜ ⎝ t ⎜ ⎝ ⎟ ⎠ . m m m Although we had previously specified x > 0 and v > 0 , Eq. (23.2.21) is seen to be a 0 x,0 valid solution of the SHO equation for any values of x0 and v x,0 . Example 23.1: Phase and Amplitude Show that x(t) = C cosω0t + Dsinω0t = Acos(ω0t +φ) , where A = (C 2 + D2)1 2 > 0 , and φ = tan−1(−D / C). 23-7 Solution: Use the identity Acos(ω0t + φ) = Acos(ω0t)cos(φ) − Asin(ω0t)sin(φ) . Thus C cos(ω0t) + Dsin(ω0t) = Acos(ω0t)cos(φ) − Asin(ω0t)sin(φ) . Comparing coefficients we see that C = Acosφ and D = − Asinφ . Therefore (C 2 + D2)1 2 A2 = A2(cos2 φ + sin2 φ) = . We choose the positive square root to ensure that A > 0 , and thus A = (C 2 + D2)1 2 (23.2.23) sinφ − D / A D tanφ = = = − , cosφ C / A C φ = tan−1(−D / C) . (23.2.24) Thus the position as a function of time can be written as x(t) = Acos(ω0t +φ) . (23.2.25) In Eq. (23.2.25) the quantity ω0t +φ is called the phase, and φ is called the phase constant. Because cos(ω0t + φ) varies between +1 and −1 , and A > 0 , A is the amplitude defined earlier. We now substitute Eq. (23.2.20) into Eq. (23.2.23) and find that the amplitude of the motion described in Equation (23.2.21), that is, the maximum value of x( ) t , and the phase are given by A = x0 2 + (v x,0 / ω0 )2 . (23.2.26) φ = tan−1(−v / ω x ) . (23.2.27) x,0 0 0 A plot of x( ) t vs. t is shown in Figure 23.4a with the values A = 3, T = π , and φ = π / 4 . Note that x(t) = Acos(ω0t + φ) takes on its maximum value when cos(ω0t + φ) = 1 . This occurs when ω0t + φ = 2π n where n = 0, ± 1, ± 2,⋅⋅⋅ . The maximum value associated with n = 0 occurs when ω0t + φ = 0 or t = −φ / ω0 . For the case shown in Figure 23.4a where φ = π / 4 , this maximum occurs at the instant t = −T / 8 . Let’s plot x(t) = Acos(ω0t + φ) vs. t for φ = 0 (Figure 23.4b). For φ > 0 , Figure 23.4a shows the plot x(t) = Acos(ω0t + φ) vs. t . Notice that when φ > 0 , x(t) is shifted to the left compared with the case φ = 0 (compare Figures 23.4a with 23.4b). The function x(t) = Acos(ω0t + φ) with φ > 0 reaches its maximum value at an earlier time than the function x(t) = Acos(ω0t) . The difference in phases for these two cases is (ω0t +φ) −ω0t = φ and φ is sometimes referred to as the phase shift. When φ < 0 , the 23-8 function x(t) = Acos(ω0t + φ) reaches its maximum value at a later time t = T / 8 than the function x(t) = Acos(ω0t) as shown in Figure 23.4c. (a) (b) (c) Figure 23.4 Phase shift of x(t) = Acos(ω0t + φ) (a) to the left by φ = π / 4 , (b) no shift φ = 0, (c) to the right φ = −π / 4 Example 23.2: Block-Spring System 23-9 ⎜ ⎟ A block of mass m is attached to a spring with spring constant k and is free to slide along a horizontal frictionless surface. At t = 0 , the block-spring system is stretched an amount x > 0 from the equilibrium position and is released from rest, v = 0 . What is 0 x,0 the period of oscillation of the block? What is the velocity of the block when it first comes back to the equilibrium position? Solution: The position of the block can be determined from Eq. (23.2.21) by substituting the initial conditions x > 0 , and v = 0 yielding 0 x,0 ⎛ k ⎞ x( ) t = x cos t , (23.2.28) 0 ⎜ ⎟ m ⎝ ⎠ and the x -component of its velocity is given by Eq. (23.2.22), ⎛ ⎞ k k v x (t) = − x0 sin . (23.2.29) t ⎜ ⎝ ⎟ ⎠ m m The angular frequency of oscillation is ω0 = k / m and the period is given by Eq. (23.2.7), 2π m T = = 2π . (23.2.30) ω0 k The block first reaches equilibrium when the position function first reaches zero. This occurs at time t1 satisfying k π π m T t1 = , t1 = = . (23.2.31) m 2 2 k 4 The x -component of the velocity at time t1 is then ⎛ ⎞ k k k k v (t1) = − x0 sin⎜ t1⎟ = − x0 sin(π / 2) = − x0 = −ω0 x0 (23.2.32) x m m m m ⎝ ⎠ Note that the block is moving in the negative x -direction at time t1; the block has moved from a positive initial position to the equilibrium position (Figure 23.4(b)). 23.3 Energy and the Simple Harmonic Oscillator 23-10 ⎜ ⎟ ⎜ ⎟ v Let’s consider the block-spring system of Example 23.2 in which the block is initially stretched an amount x0 > 0 from the equilibrium position and is released from rest, x,0 = 0 . We shall consider three states: state 1, the initial state; state 2, at an arbitrary time in which the position and velocity are non-zero; and state 3, when the object first comes back to the equilibrium position. We shall show that the mechanical energy has the same value for each of these states and is constant throughout the motion. Choose the equilibrium position for the zero point of the potential energy. State 1: all the energy is stored in the object-spring potential energy, U1 = (1/ 2) k x0 2. The object is released from rest so the kinetic energy is zero, K1 = 0 . The total mechanical energy is then E1 = U1 = 1 k x0 2 . (23.3.1) 2 State 2: at some time t , the position and x -component of the velocity of the object are given by x(t) = x0 cos k m t ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ k m x0 sin k m t ⎛ ⎝ ⎜ (23.3.2) ⎞ (t) = − ⎟ ⎠ v . x The kinetic energy is 1 2 1 2 2 ⎛ k ⎞ K2 = mv = k x0 sin t , (23.3.3) ⎜ ⎟ 2 2 m ⎝ ⎠ and the potential energy is 1 2 1 2 2 ⎛ k ⎞ U2 = k x = k x0 cos t . (23.3.4) ⎜ ⎟ 2 2 m ⎝ ⎠ The mechanical energy is the sum of the kinetic and potential energies E2 = K2 +U2 = 1 mv 2 + 1 k x2 2 x 2 ⎛ ⎞ k k ⎛ ⎛ ⎞ ⎞ 1 k x0 2 2 ⎟ ⎠ + sin2 ⎜ ⎝ (23.3.5) t t ⎜ ⎝ ⎟ ⎠ ⎜ ⎝ ⎟ ⎠ = cos 2 m m = 1 k x0 2, 2 where we used the identity that cos2 ω0t + sin2 ω0t = 1 , and that ω0 = k / m (Eq. (23.2.6)). 23-11 The mechanical energy in state 2 is equal to the initial potential energy in state 1, so the mechanical energy is constant. This should come as no surprise; we isolated the object-spring system so that there is no external work performed on the system and no internal non-conservative forces doing work. Figure 23.5 State 3 at equilibrium and in motion State 3: now the object is at the equilibrium position so the potential energy is zero, U3 = 0 , and the mechanical energy is in the form of kinetic energy (Figure 23.5). 1 2 E3 = K3 = mveq . (23.3.6) 2 Because the system is closed, mechanical energy is constant, E1 = E3. (23.3.7) Therefore the initial stored potential energy is released as kinetic energy, 1 2 1 2 k x0 = mveq , (23.3.8) 2 2 and the x -component of velocity at the equilibrium position is given by v x,eq = ± k x0 . (23.3.9) m Note that the plus-minus sign indicates that when the block is at equilibrium, there are two possible motions: in the positive x -direction or the negative x -direction. If we take x > 0 , then the block starts moving towards the origin, and v will be negative the first 0 x,eq time the block moves through the equilibrium position. We can show more generally that the mechanical energy is constant at all times as follows. The mechanical energy at an arbitrary time is given by 23-12 E = K +U = 1 mv x 2 + 1 k x2 . (23.3.10) 2 2 Differentiate Eq. (23.3.10) d 2 dE dvx dx ⎛ x ⎞ = mv x + k x = vx m + k x . (23.3.11) dt dt dt ⎝ ⎜ dt2 ⎠ ⎟ Now substitute the simple harmonic oscillator equation of motion, (Eq. (23.2.1) ) into Eq. (23.3.11) yielding dE = 0 , (23.3.12) dt demonstrating that the mechanical energy is a constant of the motion. 23.3.1 Simple Pendulum: Force Approach A pendulum consists of an object hanging from the end of a string or rigid rod pivoted about the point P . The object is pulled to one side and allowed to oscillate. If the object has negligible size and the string or rod is massless, then the pendulum is called a simple pendulum. Consider a simple pendulum consisting of a massless string of length l and a point-like object of mass m attached to one end, called the bob. Suppose the string is fixed at the other end and is initially pulled out at an angle θ0 from the vertical and released from rest (Figure 23.6). Neglect any dissipation due to air resistance or frictional forces acting at the pivot. Figure 23.6 Simple pendulum Let’s choose polar coordinates for the pendulum as shown in Figure 23.7a along with the free-body force diagram for the suspended object (Figure 23.7b). The angle θ is defined with respect to the equilibrium position. When θ > 0 , the bob is has moved to the right, and when θ < 0 , the bob has moved to the left. The object will move in a circular arc centered at the pivot point. The forces on the object are the tension in the string T = −T r ˆ and gravity mg . The gravitation force on the object has r ˆ - and θ ˆ components given by 23-13 mg = mg(cosθ r ˆ − sinθ θ ˆ) . (23.3.13) Figure 23.7 (a) Coordinate system Figure 23.7 (b) free-body force diagram Our concern is with the tangential component of the gravitational force, F θ = −mgsinθ . (23.3.14) The sign in Eq. (23.3.14) is crucial; the tangential force tends to restore the pendulum to the equilibrium value θ = 0 . If θ > 0 , F θ < 0 and if θ < 0 , F θ > 0 , where we are that because the string is flexible, the angle θ is restricted to the range −π / 2 < θ < π / 2 . (For angles θ > π / 2 , the string would go slack.) In both instances the tangential component of the force is directed towards the equilibrium position. The tangential component of acceleration is aθ = lα = l d 2θ . (23.3.15) dt 2 Newton’s Second Law, F θ = maθ , yields −mgl sinθ = ml2 d 2θ . (23.3.16) dt 2 We can rewrite this equation is the form d 2θ g = − sinθ . (23.3.17) dt 2 l This is not the simple harmonic oscillator equation although it still describes periodic motion. In the limit of small oscillations, sinθ ≅θ , Eq. (23.3.17) becomes 23-14 d 2θ g ≅− θ. (23.3.18) dt 2 l This equation is similar to the object-spring simple harmonic oscillator differential equation d 2 x k = − x . (23.3.19) dt 2 m By comparison with Eq. (23.2.6) the angular frequency of oscillation for the pendulum is approximately g ω 0 (23.3.20) l , with period 2π l T = 2π . (23.3.21) ω 0 g The solutions to Eq. (23.3.18) can be modeled after Eq. (23.2.21). With the initial conditions that the pendulum is released from rest, dθ (t = 0) = 0 , at a small angle dt θ(t = 0) = θ0 , the angle the string makes with the vertical as a function of time is given by ⎛ 2π ⎞ ⎛ g ⎞ θ(t) = θ0 cos(ω 0 t) = θ0 cos t⎠ ⎟ = θ0 cos t (23.3.22) ⎠ ⎟ .
⎝ ⎜ T ⎝ ⎜ l The z -component of the angular velocity of the bob is ⎛ ⎞ dθ g g ω z (t) = (t) = − θ0 sin⎜ t⎟ . (23.3.23) dt l l ⎝ ⎠ Keep in mind that the component of the angular velocity ω z = dθ / dt changes with time in an oscillatory manner (sinusoidally in the limit of small oscillations). The angular frequency ω 0 is a parameter that describes the system. The z -component of the angular velocity ω z (t) , besides being time-dependent, depends on the amplitude of oscillation θ0 . In the limit of small oscillations, ω0 does not depend on the amplitude of oscillation. The fact that the period is independent of the mass of the object follows algebraically from the fact that the mass appears on both sides of Newton’s Second Law and hence cancels. Consider also the argument that is attributed to Galileo: if a pendulum, consisting of two identical masses joined together, were set to oscillate, the two halves would not exert forces on each other. So, if the pendulum were split into two pieces, the 23-15 pieces would oscillate the same as if they were one piece. This argument can be extended to simple pendula of arbitrary masses. 23.3.2 Simple Pendulum: Energy Approach We can use energy methods to find the differential equation describing the time evolution of the angle θ . When the string is at an angle θ with respect to the vertical, the gravitational potential energy (relative to a choice of zero potential energy at the bottom of the swing where θ = 0 as shown in Figure 23.8) is given by U = mgl(1− cosθ ) (23.3.24) The θ -component of the velocity of the object is given by vθ = l(dθ / dt) so the kinetic ⎞ ⎠ energy is ⎛ dt ⎝l dθ ⎜ ⎟ 2 1 2 1 2 mv2 K = (23.3.25) m = . Figure 23.8 Energy diagram for simple pendulum The mechanical energy of the system is then 1 ⎛ l dθ ⎞ 2 E = K + U = + mgl (1 − cosθ ) . (23.3.26) 2 m⎝ ⎜ ⎠ ⎟ dt Because we assumed that there is no non-conservative work (i.e. no air resistance or frictional forces acting at the pivot), the energy is constant, hence dE 1 d 2θ dθ 0 = = 2 m 2l2 dθ + mgl sinθ dt dt dt 2 dt (23.3.27) ⎛ d 2θ g ⎞ = ml2 dθ + sinθ⎠ ⎟ .
dt ⎝ ⎜ dt 2 l 23-16 There are two solutions to this equation; the first one dθ / dt = 0 is the equilibrium solution. That the z -component of the angular velocity is zero means the suspended object is not moving. The second solution is the one we are interested in d 2θ g + sinθ = 0 , (23.3.28) dt 2 l which is the same differential equation (Eq. (23.3.16)) that we found using the force method. We can find the time t1 that the object first reaches the bottom of the circular arc by setting θ(t1) = 0 in Eq. (23.3.22) ⎛ g ⎞ 0 = θ0 cos t1 (23.3.29) ⎠ ⎟ . ⎝ ⎜ l This zero occurs when the argument of the cosine satisfies π t1 = (23.3.30) 2 .
g l The z -component of the angular velocity at time t1 is therefore dθ dt (t1) = − θ0 sin ⎛ ⎝ ⎜ t1 ⎞ ⎠ ⎟ = − ⎛π ⎞ ⎝ ⎜ ⎠ ⎟ = − g l g l g l g l θ0 sin θ0 . (23.3.31) 2 Note that the negative sign means that the bob is moving in the negative θ ˆ -direction when it first reaches the bottom of the arc. The θ -component of the velocity at time t1 is therefore = l dθ g ⎛ g ⎞ ⎛π ⎞ vθ (t1) ≡ v1 (t1) = −l θ0 sin t1 ⎠ ⎟ = − lg θ0 sin⎝ ⎜ ⎠ ⎟ = − lg θ0 .(23.3.32) dt l ⎝ ⎜ l 2 We can also find the components of both the velocity and angular velocity using energy methods. When we release the bob from rest, the energy is only potential energy E = U0 = mgl(1 − cosθ0 ) ≅ mgl θ 2 0 2 , (23.3.33) where we used the approximation that cosθ0 ≅ 1−θ0 2 / 2 . When the bob is at the bottom of the arc, the only contribution to the mechanical energy is the kinetic energy given by 23-17 K1 = 2 1 mv1 2 . (23.3.34) Because the energy is constant, we have that U0 = K1 or mgl θ 2 2 0 1 2 (23.3.35) mv1 = .
2 We can solve for the θ -component of the velocity at the bottom of the arc vθ ,1 = ± gl θ0 . (23.3.36) The two possible solutions correspond to the different directions that the motion of the bob can have when at the bottom. The z -component of the angular velocity is then dθ (t1) = dt v1 l = ± g l θ0 , (23.3.37) in agreement with our previous calculation. If we do not make the small angle approximation, we can still use energy techniques to find the θ -component of the velocity at the bottom of the arc by equating the energies at the two positions mgl (1 − cosθ0 ) = 2 1 mv1 2 , (23.3.38) (23.3.39) 23.4 Worked Examples Example 23.3: Rolling Without Slipping Oscillating Cylinder Attach a solid cylinder of mass M and radius R to a horizontal massless spring with spring constant k so that it can roll without slipping along a horizontal surface. At time t , the center of mass of the cylinder is moving with speed Vcm and the spring is compressed vθ , 1 = ± 2gl 1− cosθ0 ( ) . a distance x from its equilibrium length. What is the period of simple harmonic motion for the center of mass of the cylinder? Figure 23.9 Example 23.3 23-18 Solution: At time t , the energy of the rolling cylinder and spring system is Mvcm ⎝ ⎜ ⎠ ⎟ 2 E = 1 2 + 1 2 Icm ⎛ dθ ⎞ dt 2 + 1 2 kx2 . (23.4.1) where x is the amount the spring has compressed, Icm = (1/ 2)MR2 , and because it is rolling without slipping dθ Vcm = . (23.4.2) dt R Therefore the energy is 2 MVcm ⎜ ⎟ MVcm ⎞ ⎠ ⎛ ⎝ 2 2 1 2 1 4 1 2 3 4 1 2 Vcm MR2 kx2 kx2 E = (23.4.3) + + + = .
R The energy is constant (no non-conservative force is doing work on the system) so dE 3 dVcm 1 dx (2 3 M d 2 x 0 = = 2MVcm + 2 k2x = Vcm dt 2 + kx) (23.4.4) dt 4 dt dt Because Vcm is non-zero most of the time, the displacement of the spring satisfies a simple harmonic oscillator equation d 2 x dt 2 + 2k 3M x = 0 . (23.4.5) Hence the period is T = 2π ω 0 = 2π 3M 2k . (23.4.6) Example 23.4: U-Tube A U-tube open at both ends is filled with an incompressible fluid of density ρ . The cross-sectional area A of the tube is uniform and the total length of the fluid in the tube is L . A piston is used to depress the height of the liquid column on one side by a distance x0 , (raising the other side by the same distance) and then is quickly removed (Figure 23.10). What is the angular frequency of the ensuing simple harmonic motion? Neglect any resistive forces and at the walls of the U-tube. 23-19 Figure 23.10 Example 23.4 Figure 23.11 Energy diagram for water Solution: We shall use conservation of energy. First choose as a zero for gravitational potential energy in the configuration where the water levels are equal on both sides of the tube. When the piston on one side depresses the fluid, it rises on the other. At a given instant in time when a portion of the fluid of mass Δm = ρ Ax is a height x above the equilibrium height (Figure 23.11), the potential energy of the fluid is given by U = Δmgx = (ρ Ax)gx = ρ Agx2 . (23.4.7) At that same instant the entire fluid of length L and mass m = ρ AL is moving with speed v , so the kinetic energy is 1 2 1 K = mv = ρ ALv2 . (23.4.8) 2 2 Thus the total energy is E = K +U = 1 ρ ALv2 + ρ Agx2 . (23.4.9) 2 By neglecting resistive force, the mechanical energy of the fluid is constant. Therefore dE dv dx 0 = = ρ ALv + 2ρ Agx . (23.4.10) dt dt dt If we just consider the top of the fluid above the equilibrium position on the right arm in Figure 23.13, we rewrite Eq. (23.4.10) as dE dvx dx 0 = = ρ ALv + 2ρ Agx , (23.4.11) dt x dt dt where vx = dx / dt . We now rewrite the energy condition using dvx / dt = d 2 x / dt2 as 23-20 ⎛ d 2 x ⎞ 0 = vx ρ A L + 2gx . (23.4.12) ⎝ ⎜ dt2 ⎠ ⎟ This condition is satisfied when vx = 0 , i.e. the equilibrium condition or when d 2 x 0 = L + 2gx . (23.4.13) dt2 This last condition can be written as d 2 x 2g = − x . (23.4.14) dt2 L This last equation is the simple harmonic oscillator equation. Using the same mathematical techniques as we used for the spring-block system, the solution for the height of the fluid above the equilibrium position is given by x(t) = Bcos(ω0t) + C sin(ω0t) , (23.4.15) where 2g ω0 = (23.4.16) L is the angular frequency of oscillation. The x -component of the velocity of the fluid on the right-hand side of the U-tube is given by dx(t) v (t) = = −ω0 Bsin(ω0t) + ω0C cos(ω0t) . (23.4.17) x dt The coefficients B and C are determined by the initial conditions. At t = 0 , the height of the fluid is x(t = 0) = B = x0. At t = 0 , the speed is zero so vx (t = 0) = ω0C = 0 , hence C = 0 . The height of the fluid above the equilibrium position on the right hand-side of the U-tube as a function of time is thus ⎛ ⎞ 2g x(t) = x0 cos . (23.4.18) t ⎜ ⎝ ⎟ ⎠ L 23.5 Damped Oscillatory Motion Let’s now consider our spring-block system moving on a horizontal frictionless surface but now the block is attached to a damper that resists the motion of the block due to viscous friction. This damper, commonly called a dashpot, is shown in Figure 23.13. The viscous force arises when objects move through fluids at speeds slow enough so that there is no turbulence. When the viscous force opposes the motion and is proportional to the velocity, so that 23-21 fvis = −bv , (23.5.1) the dashpot is referred to as a linear dashpot. The constant of proportionality b depends on the properties of the dashpot. Figure 23.12 Spring-block system connected to a linear dashpot Choose the origin at the equilibrium position and choose the positive x -direction to the right in the Figure 23.13. Define x(t) to be the position of the object with respect to the equilibrium position. The x -component of the total force acting on the spring is the sum of the linear restoring spring force, and the viscous friction force (Figure 23.13), dx F = −k x − b (23.5.2) x dt Figure 23.13 Free-body force diagram for spring-object system with linear dashpot Newton’s Second law in the x -direction becomes dx −k x − b dt d 2 x = m dt2 . (23.5.3) We can rewrite Eq. (23.5.3) as d 2 x dt2 + b m dx dt + k x = 0 . m (23.5.4) When (b / m)2 < 4k / m , the oscillator is called underdamped, and the solution to Eq. (23.5.4) is given by x(t) = x m e−αt cos(γ t +φ) (23.5.5) 23-22 where γ = (k / m − (b / 2m)2)1 2 is the angular frequency of oscillation, α = b 2m is a parameter that measured the exponential decay of the oscillations, x m is a constant and φ is the phase constant. Recall the undamped oscillator has angular frequency = (k / m)1 2 , so the angular frequency of the underdamped oscillator can be expressed as ω0 2 −α 2)1 2 γ = (ω0 . (23.5.6) In Appendix 23B: Complex Numbers, we introduce complex numbers and use them to solve Eq.(23.5.4) in Appendix 23C: Solution to the Underdamped Simple Harmonic Oscillator Equation. The x -component of the velocity of the object is given by −αt vx (t) = dx dt = (−γ x sin(γ t +φ) −α x cos(γ t +φ))e . (23.5.7) m m The position and the x -component of the velocity of the object oscillate but the amplitudes of the oscillations decay exponentially. In Figure 23.14, the position is plotted as a function of time for the underdamped system for the special case φ = 0. For that case x(t) = x m e−αt cos(γ t) . (23.5.8) and −αt vx (t) = dx dt = (−γ x m sin(γ t) −α x m cos(γ t))e . (23.5.9) Figure 23.14 Plot of position x(t) of object for underdamped oscillator with φ = 0 Because the coefficient of exponential decay α = b 2m is proportional to the b, we see that the position will decay more rapidly if the viscous force increases. We can introduce a time constant τ = 1 α = 2m / b . (23.5.10) 23-23 When t = τ , the position is x(t = τ ) = x m cos(γτ )e−1 . (23.5.11) The envelope of exponential decay has now decreases by a factor of e−1 , i.e. the amplitude can be at most x m e−1 . During this time interval [0,τ ] , the position has undergone a number of oscillations. The total number of radians associated with those oscillations is given by γτ = (k / m − (b / 2m)2)1 2 (2m / b) . (23.5.12) The closest integral number of cycles is then n = ⎡ ⎣γτ / 2π ⎤ ⎦ = ⎡ ⎣(k / m − (b / 2m)2)1 2 (m / πb)⎤ ⎦ . (23.5.13) If the system is very weakly damped, such that (b / m)2 << 4k / m , then we can approximate the number of cycles by (k / m)1 2 (m / πb) n = ⎡ ⎣γτ 2π ⎤ ⎦ ⎡ ⎣ ⎤ ⎦ = ⎡ ⎣ω0(m / π b)⎤ ⎦ , (23.5.14) = (k / m)1 2 where ω0 is the angular frequency of the undamped oscillator. We define the quality, Q , of this oscillating system to be proportional to the number of integral cycles it takes for the exponential envelope of the position function to fall off by a factor of e−1. The constant of proportionality is chosen to be π . Thus Q = nπ . (23.5.15) For the weakly damped case, we have that Q ω0(m / b) . (23.5.16) 23.5.1 Energy in the Underdamped Oscillator For the underdamped oscillator, (b / m)2 < 4k / m , γ = (k / m − (b / 2m)2)1 2 , and α = b 2m . Let’s choose t = 0 such that the phase shift is zero φ = 0. The stored energy in the system will decay due to the energy loss due to dissipation. The mechanical energy stored in the potential and kinetic energies is then given by 1 1 2 E = kx2 + mv . (23.5.17) 2 2 23-24 where the position and the x -component of the velocity are given by Eqs. (23.5.8) and (23.5.9). The mechanical energy is then 1 1 2 −2αt E = kx cos2(γ t)e−2αt + m(−γ x sin(γ t) −α x cos(γ t)) 2 e . (23.5.18) m m m 2 2 Expanding this expression yields 1 2(γ t)e −2αt + 1 mγ 2 2 −2αt E = (k + mα 2)x cos −2αt + mγα x 2 sin(γ t)cos(γ t)e x 2 sin2(γ t)e (23.5.19) m m m 2 2 The kinetic energy, potential energy, and mechanical energy are shown in Figure 23.15. Figure 23.15 Kinetic, potential and mechanical energy for the underdamped oscillator The stored energy at time t = 0 is E(t = 0) = 1 2 (k + mα 2 )x m 2 (23.5.20) The mechanical energy at the conclusion of one cycle, with γ T = 2π , is E(t = T ) = 1 2 −2αT (k + mα 2 )x e m 2 (23.5.21) The change in the mechanical energy for one cycle is then E(t = T ) − E(t = 0) = − 1 (k + mα 2 )x 2 (1− e−2αT ) . m 2 (23.5.22) 23-25 Recall that α 2 = b2 4m2 . Therefore E(t = T ) − E(t = 0) = − 1 (k + b2 4m)x m 2(1− e−2αT ) . (23.5.23) 2 We can show (although the calculation is lengthy) that the energy dissipated by the viscous force over one cycle is given by the integral = T ⋅ v dt = −⎛ k + b2 ⎞ x m 2 (1− e−2αt ) . (23.5.24) Edis ∫Fvis 0 ⎝ ⎜ 4m⎠ ⎟ 2 By comparison with Eq. (23.5.23), the change in the mechanical energy in the underdamped oscillator during one cycle is equal to the energy dissipated due to the viscous force during one cycle. 23.6 Forced Damped Oscillator Let’s drive our damped spring-object system by a sinusoidal force. Suppose that the x component of the driving force is given by Fx (t) = F 0 cos(ωt) , (23.6.1) where F 0 is called the amplitude (maximum value) and ω is the driving angular frequency. The force varies between F 0 and −F 0 because the cosine function varies between +1 and −1. Define x(t) to be the position of the object with respect to the equilibrium position. The x -component of the force acting on the object is now the sum dx F = F 0 cos(ωt) − kx − b . (23.6.2) x dt Newton’s Second law in the x -direction becomes d 2 dx x F 0 cos(ωt) − kx − b = m . (23.6.3) dt dt2 We can rewrite Eq. (23.6.3) as d 2 x dx F 0 cos(ωt) = m + b + kx . (23.6.4) dt2 dt 23-26 We derive the solution to Eq. (23.6.4) in Appendix 23E: Solution to the forced Damped Oscillator Equation. The solution to is given by the function x(t) = x0 cos(ωt +φ) , (23.6.5) where the amplitude x0 is a function of the driving angular frequency ω and is given by F 0 / m (ω ) = . (23.6.6) x0 1/2 ((b / m)2ω 2 + (ω0 2 −ω 2)2 ) The phase constant φ is also a function of the driving angular frequency ω and is given by ⎛ (b / m)ω ⎞ φ(ω ) = tan−1 . (23.6.7) ⎜ 2 ⎟ ⎝ω 2 −ω0 ⎠ In Eqs. (23.6.6) and (23.6.7) k ω0 = (23.6.8) m is the natural angular frequency associated with the undriven undamped oscillator. The x -component of the velocity can be found by differentiating Eq. (23.6.5), dx vx (t) = (t) = −ω x0 sin(ωt +φ) , (23.6.9) dt where the amplitude x0(ω ) is given by Eq. (23.6.6) and the phase constant φ(ω ) is given by Eq. (23.6.7). 23.6.1 Resonance When b / m << 2ω0 we say that the oscillator is lightly damped. For a lightly-damped driven oscillator, after a transitory period, the position of the object will oscillate with the same angular frequency as the driving force. The plot of amplitude x0(ω ) vs. driving angular frequency ω for a lightly damped forced oscillator is shown in Figure 23.16. If the angular frequency is increased from zero, the amplitude of the x0(ω ) will increase until it reaches a maximum when the angular frequency of the driving force is the same as the natural angular frequency, ω0 , associated with the undamped oscillator. This is called resonance. When the driving angular frequency is increased above the natural angular frequency the amplitude of the position oscillations diminishes. 23-27 Figure 23.16 Plot of amplitude x0(ω ) vs. driving angular frequency ω for a lightly damped oscillator with b / m << 2ω0 We can find the angular frequency such that the amplitude x0(ω ) is at a maximum by setting the derivative of Eq. (23.6.6) equal to zero, d F 0(2ω ) ((b / m)2 − 2(ω0 2 −ω 2)) 0 = (ω ) = − . (23.6.10) x0 3/2 dt 2m 2 −ω 2)2 ((b / m)2ω 2 + (ω0 ) This vanishes when ω = (ω0 2 − (b / m)2 / 2)1/2 . (23.6.11) For the lightly-damped oscillator, ω0 >> (1/ 2)b / m , and so the maximum value of the amplitude occurs when = (k / m)1/2 ω ω0 . (23.6.12) The amplitude at resonance is then F 0 x0(ω = ω0) = (lightly damped) . (23.6.13) bω0 The plot of phase constant φ(ω ) vs. driving angular frequency ω for a lightly damped forced oscillator is shown in Figure 23.17. 23-28 Figure 23.17 Plot of phase constant φ(ω ) vs. driving angular frequency ω for a lightly damped oscillator with b / m << 2ω0 The phase constant at resonance is zero, φ(ω = ω0) = 0 . (23.6.14) At resonance, the x -component of the velocity is given by dx F 0 vx (t) = (t) = − sin(ω0t) (lightly damped) . (23.6.15) dt b When the oscillator is not lightly damped ( b / m ω0 ), the resonance peak is shifted to the left of ω = ω0 as shown in the plot of amplitude vs. angular frequency in Figure 23.18. The corresponding plot of phase constant vs. angular frequency for the non-lightly damped oscillator is shown in Figure 23.19. Figure 23.18 Plot of amplitude vs. angular frequency for lightly-damped driven oscillator where b / m ω0 23-29 Figure 23.19 Plot of phase constant vs. angular frequency for lightly-damped driven oscillator where b / m ω0 23.6.2 Mechanical Energy The kinetic energy for the driven damped oscillator is given by 1 2(t) = 1 mω 2 K(t) = mv x0 2 sin2(ωt +φ) . (23.6.16) 2 2 The potential energy is given by 1 1 2 U (t) = kx2(t) = kx0 cos2(ωt +φ) . (23.6.17) 2 2 The mechanical energy is then 1 1 1 1 2 E(t) = mv2(t) + kx2(t) = mω 2 x0 2 sin2(ωt +φ) + kx0 cos2(ωt +φ) .(23.6.18) 2 2 2 2 Example 23.5: Time-Averaged Mechanical Energy The period of one cycle is given by T = 2π / ω . Show that 1 T 1 (i) ∫ sin2(ωt +φ) dt = , (23.6.19) T 0 2 (ii) 1 T ∫ cos2(ωt +φ)dt = 1 , (23.6.20) T 0 2 1 T (iii) sin(ωt)cos(ωt) dt = 0 . (23.6.21) ∫ T 0 23-30 Solution: (i) We use the trigonometric identity sin2(ωt +φ)) = 1 (1− cos(2(ωt +φ)) (23.6.22) 2 to rewrite the integral in Eq. (23.6.19) as 1 T ∫sin2(ωt +φ)) dt = 1 ∫ T (1− cos(2(ωt +φ))dt (23.6.23) 2T T 0 0 Integration yields T =2π /ω 1 T 1 ⎛ sin(2(ωt +φ))⎞ ∫(1− cos(2(ωt +φ)) dt = − ⎝ ⎜ ⎠ ⎟ 2T 0 2 2ω T =0 (23.6.24) 1 ⎛ sin(4π + 2φ) sin(2φ)⎞ 1 = − − , ⎠ ⎟ = 2 ⎝ ⎜ 2ω 2ω 2 where we used the trigonometric identity that sin(4π + 2φ) = sin(4π )cos(2φ) + sin(2φ)cos(4π ) = sin(2φ) , (23.6.25) proving Eq. (23.6.19). (ii) We use a similar argument starting with the trigonometric identity that cos2(ωt +φ)) = 1 (1+ cos(2(ωt +φ)) . (23.6.26) 2 Then 1 T 1 T cos2(ωt +φ)) dt = (1+ cos(2(ωt +φ)) dt . (23.6.27) ∫ ∫ T 0 2T 0 Integration yields T =2π /ω 1 T 1 ⎛ sin(2(ωt +φ))⎞ (1+ cos(2(ωt +φ)) dt = + ∫ 2T 0 2 ⎝ ⎜ 2ω ⎠ ⎟ T =0 (23.6.28) 1 ⎛ sin(4π + 2φ) sin(2φ)⎞ 1 = + − .
⎠ ⎟ = 2 ⎝ ⎜ 2ω 2ω 2 (iii) We first use the trigonometric identity that 23-31 1 sin(ωt)cos(ωt) = sin(ωt) . (23.6.29) 2 Then 1 T ∫ 1 ∫ T sin(ωt)cos(ωt)dt = sin(ωt) dt T 0 T 0 (23.6.30) T 1 1 cos(ωt) = − = − (1−1) = 0.
T 2ω 0 2ωT The values of the integrals in Example 23.5 are called the time-averaged values. We denote the time-average value of a function f (t) over one period by 1 T f ≡ ∫ f (t) dt . (23.6.31) T 0 In particular, the time-average kinetic energy as a function of the angular frequency is given by K(ω ) = 1 mω 2 x0 2 . (23.6.32) 4 The time-averaged potential energy as a function of the angular frequency is given by 1 2 U (ω ) = kx0 . (23.6.33) 4 The time-averaged value of the mechanical energy as a function of the angular frequency is given by 1 1 2 1 2 E(ω ) = mω 2 x0 2 + kx0 = (mω 2 + k)x0 . (23.6.34) 4 4 4 We now substitute Eq. (23.6.6) for the amplitude into Eq. (23.6.34) yielding F 0 2 (ω0 2 + ω 2) E(ω ) = . (23.6.35) ⎛ 2 4m ⎞ 2 −ω 2 (b / m)2ω 2 +(ω0 ) ⎝ ⎠ A plot of the time-averaged energy versus angular frequency for the lightly-damped case ( b / m << 2ω0 ) is shown in Figure 23.20. 23-32 Figure 23.20 Plot of the time-averaged energy versus angular frequency for the lightly-damped case ( b / m << 2ω0 ) We can simplify the expression for the time-averaged energy for the lightly-damped case by observing that the time-averaged energy is nearly zero everywhere except where ω = ω0 , (see Figure 23.20). We first substitute ω = ω0 everywhere in Eq. (23.6.35) except the term ω0 2 −ω 2 that appears in the denominator, yielding F 0 2 (ω0 2) E(ω ) = . (23.6.36) 2m ⎛ 2 ⎞ (b / m)2ω0 2 +(ω0 2 −ω 2 ) ⎝ ⎠ We can approximate the term ω0 2 −ω 2 = (ω0 −ω )(ω0 + ω ) 2ω0(ω0 −ω ) (23.6.37) Then Eq. (23.6.36) becomes F 0 2 1 E(ω ) = (lightly damped) . (23.6.38) 2m ((b / m)2 + 4(ω0 −ω )2 ) The right-hand expression of Eq. (23.6.38) takes on its maximum value when the denominator has its minimum value. By inspection, this occurs when ω = ω0 . Alternatively, to find the maximum value, we set the derivative of Eq. (23.6.35) equal to zero and solve for ω , 23-33 0 = d d F 0 2 1 E(ω ) = dω dω 2m ((b / m)2 + 4(ω0 −ω )2 ) . (23.6.39) 4F 0 2 (ω0 −ω ) = −ω )2 2 m ((b / m)2 + 4(ω0 ) The maximum occurs when occurs at ω = ω0 and has the value mF 0 2 E(ω0) = (underdamped) . (23.6.40) 2b2 23.6.3 The Time-averaged Power The time-averaged power delivered by the driving force is given by the expression 1 T 1 T F 0 2ω cos(ωt)sin(ωt +φ) P(ω ) = ∫ Fxvx dt = −∫ 1/2 dt , (23.6.41) T 0 T 0 m (b / m)2ω 2 + (ω0 2 −ω 2)2 ( ) where we used Eq. (23.6.1) for the driving force, and Eq. (23.6.9) for the x -component of the velocity of the object. We use the trigonometric identity sin(ωt +φ) = sin(ωt)cos(φ) + cos(ωt)sin(φ) (23.6.42) to rewrite the integral in Eq. (23.6.41) as two integrals 1 T F 0 2ω cos(ωt)sin(ωt)cos(φ) P(ω ) = − dt ∫ 1/2 T 2 −ω 2)2 0 m((b / m)2ω 2 + (ω0 ) (23.6.43) 1 T F 0 2ω cos2(ωt)sin(φ) − dt.
∫ 1/2 T 2 −ω 2)2 0 m((b / m)2ω 2 + (ω0 ) Using the time-averaged results from Example 23.5, we see that the first term in Eq. (23.6.43) is zero and the second term becomes F 0 2ω sin(φ) = 2 −ω 2)2 1/2 (23.6.44) 2m((b / m)2ω 2 + (ω0 ) P(ω ) For the underdamped driven oscillator, we make the same approximations in Eq. (23.6.44) that we made for the time-averaged energy. In the term in the numerator and the 23-34 term on the left in the denominator, we set ω ω0 , and we use Eq. (23.6.37) in the term on the right in the denominator yielding F 0 2 sin(φ) P(ω ) = 1/2 (underdamped) . (23.6.45) 2m((b / m)2 + 2(ω0 −ω )) The time-averaged power dissipated by the resistive force is given by T T T 1 1 1 F 0 2ω 2 sin2(ωt +φ)dt (ω ) = (F dt = − bv 2 dt = P dis T ∫ x )dis vx T ∫ x T ∫ m2 ((b / m)2ω 2 + (ω0 2 −ω 2)2 ) 0 0 0 ,(23.6.46) F 0 2ω 2dt = .
2 2m ((b / m)2ω 2 + (ω0 2 −ω 2)2 ) where we used Eq. (23.5.1) for the dissipative force, Eq. (23.6.9) for the x -component of the velocity of the object, and Eq. (23.6.19) for the time-averaging. 23.6.4 Quality Factor The plot of the time-averaged energy vs. the driving angular frequency for the underdamped oscullator has a width, Δω (Figure 23.20). One way to characterize this width is to define Δω = ω + −ω − , where ω ± are the values of the angular frequency such that time-averaged energy is equal to one half its maximum value 1 mF 0 2 E(ω ± ) = E(ω0) = . (23.6.47) 4b2 2 The quantity Δω is called the line width at half energy maximum also known as the resonance width. We can now solve for ω ± by setting F 0 2 1 mF 0 2 E(ω ± ) = = , (23.6.48) 2m ((b / m)2 + 4(ω0 −ω ± )2 ) 4b2 yielding the condition that (b / m)2 = 4(ω0 −ω ± )2 . (23.6.49) Taking square roots of Eq. (23.6.49) yields (b / 2m) = ω0 −ω ± . (23.6.50) 23-35 Therefore ω ± = ω0 ± (b / 2m) . (23.6.51) The half-width is then Δω = ω + (b / 2m)) − (ω0 − (b / 2m)) = b / m . (23.6.52) + −ω − = (ω0 We define the quality Q of the resonance as the ratio of the resonant angular frequency to the line width, ω0 ω0 Q = = . (23.6.53) Δω b / m Figure 23.21 Plot of time-averaged energy vs. angular frequency for different values of b / m In Figure 23.21 we plot the time-averaged energy vs. angular frequency for several different values of the quality factor Q = 10, 5, and 3. Recall that this was the same result that we had for the quality of the free oscillations of the damped oscillator, Eq. (23.5.16) (because we chose the factor π in Eq. (23.5.16)). 23.7 Small Oscillations Any object moving subject to a force associated with a potential energy function that is quadratic will undergo simple harmonic motion, 1 U (x) = U0 + 2 k(x − xeq )2 . (23.7.1) where k is a “spring constant”, xeq is the equilibrium position, and the constant U0 just depends on the choice of reference point xref for zero potential energy, U (xref ) = 0 , 23-36 0 = U (xref ) = U0 + 1 k(xref − xeq )2 . (23.7.2) 2 Therefore the constant is 1 )2 = − − x . (23.7.3) U0 2 k(xref eq The minimum of the potential x0 corresponds to the point where the x -component of the force is zero, dU = 2k(x0 − x ) = 0 ⇒ x0 = x , (23.7.4) eq eq dx x = x0 corresponding to the equilibrium position. Therefore the constant is U (x0) = U0 and we rewrite our potential function as U (x) = U (x0) + 1 k(x − x0)2 . (23.7.5) 2 Now suppose that a potential energy function is not quadratic but still has a minimum at x0 . For example, consider the potential energy function ⎛ 3 2 ⎞ ⎛ x ⎞ ⎛ x ⎞ U (x) = −U1 ⎜ − ⎟ , (23.7.6) ⎜ ⎝ ⎜ x1 ⎠ ⎟ ⎝ ⎜ x1 ⎠ ⎟ ⎟ ⎝ ⎠ (Figure 23.22), which has a stable minimum at x0 . Figure 23.22 Potential energy function with stable minima and unstable maxima When the energy of the system is very close to the value of the potential energy at the minimum U (x0) , we shall show that the system will undergo small oscillations about the 23-37 minimum value x0 . We shall use the Taylor formula to approximate the potential function as a polynomial. We shall show that near the minimum x0 , we can approximate the potential function by a quadratic function similar to Eq. (23.7.5) and show that the system undergoes simple harmonic motion for small oscillations about the minimum x0 . We begin by expanding the potential energy function about the minimum point using the Taylor formula dU 1 d 2U 1 d 3U U (x) = U (x0) + (x − x0) + (x − x0)3 + ⋅⋅⋅ (23.7.7) (x − x0)2 + dx 2! dx2 3! dx3 x=x0 x=x0 x=x0 1 d 3U where (x − x0)3 is a third order term in that it is proportional to (x − x0)3 , and 3! dx3 x=x0 d 3U d 2U dU , dx2 and are constants. If x0 is the minimum of the potential dx dx3 , x=x0 x=x0 x=x0 energy, then the linear term is zero, because dU = 0 (23.7.8) dx x=x0 and so Eq. ((23.7.7)) becomes 1 d 2U 1 d 3U U (x) U (x0) + (x − x0)3 + ⋅⋅⋅ (23.7.9) 2 dx2 (x − x0)2 + 3! dx3 x=x0 x=x0 For small displacements from the equilibrium point such that is sufficiently small, the third order term and higher order terms are very small and can be ignored. Then the potential energy function is approximately a quadratic function, x − x0 1 d 2U U (x) U (x0) + (x − x0)2 = U (x0) + 1 keff (x − x0)2 (23.7.10) 2 dx2 2 x=x0 where we define keff , the effective spring constant, by d 2U ≡ . (23.7.11) keff dx2 x=x0 23-38 Because the potential energy function is now approximated by a quadratic function, the system will undergo simple harmonic motion for small displacements from the minimum with a force given by F x dU = − dx (x − x0 ) .
= −keff (23.7.12) At x = x0 , the force is zero F (x0 ) = x dU (x0 ) = 0 .
dx (23.7.13) We can determine the period of oscillation by substituting Eq. (23.7.12) into Newton’s Second Law d 2 x (x − x0 (23.7.14) −keff ) = meff dt2 where meff is the effective mass. For a two-particle system, the effective mass is the reduced mass of the system. m 1m2 = (23.7.15) meff ≡ µred , m 1 + m2 Eq. (23.7.14) has the same form as the spring-object ideal oscillator. Therefore the angular frequency of small oscillations is given by ω0 = keff meff = d 2U dx2 x=x0 meff . (23.7.16) Example 23.6: Quartic Potential A system with effective mass m has a potential energy given by ⎛ ⎛ x ⎞ 2 ⎛ x ⎞ 4 ⎞ U (x) = U0 ⎜−2 + ⎟ , (23.7.17) ⎜ ⎝ ⎜ x0 ⎠ ⎟ ⎝ ⎜ x0 ⎠ ⎟ ⎟ ⎝ ⎠ where U0 and x0 are positive constants and U (0) = 0 . (a) Find the points where the force on the particle is zero. Classify these points as stable or unstable. Calculate the value of U (x) / U0 at these equilibrium points. (b) If the particle is given a small displacement from an equilibrium point, find the angular frequency of small oscillation. Solution: (a) A plot of U (x) / U0 as a function of x / x0 is shown in Figure 23.23. 23-39 Figure 22.23 Plot of U (x) / U0 as a function of x / x0 The force on the particle is zero at the minimum of the potential energy, ⎛ 2 4 ⎞ dU ⎛ 1 ⎞ ⎛ 1 ⎞ 0 = = U0 ⎜−4 x + 4 x3 ⎟ dx ⎜ ⎝ ⎝ ⎜ x0 ⎠ ⎟ ⎝ ⎜ x0 ⎠ ⎟ ⎟ ⎠ (23.7.18) ⎛ 1 ⎞ 2 ⎛ ⎛ x ⎞ 2 ⎞ 2 = −4U0 x ⎜1− ⎟⇒ x = x0 2 and x = 0. ⎝ ⎜ x0 ⎠ ⎟⎜ ⎝ ⎝ ⎜ x0 ⎠ ⎟ ⎟ ⎠ The equilibrium points are at x = ±x0 which are stable and x = 0 which is unstable. The second derivative of the potential energy is given by d 2U ⎛ ⎛ 1 ⎞ 2 ⎛ 1 ⎞ 4 2 ⎞ = U0 ⎜−4 +12 x ⎟ . (23.7.19) dx2 ⎜ ⎝ ⎝ ⎜ x0 ⎠ ⎟ ⎝ ⎜ x0 ⎠ ⎟ ⎟ ⎠ If the particle is given a small displacement from x = x0 then ⎛ 2 4 ⎞ d 2U ⎛ 1 ⎞ ⎛ 1 ⎞ 8 ⎜−4 +12 2 ⎟ . (23.7.20) = U0 x0 = U0 2 ⎜ ⎝ ⎜ x0 ⎠ ⎟ ⎝ ⎜ x0 ⎠ ⎟ ⎟ x0 dx2 x=x0 ⎝ ⎠ (b) The angular frequency of small oscillations is given by . (23.7.21) ω0 = d 2U dx2 x=x0 mx0 / m = 8U0 2 23-40 Example 23.7: Lennard-Jones 6-12 Potential A commonly used potential energy function to describe the interaction between two atoms is the Lennard-Jones 6-12 potential U (r) = U0 ⎡ ⎣(r 0 / r)12 − 2(r 0 / r)6 ⎤ ⎦ ; r > 0 , (23.7.22) where r is the distance between the atoms. Find the angular frequency of small oscillations about the stable equilibrium position for two identical atoms bound to each other by the Lennard-Jones interaction. Let m denote the effective mass of the system of two atoms. Solution: The equilibrium points are found by setting the first derivative of the potential energy equal to zero, ⎡ 6 ⎤ dU 12 6 −7 6 −7 r 0 0 = = U0 ⎡ ⎣−12r 0 r −13 +12r 0 r ⎤ ⎦ = U012r 0 r ⎢−⎛ ⎞ +1⎥ . (23.7.23) dr ⎢ ⎣ ⎝ ⎜ r ⎠ ⎟ ⎥ ⎦ The equilibrium point occurs when r = r 0 . The second derivative of the potential energy function is d 2U 12 6 −8 = U0 ⎡ ⎣+(12)(13)r 0 r −14 − (12)(7)r 0 r ⎤ ⎦ . (23.7.24) dr 2 Evaluating this at r = r 0 yields d 2U = 72U0r 0 −2 . (23.7.25) dr 2 r=r0 The angular frequency of small oscillation is therefore ω0 = d 2U dr 2 r=r0 / m = 72U0 / mr 0 2 . (23.7.26) 23-41 Appendix 23A: Solution to Simple Harmonic Oscillator Equation In our analysis of the solution of the simple harmonic oscillator equation of motion, Equation (23.2.1), d 2 x −k x = m , (23.A.1) dt 2 we assumed that the solution was a linear combination of sinusoidal functions, x(t) = Acos(ω0 t) + Bsin(ω0 t) , (23.A.2) where ω0 = k / m . We shall now derive Eq. (23.A.2). Assume that the mechanical energy of the spring-object system is given by the constant E . Choose the reference point for potential energy to be the unstretched position of the spring. Let x denote the amount the spring has been compressed ( x < 0 ) or stretched ( x > 0 ) from equilibrium at time t and denote the amount the spring has been compressed or stretched from equilibrium at time t = 0 by x(t = 0) ≡ x0 . Let vx = dx / dt denote the x -component of the velocity at time t and denote the x -component of the velocity at time t = 0 by vx (t = 0) ≡ v x,0 . The constancy of the mechanical energy is then expressed as 1 2 1 2 E = K +U = k x + mv . (23.A.3) 2 2 We can solve Eq. (23.A.3) for the square of the x -component of the velocity, 2 2E k 2 2E ⎛ k 2 ⎞ v = − x = 1− x (23.A.4) x ⎝ ⎜ ⎠ ⎟ . m m m 2E Taking square roots, we have dx 2E k 2 = 1− x . (23.A.5) dt m 2E (why we take the positive square root will be explained below). Let a1 ≡ 2E / m and a2 ≡ k / 2E . It’s worth noting that a1 has dimensions of velocity and w has dimensions of [length]−2 . Eq. (23.A.5) is separable, 23-42 dx 2 = a1 1− a2 x dt (23.A.6) dx = a1 dt. 1− a2 x2 We now integrate Eq. (23.A.6), dx ∫ = a1 dt . (23.A.7) The integral on the left in Eq. (23.A.7) is well known, and a derivation is presented here. We make a change of variables cosθ = a2 x with the differentials dθ and dx related by −sinθ dθ = a2 dx . The integration variable is θ = cos−1 ( a2 x) . (23.A.8) Eq. (23.A.7) then becomes a1 dt . (23.A.9) This is a good point at which to check the dimensions. The term on the left in Eq. (23.A.9) is dimensionless, and the product a2 a1 on the right has dimensions of inverse time, [length]−1[length ⋅ time−1] = [time−1] , so a2 a1 dt is dimensionless. Using the trigonometric identity 1−cos2 θ = sinθ , Eq. (23.A.9) reduces to = − dt . (23.A.10) ∫ dθ ∫ a2 a1 Although at this point in the derivation we don’t know that a2 a1 , which has dimensions of frequency, is the angular frequency of oscillation, we’ll use some foresight and make the identification ω0 ≡ a2 a1 = k 2E 2E m = k m , (23.A.11) and Eq. (23.A.10) becomes θ t dθ ∫ = − ∫ ω0 dt . (23.A.12) θ =θ0 t=0 1− a1 x2 ∫ −sinθ dθ 1−cos2 θ ∫ = a2 ∫ 23-43 After integration we have θ −θ0 = −ω0 t , (23.A.13) −1 where θ0 ≡ −φ is the constant of integration. Because θ = cos ( a2 x(t)) , Eq. (23.A.13) becomes cos−1 ( a2 x(t)) = −(ω0 t +φ) . (23.A.14) Take the cosine of each side of Eq. (23.A.14), yielding 1 2E x(t) = cos(−(ω0 t +φ)) = cos(ω0 t +φ) . (23.A.15) k a2 At t = 0, 2E k x0 ≡ x(t = 0) = cosφ . (23.A.16) The x -component of the velocity as a function of time is then dx(t) 2E v (t) = = −ω0 sin(ω0 t + φ) . (23.A.17) x dt k At t = 0, 2E v ≡ v (t = 0) = −ω0 sinφ . (23.A.18) x,0 x k We can determine the constant φ by dividing the expressions in Eqs. (23.A.18) and (23.A.16), v − x,0 = tanφ. (23.A.19) ω0 x0 Thus the constant φ can be determined by the initial conditions and the angular frequency of oscillation, ⎛ v ⎞ φ = tan−1 ⎜− x,0 ⎟ . (23.A.20) ⎝ ω0 x0 ⎠ Use the identity cos(ω0t +φ) = cos(ω0t)cos(φ) − sin(ω0t)sin(φ) (23.A.21) to expand Eq. (23.A.15) yielding 23-44 2E 2E x(t) = cos(ω0t)cos(φ) − sin(ω0t)sin(φ) , (23.A.22) k k and substituting Eqs. (23.A.16) and (23.A.18) into Eq. (23.A.22) yields v x(t) = x cosω t + x,0 sinω t , (23.A.23) 0 0 ω0 0 agreeing with Eq. (23.2.21). So, what about the missing ± that should have been in Eq. (23.A.5)? Strictly speaking, we would need to redo the derivation for the block moving in different directions. Mathematically, this would mean replacing φ by π −φ (or φ −π ) when the block’s velocity changes direction. Changing from the positive square root to the negative and changing φ to π −φ have the collective action of reproducing Eq. (23.A.23). Appendix 23B: Complex Numbers A complex number z can be written as a sum of a real number x and a purely imaginary number iy where i = z = x + iy . (23.B.1) The complex number can be represented as a point in the x-y plane as show in Figure 23B.1. −1 , Figure 23B.1 Complex numbers The complex conjugate z of a complex number z is defined to be The modulus of a complex number is z = x − iy . (23.B.2) z = (zz )1 2 = ((x + iy)(x − iy))1 2 2 )1 2 = (x2 + y . (23.B.3) 23-45 z where we used the fact that i2 = −1. The modulus represents the length of the ray from the origin to the complex number z in Figure 23B.1. Let φ denote the angle that the ray with the positive x -axis in Figure 23B.1. Then x = z cosφ , (23.B.4) y = z sinφ . (23.B.5) Hence the angle φ is given by φ = tan−1( y / x) . (23.B.6) The inverse of a complex number is then 1 z x − iy = = . (23.B.7) z zz (x2 + y2) The modulus of the inverse is the inverse of the modulus; 1 1 1 = = . (23.B.8) 2)1 2 z (x2 + y z The sum of two complex numbers, z1 = x1 + iy1 and z2 = x2 + iy2 , is the complex number z3 = z1 + z2 = (x1 + x2) + i( y1 + y2) = x3 + iy3 , (23.B.9) where x3 = x1 + x2 , y3 = y1 + y2 . We can represent this by the vector sum in Figure 23B.2, Figure 23B.2 Sum of two complex numbers The product of two complex numbers is given by z3 = z1z2 = (x1 + iy1)(x2 + iy2) = (x1x2 − y1 y2) + i(x1 y2 + x2 y1) = x3 + iy3 , (23.B.10) 23-46 where x3 = x1x2 − y1 y2 , and y3 = x1 y2 + x2 y1. One of the most important identities in mathematics is the Euler formula, eiφ = cosφ + isinφ . (23.B.11) This identity follows from the power series representations for the exponential, sine, and cosine functions, n=∞ φ 2 φ3 φ 4 φ5 iφ 1 (iφ)n e = ∑ = 1+ iφ − − i + + i ..., (23.B.12) n=0 n! 2 3! 4! 5! φ 2 φ 4 cosφ = 1− + − ..., (23.B.13) 2 4! φ3 φ5 sinφ = φ − + − .... (23.B.14) 3! 5! We define two projection operators. The first one takes the complex number eiφ and gives its real part, Re eiφ = cosφ . (23.B.15) The second operator takes the complex number eiφ and gives its imaginary part, which is the real number Im eiφ = sinφ . (23.B.16) A complex number z = x + iy can also be represented as the product of a modulus z and a phase factor eiφ , z = z eiφ . (23.B.17) The inverse of a complex number is then 1 1 1 − iφ = iφ = e , (23.B.18) z z z e where we used the fact that 1 − iφ iφ = e . (23.B.19) e iφ1 z In terms of modulus and phase, the sum of two complex numbers, z1 = e and 1 iφ2 z = z e , is 2 iφ1 z1 + z2 = e + e iφ2 . (23.B.20) z1 z2 2 23-47 A special case of this result is when the phase angles are equal, φ1 = φ2 , then the sum z1 + z2 has the same phase factor e iφ1 as z1 and z2 , iφ1 iφ1 iφ1 z1 + z2 = e + e =( + )e . (23.B.21) z1 z2 z1 z2 iφ2 z z The product of two complex numbers, z1 = e iφ1 , and z2 = e is 1 2 iφ1 iφ2 iφ1+φ2 z1z2 = e e = e . (23.B.22) z1 z2 z1 z2 When the phases are equal, the product does not have the same factor as z1 and z2 , iφ1 iφ1 z1z2 = e e = e i2φ1 . (23.B.23) z1 z2 z1 z2 Appendix 23C: Solution to the Underdamped Simple Harmonic Oscillator Consider the underdamped simple harmonic oscillator equation (Eq. (23.5.4)), d 2 x b dx k + + x = 0 . (23.C.1) dt2 m dt m When (b / m)2 < 4k / m , we show that the equation has a solution of the form x(t) = x m e−αt cos(γ t +φ) . (23.C.2) Solution: Let’s suppose the function x(t) has the form zt ) x(t) = ARe(e (23.C.3) where z is a number (possibly complex) and A is a real number. Then dx = zAezt (23.C.4) dt d 2 x = z2 Aezt (23.C.5) dt2 We now substitute Eqs. (23.C.3), (23.C.4), and (23.C.5), into Eq. (23.C.1) resulting in 23-48 b k z2 Aezt + zAezt + Aezt = 0 . (23.C.6) m m Collecting terms in Eq. (23.C.6) yields ⎛ b k ⎞ z2 + z + ⎠ ⎟ Aezt = 0 (23.C.7) ⎝ ⎜ m m The condition for the solution is that b k z2 + z + = 0 . (23.C.8) m m This quadratic equation has solutions −(b / m) ± ((b / m)2 − 4k / m)1 2 z = . (23.C.9) 2 When (b / m)2 < 4k / m , the oscillator is called underdamped, and we have two solutions for z , however the solutions are complex numbers. Let γ = (k / m − (b / 2m)2)1 2 ; (23.C.10) and α = b 2m . (23.C.11) . Recall that the imaginary number i = −1 . The two solutions are then z1 = −α + iγ t and z2 = −α − iγ t . Because our system is linear, our general solution is a linear combination of these two solutions, −α−iγ t −αt x(t) = A 1e−α +iγ t + A 2e = ( A 1eiγ t + A 2e−iγ t )e , (23.C.12) where A 1 and A2 are constants. We shall transform this expression into a more familiar equation involving sine and cosine functions with help from the Euler formula, ±iγ t e = cos(γ t) ± isin(γ t). (23.C.13) Therefore we can rewrite our solution as −αt x(t) = ( A 1(cos(γ t) + isin(γ t)) + A2(cos(γ t) − isin(γ t)))e . (23.C.14) A little rearrangement yields 23-49 −αt x(t) = (( A 1 + A2)cos(γ t) + i( A 1 − A2)sin(γ t))e . (23.C.15) Define two new constants C = A 1 + A 2 and D = i( A 1 − A 2) . Then our solution looks like x(t) = (C cos(γ t) + Dsin(γ t))e−αt . (23.C.16) Recall from Example 23.5 that we can rewrite C cos(γ t) + Dsin(γ t) = x m cos(γ t +φ) (23.C.17) , where x m = (C 2 + D2)1 2 , and φ = tan−1(D / C) . Then our general solution for the underdamped case (Eq. (23.C.16)) can be written as x(t) = x m e−αt cos(γ t +φ) . (23.C.18) There are two other possible cases which we shall not analyze: when (b / m)2 > 4k / m , a case referred to as overdamped, and when (b / m)2 = 4k / m , a case referred to as critically damped. Appendix 23D: Solution to the Forced Damped Oscillator Equation We shall now use complex numbers to solve the differential equation d 2 x dx F 0 cos(ωt) = m + b + kx . (23.D.1) dt2 dt We begin by assuming a solution of the form x(t) = x0 cos(ωt +φ) . (23.D.2) where the amplitude x0 and the phase constant φ need to be determined. We begin by defining the complex function i(ωt+φ ) z(t) = x0e . (23.D.3) Our desired solution can be found by taking the real projection x(t) = Re(z(t)) = x0 cos(ωt +φ) . (23.D.4) 23-50 Our differential equation can now be written as iωt d 2 z dz F 0e = m + b + kz . (23.D.5) dt2 dt We take the first and second derivatives of Eq. (23.D.3), dz i(ωt+φ ) (t) = iω x0e = iω z . (23.D.6) dt d 2 z i(ωt+φ ) (t) = −ω 2 x0e = −ω 2 z . (23.D.7) dt2 We substitute Eqs. (23.D.3), (23.D.6), and (23.D.7) into Eq. (23.D.5) yielding iωt i(ωt+φ ) F 0e = (−ω 2m + biω + k)z = (−ω 2m + biω + k)x0e . (23.D.8) We divide Eq. (23.D.8) through by eiωt and collect terms using yielding iφ F 0 / m x0e = . (23.D.9) ((ω0 2 −ω 2) + i(b / m)ω ) where we have used ω0 2 = k / m . Introduce the complex number z1 = (ω0 2 −ω 2) + i(b / m)ω . (23.D.10) Then Eq. (23.D.9) can be written as F 0 x0eiφ = . (23.D.11) my Multiply the numerator and denominator of Eq. (23.D.11) by the complex conjugate z1 = (ω0 2 −ω 2) − i(b / m)ω yielding iφ F 0 z1 F 0 ((ω0 2 −ω 2) − i(b / m)ω ) x0e = = ≡ u + iv . (23.D.12) mz1z1 m ((ω0 2 −ω 2)2 + (b / m)2ω 2) where 2 −ω 2) F 0 (ω0 u = , (23.D.13) m ((ω0 2 −ω 2)2 + (b / m)2ω 2) F 0 (b / m)ω v = − . (23.D.14) m ((ω0 2 −ω 2)2 + (b / m)2ω 2) 23-51 Therefore the modulus x0 is given by / m 2 )1/2 F 0 2 + v x0 = (u = ((ω0 2 − ω 2 )2 + (b / m)2ω 2 ) , (23.D.15) and the phase is given by −(b / m)ω φ = tan−1(v / u) = 2 − ω 2 ) (ω0 . (23.D.16) 23-52 MIT OpenCourseWare 8.01 Classical Mechanics Spring 2022 For information about citing these materials or our Terms of Use, visit: |
11032 | https://levjake.wordpress.com/2013/11/25/the-rsk-correspondence/ | The RSK Correspondence « Jake's Blog
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This is another tableau combinatorics post. The Robinson-Schensted-Knuth correspondence, or RSK for short, is another important theorem and algorithm in tableau combinatorics; I’ll discuss it here, though I won’t try to motivate it. For now, it will be something of a curiosity. As always, my reference is Fulton’s book Young Tableaux.
First, recall the definitions of partitions, Young diagrams and Young tableaux:
Definition. A partition is a finite, weakly decreasing sequence of positive integers, say ). A Young diagram is a pictorial representation of a partition, as a grid of left-aligned boxes, with boxes in the first row, in the second, and so on.
Definition. A Young Tableau is a Young diagram with positive integers entered in the boxes, nondecreasing in both the rows and columns. A semi-standard or column-strict Young Tableau is one where the columns are strictly increasing.
For RSK, we’ll also be interested in the weight of our tableaux: if has 1s, 2s, and so on, the weight of is the sequence
So, let’s let denote the set of semi-standard Young tableaux of shape and weight . These sets show up elsewhere in algebraic combinatorics (coming soon!). Their cardinalities are known as the Kostka numbers. The RSK correspondence will be useful later on: it’ll show us that a certain pair of change-of-basis matrices coincide.
Without further ado:
Theorem (RSK correspondence). There is a bijection between the following:
Ordered pairs of tableaux , of the same shape, with weights , and
Matrices with nonnegative integer entries, with column sums given by and row sums given by
In this bijection, we refer to as the “bumping tableau” and as the “recording tableau”. We let denote the corresponding set of matrices.
Notationally, we’re giving a bijection:
where is the set of matrices with nonnegative integer matrices.
Various consequences follow from the RSK correspondence, largely because integer matrices are easier to understand and picture than Young tableaux. For example: the transpose operation gives a bijection (an involution, in fact)
,
swapping the row and column sums. What does this correspond to on the tableau side? Hopefully it corresponds to swapping the bumping and recording tableau. This will be something of a surprise, as we’ll see, since the correspondence works by an algorithm that treats the two tableaux completely differently from each other.
The RSK Correspondence
Here’s how the algorithm works: we’ll make use of the row bumping algorithm from my last post. Given a matrix with nonnegative integer entries, we’ll use row-bumping to construct a tableau . Simultaneously, we’ll “record” the algorithm step-by-step in another tableau .
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Here’s how it works: we read through the matrix, in standard English order (left to right, top to bottom). The entry tells us to bump in boxes into , each containing the entry . Each time we do this, the size of grows by one box, somewhere along its outer edge. We “record” this growth by adding one box to the recording tableau, , in the same location, and the recording box gets the entry .
Observation. By construction, and have the same shape. Moreover, the first row sum of tells us how many 1s are in ; the first column sum tells us how many 1s are in . (Similarly for the other rows and columns.)
The only mysterious part of the process is that the recording tableau is, in fact, semistandard. Note that the recording tableau receives a bunch of 1s as we go along the first row of , then a bunch of 2s, and so on, so its entries get filled in (weakly) increasingly. So, certainly the row condition is satisfied. The column-strictness condition is less obvious to check (relying on tracking what happens if we bump in a sequence of boxes with increasing, or decreasing, entries.)
As with other combinatorial theorems, a lot of the hard work went into correctly formulating the problem.
Some consequences
The general statement I gave above is certainly nice, and the algorithm is very clever! But there’s more: if we specialize it to various cases, we’ll get some cool combinatorial facts.
First, suppose we restrict the integer matrix to have row and column sums equal to 1. (This means our matrix is square, and that it should be a permutation matrix.) On the tableau side, this means we will end up with standard tableaux, where both the rows and columns are strictly increasing, and each number from to occurs exactly once. On the matrix side, there are exactly permutation matrices, so RSK specializes to the formula:
The connection to the symmetric group is very strong here. Namely, we’ll later on use standard tableaux to construct irreducible representations of , of dimension , for each with boxes. Then, the above sum becomes a combinatorial version of the statement that, for any finite group with irreducible representations ,
In the same vein of thought, if we just restrict to having row sums equal to 1, and we restrict the matrix to be , then there are such matrices. On the tableau side, we’re getting a recording tableau of weight (with s), so the recording matrix is a standard tableau. The bumping tableau can be anything with boxes, but its entries can only range from to . So, if we let be the set of SSYTs of shape and entries ranging from to , then
There are other identities and relationships we can find. For instance, since the transpose corresponds to swapping the recording and bumping tableaux, symmetric matrices correspond to individual tableaux, i.e. the set .
One last observation is that permuting the rows and columns of our integer matrix corresponds to reordering the weights. Since this clearly doesn’t change the number of admissible integer matrices, we get the following:
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Theorem. For any permutation , the sets and have the same cardinality. That is, the number of semistandard tableaux of shape and weight doesn’t depend on the ordering of .
This is quite a surprise, since SSYTs are not “symmetric” in any obvious way with respect to reordering the alphabet. It’s certainly not clear that there are as many ways to fill a tableau with, say, , (one 1, three 2s, five 3s) as (five 1s, one 2, three 3s). The 1s need to go in the top row, for example, while the 2s and 3s have other, different rules constraining their placement.
Still, it’s true, and there actually is a way to make this bijection clear, called the Bender-Knuth involution.
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Knuth Equivalence and Tableau ProductsNovember 25, 2013 In "Algebraic combinatorics"
Everything you wanted to know about symmetric polynomials, part IIINovember 28, 2013 In "Algebraic combinatorics"
Everything you wanted to know about symmetric polynomials, part INovember 26, 2013 In "Algebraic combinatorics"
Tags:RSK correspondence, tableau combinatorics
By Jake Levinsonin Algebraic combinatorics, Mathematics on November 25, 2013.
← Knuth Equivalence and Tableau ProductsEverything you wanted to know about symmetric polynomials, part I →
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Everything you wanted to know about symmetric polynomials, part III « Jake's Blogsays: November 28, 2013 at 8:36 pm
[…] The RSK Correspondence […]
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2. Everything you wanted to know about symmetric polynomials, part V « Jake's Blogsays: December 4, 2013 at 2:42 am
[…] already shown this (I hinted at it in the second post): this equality is none other than the the RSK correspondence. More specifically, the RSK correspondence is the above equality when written in the basis. To see […]
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Surface Area of Cone
This document provides instructions for an activity to calculate the lateral surface area and total surface area of a right circular cone experimentally. Students are instructed to make a co…
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MATHEMATICS CLASS : IX ACTIVITY
–
8 Page No : 1
Date :
Objective
To find the formula for the lateral surface area and total surface area of a right circular cone experimentally.
Prerequisite Knowledge
Concept of circumference, sector, formula to find the circumference of a circle.
Materials Required
A cone made of chart paper, a pair of scissors, geometry box, fevicol, cello tape. Fig. (i) Fig. (ii) Fig. (iii)
Procedure
1.
Make a cone of pink chart paper having base radius r, slant
height ‘l’
and height h. 2.
Cut the cone along slant height as shown in fig (i) as per dotted line and unroll it to get a sector, as shown in fig. (ii). Name this sector OAB. 3.
Identify the arc length AB of the sector OAB = circumference of the base of the cone. radius of the sector OAB = slant height of the cone. As shown in the fig. (ii). 4.
Cut the sector OAB into 4 small equal sectors, along dotted lines as shown in fig. (ii) and fill red colour in two sectors as shown. 5.
Arrange these small sectors to get a parallelogram as shown in fig. (iii).
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Observation
1.
Base of parallelogram = 1
length of sector OAB =
2
1
2
r
r
2
2.
Altitude of parallelogram = slant height of cone = l 3.
Area of parallelogram =
πr
× l = Curved surface area of cone. 4.
Total surface area of cone =
πrl
area of circular base of cone. =
πrl
πr
2
=
πr
(
l
r
)
Learning Outcome
Students will learn the concept to differentiate between the curved surface area and total surface area of a cone.
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11034 | https://en.wikipedia.org/wiki/List_of_parrots | Jump to content
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Contents
1 Conventions
2 Classification
3 True parrots
3.1 Family Psittacidae
3.1.1 Subfamily Psittacinae
3.1.2 Subfamily Arinae (neotropical parrots)
3.2 Family Psittaculidae
3.2.1 Subfamily Platycercinae
3.2.2 Subfamily Psittacellinae
3.2.3 Subfamily Loriinae
3.2.4 Subfamily Agapornithinae
3.2.5 Subfamily Psittaculinae
3.3 Family Psittrichasiidae
3.3.1 Subfamily Coracopsinae
3.3.2 Subfamily Psittrichasinae
4 Cockatoos
4.1 Family Cacatuidae
4.1.1 Subfamily Nymphicinae
4.1.2 Subfamily Calyptorhynchinae
4.1.3 Subfamily Cacatuinae
5 New Zealand parrots
5.1 Family Nestoridae
5.2 Family Strigopidae
6 See also
7 Notes
8 References
9 External links
List of parrots
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From Wikipedia, the free encyclopedia
Parrots, also known as psittacines (/ˈsɪtəsaɪnz/), are the 402 species of birds that make up the order Psittaciformes, found in most tropical and subtropical regions, of which 387 are extant. The order is subdivided into three superfamilies: the Psittacoidea ("true" parrots), the Cacatuoidea (cockatoos), and the Strigopoidea (New Zealand parrots). Parrots have a generally pantropical distribution with several species inhabiting temperate regions in the Southern Hemisphere as well. The greatest diversity of parrots is in South America and Australasia.
The Cacatuoidea are quite[clarification needed] distinct, having a movable head crest, a different arrangement of the carotid arteries, a gall bladder, differences in the skull bones, and lack the Dyck texture feathers that—in the Psittacoidea—scatter light to produce the vibrant colours of so many parrots. Lorikeets were previously regarded as a family, Loriidae,: 45 but are now considered a tribe (Loriini) within the subfamily Loriinae, family Psittaculidae. Some species, such as the Puerto Rican amazon (Amazona vittata) have had a population bottleneck (in this case reduced to 13 individuals in 1975) and subsequently have low genetic variability and low reproductive success, leading to complications with conservation.
No consensus existed regarding the taxonomy of Psittaciformes until recently. The placement of the Strigopoidea species has been variable in the past. They were once considered part of the Psittacoidea, but recent 21st-century studies place this group of New Zealand species as their own superfamily next to the Cacatuoidea and remaining members of the Psittacoidea. Many studies have confirmed the unique placement of this group at the base of the parrot tree. Most authors now recognize this group as a separate taxon containing two families: Nestoridae and Strigopidae. Conversely, the relationships among various cockatoo genera are largely resolved.
Conventions
[edit]
IUCN Red List categories
| Conservation status |
| EX | Extinct (15 species) |
| EW | Extinct in the wild (1 species) |
| CR | Critically endangered (17 species) |
| EN | Endangered (38 species) |
| VU | Vulnerable (54 species) |
| NT | Near threatened (59 species) |
| LC | Least concern (219 species) |
| Other categories |
| DD | Data deficient (0 species) |
| NE | Not evaluated (5 species) |
Conservation status codes listed follow the International Union for Conservation of Nature (IUCN) Red List of Threatened Species. Range maps are provided wherever possible; if a range map is not available, a description of the bird's range is provided. Ranges are based on the IUCN red list for that species unless otherwise noted. All extinct species listed went extinct after 1500 CE (recently extinct), and are indicated by a dagger symbol "†".
Classification
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The order Psittaciformes consists of 387 extant species belonging to 87 genera. The following classification is based on the most recent proposals as of 2012.
Superfamily Psittacoidea: true parrots
Family Psittacidae
Subfamily Psittacinae: two African genera, Psittacus and Poicephalus
Subfamily Arinae
Tribe Arini: eighteen genera
Tribe Androglossini: seven genera
Family Psittaculidae
Subfamily Psittrichasinae: one species, Pesquet's parrot
Subfamily Coracopsinae: one genus with several species
Subfamily Platycercinae
Tribe Pezoporini: ground parrots and allies
Tribe Platycercini: broad-tailed parrots
Subfamily Psittacellinae: one genus (Psittacella) with several species
Subfamily Loriinae
Tribe Loriini: lories and lorikeets
Tribe Melopsittacini: one genus with one species, the budgerigar
Tribe Cyclopsittini: fig parrots
Subfamily Agapornithinae: three genera
Subfamily Psittaculinae
Tribe Polytelini: three genera
Tribe Psittaculini: Asian psittacines
Tribe Micropsittini: pygmy parrots
Superfamily Cacatuoidea: cockatoos
Family Cacatuidae
Subfamily Nymphicinae: one genus with one species, the cockatiel
Subfamily Calyptorhynchinae: the black cockatoos
Subfamily Cacatuinae
Tribe Microglossini: one genus with one species, the black palm cockatoo
Tribe Cacatuini: four genera of white, pink, and grey species
Superfamily Strigopoidea: New Zealand parrots
Family Nestoridae: two genera with two living (kea and New Zealand kaka) and several extinct species of the New Zealand region
Family Strigopidae: the flightless, critically endangered kākāpō of New Zealand
True parrots
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Family Psittacidae
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Subfamily Psittacinae
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Main article: Psittacinae
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| Genus Psittacus (grey parrots) Linnaeus, 1758 – two species | Common name | Scientific name | IUCN Red List Status | Range | Picture | --- --- | Grey parrot | P. erithacus Linnaeus, 1758 | eEN IUCN | | | | Timneh parrot | P. timneh Fraser, 1844 | eEN IUCN | | | |
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| Genus Poicephalus Swainson, 1837 – ten species | Common name | Scientific name | IUCN Red List Status | Range | Picture | --- --- | Senegal parrot | P. senegalus (Linnaeus, 1766) | LC IUCN | West Africa (excluding the Maghreb) | | | Red-bellied parrot | P. rufiventris (Rüppell, 1842) | LC IUCN | Eastern Horn of Africa, eastern Kenya, and northeast Tanzania | | | Rüppell's parrot | P. rueppellii (G. R. Gray, 1849) | LC IUCN | Northern Namibia and the coast of Angola | | | Brown-necked parrot | P. fuscicollis (Kuhl, 1820) | LC IUCN | From Gambia and southern Senegal to Ghana and Togo | | | Cape parrot | P. robustus (Gmelin, 1788) | gVU IUCN | | | | Meyer's parrot | P. meyeri (Cretzschmar, 1827) | LC IUCN | | | | Red-fronted parrot | P. gulielmi (Jardine, 1849) | LC IUCN | | | | Yellow-fronted parrot | P. flavifrons (Rüppell, 1842) | LC IUCN | Western Ethiopia | | | Brown-headed parrot | P. cryptoxanthus (Peters, 1854) | LC IUCN | Southeast African coast | | | Niam-Niam parrot | P. crassus (Sharpe, 1884) | LC IUCN | Central Central African Republic | | |
Subfamily Arinae (neotropical parrots)
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Main article: Arinae
Tribe Arini
Main article: Arini (tribe)
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| Genus Cyanoliseus Bonaparte, 1854 – one species | Common name | Scientific name | IUCN Red List Status | Range | Picture | --- --- | Burrowing parrot | C. patagonus (Vieillot, 1818) | LC IUCN | Central Argentina stretching to the coasts | | |
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| Genus Enicognathus G. R. Gray, 1840 – two species | Common name | Scientific name | IUCN Red List Status | Range | Picture | --- --- | Slender-billed parakeet | E. leptorhynchus (King, 1831) | LC IUCN | Central Chilean coast (central meaning halfway between the northern and southern extremes) | | | Austral parakeet | E. ferrugineus (Müller, 1776) | LC IUCN | Eastern Patagonian coast, southern Tierra del Fuego, the Falkland Islands, South Sandwich Islands and South Georgia | | |
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| Genus Rhynchopsitta (thick-billed parrots) Bonaparte, 1854 – two species | Common name | Scientific name | IUCN Red List Status | Range | Picture | --- --- | Thick-billed parrot | R. pachyrhyncha (Swainson, 1827) | eEN IUCN | Sierra Madre Occidental | | | Maroon-fronted parrot | R. terrisi Moore, 1947 | eEN IUCN | Sierra Madre Oriental | | |
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| Genus Pyrrhura Bonaparte, 1856 – 28 species | Common name | Scientific name | IUCN Red List Status | Range | Picture | --- --- | Ochre-marked parakeet | P. cruentata (Wied-Neuwied, 1820) | gVU IUCN | Scattered populations around the southeastern coast of Brazil (north of São Paulo) | | | Maroon-bellied parakeet | P. frontalis (Vieillot, 1818) | LC IUCN | Southwest Brazil, northern Uruguay, and southern Paraguay | | | Blaze-winged parakeet | P. devillei (Massena and de Souancé, 1854) | iNT IUCN | Northern Paraguay and Southwest Brazil along the Brazil/(northern) Paraguay border | | | Crimson-bellied parakeet | P. perlata (Spix, 1824) | gVU IUCN | Central Brazil and along the Brazil/Bolivia border | | | Pearly parakeet | P. lepida (Wagler, 1832) | gVU IUCN | Northeast Brazil south of the Amazon River | | | Green-cheeked parakeet | P. molinae (Massena and de Souancé, 1854) | LC IUCN | | | | Painted parakeet | P. picta (Müller, 1776) | LC IUCN | Northeast South America, north of the Amazon river and east of the Venezuela/Colombia border | | | Sinú parakeet | P. subandina (Todd, 1917) | CR IUCN | Sinú Valley in northern Colombia near (but not bordering) the Gulf of Darién | | | Azuero parakeet | P. eisenmanni Delgado, 1985 | eEN IUCN | Southwest Azuero Peninsula | | | Venezuelan parakeet | P. emma Salvadori, 1891 | LC IUCN | Northern coast of Venezuela | | | Santarem parakeet | P. amazonum Hellmayr, 1906 | eEN IUCN | Central and southeast Amazon in Brazil | | | Madeira parakeet | P. snethlageae Joseph and Bates, 2002 | gVU IUCN | Eastern Brazil, west of the Brazil/Bolivia border, south of the Madeira River, and north of the Araguaia River | | | Bonaparte's parakeet | P. lucianii (Deville, 1851) | LC IUCN | Northwest Amazon rainforest in Brazil | | | Rose-fronted parakeet | P. roseifrons (Gray, 1859) | LC IUCN | Along the Peru/Brazil border and Peru/Bolivia border | | | Wavy-breasted parakeet | P. peruviana Hocking, Blake, and Joseph, 2002 | LC IUCN | Along the Marañón river in northern Peru and Ecuador | | | White-eared parakeet | P. leucotis (Kuhl, 1820) | gVU IUCN | Southeast Brazilian coast, north of Rio de Janeiro and south of Salvador de Bahia | | | Grey-breasted parakeet | P. griseipectus Salvadori, 1900 | eEN IUCN | Scattered populations in northern Ceará, a state in Brazil | | | Pfrimer's parakeet | P. pfrimeri de Miranda-Ribeiro, 1920 | eEN IUCN | Dry forest near the Serra Geral | | | Fiery-shouldered parakeet | P. egregia (Sclater, 1881) | LC IUCN | Venezuela west of the Rio Caroni, and Guyana east of the Mazaruni River | | | Santa Marta parakeet | P. viridicata Todd, 1913 | eEN IUCN | Sierra Nevada de Santa Marta | | | Maroon-tailed parakeet | P. melanura (Spix, 1824) | LC IUCN | Northwest South America, east of the Andes Mountains | | | El Oro parakeet | P. orcesi Ridgely & Robbins, 1988 | eEN IUCN | El Oro Province in Ecuador | | | Black-capped parakeet | P. rupicola (Tschudi, 1844) | iNT IUCN | Around the Brazil/Peru/Bolivia border within the State of Acre and west of the Andes mountains | | | White-breasted parakeet | P. albipectus Chapman, 1914 | gVU IUCN | Zamora-Chinchipe Province | | | Flame-winged parakeet | P. calliptera (Massena and de Souancé, 1854) | gVU IUCN | Andes mountains in Colombia | | | Blood-eared parakeet | P. hoematotis de Souancé, 1857 | LC IUCN | Northern coast of Venezuela | | | Rose-crowned parakeet | P. rhodocephala (Sclater and Salvin, 1871) | LC IUCN | Andes mountains in Venezuela, near Lake Maracaibo | | | Sulphur-winged parakeet | P. hoffmanni (Cabanis, 1861) | LC IUCN | Continental Divide running through Costa Rica and Panama | | |
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| Genus Anodorhynchus (blue macaws) von Spix, 1824 – three species | Common name | Scientific name | IUCN Red List Status | Range | Picture | --- --- | Glaucous macaw | A. glaucus (Vieillot, 1816) | CR IUCN | A small population in northeast Argentina near the Argentina/Brazil/Paraguay border | | | Hyacinth macaw | A. hyacinthinus (Latham, 1790) | gVU IUCN | | | | Lear's macaw | A. leari Bonaparte, 1856 | eEN IUCN | | | |
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| Genus Leptosittaca von Berlepsch and Stolzmann, 1894 – one species | Common name | Scientific name | IUCN Red List Status | Range | Picture | --- --- | Golden-plumed parakeet | L. branickii von Berlepsch and Stolzmann, 1894 | gVU IUCN | Andes mountains in Peru, Colombia, and near the Ecuador/Peru and Ecuador/Colombia borders | | |
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| Genus Ognorhynchus Bonaparte, 1857 – one species | Common name | Scientific name | IUCN Red List Status | Range | Picture | --- --- | Yellow-eared parrot | O. icterotis (Massena and de Souancé, 1854) | eEN IUCN | Along the Chocó Department/Antioquia Department border | | |
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| Genus Diopsittaca Ridgway, 1912 – two species | Common name | Scientific name | IUCN Red List Status | Range | Picture | --- --- | Southern red-shouldered macaw | D. cumanensis (Lichtenstein, 1823) | LC IUCN | Brazil, south of the Amazon rainforest | | | Northern Red-shouldered Macaw | D. nobilis (Linnaeus, 1758) | LC IUCN | Northern coasts of French Guiana and Suriname, northern Guyana, and eastern Venezuela | | |
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| Genus Guaruba Lesson, 1830 – one species | Common name | Scientific name | IUCN Red List Status | Range | Picture | --- --- | Golden parakeet | G. guarouba (Gmelin, 1788) | gVU IUCN | Southern border of the Amazon rainforest in Brazil, but regionally extinct near the coast | | |
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| Genus †Conuropsis Salvadori, 1891 – one species | Common name | Scientific name | IUCN Red List Status | Range | Picture | --- --- | Carolina parakeet | †C. carolinensis (Linnaeus, 1758) | aEX IUCN | Formerly the eastern and southern United States | | |
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| Genus Cyanopsitta Bonaparte, 1854 – one species | Common name | Scientific name | IUCN Red List Status | Range | Picture | --- --- | Spix's macaw | C. spixii (Wagler, 1832) | jEW IUCN | Central eastern Brazil | | |
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| Genus Orthopsittaca Ridgway, 1912 – one species | Common name | Scientific name | IUCN Red List Status | Range | Picture | --- --- | Red-bellied macaw | O. manilatus (Boddaert, 1783) | LC IUCN | | | |
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| Genus Ara de Lacépède, 1799 – nine species | Common name | Scientific name | IUCN Red List Status | Range | Picture | --- --- | Great green macaw | A. ambiguus (Bechstein, 1811) | EN IUCN | | | | Blue-and-yellow macaw | A. ararauna (Linnaeus, 1758) | LC IUCN | | | | Red-and-green macaw | A. chloropterus Gray, 1859 | LC IUCN | | | | Blue-throated macaw | A. glaucogularis Dabbene, 1921 | CR IUCN | | | | Scarlet macaw | A. macao (Linnaeus, 1758) | LC IUCN | | | | Military macaw | A. militaris (Linnaeus, 1766) | gVU IUCN | | | | Red-fronted macaw | A. rubrogenys de Lafresnaye, 1847 | CR IUCN | | | | Chestnut-fronted macaw | A. severus (Linnaeus, 1758) | LC IUCN | | | | Cuban macaw | †A. tricolor (Bechstein, 1811) | aEX IUCN | Formerly Cuba and Isla de la Juventud until 1885 | | |
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| Genus Primolius Bonaparte, 1857 – three species | Common name | Scientific name | IUCN Red List Status | Range | Picture | --- --- | Golden-collared macaw | P. auricollis (Cassin, 1853) | LC IUCN | | | | Blue-headed macaw | P. couloni (Sclater, 1876) | gVU IUCN | Southern Peru east of the Andes, and the along the northern Bolivia/Brazil and Bolivia/Peru borders | | | Blue-winged macaw | P. maracana (Vieillot, 1816) | iNT IUCN | | | |
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| Genus Aratinga von Spix, 1824 – six species | Common name | Scientific name | IUCN Red List Status | Range | Picture | --- --- | Golden-capped parakeet | A. auricapillus (Kuhl, 1820) | iNT IUCN | Scattered populations in southeast Brazil, north of São Paulo | | | Jandaya parakeet | A. jandaya (Gmelin, 1788) | LC IUCN | Eastern Brazil | | | Sulphur-breasted parakeet | A. maculata (Müller, 1776) | LC IUCN | Scattered populations in Pará and Amapá (states in Brazil) | | | Nanday parakeet | A. nenday (Vieillot, 1823) | LC IUCN | Central Paraguay, spilling over to the southeastern Bolivia/Brazil border | | | Sun parakeet | A. solstitialis (Linnaeus, 1758) | eEN IUCN | Northwest Roraima (a state in Brazil), and along the Brazil/Venezuela/Guyana border | | | Dusky-headed parakeet | A. weddellii (Deville, 1851) | LC IUCN | Eastern South America, east of the Andes, south of the Colombia/Peru border, and north of central Bolivia | | |
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| Genus Eupsittula Bonaparte, 1853 – six species | Common name | Scientific name | IUCN Red List Status | Range | Picture | --- --- | Aztec parakeet | E. astec (de Souancé, 1857) | LC IUCN | Middle America along the Caribbean Sea | | | Peach-fronted parakeet | E. aurea (Gmelin, 1788) | LC IUCN | Brazil (excluding northwest Brazil), eastern Bolivia, and central Paraguay | | | Caatinga parakeet | E. cactorum (Kuhl, 1820) | LC IUCN | The Caatinga region in Brazil | | | Orange-fronted parakeet | E. canicularis (Linnaeus, 1758) | LC IUCN | Middle America bordering the Pacific Ocean (excluding the Gulf of California), and introduced to Puerto Rico | | | Jamaican parakeet | E. nana (Vigors, 1830) | iNT IUCN | Jamaica | | | Brown-throated parakeet | E. pertinax (Linnaeus, 1758) | LC IUCN | Northern South America and Panama, introduced to the British Virgin Islands and Dominica | Mainland phenotype left, Antillean phenotype right | |
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| Genus Psittacara Vigors, 1837 – 12 species | Common name | Scientific name | IUCN Red List Status | Range | Picture | --- --- | Blue-crowned parakeet | P. acuticaudatus (Vieillot, 1818) | LC IUCN | Northern Venezuela, southwest Bolivia, Paraguay, northern Argentina (west of the Andes and excluding the coast), and the Caatinga | | | Socorro parakeet | P. brevipes (Lawrence, 1871) | NE IUCN [a] | Socorro Island | | | Hispaniolan parakeet | P. chloropterus de Souancé, 1856 | gVU IUCN | Scattered areas in Hispaniola | | | Red-masked parakeet | P. erythrogenys Lesson, 1844 | iNT IUCN | Ecuador and northern Peru, west of the Andes mountains | | | Cuban parakeet | P. euops (Wagler, 1832) | gVU IUCN | Scattered areas in Cuba | | | Finsch's parakeet | P. finschi (Salvin, 1871) | LC IUCN | Costa Rica, Nicaragua, and western Panama (along the Pacific Ocean) | | | Cordilleran parakeet | P. frontatus (Cabanis, 1846) | iNT IUCN | Peru west of the Andes mountains | | | Green parakeet | P. holochlorus (Sclater, 1859) | LC IUCN | Mexico east of the Sierra Madre Oriental and west of the Sierra Madre Occidental, and Honduras and Guatemala bordering the Pacific Ocean | | | Guadeloupe parakeet | †P. labati (Rothschild, 1905) | aEX IUCN | Formerly Guadeloupe until the mid-18th century | | | White-eyed parakeet | P. leucophthalmus (Müller, 1776) | LC IUCN | Central South America and the northeast coast | | | Puerto Rican parakeet | †P. maugei Souancé, 1856 | NE IUCN [b] | Formerly Puerto Rico until the mid-19th century | | | Mitred parakeet | P. mitratus (von Tschudi, 1844) | LC IUCN | Andes mountains in Bolivia, southern Peru, and northern Argentina | | | Red-throated parakeet | P. rubritorquis (Sclater, 1887) | LC IUCN | Honduras, southern Guatemala, and northern Nicaragua, excluding the coasts | | | Pacific parakeet | P. strenuus (Ridgway, 1915) | NE IUCN [c] | western Nicaragua | A green parrot | | Scarlet-fronted parakeet | P. wagleri (Gray, 1845) | iNT IUCN | Andes mountains in Colombia and Venezuela | | |
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| Genus Pionites (caiques) Heine, 1890 – four species | Common name | Scientific name | IUCN Red List Status | Range | Picture | --- --- | Green-thighed parrot | P. leucogaster (Kuhl, 1820) | eEN IUCN | North central Brazil | | | Black-legged parrot | P. xanthomerius (Sclater, 1858) | LC IUCN | Northern Bolivia, eastern Peru, and western Brazil around the Peru/Brazil and Bolivia/Brazil borders | | | Yellow-tailed parrot | P. xanthurus (Todd, 1925) | gVU IUCN | Northwest Brazil | | | Black-headed parrot | P. melanocephalus (Linnaeus, 1758) | LC IUCN | | | |
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| Genus Deroptyus Wagler, 1832 – one species | Common name | Scientific name | IUCN Red List Status | Range | Picture | --- --- | Red-fan parrot | D. accipitrinus (Linnaeus, 1758) | LC IUCN | | | |
Tribe Forpini
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| Genus Forpus Bonaparte, 1857 – eight species | Common name | Scientific name | IUCN Red List Status | Range | Picture | --- --- | Pacific parrotlet | F. coelestis (Lesson, 1847) | LC IUCN | Ecuador and northern Peru, west of the Andes mountains | | | Spectacled parrotlet | F. conspicillatus (de Lafresnaye, 1848) | LC IUCN | Northern Colombia and eastern Panama | | | Mexican parrotlet | F. cyanopygius (de Souancé, 1856) | iNT IUCN | Sierra Madre Occidental | | | Dusky-billed parrotlet | F. modestus (Cabanis, 1849) | LC IUCN | Northern Bolivia, northern and eastern Peru, southern Colombia, western Brazil, and the mouth of the Amazon river | | | Green-rumped parrotlet | F. passerinus (Linnaeus, 1758) | LC IUCN | Northern Atlantic coast of South America, and introduced to Barbados and Jamaica | | | Turquoise-winged parrotlet | Forpus spengeli (Hartlaub, 1885) | LC IUCN | Pacific coast of Colombia | | | Yellow-faced parrotlet | F. xanthops (Salvin, 1895) | gVU IUCN | Marañón Valley, Peru | | | Cobalt-rumped parrotlet | F. xanthopterygius (von Spix, 1824)[d] | LC IUCN | Eastern Brazil, central Bolivia, and Peru east of the Andes mountains | | | Riparian parrotlet | F. crassirostris (Taczanowski, 1883) | NE | Southeastern Colombia to western Brazil and northern Peru | | |
Tribe Amoropsittacini
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| Genus Psilopsiagon Ridgway, 1912 – two species | Common name | Scientific name | IUCN Red List Status | Range | Picture | --- --- | Mountain parakeet | P. aurifrons (Lesson, 1831) | LC IUCN | Andes mountains in Peru, Bolivia, and Argentina | | | Gray-hooded parakeet | P. aymara (d'Orbigny, 1839) | LC IUCN | Andes mountains in Bolivia and Argentina | | |
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| Genus Bolborhynchus Bonaparte, 1857 – three species | Common name | Scientific name | IUCN Red List Status | Range | Picture | --- --- | Rufous-fronted parakeet | B. ferrugineifrons (Lawrence, 1880) | gVU IUCN | Andes mountains in southern Colombia | | | Barred parakeet | B. lineola (Cassin, 1853) | LC IUCN | Andes mountains excluding those in Chile and Argentina, and the mountains of Central America | | | Andean parakeet | B. orbygnesius (de Souancé, 1856) | LC IUCN | Andes mountains of Bolivia and Peru | | |
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| Genus Nannopsittaca Ridgway, 1912 – two species | Common name | Scientific name | IUCN Red List Status | Range | Picture | --- --- | Manu parrotlet | N. dachilleae O'Neill, Munn, and Franke, 1991 | iNT IUCN | Southern Peru, north of the Andes mountains, along the Peru/Bolivia border | | | Tepui parrotlet | N. panychlora (Salvin and Godman, 1883) | LC IUCN | Scattered in southern Venezuela | on top | |
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| Genus Touit Gray, 1855 – eight species | Common name | Scientific name | IUCN Red List Status | Range | Picture | --- --- | Lilac-tailed parrotlet | T. batavicus (Boddaert, 1783) | LC IUCN | Northern Guyana, Suriname, and French Guiana | | | Scarlet-shouldered parrotlet | T. huetii (Temminck, 1830) | gVU IUCN | Andes mountains in Peru and Ecuador, western Venezuela, northern Guyana, and north central Brazil | | | Red-fronted parrotlet | T. costaricensis (Cory, 1913) | gVU IUCN | Mountains of Costa Rica and western Panama | | | Blue-fronted parrotlet | T. dilectissimus (Sclater and Salvin, 1871) | LC IUCN | Ecuador and Colombia west of the Andes mountains | | | Brown-backed parrotlet | T. melanonotus (Wied-Neuwied, 1820) | eEN IUCN | Southeast coast of Brazil | | | Sapphire-rumped parrotlet | T. purpuratus (Gmelin, 1788) | LC IUCN | Northeast South America, stretching to the Peru/Colombia/Ecuador border | | | Spot-winged parrotlet | T. stictopterus (Sclater, 1862) | gVU IUCN | Andes mountains in Ecuador | | | Golden-tailed parrotlet | T. surdus (Kuhl, 1820) | gVU IUCN | Coast of Brazil, excluding the northeast coast | | |
Tribe Androglossini
Main article: Androglossini
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| Genus Pionopsitta Bonaparte, 1854 – one species | Common name | Scientific name | IUCN Red List Status | Range | Picture | --- --- | Pileated parrot | P. pileata (Scopoli, 1769) | LC IUCN | Southern Paraguay, and the southeast coast of Brazil | | |
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| Genus Triclaria Wagler, 1832 – one species | Common name | Scientific name | IUCN Red List Status | Range | Picture | --- --- | Blue-bellied parrot | T. malachitacea (von Spix, 1824) | iNT IUCN | Fragmented populations near Rio de Janeiro, São Paulo, and Rio Grande do Sul | | |
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| Genus Pyrilia Bonaparte, 1856 – seven species | Common name | Scientific name | IUCN Red List Status | Range | Picture | --- --- | Bald parrot | P. aurantiocephala (Gaban-Lima, Raposo, and Hofling, 2002) | iNT IUCN | Lower Madeira and Upper Tapajós rivers | | | Orange-cheeked parrot | P. barrabandi (Kuhl, 1820) | iNT IUCN | Northwest South America east of the Andes mountains | | | Caica parrot | P. caica (Latham, 1790) | iNT IUCN | Roraima, Pará, French Guiana, Suriname, Guyana, and eastern Venezuela | | | Brown-hooded parrot | P. haematotis (Sclater and Salvin, 1860) | LC IUCN | Central America, excluding the mountain ranges | | | Rose-faced parrot | P. pulchra (von Berlepsch, 1897) | LC IUCN | Ecuador and Colombia west of the Andes mountains | | | Saffron-headed parrot | P. pyrilia (Bonaparte, 1853) | iNT IUCN | Andes mountains in Colombia and western Venezuela | | | Vulturine parrot | P. vulturina (Kuhl, 1820) | gVU IUCN | Pará south of the Amazon river | | |
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| Genus Pionus Wagler, 1832 – 8 species | Common name | Scientific name | IUCN Red List Status | Range | Picture | --- --- | Bronze-winged parrot | P. chalcopterus (Fraser, 1841) | LC IUCN | Andes mountains in Ecuador, Colombia, and Venezuela | | | Dusky parrot | P. fuscus (Müller, 1776) | LC IUCN | Northeast South America and along the Colombia/Venezuela border west of Lake Maracaibo | | | Scaly-headed parrot | P. maximiliani (Kuhl, 1820) | LC IUCN | Paraguay, southern Brazil, and east and central Bolivia | | | Blue-headed parrot | P. menstruus (Linnaeus, 1766) | LC IUCN | Northern and central South America, excluding the Andes | | | White-crowned parrot | P. senilis (von Spix, 1824) | LC IUCN | Central America bordering the Caribbean Sea and the Sierra Madre Oriental | | | White-capped parrot | P. seniloides (Massena and de Souancé, 1854) | LC IUCN | Andes mountains in Venezuela, Colombia, Ecuador, and northern Peru | | | Red-billed parrot | P. sordidus (Linnaeus, 1758) | LC IUCN | Andes mountains in Bolivia, Venezuela, Ecuador, northern Peru, and southern Colombia | | | Speckle-faced parrot | P. tumultuosus (von Tschudi, 1844) | LC IUCN | Andes mountains in southern Peru and northern Bolivia | | |
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| Genus Graydidascalus Bonaparte, 1854 – one species | Common name | Scientific name | IUCN Red List Status | Range | Picture | --- --- | Short-tailed parrot | G. brachyurus (Temminck & Kuhl, 1820) | LC IUCN | Along the Amazon river | | |
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| Genus Alipiopsitta Caparroz and Pacheco, 2006 – one species | Common name | Scientific name | IUCN Red List Status | Range | Picture | --- --- | Yellow-faced parrot | A. xanthops (von Spix, 1824) | iNT IUCN | Central Brazil south of the Amazon | | |
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| Genus Amazona (Amazon parrots) Lesson, 1830 – 35 species | Common name | Scientific name | IUCN Red List Status | Range | Picture | --- --- | Turquoise-fronted amazon | A. aestiva (Linnaeus, 1758) | LC IUCN | Paraguay, western Bolivia, northern Argentina, southern Brazil | | | Black-billed amazon | A. agilis (Linnaeus, 1758) | VU IUCN | Central Jamaica | | | White-fronted amazon | A. albifrons (Sparrman, 1788) | LC IUCN | Northwest Costa Rica, western Nicaragua, western Honduras, Western Mexico, and the Yucatán Peninsula | | | Orange-winged amazon | A. amazonica (Linnaeus, 1766) | LC IUCN | Northern South America and the southeast coast of Brazil | | | Red-necked amazon | A. arausiaca (Müller, 1776) | gVU IUCN | Central Dominica | | | Yellow-naped amazon | A. auropalliata (Lesson, 1842) | gVU IUCN | Pacific coast of Guatemala, Honduras, Nicaragua, northern Costa Rica, and southern Mexico | | | Red-lored amazon | A. autumnalis (Linnaeus, 1758) | LC IUCN | Central America, northern Colombia, and the Sierra Madre Oriental | | | Yellow-shouldered amazon | A. barbadensis (Gmelin, 1788) | gVU IUCN | Margarita Island, Blanquilla Island, Curaçao, Bonaire, and northern Venezuela | | | Red-tailed amazon | A. brasiliensis (Linnaeus, 1758) | gVU IUCN | Southern coast of São Paulo | | | Yellow-billed amazon | A. collaria (Linnaeus, 1758) | gVU IUCN | Jamaica | | | Diademed amazon | A. diadema (von Spix, 1824) | eEN IUCN | Northeast Amazonas | | | Blue-cheeked amazon | A. dufresniana (Shaw, 1812) | iNT IUCN | Suriname, French Guiana, Guyana, and east Venezuela | | | Festive amazon | A. festiva (Linnaeus, 1758) | iNT IUCN | Along the Amazon river | | | Lilac-crowned amazon | A. finschi (Sclater, 1864) | eEN IUCN | Pacific coast of Mexico | | | Mealy amazon | A. farinosa (Boddaert, 1783) | iNT IUCN | Central and South America | | | Saint Vincent amazon | A. guildingii (Vigors, 1837) | VU IUCN | Northeast St Vincent and the Grenadines | | | Imperial amazon | A. imperialis Richmond, 1899 | eEN IUCN | Dominica | | | Kawall's amazon | A. kawalli Grantsau and Camargo, 1989 | iNT IUCN | Southern Amazonas and eastern Pará | | | Cuban amazon | A. leucocephala (Linnaeus, 1758) | iNT IUCN | Cuba, the Isle of Pines, Abaco Islands, Great Inagua, and the Cayman Islands | | | Lilacine amazon | A. lilacina (Lesson, 1844) | eEN IUCN | Ecuador west of the Andes mountains | | | Martinique amazon | †A. martinicana Clark, 1905 | aEX IUCN | Formerly Martinique until 1779 | | | Scaly-naped amazon | A. mercenarius (von Tschudi, 1844) | LC IUCN | Andes mountains in Venezuela, Colombia, Ecuador, Peru, and northern Bolivia | | | Yellow-crowned amazon | A. ochrocephala (Gmelin, 1788) | LC IUCN | Northern South America, east of the Andes | | | Yellow-headed amazon | A. oratrix Ridgway, 1887 | eEN IUCN | Scattered coastal areas of southern Mexico | | | Red-spectacled amazon | A. pretrei (Temminck, 1830) | gVU IUCN | Scattered populations in central Rio Grande do Sul | | | Red-browed amazon | A. rhodocorytha (Salvadori, 1890) | eEN IUCN | Scattered populations along the southeast coastline of Brazil | | | Tucumán amazon | A. tucumana (Cabanis, 1885) | gVU IUCN | Andes mountains in northern Argentina and southern Bolivia | | | Hispaniolan amazon | A. ventralis (Müller, 1776) | gVU IUCN | Hispaniola, Grande Cayemite, Gonve, Beata, and Saona Island, and introduced to Puerto Rico, St Croix, and St. Thomas Island | | | Saint Lucia amazon | A. versicolor (Müller, 1776) | gVU IUCN | St Lucia | | | Vinaceous-breasted amazon | A. vinacea (Kuhl, 1820) | eEN IUCN | | | | Guadeloupe amazon | †A. violacea (Gmelin, 1789) | aEX IUCN | Formerly Guadeloupe until 1779 | | | Red-crowned amazon | A. viridigenalis (Cassin, 1853) | eEN IUCN | Sierra Madre Occidental | | | Puerto Rican amazon | A. vittata (Boddaert, 1783) | CR IUCN | Former range left, current range right | | | Yucatán amazon | A. xantholora (Gray, 1859) | LC IUCN | Yucatán Peninsula | | |
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| Genus Hapalopsittaca Ridgway, 1912 – four species | Common name | Scientific name | IUCN Red List Status | Range | Picture | --- --- | Rusty-faced parrot | H. amazonina (des Murs, 1845) | gVU IUCN | Andes mountains in Colombia and Venezuela | | | Fuertes's parrot | H. fuertesi (Chapman, 1912) | CR IUCN | Andes mountains in central Colombia | | | Black-winged parrot | H. melanotis (de Lafresnaye, 1847) | LC IUCN | Andes mountains in northern Bolivia and central Peru | | | Red-faced parrot | H. pyrrhops (Salvin, 1876) | gEN IUCN | Andes mountains in southern Ecuador | | |
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| Genus Brotogeris Vigors, 1825 – eight species | Common name | Scientific name | IUCN Red List Status | Range | Picture | --- --- | Yellow-chevroned parakeet | B. chiriri (Vieillot, 1818) | LC IUCN | Southern Brazil, eastern Bolivia and Paraguay, and introduced to California around Los Angeles and San Francisco, and southern Florida | | | Golden-winged parakeet | B. chrysoptera (Linnaeus, 1766) | LC IUCN | French Guiana, Guyana, Suriname, eastern Venezuela, and northeast Brazil | | | Cobalt-winged parakeet | B. cyanoptera (Pelzeln, 1870) | LC IUCN | Northwest South America, east of the Andes mountains | | | Orange-chinned parakeet | B. jugularis (Müller, 1776) | LC IUCN | Panama, Costa Rica, Nicaragua, southern El Salvador and Guatemala, northern Colombia, and northwest Venezuela | | | Grey-cheeked parakeet | B. pyrrhoptera (Latham, 1801) | eEN IUCN | Ecuador, west of the Andes mountains | | | Tui parakeet | B. sanctithomae (Müller, 1776) | LC IUCN | Northern Bolivia, eastern Peru, and along the Amazon river in Brazil | | | Plain parakeet | B. tirica (Gmelin, 1788) | LC IUCN | Southern coast of Brazil | | | White-winged parakeet | B. versicolurus (Müller, 1776) | LC IUCN | Along the Amazon river, and introduced to California around Los Angeles and San Francisco | | |
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| Genus Myiopsitta Bonaparte, 1854 – two species | Common name | Scientific name | IUCN Red List Status | Range | Picture | --- --- | Cliff parakeet | M. luchsi (Finsch, 1868) | LC IUCN | Andes mountains in Bolivia | | | Monk parakeet | M. monachus (Boddaert, 1783) | LC IUCN | Paraguay, Uruguay, and eastern Argentina, introduced to Puerto Rico, Guadeloupe, Japan, Virgin Islands, the southern and eastern United States, and western Europe | | |
Family Psittaculidae
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Subfamily Platycercinae
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Main article: Platycercinae
Tribe Platycercini (broad-tailed parrots)
Main article: Platycercini
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| Genus Platycercus (rosellas) Vigors, 1825 – six species | Common name | Scientific name | IUCN Red List Status | Range | Picture | --- --- | Western rosella | P. icterotis (Temminck & Kuhl, 1820) | LC IUCN | Southwest Australia | | | Crimson rosella | P. elegans (Gmelin, 1788) | LC IUCN | | | | Green rosella | P. caledonicus (Gmelin, 1788) | LC IUCN | Tasmania and the Bass Strait islands | | | Pale-headed rosella | P. adscitus (Latham, 1790) | LC IUCN | Cape York Peninsula and northeastern New South Wales | | | Eastern rosella | P. eximius (Shaw, 1792) | LC IUCN | Eastern Australia and Tasmania, naturalized in New Zealand | | | Northern rosella | P. venustus (Kuhl, 1820) | LC IUCN | Western half of the Gulf of Carpentaria to the Kimberley | | |
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| Genus Prosopeia (shining parrots) Bonaparte, 1854 – three species | Common name | Scientific name | IUCN Red List Status | Range | Picture | --- --- | Masked shining parrot | P. personata (Gray, 1848) | iNT IUCN | Viti Levu, the largest island of Fiji | | | Maroon shining parrot | P. tabuensis, (Gmelin, 1788) | LC IUCN | Vanua Levu and Taveuni, which are two islands of Fiji | | | Crimson shining parrot | P. splendens, (Peale, 1849) | gNT IUCN | Kadavu and Ono, which are part of the Kadavu Group in Fiji | | |
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| Genus Eunymphicus Peters, 1937 – two species | Common name | Scientific name | IUCN Red List Status | Range | Picture | --- --- | Horned parakeet | E. cornutus, (Gmelin, 1788) | gVU IUCN | New Caledonia | | | Ouvea parakeet | E. uvaeensis (Layard, EL & Layard, ELC, 1882) | eEN IUCN | Uvea, an island of Loyalty Islands, and New Caledonia | | |
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| Genus Cyanoramphus Bonaparte, 1854 – seven species | Common name | Scientific name | IUCN Red List Status | Range | Picture | --- --- | Yellow-crowned parakeet | C. auriceps (Kuhl, 1820) | iNT IUCN | New Zealand, Stewart Island, and Auckland Island | | | Malherbe's parakeet | C. malherbi de Souancé, 1857 | CR IUCN | South Island (of New Zealand) | | | Red-crowned parakeet | C. novaezelandiae (Sparrman, 1787) | iNT IUCN | New Zealand and several nearby islands, New Caledonia, and Norfolk Island | | | Society parakeet | †C. ulietanus (Gmelin, 1788) | aEX IUCN | Formerly Raiatea, an island of the Society Islands in French Polynesia until 1777 | | | Antipodes parakeet | C. unicolor (Lear, 1831) | gVU IUCN | Antipodes Islands of New Zealand | | | Black-fronted parakeet | †C. zealandicus (Latham, 1790) | aEX IUCN | Formerly Tahiti and French Polynesia until 1844 | | |
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| Genus Barnardius Bonaparte 1854 – one species | Common name | Scientific name | IUCN Red List Status | Range | Picture | --- --- | Australian ringneck | B. zonarius (Shaw, 1805) | LC IUCN | Australian mainland, excluding the northern and eastern coasts | | |
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| Genus Purpureicephalus Bonaparte 1854 – one species | Common name | Scientific name | IUCN Red List Status | Range | Picture | --- --- | Red-capped parrot | P. spurius (Kuhl, 1820) | LC IUCN | Southwestern tip of Australia | | |
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| Genus Lathamus Lesson 1830 – one species | Common name | Scientific name | IUCN Red List Status | Range | Picture | --- --- | Swift parrot | L. discolor Shaw, 1790 | CR IUCN | | | |
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| Genus Northiella Mathews 1912 – two species | Common name | Scientific name | IUCN Red List Status | Range | Picture | --- --- | Eastern bluebonnet | N. haematogaster (Gould, 1838) | LC IUCN | New South Wales and into Southern Australia and the southeastern corner of Western Australia | | | Naretha bluebonnet | N. narethae (HL White, 1921) | NE IUCN [e] | southeastern Western Australia, southwestern South Australia | | |
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| Genus Psephotus Gould 1845 – one species | Common name | Scientific name | IUCN Red List Status | Range | Picture | --- --- | Red-rumped parrot | P. haematonotus (Gould, 1838) | LC IUCN | New South Wales and southern Queensland | | |
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| Genus Psephotellus Mathews 1913 – four species | Common name | Scientific name | IUCN Red List Status | Range | Picture | --- --- | Mulga parrot | P. varius (Clark, 1910) | LC IUCN | Southern and central Australia | | | Hooded parrot | P. dissimilis (Collett, 1898) | LC IUCN | Northern Territory | | | Golden-shouldered parrot | P. chrysopterygius (Gould, 1858) | eEN IUCN | Cape York Peninsula | | | Paradise parrot | †P. pulcherrimus (Gould, 1845) | aEX IUCN | Formerly eastern Australia until 1928 | | |
Tribe Pezoporini Bonaparte, 1837
Main article: Pezoporini
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| Genus Neophema Salvadori 1891 – six species | Common name | Scientific name | IUCN Red List Status | Range | Picture | --- --- | Blue-winged parrot | N. chrysostoma (Kuhl, 1820) | VU IUCN | Western New South Wales, eastern Southern Australia, and a resident population in Tasmania and the southern coast of New South Wales | | | Elegant parrot | N. elegans (Gould, 1837) | LC IUCN | | | | Rock parrot | N. petrophila (Gould, 1841) | LC IUCN | Spencer Gulf, Peron Peninsula, and the southwest coast of Western Australia | | | Orange-bellied parrot | N. chrysogaster (Latham, 1790) | CR IUCN | Melaleuca, Tasmania and they spend winter in the southern coast of Victoria | | | Turquoise parrot | N. pulchella (Shaw, 1792) | LC IUCN | | | | Scarlet-chested parrot | N. splendida (Gould, 1841) | LC IUCN | Large patches in the southern half of Australia | | |
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| Genus Neopsephotus Mathews 1912 – one species | Common name | Scientific name | IUCN Red List Status | Range | Picture | --- --- | Bourke's parrot | N. bourkii (Gould, 1841) | LC IUCN | Western and South Australia, and central Australia | | |
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| Genus Pezoporus Illiger 1811 – three species | Common name | Scientific name | IUCN Red List Status | Range | Picture | --- --- | Night parrot | P. occidentalis (Gould, 1861) | eCR IUCN | Known locations since 1950 | | | Ground parrot | P. wallicus (Kerr, 1792) | LC IUCN | in orange | | |
Subfamily Psittacellinae
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Main article: Psittacellinae
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| Genus Psittacella (tiger parrots) Schlegel 1871 – four species | Common name | Scientific name | IUCN Red List Status | Range | Picture | --- --- | Madarasz's tiger parrot | P. madaraszi Meyer, 1886 | LC IUCN | New Guinea Highlands | | | Modest tiger parrot | P. modesta Schlegel, 1871 | LC IUCN | New Guinea Highlands | | | Painted tiger parrot | P. picta Rothschild, 1896 | LC IUCN | New Guinea Highlands on the Papua New Guinea side | | | Brehm's tiger parrot | P. brehmii Schlegel, 1871 | LC IUCN | New Guinea Highlands | | |
Subfamily Loriinae
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Main article: Loriinae
Tribe Loriini Selby, 1836
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| Genus Chalcopsitta Bonaparte 1850 – three species | Common name | Scientific name | IUCN Red List Status | Range | Picture | --- --- | Black lory | C. atra (Scopoli, 1786) | LC IUCN | West Papua | | | Brown lory | C. duivenbodei Dubois, 1884 | LC IUCN | Northern New Guinea | | | Yellow-streaked lory | C. scintillata (Temminck, 1835) | LC IUCN | New Guinea, south of the New Guinea Highlands | | |
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| Genus Eos Wagler 1832 – six species | Common name | Scientific name | IUCN Red List Status | Range | Picture | --- --- | Black-winged lory | E. cyanogenia Bonaparte, 1850 | gVU IUCN | Schouten Islands | | | Violet-necked lory | E. squamata (Boddaert, 1783) | LC IUCN | North Maluku | | | Blue-streaked lory | E. reticulata (Müller, 1841) | iNT IUCN | Banda Sea islands | | | Red-and-blue lory | E. histrio (Müller, 1776) | eEN IUCN | Talaud Islands | | | Red lory | E. bornea (Linnaeus, 1758) | LC IUCN | Seram Island, an island of the Maluka archipelago between the Banda and Ceram seas | | | Blue-eared lory | E. semilarvata Bonaparte, 1850 | LC IUCN | Central Pulua Seram, and island of the Maluka archipelago between the Banda and Ceram seas | | |
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| Genus Pseudeos Peters 1935 – two species | Common name | Scientific name | IUCN Red List Status | Range | Picture | --- --- | Dusky lory | P. fuscata (Blyth, 1858) | LC IUCN | New Guinea | | | Cardinal lory | P. cardinalis[f] (Gray, 1849) | LC IUCN | Solomon Islands | | |
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| Genus Trichoglossus Stephens 1826 – ten species | Common name | Scientific name | IUCN Red List Status | Range | Picture | --- --- | Rainbow lorikeet | T. moluccanus (Gmelin, 1788) | LC IUCN | Eastern Australian coast, introduced to Perth | | | Sunset lorikeet | T. forsteni Bonaparte, 1850 i | NT IUCN | Islands of Bali, Lombok, Sumbawa, Tanahjampea, and Kalaotoa | | | Leaf lorikeet | T. weberi (Büttikofer, 1894) | iNT IUCN | Flores | | | Marigold lorikeet | T. capistratus (Bechstein, 1811) | LC IUCN | Timor | | | Coconut lorikeet | T. haematodus (Linnaeus, 1771) | LC IUCN | Maluku, New Guinea, Solomon Islands, Vanuatu, and New Caledonia | | | Biak lorikeet | T. rosenbergii Schlegel, 1871 | gVU IUCN | Biak, and island of the Schouten Islands | | | Red-collared lorikeet | T. rubritorquis Vigors and Horsfield, 1827 | LC IUCN | Northern Australia, excluding Cape York Peninsula | | | Olive-headed lorikeet | T. euteles (Temminck, 1835) | LC IUCN | East Nusa Tenggara and Timor | | | Pohnpei lorikeet | T. rubiginosus (Bonaparte, 1850) | iNT IUCN | Pohnpei, and island of Micronesia | | | Scaly-breasted lorikeet | T. chlorolepidotus (Kuhl, 1820) | LC IUCN | Eastern Australia | | |
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| Genus Saudareos Joseph et al., 2020 – 5 species | Common name | Scientific name | IUCN Red List Status | Range | Picture | --- --- | Yellow-cheeked lorikeet | S. meyeri (Walden, 1871) | LC IUCN | Sulawesi | | | Mindanao lorikeet | S. johnstoniae (Hartert, 1903) | iNT IUCN | Mindanao, the second largest island of the Philippines | | | Sula lorikeet | S. flavoviridis Wallace, 1863 | iLC IUCN | Sula Islands | | | Ornate lorikeet | S. ornatus (Linnaeus, 1758) | LC IUCN | Sulawesi and surrounding islands | | | Iris lorikeet | S. iris (Temminck, 1835) | LC IUCN | Timor | | |
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| Genus Psitteuteles Bonaparte 1854 – one species | Common name | Scientific name | IUCN Red List Status | Range | Picture | --- --- | Varied lorikeet | P. versicolor (Lear, 1831) | LC IUCN | Northern Australia | | |
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| Genus Glossoptilus Rothschild and Hartert, 1896 – one species | Common name | Scientific name | IUCN Red List Status | Range | Picture | --- --- | Goldie's lorikeet | G. goldiei (Sharpe, 1882) | LC IUCN | New Guinea Highlands | | |
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| Genus Lorius Vigors 1825 – six species | Common name | Scientific name | IUCN Red List Status | Range | Picture | --- --- | Chattering lory | L. garrulus (Linnaeus, 1758) | gVU IUCN | North Maluku | | | Purple-naped lory | L. domicella (Linnaeus, 1758) | eEN IUCN | Seram, Ambon, and possibly Haruku and Saparua of South Maluku | | | Black-capped lory | L. lory (Linnaeus, 1758) | LC IUCN | New Guinea, excluding the New Guinea Highlands | | | Purple-bellied lory | L. hypoinochrous Gray, 1859 | LC IUCN | New Britain, New Ireland, and the eastern coastline of the Papuan Peninsula | | | White-naped lory | L. albidinucha (Rothschild and Hartert, 1924) | iNT IUCN | New Ireland | | | Yellow-bibbed lory | L. chlorocercus Gould, 1856 | LC IUCN | Southern Solomon Islands | | |
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| Genus Vini Lesson, 1833 – 11 species | Common name | Scientific name | IUCN Red List Status | Range | Picture | --- --- | Blue-crowned lorikeet | V. australis (Gmelin, 1788) | LC IUCN | Various islands east of Fiji, including Samoa, and formerly Tonga and Wallis and Futuna | | | Kuhl's lorikeet | V. kuhlii (Vigors, 1824) | eEN IUCN | Washington Island, Fanning Island, and Christmas Island of Kiribati, and Rimatara of French Polynesia | | | Stephen's lorikeet | V. stepheni (North, 1908) | gVU IUCN | Henderson Island in the Pitcairn Islands (UK) | | | Blue lorikeet | V. peruviana (Müller, 1776) | gVU IUCN | Eastern French Polynesia, and formerly central French Polynesia | | | Ultramarine lorikeet | V. ultramarina (Kuhl, 1820) | eEN IUCN | Marquesas Islands | | | Collared lory | V. solitaria (Suckow, 1800) | LC IUCN | Fiji | | | Meek's lorikeet | V. meeki (Rothschild and Hartert, 1901) | iLC IUCN | Bougainville Island and six islands of the Solomon Islands | | | Red-chinned lorikeet | V. rubrigularis (Sclater, 1881) | LC IUCN | New Britain, New Ireland, New Hanover Island, and Karkar Island | | | New Caledonian lorikeet | V. diadema (Verreaux and des Murs, 1860) | CR IUCN | Last official sighting in 1913 in New Caledonia | | | Red-throated lorikeet | V. amabilis (Ramsay, 1875) | CR IUCN | Viti Levu, Vanua Levu, Taveuni and Ovalau (islands of Fiji) | | | Palm lorikeet | V. palmarum (Gmelin, 1788) | gVU IUCN | Santa Cruz and Vanuatu | | |
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| Genus Parvipsitta Mathews, 1916 – two species[g] | Common name | Scientific name | IUCN Red List Status | Range | Picture | --- --- | Purple-crowned lorikeet | P. porphyrocephala[h] (Dietrichsen, 1837) | LC IUCN | | | | Little lorikeet | P. pusilla[i] (Shaw, 1790) | LC IUCN | | | |
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| Genus Glossopsitta Bonaparte, 1854 – one species | Common name | Scientific name | IUCN Red List Status | Range | Picture | --- --- | Musk lorikeet | G. concinna (Shaw, 1791) | LC IUCN | Southeast Australia and Tasmania | | |
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| Genus Charminetta Iredale, 1956 – one species | Common name | Scientific name | IUCN Red List Status | Range | Picture | --- --- | Pygmy lorikeet | C. wilhelminae (Meyer, 1874) | LC IUCN | New Guinea Highlands | | |
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| Genus Hypocharmosyna Salvadori, 1891 – two species | Common name | Scientific name | IUCN Red List Status | Range | Picture | --- --- | Red-fronted lorikeet | H. rubronotata | LC IUCN | Northern New Guinea and West Papua | | | Red-flanked lorikeet | H. placentis (Temminck, 1835) | LC IUCN | New Guinea (excluding the New Guinea Highlands), North Maluku, Maluku, New Britain, New Ireland, and Bougainville Island | | |
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| Genus Charmosynopsis Salvadori, 1877 – two species | Common name | Scientific name | IUCN Red List Status | Range | Picture | --- --- | Blue-fronted lorikeet | C. toxopei (Siebers, 1930) | CR IUCN | Buru of Maluku | | | Fairy lorikeet | C. pulchella (Gray, 1859) | LC IUCN | New Guinea Highlands, West Papua, and Papuan Peninsula | | |
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| Genus Synorhacma Joseph et al., 2020 – one species | Common name | Scientific name | IUCN Red List Status | Range | Picture | --- --- | Striated lorikeet | S. multistriata (Rothschild, 1911) | iLC IUCN | New Guinea Highlands | | |
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| Genus Charmosynoides Joseph et al., 2020 – one species | Common name | Scientific name | IUCN Red List Status | Range | Picture | --- --- | Duchess lorikeet | C. margarethae (Tristram, 1879) | iLC IUCN | Bougainville Island and six islands of the Solomon Islands | | |
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| Genus Charmosyna Wagler, 1832 – 3 species | Common name | Scientific name | IUCN Red List Status | Range | Picture | --- --- | Josephine's lorikeet | C. josefinae (Finsch, 1873) | LC IUCN | New Guinea Highlands | | | West Papuan lorikeet | C. papou (Scopoli, 1786) | LC IUCN | West Papua | | | Stella's lorikeet | C. stellae Meyer, 1886 | LC IUCN | West Papua | | |
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| Genus Oreopsittacus Salvadori, 1877 – one species | Common name | Scientific name | IUCN Red List Status | Range | Picture | --- --- | Plum-faced lorikeet | O. arfaki (Meyer, 1874) | LC IUCN | New Guinea Highlands | | |
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| Genus Neopsittacus Salvadori, 1875 – two species | Common name | Scientific name | IUCN Red List Status | Range | Picture | --- --- | Yellow-billed lorikeet | N. musschenbroekii (Schlegel, 1871) | LC IUCN | New Guinea Highlands | | | Orange-billed lorikeet | N. pullicauda Hartert, 1896 | LC IUCN | New Guinea Highlands | | |
Tribe Melopsittacini
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| Genus Melopsittacus Gould, 1840 – one species | Common name | Scientific name | IUCN Red List Status | Range | Picture | --- --- | Budgerigar | M. undulatus (Shaw, 1805) | LC IUCN | | | |
Tribe Cyclopsittini (fig parrots)
Main article: Cyclopsittini
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| Genus Cyclopsitta Reichenbach, 1850 – four species | Common name | Scientific name | IUCN Red List Status | Range | Picture | --- --- | Blue-fronted fig parrot | C. gulielmitertii (Schlegel, 1866) | LC IUCN | Salawati and the Bird's Head Peninsula in New Guinea | Female left, male right | | Black-fronted fig parrot | C. nigrifrons Reichenow, 1891 | LC | northern New Guinea | | | Dusky-cheeked fig parrot | C. melanogenia (Rosenberg, HKB, 1866) | LC | southern New Guinea and Aru Island | | | Double-eyed fig parrot | C. diophthalma (Hombron and Jacquinot, 1841) | LC IUCN | Aru Islands, Waigeo, New Guinea, and the eastern coast of Cape York Peninsula | | |
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| Genus Psittaculirostris G. R. Gray and J. E. Gray, 1859 – three species | Common name | Scientific name | IUCN Red List Status | Range | Picture | --- --- | Large fig parrot | P. desmarestii (Desmarest, 1826) | LC IUCN | West Papua | | | Edwards's fig parrot | P. edwardsii (Oustalet, 1885) | LC IUCN | Northern coast of Papua New Guinea | | | Salvadori's fig parrot | P. salvadorii (Oustalet, 1880) | LC IUCN | Northern coast of Papua Province | | |
Subfamily Agapornithinae
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Main article: Agapornithinae
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| Genus Agapornis (lovebirds) Selby, 1836 – nine species | Common name | Scientific name | IUCN Red List Status | Range | Picture | --- --- | Rosy-faced lovebird | A. roseicollis (Vieillot, 1818) | LC IUCN | | | | Yellow-collared lovebird | A. personatus | LC IUCN | | | | Fischer's lovebird | A. fischeri Reichenow, 1887 | iNT IUCN | | | | Lilian's lovebird | A. lilianae Shelley, 1894 | iNT IUCN | | | | Black-cheeked lovebird | A. nigrigenis Sclater, 1906 | gVU IUCN | | | | Gray-headed lovebird | A. canus (Gmelin, 1788) | LC IUCN | Coast of Madagascar and formerly Comoros | Female left, male right | | Black-winged lovebird | A. taranta (Smith-Stanley, 1814) | LC IUCN | Ethiopia and central Eritrea | | | Red-headed lovebird | A. pullarius (Linnaeus, 1758) | LC IUCN | Sub-Saharan Africa north of the Democratic Republic of the Congo, and around the Gulf of Guinea | | | Black-collared lovebird | A. swindernianus (Kuhl, 1820) | LC IUCN | | | |
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| Genus Loriculus (hanging parrots) Blyth, 1849 – 15 species | Common name | Scientific name | IUCN Red List Status | Range | Picture | --- --- | Moluccan hanging parrot | L. amabilis Wallace, 1862 | LC IUCN | North Maluku | | | Orange-fronted hanging parrot | L. aurantiifrons Schlegel, 1871 | LC IUCN | New Guinea, excluding the New Guinea Highlands | | | Sri Lanka hanging parrot | L. beryllinus (Pennant, 1781) | LC IUCN | Southern Sri Lanka | | | Sangihe hanging parrot | L. catamene Schlegel, 1871 | iNT IUCN | Sangihe Islands | | | Pygmy hanging parrot | L. exilis Schlegel, 1866 | iNT IUCN | Sulawesi | | | Wallace's hanging parrot | L. flosculus Wallace, 1864 | eEN IUCN | Flores | | | Blue-crowned hanging parrot | L. galgulus (Linnaeus, 1758) | LC IUCN | Sumatra, Borneo and the Malay Peninsula | Female left, male right | | Sula hanging parrot | L. sclateri Wallace, 1863 | LC IUCN | Taliabu, Peleng, and smaller, nearby islands | | | Great hanging parrot | L. stigmatus (Müller, 1843) | LC IUCN | Sulawesi | | | Bismarck hanging parrot | L. tener Sclater, 1877 | iNT IUCN | New Britain, New Ireland, and New Hanover | | | Philippine hanging parrot | L. philippensis (Müller, 1776) | LC IUCN | The Philippines | | | Black-billed hanging parrot | L. bonapartei Souancé, 1856 | | Sulu Archipelago of the Philippines | | | Camiguin hanging parrot | L. camiguinensis Tello, Degner, Bates & Willard, 2006 | | Camiguin of the Philippines | | | Yellow-throated hanging parrot | L. pusillus Gray, 1859 | iNT IUCN | Java and Bali | | | Vernal hanging parrot | L. vernalis (Sparrman, 1787) | LC IUCN | Indochina (excluding the Malay Peninsula), and the Eastern and Western Ghats | | |
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| Genus Bolbopsittacus Salvadori 1891 – one species | Common name | Scientific name | IUCN Red List Status | Range | Picture | --- --- | Guaiabero | B. lunulatus (Scopoli, 1786) | LC IUCN | Philippines, excluding the central islands | | |
Subfamily Psittaculinae
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Main article: Psittaculinae
Tribe Polytelini
Main article: Polytelini
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| Genus Alisterus (king parrots) Mathews, 1911 – three species | Common name | Scientific name | IUCN Red List Status | Range | Picture | --- --- | Moluccan king parrot | A. amboinensis (Linnaeus, 1766) | LC IUCN | West Papua, Maluku, North Maluku, Buru Island, and the Banggai Islands | | | Australian king parrot | A. scapularis (Lichtenstein, 1818) | LC IUCN | Eastern coast of Australia | | | Papuan king parrot | A. chloropterus (Ramsay, 1879) | LC IUCN | New Guinea, excluding West Papua and the New Guinea Highlands | | |
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| Genus Aprosmictus Gould, 1842 – two species | Common name | Scientific name | IUCN Red List Status | Range | Picture | --- --- | Jonquil parrot | A. jonquillaceus (Vieillot, 1818) | iNT IUCN | Timor | | | Red-winged parrot | A. erythropterus (Gmelin, 1788) | LC IUCN | | | |
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| Genus Polytelis Wagler, 1832 – three species | Common name | Scientific name | IUCN Red List Status | Range | Picture | --- --- | Superb parrot | P. swainsonii (Desmarest, 1826) | LC IUCN | Central New South Wales | | | Regent parrot | P. anthopeplus (Lear, 1831) | LC IUCN | Southwest Western Australia and southwest New South Wales and northwest Victoria | | | Princess parrot | P. alexandrae Gould, 1863 | iNT IUCN | Central Australia | | |
Tribe Psittaculini
Main article: Psittaculini
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| Genus Prioniturus (racket-tails) Wagler, 1842 – nine species | Common name | Scientific name | IUCN Red List Status | Range | Picture | --- --- | Montane racket-tail | P. montanus Ogilvie-Grant, 1895 | iNT IUCN | Luzon | | | Mindanao racket-tail | P. waterstradti Rothschild, 1904 | iNT IUCN | Mindanao, an island of the Philippines | | | Blue-headed racket-tail | P. platenae Blasius, 1888 | gVU IUCN | Palawan, an island chain of the Philippines | | | Green racket-tail | P. luconensis Steere, 1890 | eEN IUCN | Luzon and Marinduque | | | Blue-crowned racket-tail | P. discurus (Vieillot, 1822) | LC IUCN | Central and southern Philippines | | | Mindoro racket-tail | P. mindorensis Steere, 1890 | gVU IUCN | Mindoro, an island of the Philippines | | | Blue-winged racket-tail | P. verticalis Sharpe, 1893 | CR IUCN | Sulu Archipelago | | | Yellow-breasted racket-tail | P. flavicans Cassin, 1853 | iNT IUCN | Northern Sulawesi | | | Golden-mantled racket-tail | P. platurus (Vieillot, 1818) | LC IUCN | Sulawesi and the Sula Islands | | | Buru racket-tail | P. mada Hartert, 1900 | LC IUCN | | | |
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| Genus Eclectus Wagler, 1832 – five species | Common name | Scientific name | IUCN Red List Status | Range | Picture | --- --- | Moluccan eclectus | E. roratus (Müller, 1776) | LC IUCN | Maluku, North Maluku | | | Sumba eclectus | E. cornelia Bonaparte, 1850 | EN | Sumba | | | Tanimbar eclectus | E. riedeli AB Meyer, 1882 | VU | Tanimbar Islands | | | Papuan eclectus | E. polychloros (Scopoli, 1786) | LC | from Kai Islands and western islands of the West Papua province in the west, across the island of New Guinea to the Trobriands, D'Entrecasteaux Islands, and Louisiade Archipelago | | | Oceanic eclectus | †E. infectus Steadman, 2006 | aEX IUCN | Formerly Tonga and possibly Malakula until the late 1700s | | |
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| Genus Geoffroyus Bonaparte, 1850 – four species | Common name | Scientific name | IUCN Red List Status | Range | Picture | --- --- | Red-cheeked parrot | G. geoffroyi (Bechstein, 1811) | LC IUCN | New Guinea (excluding the New Guinea Highlands), Maluku, North Maluku, Buru Island, Aru Islands, Timor, and the Lesser Sunda Islands | | | Blue-collared parrot | G. simplex (Meyer, 1874) | LC IUCN | New Guinea Highlands and West Papua | | | Song parrot | G. heteroclitus (Hombron and Jacquinot, 1841) | LC IUCN | New Britain, New Ireland, New Hanover, and the Solomon Islands | | | Rennell parrot | G. hyacinthinus (Mayr, 1931) | LC IUCN | Rennell Island, an island of the Solomon Islands | | |
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| Genus Tanygnathus Wagler, 1832 – four species | Common name | Scientific name | IUCN Red List Status | Range | Picture | --- --- | Great-billed parrot | T. megalorynchos (Boddaert, 1783) | LC IUCN | North Maluku, Maluku, Buru Island, Tanimbar Islands, East Nusa Tenggara, and Sumba | | | Blue-naped parrot | T. lucionensis (Linnaeus, 1766) | iNT IUCN | The Philippines | | | Blue-backed parrot | T. sumatranus (Raffles, 1822) | LC IUCN | The Philippines, Sulawesi, and Taliabu | | | Black-lored parrot | T. gramineus (Gmelin, 1788) | gVU IUCN | Buru Island | | |
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| Genus Psittinus Blyth, 1842 – two species | Common name | Scientific name | IUCN Red List Status | Range | Picture | --- --- | Blue-rumped parrot | P. cyanurus (Forster, 1795) | iNT IUCN | Borneo, Sumatra, the Mentawai Islands, Bangka Island, Riau Archipelago, the Lingga Islands, and the Malay Peninsula | | | Simeulue parrot | P. abbotti Richmond, 1902 | iNT IUCN | Simeulue and the Siumat Islands | | |
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| Genus Psittacula Cuvier, 1800 – fifteen species | Common name | Scientific name | IUCN Red List Status | Range | Picture | --- --- | Newton's parakeet | †P. exsul (Newton, 1872) | aEX IUCN | Formerly Rodrigues, an island of Mauritius until 1875 | | | Echo parakeet | P. eques (Boddaert, 1783) | EN IUCN | Southwest corner of Mauritius and formerly Réunion | | | Rose-ringed parakeet | P. krameri (Scopoli, 1769) | LC IUCN | | | | Alexandrine parakeet | P. eupatria (Linnaeus, 1766) | iNT IUCN | Eastern India, Western Ghats, the Himalayas, Indochina, and introduced to Bahrain and southwest Iran | | | Seychelles parakeet | †P. wardi (Newton, 1867) | aEX IUCN | Formerly Mahé, Praslin, and Silhouette Island until 1883 | | | Plum-headed parakeet | P. cyanocephala (Linnaeus, 1766) | LC IUCN | India, Sri Lanka, and the Himalayas | | | Blossom-headed parakeet | P. roseata Biswas, 1951 | iNT IUCN | Indochina (excluding Malaysia) and the eastern Himalayas | | | Slaty-headed parakeet | P. himalayana (Lesson, 1831) | LC IUCN | the Himalayas | | | Gray-headed parakeet | P. finschii (Hume, 1874) | iNT IUCN | Indochina, excluding Thailand and Malaysia | | | Blue-winged parakeet | P. columboides (Vigors, 1830) | LC IUCN | Western Ghats | Female left, male right | | Layard's parakeet | P. calthrapae (Blyth, 1849) | LC IUCN | Southern Sri Lanka | | | Lord Derby's parakeet | P. derbiana (Fraser, 1852) | iNT IUCN | Southern China near the China/India and China/Myanmar border | | | Red-breasted parakeet | P. alexandri (Linnaeus, 1758) | iNT IUCN | Indochina (excluding central Thailand) and the Himalayas | | | Nicobar parakeet | P. caniceps (Blyth, 1846) | iNT IUCN | Nicobar Islands which is northwest of Sumatra | | | Long-tailed parakeet | P. longicauda (Boddaert, 1783) | iNT IUCN | Sumatra, coasts of Borneo, tip of the Malay Peninsula | | | Mascarene grey parakeet | †P. bensoni[j] (Holyoak, 1973) | aEX IUCN | Formerly Mauritius until 1764 | | |
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| Genus †Lophopsittacus Cuvier, 1800 – one species | Common name | Scientific name | IUCN Red List Status | Range | Picture | --- --- | Broad-billed parrot | †L. mauritianus (Owen, 1866) | aEX IUCN | Formerly Mauritius until 1693 | | |
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| Genus †Necropsittacus Milne-Edwards, 1874 – one species | Common name | Scientific name | IUCN Red List Status | Range | Picture | --- --- | Rodrigues parrot | †N. rodricanus (Milne-Edwards, 1867) | aEX IUCN | Formerly Rodrigues, an island of Mauritius, until 1761 | | |
Tribe Micropsittini
Main article: Micropsittini
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| Genus Micropsitta (pygmy parrots) Lesson, 1831 – six species | Common name | Scientific name | IUCN Red List Status | Range | Picture | --- --- | Buff-faced pygmy parrot | M. pusio (Sclater, 1866) | LC IUCN | Northern coast of New Guinea, coast of the Papuan Peninsula, and the southern islands of the Bismarck Archipelago | | | Yellow-capped pygmy parrot | M. keiensis (Salvadori, 1876) | LC IUCN | Southern New Guinea, the Aru Islands, coast of West Papua, and the Raja Ampat Islands | | | Geelvink pygmy parrot | M. geelvinkiana (Schlegel, 1871) | iNT IUCN | Biak-Supiori and Numfor (islands of the Baik Island) | | | Meek's pygmy parrot | M. meeki Rothschild and Hartert, 1914 | LC IUCN | Admiralty Islands and the St Matthias Islands | | | Finsch's pygmy parrot | M. finschii (Ramsay, 1881) | LC IUCN | | | | Red-breasted pygmy parrot | M. bruijnii (Salvadori, 1875) | LC IUCN | New Guinea Highlands, New Britain, New Ireland, Bougainville Island, Kolombangara Island, Guadalcanal, West Papua, and Maruku | | |
Family Psittrichasiidae
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Main article: Psittrichasiidae
Subfamily Coracopsinae
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Main article: Coracopsinae
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| Genus Coracopsis (vasa parrots) Wagler, 1832 – three species | Common name | Scientific name | IUCN Red List Status | Range | Picture | --- --- | Greater vasa parrot | C. vasa (Shaw, 1812) | LC IUCN | The coasts of Madagascar and Comoros (excluding Mayotte) | | | Lesser vasa parrot | C. nigra (Linnaeus, 1758) | LC IUCN | Madagascar | | | Seychelles black parrot | C. barklyi Newton, 1867 | gVU IUCN | Praslin and possibly Curieuse Island | | |
Subfamily Psittrichasinae
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Main article: Psittrichasinae
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| Genus Psittrichas Lesson, 1831 – one species | Common name | Scientific name | IUCN Red List Status | Range | Picture | --- --- | Pesquet's parrot | P. fulgidus (Lesson, 1830) | gVU IUCN | New Guinea Highlands | | |
Cockatoos
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Main article: Cockatoo
Family Cacatuidae
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Subfamily Nymphicinae
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Main article: Nymphicinae
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| Genus Nymphicus Wagler, 1832 – one species | Common name | Scientific name | IUCN Red List Status | Range | Picture | --- --- | Cockatiel | N. hollandicus (Kerr, 1792) | LC IUCN | | Male left, female right | |
Subfamily Calyptorhynchinae
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Main article: Calyptorhynchinae
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| Genus Calyptorhynchus Desmarest, 1826 – two species | Common name | Scientific name | IUCN Red List Status | Range | Picture | --- --- | Red-tailed black cockatoo | C. banksii (Latham, 1790) | LC IUCN | | | | Glossy black cockatoo | C. lathami (Temminck, 1807) | VU IUCN | | Male left, female right | |
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| Genus Zanda Mathews, 1913 – three species | Common name | Scientific name | IUCN Red List Status | Range | Picture | --- --- | Baudin's black cockatoo | Z. baudinii (Lear, 1832) | eCR IUCN | | | | Yellow-tailed black cockatoo | Z. funerea (Shaw, 1794) | LC IUCN | Z. f. funereus in red, Z. f. xanthanotus in green | | | Carnaby's black cockatoo | Z. latirostris (Carnaby, 1948) | eEN IUCN | | | |
Subfamily Cacatuinae
[edit]
Main article: Cacatuinae
Tribe Microglossini
Main article: Microglossini
| | | | | | | | | | | |
--- --- --- --- ---
| Genus Probosciger Kuhl, 1820 – one species | Common name | Scientific name | IUCN Red List Status | Range | Picture | --- --- | Palm cockatoo | P. aterrimus (Gmelin, 1788) | NT IUCN | | | |
Tribe Cacatuini
Main article: Cacatuini
| | | | | | | | | | | |
--- --- --- --- ---
| Genus Callocephalon Lesson, 1837 – one species | Common name | Scientific name | IUCN Red List Status | Range | Picture | --- --- | Gang-gang cockatoo | C. fimbriatum (Grant, 1803) | VU IUCN | | | |
| | | | | | | | | | | |
--- --- --- --- ---
| Genus Eolophus Bonaparte, 1854 – one species | Common name | Scientific name | IUCN Red List Status | Range | Picture | --- --- | Galah | E. roseicapilla (Vieillot, 1817) | LC IUCN | | | |
| | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | |
--- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- |
| Genus Cacatua Vieillot, 1817 – 12 species | Common name | Scientific name | IUCN Red List Status | Range | Picture | --- --- | White cockatoo | C. alba (Müller, 1776) | eEN IUCN | North Maluku | | | Solomons cockatoo | C. ducorpsii Pucheran, 1853 | LC IUCN | Solomon Islands | | | Sulphur-crested cockatoo | C. galerita (Latham, 1790) | LC IUCN | Native range in red, introduced range in violet | | | Tanimbar corella | C. goffiniana Roselaar and Michels, 2004 | iNT IUCN | Yamdena and Larat of the Tanimbar Islands, and introduced to the Kai Islands | | | Red-vented cockatoo | C. haematuropygia (Müller, 1776) | CR IUCN | The Philippines | | | Major Mitchell's cockatoo | C. leadbeateri (Vigors, 1831) | LC IUCN | | | | Salmon-crested cockatoo | C. moluccensis (Gmelin, 1788) | gVU IUCN | Seram, Ambon Island, Saparua and Haruku Island | | | Blue-eyed cockatoo | C. ophthalmica Sclater, 1864 | gVU IUCN | | | | Western corella | C. pastinator (Gould, 1841) | LC IUCN | Southwest Australian coast | | | Little corella | C. sanguinea Gould, 1843 | LC IUCN | Australia, excluding Cape York Peninsula and west central Australia, and introduced to Tasmania | | | Yellow-crested cockatoo | C. sulphurea (Gmelin, 1788) | CR IUCN | Native range in blue, introduced range in red | | | Long-billed corella | C. tenuirostris (Kuhl, 1820) | LC IUCN | New South Wales, excluding the eastern coast | | |
New Zealand parrots
[edit]
Main article: New Zealand parrot
Family Nestoridae
[edit]
Main article: Nestoridae
| | | | | | | | | | | | | | | | | | | | | |
--- --- --- --- --- --- --- --- --- ---
| Genus Nestor Lesson, 1830 – three species | Common name | Scientific name | IUCN Red List Status | Range | Picture | --- --- | New Zealand kākā | N. meridionalis (Gmelin, 1788) | eEN IUCN | | | | Norfolk kākā | †N. productus (Gould, 1836) | aEX IUCN | Formerly Norfolk Island and Phillip Island until 1851 | | | Kea | N. notabilis Gould, 1856 | gVU IUCN | | | |
Family Strigopidae
[edit]
Main article: Strigopidae
| | | | | | | | | | | |
--- --- --- --- ---
| Genus Strigops Gray, 1845 – one species | Common name | Scientific name | IUCN Red List Status | Range | Picture | --- --- | Kākāpō | S. habroptilus Gray, 1845 | CR IUCN | Codfish Island / Whenua Hou, Anchor Island, and Little Barrier Island | | |
See also
[edit]
Birds portal
List of macaws
List of amazon parrots
List of Nestoridae
List of Aratinga parakeets
Notes
[edit]
^ Officially Psittacara holochlorus brevipes, a subspecies of green parakeet, by the IUCN
^ Officially Psittacara chloropterus maugei, a subspecies of Hispaniolan parakeet, by the IUCN
^ Officially Psittacara holochlorus strenuus, a subspecies of green parakeet, by the IUCN
^ The species authority is Johann Baptist von Spix, 1824, by the IOC, however the IUCN considers it to be Władysław Taczanowski, 1883
^ Officially listed as Northiella haematogaster narethae, a subspecies of eastern bluebonnet, by the IUCN
^ Officially Chalcopsitta cardinalis by the IUCN
^ Synonymous with Glossopsitta according to the IUCN
^ Officially Glossopsitta porphyrocephala by the IUCN
^ Officially Glossopsitta pusilla by the IUCN
^ This parrot is considered a member of the genus Psittacula by the IOC since 2007, but as a member of the genus Lophopsittacus by the IUCN.
