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Calibration Curve Calculator | Concentration
The standard addition method
How to calculate concentration from the calibration curve?
How to use our calibration curve calculator?
How to calculate unknown concentration from the calibration curve? An example
There are many ways to calculate the concentration of an unknown sample: if your experiment has matrix effects, you can use our calibration curve calculator to find it out! We use the standard addition method to help you — if you want to learn more about this, keep on reading. Here you will find:
What a calibration curve is and its different types;
When we use the standard addition method; and
How to calculate the concentration from the calibration curve.
In addition, it will provide you with a step-by-step tutorial on how to calculate the unknown concentration based on the calibration curve. Ready?
Calibration is a measurement technique to ensure that a method/instrument provides accurate results. Chemists write entire books about this topic, but we will try to keep it simple!
Analytical chemistry needs calibration: the reliability of a method of measurement relies on the correct interpretation of the relationship between the concentration of an analyte and the signal of the instrument used. A well-calibrated environment ensures that the results of an analysis will be accurate.
The process of calibration requires an understanding of the concept of calibration curve. This curve (though it is often a straight line) is obtained by testing a certain amount of samples with known concentration with the desired instrument, and then fitting the results using the mathematical model explaining the operations of the method. A plot of the curve shows the instrumental response (the so-called analytical signal) to an analyte (the substance which is measured) and allows to predict the concentration in an unknown sample.
There are many calibration curves types, differentiated by the kind of answer expected from the model:
Linear with non-null intercept — the standard addition technique;
Linear with zero intercept — the working curve technique;
Power (the concentration is raised to the power of a given exponent); and
Polynomial (you can consider this type similar to the polynomial one).
Here we will focus only on the standard addition method, which is also implemented in our calibration curve calculator: keep on reading to see if it fits your problem!
Our calibration curve calculator uses the standard-addition method to compute the value of concentration. The standard addition calibration is used when the sample comes with a matrix that gives a constant background signal in the measurement. Think of it as other solutes if their concentrations don't change, or as the signal of the solvent. Here you will learn how to use this method!
First thing: you need to build the calibration curve. One or more standards are required. A standard is a sample with a known concentration. In order to be known, a process of validation is required; this is however a pretty complex process, and it's not relevant here.
In the absence of standards, prepare a set of samples with different concentrations (if you need help with it, check our serial dilution calculator). Make sure that the value of concentration is included in the range of the samples. Measure your samples with the desired instrument: you will obtain a set of instrumental responses. Record them several times (usually three) — this will help reduce the uncertainty associated with the measurement process.
A linear regression model is used to fit the data. The plotted data represents the instrumental response (signal) vs. the concentration. You'll obtain two parameters, and they are fitted by the function:
y = a \cdot x + b
This is the calibration curve equation: here,
a
is the angular coefficient of the line, which translates to the sensitivity of the instrument.
b
is the intercept, and it corresponds to the background signal of the matrix. The two variables
y
x
are, respectively, the instrumental response and the concentration. As you can see. the intercept corresponds to the instrumental response for null concentration (
x = 0
💡 A linear fit is a regression technique that finds the line deviating the smallest amount from any sample in a set.
That's it! Now you have a calibration curve obtained by using the standard addition method. In the next section, you'll learn how to calculate the unknown concentration from the calibration curve equation.
The standard addition method finds applications in various techniques in analytic chemistry: absorption spectrometry (which uses the Lambert-Beer law), mass spectrometry, and gas chromatography are just some examples.
In order to calculate the unknown concentration, the equation of the linear fit is transformed into the equation:
x = (y - b)/a
Here you subtract the background
b
(the effect of the matrix) from the signal
y
, and then you divide by the sensitivity of the instrument used,
a
. The result is the concentration,
x
, with units depending on the technique with which the analysis is performed.
If you already have the values of the linear fit's parameters, simply insert them in the calibration curve calculator in their respective fields.
🙋 We decided to omit units from our calculator, since the signal coming from the instrument depends on the physical phenomena employed in the analysis. Remember to be consistent — finding the units of the concentration of your unknown sample won't be hard!
Now write the signal, and find out the unknown concentration.
Do you know that you can use our calculators in "reverse" too? Just fill the concentration field, and find out the expected signal!
If you don't know the parameters of your fit but you have the data from the standard samples, you can use our linear regression calculator to find these values. Check it out!
Calculate the equation which describes the calibration curve. Let’s assume that it is
y = 0.5x + 0.1
Transform the above equation into
x = (y - 0.1)/0.5
Measure the instrumental response of the unknown sample. Let’s assume that it is
2.1
. The units vary from experiment to experiment, and from instrument to instrument: we kept things general.
You can calculate the unknown concentration by substituting the values:
x = (2.1 - 0.1)/0.5 = 2/0.5 = 4
If you want to recompute concentration (for example switching from molarity and percentage concentration), you can use our concentration calculator. If you want to calculate the concentration of a diluted solution, you can use our solution dilution calculator. To convert between concentration units, use our molality and molarity calculators!
How do I calculate an unknown concentration from the calibration curve?
Choose the right calibration technique, for example, the standard addition method.
Measure the instrumental response (signal) from your solution.
Determine the parameters for the method: background and sensitivity.
Compute the concentration by subtracting the background from the response and dividing this difference by sensitivity.
That's all! Enjoy the result!
When can I use the standard addition method?
The standard addition method is best suited for models that include a background signal coming from a matrix. This translates into the presence of an intercept in the regression curve.
What do I need to calculate the concentration from the calibration curve?
Every calibration curve is defined by a set of parameters: in the case of linear calibration curves, they are usually:
The slope of the line (its angle in relation to the horizontal axis); and
To find out these parameters, you need to measure the signal obtained from a set of samples with known concentrations.
Lastly, measure the response from the unknown sample: that's the final quantity you need to calculate the unknown concentration.
Which calibration curve to use for the absorption spectroscopy technique?
Since the absorption spectroscopy technique has a constant background, you need to consider it when you build the calibration curve: the best model for this technique is the standard addition method. Find out more about it at Omni Calculator's website!
An example of a calibration curve, where you can see the various parameters necessary to calculate the concentration of an unknown sample. The sensitivity a of the instrument corresponds to the slope of the line, while the background signal b is equal to its intercept.
Background (b)
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Empirical Analysis of Commercial Housing Sales Based on EARCH(1,1) Model
Empirical Analysis of Commercial Housing Sales Based on EARCH(1,1) Model
Since the 1980s, China’s commercial housing market has shown an unprecedented rapid development, and the commercial houses still has a high price. This paper studies the sales rate of commercial housing sales to find an appropriate model, and it analyzes the volatility of the commercial housing market to describe the sustainable development of the commercial housing market. By selecting month data of China’s commercial housing sales from January 2006 to October 2018, this paper uses EViews7.2 and the ARMA Model as the tool in order to establish EARCH(1,1) through the method of quantitative analysis. It is found that the yield of commercial housing sales has obvious cluster, asymmetry and leverage effect, and the impact of adverse news on the commercial housing market is more significant than the impact of favorable news.
Sales Volume of Commercial Housing, ARMA Model, EARCH Model, Leverage Effect
In recent years, China’s development rate has been very rapid, and the people’s living standards have also been significantly improved. China’s commercial housing market has ushered in an unprecedented period of development especially in the context of the establishment of a comprehensive well-off society. Many domestic scholars have conducted a lot of researches for the future development prospects of the commercial housing market. Wei Junjia and Zhang Chi [1] (2014) explored the commercial housing in Nanning through the method of cointegration analysis and stationarity analysis. Shi Huiling [2] (2018) used the cross-sectional data of commercial housing sales and GDP in 30 provinces to explore the relationship between commercial housing prices and economic development. Zheng Lan [3] (2006) used the VAR model to discuss the transmission effect of monetary policy in the commercial housing market. Tian Tian and Dong Weijuan [4] (2009) studied the impact of macroeconomics on the demand for commercial housing. None of the above documents have studied the volatility of commercial housing market. This paper establishes commodity housing sales rate of return sequence and also makes a series of quantitative analysis of the residual of the model in order to EARCH model. The development status of China’s commercial housing sales market is analyzed, and relevant suggestions are put forward as a reference.
2.1. ARCH Effect Test
If the random perturbation term of the subject model at-ARCH(q) [5] , the regression equation can be established:
{a}_{t}={\alpha }_{0}+{\alpha }_{1}{a}_{t-1}^{2}+\cdots +{\alpha }_{q}{a}_{t-q}^{2}
The null hypothesis and alternative hypothesis to be tested are:
\begin{array}{l}{H}_{0}:{\alpha }_{1}={\alpha }_{2}=\cdots ={\alpha }_{q}=0\\ {H}_{1}:\exists {\alpha }_{i}\ne 0\text{\hspace{0.17em}}\left(1\le i\le q\right)\end{array}
LM=n{R}^{2}~{\chi }^{2}\left(q\right)
where, n is the number of sample data; R2 is the determinant of the auxiliary regression formula. When significance level
\alpha
and degree of freedom q are known, if
LM>{\chi }_{\alpha }^{2}\left(q\right)
, the null hypothesis is not accepted, indicating that there is ARCH effect in this sequence. Otherwise, there is no ARCH effect.
2.2. EARCH Model
The EARCH model is also called exponential GARCH model, which was proposed by Nelson [6] in 1991. The conditional variance expression of the model is:
\mathrm{log}\left({h}_{t}\right)={\alpha }_{0}+\underset{j=1}{\overset{p}{\sum }}{\theta }_{j}\mathrm{log}\left({h}_{t-j}\right)+\underset{i=1}{\overset{q}{\sum }}\left[{\alpha }_{i}|\frac{{\epsilon }_{t-i}}{\sqrt{{h}_{t-i}}}|+{\phi }_{i}\frac{{\epsilon }_{t-i}}{\sqrt{{h}_{t-i}}}\right]
where the expression of
{h}_{t}
{h}_{t}={\alpha }_{0}+{\alpha }_{1}{\epsilon }_{t-1}^{2}+\cdots +{\alpha }_{q}{\epsilon }_{t-q}^{2}
\phi \ne 0
, the information function is asymmetric. When
\phi
is less than 0, leverage is significant.
The skewness measures the symmetry of the data. Normal distribution has a skewness of 0, the right is positive, and the left is negative. The calculation formula of the skewness is as follows:
S=\frac{\frac{1}{n}{\displaystyle \underset{i=1}{\overset{n}{\sum }}{\left({x}_{i}-\stackrel{¯}{x}\right)}^{\text{3}}}}{{\left(\frac{1}{n}{\displaystyle \underset{i=1}{\overset{n}{\sum }}{\left({x}_{i}-\stackrel{¯}{x}\right)}^{\text{2}}}\right)}^{\frac{3}{2}}}
The kurtosis measures the flatness of the data distribution. The large data distribution at the tail has a large kurtosis value. The normal distribution has a kurtosis value of 3. Its formula is as follows:
K=\frac{\frac{1}{n}{\displaystyle \underset{i=1}{\overset{n}{\sum }}{\left({x}_{i}-\stackrel{¯}{x}\right)}^{\text{4}}}}{{\left(\frac{1}{n}{\displaystyle \underset{i=1}{\overset{n}{\sum }}{\left({x}_{i}-\stackrel{¯}{x}\right)}^{\text{2}}}\right)}^{2}}
The standard deviation can reflect the degree of dispersion of the data set, and the calculation formula is as follows:
S=\sqrt{\frac{1}{n-1}{\displaystyle \underset{i=1}{\overset{n}{\sum }}{\left({x}_{i}-\stackrel{¯}{x}\right)}^{\text{2}}}}
3. Empirical Analysis of ARCH Model
This paper takes the data of China’s commercial housing sales from January 2006 to October 2018 as the sample time series, and conducts data analysis with the help of EViews7.2 [7] [8] . In order to study the ARCH model of commodity housing sales, the ARMA model is used to make an empirical analysis of China’s commodity housing market by determining the return rate sequence of commodity housing sales and the required sequence must be stable. The data comes from China statistical yearbook (2018).
It can be seen from Figure 1 (commercial housing sales) that the sales volume of commercial housing presents an exponential trend on the whole. The annual December is much larger than the annual January, showing seasonal influence. And the unit root test is further used (Table 1).
According to the results in Table 1, the ADF (Augmented Dickey-Fuller test statistic) value is 0.358366. At the significance level of 1%, the sequence accepts the original false, and the P value (Probable value) is 0.9805. It is believed that the unit root exists in the sequence Xt, indicating that the commercial housing sales volume sequence is a non-stationary time series.
In order to study the sequence of commercial housing sales, relevant data are
Figure 1. Time series diagrams of Xt.
processed as follows: two first-order and one seasonal difference were performed on the data, respectively eliminating the exponential trend and seasonal influence of the sequence, so that the new yield sequence obtained does not have the correlation trend and periodic influence, and at this time, the yield sequence is a stationary sequence.
It can be seen from Figure 2 that the sequence Rt is stable, and the continuity and clustering of the sequence can be observed.
According to the results in Table 2, the ADF value is −8.295501. At the significance level of 1%, the sequence does not accept the null hypothesis, and the P value is 0. It is believed that there is no unit root in the sequence Rt, indicating that the rate of return at this time is a stable time series.
Calculate each statistic for the new sequence, as known in Table 3.
Table 3 lists the results of statistics of the return rate of commercial housing sales Rt, in which the skewness is −0.488687, less than 0, indicating the existence of left-deviation phenomenon of the return rate sequence. The kurtosis value was 15.75282, while the kurtosis value of the normal distribution was 3, which was much higher than its range value, indicating that the sequence Rt had a distribution of “sharp peak and fat tail”. The value of the J-B statistic is 954.2733,
Figure 2. Time series diagrams of Rt.
Table 3. Rt sequence statistics.
and its P value is 0, indicating that the sequence Rt does not obey the normal distribution.
In order to study the relationship within the return sequence Rt, ARMA (auto-regressive moving average) model is taken as the main model, ARMA(2,0) was finally determined as the subject model by analyzing the autocorrelation and partial autocorrelation graphs of the yield Rt series, as known in Table 4.
Its parameters are estimated as follows:
{R}_{t}=-0.\text{6364}0\text{9}{R}_{t-1}-0.\text{339977}{R}_{t-2}+{a}_{t}
The sequence autocorrelation LM (Lagrange Multiplier) test for the random perturbation term at of the equation Rt obtained a P value of 0.0034, indicating the existence of autocorrelation in the sequence. Next, the author performs the autocorrelation LM test with order 7 on the sequence, as known in Table 5.
Table 4. Parameter estimation of sequence Rt.
Table 5. LM test for the sequence Rt.
It can be seen from Table 5 that the p-value of the seventh-order statistic is 0.0017, less than 0.01, indicating that the random perturbation term at sequence of the ARMA model has an autocorrelation, and it has the high order ARCH effect, so the return rate sequence Rt can build the GARCH model.
GARCH(1,1) model should be selected to fit the high-order ARCH effect in the yield sequence Rt, and other types of ARCH effect models should be fitted. Table 6 lists GARCH(1,1), GARCH-M(1,1), TARCH(1,1), EARCH(1,1), logarithmic likelihood function values, AIC values and SC values of the return sequence Rt.
According to Table 6, the rate of return sequence Rt is more suitable for EARCH(1,1) model under AIC (Akaike Information Criterion) and SC (Schwarz Criterion) criteria, and its value reaches the minimum, which is 19.68798 and 19.81525 respectively. Meanwhile, the value of the logarithmic likelihood function is the largest, and the value is −1352.471. Therefore, the fitting model EARCH(1,1) is finally determined.
It can be seen from Table 7 that all P values of the coefficients of EARCH(1,1) model are 0, indicating good fitting effect:
Mean value equation:
{R}_{t}=-0.266744{R}_{t-1}-0.312926{R}_{t-2}+{a}_{t}
\begin{array}{c}\mathrm{log}\left({h}_{t}\right)=9.414443+0.450156\mathrm{log}\left({h}_{t-1}\right)\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}-0.524599\frac{|{a}_{t-1}|}{\sqrt{{h}_{t-1}}}-0.775541\frac{{a}_{t-1}}{\sqrt{{h}_{t-1}}}\end{array}
According to the variance equation, the parameters
{\alpha }_{0}=9.41443
{\theta }_{1}=0.450156
are both greater than 0, indicating that the fluctuation of income is positive and in line with the actual situation of commercial housing sales today. the asymmetric coefficient
{\phi }_{1}=-0.775541
is less than 0, indicating that the fluctuation phenomenon of the rate of return on sales of commercial housing in China is asymmetric, when
{\alpha }_{1}+{\phi }_{1}=-1
. 30014, it shows the impact when there is good news, when
{\alpha }_{1}-{\phi }_{1}=0.250942
, it shows the impact of bad news.
3.5. Adaptability Test
The ARCH effect test is performed by the sequence in the variance equation to
Table 7. Parameter estimation of EARCH model.
find out whether this effect has been eliminated. Figure 3 is an autocorrelation and partial autocorrelation test of the standard deviation residual sequence of the EARCH(1,1) model of the sales rate of commercial housing in China.
It can be seen from Figure 3 that the P value is 0.017 when proceeding to the fifth step, and the autocorrelation hypothesis that the significance level is 0.01 standard deviation residual sequence is rejected, indicating that the residual sequence has no ARCH effect. It may be reasonable to establish an EARCH(1,1) model. Next, ARCH effect test can be performed on the residual sequence.
It can be seen from Table 8 that the P value of LM statistic is 0.7131, greater than 0.01, which further indicates that the ARCH effect no longer exists in the residual sequence. Therefore, fitting EARCH(1,1) model is appropriate.
This paper mainly studies the characteristics of the return rate of commercial housing sales and finds that the return rate sequence has ARCH effect, and EARCH(1,1) model is still effective. From 2006 to 2018, China’s commercial housing sales have witnessed unprecedented rapid development. What’s more,
Figure 3. Correlation diagram of standard deviation residual sequence.
Table 8. ARCH effect test results of sequence.
the fluctuation of the return rate of commercial housing sales is positive, and the leverage effect coefficient of model is negative, indicating that the fluctuation of China’s commercial housing sales is asymmetric and has significant leverage effect. At the same time, by analyzing the variance equation of the model, it can be seen that the rate of return on sales of commercial housing has a more significant response to adverse news. In other words, adverse factors in the market have a greater impact on commercial housing market. With the increase of favorable news, the commercial housing market will eventually level off. Through the above empirical analysis, several suggestions can be made for the development of the commercial housing market. The government can regulate and control the commercial housing market; it can increase good news that is conducive to market development. Buyers can determine the purchase intention according to the price increase. The government should adopt policy measures in the commodity housing market to steadily increase sales and consolidate the development of the commercial housing market.
Shen, S.C. and Feng, C. (2019) Empirical Analysis of Commercial Housing Sales Based on EARCH(1,1) Model. Open Journal of Statistics, 9, 299-307. https://doi.org/10.4236/ojs.2019.92021
1. Wei, J.J. and Zhang, C. (2014) Empirical Analysis on the Influencing Factors of Commercial Housing Sales in Nanning City. Economic and Social Development, 12, 16-19.
2. Shi, H.L. (2018) Empirical Analysis of Real Estate Price and Economy—Based on Sectional Data Analysis of Commodity Housing Sales and Gross Domestic Product in 30 Provinces. Era Economy and Trade, 7, 57-58.
3. Zheng, L. (2006) Research on Monetary Policy Transmission Effect in Real Estate Market Based on VAR Model. Hunan University Press, Changsha.
4. Tian, T. and Dong, W.J. (2009) Research on the Impact of Macroeconomic Changes on the Demand for Commercial Housing. Modern Economic Information, 9, 54-55.
5. Engle, R.F. (1982) Autoregressive Conditional Heteroscedaticity with Estimates of the Variance of U.K. Inflation. Econometrica, 50, 987-1007. https://doi.org/10.2307/1912773
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How to use the human punch force calculator
Formula to calculate how much force a human punch exerts
Using our human punch force calculator, you can find out how many newtons of force is in your punch and learn how many pounds of punch force a human can deliver on average.
Here, we will tell you about your potential impact force and the pressure of your punch, all based on your weight, punching speed, time of delivery, and the surface area of your punch.
Please continue reading to learn about:
How much punching force a human has;
What is the average punching force; and
How amateur punches compare to those of athletes.
In this calculator, we've often referred to force and pressure. If you want to improve your grip on these matters, check out our force and pressure conversion tools.
The force of a human punch varies drastically based on the following significant factors:
Mass of the punching and receiving body;
Velocity at which the punch hits the target;
Angle & surface area of the punch and the target;
Time it takes for the punch to reach its max kinetic energy; and
Cushion effect on both ends due to foam, padding, gloves, or wrapping.
General observation shows that an untrained person can deliver above 750 newtons of punch force with 150 psi of pressure in ideal conditions. In contrast, an elite boxer or mixed martial artist can go as high as 5000 newtons and 800 psi. So watch out! 👊
The human punch force calculator is an effective tool to help you understand your punch pressure and potential impact force, i.e., the maximum force attainable under the given conditions.
In the first field, enter your body weight, e.g., 75 kg.
The system will then calculate your potential impact force and punch pressure with the default values.
🙋 These default values are suitable for an average person who is untrained in the art of combat sport. 🥋
Let's look at the remaining fields to learn more about how we can utilize them.
The second field is the punch speed where we enter the velocity of our punch, e.g., 3.8 m/s (default).
💡 An untrained human can throw a punch as fast as 7 m/s. In comparison, professional athletes reach up to 20 m/s.
The third field is the delivery time for the punch to gain maximum kinetic energy before hitting the target, e.g., 0.3 seconds (default).
The fourth field is the surface area, required to calculate the pressure of the punch, e.g., 4 inches2 (default).
💡 The greater the surface area of the punch, the more distributed its pressure will be.
The fifth field is the impact force, which is the punch force that we wish to calculate.
Based on the previously given values, an average person with a mass of 75 kg can have a potential impact force of 950 newtons or 214 pounds-force.
And the last field is the punch pressure, which gives the pressure of the human punch force based on the given surface area. For the sampled values, we get 53 psi.
💡 An untrained person can reach up to 150 psi. Comparatively, the punch pressure of an elite combat athlete can reach around 800 psi.
Thus, the impact force result shows us how many newtons of force are in a punch, and the punch pressure expresses how much psi a punch exerts.
🔎 To see the acceleration of your punch, press the Advanced mode button, located under the calculator.
We use the following formulas to calculate how much force a human punch exerts:
For calculating the impact force, we use:
F = m \times a
Here, we know that:
a = v \div t
And for calculating the punch pressure:
P = F \div A
F
– Impact force of the punch, measured in newtons (N);
m
– Mass or weight of the punching body, in kilograms (kg);
a
– Acceleration of the punch, in meters per second squared (m/s2);
v
– Velocity of the punch, measured in meters per second (m/s);
t
– Time it takes for the punch to hit the target, in seconds (s);
P
– Pressure exerted by the punch, in pascals (Pa); and
A
– Average surface area of the punch in square meters (m2).
Let's use these formulas in an example and find out how many pounds of punch force can a human of 100-kilogram mass deliver without training:
First, let's find the acceleration of the punch. We need to divide the velocity, a.k.a, the speed of the punch, by the time it takes to deliver the punch.
Since our person is untrained, let's take an average punch speed of 3.8 m/s and a delivery time of 0.4 seconds.
Placing the values in the acceleration formula, we get:
a = 3.8 \div 0.4 = 9.5
Thus, the acceleration of our punch is 9.5 m/s2.
🙋 Generally, a person's reaction time is directly proportional to their weight.
Multiplying our given mass with this acceleration, we get the impact force of our punch:
F = 100 \times 9.5 = 950
And so, the potential impact force of the punch is 950 newtons.
To convert newtons (N) into pounds-force (lbf), we multiply our result by a constant number of 0.22481:
950 \times 0.22481 = 213.5695
Thus, an untrained human weighing 100-kilograms can deliver around 213.6 pounds of force.
💡 It requires around 300 pounds of force to crush a watermelon 🍉, whereas, for a steel can, it requires about 50 pounds🥫.
How many pounds of punch force can a human bear?
About 900 pounds of punch force can break the strongest bone in the human body, i.e., the femur. Thus, it is safe to say that anything above that may be fatal.
What is the potential punch force of a boxer weighing 80 kg?
2400 newtons or more is the potential impact force of a punch from someone skilled in combat sports like boxing.
We calculate this force using the following formula:
F – Impact force of the punch;
m – Mass of the boxer; and
a – Acceleration of the punch reaching the target.
To calculate the punch force:
Assume the acceleration of the punch is 30 m/s² for a skilled boxer.
Multiply their weight by the acceleration of the punch (80 kg × 30 m/s²).
The result is 2400 N or 540 pounds-force.
How much pounds-force will I require to squeeze a watermelon?
Between 240 to 360 pounds-force or 1000 to 1600 newtons of force is required to squeeze a watermelon. However, it depends on how ripe the watermelon is, as over ripped watermelons tend to go soft and are easily crushed.
How much force would it take to punch through someone?
Around 45,000 psi, or 800,000 newtons will be exerted if a human weighing 80 kg generates a velocity close to that of a bullet, i.e., 760 m/s in 75 milliseconds. That's how much force it would take to punch through someone.
The Duckworth Lewis calculator helps you set a fair target for the chasing team in an interrupted game of one-day cricket.
The triathlon nutrition calculator will tell you how many calories you burned during a race, and what to eat so that you don't lose strength. 🏊♀️🚴♀️🏃♂️
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Population parameter vs. statistic
What is sampling error? – definition
How to calculate the sampling error
How to use this sample margin of error calculator
Is sampling error the same as a standard error?
Calculating the sampling error is an essential task in almost every statistical study.
Whether you're making a pool about the voting intention of a country or trying to infer the average height of the American population, you need to express the error of your estimations, as it will always exist.
This tool calculates the error of your sample given the sample size and proportion or standard deviation. If you're interested in the opposite problem: the probability of finding a range of sample means or proportions, take a look at our normal probability for sampling distributions and sampling distribution of the sample proportion calculators.
Keep reading to know more about things like:
How to calculate a sampling error.
How to reduce sampling error.
A parameter is a numeric characteristic of a population. Examples are:
The mean height of American women over 20 years old.
The proportion of citizens of a country who have the intention to vote for some candidate.
The mean diastolic pressure of the citizens who don't exercise in a specific country.
On the other hand, a statistic is a point estimate or numeric characteristic of a sample. It is the data we usually can know.
These concepts are relevant to understanding the concept of sampling error.
There's not a universal consensus about the definition of sampling error. Some authors define it as the error caused by any source, i.e., sample variability, poor study design, or a nonrepresentative sampling. Other authors make a differentiation between a random part of the error (sampling error) and a nonrandom part (known as "bias" or "nonsampling error") and define the sampling error as something related only to the variability from sample to sample.
Here we'll take the last definition, which coincides with the Glossary of Statistical Terms published by professor P.B. Stark of the University of California.
🙋 The sampling error is also known as "sample error" or "sample margin of error."
Considering this, mathematically, we can define the sampling error as the difference between the population parameter and our statistic. The problem is we rarely know the population parameter value. Therefore, the actual sampling error is unknown, and we must estimate it.
Now that you know what a sampling error is let's see how to calculate it.
The equation for the sampling error depends on whether we're trying to estimate a population proportion or a population mean. In any case, it consists of the product of a critical value related to the confidence level (z-score or t-statistic) and the standard error. The standard error estimates the standard deviation of the sampling distribution of the parameter under study. The most common example is calculating the standard error of the distribution of sample means, but we can apply it to any other parameter, i.e., variance, range, standard deviation, etc.
🔎 You can learn more about the standard error in our standard error calculator.
How to calculate sampling error for a sample proportion
If you're estimating a population proportion (
p
), the sampling error (
e_p
) of your sample proportion (
\hat p
\small e_{\hat p} = z_{α/2}SE(\hat p) = z_{α/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}
n
– Sample size;
SE(\hat p)
– Standard error of the sample proportion, which equals
\sqrt{\frac{\hat p(1-\hat p)}{n}}
z_{α/2}
– Z-score, which depends on the required confidence level.
How to calculate margin of error for a sample mean
If you don't know the population standard deviation (
σ
) (as usually happens in practice), you estimate the sample margin of error using the sample standard deviation (
s
\small e_{\bar X} = t_{α/2}SE(\bar X) = t_{α/2} \frac{s}{\sqrt n}
SE(\bar X)
– Standard error of the sample mean (its formula is
s/√n
), which is an estimate of the standard deviation of the sampling distribution of the means.
t_{α/2}
– Equivalent value to
z_{α/2}
, with the difference that it considers the sample size.
If you do know (
σ
), you can calculate the sample margin of error of the sample mean (
e_{\bar X}
) with this formula:
\small e_{\bar X} = z_{α/2}SE(\bar X) = z_{α/2} \frac{σ}{\sqrt n}
💡 You can learn more about the
z_{α/2}
t_{α/2}
critical values and its calculation in our critical value and z-score calculators.
Suppose you're carrying out a study to determine the percentage (proportion) of citizens who have the intention to vote for the candidate called Don Quixote in the next elections. You've taken a random sample of 500 citizens, and 400 of them affirm to have the intention to vote for Don Quijote in the next election. What is the margin of error of your pool? Follow these steps to know it:
In the calculator, select "Sample proportion error" as the error to estimate.
Input 30 as the sample size in the second box.
Calculate the sample proportion
\hat p = 400/500 = 0.8
, and input it in the sample proportion box.
Select a confidence level. We'll take 95%, as it is the most common.
That's it. The sampling error is ±0.0351 (3.51%).
We estimated an 80% vote intention with our sample, but the sampling error lets us indicate with a 95% confidence level that the actual population vote intention is within 76.49% and 83.51%.
You can check the results using the sampling error formula:
e_p = 1.96 \times \sqrt{\frac{0.8(1-0.8)}{500}} = 0.0351
Now suppose you're investigating the caloric content of a new food product. You're interested in the average energy content of a batch of that product. You take a sample of 30 and measure how many calories they contain. The mean caloric content of the sample is 600 kcal, with a standard deviation of 70 kcal. If you want to calculate the sampling error of your energy content estimation, these are the steps:
Select "Sample mean error" as the error to estimate.
Select "Sample standard deviation" as the info you know.
Input 30 as the sample size.
Input 70 as the sample standard deviation.
Select a confidence level. We'll take 95%, as it is the most common one.
That's it. The sampling error must be ±26.1384 kcal from the mean content.
What this result indicates: We can say, with a 95% confidence level, that the mean caloric content of the studied batch lies between 626.1384 and 573.8616 kcal.
You can check the results using the sampling error formula, taking into account that, in this case,
t_{α/2} = 2.0452
e_{\bar X} = 1.96 \frac{70}{\sqrt{600}} = 26.138
What this result doesn't indicate: The result only gives information about the mean caloric content of a population of products. It provides the possible values of a sample mean and not the possible values for each individual unit of the product.
🙋 When we say the mean caloric content lies between 64.4 and 75.6 kcal, we're stating our statement as a confidence interval. Learn more about it in our confidence interval calculator.
No, sampling error is not the same as standard error, although they relate to each other.
The standard error is the estimated standard deviation of a sampling distribution.
The sampling error equals the standard error multiplied by a z-score or the t-statistic. It represents the error we incur when estimating a population parameter.
Sampling error is the same as standard error only when the z-score or the t-statistic equal 1.
No, standard error is not the same as margin of error, but they relate to each other.
The standard error is the estimated standard deviation of a statistic's sampling distribution.
A margin of error is the standard error multiplied by a z-score or the t-statistic. It represents the error we incur when estimating a population parameter.
The margin of error has a confidence interval associated with it.
The standard error is the same as the margin of error only when the z-score or the t-statistic equal 1.
How to reduce sampling error?
If we keep other factors constant, we can reduce sampling error by increasing the sample size. That occurs because sampling error is inversely proportional to the square root of the sample size.
How to calculate a 95% confidence interval from standard error?
The way how you calculate a 95% confidence interval from standard error depends on the parameter to estimate and the available information:
If you're estimating a population mean:
If you know the population standard deviation (σ), multiply the standard error (σ/√n) by the corresponding z-score, which is 1.96 for this confidence level.
If you only know the sample standard deviation (s), multiply the standard error (s/√n) by the corresponding t-value for 95%, which will depend on the sample size.
If you're estimating a population proportion, multiply the standard error (√[p̂(1 - p̂)/n] by the corresponding z-score, zα/2 = 1.96.
Error to estimate
Sample proportion error
Sample proportion (p̂)
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Support Vector Regression with R - SVM Tutorial
August 19, 2021 October 23, 2014 by Alexandre KOWALCZYK
In this article I will show how to use R to perform a Support Vector Regression.
We will first do a simple linear regression, then move to the Support Vector Regression so that you can see how the two behave with the same data.
To begin with we will use this simple data set:
I just put some data in excel. I prefer that over using an existing well-known data-set because the purpose of the article is not about the data, but more about the models we will use.
As you can see there seems to be some kind of relation between our two variables X and Y, and it look like we could fit a line which would pass near each point.
Let's do that in R !
Step 1: Simple linear regression in R
Here is the same data in CSV format, I saved it in a file regression.csv :
We can now use R to display the data and fit a line:
# Load the data from the csv file
dataDirectory <- "D:/" # put your own folder here
data <- read.csv(paste(dataDirectory, 'regression.csv', sep=""), header = TRUE)
model <- lm(Y ~ X, data)
# Add the fitted line
The code above displays the following graph:
Step 2: How good is our regression ?
In order to be able to compare the linear regression with the support vector regression we first need a way to measure how good it is.
To do that we will change a little bit our code to visualize each prediction made by our model
dataDirectory <- "D:/"
model <- lm(Y ~ X , data)
# make a prediction for each X
predictedY <- predict(model, data)
# display the predictions
points(data$X, predictedY, col = "blue", pch=4)
X_i
the model makes a prediction
\hat{Y}_i
displayed as a blue cross on the graph. The only difference with the previous graph is that the dots are not connected with each other.
In order to measure how good our model is we will compute how much errors it makes.
We can compare each
Y_i
value with the associated predicted value
\hat{Y}_i
and see how far away they are with a simple difference.
\hat{Y}_i - Y_i
is the error, if we make a perfect prediction
\hat{Y}_i
Y_i
and the error will be zero.
If we do this for each data point and sum the error we will have the sum of the errors, and if we takes the mean we will get the Mean Squared Error (MSE)
A common way to measure error in machine learning is to use the Root Mean Squared Error (RMSE) so we will use it instead.
To compute the RMSE we take the square root and we get the RMSE
Using R we can come with the following code to compute the RMSE
rmse <- function(error)
sqrt(mean(error^2))
error <- model$residuals # same as data$Y - predictedY
predictionRMSE <- rmse(error) # 5.703778
We know now that the RMSE of our linear regression model is 5.70. Let's try to improve it with SVR !
Step 3: Support Vector Regression
In order to create a SVR model with R you will need the package e1071. So be sure to install it and to add the library(e1071) line at the start of your file.
Below is the code to make predictions with Support Vector Regression:
model <- svm(Y ~ X , data)
points(data$X, predictedY, col = "red", pch=4)
As you can see it looks a lot like the linear regression code. Note that we called the svm function (not svr !) it's because this function can also be used to make classifications with Support Vector Machine. The function will automatically choose SVM if it detects that the data is categorical (if the variable is a factor in R).
The code draws the following graph:
This time the predictions is closer to the real values ! Let's compute the RMSE of our support vector regression model.
# /!\ this time svrModel$residuals is not the same as data$Y - predictedY
# so we compute the error like this
error <- data$Y - predictedY
svrPredictionRMSE <- rmse(error) # 3.157061
As expected the RMSE is better, it is now 3.15 compared to 5.70 before.
But can we do better ?
Step 4: Tuning your support vector regression model
In order to improve the performance of the support vector regression we will need to select the best parameters for the model.
In our previous example, we performed an epsilon-regression, we did not set any value for epsilon (
\epsilon
), but it took a default value of 0.1. There is also a cost parameter which we can change to avoid overfitting.
The process of choosing these parameters is called hyperparameter optimization, or model selection.
The standard way of doing it is by doing a grid search. It means we will train a lot of models for the different couples of
\epsilon
and cost, and choose the best one.
tuneResult <- tune(svm, Y ~ X, data = data,
ranges = list(epsilon = seq(0,1,0.1), cost = 2^(2:9))
print(tuneResult)
# Draw the tuning graph
plot(tuneResult)
There is two important points in the code above:
we use the tune method to train models with
\epsilon = 0, 0.1, 0.2, ... ,1
and cost =
2^2, 2^3, 2^4, ... ,2^9
which means it will train 88 models (it can take a long time)
the tuneResult returns the MSE, don't forget to convert it to RMSE before comparing the value to our previous model.
The last line plot the result of the grid search:
On this graph we can see that the darker the region is the better our model is (because the RMSE is closer to zero in darker regions).
This means we can try another grid search in a narrower range we will try with
\epsilon
values between 0 and 0.2. It does not look like the cost value is having an effect for the moment so we will keep it as it is to see if it changes.
ranges = list(epsilon = seq(0,0.2,0.01), cost = 2^(2:9))
We trained different 168 models with this small piece of code.
As we zoomed-in inside the dark region we can see that there is several darker patch.
From the graph you can see that models with C between 200 and 300 and
\epsilon
between 0.08 and 0.09 have less error.
Hopefully for us, we don't have to select the best model with our eyes and R allows us to get it very easily and use it to make predictions.
tunedModel <- tuneResult$best.model
tunedModelY <- predict(tunedModel, data)
error <- data$Y - tunedModelY
# this value can be different on your computer
# because the tune method randomly shuffles the data
tunedModelRMSE <- rmse(error) # 2.219642
We improved again the RMSE of our support vector regression model !
If we want we can visualize both our models. The first SVR model is in red, and the tuned SVR model is in blue on the graph below :
I hope you enjoyed this introduction on Support Vector Regression with R.
You can get the source code of this tutorial. Each step has its own file.
If you want to learn more about Support Vector Machines, you can now read this article:
Categories SVM in Practice, SVM in R Tags Linear Regression, R, Support Vector Regression Post navigation
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How to use our kilometer calculator?
Another length-related calculators
In the cool kilometer calculator, we help you convert such large length unit into several ones from different distance systems:
Kilometers to miles, feet, yard, inches.
Kilometers to meters, centimeters, and millimeters.
Keep reading and find out everyday conversions you might consider helpful, like how many kilometers are in a mile or how many meters are in a kilometer.
A kilometer is a distance unit that is part of the metric measurement system that represents a thousand meters:
1km = 1,000 meters
Therefore, from the length converter calculator, we get the following conversion factors:
But what if we need to convert miles to kilometers?
Then we can use the following conversion factor:
Besides, for taking the other way around, now divide one by 1.6093. If you want to convert kilometers to miles, you use:
Furthermore, you can use the imperial conversions to get yourself around feet, yards, and inches:
Don't worry about memorizing these conversion factors; that is why we created the fantastic kilometer calculator for you.
You can use our kilometer calculator in two ways:
Type the kilometer value you want to convert and see how our tool provides its equivalent in 7 different units.
Type the unit you have. It can be mm, miles, yard, among others, and our tool will indicate its value in kilometers.
Let's see an example in-between values of the metric measurement system: How many meters are in 1.92 kilometers? Also, let's express the result in yards.
We will use the conversion factor: 1km = 1,000 meters
1.92km = 1920 m
For yards, we will use two conversion factors: 1 km = 0.6214 miles and 1 mile = 1760 yards; therefore, we have:
1.92 km * \frac {0.6214mi} {1 km} * \frac {1760 yd} {1 mi} = 2099.7 yd
Besides of our kilometer calculator, here we want to mention other distance related calculators you could find useful:
1.6093 kilometers. You can also calculate it this way:
Remember 1609.3 meters build a mile.
Divide 1609.3 meters by the conversion factor for meters to kilometers.
The result is the equivalent of a mile expressed in kilometers.
61.14 miles. You can follow the next steps for obtaining the same value:
Note the conversion factor from kilometers to miles: 1.6093 miles equals one kilometer.
Divide 100 kilometers by the conversion factor: 100 / 1.6093.
The result is your kilometers value expressed in miles: 61.14 miles.
This angle conversion calculator converts between ten different angle measurement units. It explains how to convert degrees to radians, and gives you the definition of an angle, radian, and more.
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Our square meters to square feet converter is an efficient, structured tool to convert between units of area measurements. As the name indicates, it converts square meters to square feet also the other way round. But that is not al. As a bonus; you have many other units to choose from.
Continue reading, and you will learn:
How to convert square meters to square feet; and
To convert feet squared to meters squared.
Our square meters to square feet converter is an efficient tool that allows you to convert from a SI unit of area to an imperial unit of area.
Square meters or m² is a SI unit of area. We determine it by multiplying the length by the width of any surface.
Square feet or ft² is an imperial unit of area, and similar to calculating any other area, to calculate ft², we multiply the length by the width of a surface.
To use our converter, all you have to do is:
Input the area in square meters.
The result is the area in square feet.
But if you decide to use it the other way round:
Input the area in ft².
The result is the area in m².
Again, a piece of cake.
For example, you enter
3\text{ m}^2
32.2917\text{ ft}^2
One foot is precisely equal to 0.3048 meters, defined in 1959. It makes one meter equal to 3.28084 ft. Since the area is a squared number, we can conclude our sq meters to sq feet converter equations.
To convert from sq meter to sq feet:
\text{ft}^2 = \text{m}^2 × 10.7639
1\text{ m}^2= 10.7639\text{ ft}^2
Take any number in meters squared and multiply it by 10.7639, and you have your area in feet squared. For instance, you have a table with an area of
2\text{m}^2
. To convert it to ft2, multiply 2 by 10.7639. It is
21.5278\text{ ft}^2
To convert from sq feet to sq meters:
\text{ft}^2 = \text{m}^2 / 10.7639
Take any number in feet squared and divide it by 10.7639, and you have your area in meters squared. For instance, you have a rug with an area of
7\text{ft}^2
. To convert it to m2, divide 7 by 10.7639. It is
0.650321\text{ m}^2
We can now conclude that our square meters to square feet converter is a fantastic tool, but we don't stop there.
Here is a list of some wonderful area conversion tools.
To convert the area from square meters to square feet, it would be best if you remembered the conversion factor 10.7639.
To convert from ft² to m²:
Note the area in m².
Multiply it by 10.7639.
The result is ft².
To convert from m² to ft²:
Note the area in ft².
Divide it by 10.7639.
The result is m².
How many square feet are in 56m²?
There are 602.779 ft² in 56 m².
The easiest way to convert from m² to ft² is to multiply the amount in meters by 10.7639. And you have the amount in feet squared.
ft² = m² × 10.7639
ft - Feet; and
How many square meters are in 100ft²?
There are 9.2903m² in 100ft².
This answer is based on:
m² = ft² / 10.7639
This means all you have to do is divide the amount in square feet by 10.7639, and you have your area in square meters. It's easy, isn't it!
The acres to square feet converter does exactly what the title says it does.
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1 Institute of Environmental Engineering and Building Installations, Technical University of Łódź, Łódź, Poland.
2 Leather Research Institute, Łódź, Poland.
Abstract: Calculations of chemical structures and photofading of parabens (PHB—4 hydroxybenzoic acid), which are p-hydroxybenzoic acid alkyl esters were performed. These compounds are used as preservatives for the substances used in cosmetics. The reactivity of these derivatives with an oxidant—singlet oxygen—have been tested with a theoretical method of frontier orbitals. All-valence molecular orbital methods, AM1 and PM3, have been used to calculate frontier electron density for higher occupied HOMO and lower unoccupied LUMO orbitals, which might be sensitive to an electrophilic (with singleton oxygen atom 1O2) or nucleophilic ( superoxide anion radical) attack at a particular atom in a molecule. Using AM1 and PM3, we calculated the reactivity , superdelocalisability and electron density distributions. The obtained superdelocalisability rates allow you to explain the fastness values in different chemical molecules. The structure of parabens (PHB) was optimized by MM+, DM, AM1 or PM3, to achieve constant energy values at a convergence criterion of 0.01 kcal/mol. The performed calculations indicate that the electrophilic oxidation reaction should take place in the aromatic ring in the 2-position to the hydroxyl residue of PHB, whereas the superoxide radical reaction occurs mainly on the alkyl residues of the ester group. The reaction may take place according to superoxide mechanism or 1,2-addition, where the higher superdelocalisability values SN are located on neighboring atoms in aromatic systems.
Keywords: Photofading of Parabens, All-Valence MO Calculation, Orbital Frontier Electron Density, Electrophilic and Nucleophilic Attack, Theoretical Study
{f}_{r}^{E}
{f}_{r}^{N}
{\text{O}}_{2}{}^{•-}
{S}_{r}^{E\left(N\right)}
{S}_{r}^{E\left(N\right)}
{S}_{r}^{E\left(N\right)}={f}_{r}^{E\left(N\right)}/{E}_{HOMO\left(LUMO\right)}
{f}_{r}^{E\left(N\right)}
{f}_{r}^{N}
{S}_{r}^{N}
{\text{O}}_{2}{}^{•-}
{f}_{r}^{N}
{S}_{r}^{N}
{S}_{r}^{E}
{S}_{r}^{N}
{\text{O}}_{2}{}^{•-}
{Q}_{f}^{j}=2\underset{i-1}{\overset{n}{\sum }}{c}_{ji}^{2}
{Q}_{f}^{j}
{S}_{m,n}^{E}
{f}_{r}^{E}
{f}_{r}^{E}
{f}_{r}^{E}
{f}_{r}^{N}
{\text{O}}_{2}{}^{•-}
{\text{O}}_{2}{}^{•-}
{\text{O}}_{2}{}^{•-}
{\text{O}}_{2}{}^{•-}
{\text{O}}_{2}{}^{•-}
{\text{O}}_{2}{}^{•-}
Cite this paper: Wojciechowski, K. and Szuster, L. (2019) Photofading of Derivatives of Paraben (PHB) by AM1 and PM3 Methods: A Theoretical Study. Computational Chemistry, 7, 39-50. doi: 10.4236/cc.2019.72003.
[1] Tripathi, S., Vasudev, I. and Ray, A. (2008) Photofading of Azo Dyes: A Theoretical Study. Coloration Technology, 124, 151.
[2] Hihara, T., Okada, Y. and Morita, Z. (2003) Reactivity of Phenylazonaphthol Sulfonates, Their Estimation by Semiempirical Molecular Orbital PM5 Method, and the Relations between Their Reactivity and Azo-Hydrazone Tautomerism. Dyes and Pigments, 59, 201.
[3] Ozen, A.S., Aviyente, V. and Klein, R.A. (2003) Modeling the Oxidative Degradation of Azo Dyes: Density Functional Theory Study. Journal of Physical Chemistry A, 107, 4898.
[4] Morley, J.O., Guy, O.J. and Charlton, M.H. (2004) Molecular Modeling Studies on the Photochemical Stability of Azo Dyes. Journal of Physical Chemistry A, 108, 10542.
[5] Ray, A. and Deheri, G.M. (1995) CNDO/II Study on the Fading of Antraquinone Dyes. Dyes and Pigments, 27, 327.
[6] Guo, L., Meng, F.S., Gong, X.D., Xiao, X.M., Cheng, K.C. and Tian, H. (2001) Synthesis and Spectral Properties of Soluble Trimethylsilyl Substituted Metal-Phthalocyjanines. Dyes and Pigments, 49, 83.
[7] Wang, L.F., Wang, X.L. and Zhang, H.Y. (2005) A Theoretical Study on Photobleaching Mechanisms of Hypocrellins. Dyes and Pigments, 67, 161.
[8] Fukui, K., Yonezawa, T. and Nagata, C. (1957) MO-Theoretical Approach to the Mechanism of Charge Transfer in the Process of Aromatic Substitutions. Bulletin of the Chemical Society of Japan, 27, 1247.
[9] Okada, Y., Hihara, T., Hirose, M. and Morita, Z. (2010) Substituent Effects on the Photofading of Disperse Azo Dyes on Poly(ethylene terephthalate) Substrate. Coloration Technology, 126, 127.
[10] Ozen, A.S., Aviyente, V. and Klein, R.A. (2004) Modeling the Substituent Effect on the Oxidative Degradation of Azo Dyes. Journal of Physical Chemistry A, 108, 5990.
[11] Morita, Z. and Hada, S. (1999) A Semiempirical Molecular Orbital Study on the Reaction of an Aminopyrazolinyl Azo Dye with Singlet Molecular Oxygen. Dyes and Pigments, 41, 1.
[12] Fukui, K. and Yonezawa, T. (1957) C Nagata, Interrelation of Quantum-Mechanical Quantities Concerning Chemical Reactivity of Conjugated Molecules. Journal of Chemical Physics, 26, 831.
[13] Dewar, M.J.S., Zoebisch, E.G., Hearly, E.R. and Steward, J.J.P. (1985) The Development and Use of Quantum-Mechanical Molecular Models. 76. AM1: A New General Purpose Quantum Mecanical Molecular Model. Journal of American Chemical Society, 107, 3902.
[14] Dewar, M.J.S. and Thiele, W. (1975) Ground-States of Molecules .30. Mindo-3 Study of Reactions of Singlet (1-Delta-g) Oxygen with Carbon-Carbon Double-Bonds. Journal of American Chemical Society, 97, 3978.
[15] Hihara, T., Okada, Y. and Morita, Z. (2004), Relationship between Photochemical Properties and Colourfastness Due to Light-Related Effects on Monoazo Reactive Dyes Derived from H-Acid, γ-Acid, and Related Naphthalene Sulfonic Acids. Dyes and Pigments, 59, 269.
[16] Griffiths, J. and Hawkins, C. (1976) Oxidation by Singlet Oxygen of Arylazonaphthols Exhibiting Azo-Hydrazone Tautomerism. Journal of the Chemical Society, Perkin Transactions II, 747-752.
[17] Commission Regulation (EU) No. 358/2014, Official Journal of the EU, L 107 (Vol. 57), 10 April 2014.
https://green.pidc.org.tw/upload/news/OJ-JOL_2014_107_R_0002-EN-TXT
[18] Wang, X.L., Wang, L.F. and Zhang, H.Y. (2004) A Theoretical Study on Isomerisation Process of Hypocrellin A in Ground State. Dyes and Pigments, 63, 23-28.
[19] Vanbeek, H.C.A. and Heertjes, P.M. (1966) The Mechanism of the Photoreduction of Azo Dyes in the Presence of DL-Mandelic Acid and in the Absence of Oxygen. Journal of Physical Chemistry, 70, 1704-1711.
[20] Biswas, N. and Umpathy, S. (1999) Structures, Vibrational Frequencies, and Normal Modes of Substituted Azo Dyes: Infrared, Raman, and Density Functional Calculations. Journal of Physical Chemistry, 104, 2734-2745.
[21] Foote, C.S. (1968) Photosensitised Oxygenations and the Role of Singlet Oxygen. Accounts of Chemical Research, 1, 104-110.
[22] Matsuura, T. (1977) Bio-Mimetic Oxygenation. Tetrahedron, 33, 2869-2905.
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f\prime
, explaining that it is a function giving the slope of the line tangent to the graph of
at points in the domain of
. It was also interpreted as the instantaneous rate of change of
x
f \prime \left(x\right)
f \prime \left(x\right)=\underset{h\to 0}{lim}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}\frac{f\left(x+h\right)-f\left(x\right)}{h}
x
x
\left(x,f\left(x\right)\right)
\left(z,f\left(z\right)\right)
\frac{f\left(z\right)-f\left(x\right)}{z-x}
h=z-x
z=x+h
\frac{f\left(x+h\right)-f\left(x\right)}{h}
\frac{f\left(x+h\right)-f\left(x\right)}{h}
g\prime
x=c
g\prime \left(c\right)
\frac{\mathrm{df}}{\mathrm{dx}}
\frac{ⅆ}{ⅆ\phantom{\rule[-0.0ex]{0.2em}{0.0ex}}\textcolor[rgb]{0.784313725490196,0,0.784313725490196}{x}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}\textcolor[rgb]{0,0.627450980392157,0.313725490196078}{f}
g\prime \left(c\right)
\frac{\mathrm{dg}\left(c\right)}{\mathrm{dx}}
g\left(c\right)
x
\stackrel{.}{g}\left(t\right)
\frac{\mathrm{dy}}{\mathrm{dx}}
\mathrm{dy}
\mathrm{dx}
y\prime
\mathrm{dx}
f
a
a.
The differentiability of
a
f\prime \left(a\right)=\underset{h\to 0}{lim}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}\frac{f\left(a+h\right)- f\left(a\right)}{h}
a
\underset{x\to a}{lim} f\left(x\right)=f\left(a\right)
x\ne a
f\left(x\right)
= f\left(a\right)+f\left(x\right)-f\left(a\right)
=
f\left(a\right)+\frac{f\left(x\right)-f\left(a\right)}{x-a}
x-a
\underset{x→a}{lim}f\left(x\right)
=\underset{x\to a}{lim}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}\left(f\left(a\right)+\frac{f\left(x\right)-f\left(a\right)}{x-a} \left(x-a\right)\right)
=f\left(a\right)+\underset{x\to a}{lim}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}\left(\frac{f\left(x\right)-f\left(a\right)}{x-a}\right)\underset{x\to a}{lim} \left(x-a\right)
The limit of the difference quotient can be recognized as the derivative of
a
f\prime \left(a\right)
\underset{x→a}{lim}f\left(x\right)
=f\left(a\right)+f\prime \left(a\right)\cdot 0
=f\left(a\right)
a
\genfrac{}{}{0}{}{\frac{\mathrm{df}}{\mathrm{dx}}}{\phantom{x=a}}|\genfrac{}{}{0}{}{\phantom{\mathrm{f\left(x\right)}}}{x=c}=f\prime \left(c\right)=
\underset{h\to 0}{lim}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}\frac{f\left(c+h\right)-f\left(c\right)}{c}
\underset{z\to c}{lim}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}\frac{f\left(z\right)-f\left(c\right)}{z-c}
f\left(x\right)={x}^{3}
f\prime \left(c\right)
\left(c,f\left(c\right)\right)
f\left(x\right)=\frac{{x}^{2}+1}{x+2}
f\prime \left(c\right)
c≠-2
f\left(x\right)=\sqrt{x-1}
f\prime \left(c\right)
c>1
\underset{z\to 2}{lim}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}\frac{\sqrt{{z}^{2}+1}-\sqrt{5}}{z-2}
|
Travel - Whitepaper
Dtravel allows guests to pay for bookings in a wide variety of fiat currencies and cryptocurrencies, including TRVL. Payments in TRVL are encouraged and incentivized by a guest cashback of up to 25% of the total nominal fees. The exact percentage
c_g(u)
of cashback for a guest
u
depends on the guest's Boost Factor (
b(u)
), according to the equation below:
c_g(u) = b(u) * 10\%
Furthermore, hosts are encouraged and incentivized to choose to receive payments in TRVL through a host cashback of up to 25% of the total nominal fees. The exact percentage
c_h(u)
of cashback for a host
u
depends on the host's Boost Factor (
b(u)
c_h(u) = b(u) * 10\%
The cashbacks are paid in TRVL to the host's and guest's internal Dtravel wallets when the booking is completed.
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Compute Markov chain hitting times - MATLAB hittime - MathWorks América Latina
P=\left[\begin{array}{cccc}1& 0& 0& 0\\ 1/2& 0& 1/2& 0\\ 0& 0& 0& 1\\ 0& 0& 1/2& 1/2\end{array}\right].
P=\left[\begin{array}{cccc}0& 1/2& 0& 1/2\\ 2/3& 0& 1/3& 0\\ 0& 1/2& 0& 1/2\\ 1/3& 0& 2/3& 0\end{array}\right].
P=\left[\begin{array}{ccccccc}1/2& 0& 1/2& 0& 0& 0& 0\\ 0& 1/3& 0& 2/3& 0& 0& 0\\ 1/4& 0& 3/4& 0& 0& 0& 0\\ 0& 2/3& 0& 1/3& 0& 0& 0\\ 1/4& 1/8& 1/8& 1/8& 1/4& 1/8& 0\\ 1/6& 1/6& 1/6& 1/6& 1/6& 1/6& 0\\ 1/2& 0& 0& 0& 0& 0& 1/2\end{array}\right].
The expected commute time from state
\mathit{i}
\mathit{j}
is the expected time required for a Markov chain to transition from state
\mathit{i}
\mathit{j}
(the expected first hitting time ht
\left(\mathit{i},\mathit{j}\right)
), then back to state
\mathit{i}
(ht
\left(\mathit{j},\mathit{i}\right)
). Formally, the expected commute time is
\mathit{C}\left(\mathit{i},\mathit{j}\right)=
\left(\mathit{i},\mathit{j}\right)+\text{\hspace{0.17em}}
\left(\mathit{j},\mathit{i}\right)
\left\{\begin{array}{cc}{k}_{i}^{A}=0& ,i\in A\\ {k}_{i}^{A}=1+\sum _{j\notin A}^{}{P}_{ij}{k}_{j}^{A}& ,i\notin A,\end{array}
{k}_{i}^{A}
= ht(i), the expected first hitting time for the subset of states A, beginning from state i.
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The average value of
f\left(x\right)
x∈\left[a,b\right]
\frac{1}{b-a}{∫}_{a}^{b}f\left(x\right) \mathrm{dx}
Similarly, the average value of
f\left(x,y\right)
on a region
R
{A}_{R}
\frac{1}{{A}_{R}}∫{∫}_{R}f\left(x,y\right) \mathrm{dA}
\frac{∫{∫}_{R}f\left(x,y\right) \mathrm{dA}}{∫{∫}_{R}1 \mathrm{dA}}
In polar coordinates, the average value of
F\left(r,\mathrm{θ}\right)
R
{A}_{R}
\frac{1}{{A}_{R}}∫{∫}_{R}F\left(r,\mathrm{θ}\right) \mathrm{dA}
\frac{∫{∫}_{R}F\left(r,\mathrm{θ}\right) \mathrm{dA}}{∫{∫}_{R}1 \mathrm{dA}}
where now,
\mathrm{dA}
\left|\frac{∂\left(x,y\right)}{∂\left(r,\mathrm{θ}\right)}\right|=r
, the absolute value of the Jacobian of the transformation from polar to Cartesian coordinates.
Figure 6.4.1 shows the average value of
f\left(x,y\right)={x}^{2}+{y}^{2}
R
, the first-quadrant portion of the unit circle.
The average value is given by
\frac{{∫}_{0}^{1}{∫}_{0}^{{\sqrt{1-{x}^{2}}}_{}}\left({x}^{2}+{y}^{2}\right) \mathrm{dy} \mathrm{dx}}{\mathrm{π}/4}
\frac{\mathrm{π}/8}{\mathrm{π}/4}=\frac{1}{2}
However, there is as much volume above the plane
z=1/\sqrt{2}
as there is below.
local p1,p2,p3;
p1:=plot3d(x^2+y^2,x=0..1,y=0..sqrt(1-x^2),filled=true):
p2:=plot3d(1/2,x=0..1,y=0..sqrt(1-x^2)):
p3:=display(p1,p2,orientation=[-60,80,0],axes=frame,labels=[x,y,z],tickmarks=[2,2,3],scaling=constrained,lightmodel=none);
Figure 6.4.1 Average value of
{x}^{2}+{y}^{2}
The volume below
z=1/\sqrt{2}
{∫}_{0}^{1/{2}^{1/4}}{∫}_{0}^{{\sqrt{1/\sqrt{2}-{x}^{2}}}_{}}\left({x}^{2}+{y}^{2}\right) \mathrm{dy} \mathrm{dx}
\mathrm{π}/16
, exactly half the total volume. Hence, care must be taken in interpreting the meaning of the average value of a function of several variables.
Maple Tools for Calculating Average Value
The Student MultivariateCalculus package contains the FunctionAverage command that will construct and evaluate the average value of a function of two variables in either Cartesian or polar coordinates.
In addition, this command for functions of two variables has been implemented in two task templates, one for Cartesian coordinates and one for polar coordinates. In the Cartesian case, the task template iterates in the order
\mathrm{dy} \mathrm{dx}
; and in the polar case, in the order
\mathrm{dr} d\mathrm{θ}
Of course, there is always the option of implementing the relevant integrals from first principles using either the Int and int commands at top level, or the double-integral templates in the Calculus palette.
F=x y
R
, the finite region bounded by the graph of
y=x \left(1-x\right)
x
-axis. See Example 6.2.1 and Example 6.1.1.
F=2 {x}^{2}+3 {y}^{2}
R
, the finite region bounded by the graphs of
x={y}^{2}
y=3-2 x
. See Example 6.2.2 and Example 6.1.2.
F=2 x+3 y+1
R
f\left(x\right)=\mathrm{sin}\left(x\right)
g\left(x\right)=\mathrm{sin}\left(2 x\right)
0≤x≤\mathrm{π}
F=3 {x}^{2}+2 {y}^{2}+1
R
f\left(x\right)=\mathrm{arctan}\left(x+1\right)-1/2
g\left(x\right)=\mathrm{sin}\left(x\right)
x∈\left[0,\mathrm{π}/2\right]
See Example 6.2.4 and Example 6.1.4.
F=2-{\left(y-1/2\right)}^{2}
R
, the region bounded by the graphs of 1,
\mathrm{cos}\left(x\right)
y=x
0≤x≤1
F\left(r,\mathrm{θ}\right)=1+2 r \mathrm{cos}\left(\mathrm{θ}\right)+3 r \mathrm{sin}\left(\mathrm{θ}\right)
R
, the interior of the loop of the 4-leaf rose
r=\mathrm{cos}\left(2 \mathrm{θ}\right)
that straddles the positive
x
-axis. See Example 5.7.1.
F\left(r,\mathrm{θ}\right)=4-3 r \mathrm{cos}\left(\mathrm{θ}\right)+5 r \mathrm{sin}\left(\mathrm{θ}\right)
R
, the region that is inside the cardioid
r=1+ \mathrm{cos}\left(\mathrm{θ}\right)
r=1
. See Example 5.7.2.
F\left(r,\mathrm{θ}\right)={r}^{2}
R
, the region that is inside the circle
r=3 \mathrm{sin}\left(\mathrm{θ}\right)
but outside the cardioid
r=1+\mathrm{sin}\left(\mathrm{θ}\right)
F\left(r,\mathrm{θ}\right)={r}^{2}\left(2 {\mathrm{cos}}^{2}\left(\mathrm{θ}\right)+3 {\mathrm{sin}}^{2}\left(\mathrm{θ}\right)\right)
R
r=3 \mathrm{cos}\left(\mathrm{θ}\right)
r=1+\mathrm{cos}\left(\mathrm{θ}\right)
F\left(r,\mathrm{θ}\right)=r+\mathrm{cos}\left(\mathrm{θ}\right)
R
r=3 \mathrm{cos}\left(\mathrm{θ}\right)
but outside the limaçon
r=2-\mathrm{cos}\left(\mathrm{θ}\right)
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The Doppler Effect - Maple Help
Home : Support : Online Help : Math Apps : Natural Sciences : Physics : The Doppler Effect
The Doppler Effect refers to a change in the observed frequency of waves emitted by a moving source. This effect can be heard when a vehicle sounding a siren or horn approaches, passes, and drives away from you: when compared to the actual frequency being emitted, the observed frequency of the sound is higher as the vehicle approaches and lower as it recedes.
When the source of the waves is moving toward the observer, each successive wavefront is emitted from a position slightly closer to the observer than the previous one, thereby reducing the distance between successive wavefronts. The waves are bunched together, and the time between the arrival of successive wavefronts at the observer is similarly reduced, that is, the frequency is increased.
When the source of the waves is moving away from the observer, each successive wavefront is emitted from a position slightly further away from the observer than the previous one, thereby increasing the distance between successive wavefronts. The waves are spread out, and the time between the arrival of successive wavefronts at the observer is similarly increased, that is, the frequency is decreased.
For a source moving along the same line that connects the source to the observer, the relationship between the observed frequency (f) and the emitted frequency (
{f}_{0}
f = \left(\frac{c+{v}_{r}}{c+{v}_{s}}\right)\cdot {f}_{0}
c
is the speed of the waves relative to the medium
{v}_{r}
is the speed of the receiver relative to the medium (positive if the receiver is moving closer to the source, and negative otherwise)
{v}_{s}
is the speed of the source relative to the medium (positive if the source is moving away from the receiver, negative otherwise)
{f}_{0}
is the actual frequency at which the waves are being emitted
Using Sound as an Example
A Mach number (M) refers to the ratio of the speed of an object to the speed of sound in a particular fluid medium:
M = \frac{{v}_{s}}{c}
{v}_{s}
is the speed of the source relative to the medium and
c
is the speed of sound in the medium.
Subsonic Speeds,
\mathbit{M}\mathbf{ }\mathbf{<}\mathbf{ }\mathbf{1}
In this case, let's say the source of the sound is an emergency vehicle, such as an ambulance.
When the vehicle is stationary, all observers will hear the siren at the same pitch. The sound waves are being emitted at a constant frequency (
{f}_{0}
) and the wavefronts are propagating away from the source at the constant speed of sound (c), which means that the waves will reach all observers with the same frequency (equal to the actual frequency at which they are being emitted).
When the vehicle is moving, observers in front of it will hear the siren at a higher pitch because the sound waves are bunching together in front of the vehicle and thus reach these observers at a higher frequency. Meanwhile, observers behind the vehicle will hear the siren at a lower pitch because the sound waves are spreading out behind the vehicle and thus reach these observers at a lower frequency.
Breaking the Sound Barrier,
\mathbit{M}\mathbf{ }\mathbf{=}\mathbf{ }\mathbf{1}
In this case, let's say the source of the sound is a jet.
If the jet was to fly exactly at the speed of sound (which is approximately 343.2 m/s in dry air at 20°C), the wavefronts ahead of the jet would all bunch together at a single point, which means that observers in front it will hear nothing until the jet arrives and passes overhead.
Supersonic Speeds,
\mathbit{M}\mathbf{ }\mathbf{>}\mathbf{ }\mathbf{1}
In this case, the source of the sound is also a jet.
If the jet was to fly faster than the speed of sound, it would actually lead the advancing wavefronts and pass by the observer before they hear the sound it creates. Behind the jet, the sound waves would create a cone-like formation known as a Mach cone, and the intense air pressure which builds up on the leading edges of this cone would cause a shock wave known as a sonic boom just after the jet passes.
In this model, explore how sound waves change as the jet travels at subsonic or supersonic speeds. Use the slider to adjust the Mach ratio. Click the radio buttons to choose the frequency at which the sound waves are emitted by the source. Select "Add an Observer" and then click or drag on the plot to place an observer and watch how the frequency they observe changes depending on whether they are in front of or behind the source.
Ratio of the velocity of the jet,
{v}_{s}
, to the speed of sound,
c
\frac{{v}_{s}}{c}
Emitted Frequency,
{f}_{0}=
Observed Frequency, f =
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Simulate sample paths of threshold-switching dynamic regression model - MATLAB simulate - MathWorks Deutschland
Simulate Response Path from SETAR Model
Return Innovations and States
Initialize Multivariate Model Simulation from Multiple Starting Conditions
Simulate Model Containing Exogenous Regression Component
Simulate sample paths of threshold-switching dynamic regression model
Y = simulate(Mdl,numObs) returns a random numObs-period path of response series Y from simulating the fully specified threshold-switching dynamic regression model Mdl.
Y = simulate(Mdl,numObs,Name,Value) uses additional options specified by one or more name-value arguments. For example, simulate(Mdl,10,NumPaths=1000,Y0=Y0) simulates 1000 sample paths of length 10, and initializes the dynamic component of each submodel by using the presample response data Y0.
Suppose a data-generating process (DGP) is a two-state, self-exciting threshold autoregressive (SETAR) model for a 1-D response variable. Specify all parameter values (this example uses arbitrary values).
\left(-\infty ,0\right)
\left[0,\infty \right)
{\mathit{y}}_{\mathit{t}}=-1+0.1{\mathit{y}}_{\mathit{t}-1}+{\epsilon }_{1,\mathit{t}}
{\epsilon }_{1,\mathit{t}}\sim Ν\left(0,1\right)
{\mathit{y}}_{\mathit{t}}=1++0.3{\mathit{y}}_{\mathit{t}-1}+0.2{\mathit{y}}_{\mathit{t}-2}+{\epsilon }_{2,\mathit{t}}
{\epsilon }_{2,\mathit{t}}\sim Ν\left(0,{2}^{2}\right)
mdl1 = arima(Constant=c1,AR=ar1,Variance=v1,...
Generate one random response path of length 50 from the model. simulate assumes that the threshold variable is
{\mathit{y}}_{\mathit{t}-1}
, which implies that the model is self-exciting.
Plot the response path with the threshold by using ttplot.
ttplot(Mdl.Switch,Data=y)
Consider the following logistic TAR (LSTAR) model for the annual, CPI-based, Canadian inflation rate series
{\mathit{y}}_{\mathit{t}}
{\mathit{y}}_{\mathit{t}}=-5+{\epsilon }_{1,\mathit{t}}
{\epsilon }_{1,\mathit{t}}\sim Ν\left(0,0.{1}^{2}\right).
{\mathit{y}}_{\mathit{t}}={\epsilon }_{2,\mathit{t}}
{\epsilon }_{2,\mathit{t}}\sim Ν\left(0,0.{2}^{2}\right).
{\mathit{y}}_{\mathit{t}}=5+{\epsilon }_{3,\mathit{t}}
{\epsilon }_{3,\mathit{t}}\sim Ν\left(0,0.{3}^{2}\right).
{\mathit{y}}_{\mathit{t}}<2
2\le {\mathit{y}}_{\mathit{t}}<8
The transition function rate between states 1 and 2 is 3.5, and the transition function rate between states 2 and 3 is 1.5.
Create an LSTAR model representing
{\mathit{y}}_{\mathit{t}}
tt = threshold([2 8],Type="logistic",Rates=[3.5 1.5]);
Mdl = tsVAR(tt,[mdl1; mdl2; mdl3]);
Load the Canadian inflation and interest rate data set.
Extract the CPI-based inflation rate series.
numObs = length(INF_C);
Simulate ten paths from the model. Specify the threshold variable type and its data.
Y = simulate(Mdl,numObs,NumPaths=10,Type="exogenous",Z=INF_C);
Y is a numObs-by-10 matrix of simulated paths. Each column represents an independently simulated path.
In a tiled layout, plot the threshold transitions with the data by using ttplot,and plot the simulated paths to one tile.
xticklabels(dates(xticks))
title("Simulations")
Y switches between submodels according to the value of the threshold variable INF_C. Mixing is evident for observations near thresholds, such as at the inflation rates of 1964 and 1978.
Consider the model for the annual, CPI-based, Canadian inflation rate series in Simulate Multiple Paths.
Create the LSTAR model for the series.
Load the Canadian inflation and interest rate data set and extract the inflation rate series.
Simulate a length numObs path from the model. Specify the threshold variable type and its data. Return the innovations and states.
[y,e,s] = simulate(Mdl,numObs,NumPaths=10,Type="exogenous",Z=INF_C);
ylabel("Simulated Response")
This example shows how to initialize simulated paths from presample responses and initial states. The example uses arbitrary parameter values.
{y}_{t}=\left[\begin{array}{c}1\\ -1\end{array}\right]+{\epsilon }_{1,t},
{\epsilon }_{1,t}\sim N\left(\left[\begin{array}{c}0\\ 0\end{array}\right],\left[\begin{array}{cc}1& -0.1\\ -0.1& 1\end{array}\right]\right).
{y}_{t}=\left[\begin{array}{c}2\\ -2\end{array}\right]+\left[\begin{array}{cc}0.5& 0.1\\ 0.5& 0.5\end{array}\right]{y}_{t-1}+{\epsilon }_{2,t},
{\epsilon }_{2,t}\sim N\left(\left[\begin{array}{c}0\\ 0\end{array}\right],\left[\begin{array}{cc}2& -0.2\\ -0.2& 2\end{array}\right]\right).
{y}_{t}=\left[\begin{array}{c}3\\ -3\end{array}\right]+\left[\begin{array}{cc}0.25& 0\\ 0& 0\end{array}\right]{y}_{t-1}+\left[\begin{array}{cc}0& 0\\ 0.25& 0\end{array}\right]{y}_{t-2}+{\epsilon }_{3,t},
{\epsilon }_{3,t}\sim N\left(\left[\begin{array}{c}0\\ 0\end{array}\right],\left[\begin{array}{cc}3& -0.3\\ -0.3& 3\end{array}\right]\right).
{\mathit{y}}_{2,\mathit{t}-4}<-1
-1\le {\mathit{y}}_{2,\mathit{t}-4}<1
Initialize Simulation from Presample Responses
Consider simulating 5 paths initialized from presample responses. Specify a numPreObs-by-numSeries-by-numPaths array of presample responses, where:
numPreObs is the number of presample responses per series and path. You must specify enough presample observations to initialize all AR components in the VAR models and the endogenous threshold variable. The largest AR component order is 2 and the threshold variable delay is 4, therefore simulate requires numPreObs=4 presample observations per series and path.
numSeries=2, the number of response series in the system.
numPaths=5, the number of independent paths to simulate.
Y0 = zeros(delay,Mdl.NumSeries,numPaths);
Y0(:,:,j) = 10*j*ones(delay,Mdl.NumSeries);
Simulate 10 paths of length 100 from the LSETAR model from the presample. Specify the endogenous threshold variable and its delay,
{\mathit{y}}_{2,\mathit{t}-4}
Y = simulate(Mdl,numObs,NumPaths=numPaths,Y0=Y0,Index=2,Delay=4);
Y is a 100-by-2-by-5 array of simulate response paths. For example, Y(50,2,3) is the simulated response of path 3, of series Y2, at time point 50.
Plot the simulated paths for each variable on separate plots.
plot(squeeze(Y(:,1,:)))
The system quickly settles regardless of the presample.
Initialize Simulation from States
Simulate three paths of length 100, where each of the three states initialize a path. Specify state indices for initialization, and specify the endogenous threshold variable and its delay.
S0 = 1:Mdl.NumStates;
numPaths = numel(S0);
Y = simulate(Mdl,numObs,NumPaths=numPaths,S0=S0,Index=2,Delay=4);
Consider including regression components for exogenous variables in each submodel of the threshold-switching dynamic regression model in Initialize Multivariate Model Simulation from Multiple Starting Conditions.
{y}_{t}=\left[\begin{array}{c}1\\ -1\end{array}\right]+\left[\begin{array}{c}1\\ -1\end{array}\right]{x}_{1,t}+{\epsilon }_{1,t},
{\epsilon }_{1,t}\sim N\left(\left[\begin{array}{c}0\\ 0\end{array}\right],\left[\begin{array}{cc}1& -0.1\\ -0.1& 1\end{array}\right]\right).
{y}_{t}=\left[\begin{array}{c}2\\ -2\end{array}\right]+\left[\begin{array}{cc}2& 2\\ -2& -2\end{array}\right]{x}_{2,t}+\left[\begin{array}{cc}0.5& 0.1\\ 0.5& 0.5\end{array}\right]{y}_{t-1}+{\epsilon }_{2,t},
{\epsilon }_{2,t}\sim N\left(\left[\begin{array}{c}0\\ 0\end{array}\right],\left[\begin{array}{cc}2& -0.2\\ -0.2& 2\end{array}\right]\right).
{y}_{t}=\left[\begin{array}{c}3\\ -3\end{array}\right]+\left[\begin{array}{ccc}3& 3& 3\\ -3& -3& -3\end{array}\right]{x}_{3,t}+\left[\begin{array}{cc}0.25& 0\\ 0& 0\end{array}\right]{y}_{t-1}+\left[\begin{array}{cc}0& 0\\ 0.25& 0\end{array}\right]{y}_{t-2}+{\epsilon }_{3,t},
{\epsilon }_{3,t}\sim N\left(\left[\begin{array}{c}0\\ 0\end{array}\right],\left[\begin{array}{cc}3& -0.3\\ -0.3& 3\end{array}\right]\right).
{\mathit{x}}_{1,\mathit{t}}
{\mathit{x}}_{2,\mathit{t}}
{\mathit{x}}_{3,\mathit{t}}
%VARX submodels
If you do not supply exogenous data, simulate ignores the regression components in the submodels. Simulate a single path of responses, innovations, and states into a simulation horizon of length 50. Then plot each path separately.
[Y,E,SP] = simulate(Mdl,numObs);
stem(SP)
simulate requires exogenous data in order to generate random paths from the model. Simulate exogenous data for the three regressors by generating 50 random observations from the 3-D standard Gaussian distribution.
rng(1); % Reset seed for comparison
[Y,E,SP] = simulate(Mdl,numObs,X=X);
This example shows how to use Monte Carlo estimation to obtain an interval estimate of the threshold mid-level.
Consider a SETAR model for the real US GDP growth rate
{\mathit{y}}_{\mathit{t}}
with AR(4) submodels. Suppose the threshold variable is
{\mathit{y}}_{\mathit{t}}
(self exciting with 0 delay).
Create a discrete threshold transition at unknown mid-level
{\mathit{t}}_{1}
. Label the states "Recession" and "Expansion".
tt = threshold(NaN,StateNames=["Recession" "Expansion"]);
For each state, create a partially specified AR(4) model with one coefficient at lag 4. Store the state submodels in a vector.
submdl = arima(ARLags=4);
mdl = [submdl; submdl];
Each submodel has an unknown, estimable lag 4 coefficient, model constant, and innovations variance.
Create a partially specified TAR model from the threshold transition and submodel vector.
Create a fully specified threshold transition that has the same structure as tt, but set the mid-level to 0.
Load the US macroeconomic data set. Compute the real GDP growth rate as a percent.
pRGDP = 100*price2ret(rGDP);
T = numel(pRGDP);
Fit the TAR to the real GDP rate series.
EstMdl = estimate(Mdl,tt0,pRGDP,Z=pRGDP,Type="exogenous");
Simulate 100 response paths from the estimated model.
numPaths= 100;
Y = simulate(EstMdl,T,NumPaths=numPaths,Z=pRGDP,Type="exogenous");
Fit the TAR model to each simulated response path. Specify the estimated threshold transition EstMdl.Switch to initialize the estimation procedure. For each path, store the estimated threshold transition mid-level.
tMC = nan(T,1);
EstMdlSim = estimate(Mdl,EstMdl.Switch,Y(:,j),Z=Y(:,j),Type="exogenous");
tMC(j) = EstMdlSim.Switch.Levels;
tMC is a 100-by-1 vector representing a Monte Carlo sample of threshold transitions.
Obtain a 95% confidence interval on the true threshold transition by computing the 0.25 and .975 quantiles of the Monte Carlo sample.
tCI = quantile(tMC,[0.025 0.975])
tCI = 1×2
A 95% confidence interval on the true threshold transition is (0.52%, 0.98%).
Example: NumPaths=1000,Y0=Y0 simulates 1000 sample paths and initializes the dynamic component of each submodel by using the presample response data Y0.
Number of sample paths to generate, specified as a positive integer.
{z}_{t}={y}_{j,\left(t-d\right)},
Presample response data, specified as a numeric matrix or array.
To use the same presample data for each of the numPaths path, specify a numPreSampleObs-by-numSeries matrix, where numPaths is the value of NumPaths, numPreSampleObs is the number of presample observations, and numSeries is the number of response variables.
The number of presample observations numPreSampleObs must be sufficient to initialize the AR terms of all submodels. For models of type "endogenous", the number of presample observations must also be sufficient to initialize the delayed response. If numPreSampleObs exceeds the number necessary to initial the model, simulate uses only the latest observations. The last row contains the latest observations.
simulate updates Y0 using the latest simulated observations each time it switches states.
By default, simulate determines Y0 by the submodel of the initial state:
Threshold variable data for simulations of type "exogenous", specified as a numeric vector of length numObsZ or a numObsZ-by-numPaths numeric matrix.
For a numeric vector, simulate applies the same data to all simulated paths. For a matrix, simulate applies columns of Z to corresponding simulated paths.
If numObsZ exceeds numobs, simulate uses only the latest observations. The last row contains the latest observation.
simulate determines the initial state of simulations by values in the first row Z(1,:).
Threshold variable delay d in yj,t−d for simulations of type "endogenous", specified as a positive integer.
Threshold variable index j in yj,t−d for simulations of type "endogenous", specified as a scalar in 1:Mdl.NumSeries.
simulate ignores Index for univariate AR models.
S0 — Initial states
Initial states of simulations, for simulations of type "endogenous", specified as a numeric scalar or vector of length numPaths. Entries of S0 must be in 1:Mdl.NumStates.
A scalar S0 applies the same initial state to all paths. A vector S0 applies initial state S0(j) to path j.
If you specify Y0, simulate ignores S0 and determines initial states by the specified presample data.
Example: 'S0',2 applies state 2 to initialize all paths.
simulate generates innovations using the covariance specification in Mdl. For more details, see tsVAR.
If threshold levels in Mdl.Switch.Levels are t1, t2,… ,tn, simulate labels states of the threshold variable (-∞,t1), [t1,t2), … [tn,∞) as 1, 2, 3,... n + 1, respectively.
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first order Eulerian numbers
The eulerian1(n, k) command counts the number of permutations of
{\mathrm{pi}}_{1}{\mathrm{pi}}_{2}\mathrm{...}{\mathrm{pi}}_{n}
{1,2,\mathrm{...},n}
that have k ascents, namely, k places where
{\mathrm{pi}}_{j}<{\mathrm{pi}}_{j+1}
\mathrm{eulerian1}\left(n,k\right)=\left(k+1\right)\mathrm{eulerian1}\left(n-1,k\right)+\left(n-k\right)\mathrm{eulerian1}\left(n-1,k-1\right)
It also satisfies the following identities:
{x}^{n}=\textcolor[rgb]{0.564705882352941,0.564705882352941,0.564705882352941}{\sum }_{k=0}^{n}\mathrm{eulerian1}\left(n,k\right)\left(\genfrac{}{}{0}{}{x+k}{n}\right)
m!\mathrm{Stirling2}\left(n,m\right)=\textcolor[rgb]{0.564705882352941,0.564705882352941,0.564705882352941}{\sum }_{k=0}^{n}\mathrm{eulerian1}\left(n,k\right)\left(\genfrac{}{}{0}{}{k}{n-m}\right)
\mathrm{eulerian1}\left(n,k\right)
\textcolor[rgb]{0.564705882352941,0.564705882352941,0.564705882352941}{\sum }_{m=0}^{k}{\left(-1\right)}^{m}\left(\genfrac{}{}{0}{}{n+1}{m}\right){\left(k+1-m\right)}^{n}
\textcolor[rgb]{0.564705882352941,0.564705882352941,0.564705882352941}{\sum }_{m=0}^{n}{\left(-1\right)}^{n-m-k}m!\left(\genfrac{}{}{0}{}{n-m}{k}\right)\mathrm{Stirling2}\left(n,m\right)
\mathrm{with}\left(\mathrm{combinat}\right):
\mathrm{Matrix}\left([\mathrm{seq}\left([\mathrm{seq}\left(\mathrm{eulerian1}\left(n,k\right),k=0..5\right)],n=0..5\right)]\right)
[\begin{array}{cccccc}\textcolor[rgb]{0,0,1}{1}& \textcolor[rgb]{0,0,1}{0}& \textcolor[rgb]{0,0,1}{0}& \textcolor[rgb]{0,0,1}{0}& \textcolor[rgb]{0,0,1}{0}& \textcolor[rgb]{0,0,1}{0}\\ \textcolor[rgb]{0,0,1}{1}& \textcolor[rgb]{0,0,1}{0}& \textcolor[rgb]{0,0,1}{0}& \textcolor[rgb]{0,0,1}{0}& \textcolor[rgb]{0,0,1}{0}& \textcolor[rgb]{0,0,1}{0}\\ \textcolor[rgb]{0,0,1}{1}& \textcolor[rgb]{0,0,1}{1}& \textcolor[rgb]{0,0,1}{0}& \textcolor[rgb]{0,0,1}{0}& \textcolor[rgb]{0,0,1}{0}& \textcolor[rgb]{0,0,1}{0}\\ \textcolor[rgb]{0,0,1}{1}& \textcolor[rgb]{0,0,1}{4}& \textcolor[rgb]{0,0,1}{1}& \textcolor[rgb]{0,0,1}{0}& \textcolor[rgb]{0,0,1}{0}& \textcolor[rgb]{0,0,1}{0}\\ \textcolor[rgb]{0,0,1}{1}& \textcolor[rgb]{0,0,1}{11}& \textcolor[rgb]{0,0,1}{11}& \textcolor[rgb]{0,0,1}{1}& \textcolor[rgb]{0,0,1}{0}& \textcolor[rgb]{0,0,1}{0}\\ \textcolor[rgb]{0,0,1}{1}& \textcolor[rgb]{0,0,1}{26}& \textcolor[rgb]{0,0,1}{66}& \textcolor[rgb]{0,0,1}{26}& \textcolor[rgb]{0,0,1}{1}& \textcolor[rgb]{0,0,1}{0}\end{array}]
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Some of the following algebraic fractions have common denominators and some do not. Add or subtract the expressions and, if possible, simplify.
\frac { 3 } { ( x - 4 ) ( x + 1 ) } + \frac { 6 } { x + 1 }
\frac{3}{(x-4)(x+1)}+\frac{6}{x+1}\cdot \frac{x-4}{x-4}
\frac{3+6(x-4)}{(x-4)(x+1)}
\frac{6x-21}{(x-4)(x+1)}
\frac { 5 } { 2 ( x - 5 ) } + \frac { 3 x } { x - 5 }
\frac { x } { x ^ { 2 } - x - 2 } - \frac { 2 } { x ^ { 2 } - x - 2 }
Try factoring after subtracting.
\frac { x + 2 } { x ^ { 2 } - 9 } - \frac { 1 } { x + 3 }
Factor to find the common denominator, then use the strategy from part (a) to simplify.
\frac{5}{x^2-9}
-\frac{1}{x^2-9},
check the step where you subtracted.
−(x−3)=x+3
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Floating-point arithmetic - Wikipedia @ WordDisk
{\displaystyle {\text{significand}}\times {\text{base}}^{\text{exponent}},}
{\displaystyle 1.2345=\underbrace {12345} _{\text{significand}}\times \underbrace {10} _{\text{base}}\!\!\!\!\!\!^{\overbrace {-4} ^{\text{exponent}}}.}
This article uses material from the Wikipedia article Floating-point arithmetic, and is written by contributors. Text is available under a CC BY-SA 4.0 International License; additional terms may apply. Images, videos and audio are available under their respective licenses.
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Steel_square Knowpia
The steel square is a tool used in carpentry. Carpenters use various tools to lay out structures that are square (that is, built at accurately measured right angles), many of which are made of steel, but the name steel square refers to a specific long-armed square that has additional uses for measurement, especially of various angles. It consists of a long, wider arm and a shorter, narrower arm, which meet at an angle of 90 degrees (a right angle). Today the steel square is more commonly referred to as the framing square or carpenter's square, and such squares are no longer invariably made of steel (as they were many decades ago); they can also be made of aluminum or polymers, which are light and resistant to rust.
Carpentry hand tool
The longer wider arm is two inches (51 mm) wide, and is called the blade; the shorter narrower arm, is one and a half inches (38 mm) wide, and is called the tongue. The square has many uses, including laying out common rafters, hip rafters and stairs.[1] It has a diagonal scale, board foot scale and an octagonal scale. On the newer framing squares there are degree conversions for different pitches and fractional equivalents.
Framing squares may also be used as winding sticks.
Blade and tongueEdit
In traditional timber frame joinery, mortises and tenons were typically two inches (51 mm) wide and two inches (51 mm) from the edge of the timber when working with softwoods, giving rise to the width of the blade. Likewise, mortises and tenons were traditionally one and a half inches (38 mm) wide when working in hardwoods, explaining the width of the tongue. This allowed for quick layouts of mortise and tenon joints when working both hard and softwoods.
A steel square is self-proving and self-calibrating in that you can lay out a perpendicular line, flip the square over, and determine the size and direction of the error. The error can be corrected by opening or closing the angle with a center punch.[2]
Stair framingEdit
Theoretical rise and run of stringer, placement of square, marking of tread and rise, dropping the stringer, ABC=90°, total rise of stringer = 2R-T, total run of stringer = 2AB.
Stairs usually consist of three components. They are the stringer, the tread and the riser. The stringer is the structural member that carries the load of the staircase, the tread is the horizontal part that is stepped on, and the riser board is the vertical part which runs the width of the structure. There are many types of stairs: open, closed, fully housed, winding, and so on, to mention a few of them.
Laying out a staircase requires rudimentary math. There are numerous building codes to which staircases must conform. In an open area the designer can incorporate a more desirable staircase. In a confined area this becomes more challenging. In most staircases there is one more rise than there are treads.
The rise (vertical measurement), and the run (horizontal measurement). Note that the stringer will rest partially on the horizontal surface.
This is a two-by-twelve piece of lumber. A framing square is placed on the lumber so that the desired rise and tread marks meet the edge of the board. The outline of the square is traced. The square is slid up the board until the tread is placed on the mark and the process is repeated.
The board is cut along the dotted lines, and the top plumb cut and the bottom level cut are traced by holding the square on the opposite side.
The stringer in this example has two pieces of tread stock. This allows for a slight overhang. There is also a space in between the boards. The bottom of the stringer must be cut to the thickness of the tread. This step is called dropping the stringer. After one stringer is cut this piece becomes the pattern that is traced onto the remaining stringers.
Roof framingEdit
Blade of steel square
Common rafter length per foot run 21.63" 19.21" 16.97" 14.42"
Hip or valley rafter length per foot run 24.74" 22.65" 20.78" 18.76"
Difference in length of jacks 16 inch centers 28.88" 25.63" 22.63" 19.25"
Side cut length of jack rafters 6.69" 7.5" 8.5" 10.00"
Side cut of hip rafter or valley rafter 8.25" 9.0" 9.81" 10.88"
This table shows five different types of rafter calculations and one table for marking an angle called the side cut or cheek cut.
There is a table of numbers on the face side of the steel square; this is called the rafter table. The rafter table allows the carpenter to make quick calculations based on the Pythagorean theorem. The table is organized by columns that correspond to various slopes of the roof. Each column describes a different roof inclination (pitch) and contains the following information:
This is a common rafter with the two different cuts. The plumb cut fits in the ridge board and the Bird's mouth fits on the wall plate.
Common rafter per foot of run. The common rafter connects the peak of a roof (the ridge) to the base of a roof (the plate). This number gives the length (hypotenuse) of the common rafter per twelve units of horizontal distance (run).
Hip or valley rafter per foot of run. The hip or valley rafter also connects the ridge to the plate, but lies at a 45-degree angle to the common rafter. This number gives the length of the hip or valley rafter per seventeen units of run.
Difference in lengths jacks. The jack rafters lie in the same plane as the common rafter but connect the top plate (the wall) or ridge board to the hip or valley rafter respectively. Since the hip or valley rafter meets the ridge board and the common rafter at angles of 45 degrees, the jack rafters will have varying lengths when they intersect the hip or valley. Depending on the spacing of the rafters, their lengths will vary by a constant factor—this number is the common difference.
This angle can be cut on the fly by aligning this given number on the blade of the steel square and the twelve-inch mark on the tongue, and drawing a line along the tongue.
Cutting hip and valley cripple rafters are all cut in a similar way.
The relationship between hip, jack and common rafters, and how they tie into the ridge and bottom plate. The rafters are fastened to the horizontal ridge board at the peak of the roof.
The side cut is the beveled angle of the hip or valley rafter that fits into the ridge board in this image.
Common and jack rafters all use twelve as the common reference to mark the plumb cut.
Hip and jack rafters use twelve as a common reference while aligning the desired pitch in the side cut column.
Hip and valley rafters use seventeen as the common reference for marking the plumb cut of a rafter.
Octagon scaleEdit
The octagon scale allows the user to inscribe an octagon inside a square, given the length of the side of the square. The markings indicate half the length of the octagon's sides, which can be set to a compass or divider. Arcs drawn from the midpoints of the square's sides will intersect the square at the vertices of the planned octagon. All that remains is to cut four triangular sections from the square.
Octagon table located on the front side of the steel square.
Octagon table viewed from an aluminum square.
Diagonal scaleEdit
Knee bracing is a common feature in timber framing to prevent racking under lateral loads. The diagonal scale is useful for determining the length of the a knee brace desired for a given distance from the joint between the post and beam.
This is the location of the diagonal scale on the square.
The diagonal scale gives the diagonal, or the hypotenuse, for the different legs of the triangle for which a brace is to be cut.
Calculators in roof framingEdit
In addition to use the square tool, construction calculators are also used to verify and determine roofing calculations. Some are programmed to calculate all side cuts for hip, valley and jack regular rafters to be exactly 45° for all rafter pitches. The rafter table is expressed in inches, and the higher the numerical value of the pitch, the greater the difference between side cut angles within a given pitch. Only a level roof, or a 0 pitch will require a 45° angle side cut (cheek cut) for hip and jack rafters.
Side cut hip/valley rafter tableEdit
If a right triangle has two angles that equal 45° then the two sides are equidistant. The rafter is the hypotenuse and the legs or catheti of the triangle are the top wall plates of the structure. The side cut is located at the intersection of the given pitch column and the side cut of the hip/valley row. The regular hip/valley rafter runs at a 45° angle to the main roof and the unit of measurement is 16.97 inches of run. Regular hip/valley and jack rafters have different bevel angles within any given pitch and the angle decreases as the pitch increases.
{\displaystyle c={\sqrt {12^{2}+12^{2}}}\,}
{\displaystyle L={\sqrt {c^{2}+P^{2}}}\,}
{\displaystyle tangent={\frac {c}{L}}\,}
P= pitch
L= rafter length
Z= difference in L of jack rafter 16" OC
The side cut of the hip/valley rafter = (Tangent)(12) = side cut in inches. The side cuts in the rafter table are all in a base 12. The arc tan can be determined from any given pitch. Most power tools and angle measuring devices use 90° as 0° in construction. The complementary angles of the arc tan are used with tools like the speed square.
Side cut of jack raftersEdit
The side cut is located at the intersection of the side cut of jack rafters row and the pitch column on the Steel square. There is a row for the difference in length of jacks, 16 and 24 inch centers on the blade. The tangents are directly proportional for both centers.
{\displaystyle tangent={\frac {16}{z}}\,}
The tangent is in a base 12. The tangent x 12 = side cut of jack rafters. This corresponds to the side cut on the Steel square. The complementary angles of the arc tan are used on most angle measuring devices in construction. The tangent of hip, valley, and jack rafters are less than 1.00 in all pitches above 0°. An eighteen pitch has a side cut angle of 29.07° and a two pitch has a side cut angle of 44.56° for jack rafters. This is a variation of 15.5° between pitches.
Side cut angles versus pitch
This is a reference table for side cuts versus pitch. (only valid for 90 degrees eave angle) :
Pitch expressed in rise units / run units
Pitch 18/12 ==> 60,86 deg
Pitch 9/12 ==> 51,25 deg
Plumb cut of jack and common raftersEdit
The plumb cut for jack and common rafters are the same angles. The level cut or seat cut is the complementary angle of the plumb cut. The notch formed at the intersection of the level and plumb cut Is commonly referred to as the bird's mouth .
{\displaystyle tangent={\frac {P}{b}}.\,}
Plumb cut of hip/valley raftersEdit
The plumb cut of the hip/valley rafter is expressed in the formula. The level cut is the complementary angle or 90° minus the arc tan.
{\displaystyle tangent={\frac {P}{\sqrt {a^{2}+b^{2}}}}.\,}
Irregular hip/valley raftersEdit
The only Framing Square that has tables for unequal pitched roofs is the Chappell Universal Square, (patent #7,958,645). There is also a comprehensive rafter table for 6 & 8 sided polygon roofs (first time ever on a framing square). The traditional steel square's rafter table (patented April 23,1901) is limited in that it does not have tables that allow for work with unequal pitched roofs. Irregular hip/valley rafters are characterized by plan angles that are not equal or 45°. The top plates can be 90° at the outside corners or various other angles. There are numerous irregular h/v roof plans.
irregular hip/valley gable roof plan.
irregular roof plan.
Intersecting irregular hip/valley gable roof plan.
Carpenter's squareEdit
In carpentry, a square is a guide for establishing right angles (90° angles) or mitre angles, usually made of metal. There are various types of square, such as speed squares, try squares and combination squares.
^ Elliot, J. Hamilton (1910), Cobleigh, Rolfe (ed.), "Use of the Steel Square", Handy Farm Devices and How to Make Them, Orange Judd and Company.
^ "Framing Square Test & Repair". Inspectapedia. Retrieved 10 December 2021.
Wikimedia Commons has media related to Steel squares.
Hodgson, Frederick T. (1880). The Carpenters' Steel Square and Its Uses. The Industrial Publication Company. LCCN 06036488.
Ulrey, Harry F. (1972). Carpenters and Builders Library. Vol. No.3. Theodore Audel. LCCN 74099760.
Schuttner, Scott (1990). Basic Stairbuilding. Taunton Press. ISBN 0-942391-44-6.
Falconer, John (1925). Ednie, John (ed.). The Steel Square. Carpentry and Joinery. Vol. V. Gresham.
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Introduction to Thermodynamics | Boundless Chemistry | Course Hero
Thermodynamics is the study of heat energy and other types of energy, such as work, and the various ways energy is transferred within chemical systems. "Thermo-" refers to heat, while "dynamics" refers to motion.
\Delta E=\Delta E_{sys}+\Delta E_{surr}=0
If you've ever witnessed a video of a space shuttle lifting off, the chemical reaction that occurs also releases tremendous amounts of heat and light. Another useful form of the first law of thermodynamics relates heat and work for the change in energy of the internal system:
\Delta E_{sys}=Q+W
Both heat and work refer to processes by which energy is transferred to or from a substance.
Define heat and work
When energy is exchanged between thermodynamic systems by thermal interaction, the transfer of energy is called heat.
Heat and work are related: work can be completely converted into heat, but the reverse is not true: heat cannot be completely converted to work.
thermochemistry: The study of the energy and heat associated with chemical reactions and/or physical transformations.
When energy is exchanged between thermodynamic systems by thermal interaction, the transfer of energy is called heat. The units of heat are therefore the units of energy, or joules (J). Heat is transferred by conduction, convection, and/or radiation.
Heat is transfer by conduction occurs when an object with high thermal energy comes into contact with an object with low thermal energy. Heat transfer by convection occurs through a medium. For example, when heat transfers from the hot water at the bottom of the pot to the cooler water at the top of the pot. Lastly, heat can also be transferred by radiation; a hot object can convey heat to anything in its surroundings via electromagnetic radiation.
When a high temperature body is brought into contact with a low temperature body, the temperatures equilibrate: there is heat flow from higher to lower temperature, like water flowing downhill, until the temperatures of the bodies are equivalent. The high temperature body loses thermal energy, and the low temperature body acquires this same amount of thermal energy. The system is then said to be at thermal equilibrium.