References
[edit]
^ "Psittacine". American Heritage Dictionary of the English Language (4th ed.). Houghton Mifflin Company. 2000. Archived from the original on 27 August 2007. Retrieved 12 August 2016.
^ "Psittacine". Merriam-Webster Online Dictionary. Merriam-Webster, Inc. Retrieved 12 August 2016.
^ a b c d e Joseph, L.; Toon, A.; Schirtzinger, E. E.; Wright, T. F.; Schodde, R. (2012). "A revised nomenclature and classification for family-group taxa of parrots (Psittaciformes)" (PDF). Zootaxa. 3205 (3205): 26–40. doi:10.11646/zootaxa.3205.1.2. Archived from the original (PDF) on 2013-12-11. Retrieved 2016-08-13.
^ Forshaw, J. M.; Cooper, W. T. (1981). Parrots of the World (2nd ed.). London, England: David & Charles, Newton Abbot. ISBN 978-0-7153-7698-0.
^ Cooke, F.; Bruce, J. (2004). The Encyclopedia of Animals: a complete visual guide (1st ed.). Berkeley, California: University of California Press. p. 296. ISBN 978-0-520-24406-1.
^ Heatley, J. J.; Cornejo, J. (2015). "Psittaciformes". In Miller, R. E.; Fowler, M. E. (eds.). Zoo and Wild Animal Medicine. Vol. 8. St. Louis, Missouri: Elsevier Saunders. p. 172. ISBN 978-1-4557-7397-8.
^ Burtt, E. H.; Schroeder, M. R.; Smith, L. A.; Sroka, J. E.; McGraw, K. J. (2010). "Colourful parrot feathers resist bacterial degradation". Biology Letters. 7 (2): 214–216. doi:10.1098/rsbl.2010.0716. PMC 3061162. PMID 20926430.
^ Forshaw, J. M.; Cooper, W. T. (1978) . Parrots of the World (2nd ed.). Melbourne, Australia: Landsdowne Editions. ISBN 978-0-7018-0690-3.
^ a b c d e Wright, T. F.; Schirtzinger, E. E.; Matsumoto, T.; Eberhard, J. R.; Graves, G. R.; Sanchez, J. J.; Capelli, S.; Muller, H.; Scharpegge, J.; Chambers, G. K.; Fleischer, R. C. (2008). "A Multilocus Molecular Phylogeny of the Parrots (Psittaciformes): Support for a Gondwanan Origin during the Cretaceous". Molecular Biology and Evolution. 25 (10): 2141–2156. doi:10.1093/molbev/msn160. PMC 2727385. PMID 18653733.
^ a b c d de Kloet, R. S.; de Kloet, S. R. (2005). "The evolution of the spindlin gene in birds: Sequence analysis of an intron of the spindlin W and Z gene reveals four major divisions of the Psittaciformes". Molecular Phylogenetics and Evolution. 36 (3): 706–721. Bibcode:2005MolPE..36..706D. doi:10.1016/j.ympev.2005.03.013. PMID 16099384.
^ a b c Tokita, M.; Kiyoshi, T.; Armstrong, K. N. (2007). "Evolution of craniofacial novelty in parrots through developmental modularity and heterochrony". Evolution and Development. 9 (6): 590–601. doi:10.1111/j.1525-142X.2007.00199.x. PMID 17976055. S2CID 46659963.
^ Brock, M. K.; White, B. N. (1992). "Application of DNA fingerprinting to the recovery program of the endangered Puerto Rican parrot" (PDF). Proceedings of the National Academy of Sciences. 89 (23): 11121–11125. Bibcode:1992PNAS...8911121B. doi:10.1073/pnas.89.23.11121. PMC 50501. PMID 1454788.
^ Sibley, C. G.; Ahlquist, J. E. (1991). Phylogeny and Classification of Birds. Yale University Press.
^ Christidis, L.; Boles, W. E. (2008). Systematics and Taxonomy of Australian Birds. Canberra, Australia: CSIRO Publishing. p. 200. ISBN 978-0-643-06511-6.
^ Livezey, B. C.; Zusi, R. L. (2007). "Higher-order phylogeny of modern birds (Theropoda, Aves: Neornithes) based on comparative anatomy: II. – Analysis and discussion". Zoological Journal of the Linnean Society. 149 (1): 1–94. doi:10.1111/j.1096-3642.2006.00293.x. PMC 2517308. PMID 18784798.[dead link]
^ Homberger, D. G. (2006). "Classification and the status of wild populations of parrots". In Luescher AU (ed.). Manual of parrot behavior. Ames, Iowa: Blackwell Publishing. pp. 3–11. ISBN 978-0-8138-2749-0.
^ a b White, N. E.; Phillips, M. J.; Gilbert, M. T. P.; Alfaro-Núñez, A.; Willerslev, E.; Mawson, P. R.; Spencer, P. B. S.; Bunce, M. (2011). "The evolutionary history of cockatoos (Aves: Psittaciformes: Cacatuidae)". Molecular Phylogenetics and Evolution. 59 (3): 615–622. Bibcode:2011MolPE..59..615W. doi:10.1016/j.ympev.2011.03.011. PMID 21419232.
^ Astuti, D. (2004). A phylogeny of cockatoos (Aves: Psittaciformes) inferred from DNA sequences of the seventh intron of nuclear β-fibrinogen gene (PDF) (Thesis). Graduate School of Environmental Earth Science, Hokkaido University, Japan. pp. 1–3.
^ Adams, M.; Baverstock, P. R.; Saunders, D.A.; Schodde, R.; Smith, G. T. (1984). "Biochemical systematics of the Australian cockatoos (Psittaciformes: Cacatuinae)". Australian Journal of Zoology. 32 (3): 363–377. doi:10.1071/ZO9840363.
^ "Introduction". IUCN Red List of Threatened Species. Retrieved 12 August 2016.
^ Schweizer, M.; Seehausen, O.; Güntert, M.; Hertwig, S. T. (2009). "The evolutionary diversification of parrots supports a taxon pulse model with multiple trans-oceanic dispersal events and local radiations". Molecular Phylogenetics and Evolution. 54 (3): 984–994. doi:10.1016/j.ympev.2009.08.021. PMID 19699808. S2CID 1831016.
^ Joseph, L.; Toon, A.; Schirtzinger, E. E.; Wright, T. F. (2011). "Molecular systematics of two enigmatic genera Psittacella and Pezoporus illuminate the ecological radiation of Australo-Papuan parrots (Aves: Psittaciformes)". Molecular Phylogenetics and Evolution. 59 (3): 675–684. Bibcode:2011MolPE..59..675J. doi:10.1016/j.ympev.2011.03.017. PMID 21453777.
^ Schweizer, M.; Seehausen, O.; Hertwig, S. T. (2011). "Macroevolutionary patterns in the diversification of parrots: effects of climate change, geological events and key innovations". Journal of Biogeography. 38 (11): 2176–2194. Bibcode:2011JBiog..38.2176S. doi:10.1111/j.1365-2699.2011.02555.x. S2CID 85625053.
^ "Platycercus/Barnardius – Rosellas". Avian Web. 2006. Retrieved 9 August 2016.
^ Forshaw, Joseph M. (2006). Parrots of the World; an Identification Guide. Princeton University Press. ISBN 978-0-691-09251-5.
^ Lendon, A. H. (1973). Australian Parrots in Field and Aviary (2nd ed.). Sydney, Australia: Angus and Robertson. ISBN 978-0-207-12424-2.
^ Falla, R. A.; Sibson, R. B.; Turbot, E. G. (1966). A Field guide to the birds of New Zealand. London, England: Houghton Mifflin and Company. ISBN 978-0-00-212022-7.
^ BirdLife International. "Northern Rosella Platycercus venustus". IUCN Redlist for Birds. Retrieved 9 August 2016.
^ Gill, F.; Donsker, D. (2016). "Parrots and cockatoos". IOC World Bird List. 6.4. International Ornithological Congress. doi:10.14344/IOC.ML.6.4. Retrieved 26 December 2016.
^ International, B. (2016). "Lophopsittacus bensoni". IUCN Red List of Threatened Species. 2016 e.T22728844A94998578. doi:10.2305/IUCN.UK.2016-3.RLTS.T22728844A94998578.en.{{cite iucn}}: uses deprecated |page= identifier (help)
External links
[edit]
eBird – Enter the name of a parrot species to find detailed information about their current range as well as sightings across the world
IOC World Bird List – A comprehensive list of parrots with up-to-date information on taxonomy, including synonyms and authority
Integrated Taxonomic Information System – A list providing taxonomic information on all subordinate taxa of Psittaciformes, including subspecies
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11035 | https://www.wyzant.com/resources/answers/859201/find-the-probability-of-selecting-two-red-cards-when-two-cards-are-drawn-wi | find the probability of selecting two red cards when two cards are drawn without replacement from a standard deck of cards | Wyzant Ask An Expert
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Probability
Ann G.
asked • 07/20/21
find the probability of selecting two red cards when two cards are drawn without replacement from a standard deck of cards
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52 cards in a deck, 26 red cards.
26/52 = 1/2 chance first draw is red card, leaving 51 cards, of which 25 are red, so
25/51 chance second draw is a red card given the first one drawn is red.
probability both of the two cards drawn are red is 1/2 25/51 = 25/102.
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11036 | https://ovarianresearch.biomedcentral.com/articles/10.1186/s13048-015-0205-8 | Progesterone administration for luteal phase deficiency in human reproduction: an old or new issue? | Journal of Ovarian Research | Full Text
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Journal of Ovarian Research
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Progesterone administration for luteal phase deficiency in human reproduction: an old or new issue?
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Published: 19 November 2015
Progesterone administration for luteal phase deficiency in human reproduction: an old or new issue?
Stefano Palomba1,
Susanna Santagni1&
Giovanni Battista La Sala2
Journal of Ovarian Researchvolume 8, Article number:77 (2015) Cite this article
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Abstract
Luteal phase deficiency (LPD) is described as a condition of insufficient progesterone exposure to maintain a regular secretory endometrium and allow for normal embryo implantation and growth. Recently, scientific focus is turning to understand the physiology of implantation, in particular the several molecular markers of endometrial competence, through the recent transcriptomic approaches and microarray technology. In spite of the wide availability of clinical and instrumental methods for assessing endometrial competence, reproducible and reliable diagnostic tests for LPD are currently lacking, so no type-IA evidence has been proposed by the main scientific societies for assessing endometrial competence in infertile couples. Nevertheless, LPD is a very common condition that may occur during a series of clinical conditions, and during controlled ovarian stimulation (COS) and hyperstimulation (COH) programs. In many cases, the correct approach to treat LPD is the identification and correction of any underlying condition while, in case of no underlying dysfunction, the treatment becomes empiric. To date, no direct data is available regarding the efficacy of luteal phase support for improving fertility in spontaneous cycles or in non-gonadotropin induced ovulatory cycles. On the contrary, in gonadotropin in vitro fertilization (IVF) and non-IVF cycles, LPD is always present and progesterone exerts a significant positive effect on reproductive outcomes. The scientific debate still remains open regarding progesterone administration protocols, specially on routes of administration, dose and timing and the potential association with other drugs, and further research is still needed.
Background
Multiple uses of progesterone and progestogens for women’s health in clinical practice are recognized. In obstetric care they are used for treating abortion and preterm labor prevention, and in gynecology to balance estrogens in hormonal replacement therapies or in oral contraception, and as palliative care for the treatment of gynecological malignancies. On the other hand, the role of progesterone for luteal phase support still represents a controversial topic, due to its wide and empirically transverse clinical use, from natural ovulatory cycles to assisted reproductive technologies (ARTs). Most of these controversies derive from too much confusion existing in the knowledge of corpus luteum function, progesterone production, implantation window and endometrial competence and, consequently, about luteal phase deficiency (LPD) and support.
LPD is described as a condition of insufficient progesterone exposure to maintain a normal secretory endometrium and allow for normal embryo implantation and growth . This definition has been sustained during years from its first description .
After several years where endometrium has received much less attention in reproductive studies compared to ovary and embryo development, the endometrium is now receiving the research attention it deserves, due to its physiological importance in reproduction . In fact, embryo implantation represents a critical step of the reproductive process consisting of a unique biological phenomenon . Successful implantation requires a receptive endometrium, a functional embryo at the blastocyst stage and a synchronized dialog between maternal and embryonic tissues . The human endometrium undergoes a complex series of organized proliferative and secretory changes in each menstrual cycle and exhibits only a short period of receptivity, known as the ‘implantation window’ . Endometrial receptivity during the implantation window requires a close cooperation of an extremely large number of different factors; unfortunately, the individual role of each factor in the network of endometrial development is still not completely understood .
The aim of this descriptive review will be to provide clinicians with a brief document based on the best and current scientific evidences, summarizing the actual knowledge regarding the role of progesterone in implantation physiology, and to show the potential and correct use of progesterone and progestogens for treatment of LPD in the different clinical settings. To obtain evidence-based data, we performed a systematic search for studies (articles and/or abstracts), without English language limitation, collecting and analyzing the published articles in literature until July 2015. We searched on Medline (through PubMed), with the combination of the following medical subject headings or keywords: “luteal phase deficiency”, “window of implantation”, “endometrial competence”, “progesterone AND luteal phase support”, “assisted reproductive technologies AND progesterone”, “controlled ovarian hyperstimulation AND luteal phase support” and “frozen-thawed cycles AND luteal phase support”. Additional literature searches were performed on the references from the identified studies. We gave priority to meta-analysis, systematic reviews and randomized clinical trials (RCTs), based on the personal evaluation of each author. When meta-analytic data or data from RCTs were lacking, prospective non-randomized and then cohort studies were included in the final analysis.
Implantation physiology
The uterus differentiates into an altered state when implantation-competent blastocysts are ready to initiate implantation. This state is called uterine receptivity for implantation, i.e. the implantation window, and lasts for a limited time. At this stage, the uterine environment is conducive to blastocyst growth, attachment and the subsequent events of implantation. The main hormones involved in uterine receptivity are the ovarian steroids, progesterone and estrogen. Uterine sensitivity to implantation is classified into pre-receptive, receptive and non-receptive (refractory) phases. In humans, the endometrium is classified histologically and functionally into proliferative and secretory phases during the average 28–30 day menstrual cycle. During the secretory phase, the uterus is considered pre-receptive for the first ~7 days following ovulation; it then becomes receptive during the mid-secretory phase, which spans 7–10 days after ovulation; the non-receptive (refractory) phase comprises the rest of the secretory phase .
Molecular and genetic evidence indicates that locally produced signaling molecules, including cytokines, growth factors, homeobox transcription factors, lipid mediators and morphogens, together with ovarian hormones serve as autocrine, paracrine and juxtacrine factors to obtain uterine receptivity . Therefore, the implantation mechanism includes the cooperation of several factors, such as the integrins on both blastocyst and receptive endometrium surfaces, progesterone which stimulates the endometrium receptivity and the decidulization of stroma and a series of cytokines, such as IL-1 and EGF, which participate in the regulation of endometrial immune tolerance, expression of integrin nd prostaglandins production [7, 8].
The corpus luteum is derived from the transformation of granulosa and theca cells into luteal cells in response to the mid-cycle surge of gonadotropins or to an exogenous human chorionic gonadotropin (hCG) bolus administration. The most important function of the corpus luteum is progesterone secretion which is necessary to obtain a secretory transformation of the endometrium and to sustain the early pregnancy. Progesterone prepares the endometrium for pregnancy by stimulating proliferation in response to hCG. This occurs in the luteal phase of the menstrual cycle.
During the midluteal phase of a natural cycle, progesterone is involved in the modulation process of the expression of ultrastructural hallmarks of secretory transformation, such as giant mitochondria, subnuclear glycogen deposits, pinopodes and nucleolar channel system (NCS) . With the aim of inducing endometrial competence/receptivity, progesterone can also act by stimulating the immune system to produce non-inflammatory T-helper 2 cytokines and C3-C4, as shown in patients with autoimmune diseases [10, 11], increasing nitric oxide production, with improvement of the blood flow and oxygen to the endometrium . Progesterone is also able to reduce the contractility of the myometrium at the time of the implantation .
The crucial role of the corpus luteum in human reproduction in the maintenance of early pregnancy is demonstrated by the harmful effects of a lutectomy during the first weeks of a pregnancy. Initial studies on corpus luteum function demonstrated that, in patients at 7 weeks of pregnancy, after a tubal ligation, the plasma progesterone levels were normal and pregnancy continued but, after a tubal ligation plus lutectomy, the plasma progesterone levels dropped to near zero within 5 days and the pregnancy was aborted . In this week of gestation, after tubal ligation and lutectomy, progesterone replacement was effective in maintaining the pregnancy. At and after 8 weeks of amenorrhea, tubal ligation plus lutectomy decreased plasma progesterone levels until day 4 but these then increased to pre-treatment concentrations with successful continuation of pregnancy .
This “genuine” endometrial competence, however, can be negatively affected in clinical practice by several functional diseases, such as polycystic ovarian syndrome (PCOS) and obesity and anatomic abnormalities, such as congenital uterine anomalies , endometrial polyps , myomas (submucosal) , hydrosalpinx and endometriosis . In fact, the treatments of these diseases have been discussed as potential therapies to improve fertility and clinical pregnancy rate in spontaneous and/or in in-vitro fertilization (IVF) cycles. The pregnancy rate was found to be two-fold higher [relative risk (RR) 2.0, 95% confidence interval (CI) 1.1 to 3.9] after myomectomy of submucosal myomas or after polipectomy of 3–24 mm endometrial polyps (RR 2.1, 95% CI 1.5 to 2.9) . Laparoscopic salpingectomy for hydrosalpinges versus no-surgical treatments improved the clinical pregnancy rate by more than two-fold odds ratio (OR) 2.49, 95% CI 1.60 to 3.86) [. Finally, compared with diagnostic laparoscopy, laparoscopic surgery for endometriosis was associated with a significant improvement of the clinical pregnancy rates (OR 1.89, 95% CI 1.25 to 2.86) .
Endometrial competence evaluation
Endometrial competence can be assessed using different methods, such as histology, timing, ultrasound and biomarkers.
In consideration of the LPD definition, histological evaluation of the endometrium has been considered the gold standard procedure to study endometrial competence. The histological method is based on a careful examination of an endometrial biopsy, and on the definition of the histologic characteristics of a secretory endometrium and describing the temporal responses to progesterone, i.e. endometrial dating. An endometrium is considered out of phase when there is a lag of more than 2 days . Over the years, the histological method has developed more specific instruments, such as scanning electron microscopy [23, 24] and the immunohistochemical analysis of estrogen and progesterone receptors . However, randomized clinical trials (RCTs) suggest that endometrial biopsy is an inaccurate tool for differentiating fertile women from women with LPD and infertility .
The timing as criteria of endometrial competence is based on LH peak, ultrasound demonstration of ovulation, basal body temperature shift and menstrual flow after the biopsy but it also presents a wide inter- and intra-subject variability [1, 26].
Ultrasound assessment is still used to evaluate endometrial competence though no significant correlation has been seen between endometrial measurement (>7 mm) and the pregnancy rate; moreover, the triple-layered pattern is visible in the same rate in infertile and fertile women (91% vs. 90%, respectively) . In order to increase accuracy of the ultrasound method and to find a predictor sign of pregnancy during IVF-treatment, the power-Doppler was proposed as the instrument for evaluating endometrial competence, but no statistically significant differences in the vascularization indexes of endometrial and sub-endometrial blood flows were found between pregnant and non-pregnant groups .
Several molecular markers of endometrial competence have been discovered in blood and uterine fluid analysis and endometrial biopsy, such as estradiol, progesterone, pinopodes, glycodelin, IL-1 system, cytokines, integrins and HOXA genes . Serum progesterone assay has been proposed as surrogate of endometrial competence but it has been observed that no minimum progesterone levels can define a “fertile” luteal phase because progesterone concentrations may fluctuate up to eightfold within 90 min and its levels peak 6 to 8 days after the ovulation .
Among the over one hundred different genes identified, all potential biomarkers of the implantation window, few studies exist with adequate power and validity based on study design to help determine which biomarkers have the greatest value and consistency; therefore, no reliable methods to assess “receptivity” have been established or adequately validated . Nevertheless, the transcriptomic approaches and microarray technology make it possible to identify biomarkers of endometrial receptivity and to report modifications related to the gene expression profile associated with the transition of the human endometrium from a pre-receptive to a receptive state during a natural cycle and, most of all, to reveal either moderate or strong alterations of endometrial receptivity under controlled ovarian stimulation (COS) protocols . In fact, a study analyzed the endometrial gene expression profiles by paired samples from the same patients during the pre-receptive to the receptive transition both in a natural and a subsequent stimulated cycle . This study has shown that COS regimens affect the transcriptomic pattern of endometrial cells in comparison with natural cycles; in particular, there were numerous differences in the main systems involved in the implantation process, such as the TGFb signaling pathway, the complement and coagulation cascades and leukocyte transendothelial migration . This information could open new perspectives, particularly in patients with multiple implantation failures . A recent study revealed that some cases of repeated implantation failure could be associated with an aberrant gene expression profile, particularly of transcripts related to the immune function and complement activation, and altered progesterone signaling might be an underlying mechanism for such endometrial gene expression deregulation in women with repeated implantation failure . Other research has been focused on the family of fibroblast growth factors (FGFs), which affects embryo implantation and supports improved endometrial trophoblastic interaction . Compared with the fertile group, FGF-1 is not expressed strongly enough in the failed IVF patients, which may have caused a lack of endothelial cell migration, important for implantation, stopped the process of blood vessel formation or induced early vascularization of implantation problems .
However, in the current state, no type-IA evidence has been proposed by the main scientific societies for assessing endometrial competence in infertile couples. In fact, the Royal College of Obstetricians & Gynecologists (RCOG) guidelines for infertility do not recommend the use of basal body temperature charts nor an endometrial biopsy to evaluate the luteal phase in routine infertility investigation . The blood test measurement of serum progesterone in the mid-luteal phase of the cycle is recommended but with type-B evidence . At the same time, according to the recommendation of the American Society for Reproductive Medicine (ASRM), no diagnostic test for luteal phase insufficiency has been proven to be reliable in a clinical setting .
Luteal phase support in clinical practice
The discussion regarding the clinical role of LPD is mainly due to the lack of reproducible and reliable diagnostic tests. Nevertheless, LPD is a common condition, that occurs during a series of clinical patterns characterized by low follicular-stimulating hormone (FSH) levels, altered follicular FSH/luteinizing hormone (LH) ratio and/or abnormal FSH and LH pulsatility, such as functional hypothalamic amenorrhea, thyroid and prolactin disorders , obesity and PCOS and during COS for IVF cycles . Therefore, the correct approach to the treatment of LPD is the identification and correction of any underlying condition. On the other hand, in case of no underlying dysfunction, the treatment becomes empiric .
Notwithstanding the empiricism of the LPD treatment, there are different clinical conditions where luteal phase support may be more or less useful, and can drive the choice for a specific luteal phase support.
Spontaneous ovulatory cycles
In spontaneous cycles there are no direct data on the efficacy of luteal phase support for improving fertility. A recent Cochrane meta-analysis on miscarriage prevention showed no statistically significant difference between women receiving progestogens (either natural and synthetic) and those receiving only placebo or no treatment (OR 0.99, 95 % CI 0.78 to 1.24) . The same meta-analysis found a statistically significant reduction (OR 0.39, 95 % CI 0.21 to 0.72) in miscarriage rate after progestogen administration only in women with recurrent miscarriage, i.e. three or more consecutive miscarriages. Moreover, these findings deserve further studies because the trials included were of poorer methodological quality . Also for the treatment of threatened miscarriage, there is insufficient evidence to support the routine use of progestogens (whether natural or synthetic) . In conclusion, in natural and unstimulated cycles, no treatment for LPD has been shown to improve pregnancy outcomes .
Non-gonadotropin induced ovulatory cycles
The non-gonadotropin induced ovulatory cycles mainly use clomiphene citrate (CC), while aromatase inhibitors (AIs) and metformin are still considered experimental treatments and to be employed in selected cases, respectively.
CC is a nonsteroidal selective estrogen modulator with both estrogenic and anti-estrogenic properties that interferes with the normal feedback mechanisms and leads to increased and prolonged FSH and LH secretion, which in turn stimulates follicular development. Cumulative CC administrations induce weak and prevalent anti-estrogenic effects on sensitive tissues, such as endometrium and/or ovary-related luteal phase defect and/or folliculogenesis alterations . The competitive binding of the CC with estrogen receptors makes the estrogenic sensitive organs less-sensitive to endogenous and exogenous steroid . Moreover, CC administration appears to be related to a reduced perifollicular vascularization and to a rate of high grade follicles significantly lower than in healthy controls . Also the vascularization of corpus luteum is significantly lower and with higher resistance in CC cycles, considering this hypo-vascularization of corpus luteum as a possible LPD cause .
To this regard, several data are available in literature about luteal phase progesterone support after ovulation induction in intrauterine insemination (IUI) cycles. A recent meta-analysis concluded that progesterone support did not benefit the clinical pregnancy rate in patients undergoing ovulation induction with CC (RR 0.89, 95 % CI 0.47 to 1.67) nor a significant difference in miscarriage per cycle between the two groups (OR 1.03, 95 % CI 0.52 to 2.04) . Of note, no heterogeneity in the findings obtained was detected and there were no data on live birth to perform data synthesis. The same results were obtained in another recent meta-analysis . In CC-stimulated cycles, progesterone administration did not improve the reproductive outcomes, such as the live birth rate (RR 1.30, 95 % CI 0.68 to 2.50) and the clinical pregnancy rate (RR 1.17, 95 % CI 0.82 to 1.65) and no differences were observed regarding miscarriage rate between the progesterone-treated and not-treated groups in the CC stimulation protocol (RR 1.14, 95% CI 0.63 to 2.06) .
In CC-stimulated cycles, the combined administration of estrogens and progesterone was tested in clinical studies showing not univocal results. In fact, endometrial estrogen receptors are blocked by competitive binding of the CC. Thus, they could be not stimulating by estrogens. Moreover, a RCT demonstrated that ethinylestradiol administration from day 8 for 5 days (0.05 mg daily) significantly improves the efficacy of progesterone support administered intramuscularly . Similarly, another RCT showed a significant reproductive benefit of oral estradiol administration (3 mg daily in two administrations from cycle day 8 until ovulation) followed by vaginal progesterone in CC cycles . Finally, a very recent randomized, double-blind, placebo-controlled trial compared in PCOS women the clinical pregnancy rate between two groups treated with CC plus ethinyl estradiol and CC plus placebo, respectively. The study resulted in an increase of clinical pregnancy rate in the CC + EE group (29 % vs 10 %, p = 0.02) even if there was no statistically significant difference in the ongoing pregnancy rate between the two groups . However, an interesting recent retrospective cohort analysis revealed that luteal phase progesterone supplementation in CC-IUI cycles can improve endometrial receptivity with an effect closely related to the endometrial thickness . Patients who appeared to receive the greatest benefit of progesterone supplementation had an endometrial thickness of 6–8 mm; their clinical pregnancy rate was found to be improved two-fold (OR 2.04; 95 % CI 1.01 to 4.14) . These patients seem to have an endometrium still receptive to progesterone administration, whereas patients with an endometrial thickness less than 6 mm are not responsive to progesterone supplementation for CC-related receptors depletion/inhibition and patients with an endometrial thickness greater than 8 mm represent a group with good reproductive prognosis in which progesterone supplementation is unable to provide further reproductive improvement.
Metformin improves the regular menstrual cycles and increases ovulation rate in patients with PCOS, although the efficacy of the drug is extremely variable both between different PCOS populations and within the same population . The efficacy of metformin in inducing ovulation in patients with PCOS is probably due to a direct action of the drug on the ovary; in fact, the ovulatory response to the drug seems to be related more to local drug sensitivity or resistance than to improvements in the systemic hormonal and/or metabolic environments , as shown by the analysis of follicular fluid that seemed to confirm that metformin acts directly on the ovary, improving local levels of androgens, ovarian insulin resistance and the levels of several growth factors . As regards the use of metformin as ovulation inductor, unfortunately few and no direct data exist concerning the need of a luteal phase support. Under metformin treatment, the hormonal pattern and the ovarian dynamics of the ovulatory cycles were found to be similar to spontaneous cycles as observed in normo-ovulating controls . In addition, the preovulatory follicles obtained under metformin treatment had a vascularization similar to that observed during natural cycles of healthy women, and the rate of high grade preovulatory follicles observed in women with PCOS who ovulated with metformin was not significantly different from controls . Finally the vascularization around the corpus luteum was found to be similar between metformin and spontaneous cycles, confirming the beneficial effects of metformin on the corpus luteum function .
Letrozole is an aromatase inhibitor recently employed as ovulation inducer. A recent large multicenter randomized double-blind parallel controlled trial, published in 2014, demonstrated the superiority of letrozole as first-line therapy for anovulatory infertility in women with PCOS when compared with CC . A systematic review of RCTs with meta-analysis concluded that letrozole is associated with significantly higher live birth rates than with CC (OR 1.64, 95 % CI 1.32 to 2.04) and with significantly higher clinical pregnancy rate compared to CC (OR 1.40, 95% CI 1.18 to 1.65) even if the quality of the evidence was rated as low . Letrozole acts by inhibiting the aromatase, a cytochrome P450 enzyme complex which is responsible for androgen to estrogen conversion, so it induces a hypoestrogenic state that stimulates, through activation of hypothalamic-pituitary axis, increased FSH secretion and ovarian follicle development . It does not exert an antagonist effect on endometrial estrogen receptors; on the other hand suppression of the systemic estrogen levels and in peripheral tissues can result in up-regulation of the estrogen receptors in the endometrium, leading to rapid endometrial growth once follicle development starts and estrogen secretion is restored .
Controlled ovarian stimulation (COS) with gonadotropins for non-IVF cycles
In non-IVF cycles with gonadotropins COS, two recent meta-analyses [45, 46] have demonstrated the benefit in improving reproductive outcomes with vaginal progesterone use as luteal phase support. Specifically, in the first meta-analysis, the clinical pregnancy rate was increased to 70% (OR 1.77, 95 % CI 1.20 to 2.60), as well as the likelihood of live birth per cycle (OR 2.63, 95 % CI 1.42 to 4.80), in patients receiving progesterone support . In the second meta-analysis, patients treated with FSH showed higher biochemical pregnancy (RR 1.81, 95 % CI 1.36 to 2.43), clinical pregnancy (RR 1.57, 95 % CI 1.15 to 2.15) and live birth (RR 2.28, 95 % CI 1.49 to 3.51) rates on receiving progesterone supplementation . No differences were observed regarding miscarriage rate between the two study groups (progesterone treated and not-treated groups) . In both meta-analyses [45, 46], data heterogeneity was low and not significant.
To the regard of the optimal progesterone dosage to use in non-IVF COS cycles with gonadotropins, very few clinical evidences are available in literature. A very recent RCT, evaluating two doses of vaginal progesterone for IUI cycles in terms of pregnancy rates, demonstrated that 300 mg of intravaginal micronized progesterone should be the maximum dosage for luteal phase when compared with 600 mg .
Controlled ovarian hyperstimulation (COH) with gonadotropins for IVF cycles
To explain in detail the role of progesterone and/or progestogens for the luteal phase support in gonadotropins COH for IVF cycles, the issue should be approached considering conventional and new protocols of gonadotropins COH separately.
Conventional COH protocols
Conventional COH protocols with gonadotropins have seen the transition over time from use of depot GnRH-agonist plus high-dose gonadotropins (225 IU/daily) in step-down or chronic regimen to daily GnRH-agonist plus low-doses gonadotropin (150 IU/day) in the patients’ response adjusted regimen, i.e. a “mild stimulation”. Ovarian hyper-stimulation was seen to affect embryo quality as assessed by morphology as well as the chromosomal constitution of the embryos [56–58], as result of interference with the natural selection of good-quality oocytes or the exposure of growing follicles to the potentially negative effects of ovarian stimulation . Mild stimulation approaches, aiming at a more physiological response, seem to improve embryo quality . In fact, a meta-analysis combining the results of three separate RCTs suggests that the retrieval of a reduced number of oocytes following mild stimulation is associated with a higher implantation rate compared with patients where the same number of oocytes is retrieved following conventional stimulation . Mild stimulation approaches might improve also embryo implantation rates .
In both these conventional COH for IVF cycles, an LFD is always present and it is due to supra-physiological estradiol levels which inhibit LH release, as negative feed-back mechanism, and alter endometrial responsiveness to progesterone . Moreover, the use of GnRH agonist and antagonist implies luteolysis and decreased LH pulsatility because of a competitive receptor’s blockage.
As result, in conventional IVF cycles, after hCG trigger, hCG levels rapidly increase until maximum levels at the time of oocyte retrieval, before falling to the lowest levels approximately 5 days after the retrieval . The trend of progesterone follows that of hCG more slowly, with a peak near to 5 days after oocyte retrieval, then decreases rapidly . Therefore, it creates a window during which progesterone lacks hCG stimulation to reach the threshold of 80–100 nmol/L, necessary to maintain the pregnancy . The luteal support is required in this time interval.
The main guidelines of the scientific society agree to recommend exogenous progesterone supplementation in assisted reproductive technologies (ARTs) cycles [68, 69]. A recent Cochrane meta-analysis regarding luteal phase support for ARTs cycles confirmed that progesterone exerts a significant positive effect on clinical pregnancy (OR 1.89, 95 % CI 1.30 to 2.75) and on pooled live birth or ongoing pregnancy (OR 1.77, 95 % CI 1.09 to 2.86) rate . However, when the analysis was restricted to live birth rate alone, evidence suggested no significant effect of the progesterone administration (OR 4.21, 95 % CI 0.93 to 19.18) . Surprisingly, in a previous Cochrane review the same authors concluded that progesterone had a significant positive effect on live birth rate compared to placebo, although in both reviews the effect of progesterone on the live birth derived from the results of a single study. In the most recent Cochrane , no significant data heterogeneity was detected, although the quality of the evidence was considered as very low to low. On the other hand, even if the hCG administration for luteal phase support is also effective, i.e. no differences between progesterone and hCG in rates of live birth or ongoing pregnancy (OR 0.95, 95% CI 0.65 to 1.38), its use increases the ovarian hyperstimulation syndrome (OHSS) risk by more than four-fold compared to placebo (OR 4.28, 95 % CI 1.91 to 9.60) and compared to the progesterone group (OR for OHSS of progesterone group vs. hCG group: 0.50, 95 % CI 0.34 to 0.76) .
Progesterone has different pharmacokinetic and pharmacodynamics properties when used in different routes of administration. In fact, intramuscular progesterone in oil formulation is related to higher serum progesterone concentrations and lower endometrial concentrations (unlike vaginal route) . In addition, in consideration of a half-life longer than one day, a depot effect with a continuous release over time without intermittent peaks (as in the case of vaginal route) has been observed for intramuscular progesterone . These pharmacokinetic properties of intramuscular route allow a wider implantation window and less endometrial contractions per minutes on the day of the embryo transfer (1 vs. more than 4 endometrial waves) . Unfortunately, intramuscular progesterone in oil formulation is available only in a few countries and represents a very painful route of administration. Similarly, the other available options for delivering exogenous progesterone have other restrictions, such as a low absorption rate for oral administration, due to the first liver-pass effect and potential side effects due to its metabolites, or the need for daily repeated administrations of vaginal progesterone, which can result in an uncomfortable route and with absorption variability due to leakage . Different routes of progesterone administration have been compared in clinical studies. An initial meta-analysis found no significant differences in treatment outcomes between vaginal and intramuscular progesterone administration. More recently, also a new systematic review and meta-analysis showed that there were no statistically significant differences in live birth rate between intramuscular vs. oral (OR 0.71, 95 % CI 0.14 to 3.66) and vs. vaginal or rectal (OR 1.31, 95 % CI 0.84 to 2.05) route. The same statistical inconsistence remained between the different routes of administration also when other intermediate endpoints, such as the clinical pregnancy rate, were evaluated . Moreover, results were obtained after data extrapolation from few studies on little study’ populations .
Regarding to the dose of progesterone, the most recent Cochrane review demonstrated no differences in live birth or ongoing pregnancy rate (OR 0.97, 95 % CI 0.84 to 1.11) between standard (90 mg/day) or high (equal or more than 100 mg/day) doses of vaginal progesterone without heterogeneity among studies . Moreover, the RCTs synthesized included different formulations and were of suboptimal quality. Both histopathological and clinical studies suggested that vaginal progesterone in doses lower than 300 mg/day are less efficacy to intramuscular progesterone (100 mg daily) whereas no difference in efficacy was demonstrated when doses of 600 mg/day of vaginal progesterone were used . In addition, another recent RCT comparing low-dose of micronized vaginal progesterone (200 mg twice daily) versus high-dose (200 mg three times daily) confirmed that the first regimen is less effective in term of clinical and ongoing pregnancy . Moreover, a noninferiority RCT found no statistical differences in clinical pregnancy rate, implantation rate and miscarriage rate between intramuscular (50 mg/day) and vaginal progesterone, when administrated as micronized progesterone bioadhesive gel both at standard (90 mg/day) and high-dosage (90 mg twice a day) . The efficacy of micronized progesterone in bioadhesive gel formulation (90 mg/day) was also confirmed in a multicentre RCTs demonstrating no statistical differences in live-birth rate and pregnancy rates when compared with other vaginal formulations (100 mg), although used at higher dosages [81, 82].
The goals for new progesterone treatment seem to be represented by aqueous subcutaneous progesterone. Two large multicentre RCTs demonstrated that the 25 mg subcutaneous progesterone administered once daily is effective and well tolerated [83, 84]. Both 25 and 50 mg daily doses of aqueous subcutaneous progesterone led to similar secretory transformation of endometrium in reproductive-aged women who were down-regulated with GnRH agonist and treated with estradiol suggesting that 25 mg should be considered the lowest effective dose. The subcutaneous administration maximizes the kinetic profile and the physiological response, i.e. lower dose needed, minimizing side effects [85, 86]. In fact, it reduces patient discomfort (fewer skin reactions) and it may be appealing to women who prefer to avoid the intramuscular and vaginal routes of administration . This dose also relates to the daily production of progesterone, which has been estimated to be 25 mg.
Other compounds were assessed alone or in combination with progesterone for luteal phase support in conventional IVF cycles. As explained earlier, hCG should be avoided as it leads to a higher risk of OHSS (see above) and the addition of estrogens to progesterone did not show a better effect than progesterone alone in the rates of ongoing pregnancies (OR 1.12, 95 % CI 0.91 to 1.38) and live births (OR 1.32, 95 % CI 0.93 to 1.86) . The only significant result was found in the comparison of progesterone versus progesterone plus GnRH agonists . Specifically, the addition of GnRH agonist to progesterone had a significant greater effect than progesterone alone in live birth rate (OR 0.40, 95 % CI 0.26 to 0.61), clinical pregnancy rate (OR 0.71, 95% CI 0.59 to 0.87) and pooled live birth/ongoing pregnancy rate (OR 0.67, 95 % CI 0.56 to 0.81) . Although a high statistical heterogeneity was detected in this analysis, the direction of the effect was considered consistent. The effectiveness of the GnRH agonist administration in luteal phase support has been confirmed both in GnRH agonist and GnRH antagonist COHs . Moreover, the luteal support with GnRH agonist requires repeated daily administrations. In fact, due to the pharmacokinetics of the GnRH agonist-induced LH releases (see below), a clear-cut positive correlation between the frequency of GnRH agonist administration (i.e., buserelin) and serum progesterone levels was observed [88, 89].
In literature, there is a debate regarding also the timing of progesterone administration, once a decreased likelihood of pregnancy has been observed if progesterone is initiated both before oocyte retrieval or 6 days after oocyte aspiration . A systematic review of 5 RCTs with a total of 872 patients and 1010 cycles shows the presence of a window for progesterone start, occurring between the evening of oocyte retrieval and day 3 after oocyte retrieval, when the embryo-to-endometrial synchrony and exogenous luteal phase support seem to be optimized . Of these RCTs, only one reported live birth rates, finding no differences in live birth between patients randomized to receive progesterone 36 h before oocyte retrieval, the evening of the oocyte aspiration or day 3 after oocyte retrieval even if this study was not powered to detect a difference in live birth rate. In fact, all studies reported the clinical pregnancy rate as a primary outcome. A lower clinical pregnancy rate was detected in patients starting progesterone 12 h before oocyte retrieval compared with those patients starting progesterone the evening of oocyte retrieval (12.9% vs. 24.6%, respectively; P = 0.01) . Similarly, a lower clinical pregnancy rate was observed in patients undergoing fresh autologous IVF and starting progesterone 6 days after oocyte retrieval when compared with starting progesterone 3 days after oocyte retrieval (44.8% vs. 61.0%, respectively; P = 0.05) . Finally, the last three studies that compared the clinical pregnancy rate in patients starting progesterone the evening of oocyte retrieval versus 2 days after and 3 days after oocyte retrieval did not detect significant differences between the groups [91, 94, 95].
The optimal duration of progesterone supplementation after IVF cycles has also been the object of debate. A meta-analysis assessed the effects of different lengths of progesterone treatment . No statistical difference in live birth (RR 0.95, 95 % CI 0.86 to 1.05), miscarriage (RR 1.01, 95 % CI 0.74 to 1.38) and ongoing pregnancy (RR 0.97, 95 % CI 0.90 to 1.05) rate between an early progesterone cessation at the first positive pregnancy test and a progesterone continuation until 6th-7th weeks of pregnancy was observed . No or low heterogeneity was observed among studies in the main outcome measures .
Non-conventional COH protocols
In order to optimize the safety of the COH in selected populations, such as cancers patients and donors, new COH protocols have been initially proposed, and further extended to hyper- and normo-responders to achieve the “OHSS-free clinic” dream . The “non-conventional” COH protocols consist of using GnRH antagonist for avoiding LH surge and standard/high gonadotropin dosage [98, 99]. In case of no or low OHSS risk, multiple ovulation triggering can be induced by conventional hCG administration [98–103]. On the other hand, in case of moderate and/or high OHSS risk, GnRH agonist superovulation triggering followed by elective [100–103] or non-elective cryopreservation programs (segmentation of IVF cycles) [98, 99] can be an option. The “IVF cycle segmentation” consists in the cryopreservation of all embryos produced and their replacement in a receptive non-stimulated endometrium, such as in a natural cycle, or after artificial endometrial preparation . This concept is also supported by positive reproductive outcomes of existing RCTs in favor of a strategy of frozen embryo transfer, although same aspects remain unclear as well as the higher incidence of “large baby syndrome” . Thus, larger trials are needed before a critical change in clinical practice can be widely accepted .
From the first study analyzing the potential role of GnRH agonist to induce final follicular maturation, the concern of a deep LPD in these IVF cycles and the need of an intensive luteal phase support emerged . The use of standard treatment to support the luteal phase after GnRH agonist triggering is considered inadequate in consideration of the lower conception rates observed . In fact, GnRH agonist triggering shows a combined negative effect on the function of the corpus luteum and on the function of the endometrium [109, 110].
Relevant differences exist regarding the duration and profile of the GnRH agonist-induced surge of gonadotropins when compared with that of the natural cycle [111, 112]. The GnRH agonist-induced LH surge consists of two phases: a short ascending limb of about 4 h and long descending limb of about 20 h for a total length of 24–36 h. On the other hand, the mid-cycle LH-surge of a natural cycle is characterized by three phases: a rapidly ascending phase lasting 14 h, a plateau of 14 h and a descending phase of 20 h, for a total length of 48 h . Therefore, the total amount of LH released during the surge is significantly reduced when GnRH agonist is used to trigger ovulation compared with that in natural cycle or with hCG triggering . Moreover, the LPD after GnRH agonist triggering is not due to low serum LH levels but to the rapid half-life of LH. In fact, LH levels of 4–10 IU/L are sufficient to induce a physiologic peak of progesterone (25 nmol/L) , and even if the triggering with hCG induces a release in progesterone higher than 100–250 nmol/L, the serum progesterone levels of 80–100 nmol/L, as obtained after GnRH agonist triggering, should be sufficient to support the luteal phase [67, 114]. On the other hand, the half-life of LH is approximately 21 min vs. the half-life of hCG that is 12 h . Thus, the mean duration of a non-supplemented luteal phase after GnRH trigger may be as short as 4 days, compared with 13 days after hCG trigger .
Finally, biological changes in endometrium related to GnRH agonist use were also observed. In particular, differences were seen in endometrial gene expression, related to the type of ovulation trigger and luteal support. The mode of triggering is reflected in the transcriptome of the somatic cells of the follicle since they differentially expressed genes like ANGPT1 and SEMA3A in mural granulosa cells. This suggests an impaired induction of angiogenesis in the GnRHa-triggered as compared with the hCG-triggered, which in association with the lower postovulatory LH activity, and may explain the insufficient luteal phase observed after COH and GnRHa trigger . A RCT showed that the gene expression after GnRH-a trigger and the modified luteal support adding LH/hCG activity more closely resembles the pattern seen with the use of hCG for trigger and of a standard luteal support with vaginal progesterone .
The final result is a deeper LPD in GnRH agonist triggering cycles compared with hCG triggering cycles. Actually, the scientific debate still remains open regarding the role of GnRH agonist in trigger ovulation because the most recent and updated Cochrane review about GnRH agonist vs. hCG for oocyte triggering in antagonist ARTs largely confirmed the same conclusions as the previous one . In particular, the Authors concluded that GnRH agonist, as a final oocyte maturation trigger in fresh autologous cycles, is associated with a lower live birth rate, a lower ongoing pregnancy rate (beyond 12 weeks of amenorrhea) and a higher rate of early miscarriage (less than 12 weeks) . Thus, GnRH agonist triggering could be useful and employed only for women who choose to avoid fresh transfers, women who donate oocytes to recipients or women who wish to freeze their eggs for later use . In reality, a data synthesis of reproductive results may be not feasible since the studies included were not comparable due to difference in luteal phase support protocols .
New and different regimens have been proposed for a greater luteal phase support in GnRH-a triggering: the intensive or “American” approach which consists of an aggressive steroidal support (intramuscular or vaginal progesterone plus transdermal estradiol) with adjuvant low-dose hCG trigger only in selected cases, such as women with peak serum E 2 less than 4000 pg/ml on the day of trigger, and the moderate or the “European” approach which promotes the production of endogenous steroids by the corpus luteum via exogenous hCG supplementation, immediately after the oocyte retrieval, at dose low enough to avoid the development of OHSS . Youssef et al. highlighted that the modified luteal phase support with LH/hCG (the European concept) is associated with pregnancy rates almost comparable with those of hCG triggering cycles, albeit still significantly lower. Another regimen suggested for a more sustained luteal support is the use of one bolus of 1500 IU hCG concomitant with GnRH-a (dual trigger) 34–36 h before oocyte retrieval [122, 123]. With the dual trigger, acceptable rates of implantation, clinical pregnancy, ongoing pregnancy rates, and early pregnancy loss has been achieved in high responders [122, 123], even if the incidence of clinically significant OHSS was not eliminated, but rather reduced to 0.5% . In fact, the minimal hCG activity needed for luteal phase support without inducing late-onset OHSS is not known. In a RCT, two cases of moderate OHSS out of 125 patients (considered normal responders) treated with GnRH agonist triggering plus 1500 IU hCG on the day of oocyte retrieval and an additional bolus of 1500 IU of hCG 5 days after the oocyte retrieval were reported . Thus, that protocol for luteal phase should be avoided because of a persistent OHSS risk . Finally, a case of recurrent empty follicle syndrome has recently been described, successfully treated by ovulation trigger with GnRH-a 40 h and hCG added 34 h prior to oocyte retrieval (double triggering) . This new regimen is based on the concept of prolonging the time interval between ovulation triggering with GnRH-a and oocyte retrieval with the consequent simultaneous induction of an FSH surge; thus the "double trigger" could overcome any existing impairments in granulosa cell function, oocyte meiotic maturation or cumulus expansion, resulting in successful aspiration of mature oocytes, pregnancy and delivery . In line with these results, the double triggering was later offered also to two groups of patients demonstrating abnormal final follicular maturation despite normal response to COH, those with low (<50%) number of oocytes retrieved per number of dominant follicles (i.e. > 14 mm in diameter) on the day of hCG administration and those with low proportion of mature/metaphase-II (MII) oocytes (<66%) per number of oocytes retrieved . In both groups, with double triggering, patients showed significantly higher number of oocytes retrieved, higher number of mature oocytes per number of oocytes retrieved, with a tendency toward a higher number of top-quality embryos, as compared to the hCG-only trigger cycles [128, 129].
Summarizing, available evidence supports the use of GnRH agonist trigger as a helpful approach in women with moderate and/or high risk for OHSS. In fact, in women with an extreme response to stimulation a GnRH agonist trigger will be followed by a "freeze-all" strategy to eliminate OHSS risk, while in patients with a low-moderate OHSS risk the GnRH-agonist trigger followed by intensive luteal phase support could allow fresh transfer with good, albeit inferior, reproductive outcome and a significantly reduced risk of OHSS. Finally, in low OHSS risk patients, a GnRH agonist trigger with modified luteal support (hCG rescue bolus after oocyte retrieval) could be an alternative to conventional hCG trigger, given the excellent pregnancy rates while GnRH-a and hCG may be offered concomitantly, 35–37 h prior to oocyte retrieval (i.e. dual trigger) or 40 h and 34 h prior to oocyte retrieval respectively, i.e. double trigger, in women with abnormal final follicular maturation .
Among these women, GnRH agonist trigger also introduces the possibility of an “exogenous progesterone-free” luteal phase, that relies solely on endogenous progesterone production from the corpus luteum driven by small boluses of LH activity (hCG) administered during luteal phases [130, 131]. This innovative option for luteal phase support needs to be corroborated in future large trials but it could bring an end to inconvenient vaginal discharge and/or to painful intramuscular injections of progesterone .
Donor and/or frozen-thawed cycles
Due to increased mean age of maternity in developed countries, IVF using donor oocytes is an increasingly used infertility treatment option for women with irreversible ovarian function loss or primary ovarian failure showing a pregnancy and implantation rates in oocyte donation cycles higher than in standard IVF/ICSI cycles . In these cases, the support of the luteal is an essential prerequisite. In addition, as explained above, more and more frozen-thawed embryo transfers (FET) are employed with the aim of avoiding OHSS in high-risk patients. FET makes it possible for the embryos generated by IVF and ICSI to be stored and utilized later, with a segmentation of the IVF cycle, separating the ovarian stimulation and trigger from the fresh transfer, and vitrifying all embryos for subsequent transfer in frozen thawed cycles.
Moreover, not only for reducing the OHSS risk, in recent years the amount of FET cycles has dramatically increased due to the trend towards transferring fewer embryos after a fresh IVF cycle, and as a result of improved laboratory techniques [134, 135] until to the emerging “freeze-all” policy as an alternative to fresh embryo transfer (ET) to improve IVF outcomes . A recent prospective, observational, cohort study has suggested that the “freeze-all” policy is advantageous when compared with fresh ETs . Clinical (RR 1.29, 95 % CI 1.05 to 1.59) and ongoing pregnancy (RR 1.28, 95 % CI 1.01 to 1.62) rates were significantly better after FET . In addition, a recent meta-analysis of observational studies has shown that singleton pregnancies after FET are associated with a lower risk of obstetric outcomes, such as small for gestational age baby (RR 0.45, 95 % CI 0.30 to 0.66) and preterm birth (RR 0.84, 95 % CI 0.78 to 0.90) .
Unlike the complex ovarian stimulation protocols of IVF and ICSI cycles, in the FET cycles, the primary target is limited to adequate preparation of the endometrium to receive the thawed embryos. The endometrial preparation can be achieved by means of two different methods, a natural cycle (NC) which consists of using endogenous sex steroids production from developing follicles in patients with regular ovulatory cycles and the artificial cycles (AC), where the endocrine preparation of the endometrium is achieved by sequential administration of estrogens and progestogens to mimic a normal cycle, more suitable for women without regular ovulatory cycles. The entire success of the NC-FET cycle depends on the correct identification of ovulation and calculation of the likely subsequent period of optimal endometrial receptivity [138, 139]. The identification of ovulation is based on detecting the spontaneous LH surge, therefore LH levels need to be regularly (daily) monitored in blood or urine samples, because it is assumed that ovulation will occur 36 – 40 h after LH peak . Unfortunately, the LH surges in urine lag up to 21 h behind the appearance of the blood surge, creating difficulties in correct assessing of data. To overcome these problems, a modified-NC (mNC) has been proposed, which consists of hCG triggering when dominant follicle is of optimal size (>17 mm), after an ultrasound evaluation to ensure the appropriate timing of hCG administration.
In the AC-FET the timing of embryo thawing and transfer is planned according to the moment of progesterone supplementation but the sequential administration of exogenous sex steroids does not guarantee complete pituitary suppression (at least 5% of early luteinizations, ), putting the endometrium at risk of early exposure to progesterone. Also for the AC cycle, a modified AC-FET (mAC) has been proposed consisting in the co-treatment with GnRH agonist in order to down-regulate the pituitary and prevent early follicular luteinizations .
A recent meta-analysis regarding the optimal endometrial preparation method in FET revealed no significant advantage of one specific approach to prepare the endometrium for FET in terms of clinical pregnancy rates or live birth rates . In particular, no significant benefit of true NC-FET versus modified NC-FET with regard to clinical pregnancy (OR 0.91, 95 % CI 0.74 to 1.10), ongoing pregnancy (OR 1.0, 95 % CI 0.66 to 1.60) or live birth (OR 1.0, 95 % CI 0.63 to 1.60) was observed. Also luteal phase support was evaluated as potential influencing factor on pregnancy rate but the meta-analysis concludes that currently there is too little evidence supporting a positive effect of luteal phase support in patients undergoing NC-FET . More recently, a retrospective study evaluating 228 patients showed that progesterone for luteal phase supplementation decreases miscarriage rate and improves live birth rate in NC-FET cycles but, due to different study designs, the overall conflicting results of the available studies, do not make it possible to conclude a positive effect of luteal phase support in patients undergoing NC-FET .
Other issues of discussion concerned the different timing of progesterone starting and routes of administration in donor and/or FET cycles. A Cochrane review analyzed the most effective endometrial preparation for women undergoing embryo transfer with frozen embryos and with embryos derived from donated oocytes. Starting progesterone on the day of oocyte aspiration (OR 1.92, 95 % CI 1.08 to 3.42) or on the day after (OR 1.81, 95% CI 1.01 to 3.24) led to higher pregnancy rate than when progesterone was started the day before oocyte retrieval . However, no difference was found between starting progesterone the day of oocyte aspiration or the day after (OR 0.94, 95% CI 0.53 to 1.68) [95, 145]. Finally, no differences were found (OR 0.75, 95% CI 0.24 to 2.34) between starting one or two days before the oocyte retrieval when the transfer is performed on blastocyst stage .
As regards progesterone route of administration, no statistical significant difference was detected in live-birth and clinical pregnancy rate between vaginal and intramuscular administration . On the contrary, in a retrospective cohort study women supplemented with vaginal progesterone gel had significantly lower rate of clinical pregnancy (36.9% vs. 51.1%, respectively) and live birth (24.4% vs. 39.1%, respectively) compared with intramuscular progesterone. The benefit of intramuscular route could be explained with the higher and continuous serum progesterone levels that act relaxing uterus and reducing the frequency of endometrial waves in the luteal phase that are associated with lower pregnancy rates . Moreover, the booster injection of intramuscular progesterone (50 mg IM, once every 3 days) did not improved the pregnancy rates of patients who received vaginal progesterone (100 mg, three times daily) .
The worst results in vaginal progesterone group probably depend on the dosage of vaginal progesterone. In a retrospective study patients treated with vaginal progesterone gel twice daily had a lower risk of pregnancy loss compared with women treated once a day (43.7% vs. 67.4%, respectively) resulting in a delivery rate (20.5% vs. 8.7%, respectively) more than two-fold higher . However, considering recipients who received vitrified blastocysts on day 6, no statistical differences were seen in implantation and pregnancy rate between vaginal and intramuscular progesterone luteal phase support .
Very few data are available regarding the best dosage of progesterone administration. A recent study seems to demonstrate that the optimal dose of intramuscular progesterone to for luteal phase support ranges between 50 and 100 mg/day . In fact, that dose lets to reach progesterone blood levels similar to those observed during the mid-luteal phase of a spontaneous ovulatory cycle. Moreover, an increase in progesterone dosage could be useful in donor and/or frozen-thawed AC cycles when the endometrial pattern at ultrasound is non-homogeneous hyperechoic in mid-luteal phase .
Conclusions
Current data do not completely explain the clinical significance of LPD and its treatment. Several mechanisms of the implantation physiology and of “implantation window” are still not completely understood and are difficult to study owing to the lack of adequate tools for assessing endometrial receptivity. Future perspectives should concern endometrial competence evaluation, with the aim of revealing its physiologic mechanisms and to identify a correct diagnostic test for LPD in order to recognize it when it occurs in infertile couples.
In fact, LPD, is not univocally considered an infertility factor, and, even if it can be associated with infertility, it should be treated only in selected cases. The correct approach to the treatment of LPD is the identification and correction of any underlying condition, such as hypothalamic amenorrhea, thyroid and prolactin disorders, obesity and PCOS and during COS for IVF cycles. The available scientific literature is in agreement that in all gonadotropin COS/COH cycles for IVF and non-IVF cycles, progesterone supplementation is needed for luteal phase support, showing an improvement of the main reproductive outcomes. The protocols of luteal phase support with progesterone change according to the type of protocol of ovarian stimulation, i.e. conventional or non-conventional COH. Among different routes of administrations, there are no statistically significant differences between intramuscular progesterone and the others routes (vaginal, oral and rectal) for IVF cycles, fresh and FET, and the optimal timing and doses are not yet known.
Much of the current scientific evidence is based on reviews and meta-analyses of observational studies and on few RCTs, thus the future perspectives should take into consideration the implementation of randomized trials and the evaluation of the promising efficacy of subcutaneous progesterone at high-doses in the new COH cycles in which GnRH agonist are used to trigger multiple ovulation. Thus, at the moment, the scientific debate still remains open regarding progesterone administration protocols, in particular routes of administration, dose and timing and the potential association with other drugs, and further research is still needed.
Abbreviations
ART:
Assisted reproduction technology
LPD:
Luteal phase deficiency
HCG:
Human chorionic gonadotropin
NCS:
Nucleolar channel system
PCOS:
Polycystic ovarian syndrome
IVF:
In-vitro fertilization
COS:
Controlled ovarian stimulation
FGF:
Fibroblast growth factors
FSH:
Follicular stimulating hormone
LH:
Luteinizing hormone
CC:
Clomiphene citrate
AI:
Aromatase inhibitor
IUI:
Intrauterine insemination
OHSS:
Ovarian hyperstimulation syndrome
COH:
Controlled ovarian hyperstimulation
GnRH:
Gonadotropin releasing hormone
ICSI:
Intracytoplasmic sperm injection
FET:
Frozen-thawed embryo transfer
NC:
Natural cycle
AC:
Artificial cycle
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Centre of Reproductive Medicine and Surgery, Arcispedale Santa Maria Nuova - IRCCS, Viale Risorgimento 80, 42123, Reggio Emilia, Italy
Stefano Palomba&Susanna Santagni
Centre of Reproductive Medicine and Surgery, Arcispedale Santa Maria Nuova - IRCCS, University of Modena and Reggio Emilia, Via Università 4, 41100 Viale Risorgimento 80, 42123, Modena, Italy
Giovanni Battista La Sala
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1. Stefano PalombaView author publications Search author on:PubMedGoogle Scholar
2. Susanna SantagniView author publications Search author on:PubMedGoogle Scholar
3. Giovanni Battista La SalaView author publications Search author on:PubMedGoogle Scholar
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Correspondence to Stefano Palomba.
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The authors declare that they have no financial and non-financial competing interests in relation to this manuscript.
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SP substantially contributed to conceive, design and draft the manuscript, revising it critically for important intellectual content; SS substantially contributed to collect data, analyze the results, and draft the manuscript; and GBLS contributed to design the study and to revise critically the work. All authors gave final approval of the version to be published and agree to be accountable for all aspects of the work in ensuring that questions related to the accuracy or integrity of any part of the work.
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Palomba, S., Santagni, S. & La Sala, G.B. Progesterone administration for luteal phase deficiency in human reproduction: an old or new issue?. J Ovarian Res8, 77 (2015).
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Published: 19 November 2015
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Keywords
ART
Endometrium
Implantation
Luteal phase deficiency
Progesterone
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Abstract
Background
Implantation physiology
Endometrial competence evaluation
Luteal phase support in clinical practice
Spontaneous ovulatory cycles
Non-gonadotropin induced ovulatory cycles
Controlled ovarian stimulation (COS) with gonadotropins for non-IVF cycles
Controlled ovarian hyperstimulation (COH) with gonadotropins for IVF cycles
Donor and/or frozen-thawed cycles
Conclusions
Abbreviations
References
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Check JH, Dietterich C, Cohen R, Choe JK, Amui J, Brasile D. Increasing the dosage of progesterone (P) supplemention from the mid-luteal phase in women not attaining a mid-luteal homogeneous hyperechogenic (HH) pattern with sonography improves pregnancy rates (PRS) following frozen embryo transfer (ET). Clin Exp Obstet Gynecol. 2010;37:13–4.
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11037 | https://uploads-ssl.webflow.com/605fe570e5454a357d1e1811/60a030fa3997b95bed93724a_AP-Physics-I-Momentum.pdf | AP Physics Study Guide Momentum From Simple Studies, & @simplestudiesinc on Instagram All images are from the Openstax college physics textbook Momentum is defined as the product of a system’s mass multiplied by its velocity ● 𝑝= 𝑚𝑣 ○ The greater an object’s mass or velocity, the greater its momentum ○ Momentum is a vector having the same direction as the velocity Newton’s second law of motion in terms of momentum states that the net external force equals the change in momentum of a system divided by the time over which it changes ● 𝐹 𝑛𝑒𝑡= 𝛥𝑝/𝛥𝑡 ● It can be applied to systems where the mass is changing as well as to systems of constant mass Impulse is the change in momentum ● 𝛥𝑝= 𝐹𝑛𝑒𝑡𝛥𝑡 ○ A small force could cause the same impulse as a large force, but it would have to act for a much longer time ● Our definition of impulse includes an assumption that the force is constant over the time interval ○ However, forces are usually not constant ○ It is possible to find an average effective force that produces the same result as the the corresponding time-varying force The conservation of momentum principle states that when the net external force is zero, the total momentum of the system is conserved or constant ● 𝑝𝑡𝑜𝑡= 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 ● 𝑝𝑡𝑜𝑡= 𝑝′𝑡𝑜𝑡(isolated system) ○ An isolated system is defined to be one in which the net external force is zero An elastic collision is one that also conserves internal kinetic energy ● Internal kinetic energy is the sum of the kinetic energies of the objects in the system An inelastic collision is one in which the internal kinetic energy changes (it is not conserved) ● A collision in which the objects stick together is called “perfectly inelastic collision” ○ It reduces internal kinetic energy more than does any other type of elastic collision Two-dimensional collisions might cause objects to rotate before or after the collision ● To avoid rotation, we consider only the scattering of point masses - structureless particles that cannot rotate or spin ○ The conservation of momentum for the x-axis is 𝑝1𝑥+ 𝑝2𝑥= 𝑝′1𝑥+ 𝑝′2𝑥 ■ This then leads to 𝑚1𝑣1 = 𝑚1𝑣′1 𝑐𝑜𝑠𝜃1 + 𝑚2𝑣′2 𝑐𝑜𝑠𝜃2 ○ The conservation of momentum for the y-axis is 𝑝1𝑦+ 𝑝2𝑦= 𝑝′1𝑦+ 𝑝′2𝑦 ■ This then leads to 0 = 𝑚1𝑣′1 𝑠𝑖𝑛𝜃1 + 𝑚2𝑣′2 𝑠𝑖𝑛𝜃2 |
11038 | https://www.youtube.com/watch?v=TOReHEHhR6o | Math Olympiad – Topic: Geometry – Isosceles Triangle with the Apex Angle of 20° - Exercises – p24
Math Olympiad by MG Niasar
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Posted: 18 Apr 2025
A classic geometry problem on isosceles and congruent triangles, designed to challenge young students.
Transcript:
Hi, in this video we have a very famous geometric question that is about isocelis triangle with the aex angle of 20°. Usually this question is used to tease a students at the sixth or seventh grades after they study the topic of concurrent triangles and also similar triangles. So we have isocles triangle ABC is equal. This angle is 20°. We draw a line BD such that this angle is 20°. We draw a line C such that this angle is 30°. The question is what is the angle E DB? This red one. All right. So maybe spend 10 to 20 minutes try to solve this exercise and after that come back and see the solution. Actually there are many different solutions. Hopefully you will find one of them or you will come back and see what I will show you. Let us start. Okay. So we have the triangle here. Just by looking at this exercise, we already can obtain certain information. For example, this is an isoseris triangle. It means A is equal A C. And that means these two angle B and C they are equal. So if the AEX is 20° and these two are equal each of them will be 80° because 80 2 is 160 + 20 is 180. Okay, that is the first conclusion. because this angle B is 80°. Okay. And right now we know that here is 20°. So this one will be 60°. Okay. This is also 80° and this is 30. So that means this part E CB is going to be 50°. So this is 50. This B angle is 80. 50 + 80 is 130. That means this angle will also be 50. So EC this triangle is an isoselles triangle. That means EB is equal C. Okay. This is what I have written here. All right. So now in order to solve the exercise we are going to select point F on the side AC such that BF is equal to BC because BF is equal BC. FBC is going to be an isocles triangle and obviously since this one is already 80° this angle will also be 80° 80. So this angle will be 20°. Okay. So this is 20°. This is also 20° and we know that the total is 80. So this angle will be 40°. Okay. So this angle is 40° and we know that here we have 80. Therefore this angle will also be 40°. So let's see what I have written here. BF is equal BC that is okay. This is an isoelis that is 80 and that is 20 which means these two angles are going to be 40°. And based on this we can conclude that BF is equal to DF. So actually this triangle is also an isoceteris triangle. We said that B is equal to BC. Also we ourself we selected that BF should be equal to BC. Therefore BE is also equal to BF. So this one and this one they are total angle is 80°. This part is 20°. This angle E BF is going to be 60°. So here the triangle E BFF it is an isocles triangle because we have shown that B is equal to BF but also the angle in between them is 60° that mean EBF is going to be an equilateral triangle. So this triangle is an equilateral which means all the angles are going to be 60°. Very important information. PF is equal to DF. We have already shown it but we said that this is an equilateral EF is also equal to BF. Therefore we can say that DF is equal to EF. Okay. So this triangle is going to be also an isocetis triangle. Everything become isoceleris now in this exercise. Okay. So now everything is actually very simple. We know that this is 80. is 60. So, E FD this angle is going to be 180 - 60 - 80 which is 40°. So, this angle is 40° is the aex angle of this isocleser triangle EDF. Therefore, we can immediately conclude that this whole angle will be 180 minus 40. this one which becomes 140 / 2. So, ED DF is 140 / 2 which becomes 70°. So, this total angle is 70°. But we know that this little part is 40°. Therefore, E DB will become 30° and that means alpha is 30°. So, this is actually one solution which is very straightforward. we had to only select one point on one side and then everything will reveal itself. Okay. So we also have other solutions for this exercise. Many different solutions actually here I will give you some hint if you want to try it yourself. For example, you can show that E B D this triangle is similar to E A C because this angle for both of them are 20° and you can show the ratio of B to BD will be equal to AE to A C and therefore these two triangle will be similar and from there you can conclude that this angle is equal to this angle becomes Alternatively, you can actually introduce new lines. For example, here we said that this angle will be 60 because totally is 80 minus 20 is 60. We are going to make this blue line CF such that this angle is also 60°. We continue this because this is 60 and this is 60. This will be equilateral. You can show also this top one is an equilateral. And after that try to show E DG is concurrent to E DF. And if you show that that means these two angles are equal because total of them is 60. Half of it will be 30°. This is another solution. There are actually several more solutions that you can show uh the angle alpha is equal to 30°. All right. So this is end of the video. Let's see what we have learned. In geometric questions sometimes you need to introduce new lines, new points, circles maybe. So after introducing these new lines or shape you have to gather more information from the problem and put them together and basically solve the exercise. All right. So see you next time. Bye. |
11039 | https://www.geogebra.org/m/kxyfntj9 | Exploration: Circumference and Area of a Circle APS – GeoGebra
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Exploration: Circumference and Area of a Circle APS
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This applet illustrates formulas for the circumference by rolling a circle along a line, then measuring the length in units of the diameter, or the radius. Check the "Show rolling circle" box, then roll the circle with the slider labeled "Roll circle". Check "Show C measured with r, d." This leads to a formula for the area of a circle, too. Cut it apart into sectors, and rearrange into a "parallelogram". Check the last three boxes to see this.
Formulas: Circumference = diameter (that is, ) Circumference = 2 radius (that is, ) Area = Circumference radius Area = radius squared (that is, )
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11040 | https://courses.lumenlearning.com/uvu-introductoryalgebra/chapter/3-2-5-roots-on-variables/ | Chapter 3: Variables, Expressions and Equations
3.2.3: Radical Expressions
Learning Outcomes
Simplify square roots with variables
Recognize that by definition [latex]\sqrt{x^{2}}[/latex] is always nonnegative
Key words
Radical expression: an expression that contains radicals
Radical expressionsare expressions that contain radicals. Radical expressions come in many forms, from simple and familiar, such as[latex]\sqrt{16}[/latex], to more complicated, as in [latex]\sqrt[3\;]{250{{x}^{4}}y}[/latex]. In this section we will discover how to simplify expressions containing square roots.
Simplifying Square Roots
We have already investigated simplifying a radical expression with integers by using factoring and the product property of square roots. For example, [latex]\sqrt{45}=\sqrt{9\cdot 5}=\sqrt{9}\cdot\sqrt{5}=3\sqrt{5}[/latex]. We can use this same method to simplify a radical term that contains variables. However, we have to be careful.
Consider the expression [latex]\sqrt{{{x}^{2}}}[/latex]. This looks like it should be equal to [latex]x[/latex], right? Let’s test some values for[latex]x[/latex]and see what happens.
Suppose, [latex]x=5[/latex]. Then [latex]\sqrt{{{x}^{2}}}=\sqrt{{{5}^{2}}}=\sqrt{25}=5=x[/latex]. In this case, [latex]\sqrt{{{x}^{2}}}=x[/latex].
Now suppose [latex]x=-3[/latex]. Then [latex]\sqrt{{{x}^{2}}}=\sqrt{{{(-3)}^{2}}}=\sqrt{9}=3\neq x[/latex]. In this case, [latex]\sqrt{{{x}^{2}}}\neq x[/latex].
The table shows more examples.
| [latex]x[/latex] | [latex]x^{2}[/latex] | [latex]\sqrt{x^{2}}[/latex] | [latex]\left|x\right|[/latex] |
--- --- |
| [latex]−5[/latex] | [latex]25[/latex] | [latex]5[/latex] | [latex]5[/latex] |
| [latex]−2[/latex] | [latex]4[/latex] | [latex]2[/latex] | [latex]2[/latex] |
| [latex]0[/latex] | [latex]0[/latex] | [latex]0[/latex] | [latex]0[/latex] |
| [latex]6[/latex] | [latex]36[/latex] | [latex]6[/latex] | [latex]6[/latex] |
| [latex]10[/latex] | [latex]100[/latex] | [latex]10[/latex] | [latex]10[/latex] |
Notice that in cases where [latex]x[/latex] is a negative number, [latex]\sqrt{x^{2}}\neq{x}[/latex]! (This happens because the process of squaring the number loses the negative sign, since a negative times a negative is a positive.) However, in all cases [latex]\sqrt{x^{2}}=\left|x\right|[/latex]. We need to consider this fact when simplifying radicals that contain variables, because by definition [latex]\sqrt{x^{2}}[/latex] is always nonnegative.
When we square any exponential term, we multiply the exponent by 2. For example, [latex]\left ( y^3\right )^2=y^{3\cdot 2}=y^6[/latex]. This means that in order to take the square root of an exponential term, the exponent must be even.
Let’s consider [latex]\sqrt{y^6}[/latex]. We can write [latex]y^6[/latex] as [latex]\left (y^3\right )^2[/latex]. Then [latex]\sqrt{y^6}=\sqrt{\left (y^3\right )^2}=\left | x^3 \right |[/latex]. We need the absolute value because the square root must be non-negative.
Notice that when we square an exponential term, we multiply the exponent by [latex]2[/latex]. Since taking the square root “undoes” squaring, we can divide the exponent by 2 t take the square root, provided the exponent is even.
the Square Root Of variables
[latex]\sqrt{x^{2}}=\left|x\right|[/latex]
[latex]\sqrt{x^{2n}}=\left | x^n \right |[/latex]
Now that we know this, we can simplify a radical expression by using factoring and the product property of square roots.
The goal is to find factors under the radical that are perfect squares so that we can simplify.
Example
Simplify. [latex]\sqrt{9{{x}^{6}}}[/latex]
Solution
Write the radicand as perfect square factors. Note how we use the power rule for exponents to write [latex]x^6[/latex] as a perfect square: [latex]\left ({x^3}\right )^2[/latex]
[latex]\sqrt{{{3}^{2}}\cdot {{\left( {{x}^{3}} \right)}^{2}}}[/latex]
Separate into individual radicals using the product property.
[latex]\sqrt{{{3}^{2}}}\cdot \sqrt{{{\left( {{x}^{3}} \right)}^{2}}}[/latex]
Take the square roots, remembering that [latex]\sqrt{{{x}^{2}}}=\left|x\right|[/latex].
[latex]3\left|{{x}^{3}}\right|[/latex]
Answer
[latex]\sqrt{9{{x}^{6}}}=3\left|{{x}^{3}}\right|[/latex] We have to keep the absolute value since square roots are never negative.
Let’s try to simplify another radical expression.
Example
Simplify. [latex]\sqrt{100{{x}^{2}}{{y}^{4}}}[/latex]
Solution
Find perfect squares under the radical: exponents that are even.
[latex]\sqrt{10^2\cdot {x}^{2}\cdot {y}^{4}}[/latex]
Separate the perfect squares into individual radicals.
[latex]\sqrt{100}\cdot \sqrt{{x}^{2}}\cdot \sqrt{({y}^{2})^{2}}[/latex]
Simplify each radical by taking the square root. Remember to put absolute values on variable terms.
[latex]10\cdot\left|x\right|\cdot \left |{y}^{2}\right |[/latex]
Simplify. Since [latex]y^2\geq 0[/latex], [latex]\left |{y}^{2}\right | = y^2[/latex].
[latex]10\left|x\right|y^{2}[/latex]
Answer
[latex]\sqrt{100{{x}^{2}}{{y}^{4}}}=10\left| x \right|{{y}^{2}}[/latex]
Remember that we can check always check our answer by squaring it to be sure it equals [latex]100{{x}^{2}}{{y}^{4}}[/latex].
Example
Simplify. [latex]\sqrt{49{{x}^{10}}{{y}^{8}}}[/latex]
Solution
Look for perfect squares: numbers and variables.
49 is a perfect square since [latex]7^2=49[/latex]; [latex]x^{10}[/latex] and [latex]y^8[/latex] are perfect squares since their exponents are even. [latex]x^{10}=\left (x^5\right )^2[/latex] and [latex]y^8=\left (y^4\right )^2[/latex].
Separate the squared factors into individual radicals.
[latex]\sqrt{7^2}\cdot\sqrt{({x^5})^2}\cdot\sqrt{({y^4})^2}[/latex]
Take the square root of each radical using the rule that [latex]\sqrt{{{x}^{2}}}=\left|x\right|[/latex].
[latex]7\cdot\left|{{x}^{5}}\right|\cdot\left |{{y}^{4}}\right |[/latex]
Simplifly: [latex]\left |{{y}^{4}}\right |=y^4[/latex], since [latex]y^4\geq 0[/latex].
[latex]7\left|{{x}^{5}}\right|{{y}^{4}}[/latex]
Answer
[latex]\sqrt{49{{x}^{10}}{{y}^{8}}}=7\left|{{x}^{5}}\right|{{y}^{4}}[/latex]
In order to check this calculation, we could square [latex]7\left|{{x}^{5}}\right|{{y}^{4}}[/latex], hoping to arrive at [latex]49{{x}^{10}}{{y}^{8}}[/latex]. And, in fact, we would get this expression if we evaluated [latex]{\left({7\left|{{x}^{5}}\right|{{y}^{4}}}\right)^{2}}[/latex].
Try It
Simplify. [latex]\sqrt{81{{x}^{6}}{{y}^{4}}}[/latex]
Show Answer
[latex]9\left | x^3\right | y^2[/latex]
Try It
Simplify. [latex]\sqrt{144{{x}^{14}}{{y}^{12}}}[/latex]
Show Answer
[latex]12\left | x^7\right | y^6[/latex]
So far we have seen examples that have perfect squares under the radicals: the exponents have all been even. If we have an odd exponent then it is not a perfect square. For example, [latex]x^5[/latex] is not a perfect square. However, it contains perfect square factors of [latex]x^2[/latex] and [latex]x^4[/latex].
[latex]x^5=x\cdot x\cdot x\cdot x\cdot x=x^2\cdot x^2\cdot x=x^4\cdot x[/latex]
Each pair of factors makes a perfect square. This means that we can always write a variable with an odd exponent as the variable to one power less times the variable. For example, [latex]x^7=x^6\cdot x,\;y^9=y^8\cdot y,\; z^21=z^{20}\cdot z[/latex].
perfect square factors of powers of variables
[latex]x^n=x^{n-1}\cdot x[/latex]
This means that we can now simplify more radical expressions.
Example
Simplify. [latex]\sqrt{{{a}^{3}}{{b}^{5}}{{c}^{2}}}[/latex]
Solution
Factor to find variables with even exponents: [latex]\sqrt{{{a}^{2}}\cdot a\cdot {{b}^{4}}\cdot{b}\cdot{{c}^{2}}}[/latex]
Separate the perfect square factors into individual radicals: [latex]\sqrt{a^2}\cdot\sqrt{b^4}\cdot\sqrt{c^2}\cdot \sqrt{a\cdot b}[/latex]
Take the square root of each radical with a perfect square radicand. Remember that [latex]\sqrt{{{a}^{2}}}=\left| a \right|[/latex]: [latex]\left| a \right|\cdot \left |{b^2}\right |\cdot\left|{c}\right|\cdot\sqrt{a\cdot b}[/latex]
Simplify: [latex]\left| a\right |{b^2}\left | c \right|\sqrt{ab}[/latex]
Answer
[latex]\sqrt{{{a}^{3}}{{b}^{5}}{{c}^{2}}} \left| a\right |{b^2}\left | c \right|\sqrt{ab}[/latex]
Try It
Simplify. [latex]\sqrt{{{x}^{4}}{{y}^{9}}{{z}^{3}}}[/latex]
Show Answer
[latex]x^2 y^4 \left |z \right |\sqrt{yz}[/latex]
Try It
Simplify. [latex]\sqrt{{{m}^{9}}{{n}^{7}}{{p}^{11}}}[/latex]
Show Answer
[latex]m^4 \left | n^3\right | \left |p^5 \right |\sqrt{mnp}[/latex]
Candela Citations
CC licensed content, Original
Radical definition; Examples; Try Its. Authored by: Hazel McKenna. Provided by: Utah Valley University. License: CC BY: Attribution
Revision and Adaptation. Provided by: Lumen Learning. License: CC BY: Attribution
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Revised and adapted: Unit 16: Radical Expressions and Quadratic Equations, from Developmental Math: An Open Program. Provided by: Monterey Institute of Technology and Education. License: CC BY: Attribution
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CC licensed content, Original
Radical definition; Examples; Try Its. Authored by: Hazel McKenna. Provided by: Utah Valley University. License: CC BY: Attribution
Revision and Adaptation. Provided by: Lumen Learning. License: CC BY: Attribution
CC licensed content, Shared previously
Revised and adapted: Unit 16: Radical Expressions and Quadratic Equations, from Developmental Math: An Open Program. Provided by: Monterey Institute of Technology and Education. License: CC BY: Attribution
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11041 | https://math.stackexchange.com/questions/2404547/remainder-of-fracpxx2-1 | Stack Exchange Network
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Remainder of $\frac{P(X)}{X^2-1}$
Given that P(X) a polynomial of degree at least $2$, that
$\frac{P(X)}{X-1}=q_1(X)$ with a remainder of $3$, that
$\frac{P(X)}{X+1}=q_2(X)$ with a remainder of $-5$, what is the remainder when $P(X)$ is divided by $X^2-1$?
What I have done:
$P(X)=q_1(X)(x-1)+3$
$P(X)=q_2(X)(x+1)-5$
$P(1)=3=2q_2(1)-5$ which gives $q_2(1)=4$
$P(-1)=-5=-2q_1(-1)+3$ which gives $q_1(-1)=4$
And that's about it. I'm not sure how I should approach this kind of exercise, can I have some hints on how to get it done?
2 Answers 2
You know that $P(x)=(x^2-1)Q(x)+ax+b$ for some numbers $a$ and $b$, and from $P(1)=3$ and $P(-1)=-5$, you can solve for $a$ and $b$.
$$\left{\begin{array}{rcr} P(x) &\equiv& 3\pmod{x-1}\ P(x) &\equiv& -5\pmod{x+1}\end{array}\right.$$
imply that $-1$ is a root of $P(x)+5$ and $1$ is a root of $P(x)-3$. By imposing the first constraint
$$ P(x) = -5+(x+1)\,f(x) $$
and by imposing the second constraint $-8+(1+1)\,f(1) = 0$, or $f(1)=4$, from which
$$ P(x) = -5+(x+1)\left[4+(x-1)\,g(x)\right] =\color{red}{4x-1}+(x^2-1)g(x).$$
You may notice a strong similarity with one of the usual proofs of the Chinese remainder theorem: well, this is not a coincidence. And the given problem can be read as an interpolation problem, too: which linear polynomials equal $3$ at $x=1$ and $-5$ at $x=-1$? Just $4x-1$, clearly, it is enough to draw a line through $(1,3)$ and $(-1,-5)$.
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11042 | https://www.amboss.com/us/knowledge/esophageal-cancer/ | Expand all sections
Esophageal cancer
Summarytoggle arrow icon
Esophageal cancer (EC) is the eighth most common type of cancer worldwide and affects men more than women (3:1 ratio). The two main forms are esophageal adenocarcinoma and squamous cell carcinoma. Esophageal adenocarcinomas are among the neoplasms with the fastest increasing incidence in northern and western Europe and North America, while squamous cell carcinoma is the most common form worldwide. Adenocarcinoma, which usually affects the lower third of the esophagus, may be preceded by gastroesophageal reflux disease and Barrett esophagus. Other risk factors include smoking and obesity. Risk factors for squamous cell carcinoma include carcinogen exposure (e.g., in the form of alcohol and tobacco) and a diet high in nitrosamines but low in fruits and vegetables. Locally advanced disease is common at the time of diagnosis because EC is typically asymptomatic early in the disease course. Symptomatic patients may experience weight loss, dyspepsia, progressive dysphagia, cervical adenopathy, hoarseness or persistent cough, and signs of upper gastrointestinal bleeding, such as hematemesis, melena, or anemia. Esophagogastroduodenoscopy (EGD) is used to directly visualize the lesion and obtain a biopsy sample for histopathological confirmation. Staging of the tumor involves CT scan of the chest and abdomen, a PET scan, and often transesophageal endoscopic ultrasound (EUS). Curative surgical resection may be considered for locally invasive cancers, but EC is unresectable in approximately 60% of patients at the time of diagnosis. For patients with unresectable disease, treatment options include chemotherapy, radiation, and palliative stenting. Prognosis is generally poor because of the aggressive nature of EC and oftentimes late diagnosis.
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Epidemiologytoggle arrow icon
Adenocarcinoma is more common in the US of America.
Epidemiological data refers to the US, unless otherwise specified.
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Etiologytoggle arrow icon
Adenocarcinoma
The most important risk factors for esophageal adenocarcinoma are gastroesophageal reflux and associated Barrett esophagus.
Squamous cell carcinoma (SCC)
The primary risk factors for squamous cell esophageal cancer are alcohol consumption, smoking, and dietary factors (e.g., diet low in fruits and vegetables).
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Classificationtoggle arrow icon
Siewert classification of adenocarcinoma of the esophagogastric junction
| Overview of Siewert classification | | |
---
| Type | Localization | Surgical approaches |
| Siewert type I | Center of the tumor located 1–5 cm above the Z line (associated with Barrett esophagus) | Transthoracic esophagectomy Partial gastrectomy Lymphadenectomy |
| Siewert type II | Center of the tumor located from 1 cm above to 2 cm below the Z line | Transhiatal esophagectomy Total gastrectomy Extended lymphadenectomy |
| Siewert type III | Center of the tumor located 2–5 cm below the Z line (subcardial gastric cancer) |
Siewert type I
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Clinical featurestoggle arrow icon
Early stages
Advanced stages
Initially, esophageal cancer is often asymptomatic. It typically becomes symptomatic at advanced stages.
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Diagnosistoggle arrow icon
Esophagogastroduodenoscopy (EGD) with biopsy is the best initial and confirmatory test in patients with suspected esophageal cancer.
EGD
Barium swallow
Staging investigations
Consider the following studies in consultation with a multidisciplinary team.
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11044 | https://www.amathsdictionaryforkids.com/qr/u/unitPrice.html | unit price ~ A Maths Dictionary for Kids Quick Reference by Jenny Eather
AaBbCcDdEeFfGgHhIiJjKkLlMm
NnOoPpQqRrSsTtUuVvWwXxYyZz
| Uu |
| unit price • a unit price compares the price of something to a particular unit of measurement, for example, cost per kilogram or cost per litre or gallon, so you can see which is the cheapest. EXAMPLES: |
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11045 | https://math.stackexchange.com/questions/868675/line-not-intersecting-circle-maximum-value-of-expression-involving-radius | algebra precalculus - Line not intersecting circle, maximum value of expression involving radius - Mathematics Stack Exchange
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Line not intersecting circle, maximum value of expression involving radius
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If line y+x=2 y+x=2 do not intersect any member of circles x 2+y 2−a x=0 x 2+y 2−a x=0 at two distinct points where a is parameter, then maximum value of |a+4||a+4|.
My try: Since the line does not intersect the circle at 2 distinct points therefore the distance of line from center of circle is greater than or equal to the radius of the circle
Circle is x 2+y 2−a x=0 x 2+y 2−a x=0 therefore its radius is a a and center is (a,0)(a,0)
Distance of line x 2+y 2−a x=0 x 2+y 2−a x=0 from center of circle is given by |a−2|√2|a−2|2√
Therefore we get the inequality |a−2|√2≥a
|a−2|2–√≥a
But I can't find a way to solve it further to obtain the maximum value of |a + 4|
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edited Jun 12, 2020 at 10:38
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asked Jul 16, 2014 at 5:53
kartikeykant18kartikeykant18
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Try to find values of a a such that line and circle are tangent.Gerry Myerson –Gerry Myerson 2014-07-16 07:18:10 +00:00 Commented Jul 16, 2014 at 7:18
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Use substitution to solve the system of equations: {x 2+y 2−a x=0 y+x=2⟹x 2+(2−x)2−a x=0⟹x 2+(4−4 x+x 2)−a x=0⟹2 x 2+(−a−4)x+4=0
{x 2+y 2−a x=0 y+x=2⟹x 2+(2−x)2−a x=0⟹x 2+(4−4 x+x 2)−a x=0⟹2 x 2+(−a−4)x+4=0
But since there is at most one intersection point, we know that this equation must have at most one solution so that the discriminant of the LHS is either negative or zero: (−a−4)2−4(2)(4)≤0(−(a+4))2−32≤0(a+4)2≤32√(a+4)2≤√32|a+4|≤4√2
(−a−4)2−4(2)(4)(−(a+4))2−32(a+4)2(a+4)2−−−−−−−√|a+4|≤0≤0≤32≤32−−√≤4 2–√
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answered Jul 16, 2014 at 8:03
AdrianoAdriano
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Was the expression I got difficult to solve ?kartikeykant18 –kartikeykant18 2014-07-16 08:37:33 +00:00 Commented Jul 16, 2014 at 8:37
I'm not entirely sure how you came up with that expression. For one thing, the circle's centre is actually at (a 2,0)(a 2,0) and its radius is actually a 2 a 2.Adriano –Adriano 2014-07-16 08:54:12 +00:00 Commented Jul 16, 2014 at 8:54
oh yes! I made this mistake yet another time....kartikeykant18 –kartikeykant18 2014-07-16 11:35:35 +00:00 Commented Jul 16, 2014 at 11:35
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11046 | https://www.nagwa.com/en/videos/402173903696/ | Pop Video: Dot Products and Duality | Nagwa
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Pop Video: Dot Products and Duality
Grant Sanderson • 3Blue1Brown • Boclips
Dot Products and Duality
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Traditionally, dot products are something that’s introduced really early on in a linear algebra course, typically right at the start. So it might seem strange that I’ve pushed them back this far in the series. I did this because there’s a standard way to introduce the topic which requires nothing more than a basic understanding of vectors. But a fuller understanding of the role the dot products play in math can only really be found under the light of linear transformations. Before that though, let me just briefly cover the standard way that products are introduced, which I’m assuming is at least partially review for a number of viewers.
Numerically, if you have two vectors of the same dimension, two lists of numbers with the same length, taking their dot product means pairing up all of the coordinates, multiplying those pairs together, and adding the result. So the vector one, two dotted with three, four would be one times three plus two times four. The vector six, two, eight, three dotted with one, eight, five, three would be six times one plus two times eight plus eight times five plus three times three.
Luckily, this computation has a really nice geometric interpretation. To think about the dot product between two vectors 𝐕 and 𝐖, imagine projecting 𝐖 onto the line that passes through the origin and the tip of 𝐕. Multiplying the length of this projection by the length of 𝐕, you have the dot product 𝐕 dot 𝐖. Except when this projection of 𝐖 is pointing in the opposite direction from 𝐕, that dot product will actually be negative.
So when two vectors are generally pointing in the same direction, their dot product is positive. When they’re perpendicular, meaning the projection of one onto the other is the zero vector, their dot product is zero. And if they’re pointing generally the opposite direction, their dot product is negative.
Now, this interpretation is weirdly asymmetric; it treats the two vectors very differently. So when I first learned this, I was surprised that order doesn’t matter. You could instead project 𝐕 onto 𝐖; multiply the length of the projected 𝐕 by the length of 𝐖 and get the same result. I mean, doesn’t that feel like a really different process? Here’s the intuition for why order doesn’t matter: if 𝐕 and 𝐖 happened to have the same length, we could leverage some symmetry, since projecting 𝐖 onto 𝐕 then multiplying the length of that projection by the length of 𝐕 is a complete mirror image of projecting 𝐕 onto 𝐖 then multiplying the length of that projection by the length of 𝐖.
Now, if you scale one of them, say 𝐕 by some constant like two, so that they don’t have equal length, the symmetry is broken. But let’s think through how to interpret the dot product between this new vector two times 𝐕 and 𝐖. If you think of 𝐖 is getting projected onto 𝐕, then the dot product two 𝐕 dot 𝐖 will be exactly twice the dot product 𝐕 dot 𝐖. This is because when you scale 𝐕 by two, it doesn’t change the length of the projection of 𝐖, but it doubles the length of the vector that you’re projecting onto.
But, on the other hand, let’s say you’re thinking about 𝐕 getting projected onto 𝐖. Well, in that case, the length of the projection is the thing to get scaled when we multiply 𝐕 by two. The length of the vector that you’re projecting onto stays constant. So the overall effect is still to just double the dot product. So, even though symmetry is broken in this case, the effect that this scaling has on the value of the dot product is the same under both interpretations.
There’s also one other big question that confused me when I first learned this stuff. Why on earth does this numerical process of matching coordinates, multiplying pairs, and adding them together have anything to do with projection? Well, to give a satisfactory answer and also to do full justice to the significance of the dot product, we need to unearth something a little bit deeper going on here, which often goes by the name “duality.” But before getting into that, I need to spend some time talking about linear transformations from multiple dimensions to one dimension, which is just the number line. These are functions that take in a 2D vector and spit out some number. But linear transformations are, of course, much more restricted than your run-of-the-mill function with a 2D input and a 1D output. As with transformations in higher dimensions, like the ones I talked about in chapter 3, there are some formal properties that make these functions linear. But I’m going to purposely ignore those here so as to not distract from our end goal and instead focus on a certain visual property that’s equivalent to all the formal stuff.
If you take a line of evenly spaced dots and apply a transformation, a linear transformation will keep those dots evenly spaced, once they land in the output space, which is the number line. Otherwise, if there’s some line of dots that gets unevenly spaced, then your transformation is not linear. As with the cases we’ve seen before, one of these linear transformations is completely determined by where it takes 𝑖-hat and 𝑗-hat. But this time, each one of those basis vectors just lands on a number. So when we record where they land as the columns of a matrix, each of those columns just has a single number. This is a one-by-two matrix.
Let’s walk through an example of what it means to apply one of these transformations to a vector. Let’s say you have a linear transformation that takes 𝑖-hat to one and 𝑗-hat to negative two. To follow where a vector with coordinates, say, four, three, ends up, think of breaking up this vector as four times 𝑖-hat plus three times 𝑗-hat. A consequence of linearity is that after the transformation, the vector will be four times the place where 𝑖-hat lands, one, plus three times the place where 𝑗-hat lands, negative two, which in this case implies that it lands on negative two. When you do this calculation purely numerically, it’s matrix-vector multiplication.
Now, this numerical operation of multiplying a one-by-two matrix by a vector feels just like taking the dot product of two vectors. Doesn’t that one-by-two matrix just look like a vector that we tipped on its side? In fact, we could say right now that there’s a nice association between one-by-two matrices and 2D vectors, defined by tilting the numerical representation of a vector on its side to get the associated matrix or to tip the matrix back up to get the associated vector.
Since we’re just looking at numerical expressions right now, going back and forth between vectors and one-by-two matrices might feel like a silly thing to do. But this suggests something that’s truly awesome from the geometric view. There’s some kind of connection between linear transformations that take vectors to numbers and vectors themselves.
Let me show an example that clarifies the significance and which just so happens to also answer the dot product puzzle from earlier. Unlearn what you have learned and imagine that you don’t already know that the dot product relates to projection. What I’m gonna do here is take a copy of the number line and place it diagonally and space somehow with the number zero sitting at the origin. Now think of the two-dimensional unit vector, whose tips sit where the number one on the number line is. I want to give that guy a name, 𝐮-hat. This little guy plays an important role in what’s about to happen, so just keep them in the back of your mind. If we project 2D vectors straight onto this diagonal number line, in effect, we’ve just defined a function that takes 2D vectors to numbers. What’s more, this function is actually linear since it passes our visual test that any line of evenly spaced dots remains evenly spaced once it lands on the number line.
Just to be clear, even though I’ve embedded the number line in 2D space like this, the output of the function are numbers, not 2D vectors. You should think of a function that takes in two coordinates and outputs a single coordinate. But that vector 𝐮-hat is a two-dimensional vector living in the input space. It’s just situated in such a way that overlaps with the embedding of the number line.
With this projection, we just defined a linear transformation from 2D vectors to numbers, so we’re gonna be able to find some kind of one-by-two matrix that describes that transformation. To find that one-by-two matrix, let’s zoom in on this diagonal number line setup and think about where 𝑖-hat and 𝑗-hat each land, since those landing spots are gonna be the columns of the matrix.
This part is super cool; we can reason through it with a really elegant piece of symmetry. Since 𝑖-hat and 𝐮-hat are both unit vectors, projecting 𝑖-hat onto the line passing through 𝐮-hat looks totally symmetric to projecting 𝐮-hat onto the 𝑥-axis. So when we ask: what number does 𝑖-hat land on when it gets projected? The answer is gonna be the same as whatever 𝐮-hat lands on when its projected onto the 𝑥-axis. But projecting 𝐮-hat onto the 𝑥-axis just means taking the 𝑥-coordinate of 𝐮-hat. So, by symmetry, the number where 𝑖-hat lands when it’s projected onto that diagonal number line is gonna be the 𝑥-coordinate of 𝐮-hat. Isn’t that cool?
The reasoning is almost identical for the 𝑗-hat case. Think about it for a moment. For all the same reasons, the 𝑦-coordinate of 𝐮-hat gives us the number where 𝑗-hat lands when it’s projected onto the number line copy.
Pause and ponder that for a moment; I just think that’s really cool. So the entries of the one-by-two matrix describing the projection transformation are going to be the coordinates of 𝐮-hat. And computing this projection transformation for arbitrary vectors in space, which requires multiplying that matrix by those vectors, is computationally identical to taking a dot product with 𝐮-hat. This is why taking the dot product with a unit vector can be interpreted as projecting a vector onto the span of that unit vector and taking the length.
So what about non-unit vectors? For example, let’s say we take that unit vector 𝐮-hat, but we scale it up by a factor of three. Numerically, each of its components gets multiplied by three. So looking at the matrix associated with that vector, it takes 𝑖-hat and 𝑗-hat to three times the values where they landed before. Since this is all linear, it implies more generally, that the new matrix can be interpreted as projecting any vector onto the number line copy and multiplying where it lands by three. This is why the dot product with a non-unit vector can be interpreted as first projecting onto that vector then scaling up the length of that projection by the length of the vector.
Take a moment to think about what happened here. We had a linear transformation from 2D space to the number line, which was not defined in terms of numerical vectors or numerical dot products; it was just defined by projecting space onto a diagonal copy of the number line. But because the transformation is linear, it was necessarily described by some one-by-two matrix. And since multiplying a one-by-two matrix by a 2D vector is the same as turning that matrix on its side and taking a dot product, this transformation was inescapably related to some 2D vector.
The lesson here is that anytime you have one of these linear transformations, whose output space is the number line, no matter how it was defined there’s gonna be some unique vector 𝐕 corresponding to that transformation, in the sense that applying the transformation is the same thing as taking a dot product with that vector.
To me, this is utterly beautiful. It’s an example of something in math called “duality.” Duality shows up in many different ways and forms throughout math, and it’s super tricky to actually define. Loosely speaking, it refers to situations where you have a natural, but surprising correspondence between two types of mathematical thing. For the linear algebra case that you just learned about, you’d say that the dual of a vector is the linear transformation that it encodes. And the dual of a linear transformation from some space to one dimension is a certain vector in that space.
So, to sum up, on the surface, the dot product is a very useful geometric tool for understanding projections and for testing whether or not vectors tend to point in the same direction. And that’s probably the most important thing for you to remember about the dot product. But at deeper level, dotting two vectors together is a way to translate one of them into the world of transformations. Again, numerically, this might feel like a silly point to emphasize; it’s just two computations that happen to look similar. But the reason I find this so important, is that throughout math, when you’re dealing with a vector, once you really get to know its personality sometimes you realize that it’s easier to understand it not as an arrow in space, but as the physical embodiment of a linear transformation. It’s as if the vector is really just a conceptual shorthand for certain transformation, since it’s easier for us to think about arrows and space rather than moving all of that space to the number line.
In the next video, you’ll see another really cool example of this duality in action as I talk about the cross product.
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11048 | https://www.acog.org/education-and-events/creog/curriculum-resources/-/media/0fb3d64d83c34fcea19e15d43b0f6e45.ashx | Original Research
Estimating Gestational Age From Ultrasound Fetal Biometrics
Daniel W. Skupski, MD , John Owen, MD , MSPH , Sungduk Kim, PhD , Karin M. Fuchs, MD ,Paul S. Albert, PhD , and Katherine L. Grantz, MD , MS , for the Eunice Kennedy Shriver National Institute of Child Health and Human Development Fetal Growth Studies
OBJECTIVE: To compare the accuracy of a new formula with one developed in 1984 (and still in common use) and to develop and compare racial and ethnic-specific and racial and ethnic-neutral formulas.
METHODS: The Eunice Kennedy Shriver National Insti-tute of Child Health and Human Development (NICHD) Fetal Growth Studies-Singletons was a prospective cohort study that recruited women in four self-reported racial–ethnic groups—non-Hispanic black, Hispanic, non-Hispanic white, and Asian—with singleton gestations from 12 U.S. centers (2009–2013). Women with a certain last menstrual period confirmed by first-trimester ultrasonogram had longitudinal fetal measurements by credentialed study ultrasonographers blinded to the gestational age at their five follow-up visits. Regression analyses were performed with linear mixed models to develop gestational age estimating formulas. Repeated cross-validation was used for valida-tion. The estimation error was defined as the mean squared difference between the estimated and observed gestational age and was used to compare the formulas’ accuracy.
RESULTS: The new formula estimated the gestational age ( 62 SD) within 67 days from 14 to 20 weeks of gestation, 610 days from 21 to 27 weeks of gestation, and 617 days from 28 to 40 weeks of gestation. The new formula performed significantly better than a for-mula developed in 1984 with an estimation error of 10.4 compared with 11.2 days from 21 to 27 weeks of gestation and 17.0 compared with 19.8 days at 28–40 weeks of gestation, respectively. Racial and ethnic-specific formulas did not outperform the racial and ethnic-neutral formula.
CONCLUSION: The NICHD gestational age estimation formula is associated with smaller errors than a well-established historical formula. Racial and ethnic-specific formulas are not superior to a racial–ethnic-neutral one.
(Obstet Gynecol 2017;130:433–41) DOI: 10.1097/AOG.0000000000002137
Accurate gestational dating is imperative for opti-mal maternal and neonatal outcomes. The addi-tion of ultrasound fetal biometric measurements to a woman ’s reported last menstrual period (LMP) allows for improvement in gestational dating. 1 Meth-ods for estimating gestational age from ultrasound fetal biometric measurements were reported decades ago, mostly in racially homogenous populations. 2Indeed, formulas from early reports are still used today to estimate gestational age. 3 Formula accuracy
From the Division of Maternal Fetal Medicine, Department of Obstetrics and Gynecology, New York Presbyterian Hospital, Queens, New York; the Division of Maternal-Fetal Medicine, Department of Obstetrics and Gynecology, University of Alabama at Birmingham, School of Medicine, Birmingham, Alabama; the Division of Intramural Population Health Research, Eunice Kennedy Shriver
National Institute of Child Health and Human Development, National Insti-tutes of Health, Bethesda, Maryland; and the Department of Obstetrics and Gynecology, Columbia University Medical Center, New York, New York. Supported by funding through the Eunice Kennedy Shriver National Institute of Child Health and Human Development intramural program (contracts HHSN275200800013C, HHSN275200800002I, HHSN27500006, HHSN275200800003IC, HHSN275200800014C, HHSN275200800012C, HHSN275200800028C, and HHSN275201000009C) and included funding through the American Recovery and Reinvestment Act of 2009. Presented at the Annual Meeting of the Society for Maternal-Fetal Medicine, February 4 –7, 2016, Atlanta, Georgia. The authors thank GE Healthcare Women ’s Health Ultrasound for their support and training on the Voluson product over the course of this study. The authors also thank the principal investigators at the clinical sites for their support and the site sonographers for their good and careful work. For a list of principal inves-tigators who participated in this study, see Appendix 1 online at lww.com/AOG/A972. Each author has indicated that he or she has met the journal ’s requirements for authorship. Corresponding author: Daniel W. Skupski, MD, Obstetrics and Gynecology, New York Presbyterian Queens, 56-45 Main Street, Room M-365, Flushing, NY 11355; email: dwskupsk@med.cornell.edu.
Financial Disclosure
The authors did not report any potential conflicts of interest. © 2017 by The American College of Obstetricians and Gynecologists. Published by Wolters Kluwer Health, Inc. All rights reserved.
ISSN: 0029-7844/17
VOL. 130, NO. 2, AUGUST 2017 OBSTETRICS & GYNECOLOGY 433
Copyright ª by The American College of Obstetricians and Gynecologists. Published by Wolters Kluwer Health, Inc. Unauthorized reproduction of this article is prohibited. was initially evaluated and then further refined using pregnancies conceived by in vitro fertilization as the referent standard. 4 Ultrasound imaging has continued to advance since that time 5,6 and may allow for improved gestational age estimation through better image quality and more precise caliper placement. The recent performance and completion of the Eunice Kennedy Shriver National Institute of Child Health and Human Development (NICHD) Fetal Growth Studies provides a unique opportunity to readdress the issue of gestational age estimation in a larger and diverse cohort, one that included a cohort of women of four different racial –ethnic groups, conditions that were optimal for fetal growth, a standardized ultrasound protocol, and extensive training and quality control of images and measurements. 7,8 In addition, maternal health characteristics have changed in the last 30 years or more such as age and obesity rates and a reevalu-ation seems prudent. Because there is no consensus in the obstetric community on whether racial and ethnic-specific formulas are superior to racial and ethnic-neutral formulas, the NICHD Fetal Growth Studies provide this opportunity as well. Our objectives were to develop a gestational age estimation model using ultrasound fetal parameters in women who participated in the NICHD Fetal Growth Studies, to compare this formula with one in common use 3 and to assess the accuracy of racial and ethnic-specific formulas.
MATERIALS AND METHODS
The NICHD Fetal Growth Studies-Singletons was a prospective cohort study that recruited women of four self-reported racial or ethnic groups (non-His-panic black, Hispanic, non-Hispanic white, and Asian), who had low-risk pregnancies with optimal conditions for fetal growth. To be included, women had to have a certain LMP and a first-trimester ultrasonogram that confirmed their pregnancy dating. The ultrasound estimate of gestation had to be between 8 0/7 weeks and 13 6/7 weeks and match the LMP-based gestational age within 5 days for women between 8 0/7 weeks and 10 6/7 weeks, within 6 days for those between 11 0/7 weeks and 12 6/7 weeks, and within 7 days for participants between 13 0/7 weeks and 13 6/7 weeks. 1 If these criteria were not met, the patient was excluded. Cycle length and other hormonal contraception were not assessed and were not exclusions to participation. The project gestational age was then based on the menstrual date. Women were screened at 8 0/7 weeks of gestation to 13 6/7 weeks of gestation for maternal health status associated with normal fetal growth (aged 18 –40 years; body mass index [BMI, calculated as weight (kg)/[height (m)] 2 ] 19.0 –29.9; healthy lifestyles and living conditions [see the exclusion criteria subsequently]; low-risk medical and obstetric history). Body mass index was calculated from self-reported prepregnancy height and weight and was confirmed by measurement at the enrollment visit. Details of the study methodology have been published elsewhere. 7A cohort of women with BMI 30.0 –45.0 mg/kg 2 was recruited also, but is not detailed in the original report. 7 A separate report comparing the obese cohort with the nonobese cohort is in development, but this comparison is not shown in our current study on gestational age assessment. The exclusion criteria in the NICHD Fetal Growth Studies-Singletons were extensive: current pregnancy with multifetal gestation, history of preterm low birth weight (less than 2,500 g), or macrosomic (greater than 4,000 g) neonate; history of stillbirth or neonatal death; medically assisted conception; cigarette smoking or illicit drug use in the past 6 or 12 months, respectively; one or more daily alcoholic drinks; previous fetal congenital malformation; history of noncommunicable diseases (asthma requiring weekly medication, autoimmune disorders, cancer, diabetes mellitus, epilepsy or seiz-ures requiring medication, hematologic disorders, hypertension, psychiatric disorders, renal disease, thyroid disease); or history of gravid diseases (gesta-tional diabetes, severe preeclampsia or eclampsia, or hemolysis, elevated liver enzymes, low platelet count syndrome). Women who participated but who had complications in the current pregnancy or adverse outcomes for the neonate were excluded (post hoc) in the original report. 7 Sample size was determined so that the 5th and 95th percentiles for each gestational week for each racial or ethnic group could be pre-cisely estimated over gestation and assuming a 10% attrition and 30% post hoc exclusion rate for women with pregnancy or neonatal complications. We esti-mated needing 600 women in each group to reach our estimate of 320 women (after post hoc exclusions); our high completion and lower than expected exclu-sion rate resulted in larger cohorts. Women were recruited from 12 participating U.S. clinical sites from July 2009 through January 2013. After an ultrasonogram at 8 0/7 weeks to 13 6/7 weeks of gestation, women underwent serial ultra-sound assessment of fetal biometry, five targeted visits approximately once every 4 to 8 weeks, in one of four ultrasound schedules, assigned randomly, so that each gestational week from 14 to 40 was represented without exposing individual women to weekly
434 Skupski et al NICHD Gestational Age Estimation OBSTETRICS & GYNECOLOGY
Copyright ª by The American College of Obstetricians and Gynecologists. Published by Wolters Kluwer Health, Inc. Unauthorized reproduction of this article is prohibited. ultrasonography. All ultrasound examinations were performed using Voluson E8 machines. All study ultrasonographers underwent ante hoc training and credentialing, and their measurement techniques were subject to rigorous quality assurance. 9 Ultrasonogra-phers were blinded to the project gestational age, to the results of all prior scans, and to the gestational age of each measurement calculated in real time on the ultrasound machine to avoid measurement bias. Insti-tutional review board approval was obtained for the NICHD and all participating clinical institutions, and the data and imaging coordinating centers. Women were enrolled from four racial –ethnic groups, 611 non-Hispanic black, 649 Hispanic, 614 non-Hispanic white, and 460 Asian or Pacific Islander women. Protocol completion rates were 92%, 93%, 93%, and 90%, respectively. Using either a transabdominal or transvaginal approach, biparietal diameter was measured at the level of the thalami and cavum septa pellucida or the cerebral peduncles as the linear distance from the outer edge of the proximal to the inner edge of the distal skull; head circumference was measured at the same level (and often on the same images) using the ellipse function around the outer perimeter of the skull. Abdominal circumference was measured using the ellipse function circumscribing the actual or projected skin line in the transverse plane at the level of the stomach and the junction of the umbilical vein and portal sinus. Femur length was measured as the linear distance between the midpoints of each end of the calcified femoral diaphysis. The nonobese cohort was combined with the obese cohort for the current study of gestational age assessment. A total of 468 obese women were included, 192 non-Hispanic black, 132 Hispanic, 137 non-Hispanic white, and 7 Asian or Pacific Islander. The obese cohort had the same eligibility criteria with the addition of the following exclusion criteria, which are different and not as extensive as the exclusion criteria of the singleton cohort: autoimmune diseases, cancer, chronic hypertension requiring two or more medications, chronic renal disease, diabetes while not pregnant, human immunodeficiency virus, or psychiatric disorders. These additional exclusion criteria were added because major chronic conditions such as hypertension and diabetes disproportionately affect obese women, making it hard to determine whether the elevated risks of large for gestational age or macrosomia (effects on fetal growth) are the result of those complications or the excess maternal fatness. Our study ’s obese cohort of women was free of these major pre-existing chronic conditions, which helps to differentiate morbidity from obesity-related fetal effects. This will be detailed in a separate report. All participating women from the nonobese and obese cohorts were included in this analysis of gestational age estimation; there were no post hoc exclusions, which is different than the original fetal growth study report. 7Gestational age estimation models were devel-oped using a backward elimination regression tech-nique that initially contained all biometric measurements (biparietal diameter, head circumfer-ence, abdominal circumference, and femur length), including first-order, quadratic, and interaction terms, then removing the least significant terms until only terms that were significant at the .05 level remained. We used linear mixed models to account for the intrapatient correlation associated with repeated measurements. The final chosen model (hereafter referred to as the NICHD model) was fit using all patients in the database and all gestational ages from 14 to 40 weeks. This model was validated using the technique of repeated cross-validation, a paradigm that includes development of the model on a random selection of 50% of the data set and validation of the model on the remaining 50%. For each formula, a single scan of each patient was used and chosen randomly on half of the population and tested on the other half of the population. We performed this 50% test –50% training procedure 1,000 times. Using the coefficients of this validated model, we derived a for-mula for estimating gestational age based on the statistically significant terms of biometric parameters (hereafter referred to as the NICHD formula). The estimation error was defined as the mean squared difference between estimated and observed (project) gestational age among test set samples and was used to assess formula performance in several ways. First we evaluated the NICHD formula ’s SD of estimation in the second compared with third trimes-ters. Second, we compared the NICHD formula ’s SD of estimation with that of the 1984 Hadlock formula, which is in common use today. 3 Finally, using the same statistical techniques described previously, we developed racial and ethnic-specific formulas for each of the four racial –ethnic groups and compared the performance of the NICHD formula ’s SD of estima-tion with racial and ethnic-specific formulas. The SDs were multiplied by 1.96 to obtain the estimation error with the interpretation that in 95% of patients, the difference between the formula ’s estimated gestational age and the project gestational age is within plus or minus this value (ie, gestational weeks or days).
VOL. 130, NO. 2, AUGUST 2017 Skupski et al NICHD Gestational Age Estimation 435
Copyright ª by The American College of Obstetricians and Gynecologists. Published by Wolters Kluwer Health, Inc. Unauthorized reproduction of this article is prohibited. Prior literature has focused on three smaller gestational age windows as separate entities because of the different variation in the accuracy of estimation among these windows. 1,3,4,10 –12 Therefore, we also analyzed the difference in SD of estimation error among gestational age windows (14 –20, 21 –27, and 28 –40 weeks). The clinical effect of more accurate gestational dating was assessed by comparing the number of women who would be outside of a prescribed range of error for gestational age prediction using either formula for each of the three gestational age windows studied (14 –20, 21 –27, 28 –40 weeks). The designated
“acceptable ” range of error is based on prior litera-ture 1 and was as follows: 14 –20 weeks 67 days, 21 –27 weeks 610 days, and 28 –40 weeks 614 days. We used a generalized estimating equation to account for multiple measurements per individual and compare rates of estimated gestational ages out-side of prescribed ranges for both the Hadlock and NICHD formulas. This allowed an assessment of any statistically significant differences between the two formulas. All analyses were implemented using SAS 9.4 or R 3.1.2.
RESULTS
The study population included 803 non-Hispanic black (28.6%), 781 Hispanic (27.9%), 751 non-Hispanic white (26.8%), and 467 Asian (16.7%), totaling 2,802 women. Table 1 includes demographic data of the population. Table 2 presents the selected fetal anthropometric parameters and their associated estimated regression coefficients obtained from model selection. The r2 for the new NICHD formula was 0.975. The concordance among study site ultrasonog-raphers for all measurements was in excess of r50.99, demonstrating no significant difference among study sites. The equation is: gestational age (weeks) 510.6 –
0.168 3BPD+0.045 3HC+0.03 3AC+0.058 3FL+ 0.002 3BPD 2 +0.002 3FL 2 +0.0005 3(BPD 3AC) 2
0.005 3(BPD 3FL) 20.0002 3(HC 3AC)+0.0008 3
(HC 3FL)+0.0005 3(AC 3FL), in which BPD 5bi-parietal diameter, HC 5head circumference, AC 5abdominal circumference, and FL 5femur length. Figure 1 demonstrates the observed gestational age based on the LMP (confirmed by first-trimester ultrasonogram) plotted against the gestational age estimated by the NICHD formula applied to our pop-ulation for gestational weeks 14 –40. The figure shows that the accuracy of estimation of the formula is higher early in gestation and decreases with advancing gestation.
Table 1. Characteristics of the Study Population (N 52,802) Characteristic Value
Age (y) 28.16 65.49 Self-reported height (cm) 162.60 67.08 Self-reported weight (kg) 67.34 614.98 BMI (kg/m 2)25.41 65.13 Gestational age at ultrasound visit (wk) Visit 0 (enrollment) 12.7 60.95 Visit 1 19.8 62.44 Visit 2 27.1 62.08 Visit 3 32.0 61.55 Visit 4 35.9 61.36 Visit 5 38.6 60.83 Parity 01,319 (47.1) 1943 (33.7) 2 or more 540 (19.3) Race–ethnicity Non-Hispanic white 751 (26.8) Non-Hispanic black 781 (27.9) Hispanic 803 (28.7) Asian 467 (16.7) Marital status Never married 632 (22.6) Living as married or married 2,082 (74.4) Divorced, separated, or widowed 84 (3.0) Education No college 841 (30.0) Some college 850 (30.3) College graduate 645 (23.0) Postgraduate degree 466 (16.6) Annual family income Less than $30,000 707 (29.3) $30,000–49,999 452 (18.7) $50,000–74,999 311 (12.9) $75,000–99,999 305 (12.6) $100,000 or greater 640 (26.5) Health insurance Private, managed care 1,459 (56.5) Medicaid; other 1,064 (41.2) Self-pay 61 (2.4) Neonatal sex Male 1,309 (51.2) Female 1,249 (48.8) Gestational age at delivery (wk) 39.17 61.91 Total no. of images All images together, counts 54,012 BPD, counts 13,461 HC, counts 13,470 AC, counts 13,540 FL, counts 13,541 No. of images per patient All images together 19.28 65.40 BPD 4.80 61.37 HC 4.81 61.37 AC 4.83 61.35 FL 4.83 61.36 BMI, body mass index; BPD, biparietal diameter; HC, head circumference; AC, abdominal circumference; FL, femur length. Data are mean 6SD, n (%), or n.
436 Skupski et al NICHD Gestational Age Estimation OBSTETRICS & GYNECOLOGY
Copyright ª by The American College of Obstetricians and Gynecologists. Published by Wolters Kluwer Health, Inc. Unauthorized reproduction of this article is prohibited. Figure 2 shows the gestational ages estimated by the NICHD formula compared with the Hadlock formula. The figure suggests that early in gestation, the two approaches give similar estimates and demonstrate increased differences in estimates with advancing gestation. Table 3 presents the error of estimation results for the NICHD formula for different gestational age windows. The estimation of gestational age in weeks is more precise over the interval of 14 –20 weeks compared with a later interval of 21 –27 weeks. Fur-thermore, the accuracy is lower from 28 to 40 weeks of gestation as compared with the other two win-dows. This indicates more accurate estimation earlier in gestation. The NICHD formula ’s performance was similar to Hadlock at 14 –20 weeks of gestation, whereas it was significantly more accurate after 20 weeks of gestation. The accuracy of the NICHD for-mula is 67 days from 14 to 20 weeks of gestation,
610 days from 21 to 27 weeks of gestation, and 617 days from 28 to 40 weeks of gestation. The new for-mula performed better than a formula developed in 1984 with an estimation error of 10.4 compared with 11.2 days from 21 to 27 weeks of gestation and 17.0 compared with 19.8 days at 28 –40 weeks of gesta-tion, respectively. Table 3 also shows the comparison of racial and ethnic-specific formulas and the NICHD formula applied to each racial –ethnic group. Results were similar overall (gestational weeks 14 –40) and for each smaller gestational age window (14 –20 weeks, 21 –27 weeks, 28 –40 weeks).