An illustration of thermal equilibrium: The can of cola and ice cube start at different temperatures. When they come into contact, heat is transferred from the cola can to the ice cube until both bodies reach thermal equilibrium.
Work is the transfer of energy by any process other than heat. Like heat, the unit measurement for work is joules (J). There are many forms of work, including but not limited to mechanical, electrical, and gravitational work. For our purposes, we are concerned with P-V work, which is the work done in an enclosed chemical system. In this type of system, work is defined as the change in the volume (V) in liters within the system multiplied by a pressure (P). Assuming the system is at constant pressure, this equates to the following:
W=P\Delta V
Most often, we are interested in the work done by expanding gases. Assuming the gases are ideal, we can apply the ideal gas law to the above equation to get the following:
W=P\Delta V=nR\Delta T
Heat and work are related. Work can be completely converted into heat, but the reverse is not true: heat energy cannot be wholly transformed into work energy. Scientists and engineers have been able to exploit the principles of thermochemistry to develop technologies ranging from hot/cold packs to gasoline powered combustion engines.
For a closed system, the change in internal energy (∆U) is related to heat (Q) and work (W) as follows:
\Delta U = Q+W
This means that the total energy within a system is affected by the sum of two possible energy transfers: heat and work.
Energy, heat, and work in Chemistry. Provided by: Steve Lower's Website. Located at: http://www.chem1.com/acad/webtext/energetics/CE-1.html#SEC3. License: CC BY-SA: Attribution-ShareAlike
Work (thermodynamics). Provided by: Wikipedia. Located at: http://en.wikipedia.org/wiki/Work_(thermodynamics). License: CC BY-SA: Attribution-ShareAlike
Provided by: Steve Lower's Website. Located at: http://www.chem1.com/acad/webtext/pre/enheat.html. License: CC BY-SA: Attribution-ShareAlike
Heat. Provided by: Wikipedia. Located at: http://en.wikipedia.org/wiki/Heat. License: CC BY-SA: Attribution-ShareAlike
Thermochemistry. Provided by: Wikipedia. Located at: http://en.wikipedia.org/wiki/Thermochemistry. License: CC BY-SA: Attribution-ShareAlike
OpenStax College, Heat. October 20, 2012. Provided by: OpenStax CNX. Located at: http://cnx.org/content/m42223/latest/. License: CC BY: Attribution
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How to use the round pen calculator
How to manually calculate the round pen size
Our round pen calculator helps you calculate the round pen size you need based on the diameter or circumference of the designated land. This tool determines how many round pen panels you need based on your desired panel length.
This article briefly discusses how to build a round pen. More importantly, it focuses on how many round pen panels you need based on the required size of the pen.
Here's an overview of how we can build a round pen:
Choose your pen's appropriate diameter and height based on the animals you wish to keep.
For example, if you plan to train a horse, you should have a round pen at least 60 feet in diameter.
Make sure to build the pen on firm ground that isn't prone to flooding. Clean and level the land before installing the enclosure. Keeping in mind the safety and comfort of the animal, have suitable amounts of sand, wood shavings or soft dirt for the ground, and cement for fixing posts.
Decide between steel or wood for the pen. Wood may be more economical, but you cannot move it once fixed. In comparison, steel may be more expensive but easier to assemble and more manageable for increasing or decreasing the size of the pen.
🔎 To learn about different types of steel and their weight, check out our steel weight calculator.
Determine the required number of posts and panels based on your preferred panel lengths and pen size. The number of posts will be a multiple of the number of panels.
You can calculate your round pen panel quantity using our given calculator.
Starting from the entrance, mark the ground for posts. Dig the holes about 2 feet deep. Attach the panels to the posts and fix the posts in the ground using cement.
Once you've attached the gate, you're ready to bring the animals in.
Here's how you can use our round pen calculator:
Enter your preferred pen diameter, e.g., 80 ft.
The calculator will give you the pen circumference of 251.3 ft.
🔎 Alternatively, you can also enter the pen's circumference in our calculator to obtain its diameter or use our circumference to diameter conversion calculator.
Enter your preferred panel length. A typical panel size is 8 feet.
💡 Depending on the requirement, a standard round pen's panel length can be anywhere between 4 and 20 feet wide.
Once you enter your preferred panel length, you'll get the number of panels required to build this round pen, which in this case is 31 panels.
You can manually calculate your round pen panels quantity using the following formula:
P = D \times \pi \div L
P
– Number of panels required to build the round pen;
D
– Diameter of the round pen;
\pi
– Math constant with a value of 3.1416 approx.; and
L
– Preferred length of the panel.
Here's an example of calculating the required number of panels for a round pen, where the pen's diameter should be 100 feet and our preferred panel length is 15 feet.
P = 100 \times \pi \div 15 ≈ 21
Thus, we require 21 panels of 15 feet each for building a round pen of 100 feet in diameter.
For a round pen, how many 12 foot panels do I need?
16. That's how many 12-foot round pen panels you need for a standard round pen with a diameter of 60 feet.
You can calculate the round pen panels that are required by using this formula:
Number of panels = Pen diameter × π / Preferred panel length.
How big are round pens for horses?
A round pen with a diameter of 60 feet and a height of 5 feet is sufficient for keeping regular horses. For riding and training wild horses, such as stallions, a round pen of about 80 feet in diameter and a height of 6 feet is more suitable.
How do I calculate the number of panels for a round pen?
To calculate the number of panels for a round pen:
Measure the diameter of your pen in feet. As an example, let's say it's 45 ft.
Multiply the result by π to find its circumference, e.g., 45 × 3.1416 = 141.4 ft.
Divide the circumference by your desired panel length (in feet). So that's 141.4 / 12 = 12.
Thus, with a 45 feet diameter, you'll need 12 panels, each 12 feet long.
How many 10′ panels in a 50′ round pen?
16. That's how many round pen panels you need for a round pen of 50′ diameter.
You can calculate the number of panels using the following formula:
Pen circumference
Find the weight of any volume of glass material made from the most common glass types in production today using this glass weight calculator.
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First-price sealed-bid auction - Wikipedia
A first-price sealed-bid auction (FPSBA) is a common type of auction. It is also known as blind auction.[1] In this type of auction, all bidders simultaneously submit sealed bids so that no bidder knows the bid of any other participant. The highest bidder pays the price that was submitted.[2]: p2 [3]
2 Incentive-compatible variant
3 Comparison to second-price auction
4 Comparison to other auctions
In a FPSBA, each bidder is characterized by their monetary valuation of the item for sale.
Suppose Alice is a bidder and her valuation is
{\displaystyle a}
. Then, if Alice is rational:
She will never bid more than
{\displaystyle a}
, because bidding more than
{\displaystyle a}
can only make her lose net value.
If she bids exactly
{\displaystyle a}
, then she will not lose but also not gain any positive value.
If she bids less than
{\displaystyle a}
, then she may have some positive gain, but the exact gain depends on the bids of the others.
Alice would like to bid the smallest amount that can make her win the item, as long as this amount is less than
{\displaystyle a}
. For example, if there is another bidder Bob and he bids
{\displaystyle y}
{\displaystyle y<a}
, then Alice would like to bid
{\displaystyle y+\varepsilon }
{\displaystyle \varepsilon }
is the smallest amount that can be added, e.g. one cent).
Unfortunately, Alice does not know what the other bidders are going to bid. Moreover, she does not even know the valuations of the other bidders. Hence, strategically, we have a Bayesian game - a game in which agents do not know the payoffs of the other agents.
The interesting challenge in such a game is to find a Bayesian Nash equilibrium. However, this is not easy even when there are only two bidders. The situation is simpler when the valuations of the bidders are i.i.d. random variables, i.e.: there is a known prior distribution and the valuations of the bidders are all drawn from the same distribution.[4]: 234–236
Suppose there are two bidders, Alice and Bob, whose valuations
{\displaystyle a}nd
{\displaystyle b}
are drawn from a Continuous uniform distribution over the interval [0,1]. Then, it is a Bayesian-Nash equilibrium when each bidder bids exactly half his/her value: Alice bids
{\displaystyle a/2}
and Bob bids
{\displaystyle b/2}
PROOF: The proof takes the point-of-view of Alice. We assume that she knows that Bob bids
{\displaystyle f(b)=b/2}
, but she does not know
{\displaystyle b}
. We find the best response of Alice to Bob's strategy. Suppose Alice bids
{\displaystyle x}
. There are two cases:
{\displaystyle x\geq f(b)}
. Then Alice wins and enjoys a net gain of
{\displaystyle a-x}
. This happens with probability
{\displaystyle f^{-1}(x)=2x}
{\displaystyle x<f(b)}
. Then Alice loses and her net gain is 0. This happens with probability
{\displaystyle 1-f^{-1}(x)}
All in all, Alice's expected gain is:
{\displaystyle G(x)=f^{-1}(x)\cdot (a-x)}
. The maximum gain is attained when
{\displaystyle G'(x)=0}
. The derivative is (see Inverse functions and differentiation):
{\displaystyle G'(x)=-f^{-1}(x)+(a-x)\cdot {1 \over f'(f^{-1}(x))}}
and it is zero when Alice's bid
{\displaystyle x}
{\displaystyle f^{-1}(x)=(a-x)\cdot {1 \over f'(f^{-1}(x))}}
Now, since we are looking for a symmetric equilibrium, we also want Alice's bid
{\displaystyle x}
{\displaystyle f(a)}
{\displaystyle f^{-1}(f(a))=(a-f(a))\cdot {1 \over f'(f^{-1}(f(a)))}}
{\displaystyle a=(a-f(a))\cdot {1 \over f'(a)}}
{\displaystyle af'(a)=(a-f(a))}
The solution of this differential equation is:
{\displaystyle f(a)=a/2}
{\displaystyle v_{i}}
- the valuation of bidder
{\displaystyle i}
{\displaystyle y_{i}}
- the maximum valuation of all bidders except
{\displaystyle i}
{\displaystyle y_{i}=\max _{j\neq i}{v_{j}}}
Then, a FPSBA has a unique symmetric BNE in which the bid of player
{\displaystyle i}s given by:[5]: 33–40
{\displaystyle E[y_{i}|y_{i}<v_{i}]}
Incentive-compatible variant[edit]
The FPSBA is not incentive-compatible even in the weak sense of Bayesian-Nash-Incentive-Compatibility (BNIC), since there is no Bayesian-Nash equilibrium in which bidders report their true value.
However, it is easy to create a variant of FPSBA which is BNIC, if the priors on the valuations are common knowledge. For example, for the case of Alice and Bob described above, the rules of the BNIC variant are:
The highest bidder wins;
The highest bidder pays 1/2 of his/her bid.
In effect, this variant simulates the Bayesian-Nash equilibrium strategies of the players, so in the Bayesian-Nash equilibrium, both bidders bid their true value.
This example is a special case of a much more general principle: the revelation principle.
Comparison to second-price auction[edit]
The following table compares FPSBA to sealed-bid second-price auction (SPSBA):
Winner: Agent with highest bid Agent with highest bid
Winner pays: Winner's bid Second-highest bid
Loser pays: 0 0
Dominant strategy: No dominant strategy Bidding truthfully is dominant strategy[6]
Bayesian Nash equilibrium[7] Bidder
{\displaystyle i}
{\displaystyle {\frac {n-1}{n}}v_{i}}
{\displaystyle i}
truthfully bids
{\displaystyle v_{i}}
Auctioneer's revenue[7]
{\displaystyle {\frac {n-1}{n+1}}}
{\displaystyle {\frac {n-1}{n+1}}}
The auctioneer's revenue is calculated in the example case, in which the valuations of the agents are drawn independently and uniformly at random from [0,1]. As an example, when there are
{\displaystyle n=2}
agents:
In a first-price auction, the auctioneer receives the maximum of the two equilibrium bids, which is
{\displaystyle \max(a/2,b/2)}
In a second-price auction, the auctioneer receives the minimum of the two truthful bids, which is
{\displaystyle \min(a,b)}
In both cases, the auctioneer's expected revenue is 1/3.
This fact that the revenue is the same is not a coincidence - it is a special case of the revenue equivalence theorem. This holds only when the agents' valuations are statistically independent; when the valuations are dependent, we have a common value auction, and in this case, the revenue in a second-price auction is usually higher than in a first-price auction.
The item for sale may not be sold if the final bid is not high enough to satisfy the seller, that is, the seller reserves the right to accept or reject the highest bid. If the seller announces to the bidders the reserve price, it is a public reserve price auction.[8] In contrast, if the seller does not announce the reserve price before the sale but only after the sale, it is a secret reserve price auction.[9]
Comparison to other auctions[edit]
A FPSBA is distinct from the English auction in that bidders can only submit one bid each. Furthermore, as bidders cannot see the bids of other participants, they cannot adjust their own bids accordingly.[3]
FPSBA has been argued to be strategically equivalent to the Dutch auction.[2]: p13
What are effectively FPSBA are commonly called tendering for procurement by companies and organizations, particularly for government contracts and auctions for mining leases.[3] FPSBA are thought to lead to low procurement costs through competition and low corruption through increased transparency, even though they may entail a higher ex-post extra cost of the completed project and extra time to complete it.[10]
A Generalized first-price auction is a non-truthful auction mechanism for sponsored search (aka position auction).
A generalization of both 1st-price and 2nd-price auctions is an auction in which the price is some convex combination of the 1st and 2nd price.[11]
^ Shor, Mikhael, "blind auction" Dictionary of Game Theory Terms
^ a b Krishna, Vijay (2002), Auction Theory, San Diego, USA: Academic Press, ISBN 978-0-12-426297-3
^ a b c McAfee, Dinesh Satam; McMillan, Dinesh (1987), "Auctions and Bidding" (PDF), Journal of Economic Literature, American Economic Association (published June 1987), vol. 25, no. 2, pp. 699–738, JSTOR 2726107, retrieved 2008-06-25
^ Vazirani, Vijay V.; Nisan, Noam; Roughgarden, Tim; Tardos, Éva (2007). Algorithmic Game Theory (PDF). Cambridge, UK: Cambridge University Press. ISBN 0-521-87282-0.
^ Daron Acemoglu and Asu Ozdaglar (2009). "Networks Lectures 19-21: Incomplete Information: Bayesian Nash Equilibria, Auctions and Introduction to Social Learning". MIT. Retrieved 8 October 2016. {{cite web}}: CS1 maint: uses authors parameter (link)
^ Hence a second-price auction is a truthful mechanism.
^ a b Calculated for
{\displaystyle n}
bidders whose valuations are drawn independently and uniformly at random from [0,1]
^ Elyakime, B.; Laffont, J.J.; Loisel, P.; Vuong, Q. (1994). "First-Price Sealed-Bid Auctions with Secret Reservation Prices". Annales d'Économie et de Statistique. 34 (34): 115–141. doi:10.2307/20075949. JSTOR 20075949.
^ Decarolis, Francesco (2014). "Awarding Price, Contract Performance, and Bids Screening: Evidence from Procurement Auctions". American Economic Journal: Applied Economics. 6 (1).
^ Güth, W.; van Damme, E. (1986-09-01). "A comparison of pricing rules for auctions and fair division games". Social Choice and Welfare. 3 (3): 177–198. doi:10.1007/bf00433534. ISSN 0176-1714.
Hammami, Farouk; Rekik, Monia; Coelho, Leandro C. (2019). "Exact and heuristic solution approaches for the bid construction problem in transportation procurement auctions with a heterogeneous fleet". Transportation Research Part E: Logistics and Transportation Review. 127: 150–177. doi:10.1016/j.tre.2019.05.009. Combinatorial auctions for transportation services procurement with first-price sealed-bid rules.
Nash equilibrium in first price auction - in math.stackexchange.com.
Retrieved from "https://en.wikipedia.org/w/index.php?title=First-price_sealed-bid_auction&oldid=1087716979"
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SVMs - An overview of Support Vector Machines - SVM Tutorial
Today we are going to talk about SVMs in general.
I recently received an email from a reader of my serie of articles about the math behind SVM:
I felt I got deviated a lot on Math part and its derivations and assumptions and finally got confused what exactly SVM is ? And when to use it and how it helps ?
Here is my attempt to clarify things.
What exactly is SVM ?
SVM is a supervised learning model
It means you need a dataset which has been labeled.
Exemple: I have a business and I receive a lot of emails from customers every day. Some of these emails are complaints and should be answered very quickly. I would like a way to identify them quickly so that I answer these email in priority.
Approach 1: I can create a label in gmail using keywords, for instance "urgent", "complaint", "help"
The drawback of this method is that I need to think of all potential keywords that some angry users might use, and I will probably miss some of them. Over time, my keyword list will probably become very messy and it will be hard to maintain.
Approach 2: I can use a supervised machine learning algorithm.
Step 1: I need a lot of emails, the more the better.
Step 2: I will read the title of each email and classify it by saying "it is a complaint" or "it is not a complaint". It put a label on each email.
Step 3: I will train a model on this dataset
Step 4: I will assess the quality of the prediction (using cross validation)
Step 5: I will use this model to predict if an email is a complaint or not.
In this case, if I have trained the model with a lot of emails then it will perform well. SVM is just one among many models you can use to learn from this data and make predictions.
Note that the crucial part is Step 2. If you give SVM unlabeled emails, then it can do nothing.
SVM learns a linear model
Now we saw in our previous example that at the Step 3 a supervised learning algorithm such as SVM is trained with the labeled data. But what is it trained for? It is trained to learn something.
What does it learn?
In the case of SVM, it learns a linear model.
What is a linear model? In simple words: it is a line (in complicated words it is a hyperplane).
If your data is very simple and only has two dimensions, then the SVM will learn a line which will be able to separate the data.
The SVM is able to find a line which separates the data
If it is just a line, why do we talk about a linear model?
Because you cannot learn a line.
So instead of that:
1) We suppose that the data we want to classify can be separated by a line
2) We know that a line can be represented by the equation
y =\mathbf{w}\mathbf{x}+b
(this is our model)
3) We know that there is an infinity of possible lines obtained by changing the value of
\mathbf{w}
b
4) We use an algorithm to determine which are the values of
\mathbf{w}
b
giving the "best" line separating the data.
SVM is one of these algorithms.
Algorithm or model?
At the start of the article I said SVM is a supervised learning model, and now I say it is an algorithm. What's wrong? The term algorithm is often loosely used. For instance, you will sometime read that SVM is a supervised learning algorithm. This is not true if you consider that an algorithm is a set of actions to perform to obtain a specific result. Sequential minimal optimization is the most used algorithm to train SVM, but you can train an SVM with another algorithm like Coordinate descent. However, most people are not interested in details like this, so we simplify and say that we use the SVM "algorithm" (without saying in details which one we use).
SVM or SVMs?
Sometime, you will see people talk about SVM, and sometime about SVMs.
As often Wikipedia is quite good at stating things clearly:
In machine learning, support vector machines (SVMs) are supervised learning models with associated learning algorithms that analyze data used for classification and regression analysis. (Wikipedia)
So, we now discover that there are several models, which belongs to the SVM family.
Wikipedia tells us that SVMs can be used to do two things: classification or regression.
SVM is used for classification
SVR (Support Vector Regression) is used for regression
So it makes sense to say that there are several Support Vector Machines. However, this is not the end of the story !
In 1957, a simple linear model called the Perceptron was invented by Frank Rosenblatt to do classification (which is in fact one of the building block of simple neural networks also called Multilayer Perceptron).
A few years later, Vapnik and Chervonenkis, proposed another model called the "Maximal Margin Classifier", the SVM was born.
Then, in 1992, Vapnik et al. had the idea to apply what is called the Kernel Trick, which allow to use the SVM to classify linearly nonseparable data.
Eventually, in 1995, Cortes and Vapnik introduced the Soft Margin Classifier which allows us to accept some misclassifications when using a SVM.
So just when we talk about classification there is already four different Support Vector Machines:
The original one : the Maximal Margin Classifier,
The kernelized version using the Kernel Trick,
The soft-margin version,
The soft-margin kernelized version (which combine 1, 2 and 3)
And this is of course the last one which is used most of the time. That is why SVMs can be tricky to understand at first, because they are made of several pieces which came with time.
That is why when you use a programming language you are often asked to specify which kernel you want to use (because of the kernel trick), and which value of the hyperparameter C you want to use (because it controls the effect of the soft-margin).
In 1996, Vapnik et al. proposed a version of SVM to perform regression instead of classification. It is called Support Vector Regression (SVR). Like the classification SVM, this model includes the C hyperparameter and the kernel trick.
I wrote a simple article, explaining how to use SVR in R.
If you wish to learn more about SVR, you can read this good tutorial by Smola and Schölkopft.
Maximal Margin Classifier (1963 or 1979)
Kernel Trick (1992)
Soft Margin Classifier (1995)
Support Vector Regression (1996)
If you want to know more, you can learn this very detailed overview of the history.
Other type of Support Vector Machines
Because SVMs have been very successful at classification, people started to think about using the same logic for other type of problems, or to create derivation. As a result there exists now several different and interesting methods in the SVM family:
Structured support vector machine which is able to predict structured objects
Least square support vector machine used for classification and regression
Support vector clustering used to perform cluster analysis
Transductive Support Vector Machine used for semi-supervised learning
Ranking SVM used to sort results
One class support vector machine used for anomaly detection
We have learned that it is normal to have some difficulty to understand what SVM is exactly. This is because there are several Support Vector Machines used for different purposes. As often, history allows us to have a better vision of how the SVM we know today has been built.
I hope this article give you a broader view of the SVM panorama, and will allow you to understand these machines better.
If you wish to learn more about how SVM work for classification, you can start reading the math series:
SVM - Understanding the math
Part 1: What is the goal of the Support Vector Machine (SVM)?
Part 2: How to compute the margin?
Part 3: How to find the optimal hyperplane?
Part 4: Unconstrained minimization
Part 6: Duality and Lagrange multipliers
Categories SVM Tutorial Tags SVM, SVMs, SVR Post navigation
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System Identification Toolbox™ offers three commands for nonlinear state estimation:
x\left[k\right]=f\left(x\left[k-1\right],u\left[k-1\right]\right)+w\left[k-1\right]
x\left[k\right]=f\left(x\left[k-1\right],w\left[k-1\right],u\left[k-1\right]\right)
\begin{array}{l}{\underset{}{\overset{˙}{x}}}_{1}={x}_{2}\\ {\underset{}{\overset{˙}{x}}}_{2}=\left(1-{x}_{1}^{2}\right){x}_{2}-{x}_{1}\end{array}
\underset{}{\overset{˙}{x}}={f}_{c}\left(x\right)
x=\left[{x}_{1};{x}_{2}\right]
{T}_{s}
\underset{}{\overset{˙}{x}}={f}_{c}\left(x\right)
\underset{}{\overset{˙}{x}}\approx \frac{x\left[k+1\right]-x\left[k\right]}{{T}_{s}}
\begin{array}{ll}x\left[k+1\right]& =x\left[k\right]+{f}_{c}\left(x\left[k\right]\right)\phantom{\rule{0.2777777777777778em}{0ex}}{T}_{s}\\ & =f\left(x\left[k\right]\right)\end{array}
{T}_{s}
{T}_{s}
{T}_{s}
{T}_{s}=0.05s
y\left[k\right]=h\left(x\left[k\right],u\left[k\right]\right)+v\left[k\right]
y\left[k\right]=h\left(x\left[k\right],v\left[k\right],u\left[k\right]\right)
{x}_{1}
y\left[k\right]={x}_{1}\left[k\right]\phantom{\rule{0.2777777777777778em}{0ex}}\left(1+v\left[k\right]\right)
{x}_{1}
{x}_{2}
Create this state transition function and save it in a file named vdpMeasurementNonAdditiveNoiseFcn.m. You provide this function to the unscented during object construction.
The top plot shows the true, estimated, and the measured value of the first state. The filter utilizes the system model and noise covariance information to produce an improved estimate over the measurements. The bottom plot shows the second state. The filter is successful in producing a good estimate.
1\sigma
{x}_{1}\left[k\right]*\left(1+v\left[k\right]\right)
1\sigma
Unscented and extended Kalman filters aim to track the mean and covariance of the posterior distribution of the state estimates by different approximation methods. These methods may not be sufficient if the nonlinearities in the system are severe. In addition, for some applications, just tracking the mean and covariance of the posterior distribution of the state estimates may not be sufficient. Particle filters can address these problems by tracking the evolution of many state hypotheses (particles) over time, at the expense of higher computational cost. The computational cost and estimation accuracy increases with the number of particles.
The particleFilter command in System Identification Toolbox implements a discrete-time particle filter algorithm. This section walks you through constructing a particle filter for the same van der Pol oscillator used earlier in this example, and highlights the similarities and differences with the unscented Kalman filter.
There are differences between the state transition function you supply to unscentedKalmanFilter and particleFilter. The state transition function you used for unscented Kalman filter just described propagation of one state hypothesis to the next time step, instead of a set of hypotheses. In addition, the process noise distribution was defined in ProcessNoise property of the unscentedKalmanFilter, just by its covariance. Particle filter can consider arbitrary distributions that may require more statistical properties to be defined. This arbitrary distribution and its parameters are fully defined in the state transition function you provide to the particleFilter.
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dBm to Watts Calculator | Convert dBm to Watts
Formula for dBm to watts conversion
dBm to watts conversion example
How to use the dBm to watts calculator
Omni's dBm to watts calculator allows you to switch between decibels per milliwatt (dBm) and watts (W)/milliwatts (mW).
What is the difference between dBm and dB?
How to convert from dBm to watts (W) or dBm to mW and vice-versa?
You will also learn how to use the dBm to watts calculator.
If you are interested in converting between other power units, our power converter might be the right tool for you.
Decibels per milliwatt or dBm is a measure of power level. We often express sound intensity level or the power level of an electrical signal using a relative unit of measurement called the decibel or dB. For example, we can write power level in decibels as the ratio:
\!\small \text{P (dB)} = 10\ \text{log}_{10} \left (\frac{P_2}{P_1} \right),
P_1
P_2
are the power in watts, with
P_1
usually being the reference power.
When we express power level in decibels (dB) by using
P_1 = 1.0\ \text{mW}
as a reference, we indicate the power level as dBm.
🙋 The terms decibels (dB) or decibel milliwatts (dBm) do not measure power. Instead, they give a relative comparison (or ratio) between two power values.
To convert from dBm to watts, we will use the following formula:
\!\small P (\text W) = 1 \text W \times \frac{10^ \frac{P(\text{dBm})}{10}}{1000}
P(\text W)
– Power in watts (W); and
P(\text{dBm})
– Power in decibel-milliwatts (dBm).
Rearranging the above equation, we will get the formula for converting from watts to dBm:
\!\footnotesize \begin{align*} P (\text{dBm}) &= 10 \times \text{log}_{10} \left ( \frac{1000 \times P (\text W)}{ 1 \text W} \right )\\\\ & = 10 \times \text{log}_{10} \left ( \frac{P (\text W)}{ 1 \text W} \right ) + 30 \end{align*}
As an example, let us convert 50 dBm to watts. We will use the dBm to watts formula:
\small \begin{align*} P (\text W) &= 1\ \text W \times \frac{10^ \frac{P(\text{dBm})}{10}}{1000}\\\\ &= 1\ \text W \times \frac{10^ \frac{50}{10}}{1000}\\\\ &= 1\ \text W \times \frac{10^5}{1000}\\\\ &= 100\ \text W \end{align*}
Hence, 50 dBm is equivalent to 100 watts. The following dBm to watts conversion table shows a chart that you can use for converting certain specific values.
dBm to watts conversion table.
In the next section, we will use the dBm to watts conversion calculator for converting between the two.
Now let us see how we can use the dBm to watts calculator to convert 50 dBm to watts.
Using the drop-down menu, choose the conversion type as dBm to watts.
Enter the power in dBm, i.e., 50 dBm.
The tool will convert the power in dBm to power in watts, i.e., 100 watts. If you want a conversion from dBm to mW, use the drop-down menu to choose dBm to milliwatts as the conversion type.
You can also use this calculator to convert from watts/milliwatts to dBm.
We also have a tool that converts signal distortion from percent to decibel and vice versa, so make sure you check out our total harmonic distortion calculator as well!
How do I convert dBm to watts?
To convert dBm to watts, follow the given instructions:
Divide the power in dBm by 10.
Calculate 10 to the power of the result from step 1.
Now divide the value from step 2 by 1000, and you will get the value of power in watts.
What is the dBm equivalent for 1 watt?
30 dBm. Using the formula P(dBm) = 10 × log₁₀(1000 × P(W)/1W), we can calculate for 1 W, P(dBm) = 10 ×log₁₀(1000 × 1W/1W) = 30 dBm.
How many dBm is 5 watts?
36.99 dBm. We know that the watts to dBm formula is P(dBm) = 10 × log₁₀(P(W)/1W) +30. Hence, we can write P(dBm) = 10 × log₁₀(5) + 30 or P(dBm) = 10 × 0.699 + 30 = 36.99.
0.00631 watts. For converting 8 dBm to watts, we will use the formula P(W) = 1W × (10ᴾ⁽ᵈᴮᵐ⁾/¹⁰)/1000:
Calculating 10 raised to the power of 8/10 = 0.8, we get 6.31.
Dividing the result by 1000, we will get the power in watts, i.e., 0.00631 watts.
🤔 Choose the conversion type!💡
Use our **lbs to g converter** to quickly convert weight (mass) in pounds (lbs) to gram (g).
This lbs to stone converter allows you to effortlessly change a value from lbs to stone.
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Considering High Availability | Seeking Consistency
High Availability (HA) is a fantastic topic. It's often an overlooked topic in the world of performance. It's not as sexy as low latency nor as impressive as high throughput. Yet so many business processes can be significantly adversely impacted by an unavailable application. This is especially true in auto pricing systems employed by banks. Downtime for them means they're not competing in the market and they're not reaching their customer base. This not only represents a missed opportunity but in the more severe cases, loss of business and reputational damage.
Let me share some thoughts on how to think about HA.
Disclaimer: This is a simplified view of the world and considers a single failure in a pair of nodes running on separate servers only. This does not discuss network availability, platform issues such as garbage collection, persistent stores such as databases, or the fault tolerance of the consumers with regards to stream recovery.
Overview of available high availability mechanisms
Generally people speak of HA in terms of 4 policies:
hot-active/cold-passive
hot-active/warm-passive
hot-active/warm-active
hot-active/hot-active
The active/passive part denotes whether the node is doing processing or not. The hot/warm/cold part denotes how active a node is. Cold are inactive nodes that are not running. Warm nodes are running but may or may not be active. Hot nodes are always active.
hot-active/cold-passive: Upon failure the secondary is started, which in addition to booting up will need to load the last good state of the primary process before it can become active. This tends to result in more latent failovers but is one of the simplest HA mechanisms to get going with and frankly, is often one of the most reliable.
hot-active/warm-passive: When failover occurs the secondary must load the last good state of the failed primary process before it can become the active. Failover occurs with less latency than the previous policy because the secondary has already booted up.
hot-active/warm-active: The secondary is running and doing processing but either emitting no output or being ignored by consumers until a failover. When failover occurs the secondary node already has the last good state (thanks to being active) and so only needs to become hot, either by starting to emit output or by the consumers recognising the secondary as the hot node. Again, this improves upon latency of the previous policy because there's no need to boot up or load the last good state.
hot-active/hot-active: Often this accompanied by either load balancing or deduplication on the consumer. The idea is that failover occurs immediately with the lowest latency possible. Users typically shouldn't notice any outage. Imagine you've sent a request, both nodes receive it, process it and emit a response. The requester must then deduplicate the responses, this is fine so long as the responses are always identical. If the response is formed by an aggregation, random or time sensitive computation then this can become difficult and consistency of the response needs considering when deduplicating (My favourite paper that help reason about this is: High-Availability Algorithms for Distributed Stream Processing)
Detecting failure and triggering failover
Detecting a failure is a more significant topic than I've the patience to go into, but fundamentally there are a few simple ways to detect a failure.
Is the process running? A monitoring process checks the process list periodically, if the application is not running, there's been a failure.
Is the process heartbeating? The process on a periodic interval will send a heartbeat, this may take the form of a message that's broadcast, a file that's touched or even a database table that is updated. The monitor waits for a sensible period longer than the heartbeat interval (to allow for natural variance in sending and receiving the heartbeat) before marking the process as failed.
Is the process in a good state? This is much easier to say than it is to do. A simple (read as naive) way to accomplish this is to allow the process to send a status message that contains it's own assessment of factors contributing to it's health. This could be anything from a flag to a list of logical status' of all subsystems. This does not allow you to detect deeper errors (i.e. a subsystem has failed but the process fails to determine this). Often the simplest way of determining if a process is in a good state is by humans inspecting the process and it's output such as logs.
Most of the systems I have worked on require the standby node(s) to observe the active, and upon detecting the failure makes a decision on how to failover then triggers it automatically. That said, this is not the only way to approach this. I see there being three main triggering mechanisms:
Humans. Don't under estimate the value in having a human inspect the environment, determine it's health then trigger a failover manually (uh, in this world where I'm (optimistically) imagining humans without their propensity to err).
Supervisor/Monitor processes. These processes are dedicated to watching the health of the HA group nodes. When it detects an unhealthy node the common approach is to kill it off and either start the secondary (hot/cold) or to send a signal to the secondary node to take over (hot/warm). In the hot/warm case this means that the process must have been developed with a component that can understand and act on the signal.
Automatic standby failover. Here the standby processes monitor the active and take over on failure. This typically combines the a detection and a triggering mechanism and requires your application to have been developed with an embedded component to support both. This isn't suitable for hot/cold policies since the standby isn't running.
Reasoning about the performance of your high availability mechanism
Non-functional requirements for HA are usually expressed as.
The system must not be unavailable for more than a
x
period in a
y
People often speak of a a percentage of time available, e.g. 99.9% (also known as three nines - wikipedia) but this boils down to the same expression. Occasionally the more liberal requirement is stated:
The system must recover from a failure within
z
These requirements relate to one another such that in any
period the number of failures a system can experience,
\epsilon
\epsilon=\frac{x}{z}
e.g. if you cannot be unavailable for more than 10 minutes in a 60 minute period and your system can recover within 2 minutes this means the total number of failures you can handle in any 60 minute interval is:
\epsilon=\frac{10}{2} = 5
The time it takes to failover,
z
can be further decomposed into the time it takes to detect a failure,
d
and the time it takes to failover,
f
If you're detecting failure using a process list then the interval you should refresh your list at is
d
. If you're using a heartbeat then your last heartbeat must have been received within the period
(this means your heartbeat interval itself will be
< d
, I'd recommend the interval should typically be
\frac{d}{2}
The time it takes to failover depends on the HA policy. If we decompose this into the time it takes to boot up,
b
and the time it takes to initialise the last known good state of the active,
i
. Then the time it takes to failover
z
z=d+b+i
Then the following summaries can be considered true
hot-active/cold-passive;
d>0
i>0
b>0
hot-active/warm-passive;
d>0
i>0
b=0
hot-active/warm-active;
d>0
b=i=0
hot-active/hot-active;
d=b=i=0
Note the hot-active/hot-active policy where I've made the assumption that responses to a request are always identical and the consumers can safely deduplicate them. If we consider the case where they may not be identical then a load balancer or queue is required in front of the HA group. With a load balancer
d\not=0
. In this case you will need to consider how long it takes for the load balancer to detect a failure and remove it from it's routes.
This could mean your value of
z
(the time it takes to failover) in hot-active/hot-active may be higher or similar to an hot-active/warm-active policy. In spite of this the hot-active/hot-active policy is very good when your system would otherwise be overloaded and thus become unavailable since the hot-active/warm-active would be insufficient to cope with this scenario.
In addition to knowing how long your system may be unavailable over a period you should consider how many simultaneous failures your system can tolerate, this capability should be stated:
The system remains highly available for
n
simultaneous failures
So far what I've spoke about assumes that it can cope with a single process failure. I have not considered more practical elements such as how the network or network topology affects the HA. Here for instance a using a single network switch can violate this requirement - if it blows then your entire system blows.
Nor have I discussed how to cope with failure in the monitoring processes or how to consider disaster recovery (where your entire server site is unavailable for one reason or another).
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Water | Boundless Biology | Course Hero
Water's polarity is responsible for many of its properties including its attractiveness to other molecules.
Describe the actions that occur due to water's polarity
One of water's important properties is that it is composed of polar molecules. The two hydrogen atoms and one oxygen atom within water molecules (H2O) form polar covalent bonds. While there is no net charge to a water molecule, the polarity of water creates a slightly positive charge on hydrogen and a slightly negative charge on oxygen, contributing to water's properties of attraction. Water's charges are generated because oxygen is more electronegative, or electron loving, than hydrogen. Thus, it is more likely that a shared electron would be found near the oxygen nucleus than the hydrogen nucleus. Since water is a nonlinear, or bent, molecule, the difference in electronegativities between the oxygen and hydrogen atoms generates the partial negative charge near the oxygen and partial positive charges near both hydrogens.
Nonpolar Molecules: Oil and water do not mix. As this macro image of oil and water shows, oil does not dissolve in water but forms droplets instead. This is due to it being a nonpolar compound.
As a result of water's polarity, each water molecule attracts other water molecules because of the opposite charges between them, forming hydrogen bonds. Water also attracts, or is attracted to, other polar molecules and ions, including many biomolecules, such as sugars, nucleic acids, and some amino acids. A polar substance that interacts readily with or dissolves in water is referred to as hydrophilic (hydro- = "water"; -philic = "loving"). In contrast, nonpolar molecules, such as oils and fats, do not interact well with water, as shown in. These molecules separate from it rather than dissolve in it, as we see in salad dressings containing oil and vinegar (an acidic water solution). These nonpolar compounds are called hydrophobic (hydro- = "water"; -phobic = "fearing").
Water's States: Gas, Liquid, and Solid
Water's High Heat Capacity
\displaystyle{\text{C}=\frac{\text{Q}}{\Delta \text{T}}}.
Evaporation of water requires a substantial amount of energy due to the high heat of vaporization of water.
Explain how heat of vaporization is related to the boiling point of water
The dissociation of liquid water molecules, which changes the substance to a gas, requires a lot of energy.
The boiling point of water is the temperature in which there is enough energy to break the hydrogen bonds between water molecules.
Evaporation of sweat (mostly water) removes heat from the surface of skin, cooling the body.
heat of vaporization: The energy required to transform a given quantity of a substance from a liquid into a gas at a given pressure (often atmospheric pressure).
Water's Heat of Vaporization
Water in its liquid form has an unusually high boiling point temperature, a value close to 100°C. As a result of the network of hydrogen bonding present between water molecules, a high input of energy is required to transform one gram of liquid water into water vapor, an energy requirement called the heat of vaporization. Water has a heat of vaporization value of 40.65 kJ/mol. A considerable amount of heat energy (586 calories) is required to accomplish this change in water. This process occurs on the surface of water. As liquid water heats up, hydrogen bonding makes it difficult to separate the water molecules from each other, which is required for it to enter its gaseous phase (steam). As a result, water acts as a heat sink, or heat reservoir, and requires much more heat to boil than does a liquid such as ethanol (grain alcohol), whose hydrogen bonding with other ethanol molecules is weaker than water's hydrogen bonding. Eventually, as water reaches its boiling point of 100° Celsius (212° Fahrenheit), the heat is able to break the hydrogen bonds between the water molecules, and the kinetic energy (motion) between the water molecules allows them to escape from the liquid as a gas. Even when below its boiling point, water's individual molecules acquire enough energy from each other such that some surface water molecules can escape and vaporize; this process is known as evaporation.
Humidity, Evaporation, and Boiling: (a) Because of the distribution of speeds and kinetic energies, some water molecules can break away to the vapor phase even at temperatures below the ordinary boiling point. (b) If the container is sealed, evaporation will continue until there is enough vapor density for the condensation rate to equal the evaporation rate. This vapor density and the partial pressure it creates are the saturation values. They increase with temperature and are independent of the presence of other gases, such as air. They depend only on the vapor pressure of water.
The fact that hydrogen bonds need to be broken for water to evaporate means that a substantial amount of energy is used in the process. As the water evaporates, energy is taken up by the process, cooling the environment where the evaporation is taking place. In many living organisms, including humans, the evaporation of sweat, which is 90 percent water, allows the organism to cool so that homeostasis of body temperature can be maintained.
Water's Cohesive and Adhesive Properties
Acids dissociate into H+ and lower pH, while bases dissociate into OH- and raise pH; buffers can absorb these excess ions to maintain pH.
Buffers are solutions that contain a weak acid and its a conjugate base; as such, they can absorb excess H+ ions or OH- ions, thereby maintaining an overall steady pH in the solution.
pH is equal to the negative logarithm of the concentration of H+ ions in solution: pH = - log[H+].
2\text{H}_2\text{O} \leftrightharpoons \text{H}_3\text{O}^+ + \text{OH}^-
The stronger the acid, the more readily it donates H+. For example, hydrochloric acid (HCl) is highly acidic and completely dissociates into hydrogen and chloride ions, whereas the acids in tomato juice or vinegar do not completely dissociate and are considered weak acids; conversely, strong bases readily donate OH– and/or react with hydrogen ions. Sodium hydroxide (NaOH) and many household cleaners are highly basic and give up OH– rapidly when placed in water; the OH- ions react with H+ in solution, creating new water molecules and lowering the amount of free H+ in the system, thereby raising the overall pH. An example of a weak basic solution is seawater, which has a pH near 8.0, close enough to neutral that well-adapted marine organisms thrive in this alkaline environment.
Structural Biochemistry/Unique Properties/Cohesive Behavior/Melting Point and Boiling Point. Provided by: Wikibooks. Located at: http://en.wikibooks.org/wiki/Structural_Biochemistry/Unique_Properties/Cohesive_Behavior/Melting_Point_and_Boiling_Point. License: CC BY-SA: Attribution-ShareAlike
heat of vaporization. Provided by: Wikipedia. Located at: http://en.wikipedia.org/wiki/heat%20of%20vaporization. License: CC BY-SA: Attribution-ShareAlike
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Carrying Capacity Calculator | Population Math
How does a population grow? – the logistic model
Definition of carrying capacity in biology
How do I calculate the carrying capacity?
Some practical examples of carrying capacity: rabbits and bacteria
Did humanity pass the carrying capacity of Earth?
Math underlines the dynamics of populations: learn one of the fundamental laws of ecology with our carrying capacity calculator.
Why can we find math in biology?
What are the laws of population growth?
How to calculate the carrying capacity from the logistic equation.
Some examples of carrying capacity.
Plus much more. Explore the point of contact between math and biology with Omni!
To enter this topic, we need to introduce a friend: Thomas Robert Malthus. At the end of the 18th century, he single-handedly revolutionized the field of mathematical biology by finding and modeling the correlation between population growth and exponential functions.
The connection formally made it possible to describe the growth of a population with a dynamical system.
🔎 A dynamical system, in science, is a set of equations describing the trajectory of a point in a specific space. Often, dynamical systems use relatively simple equations with an emerging complexity that can lead to the rise of chaotic phenomena.
In particular, the logistic equation governs the number of individuals in a population. Please take a look at it: it is surprisingly simple!
\frac{dN}{dt} = r\times N \times \left(1-\frac{N}{K}\right)
We could write a couple of books about this equation, but we will keep it short. Let's identify the elements of the logistic equation:
N
is the number of individuals at a given time;
r
is the rate of change per individual; and
K
is the carrying capacity.
The logistic equation is a differential equation: the left-hand side is a derivative, particularly the derivative of the number of individuals with respect to time. In other words, this derivative is the rate of increase of the population. If you think of the curve describing the population over time, the equation above is nothing but its slope.
🔎 Mathematical corner! The expression
\tfrac{dN}{dt}
is a derivative. We read it as "derivative of
N
t
", and is only a notation, where
dN
dt
are infinitely small intervals of the relative quantities.
In the graph below, you can appreciate the behavior of that function.
Derivative of the logistic function.
Note that the population starts and ends changing at an extremely low pace, while it has a maximum between the beginning and when it reaches the carrying capacity.
You can find it interesting to see how the population effectively changes with time: that's why we solved the differential equation for you (it is slightly tricky and out of scope):
N = \frac{1}{(1 + ((K - 1)\times e^{-r\times t}))}
We assumed that the initial population is
N_0 =1
, but this doesn't make much of a difference in this context. Now we can plot it!
A population grows following the logistic curve. The dashed red line indicates the carrying capacity.
As you can see, the population (on the vertical axis) starts growing slowly, then accelerates: it makes sense since there are more individuals capable of reproducing.
However, the curve slows down, and reaches a plateau: the population met the carrying capacity. We will talk in detail about the carrying capacity in the next section. For now, we need to focus for a moment on
r
, the rate of change per individual. The value of
r
defines the variation of the population for each individual. If there is one birth and one death for each individual, then
r=0
. If births outnumber deaths, the population grows.
The carrying capacity definition is the maximum size of a population sustainable by a specific environment. When a population reaches the carrying capacity, the net growth rate is
0
: the number of births equals the number of deaths (and the other factors affecting the number of individuals balance each other).
The population plateaus because the environment can't support more than that number. A temporary increase would necessarily result in a decrease that would bring the number back toward the carrying capacity: we say that the population is controlled by a limiting factor.
The real world is a world of limitations. Space, food, resources: you name it; they all contribute to keeping the populations down to manageable numbers, and with a reason: the characteristic times of populations growth are centuries if not more, generally outliving the lifespan of a single individual. These extended times allow for careful tailoring of the populations. Only catastrophic events tilt the balance.
🔎 Why do we call the carrying capacity
K
? It comes from the German Kapazitätsgrenze, which means "capacity limit".
A combination of the growth rate per individual and the carrying capacity defines the story of a population, from its explosive growth to the eventual plateau and stabilization. What if other factors come and influence the growth of a population? A pandemic, or the discovery of new fertile grounds, are a couple of examples? At that point, both
r
K
To calculate the carrying capacity, we start from the differential form of the logistic equation.
\frac{dN}{dt} = r\times N \times \left(1-\frac{N}{K}\right)
Take the derivative and substitute it with its value at the desired point in the population growth: we can call it
C_p
. Now we can invert the equation to find the definition of carrying capacity:
K= \frac{N}{1-\frac{C_p}{r\times N}}
K
acts as the attractor of the dynamical system. Every trajectory (the story of a population given a set of initial conditions) converges on the carrying capacity.
Rabbits! Bunnies are the perfect example for studies in population dynamics since they reproduce... like rabbits 😁
Rabbits were a species unknown in Australia until Thomas Austin introduced a few pairs (12) in 1859 because rabbits reminded him of England. Australia didn't have a natural predator with a specialization in cottontails, which literally recreated the "it's free real estate" meme. Rabbit Inc. was in for serious business, and in six years, the population reached
22
million individuals.
You can pinpoint this tumultuous growth stage in the first part of the population curve: the exponential growth. What happened after?
We know the parameters of this frantic growth:
r=2.3
: The next generation had almost two and a half new rabbits for each rabbit. For comparison, the US population had
r=0.4
6
years in, the population was
22,000,000
, and the rate of change of the population
C_p
C_p=49.12\times10^6
rabbits per year.
Can we calculate the carrying capacity of Australia for an uncontrolled rabbit population?
\begin{align*} K &= \frac{N}{1-\frac{C_p}{r\times N}}\\\\ &= \frac{2.2\times10^7}{1-\frac{49.12\times10^6}{2.3\times 2.2\times10^7}}\\\\ &=7.52\times10^8 \end{align*}
The result is in accordance with the outcome of the rabbit baby boom. In 1930, their population reached an estimated
750
million individuals, a number at which it stopped growing: it touched the carrying capacity.
The story ends in tragedy, though, at least for the rabbits. After this explosive growth, the Australian government decided to introduce a deadly virus for rabbits, myxomatosis. After the infection, in 1950, the population quickly decreased by over
90\%
. Moral of the story? Don't be like Thomas Austin and keep your country's animals in another country!
Let's take another example. Imagine a bacteria population (the good old E. coli) growing in a welcoming Petri dish. The intrinsic growth rate is
r=2.0
(perfect mitosis and no deaths). Considering that at the time we are measuring the population, we find that:
The population is
\log{N}=4.75\ \text{CFU/ml}
It is growing with a rate of change of
C_p= 5.05\ \text{CFU/ml}
What's a CFU? CFUs are a widely used unit in laboratories worldwide, the colony-forming unit. It only includes the viable cells, which we consider in a population growth model.
What is the carrying capacity of our Petri dish? Input these values in our calculator (we purposefully left out the measurement units): you will find out that the carrying capacity is
\log{K}=4.75\ \text{CFU/ml}
. Once our bacteria reach that point, the nutrients and the space in the dish would not allow for growth anymore, but for a certain period of sustained population. Without intervention, a decrease would follow.
If you are interested in the math behind bacterial growth, try our doubling time calculator and our generation time calculator!
Science is at the basis of human societies, whether we know it or not. Take, for example, the Haber process, which is crucial to fertilizer production. Its development allowed for the population boom of the 20th century: to reach one billion, humanity took tens of thousands of years; it took barely more than a century to reach two billion. From 1900 to 2000, however, the population reached six billion, and it is still growing.
The Haber process is often remembered as the most important discovery of the 20th century. However, shadows cover the ability to turn air into bread: its discoverer also contributed to the development of chemical warfare. Germany's increased agricultural productivity is thought to be a contributing factor to the events of both world wars.
Did humanity reach the Earth's carrying capacity? We don't know. Estimates vary from a mere seven billion (which we passed) to a reasonable 9-10 billion. In any case, there will be a limit to humanity's growth, and it's nearing. Reaching that limit at the speed our species is expanding today would have dramatic consequences, especially dire famines and conflicts (a Malthusian catastrophe or trap: you can read more on this page). We can hope the growth will slow before this, reaching that carrying capacity plateau slowly.
What is the carrying capacity in biology?
The carrying capacity is the maximum number of individuals sustainable by a specific environment. This quantity corresponds to a plateau in the population reached after a period of growth or shrinkage. In the logistic model, only a few factors affect the carrying capacity:
The intrinsic growth rate r; and
The rate of change of the population at a given time.
What is the carrying capacity of Earth for a human population?
Between 7 and 11 billion. Scientists don't know precisely the limit to humanity's growth, but they agree that it is close to the current number of individuals hosted on the Planet. If we've already passed Earth's carrying capacity, we will face a phase of rapid shrinking of the population; otherwise, we will meet a plateau (expected for the second half of the century).
To calculate the carrying capacity of a population, we can start from the differential form of the logistic equation:
dN/dt = r × N × (1 - N/K)
r – Intrinsic rate of change;
dN/dt – Change in the number of individuals;
N – Number of individuals; and
By defining the rate of change when the population is N as Cₚ, we can compute the carrying capacity as:
K = N/(1 - (Cₚ/r × N))
What is the carrying capacity of a population of 100 individuals growing at a rate of 25 individuals per year, and with an intrinsic growth rate r=0.27?
1,350 individuals. To calculate the carrying capacity, first identify the given data:
N = 100; and
Cₚ = 25.
This combination identifies a particular moment of the growth of a population. The rate of change is rather high, meaning that we are far from the carrying capacity. Let's use the formula to calculate K:
K = N/(1 - (Cₚ/r × N)) = 100 × 1/(1 - (25/(0.27 × 100)) = 1,350.
What happens if a population passes the carrying capacity?
Since the carrying capacity is a property of the environment and the population together, if the latter passes this critical value, the former will react accordingly. A scenario where a population exceeds the carrying capacity can bring:
More damaging spread of infectious disease;
Degradation of the environment;
Reduced space and territory; and
Unbalance in the population of other species.
If left untouched, nature is usually good at finding a balance and preventing that scenario.
Change in population (Cₚ)
Intrinsic rate of change (r)
Dog age calculator finds the human age equivalent of your dog's age.
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What is the sampling distribution of the sample proportion?
How to use this sampling distribution of the proportion calculator: an example
How to find the sample proportion?
How to find the standard deviation of sample proportion?
This sampling distribution of the sample proportion calculator finds the probability that your sample proportion lies within a specific range: P(p₁ < p̂ < p₂), P(p₁ > p̂), or P(p₁ < p̂). Using the sample size (n), population proportion (p), and the specified proportions range, it calculates the probability.
Population proportion is vital in many scientific fields, as many research questions of interest involve this parameter. If you're interested in the opposite problem: finding a range of possible values of the population proportion given a confidence level, check our sampling error calculator.
What is the distribution of the sample proportion?
How to find the sample proportion.
How to use our calculator for the sampling distribution of the sample proportion.
To know what is the distribution of the sample proportion, we first have to talk about proportions. Proportions are one of the more predominant parameters in research. In plain words, they indicate the percentage of individuals with a specific characteristic. Examples are:
The proportion of citizens who approve of how the president governs his country.
Testing whether introducing physical activity diminishes cardiovascular disease incidence.
The proportion of TVs of a particular model who fail after one year.
This parameter is straightforward to calculate. Divide the number of individuals who possess the characteristic by the total number of individuals, and you'll get it. As in the case of the mean, we usually use sample proportions to estimate population proportions.
If we took samples of and plotted each of the sample proportions encountered, we would end up with a sampling distribution, similar to what happens with the sampling distribution of the mean. The resulting distribution is called the sampling distribution of the sample proportion and is a graphical representation of the possible values of the population proportion.
Nevertheless, there are fundamental differences with the sampling distribution of the mean. The sample proportion is a discrete variable and not a continuous variable. Therefore, the sample proportion values are infinite, unlike the sample mean case.
The sampling distribution of the sample proportion doesn't follow a normal distribution but a binomial distribution, which depends on the population proportion and the sample size. The good part is that, in most cases, we can approximate that discrete binomial distribution as a continuous normal distribution and use the widely known methods to find probabilities.
The approximation results in a normal distribution with mean
µ_{\hat p}
σ_p̂
, expressed in terms of the sample size
n
and the population proportion
p
\footnotesize µ_{\hat p} = p
\footnotesize σ_{\hat p} = \sqrt{\frac{p(1-p)}{n}}
We use those values to find the z-scores and finally find the probability values, as we would usually do with a population mean:
\footnotesize Z = \frac{p₀-µ_{\hat p}}{σ_p̂} = \frac{p₀-p}{σ_{\hat p}}
🙋 In which cases does this approximation apply?
This approximation is inadequate for small sample sizes and p values near 0 or 1. The following is a mathematical criterion for determining if the approximation holds well:
np(p-1) ≥ 15
The above equations translate into the fact that there have to be at least 15 successes and 15 failures for the approximation to be reasonably good.
The following image shows how proportions far from 0 and 1 and larger sample sizes approximate a normal distribution.
Probability mass function of three different binomial distributions. Attribution: Tayste, Public domain, via Wikimedia Commons.
Let's look at an example of using these formulas and our sampling distribution of the sample proportion calculator.
Suppose you know that 70% percent of the American population approves the management of the current president. If you conduct a pool and take a random sample of 500 people, what is the probability of finding at least 325 who support him?
Determine the proportion values.
p₁ = 320/500 = 60% = 0.65
Input the
p
n
values in the calculator.
As we're interested in the values greater than p₁, select that you want "P(p̂ > p₁)".
Input p₁ = 0.65.
That's it. The answer is 0.992651, so the probability of finding at least this proportion of people who approve him is about 99.3%. In statistical notation: P(p̂ > 0.65) = 99.3%.
You can also check the results using the formulas shown before:
\footnotesize \begin{align*} σ_{\hat p} &= \sqrt{\frac{p(1-p)}{n}}\\\\ &= \sqrt{\frac{0.7(1-0.7)}{500}} = 0.0204939 \end{align*}
\footnotesize \begin{align*} Z &= \frac{p₁-p}{σ_{\hat p}}\\\\ &= \frac{0.65-0.7}{0.0204939} = -2.43975 \end{align*}
\footnotesize \begin{align*} P (p̂ > 0.65) &= P ( Z > −2.43975)\\ &= 0.992651 \end{align*}
P ( Z > −2.43975)
is not straightforward to calculate. To do it, you need a table to convert from z-score to P values, or you can use our p-value calculator (better and easier).
💡 If you want to calculate the probability of getting a
\hat p
between two values, you have to select the option "P(p₁ < p̂ < p₂)" in the "What probability do you want" box. In the previous example, if you wanted to calculate the probability of finding a proportion of approvers between 0.65 and 0.75, your calculator should look this way:
Do you have doubts about the calculations of the sampling distribution of the sample proportion? Maybe the following FAQs can help you.
Follow these steps to find the sample proportion:
Determine the number of successes in your sample.
Divide the number of successes by the sample size. This result represents the fraction or percentage of successes in your sample.
The process to find the standard deviation of the sample proportion depends on the available information:
If you know the population proportion (p) and the sample size (n), input those values in the sample proportion standard deviation formula: √[p (p - 1)/n].
If you only know the sample proportion (p̂ ) and the sample size (n), input those values in the standard error of the sample proportion formula: √[p̂ (p̂ - 1)/n], which is an estimate of the sample proportion standard deviation.
What is the difference between population proportion and sample proportion?
The population proportion (p) is the number of successes (X) of the entire population divided by the population size (N): p = X/N.
The sample proportion (p̂ ) is the number of successes found in the sample (x) divided by the sample size (n): p̂ = x/n.
What is the probability of getting a sample proportion higher than the population proportion?
The probability of getting a sample proportion higher than the population proportion is 50%, as long as the sampling distribution of the proportion is symmetrical. The larger the sample size or the farthest the proportion value is from 0 or 1, the more symmetric the distribution.
P(p₁ < p̂ < p₂)
Z-score of p₁
Z-score of p₂
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Modeling Low-Pressure Fluid Transportation Systems - MATLAB & Simulink
How Fluid Transportation Systems Differ from Power and Control Systems
Available Blocks and How to Use Them
Low-Pressure Fluid Transportation System
In hydraulics, the steady uniform flow in a component with one entrance and one exit is characterized by the following energy equation
-\frac{{\stackrel{˙}{W}}_{s}}{\stackrel{˙}{m}g}=\frac{{V}_{2}^{2}-{V}_{1}^{2}}{2g}+\frac{{p}_{2}-{p}_{1}}{\rho g}+{z}_{2}-{z}_{1}+{h}_{L}
{\stackrel{˙}{W}}_{s}
Work rate performed by fluid
\stackrel{˙}{m}
V2 Fluid velocity at the exit
V1 Fluid velocity at the entrance
p1, p2 Static pressure at the entrance and the exit, respectively
z1, z2 Elevation above a reference plane (datum) at the entrance and the exit, respectively
hL Hydraulic loss
Subscripts 1 and 2 refer to the entrance and exit, respectively. All the terms in Equation 1 have dimensions of height and are named kinematic head, piezometric head, geometric head, and loss head, respectively. For a variety of reasons, analysis of hydraulic power and control systems is performed with respect to pressures, rather than to heads, and Equation 1 for a typical passive component is presented in the form
\frac{\rho }{2}{V}_{1}^{2}+{p}_{1}+\rho g{z}_{1}=\frac{\rho }{2}{V}_{2}^{2}+{p}_{2}+\rho g{z}_{2}+{p}_{L}
V1, p1, z1 Velocity, static pressure, and elevation at the entrance, respectively
V2, p2, z2 Velocity, static pressure, and elevation at the exit, respectively
pL Pressure loss
\frac{\rho }{2}{V}^{2}
is frequently referred to as kinematic, or dynamic, pressure, and
\rho gz
as piezometric pressure. Dynamic pressure terms are usually neglected because they are very small, and Equation 2 takes the form
{p}_{1}+\rho g{z}_{1}={p}_{2}+\rho g{z}_{2}+{p}_{L}
The size of a typical power and control system is usually small and rarely exceeds 1.5 – 2 m. To add to this, these systems operate at pressures in the range 50 – 300 bar. Therefore,
\rho gz
terms are negligibly small compared to static pressures. As a result, Simscape™ Fluids™ components (with the exception of the ones designed specifically for low-pressure simulation, described in Available Blocks and How to Use Them) have been developed with respect to static pressures, with the following equations
\begin{array}{l}p={p}_{L}=f\left(q\right)\\ q=f\left({p}_{1},{p}_{2}\right)\end{array}
p Pressure difference between component ports
q Flow rate through the component
Fluid transportation systems usually operate at low pressures (about 2-4 bar), and the difference in component elevation with respect to reference plane can be very large. Therefore, geometrical head becomes an essential part of the energy balance and must be accounted for. In other words, the low-pressure fluid transportation systems must be simulated with respect to piezometric pressures
{p}_{pz}=p+\rho gz
, rather than static pressures. This requirement is reflected in the component equations
\begin{array}{l}p={p}_{L}=f\left(q,{z}_{1},{z}_{2}\right)\\ q=f\left({p}_{1},{p}_{2},{z}_{1},{z}_{2}\right)\end{array}
Equations in the form Equation 5 must be applied to describe a hydraulic component with significant difference between port elevations. In hydraulic systems, there is only one type of such components: hydraulic pipes. The models of pipes intended to be used in low pressure systems must account for difference in elevation of their ports. The dimensions of the rest of the components are too small to contribute noticeably to energy balance, and their models can be built with the constant elevation assumption, like all the other Simscape Fluids blocks. To sum it up:
You can build models of low-pressure systems with difference in elevations of their components using regular Simscape Fluids blocks, with the exception of the pipes. Use low-pressure pipes, described in Available Blocks and How to Use Them .
When modeling low-pressure systems, you must use low-pressure pipe blocks to connect all nodes with difference in elevation, because these are the only blocks that provide information about the vertical locations of the system parts. Nodes connected with any other blocks, such as valves, orifices, actuators, and so on, will be treated as if they have the same elevation.
When modeling low-pressure hydraulic systems, use the pipe blocks from the Low-Pressure Blocks library instead of the regular pipe blocks. These blocks account for the port elevation above reference plane and differ in the extent of idealization, just like their high-pressure counterparts:
Resistive Pipe LP — Models hydraulic pipe with circular and noncircular cross sections and accounts for friction loss only, similar to the Resistive Tube block, available in the Simscape Foundation library.
Resistive Pipe LP with Variable Elevation — Models hydraulic pipe with circular and noncircular cross sections and accounts for friction losses and variable port elevations. Use this block for low-pressure system simulation in which the pipe ends change their positions with respect to the reference plane.
Hydraulic Pipe LP — Models hydraulic pipe with circular and noncircular cross sections and accounts for friction loss along the pipe length and for fluid compressibility, similar to the Hydraulic Pipeline block in the Pipelines library.
Hydraulic Pipe LP with Variable Elevation — Models hydraulic pipe with circular and noncircular cross sections and accounts for friction loss along the pipe length and for fluid compressibility, as well as variable port elevations. Use this block for low-pressure system simulation in which the pipe ends change their positions with respect to the reference plane.
Segmented Pipe LP — Models circular hydraulic pipe and accounts for friction loss, fluid compressibility, and fluid inertia, similar to the Segmented Pipe block in the Pipelines library.
Use these low-pressure pipe blocks to connect all Hydraulic nodes in your model with difference in elevation, because these are the only blocks that provide information about the vertical location of the ports. Nodes connected with any other blocks, such as valves, orifices, actuators, and so on, will be treated as if they have the same elevation.
The additional models of pressurized tanks available for low-pressure system simulation include:
Constant Head Tank — Represents a pressurized hydraulic reservoir, in which fluid is stored under a specified pressure. The size of the tank is assumed to be large enough to neglect the pressurization and fluid level change due to fluid volume. The block accounts for the fluid level elevation with respect to the tank bottom, as well as for pressure loss in the connecting pipe that can be caused by a filter, fittings, or some other local resistance. The loss is specified with the pressure loss coefficient. The block computes the volume of fluid in the tank and exports it outside through the physical signal port V.
Variable Head Tank — Represents a pressurized hydraulic reservoir, in which fluid is stored under a specified pressure. The pressurization remains constant regardless of volume change. The block accounts for the fluid level change caused by the volume variation, as well as for pressure loss in the connecting pipe that can be caused by a filter, fittings, or some other local resistance. The loss is specified with the pressure loss coefficient. The block computes the volume of fluid in the tank and exports it outside through the physical signal port V.
Variable Head Two-Arm Tank — Represents a two-arm pressurized tank, in which fluid is stored under a specified pressure. The pressurization remains constant regardless of volume change. The block accounts for the fluid level change caused by the volume variation, as well as for pressure loss in the connecting pipes that can be caused by a filter, fittings, or some other local resistance. The loss is specified with the pressure loss coefficient at each outlet. The block computes the volume of fluid in the tank and exports it outside through the physical signal port V.
Variable Head Three-Arm Tank — Represents a three-arm pressurized tank, in which fluid is stored under a specified pressure. The pressurization remains constant regardless of volume change. The block accounts for the fluid level change caused by the volume variation, as well as for pressure loss in the connecting pipes that can be caused by a filter, fittings, or some other local resistance. The loss is specified with the pressure loss coefficient at each outlet. The block computes the volume of fluid in the tank and exports it outside through the physical signal port V.
The following illustration shows a simple system consisting of three tanks whose bottom surfaces are located at heights H1, H2, and H3, respectively, from the reference plane. The tanks are connected by pipes to a hydraulic manifold, which may contain any hydraulic elements, such as valves, orifices, pumps, accumulators, other pipes, and so on, but these elements have one feature in common – their elevations are all the same and equal to H4.
\begin{array}{ll}{p}_{i}=\rho g{F}_{i}\hfill & \text{for }i=1,2,3\hfill \end{array}
The components inside the manifold can be simulated with regular Simscape Fluids blocks, like you would use for hydraulic power and control systems simulation. The pipes must be simulated with one of the low-pressure pipe models: Resistive Pipe LP, Hydraulic Pipe LP, or Segmented Pipe LP, depending on the required extent of idealization. Use the Constant Head Tank or Variable Head Tank blocks to simulate the tanks. For details of implementation, see the Water Supply System and the Three Constant Head Tanks examples.
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Fumseck, the Solarized WordPress theme — Guillaume Paumier
Roles: Designer, Developer
Topics: Latin Modern, Solarized, WordPress
I've never been entirely satisfied with any of the themes that I've tested and used over the years. At some point, I realized that the only way I was going to be happy with both the design and the features was by coding the theme myself. The problem, of course, was that I had never coded in PHP (the scripting language used by WordPress), and that my knowledge of CSS was very basic, although they were both fields I wanted to dive into eventually.
In Fall 2013, I decided it was time, and got to work. I did some research and read many web articles I had saved in preparation for this project, notably on web technologies, responsiveness and the latest good practices. In the end, there wasn't much that couldn't be understood with a general knowledge of programming and a good look at the reference documentation.
“Fumseck” is the name of Professor Dumbledore's phoenix Fawkes in the French version of the Harry Potter book series. Choosing this name was a tongue-in-cheek reference to the fact that this site had been moribund for a while and that I wanted it to "rise from its ashes".
Fumseck is used in conjunction with "Batbelt", a plugin for WordPress that I've written to bundle together customizations I've made to this site that aren't directly related to its appearance. For example, it's the plugin that implements the Project custom content type (used on this page) and custom taxonomies such as Locations. The Fumseck theme is then in charge of defining the appearance of those customizations, in addition to the rest of the site.
An identity in colors and typefaces
I fell in love with
\LaTeX
, the document typesetting system, when I started using it for reports and theses in college. I was immediately attracted by its elegance and simplicity. I also found the idea of compiling a document really cool.
I wanted this site to reflect who I was, and when I found out that Computer Modern, the family of typefaces designed by Donald Knuth for
\TeX
, had been ported to other font formats by the GUST e-foundry (under the name "Latin Modern"), I knew those were the Chosen Ones.
I also loved the fact that I could use different typefaces from the same family, meaning I could, for example, mix a Serif and a Sans serif typeface that would nonetheless be consistent with each other. Additionally, Latin Modern fonts come with true variations like bold and italics. And have I mentioned how beautiful they look?
If you're using a reasonably modern browser, you're seeing those typefaces right now, loaded as web fonts if you don't already have them installed on your computer. I've used a large base font size because I dislike having to zoom in on so many web pages, even if I have excellent vision.
The color scheme was the other main choice to make regarding the overall appearance of the site. There wasn't much to think about there, though. I knew from the start that I wanted to use Ethan Schoonover's Solarized palette, which I've been using in a variety of documents since I discovered it in 2011.
Solarized is an awesome color scheme with a lot of cool features, notably a soft contrast and the ability to switch between a light and a dark background with minimal changes. I have plans to create an alternate style sheet based on the dark version of the color scheme, and perhaps a high-contrast version as well: I like the soft tones of Solarized, but the contrast may be insufficient for some people.
Last, since I post content in both English and French, I wanted the browsing experience to be as enjoyable in either language. Not all the content is available in both languages, but I wanted the interface to be fully localized. The theme is therefore written in English, but fully translated and localized into French.
Discovering and experimenting with SASS
One of the first things I did for this project was to learn how to use Sass, the CSS extension language, and its associated framework, Compass. I read the book Sass and Compass for Designers by Ben Frain in two days, and started writing Sass files right away. I loved the concept of a CSS preprocessor, because of the modularity and reusability it provided.
Then, I moved on to coding the PHP templates for the different types of pages. Creating the responsive grids and layout was made easy with the Susy framework.
The layout for image posts was a bit challenging, because I wanted to take advantage of the room available on wide viewports while keeping a logical information hierarchy on smaller screens. I also wanted to extract and display some photo-specific information, like photograph metadata (f-number, exposure, etc.).
The layout for image posts was also challenging in that it needed to accommodate both landscape- and portrait-oriented pictures. I experimented a bit with responsive images and played around with <picture> and src-N, until it appeared that there was still no foreseeable consensus regarding their implementation in web browsers.
Bootstrap and the path to LESS CSS
I had completed much of the PHP templates and a reasonable amount of styling when I started working on the header and menus. I thought at first that I'd use a couple of JavaScript libraries to handle the interactive behavior, but I finally settled on moving to Bootstrap because it provided most of what I needed. I was already using the icons from FontAwesome, so it made sense to use tools from the same family.
Bootstrap is written in LESS CSS, another CSS extension language. I didn't want to maintain CSS in two different languages, so I converted all my SASS files to LESS. I didn't know the LESS syntax at first, but it turned out to be very similar to that of SASS, so switching to it was relatively painless.
I converted the Susy-based grids to Bootstrap's layout system, at first by adding the relevant CSS classes directly to the HTML. This solution felt messy, though, and after a little investigation, I managed to move the Bootstrap grid elements to the CSS by importing the definitions and compiling the stylesheet.
More PHP templates
Once that conversion was done, I finished the header and continued to add the remaining PHP templates, like archive pages and custom page templates. I did a lot of fixing and tweaking, added navigation links, menus, and a custom language switcher for WPML, the WordPress plugin I use to provide a proper multilingual site. I also looked into theme localization, .mo and .po files, and translated and localized the theme's interface.
I usually use a GPS logger when taking pictures, so that I can then correlate the files and geolocalize the pictures by embedding the GPS geographic coordinates into the photo files. To reuse that data automatically, I implemented a feature in the theme that extracts those coordinates, converts them to a human-readable format and displays them on the photo's page; there is also a link to the location on OpenStreetMap based on another coordinates format.
After some more tweaking and a lot of testing, I eventually enabled Fumseck on this website in mid-December 2013, and it's been working great since then.
Fumseck was my first WordPress theme written from scratch; I had barely done any PHP coding before, and my CSS skills were very limited. There are still plenty of small things I want to fix or improve (such as a better interface for the comments, instead of the one provided by WordPress) but I'm really happy with the result.
I feel this theme is a genuine reflection of my personality and my tastes, and I really enjoy publishing content and navigating the site. I hope you'll find the browsing and reading experience as enjoyable as I do.
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How to calculate ICC T20 rankings — points gained per match
Example: Using the ICC calculator for T20 points calculation
The ICC calculator determines where your cricket team stands in the International Cricket Council (ICC) T20 team rankings after their current win or loss. The points calculation for ICC T20 is based on the team's current standings before the match and who they are playing. The rankings are calculated over 12 months, and as the games get older, their weighting reduces over time.
The rankings are crucial for a national team to secure a spot in the World Cup main event or the qualifiers. The top 8 teams on the cut-off date set by the ICC qualify directly for the main event, whereas the next set of 10 or 12 teams plays against each other for a few qualifying spots. This calculator focuses on one match per time, and it will tell you how many points a team will gain for the match outcome.
Depending on those points, the teams move up or down the rankings ladder. With the next ICC World T20 cup scheduled for this year (2022), the rankings are crucial to making the qualifiers and super 12s. You can start by entering some numbers or continue reading to understand how to calculate ICC T20 rankings.
The rules to calculate ICC T20 rankings depend mainly on the team's current standings. The winning margin of runs or wickets does not affect whatsoever the points gained by the team for the win. First, the data for wins and losses is collected over a duration, mostly 12 months. Then, the points accumulated for each win or loss are calculated. The total sum of points is divided by the number of matches played to obtain the rating value used to rank the teams. As of 2022, there are 91 teams in the ranking table. The points gained for each match depend on the following factors:
Pre-match ratings of both sides;
Difference between the pre-match ratings; and
Standing of winning team.
Say two teams, A and B, are playing against each other; firstly, the rating before the match is recorded. The rating is compared to determine the difference between the ratings. Here, the weaker team is one with a lesser rating and vice versa. The points are as follows:
Opponent's rating - 50
Weaker team
To calculate the points per match:
Enter the rating for Team 1.
Fill in the rating for Team 2.
Select the result of the match from the list.
Based on the match result, the ICC calculator will return the points gained by each team.
Find the number of points gained by England and Australia playing against each other in a T20i match. Take the team ratings for England and Australia as 248 and 275, respectively.
Enter the rating for Team 1, England, as 248.
Fill in the rating for Team 2, Australia, as 275.
Select the result of the match from the list as Team 1 wins.
As per the ICC T20 points calculation:
\qquad \scriptsize \begin{align*} \text{Team 1 gains} &= 275 + 50 = 325\\ \text{Team 2 gains} &= 248 - 50 = 198 \end{align*}
You can also calculate the team ratings before and after the match using the advanced mode.
What are the factors affecting ICC T20 team rankings?
The factors that affect ICC T20 team rankings are:
Rating of both teams involved in the match;
Difference between the ratings of the two teams; and
The victor side.
If the team with a higher rating loses the match, the points gained by their opponent will be more. This system rewards the weaker teams when they win against the stronger side.
How do I calculate ratings for winning a T20 match?
You can calculate the rating gained or lost per match by:
Find the number of points gained.
Add the points to the current tally.
Divide the points by the number of matches played.
If the team loses the match, the points gained will be less, therefore reducing the average or team rating and vice versa.
Which team is the number 1 in men's ICC T20 rankings?
As of February 2022, England leads the men's ICC T20 rankings having 9354 points from 34 matches resulting in a team rating of 275. Cricket teams from India and Pakistan hold the following positions with 9627 and 12207 points with 267 and 265 ratings, respectively.
Which team is the number 1 in women's ICC T20 rankings?
As of February 2022, Australia leads the women's ICC T20 rankings having 5824 points from 20 matches resulting in a team rating of 291. Cricket teams from England and India hold the following positions with 7157 and 6081 points with 286 and 264 ratings, respectively.
This brand new ACFT calculator determines whether you're ready to take the new Army Combat Fitness Test and get into your dream unit.
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What is normal approximation to binomial distribution?
How to calculate normal approximation — normal approximation calculator with steps
How do I calculate normal approximation to binomial distribution?
Normal approximation to the binomial — Example #1
The normal approximation calculator (more precisely, normal approximation to binomial distribution calculator) helps you to perform normal approximation for a binomial distribution.
The normal approximation of binomial distribution is a process where we apply the normal distribution curve to estimate the shape of the binomial distribution. The fundamental basis of the normal approximation method is that the distribution of the outcome of many experiments is at least approximately normally distributed. If you are not familiar with that typically bell-shaped curve, check our normal distribution calculator.
You need to set the following variables to run the normal approximation to binomial calculator. Before using the tool, however, you may want to refresh your knowledge of the concept of probability with our probability calculator.
First, tell the normal approximation calculator about the probabilistic problem.
Number of occurrences or trials (
N
Probability of success (
p
) or the probability of failure (
q = 1-p
Number of successes (
n
Select the probability you would like to approximate at the event restatement.
P(x = n)
— the probability for an exact discrete value of
n
P(x > n)
— the probability for events corresponding to a value greater than
P(x ≤ n)
— the probability for events occurring at most
n
P(x < n)
— the probability for events corresponding to a value lesser than
P(x ≥ n)
— the probability for events occurring at least
n
After specifying the problem, you can immediately read both the final and partial results.
μ
The variance (
σ^2
The standard deviation (
σ
The continuity correction;
The Z-score;
The Z-value; and
The approximated probability.
If you want to compute the normal approximation to binomial distribution by hand, follow the below steps.
Find the sample size (the number of occurrences or trials,
N
) and the probabilities
p
q
— which can be the probability of success (
p
) and probability of failure (
q = 1 - p
), for example.
Check if you can apply the normal approximation to the binomial. If
N \times p
N \times q
are both larger than
5
, then you can use the approximation without worry.
Find the mean (
μ
) by multiplying
n
p
μ = N \times p
Compute the variance (
σ^2
N
p
q
σ^2 = N \times p \times q
Determine the standard deviation (
\text{SD}
σ
) by taking the square root of the variance:
\sqrt{N \times p \times q}
State the problem (the number of successes,
n
) using the continuity correction factor according to the below table. This is necessary because the normal distribution is a continuous probability distribution, and the binomial distribution is a discrete probability distribution).
n-0.5 < x < n+0.5
x ≤ n
x < n + 0.5
x < n − 0.5
x ≥ n
x > n − 0.5
x > n + 0.5
Find the Z-score with
(x - μ) / σ
Check the Z-value in the Z-table.
Determine the probability associated with the Z-values according to the below table, which will be the normal approximation of binomial distribution.
difference of Z-values for n+0.5 and n-0.5
1 − Z-value
Assume you have a fair coin and want to know the probability that you would get 40 heads after tossing the coin 100 times.
Gather information from the above problem.
N = 100
(number of occurrences or trials);
n = 40
(number of successes); and
p = 0.5
(probability of success on a given trial).
Verify that the sample size is large enough to use the normal approximation.
N \times p = 50 ≥ 5
N \times (1-p) = 50 ≥ 5
State the problem using the continuity correction factor.
x ≤ 40 \therefore P(x < 40.5)
μ
σ
) of the binomial distribution.
μ = N \times p = 100 \times 0.5 = 50
σ² = N \times p \times (1-p) = 100 \times 0.5 \times (1-0.5) = 25
σ = \sqrt{25} = 5
Find the Z-score using the mean and standard deviation.
z = (x - μ) / σ = (40.5 - 50) / 5 = -9.5 / 5 = -1.9
Find the Z-value and determine the probability.
z = 0.4713
Thus, the probability that a coin lands on heads less than or equal to 40 times during 100 flips is
0.0287
2.8717\%
Assume you have reliable data stating that 60% of the working people commute by public transport to work in a given city. If a random sample of size 30 is selected (all working persons), what is the probability that precisely 10 persons will travel from those by public transport?
Gather information from the above statement.
N = 30
n = 10
p = 0.6
N \times p = 18 ≥ 5
N \times (1-p) = 12 ≥ 5
As these numbers are nice and large, we're good to go!
x = 10 \therefore P(9.5 < x < 10.5)
\mu
\sigma
μ = N \times p = 30 \times 0.6 = 18
σ^2 = N \times p \times (1-p) = 30 \times 0.6 \times (1-0.6) = 7.2
σ = \sqrt{7.2} = 2.6833
Find the two z-scores using the mean and standard deviation.
z_1 = (x - μ) / σ = (9.5 - 18) / 2.68 = −8.5 / 2.68 = -3.168
z_2 = (x - μ) / σ = (10.5 - 18) / 2.68 = −7.5 / 2.68 = -2.795
P = z_2 - z_1 = 0.0026 - 0.0008 = 0.0018
Thus, the probability that precisely 10 people travel by public transport out of the 30 randomly chosen people is
0.0018
0.18\%
Can I use normal approximation if the product of the trials and the probability of the event is less than five?
No. The number of trials (or occurrences, N) relative to its probabilities (p and 1−p) must be sufficiently large (N×p ≥ 5 and N×(1−p) ≥ 5) to apply the normal distribution in order to approximate the probabilities related to the binomial distribution.
The normal approximation to the binomial distribution is a process by which we approximate the probabilities related to the binomial distribution.
What is the z-value of 60.5 occurrences when the mean is 50 and standard deviation is 5?
The z-value is 2.3 for the event of 60.5 (x = 60.5) occurrences with the mean of 50 (μ = 50) and standard deviation of 5 (σ = 5). The computation takes the form of (x – μ) / σ = (60.5 – 50) / 5 = 11.5 / 5 = 2.3).
What are the main steps for the normal approximation to binomial distribution?
You should take the following steps to proceed with the normal approximation to binomial distribution.
Find the number of occurrences or trials (N) with its probabilities (p).
Check if the number of trials is sufficiently high (N×p ≥ 5 and N×(1-p) ≥ 5).
Apply a continuity correction by adding or subtracting 0.5 from the discrete x-value.
Find the mean (μ) and standard deviation (σ).
Find the z-score (z = (x – μ) / σ).
Find the probability associated with the z-score.
Probability of success (0<p<1)
Probability of failure (0<q<1)
Number of occurrences (N)
Number of successes (n)
Event restatement
P(x ≥ n)
Problem statement P(x ≥ 40)
Continuity correction P(x > 39.5)
Z-score P(z ≥ -2.1)
Z-value 0.0179
Approximated probability 1 - 0.0179 = 0.9821
The approximated probability for the event to happen is 0.9821 or 98.2136%.
The IQR calculator allows you to find the interquartile range of up to 50 values.
IQR Calculator - Interquartile Range
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Actual cash value vs replacement value 🤔
How to calculate depreciation on appliances
Example: Using the appliance depreciation calculator
Appliance depreciation table
The appliance depreciation calculator estimates the actual cash value of any home appliances that you own. Whether you are thinking about replacing your old appliances like a washing machine or dealing with a home insurance policy that offers replacement cash value or actual cash value, this calculator has got you covered!
The following article will explain the difference between replacement cost and actual cash value. There's also an appliance depreciation table at the end to give you an idea of how much appliances depreciate each year. Not just for appliances, you can also use this tool to estimate the depreciation cost of other things, like the roof of your house or a window. You can start with entering some numbers in the calculator or continue reading to understand appliance depreciation and how do you calculate depreciation.
The devices or appliances we own get older as time passes. With regular usage of said appliances, there's wear and tear that causes damage to it, and the value of our devices reduces. This reduction in the value of the appliances is appliance depreciation.
The appliance depreciation has four components:
Depreciation rate (DR) — Rate at which the value of appliance is getting depreciated;
Age of item — Number of years in use or since purchase;
Actual cash value (ACV) — Current valuation or the replacement cash value minus the depreciated value of the appliance; and
Replacement cash value (RCV) — Amount it would take to replace the appliance at present.
The depreciation formula is:
\scriptsize \text{ACV} = \text{RCV} - (\text{DR}/100 \times \text{RCV} \times \text{Age})
While buying insurance cover for your home appliances or personal property, please read the terms and conditions for these terms. It is often the case that the insurance provider only pays the actual cash value of the objects rather than the replacement cash value. There are also different policy types based on the above parameters.
Please read your policy declarations carefully to check if it covers the replacement cash value to receive maximum benefits in any accidents, thefts, or other unpleasant circumstances. Also, look for additional factors like deductibles and limits.
The policy that covers replacement cash value will pay you the value of goods without any deduction for depreciation, i.e., the insurer will likely reimburse you a larger sum that you can use to buy yourself the covered personal property directly. However, if the policy covers the actual cash value, the money you'll receive upon claims is the present value of the appliance considering the depreciation. You may have to make the out-of-pocket expense to cover the rest of the cost to replace the device.
You can calculate the appliance depreciation using the above equation. Follow the steps below:
Select the appliance from the list or enter the depreciation rate directly.
Fill in the age of the item.
Insert the replacement cash value.
The appliance depreciation calculator will return the appliance's actual cash value.
Find the actual cash value for a 4-year old space heater presently available for $75.
To find the depreciation cost for a space heater:
Select the appliance from the list as space heater.
Fill in the age of the item as 4 years.
Insert the replacement cash value as $75.
The actual cash value is:
\qquad \scriptsize \begin{align*} \text{ACV} &= 75 - (6.67/100 \times 75 \times 4) \\ &= \text{\textdollar}54.99 \end{align*}
The table below has the depreciation rate of most of the appliances:
Refrigerator - regular
Stoves, ranges – electric
Vaccum cleaner (large)
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To calculate depreciation on appliances:
Multiply the age of the appliance by the replacement cash value.
Multiply this product with the depreciation rate to obtain the depreciated value of appliances.
Subtract the depreciation value from the replacement cash value to find out the actual cash value of the appliance.
The replacement cash value is the current cost of the product in the market, i.e., the cost required to replace the item you own with a new one at present. The actual cash value is the value of an appliance after considering depreciation due to wear and tear. The actual cash value is the difference between the replacement cost and the depreciation cost.
What do you mean by depreciation rate?
The depreciation rate tells you how much the appliance depreciate each year. It is the set percentage value or rate at which an item loses its value. For instance, the depreciation rate of a $50 water heater is 10%. Every year the water heater would lose 10% of $50 = $5 in its value. This value is known as the actual cash value of the appliance.
What are the factors affecting actual cash value of an appliance?
The actual cash value of an appliance primarily depends upon the depreciation rate – the rate at which the value of the appliance is getting depreciated – and the age of item, which is the number of years in use or since the purchase made.
Appliance wattageBack to schoolBalloon arch… 17 more
Our winch size calculator can help you find the winch size you may need for your vehicle in any situation.
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Hedging an Option Using Reinforcement Learning Toolbox - MATLAB & Simulink - MathWorks España
{\mathit{S}}_{\mathit{i}}^{}\text{\hspace{0.17em}}
\mathit{i}
{\mathit{A}}_{\mathit{i}}^{}\text{\hspace{0.17em}}
is the action taken at
\mathit{i}
{\mathit{R}}_{\mathit{i}+1}^{}
is the resulting reward at time
\mathit{i}+1
{\mathit{R}}_{\mathit{i}+1}={\mathit{V}}_{\mathit{i}+1}-{\mathit{V}}_{\mathit{i}}
{\mathit{H}}_{\mathit{i}+1}
{\mathit{S}}_{\mathit{i}+1}
{\mathit{S}}_{\mathit{i}}
{\kappa |{\mathit{S}}_{\mathit{i}+1}\left({\mathit{H}}_{\mathit{i}+1}-{\mathit{H}}_{\mathit{i}}\right)|\text{\hspace{0.17em}}\text{\hspace{0.17em}}}_{}
{\mathit{R}}_{\mathit{i}}:\mathrm{Reward}
{\mathit{V}}_{\mathit{i}}:\mathrm{Value}\text{\hspace{0.17em}}\mathrm{of}\text{\hspace{0.17em}}\mathrm{option}
{\mathit{S}}_{\mathit{i}}:\mathrm{Spot}\text{\hspace{0.17em}}\mathrm{price}\text{\hspace{0.17em}}\mathrm{of}\text{\hspace{0.17em}}\mathrm{underlying}\text{\hspace{0.17em}}\mathrm{asset}
{\mathit{H}}_{\mathit{i}}:\mathrm{Holding}
\kappa :\mathrm{Transaction}\text{\hspace{0.17em}}\mathrm{costs}
A final reward at the last time step liquidates the hedge that is
{\kappa |{\mathit{S}}_{\mathit{n}}\left({\mathit{H}}_{\mathit{n}}\right)|.\text{\hspace{0.17em}}\text{\hspace{0.17em}}}_{}
In this implementation, the reward (
{\mathit{R}}_{\mathit{i}}
) is penalized by the square of the reward multiplied by a constant to punish large swings in the value of the hedged position:
{\mathit{R}}_{\mathit{i}+1}={\mathit{R}}_{\mathit{i}+1}-\mathit{c}{\left({\mathit{R}}_{\mathit{i}+1}\right)}^{2}
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How do I calculate the discharging time of a battery?
How do I measure a lead-acid battery capacity?
How do I check the capacity of a lithium-ion battery after use?
How to use battery size calculator?
Omni's battery size calculator (or remaining battery capacity calculator) explains in detail how to check the battery capacity for both lithium-ion and lead-acid batteries.
Our tool has many uses — whether you want to know how much longer your drone will fly after already using it for a few hours, or if you want to compare lead-acid and lithium-ion batteries in terms of their battery capacity, the battery size calculator does it all!
You just need to divide the battery's ampere-hour or amp-hour (Ah) rating with the desired load current to estimate the discharging time or run time:
discharging time = amp-hour / load current
The amp-hour rating helps us determine the maximum steady current supplied in an hour. Suppose your battery has a 20 Ah rating on it. Using the battery runtime formula, we find out that it can supply 20 A for an hour, 10 A for 2 hours, 5 A for 4 hours, or 1 A for 20 hours.
Do you know that you shouldn't discharge batteries completely? Learn how to estimate how long you can safely use a battery with our battery life calculator!
Now that you know how to calculate the runtime/discharge time of any type of battery, let's learn how to measure the capacity of lead-acid batteries after putting them to use.
We calculate the remaining capacity of a lead-acid battery using the following formula:
B_\text{Pb} = \frac{100\cdot I_\text{L}\cdot t}{(100 - Q)(0.02 t + 0.6)},
B_\text{Pb}
— Remaining capacity of the lead-acid battery (
\text{Pb}
because it's the chemical symbol for lead);
I_\text{L}
— Load current;
t
— Duration for which the power is supplied to the load; and
Q
— Percentage of charge that should remain after the battery is used.
For a lithium-ion battery, the formula for the battery capacity is:
B_\text{Li} = \frac{100 \cdot I_\text{L} \cdot t}{(100 - Q)},
B_\text{Li}
— Remaining capacity of the lithium-ion battery (
\text{Li}
because it's the chemical symbol for lithium);
I_\text{L}
t
— Duration for which the electrical power is supplied to the load; and
Q
Do you want to estimate the full capacity of your battery? Check our battery capacity calculator!
🙋 Lithium-ion batteries have a higher energy density than lead-acid batteries. So, we can store more energy in a lithium-ion battery using the same physical space. Therefore, a lithium-ion battery can supply more power than a lead-acid battery of the same size!
You can easily find out the remaining battery capacity using our tool by following these simple steps:
Enter the wattage of the appliance in the input box for Application load.
Select the type of your battery from the drop-down list beside Battery type.
Type the voltage rating specified on your battery in the input box for Voltage.
Enter the number of hours (or you can select the duration in other units) you intend to use the battery for.
Input the percentage of charge you want to remain in the battery after use.
Voila! The calculator tells you the Load current and Remaining capacity or the battery size!
⚠️ You shouldn't discharge lead-acid and lithium-ion batteries completely. Discharge lead-acid batteries up to 50% and lithium-ion batteries up to 20% to avoid any irreversible damage and for improved cell life.
Example: To find the remaining charge in your UPS after running a desktop computer of 200 W for 10 minutes:
Enter 200 for the Application load, making sure W is selected for the unit.
Usually, a UPS uses a lead-acid battery. The Battery type is Lead-acid by default. So you don't need to choose the type manually in this case.
Enter 12 for the Voltage as the lead-acid battery provides a total input voltage of 12 V.
Select minutes as the unit from the drop-down list beside the input box for Required duration.
Input 10 in the Required duration input box.
Type 50 in the Remaining charge box.
The calculator displays both the load current 16.7 A and the remaining capacity or the battery size 9.2 Ah!
Read on to find answers to some of the questions on a battery's capacity and charging. Also, send us your questions on battery capacity in the comment box displayed when you rate the calculator.
What is the difference between ampere and ampere-hour?
Ampere (A) is the SI base unit of the electrical current while ampere-hour or amp-hour (Ah) denotes the electric current produced or consumed in an hour. We use amp-hour to measure the capacity of a battery. We also use watt-hour to measure battery capacity.
The reserve capacity (RC) of a battery tells us how many minutes it can supply a current of 25 A at a potential difference above 10.5 V.
How do I calculate the battery size for inverters?
You can calculate the battery size for inverters using the formula B = P × t / Vdc, where B is the battery capacity in ampere-hour, P is the inverter's power rating, t is the duration of power supply in hours, and Vdc is the DC voltage of the inverter.
How do I calculate the lithium-ion battery size for a varying load?
To calculate the battery size for a varying load which requires I1 in the interval t1 and I2 in the remaining time:
Estimate the average load current — Iav = (I1 × t1 / t) + (I2 × [t - t1 / t]).
Substitute I = Iav in the equation for battery capacity of lithium-ion. BLi = 100 × I × t / (100 - q) where B is the battery capacity, I is the load current, t is the duration of power supply, and q is the percentage of charge which should remain in the battery after the discharge.
How long will a 100 Ah battery run an appliance that requires 1,000 W?
You can calculate the run-time using the formula, t = (amp-hour × V)/P, where amp-hour is the battery's maximum capacity, V is the voltage of the power supply, and P is the appliance's wattage.
In the US, the household power supply's voltage is 120 V. Therefore, a 100 Ah battery can supply power for 12 hours in the US for a 1000W-appliance.
How do I calculate the charging time of a battery?
If we ignore the losses during charging, divide the battery capacity by the charging current to obtain the battery's charging time in hours.
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Schulze STV - Wikipedia
Proportional-representation ranked voting system
Schulze STV is a draft single transferable vote (STV) ranked voting system designed to achieve proportional representation.[1][2] It was invented by Markus Schulze, who developed the Schulze method for resolving ties using a Condorcet method. Schulze STV is similar to CPO-STV in that it compares possible winning candidate pairs and selects the Condorcet winner. It is not used in parliamentary elections.
The system is based on Schulze's investigations into vote management and free riding.[jargon explanation needed][3] When a voter prefers a popular candidate, there is a strategic advantage to first choosing a candidate who is unlikely to win (Woodall free riding) or omitting his preferred candidate from his rankings (Hylland free riding). According to Schulze, vote management is party coordination of free riding.
Although Schulze STV is resistant to both types of free riding, Hylland free riding is impossible to eliminate. Schulze created a criterion known as "weak invulnerability to Hylland free riding".[This quote needs a citation] A method meets this criterion if it is invulnerable to Hylland free riding, except in cases where the Droop proportionality criterion would be violated. Schulze STV meets this criterion.
1.1 Resistance to vote management
2 STV and Schulze STV
Each voter ranks candidates in their order of preference. In a hypothetical election, three candidates vie for two seats; Andrea and Carter represent the Yellow Party, and Brad represents the Purple Party. Andrea is a popular candidate, and has supporters who are not Yellow Party supporters. It is assumed that the Yellow Party can influence their own supporters, but not Andrea's.
There are 90 voters, and their preferences are
Andrea (Y)
Brad (P)
Carter (Y)
In the STV system, the initial tallies are:
Andrea (Y): 50
Carter (Y): 13
Brad (P): 27
The quota is determined according to
{\displaystyle ({\rm {\mbox{valid votes cast}}})/({\rm {\mbox{seats to fill}}}+1)=90/(2+1)=30.}
Andrea is declared elected and her surplus,
{\displaystyle {\rm {\mbox{Andrea's votes}}}-{\rm {\mbox{quota}}}=50-30=20}
, is distributed with
{\displaystyle {\mbox{Round}}\left({{\mbox{votes for next preference belonging to the original candidate}} \over {\mbox{total votes for the original candidate}}}\times {\mbox{surplus votes for original candidate}}\right).}
Carter (Y):
{\displaystyle 13+{\mbox{Round}}\left({\frac {26+12}{50}}\times 20\right)=13+15=28}
Brad (P):
{\displaystyle 27+{\mbox{Round}}\left({\frac {12}{50}}\times 20\right)=27+5=32}
Brad is also elected.
The Schultze STV system has three possible outcomes (sets of winners) in the election: Andrea and Carter, Andrea and Brad, and Carter and Brad. In this system, any candidate with more than the Droop quota of first choices will be elected. Andrea is certain to be elected, with two possible outcomes: Andrea and Carter, and Andrea and Brad.
Resistance to vote management[edit]
In vote management, a party instructs its voters not to rank a popular party candidate first. If the Yellow Party's leaders instruct their supporters to choose Carter first (followed by Andrea), the balloting changes. Unlike STV, however, Schulze STV resists vote management.
STV and Schulze STV[edit]
Ordinary STV, being based on IRV, can be sensitive to in what order the losers are eliminated: what candidates win may depend on whether a loser has fewer first preferences than another loser. By using Condorcet pair comparisons, Schulze STV (like CPO-STV) greatly reduces this defect.
The number of pair comparisons is smaller than CPO-STV, since Schulze STV only compares outcomes differing by one candidate; CPO-STV compares all possible pairs.
Potential for tactical voting[edit]
Proportional representation systems are much less susceptible to tactical voting systems than single-winner systems such as the first past the post system and instant-runoff voting (IRV), if the number of seats to be filled is sufficiently large. Schulze STV has additional resistance to forms of tactical voting which are specific to single transferable voting methods.
All forms of STV that reduce to IRV in single winner elections fail the monotonicity criterion. This means that it is sometimes possible to benefit a candidate by ranking them lower than one's true order of preference, or to harm a candidate by ranking them higher. This isn't the case for Schulze STV. When some voters rank candidate
{\displaystyle b}
higher without changing the order in which they rank the other candidates relatively to each other, then the strength of the vote management of the candidates
{\displaystyle a_{1},\ldots ,a_{M}}
against candidate
{\displaystyle b}
can't increase. I. e. the strength of any vote management and the strength of beatpaths is monotonic in
{\displaystyle b}
and the monotonicity follows from that of the underlying Schulze method.
As Schulze STV reduces to the Schulze method in single winner elections, it fails the participation criterion, the later-no-harm criterion and the later-no-help criterion, whereas traditional forms of STV (that reduce to IRV in single winner elections) fulfill later-no-help and later-no-harm.
STV methods which make use of Meek's or Warren's method are resistant to Woodall Free Riding, but are still vulnerable to Hylland Free Riding. Schulze's method is not vulnerable to Hylland Free Riding, except where necessary in order to meet the Droop proportionality criterion.[1]
A method which doesn't meet the Droop proportionality criterion has the potential to give disproportional results, unless it meets a similar proportionality criterion. Thus, Schulze STV can be considered invulnerable to Hylland Free Riding to as great an extent possible, subject to actually being a proportional representation method.
Schulze STV is no more complicated for the voter than other forms of STV; the ballot is the same, and candidates are ranked in order of preference. In calculating an election result, however, Schulze STV is significantly more complex than STV. Although it is less complex than CPO-STV, computer calculation would be required for large-scale elections. Computing the result would still be difficult in some cases, since Schulze STV does not have polynomial runtime.