Table 2. Regression Model for Estimation of Gestational Age Variable Estimate SE R2 R2
Intercept 10.5992 0.1815 0.9754 0.956731 BPD 20.1683 0.0194 HC 0.0452 0.0058 AC 0.0302 0.0037 FL 0.0576 0.0168 BPD 2 0.0025 0.0004 FL 2 0.0017 0.0007 BPD 3AC 0.0005 0.0002 BPD 3FL 20.0052 0.0012 HC 3AC 20.0003 0.0001 HC 3FL 0.0008 0.0003 AC 3FL 0.0006 0.0001 SE, standard error; BPD, biparietal diameter; HC, head circumfer-ence; AC, abdominal circumference; FL, femur length; GA, gestational age. The equation is: GA (weeks) 510.6–0.168 3BPD+0.045 3HC +0.03 3AC+0.058 3FL+0.002 3BPD 2+0.002 3FL 2+0.0005 3
(BPD 3AC) 20.005 3(BPD 3FL) 20.0002 3(HC 3AC)+0.0008 3
(HC 3FL)+0.0005 3(AC 3FL). The regression equation was obtained using linear mixed modeling to account for the use of multiple anthropometric measure-ments per fetus.
R2is based on randomly selected one sample for each participant. Calculated from whole dataset from fitted model based on one sample for each participant.
Fig. 1. Estimated gestational age compared with observed gestational age from 15 to 40 weeks. Use of the Eunice Kennedy Shriver National Institute of Child Health and Human Development (NICHD) model produced the esti-mated gestational age. Observed gestational age is that obtained from the last menstrual period.
Skupski. NICHD Gestational Age Estimation. Obstet Gynecol 2017.
Fig. 2. Estimated gestational age. Eunice Kennedy Shriver
National Institute of Child Health and Human Develop-ment (NICHD) model compared with the Hadlock model.
Skupski. NICHD Gestational Age Estimation. Obstet Gynecol 2017.
VOL. 130, NO. 2, AUGUST 2017 Skupski et al NICHD Gestational Age Estimation 437
Copyright ª by The American College of Obstetricians and Gynecologists. Published by Wolters Kluwer Health, Inc. Unauthorized reproduction of this article is prohibited. Figure 3 and Table 4 present the error of estima-tion for each gestational week when applying the NICHD and Hadlock formulas to our population. This can be used to establish gestational age when LMP differs from the ultrasound biometry outside of the estimated error range. The results are better with the NICHD compared with the Hadlock formula after 35 weeks of gestation. Before 35 weeks of gestation, the Hadlock and NICHD formulas are very similar. Table 5 presents an example of the effect of more accurate gestational dating using the NICHD formula compared with the Hadlock formula. There is a signif-icantly smaller proportion of patients outside the prescribed range of acceptable error using the NICHD formula at 21 –27 weeks and 28 –40 weeks of gestation.
DISCUSSION
We have developed a model for gestational age estimation and compared this new formula with a commonly used formula (Hadlock 1984). 3 We found the new formula to have improved estimation in the second and third trimesters and used a robust method for evaluating racial and ethnic-specific for-mulas and found that they are not superior to the new formula. Similar to prior literature, gestational age estima-tion is better with the NICHD formula from 14 to 20 weeks of gestation compared with later windows, which indicates less biological variation earlier in gestation. 2–4 The accuracy of the NICHD formula is
67 days from 14 to 20 weeks of gestation, 610 days from 21 to 27 weeks of gestation, and 617 days from 28 to 40 weeks of gestation, although we also provide individual week errors, which can be used to establish gestational age when LMP differs from the ultrasound biometry outside of the estimated error range (Table 4). The large population also allows us to support a recommendation to use ultrasonography to establish gestational age when fetal biometry at 14 –15 weeks of gestation is 8 days or more different than the LMP. This type of recommendation has previously been lacking in the literature. 13 We believe several factors are responsible for the observed improvements. The rigorous credentialing process for ultrasonographers, including masking gestational age at each examination, provided an unbiased method for obtaining each measurement and allowed minimization of random and systematic error. 4,14 Also, the ongoing quality improvement program provided sustained use of a consistent mea-surement method. 9,15 Modern ultrasound equipment with better image quality also possibly improved the
Table 3. Error of Estimation in Days for Different Gestational Age Windows, Comparing the Eunice Kennedy Shriver National Institute of Child Health and Human Development Model, the Hadlock Model 4 Normal and Obese Cohorts, and Four Racial- and Ethnic-Specific Models GA (wk) NICHD Model (Entire Cohort) Hadlock Model (Entire Cohort) NICHD Model Hadlock Model Non-Hispanic Black Model Hispanic Model Non-Hispanic White Model Asian Model Normal Obese Normal Obese
14–40 14.66 16.82 14.54 15.22 14.54 15.22 14.51 15.06 14.32 13.80 14–20 7.15 7.08 6.94 8.06 6.94 8.06 6.81 7.28 7.05 7.55 21–27 10.36 11.22 10.29 10.74 10.29 10.74 10.11 11.07 9.79 9.75 28–40 17.01 19.75 16.88 17.61 16.88 17.61 16.96 17.34 16.68 15.92 GA, gestational age; NICHD, Eunice Kennedy Shriver National Institute of Child Health and Human Development. The numbers were generated by multiplying the SD times 1.96. In 95% of cases, the error of the predictor is within plus or minus this value.
Fig. 3. Plot comparing the Eunice Kennedy Shriver
National Institute of Child Health and Human Develop-ment (NICHD) model with the Hadlock model for estima-tion interval ( 6estimated error in days) for each week of gestation.
Skupski. NICHD Gestational Age Estimation. Obstet Gynecol 2017.
438 Skupski et al NICHD Gestational Age Estimation OBSTETRICS & GYNECOLOGY
Copyright ª by The American College of Obstetricians and Gynecologists. Published by Wolters Kluwer Health, Inc. Unauthorized reproduction of this article is prohibited. accuracy. Improved image quality in all clinical situa-tions (eg, previous abdominal scar, increased BMI) has been a focus of ultrasound vendors for many years. 16 –18 The results of Table 5 demonstrate the clinical importance of more accurate estimation. Because the SDs of estimated errors between the two groups can-not be compared using inferential statistics, and thus
P values cannot be displayed, we created Table 5 to demonstrate that the clinical effect of improved esti-mation with the NICHD formula is significant. 19 These findings are particularly important for women who present late to prenatal care, when ultrasonogra-phy has a wider range of error. Currently, 5 –15% of women fall outside the prescribed range of acceptable error. Using the NICHD formula, 2 –5% of women would have more accurate dating, thus preventing interventions for preterm labor or postterm preg-nancy (Table 5). Up to 25% of the entire population in the United States undergoes induction of labor, 20 providing a significant potential to prevent morbidity from unnecessary interventions. The error range for the third trimester (+17 days) argues that caution in gestational dating is necessary late in gestation. We believe the finding that the racial and ethnic-specific formulas were similar to a racial and ethnic-neutral formula is the result of the adequately represented racial –ethnic diversity in the population from which our formula was developed. 21 –23 In addition, this fact simplifies the workflow in ultra-sound units because a single formula for gestational age estimation is accurate for all races and ethnicities. The NICHD Fetal Growth Studies showed previously that there is a significant difference in many fetal measurements when comparing different racial –ethnic groups. 7 The results of the current study found no difference among different racial –ethnic groups. This discrepancy is explained by the fact that growth trajectories, as assessed in the parent trial, and the relationship between cross-sectional measures and gestational age, as assessed in the current trial, reflect different biological processes.
Table 4. Error of Estimation for Gestational Age by Each Gestational Week, Comparing the Eunice Kennedy Shriver National Institute of Child Health and Human Development and Hadlock Models Gestational Age (wk) NICHD Hadlock Error of Estimation (wk) Error of Estimation (d) † Error of Estimation (wk) Error of Estimation (d) †
15 to less than 16 0.96 71.05 716 to less than 17 0.92 60.92 617 to less than 18 1.02 70.99 718 to less than 19 1.04 70.98 719 to less than 20 1.07 71.07 720 to less than 21 1.19 81.27 921 to less than 22 1.31 91.41 10 22 to less than 23 1.22 91.38 10 23 to less than 24 1.37 10 1.58 11 24 to less than 25 1.50 11 1.73 12 25 to less than 26 1.60 11 1.73 12 26 to less than 27 1.67 12 1.78 12 27 to less than 28 1.72 12 1.76 12 28 to less than 29 1.72 12 1.70 12 29 to less than 30 1.89 13 1.85 13 30 to less than 31 2.00 14 1.90 13 31 to less than 32 2.16 15 2.09 15 32 to less than 33 2.26 16 2.13 15 33 to less than 34 2.35 16 2.34 16 34 to less than 35 2.39 17 2.46 17 35 to less than 36 2.52 18 2.73 19 36 to less than 37 2.73 19 3.10 22 37 to less than 38 2.60 18 3.34 23 38 to less than 39 2.92 20 4.00 28 39 to less than 40 3.23 23 4.72 33 40 to less than 41 3.14 22 4.94 35 NICHD, Eunice Kennedy Shriver National Institute of Child Health and Human Development. Estimated SD 31.96. †Estimated SD 31.96 37.
VOL. 130, NO. 2, AUGUST 2017 Skupski et al NICHD Gestational Age Estimation 439
Copyright ª by The American College of Obstetricians and Gynecologists. Published by Wolters Kluwer Health, Inc. Unauthorized reproduction of this article is prohibited. Our study has strengths. We included a large number of women in multiple racial –ethnic groups in many centers. We controlled for bias inherent in using repeated measurements with the use of linear mixed regression models. We used statistical methodology to control for the problem of developing the predictive model in the same population in which we then tested the model —cross-validation with 1,000 iterations of a 50% test –50% training algorithm. The richness of our data allows us to suggest that our results will be more generally applicable to the U.S. obstetric population than any other formula. Our study has limitations. Our study population did not have a known date of conception such as with in vitro fertilization that allows us to pinpoint the
“true ” gestational age. Even with a known date of conception, as is seen with in vitro fertilization pregnancies, ultrasonography is associated with an unavoidable range of error, which is similar to the range of error seen in our study. 4 In addition, both the NICHD and Hadlock formulas were subject to the same estimation errors because their estimates were both compared with the same project gestational age. So, compared with a commonly used formula, the NICHD formula is associated with significantly small-er error ranges when gestational age is estimated beyond 20 weeks. We cannot comment on gestational age assessment in pregnancies conceived by assisted reproductive technologies because they were not included in our cohort. Using the NICHD formula has the potential to decrease unnecessary obstetric interventions with associated decreases in morbidity and cost savings. Given the large number of annual births in the United States and the significant rate of interventions, these improvements could have important implications at a population level.
REFERENCES
Methods for estimating the due date. Committee Opinion No. 700. American College of Obstetricians and Gynecologists. Obstet Gynecol 2017;129:e150 –4. 2. Harrison RF, Roberts AP, Campbell S. A critical evaluation of tests used to assess gestational age. Br J Obstet Gynaecol 1977; 84:98 –107. 3. Hadlock FP, Deter RL, Harrist RB, Park SK. Estimating fetal age: computer-assisted analysis of multiple fetal growth param-eters. Radiology 1984;152:497 –501. 4. Chervenak FA, Skupski DW, Romero R, Myers MK, Smith-Levitin M, Rosenwaks Z, et al. How accurate is fetal biometry in the assessment of fetal age? Am J Obstet Gynecol 1998;178:678 –87. 5. Morales-Roselló J, Hervás-Marin D, Stirrup O, Perales-Marín A, Khalil A. International standards for fetal growth: relevance of advances in ultrasound technology. Ultrasound Obstet Gynecol 2015;46:631 –2. 6. Papageorghiou AT, Ohuma EO, Altman DG, Todros T, Cheikh Ismail L, Lanbert A, et al. International standards for fetal growth based on serial ultrasound measurements: the Fetal Growth Longitudinal Study of the INTERGROWTH-21st Project. Lancet 2014;384:869 –79. 7. Buck-Louis G, Grewal J, Albert PS, Sciscione A, Wing DA, Grobman WA, et al. Racial/ethnic standards for fetal growth: the NICHD Fetal Growth Studies. Am J Obstet Gynecol 2015; 213:449.e1 –41. 8. Grantz KL, Grewal J, Albert PS, Wapner R, D ’Alton ME, Sciscione A, et al. Dichorionic twin trajectories: the NICHD Fetal Growth Studies. Am J Obstet Gynecol 2016;215:221. e1 –16. 9. Hediger ML, Fuchs KM, Grantz KL, Grewal J, Kim S, Gore-Langton RE, et al. Ultrasound quality assurance for singletons in the National Institute for Child Health and Human Develop-ment Fetal Growth Studies. J Ultrasound Med 2016;35:1725 –33. 10. Sabbagha RE, Tamura RK, Socol ML. The use of ultrasound in obstetrics. Clin Obstet Gynecol 1982;25:735 –52. 11. Romero R, Quintero R, Brekus C. Assessment of gestational age. In: Divon M, editor. Abnormal fetal growth. New York (NY): Elsevier Science; 1991. 12. Kalish RB, Thaler HT, Chasen ST, Gupta M, Berman SJ, Rosenwaks Z, et al. First- and second-trimester ultrasound assess-ment of gestational age. Am J Obstet Gynecol 2004;191:975 –8. 13. Spong CY. Redefining ‘term ’ pregnancy: recommendations from the defining ‘term ’ pregnancy workgroup. JAMA 2013; 309:2445 –6. 14. Perni SC, Chervenak FA, Kalish RB, Magherini-Rothe S, Predanic M, Streltzoff J, et al. Intraobserver and interobserver reproducibility of fetal biometry. Ultrasound Obstet Gynecol 2004;24:654 –8. 15. Boamah EA, Asante K, Ae-Ngibise K, Kinney PL, Jack DW, Manu G, et al. Gestational age assessment in the Ghana Randomized Air Pollution and Health Study (GRAPHS): ultrasound capacity building, fetal biometry protocol devel-opment, and ongoing quality control. JMIR Res Protoc 2014;3:e77.
Table 5. Comparison of the Error in Gestational Age Estimation of the Eunice Kennedy Shriver National Institute of Child Health and Human Development and Hadlock Formulas Formula In Range Out of Range P
14–20 wk Hadlock 1,352 (94.5) 79 (5.5) .77 NICHD 1,350 (94.3) 81 (5.7) 21–27 wk Hadlock 2,312 (92.2) 196 (7.8) ,.001 NICHD 2,376 (94.7) 132 (5.3) 28–40 wk Hadlock 5,712 (84.4) 1,058 (15.6) ,.001 NICHD 6,076 (89.7) 694 (10.3) NICHD, Eunice Kennedy Shriver National Institute of Child Health and Human Development. Data are n (%) unless otherwise specified. In range, 14–20 weeks 67 days, 21–27 weeks 610 days, 28–40 weeks 614 days. Out of range, when measurements are outside these ranges.
440 Skupski et al NICHD Gestational Age Estimation OBSTETRICS & GYNECOLOGY
Copyright ª by The American College of Obstetricians and Gynecologists. Published by Wolters Kluwer Health, Inc. Unauthorized reproduction of this article is prohibited. 16. Blackwell R. New developments in equipment. Clin Obstet Gynaecol 1983;10:371 –94. 17. Malone FD, Athanassiou A, Nores J, D ’Alton ME. Effect of ISDN bandwidth on image quality for telemedicine transmis-sion of obstetric ultrasonography. Telemed J 1998;4:161 –5. 18. D ’Alton ME, Cleary-Goldman J. Education and quality review for nuchal translucency ultrasound. Semin Perinatol 2005;29:380 –5. 19. Liang K-Y, Zeger SL. Longitudinal data analysis using generalized linear models. Biometrika 1986;73:13 –22. 20. Chauhan SP, Ananth CV. Induction of labor in the United States: a critical appraisal of appropriateness and reducibility. Semin Perinatol 2012;36:336 –43. 21. Schofield LS. Correcting for measurement error in latent variables used as predictors. Ann Appl Stat 2015;9: 2133 –2152. 22. Al-Gindan YY, Hankey CR, Govan L, Gallagher D, Heymsfield SB, Lean ME. Derivation and validation of simple anthropometric equations to predict adipose tissue mass and total fat mass with MRI as the reference method. Br J Nutr 2015;114:1852 –67. 23. Al-Gindan YY, Hankey C, Govan L, Gallagher D, Heymsfield SB, Lean ME. Derivation and validation of simple equations to predict total muscle mass from simple anthropometric and demographic data. Am J Clin Nutr 2014;100:1041 –51.
VOL. 130, NO. 2, AUGUST 2017 Skupski et al NICHD Gestational Age Estimation 441
Copyright ª by The American College of Obstetricians and Gynecologists. Published by Wolters Kluwer Health, Inc. Unauthorized reproduction of this article is prohibited. |
11049 | https://www.youtube.com/watch?v=V8kAUmbgYGo | Triangular numbers induction proof
SB MathsYT
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Posted: 22 Jul 2021
In today's maths video we look at a proof exercise in which we use induction. This proof involves summations which appear alot on A Level Further maths papers here in the uk.
I have plenty of videos using this technique and will have more coming up in the future so be sure to subscribe to see those.
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Transcript:
we're going to do an induction proof and what we're going to look to prove is that the triangular numbers um are given the nth triangle number is given by a certain sum okay and we're going to represent that as such so t n is equal to n over two multiplied by m plus one okay so what are the triangular numbers so we start with one and then we add on two then we add on three then we add on four then we have five and so on and they form this triangular shape okay so the first triangular number is one so what we do for an induction proof is we check that t one so we know t one is equal to one and then one half one plus one is equal to one um so this works for n equals one okay and then next up we we do an anchor step so we're going to assume uh for some [Music] n equals k that the t k is equal to k over 2 k plus one okay and then we're going to consider k plus one uh consider n equals k plus one okay and this is the the formula that but not the formula the the recipe i guess we use for all induction proofs um so we know that t k plus one t k plus one is equal to t k plus one more term and and that term is k plus one okay so you see because on the first term we added one on the second term we're adding two on the third term we're in three so plus k plus one we've assumed that t k is equal to k over two k plus one and then we've got plus k plus one and we're gonna factor out this k plus one so that's k over two uh plus one and then we're going to factor out this half so that's equal to one half k plus one k plus two but we notice here the k plus 2 is k plus 1 plus 1. so this formula works for k plus 1 given it works for k and since we've shown it works for one it therefore works for two three four five six seven eight so on and that we can continue that logic forever um and therefore we've done our proof you'd be required on an example to write some sort of um closing statement just to just to justify it's just a couple of lines um and explain your steps as you go along but obviously i don't have much room here so i just did the did the math um a couple of questions on this method so this step here i believe is is obvious but how do i know to factor out this k plus one well it just seems as a k plus one in each term seem like the the logical thing to do you're just trying to rearrange this into the into the correct form and then um i notice that once you take this half out you're gonna get a two here which is what we want to get this result here okay so just a bit of uh trial and error really but the more you do with these especially these type because this is despite being a question about trying the numbers it's a question about sums the other way this question could be formulated is to prove for all n that uh this sum is equal to one-half and n plus one okay it's exactly the same question you can consider why that is um and on exams you might get a lot of summations and be given what what they and be given a formula to find those summations and this is the the procedure you have to follow sometimes the manipulation is a little bit trickier um but yeah get make sure you get lots of practice in if you did struggle with this example thank you very much for watching this video if you did enjoy it found any utility in it please be sure to subscribe to the channel any support is much appreciated i hope to see you in the next one have a good day |
11050 | https://pubs.rsc.org/en/content/articlehtml/2023/ra/d2ra07642j | View PDF VersionPrevious ArticleNext Article
DOI: 10.1039/D2RA07642J
(Review Article)
RSC Adv., 2023, 13, 3843-3876
A comprehensive review on the electrochemical parameters and recent material development of electrochemical water splitting electrocatalysts†
Received
1st December 2022
, Accepted 18th January 2023
First published on 26th January 2023
Abstract
Electrochemical splitting of water is an appealing solution for energy storage and conversion to overcome the reliance on depleting fossil fuel reserves and prevent severe deterioration of the global climate. Though there are several fuel cells, hydrogen (H2) and oxygen (O2) fuel cells have zero carbon emissions, and water is the only by-product. Countless researchers worldwide are working on the fundamentals, i.e. the parameters affecting the electrocatalysis of water splitting and electrocatalysts that could improve the performance of the hydrogen evolution reaction (HER) and oxygen evolution reaction (OER) and overall simplify the water electrolysis process. Noble metals like platinum for HER and ruthenium and iridium for OER were used earlier; however, being expensive, there are more feasible options than employing these metals for all commercialization. The review discusses the recent developments in metal and metalloid HER and OER electrocatalysts from the s, p and d block elements. The evaluation perspectives for electrocatalysts of electrochemical water splitting are also highlighted.
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| Asha Raveendran | Asha Raveendran is pursuing her PhD degree in the Department of Chemistry, National Institute of Technology Puducherry, Karaikal, India. After obtaining Master's degree from Central University of Kerala, India, she joined Nano Electrochemistry Lab (NEL) under the guidance of Dr Ragupathy Dhanusuraman. Her research is on the design, synthesis, and fundamental electrochemical studies of 2D materials as well as conducting polymers based nanoelectrocatalysts for energy conversion studies like electrochemical water splitting, direct alcohol fuel cell applications and energy storage systems, such as supercapacitors. |
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| Mijun Chandran | Mijun Chandran completed his undergraduate studies (BSc) in Chemistry from Kannur University in 2013. He received his MSc degree (2015) and PhD (2021) in Chemistry from Central University of Kerala, India, under the supervision of Dr M. Bhagiyalakshmi. His research focuses on the development and characterization of 2-dimensional MXene-based electrocatalysts for supercapacitors, direct alcohol fuel cell and biosensor applications. He has also worked as a DST-SERB Junior Research fellow entitled “Exploration on the bio-electrocatalytic system for converting CO2 to formic acid/methanol in CO2 capture and utilisation processes”. He is currently working as Technical Assistant at Department of Chemistry, Central University of Tamil Nadu, India. |
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| Ragupathy Dhanusuraman | Ragupathy Dhanusuraman is Associate Professor at National Institute of Technology Puducherry, Karaikal, India. Previously, he worked as an Assistant Professor at Department of Chemical System Engineering, Keimyung University, Daegu, South Korea (2012). Between 2010 and 2012, he pursued as Postdoctoral Fellow and Post-Doctoral Researcher at Kyungpook National University & Pusan National University. He has 91 Peer-reviewed research articles, 4 Patents, and 5 book chapters and has received “Young Scientist Award – 2014” from Department of Science & Technology, Government of India, New Delhi. His recent research includes Nanotechnology for Energy, Catalytic and Biomedical Applications. |
1. Introduction
One of the significant challenges we face, inevitable in the 21st century, is the intensifying energy demand, depletion of fossil fuels and detrimental effects of the utilization of non-renewable resources on our environment.1–3 Renewable energy resources like solar, tidal and wave power are alternative energy sources for meeting energy demands, but they are erratic, owing to inconsistencies due to seasonal and regional variability.4,5 Electrochemical water splitting is considered friendly to the environment and a promising route to address the intermittency problem by converting electrical energy from renewable resources into chemical energy resulting in the electrocatalytic generation of H2.6 Hydrogen is termed as the future fuel7 because of its high gravimetric energy density (120 MJ kg−1) compared to gasoline (44 MJ kg−1), excellent energy conversion efficiency, environmental compatibility and zero carbon dioxide emission with a byproduct of just water.8–12 Moreover, Hydrogen has been applied in the synthesis of ammonia (Haber's Process), methanol synthesis, hydrocracking of crude oil, production of hydrochloric acid, and hydrogenation processes of oils and fats.13,14 As no natural hydrogen exists on the earth, hydrogen production is currently via steam reforming hydrocarbons at high temperatures and pressure, which inevitably leads to consumption of limited fossil fuels and carbon dioxide emission.15 Moreover, the H2 obtained by this method is accompanied by the oxides of C, N and S, which would poison the surface of the catalyst, reducing their cycle life.16,17 Other methods include photoelectrochemical water splitting, which utilizes photons to produce H2.18,19 Though greener and evolve pure H2, they make insufficient quantity per unit time attributable to their low solar to hydrogen (STH) conversion efficiencies and, as a result, cannot be replaced for bulk and immediate generation.20,21 Hydrolysis of metal hydrides and reactive metal can be utilized to produce large quantities of H2 rapidly.22,23 Nonetheless, their precursors are frequently toxic metals and are synthesized through fine chemical industries polluting the environment and cannot be opted as a greener method of production.24–27 Thus, water electrolysis is the only environmental friendly method to generate the immediate and large scale H2, and studies on improving the water electrolyzers performance by developing cost-effective electrocatalysts with excellent efficiency for the decomposition of water is a raging hotspot among researchers.28–30
2. Electrochemistry involved in water electrolysis
The electrolytic cell was first proposed in 1789 and composed of three parts: cathode, anode and an electrolyte.31 The reaction involved in electrochemical water splitting;
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One mole of hydrogen and half a mole of oxygen gas are produced during the electrolysis of one mole of water. The process utilizes the thermodynamic potentials and the first law of thermodynamics. The pertinent values are taken from a table of thermodynamic properties, assuming the process at 298 K and one pressure atmosphere (Table 1).32
Table 1 Thermodynamic quantites of H2, O2 and H2O
| Quantity | H2O | H2 | 0.5O2 | Change |
| Enthalpy | −285.85 kJ | 0 | 0 | ΔH = 285.83 kJ |
| Entropy | 69.91 J k−1 | 130.68 J k−1 | 0.5 × 205.14 J k−1 | TΔS = 48.7 kJ |
Energy for the dissociation and the expansion of the gases that are produced must be supplied by the process, and they are accounted for in the change in enthalpy shown in the table above. Considering the system works at 298 K and one-atmosphere pressure;
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| W = PΔV = (101.3 × 103 Pa)(1.5 moles)(22.4 × 10−3 m3 mol−1)(298 K/273 K) = 3715 J | (2) |
As the enthalpy H= U + PV, the change in internal energy U is then
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| ΔU = ΔH − PΔV = 285.83 kJ − 3.72 kJ = 282.1 kJ | (3) |
The expansion of the gases produced occurs along with this change in internal energy; hence change in enthalpy represents the energy required to complete the electrolysis. Therefore
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| ΔG0 = ΔH − TΔS = 285.83 kJ − 48.7 kJ = 237.1 kJ | (4) |
Water splitting is processed via two half-cell reaction; water reduction & water oxidation. Hydrogen evolution reaction (HER) is the reduction of protons at the cathode;
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| 2H+(aq) + 2e− → H2(g) | (5) |
and oxygen evolution reaction (OER) is oxidation of water.33
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| 2H2O(l) → O2(g) + 4H+(aq) + 4e− | (6) |
Subsequently water is highly stable and weakly conductive, its conductivity is enhanced by electrolytes like sulphuric acid, potassium hydroxide and potassium phosphate buffer making the medium acidic, basic and neutral.34
In acidic medium;35
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| Cathode: 4H+ + 4e− → 2H2, E° = 0.0 V | (7) |
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| Anode: 2H2O → O2 + 4H+ + 4e−, E° = +1.23 V | (8) |
In basic medium;
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| Cathode: 4H2O + 4e− → 2H2 + 4OH−, E° = −0.828 V | (9) |
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| Anode: 4OH− → O2 + 2H2O + 4e−, E° = +0.401 V | (10) |
In neutral medium
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| Cathode: 4H2O + 4e− → 2H2 + 4OH−, E° = +0.413 V | (11) |
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| Anode: 2H2O → O2 + 4H+ + 4e−, E° = +0.817 V | (12) |
Water electrolysis conducted in utmost pH conditions has its pros and cons. As plenty of protons are accessible for immediate charge and discharge in acidic medium, the efficiency of HER is better in acidic medium.36,37 Nonetheless, OER can be catalysed only by oxides of precious and expensive iridium (Ir) or ruthenium (Ru) alloys at minimum overpotential at this medium. It is crucial to note that only a small number of electrocatalytically active oxides exhibit metallic-type conductivity. With a relatively high electrical conductivity of 104 cm−1, both IrO2 and RuO2 exhibit metallic behaviour. This is due to the lack of a band gap caused by the relatively small electronegativity difference between oxygen and metal (less than 1.3 units). It is owing to weak oxygen binding and comparatively stronger hydroxyl (HO) binding on the surface, both oxides exhibit lower over-potentials of O2-evolution than other metal anodes. Additionally, the –O, –OH and –OOH binding energies on the (1 1 0) surfaces satisfy universal linear relations comparable to those observed on metallic surface, such as RuO2 and IrO2.38,39 Thus OER in acidic medium remains a bottleneck for electrochemical researchers. The efficiency of OER is observed to increase drastically in alkaline medium utilizing oxides/hydroxides of less expensive 3d transition metals; the H2 production is, however, low as there is no instantaneous supply of H+ to cathode at high pH.40,41
Moreover, water splitting occurs at considerable cell potential (1.8–2.0 V) for commercial electrolysers, about 0.55 to 0.77 V higher than the theoretical value (1.23 V).42 This excess potential is obligatory to overcome the energy barrier of the reaction system, and it is referred to as overpotential. Overpotential may be due to the contact and solution resistance by the electrode, electrolyte, their distance, and circuit and by the activation hindrance from the kinetic steps in both cathode and anode. Thus to practically carry out the process of water splitting, potential higher than thermodynamic potential is required. A decrease in overpotential can be achieved by utilizing highly efficient, sustainable and eco-friendly HER and OER electrocatalysts & other resistances can be minimalized by augmenting the electrolyzer design, thus making electrochemical water splitting greener and affordable.43,44
3. Hydrogen evolution reaction (HER)
Despite the pH of the medium, the primary step in the multi-step HER is the adsorption of hydrogen on the surface of the electrode by the discharged protons. The reaction is stated as the Volmer reaction. The second step involves the desorption of H2 from the cathode either via electrochemical (Heyrovsky step) or via the chemical route (Tafel step), as observed in (Fig. 1).45 Between the two, the Tafel mechanism is more facile.
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| | Fig. 1 A schematic illustration of a potential HER reaction pathway on cathode.48 | |
HER in acidic media ( stands for an active center on the electrocatalyst):46,47
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| H+ + e− → H(Volmer) | (13) |
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| H+ + e− + H → H2(Heyrovsky) | (14) |
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| (or) 2H → H2(Tafel) | (15) |
HER in alkaline media:
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| H2O + e− → H + OH−(Volmer) | (16) |
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| H2O + e− → H2 + OH−(Heyrovsky) | (17) |
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| (or) 2H → H2(Tafel) | (18) |
The mechanism by which HER proceed is studied by the Tafel slope value is based on the coverage of the Hads. When the Hads surface coverage is low, Hads react with a proton and electron simultaneously since there are active sites adjacent to the adsorbed hydrogen on the cathode surface leading to Heyrovsky reaction. If the surface coverage is high, the adjacent two Hads combine, and H2 is generated.48,49 Tafel slope values are expressed as follows:
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| | (19) |
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where b stands for the Tafel slope (bv: Volmer, bH: Heyrovsky, bT: Tafel) α is the symmetry value 0.5, R is the ideal gas constant, T is the temperature, and F is the Faraday constant. The quicker the kinetics of the reaction, the lower would the Tafel slope value.50 The observed slope values are 11 mV dec−1 for the Volmer step, 25–35 mV dec−1 for the Tafel reaction and slope values greater than 39 mV dec−1 are observed in the slower Heyrovsky reaction. Ultimately, the free energy for the adsorption of hydrogen (ΔGH) can be a favourable indicator for HER performance of an electrocatalyst. For instance, electrocatalysts with ΔGH value approx. zero, Pt-based exhibits the most efficient activity for HER.
If ΔGH is too negative, the adsorbed H have difficulty desorbing from the electrocatalyst surface leading to slow Tafel or Heyrovsky steps. If ΔGH is too positive due to the poor interaction of Hydrogen and electrocatalyst, on contrary is the Volmer step. Optimizing the design of the HER electrocatalyst depends on adjusting the value of ΔGH.51
The HER activities of different metals are reflected in the Volcano relationship, which is experimentally plotted ΔGH calculated from density functional theory (DFT) versus the logarithm of their corresponding exchange current densities (logj0) of a wide range of catalysts. The so-called volcano diagram in (Fig. 2) is a simple approach to illustrate and compare and optimize the activity and design of different metals. Thus a good HER catalyst should have numerous active sites with optimal electron density for moderate bonding strength with adsorbed H(Sabatier principle), minimal charge transfer resistance between electrode–electrolyte interfaces, and should be unwavering in electrolyte medium.52,53
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| | Fig. 2 Hydrogen evolution reaction (HER) volcano plot on metal electrodes under low pH environment. For each metal surface, the logarithm of the exchange current density j0 is plotted against M–H bond energy. Reproduced with permission2 Copyright 2010, American Chemical Society. | |
4. Oxygen evolution reaction (OER)
In comparison to HER, Volcano relationship reflects a sluggish reaction involving a multistep four-electron transfer process at the anode. In an acidic electrolyte, H2O is oxidized into hydrogen and oxygen, while in alkaline or neutral electrolyte, the hydroxyl ions are oxidized as oxygen and water. The kinetics with which the reaction proceeds are entirely dependent on the material by which the reaction is catalysed. Ir and Ru based catalysts show better efficiency in acidic than in alkaline medium. Transition metals like iron (Fe), nickel (Ni) and cobalt (Co) based catalysts catalyze OER better in alkaline medium.54,55 OER mechanisms in both acidic and basic mediums were reviewed in detail in 1986 by Matsumoto and Sato56 including the Yeager and wade, Krasil shchkov, Bockris57 and Hackerman58 pathway along with the renowned electrochemical oxide and oxide pathway. In basic medium, all the listed mechanisms commence with the coordination of hydroxide to the active site as their elementary step and would proceed through different steps. The overpotential is higher due to kinetic barriers in each steps of the reaction mechanism.58,59
OER in acidic media;
(i) Electrochemical oxide path
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| +H2O → OH + H+ + e− | (22) |
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| OH → O + H+ + e− | (23) |
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| O + H2O → OOH + H+ + e− | (24) |
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| O2 → + O2 | (25) |
(ii) Oxide path
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| + H2O → OH + H+(aq) + e− | (26) |
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| 2OH → O + + H2O | (27) |
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| 2O → 2 + O2(g) | (28) |
(iii) Krasil' shchkov path
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| + H2O → OH + H+(aq) + e− | (29) |
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| OH → S–O− + H+(aq) | (30) |
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| O− → O + e− | (31) |
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| 2O → 2 + O2(g) | (32) |
(iv) Wade and Hackerman's path
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| 2 + 2H2O → O + H2O + 2H+(aq) + 2e− | (33) |
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| O + 2OH− → H2O + 2 + O2(g) + 2e− | (34) |
OER in basic media;
(i) Electrochemical oxide path
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| + OH− → OH + e− | (35) |
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| OH + OH−(aq) → O + H2O + e− | (36) |
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| 2O → 2 + O2(g) | (37) |
(ii) Oxide path
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| + OH−(aq) → OH + e− | (38) |
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| 2OH → O + + H2O | (39) |
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| 2O → 2 + O2(g) | (40) |
(iii) Krasil' shchkov path
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| + OH−(aq) → OH + e− | (41) |
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| |
| OH + OH−(aq) → O− + H2O | (42) |
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| |
| O− → O + e− | (43) |
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| O + O → 2 + O2(g) | (44) |
(iv) Yeager's path
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| + OH−(aq) → OH + e− | (45) |
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| |
| zOH → z+1OH + + e− | (46) |
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| |
| 2z+1OH + 2OH−(aq) → 2 + 2H2O + O2(g) | (47) |
(v) Bockris path
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| + OH−(aq) → OH + e− | (48) |
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| OH + OH−(aq) → H2O2 + e− | (49) |
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| H2O2 + OH−(aq) → OOH− + H2O | (50) |
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| H2O2 + OOH− → 2 + OH−(aq) + O2(g) | (51) |
The mechanisms mentioned above match the experimental results of both hydroxide and oxide catalysts but shows inconsistencies in perovskite materials.60 18O isotope labelling and DFT studies are conducted to address these inconsistencies, and it was observed that some material prefer adsorbate evolution mechanism & some others prefer Lattice oxygen participation mechanism (LOM).61,62
Steps involved in AEM;
| |
| |
| + H2O → OH + H + e− | (52) |
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| OH → O + H + e− | (53) |
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| O + H2O → OOH + H + e− | (54) |
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| OOH → O2(g) + H + e− | (55) |
Steps involved in LOM (Vo: oxygen vacancy due to lattice oxygen atom participation in OER)
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| |
| OH → (Vo + OO) + H + e− | (56) |
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| |
| (Vo + OO) + H2O(l) → O2(g) + (Vo + OH) + H+ + e− | (57) |
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| |
| | (58) |
| |
| |
| | (59) |
The usage of parenthesis is to indicate adsorbate of the same supercell. AEM and LOM pathways compete with each other and are based on the material used to synthesise of the electrocatalyst. In all the mechanisms, the active sites undergo cycles of oxidation & reduction in both acidic and basic mediums of OER, as shown in (Fig. 3). Thus metals with stable & variable oxidation states can behave as electrocatalysts for OER. Nonetheless, for all the possible OER mechanisms, the formation of intermediates O, OH and OOH on the surface of the electrocatalyst is a prerequisite and behaves as the rate-determining step.63
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| | Fig. 3 The mechanisms governing the oxygen evolution reaction in acidic (blue line) and alkaline electrolytes, respectively. The corresponding black and green lines represent the two likely intermediates, M–OOH and M–O. Reproduced with permission64 Copyright 2020, Elsevier. | |
The free energy for the adsorption value of ΔGOOH − ΔGOH of an electrocatalyst has an impact on the OER activity as they provide the thermodynamic information between the intermediates as well as the electrocatalyst (too strong & weak bonds cause sluggish reactions), while the Tafel slope gives the kinetic information of the catalyst.64,65 Evolution of oxygen gas is generally from the oxide surface of the catalyst and not the bare surface, and thus the OER mechanism varies for different oxides having different surface morphology. Oxides with similar compositions portray different kinetics due to differences in preparation methods leading to different structures and thickness of their oxide layers.
From the volcano plot portraying the activity O2 production from the surface of first row transition metal oxides in (Fig. 4), due to excellent electrical conductivity and low redox potential, metal oxides of noble Ru and Ir are on top of the plot. The drawback, however is their limited HER activity and being cost ineffective.56,66 Researchers are looking into affordable and sustainable metals electrocatalysts that demonstrate extraordinary activity for both HER and OER accompanied by stability and the list of elements suitable for water splitting are portrayed in (Fig. 5).67
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| | Fig. 4 Volcano plot demonstrates the relationship between the rate of O2 oxidation on transition metal oxide surfaces and change in enthalpy in acidic () and basic () solutions. Measurement of overpotential in relation to 0.1 mA cm2 current density. Reproduced with permission2 Copyright 2010, American Chemical Society. | |
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| | Fig. 5 Endangered species of elements and their potential application as electrocatalysts for water splitting, according to the aster diagram. Reproduced with permission67 Copyright 2022, Elsevier. | |
5. Parameters for the evaluation of electrochemical water splitting
The main evaluation parameters for electrochemical splitting in (Fig. 6) include overpotential at a defined current density, electrochemical surface area (ECSA), Tafel slope, exchange current density, turnover frequency (TOF), faradaic efficiency, mass and specific activities. Tafel slope, exchange current density and TOF gives the kinetic activity i.e. how fast the reaction proceeds. All parameters however have both their merits and demerits.68,69
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| | Fig. 6 The evaluation parameters of electrochemical water splitting. | |
5.1. Overpotential
As mentioned earlier, an extra potential is required to overcome the intrinsic kinetic hindrance in electrochemical water splitting (in both HER and OER) known as overpotential and is symbolized as η.58 The reversible thermodynamic potential for HER and OER are 0 and 1.23 V vs. the overpotential at desired current density without iR compensation are given by the equations; ηHER = ERHE − 0 V and ηOER = ERHE − 1.23 V.70–72 In commercial water electrolysis, the electrochemical setup works in a constant cell voltage which varies from 1.6 V to 2.0 V based on the electrocatalyst of HER and OER as well as the electrolyte pH. The accepted benchmark for current density is 10 mA cm−2 for comparing HER and OER electrocatalyst in mediums of different pH. The numerical figure of merit is the current density predicted for 10% competent solar to fuel conversion device, under 1 sun illumination.41,73 This benchmark as well as detailed convention for measuring the ECSA via CV as well as electrochemical impedance spectroscopy (EIS), faradaic efficiency using rotating ring disc voltammetry, catalytic activity using rotating disc voltammetry along with the long term cyclic stability of the catalysts in controlled electrolysis conditions was proposed by Jaramillo and co-workers.74
There are different types of overpotential in water splitting i.e., activation overpotential, ohmic overpotential, concentration overpotential, bubble overpotential and Kinetic overpotential of HER and OER.2
5.1.1. Activation overpotential. The kinetics of electrode at the reaction site determine the overpotential caused by the activation energy required for charge transfer, which is known as the activation overpotential. Thus it accounts for the electrochemical kinetics, electron migration, and proton migration. The Tafel equation can be used to calculate the anode activation overpotential (ηa) and cathode activation overpotential (ηC).
| |
| |
| ηa,a = aa + balnj | (60) |
| |
| |
| ηc,a = ac + bclnj | (61) |
where ac and aa are the constants of the Tafel equation at the cathode and anode respectively, j is the current density (A m−2), and ba and bc are the Tafel slopes of the Tafel equation at the anode and cathode, respectively. By choosing an appropriate electrocatalyst, one could decrease the activation overpotential which is the inherent property of the electrocatalyst.42,75
5.1.2. Concentration overpotential. Concentration overpotential is caused by a rapid reduction in concentration at the interfaces due to concentration variance between ions in the bulk of the solution and on the electrode surface.
The expression for j under pure mass transfer control can be written as
| |
| |
| | (62) |
where jL is the limiting current density and conc. is the concentration overpotential, the current at which the reactant's surface concentration falls to the limiting case of zero.76 The anode and cathode concentration overpotentials make up this phenomenon, which is caused by mass transfer restrictions at higher current densities:
| |
| |
| ηcon = ηacon + ηccon | (63) |
where ηacon is the anode concentration overpotential, and ηccon is the cathode concentration overpotential, respectively. Due to the presence of a diffusion layer, it can be considerably decreased by agitating the solution.
5.1.3. Ohmic overpotential. The resistance between the surfaces and interfaces of the measuring system causes an additional voltage drop, resulting in the junction overpotential or resistance overpotential. Ohmic overpotential adheres to Ohm's law ηohmic = iR, where I is the applied current. According to this relationship, the ohmic drop changes linearly with the applied current. The ohmic overpotential can be expressed as
| |
| |
| ηohmic = ηabp + ηcbp + ηae + ηce + ηm | (64) |
where ηabp is the ohmic losses of the anode bipolar plates and ηcbp, the ohmic losses of the cathode bipolar plates, respectively. The ohmic losses of the anode backing layers and the ohmic losses of the cathode electrode backing layers are denoted by ηae and ηce. The overpotential due to the polymer electrolyte membrane is ηm.
Resistance overpotential could be excluded by performing ohmic drop compensation, also known as iR compensation which is available now in many electrochemical workstation.35 It could also be done manually by multiplying the resultant current density with uncompensated resistance (Ru) which gives a potential E. Thus iR drop should be subtracted from the experimental potential. Commercial water electrolyzers doesnot operate to the reported iR corrected overpotential in literature, thus its mandatory to report both iR uncompensated as well as compensated overpotential at the same current density.51,77,78
5.1.3.1. Proton transport and mass transport losses. The main losses occur in the middle of the operating range of PEM cell and are caused by the cell resistance of electrolyzer (Rcell) produced by electronic and ionic conduction. Application of Ohm's law is used to calculate the overall ohmic losses as mentioned above. Due to the high conductivity of frequently used materials, it is customarily believed that electron transport is significantly faster than protonic transport and that only ohmic losses resulting from proton transport through the PEM (the protons produced at the anode passes through the membrane and are transported to the cathode where they are reduced by electron) are taken into consideration.79 The ohmic overpotential with regard to proton transport across the membrane is given by;
| |
| |
| | (65) |
where x is the location in the membrane measured from the electrode-membrane interface, tm is the thickness of the membrane and the conductivity of the membrane is given as km which is further evaluated as
| |
| |
| | (66) |
where λm is the water content in the membrane and the temperature of the electrolyzer is given as T.
Since electrolyzers typically don't operate at high current densities, many authors neglect to take these mass transport limitations into account when calculating the major losses (such as the concentration losses), which usually occur at very high current densities.80 The mass transport losses are governed by the Nernst equation, which states that as the concentration of the product species at the reaction interface increases, the overpotential caused by mass transport limitation also increases. When the current density is high enough to prevent reactants from reaching active sites due to an excess of reacting molecules, mass transport losses take place, slowing down the rate of reaction. The Nernst equation can be used to calculate the mass transport overpotential (also known as the diffusion overpotential ηDiff).
| |
| |
| | (67) |
where C0 is taking working concentration as reference concentration whereas C is the hydrogen or oxygen concentration at the electrode-membrane interface. Until operating current densities are moderate, the mass transport losses are not significant. These losses can be neglected with no appreciable errors in voltage prediction at current densities of 1.6 A cm−2.81
5.1.4. Bubble overpotential. The bubbles (H2/O2) generated on the electrode's surface should not be avoided. While lot of the bubbles get away from the surface of the electrode & some cannot, as a result of which direct loss of effective active area, i.e. water is electrolyzed, and bubbles cannot be removed from the electrolytic system quickly; instead, they accumulate on the electrode surface or disperse in the electrolyte. The phenomenon, known as the “bubble effect,” would result in a high overpotential and a significant ohmic voltage drop.82
Current density for gas evolution is given as
| |
| |
| | (68) |
where A is the geometric surface area and I is current intensity. According to the Tafel equation, the reaction overpotential is
| |
| |
| | (69) |
Since the active surface area is reduced by the bubble coverage, the overpotential (ηθ) and the real current density iθ
| |
| |
| | (70) |
| |
| |
| | (71) |
where θ is the bubble coverage ratio on the electrode surface, 0 < θ < 1. Severe bubble coverage decreases the electrode's effective active area. As a result, according to the Tafel equation, iθ grows to be larger than i, and the reaction overpotential also increases.42
5.1.5. Kinetic overpotentials of HER and OER. An exponential relationship between the overpotential and the rate, or current density, j, of the desired reaction results from assuming an Arrhenius-type relationship for the interfacial charge-transfer kinetics.
| |
| |
| | (72) |
The exchange current density which is the current density that flows equally in both forward and backward direction at equilibrium is represented as j0, gas constant, Faraday constant and absolute temperature is represented as F, R and T respectively. α signifies the portion of the overpotential that lowers the reaction's kinetic barrier at the electrode/electrolyte interface.83 For every electrode and chemical reaction of interest, j0 and α are different because they depend on the materials. As a result, the overpotentials at all of the electrodes in an electrolyzer's electrolysis cell add up to form the device's overall kinetic overpotential. In comparison to electrocatalysts in contact with neutral electrolytes, the kinetic overpotentials for the water-splitting half-reactions are typically lower for electrocatalysts in contact with strongly acidic or strongly alkaline electrolytes. Electrocatalysts for the HER exhibit much lower overpotentials in acid than at neutral pH because it is simpler to reduce a positively charged proton than to reduce a neutral species, such as water. Electrocatalysts for the OER, however, display much lower overpotentials in alkaline media than at neutral pH because oxidation of a negatively charged hydroxide ion is easier than oxidation of a neutral species, water.84 The polarization curve obtained by plotting current density vs. overpotential is used to calculate the overpotential of a given current density, if value of the overpotential is lower, higher is the performance of the electrocatalyst. As the kinetics of HER are faster than that of OER, the activation overpotential has priority over the other overpotential as is referred to as onset overpotential. In the case of OER, other overpotentials are also important. Since, the proposed mechanisms of OER in both acidic and basic medium are multistep, Man et al. along with others proposed an expression for calculating the theoretical overpotential at U = 0 vs. SHE by studying the thermodynamics of the OER mechanism.59
| |
| |
| | (73) |
This theoretical value had huge difference compared to the experimental values of the standard free energy change related with the elementary step of conversion of peroxide form oxide. Thus, it's clear the kinetic constraints are not taken into consideration while studying thermodynamics of the overpotential. Due to the reason stated above, instead of an onset overpotential, overpotential (η10) at a fixed current density like 10 mA cm−2 is considered a quantitative parameter to evaluate an electrocatalyst for both HER and OER.85,86 Materials having intense redox peaks i.e. current density greater than 10 mA cm−2 and catalysts like layered double hydroxides which exhibit high performance, i.e. current density more than 500 mA cm−2, overpotentials at higher current densities such as 50 mA cm−2 and 100 mA cm−2 are used as alternative parameters.87–89 Overpotential of a catalyst is mass dependent i.e. it varies with different mass loading.90–92
5.2. Tafel slope and exchange current density
Tafel slope and exchange current density are the two kinetic parameters attained from Tafel plot. Tafel plot is retrieved by replotting the polarization curves i.e. LSV (linear sweep voltammogram) as log current density (j) versus overpotential. It is defined as the dependence of the current density with the iR compensated overpotential and is derived from the Butler–Volmer equation which gives the relationship between the overpotential and current at the electrical interface.93 It is represented as follows:
| |
| |
| | (74) |
Here, I stands for the current density, I0 is the exchange current density i.e. the current at equilibrium potential, n is the number of electrons involved in the reaction (2 in case of HER and 4 in case of OER), F is the Faraday constant (96485C), T is the absolute temperature in Kelvin, η is the overpotential, αA and αC are the charge transfer coefficients for anodic reaction (OER) and cathodic reaction (HER) and R is the ideal constant. Tafel equations of anodic and cathodic polarization curves are the result of the high overpotential approximation of the Butler–Volmer equation.94 They are stated as follows:
| |
| |
| | (75) |
| |
| |
| | (76) |
Both equations are in the form of y = b + mx and would give linear line on plotting logI vs. η and their slope for anodic and cathodic polarizations are
| |
| |
| | (77) |
| |
| |
| | (78) |
Tafel slope is related inversely to charge transfer coefficient (α) and number of electrons transferred, information about the charge rate kinetics and the reaction mechanism can be deduced. The smaller Tafel slope value indicates that increasing the same current density would need smaller overpotential suggesting a faster reaction rate.53 Usually the LSV procured at lowest possible scan rate gives a Tafel slope value with negligible experimental inaccuracies. The scan rate used for acquiring a Tafel plot would cause an issue only if the catalyst exhibit very high capacitive behaviour. This ultimately causes large error in the determination of the exchange current density. Exchange current density is obtained via the extrapolation of the linear fit which intersects at a point where equilibrium potential (zero potential) of the electrocatalytic process meet the respective current density in logarithmic scale. This current density is referred to as exchange current density. Higher the exchange current density, better is the electrocatalyst, it gives an insight on the intrinsic rate of reaction transfer between electrode and electrolyte under equilibrium conditions and is dependent on the electrocatalyst material, analyte in solution as well as the temperature.95,96 The electron transfer efficiency across the catalytic interface will be facile when exchange current density is high, which means very small activation energy is required. Hence the overpotential would be low. In case of electrochemical water splitting, at the potentials of interceptions of 0 and 1.23 V vs. RHE, the exchange current densities for HER and OER are calculated. Thus for electrocatalyst to be ideal for water splitting, they should have low overpotential, small Tafel slope value and a large exchange current density.97
As mentioned earlier in this review, Tafel slope is also correlated to the mechanism of HER, Tafel mechanism have lower Tafel slope value compared to Heyrovsky.98 For Tafel mechanism, the surface active sites separation should not be farther than van der Waals radius of the two adsorbed hydrogen atoms; i.e. smaller the space between them, greater would be the chance of chemical desorption. Comparatively, the Volmer–Heyrovsky mechanism takes place when the electroactive sites are relatively less available and the distance between adjacent sites is greater than the van der Waals radius of the adsorbed two hydrogen atoms. Due to this reason it unlikely for Tafel mechanism to occur. In case of OER, there are a number of mechanisms having different possible elementary steps with any of them being the rate determining step that changes as the pH of the medium changes. Though it is difficult to conclude the mechanism undertaken based on the Tafel slope value, it would be possible to predict the number of electrons transferred. This can be clearly understood from the Tafel slope values of IrO2 and RuO2 that are in between 30–50 mV dec−1 (acidic medium) and 60–120 mV dec−1 in case of the oxides/hydroxides of first row 3d transition metals like Fe, Co and Ni (basic medium). This is because the oxidation state of IrO2 and RuO2 changes from 4+ to 6+ by releasing two electrons per site during OER whereas the oxidation state of M(OH)2 (M = Fe, Co and Ni) changes from +2 to +3 by releasing one electron per site.71
There are different ways in extracting Tafel plots for both HER and OER. The electrochemical techniques used are voltammetry, chronopotentiometry, chronoamperometry and EIS. Tafel plots from voltammograms are in fact not the real steady state polarization curve as the overpotential and current obtained are not retrieved at steady state, instead they were changing as the time goes on increasing. These plots also retained the current contribution from the interface capacitance to the evolution of gas. Due to these demerits, researchers are giving importance to study steady-state polarization curves utilizing static electrochemical techniques like potentiometry and amperometry. After allowing sufficient time for the electrocatalytic interface to reach steady state, the current density and potential are measured for plotting the Tafel curves. The Tafel plot extracted from the steady state methods didn't alter the Tafel slope values, but a significant change in the exchange current density was observed due to the exclusion of the capacitance current in the chronoamperometric measurements.71
Based on the review by Fabbri et al., Tafel plots of IrTiO2 electrode obtained by voltammogram recorded at 5 mV s−1 in (Fig. 7) showed to be in agreement with amperometric measurements.58 Though these steady state electrochemical technique seem accurate in comparison to voltammetry, they too have issues. The current may be free of capacitance current and is obtained in steady state of potential, yet the different types of resistances like intrinsic resistance, solution resistance, electrode–electrolyte resistance and other resistances that occur due to the hetero-junction in the series were not eliminated. Hence the intrinsic and the real activity, the kinetics of the electrocatalyst is not correctly studied from the Tafel parameters gained via this method.
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| | Fig. 7 IrTiO2 electrode's Tafel curves were obtained using both chronoamperometry in 0.1 M perchloric acid and CV at 5 mV s−1. Reproduced with permission from ref. 58 Copyright 2014, Royal Society of Chemistry. | |
Tafel parameters from EIS technique was studied extensively by Hu and co-workers in 2012 who proposed a method to neglect the series resistance (Ru) and catalysts intrinsic resistance.99 EIS measurements were performed at different overpotential in intervals varying with catalyst activity range with an AC sine wave of amplitude ranging 10 to 50 mV from the operational overpotential. The Nyquist plot of log charge transfer resistance (Rct) vs. the overpotential gives the Tafel slope reflecting the kinetics of the catalyst excluding the influence of series resistance as it depends only on the charge transfer ability of the catalyst. S. Anantharaj et al. plotted the Tafel curves of commercial Pt/C 20 wt% in all three methods as seen in (Fig. 8). The Tafel slope observed in LSV recorded in 0.5 M H2SO4 at scan rate of 5 mV s−1 is 43 mV dec−1 and that derived chronoamperometry & EIS under the similar experimental circumstances are 33 and 28 mV dec−1. The theoretical Tafel slope value of Pt/C is 30 mV dec−1 and among the three, EIS derived Tafel curve exhibited the closest value. Thus EIS method gives the exact information of the electrocatalytic activity in terms of Tafel slope and is superior among the three. The precise exchange current density of the catalyst is established by the 1/Rct obtained at the onset overpotential. Based on the necessities and conditions, any one of these approaches can be opted for finding the Tafel slope of an electrocatalyst. The important steps to be followed regardless the method opted for Tafel analysis are; the potential of the Tafel region should be greater than at least 0.12 V from the onset potential of HER and OER, minimum 2 to 3 decades should be present within the linear portion of the slope being measured. The third condition applicable for reversible/quasi reversible redox reactions i.e. attaining a minimum of 1/10 of the limiting current isn't relevant in the case of HER and OER as they are irreversible redox reactions governed by the charge transfer.68
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| | Fig. 8 (a) Linear sweep voltammograms recorded at 5 mV s−1 in 0.5 M H2SO4 exhibits the Tafel plot of Pt/C 20 wt% catalyst (b) & (c) are Tafel plots of the same obtained using Chronoamperometry & EIS methods, respectively,under similar experimental circumstances. Reproduced with permission from ref. 68 Copyright 2018, Royal Society of Chemistry. | |
5.3. Stability
Stability of an electrocatalyst is a crucial parameter determining its promotability to commercial level large scale operations by assessing their activity with a long term operation period. The stability of HER and OER is studied by the cycling of the catalyst within the potential window via two techniques: cyclic voltammetry (CV) or linear sweep voltammogram (LSV) at higher scan rate, also known as accelerated degradation test & prolonged potentiostatic or galvanostatic electrolysis measurements.34 The number of cycles determining the stability of the catalytic material is different in case of OER and HER. In HER, over thousands of cycles can be carried out as the polarization curve starts from 0 V vs. NHE. The accelerated degradation test beyond 1000 is unlikely to be observed in case of OER, as reports on OER catalysts having extreme stability (beyond 250–1000 cycles) are rare. The shift in both onset potential as well as overpotential at a defined current density of 10 mA cm−2 are indicative parameter of electro catalyst. It is considered apt if the overpotential does not increase more than 30 mV and if its final activity degradation after running the stability tests does not exceed 5%. A stable current density (e.g.: 10 mA cm−2 and 50 mA cm−2 or 100 mA cm−2 for high performance/catalysts with strong redox peaks) for over 12 h in chronoamperometric measurements or negligible increase in the overpotential at a current density of 10 mA cm−2 for more than 12 h in chronoamperometric measurements for an electrocatalyst is also another widely accepted parameter for analyzing the stability. The substrate, electrode being used, the catalyst fixation method all contribute to the stabilization of the electrocatalyst being studied.98
5.4. Turnover frequency (TOF)
TOF is another kinetic parameter that tells how fast the electrocatalyst can catalyze the electrochemical reaction at a defined overpotential. It is defined as the number of moles of O2 or H2 gas evolved at each available catalytic site. Both HER and OER undergo pseudo first order kinetics, thus both of them have TOF value in terms of time. TOF is independent of mass loading but is highly dependent on the high coverage, i.e. shows linear relationship only if the coverage is below 100%.99–101 Higher the TOF value better the catalyst.
| |
| |
| | (79) |
Here, j stands for current density (A cm−2), n is the number of electrons transferred (2 for HER and 4 for OER), NA is Avogadro number, F is the Faraday constant, Γ is the total/surface concentration of active sites of the catalysts or the number of participating atoms in electrocatalyst material. The challenges of estimating surface concentration render down the usage of this method as kinetic parameter. There are different methods in determining the total/surface concentration of catalysts. In the first, the surface concentration can be calculated from the redox peaks in the cyclic voltammogram after the activation of the catalyst by CV cycling. The TOF values of 3d transition metals in OER are determined from the redox peak in the CV. The surface active concentration of the metal sites is calculated from the region beneath the redox peak for the formation of oxyhydroxide from oxide/hydroxide. Prior to the onset of OER, metal sites that are undergoing in situ oxidation and reduction are thought to be involved in the catalysis.102,103 The total concentration of atoms can also be calculated using Avogadro number method utilizing the average particle diameter of the catalyst. When the catalyst surface is even and smooth or have sheet like morphology, one can use another method where it's assumed to have monolayer.36 All the three listed method have their demerits. Calculating TOF value from CV could lead to potential error if the entire catalyst is not activated, if the total current is not only contributed by the faradaic reaction but also have non-faradaic component. The problem can be rectified by using the ORR current measured during RRDE (rotating ring disk electrode) experimental in correspondence to OER at the disk and concurrent ORR at the ring is used. The current measured at ORR at ring electrode is solely due to O2 generated from the disk electrode giving more accurate TOF value. In case of HER, techniques like CO stripping, hydrogen under potential deposition (H-UPD), peroxide reactions are used. In the second method the atoms in the core are also taken into consideration though they donot contribute to the catalytic activity thus the exact catalytic property is not reflected and the TOF calculated differs by the order 102 to 103. The third method would lead to error when the material is uneven or may degrade under severe electrochemical conditions. Samar K et al. utilised the galvanostatic Tafel plots to calculate the relevant turnover frequency (TOF) values,104 using the formula
| |
| |
| TOF = I/2mF | (80) |
where current value was acquired from the chronoamperometry electrolysis was represented as I, n is the number of active sites (mol), F is the Faraday constant (C mol−1), 2 represents the number of electrons needed to create one mole of hydrogen; the number of active sites m is calculated as follows;
| |
| |
| | (81) |
Q is derived by integrating the current vs. time plot produced from the trace of the cyclic voltammogram
| |
| |
| | (82) |
The method adopted for TOF calculation should be appropriate for the nature of the catalyst.105
5.5. Faradaic efficiency
It is the efficiency of electrocatalyst for electron transfer provided by the external circuit across the electrode-electrolyte surface which promote the electrochemical reaction i.e. HER and OER. There are two different ways in determining it, either via gas chromatography (GC) for both HER and OER and rotating ring disk electrode (RRDE) for OER to measure the amount of O2. If the reaction course is studied using GC, one could gather information on the how the reaction rate of gas evolution reaction is affected with the change in applied potential by associating the results with the calculated theoretical amount of the evolved gas at certain time interval.
To analysis the FE of an electrocatalyst of OER in both acidic and basic medium, RRDE method is used. In this technique, the O2 molecule evolved at the disk electrode are trapped and reduced at which the potential is swept at a very low scan rate of 5 mV s−1 by keeping potential of the ring electrode constant where complete reduction of O2 to water is possible. The potential of Pt ring is set at constant potential of ORR which varies depending on the pH of the solution. The setup includes glassy carbon disk and Pt ring/Au disk and Pt ring electrode. Faradaic efficiency of an OER electrocatalyst using RRDE is determined using the equation:
| |
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| FE = jR × nD/(jD × nR × NCL) | (83) |
In the above equation, jR is ring current density, nD is number of electron transferred at the disk electrode (4), jD is the current density of disk electrode and nR is number of electrons transferred for reduction of O2 (4, since the ring electrode is Pt and thus electron pathway probable for ORR on Pt surface) and NCL is collection efficiency of ring electrode. FE along with other parameters are mandatory for the screening of HER/OER/water splitting electrocatalyst as it reflects the selectivity of the electrocatalyst. FE is exempted from experimental inaccuracies as they are unaffected by the shape, size, and morphology of the catalyst.105 Quite recently, researchers have tried fluorescence spectroscopy for the measurement of gas evolved, at periodic intervals during galvanostatic or potentiostatic electrolysis instead of GC for both HER and OER to determine FE of their electrocatalysts. Here the oxygen evolved is excited from doublet to singlet state and relaxed back by fluorescence. The intensity of the fluorescent light is the direct measure of the amount of O2 evolved. The majority of faradaic loss are contributed by the formation of by products or due to heat loss.35,53 faradaic efficiency can be defined as the ratio of experimentally evolved volume of gas value (hydrogen or oxygen) to theoretically calculated volume of gas value.
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Assuming a 100% faradaic efficiency, the theoretical volume of gas can be calculated using the Faraday's second law, based on the current density, electrolysis time, and electrode area, as shown in eqn (85). when the water–gas displacement method or a gas chromatography analysis can be used to determine the practical amount that the experiment produced.
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The theoretical hydrogen yield VH2, the ideal gas expression (VM = R(273 + T)/P) is represented as VM, ideal gas constant (0.0821 atm K−1 mol−1), pressure (atm), temperature is denoted as R, P, t respectively; Faraday's constant (96485C mol−1) and applied current (A) is represented F and I. By computing the electrolyte volume reduction during the electrolysis process at the operating temperature and pressure, the volume of hydrogen can be determined. Due to the hydrogen gas being assumed to be a perfect gas to facilitate calculations, the real volume (VH2 (produced)) will be as described in the following equation under normal circumstances.
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where VH2(measured) is the volume obtained via alkaline water displacement, the operating temperature and the standard temperatures in Kelvin is depicted as Tstandard and Tmeasured respectively.106,107 The methods utilized for the calculation of FE depends on the demands of the catalysts. An electrocatalyst would become suitable for commercialization for water splitting if the FE is atleast 90%.
5.6. Mass and specific activities
The catalytic activity of the electro catalyst are also given by two other parameters; mass and specific activity.58,108 Similar to TOF, both are acquired at a particular overpotential. Mass activity is the overpotential at the current density normalized by mass and is expressed as A g−1 while the overpotential at ECSA or BET normalized current density is specific activity. At present, current density normalized by geometrical area are being used but they reflect the area of electrodes with planar and smooth surfaces only. This does not become applicable for electro catalysts with uneven or roughened surface area such as in situ grown catalysts or electrodes modified with nano powders. N2 gas adsorption and desorption measurements are used in BET (Brunauer–Emmett–Teller) adsorption isotherm studies to determine the real surface area. The measurements were made such that N2 are adsorbed and desorbed on all the sites, yet all these sites cannot be electrochemically active. The results acquired by this method have high chances of having inaccuracies, thus ECSA normalized activity is gaining attention over BET. ECSA compared to geometrical surface area is more accurate as they give an insight about the inherent catalytic property of the material of the catalyst. Their disadvantage is that there are number of methods for ECSA determination and the measured values varies from method to method. Thus ECSA normalized activity alone cannot be considered as an ideal parameter, but it along with other essential parameters can be used for the evaluation of the catalyst for water splitting.
6. Periodic classification of metal electrocatalysts of electrochemical water splitting
In this section of the review, elements acting as dopants, electrocatalysts exhibiting electrocatalytic activity for HER, OER and overall water splitting are discussed as per their periodic classifications below.
6.1. s- Block elements
All elements of s block are metals, have low ionization enthalpies, are highly reactive and readily form ionic compounds. They have the ability to be easily oxidized in the presence of air and form oxides and have highly negative standard potential, thus acting as strong reducing agents. Owing to these properties, alkali metals and alkali earth metals have been used in many energy storage and conversion applications. Lithium plays essential and integral part in today's technology due to their small ionic size and negative electrode potential, increasing their ability to store more energy and hence act as ionic mediator in LIBs, Li–S etc.109 For electrochemical water splitting, Li incorporation of lithium as a dopant is seen to boost electrocatalysis. The super Lewis acidity of Li+ accelerated the hydrogen evolution activity in case of Li+–Ni(OH)2–Pt.110 Approximately a 10 fold of total increase in activity was seen due to the generation of hydrogen intermediates which increased due to the Li+- induced destabilization of HO–H bond. The high HER activity of LixMoS2 is because of key role played by Li adsorption. Upon the intercalation of Li, a phase transition from 2H- to 1T′-MoS2 occurs; altering conducting properties. The adsorption of Li direct ΔGH in a favorable direction, promoting HER active edges sites.111 LiCoO2 is a non-precious metal-based catalyst and a prototype cation-exchanged layered metal oxide which is being used as a well-known cathode material for Li-ion batteries, which has its application in OER.112 Study conducted by Shao-Horn et al. on lithium cobalt oxide and phosphate (LiCoO3 and LiCoPO4) water splitting activity in neutral medium, further proved the importance of Li in the electrocatalysis of water splitting, their structure as well as their electrocatalytic activities are observed in (Fig. 9) and (Fig. 10) respectively. Upon cycling in neutral pH LiCoO2 and LiCoPO4 underwent deactivation and activation respectively coincidental with the morphological distortion leading to spinel like surface for LiCoO2 and amorphous surface for LiCoPO4.113
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| | Fig. 9 The surface regions of LiCoO2 and LiCoPO4 are shown in HRTEM images and (right) FFTs in the following order: (a) pristine LiCoO2; (b) cycled LiCoO2 in 0.1 M KPi (pH 7) (c) cycled LiCoO2 in 0.1 M KOH (pH 13); (d) pristine LiCoPO4; (e) cycled LiCoPO4 in 0.1 M KPi (pH 7); (f) cycled LiCoPO4 in 0.1 M KOH (pH 13). The interface between amorphous and crystalline areas is denoted by white dashed lines. The electrodes underwent 100 uninterrupted cycles at a scan rate of 10 mV s−1 at 900 rpm between 1.2 and 1.8 V vs. RHE in 0.1 M KPi at pH 7 and between 1.0 and 1.7 V vs. RHE in 0.1 M KOH at pH 13. Reproduced with permission from ref. 113 Copyright 2018, American Chemical Society. | |
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| | Fig. 10 (a) LiCoO2 and (b) LiCoPO4 electrodes were subjected to 10 mV s−1, 900 rpm CV scans in 0.1 M KPi electrolyte with a pH of 7.0. Dashed arrows indicate the progression of consecutive cycling, and the number labels indicate the number of cycles 0.1 M KOH at pH 13 was used for the CV scans of electrodes (c) LiCoO2 and (d) LiCoPO4 at a scan rate of 10 mV s−1 at 900 rpm. The results of 100 uninterrupted cycles are displayed CV scans in (e) 0.1 M KPi at pH 7 and (f) 0.1 M KOH at pH 13 for LiCoO2 (100th cycle), LiCoPO4 (100th cycle), and electrodeposited Co-Pi. Reproduced with permission from ref. 113. Copyright 2018, American Chemical Society. | |
Synthesis temperature of spinel-type lithium cobalt oxide also attribute to their activity towards OER. LT-LiCoO2, synthesized at 400 °C has higher activity than of layered lithium cobalt oxide synthesized at 800 °C (designated as HT-LiCoO2) for the oxygen evolution reaction.114 Zhiyi Lu et al.115 reported that electrochemical tuning (delithiation) of LiCoO2 improved oxygen evolution reaction. The monoclinic phase formation of Li0.5CoO2, improved the electronic structure altering the catalytic reaction. Compared to commercial Ir/C catalyst, De-LiCo0.33Ni0.33Fe0.33O2 showed optimum OER activity.116 Jianghao Wang et al. studies showcased ultrathin LiCoO2 nanosheets which displayed excellent activity due to high electron conduction and superior electrophilicity to the adsorbed oxygen. The mass activity of ultrathin LiCoO2 was 95.6 A g−1 at an over-potential of 410 mV, relatively greater than bulk LiCoO2.117 On the contrary, OER activity is quite low for Li1−xNiO2 since the intermixing of cations hinder the preparation of stoichiometric LiNiO2 with well-ordered Li+and Ni III into alternate (111) planes of the cubic close-packed oxygen sub lattices leading to the production of non-stoichiometric Li1−xNi1+xO2 which also include Ni II. Asha Gupta et al. reported that Al III doping helped stabilize the Ni III and via both conventional solid-state synthesis & solution-combustion synthesis, LiNi1−xAlxO2 with 0 ≤ x ≤ 0.4 was synthesized with/without Li2O2 (strong oxidizer) in the precursor which curbed the intermixing of Ni II in the Li+layer, improving the OER activity.114 The activity of LiNiO2 can be improved by electrochemical delithiation which assists the formation of superoxo/peroxo-like (O2)n− species like NiOO.The adaptive heterojunction formed by OER-induced surface reconstruction is growth of NiOOH on delithiated LiNiO2.112 The lithium vacancies in the delithiated LiNiO2 augment the electronic structure of the surface NiOOH producing stable NiOO species at this point, allowing for improved OER activity. Due to its inferior activity in terms of electrochemical water splitting, Mg as catalyst is least understood. Recently, Zheng Li et al.118 investigated how Mg doping increased the performance of a Ru-based electrocatalyst for acidic oxygen evolution process. It was observed that at different annealing temperatures, Mg-MOF-74 was able to generate metal oxide particles. Thus they were considered a good precursor for the introduction of Mg into RuO2 to decrease the Ru loading. Mg–RuO2 facilitates better oxygen evolution reaction (OER) in 0.5 M H2SO4 solution, with a low overpotential of 228 mV at 10 mA cm−2 and excellent stability in durability tests of 10000 cyclic voltammograms (CV) cycles as well as exhibited good lifetime for 30 h.
Introduction of beryllium, accelerated the hydrogen generation performance of Ru in both neutral and alkaline medium with the durability up to 50 h.119 However, they had poor activity in acidic medium. From density functional theory (DFT) calculations, it was observed charge on Be incorporation, charge transfer for electrocatalysis as well as improved water adsorption and lowered the energy barrier for dissociation of water, which are advantageous for better hydrogen production from water.
Various calcium based electrocatalysts have also been reported in water splitting especially in oxygen evolution reaction such as calcium copper titanate surface induced graphene oxide, reduced graphene oxide supported cobalt–calcium phosphate composite, F.
Fe-doped calcium cobaltites, layered calcium cobalt oxide, Cu-doping on the activity of calcium cobaltite, iron phosphate modified calcium iron oxide, Fe substitution in lanthanum calcium cobalt oxide are few among those reported.
Strontium is one of the most commonly used s block elements in electronic devices like in light emitting diodes (LED). This metal has shown their capability to act water splitting electrocatalyst while being part of perovskite material. Perovskites are mixed oxides with the general formula ABO3 and are effectively being used as OER electrocatalysts in both acidic and basic medium. Some of the low cost and effectual OER catalysts are Ba0.5Sr0.5Co0.8Fe0.2O3−δ and La1−xSrxCoO0.2Fe0.8O3−δ.120–122 Strontium cobaltite SrCoO3 due to their unique electronic structure exhibit excellent electrochemical property and is on the top of OER volcano chart.123 SrSc0.025Nb0.025Co0.95O3d (SSNC1) and SrSc0.175Nb0.025Co0.8O3d (SSNC2) as a series of extremely active and durable electrocatalysts for the OER in alkaline solution were also produced by Zhou et al. using scandium and niobium cation (Sc3+ and Nb5+) doped A-site strontium cobalt.124 In comparison to gold-standard of OER electrocatalysts, these A-site doped perovskites not only demonstrated up to a 50-fold increase in intrinsic activity, but also outstanding durability. Tafel slope of these catalysts rose to 70 mV dec−1 at lower voltage ranges (1.6 V), while the slopes of SrCo0.8Fe0.2O3d and RuO2 grew to 145 and 150 mV dec−1 correspondingly. Strontium palladium perovskite Sr2PdO3 when synthesized via citrate-combustion methodology portrayed excellent operation stability with the self-activation property and thus is employed as efficient electrocatalyst for OER as well as HER thus encouraging its use in fuel cells with water splitting application, where the catalyst remained stable for 3 h for OER activity.125 Mengran Li et al.126 reported that doping of Sr+ into A-site of the La(Ni,Fe)O3−δ optimizes the electronic structure, tunes the crystalline structure and enhance the number of OER active sites on the surfaces such as hydroxylation, oxygen vacancy defects, or transition metal ions with higher valences for Co or Fe-based perovskite OER catalysts. Excellent OER is achieved on the substitution of La3+ with Sr2+ by improving number of Ni III OER active species, elevating Ni/Fe surface ratio and increasing O22−/O− oxygen species.
From AES and DFT studies of La2−xSrxNiO4−δ (LSNO), Dongkyu Lee et al. concluded that Sr substitution leads to reorientation to (001) tetra. (out-of-plane) from the (100) tetra. (in-plane) orientation as a function of the Sr content in (Fig. 11).127 DFT modelling also indicates a trend in adsorption energies of the LSNO system, the oxygen surface binding decreases with the increase in content of Sr.
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| | Fig. 11 Dependency on content of Sr (a) strain energies (gray) of the La2−xSrxNiO4±δ (LSNO) thin films with 0.0 ≤ xSr ≤ 1.0 derived using the surface energies and strain energy density equation, and (b) oxygen molecule adsorption energies on the (100)tetra. (blue) and (001)tetra. (red) surfaces of the LSNO thin films were acquired from density functional theory (DFT). Adsorption sites of surface oxygen on (c) the (100) tetra. slabs and (d) the (001)tetra. slabs. Anisotropy in LSNO thin films is represented by both surface and adsorption energies. AES data of the La2−xSrxNiO4±δ (LSNO) thin films with 0.0 ≤ xSr ≤ 1.0 annealed at 550 °C in an 1 atm partial pressure of oxygen. (e) LaMNN cation variation (RSF: 0.059), (f) SrLMM cation variation (RSF: 0.027), and (g) NiLMM cation variation (RSF: 0.277) as a function of Sr contents. The shift in the spectra of La and Sr cations scales with the concentration of La and Sr, indicating that the surface chemistry is characteristic of the surface cationic ratio desired. The increase in NiO6 octahedra population (normalised by surface area) with increasing Sr content is represented by a change in the Ni cation spectra which indicate a change in the thin film orientation. Reproduced with permission from ref. 127 Copyright 2014, Royal Society of Chemistry. | |
Barium based metal oxides are currently being studied for water splitting application. Through thermal reduction induced phase transformation, BaMoO3 perovskite is derived from Scheelite BaMoO4 which promotes alkaline HER activity.128 Compared to many bulk perovskite composed of Ni, Co, Fe; perovskite BaMoO3 have lower overpotential at current density of 10 mA cm−2. The higher electrical conductivity is due to interconnected MoO6 octahedra in the cubic-structured BaMoO3 perovskite compared to disconnected MoO4 tetrahedra in the tetragonal-structured BaMoO4 scheelite increasing the number of oxygen surface vacancies favoring the ensuing of the Volmer step of HER as shown in (Fig. 12). The precursor Barium salicylate is employed in the synthesis of BaMnO3 which enhanced the O2 evolution reaction.129 The nanoparticle size, shape as well as uniformity is affected on changing the precursor. Maximum TON and TOF accompanied with the high amount of O2 evolution is related to the nanoplate disc utilizing barium salicylate. Owing to the high charge transferability, May et al. reported Barium-based strontium-doped cobalt iron perovskite oxide Ba0.5Sr0.5Co0.8Fe 0.2O3−δ (BSCF82) to be an excellent catalyst for oxidation of water. The amorphous nature of the catalyst lead to high current for oxygen evolution as well as in double layer capacitance due to the lower O–P band centres w.r.t the Fermi level. On continuing the cyclic voltammetry, changes occurred in the surface morphology of BSCF82 which increased the OER activity.130 Barium based oxygen deficient bi-functional catalyst BaTiO3−x reported by Chen et al.131 was synthesised via sol–gel method executed better performance than noble and other precious metals synthesised through the same procedure. BaTiO3−x exhibited lower onset potential and greater current densities at low potentials <1.6 V than IrO2.
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| | Fig. 12 In 1.0 M KOH, the electrocatalytic behaviour of BaMoO3 perovskite toward the HER was investigated. (a) BaMoO4 and BaMoO3 LSV curves, where currents are normalised to the geometric area of GCE (in mA cmgeo−2) and for reference, the GCE's LSV curve is shown in (a). The inset shows an enlarged version of the low current density region (b) LSV curves of BaMoO4 and BaMoO3, where currents are normalised to the BET surface area to calculate specific activity (in mA cmoxide−2). (c)BaMoO4 and BaMoO3's electrical conductivity was measured in ambient conditions. Schematic representations of the scheelite-to-perovskite phase transition and the shift from insulating to metallic conduction are shown in the inset. Reproduced with permission from ref. 128 Copyright 2020, Elsevier. | |
6.2. p- Block elements
p- Block comprises of a wide range of elements including metals, metalloids and nonmetal. They possess a relatively higher ionization energy and besides the group oxidation state, they show a number of other oxidation states. In this part of review, we would be discussing the recent development of most of the electrocatalysts based on metal as well as metalloid p block elements for OER, HER and electrochemical water splitting. The metal p block elements discussed here are Al, Ga, In, Sn and Bi.
Aluminium due to their nontoxic nature are considered as a good candidate in both acid and alkaline medium. They are inexpensive and are found abundant in the crust and are also observed to produce upto 11 wt% in the electrolysis of water. Aluminium doping onto transition metals like Co and Ni enable them to attain extra electrons from Al promoting interaction between reactant as well as transition metal sites. Al doped CoP nanoarray on carbon cloth (AlCoP/CC)132 exhibiting Pt like super high activity and good stability in acidic medium. From DFT calculations it was concluded that doping of Al lead to weakening of interaction between Co and H causing more thermo-neutral hydrogen adsorption free energy (ΔGH). AlCoP/CC showed excellent activity in basic medium too as only 1.56 V was required to reach 10 mA cm−2 and the lifetime as HER electrocatalyst was studied up to 40 h. Gerken et al. and Chen et al. reported amorphous ternary metal oxide containing aluminium Al0.2Fe0.2Ni0.6Ox and Al0.4Fe0.2Ni0.4Ox had strong OER activity among various ternary metal-oxide combinations and the durability of Al0.4Fe0.2Ni0.4Ox was reported for 24 h.133 Phosphorus and aluminum co-doped porous nio nanosheets studied by Zhao Li134 identified that varying Al concentrations and comparing by LSV measurements, Al doping had significant effect and improve electrocatalytic activity. The elemental ratio of Ni:Al = 2:1 of PA-NiO had the highest catalytic activity OER and HER in series samples. For both HER/OER aluminum doped nickel-molybdenum oxide,135 the XPS results demonstrated Al acts as electron modulator by manipulating the electron density around Mo, inducing unsaturated coordination electron pair and causing electronic structural change. Al–NiMoO4-rods had overpotential of 259 mV to reach 10 mA cm−2 and low Tafel slope of 117 mV dec−1 for OER and overpotential of 131 mV to reach 10 mA cm−2 with low Tafel slope of 82 mV dec−1 for HER and the lifetime of catalyst was aquired up to 30 h. Hang Zhang et al. showed that the modification of electronic states of Ru nanosheets with Al in (Fig. 13) induced delocalize electrons, exposed more active sites due to formed holey structure.136 Metallic Al decoration on MoS2 ultrathin nanosheets have overpotential at 248 mV cm−2 in acid media and 198 mV cm−2 in basic media for HER as presence of Al reduces the band gap of MoS2 nanosheets, improving intrinsic conductivity and the improved is active surface area probability due to the activation of the inert surface of MoS2 by Al decoration.132 Inverse spinel NiFeAlO4 behaving as oxygen evolution electrocatalyst was reported by Jamie Y. C et al.137 revealed the mechanistic role of Al III is modulation of the redox properties of Ni and its ability to activate a water molecule for O–O bond formation owing to low overpotential and durability up to 15 h. DFT studied Co3−xAlxO4 Nanosheets for Water Oxidation rationalized Al-introduction lead occupancy of Co2+ Td sites by Al3+ atoms, decreasing the reaction barriers of intermediates and catalyst surface during the OER processes thereby enhancing the intrinsic activity.138 Co1.75Al1.25O4 is considered one of the best Co-based OER electrocatalysts to date with low overpotential of 248 mV to deliver a current density of 10 mA cm−2 and the authors studied the stability of the catalyst stability up to 30 h. Thus synthesis of Al incorporated transition metals, metal oxides is facial for obtaining high performance functional materials.
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| | Fig. 13 Ru nanosheets with and without Al doping, calculated using DFT: optimised 2D Ru structure in (a); charge-density distribution on the surface with hydrogen adsorption sites indicated in (b); DOSs in (c and d) (inset: enlarged images for states near the Fermi level); and (e) HER free-energy diagrams. Electrochemical characteristics (f) LSV curves; (g) the necessary overpotential for 10 and 50 mA cm−2; (h) Tafel slopes; (i) the EIS at a 150 mV overpotential; (j) the CO stripping measurement; and (k) the LSV curves for the durability assessment. Inset: overpotential for 10 A cm−2 measured every 500 cycles. Reproduced with permission from ref. 136 Copyright 2019, American Chemical Society. | |
Gallium has attracted a lot of interest due to their low melting point and toxicity. Upon exposure of air, gallium can be self passivated by formation of ultrathin Ga oxide layer, this behaviour is similar to Al. Gallium oxide compared to iron oxide has better electrical conductivity, and α-Ga2O3 exhibited good water splitting ability. Gallium oxide nano fibers due to abundances their Lewis acid sites exhibit electrocatalytic behaviour.139 Their low onset potential of −0.34 V (vs. RHE), Tafel slope of this catalyst was 70 mV dec−1 and current density increment better than Pt/C, this catalyst in the absence of carbon additive and heteroatom doping show high performance. Ga intermetallic compounds like AuGa2 and AgGa2 obtained by the ultrasonic irradiation of Ga particles into aqueous solution or organic liquids possess positive reduction potential compared to metallic Ga. Suh-Ciuan Lim et al. reported popcorn-shaped GaPt3 as excellent HER catalyst synthesised via hot-solvent synthesis method.140 They exhibited exceptional durability and an overwhelmingly low overpotential (<80 mV). The same group also reported another Ga based HER electrocatalyst Gallium–Palladium (GaPd2)141 with high surface area and catalytic activity due to synergistic effect by the alloying of Pd–Ga with durability up to 24 h. Ni Wang et al. has stated the hydrogen evolution activity of gallium hydride and ligand-centered reduction.142 Their group was able to study the catalytic proton reduction cycle via mechanistic studies, identified the key intermediates and concluded that Ga III behave as the hydride binding site, whereas the porphyrin ligand behave as the electron transfer site enabling the redox inert metal ions to be catalytically active in the redox reactions. Tao Jing and group studied the Ga based electrocatalyst for oxygen evolution reaction via density functional theory studies.143
Using epitaxial graphene, graphene-like 2D gallium nitride (g-GaN) was prepared by migration-enhanced encapsulated growth method. All the three electrocatalysts, Fe/g-GaN, Ni/g-GaN, and Au/g-GaN are promising electrocatalysts for the OER. However among them Ni/g-GaN showed the least overpotential at 0.26 V. The g-GaN monolayer behave as excellent substrates and improved catalytic activity is related to the d-band center of active atoms. Introduction of Ga to the FeCoOx by Qiaoqiao Zhang via facile sol gel strategy produced amorphous iron–cobalt–gallium oxide (FeCoGaOx) as OER catalyst. The charge transfer kinetics improved due to the inclusion of third metal Ga, thus accelerating the water oxidation process. In 1.0 M KOH, FeCoGaOx on glassy carbon, platinum, and nickel foam had overpotential of 245, 275, and 215 mV respectively.144 Ga doped selenides are rare and recently shun Zhang reported Ga doped CoSe2 nanosheets for enhancing OER activity.145 Ga-doped CoSe2 was synthesised via magnetron sputtering and alloying on stainless steel mesh (SSM) and from XPS data it was revealed that electron drawing from Co to Ga, modulating the electronic structure of cobalt inducing electron deficiency in Co. This deficiency facilitated the transition of Co2+ in CoSe2 to Co3+/Co4+ during the OER process improving the intrinsic OER activity. Recently indium doped transition metal compounds were detected have improved catalytic performance due to adjustment in the strength of adsorption with the reactant, indium based spinels are quite rare and Jiajun Wang recently prepared cubic spinel CoIn2Se4 based on the polyol solution process in an environmental friendly manner where in the crystal site Co2+ occupied the tetrahedral and In3+ occupy the octahedral site as in (Fig. 14).146
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| | Fig. 14 (a) and (b) Represents the CoIn2Se4, In2Se3, CoSe2 and Ir/C electrocatalysts respectively (c) the determined capacitive currents plotted as a function of scan rate for CoIn2Se4, and CoSe2, (d) LSV curves of CoIn2Se4 before and after 1000 CV cycles (inset is the time-dependent current density curves on CoIn2Se4 and Ir/C), (e) ORR polarization curves and (f) the Eonset and E1/2 of CoIn2Se4, In2Se3, CoSe2 and Pt/C. Reproduced with permission from ref. 146 Copyright 2020, American Chemical Society. | |
Their boosted activity was contributed by charge transfer ability and numerous active site in alkaline electrolyte. Gd2O3In2O3–ZnO147 based ternary oxides in short GIZO was studied for overll water splitting activity i.e. both HER and OER. Firstly, the water molecule is reduced on the GIZO surface and liberated Hads, which proceeds to generate H2. For OER and HER; the Tafel slopes values were 121 and 64 mV dec−1 and the overpotential values at ±10 mA cm−2 are 282 and 271 mV respectively with the lifetime of catalyst of 10 h. One step solvent mediated process utilizing different solvents such as pure ethylene glycol, ethylene glycol/water, and ethylene glycol/ethanol mixtures was used to synthesis hollow and porous In2S3 spheres for overall electrocatalytic water splitting. The synthesised In2S3 exhibited overpotential at 230 mV for oxygen evolution reaction and 239 mV for hydrogen evolution reaction. Family of III–VI layered compounds, MX (M = Ga, In; X = S, Se, Te), are promising candidates for hydrogen production are held together via van der Waals interaction among stacked quaternary layers. InSe is seen to exhibit electrocatalytic property in both acidic and basic medium with overpotential of 549 and 471 mV, respectively.148 Jinjie Qian et al. reported (InOF-16, [(Me2NH2)2][In3(BTC)5Co2(DMF)6]·solvent) calcinated heterometallic oxides and/or carbides from a new bimetallic metal organic framework for increased electrocatalytic oxygen evolution activity. The MOF was studied in different thermal treatment i.e. InOF-16-Ar/O2-T (T = 550, 700 and 1000 °C).149 The composition was studied in different environment too.
In pure oxygen environment, there was low OER activity due to electrochemically inactive indium oxide and only binary oxides In(III)2O3 and Co(II,III)3O4 was be obtained. Superior performance was observed in argon atmosphere indicating In(III)2O3 and Co(I)3 In(0)C0.75 nanoparticles (NPs) are well embedded within these graphitic carbon material. The catalyst procured at 1000 °C, InOF-16-Ar/O2-1000 °C present best OER activities at 10 mA cm−2 with much lower overpotential compared to the non-precious OER catalysts. Indium tin oxide (ITO) electrodes are among the important electrodes for electrochemical measurements and is used as support for apprehending the NPs electrocatalytic activities such as HER and OER. Paolo Ciocci150 utilized an in situ optical microscopy approach to differentiate electrochemically active regions of ITO electrodes for hydrogen evolution. The electrochemical behaviour was studied by opto-electrochemical monitoring in a scanning electrochemical cell microscopy, SECCM in 5 mM H2SO4. The optical responses and the electrochemical behavior altered with the sample resistance however the onset potential of the HER remains mainly the same.
Doping of cost effective metal Tin (Sn) onto various electrocatalysts has also caught the attention of researchers in water splitting. Juan Jian reported the effects of doping tin onto Ni(OH)2 on water splitting performance.151 Compared to the undoped Ni(OH)2, Sn4+-doped Ni(OH)2, Sn–Ni(OH)2 processed via one step hydrothermal process had hugely enhanced catalytic activity for both HER and OER. Based on both computational as well as experimental data, Sn4+ was observed to act as active site for Ni sites for OER, while in HER Sn itself act active reaction site. Though Sn4+ has different effect for both hydrogen and oxygen evolution reaction, overall the catalyst act as overall water splitting. The overpotentials of Sn–Ni(OH)2 are 312 mV (OER) and 298 mV (HER), which are lower than the corresponding 396 and 427 mV of Ni(OH)2, respectively at 100 mA cm−2. Jing Y carried out in situ growth of Sn-doped Ni3S2 ultrathin nanosheets on Ni foam and studied their electrocatalytic activity. Its activity was studied in both acidic and basic media.152 Comparing the Nyquist plots of both doped and undoped Ni3S2, the electrochemical impedance was reduced in the case of doped catalyst. Tin due to its weak hydrogen binding and ability to maintain a good balance between adsorption and desorption of hydrogen atoms, considered a good choice to alloy with Ni. Sn doped Ni3S2 had lower Tafel slope value indicating accelerated kinetics and lower overpotential demonstrating they possess superior HER activity. Ni–Sn@C core/shell, though had lower stability in acid medium, their large active surface area and high conductivity led to good electrocatalytic activity toward HER.153 Ternary alloy including Sn as one of the metals for electrochemical water splitting application are also studied. Yuchan Liu successfully synthesised Ni–Co–Sn alloy electrodes154 via galvanostatic deposition onto copper foil and the synthesised catalyst exhibited both HER and OER activities as shown in (Fig. 15). The good performance is due to the synergistic effect of tin incorporation onto Ni–Co as well as the numerous active sites. The Tafel slope values are 63 mV dec−1 and 62 mV dec−1, for HER and OER activities and from chronoamperometric measurements, current density was maintained up to 92% even after 10 h. The bifunctional catalyst better activity with a small cell voltage of 1.76 V at 50 mA cm−2 for overall water splitting. Fang Liu et al. studied an efficient way to optimize the hydrogen evolution reaction activity by doping a main group metal like Sn onto CoS2 causing the activation of their sulphur sites.155 DFT studies evidences improved performance with long term durability which is attributed to increased intrinsic activity and the optimized ΔGH by activating the inert S sites. Ravula et al. reported SnO2 enhanced the HER activity of MoS2 in the efficient HER electrocatalyst MoS2/SnO2/reduced graphene oxide (rGO) multilayer hybrid sheets.156 Yuksel Akinay et al. have reported 3D Sn doped Sb2O3 (ref. 157) as electrocatalysts for hydrogen evolution reaction in acidic medium & demonstrated they show excellent activity & are also stable electrocatalyst.
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| | Fig. 15 (a) Polarization curves of Cu substrate, commercial Pt/C, and Ni–Co–Sn catalysts with a scan rate of 2 mV s−1 in 1 M NaOH for HER (b) Ni–Co–Sn catalyst, Cu substrate, and commercial Ir/C polarisation curves in 1.0 M NaOH for OER at a scan rate of 2 mV s−1. (c) The double-layer capacitances for OER catalysts with pH values of 5, 6, and 7. Reproduced with permission from ref. 154 Copyright 2019, Elsevier. | |
Bismuth based catalyst owing to their cost effectiveness, high electrochemical active sites, environmental friendliness and good electrocatalytic activity towards water splitting has been investigated for numerous applications. Though Bi based catalysts are considered to showcase poor HER activity, the activity can be improved by the synergistic effect with oxy/hydroxide phases of metals. Rahmad Syah et al. incorporated metallic Bi with Bi2O3 (ref. 158) residuals on Ni and observed high catalytic performance. Evaluation of the EIS techniques indicated low Tafel slope value 55 mV dec−1 indicating rapid kinetics with low overpotential of 250 mV dec to the synergetic effect of metallic Bismuth and oxidized Bismuth in the crystal structure. The ion diffusion was readily reduced due to the 3D porous structure leading to high electron transport. Shuai Wang et al. reported succulent-like binary metal sulphide heterostructure activated by bismuth, Bi2S3/Ni3S2/NF prepared by mild solvothermal method as OER electrocatalyst in alkaline medium. Bi2S3/Ni3S2/NF, as a low overpotential of 268 mV with low Tafel slope (82 mV dec−1) and good stability for over 12 h at high current density.159 Various perovskite-based bifunctional catalysts are synthesised utilizing alkaline earth metals (Ba, Sr, Ca, and Mg)-doped bismuth iron oxides Bi0.6M0.4FeO3 (BFO).160 Among them, Ca(Bi0.6Ca0.4FeO3 (BCFO)) exhibited good OER performance in alkaline media, as BCFO160 have current density 6.93 mA cm−2 at a fixed overpotential of 0.42 V which is double compared to other catalysts having current density 3.06 to 4.04 mA cm−2. Bismuth molybedate, Bi2MoO6, have layered structure consisting perovskite layer (An−1BnO3n+1)2−, A = Ba, Bi, Pb, etc.; B = Ti, Nb, W, Mo, etc. and bismuth oxide layers (Bi2O2)2+ and is considered an advanced material. Sakila Khatun et al. reported the development of Bi2MoO6 along with a series of bismuth iron molybedate comprising of major component Bi2MoO6 and minor component monoclinic Bi3(FeO4)(MoO4)2 existing as solid solution and studied their performance in both OER and HER which exhibited overall ater splitting at 1.67 V @ 10 mA cm−2.161 The optimum blending of these components in solid solution exhibited overpotential at current density 10 mA cm−2 and 50 mA cm−2 at 266 and 288 mV respectively for OER and overpotential at 10 mA cm−2 at 207 mV for HER portraying the remarkable activity. Chenguang Duan et al. reported high catalytic activity of 2D Bi for hydrogen evolution reaction.162 The 2D Bi nanosheets are synthesised via sonication assisted liquid phase exfoliation method and their 2D structure has led to Bi nanosheets having shorter charge diffusion distance and more active sites compared to bulk counterpart. Pillai et al. calculated that adsorption energy of the hydrogen atom and oxygen atom on the surface of Bi were −1.418 and −3.963 eV respectively. The small atomic hydrogen adsorption value indicate that HER occur on the 2D Bi surface.
The metalloids of p Block elements being discussed here are B, Si, Ge, As, Sb, Te.
Transition metal borides in (Fig. 16) for electrochemical water splitting was studied as early as 2009 by Daniel Nocera and group by investigating the electrocatalytic properties of Co and Ni borates in near neutral medium. Followed by Hu and Patel for development of Mo and Co borides.163–167 Borates are referred to electrocatalysts prepared via electrodepostion and borides are electrocatalysts prepared by other techniques. Over the years many methods to improve the performance by usage of substrate like metal foams and plates, carbon and graphene based structures, nano structuring i.e. metal core-oxide shell, ternary alloys: Ni–Fe-b, Co–Mo–B, Co–Fe–B, Co–Ni–B, Co–W-b etc., quaternary alloy; Fe–Co–Ni–B, V–Co–Ni–B, alloying with P and different synthesis strategies; electrodeposition: amorphous coating of metal borates, solid state: molybdenum based borides, CVD/boronisation: thin films, boronizing metal surfaces, electro less deposition; nanostructured coating of Co and Ni borides. Chemical reduction: amorphous nanoparticles of Co and Ni borides have been studied. Though a number of reports and dedicated reviews are there for metal borides and borates, in this review, the most recent works will be discussed.
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| | Fig. 16 Performance of various metal boride/borate electrocatalysts was compared at pH 9.2. The graph also displays the most efficient catalysts from different nonprecious metal catalyst families and precious metal-based catalysts. (Acronyms used—C: Vulcan carbon; CC: carbon cloth; CP: carbon paper; NF: Ni foam; Ni: Ni foil/plate; Ti: Ti mesh/foil.)48 | |
Xuetao, Liu et al. successfully synthesised boron-doped cobalt–iron bimetal phosphides nanosheets by hydrothermal boronation of bimetallic metal–organic frameworks precursors with the subsequent low-temperature phosphidation for enhanced oxygen evolution.168 The cooperative effect between Co and Fe with B and P influenced the electronic configuration surrounding the metal centres. The catalyst exhibited excellent durability with low Tafel slope value of 49.5 mV dec−1 and overpotential of 294 mV at 10 mA cm−2 in 1.0 M KOH. Tungsten and phosphorus co-doped amorphous FeB in alkaline medium169 is considered to be among the efficient boride based OER catalysts due to low overpotential 209 mV of 10 mA cm−2 and durability for over 25 h. This study led by Zhijie Chen also offers an approach optimize metal borides such as NiBx, CoBx, and MoBx as precatalysts for OER. B doping onto 2D MXene nanosheets was seen to improve the adsorption kinetics of intermediate H with reduction of charge transfer resistance towards HER based on quantum mechanical first-principles calculations.170 Munkhjargal Bat-Erdene incorporated Ru nanoparticles onto Boron doped Ti3C2Tx MXene, Ru@B-Ti3C2Tx as HER catalysts exhibited low overpotential of 62.9 and 276.9 mV at the current density 10 and 100 mA cm−2 with over 1000 cycles for stability. Dong et al. reported boron carbonitrides with tunable B/N Lewis acid/base sites for overall water splitting. B-BCN had onset potential at 1.4 V for OER with Tafel slope 65.2 mV dec−1.171 N-BCN had low onset potential of −50 mV for HER with a small Tafel slope of 61.3 mV dec−1 with the lifetime of 48 h and the overall water splitting occurs at 1.62 @ 10 mA cm−2 & 1.79 @ 100 mA cm−2. B-BCN//N-BCN exhibited onset potential of 1.52 V for overall water splitting. Thus BCN with different B/N contents exhibit electrocatalytic activity towards water electrolysis. B doped Ni (B-Ni) in basic medium exhibited bifunctional electro activity for both HER and OER and showcased excellent water splitting performance with 1.62 and 1.74 V (vs. RHE) to attain 10 and 100 mA cm−2, respectively.172 Due to electronegativity difference, doped B extract electron from Ni sites increasing their charge density and resulting in loss of charge density on the Ni atoms, this charge distribution favours HER processes. In-Kyoung Ahn et al. reported boron-doped nickel-iron layered double hydroxide to facilitate charge transfer in oxygen evolution electrocatalysts.173 B:NiFe LDH had 281 mV at 10 mA cm−2 which was further reduced to 229 mV through the electrochemical oxidation, maximizing the doing effect and activating the catalyst. EIS measurements confirmed that doping as well as activation reduced the charge transfer resistance. Boron doped FeNi solid solution grafted onto activated hydrophilic sponge (B-FeNi@HS) behave as a as a water splitting bifunctional catalyst.174 B-FeNi@HS exhibit good activity with overpotential at 10 mA cm−2 for HER and OER being 54 and 171 mV respectively. Hong ling Li fabricated and studied fabricate hybrid boron carbon oxynitride nano fibrous ((BCNONF)175 and they exhibit high performance and greater durability in alkaline electrolyte up to 10 h and overall water splitting 1.79 @ 10 mA cm−2.
Silicon is among the most abundant element in the Earth's crust. Silicon though compared to carbon has the same electronic configuration, carbides are widely explored in water splitting applications where as silicides are not much investigated. Mono silicide,176 RuSi which is recently reported showcased good HER activity, as interstitial silicon activated the recessive active Ru site on the catalyst surface. In the detailed study of transition metal monosilicides including silicides of Ti, Mn, Ru, Ni, Co, Pd, Fe and Rh as promising new HER catalysts by Yuan He, almost all of them exhibited excellent activity due to their large ECSA i.e. up to 211.5 cm2 and turn over frequency up to 118 s−1 at 100 mV. Yujia Huang176 reported for the first time reduced graphene oxide (rGO)/SiO2 ceramic composite as HER catalyst. The incorporation of SiO2 into the rGO layers avoided the aggregation of the rGO layers, prevent electrolyte diffusion and blockage of gas. The close contact between SiO2 and rGO improved the durability as well as electrical conductivity. Si based intermetallics are also investigated for OER activity. Porous Fe5Si3 has superior OER performance in extremely acidic condition showcasing their superior corrosion resistance.177 Amar A. Bhardwaj and group showed that Chlorine Evolution reaction can be suppressed greatly while still allowing OER to take place in acidic and pH neutral seawater, by the deposition of catalytically inert SiOx over layer planar Pt electrocatalyst, SiOx encapsulated electrocatalyst could be improved by enhancing the adhesion properties of the interface thus regulating the H2O/Cl− transport selectivity. Mesoporous silica (SBA-15) are also considered for evaluation for HER activity. Elham Chiani178 demonstrated that PdCu bimetallic nanoparticles decorated on ordered mesoporous silica (SBA-15)/MWCNTs showed excellent HER activity with high exchange current density of 165.24 mA cm−2 at −360 mV and low Tafel slope of 45 mV dec−1. The presence of mesoporous SBA-15 contributes to the improvement of the activity by ensuring abundant active sites for Pd and Cu NPs dispersion.
Another new candidate in the field of electrocatalysis is graphene like germanium carbide (g-GeC) that are being considered due to their adjustable electronic properties and large surface. Xin Chen et al. studied OER and HER activity of TM doped g-GeC (M-GeC). For OER, among the 3d, 4d, 5d transition metals, Pd-GeC and Ni-GeC exhibited highest activity with low overpotential values of 0.41 and 0.55 V, respectively.179 While for HER, Mn-GeC with ΔGH equal to −0.04 eV possesses excellent electrocatalytic activity. Meihong Fan et al. based on DFT calculations and experimental study, explained how adsorption of germanium on ruthenium catalyst promote Hydrogen evolution activity.180 Preferential adsorption sites are altered from the hollow to top of Ru catalyst due to germanium. J. Niklas Hausmann reported intermetallic Fe6Ge5 synthesised via high temperature method, which consists of polyhedral of five different Fe atoms forming square pyramids whereas Ge atoms form pentagonal prisms with distorted octahedra.181 Fe6Ge5 behaves as a novel OER precatalyst in alkaline media with overpotential of 272 mV at 100 mA cm−2 and Tafel slope value of 32 mV dec−1.
The unique morphology and electronic structure of the natural mineral orthorhombic arsenic has attracted attention. Owing to their black colour they are represented as b-As.182 They have puckered honeycomb structure with semiconductor properties whose electronic structures are thickness related. Single and few layered b-As promote electron mobility in HER due to high carrier transport ability. Though pristine b-As has poor hydrogen atom binding and show weak HER properties, doping of atoms like N, S, C, O and P enhance their performance. Among them O is easily embedded into b-As lattice and reduce |ΔGH| to as low as 0.044 eV close to Pt with |ΔGH| = −0.090 eV. Joseph A. Gauthier et al. studied various transition metal arsendies and concluded CoAs and MoAs to be active arsenide materials for HER.183 MoAs exhibited weak hydrogen binding whereas CoAs bonds to hydrogen too strongly, both the catalyst exhibit moderate overpotential on the basis of normalized electrochemical surface area basis. Doping of grey arsenene (g-As) along with pristine black (b-As) with heteroatoms184 was studied for both HER and OER activities by Sengpajan Santisouk. Among them single O-doped g-AsO1 as well as bAsO2 exhibited OER activity whereas double C-doped g-AsC2 exhibited HER activity.
Elemental antimony nanoparticles as well as conductive antimony tin oxide (ATO) are used in electrochemistry Li ion and Na ion batteries. Raspberry like structures of SbPt synthesised via arc melting and sintering was observed to have ECSA 1.25 times greater than Pt black in (Fig. 17). Existence of Sb have synergistic effects from the hybridization of 3d orbital prevents the adsorption of hydrogen improving the catalytic performance.185 Song Lu et al. studied the DFT calculations to analyze the efficiency of single transition metal embedded antimonene monolayers as HER and OER catalyst. Transition metals like Sc, Ti, V, Cr, Mn, Fe, Co, Ni, Cu, Zn, Ru, Rh, Pd, Ag, Cd, Ir, Pt and Au are embedded onto the single Sb vacancy of the antimonene monolayer to form SACs (Single Atom Catalyst). Based on calculations, Ir@Sb has best electrocatalytic activity toward the HER as ΔGH value is 0.009 eV and, Pt@Sb had best OER performance having the least overpotential of 0.48 V which lower than state of the art catalyst IrO2 with overpotential 0.56 V.186
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| | Fig. 17 At pH 9.2, the effectiveness of various metal boride/borate electrocatalysts was evaluated. Additionally, the most effective catalysts from various non-precious metal catalyst families and precious metal-based catalysts are shown in the graph. Reproduced with permission from ref. 185 Copyright 2021, Elsevier. | |
NiTe and CoTe187 are metal tellurides studied for bifunctional catalysts, OER and HER in metal chalcogenides due to their conductivity, specific area and chemical stability. It was explored that transition metal based catalysts when tellurized improves the intrinsic water splitting activity of the material by increasing the covalent character of the metal anion bond. Among all chalcogenides; tellurides are less toxic and has less electronegativity thus increasing their covalent character when bonded to metal, lowering redox potentials of metals generating active sites for catalysis. For HER, Nickel tellurides has overpotential of 679 mV and a Tafel slope of 151 mV dec−1 whereas CoTe nanostructures over carbon fiber paper show a low overpotential of just 230 mV at 100 mA cm−2 and CoTe2 with low overpotential of 246 mV at 10 mA cm−2.
CuTe-NS are deposited over conductive NiF substrate188 synthesised via hydrothermal approach demonstrate decent HER activity with low Tafel slope of 36 mV dec−1. Hagyeong Kwon recently reported nanoporous silver telluride on the basal plane of layered metallic support, WTe2.189 AgTe compared to other Ag related catalyst which have poor catalytic activity, reveal higher HER performance and better durability overcoming the problems faced by many state of the art catalysts.
6.3. d- Block elements
d- Block elements consist of first transition series (from Sc to Cu), the second transition series (from Y to Ag), and the third transition series (the element La and the elements from Hf to Au). They form stable complexes, have high melting and boiling point, exhibit variable doxidation states and form compounds with profound catalytic activity. As there are number of elements in d block, in this part of the review, the elements partaking in water electrolysis, HER and OER are divided into groups based on their properties. The groups are transition metal carbides/nitrides, transition metal phosphides, transition metal sulfides, and selenides and noble metal-based electrocatalysts that are recently reported will be discussed.
6.3.1. Transition metal carbides and nitrides. Transition metals carbides and nitrides are kind of interstitial compounds which, due to their noble metal like behaviour and electronic structure, are widely studied in water splitting. The interstitial sites of the parent metal are occupied by the carbon and nitrogen atoms. Their properties are a combination of covalent, ionic and metallic. Covalent character ensures tolerance to stress, and metallic character displays high electrical conductivity, whereas electronic structures like noble metals are due to the ionic character of bonding. The properties of monometallic carbides or nitrides are not up to that of state of the art catalysts. Thus further improvement can be achieved by modifying size, doping of heteroatom, usage of substrate etc.190,191 Recently, Mo2C was doped with different transition metals such as Cr, Fe, Co and Ni utilizing precursors, Anderson-type polyoxometalates (POMs) and their HER activities were studied. It was observed among the prepared four catalysts, Ni–Mo2C exhibited the highest activity followed by Co–Mo2C and Fe–Mo2C and the least activity was shown by Cr–Mo2C.192 POMs are a distinct class of nanosized clusters that may be used as a preliminary step for the synthesis of nanoscale material. Additionally, high-negative-charge polyoxoanions can readily combine with nearly all transition-metal elements in a single molecule, suggesting that they can be used as an ideal molecular platform for the design and preparation of multi-metallic composited electrocatalysts. In order to achieve 50 mA cm−2 current for HER and OER, respectively, Zhang et al.193 used a low-cost and effective catalyst of Co2P@Co2P/Co-POM/NF that delivered a minor overpotential of 130 and 336 mV with Tafel slope of 135 mV dec−1 and 57 mV dec−1 at a low cell voltage of 1.6 V @ 10 mA cm−2. Using the Anderson-type (NH4)4[Co(II)Mo6O24H6]·6H2O polyoxometalate as a bimetal precursor, Hou et al. recently demonstrated the synthesis of oxygenated CoMoS with low overpotentials of 97 mV and 272 mV for HER and OER at 10 mA cm−2 with Tafel slope of 70 and 45 mV in alkaline solution at low cell voltage of 1.6 V with the durability of 10 h.194 Samar K et al. K2[Ni(H2O)6]2[V10O28]·4H2O (1) synthesised is the first homogeneous POM catalyst with nickel that is used for electrochemical hydrogen evolution in acidic medium195 with overpotential of 127 mV, turn over frequency of 2.1 s−1 with durability of 5 h at 1 mA cm−2. For heterogenous water oxidation, surface-modifying agent APTS was used to immobilize the high nuclearity 16-Co(II)-containing POM over commercial Ni foam, and the synthesised electrode Co16-GeW9@NiF195,196 exhibited low overpotential of 370 mV at 10 mA cm−2 and a Tafel slope of 84 mV dec−1, in basic pH with stability of 8 h. K5[Bi(H2O)2SiW11O39]·13H2O (K51·13H2O). [Bi(OH2)2]3± functionalized Keggin polyoxometalate (POM) compound was fabricated with ZIF-8, a metal–organic framework containing zinc imidazole; H51@ZIF8 exhibited OER activity with 375 mV, the overpotential and Tafel slope is 188 mV dec−1 in 0.1 M KOH with stability studies carried up to 3 h.197 The overpotential value @ 10 mA cm−2 and Tafel slopes are found to be 726.04 mV and 173.80 mV dec−1 respectively for the anti-Lipscomb (Krebs type) compound incorporated with cobalt; Na6[{CoII(H2O)3}2{WVI(OH)2}2{(BiIIIWVI9 O33)2}]·8H2O suitable for HER activity to support up to 10 hours of chronoamperometry.198 Core shell-type metal–organic framework (MOF) derived Co-NC (Co-NC-POM) coated Keggin-type polyoxometalate (POM) nanoscale particles were prepared using a simplistic coprecipitation method by Huimin et al.199 where the as-synthesised Co-NC-POM hybrids exhibited excellent electrocatalytic performance for both HER and OER due to their active sites, distinctive structure, and defects as well as synergy effect. Co-NC-POM has lower overpotential and Tafel value of 133 mV at 10 mA cm−2 and 124 mV dec−1 for HER activity while for OER the catalyst also provides low overpotential of 400 mV with Tafel slope of 94 mV dec−1 with excellent stability up to 24 h and low voltage for overall water splitting 1.60 V. Due to their numerous active sites, atomic-level thickness and substantial surface area, the 2D nanosheets of metal–organic frameworks (MOFs) have recently become recognized as a promising material. Investigated in 1 M KOH, the OER performance of Te, Cl–NiFe MOF on nickel foam with overpotential of 224 mV at 30 mA cm−2 and Tafel slope value of 37.6 mV dec−1.200 The incorporation of Te in NiFe MOF improved charge transfer and allow for the adsorption of oxygen on the surface where as Cl helped to increase the number of active sites on the NiFe MOF surface. G/THTA-Co G/H3 [Co3(tht)(tha)], pristine MOF based electrocatalyst in acidic medium exhibited HER activity provided low overpotential of 230 mV with Tafel slope value of 70 mV dec−1 with stability of up to 4 h.201 MOF based electrocatalyst CoP/Co-MOF on carbon cloth exhibited overpotential of 49 mV and Tafel slope of 63 mV dec−1 in 1 M PBS, overpotential of 34 mV and Tafel slope of 56 mV dec−1 in 1 M KOH, overpotential 27 mV and Tafel slope 43 mV dec−1 in 0.5 M H2SO4 while MOF-derived electrocatalysts CoP/Mo2C-NCG with overpotential of 55.7 mV and Tafel slope of 49 mV dec−1 in 0.5 M H2SO4 and overpotential of 67.2 mV and 66 mV dec−1 in 1 M KOH. Co-Zn/PNC shows both HER and OER with overpotentials 180 and 348 mV and Tafel slopes 100 mV and 84 mV respectively with overall water splitting at 1.63 V @ 10 mA cm−2.202 Li et al. used the semi-MOF precursor NiCoFe-MOF-74 to create NiCo/Fe3O4 heteroparticles encased in MOF-74 (NiCo/Fe3O4/MOF-74) using a partial pyrolysis technique. Strong evidence of an interaction between NiCo and Fe3O4 was revealed by DFT calculations. A high activity with an overpotential of 238 mV at 10 mA cm2 and Tafel slope 29 mV dec−1 and exceptional long-term stability of 36 h with overall water splitting 1.47 V were observed when the NiCo/Fe3O4/MOF-74 was tested for the OER.203
Porous molybdenum carbide nano-octahedrons was synthesised by Lou et al. using MOF based material (NENU-5; [Cu2(BTC)4/3(H2O)2]6[H3PMo12O40]) which shows excellent activity in both acidic and basic medium. They have low Tafel slope value 59 mV dec−1 with overpotential of mV at 10 mA cm−2.204 Both HER and OER activity is observed in Co doped Mo2C nanosheet (Co SAs/Mo2C), the Co-anchoring is carried out by sacrificial Zn strategy.205 α-Mo2C synthesised via simple urea-glass route of molybdenum carbide, due to small particle size (2–17 nm) exhibit enhanced HER performance in basic medium than in acidic medium.206 Wang et al. developed 3D molybdenum carbide (Mo2CTx) microflowers207 in (Fig. 18) with outstanding OER activity having an overpotential 180 mV to reach 10 mA cm−2.208
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| | Fig. 18 (a) Mo2CTx electrocatalytic OER schematic illustration. (b) Image of the Mo2CTx MFs obtained using scanning electron microscopy. (c) A comparison of different Mo2CTx samples' Linear Sweep Voltammogram curves Reproduced with permission208 Copyright 2021, Elsevier. | |
Streb et al. using Mn/V oxide as precursor for metal oxides/carbides-CNT hybrid that it demonstrates higher activity and durability for OER activity.209 Titanium carbide (MXene) owing to their electronegative surfaces as well as large surface area has ability to be potential carrier material of MOF and FeNi-LDH. MOF/Ti3C2Tx and FeNi-LDH/Ti3C2-MXene exhibit good activity OER. Yang et al. introduced Mo into Nickel carbide to form Mo6Ni6C catalyst which shows good OER activity as at the atomic level, Mo6+ has integrated with the Ni species and the carbon element has been insert into MoNi array.210 Doping of Fe led to tuning of electronic properties of Ni3C enhancing its catalytic reactivity in both HER and OER. Among transition metal nitrides, molybdenum and tungsten based nitrides are popular than other TMNs. 2D nitrogen-rich hexagonal W2N3 (h-W2N3)211 developed and studied by Zhou and group showed HER activity with overpotential −98.2 mV at 10 mA cm−2 in acidic medium in seen (Fig. 19).212
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| | Fig. 19 (a) Schematic diagram depicts the preparation of 2D h-W2N3. (b) Images of 2D h-W2N3 nanosheets from STEM, AFM, and EDS. 2D h-W2N3, bulk-W2N3, and 20% Pt/C polarisation curves are shown in (c). Reproduced with permission212 Copyright 2021, Elsevier. | |
NiMoN nanowires reported by Hao et al.213 had remarkable HER and OER activity, with overpotential of 38 mV at 10 mA cm−2, lower compared Pt/C in HER and 290 mV at 50 mA cm−2, lower than Ru2O for OER, durability was studied up to 5 h and 1.498 @ 20 mA cm−2, 1.532 @ 50 mA cm−2, 1.559 @ 100 mA cm−2. Ni2Mo3N synthesised via urea glass route shows OER activity in alkaline medium with overpotential at 270 mV. Treating Co3Mo3N in ammonia at low temperature, close packed Co0.6Mo0.14N2 was procured by Khalifah and group214 with small Tafel slope and low overpotential, showing HER activity. By facile pyrolysis, among the series of bimetal embedded nitrogen-doped carbon framework synthesised, Co0.75Fe0.25-NC displayed low value of Tafel slope and small onset for both HER and OER. Ni3FeN supported Fe3Pt fabricated by Good enough et al. has more activity than OER catalyst (Ir/C).215 Zeng et al. synthesised plasmonic material TiOxNy nanoshells supported on Fe2N that have holes due to movement of hot electrons upon laser radiation which capture electrons from the reaction intermediate and elevate the OER activity.216 A series of VN support was synthesised by simple ammonolysis and Wang & group concluded that NiCo2O4 supported on VN due to the facile electron transfer from VN to NiCo2O4 surface shows more durability and catalytic activity than NiCo2O4 supported on carbon black.217 The same group studied MoS2 supported on VN which show superior HER performance.218
6.3.2. Transition metal phosphides. Transition metal phosphides are other promising class of electrocatalysts for electrochemical water splitting. They are being used extensively catalysts in hydro processing reactions like hydrodesulphurization, that involves hydrogen adsorption and dissociation steps, hence embarking potential adequacy as electrocatalyst for HER. As protons are not easily available in alkaline medium, the electrocatalytic activity of TMP was found to be unsatisfactory. Attempts to boost the catalytic activity is by doping via heteroatom, by bimetallic alloys enhancing surface area by architecture engineering or by usage of substrates. Among non-metal dopants sulphur is the most common, its notices that even the amount sulphur being doped has a drastic impact on the performance of the electrocatalyst. Varying amount of sulphur doped in nickel phosphide nanoplate arrays on carbon paper was evaluated for HER activity by Xing and group,219 it was observed 6 wt% of sulphur gave the maximum catalyst reactivity which have overpotential of 56 mV at 10 mA cm−2 and 104 mV at 100 mA cm−2. Doping of Zn along with sulphur has also improved overall water splitting activity both n neutral and alkaline medium and was extensively studied by Li et al. To drive HER process, the catalyst Zn0.075S–Co0.925P NRCs/CP in neutral medium (1.0 M PBS) has overpotential of 67 mV at 10 mA cm−2 & 37 mV at 10 mA cm−2 in 1.0 M KOH with durability of 20 h and overall water splitting of 1.70 V.220 The structural analysis accompanied the DFT calculation gave an insight on how doping of both Zn and S, optimized the electronic structure and increased the active sites. Boron doped TMP recently gained attention as HER electro activity as Boron being less electronegative than P The XPS studies of B-CoP anchored on MWCNT221 show that the electron density of the bonded Co is only weakly withdrawn by B than P, confirmed by the shift of the binding energies by 0.3 eV of both Co 2p3/2 and Co 2p1/2 making the oxidation state of B-CoP. Compared to undoped NiMoP2/Ni their oxygen doped counterpart had lower overpotential as ΔGH value (DFT calculations) approached zero on doping making the HER kinetic facile.221,222 Ni2P when doped with oxygen and iron exhibited excellent performance for HER and OER at high pH.223 The electrochemical surface oxidation of Ni2P in 0.1 M potassium hydroxide led to the formation of a interface between Ni2P and amorphous nihydr(oxy)oxides (sc-Ni2Pδ−/NiHO) forming negatively charged Ni2P, confirmed by XANES and XPS studies. sc-Ni2Pδ−/NiHO compared to Ni2P exhibited higher HER activity and lifetime of 100 h.224 Fe doped CoP holey nanosheets in alkaline medium behave as bifunctional catalysts for both OER and HER as seen in (Fig. 20) with faradaic efficiency of 96.4% and 95.4% and overpotential 79 and 220 mV to achieve 10 mA cm−2, for hydrogen evolution reaction and oxygen evolution reaction respectively along with the overall water splitting at 1.55 V @ 10 mA cm−2 and lifetime of 20 h.225 Moreover there was no attenuation after running chronopotentiometric tests for one day which is viable quality for OER catalysts.
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| | Fig. 20 Fe doped CoP electrocatalyst TEM (a), electrocatalyst Tafel plots (b), and LSV curves before and after a 24 hours test for Fe CoP HNSs (c). Reproduced with permission225 Copyright 2021, Elsevier. | |
Da Huo226 recently published their work on carbon encapsulated nickel phosphide as OER catalyst. A safe and simple strategy was considered, i.e. by pyrolysis of the precursor nickel triphenylphosphine which generate current density of 10 mA cm−2 at overpotential of 0.26 V and stability was ensured as there was no current change even after polarization for over 25 h. Core shell Fe2O3@C@CoP nanocomposite227 synthesised by Xia Zhong proposed that phosphorus vacancies promote Co O C bonds i.e. direct coupling of cobalt of CoP with the functional group oxygen from the carbon layer thereby enhancing OER performance having a small Tafel slope of 55 mV dec−1 with low overpotential of 230 mV and durability of 25 h. The electro activity is seen to increase even by doping with only trace amount of transition metal like Co and Ni to TMP which was evaluated by Zhang et al.228 Yihao Liu and group studied the pre and post electrolysis characterisation of Mn doped CoP and noticed that during OER Mn–CoP is converted to Mn–CoOOH hexagonal nanosheets which serve as active species.229 By doping Mn gap the gap states of the Fermi level of active O sites allowing the deprotonation of OH thereby decreasing the RDS energy barrier. Doping of around 0.02% Ni or Co showed significant improvement in MoP as HER electrocatalyst in acidic medium with overpotential 102 mV at −10 mA cm−2 whereas MoP had around twice the overpotential i.e. 210 mV.228 3D core–shell CoP@CoFe-LDH/NF heterostructures exhibited enhanced the OER performance with small Tafel slope value of 69.2 mV dec−1 and high TOF value of 0.093 s−1 at 300 mV and durability of 20 h.230 The nanosheets are interconnected improving the number of active sites doping of TMP with noble metals have significantly bettered the activity and reduced the entire reliance on precious metals. Only 3.85 wt% ruthenium doped Ni5P4 synthesised via impregnation of Ru3+ into Ni(OH)2 followed by phosphidation as HER electrocatalyst with Tafel slope of 52 mV dec−1 and overpotential of 54 mV at −10 mA cm−2 with excellent durability of 120 h.231 The most recent work involves doping of Ru onto NiCoP inducing numerous Ni and Co vacancies leading to abundant number of active sites and increased conductivity after electrochemical activation. Dr Jun Kim and group232 proved Ru doped NiCoP showed excellent OER catalytic performance with Tafel slope value of 42.7 mVdec−1. Multicomponent heterostructures Co–Mo–P/Zn–Co–S in alkaline medium act as bifunctional catalyst for HER and OER displaying outstanding activity that outperforms the state of the art catalyst because of their large surface area, synergistic effect of the hybrid material and augmented electron and mass transfer.
6.3.3. Transition metal sulphides and selenides. The intrinsic electro activity, excellent performance due their metallic/semiconducting nature, variety of crystal phases, flexibility for incoporperation of dopants make transition metal disulphides (TMDs) excellent HER and OER electrocatalysts. The efficient HER electrocatalyst among MoS2, WS2, NbS2 and Nb1.35S2 was studied by Jieun et al. and the disulphide with excess niobium showcase excellent activity and lowest onset potential.233 MoS2 & WS2 deposition on sulphur doped graphene had good HER activity and Antonia et al.234 confirmed that presence of the doped graphene was responsible for the efficiency. Among the many single atom dopants of MoS2, ruthenium doped MoS2 exhibit the highest HER activity. HER performance was also studied for gold encapsulated MoS2 shells and it was concluded the core size of Au matters. Low Tafel slope value and overpotential was exhibited by the largest Au core.235 Whereas Co doped WS2 (ref. 236) also showed HER activity and the presence of surface macropores due to the use of hydrogen peroxide, helped facilitate the proton and electrolyte diffusion. By supporting CoS2 nanowires on graphdiyne, the unsaturated sulphurs in (211) plane of CoS2 becomes activated, diminishing the energy barriers of electron transfer caused by the synergistic effect between the planar unsaturated sulphur and the sp1 hybridized carbons of graphdiyne.237 Graphdiyne-CoS2 heterojunction nanocomposite with low overpotential of 97 mV at 10 mA cm−2 exhibit superior HER catalytic performance. Nickel based sulphide being easy to prepare, cheap and having electroactive sites are considered promising alternatives for noble metal catalysts. Its conductivity and metallicity is due to continuous Ni–Ni bonds. Recently, by rational cationic-doped strategy, Zn was doped on Ni3S2 nanosheet arrays supported on Ni foam (Zn–Ni3S2/NF)238 and it was revealed that the electronic structure is modulated on doping with Zn altering the ΔGH thus exhibiting HER catalytic performance the interaction involved contributing to this electro activity is further confirmed XPS and Raman spectrum. XiaoqingMao measured the electrochemical tests for oxygen evolution and concluded that Co and Fe doped Ni3S4 shows better performance that than single cation doped and pure Ni3S4 with the overall water splitting of 1.47 V @ 10 mA cm−2.239 Among four different morphologies, Gong et al. concluded that 3D NiCo2S4 nanostructures had the excellent OER catalytic activity.240 Mn-doped NiCo2S4,241 sea urchin-shaped nanospheres catalyst grown on nickel foam also exhibit OER activity and have overpotential 40 mV less than NiCo2S4/NF and Tafel slope lower than pure NiCo2S4. Mingjin Cui synthesised and studied the electrocatalytic OER properties of quinary high entropy metal sulfide (CrMnFeCoNi)Sx in 1 M KOH.242 The catalyst was observed to outperform most MxSy materials because of the synergistic effect of the metals caused by the transfer of charge state between the different metallic sites, established from computational results. The quinary HEMS nanoparticles had low overpotential of 295 mV at a current density of 100 mA cm−2 and excellent stability. Yu Ma designed a 2D bimetallic sulfoselenide nanosheets (NiSe0.1MoS6.4/NF-P123) via one step hydrothermal method. Inorder to tailor the nanosheet, morphology-controlling reagents such as Mo7O246− and Pluronic P123 are used.242,243 The synergistic effect of Ni and Mo accompanied by presence of S and Se contribute to electronic properties and elevate HER kinetics and behave as outstanding non-metal based HER electrocatalysts with the excellent stability up to 18 h.
Selenium centres have localized polarization induced partial negative charges enabling them to capture protons and improve HER. Moreover the coordination step with hydroxides compared to P and S site; Se site is less affected, allowing the dioxygen molecule delivery to be quicker, making them more efficient than sulphides and phosphides. The primary preparation method to synthesis Se based electrocatalyst is to use hydrothermal or solvothermal method for precursor followed by selenization by selenium powder or its composite and the different mono/bi and multimetallic transition metal selenides, their HER and OER aspects was reported by Xia & group. Increasing the Se content on NiSe2 reduced the overpotential and lower the Tafel slope value in case of NiSe2 nanosheet for HER.244 The morphology optimization also counts as NiSe2 nanowire had higher HER activity over NiSe2 nanoparticle. Hui Lui doped P anion onto NiSe2 nanosheet in (Fig. 21), studied their HER catalytic kinetics in alkaline medium and noticed improved activity with Tafel slope of 61.3 mV dec−1 and a low overpotential of 86 mV at 10 mA cm−2 with overall water splitting at 1.62 V @ 10 mA cm−2 and durability of 4 h.245 Bimetallic NiFe selenide based metal–organic framework is also studied for their efficiency in oxygen evolution reaction. The synergistic effects of Ni and Fe increased the OER electrocatalytic activity Fe2NiSe4 phase.246 NiFe–Se/CFP compared with monometallic FeSe2/CFP and NiSe2/CFP have lower overpotential of 281 mV at 10 mA cm−2 and shows a small Tafel slope of 40.93 mV dec−1. Heterostructural NiSe2/FeSe2 nanoparticles surpass the OER performance of the state of the art catalyst RuO2 with overpotential of 274 mV at 40 mA cm−2 and Tafel slope value of 57.07 mV dec−1, owing to the heterointerface interaction between NiSe2 and FeSe2 which reduces the Gibbs free energy of intermediates having oxygen from 3.15 eV for NiSe2 to 2.14 eV for NiSe2/FeSe2 heterostructures which is confirmed from the DFT calculations. Yuan Huang et al. reported phosphorus doping of the fe electronic structure in iron–nickel selenide nanosheets as OER electrocatalysts.247
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| | Fig. 21 (a) Polarization curves of P-NiSe2/CC, NiSe2/CC, Pt/C, and CC electrodes in 1.0 M KOH, (b) corresponding Tafel plots of P-NiSe2/CC, NiSe2/CC, and Pt/C electrodes, (c) Nyquist plots of P-NiSe2/CC and NiSe2/CC electrodes and inset has equivalent circuit, (d) the Cdl values of P-NiSe2/CC and NiSe2/CC electrodes at scanning rate of 5, 25, 50, 75 and 100 mV s. (e) Based on the electrochemical active surface area, the specific activity of P-NiSe2/CC and NiSe2/CC electrodes are depicted (f) durability testing showing the P-NiSe2/CC electrode's first and 3000th polarization curves; inset shows a time-dependent current density curve recorded on the P-NiSe2/CC electrode for 55 hours at a constant overpotential of 10 mV (vs. RHE). (g) Fabrication process schematic illustration. Reproduced with permission.245 Copyright 2021, Elsevier. | |
The number of active sites increased on doping with P as P-Ni0.75Fe0.25Se2 had higher electrochemical double layer capacitance (Cdl) value of 6.25 mF cm−2 when compared to Ni0.75Fe0.25Se2 with 4.13 mF cm−2 and the RuO2 with 5.92 mF cm−2. The OER activity also follow the similar trend as that of the capacitances making them ideal electrocatalyst. Shuyuan Pan et al. synthesised heterostructured electrocatalyst having boosted HER activity comprising of metallic Rh and Rh3Se4 prepared by partial selenization of Rh/C at 300C with overpotential 32 mV and 29 mV at 10 mA cm−2 in both acidic and alkaline environments respectively.248 Rhodium increased the electric density near Fermi level thereby improving the electronic conductivity.
6.3.4. Noble metals and their composites. Electrocatalyst based on platinum, palladium, ruthenium, rhodium and iridium noble metals are commonly acknowledged as outstanding electrocatalyst for electrochemical water splitting. While Ru/Ir oxides are excellent OER catalyst and HER activity is showed by Pt and Pt based nanomaterials. However because to their expensive amount and lack of abundance, cost effective approaches such as substitution of these noble metals, doping of low amount of noble metals which act as active centres into non-noble electrocatalyst are being taken into consideration. During proton exchange membrane fuel cell operation, the catalyst deteriorates gradually, preventing sudden proton exchange membrane fuel cell failure. since the anode is exposed to high potentials (>1.5 V), the stability of the electrocatalyst at the anode is crucial. In a few studies, the state-of-the-art iridium dissolution during anodic treatment has been documented. The mechanism of iridium dissolution has been proposed by Cherevko et al.249 Because soluble iridium (III) complexes, such as Ir(OH)3 or Ir2O3, are well known for their instability and have a relatively high solubility at low pH, were accounted for the dissolution of iridium catalyst. The authors claim that the formation pathway of Ir(III) complexes i.e. the dissolution of hydrous iridium oxide in the region of OER is explained below.