^ a b Markus Schulze, Free Riding and Vote Management under Proportional Representation by Single Transferable Vote
^ Markus Schulze, Implementing the Schulze STV Method
^ Markus Schulze, Free Riding, Voting matters, issue 18, pages 2–8, June 2004
The Schulze Method of Voting (section 9) by Markus Schulze
Personalisierung der Verhältniswahl durch Varianten der Single Transferable Vote (in German) by Martin Wilke
A voting web application that uses Schulze STV
Retrieved from "https://en.wikipedia.org/w/index.php?title=Schulze_STV&oldid=1078007228"
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Kg to mL conversion formula
Kg to mL conversion example
How to use the kg to mL converter?
Other mass to volume converters
Our kg to mL converter can help you convert between the mass of any substance in kg to its volume in mL.
Continue reading to know the relation between mass and volume. You will also find answers to questions like:
How to convert kg to mL?
How many mL in a kg of water or milk?
First, let us start with the basics of kg to mL conversion.
As you already know that kg or kilogram is a unit of mass, i.e., the quantity of matter in any object. Whereas, mL or milliliter is a unit of volume, i.e., the space occupied by matter.
To convert between mass and volume, we can use a very simple formula:
\rm Volume = \frac{Mass} {Density}
In the above equation, the mass, density, and volume of the substance are expressed in the S.I units, i.e., kg, kg/m3, and m3, respectively.
Now, using the fact that 1 m3 = 106 mL, we can write the formula for kg to mL conversion as:
\rm Volume (mL) = \frac{Mass (kg)}{Density (kg/m^3)} \times 10^6~ \rm{mL/m^3}
Let us consider a simple example of converting 1 kg of milk to ml. We know that the density of milk is about 1030 kg/m3.
Hence, we can calculate its volume in mL as:
\begin{align*} \text{Volume} (\rm{mL}) &= \frac{1~\rm{kg}} {1030~ \rm{kg/m^3}} \times 10^6~ \rm{mL/m^3} \\ & = 970.9~ \rm{mL} \end {align*}
i.e., 1 kg of milk is equivalent to about 970.9 mL.
As you can see converting from kg to mL is very easy if you know the density of the substance of interest.
Now let us see how we can use the kg to mL converter for the same problem:
Using the drop-down menu, select the category as "food" and "milk".
The density field will auto-populate. You can also enter the density if you know its value.
Input the weight/mass as 1 kg. The tool will display the volume of milk as 970.9 mL.
If you like our kg to mL converter, make sure to check out our other tools that deal with mass to volume conversions:
How do I convert kg to mL?
To convert kg to mL for any substance, follow the given instructions:
Find out the density of the substance in kg/m3.
Now divide its mass in kg by its density in kg/m3 to get its volume in m3.
Multiply the volume in m3 by 106 to get the volume in mL.
How many mL in a kg of water?
About 1,000 mL. The density of water is about 1000 kg/m3. Hence, the volume of 1 kg of water is about 10-3 m3 or 10-3 × 106 = 1,000 mL.
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Magic Number Calculator | Let's Play Ball!
Reviewed by Łucja Zaborowska, MD, PhD candidate and Jack Bowater
How to use the magic number calculator?
How to calculate the magic number in baseball using formula
What is the tragic number?
Using our magic number calculator, you can determine how many games your team needs to win or the current opponent team needs to lose for you to secure the title or the next playoff spot.
This article will teach you how to calculate the magic number in baseball and discuss what this magic number is. You will observe how it resembles and differs from the winning percentage .
To determine a batter's effectiveness at reaching base, check out our on base percentage calculator;
For their offensive skills, the batting average calculator may prove to be helpful; and
For their value of walks plus hits per inning pitched, check out our WHIP calculator.
A magic number represents the required number of wins a team needs (or the same combination of losses by its present competitor) to achieve the title or the next playoff spot. Every time a team wins, its magic number drops by one, which means it's closer to the final goal.
🙋 Every baseball team starts with 163 as their magic number. There are a total of 162 games played, with one more being played to break a tie.
Our tool helps you calculate the MLB magic number as follows:
Enter the number of games your team has won in the first field, e.g., 72 games.
Enter your current opponent's losses in the second field, e.g., 50 games.
You get your magic number as a result, i.e., 41 games.
That's how you calculate the baseball magic number.
💡 When a team is up against multiple competitors, find the magic number against their closest competitor.
Here's the formula that we use to calculate the MLB magic number:
M_N = T_G - W_T - L_O + 1
M_N
- The magic number of a baseball team;
T_G
- The total number of games to be played;
W_T
- The number of wins by your team; and
L_O
- The number of losses by your opponent.
Let's take an example and find the magic number of team A, with 61 wins, while team B has 56 losses.
M_N = 162 - 61 - 56 + 1 = 46
Thus, team A's magic number is 46. They need to win 46 more games, unless team B loses 46 games, to proceed to the subsequent playoffs.
Tragic numbers, also called elimination numbers, or E, are the opposite of what magic numbers are. We calculate them by subtracting the losses of our team and winnings of the opponent team from the total number of games to be played, plus one, i.e., 163. This number determines how many losses will eliminate our team.
How do I find the magic number in baseball?
To find the magic number in baseball:
Take the total number of games, e.g., 162.
Subtract the total wins of the first team, e.g., 55.
Subtract the losses of the second team, e.g., 57.
Add 1 to the result and you have your first team's magic number, i.e., 51.
What does MLB, NLB, ALDS, ALCS, NLDS, and NLCS stand for?
MLB stands for major league baseball;
NLB stands for national league baseball;
ALDS stands for American league division series;
ALCS stands for American league championship series;
NLDS stands for national league division series; and
NLCS stands for the national league championship series.
What is a wild card team in baseball?
A wild card team is the one that doesn't win their division but has the best record after the winning team and thus gets a chance to participate in the playoffs. There are two wild card teams selected from each league.
Your team's wins
Opponent's losses
Every MLB team starts with a magic number of 163. ⚾
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Detecting memory corruption with dog tags | Seeking Consistency
There's a fantastic technique I've used in my C days that attempted to detect memory corruption. It's a cheap and usually effective technique because it relies on common causes of corruption, namely off by one or any bound writes into memory beyond your allocated limit or even pointer arithmetic gone a-rye.
It works by whenever you allocate memory you write a tag at the start and end. This can be anything but should be big enough such that it's unlikely to ever end up in legitimate memory when the buffer is over ran. Therefore your allocate a buffer that is
desiredSize + 2(tagSize)
. When your allocation function returns, it returns the pointer to
actualPointer + tagSize
#define TAG_TYPE int
#define TAG_SIZE sizeof(TAG_TYPE)
#define VALID_START_TAG 0xDEADBEEF
#define VALID_END_TAG 0xCAFEBABE
#define TAG_INVALID_VALUE 0xBADC0DE5
#define PTR_START_TAG(memory) (TAG_TYPE *)(memory->ptr - TAG_SIZE)
#define PTR_END_TAG(memory) (TAG_TYPE *)(memory->ptr + memory->size)
At any time you can check whether the memory is corrupt by inspecting that the front (
fakePointer - tagSize
) and rear (
fakePointer + size
) tags are correct.
When you free your memory you can again check that they are correct and if they are not warn the user. You also invalidate the tags so that a later allocation wont accidentally contain the tags by chance.
I used a struct above for convenience.
Full code available at https://gist.github.com/kay/5658481
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Section 3-9 - IndeterminateForms - Maple Help
Home : Support : Online Help : Study Guides : Calculus : Chapter 3 - Applications of Differentiation : Section 3-9 - IndeterminateForms
Section 3.9: Indeterminate Forms and L'Hôpital's Rule
The seven collections of symbols on the left in Table 3.9.1 are called indeterminate forms, insofar as they arise when calculating the limits of certain expressions. The stratagems on the right in the table extend the set of rules for computing limits without having to invoke Definition 1.2.1.
Limit Stratagem
\frac{0}{0}
\frac{\infty }{\infty }
0\cdot \infty
Rewrite as fraction and use L'Hôpital's rule
\infty -\infty
Convert to quotient via common denominator, rationalization, or factoring;
then use L'Hôpital's rule
{0}^{0}
{\infty }^{0}
{1}^{\infty }
Take the limit of the log and exponentiate the result.
Equivalently, write
u{\left(x\right)}^{v\left(x\right)}
{e}^{v \mathrm{ln}\left(u\right)}
Table 3.9.1 Indeterminate forms and appropriate approaches to their limits
Recall from Table 1.3.1 that the limit of a quotient is the quotient of the limits when the limits exist and the limit of the denominator is not zero. If, in computing the limit of the fraction
f\left(x\right)/g\left(x\right)
one of the forms
0/0
\infty /\infty
results, no formal technique in Chapter 1 applies.
Recall from Table 1.3.1 that the limit of a product is the product of the limits, provided both limits exist. If, in computing the limit of the product
f\left(x\right)\cdot g\left(x\right)
0\cdot \infty
Recall from Table 1.3.1 that the limit of a difference is the difference of the limits, provided both limits exist. If, in computing the limit of the difference
f\left(x\right)-g\left(x\right)
\infty -\infty
The three indeterminate forms at the end of Table 3.9.1 arise when taking the limit of expressions of the form
u{\left(x\right)}^{v\left(x\right)}
. They do not arise if one of the functions
u
v
is strictly constant. For example, Maple evaluates
{0}^{0}
to 1, and
\underset{x→\infty }{lim}{1}^{x}
\textcolor[rgb]{0,0,1}{1}
Theorem 3.9.1 - L'Hôpital's Rule
a∈I
, an open interval
f
g
differentiable on
I
g\prime \left(x\right)≠0
I
, except possibly at
x=a
\underset{x→a}{lim}f\left(x\right)=0
\underset{x→a}{lim}g\left(x\right)=0
\underset{x→a}{lim}f\left(x\right)=±\infty
\underset{x→a}{lim}g\left(x\right)=±\infty
\underset{x→a}{lim}\frac{f\left(x\right)}{g\left(x\right)}
exists or is
±\infty
\underset{x→a}{lim}\frac{f\left(x\right)}{g\left(x\right)}=\underset{x→a}{lim}\frac{f \prime \left(x\right)}{g\prime \left(x\right)}
Condition 4 of Theorem 3.9.1 states that L'Hôpital's rule deals with the two indeterminate forms
0/0
\infty /\infty
. If the rule is applied to a form other than one of these, an incorrect limit will be obtained.
\underset{x→a}{lim}\left(f\left(x\right)-g\left(x\right)\right)
results in the indeterminate form
\infty -\infty
, the algebra in Table 3.9.2 provides a formal way of converting the difference to an expression amenable to L'Hôpital's rule.
f-g\equiv \frac{\frac{1}{g}-\frac{1}{f}}{\frac{1}{f g}}
Table 3.9.2 Converting the form
\infty -\infty
0/0
f→\infty
g→\infty
, the numerator of the fraction on the right in Table 3.9.2 tends to
0-0=0
. Similarly, the denominator on the right tends to 0 also. Hence, Table 3.9.2 converts the form
\infty -\infty
0/0
for which L'Hôpital's rule is applicable. However, the algebra involved is cumbersome, and any ad hoc manipulations that can replace the formalism of Table 3.9.2 should be considered.
Note on the pronunciation of "L'Hôpital"
There is variation in the spelling of this French name, some authors using the upper-case "L", some using the lower-case. Some texts even use the alternate spelling L'Hospital, causing students to pronounce the "s".
Consultation with a French mathematician yielded the following conclusions. Always use the upper-case "L". Never spell the name with an "s". Never pronounce the name with an "s". A typical North American pronunciation would be lōp-ē-tăhl'. The internet provides sites at which the French pronunciation can be heard.
\underset{x→0}{lim}\frac{\mathrm{sin}\left(x\right)}{x}
, then detail an applicable strategy taken from Table 3.9.1.
\underset{x→0}{lim}\frac{{\mathrm{sin}}^{2}\left(x\right)}{x}
\underset{x→1}{lim}\frac{{x}^{2}-1}{x-1}
\underset{x→0}{lim}\frac{\sqrt{x+1}-1}{x}
\underset{x→\infty }{lim}\frac{{e}^{x}}{{x}^{2}}
\underset{x→\infty }{lim}\frac{2 {x}^{2}-1}{3 {x}^{2}+5 x}
\underset{x→0+}{lim}x \mathrm{ln}\left(x\right)
\underset{x→0}{lim}\left(\mathrm{csc}\left(x\right)-\mathrm{cot}\left(x\right)\right)
\underset{x→0+}{lim}{x}^{x}
\underset{x→0+}{lim}{\left(-\mathrm{ln}\left(x\right)\right)}^{x}
\underset{x→\infty }{lim}{\left(\frac{x}{1+x}\right)}^{x}
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How do I calculate the percent composition of a compound?
How to find the molar mass from the percent composition of a compound?
How to use this percent composition calculator?
This percent composition calculator is a simple tool to evaluate the percentage of each element by weight (or mass) in a compound from its chemical formula. This article will explore percent composition, what is needed to determine percent composition, how to calculate it using the percent composition formula, and some frequently asked questions.
The percent composition of a compound expresses the weight (or mass) of each element in the compound as a percentage of the total weight of the compound. The formula for percent composition is given by:
element wgt% = (element mass/compound mass) × 100
element wgt% – Percentage of the element present in the compound by mass;
element mass – Mass of the element in one mole of the compound; and
compound mass – Mass of one mole of the compound or its molar mass.
Before you learn how to find the percent composition of a substance in the next section, use our mole calculator and Avogadro's number calculator to understand the concept of moles as a unit of measurement.
To determine the percent composition of a substance, follow these simple steps:
Determine the molar mass of the substance either from its molecular weight or from its mass and number of moles.
Compute the mass of each element in one mole of the compound by multiplying their atomic mass with the number of atoms in one molecule of the compound.
Calculate percent composition of each element as (mass of the element in 1 mol of compound/molar mass of compound) × 100.
Verify your calculations with our percent composition calculator.
For example, let's calculate the mass percent composition of sulfuric acid,
\text{H}_2\text{SO}_4
, using the percent composition equation.
Molar mass of sulfuric acid:
Start by finding the molecular weight of
\text{H}_2\text{SO}_4
, given by multiplying the number of atoms of each element in the molecule by its atomic mass:
2 × (1.008) + (32.06) + 4 × (15.999) = 98.07 amu.
Since the molar mass of a compound in g/mol is numerically equivalent to its molecular weight in amu, we get molar mass of sulfuric acid as 98.07 g/mol.
Mass of each element in one mole of sulfuric acid:
Mass of hydrogen = 2 × (1.008) = 2.016 g/mol
Mass of sulfur = 1 × 32.06 = 32.06 g/mol
Mass of oxygen = 4 × 15.999 = 63.996 g/mol
Percent composition of each element in sulfuric acid:
Hydrogen% = (2.016/98.07) × 100 = 2.0556%
Sulfur% = (32.06/98.07) × 100 = 32.69%
Oxygen% = (63.996/98.07) × 100 = 65.25%
Now, do you understand how to calculate the percentage of an element in a compound?
To calculate the molar mass of a compound from its percent composition, follow these simple steps:
Divide the percent composition of an element in the compound by 100 to get the value in decimals.
Divide the mass of the element in one mole of the compound by this decimal value to obtain the molar mass of the compound.
You can verify your result yourself by repeating these steps with the percent composition of other elements too!
Our percent composition calculator is a simple tool to use. It can support up to five elements in a compound.
In the field labeled Number of atoms of first element, you can select the element from the units drop-down list (five of the most common elements in compounds are the default).
Enter the number of atoms of the selected element present in one molecule of the compound. For example, if the compound is
\text{H}_2\text{O}
, enter 2 atoms for hydrogen and 1 for oxygen.
Repeat these steps for every element in the compound.
The calculator will automatically calculate the percent composition by mass using the percent composition equation and display them in the corresponding fields in the Composition of each element group.
For the element fields you don't use, let their value stay at the default value of 0. Empty fields halt the calculation of molecular weight.
We've put together a collection of related calculators for you to explore and learn further:
How much carbon is present in 100 g of sugar?
There are 42.1 grams of carbon in 100 grams of sugar. The molecular formula for sugar is C₁₂H₂₂O₁₁. From this, we can determine that the percent composition of carbon is 42.11%, hydrogen is 6.479%, and oxygen is 51.41%. Therefore, in 100 grams of sugar, we shall find:
42.1 grams of carbon;
Around 6.5 grams of hydrogen; and
51.4 grams of oxygen.
What is the mass percentage of hydrogen in water?
11.19%. To replicate this result, follow these simple steps:
Determine the molecular weight of water as the sum of the atomic mass of all atoms in one molecule. 2 × 1.008 + 15.999 = 18.015 amu.
Obtain the molar mass of water as the numerical equivalent of its molecular weight, when expressed in g/mol. Molar mass of water = 18.015 g/mol.
Compute the mass of hydrogen in one mole of water as the numerical equivalent of the atomic mass of hydrogen atoms in one water molecule, when expressed in g/mol. Mass of hydrogen in 1 mol water = 2 × 1.008 = 2.016 g/mol.
Calculate the mass percent composition of hydrogen in water as (mass of hydrogen in 1 mol water/molar mass of water) × 100 = (2.016/18.015) × 100 = 11.19%.
How to find the molecular formula from the percent composition of a compound?
To find the molecular formula of a compound from its percent composition, you need to know its molar mass:
Determine the molecular weight of the compound in amu from its molar mass in g/mol since they are numerical equivalent.
Calculate the weight of each element in the compound from the percent composition, using mass of an element per molecule = (element percent composition element × molecular weight)/100.
Find the number of atoms of each element in the compound using their atomic mass. Number of atoms of each element per molecule = mass of each element per molecule/element atomic mass.
Formulate the molecular formula of the compound using the element symbols and the number of atoms in one molecule.
What is the percent composition of NaHCO₃?
Total molecular weight : 84 amu
Number of atoms of each element in the compound
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Mw
Adnan Barkat; Farhan Javed; Yen Joe Tan; Aamir Ali; Muhammad Tahir Javed ...
Mw
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Focus Section: Big Data Problems in Seismology
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Logic | Introduction to College Mathematics | Course Hero
2\text{ }{\in}\text{ }\mathbb{Z}
(2 is an element of the set of integers (or more simply, 2 is an integer).)
\sqrt{2}\text{ }{\notin}\text{ }\mathbb{Z}
(The square root of 2 is not an integer.)
\mathbb{N}\text{ }{\subseteq}\text{ }\mathbb{Z}
(The set of natural numbers is a subset of the set of integers.)
\sqrt{2}\text{ }{\notin}\text{ }\mathbb{R}
(The square root of 2 is not a real number.)
\mathbb{Z}\subseteq\mathbb{N}
(The set of integers is a subset of the set of natural numbers.)
{0,1,2}\cap\mathbb{N}=\varnothing
(The intersection of the set {0,1,2} and the natural numbers is the empty set.)
\mathbb{Z}
(The set of integers)
42\text{ }{\in}\text{ }\mathbb{Z}
(42 is an element of the set of integers.)
\mathbb{N}\subseteq\varnothing
{0,-1,-2}\cap\mathbb{N}=\varnothing
P : The solutions of the equation
\displaystyle{a}x^2+bx+c=0\text{ are }x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}
Q : If a right triangle has legs of lengths a and b and hypotenuse of length
a^2+b^2={c}^2
.. Here is a very famous statement, so famous, in fact, that it has a name. It is called Fermat’s last theorem after Pierre Fermat, a seventeenth-century French mathematician who scribbled it in the margin of a notebook.
R : For all numbers a, b, c, n ∈
\mathbb{N}
with n > 2, it is the case that
a^n+b^n\neq{c}^n
. Fermat believed this statement was true. He noted that he could prove it was true, except his notebook’s margin was too narrow to contain his proof. It is doubtful that he really had a correct proof in mind, for after his death generations of brilliant mathematicians tried unsuccessfully to prove that his statement was true (or false). Finally, in 1993, Andrew Wiles of Princeton University announced that he had devised a proof. Wiles had worked on the problem for over seven years, and his proof runs through hundreds of pages. The moral of this story is that some true statements are not obviously true.
For another example, starting with the false statement “
2\in\varnothing
” we get the true statement “It is not true that
2\in\varnothing
.” We use the symbol ∼ to stand for the words “It’s not true that,” so ∼ P means “It’s not true that P.” We often read ∼ P simply as “not P.” Unlike ∧ and ∨, which combine two statements, the symbol ∼ just alters a single statement. Thus its truth table has just two lines, one for each possible truth value of P.
{P}\Rightarrow{Q}\begin{cases}{\text{If }P\text{, then }Q.}\\{Q\text{ if }P.}\\{Q\text{ whenever }P.}\\{Q\text{, provided that }P.}\\{\text{Whenever }P\text{, then also }Q.}\\{P\text{ is a sufficient condition for }Q.}\\{\text{For }Q\text{, it is sufficient that }P.}\\{Q\text{ is a necessary condition for }P.}\\{\text{For }P\text{, it is necessary that }Q.}\\{P\text{ only if }Q.}\end{cases}
However, the contrapositive of P ⇒ Q, ~Q ⇒ ~P, is equivalent to P ⇒ Q. Similarly, the inverse of P ⇒ Q, which is ~P ⇒ ~Q, is equivalent to the converse Q ⇒ P. In "Truth Tables for Statements," we will learn how to show these equivalences using a truth table.
{P}\iff{Q}\begin{cases}{P\text{ if and only if }Q.}\\{P\text{ is a necessary and sufficient condition for }Q.}\\{\text{For }P\text{ it is necessary and sufficient that }Q.}\\{\text{If }P\text{, then }Q\text{, and conversely.}}\end{cases}
\text{Contrapositive law}\begin{cases}P\Rightarrow{Q}=(\sim{Q})\Rightarrow(\sim{P})\end{cases}
\text{DeMorgan's laws}\begin{cases}{\sim(P\land{Q})=\sim{P}\lor\sim{Q}}\\{\sim(P\lor{Q})=\sim{P}\land\sim{Q}}\end{cases}
\text{Commutative laws}\begin{cases}{(P\land{Q})={P}\land{Q}}\\{(P\lor{Q})={P}\lor{Q}}\end{cases}
\text{Distributive laws}\begin{cases}{{P}\land(Q\lor{R})=({P}\land{Q})\lor(P\land{R})}\\{P\lor(Q\land{R})=({P}\lor{Q})\land(P\lor{R})}\end{cases}
\text{Associative laws}\begin{cases}{P\land(Q\land{R})=(P\land{Q})\land{R}}\\{P\lor(Q\lor{R})=(P\lor{Q})\lor{R}}\end{cases}
(from "Logical Equivalence") can be viewed as rules that tell us how to negate the statements P ∧Q and P ∨Q. Here are some examples that illustrate how DeMorgan’s laws are used to negate statements involving “and” or “or.”
Now let’s move on to a slightly different kind of problem. It’s often necessary to find the negations of quantified statements. For example, consider ∼(∀x ∈
\mathbb{N}
, P(x)). Reading this in words, we have the following:
This means P(x) is false for at least one x. In symbols, this is ∃ x ∈
\mathbb{N}
, ∼P(x). Thus ∼ (∀x ∈
\mathbb{N}
, P(x)) = ∃ x ∈
\mathbb{N}
, ∼P(x). Similarly, you can reason out that ∼ (∃ x ∈
\mathbb{N}
, P(x)) = ∀x ∈
\mathbb{N}
, ∼P(x). In general:
It is important to be aware of the reasons that we study logic. There are three very significant reasons. First, the truth tables we studied tell us the exact meanings of the words such as “and,” “or,” “not,” and so on. For instance, whenever we use or read the “If..., then” construction in a mathematical context, logic tells us exactly what is meant. Second, the rules of inference provide a system in which we can produce new information (statements) from known information. Finally, logical rules such as DeMorgan’s laws help us correctly change certain statements into (potentially more useful) statements with the same meaning. Thus logic helps us understand the meanings of statements and it also produces new meaningful statements.
Logic is the glue that holds strings of statements together and pins down the exact meaning of certain key phrases such as the “If..., then” or “For all” constructions. Logic is the common language that all mathematicians use, so we must have a firm grip on it in order to write and understand mathematics.
Logic. Authored by: Richard Hammack. Located at: http://www.people.vcu.edu/~rhammack/BookOfProof/. Project: BOOK OF PROOF. License: CC BY-ND: Attribution-NoDerivatives
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Created by Steven Wooding and Rijk de Wet
Peter Backus “Why I don't have a girlfriend: An application of the Drake Equation to love in the UK“ (January 2010)
The Drake equation for love calculator
Finding love and aliens with maths
Is love more rare than aliens?
Love is out there — just ask Peter
Mathematical tips for finding love
7.9 billion people in the world and you're single on Valentine's Day. What are the odds of that? Well, roughly 40% of the UK were partnerless in 2020, according to the latest data from the Office for National Statistics (ONS) and Scotland's 2011 census. It's safe to say your chances of being single are surprisingly high. But is there a scientific way to determine it? And can you increase your chances of finding love?
Peter Backus, an economist of the University of Warwick, set out to discover exactly that when he applied the famous Drake equation (for finding alien civilisations) to the dating world.
Our Drake equation for love calculator applies Backus's formula to tell you how many potential partners are out there and your chances of finding them — much like the Drake equation does, but for love instead of aliens!
Can you mix aliens, maths, and romance? Only one way to find out!
Soon, this could be you!
First, let's show you how to use our calculator and get your result. In the next section, we'll dig into the maths behind the calculator.
The calculator is really easy to use — just answer a few questions about yourself and your ideal partner, and it will work out how many matches are out there and your chances of finding them. Let's go through the calculator and get your results.
Location 🏙️ — Select the city where you live or one near you. This defines the pool of people the calculator will select from. If you are not near one of the listed cities, choose the "UK" option instead, or use the advanced mode (see below for details).
Your attractiveness — Rate how attractive you are on a scale of 1 to 5 (5 being very attractive). It's best to be honest when answering this question. If you're unsure, ask a good and honest friend, or a stranger at the nearest coffee shop.
Social skills — Say how well you get on with other people, on a scale of 1 (difficult) to 5 (very easy). Again, be honest and/or seek the opinion of those around you.
Gender ♂♀ — Are you looking for a boyfriend or a girlfriend? You can also say you don't mind which gender.
Minimum age — What is the minimum age of a partner that would be a good fit for you?
Maximum age — What is the maximum age of a person you are looking for?
University educated 🎓 — Should the person you're looking for have a degree or not? You can also say you don't mind either way.
Attractiveness of people — Thinking about people that match your preferences so far, what proportion do you find attractive? Select one of the three existing options, or select "Custom value" to enter this factor as a percentage.
Once you've entered all the inputs of the calculator, it will show you the number of potential partners that are out there. It also tells you the percentage chance of finding love based on the number of people in your chosen city or the whole of the UK. For a bit of fun, we compare your chances with the existence of an alien civilisation within 1000 light-years of Earth. 👽
For those who want to specify your own factors to use in the Drake equation for love, click on the advanced mode button. Doing this displays the following inputs:
Population of the UK — This is the base population figure that the calculator uses. You could change this value to reflect the population of a different country or region.
Gender — The decimal fraction of the population who is the gender you are looking for.
Location — The faction of the population of the correct gender who live near you.
Age — The proportion of these people who are age-appropriate for you.
University degree — The proportion of these people that have a university degree. If you don't mind about this option, enter 1.
Single rate — The fraction of these people who are single and ready to mingle.
Once you have filled in all of these fields, the calculator will display the results using these inputs. Note that this result still takes into account your answers for your attractiveness, social skills, and your perceived attractiveness of people.
To switch back to basing the results on the standard mode of the calculator, remove one of the advanced inputs.
💡 If you have changed the gender or location factors, you'll need to re-select these options.
The original Drake equation (as formulated by the British astronomer Dr Frank Drake in 1961) uses Bayesian probability to calculate the number of alien civilisations in our galaxy with whom we could communicate. It famously states:
G = R\cdot f_p\cdot n_e\cdot f_\ell\cdot f_i\cdot f_c \cdot L
The result of this equation,
G
, is the number of contactable alien civilisations in our galaxy. The other values represent increasingly specific conditions (like
f_\ell
, the fraction of Earth-like planets that develop some form of life). This way, the number of alien civilisations that ought to exist is narrowed down.
Using his formula, Dr. Drake estimated that 10,000 alien civilisations may exist in our galaxy alone. While that sounds like an exciting number, there are more than 200 billion stars in our Milky Way. So, the odds of pointing your telescope at a random star and seeing E.T. wave back at you is around 0.000005%. And that's if Dr. Drake's parameter estimates were even correct.
Many years after Dr Drake's equation became famous, Peter Backus realised that a girlfriend might just be even more rare than an alien civilisation. Using the mathematical principles behind the Drake equation, Backus estimated the size of his dating pool and published his findings in a darkly humourous paper. His girlfriend-finding equation is:
G = R \cdot f_G \cdot f_L \cdot f_A \cdot f_U \cdot f_B \cdot L
Each variable whittles away at the population based on Backus's own criteria for a life partner — with the same maths as the Drake equation. The variables in the equation are:
G
Your number of potential partners.
R
The UK's population growth.
f_G
The fraction of the UK's population who are the appropriate gender.
f_L
The fraction of men/women (as per your preference) living in your city.
f_A
The fraction of men/women in your city who are in your preferred age range.
f_U
The fraction of age-appropriate men/women in your city with a university education (optional).
f_B
The fraction of university-educated, age-appropriate men/women in your city who you find physically attractive.
L
Your own age.
By looking at data for the empirical parameters and estimating the rest, Backus found that 10,510 people meet his criteria. You might think it doesn't sound too bad — until he realised he didn't account for whether these people find him attractive, whether they are single, and whether they'd get along. After multiplying these extra fractions (
f_Q
f_S
f_C
respectively) at the end of his equation for love, the number of potential girlfriends dropped down to 26. In the whole of London. "Not great," as he had put it himself.
💡 Dr. Drake (who is still alive today) has yet to make any comments on Backus's adaptation of his work.
In the rom-com How to Be Single, Allison Brie follows a similar method as Backus.
As he was so disheartened to find out, Backus's odds of finding a girlfriend were extremely low. And yet, a mere three years after publishing his paper, Backus got married to a friend of a friend he met at a dinner party. One of his 26 viable partners was only two degrees of separation away!
Girlfriends (unlike aliens) aren't science fiction. We don't know whether Backus's wife Rose matches all his criteria, but one thing is certain: if this pessimistic mathematician can find love against the odds he calculated himself, then so can you!
Love can be anywhere, and more people find it every day. There were more than 235,000 new marriages in the UK in 2018, and around 51% of the nation were in a marriage or civil partnership in 2020.
There are plenty of ways to beat the odds presented by the Drake equation for love, as Backus told TODAY himself.
🙋 "Keep looking, and spend a lot of time hanging out in places where other people who satisfy your criteria hang out. Go to bars, go to music shows, go to places where people hang out, and increase the probability. That's what the equation shows." — Peter Backus
In response to the bleak outlook of Backus's paper, love gurus and mathematicians alike shared their tips to find love:
Relax your criteria. Based on her statistical studies, Dr Hanna Fry suggests that Backus underestimated many values in the equation when he reached his conclusion. She claims that if Backus was less picky (e.g., if he increased
f_B
from his original 5% to 20%), he would increase his original estimate from 26 to 832.
Dr Fry had also found that the happiest and longest lasting marriages are ones with "low negativity thresholds" — where conflict and anger subsides quickly. For this reason, the mathematician Bobby Seagull encourages stress-testing your potential partners and eliminate any that are easy to anger or argumentative. Perhaps bring up the latest about Boris?
Put yourself out there. As Backus suggested, the more you expose yourself to the dating world, the greater your chances of meeting someone you can grow old with. It's just statistics — and common sense. And while Tinder has its hidden algorithms, there's no mathematical way to predict who you'll click with — you just have to take the plunge and go looking for someone yourself.
💡 Backus and his peers presumably have no tips for finding aliens, though.
Steven Wooding and Rijk de Wet
What are your chances of finding love 💕 in time for this year’s Valentine’s?
Let's use the Drake equation (a concept from cosmology) to calculate how many potential perfect partners are out there for you!
Location 🏙️
Your attractiveness (fQ)
Social skills (fC)
Gender ♂♀
University educated 🎓
% you find attractive
Use advanced mode to directly modify the factors of the Drake equation for love.
You can also use our Dating Theory calculator to see how many dates to go on.
Disclaimer: Some parameters used in the equation are subjective and may not correspond well with reality. Also, where there are gaps in data coverage, general averages are used. May the odds be ever in your favour.
AudiobooksBeach Price IndexHammock hang… 11 more
Ring size converter allows you to calculate the equivalent of your ring size in different measures 💍
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How to convert cups to oz?
The must have volume converters
Our cups to oz converter is a one-stop tool for you if you are looking to convert units of volume. The tool allows you to convert cups to oz and vice versa.
The conversion from cups to ounces can come in handy for many measurements. Whether you are making Sunday morning pancakes or baking a pizza for your next pizza party, you won't have to wonder how to convert oz to cups. We've got your back.
The cups to oz converter is a one-stop tool for you to convert the volume of any liquid. Whether it's water, oil, or milk, you can enter the amount in one unit and get the answer in the other.
Yup! it is that simple.
If you want to follow some steps on how to use the calculator, which won't be many, here is how you use it:
Input the value of the ingredients in cups.
As a result, the tool will display the amount in US fluid ounces (oz).
But wait, you can do that in reverse too. Enter the amount in oz and get the result in cups.
And to top it all, you have a list of other volume units to switch.
The volume conversion between various measurement units is independent of density, or any other factor and hence requires simple conversion methods.
Let's look at two ways to convert between cups and ounces.
Convert cups to oz
When you have a quantity measured in cups, and you wish to convert it to ounces, you may use the following formula:
\text{oz} = \text{cups} × 8
So, all you have to do is multiply the amount in cups by 8. You have your answer in US fluid ounces.
This implies that 1 cup is equal to 8 fl. oz..
For instance, you have 3 cups of oil, and you are curious to know how much is it in ounces. Multiply 3 by 8, and you have 24 fl. oz. of oil.
The next scenario is when you a quantity measured in ounces but you would like to know it in terms of cups. You may use the formula:
\text{cups} = \text{oz} / 8
You see, nothing is left to guessing. You have a fool-proof method to convert ounces to cups.
For example, you have
32\ \text{oz}
of cream and want to make sure if it is equal to
4\ \text{cups}
or not. You can easily divide 32 by 8 and confirm that 32 oz is indeed equal to 4 cups.
How many cups are there in 62 oz?
There are 7.7 cups in 62 oz.
The conversion from ounces (oz) to cups is so easy, you can do it at your fingertips.
cups = oz / 8
So, if you divide 62 by 8, you get your amount in cups, i.e, 7.7 cups.
How do I convert 3 cups to ounces?
3 cups are equal to 24 ounces.
The simplest way to convert cups to ounces is to multiply the amount in cups by 8, and you have your result in ounces.
Hence the formula is:
oz = cups × 8
You can even do this calculation the other way round if you have a quantity in ounces and convert it to cups. All you have to do is divide the amount in ounces by 8.
Are there 16 ounces in a cup?
No, there are 8 ounces in a cup. That means 16 ounces are actually equal to 2 cups.
The formula to convert from oz(ounces) to cups is:
No matter what quantity you have in fluid ounces you can convert it to cups by diving it by 8.
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2022 A small closed convex projective 4-manifold via Dehn filling
Gye-Seon Lee, Ludovic Marquis, Stefano Riolo
Gye-Seon Lee,1 Ludovic Marquis,2 Stefano Riolo3
1Department of Mathematics, Sungkyunkwan University, Suwon, South Korea
2Univ Rennes, CNRS, IRMAR - UMR 6625, F-35000 Rennes, France
3Dipartimento di Matematica, Università di Pisa, Italy
In order to obtain a closed orientable convex projective
4
-manifold with small positive Euler characteristic, we build an explicit example of convex projective Dehn filling of a cusped hyperbolic
4
-manifold through a continuous path of projective cone-manifolds.
Gye-Seon Lee. Ludovic Marquis. Stefano Riolo. "A small closed convex projective 4-manifold via Dehn filling." Publ. Mat. 66 (1) 369 - 403, 2022. https://doi.org/10.5565/PUBLMAT6612215
Received: 17 June 2020; Accepted: 6 April 2021; Published: 2022
Primary: 22E40 , 53A20 , 53C15 , 57M50 , 57N16 , 57S30
Keywords: cone-manifold , Dehn filling , Euler characteristic , Hilbert geometry , hyperbolic 4-manifold , real projective structure
Gye-Seon Lee, Ludovic Marquis, Stefano Riolo "A small closed convex projective 4-manifold via Dehn filling," Publicacions Matemàtiques, Publ. Mat. 66(1), 369-403, (2022)
|
Reviewed by Hanna Pamuła, PhD candidate, Steven Wooding, Bogna Szyk and Rijk de Wet
About the picture frame calculator
How to measure a picture frame using our picture frame size calculator
Picture frame calculation formulas
Examples for manually calculating custom picture frame sizes
Collage picture frame sizes
Picture frame size standards
Our picture frame calculator can help you calculate your favorite photos' required frame size. It also calculates the picture mat size and the inner and outer frame dimensions to help you make the best custom picture frames.
There is so much happening around us. Times are changing, and we are growing every day. Amidst all this hustle and bustle, we shouldn't lose sight of what's essential and meaningful in our lives.
Only by capturing every moment and making the best of every arising opportunity can we truly live the best version of our lives.
And in those times, we should frame our experiences so we can share them with those who truly matter to us.
Thus, with our picture frame calculator, you'll be able to find the perfect frame sizes to preserve those beautiful moments and walk down memory lane every time you see them.
Our picture frame calculator helps you determine the most suitable frame sizes for your valued photo collection.
Upon entering the values, you will also be able to view the picture frames digitally in proper proportions to visualize how they'll look with your wedding 🤵👰 photos, or from when you took a blink-free photo to everyone's surprise, on vacation with your family 🏖️. Oh! And don't forget to think of some from your late-night pizza parties 🍕.
Our picture frame calculator has the following salient features:
🔲 You can get the area of a picture frame in the format of its inner and outer dimensions by entering the picture width, height and the border width. This border width is the picture framing mold or the molding width.
You can also add or calculate the picture mat size 🔳 to give it a more polished look. Mat size is the whitespace between your picture and the frame, also called picture frame spacing.
And lastly, you can find the molding length to help measure custom frame sizes. The molding length is the total size of the picture framing mold.
Using these features, you'll finally be able to make your frames and display your favorite pictures or artwork illustrations with ease.
You can calculate the picture frame size and mat for your photo frame in a few easy steps:
Choose your unit of measurement and input your molding width for your frame in the border width field, e.g., 1 inch.
Then enter the height of your picture in the picture height field, e.g., 12 inches.
And the width of your picture in picture width field, e.g., 12 inches.
Image describing picture height and width.
And you'll get the area of your picture frame in the form of its inner and outer dimensions, i.e., 12 and 14 inches respectively, along with the total molding length required for the frame, i.e., 56 inches.
💡 The total length of the frame, a.k.a mold size, can vary based on joint types and miter angles. Our calculator computes picture frame spacing based on 45° miter angles with 4 corners.
Enter your desired amount of whitespace between the picture and the frame in the mat height and width fields, e.g., 6 inches. Or input your custom frame dimensions to find the required mat size.
Image describing mat height and width.
Now click the advanced mode for further frame adjustments.
Here you will see the rabbet width and a fudge factor.
The rabbet width is the frame area where your picture and mat fit when mounted. In other words, rabbet width is the difference between the viewable area and the actual size of your picture and mat in the frame.
And the fudge factor adds a little extra space inside the frame to account for marginal errors.
The standard fudge factor is usually around 0.05 inches.
Once you've updated these values, the calculator will automatically update the output fields, i.e., the inner and outer frame dimensions, and the required molding length, that is our picture frame's perimeter.
Image describing rabbet and frame dimensions.
Next, we will see how to calculate some of these values using different formulas.
We use the following formulas for calculating picture frames.
Calculating picture frame's molding length, a.k.a its perimeter:
F _{\text{perimeter}} = ( F _{\text{height}} + F _{\text{width}} ) × 2
Calculating picture frame's outer dimensions, a.k.a., its external height and width:
F _{\text{height}} = P _{\text{height}} + ( M _{\text{height}} + B ) × 2
F _{\text{width}} = P _{\text{width}} + ( M _{\text{width}} + B ) × 2
Calculating picture frame's inner dimensions, a.k.a its internal height and width:
^iF _{\text{height}} = P _{\text{height}} + ( M _{\text{height}} - R ) × 2
^iF _{\text{width}} = P _{\text{width}} + ( M _{\text{width}} - R ) × 2
F _{\text{perimeter}}
- Perimeter of the picture frame;
F _{\text{height | width}}
- Outer height or width of the frame;
^iF _{\text{height | width}}
- Inner height or width of the frame;
P _{\text{height}}
- Height of the picture or artwork;
P _{\text{width}}
- Width of the picture or artwork;
M _{\text{height}}
- Top and bottom mat height in the picture frame;
M _{\text{width}}
- Left and right mat width in the picture frame;
B
- Border or molding width of the picture frame; and
R
- Rabbet width in the picture framing mold.
🔎 Note: if you want the mat height
M _{\text{height}}
or width
M _{\text{width}}
to respectively have different top and bottom or left and right values, you can rewrite them as
\frac{M _{\text{top}} + M _{\text{bottom}}}{2}
\frac{M _{\text{left}} + M _{\text{right}}}{2}
Now we know how to measure a picture frame. Let's take a look at some examples in the next section.
Suppose we want to make a photo frame for a picture of 4 x 6 inches. Where we want the frame border to be 1 inch.
P _{\text{height}} = 4
P _{\text{width}} = 6
B = 1
Let's find the outer dimensions of our frame by placing the given values in our formulas.
F _{\text{height}} = 4 + ( 0 + 1 ) × 2
F _{\text{width}} = 6 + ( 0 + 1 ) × 2
M
with 0, because we don't have a mat.
Thus, our required frame height is 6 inches, and the frame width is 8 inches.
Now let's find the required molding length of our picture frame.
F _{\text{perimeter}} = ( 6 + 8 ) × 2
Thus we need a molding length of 28 inches to create a picture frame of the required specifications.
Let's take another example where we want to find the visible mat size for a picture frame of 16 x 20 inches with a border of 1 inch and a rabbet of 1/4 inches, to display a picture of 8 x 11 inches.
F _{\text{height}} = 16
F _{\text{width}} = 20
B = 1
R = 1/4
P_{\text{height}} = 8
P _{\text{width}} = 11
We can obtain the inner frame size from the outer frame as follows:
^iF _{\text{height}} = F _{\text{height}} - B × 2
^iF _{\text{width}} = F _{\text{width}} - B × 2
^iF _{\text{height}} = 14
i.e., 16 - 1 × 2
^iF _{\text{width}} = 18
These values are generally given on the frames.
Rearranging the inner frame formula, we get:
M _{\text{height}} = \frac {^iF _{\text{height}} - P _{\text{height}}}{2} + R
M _{\text{width}} = \frac {^iF _{\text{width}} - P _{\text{width}}}{2} + R
Placing the values in our formula, we get the required mat height and width for each side.
M _{\text{height}} = \frac {14 - 8}{2} + \frac{1}{4}
M _{\text{height}} = \frac {18 - 11}{2} + \frac{1}{4}
Thus, our visible mat height is 3.25 inches and mat width is 3.75 inches.
🙋 For additional help, check out our image ratio and resolution scale calculator to help match the size of your digital images with your picture frames.
Can a picture 📷 really be worth a thousand words? Absolutely, when you have beautiful memories associated with them. And to top it off, what better way to compliment those feelings than by expressing them in personalized collage picture frames.
We've added a fun section for you to try out in our calculator. Head over to Find the frame size for your: and select collage. There you can enter up to 5 picture sizes to design your collage. Also, once you've designed your collage, you can switch back to picture mode to enter each picture's size to get your framing details.
To give you some idea for your picture frame sizes 🖼️, here are some height x width standards.
Just as reels 🎞 and black & white pictures 👥 are considered vintage, someday, these colored picture frames in golden ratio on our walls and bedside tables may also become obsolete, getting replaced by digital picture frames. The artwork in our living room will have animated displays, where we'll remotely upload our latest collection to share with our viewers. We may even have paid subscriptions to beautiful sceneries or patterns ⚜️ working as our accent walls, and casually trade digital artwork as NFTs (non-fungible tokens).
But until then, having printed albums and static photo frames is our best chance at preserving the beautiful experiences and memories 💕 we capture in our lifetimes.
Find the frame size for your:
Picture mat dimensions
Glaze size, i.e., picture + mat size
Glaze height
Glaze width
Inner frame height
Inner frame width
Outer frame height
Outer frame width
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Lumens to Watts Calculator - LED, Bulbs, Tubes
Features of lumens to watts calculator
How to calculate lumens to watts using this calculator
Light bulbs efficiency comparison chart
How to calculate the wattage of LED lights and other bulbs manually
How to compare wattage of LED to incandescent and other bulbs
You can use our lumens to watts calculator to calculate the efficiency of different bulbs and low wattage LEDs based on their brightness to power ratio, known as luminous efficacy, where the brightness of the light is in lumens and its power, in wattage.
You can then compare your LED bulb wattage with other light bulbs to observe how much electricity you'll save. And to further reduce your electricity bill, you may try our outstanding appliance wattage calculator.
Furthermore, we will tell you how to calculate lumens to watts using a formula and show you a comparison chart of LED wattage consumption against other bulbs.
We can use this lumens to watts calculator to find the required LED bulb wattage, i.e., the power needed for our light bulb to produce the specified amount of visible light, based on its luminous efficacy.
You can also use it to find the luminous flux, i.e., the brightness of light that our bulb produces with a given amount of power and luminous efficacy.
Finally, we can find the luminous efficacy, i.e., how efficiently our light bulb produces visible light based on its power to Lumen ratio.
Here's how we use the lumens to wattage calculator for LED lights.
Enter the luminous efficacy of your light source. The default is at 100 lm/W, primarily for LEDs.
🔎 You can find your value from our light bulbs efficiency chart below if you have a different light source.
Enter the brightness of your LED in lumens or nits, e.g., 1200 lumens or 350.25 nits. It is usually mentioned on the packaging of the bulb or you know the lumens of the old bulb you want to replace.
💡 Nit generally measures the brightness of digital displays, such as mobiles, computers, or TVs. For reference, the sun at noon measures about 1.6 billion nits.
Once you have entered your bulbs luminous efficacy and brightness, you'll determine how much power or wattage your light source requires.
Thus, with a luminous efficacy of 100 lm/W, for producing 1200 lumens, the required power is 12 watts.
The following table shows a range of luminous efficacies for different light bulbs:
Luminous efficacies of different light bulbs.
Tungsten Incandescent bulbs
Because different manufacturers have different standards, there are variations in lumens to watts ratio for the same light sources.
You can also try our lighting calculator to determine how much lighting you'll require based on the dimensions of your surface.
Based on their brightness and efficiency, we use the following formula when we want to find the required wattage of our low-wattage LED bulbs:
P _{(\text W)} = \varPhi _{V (\text{lm})} / \eta _{(\text{lm/W})}
P
– Power required, in watts;
\varPhi _{V}
– Luminous flux or the brightness in lumens (lm); and
\eta
– Luminous efficacy or the efficiency of producing visible light in lumens per watt (lm/W).
As an example, let's find the required wattage of an LED that has a luminous efficacy of 120 lm/W, and produces 1200 lumens:
P _{(\text W)} = 1200 / 120 = 10
Thus, our LED requires 10 watts for producing 1200 lumens.
💡 Standard LED luminous efficacy is between 80 and 120. However, some specific LEDs take this luminous efficacy up to 200 lm/W.
Let's take another example where we wish to find the luminous efficacy of our bulb, which consumes 15 watts and produces 1000 lumens:
In order to find its luminous efficacy, we rearrange the formula as follows:
\eta _{(\text{lm/W})} = \varPhi _{V (\text{lm})} / P _{(\text W)}
\eta = 1000 / 15
Thus, our bulb's luminous efficacy is 66.67 lm/W.
In case there is no mention of wattage on a device, we multiply its voltage by the electric current in amperes, i.e.,
\text{Wattage = Voltage × Ampere}
Here's an LED wattage chart comparing a low-wattage LED bulb of 12 W at 100 lm/W against other bulb types with different luminous efficacies.
Power consumption of LED vs other light bulbs.
Tungsten Incandescent bulb
Requires almost 7 times more power
Requires almost 1.7 times more power
Sulfur plasma light
✅ The chart tells us that the equivalent wattage of an LED to an incandescent light bulb is almost 7 times less for producing the same amount of light.
Disclaimer: This LED wattage chart only represents approximate values, as different manufacturers have different standards.
How do I calculate the wattage of an LED strip?
You can multiply the required voltage by amperes (current) to obtain the wattage.
Or you can divide the lumens, i.e., the brightness of the LED strip, by its luminous efficacy to obtain the wattage.
η = lm / W
η – Luminous efficacy;
lm – Brightness in lumens; and
W – Power in wattage.
How do I convert 500 watts to lumens?
To convert 500 watts to lumens:
Find the type of bulb to determine its luminous efficacy.
Then multiply the wattage of the bulb with its luminous efficacy to obtain lumens.
For example, if our 500 watts bulb has a luminous efficacy of 40 lm/W: The amount of light it can produce will be 500 × 40 = 20,000 lumens.
Can we use higher wattage LED bulbs?
Yes, as long as your LED wattage doesn't exceed the maximum recommended power load on your fixture or wiring, you can use higher wattage LED bulbs. Be careful not to overload, as that may damage your wiring due to excessively generated heat and cause short-circuiting before catching fire.
How does LED wattage compare to incandescent light bulbs?
The wattage consumption of LED is roughly 7 to 10 times less for producing the same amount of light compared to incandescent light bulbs.
Divide the lumens, i.e., the brightness of a light bulb, by its luminous efficacy to obtain this wattage:
watts = lumens / luminous efficacy
Why are bulb lighting colors in Kelvin?
The bulb light colors artificially mimic the Kelvin temperature of a metal object when it’s heated, ranging from 1000 K to 10,000 K.
Lower numbers have warmer tones, while higher numbers have cooler tones.
We generally use between 2700 K to 4100 K lights in our daily lives.
Bulb's efficiency for producing visible light
Luminous efficacy η
Bulb's brightness
The jacket size calculator will help you pick a jacket or coat that fits you perfectly.
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Section 6-6 - Rationalizing Substitutions - Maple Help
Home : Support : Online Help : Study Guides : Calculus : Chapter 6 - Techniques of Integration : Section 6-6 - Rationalizing Substitutions
Section 6.6: Rationalizing Substitutions
Table 6.6.1 lists two common types of substitutions that turn an integrand into a rational function of a single variable, in which case the method of partial fractions would apply.
\sqrt[n]{g\left(x\right)}
u=\sqrt[n]{g\left(x\right)}
u=g\left(x\right)
{u}^{n}=g\left(x\right)
\sqrt[n]{g\left(x\right)}
\sqrt[m]{g\left(x\right)}
u=\sqrt[k]{g\left(x\right)}
k
= least common multiple of
m
n
Rational function of
\mathrm{sin}\left(x\right)
\mathrm{cos}\left(x\right)
z=\mathrm{tan}\left(\frac{x}{2}\right),\mathrm{dx}=\frac{2 \mathrm{dz}}{1+{z}^{2}},\mathrm{sin}\left(x\right)=\frac{2 z}{1+{z}^{2}},\mathrm{cos}\left(x\right)=\frac{1-{z}^{2}}{1+{z}^{2}}
Table 6.6.1 Common rationalizing substitutions
The consequences of the substitution
z=\mathrm{tan}\left(x/2\right)
listed in Table 6.6.1 hinge on Figure 6.6.1 and the calculations to its right.
z=\mathrm{tan}\left(x/2\right)
and Figure 6.6.1 it follows that
\mathrm{sin}\left(\frac{x}{2}\right)=\frac{z}{\sqrt{1+ {z}^{2}}}
\mathrm{cos}\left(\frac{x}{2}\right)=\frac{1}{\sqrt{1+ {z}^{2}}}
From the trig identities
\mathrm{sin}\left(x\right)=2 \mathrm{sin}\left(\frac{x}{2}\right)\mathrm{cos}\left(\frac{x}{2}\right)
\mathrm{cos}\left(x\right)={\mathrm{cos}}^{2}\left(\frac{x}{2}\right)-{\mathrm{sin}}^{2}\left(\frac{x}{2}\right)
\mathrm{sin}\left(x\right)=\frac{2 z}{1+ {z}^{2}}
\mathrm{cos}\left(x\right)=\frac{1-{z}^{2}}{1+ {z}^{2}}
x=2 \mathrm{arctan}\left(z\right)
\mathrm{dx}=\frac{2 \mathrm{dt}}{1+{z}^{2}}
z=\mathrm{tan}\left(\frac{x}{2}\right),z∈\left(-\mathrm{π},\mathrm{π}\right)
∫\frac{x}{\sqrt{x-1}} \mathit{ⅆ}x
∫\frac{1}{{x}^{1/3}+{x}^{1/4}} \mathit{ⅆ}x
∫\frac{1}{2 \mathrm{sin}\left(x\right)+3 \mathrm{cos}\left(x\right)} \mathit{ⅆ}x
∫\frac{1}{2 \mathrm{sin}\left(x\right)+3 \mathrm{cos}\left(x\right)} \mathit{ⅆ}x
without making the rationalizing substitution
z=\mathrm{tan}\left(x/2\right)
Obtain a continuous antiderivative for
1/\left(7+2 \mathrm{sin}\left(x\right)+3 \mathrm{cos}\left(x\right)\right)
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cubefree - Maple Help
Home : Support : Online Help : Mathematics : Numbers : Type Checking : cubefree
check for a square-free integer
type/cubefree
check for a cube-free integer
type( expr, squarefree )
type( expr, cubefree )
The type( expr, squarefree ) command returns true if expr is a square-free integer, and false otherwise. An integer is square-free if it is not divisible by the square of any prime number.
An integer is square-free if, and only if, it is equal to its radical.
The type( expr, cubefree ) command returns true if expr is a cube-free integer, and false otherwise. An integer is cube-free if it is not divisible by the cube of any prime number.
\mathrm{type}\left(6,'\mathrm{squarefree}'\right)
\textcolor[rgb]{0,0,1}{\mathrm{true}}
\mathrm{type}\left(12,'\mathrm{squarefree}'\right)
\textcolor[rgb]{0,0,1}{\mathrm{false}}
\mathrm{type}\left(12,'\mathrm{cubefree}'\right)
\textcolor[rgb]{0,0,1}{\mathrm{true}}
\mathrm{type}\left(54,'\mathrm{cubefree}'\right)
\textcolor[rgb]{0,0,1}{\mathrm{false}}
The asymptotic natural density of square-free integers is equal to
\frac{6}{{\mathrm{\pi }}^{2}}=\frac{1}{\mathrm{zeta}\left(2\right)}
N≔100000:
\frac{\mathrm{nops}\left(\mathrm{select}\left(\mathrm{type},[\mathrm{seq}]\left(1..N\right),'\mathrm{squarefree}'\right)\right)}{N}
\frac{\textcolor[rgb]{0,0,1}{30397}}{\textcolor[rgb]{0,0,1}{50000}}
\mathrm{evalf}\left(%\right)
\textcolor[rgb]{0,0,1}{\mathrm{lastexception}}
\mathrm{evalf}\left(\frac{6}{{\mathrm{\pi }}^{2}}\right)
\textcolor[rgb]{0,0,1}{0.6079271016}
Similarly, the asymptotic natural density of the cube-free integers is
\frac{1}{\mathrm{zeta}\left(3\right)}
N≔10000:
\frac{\mathrm{nops}\left(\mathrm{select}\left(\mathrm{type},[\mathrm{seq}]\left(1..N\right),'\mathrm{cubefree}'\right)\right)}{N}
\frac{\textcolor[rgb]{0,0,1}{8319}}{\textcolor[rgb]{0,0,1}{10000}}
\mathrm{evalf}\left(%\right)
\textcolor[rgb]{0,0,1}{\mathrm{lastexception}}
\mathrm{evalf}\left(\frac{1}{\mathrm{\zeta }\left(3\right)}\right)
\textcolor[rgb]{0,0,1}{0.8319073727}
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40 CFR § 1066.635 - NMOG determination. | CFR | US Law | LII / Legal Information Institute
Subpart G - Calculations
40 CFR § 1066.635 - NMOG determination.
For vehicles subject to an NMOG standard, determine NMOG as described in paragraph (a) of this section. Except as specified in the standard-setting part, you may alternatively calculate NMOG results based on measured NMHC emissions as described in paragraphs (c) through (f) of this section.
(a) Determine NMOG by independently measuring alcohols and carbonyls as described in 40 CFR 1065.805 and 1065.845. Use good engineering judgment to determine which alcohols and carbonyls you need to measure. This would typically require you to measure all alcohols and carbonyls that you expect to contribute 1% or more of total NMOG. Calculate the mass of NMOG in the exhaust, mNMOG, with the following equation, using density values specified in § 1066.1005(f):
\begin{array}{c}{m}_{\mathrm{NMOG}}={m}_{\mathrm{NMHC}}-{\rho }_{\mathrm{NMHC}}·\sum _{i=1}^{N}\frac{{m}_{\mathrm{OHCi}}}{{\rho }_{\mathrm{OHCi}}}·{\mathrm{RF}}_{\mathrm{OHCi}\left[\text{THC-FID}\right]}+\sum _{i=1}^{N}{m}_{\mathrm{OHCi}}\\ \text{Eq. 1066.635-1}\end{array}
mNMHC = the mass of NMHC and all oxygenated hydrocarbon (OHC) in the exhaust, as determined using Eq. 1066.605-2. Calculate NMHC mass based on ρNMHC.
rNMHC = the effective C1-equivalent density of NMHC as specified in § 1066.1005(f).
mOHCi = the mass of oxygenated species i in the exhaust calculated using Eq. 1066.605-2.
rOCHi = the C1-equivalent density of oxygenated species i.
RFOHCi[THC-FID] = the response factor of a THC-FID to oxygenated species i relative to propane on a C1-equivalent basis as determined in 40 CFR 1065.845.
(b) The following example shows how to determine NMOG as described in paragraph (a) of this section for (OHC) compounds including ethanol (C2H5OH), methanol (CH3OH), acetaldehyde (C2H4O), and formaldehyde (CH2O) as C1-equivalent concentrations:
mNMHC = 0.0125 g
mCH3OH = 0.0002 g
mC2H5OH = 0.0009 g
mCH2O = 0.0001 g
mC2H4O = 0.00005 g
RFCH3OH[THC-FID] = 0.63
RFC2H5OH[THC-FID] = 0.75
RFCH2O[THC-FID] = 0.00
RFC2H4O[THC-FID] = 0.50
rNMHC-liq = 576.816 g/m 3
rCH3OH = 1332.02 g/m 3
rC2H5OH = 957.559 g/m 3
rCH2O = 1248.21 g/m 3
rC2H4O = 915.658 g/m 3
\begin{array}{c}{m}_{\mathrm{NMOG}}=0.0125-576.816·\left(\genfrac{}{}{0}{}{\frac{0.0002}{1332.02}·0.63+\frac{0.0009}{957.559}·0.75+}{\frac{0.0001}{1248.1}+0.00+\frac{0.00005}{915.658}+0.5}\right)+\\ 0.0002+0.0009+0.0001+0,00005\\ {m}_{\mathrm{NMOG}}=0.013273\end{array}
(c) For gasoline containing less than 25% ethanol by volume, you may calculate NMOG from measured NMHC emissions as follows:
(1) For hot-start and hot-running test cycles or intervals other than the FTP, you may determine NMOG based on the NMHC emission rate using the following equation:
\begin{array}{c}{e}_{\mathrm{NMOGh}}={e}_{\mathrm{NMHCh}}·1.03\\ \text{Eq. 1066.635-2}\end{array}
eNMOGh = mass emission rate of NMOG from the hot-running test cycle.
eNMHCh = mass emission rate of NMHC from the hot-running test cycle, calculated using rNMHC-liq.
eNMHCh = 0.025 g/mi
eNMOGh = 0.025 · 1.03 = 0.026 g/mi
(2) You may determine weighted composite NMOG for FTP testing based on the weighted composite NMHC emission rate and the volume percent of ethanol in the fuel using the following equation:
\begin{array}{c}{e}_{\mathrm{NMOGcomp}}={e}_{\mathrm{NMHCcomp}}·\left(1.0302+0.0071·{\mathrm{VP}}_{\mathrm{EtOH}}\right)\\ \text{Eq. 1066.635-3}\end{array}
eNMOGcomp = weighted FTP composite mass emission rate of NMOG.
eNMHCcomp = weighted FTP composite mass emission rate of NMHC, calculated using rNMHC-liq.
VPEtOH = volume percentage of ethanol in the test fuel. Use good engineering judgment to determine this value either as specified in 40 CFR 1065.710 or based on blending volumes, taking into account any denaturant.
eNMHCcomp = 0.025 g/mi
VPEtOH = 10.1%
eNMOGcomp = 0.025 · (1.0302 + 0.0071 · 10.1) = 0.0275 g/mi
(3) You may determine NMOG for the transient portion of the FTP cold-start test for use in fuel economy and CREE calculations based on the NMHC emission rate for the test interval and the volume percent of ethanol in the fuel using the following equation:
\begin{array}{c}{e}_{\mathrm{NMOG}-\mathrm{FTPct}}={e}_{\mathrm{NMH}-\mathrm{FTPct}}·\left(1.0246+0.0079·{\mathrm{VP}}_{\mathrm{EtOH}}\right)\\ \text{Eq. 1066.635-4}\end{array}
eNMOG-FTPct = mass emission rate of NMOG from the transient portion of the FTP cold-start test (generally known as bag 1).
eNMHC-FTPct = mass emission rate of NMHC from the transient portion of the FTP cold-start test (bag 1), calculated using rNMHC-liq.
eNMHC-FTPct = 0.052 g/mi
eNMOG-FTPct = 0.052 · (1.0246 + 0.0079 · 10.1) = 0.0574 g/mi
(4) You may determine NMOG for the stabilized portion of the FTP test for either the cold-start test or the hot-start test (bag 2 or bag 4) for use in fuel economy and CREE calculations based on the corresponding NMHC emission rate and the volume percent of ethanol in the fuel using the following equation:
\begin{array}{c}{e}_{\mathrm{NMOG}-\mathrm{FTPcs}-\mathrm{hs}}={e}_{\mathrm{NMHC}-\mathrm{FTPcs}-\mathrm{hs}}·\left(1.1135+0.001·{\mathrm{VP}}_{\mathrm{EtOH}}\right)\\ \text{Eq. 1066.635-5}\end{array}
eNMOG-FTPcs-hs = mass emission rate of NMOG from the stabilized portion of the FTP test (bag 2 or bag 4).
eNMHC-FTPcs-hs = mass emission rate of NMHC from the stabilized portion of the FTP test (bag 2 or bag 4), calculated using rNMHC-liq.
(5) You may determine NMOG for the transient portion of the FTP hot-start test for use in fuel economy and CREE calculations based on the NMHC emission rate for the test interval and the volume percent of ethanol in the fuel using the following equation:
\begin{array}{c}{e}_{\mathrm{NMOG}-\mathrm{FTPht}}={e}_{\mathrm{NMHC}-\mathrm{FTPht}}·\left(1.0195+0.0031·{\mathrm{VP}}_{\mathrm{EtOH}}\right)\\ \text{Eq. 1066.635-6}\end{array}
eNMOG-FTPht = mass emission rate of NMOG from the transient portion of the FTP hot-start test (bag 3).
eNMHC-FTPht = mass emission rate of NMHC from the transient portion of the FTP hot-start test (bag 3), calculated using rNMHC-liq.
(6) For PHEVs, you may determine NMOG based on testing over one full UDDS using Eq. 1066.635-3.
(d) You may take the following alternative steps when determining fuel economy and CREE under 40 CFR part 600 for testing with ethanol-gasoline blends that have up to 25% ethanol by volume:
(1) Calculate NMOG by test interval using Eq. 1066.635-3 for individual bag measurements from the FTP.
(2) For HEVs, calculate NMOG for two-bag FTPs using Eq. 1066.635-3 as described in 40 CFR 600.114.
(e) We consider NMOG values for diesel-fueled vehicles, CNG-fueled vehicles, LNG-fueled vehicles, and LPG-fueled vehicles to be equivalent to NMHC emission values for all test cycles.
(f) For all fuels not covered by paragraphs (c) and (e) of this section, manufacturers may propose a methodology to calculate NMOG results from measured NMHC emissions. We will approve adjustments based on comparative testing that demonstrates how to properly represent NMOG based on measured NMHC emissions.
[79 FR 23823, Apr. 28, 2014, as amended at 80 FR 9122, Feb. 19, 2015; 81 FR 74212, Oct. 25, 2016]
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Animation and Solution of Double Pendulum Motion - MATLAB & Simulink Example
Step 1: Define Displacement, Velocity, and Acceleration of Double Pendulum Masses
Step 2: Define Equations of Motion
Step 3: Evaluate Forces and Reduce System Equations
Step 4: Solve System Equations
Step 5: Create Animation of Oscillating Double Pendulum
This example shows how to model the motion of a double pendulum by using MATLAB® and Symbolic Math Toolbox™.
Solve the motion equations of a double pendulum and create an animation to model the double pendulum motion.
The following figure shows the model of a double pendulum. The double pendulum consists of two pendulum bobs and two rigid rods.
Describe the motion of the double pendulum by defining the state variables:
the angular position of the first bob
{\theta }_{1}\left(\mathit{t}\right)
the angular position of the second bob
{\theta }_{2}\left(\mathit{t}\right)
Describe the properties of the double pendulum by defining the variables:
the length of the first rod
{\mathit{L}}_{1}
the length of the second rod
{\mathit{L}}_{2}
the mass of the first bob
{\mathit{m}}_{1}
the mass of the second bob
{\mathit{m}}_{2}
the gravitational constant
\mathit{g}
For simplicity, ignore the masses of the two rigid rods. Specify all variables by using syms.
syms theta_1(t) theta_2(t) L_1 L_2 m_1 m_2 g
Define the displacements of the double pendulum in Cartesian coordinates.
x_1 = L_1*sin(theta_1);
y_1 = -L_1*cos(theta_1);
x_2 = x_1 + L_2*sin(theta_2);
y_2 = y_1 - L_2*cos(theta_2);
Find the velocities by differentiating the displacements with respect to time using the diff function.
vx_1 = diff(x_1);
vy_1 = diff(y_1);
Find the accelerations by differentiating the velocities with respect to time.
ax_1 = diff(vx_1);
ay_1 = diff(vy_1);
Define the equations of motion based on Newton's laws.
First, specify the tension of the first rod as
{\mathit{T}}_{1}
, and the tension of the second rod
{\mathit{T}}_{2}
syms T_1 T_2
Next, construct free-body diagrams of the forces that act on both masses.
Evaluate the forces acting on
{\mathit{m}}_{1}
. Define the equations of motion of the first bob by balancing the horizontal and vertical force components. Specify these two equations as symbolic equations eqx_1 and eqy_1.
eqx_1 = m_1*ax_1(t) == -T_1*sin(theta_1(t)) + T_2*sin(theta_2(t))
eqx_1 =
-{m}_{1} \left({L}_{1} \mathrm{sin}\left({\theta }_{1}\left(t\right)\right) {\left(\frac{\partial }{\partial t}\mathrm{ }{\theta }_{1}\left(t\right)\right)}^{2}-{L}_{1} \mathrm{cos}\left({\theta }_{1}\left(t\right)\right) \frac{{\partial }^{2}}{\partial {t}^{2}}\mathrm{ }{\theta }_{1}\left(t\right)\right)={T}_{2} \mathrm{sin}\left({\theta }_{2}\left(t\right)\right)-{T}_{1} \mathrm{sin}\left({\theta }_{1}\left(t\right)\right)
eqy_1 = m_1*ay_1(t) == T_1*cos(theta_1(t)) - T_2*cos(theta_2(t)) - m_1*g
eqy_1 =
{m}_{1} \left({L}_{1} \mathrm{sin}\left({\theta }_{1}\left(t\right)\right) \frac{{\partial }^{2}}{\partial {t}^{2}}\mathrm{ }{\theta }_{1}\left(t\right)+{L}_{1} \mathrm{cos}\left({\theta }_{1}\left(t\right)\right) {\left(\frac{\partial }{\partial t}\mathrm{ }{\theta }_{1}\left(t\right)\right)}^{2}\right)={T}_{1} \mathrm{cos}\left({\theta }_{1}\left(t\right)\right)-g {m}_{1}-{T}_{2} \mathrm{cos}\left({\theta }_{2}\left(t\right)\right)
{\mathit{m}}_{2}
. Define the equations of motion of the second bob by balancing the horizontal and vertical force components. Specify these two equations as symbolic equations eqx_2 and eqy_2.
eqx_2 = m_2*ax_2(t) == -T_2*sin(theta_2(t))
-{m}_{2} \left({L}_{1} \mathrm{sin}\left({\theta }_{1}\left(t\right)\right) {\left(\frac{\partial }{\partial t}\mathrm{ }{\theta }_{1}\left(t\right)\right)}^{2}+{L}_{2} \mathrm{sin}\left({\theta }_{2}\left(t\right)\right) {\left(\frac{\partial }{\partial t}\mathrm{ }{\theta }_{2}\left(t\right)\right)}^{2}-{L}_{1} \mathrm{cos}\left({\theta }_{1}\left(t\right)\right) \frac{{\partial }^{2}}{\partial {t}^{2}}\mathrm{ }{\theta }_{1}\left(t\right)-{L}_{2} \mathrm{cos}\left({\theta }_{2}\left(t\right)\right) \frac{{\partial }^{2}}{\partial {t}^{2}}\mathrm{ }{\theta }_{2}\left(t\right)\right)=-{T}_{2} \mathrm{sin}\left({\theta }_{2}\left(t\right)\right)
eqy_2 = m_2*ay_2(t) == T_2*cos(theta_2(t)) - m_2*g
{m}_{2} \left({L}_{1} \mathrm{cos}\left({\theta }_{1}\left(t\right)\right) {\left(\frac{\partial }{\partial t}\mathrm{ }{\theta }_{1}\left(t\right)\right)}^{2}+{L}_{2} \mathrm{cos}\left({\theta }_{2}\left(t\right)\right) {\left(\frac{\partial }{\partial t}\mathrm{ }{\theta }_{2}\left(t\right)\right)}^{2}+{L}_{1} \mathrm{sin}\left({\theta }_{1}\left(t\right)\right) \frac{{\partial }^{2}}{\partial {t}^{2}}\mathrm{ }{\theta }_{1}\left(t\right)+{L}_{2} \mathrm{sin}\left({\theta }_{2}\left(t\right)\right) \frac{{\partial }^{2}}{\partial {t}^{2}}\mathrm{ }{\theta }_{2}\left(t\right)\right)={T}_{2} \mathrm{cos}\left({\theta }_{2}\left(t\right)\right)-g {m}_{2}
Four equations of motion describe the kinematics of the double pendulum. Evaluate the forces acting on the rods and reduce the set of four equations to two equations.
The equations of motion have four unknowns:
{\theta }_{1}
{\theta }_{2}
{\mathit{T}}_{1}
{\mathit{T}}_{2}
. Evaluate the two unknowns
{\mathit{T}}_{1}
{\mathit{T}}_{2}
from eqx_1 and eqy_1. Use solve function to find
{\mathit{T}}_{1}
{\mathit{T}}_{2}
Tension = solve([eqx_1 eqy_1],[T_1 T_2]);
Substitute the solutions for
{\mathit{T}}_{1}
{\mathit{T}}_{2}
into eqx_2 and eqy_2.
eqRed_1 = subs(eqx_2,[T_1 T_2],[Tension.T_1 Tension.T_2]);
eqRed_2 = subs(eqy_2,[T_1 T_2],[Tension.T_1 Tension.T_2]);
The two reduced equations fully describe the pendulum motion.
Solve the system equations to describe the pendulum motion.
First, define the values for the masses in
\mathrm{kg}
, the rod lengths in
\mathrm{m}
, and the gravity in
\mathrm{m}/{\mathrm{s}}^{2}
(SI units). Substitute these values into the two reduced equations.
L_1 = 1;
L_2 = 1.5;
eqn_1 = subs(eqRed_1)
eqn_1 =
\begin{array}{l}\mathrm{cos}\left({\theta }_{1}\left(t\right)\right) {\sigma }_{1}-\frac{3 \mathrm{sin}\left({\theta }_{2}\left(t\right)\right) {\left(\frac{\partial }{\partial t}\mathrm{ }{\theta }_{2}\left(t\right)\right)}^{2}}{2}-\mathrm{sin}\left({\theta }_{1}\left(t\right)\right) {\left(\frac{\partial }{\partial t}\mathrm{ }{\theta }_{1}\left(t\right)\right)}^{2}+\frac{3 \mathrm{cos}\left({\theta }_{2}\left(t\right)\right) \frac{{\partial }^{2}}{\partial {t}^{2}}\mathrm{ }{\theta }_{2}\left(t\right)}{2}=-\frac{2 \mathrm{sin}\left({\theta }_{2}\left(t\right)\right) \left({\mathrm{cos}\left({\theta }_{1}\left(t\right)\right)}^{2} {\sigma }_{1}+{\mathrm{sin}\left({\theta }_{1}\left(t\right)\right)}^{2} {\sigma }_{1}+\frac{49 \mathrm{sin}\left({\theta }_{1}\left(t\right)\right)}{5}\right)}{\mathrm{cos}\left({\theta }_{1}\left(t\right)\right) \mathrm{sin}\left({\theta }_{2}\left(t\right)\right)-\mathrm{cos}\left({\theta }_{2}\left(t\right)\right) \mathrm{sin}\left({\theta }_{1}\left(t\right)\right)}\\ \\ \mathrm{where}\\ \\ \mathrm{ }{\sigma }_{1}=\frac{{\partial }^{2}}{\partial {t}^{2}}\mathrm{ }{\theta }_{1}\left(t\right)\end{array}
\begin{array}{l}\mathrm{cos}\left({\theta }_{1}\left(t\right)\right) {\left(\frac{\partial }{\partial t}\mathrm{ }{\theta }_{1}\left(t\right)\right)}^{2}+\frac{3 \mathrm{cos}\left({\theta }_{2}\left(t\right)\right) {\left(\frac{\partial }{\partial t}\mathrm{ }{\theta }_{2}\left(t\right)\right)}^{2}}{2}+\mathrm{sin}\left({\theta }_{1}\left(t\right)\right) {\sigma }_{1}+\frac{3 \mathrm{sin}\left({\theta }_{2}\left(t\right)\right) \frac{{\partial }^{2}}{\partial {t}^{2}}\mathrm{ }{\theta }_{2}\left(t\right)}{2}=\frac{2 \mathrm{cos}\left({\theta }_{2}\left(t\right)\right) \left({\mathrm{cos}\left({\theta }_{1}\left(t\right)\right)}^{2} {\sigma }_{1}+{\mathrm{sin}\left({\theta }_{1}\left(t\right)\right)}^{2} {\sigma }_{1}+\frac{49 \mathrm{sin}\left({\theta }_{1}\left(t\right)\right)}{5}\right)}{\mathrm{cos}\left({\theta }_{1}\left(t\right)\right) \mathrm{sin}\left({\theta }_{2}\left(t\right)\right)-\mathrm{cos}\left({\theta }_{2}\left(t\right)\right) \mathrm{sin}\left({\theta }_{1}\left(t\right)\right)}-\frac{49}{5}\\ \\ \mathrm{where}\\ \\ \mathrm{ }{\sigma }_{1}=\frac{{\partial }^{2}}{\partial {t}^{2}}\mathrm{ }{\theta }_{1}\left(t\right)\end{array}
The two equations are nonlinear second-order differential equations. To solve these equations, convert them to first-order differential equations by using the odeToVectorField function.
[V,S] = odeToVectorField(eqn_1,eqn_2);
The elements of the vector V represent the first-order differential equations that are equal to the time derivative of the elements of S. The elements of S are the state variables
{\theta }_{2}
\mathit{d}{\theta }_{2}/\mathrm{dt}
{\theta }_{1}
\mathit{d}{\theta }_{1}/\mathrm{dt}
. The state variables describe the angular displacements and velocities of the double pendulum.
\left(\begin{array}{c}{\theta }_{2}\\ {\mathrm{Dtheta}}_{2}\\ {\theta }_{1}\\ {\mathrm{Dtheta}}_{1}\end{array}\right)
Next, convert the first order-differential equations to a MATLAB function with the handle M.
M = matlabFunction(V,'vars',{'t','Y'});
Define the initial conditions of the state variables as [pi/4 0 pi/6 0]. Use the ode45 function to solve for the state variables. The solutions are a function of time within the interval [0 10].
initCond = [pi/4 0 pi/6 0];
sols = ode45(M,[0 10],initCond);
Plot the solutions of the state variables.
plot(sols.x,sols.y)
legend('\theta_2','d\theta_2/dt','\theta_1','d\theta_1/dt')
title('Solutions of State Variables')
ylabel('Solutions (rad or rad/s)')
Create the animation of the oscillating double pendulum.
First, create four functions that use deval to evaluate the coordinates of both pendulums from the solutions sols.
x_1 = @(t) L_1*sin(deval(sols,t,3));
y_1 = @(t) -L_1*cos(deval(sols,t,3));
x_2 = @(t) L_1*sin(deval(sols,t,3))+L_2*sin(deval(sols,t,1));
y_2 = @(t) -L_1*cos(deval(sols,t,3))-L_2*cos(deval(sols,t,1));
Next, create a stop-motion animation object of the first pendulum bob by using the fanimator function. By default, fanimator creates an animation object with 10 generated frames per unit time within the range of t from 0 to 10. Plot the coordinates by using the plot function. Set the x-axis and y-axis to be equal length.
fanimator(@(t) plot(x_1(t),y_1(t),'ro','MarkerSize',m_1*10,'MarkerFaceColor','r'));
Next, add the animation objects of the first rigid rod, the second pendulum bob, and the second rigid rod.
fanimator(@(t) plot([0 x_1(t)],[0 y_1(t)],'r-'));
fanimator(@(t) plot(x_2(t),y_2(t),'go','MarkerSize',m_2*10,'MarkerFaceColor','g'));
fanimator(@(t) plot([x_1(t) x_2(t)],[y_1(t) y_2(t)],'g-'));
fanimator(@(t) text(-0.3,0.3,"Timer: "+num2str(t,2)));
Use the command playAnimation to play the animation of the double pendulum.
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Section 5-5 - Surface of Revolution - Maple Help
Home : Support : Online Help : Study Guides : Calculus : Chapter 5 - Applications of Integration : Section 5-5 - Surface of Revolution
Section 5.5: Surface Area of a Surface of Revolution
Table 5.5.1 summarizes the contents of the "formula"
2 \mathrm{π} {∫}_{a}^{b}\mathrm{ρ} \mathit{ⅆ}s
that gives the surface area of a surface of revolution generated by rotating a curve about either a horizontal axis (
y=c
) or a vertical axis (
x=c)
. The distance of an arc-length element
\mathrm{ds}
from the axis of rotation is
\mathrm{ρ}
, so for curves defined either explicitly or parametrically, Table 5.5.1 lists the appropriate expressions for
\mathrm{ρ}
\mathrm{ds}
y=c
x=c
y=f\left(x\right)
\mathrm{ρ}=\left|y-c\right|
\left|f\left(x\right)-c\right|
\mathrm{ds}=\sqrt{1+{\left(\frac{\mathrm{df}}{\mathrm{dx}}\right)}^{2}}\mathrm{dx}
\mathrm{ρ}=\left|x-c\right|
\mathrm{ds}=\sqrt{1+{\left(\frac{\mathrm{df}}{\mathrm{dx}}\right)}^{2}}\mathrm{dx}
x=g\left(y\right)
\mathrm{ρ}=\left|y-c\right|
\mathrm{ds}=\sqrt{1+{\left(\frac{\mathrm{dg}}{\mathrm{dy}}\right)}^{2}}\mathrm{dy}
\mathrm{ρ}=\left|x-c\right|
\left|g\left(y\right)-c\right|
\mathrm{ds}=\sqrt{1+{\left(\frac{\mathrm{dg}}{\mathrm{dy}}\right)}^{2}}\mathrm{dy}
x=x\left(t\right)
y=y\left(t\right)
\mathrm{ρ}=\left|y-c\right|
\left|y\left(t\right)-c\right|
\mathrm{ds}=\sqrt{{\left(\frac{\mathrm{dx}}{\mathrm{dt}}\right)}^{2}+{\left(\frac{\mathrm{dy}}{\mathrm{dt}}\right)}^{2}}\mathrm{dt}
\mathrm{ρ}=\left|x-c\right|
\left|x\left(t\right)-c\right|
\mathrm{ds}=\sqrt{{\left(\frac{\mathrm{dx}}{\mathrm{dt}}\right)}^{2}+{\left(\frac{\mathrm{dy}}{\mathrm{dt}}\right)}^{2}}\mathrm{dt}
Table 5.5.1 Surface area of a surface of revolution:
2 \mathrm{π}{∫}_{a}^{b}\mathrm{ρ} \mathit{ⅆ}s
The arc-length element appears in the surface-area integrals because the surface area is defined as the limiting sum of the areas of approximating frustums of a cone. (A frustum of a cone is a segment cut off by two planes parallel to the base.) If
r
R
are respectively the smaller and larger radii of a frustum whose generator (lateral height) has length
L
, then the surface area of the frustum is
\mathrm{π} L \left(r+R\right)
. As the number of frustums increases,
\left(r+R\right)/2
\mathrm{ρ}
L
\mathrm{ds}
. Hence, the surface-area element becomes
2 \mathrm{π} \mathrm{ρ} \mathrm{ds}
, and the integrals in Table 5.5.1 follow.
Calculate the surface area of the surface of revolution formed when the graph of
y={x}^{2},x∈\left[0,1\right]
x
y={x}^{2},x∈\left[0,1\right]
, is rotated about the line
y=2
y={x}^{2},x∈\left[0,1\right]
y
y={x}^{2},x∈\left[0,1\right]
x=2
y={e}^{-x},x∈\left[0,2\right]
y
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Ideal source of hydraulic energy, characterized by pressure - MATLAB - MathWorks France
Hydraulic Pressure Source
Ideal source of hydraulic energy, characterized by pressure
The Hydraulic Pressure Source block represents an ideal source of hydraulic energy that is powerful enough to maintain specified pressure at its outlet regardless of the flow rate consumed by the system. Block connections T and P correspond to the hydraulic inlet and outlet ports, respectively, and connection S represents a control signal port. The pressure differential across the source
p={p}_{P}-{p}_{T}
where pP, pT are the gauge pressures at the source ports, is directly proportional to the signal at the control port S. The entire variety of Simulink® signal sources can be used to generate desired pressure variation profile.
The block positive direction is from port P to port T. This means that the flow rate is positive if it flows from P to T. The power generated by the source is negative if the source delivers energy to port P.
Hydraulic Constant Pressure Source | Hydraulic Pressure Sensor
|
Geometry/Differential Geometry/Introduction - Wikibooks, open books for an open world
Geometry/Differential Geometry/Introduction
< Geometry | Differential Geometry
Differential geometry studies geometry by considering differentiable paramaterizations of curves, surfaces, and higher dimensional objects. Prerequisites include vector calculus, linear algebra, analysis, and topology.
One goal of differential geometry is to classify and represent differentiable curves in ways which are independent of their paramaterization. For example consider the curve represented by
{\displaystyle y=3x}
{\displaystyle (x,y)=(t,3t)}
{\displaystyle (x,y)=(3t,9t)}
are different paramterizations, they both represent the same curve. More generally, we consider the slope of the curve
{\displaystyle 3={\frac {dy}{dx}}={\frac {dy}{dt}}\cdot {\frac {1}{\frac {dx}{dt}}}={\frac {3}{1}}}
We call this type of curve a line. We can even rotate, and move it around, but it is still a line. The goal of Differential Geometry will be to similarly classify, and understand classes of differentiable curves, which may have different paramaterizations, but are still the same curve.
By adding sufficient dimensions, any equation can become a curve in geometry. Therefore, the ability to discern when two curves are unique also has the potential for applications in distinguishing information from noise. There may be multiple ways of receiving the same information--in different paramterizations, but we want to distinguish if the information is actually unique.
Retrieved from "https://en.wikibooks.org/w/index.php?title=Geometry/Differential_Geometry/Introduction&oldid=3154731"
Book:Geometry
|
[short-fuse] Percent efficacy of J&J vaccine? | Metaculus
[short-fuse] Percent efficacy of J&J vaccine?
The single-dose adenovirus-vectored vaccine Ad26.COV2.S, which encodes the spike protein of SARS-CoV-2, is currently undergoing Phase III testing with the support of Johnson & Johnson. This randomized double-blind Phase III trial, known as ENSEMBLE, has enrolled 45,000 adult participants in multiple countries.
Johnson & Johnson has already completed a Phase I/II trial on its single-dose candidate. It says that it expects to report interim efficacy data from an analysis of the Phase III trial in early February 2021.
The primary outcome of the ENSEMBLE study is "Number of Participants with First Occurrence of Molecularly Confirmed Moderate to Severe/Critical Coronavirus Disease (COVID-19) with Seronegative Status". This primary endpoint is the same as was the case for the Pfizer and Moderna Phase III trials, which reported median vaccine efficacies of 95.0% and 94.1%, respectively. The AstraZeneca/Oxford vaccine, which is similar to the Johnson & Johnson vaccine in that it is also an adenoviral-vectored vaccine, has reported efficacy of 70.4%, though there are many issues with this final figure.
What will be the absolute efficacy of the single-dose Johnson & Johnson Ad26.COV2.S vaccine candidate according to the results of Phase III testing?
In this study, confirmed cases of COVID-19 are defined as molecularly confirmed (PCR positive) symptomatic cases of COVID-19 from 14 days post-vaccination.
This question resolves as the median estimate of the absolute vaccine efficacy of Ad26.COV2.S,
\frac{ARU-ARV}{ARU} \times 100
, where ARU is the disease attack rate in the unvaccinated group and ARV is the disease attack rate in the vaccinated group. The disease attack rate is the proportion of virologically confirmed (PCR positive) symptomatic cases of COVID-19 (WHO, 2017). Such a finding should be presented in a peer-reviewed research article that is authored and/or supported by Johnson & Johnson. If no median estimate is provided, we will resolve this question on the mean absolute vaccine efficacy.
This question will close retroactively to the date when the interim results expected in early February are released. However, it will not resolve on the basis of that efficacy data — rather, it will resolve on the basis of the final phase III efficacy data in a published peer-reviewed article.
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Vertical coverage diagram - MATLAB radarvcd - MathWorks Deutschland
radarvcd
Vertical Coverage Pattern with Specified Antenna Pattern
Plot Vertical Coverage Diagram For User-Specified Antenna
SurfaceRelativePermittivity
SurfaceHeightStandardDeviation
ElevationBeamwidth
AntennaPattern
PatternAngles
ElevationStepSize
Vertical Coverage Pattern
Vertical coverage diagram
[vcp,vcpangles] = radarvcd(freq,rfs,anht)
[vcp,vcpangles] = radarvcd(___,Name,Value)
radarvcd(___)
[vcp,vcpangles] = radarvcd(freq,rfs,anht) calculates the vertical coverage pattern of a narrowband radar antenna. The Vertical Coverage Pattern is the range of the radar vcp as a function of elevation angle vcpangles. The vertical coverage pattern depends on three parameters: maximum free-space detection range of the radar rfs, the radar frequency freq, and the antenna height anht.
[vcp,vcpangles] = radarvcd(___,Name,Value) allows you to specify additional input parameters using name-value arguments. You can specify multiple name-value arguments in any order.
radarvcd(___) displays the vertical coverage diagram for a radar system. The plot is the locus of points of maximum radar range as a function of target elevation. This plot is also known as the Blake chart. To create this chart, radarvcd invokes the function blakechart using default parameters. To produce a Blake chart with different parameters, first call radarvcd to obtain vcp and vcpangles. Then, call blakechart with user-specified parameters. This syntax can use any of the previous syntaxes.
Set the frequency to 100 MHz, the antenna height to 10 m, and the free-space range to 200 km. The antenna pattern is a sinc function with 45° half-power width. The surface height standard deviation is set to
1/2\sqrt{2}
m. The antenna tilt angle is set to 0°, and the field polarization is horizontal.
ntn = phased.SincAntennaElement('Beamwidth',45);
pat = ntn(freq,pat_angles');
[vcp,vcpangles] = radarvcd(freq,rng_fs,ant_height,...
'PatternAngles',pat_angles,...
'TiltAngle',tilt_ang,'SurfaceHeightStandardDeviation',1/(2*sqrt(2)));
Call radarvcd with no output arguments to display the vertical coverage pattern.
radarvcd(freq,rng_fs,ant_height,...
Alternatively, use the radarvcd output arguments and the blakechart function to display the vertical coverage pattern to a maximum range of 400 km and a maximum height of 50 km. Customize the Blake chart by changing the color.
blakechart(vcp,vcpangles,400,50, ...
'FaceColor',[0.8500 0.3250 0.0980],'EdgeColor',[0.8500 0.3250 0.0980])
Plot the range-height-angle curve (Blake chart) for a radar with a user-specified antenna pattern.
Define a sinc-function antenna pattern with a half-power beamwidth of 90 degrees. The radar transmits at 100 MHz.
Specify a free-space range of 200 km. The antenna height is 10 meters, the antenna tilt angle is zero degrees, and the surface roughness is one meter.
Create the radar range-height-angle plot.
'TiltAngle',tilt_ang,...
freq — Radar frequency
real-valued scalar less than 10 GHz
Radar frequency, specified as a real-valued scalar less than 10 GHz (1010 Hz).
rfs — Free-space range
Free-space range, specified as a positive scalar or vector. rfs is the calculated or assumed free-space range for a target or for a one-way RF system at which the field strength would have a specified value. Range units are set by the RangeUnit name-value argument.
anht — Radar antenna height
Radar antenna height, specified as a real-valued scalar. The height is referenced to the surface. Height units are set by the HeightUnit name-value argument.
Example: 'HeightUnit','km'
Radar range units denoting kilometers, nautical miles, miles, feet, meters, or kilofeet. This argument specifies the units for the free-space range argument rfs and the output vertical coverage pattern vcp.
HeightUnit — Antenna height units
'm' (default) | 'nmi' | 'mi' | 'km' | 'ft' | 'kft'
Antenna height units denoting meters, nautical miles, miles, kilometers, feet, or kilofeet. This argument specifies the units for the antenna height anht and the 'SurfaceRoughness' argument.
Polarization — Transmitted wave polarization
'H' (default) | 'V'
Transmitted wave polarization, specified as 'H' for horizontal polarization or 'V' for vertical polarization.
SurfaceRelativePermittivity — Complex permittivity of reflecting surface
frequency dependent model (default) | complex-valued scalar
Complex permittivity (dielectric constant) of the reflecting surface, specified as a complex-valued scalar. The default value of this argument depends on the value of freq. radarvcd uses a seawater model that is valid for frequencies up to 10 GHz.
SurfaceHeightStandardDeviation — Standard deviation of surface height
Standard deviation of surface height, specified as a nonnegative real-valued scalar. A value of 0 indicates a smooth surface. Use 'HeightUnit' to specify the units of height.
The surface height standard deviation relates to the crest-to-trough "surface roughness" height through
Surface roughness = 2 × √2 × Surface height standard deviation.
SurfaceSlope — Surface slope
Surface slope in degrees, specified as a nonnegative scalar. This value is expected to be 1.4 times the RMS surface slope. Given the condition that
2 × GRAZ/β0 < 1,
where GRAZ is the grazing angle of the geometry specified in degrees and β0 is the surface slope, the effective surface height standard deviation in meters is calculated as
Effective HGTSD = HGTSD × (2 × GRAZ/β0)1/5.
This calculation better accounts for shadowing. Otherwise, the effective height standard deviation is equal to HGTSD. This argument defaults to 0, indicating a smooth surface.
VegetationType — Vegetation type
'None' (default) | 'Trees' | 'Brush' | 'Weeds' | 'Grass'
Surface vegetation type, specified as 'Trees', 'Weeds', and 'Brush' are assumed to be dense vegetation. 'Grass' is assumed to be thin grass. Use this argument when using the function on surfaces different from the sea.
ElevationBeamwidth — Half-power elevation beamwidth
10 (default) | scalar between 0° and 90°
Half-power elevation beamwidth in degrees, specified as a scalar between 0° and 90°. The elevation beamwidth is used in the calculation of a sinc antenna pattern. The default antenna pattern is symmetric with respect to the beam maximum and is of the form sin(u)/u. The parameter u is given by u = k sin(θ), where θ is the elevation angle in radians and k is given by k = x0 / sin(π × ELBW/360), where ELBW is the half-power elevation beamwidth and x0 ≈ 1.3915573 is a solution of sin(x) = x/√2.
AntennaPattern — Antenna elevation pattern
Antenna elevation pattern, specified as a real-valued column vector. Values for 'AntennaPattern' must be specified together with values for 'PatternAngles'. Both vectors must have the same size. If both an antenna pattern and an elevation beamwidth are specified, radarvcd uses the antenna pattern and ignores the elevation beamwidth value. This argument defaults to a sinc antenna pattern.
Example: cosd([–90:90])
PatternAngles — Antenna pattern elevation angles
Antenna pattern elevation angles specified as a real-valued column vector. The size of the vector specified by PatternAngles must be the same as that specified by AntennaPattern. Angle units are expressed in degrees and must lie between –90° and 90°. In general, the antenna pattern should fill the whole range from –90° to 90° for the coverage to be computed properly.
TiltAngle — Antenna tilt angle
Antenna tilt angle, specified as a real-valued scalar. The tilt angle is the elevation angle of the antenna with respect to the surface. Angle units are expressed in degrees.
Effective Earth radius in meters, specified as a positive scalar. The effective Earth radius is an approximation used for modeling refraction effects in the troposphere. The default value calculates the effective Earth radius using a refraction gradient of -39e-9, which results in approximately 4/3 of the real Earth radius.
MaxElevation — Maximum elevation angle
Maximum elevation angle, specified as a real-valued scalar. The maximum elevation angle is the largest angle for which the vertical coverage pattern is calculated. Angle units are expressed in degrees.
MinElevation — Minimum elevation angle
Minimum elevation angle, specified as a real-valued scalar. The minimum elevation angle is the smallest angle for which the vertical coverage pattern is calculated. Angle units are expressed in degrees.
ElevationStepSize — Elevation angle increment
Elevation angle increment, specified as a positive scalar in degrees. The elevation vector goes from the minimum value specified in MinElevation and the maximum value specified in MaxElevation in increments of ElevationStepSize. The default value of this argument is given by
Δ = 885.6/(π × fMHz × ha,ft),
where fMHz is the frequency in MHz and ha,ft is the antenna height in feet.
Vertical coverage pattern, returned as a real-valued column vector or matrix. The vertical coverage pattern is the actual maximum range of the radar. Each row of the vertical coverage pattern corresponds to one of the angles returned in vcpangles The columns of vcp correspond to the ranges specified in rfs.
Vertical coverage pattern angles, returned as a column vector. The angles range from –90° to 90°. Each entry of vcpangles specifies the elevation angle at which the vertical coverage pattern is measured.
The maximum detection range of a radar antenna can differ depending on placement. Suppose you place a radar antenna near a reflecting surface, such as the earth's land or sea surface and computed maximum detection range. If you then move the same radar antenna to free space far from any boundaries, it results in a different maximum detection range. This is an effect of multipath interference that occurs when waves, reflected from the surface, constructively add to or nullify the direct path signal from the radar to a target. Multipath interference gives rise to a series of lobes in the vertical plane. The vertical coverage pattern is the plot of the actual maximum detection range of the radar versus target elevation and depends upon the maximum free-space detection range and target elevation angle. See Blake [1].
The vertical coverage pattern is generally considered to be valid for antenna heights that are within a few hundred feet of the surface and with targets at altitudes that are not too close to the radar horizon.
blakechart | el2height | height2el | height2range | height2grndrange | landroughness | range2height | refractionexp | searoughness
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Blast wave - Wikipedia
Increased fluid pressure and flow from an explosion
In fluid dynamics, a blast wave is the increased pressure and flow resulting from the deposition of a large amount of energy in a small, very localised volume. The flow field can be approximated as a lead shock wave, followed by a self-similar subsonic flow field. In simpler terms, a blast wave is an area of pressure expanding supersonically outward from an explosive core. It has a leading shock front of compressed gases. The blast wave is followed by a blast wind of negative gauge pressure, which sucks items back in towards the center. The blast wave is harmful especially when one is very close to the center or at a location of constructive interference. High explosives that detonate generate blast waves.
3.1 Mach stem formation
5.2 Effects of blast loads on buildings
High-order explosives (HE) are more powerful than low-order explosives (LE). HE detonate to produce a defining supersonic over-pressurization shock wave. Several sources of HE include trinitrotoluene, C-4, Semtex, nitroglycerin, and ammonium nitrate fuel oil (ANFO). LE deflagrate to create a subsonic explosion and lack HE’s over-pressurization wave. Sources of LE include pipe bombs, gunpowder, and most pure petroleum-based incendiary bombs such as Molotov cocktails or aircraft improvised as guided missiles. HE and LE induce different injury patterns. Only HE produce true blast waves.[1]
The classic flow solution—the so-called Taylor–von Neumann–Sedov blast wave solution—was independently devised by John von Neumann[2][3] and British mathematician Geoffrey Ingram Taylor[4][5] during World War II. After the war, the similarity solution was published by three other authors—L. I. Sedov,[6] R. Latter,[7] and J. Lockwood-Taylor[8]—who had discovered it independently.[9]
Since the early theoretical work, both theoretical and experimental studies of blast waves have been ongoing.[10][11]
A Friedlander waveform is the simplest form of a blast wave.
The simplest form of a blast wave has been described and termed the Friedlander waveform.[12] It occurs when a high explosive detonates in a free field, that is, with no surfaces nearby with which it can interact. Blast waves have properties predicted by the physics of waves. For example, they can diffract through a narrow opening, and refract as they pass through materials. Like light or sound waves, when a blast wave reaches a boundary between two materials, part of it is transmitted, part of it is absorbed, and part of it is reflected. The impedances of the two materials determine how much of each occurs.
The equation for a Friedlander waveform describes the pressure of the blast wave as a function of time:
{\displaystyle P(t)=P_{s}e^{-{\frac {t}{t^{*}}}}\left(1-{\frac {t}{t^{*}}}\right).}
where Ps is the peak pressure and t* is the time at which the pressure first crosses the horizontal axis (before the negative phase).