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| 2IrO2(OH) + 2H2O → 2Ir(OH)3 + O2 | (87) |
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| 2IrO2(OH) → 2HIrO2 + O2 | (88) |
The entire oxygen evolution reaction pathway, forming a closed cycle, can be rewritten as follows for the second reaction:
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| HIrO2 → IrO2 + H+ + e− | (89) |
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| IrO2 + H2O → IrO2(OH) + H+ + e− | (90) |
| |
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| 2IrO2(OH) → 2HIrO2 + O2 | (91) |
Additionally, the oxidizing ability of iridium is correlated with the rate at which anode catalyst dissolves. The catalyst dissolution process may be slowed down by a lower oxidation state caused by interactions between the metal and the metal oxide support. Under high current densities (4 A cm2), it has been noted that the dissolution and migration of iridium particles worsen significantly and significant portion of the iridium catalyst particles was seen to diffuse into the membrane after a lengthy test.250
Doping of ultralow Ru into Co MoF followed by annealing in air led to conversion of hollow Ru–Co MOF derivative into (Ru–Co)Ox nanosheets which consist of Co3O4 and RuO2.251 The nanosheets provide active sites and Ru induces modulation in electronic structures that helps strengthen the adsorption of Oads and weaken the Hads adsorption resulting in outstanding OER and HER activity. Li et al. introduced a single atom Ru into MOF and observed that the electronic structure was tailored upon the doping. H and H2O, binding strength are altered owing to the excellent HER activity. Ir4+ doped NiFe layered double hydroxide (LDH) as efficient HER electrocatalyst was reported by Chen and coworkers, and the improved dissolution kinetics of water molecules is was found to contributed by doped Ir.252 Fan et al. fabricated atomic Ir doped NiCo LDH and the EPR (electron paramagnetic resonance) spectra confirmed the elevation in production of oxygen vacancies after incorporation of Ir.253 There was also increase in active sites, promoting both HER and OER performances. Bifunctional catalyst Ir doped NiV(OH)2 developed by Du et al., carried out both theoretical and experimental calculations and concluded that Ir, V and O modified the electronic structure as well as adsorption energies of intermediated and water molecules acting as HER and OER electrocatalysts.254 Zhu et al. observed that rhodium atom improves electrical conductivity of CoFe LDH by reducing energy barriers and induce Fe vacancies which favor both HER and OER activity, as they have overpotential of 28 mV for the hydrogen evolution reaction and 245 mV for the oxygen evolution reaction at 10 mA cm−2.255
2 Dimensional Ru nanosheets were developed by Kong et al.256 via simple solvothermal method that exhibited overall water splitting. 2D nanostructures are highly appreciated as catalysts as there is exposure of nearly all atoms which act as active sites and favor interfacial charge transfer. Compared to rhodium nanocrystal with tetrahedra and concave tetrahedra, rhodium nanosheets delivered better electrocatalytic activity towards both HER and OER. This is because the electrochemical surface area (ECSA) of Rh tetrahedra is 32.5 m2 g−1, Rh concave tetrahedra is 18.9 m2 g−1, while Rh nanosheets have the highest at 65.3 m2 g−1. 2D PdCu nanosheets facilitate electron transfer due to 3D their architecture and synergistic effect between Pd & Cu, thus exhibiting HER performance.257 Multi metallic ultrathin nanorings have seen to have improved electron kinetics and efficiency and thus Guo and coworkers demonstrated that PtPd based multi metallic nanorings have excellent OER activity. Pt deposition on 2D Pd3Pb enhanced the electrocatalytic HER performance in (Fig. 22) due to the improved H binding on Pt attributed charge transfer between Pd3Pb.258,259 2D Ir nanosheets developed by Yamauchi et al. demonstrated better oxygen evolution activity having an overpotential of 240 mV at 10 mA cm−2.260 Table S1† shows the comparison of bi-functional OER and HER activities of various s, p & d block metal catalyst & noble metals based on performance and lifetime.
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| | Fig. 22 Atomic-resolution HAADF-STEM images of the (A) Pd3Pb and (B) AL-Pt/Pd3Pb. (C) Pt and Pd3Pb charge density differences in AL-Pt/Pd3Pb. (D) The calculated average binding energy of H at various hydrogen coverages on Pt(100), strained Pt(100), and Pt/Pd3Pb(100) Reproduced with permission258 Copyright 2019, American Chemical Society. | |
7. Conclusion
Fuel cell technology is a more environmentally friendly method of generating electrical energy from various fuels, including hydrogen, oxygen, methanol, formic acid, borohydride, etc. The greenest approach to creating electrical power is via a simple process employing H2 and O2 as fuel. The electrochemical redox process generates water as a byproduct. If the H2 provided is as pure as feasible, the fuel cell's efficiency can be as high as attainable. The current technique for H2 production of coal reforming technology isn't the best option for getting the cleanest H2. Electrochemical water splitting is a better method of generating vast amounts of hydrogen in a short period of time with the most outstanding possible purity, thereby minimizing the impact on the environment. The key to improving water electrolysis efficiency and lowering costs is the creation of high-efficiency, low-cost electrocatalysts with low overpotential and minimum value of Tafel slope for HER & OER. The state-of-the-art electrocatalyst Pt is for HER, and in the case of OER, the state-of-the-art catalysts are iridium or ruthenium. Their compounds were considered and continue to be ideal catalysts. PGM catalysts have demonstrated good electrocatalytic activity in both HER and OER to date. Their morphology also matters as noble metals with nanostructure and controlled crystal form has been widely studied to get higher performance to optimize the usage of these rare resources. Modifying, composing, or alloying PGMs with specific materials on purpose has led to enhanced electrochemical activity. Recently there has been a greater emphasis on using inexpensive non-noble metals to replace PGM catalysts. Fabrication of non-noble metal electrocatalysts with nanostructures with different morphologies to increase their surface area and conductivity also accompanied by support and different precursor templates are extensively being studied. In this review, other mechanisms of HER and OER, parameters involved in electrochemical studies, and many of the metal and metalloids based electrocatalysts that have been recently reported are detailed. Their electrochemical activities in different electrolytes in accordance with their physiochemical characterizations are briefly explained. There are several challenges while switching from conventional combustion engines to fuel cells where hydrogen is considered a fuel. Most HER/OER electrocatalysts have been investigated in current density ranging from 10 to 100 mA cm−2. However, their operation in high current densities like 2 A cm−2 in water electrolyzer is still challenging, and these conditions require much more advanced electrochemical techniques besides CV and Tafel slopes. Another major problem is that O2 may diffuse into the cathode chamber, combine with the generated hydrogen and be converted back to the water. This worsens if operated at high current densities. Another reason for opting for low current density is to decrease the ohmic resistance of the diaphragm and liquid electrolyte. Experimental studies convey that HER activity is excellent in acidic medium whereas OER activity in alkaline medium. Still, combining their activity is drastically lower with weak stability and insufficient corrosion resistance. Many groups are working to rectify this issue and develop a versatile material in all pH.
Thus, many efforts are put across the globe to develop new materials to improve the performance of the already developed catalysts for HER and OER processes. The investigations are directed to the synthesis of inexpensive, sustainable and environmental friendly bifunctional catalysts that could carry out both HER and OER activities in acidic, alkaline, and near-neutral mediums and their structural changes, chemical adsorption processes, and coordination environment can be studied by in situ techniques such as XPS, in situ operando Raman, XAS which help in the upcoming studies.
Conflicts of interest
The authors declare no conflict of interest.
Acknowledgements
The author wishes to acknowledge the basic support from the National Institute of Technology Puducherry, Karaikal, India.
Notes and references
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Factoring
Factoring is an essential part of any mathematical toolbox. To factor, or to break an expression into factors, is to write the expression (often an integer or polynomial) as a product of different terms. This often allows one to find information about an expression that was not otherwise obvious.
Contents
[hide]
1 Differences and Sums of Powers
2 Vieta's/Newton Factorizations
3 Other Useful Factorizations
4 Practice Problems
5 Other Resources
Differences and Sums of Powers
Using the formula for the sum of a geometric sequence, it's easy to derive the general formula for difference of powers:
If , this creates the difference of squares factorization,
This leads to the difference of cubes factorization,
In addition, if is odd:
This also leads to the formula for the sum of cubes,
Another way to discover these factorizations is the following: the expression is equal to zero if . If one factorizes a product which is equal to zero, one of the factors must be equal to zero, so must have a factor of . Similarly, we note that the expression when is odd is equal to zero if , so it must have a factor of . Note that when is even, , rather than 0, so this gives us no useful information.
Vieta's/Newton Factorizations
These factorizations are useful for problems that could otherwise be solved by Newton's sums or problems that give a polynomial and ask a question about the roots. Combined with Vieta's formulas, these are excellent factorizations that show up everywhere.
Other Useful Factorizations
Binomial theorem
Simon's Favorite Factoring Trick
Sophie Germain Identity
Factor Theorem
Practice Problems
Prove that is never divisible by 121 for any positive integer .
Prove that is divisible by 7. - USSR Problem Book
Factor .
Factor into two polynomials with real coefficients.
Given that , prove that .
Other Resources
More Common Factorizations.
Retrieved from "
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Table of Contents
Table of Contents
What Is Percentage Change?
How It Works
Formula and Calculation
Uses
Example
FAQs
The Bottom Line
How to Calculate a Percentage Change
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Definition
A percentage change represents the relative change in the value of an item.
What Is Percentage Change?
Percentage change is a calculation used to assess the relative performance of a stock or other investment over a certain period.
Investors commonly review the percentage change of their portfolio compared to the percentage change of similar investments over the same timeframe.
Percentage change can also be used to assess the change in the financial value of any item—from the increase in the price of eggs from one month to the next, to how much a car costs today compared to what it cost a year ago.
Key Takeaways
To find the percentage change of a security, investors can subtract the previous price from the current price, divide the result by the earlier price, and multiply by 100.
Businesses use percentage change to analyze revenue growth year over year.
Monitoring percentage changes over time helps investors determine if their portfolios need rebalancing.
Percentage change is useful for assessing the change in value of many things, in addition to investments.
How Percentage Changes Work
Percentage changes can be calculated for a variety of variables for any items that you wish to measure over time.
For example, people might be interested in the percentage change in the amount of rainfall in their area from season to season. Or a baseball fan might want to keep track of the percentage change in the number of home runs that their favorite team hits yearly.
In finance, percentage change is often used to track the prices of large market indexes like the S&P 500 and Dow Jones Industrial Average (DJIA) and individual securities. It's also helpful when comparing the fluctuating values of different nations' currencies.
Percentage change is also a widely used metric in business, such as when a company illustrates its revenue growth year over year (YOY) in its balance sheet.
Starbucks Financial Results
Often, if there is a significant percentage change, the company will try to explain why. For example, for the third quarter of 2020, Starbucks reported a 38% drop in net revenues over the same quarter in 2019 "due to the adverse impact of COVID-19."
Subsequent quarterly reports showed the gradual recovery of Starbucks' revenues and positive percentage changes in net revenues, as the business disruptions caused by COVID-19 diminished.
Moving on, Starbucks' 2024 announcement of Q4 and full fiscal year 2024 results noted total net revenues for the 2024 and 2023 fiscal years as $36.2 billion and $35.98 billion, respectively. The increase year-over-year was a positive percentage change of 0.6%.
Fast Fact
Percentage change, like many other formulas used in finance, can be calculated using spreadsheets, such as Microsoft Excel or Google Sheets.
Formula and Calculation of Percentage Change
To calculate a percentage increase, first determine the difference (increase) between the two numbers you are comparing:
Increase=New Number−Original Number
Next, divide the increase by the original number and multiply the answer by 100:
Percentage Increase=(Original NumberIncrease)×100.
This expresses the change as a percentage—i.e., the percentage change.
Similarly, to calculate a percentage decrease, determine the difference (decrease) between the two numbers you are comparing.
Decrease=Original Number−New Number
Next, divide the decrease by the original number and multiply the answer by 100.
Percentage Decrease=(Original NumberDecrease)×100
Again, the result expresses the change as a percentage—i.e., the percentage change.
Tip
If you only want to remember one formula, use the one for a positive increase. The result will be either positive or negative (if you use both formulas, the result is always positive). That will tell you whether the percentage change is an increase (positive) or a decrease (negative).
Uses of Percentage Change
Investors, companies, and entire industries can benefit from analyzing how prices or other measures rise or fall from one period to the next.
Here are some ways that investors can use percentage change calculations to their advantage:
Measure individual investment returns:The return on investment (ROI) for various assets, such as stocks, bonds, or mutual funds, is expressed in terms of percentage changes. This helps investors evaluate the performance of their assets over a particular time frame and compare them to other possible investments.
Evaluate portfolios: In addition to computing the percentage changes of individual assets within a portfolio, investors can compute the change for their entire portfolio. This can be useful in determining whether their current asset allocation is delivering the returns they need or whether they should consider rebalancing.
Analyze price movements: Percentage changes in the prices of stocks, commodities, or other financial instruments can also help identify trends, volatility, and potential trading opportunities.
Compare against benchmarks:Investors can check the percentage changes in their investments against the percentage changes in a corresponding stock index or other benchmark to see whether their holdings are outperforming or underperforming the market. If they are significantly underperforming the benchmark, you should find out why.
Manage risk:Investors who use stop-loss orders to protect themselves from steep declines in the value of an investment can set their targets based on percentage changes.
Example of Calculating Percentage Change
Consider Grace, who bought shares of a stock at $35 per share on January 1 of last year. A year later, the stock was worth $45.50 per share. By what percentage did Grace's share value increase?
To answer this question, first calculate the increase in price between the new and old numbers: $45.50 - $35 = $10.50. Then divide the increase by the original price:
3510.5=0.3 Finally, to get the percentage, multiply the answer by 100 (or simply move the decimal point two spaces to the right).
0.3×100=30 Grace's stock increased by 30%.
How Do I Calculate Percent Change?
If you are tracking a price increase, use the formula: (New Price - Old Price) ÷ Old Price, and then multiply that number by 100. Conversely, if the price decreased, use the formula (Old Price - New Price) ÷ Old Price and multiply that number by 100.
What Is a Balance Sheet, and How Does It Relate to Percentage Changes?
A balance sheet is a financial statement that companies use to report assets, liabilities, and shareholder equity. Balance sheets provide a snapshot of a company's finances for a specific period, such as a quarter or a fiscal year. Many companies choose to analyze their balance sheet by examining the percentage change in specific account balances from one period to the next. For example, a company can check its immediate liquidity trend by examining the percentage change in cash on hand for the last several years.
How Is Percentage Change Used in Finance?
Percentage change is often used by investors to track the increase or decrease in the value of a security over time and then to compare that against the performance of a relevant index. It is also used to compare the changing values of different national currencies as well as to measure the appreciation of physical assets like real estate.
The Bottom Line
Percentage change is a simple calculation with many uses in finance and business.
If investors want a quick read on how a particular investment of theirs is performing, they should compare its percentage change in value for the past several years to discern a trend upward or downward.
In addition, they can compare it to that of similar investments, such as a benchmark index.
Article Sources
Investopedia requires writers to use primary sources to support their work. These include white papers, government data, original reporting, and interviews with industry experts. We also reference original research from other reputable publishers where appropriate. You can learn more about the standards we follow in producing accurate, unbiased content in our editorial policy.
Starbucks. "Starbucks Reports Q3 Fiscal 2020 Results."
Starbucks. "Starbucks Reports Record Q4 and Full Year 2021 Results."
Starbucks. "Starbucks Reports Q4 and Full Fiscal Year 2024 Results."
Reed.edu. "Percentage Change and Percentage Point Change: A Primer."
Harvard Business School Online. "How to Prepare a Balance Sheet: 5 Steps for Beginners."
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How to find a point where a line intersects an ellipse in 2D (C#)
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I need to find a point where a line (its origin is ellipse' center) intersects an ellipse in 2D... I can easily find a point on a circle, because I know an angle F and the circle' radius (R):
x = x0 + R cosF y = y0 + R sinF
However I just can't figure how am I supposed to deal with an ellipse... I know it's dimensions (A & B), but what is the way of finding parameter T?!
x = x0 + A cosT y = y0 + B sinT
From what I understand the parameter T (T angle) is not far from the F angle (approximately +-15 degrees in some cases), but I just can't figure how to calculate it!!!
If there is a kind hearted soul, please help me with this problem...
c#
math
geometry
trigonometry
ellipse
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asked May 21, 2012 at 20:59
I_really_love_MSVCI_really_love_MSVC
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1
I see nothing programming related about this question. Probably belongs here: math.stackexchange.com
Marlon
– Marlon
2012-05-21 21:15:08 +00:00
Commented May 21, 2012 at 21:15
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5 Answers 5
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8
The standard equation of an ellipse, stationed at 0,0, is:
1 = (x)^2 / (a) + (y)^2 / (b)
Where a is 1/2 the diameter on the horizontal axis, and b is 1/2 the diameter on the vertical axis.
you have a line, assuming an equation:
y = (m)(x - x0) + y0
So, let us plug-and-play!
1 = (x)^2 / (a) + (m(x - x0) + y0)^2 / (b) 1 = x^2 / a + (mx + (y0 - mx0))^2 / b 1 = x^2 / a + (m^2 x^2 + 2mx(y0 - mx0) + (y0 - mx0)^2) / b 1 = x^2 / a + (m^2 x^2) / b + (2mx(y0 - mx0) + (y0^2 - 2y0mx0 + m^2x0^2)) / b 1 = ((x^2 b) / (a b)) + ((m^2 x^2 a) / (a b)) + (2mxy0 - 2m^2xx0)/b + (y0^2 - 2y0mx0 + m^2x0^2)/b 1 = ((bx^2 + am^2x^2)/(ab)) + (x(2my0 - 2m^2x0))/b + (y0^2 - 2y0mx0 + m^2x0^2)/b 0 = x^2((b + am^2)/(ab)) + x((2my0 - 2m^2x0)/b) + (((y0^2 - 2y0mx0 + m^2x0^2)/b) - 1)
That last equation follows the form of a standard quadratic equation.
So just use the quadratic formula, with:
((b + am^2)/(ab)) ((2my0 - 2m^2x0)/b) and (((y0^2 - 2y0mx0 + m^2x0^2)/b) - 1)
to get the X values at the intersections; Then, plug in those values into your original line equation to get the Y values.
Good luck!
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answered May 21, 2012 at 21:18
SergeSerge
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public Hits EllipseLineIntersection ( float rx , float ry , float2 p1 , float2 p2 ) { Hits hits = default(Hits); float2 p3, p4; Rect rect = default(Rect); { rect.xMin = math.min(p1.x,p2.x); rect.xMax = math.max(p1.x,p2.x); rect.yMin = math.min(p1.y,p2.y); rect.yMax = math.max(p1.y,p2.y); } float s = ( p2.y - p1.y )/( p2.x - p1.x ); float si = p2.y - ( s p2.x ); float a = ( ryry )+( rxrx ss ); float b = 2f rxrx si s; float c = rxrx sisi - rxrx ryry; float radicand_sqrt = math.sqrt( ( bb )-( 4f a c) ); p3.x = ( -b - radicand_sqrt )/( 2fa ); p4.x = ( -b + radicand_sqrt )/( 2fa ); p3.y = sp3.x + si; p4.y = sp4.x + si; if( rect.Contains(p3) ) hits.Push( p3 ); if( rect.Contains(p4) ) hits.Push( p4 ); return hits; } public struct Hits { public byte count; public T point0, point1; public void Push ( T val ) { if( count==0 ) { point0 = val; count ++; } else if( count==1 ) { point1 = val; count ++; } else print("This structure can only fit 2 values"); } }
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answered May 13, 2020 at 20:51
Andrew ÅukasikAndrew Åukasik
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3 Comments
Sierramike
Sierramike
I did try your code and it really fits my needs, I was fearing it calculates "line" intersection but it actually calculates "segment" intersection, which is exactly why I was looking for. But unfortunately, it doesn't detect tangent intersection, e.g. EllipseLineIntersection(100, 50, new PointF(-100, 50), new PointF(100, 50)) will return 0 hits.
Sierramike
Sierramike
Unfortunately I found another case where it's not working : EllipseLineIntersection(100, 50, new PointF(-200, 0), new PointF(200, 0)) should cross the Ellipse horizontally on the exact big radius, and should return (-100, 0) and (100, 0) as crossing points, but it returns 0 hits. So I'm afraid to say this method is not reliable.
Sierramike
Sierramike
Ok, I figured out my issue. I don't know what is float2 and Rect, I'm using regular C# and I converted this code to do so, using PointF and RectangleF. Unfortunately, if the RectangleF had a height of 0 or a width of 0 (meaning the crossing line is totaly vertical or totaly horizontal), rect.Contains(p) is not working. I modified the code and will post it as another answer.
3
Don't do it this way. Instead check the equation that forms an ellipse and that forming a line and solve the set:
The ellipse: (x/a)^2 + (y/b)^2 = 1 Your line: y = cx
You know a, b and c, so finding a solution is going to be easy. You'll find two solutions, because the line crosses the ellipse twice.
EDIT: Note I moved your ellipse's center to (0,0). It makes everything easier. Just add (x0,y0) to the solution.
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edited May 21, 2012 at 21:11
answered May 21, 2012 at 21:05
zmbqzmbq
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2 Comments
tdammers
tdammers
There may also be zero solutions (no intersection at all), or two identical ones (line is a tangent).
zmbq
zmbq
How can the line not cross the ellipse twice? The line passes through the ellipse's center.
1
I wrote a C# code for your problem and I hope you can find it helpful. the distance function inside this code calculates euclidean distance between two points in space.
wX denotes horizontal radios of ellipse and wY denotes vertical radios.
private PointF LineIntersectEllipse(PointF A, PointF B, float wX, float wY) { double dx = B.X - A.X; double dy = B.Y - A.Y; double theta = Math.Atan2(dy, dx); double r = distance(A, B) - ((wX wY) / Math.Sqrt(Math.Pow(wY Math.Cos(theta), 2) + Math.Pow(wX Math.Sin(theta), 2))); return PointF((float)(A.X + r Math.Cos(theta)), (float)(A.Y + r Math.Sin(theta))); }
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edited Sep 15, 2016 at 10:35
ideasman42
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answered Sep 6, 2014 at 19:36
Danial EsmaeiliDanial Esmaeili
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Andrew Åukasik
Andrew Åukasik
Is it just me or it breaks once you move point B?
0
Andrew Åukasik posted a good and useful answer, however it is not using regular C# types. As I wrote in the comments, I converted the code using System.Drawing objects PointF and RectangleF. I found out that if the points given as parameters are aligned as a vertical or horizontal line, then "rect" will have a width or a height equal to 0. Then, rect.Contains(point) will return false even if the point is on this line.
I also modified the "Hits" structure to check if the point pushed is not already existing, which is the case if the line is perfectly tangent, then p3 and p4 will have same coordinates, as the exact tangent point is the only crossing point.
Here is the new code taking care of all the cases :
public static Hits EllipseLineIntersection0(float rx, float ry, PointF p1, PointF p2) { Hits hits = default(Hits); PointF p3 = new PointF(); PointF p4 = new PointF(); var rect = default(RectangleF); rect.X = Math.Min(p1.X, p2.X); rect.Width = Math.Max(p1.X, p2.X) - rect.X; rect.Y = Math.Min(p1.Y, p2.Y); rect.Height = Math.Max(p1.Y, p2.Y) - rect.Y; float s = (p2.Y - p1.Y) / (p2.X - p1.X); float si = p2.Y - (s p2.X); float a = (ry ry) + (rx rx s s); float b = 2f rx rx si s; float c = rx rx si si - rx rx ry ry; float radicand_sqrt = (float)Math.Sqrt((b b) - (4f a c)); p3.X = (-b - radicand_sqrt) / (2f a); p4.X = (-b + radicand_sqrt) / (2f a); p3.Y = s p3.X + si; p4.Y = s p4.X + si; if (rect.Width == 0) { if (p3.Y >= rect.Y && p3.Y <= rect.Y + rect.Height) hits.Push(p3); if (p4.Y >= rect.Y && p4.Y <= rect.Y + rect.Height) hits.Push(p4); } else if (rect.Height == 0) { if (p3.X >= rect.X && p3.X <= rect.X + rect.Width) hits.Push(p3); if (p4.X >= rect.X && p4.X <= rect.X + rect.Width) hits.Push(p4); } else { if (rect.Contains(p3)) hits.Push(p3); if (rect.Contains(p4)) hits.Push(p4); } return hits; } public struct Hits { public byte Count; public T P0, P1; public void Push(T val) { if (Count == 0) { P0 = val; Count++; } else if (Count == 1) { if (!P0.Equals(val)) { P1 = val; Count++; } } else throw new OverflowException("Structure Hits can only fit 2 values."); } }
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answered Apr 17, 2022 at 20:18
SierramikeSierramike
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Silverlan
Silverlan
There is a division by 0 if p1 and p2 have the same x axis value.
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Thermodynamics Formula Sheet - Key Expressions and Constants
A useful formula sheet for thermodynamics
Original title:
Thermodynamics formula sheet
Course
Engineering Physics A: Mechanics, Waves, Electromagnetism and Thermal Physics (PHYS101)
263 documents
University
University of Canterbury
Academic year:2022/2023
Uploaded by:
Anonymous Student
University of Canterbury
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Thermodynamics: Useful Information and Expressions
R = 8 J/(mol K) kB = 1. 38 × 10 − 23 J/K NA = 6. 023 × 1023 molecules/mole Density of water = 1. 00 × 103 kg/m 3 Specific heat of water = 4. 2 × 103 J/(kg K) Specific heat of ice = 2. 1 × 103 J/(kg K) Specific heat of steam = 2. 0 × 103 J/(kg K) Latent heat of fusion of water = 330 × 103 J/kg Latent heat of vaporisation of water = 2260 × 103 J/kg Atmospheric pressure = 1 × 105 Pa Atmospheric density = 1 kg/m 3 Acceleration due to gravity g = 9 m/s 2 Stefan Boltzmann constant σ = 5. 67 × 10 − 8 W/(m 2 K 4 ) Sphere: A = 4πr 2 V = 43 πr 3 Density: ρ = m V Kinetic Energy: K = 12 mv 2 Potential Enery: U = mgh
Heat Q = mc∆T and Q = mL Linear expansion ∆L = αL∆T Thermal conductivity P = kA dT dx
Radiation P = εσAT 4 Ideal gas law P V = nRT = N kB T = 1 3 N mv 2 rms
Molar specific heats for a monatomic gas CV = 32 R, CP = 52 R First law of thermodynamics ∆Eint = Q + W Adiabatic process P V γ = constant, T V γ− 1 = constant, γ = Cp/CV
Gas velocity vrms =
√
3 RT
M
Work dW = −P dV Isothermal process W = nRT ln
( V
i Vf
)
Isobaric process Q = nCp∆T Isovolumetric process Q = nCV ∆T Monatomic gas Eint = 32 N kB T Diatomic gas Eint = 52 N kB T
Efficiency of an engine ε = QWH = 1 − | |QQHC | |
Coefficient of performance COPcooling = |Q WC | or COPheating = |Q WH |
Entropy ∆S =
∫ dQ rev T Carnot εC = 1 − TTCH
Thermodynamics Formula Sheet - Key Expressions and Constants
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Thermodynamics Formula Sheet - Key Expressions and Constants
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Thermo dynamics:Useful Information and Expressions
R=8.314 J/(mol K)
k B=1.38×10−23 J/K
N A=6.023×10 23 molecules/mole
Density of w ater=1.00×10 3 kg/m 3
Sp ecific heat of w ater=4.2×10 3 J/(kg K)
Sp ecific heat of ice=2.1×10 3 J/(kg K)
Sp ecific heat of steam=2.0×10 3 J/(kg K)
Laten t heat of fusion of w at er=330×10 3 J/kg
Laten t heat of v ap orisation of w ater=2260×10 3 J/kg
A tmosph eric pressure=1.013×10 5 P a
A tmosph eric densit y=1.21 kg/m 3
Acceleration due to gravit y g=9.80 m/s 2
Stefan Boltzmann constant σ=5.67×10−8 W/(m 2 K 4)
Sphere:A=4 π r 2 V=4
3 π r 3
Density:ρ=m
V
Kinetic Energy:K=1
2 mv 2
P otential Enery:U=mg h
Heat Q=mc∆T and Q=mL
Linear expansion∆L=αL∆T
Thermal conductivity P=k A dT
dx
Radiation P=εσ AT 4
Ideal gas la w P V=nRT=N k B T=1
3 N mv 2
rms
Molar sp ecific heats for a monatomic gas C V=3
2 R,C P=5
2 R
First law of thermodynamics∆E int=Q+W
Adiabatic pro cess P V γ=constan t,T V γ−1=constan t,γ=C p/C V
Gas v elo cit y v rms=r 3 RT
M
W ork dW=−P dV
Isothermal pro cess W=nRT lnV i
V f
Isobaric pro cess Q=nC p∆T
Iso volumetric pro cess Q=nC V∆T
Monatomic gas E int=3
2 N k B T
Diatomic gas E int=5
2 N k B T
Efficiency of an engine ε=W
Q H
=1−|Q C|
|Q H|
Co efficien t of p erformance COP coolin g=|Q C|
W or COP heating=|Q H|
W
En tr op y∆S=Z dQ rev
T
Carnot ε C=1−T C
T H
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11056 | https://math.stackexchange.com/questions/37157/recurrence-relation-linear-second-order-constant-coefficients | Stack Exchange Network
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Recurrence relation (linear, second-order, constant coefficients)
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Asked
Modified 5 years, 2 months ago
Viewed 9k times
4
$\begingroup$
Q1. Find the general solution to the difference equation
$$ a_{n} - 4a_{n-1} + 3a_{n-2} = 6 $$
Q2. Solve the difference equation
$$ a_{n} - a_{n-1} - 2a_{n-2} = 0, a_0 = 2, a_1 = 4 $$
I am completely lost in solving recurrence relation questions. Can anyone guide me in steps to solve the following 2 questions?
discrete-mathematics
recurrence-relations
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edited Jun 27, 2015 at 9:49
Américo Tavares
39.2k1414 gold badges110110 silver badges252252 bronze badges
asked May 5, 2011 at 10:03
ilovetolearnilovetolearn
1,60355 gold badges2020 silver badges3838 bronze badges
$\endgroup$
4
$\begingroup$ Re Q2, you could try to compute the first terms, to see if a pattern emerges. $\endgroup$
Did
– Did
2011-05-05 10:08:36 +00:00
Commented May 5, 2011 at 10:08
$\begingroup$ Don't you have a guide for that in your textbooks? Those are pretty standard. $\endgroup$
Raskolnikov
– Raskolnikov
2011-05-05 10:10:20 +00:00
Commented May 5, 2011 at 10:10
$\begingroup$ @Raskolnikov I dont understand the examples on the textbook. $\endgroup$
ilovetolearn
– ilovetolearn
2011-05-05 10:13:28 +00:00
Commented May 5, 2011 at 10:13
1
$\begingroup$ For linear recurrence relations, one can get some insight but making the ansatz $a_n = \lambda^n$ and then figuring out which $\lambda$s fulfill the equation (then one can take an arbitrary linear combination of these solutions in order to get the initial conditions right). $\endgroup$
Fabian
– Fabian
2011-05-05 10:13:31 +00:00
Commented May 5, 2011 at 10:13
Add a comment |
4 Answers 4
Reset to default
7
$\begingroup$
The general approach to solving recurrence relations is the following: given a recurrence relation $$a_n+\alpha_1a_{n-1}+...+\alpha_ra_{n-r}=\beta(n) \; .$$
I) First you solve the homogeneous part $a_n^{(h)}+\alpha_1a_{n-1}^{(h)}+...+\alpha_ra_{n-r}^{(h)}=0$:$\quad$ a) By solving the characteristic equation: $x^r+\alpha_1x^{r-1}+...+\alpha_r=0$ you get $x_1,...,x_r$ the roots of the equation.$\quad$ b) If all $x_1,...,x_r$ are different then the $a_n^{(h)}=A_1x_1^n+...+A_rx_r^n$, where $A_1,...,A_r$ are the coefficients.$\quad$ c) If you have a root $x_j$ is of multiplicity $k$ then you replace the term $A_jx_j^n$ with $(B_1n^{k-1}+...+B_k)x_j^n$
II) Now you find a particular solution to the equation $a_n^{(p)}+\alpha_1a_{n-1}^{(p)}+...+\alpha_ra_{n-r}^{(p)}=\beta(n)$. Each $\beta(n)$ needs a different approch to guess $a_n^{(p)}$. For example, if $\beta(n)=k^nf(n)$, where $k$ is a number and $f(n)$ is a polynomial in $n$, then $a_n^{(p)}=k^nn^sg(n)$ where $k$ is the same $k$, $s$ is the multiplicity of $k$ as a root of the characteristic equation in the homogeneous part and $g(n)$ is a polynomial of the same degree as $f(n)$.
Finally, $a_n=a_n^{(h)}+a_n^{(p)}$
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edited May 5, 2011 at 11:12
Raskolnikov
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answered May 5, 2011 at 10:58
Dennis GulkoDennis Gulko
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$\endgroup$
3
$\begingroup$ can you explain the purpose of finding general solution and particular solutions for linear difference equation? $\endgroup$
ilovetolearn
– ilovetolearn
2011-05-05 11:01:15 +00:00
Commented May 5, 2011 at 11:01
$\begingroup$ @liang The purpose is that you find a general solution for a simpler equation and a particular solution for the complicated equation to arrive at the general solution for the complicated equation. $\endgroup$
Phira
– Phira
2011-05-05 11:17:17 +00:00
Commented May 5, 2011 at 11:17
$\begingroup$ @liangteh: In addition to what user9325 said, you can check that the set of all sequences $a_n$ that obey to a homogeneous recurrence relation form a vector space over $\mathbb{C}$. Hence, the set of all solutions of a non-homogeneous recurrence relation forms an affine vector space. $\endgroup$
Dennis Gulko
– Dennis Gulko
2011-05-05 11:32:46 +00:00
Commented May 5, 2011 at 11:32
Add a comment |
3
$\begingroup$
The general second order homogeneous linear recurrence/difference equation with constant coefficients
$$x_{n}+c_{1}x_{n-1}+c_{2}x_{n-2}=0\qquad (1)$$
has two fundamental solutions $(\lambda _{1}^{n})_{n\geq 0}$ and $(\lambda _{2}^{n})_{n\geq 0}$, where $\lambda _{1},\lambda _{2}$ are the two zeroes of the characteristic polynomial
$$\lambda ^{2}+c_{1}\lambda +c_{2}\qquad (2)$$
Let's confirm.
$$\begin{eqnarray} \lambda _{1}^{n}+c_{1}\lambda _{1}^{n-1}+c_{2}\lambda _{1}^{n-2} &=&\lambda _{1}^{n-2}\left( \lambda _{1}^{2}+c_{1}\lambda _{1}+c_{2}\right) \equiv 0 \ && \ \lambda _{2}^{n}+c_{1}\lambda _{2}^{n-1}+c_{2}\lambda _{2}^{n-2} &=&\lambda _{2}^{n-2}\left( \lambda _{2}^{2}+c_{2}\lambda _{2}+c_{2}\right) \equiv 0 \end{eqnarray}$$
If $\lambda _{1}\neq \lambda _{2}$ the general solution of $(1)$ is a linear combination of $\lambda _{1}^{n}$ and $\lambda _{2}^{n}$
$$x_{n}=A\lambda _{1}^{n}+B\lambda _{2}^{n}\qquad (3)$$
as you can confirm substituting $(3)$ in $(1)$. Let's apply this result to your second difference equation $a_{n}-a_{n-1}-2a_{n-2}=0$. The characteristic polynomial $\lambda ^{2}-\lambda -2$ has the zeroes $\lambda _{1}=-1,\lambda _{2}=2$.
Thus $(3)$ becomes
$$a_{n}=A(-1)^{n}+B2^{n}$$
The constants $A$ and $B$ are determined by using the initial conditions $a_{0}=2$, $a_{1}=4$.
Added: Determination of $A$ and $B$:
$$a_{0}=A(-1)^{0}+B2^{0}=A+B=2\Leftrightarrow B=2-A$$
Hence
$$a_{n}=A(-1)^{n}+\left( 2-A\right) 2^{n}$$
and
$$a_{1}=A(-1)^{1}+\left( 2-A\right) 2^{1}=-A+4-2A=-3A+4=4\Leftrightarrow A=0.$$
Since $A=0$ and $B=2-A=2$ the solution is
$$a_{n}=2\cdot 2^{n}=2^{n+1}\qquad n\geq 2$$
As for your first equation it is not homogenous because the RHS is not zero. For solving it, please see the explanation by Dennis Gulko in his answer.
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edited May 5, 2011 at 12:23
answered May 5, 2011 at 11:49
Américo TavaresAmérico Tavares
39.2k1414 gold badges110110 silver badges252252 bronze badges
$\endgroup$
9
$\begingroup$ @Américo Tavares: why one gets a cubic characteristic polynomial for the first equation? $\endgroup$
Fabian
– Fabian
2011-05-05 11:56:18 +00:00
Commented May 5, 2011 at 11:56
$\begingroup$ @Fabian: It was my big mistake. I will correct it. Many thanks! $\endgroup$
Américo Tavares
– Américo Tavares
2011-05-05 11:58:24 +00:00
Commented May 5, 2011 at 11:58
$\begingroup$ my try at Q2 after the explanations. Please pardon the mistake if there are any. characteristic equation: $$ x^2 - x - x = 0 $$ $$ (x+1)(x-2) = 0 $$ $$ x = -1, 2 $$ General solution $$ A_n = A_1(-1)^n + A_2(2)^n $$ Now $$ a_0 = 2, a_1 = 4 $$ $$ a_0 = A_1 (-1)^0 + A_2(2)^0 $$ $$ A_1 + A_2 = 2 $$ $$ A_1 = A_1(-1)^1 + A_2(2)^1 $$ $$ 4+2 = -K_1 + 2K_2 + K_1 + K_2 $$ $$ 6 = 3K_2 $$ So $$ K_1 = 0, K_2 = 2 $$ Hence $$ A_n = 0(-1)^n + 2(2)^n $$ $$ A_n = 2(2)^n $$ Please pardon if there any mistakes $\endgroup$
ilovetolearn
– ilovetolearn
2011-05-05 12:12:43 +00:00
Commented May 5, 2011 at 12:12
$\begingroup$ The independant term is -2. The characteristic equation is $x^2-x-2=0$, which is equivalent to $(x+1)(x-2)=0$. I used the variable $\lambda$ instead. $\endgroup$
Américo Tavares
– Américo Tavares
2011-05-05 12:15:47 +00:00
Commented May 5, 2011 at 12:15
$\begingroup$ @Américo Tavares: Noted. Your explanation is quite similar to the textbook examples. $\endgroup$
ilovetolearn
– ilovetolearn
2011-05-05 12:24:18 +00:00
Commented May 5, 2011 at 12:24
| Show 4 more comments
2
$\begingroup$
Consider the recurrence relation for $n\ge 1$ given by \begin{equation} \tag{1} \label{1} A_{n+1}=\alpha A_n+\beta A_{n-1}+f_n, \end{equation} where $\alpha$ and $\beta$ are constants (not dependent on $n$), and $f_n$ is a given function of $n$. Eqn.~\eqref{1} is to be solved either for specified $A_0$, $A_1$, or two constraints of the form \begin{align}\tag{2}\label{2} \sum_{n=1}^{\infty} c_nA_n&=h_n, \ \sum_{n=1}^{\infty} d_nA_n&=g_n. \end{align}
We can write Eqn.~\eqref{1} as \begin{equation} \tag{3} \label{3} \begin{bmatrix} A_{n+1} \ A_n \end{bmatrix}=T\begin{bmatrix} A_n \ A_{n-1} \end{bmatrix}+\begin{bmatrix} f_n \ 0 \end{bmatrix}, \end{equation} where \begin{equation} T=\begin{bmatrix} \alpha & \beta \ 1 & 0 \end{bmatrix}. \end{equation}
First consider the case where the eigenvalues of $T$ are distinct, and given by \begin{align} \tag{4} \label{4} \lambda_1&=\frac{\alpha-\sqrt{\alpha^2+4\beta}}{2}, \ \lambda_2&=\frac{\alpha+\sqrt{\alpha^2+4\beta}}{2}. \end{align} We deal with the case of repeated eigenvalues where $\beta=-\alpha^2/4$, and $\lambda_1=\lambda_2=\alpha/2$ later by taking appropriate limits. Since the eigenvalues are distinct, $T$ is diagonalizable, and can be written as \begin{equation} \tag{5} \label{5} T=\lambda_1 P_1+\lambda_2 P_2, \end{equation} where $P_1$ and $P_2$ are projections given by \begin{align} P_1&=\frac{T-\lambda_2 I}{\lambda_1-\lambda_2}, \ P_2&=\frac{T-\lambda_1 I}{\lambda_2-\lambda_1}. \end{align} Thus, we have \begin{equation} \tag{6} \label{6} T^n=\lambda_1^n P_1+\lambda_2^n P_2. \end{equation}
We now have \begin{align} \tag{7} \label{7} \begin{bmatrix} A_{n+1} \ A_n \end{bmatrix}&=T^n\begin{bmatrix} A_1 \ A_0 \end{bmatrix}+\sum_{k=1}^{n-1}T^{n-k}\begin{bmatrix} f_k \ 0 \end{bmatrix} +\begin{bmatrix} f_n \ 0 \end{bmatrix} \notag \ &=(\lambda_1^n P_1+\lambda_2^n P_2)\begin{bmatrix} A_1 \ A_0 \end{bmatrix}+\sum_{k=1}^{n-1}(\lambda_1^{n-k} P_1+\lambda_2^{n-k} P_2) \begin{bmatrix} f_k \ 0 \end{bmatrix}+\begin{bmatrix} f_n \ 0 \end{bmatrix}, \end{align} which leads to the explicit formula \begin{equation} \tag{8} \label{8} A_n=\frac{1}{\lambda_2-\lambda_1}\left[(\lambda_2^n-\lambda_1^n)A_1-\lambda_1\lambda_2\left(\lambda_2^{n-1}-\lambda_1^{n-1}\right)A_0 +\sum_{k=1}^{n-1} \left(\lambda_2^{n-k}-\lambda_1^{n-k}\right)f_k\right]. \end{equation} If $f_n=f_0$ is a constant, Eqn.~\eqref{8} simplifies to \begin{equation} \tag{9} \label{9} A_n=\frac{1}{\lambda_2-\lambda_1}\left[(\lambda_2^n-\lambda_1^n)A_1-\lambda_1\lambda_2\left(\lambda_2^{n-1}-\lambda_1^{n-1}\right)A_0 +\left(\frac{\lambda_2(\lambda_2^{n-1}-1)}{\lambda_2-1}-\frac{\lambda_1(\lambda_1^{n-1}-1)}{\lambda_1-1}\right)f_0\right]. \end{equation} If $\lambda_1=1$, say, then by taking the limit in the above equation, we get \begin{equation} \tag{10} \label{10} A_n=\frac{1}{\lambda_2-1}\left[(\lambda_2^n-1)A_1-\lambda_2\left(\lambda_2^{n-1}-1\right)A_0 +\left(\frac{\lambda_2(\lambda_2^{n-1}-1)}{\lambda_2-1}-(n-1)\right)f_0\right]. \end{equation} In case the constraints in Eqn.~\eqref{2} are imposed in place of prescribed values for $A_0$ and $A_1$, then we determine $A_0$ and $A_1$ by substituting the solution given by Eqn.~\eqref{8} into Eqns.~\eqref{2}.
The solution for the case where the eigenvalues of $T$ are repeated is obtained by taking the limit $\lambda_2\rightarrow \lambda_1$ in Eqn.~\eqref{8}, and setting $\lambda_1=\alpha/2$ as \begin{equation} \tag{11} \label{11} A_n=\left(\frac{\alpha}{2}\right)^{n-1}\left[nA_1-\frac{\alpha(n-1)A_0}{2}\right]+\sum_{k=1}^{n-1}(n-k)f_k\left(\frac{\alpha}{2}\right)^{n-k-1}. \end{equation} If $f_n=f_0$ is a constant, Eqn.~\eqref{11} simplifies to \begin{align} A_n&=\left(\frac{\alpha}{2}\right)^{n-1}\left[nA_1-\frac{\alpha(n-1)A_0}{2}\right]+\frac{4\left[2^n+(n-1)\alpha^n-2n\alpha^{n-1}\right]f_0}{(\alpha-2)^22^n}, && (\alpha\ne 2), \ &=nA_1-(n-1)A_0+\frac{n(n-1)f_0}{2}, && (\alpha=2). \end{align}
As examples, consider the following:
The Fibonacci sequence defined by $A_{n+1}=A_n+A_{n-1}$ with $A_0=0$, $A_1=1$, so that $\alpha=1$, $\beta=1$, $\lambda_1=(1-\sqrt{5})/2$, $\lambda_2=(1+\sqrt{5})/2$, $f_n=0$, and \begin{equation} A_n=\frac{\lambda_2^n-\lambda_1^n}{\lambda_2-\lambda_1}. \end{equation}
The recurrence \begin{equation} A_{n+1}=4A_n-4A_{n-1}+4^{n-1}, \end{equation} with $A_0=0$, $A_1=1$, so that $\alpha=4$, $\beta=-4$, $\lambda_1=\lambda_2=2$, $f_n=4^{n-1}$. From Eqn.~\eqref{11}, we get \begin{equation} A_n=n2^{n-1}+2^{n-2}(2^n-n-1). \end{equation}
The solution to your Q1 is given by Eqn.~\eqref{10} with $\lambda_2=3$ and $f_0=6$.
The above procedure can obviously be generalized to higher-order recurrences, by considering the eigenvalues of $T$ to be distinct, and then deriving various subcases for repeated eigenvalues by taking appropriate limits as in the development above.
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edited Jul 21, 2020 at 4:29
answered Jul 20, 2020 at 11:44
JogJog
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Simple, general way of disposing such equations is using generating functions. Define: $$ A(z) = \sum_{n \ge 0} a_n z^n $$ Write the recurrence so it hasn't subtractions in indices: $$ a_{n + 2} - 4 a_{n + 1} + 3 a_n = 6 $$ Multiply by $z^n$, sum over $n \ge 0$ an recognize the resulting terms: $$ \frac{A(z) - a_0 - a_1 z}{z^2} - 4 \frac{A(z) - a_0}{z} + 3 A(z) = 6 \frac{1}{1 - z} $$
Substituting your initial values, and solving for $A(z)$, writing that as partial fractions: $$ A(z) = \frac{3}{(1 - z)^3} + \frac{5}{2 (1 - z)} + \frac{5}{2 (1 - 3 z)} $$ Using the generalized binomial theorem you can read off the coefficients: \begin{align} a_n &= 3 \binom{-3}{n} (-1)^n + \frac{5}{2} + \frac{5}{2} \cdot 3^n \ &= 3 \binom{n + 3 - 1}{3 - 1} + \frac{5}{2} + \frac{5}{2} \cdot 3^n \ &= \frac{3 (n + 2) (n + 1)}{2} + \frac{5}{2} + \frac{5}{2} \cdot 3^n \ &= \frac{5 \cdot 3^n + 3 n^2 + 9 n + 11}{2} \end{align} The other one I leave as a exercise for the gentle reader.
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answered Apr 13, 2014 at 17:17
vonbrandvonbrand
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11057 | https://www.webqc.org/balanced-equation-Al+O2=Al2O3 | Printed from
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| Balanced equation: 4 Al + 3 O2 = 2 Al2O3Reaction type: synthesis | Reaction stoichiometry | Limiting reagent | --- | | Compound | Coefficient | Molar Mass | Moles | Weight | | Reagents | | Al | 4 | 26.98 | | | | O2 | 3 | 32.00 | | | | Products | | Al2O3 | 2 | 101.96 | | | Units: molar mass - g/mol, weight - g. | Word equation | | 4 Aluminium + 3 Oxygen = 2 Aluminium oxide | | Balancing step by step using the inspection method | | Let's balance this equation using the inspection method.First, we set all coefficients to 1:1 Al + 1 O2 = 1 Al2O3For each element, we check if the number of atoms is balanced on both sides of the equation.Al is not balanced: 1 atom in reagents and 2 atoms in products.In order to balance Al on both sides we:Multiply coefficient for Al by 22 Al + 1 O2 = 1 Al2O3O is not balanced: 2 atoms in reagents and 3 atoms in products.In order to balance O on both sides we:Multiply coefficient for O2 by 3Multiply coefficient for Al2O3 by 22 Al + 3 O2 = 2 Al2O3Al is not balanced: 2 atoms in reagents and 4 atoms in products.In order to balance Al on both sides we:Multiply coefficient for Al by 24 Al + 3 O2 = 2 Al2O3All atoms are now balanced and the whole equation is fully balanced: 4 Al + 3 O2 = 2 Al2O3 | | Balancing step by step using the algebraic method | | Let's balance this equation using the algebraic method.First, we set all coefficients to variables a, b, c, d, ...a Al + b O2 = c Al2O3Now we write down algebraic equations to balance of each atom:Al: a 1 = c 2O: b 2 = c 3Now we assign a=1 and solve the system of linear algebra equations:a = c 2b 2 = c 3a = 1Solving this linear algebra system we arrive at:a = 1b = 0.75c = 0.5To get to integer coefficients we multiply all variable by 4a = 4b = 3c = 2Now we substitute the variables in the original equations with the values obtained by solving the linear algebra system and arrive at the fully balanced equation: 4 Al + 3 O2 = 2 Al2O3 | Direct link to this balanced equation: Please tell about this free chemistry software to your friends! |
| Instructions on balancing chemical equations: Enter an equation of a chemical reaction and click 'Balance'. The answer will appear below Always use the upper case for the first character in the element name and the lower case for the second character. Examples: Fe, Au, Co, Br, C, O, N, F. Compare: Co - cobalt and CO - carbon monoxide To enter an electron into a chemical equation use {-} or e To enter an ion, specify charge after the compound in curly brackets: {+3} or {3+} or {3}.Example: Fe{3+} + I{-} = Fe{2+} + I2 Substitute immutable groups in chemical compounds to avoid ambiguity. For instance equation C6H5C2H5 + O2 = C6H5OH + CO2 + H2O will not be balanced, but PhC2H5 + O2 = PhOH + CO2 + H2O will Compound states [like (s) (aq) or (g)] are not required. If you do not know what products are, enter reagents only and click 'Balance'. In many cases a complete equation will be suggested. Reaction stoichiometry could be computed for a balanced equation. Enter either the number of moles or weight for one of the compounds to compute the rest. Limiting reagent can be computed for a balanced equation by entering the number of moles or weight for all reagents. The limiting reagent row will be highlighted in pink. Examples of complete chemical equations to balance: Fe + Cl2 = FeCl3 KMnO4 + HCl = KCl + MnCl2 + H2O + Cl2 K4Fe(CN)6 + H2SO4 + H2O = K2SO4 + FeSO4 + (NH4)2SO4 + CO C6H5COOH + O2 = CO2 + H2O K4Fe(CN)6 + KMnO4 + H2SO4 = KHSO4 + Fe2(SO4)3 + MnSO4 + HNO3 + CO2 + H2O Cr2O7{-2} + H{+} + {-} = Cr{+3} + H2O S{-2} + I2 = I{-} + S PhCH3 + KMnO4 + H2SO4 = PhCOOH + K2SO4 + MnSO4 + H2O CuSO45H2O = CuSO4 + H2O calcium hydroxide + carbon dioxide = calcium carbonate + water sulfur + ozone = sulfur dioxide Examples of the chemical equations reagents (a complete equation will be suggested): H2SO4 + K4Fe(CN)6 + KMnO4 Ca(OH)2 + H3PO4 Na2S2O3 + I2 C8H18 + O2 hydrogen + oxygen propane + oxygen Understanding chemical equations A chemical equation represents a chemical reaction. It shows the reactants (substances that start a reaction) and products (substances formed by the reaction). For example, in the reaction of hydrogen (H₂) with oxygen (O₂) to form water (H₂O), the chemical equation is: H2 + O2 = H2O However, this equation isn't balanced because the number of atoms for each element is not the same on both sides of the equation. A balanced equation obeys the Law of Conservation of Mass, which states that matter is neither created nor destroyed in a chemical reaction. Balancing with inspection or trial and error method This is the most straightforward method. It involves looking at the equation and adjusting the coefficients to get the same number of each type of atom on both sides of the equation. Best for: Simple equations with a small number of atoms. Process: Start with the most complex molecule or the one with the most elements, and adjust the coefficients of the reactants and products until the equation is balanced. Example:H2 + O2 = H2O 1. Count the number of H and O atoms on both sides. There are 2 H atoms on the left and 2 H atom on the right. There are 2 O atoms on the left and 1 O atom on the right. 2. Balance the oxygen atoms by placing a coefficient of 2 in front of H2O: H2 + O2 = 2H2O 3. Now, there are 4 H atoms on the right side, so we adjust the left side to match: 2H2 + O2 = 2H2O 4. Check the balance. Now, both sides have 4 H atoms and 2 O atoms. The equation is balanced. Balancing with algebraic method This method uses algebraic equations to find the correct coefficients. Each molecule's coefficient is represented by a variable (like x, y, z), and a series of equations are set up based on the number of each type of atom. Best for: Equations that are more complex and not easily balanced by inspection. Process: Assign variables to each coefficient, write equations for each element, and then solve the system of equations to find the values of the variables. Example: C2H6 + O2 = CO2 + H2O 1. Assign variables to coefficients: a C2H6 + b O2 = c CO2 + d H2O 2. Write down equations based on atom conservation: 2 a = c 6 a = 2 d 2 b = 2c + d 3. Assign one of the coefficients to 1 and solve the system. a = 1 c = 2 a = 2 d = 6 a / 2 = 3 b = (2 c + d) / 2 = (2 2 + 3) / 2 = 3.5 4. Adjust coefficient to make sure all of them are integers. b = 3.5 so we need to multiply all coefficient by 2 to arrive at the balanced equation with integer coefficients: 2 C2H6 + 7 O 2 = 4 CO2 + 6 H2O Balancing with oxidation number method Useful for redox reactions, this method involves balancing the equation based on the change in oxidation numbers. Best For: Redox reactions where electron transfer occurs. Process: identify the oxidation numbers, determine the changes in oxidation state, balance the atoms that change their oxidation state, and then balance the remaining atoms and charges. Example: Ca + P = Ca3P2 1. Assign oxidation numbers: Calcium (Ca) has an oxidation number of 0 in its elemental form. Phosphorus (P) also has an oxidation number of 0 in its elemental form. In Ca3P2, calcium has an oxidation number of +2, and phosphorus has an oxidation number of -3. 2. Identify the changes in oxidation numbers: Calcium goes from 0 to +2, losing 2 electrons (oxidation). Phosphorus goes from 0 to -3, gaining 3 electrons (reduction). 3. Balance the changes using electrons: Multiply the number of calcium atoms by 3 and the number of phosphorus atoms by 2. 4. Write the balanced Equation: 3 Ca + 2 P = Ca3P2 Balancing with ion-electron half-reaction method This method separates the reaction into two half-reactions – one for oxidation and one for reduction. Each half-reaction is balanced separately and then combined. Best for: complex redox reactions, especially in acidic or basic solutions. Process: split the reaction into two half-reactions, balance the atoms and charges in each half-reaction, and then combine the half-reactions, ensuring that electrons are balanced. Example: Cu + HNO3 = Cu(NO3)2 + NO2 + H2O 1. Write down and balance half reactions: Cu = Cu{2+} + 2{e} H{+} + HNO3 + {e} = NO2 + H2O 2. Combine half reactions to balance electrons. To accomplish that we multiple the second half reaction by 2 and add it to the first one: Cu + 2H{+} + 2HNO3 + 2{e} = Cu{2+} + 2NO2 + 2H2O + 2{e} 3. Cancel out electrons on both sides and add NO3{-} ions. H{+} with NO3{-} makes HNO3 and Cu{2+} with NO3{-} makes Cu(NO3)3: Cu + 4HNO3 = Cu(NO3)2 + 2NO2 + 2H2O Learn to balance chemical equations: Practice what you learned: Practice balancing chemical equations Related chemical tools: Molar mass calculator pH solver |
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11058 | https://askfilo.com/user-question-answers-smart-solutions/multiple-choice-questions-on-vectors-93-the-dot-product-of-3336343432363433 | Question asked by Filo student
Multiple Choice Questions on Vectors
Multiple Choice Questions on Vectors
The dot-product of two non-zero
Views: 5,843 students
Updated on: Sep 20, 2025
Text SolutionText solutionverified iconVerified
Solutions to Multiple Choice Questions on Vectors
93. Dot product equals product of magnitudes only if angle between them is 0° (cos 0° = 1).
Answer: A) 0°
94. Dot product = |A||B|cos θ = (1/2) |A||B| implies cos θ = 1/2. This happens at 60°.
Answer: C) 60°
95. Dot product A.A = |A|^2 = A².
Answer: D) A.A = A²
96. Dot product of unit vectors i and j is zero (they are perpendicular).
Answer: B) i.j = 0
97. Dot product of two anti-parallel unit vectors equals -1 because cos 180° = -1.
Answer: C) -1
98. i . (j x k) is scalar triple product of unit vectors, equals 1.
Answer: B) 1
99. If A ⊥ B then A.B=0.
Calculate A(A+B) = A.A + A.B = A² + 0 = A².
Answer: B) A(A+B) = A²
100. Angle θ between vectors given by:
cosθ=∣A∣∣B∣A⋅B
Calculate dot product:
(−2)(1)+3(2)+1(−4)=−2+6−4=0
Magnitude of each vector:
∣A∣=(−2)2+32+12=4+9+1=14
∣B∣=12+22+(−4)2=1+4+16=21
So,
cosθ=0/(14×21)=0
Hence,
θ=90∘
Answer: C) 90°
101. For vector 4i-3j, a perpendicular vector satisfies dot product zero:
Let vector be xi + yj:
4x−3y=0⟹4x=3y⟹y=(4/3)x
Check options:
Since 7k is perpendicular (k is perpendicular to i and j directions), it also qualifies.
Among given options, B) is in xy-plane and perpendicular.
Answer: B) 3i + 4j
102. For vectors to be perpendicular:
(3)(2)+(−2)(6)+(1)(m)=0
6−12+m=0⟹m=6
Answer: D) m = 6
103. Pairs represent orthogonal vectors if dot product = 0.
Check options (interpreting typographical errors as best as possible):
Dot product = 72 + 7(-3) = 14 -21 = -7 ≠ 0
Dot = 23 + 3(-2) = 6 -6 = 0
Answer: D) (2i + 3j) and (3i - 2j)
104. Vector B perpendicular to A = A cos θ i + A sin θ j:
Perpendicular vector is obtained by rotating by 90°, so:
B = B sin θ i - B cos θ j
Answer: D) B sin θ i - B cos θ j
105. Angle between (i + j) and (j + k):
Dot product = 01 + 11 + 01 = 1
Magnitude of (i+j) = √1² +1² = √2
Magnitude of (j+k) = √1² +1² = √2
Cos θ = 1 / (√2 √2) = 1/2
θ = 60°
Answer: C) 60°
106. AxB = determinant:
|i j k|
|2 1 0|
|1 3 0|
For cross product in k direction:
(2)(3) - (1)(1) = 6 - 1 = 5
So,
AxB = 5k
Answer: A) 5k
107. If AxB points along x-axis, then A and B lie in yz plane because cross product is perpendicular to both.
Answer: B) yz plane
108. If A = Ai and B = Bj,
A×B=ABk (since i x j = k)
Answer: C) AB k
109. Cross product of two parallel vectors is zero.
Answer: C) 0
110. A = j + k, B = -j - k => B = -A
Cross product of parallel or anti-parallel vectors is zero.
Answer: A) Of zero magnitude
111. Cross product of vector with itself is zero (since angle = 0° and sine 0 = 0).
Answer: A) 0
112. (A - B) and (A x B) are perpendicular since cross product is perpendicular to both A and B, so (A - B) is linear combination of A and B.
Angle = π/2 rad
Answer: B) π/2 rad
113. Magnitude of AxB equals area of parallelogram formed by A and B.
Answer: C) Parallelogram
114. Aβ sin θ x Aβ sin θ interpreted as (A B sin θ) x (A B sin θ) = (A B sin θ)^2
Answer: A) A² B² sin² θ
115. Area of parallelogram formed by A and B = AB sin θ.
Answer: B) AB sin θ
116. Cross product of unit vector with itself is zero.
Answer: C) 0
117. i . (j x k) = scalar triple product = 1
Answer: B) 1
118. i . (j x k) + j . (k x i) = 1 + 1 = 2
Answer: B) 2
119. i . (j x k) + j . (i x j) + k . (j x k) = 1 + 0 + 0 = 1
(Recall j . (i x j) = 0 and k . (j x k) = 0)
Answer: B) 1
120. (i x i) x i + (j x j) x j + (k x k) x i = 0 + 0 + 0 = 0
Answer: D) None of these
121. Cross product of two perpendicular vectors = AB n (magnitude AB; direction perpendicular)
Answer: B) AB n
122. Self-cross product angle is 0°.
Answer: A) 0°
123. If A and B are perpendicular, dot product AB = 0.
Answer: A) AB = 0
124. Magnitude of both dot and cross product are equal if vectors are at 45° because:
Dot product magnitude = AB cos θ
Cross product magnitude = AB sin θ
Set equal: cos θ = sin θ → θ = 45°
Neither parallel (0°) nor perpendicular (90°).
Answer: Neither option A nor B exactly, but correct condition is 45° angle.
125. Magnitude of dot and cross products equal when angle θ satisfies:
|A||B| cos θ = |A||B| sin θ
cos θ = sin θ
θ = 45°
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| Question Text | ### Multiple Choice Questions on Vectors 93. The dot-product of two non-zero vectors equals the product of their magnitudes only if the angle between them is: - A) 0° - B) 90° - C) 120° - D) 180° 94. For which of the following pair of angles, the dot product of two vectors is equal to one-half of the product of the magnitudes of two vectors? - A) 30° - B) 45° - C) 60° - D) 90° 95. For the vector A: - A) A.A = 0 - B) A.A = 1 - C) A.A = A - D) A.A = A² 96. Which one of the followings is correct? - A) i.j = k - B) i.j = 0 - C) i.j = 1 - D) i.j = -k 97. Dot product of two anti-parallel unit vectors is: - A) 0 - B) 1 - C) -1 - D) ∞ 98. |i.(j x k)| = ? - A) 0 - B) 1 - C) -1 - D) 2i 99. If A is perpendicular to B, then: - A) AB = 1 - B) A(A+B) = A² - C) AB = AB - D) A(A+B) = A² + AB 100. The angle between two vectors -2i + 3j + k and i + 2j - 4k is: - A) 0° - B) 60° - C) 90° - D) 180° 101. Consider a vector 4i - 3j. Another vector that is perpendicular to it is: - A) 4i + 3j - B) 3i + 4j - C) 3i - 4j - D) 7k 102. Vectors 3i - 2j + k and 2i + 6j + mk will be perpendicular if: - A) m = -1 - B) m = 3 - C) m = -6 - D) m = 6 103. Which of the following pairs represent orthogonal vectors? - A) (2i + 3i) and (4î - 2) - B) (5i + 7j) and (4î - 2) - C) (7i + 7j) and (21 - 3j) - D) (2i + 3)) and (3i - 2) 104. Let A = A cos θ i + A sin θ j be any vector. Another vector B, which is at right angle to A can be expressed as: - A) B cos θ i + B sin θ j - B) B cos θ i - B sin θ j - C) B sin θ i + B cos θ j - D) B sin θ i - B cos θ j 105. The angle between i + j and j + k is: - A) 30° - B) 45° - C) 60° - D) 90° 106. If A = 2i + j and B = i + 3j then A x B = ? - A) 5k - B) -5k - C) 7k - D) -7k 107. If A x B points along x-axis then the vector A and B must be in: - A) xy plane - B) yz plane - C) zx plane - D) In space 108. If A = Ai and B = Bj then A x B = ? - A) AB - B) AB sine k - C) AB k - D) 0 109. A x B for two non-zero parallel vectors will be equal to: - A) AB sine - B) AB sine n - C) 0 - D) 0 110. If A = j + k and B = -j - k, then cross-product of A and B is: - A) Of zero magnitude - B) Of non-zero magnitude - C) Of negative value - D) Of maximum value 111. The cross product of a vector with itself has the magnitude: - A) 0 - B) 1 - C) -1 - D) A² 112. What is the angle between (A - B) and (A x B)? - A) 0 rad - B) π/2 rad - C) π rad - D) 3π/2 rad 113. The magnitude of A x B is equal to the area of adjacent sides of: - A) Rectangle - B) Triangle - C) Parallelogram - D) Parallelepiped 114. Aβ sin θ x Aβ sin θ is: - A) A²B² sin² θ - B) A²B² - C) A²B² θ - D) 0 115. Area of parallelogram formed by A and B as its two adjacent sides is given as: - A) AB cos θ - B) AB sin θ - C) AB tan θ - D) AB 116. In case of unit vectors i x i = j x j = k x k = ? - A) 1 - B) -1 - C) 0 - D) 0 117. The magnitude of i . (j x k) is equal to: - A) 0 - B) 1 - C) -1 - D) 0 118. i . (j x k) + j . (k x i) = ? - A) 1 - B) 2 - C) -1 - D) -2 119. i . (j x k) + j . (i x j) + k . (j x k) = ? - A) 0 - B) 1 - C) 2 - D) 3 120. (i x i) x i + (j x j) x j + (k x k) x i = ? - A) i - B) j - C) k - D) None of these 121. Cross product of two perpendicular vectors is equal to: - A) AB - B) AB n 122. In self-cross product, the angle is: - A) 0° - B) 90° 123. If A and B are perpendicular then: - A) AB = 0 - B) AB = 1 124. The magnitude of both dot and cross product of two vectors A and B - A) A and B are parallel to each other - B) A and B are perpendicular to each other 125. The magnitude of dot and cross product of two vectors are equal when |
| Updated On | Sep 20, 2025 |
| Topic | All topics |
| Subject | Smart Solutions |
| Class | Class 12 Passed |
| Answer Type | Text solution:1 |
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11059 | https://math.stackexchange.com/questions/4301417/deducing-a-lower-bound-for-the-number-of-sign-changes | real analysis - Deducing a lower bound for the number of sign changes - Mathematics Stack Exchange
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Deducing a lower bound for the number of sign changes
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Let Z(t)Z(t) denote Hardy's Z function (just treat it as any continuous real function defined on positive real numbers). It is known that for all large T T there exists absolute constants A,B>0 A,B>0 such that
∫2 T T|Z(t)|d t≥A T(1)(1)∫T 2 T|Z(t)|d t≥A T
∣∣∣∫2 T T Z(t)d t∣∣∣≤B T 3/4(2)(2)|∫T 2 T Z(t)d t|≤B T 3/4
In §2.1 of Ivić's The Theory of Hardy's Z-Function, the author states without justification that (for large T T)
This argument actually shows that there is always a zero of Z(t)Z(t) in [T,T+C T 3/4][T,T+C T 3/4] for suitable C>0 C>0.
Then, the author says that one can deduce from this that there exists K>0 K>0 such that N 0(T)>K T 1/4 N 0(T)>K T 1/4 for T T large enough, where N 0(T)N 0(T) denotes the number of zeros of Z(t)Z(t) on [0,T][0,T].
However, I am only able to deduce an inferior estimate: Suppose Z(t)Z(t) does not change sign in [T,2 T][T,2 T] for all large T T. Then
A T≤∫2 T T|Z(t)|d t≤B T 3/4 A T≤∫T 2 T|Z(t)|d t≤B T 3/4
which leads to a contradiction. This means there exists a constant M>0 M>0 such that N 0(T)≥M log T N 0(T)≥M logT for all sufficiently large T T.
I wonder how the author is able to reduce the range of potential zero into [T,T+C T 3/4][T,T+C T 3/4] and obtain such a strong lower bound.
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edited Nov 9, 2021 at 18:26
TravorLZHTravorLZH
asked Nov 9, 2021 at 16:33
TravorLZHTravorLZH
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I don't feel that the given information is enough to imply these localized sign changes. (Imagine that Z(t)Z(t) were incredibly small yet positive between T T and 2–√⋅T 2⋅T, then macroscopic and oscillating between 2–√⋅T 2⋅T and 2 T 2 T, and then repeated periodically in the multiplicative sense.) Perhaps Ivić literally means that the argument/proof implies this, not the final statements. Does he prove a more localized version of (1) perhaps?Greg Martin –Greg Martin 2021-11-09 17:13:02 +00:00 Commented Nov 9, 2021 at 17:13
@GregMartin I insist that the conditions I provided in the question is sufficient to justify a sign change. You might want to have a look at my revised question description.TravorLZH –TravorLZH 2021-11-09 18:27:35 +00:00 Commented Nov 9, 2021 at 18:27
@GregMartin Yeah, it turns out you are correct. The estimates on the integral over [T,2 T][T,2 T] alone is not enough to conclude anything on the sign changes within [T,T+C T 3/4][T,T+C T 3/4].TravorLZH –TravorLZH 2021-11-11 06:30:23 +00:00 Commented Nov 11, 2021 at 6:30
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Let 1/4<a<1 1/4<a<1 say.
Now ∫T+C T a T|Z(t)|d t≥|∫T+C T a T ζ(1/2+i t)|d t=|(∫2+i T 1/2+i T+∫2+i(T+C T a)2+i T+∫1/2+i(T+C T a)2+i(T+C T a))ζ(s)d s|∫T T+C T a|Z(t)|d t≥|∫T T+C T a ζ(1/2+i t)|d t=|(∫1/2+i T 2+i T+∫2+i T 2+i(T+C T a)+∫2+i(T+C T a)1/2+i(T+C T a))ζ(s)d s|
We have that |∫2+i T 1/2+i T ζ(s)d s|+|∫1/2+i(T+C T a)2+i(T+C T a)ζ(s)d s|=O(T 1/4)|∫1/2+i T 2+i T ζ(s)d s|+|∫2+i(T+C T a)1/2+i(T+C T a)ζ(s)d s|=O(T 1/4)
(from the convexity estimate of ζ ζ - and we can do even better using sharper available estimates )
and |∫2+i(T+C T a)2+i T ζ(s)d s|=C T a+O(1)|∫2+i T 2+i(T+C T a)ζ(s)d s|=C T a+O(1) from the Dirichlet series expansion, giving:
∫T+C T a T|Z(t)|d t≥C T a+O(T 1/4)∫T T+C T a|Z(t)|d t≥C T a+O(T 1/4)
On the other hand,the proof in Ivic shows that |∫T+C T a T Z(t)d t|=O(T 3/4)|∫T T+C T a Z(t)d t|=O(T 3/4) (where the O O doesn't depend on C,a C,a as long as say C T a<T C T a<T), so it follows that for a≥3/4 a≥3/4 and for C C large enough (hence for T T large enough so C T a|∫T+C T a T Z(t)d t|∫T T+C T a|Z(t)|d t>|∫T T+C T a Z(t)d t| implying the existence of a zero there
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edited Nov 9, 2021 at 19:01
answered Nov 9, 2021 at 18:50
ConradConrad
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I see, so it's Ivić did not ask reader to use the exactly same argument to deduce the existence of a zero. I should have modified it a bit before asking this question, but huge appreciation anyway!TravorLZH –TravorLZH 2021-11-09 19:07:53 +00:00 Commented Nov 9, 2021 at 19:07
By the way, I wonder what is the known best lower bound obtained by this trick (if there is one). Maybe we could generalize the problem to contrasting the behavior of Z(t)Z(t) and |Z(t)||Z(t)| when being integrated over [T,T+Q][T,T+Q] for some suitable Q Q to be chosen later.TravorLZH –TravorLZH 2021-11-09 19:09:24 +00:00 Commented Nov 9, 2021 at 19:09
1 Yes, Ivic's equation 2.2 (easily as shown above) holds for integrals from T T to T+C T a T+C T a for a<1 a<1 for which an estimate of the type |ζ(s)|<<|s|a−ϵ,R s≥1/2|ζ(s)|<<|s|a−ϵ,ℜ s≥1/2 is available; there are a lot of results along these line obtained by Moser and then Karatsuba (see Karatsuba's book Complex Analysis in Number Theory Ch 2 and references there) where clever exponential sum estimates for |∫T+C T a T Z(t)d t||∫T T+C T a Z(t)d t| (which is the hard part) much improve both the exponent a a and the estimate for the number of zeroes on the critical line between T T and T+C T a T+C T a Conrad –Conrad 2021-11-09 19:13:55 +00:00 Commented Nov 9, 2021 at 19:13
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To generalize Conrad's argument, one is motivated to define
I(T,Δ)=∣∣∣∫T+Δ T Z(t)d t∣∣∣I(T,Δ)=|∫T T+Δ Z(t)d t|
and
J(T,Δ)=∫T+Δ T|Z(t)|d t=∫T+Δ T∣∣∣ζ(1 2+i t)∣∣∣d t J(T,Δ)=∫T T+Δ|Z(t)|d t=∫T T+Δ|ζ(1 2+i t)|d t
Undeniably, it is possible to attack them via approximate functional equation, but I personally found the Phragmén-Lindelöf corollary a simpler result to use (see chapter 4 of Titchmarsh's The theory of the Riemann zeta-function):
ζ(σ+i t)≪ε|t|(1−σ)/2+ε ζ(σ+i t)≪ε|t|(1−σ)/2+ε
Applying this and a few other approximation lemmas in chapter 4 of the same book, one obtains that when Δ≤T Δ≤T,
I(T,Δ)≪δ T 3/4+δ(1)(1)I(T,Δ)≪δ T 3/4+δ
for all δ>0 δ>0. In addition, we have
J(T,Δ)≫Δ(2)(2)J(T,Δ)≫Δ
whenever Δ≫ε T 1/4+ε Δ≫ε T 1/4+ε. This indicates that if we were plug Δ=T 3/4+η Δ=T 3/4+η into (2) and set δ=ε/2 δ=ε/2 in (1), then we can obtain the following density estimate:
For every η>0 η>0, there exists T 0=T 0(η)>0 T 0=T 0(η)>0 such that there is always a zero of Z(t)Z(t) lying within [T,T+T 3/4+η][T,T+T 3/4+η].
Though weaker than the original question's density estimate, this result still allows us to give a rough lower bound for N 0(T)N 0(T):
N 0(T)≫η T 1/4−η(3)(3)N 0(T)≫η T 1/4−η
In §10.5 of Titchmarsh's book, the author studied the Δ=T Δ=T case and showed that there is always a zero in [T,2 T][T,2 T], and it is not difficult to port these arguments to deduce (3). I have written a detailed account on this in Chinese available on Zhihu.
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answered Nov 16, 2021 at 1:37
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11060 | https://www.cs.dartmouth.edu/~ac/Teach/CS105-Winter05/Notes/nanda-scribe-3.pdf | CS 105: Algorithms (Grad) Subset Sum Problem Soumendra Nanda March 2, 2005 1 What is the Subset Sum Problem?
An instance of the Subset Sum problem is a pair (S, t), where S = {x1, x2, ..., xn} is a set of positive integers and t (the target) is a positive integer. The decision problem asks for a subset of S whose sum is as large as possible, but not larger than t.
This problem is NP-complete.
This problem arises in practical applications. Similar to the knapsack problem we may have a truck that can carry at most t pounds and we have n different boxes to ship and the ith box weighs xi pounds.
The naive approach of computing the sum of the elements of every subset of S and then selecting the best requires exponential time. Below we present an exponential time exact algorithm.
2 An Exact Algorithm for the Subset-Sum Problem In iteration i, we compute the sums of all subsets of {x1, x2, ..., xi}, using as a starting point the sums of all subsets of {x1, x2, ..., xi−1}. Once we find the sum of a subset S’ is greater than t, we ignore that sum, as there is no reason to maintain it. No superset of S’ can possibly be the optimal solution.
Notation: If L is a list of positive integers and x is another positive integer, then we let L + x denote the list of integers derived from L by increasing each element of L by x. For example, if L = ⟨1, 2, 4⟩then L + 2 = ⟨3, 4, 6⟩ MergeLists(L, L′) returns the sorted list that is the merge of two sorted input lists L and L’ Algorithm 1: Exact-Subset-Sum(S, t) n ← −|S| 1 L0 ← −⟨0⟩ 2 for i = 1 to n do 3 Li ← −MergeLists(Li−1, Li−1 + xi) 4 remove from Li every element greater than t 5 return the largest element in Ln 6 Analysis: It can be shown by induction that above algorithm is correct. Exact-Subset-Sum(S, t) is an exponential time algorithm in general since the length of Li can be as high as 2i.
Example: Let S = {1, 4, 5} then.......
L0 = {0} L1 = {0, 1} L2 = {0, 1, 4, 5} L3 = {0, 1, 4, 5, 6, 9, 10} So if the target was 9 we would have removed 10 from the last list.
Page 1 of 4 CS 105: Algorithms (Grad) Subset Sum Problem Soumendra Nanda March 2, 2005 3 FPTAS for the Subset-Sum Problem 3.1 What is a PTAS / FPTAS?
A polynomial time approximation scheme (PTAS) is an algorithm that takes as input not only an instance of the problem but also a value ϵ > 0 and approximates the optimal solution to within a ratio bound of 1 + ϵ. For any choice of ϵ the algorithm has a running time that is polynomial in n, the size of the input.
Example: a PTAS may have a running time bound of O(n2/ϵ) A fully polynomial-time approximation scheme (FPTAS) is a PTAS with a running time that is polynomial not only in n but also in 1/ϵ.
Example: a PTAS with a running time bound of O((1/ϵ)2n3) is an FPTAS 3.2 Trim Subroutine for Approximate Subset Sum Algorithm Before we explain the approximation algorithm, we explain how to trim our list Li using the parameter δ. If several values in L are close to each other, maintain only one of them,i.e we trim each list Li after it is created.
Given a parameter δ, where 0 < δ < 1 , element z approximates element y if y/(1 + δ) ≤z ≤y.
To trim a list Li by δ, remove as many elements as possible such that every element that is removed is approximated by some remaining element in the list.
Example: If δ = 0.1 and L = ⟨10, 11, 12, 15, 20, 21, 22, 23, 24, 29⟩ will be trimmed to L = ⟨10, 12, 15, 20, 23, 29⟩ since 11 approximates 12, 20 approximates 21 and 22 and similarly 23 approximates 24.
Algorithm 2: Trim(L, δ) m ← −|L| 1 L′ ← −⟨0⟩ 2 last ← −y1 3 for i = 2 to n do 4 if yi > last.(1 + δ) then 5 append yi onto the end of L′ 6 last ← −yi 7 return L′ 8 The running time for above algorithm is Θ(m). We now present the approximate subset sum algorithm that uses Trim and MergeLists.
Page 2 of 4 CS 105: Algorithms (Grad) Subset Sum Problem Soumendra Nanda March 2, 2005 3.3 Approximate Subset Sum Algorithm Algorithm 3: Approx-Subset-Sum(S, t, ϵ) n ← −|S| 1 L0 ← −⟨0⟩ 2 for i = 1 to n do 3 Li ← −MergeLists(Li−1, Li−1 + xi) 4 Li ← −Trim(Li, ϵ/2n) 5 remove from Li every element greater than t; 6 return the largest element in Ln 7 3.4 Analysis of Approximate Subset Sum Algorithm Theorem: For 0 < ϵ < 1, Approx-Subset-Sum(S, t, ϵ) is a FPTAS for the subset sum problem.
Proof: We need to show that 1. The solution returned is within a factor of 1 + ϵ of the optimal solution.
2. The running time is polynomial in both n and 1/ϵ Let z∗be the value returned by Approx-Subset-Sum(S, t, ϵ) and let y∗be an optimal solution.
As z∗≤y∗, we need to show that y∗ z∗≤1 + ϵ For every element y ≤t that is the sum of a subset of the first i numbers in S, there is a z ∈Li, such that y (1 + ϵ/2n)i ≤z ≤y Taking i = n and using the fact that z∗is the largest element in Ln, y∗ z∗≤(1 + ϵ/2n)n ≤eϵ/2 ≤1 + ϵ Therefore, y∗ z∗≤1 + ϵ (1) Now lets look at the number of elements in Li: After trimming, successive elements z and z′ in Li must differ by at least 1 + ϵ/2n The number of elements in Li is at most 2 + log1+ϵ/2n t = 2 + ln t ln(1 + ϵ/2n) Page 3 of 4 CS 105: Algorithms (Grad) Subset Sum Problem Soumendra Nanda March 2, 2005 ≤2 + 4n ln t ϵ (2) This implies that this above value is polynomial in size of the input(i.e log t plus some polyno-mial in n) and in 1/ϵ.
Therefore since the running time of Approx-Subset-Sum is polynomial in size of the length of Li, from (1) and (2) we have shown that Approx-Subset-Sum is an FPTAS.
Page 4 of 4 |
11061 | https://www.unitconverters.net/inductance/henry-to-millihenry.htm | Home / Inductance Conversion / Convert Henry to Millihenry
Convert Henry to Millihenry
Please provide values below to convert henry [H] to millihenry [mH], or vice versa.
Henry to Millihenry Conversion Table
| Henry [H] | Millihenry [mH] |
--- |
| 0.01 H | 10 mH |
| 0.1 H | 100 mH |
| 1 H | 1000 mH |
| 2 H | 2000 mH |
| 3 H | 3000 mH |
| 5 H | 5000 mH |
| 10 H | 10000 mH |
| 20 H | 20000 mH |
| 50 H | 50000 mH |
| 100 H | 100000 mH |
| 1000 H | 1000000 mH |
How to Convert Henry to Millihenry
1 H = 1000 mH
1 mH = 0.001 H
Example: convert 15 H to mH:
15 H = 15 × 1000 mH = 15000 mH
Convert Henry to Other Inductance Units
Henry to Exahenry
Henry to Petahenry
Henry to Terahenry
Henry to Gigahenry
Henry to Megahenry
Henry to Kilohenry
Henry to Hectohenry
Henry to Dekahenry
Henry to Decihenry
Henry to Centihenry
Henry to Microhenry
Henry to Nanohenry
Henry to Picohenry
Henry to Femtohenry
Henry to Attohenry
Henry to Weber/ampere
Henry to Abhenry
Henry to EMU Of Inductance
Henry to Stathenry
Henry to ESU Of Inductance
All Converters
Common ConvertersEngineering ConvertersHeat ConvertersFluids ConvertersLight ConvertersElectricity Converters
ChargeLinear Charge DensitySurface Charge DensityVolume Charge DensityCurrentLinear Current DensitySurface Current DensityElectric Field StrengthElectric PotentialElectric ResistanceElectric ResistivityElectric ConductanceElectric ConductivityElectrostatic CapacitanceInductance
Magnetism ConvertersRadiology ConvertersCommon Unit Systems |
11062 | https://www.researchgate.net/publication/261460614_Mathematical_computer_games_based_on_modular_arithmetic | Published Time: 2012-09-01
(PDF) Mathematical computer games based on modular arithmetic
Conference Paper PDF Available
Mathematical computer games based on modular arithmetic
September 2012
DOI:10.1109/ICPCI.2012.6486317
Conference: Problems of Cybernetics and Informatics (PCI), 2012 IV International Conference
Authors:
Fidan Nuriyeva
Dokuz Eylül University
Asli Guler
Yaşar University
Deniz Tanir
Kafkas University
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References (2)
Abstract
In this paper, computer aided games, which motivate and bring out the mathematical abilities of students, have been studied by inspiring “Bachet's Game”. The winning strategies of the games are based on modular arithmetic. Winning algorithms and mathematical models are included in the paper. Levels of the logical and math games which have been proved to be beneficial for students in order to gain ability of appropriate selection and application of analytical technics and strategies in terms of problems could be applied in compliance with age group of students and be permanent during the academic life of the student as a result of its character.
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zerbaljan National
ademy of Sclences Ministry of Education Problem Council of
of Azerbaijan Repubtic Cybernetics and Information
Technology
Ministry of Communications
and Information Technologies
of Azerbaijan Republic
ıstitute of Cybernetics
ıf Aze.rbajjan National
Acaderny of Sciences
Institute of Information
Technology
of
Azerbaijan
National Academy of Sclences
Baku State University,
Research Institute of
Applied Mathematics of
Baku State University
National Academy
of Avlation
Azerbaljan State
Oil Academy
Azerpaljan
Technicaf University
{il'
\ 1993
E.
:7tr 1
t:4.N ••
Azerbaljan
University of Qafqaz University
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and
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Baku, Azerbaijan
IV International Conference"Problems
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www.pci2012.science.azJ2115.pdf
Mathematical Computer Games Based on Modular
Aritlımetic
__ L_~( __ ~( ~( ~_
Fidan Nuriyeva', Asli Guler", Deniz Tanır)
I.JEge University,ızmir,Turkey
2 Yasar University,İzmir, Turkey
i
nuriyevajidan@gmail.com,
2
asli.guler@yasar.edu.tr,3deniz _tanir _35@ho/mail.com
Abstract=- In this paper,computer aided games,whicb motivate
and bring out the mathematical abilities of students,have been
studied by inspiring"Bachct's Gamc",The winning stratcgies of
the games are based on modular
arithrnetic.
Winning algo rtthms
and mathematical models are included in the puper.
Levels of the logical and math games which have been proved to
be beneficial for students in order to gain ability of appropriare
selection and application of analytical technics and stratcgies in
terms of problems could be applied in compliance with age group
ol
students and be permanent during the academic life of the
student as a result of its charactcr.
Keywords-s-Computer g anıe, k/allı g anıe, Mind game,
Modular
aritlımetic;Winniııg strategies,Logical straıegy gome,Computer
g
ame
programming.
ı.
INTRODUCTION
Inspiring"Bachet' s Game",same games have been
produced in asimilar way:'N!:o-dular arithmetic has been used
for search ing mathernatical structures of the games.In order
to develop winning strategies and
algorithrns,
it is benefited
from techniques of Artifıcial Intelligence and Theory of
Algorithms [2,3]. The winning strategies of the games are
based
on
modular
arithmetic.Mathematica!mode!s of tlıe
games have been examined in detai!and winning strategies
(algorithms)have been proposed for different
cases.
Leve!s of
the logical and math ganıes which have been proved to be
beneficial
for students in order to gain ability of appropriate
seleetion and application of analyrical technics and strategies in
terms of problems could be applied in compliance with age
group of students and be perrnanent during the academic life of
the student as a result of its character.
II.
yYpES OF BACHET'S GAME
Bachet's
Garne is a mathemarical game
whose
winning
strategy is based on modular arithrnetic.The ganıe is played in
a competitive environment. There are
N
items on a table,and a
player makes a move by removing at most
S
items from the
table. The player who removes the last
item
loses the game.
The gaıne is diversified by values of
N
and S .
A."Fifteen Items"Game
Fifteen items are given.Players can remove one or two
items in
tum.
The player
who removes
the
last
item loses the
game.
Le! us arrange the items in groups as below.
+.IEEE
Figure I.Grouping of the items
As it is seen in "Fig.1", in order that the first player wins
the game,he should remove two items in his fırst turn.Therı,
he should make the number of items,that is removed by the
opponent,
up
to three in the latter turns.
In
otlıer words,if his
opponent removes one item,he should remove two items,or if
the opponent removes two items,he should remove one item.
The second player
must
remove the last
item,
and lose tlıe
gaıne in tlıis way.
i)
Generalizatian of "Fifteen Items"Game
Given n iterns,Two players can remove s items
(L:S;
s
:s;
nı,
m
<
n)
respectively.The player who removes the
last item
loses,
Let us arrange the iterns ın groups in a similar way as
"Fifteen items" game.
n
~--------~~~----------
("
'\
(-....-)(-...-)...(-...-J-
~~~ ~
k "'-
m
T
!
m+l
m+i)
y
i
Figure 2. Grouping ofiteıns in the general game
As it is seen in "Fig. 2", the first player should remove k
items in his
fırst
tum in order to win.
In
latter turns, the first
player
should remove
(nı
+
ı-
s)
items when the second player
removes
s
items
(s :;;In).
Therefore,the second player has to
remove the last item and loses,
We can see that this strategy is also valid for "fifteen iterns"
garne. Since
n
=
15,
in
=
2 in the game,as in (1),
k
=
2 is
obtained.
k=(n-ı)'
n-IJ=ı4-'~J'3=14-4'3=2 (1)
Lm+l L
3
2) Anather Version o/the Game
194
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IV International Conference"Probtems ofCybernettes an d
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(pcrzoız),
Septemoer
IZ-l4, ZOLZ
www.pci20ı2.scienee.az/2I1S.pdf
In the above"n: m"game,the player who removes the last
item loses the game.Let us consider the game with a eondition
that if the player removes the last itern,he wins.In other
words,
given n
iterns, rwo
players can remove
s
items
(ı~ s ~
m,m <n) respectively.The player who
removes
the
last İtem
wins.
As
it
seen in "Fig. 3", we arrange the items in groups,
n
------------~------------
(
"\
(_...-J
'---Y---J
m+! ~
-
[-
...
-)[-
...
-)
'---Y---J '---Y---J '---Y---J
k... m+l m+!
y
i
Figure
3.
Grouping of items in
the new version
The first player should remove
k
items,
k
S
m,
by result(3)
in order to win.In latter tums,the first player should remove
( m
+
i-
s)
items when the second player
removes s
iterns
( s~m).
Therefore,there ternain
(m
+
i)
items on the table at
step
(I
+1) according to value
i
calculated by equation(2). The
second
player
can remove at most m items;so the first player
wins the game by removing the rest ofthem.
L
n
-1
J
1=
m+]
k
==
n
(mod
(m
+
l))
B"Hundred: ten"Game
The game is played with
two
people. The
fırst
player tells
a
number between
i
and
ı
o.
The second player tells a new
number by adding at least
i
at most
ıo
to the previous number.
The game continues like this;and the player who tells the
number
ı
00 wins the game.
The strategies mentioned before can be applied for this
game as welL.
Here,
11
=
i
00,
m
=
Lo and tn +
i
=
IL.
100-1l=89
89-11=78
78-11=67
67-ll=56
56-11=45
45-11=34
34-J J=23
23-11=12
(4)
(5)
(6)
(7)
(8)
(9)
(10)
(ll)
.IEEE
-,'-';~-,~i+-:• ~~~~
_ ~",-'::::.:;,,""T
12-11
=
1 (12)
The fırst player should teli the number
i
(one)to win this
game.Then,he should make the number s, that is told by the
opponent,( s:5 10),up to II in the latter turns.in this way,he
tells 12,23,34, 45,56, 67,78,89 and
ı
00 at last.Therefore,
the number
i
00 is told by the fırst player.
i)
Generalizatian of"Hundred: ten"Game
The game İs played with two people.The fırst player tells
a number between 1 and m. The second player
tells
a new
number by adding at
least
i
at most m to the previous
number.The game contiııues like this;and the player who tells
the number
n
wins the game.
We use the formula(13)andfınd the number k.
k
=
n
-l~J(m+
i)
111+1
(13)
The fırst pIayer should
tell
the number
k
at the beginning.
Then,he should make the number s, that is (old by the
opponent,
(s:5
nı),up to
171+1
in the latter turns.in other
words,he telIs
(m
+
ı-
s )
at each turn,so the number
n
is told
by the fırst player.
llL.DESIGNING COMPUTER PROGRAMS FOR BACHET'S
GAME
(2)
Inspiring so-called"Bachet's Game',some games have
been produced in a similar way,then computer programs have
been proposed by making different generalizations of the
ganıes
.
Two of the
produced
programs are run by being
loaded to computer and the other one is pIayed on the Internet.
C# has been used for the first
two
of the games(ITEM and
NUMBER),and the other game(MODAR)has been designed
in JavaScript and HtmL.
A.MODAR Game
MODAR ganıe takes its name from the union of thefırst
Ietters of the word s "Modular" and "Arithmetics", This game
has
been
designed for ten
items
by JavaScript and Html
with
the name MODAR,and it can be played on
[5J.
MODAR game has been generated by using web
progranuning languages h0111
and javascript.
The game
can be
played
againsı
the computer or a human opponerıt.When
playing against the computer,there are three levels of the
game.The computer uses all winning strategies at the most
difficult level.There are ten balls in the game.Choosing the
first player the game starts.The players can remove one or two
ba1ls respecıively.The player who removes the last balı wins
the game.
B.Programming Generalized Fifieen Iıems Game
by
C#
There are input units where we can change the total number
of objects and the number of items that can be rernoved by
players in our program.If desired,the computer or we can start
the game,The starting view
of the
program is
like in
"Fig.4".
(3)
195
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Internasional
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andInformaıtcs=
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12-14 2012
www.pci2012.science.a7J2IlS.pdf '
uınlrollY19 Panel
T
ota!
objed Number
MilXimum
tak"" dem.
Figure 4.The
starung
view of the program
When pressing the Start bu tton,there will be definite
number of
squares
on the screen as in
"Fig. 5"
that
we
have
determined at the beginrıing.At each step,we alsa determine
that how maııy items we remove,theıı this number of squares
wiU disappear from the screen.
ii·
Figure S.The view of items
We can reach the information about how the game
is played
by pressing the Help button.
If we lose the game a message box will be appeared on the
screen.
The computer plays according to the winning
strategies
when playing against the computer.
C. Programıning Generalized
"Hundred.Ten"
Game by Ct:
In the program,
we determine
the target number and the
inereasing
number
ourselves by writing thern into "New
Target"and "New Increasing"boxes.
The opponent(computer)or we can start the game.The
starting view of the program is like in"Fig.6".
We can reach the information about how the game is played
by pressing the"Help"button.
When we press the "Start"button,the gaıne starts.
We write the number of iterns,that we wlll remove,into
"Increasing"box,the gaıne continues like this.
+,IEEE
Kontrol
Por~ı; Geme Panel
Ta!'}ll:t
Increasr,g
1---..;1
ı..._,... __ .
..:.......-1
New Increasing:
Figure 6.The starting vıew of the program
IV.
CONCLUSION
A combinatorio game has been considered in this paper,
moreover its generalizations
and different versions have been
analysed by
using
modular aritlımetic.New
logical
games, that
are not at the market,have been gerıerated, solution algorithms
have been
proposed
and computer programs have been
designed. it is aimed that these programs will provide students
to use mathemaıics,mind and game strategies by several math
games.
lt
has been seen that generating different new ga:mes
is
possible by changing the corıditions in mentioned games.Some
strategies have been proposed and algorithms have been
desigrıed for solutions,Furthermore,the algorithm of a game
that runs
at an
interactive web environment has
been
written by
javascript language.
lt has been shown that how mathematics can be used
effectively in garnes and a new activity has been added to
activities that can
rnake
students like mathematics.
The computer programs can be used for education in
schools.Mind games,whose main purpose is to help bringing
up individuals who think,questionize,reason and analyse,
provide students to meet analytic thinking techniques and to
improve these skills when spending enjoyable time.
REFERENCES
[ı
i
Alizade R.,
Ufuktepe.
U, (2006), Finite
Mathematics.
Olyrnpiad
Questions and Solunons,TUBlTAK Publishing,Ankara,273 p.
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11063 | https://diversification.com/term/marginal-rate-of-technical-substitution | ← Back to M Definitions
Marginal rate of technical substitution
What Is Marginal Rate of Technical Substitution?
The marginal rate of technical substitution (MRTS) is an economic concept that quantifies the rate at which one input, typically labor, can be substituted for another input, such as capital, while maintaining a constant level of output. It is a fundamental element within production theory, a branch of microeconomics that examines how firms make decisions about resource allocation to produce goods and services. The MRTS helps businesses determine the most efficient combination of factors of production to achieve a specific production target. Understanding the MRTS is crucial for optimizing the use of inputs and achieving cost minimization in a firm's operations.
History and Origin
The concept of the marginal rate of technical substitution is deeply rooted in neoclassical economics, which developed in the late 19th and early 20th centuries. This school of thought focuses on how individuals and firms make rational decisions to allocate limited resources, with an emphasis on supply and demand dynamics24. The development of tools like isoquants, which graphically represent combinations of inputs yielding the same output, provided a framework for understanding input substitution.
However, the theoretical underpinnings of production functions and the nature of capital itself faced significant scrutiny during the "Cambridge Capital Controversies" of the 1950s and 1960s. This debate primarily involved economists from the University of Cambridge, UK, and the Massachusetts Institute of Technology (MIT) in Cambridge, Massachusetts. The controversies challenged the neoclassical assumption of capital as a homogeneous, quantifiable aggregate that could be easily measured and substituted, which is foundational to concepts like MRTS. Economists like Joan Robinson and Piero Sraffa argued against this simplified view, highlighting complexities in the measurement and role of capital goods in aggregate production and distribution22, 23. This intellectual dispute, while not directly refuting the concept of MRTS at the micro-level, underscored the limitations of applying such concepts to broader macroeconomic analysis20, 21.
Key Takeaways
The Marginal Rate of Technical Substitution (MRTS) measures how much one input can decrease as another increases, keeping output constant.
It is represented by the slope of an isoquant curve.
The MRTS typically diminishes as more of one input is substituted for another, reflecting that inputs are not perfect substitutes.
Firms use MRTS to identify the most optimal input mix for a given level of production, contributing to economic efficiency.
The concept is fundamental in the theory of the firm and production economics.
Formula and Calculation
The marginal rate of technical substitution (MRTS) between two inputs, typically labor (L) and capital (K), is calculated as the ratio of the marginal product of labor (MPL) to the marginal product of capital (MPK). It represents the rate at which capital can be reduced for every additional unit of labor employed, without changing the total output.
The formula for MRTS is:
MRTSL,K=−ΔLΔK=MPKMPL
Where:
( MRTS_{L,K} ) = Marginal Rate of Technical Substitution of labor for capital
( \Delta K ) = Change in the quantity of capital
( \Delta L ) = Change in the quantity of labor
( MP_L ) = Marginal product of labor (the additional output produced by one more unit of labor)
( MP_K ) = Marginal product of capital (the additional output produced by one more unit of capital)
The negative sign indicates that as one input increases, the other must decrease to maintain the same output level, reflecting the downward slope of the isoquant curve.
Interpreting the Marginal Rate of Technical Substitution
Interpreting the marginal rate of technical substitution involves understanding the trade-offs a firm faces when adjusting its production inputs. A high MRTS indicates that a large amount of one input (e.g., capital) can be given up for a small increase in another input (e.g., labor) while maintaining the same output. Conversely, a low MRTS suggests that only a small amount of one input can be substituted for a relatively large increase in the other.
As a firm substitutes more of one input for another, the MRTS typically diminishes. This phenomenon, known as the law of diminishing returns, implies that as additional units of a variable input are added to fixed inputs, the additional output generated from each new unit of the variable input will eventually decrease19. For example, as more workers are added to a fixed amount of machinery, each additional worker contributes less to total output than the one before them. This diminishing MRTS causes isoquant curves to be convex to the origin18. Firms use this information to find the most cost-effective combination of inputs, often illustrated at the point where an isoquant curve is tangent to an isocost line16, 17.
Hypothetical Example
Consider a hypothetical T-shirt manufacturing company, "ShirtCo," that produces 1,000 T-shirts per day. ShirtCo uses two primary inputs: labor (number of employees) and capital (number of specialized sewing machines).
Initially, ShirtCo uses 50 employees and 10 sewing machines to produce 1,000 T-shirts.Due to rising labor costs, ShirtCo wants to explore substituting some labor with additional capital, while still producing 1,000 T-shirts.
Scenario 1: ShirtCo reduces its employees by 5 (to 45) and finds it needs to add 1 more sewing machine (to 11) to maintain the output of 1,000 T-shirts.In this case:( \Delta L = -5 )( \Delta K = +1 )The MRTS (labor for capital) at this point is approximately ( - \frac{1}{-5} = 0.2 ). This means for every 5 employees ShirtCo reduces, it needs 1 additional sewing machine to keep output constant.
Scenario 2: ShirtCo further reduces employees by another 5 (to 40). At this point, to maintain 1,000 T-shirts, it finds it needs to add 2 more sewing machines (to 13) because the marginal product of each additional machine has started to decrease. In this case: ( \Delta L = -5 ) ( \Delta K = +2 ) The MRTS has now changed to approximately ( - \frac{2}{-5} = 0.4 ).
This example illustrates the diminishing marginal rate of technical substitution. As ShirtCo substitutes more capital for labor, it requires increasingly more units of capital to replace the same amount of labor, reflecting that these inputs are not perfect substitutes for each other15. By analyzing this, ShirtCo can identify the most optimal input mix that minimizes its production costs for 1,000 T-shirts.
Practical Applications
The marginal rate of technical substitution (MRTS) is a critical concept with several practical applications in business and economic analysis, particularly within the framework of production theory.
Production Planning and Efficiency: Firms utilize MRTS to make informed decisions about their mix of factors of production, such as labor and capital. By understanding the rate at which one input can be substituted for another without affecting output, businesses can achieve the lowest possible production cost for a given output level, a process known as cost minimization13, 14. This is essential for maintaining competitiveness and maximizing profits.
Technological Change Analysis: When new technologies emerge, they often alter the marginal productivities of labor and capital, thereby changing the MRTS. For example, automation can significantly increase the marginal product of capital, leading firms to substitute capital for labor. Economists and policymakers monitor aggregate data, such as the Federal Reserve Board's "Industrial Production and Capacity Utilization" (G.17) report, to observe broad trends in industrial output and the shifting intensity of capital and labor utilization across sectors11, 12.
Resource Allocation and Policy: Governments and international organizations also consider the principles underlying MRTS when formulating economic policies. For instance, policies related to labor laws, investment incentives, or technological advancements can influence the relative costs and productivities of inputs, impacting how firms combine labor and capital in their production processes10. The overall economic efficiency of an economy depends on how effectively resources are allocated and utilized.
Limitations and Criticisms
While the marginal rate of technical substitution (MRTS) is a valuable tool in production theory, it has certain limitations and has faced criticisms, particularly stemming from broader debates in neoclassical economics.
One primary limitation is the assumption that inputs are continuously divisible and perfectly substitutable to a certain extent. In reality, substituting discrete units of capital or labor may not always be smooth or perfectly interchangeable without affecting quality or efficiency9. For instance, replacing a specific skilled worker with a machine might not yield the same output quality, or it might require a completely different production process.
Furthermore, the concept's reliance on aggregate measures of capital and its treatment within neoclassical models has been a point of contention. The "Cambridge Capital Controversies" challenged the idea that capital could be measured as a homogeneous quantity independent of its rate of return or the distribution of income. Critics argued that aggregating diverse capital goods into a single "quantity of capital" could lead to theoretical inconsistencies, which in turn could affect the interpretation of concepts like the MRTS when applied at a macro level7, 8. This debate highlighted that simplistic assumptions about factors of production might not capture the full complexity of real-world production processes.
Additionally, the MRTS focuses on a constant output level. In dynamic business environments, firms often aim to increase output or respond to changing supply and demand conditions, making a static analysis of substitution less comprehensive. The theory of production, which includes the MRTS, typically assumes fixed technology; however, technological advancements can significantly shift the underlying production function, altering the trade-offs between inputs in ways not fully captured by a static MRTS calculation6.
Marginal Rate of Technical Substitution vs. Marginal Rate of Substitution
The marginal rate of technical substitution (MRTS) and the marginal rate of substitution (MRS) are both concepts that involve trade-offs, but they apply in different economic contexts:
| Feature | Marginal Rate of Technical Substitution (MRTS) | Marginal Rate of Substitution (MRS) |
---
| Context | Production theory (firm behavior) | Consumer theory (individual behavior) |
| Focus | How inputs are substituted to maintain a constant output level | How goods are substituted to maintain a constant level of consumer utility |
| Inputs/Goods | Typically labor and capital (factors of production) | Two consumer goods or services |
| Curve Represented | Isoquant curve | Indifference curve |
| Decision-Maker | Producer/Firm | Consumer |
The key distinction lies in the economic agent and their objective. MRTS helps producers determine the most optimal input mix for efficient production, whereas MRS helps consumers allocate their budgets to maximize satisfaction from different goods. While both exhibit the property of diminishing returns (or diminishing MRS for consumers) as more of one item is substituted for another, their application and implications are distinct to their respective fields of microeconomic analysis5.
FAQs
What does a diminishing marginal rate of technical substitution mean?
A diminishing marginal rate of technical substitution means that as a firm substitutes more and more of one input (e.g., labor) for another (e.g., capital) while keeping output constant, it takes progressively larger amounts of the increasing input to replace each unit of the decreasing input. This happens because inputs are not perfect substitutes, and as a firm uses more of one factor, its marginal product tends to decrease4.
How does MRTS relate to the isoquant curve?
The MRTS is the absolute value of the slope of an isoquant curve at any given point. An isoquant curve shows all combinations of two inputs (e.g., labor and capital) that produce the same level of output. The slope indicates the rate at which one input can be substituted for the other while remaining on the same output level.
Why is MRTS important for businesses?
MRTS is important for businesses because it helps them identify the most cost-effective and efficient combinations of inputs to produce a desired quantity of output. By understanding these trade-offs, firms can optimize their resource allocation, reduce production costs, and improve their overall economic efficiency2, 3.
Can MRTS be constant?
While MRTS typically diminishes, it can be constant in specific theoretical cases where two inputs are perfect substitutes. In such a scenario, the isoquant curve would be a straight line, indicating that one input can be substituted for another at a fixed rate without any change in the marginal productivity of the inputs1. However, this is less common in real-world production.
Related Definitions
Bill of materialsAdjusted growth rate effectAdjusted cumulative tax rateAdjusted fill rate
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11064 | https://www.tiktok.com/@professoralgebro/video/7275875335353158958 | Three consecutive integers, Professor AlgeBro, #Math #Education #FYP #... | TikTok
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ayden
27,28,29. Divide 84 by 3 and whatever number it is it’s the next smaller and bigger number
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Kiss it 💋
Thank you so much I have a test tomorrow and I didn’t understand were you randomly got 3 numbers I was so lost
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toemuncher17
i just learned this today in class
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dontbemadtho
i think of it like the answer is less then 90 but more than 75 all 3 are less than 30 but more than 25. I just go from there
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GoldCoastKiss
All you have to do is subtract 3 from 84 then divide 81 by 3. That is 27. Add 28 + 29 and it is 84. Much easier.
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Rach
How do you know to use the +1 and then +2? I’m confused.
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thank you bro
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old evil raven
28, 28, 28
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Jüice
Why are we dividing by 3?
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misssassyspoons
27,28, and 29
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ProfessorAlgeBro
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@Sum_12 here you go..
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professor alegebro 😭😭😭 immediate follow
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11065 | https://ponce.sdsu.edu/onlinestandardsuppressedrectangular.php | Calculation of discharge through a standard suppressed rectangular weir, Victor Miguel Ponce, San Diego State University
online_standard_suppressed_rectangular: Discharge through a standard suppressed rectangular weir Formula: Q = 3.33 L H 3/2 [in U.S. Units] Head H on the weir, in m [or ft] Length L of the weir, in m [or ft] Height P to the weir crest, in m [or ft] Discharge Qin L/s [or cfs] See USBR Manual for general methodology INPUT DATA:[Description] Select Units: Head H : [H> 0 ] Length L : [L> 0 ] Height P : [H> 0 ] OUTPUT: Warning:Ratio P/H = 0 < 2. Warning:Ratio L/H = 0 ≤ 3. Discharge Q: 0 Your request was processed at 07:17:21 pm on September 28th, 2025[ 250928 19:17:21]. Thank you for running online_standard_suppressed_rectangular Please call again. |
11066 | https://www.ncss.com/wp-content/themes/ncss/pdf/Procedures/NCSS/NP_Charts.pdf | NCSS Statistical Software NCSS.com 257-1 © NCSS, LLC. All Rights Reserved. Chapter 257 NP Charts Introduction This procedure generates the NP control chart for the number nonconforming of a sample. The format of the control charts is fully customizable. This procedure permits the defining of stages. For the NP chart, the value for P can be entered directly or NP can be estimated from the data, or a sub-set of the data. A list of out-of-control points can be produced in the output, if desired, and proportion nonconforming values may be stored to the spreadsheet. NP Control Charts NP charts are used to monitor the number of nonconforming units of a process based on samples taken from the process at given times (hours, shifts, days, weeks, months, etc.). Typically, an initial series of samples is used to estimate the average number of nonconforming units per sample. The estimated average is then used to produce control limits for the number of nonconforming units. During this initial phase, the process should be in control. If points are out-of-control during the initial (estimation) phase, the assignable cause should be determined, and the sample should be removed from estimation. Once the control limits have been established for the NP chart, these limits may be used to monitor the number nonconforming going forward. When a point is outside these established control limits it indicates that the number of nonconforming units of the process is out-of-control. An assignable cause is suspected whenever the control chart indicates an out-of-control process. NCSS Statistical Software NCSS.com NP Charts 257-2 © NCSS, LLC. All Rights Reserved. The NP Chart versus the P Chart The NP chart is very similar to the P chart. Rather than focusing on the proportion of nonconforming units, as does the P chart, the NP chart focuses on the average number of non-conforming units. As such, choice of the P or NP chart is simply a matter of preference, as each is a scaled version of the other. One case in which the P chart may be recommended over the NP chart is the case where the sample size varies across samples, since the P chart is easier to interpret for this scenario. NP Chart Formulas Suppose we have k samples, each of size n. Let Di represent the number of nonconforming units in the ith sample. Formulas for the Points on the Chart Each point on the chart is given by Di, the number of nonconforming units in the ith sample. Estimating the NP Chart Center Line In the NP Charts procedure, the center line proportion may be input directly, or it may be estimated from a series of samples. If it is estimated from the samples the formula for the center line is 𝑛𝑛𝑝𝑝̅, where 𝑝𝑝̅ = ∑ 𝐷𝐷𝑖𝑖 𝑘𝑘 𝑖𝑖=1 𝑘𝑘𝑘𝑘 = ∑ 𝑝𝑝𝑖𝑖 𝑘𝑘 𝑖𝑖=1 𝑘𝑘 NP Chart Limits The lower and upper control limits for the P chart are calculated using the formulas 𝐿𝐿𝐿𝐿𝐿𝐿= 𝑛𝑛𝑝𝑝̅ −𝑚𝑚ඥ𝑛𝑛𝑝𝑝̅(1 −𝑝𝑝̅) 𝑈𝑈𝑈𝑈𝑈𝑈= 𝑛𝑛𝑝𝑝̅ + 𝑚𝑚ඥ𝑛𝑛𝑝𝑝̅(1 −𝑝𝑝̅) where m is a multiplier (usually set to 3) chosen to control the likelihood of false alarms (out-of-control signals when the process is in control). NCSS Statistical Software NCSS.com NP Charts 257-3 © NCSS, LLC. All Rights Reserved. Runs Tests The strength of control charts comes from their ability to detect sudden changes in a process that result from the presence of assignable causes. Unfortunately, the NP chart is poor at detecting drifts (gradual trends) or small shifts in the process. For example, there might be a positive trend in the last ten samples, but until a number of nonconforming units goes above the upper control limit, the chart gives no indication that a change has taken place in the process. Runs tests can be used to check control charts for unnatural patterns that are most likely caused by assignable causes. Runs tests are sometimes called “pattern tests”, “out-of-control” tests, or “zones rules”. While runs tests may be helpful in identifying patterns or smaller shifts in the proportion, they also increase the likelihood of false positive indications. The rate of false positives is typically measured using the average run length (the average length of a run before a false positive is indicated). When several runs tests are used the average run length of the control chart becomes very short. In order to perform the runs tests, the control chart is divided into six equal zones (three on each side of the center line). Since the control limit is three sigma limits (three standard deviations of the proportion) in width, each zone is one sigma wide and is labeled A, B, or C, with the C zone being the closest to the center line. There is a lower zone A and an upper zone A. The same is true for B and C. The runs tests look at the pattern in which points fall in these zones. The runs tests used in this procedure are described below. Test 1: Any Single Point Beyond Zone A This runs test simply indicates a single point is beyond one of the two three-sigma limits. Test 2: Two of Three Successive Points in Zone A or Beyond This usually indicates a shift in the process average. Note that the two points have to be in the same Zone A, upper or lower. They cannot be on both sides of the center line. The third point can be anywhere. Test 3: Four of Five Successive Points in Zone B or Beyond This usually indicates a shift in the process average. Note that the odd point can be anywhere. Test 4: Eight Successive Points in Zone C or Beyond All eight points must be on one side of the center line. This is another indication of a shift in the process average. NCSS Statistical Software NCSS.com NP Charts 257-4 © NCSS, LLC. All Rights Reserved. Test 5: Fifteen Successive Points Fall in Zone C on Either Side of the Center Line Although this pattern might make you think that the variation in your process has suddenly decreased, this is usually not the case. It is usually an indication of stratification in the sample. This happens when the samples come from two distinct distributions having different means. Perhaps there are two machines that are set differently. Try to isolate the two processes and check each one separately. Test 6: Eight of Eight Successive Points Outside of Zone C This usually indicates a mixture of processes. This can happen when two supposedly identical production lines feed a single production or assembly process. You must separate the processes to find and correct the assignable cause. There are, of course, many other sets of runs tests that have been developed. You should watch your data for trends, zig-zags, and other nonrandom patterns. Any of these conditions could be an indication of an assignable cause and would warrant further investigation. Issues in Using Control Charts There are several additional considerations surrounding the use of control charts that will not be addressed here. Some important questions are presented below without discussion. For a full treatment of these issues, you should consider a statistical quality control text such as Ryan (2011) or Montgomery (2013). Sample Size How many items should be sampled for each sample? How does the sample size affect my use of control charts? What about unequal sample sizes? Dealing with Out-of-Control Points How do you deal with out-of-control points once they have been detected? Should they be included or excluded in the process proportion estimate? Control Limit Multiplier Three-sigma limits are very common. When should one consider a value other than three? Startup Time How many samples should be used to establish control for my process? NCSS Statistical Software NCSS.com NP Charts 257-5 © NCSS, LLC. All Rights Reserved. Data Structure In this procedure, the data are entered in two columns. One column contains the sample size for each sample, the other column contains the number of nonconforming units. Example Dataset N Nonconf 50 2 50 8 50 6 50 3 50 4 50 2 50 7 50 1 50 9 50 7 50 3 50 5 . . . . . . NCSS Statistical Software NCSS.com NP Charts 257-6 © NCSS, LLC. All Rights Reserved. Attribute Chart Format Window Options This section describes the specific options available on the Attribute Chart Format window, which is displayed when the Attribute Chart Format button is clicked. Common options, such as axes, labels, legends, and titles are documented in the Graphics Components chapter. Attribute Chart Tab Symbols Section You can modify the attributes of the symbols using the options in this section. Lines Section You can specify the format of the various lines using the options in this section. Note that when shading is desired, the fill will be to the bottom for single lines (such as the mean line), and between the lines for pairs of lines (such as primary limits). Titles, Legend, Numeric Axis, Group Axis, Grid Lines, and Background Tabs Details on setting the options in these tabs are given in the Graphics Components chapter. The legend does not show by default but can easily be included by going to the Legend tab and clicking the Show Legend checkbox. NCSS Statistical Software NCSS.com NP Charts 257-7 © NCSS, LLC. All Rights Reserved. Example 1 – NP Chart Analysis (Phase I) This section presents an example of how to run an initial NP Chart analysis to establish control limits. In this example, a specific location of an automotive connection will be examined for a leak. Seventy connections are to be examine each day for 40 days to establish control of the process. The data used are in the Leak dataset. Setup To run this example, complete the following steps: 1 Open the Leak example dataset • From the File menu of the NCSS Data window, select Open Example Data. • Select Leak and click OK. 2 Specify the NP Charts procedure options • Find and open the NP Charts procedure using the menus or the Procedure Navigator. • The settings for this example are listed below and are stored in the Example 1 settings file. To load these settings to the procedure window, click Open Example Settings File in the Help Center or File menu. Variables Tab __________________________________________________________________________________________________ Sample Size Variable ...................................... Size Number Nonconforming Variable ..................... Leak 3 Run the procedure • Click the Run button to perform the calculations and generate the output. Chart Summary (Estimation and Control Limits) Chart Summary (Estimation and Control Limits) for Samples 1 to 40 ──────────────────────────────────────────────────────────────────────── Number of Samples: 40 Sample Size Variable: Size Number Nonconforming Variable: Leak ──────────────────────────────────────────────────────────────────────── Number Nonconforming ────────────────────────────── Sample Size Control Limits ───────────── Proportion ────────────── Average Total Nonconforming Average Total Lower Upper ───────────────────────────────────────────────────────────────────────────────────────────────────────── 70 2800 0.1060714 7.425 297 0 15.153957 ──────────────────────────────────────────────────────────────────────── This section displays the calculation values and limits that are to be used in the NP chart. The formulas for each calculation are described in the NP chart formula section toward the beginning of this chapter. NCSS Statistical Software NCSS.com NP Charts 257-8 © NCSS, LLC. All Rights Reserved. Average Sample Size This is the average of all the sample sizes. Total Sample Size This is the sum of all the sample sizes. Proportion Nonconforming This is the proportion of nonconforming units. Average Number Nonconforming This is the average number of nonconforming units. Total Number Nonconforming This is the sum of all the numbers of nonconforming units. Number Nonconforming Control Limits These are the lower and upper control limits for the NP chart. NP Chart NP Chart for Samples 1 to 40 ──────────────────────────────────────────────────────────────────────── The NP chart shows the sequence of numbers of leaking connections. The chart shows that samples 32 and 33 merit further investigation for an assignable cause. After reviewing the records, the investigators found that the 32nd and 33rd samples came during and immediately after the training of a new connection installer. NCSS Statistical Software NCSS.com NP Charts 257-9 © NCSS, LLC. All Rights Reserved. Out-of-Control List Out-of-Control List for Samples 1 to 40 ──────────────────────────────────────────────────────────────────────── Number Row Label Nonconforming Reason ───────────────────────────────────────────────────────────────────────── 32 32 24 beyond control limits 33 33 19 beyond control limits 34 34 5 2 of 3 in zone A 37 37 2 2 of 3 in zone A ──────────────────────────────────────────────────────────────────────── This report provides a list of the samples that failed one of the runs tests (including points outside the control limits). The report shows that samples 32 and 33 are beyond the control limits. Because the 32nd and 33rd samples came during and immediately after the training of a new connection installer, the investigators determine that the chart limits should be revised to exclude samples 32 and 33. Samples 34 and 37 show runs in the data, but they do seem to signal an important change in the process. NCSS Statistical Software NCSS.com NP Charts 257-10 © NCSS, LLC. All Rights Reserved. Example 2 – NP Chart Revised (Phase I) This section presents a continuation of the previous example. In this example the limits are revised to exclude the data from samples 32 and 33, due to assignable cause. Setup To run this example, complete the following steps: 1 Open the Leak example dataset • From the File menu of the NCSS Data window, select Open Example Data. • Select Leak and click OK. 2 Specify the NP Charts procedure options • Find and open the NP Charts procedure using the menus or the Procedure Navigator. • The settings for this example are listed below and are stored in the Example 2 settings file. To load these settings to the procedure window, click Open Example Settings File in the Help Center or File menu. Variables Tab _________________________________________________________________________________________________ Sample Size Variable ...................................... Size Number Nonconforming Variable ..................... Leak Specification Method ....................................... Keep Rows Variable Use all rows where variable ............................. Keep =..................................................................... 1 3 Run the procedure • Click the Run button to perform the calculations and generate the output. Output Chart Summary (Estimation and Control Limits) for Samples 1 to 40 where Keep = 1 ──────────────────────────────────────────────────────────────────────── Number of Samples: 38 Sample Size Variable: Size Number Nonconforming Variable: Leak ──────────────────────────────────────────────────────────────────────── Number Nonconforming ────────────────────────────── Sample Size Control Limits ───────────── Proportion ────────────── Average Total Nonconforming Average Total Lower Upper ────────────────────────────────────────────────────────────────────────────────────────────────────────── 70 2660 0.09548872 6.684211 254 0 14.060761 ──────────────────────────────────────────────────────────────────────── NCSS Statistical Software NCSS.com NP Charts 257-11 © NCSS, LLC. All Rights Reserved. NP Chart for Samples 1 to 40 ──────────────────────────────────────────────────────────────────────── Out-of-Control List for Samples 1 to 40 ──────────────────────────────────────────────────────────────────────── Number Row Label Nonconforming Reason ───────────────────────────────────────────────────────────────────────── 10 10 15 beyond control limits 32 32 24 beyond control limits 33 33 19 beyond control limits 34 34 5 2 of 3 in zone A ──────────────────────────────────────────────────────────────────────── The limits in this report and chart are now based on all samples except samples 32 and 33. The values for samples 32 and 33 are included in the chart, but they are not included in the calculations. Because two of the more extreme values were removed, the control limits are narrower than those of Example 1. Although sample 10 now gives an out-of-control signal, the investigators did not find an assignable cause and therefore chose to leave sample 10 in the calculations of the center line and limits. Sample 34 gives an out-of-control signal due to samples 32 and 33 being beyond zone A. The NP chart can be additionally enhanced by labeling the points that were not included in the calculations. To do this, set the Point Label Variable to Label. NCSS Statistical Software NCSS.com NP Charts 257-12 © NCSS, LLC. All Rights Reserved. Chart Section for Samples 1 to 40 ───────────────────────────────────────────────────────────────────────── NCSS Statistical Software NCSS.com NP Charts 257-13 © NCSS, LLC. All Rights Reserved. Example 3 – NP Chart Analysis (Phase II) Continuing with Examples 1 and 2, the investigators obtain samples for an additional 20 days. They wish to use the limits based on the first 40 samples, excluding samples 32 and 33. Setup To run this example, complete the following steps: 1 Open the Leak example dataset • From the File menu of the NCSS Data window, select Open Example Data. • Select Leak and click OK. 2 Specify the NP Charts procedure options • Find and open the NP Charts procedure using the menus or the Procedure Navigator. • The settings for this example are listed below and are stored in the Example 3 settings file. To load these settings to the procedure window, click Open Example Settings File in the Help Center or File menu. Variables Tab __________________________________________________________________________________________________ Sample Size Variable ...................................... SizeCont Number Nonconforming Variable ..................... LeakCont Specification Method ....................................... Keep Rows Variable Use all rows where variable ............................. KeepCont =..................................................................... 1 Point Label Variable ......................................... Label Report Options (in the Toolbar) _________________________________________________________________________________________________ Variable Labels ................................................ Column Names 3 Run the procedure • Click the Run button to perform the calculations and generate the output. NCSS Statistical Software NCSS.com NP Charts 257-14 © NCSS, LLC. All Rights Reserved. Output Chart Summary (Estimation and Control Limits) for Samples 1 to 60 where KeepCont = 1 ──────────────────────────────────────────────────────────────────────── Number of Samples: 38 Sample Size Variable: SizeCont Number Nonconforming Variable: LeakCont ──────────────────────────────────────────────────────────────────────── Number Nonconforming ────────────────────────────── Sample Size Control Limits ───────────── Proportion ────────────── Average Total Nonconforming Average Total Lower Upper ────────────────────────────────────────────────────────────────────────────────────────────────────────── 70 2660 0.09548872 6.684211 254 0 14.060761 ──────────────────────────────────────────────────────────────────────── NP Chart for Samples 1 to 60 ──────────────────────────────────────────────────────────────────────── Out-of-Control List for Samples 1 to 60 ──────────────────────────────────────────────────────────────────────── Number Row Label Nonconforming Reason ───────────────────────────────────────────────────────────────────────── 10 10 15 beyond control limits 32 32 24 beyond control limits 33 33 19 beyond control limits 34 34 5 2 of 3 in zone A ──────────────────────────────────────────────────────────────────────── The estimation and limits section shows the same results as Example 2, since the calculations are again based only on the first 40 samples, excluding 32 and 33. The plot shows a stable continuation of the process from samples 41 to 60. NCSS Statistical Software NCSS.com NP Charts 257-15 © NCSS, LLC. All Rights Reserved. Example 4 – NP Chart with Additional Formatting This example uses the same setup as Example 3, except that a variety of improvements are made in the plot format. These improvements are made by clicking the NP Chart format button on the NP Chart tab. The settings for this example are stored in the Example 4 settings file. To load these settings to the procedure window, click Open Example Settings File in the Help Center or File menu. NP Chart NP Chart for Samples 1 to 60 ──────────────────────────────────────────────────────────────────────── As shown here, a variety of enhancements can be made to the formatting of the control chart to make the chart as easy to read as possible. NCSS Statistical Software NCSS.com NP Charts 257-16 © NCSS, LLC. All Rights Reserved. Example 5 – NP Chart with Stages This section presents a continuation of Examples 1, 2, and 3. After the first 60 days, a new connection process is implemented to reduce the number of leaks. The investigators obtained samples of the process for 60 additional days, using the first 30 samples of the additional 60 days (61 to 90) to determine the in-control limits of the updated process. The investigators wish to view the entire progress of the process, beginning with day 1. This example shows the use of stages to monitor a process with a change in center line and limit calculations. Setup To run this example, complete the following steps: 1 Open the Leak example dataset • From the File menu of the NCSS Data window, select Open Example Data. • Select Leak and click OK. 2 Specify the NP Charts procedure options • Find and open the NP Charts procedure using the menus or the Procedure Navigator. • The settings for this example are listed below and are stored in the Example 5 settings file. To load these settings to the procedure window, click Open Example Settings File in the Help Center or File menu. Variables Tab _____________________________________________________________________________________________________ Sample Size Variable ...................................... SizeStage Number Nonconforming Variable ..................... LeakStage Number of Stages ............................................ Multiple Stages Stage Variable ................................................. Stage Specification Method ....................................... Keep Rows Variable Use all rows where variable ............................. KeepStage =..................................................................... 1 Point Label Variable ......................................... Label 3 Run the procedure • Click the Run button to perform the calculations and generate the output. NCSS Statistical Software NCSS.com NP Charts 257-17 © NCSS, LLC. All Rights Reserved. Chart Summary (Estimation and Control Limits) Chart Summary (Estimation and Control Limits) for Samples 1 to 60 where KeepStage = 1 ──────────────────────────────────────────────────────────────────────── Number of Samples: 38 Sample Size Variable: SizeStage Number Nonconforming Variable: LeakStage ──────────────────────────────────────────────────────────────────────── Number Nonconforming ────────────────────────────── Sample Size Control Limits ───────────── Proportion ────────────── Average Total Nonconforming Average Total Lower Upper ────────────────────────────────────────────────────────────────────────────────────────────────────────── 70 2660 0.09548872 6.684211 254 0 14.060761 ──────────────────────────────────────────────────────────────────────── Chart Summary (Estimation and Control Limits) for Samples 61 to 120 where KeepStage = 1 ──────────────────────────────────────────────────────────────────────── Number of Samples: 30 Sample Size Variable: SizeStage Number Nonconforming Variable: LeakStage ──────────────────────────────────────────────────────────────────────── Number Nonconforming ───────────────────────────── Sample Size Control Limits ───────────── Proportion ────────────── Average Total Nonconforming Average Total Lower Upper ──────────────────────────────────────────────────────────────────────────────────────────────────────── 70 2100 0.05857143 4.1 123 0 9.993956 ──────────────────────────────────────────────────────────────────────── This section displays the estimation for each stage separately. The average number nonconforming has decreased by about 40%. NCSS Statistical Software NCSS.com NP Charts 257-18 © NCSS, LLC. All Rights Reserved. NP Chart NP Chart for Samples 1 to 120 ──────────────────────────────────────────────────────────────────────── The NP chart shows the substantial decrease in the number of leaking connections in the second stage. The center line and limits of the second stage are estimated from samples 61 to 90. Out-of-Control List Out-of-Control List for Samples 1 to 60 ──────────────────────────────────────────────────────────────────────── Number Row Label Nonconforming Reason ───────────────────────────────────────────────────────────────────────── 10 10 15 beyond control limits 32 32 24 beyond control limits 33 33 19 beyond control limits 34 34 5 2 of 3 in zone A ──────────────────────────────────────────────────────────────────────── With no new out-of-control points, the out-of-control list is the same as that of the previous examples. There is not an Out-of-Control List for samples 61 to 120 since there are no out-of-control points in this group. |
11067 | https://www.emed.theclinics.com/article/S0733-8627(17)30075-5/fulltext | Airway Management in Trauma - Emergency Medicine Clinics
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Review articleVolume 36, Issue 1p61-84 February 2018 Open access
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Airway Management in Trauma
George Kovacs, MD, MHPE, FRCPC
George Kovacs, MD, MHPE, FRCPC
Correspondence
Corresponding author. Charles V. Keating Emergency & Trauma Centre, QEII Health Sciences Centre, 1799 Robie Street, Halifax, Nova Scotia B3H 3G1, Canada.
gkovacs@dal.ca
Affiliations
Department of Emergency Medicine, Division of Medical Education, Dalhousie University, 3rd Floor, HI Site, Suite 355, Room 364D, Halifax, Nova Scotia B3H 3A7, Canada
Department of Anaesthesia, Division of Medical Education, Dalhousie University, 3rd Floor, HI Site, Suite 355, Room 364D, Halifax, Nova Scotia B3H 3A7, Canada
Department of Medical Neurosciences, Division of Medical Education, Dalhousie University, 3rd Floor, HI Site, Suite 355, Room 364D, Halifax, Nova Scotia B3H 3A7, Canada
Charles V. Keating Trauma & Emergency Centre, QEII Health Sciences Centre, 1799 Robie Street, Halifax, Nova Scotia B3H 3G1, Canada
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a,b,c,dgkovacs@dal.ca ∙ Nicholas Sowers, MD, FRCPC
Nicholas Sowers, MD, FRCPC
Affiliations
Department of Emergency Medicine, Division of Medical Education, Dalhousie University, 3rd Floor, HI Site, Suite 355, Room 364D, Halifax, Nova Scotia B3H 3A7, Canada
Charles V. Keating Trauma & Emergency Centre, QEII Health Sciences Centre, 1799 Robie Street, Halifax, Nova Scotia B3H 3G1, Canada
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a,d
Affiliations & Notes Article Info
a Department of Emergency Medicine, Division of Medical Education, Dalhousie University, 3rd Floor, HI Site, Suite 355, Room 364D, Halifax, Nova Scotia B3H 3A7, Canada
b Department of Anaesthesia, Division of Medical Education, Dalhousie University, 3rd Floor, HI Site, Suite 355, Room 364D, Halifax, Nova Scotia B3H 3A7, Canada
c Department of Medical Neurosciences, Division of Medical Education, Dalhousie University, 3rd Floor, HI Site, Suite 355, Room 364D, Halifax, Nova Scotia B3H 3A7, Canada
d Charles V. Keating Trauma & Emergency Centre, QEII Health Sciences Centre, 1799 Robie Street, Halifax, Nova Scotia B3H 3G1, Canada
Footnotes:
Disclosure Statement: The authors have nothing to disclose.
DOI: 10.1016/j.emc.2017.08.006 External LinkAlso available on ScienceDirect External Link
Copyright: © 2017 The Authors. Published by Elsevier Inc.
User License: Creative Commons Attribution – NonCommercial – NoDerivs (CC BY-NC-ND 4.0) | Elsevier's open access license policy
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Outline
Outline
Keywords
Key points
Introduction
Does early definitive trauma airway management save lives?
Trauma and the difficult airway
Airway management trauma scenarios
The awake intubation
Rapid sequence intubation
Front of neck airway to secure the airway
Summary
References
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Keywords
Key points
Introduction
Does early definitive trauma airway management save lives?
Trauma and the difficult airway
Airway management trauma scenarios
The awake intubation
Rapid sequence intubation
Front of neck airway to secure the airway
Summary
References
Article metrics
Related Articles
Keywords
Airway
Trauma
Airway management
Key points
•
Airway management in trauma presents numerous unique challenges.
•
A safe approach to airway management in trauma requires recognition of these anatomic and physiologic challenges.
•
An approach to airway management for these complicated patients is presented based on an assessment of anatomic challenges and optimizing physiologic parameters.
Introduction
The “ABCs” of trauma resuscitation were born from the assumption that correcting hypoxemia and hypotension reduces morbidity and mortality. Definitive care for severely injured or polytrauma patients includes the ability to provide advanced airway management in a variety of settings: in the emergency department, 20% to 30% intubations are for trauma.1,2
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Airway management in the trauma patient presents numerous unique challenges beyond placement of an endotracheal tube (ETT), with outcomes dependent on the provider’s ability to predict and anticipate difficulty and have a safe and executable plan.
Does early definitive trauma airway management save lives?
Despite significant advances in prehospital care, injury prevention, and the development of trauma systems, early mortality from trauma has essentially remained unchanged.3
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R. Adams Cowley, founder of Baltimore’s Shock Trauma Institute, defined the “golden hour” as a window to arrest the physiologic consequences of severe injury by rapidly transporting trauma patients to definitive care.4,5
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The “stay and play” versus “scoop and run” approach to prehospital trauma care has been a topic of debate since the early 1980s.6,7
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Specific to airway management, there is evidence to support the argument that advanced airway management can be performed in the prehospital setting without delaying transfer to a trauma center.8,9
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In the hospital setting, delayed intubation is associated with increased mortality in noncritically injured trauma patients.11
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Conversely, there is a growing body of evidence that prehospital advanced airway management may increase mortality for trauma patients in some circumstances.8,12–14
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“No conclusion could be reached regarding prehospital intubation for patients with traumatic brain injury, with or without RSI [rapid sequence intubation]. Diversity of patient population, differing airway algorithms, various experience among emergency medical service personnel in ETI, and differing reporting make consensus difficult.”15
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It may be that the technical, procedure-focused management imperative of “getting the tube” is diverting attention away from the physiologic principles of oxygen delivery. Translated physiologically, the ABC priorities of trauma resuscitation are “stop the bleeding, maintain perfusion and oxygenate.” Lifesaving oxygenation maneuvers may include a jaw thrust, temporary bag-mask ventilation (BMV), placement of a supraglottic airway device, or ETI. Advanced does not necessarily mean better.
Trauma and the difficult airway
A “difficult airway” is defined as difficulty with laryngoscopy and intubation, BMV, supraglottic device ventilation, and/or front of neck airway (FONA) access.16,17
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The pathophysiology of trauma adds an additional layer of complexity and difficulty (Table 1).
| Difficult Airway | Trauma Related Difficulty | Approach |
---
| Difficult laryngoscopy and intubation |
| Limited mouth opening/jaw displacement | Collar/improper MILS Trismus | Open collar/ear-muff MILS |
| Inability to position | MILS | ELM/bougie/VL |
| Blood/vomitus | Facial injuries/full stomach, delayed gastric emptying | 2 suctions/SALAD approach FONA |
| Penetrating or blunt neck trauma | Disrupted or distorted airway | Awake primary FIE; if not feasible RSI VL-assisted FIE |
| Difficult BVM |
| Limited jaw thrust | Mandibular fractures | Early SGA use |
| Poor seal | Facial injuries with swelling, disruption | Early SGA use |
| Blood/vomitus | Facial injuries/full stomach, delayed gastric emptying | 2 suctions/SALAD approach FONA |
| Penetrating or blunt neck trauma | Distorting subcutaneous emphysema, disrupted airway | Passive oxygen delivery/minimize PPV |
| Difficult SGA use |
| Blood/vomitus | Facial injuries/full stomach, delayed gastric emptying | 2 suctions/SALAD approach FONA |
| Penetrating or blunt neck trauma | Distorted/disrupted airway | Direct visualization FIE/FONA, low tracheotomy |
| FONA |
| Penetrating or blunt neck trauma | Distorted/disrupted airway CTM not accessible or injury at or below CTM | Low tracheotomy |
Table 1
Predictors of difficult airway management in trauma
Abbreviations: BVM, bag-valve-mask; CTM, cricothyroid membrane; ELM, external laryngeal manipulation; FIE, flexible intubating endoscope; FONA, front of neck airway; MILS, manual inline stabilization; PPV, partial-pressure ventilation; RSI, rapid sequence intubation; SALAD, suction-assisted laryngoscopy airway decontamination; SGA, supraglottic airway; VL, video laryngoscopy.
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The “physiologically difficult airway” is used to describe nonanatomic patient factors that can influence the outcome of airway management. Uncorrected hypoxemia, hypocapnia, and hypotension can have devastating consequences in the peri-intubation period. All trauma patients should have both anatomic and physiologic factors considered, planned for, and ideally corrected as part of their airway plan.22
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In patients in whom both ETI and rescue oxygenation (bag-mask or supraglottic airway ventilation) are anticipated to be difficult, most existing airway algorithms recommend an “awake” intubation approach, in which the patient maintains spontaneous respiration throughout the procedure. There are a variety of reasons why awake intubation is uncommonly used for the trauma airway, and these are discussed later in this text.
Although the difficult airway is defined with reference to an experienced airway provider with an array of available recourses, other context-related challenges, including human factors, environment, clinician experience, and skill will invariably influence outcomes. Understanding when and why trauma patients may encounter difficulty in airway management can help guide the logistical and mental exercise of developing specific mitigating strategies and contingency planning. A call for help should always be viewed as a patient-focused measure, not a sign of provider weakness.
Airway management trauma scenarios
The Head-Injured Patient
Traumatic brain injury (TBI) is the most common cause of mortality in trauma patients. Airway management in this cohort of patients is often performed for airway protection. Given the relatively high incidence of peri-intubation desaturation, hypocapnea, and hypotension in emergency intubations, the benefit of ETI for airway protection to prevent aspiration must be weighed against the risk of the occurrence of physiologic adverse events known to increase morbidity and mortality in TBI patients.23–26
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If intubating for the purpose of airway protection, it is usually less time sensitive and should not be rushed. Every precaution should be taken to adequately preoxygenate and resuscitate first.
Apnea resulting from head injury requires immediate intervention. There are 3 mechanisms by which apnea may occur in TBI:
1.
Severe or catastrophic brain injury
2.
Impact brain apnea (IBA)
3.
Loss of consciousness with resultant functional airway obstruction
Severe or catastrophic brain injury is usually nonsurvivable, and associated with early death. Predictions of outcome are usually not made until the patient has undergone a full trauma resuscitation, which often includes ETI. Contrastingly, IBA and functional airway obstruction may be correctable with simple airway opening maneuvers, with or without brief ventilation support. IBA from head trauma results in a primary respiratory arrest without significant parenchymal injury to the brain.27
27.
Wilson, M.H. ∙ Hinds, J. ∙ Grier, G. ...
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In contrast to patients with head injury with functional airway obstruction, patients with IBA do not respond to simple airway opening maneuvers alone, and may require brief ventilation support to prevent secondary hypoxic insult. With appropriate treatment, prognosis is generally good.
Head-injured patients with a decreased level of consciousness frequently receive prehospital advanced airway management.10
10.
Lockey, D.J. ∙ Healey, B. ∙ Crewdson, K. ...
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In one series, 30% to 40% of patients are assessed as having partial or complete airway obstruction on EMS arrival.10
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Lockey, D.J. ∙ Healey, B. ∙ Crewdson, K. ...
Advanced airway management is necessary in prehospital trauma patients
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A proportion of these patients will respond to basic maneuvers, and those who do not usually have more severe, less survivable injuries. This observation in part explains the comparatively poor survival rates for trauma patients who are intubated in the prehospital setting
Management pearls for the patient with traumatic brain injury (TBI)
•
Hypoxemia and hypotension during airway management significantly worsens outcomes in patients with TBI.
•
Airway management for airway protection should proceed only after adequate measures have been taken to prevent intubation related physiologic disturbances.
•
Postintubation hypocapnia is also associated with poor outcomes in patients with TBI and often the result of adrenaline induced overzealous postintubation ventilation.
•
Postinjury apnea requiring ventilation support does not necessarily predict poor outcome.
.
Airway Management in Patients with Suspected Cervical Spine Injuries
Trauma resuscitations typically proceed under the assumption that the patient has an unstable cervical spine (c-spine) injury until proven otherwise. In the prehospital environment, trauma patients are often placed in a cervical spine collar and secured to a rigid backboard with blocks. Although a long-standing tradition in emergency medicine and trauma care, there is very limited published evidence to support the notion that cervical spine collars and immobilization prevent secondary spinal cord injury.28,29
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Although the incidence of c-spine injuries is relatively low (occurring in approximately 2% in the general trauma population and 6%–8% in patients with head and facial trauma), practitioners often operate with deep concern that intubation may cause secondary spinal cord injury, making it one of the most frequently encountered reasons for difficulty in trauma airway management.30–33
30.
Mulligan, R.P. ∙ Friedman, J.A. ∙ Mahabir, R.C.
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A frequently studied outcome is the amount of translational or angular movement of the cervical spine caused by airway manipulation. Although it appears that spinal movement occurs to a variable degree depending on the airway technique used, it is unclear whether or not this results in any important differences in clinical outcomes.34
34.
Kill, C. ∙ Risse, J. ∙ Wallot, P. ...
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In cadaveric studies of unstable c-spine injuries, movement occurring with both direct laryngoscopy (DL) and indirect laryngoscopy do not significantly exceed the physiologic values observed with intact spines.35,36
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Despite the need to be cautious, even in patients with known cervical spine injuries, secondary neurologic deterioration is rare, with a reported incidence of 0.03%.33,37–39
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Trauma patients with suspected spinal injury are typically fully supine, inhibiting the practitioner’s ability to optimally position the patient for DL. Manual inline stabilization (MILS) worsens the view obtained with DL in up to 50% of cases.40
40.
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Effect of manual in-line stabilization of the cervical spine in adults on the rate of difficult orotracheal intubation by direct laryngoscopy: a randomized controlled trial
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Minimizing challenges with laryngoscopy and intubation mandates proper application of MILS, whereby the provider tasked with this role immobilizes the head and neck without immobilizing the mandible (Fig.1). C-spine collars and improperly applied MILS will restrict mouth opening and tongue and mandibular displacement required for optimal laryngoscopy. Despite properly applied MILS, the provider should still expect a higher occurrence of a poor view with DL, longer intubation times, and more frequent failed intubation attempts.40
40.
Thiboutot, F. ∙ Nicole, P.C. ∙ Trépanier, C.A. ...
Effect of manual in-line stabilization of the cervical spine in adults on the rate of difficult orotracheal intubation by direct laryngoscopy: a randomized controlled trial
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This scenario is often easily managed by applying external laryngeal manipulation or use of a bougie.
Figure viewer
Fig.1(A) MILS applied incorrectly limiting mandibular range of motion (ROM). (B) MILS applied correctly with hands over the ears (ear-muff approach) not limiting mandibular ROM.
Another theoretic concern is that application of MILS results in the need for an increased applied force during laryngoscopy, which paradoxically may lead to more movement during intubation than occurs without MILS.40,41
40.
Thiboutot, F. ∙ Nicole, P.C. ∙ Trépanier, C.A. ...
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Recognizing our inability to correct the fundamental geometric challenge of DL, the provider may opt to use a “look-around-the-corner” indirect device, such as a video laryngoscope with a hyperangulated blade. The 3 classes of video laryngoscopes are described in Box 1.42
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Box 1
Classes of video laryngoscopes
Data from Kovacs G, Law JA. Lights camera action: redirecting videolaryngoscopy. EMCrit. 2016. Available at: Accessed February 25, 2017.
1.
Macintosh video laryngoscopy (VL; also known as standard geometry blade) for example, C-MAC (Mac Blade; Karl Storz, Tuttlingen, Germany), McGrath Mac (Mac blade; Medtronic, Minneapolis, MN), GlideScope Titanium Mac (GlideScope, Verathon, WA), Venner APA (Mac blade; Venner Medical, Singapore, Republic of Singapore).
2.
Hyperangulated VL (also known as indirect VL), for example, C-MAC (D-Blade), McGrath Mac (X blade) standard GlideScope, KingVision (nonchanneled blade; Ambu, Ballerup, Denmark).
3.
Channeled blade VL, for example, King Vision, Pentax AWS (Pentax, Tokyo, Japan), Airtraq (Teleflex Medical, Wayne, PA).
Data from Kovacs G, Law JA. Lights camera action: redirecting videolaryngoscopy. EMCrit. 2016. Available at: Accessed February 25, 2017.
It would seem intuitive that because indirect hyperangulated video laryngoscopy (VL) consistently provides an improved glottic view and that c-spine immobilization consistently impairs the glottic view with DL, that VL is the better choice for trauma patients.43–45
43.
Bathory, I. ∙ Frascarolo, P. ∙ Kern, C. ...
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However, having a good view with VL does not mean that easy ETI will follow.46
46.
Gu, Y. ∙ Robert, J. ∙ Kovacs, G. ...
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When using a hyperangulated video laryngoscope, a deliberate restricted view may be desired to facilitate the often seemingly frustrating paradox of having a great view of the glottis but not being able to deliver the ETT.42,46,47
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Literature comparing intubation devices in c-spine immobilized patients has yielded inconsistent findings, and no consensus as to the optimal approach.38,44,45,48,49
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A recent meta-analysis by Suppan and colleagues45
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Suppan, L. ∙ Tramèr, M.R. ∙ Niquille, M. ...
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reported more failed intubations for DL compared with several alternative intubating devices in patients with c-spine immobilization. Although the investigators acknowledge the weaknesses of available literature, they note there was no statistically significant difference in first-attempt success between the more commonly used VL devices (GlideScope, C-MAC) and DL.45
45.
Suppan, L. ∙ Tramèr, M.R. ∙ Niquille, M. ...
Alternative intubation techniques vs Macintosh laryngoscopy in patients with cervical spine immobilization: systematic review and meta-analysis of randomized controlled trials
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It is less likely that there is a “the right device” for the unstable c-spine and more important is the right experienced practitioner, using a device with which he or she is the most comfortable.50,51
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Airway management in the patient with a possible c-spine injury must strike a balance between minimizing movement and the need to quickly and successfully intubate on first attempt, thereby minimizing the harm of hypoxemia that may be associated with multiple attempts at intubation.52
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It seems reasonable to consider that if the patient’s spinal cord has survived the massive forces of the crash, as well as repositioning during extrication and immobilization, that the chances that movement occurring during controlled airway management will result in cord injury is extremely low. As suggested by Aprahamian and colleagues,33,53
33.
Manoach, S. ∙ Paladino, L.
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the primary benefit of a rigid cervical collar is to serve as a reminder about the potential existence of an unstable c-spine injury
Management pearls for patients with unstable cervical spine injuries
•
Imaging should not delay airway management and assume all trauma patients have unstable cervical spines.
•
The provider should optimally use the intubation device he or she is most experienced with.
•
Be prepared for a poor view with direct laryngoscopy (DL) and always have a bougie ready for use.
•
Rigid cervical collars must be opened or removed and replaced by properly applied manual inline stabilization (MILS).
•
Properly applied MILS should avoid immobilization of the mandible.
•
If using a hyperangulated video laryngoscope, a deliberate restricted glottic view may facilitate difficult ETT advancement.
.
The Contaminated Airway
The presence of airway contamination with either blood or vomitus has been shown to decrease the rate of first-attempt intubation success, regardless of the device used.54
54.
Sakles, J.C. ∙ Corn, G.J. ∙ Hollinger, P. ...
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Blood and vomit in the airway can lead to early and late complications related to difficult airway management and/or aspiration. The bloody airway is not uncommon in trauma patients with injuries to the face and/or neck and may range in severity from scant bleeding, which is easily managed, to significant hemorrhage. The combination of altered levels of consciousness, diminished protective airway reflexes, delayed gastric emptying, and full stomachs place trauma patients at high risk of vomiting and aspiration during airway management. Management of contaminated airway must begin with the expectation that the degree of blood, vomit, and secretions appreciated externally represents only a fraction of what may be encountered on initiation of an RSI. As such, providers must ensure that adequate suction is available (at least 2 large rigid suction catheters). Consideration must be given to positioning, placing the patient in reverse Trendelenburg or, if safe to do so, seated upright or even leaning forward to allow drainage of blood and secretions. For c-spine immobilized patients, suction must be immediately within reach, and restraints securing the patient to the bed should be avoided. During the preoxygenation phase, positive-pressure ventilation (PPV) should be used only if necessary balanced against the patient’s oxygenation status, as ventilatory pressures of 20 cm H2O or more are likely to ventilate the stomach, increasing the risk of regurgitation and aspiration.
When blood or vomitus is overwhelming suction capabilities, the provider may place either one rigid suction or an ETT in the upper esophagus to divert the offending contaminants. The ETT or rigid suction may then be stabilized to the left of the laryngoscope and the second suction used during laryngoscopy in search of the epiglottis (Fig.2). Often the epiglottis may be “lifted” (more easily accomplished in a reverse Trendelenburg) out of the contaminant during laryngoscopy, providing an anatomic reference for placing a bougie.
Figure viewer
Fig.2 Suction in upper esophagus stabilized to left of laryngoscope (SALAD approach).
(Courtesy of Ruben Strayer, MD.)
Most of the literature comparing DL with VL in the bloody or vomitus-filled airway is simulation-based, and concern exists about the vulnerability of VL camera lens in the contaminated airway.55,56
55.
Ohchi, F. ∙ Komasawa, N. ∙ Mihara, R. ...
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Recently, Sakles and colleagues54
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retrospectively reviewed more than 4600 intubations and demonstrated that, although airway contamination was associated with a decreased first-attempt success rate, this was irrespective of the choice of GlideScope or DL as the first-attempt device used. The use of DL or Macintosh VL, in which a direct approach can be used if the camera is obscured with the aid of a bougie, may be preferred approaches.
Although not studied in a clinical setting, the Ducanto suction-assisted laryngoscopy airway decontamination (SALAD) approach has gained acceptance as a method to manage the soiled airway.57,58
57.
DuCanto, J. ∙ Serrano, K. ∙ Thompson, R.
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In the uncommon circumstance in which blood or vomit is overwhelming these management strategies, intubation is not possible and the patient is critically desaturating, rescue oxygenation with a BVM (bag-valve-mask) or SGA (supraglottic airway) is unlikely to work and an FONA approach is indicated
Ducanto suction-assisted laryngoscopy airway decontamination approach to managing massive airway contamination
•
Use rigid large-bore suction to initially decontaminate
•
Perform laryngoscopy keeping blade superior against tongue away from fluid
•
Advance suction tip into upper esophagus then wedge in place to left of the laryngoscope
•
Use second suction as needed
•
Rotate laryngoscope blade 30 degrees to the left to open blade channel
•
Place endotracheal tube (ETT), inflate the cuff
Management pearls for the patient with the contaminated airway
•
Have at least 2 large-bore rigid suction catheters.
•
Consider alternative options for hemorrhage control (sutures, packing, epistaxis kit).
•
Minimize positive-pressure ventilation (PPV) and use a monometer for provider feedback when mask ventilation is indicated.
•
Look for epiglottis as an important landmark for glottis and have a bougie prepared for use with DL.
•
If a VL is considered the best option, Macintosh VL may be the preferred device, as it may be used directly if contamination obstructs camera.
•
Consider esophageal ETT diversion connected to suction.
•
Suction-assisted laryngoscopy airway decontamination (SALAD) approach.
•
If intubation fails and patient is desaturating, front of neck airway (FONA) rescue oxygenation approach is indicated.
.
The Uncooperative or Agitated Patient
Uncooperative, violent, or agitated patients can encumber adequate assessment, leading to missed injuries and inadequate resuscitation. Agitation can be multifactorial and may be the result of head injury, hypoperfusion, hypoxemia, or intoxication. It may not be clear why a patient is agitated and providers must determine if the patient is agitated AND injured or agitated BECAUSE the patient is injured.
The EAST guidelines recommend that aggressive behavior refractory to initial pharmacologic intervention is a discretionary indication for intubation; specifically that if a patient’s level of agitation prevents assessment and resuscitation, intubation and sedation should follow.15
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Mayglothling, J. ∙ Duane, T.M. ∙ Gibbs, M. ...
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reviewed 1078 trauma patients intubated for discretionary indications (eg, agitation, alcohol intoxication) and found that 62% of patients, once investigated, had a significant head injury. Importantly, there was no significant difference in complications associated with acute airway management in patients intubated for discretionary indications, as compared with those intubated for higher acuity reasons.59
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In severely agitated patients, RSI is at times undertaken before optimal hemodynamic resuscitation and preoxygenation has been achieved. Patients rendered apneic as part of an RSI without adequate preoxygenation are at high risk of desaturation. The use of ketamine to facilitate cooperation and allow interventions including preoxygenation has been described as “delayed-sequence intubation” by Weingart and colleagues.60
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If given slowly, a dissociative intravenous dose of 1 to 1.5 mg/kg poses little risk of respiratory depression. However, the use of any sedative, particularly in the presence of other intoxicating ingestions, may inhibit airway reflexes. Concerns that ketamine may raise intracranial pressure and worsen outcomes in TBI is not supported by evidence
Management pearls in the agitated trauma patient
•
Agitation may be a symptom of traumatic pathology.
•
Agitated patients may require facilitated cooperation to ensure adequate preoxygenation.
•
Ketamine is an appropriate agent to facilitate cooperation in agitated patients in preparation for airway management.
•
Always be prepared to provide definitive airway intervention before administering sedation.
.61,62
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Maxillofacial Injuries
Maxillofacial fractures may present dramatically and affect airway management in one of several ways.63
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Posterior displacement from fractured maxillofacial segments may cause soft tissue collapse and occlude the airway, which may be worsened by the presence of c-spine collar.64,65
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Bleeding may be significant and cause airway management challenges, as previously discussed. In the supine position, the pooling of blood in the oropharynx may stimulate a gag response or vomiting, which in turn may worsen bleeding. Although patients with mandibular fractures in 2 or more locations may be easier to intubate due to increased mobility of the mandible and attached soft tissues, associated condylar fractures may cause a mechanical obstruction limiting mouth opening, making laryngoscopy and intubation difficult.64,66
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Maxillofacial fractures may also cause trismus that may resolve with the neuromuscular blockade; however, differentiating this from a mechanical obstruction before intubation is required and is often difficult.
Airway management begins with careful consideration to patient positioning. Awake, neurologically intact patients without neck pain should be allowed to position themselves however they are most comfortable to control tissue obstruction and allow drainage of blood and secretions. They may be given a rigid suction catheter to use themselves, which is more often tolerated, effective, and less likely to stimulate a gag and resultant vomiting. Adherence to protocols requiring rigid spinal immobilization and supine positioning may result in catastrophe.
The provider should presume that preoxygenation in patients with facial trauma may be difficult, and that reoxygenation with mask ventilation during RSI if the first attempt is unsuccessful may be difficult or impossible. Distortion of facial structures may make obtaining a seal with a BVM device difficult and patients may poorly tolerate PPV, as disruption of tissues may result in worsening bleeding and in cases of associated lower airway trauma, significant subcutaneous emphysema. Practitioners must proceed with the assumption that structural collapse of the airway may occur during an RSI.
The choice of approach is based on the patient’s ability to maintain a patent airway and their oxygenation status. For a “have no time” scenario (obstructing and hypoxemic), the primary approach may require a FONA, facilitated by a dissociative ketamine dosing. Alternatively, a “double set-up” may be used: RSI with a single attempt at oral intubation followed immediately by FONA rescue if needed (Fig.3).
Figure viewer
Fig.3 Approach to maxillofacial trauma. OR, operating room. a Facilitated cooperation using ketamine.
(Adapted from Mercer SJ, Jones CP, Bridge M, et al. Systematic review of the anaesthetic management of non-iatrogenic acute adult airway trauma. Br J Anaesth 2016;117(Suppl 1):i55; with permission.)
If the patient is maintaining adequate oxygenation, the clinician should proceed with a focused physical examination to assess the specific pattern of facial injury and plan accordingly. For example, swelling and tenderness at the temporomandibular joints suggests the presence of condylar factures and the possibility of mechanical trismus. Because anticipated difficulty in patients with facial trauma may involve challenges with intubation, mask ventilation, and possibly supraglottic airway rescue, preservation of spontaneous respiration during attempts to secure the airway should be considered. These patients may be best served when feasible by an “awake” approach with a laryngoscope or flexible intubating endoscope (FIE) or in selected cases a primary FONA
Management pearls for the patient with facial injuries
•
These patients require careful assessment of damaged anatomy recognizing the unique airway complications associated with facial fractures.
•
Both laryngoscopy and mask ventilation may be challenging and a double set-up should be prepared for when rapid sequence intubation (RSI) is the chosen approach.
•
An awake approach, although not always practical, should be considered.
•
Management of aggressive bleeding should be anticipated.
•
Allow patients to assume a position of comfort when safe to do so.
.
The Traumatized Airway
Airway management for the patient with a primary injury to the larynx or trachea is a high-stakes scenario, in which the loss of a stable airway can happen rapidly and with little warning. Suspicion of a traumatized airway should initiate a call for help to an experienced colleague.
Primary airway trauma is relatively uncommon in the civilian urban setting, with a reported incidence of less than 1% (0.4% for blunt and 4.5% for penetrating injuries).67
67.
Kummer, C. ∙ Netto, F.S. ∙ Rizoli, S. ...
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Accordingly, practitioners have infrequent or limited experience in managing these patients and existing management guidelines for care of these patients are mostly based on expert opinion.68
68.
Mercer, S.J. ∙ Jones, C.P. ∙ Bridge, M. ...
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Clinical findings suggestive of significant laryngotracheal airway injury include dysphagia, hoarseness, stridor, bleeding in the upper airway, subcutaneous emphysema, expanding hematoma, or in open penetrating injuries, obvious disruption of the larynx or trachea. If airway injury is suspected, aggressive PPV should be avoided. PPV in the setting of airway disruption can create or worsen pneumothorax, pneumomediastinum, or subcutaneous emphysema. Massive subcutaneous emphysema can distort airway anatomy, further complicating management. A potentially catastrophic complication is the conversion of a partial tracheal transection into a complete transection with the force of blindly passing an ETT or a bougie, particularly if relying on distal “hold-up” to confirm placement.68,69
68.
Mercer, S.J. ∙ Jones, C.P. ∙ Bridge, M. ...
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For patients with known or suspected airway injury, the safest way to facilitate ETI is placement of the ETT under direct visualization, ideally from the oropharynx to the carina using an FIE.68,70
68.
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Horton, C.L. ∙ Iii, C.A.B. ∙ Raja, A.S. ...
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Mercer and colleagues68
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Mercer, S.J. ∙ Jones, C.P. ∙ Bridge, M. ...
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described an approach to managing patients with suspected laryngotracheal injury (Fig.4). In the “have time” scenario in the patient who is oxygenating with minimal assistance, awake, and cooperative, the airway can be approached either from above after adequate topicalization, using a FIE, or infraglottically with a FONA approach (most commonly an awake tracheostomy) depending on the location of the injury.
Figure viewer
Fig.4 Approach to laryngotracheal trauma. a Facilitated cooperation using ketamine.
(Adapted from Mercer SJ, Jones CP, Bridge M, et al. Systematic review of the anaesthetic management of non-iatrogenic acute adult airway trauma. Br J Anaesth 2016;117(Suppl 1):i56; with permission.)
Careful titration of ketamine is often desirable to improve comfort and cooperation while maintaining airway reflexes. Use of an FIE allows both a diagnostic and therapeutic advantage: visualizing the specific pattern of injury, and facilitating careful ETT placement, while avoiding conversion to a complete transection. If an area of partial injury is identified, the FIE can be advanced distally and used as a guide to ensure safe advancement of the ETT beyond the site of injury.
In a “have no time” situation with a deteriorating patient in whom an RSI is considered the only viable option, a double set-up is mandatory, recognizing that the level of airway breach will influence whether FONA can occur through the cricothyroid membrane or if a tracheostomy is required. One option to improve visual navigation past the airway injury is to use a VL-assisted flexible endoscopic intubation.71
71.
Sowers, N. ∙ Kovacs, G.
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In this scenario, an RSI is initiated in the usual fashion and a VL is used to control soft tissues and obtain a view of the glottic inlet. A flexible intubating endoscope is then advanced through the vocal cords with the visual aid of VL distally to the carina facilitating inspection of the airway, and the ETT is advanced over the FIE into position
Management pearls for the patient with a primary airway injury
•
Decompensation in the patient with a traumatized airway may be rapid and catastrophic.
•
PPV should be avoided if possible.
•
An awake approach with appropriate topicalization is the preferred approach.
•
If an RSI is chosen, a double set-up with a FONA plan for accessing the trachea based on the level of the airway breach.
•
ETT placement should ideally be performed with visualization of the airway using a flexible intubating endoscope (FIE).
•
Advanced techniques using FIE either primarily in an awake patient or assisted by VL when an RSI is chosen are recommended when resources and skill are available.
.
The awake intubation
A successful awake intubation is dependent on careful patient selection, and, in particular, identifying anatomic, pathologic, or physiologic features that would make RSI problematic. There are several specific patient populations, including the burn patient and the patient with penetrating neck injuries, in whom an awake intubation may be the approach of choice, as a strategy to mitigate both predicted difficulty and anticipated dynamic changes in airway anatomy and physiology.
The awake intubation is a “have time” approach, involving placement of an ETT following adequate topicalization in a patient who is able to maintain spontaneous respirations. It is not device-specific and can be performed using DL, VL, or an FIE. Success with awake intubation is dependent on meticulous airway topicalization, and in general requires an awake and cooperative patient.72,73
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Higgs, A. ∙ Cook, T.M. ∙ McGrath, B.A.
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The use of sedation is not routine, and has been associated with increased awake intubation failures.74
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Specifically, sedation should never be used in place of adequate airway topicalization. A difficult airway paradox exists here: patients identified as difficult are selected to undergo a technically more challenging awake approach, a procedure that is performed infrequently by most emergency physicians. There is no simple answer to this resource, skill availability dilemma. It is our opinion that physicians who are responsible for acute airway management should acquire and maintain the skills required for awake intubation, as it can be a lifesaving approach in a specific subset of dynamic airway situations.
Rapid sequence intubation
RSI involves the rapid administration of an induction agent and a neuromuscular blocking agent in quick succession to facilitate ETT placement in a patient who is presumed to have a full stomach. RSI is the most common approach for airway management in trauma.75,76
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Oxygenation with or without ventilation during the procedure (referred to by some as a “modified” RSI) is considered standard by most acute care practitioners.77–79
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Historically, the application of cricoid pressure (CP) to prevent passive aspiration has been considered an essential component of an RSI. However, its routine use remains controversial, with some evidence suggesting it may make various aspects of airway management more challenging. If cricoid pressure is being applied and the practitioner experiences difficulty with laryngoscopy, intubation, or ventilation, CP should be immediately discontinued.80,81
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The “rapid” part of an RSI refers to the delivery of the induction drug and neuromuscular blocking agent, and is not meant to imply a hurried or rushed process. RSI in underresuscitated patients may result in unintended poor outcomes, including critical hypoxemia and circulatory collapse.82,83
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The term “resuscitative sequence intubation” has been suggested as a more representative term used to describe the preparation and optimization of the patient’s physiologic status before definitive airway management.84
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In general, checklists should be simple, use terminology that is clearly understood by the entire team, and can be performed rapidly (Fig.5).
Figure viewer
Fig.5 Intubation checklist example. BP, blood pressure; BVM: bag-valve mask; CXR, chest radiograph; DL: direct laryngoscope; ETI: endotracheal intubation; ETT: endotracheal tube; HFNO: high flow nasal oxygen; LPM: leters per minute; MILS: manual in-line stabilization; NIV: non-invasive ventilation; NMBA: neuromuscular blocking agent; NRBM: non-rebreather mask; OG: orogastric; pulse ox, pulse oximetry. ReOx: reoxygenation; SGA: supraglottic airway; SI: shock index (SBP/HR); VL: video laryngoscope.
(FromSaferAirway.org Toolkit (Emergency Medicine Associates, Germantown, Maryland); with permission.)
Optimizing Hemodynamics
Hemodynamic instability in trauma is most commonly caused by hypovolemia due to hemorrhage. Intravascular depletion shifts the gradient between right atrial pressure and mean systemic pressure, reducing preload and subsequently mean arterial pressure. The transition to PPV, either through a closed system BVM with an attached with positive end-expiratory pressure (PEEP) valve or with mechanical ventilation once intubated, abruptly increases the intrathoracic pressure, further shifting this gradient.
Postintubation hypotension is defined as a systolic blood pressure (SBP)<90 mm Hg or a mean arterial pressure <65 mm Hg within 30 minutes of intubation.92
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Early use of blood products is recommended when the etiology of shock in trauma is believed to be from hemorrhage. Vasopressors have traditionally been avoided in trauma based on concerns that they may worsen bleeding. During the peri-intubation phase, early aggressive use of blood products and hemorrhage control measures (splinting, pelvic binding) should remain the mainstay of resuscitation efforts. Careful use of vasopressors should be strongly considered in at-risk head-injured patients to prevent or manage postintubation hypotension. Sustained vasopressor use by infusion beyond the immediate peri-intubation period may be required to mitigate the effects of PPV, postparalysis sedation, and ongoing losses unresponsive to fluid and blood product replacement.
All commonly used induction agents will cause hypotension, particularly when a full dose is administered to a volume-constricted patient. Dosage recommendations for RSI induction agents are largely based on patients without hemodynamic instability. As such, in trauma patients with hypotension or a shock index greater than 0.8, it seems prudent that the dose of the induction agent be reduced by at least 50%.83,91,95,99
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In light of this, it is probably true to suggest that the dose of the drug is more important than the choice of drug.
Paralysis should have no direct effect on the patient’s hemodynamic status, and in hypoperfused states their dose should be increased. There is some debate regarding which neuromuscular blocking agent is superior for an RSI: both succinylcholine and rocuronium may be safely used and can provide good intubating conditions.101
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It should be emphasized, however, that administering too small a dose (<1.0 mg/kg) of rocuronium, particularly in low-flow states, will result in inadequate intubating conditions and a larger dose is recommended (1.2–1.6 mg/kg).101,102
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Although clinicians may be weary of the prolonged effect of high-dose rocuronium in potentially difficult airway cases, extended, deep paralysis is in fact desirable in this circumstance, as it helps to create optimal conditions for laryngoscopy, BVM and SGA ventilation, and FONA. Having the sick trauma patient wake up to a state that he or she will be able rescue his or her own airway is simply not realistic, and will make efforts to secure the airway even more difficult
Management of peri-intubation hemodynamic instability
•
Resuscitation using blood products (packed red blood cells/massive transfusion) should be done early in the preintubation phase of trauma management.
•
In selected scenarios consider the use of vasopressors during the peri-intubation phase.
•
Reduce the dose of all induction agents by at least 50% and increase the dose of the paralytic.
.
Avoiding Hypoxemia
Hypoxemia during emergency department (ED) RSI is common, occurring in more than one-third of cases.24
24.
Bodily, J.B. ∙ Webb, H.R. ∙ Weiss, S.J. ...
Incidence and duration of continuously measured oxygen desaturation during emergency department intubation
Ann Emerg Med. 2015; 1-7
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Proceeding with an RSI in a patient who is already hypoxemic can result in catastrophic complications. Patients with preintubation oxygen saturations of less than 95% are at risk of abrupt desaturation within 90 seconds of the onset of apnea.103
103.
Weingart, S.D. ∙ Levitan, R.M.
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Ann Emerg Med. 2012; 59:165-175
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Although recent literature has focused on extending safe apnea time with passive high-flow nasal oxygenation (HFNO) with or without noninvasive ventilation, the most important determinant of time to desaturation remains preoxygenation status.104
104.
Wong, D.T. ∙ Yee, A.J. ∙ May Leong, S. ...
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Although high oxygen saturation is reassuring, it is not a true measurement of oxygen reserve. Furthermore, oxygen saturation alone provides little information about the true safe apnea time, which is defined by the relationship between oxygen reserve and the rate of oxygen consumption. Effective management of the preoxygenation phase requires both increasing the patient’s oxygen reserve while simultaneously decreasing the rate of oxygen consumption through aggressive resuscitation. Increasing the oxygen reserve has 3 components: denitrogenation of the lungs, recruitment of alveoli with PEEP, and apneic oxygenation.
Denitrogenation is the primary physiologic mechanism of preoxygenation and is dependent on delivery of a high concentration of oxygen, resulting in a 10-fold increase available oxygen. Increased functional residual capacity (FRC) provides an ongoing reservoir to keep hemoglobin saturated. Delivery of close to 100% oxygen for approximately 4 minutes is required to denitrogenate normal lungs and may be most easily accomplished with high-flow, “flush-rate” (40 LPM) oxygen using a conventional non-rebreather mask.105
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Driver, B.E. ∙ Prekker, M.E. ∙ Kornas, R.L. ...
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In patients with a high minute ventilation or shunt physiology, a closed system BVM with a PEEP valve and a tight-fitting mask is the preferred preoxygenation technique. If respiratory effort support is required, this is best achieved using a mechanical (NIV) as opposed to manual ventilator (BVM).
With atelectasis, pulmonary contusion, and other lung pathologies, the FRC is diminished and the resultant shunt physiology renders preoxygenation with a high Fi o 2 less effective in preventing desaturation.106–108
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Alveolar recruitment using an oxygen delivery device with PEEP is necessary to help mitigate the negative effects of shunt physiology. A PEEP valve should be considered standard when using a BVM for preoxygenation, as it will prevent entrainment in BVMs without a dedicated expiratory valve. A PEEP valve attached to a BVM and applied over conventional nasal prongs at high flow (≥15 LPM) in a spontaneously breathing patient will produce continuous positive airway pressure–like conditions. Assisted ventilations are best delivered using a dedicated noninvasive ventilator; however, can be performed with the same BVM/PEEP, nasal cannula combination. It is advisable to use a pressure manometer attached to the BVM when using this combination to minimize high pressures that may result in gastric distention and aspiration.
Alveolar oxygen delivery that continues without respiratory effort is referred to as apneic oxygenation (AO). AO is facilitated by the pressure gradient between the oropharynx and the alveoli created by the differential uptake of oxygen and delivery of CO2 to and from the alveoli, resulting in the passive transfer of oxygen.104,107
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By continuously replenishing the FRC oxygen stores, apneic oxygenation using conventional or specialized nasal prongs to administer high-flow oxygen (10–70 LPM) may extend the safe apnea time after an RSI.103,104
103.
Weingart, S.D. ∙ Levitan, R.M.
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Numerous studies have evaluated to effectiveness of HFNO as an adjunct to preoxygenation for RSI, and the balance of evidence suggest the procedure as both safe and effective.104
104.
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Of note, the ability to extend the safe apnea time must not allow providers to become cavalier and should not encourage prolonged intubation attempts.106,109
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By the time peripheral oxygen saturation begins to fall, cerebral hypoxemia has already occurred, a phenomenon known as “pulse-ox lag.”103
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Effective situation awareness is required, even (and perhaps especially) when supporting gas exchange by way of passive oxygenation, to stay within the “safe apnea” zone
Preoxygenation: “the rule of 2s”
•
Elevate the head (ear to sternum) and the bed greater than 20° (reverse Trendelenburg).
•
Two sources of oxygen for all critically ill patients: high-flow nasal prongs ≥15 L/min and NRB/bag-mask ventilation ≥15 L/min.
•
Two approaches for obstruction: OPA with a jaw thrust for soft tissue obstruction.
•
Two attachments for your BVM: positive end-expiratory pressure valve and pressure manometer.
•
Two hands on all face masks: to ensure closed system oxygenation and ventilation and perform an aggressive jaw thrust.
•
Two providers: the most experienced obtaining a tight mask seal and aggressive jaw thrust giving feedback to the provider squeezing the bag to avoid overventilation and hyperventilation.
Abbreviations: BVM, bag-valve-mask; NRB, non-rebreather mask; OPA, oropharyngeal airway.
.
Front of neck airway to secure the airway
Numerous methods are used to access the trachea infraglottically, and the terminology describing the procedure is as variable as the techniques available to do so. Perhaps most accurate is the new term “front of neck airway” (FONA), which eliminates ambiguous verbiage like “surgical airway” or “airway rescue.”110
110.
Pracy JP, Brennan L, Cook TM, et al. Surgical intervention during a can’t intubate can’t oxygenate ( CICO ) event: emergency front-of-neck airway ( FONA )?. Br J Anaesth 2016;117(4):426–8
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Although rare (0.05%–1.7% of ED-based intubations), the decision to perform FONA must begin during the initial assessment of the patient’s airway, long before the “cannot intubate cannot oxygenate” (CICO) scenario is encountered.111
111.
Law, J.A. ∙ Broemling, N. ∙ Cooper, R.M. ...
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This begins with a pre-procedure briefing with team members to define clear triggers for moving ahead with the procedure. Palpation of the anterior neck, and perhaps even marking the anticipated location of the cricothyroid membrane should be done routinely for all emergency airway cases and equipment should be both familiar to team members and immediately available.
Cognitive and team-based preparation is vital, as the decision to proceed with a FONA is often delayed until critical hypoxemia has occurred.111
111.
Law, J.A. ∙ Broemling, N. ∙ Cooper, R.M. ...
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It is widely speculated that the most significant delay is often the result of hesitant decision-making and a reluctance to perform a rarely encountered procedure.112
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Open discussion of the emergency surgical airway as a potential outcome familiarizes the team and normalizes the procedure, shifting from the negative connotation of the “failed airway” to the recognition of the ultimately inevitable surgical airway.113
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By normalizing the procedure, at least cognitively, we aim to reduce the psychological distress associated with it.
No specific oxygen saturation level should be used as a trigger to perform FONA, recognizing that a failed oxygenation situation is dynamic and characterized by a rapidly falling oxygen saturation despite maximal efforts to reoxygenate the patient.111
111.
Law, J.A. ∙ Broemling, N. ∙ Cooper, R.M. ...
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Failed intubation followed by difficult mask ventilation with falling saturations represents an impending FONA that should be initiated after a single rescue attempt with supraglottic airway device. In rare situations, FONA may be the first and only invasive airway technique attempted, even in the setting of normal oxygen saturation; for example, massive facial trauma disrupting all recognizable airway landmarks. This “surgically inevitable” airway needs to be identified and declared early, such that time is not wasted on fruitless efforts to intubate “from above.”
There are several options for the emergency FONA and there remains some controversy regarding the preferred approach. Historically, methods such as transtracheal jet ventilation and percutaneous cricothyroidotomy with a Seldinger technique, have been advocated. However, recent reviews have shown that complications associated with jet ventilation are unacceptably high and wire-guided approaches are not as easy or successful as once believed.114–117
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Cook, T.M. ∙ Woodall, N. ∙ Frerk, C.
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There has been a recent push to adopt a modified open technique using a scalpel, finger, and bougie (the “scalpel-bougie” technique). The 2015 Difficult Airway Society guidelines recommend that all clinicians responsible for airway management be able to perform a FONA, and that the scalpel-bougie technique is the technique of choice.89
89.
Frerk, C. ∙ Mitchell, V.S. ∙ Mcnarry, A.F. ...
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The scalpel-bougie technique has several advantages for the emergency clinician: it relies on gross motor skills (which are more likely to be preserved during periods of acute stress), uses familiar equipment (scalpel, bougie, and a #6 ETT), and has a minimal number of steps.
The body of evidence for this rarely needed procedure is (and will likely remain) limited, yet all clinicians need to be mentally and technically prepared to rapidly perform front of neck access to secure the airway. Success for high-acuity, low-opportunity events like this requires frequent, deliberate practice using simulation.118
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Petrosoniak, A. ∙ Hicks, C.M.
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FONA trainers need not be expensive or complicated; motor habit can be developed using simple models with Venturi oxygen tubing or an empty roll of bathroom tissue.
Summary
Effective trauma care requires a team approach, with resuscitation priorities clearly communicated and interventions guided by the physiologic priorities that ensure adequate oxygen delivery. Although ensuring oxygenation and ventilation are priorities, airway management as the technical imperative of putting the “tube in the hole” must not overshadow other resuscitative elements.
Providing advanced airway management is part of the A and B and C parallel resuscitative priorities of trauma care. Safely managing both the anticipated and unanticipated difficult airway requires technical expertise; however, decisions of when and how to intervene are equally important determinants of outcome. Airway management in trauma begins as soon as patient contact is made and rarely starts with placement of an ETT. Gaining intravenous access, beginning fluid resuscitation, and applying oxygen may be lifesaving and/or bridging interventions that allow for the safe execution of downstream more definitive procedures. Whether the team is a doctor, nurse, and transporting paramedics or a group of 10, success is dependent on a shared understanding of the importance of resuscitation before intubation and clear communication of what and when various airway interventions will be performed. Then, securing the airway will have the best chance of making a positive difference in trauma patient outcomes.
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The Internet Encyclopedia of Philosophy (IEP) was founded in 1995 to provide open access to detailed, scholarly information on key topics and philosophers in all areas of philosophy. The Encyclopedia's articles are written with the intention that most of the article can be understood by advanced undergraduates majoring in philosophy and by other scholars who are not working in the field covered by that article. The IEP articles are written by experts but not for experts in analogy to the way the Scientific American magazine is written by scientific experts but not primarily for scientific experts.
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Anatomy and Physiology 2e
23.2 Digestive System Processes and Regulation
Anatomy and Physiology 2e
23.2
Digestive System Processes and Regulation
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Learning Objectives
By the end of this section, you will be able to:
Discuss six fundamental activities of the digestive system, giving an example of each
Compare and contrast the neural and hormonal controls involved in digestion
The digestive system uses mechanical and chemical activities to break food down into absorbable substances during its journey through the digestive system. Table 23.3 provides an overview of the basic functions of the digestive organs.
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Visit this site for an overview of digestion of food in different regions of the digestive tract. Note the route of non-fat nutrients from the small intestine to their release as nutrients to the body.
Functions of the Digestive Organs
| Organ | Major functions | Other functions |
---
| Mouth | Ingests food Chews and mixes food Begins chemical breakdown of carbohydrates Moves food into the pharynx Begins breakdown of lipids via lingual lipase | Moistens and dissolves food, allowing you to taste it Cleans and lubricates the teeth and oral cavity Has some antimicrobial activity |
| Pharynx | Propels food from the oral cavity to the esophagus | Lubricates food and passageways |
| Esophagus | Propels food to the stomach | Lubricates food and passageways |
| Stomach | Mixes and churns food with gastric juices to form chyme Begins chemical breakdown of proteins Releases food into the duodenum as chyme Absorbs some fat-soluble substances (for example, alcohol, aspirin) Possesses antimicrobial functions | Stimulates protein-digesting enzymes Secretes intrinsic factor required for vitamin B12 absorption in small intestine |
| Small intestine | Mixes chyme with digestive juices Propels food at a rate slow enough for digestion and absorption Absorbs breakdown products of carbohydrates, proteins, lipids, and nucleic acids, along with vitamins, minerals, and water Performs physical digestion via segmentation | Provides optimal medium for enzymatic activity |
| Accessory organs | Liver: produces bile salts, which emulsify lipids, aiding their digestion and absorption Gallbladder: stores, concentrates, and releases bile Pancreas: produces digestive enzymes and bicarbonate | Bicarbonate-rich pancreatic juices help neutralize acidic chyme and provide optimal environment for enzymatic activity |
| Large intestine | Further breaks down food residues Absorbs most residual water, electrolytes, and vitamins produced by enteric bacteria Propels feces toward rectum Eliminates feces | Food residue is concentrated and temporarily stored prior to defecation Mucus eases passage of feces through colon |
Table
23.3
Digestive Processes
The processes of digestion include six activities: ingestion, propulsion, mechanical or physical digestion, chemical digestion, absorption, and defecation.
The first of these processes, ingestion, refers to the entry of food into the alimentary canal through the mouth. There, the food is chewed and mixed with saliva, which contains enzymes that begin breaking down the carbohydrates in the food plus some lipid digestion via lingual lipase. Chewing increases the surface area of the food and allows an appropriately sized bolus to be produced.
Food leaves the mouth when the tongue and pharyngeal muscles propel it into the esophagus. This act of swallowing, the last voluntary act until defecation, is an example of propulsion, which refers to the movement of food through the digestive tract. It includes both the voluntary process of swallowing and the involuntary process of peristalsis. Peristalsis consists of sequential, alternating waves of contraction and relaxation of alimentary wall smooth muscles, which act to propel food along (Figure 23.5). These waves also play a role in mixing food with digestive juices. Peristalsis is so powerful that foods and liquids you swallow enter your stomach even if you are standing on your head.
Figure
23.5
Peristalsis
Peristalsis moves food through the digestive tract with alternating waves of muscle contraction and relaxation.
Digestion includes both mechanical and chemical processes. Mechanical digestion is a purely physical process that does not change the chemical nature of the food. Instead, it makes the food smaller to increase both surface area and mobility. It includes mastication, or chewing, as well as tongue movements that help break food into smaller bits and mix food with saliva. Although there may be a tendency to think that mechanical digestion is limited to the first steps of the digestive process, it occurs after the food leaves the mouth, as well. The mechanical churning of food in the stomach serves to further break it apart and expose more of its surface area to digestive juices, creating an acidic “soup” called chyme. Segmentation, which occurs mainly in the small intestine, consists of localized contractions of circular muscle of the muscularis layer of the alimentary canal. These contractions isolate small sections of the intestine, moving their contents back and forth while continuously subdividing, breaking up, and mixing the contents. By moving food back and forth in the intestinal lumen, segmentation mixes food with digestive juices and facilitates absorption.
In chemical digestion, starting in the mouth, digestive secretions break down complex food molecules into their chemical building blocks (for example, proteins into separate amino acids). These secretions vary in composition, but typically contain water, various enzymes, acids, and salts. The process is completed in the small intestine.
Food that has been broken down is of no value to the body unless it enters the bloodstream and its nutrients are put to work. This occurs through the process of absorption, which takes place primarily within the small intestine. There, most nutrients are absorbed from the lumen of the alimentary canal into the bloodstream through the epithelial cells that make up the mucosa. Lipids are absorbed into lacteals and are transported via the lymphatic vessels to the bloodstream (the subclavian veins near the heart). The details of these processes will be discussed later.
In defecation, the final step in digestion, undigested materials are removed from the body as feces.
Aging and the...
Digestive System: From Appetite Suppression to Constipation
Age-related changes in the digestive system begin in the mouth and can affect virtually every aspect of the digestive system. Taste buds become less sensitive, so food isn’t as appetizing as it once was. A slice of pizza is a challenge, not a treat, when you have lost teeth, your gums are diseased, and your salivary glands aren’t producing enough saliva. Swallowing can be difficult, and ingested food moves slowly through the alimentary canal because of reduced strength and tone of muscular tissue. Neurosensory feedback is also dampened, slowing the transmission of messages that stimulate the release of enzymes and hormones.
Pathologies that affect the digestive organs—such as hiatal hernia, gastritis, and peptic ulcer disease—can occur at greater frequencies as you age. Problems in the small intestine may include duodenal ulcers, maldigestion, and malabsorption. Problems in the large intestine include hemorrhoids, diverticular disease, and constipation. Conditions that affect the function of accessory organs—and their abilities to deliver pancreatic enzymes and bile to the small intestine—include jaundice, acute pancreatitis, cirrhosis, and gallstones.
In some cases, a single organ is in charge of a digestive process. For example, ingestion occurs only in the mouth and defecation only in the anus. However, most digestive processes involve the interaction of several organs and occur gradually as food moves through the alimentary canal (Figure 23.6).
Figure
23.6
Digestive Processes
The digestive processes are ingestion, propulsion, mechanical digestion, chemical digestion, absorption, and defecation.
Some chemical digestion occurs in the mouth. Some absorption can occur in the mouth and stomach, for example, alcohol and aspirin.
Regulatory Mechanisms
Neural and endocrine regulatory mechanisms work to maintain the optimal conditions in the lumen needed for digestion and absorption. These regulatory mechanisms, which stimulate digestive activity through mechanical and chemical activity, are controlled both extrinsically and intrinsically.
Neural Controls
The walls of the alimentary canal contain a variety of sensors that help regulate digestive functions. These include mechanoreceptors, chemoreceptors, and osmoreceptors, which are capable of detecting mechanical, chemical, and osmotic stimuli, respectively. For example, these receptors can sense when the presence of food has caused the stomach to expand, whether food particles have been sufficiently broken down, how much liquid is present, and the type of nutrients in the food (lipids, carbohydrates, and/or proteins). Stimulation of these receptors provokes an appropriate reflex that furthers the process of digestion. This may entail sending a message that activates the glands that secrete digestive juices into the lumen, or it may mean the stimulation of muscles within the alimentary canal, thereby activating peristalsis and segmentation that move food along the intestinal tract.
The walls of the entire alimentary canal are embedded with nerve plexuses that interact with the central nervous system and other nerve plexuses—either within the same digestive organ or in different ones. These interactions prompt several types of reflexes. Extrinsic nerve plexuses orchestrate long reflexes, which involve the central and autonomic nervous systems and work in response to stimuli from outside the digestive system. Short reflexes, on the other hand, are orchestrated by intrinsic nerve plexuses within the alimentary canal wall. These two plexuses and their connections were introduced earlier as the enteric nervous system. Short reflexes regulate activities in one area of the digestive tract and may coordinate local peristaltic movements and stimulate digestive secretions. For example, the sight, smell, and taste of food initiate long reflexes that begin with a sensory neuron delivering a signal to the medulla oblongata. The response to the signal is to stimulate cells in the stomach to begin secreting digestive juices in preparation for incoming food. In contrast, food that distends the stomach initiates short reflexes that cause cells in the stomach wall to increase their secretion of digestive juices.
Hormonal Controls
A variety of hormones are involved in the digestive process. The main digestive hormone of the stomach is gastrin, which is secreted in response to the presence of food. Gastrin stimulates the secretion of gastric acid by the parietal cells of the stomach mucosa. Other GI hormones are produced and act upon the gut and its accessory organs. Hormones produced by the duodenum include secretin, which stimulates a watery secretion of bicarbonate by the pancreas; cholecystokinin (CCK), which stimulates the secretion of pancreatic enzymes and bile from the liver and release of bile from the gallbladder; and gastric inhibitory peptide, which inhibits gastric secretion and slows gastric emptying and motility. These GI hormones are secreted by specialized epithelial cells, called endocrinocytes, located in the mucosal epithelium of the stomach and small intestine. These hormones then enter the bloodstream, through which they can reach their target organs.
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11071 | https://www.youtube.com/watch?v=FQTfWcpO3CQ | Parabola - Axis of Symmetry - Using X Intercepts and Using 2 Points With Same Y Coordinate
Radford Mathematics
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Posted: 1 Nov 2021
We learn how to find the equation of a parabola's axis of symmetry using its x intercepts, also called its zero or roots. The method is explained with several examples.
If a parabola cuts the x axis at x = p and x = q then the axis of symmetry has equation:
x = (p+q)/2
We also see that the axis of symmetry can be found using any two points on the parabola, provided they have the same y coordinate.
Transcript:
in this video we learn how to find the equation of the axis of symmetry of a parabola that is a quadratic functions curve using its x-intercepts which we also call zeros and we also call them roots and at the end of this video we'll even extend the method we learn to finding the equation of the axis of symmetry using any two points on a parabola provided they have the same y-coordinate that being said let's get started to learn the method i'm going to work through the examples we see here here's the idea looking at the first parabola we have here its axis of symmetry splits the entire parabola in two and looking at this parabola we can see that it crosses the x-axis twice here at one and here at five and since the axis of symmetry runs straight down the middle of these two values we can use them to find its equation and here's how the equation of this axis of symmetry is x equals to the average of the two x values at which the curve crosses the x-axis in other words it's equal to one plus five over two and that's equal to 6 over 2. finally this parabola's axis of symmetry has equation x equals to 3. and we're done and i can even label that x equals to 3. there we go looking at the next parabola we have here we can see that it crosses the x-axis at negative three and at one and so using the same method we can state that the axis of symmetry which i'm drawing right now there we go has equation x equals to negative three plus one over two that leads us to negative two over two finally x equals to negative one and once more i go ahead and box that result and i'll label the axis of symmetry x equals to negative one looking at one more example here we can see that this parabola crosses the x-axis at one and at seven and so without much effort we can find the equation of the axis of symmetry that i'm drawing right now indeed its axis of symmetry has equation x equals to one plus seven over two that's equal to eight over two finally x equals to four done and again i'll label the line x equals to four and the method we've just seen can be generalized with the following formula let me just make a bit of space like so we can go ahead and state that given a parabola that crosses the x-axis which i'm drawing right now that's my x-axis in such a way that the parabola crosses the x-axis at two values say p and q then its axis of symmetry which i'm drawing in a dotted line right now has equation x equals to p plus q over 2. and so if you want to memorize this with a formula this would be the one to go for and this method can in fact be extended indeed consider the following parabola i'm just drawing my xy grid here there we go that's my x and that's my y and let's say we have a parabola doing something like this and now let's say that i know the coordinates of two points along this parabola the first one is here with coordinates 1 6 and the second one is right here with coordinates five six now what's important to notice here is that both of these points have the same y-coordinate consequently they both lie on the horizontal line i'm drawing right now whose equation would be y equals to six now provided we have two points on a parabola who have the same y-coordinate we can calculate the equation of the axis of symmetry that i'm drawing right now by calculating the average of the two x-coordinates of these points so in this case the average of one and five and so for this parabola we could state that the equation of its axis of symmetry is x equals to one plus five over two that's equal to six over two finally x equals to three and there we go and i can label that on my graph as well x equals to three and this technique will always work provided the two points we're using have the same y-coordinate which was the case here indeed both of the points had a y-coordinate of 6 but it was also the case in the three examples we saw before indeed all of these zeros or x-intercepts or even roots we had seen all lay on the x-axis and consequently all had the same y-coordinate and so we see that all we need to find the equation of the axis of symmetry of a parabola are two points on the parabola with the same y-coordinate and those points can be zeros like what we saw initially or they can be two points like the ones we saw here and there we go we now know another method for calculating the equation of the axis of symmetry of a parabola and that's it for this tutorial |
11072 | https://www.sciencedirect.com/science/article/pii/S246802572030220X | Skip to article
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Green Energy & Environment
Volume 7, Issue 4, August 2022, Pages 772-781
Research paper Improved methanol synthesis performance of Cu/ZnO/Al2O3 catalyst by controlling its precursor structure
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Abstract
Methanol, a versatile chemical, fuel additive and potential H2 carrier, has attracted great attention. Despite of the wide industrialization, improvement of Cu-based methanol-synthesis catalysts is highly anticipated. Accordingly, a series of Cu/ZnO/Al2O3 with designed precursor structures were prepared, and its structurefunction relationship was investigated to make progress on this area. Results showed the catalyst derived from highly zinc-substituted malachite demonstrated the best catalytic performance in this work. It was found that the well-behaved catalyst possessed relatively high Cu specific surface area and exposed Cu concentration, and the well Cu/ZnO synergy. CuZn alloy was found by In-situ XRD tests, and its effect on the catalyst's thermostability was discussed. Fractional precipitation, which facilitated the Cu2+ substitution by Zn2+ in malachite lattice, could be an efficient preparation method of the Cu/ZnO/Al2O3 catalyst.
Graphical abstract
Fractional precipitation promoted zinc substitution in the malachite lattice. Derived from highly zinc-substituted malachite, sample I demonstrated better catalytic activity and thermostability during the methanol synthesis reaction from syngas.
Keywords
Zinc malachite
Cu2+ substitution
Hydrotalcite-like compounds
Fractional precipitation
In-situ XRD
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© 2020 Institute of Process Engineering, Chinese Academy of Sciences. Publishing services by Elsevier B.V. on behalf of KeAi Communications Co., Ltd. |
11073 | https://www.studypug.com/calculus-help/arc-length-and-surface-area-of-parametric-equations | Master Parametric Arc Length Formula: Calculus Essentials | StudyPug
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Now Playing:Arc length and surface area of parametric equations – Example 0a
Intros
Overview:
Overview: Arc Length of Parametric Equations
Overview: Surface Area of Parametric Equations
Examples
The Length of a Curve
Find the length of each of the given parametric equations:
1. x=e t sint x=e^t \sin t x=e t sin t y=e t cost y=e^t \cos t y=e t cos t where 0≤t≤2 π 0 \leq t \leq 2\pi 0≤t≤2 π
2. x=cos(θ) x=\cos (\theta) x=cos(θ) y=sin(θ)y=\sin (\theta)y=sin(θ) where 0≤θ≤π 0 \leq \theta \leq \pi 0≤θ≤π
The Surface Area of a Curve rotating about the x-axis
Find the surface area for each of the given parametric equations by rotating about the x x x-axis:
1. x=4 t−t 2 x=4t-t^2 x=4 t−t 2 y=2 t y=2t y=2 t where 0≤t≤3 0 \leq t \leq 3 0≤t≤3
2. x=r(θ−sinθ)x=r(\theta - \sin \theta) x=r(θ−sin θ) y=r(1−cosθ)y=r(1- \cos \theta )y=r(1−cos θ) where 0≤θ≤2 π,r>0 0 \leq \theta \leq 2\pi , \; r > 0 0≤θ≤2 π,r>0
Applications related to Circles and Spheres
You are given the parametric equations x=r cos(t)x=r\; \cos(t)x=r cos(t), y=r sin(t)y=r\;\sin(t)y=r sin(t) where 0≤t≤2 π 0 \leq t \leq 2\pi 0≤t≤2 π. Show that the circumference of a circle is 2 π r 2\pi r 2 π r
You are given the parametric equations x=r cos(t)x=r\; \cos(t)x=r cos(t), y=r sin(t)y=r\;\sin(t)y=r sin(t) where 0≤t≤π 0 \leq t \leq \pi 0≤t≤π. Show that the surface area of a sphere is 4 π r 2 4\pi r^2 4 π r 2
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Now Practicing:Arc Length And Surface Area Of Parametric Equations 1a
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Arc length and surface area of parametric equations
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Notes
In this lesson, we will learn how to find the arc length and surface area of parametric equations. To find the arc length, we have to integrate the square root of the sums of the squares of the derivatives. For surface area, it is actually very similar. If it is rotated around the x-axis, then all you have to do is add a few extra terms to the integral. Note that integrating these are very hard, and would require tons of trigonometric identity substitutions to make it simpler. We will first apply these formulas to some of the questions below. Then we will look at a case where using these formulas will give us much more simplified formulas in finding the arc length and surface areas of circles and spheres.
Let the curve be defined by the parametric equations x=f(t)x=f(t)x=f(t), y=g(t)y=g(t)y=g(t) and let the value of t t t be increasing from α\alpha α to β\beta β. Then we say that the formula for the length of the curve is:
L=∫α β(d x d t)2+(d y d t)2 d t L=\int_{\alpha}^{\beta}\sqrt{(\frac{dx}{dt})^2+(\frac{dy}{dt})^2}dt L=∫α β(d t d x)2+(d t d y)2d t
The formula to find the surface area is very similar.
If the curve is rotating around the x x x-axis, where f′,g′f', g'f′,g′ are continuous and g(t)≥0 g(t) \geq 0 g(t)≥0, then the formula for the surface area of the curve is
S A=∫α β 2 π y(d x d t)2+(d y d t)2 d t SA=\int_{\alpha}^{\beta} 2\pi y\sqrt{(\frac{dx}{dt})^2+(\frac{dy}{dt})^2}dt S A=∫α β2 π y(d t d x)2+(d t d y)2d t
Concept
Introduction to Arc Length and Surface Area of Parametric Equations
Welcome to our exploration of parametric equations and their applications in calculating arc length and surface area! These concepts are crucial in advanced mathematics and have wide-ranging applications in physics, engineering, and computer graphics. Our introduction video serves as an excellent starting point, providing a visual and intuitive understanding of these complex ideas. As we delve into this topic, you'll discover how parametric equations allow us to describe curves and surfaces in a more flexible way than traditional functions. We'll learn to compute arc length, which measures the distance along a curve, and surface area, which quantifies the extent of a surface in three-dimensional space. These tools are invaluable for analyzing shapes and movements in the real world. Whether you're a budding mathematician or an aspiring engineer, mastering these concepts will open up new possibilities in your studies and future career. Let's embark on this exciting journey together!
FAQs
What is the formula for the arc length of a parametric curve?
The formula for the arc length of a parametric curve is:
L = a b ((dx/dt)² + (dy/dt)²) dt
Where L is the arc length, a and b are the lower and upper bounds of the parameter t, and x(t) and y(t) are the parametric equations.
How do you calculate the surface area of a solid formed by rotating a parametric curve around the x-axis?
The formula for the surface area of a solid formed by rotating a parametric curve around the x-axis is:
S = 2π y(t) ((dx/dt)² + (dy/dt)²) dt
Where S is the surface area, y(t) is the y-coordinate of the parametric curve, and dx/dt and dy/dt are the derivatives of x and y with respect to t.
What are some practical applications of arc length and surface area calculations using parametric equations?
Practical applications include:
Analyzing trajectories of projectiles in physics
Designing roller coasters and amusement park rides
Creating 3D models in computer graphics and animation
Optimizing car body shapes for aerodynamics in the automotive industry
Calculating material requirements in architecture and engineering
What are some common challenges in calculating arc length and surface area?
Common challenges include:
Dealing with complex integrals
Applying appropriate trigonometric substitutions
Setting up correct limits of integration for surface area problems
Recognizing when to use symmetry or other simplification techniques
How can students improve their skills in solving parametric equation problems?
Students can improve their skills by:
Practicing regularly with a variety of problems
Studying worked examples and understanding each step
Using visualization tools and graphing calculators
Applying concepts to real-world scenarios
Seeking help from instructors or online resources when stuck
Prerequisites
Understanding the arc length and surface area of parametric equations is a crucial concept in advanced calculus. However, to fully grasp this topic, it's essential to have a solid foundation in certain prerequisite areas. Two key prerequisites that play a vital role in comprehending this subject are defining curves with parametric equations and the magnitude of a vector.
Firstly, a strong understanding of defining curves with parametric equations is crucial. This concept forms the basis for working with parametric equations in more complex scenarios. When dealing with arc length and surface area, you'll frequently encounter curves described parametrically. Knowing how to interpret and manipulate these equations is fundamental. For instance, the ability to distinguish between Cartesian equations and parametric equations becomes particularly important when calculating arc lengths, as the formulas differ depending on the equation type.
Secondly, familiarity with the magnitude of a vector is equally important. This concept is directly applicable when working with parametric equations, especially in the context of arc length and surface area calculations. The magnitude of a vector, particularly the magnitude of a velocity vector, is a key component in determining arc length. It helps in understanding the rate at which a curve is being traced and, consequently, its length.
When calculating arc length for parametric equations, you'll often need to integrate the magnitude of the velocity vector. This process combines your knowledge of parametric equations with vector operations, highlighting the interconnectedness of these prerequisite topics. Similarly, for surface area calculations, understanding vector magnitudes aids in comprehending the surface element and its contribution to the total area.
Moreover, these prerequisites provide the mathematical language and tools necessary for more advanced concepts. For example, the ability to work with parametric equations allows you to represent complex curves and surfaces that might be difficult or impossible to describe with standard Cartesian equations. This flexibility is crucial when dealing with real-world applications of arc length and surface area in fields like physics, engineering, and computer graphics.
In conclusion, mastering defining curves with parametric equations and understanding the magnitude of a vector are not just academic exercises. They are fundamental building blocks that enable you to tackle more complex problems involving arc length and surface area of parametric equations. By solidifying your knowledge in these areas, you'll be better equipped to understand, analyze, and solve advanced calculus problems, paving the way for a deeper appreciation of the subject and its wide-ranging applications.
Fundamental theorem of calculus
Integration using trigonometric identities
Defining curves with parametric equations
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11074 | https://books.google.com/books/about/Plane_Trigonometry.html?id=grQUWYejIq4C | Plane Trigonometry - Sidney Luxton Loney - Google Books
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Plane Trigonometry ================== Sidney Luxton Loney University Press, 1893 - Plane trigonometry - 480 pages Preview this book »
Selected pages
Title Page
Table of Contents
Index
Contents
Measure2
Trigonometrical Ratios for angles less than a right19
Simple problems in Heights and Distances40
CHAP49
the practical applications of Trigonometry It is60
General expressions for all angles having a given76
and90
L97
Inverse circular functions271
Expansions of sin ne cos ne and tan295
260302
271319
280329
288340
Exponential and Logarithmic Series349
Exponential Series for Complex Quantities363
More
Ratios of multiple and submultiple angles105
Principle of Proportional Parts159
Sides and Angles of a triangle175
Solution of triangles188
Heights and Distances209
CHAP217
Properties of a triangle226
Quadrilaterals249
Trigonometrical ratios of small angles260
Two important limits365
Logarithms of complex quantities382
Summation of Series404
Factors of x2 2xn cos ne+1416
Principle of Proportional Parts452
Errors of observation461
Miscellaneous Propositions468
461xxvi
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Plane Trigonometry (Classic Reprint)
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Common terms and phrases
A₁angle AOPangle increasesangle of elevationcentrecirclecircumcentrecircumcirclecoefficientcos²cos³coseccoshcosinecotangentdecreasesdenoteddescribed an anglediffdistanceequalEXAMPLESexpressionfeetFind the anglefind the heightfind the valuesflagstaffformulae of ArtHencehorizontal planeincircleinfiniteinscribedintegerlast articlelengthLet the revolvinglogalogarithmlogemilesmultiplenearlynumber of radiansorthocentrepedal triangleperpendicular to OAplaces of decimalspositive or negativeprovequadrilateralradiusregular polygonrespectivelyrevolving lineright anglesec²secantshewshewnSimilarlysin²sin³sinesinhtan²tangentTheoremtowertriangle ABCtrigonometrical functionstrigonometrical ratiosunityzeroβίηθηπθαπ π
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Page 36 - ... the three angles of a triangle are together equal to two right angles, although it is not known to all.
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Page 17 - Radian is the angle subtended, at the centre of a circle, by an arc equal in length to the radius...
Appears in 220 books from 1865-2008
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Page 44 - A person standing on the bank of a river observes that the angle subtended by a tree on the opposite bank is 50°; walking 40 ft.
Appears in 20 books from 1891-2006
Page 13 - All the interior angles of any rectilineal figure, together with four right angles, are equal to twice as many right angles as the figure has sidef.
Appears in 47 books from 1716-2001
Page 1 - Every circumference of a. circle, whether the circle be large or small, is supposed to be divided into 360 equal parts called degrees. Each degree is divided into 60 equal parts called minutes, and each minute into 60 equal parts called seconds.
Appears in 219 books from 1734-1996
Page 10 - If the radius of the earth be 4000 miles, what is the length of its circumference?
Appears in 8 books from 1845-1911
Page 42 - From the top of a cliff 150 ft. high the angles of depression of the top and bottom of a tower are 30° and 60°, respectively.
Appears in 36 books from 1848-2008
Page 220 - A ladder placed at an angle of 75° just reaches the sill of a window at a height of 27 feet above the ground on one side of a street. On turning the ladder over without moving its foot, it is found that when it rests against a wall on the other side of the street it is at an angle of 15° with the ground.
Appears in 8 books from 1876-1893
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Bibliographic information
Title Plane Trigonometry
Plane Trigonometry, Sidney Luxton Loney
AuthorSidney Luxton Loney
Publisher University Press, 1893
Original from the University of Michigan
Digitized Nov 23, 2005
Length 480 pages
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Automata and Computability ========================== Dexter C. Kozen Springer Science & Business Media, Jun 29, 2007 - Computers - 400 pages The aim of this textbook is to provide undergraduate students with an introduction to the basic theoretical models of computability, and to develop some of the model's rich and varied structure. Students who have already some experience with elementary discrete mathematics will find this a well-paced first course, and a number of supplementary chapters introduce more advanced concepts. The first part of the book is devoted to finite automata and their properties. Pushdown automata provide a broader class of models and enable the analysis of context-free languages. In the remaining chapters, Turing machines are introduced and the book culminates in discussions of effective computability, decidability, and Gödel's incompleteness theorems. Plenty of exercises are provided, ranging from the easy to the challenging. As a result, this text will make an ideal first course for students of computer science. More » Preview this book »
Selected pages
Title Page
Index
References
Contents
Course Road map and Historical Perspective 3
Strings and Sets 7
Finite Automata and Regular Sets 14
More on Regular Sets 19
Nondeterministic Finite Automata 25
The Subset Construction 32
Pattern Matching 40
Pattern Matching and Regular Expressions 44
More on Turing Machines 215
Equivalent Models 221
Universal Machines and Diagonalization 228
Decidable and Undecidable Problems 235
Reduction 239
Rices Theorem 245
Undecidable Problems About CFLs 249
Other Formalisms 256
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Regular Expressions and Finite Automata 49
Kleene Algebra and Regular Expressions 55
Homomorphisms 61
Limitations of Finite Automata 67
Using the Pumping Lemma 72
DFA State Minimization 77
A Minimization Algorithm 84
MyhillNerode Relations 89
The MyhillNerode Theorem 95
Collapsing Nondeterministic Automata 100
Automata on Terms 108
The MyhillNerode Theorem for Term Automata 114
TwoWay Finite Automata 119
2DFAs and Regular Sets 124
ContextFree Grammars and Languages 129
Balanced Parentheses 135
21 Normal Forms 140
The Pumping Lemma for CFLs 148
Pushdown Automata 157
Final State Versus Empty Stack 164
PDAs and CFGs 167
Simulating NPDAs by CFGs 172
Deterministic Pushdown Automata 176
Parsing 181
The CockeKasamiYounger Algorithm 191
The ChomskySchutzenberger Theorem 198
Parikhs Theorem 201
Turing Machines and Effective Computability 206
The ACalculus 262
While Programs 269
Beyond Undecidability 274
Godels Incompleteness Theorem 282
Proof of the Incompleteness Theorem 287
Godels Proof 292
Exercises 299
Homework 1 301
Homework 2 302
Homework 3 303
Homework 4 304
Homework 5 306
Homework 6 307
Homework 7 308
Homework 8 309
Homework 9 310
Homework 10 311
Homework 11 312
Homework 12 313
Finite Automata and Regular Sets 315
Pushdown Automata and ContextFree Languages 333
Turing Machines and Effective Computability 340
Hints for Selected Miscellaneous Exercises 351
Solutions to Selected Miscellaneous Exercises 357
References373
Notation and Abbreviations 381
Index389
Copyright
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Common terms and phrases
2DFAaccepts by emptyalgorithmautomatonaxiomsbinarybisimulationChomskycomputationconcatenationconfigurationcongruencecontext-free languagecorrespondingDCFLdefineddenotederivatione-transitionsempty stackencodingexampleexistsexternal queuefinite automatafinite controlfinite setformalGivegivengrammarGreibach normal formhalting problemHomeworkhomomorphisminduction hypothesisinfiniteinput alphabetinput stringinput symbolKleene algebraleft endmarkerleftmostlengthloopmarkedMiscellaneous ExerciseMyhill-Nerode relationMyhill-Nerode theoremnatural numbersnondeterministicnondeterministic finite automatonnonterminalnormal formNPDAnumber theoryoperatorparse treepebbleproductionsproofprovableProvepumping lemmar.e. setsrecursive setregular expressionregular setrejectRice's theoremscanningsentential formset of stringssimulatestack symbolstepsubstringSupplementary Lecturetapetransition functionTuring machineunaryundecidable
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Page 376 - SA Greibach, A new normal form theorem for context-free phrase structure grammars.
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Page 379 - MY Vardi. A note on the reduction of two-way automata to one-way automata. Information Processing Letters, 30(5):261-264, 1989.
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Page 373 - RC Backhouse. Closure Algorithms and the StarHeight Problem of Regular Languages. PhD thesis, Imperial College, 1975.
Appears in 7 books from 1991-2002
Page 379 - ... Syntactic Analysis and the Pushdown Store," Proceedings of Symposia on Applied Mathematics (Vol. 12). Providence, RI: American Mathematical Society, 1961. Theorem 3.4.1 on the equivalence of context-free languages and pushdown automata was proved independently by Schutzenberger, Chomsky, and Evey. MP SCHUTZENBERGER, "On Context-free Languages and Pushdown Automata," Information and Control, 6, no. 3 (1963), 246-264. N. CHOMSKY, "Context-free Grammar and Pushdown Storage," Quarterly Progress Report,...
Appears in 6 books from 1969-1998
Page 379 - A. YASUHARA, Recursive Function Theory and Logic, Academic Press, New York, 1971.
Appears in 8 books from 1951-2002
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References to this book
Engineering a Compiler
Keith D. Cooper,Linda Torczon
Limited preview - 2004
Graph Algebras and Automata
Andrei Kelarev
Limited preview - 2003
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Bibliographic information
Title Automata and Computability
Undergraduate Texts in Computer Science
AuthorDexter C. Kozen
Edition illustrated
Publisher Springer Science & Business Media, 2007
ISBN 0387949070, 9780387949079
Length 400 pages
SubjectsComputers
›
Programming
›
Algorithms
Computers / Information Technology
Computers / Machine Theory
Computers / Programming / Algorithms
Computers / User Interfaces
Export CitationBiBTeXEndNoteRefMan
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11076 | https://brainly.com/question/17209918 | [FREE] What is the total number of common tangents that can be drawn to the circles - brainly.com
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What is the total number of common tangents that can be drawn to the circles
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Explanation
The circles are concentric. I.e there is no point where the circumferences of both circles ever touch each other. Hence there can be no tangent line possible that can be common to both circles at the same time.
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The total number of common tangents between two circles depends on their positions. They can have 0, 1, 2, 3, or 4 tangents based on whether they are concentric, touching each other, intersecting, or distant. Therefore, the analysis of their relationship is crucial to determining the exact number.
Explanation
To determine the total number of common tangents that can be drawn to two circles, we need to consider their relative positions. There are several configurations that can occur between two circles:
Externally tangent circles (touching at one point): 3 common tangents (2 external and 1 internal)
Two circles intersecting at two points: 2 common tangents (both external)
One circle inside another (non-intersecting): 1 common tangent (external)
Concentric circles (one circle inside the other without touching): 0 common tangents
To find how many common tangents exist in any specific case, you would analyze the positions of the circles and their sizes. For instance, if two circles have equal radii, they can be concentric or externally tangent based on their centers' distance.
In general, the number of tangents can be summarized as follows:
If circles are distant enough, they can share 4 tangents (2 internal and 2 external).
If they touch externally, there are 3.
If they intersect, there are 2.
If one is inside the other without touching, there is 1.
If they are concentric, there are 0.
Thus, to get the complete picture when presented with specific circles, checking their relationship is essential to determine the correct number of tangents.
Examples & Evidence
For example, if two circles are concentric (one inside another without touching), there would be 0 common tangents. If they intersect at two points, there would be 2 common tangents. And if they are externally tangent, there are 3 common tangents, showing how their arrangement affects the number of lines that can touch both circles without crossing.
This information is based on established principles of geometry and the properties of circles regarding tangents and relationships between them.
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Community Answer 5.0 1 What is the total number of common tangents that can be drawn to the circles? A. 3 B. 2 C. 1 D. 0
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QUESTION 3 VERTICAL PROJECTILE MOTION (PRACTICE EXAM 5) The velocity - time graph below shows the motion of a ball that is thrown vertically upwards from the balcony of a building. It takes 0,5 for the ball to reach the highest point above the balcony, after which it falls past the balcony and strikes the ground. Ignore the effects of friction. 3.1 State the
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Question: QUESTION 3 VERTICAL PROJECTILE MOTION (PRACTICE EXAM 5) The velocity - time graph below shows the motion of a ball that is thrown vertically upwards from the balcony of a building. It takes 0,5 for the ball to reach the highest point above the balcony, after which it falls past the balcony and strikes the ground. Ignore the effects of friction. 3.1 State the
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The gradient of v t graph is the acceleration and is denoted by tan(θ) = height / base
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QUESTION 3 VERTICAL PROJECTILE MOTION (PRACTICE EXAM 5) The velocity - time graph below shows the motion of a ball that is thrown vertically upwards from the balcony of a building. It takes 0,5 for the ball to reach the highest point above the balcony, after which it falls past the balcony and strikes the ground. Ignore the effects of friction. 3.1 State the numerical value of: 3.1.1 The gradient of the above velocity - time graph. Provide a reason for your answer. (2) 3.1.2 Time, T x a as shown on the graph. (1) 3.2 Use ONLY the graph (and NOT equations of motion) to determine the maximum height the ball reaches above the balcony. (3) 3.3 Calculate, using equations of motions and data from the graph, the: 3.3.1 Velocity, x, with which the ball strikes the ground (3) 3.3.2 Height of the balcony above the ground (3) 3.4 Sketch an acceleration versus time graph for the motion of the hall. (2)
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11078 | https://essd.copernicus.org/articles/16/1137/2024/essd-16-1137-2024.pdf | Earth Syst. Sci. Data, 16, 1137–1149, 2024 © Author(s) 2024. This work is distributed under the Creative Commons Attribution 4.0 License.
An extensive spatiotemporal water quality dataset covering four decades (1980–2022) in China Jingyu Lin1,2, Peng Wang3, Jinzhu Wang4, Youping Zhou5, Xudong Zhou6, Pan Yang1,2, Hao Zhang7, Yanpeng Cai1,2, and Zhifeng Yang1,2 1Guangdong Provincial Key Laboratory of Water Quality Improvement and Ecological Restoration for Watersheds, School of Ecology, Environment and Resources, Guangdong University of Technology, Guangzhou 510006, China 2Southern Marine Science and Engineering Guangdong Laboratory (Guangzhou), Guangzhou 511458, China 3Coastal and Ocean Management Institute, College of the Environment and Ecology, Xiamen University, Xiamen 361102, China 4School of Life and Environmental Sciences, Deakin University, Burwood, Vic 3125, Australia 5Department of Ocean Science and Engineering, Southern University of Science and Technology (SUSTech), Shenzhen 518055, China 6Global Hydrological Prediction Center, Institute of Industrial Science, The University of Tokyo, Tokyo 153-8505, Japan 7CAS Key Laboratory of Tropical Marine Bio-Resources and Ecology, South China Sea Institute of Oceanology, Chinese Academy of Sciences, Guangzhou 510301, China Correspondence: Yanpeng Cai (yanpeng.cai@gdut.edu.cn) Received: 20 April 2023 – Discussion started: 15 May 2023 Revised: 31 October 2023 – Accepted: 21 November 2023 – Published: 27 February 2024 Abstract. Water quality data represent a critical resource for evaluation of the well-being of aquatic ecosystems and assurance of clean water sources for human populations. While the availability of water quality datasets is growing, the absence of a publicly accessible national water quality dataset for both inland and the ocean in China has been notable. To address this issue, we utilized R and Python programming languages to collect, tidy, reorganize, curate, and compile three publicly available datasets, thereby creating an extensive spatiotemporal repository of surface water quality data for China. Distinguished as the most expansive, clean, and easily acces-sible water quality dataset in China to date, this repository comprised over 330 000 observations encompassing daily (3588), weekly (217 751), and monthly (114 954) records of surface water quality covering the period from 1980 to 2022. It spanned 18 distinct indicators, meticulously gathered at 2384 monitoring sites, which were fur-ther categorized as daily (244 sites), weekly (149 sites), and monthly (1991 sites), ranging from inland locations to coastal and oceanic areas. This dataset will support studies relevant to the assessment, modeling, and projec-tion of water quality, ocean biomass, and biodiversity in China, and therefore make substantial contributions to both national and global water resources management.
This water quality dataset and supplementary metadata are available for download from the figshare repository at (Lin et al., 2023b).
Published by Copernicus Publications.
1138 J. Lin et al.: An extensive spatiotemporal water quality dataset covering four decades in China 1 Introduction The implications of the 2030 Agenda for Sustainable De-velopment necessitate the utilization of high-quality moni-toring data for the purpose of gauging progress and facil-itating evidence-based policymaking (Allen, 2021). Water, constituting the foundational pillar of sustainable develop-ment (WWAP, 2019), bears a profound interconnection with numerous targets within the Sustainable Development Goals (SDGs), notably SDG 6 (Sadoff et al., 2020), which endeav-ors to ensure the universal availability and sustainable man-agement of water and sanitation, and SDG 14, which focuses on the conservation and sustainable utilization of oceans, seas, and marine resources. With the campaign of ecologi-cal civilization and a series of marine policies (e.g., Maritime Power and Strategy, Chen et al., 2019), China is committed to the preservation of water resources while simultaneously ad-vancing resource management methodologies. To effectively accomplish the United Nations SDGs and align with China’s extensive policy frameworks, it is crucial to systematically compile water-related data across both inland and coastal and oceanic domains (Dai et al., 2022; Plagányi et al., 2023).
Within the context of the Source-to-Sea (S2S) aquatic contin-uum, water quality data emerge as a pivotal factor in discern-ing pollution levels (Regnier et al., 2022). This information plays a critical role in the preservation of water resources and the provision of sanitation services (WWAP, 2023).
Water quality refers to the selected physical, chemical, and biological characteristics of water that determine its suitabil-ity for a particular use (World Health Organization, 2017; Johnson et al., 1997). There are some key properties widely recognized for measuring water quality. In terms of physi-cal characteristics, key considerations include the color, tem-perature (TEMP), sediment content, turbidity, electrical con-ductivity, and the concentration of total suspended solids (TSSs) (Oteng-Peprah et al., 2018). Chemical constituents play a significant role in the determination of water qual-ity. These encompass parameters such as the potential of hy-drogen (pH), acidity levels, and indicators reflecting nutri-ent levels, including ammonia nitrogen (NH4N), nitrite ni-trogen (NO2N), and nitrate nitrogen (NO3N), and various forms of phosphorus such as dissolved inorganic phospho-rus (DIP) and total phosphorus (TP). Additionally, the con-centration of oxygen required for microorganisms to decom-pose organic matter is highly considered, which includes bio-chemical oxygen demand (BOD), chemical oxygen demand (COD), and dissolved oxygen (DO) (Hassan Omer, 2020).
Biological indicators provide insights into the presence, con-dition, and abundance of various living organisms within wa-ter bodies, such as bacteria, algae, and pathogens. Overall, these indicators are crucial for assessing water quality and ensuring the health of aquatic ecosystems and human popu-lations that rely on clean water sources.
Sustaining elevated water quality standards stands as an imperative requisite for the perpetuity of diverse spheres, en-compassing natural ecosystems, public health, and socioe-conomic systems. Contaminants such as excessive nutrients that enter water bodies can have detrimental effects on the integrity, functioning, and biodiversity of both riverine and oceanic ecosystems which provide a habitat for a diverse array of flora and fauna (Morin and Artigas, 2023). For in-stance, the influx of pesticides into aquatic systems has been unequivocally associated with the diminishment of aquatic species and perturbations in food chains (Stehle and Schulz, 2015). Consequently, the unwavering adherence to stringent water quality standards emerges as an imperative measure for ameliorating the adversative consequences, thereby safe-guarding fragile habitats, and preserving ecological equi-librium (Hering et al., 2010). Furthermore, the assurance of clean water represents a fundamental safeguard against the outbreak of waterborne maladies (Gleick and Palaniap-pan, 2010), with direct implications for the preservation of public health (Prüss-Ustün et al., 2014) and the concomi-tant mitigation of healthcare expenditures. Maladies such as cholera, typhoid, and hepatitis find direct causation in the in-adequacy of water quality (Leju Celestino Ladu et al., 2018).
Lastly, impaired water quality can have severe economic con-sequences, including reduced agricultural productivity, in-creased costs of water treatment, and damage to tourism industries reliant on pristine water bodies (United Nations, 2018).
The recognition of the significance of the water quality to nature, society, food, and security has accelerated the rise and availability of local, national, and global water quality datasets. For example, local water quality datasets include the dataset Water QUAlity, DIscharge and Catchment At-tributes providing data for 1386 German catchments for the purpose of studying the species of nitrogen, phosphorus, and organic carbon (Ebeling et al., 2022); a set of water chem-istry measurements including carbon species, dissolved nu-trients, and major ions to describe the biogeochemical con-ditions of permafrost-affected Arctic watersheds (Shogren et al., 2022); and a catchment-wide biogeochemical monitoring platform for capturing water temperature, pH, alkalinity, sus-pended solid, chlorophyll concentrations, and nutrient and cation data of the Thames basin in the United Kingdom to promote drinking water resource management (Bowes et al., 2018). The Water Quality Portal (WQP) comprises thousands of water quality variables encompassing physical conditions, chemical and bacteriological water analyses, chemical analy-ses of fish tissue, taxon abundance data, toxicity data, habitat assessment scores, and biological index scores, which was widely applied to several domains (e.g., to examine water clarity in lakes and reservoirs; Read et al., 2017). Aggregat-ing five large water quality datasets, the Global River Wa-ter Quality Archive (GRQA) has significantly expanded both the geographic and historical reach of existing water quality datasets by incorporating 42 parameters related to nutrient species, carbon content, sediment composition, and oxygen levels (Virro et al., 2021).
Earth Syst. Sci. Data, 16, 1137–1149, 2024 J. Lin et al.: An extensive spatiotemporal water quality dataset covering four decades in China 1139 Despite significant advances in open-data science for wa-ter quality research globally, Asia lags far behind other re-gions in this regard (Virro et al., 2021; Lin et al., 2023a).
As the largest country in East Asia, China’s water quality data are notably limited in the comprehensive global dataset, with a notable absence of data from coastal and oceanic re-gions. The publicly available data consist of a total of only 3595 daily observations from 24 sites, spanning from 1980 to 2009, as documented in GRQA. This is far from adequate for water quality analysis and modeling. Additionally, the water data available from open-data centers are stored in a user-unfriendly format that requires significant additional ef-forts to make them credible, editable, and reusable. For ex-ample, monthly water quality data spanning from 2006 to 2022 are presented as reports with figures derived from statis-tical analysis, instead of providing more reliable monitoring data. Although some studies have employed national-scale water quality data for assessment and modeling in China (Ma et al., 2020a, b; Huang et al., 2021; Zhang et al., 2022), these datasets are not publicly available due to licensing re-strictions and/or government sanctions (Lin et al., 2023a). To date, there is no clean and publicly accessible national water quality dataset that covers the entirety of China.
Therefore, there is a pressing need to reorganize, curate, and manage the continuous, long-time-series, standardized, well-organized, and consistent water quality datasets from inland to coastal and oceanic areas within China. These datasets represent invaluable resources to support researchers and decision-makers (Van Vliet et al., 2023). They enable an in-depth examination of water quality status, encompassing the entire spectrum from riverine environments to the vast expanse of the oceans. Furthermore, they provide the means to model various dimensions of water quality indicators and forecast the ramifications of emergent water pollution phe-nomena (i.e., coastal eutrophication and oceanic harmful al-gal blooms due to additional nitrogen input from land and re-leases of radionuclides from inland redundant nuclear power plant accidents). They are also valuable for the effective management of water resources to support the United Na-tions Water Action Decade (2018–2028) and Ocean Decade (2021–2030; Folke et al., 2021). Our water quality dataset is thus initiated to meet the huge demand for Chinese water quality data, to boost national water data sharing, and to ad-vance global water-related research and applications. It aims to collect non-sensitive and publicly available water quality data, to apply consistency to the formatting and curation, and to establish a standardized set of metadata for different water quality aspects.
2 Data and methods 2.1 Openly accessible data sources The Chinese surface water quality dataset presented herein is derived from three publicly accessible online data sources.
Details of these original datasets are provided in Table 1.
2.1.1 GRQA As the most comprehensive water quality dataset, GRQA has incorporated inland water quality data from five existing sources, including the Canadian Environmental Sustainabil-ity Indicators program, Global Freshwater Quality Database, GLObal RIver Chemistry database, European Environment Agency, and USGS WQP for selected 42 water quality pa-rameters (e.g., nutrients, carbon, oxygen, and sediments; Read et al., 2017; Virro et al., 2021) with globally 93 057 sites in total spanning from 1898 to 2020 (Table 1).
2.1.2 CNEMC As the most advanced and complete environmental data cen-ter, the China National Environmental Monitoring Centre (CNEMC) is an online information system managed by the agency of the China Ministry of Ecology and Environment.
The CNEMC was established in 1979 to monitor all envi-ronmental aspects (e.g., quality of air, water, soil), to provide publicly online data, to assess environmental impacts, and to report on the status of water environments for local and national governments. Water quality data available from this center included yearly water quality reports spanning from 2006 to 2022 ( 6.shtml, last access: 23 January 2024), 7 d moving average (weekly) inland water quality data stored in an individual WORD file or PDF file named by year with week number spanning from 2007 to 2018 (Table 1), real-time water qual-ity data for 11 indicators (TEMP, electrical conductivity, pH, DO, turbidity, CODMn, NH4H, TP, TN, chlorophyll, and al-gal density) with a frequency of 4 h ( 8070/GJZ/Business/Publish/Main.html, last access: 23 Jan-uary 2024), and real-time water quality data with a frequency of 1 month for 25 indicators (TEMP, electrical conductivity, pH, DO, turbidity, CODMn, BOD, BOD5, NH4H, TP, TN, fluoride, Cu, Zn, Se, As, Hg, Cd, Cr, Pb, cyanide, volatile phenol, total petroleum hydrocarbons (TPH), anionic surfac-tant, and sulfide) with data licensing and sharing restrictions.
In this paper, we provide the digital weekly water quality data which are publicly available.
These weekly water quality data were collected and con-structed by following the standards from the Environmen-tal Quality Standards for Surface Water (GB3838-2002).
Water samples were automatically collected at six intervals throughout the day, with a sampling frequency of one sample every 4 h (00:00–04:00, 04:00–08:00, 08:00–12:00, 12:00– 16:00, 16:00–20:00, 20:00–24:00 UTC+8). The weekly wa- Earth Syst. Sci. Data, 16, 1137–1149, 2024 1140 J. Lin et al.: An extensive spatiotemporal water quality dataset covering four decades in China Table 1. Source datasets for compiling China water quality dataset.
Name Data sources Timestep Original observations Timeframe Number of Number of sites (source/China) parameters (source/China) Global daily water quality data Global River Wa-ter Quality Archive (GRQA) Daily 17 000 000/3595 1898–2020 42 93 057/244 National weekly water quality data China National Envi-ronmental Monitoring Centre (CNEMC) Weekly (7 d moving av-erage) 225 336/225 336 2007–2018 4 150/150 National monthly water quality data National Marine Envi-ronmental Monitoring Center (NMEMC) Monthly 116 304/116 304 2017–2022 6 1991/1991 ter quality dataset was derived through the computation of daily averages encompassing Monday through Sunday. This process yielded a single numerical value that served as a rep-resentative of a set of valid data samples. Specifically, a min-imum of four data samples were aggregated to calculate the daily average, and five daily average data points were used to compute the weekly average.
2.1.3 NMEMC The National Marine Environmental Monitoring Center (NMEMC), which has been maintained by the China Min-istry of Ecology and Environment since 2018, is an agency with a history of 60 years specializing in marine ecologi-cal and environmental monitoring and protection. Monthly coastal and oceanic water quality data were accessible via (last access: 26 Octo-ber 2022) that were recorded from the year 2017 to 2023 and were updated until the present. Weekly water quality reports of some important beaches along the coastal areas of China for the period 2019–2023 were available via http: //www.nmemc.org.cn/hjzl/hsycszzb/index.shtml (last access: 23 January 2024) and annual average ocean ecological environment bulletins (last access: 23 January 2024). Observation data were only available for monthly coastal and oceanic water quality data.
Guidelines in the Specification for Offshore Environmental Monitoring (HJ 442-2008) directed the methodologies, cri-teria, and quality assurance measures for monthly sampling of oceanic water quality. Employing Niskin and Go-Flo wa-ter samplers, samples were collected multiple times annually, typically during the months of April through December, as il-lustrated in Fig. 1. The acquisition of this dataset entailed the collection of various quality control samples, including ma-trix spikes, blanks, parallels, and quality control check sam-ples, which underwent meticulous collection and subsequent intra-laboratory comparison.
Figure 1. Sampling frequency for oceanic water quality.
2.2 Procedure for downloading and preprocessing source data 2.2.1 Data capturing We extracted those sites located in China based on the geopo-litical map after importing all coordinate data of the GRQA dataset into ArcGIS 10.8. Afterwards, metadata information of countries and/or regions from GRQA were tidied and re-named for consistency. For instance, regions identified as “HK”, “Macao”, and “Taiwan” were renamed as “China”.
Therefore, we obtained daily water quality data in China from GRQA, which consisted of 244 sites for 15 selected wa-ter quality indicators (i.e., BOD, DO, COD, DIP, dissolved oxygen saturation (DOSAT), NH4N, NO2N, NO3N, pH, to-tal dissolved phosphorus (TDP), TEMP, TP, TSSs, dissolved organic carbon (DOC), and total organic carbon (TOC)).
Weekly water quality data were tidied up from the report collection derived from index.shtml (last access: 26 October 2022). To obtain all these files automatically, we inspected the elements of the website to locate the key nodes where the href attribute speci-fied the URL of the page the link goes to for each report. Sub-sequently, a series of packages (i.e., rvest, RSelenium, XML, purrr) in R language were used to request remote URL and scrape the hyperlinks. A collection of hyperlinks were listed to download the original reports using the downloader pack-Earth Syst. Sci. Data, 16, 1137–1149, 2024 J. Lin et al.: An extensive spatiotemporal water quality dataset covering four decades in China 1141 age. A total of 500 reports were identified, all of which were in WORD file format (i.e., DOC, DOCX, and PDF). These reports were originally designated with file names that com-bined the year and the week number. Upon closer examina-tion of the front-page summaries in each report, it came to our attention that certain original report file names exhibited inconsistencies with the actual content within. An illustrative example was the report labeled as “2010 – 1st week”, which erroneously contained observations from the 37th week of the same year. A comparable situation arose with the reports for the 53rd week in the years 2011 and 2013, as revealed through an individual cross-referencing of file names and report summaries. After the identification of these duplica-tions, the affected files were expunged from the collection.
Subsequently, a conversion process was undertaken to trans-form each of these files into editable CSV files. These CSV files were then amalgamated into a unified worksheet file, comprising 11 columns. These columns encompassed a se-rial number, information on the watersheds (MonitoringLo-cationDescriptionText), the site name (MonitoringLocation-Name), the monitoring site type (e.g., river, lake, and reser-voir; MonitoringLocationType), indicator values, the water quality index for the current week, the water quality index for the previous week, and descriptions on major pollutants.
The dataset excluded the columns related to the water qual-ity index and major pollutants, as these columns contained mainly descriptive text intended to summarize information about water quality. The column containing the serial num-ber was also excluded. The indicators featured in this dataset included DO, CODMn, NH4N, and pH.
We have collected the monthly coastal and oceanic water quality data from the NMEMC manually for the years 2017, 2018, 2019, 2020, 2021, and 2022. All data were stored as CSV files and were appended into a single worksheet file, which consisted of 14 columns (i.e., oceanic name (MonitoringLocationDescriptionText), province (ProvinceName), city (CityCode), code of the monitoring site (Source_MonitoringLocationCode), longi-tude (LongitudeMeasure_WGS84), latitude (LatitudeMea-sure_WGS84), monitoring date (MonitoringDate), values of the indicators, water quality index for the current month).
The column of the water quality index was removed. Indi-cators of the coastal and oceanic water quality data included COD, dissolved inorganic nitrogen (DIN), DO, DIP, pH, and TPH.
2.2.2 Coordinates of the monitoring sites Information of longitude and latitude is fundamental for identifying the location of a monitoring site. This informa-tion was used to export spatial point data and was overlapped with other maps to obtain metadata information.
For daily water quality data, the longitude and latitude in-formation was given by the GRQA dataset. The site location for weekly water quality data was coded as plain text of the administrative address, lacking geographic coordinates (i.e., longitude, latitude). We first used geocoding API methods to find the address for a given place, thereby transforming the address into a corresponding geographic entity. Afterwards, we validated each of them by overlapping with the layers of watersheds and rivers according to the official maps ob-tained from the National Geomatics Center of China (http:// www.ngcc.cn/ngcc/html/1/391/392/16114.html, last access: 26 October 2022). All sites were confirmed to be located at the outlet of a river reach. As the geographic coordinates for the station labeled “Xuqiao” were unidentifiable from the in-formation provided in the original files, the data associated with this station were excluded from the dataset.
General information for the monthly coastal and oceanic water quality data could be found via the NMEMC. How-ever, there were some information inconsistencies in longi-tude and latitude for the same station or place. For example, the station with code number FJD10003 was recorded with 120.57° E and 26.84° N in the year 2021 but with 120.58° E and 26.84° N in 2022. In addition, some stations with the same longitude and latitude may have different code num-bers. Therefore, we first grouped them by code numbers and computed the average value of the longitude and latitude of that station to replace the initial value. Subsequently, we re-moved the column of the code number to avoid the same stations. Finally, we dropped the duplicated rows to get the unique stations.
All the transferred longitude and latitude information was merged into a single table and then imported into ArcGIS 10.8 as point shapefile in World Geodetic System 1984 (WGS84). After overlapping with the city-level adminis-trative map and watersheds delineation map obtained from the National Geomatics Center of China, we derived other metadata information such as city, sub-watersheds (Mon-itoringLocationTypeName), etc. We referred to the China Area Code and Zip Code, Version 2021, for the province (ProvinceCode) and city (CityCode).
2.2.3 Data cleaning and technical validation We undertook a comprehensive standardization process across all the aforementioned data providers. This harmo-nization encompassed the transformation of downloaded time series into a uniform file format, shifting from CSV files to R time series. Additionally, we ensured consistency in in-dicator selection, units, data structure, identification of miss-ing values, and language.
Given the limited availability of indicators within the (sub)datasets, all of them were incorporated into our wa-ter quality dataset. This inclusive selection comprised both physical parameters (e.g., TEMP, TSSs) and chemical pa-rameters (e.g., pH, BOD, COD, CODMn, DO, DOSAT, DIN, NH4N, NO2N, NO3N, TDP, DIP, TP, TPH, DOC, TOC). We adopted GRQA as a reference for indicator ab-breviations, with the aim of facilitating international com- Earth Syst. Sci. Data, 16, 1137–1149, 2024 1142 J. Lin et al.: An extensive spatiotemporal water quality dataset covering four decades in China patibility when appending to global datasets. It is note-worthy that, except for temperature (°C), pH, and DOSAT (%), the original unit of measurements for all indicators in the (sub)datasets was milligrams per liter (mg L−1), and we retained this unit uniformity for consistency.
Eight columns (i.e., MonitoringLocationIdentifier, Longi-tudeMeasure_WGS84, LatitudeMeasure_WGS84, Monitor-ingDate (with the format % d/% m/% y), IndicatorsName, Value, Unit, SourceProvider) were then included for structur-ing the full dataset. A column for MonitoringLocationIden-tifier was created as an index to connect with the metadata file.
Some observations for different indicators were merged into a single column when converting the PDF file to editable files for weekly water quality data. Those columns were se-lected to be divided and tidied up into several columns via regular expression automatically and validation manually.
Three additional columns were added to indicate the specific year (column MonitoringYear), week number (column Mon-itoringWeek), and monitoring date (column MonitoringDate) for the weekly water quality data. The specific years and week numbers were subtracted from the file names. The col-umn of MonitoringDate for that specific week was estimated using R according to the international standard ISO 8601 that Monday was considered the first day of a week. They were validated with the descriptive text on the cover of each report that was deleted later from the weekly water quality dataset.
The column of MonitoringDate from ocean water quality data was assumed to occur on the first day of that month to keep consistency in the date format of other datasets.
In addition, duplicated rows were identified and removed by using a distinct function in R based on the unique site, indicators, monitoring week/date, and values from the (sub)datasets that included 1776 site pairs from the weekly water quality dataset due to the file inconsistencies men-tioned in Sect. 2.2.1. Negative values (with seven observa-tions) were omitted from the weekly water quality dataset.
No duplicated rows and negative values were identified from the monthly water quality datasets. In cases where seven sites provided two daily observations but lacked specific timestamp information from the GRQA, we substituted these records with the average value calculated of the two obser-vations. Missing (e.g., noted as “–”) and empty data were replaced with NA, and were omitted from the dataset. Val-ues that fell below known detection limits were denoted as “< DL” in the monthly water quality datasets, which contain 3490 data points. COD, DO, DIN, DIP, and TPH detection limits were 0.15, 0.32, 0.001, 0.001, and 0.001 mg L−1, re-spectively. The descriptions in the stations that were origi-nally in Chinese were replaced with Hanyu Pinyin.
2.3 Methods for quality assurance Since data quality will generate bias and uncertainty for the results despite conducting imputation (Tiyasha et al., 2020), it was a necessary step to conduct data quality assurance to determine the shortcomings, errors, and issues in the research results, and ensure a robust study for different data users (Koelmans et al., 2019). In this paper, we used data avail-ability and outliers for identifying quality assurance charac-teristics.
2.3.1 Availability Data availability was characterized to assess the available records, both spatially and temporally (Cai and Zhu, 2015).
For each time series, we first counted the length of the records (LengthofData) to illustrate the general temporal coverage. Then, we assessed the data intensity, computed as the ratio between the length of the time series and the length of the time series without missing values. Furthermore, we used overall availability, longest availability, and continuity to measure the characteristics of availability following the methods from Crochemore et al. (2019).
2.3.2 Outlier detection and treatment Outliers were detected by using the interquartile range (IQR) method. IQR is the range between the first (Q1) and third (Q3) quartile. Data points that fell below Q1–1.5 × IQR and above Q3 + 1.5 × IQR were considered outliers. Since it was difficult to determine whether an outlier is an error caused by faulty equipment or data entry errors or not, no observations were omitted from the original datasets.
3 Data records 3.1 General information of metadata All data were constructed in the form of CSV, while site information was provided with the point shapefile (.shp) map (available for download at Lin et al., 2023b). Referring to the inventory information of WQP, descriptions of the metadata for each time series of the water quality dataset are explained in Table 2.
3.2 Spatiotemporal distribution of monitoring sites After conducting cross-validation, it was observed that there was no spatial convergence among monitoring sites from dif-ferent data sources (Fig. 2). The dataset contained a large number of monitoring sites for the coastal and oceanic areas obtained from NMENC (Fig. 2). Most GRQA sites were lo-cated in tributaries, while the CNEMC provided most of the sites from the mainstream.
Our dataset encompassed monitoring site records span-ning from 1980 to 2022 (Fig. 3). The number of sites for daily, weekly, and monthly observations was 244, 149, and 1991 respectively. Overall, the number of monitoring sites Earth Syst. Sci. Data, 16, 1137–1149, 2024 J. Lin et al.: An extensive spatiotemporal water quality dataset covering four decades in China 1143 Table 2. Metadata information for water quality data.
Field name General introduction Descriptions Data type ID / Identifier for each time series Int WaterDataType Water data type within a broader aspect “W2” stands for water quality data String MonitoringLocationIdentifier Identifier for monitoring site Identifiers for the stations Int MonitoringLocationDescriptionText Given by the data source String MonitoringLocationName Given by the data source Name of the station String MonitoringLocationType Indicate the type of monitoring site River, Lake, Reservoir, Ocean String MonitoringLocationTypeCode Use code to indicate the type River (R), Lake (L), Reservoir (V), Ocean (C) Character MonitoringLocationTypeName Specify the name of the monitoring site In which rivers, which lakes String Source_MonitoringLocationCode Location code from the original datasets String LongitudeMeasure_WGS84 Float LatitudeMeasure_WGS84 Float ProvinceName The acronym of a specific province String ProvinceCode China area code and zip code Int CityCode China area code and zip code Int IndicatorsName String IndicatorsUnit String ResolutionCode Use numbers to identify the temporal resolution Int ResolutionName Temporal resolution String CountryCode Int StartDate Date EndDate Date LengthofData The count of observations in each time series Int DataIntensity Ratio between the length of the time series and the length of the time series without missing values Float OverallAvailability Length of the observation series, as a fraction of the dataset longest period Refers to Crochemore et al. (2019) Float LongestAvailability Length of the longest observation series without gaps, as a fraction of the dataset longest period Refers to Crochemore et al. (2019) Float Continuity Ratio between longest availability and overall availability Refers to Crochemore et al. (2019) Float SourceProvider Data source String SourceProviderID To separate the type of data source Classified as authoritative and non-authoritative String with records exhibited a slight increase before 2016, fol-lowed by a significant surge after 2016. Notably, GRQA pre-dominantly contributes observations from monitoring sites prior to 2006, with an average of 133 observations obtained from approximately 13 sites per year, as illustrated in Fig. 3a and b. By contrast, CNEMC provides data from monitoring sites between 2007 and 2018, averaging around 126 sites per year, while NMEMC covers the period from 2017 to 2022 with an average of approximately 1249 sites per year. De-spite CNEMC providing fewer monitoring sites, it consists of a comparable number of observations with an average of approximately 18 145 observations per year compared to NMEMC with an average of 19 159 observations. CNEMC and NMEMC datasets offer a greater number of records in comparison with GRQA. Temporal overlaps between vari-ous sources were identified on two occasions. The first in-stance transpired during the years 2007–2009, involving data from the GRQA and the CNEMC. The second temporal over-lap was documented between CNEMC and NMEMC for the years 2017–2018.
Earth Syst. Sci. Data, 16, 1137–1149, 2024 1144 J. Lin et al.: An extensive spatiotemporal water quality dataset covering four decades in China Figure 2. Spatial distribution of water quality monitoring sites from different sources with drainages in China.
Figure 3. Distribution of monitoring sites (a) and observations (b) from different sources over time.
3.3 Characteristics of time series The study identified four distinct types of monitoring site, comprising rivers, lakes, reservoirs, and coast/ocean (Ta-ble 3). The majority of the monitoring sites were located by the coast/ocean, with 1991 sites, followed by 365 sites in rivers that encompassed most of the indicators. Rivers from CNEMC demonstrated a considerable number of ob-servations for CODMn, DO, NH4N, and pH indicators, while COD, DIN, DIP, DO, pH, and TPH indicators have the most observations in the ocean. Despite having fewer sites and ob-servations for most indicators, rivers had a longer time series period compared to other types. Indicators of COD, DIP, and TPH exhibited some values that fell below the detection lim-its.
Availability (Fig. 4a) and continuity (Fig. 4b) plots were used to examine the temporal fragmentation of the time se-ries. Some dominant indicators (i.e., CODMn, DO, NH4N, pH) were selected for presentation in Fig. 4. Our analysis re-vealed that observations from inland exhibited significantly higher availability and continuity than those from ocean ar-eas. Specifically, for weekly water quality data, data avail-ability for all indicators ranged from 40 % to 80 % (Fig. 4a), indicating good data availability. By contrast, observations from the ocean showed moderate availability while exhibit-ing low data continuity for most observations.
The presentation of outlier proportions is documented in Table 3. Among all indicator types, TP and NH4N exhibited a higher proportion of outliers (Table 3). After the removal of outliers detected through the IQR test, boxplots were con-Earth Syst. Sci. Data, 16, 1137–1149, 2024 J. Lin et al.: An extensive spatiotemporal water quality dataset covering four decades in China 1145 Table 3. Statistics for different types of the monitoring sites and indicators.
Location Sites in Indicator Indicator Sites Observations Start End Below Outliers Sources type total number name date date limits (n) (%) (n) Coast and ocean 1991 6 COD 1991 19 367 May 2017 Aug 2022 94 4.88 NMEMC DIN 1991 19 369 May 2017 Aug 2022 / 8.99 NMEMC DIP 1991 19 369 May 2017 Aug 2022 939 6.76 NMEMC DO 1991 18 143 May 2017 Aug 2022 / 2.78 NMEMC pH 1991 19 338 May 2017 Aug 2022 / 3.69 NMEMC TPH 1991 19 368 May 2017 Aug 2022 2453 2.88 NMEMC River 366 15 BOD 10 432 7 Jan 1980 27 Nov 1997 / 6.71 GRQA COD 10 235 3 Jan 1988 27 Nov 1997 / 6.81 GRQA CODMn 122 45 491 29 Oct 2007 24 Dec 2018 / 4.59 CNEMC DIP 3 9 6 Aug 1981 27 Nov 1983 / 0.00 GRQA DO 135 45 932 7 Jan 1980 24 Dec 2018 / 3.99/3.59 CNEMC (45 459)/GRQA (473) DOC 5 16 22 Jul 1981 21 May 2008 / 0.00 GRQA DOSAT 24 31 14 Jan 1986 11 Feb 1999 / 3.23 GRQA NH4N 123 45 567 24 Feb 1983 24 Dec 2018 / 12.28/0.00 CNEMC (45 562)/GRQA (5) NO2N 13 334 6 Aug 1981 10 Nov 1997 / 7.19 GRQA NO3N 119 388 22 Jul 1981 5 Sep 2009 / 6.96 GRQA pH 251 46 181 21 Jan 1980 24 Dec 2018 / 0.50/0.99 CNEMC (45 571)/GRQA (610) TDP 3 16 12 Apr 1994 21 Oct 1996 / 0.00 GRQA TEMP 92 520 6 Feb 1980 5 Apr 2009 / 0.00 GRQA TOC 1 1 30 Aug 1994 30 Aug 1994 / 0.00 GRQA TP 10 196 7 Jan 1985 17 Oct 1996 / 15.31 GRQA TSSs 12 329 8 Jan 1980 22 Sep 1997 / 9.73 GRQA Lake 22 4 CODMn 22 6657 29 Oct 2007 24 Dec 2018 / 10.64 CNEMC DO 22 6656 29 Oct 2007 24 Dec 2018 / 2.48 CNEMC NH4N 22 6667 29 Oct 2007 24 Dec 2018 / 6.90 CNEMC pH 22 6661 29 Oct 2007 24 Dec 2018 / 0.05 CNEMC Reservoir 5 4 CODMn 5 2231 29 Oct 2007 24 Dec 2018 / 8.70 CNEMC DO 5 2276 29 Oct 2007 24 Dec 2018 / 1.36 CNEMC NH4N 5 2268 29 Oct 2007 24 Dec 2018 / 11.02 CNEMC pH 5 2252 29 Oct 2007 24 Dec 2018 / 0.27 CNEMC Figure 4. Overall availability (a) and continuity (b) for KMnO4 chemical oxygen demand (CODMn), dissolved oxygen (DO), ammonia nitrogen (NH4N), and pH.
structed for each indicator, illustrating a prominent positive skew in their distributions (Fig. 5). However, in the case of the TOC indicator, the generation of a boxplot was not in-formative due to the presence of only a single data point (Table 3), and as such, it was omitted from presentation in this context. This skewness behavior was consistent with the characteristics observed in the GRQA dataset. Conversely, indicators of DO and pH demonstrated a significant normal distribution across all three data sources.
4 Applications Given the amount of metadata information included in our inventory and the observations, this database will be par- Earth Syst. Sci. Data, 16, 1137–1149, 2024 1146 J. Lin et al.: An extensive spatiotemporal water quality dataset covering four decades in China Figure 5. Boxplots for all indicators with (a) biochemical oxygen demand (BOD), (b) chemical oxygen demand (COD), (c) KMnO4 chemical oxygen demand (CODMn), (d) dissolved inorganic nitrogen (DIN), (e) dissolved inorganic phosphorus (DIP), (f) dissolved oxygen (DO), (g) dissolved organic carbon (DOC), (h) dissolved oxygen saturation (DOSAT), (i) ammonia nitrogen (NH4N), (j) nitrite nitrogen (NO2N), (k) nitrate nitrogen (NO3N), (l) potential of hydrogen (pH), (m) total dissolved phosphorus (TDP), (n) temperature (TEMP), (o) total phosphorus (TP), (p) total petroleum hydrocarbons (TPH), and (q) total suspended solids (TSSs). Outliers determined by the interquartile range (IQR) have been removed. The unit of indicators except TEMP (°C), pH (%), and DOSAT (%) is mg L−1.
ticularly useful and important for researchers and decision-makers in the fields of hydrology, environmental research, water resources management, ecological studies, climate change, policy development, public health, and oceanogra-phy. For example, the indicator of NH4N can be used by hydrologists to develop predictive models, calibrate nitrogen models, and generate projections within China. The inland, coastal, and oceanic water quality data can be connected to display the dynamics of water quality from land to ocean, thereby routing the import, transport, and export of pollu-tants. Researchers can use these data to analyze long-term trends and variations in surface water quality, which can be vital for understanding the impact of various factors such as climate change, pollution, and land use on aquatic ecosys-tems. Water resource managers can utilize this repository to assess the quality of water in different regions, helping to make informed decisions about water allocation, treatment, and conservation strategies. Policymakers can rely on this repository to support evidence-based policy development re-lated to water quality standards and regulations. Health offi-cials can use these data to monitor the safety of water sources and assess potential health risks associated with waterborne contaminants. The high intensity of coastal and oceanic wa-ter quality data can be used to indicate coastal and oceanic water environments for the food web (i.e., living conditions of plankton). For instance, phytoplankton and zooplankton communities are sensitive to the changes in water quality, and respond to low DO levels, high nutrient levels (i.e., DIN), and toxic contaminants (i.e., TPH). Therefore, this spatial contin-uous coastal and oceanic water quality dataset is helpful for characterizing the patterns of spatiotemporal distributions of plankton, assessing the status and trends of biodiversity, and predicting the population succession in the changing ocean world.
Certain studies have previously utilized specific segments of the original dataset. For instance, researchers have em-ployed the weekly water quality data to examine the char-acteristics, trends, and seasonality of water quality in the Yangtze River (Di et al., 2019; Duan et al., 2018). It should be noted, however, that the complete dataset presented in this study has not been employed in any research thus far, which may limit the reliability of the dataset. In future, we plan to employ this dataset in upcoming research projects, where we will rigorously test its reliability.
5 Data availability All data records can be found via the figshare repository at (Lin et al., 2023b).
6 Conclusions This water quality dataset was developed with the express purpose of addressing the substantial demand for Chinese water quality data, facilitating the enhancement of national water data sharing initiatives, and fostering advancements in global water-related research and applications. It pro-vided a clean, editable, and sharable national water qual-ity dataset within China, compiling three publicly available (sub)datasets from GRQA, CNEMC, and NMEMC. The cur-rent dataset included water quality data at 2384 sites (daily Earth Syst. Sci. Data, 16, 1137–1149, 2024 J. Lin et al.: An extensive spatiotemporal water quality dataset covering four decades in China 1147 records at 244 sites, weekly at 149 sites, and monthly at 1991 sites) for the period of 1980–2022, with over 330 000 obser-vations for 18 indicators across inland, coastal, and oceanic domains. The predominant share of observations, compris-ing approximately 98.9 %, originates from the CNEMC and NMEMC, significantly expanding the global water quality dataset with a notable emphasis on the Asian region.
This database will be particularly useful and important for researchers and decision-makers in the fields of hydrology, environmental management, and oceanography for advanc-ing the assessment, modeling, and projection of water qual-ity, ocean biomass, and biodiversity in China. Considering the extensive coverage of oceanic monitoring sites within this dataset, it has made a substantial contribution to the dis-semination of coastal and oceanic water quality data, offer-ing a comprehensive depiction of the aquatic environment, and facilitating researchers in conducting in-depth investiga-tions into ocean ecosystem. Due to its comprehensive tempo-ral coverage of riverine water quality data, this dataset pre-sented a valuable adjunct for research requiring large-sample datasets and continuous information, especially for water-shed modeling, such as modeling and projection of water pollutants.
This water quality dataset will be regularly updated to incorporate any new publicly released government data in China, ensuring prompt availability to the community for their immediate use. Considering the existing absence of bi-ological parameters within the global water quality dataset, we have the intention to proactively incorporate relevant bi-ological parameters in the event of new government data re-leases. This dataset also introduces the metadata framework for forthcoming national datasets, a comprehensive collec-tion of water-related data throughout China that aims at pro-viding free, clean, non-sensitive, coherent, and reliable wa-ter data within China for global researchers to support the national water resources management and further promote Asian water data sharing in the future.
Author contributions. YC and ZY were involved in planning and supervised the work. JL, PW, and JW designed the code. JL car-ried out the data processing with contributions from PW and JW.
JL mapped the monitoring sites and developed the outlier detec-tion strategy. YZ helped improve the grammar and flow of the manuscript. JL prepared the manuscript and JW, YZ, XZ, PY, and HZ provided critical feedback and helped shape the research, anal-ysis, and manuscript.
Competing interests. The contact author has declared that none of the authors has any competing interests.
Disclaimer. Publisher’s note: Copernicus Publications remains neutral with regard to jurisdictional claims made in the text, pub-lished maps, institutional affiliations, or any other geographical rep-resentation in this paper. While Copernicus Publications makes ev-ery effort to include appropriate place names, the final responsibility lies with the authors.
Financial support. This work was supported by the Program for Guangdong Introducing Innovative and Entrepreneurial Teams (grant no. 2019ZT08L213) and the National Natural Science Foun-dation of China (grant nos. U20A20117, 52200213, and 52239005).
Review statement. This paper was edited by Yue Qin and re-viewed by four anonymous referees.
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Earth Syst. Sci. Data, 16, 1137–1149, 2024 |
11079 | https://danceswithcode.net/engineeringnotes/rotations_in_2d/rotations_in_2d.html | 2D Rotations
Home
Rotating Points in Two-Dimensions
D. Rose - December, 2014
Abstract
This tutorial describes the efficient way to rotate points around an arbitrary center on a two-dimensional (2D) Cartesian plane. This is a very common operation used in everything from video games to image processing. Sample code is provided in Java.
The Naïve Solution (don’t do this)
Let’s try a concrete example. Take the point (4,3) and rotate it 30 degrees around the origin in a counterclockwise direction. If you’ve taken high school trigonometry, you might come up with a solution like this:
Step 1:Convert the point from Cartesian to polar coordinates. Polar coordinates define the location of a point by its distance from the origin (r) and angle from the x-axis (θ).
The distance from the origin can be found using the Pythagorean Theorem: r 2 = x 2+y 2. If you plug in (4,3) for (x,y), you find that r = 5.
The angle can be found using trigonometry: θ = tan-1(y/x). If you plug in (4,3) for (x,y), you find that θ=36.87°. (Note: If you’re writing a computer program, you need to use the function atan2() instead of atan(). The atan2() function accepts the x and y arguments separately, and so produces the correct answer in all four quadrants.)
Step 2:Rotate the point by 30°. You do this by adding 36.87°+30°, to get a rotated angle of 66.87°.
Step 3:Convert back to Cartesian coordinates.
The x coordinate of the rotated point is r cos(θ), and the y coordinate is r sin(θ). If you plug in 5 and 66.87 for r and θ, you find that the rotated point (x 1,y 1) = (1.964, 4.598).
This is the correct answer, and it works for points in all quadrants and all rotation angles, including negative angles. But it requires five multiplications, three trig functions, a square root and an addition, which is way more computation than necessary. If you’re rotating a 2-million-pixel HDTV image, or trying to render your X-Wing fighter at 60 frames per second, you want to be as efficient as possible.
The Right Way
Equations 1 and 2 show the right way to rotate a point around the origin:
x 1 = x 0 cos(θ) – y 0 sin(θ)
(Equation 1)
y 1 = x 0 sin(θ) + y 0 cos(θ)
(Equation 2)
If we plug in our example point of (x 0, y 0) = (4, 3) and θ = 30°, we get the answer (x 1, y 1) = (1.964, 4.598), the same as before.
At first glance this may not seem to be that much faster than the naïve method, since it takes four trig functions, four multiplications, and two additions. The trick is that the trig functions only have to operate on the rotation angle, which is constant for all points. You therefore only have to compute them once, no matter how many points you are rotating. The per-point computational load is therefore only four multiplies and two additions—less than half that of the naïve method.
Rotating Points around an Arbitrary Center
But what if we want to rotate our point around something other than the origin? We simply modify equation 1 and 2 as follows:
x 1 = (x 0 – x c)cos(θ) – (y 0 – y c)sin(θ) + x c
(Equation 3)
y 1 = (x 0 – x c)sin(θ) + (y 0 – y c)cos(θ) + y c
(Equation 4)
where:
(x 0, y 0)= Point to be rotated
(x c, y c)= Coordinates of center of rotation
θ= Angle of rotation (positive counterclockwise)
(x 1, y 1)= Coordinates of point after rotation
Note that if (x c, y c) = (0, 0), then equations 3 and 4 simplify to become equations 1 and 2, which is what we expect.
Sample Code:
This listing shows a Java implementation of equations 3 and 4. Note that the sine and cosine calculations are performed only once, outside the main loop. This saves significant computation time.
void rotatePoints(
double[] x, //X coords to rotate - replaced on return
double[] y, //Y coords to rotate - replaced on return
double cx, //X coordinate of center of rotation
double cy, //Y coordinate of center of rotation
double angle) //Angle of rotation (radians, counterclockwise)
{
double cos = Math.cos(angle);
double sin = Math.sin(angle);
double temp;
for( int n=0; n<x.length; n++ ){
temp = ((x[n]-cx)cos - (y[n]-cy)sin) + cx;
y[n] = ((x[n]-cx)sin + (y[n]-cy)cos) + cy;
x[n] = temp;
}
return;
}
Here is an example of how the rotatePoints() method would be called:
```
double[] x = {0, 0, 0, 0, 0, 10, 20, 10}; //Create some test data
double[] y = {0, 10, 20, 30, 40, 40, 30, 20}; //Create some test data
double cx = 20; //X-coord of center of rotation
double cy = 30; //Y-coord of center of rotation
double angle = 45 Math.PI/180; //convert 45 degrees to radians
//Rotate the points
rotatePoints( x, y, cx, cy, angle );
//Display the results
for( int n=0; n<x.length; n++ ){
System.out.println( ""+x[n]+","+y[n] );
}
```
The following figure shows the eight points from the example code before and after the rotation:
Common Errors
Here are a few things to be conscious of when implementing this code :
This function is designed to perform the rotation “in-place.” That is, the coordinate data in the x[] and y[] input arrays is replaced by their rotated equivalents when the rotatePoints() method returns. If you want to save the original points, you need to copy them first, or modify the method to write the results to separate arrays.
Programming languages expect angles in radians. Neglecting to convert from degrees to radians is a common error.
These examples assume a coordinate system where x points to the right and y points up. Unfortunately, most computer graphics systems use a convention where x points to the right and y points down. This has the effect of reversing the direction of rotation. If you use these equations in a y-down coordinate system, then a positive rotation angle will produce a clockwise rotation, rather than counterclockwise. You can correct for this by negating the sign of the rotation angle.
Comments and error reports may be sent to the following address. We may post comments of general interest. Be sure to identify the page you are commenting on.
.
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11080 | https://mathsodology.wordpress.com/2020/11/04/s1-2-1-2-examples-ungrouped/ | S1 §2.1.2 Example 9 (Mode, median, mean for ungrouped discrete data) – Mathsodology
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S1 §2.1.2 Example 9 (Mode, median, mean for ungrouped discrete data)
Mathsodology2.1 Measures of central tendency: Mode, Median, MeanNovember 4, 2020 December 7, 2020 1 Minute
Example 9. Rebecca records the shirt collar size, , of the male students in her year. The results are shown in the table.
Shirt collar size15 15.5 16 16.5 17
Number of students3 17 29 34 12
Find for this data:
(a) the mode
(b) the median
(c) the mean
(d) Explain why a shirt manufacturer might use the mode when planning production numbers.
Edexcel 2008 Specification, S1 Ch2 Example 13; 2017 Specification S1 Ch2 Example 2.
Solution.(a) Mode = 16.5
(b) There are students/observations. The median is the -th value. There are 20 observations up to 15.5 and 49 observations up to 16. Thus, the median is 16.
(c) The mean is given by
(d) While the mean is not an actual size of the shirt, the mode is an actual size and gives the information on the most common size.
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Comment 1: For data given in a frequency table, the mean can be calculated by:
This is in fact the same formula as .
Comment 2: For the median, the cumulative frequency table is often useful.
Number of studentsCumulative frequency
15 3 3
15.5 17 20
16 29 49
16.5 34 83
17 12 95
Since the median is the 48th value, we immediately see that it belongs to the class with .
The cumulative frequency table is not absolutely necessary for the discrete data but it becomes a very useful tool for continuous data, as we will see in S1 §2.2 Other measures of location and S1 §3.3 Cumulative frequency.
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S1 §2.1.25 Exercise Q20 (Mode, median, mean for ungrouped discrete data)November 25, 2020 In "2.1 Measures of central tendency: Mode, Median, Mean"
S1 §2.1.1 Mode, Median, Mean – Measures of central tendency (Theory Explained)November 4, 2020 In "2.1 Measures of central tendency: Mode, Median, Mean"
S1 §2.1.23 Exercise Q18 (Mode, median, mean for ungrouped discrete data)November 23, 2020 In "2.1 Measures of central tendency: Mode, Median, Mean"
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Table of Contents
Abstract
Introduction
Diagnosis and Risk Factors of Preeclampsia
Maternal Antiangiogenic State in Preeclampsia
Mechanisms of Endothelial Dysfunction by Inhibiting the VEGF Signal Pathway
Mechanism of Renal Injury by Inhibiting the VEGF Signal Pathway
Complement System and Angiogenic Imbalance
Regulating sFlt1 Production by Trophoblastic Cells
Preeclampsia as a Systemic Vascular Disorder of Pregnancy
Novel Clinical and Therapeutic Strategies from the Viewpoints of Maternal Angiogenic Imbalance
Prevention of Preeclampsia
Preeclampsia and Future Risk of Cardiovascular Disease
Conclusions
Author Contributions
Funding
Conflicts of Interest
Abbreviations
References
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Open AccessReview
Preeclampsia: Maternal Systemic Vascular Disorder Caused by Generalized Endothelial Dysfunction Due to Placental Antiangiogenic Factors
by
Takuji Tomimatsu
Takuji Tomimatsu
SciProfilesScilitPreprints.orgGoogle Scholar
1,,
Kazuya Mimura
Kazuya Mimura
SciProfilesScilitPreprints.orgGoogle Scholar
1,
Shinya Matsuzaki
Shinya Matsuzaki
SciProfilesScilitPreprints.orgGoogle Scholar
1,
Masayuki Endo
Masayuki Endo
SciProfilesScilitPreprints.orgGoogle Scholar
1,
Keiichi Kumasawa
Keiichi Kumasawa
SciProfilesScilitPreprints.orgGoogle Scholar
2 and
Tadashi Kimura
Tadashi Kimura
SciProfilesScilitPreprints.orgGoogle Scholar
1
1
Department of Obstetrics and Gynecology, Osaka University Graduate School of Medicine, Osaka 565-0871, Japan
2
Department of Obstetrics and Gynecology, Tokyo University Graduate School of Medicine, Tokyo 113-0033, Japan
Author to whom correspondence should be addressed.
Int. J. Mol. Sci. 2019, 20(17), 4246;
Submission received: 29 June 2019
/
Revised: 20 August 2019
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Accepted: 28 August 2019
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Published: 30 August 2019
(This article belongs to the Special Issue Endothelial Dysfunction: Pathophysiology and Molecular Mechanisms)
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Abstract
Preeclampsia, a systemic vascular disorder characterized by new-onset hypertension and proteinuria after 20 weeks of gestation, is the leading cause of maternal and perinatal morbidity and mortality. Maternal endothelial dysfunction caused by placental factors has long been accepted with respect to the pathophysiology of preeclampsia. Over the past decade, increased production of placental antiangiogenic factors has been identified as a placental factor leading to maternal endothelial dysfunction and systemic vascular dysfunction. This review summarizes the recent advances in understanding the molecular mechanisms of endothelial dysfunction caused by placental antiangiogenic factors, and the novel clinical strategies based on these discoveries.
Keywords:
preeclampsia; endothelial dysfunction; angiogenic imbalance; systemic vascular dysfunction; arterial stiffness; vascular endothelial growth factor; placental growth factor; soluble fms-like tyrosine kinase 1
1. Introduction
Preeclampsia, a systemic vascular disorder of pregnancy characterized by hypertension in association with proteinuria, affects 5% to 10% of all pregnancies. This condition can affect virtually every organ system, causing preeclampsia-related adverse complications such as seizures (eclampsia), HELLP (hemolysis, elevated liver enzymes, and low platelets) syndrome, abruptio placentae, and fetal growth restriction. Currently, the only effective treatment is termination of pregnancy, which may also add substantial risks to the neonate if the fetus is delivered prematurely. Although the clinical symptoms of preeclampsia completely resolve after delivery, recent evidence has demonstrated significant association between history of preeclampsia and future risks of cardiovascular diseases [1,2].
Generalized maternal endothelial dysfunction due to placental factors has been considered to play an important role in the pathogenesis of preeclampsia. Many serum markers indicating endothelial activation increase , and flow-mediated dilation (FMD), the gold standard for evaluating endothelial function , is impaired in patients with preeclampsia . Over the past decade, excess placental antiangiogenic factor, soluble fms-like tyrosine kinase 1(sFlt1), has been shown to antagonize vascular endothelial growth factor (VEGF) and placental growth factor (PlGF), and to induce generalized endothelial dysfunction in these women [6,7,8,9]. This discovery generated great enthusiasm for sFlt1 as the most promising placental factor and led to the development of novel clinical strategies for managing preeclampsia. In this review, we aimed to summarize the recent advances in understanding the pathophysiology of preeclampsia, and novel clinical strategies from the viewpoints of maternal endothelial dysfunction caused by placental antiangiogenic factors.
2. Diagnosis and Risk Factors of Preeclampsia
Preeclampsia is defined as an increase in systolic blood pressure ≥140/90 mmHg in previously normotensive women as well as proteinuria ≥300 mg in a 24-h collection, or 0.3 g/g by urine protein/creatinine ratio, or +1 by urine dipstick if it is the sole available test, occurring after 20 weeks of gestation. These diagnostic criteria do not require proteinuria for diagnosis in the presence of other organ damage such as thrombocytopenia, renal failure, liver involvement, cerebral symptoms, and pulmonary edema .
Several risk factors for preeclampsia clearly indicate the presence of genetic predisposition. The incidence of preeclampsia is greatly affected by race and ethnicity. African-American women have a higher risk of developing preeclampsia than white women . Population based cohort studies revealed that the daughters and sons of women who had preeclampsia during pregnancy had a higher risk of developing preeclampsia themselves and fathering a preeclamptic pregnancy, respectively [12,13]. Recently, a large clinical genome-wide association study revealed significant association between single-nucleotide polymorphism near the FLT1 locus (rs4769613) on chromosome 13 in the fetal genome and the development of preeclampsia [14,15]. Trisomy 13 (chromosome 13 contains the FLT1 locus) is associated with increased maternal levels of sFlt1 and a high risk of preeclampsia .
Multifetal and molar pregnancies are associated with an increased risk of preeclampsia, presumably due to increased levels of sFlt1. We have reported an interesting case of mirror syndrome (maternal preeclampsia-like symptoms and fetal hydrops) caused by severe fetal anemia due to parvovirus B19 infection. Both maternal and fetal symptoms resolved immediately after intrauterine transfusion, along with the normalization of the increased levels of sFlt1 . These findings may strengthen the importance of the central contribution of maternal antiangiogenic state in the pathogenesis of preeclampsia.
Reduced paternal antigen exposure such as nulliparity, shorter periods of sexual cohabitation, and changing paternity demonstrate increased risks for developing preeclampsia, indicating immunological contribution to the pathogenesis [20,21]. Some maternal conditions, such as advanced maternal age, obesity, diabetes mellitus, chronic hypertension, antiphospholipid syndrome (APS), chronic kidney disease (CKD), and systemic lupus erythematosus (SLE), are also associated with an increased risk of preeclampsia [1,2]. These maternal conditions have been reported to be associated with endothelial dysfunction, which may contribute to increased risk of preeclampsia . In addition, it has been reported that pregnant women with chronic hypertension and pregnant women with diabetes showed significantly higher sflt1 level, and pregnant women with obesity had significantly lower PlGF level .
3. Maternal Antiangiogenic State in Preeclampsia
Maternal endothelial dysfunction caused by placental factors and the two-stage theory have long been accepted with regard to the pathophysiology of preeclampsia (Figure 1) . The initial step is considered to start from insufficient cytotrophoblast invasion of spiral arteries (abnormal placentation) [24,25,26,27]. Predisposing immunological, genetic, and preexisting maternal risk factors may affect this abnormal placentation. In preeclampsia, failure of the physiological remodeling of decidual vessels results in reduced placental perfusion , which has been believed to release placental factors into the maternal circulation.
Figure 1.
Two-stage theory of the pathophysiology of preeclampsia. Predisposing immunological, genetic, and preexisting maternal risk factors may affect abnormal cytotrophoblast invasion of spiral arteries (abnormal placentation) (First stage). The reduced uteroplacental perfusion induces placental release of antiangiogenic factors (soluble fms-like tyrosine kinase 1 (sFlt1)) into the maternal circulation, which antagonizes proangiogenic factors, leading to endothelial dysfunction and systemic vascular dysfunction (Second stage). Preexisting maternal conditions such as chronic hypertension, systemic lupus erythematosus (SLE), and obesity is also associated with endothelial dysfunction.
In 2003, Maynard et al. discovered increased levels of placental sFlt1 in the serum of women with preeclampsia. They also reported that administration of sFlt1 to pregnant rats generated preeclampsia-like symptoms. Subsequently, Levine et al. showed that serum levels of sFlt1 correlates with disease severity and declines after resolution. These experimental and epidemiological studies as well as several following landmark reports [8,9], generated compelling evidence that placental sFlt1 is one of the most important placental factors leading to maternal endothelial dysfunction .
sFlt1 is a splice variant of the VEGFR1 (Flt1) containing only the extracellular ligand-binding domain of VEGFR1 . As VEGFR1 (Flt1) interacts with VEGF and PlGF , sFlt1 inhibits proangiogenic signaling by antagonizing VEGF and PlGF. Importantly, sFlt1 is mainly produced by the trophoblastic cells, and released into the maternal circulation during pregnancy (Figure 2).