Blast waves will wrap around objects and buildings.[13] Therefore, persons or objects behind a large building are not necessarily protected from a blast that starts on the opposite side of the building. Scientists use sophisticated mathematical models to predict how objects will respond to a blast in order to design effective barriers and safer buildings.[14]
Mach stem formation[edit]
A blast wave reflecting from a surface and forming a mach stem.
Mach stem formation occurs when a blast wave reflects off the ground and the reflection catches up with the original shock front, therefore creating a high pressure zone that extends from the ground up to a certain point called the triple point at the edge of the blast wave. Anything in this area experiences peak pressures that can be several times higher than the peak pressure of the original shock front.
Constructive and destructive interference[edit]
An example of constructive interference.
In physics, interference is the meeting of two correlated waves and either increasing or lowering the net amplitude, depending on whether it is constructive or destructive interference. If a crest of a wave meets a crest of another wave at the same point then the crests interfere constructively and the resultant crest wave amplitude is increased; forming a much more powerful wave than either of the beginning waves. Similarly two troughs make a trough of increased amplitude. If a crest of a wave meets a trough of another wave then they interfere destructively, and the overall amplitude is decreased; thus making a wave that is much smaller than either of the parent waves.
The formation of a mach stem is one example of constructive interference. Whenever a blast wave reflects off of a surface, such as a building wall or the inside of a vehicle, different reflected waves can interact with each other to cause an increase in pressure at a certain point (constructive interference) or a decrease (destructive interference). In this way the interaction of blast waves is similar to that of sound waves or water waves.
Blast waves cause damage by a combination of the significant compression of the air in front of the wave (forming a shock front) and the subsequent wind that follows.[15] A blast wave travels faster than the speed of sound and the passage of the shock wave usually lasts only a few milliseconds. Like other types of explosions, a blast wave can also cause damage to things and people by the blast wind, debris, and fires. The original explosion will send out fragments that travel very fast. Debris and sometimes even people can get swept up into a blast wave, causing more injuries such as penetrating wounds, impalement and broken bones. The blast wind is the area of low pressure that causes debris and fragments to rush back towards the original explosions. The blast wave can also cause fires or secondary explosions by a combination of the high temperatures that result from detonation and the physical destruction of fuel-containing objects.
In response to an inquiry from the British MAUD Committee, G. I. Taylor estimated the amount of energy that would be released by the explosion of an atomic bomb in air. He postulated that for an idealized point source of energy, the spatial distributions of the flow variables would have the same form during a given time interval, the variables differing only in scale. (Thus the name of the "similarity solution.") This hypothesis allowed the partial differential equations in terms of r (the radius of the blast wave) and t (time) to be transformed into an ordinary differential equation in terms of the similarity variable
{\displaystyle {\frac {r^{5}\rho _{o}}{t^{2}E}}}
{\displaystyle \rho _{o}}
is the density of the air and
{\displaystyle E}
is the energy that's released by the explosion.[16][17][18] This result allowed G. I. Taylor to estimate the yield of the first atomic explosion in New Mexico in 1945 using only photographs of the blast, which had been published in newspapers and magazines.[9] The yield of the explosion was determined by using the equation:
{\displaystyle E=\left({\frac {\rho _{o}}{t^{2}}}\right)\left({\frac {r}{C}}\right)^{5}}
{\displaystyle C}
is a dimensionless constant that is a function of the ratio of the specific heat of air at constant pressure to the specific heat of air at constant volume. The value of C is also affected by radiative losses, but for air, values of C of 1.00-1.10 generally give reasonable results. In 1950, G. I. Taylor published two articles in which he revealed the yield E of the first atomic explosion,[4][5] which had previously been classified and whose publication therefore a source of controversy.[citation needed]
While nuclear explosions are among the clearest examples of the destructive power of blast waves, blast waves generated by exploding conventional bombs and other weapons made from high explosives have been used as weapons of war due to their effectiveness at creating polytraumatic injury. During World War II and the U.S.’s involvement in the Vietnam War, blast lung was a common and often deadly injury. Improvements in vehicular and personal protective equipment have helped to reduce the incidence of blast lung. However, as soldiers are better protected from penetrating injury and surviving previously lethal exposures, limb injuries, eye and ear injuries, and traumatic brain injuries have become more prevalent.
Effects of blast loads on buildings[edit]
Structural behaviour during an explosion depends entirely on the materials used in the construction of the building. Upon hitting the face of a building, the shock front from an explosion is instantly reflected. This impact with the structure imparts momentum to exterior components of the building. The associated kinetic energy of the moving components must be absorbed or dissipated in order for them to survive. Generally, this is achieved by converting the kinetic energy of the moving component to strain energy in resisting elements.[19]
Typically the resisting elements, such as windows, building facades and support columns fail, causing partial damage through to progressive collapse of the building.
The so-called Sedov-Taylor solution (see § Bombs) has become useful in astrophysics. For example, it can be applied to quantify an estimate for the outcome from supernova-explosions. The Sedov-Taylor expansion is also known as the 'Blast Wave' phase, which is an adiabatic expansion phase in the life cycle of supernova. The temperature of the material in a supernova shell decreases with time, but the internal energy of the material is always 72% of E0, the initial energy released. This is helpful for astrophysicists interested in predicting the behavior of supernova remnants.
Blast waves are generated in research environments using explosive or compressed-gas driven shock tubes in an effort to replicate the environment of a military conflict to better understand the physics of blasts and injuries that may result, and to develop better protection against blast exposure.[20] Blast waves are directed against structures (such as vehicles),[21] materials, and biological specimens[22] or surrogates. High-speed pressure sensors and/or high speed cameras are often used to quantify the response to blast exposure. Anthropomorphic test devices (ATDs or test dummies) initially developed for the automotive industry are being used, sometimes with added instrumentation, to estimate the human response to blast events. For examples, personnel in vehicles and personnel on demining teams have been simulated using these ATDs.[23]
Combined with experiments, complex mathematical models have been made of the interaction of blast waves with inanimate and biological structures.[24] Validated models are useful for "what if" experiments – predictions of outcomes for different scenarios. Depending on the system being modeled, it can be difficult to have accurate input parameters (for example, the material properties of a rate-sensitive material at blast rates of loading). Lack of experimental validation severely limits the usefulness of any numerical model.
Chapman–Jouguet condition
Zeldovich–Taylor flow
^ Explosions and Blast Injuries: A Primer for Clinicians (PDF) (Report). Centers for Disease Control (CDC). Archived (PDF) from the original on 4 March 2022. Retrieved 7 March 2022.
^ Bethe,H.A., et al, BLAST WAVE, Los Alamos Report LA-2000, Ch. 2, (1947). read online
^ a b Taylor, Sir Geoffrey Ingram (1950). "The Formation of a Blast Wave by a Very Intense Explosion. I. Theoretical Discussion". Proceedings of the Royal Society A. 201 (1065): 159–174. Bibcode:1950RSPSA.201..159T. doi:10.1098/rspa.1950.0049. S2CID 54070514.
^ a b Taylor, Sir Geoffrey Ingram (1950). "The Formation of a Blast Wave by a Very Intense Explosion. II. The Atomic Explosion of 1945". Proceedings of the Royal Society A. 201 (1065): 175–186. Bibcode:1950RSPSA.201..175T. doi:10.1098/rspa.1950.0050.
^ Sedov, L. I., "Propagation of strong shock waves," Journal of Applied Mathematics and Mechanics, Vol. 10, pages 241 - 250 (1946); in Russian: Седов Л. И. "Распространение сильных взрывных волн," Прикладная математика и механика, т. X, № 2, С. 241-250.
^ a b Batchelor, George, The Life and Legacy of G. I. Taylor, [Cambridge, England: Cambridge University Press, 1996], pages 202 - 207.
^ Dewey JM. 53 years of blast wave research, a personal history. 21st International Symposium on Military and Blast, Israel, 2010
^ Rinehart EJ, et al. DTRA weapons effects testing: a thirty year perspective. 21st International Symposium on Military and Blast, Israel, 2010 read online Archived 13 March 2012 at the Wayback Machine
^ Dewey JM. THE SHAPE OF THE BLAST WAVE: STUDIES OF THE FRIEDLANDER EQUATION. Presented at the 21st International Symposium on Military Aspects of Blast and Shock, Israel 2010 read online
^ Remmenikov AM. Modelling blast loads on buildings in complex city geometries. Computers and Structures, 2005, 83(27), 2197-2205. read online
^ for example,Cullis IG. Blast waves and how they interact with structures. J.R. Army Med Corps 147:16-26, 2001
^ Neff M. A visual model for blast waves and fracture. Master's Thesis, University of Toronto, Canada, 1998
^ Discussion of similarity solutions, including G. I. Taylor's: Buckingham Pi theorem
^ Dusenberry, Donald. 'Handbook for Blast Resistant Design of Buildings', 2010, pages 8-9.
^ Rinehart, Dr. E. J., Henny, Dr. R. W., Thomsen, J. M., Duray, J. P. DTRA Weapons Effects Testing: A Thirty Year Perspective. Applied Research and Associates, Shock Physics Division
^ for example, Bauman, R. A., Ling, G., Tong, L., Januszkiewicz, A., Agoston, D., Delanerolle, N., Kim, Y., Ritzel, D., Bell, R., Ecklund, J., Armonda, R., Bandak, F., Parks, S. An Introductory Characterization of a Combat-Casualty-Care Relevant Swine Model of Closed Head Injury Resulting from Exposure to Explosive Blast. Journal of Neurotrauma, June 2009, Mary Ann Liebert, Inc.
^ Cernak, I. The importance of systematic response in the pathobiology of blast-induced Neurotrauma. Frontiers in Neurology December, 2010.
^ Makris, A. Nerenberg, J., Dionne, J. P., Bass, C. R., Chichester. Reduction of Blast Induced Head Acceleration in the Field of Anti-Personnel Mine Clearance. Med-Eng Systems Inc.
^ for example, Stuhmiller JH. Mathematical Modeling in Support of Military Operational Medicine Final Report J3150.01-06-306 prepared for the U.S. Army Medical Research and Materiel Command Fort Detrick, Maryland 21702-5012 OMB No. 0704-0188, July, 2006.
"The formation of a blast wave by a very intense explosion" G. I. Taylor's solution
Retrieved from "https://en.wikipedia.org/w/index.php?title=Blast_wave&oldid=1086109969"
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Model fitting using MCMC - The basic framework | Scalismo
In this tutorial we show how Bayesian model fitting using Markov Chain Monte Carlo can be done in Scalismo. To be able to focus on the main components of the framework instead of technical details, we start in this tutorial with a simple toy example, namely 1D Bayesian linear regression. The application to 3D shape modelling is discussed in depth in the next tutorial.
Problem setting
In a Bayesian linear regression an outcome variable
y
is modelled a linear function of the explanatory variable
x
. The normal linear model assumes that the distribution of
y
is a normal distribution with a mean
a \cdot x + b
\sigma^2
y \sim N(a \cdot x + b, \sigma^2 ).
In the following we will denote the unknown parameters
a
b
\sigma^2
\theta
\theta = (a, b, \sigma^2)
. The inference problem is to estimate the parameters
\theta
, given observations
X=(x_1, \ldots, x_n)
Y=(y_1, \ldots, y_n)
. This is done by computing the posterior distribution:
p(\theta | Y, X) = \frac{p(Y | \theta, X)p(\theta)}{\int P(Y | \theta, X)p(\theta) \, d\theta}
The likelihood term
p(Y | \theta, X)
is given by the normal distribution
N(a \cdot x + b,\sigma^2)
define above. Hence the likelihood of observing the data
X, Y
\prod_{i=1}^n p(y_i | \theta, x_i) = \prod_{i=1}^n N(y_i | a \cdot x_i + b, \sigma^2)
As prior distribution
p(\theta)
a \sim N(0, 5) \\ b \sim N(0, 10) \\ \sigma^2 \sim logNormal(0, 0.25)
Metropolis Hastings Algorithm
The way we approach such an inference problem in Scalismo is by using the Metropolis-Hastings algorithm. The Metropolis-Hastings algorithm allows us to draw samples from any distribution, given that the unnormalized distribution can be evaluated point-wise. This requirement is easy to fulfill for all shape modelling applications.
For setting up the Metropolis-Hastings algorithm, we need two things:
The (unnormalized) target distribution, from which we want to sample. In our case this is the posterior distribution
p(\theta \mid Y, X))
. In Scalismo the corresponding class is called the DistributionEvaluator.
A proposal distribution
Q(\theta' \mid \theta)
, which generates for a given sample
\theta
a new sample
\theta'
The Metropolis Hastings algorithm introduces an ingenious scheme for accepting and rejecting the samples from this proposal distribution, based on their probability under the target density, such that the resulting sequence of samples is guaranteed to be distributed according to the target distribution. In practice, the algorithm works as follows: It uses the proposal generator to perturb a given sample
\theta
to obtain a new sample
\theta'
. Then it checks, using the evaluator, which of the two samples,
\theta
\theta'
is more likely and uses this ratio as a basis for rejecting or accepting the new sample.
Implementation in Scalismo
import breeze.stats.meanAndVariance
To make the setup simple, we generate artificial data, which follows exactly our assumptions. In this way we will be able to see how well we estimated the parameters.
val sigma2 = 0.5
val errorDist = breeze.stats.distributions.Gaussian(0, sigma2)
val data = for (x <- 0 until 100) yield {
(x.toDouble, a * x + b + errorDist.draw())
Before we discuss the two main components, the Evaluator and Proposal generator in detail, we first define a class for representing the parameters
\theta = (a, b, \sigma^2)
case class Parameters(a : Double, b: Double, sigma2 : Double)
We introduce a further class to represent a sample from the chain. A sample is simply a set of parameters together with a tag, which helps us to keep track later on, which proposal generator generated the sample:
case class Sample(parameters : Parameters, generatedBy : String)
In Scalismo, the target density is represented by classes, which we will refer to as Evaluators. Any Evaluator is a subclass of the class DistributionEvalutor, which is defined in Scalismo as follows:
trait DistributionEvaluator[A] {
/** log probability/density of sample */
def logValue(sample: A): Double
Note: This trait is already defined in Scalismo, don't paste it into your code.
We see that the only thing we need to define is the log probability of a sample.
In our case, we will define separate evaluators for the prior distribution
p(\theta)
p(Y | \theta, X)
The likelihood function, defined above, can be implemented as follows:
case class LikelihoodEvaluator(data : Seq[(Double, Double)]) extends DistributionEvaluator[Sample] {
override def logValue(theta: Sample): Double = {
val likelihoods = for ((x, y) <- data) yield {
val likelihood = breeze.stats.distributions.Gaussian(
theta.parameters.a * x + theta.parameters.b, theta.parameters.sigma2)
likelihood.logPdf(y)
likelihoods.sum
Notice that we work in Scalismo with log probabilities, and hence the product in above formula becomes a sum.
In a similar way, we encode the prior distribution:
object PriorEvaluator extends DistributionEvaluator[Sample] {
val priorDistA = breeze.stats.distributions.Gaussian(0, 1)
val priorDistB = breeze.stats.distributions.Gaussian(0, 10)
val priorDistSigma = breeze.stats.distributions.LogNormal(0, 0.25)
priorDistA.logPdf(theta.parameters.a)
+ priorDistB.logPdf(theta.parameters.b)
+ priorDistSigma.logPdf(theta.parameters.sigma2)
The target density (i.e. the posterior distribution) can be computed by taking the product of the prior and the likelihood.
val posteriorEvaluator = ProductEvaluator(PriorEvaluator, LikelihoodEvaluator(data))
Note that the posteriorEvaluator represents the unnormalized posterior, as we did not normalize by the probability of the data
p(y)
In Scalismo, a proposal generator is defined by extending the trait ProposalGenerator, which is defined as follows
trait ProposalGenerator[A] {
/** draw a sample from this proposal distribution, may depend on current state */
def propose(current: A): A
In order to be able to use a proposal generator in the Metropolis-Hastings algorithm, we also need to implement the trait TransitionProbability:
trait TransitionProbability[A] extends TransitionRatio[A] {
/** rate of transition from to (log value) */
def logTransitionProbability(from: A, to: A): Double
Note: The above traits are already defined in Scalismo, don't paste them into your code.
We use here one of the simples possible proposals, namely a random walk proposal. This is a proposal which updates the current state by taking a step of random length in a random direction. For simplicity, we update all three parameters together:
case class RandomWalkProposal(stepLengthA: Double, stepLengthB : Double, stepLengthSigma2 : Double)(implicit rng : scalismo.utils.Random)
extends ProposalGenerator[Sample] with TransitionProbability[Sample] {
val newParameters = Parameters(
a = sample.parameters.a + rng.breezeRandBasis.gaussian(0, stepLengthA).draw(),
b = sample.parameters.b + rng.breezeRandBasis.gaussian(0, stepLengthB).draw(),
sigma2 = sample.parameters.sigma2 + rng.breezeRandBasis.gaussian(0, stepLengthSigma2).draw(),
Sample(newParameters, s"randomWalkProposal ($stepLengthA, $stepLengthB)")
override def logTransitionProbability(from: Sample, to: Sample) : Double = {
val stepDistA = breeze.stats.distributions.Gaussian(0, stepLengthA)
val stepDistB = breeze.stats.distributions.Gaussian(0, stepLengthB)
val stepDistSigma2 = breeze.stats.distributions.Gaussian(0, stepLengthSigma2)
val residualA = to.parameters.a - from.parameters.a
val residualB = to.parameters.b - from.parameters.b
val residualSigma2 = to.parameters.sigma2 - from.parameters.sigma2
stepDistA.logPdf(residualA) + stepDistB.logPdf(residualB) + stepDistSigma2.logPdf(residualSigma2)
Remark: the second constructor argument implicit rng : scalismo.utils.Random is used to automatically pass the globally defined random generator object to the class. If we always use this random generator to generate our random numbers, we can obtain reproducible runs, by seeding this random generator at the beginning of our program.
Let's define two random walk proposals with different step length:
val smallStepProposal = RandomWalkProposal(0.01, 0.01, 0.01)
val largeStepProposal = RandomWalkProposal(0.1, 0.1, 0.1)
Varying the step length allow us to sometimes take large step, in order to explore the global landscape, and sometimes smaller steps, to explore a local environment. We can combine these proposal into a MixtureProposal, which chooses the individual proposals with a given probability. Here We choose to take the large step 20% of the time, and the smaller steps 80% of the time:
val generator = MixtureProposal.fromProposalsWithTransition[Sample](
(0.8, smallStepProposal),
(0.2, largeStepProposal)
Now that we have all the components set up, we can assemble the Markov Chain.
To run the chain, we obtain an iterator, which we then consume to drive the sampling generation. To obtain the iterator, we need to specify the initial sample:
val initialSample = Sample(Parameters(0.0, 0.0, 1.0), generatedBy="initial")
val mhIterator = chain.iterator(initialSample)
Our initial parameters might be far away from a high-probability area of our target density. Therefore it might take a few hundred or even a few thousand iterations before the produced samples start to follow the required distribution. We therefore have to drop the samples in this burn-in phase, before we use the samples:
val samples = mhIterator.drop(5000).take(15000).toIndexedSeq
As we have generated synthetic data, we can check if the expected value, computed from this samples, really corresponds to the parameters from which we sampled our data:
val meanAndVarianceA = meanAndVariance(samples.map(_.parameters.a))
println(s"Estimates for parameter a: mean = ${meanAndVarianceA.mean}, var = ${meanAndVarianceA.variance}")
val meanAndVarianceB = meanAndVariance(samples.map(_.parameters.b))
println(s"Estimates for parameter b: mean = ${meanAndVarianceB.mean}, var = ${meanAndVarianceB.variance}")
val meanAndVarianceSigma2 = meanAndVariance(samples.map(_.parameters.sigma2))
println(s"Estimates for parameter sigma2: mean = ${meanAndVarianceSigma2.mean}, var = ${meanAndVarianceSigma2.variance}")
In the next tutorial, we see an example of how the exact same mechanism can be used for fitting shape models. Before we discuss this, we should, however, spend some time to discuss how the chain can be debugged in case something goes wrong. You can safely skip this section and come back to it later if you first want to see a practical example.
Debugging the Markov chain
Sometimes a chain does not work as expected. The reason is usually that our proposals are not suitable for the target distribution. To diagnose the behaviour of the chain we can introduce a logger. To write a logger, we need to extend the trait AcceptRejectLogger, which is defined as follows:
trait AcceptRejectLogger[A] {
def accept(current: A, sample: A, generator: ProposalGenerator[A], evaluator: DistributionEvaluator[A]): Unit
def reject(current: A, sample: A, generator: ProposalGenerator[A], evaluator: DistributionEvaluator[A]): Unit
The two methods, accept and reject are called whenever a sample is accepted or rejected. We can overwrite these methods to implement our debugging code.
The following, very simple logger counts all the accepted and rejected samples and computes the acceptance ratio. This acceptance ratio is a simple, but already useful indicator to diagnose if all proposal generators function as expected.
evaluator: DistributionEvaluator[Sample]
def acceptanceRatios() : Map[String, Double] = {
val acceptanceRatios = for (generatorName <- generatorNames ) yield {
To use the logger, we simply rerun the chain, but pass the logger now as a second argument to the iterator method:
val mhIteratorWithLogging = chain.iterator(initialSample, logger)
val samples2 = mhIteratorWithLogging.drop(5000).take(15000).toIndexedSeq
We can now check how often the individual samples got accepted.
println("acceptance ratio is " +logger.acceptanceRatios())
We see that the acceptance ratio of the random walk proposal, which takes the smaller step is quite high, but that the larger step is often rejected. We might therefore want to reduce this step size slightly, as a proposal that is so often rejected is not very efficient.
In more complicated applications, this type of debugging is crucial for obtaining efficient fitting algorithms.
Implementation in Scalismo
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Governance Token - Fishing Master
MASTER is the native token of the GameFiMaster family which consists of Fishing Master and more games. MASTER can be used for initiating proposals, voting and sharing the project revenue. The GameFiMaster smart contract will periodically buy back and burn MASTER with the profit of the products.
CMC: https://coinmarketcap.com/currencies/fishing-master/
TGE Estimate: 2022Q2 Landing on Ethereum Mainnet.
There is no fundraising planned for the MASTER token.
Linearly vested in 36 epochs
Game farming for 36 epochs
Based on MM demand
An epoch is 28 days.
Token to Players
20% of MASTER issuance will be released to players through in-game farming, with 95% vested for 36 epochs starting from TGE and 5% unlocked immediately.
Please note that all rules and formulas are subject to change before the final announcement of TGE.
Pre-TGE Airdrop Allocation
On TGE, 439,900 MASTER will be airdropped to the valid users who played the game since Open Beta launched. The allocation is performance-based. The time of snapshot is TBD.
Based on Bullet firing amount, Blindbox spending and net deposit endurance.
Bullet firing amount: F The total Bullet fired amount of the account in game sessions since registration.
Blindbox spending: S The total FMC spending amount of the account on Blindbox since registration. The p is an undecided coefficient on S.
Net deposit: N_i The net deposit of the account during T_i. If net deposit is negative, it will cause a negative effect on the calculation.
Endurance: T_i The length of an interval(i) of certain net deposit of the account.
Airdrop Weight: W The weight of the account in the airdrop allocation. If the account's sum of net deposit x endurance is zero or negative, then W = 0.
W=(F+p*S)*\sum_{i=1}^n(N_i*T_i)
The threshold to be a valid user of this event is having 10,000 Bullets fired in record.
The minimum effective airdrop is 0.0001 MASTER. The amount below this figure will be abandoned.
Tokens will be instantly released for claiming by users via the official panel on TGE.
Post-TGE Farming Allocation
After TGE, 8,358,100 MASTER will be released over time to the valid users played the game. The allocation is performance-based. The event lasts 36 epochs. The first release is the 28th day after TGE.
Based on Bullet firing amount, Blindbox spending and average net deposit during an epoch.
Bullet firing amount: F_k The total Bullet fired amount of the account in game sessions during the epoch(k).
Blindbox spending: S_k The total FMC spending amount of the account on Blindbox during the epoch(k). The q is an undecided coefficient on S_k.
Average Net deposit: N_avg The average net deposit of the account during the epoch(k). The average net deposit is the arithmetic mean of the net deposit for 4 weeks' snapshots.
Farming Weight: W_k The weight of the account in the farming allocation of the epoch(k). If the account's average net deposit is zero or negative, then W = 0.
W_k=(F_k+q*S_k)*N_{avg}
The threshold to be a valid user of this event is having 10,000 Bullets fired during an epoch.
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Home : Support : Online Help : Connectivity : Calling External Routines : ExternalCalling : C Application Programming Interface : RTableCreate
RTableCreate(kv, rts, pdata, bounds)
pointer to array data (optional)
RTableCreate creates a new rtable with the settings specified in rts.
If pdata is NULL, then a data block is allocated and initialized to rts->fill. When specifying a previously created block of data (that is, when pdata is not NULL), it is important that rts->foreign is set to TRUE. Size, storage, data_type, order, and indexing functions must all be considered when managing your data block. It is recommended that you let Maple create the data block, then use RTableDataBlock to create a pointer to it.
ALGEB M_DECL MyIdentity( MKernelVector kv, ALGEB *args )
M_INT argc, n, i;
INTEGER32 *data;
rts.subtype = RTABLE_MATRIX;
rts.data_type = RTABLE_INTEGER32;
rt = RTableCreate(kv,&rts,NULL,bounds);
data = (INTEGER32*)RTableDataBlock(kv,rt);
data[MATRIX_OFFSET_FORTRAN_RECT(i,i,n,n)] = 1;
\mathrm{with}\left(\mathrm{ExternalCalling}\right):
\mathrm{dll}≔\mathrm{ExternalLibraryName}\left("HelpExamples"\right):
\mathrm{eye}≔\mathrm{DefineExternal}\left("MyIdentity",\mathrm{dll}\right):
M≔\mathrm{eye}\left(3\right)
\textcolor[rgb]{0,0,1}{M}\textcolor[rgb]{0,0,1}{≔}[\begin{array}{ccc}\textcolor[rgb]{0,0,1}{1}& \textcolor[rgb]{0,0,1}{0}& \textcolor[rgb]{0,0,1}{0}\\ \textcolor[rgb]{0,0,1}{0}& \textcolor[rgb]{0,0,1}{1}& \textcolor[rgb]{0,0,1}{0}\\ \textcolor[rgb]{0,0,1}{0}& \textcolor[rgb]{0,0,1}{0}& \textcolor[rgb]{0,0,1}{1}\end{array}]
\mathrm{rtable_options}\left(M,\mathrm{datatype}\right)
{\textcolor[rgb]{0,0,1}{\mathrm{integer}}}_{\textcolor[rgb]{0,0,1}{4}}
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Section 45.6 (0FGP): Projective space bundle formula—The Stacks project
Section 45.6: Projective space bundle formula (cite)
45.6 Projective space bundle formula
Let $k$ be a base field. Let $X$ be a smooth projective scheme over $k$. Let $\mathcal{E}$ be a locally free $\mathcal{O}_ X$-module of rank $r$. Our convention is that the projective bundle associated to $\mathcal{E}$ is the morphism
\[ \xymatrix{ P = \mathbf{P}(\mathcal{E}) = \underline{\text{Proj}}_ X(\text{Sym}^*(\mathcal{E})) \ar[r]^-p & X } \]
over $X$ with $\mathcal{O}_ P(1)$ normalized so that $p_*(\mathcal{O}_ P(1)) = \mathcal{E}$. Recall that
\[ [\Gamma _ p] \in \text{Corr}^0(X, P) \subset \mathop{\mathrm{CH}}\nolimits ^*(X \times P) \otimes \mathbf{Q} \]
See Example 45.3.2. For $i = 0, \ldots , r - 1$ consider the correspondences
\[ c_ i = c_1(\text{pr}_2^*\mathcal{O}_ P(1))^ i \cap [\Gamma _ p] \in \text{Corr}^ i(X, P) \]
We may and do think of $c_ i$ as a morphism $h(X)(-i) \to h(P)$.
Lemma 45.6.1 (Projective space bundle formula). In the situation above, the map
\[ \sum \nolimits _{i = 0, \ldots , r - 1} c_ i : \bigoplus \nolimits _{i = 0, \ldots , r - 1} h(X)(-i) \longrightarrow h(P) \]
is an isomorphism in the category of motives.
Proof. By Lemma 45.5.3 it suffices to show that our map defines an isomorphism on Chow groups of motives after taking the product with any smooth projective scheme $Z$. Observe that $P \times Z \to X \times Z$ is the projective bundle associated to the pullback of $\mathcal{E}$ to $X \times Z$. Hence the statement on Chow groups is true by the projective space bundle formula given in Chow Homology, Lemma 42.36.2. Namely, pushforward of cycles along $[\Gamma _ p]$ is given by pullback of cycles by $p$ according to Lemma 45.3.6 and Chow Homology, Lemma 42.59.5. Hence pushforward along $c_ i$ sends $\alpha $ to $c_1(\mathcal{O}_ P(1))^ i \cap p^*\alpha $. Some details omitted. $\square$
In the situation above, for $j = 0, \ldots , r - 1$ consider the correspondences
\[ c'_ j = c_1(\text{pr}_1^*\mathcal{O}_ P(1))^{r - 1 - j} \cap [\Gamma _ p^ t] \in \text{Corr}^{-j}(P, X) \]
For $i, j \in \{ 0, \ldots , r - 1\} $ we have
\[ c'_ j \circ c_ i = \text{pr}_{13, *}\left( c_1(\text{pr}_2^*\mathcal{O}_ P(1))^{i + r - 1 - j} \cap (\text{pr}_{12}^*[\Gamma _ p] \cdot \text{pr}_{23}^*[\Gamma _ p^ t]) \right) \]
The cycles $\text{pr}_{12}^{-1}\Gamma _ p$ and $\text{pr}_{23}^{-1}\Gamma _ p^ t$ intersect transversally and with intersection equal to the image of $(p, 1, p) : P \to X \times P \times X$. Observe that the fibres of $(p, p) = \text{pr}_{13} \circ (p, 1, p) : P \to X \times X$ have dimension $r - 1$. We immediately conclude $c'_ j \circ c_ i = 0$ for $i + r - 1 - j < r - 1$, in other words when $i < j$. On the other hand, by the projective space bundle formula (Chow Homology, Lemma 42.36.2) the cycle $c_1(\mathcal{O}_ P(1))^{r - 1} \cap [P]$ maps to $[X]$ in $X$. Hence for $i = j$ the pushforward above gives the class of the diagonal and hence we see that
\[ c'_ i \circ c_ i = 1 \in \text{Corr}^0(X, X) \]
for all $i \in \{ 0, \ldots , r - 1\} $. Thus we see that the matrix of the composition
\[ \bigoplus h(X)(-i) \xrightarrow {\bigoplus c_ i} h(P) \xrightarrow {\bigoplus c'_ j} \bigoplus h(X)(-j) \]
is invertible (upper triangular with $1$s on the diagonal). We conclude from the projective space bundle formula (Lemma 45.6.1) that also the composition the other way around is invertible, but it seems a bit harder to prove this directly.
Lemma 45.6.2. Let $p : P \to X$ be as in Lemma 45.6.1. The class $[\Delta _ P]$ of the diagonal of $P$ in $\mathop{\mathrm{CH}}\nolimits ^*(P \times P)$ can be written as
\[ [\Delta _ P] = \left(\sum \nolimits _{i = 0, \ldots , r - 1} {r - 1 \choose i} c_{r - 1 - i}(\text{pr}_1^*\mathcal{S}^\vee ) \cap c_1(\text{pr}_2^*\mathcal{O}_ P(1))^ i\right) \cap (p \times p)^*[\Delta _ X] \]
where $\mathcal{S}$ is the kernel of the canonical surjection $p^*\mathcal{E} \to \mathcal{O}_ P(1)$.
Proof. Observe that $(p \times p)^*[\Delta _ X] = [P \times _ X P]$. Since $\Delta _ P \subset P \times _ X P \subset P \times P$ and since capping with Chern classes commutes with proper pushforward (Chow Homology, Lemma 42.38.4) it suffices to show that the class of $\Delta _ P \subset P \times _ X P$ in $\mathop{\mathrm{CH}}\nolimits ^*(P \times _ X P)$ is equal to
\[ \left(\sum \nolimits _{i = 0, \ldots , r - 1} {r - 1 \choose i} c_{r - 1 - i}(q_1^*\mathcal{S}^\vee ) \cap c_1(q_2^*\mathcal{O}_ P(1))^ i\right) \cap [P \times _ X P] \]
where $q_ i : P \times _ X P \to P$, $i = 1, 2$ are the projections. Set $q = p \circ q_1 = p \circ q_2 : P \times _ X P \to X$. Consider the maps
\[ q_1^*\mathcal{S} \otimes q_2^*\mathcal{O}_ P(-1) \to q^*\mathcal{E} \otimes q^*\mathcal{E}^\vee \to \mathcal{O}_{P \times _ X P} \]
where the final arrow is the pullback by $q$ of the evaluation map $\mathcal{E} \otimes _{\mathcal{O}_ X} \mathcal{E}^\vee \to \mathcal{O}_ X$. The source of the composition is a module locally free of rank $r - 1$ and a local calculation shows that this map vanishes exactly along $\Delta _ P$. By Chow Homology, Lemma 42.44.1 the class $[\Delta _ P]$ is the top Chern class of the dual
\[ q_1^*\mathcal{S}^\vee \otimes q_2^*\mathcal{O}_ P(1) \]
The desired result follows from Chow Homology, Lemma 42.39.1. $\square$
Comment #6307 by Qingyuan Jiang on July 03, 2021 at 07:31
Since this chapter is using the contravariant convention for motives, should the
h(X)(i)
in the formula of Lemma 45.6.1 be
h(X)(-i)
h(X) \otimes L^i
L
as in Remark 45.4.5)? Similarly for the formula about the composition
\oplus c_i
\oplus c_j'
Good catch! Thanks! I have fixed this here.
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e
(typically a module) can be converted to a package by using the call convert( e, 'package' ). The resulting package is returned.
\mathrm{addSquares}\left(2,3\right)
\textcolor[rgb]{0,0,1}{\mathrm{addSquares}}\textcolor[rgb]{0,0,1}{}\left(\textcolor[rgb]{0,0,1}{2}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{3}\right)
m:-\mathrm{addSquares}\left(2,3\right)
\textcolor[rgb]{0,0,1}{13}
\mathrm{with}\left(m\right)
\mathrm{type}\left(m,'\mathrm{package}'\right)
\textcolor[rgb]{0,0,1}{\mathrm{false}}
\mathrm{convert}\left(m,'\mathrm{package}'\right):
\mathrm{type}\left(m,'\mathrm{package}'\right)
\textcolor[rgb]{0,0,1}{\mathrm{true}}
\mathrm{with}\left(m\right);
\mathrm{addSquares}\left(2,3\right)
[\textcolor[rgb]{0,0,1}{\mathrm{addSquares}}]
\textcolor[rgb]{0,0,1}{13}
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GetColumns - Maple Help
Home : Support : Online Help : Connectivity : Database Package : Connection : GetColumns
get a list of columns from the database
connection:-GetColumns( opts )
(optional) equation(s) of the form option=value where option is one of catalog, schema, table, column, or output
GetColumns returns an Array in which each row contains one column from the database connected to using connection. A column in the Array is data associated with the column. The output option controls the columns and their order in the returned Array.
output = list containing one or more of the literal symbols: name, catalog, schema, table, and type
The output option specifies the columns of information to return and their order. By default, only the name is returned. Some databases do not support all these fields.
- name returns the name of the column.
- schema returns the name of the schema used in the table.
- table returns the name of the table that contains the column.
- type returns the SQL type of the column.
Return only columns from databases in which the schema matches the specified pattern. The pattern can consist of any valid schema name characters and the special characters % and _. A % matches any string and an _ matches any character. To match a literal _ or %, you must prepend the escape character \. For example, to match a _, use \_. To return columns from databases that do not use a schema, specify the empty string (""). By default, no restrictions are applied.
Return only columns for which the table matches the specified pattern. The pattern can consist of any valid table name characters and the special characters % and _. A % matches any string and an _ matches any character. To match a literal _ or %, you must prepend the escape character \. For example, to match a _, use \_. By default, no restrictions are applied.
column = string
Return only columns for which the name matches the specified pattern. The pattern can consist of any valid table name characters and the special characters % and _. A % matches any string and an _ matches any character. To match a literal _ or %, you must prepend the escape character \. For example, to match a _, use \_. By default, no restrictions are applied.
\mathrm{driver}≔\mathrm{Database}[\mathrm{LoadDriver}]\left(\right):
\mathrm{conn}≔\mathrm{driver}:-\mathrm{OpenConnection}\left(\mathrm{url},\mathrm{name},\mathrm{pass}\right):
\mathrm{conn}:-\mathrm{GetColumns}\left('\mathrm{output}'=['\mathrm{table}','\mathrm{name}']\right)
[\begin{array}{cc}\textcolor[rgb]{0,0,1}{"Table1"}& \textcolor[rgb]{0,0,1}{"size"}\\ \textcolor[rgb]{0,0,1}{"Table1"}& \textcolor[rgb]{0,0,1}{"number"}\\ \textcolor[rgb]{0,0,1}{"Table2"}& \textcolor[rgb]{0,0,1}{"mass"}\\ \textcolor[rgb]{0,0,1}{"Table2"}& \textcolor[rgb]{0,0,1}{"density"}\\ \textcolor[rgb]{0,0,1}{"Table3"}& \textcolor[rgb]{0,0,1}{"distance"}\end{array}]
\mathrm{conn}:-\mathrm{GetColumns}\left('\mathrm{table}'="Table1",'\mathrm{output}'=['\mathrm{name}']\right)
[\begin{array}{c}\textcolor[rgb]{0,0,1}{"size"}\\ \textcolor[rgb]{0,0,1}{"number"}\end{array}]
\mathrm{conn}:-\mathrm{GetColumns}\left('\mathrm{column}'="%a%",'\mathrm{output}'=['\mathrm{name}','\mathrm{table}','\mathrm{type}']\right)
[\begin{array}{ccc}\textcolor[rgb]{0,0,1}{"mass"}& \textcolor[rgb]{0,0,1}{"Table1"}& \textcolor[rgb]{0,0,1}{"DOUBLE"}\\ \textcolor[rgb]{0,0,1}{"distance"}& \textcolor[rgb]{0,0,1}{"Table3"}& \textcolor[rgb]{0,0,1}{"DOUBLE"}\end{array}]
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Balloon Arch Calculator | Venue Decorations
Do you need a balloon arch?
How to use our balloon arch calculator
The price of a balloon arch
If you are planning a fancy party with a LOT of balloons involved, our balloon arch calculator will help you calculate how many balloons for a balloon arch — the perfect one! 🎈
Learn how to make a balloon arch with the just-right number of balloons: keep reading to learn the different types of arches and how to calculate how many balloons you need!
How to make a balloon arch;
The types of balloon arches; and
How to calculate the perfect number of balloons for your decorations.
If you are hosting an important event, from a graduation 🎓 to a wedding 💒, and you want to spotlight a certain part of the venue, you may need a balloon arch.
Who doesn't love balloons? Ok, maybe not everyone loves balloons, but if they don't explode or fly away, they look nice! That's why a balloon arch may be the best choice to create that ethereal doorframe where you can take the best pictures of the event.
But it's not that easy!
To make a balloon arch, you first have to choose how you want it to look: tall, short or balanced? This will help our balloon arch calculator give you the correct answer. Choose the size of your arch: the essential measurements are the height and the width.
You then have to choose how the style of your arch: a single row of balloons or a complex garland?
And then some other details: do you want air-filled balloons or helium-filled ones? And what about the size and color?
Aren't balloon arches beautiful? But don't let any of the balloons float away!
If you plan to make an air-filled balloon arch, you will need a rather sturdy frame to hang them. A metal wire or other appropriate support you can easily find online will do the job. On the other hand, helium will make balloons float: you won't need any support!
🔎 A string of helium-filled balloons would create a curve called catenary curve: find out more about them in our catenary curve calculator.
If you decided to make a single row "pearls' string" arch, you will have to secure each balloon apart from the other. If you opted for a fancy garland, choose how many balloons you want to use for each "turn", and tie them together.
Once you get your balloons, it's time to inflate them. We strongly suggest you buy a pump: you may end up with almost 200 balloons, and trust us, you don't want to blow them up yourself!
The critical quantity that you need to know before buying too many balloons is the length of the balloon arch. To calculate it, first choose the height
h
w
of the arch. Then, follow the formulas to find it:
L = \begin{cases} h+w & \text{ if }\ w>h\\ 1.5\cdot h + w & \text{ if } \ w\simeq h\\ 2\cdot h+w& \text{ if }\ w<h \end{cases}
To calculate the number of balloons you need for an arch, we apply a simple formula, which depends on the diameter
d
of the balloons. It's not really scientific, but it works!
To find the perfect number of balloons for your arch, round the results to the nearest integer. If you choose a single row arch:
N=\frac{L}{d}\cdot 0.87
This means that to cover a meter of arch with balloons with diameter
11"
, we need:
N=\frac{39.37"}{11"}\cdot 0.87=3.11\approx 3
3
balloons for each meter of length of the arch, around
1
balloon per foot.
If you chose to make a garland arch, you can use two formulas according to the number
n
of balloons in each turn:
N = \begin{cases} 4.8\cdot\frac{L}{d}& \text{if }\ n=4\\ 6.3\cdot\frac{L}{d}& \text{if }\ n=5 \end{cases}
Again, remember to round your result: half balloons are not good balloons!
To find how many balloons for a balloon arch, define the type of arch in our calculator:
Input the height and width of the arch
Select if the arch is single row or garland.
Then insert the measurements of your arch, and select the diameter of the balloons. And... that's it!
Let's say that you want a
10
feet tall,
6
feet wide arch, and you chose a fancy five balloons garland. Since the arch's height is bigger than its width, we calculate the length using:
L=2\cdot h + w= 2\cdot 10 + 6=26\ \text{feet}
Do you want to make it with
11"
balloons? Easy: insert the value in the Diameter field, and remember to choose the correct units in our calculator: we will do the math for you:
N=6.3\cdot \frac{26}{11/12}= 178.7 \approx 179
Note that we applied a conversion from inches to feet.
That's quite a lot of balloons! That's why you need that pump. 😉
🙋 If you want to fill the gaps between the balloons in a garland, buy some small balloons (for example,
5"
) and put them in the holes: the overall effect will be fabulous!
We added a quick functionality that allows you to calculate the budget for a balloon arch! The price obviously varies (and usually, the more balloons you buy in a single batch, the less you pay for them): we gave you some indicative values that can help you estimate your expenses!
🙋 Select the desired type of balloons in the type of balloons field of our balloon arch calculator: it's that easy! We chose the price of $0.10 for a normal balloon, $0.30 for a metalized one, and $0.70 for a personalized one.
We have to say this: balloons look nice, but they are bad for the environment. Apart from being single-use (making a lot of waste), they pose a great risk to the wildlife, in particular if they float around.
And trust us — they float a lot: the author once found the remains of a bright green balloon a few meters shy of 3 km above sea level, in the deep of the mountains!
Remember to dispose of them correctly and not to let them escape. They look nice to us, but they might like food for many animals. And if you care for the environment, search for more friendly alternatives: paper makes beautiful decorations, or fly some kites! You will be as happy, and nature will thank you. 🌎
How to build a balloon arch?
Choose the type of balloon arch, then use our balloon arch calculator at omnicalculator.com to find out how many balloons you need for the perfect balloon arch!
Then arm yourself with patience, and start gluing together all of the balloons: are you going to make them in a single row? Simply tie them to their support. Are you going to make a garland? Tie as many as you need together and create a crowded decoration!
How many balloons do I need for a garland arch 10 feet high and 7 feet wide?
To build such an arch, you will need about 140 balloons. Use the formula:
N = 4.8 × (L/d)
N is the number of balloons;
L is the length of the arch; and
d the diameter of the balloons.
If you need more balloons in each turn of your garland, change the factor 4.8 to 6.3!
Do I need a balloon arch?
No! Balloons are bad for the environment, and you can find many other alternatives for your decorations. Go for paper flowers and garlands, kites if you are outside: bring nature to the party with you!🌸
Can I use helium for a balloon arch?
Yes, you can use helium for your balloon arch. In that case, the arch will float, and you won't need to support it in any way. But remember: helium costs a lot (there is a shortage), and the balloons float around. Stick to air: better for your pockets and the environment!
Size of the balloon arch
How packed is your arch?
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Effect of Residual Stress and Heterogeneity on Circumferential Stress in the Arterial Wall | J. Biomech Eng. | ASME Digital Collection
S. J. Peterson,
Department of Mechanical Engineering, Washington University, St. Louis, MO 63130
e-mail: rjo@mecf.wustl.edu
Contributed by the Bioengineering Division of THE AMERICAN SOCIETY OF MECHANICAL ENGINEERS. Manuscript received by the Bioengineering Division June 6, 1999; revision received March 22, 2000. Associate Technical Editor: J. D. Humphery.
Peterson , S. J., and Okamoto , R. J. (March 22, 2000). "Effect of Residual Stress and Heterogeneity on Circumferential Stress in the Arterial Wall ." ASME. J Biomech Eng. August 2000; 122(4): 454–456. https://doi.org/10.1115/1.1288210
Quantifying the stress distribution through the arterial wall is essential to studies of arterial growth and disease. Previous studies have shown that both residual stress, as measured by opening angle, and differing material properties for the media-intima and the adventitial layers affect the transmural circumferential stress
σθ
distribution. Because a lack of comprehensive data on a single species and artery has led to combinations from multiple sources, this study determined the sensitivity of
σθ
to published variations in both opening angle and layer thickness data. We fit material properties to previously published experimental data for pressure–diameter relations and opening angles of rabbit carotid artery, and predicted
σθ
through the arterial wall at physiologic conditions. Using a one-layer model, the ratio of
σθ
at the internal wall to the mean
σθ
decreased from 2.34 to 0.98 as the opening angle increased from 60 to 130 deg. In a two-layer model using a 95 deg opening angle, mean
σθ
in the adventitia increased (112 percent for 25 percent adventitia) and mean
σθ
in the media decreased (47 percent for 25 percent adventitia). These results suggest that both residual stress and wall layers have important effects on transmural stress distribution. Thus, experimental measurements of loading curves, opening angles, and wall composition from the same species and artery are needed to accurately predict the transmural stress distribution in the arterial wall. [S0148-0731(00)02204-4]
biomechanics, blood vessels, stress analysis, physiological models, Arteries, Residual Stress, Stress Analysis
Hoop stress, Stress, Stress analysis (Engineering), Physiology, Pressure
Elastic Properties of Arteries: A Nonlinear Two-Layer Cylindrical Model
Alternate Method for the Analysis of Residual Strain in the Arterial Wall
Rhodin, J., 1979, “Architecture of the Vessel Wall,” in: Handbook of Physiology, Section 2, The Cardiovascular System, Berne, R. M., ed., American Physiological Society, pp. 1–31.
Fung, Y. C., 1993, Biomechanics: Mechanical Properties of Living Tissues, Springer-Verlag, New York.
Strain Distribution in the Layered Wall of the Esophagus
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What are crushed stones?
How to use crushed stones of different sizes
How to use our crushed stone calculator for yards
Calculate how much crushed stone do I need manually
Using our crushed stone calculator, you can find the quantity and weight of the crushed stone or gravel that you'll need to layout or fill an area.
The following article discusses what crushed stones are, their different sizes and how we can use these crushed stones.
You're then given a guide on how to use the crushed stone calculator for calculating crushed stone in tons or yards while also sharing the formula for manually calculating the crushed stone in yards, then helping you convert that into tons.
Using our tool, you'll be able to calculate how much crushed stone you need for your next project. 😊
🔎 Check out our cubic yards to tons converter for a more precise volume to weight conversion of different materials.
When processed together, several different rock fragments make up crushed stones, also called crush for simplicity. The most commonly found rocks of mixed qualities and quantities in crushed stones are:
Argillite;
Slates;
Traprocks;
Quartzite; and
Volcanic cinder.
Using crushers, we can crush stones in various forms and sizes. Different crush sizes can have different uses.
Here's a list of standard crush sizes with their uses:
Stone dust – Used in concrete block fabrications, filling walls, patching holes, and building smooth pavements.
Up to 1/4″ – Used in barns, ground fillers, horse stables, tennis courts, and walkways.
Up to 1/2″ – Mixed in asphalt and concrete for roads, water tank beddings, filtration, and various environmental applications, such as landscaping and building.
Up to 3/4″ – Used in bedding for large pipes, drainage, driveways, farm roads, french drains, path fillers, and slab bases.
Up to 1″ – Best suited for pavements, road bases, and jogging tracks.
Up to 11/4″ – Used as a base material for driveways, rough terrain pads, and temporary paths for heavy-duty trucks at construction sites.
Up to 2″ – Drainage, dry wells, septic systems, and railroad tracks.
Up to 4″ – Drain outfall areas, heavy-duty haul roads, septic tanks, and slopes.
That's how we use crushed stones of varying sizes.
Here's how you can use the crushed stone calculator:
Select your surface shape, e.g., Rectangular.
In surface length, enter the length of your surface, e.g., 12 feet.
In surface width, enter the width of your surface, e.g., 8 feet.
Now, enter in depth, how deep do you want to fill the layout of your surface, e.g., 3 inches.
The waste factor tells how much crushed stone wastes during the layout process, e.g., 10% (default).