Figure 2.
Mechanisms of endothelial dysfunction leading to systemic vascular dysfunction in preeclampsia. Excessive soluble fms-like tyrosine kinase 1 (sFlt1) antagonizes VEGF or PlGF, or both, and causes endothelial dysfunction, including a decrease in vasodilators such as nitric oxide (NO) and prostacyclin (PGI2) and an increase in vasoconstrictors such as endothelin-1 (ET-1). VEGF: vascular endothelial growth factor; PLGF: placental growth factor; VEGFR1: VEGF receptor 1 (also known as Flt1).
4. Mechanisms of Endothelial Dysfunction by Inhibiting the VEGF Signal Pathway
The precise mechanisms of endothelial dysfunction in women with preeclampsia due to high levels of circulating sFlt1 remains unclear. Direct administration of VEGF augments the release of nitric oxide (NO) from the vascular endothelium and causes nitric oxide-dependent hypotension in vivo . VEGF has been shown to stimulate NO production via upregulation of nitric oxide synthase (NOS) expression in endothelial cells [35,36]. This vasodilation effect of VEGF may be mediated by both VEGFR1 (Flt1) and VEGFR2 (KDR/FlK1) receptors, but VEGFR2 is the predominant receptor mediating this effect . VEGF, but not PlGF, was also shown to induce prostacyclin (PGI2) synthesis [38,39]. NO activates soluble guanylate cyclase (sGC), leading to cGMP synthesis. PGI2 activates adenylyl cyclase (AC) and increases cAMP synthesis. Both cGMP and cAMP lead to decreased intracellular Ca2+ concentrations, which induce smooth muscle relaxation and vasodilation .
The recent introduction of VEGF inhibitor therapies in cancer patients and its preeclampsia-like adverse effects, particularly hypertension and renal injury, have attracted much attention and made doubly sure that inhibiting the VEGF signal pathway is central to the pathophysiology in preeclampsia [41,42,43,44,45,46]. VEGF inhibitor therapies in cancer patients also added several interesting insights into the mechanisms of preeclampsia. Of these, several lines of evidence reported a dose-dependent activation of the endothelin-1 (ET-1) in response to VEGF inhibitor therapies [47,48,49]. Considering that ET-1 is the most potent vasoconstrictor and increased levels of ET-1 have also been reported in women with preeclampsia [50,51], it seems plausible that ET-1 is involved in pathogenesis. Although the precise mechanism leading to increased levels of ET-1 by inhibiting the VEGF signal pathway remains unclear, it has been reported that VEGF enhances prepro-ET-1 mRNA expression and induces endothelin-converting enzyme-1 (ECE-1), which is a key enzyme in endothelin processing [52,53].
5. Mechanism of Renal Injury by Inhibiting the VEGF Signal Pathway
VEGF is synthesized by podocytes within the glomerulus where it maintains fenestrated endothelium . Inhibiting the VEGF signal pathway causes endothelial swelling, termed glomerular endotheliosis, which is the renal lesion frequently seen in women with preeclampsia [55,56]. The importance of NO in maintaining normal renal function has also been well documented . Inhibiting the VEGF signal pathway reduces NO production due to the decreasing expression of endothelial and neuronal NOS in the kidney . In addition, it has been reported that proteinuria and glomerular endotheliosis caused by VEGF inhibitor therapies were prevented by the endothelin receptor blocker , indicating the involvement of the endothelin system in the mechanism of renal injury by inhibiting the VEGF signal pathway.
6. Complement System and Angiogenic Imbalance
There has been compelling evidence that complement activation is implicated in the pathogenesis of preeclampsia [60,61]. Clinical similarities between atypical hemolytic uremic syndrome (aHUS), a disease of excessive activation of the alternative complement pathway, and HELLP syndrome, a severe variant of preeclampsia, along with the findings of several studies [62,63], have strengthened the role of complement activation in the pathogenesis of preeclampsia. Elevated levels of urinary C5b-9 in women with preeclampsia have been shown to be a useful biomarker that differentiates preeclampsia from other hypertensive disorders [64,65]. Immunohistochemical studies using renal biopsy specimens from women with preeclampsia revealed increased renal C4d-a and C1q-positive glomeruli, suggesting the importance of the classical complement pathway in the pathogenesis . Increased C4 deposits in the glomeruli were also shown in the sFlt1-injected pregnant mouse model, indicating that angiogenic dysregulation may play a role in complement activation within the kidney . It has also been reported that aHUS was also induced in cancer patients under VEGF inhibitor therapies . Recently, a possible mechanism linking the complement system to angiogenic imbalance was shown, in which inhibiting the VEGF signal pathway decreases local complement inhibitor synthesis in renal glomeruli, potentially making these sites vulnerable to complement activation . Another study reported that human extravillous trophoblast cell line HTR-8/Svneo treated with C5a expressed significantly increased mRNA levels of sFlt1 and decreased mRNA levels of PlGF .
7. Regulating sFlt1 Production by Trophoblastic Cells
The mechanisms of placental sFlt1 upregulation are largely unknown. Alternative splicing of the pre-mRNA encoding FLT1 results in the production of sFlt1 containing only the extracellular ligand-binding domain of Flt1 but lacking the intracellular and membrane-spanning domains . It is believed that hypoxic environment caused by abnormal placentation stimulates sFlt1 production . In accordance with this notion, elevated expression of transcription factor hypoxia-inducible factor 1α (HIF1α) was shown to contribute to sFlt1 upregulation in in vivo and in vitro models of human placenta . Inhibition of complement activation has been shown to block the increase of sFlt1 in pregnant mice . Mitochondrial dysfunction leading to reactive oxygen species generation and oxidative stress may contribute to sFlt1 production . Recently, the upregulation of VEGF in maternal decidual cell was advocated as a trigger of sFlt1 production by trophoblastic cells .
8. Preeclampsia as a Systemic Vascular Disorder of Pregnancy
Systemic vascular dysfunction is considered as a final step in the pathophysiology of preeclampsia (Figure 1). During normal pregnancy, maternal vascular resistance decreases, resulting in slightly decreased blood pressure [75,76,77,78,79,80]. In women with preeclampsia, it has been thought that these adaptations do not occur sufficiently due to systemic vascular disorder with generalized endothelial dysfunction. Although the precise mechanism of systemic vascular disorder caused by endothelial dysfunction remains elusive, abnormalities in matrix metalloproteinases (MMPs) and increased collagen deposition in extracellular matrix (ECM) are considered to play significant roles in inadequate vascular remodeling leading to systemic vascular dysfunction .
Recently, noninvasive assessment of vascular function has directly revealed the presence of systemic vascular dysfunction in women with preeclampsia. FMD has been shown to increase during pregnancy . In women with preeclampsia, significantly lower FMD was found both before and after the development of the disease as well as 3 years after delivery . VEGF inhibitor therapies with bevacizumab have also been shown to result in reduced endothelium-mediated vasodilation . Pulse wave analysis (PWA) measures the composite stiffness of the conduit and resistance artery [85,86,87]. PWA indices (augmentation index and central systolic pressure) decline markedly during pregnancy [88,89,90]. In women with preeclampsia, these indices are significantly increased [91,92]. Abnormal PWA measures have also been observed before the onset of disease [93,94], as well as 6–24 months postpartum . Increased PWA measures were also shown to be more relevant to intrauterine fetal growth than conventional brachial blood pressure [96,97]. Pulse wave velocity (PWV) has been considered to provide information regarding the stiffness of conduit arteries [98,99], and is elevated in pregnant women with preeclampsia [100,101].
The results from these vascular function tests provided a comprehensive picture of a systemic vascular dysfunction due to endothelial dysfunction, the final step in the pathophysiology of preeclampsia (Figure 1). It is expected that combining vascular function tests with the assessment of angiogenic imbalance might improve prediction accuracy of preeclampsia-related adverse complications. In addition, these tests revealed the presence of vascular dysfunction even after the resolution of clinical symptoms of preeclampsia, indicating its possible association with future cardiovascular disease risks.
9. Novel Clinical and Therapeutic Strategies from the Viewpoints of Maternal Angiogenic Imbalance
9.1. Prediction of Disease
Identifying women at risk for preeclampsia in the first trimester is now an area of important clinical research, as low-dose aspirin started before 16 weeks of gestation was reported to be associated with a significant decrease in the prevalence of preeclampsia in women identified to be at high-risk based on the conventional maternal clinical risk factors, such as nulliparity, a history of preeclampsia, and chronic hypertension .
Although extensive investigations revealed that the value of sFlt1 levels in the first trimester showed no clear association with the development of preeclampsia , PlGF levels in the first trimester have been shown to have consistent and promising results in the prediction of preeclampsia . Recently, screening performance for preeclampsia in first trimester based on an algorithm combining PlGF levels with maternal clinical factors, mean arterial pressure (MAP), and uterine artery pulsatility index (UtA-PI) was reported to be by far superior to the screening performance based on conventional maternal clinical risk factors . Using this algorithm, a recent large clinical trial (ASPRE trial) confirmed that first-trimester screening combining with low-dose aspirin administration resulted in a substantial decrease in the incidence of preterm preeclampsia (odds ratio, 0.38; 95% confidence interval (CI), 0.20–0.74; p = 0.004) .
9.2. Prediction of Adverse Maternal and Perinatal Complications
Clinically, it is not unusual for pregnant women without hypertension or without proteinuria to develop preeclampsia-related adverse complications, such as eclampsia [107,108,109]. Conversely, a significant portion of women who meet the diagnostic criteria for preeclampsia do not show any adverse complications and are able to carry a pregnancy to nearly full term. Therefore, the association of maternal angiogenic imbalance with the occurrence of preeclampsia-related adverse complications, rather than with the development of preeclampsia, has been vigorously investigated.
In pregnant women with suspected preeclampsia, the severity of the maternal antiangiogenic state predicted preeclampsia-related adverse complications more accurately than the highest systolic blood pressure, a hallmark of the diagnostic criteria for preeclampsia [110,111]. In women with suspected preeclampsia presenting at <34 weeks, an sFlt1/PlGF ratio ≥85 predicted preterm delivery within 2 weeks with a hazard ratio of 15.2 . Furthermore, a secondary analysis of this study revealed that patients who meet diagnostic criteria for preeclampsia but had a normal angiogenic profile showed no preeclampsia-related adverse maternal and fetal complications . In women with suspected preeclampsia presenting at ≤36 weeks, an sFlt-1/PlGF ratio <38 showed an extremely high negative predictive value of 99.3% on the occurrence of preeclampsia-related adverse complications within 1 week . Furthermore, a secondary analysis of this study revealed that patients with an sFlt1/PlGF ratio ≥38 showed significantly shorter remaining time to delivery and a higher rate of preterm delivery, irrespective of the development of preeclampsia . In 90% of women with suspected or confirmed preeclampsia with an sFlt1/PlGF ratio ≤38, the ratio was largely stable and did not increase further up to 100 days . Recently, a randomized controlled trial confirmed that implementing PlGF measurement in managing women with suspected preeclampsia significantly improved maternal outcome .
9.3. Assessing Angiogenic Imbalance in Differential Diagnosis
Assessing angiogenic imbalance is reported to be useful in differentiating preeclampsia from other diseases with preeclampsia-like symptoms, including chronic kidney disease (CKD) , gestational thrombocytopenia , and chronic hypertension . In case of a flare of SLE during pregnancy, assessing angiogenic imbalance can lead to appropriate treatment (prednisolone escalation), instead of unnecessary iatrogenic preterm deliveries [120,121]. In pregnant women with SLE, APS, or both, circulating angiogenic factors measured during early gestation have a high negative predictive value of 93% in ruling out the development of severe adverse outcomes, including early-onset preeclampsia, fetal/neonatal death, and iatrogenic preterm delivery before 30 weeks of gestation .
9.4. Therapeutic Potential of Modulating Angiogenic Factors
A large number of basic research studies have suggested the therapeutic potential of modulating angiogenic factors [123,124,125,126,127,128,129,130], including administration of recombinant VEGF or PlGF , and reducing sFlt1 levels via RNA interference (RNAi) . However, currently, only one pilot human trial aimed at direct modulation of maternal angiogenic imbalance has reported a clinical benefit, in which removal of sFlt1 by dextran sulfate apheresis resulted in the stabilization of blood pressure and prolongation of pregnancy in women with very preterm (<32 weeks) preeclampsia .
10. Prevention of Preeclampsia
10.1. Low-Dose Aspirin
The role of aspirin in the primary or secondary prevention of preeclampsia has long been an important clinical concern. In preeclampsia, thromboxane A2 (TXA2: platelet activator and vasoconstrictor) production by the platelets increases, whereas PGI2 production by the endothelium decreased . Both TXA2 and PGI2 are synthesized from arachidonic acid by the action of cyclooxygenase (COX) . Although low-dose acetylsalicylic acid (aspirin) blocks COX irreversibly, the endothelium recovers PGI2 production by de novo synthesis of COX . However, the platelets, where TXA2 is synthesized, cannot synthesize COX as the platelets are anuclear. In accordance with this notion, several studies reported that low-dose aspirin reduced TXA2 production without altering the PGI2 production [135,136], leading to normal TXA2/PGI2 balance by two weeks of treatment . Aspirin is also shown to have angiogenic properties by blocking sFlt1 production in human trophoblasts and by increasing PlGF production in BeWo trophoblast cells .
Although the first clinical trials showed significant efficacy of aspirin in preventing preeclampsia in 1985 , subsequent large-scale clinical studies have shown limited or no clinical benefit of low-dose aspirin for the prevention of preeclampsia [141,142,143]. In 2010, Bujold et al. published a meta-analysis of double-blind randomized trials and suggested a greater benefit when aspirin treatment was started before 16 weeks of gestation (relative risk (RR), 0.47; 95% CI 0.34–0.65) in high-risk patients . As mentioned above, a recent multicenter, double blind, randomized, placebo-controlled trial (ASPRE trial) evaluated the effect of low-dose aspirin, administered from 11 to 14 weeks of gestation until 36 weeks of gestation, among women who were identified as high risk following first-trimester screening. This trial revealed a substantial reduction in the incidence of preterm preeclampsia (odds ratio, 0.38; 95% CI, 0.20–0.74; p = 0.004) .
10.2. Low-Dose Aspirin plus Heparin
Antiphospholipid syndrome (APS) is an autoimmune disease that causes an increased risk of thrombotic or adverse obstetrical events in patients with persistent antiphospholipid antibodies . In pregnant women with APS, of whom a third of these women develop preeclampsia, treatment with low-dose aspirin plus heparin has long been the most efficacious regimen, with significant improvement of maternal and perinatal outcomes . In a mouse model of APS, one of the beneficial mechanisms of heparin in women with APS was shown to be mediated by inhibition of the complement system , which also can be beneficial in women with preeclampsia.
For preventing preeclampsia in patients with and without APS, a recent meta-analysis reported that heparin improved the efficacy of low-dose aspirin alone (RR, 0.54; 95% CI 0.31–0.92) . However, subsequent large multicenter trials did not show the potential efficacy of heparin for the prevention of preeclampsia in high-risk patients without APS, indicating that heparin may benefit only a subset of patients [147,148]. Moreover, several studies reported the contradictory but consistent observations that heparin increases both circulating PlGF and sFlt1 levels, suggesting that heparin does not have ideal properties for restoring angiogenic imbalance observed in women with preeclampsia [149,150,151].
11. Preeclampsia and Future Risk of Cardiovascular Disease
As mentioned above, it is now established that preeclampsia is associated with the future risks for cardiovascular diseases [152,153,154,155]. Women with a history of preeclampsia are at increased risks for future cardiovascular diseases, such as chronic hypertension (RR, 3.7; 95% CI, 2.70–5.05) , and heart failure [RR, 4.19 (2.09–8.38)] , stroke (RR, 1.81 (1.29–2.55)) , coronary heart disease (RR, 2.50 (1.43–4.37)) , and cardiovascular mortality (RR, 2.21 (1.83–2.66)) when compared to women without a history of preeclampsia.
Although the precise mechanisms remain unclear, several recent studies revealed that the persistence of maternal endothelial and vascular dysfunction after delivery by using several non-invasive vascular function tests or echocardiography [95,156,157]. Although the sFlt1 level has been reported to decrease to less than 1% of its pre-delivery value within 24 h of the delivery , the levels of sFlt1 and the sFlt1/PlGF ratio were still higher at 1 year postpartum in women with preeclampsia . Elevated levels of sFlt1 were also reported at 5–8 years postpartum in women with preeclampsia .
Recently, the clinical significance of sFlt1 in cardiac functions has been advocated. Increased levels of sFlt1 were related to the development of acute heart failure in patients with acute myocardial infarction . In women with preeclampsia, the extent of subclinical cardiac dysfunction correlates with the circulating levels of sFlt1, and women with peripartum cardiomyopathy showed abnormally increased level of sFlt1 even 4–6 weeks postpartum . In addition, systemic administration of sFlt1 in a mouse model of peripartum cardiomyopathy induced substantial cardiac dysfunction .
Although a history of preeclampsia is now recognized as a women-specific risk factor for cardiovascular disease in later life, it is still unclear how the cardiovascular health of these women should be improved. Presently, several guidelines suggest management for monitoring of hypertension, hyperlipidemia, and diabetes, and provision of healthy lifestyle advice for women with a history of preeclampsia . Further studies are needed to define the appropriate monitoring and intervention strategies for these women.
12. Conclusions
Our understanding of the pathophysiology of preeclampsia has advanced considerably in the past decade. Maternal systemic vascular dysfunction caused by generalized endothelial dysfunction due to placental antiangiogenic factors has emerged as one of the most important mechanisms, and clinical strategies from the viewpoints of maternal angiogenic imbalance has been increasingly incorporated in the clinical practice. In the future, novel clinical and therapeutic strategies aimed at restoring angiogenic imbalance is expected to ameliorate complications and prolong gestation in women with preeclampsia. In addition, elucidation of pathophysiology and establishment of effective screening and prevention strategies will guide clinicians to reduce future risks of cardiovascular disease in women with preeclampsia.
Author Contributions
All authors contributed to writing of this review.
Funding
This work was supported by Grant-in-Aid for General Scientific Research (T17K112330) from the Ministry of Education, Culture, Sports, Science and Technology of Japan, and by AMED under Grant Number JP19gk0110030h0002.
Conflicts of Interest
The authors declare no conflict of interest.
Abbreviations
| | |
--- |
| FMD | Flow-mediated dilation |
| sFlt1 | Soluble fms-like tyrosine kinase 1 |
| VEGF | Vascular endothelial growth factor |
| PlGF | Placental growth factor |
| sVEGFR1 | Soluble vascular endothelial growth factor receptor 1 |
| NO | Nitric oxide |
| NOS | Nitric oxide synthase |
| PGI2 | Prostacyclin |
| TXA2 | Thromboxane A2 |
| COX | Cyclooxygenase |
| sGC | Soluble guanylate cyclase |
| AC | Adenylyl cyclase |
| ET-1 | Endothelin 1 |
| ECE-1 | Endothelin-converting enzyme 1 |
| aHUS | Atypical hemolytic uremic syndrome |
| IUGR | Intrauterine growth restriction |
| APS | Antiphospholipid syndrome |
| CKD | Chronic kidney disease |
| SLE | Systemic lupus erythematosus |
| PWA | Pulse wave analysis |
| PWV | Pulse wave velocity |
| RNAi | RNA interference |
| MAP | Mean arterial pressure, and |
| UtA-PI | Uterine artery pulsatility index |
| MMPs | Matrix metalloproteinases |
| ECM | Extracellular matrix |
| HIF1α | Hypoxia-inducible factor 1α |
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Figure 1.
Two-stage theory of the pathophysiology of preeclampsia. Predisposing immunological, genetic, and preexisting maternal risk factors may affect abnormal cytotrophoblast invasion of spiral arteries (abnormal placentation) (First stage). The reduced uteroplacental perfusion induces placental release of antiangiogenic factors (soluble fms-like tyrosine kinase 1 (sFlt1)) into the maternal circulation, which antagonizes proangiogenic factors, leading to endothelial dysfunction and systemic vascular dysfunction (Second stage). Preexisting maternal conditions such as chronic hypertension, systemic lupus erythematosus (SLE), and obesity is also associated with endothelial dysfunction.
Figure 2.
Mechanisms of endothelial dysfunction leading to systemic vascular dysfunction in preeclampsia. Excessive soluble fms-like tyrosine kinase 1 (sFlt1) antagonizes VEGF or PlGF, or both, and causes endothelial dysfunction, including a decrease in vasodilators such as nitric oxide (NO) and prostacyclin (PGI2) and an increase in vasoconstrictors such as endothelin-1 (ET-1). VEGF: vascular endothelial growth factor; PLGF: placental growth factor; VEGFR1: VEGF receptor 1 (also known as Flt1).
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Tomimatsu, T.; Mimura, K.; Matsuzaki, S.; Endo, M.; Kumasawa, K.; Kimura, T.
Preeclampsia: Maternal Systemic Vascular Disorder Caused by Generalized Endothelial Dysfunction Due to Placental Antiangiogenic Factors. Int. J. Mol. Sci. 2019, 20, 4246.
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Tomimatsu T, Mimura K, Matsuzaki S, Endo M, Kumasawa K, Kimura T.
Preeclampsia: Maternal Systemic Vascular Disorder Caused by Generalized Endothelial Dysfunction Due to Placental Antiangiogenic Factors. International Journal of Molecular Sciences. 2019; 20(17):4246.
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Tomimatsu, Takuji, Kazuya Mimura, Shinya Matsuzaki, Masayuki Endo, Keiichi Kumasawa, and Tadashi Kimura.
2019. "Preeclampsia: Maternal Systemic Vascular Disorder Caused by Generalized Endothelial Dysfunction Due to Placental Antiangiogenic Factors" International Journal of Molecular Sciences 20, no. 17: 4246.
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11082 | https://ocw.mit.edu/courses/18-782-introduction-to-arithmetic-geometry-fall-2013/f37bbbd7fb8c7ee4e077cc5aa1d2a220_MIT18_782F13_lec24.pdf | 18.782 Introduction to Arithmetic Geometry Fall 2013 Lecture #24 12/03/2013 24.1 Isogenies of elliptic curves Definition 24.1. Let E1/k and E2/k be elliptic curves with distinguished rational points O1 and O2, respectively. An isogeny ϕ: E1 →E2 of elliptic curves is a surjective morphism that maps O1 to O2.
¯ As an example, the negation map that send P ∈E(k) to its additive inverse is an isogeny from E to itself; as noted in Lecture 23, it is an automorphism, hence a surjective morphism, and it clearly fixes the identity element (the distinguished rational point O).
Recall that a morphism of projective curves is either constant or surjective, so any nonconstant morphism that maps O1 to O2 is automatically an isogeny. The composition of two isogenies is an isogeny, and the set of elliptic curves over a field k and the isogenies between them form a category; the identity morphism in this category is simply the identity map from an elliptic curve to itself, which is is clearly an isogeny. Given that the set of rational points on an elliptic curve form a group, it would seem natural to insist that, as morphisms in the category of elliptic curves, isogenies should preserve this group structure.
But there is no need to put this requirement into the definition, it is necessarily satisfied.
Theorem 24.2. Let E1/k and E2/k be elliptic curves and let ϕ: E1 →E2 be an isogeny defined over k. Then ϕ is a group homomorphism from E1(L) to E2(L), for any algebraic extension L/k.
Proof. This is essentially immediate (just consider the pushforward map on divisors), but let us spell out the details.
By base extension to L, it suffices to consider the case L = k. For i = 1, 2, let Oi be the distinguished rational point of Ei and let φ 0 i : Ei(k) →Pick Ei be the group isomorphism that sends P ∈Ei(k) to the divisor class [P −Oi]. Let ϕ : Pic0 k E1 →Pic0 k E2 be the pushforward ∗ map on divisor classes of degree zero. For any P ∈E1(k) we have ϕ ([P]) = [ϕ(P)], since ∗ P and ϕ(P) both have degree one, and ϕ (φ1(P)) = ϕ ([P −O1]) = [ϕ (P −O1)] = [ϕ(P) −ϕ(O1)] = [ϕ(P) −O2] = φ ∗ ∗ ∗ 2(ϕ(P)).
For any P, Q ∈E1(k) with P ⊕Q = R we have P ⊕Q = R φ1(P) + φ1(Q) = φ1(R) ϕ (φ ∗ 1(P) + φ(Q)) = ϕ (φ ∗ !(R)) ϕ (φ )) ∗ 1(P)) + ϕ (φ Q ∗ 1( = ϕ (φ ∗ 1(R)) φ2(ϕ(P)) + φ2(ϕ(Q)) = φ2(ϕ(R)) ϕ(P) ⊕ϕ(Q) = ϕ(R), where ⊕denotes the group operation on both E1(k) and E2(k).
Now that we know that an isogeny ϕ: E1 →E2 is a group homomorphism, we can speak of its kernel ϕ−1(O2). One can view ker ϕ as a set of closed points of E1/k, but it is more ¯ useful to view it as a subgroup of E1(k).
Andrew V. Sutherland 1 Definition 24.3. Let ϕ: E1 →E2 be an isogeny of elliptic curves over k. The kernel of ϕ, ¯ ¯ denoted ker ϕ is the kernel of the group homomorphism ϕ: E1(k) →E2(k).
Recall the translation-by-Q automorphism τQ : E →E that sends P to P ⊕Q. The ¯ induced map τQ ∗ ¯ ¯ : k(E) →k(E) is an automorphism of the function field k(E).
Lemma 24.4. Let ϕ: E1 →E2 be an isogeny of elliptic curves. For each P ∈ker ϕ, the automorphism τ ∗ ¯ fixes ϕ∗¯ ¯ P (k(E2)), and the map ker ϕ →Aut(k(E1)/ϕ∗(k(E2))) defined by P 7→τP ∗is an injective group homomorphism.
Proof. Let P ∈ ¯ ker ϕ and let f ∈k(E2). Then τP ∗(ϕ∗(f))(Q) = (f ◦ϕ ◦τP )(Q) = f(ϕ(P ⊕Q)) = f(ϕ(Q)) = (f ◦ϕ)(Q) = ϕ∗(f)(Q), ¯ since ϕ is a group homomorphism and P lies in its kernel. Thus τP fixes ϕ∗(k(E2)).
For any P, Q ∈ker ϕ and f ∈¯ k(E1) we have τP ∗ ⊕Q(f) = f ◦τP⊕Q = f ◦τQ ◦τP = τP ∗(f ◦τQ) = τP ∗(τQ ∗(f)), so τP ∗ Q = τP ∗◦τQ ∗, and the map P 7→τP ∗is a group homomorphism. It is clearly injective, ⊕ since if P = Q then P ⊖Q = O and τ 1 P ∗ ⊖Q = τP ∗◦(τQ ∗)−is not the identity map (apply it ¯ to any nonconstant f ∈k(E1)).
Corollary 24.5. For any isogeny ϕ: E1 →E2 of elliptic curves, # ker ϕ divides deg ϕ.
In particular, the kernel of an isogeny is finite.
¯ ¯ Proof. By definition, deg ϕ = [k(E1) : ϕ∗(k(E2))], and we know from Galois theory that the order of the automorphism group of a finite extension divides the degree of the extension.
¯ ¯ Since ker ϕ injects into Aut(k(E1)/ϕ∗(k(E2))), its order must divide deg ϕ.
Remark 24.6. In fact, the homomorphism in Lemma 24.4 is an isomorphism, and the corollary implies that when ϕ is separable we have # ker ϕ = deg ϕ; see [1, III.4.10].
24.2 Torsion points on elliptic curves Definition 24.7. Let E/k be an elliptic curve and let n be a positive integer.
The ¯ ¯ multiplication-by-n map [n]: E(k) →E(k) is the group homomorphism defined by nP = P ⊕P ⊕· · · ⊕P.
The points P ∈ ¯ E(k) for which nP = O are called n-torsion points. They form a subgroup ¯ of E(k) denoted E[n].
If ϕ: E1 →E2 is an isogeny, then we know from Corollary 24.5 that n = deg ϕ is a multiple of the order of ker ϕ. It follows that every point in ker ϕ is an n-torsion point. By definition, [n] is a group homomorphism. We now show that [n] is an isogeny.
Theorem 24.8. The multiplication-by-n map on an elliptic curve E/k is an isogeny.
̸ ̸ 2 Proof assuming char(k) = 2: The case n = 1 is clear, and for n = 2 the map P 7→P ⊕P is a rational map, hence a morphism (by Theorem 18.6, a rational map from a smooth projective curve is a morphism), since it can be defined in terms of rational functions of ¯ the coordinates of P via the algebraic formulas for the group operation on E(k). More generally, given any morphism φ: E →E, plugging the coordinate functions of φ into the formulas for the group law yields a morphism that sends P to φ(P) ⊕P. It follows by induction that [n] is a morphism, and it clearly fixes the identity element O.
It remains to show that [n] is surjective. For this it suffices to show that it does not map every point to O, since a morphism of smooth projective curves is either surjective or ¯ constant (by Corollary 18.7). We have already seen that there are exactly 4 points in E(k) that are fixed by the negation map, three of which have order 2 (in short Weierstrass form, these are the point at infinity and the 3 points whose y-coordinate is zero). For n odd, [n] cannot map a point of order 2 to O, so [n] is surjective for n odd. For n = 2km with m odd we may write [n] = ◦· · · ◦ ◦[m]. We already know that [m] is surjective, so it suffices ¯ to show that is. But cannot map any of the infinitely many points in E(k) that are not one of the 4 points fixed by the negation map to O, so must be surjective.
Remark 24.9. Note that in characteristic 2 there are not four 2-torsion points, in fact there may be none. But one can modify the proof above to use 3-torsion points instead.
Corollary 24.10. Let E/k be an elliptic curve. For any positive integer n, the number of ¯ n-torsion points in E(k) is finite.
Remark 24.11. In fact one can show that the number of n-torsion points divides n2, and for n not divisible by char(k), is equal to n2.
24.3 Torsion points on elliptic curves over Q Let E be an elliptic curve Q, which we may assume is given by a short Weierstrass equation E : y2 = x3 + a4x + a6, with a4, a6 ∈Q. Let d be the LCM of the denominators of a4 and a6. After multiplying both sides by d6 and replacing y by d3ny and x by d2nx, we may assume a4, a6 ∈Z. Since E is non-singular, we must have ∆= ∆(E) := −16(4a3 4 + 27a2 6) = 0.1 For each prime p the equation for E also defines an elliptic curve over Qp. For the sake of simplicity we will focus our attention on primes p that do not divide ∆, but everything we do below can be extended to arbitrary p (as we will indicate as we go along). Let E0 denote the elliptic curve over Qp obtained by base extension from Q to Qp. Let E/Fp denote the curve over Fp obtained by reducing the equation for E modulo p. Here we are assuming ∆̸≡0 mod p so that the reduced equation has no singular points, meaning that E is an elliptic curve. We say that E has good reduction at p when this holds.
The reduction map E0(Qp) →E(Fp) is a group homomorphism, and we define E1(Qp) to be its kernel; these are the points that reduce to (0 : 1 : 0) modulo p. In fact, E1(Qp) can be defined as the kernel of the reduction map regardless of whether E has good reduction at p or not and one can show that the points in E1(Qp) still form a group.
The points in E1(Qp) are precisely the points in E0(Qp) that can be represented as (x : y : z), with vp(x), vp(z) > 0 and vp(y) = 0; equivalently, the points with vp(x/y) > 0 1The leading factor of −16 appears for technical reasons that we won’t explain here, but it is useful to have a factor of 2 in ∆because a short Weierstrass equation always has singular points in characteristic 2.
̸ ̸ 3 (note that vp(x/y) does not depend on how the coordinates are scaled). For all positive integers n we thus define En(Qp) = (x : y : z) ∈E0(Qp) : vp(x/y) ≥n Q , and note that this agrees with our previous definition of E1( p).
Lemma 24.12. For n > 0, each of the sets En+1(Qp) is an index p subgroup of En(Qp).
Proof. Containment is clear from the definition, but we need to show that the sets En(Qp) are actually groups. For O = (0 : 1 : 0) we have vp(x/y) = ∞, so O ∈En(Qp) for all n. Any affine point P ∈En(Qp) −En+1(Qp) has vp(x/y) = n, and and after dividing through by z can be written as (x : y : 1) with vp(y) < 0. Since a4, a6 ∈Zp, the equation y2 = x3+a4x+a6 implies 3vp(x) = 2vp(y), so n = vp(x/y) = vp(x) −vp(y) = −vp(y)/3, and therefore vp(y) = n −3n and vp(x) = −2n. After multiplying through by p3n we can write P = (p x 3 0 : y0 : p n) with x0, y0 ∈Z× p . We then have p3ny2 3 0 = p nx3 0 + a4p7nx0 + a 9 6p n y2 0 = x3 0 + a 4n 6n 4p x0 + a6p .
After reducing mod p we obtain an affine point (x0 : y0 : 1) whose coordinates are all nonzero and which lies on the singular variety C0/Fp defined by y2z = x3, which also contains the reduction of O = (0 : 1 : 0). If we consider the image of the group law on E0(Qp) on C0(Fp), we still have an operation defined by the rule that three colinear points sum to zero. We claim that this makes the set S of nonsingular points in C0(Fp) into a group of order p. To show this, we first determine S. We have (∂/∂x)(y2z −x3) = −3x2, (∂/∂y)(y2z 2 −x3) = 2yz, (∂/∂z)(y z −x3) = y2.
It follows that a point in C0(Fp) is singular if and only if its y-coordinate is zero.
In particular all of the reductions of points in En(Qp) are non-singular, for any n ≥1. Every non-singular point in C0(Fp) can be written as (x : 1 : x3), and this gives a bijection from F 3 p to S defined by x 7→(x : 1 : x ). Thus the set S has order p and it contains the identity element. It is clearly closed under negation, and we now show it is closed under addition.
If P and Q are two elements of S not both equal to (0 : 1 : 0), then at least one of them has non-zero z-coordinate and the line L defined by P and Q can be written in the form z = ax + by. Plugging this into the curve equation gives y2(ax + by) = x3, and it is then clear that the third point R in C0 ∩L must have nonzero y-coordinate, since y0 = 0 ⇒x0 = 0 ⇒z0 = 0 for any (x0 : y0 : z0) ∈C0 ∩L. Since P and Q are both in C0(Fp), so is R, thus R lies in S, as does its negation, which is P ⊕Q. Therefore the reduction map En(Q ) →C (F ) defines a group homomorphism from En p 0 p (Qp) to S, and its kernel is En+1(Qp), an index p subgroup of En(Qp).
4 Definition 24.13. The infinite chain of groups E0(Qp) ⊃E1(Qp) ⊃E2(Qp) ⊃· · · is called the p-adic filtration of E/Q.
Theorem 24.14. Let E/Q be an elliptic curve and let p be a prime not dividing ∆(E).
The p-adic filtration of E satisfies (1) E0(Qp)/E1(Qp) ≃E(Fp); (2) En(Qp)/En+1(Qp) ≃E(Fp) ≃Z/pZ for all n > 0; (3) ∩nEn = {O}.
Proof. The group E1(Qp) is, by definition, the kernel of the reduction map from E0(Qp) to E(Fp). To prove (1) we just need to show that the reduction map is surjective.
Let P = (a1 : a2 : a3) be a point in E(Fp) with the ai ∈Z/pZ. The point P is non-singular, so at least one of the partial derivatives of the curve equation f(x1, x2, x3) = 0 for E does not vanish. Without loss of generality, suppose ∂f/∂x1 does not vanish at P. If we ˆ pick an arbitrary point P = (a ˆ1 : a ˆ2 : a ˆ3) with coefficients a ˆi ∈Zp such that a ˆi ≡ai mod p, we can apply Hensel’s to the polynomial g(t) = f(t, a ˆ2, a ˆ3) using a1 ∈Z/pZ as our initial solution, which satisfies g′(ai) = 0. This yields a point in E0(Qp) that reduces to P, thus the reduction map is surjective, which proves (1).
Property (2) follows from the lemma above. For (3), note that E1(Qp) contains only points with nonzero y-coordinate, and the only such point with vp(x/y) = ∞is O; every other other point (x : y : z) ∈E1(Qp) lies in En(Qp) −En+1(Qp), where n = vp(x/y).
Remark 24.15. Theorem 24.14 can be extended to all primes p by replacing E(Fp) in (1) with the set S of non-singular points on the reduction of E modulo p. As in the proof of Lemma 24.12, one can show that S always contains O and is closed under the group operation, but there are now three different group structures that can arise: 1. A cyclic group of order p isomorphic to the additive group of Fp; in this case E is said to as additive reduction at p.
2. A cyclic group of order p−1 isomorphic to the multiplicative group of Fp; in this case E is said to have split multiplicative reduction at p.
3. A cyclic group of order p + 1 isomorphic to the subgroup of the multiplicative group of a quadratic extension of Fp consisting of elements of norm one; in this case E is said to have non-split multiplicative reduction at p.
Parts (2) and (3) of the theorem remain true for all primes p (as we will now assume).
Corollary 24.16. Suppose P = (x : y : 1) is an affine point in E0(Qp) with finite order prime to p. Then x, y ∈Zp.
Proof. Suppose not.
Then both x and y must have negative p-adic valuations in order to satisfiy y2 = x3 + a4x + a6 with a4, a6 ∈Zp, and we must have 2vp(x) = 3vp(y), so vp(x/y) > 0. Let n = vp(x/y). Then P ∈En(Qp) −En+1(Qp), and the image of P in En(Qp)/En+1(Qp) ≃Z/pZ is not zero, hence it has order p. The order m of P is prime to p, so the image of mP in En(Qp)/En+1(Qp) is also nonzero. Thus mP ̸∈En+1(Qp), but this is a contradiction, because mP = O ∈En+1(Qp).
̸ 5 Lemma 24.17. Suppose P1, P2, P3 are colinear points in En(Qp), for some n > 0, with Pi = (xi : 1 : zi). Then vp(x1 + x2 + x3) ≥5n.
Proof. We have already seen that for P ∈En(Q ) we have x n 3n i p i ∈p Zp and zi ∈p Zp.
Fixing y = 1, if P1 = P2 then the equation of the line through P1 and P2 in the x-z plane has the form z = αx + β with α = (z2 −z1)/(x2 −x1). Using the curve equation z = x3 + a3xz2 + z3 (with y = 1), we can rewrite α as z2 α = −z1 x2 −x1 z2 −z1 1 −a z2 4 2 −a 2 6(z = x2 −x1 · 2 + z1z2 + z2 1) 1 −a4z2 2 3 −a 2 6(z2 + z1z2 + z2 1) (z2 = −a6z2) −(z1 −a4x1z2 1 −a6z3 1) −a4x1z2 2 (x1 −x2)(1 −a 2 4z2 −a6(z2 2 + z1z2 + z2 1)) (x3 + a x z2) −x3 2 − 2 2 4 = 2 1 a4x1z2 (x1 −x2)(1 −a4z2 2 2 −a 2 6(z2 + z1z2 + z2 1)) (x2 −x1)(x2 + x 2 1x2 + x = 1 + a4z2 2) (x −x )(1 −a z2 −a (z2 2 1 2 4 2 6 2 + z1z2 + z1)) x2 2 + x1x2 + x2 1 + a4z2 = 2 .
1 −a z2 −a (z2 + z z + z2 4 2 6 2 1 2 1) The key point is that the denominator of α is then a p-adic unit. It follows that α ∈p2nZp, and then β = z1 −αx1 ∈p3nZp. Substituting z = αx + β into the curve equation gives αx + β = x3 + a4x(αx + β)2 + b(αx + β)3.
We know that x1, x2, x3 are the three roots of the cubic defined by the equation above, thus x1 + x2 + x3 is equal to the coefficient of the quadratic term divided by the coefficient of the cubic term. Therefore 2a4αβ + 3a6α2β x1 + x2 + x3 = ∈p5n 1 + a4α2 + a6β3 Zp.
The case P1 = P2 is similar.
Corollary 24.18. The group E1(Qp) is torsion-free.
Proof. By the previous corollary we only need to consider the case of a point of order p. So suppose pP = 0 for some P ∈En(Qp) −En+1(Qp). Consider the map En(Qp) →pnZp/p5nZp that sends P := (x : 1 : z) to the reduction of x in pnZp/p5nZp. By the lemma above, this is a homomorphism of abelian groups, so it sends pP to the reduction of px ∈pn+1Zp−pn+2Zp modulo p5nZp, which is not zero, a contradiction.
Corollary 24.19. Let E/Q be an elliptic curve and let p be a prime of good reduction. The torsion subgroup of E(Q) injects into E(Fp). in particular, the torsion subgroup is finite.
Proof. The group E(Q) is isomorphic to a subgroup of E0(Qp) and E(Fp) = E0(Qp)/E1(Qp).
But E1(Qp) is torison free, so the torsion subgroup of E(Q) injects into E(Fp).
̸ 6 Now that we know that each elliptic curve over Q has a finite number of rational torsion points, one might ask whether there is a uniform upper bound that applies to every elliptic curve over Q. It’s not a priori clear that this should be the case; one might suppose that by varying the elliptic curve we could get an arbitrarily large number of rational torsion points. But this is not the case; an elliptic curve over Q can have at most 16 rational points of finite order. This follows from a celebrated theorem of Mazur that tells us exactly which rational torsion subgroups can (and do) arise for elliptic curves defined over Q.
Theorem 24.20 (Mazur). Let E/Q be an elliptic curve. The torsion subgroup of E(Q) is isomorphic to one of the fifteen subgroups listed below: Z/nZ (n = 1, 2, 3, . . . , 9, 10, 12), Z/2Z × Z/2nZ (n = 1, 2, 3, 4).
The proof of this theorem is well beyond the scope of this course.2 However, as a further refinement of our results above, we can prove the Nagell-Lutz Theorem.
Theorem 24.21 (Nagell-Lutz). Let P = (x0 : y0 : 1) be an affine point of finite order on the elliptic curve y2 = x3 +a4x+a6 over Q, with a4, a6 ∈Z. Then x0, y0 Z, and if y0 = 0 then y2 divides 4a3 2 ∈ 0 4 + 27a6.
Proof. For any prime p, if vp(x0) < 0 then 2vp(y0) = 3vp(x0) and vp(x0/y0) > 0. It follows that P ∈E1(Q), but then P cannot be a torsion point. So vp(x) ≥0 for all primes p. Thus x0 is an integer, and so is y2 0 = x3 0 + a4x0 + a6, and therefore y.
If P has order 2 then y0 = 0; otherwise, the x-coordinate of 2P is an integer equal to λ2 −2x , where λ = (3x2 0 0 + a4)/(2y0) is the slope of the tangent at P. Thus 4y2 0 and therefore y2 divides λ2 2 2 3 0 = (3x0 + a4) , as well as x0 + a4x0 + a6. We now note that (3x2 0 + 4a4)(3x2 0 + a4)2 = 27x6 0 + 54a4x4 0 + 27a2 4x2 0 + 4a3 4 (3x2 0 + 4a4)(3x2 0 + a 2 4) = 27(x3 0 + a4x0)2 + 4a3 4 0 ≡27a2 6 + 4a3 4 mod y2 0, since (x3 0 + a4x0) ≡−a6 mod y2 0, thus y0 divides 4a3 4 + 27a2 6.
The Nagell-Lutz theorem gives an effective method for enumerating all of the torsion points in E(Q) that is quite practical when the coefficients a4 and a6 are small. By factoring D = 4a3 4 + 27a2 6, one can determine all the squares y2 0 that divide D. By considering each of these, along with y0 = 0, one then checks whether there exists an integral solution x0 to y2 0 = x3 0 + ax0 + a6 (note that such an x0 must be a divisor of a6 −y2 0).
This yields a list of candidate torsion points P = (x0 : y0 : 1) that are all points in E(Q), but do not necessarily all have finite order. To determine which do, one computes multiples nP for increasing values of n (by adding the point P at each step, using the group law on E), checking at each step whether nP = O. If at any stage it is found that the affine coordinates of nP are not integers then nP, and therefore P, cannot be a torsion point, and in any case we know from Mazur’s theorem that if nP = O for any n ≤12 then P is not a torsion point; alternatively, we also know that n must divide #E(Fp), where p is the least prime that does not divide ∆(E).
However, this method is not practical in general, both because it requires us to factor D, and because D might have a very large number of square divisors (if D is, say, the product 2There was recently a graduate seminar at Harvard devoted entirely to the proof of Mazur’s theorem; see for notes and references.
̸ ̸ 7 of the squares of the first 100 primes, then we have 2100 values of y0 to consider). But Cororllary 24.19 gives us a much more efficient alternative that can be implemented to run in quasi-linear time (roughly proportional to the number of bits it takes to represent a4 and a6 on a computer).
We first determine the least odd prime p that does not divide D (we don’t need to factor D to do this), and then exhaustively compute the set E(F √ p), which clearly has cardinality at most 2p (in fact, at most p + 1 + 2 p). We then use Hensel’s lemma to “lift” these points to Z by computing p-adic approximations of their coordinates to sufficiently high precision; the fact that y2 0 must divide D gives us an upper bound on x0 and y0. This yields a small set of candidate points P = (x0 : y0 : 1) with x0, y0 ∈Z that satisfy the curve equation modulo some power of p. For each candidate P we then check whether P ∈E(Q) by plugging its coordinates into the curve equation (it need not be satisfied, P could be the reduction of a Qp-point that doe not lie in E(Q)). If P ∈E(Q), we then check that it is actually a torsion point by computing nP = E(Q), where n is the order of the reduction of P in E(Fp).
References J. H. Silverman, The arithmetic of elliptic curves, Springer, 2009.
8 MIT OpenCourseWare ,QWURGXFWLRQWR$ULWKPHWLFHRPHWU\ )DOO 201 For information about citing these materials or our Terms of Use, visit: |
11083 | https://labobineuse.com/en/blogs/bobi-blog/la-fameuse-regle-de-3?srsltid=AfmBOooD7D5Dbyva8YKCKVm5TI-uHymsVgKE08IWHxMI495l7edZLVAI | Language
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The famous rule of 3!
If you've ever been to our store and talked about sizing or size adaptation with one of us, we've probably told you about the rule of 3! It's almost become a passion for us!
What is the rule of 3?
The rule of three (also called the cross product rule) is a simple method in mathematics that allows you to solve proportion problems. It is useful when you want to find a missing value knowing that two things are proportional to each other.
How It Works
If we have two quantities that are proportional, we can use the rule of three to calculate a missing quantity.
For example, let's say we know that:
In proportion, it is written like this:
To find x , we apply the rule of three with this formula:
Practical Example
Let's imagine that we are looking for how many stitches we will need to cast on to make a blanket 60 cm wide, knowing that we have cast on 30 stitches to make our 10 cm wide sample.
We use the rule of three to find x (the number of stitches):
So, we will need to cast on 180 stitches to get 60 cm.
How to Apply the Rule of Three
It's really simple and effective for finding the right size of patterns according to our measurements, knitting simple projects according to our samples and adapting patterns to the right size (provided you can adapt the decreases/increases, which is not so easy!)
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The famous rule of 3!
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Arithmetic Mean or Average on GMAT
Quick Summary
Average equals sum of terms divided by number of terms. Example: 5, 7, and 9 sum to 21, so the average is 21 ÷ 3 = 7. Adding k to each term raises the average by k; multiplying/dividing all terms by k multiplies/divides the average by k.
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Overview
The concept of averages, or the arithmetic mean, may appear simple but is one of the most widely tested ideas in aptitude and competitive exams. It goes far beyond dividing a sum by the number of terms. Averages test your ability to see patterns, apply rules consistently, and avoid hidden traps. For instance, when every term in a set changes by the same value, the mean changes by exactly that value. Similarly, when every term is multiplied or divided by a number, the mean transforms in the same way. This makes averages a powerful tool for quick mental calculations. At times, students confuse mean with median or mode, but there is no fixed hierarchy between them. Depending on the data, any one of them could be the largest or the smallest. Building confidence with such subtleties is vital for GMAT preparation, and solving timed questions helps master them.
Arithmetic Mean or Average on GMAT
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Understanding the Arithmetic Mean or Average
The mean, or average, is defined as the sum of all terms in a set divided by the number of terms.
Example:
{10, 7, 5, -20, 10, 15, 25, 50, 100, 200}
Sum of terms = 402
Number of Terms = 10
Mean = 40.2
Average or the arithmetic means is a measure of central tendency that often summarizes an entire dataset in one value.
Key Properties of Averages
The mean may or may not be an actual term in the set.
If every term in the set increases or decreases by the same number X, the mean also increases or decreases by X.
If every term is multiplied or divided by X, the mean also gets multiplied or divided by X.
The mean has no fixed relationship with median or mode; any of them could be higher or lower depending on the data.
Why Averages Matter
Averages are not just calculations; they represent insight. Questions often test whether you can apply the properties logically, rather than perform lengthy arithmetic. With practice, averages become a tool for elegant shortcuts that save time on the test. Consistent effort through GMAT practice tests ensures that these patterns become second nature.
Parting Thought
Average is a quiet reminder that balance is built one number at a time. Each small change shifts the center, just as each study session nudges your preparation toward clarity. The mean rewards consistency, not bursts of effort. It teaches you to integrate mistakes without fear, to learn, and to move on. Keep adding steady, honest work to your set. Over days and weeks, your center will move. What you practice repeatedly becomes who you are on test day, truly.
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11085 | https://math.stackexchange.com/questions/3030102/what-is-the-coefficient-of-x11-in-3x-919 | exponential function - What is the coefficient of $x^{11}$ in $(3x-9)^{19}$? - Mathematics Stack Exchange
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What is the coefficient of x 11 x 11 in (3 x−9)19(3 x−9)19?
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I am currently studying for finals, and I do not know how to do this problem from my study guide. I have tried to watch a few YouTube videos and I know that I will end up with 3 x 11×(−9)8 3 x 11×(−9)8, but from here I do not know where to go. The YouTube videos I have watched have been unclear, and I do not have a chance to ask my professor until just a little before the exam. If anyone has any tips on where to go from here I would greatly appreciate it!
exponential-function
binomial-coefficients
binomial-theorem
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edited Dec 7, 2018 at 17:10
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asked Dec 7, 2018 at 16:36
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Don't forget about the binomial coefficient.John Wayland Bales –John Wayland Bales 2018-12-07 16:40:44 +00:00 Commented Dec 7, 2018 at 16:40
Are you aware of the BInomial expansion (a x−b)n=∑n r=0(n r)(a x)n−r⋅(−1)r⋅b r(a x−b)n=∑r=0 n(n r)(a x)n−r⋅(−1)r⋅b r?Prakhar Nagpal –Prakhar Nagpal 2018-12-07 16:41:27 +00:00 Commented Dec 7, 2018 at 16:41
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(3 x−9)19=(3 x−9)(3 x−9)⋯(3 x−9)(3 x−9)19=(3 x−9)(3 x−9)⋯(3 x−9)
Notice that on the right hand side there exist 19 19 products. A term can be formed by choosing either 3 x 3 x or −9−9 from each of these 19 19 products.
If your goal is to find the coefficient of x 11 x 11, you have to choose 3 x 3 x for a 11 11 times from the available 19 19 products, and then choose −9−9 for 8 8 times from the remaining 8 8 products.
If you choose 3 x 3 x for 11 11 times, you end up with (3 x)11(3 x)11, and you can do this in exactly (19 11)(19 11) ways. (because you have 19 products available and you're considering only 11 of them for 3 x.)(because you have 19 products available and you're considering only 11 of them for 3 x.)
For each of the above (19 11)(19 11) ways, you can choose −9−9 from the remaining 8 8 products, this gives (−9)8(−9)8
So the overall term would be (19 11)(3 x)11(−9)8(19 11)(3 x)11(−9)8
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edited Dec 7, 2018 at 17:12
answered Dec 7, 2018 at 17:00
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Thank you this makes sense! I completely forgot about the choose bmurf17 –bmurf17 2018-12-07 19:04:41 +00:00 Commented Dec 7, 2018 at 19:04
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HINT: The Binomial expansion :
(a+b)n=∑k=0 n(n k)a k b n−k.(a+b)n=∑k=0 n(n k)a k b n−k.
Substitute a=3 x a=3 x and b=−9 b=−9 . To find out the coefficient of x 11 x 11,put k=11 k=11 in (n k)a k b n−k(n k)a k b n−k
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answered Dec 7, 2018 at 16:48
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Note that by expansion of a binomial to the n t h n t h power, you get
(a+b)n=∑r=0 n(n r)a n−r b r=a n+a n−1 b+a n−2 b 2+…+a n−r b r+…+a b n−1+b n(a+b)n=∑r=0 n(n r)a n−r b r=a n+a n−1 b+a n−2 b 2+…+a n−r b r+…+a b n−1+b n
where (n r)(n r) corresponds to the binomial coefficients, or Pascal Numbers. (Never forget those for coefficients.)
For (3 x−9)19(3 x−9)19, the first term has x 19 x 19, the second term has x 18 x 18, etc. So, x 11 x 11 corresponds to the ninth term. Hence, you get
(19 19−8)(3 x)11(−9)8(19 19−8)(3 x)11(−9)8
(19 11)(3 x)11 9 8(19 11)(3 x)11 9 8
Simplifying from here, you can solve for the ninth term and find the coefficient of x 11 x 11.
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edited Dec 7, 2018 at 17:02
answered Dec 7, 2018 at 16:44
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You are part way there.
If you were to write out (3 x−9)19(3 x−9)19 in full, it would be a product of nineteen (3 x−9)(3 x−9). To get an x 11 x 11 you would need to pick 11 of these (3 x−9)(3 x−9).
There are (19 11)(19 11) ways of doing that, so the term in x 11 x 11 is
(19 11)(3 x)11(−9)8(19 11)(3 x)11(−9)8
Thus the coefficient of x 11 x 11 is
(19 11)(3)11(−9)8=(19 11)3 27(19 11)(3)11(−9)8=(19 11)3 27
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answered Dec 7, 2018 at 17:04
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11086 | https://davidaltizio.web.illinois.edu/Divisors%20and%20Divisibility%20Overview.pdf | Randolph High School Math League 2013-2014 Page 1 Introduction to Number Theory 1 What is Number Theory?
Number Theory is a branch of mathematics that explores the integers and their properties. This is a much different way to approach mathematics, as previously the problems many of you have experienced deal with real numbers, a more general case. Number theory uses its own special tools in order to restrict the set of solutions within the set of integers.
2 Basic Terminology These are all things you should be familiar with from previous mathematics courses, but in the interest of being self-contained I will review them.
• It’s sort of difficult to define integers in themselves, but in this case the Wikipedia definition suffices: “An integer is a number that can be written without a fractional or decimal component.” Some examples of integers are 5, −17, and 3628800. It is important to know that while the set of integers is closed1 under addition, subtraction, or multiplication, it is NOT under division!
• An integer a is said to be a multiple of another integer b if there exists an integer k such that a = kb. The integer b here is said to be called a factor or a divisor2 of a. Furthermore, a is said to be divisible by b.
If an integer has no positive divisors other than 1 and itself, it is said to be prime; otherwise, it is said to be composite (with the exception of 1, of course.) 3 Divisibility Some of the simplest number theory problems ask you to manipulate divisibility rules in order to solve a question.
In any case, don’t be afraid to churn out a little bit of basic algebra and/or casework, as there may be multiple solutions to the problem.
Example 1. What is the value of the digit X such that the three-digit number 3X4 is divisible by 9?
Solution. The divisibility rule states that the sum of the digits of any number must be divisible by 9 in order for the original number to be divisible by 9. The value of X such that 7 + X is divisible by 9 must be X = 2 .
■ Example 2. The four-digit positive integer 3BAA is divisible by 12. What is the sum of all possible values of A + B?
Solution. It suffices to examine the divisibility rules for 3 and 4. In order for the integer to be divisible by 4, AA must also be divisible by 4, so A = 4 or A = 8. If A = 4, then the sum of the digits of the number is 11 + B, and since this sum must be divisible by 3, B = 1, 4, 7 all work. Similarly, if A = 8, then the sum of the digits of the number is 19 + B, and the values of B that make this sum divisible by 3 are B = 2, 5, 8. Hence the six ordered pairs (A, B) = (4, 1), (4, 4), (4, 7), (8, 2), (8, 5), (8, 8) all work, and the sum of all possible values of A + B is 63 .
■ 4 Prime Factorization Any integer N can be written as the product of the primes it is divisible by. The prime factorization of N is N = Y p∈P pei = 2e1 · 3e2 · 5e3 · . . . , where P is the set of positive primes and {ei} is a sequence of integers determining how many times the ith prime number can be divided out of N. For example, 144 = 22 · 32 and 7! = 24 · 32 · 5 · 7.
Often times, the crux of a number theory problem is determining how to deal with its prime factorization, albeit indirectly.
1A set A is said to be closed under an operation if said operation takes members from the set to produce a member from that same set. For example, multiplying two integers produces another integer.
2While both of these terms are usually interchangable, usually the latter one is more often used.
Randolph High School Math League 2013-2014 Page 2 Example 3. What is the smallest positive integer that, when multiplied by 60, results in a perfect cube?
Solution. As the title of this section suggests, we look at the prime factorization of 60, 22 · 3 · 5. Note that in order for an integer to be a perfect cube, all of the exponents in its prime factorization must be multiples of 3. Therefore, in order to do that, we need to add one factor of 2, two factors of 3, and two factors of 5. Hence the requested number is 2 · 32 · 52 = 450 .
■ 5 GCD and LCM The greatest common divisor of a set of integers A is the largest positive integer n that divides evenly into every element in A. For example, gcd(15, 20) = 5 and gcd(12, 18) = 6. Similarly, the least common multiple of a set of integers B is the smallest positive integer N such that N is evenly divisible by every integer in B. For example, lcm(15, 20) = 60 and lcm(12, 18) = 36.
There is a general expression for the greatest common divisor and least common multiple of two integers, and its roots lay within the idea of prime factorization: Important: Let M and N be positive integers. Consider the set of primes p1, p2, . . . , pk that divide either M or N. Let M = pe1 1 · pe2 2 · . . . · pek k , N = pf1 1 · pf2 2 · . . . · pfk k .
Then gcd(M, N) = pmin{e1,f1} 1 · pmin{e2,f2} 2 · . . . · pmin{ek,fk} k and lcm(M, N) = pmax{e1,f1} 1 · pmax{e2,f2} 2 · . . . · pmax{ek,fk} k .
This is easily justified by the definition of GCD and LCM; I leave it as an exercise to the interested individual.
Example 4. What is the greatest common divisor of 60 and 150?
Solution. It is easy to see that 60 = 22 · 3 · 5 and 150 = 2 · 3 · 52. In order for an integer to divide both 60 and 150, it must have at most one factor of 2, at most one factor of 3, and at most one factor of 5.3 Therefore the largest possible integer that divides both 60 and 150 is 2 · 3 · 5 = 30 .
■ Of course, this technical definition is not always necessary. Sometimes, it is important to simply understand what the definitions of gcd and lcm actually imply.
Example 5 (Math League HS 2000-2001). With each entry I submit, I have to write a different pair of positive integers whose greatest common factor is 1 and whose sum is 2000. (Pairs differing only in the order of addition are counted as 1 pair, not two different pairs.) For example, I submitted the pair (1, 1999) with my first entry.
With these restrictions, at most how many entries can one person submit?
Solution. Instead of focusing on the pairs where the greatest common divisor is 1, we instead focus on the opposite set: those pairs in which the two integers share a common factor.4 Note that by the definition of greatest common divisor, x and y are both divisible by gcd(x, y). Therefore, the sum of these integers must be divisible by gcd(x, y) as well. Since 2000 = 24 · 53, in order for gcd(x, y) to be greater than 1, x and y must be both divisible by either 2 or 5 (or both).
We now proceed to count the number of unordered pairs of integers (x, y) that satisfy the above condition. The pairs in which x and y are both divisible by 2 are (2, 1998), (4, 1996), . . . , (1000, 1000). (By symmetry, we only need to go up to x = 1000, due to the condition that the pair (x, y) is identical to the pair (y, x).) Similarly, the pairs in which x and y are both divisible by 5 are (5, 1995), (10, 1990), . . . , (1000, 1000). However, if x and y are divisible by both 2 and 5, then the pair (x, y) is counted in both lists, so we need to subtract those pairs once in order to not overcount!5 These pairs are (10, 1990), (20, 1980), . . . , (1000, 1000). By simple counting arguments, we can determine that there are 500, 200, and 100 pairs in the three lists respectively, for a total of 500 + 200 −100 = 600 unordered pairs that satisfy the condition. Therefore, since there are 1000 possible unordered pairs, I can submit at most 1000 −600 = 400 entries in total.
■ 3Try and see why this reasoning works and how it derives the boxed formula shown above!
4This technique is more broadly referred to as complementary counting, which will be discussed more in-depth in the lecture on counting and probabilty.
5This technique is called the Principle of Inclusion-Exclusion, another idea prevalent in combinatorics.
Randolph High School Math League 2013-2014 Page 3 6 Practice Problems 1. What is the largest prime factor of 143 000 000?
2. Three positive integers are each greater than 1, have a product of 27000, and are pairwise relatively prime.
What is their sum?
3. Let (a, b) denote the greatest common factor of a and b, and let [a, b] denote the least common multiple of a and b. Evaluate the following expressions: (a) (15, 20) (b) [66, 121] (c) (126, 184) ⋆(d) (2143567, 2143369) 4. The six-digit integer 12345X is divisible by 11. What is the value of the digit X?
5. When 270 is divided by the odd number x, the quotient is a positive prime (and the remainder is 0). What is the value of x?
6. What is the smallest positive integer that is neither prime nor square and that has no prime factor less than 50?
7. What is the greatest possible sum of two multiples of 12, each less than 100, whose greatest common factor is 24?
8. A standard six-sided die is rolled, and P is the product of the five numbers that are visible. What is the largest number that is certain to divide P?
9. Which of the following numbers is a perfect square?
(A) 14!15!
2 (B) 15!16!
2 (C) 16!17!
2 (D) 17!18!
2 (E) 18!19!
2 10. A store sold 72 decks of cards for $a67.9b. Find a + b.
11. A monoprime is a positive even number that isn’t the product of two even numbers. An irregular number is an integer which can be written as a product of two (possibly equal) monoprimes in more than one way.
What is the smallest irregular number?
12. The product of any two of the positive integers 30, 72, and N is divisible by the third. What is the smallest possible value of N?
13. The product of two positive integers is 9984, and the greatest common factor of these integers equals the difference between them. What are the two integers?
14. Find the largest natural number n below 50 such that LCM(n, n + 1, . . . , 50) = LCM(1, 2, . . . , 50), where LCM stands for least common multiple.
15. For positive integers n ≥2, define g(n) to be one more than the largest proper divisor of n. Hence g(35) = 8, since the proper divisors of 35 are 1, 5, and 7. For how many n in the range 2 ≤n ≤100 do we have g(g(n)) = 2?
⋆16. Let [r, s] denote the least common multiple of positive integers r and s. Find the number of ordered triples (a, b, c) of positive integers for which [a, b] = 1000, [b, c] = 2000, and [c, a] = 2000.
⋆17. Let x and y be positive integers such that 7x5 = 11y13.
The minimum possible value of x has a prime factorization acbd. What is a + b + c + d?
⋆18. An infinite sequence of positive integers {an} is such that for any two positive integers i ̸= j, gcd{ai, aj} = gcd{i, j}.
Show that ai = i for all i. |
11087 | https://peer.asee.org/integration-of-the-boiling-experiments-in-the-undergraduate-heat-transfer-laboratory.pdf | Session 3226 Integration of Boiling Experiments in the Undergraduate Heat Transfer Laboratory Hosni I. Abu-Mulaweh, Josué Njock Libii Engineering Department Indiana University-Purdue University at Fort Wayne Fort Wayne, IN 46835, USA Abstract This paper presents three boiling experiments that can be integrated in the undergraduate heat transfer laboratory. The objective of these experiments is to enhance the understanding of boiling process by undergraduate mechanical engineering students. These experiments expose the students to several important concepts in boiling, such as subcooled boiling, modes of pool boiling, and Leidenfrost Phenomenon. The experimental setup and apparatus required to carry out these experiments is simple. It includes a metallic plate such as brass, stainless steel or aluminum, a heating source such as a Hot Plate, thermocouples, a stopwatch, a liquid dropper, and a camera. These equipments are inexpensive and available in almost all undergraduate heat transfer laboratories.
I. Introduction Boiling and condensing processes play an important role in a large number of practical engineering applications, such as the production of electrical power from vapor cycles, production of refrigeration, and the design of petrochemical processes (such as the refining of petroleum and the manufacture of chemicals). Boiling and condensing are vapor-liquid phase change processes where fluid motion is involved. Due to this fact, boiling and condensing are classified as convective mechanisms. However, there are major differences between these mechanisms and single phase convective heat transfer. This is because there are significant differences between the various fluid properties in the two phases, such as conductivity, specific heat, and density. Also, there is a consumption or release of latent heat hfg which influences the heat transfer rates greatly during phase change.
Both boiling and evaporation are liquid-to-vapor phase change processes, but there are major differences between the two. Evaporation process occurs at the liquid-vapor interface when the vapor pressure pv is less than the saturation pressure psat of the liquid at a given temperature. And evaporation does not involve bubble formation or bubble motion. Examples of evaporation are the evaporation of sweat to cool the human body and the drying of fruits and cloths. On the other hand, boiling occurs at the solid-liquid interface when the temperature of the surface is maintained at a temperature Ts that exceeds the saturation temperature Tsat corresponding to the pressure of the liquid that is in contact with the surface. The boiling process is characterized by Page 5.382.1 the rapid formation of vapor bubbles at the solid-liquid interface. When the vapor bubbles reach a certain size they start to detach from the surface and attempt to rise to the free surface of the liquid. Bubbles are formed, during the boiling process, as a result of the surface tension F at the liquid-vapor interface due to the attraction force on molecules at the interface toward the liquid phase.
Boiling is classified as pool boiling or flow boiling (forced convection boiling) depending on the absence or presence of fluid motion, respectively. In the case of pool boiling, the fluid is stationary, and its motion near the surface is due to natural convection and to the motion of the bubbles caused by their growth and detachment. Whereas, in flow boiling, the fluid is set in motion by external means such as a pump, as well as by natural convection and the motion of the bubbles. In addition, boiling is classified as subcooled boiling or saturated boiling, depending on the liquid temperature. Boiling is referred to as subcooled when the temperature of the liquid is below the saturated temperature Tsat (i.e., the liquid is subcooled) and it is considered saturated when the temperature of the liquid is equal to the saturated temperature Tsat (i.e., the liquid is saturated). It should be noted that the three experiments presented in this paper examine pool boiling under subcooled conditions.
Depending on the value of the excess temperature )Te which represents the excess of the surface temperature above the saturation temperature of the liquid ()Te = Ts - Tsat ), pool boiling takes different forms. These forms or regimes are natural convection boiling, nucleate boiling, transition boiling, and film boiling. One of the three experiments suggested in this paper is to observe the different mechanisms of pool boiling in these different regimes. In another experiment, the total evaporation time of droplets of water deposited on a hot surface will be measured at different surface temperatures. The trends will be compared with the boiling curve (see section B). In the third experiment, the Leidenfrost point, where the heat flux reaches a minimum, will be determined for different liquids.
The objective of this paper is to develop laboratory experiments to enhance the learning of basic boiling concepts by undergraduate mechanical engineering students. The equipments required to achieve this goal are inexpensive and available in almost all undergraduate heat transfer laboratories.
II. Experimental Setup and Equipment The experimental apparatus is relatively simple and inexpensive. A brass plate (140 mm long, 76 mm wide and 12.7 mm thick) is placed horizontally on a Scientific Hot Plate. The brass plate can be heated and maintained at a constant temperature by adjusting the Scientific Hot Plate to the appropriate heating level. The temperature of the brass plate is measured by a thermocouple.
The measuring junction of the thermocouple is inserted from the side of the brass plate into a small hole and it reaches the center of the plate. The output of the thermocouple is read using an Omega Microprocessor Thermometer. A depression in the form of a shallow spherical cap (35 mm in diameter and 3.4 mm deep) has been machined into the brass plate in order to provide a seat onto which droplets of liquid could be deposited for observation. A camera was used to take pictures of the different boiling forms and regimes. A stopwatch was also used to measure the Page 5.382.2 time that the liquid droplet would take to evaporate. All observations and measurements were made only after the system had reached steady state conditions.
III. The Experiments A. Observations of the Different Regimes in Pool Boiling of Subcooled Liquid The objective of this experiment was to observe the different behaviors associated with the different regimes of pool boiling (i.e., natural convection boiling, nucleate boiling, transition boiling, and film boiling) of a droplet of subcooled water on a brass plate that is heated to a temperature that exceeds the saturated temperature of water.
The horizontal brass plate is heated to the desired temperature Ts by adjusting the electrical energy input to the Scientific Hot Plate. When steady state is reached, a droplet of deionized water at room temperature is placed in the grove on the heated brass plate using a liquid dropper.
The behavior of the water droplet during the boiling process is observed carefully and recorded.
Also, a picture of the droplet during the boiling process can be taken for the record. This procedure is repeated several times, by readjusting the electrical energy input to the Scientific Hot Plate, to cover all the regimes of pool boiling. It should be noted that the temperature of the brass plate Ts dropped about 1 to 1.5oC after the water droplet is deposited on the surface. This is because of the heat transfer (the heat loss) from the surface of the heated brass plate to the water droplet during the evaporation of the droplet. But once the evaporation of the water droplet is completed, the temperature of the plate Ts increases to the original temperature value.
It was observed that the water droplet assumed completely different shapes during the boiling process in each regime and the time of evaporation was different from one regime to another and within the regime itself. The results were carefully analyzed and grouped into seven different zones of behavior that are summarized in Table 1. The reason for arranging these observation in seven zones and not four, the number of the different regimes in pool boiling (natural convection boiling, nucleate boiling, transition boiling, and film boiling) is that the characteristics sometimes are different within one regime. Zones I and II correspond to natural convection boiling and nucleate boiling, respectively. Zone III is the neighborhood where minimum evaporation time of the liquid droplet is attained and at which there is maximum heat flux qs (i.e., maximum heat transfer rate from the heated surface to the liquid droplet). The transition boiling regime is divided into three zones (zones IV, V, and VI). This is because the behavior of the evaporation of the liquid droplet was not exactly the same throughout the transition regime as Table 1 illustrates. And finally, zone VII corresponds to the film boiling regime. Figures 1, 2, 3, and 4 present pictures of the droplet behavior during the different regimes of pool boiling.
The pictures (a) and (b) in Fig. 1 show the pool boiling process of a subcooled water droplet in the natural convection boiling regime at )Te = 7oC. Figure 2 shows the pool boiling process of a subcooled water droplet in the nucleate boiling regime at )Te = 16oC. The transition boiling at )Te = 120oC and film boiling at )Te = 161oC are shown in Figs. 3 and 4, respectively. The description of the behavior of the water droplet during the boiling process shown in these pictures is summarized in Table 1.
Page 5.382.3 Table 1 Characteristic Zones and their Descriptions.
Observed Zones Excess Temperature Range ( oC) Observed Behavior I 0 < )Te < 10 Many small bubbles form instantly at the base. Two main features are observed here: 1) a single bubble remains by itself for a short while; and 2) after this single bubble disappears, there is a period when no bubbles can be seen in the drop at all. Finally, the drop shrinks until it disappears. II 10 < )Te < 44 Many bubbles form at the base of the drop. The drop spreads out, swells up and breaks up into two or more patches. The bubbles disappear. Thereafter, individual patches shrink in size continuously until they disappear.
Increases in temperature appear to speed up the processes described in this zone.
III 44 < )Te < 50 The drop breaks up very quickly and its evaporation is so rapid that it seems instantaneous.
IV 50 < )Te < 70 The drop breaks up so quickly that it starts to shatter into small droplets. Evaporation is very rapid. Shattering becomes even more evident at higher temperatures.
V 70 < )Te < 120 The drop shatters instantly into crystallized balls that are well defined. One of the balls is typically larger than the rest. The balls jump from one spot to another repeatedly ( like pop corn). The large ball gets larger with increasing temperatures and the smaller balls become fewer and fewer. At still higher temperatures, however, the large ball, once formed, breaks up into smaller balls whose sizes vary with the temperature.
VI 120 < )Te < 130 One single ball is formed. It takes it a while to evaporate.
However, just before it evaporates completely, it would shatter into tiny balls.
VII 130 < )Te < 200 One single ball is formed. It takes it a while to evaporate.
The ball gets smaller and smaller until it evaporates completely. No shattering is observed at all.
Page 5.382.4 (a ) (b ) Fig. 1 Natural Convection Boiling Regime Fig. 2 Nucleate Boiling Regime Fig. 3 Transition Boiling Regime Fig. 4 Film Boiling Regime Page 5.382.5 B. Measurements of Total Evaporation Time of a Droplet of Subcooled Water in Pool Boiling The objective of this experiment was to experimentally determine the total evaporation time of a droplet of deionized water in the different pool boiling regimes. The mass of the individual droplet was measured to be m = 32 mg. This value represents the average of 10 different measurements using a Denver Instrument M-310 digital scale. The repeatability of the mass measurements was determined to be within 2.5 mg (7.8%).
The experimental procedure for this experiment is relatively simple. The horizontal brass plate is heated to the desired temperature by the Scientific Hot Plate. Once steady state condition is reached, the water droplet is then deposited onto the heated plate using a liquid dropper. The total evaporation time (i.e., the time elapsed from the instant at which the water droplet is deposited on the heated plate to the instant at which the droplet is evaporated completely) was measured using a stopwatch. Each measurement was repeated 5 times and then an average of the total evaporation times was calculated. The repeatability of the time measurements was determined to be within 9%. It should be noted that the measurements of the total evaporation time were carried out for three regimes only: the natural convection boiling regime, nucleate boiling regime, and film boiling regime. It was not possible to measure the total evaporation time of the droplet in the transition boiling regime. This is because the ball that formed during the transition boiling shatters to smaller balls and some of these balls jump off the heated brass plate. The transition boiling regime is also known as the unstable boiling or partial film boiling regime. In this boiling regime, an unstable condition exists in which the process oscillates between nucleate boiling and film boiling.
The variation of the total evaporation time with the excess temperature )Te in the natural convection and nucleate boiling regimes is illustrated in Fig. 5. The figure clearly shows that the total evaporation time in these two regimes decreases as the excess temperature )Te increases.
This is because in these two regimes the heat flux qs increases with increasing excess temperature )Te (i.e., the rate of heat transfer from the heated plate to the water droplet increases as the surface temperature Ts increases) as the general boiling curve in reference shows. The measured results of the total evaporation time in these two regimes agree favorably with the results reported by Cumo, et al. .
Figure 6 shows the effect of the excess temperature )Te on the total evaporation time in the film boiling regime. It can be seen from the figure that the total evaporation time decreases as the excess temperature )Te increases. Again, this is because in the film boiling regime, the heat flux qs increases as the excess temperature increases (i.e., the rate of heat transferred from the hot surface increases as the surface temperature Ts increases in this regime) as the general boiling curve in reference shows. In the film boiling regime, the heat flux qs increases with increasing excess temperature )Te as a result of heat exchange between the heated plate and the liquid droplet through the vapor film due to radiation, which becomes significant at high temperatures ()Te $150oC). These measured results of the total evaporation time in the film boiling regime agree favorably with the results reported by Cumo, et al. in this regime. Page 5.382.6 Page 5.382.7 C. The Leidenfrost Phenomenon: Film Boiling of Liquid Droplets on a Flat Plate The film boiling of small droplets of liquid on a heated surface is commonly known as the Leidenfrost Phenomenon after J. G. Leidenfrost who first studied the process in 1756. The objective of this experiment is to determine the Leidenfrost point, defined as the excess temperature at which the droplet evaporation time is greatest, for different liquids, such as water, ethanol, benzene. It was reported by Gottfried et al. that the Leidenfrost point is well defined for organic liquids (ethanol and benzene) and it is at excess temperature )Te of 100 - 105oC , while for water it varies between excess temperatures )Te of 150oC and 210oC. This is because for water the Leidenfrost point depends on the surface and the method of depositing the droplet on the surface. They also reported that the Leidenfrost point is independent of droplet size.
However, the experimental study of Cumo, et al. indicate that the Leidenfrost point for water droplet on a hot plate is about an excess temperature )Te of 140oC. And in our experimental study, the Leidenfrost point for a dionized water droplet on a heated brass plate was observed to be about an excess temperature )Te of 130oC. The boiling process of liquid droplets on a hot surface was extensively examined in the past (see for example, Wachters et al. and Wachters and Westerling , Wachters and van Andel ). In this experiment, the students will try to determine Leidenfrost point for droplets of water and a couple of organic liquids on a hot horizontal plate, and then compare their results with those of reported in the literature. Also, they will examine the fact that the Leidenfrost point is independent of the droplet size.
IV. Impact of the Project The three boiling experiments discussed above were introduced to the students of the undergraduate heat transfer laboratory (ME 322) in spring 2000. Upon the completion of these experiments, the students were asked to submit some feedback regarding the integration of these experiments into the undergraduate heat transfer laboratory. Almost all of the students favorably agreed that the inclusion of these three boiling experiments will widen the scope of the undergraduate heat transfer laboratory. They also have indicated that these experiments have enhanced their understanding of the boiling phenomenon.
Bibliography 1. Incropera, F. P. and DeWitt, D. P. (1996), “Fundamentals of Heat and Mass Transfer,” John Wiley & Sons.
2. Cumo, M., Farello, G.E., and Ferrari, G. (1969), “Notes on Droplet Heat Transfer,” Chemical Engineering Progress Symposium Series, Heat Transfer, Vol. 65, pp. 175-187.
3. Gottfried, B. S., Lee, C. J., and Bell, K. J. (1966), “The Leidenfrost Phenomenon: Film Boiling of Liquid Droplets on a Flat Plate,” International Journal of Heat and Mass Transfer, Vol. 9, pp. 1167-1187.
4. Wachters, L. H. J., Bonne, H., and van Nouhuis, H. J. (1966), “The Heat Transfer from a Hot Horizontal Plate to sessile Water Drops in the Spherodial State,” Chemical Engineering Science, Vol. 21, pp. 923-936.
5. Wachters, L. H. J. and Westerling, N. A. J. (1966), “The Heat Transfer from a Hot Wall to Impinging Water Drops in the Spherodial State,” Chemical Engineering Science, Vol. 21, pp. 1047-1056.
6. Wachters, L. H. J. and van Andel, E. (1966), “Boiling on a horizontal Plate with Periodically Varying Heat Flux,” Chemical Engineering Science, Vol. 21, pp. 937-940.
Page 5.382.8 HOSNI I. ABU-MULAWEH Hosni I. Abu-Mulaweh is an Assistant Professor of Mechanical Engineering at Indiana University-Purdue University, Fort Wayne, Indiana. He earned his B.S. , M.S., and Ph.D. in Mechanical Engineering from the University of Missouri-Rolla, Rolla, Missouri. His areas of interest are Heat Transfer, Thermodynamics, and Fluid Mechanics.
JOSUÉ NJOCK LIBII Josué Njock Libii is an Associate Professor of Mechanical Engineering at Indiana University- Purdue University, Fort Wayne, Indiana. He earned a B.S.E in Civil Engineering, an M.S.E. in Engineering Mechanics, and a Ph.D. in Engineering Mechanics from the University of Michigan, Ann Arbor, Michigan. His areas of interest are Fluid Mechanics and its application and uses in related fields. Page 5.382.9 |
11088 | https://blog.csdn.net/u013172930/article/details/143604418 | 一元二次函数的最值公式_一元二次函数最值-CSDN博客
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一元二次函数的最值公式
最新推荐文章于 2024-12-30 19:50:26 发布
原创 于 2024-11-07 18:15:35 发布·3.8k 阅读
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#一元二次函数
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一元二次函数的标准形式为：
f(x)=ax2+bx+c f(x) = ax^2 + bx + c f(x)=a x 2+b x+c
其中，aa a、bb b、cc c 是常数，且 a≠0a \neq 0 a=0。
一元二次函数的最值公式
一元二次函数的最值与二次项系数 aa a 的符号有关：
当 a>0a > 0 a>0 时，抛物线开口向上，函数有最小值。
当 a<0a < 0 a<0 时，抛物线开口向下，函数有最大值。
最值的具体公式
对于一元二次函数 f(x)=ax2+bx+cf(x) = ax^2 + bx + c f(x)=a x 2+b x+c，其在 xx x 轴上的对称轴为：
x=−b2a x = -\frac{b}{2a} x=−2 a b
函数在该点的取值（即最值）为：
f(−b2a)=a(−b2a)2+b(−b2a)+c f\left(-\frac{b}{2a}\right) = a\left(-\frac{b}{2a}\right)^2 + b\left(-\frac{b}{2a}\right) + c f(−2 a b)=a(−2 a b)2+b(−2 a b)+c
化简后，最值可以表示为：
f(−b2a)=4ac−b24a f\left(-\frac{b}{2a}\right) = \frac{4ac - b^2}{4a} f(−2 a b)=4 a 4 a c−b 2
总结
当 a>0a > 0 a>0 时，最小值为 4ac−b24a\frac{4ac - b^2}{4a}4 a 4 a c−b 2。
当 a<0a < 0 a<0 时，最大值为 4ac−b24a\frac{4ac - b^2}{4a}4 a 4 a c−b 2。
最值对应的 xx x 值是 x=−b2ax = -\frac{b}{2a}x=−2 a b。
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一元 二次方程 形态：ax²+bx+c=0 (a≠0) 求根 公式： x=(-b±(b²-4ac)^(1/2))/(2a) 当b²-4ac>0时有两个不相等的实根 当b²-4ac=0时有两个相同的实根 当b²-4ac（虚根） 注：在正式学习高数之前，对于复数只需要知道i²=-1即可 韦达定理（x1、x2为方程的两个根）：x1+x2=-(b/a),x1
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线性回归是 最 简单的一个函数拟合过程,一元 线性回归 公式 为y=ax+b。我们做拟合,首先需要定义一个损失函数。一般常用的损失函数有: 0-1损失函数和绝对 值 损失函数 0-1损失是指,预测 值 和目标 值 不相等为1,否则为0: 感知机就是用的这种损失函数。但是由于相等这个条件太过严格,因此我们可以放宽条件,即满足 ...
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动态演示 二次函数 最 值 及恒成立问题，包括轴动区间定、轴定区间动求 最 值；二次函数 在区间上恒大于零等问题。
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数学规律，一元 二次方程求 最 大 值。
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文章目录abstract 一元 二次方程求根 公式 一元 二次函数 最 简单类型 一元 二次函数 抛物线形状函数平移规律 一元 二次函数 的三种形式👺形式转换配方👺解 一元 二次方程配方判别式 一元 二次式或函数的配方顶点(对称轴和 最 值)例根和零点根与系数的关系(韦达定理)👺推导1推导2 公式 中的分母应用例refs abstract 一元 二次函数 f(x)=ax2+bx+cf(x)=ax^{2}+bx+cf(x)=ax2+bx+c性质,其核心 公式 如下,便于查阅 配方 公式:ax2+bx+cax^{2}+bx+cax2+bx+c=a(x+b2
指数函数中x的取 值 范围_学习随笔高中数学中 最 值 与圆的求解
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对于 一元 二次函数 baiy=ax²+bx+c(a≠0)来说: 当x=-b/2a 时,du有 最 zhi 值;且 最 值 公式 dao为:(4ac—b^2)/4a 当a>0时, 为 最 小 值, 当a<0时, 为 最 大 值。 一元 二次方程成立必须同时满足三个条件: 1、是整式方程,即等号两边都是整式,方程中如果有分母;且未知数在分母上,那么这个方程就是分式方程...
初三数学 二次函数 最 值 问题及压轴题.pdf
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其次，对于含有绝对 值、分式、不等式等复杂条件的 最 值 问题，通常需要结合一次函数、一元 二次不等式等知识进行分析。例如，当求解“在满足某个条件的情况下，二次函数 的 最 大 值 或 最 小 值”时，可能需要先确定自变量的...
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本知识点主要集中在 一元 二次函数 的 最 值 问题上，涉及了函数对称轴与给定区间的关系以及如何通过分析这些关系来确定函数的 最 大 值 和 最 小 值。 首先，一元 二次函数 的 最 值 问题与函数的开口方向和对称轴的位置紧密相关。...
二次函数 的 最 值 问题.doc
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首先，我们可以通过配方法或者利用 一元 二次方程的顶点 公式 找到对称轴，然后根据对称轴的位置和区间的相对关系，确定 最 大 值 和 最 小 值。 在例2中，已知函数在特定区间上的 最 大 值 和 最 小 值，我们可以反推出 a a, ( ...
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在C++中，求解 一元 二次方程的 最 值 通常涉及到找到该方程的根，因为 二次函数 在其顶点处取得 最 小 值（如果开口向上）或 最 大 值（如果开口向下）。如果你有一个一般形式的二次方程，例如 ax^2 + bx + c = 0，你可以使用...
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如何使用均 值 定理求函数的 最 值 河北省武邑县职教中心【摘要】摘要：均 值 定理是高中数学中重要的内容，在高考中占有很重要的地位，成为高考的高频考点.它们总能在高考的舞台上与其姊妹知识合理、巧妙、有机地结合在一起进行联合演出，成为检查学生知识掌握情况和提升学生综合应用能力的训练战场.本文重点介绍了使用均 值 定理求 最 值 的常见题型及使用方法，以供参考.【期刊名称】河北理科教学研究【年(卷),期】2016(000)...
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我们知道：一元 二次方程：y = f(x) = x^2 + 2 x + 1 = (x + 1)^2，当x = -1时，f(x)达到 最 小 值 0。 对方程求导数y’ = f’(x) = 2 x + 2。也就是说y’随着x的变化而变化，当x = -1时，导数为0，而在这个时候，f(x)也恰好达到 最 小 值。 另外，当x = -4时，y = 9，导数y’ = -6。红色的直线是x = -4时, f(x)的切线。沿着导数y’的反方向移动可以到达x = -1，即到达f(x)的 最 小 值 0。 相反，如果x = 2，y = 9
求 一元 二次方程求根 公式 与韦达定理.
kwame211的博客
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一元 二次方程ax^2+bx+c=0中, 一元 二次方程求根 公式： 两根x1,x2= [-b±√(b^2-4ac)]/2a 韦达定理： 两根x1,x2有如下关系: x1+x2=-b/a,x1x2=c/a
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目录
一元二次函数的最值公式
最值的具体公式
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上一篇： 《XGBoost算法的原理推导》12-15目标函数分解成叶子节点的形式 公式解析 下一篇： 《XGBoost算法的原理推导》12-22计算信息增益（Gain）的公式 公式解析
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From Wikipedia, the free encyclopedia
Abstract strategy board game
"Omok" redirects here. For the language, see Omok language.
Gomoku
| | |
| Genres | Board game Abstract strategy game |
| Players | 2 |
| Setup time | Minimal |
| Chance | None |
| Skills | Strategy, tactics |
Gomoku, also called five in a row, is an abstract strategy board game. It is traditionally played with Go pieces (black and white stones) on a 15×15 Go board while in the past a 19×19 board was standard. Because pieces are typically not moved or removed from the board, gomoku may also be played as a paper-and-pencil game. The game is known in several countries under different names.
Rules
[edit]
Players alternate turns placing a stone of their color on an empty intersection. Black plays first. The winner is the first player to form an unbroken line of five stones of their color horizontally, vertically, or diagonally. In some rules, this line must be exactly five stones long; six or more stones in a row does not count as a win and is called an overline. If the board is completely filled and no one has made a line of 5 stones, then the game ends in a draw.
Origin
[edit]
Historical records indicate that the origins of gomoku can be traced back to the mid-1700s during the Edo period. It is said that the 10th generation of Kuwanaya Buemon, a merchant who frequented the Nijō family, was highly skilled in this game, which subsequently spread among the people. By the late Edo period, around 1850, books had been published on gomoku. The earliest published book on gomoku that can be verified is the Gomoku Jōseki Collection (五石定磧集) in 1856.
The name "gomoku" is from the Japanese language, in which it is referred to as gomokunarabe (五目並べ). Go means five, moku is a counter word for pieces and narabe means line-up. The game is popular in China, where it is called Wuziqi (五子棋). Wu (五 wǔ) means five, zi (子 zǐ) means piece, and qi (棋 qí) refers to a board game category in Chinese. The game is also popular in Korea, where it is called omok (오목 [五目]) which has the same structure and origin as the Japanese name.
In the nineteenth century, the game was introduced to Britain where it was known as Go Bang, said to be a corruption of the Japanese word goban, which was itself adapted from the Chinese k'i pan (qí pán) "go-board."
First-player advantage
[edit]
Gomoku has a strong advantage for the first player when unrestricted.
Championships in gomoku previously used the "Pro" opening rule, which mandated that the first player place the first stone in the center of the board. The second player's stone placement was unrestricted. The first player's second stone had to be placed at least three intersections away from the first player's first stone. This rule was used in the 1989 and 1991 world championships. When the win–loss ratio of these two championships was calculated, the first player (black) won 67 percent of games.
This was deemed too unbalanced for tournament play, so tournament gomoku adopted the Swap2 opening protocol in 2009. In Swap2, the first player places three stones, two black and one white, on the board. The second player then selects one of three options: play as black, play as white and place another white stone, or place two more stones, one white and one black, and let the first player choose the color.
The win ratio of the first player has been calculated to be around 52 percent using the Swap2 opening protocol, greatly balancing the game and largely evening out the first-player advantage.
Variants
[edit]
Freestyle gomoku
[edit]
Freestyle gomoku has no restrictions on either player and allows a player to win by creating a line of five or more stones, with each player alternating turns placing one stone at a time.
Swap after 1st move
[edit]
The rule of "swap after 1st move" is a variant of the freestyle gomoku rule, and is mostly played in China. The game can be played on a 19×19 or 15×15 board. As per the rule, once the first player places a black stone on the board, the second player has the right to swap colors. The rest of the game proceeds as freestyle gomoku. This rule is set to balance the advantage of black in a simple way.
Renju
[edit]
Black (the player who makes the first move) has long been known to have an advantage, even before L. Victor Allis proved that black can force a win (see below). Renju attempts to mitigate this imbalance with extra rules that aim to reduce black's first player advantage.
It is played on a 15×15 board, with the rules of three and three, four and four, and overlines applied to Black only.
The rule of three and three bans a move that simultaneously forms two open rows of three stones (rows not blocked by an opponent's stone at either end).
The rule of four and four bans a move that simultaneously forms two rows of four stones (open or not).
Overlines prevent a player from winning if they form a line of 6 or more stones.
Renju also makes use of various tournament opening rules, such as Soosõrv-8, the current international standard.
Caro
[edit]
In Caro, (also called gomoku+, popular among Vietnamese), the winner must have an overline or an unbroken row of five stones that is not blocked at either end (overlines are immune to this rule). This makes the game more balanced and provides more power for White to defend.
Omok
[edit]
Omok is similar to Freestyle gomoku; however, it is played on a 19×19 board and includes the rule of three and three.
Ninuki-renju
[edit]
Also called Wu, Ninuki Renju is a variant which adds capturing to the game; A pair of stones of the same color may be captured by the opponent by means of custodial capture (sandwiching a line of two stones lengthwise). The winner is the player either to make a perfect five in a row, or to capture five pairs of the opponent's stones. It uses a 15x15 board and the rules of three and three and overlines. It also allows the game to continue after a player has formed a row of five stones if their opponent can capture a pair across the line.
Pente
[edit]
Pente is related to Ninuki-Renju, and has the same custodial capture method, but is most often played on a 19x19 board and does not use the rules of three and three, four and four, or overlines.
Tournament opening rules
[edit]
Tournament rules are used in professional play to balance the game and mitigate the first player advantage. The tournament rule used for the gomoku world championships since 2009 is the Swap2 opening rule. For all of the following professional rules, an overline (six or more stones in a row) does not count as a win.
Pro
[edit]
The first player's first stone must be placed in the center of the board. The second player's first stone may be placed anywhere on the board. The first player's second stone must be placed at least three intersections away from the first stone (two empty intersections in between the two stones).
Long Pro
[edit]
The first player's first stone must be placed in the center of the board. The second player's first stone may be placed anywhere on the board. The first player's second stone must be placed at least four intersections away from the first stone (three empty intersections in between the two stones).
Swap
[edit]
The tentative first player places three stones (two black, and one white) anywhere on the board. The tentative second player then chooses which color to play as. Play proceeds from there as normal with white playing their second stone.
Swap2
[edit]
The tentative first player places three stones on the board, two black and one white. The tentative second player then has three options:
They can choose to play as white and place a second white stone
They can swap their color and choose to play as black
Or they can place two more stones, one black and one white, and pass the choice of which color to play back to the tentative first player.
Because the tentative first player doesn't know where the tentative second player will place the additional stones if they take option 3, the swap2 opening protocol limits excessive studying of a line by only one of the players.
Theoretical generalizations
[edit]
m,n,k-games are a generalization of gomoku to a board with m×n intersections, and k in a row needed to win. Connect Four is (7,6,4) with piece placement restricted to the lowest unoccupied place in a column.
Connect(m,n,k,p,q) games are another generalization of gomoku to a board with m×n intersections, k in a row needed to win, p stones for each player to place, and q stones for the first player to place for the first move only. In particular, Connect(m,n,6,2,1) is called Connect6.
Example game
[edit]
This game on the 15×15 board is adapted from the paper "Go-Moku and Threat-Space Search".
The opening moves show clearly black's advantage. An open row of three (one that is not blocked by an opponent's stone at either end) has to be blocked immediately, or countered with a threat elsewhere on the board. If not blocked or countered, the open row of three will be extended to an open row of four, which threatens to win in two ways.
White has to block open rows of three at moves 10, 14, 16 and 20, but black only has to do so at move 9.
Move 20 is a blunder for white (it should have been played next to black 19). Black can now force a win against any defense by white, starting with move 21.
There are two forcing sequences for black, depending on whether white 22 is played next to black 15 or black 21. The diagram on the right shows the first sequence. All the moves for white are forced. Such long forcing sequences are typical in gomoku, and expert players can read out forcing sequences of 20 to 40 moves rapidly and accurately.
The diagram on the right shows the second forcing sequence. This diagram shows why white 20 was a blunder; if it had been next to black 19 (at the position of move 32 in this diagram) then black 31 would not be a threat and so the forcing sequence would fail.
World championships
[edit]
World Gomoku Championships have occurred 2 times in 1989, 1991.
Since 2009 tournament play has resumed, with the opening rule changed to swap2.
List of the tournaments occurred and title holders follows.
World Championship
| Title year | Hosting city, country | Gold | Silver | Bronze | Opening rule |
| 1989 | Kyoto, Japan | Sergey Chernov | Yuriy Tarannikov | Hirouji Sakamoto | Pro |
| 1991 | Moscow, Soviet Union | Yuriy Tarannikov | Ando Meritee | Sergey Chernov | Pro |
| 2009 | Pardubice, Czech Republic | Artur Tamioła | Attila Demján | Pavel Laube | Swap2 |
| 2011 | Huskvarna, Sweden | Attila Demján | Artur Tamioła | Michał Żukowski | Swap2 |
| 2013 | Tallinn, Estonia | Attila Demján | Pavel Laube | Mikhail Kozhin | Swap2 |
| 2015 | Suzdal, Russia | Rudolf Dupszki | Gergő Tóth | Mikhail Kozhin | Swap2 |
| 2017 | Prague, Czech Republic | Zoltán László | Rudolf Dupszki | Denis Osipov | Swap2 |
| 2019 | Tallinn, Estonia | Martin Muzika | Oleg Bulatowsky | Michał Żukowski | Swap2 |
| 2023 | Budapest, Hungary | Pavel Laube | Adrian Fitzermann | Martin Muzika | Swap2 |
| 2025 | Brno, Czech Republic | Pavel Laube | Martin Muzika | Adrian Fitzermann | Swap2 |
Team World Championship
| Title year | Hosting city, country | Gold | Silver | Bronze | Opening rule |
| 2016 | Tallinn, Estonia | Poland Michał Żukowski Michał Zajk Łukasz Majksner Piotr Małowiejski | Czech Republic Pavel Laube Igor Eged Štěpán Tesařík Marek Hanzl | Chinese Taipei Lu Wei-Yuan Chen Ko-Han Chang Yi-Feng Sung Pei-Jung | Swap2 |
| 2018 | Płock, Poland | Russia-1 Edvard Rizvanov Denis Osipov Ilya Muratov Maksim Karasev Mikhail Kozhin | Hungary Zoltán László Gergő Tóth Márk Horváth Gábor Gyenes Attila Hegedűs | Poland Łukasz Majksner Michał Żukowski Michał Zajk Marek Gorzecki Paweł Tarasiński | Swap2 |
| 2020 | | | | | | --- --- | | Cancelled due to the COVID-19 pandemic | | | | | | | | |
| 2022 | | | | | --- | Not organised | | | | | | | |
| 2024 | Xintai, China | | | | | | --- --- | | Organised but not played | | | | | | | |
Computers and gomoku
[edit]
Researchers have been applying artificial intelligence techniques to playing gomoku for several decades. Joseph Weizenbaum published a short paper in Datamation in 1962 entitled "How to Make a Computer Appear Intelligent" that described the strategy used in a gomoku program that could beat novice players. In 1994, L. Victor Allis raised the algorithm of proof-number search (pn-search) and dependency-based search (db-search), and proved that when starting from an empty 15×15 board, the first player has a winning strategy using these searching algorithms. This applies to both free-style gomoku and standard gomoku without any opening rules. It seems very likely that black wins on larger boards too. In any size of a board, freestyle gomoku is an m,n,k-game, hence it is known that the first player can force a win or a draw. In 2001, Allis's winning strategy was also approved for renju, a variation of gomoku, when there was no limitation on the opening stage.
However, neither the theoretical values of all legal positions, nor the opening rules such as Swap2 used by the professional gomoku players have been solved yet, so the topic of gomoku artificial intelligence is still a challenge for computer scientists, such as the problem on how to improve the gomoku algorithms to make them more strategic and competitive. Most state-of-the-art gomoku algorithms are based on the alpha-beta pruning framework.[citation needed]
Reisch proved that Generalized gomoku is PSPACE-complete. He also observed that the reduction can be adapted to the rules of k-in-a-Row for fixed k. Although he did not specify exactly which values of k are allowed, the reduction would appear to generalize to any k ≥ 5.
There exist several well-known tournaments for gomoku programs since 1989. The Computer Olympiad started with the gomoku game in 1989, but gomoku has not been in the list since 1993. The Renju World Computer Championship was started in 1991, and held for 4 times until 2004. The Gomocup tournament has been held annually since 2000, with more than thirty participants from about ten countries taking part. The Hungarian Computer Go-Moku Tournament was played twice in 2005. There were also two Computer vs. Human tournaments played in the Czech Republic, in 2006 and 2011. Not until 2017 did the computer programs prove capable of outperforming the world human champion in public competitions. In the Gomoku World Championship 2017, there was a match between the world champion program Yixin and the world champion human player Rudolf Dupszki. Yixin won the match with a score of 2–0.
In popular culture
[edit]
Gomoku was featured in a 2018 Korean film by Baek Seung-Hwa starring Park Se-wan titled Omok Girl. The film follows Baduk Lee (Park Se-wan), a former go prodigy who retired after a humiliating loss on time. Years later, Baduk Lee works part time at a go club, where she meets Ahn Kyung Kim, who introduces her to an Omok (Korean gomoku) tournament. Lee is initially uninterested and considers Omok a children's game, but after her roommate loses money on an impulse purchase, she enters the tournament for the prize money and loses badly, being humiliated once again. Afterwards, she begins training to redeem herself and becomes a serious omok player.
In the video game Vintage Story, omok boards and pieces (made of gold and lead) can occasionally be found in ruins or as part of luxury traders' inventory. The board and pieces are functional, allowing players to have actual omok matches. In-universe, omok is so far the only game surviving from the times before the Rot.
See also
[edit]
Renju
Pente
Pegity
Connect6
Connection game
Reversi
References
[edit]
^ "Gomoku - Japanese Board Game". Japan 101. Archived from the original on 2014-03-26. Retrieved 2013-06-25.
^ "Game Theory | GomokuWorld.com". gomokuworld.com. Archived from the original on 2021-07-22. Retrieved 2021-07-28.
^ Lasker, Edward (1960). Go and go-moku: the oriental board games (2nd rev. ed.). New York: Dover. ISBN 9780486206134. {{cite book}}: ISBN / Date incompatibility (help)
^ "The rules and the history of Renju and other five-in-a-row games." Luffarschack, renju.se/rif/r1rulhis.htm. Accessed 28 July 2021.
^ "Game Theory | GomokuWorld.com". gomokuworld.com. Archived from the original on 2021-07-22. Retrieved 2021-07-22.
^ a b c "The Renju International Federation portal - RenjuNet". www.renju.net. Archived from the original on 2023-02-10. Retrieved 2023-02-10.
^ "About the origin and rules of renju". Nihon Renju-sha (in Japanese). 2022-09-19. Archived from the original on 2023-04-03. Retrieved 2023-04-26.
^ "Origins of renju". www.success-simulation.com. 1999-10-01. Archived from the original on 2022-03-31. Retrieved 2023-04-26.
^ "The Renju International Federation portal - RenjuNet". www.renju.net. Retrieved 2023-04-30.
'^ OED citations: 1886 GUILLEMARD Cruise 'Marchesa I. 267 Some of the games are purely Japanese..as go-ban. Note, This game is the one lately introduced into England under the misspelt name of Go Bang. 1888 Pall Mall Gazette 1. Nov. 3/1 These young persons...played go-bang and cat's cradle.
The board below shows the three types of winning arrangements as they might appear on an 8x8 Petteia board. Obviously the cramped conditions would result in a draw most of the time, depending on the rules. Play would be easier on a larger Latrunculi board of 12x8 or even 10x11.
.
^ a b "BoardGameGeek". boardgamegeek.com. Archived from the original on 2021-07-22. Retrieved 2021-01-26.
^ a b "Game database | GomokuWorld.com". gomokuworld.com. Retrieved 2021-01-26.
^ a b "The Renju International Federation portal - RenjuNet". Renju.net. Archived from the original on 2021-07-22. Retrieved 2012-10-03.
^ a b "Gomoku - swap2 rule". renju.net. Retrieved 2016-11-09.
^ a b c d e "Opening rules | GomokuWorld.com". gomokuworld.com. Archived from the original on 2021-07-22. Retrieved 2021-07-07.
^ a b c "History | GomokuWorld.com". gomokuworld.com. Archived from the original on 2021-07-07. Retrieved 2021-01-26.
^ "Swap after 1st move rule". www.wuzi8.com (in Chinese). Archived from the original on 2021-12-08. Retrieved 2023-04-28.
^ "The Renju International Federation portal - RenjuNet". Archived from the original on 11 July 2021. Retrieved 2021-07-22.
^ "The Renju International Federation portal - RenjuNet". renju.net. Archived from the original on 2021-07-22. Retrieved 2021-07-22.
^ "Caro (aka Gomoku)". LearnPlayWin. Archived from the original on 2021-07-22. Retrieved 2021-07-22.
^ "Omok: A Korean Game of Five Stones". KPOP Jacket Lady. 2016-10-06. Archived from the original on 2021-07-22. Retrieved 2021-07-22.
^ Sungjin, Nam. "Omok." Encyclopedia of Korean Folk Culture, National Folk Museum of Korea, . Accessed 22 July 2021.
^ "Rules of Pente, Keryo-Pente and Ninuki". Renju. Archived from the original on 2021-07-22. Retrieved 2021-07-22.
^ "Pente". www.mindsports.nl. Archived from the original on 2021-07-01. Retrieved 2021-07-22.
^ "Gomoku – pro rule". www.renju.net. Retrieved 2021-07-28.
^ Allis, L. V., Herik, H. J., & Huntjens, M. P. H. (1993). Go-moku and threat-space search. University of Limburg, Department of Computer Science.
^ How to Make a Computer Appear Intelligent, Datamation, February, 1962
^ L. Victor Allis (1994). Searching for Solutions in Games and Artificial Intelligence. Ph.D. thesis, University of Limburg, The Netherlands. pp. 121–154. CiteSeerX 10.1.1.99.5364. ISBN 90-900748-8-0.
^ J. Wágner and I. Virág (Mar 2001). "Solving Renju". ICGA Journal. 24 (1): 30–35. doi:10.3233/ICG-2001-24104. S2CID 207577292.
^ Stefan Reisch (1980). "Gobang ist PSPACE-vollständig (Gomoku is PSPACE-complete)". Acta Informatica. 13: 59–66. doi:10.1007/bf00288536. S2CID 21455572.
^ Demaine, Erik; Hearn, Robert (2001). "Playing Games with Algorithms: Algorithmic Combinatorial Game Theory". arXiv:cs/0106019v2.
^ "Go-Moku (ICGA Tournaments)". game-ai-forum.org. Retrieved 2016-06-02.
^ "Renju Computer World Championship". 5stone.net. Retrieved 2016-06-02.
^ "4-th World Championship among Computer programs". Nosovsky Japanese Games Home Page. Retrieved 2016-06-03.
^ "Gomocup - The Gomoku AI Tournament". Gomocup. Archived from the original on 2016-06-04. Retrieved 2016-06-02.
^ "Hungarian Computer Gomoku Tournament 2005 | GomokuWorld.com". gomokuworld.com. Retrieved 2016-06-02.
^ "2nd Hungarian Computer Go-Moku Open Tournament". sze.hu. Retrieved 2016-06-03.
^ "The 1st tournament AI vs. Human (November the 11th, 2006) | Gomocup". gomocup.org. Retrieved 2016-06-02.
^ "AI vs. Člověk 2011 | Česká federace piškvorek a renju". piskvorky.cz. Retrieved 2016-06-02.
^ "Rudolf Dupszki versus Yixin". AIEXP.
^ "Rudolf Dupszki vs. Yixin 2017". Facebook.
^ Seung-hwa, Baek, writer. Omok Girl. Performance by Park Se-wan, SK Telecom, 2018.
Further reading
[edit]
Five-in-a-Row (Renju) For Beginners to Advanced Players ISBN 4-87187-301-3
External links
[edit]
Gomoku World
Renju International Federation website
Gomocup tournament
| | | |
---
| Variants | 3D tic-tac-toe Gomoku Notakto + Treblecross Number Scrabble Order and Chaos Pente Quantum tic-tac-toe Renju SOS Ultimate tic-tac-toe Wild tic-tac-toe | |
| Related concepts | m,n,k-game nd game Kaplansky's game Harary's generalized tic-tac-toe Hales–Jewett theorem Strategy-stealing argument Futile game Paper-and-pencil game |
| Similar games | Achi Connect Four Connect6 Gobblet Nine Holes Nine men's morris OXO Pentago Quarto Score Four + Tic-Stac-Toe Teeko Three men's morris Toss Across |
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11090 | https://pmc.ncbi.nlm.nih.gov/articles/PMC9491185/ | Diagnosis, Management, and Future Control of Cholera - PMC
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Diagnosis, Management, and Future Control of Cholera
Fahima Chowdhury
Fahima Chowdhury
a International Center for Diarrheal Disease Research, Bangladesh, Dhaka, Bangladesh
b Menzies Health Institute Queensland, Griffith University, Gold Coast, Southport, Queensland, Australia
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a,b,✉, Allen G Ross
Allen G Ross
c Rural Health Research Institute, Charles Sturt University, Orange, New South Wales, Australia
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c, Md Taufiqul Islam
Md Taufiqul Islam
a International Center for Diarrheal Disease Research, Bangladesh, Dhaka, Bangladesh
b Menzies Health Institute Queensland, Griffith University, Gold Coast, Southport, Queensland, Australia
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a,b, Nigel A J McMillan
Nigel A J McMillan
b Menzies Health Institute Queensland, Griffith University, Gold Coast, Southport, Queensland, Australia
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b, Firdausi Qadri
Firdausi Qadri
a International Center for Diarrheal Disease Research, Bangladesh, Dhaka, Bangladesh
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a International Center for Diarrheal Disease Research, Bangladesh, Dhaka, Bangladesh
b Menzies Health Institute Queensland, Griffith University, Gold Coast, Southport, Queensland, Australia
c Rural Health Research Institute, Charles Sturt University, Orange, New South Wales, Australia
The authors declare no conflict of interest.
✉
Corresponding author.
Collection date 2022 Sep.
Copyright © 2022 American Society for Microbiology.
All Rights Reserved.
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PMCID: PMC9491185 PMID: 35726607
SUMMARY
Cholera, caused by Vibrio cholerae, persists in developing countries due to inadequate access to safe water, sanitation, and hygiene. There are approximately 4 million cases and 143,000 deaths each year due to cholera. The disease is transmitted fecally-orally via contaminated food or water. Severe dehydrating cholera can progress to hypovolemic shock due to the rapid loss of fluids and electrolytes, which requires a rapid infusion of intravenous (i.v.) fluids. The case fatality rate exceeds 50% without proper clinical management but can be less than 1% with prompt rehydration and antibiotics. Oral cholera vaccines (OCVs) serve as a major component of an integrated control package during outbreaks or within zones of endemicity. Water, sanitation, and hygiene (WaSH); health education; and prophylactic antibiotic treatment are additional components of the prevention and control of cholera. The World Health Organization (WHO) and the Global Task Force for Cholera Control (GTFCC) have set an ambitious goal of eliminating cholera by 2030 in high-risk areas.
KEYWORDS:Vibrio cholerae, WaSH, oral cholera vaccine, treatment
INTRODUCTION
Vibrio cholerae is a major cause of severe dehydrating diarrheal disease and remains a major public health problem in low- and middle-income countries (LMICs) where water, sanitation, and hygiene (WaSH) are inadequate (1). Severe cholera can rapidly lead to hypovolemic shock after the onset of diarrhea and vomiting. However, with appropriate oral and intravenous (i.v.) fluid therapy, clinicians can lower the case fatality rate from over 50% to less than 1% (2). There are two fecal-oral mechanisms of V. cholerae transmission: one involves direct spread from person to person as a consequence of eating bacterium-contaminated food or water, and the other involves drinking environmentally polluted water from ponds, lakes, or rivers (3, 4). Fomites and flies may help disseminate V. cholerae by acting as mechanical vectors (3).
The World Health Organization (WHO) approximates 1.4 million to 4.0 million cases and 21,000 to 143,000 deaths globally due to the disease per annum (5). A zone where cholera is endemic is defined as an area with confirmed cases detected over the past 3 years (5). Cholera is endemic in Asia, Latin and Central America, and sub-Saharan Africa (1, 5, 6). V. cholerae originated in the Ganges River Delta and remains highly prevalent in Asia and Africa (7). Outbreaks of cholera are unpredictable and can occur in both zones where cholera is endemic and those where it is nonendemic depending on environmental conditions, with refugee camps being particularly susceptible (8). V. cholerae is always evolving, with new phenotypes and genotypes emerging with outbreaks and as a result of increased antibiotic resistance (9). The WHO’s Global Task Force for Cholera Control (GTFCC) recommends access to both oral cholera vaccination and improved WaSH to avert cholera transmission as well as continued diarrheal disease surveillance (10).
EPIDEMIOLOGY
Geographical Distribution and Burden of Cholera
In resource-poor countries of endemicity, data from the last decade have shown that the burden of cholera has increased, and it has become an important public health problem (8). Figure 1 illustrates the global geographical distribution of cholera between 2016 and 2019 (11–14). Cholera has been historically endemic in the Asian subcontinent (e.g., Indonesia, India, Bangladesh, Vietnam, Thailand, Pakistan, Nepal, and Iraq) but is now endemic in Africa (e.g., South Africa, Mozambique, Botswana, Zambia, Sierra Leone, Nigeria, Angola, the Democratic Republic of the Congo [DRC], Yemen, Zimbabwe, the United Republic of Tanzania, and Guinea), Latin America (Brazil, Peru, Chile, Columbia, and Ecuador), and the Caribbean (Haiti, Cuba, and the Dominican Republic) (11–14). Due to weak or absent surveillance systems, many countries do not report cholera cases or deaths (3). Some countries are cautious in announcing cholera outbreaks to avoid economic losses in tourism and exports and to prevent general panic in society. But early reporting of cholera outbreaks has resulted in shorter durations of epidemics (15). Since 2006, 52 developing countries have reported increasing numbers of cholera cases (6).
FIG 1.
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Global distribution of cholera between 2016 and 2019. ★, the imported cases were from elsewhere in the country where cholera is not common. The red zones represent the affected countries where cholera is endemic during the 3-year period.
In the Bengal Delta region, cholera epidemics typically follow a seasonal pattern, with one peak in spring (March to May) and another following the rainy season (September to November) (16–20). In Africa, epidemics occur in different regions during the rainy season, with recent outbreaks being reported in Zanzibar, the Eastern DRC, Angola, and West Africa. Haiti experienced a recent outbreak from 2017 to 2018 due to Hurricane Matthew, and the WHO reported 800,000 cholera cases and approximately 10,000 deaths from cholera since the outbreak (3). Mathematical modeling of cholera transmission suggests that outbreaks are reliant on variations in the environment and herd protection (21).
Risk Factors
The risk factors for acquiring cholera are associated with poverty, including inadequate sanitation, contaminated drinking water, and poor food hygiene (e.g., street foods). Handwashing with soap before and after meals and after defecation is associated with a reduced risk (22).
There are some biological factors that have been recognized as risk factors for cholera, and these are female gender, blood group O, and retinol deficiency, and hypochlorhydria (i.e., in people who take histamine receptor blockers, antacids, and proton pump inhibitors) also increases the risk of contracting cholera (23–28). Moreover, Helicobacter pylori infection and gastrectomy are two major factors contributing to severe disease (1). Malnutrition increases susceptibility, especially among young children (1). The incidence of cholera in countries of endemicity is highest among children less than 5 years of age as they have low levels of acquired immunity compared to adults (29).
Secretory immunoglobulin A (SIgA), which is secreted in breast milk, protects against severe cholera, and thus, exclusive breastfeeding for the first 6 months of life is recommended for women living in communities of endemicity (30). Long palate, lung, and nasal epithelium clone protein 1 (LPLUNC1) is expressed in Paneth cells of the intestinal mucosa and is strongly associated with modulating host inflammatory responses to V. cholerae infection and disease severity (25). Concomitant infection of the gut with other bacteria like enterotoxigenic Escherichia coli (ETEC) or parasites increases the chance of V. cholerae infection (31, 32).
A household contact study of cholera patients in Bangladesh showed that first-degree relatives (parents, offspring, and siblings) have a greater chance of acquiring cholera than second-degree relatives (grandchildren, grandparents, uncles, and aunts) staying in the same household (15).
PATHOLOGY
Microbiology and Pathogenesis
V. cholerae is a Gram-negative, comma-shaped bacterium that is classified serologically into over 200 serogroups. Of them, V. cholerae O1 and O139 have caused recent cholera epidemics (33). Based on biochemical structure, V. cholerae O1 is categorized into two biotypes, classical and El Tor, and, more recently, the altered El Tor biotype. Furthermore, each biotype is differentiated into three serotypes, Ogawa, Inaba, and the rare Hikojima type (34).
V. cholerae releases a heat-labile exotoxin based on the AB5 multimeric protein, named cholera toxin (CT), which adheres to the small intestinal mucosa of the gut. CT consists of two subunits (one A subunit [CTA] and five B subunits [CTB]), which cause fluid and electrolyte loss. The B subunit fixes onto eukaryotic cells, whereas the A subunit is shifted into the cell, which assists in increasing cyclic AMP (cAMP) and subsequently leads to secretory diarrhea, which causes severe dehydration. The toxin-coregulated pilus (TcpA), which is required for colonization, serves as a receptor for the cholera toxin phage (CTXphi), which is the V. cholerae-specific filamentous bacteriophage and carries the gene for CT (35).
V. cholerae exists in stagnant water or marine environmental reservoirs where the water source is infected with human and/or animal waste. V. cholerae in freshly shed stool appears to be hyperinfectious for 24 h after release into the environment (36). V. cholerae flourishes in 30°C water with 15% salinity and a pH of 8.5 (37). During spring and after the monsoon season, more adverse environmental conditions can trigger an outbreak of V. cholerae. Real-time recording of climatic parameters along with active surveillance systems can assist public health warning systems in minimizing risk factors (37). The intake of polluted water and contaminated food are the main sources of infection, although acidic gastric enzymes kill most of the bacteria, and the rest colonize the small intestine. The infective dose of V. cholerae is 10 3 to 10 8 CFU when ingested with water, but a minimum dose of ~10 2 to 10 4 CFU can cause diarrhea when ingested with food. ( A high infectious dose, 10 8 CFU, is required to cause diarrhea in healthy individuals, while a very small inoculum of 10 5 CFU can cause diarrhea in individuals with low levels of gastric acid (38). The incubation period of V. cholerae may range from 12 h to 5 days (39).
Molecular Epidemiology of V. cholerae
One focus of molecular epidemiological analysis of V. cholerae is the CTX phage, which contains the CT genes. The cholera toxin-encoding genes ctxAB carried by toxigenic V. cholerae have undergone numerous genetic mutations, which include two main components of CTX phage (CTX cla and CTX-1) and repeat sequence 1 (RS1), along with point mutations in ctxB (40). Since 1817, a total of seven cholera pandemics have occurred worldwide, with the most recent pandemic continuing until the present (3). It appears that the O1 classical biotype was the causative agent of the first five pandemics (1817 to 1896). Following this, the O1 classical biotype CTX cla was responsible for the sixth pandemic (1899 to 1923) (40, 41). During the pre-seventh-pandemic period (1923 to 1961), only a few sporadic outbreaks were reported with the El Tor biotype (41). The current cholera pandemic (7th cholera pandemic [7CP]) began in 1961 in Indonesia, with the El Tor biotype as the causative agent, and spread to South Asia after 2 years and then to Africa (1970), South America (1990), and the Caribbean (2010).
A new V. cholerae serogroup, O139, emerged in the Indian subcontinent in 1992 and transmitted throughout the Asian subcontinent by the mid-2000s. In 2004, another new V. cholerae O1 type was isolated in Asia and Africa as a hybrid El Tor biotype encoded with classical CT (40, 41).
Eight diverse phyletic lineages (L1 to L8) have been classified by genomic analysis, based on single nucleotide polymorphisms (SNPs) from the O1 and O139 serogroups. The classical (L1) and El Tor (L2) biotypes were separated into two distinctly evolved lineages. The L1 and L3 to -6 lineages represent the first six pandemics, while the L2 lineage, also known as 7 P V. cholerae El Tor (7PET), is responsible for 7CP having three waves and several phylogenetic sublineages of transmission events (T1 to -13 and Latin American transmission 1 [LAT-1] to LAT-3). T1 to T12 originated from Africa, while T13 originated from East Africa and Yemen (40, 41).
The wave 1 (T1 to -6 and LAT-1 and -2) isolates were composed of CTX-1, the repressor gene (rstR) of the El Tor biotype (rstR El Tor) on chromosome 1, and CT genotype 3 (ctxB3) of V. cholerae O1 added with the toxin-linked cryptic (TLC) component and persisted from 1961 to 1999. Wave 1 spread from Indonesia to Southeast Asia, Mozambique, Angola, the Middle East, East Europe, Ethiopia, Angola, the U.S. Gulf Coast, and Latin America. Wave 2 (T7 and -8) isolates were composed of CT genotype 1 (ctxB1), a recurrence of ctx-2 on chromosome 2, ctx-1, and RS1 on chromosome 1. The wave was prominent between 1978 and 1984. It originated in India and spread to East Asia and Africa. V. cholerae O139 contains ctxB4 to ctxB6 and was prevalent from 1999 to 2005 in Bangladesh. Wave 3 (T9 to -13 and LAT-3) isolates are composed of rstR El Tor and TLC:RS1:CTX3 to TLC:RS1:CTX6 in CT genotype 1 (1991 to 2010) and carry the integrating and conjugative element (ICE) (specifically, ICEVchInd5) of the SXT/R391 type (encoding resistance to chloramphenicol, streptomycin, tetracycline, sulfamethoxazole, and trimethoprim). Wave 3 isolates are divided into three or more types. Wave 3 originated in India in 2006 and caused outbreaks in Haiti, Yemen, and Mariupol, Ukraine, in 2019 (40).
Clinical and Metabolic Manifestations
Cholera infection may be asymptomatic, mild, moderate, or severe (42). Diarrhea in cholera patients is usually painless and may contain bile or fecal matter in the early stages of infection. “Rice water stools” are unique to cholera patients. They are starchy in color, look more like water that contains uncooked rice or has been used to wash rice, and have a fishy odor (42, 43). An adult cholera patient can produce up to 1,000 mL per h of loose watery stool, leading to hypovolemia, shock, and death, termed “cholera gravis.” The rate of excretion rate stool in children with severe cholera is generally between 10 and 20 mL/kg/h (44). This stool contains potassium, sodium, and bicarbonate. Signs of dehydration (e.g., sunken eyes, tears, dry mouth, thirst, rapid pulse, lethargy, cold skin, loss of skin elasticity, or crumpled hands and feet) are present due to profuse watery diarrhea (Table 1). Deep and rapid respiration due to hyperventilation (Kussmaul breathing) and acidosis (loss of bicarbonate in the stool) are some of the striking features of severe cholera. Symptomatic cholera patients can shed bacteria in their stool for 2 days to 2 weeks from the onset of infection (Fig. 2), whereas asymptomatic carriers shed for only a few days (<7 days) (15). In settings where cholera is endemic, spatiotemporal analysis has shown that index cholera cases can be very infectious during the first 5 days of infection and can spread the bacteria within a 200-m radius of their home. Household contacts have a 100-times-higher risk of acquiring cholera than those outside the radius (45).
TABLE 1.
Clinical assessment of dehydrationd
| Condition | Assessment of dehydration severity |
--- |
| No signs of dehydration | Some dehydration | Severe dehydration |
| Physical appearance | Well, alert | Restless/irritablea | Lethargic/unconsciousa |
| Eyes | Normal | Sunken | Very sunken and dry |
| Tearsb | Present | Absent | Absent |
| Mouth and tongue | Moist | Dry | Very dry |
| Thirst | Drinks normally, not thirsty | Thirsty, drinks eagerlya | Drinks poorly/unable to drinka |
| Skin pinchc | Goes back quickly | Goes back slowlya | Goes back very slowlya |
| Radial pulse | Normal | Rapid, low vola | Weak or absenta |
| Diagnosis | No signs of dehydration | If the patient has 2 or more signs, including at least 1 signa of mandatory criteria, there is some dehydration | If the patient has 2 or more signs, including at least 1 signa of mandatory criteria, there is severe dehydration |
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a
Mandatory criteria for the diagnosis of different grades of dehydration.
b
Tear sign is applicable only to infants and younger children.
c
The skin pinch is less useful in severely malnourished children or overweight patients.
d
Adapted from WHO guidelines on the management of patients with cholera (58, 144).
FIG 2.
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Life cycle of V. cholerae.
Approximately 5 to 10% of patients develop severe dehydration from an increased frequency of profuse watery stool and excessive vomiting, which can rapidly deplete a large volume of water from the body, leading to kidney failure, shock, sepsis, and even death within a few hours if left untreated (1). Electrolyte imbalance is a common complication of cholera, which includes hyponatremia or hypernatremia, hypocalcemia, and hypokalemia (42). Renal failure due to decreased urinary output and aspiration pneumonia are also common in children (46). The accumulation of fluid in the intestinal lumen (cholera sicca) is very uncommon. Children under 5 years of age may develop chronic enteropathy and malnutrition. Inadequate rehydration may cause metabolic abnormalities among patients suffering from severe dehydration (42). Reduced food intake during acute illness may lead to hypoglycemia, a lethal complication that is more common in children (1).
DIAGNOSIS
The diagnosis of cholera is frequently based on clinical signs and symptoms (Table 1) in resource-limited areas of endemicity where laboratory facilities are not available (42). In nonendemic settings, cholera is suspected if a patient has severe dehydration or someone has died from acute watery diarrhea (AWD). But in a cholera epidemic setting, if a patient (>5 years of age) has AWD more than three times with or without vomiting within 24 h, cholera is indicated (47).
Generally, in the developed laboratory setting, for the diagnosis of cholera, stool or rectal swab culture is the gold-standard reference method and costs approximately $10 (40). The specimens are placed into an enrichment broth made of alkaline peptone water, which enhances the sensitivity of the culture, and are later subcultured on selective thiosulfate citrate bile salt (TCBS) agar or taurocholate tellurite gelatin agar (TTGA), which is the ideal culture medium (48). Hence, performing stool culture requires trained personnel and a laboratory facility (49). Cary-Blair medium is commonly used as the medium for transport from field settings to the laboratory (40).
Different methods (e.g., biochemical, immunochemical, or molecular) with specific antiserum or monoclonal antibodies are used to characterize the biochemical properties of V. cholerae O1 such as classical, El Tor, or altered variants (43). Dark-field microscopy can be used to rapidly detect V. cholerae in stool samples before culture (50). However, more than half of the dark-field-microscopy-negative samples are found to be positive for V. cholerae when cultured (51). V. cholerae diagnosis using PCR is highly sensitive, but this technique needs an enhanced laboratory capacity, which is often lacking in most LMICs (43, 46). PCR can be used to detect molecular markers of certain phenotypes with target genes such as ctxA, tcpA, and ompW (52). Even though PCR costs are approximately the same as those of stool culture and PCR requires a specific laboratory setup, the results can be obtained much earlier than with stool culture.
In rural or underdeveloped health care settings where there is a scarcity of culture medium/PCR and/or trained personnel (5), rapid diagnostic tests (RDTs) of stool samples cost only $2, and these can be performed without special training. RDTs can also provide an early warning for public health experts when a cholera outbreak is imminent. Different categories of RDTs are available on the market, with a wide range of sensitivities and specificities. Monoclonal antibodies in Crystal VC (Span Diagnostics Ltd., Surat, India) can easily detect the lipopolysaccharide (LPS) antigens of both V. cholerae O1 and O139 serogroups. Crystal VC has shown 97% sensitivity and 76% specificity (53). Cholkit is available in Bangladesh, with acceptable sensitivity (76%) and good specificity (90%), and is less expensive (less than $2) than the existing RDTs (54). However, this kit detects only the V. cholerae O1 serogroup and hence needs to be used in settings where cholera is endemic and where this serogroup is predominant. Currently, V. cholerae O1 is the predominant serogroup, although V. cholerae O139 strains are occasionally isolated in Bangladesh (3, 55).
CLINICAL MANAGEMENT
Fluid Replacement
Early diagnosis and rapid management of dehydration are crucial for increasing positive outcomes. Most cholera cases are usually mild to moderate and easily managed with an oral rehydration solution (ORS) ( Currently, the WHO and UNICEF recommend low-osmolarity ORS, which contains sodium, chloride, potassium, citrate, and anhydrous glucose prepared in 1,000 mL of sterile water (Table 2). This improved ORS recipe is safe, lowers hypertonicity, and decreases stool output (56). ORS can also be prepared at home by mixing 1/2 teaspoon of salt and 6 teaspoons of sugar in 1,000 mL of sterile water. Rice-based saline (i.e., rice powder) is also used for those above 6 months of age but is more difficult to prepare (57). After each purging of watery stool, ORS is given according to different age groups to counterbalance the amount of stool loss (Fig. 3). For young children up to 2 years of age, breastfeeding is vital, along with fluid replacement (8). Many countries have introduced low-osmolarity ORS plus zinc for the treatment of cholera.
TABLE 2.
Compositions of different rehydration solutionsa
| Solution | Content (mmol/L) |
--- |
| Na+ | K+ | Cl− | HCO 3− | Carbohydrate |
| i.v. lactated Ringer’s solution | 130 | 4 | 109 | 28 | |
| i.v. normal saline, 0.9% | 154 | 0 | 154 | 0 | |
| i.v. cholera saline (Dhaka solution) | 133 | 13 | 98 | 48 | 140 |
| ORS (standard) | 90 | 20 | 80 | 10 (citrate) | 111 |
| ORS (WHO 2002)a | 75 | 20 | 65 | 10 (citrate) | 75 (glucose) |
| Rice-based ORS (e.g., Cera ORS 75) | 75 | 20 | 65 | 10 (citrate) | 27 g rice syrup solids |
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a
The data on the compositions of rehydration solutions are adapted from standard guidelines (56, 58, 145).
FIG 3.
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Management of cholera based on the severity of dehydration. These guidelines have been adapted from WHO guidelines for the treatment of diarrhea (58).
Treatment with i.v. fluids can be considered if vomiting continues more than three times in 1 h or if ORS is not improving the patient’s condition. Instant corrections of fluid and electrolyte deficits are the bases of rehydration therapy in order to counterbalance ongoing losses. Severe cholera cases require i.v. fluid (25). The WHO recommends Ringer’s lactate solution (Na+ at 130 mmol/L, K+ at 4 mmol/L, Cl− at 109 mmol/L, HCO 3− at 28 mmol/L, and Ca+ at 1.5 mmol/L [osmolarity of 273 mmol/L and pH of 6.5]) over normal saline (Na+ at 154 mmol/L and K+ at 154 mmol/L [osmolarity of 308 mmol/L and pH of 4.5]) as it contains more potassium and bicarbonate. The International Centre for Diarrhoeal Disease Research, Bangladesh (icddr,b), and other governing bodies of countries where cholera is endemic prefer to use the “Dhaka solution” or “cholera saline” (Na+ at 133 mmol/L, K+ at 13 mmol/L, Cl− at 154 mmol/L, HCO 3− at 48 mmol/L, carbohydrate at 140 mmol/L) (Table 2), which contains glucose as well as more potassium and bicarbonate than Ringer’s lactate solution and can reduce adverse outcomes such as electrolyte imbalances (58, 59).
Severely dehydrated patients need an instant i.v. fluid infusion at a bolus dose of 100 mL/kg of body weight over 3 h and a one-third infusion in the first 30 min (59). Patients aged 1 year and above need 30 mL/kg in the first 30 min and 70 mL/kg in the next 2 1/2 h. The total duration of rehydration is 6 h for children less than 1 year of age (Fig. 3). Severe cholera patients should be kept on a cholera cot (e.g., a cot with a hole and an underlying bucket) to monitor the ongoing fluid loss so that the actual amount of fluid can be replaced. When a patient is able to drink, ORS is started again for hydration (5). For all patients, the fluid infusion should be given in repetitive measures if danger signs (hypovolemia, low radial pulse, and deep breathing, etc.) appear even after starting an i.v. infusion as a bolus therapy (Fig. 4). Malnourished children require a high-energy diet after correction of fluid deficiency to prevent hypoglycemia, hyponatremia, and hypokalemia (58).
FIG 4.
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Flow diagram of the course of a cholera patient from admission to recovery.
Antibiotics and Antimicrobial Resistance
The WHO has recommended antibiotics for severe cholera patients irrespective of age and for patients who require hospitalization (5, 60). Several studies have shown that antibiotics shorten the length of infectious diarrhea (from 5 days to 1 to 2 days) as well as decrease the volume of stool output by up to 50% (1, 43, 61). Tetracycline, fluoroquinolones, co-trimoxazole, doxycycline, ciprofloxacin, trimethoprim-sulfamethoxazole, erythromycin, and azithromycin are the most commonly used antimicrobials for treating cholera patients, but resistance has become a global concern (43). The choice of antibiotics depends on local antibiotic susceptibility patterns. In most countries, doxycycline is recommended as a first-line treatment as a 300-mg single oral dose for adults (including pregnant women) and as a 2- to 4-mg/kg single oral dose for children. If resistance to doxycycline is documented, azithromycin and ciprofloxacin are alternative options. For children under 12 years of age, azithromycin is prescribed as a 20-mg/kg (maximum, 1-g) single oral dose, and ciprofloxacin is prescribed as a 20-mg/kg (maximum, 1-g) single oral dose. A single oral dose of azithromycin (1 g) or ciprofloxacin (1 g) is prescribed to adults suffering from severe cholera (
Antimicrobial resistance (AMR) against trimethoprim-sulfamethoxazole, nalidixic acid, furazolidone, tetracycline, and ciprofloxacin developed between 1990 and 2010 (62). Hence, frequent performance of antimicrobial testing on V. cholerae clinical isolates is recommended for the treatment of cholera patients in settings where cholera is endemic as multidrug resistance (MDR) has developed due to frequent chromosomal mutation. Over the past decade, whole-genome sequencing of V. cholerae isolates has shown mobile genetic elements (MGEs) from other bacterial species, plasmids, conjugative elements, superintegrons, and supplemental sequences, all of which could lead to AMR (63). Studies from Africa showed increased resistance of V. cholerae O1 serotype strains to co-trimoxazole, chloramphenicol, ampicillin, and tetracycline (64, 65). Countries such as India and Nepal have also reported V. cholerae AMR to nalidixic acid, ciprofloxacin, tetracycline, and furazolidone, and there is an increasing number of MDR V. cholerae El Tor strains (66, 67).
Between 2009 and 2014, 17% of V. cholerae strains were resistant to third-generation cephalosporins, and 93% MDR strains were detected, exhibiting resistance to streptomycin, nalidixic acid, tetracycline, and trimethoprim-sulfamethoxazole (68, 69). An alarming fact is that the isolates were also becoming resistant to ciprofloxacin and azithromycin as these antibiotics were the drugs of choice for the management of cholera cases (70).
Self-medication with antibiotics has been reported in 83% (120/144) of infected and symptomatic household contacts in Bangladesh (71). Antibiotics are easily obtainable without a doctor’s prescription, and the habit of frequent and incomplete courses is quite common in developing countries (72, 73). Most recently, a randomized controlled trial of electronic decision-based diarrheal management through mobile phones using the mHealth Diarrhea Management (mHDM) platform in 10 district hospitals in Bangladesh showed a 10% reduction in the ordering of antibiotics in hospitalized patients compared to paper-based decisions in clinical facilities. Additionally, that study also found a remarkable decrease in prescribing of nonindicated antibiotics using mHDM among patients <5 years (28.5%) and >18 years (11.8%) of age (74). Such findings may help with antimicrobial stewardship in Bangladesh.
V. cholerae susceptibility to antibiotics is usually tested by two methods, disk diffusion and MIC methods for antibacterial properties (75, 76). To better understand the pattern of declining sensitivity of V. cholerae to antibiotics and its alarming resistance surge in the world, several molecular techniques are being conducted globally. National surveillance systems to identify changing sensitivity patterns are being used in order to identify the most appropriate drugs for cholera management (77). Moreover, the advantage of stool culture with sensitivity tests relative to PCR or RDTs is the added ability to perform antimicrobial susceptibility testing (AST) and monitor resistance patterns.
Antibiotic Prophylaxis
Findings from a systematic review have revealed that household contacts of cholera cases obtained protection against the disease when antibiotics were given to them as prophylaxis (78). In another study, one dose of doxycycline (300 mg) was administered orally to all prisoners and prison staff to prevent the spread of cholera in a prison in Cameroon in 2004, and no new cholera cases were reported over the next 4 months (79). However, mass chemoprophylaxis with an antibiotic is not recommended by the WHO for cholera control as it can lead to antibiotic resistance, but selective chemoprophylaxis may be provided for the prophylactic treatment of close contacts of a cholera patient (80). More studies are needed to inform the future use of targeted chemoprophylaxis in high-risk subpopulations.
Micronutrients
Zinc supplementation in children <5 years of age can also reduce the length and stool volume of diarrhea. Studies in various countries have shown that the addition of zinc to ORS reduces the severity of diarrhea and subsequently limits the use of antimicrobials (81). Zinc inhibits basolateral potassium channels by blocking cAMP-induced chloride-dependent fluid secretion. Zinc also regenerates the intestinal epithelium and increases the secretion of enzymes and the absorption of water and electrolytes, thus enhancing the immune response (82, 83). Vitamin A supplementation is suggested for children 6 months to 5 years of age to avert further occurrences of diarrhea (84). A high-calorie diet may reduce hypokalemia, hypoglycemia, and malnutrition, even when diarrhea is present (85).
FUTURE TREATMENTS
Supplementation with probiotics is relatively new for cholera management (86). Cholera toxin alters the gut microbiota (87). Probiotics can restore the gut microbiome and thus can potentially be used to clinically manage cholera. The use of probiotics in cholera management may limit AMR by reducing antibiotic use. Studies have shown the important contribution of the gut microbiome to fighting cholera and other diarrheal diseases (82, 83). There are several bacterial species in the gut that have been found to suppress cholera infection. Ruminococcus obeum has a positive correlation with cholera recovery (88). Another study showed that Lactobacillus rhamnosus strain GG (ATCC 53103) and Bifidobacterium longum 46 (DSM 14583) in coculture with V. cholerae were capable of removing CT from its environment (89). The therapeutic use of lytic bacteriophages, which is known as “phage therapy,” is another novel approach to the treatment of cholera (90). Phages are efficient at killing MDR bacteria. Studies conducted in infant mice and rabbits found that a combination of three isolated V. cholerae-specific virulent phages (ICP1, ICP2, and ICP3) was able to decrease the bacterial load of V. cholerae and prevent cholera-like diarrhea (91). Experimental studies targeting the inhibitor of cystic fibrosis channel transmembrane (CFTR), inhibitors of virulence factors, and a monosialoganglioside (GM1) antagonist found that they reduced intestinal secretion induced by CT of V. cholerae (92).
INTEGRATED CONTROL
Water, Sanitation, and Hygiene
Water, sanitation, and hygiene (WaSH) are of the utmost importance for the prevention and control of cholera and other enteric infections in the developing world (88, 89). Treatment of water at the source will not prevent the disease as contamination may occur at any time from collection to consumption. Water storage can also lead to contamination (93). When an outbreak is declared, interdisciplinary action is crucial for containing the pathogen, along with prioritizing which components of WaSH need to be implemented (93). Behavioral change practices (e.g., handwashing, use of soapy water, chlorination of household drinking water) have been shown to improve the protective efficacy of oral cholera vaccination when administered together (94). The Philippines, Thailand, and Vietnam have initiated cholera control strategies comprising surveillance, health promotion, and the supply of safe water. But in the cross-border areas of Malaysia, great challenges remain where the scarcity of sterile water, poor hygiene, and open defecation are noticeable concerns. Vietnam has reported zero cholera cases since 2012, and the Philippines has implemented a zero-open-defecation program along with other WaSH interventions to control the disease (95). Countries with high cholera disease burdens, such as Bangladesh, Pakistan, Nepal, and India, have minimal piped water in many localities, limited access to potable water in rural areas, shared latrines among the urban poor, open defecation in rural areas, and poor compliance with handwashing (96). Reports on evidence of household water treatment reducing cholera incidence have been published, but it is not effective in high-risk households due to financial constraints, poor education, and practices resulting in the low uptake of interventions (96, 97). Few high-risk households (12% in Nigeria, 19% in Nepal, and 24% in Haiti) have reported using water treatment (e.g., filtration, boiling, UV purification, or chlorine disinfectant use) (97). Without the support of national and local governments and nongovernment organizations, such implementations are impossible to sustain. Additionally, effective sewage systems with safe waste disposal and mechanisms to prevent untreated waste from reentering the environment are essential for controlling cholera (97). A strategy implemented in Mozambique, following the impact of Cyclone Idai in March 2019, was successful in controlling an outbreak of cholera. This included the establishment of a real-time surveillance system, WaSH, and vaccination (98). The WHO also recommends reinforcement and access to improved potable water, standard sanitation laws for food industries, execution of handwashing practices with soap, and safe handling of food as part of the cholera prevention strategy (5).
Inexpensive WaSH interventions such as solar power water purifiers, handwashing facilities, soapy water dispensers, Aquatabs (hypochlorous acid) for water purification, and safe storage containers (Fig. 5) have been modestly effective in lowering the numbers of cholera cases in community trials (99). During outbreaks, emergency WaSH interventions, including inexpensive WaSH strategies, health education, and media coverage, are crucial for reducing mortality as well as preventing further outbreaks (99). Community engagement plays an important role in long-term behavior changes and the prevention of cholera. Awareness campaigns using advertisements on the radio, television, billboards, and text messages can be used to educate people on when to seek medical attention and the use of ORS.
FIG 5.
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Low-cost WaSH interventions for the control of cholera during an outbreak.
Vaccination
In areas where cholera is endemic, the WHO recommends cholera vaccination as part of the national cholera control program along with WaSH (10). Several studies have shown that the early introduction of vaccines during an outbreak offers 79% protection against cholera (100, 101), and even one dose of the cholera vaccine significantly reduces the risk (5, 102–104).
Broadly speaking, there are four different types of cholera vaccines (Table 3): (i) killed whole-cell (WC) monovalent V. cholerae O1 (classical Inaba, and Ogawa, and El Tor Inaba) vaccines with a recombinant B subunit of cholera toxin, (ii) killed WC modified bivalent V. cholerae O1 and V. cholerae O139 vaccines without the B subunit, (iii) live attenuated oral cholera vaccines (OCVs), and (iv) parenteral cholera vaccines (5, 105, 106). Presently, only two OCVs are available worldwide: (i) the killed WC monovalent vaccine with a recombinant B subunit of cholera toxin (Dukoral) and (ii) killed and modified WC (O1 and O139) vaccines without the B subunit (Shanchol, Euvichol, Euvichol Plus, and mORC-Vax ).
TABLE 3.
Comparison of oral cholera vaccinesa
| Vaccine | Description for vaccine |
--- |
| Dukoral | OraVacs | Shanchol | Euvichol/Euvichol Plus | mORC-Vax | Cholvax | Vaxchora | Hillchol |
| Generic name | WC-rBSb | WC-rBSb | Modified bivalent WC | Modified bivalent WC | Modified bivalent WC | Modified bivalent WC | CVD 103-HgR | |
| Type | Oral killed; buffer is needed | Oral killed; capsule | Oral killed; no buffer | Oral killed; no buffer | Oral killed; no buffer | Oral killed; no buffer | Oral live attenuated; buffer is needed | Oral killed; no buffer |
| Composition | Monovalent; killed whole-cell vaccine O1 serogroup (classical Inaba, and Ogawa, and El Tor Inaba) and recombinant cholera toxin B subunit | Monovalent; killed whole-cell vaccine O1 serogroup (classical Inaba and El Tor Inaba) and recombinant cholera toxin B subunit | Bivalent; killed modified whole-cell bivalent (O1 and O139) vaccine without the B subunit | Bivalent; killed modified whole-cell bivalent (O1 and O139) vaccine without the B subunit | Bivalent; killed modified whole-cell bivalent (O1 and O139) vaccine without the B subunit | Bivalent; killed modified whole-cell bivalent (O1 and O139) vaccines without the B subunit | Monovalent; live attenuated serogroup O1 classical Inaba strain 569B | Single strain; whole-cell recombinant Hikojima strain MS1568 |
| Age range (yrs) | ≥2 | ≥2 | ≥1 | ≥1 | ≥1 | ≥1 | 18–64 | 1–45 |
| Dose regimen(s) | 2 doses given 1–6 wks apart, 3 doses for children aged 2–5 yrs | | 2 doses, 14 days apart | 2 doses, 14 days apart | 2 doses, 14 days apart | 3 doses, days 0, 7, and 28 | Single dose | 2 doses, 14 days apart |
| Manufacturer | Developed by SBL, Sweden; now Valneva, France | Shanghai United Cell Biotechnology, China | Shantha Biotechnics, India | Eubiologics, South Korea | Vabiotech, Hanoi, Vietnam | Incepta Vaccine Ltd., Bangladesh | PaxVax Inc., USA | Incepta Vaccine Ltd., Bangladesh |
| First licensing country(ies), yr(s) | Sweden, 1991 | China and Philippines | India, 2009 | South Korea, 2004 | Vietnam, 1997, 2009 | Bangladesh, 2019 | USA, 2016 | Under development |
| WHO prequalification date (mo and yr) | October 2001 | No | September 2011 | December 2014/2017 | No | No | No | No |
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a
Information on the different oral cholera vaccines is adapted from references 5 and 106.
b
Whole-cell recombinant cholera B subunit.
Killed whole-cell cholera vaccines.
(i) Monovalent WC V. cholerae O1 oral cholera vaccine with a recombinant B subunit of cholera toxin.
The killed WC monovalent OCV was first manufactured in Sweden and gained licensure in 1991. Currently, it is available in more than 60 countries (5). It is found under the trade name Dukoral, manufactured by Valneva, France, and achieved WHO prequalification in October 2001 (106). The vaccine was formulated in combination with a recombinant B subunit of cholera toxin and formalin- or heat-killed WC V. cholerae O1 (classical Inaba and Ogawa and El Tor Inaba). The B unit of V. cholerae toxin is analogous to the heat-labile toxin (LT) of enterotoxigenic E. coli (ETEC) in its composition and functional capability and elicits cross-protection against ETEC (5). The vaccine is administered as a three-dose regimen 1 week apart for those aged 2 to 6 years and as a two-dose regimen at least 1 week apart for individuals >6 years of age. The added B subunit in the vaccine can be neutralized by gastric acids, and hence, to protect its functionality, the vaccine is required to be administered along with a buffer solution. The Dukoral vaccine does not defend against the V. cholerae O139 serogroup or other types of vibrios and has also been reported to elicit higher immune responses in cholera-naive populations than in populations where cholera is endemic (5, 105, 106).
OraVacs is a monovalent, killed WC vaccine containing the O1 serogroup (classical or El Tor biotype) and a recombinant cholera toxin B (rCTB) subunit and is indicated for traveler’s diarrhea. A clinical trial of this vaccine revealed that it is safe and immunogenic for children older than 2 years of age and adults. However, no efficacy trial data are available (106). This vaccine is manufactured by Shanghai United Cell Biotechnology, China, and is licensed in China and the Philippines (Table 3).
(ii) Bivalent modified WC V. cholerae O1 and V. cholerae O139 vaccines.
In the mid-1980s, scientists in Vietnam developed a modified killed WC vaccine comprising the V. cholerae O1 serogroup excluding CTB, ORC-Vax, with technology transfer from Sweden (105), and including both V. cholerae O1 and O139; hence, this vaccine was called the bivalent killed WC vaccine and was first licensed in Vietnam in 1997 (5, 105, 106). It was manufactured by Vabiotech (Hanoi, Vietnam), and it was included in the national Expanded Program for Immunization (EPI) and used during cholera outbreaks in that country. However, this vaccine was not WHO prequalified as Vietnam’s National Regulatory Agency (NRA) was not recognized by the WHO (105). Hence, to comply with international guidelines, the vaccine’s manufacturing technology was transferred to the International Vaccine Institute (IVI) (Seoul, South Korea) in 2004 from Vabiotech. This reformulated vaccine was named mORC-Vax and is now licensed in Vietnam. Several clinical trials have evaluated the two-dose schedule of this reformulated vaccine and found it to be safe and to produce antibody responses to O1, but immunogenicity was less pronounced against the O139 serotype (107, 108).
The IVI transferred the modified technology to Shantha Biotechnics (Hyderabad, India) to distribute mORC-Vax internationally. The vaccine was reformulated once again to WHO standards and met the requirements of safety and immunogenicity following several clinical trials. The modified version of the vaccine was called Shanchol (Shantha Biotechnics-Sanofi Pasteur), and it was licensed in 2009 and subsequently WHO prequalified in 2011 (105, 106). This final version of the vaccine was found to have fewer adverse reactions and higher levels of antibody responses in India, Bangladesh, Vietnam, and Ethiopia (107–110). In Kolkata, India, and Bangladesh, large phase II and III trials were conducted, and it was proven to be safe and efficacious. During the 2-, 3-, and 5-year follow-up periods, the vaccine provided 67%, 66%, and 65% protection against V. cholerae O1, respectively (111–113). A single dose of the Shanchol OCV is also effective in individuals above 5 years of age in settings where cholera is highly endemic (114, 115). The vaccine is safe and stable at elevated temperatures and is presently stockpiled (116). To meet the increasing demand for the OCV internationally, the IVI transferred the vaccine manufacturing technology to Eubiologics (Seoul, South Korea) to manufacture Euvichol, similar to Shanchol. The Euvichol (glass vial) vaccine was licensed in 2004 by the Government of South Korea, and the vaccine was prequalified by the WHO in 2015 (117). Clinical trials were conducted in South Korea and the Philippines, and the study outcomes revealed that robust antibody responses after two doses of Euvichol were comparable to those elicited by Shanchol in adults (82% versus 76%) and children (87% versus 89%) (117). Euvichol Plus (plastic vial) was developed at a lower cost, is easier to store, and received WHO prequalification in 2017 (105, 106). Euvichol is now stockpiled.
Live attenuated oral cholera vaccine.
In the 1970s, researchers from the Center for Vaccine Development (CVD) at the University of Maryland School of Medicine manufactured a live attenuated OCV with a V. cholerae O1 strain (106, 118). The live attenuated oral vaccine (like other live vaccines) provides greater efficacy, a more rapid immune response, and greater long-term protection among naive populations than among populations in areas where cholera is endemic (106, 119). To date, four live attenuated oral cholera vaccines have been produced, one of which is currently available, CVD 103-HgR (119).
CVD 103-HgR is the first live attenuated single-dose OCV (monovalent classical Inaba O1). Commercially, this vaccine was produced by the Swiss Serum and Vaccine Institute (SSVI), Berne, with the market names Orochol and Mutacol. Unfortunately, these vaccines were withdrawn from the market in 2003 for economic reasons (120, 121). In 2016, PaxVax Inc., a pharmaceutical company in the United States, acquired the licensure of CVD 103-HgR with the market name Vaxchora for U.S. adult travelers (122). Single-dose administration of Vaxchora showed a >90% vibriocidal seroconversion rate and gave protection among U.S. adult study participants following vaccination (119, 120, 122, 123). It had 79% protective efficacy when administered during a mass cholera vaccination campaign following an epidemic in Micronesia (119). Moreover, recently, satisfactory antibody responses and very few adverse events were seen in U.S. adults, children, and HIV patients (119, 122).
Licensures of other live attenuated OCVs are currently under development, such as Peru-15, composed of the O1 El Tor Inaba strain (Cholera Garde; Harvard Medical School, USA); OCV VA 1.4 (Government of India); a live attenuated El Tor Ogawa strain (638); IEM 108 (China); and HaitiV (United States) (106, 124, 125).
Parenteral vaccines.
Several parenteral cholera vaccines have been developed, among which are a killed WC vaccine, a purified lipopolysaccharide vaccine, a killed WC vaccine in combination with different adjuvants, and a polysaccharide-cholera toxin conjugate vaccine (126, 127). Only the killed WC vaccine was broadly available for several years but was not recommended by the WHO as this vaccine provided only a short duration of protective efficacy (43% protection for 3 months), induced a high level of adverse events (106, 126, 127), and was not recommended for pregnant women (126, 128). Most recently, a preclinical study with a newly developed parenteral cholera conjugate vaccine composed of Inaba O-specific polysaccharide (OSP) and the recombinant tetanus toxoid showed significant boosting of vibriocidal immune responses in mice following a single dose (CVD 103-HgR) (129). The parenteral cholera vaccine does not protect against cholera caused by V. cholerae O139 and did not show great effectiveness during cholera outbreaks. It also has a high-adverse-event profile, especially after intramuscular (i.m.) or subcutaneous (s.c.) administration, including local pain, erythema, induration, fever, malaise, and headache in most individuals, in comparison to oral cholera vaccines. OCVs are also more feasible to administer in an emergency setting than any parenteral vaccine and are also safe for pregnant women (126).
New vaccines under development.
Incepta Vaccine Ltd., a Bangladeshi company, has developed two OCVs, Cholvax and Hillchol. The technology for Cholvax vaccine production was transferred from the IVI (106). Cholvax has the same formulation as that of the Shanchol vaccine regarding strains and other formulations required for maintaining international good manufacturing practice (GMP) standards and WHO production guidelines (130). The manufacturing process for the Cholvax vaccine is also less expensive than those for the other available WHO-prequalified OCVs, and Cholvax has been found to be safe, immunogenic, and noninferior to the Shanchol vaccine (106, 131). Cholvax has been approved by the Directorate General of Drug Administration (DGDA) and was licensed in Bangladesh in 2019. The annual capacity for the production of Cholvax is 20 million to 40 million doses, which will be helpful to reduce the cholera burden in Bangladesh (106).
Hillchol is a formaldehyde-inactivated WC single-strain vaccine that originated from recombinant Hikojima V. cholerae strain MS1568 and was generated from an El Tor V. cholerae O1 parent/ancestor Inaba strain (Phil6973). This strain contains 50% LPS of each of the Ogawa and Inaba serotypes (132). The new monovalent Hillchol vaccine is manufactured with a solitary inactivation process and is anticipated to have a lower price than other current OCVs. The Hillchol vaccine was found to be safe, immunogenic, and noninferior to Shanchol among study participants (healthy adults and older and younger children) in Bangladesh (133).
The amended heat-stable Hillchol-B vaccine is composed of WC, formalin-inactivated Ogawa and Inaba strains in combination with rCTB in an enteric-coated capsule and can be easily administered during cholera outbreaks and among travelers to areas where cholera is endemic. Very recently, this vaccine was named the DuoChol capsule (124). Another OCV is in preclinical development in Sweden and is composed of formalin-killed cocultured isogenic El Tor Ogawa and Inaba serogroups (112). Eubiologics and the IVI are developing a formalin-killed classical Ogawa (Cairo50) and El Tor Inaba (Phil6973) vaccine, which has completed preclinical trials and is now set to move forward with a clinical trial in South Korea (124).
Vaccine enhancement in vulnerable populations.
Many low-income countries where cholera is endemic have reported a high cholera prevalence among young children. Current vaccines have shown a minimal level of protection and a short duration of protection in those aged 2 to 5 years (10, 126). Many approaches have been suggested to improve vaccine efficacy for this age cohort. Different regimes of vaccine administration, such as 3 doses at 4-week intervals or 2 doses at 8-week intervals, have been suggested to enhance immunogenicity in children. One study showed that withholding breastfeeding 3 h before vaccination increases vaccine efficacy along with supplementation with 20 mg of zinc per day for 42 days (134). Blood group, gut microbiota, malnutrition, environmental enteropathy, and the presence of multiple copathogens may also have a strong association with lower immunity in children (119).
The WHO recommends OCVs for pregnant and lactating women as these vaccines have potential benefits that outweigh the negligible risks (5). During pregnancy, severe dehydration can lead to premature delivery, miscarriage, and fetal death. OCVs were found to be safe, with no adverse fetal outcomes observed in several studies (135–138). The WHO now recommends the use of OCVs for pregnant mothers in areas where cholera is endemic to prevent severe dehydrating cholera that may harm the fetus.
Herd immunity and vaccines.
It has been shown that inactivated OCVs can give significant herd protection in various study settings, which were analyzed using geographic information system (GIS) tools (138). The herd-protective effects of OCVs were measured using various study designs (individually or cluster randomized trials and cohort or case-control studies), and significant herd protection (both direct and indirect) against cholera was seen among unvaccinated persons and in the community (138). Mathematical modeling of cholera transmission (139) using Matlab showed that 93% of cholera infections in Bangladesh can be prevented with 50% OCV coverage (140). The model predicts an 89% drop in the incidence of cholera in the population that is not vaccinated. Study findings in Zanzibar showed that after mass cholera vaccination, OCVs also gave herd immunity (both direct and indirect) in this African setting (101). A large feasibility trial in Bangladesh revealed that children less than 3 years of age had a 47% reduction in the incidence of cholera if their mothers were given OCVs (139). Another analysis by Ali et al. showed that the chance of having cholera in the unvaccinated adult population was reduced by 14% with a 10% rise in OCV coverage in all age strata with a killed oral vaccine (138).
Challenges with oral cholera vaccines.
In 2013, after the formation of the global cholera vaccine stockpile, the logistic complications of vaccine delivery and the consistent scarcity of the vaccine supply hampered successful OCV implementation. The main challenge is to deliver two doses of the vaccine at a 14-day interval in rural or urban field settings (141). Recently, a heat stability study conducted on Shanchol showed that the vaccine is thermostable (116). Fortunately, Shanchol can be used at an ambient temperature (up to 42°C) for up to 1 month, but the storage temperature should be between 2°C and 8°C according to WHO guidelines (5).
Future vaccine strategies.
A single dose of an OCV induces a vibriocidal response among exposed populations, as observed in previous clinical trials (109, 111, 114, 142). A single dose of an OCV was shown to be efficacious (57%) among those above 5 years of age; however, no protective efficacy was observed for those below 5 years of age (114). A possible reason is a lack of acquired immunity. Thus, the memory B cell response against cholera-specific antigen develops over time due to recurrent natural exposure to V. cholerae (106, 142).
Studies on the use of booster doses of OCVs have been carried out in Bangladesh, which showed that children who received a single dose of an OCV 3 years earlier showed significantly increased vibriocidal antibody responses after receiving one booster dose of an OCV compared to those who did not receive an OCV earlier (142). These results suggest that boosting with one dose of an OCV augmented the immune responses in children, although more studies are required to adjust the primary booster doses of OCVs as well as to determine the duration between the prime and booster doses (142). Nevertheless, one dose of an OCV was also found to be protective during an outbreak among people (5 years of age and older) in Zambia who had less exposure to cholera (100).
Large campaigns of two doses of an OCV were carried out in 2017 among the Rohingya population (forcibly displaced Myanmar Nationals) in Cox’s Bazar in Bangladesh. A trial was conducted to assess the immunological parameters before and after vaccination. The study revealed a significant increase in vibriocidal antibody titers 14 days following the first dose of the OCV (143). Similarly, another study conducted during a humanitarian crisis in South Sudan showed that only one dose of an OCV was immunogenic and induced short-term antibodies (106, 115).
CHOLERA ELIMINATION
The GTFCC of the WHO has launched an initiative, entitled Ending Cholera: a Global Roadmap to 2030, aiming for at least a 90% mortality reduction in 47 countries of endemicity. The global roadmap aligns health and WaSH resources and targets areas most in need, saving lives, enhancing equity, and reducing the significant economic burden of cholera as well as other enteric diseases. The global roadmap focuses on three strategic priorities to control cholera. The first strategy includes the rapid detection of cholera cases and early responses to outbreaks through an early-warning surveillance system (EWARS) with an enhanced laboratory culture capacity along with dedicated health care facilities to treat cholera. The second strategy focuses on averting cholera recurrence in identified hot spots by improving WaSH and the delivery of OCVs. The third strategy is to develop a well-organized and efficient network to provide financial support to countries where cholera is endemic and bring national and international collaborators together to promote intersectorial coordination, supply mobilization, technical assistance, and strong cooperation to control cholera (
Bangladesh is one of 20 countries of endemicity targeting cholera elimination. The incidence rate for cholera is 1.64 per 1,000 population annually. Bangladesh formulated the National Cholera Control Plan (NCCP) ( in 2019, which is the guiding document to ensure OCV delivery to the target population according to a GTFCC strategic approach. A large OCV campaign was demonstrated among 1.2 million people as part of the NCCP in February 2020 in the capital city of Dhaka, comprising six high-risk cholera-prone areas (
Along with Bangladesh, Uganda, Zambia, and Zanzibar are also in the process of trying to meet the Ending Cholera: a Global Roadmap to 2030 objectives. The Zambia government is working toward improving the WaSH sector in compliance with the roadmap supported by the GTFCC ( Zanzibar is implementing the Zanzibar Comprehensive Cholera Elimination Plan (ZACCEP) ( which is a 10-year program to eliminate the indigenous transmission of cholera and promote a healthy and clean environment. Uganda has incorporated a 5-year plan to reduce cholera by 50% by 2022 through vaccination and the implementation of WaSH among 300,000 persons in the first 3 years, as reported at the 5th annual meeting of the GTFCC ( Developed countries have eliminated cholera largely through improved sewage systems and clean water supplies. However, in the 21st century, high-risk populations and those in LMICs still do not have access to safe drinking water and formal sewage systems (97).
CONCLUSIONS
V. cholerae causes periodic cholera epidemics in several regions around the globe. The disease requires immediate treatment as it can cause death within hours in patients with moderate to severe cholera. With the development of i.v. fluids, ORS, and Zn therapy, progression to severe dehydration and mortality has been remarkably reduced. Antibiotics, micronutrients, and probiotics can further assist in recovery. In the developing world, treatment challenges are primarily due to delays in receiving prompt medical attention by health care professionals. The use of cholera RDTs as a point-of-care diagnostic during an outbreak along with PCR requires significant laboratory investment and skilled personnel. The ambitious global roadmap to end cholera by 2030 requires countries of endemicity to use evidence-based solutions to make this goal a reality ( To control cholera in endemic or outbreak situations at the domestic and communal levels, an OCV is considered an essential component of an integrated control package along with WaSH. Extensive and robust cholera surveillance, rapid diagnostics, treatment, and health education will be required for sustained control. Finally, to make all of this possible there must be political will at all levels of government. Without such support, the WHO 2030 targets will not be met.
ACKNOWLEDGMENTS
This article received no funding from any donor agency in the public, profitable, or nonprofitable sector.
The icddr,b is thankful to the Governments of Bangladesh, Canada, Sweden, and the United Kingdom for providing core/unrestricted support.
There is no conflict of interest.
Biographies
Fahima Chowdhury, Associate Scientist and Project Coordinator, Mucosal Immunology and Vaccinology Unit, Infectious Diseases Division, icddr,b, is currently doing her Ph.D. under the Faculty of Medicine at Griffith University, Australia, under the supervision of Professor Nigel McMillan, Professor Allen Ross, and Dr. Firdausi Qadri. Dr. Fahima is working on different immune response studies in cholera patients and recently conducted an immunogenicity trial among forcefully displaced Myanmar Nationals (FDMNs). She was actively involved in large field trials of oral cholera vaccines (OCVs). Currently, as a PI, she completed clinical trials of locally produced vaccines such as cholera and hepatitis B vaccines. Very recently, she has been conducting a study among hospitalized COVID-19 patients to detect virus-specific antibodies and measure the immune responses in vitro and in vivo as well as antibody responses after different COVID-19 vaccines. She had published around 95 papers in peer-reviewed journals, including 16 papers as a first author.
Allen G. Ross is Professor of Medicine and Executive Director of the Rural Health Research Institute with Charles Sturt University, Orange, NSW, Australia. Professor Ross’s expertise and research interests lie in the realm of graduate education, global health, medical microbiology, tropical infectious diseases, enteric diseases, disease control, pandemic planning, and vaccination. Professor Ross completed a bachelor of science in Biology (B.Sc., 1990) and a master of Science (M.Sc., 1994) in Human Biology in Canada before proceeding to Brisbane, Australia, where he completed a doctorate in Tropical Health with distinction (Ph.D., 1998); a bachelor of Medicine, bachelor of Surgery (M.B.B.S., 2005); a doctorate of Medicine (M.D., 2010); and a doctorate of Medical Science (D.Sc., 2017). He is a Fellow of the Royal College of Physicians of Edinburgh in the United Kingdom, Fellow of the Royal College of Pathology in the United Kingdom, Fellow of the Australasian College of Tropical Medicine, and Fellow of the Royal Society of Public Health in the United Kingdom.
Md Taufiqul Islam, Deputy Project Coordinator, Mucosal Immunology and Vaccinology Unit, Infectious Diseases Division, icddr,b, completed his M.B.B.S. from Mymensingh Medical College Hospital, University of Dhaka, in 2008. Subsequently, he obtained his M.P.H. degree in 2016. He was enrolled in the Ph.D. program at Griffith University, Australia, in 2020. His area of work includes the introduction of an oral cholera vaccine in Bangladesh and a single-dose oral cholera vaccine study in urban Dhaka. Dr. Islam has contributed to preparing the Cholera Vaccine Investment Strategy, which is an essential instrument to enhance the cholera control program in Bangladesh. Evaluation of the safety of using whole-cell oral cholera vaccines in pregnant women is another important scientific involvement. Moreover, his area of work has been expanded by being involved in the vaccine delivery and surveillance network in the complex humanitarian crisis in Bangladesh. He is also actively working in the program for the control of endemic cholera in Bangladesh to achieve the target of End Cholera 2030, the global roadmap.
Nigel A. J. McMillan, Griffith Centre for Cell and Gene Medicine, Menzies Health Institute, Queensland, Griffith University, obtained his Ph.D. at the University of Otago, New Zealand, in 1991. He is currently Professor of Infectious Diseases at Griffith University and explores the links between viruses and cancer. This includes developing novel nanoparticle delivery systems for RNAi and CRISPR. His laboratory was the first to report a cure for cancer using CRISPR-loaded nanoparticles in 2019, and in 2021, his laboratory published the first cure for SARS-CoV-2 infection in animal models using siRNA.
Firdausi Qadri, Ph.D., is a Senior Scientist, Infectious Diseases Division, at icddr,b, Dhaka, Bangladesh. She also heads the Mucosal Immunology and Vaccinology Unit. Her work includes basic and applied immunology of infectious diseases but also clinical and large field-based studies on enteric vaccines and has coauthored 406 peer-reviewed publications, including reviews and book chapters. Special interests are infections caused by V. cholerae, Shigella, ETEC, Salmonella spp., and Helicobacter pylori under grants from Sida, the NIH, BMGF, USAID, the University of Oxford, and the European Union. The results have produced an impact on the field of enteric diseases, specifically in the areas of immunological, genetic, and genomic mechanisms, proteomics, metagenomics, diagnostics, and vaccine development. She has been elected fellow of many societies, including ASM, AAM, TWAS, IDSA, BAS, and INSA, and serves on advisory boards, including the ISDB science, biotechnology, and innovation board.
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11091 | https://courses.lumenlearning.com/chemistryformajors/chapter/stoichiometry-of-gaseous-substances-mixtures-and-reactions/ | Stoichiometry of Gaseous Substances, Mixtures, and Reactions
Learning Outcomes
Use the ideal gas law to compute gas densities and molar masses
Perform stoichiometric calculations involving gaseous substances
State Dalton’s law of partial pressures and use it in calculations involving gaseous mixtures
The study of the chemical behavior of gases was part of the basis of perhaps the most fundamental chemical revolution in history. French nobleman Antoine Lavoisier, widely regarded as the “father of modern chemistry,” changed chemistry from a qualitative to a quantitative science through his work with gases. He discovered the law of conservation of matter, discovered the role of oxygen in combustion reactions, determined the composition of air, explained respiration in terms of chemical reactions, and more. He was a casualty of the French Revolution, guillotined in 1794. Of his death, mathematician and astronomer Joseph-Louis Lagrange said, “It took the mob only a moment to remove his head; a century will not suffice to reproduce it.”
As described in an earlier chapter of this text, we can turn to chemical stoichiometry for answers to many of the questions that ask “How much?” We can answer the question with masses of substances or volumes of solutions. However, we can also answer this question another way: with volumes of gases. We can use the ideal gas equation to relate the pressure, volume, temperature, and number of moles of a gas. Here we will combine the ideal gas equation with other equations to find gas density and molar mass. We will deal with mixtures of different gases, and calculate amounts of substances in reactions involving gases. This section will not introduce any new material or ideas, but will provide examples of applications and ways to integrate concepts we have already discussed.
Gas Density and Molar Mass
The ideal gas law described previously in this chapter relates the properties of pressure P, volume V, temperature T, and molar amount n. This law is universal, relating these properties in identical fashion regardless of the chemical identity of the gas:
PV=nRTPV=nRT
The density d of a gas, on the other hand, is determined by its identity. As described in another chapter of this text, the density of a substance is a characteristic property that may be used to identify the substance.
d=mvd=mv
Rearranging the ideal gas equation to isolate V and substituting into the density equation yields
d=mPnRT=(mn)PRTd=mPnRT=(mn)PRT
The ratio m/n is the definition of molar mass, ℳ:
ℳ=mnM=mn
The density equation can then be written
d=MPRTd=MPRT
This relation may be used for calculating the densities of gases of known identities at specified values of pressure and temperature as demonstrated in Example 1.
Example 1: Measuring Gas Density
What is the density of molecular nitrogen gas at STP?
Solution
Show Solution
The molar mass of molecular nitrogen, N2, is 28.01 g/mol. Substituting this value along with standard temperature and pressure into the gas density equation yields
d=ℳPRT=(28.01g/mol)(1.00atm)(0.721L∗atm∗mol−1K−1)(273K)=1.25g/Ld=MPRT=(28.01g/mol)(1.00atm)(0.721L∗atm∗mol−1K−1)(273K)=1.25g/L
Check Your Learning
What is the density of molecular hydrogen gas at 17.0 °C and a pressure of 760 torr?
Show Solution
d = 0.0847 g/L
When the identity of a gas is unknown, measurements of the mass, pressure, volume, and temperature of a sample can be used to calculate the molar mass of the gas (a useful property for identification purposes). Combining the ideal gas equation
PV=nRTPV=nRT
and the definition of molarity
ℳ=mnM=mn
yields the following equation:
ℳ=mRTPVM=mRTPV
Determining the molar mass of a gas via this approach is demonstrated in Example 2.
Example 2: Empirical/Molecular Formula Problems Using the Ideal Gas Law and Density of a Gas
Cyclopropane, a gas once used with oxygen as a general anesthetic, is composed of 85.7% carbon and 14.3% hydrogen by mass. Find the empirical formula. If 1.56 g of cyclopropane occupies a volume of 1.00 L at 0.984 atm and 50 °C, what is the molecular formula for cyclopropane?
Show Solution
Strategy: First solve the empirical formula problem using methods discussed earlier. Assume 100 g and convert the percentage of each element into grams. Determine the number of moles of carbon and hydrogen in the 100-g sample of cyclopropane. Divide by the smallest number of moles to relate the number of moles of carbon to the number of moles of hydrogen. In the last step, realize that the smallest whole number ratio is the empirical formula:
85.7 g C×1 mol C12.01 g C=7.136 mol C7.1367.136=1.00 mol C85.7 g C×1 mol C12.01 g C=7.136 mol C7.1367.136=1.00 mol C
14.3 g H×1 mol H1.01 g H=14.158 mol H14.1587.136=1.98 mol H14.3 g H×1 mol H1.01 g H=14.158 mol H14.1587.136=1.98 mol H
Empirical formula is CH2 [empirical mass (EM)(EM) of 14.03 g/empirical unit].
Next, use the provided values for mass, pressure, temperature and volume to compute the molar mass of the gas:
ℳ=mRTPV=(1.56 g)(0.0821L⋅atm⋅mol−1K−1)(323K)(0.984atm(1.00L)=42.0 g/molM=mRTPV=(1.56 g)(0.0821L⋅atm⋅mol−1K−1)(323K)(0.984atm(1.00L)=42.0 g/mol
Comparing the molar mass to the empirical formula mass shows how many empirical formula units make up a molecule:
ℳEM=42.0g/mol14.0g/mol=3MEM=42.0g/mol14.0g/mol=3
The molecular formula is thus derived from the empirical formula by multiplying each of its subscripts by three:
(CH2)3=C3H6(CH2)3=C3H6
Check Your Learning
Acetylene, a fuel used welding torches, is comprised of 92.3% C and 7.7% H by mass. Find the empirical formula. If g of acetylene occupies of volume of 1.00 L at 1.15 atm and 59.5 °C, what is the molecular formula for acetylene?
Show Solution
Empirical formula, CH; Molecular formula, C2H2
Example 3: Determining the Molar Mass of a Volatile Liquid
The approximate molar mass of a volatile liquid can be determined by:
Heating a sample of the liquid in a flask with a tiny hole at the top, which converts the liquid into gas that may escape through the hole
Removing the flask from heat at the instant when the last bit of liquid becomes gas, at which time the flask will be filled with only gaseous sample at ambient pressure
Sealing the flask and permitting the gaseous sample to condense to liquid, and then weighing the flask to determine the sample’s mass (see Figure 1)
Using this procedure, a sample of chloroform gas weighing 0.494 g is collected in a flask with a volume of 129 cm3 at 99.6 °C when the atmospheric pressure is 742.1 mm Hg. What is the approximate molar mass of chloroform?
Show Solution
Since ℳ=mnM=mn and n=PVRT,n=PVRT, substituting and rearranging gives ℳ=mRTPV,M=mRTPV, then
ℳ=mRTPV=(0.494 g)×0.08206 L⋅ atm/mol K×372.8 K0.976 atm×0.129 L=120 g/molM=mRTPV=(0.494 g)×0.08206 L⋅ atm/mol K×372.8 K0.976 atm×0.129 L=120 g/mol
Check Your Learning
A sample of phosphorus that weighs 3.243 × 10-2 g exerts a pressure of 31.89 kPa in a 56.0-mL bulb at 550 °C. What are the molar mass and molecular formula of phosphorus vapor?
Show Solution
124 g/mol P4
The Pressure of a Mixture of Gases: Dalton’s Law
Unless they chemically react with each other, the individual gases in a mixture of gases do not affect each other’s pressure. Each individual gas in a mixture exerts the same pressure that it would exert if it present alone in the container (Figure 2). The pressure exerted by each individual gas in a mixture is called its partial pressure. This observation is summarized by Dalton’s law of partial pressures: The total pressure of a mixture of ideal gases is equal to the sum of the partial pressures of the component gases:
PTotal=PA+PB+PC+…=ΣiPiPTotal=PA+PB+PC+…=ΣiPi
In the equation PTotal is the total pressure of a mixture of gases, PA is the partial pressure of gas A; PB is the partial pressure of gas B; PC is the partial pressure of gas C; and so on.
The partial pressure of gas A is related to the total pressure of the gas mixture via its mole fraction, a unit of concentration defined as the number of moles of a component of a solution divided by the total number of moles of all components):
PA=XA×PTotal where XA=nAnTotalPA=XA×PTotal where XA=nAnTotal
where PA, XA, and nA are the partial pressure, mole fraction, and number of moles of gas A, respectively, and nTotal is the number of moles of all components in the mixture.
Example 4: The Pressure of a Mixture of Gases
A 10.0-L vessel contains 2.50 × 10-3 mol of H2, 1.00 × 10-3 mol of He, and 3.00 × 10-4 mol of Ne at 35 °C.
What are the partial pressures of each of the gases?
What is the total pressure in atmospheres?
Show Solution
The gases behave independently, so the partial pressure of each gas can be determined from the ideal gas equation, using P=nRTV:P=nRTV:
PH2=(2.50×10−3mol)(0.08206Latmmol−1K−1)(308 K)10.0L=6.32×10−3atmPH2=(2.50×10−3mol)(0.08206Latmmol−1K−1)(308 K)10.0L=6.32×10−3atm
PHe=(1.00×10−3mol)(0.08206 L atmmol−1K−1)(308 K)10.0L=2.53×10−3atmPHe=(1.00×10−3mol)(0.08206 L atmmol−1K−1)(308 K)10.0L=2.53×10−3atm
PNe=(3.00×10−4mol)(0.08206Latmmol−1K−1)(308 K)10.0L=7.58×10−4atmPNe=(3.00×10−4mol)(0.08206Latmmol−1K−1)(308 K)10.0L=7.58×10−4atm
The total pressure is given by the sum of the partial pressures:
PT=PH2+PHe+PNe=(0.00632+0.00253+0.00076)atm=9.61×10−3atmPT=PH2+PHe+PNe=(0.00632+0.00253+0.00076)atm=9.61×10−3atm
Check Your Learning
A 5.73-L flask at 25 °C contains 0.0388 mol of N2, 0.147 mol of CO, and 0.0803 mol of H2. What is the total pressure in the flask in atmospheres?
Show Solution
1.137 atm
Here is another example of this concept, but dealing with mole fraction calculations.
Example 5: The Pressure of a Mixture of Gases
A gas mixture used for anesthesia contains 2.83 mol oxygen, O2, and 8.41 mol nitrous oxide, N2O. The total pressure of the mixture is 192 kPa.
What are the mole fractions of O2 and N2O?
What are the partial pressures of O2 and N2O?
Show Solution
The mole fraction is given by XA=nAnTotalXA=nAnTotal and the partial pressure is PA = XA × PTotal.
For O2,
XO2=nO2nTotal=2.83 mol(2.83+8.41)mol=0.252XO2=nO2nTotal=2.83 mol(2.83+8.41)mol=0.252
and
PO2=XO2×PTotal=0.252×192 kPa=48.4 kPaPO2=XO2×PTotal=0.252×192 kPa=48.4 kPa
For N2O,
XN2=nN2nTotal=8.41 mol(2.83+8.41)mol=0.748XN2=nN2nTotal=8.41 mol(2.83+8.41)mol=0.748
and
PN2=XN2×PTotal=0.748×192 kPa=143.6kPaPN2=XN2×PTotal=0.748×192 kPa=143.6kPa
Check Your Learning
What is the pressure of a mixture of 0.200 g of H2, 1.00 g of N2, and 0.820 g of Ar in a container with a volume of 2.00 L at 20 °C?
Show Solution
1.87 atm
Collection of Gases over Water
A simple way to collect gases that do not react with water is to capture them in a bottle that has been filled with water and inverted into a dish filled with water. The pressure of the gas inside the bottle can be made equal to the air pressure outside by raising or lowering the bottle. When the water level is the same both inside and outside the bottle (Figure 3), the pressure of the gas is equal to the atmospheric pressure, which can be measured with a barometer.
However, there is another factor we must consider when we measure the pressure of the gas by this method. Water evaporates and there is always gaseous water (water vapor) above a sample of liquid water. As a gas is collected over water, it becomes saturated with water vapor and the total pressure of the mixture equals the partial pressure of the gas plus the partial pressure of the water vapor. The pressure of the pure gas is therefore equal to the total pressure minus the pressure of the water vapor—this is referred to as the “dry” gas pressure, that is, the pressure of the gas only, without water vapor. The vapor pressure of water, which is the pressure exerted by water vapor in equilibrium with liquid water in a closed container, depends on the temperature (Figure 4); more detailed information on the temperature dependence of water vapor can be found in Table 1, and vapor pressure will be discussed in more detail in the next chapter on liquids.
Figure 4. This graph shows the vapor pressure of water at sea level as a function of temperature.
Table 1. Vapor Pressure of Ice and Water in Various Temperatures at Sea Level
| Temperature (°C) | Pressure (torr) | Temperature (°C) | Pressure (torr) | Temperature (°C) | Pressure (torr) |
| –10 | 1.95 | 18 | 15.5 | 30 | 31.8 |
| –5 | 3.0 | 19 | 16.5 | 35 | 42.2 |
| –2 | 3.9 | 20 | 17.5 | 40 | 55.3 |
| 0 | 4.6 | 21 | 18.7 | 50 | 92.5 |
| 2 | 5.3 | 22 | 19.8 | 60 | 149.4 |
| 4 | 6.1 | 23 | 21.1 | 70 | 233.7 |
| 6 | 7.0 | 24 | 22.4 | 80 | 355.1 |
| 8 | 8.0 | 25 | 23.8 | 90 | 525.8 |
| 10 | 9.2 | 26 | 25.2 | 95 | 633.9 |
| 12 | 10.5 | 27 | 26.7 | 99 | 733.2 |
| 14 | 12.0 | 28 | 28.3 | 100.0 | 760.0 |
| 16 | 13.6 | 29 | 30.0 | 101.0 | 787.6 |
Example 6: Pressure of a Gas Collected Over Water
If 0.200 L of argon is collected over water at a temperature of 26 °C and a pressure of 750 torr in a system like that shown in Figure 3, what is the partial pressure of argon?
Show Solution
According to Dalton’s law, the total pressure in the bottle (750 torr) is the sum of the partial pressure of argon and the partial pressure of gaseous water:
PT=PAr+PH2OPT=PAr+PH2O
Rearranging this equation to solve for the pressure of argon gives:
PAr=PT−PH2OPAr=PT−PH2O
The pressure of water vapor above a sample of liquid water at 26 °C is 25.2 torr (Water Properties), so:
PAr=750 torr−25.2 torr=725 torrPAr=750 torr−25.2 torr=725 torr
Check Your Learning
A sample of oxygen collected over water at a temperature of 29.0 °C and a pressure of 764 torr has a volume of 0.560 L. What volume would the dry oxygen have under the same conditions of temperature and pressure?
Show Solution
0.537 L
Chemical Stoichiometry and Gases
Chemical stoichiometry describes the quantitative relationships between reactants and products in chemical reactions.
We have previously measured quantities of reactants and products using masses for solids and volumes in conjunction with the molarity for solutions; now we can also use gas volumes to indicate quantities. If we know the volume, pressure, and temperature of a gas, we can use the ideal gas equation to calculate how many moles of the gas are present. If we know how many moles of a gas are involved, we can calculate the volume of a gas at any temperature and pressure.
Avogadro’s Law Revisited
Sometimes we can take advantage of a simplifying feature of the stoichiometry of gases that solids and solutions do not exhibit: All gases that show ideal behavior contain the same number of molecules in the same volume (at the same temperature and pressure). Thus, the ratios of volumes of gases involved in a chemical reaction are given by the coefficients in the equation for the reaction, provided that the gas volumes are measured at the same temperature and pressure.
We can extend Avogadro’s law (that the volume of a gas is directly proportional to the number of moles of the gas) to chemical reactions with gases: Gases combine, or react, in definite and simple proportions by volume, provided that all gas volumes are measured at the same temperature and pressure. For example, since nitrogen and hydrogen gases react to produce ammonia gas according to N2(g)+3H2(g)→2NH3(g)N2(g)+3H2(g)→2NH3(g), a given volume of nitrogen gas reacts with three times that volume of hydrogen gas to produce two times that volume of ammonia gas, if pressure and temperature remain constant.
The explanation for this is illustrated in Figure 5. According to Avogadro’s law, equal volumes of gaseous N2, H2, and NH3, at the same temperature and pressure, contain the same number of molecules. Because one molecule of N2 reacts with three molecules of H2 to produce two molecules of NH3, the volume of H2 required is three times the volume of N2, and the volume of NH3 produced is two times the volume of N2.
Figure 5. One volume of N2 combines with three volumes of H2 to form two volumes of NH3.
Example 7: Reaction of Gases
Propane, C3H8(g), is used in gas grills to provide the heat for cooking. What volume of O2(g) measured at 25 °C and 760 torr is required to react with 2.7 L of propane measured under the same conditions of temperature and pressure? Assume that the propane undergoes complete combustion.
Show Solution
The ratio of the volumes of C3H8 and O2 will be equal to the ratio of their coefficients in the balanced equation for the reaction:
C3H8(g)+5O2(g)⟶3CO2(g)+4H2O(l)1 volume+5 volumes3 volumes+4 volumesC3H8(g)+5O2(g)⟶3CO2(g)+4H2O(l)1 volume+5 volumes3 volumes+4 volumes
From the equation, we see that one volume of C3H8 will react with five volumes of O2:
2.7LC3H8×5 LO21LC3H8=13.5 LO22.7LC3H8×5 LO21LC3H8=13.5 LO2
A volume of 13.5 L of O2 will be required to react with 2.7 L of C3H8.
Check Your Learning
An acetylene tank for an oxyacetylene welding torch provides 9340 L of acetylene gas, C2H2, at 0°C and 1 atm. How many tanks of oxygen, each providing 7.00 × 103 L of O2 at 0 °C and 1 atm, will be required to burn the acetylene?
2C2H2+5O2→4CO2+2H2O2C2H2+5O2→4CO2+2H2O
Show Solution
3.34 tanks (2.34 × 104 L)
Example 8: Volumes of Reacting Gases
Ammonia is an important fertilizer and industrial chemical. Suppose that a volume of 683 billion cubic feet of gaseous ammonia, measured at 25 °C and 1 atm, was manufactured. What volume of H2(g), measured under the same conditions, was required to prepare this amount of ammonia by reaction with N2?
N2(g)+3H2(g)→2NH3(g)N2(g)+3H2(g)→2NH3(g)
Show Solution
Because equal volumes of H2 and NH3 contain equal numbers of molecules and each three molecules of H2 that react produce two molecules of NH3, the ratio of the volumes of H2 and NH3 will be equal to 3:2. Two volumes of NH3, in this case in units of billion ft3, will be formed from three volumes of H2:
683 billion ft3NH3×3 billion ft3H22 billion ft3NH3=1.02×103billion ft3H2683 billion ft3NH3×3 billion ft3H22 billion ft3NH3=1.02×103billion ft3H2
The manufacture of 683 billion ft3 of NH3 required 1020 billion ft3 of H2. (At 25 °C and 1 atm, this is the volume of a cube with an edge length of approximately 1.9 miles.)
Check Your Learning
What volume of O2(g) measured at 25 °C and 760 torr is required to react with 17.0 L of ethylene, C2H4(g), measured under the same conditions of temperature and pressure? The products are CO2 and water vapor.
Show Solution
51.0 L
Example 9: Volume of Gaseous Product
What volume of hydrogen at 27 °C and 723 torr may be prepared by the reaction of 8.88 g of gallium with an excess of hydrochloric acid?
2Ga(s)+6HCl(aq)→2GaCl3(aq)+3H2(g)2Ga(s)+6HCl(aq)→2GaCl3(aq)+3H2(g)
Show Solution
To convert from the mass of gallium to the volume of H2(g), we need to do something like this:
The first two conversions are:
8.88g Ga×1mol Ga69.723g Ga×3 molH22mol Ga=0.191mol H28.88g Ga×1mol Ga69.723g Ga×3 molH22mol Ga=0.191mol H2
Finally, we can use the ideal gas law:
VH2=(nRTP)H2=0.191mol×0.08206 Latmmol−1K−1×300 K0.951atm=4.94 LVH2=(nRTP)H2=0.191mol×0.08206 Latmmol−1K−1×300 K0.951atm=4.94 L
Check Your Learning
Sulfur dioxide is an intermediate in the preparation of sulfuric acid. What volume of SO2 at 343 °C and 1.21 atm is produced by burning l.00 kg of sulfur in oxygen?
Show Solution
1.30 × 103 L
Greenhouse Gases and Climate Change
The thin skin of our atmosphere keeps the earth from being an ice planet and makes it habitable. In fact, this is due to less than 0.5% of the air molecules. Of the energy from the sun that reaches the earth, almost 1313 is reflected back into space, with the rest absorbed by the atmosphere and the surface of the earth. Some of the energy that the earth absorbs is re-emitted as infrared (IR) radiation, a portion of which passes back out through the atmosphere into space. However, most of this IR radiation is absorbed by certain substances in the atmosphere, known as greenhouse gases, which re-emit this energy in all directions, trapping some of the heat. This maintains favorable living conditions—without atmosphere, the average global average temperature of 14 °C (57 °F) would be about –19 °C (–2 °F). The major greenhouse gases (GHGs) are water vapor, carbon dioxide, methane, and ozone. Since the Industrial Revolution, human activity has been increasing the concentrations of GHGs, which have changed the energy balance and are significantly altering the earth’s climate (Figure 6).
There is strong evidence from multiple sources that higher atmospheric levels of CO2 are caused by human activity, with fossil fuel burning accounting for about 3434 of the recent increase in CO2. Reliable data from ice cores reveals that CO2 concentration in the atmosphere is at the highest level in the past 800,000 years; other evidence indicates that it may be at its highest level in 20 million years. In recent years, the CO2 concentration has increased from historical levels of below 300 ppm to almost 400 ppm today (Figure 7).
Figure 8. Susan Solomon’s research focuses on climate change and has been instrumental in determining the cause of the ozone hole over Antarctica. (credit: National Oceanic and Atmospheric Administration)
Portrait of a Chemist: Susan Solomon
Atmospheric and climate scientist Susan Solomon (Figure 8) is the author of one of The New York Times books of the year (The Coldest March, 2001), one of Time magazine’s 100 most influential people in the world (2008), and a working group leader of the Intergovernmental Panel on Climate Change (IPCC), which was the recipient of the 2007 Nobel Peace Prize. She helped determine and explain the cause of the formation of the ozone hole over Antarctica, and has authored many important papers on climate change. She has been awarded the top scientific honors in the US and France (the National Medal of Science and the Grande Medaille, respectively), and is a member of the National Academy of Sciences, the Royal Society, the French Academy of Sciences, and the European Academy of Sciences. Formerly a professor at the University of Colorado, she is now at MIT, and continues to work at NOAA.
Key Concepts and Summary
The ideal gas law can be used to derive a number of convenient equations relating directly measured quantities to properties of interest for gaseous substances and mixtures. Appropriate rearrangement of the ideal gas equation may be made to permit the calculation of gas densities and molar masses. Dalton’s law of partial pressures may be used to relate measured gas pressures for gaseous mixtures to their compositions. Avogadro’s law may be used in stoichiometric computations for chemical reactions involving gaseous reactants or products.
Key Equations
PTotal=PA+PB+PC+…=ΣiPiPTotal=PA+PB+PC+…=ΣiPi
PA=XA⋅PTotalPA=XA⋅PTotal
XA=nAnTotalXA=nAnTotal
Try It
What is the density of laughing gas, dinitrogen monoxide, N2O, at a temperature of 325 K and a pressure of 113.0 kPa?
Calculate the density of Freon 12, CF2Cl2, at 30.0°C and 0.954 atm.
Which is denser at the same temperature and pressure, dry air or air saturated with water vapor? Explain.
A cylinder of O2(g) used in breathing by emphysema patients has a volume of 3.00 L at a pressure of 10.0 atm. If the temperature of the cylinder is 28.0 °C, what mass of oxygen is in the cylinder?
What is the molar mass of a gas if 0.0494 g of the gas occupies a volume of 0.100 L at a temperature 26 °C and a pressure of 307 torr?
What is the molar mass of a gas if 0.281 g of the gas occupies a volume of 125 mL at a temperature 126 °C and a pressure of 777 torr?
How could you show experimentally that the molecular formula of propene is C3H6, not CH2?
The density of a certain gaseous fluoride of phosphorus is 3.93 g/L at STP. Calculate the molar mass of this fluoride and determine its molecular formula.
Consider this question: What is the molecular formula of a compound that contains 39% C, 45% N, and 16% H if 0.157 g of the compound occupies l25 mL with a pressure of 99.5 kPa at 22 °C?
Outline the steps necessary to answer the question.
Answer the question.
A 36.0–L cylinder of a gas used for calibration of blood gas analyzers in medical laboratories contains 350 g CO2, 805 g O2, and 4,880 g N2. What is the pressure in the flask in atmospheres, in torr, and in kilopascals?
A cylinder of a gas mixture used for calibration of blood gas analyzers in medical laboratories contains 5.0% CO2, 12.0% O2, and the remainder N2 at a total pressure of 146 atm. What is the partial pressure of each component of this gas? (The percentages given indicate the percent of the total pressure that is due to each component.)
A sample of gas isolated from unrefined petroleum contains 90.0% CH4, 8.9% C2H6, and 1.1% C3H8 at a total pressure of 307.2 kPa. What is the partial pressure of each component of this gas? (The percentages given indicate the percent of the total pressure that is due to each component.)
A mixture of 0.200 g of H2, 1.00 g of N2, and 0.820 g of Ar is stored in a closed container at STP. Find the volume of the container, assuming that the gases exhibit ideal behavior.
Most mixtures of hydrogen gas with oxygen gas are explosive. However, a mixture that contains less than 3.0 % O2 is not. If enough O2 is added to a cylinder of H2 at 33.2 atm to bring the total pressure to 34.5 atm, is the mixture explosive?
A commercial mercury vapor analyzer can detect, in air, concentrations of gaseous Hg atoms (which are poisonous) as low as 2 × 10-6 mg/L of air. At this concentration, what is the partial pressure of gaseous mercury if the atmospheric pressure is 733 torr at 26 °C?
A sample of carbon monoxide was collected over water at a total pressure of 756 torr and a temperature of 18 °C. What is the pressure of the carbon monoxide? (See Table 9.2 for the vapor pressure of water.)
In an experiment in a general chemistry laboratory, a student collected a sample of a gas over water. The volume of the gas was 265 mL at a pressure of 753 torr and a temperature of 27 °C. The mass of the gas was 0.472 g. What was the molar mass of the gas?
Joseph Priestley first prepared pure oxygen by heating mercuric oxide, HgO: 2HgO(s)→2Hg(l)+O2(g)2HgO(s)→2Hg(l)+O2(g)
Outline the steps necessary to answer the following question: What volume of O2 at 23° C and 0.975 atm is produced by the decomposition of 5.36 g of HgO?
Answer the question.
Cavendish prepared hydrogen in 1766 by the novel method of passing steam through a red-hot gun barrel:4H2O(g)+3Fe(s)→Fe3O4(s)+4H2(g)4H2O(g)+3Fe(s)→Fe3O4(s)+4H2(g)
Outline the steps necessary to answer the following question: What volume of H2 at a pressure of 745 torr and a temperature of 20 °C can be prepared from the reaction of 15.O g of H2O?
Answer the question.
The chlorofluorocarbon CCl2F2 can be recycled into a different compound by reaction with hydrogen to produce CH2F2(g), a compound useful in chemical manufacturing: CCl2F2(g)+4H2(g)→CH2F2(g)+2HCl(g).CCl2F2(g)+4H2(g)→CH2F2(g)+2HCl(g).
Outline the steps necessary to answer the following question: What volume of hydrogen at 225 atm and 35.5 °C would be required to react with 1 ton (1.000 × 103 kg) of CCl2F2?
Answer the question.
Automobile air bags are inflated with nitrogen gas, which is formed by the decomposition of solid sodium azide (NaN3). The other product is sodium metal. Calculate the volume of nitrogen gas at 27 °C and 756 torr formed by the decomposition of 125 g of sodium azide.
Lime, CaO, is produced by heating calcium carbonate, CaCO3; carbon dioxide is the other product.
Outline the steps necessary to answer the following question: What volume of carbon dioxide at 875° and 0.966 atm is produced by the decomposition of 1 ton (1.000 × 103 kg) of calcium carbonate?
Answer the question.
Before small batteries were available, carbide lamps were used for bicycle lights. Acetylene gas, C2H2, and solid calcium hydroxide were formed by the reaction of calcium carbide, CaC2, with water. The ignition of the acetylene gas provided the light. Currently, the same lamps are used by some cavers, and calcium carbide is used to produce acetylene for carbide cannons.
Outline the steps necessary to answer the following question: What volume of C2H2 at 1.005 atm and 12.2 °C is formed by the reaction of 15.48 g of CaC2 with water?
Answer the question.
Calculate the volume of oxygen required to burn 12.00 L of ethane gas, C2H6, to produce carbon dioxide and water, if the volumes of C2H6 and O2 are measured under the same conditions of temperature and pressure.
What volume of O2 at STP is required to oxidize 8.0 L of NO at STP to NO2? What volume of NO2 is produced at STP?
Consider the following questions:
What is the total volume of the CO2(g) and H2O(g) at 600 °C and 0.888 atm produced by the combustion of 1.00 L of C2H6(g) measured at STP?
What is the partial pressure of H2O in the product gases?
Methanol, CH3OH, is produced industrially by the following reaction:
CO(g)+2H2(g)copper catalyst 300∘ C,300 atm→CH3OH(g)CO(g)+2H2(g)copper catalyst 300∘ C,300 atm−−−−−−−−−−−−−−−−−−→CH3OH(g)
Assuming that the gases behave as ideal gases, find the ratio of the total volume of the reactants to the final volume.
What volume of oxygen at 423.0 K and a pressure of 127.4 kPa is produced by the decomposition of 129.7 g of BaO2 to BaO and O2?
A 2.50-L sample of a colorless gas at STP decomposed to give 2.50 L of N2 and 1.25 L of O2 at STP. What is the colorless gas?
Ethanol, C2H5OH, is produced industrially from ethylene, C2H4, by the following sequence of reactions:
3C2H4+2H2SO4→C2H5HSO4+(C2H5)2SO43C2H4+2H2SO4→C2H5HSO4+(C2H5)2SO4 C2H5HSO4+(C2H5)2SO4+3H2O→3C2H5OH+2H2SO4C2H5HSO4+(C2H5)2SO4+3H2O→3C2H5OH+2H2SO4
What volume of ethylene at STP is required to produce 1.000 metric ton (1000 kg) of ethanol if the overall yield of ethanol is 90.1%?
One molecule of hemoglobin will combine with four molecules of oxygen. If 1.0 g of hemoglobin combines with 1.53 mL of oxygen at body temperature (37 °C) and a pressure of 743 torr, what is the molar mass of hemoglobin?
A sample of a compound of xenon and fluorine was confined in a bulb with a pressure of 18 torr. Hydrogen was added to the bulb until the pressure was 72 torr. Passage of an electric spark through the mixture produced Xe and HF. After the HF was removed by reaction with solid KOH, the final pressure of xenon and unreacted hydrogen in the bulb was 36 torr. What is the empirical formula of the xenon fluoride in the original sample? (Note: Xenon fluorides contain only one xenon atom per molecule.)
One method of analyzing amino acids is the van Slyke method. The characteristic amino groups (-NH2) in protein material are allowed to react with nitrous acid, HNO2, to form N2 gas. From the volume of the gas, the amount of amino acid can be determined. A 0.0604-g sample of a biological sample containing glycine, CH2(NH2)COOH, was analyzed by the van Slyke method and yielded 3.70 mL of N2 collected over water at a pressure of 735 torr and 29 °C. What was the percentage of glycine in the sample?
CH2(NH2)CO2H+HNO2→CH2(OH)CO2H+H2O+N2CH2(NH2)CO2H+HNO2→CH2(OH)CO2H+H2O+N2
Selected Answers
2. ρ=PMRT=0.954atm[12.011+2(18.9954)+2(35.453)]gmol−10.08206 Latmmol−1K−1×303.15K=4.64 gL−1ρ=PMRT=0.954atm[12.011+2(18.9954)+2(35.453)]gmol−10.08206 Latmmol−1K−1×303.15K=4.64 gL−1
4. massO2=(31.9988 gmol−1)(10.0atm)(3.00L)(0.08206Latmmol−1K−1)(301.15K)=38.8 gmassO2=(31.9988 gmol−1)(10.0atm)(3.00L)(0.08206Latmmol−1K−1)(301.15K)=38.8 g
From the ideal gas law, PV = nRT, set n=massmolar massn=massmolar mass and solve the molar mass. molar mass=(0.281 g)(0.08206Latmmol−1K−1)(399.15K)(777torr760torratm−1)(0.125L)=72.0 gmol−1molar mass=(0.281 g)(0.08206Latmmol−1K−1)(399.15K)(777torr760torratm−1)(0.125L)=72.0 gmol−1
M=mRTPVD=mVM=DRTPM=mRTPVD=mVM=DRTP M=3.93 gL−1(0.08206 Latmmol−1K−1)(273.15K)1.00atm=88.1 gmol−1M=3.93 gL−1(0.08206 Latmmol−1K−1)(273.15K)1.00atm=88.1 gmol−1 ℳphosphorous = 30.97376 g/mol
ℳfluorine = 18.998403 g/mol
molecular formula: phosphorous: 30.97376flourine:3(18.998403)_87.968969molecular formula: phosphorous: 30.97376flourine:3(18.998403)–––––––––––––––87.968969
The molecular formula is PF3.
To find this answer you can either use trial and error, or you can realize that since phosphorus is in group 5, it can fill its valence shell by forming three bonds. Fluorine, being in group 7, needs to form only one bond to fill its shell. Thus it makes sense to start with PF3 as a probable formula.
Calculate the moles of each gas present and from that, calculate the pressure from the ideal gas law. Assume 25°C. The calibration gas contains:
350gCO244.0098gmol−1CO2=7.953 mol CO2350gCO244.0098gmol−1CO2=7.953 mol CO2
805gO231.9988gmol−1O2=25.157 mol O2805gO231.9988gmol−1O2=25.157 mol O2
4880gN228.01348gmol−1N2=174.202 mol N24880gN228.01348gmol−1N2=174.202 mol N2
Total moles = 7.953 + 25.157 + 174.202 = 207.312 mol
P=nRTV=207.312mol×0.08206Latmmol−1K−1×298.15K36.0L=141 atmP=nRTV=207.312mol×0.08206Latmmol−1K−1×298.15K36.0L=141 atm
Since these are percentages of the total pressure, the partial pressure can be calculated as follows:
CH4: 90% of 307.2 kPa = 0.900 × 307.2 = 276 kPa
C2 H6: 8.9% of 307.2 kPa = 0.089 × 307.2 = 27 kPa
C3 H8: 1.1% of 307.2 kPa = 0.011 × 307.2 = 3.4 kPa
14. The oxygen increases the pressure within the tank to (34.5 atm – 33.2 atm =) 1.3 atm. The percentage O2 on a mole basis is 1.334.5×100%=3.77%.1.334.5×100%=3.77%. The mixture is explosive. However, the percentage is given as a weight percent. Converting to a mass basis increases the percentage of oxygen even more, so the mixture is still explosive.
The vapor pressure of water at 18 °C is 15.5 torr. Subtract the vapor pressure of water from the total pressure to find the pressure of the carbon monoxide:
PT = Pgas + Pwater
Rearrangement gives: PT – Pwater =Pgas
756 torr – 15.5 torr = 740 torr
The answers are as follows:
Determine the moles of HgO that decompose; using the chemical equation, determine the moles of O2 produced by decomposition of this amount of HgO; and determine the volume of O2 from the moles of O2, temperature, and pressure.
5.36g HgO×1 mol HgO(200.59+15.9994)g HgO=0.0247 mol HgO0.0247mol HgO×1 molO22mol HgO=0.01235 molO25.36g HgO×1 mol HgO(200.59+15.9994)g HgO=0.0247 mol HgO0.0247mol HgO×1 molO22mol HgO=0.01235 molO2
PV = nRT
P = 0.975 atm
T = (23.0 + 273.15) K
V=nRTP=0.01235mol(0.08206 Latmmol−1K−1)(296.15K)0.975atm=0.308 LV=nRTP=0.01235mol(0.08206 Latmmol−1K−1)(296.15K)0.975atm=0.308 L
The answers are as follows:
Determine the molar mass of CCl2F2. From the balanced equation, calculate the moles of H2 needed for the complete reaction. From the ideal gas law, convert moles of H2 into volume.
Molar mass of CCl2 F2 = 12.011 + 2 × 18.9984 + 2 × 35.4527 = 120.913 g/mol
molH2=1.000×106g×1 molCCL2F2120.913 g×4 molH21 molCCl2F2=3.308×104molmolH2=1.000×106g×1 molCCL2F2120.913 g×4 molH21 molCCl2F2=3.308×104mol
V=nRTP=(3.308×104mol)(0.08206 Latmmol−1K−1)(308.65K)225atm=3.72×103LV=nRTP=(3.308×104mol)(0.08206 Latmmol−1K−1)(308.65K)225atm=3.72×103L
22. The answers are as follows:
Balance the equation. Determine the grams of CO2 produced and the number of moles. From the ideal gas law, determine the volume of gas.
CaCO3(s)→CaO(s)+CO2(g)CaCO3(s)→CaO(s)+CO2(g)
massCO2=1.00×106g×1 molCaCO2100.087 g×44.01 gCO21 molCO2×1 molCO21 molCaCO2=4.397×105gmassCO2=1.00×106g×1 molCaCO2100.087 g×44.01 gCO21 molCO2×1 molCO21 molCaCO2=4.397×105g
molCO2=4.397×105g44.01 gmol−1=9991 molmolCO2=4.397×105g44.01 gmol−1=9991 mol
V=nRTP=(9991 mol)(0.08206 L atmmol−1K−1)(875 K)0.966 atm=7.43×105LV=nRTP=(9991 mol)(0.08206 L atmmol−1K−1)(875 K)0.966 atm=7.43×105L
2C2H6(g)+7O2(g)→4CO2(g)+6H2O(g)2C2H6(g)+7O2(g)→4CO2(g)+6H2O(g)
From the balanced equation, we see that 2 mol of C2H6 requires 7 mol of O2 to burn completely. Gay-Lussac’s law states that gases react in simple proportions by volume. As the number of liters is proportional to the number of moles,
12.00 L2 molC2H6=V(O2)7 molO212.00 L2 molC2H6=V(O2)7 molO2
V(O2)=12.00 L×72=42.00 LV(O2)=12.00 L×72=42.00 L
26. The answers are as follows:
(a) The scheme to solve this problem is:
volume C2H6(g)ideal gasequation→mol C2H6(g)reactionstoichiometry→mol CO2+H2Oideal gasequation→volume CO2+H2Ovolume C2H6(g)ideal gasequation→mol C2H6(g)reactionstoichiometry→mol CO2+H2Oideal gasequation→volume CO2+H2O
C2H6(g)+312O2(g)→2CO2(g)+3H2O(g)1.n(C2H6)=PVRT=1.00atm×1.00L0.08206Latmmol−1K−1(273.15K)=0.0446 mol2.0.0446molC2H6×5 mol products1molC2H6=0.223 mol products3.V=nRT=(0.223mol)(0.08206 Latmmol−1K−1)(873.15K)0.888atm=18.0 LC2H6(g)+312O2(g)→2CO2(g)+3H2O(g)1.n(C2H6)=PVRT=1.00atm×1.00L0.08206Latmmol−1K−1(273.15K)=0.0446 mol2.0.0446molC2H6×5 mol products1molC2H6=0.223 mol products3.V=nRT=(0.223mol)(0.08206 Latmmol−1K−1)(873.15K)0.888atm=18.0 L
(b) First, calculate the mol H2O produced:
0.0446 molC2H6×3 mol products1 molC2H6=0.1338 mol0.0446 molC2H6×3 mol products1 molC2H6=0.1338 mol
Second, calculate the pressure of H2O:
P=nRTV=(0.1338mol)(0.8206Latmmol−1K−1)(873.15K)18.0L=0.533 atm
First, we must write a balanced equation to establish the stoichiometry of the reaction:
2BaO2→2BaO+O2
We are given the mass of BaO2 that decomposes, so the scheme for solving this problem will be:
Mass (BaO2) = 137.33 + 2(15.9994) = 169.33 g/mol
n(O2)=129.7 gBaO2×1 molBaO2169.33 gBaO2×1 molO22 molBaO2=0.3830 molO2
V(O2)=nRTP=0.3830 mol(8.314 L kPamol−1K−1)(423.0 K)127.4 kPa=10.57 LO2
At 90.1% conversion, a 1.000 × 106 g final yield would require a (1.000×1060.901)=1.1099×106g theoretical yield.
3C2H4 produces 3C2H5OH, giving a 1:1 ratio:
mol(C2H4)=1.1099×106gC2H5OH×1molC2H2OH46.069gC2H5OH×1 molC2H41molC2H2OH=2.409×104mol
V (C2 H4) = 22.4 L/mol × 2.409 × 104 mol = 5.40 × 105 L
The reaction is:
XeFx+x2H2→Xe+xHF
The pressure of H2 that reacts is 48 torr – 24 torr = 24 torr
The number of moles of gas is proportional to the partial pressures. The reaction used 24 torr of XeFx and 24 torr of H2 so: x2=1 and x = 2
The formula for the xenon compound is XeF2.
Immediately after the H2 is added (before the reaction):
PTotal=PXeF2+PH2PH2=PTotal−PXeF2=72 torr−24 torr=48torr
After the reaction:
PXe=24 torr (1 molXeFx→1 mol Xe)
And the partial pressure of unreacted H2 is:
PH2=PTotal−PXe=48 torr−24 torr=24 torr
Glossary
Dalton’s law of partial pressures: total pressure of a mixture of ideal gases is equal to the sum of the partial pressures of the component gases.
mole fraction: concentration unit defined as the ratio of the molar amount of a mixture component to the total number of moles of all mixture components
partial pressure: pressure exerted by an individual gas in a mixture
vapor pressure of water: pressure exerted by water vapor in equilibrium with liquid water in a closed container at a specific temperature
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11092 | https://www.ahajournals.org/doi/10.1161/cir.0000000000000476 | Meal Timing and Frequency: Implications for Cardiovascular Disease Prevention: A Scientific Statement From the American Heart Association | Circulation
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Reference #1
Review Article
Originally Published 30 January 2017
Free Access
Meal Timing and Frequency: Implications for Cardiovascular Disease Prevention: A Scientific Statement From the American Heart Association
Marie-Pierre St-Onge, PhD, FAHA, Chair, Jamy Ard, MD, Monica L.Baskin, PhD, Stephanie E.Chiuve, ScD, Heather M.Johnson, MD, FAHA, Penny Kris-Etherton, PhD, RD, FAHA, and Krista Varady, PhDOn behalf of the American Heart Association Obesity Committee of the Council on Lifestyle and Cardiometabolic Health; Council on Cardiovascular Disease in the Young; Council on Clinical Cardiology; and Stroke CouncilAuthor Info & Affiliations
Circulation
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Contents
Abstract
Additional Methods
Breakfast Skipping and Cardiometabolic Risk: Epidemiological Findings
Breakfast Skipping and Cardiometabolic Risk: Clinical Intervention Findings
Intermittent Fasting, Meal Frequency, and Cardiometabolic Risk: Epidemiological Findings
Intermittent Fasting, Meal Frequency, and Cardiometabolic Risk: Clinical Intervention Findings
Meal Timing and Cardiometabolic Risk: Observational Findings
Meal Timing and Cardiometabolic Risk: Clinical Intervention Findings
Research Gaps in Defining Meals and Eating Occasions
Clinical Implications
Recommendations for Special Populations
References
eLetters
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Abstract
Eating patterns are increasingly varied. Typical breakfast, lunch, and dinner meals are difficult to distinguish because skipping meals and snacking have become more prevalent. Such eating styles can have various effects on cardiometabolic health markers, namely obesity, lipid profile, insulin resistance, and blood pressure. In this statement, we review the cardiometabolic health effects of specific eating patterns: skipping breakfast, intermittent fasting, meal frequency (number of daily eating occasions), and timing of eating occasions. Furthermore, we propose definitions for meals, snacks, and eating occasions for use in research. Finally, data suggest that irregular eating patterns appear less favorable for achieving a healthy cardiometabolic profile. Intentional eating with mindful attention to the timing and frequency of eating occasions could lead to healthier lifestyle and cardiometabolic risk factor management.
The patterns of meal and snack eating behavior in American adults have changed over the past 40 years. Based on NHANES (National Health and Nutrition Examination Survey) data from 1971 to 1974 to 2009 to 2010 (n=62 298), women 20 to 74 years of age reported a decrease in 24-hour meal-derived total energy intake (TEI) from 82% in the 1970s to 77% in 2009 to 2010 and an increase in the proportion of TEI consumed from snacks from 18% to 23%.1 Similar trends were reported among men. The proportion of men and women who reported consuming all 3 standard meals declined over this period (from 73% to 59% in men; from 75% to 63% in women),1 reflecting changes in eating patterns rather than changes in eating frequency. Indeed, the traditional breakfast-lunch-dinner pattern was not observed in a population of healthy, non–shift-working adults.2 In that study, the number of eating occasions, defined as consumption of any food or beverage providing at least 5 kcal, was ≈4.2 times a day in the lowest decile and 10.5 times a day for the top decile. There were only 5 hours during the 24-hour day when <1% of all eating occasions occurred: between 1 and 6 am. This study clearly demonstrated that adults in the United States eat around the clock. Because feeding and fasting entrain clock genes, which regulate all aspects of metabolism, meal timing can have serious implications for the development of cardiovascular disease (CVD), type 2 diabetes mellitus, and obesity.3,4
The circadian rhythms of the body are controlled by the central clock located in the suprachiasmatic nucleus of the hypothalamus but also by clocks of peripheral organs. Although the master clock is strongly entrained by light, clocks of peripheral organs are additionally responsive to food supply, and temporal restriction of food can reset clock gene rhythms. In mice, food given in the normal sleeping period can uncouple peripheral clocks from the master clock.5 In fact, time-restricted feeding in mice alters the robustness and coherence of rhythmic gene transcripts,6 which may be relevant for cardiometabolic health. Indeed, the CLOCK and BMAL1 genes are implicated in regulating genes involved in lipid metabolism.7,8 Polymorphisms in the CLOCK gene correlated with the development of metabolic syndrome; those in BMAL1, with type 2 diabetes mellitus and hypertension. Therefore, time of eating and nutrient delivery may have cardiometabolic health implications via alterations in peripheral clocks, most notably that of the liver.
In this statement, we review epidemiological and clinical evidence linking various eating patterns with cardiometabolic health markers in adults. We focus on patterns of food consumption as they relate to meal times and number of eating occasions/eating frequency rather than dietary profiles such as nutrient intakes. However, some attention will be given to diet quality to the extent that food intake patterns influence overall diet macronutrient composition. A comprehensive literature search strategy was used to identify relevant articles for this review. We searched electronic databases (eg, MEDLINE, PubMed, PubMed Central) and reference lists from retrieved articles and consulted expert colleagues. For electronic searches, we crossed various key words related to cardiometabolic health markers: lipid profile, fasting glucose, fasting insulin, homeostatic model assessment (HOMA), insulin resistance (IR), insulin sensitivity, blood pressure, body weight, overweight, obesity, fat mass, visceral fat, adiposity, lean body mass, and weight loss. Key words relevant to eating patterns under review included breakfast skipping, intermittent fasting, meal frequency, and meal timing.
Additional Methods
A systematic search in MEDLINE PubMed was performed with the use of various combinations of the following search terms: alternate day fasting, intermittent fasting, fasting, intermittent energy restriction, meal frequency, meal skipping, meal timing, late day eating, late day meals, evening eating, evening meals, obesity, body weight, weight loss, cardiovascular risk, coronary heart disease (CHD), cholesterol, plasma lipids, lipid profile, blood pressure, glucose, insulin, and insulin resistance. Articles were excluded if they did not include original data; if they were editorials, letters, comments, or conferences proceedings; or if they did not meet the inclusion criteria described below. References of the retrieved articles were also screened for additional studies. Inclusion criteria were as follows: (1) randomized, controlled trials and nonrandomized trials; (2) cohort and observational studies; (3) sample size ≥7 subjects per study arm for intervention studies; (4) primary end points of body weight or ≥1 relevant cardiovascular risk parameters; (5) age between 18 and 75 years; (6) nonsmokers; and (7) sedentary or moderately active individuals. Exclusion criteria included (1) trials that included dietary supplements, pharmacological substances, or exercise; (2) individuals with type 2 diabetes mellitus; and (3) very active individuals or athletes. This search was limited to clinical trials with human subjects reported in the English language.
Breakfast Skipping and Cardiometabolic Risk: Epidemiological Findings
It is commonly reported that “breakfast is the most important meal of the day.”9 However, ≈20% to 30% of US adults do not eat breakfast, and breakfast consumption has declined in recent decades.10–14 Of all meals, the prevalences of breakfast and lunch consumption have seen the greatest decline over the past 40 years.1 The decline in breakfast consumption has paralleled the increase in obesity prevalence, fostering studies of the association between breakfast consumption, cardiometabolic health risks, and chronic disease.15–21 The definition of breakfast varies across studies.9 However, 2 common definitions are the following: (1) the first meal of the day eaten before or at the start of daily activities within 2 hours of waking, typically no later than 10 am, and consisting of a calorie level of 20% to 35% of total daily energy needs18,20,22,23 and (2) the consumption of food or beverage (excluding water) between 5 and 9 am.12 Globally, adult population predictors of skipping breakfast have consistently included younger age, current tobacco use, late dinner, higher alcohol consumption, higher daily energy intake, and infrequent exercise.13,20,24–34
Breakfast Skipping and Diet Quality
There is an association between skipping breakfast and low nutritional adequacy of adult diets.13,35 A cross-sectional survey of the Bogalusa Heart Study (n=504; 58% women; 70% white; mean age, 23 years [range, 19–28 years]) demonstrated that 74% of breakfast skippers did not meet two thirds of the Recommended Dietary Allowance for vitamins and minerals compared with 41% of those who consumed breakfast.36 Using data from NHANES 1999 to 2002, investigators demonstrated that young adults (n=2615; age, 20–39 years) who reported skipping breakfast (defined as the absence of any food/beverage, excluding water) had greater TEI from added sugars, a lower mean adequacy ratio for nutrient intake, and a lower Healthy Eating Index.37
Breakfast Skipping and CVD Risk Factors: Obesity and Weight Gain
An abundance of data support an association between breakfast skipping and adiposity, which has led to recommendations to consume breakfast as a possible strategy to achieve a healthy body weight38 and successful weight loss maintenance.39 This association between breakfast skipping and higher body mass index (BMI) has been reported globally, primarily in cross-sectional studies.33,40–51 For example, in NHANES 1999 to 2002, young adults (n=5316; age, 20–39 years) who reported consuming ready-to-eat cereal were 31% less likely to be overweight/obese and 39% less likely to have abdominal obesity compared with breakfast skippers.52 Additionally, a meta-analysis of 19 studies in the Asian and Pacific regions (n=19 108 participants) demonstrated a pooled odds ratio (OR) of 1.75 (95% confidence interval [CI], 1.57–195) for prevalence of overweight or obesity among the lowest compared with the highest breakfast consumption frequency.53
The association between breakfast consumption and lower risk of obesity and weight gain is supported further by results from several large, long-term, prospective, observational studies. The prospective design minimizes the potential for reverse causation bias that hinders cross-sectional analyses. Among 6764 adults in the United Kingdom, a 1% increase in reported TEI at breakfast was associated with a relatively lower weight gain (−0.021 kg over an average of 3.7 years of follow-up). After adjustment for sociodemographic and dietary factors, the percentage of TEI consumed at breakfast was inversely associated with weight gain.54 Among young adults (n=3598; baseline age, 18–30 years), daily breakfast eaters gained 1.91 kg less than infrequent breakfast consumers (<4 d/wk) over 18 years.55 In men, breakfast consumers were less likely to gain ≥5 kg over 10 years, independently of lifestyle and BMI at baseline (hazard ratio [HR], 0.87; 95% CI, 0.82–0.93).56 Interestingly, the inverse association between breakfast consumption and weight gain was greater in normal-weight men (adjusted HR, 0.78; 95% CI, 0.70–0.87) compared with overweight men (HR, 0.92; 95% CI, 0.85–1.00).56
Breakfast Skipping and Other CVD Risk Factors
In cross-sectional studies, daily breakfast eaters were less likely to have CVD risk factors, including elevated serum low-density lipoprotein (LDL) cholesterol, low serum high-density lipoprotein (HDL) cholesterol, and elevated blood pressure.52,57 Conversely, among 415 healthy Korean adults, rare breakfast eaters (eating breakfast 1 of 3 days) were less likely to have elevated serum triglycerides (≥150 mg/dL).35 Notably, in the Korean study, the percent of TEI from carbohydrates was lower and the percent from fat was higher in rare breakfast eaters, which may explain this observation.
Skipping breakfast has been associated with markers of impaired glucose metabolism, including elevated hemoglobin A 1c,44 higher fasting plasma glucose58 and all-day postprandial hyperglycemia,59,60 and a higher rate of impaired fasting glucose.28 In contrast, HOMA-IR did not differ statistically between breakfast skippers and eaters.57 Furthermore, breakfast skipping has been associated with a greater risk of clinically diagnosed diabetes mellitus, which is a cardiovascular risk equivalent, in 3 recent long-term, prospective studies. In a prospective study among men in the Health Professionals Follow-Up Study (n=29 205; age, 40–75 years at baseline), skipping breakfast (defined as not eating anything before lunch) was associated with a 21% higher risk of developing type 2 diabetes mellitus (relative risk, 1.21; 95% CI, 1.07–1.35) after adjustment for BMI, age, dietary quality, and other potential confounders.61 Among women in the Nurses’ Health Study (n=46 289; mean baseline age, 64.7 years), skipping breakfast even once per week was associated with a 28% higher risk of incident type 2 diabetes mellitus (HR, 1.28; 95% CI, 1.14–1.44) after adjustment for numerous confounders.29 In a prospective analysis among 4631 middle-aged Japanese workers (age, 35–66 years at baseline), participants who skipped breakfast ≥2 d/wk had a 73% greater risk of diabetes mellitus (HR, 1.73; 95% CI, 1.24–2.42) compared with participants who consumed breakfast 6 or 7 d/wk over 8.9 years of follow-up after adjustment for confounders.62 Finally, in a prospective analysis of young adults (n=3598; baseline age, 18–30 years), daily breakfast consumers had a multivariable-adjusted HR for diabetes mellitus of 0.66 (95% CI, 0.51–0.86) compared with infrequent breakfast eaters (0–3 d/wk) over 18 years of follow-up.55 This association was mediated by BMI (HR, 0.81; 95% CI, 0.63–1.05). Furthermore, daily breakfast was not associated with a lower risk of diabetes mellitus among black women. The multivariable HR for diabetes mellitus comparing daily with infrequent breakfast consumption was 1.00 (95% CI, 0.66–1.50) in black women but 0.54 (95% CI, 0.39–0.75) in black men, white men, and white women. Additionally, in CARDIA (Coronary Artery Risk Development in Young Adults), daily breakfast eaters also had a lower risk of hypertension (HR, 0.74; 95% CI, 0.63–0.86) and metabolic syndrome (HR, 0.63; 95% CI, 0.54–0.75) compared with infrequent breakfast eaters (0–3 times a week), and these associations remained significant after adjustment for baseline measures of adiposity.55 Thus, data from large, prospective studies, supported by data from cross-sectional studies, suggest that breakfast consumption may play an important role in the prevention of cardiometabolic outcomes.
Breakfast Skipping and CVD Risk
To the best of our knowledge, only 2 prospective studies have examined the association between breakfast skipping and risk of CVD.62,63 Over 16 years of follow-up, men who reported usually skipping breakfast had a 27% (relative risk, 1.27; 95% CI, 1.06–1.53) higher risk of CHD (defined as a nonfatal myocardial infarction or fatal CHD) compared with men who did not skip breakfast after adjustment for age, demographic factors, and dietary and lifestyle factors.63 In the second study, conducted in Japan, infrequent breakfast consumption was associated with a greater risk of CVD, specifically greater risk of hemorrhagic stroke, after adjustment for age, sex, dietary and lifestyle factors, perceived mental stress, living alone, physical labor, and public health center area.64 Individuals who consumed breakfast 0 to 2 times a week compared with those with daily breakfast eating had an HR of 1.14 (95% CI, 1.01–1.27) for total CVD, 1.18 (95% CI, 1.04–1.34) for total stroke, and 1.36 (95% CI, 1.10–1.70) for cerebral hemorrhage. In contrast, infrequent breakfast intake was not associated with greater risk of subarachnoid hemorrhage, cerebral infarction, or CHD in this population. Compared with the prevalence in Western countries, the prevalence of stroke, especially cerebral hemorrhage, is higher and the prevalence of CHD is lower in Japan, which may be attributable to differential CVD risk factor patterns between these countries.65,66 The authors speculated that the low prevalence of CHD may explain the lack of association between breakfast skipping and CHD risk in Japan.64
Epidemiological studies provide strong evidence of a relation between breakfast skipping and cardiometabolic risk. These include greater risk of overweight and obesity, metabolic risk profile, diabetes mellitus, CVD, and hypertension. These risks seem to be independent of differences in diet quality between breakfast eaters and nonconsumers. Notably, it is possible that reverse causation is responsible in part for the findings in the cross-sectional analyses. Although findings from observational studies cannot establish causality, large, prospective studies with long-term follow-up and the assessment of clinical end points, including CVD and diabetes mellitus, can provide important insight into these associations. These are particularly meaningful when supported by evidence from experimental studies and clinical interventions.
Breakfast Skipping and Cardiometabolic Risk: Clinical Intervention Findings
Few randomized, controlled trials have studied the impact of breakfast consumption on body weight. One of the pioneering studies in this field randomized breakfast eaters (≥4 times per week) and skippers to either maintaining their breakfast consumption patterns or switching to the alternative pattern while following a 1200-kcal/d diet for 12 weeks.67 The breakfast skippers randomized to breakfast consumption lost 7.7 kg, whereas those maintained on a no-breakfast diet lost 6.0 kg. In comparison, breakfast consumers maintained on a breakfast-eating weight loss diet lost 6.2 kg compared with 8.9 kg for those randomized to skip breakfast. It was therefore the change in breakfast eating habits, rather than breakfast consumption or skipping per se, that resulted in greater weight loss. However, behavioral data demonstrated that eating breakfast reduced overall dietary fat intake and minimized impulsive snacking, which are critical for successful weight reduction.67 A subsequent clinical trial randomized >300 overweight and obese adults (age, 20–65 years) stratified by baseline breakfast habits to consume breakfast or to skip breakfast for 16 weeks.68 No other dietary advice was provided. In contrast to the prior study, treatment assignment did not have a significant effect on weight. A 4-week trial randomized a total of 36 obese men and women to a higher-fiber breakfast (hot oat cereal), a nonfiber breakfast (frosted ready-to-eat cereal), or a no-breakfast arm.69 The no-breakfast group had greater weight loss (−1.18 kg) compared with the breakfast groups; however, breakfast skippers had an increase in serum total cholesterol.69 Another small, randomized, crossover trial of 10 normal-weight women also demonstrated that omitting breakfast for 2 weeks resulted in higher fasting total and LDL cholesterol.70
In an open-label trial, 93 overweight and obese women with metabolic syndrome (age, 30–57 years; BMI, 32.2±1.2 kg/m 2) were randomized to compare change in weight and metabolic outcomes on a high-calorie breakfast versus a high-calorie dinner.71 Both groups consumed a 500-kcal lunch. However, the breakfast group consumed a 700-kcal breakfast and 200-kcal dinner. In contrast, the dinner group consumed a 200-kcal breakfast and 700-kcal dinner. After 12 weeks, although body weight, waist circumference, fasting glucose, and insulin were reduced in both groups, they were all significantly lower in the breakfast group. This suggests that a higher calorie intake during breakfast (earlier in the day) may influence weight and glucose metabolism. Additionally, despite weight loss in both groups, mean triglyceride levels decreased by 33% in the high-calorie breakfast group but increased by 14% in the dinner group, raising concerns about nocturnal postprandial lipid metabolism. However, despite the short trial duration, 17% of the breakfast group and 23% of the dinner group dropped out before completion.
Other studies suggest that the impact of breakfast skipping may differ by weight status. In obese participants randomized to consume at least 700 kcal before 11 am (n=11) or to fast until noon (n=12), participants compensated for the lack of morning energy intake, and there was no impact of the intervention on body weight.72 However, morning physical activity was reduced in the fasting group, and insulin sensitivity was reduced relative to the breakfast consumers. The lipid profile was unaffected by breakfast consumption patterns. This contrasts with an identical intervention study in which lean individuals did not compensate for the reduced morning caloric intakes.73 However, in that study also, body weight was unaffected by the intervention. Similar findings with respect to lipid profile and insulin sensitivity were reported. Despite the epidemiological association of breakfast skipping and higher BMI, the few clinical trial interventions have significant limitations and conflicting outcomes that prohibit evidence-based recommendations on daily breakfast consumption to promote weight loss solely.
In a small crossover study of individuals (n=26) with type 2 diabetes mellitus who regularly ate breakfast at baseline (age, 30–70 years; BMI, 22–35 kg/m 2; hemoglobin A 1c, 7%–9%), randomization to 2 days of consumption versus omitting breakfast suggested that skipping breakfast increases postprandial hyperglycemia after lunch and dinner in association with lower intact glucagon-like peptide-1 and impaired insulin response.74 In clinical intervention studies, skipping breakfast has been associated with adverse metabolic and behavioral responses.75,76 Adverse effects include higher blood glucose and serum insulin responses, higher plasma free fatty acids, and higher hunger and desire-to-eat ratings that likely result in a compensatory increase in energy intake later in the day.75
In summary, the limited evidence of breakfast consumption as an important factor in combined weight and cardiometabolic risk management is suggestive of a minimal impact. There is increasing evidence that advice related to breakfast consumption does not improve weight loss, likely because of compensatory behaviors during the day. On the other hand, breakfast consumption can contribute to a healthier eating pattern that leads to slight improvements in cardiometabolic risk profile. Additional, longer-term studies are needed in this field because most metabolic studies have been either single-day studies or of very short duration.
Summary
On the basis of the combined epidemiological and clinical intervention data, daily breakfast consumption among US adults may decrease the risk of adverse effects related to glucose and insulin metabolism. In addition, comprehensive dietary counseling that supports daily breakfast consumption may be helpful in promoting healthy dietary habits throughout the day.
Intermittent Fasting, Meal Frequency, and Cardiometabolic Risk: Epidemiological Findings
Intermittent Fasting
Observational data on the long-term relationship between intermittent fasting and risk of cardiometabolic disease are limited. The Intermountain Heart Collaborative Study includes a large proportion of patients who identify themselves as having a Latter-Day Saint religious preference, in which routine fasting (1 time per month for 24 hours) is common.77 A meta-analysis of 2 small studies within this cohort including patients from 2004 to 2006 (n=448) and from 2007 to 2008 (n=200) found that patients who fasted routinely had an OR of 0.65 (95% CI, 0.46–0.94) for coronary artery disease compared with individuals who did not fast routinely after adjustment for confounders. In the same study, the OR comparing routine fasting and nonfasting was 0.56 (95% CI, 0.36–0.88) for diabetes mellitus. However, these estimates were adjusted for age and sex only. Further studies are needed with larger sample sizes, adjustment for confounding by other lifestyle behaviors, and prospective data collection to determine the long-term relationship between routine fasting and disease outcomes. Furthermore, epidemiological studies involving participants with more frequent fasting days are needed.
Meal Frequency
Greater eating frequency has been associated with lower risk of obesity in several cross-sectional studies within free-living populations. In a cross-sectional analysis within the prospective SEASONS study (Seasonal Variation of Blood Cholesterol Study in Worcester County, Massachusetts; n=499; 50.3% men; mean age, 48 years), individuals who ate ≥4 times a day had a significantly lower risk of obesity (OR, 0.55; 95% CI, 0.33–0.91) compared with individuals who ate ≤3 times a day, after adjustment for age, sex, physical activity, and TEI.48 In the Malmo Diet and Cancer study (n=1355 men, 1654 women; age, 47–68 years), meal frequency was associated with an increased risk of obesity in men but not women.78 Compared with men who ate ≥6 times per day, men who ate ≤3 times a day were more likely to be obese (OR, 2.42; 95% CI, 1.02–5.73) and to have an increased waist circumference (≥102 cm; OR, 2.09; 95% CI, 1.03–4.27) after adjustment for TEI, lifestyle, and diet. However, the cross-sectional nature of these studies precludes us from establishing the causality or temporality of this association.
In a prospective analysis of adult men, increasing the number of eating occasions beyond 3 meals a day was associated with greater risk of gaining ≥5 kg over 10 years (HR, 1.15; 95% CI, 1.06–1.25 for ≥2 versus 0 additional eating occasions).56 Additional prospective studies are needed to better understand the role of eating frequency on adiposity and long-term weight gain or loss.
In a large, cross-sectional study conducted within the Norfolk cohort of EPIC (European Prospective Investigation Into Cancer; n=14 666; age, 45–75 years), greater frequency of eating was associated with lower mean concentrations of total and LDL cholesterol.79 Individuals who reported eating ≥6 times a day had mean total cholesterol levels that were ≈0.15 mmol/L lower than in individuals who ate 1 or 2 times a day, independently of TEI, age, BMI, smoking, physical activity, and nutrients. Eating frequency was not associated with HDL cholesterol.
Among men, infrequent meal frequency was associated with greater risk of type 2 diabetes mellitus over 16 years of follow-up.61 Compared with men who ate 3 meals a day, men who ate 1 to 2 times a day had an HR for diabetes mellitus of 1.26 (95% CI, 1.09–1.46) after adjustment for age, lifestyle, diet, and other potential confounders. This risk was not altered after adjustment for BMI (HR, 1.25; 95% CI, 1.08–1.45). In contrast, more frequent eating (≥4 times a day) was not associated with risk of diabetes mellitus independently of BMI. In contrast, eating frequency was not associated with risk of diabetes mellitus among women in the Nurses’ Health Study over 6 years of follow-up.29 Compared with women who ate 3 times a day, the HRs for diabetes mellitus were 1.09 (95% CI, 0.84–1.41) for women who ate 1 to 2 times a day, 1.13 (95% CI, 1.00–1.27) for women who ate 4 to 5 times a day, and 0.99 (95% CI, 0.81–1.21) for women who ate ≥6 times a day.
To the best of our knowledge, there has been only 1 prospective, cohort study that quantified the association between eating frequency and risk of CHD.63 Compared with men who ate 3 times a day, the HRs for CHD were 1.10 (95% CI, 0.92–1.32) for men who ate 1 to 2 times a day, 1.05 (95% CI, 0.94–1.18) for men who ate 4 to 5 times a day, and 1.26 (95% CI, 0.90–1.77) for men who ate ≥6 times a day after adjustment for TEI, diet quality, lifestyle, and other CVD risk factors.
Epidemiological studies of eating frequency lead to different conclusions, depending on the outcome of interest. The relation between eating frequency and obesity is mixed but seems to be more consistent with respect to CVD risk factors and diabetes mellitus. In those cases, greater eating frequency seems to be related to improved risk status.
Intermittent Fasting, Meal Frequency, and Cardiometabolic Risk: Clinical Intervention Findings
Intermittent Fasting
Intermittent-fasting diets have increased in popularity over the past decade, particularly with respect to clinical intervention studies. The 2 most common forms of intermittent fasting include alternate-day fasting and periodic fasting. Alternate-day fasting involves a “fast day,” when individuals consume ≤25% of baseline energy needs during a 24-hour period, alternated with a “feast day,” when ad libitum food consumption is permitted for 24 hours. Periodic fasting, on the other hand, requires participants to fast only 1 or 2 d/wk and allows 5 to 6 days of ad libitum food consumption per week.
Changes in body weight by alternate-day fasting and periodic-fasting regimens are displayed in Table 1. Body weight decreased significantly in all studies by 3% to 8% after 3 to 24 weeks of treatment.80–89 Studies that provided food on the fast day produced the greatest weight loss. For instance, overweight participants lost 8% of their body weight over an 8-week period when provided with a 320- to 380-kcal meal replacement shake on each fast day during an alternate-day fasting protocol.82 Similar decreases in body weight (4%–7%) were demonstrated in the other 8-week alternate-day fasting studies that provided food on the fast day.83–88 The frequency of weekly fast days also appears to affect the degree of weight loss. Not surprisingly, participants lost weight faster in the alternate-day fasting studies that required 3 to 4 days of fasting a week compared with periodic-fasting studies, which require participants to fast only 1 to 2 d/wk. On average, alternate-day fasting produces a 0.75-kg/wk reduction in body weight,80–87 whereas periodic fasting produces a 0.25-kg/wk reduction in body weight.88,89
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Table 1. Intermittent Fasting: Effect on CHD Risk Parameters
| Reference | Duration, wk | Subjects | Intervention | Weight, % Change | CHD Risk Parameters, % Change |
--- --- --- |
| TC | LDL | HDL | TG | SBP | DBP | Glucose | Insulin | HOMA-IR |
| Alternate-day fasting (fasting 3–4 d/wk) |
| Heilbronn et al,80 2005 | 3 | n=16, M and FAge, 23–53 y OverweightRace/ethnicity: NR | Fast day: 0% intakeFeed day: ad libitum intakeFood not provided | ↓3 | … | … | ↑VNR in women only | ↓VNR | … | … | ↓1 | ↓57 | … |
| Eshghinia and Mohammadzadeh,81 2013 | 6 | n=15, FAge, 34±6 y ObeseRace/ethnicity: NR | Fast day: 30% intakeFeed day: ad libitum intakeFood not provided | ↓7 | ↓6 | ↓12 | ↑19 | ↓11 | ↓8 | ↓10 | ↓6 | … | … |
| Johnson et al,82 2007 | 8 | n=10, M and FAge: NR ObeseRace/ethnicity: NR | Fast day: 20% intakeFeed day: ad libitum intakeFood provided on fast day | ↓8 | ↓9 | ↓10 | ↑4 | ↓42 | … | … | ↑6 | ↓37 | ↓33 |
| Varady et al,83 2009 | 8 | n=16, M and FAge, 46±2 y ObesePrediabetic Race/ethnicity: 6–H8–B2–W | Fast day: 25% intakeFeed day: ad libitum intakeFood provided on fast day | ↓6 | ↓21 | ↓32 | ↑9 | ↓35 | ↓6 | ↓6 | ↓4 | ↓20 | ↓19 |
| Klempel et al,84 2012 | 8 | n=32, FAge, 42±2 y ObeseRace/ethnicity: 8–H24–B0–W | Fast day: 25% intake, high fatFeed day: ad libitum intake, high fatFast day: 25% intake, low fatFeed day: ad libitum intake, low fatFood provided all days in all groups | ↓5↓4 | ↓13↓16 | ↓18↓25 | ↓1↑5 | ↓14↓24 | ↓2↓2 | ↓3↑3 | ↓2↓2 | … | … |
| Alternate-day fasting (fasting 3–4 d/wk) Continued |
| Hoddy et al,85 2014 | 8 | n=74, M and FAge, 45±3 yObeseRace/ethnicity: NR | Fast day: 25% intake as lunchFeed day: ad libitum intakeFast day: 25% intake as dinnerFeed day: ad libitum intakeFast day: 25% intake as small mealsFeed day: ad libitum intakeFood provided on fast day in all groups | ↓4↓4↓4 | ↓1↓30 | ↓100 | ↓40↓2 | ↓6↓8↓1 | ↓2↓4↓5 | ↓1↓4↓1 | ↓2↓1↓1 | 0↓18↓12 | ↓10↓26↓19 |
| Bhutani et al,86 2013 | 12 | n=32, M and FAge, 43±3 y ObeseRace/ethnicity: 18–H41–B21–W3–Other | Fast day: 25% intakeFeed day: ad libitum intakeFood provided on fast dayControl: ad libitum fed every dayFood not provided | ↓40 | ↑7↑1 | ↓1↑3 | 0↑8 | ↑3↑5 | ↓3↓2 | ↓2↓2 | ↓3↑2 | ↓11†↑1 | ↓9†↑2 |
| Varady et al,87 2013 | 12 | n=32, M and FAge, 47±4 y Normal weight and overweight PrediabeticRace/ethnicity: 3–H13–B14–W | Fast day: 25% intakeFeed day: ad libitum intakeFood provided on fast dayControl: ad libitum fed every dayFood not provided | ↓7↓1 | ↓13↓4 | ↓16↓7 | ↓4↑2 | ↓2↑9 | ↓6↑1 | ↓6↑1 | ↓8↑2 | ↓31†↑2 | ↓28†↑2 |
| Periodic fasting (fasting 1–2 d/wk) |
| Klempel et al,88 2012 | 8 | n=54, FAge, 48±2 y Obese PrediabeticRace/ethnicity‡: 9–H34–B6–W5–A | 1 d/wk: 0% intake6 d/wk: 70% intake, liquid dietFood provided1 d/wk: 0% intake6 d/wk: 70% intake, food dietFood not provided | ↓4↓3 | ↓19↓8 | ↓20↓7 | ↓5↓2 | ↓17↓3 | ↓2↓5 | ↓50 | ↓3↓2 | ↓21↓13 | ↓23↓12 |
| Harvie et al,89 2011 | 24 | n=53, FAge, 30–45 y Overweight and obeseRace/ethnicity: 2–B103–W2–Other | 2 d/wk: 25% intake5 d/wk: ad libitum intakeFood not provided | ↓7 | ↓6 | ↓10 | 0 | ↓16 | ↓3 | ↓6 | ↓2 | ↓29 | ↓27 |
Expand Table
A indicates Asian; B, black, African American, or Afro-Caribbean; CHD, coronary heart disease; DBP, diastolic blood pressure; F, female; H, Hispanic; HDL, high-density lipoprotein cholesterol; HOMA-IR, homeostatic model assessment of insulin resistance; LDL, low-density lipoprotein cholesterol; M, male; NR, not reported; SBP, systolic blood pressure; TC, total cholesterol; TG, triglycerides; VNR, value not reported; and W, white.
Posttreatment value significantly different from baseline value (P<0.05).
†
Significantly different from control group (P<0.05).
‡
Reflects 30 of 32 completing the study and included in the analyses.
The impact of intermittent fasting on total and LDL cholesterol concentrations is variable (Table 1). Although some trials report reductions in total cholesterol ranging from 6% to 21% and LDL cholesterol ranging from 7% to 32%,82–84,87–89 others report no effect.81,85,86 The variation between studies cannot be clearly explained because the extent of weight loss is similar between studies.81–89 On closer examination, however, it is possible that baseline cholesterol levels may play a role. Significant reductions in these lipid risk factors have been observed only in studies in which participants had mildly elevated cholesterol (ie, LDL cholesterol >110 mg/dL).82–84,87–89 HDL cholesterol concentrations remained unchanged in most of the studies reviewed here (Table 1).
Triglyceride concentrations decreased in the majority of intermittent-fasting studies, with reductions ranging from 16% to 42% (Table 1). The greatest decreases in triglycerides were generally observed in studies with the greatest weight loss. For instance, in studies that achieved 1-kg/wk weight loss, triglycerides decreased by ≈30% to 40%,82,83 whereas in the studies that achieved 0.25- to 0.5-kg/wk weight loss,84,87–89 triglycerides decreased by ≈10% to 20%. Thus, both alternate-day fasting and periodic-fasting regimens appear to be effective in lowering triglyceride levels, but the effect is dependent on the amount of weight lost.
Systolic and diastolic blood pressures decreased only in the intermittent-fasting studies that achieved 6% to 7% weight loss.81,83,87,89 In these trials, systolic blood pressure reductions ranged from 3% to 8% and diastolic blood pressure reductions ranged from 6% to 10% after 6 to 24 weeks of treatment.81,83,87,89 Participants in these studies81,83,87,89 all had borderline prehypertension, suggesting that these diets may prevent the progression of prehypertension to hypertension. Longer-term and larger-scale trials are needed to confirm this interesting preliminary finding.
Intermittent fasting appears to have no effect on fasting glucose concentrations in healthy individuals (Table 1). On the other hand, the diet seems to have a minor beneficial effect in those with prediabetes, with decreases in fasting glucose ranging from 3% to 6%.83,87,88 The greatest decrease in fasting glucose was observed in individuals with prediabetes randomized to an alternate-day fasting group versus a no-intervention control group.87 After 12 weeks, fasting glucose was reduced by 6% in the alternate-day fasting group relative to the control group.87 Reductions in glucose (3%–4%) in patients with prediabetes were also demonstrated in 2 other 8-week studies of alternate-day fasting83 and periodic fasting.88
In contrast to the variable findings for glucose, fasting insulin decreased in all but 1 study (Table 1). Decreases in fasting insulin ranged from 11% to 57% after 3 to 24 weeks of following the intermittent-fasting dietary pattern.80,82,83,85–89 These reductions in insulin were not dependent on the prediabetes status of participants but rather appeared to be most strongly related to the degree of imposed energy restriction yet were somewhat independent of weight loss. Trials with the largest decrease in insulin had an average daily restriction of 40% to 50% of baseline energy needs.80,82 For instance, in 1 study,80 participants underwent 50% average daily restriction via complete fasting on fast days, alternating with ad libitum intake on the day of feeding. Although weight was reduced by only 3% from baseline, a 57% decrease in fasting insulin was observed.80 Likewise, in a study in which participants underwent 40% average daily restriction,82 a 37% reduction in fasting insulin levels was observed. In a comparison of periodic fasting and alternate-day fasting, it would appear that both dietary patterns produce sizeable reductions in insulin concentrations. Improvements in this CVD risk parameter may be observed with fasting 1 to 2 d/wk, as with periodic fasting, or fasting 3 to 4 d/wk, as with alternate-day fasting.
Intermittent-fasting protocols produce fairly consistent reductions in IR after 8 to 24 weeks of treatment (Table 1). In all studies, changes in IR were quantified by the HOMA-IR method.82,83,86–89 The most pronounced decreases in IR occurred with the greatest weight loss. For example, in 1 alternate-day fasting study,82 participants lost 8% of body weight, and this corresponded to the largest decline in HOMA-IR (33%). In another alternate-day fasting study in which the reduction in body weight was half as large (4%),86 moderate reductions in HOMA-IR (9%) were observed. Notable reductions in IR were also demonstrated by periodic-fasting protocols. For example, HOMA-IR decreased by 23% after 8 weeks of fasting 1 d/wk.88 Likewise, a 27% reduction in HOMA-IR was demonstrated with 24 weeks of fasting on 2 d/wk.89 Together, these studies show that intermittent-fasting protocols that produce at least 4% weight loss may be helpful in decreasing IR in obese patients. Whether these findings can be reproduced when more robust measures of insulin sensitivity, that is, the hyperinsulinemic-euglycemic clamp, are implemented warrants investigation.
Summary
There is evidence that both alternate-day fasting and periodic fasting may be effective for weight loss, although there are no data that indicate whether the weight loss can be sustained long term. In addition, both eating patterns may be useful for lowering triglyceride concentrations but have little or no effect on total, LDL, or HDL cholesterol concentrations. These protocols may also be beneficial for lowering blood pressure, but a minimum weight loss of 6% may be required to see an effect. Intermittent fasting may also be effective for decreasing fasting insulin and IR, but fasting glucose remains largely unchanged. Future work in this area should aim to examine whether these effects still persist in longer-term (>52 weeks) randomized, controlled trials.
Meal Frequency
To date, 9 trials90–98 have been performed that examined the impact of meal frequency without calorie restriction on CHD risk. These trials implemented the following meal frequency regimens: 1, 3, 6, 9, 12, and 17 meals a day. The effects of these diets on key cardiometabolic risk factors are displayed in Table 2.
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Table 2. Meal Frequency Regimens Without Calorie Restriction: Effect on CHD Risk Parameters
| Reference | Duration, wk | Subjects | Intervention | Weight, % Change | CHD Risk Parameters, % Change |
--- --- --- |
| Baseline | Treatment | TC | LDL | HDL | TG | DBP | SBP | Glucose | Insulin | HOMA-IR |
| 1 Meal per day |
| Stote et al,90 2007 | 8 | n=15, M and FAge, 45±1 y Normal weightRace/ethnicity: NR | Not stated | 15% Pro35% Fat50% CarbFood provided | ↓1 | ↑19 | ↑25 | ↑17 | ↓4 | ↑1 | ↑1 | … | … | … |
| 3 Meals per day |
| McGrath and Gibney,91 1994 | Not stated | n =12, MAge: not stated Normal weightRace/ethnicity: NR | 13% Pro39% Fat48% Carb | 14% Pro41% Fat45% CarbFood not provided | 0 | ↑1 | ↑3 | 0 | ↓19 | … | … | … | … | … |
| Jenkins et al,92 1989 | 2 | n=7, MAge, 31–51 yBMI: not statedRace/ethnicity: NR | Not stated | 15% Pro33% Fat52% CarbFood provided | ↓1 | ↓4 | ↑2 | ↓4 | ↓39 | … | … | No ΔVNR | … | … |
| Murphy et al,93 1996 | 2 | n =11, FAge, 22±1 yBMI: not statedRace/ethnicity: NR | Not stated | 20% Pro40% Fat40% CarbFood not provided | 0 | ↑3 | 0 | ↑7 | ↓2 | … | … | No ΔVNR | No ΔVNR | … |
| Arnold et al,94 1993 | 2 | n=19, M and FAge, 32±2 y Normal weightRace/ethnicity: NR | 15% Pro28% Fat57% Carb | 15% Pro28% Fat57% CarbFood not provided | 0 | ↓4 | ↓7 | ↑1 | ↑3 | No ΔVNR | No ΔVNR | No ΔVNR | No ΔVNR | … |
| Arnold et al,95 1994 | 4 | n=16, M and FAge, 50±2 yBMI: Not statedRace/ethnicity: NR | Not stated | 17% Pro35% Fat48% CarbFood not provided | 0 | ↓1 | ↑4 | ↑1 | ↓23 | No ΔVNR | No ΔVNR | No ΔVNR | No ΔVNR | … |
| 3 Meals per day Continued |
| Arciero et al,96 2013 | 4 | n=10, M and FAge, 46±2 y OverweightRace/ethnicity: NR | Not stated | 35% Pro21% Fat44% CarbFood not provided | ↓1 | … | … | … | … | … | … | No ΔVNR | No ΔVNR | No ΔVNR |
| Stote et al,90 2007 | 8 | n=15, M and FAge, 45±1 y Normal weightRace/ethnicity: NR | Not stated | 15% Pro35% Fat50% CarbFood provided | ↑1 | ↑1 | ↑4 | ↑6 | ↓4 | ↓3 | ↓5 | … | … | … |
| 6 Meals per day |
| Farshchi et al,97 2004 | 2 | n=9, FAge, 18–42 y Normal weightRace/ethnicity: NR | 18% Pro35% Fat47% Carb | 14% Pro40% Fat46% CarbFood not provided | ↓1 | ↓1 | ↓8 | ↑8 | ↑5 | — | — | ↓6% | ↓11% | ↓11% |
| Farshchi et al,98 2005 | 2 | n=10, FAge, 32–47 y ObeseRace/ethnicity: NR | 17% Pro38% Fat45% Carb | 17% Pro38% Fat45% CarbFood not provided | 0 | ↓3 | ↓6 | ↑1 | ↓4 | … | … | No ΔVNR | No ΔVNR | … |
| Arciero et al,96 2013 | 4 | n=10, M and FAge, 46±2 y OverweightRace/ethnicity: NR | Not stated | 35% Pro21% Fat44% CarbFood not provided | ↓1 | … | … | … | … | … | … | No ΔVNR | No ΔVNR | No ΔVNR |
| McGrath and Gibney,91 1994 | Not stated | n=11, MAge: not stated Normal weightRace/ethnicity: NR | 13% Pro39% Fat48% Carb | 14% Pro41% Fat45% CarbFood not provided | 0 | ↓8 | ↓12 | ↑1 | ↓15 | … | … | … | … | … |
| 9 Meals per day |
| Arnold et al,94 1993 | 2 | n=19, M and FNormal weightRace/ethnicity: NR | 15% Pro28% Fat57% Carb | 15% Pro28% Fat57% CarbFood not provided | 0 | ↓10 | ↓14 | ↓3 | ↑1 | No ΔVNR | No ΔVNR | No ΔVNR | No ΔVNR | … |
| Arnold et al,95 1994 | 4 | n=16, M and FBMI: not statedRace/ethnicity: NR | Not stated | 17% Pro35% Fat48% CarbFood not provided | 0 | 0 | ↑6 | ↓1 | ↓21 | No ΔVNR | No ΔVNR | No ΔVNR | No ΔVNR | … |
| 12 Meals per day |
| Murphy et al,93 1996 | 2 | n=11, FBMI: not statedRace/ethnicity: NR | Not stated | 20% Pro40% Fat40% CarbFood provided | 0 | ↓3 | ↓1 | ↓2 | ↑3 | … | … | No ΔVNR | No ΔVNR | … |
| 17 Meals per day |
| Jenkins et al,92 1989 | 2 | n=7, M1–51 yBMI: not statedRace/ethnicity: NR | Not stated | 15% Pro33% Fat52% CarbFood provided | ↓1 | ↓16 | ↓18 | ↓2 | ↓26 | … | … | No ΔVNR | … | … |
Expand Table
BMI indicates body mass index; Carb, percent energy from carbohydrates; CHD, coronary heart disease; DBP, diastolic blood pressure; F, female; Fat, percent energy from fat; HDL, high-density lipoprotein cholesterol; HOMA-IR, homeostatic model assessment of insulin resistance; LDL, low-density lipoprotein cholesterol; M, male; NR, not reported; Pro, percent energy from protein; SBP, systolic blood pressure; TC, total cholesterol; TG, triglycerides; and VNR, value not reported.
Posttreatment value significantly different from baseline value (P<0.05).
Because calorie restriction was not imposed in any of the trials reviewed,90–98 body weight remained unchanged for all studies (Table 2). We intentionally chose not to include meal frequency studies that also applied calorie restriction because the effect of meal frequency on weight status would be difficult to isolate from that of weight loss. One study assessed the impact of meal frequency on energy expenditure in overweight women in a metabolic chamber.99 Eight to 10 women participated in 3 separate studies testing consumption of 2 meals served at 11 am and 7 pm or 6 meals served every 2 hours between 9 am and 7 pm. In 1 study, additional foods could be consumed outside of those times, and in the third study, the 6-meal pattern was compared with 4 meals served at 1, 2, 5, and 7 pm. The third study also allowed additional food intake. No differences were observed between meal patterns in any of the 3 studies on total energy expenditure or energy balance. However, nighttime energy expenditure was higher in the first study when participants were given 2 meals compared with 6 meals. Nevertheless, these studies do not suggest a role of meal frequency on energy metabolism.
One study examined the effects of low versus high eating frequency on appetite (n=12 men and women; age, 18–50 years; BMI ≥18 kg/m 2).100 Participants consumed 3 or 8 equally spaced meals daily for 21 days in a randomized, crossover design. Body weight and waist-to-hip ratio did not change during the study. After the 21-day period, appetite was assessed over a 4-hour window. In the low eating frequency condition, 1 meal was served, whereas in the high eating frequency condition, 2 meals were served. In each condition, participants consumed 33% of their energy requirements over the testing period. Mean hunger ratings were significantly lower in the low compared with the high eating frequency condition, but the area under the curve for the entire testing session did not differ between conditions. The mean and area under the curve over the entire session for desire-to-eat ratings were lower in the low eating frequency condition compared with the high eating frequency condition. The mean composite score for appetite and the area under the curve for the entire testing period were higher in the high eating frequency condition compared with the low eating frequency condition. This small study does not support the conclusion that increasing eating frequency reduces appetite.
The effects of various meal frequency strategies on total and LDL cholesterol are reported in Table 2. The greatest reductions in total cholesterol (16%) and LDL cholesterol (18%) concentrations were observed when normocholesterolemic men consumed 17 meals a day for 2 weeks.92 Consuming 9 meals a day reduced total and LDL cholesterol by 10% and 14%,94 respectively; consuming 6 meals a day lowered total cholesterol by 1% to 8% and LDL cholesterol by 6% to 8%91,97,98; and consuming 3 meals a day resulted in little or no change.90–96 Interestingly, in a controlled feeding study,90 participants who ate all of their daily energy needs in 1 sitting had increases in total and LDL cholesterol levels of 19% and 25%, respectively, after 8 weeks. Taken together, these findings indicate that increasing meal frequency may reduce cholesterol levels, whereas decreasing meal frequency may have unfavorable effects. However, it is important to note that macronutrient distribution did not change from baseline to after treatment in any of the studies reviewed here.90–98 Average dietary fat intake ranged from 30% to 40% of daily energy needs, and in most studies, participants were consuming at least 35% of energy as fat and at least 10% of energy from saturated fat.70,90,91,93,95,97,98 We also observed that diets providing >35% of energy as fat may negate the improvements observed as a result of increased meal frequency. For example, in a study providing 9 meals a day in the context of a low-fat diet (28% of energy as fat),94 total and LDL cholesterol concentrations were reduced by 10% and 14%, respectively. In contrast, when the background diet provided 35% of energy as fat,95 there was no effect of increasing meal frequency to 9 meals a day. Thus, the increased meal frequency pattern may be more effective at lowering total and LDL cholesterol when combined with lower-fat diets (<30% energy as fat). The reason for this is not apparent.
The impact of meal frequency on HDL cholesterol concentrations has also been assessed (Table 2). In a randomized, crossover trial,90 HDL cholesterol was augmented by 17% from baseline when participants ate all of their energy needs as 1 meal a day for 8 weeks but remained unchanged when 3 meals a day were consumed. These modulations in HDL cholesterol appear to be independent of dietary macronutrient composition because dietary cholesterol and fatty acid intakes were held constant. Increasing meal frequency to 6, 9, 12, or 17 meals a day did not affect HDL cholesterol concentrations.91–98 These findings suggest that increasing meal frequency under isocaloric conditions has little effect on HDL cholesterol levels. The reason why increases in HDL cholesterol were observed with a reduction in meal frequency to 1 meal a day remains unclear.90 Additional studies are needed to confirm this observation.
Studies suggest that altering meal frequency has no significant impact on triglyceride concentrations (Table 2).90–98 Although several studies report large numeric reductions in triglyceride levels,91,92,94,95 none of these reductions were significantly different when posttreatment values were compared with baseline values. This lack of a significant difference in triglyceride concentrations is most likely attributable to the large variability in subject responses reported in each trial.91,92,94,95 Moreover, it is worth noting that these studies have not compared the effects of the intervention relative to the control but rather performed within-group comparison.
Only 1 study reported a significant impact of meal frequency on blood pressure (Table 2).90 Results indicate that consuming 1 meal a day increases both systolic and diastolic blood pressures (1% increase from baseline) after 8 weeks of treatment. When these same individuals crossed over to a regimen of 3 meals a day, systolic and diastolic blood pressures decreased by 6% and 4%, respectively.90 In other studies that examined the impact of 3 or 9 meals a day on blood pressure,94,95 no change was noted after 2 to 4 weeks of treatment. Whether blood pressure would be affected with longer treatment duration remains uncertain.
Fasting glucose and insulin concentrations were examined in the majority of studies reviewed here (Table 2). Results from these small, uncontrolled, short-term trials indicate that consuming a few meals per day (ie, 1 or 3 meals a day)90–96 or several meals per day (ie, 6, 9, 12, or 17 meals a day)91–98 has no effect on either fasting glucose or insulin in the absence of weight loss.
Preliminary studies suggest that consuming 3 or 6 meals a day under isocaloric conditions for 2 to 4 weeks has no effect on IR.96,97 These results are not surprising because glucose and insulin did not change during these trials.
These findings should be interpreted with caution, however, because each of these trials has several limitations. First, all of these trials lacked a control group, and very few studies implemented a controlled feeding protocol.70,90–98 Second, sample sizes of the various study arms were quite small (n=7–19),90,92 and the studies may have lacked power to detect significant differences within or between groups. In view of these limitations, larger-scale and longer-term controlled trials will be required before any definitive conclusions can be reached on the impact of meal frequency on these cardiometabolic risk parameters. Further insight into the effects of meal frequency on the long-term risk of clinical events such as CVD or diabetes mellitus may be gained through observational studies in free-living individuals. Moreover, the impact of meal frequency on TEI should be examined because increasing meal frequency may lead to undesirable increases in intakes that could lead to weight gain.
Summary
Altering meal frequency under isocaloric conditions may not be useful for decreasing body weight or improving traditional cardiometabolic risk factors.
Meal Timing and Cardiometabolic Risk: Observational Findings
Late-night eating has been associated with a greater risk of poor cardiometabolic health in several cross-sectional studies. Late-night eating was associated with an OR for obesity of 1.62 (95% CI, 1.10–2.39) compared with no late-night eating among 3610 Swedish men and women.50 In a small, cross-sectional study among 239 US adults, individuals who consumed ≥33% of their TEI in the evening had twice the risk of being obese (OR, 2.00; 95% CI, 1.03–3.89) compared with individuals who consumed <33% of their TEI at night.101 Additionally, the combination of late-night eating and skipping breakfast was associated with a greater risk of having the metabolic syndrome among Japanese adults (n=60 800; age, 20–75 years).102 Compared with individuals with healthy eating patterns, those who ate late at night, defined as eating dinner within 2 hours of bedtime, and skipped breakfast had an OR for the metabolic syndrome of 1.17 (95% CI, 1.08–1.28). The associations of eating frequency and timing with inflammation and IR biomarkers were assessed in female participants in NHANES 2009 to 2010 (n=2212; mean age, 46.8 years).103 Independent variables included eating frequency (number of eating occasions per day), percent of TEI eaten between 5 pm and midnight, and nighttime fasting duration. Each 10% increase in the proportion of TEI consumed in the evening was associated with a 3% increase in C-reactive protein concentrations, whereas having 1 additional eating occasion per day was associated with an 8% decrease in C-reactive protein. There was no relation with nighttime fasting duration. None of the variables were associated with HOMA-IR. Interestingly, favorable effects of nighttime fasting on inflammation and IR were observed only in women who stopped eating before 6 pm, and lengthening nighttime fast duration by skipping breakfast did not have the same favorable effect. An important limitation to consider with these studies is that the definition of late-night eating differs among studies. Furthermore, the causality of these associations between eating patterns and CVD risk factors is unclear because of the cross-sectional nature of the studies.
To date, only 1 prospective, observational, epidemiological study has reported the association between late-night eating and risk of CHD.63 In the Health Professional Follow-Up Study, men were asked to respond to the following question on a biennial questionnaire: “Please indicate the time of day that you usually eat (mark all that apply): before breakfast, breakfast, between breakfast and lunch, lunch, between lunch and dinner, dinner, between dinner and bedtime, after going to bed.” Nighttime eating was defined as a positive response to eating after going to bed. Men with nighttime eating had a relative risk of CVD of 1.55 (95% CI, 1.05–2.29) compared with men who did not eat during the night after adjustment for demographic data, diet, lifestyle, and CHD risk factors. This association was mediated by BMI, hypercholesterolemia, hypertension, and diabetes mellitus. However, this eating behavior may be considered pathological and does not represent the more common late-night eating pattern of food consumption before bedtime. In this population, only 1% of the men reported late-night eating (n=29 cases).
Meal Timing and Cardiometabolic Risk: Clinical Intervention Findings
Very few interventions have focused on meal timing and cardiometabolic risk, and most have been single-day studies with few participants. For example, 1 study tested the impact of a late dinner, given at 11 pm, relative to 6 pm dinner on glucose handling after breakfast the following day.104 On the day before testing, standard meals were served at 7 am, noon, and 6 or 11 pm. Breakfast on test day was served at 8:30 am. Orocecal transit time was longer after breakfast consumption after a late dinner, and unabsorbed dietary carbohydrates were lower compared with the usual dinnertime. In addition, in these 12 young, healthy, normal-weight women, glucose was higher after breakfast in the late dinner condition compared with the usual dinner condition. The authors concluded that eating a late dinner could increase the risk of type 2 diabetes mellitus because high postprandial glucose is a risk factor for this disorder. Conversely, a high-energy breakfast was also tested in women with polycystic ovary syndrome over a 12-week period.105 Young, normal-weight women were randomized to consume an 1800-kcal diet with energy distribution among meals being either 54%, 35%, and 11% or 11%, 35%, and 54% TEI distribution at breakfast, lunch, and dinner, respectively. Women were instructed to consume breakfast between 6 and 9 am, lunch between noon and 3 pm, and dinner between 6 and 9 pm. Compliance was assessed from biweekly 3-day diet records. There was no difference in change in BMI, waist circumference, and adiposity between groups. However, metabolic risk factors changed differently between groups. Women in the group who ate a large breakfast had reductions in fasting plasma glucose and insulin, along with a decrease in HOMA-IR and an increase in the insulin sensitivity index. There was no change in those parameters in women in the group who ate a large dinner. Oral glucose tolerance test glucose and insulin areas under the curve were also reduced in the breakfast group to a greater extent than in the dinner group. The authors concluded that the improvement in IR in the group consuming a large breakfast implies an eating schedule that is synchronized with the circadian pacemaker but that consuming a larger dinner does not worsen IR in these women.
This beneficial effect of a larger breakfast on the cardiometabolic risk profile is not universally observed, however. During Ramadan fasting, individuals eat breakfast 30 minutes before sunrise and eat dinner after sunset. Adult men with the metabolic syndrome were studied during this month, in which 58% of their TEI was consumed in the evening.106 Self-reported TEI was reduced by ≈56 kcal/d, and body weight and waist circumference decreased by 2.4% during the month of Ramadan. Fasting plasma glucose and systolic and diastolic blood pressures decreased, and HDL cholesterol and insulin sensitivity increased, but fasting plasma insulin, triglycerides, and HOMA-IR did not change. The authors concluded that Ramadan fasting can improve some metabolic risk factors, but several limitations in this study are worth noting. There was no control group, and data were not adjusted for change in body weight or waist circumference. Moreover, discordant results were observed on 2 measures of IR. Therefore, a definitive conclusion cannot be drawn.
Other studies have focused on the amount and type of carbohydrates and cardiometabolic risk factors.107,108 In a small study (n=6) of young, normal-weight adults, diets with low (34) versus high (84) glycemic index provided either mostly in the morning (60%, 20%, and 20% TEI at breakfast, lunch, and dinner, respectively) or in the evening (20%, 20%, and 60% TEI at breakfast, lunch, and dinner, respectively) were compared for their short-term effects on glycemia with a 2×2 factorial design.107 Meals were served at 9:30 am, 1:30 pm, and 8:30 pm, and blood samples were collected before and for 2 hours after each meal. Glucose and insulin areas under the curve were highest for the diet with a high glycemic index with the greater percent of energy provided in the evening. This was significant compared with all other diets for glucose and relative to the diets with low glycemic index for insulin. Triglycerides and nonesterified fatty acids were not affected by the glycemic index of the diets or the shift in caloric load between morning and evening. The authors suggested that eating patterns with the largest energy load in the evening may contribute to metabolic syndrome through deterioration of postprandial glucose and insulin and that avoiding large meals with a high glycemic index in the evening could improve glycemic profile.
One longer-term study tested whether changing both the carbohydrate content and the energy content of meals affected weight loss and weight loss maintenance in obese adult men and women.108 Participants were instructed to follow a low-calorie diet with either a low- or high-carbohydrate (10% versus 20% of TEI, respectively) and –energy content breakfast for 16 weeks; follow-up continued after intervention for an additional 16 weeks. The low-carbohydrate, low-energy breakfast diet provided 300 kcal at breakfast and 600 to 700 kcal at dinner, whereas the high-carbohydrate, high-energy breakfast diet provided 600 kcal at breakfast and 300 to 400 kcal at dinner. The higher energy content of breakfast was achieved by including dessert for the high-carbohydrate group (chocolate, ice cream, cookies, cake, etc). During follow-up, participants were encouraged to continue their intervention dietary pattern. Weight loss was similar between groups during the intervention phase, but the high-carbohydrate group continued to lose weight during follow-up, whereas the low-carbohydrate group gained weight. This resulted in significantly different body weights between groups at the end of the follow-up period. Similar results were observed for glucose, insulin, and HOMA-IR. After the intervention, triglycerides were significantly lower and HDL cholesterol was significantly higher in the high-carbohydrate group. After follow-up, those differences remained, and total cholesterol and LDL cholesterol were also reduced in the high-carbohydrate compared with the low-carbohydrate group. In line with the greater weight loss observed in the high-carbohydrate group, craving scores were lower and the ghrelin nadir in response to a breakfast meal challenge decreased to a greater extent in the high-carbohydrate compared with the low-carbohydrate group at follow-up. This study had unusually high weight loss for an outpatient intervention study (≈21 kg), and the level of carbohydrate intake in both groups was extremely low (<100 g, on average).
Data from the above studies suggest that consuming a larger percentage of energy later in the day may lead to an adverse cardiometabolic risk profile, but these studies have major limitations. However, a separate study showed that providing 35 g of available starch from brown beans versus white wheat bread at dinner leads to a 23% lower glucose response and 16% lower insulin response after breakfast the following day.109 Interleukin-6 and interleukin-18 were also lower after breakfast consumed the day after the brown bean evening meal.
Another study tested whether late-night eating, without altering meal frequency, alters energy and glucose metabolism.110 Eleven young, normal-weight women consumed a 200-kcal snack (45% carbohydrates, 50% fat) at 10 am or 11 pm for 13 days in a randomized, crossover study. Body weight did not differ between phases. There were no differences in HDL cholesterol, triglycerides, nonesterified fatty acids, adiponectin, leptin, and oral glucose tolerance test glucose and insulin areas under the curve between snack patterns. Despite these results, the authors concluded that having late snacks could increase the risk of obesity and CVD. However, given the lack of significant difference between groups in metabolic risk factors, the data do not fully support this conclusion.
More extreme patterns of eating during nighttime hours were tested in a shift-work paradigm.111,112 In 1 study, 11 female nurses were tested over 3 days of night-shift work.111 Glycemic response to standard meals containing 440 kcal provided at 7:30 pm, 11:30 pm, and 3:30 am was assessed over 4 hours. Basal glucose was similar before each meal, but postprandial concentrations were higher after the meal consumed at 11:30 pm. Basal insulin was highest before the 11:30 pm meal and lowest at 3:30 am; postprandial values followed the same pattern. The authors concluded that glucose tolerance was lower around midnight, as the circadian oscillations would predict, and proposed that the rise in CHD risk in night workers could be attributable to deregulated coupling of food intake to the circadian system. Furthermore, they posited that food quality and timing could play a role in metabolic responses to meals. The other study was performed in 8 young, lean men.112 Participants were randomized to simulated day or night work for a single shift in a crossover design. The day shift occurred at noon to 8 pm with meals provided at 1 pm and 7 pm and a snack at 4 pm; the night shift was performed at midnight to 8 am with meals at 1 and 7 am and a snack given at 4 am. Diets were identical in both phases. Glucose and triglyceride concentrations were higher during the night shift compared with the day shift. The authors reported relative lipid intolerance throughout the night shift with metabolic recovery of glucose tolerance toward the end of the night shift and suggested that restricting fat intake throughout the night would be beneficial. However, this was not directly tested in this study.
Even less extreme shifts in food intake could influence cardiometabolic risk status. A 2-week intervention study tested the effects of early (1 pm) or late (4:30 pm) lunch on glucose tolerance in young, normal-weight women (n=32; mean age, 24 years; BMI, 22.9 kg/m 2).113 Postprandial glucose area under the curve was increased by 46% relative to the early lunch condition. In addition, consumption of a late lunch blunted the daily cortisol profile. The authors concluded that delaying the timing of an identical meal for 1 week reduced glucose tolerance. However, because breakfast consumption occurred at a fixed time, 8 am throughout the study, these results may be attributable to the longer intermeal interval in the late lunch condition relative to early lunch. It is unknown whether similar results would be observed if meals were consumed at similar intervals from one another, and the contributions of fasting duration and circadian cycle to these markers could not be distinguished.
Given the seemingly negative health effects of eating late during the day, studies have examined whether restricting food intake to earlier daytime hours would affect cardiometabolic health markers. In 1 study, men were asked not to eat between 7 pm and 6 am daily for 2 weeks or to continue their regular eating pattern (n=27; mean age, 20.9 years; BMI, 24.4 kg/m 2).114 TEI was reduced when night eating was not permitted, resulting in a 0.4-kg weight loss compared with a 0.6-kg weight gain during the control period. Similarly, in a small pilot study (n=8) in which food intake was limited to a self-determined 10- to 12-hour window during the day, body weight decreased by 3.3 kg over 16 weeks.2 This study did not include a control group. Nevertheless, both studies suggested that restricting the period of time during which food consumption occurs could be a promising method for weight management.2,114
Summary
The impact of meal timing, particularly related to the evening meal, deserves further study. Epidemiological findings suggest a potential detrimental effect of late meals on cardiometabolic health, but clinical intervention studies, which would address causality, have been limited in scope and too diverse to draw definitive conclusions and make recommendations. Moreover, the potential benefit of increased meal frequency should be evaluated in the context of timing and duration of the daily prandial period.
Research Gaps in Defining Meals and Eating Occasions
Studies have suggested that eating late in the day or skipping breakfast may be associated with weight gain and obesity, whereas intermittent fasting and high meal frequency may have beneficial cardiometabolic health effects.48,55 However, research in this field has been impeded by the lack of consensus on the definition of a meal, snack, and meal timing. Two common definitions have been identified to distinguish between meals and snacks based on (1) participant identification of a meal, labeled as breakfast, brunch, lunch, dinner, and supper, and snack, labeled as snack, morning or afternoon tea, or beverage break, and (2) time-of-day reports, with a meal being the largest eating occasion within each time period: 6 to 10 am, noon to 3 pm, and 7 to 9 pm, with all other eating occasions being labeled as snacks.115 Another definition that has been used to distinguish between meals and snacks is the contribution of the eating occasion to the TEI, with a meal providing ≥15% of TEI, regardless of the time of day or type of food or beverage consumed.116 Eating occasions contributing to <15% of TEI would be labeled as snacks. This cut point to distinguish between meals and snacks was based on US national data of usual energy intakes from self-defined meals and snacks. Another definition used by these investigators is based on time of day: A meal is any eating occasion that occurs between 6 and 10 am, noon and 3 pm, and 6 and 9 pm; snacks are eating occasions that occur outside of these time periods.116 This is distinct from the previous definition (No. 2), in which only 1 meal, defined as the largest eating occasion, can occur within each time period.
There are concerns about using time of day as a means of defining meals and snacks because this is restrictive to specific cultures and may not be appropriate for some population subgroups such as night or shift workers. For example, 1 definition uses the time period between 6 and 9 pm to categorize evening meals in a British population.116 Meanwhile, median time of the dinner meal in a Spanish population was reported to be 9:30 pm.117 Therefore, a socially and cross-culturally appropriate definition for meals and snacks should not contain predefined time periods to designate between these eating occasions. The use of eating occasion, which encompasses meals and snacks, is a preferable term. Several definitions of eating occasions have been put forth. For example, eating occasions could be separate occurrences if they are at least 15, 30, or 60 minutes apart or have a minimum energy content of 210 kJ and are 15, 30, or 60 minutes apart.115 Another definition is one in which at least 210 kJ is consumed with distinct eating episodes ≥45 minutes apart.118 When 6 definitions of eating occasions based on time between occurrences of 15, 30, and 60 minutes and inclusion of, or no regard to, minimum energy provision of 210 kJ were compared, the definition of eating occasions as eating episodes providing at least 210 kJ occurring 15 minutes apart best predicted the variance in TEI.115
If meals and snacks must be distinguished, then a definition based on the contribution of the eating occasion to TEI may be used. When meals are defined as eating occasions providing ≥15% of TEI, meal frequency has been positively associated with the Mediterranean Diet Score, an index of the healthfulness of a diet, in both men and women, but when defined by time periods, this was significant only in women.116 Snacking, on the other hand, was associated with unfavorable dietary intake patterns for both men and women, regardless of the definition used.
On the basis of the current information, we propose that eating occasions be defined as any eating/drinking episode providing at least 210 kJ and that 15 minutes should be the minimum amount of time elapsed between separate occasions. Distinguishing between meals and snacks should be left to the participant’s discretion. This will provide a definition that accommodates different social norms and cultural behaviors. However, we understand that information on participant-based categorization of meals and snacks is not always collected in surveys and various studies. In such instances, a definition of breakfast could be adapted and generalized to meals.9 Breakfast has been defined as the first meal of the day that breaks the fast after the longest period of sleep, occurs within 2 to 3 hours of awakening, and contains foods and beverages from at least 1 food group.9 A high-quality breakfast would provide 15% to 25% of total energy requirements with 10% of the daily value for as many nutrients as possible and 20% for nutrients of concern, identified as calcium, potassium, vitamin D, and fiber. With this scenario, we propose that an established definition of breakfast9 be adopted and all other meals be defined as eating occasions providing ≥15% of TEI. Eating occasions providing <15% of TEI would be categorized as snacks.
Clinical Implications
The data reviewed here suggest that even when considering a wide range of definitions for meals and snacks, irregular patterns of TEI appear less favorable for the maintenance of body weight and optimal cardiometabolic health. Ultimately, clinicians may be able to use this information to suggest to patients that a more intentional approach to eating that focuses on the timing and frequency of meals and snacks could be the basis of a healthier lifestyle and improved risk factor management. An intentional approach to eating requires eating at planned intervals to distribute TEI throughout the day (Table 3). This may be challenging for many because time constraints limit meal planning and preparation, leading to increased use of convenience food items (eg, fast food, vending machines) and haphazard eating supported by an environment with readily accessible food options.119–121 What results is a poor-quality diet with large portions that are often energy dense but nutrient poor.
Although more direct translational research is still needed, these data suggest that intervening on meal timing and frequency may be beneficial. By focusing on meal frequency and timing as an intervention target, patients may directly address poor dietary quality without the need to deal with calorie restriction to promote weight loss. Ultimately, the clinician’s goal may be to help the patient spread energy intake over a defined portion of the day in a more balanced way rather than limited to 1 segment of the day or continuously over long periods of time (ie, grazing). This does not mean that TEI and macronutrient balance can be ignored but simply that the frequency and timing of intake are the basis for building the structure for intentional eating.
One of the challenges of translating the findings reviewed here into clinical practice is the limited value of some of the definitions of meals and snacks. In Research Gaps in Defining Meals and Eating Occasions, we review and propose definitions for meals and snacks to be used for research purposes. However, from a practical perspective, people rarely think of meals or snacks as being defined by the amount of calories provided within the foods they consume. Studies show that cues such as time of day, portion size, food type, and preparation are more likely to be the bases for categorizing an eating episode as a meal or a snack.122,123 Furthermore, most people do not consider energy-containing beverages, with the exception of meal replacements, as stand-alone snacks or meals, although they would meet our proposed definition of such because of their energy content.124 Consequently, it may be best to frame discussions with patients by using language and references that the individual patient can identify with when counseling on the timing and frequency of eating episodes.
Another major challenge of addressing meal timing and frequency that is highlighted in the research reviewed here is the understanding that eating occasions are influenced by previous eating episodes, and thus, making adjustments in 1 period of the day or meal/snack timing has an influence on other opportunities for consumption. Meal timing and frequency may be as much about the amount of time between eating episodes as about the types and amounts of energy consumed at each eating episode. Many of the studies reviewed included controlled calorie conditions without opportunities for ad libitum intake. When calorie intake is not controlled, it is plausible to hypothesize that a longer interval between eating episodes leads to greater consumption at the next available eating opportunity. However, it is clear from the data that longer periods of time between eating episodes during the day may have different implications from longer overnight fast periods.
Although most of the research reviewed in this document was not in the setting of weight reduction interventions, given that 1 in 3 adults in the United States is obese, understanding how meal frequency and timing influence energy balance may be helpful clinically.125 Intermittent fasting approaches appear to be feasible to help patients lower calorie intake consistently. More data are needed on longer-term outcomes for weight reduction and risk factor modification. Additionally, the use of intermittent fasting for weight loss maintenance is a strategy that should be tested empirically. In the long run, approaches such as intermittent fasting may simply provide another way to help patients be more intentional about consuming energy within a given time period.
Focusing on meal timing and frequency as a starting point to address obesity and positive energy balance appears to be beneficial for several reasons on the basis of the data reviewed here. First, the data suggest that making changes that promote more regular intake of energy during the day, perhaps with a greater proportion of calories earlier in the day, has positive effects on risk factors for heart disease and diabetes mellitus. Several of the studies suggest that this may also have positive effects on body weight, but randomized, controlled trials are needed to test this hypothesis. Second, this approach may be a better way to focus on improving dietary quality because added eating episodes can be used to introduce a wider variety of healthful food options. Finally, if and when the individual is ready to reduce calories, having an intake structure on which to superimpose calorie restriction may improve the patient’s ability to adhere to the weight loss plan. Given the fact that eating episodes are influenced by each other, portion control may be achieved more easily in the context of planned meals and snacks timed throughout the day to help manage hunger.
Eating frequency may also be influenced by eating speed, which was not evaluated in the present document. A meta-analysis on this topic has reported a small to medium effect of eating rate on energy intakes at a meal, with lower intakes when eating speed is slow.126 The difference in energy intakes was proportional to the difference in eating speed. However, there was no effect of eating rate on hunger after the meal and no difference in hunger ratings between meals of fixed content eaten at different speeds. In a fixed-meal study, meal duration (30 versus 10 min) had no effect on appetite and hunger hormones or feelings of hunger or fullness after the meal.127 Intakes throughout the rest of the day also were not different between eating speed conditions. Given that high heterogeneity between studies has been reported126 and that no studies have assessed long-term effects of eating speed on cardiometabolic risk factors, additional studies are needed in this area.
Recommendations for Special Populations
Although the research reviewed here generally supports an intentional eating approach with planned distribution of daily TEI for both weight management and cardiometabolic health, this body of work is limited in its investigation of vulnerable populations. There are established racial/ethnic disparities in the prevalence of obesity128 and CVD.129,130 With respect to obesity, black women, Mexican American women, Native Americans, and Pacific Islanders are disproportionately affected.128 Similarly, death rates from heart disease are up to 3 times greater for blacks in the United States,129,131 and hypertension among blacks is among the highest worldwide.129 Despite the increased risk among these groups, the majority of studies available for review here did not include these populations or failed to report the racial/ethnic distribution of their samples. With known cultural variations in dietary intake, including TEI, meal frequency, and meal timing, adequate representation of racial/ethnic groups at higher risk is needed to increase confidence in generalizing findings from earlier studies. Furthermore, future research should include racially/ethnically diverse samples, a full description of the sample in publications, and subgroup analysis (as appropriate) and should specifically discuss limitations of findings relative to the level of diversity in their sample.132
There has also been limited published research investigating meal patterns and frequency among children and adolescents and body weight and other metabolic risk factors, although there has been tremendous focus on childhood obesity and early predictors of CVD risk.133 Evidence from this scant body of literature primarily mirrors findings in adult populations. Namely, there is support for regular and more frequent main meals to lower obesity risk134,135 among children and adolescents. Youths with less frequent eating are more likely to have greater body weight, even after adjustment for other confounders.134 Likewise, skipping meals is associated with higher metabolic risk (eg, higher BMI, waist circumference, fasting serum insulin, and fasting plasma glucose; decreased fasting plasma HDL).134,136 Again, more research with diverse samples of children and adolescents is needed to increase confidence in the findings of the aforementioned studies.
Finally, there is a critical need to better understand the implications of dietary patterns and frequency on CVD risk among older adults. In 2014, 46.2 million individuals ≥65 years of age were living in the United States (14.5% of the population). Over the next 25 years, it is expected that older Americans will grow to 22% of the US population. The nutritional needs of the elderly are complicated by the need to moderate caloric intake for weight management and CVD prevention while balancing the need to address nutritional deficiencies associated with diets of older adults.137 For example, a recent research review explored the potential risks and benefits of calorie restriction among the elderly and urged caution as a result of mixed results from clinically based efficacy trials and a lack of effectiveness and community-based studies.138 Most CVD prevention research has focused on younger adults; however, with the growing number of older adults and increases in the average life expectancy, more research on the nutrition behaviors of this group is warranted.
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Table 3. Intentional Approach to Eating
Develop an intentional approach to eating that focuses on the timing and frequency of meals and snacks as the basis of a healthier lifestyle and improved risk factor management
Understand the patient’s frame of reference in how he or she may define meals and snacks
Recommend distributing calories over a defined portion of the day
Recommend eating a greater share of the total calorie intake earlier in the day to have positive effects on risk factors for heart disease and diabetes mellitus
Promote consistent overnight fast periods
Link eating episodes to influence subsequent energy intake (eg, place snacks strategically before meals that might be associated with overeating)
Include intermittent fasting approaches as an option to help lower calorie intake and to reduce body weight
Use added eating episodes to introduce a wider variety of healthful food options and to displace less healthful foods
Use planned meals and snacks timed throughout the day to help manage hunger and to achieve portion control
Expand Table
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b [...] a 0.75-kg/wk reduction in body weight,
c [...] Heilbronn et al,
d [...] the intermittent-fasting dietary pattern.
e [...] of 40% to 50% of baseline energy needs.
f [...] For instance, in 1 study,
g [...] decrease in fasting insulin was observed.
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b [...] a 0.75-kg/wk reduction in body weight,
c [...] Eshghinia and Mohammadzadeh,
d [...] others report no effect.
e [...] of weight loss is similar between studies.
f [...] studies that achieved 6% to 7% weight loss.
g [...] 6% to 10% after 6 to 24 weeks of treatment.
h [...] Participants in these studies
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b [...] during an alternate-day fasting protocol.
c [...] a 0.75-kg/wk reduction in body weight,
d [...] Johnson et al,
e [...] and LDL cholesterol ranging from 7% to 32%,
f [...] of weight loss is similar between studies.
g [...] (ie, LDL cholesterol >110 mg/dL).
h [...] triglycerides decreased by ≈30% to 40%,
i [...] the intermittent-fasting dietary pattern.
j [...] of 40% to 50% of baseline energy needs.
k [...] underwent 40% average daily restriction,
l [...] IR were quantified by the HOMA-IR method.
m [...] example, in 1 alternate-day fasting study,
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b [...] studies that provided food on the fast day.
c [...] a 0.75-kg/wk reduction in body weight,
d [...] Varady et al,
e [...] and LDL cholesterol ranging from 7% to 32%,
f [...] of weight loss is similar between studies.
g [...] (ie, LDL cholesterol >110 mg/dL).
h [...] triglycerides decreased by ≈30% to 40%,
i [...] studies that achieved 6% to 7% weight loss.
j [...] 6% to 10% after 6 to 24 weeks of treatment.
k [...] Participants in these studies
l [...] in fasting glucose ranging from 3% to 6%.
m [...] 8-week studies of alternate-day fasting
n [...] the intermittent-fasting dietary pattern.
o [...] IR were quantified by the HOMA-IR method.
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Klempel MC, Kroeger CM, Varady KA. Alternate day fasting (ADF) with a high-fat diet produces similar weight loss and cardio-protection as ADF with a low-fat diet. Metabolism. 2013;62:137–143. doi: 10.1016/j.metabol.2012.07.002.
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b [...] studies that provided food on the fast day.
c [...] a 0.75-kg/wk reduction in body weight,
d [...] Klempel et al,
e [...] and LDL cholesterol ranging from 7% to 32%,
f [...] of weight loss is similar between studies.
g [...] (ie, LDL cholesterol >110 mg/dL).
h [...] achieved 0.25- to 0.5-kg/wk weight loss,
85.
Hoddy KK, Kroeger CM, Trepanowski JF, Barnosky A, Bhutani S, Varady KA. Meal timing during alternate day fasting: Impact on body weight and cardiovascular disease risk in obese adults [published correction appears in Obesity (Silver Spring). 2015;23:914]. Obesity (Silver Spring). 2014;22:2524–2531. doi: 10.1002/oby.20909.
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b [...] studies that provided food on the fast day.
c [...] a 0.75-kg/wk reduction in body weight,
d [...] Hoddy et al,
e [...] others report no effect.
f [...] of weight loss is similar between studies.
g [...] the intermittent-fasting dietary pattern.
86.
Bhutani S, Klempel MC, Kroeger CM, Trepanowski JF, Varady KA. Alternate day fasting and endurance exercise combine to reduce body weight and favorably alter plasma lipids in obese humans. Obesity (Silver Spring). 2013;21:1370–1379. doi: 10.1002/oby.20353.
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b [...] studies that provided food on the fast day.
c [...] a 0.75-kg/wk reduction in body weight,
d [...] Bhutani et al,
e [...] others report no effect.
f [...] of weight loss is similar between studies.
g [...] the intermittent-fasting dietary pattern.
h [...] IR were quantified by the HOMA-IR method.
i [...] in body weight was half as large (4%),
87.
Varady KA, Bhutani S, Klempel MC, Kroeger CM, Trepanowski JF, Haus JM, Hoddy KK, Calvo Y. Alternate day fasting for weight loss in normal weight and overweight subjects: a randomized controlled trial. Nutr J. 2013;12:146. doi: 10.1186/1475-2891-12-146.
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b [...] studies that provided food on the fast day.
c [...] a 0.75-kg/wk reduction in body weight,
d [...] Varady et al,
e [...] and LDL cholesterol ranging from 7% to 32%,
f [...] of weight loss is similar between studies.
g [...] (ie, LDL cholesterol >110 mg/dL).
h [...] achieved 0.25- to 0.5-kg/wk weight loss,
i [...] studies that achieved 6% to 7% weight loss.
j [...] 6% to 10% after 6 to 24 weeks of treatment.
k [...] Participants in these studies
l [...] in fasting glucose ranging from 3% to 6%.
m [...] versus a no-intervention control group.
n [...] group relative to the control group.
o [...] the intermittent-fasting dietary pattern.
p [...] IR were quantified by the HOMA-IR method.
88.
Klempel MC, Kroeger CM, Bhutani S, Trepanowski JF, Varady KA. Intermittent fasting combined with calorie restriction is effective for weight loss and cardio-protection in obese women. Nutr J. 2012;11:98. doi: 10.1186/1475-2891-11-98.
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b [...] studies that provided food on the fast day.
c [...] a 0.25-kg/wk reduction in body weight.
d [...] Klempel et al,
e [...] and LDL cholesterol ranging from 7% to 32%,
f [...] of weight loss is similar between studies.
g [...] (ie, LDL cholesterol >110 mg/dL).
h [...] achieved 0.25- to 0.5-kg/wk weight loss,
i [...] in fasting glucose ranging from 3% to 6%.
j [...] and periodic fasting.
k [...] the intermittent-fasting dietary pattern.
l [...] IR were quantified by the HOMA-IR method.
m [...] by 23% after 8 weeks of fasting 1 d/wk.
89.
Harvie MN, Pegington M, Mattson MP, Frystyk J, Dillon B, Evans G, Cuzick J, Jebb SA, Martin B, Cutler RG, Son TG, Maudsley S, Carlson OD, Egan JM, Flyvbjerg A, Howell A. The effects of intermittent or continuous energy restriction on weight loss and metabolic disease risk markers: a randomized trial in young overweight women. Int J Obes (Lond). 2011;35:714–727. doi: 10.1038/ijo.2010.171.
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b [...] a 0.25-kg/wk reduction in body weight.
c [...] Harvie et al,
d [...] and LDL cholesterol ranging from 7% to 32%,
e [...] of weight loss is similar between studies.
f [...] (ie, LDL cholesterol >110 mg/dL).
g [...] achieved 0.25- to 0.5-kg/wk weight loss,
h [...] studies that achieved 6% to 7% weight loss.
i [...] 6% to 10% after 6 to 24 weeks of treatment.
j [...] Participants in these studies
k [...] the intermittent-fasting dietary pattern.
l [...] IR were quantified by the HOMA-IR method.
m [...] with 24 weeks of fasting on 2 d/wk.
90.
Stote KS, Baer DJ, Spears K, Paul DR, Harris GK, Rumpler WV, Strycula P, Najjar SS, Ferrucci L, Ingram DK, Longo DL, Mattson MP. A controlled trial of reduced meal frequency without caloric restriction in healthy, normal-weight, middle-aged adults. Am J Clin Nutr. 2007;85:981–988.
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b [...] Stote et al,
c [...] Stote et al,
d [...] not imposed in any of the trials reviewed,
e [...] a day resulted in little or no change.
f [...] in a controlled feeding study,
g [...] in any of the studies reviewed here.
h [...] at least 10% of energy from saturated fat.
i [...] ). In a randomized, crossover trial,
j [...] frequency to 1 meal a day remains unclear.
k [...] Table 2).
l [...] Table 2).
m [...] decreased by 6% and 4%, respectively.
n [...] few meals per day (ie, 1 or 3 meals a day)
o [...] implemented a controlled feeding protocol.
p [...] study arms were quite small (n=7–19),
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c [...] McGrath and Gibney,
d [...] not imposed in any of the trials reviewed,
e [...] by 1% to 8% and LDL cholesterol by 6% to 8%
f [...] a day resulted in little or no change.
g [...] in any of the studies reviewed here.
h [...] at least 10% of energy from saturated fat.
i [...] not affect HDL cholesterol concentrations.
j [...] Table 2).
k [...] numeric reductions in triglyceride levels,
l [...] subject responses reported in each trial.
m [...] few meals per day (ie, 1 or 3 meals a day)
n [...] per day (ie, 6, 9, 12, or 17 meals a day)
o [...] implemented a controlled feeding protocol.
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b [...] Jenkins et al,
c [...] Jenkins et al,
d [...] not imposed in any of the trials reviewed,
e [...] men consumed 17 meals a day for 2 weeks.
f [...] a day resulted in little or no change.
g [...] in any of the studies reviewed here.
h [...] not affect HDL cholesterol concentrations.
i [...] Table 2).
j [...] numeric reductions in triglyceride levels,
k [...] subject responses reported in each trial.
l [...] few meals per day (ie, 1 or 3 meals a day)
m [...] per day (ie, 6, 9, 12, or 17 meals a day)
n [...] implemented a controlled feeding protocol.
o [...] study arms were quite small (n=7–19),
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Circulation
Volume 135 • Number 9 • 28 February 2017
Pages: e96 - e121
PubMed: 28137935
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Keywords
AHA Scientific Statements
cardiovascular diseases
lipids
meals
obesity
prevention and control
risk factors
Subjects
Statements and Guidelines
Authors
Affiliations Expand All
Marie-Pierre St-Onge, PhD, FAHA, Chair
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Jamy Ard, MD
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Monica L.Baskin, PhD
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Stephanie E.Chiuve, ScD
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Heather M.Johnson, MD, FAHA
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Penny Kris-Etherton, PhD, RD, FAHA
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Krista Varady, PhD
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On behalf of the American Heart Association Obesity Committee of the Council on Lifestyle and Cardiometabolic Health; Council on Cardiovascular Disease in the Young; Council on Clinical Cardiology; and Stroke Council
Notes
The American Heart Association makes every effort to avoid any actual or potential conflicts of interest that may arise as a result of an outside relationship or a personal, professional, or business interest of a member of the writing panel. Specifically, all members of the writing group are required to complete and submit a Disclosure Questionnaire showing all such relationships that might be perceived as real or potential conflicts of interest.
This statement was approved by the American Heart Association Science Advisory and Coordinating Committee on September 7, 2016, and the American Heart Association Executive Committee on October 25, 2016. A copy of the document is available at by using either “Search for Guidelines & Statements” or the “Browse by Topic” area. To purchase additional reprints, call 843-216-2533 or e-mail kelle.ramsay@wolterskluwer.com.
The American Heart Association requests that this document be cited as follows: St-Onge M-P, Ard J, Baskin ML, Chiuve SE, Johnson HM, Kris-Etherton P, Varady K; on behalf of the American Heart Association Obesity Committee of the Council on Lifestyle and Cardiometabolic Health; Council on Cardiovascular Disease in the Young; Council on Clinical Cardiology; and Stroke Council. Meal timing and frequency: implications for cardiovascular disease prevention: a scientific statement from the American Heart Association. Circulation. 2017;135:e96–e121. doi: 10.1161/CIR.0000000000000476.
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Disclosures
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Writing Group Disclosures
| Writing Group Member | Employment | Research Grant | Other Research Support | Speakers’ Bureau/Honoraria | Expert Witness | Ownership Interest | Consultant/Advisory Board | Other |
--- --- --- ---
| Marie-Pierre St-Onge | Columbia University | NIH† | None | None | None | None | None | None |
| Jamy Ard | Wake Forest Baptist Medical Center | None | None | None | None | None | None | None |
| Monica L. Baskin | University of Alabama at Birmingham | None | None | None | None | None | None | None |
| Stephanie E. Chiuve | Brigham and Women’s Hospital | None | None | None | None | None | None | None |
| Heather M. Johnson | University of Wisconsin School of Medicine and Public Health | None | None | None | None | None | None | None |
| Penny Kris-Etherton | Pennsylvania State University | None | None | None | None | None | None | None |
| Krista Varady | University of Illinois at Chicago | University of Illinois at Chicago, Campus Research Board grant; American Diabetes Association grants (1-16-ICTS-114, 1-16-ICTS-022) | None | North American Menopause Society; American Institute Cancer Research | None | None | Nestle | None |
This table represents the relationships of writing group members that may be perceived as actual or reasonably perceived conflicts of interest as reported on the Disclosure Questionnaire, which all members of the writing group are required to complete and submit. A relationship is considered to be “significant” if (a) the person receives $10 000 or more during any 12-month period, or 5% or more of the person’s gross income; or (b) the person owns 5% or more of the voting stock or share of the entity, or owns $10 000 or more of the fair market value of the entity. A relationship is considered to be “modest” if it is less than “significant” under the preceding definition.
Modest.
†
Significant.
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Reviewer Disclosures
| Reviewer | Employment | Research Grant | Other Research Support | Speakers’ Bureau/Honoraria | Expert Witness | Ownership Interest | Consultant/Advisory Board | Other |
--- --- --- ---
| Marco Bertolotti | Universita degli Studi di Modena e Reggio Emilia (Italy) | None | None | None | None | None | None | None |
| Richard Milani | Ochsner Clinic Foundation | None | None | None | None | None | None | None |
| Paul Nestel | Baker IDI Heart & Diabetes Institute (Australia) | None | None | None | None | None | None | None |
This table represents the relationships of reviewers that may be perceived as actual or reasonably perceived conflicts of interest as reported on the Disclosure Questionnaire, which all reviewers are required to complete and submit. A relationship is considered to be “significant” if (a) the person receives $10 000 or more during any 12-month period, or 5% or more of the person’s gross income; or (b) the person owns 5% or more of the voting stock or share of the entity, or owns $10 000 or more of the fair market value of the entity. A relationship is considered to be “modest” if it is less than “significant” under the preceding definition.
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Vassilios S Vassiliou,
Demosthenes Panagiotakos,
Diet and nutrition in cardiovascular disease prevention: a scientific statement of the European Association of Preventive Cardiology and the Association of Cardiovascular Nursing & Allied Professions of the European Society of Cardiology, European Journal of Preventive Cardiology, (2025).
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Isabella Mendes,
Josefina Bressan,
Arieta Carla Gualandi Leal,
Andreia Queiroz Ribeiro,
Adriano Marçal Pimenta,
Helen Hermana Miranda Hermsdorff,
Interaction between skipping breakfast and depression on the incidence of metabolic syndrome among Brazilian adult graduates: six-year follow-up of the Cohort of Universities of Minas Gerais (CUME Study), Nutrition, Metabolism and Cardiovascular Diseases, (104202), (2025).
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Table 1. Intermittent Fasting: Effect on CHD Risk Parameters
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Table 2. Meal Frequency Regimens Without Calorie Restriction: Effect on CHD Risk Parameters
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Table 3. Intentional Approach to Eating
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b [...] a 0.75-kg/wk reduction in body weight,
c [...] Heilbronn et al,
d [...] the intermittent-fasting dietary pattern.
e [...] of 40% to 50% of baseline energy needs.
f [...] For instance, in 1 study,
g [...] decrease in fasting insulin was observed.
81.
Eshghinia S, Mohammadzadeh F. The effects of modified alternate-day fasting diet on weight loss and CAD risk factors in overweight and obese women. J Diabetes Metab Disord. 2013;12:4. doi: 10.1186/2251-6581-12-4.
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a [...] 3% to 8% after 3 to 24 weeks of treatment.
b [...] a 0.75-kg/wk reduction in body weight,
c [...] Eshghinia and Mohammadzadeh,
d [...] others report no effect.
e [...] of weight loss is similar between studies.
f [...] studies that achieved 6% to 7% weight loss.
g [...] 6% to 10% after 6 to 24 weeks of treatment.
h [...] Participants in these studies
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Johnson JB, Summer W, Cutler RG, Martin B, Hyun DH, Dixit VD, Pearson M, Nassar M, Telljohann R, Maudsley S, Carlson O, John S, Laub DR, Mattson MP. Alternate day calorie restriction improves clinical findings and reduces markers of oxidative stress and inflammation in overweight adults with moderate asthma [published correction appears in Free Radic Biol Med. 2007;43:1348]. Free Radic Biol Med. 2007;42:665–674. doi: 10.1016/j.freeradbiomed.2006.12.005.
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a [...] 3% to 8% after 3 to 24 weeks of treatment.
b [...] during an alternate-day fasting protocol.
c [...] a 0.75-kg/wk reduction in body weight,
d [...] Johnson et al,
e [...] and LDL cholesterol ranging from 7% to 32%,
f [...] of weight loss is similar between studies.
g [...] (ie, LDL cholesterol >110 mg/dL).
h [...] triglycerides decreased by ≈30% to 40%,
i [...] the intermittent-fasting dietary pattern.
j [...] of 40% to 50% of baseline energy needs.
k [...] underwent 40% average daily restriction,
l [...] IR were quantified by the HOMA-IR method.
m [...] example, in 1 alternate-day fasting study,
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Varady KA, Bhutani S, Church EC, Klempel MC. Short-term modified alternate-day fasting: a novel dietary strategy for weight loss and cardioprotection in obese adults. Am J Clin Nutr. 2009;90:1138–1143. doi: 10.3945/ajcn.2009.28380.
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a [...] 3% to 8% after 3 to 24 weeks of treatment.
b [...] studies that provided food on the fast day.
c [...] a 0.75-kg/wk reduction in body weight,
d [...] Varady et al,
e [...] and LDL cholesterol ranging from 7% to 32%,
f [...] of weight loss is similar between studies.
g [...] (ie, LDL cholesterol >110 mg/dL).
h [...] triglycerides decreased by ≈30% to 40%,
i [...] studies that achieved 6% to 7% weight loss.
j [...] 6% to 10% after 6 to 24 weeks of treatment.
k [...] Participants in these studies
l [...] in fasting glucose ranging from 3% to 6%.
m [...] 8-week studies of alternate-day fasting
n [...] the intermittent-fasting dietary pattern.
o [...] IR were quantified by the HOMA-IR method.
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Klempel MC, Kroeger CM, Varady KA. Alternate day fasting (ADF) with a high-fat diet produces similar weight loss and cardio-protection as ADF with a low-fat diet. Metabolism. 2013;62:137–143. doi: 10.1016/j.metabol.2012.07.002.
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a [...] 3% to 8% after 3 to 24 weeks of treatment.
b [...] studies that provided food on the fast day.
c [...] a 0.75-kg/wk reduction in body weight,
d [...] Klempel et al,
e [...] and LDL cholesterol ranging from 7% to 32%,
f [...] of weight loss is similar between studies.
g [...] (ie, LDL cholesterol >110 mg/dL).
h [...] achieved 0.25- to 0.5-kg/wk weight loss,
85.
Hoddy KK, Kroeger CM, Trepanowski JF, Barnosky A, Bhutani S, Varady KA. Meal timing during alternate day fasting: Impact on body weight and cardiovascular disease risk in obese adults [published correction appears in Obesity (Silver Spring). 2015;23:914]. Obesity (Silver Spring). 2014;22:2524–2531. doi: 10.1002/oby.20909.
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a [...] 3% to 8% after 3 to 24 weeks of treatment.
b [...] studies that provided food on the fast day.
c [...] a 0.75-kg/wk reduction in body weight,
d [...] Hoddy et al,
e [...] others report no effect.
f [...] of weight loss is similar between studies.
g [...] the intermittent-fasting dietary pattern.
86.
Bhutani S, Klempel MC, Kroeger CM, Trepanowski JF, Varady KA. Alternate day fasting and endurance exercise combine to reduce body weight and favorably alter plasma lipids in obese humans. Obesity (Silver Spring). 2013;21:1370–1379. doi: 10.1002/oby.20353.
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b [...] studies that provided food on the fast day.
c [...] a 0.75-kg/wk reduction in body weight,
d [...] Bhutani et al,
e [...] others report no effect.
f [...] of weight loss is similar between studies.
g [...] the intermittent-fasting dietary pattern.
h [...] IR were quantified by the HOMA-IR method.
i [...] in body weight was half as large (4%),
87.
Varady KA, Bhutani S, Klempel MC, Kroeger CM, Trepanowski JF, Haus JM, Hoddy KK, Calvo Y. Alternate day fasting for weight loss in normal weight and overweight subjects: a randomized controlled trial. Nutr J. 2013;12:146. doi: 10.1186/1475-2891-12-146.
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b [...] studies that provided food on the fast day.
c [...] a 0.75-kg/wk reduction in body weight,
d [...] Varady et al,
e [...] and LDL cholesterol ranging from 7% to 32%,
f [...] of weight loss is similar between studies.
g [...] (ie, LDL cholesterol >110 mg/dL).
h [...] achieved 0.25- to 0.5-kg/wk weight loss,
i [...] studies that achieved 6% to 7% weight loss.
j [...] 6% to 10% after 6 to 24 weeks of treatment.
k [...] Participants in these studies
l [...] in fasting glucose ranging from 3% to 6%.
m [...] versus a no-intervention control group.
n [...] group relative to the control group.
o [...] the intermittent-fasting dietary pattern.
p [...] IR were quantified by the HOMA-IR method.
88.
Klempel MC, Kroeger CM, Bhutani S, Trepanowski JF, Varady KA. Intermittent fasting combined with calorie restriction is effective for weight loss and cardio-protection in obese women. Nutr J. 2012;11:98. doi: 10.1186/1475-2891-11-98.
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a [...] 3% to 8% after 3 to 24 weeks of treatment.
b [...] studies that provided food on the fast day.
c [...] a 0.25-kg/wk reduction in body weight.
d [...] Klempel et al,
e [...] and LDL cholesterol ranging from 7% to 32%,
f [...] of weight loss is similar between studies.
g [...] (ie, LDL cholesterol >110 mg/dL).
h [...] achieved 0.25- to 0.5-kg/wk weight loss,
i [...] in fasting glucose ranging from 3% to 6%.
j [...] and periodic fasting.
k [...] the intermittent-fasting dietary pattern.
l [...] IR were quantified by the HOMA-IR method.
m [...] by 23% after 8 weeks of fasting 1 d/wk.
89.
Harvie MN, Pegington M, Mattson MP, Frystyk J, Dillon B, Evans G, Cuzick J, Jebb SA, Martin B, Cutler RG, Son TG, Maudsley S, Carlson OD, Egan JM, Flyvbjerg A, Howell A. The effects of intermittent or continuous energy restriction on weight loss and metabolic disease risk markers: a randomized trial in young overweight women. Int J Obes (Lond). 2011;35:714–727. doi: 10.1038/ijo.2010.171.
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a [...] 3% to 8% after 3 to 24 weeks of treatment.
b [...] a 0.25-kg/wk reduction in body weight.
c [...] Harvie et al,
d [...] and LDL cholesterol ranging from 7% to 32%,
e [...] of weight loss is similar between studies.
f [...] (ie, LDL cholesterol >110 mg/dL).
g [...] achieved 0.25- to 0.5-kg/wk weight loss,
h [...] studies that achieved 6% to 7% weight loss.
i [...] 6% to 10% after 6 to 24 weeks of treatment.
j [...] Participants in these studies
k [...] the intermittent-fasting dietary pattern.
l [...] IR were quantified by the HOMA-IR method.
m [...] with 24 weeks of fasting on 2 d/wk.
90.
Stote KS, Baer DJ, Spears K, Paul DR, Harris GK, Rumpler WV, Strycula P, Najjar SS, Ferrucci L, Ingram DK, Longo DL, Mattson MP. A controlled trial of reduced meal frequency without caloric restriction in healthy, normal-weight, middle-aged adults. Am J Clin Nutr. 2007;85:981–988.
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f [...] in a controlled feeding study,
g [...] in any of the studies reviewed here.
h [...] at least 10% of energy from saturated fat.
i [...] ). In a randomized, crossover trial,
j [...] frequency to 1 meal a day remains unclear.
k [...] Table 2).
l [...] Table 2).
m [...] decreased by 6% and 4%, respectively.
n [...] few meals per day (ie, 1 or 3 meals a day)
o [...] implemented a controlled feeding protocol.
p [...] study arms were quite small (n=7–19),
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f [...] a day resulted in little or no change.
g [...] in any of the studies reviewed here.
h [...] at least 10% of energy from saturated fat.
i [...] not affect HDL cholesterol concentrations.
j [...] Table 2).
k [...] numeric reductions in triglyceride levels,
l [...] subject responses reported in each trial.
m [...] few meals per day (ie, 1 or 3 meals a day)
n [...] per day (ie, 6, 9, 12, or 17 meals a day)
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g [...] in any of the studies reviewed here.
h [...] not affect HDL cholesterol concentrations.
i [...] Table 2).
j [...] numeric reductions in triglyceride levels,
k [...] subject responses reported in each trial.
l [...] few meals per day (ie, 1 or 3 meals a day)
m [...] per day (ie, 6, 9, 12, or 17 meals a day)
n [...] implemented a controlled feeding protocol.
o [...] study arms were quite small (n=7–19),
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g [...] at least 10% of energy from saturated fat.
h [...] not affect HDL cholesterol concentrations.
i [...] Table 2).
j [...] few meals per day (ie, 1 or 3 meals a day)
k [...] per day (ie, 6, 9, 12, or 17 meals a day)
l [...] implemented a controlled feeding protocol.
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g [...] in any of the studies reviewed here.
h [...] of a low-fat diet (28% of energy as fat),
i [...] not affect HDL cholesterol concentrations.
j [...] Table 2).
k [...] numeric reductions in triglyceride levels,
l [...] subject responses reported in each trial.
m [...] of 3 or 9 meals a day on blood pressure,
n [...] few meals per day (ie, 1 or 3 meals a day)
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e [...] a day resulted in little or no change.
f [...] in any of the studies reviewed here.
g [...] at least 10% of energy from saturated fat.
h [...] diet provided 35% of energy as fat,
i [...] not affect HDL cholesterol concentrations.
j [...] Table 2).
k [...] numeric reductions in triglyceride levels,
l [...] subject responses reported in each trial.
m [...] of 3 or 9 meals a day on blood pressure,
n [...] few meals per day (ie, 1 or 3 meals a day)
o [...] per day (ie, 6, 9, 12, or 17 meals a day)
p [...] implemented a controlled feeding protocol.
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c [...] Arciero et al,
d [...] not imposed in any of the trials reviewed,
e [...] a day resulted in little or no change.
f [...] in any of the studies reviewed here.
g [...] not affect HDL cholesterol concentrations.
h [...] Table 2).
i [...] few meals per day (ie, 1 or 3 meals a day)
j [...] per day (ie, 6, 9, 12, or 17 meals a day)
k [...] for 2 to 4 weeks has no effect on IR.
l [...] implemented a controlled feeding protocol.
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c [...] not imposed in any of the trials reviewed,
d [...] by 1% to 8% and LDL cholesterol by 6% to 8%
e [...] in any of the studies reviewed here.
f [...] at least 10% of energy from saturated fat.
g [...] not affect HDL cholesterol concentrations.
h [...] Table 2).
i [...] per day (ie, 6, 9, 12, or 17 meals a day)
j [...] for 2 to 4 weeks has no effect on IR.
k [...] implemented a controlled feeding protocol.
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c [...] not imposed in any of the trials reviewed,
d [...] by 1% to 8% and LDL cholesterol by 6% to 8%
e [...] in any of the studies reviewed here.
f [...] at least 10% of energy from saturated fat.
g [...] not affect HDL cholesterol concentrations.
h [...] Table 2).
i [...] per day (ie, 6, 9, 12, or 17 meals a day)
j [...] implemented a controlled feeding protocol.
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11093 | https://fiveable.me/ap-physics-2-revised/unit-10/1-electric-charge-and-electric-force/study-guide/E6OYkOGeroCXwgw1 | printables
🧲AP Physics 2 (2025) Unit 10 Review
10.1 Electric Charge and Electric Force
🧲AP Physics 2 (2025) Unit 10 Review
10.1 Electric Charge and Electric Force
Written by the Fiveable Content Team • Last updated September 2025
Verified for the 2026 exam
Verified for the 2026 exam•Written by the Fiveable Content Team • Last updated September 2025
🧲AP Physics 2 (2025)
Unit & Topic Study Guides
10.1 Electric Charge and Electric Force
10.2 The Process of Charging
10.3 Electric Fields
10.4 Electric Potential Energy
10.5 Electric Potential
10.6 Capacitors
10.7 Conservation of Electric Energy
The Electric Force Between Charged Objects
Electric force is one of the fundamental forces in nature that governs how charged particles interact with each other. This force operates according to specific patterns based on the charges involved.
When objects have the same charge sign (both positive or both negative), they repel each other
When objects have opposite charge signs (one positive, one negative), they attract each other
The force acts along the line connecting the centers of the charged objects
Electric forces follow Coulomb's Law: F=kr2q1q2 where k is Coulomb's constant, q1 and q2 are the charges, and r is the distance between them
In our everyday experience, electric forces are responsible for many common phenomena, from static electricity to the rigidity of solid objects. When you touch a doorknob after walking across a carpet, the small shock you feel is due to electric forces between charged particles.
Although electric forces operate at the atomic level, we typically describe macroscopic interactions using simplified models like contact forces (normal force, friction, tension) rather than calculating all the individual electric interactions between atoms.
more resources to help you study
practice questionscheatsheetscore calculator
Comparing Electric and Gravitational Forces
Both electric and gravitational forces are fundamental forces in nature, but they have important differences:
Electric forces can be either attractive or repulsive, while gravitational forces are always attractive
For objects with both mass and charge, the electric force is typically much stronger than the gravitational force at small scales
The ratio of electric to gravitational force between two protons is approximately 1036 (extremely large!)
Despite their relative weakness, gravitational forces dominate at astronomical scales because:
Large objects tend to be electrically neutral (equal amounts of positive and negative charge)
Electric forces between neutral objects largely cancel out
Gravitational forces always add up and never cancel each other
This explains why planets orbit stars due to gravity rather than electric forces, even though electric forces are inherently stronger.
Electric Permittivity
Electric permittivity describes how a material responds to an electric field and affects how electric charges interact within that material.
When an electric field is applied to a material, it can cause polarization—a slight separation of positive and negative charges within the material's atoms or molecules. This polarization affects how electric forces propagate through the material.
Key aspects of permittivity include:
Free space (vacuum) has a constant permittivity value of ε0=8.85×10−12 F/m
Materials have different permittivity values than free space, often expressed as a relative permittivity εr (ratio to free space)
The permittivity affects the strength of electric forces within the material according to Coulomb's Law: F=4πε1r2q1q2
The permittivity of a material depends on its molecular structure and how easily its electrons can rearrange in response to an electric field:
Conductors (like metals) have free electrons that can move easily throughout the material
Insulators (like rubber or plastic) have tightly bound electrons that resist movement
Dielectrics (like glass or ceramic) fall between conductors and insulators and can become polarized
🚫 Boundary Statement
Calculations of electric force are limited to four or fewer interacting charged objects or systems, with the exception of highly symmetrical situations where analyzing the resulting force from more charges is allowed.
Practice Problem 1: Electric Force Calculation
Two point charges are placed 0.3 meters apart. The first charge is +5.0 μC and the second charge is -2.0 μC. Calculate the magnitude and direction of the electric force between them. (Coulomb's constant k = 9.0 × 10^9 N·m²/C²)
Solution
To solve this problem, we need to use Coulomb's Law:
F=kr2∣q1q2∣
Where:
k=9.0×109 N·m²/C²
q1=+5.0×10−6 C
q2=−2.0×10−6 C
r=0.3 m
Substituting these values:
F=(9.0×109)×(0.3)2∣5.0×10−6×(−2.0×10−6)∣
F=(9.0×109)×0.0910.0×10−12
F=(9.0×109)×(1.11×10−10)
F=1.0×100 N = 1.0 N
Since one charge is positive and one is negative, the force is attractive, meaning the charges pull toward each other.
Practice Problem 2: Comparing Electric and Gravitational Forces
Calculate the ratio of the electric force to the gravitational force between a proton and an electron. (Given: proton charge = +1.6 × 10^-19 C, electron charge = -1.6 × 10^-19 C, proton mass = 1.67 × 10^-27 kg, electron mass = 9.11 × 10^-31 kg, G = 6.67 × 10^-11 N·m²/kg², k = 9.0 × 10^9 N·m²/C²)
Solution
We need to calculate both forces and find their ratio.
Electric force using Coulomb's Law: Fe=kr2∣q1q2∣
Fe=kr2∣qp×qe∣=kr2∣(1.6×10−19)(−1.6×10−19)∣=kr22.56×10−38
Gravitational force using Newton's Law of Gravitation: Fg=Gr2m1m2
Fg=Gr2mp×me=Gr2(1.67×10−27)(9.11×10−31)=Gr21.52×10−57
The ratio of electric to gravitational force: FgFe=G(1.52×10−57)/r2k(2.56×10−38)/r2=Gk×1.52×10−572.56×10−38
FgFe=6.67×10−119.0×109×1.52×10−572.56×10−38
FgFe=1.35×1020×1.68×1019=2.27×1039
This enormous ratio (approximately 10^39) demonstrates that the electric force between fundamental particles is vastly stronger than the gravitational force.
Frequently Asked Questions
What is electric charge and how does it work?
Electric charge is a basic property of matter that comes in two signs: positive (+) and negative (−). Electrons carry −e and protons +e (e ≈ 1.6×10⁻¹⁹ C). Like charges repel and opposite charges attract. For point charges the electrostatic force between two charges q₁ and q₂ separated by r is given by Coulomb’s law: |FE| = k|q₁q₂|/r² (k = 1/4πϵ₀). The force acts along the line connecting the charges; its direction depends on their signs. Electric permittivity (ϵ) of a medium tells you how the medium reduces the effective force compared to vacuum (ϵ₀). On the AP exam you’ll use these ideas to describe and calculate forces (CED allows up to 4 charges or high-symmetry cases) and compare electrostatic vs. gravitational forces. For a focused review, see the Topic 10.1 study guide ( the Unit 10 overview ( and practice problems (
Why do some objects have positive charge and others have negative charge?
Charge sign comes down to which elementary charge is in excess. Protons (in nuclei) each carry +e and electrons carry −e (CED 10.1.A.1.ii–iii). Normally objects are neutral because they have equal numbers of protons and electrons. If an object gains extra electrons it becomes negatively charged; if it loses electrons it becomes positively charged. Protons don’t move in ordinary materials, so charging usually happens by moving electrons (see Topic 10.2: the process of charging). The choice of “positive” vs “negative” is a historical convention, but the physics follows: Coulomb’s law (CED 10.1.A.2) tells you the force magnitude depends on |q1q2|/r^2 and the force direction depends on the signs (like charges repel, opposite attract, CED 10.1.A.3). For more on charging mechanisms and practice problems, check the Topic 10.1 study guide ( and the unit page (
Can someone explain Coulomb's law in simple terms because I'm really confused?
Coulomb’s law tells you how strong the electric force is between two point charges and which way it points. In simple terms: the force magnitude is proportional to the product of the two charges and falls off like 1/r^2 (inverse-square). Mathematically: |FE| = (1/4πε0) |q1 q2| / r^2 = k |q1 q2| / r^2. Same-sign charges repel; opposite-sign charges attract, and the force acts along the line joining them. Treat objects as point charges when their size is negligible (CED 10.1.A.1.iv). Use vector form and the superposition principle to add forces from multiple charges (CED boundary: up to four charges in AP problems). Remember ε0 is vacuum permittivity and appears when you need k = 1/(4πε0) (CED 10.1.C). This is a core AP 2 topic—practice calculating magnitudes, directions, and using superposition for up to four charges. For a quick review see the Topic 10.1 study guide ( the unit overview ( and try practice problems (
What's the difference between electric force and gravitational force?
Electric and gravitational forces are both inverse-square long-range forces, but they differ in key ways you need for AP Physics 2. Electric force (Coulomb’s law: FE = k|q1 q2|/r^2) can be attractive or repulsive depending on charge signs; gravitational force (Newton’s law: FG = G m1 m2 / r^2) is always attractive (CED 10.1.A.2, 10.1.B.1). For two charged masses, the electrostatic force is usually enormously larger than the gravitational force between them (CED 10.1.B.2), but gravity dominates at large (astronomical) scales because most large systems are electrically neutral (CED 10.1.B.3). Direction: electric force is along the line joining charges and set by sign (repel/attract); gravity is always toward the other mass. On the AP exam you may be asked to compare magnitudes, signs, or use Coulomb’s law vs Newton’s law in calculations—keep units and inverse-square dependence in mind. For extra review and practice problems on Topic 10.1, see the Fiveable study guide ( and Unit 10 overview (
How do I know if two charges will attract or repel each other?
Check the signs of the charges and use Coulomb’s law for the force’s magnitude and direction. If both charges have the same sign (both + or both −) they repel; if they have opposite signs (+ and −) they attract (CED 10.1.A.3.i–ii). For point charges the magnitude is |FE| = k |q1 q2| / r^2 and the force acts along the line joining them; the sign of q1q2 determines whether that line points toward the other charge (attraction) or away (repulsion) (CED 10.1.A.2–3). Remember: Coulomb’s law gives magnitude; use the charges’ signs to set direction. On the AP exam you’ll be expected to apply this (often with up to four charges and superposition) in Unit 10 problems. For a quick review, see the Topic 10.1 study guide ( and try practice problems (
What does it mean when they say charge is quantized and what is elementary charge?
When physicists say charge is quantized they mean any object's net electric charge comes in whole-number multiples of a single unit—you can't have 1.3 times that smallest unit. Mathematically: Q = n·e, where n is an integer and e is the elementary charge. The elementary charge e ≈ 1.602×10^−19 C. By the CED, an electron carries charge −e, a proton +e, and a neutron 0 (10.1.A.1.ii–iii). Coulomb’s law then uses these charges to find forces (10.1.A.2). For AP problems treat e as the indivisible unit (quarks have fractional charges but aren’t observed free in typical AP contexts). If you want more practice or a quick refresher on this topic, check the Topic 10.1 study guide ( the unit overview ( or try practice problems (
I don't understand what a point charge is - can someone explain?
A point charge is just a simplifying model: you treat a charged object as if all its charge sits at a single point in space because the object's size is negligible compared to the distances you care about (CED 10.1.A.1.iv). That lets you use Coulomb’s law easily: FE = k|q1 q2|/r^2 with r the distance between the points and the force along the line joining them (10.1.A.2–3). Use the point-charge model when the separation between charges is much larger than the physical size of the objects (e.g., charged sphere looked at from far away). Remember the elementary charge e is the smallest indivisible charge (10.1.A.1.ii–iii). For multiple point charges, apply superposition: calculate each Coulomb force and add vectorially (CED keywords). AP exams expect you to use point-charge approximations in Coulomb’s-law problems (unit 10 practice items). For a quick refresher, see the Topic 10.1 study guide ( the full unit overview ( and tons of practice problems (
Why is the electric force so much stronger than gravity but we don't notice it in everyday life?
Electric forces are enormously stronger than gravity for individual particles because the constants in the two laws differ hugely. Coulomb’s law uses k ≈ 9×10^9 N·m^2/C^2 while Newton’s gravity uses G ≈ 6.7×10^-11 N·m^2/kg^2, and for a proton and electron the electric attraction is about 10^39 times larger than their gravitational attraction. So at the particle level electric interactions dominate. We don’t notice that huge strength every day because ordinary matter is electrically neutral: positive charges (protons) and negative charges (electrons) cancel out at macroscopic scales, so net electric forces largely cancel while gravity always adds up (always attractive). Also charges in materials rearrange (polarization, conductors vs insulators) so local electric fields get screened. That’s why gravity, though weak, governs planets and stars while electric forces control atomic and molecular structure and contact forces (normal, friction) you experience (CED Topic 10.1, especially 10.1.A and 10.1.B). For a quick review, check the Topic 10.1 study guide ( and try practice problems (
What happens to the electric force when you double the distance between two charges?
By Coulomb’s law (CED 10.1.A.2), the magnitude of the electrostatic force between two point charges is F = k|q1 q2|/r^2. If you double the separation r → 2r, the force changes by the factor 1/(2^2) = 1/4. So the force becomes one-quarter as large. The direction still depends on the signs of the charges: like charges repel, opposite charges attract (CED 10.1.A.3). On the AP exam you should be able to state and use this inverse-square dependence in calculations (the CED limits calculations to four or fewer charges). For a quick review, see the Topic 10.1 study guide on Fiveable ( and try practice problems ( to get comfortable applying the 1/r^2 relationship.
How does electric permittivity affect the force between charges?
Permittivity shows up in Coulomb’s law and controls how strongly two charges interact. In vacuum the constant ε0 (≈ 8.85×10⁻¹² F/m) appears in the Coulomb constant k = 1/(4π ε0), so larger ε0 → smaller k → weaker force. In materials you use ε = κ ε0 (κ = relative permittivity or dielectric constant). A medium with κ > 1 becomes polarized by an external field, which reduces the effective field between charges and therefore reduces the magnitude of the Coulomb force by a factor of 1/κ compared to vacuum. Conductors and high-κ dielectrics polarize more easily; insulators polarize less. For AP Physics 2 you should be able to state this qualitative effect, use ε0 in Coulomb’s law, and relate relative permittivity (dielectric constant) to reduced force (Topic 10.1 and 10.1.C; see the Topic 10.1 study guide ( For extra practice, check the unit review ( and many practice problems (
What's the difference between conductors and insulators in terms of how electrons move?
Conductors (like metals) have lots of free charge carriers—electrons that can move easily through the material. That means when you put a conductor near other charges or an external field, those free electrons rearrange quickly, letting charge flow and cancel the internal electric field in electrostatic equilibrium. Insulators (like rubber or glass) have electrons tightly bound to atoms or molecules, so they can’t move freely from place to place. Instead, an external field only slightly shifts bound electrons relative to their nuclei (electric polarization), changing the material’s permittivity but not producing macroscopic charge flow. This distinction is exactly what Topic 10.1.C emphasizes: conductors allow easy movement of charge carriers; insulators do not. For more review, see the Topic 10.1 study guide ( and practice problems (
I missed the lab on static electricity - what were we supposed to observe about charged objects?
In the static electricity lab you should’ve seen charged objects either attract or repel depending on sign—like two rubbed balloons repelling (both negative) or a rubbed balloon sticking to a neutral wall (induced polarization). You’d observe charging by friction (transfer of electrons), conduction (touching), and induction (bringing a charged rod near a conductor and grounding). Insulators hold localized charge; conductors let charge move. Small sparks or hair standing up show large local electric forces. These behaviors connect to Coulomb’s law (force ∝ |q1 q2|/r²) and the idea that direction depends on charge sign (CED 10.1.A.2–A.3). For exam prep, practice calculating forces (superposition, up to four point charges allowed) and think about permittivity/dielectrics if materials are involved. Review the Topic 10.1 study guide ( and try problems from Fiveable’s practice set (
How do you calculate the net force when you have multiple charges acting on one object?
Use Coulomb’s law for each pair and the superposition principle: each other charge exerts a force F = k|q q0|/r^2 (k = 8.99×10^9 N·m^2/C^2) on the charge you care about. The steps: 1. For every other charge i, compute the magnitude Fi = k|qi q0|/r_i^2 and decide direction (repel if same sign, attract if opposite) along the line joining them (CED: 10.1.A.2–3). 2. Treat each Fi as a vector. If charges aren’t colinear, resolve each Fi into x and y components (Fx = Fi cosθ, Fy = Fi sinθ). 3. Sum components: Fnet,x = ΣFx_i, Fnet,y = ΣFy_i. 4. Get magnitude and direction: |Fnet| = sqrt(Fnet,x^2 + Fnet,y^2), θ = arctan(Fnet,y/Fnet,x). AP note: you’ll only be asked up to four interacting point charges except in high-symmetry cases (CED boundary). For worked examples and practice, see the Topic 10.1 study guide ( and more practice problems (
Why do we use the constant k in Coulomb's law and what does it actually represent?
We use k in Coulomb’s law as a constant that sets the scale of the electrostatic force in SI units. Mathematically k = 1/(4πϵ0), where ϵ0 is the permittivity of free space. Its numeric value is about 8.99×10^9 N·m^2/C^2, so Coulomb’s law becomes |FE| = k |q1 q2|/r^2. Physically k (or ϵ0) tells you how easily electric field lines spread out in a medium: larger ϵ means field lines are “diluted” more and forces are weaker. In materials you replace ϵ0 with ϵ = κϵ0 (κ = dielectric constant), so the medium’s permittivity appears directly in the force. Remember the law is inverse-square and direction depends on the signs of charges (attract vs repel). For AP Physics 2 practice, see the Topic 10.1 study guide ( and hundreds of practice questions (
What does it mean when a material gets polarized by an electric field?
Polarization means an external electric field shifts charges inside a material so opposite signs separate slightly—without adding net charge. In atoms or molecules electrons move a bit relative to positive nuclei (induced dipoles); in an insulator charges stay bound and just shift, while in a conductor free charges move and can pile up at surfaces. That local separation reduces the net field inside the material and is why materials have different electric permittivities (a measure of how easily they polarize). On the AP, this is covered in 10.1.C: electric polarization is modeled as induced rearrangement of electrons and helps explain dielectric behavior and why the permittivity of matter differs from ε0. If you want a short recap tied to the CED, check the Topic 10.1 study guide ( the full unit overview ( and try practice problems ( to see polarization in capacitor/dielectric questions on the exam. |
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Can an asymptotically slower algorithm beat a naive and asymptotically faster algorithm in a record-setting computation? We show how this was done. We will look at three computational problems themed around elementary arithmetic to see how problems from antiquity continue to pose interesting challenges to modern computer science. The third problem will involve multiplication tables, the same table that every elementary school student is required to learn. We will evaluate M(n), the function that counts the distinct numbers in an n ✕ n multiplication table. The 10 ✕ 10 table has 42 distinct numbers in it; M(10) = 42. We modify an algorithm from antiquity, the sieve of Eratosthenes, to compute M(2^{30} - 1) = 204505763483830092.
Join us on Monday, January 22, for this exciting presentation from Jonathan Webster, associate professor of mathematics and actuarial science at Butler University. The presentation will begin at 4:10pm in Hayes Hall 109. We hope to see you there!
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11095 | https://www.inchcalculator.com/convert/micrometer-to-meter/ | Inch Calculator Skip to Content
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Micrometers to Meters Converter
Enter the length in micrometers below to convert it to meters.
Have a Question or Feedback?
Result in Meters:
1 µm = 1.0E-6 m
Hint: use a scientific notation calculator to convert E notation to decimal
Learn how we calculate this below
Do you want to convert meters to micrometers?
On this page:
Micrometers to Meters Converter
How to Convert Micrometers to Meters
What Is a Micrometer?
What Is a Meter?
Micrometer to Meter Conversion Table
References
By
Joe Sexton
Joe is the creator of Inch Calculator and has over 20 years of experience in engineering and construction. He holds several degrees and certifications.
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Sexton, J. (n.d.). Micrometers to Meters Converter. Inch Calculator. Retrieved September 24, 2025, from
How to Convert Micrometers to Meters
To convert a measurement in micrometers to a measurement in meters, divide the length by the following conversion ratio: 1,000,000 micrometers/meter.
Since one meter is equal to 1,000,000 micrometers, you can use this simple formula to convert:
meters = micrometers ÷ 1,000,000
The length in meters is equal to the length in micrometers divided by 1,000,000.
For example, here's how to convert 5,000,000 micrometers to meters using the formula above.
meters = (5,000,000 µm ÷ 1,000,000) = 5 m
Micrometers and meters are both units used to measure length. Keep reading to learn more about each unit of measure.
What Is a Micrometer?
One micrometer is equal to one-millionth (1/1,000,000) of a meter, which is defined as the distance light travels in a vacuum in a 1/299,792,458 second time interval.
The micrometer, or micrometre, is a multiple of the meter, which is the SI base unit for length. In the metric system, "micro" is the prefix for millionths, or 10-6. A micrometer is sometimes also referred to as a micron. Micrometers can be abbreviated as µm; for example, 1 micrometer can be written as 1 µm.
To get an idea of the actual physical length of a micrometer, one human hair is 40-50 µm thick, demonstrating how small this unit of measure is.
Learn more about micrometers.
What Is a Meter?
According to the most recent 2019 definition, the meter is defined as the distance traveled by light in vacuum during a time interval with a duration of 1/299,792,458 of a second. One meter is equal to 100 centimeters, 3.28084 feet, or 39.37 inches.
The meter, or metre, is the SI base unit for length in the metric system. Meters can be abbreviated as m; for example, 1 meter can be written as 1 m.
Learn more about meters.
We recommend using a ruler or tape measure for measuring length, which can be found at a local retailer or home center. Rulers are available in imperial, metric, or a combination of both values, so make sure you get the correct type for your needs.
Need a ruler? Try our free downloadable and printable rulers, which include both imperial and metric measurements.
Micrometer to Meter Conversion Table
Table showing various micrometer measurements converted to meters.
| Micrometers | Meters |
| 1 µm | 0.000001 m |
| 2 µm | 0.000002 m |
| 3 µm | 0.000003 m |
| 4 µm | 0.000004 m |
| 5 µm | 0.000005 m |
| 6 µm | 0.000006 m |
| 7 µm | 0.000007 m |
| 8 µm | 0.000008 m |
| 9 µm | 0.000009 m |
| 10 µm | 0.00001 m |
| 100 µm | 0.0001 m |
| 1,000 µm | 0.001 m |
| 10,000 µm | 0.01 m |
| 100,000 µm | 0.1 m |
| 1,000,000 µm | 1 m |
References
International Bureau of Weights and Measures, The International System of Units, 9th Edition, 2019,
More Micrometer & Meter Conversions
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11096 | https://www.youtube.com/watch?v=6rgccmObdc0 | How to use Descartes Rule of signs to find positive & negative zeros
Brian McLogan
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Posted: 19 Sep 2012
👉 Learn about Descartes' Rule of Signs. Descartes' rule of the sign is used to determine the number of positive and negative real zeros of a polynomial function. Knowing the number of positive and negative real zeros enables also to also know the number of complex zeros of a complex number.
Descartes' rule of signs states that the number of positive real zeroes in a polynomial function f(x) is the same or less than by an even number as the number of changes in the sign of the coefficients of the terms of the polynomial function. The number of negative real zeroes of the f(x) is the same as the number of changes in sign of the coefficients of the terms of f(-x) or less than this by an even number.
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Transcript:
okay so by using the carts real signs what it says is if I take F ofx I need to determine the variations of the sign of the coefficients so what I look is I have a NE 5x to a positive X right or to a positive x^ s so is that a variation of a sign yes then I go from a positive again to a negative then a negative positive to a negative to a positive so you can see I have three variations in my sign correct so therefore I have a possible of three positive real zeros now remember this is only talking about real zeros all right real zeros not imaginary real so therefore I have three positive real zeros or minus that by an even integer or one so you take three and then subtract two which would be your even integer one or it could be if you have multiple it could keep on going down let's say you had six it could be 6 4 2 or zero all right so I have either three positive real zeros or one positive real zero all right so you take the number of variations and then subtract it by your even integer now that's for that's for real zeros if you guys look at this FX equal x Cub my linear factors how many possible zeros do I have real and complex three so is it possible for all all my zeros to be positive yes right but let's look at what about the negative are there any negative real zeros so to do that you're going to do the same test but you're going to test it for f ofx so remember this is the evaluation tool so you do5 you plug INX so all I do is instead of using X I plug in negative xgx raised to an odd power still going to give Mex um then times A5 is going to give me a postive 5x Cub x^2 ATX raised to an even power which is going to give me now a positive x squared negative a negative would be a positive X POS 5 do you guys see any variation in sign no no so is there any negative real zeros no so guess what all my zeros are going to be positive so here's why it's helpful if you're going to be doing synthetic division right synthetic division and let's say you don't have a graph calculator and you had to do synthetic division by using the Rational Zero test you had to do p over Q right so you're going to do plus orus 5 over 5 plus or - 5 over 1 plus or - 1 over 5 plus or - 1 over 1 which is again is 1 are you going to have to test any negative numbers no no they're all going to be positive right so there you go it's another just tool or helpful um yeah helpful you know process to help you identify the POS or the real Zer if they're positive or negative okay that's really about it that's all you guys have to do so you will have a question on your test though that is just going to say hey |
11097 | https://incoherency.co.uk/jesref/thermal-expansion.html | Coefficients of thermal expansion
jesref
Coefficients of thermal expansion
This table lists coefficients of thermal expansion for common materials.
| Material | 10-6/°C | μm/mm at 100°C change |
---
| Aluminium | 23 | 2.3 |
| Brass | 20 | 2.0 |
| Copper | 18 | 1.8 |
| Stainless steel | 10 - 16 | 1.0 - 1.6 |
| Mild steel | 10 | 1.0 |
| Cast iron | 10 | 1.0 |
Source: www.EngineeringToolBox.com
For example, the value for aluminium is 2310-6/°C, so if an aluminium bar 100 mm long is heated by 100°C, then it will expand in length by 2310-6100100 = 0.23mm. |
11098 | https://eyewiki.org/Scleral_Buckling_for_Rhegmatogenous_Retinal_Detachment | Please note: This website includes an accessibility system. Press Control-F11 to adjust the website to people with visual disabilities who are using a screen reader; Press Control-F10 to open an accessibility menu.
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Scleral Buckling for Rhegmatogenous Retinal Detachment
From EyeWiki
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All content on Eyewiki is protected by copyright law and the Terms of Service. This content may not be reproduced, copied, or put into any artificial intelligence program, including large language and generative AI models, without permission from the Academy.
Article initiated by:
Charles P. "Pat" Wilkinson, MD
All contributors:
Charles P. "Pat" Wilkinson, MD, Brad H. Feldman, M.D., Leo A. Kim, MD, PhD, Joobin Khadamy, MD, FEBO, FEBOS-CR, Anna Murchison, MD, MPH, Grant A. Justin, MD, Neelakshi Bhagat, MD, FACS, Jennifer I Lim MD, Peter A.Karth, MD, Nimesh Patel, MD, Boonkit Purt, MD
Assigned editor:
Boonkit Purt, MD, Nimesh Patel, MD
Review:
Assigned status Up to Date
by Nimesh Patel, MD on June 23, 2025.
| |
| add |
| Contributing Editors: | add |
Scleral buckling is an ophthalmic surgical technique that has been successfully employed as a primary or adjuvant procedure to repair rhegmatogenous retinal detachments for over 60 years. In the past two decades, pneumatic retinopexy and vitrectomy have been added to the retina surgeons' reattachment armamentarium. Although considerable debate persists regarding the optimal form of treatment for many types of retinal detachments, scleral buckling is declining in popularity, particularly in regard to pseudophakic cases. Still, it remains a valuable procedure in many instances, and scleral buckling techniques should continue to be part of retina surgical education in the years ahead.
Contents
1 Disease Entity
2 Risk Factors
3 Pathophysiology
4 Primary prevention
5 History
6 Diagnosis
7 Differential diagnosis
8 Management
8.1 Surgery
8.1.1 Principles of scleral buckling
8.1.1.1 Scleral buckle configuration
9 The scleral buckling operation
9.1 Prep and Drape
9.2 Conjunctival incision and isolation of rectus muscles
9.3 Localization of retinal breaks
9.4 Thermal treatment of retinal breaks
9.5 Use of buckling materials
9.5.1 Segmental episcleral buckles
9.5.2 Encircling episcleral buckles
9.5.3 Modification of routine buckling
9.5.4 Intrascleral buckles
9.6 Management of subretinal fluid
9.6.1 Non-drainage technique
9.6.2 Drainage procedures
9.7 Adjustment of scleral buckle
9.8 Accessory techniques
9.8.1 Intravitreal injection of balanced salt solution
9.8.2 Intravitreal gas injection.
9.8.3 Gas-fluid exchange
9.8.4 Laser therapy
9.9 Closure of incisions
9.10 Results of scleral buckling
9.11 Complications of scleral buckling
9.11.1 Common complications of scleral buckling
9.11.1.1 Selected Intraoperative Complications
9.11.1.1.1 Corneal Complications
9.11.1.1.2 Pupillary Complications
9.11.1.1.3 Scleral perforation with suture needles
9.11.1.1.4 Complications of draining subretinal fluid
9.11.1.1.5 “Fishmouthing” of retinal breaks.
9.11.1.1.6 Complications of intravitreal gas injections
9.11.1.2 Selected Postoperative Complications
9.11.1.2.1 Increased intraocular pressure
9.11.1.2.2 Endophthalmitis and scleral abscess.
9.11.1.2.3 Choroidal Detachment
9.11.1.2.4 Later periocular infection and implant extrusion.
9.11.1.2.5 Cystoid macular edema (CME)
9.11.1.2.6 Epimacular proliferation
9.11.1.2.7 Proliferative vitreoretinopathy (PVR)
9.11.1.2.8 Recurrent retinal detachment.
9.11.1.2.9 Altered refractive error.
9.11.1.2.10 Muscle imbalance
10 Additional Resources
Disease Entity
The essential requirements for a rhegmatogenous retinal detachment include a retinal break (rhegma = rent or rupture) and vitreous liquefaction sufficient to allow fluid in the vitreous cavity to pass through the break(s) into the subretinal space. The usual pathological sequence that results in retinal detachment is vitreous liquefaction followed by a posterior vitreous detachment (PVD), which in turn causes retinal tears at the sites of significant vitreoretinal adhesions (Figure 1). In a smaller but significant percentage of cases, a complete PVD does not occur, and vitreoretinal traction may occur in regions of the retina that are near breaks which were unrelated to the PVD process. All ocular conditions that are associated with an increased prevalence of vitreous liquefaction and PVD or with an increased number or extent of vitreoretinal adhesions are associated with a higher incidence of retinal detachment, including trauma.
The majority of eyes with retinal breaks do not develop retinal detachment because normal physiological forces keep the retina in place. However, the combination of retinal breaks, vitreous liquefaction and detachment, traction on the retina (vitreoretinal traction), and intraocular fluid currents associated with movement of liquid vitreous and subretinal fluid can overwhelm these "attachment factors," causing retinal detachment.
Risk Factors
Major risk factors have in common an increased incidence of retinal break(s), vitreous liquification and detachment, and abnormal vitreoretinal adhesions. Specific entities include myopia, surgical and non-surgical trauma including complicated cataract surgery with posterior capsular rupture and vitreous presentation, lattice degeneration, infectious retinitis, and hereditary vitreoretinal disorders.
Pathophysiology
As noted above, retinal detachment occurs when the combination of factors that promote retinal detachment overwhelms the normal attachment forces. This is due to a combination of retinal breaks, vitreous changes inducing a retinal break and vitreoretinal traction, and intraocular fluid currents.
Retinal breaks are traditionally classified as holes, tears, or dialyses. Retinal holes are full-thickness retinal defects that are typically not associated with persistent vitreoretinal traction in their vicinity. They usually occur as a result of localized atrophic intraretinal abnormalities. Retinal tears are usually produced by an acute PVD due to excess vitreoretinal traction at sites of significant vitreoretinal adhesions. Vitreous traction usually persists at the edge of a tear, which promotes progression of the retinal detachment. Dialyses are circumferential retinal breaks that occur at the ora serrata. Although most are associated with blunt ocular trauma, dialyses can occur spontaneously.
Aging of the human vitreous (synchysis senilis) is characterized by liquefaction of the vitreous gel and progressively enlarging pools of fluid (lacunae) within the gel. These optically empty liquid spaces coalesce with aging. Extensive liquefaction within the vitreous cavity leads to a reduction in both the shock-absorbing capabilities and the stability of the gel. Posterior vitreous detachment (PVD) usually occurs as an acute event after liquefaction of the vitreous gel reaches a critical degree. The precipitating event is probably a break in the posterior cortical vitreous in the region of the macula.2 This is followed by the immediate passage of intravitreal fluid into the space between the cortical vitreous and retina. Characteristically, this rapid movement of fluid and the associated collapse of the remaining structure of the gel result in extensive separation of the vitreous gel and retina posterior to the vitreous base, especially in the superior quadrants. Partial PVDs usually progress rapidly (within days) to become complete, although they do not always separate from the entire posterior retina. Vitreoretinal traction has a number of causes, which range from simple action of gravitational force on the vitreous gel to prominent transvitreal fibrocellular membranes. Gravitational force is important and probably accounts for the high percentage of superior retinal tears (80%). However, rotational eye movements, which exert strong forces on all vitreoretinal adhesions, are probably more important causes of ongoing vitreoretinal traction. When the eye rotates, the inertia of the detached vitreous gel causes it to lag behind the rotation of the eye wall and, therefore, the attached retina. The retina at the site of a vitreoretinal adhesion exerts force on the vitreous gel, which causes the adjacent vitreous to rotate. The vitreous gel, because of its inertia, exerts an equal and opposite force on the retina, which can cause a retinal break or separate the neural retina farther from the pigment epithelium if subretinal fluid is already present (Figure 2). When the rotational eye movement stops, the vitreous gel continues its internal movement and exerts vitreoretinal traction in the opposite direction.
Continuous flow of liquid vitreous through a retinal break into the subretinal space is necessary to maintain a rhegmatogenous retinal detachment, because subretinal fluid is absorbed continually from the subretinal space via the RPE. Trans-break flow is encouraged by vitreoretinal traction, which tends to elevate the retina from the RPE. Rotary eye movements cause liquid currents in the vitreous to push against the gel adjacent to the retinal break and to dissect beneath the edge of a retinal break into the subretinal space (Figure 2) Subsequent eye movements also have an inertia effect on the subretinal fluid that favors extension of the retinal detachment.
Primary prevention
Attempts to prevent retinal detachment have been advocated for many decades, but the value of such interventions remains questionable in most instances (This subject is discussed in more detail in Retinal detachment). The American Academy of Ophthalmology Preferred Practice Pattern on the topic contains only a single evidence-based recommendation for prophylactic therapy. This is in regard to the proven value of treating symptomatic retinal tears associated with persistent vitreoretinal traction.
History
The early symptoms of acute retinal detachment are the same as those of an acute PVD: the sudden onset of many tiny dark floating objects, frequently associated with significant photopsia (flashes). Photopsia may be brief but are usually dramatic, in the temporal visual field, and are best seen in the dark and associated with eye movement.
Symptomatic loss of visual field does not usually occur until sufficient subretinal fluid has passed through the retinal break(s) to cause the retinal detachment to progress posterior to the equator. The majority of retinal breaks are located at the equator or more anteriorly; subretinal fluid initially accumulates in the retinal periphery, where it causes a corresponding loss of peripheral vision in the area that is inverse (and reverse) to the location of the retinal detachment. The loss of peripheral vision (a “curtain effect”) increases as the detachment enlarges; central visual acuity is lost when subretinal fluid passes beneath the fovea. Frequently, patients do not notice any symptoms until the macula becomes involved.
Diagnosis
If the retina can be well visualized, the diagnosis of rhegmatogenous retinal detachment is made on the basis of clinical examination. In eyes with opaque media, the presence of a retinal detachment is usually determined ultrasonographically; the location and identification of the causative retinal breaks are based upon the configuration of the detachment as well as on the patient’s history and associated findings. The vast majority of retinal detachments are diagnosed easily with a binocular stereoscopic evaluation of the entire retina. Areas of retinal detachment are recognized by elevation of the neural retina from the RPE and loss of pigment epithelial and choroidal detail beneath the elevated retina. Retinal breaks also maybe discovered by direct ophthalmoscopic examination. Indentation of the peripheral retina (scleral depression) is employed to facilitate visualization of the anterior retina dynamically and at different angles, and this improves abilities to identify all retinal defects.
Retinal breaks associated with small amounts of subretinal fluid may be difficult to detect; however, the diagnosis becomes more obvious as the retinal detachment increases in size. A stereoscopic vitreoretinal examination typically reveals an elevated sensory retina in the area of detachment, but the important identification of all retinal breaks may remain difficult (especially critical for scleral buckling surgery)—it is considerably easier to diagnose the retinal detachment than to detect all retinal breaks. Due to the effects of gravity, the topography of a retinal detachment is of major value in the prediction of the most likely locations of retinal breaks.10 Retinal breaks are usually present superiorly within the area of detachment. Thus, if a retinal detachment involves one upper quadrant or both the superior and inferior quadrants on one side of the vertical meridian, the responsible retinal break is likely to be near the superior edge of the detachment. Retinal detachments that involve the inferior quadrants tend to follow the same rules, however the progression of the detachment is often much slower, and symmetrical spread of subretinal fluid may occur on both sides of the break. Therefore, detachments that involve one or both inferior quadrants may have a break near a superior margin of the detachment or in the meridian that bisects the area of detachment. Nevertheless, because multiple retinal breaks are common, the entire periphery of the detached retina must be meticulously examined.
Differential diagnosis
Retinal detachments that occur as a result of retinal breaks must be distinguished from several conditions in which retinal blood vessels are clearly separated from the pigment epithelium. These include retinal detachments from other causes and retinoschisis. Choroidal lesions that elevate the overlying retina and intravitreal pathology that simulates an elevated retina may also be confused with retinal detachment.
The distinction between different types of retinal detachment can be difficult in eyes with small or undetectable retinal breaks and features associated with intraocular proliferation or exudation. In some cases, both a rhegmatogenous and a traction or exudative component may be important in the pathogenesis of the detachment. This is particularly common in eyes with proliferative diabetic retinopathy and retinal detachment. Pure traction detachments usually have a concave surface, and the shape, location, and extent of the detachment can be accounted for by the evident vitreous traction (see Figure). Diabetic retinal detachments with a rhegmatogenous component are usually more extensive and often have a convex contour. Exudative detachments from a variety of causes are characterized by shifting subretinal fluid, which assumes a dependent position beneath the retina. In most cases, the fluid is located inferiorly and its cause within or beneath the retina may be apparent or quite subtle.
Management
Rhegmatogenous retinal detachment are indications for vitreoretinal surgery. The three major surgical methods are scleral buckling, vitrectomy, and pneumatic retinopexy (PR). Combinations of two or three of these are frequently employed.
This manuscript will discuss scleral buckling, and EyeWiki manuscripts regarding the alternatives are available.
Surgery
Localized indentation of the sclera, choroid, and pigment epithelium beneath a retinal break alters the anatomical and physiological factors associated with the production of a retinal detachment. The fundamental goal of scleral buckling is the functional closure of all retinal breaks, so that normal physiological forces can maintain a permanent state of attachment. Drainage of subretinal fluid and scleral buckling will usually close the responsible break(s) immediately. In a non-drainage procedure, functional closure of retinal breaks can result from several beneficial effects of a scleral buckle, including (1) reduction of vitreoretinal traction by displacing the eye wall and retina centrally; (2) displacement of subretinal fluid away from the location of the retinal break and scleral buckle; (3) postoperative increase in the height of the scleral buckle; (4) approximation of the retinal break and adjacent vitreous gel; (5) increase in resistance to fluid flow through the retinal break, with consequent increase in the relative reattachment forces; and (6) alteration in the concave shape of the eyeball, resulting in a change in the effect of intraocular currents that encourage liquid vitreous to enter the subretinal space. These effects are non-exclusive and are probably synergistic, and they are also important in drainage cases. Although contemporary scleral buckling procedures routinely include the creation of a chorioretinal burn with cryotherapy or laser to induce adhesion from reactive scaring, such an adhesion is not always necessary to maintain retinal reattachment.
Principles of scleral buckling
The most important skill required in surgery for retinal detachment is the ability to detect all retinal breaks and additional areas of vitreoretinal pathology. Scleral buckling is performed to produce functional closure of retinal breaks responsible for retinal detachment and to reduce the chances of recurrent detachment. Various kinds and shapes of silicone rubber elements are used, including segments of silicone sponge as well as solid silicone shaped into bands for encircling the eye and into additional forms to augment the width and height of the buckle in selected areas. The specific configuration of the scleral buckle depends upon a number of factors. Following localization and treatment of retinal breaks and areas of vitreoretinal degeneration, the silicone buckling element is secured to the scleral surface, usually with sutures or scleral tunnels. Drainage of subretinal fluid is often performed. Intravitreal gas or air injection is sometimes employed in conjunction with scleral buckling. Problems encountered at any point of the procedure may require modifications in technique, often leading to a vitrectomy surgery.
Scleral buckle configuration
The location, number, size, and types of retinal breaks are important variables affecting the selection of a specific buckling technique. Similarly, the presence of vitreoretinal degeneration, with or without retinal breaks, and of significant vitreoretinal traction unassociated with retinal breaks should be considered in the preoperative assessment. If retinal breaks, vitreoretinal degenerative disorders, and significant vitreoretinal traction are present in multiple quadrants, a circumferential buckle is usually favored. A single retinal break unassociated with additional significant problems may be managed with an isolated segmental buckle, if not with pneumatic retinopexy.
The anterior-posterior dimensions of retinal break(s) and areas of significant vitreoretinal degeneration and vitreoretinal traction are also important considerations in planning a buckling procedure. Scleral buckles should support all edges of the retinal breaks and associated areas of vitreoretinal degeneration. In general, the buckling effect should extend into the zone of the vitreous base to eliminate current and future traction forces..
The internal changes caused by scleral buckling are determined by the size, shape, and consistency of the buckling material, the width of the suture bites placed to attach the silicone rubber to the sclera, the tightness of the tied sutures, and the extent to which an encircling element is tightened. A "high" scleral buckle is associated with a significant displacement of intraocular volume. In order to avoid large increases in intraocular pressure, drainage of subretinal fluid, paracentesis, or removal of liquid vitreous is usually necessary, particularly in eyes with compromised aqueous outflows.
The scleral buckling operation
The procedure involves routine prepping and draping, conjunctival incision, identification and treatment of all retina breaks and areas of vitreoretinal degeneration, suturing buckling material to the sclera, and additional techniques as indicated.
Prep and Drape
For scleral buckling, as with all vitreoretinal surgery, the face, eyelids, and conjunctiva are prepared with appropriate antiseptic techniques, and the drapes are applied. A plastic adhesive drape is placed over the opened lids so that the drape adheres to the margins of the eyelids that are held securely in place with the eyelid speculum.
Conjunctival incision and isolation of rectus muscles
A 360 degree limbal conjunctival peritomy incision is usually made for encircling procedures, but a less extensive incision is performed for limited buckling procedures. A five clock-hour peritomy is usually sufficient for a quadrantic detachment when a radial or short circumferential buckle is planned, and only the two rectus muscles bordering the involved quadrant are isolated in this situation. Traction sutures are placed beneath the insertions of the exposed rectus muscles to facilitate positioning the globe. Rather heavy tier material, such as 0-0 silk, is recommended. After the traction sutures are placed, the sclera should be examined. It can be exposed for inspection by rotating the globe with two adjacent traction sutures and pushing the intermuscular fascia posteriorly with a cotton-tipped applicator. The surgeon should look for abnormalities such as anomalous vortex veins or scleral thinning. Thinning is usually seen as abnormal graying (lines or patches) in the equatorial or preequatorial area, particularly superior temporally. Thinning lines are called radial staphylomas, scleral cracks, or scleral dehiscences.
After examination of the sclera, the surgeon should carefully examine the retina 360 degrees with binocular indirect ophthalmoscopy and scleral indentation. Important lesions that were not noted preoperatively may become apparent, and occasionally this examination reveals significant changes that have developed since the preoperative examination.
Localization of retinal breaks
A localizing mark is made with a scleral marking device on the point of sclera overlying an edge of the retinal break(s). The tip of the scleral marker is firmly pressed against the eye for a few seconds, and the pressure creates a temporary black mark on the sclera. Alternatively, a flat diathermy probe can be used, yielding a light burn. The localization site is immediately dried and touched with a marking pen, because the pressure effect or light scleral burn disappear quickly. If a break is large, marks should ideally be placed at the posterior, anterior, and lateral margins. For smaller, routine tears, some surgeons prefer to localize with a single mark placed at the center of the anterior edge of the tear, preferring this location because (1) the anterior edge is the usual site of persistent vitreoretinal traction, (2) the anterior edge is always easier to mark than the posterior border, (3) a buckling effect extending from the anterior edge into the area of the vitreous base is desirable, and (4) it is relatively easy to estimate the amount of buckling effect required more posteriorly to support the respective break(s). Other surgeons prefer a single mark placed at the center of the posterior edge of the tear to help ensure that no part of the tear falls too far back on the posterior slope of the buckle. When positioning the buckle relative to the mark, the surgeon must keep in mind whether the posterior or anterior edge of the tear was marked.
Accurate marking is easily performed if the detachment is sufficiently flat to allow the retinal pigment epithelium to be pressed inward against the sensory retina. If the detachment is highly bullous, this is not possible and the surgeon must try to compensate for the parallax effect. With bullous detachments, the tendency is to localize the break too far posteriorly. If the surgeon is aware of the problem, an adequate anterior compensation can be made.
NOTE: Marking of the breaks is often done as the step after thermal treatment as opposed to prior to it, however this is at the surgeons discretion.
Thermal treatment of retinal breaks
A variety of methods have been employed to irritate the choroid and pigment epithelium so that chorioretinal adhesions create scar tissue to seal retinal breaks. The three most popular techniques, which have persisted for many years, are diathermy, cryotherapy, and laser photocoagulation.
Diathermy is no longer employed by most surgeons. It should only be applied through thinned sclera, beneath partial-thickness scleral flaps, to avoid full-thickness destruction of sclera and achieve a more uniform intraocular reaction. The surgeon may apply the diathermy while inspecting the retina with the ophthalmoscope, limiting the intensity to that required to produce a moderately white lesion in the retina. Alternatively, an experienced surgeon can apply diathermy without ophthalmoscopic control, predicting the probable ophthalmoscopic result on the basis of the scleral reaction. Diathermy burns are spaced approximately 1.5 mm apart, producing intermittent pigmentary changes in the fundus that have been referred to as “leopard skin.”
Cryotherapy has gained wide acceptance since its reintroduction in 1964. A survey of retinal surgeons revealed that the vast majority of surgeons now employ it for most buckling cases [reference needed]. Although it is possible to apply cryotherapy beneath partial-thickness scleral flaps, it is not necessary to do so. The goal of cryotherapy is to apply contiguous lesions completely surrounding each retinal break and areas of vitreoretinal degeneration. A single row of confluent freeze spots is generally sufficient, although more than one row may be required anteriorly to extend the future adhesion into the vitreous base.
Cryotherapy should be applied while observing the retina with the ophthalmoscope. A prominent white change develops at the area of retinal freezing. The freezing is generally terminated soon after the surgeon observes the appearance of an ice ball approaching the neurosensory retina. Adequate freezing of the retinal pigment epithelium is discerned by a change of color to dull orange or gray. If portions of the break remain highly elevated during scleral depression with the cryo probe, treatment can be postponed until some of the subretinal fluid is drained.
A challenge with using cryotherapy is that one cannot determine what areas of the retina have been treated for many minutes to days following treatment. This can lead to over treatment if sequential freezes overlap significantly, and this excessive therapy should be avoided if possible. However, if the burns are not confluent in appropriate areas, an inadequate adhesion will result. Thus the surgeon must form an optimal visual image of the precise limits of prior applications of cryotherapy, and considerable experience is usually required to master this technique. Cryotherapy burns should extend only to the edges of medium and large retinal breaks from retina surrounding them. If the pigment epithelium lying beneath the break is included in the cryo burn, as it often is, an intravitreal dispersion of pigmented cells capable of proliferation can occur. For the same reason, scleral depression of a treated area at the edge of a break should not be repeated after the treatment, and localization of retinal breaks should ideally precede cryotherapy.
Laser photocoagulation is not usually used to induce adhesive lesions in scleral buckling surgery for detached retina, because the retinal pigment epithelium is not sufficiently close to the retina to cause a burn. Although the surgeon can close retinal breaks with scleral buckling and/or intravitreal gas and then attempt to apply laser photocoagulation, this is usually easier a day or so following the operation when the retina is quite flat. Attached retinal breaks may sometimes be lasered intraoperatively with the laser indirect ophthalmoscope.
Use of buckling materials
Various materials have been used for scleral buckling, including fascia lata, palmaris tendon, plantaris tendon, knee cartilage, donor sclera, dura mater, polyviol, polyethylene, encircling nonabsorbable and absorbable sutures, gelatin, hydrogel, and silicone. The latter is far the most popular. Hydrogel material is no longer used, due to problems with the material over time. Silicone is a soft, synthetic rubber material that is nontoxic and nonallergenic. It is produced in a variety of molded shapes that can be modified by the surgeon and used in either solid or sponge form. Implants can be placed within the sclera or on its surface. Although they are not technically “implants” in the latter position, the term has persisted for decades and will be employed in this monograph. Episcleral implants are currently employed in the vast majority of cases. These can be segmental (radial or circumferential) or encircling in configuration.
Segmental episcleral buckles
Segmental silicone exoplants are secured to the sclera with 5-0 non-absorbable synthetic (commonly nylon) suture attached to spatula needles with cutting tips, tied in a horizontal mattress mattress configuration. One-half to two-thirds thickness intrascleral passes at least 6mm long are usually attempted for sponges and broad exoplants, whereas shorter intrascleral passes can be used to support encircling bands. The surgeon applies focal pressure with a cotton applicator near the intended suture site to prevent buckling of the sclera in front of the needle as it is passed through the sclera. Alternatively, a nearby muscle insertion can be grasped with a forceps to provide increased stability and to elevate intraocular pressure. The location of the needle tip should be visualized at all times during its passage through the sclera. Localized scleral buckles may be radially or circumferentially oriented, and a combination of the two may be considered if more extensive buckling is required. Radial scleral buckles provide focal support for a retinal tear and minimize the development of radial retinal folds commonly associated with circumferential buckles, which shorten the circumference of the eye wall but not the circumference of the retina. Circumferential buckles provide a zone of support oriented parallel to the region where vitreous traction is usually most severe, and they are an efficient means of supporting multiple areas of vitreoretinal pathology. When tightened, the sutures indent the exoplant and underlying eye wall. Using an exoplant of a given shape, the height of the buckling effect is determined by the distance between the suture bites and the tightness of the sutures when they are tied. Calipers may be used to measure the distance between the suture bites, and in general these should be 2-3 mm wider than the exoplant for a modest buckling effect and 4-8 mm wider for a relatively high buckle.
When a radially oriented segmental buckle is needed, intrascleral suture limbs are placed parallel to the meridian of the retinal break and equidistant from its edges, and the needle is passed so that the knot can be tied posteriorly (Figure 5-7). The size and type of the retinal break usually dictate the width and length of the exoplant. In general, the width of the silicone element should be at least as wide as the edges of the break marked on the sclera, and the buckle length should support both the posterior end of the break and the vitreous base anterior to the break.
If a circumferentially oriented segmental buckle is used, suture limbs are placed parallel to the limbus. The anterior bite is usually placed just anterior to the posterior margin of the vitreous base, a location estimated as lying about 2-3 mm posterior to an imaginary line drawn between the muscle insertions (and forming a portion of the spiral of Tillaux). A silicone element of sufficient width to support both the anterior and posterior edges of the retinal break(s) or other pathology is used, and the posterior suture bite is placed in a position that will produce an optimal buckling effect.
Encircling episcleral buckles
If a 360 degree encircling circumferential buckle of modest width and height is desired, an encircling #240 , #41, #42, or other silicone band is passed around the circumference of the globe and beneath the rectus muscles. The band is traditionally anchored with a single mattress suture with bites parallel to the limbus placed in the center of each quadrant. Although they are an elegant and effective means of securing a band, scleral tunnels are usually not employed by most surgeons. Suture bites that straddle a silicone band should be placed just far enough apart to allow the band to move freely beneath the suture, and this distance equals the width of the band plus two-times its thickness (Figure 5-8). Narrower bites inhibit circumferential movement of the band, particularly if the sutures are pulled tight. Wider bites will allow the band to move anterior to its desired location when its ends are joined. In its proper position, the band is usually intended to support breaks in the region of the posterior edge of the vitreous base, and these are marked and treated before suture placement. In quadrants without retinal breaks, the vitreous base margin can be marked, or its location can be estimated and the anterior suture bite placed about 2-3 mm posterior to the imaginary line mentioned above. If a grooved segment of silicone tire is required because of a need for more extensive augmentation of the buckle, additional sutures may be required, depending upon the characteristics of the specific case. If a high broad 360-degree encircling scleral buckle is required, two broad mattress sutures are placed in each quadrant to accommodate a silicone tire and an overlying silicone band (Figure 5-9). The anterior suture bite is placed at the estimated location of the ora serrata, and the posterior bite is placed far posteriorly, at a spot dictated by the width of the tire, the desired amount of indentation, and the location of tears. Frequently, this distance is equal to twice the distance from the anterior bite to the marked posterior edge of the retinal break(s). If a vortex vein must be avoided with the posterior suture bite, two small circumferential passes can be made on either side of the vessel. When using segmental scleral buckles produced by circumferentially oriented silicone materials, suture bites are always placed in the location overlying the responsible retinal break(s), because the maximum buckle height is produced in this spot. However, if an increased buckling effect is needed in only a small area in a case in which an encircling band is employed, anchoring sutures are placed away from the important pathology, and a radially oriented piece such as #103 or #106 element is placed beneath the band and usually not sutured. The ends of an encircling band are joined with a silicone Watzke silicone sleeve, tantelum clip, or suture. Tantalum clips appear to be less popular than the sleeve, primarily because of the extra time required for their use, particularly if later adjustment of the length of the band is required. Some degree of twisting of the band near the elastic sleeve can occur when the band is tightened, and this can be avoided by grasping each end near the sleeve and by keeping the ends flat against the periscleral portion already in place. The ends should be rejoined if twisting causes a narrow edge of the band to indent the sclera, as this can lead to later intrusion problems. Since even minimal twisting of the sleeve can affect the inner morphology of the buckling effect, the sleeve should be located in a quadrant relatively free of significant vitreoretinal pathology. Alternatively, a 5-0 suture tied in a loop around the overlapped ends of the encircling band minimizes twisting while allowing adjustment. To increase the tension on the band, traction is placed on the two cut ends of the band while a needle holder grasps the suture and allows it to slip. To decrease tension on the band, traction is placed on segments of the band on either side of the loop..
Modification of routine buckling
The most common problem requiring a modification in technique is the presence of thin sclera (Figure 5-4). When this is encountered, scleral suture bites must be placed in positions that are less potentially hazardous. In most such situations, this can be accomplished with suture passes on either side of the ectatic sclera and/or with the use of a wider piece of silicone buckling material. More exotic means of managing thin sclera with donor sclera, tissue glue, etc, have been described, but pneumatic retinopexy with or without drainage of subretinal fluid or vitrectomy without scleral buckling are usually employed if scleral suturing appears to be impossible.
Intrascleral buckles
Intrascleral buckles involve lamellar dissection to create a partial thickness scleral bed. They usually involve only a limited part of the circumference of the globe, but can vary in length from one to twelve hours of the clock and in width from 4 to 12 mm. An encircling band is usually used, attached directly to the surface of the sclera in the areas not undermined, and the ends are joined together with only moderate tension. The implant is placed in the bed of the undermining, and the flaps are then closed over the top with sutures. This “trap-door procedure” creates a satisfactory buckle (Figure 5-10). This technique is not commonly performed.
Management of subretinal fluid
Decisions regarding the drainage of subretinal fluid are among the most difficult associated with scleral buckling procedures, and considerable differences in opinion exist. Drainage is almost never performed if responsible breaks can be easily and almost completely approximated to the pigment epithelium with a non-drainage technique. Drainage is almost always performed if a high and broad encircling scleral buckle is required. However, in most cases the criteria for drainage or non-drainage are less obvious. In eyes in which drainage was considered to be neither clearly unnecessary nor mandatory, a small randomized trial demonstrated comparable results with both techniques. In most cases, the decision regarding drainage depends upon the size and configuration of retinal tears, the amount of traction, the appearance after the scleral buckle has been elevated beneath the retinal break(s), and the experience of the surgeon regarding the amount of subretinal fluid that can be allowed to remain between the crest of the buckle and the break(s). Most non-drainage procedures are effective if the crest of the buckle is within 3 mm of the respective retinal break. Although this may be relatively easy to accomplish when buckling a single break, the need for more extensive buckling of multiple breaks makes this a more difficult goal and favors a drainage procedure. Because of complications associated with drainage of subretinal fluid, it is avoided unless considered to be necessary for surgical success. If failure to drain results in persistent subretinal fluid postoperatively, an intravitreal gas injection performed in the office can frequently cause the fluid to settle. Nevertheless, most surgeons perform trans-scleral drainage of subretinal fluid in approximately 75% of retinal detachment cases managed with scleral buckling [reference needed]. However, with increasing popularity of primary vitrectomy for retinal detachment, many surgeons now consider vitrectomy in cases that are likely to require drainage and use drainage in a smaller percentage of scleral buckle cases. Still, drainage and non-drainage techniques play a major role in the management of a routine series of cases, and familiarity with both techniques is essential.
Non-drainage technique
The most common indication for a non-drainage technique is a retinal detachment due to a single break that can be approximated close to the pigment epithelium by scleral depression. Following the treatment of the break with cryotherapy and placement of appropriate sutures, the scleral buckle is elevated to the desired height, and the proximity of the retinal break to the surface of the buckle is re-evaluated. If the break is not perfectly positioned, the scleral buckle must be adjusted. The breadth of the buckling effect can be extended with additional sutures placed anterior or posterior to those already holding the buckle. If the posterior edge of the break is not positioned on the center of the buckle crest, at least one arm of the mattress suture must be repositioned. Assuring perfusion of the central retinal artery is critical if non-drainage techniques are employed. Unless fluid has been removed from the eye, placement of a scleral buckle will usually raise intraocular pressure so high that the central retinal artery is no longer perfused. This is temporarily tolerable but pressure must be lowered and perfusion restored approximately five minutes. Whether or not subretinal fluid is drained, pulsating perfusion of the central retinal artery must be confirmed by the end of the case. It can be difficult to determine if the central retinal artery is patent. If pulsations of the central retinal artery are observed following segmental buckling, perfusion is usually sufficient. If pulsations are not visualized and perfusion is questioned, additional digital pressure should be applied to the globe to elicit pulsations. If these do not occur, the arterial flow into the eye has probably ceased, and intraocular pressure must be reduced if pulsations do not begin soon. Intraocular pressure can be reduced by drainage of subretinal fluid, paracentesis, aspiration of fluid vitreous, and/or reduction in height or extent of the buckle. Paracentesis is usually performed in non-drainage procedures to achieve timely reduction of intraocular pressure after placement of the buckle, although this is usually not necessary if only a single radial buckle is placed. If repeated paracenteses cannot reduce intraocular pressure sufficient to reopen the central retinal artery, pressure can also be reduced by aspirating fluid from the posterior vitreous cavity. However, this maneuver is associated with many more potential complications than paracentesis. If optimal buckle size and suture width have been employed, the height of the buckle can be increased by further tightening of mattress sutures after intraocular pressure has returned to relatively normal levels. Sutures are initially tied temporarily to facilitate this adjustment. Perfusion or at least pulsation of the central retinal artery must be documented each time that sutures are tightened, and this is particularly important in eyes with reduced outflow facility.
Drainage procedures
Drainage of subretinal fluid is performed at a site determined by the configuration of the retinal detachment. Sufficient subretinal fluid is necessary to allows safe drainage. Factors considered in the selection of a drainage site include (1) the distribution of subretinal fluid when the eye is in a position at which drainage will be performed, (2) the location and size of the retinal break(s), (3) the location and configuration of the buckle, (4) the vascularity of the choroid, (5) features of vitreoretinal and epiretinal membrane traction, and (6) the ease of exposure of the proposed drainage site. The optimal locations for drainage are usually just above or below the lateral rectus muscle, because major choroidal vessels are avoided, and exposure of sclera is excellent. Choroidal vessels are also avoided by draining on either side of the three remaining rectus muscles, but exposure is frequently more difficult. If possible, drainage is usually performed some distance from retinal breaks, especially large retinal breaks, so that passage of vitreous through the break(s) and out of the eye can be minimized. The scleral depression effect provided by the buckle or by a cotton tipped applicator can help prevent this occurrence. In unusual situations in which a large buckling effect is required and very little subretinal fluid exists, drainage can be performed immediately beneath a large tear to allow both subretinal and intravitreal fluids to exit the globe. Drainage is performed prior to tightening the scleral buckling elements onto the eye, since a high intraocular pressure at the time of drainage increases the risk of complications. A site is usually selected at or slightly anterior to the equator, and a location that will ultimately be closed by the exoplant is preferred (usually in the bed of the exoplant). This avoids the need for a preplaced suture at the sclerotomy site, and it facilitates subsequent management of drainage complications, as noted later. In the situation in which drainage cannot be performed optimally at a site intended to be covered by the buckle, a preplaced suture is employed to close the scleral incision following drainage. This suture is placed after the sclera incision is made but before the choroid is penetrated. A 3-4 mm incision through sclera is performed so that the center of the sclerotomy will be at the appropriate location. All scleral fibers are carefully divided until subtle prolapse of uveal tissue is observed. The choroid may then be closely inspected for prominent choroidal vessels. If large visible vessels cannot be avoided during planned penetration, a second site nearby is selected, and another scleral incision is performed. If the area of exposed choroid is free of prominent vessels, it is sometimes treated lightly with a flat diathermy probe. This causes minimal retraction of the edges of the sclera to improve visualization, and it may reduce the risk of hemorrhage. All significant traction upon the eye is eliminated to reduce intraocular pressure as much as possible. The choroid is then penetrated with a sharp-tipped conical penetrating diathermy electrode or suture needle. Modest pressure is used to insert the device perpendicular to the surface of the sclera until the subretinal space is entered. If the conical diathermy electrode is used, this event is usually heralded by a sudden subtle "pop" which is usually perceived by touch or observation. Because of the tapered shape of the electrode, significant amounts of subretinal fluid do not exit the eye until this device is very slowly withdrawn from the eye. The penetrating diathermy electrode is relatively blunt, compared to a suture needle, and penetration of a soft eye with a congested choroid may be somewhat difficult. This is managed by modest elevation of intraocular pressure with traction upon the muscle fixation sutures. An oblique or tangential path of penetration is recommended by some authors to avoid perforating the retina, but this can result in a flap valve of the choroid, which can limit drainage. If a proper drainage site has been selected, penetration of the retina with the tapered diathermy is exceptionally rare, because it is removed prior to the release of significant amounts of subretinal fluid. Lasers have also been employed to drain subretinal fluid, but the expense and time required to use them do not appear to be balanced by a significant reduction in the rate of complications. As the globe softens during drainage, intraocular pressure is very slowly increased to encourage further drainage and to avoid complications associated with hypotony. If a large tear is present, the sclera or the buckle and sclera overlying the break are indented with a cotton applicator. This maintains intraocular pressure and inhibits passage of intravitreal fluid to the subretinal space. Pressure can also be increased by placing applicators on either side of the sclerotomy site and gently pushing them toward the center of the eye, and these maneuvers also tend to keep the sclerotomy open. Relatively normal intraocular pressure can also be maintained by indenting the sclera at a location far from the sclerotomy site with numbers of cotton applicators. The drainage site is not touched as long as fluid flows through it. Sudden and significant increases in intraocular pressure are avoided to reduce chances of incarceration of the retina in the sclerotomy, and any sudden cessation of drainage requires immediate closure of the sclerotomy and internal examination of the sclerotomy site with the indirect ophthalmoscope. The appearance of pigment granules suspended in the draining subretinal fluid usually indicates that the last of the subretinal fluid is exiting the eye. When drainage ceases, the sclerotomy site is closed by temporarily tying the sutures over an exoplant or by pulling together the ends of an encircling band. If the locations of buckling material will not adequately close the sclerotomy, the scleral incision is closed with the preplaced suture prior to significant elevation of intraocular pressure with the buckle. The eye is quickly inspected following closure of the sclerotomy site and a preliminary adjustment of the scleral buckle. The site of drainage is first evaluated for signs of subretinal bleeding, retinal incarceration, and iatrogenic hole formation, and management of these relatively unusual introperative problems is briefly discussed later. The amount of persistent subretinal fluid is then determined, and the need for further drainage is considered. Significant subretinal fluid is allowed to persist if the optimal amount of buckling nearly closes the retinal break(s). If drainage of additional subretinal fluid is required, the initial sclerotomy site must be closely evaluated with mobile scleral depression, in which a cotton-tipped applicator is rolled circumferentially beneath the area of drainage. If the pigment epithelium is clearly not in contact with the retina, the sclerotomy site can be reopened by reducing intraocular pressure and/or removing the portion of the exoplant that covers the scleral incision. Additional drainage usually occurs spontaneously, or it can be initiated by gently manipulating the edges of the sclerotomy with applicators or a forceps. In some cases, particularly those with exceptionally viscous subretinal fluid, the retina may flatten completely at the site of the sclerotomy while large amounts of subretinal fluid persist elsewhere. In this situation, additional sclerotomies must be performed if additional drainage is required to produce an adequate buckling effect. An alternative and increasingly popular method of draining subretinal fluid is to insert a small 25-30-gauge needle into the subretinal space using direct visualization with the indirect ophthalmoscope or operating microscope. The needle is usually attached to a tuberculin syringe from which the plunger has been removed. Digital pressure is exerted on the eye or significant intraocular pressure is maintained with traction sutures as the subretinal fluid passively exits the eye. The needle is dynamically positioned to remain within subretinal fluid, and it is slowly retracted as the retina approximates its tip.
Adjustment of scleral buckle
Following drainage of appropriate amounts of subretinal fluid, an optimal scleral buckling effect is created by adjusting the scleral sutures and the length of the encircling band. Broad sutures over the portion of the buckle supporting large retinal breaks are first temporarily tied in a manner intended to provide optimal width and height. If intraocular pressure remains low and the breaks are in optimal position, the sutures are permanently tied. The height of the buckle is adjusted if it is inadequate or excessive. If a fold of retina continues to communicate with an open retinal break ("fish-mouth phenomenon") following buckle adjustment, a variety of manipulations can be used to solve the dilemma, as noted later. If an encircling band has been used in combination with a wider circumferential tire of hard silicone, its ends are overlapped to provide a modest buckling effect supporting the posterior edge of the vitreous base in areas not occupied by the tire. If the encircling band is used without a tire, it is adjusted to create an indentation of somewhat greater height. The degree to which the band should be tightened depends upon the intraocular pressure, the nature and extent of vitreoretinal pathology, and the necessity of a subsequent intravitreal gas injection. If the intraocular pressure remains quite soft following appropriate adjustment of the scleral buckle and retinal breaks are flat, balanced salt solution should be injected into the vitreous cavity via the pars plana. Attempts to restore normal pressure with further increases in buckle height can lead to a number of postoperative problems. Indications for gas injections are described below. The need to replace buckling materials or sutures is quite unusual if appropriate localization of retinal breaks has been performed. However, augmentation of the buckle in certain areas is frequently desired. This can be accomplished with suture adjustments if wide elements are already in place. If an encircling band has been used, and an augmentation of the buckle is needed in an area supported only by the band, hard silicone pieces of an appropriate size may be placed beneath the band, and these usually are not sutured.
Accessory techniques
Although scleral buckles are successful in producing a functional closure of retinal breaks in most cases, their effectiveness can be enhanced with accessory techniques, including intravitreal fluid and gas injections, gas-fluid exchanges, and postoperative laser photocoagulation. The first option, injection of balanced salt solution, is employed primarily to restore normal intraocular pressure to a hypotonous eye. Intravitreal gas injection and air fluid exchange are usually used to assist in the internal closure of retinal breaks. Laser photocagulation is usually not used at the time of scleral buckling but can be employed to create or augment a chorioretinal adhesion postoperatively.
Intravitreal injection of balanced salt solution
The solution is drawn into a syringe, and all air is carefully eliminated. A 0.5-inch, 30-gauge disposable needle is recommended. As the surgeon grasps the sclera with a twist pick or forceps, the needle is introduced into the vitreous cavity 3mm (pseudophakic/aphakic eye) or 4 mm (phakic eye) posterior to the limbus. The tip of the needle is directed toward the geometric center of the globe. In a very soft eye, the tip of the needle may occasionally elevate the pars plana epithelium without perforation. Injection in that situation would produce detachment of the pars plana and retina. This can be avoided by direct visualization of the tip of the needle through the pupil and, when the needle is definitely within the vitreous cavity, the injection can be safely carried out. In phakic eyes, care must be taken to keep the tip of the needle near the middle of the vitreous cavity to avoid touching the lens and causing a subsequent cataract. After injection of the required volume to restore normal intraocular pressure, the needle is withdrawn. The self-sealing wound does not require suturing.
Intravitreal gas injection.
Gas injections to internally tamponade retinal breaks are commonly performed in association with scleral buckling procedures. Gas is usually injected after the breaks have been treated and well positioned on the scleral buckle. The type and volume of injected gas depend upon the available potential space within the vitreous cavity as well as the size of retinal break(s) and the desired duration of tamponade. The injection technique is similar to pneumatic retinopexy. In the vast majority of cases in which the buckle is in appropriate position, an effective tamponade is necessary for only 24-48 hours. Keeping in mind a 0.30 ml bubble will maintain contact with a 90-degree arc of the retina. However, injection of even 0.3 ml of gas into an eye with normal intraocular pressure will cause a marked increase in intraocular pressure and a transient occlusion of the central retinal artery. Therefore, the eye must be quite soft prior to an injection of a large gas bubble, or a smaller volume of an expansile gas is employed. In cases in which a longer tamponade effect is desired, more insoluble gases such as sulfur hexafluoride (SF6) and perfluoropropane (C3F8) are used. These possess two potentially favorable characteristics: expansile qualities if injected as pure gas, and longer duration in the eye. Numerous techniques of gas injection have been described. The injection is performed 3-4mm posterior to the limbus. The site of injection is made uppermost so the bubble will tend to remain at the site of the needle tip, to avoid formation of multiple small bubbles. Using indirect ophthalmoscopy, passage of the needle tip through the pars plana epithelium may be confirmed. The needle is then withdrawn enough to leave a short length of the needle in the eye. The predetermined volume of gas is injected moderately rapidly. The optic nerve is then inspected to document perfusion of the retinal vessels. If pulsations are visualized in a patient with normal blood pressure, no tension-lowering manipulations are performed. If pulsations are not observed and cannot be produced with digital pressure, the intraocular pressure is lowered with a paracentesis if pulsations have not resumed after several minutes. The height of the buckle can also be reduced or some of the gas can be removed if necessary to restore patency of the central retinal artery. Of note, in pseudophakic patients, there is a risk that the previously injected gas bubble will be pulled forward into the anterior chamber during an anterior paracentesis which is performed after gas injection; anterior paracentesis is preferably performed prior to gas injection.
Gas-fluid exchange
When a relatively large volume of gas is required, and intraocular pressure cannot be adequately lowered by any other means, fluid can be removed from the vitreous cavity prior to gas injection. This is commonly performed during vitreous surgery but rarely during routine scleral buckling operations. If a total posterior vitreous detachment has been documented preoperatively, and the retina is relatively flat, fluid can be aspirated from the space behind the posterior hyaloid with a 25-gauge needle inserted via the pars plana. It is helpful to have a very small volume of balanced salt solution in the syringe. When the needle is in place, 0.1-ml solution should be injected. This displaces vitreous gel at the tip of the needle and thereby facilitates the aspiration of fluid vitreous. Alternatively, a vitrectomy-cutting instrument can be used under indirect ophthalmoscopic control. This alternative has the advantage of reducing possible vitreoretinal traction but involves extra cost and time for setting up equipment. Gas is injected after the eye has been softened by either technique.
Laser therapy
In cases in which cryotherapy can not be appropriately performed or excessive treatment is feared, retinal breaks can be treated with laser therapy after the retina is totally reattached. Some surgeons prefer this technique as a routine. However, a thin film of persistent subretinal fluid following routine drainage may result in excessive laser energy being required to produce a visible burn, and it may be easier to perform laser treatment a day or two following the primary procedure. In addition, in selected cases, recurrent retinal detachments following routine scleral buckling can be repaired by reattaching the retina with a gas bubble followed by later laser photocoagulation.
Closure of incisions
After placement and adjustment of the scleral buckle have been completed, any excess silicone or relatively sharp implant edges should be carefully trimmed, especially in very anterior buckles. Irrigation of the operative field with an antibiotic solution is usually performed. Peritomies are closed with interrupted or running absorbable sutures. Some postoperative suture discomfort can be eliminated by burying the suture knots.
Results of scleral buckling
A single scleral buckling procedure is usually successful in 85% - 90% of cases [reference needed]. If postoperative retinal detachment is due to a new tear or an inadequate buckling effect unassociated with proliferative vitreoretinopathy (PVR), modification of the scleral buckle and creation of a chorioretinal adhesion will often reattach the retina. It remains difficult to provide precise success figures because of the impact of case selection and the lack of specific details regarding reasons for failure. The majority of macula-on detachments do not lose visual acuity, but a return to normal in eyes with preoperative macular involvement sometimes may occur. Barring complications, postoperative vision is primarily determined by the level of preoperative visual acuity, mostly dependent on pre-operative macula status.
Complications of scleral buckling
Common complications of scleral buckling
Complications have traditionally been divided into intraoperative and postoperative categories. Common intraoperative complications can make the operation more difficult and interfere with the ability to achieve surgical goals in an efficient fashion, and familiarity with their management is essential. Complications that follow scleral buckling procedures include factors that affect anatomical and/or visual outcomes and patient satisfaction.
Selected Intraoperative Complications
Complications during scleral buckling procedures are various and a select few will be discussed here.
Corneal Complications
Corneal epithelial edema or trauma to the epithelium impairs the clarity of the cornea and impedes optimal visualization of the retina at surgery. Several factors are responsible for a majority of the cases. It is routine in some operating rooms to instill preoperative topical anesthetics prior to dilating drops, and this practice should be avoided, because topical anesthetics adversely affect the corneal epithelium. For the same reason, concentrated detergents should not be used in the surgical prep. Prolonged elevated intraocular pressure during surgery may result in edema of the epithelium, especially in elderly adults with an aging endothelium and in diabetic patients. Some cases respond well to mechanical expression of the edema fluid by pressing firmly on a dry, cotton-tipped applicator as it is rolled across the cornea. This maneuver expresses the edema but does not remove the epithelium. In persistent cases of epithelial edema, the central portion of the epithelium can be removed with gentle strokes of a tilted knife blade, avoiding damage to Bowman’s membrane.
Pupillary Complications
Intraoperative miosis is due most commonly to hypotony. Intravitreal gas that comes in contact with the iris can also cause miosis. If the pupil does not dilate with cycloplegic and mydriatic drops, intracameral epinephrine and/or iris retractors may be considered.
Mersilene sutures traversed through the sclera, choroid, and retina are visible in this image. Surrounding the surgical site, evidence of chorioretinal atrophy is apparent. This photograph captures the aftermath of an encircling episcleral buckle procedure performed 18 years ago, shedding light on the fact that it may have no long-term severe consequences. (Courtesy of J. Khadamy)
Scleral perforation with suture needles
Penetration of sclera with a suture needle most commonly occurs with a long deep intrascleral pass of the suture needle combined with thinned sclera. Penetration of the globe is usually heralded by the sudden appearance of subretinal fluid at either end of the suture bite. Occasionally, pigment granules and blood are also observed. When an inadvertent penetration is recognized, all traction upon the eye should be immediately released, and the retina should be inspected with the indirect ophthalmoscope. The suture is temporarily left in place to minimize further drainage. Choroidal hemorrhage is the most common complication of inadvertent penetration, occurring in approximately a fourth of cases. Usually it is minimal, but if active bleeding is observed, the pressure in the eye should be immediately raised with scleral depression over the bleeding site. If the macula was involved in the detachment, blood can ultimately settle in that location and severely compromise visual acuity. This is best prevented by draining no additional subretinal fluid after the penetration. Postoperatively the patient is positioned so that the blood will settle away from the macula. If an inadvertent penetration causes an iatrogenic hole in attached or detached retina, or if an iatrogenic hole cannot be ruled out, the site is usually treated with cryotherapy and supported on a modified scleral buckle. If it is near a retinal tear, a wider piece of silicone is sutured in place to support both the break(s) and the penetration site.
Complications of draining subretinal fluid
This step of a scleral buckling procedure is usually considered the most hazardous. Although drainage complications are not statistically associated with subsequent anatomic failure, they can require a modification of the surgical plan and significantly compromise postoperative visual acuity. The three classic complications associated with drainage of subretinal fluid are: hemorrhage, retinal incarceration, and iatrogenic retinal holes. The first of these is the most common, occurring in approximately 3%-4% of cases. Most hemorrhage is confined to a small area surrounding the sclerotomy site, and additional surgical maneuvers are not required. More significant hemorrhages are managed in a fashion similar to that described following inadvertent penetration of the globe with a suture needle. Retinal incarceration is observed following drainage in 1%-3% of cases. This usually occurs soon after penetration of the choroid, and it is associated with a sudden cessation of drainage. Increased intraocular pressure contributes to this problem. When retinal incarceration is suspected, all traction upon the eye is released, and the fundus is quickly examined with indirect ophthalmoscopy. Usually the incarceration is mild, with only a localized depression in the retina surrounded by radiating striae. The drainage site should be closed immediately with the exoplant or with a suture, and then the retina should be more thoroughly evaluated with scleral depression. If an iatrogenic retinal break is discovered, it is usually treated with light cryotherapy. Incarceration sites are generally supported on a buckle in a manner similar to that described regarding inadvertent suture penetrations. In the absence of incarceration of the retina, drainage of subretinal fluid rarely causes iatrogenic retinal breaks, but the drainage site is carefully evaluated by indirect ophthalmoscopy following completion of drainage to rule this out.
“Fishmouthing” of retinal breaks.
This radial folding of the retina at the site of a large horseshoe tear is sometimes observed following a circumferential scleral buckling procedure. The open break may prevent reattachment if it is not functionally closed. Increasing the height of the circumferential buckle is tempting but counterproductive, as it worsens the radial folding. The best way to manage this phenomenon is to reduce the height of the buckle beneath the break(s) and to inject an intravitreal gas bubble.
Complications of intravitreal gas injections
The two common dilemmas that arise following intravitreal gas injections are difficulties in visualization and elevation of intraocular pressure, and the former can aggravate the management of the latter. Although the minification of details and altered reflexes due to intravitreal gas bubbles makes visualization more difficult, it is sometimes possible to see through the center of a bubble 1 cc or more in volume. Alternatively, the eye can be tilted so that the bubble will float out of the way. A more difficult problem occurs when a multitude of tiny bubbles prevent a meaningful view, and the frequency of this problem is reduced by employing optimal injection techniques. If it occurs, the mass of bubbles will frequently float out of the way with appropriate tilting of the eye. Otherwise, a coalescence of bubbles can be encouraged by a series of jerks of the globe and the passage of time. The optic nerve must always be visualized following an intravitreal injection of gas. Occasionally extreme repositioning of the patient is required to be certain that the optic nerve and retina are perfused. If perfusion is questionable, increased digital pressure should be placed upon the eye to elicit pulsations. If these do not occur and if they are not visualized after several minutes, pressure-lowering maneuvers should be immediately instituted. The common options include paracentesis and lowering scleral buckle height.
Selected Postoperative Complications
The most important complication following scleral buckling procedures is anatomical failure, but several other problems can occur with significant frequency.
Increased intraocular pressure
During the early postoperative period, corneal epithelial edema, pain, or pulsation of the central retinal artery can occur due to increased intraocular pressure. Patients likely to develop the complication include narrow-angle glaucoma suspects, elderly patients with enlarged lenses, those who have subluxated lenses, patients with pre-existing open-angle glaucoma, those who received relatively large injections of intraocular gas or intracameral injections of sodium hyaluronate, and especially those with prominent postoperative choroidal detachment. Moderate elevations to 30 mm Hg are frequently transient and usually do not require treatment. Higher pressures may be due to some degree of angle closure. The most common causative mechanism is anterior displacement of the lens and iris by the presence of choroidal detachment or swelling and subsequent compromise of the filtration angle. The condition is usually self-limiting and responds well to intravenous or oral acetazolamide. Topical corticosteroids can be useful to discourage the development of peripheral anterior synechiae. If the administration of acetazolalamide, topical glaucoma drops, hyperosmotic agents, systemic corticosteroids, and/or cycloplegic drops fails to control the pressure, and the filtration angle is obscured by peripheral iris, it may be necessary to drain the choroidal detachment via posterior sclerotomies to reopen the angle and prevent permanent anterior synechiae.
Endophthalmitis and scleral abscess.
Bacterial endophthalmitis is an exceptionally rare but potentially devastating complication following retinal detachment surgery, but is typically only associated with entering the globe. The earliest clinical symptoms and signs, developing by the third to fifth postoperative day, are pain, chemosis, lid edema, the appearance of localized vitreous opacification or petechial hemorrhages, and acute subretinal exudate at the level of the scleral buckle, noticeable in an examination by indirect ophthalmoscopy. The prompt recognition of such symptoms and signs, even if relatively subtle, is critical to proper management of endophthalmitis. The contemporary management of genuine endophthalmitis is beyond the scope of this chapter. Scleral abscesses usually present in a similar fashion clinically, but they are usually a sterile vitreous inflammatory reaction combined with evidence of severe presumed infection of the sclera. When this rare syndrome is recognized, the buckling material is removed and prompt and intense periocular, and sometimes intraocular, antibiotic therapy is instituted.
Choroidal Detachment
Some degree of choroidal detachment develops in approximately 5% to 10% of scleral buckling procedures. The fluid is assumed to be a transudation from the choroid. Its formation is related in part to advanced age and in part to surgical factors, such as thermal treatment, trauma to the choroid, hypotony, and obstruction of vortex vein outflow with resultant increased intravascular pressure throughout the choroidal vascular system. Although care is taken at surgery to minimize procedural factors, choroidal detachment is not entirely avoidable. In a majority of cases, the overlying retina remains in satisfactory apposition to the treated retinal pigment epithelium, and because of eventual spontaneous resolution of the choroidal detachment, the prognosis is usually good. Choroidal detachments that cause angle closure and glaucoma are a more serious problem, and their management was discussed above. Massive detachments of the choroid may actually touch in the central vitreous (“kissing choroidals”). This problem may require surgical drainage to prevent the possibility of adhesion of retina to retina with subsequent tractional retinal redetachment. Hemorrhagic choroidal detachments are less frequent than the common serous detachment, and the prognosis is poorer. The majority of cases follow a self-limiting course to spontaneous resolution, but some have a considerable chronic inflammatory response and are associated with failure of the retina to reattach. In some cases, the hemorrhage passes through the retina to produce a dense vitreous hemorrhage.
Later periocular infection and implant extrusion.
The various implant materials and sutures used in retinal detachment surgery are foreign bodies and can become nidi for infection. The organisms responsible are frequently coagulase-positive staphylococci but may be Gram-negative bacteria. The incidence varies, depending on the surgical technique. The major clinical indication is pain, which should be considered a symptom of periocular infection until proved otherwise. Other signs are localized inflammation of the conjunctiva, subconjunctival hemorrhage, point tenderness, purulent discharge, and occasionally a draining of the sinus tract through the conjunctiva. Fortunately, associated intraocular involvement is exceptionally rare. For extraocular infections occurring within the first few weeks of surgery, medical treatment may be administered to suppress the infection and gain sufficient time for the retina to firmly reattach before the scleral buckle is removed. For infections diagnosed after a few weeks, treatment consists of removal of the scleral buckle, identification of the organism, and appropriate antibiotics. Any evidence of intraocular extension of infection, however, demands immediate removal of the buckle. When the retina appears to be reattached and the chorioretinal adhesion mature, removal of the buckle only infrequently leads to redetachment. However, if this occurs, a second procedure may be performed as soon as all evidence of residual infection has disappeared. Late exposure of implant materials occurs as a consequence of disintegration of overlying tissues associated with infection, inadequate coverage of scleral implants at surgery, or migration of the implant. Segments of buckle material may erode through overlying Tenon’s capsule and conjunctiva (Figure 5-20). Segmental episcleral buckles have a greater tendency to erode than narrower encircling bands. Although clinical signs of bacterial infection may or may not be present, virtually all such cases can be assumed to be infected. Providing the retina is well attached, there is no advantage in attempting to modify the scleral buckle or to patch over the exposed portion. Most exposed buckles eventually have to be removed, and indications for removal include clinical infection, pain, or a cosmetic problem.
Cystoid macular edema (CME)
Cystoid macular edema (CME) is a common complication of ocular inflammatory diseases and surgery. Abnormal leakage from perifoveal capillaries causes intraretinal edema that accumulates in a characteristic cystoid pattern. Three to six weeks following successful scleral buckling, some degree of CME can develop in as many as a third of patients. There is no strong relationship between the incidence of CME and preoperative macular detachment, drainage of subretinal fluid, or the type of scleral buckle, but the problem is more common in pseudophakic cases.
The effect of CME upon final visual acuity is uncertain, primarily because of a lack of studies correlating preoperative vision with a variety of additional variables associated with scleral buckling procedures. Nevertheless, CME contributes to a loss of visual acuity in eyes without preoperative macular detachment, and it probably limits recovery of vision in eyes in which the macula was detached. Eyes with objective signs of intraocular inflammation are treated with topical and periocular corticosteroids. Eyes without visible intraocular inflammation are treated with topical non-steroidal anti-inflammatory agents and/or a trial of topical steroids.
Epimacular proliferation
Epiretinal membranes that distort or cover the macula are a relatively common cause of disappointing visual acuity following successful scleral buckling surgery. The reported incidence of this problem varies considerably due to varying criteria employed for its diagnosis. Macular puckers have been reported in from 2% to 17% of successfully buckled cases. Although the precise cause of macular pucker following reattachment surgery is unknown, it is likely that many develop from pigment epithelial cells that pass through the retinal break(s) into the vitreous cavity, not from the buckling procedure itself. The cells then become attached to the surface of the retina in the posterior pole and subsequently proliferate and contract. The development of macular puckers may be associated with a variety of factors, including vitreous hemorrhage and intraocular inflammation associated with preoperative problems and/or operative techniques. Removal of epiretinal membranes with vitrectomy techniques is the only effective means of treating macular puckers. Surgery is advisable in cases of significant epimacular fibrosis in which relatively poor postoperative visual acuity is associated with a history suggesting potential for relatively good macular function.
Proliferative vitreoretinopathy (PVR)
Proliferative vitreoretinopathy (PVR) is the only common cause of ultimate failure following retinal reattachment surgery, regardless of the employed surgical method. Unless this occurs, the vast majority of initial failures can be successfully repaired. This cell-mediated process is associated with the production of fibrocellular membranes on the posterior vitreous surface and on both surfaces of the retina. Subsequent contraction of the membranes causes significant shortening of the retina, which prevents reattachment or causes recurrence of detachment, even if all retinal breaks are closed.
Factors which cause a significant breakdown in the blood-aqueous barrier and which allow an increased number of pigment epithelial cells to enter the vitreous cavity are also associated with an increased incidence of PVR. Preoperative intraocular inflammation is clearly related to the incidence of postoperative PVR. Vitrectomy techniques are routinely employed to reattach retinas associated with extensive intravitreal and periretinal membranes. In combination with vitrectomy, an encircling scleral buckle is also an important component of surgery for the repair of retinal detachments associated with PVR. At the time of vitreous surgery, previously placed segmental buckles are usually replaced or augmented with encircling procedures, and selected previously-placed encircling buckles may be modified in an effort to counteract significant peripheral traction forces.
Recurrent retinal detachment.
Recurrent or persistent retinal detachment is the most significant complication of scleral buckling, and the severity of this problem is related to its cause. Retinal detachment following scleral buckling surgery occurs in from 5% to 20% of primary operations, and the vast majority are associated with open retinal breaks. If postoperative retinal detachment is due to a new tear or an inadequate buckling effect unassociated with PVR, modification of the scleral buckle and creation of a chorioretinal adhesion will usually reattach the retina. In selected cases, intraocular injection of a gas bubble with or without supplemental cryopexy or laser, and appropriate positioning may be sufficient to achieve reattachment and avoid revision of the buckle. If extensive PVR is responsible for the surgical failure, vitrectomy techniques are usually required for a successful reoperation.
Altered refractive error.
Scleral buckling techniques with an encircling component usually cause a myopic change in the refractive error because of their effect upon axial length. This is a significant drawback of this procedure, especially in pseudophakic patients who have a desirable refractive status. An average increase in axial length of approximately 1 mm induces an average myopic shift of approximately -2.50 diopters. The effect of radial scleral buckles is usually less significant. Significant astigmatic changes are very unusual unless the buckles are quite anterior, but some degree of astigmatism is occasionally caused by indenting the sclera near the ora serrata. Although changes in refractive error may be relatively unimportant in eyes with poor postoperative visual acuity, the induction of significant anisometropia can be devastating to some patients, particularly those with excellent postoperative visual acuity and an emmetropic fellow eye. Avoiding anisometropia is an important goal of some alternatives to scleral buckling.
Muscle imbalance
Some degree of extraocular muscle imbalance occurs in a significant number of patients undergoing scleral buckling procedures. Most of these abnormalities are temporary and due to intraoperative muscle damage. Nevertheless, some degree of permanent muscle imbalance is not uncommon. Causes of strabismus include the following: (1) abnormal adhesions between the muscle and the sclera or Tenon's capsule, (2) injury to the muscle from surgical trauma, (3) mechanical disturbances due to the location and shape of buckling materials, and (4) problems associated with disinsertion or repositioning of a muscle. Factors associated with postoperative muscle imbalance include placement of a buckle beneath a muscle, size of buckling material beneath a muscle, and reoperations. Therapy for patients with good bilateral vision usually includes an attempt to prescribe prisms to restore fusion. If this is unsuccessful, surgery is considered. Avoiding postoperative muscle imbalance is a major benefit attributed to alternative reattachment procedures.
Additional Resources
American Academy of Ophthalmology: The repair of rhegmatogenous retinal detachment. Information Statement. Ophthalmology 1990;97:1562–1572.
Brinton DA, Wilkinson CP. Retinal Detachment. Principles and practice. 3rd Edition. 2009; New York, Oxford University Press, pp149-180.
Griffith RD, Ryan EA, Hilton GF: Primary retinal detachments without apparent breaks. Am J Ophthalmol 1976;81:420–427.
Hilton, GF, Grizzard WS, Avins LR, et al: The drainage of subretinal fluid: a randomized controlled clinical trial. Retina 1981;1:271–280.
Wilkinson CP. Scleral buckling techniques: A simplified approach. In Guyer DR, Yannuzzi LA, Chang S, et al (eds) Retina-Vitreous-Macula 1999; Philadelphia, WB Saunders, pp 1248-1271.
Williams GA, Aaberg Jr TA: Techniques of scleral buckling. In: Ryan SJ Wilkinson CP (eds) Retina (Volume 3, 4th Edition). Philadelphia ; Elsevier Mosby 2006, pp 2035-207
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Headaches in Over 12s: Diagnosis and Management [NICE CG150]
CPG
Date: Friday, December 17, 2021
Overview
Evaluates the diagnosis and management of tension-type headache, migraine, cluster headache and medication overuse headache in adults and young people (aged 12 years and older).
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Resource Details
Date:December 17, 2021 Contact: practice@apta.org Content Type: CPG
Symptoms & Conditions
Headaches
Patient Age
Adult (25-44), Aged (65+), Adolescent (13-18), Young Adult (19-24), 80 and over (80+), Middle Aged (45-64)
Author(s)
National Institute for Health and Care Excellence
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