Once you've entered the above information, you get the following:
The required quantity of crushed stone for your layout, i.e., 0.98 cubic yards.
The estimated weight of this crushed stone, i.e., 1.5 tons.
Next, let's look into how we calculate the crushed stone quantity and weight manually.
Here are the formulas that we use to calculate how much crushed stone we need manually:
For square or rectangular surfaces:
S = (L \times W \times D) \div 27
And for circular or elliptical surfaces:
S = (\pi \times \frac{L}{2} \times \frac{W}{2} \times D) \div 27
S
– The required amount of crushed stones in cubic yards;
L
– Length in feet, where we want to layout the crush;
W
– Width in feet, of the same surface for laying out the crush;
D
– The depth in feet for the area which we want to fill with the crush;
\pi
– Math constant with a value of 3.1416 approx; and
27
– Conversion factor from cubic feet to cubic yards.
🙋 When calculating the required crushed stones, we add a standard 10% to the total as a waste factor.
To consider the waste factor in our total, we divide the result by 10 and sum it to the original value:
S_T = S + (S \div 10)
💡 We can convert the resulting crushed stone volume from yards to its approximate weight in tons by multiplying the result by 1.5.
Let's take an example of finding out how much crushed stone we require for our patio of 15′ × 11′, with a depth of 2″.
S = (15 \times 11 \times 0.167) \div 27 = 1.02
So, we require about 1.02 cubic yards without considering the waste factor.
To consider the waste factor, we add an additional 10% to our total:
S_T = 1.02 + (1.02 \div 10) = 1.12
And thus, the total amount of required crushed stone is about 1.12 cubic yards.
Multiplying the result by 1.5, we get the required weight of our crushed stone as 1.7 US tons.
Also, check out our simple gravel driveway calculator to determine the required amount of gravel for your driveway.
How much crushed stone do I require for a 10′ × 20′ × 4″ volume?
About 2.47 cubic yards without considering the waste factor, and 2.72 cubic yards with a waste factor of 10%. That's 3.7 and 4.1 US tons, respectively.
To calculate crushed stone in yards:
Multiply the together the length, width, and depth measured in feet.
Multiply by 27 to convert cubic feet into cubic yards.
Multiply by 1.5 to calculate the weight of the crushed stone in tons.
How do I calculate crushed stone in yards?
To calculate crushed stones in cubic yards for a rectangular surface, use the following formula:
S = (L × W × D) × 27
S – Required amount of crushed stones;
L – Surface length of where you want to fill the crushed stones;
W – Surface width where you want the crushed stones; and
D – Surface depth, to fill it with the crushed stones.
What is 3/4″ crushed stone used for?
Three-quarter inch crushed stones are most suitable for:
Bedding for large pipes;
Water drainage;
Constructing driveways;
Laying out on farm roads;
Creating french drains;
Used as path fillers; and
Slab bases.
How big is a yard of crushed stone?
A yard is equal to 3 feet. So a 1 cubic yard is 3 × 3 × 3 feet or 27 cubic feet.
One cubic yard of crushed stone or gravel weighs around 1.5 US tons, which can fill a volume of 3 cubic feet.
⛰️ Required crushed stone
Check out 3 similar driveway calculators 🚧
AsphaltConcrete driveway costGravel driveway
This siding calculator finds the square footage and cost of siding for your house.
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Goomba structure - Ukikipedia
Goomba structure
A Goomba Structure is created by mass cloning a large number of Goombas, placing them in desirable positions for Mario to utilize them. This can only be done in stages where cloning is possible and Goomba spawners are present.
As Mario is unable to release the clone normally without interacting with the Goomba's hitbox, numerous glitches that make Mario release the clone remotely is required. Hat-in-hand glitch, for instance, allows Mario to release Goombas at the HOLP since the object in front of Mario does not render, leaving the HOLP unaltered. Pause Buffered Hitstun allows Mario to release a Goomba clone since Mario is invisible in every other frame.
1 Goomba Cluster
2 Goomba Bridge
3 Goomba Staircase
4 Goomba Oasis
5 Goomba Ladder
Goomba Cluster
A Goomba Cluster is created by releasing cloned Goombas in different angles at the same HOLP. Because the Goombas are facing different directions, they have slightly different hitboxes, allowing them to be individually stomped.
Goomba Clusters are typically used to build Goomba Bridges and Goomba Staircases, both of which require a staircase of length
{\displaystyle n}
to set the HOLP to where Goomba
{\displaystyle n+1}
must be placed. Without a Goomba Cluster, an entire bridge of length
{\displaystyle n}
must be built twice to build a bridge of length
{\displaystyle n+1}
, causing both time and the number of slots to grow with order
{\displaystyle 2^{N}}
However, with a Goomba Cluster, all of the Goombas at positio{\displaystyle n}
can be placed once, removing the need to keep setting the HOLP. Each Goomba position on the bridge still requires a HOLP update, however, the HOLP must only be set there once. Therefore, at positio{\displaystyle n}
, the number of Goombas required is the number of Goomba positions after that position, which is
{\displaystyle N+1-n}
Since there is a position containing
{\displaystyle n}
Goombas for every
{\displaystyle n\in [1,N]}
, the total number of Goombas required is
{\displaystyle {\frac {N(N+1)}{2}}={\frac {N^{2}}{2}}+{\frac {N}{2}}}
, something that grows much more efficiently on the order of
{\displaystyle N^{2}}
Goomba Bridge
A Goomba Bridge is a type of structure created by cloning Goombas, allowing Mario to navigate in the air to certain locations that are hard to reach. The bridge can be as small as just one Goomba, to a massive bridge involving Goombas of two digits. Creating a bridge allows Mario to move across certain gaps that would otherwise require long jumping or triple jump dives.
The most efficient way to create Goomba Bridges is setting up the HOLP first and then release Goomba clones remotely at different angles to create a Goomba Cluster, and then step on that cluster to set a different HOLP further away, and then repeat. Before the discovery of Goomba Clusters, the old method involves cloning the Goombas one by one, which requires many trips back and forth and takes an exponential amount of Goombas to complete.
A Goomba Staircase is similar to a Goomba Bridge, except that the main purpose is for Mario to gain height instead of crossing a gap. Massive staircases can be constructed for the purpose of collecting the items to finish the objective high in the air, such as the secrets in Mario Wings to the Sky, or to reach the higher platforms in Bowser in the Sky.
Creating Goomba Staircases is most efficiently done by setting up the HOLP and create a Goomba Cluster for each step.
Goomba Oasis
A goomba oasis in BoB
A Goomba Oasis is a type of flat structure created by cloned Goombas via transport cloning. The clones are located midway between Mario and the HOLP laterally and at the HOLP's height. This allows many Goombas clones to stay in midair without having to construct a massive Goomba Bridge, as the game has limited memory for the object slots.
Goomba Ladder
A Goomba Ladder is a structure comprising of cloned Goombas going straight up, forming a ladder. The name is often used for any Goomba structure, but this use is a misnomer.
It can be created by releasing the cloned Goomba by dropping instead of throwing, making them appear at Mario's height but at HOLP laterally. This allows Mario to bounce on each Goomba to gain height. The disadvantage is that the ladder can only be as high as the course allows it, since the Goombas cloned with this method cannot go above the height that Mario can stand on at any point. Currently, there is no use of this structure in challenges.
Retrieved from "https://ukikipedia.net/mediawiki/index.php?title=Goomba_structure&oldid=16015"
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Leaderboard Incentive Program - Fishing Master
This program has launched on Dec 13, 2021.
Check this article to learn more: https://mirror.xyz/0xFD1972ADf9BB1985CaC4616240ed23518989F614/Gtyst1v2K11kf-3tBR5_Mlcj9fd53HA-4Bh6FUmP0H4
Every Monday, the game will update the leaderboard which shows the ranking of all players based on the amount of Bullets fired in the past seven days. The top players will be rewarded with Bullets.
One week as one round.
Time of Snapshot
Every Monday UTC 00:00:00.
Set by the administrator for each round.
See Pool Schedule.
Range of Winners
Top 50% of players in the ranking of weekly firing amount.
Minimum Threshold to Participate
10,000 Bullets fired during the week.
The reward is distributed based on each account's weight. Weight is determined by the account's Bullet firing amount for the week and its net deposit in the game.
weight = B_{weekshotcost} * ( B_{charge} - B_{withdraw} )
B_weekshotcost
The amount of Bullets fired by the user for the week.
B_charge
The amount of Bullets filled by the user totally.
B_withdraw
The amount of Bullets sold by the user totally.
Specially, if Bcharge - Bwithdraw < 1 then:
weight = B_{weekshotcost} * 1
Tips on How to Earn More Reward
First, please pay attention to the up left corner of the Main Menu screen. To make sure yourself a winner, you should keep your ranking inside top 50%.
The meter updates every 10 minutes.
Second, be careful to cash out Bullets because your account weight will shrink.
You must claim your reward manually.
During the test period, Claiming is suspended but all your reward is kept safely. Claiming is about to open on Jan 10.
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Rotating wave approximation - Wikipedia
Find sources: "Rotating wave approximation" – news · newspapers · books · scholar · JSTOR (August 2013) (Learn how and when to remove this template message)
The rotating wave approximation is an approximation used in atom optics and magnetic resonance. In this approximation, terms in a Hamiltonian which oscillate rapidly are neglected. This is a valid approximation when the applied electromagnetic radiation is near resonance with an atomic transition, and the intensity is low.[1] Explicitly, terms in the Hamiltonians which oscillate with frequencies
{\displaystyle \omega _{L}+\omega _{0}}
are neglected, while terms which oscillate with frequencies
{\displaystyle \omega _{L}-\omega _{0}}
are kept, where
{\displaystyle \omega _{L}}
is the light frequency and
{\displaystyle \omega _{0}}
is a transition frequency.
The name of the approximation stems from the form of the Hamiltonian in the interaction picture, as shown below. By switching to this picture the evolution of an atom due to the corresponding atomic Hamiltonian is absorbed into the system ket, leaving only the evolution due to the interaction of the atom with the light field to consider. It is in this picture that the rapidly oscillating terms mentioned previously can be neglected. Since in some sense the interaction picture can be thought of as rotating with the system ket only that part of the electromagnetic wave that approximately co-rotates is kept; the counter-rotating component is discarded.
The rotating wave approximation is closely related to, but different from, the secular approximation.[2]
1.1 Making the approximation
For simplicity consider a two-level atomic system with ground and excited states
{\displaystyle |{\text{g}}\rangle }
{\displaystyle |{\text{e}}\rangle }
, respectively (using the Dirac bracket notation). Let the energy difference between the states be
{\displaystyle \hbar \omega _{0}}
{\displaystyle \omega _{0}}
is the transition frequency of the system. Then the unperturbed Hamiltonian of the atom can be written as
{\displaystyle H_{0}={\frac {\hbar \omega _{0}}{2}}|{\text{e}}\rangle \langle {\text{e}}|-{\frac {\hbar \omega _{0}}{2}}|{\text{g}}\rangle \langle {\text{g}}|}
Suppose the atom experiences an external classical electric field of frequency
{\displaystyle \omega _{L}}
{\displaystyle {\vec {E}}(t)={\vec {E}}_{0}e^{-i\omega _{L}t}+{\vec {E}}_{0}^{*}e^{i\omega _{L}t}}
; e.g., a plane wave propagating in space. Then under the dipole approximation the interaction Hamiltonian between the atom and the electric field can be expressed as
{\displaystyle H_{1}=-{\vec {d}}\cdot {\vec {E}}}
{\displaystyle {\vec {d}}}
is the dipole moment operator of the atom. The total Hamiltonian for the atom-light system is therefore
{\displaystyle H=H_{0}+H_{1}.}
The atom does not have a dipole moment when it is in an energy eigenstate, so
{\displaystyle \left\langle {\text{e}}\left|{\vec {d}}\right|{\text{e}}\right\rangle =\left\langle {\text{g}}\left|{\vec {d}}\right|{\text{g}}\right\rangle =0.}
This means that defining
{\displaystyle {\vec {d}}_{\text{eg}}\mathrel {:=} \left\langle {\text{e}}\left|{\vec {d}}\right|{\text{g}}\right\rangle }
allows the dipole operator to be written as
{\displaystyle {\vec {d}}={\vec {d}}_{\text{eg}}|{\text{e}}\rangle \langle {\text{g}}|+{\vec {d}}_{\text{eg}}^{*}|{\text{g}}\rangle \langle {\text{e}}|}
{\displaystyle ^{*}}
denoting the complex conjugate). The interaction Hamiltonian can then be shown to be
{\displaystyle H_{1}=-\hbar \left(\Omega e^{-i\omega _{L}t}+{\tilde {\Omega }}e^{i\omega _{L}t}\right)|{\text{e}}\rangle \langle {\text{g}}|-\hbar \left({\tilde {\Omega }}^{*}e^{-i\omega _{L}t}+\Omega ^{*}e^{i\omega _{L}t}\right)|{\text{g}}\rangle \langle {\text{e}}|}
{\displaystyle \Omega =\hbar ^{-1}{\vec {d}}_{\text{eg}}\cdot {\vec {E}}_{0}}
is the Rabi frequency and
{\displaystyle {\tilde {\Omega }}\mathrel {:=} \hbar ^{-1}{\vec {d}}_{\text{eg}}\cdot {\vec {E}}_{0}^{*}}
is the counter-rotating frequency. To see why the
{\displaystyle {\tilde {\Omega }}}
terms are called counter-rotating consider a unitary transformation to the interaction or Dirac picture where the transformed Hamiltonian
{\displaystyle H_{1,I}}
{\displaystyle H_{1,I}=-\hbar \left(\Omega e^{-i\Delta \omega t}+{\tilde {\Omega }}e^{i(\omega _{L}+\omega _{0})t}\right)|{\text{e}}\rangle \langle {\text{g}}|-\hbar \left({\tilde {\Omega }}^{*}e^{-i(\omega _{L}+\omega _{0})t}+\Omega ^{*}e^{i\Delta \omega t}\right)|{\text{g}}\rangle \langle {\text{e}}|,}
{\displaystyle \Delta \omega \mathrel {:=} \omega _{L}-\omega _{0}}
is the detuning between the light field and the atom.
Making the approximation[edit]
Two-level-system on resonance with a driving field with (blue) and without (green) applying the rotating-wave approximation.
This is the point at which the rotating wave approximation is made. The dipole approximation has been assumed, and for this to remain valid the electric field must be near resonance with the atomic transition. This means that
{\displaystyle \Delta \omega \ll \omega _{L}+\omega _{0}}
and the complex exponentials multiplying
{\displaystyle {\tilde {\Omega }}}
{\displaystyle {\tilde {\Omega }}^{*}}
can be considered to be rapidly oscillating. Hence on any appreciable time scale, the oscillations will quickly average to 0. The rotating wave approximation is thus the claim that these terms may be neglected and thus the Hamiltonian can be written in the interaction picture as
{\displaystyle H_{1,I}^{\text{RWA}}=-\hbar \Omega e^{-i\Delta \omega t}|{\text{e}}\rangle \langle {\text{g}}|-\hbar \Omega ^{*}e^{i\Delta \omega t}|{\text{g}}\rangle \langle {\text{e}}|.}
Finally, transforming back into the Schrödinger picture, the Hamiltonian is given by
{\displaystyle H^{\text{RWA}}={\frac {\hbar \omega _{0}}{2}}|{\text{e}}\rangle \langle {\text{e}}|-{\frac {\hbar \omega _{0}}{2}}|{\text{g}}\rangle \langle {\text{g}}|-\hbar \Omega e^{-i\omega _{L}t}|{\text{e}}\rangle \langle {\text{g}}|-\hbar \Omega ^{*}e^{i\omega _{L}t}|{\text{g}}\rangle \langle {\text{e}}|.}
Another criterion for rotating wave approximation is the weak coupling condition, that is, the Rabi frequency should be much less than the transition frequency.[1]
At this point the rotating wave approximation is complete. A common first step beyond this is to remove the remaining time dependence in the Hamiltonian via another unitary transformation.
Given the above definitions the interaction Hamiltonian is
{\displaystyle {\begin{aligned}H_{1}=-{\vec {d}}\cdot {\vec {E}}&=-\left({\vec {d}}_{\text{eg}}|{\text{e}}\rangle \langle {\text{g}}|+{\vec {d}}_{\text{eg}}^{*}|{\text{g}}\rangle \langle {\text{e}}|\right)\cdot \left({\vec {E}}_{0}e^{-i\omega _{L}t}+{\vec {E}}_{0}^{*}e^{i\omega _{L}t}\right)\\&=-\left({\vec {d}}_{\text{eg}}\cdot {\vec {E}}_{0}e^{-i\omega _{L}t}+{\vec {d}}_{\text{eg}}\cdot {\vec {E}}_{0}^{*}e^{i\omega _{L}t}\right)|{\text{e}}\rangle \langle {\text{g}}|-\left({\vec {d}}_{\text{eg}}^{*}\cdot {\vec {E}}_{0}e^{-i\omega _{L}t}+{\vec {d}}_{\text{eg}}^{*}\cdot {\vec {E}}_{0}^{*}e^{i\omega _{L}t}\right)|{\text{g}}\rangle \langle {\text{e}}|\\&=-\hbar \left(\Omega e^{-i\omega _{L}t}+{\tilde {\Omega }}e^{i\omega _{L}t}\right)|{\text{e}}\rangle \langle {\text{g}}|-\hbar \left({\tilde {\Omega }}^{*}e^{-i\omega _{L}t}+\Omega ^{*}e^{i\omega _{L}t}\right)|{\text{g}}\rangle \langle {\text{e}}|,\end{aligned}}}
as stated. The next step is to find the Hamiltonian in the interaction picture,
{\displaystyle H_{1,I}}
. The required unitary transformation is
{\displaystyle U=e^{iH_{0}t/\hbar }=e^{i\omega _{0}t|{\text{e}}\rangle \langle {\text{e}}|}=|{\text{g}}\rangle \langle {\text{g}}|+e^{i\omega _{0}t}|{\text{e}}\rangle \langle {\text{e}}|}
where the last step can be seen to follow e.g. from a Taylor series expansion with the fact that
{\displaystyle |{\text{g}}\rangle \langle {\text{g}}|+|{\text{e}}\rangle \langle {\text{e}}|=1}
, and due to the orthogonality of the states
{\displaystyle |{\text{g}}\rangle }
{\displaystyle |{\text{e}}\rangle }
. The substitution for
{\displaystyle H_{0}}
in the second step being different from the definition given in the previous section can be justified either by shifting the overall energy levels such that
{\displaystyle |{\text{g}}\rangle }
{\displaystyle 0}
{\displaystyle |{\text{e}}\rangle }
{\displaystyle \hbar \omega _{0}}
, or by noting that a multiplication by an overall phase (
{\displaystyle e^{i\omega _{0}t/2}}
in this case) on a unitary operator does not affect the underlying physics. We now have
{\displaystyle {\begin{aligned}H_{1,I}&\equiv UH_{1}U^{\dagger }\\&=-\hbar \left(\Omega e^{-i\omega _{L}t}+{\tilde {\Omega }}e^{i\omega _{L}t}\right)e^{i\omega _{0}t}|{\text{e}}\rangle \langle {\text{g}}|-\hbar \left({\tilde {\Omega }}^{*}e^{-i\omega _{L}t}+\Omega ^{*}e^{i\omega _{L}t}\right)|{\text{g}}\rangle \langle {\text{e}}|e^{-i\omega _{0}t}\\&=-\hbar \left(\Omega e^{-i\Delta \omega t}+{\tilde {\Omega }}e^{i(\omega _{L}+\omega _{0})t}\right)|{\text{e}}\rangle \langle {\text{g}}|-\hbar \left({\tilde {\Omega }}^{*}e^{-i(\omega _{L}+\omega _{0})t}+\Omega ^{*}e^{i\Delta \omega t}\right)|{\text{g}}\rangle \langle {\text{e}}|\ .\end{aligned}}}
Now we apply the RWA by eliminating the counter-rotating terms as explained in the previous section, and finally transform the approximate Hamiltonian
{\displaystyle H_{1,I}^{\text{RWA}}}
back to the Schrödinger picture:
{\displaystyle {\begin{aligned}H_{1}^{\text{RWA}}&=U^{\dagger }H_{1,I}^{\text{RWA}}U\\&=-\hbar \Omega e^{-i\Delta \omega t}e^{-i\omega _{0}t}|{\text{e}}\rangle \langle {\text{g}}|-\hbar \Omega ^{*}e^{i\Delta \omega t}|{\text{g}}\rangle \langle {\text{e}}|e^{i\omega _{0}t}\\&=-\hbar \Omega e^{-i\omega _{L}t}|{\text{e}}\rangle \langle {\text{g}}|-\hbar \Omega ^{*}e^{i\omega _{L}t}|{\text{g}}\rangle \langle {\text{e}}|.\end{aligned}}}
The atomic Hamiltonian was unaffected by the approximation, so the total Hamiltonian in the Schrödinger picture under the rotating wave approximation is
{\displaystyle H^{\text{RWA}}=H_{0}+H_{1}^{\text{RWA}}={\frac {\hbar \omega _{0}}{2}}|{\text{e}}\rangle \langle {\text{e}}|-{\frac {\hbar \omega _{0}}{2}}|{\text{g}}\rangle \langle {\text{g}}|-\hbar \Omega e^{-i\omega _{L}t}|{\text{e}}\rangle \langle {\text{g}}|-\hbar \Omega ^{*}e^{i\omega _{L}t}|{\text{g}}\rangle \langle {\text{e}}|.}
^ a b Wu, Ying; Yang, Xiaoxue (2007). "Strong-Coupling Theory of Periodically Driven Two-Level Systems". Physical Review Letters. 98 (1): 013601. Bibcode:2007PhRvL..98a3601W. doi:10.1103/PhysRevLett.98.013601. ISSN 0031-9007. PMID 17358474.
^ Mäkelä, H.; Möttönen, M. (13 November 2013). "Effects of the rotating-wave and secular approximations on non-Markovianity". Physical Review A. 88 (5): 052111. doi:10.1103/PhysRevA.88.052111.
Retrieved from "https://en.wikipedia.org/w/index.php?title=Rotating_wave_approximation&oldid=1082671388"
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Contrast with documentation generationEdit
Creation of macrosEdit
Program as a webEdit
Order of human logic, not that of the compilerEdit
Record of the train of thoughtEdit
Remarkable examplesEdit
Literate programming practicesEdit
Mercury Python Python, TypeScript JSON format specification for ipynb Mercury turns Jupyter Notebook into interactive computational documents. They can be published as web application, dashboards, reports, REST API or slides. The excuted document can be exported as standalone HTML or PDF file. Documents can be scheduled for automatic execution. The document presence and widgets are controlled with YAML header in the first cell of the notebook.
{\displaystyle {\begin{aligned}&comp::(\beta \to \gamma )\to (\alpha \to \beta )\to (\alpha \to \gamma )\\&(g\operatorname {comp} f)x=g(fx)\end{aligned}}}
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1 UFR des Sciences Biologiques, Université de Péléforo Gon Coulibaly, Korhogo, Cote d’Ivoire.
2 Université de Rennes 1, Institut des Sciences Chimiques de Rennes (ISCR), UMR CNRS 6226, Groupe ICMV, Bat. 10A, Campus de Beaulieu, Rennes Cedex, France.
3 Laboratoire de Thermodynamique et de Physico-Chimie du Milieu, UFR SFA, Université Nangui Abrogoua, Abidjan, Côte-d’Ivoire.
Abstract: This theoretical chemical reactivity study was conducted using the Density Functional Theory (DFT) method, at computational level B3LYP/6-31G (d). It involved a series of six (06) 5-arylidene rhodanines and allowed to predict the chemical reactivity of these compounds. DFT global chemical reactivity descriptors (HOMO and LUMO energies, chemical hardness, softness, electronegativity) were examined to predict the relative stability and reactivity of rhodanin derivatives. Thus, the compound 6 which has an energy gap between the orbitals of ΔEgap = 3.004 eV is the most polarizable, the most reactive, the least stable, the best electron donor and the softest molecule. Calculation of the local indices of reactivity as well as dual descriptors revealed that the sulfur heteroatom of the Rhodanine ring is the privileged site of electrophilic attack in a state of sp3 hybridization and privileged site of nucleophilic attack in a state of sp2 hybridization.
Keywords: Rhodanine Derivatives, Global Descriptors, Local Descriptors, Dual Descriptors
I=-{E}_{\text{HOMO}}
A=-{E}_{\text{LUMO}}
\chi =-\mu =-1/2\left({E}_{\text{LUMO}}+{E}_{\text{HOMO}}\right)
\eta =\left({E}_{\text{LUMO}}-{E}_{\text{HOMO}}\right)/2
\omega =\frac{{\chi }^{2}}{2\eta }
\sigma =1/\eta
\left({f}_{k}^{+},{f}_{k}^{-}\right)
\left({\sigma }_{k}^{+},{\sigma }_{k}^{-}\right)
\left({\omega }_{k}^{+},{\omega }_{k}^{-}\right)
{f}_{k}^{+}
{f}_{k}^{-}
{\sigma }_{k}^{+}
{\omega }_{k}^{+}
{\sigma }_{k}^{-}
{\omega }_{k}^{-}
{f}_{k}^{+}={q}_{k}\left(N+1\right)-{q}_{k}\left(N\right)
{f}_{k}^{-}={q}_{k}\left(N\right)-{q}_{k}\left(N-1\right)
{\sigma }_{k}^{+}=\sigma {f}_{k}^{+}
{\sigma }_{k}^{-}=\sigma {f}_{k}^{-}
{\omega }_{k}^{+}=\omega {f}_{k}^{+}
{\omega }_{k}^{-}=\omega {f}_{k}^{-}
{q}_{k}\left(N\right)
{q}_{k}\left(N+1\right)
{q}_{k}\left(N-1\right)
\Delta f={f}_{k}^{+}-{f}_{k}^{-}
\Delta \sigma ={\sigma }_{k}^{+}-{\sigma }_{k}^{-}
\Delta \omega ={\omega }_{k}^{+}-{\omega }_{k}^{-}
Cite this paper: Coulibaly, W. , N’dri, J. , Koné, M. , Dago, C. , Ambeu, C. , Bazureau, J. and Ziao, N. (2019) Studies of the Chemical Reactivity of a Series of Rhodanine Derivatives by Approaches to Quantum Chemistry. Computational Molecular Bioscience, 9, 49-62. doi: 10.4236/cmb.2019.93005.
[1] Souness, E., Aldous, D. and Sargent, C. (2004) Immunosuppressive and Anti-Inflammatory Effects of Cyclic AMP Phosphodiesterase (PDE) Type 4 Inhibitors. Immunopharmacology, 47, 127-162.
[2] Achson, A. (2009) An Introduction to the Chemistry of Heterocyclic Compounds. 3rd Edition, Wiley, Hoboken.
[3] Dua, R., Shrivastava, S., Sonwane, S.K. and Srivastava, S.K. (2011) Pharmacological Significance of Synthetic Heterocycles Scaffold: A Review. Adv. Biol. Res. (Rennes), 5, 120-144.
[4] Coulibaly, W.K. (2012) Conception, synthèse de bis-(5-arylidène-imidazolinone) et de bis-(5-arylidène thiazolidinone) dérivés du modèle de la pentamidine (pour la malaria), et leurs évaluations biologiques. Thèse de Doctorat, No. 4549, 1-236.
[5] Jain, A.K., Vaidya, A., Ravichandran, V., Kasshaw, S.K. and Agrawal, R.K. (2012) Recent Developments and Biological Activities of Thiazolidinone Derivatives: A Review. Bioorganic & Medicinal Chemistry, 20, 3378-3395.
[6] Kaur Manjal, S., et al. (2017) Synthetic and Medicinal Perspective of Thiazolidinones: A Review. Bioorganic Chemistry, 75, 406-423.
[7] Foye, W.O. and Tovivich, P.J. (1977) N-Glucopyranosyl-5-Aralkylidenerhodanines: Synthesis and Antibacterial and Antiviral Activities. Journal of Pharmaceutical Sciences, 66, 1607-1611.
[8] Alegaon, S.G., Alagawadi, K.R., Sonkusare, P.V., Chaudhary, S.M., Dadwe, D.H. and Shah, A.S. (2012) Novel Imidazo [2,1-b] [1,3,4] Thiadiazole Carring Rhodanine-3-Acetic Acid as Potential Antitubercular Agents. Bioorganic & Medicinal Chemistry Letters, 22, 1917-1921.
[9] Radi, M., Botta, L., Casaluce, G., Bernardini, M. and Botta, M. (2010) Practical One-Pot Two-Step Protocol for the Microwave-Assisted Synthesis of Highly Functionalized Rhodanine Derivatives. Journal of Combinatorial Chemistry, 12, 200-205.
[10] Benson, S.W. and Buss, J.H. (1958) Additivity Rules for the Estimation of Molecular Properties. Thermodynamic Properties. The Journal of Chemical Physics, 29, 546-572.
[11] Kumar, V., Khandare, D.G., Chatterjee, A. and Banerjee, M. (2013) Protic Ionic Liquids as Recyclable Solvents for the Acid Catalysed Synthesis of Diphenylmethyl Thioethers. Tetrahedron Letters, 54, 5505-5509.
[12] Deohate, P.P., Deohate, J.P. and Berad, B.N. (2004) Synthesis of Some Novel 1,2,4-Dithiazolidines and Their Antibacterial and Antifungal Activity. Asian Journal of Chemistry, 16, 255-260.
[13] Rao, V.J., Reddy, T.N., Ravinder, M., Bagul, P., Ravikanti, K., Bagul, C., Nanubolu, J.B., Srinivas, K. and Banerjee, S.K. (2014) Synthesis and Biological Evaluation of New Epalrestat Analogues as Aldose Reductase Inhibitors (ARIs). European Journal of Medicinal Chemistry, 71, 53-66.
[14] Sha, X., Zhang, H. and Huang, W. (2009) Design and Synthesis of Thiazolidinedione Derivatives as Selective PPARγ Modulators. Letters in Drug Design & Discovery, 6, 224-227.
[15] Bozdag, D.O., Verspohl, E.J., Das, E., Kaup, R.M., Bauer, K., Sarikaya, M., Evi, B. and Ertan, R. (2008) Synthesis and Biological Activity of Some New Flavonyl-2,4-Thiazolidinediones. Bioorganic & Medicinal Chemistry, 16, 6747-6751.
[16] Rao, B.A., Rajpoot, R.S., Naidu, V.G.M., Srinivas, K., Ramakrishna, S. and Rao, V.J. (2011) Synthesis and in Vitro Antitumor Activity of New Nicotinyl-Rhodanine Derivatives. International Journal of Pharma and Bio Sciences, 2, 191-202.
[17] Patel, N.B. and Shaikh, F.M. (2010) New 4-Thiazolidinones of Nicotinic Acid with 2-Amino-6-Methylbenzothiazole and Their Biological Activity. Scientia Pharma-ceutica, 78, 753-765.
[18] Foresman, J.B. and Frisch, A. (1996) Exploring Chemistry with Electronic Structure Methods. Gaussian Inc., Pittsburgh.
[19] Kurt, M., Sertbakan, T.R. and Ozduran, M. (2008) Spectrochim: An Experimental and Theoretical Study of Molecular Structure and Vibrational Spectra of 3- and 4-Pyridineboronic Acid Molecules by Density Functional Theory Calculations. Spectrochimica Acta Part A: Molecular and Biomolecular Spectroscopy, 70, 664-673.
[20] Abdel-Rahman, L.H., El-Khatib, R.M., Nassr, L.A.E., Abu-Dief, A.M., Ismael, M. and Seleem, A.A. (2014) Metal Based Pharmacologically Active Agents: Synthesis, Structural Characterization, Molecular Modeling, CT-DNA Binding Studies and in Vitro Antimicrobial Screening of Iron (II) Bromosalicylidene Amino Acid Chelates. Spectrochimica Acta Part A: Molecular and Biomolecular Spectroscopy, 117, 366-378.
[21] Abdel-Rahman, L.H., Abu-Dief, A.M., Adam, M.S.S. and Hamdan, S.K. (2016) Some New Nano-Sized Mononuclear Cu(II) Schiff Base Complexes: Design, Characterization, Molecular Modeling and Catalytic Potentials in Benzyl Alcohol Oxidation. Catalysis Letters, 146, 1373-1396.
[22] Abdel Rahman, L.H., Abu-Dief, A.M., Moustafa, H. and Hamdan, S.K. (2017) Ni(II) and Cu(II) Complexes with ONNO Asymmetric Tetradentate Schiff Base Ligand: Synthesis, Spectroscopic Characterization, Theoretical Calculations, DNA Interaction and Antimicrobial Studies. Applied Organometallic Chemistry, 31, e3555.
[24] Parr, R.G. and Yang, W. (1995) Density-Functional Theory of the Electronic Structure of Molecules. Annual Review of Physical Chemistry, 46, 701-728.
[25] Parr, R.G. and Yang, W. (1989) Density Functional Theory of Atoms and Molecules. Oxford University Press, Oxford.
[26] Fukui, K., Yonezawa, Y. and Shingu, H. (1952) A Molecular Orbital Theory of Reactivity in Aromatic Hydrocarbons. The Journal of Chemical Physics, 20, 722-725.
[28] Ayers, P.W. and Levy, M. (2000) Density Functional Approach to the. Frontier-Electron Theory of Chemical Reactivity. Theoretical Chemistry Accounts, 103, 353-360.
[29] Lee, W.Y. and Parr, R.G. (1988) Development of the Colle-Salvetti Correlation-Energy Formula into a Functional of the Electron Density. Physical Review B, 37, 785-789.
[31] Frisch, M.J., Trucks, G.W., Schlegel, H.B. and Scuseria, G.E. (2009) Gaussian 09, Revision A 02. Gaussian, Inc., Wallingford.
[32] Kapp, J., Remko, M. and Schleyer, P.V.R. (1996) H2XO and (CH3)2XO Compounds (X = C, Si, Ge, Sn, Pb): Double Bonds vs. Carbene-Like Structures Can the Metal Compounds Exist at All? Journal of the American Chemical Society, 118, 5745-5751.
[33] Johnson, B.G., Gill, P.M. and Pople, J.A. (1993) The Performance of a Family of Density Functional Methods. The Journal of Chemical Physics, 98, 5612-5626.
[34] XLSTAT Version 2014.5.03 Copyright Addinsoft 1995-2014 (2014) XLSTAT and Addinsoft Are Registered Trademarks of Addinsoft.
[35] Mulliken, R.S. (1962) Criteria for the Construction of Good Self-Consistent-Field Molecular Orbital Wave Functions, and the Significance of LCAO-MO Population Analysis. The Journal of Chemical Physics, 36, 3428-3439.
[36] Koopmans, T. (1934) úber die Zuordnung von Wellenfunktiomen und Eigenwerten zu den einzelnen Elektronen eines Atoms. Physica, 1, 104-113.
[38] Yang, W. and Mortier, W. (1986) The Use of Global and Local Molecular Parameters for the Analysis of the Gas-Phase Basicity of Amines. Journal of the American Chemical Society, 108, 5708-5711.
[39] Yang, W. and Parr, R.G. (1985) Chemistry. Hardness, Softness, and the Fukui Function in the Electronic Theory of Metals and Catalysis. Proceedings of the National Academy of Sciences of the United States of America, 82, 6723-6726.
[40] Chattaraj, P.K., Maiti, B. and Sarkar, U. (2003) Philicity: A Unified Treatment of Chemical Reactivity and Selectivity. The Journal of Physical Chemistry, 107, 4973-4975.
[41] Fuentealba, P. and Contreras, R. (2002) Fukui Function in Chemistry. In: Sen, K.D., Ed., Reviews in Modern Quantum Chemistry: A Celebration of the Contributions of Robert G Parr, World Scientific, River Edge, 1013-1052.
[43] Morell, C., Grand, A. and Toro-Labbé, A. (2005) New Dual Descriptor for Chemical Reactivity. The Journal of Physical Chemistry, 109, 205-212.
[44] Morell, C., Grand, A. and Toro-Labbé, A. (2006) Theoretical Support for Using the Delta f(r). Descriptor. Chemical Physics Letters, 425, 342-346.
[45] Abdel-Baset, H.M., Hanan, G.E., El-Okr, M.M. and Ibrahim, M.A. (2015) Electrostatic Potential Maps of Fullerene C60 and Some of Its Specific Derivatives: Density Functional Theory Approach. Journal of Nanomaterials Molecular Nanotechnology, 4, 2.
[46] Hong, L.X., Ling, C.H., Zhou, Z.R. and Zhou, Z.X. (2015) Molecular and Biomolecular Spectroscopy. Spectrochimica Acta Part A, 137, 321-327.
[47] Woodward, R.B. and Hoffmann, R. (1970) The Conservation of Orbital Symmetry. Chemie, Weinheim.
[49] Lewis, D.F.V., Ioannides, C. and Parke, D.V. (1994) Interaction of a Series of Nitriles with the Alcohol-Inducible Isoform of P450: Computer Analysis of Structure-Activity Relationships. Xenobiotica, 24, 401-408.
[50] Uesugi, Y., Mizuno, M., Shimojima, A. and Takahashi, H. (1997) Transient Resonance Raman and Ab-Initio Mo Calculation Studies of the Structures and Vibrational Assignments of the t1 State and the Anion Radical of Coumarin and Its Isotopically Substituted Analogues. The Journal of Physical Chemistry, 101, 268-274.
[51] Sinha, L., Prasad, O., Narayan, V., Shukla, S. and Raman, R. (2011) FT-IR Spectroscopic Analysis and First-Order Hyperpolarisability of 3-benzoyl-5-chlorouracil by First Principles. Molecular Simulation, 37, 153-163.
[52] Pearson, R.G. (1963) Hard and Soft Acids and Bases. Journal of the American Chemical Society, 85, 3533-3539.
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Recover bits from VHT-Data field - MATLAB wlanVHTDataRecover - MathWorks Deutschland
wlanVHTDataRecover
Recover Bits from VHT Signal Transmitted Through 2x2 Fading Channel
Recover Bits from VHT Signal Transmitted through MU-MIMO Channel
Recover Bits from VHT-Data Field Using Zero-Forcing Equalization
rxDataSig
cfgVHT
numSTS
crcBits
Recover bits from VHT-Data field
dataBits = wlanVHTDataRecover(rxDataSig,chEst,noiseVarEst,cfgVHT)
dataBits = wlanVHTDataRecover(rxDataSig,chEst,noiseVarEst,cfgVHT,userIdx)
dataBits = wlanVHTDataRecover(rxDataSig,chEst,noiseVarEst,cfgVHT,userIdx,numSTS)
dataBits = wlanVHTDataRecover(___,Name,Value)
[dataBits,crcBits] = wlanVHTDataRecover(___)
[dataBits,crcBits,eqSym] = wlanVHTDataRecover(___)
[dataBits,crcBits,eqSym,cpe] = wlanVHTDataRecover(___)
dataBits = wlanVHTDataRecover(rxDataSig,chEst,noiseVarEst,cfgVHT) recovers dataBits, a column vector of bits, from rxDataSig, the received VHT-Data field of a very-high-throughput (VHT) single-user transmission. The function recovers dataBits by using chEst, a channel estimate for the occupied subcarriers, noiseVarEst, an estimate of noise variance, and cfgVHT, a configuration object that contains VHT transmission parameters.
For more information about the VHT-Data field, see VHT-Data Field.
dataBits = wlanVHTDataRecover(rxDataSig,chEst,noiseVarEst,cfgVHT,userIdx) recovers dataBits for one user, specified by user index userIdx, in a VHT multi-user transmission.
dataBits = wlanVHTDataRecover(rxDataSig,chEst,noiseVarEst,cfgVHT,userIdx,numSTS) recovers dataBits for one user in a VHT multi-user transmission for numSTS, the number of space-time streams in the transmission.
dataBits = wlanVHTDataRecover(___,Name,Value) specifies algorithm options by using one or more name-value pair arguments, in addition to any input argument combination from previous syntaxes. For example, 'LDPCDecodingMethod','layered-bp' specifies the layered belief propagation low-density parity-check (LDPC) decoding algorithm.
[dataBits,crcBits] = wlanVHTDataRecover(___) returns the VHT-SIG-B checksum bits, crcBits, using any input argument combination from previous syntaxes.
[dataBits,crcBits,eqSym] = wlanVHTDataRecover(___) returns eqSym, the equalized OFDM symbols that comprise the data subcarriers of the VHT-Data field, using any input argument combination from the previous syntaxes.
[dataBits,crcBits,eqSym,cpe] = wlanVHTDataRecover(___) returns cpe, the common phase error between the received and expected OFDM symbols, using any input argument combination from the previous syntaxes.
Recover bits from the VHT-Data field of a VHT waveform transmitted though a 2x2 fading channel by using channel estimation on the VHT long training field (VHT-LTF).
Configure a VHT transmission with a channel bandwidth of 160 MHz, two transmit antennas, and two transmission paths.
cfgVHT = wlanVHTConfig('ChannelBandwidth','CBW160','NumTransmitAntennas',2,'NumSpaceTimeStreams',2,'APEPLength',512);
Generate VHT-LTF and VHT-Data fields signals.
psduLength = 8*cfgVHT.PSDULength;
bits = randi([0 1],psduLength,1);
txLTF = wlanVHTLTF(cfgVHT);
txDataSig = wlanVHTData(bits,cfgVHT);
Pass the transmitted waveform through a 2x2 quasi-static fading channel with additive white Gaussian noise (AWGN).
H = complex(randn(2,2),randn(2,2))/sqrt(2);
rxLTF = awgn(txLTF*H,snr);
rxDataSig = awgn(txDataSig*H,snr);
Calculate the received signal power and estimate the noise variance.
powerDB = 10*log10(var(rxDataSig));
noiseVarEst = mean(10.^(0.1*(powerDB-snr)));
Perform channel estimation based on the VHT-LTF.
sym = wlanVHTLTFDemodulate(rxLTF,cfgVHT,1);
chEst = wlanVHTLTFChannelEstimate(sym,cfgVHT);
Recover the bits from the received VHT-Data field and confirm that the received bits match the transmitted bits.
dataBits = wlanVHTDataRecover(rxDataSig,chEst,noiseVarEst,cfgVHT);
numErr = biterr(bits,dataBits)
Recover bits from the VHT-Data field of a VHT multi-user transmission recovered from a fading MU-MIMO channel by using channel estimation on the VHT-LTF.
This example can return high bit error rates because the transmission does not include precoding to mitigate the interference between space-time streams. However, the example shows a typical VHT signal recovery workflow and appropriate syntax use for the functions involved.
Configure a VHT transmission with a channel bandwidth of 160 MHz, two users, and four transmit antennas. Assign one space-time stream to the first user and three space-time streams to the second user.
numSTS = [1 3];
cfgVHT = wlanVHTConfig('ChannelBandwidth',cbw,'NumUsers',2, ...
'NumTransmitAntennas',4,'NumSpaceTimeStreams',numSTS);
Generate a payload of bits for each user. This payload must be in the a 1-by-N cell array, where N is the number of users.
numUsers = cfgVHT.NumUsers;
bits = cell(1,2);
for nu = 1:numUsers
bits{nu} = randi([0 1],psduLength(nu),1);
Generate VHT-LTF and VHT-Data field signals.
txDataSym = wlanVHTData(bits,cfgVHT);
Pass the VHT-Data field signal for the first user through a 4x1 channel because this signal consists of a single space-time stream. Pass the VHT-Data field for the second user data through a 4x3 channel because this signal consists of three space-time streams. Apply AWGN to each signal, assuming an SNR of 15 dB.
H{1} = complex(randn(4,1),randn(4,1))/sqrt(2);
number = zeros(2,1);
ratio = zeros(2,1);
rxDataSym = awgn(txDataSym*H{userIdx},snr,'measured');
Apply the same channel processing to the VHT-LTF for each user.
rxLTF = awgn(txLTF*H{userIdx},snr,'measured');
Calculate the received signal power for each user and estimate the noise variance.
powerDB = 10*log10(var(rxDataSym));
Estimate the channel characteristics by using the VHT-LTF.
demod = wlanVHTLTFDemodulate(rxLTF,cbw,numSTS);
chEst = wlanVHTLTFChannelEstimate(demod,cbw,numSTS);
Recover the bits from the received VHT-Data field for each user and determine the bit error rate by comparing the recovered bits with the original payload bits.
dataBits = wlanVHTDataRecover(rxDataSym,chEst,noiseVarEst,cfgVHT,userIdx);
[number(userIdx),ratio(userIdx)] = biterr(bits{userIdx},dataBits);
disp(number(userIdx))
disp(ratio(userIdx))
Recover bits from the VHT-Data field signal of a VHT transmission recovered from a SISO AWGN channel by using a zero-forcing equalization algorithm.
Configure a VHT transmission and generate the VHT-Data field for a random payload of bits.
cfgVHT = wlanVHTConfig('APEPLength',512);
Pass the transmission through an AWGN channel.
rxDataSig = awgn(txDataSig,snr);
Recover the payload bits using a perfect channel estimate of all ones and zero-forcing equalization.
chEst = ones(242,1);
[dataBits,crcBits,eqSym,cpe] = wlanVHTDataRecover(rxDataSig,chEst,noiseVarEst,cfgVHT,'EqualizationMethod','ZF');
Verify that the recovered signal contains no bit errors.
number = biterr(bits,dataBits)
Display the VHT-Data field CRC checksum bits.
disp(crcBits')
Calculate and display the maximum common phase error.
max(abs(cpe))
rxDataSig — Received VHT-Data field
Received VHT-Data field, specified as a complex-valued array of size NS-by-NR.
NS is an integer greater than or equal to the number of time-domain samples.
The function processes one PPDU data field per entry. If you specify NS as a value greater than the field length, the function does not process additional samples at the end of rxDataSig. To process a concatenated stream of PPDU data fields, you must call the function multiple times.
chEst — Channel estimate for occupied subcarriers
Channel estimate for occupied subcarriers, specified as a complex-valued array of size NST-by-NSTS-by-NR.
NST is the number of occupied subcarriers, which depends on the ChannelBandwidth property of the cfgVHT input in accordance with this table.
Value of ChannelBandwidth Property
NSTS is the number of space-time streams, which must match the NumSpacetimeStreams property of the cfgVHT input. For multi-user transmissions, NSTS is the total number of space-time streams for all users.
cfgVHT — VHT transmission configuration
VHT transmission configuration, specified as a wlanVHTConfig object.
userIdx — User index
integer in the interval [1, NUsers]
User index, specified as an integer in the interval [1, NUsers], where NUsers is the total number of users in the transmission.
numSTS — Number of space-time streams
integer in the interval [1, 4] | row vector of integers in the interval [1, 4]
Number of space-time streams.
For a single-user transmission, specify this input as an integer in the interval [1, 4]
for a multi-user transmission, specify this input as a row vector of integers in the interval [1, 4] of length NUsers, where NUsers is the total number of users in the transmission.
Example: [1 3 2] indicates the number of space-time streams in a three-user transmission. In this case, the transmission allocates one, three, and two space-time streams to the first, second, and third users, respectively.
Equalization method, specified as the comma-separated pair consisting of 'EqualizationMethod' and one of these values.
Specify this argument as 'ZF' when either of these conditions applies.
The NumSpaceTimeStreams property of the cfgVHT input is 1.
The NumSpaceTimeStreams and STBC properties of the cfgVHT input are 2 and 1 (true), respectively.
When you specify this input as 'norm-min-sum' or 'offset-min-sum', the function sets input log-likelihood ratio (LLR) values that are greater than 1e10 or less than -1e10 to 1e10 and -1e10, respectively. The function then uses these values when executing the LDPC decoding algorithm.
To enable this argument, set the ChannelCoding property of the cfgVHT input to 'LDPC' for the user corresponding to the userIdx input.
When you set this value to 0 (false), LDPC decoding completes the number of iterations specified by 'MaximumLDPCIterationCount' regardless of parity check status.
dataBits — Bits recovered from VHT-Data field
Bits recovered from the VHT-Data field, returned as a column vector of length 8×LPSDU, where LPSDU is the length of the PSDU in bytes.
crcBits — VHT-SIG-B checksum bits
VHT-SIG-B checksum bits, returned as a binary-valued column vector of length 8.
eqSym — Equalized OFDM symbols
Equalized OFDM symbols comprising the VHT-Data field, returned as a complex-valued array of size NSD-by-NSym-by-NSS.
NSD is the number of data subcarriers
NSYM is the number of OFDM symbols in the VHT-Data field
NSS is the number of spatial streams. When the STBC property of the cfgVHT input is 0 (false), NSS is equal to NSTS, the number of space-time streams in the transmission. When the STBC property of the cfgVHT input is 0 (false), NSS is equal to NSTS/2.
Common phase error between the received and expected OFDM symbols, in radians, returned as a real-valued column vector. The length of this output is NSym, the number of OFDM symbols in the VHT-Data field.
For a detailed description of the VHT-Data field, see section 21.3.10 of IEEE® Std 802.11™-2016. The VHT Data field consists of four subfields.
c=\left({c}_{0},{c}_{1},\dots ,{c}_{n-1}\right)
, the input to the LDPC decoder is the LLR given by
L\left({c}_{i}\right)=\mathrm{log}\left(\frac{\mathrm{Pr}\left({c}_{i}=0|\text{channel output for }{c}_{i}\right)}{\mathrm{Pr}\left({c}_{i}=0|\text{channel output for }{c}_{i}\right)}\right)
L\left({r}_{ji}\right)=2\text{\hspace{0.17em}}\text{atanh}\text{\hspace{0.17em}}\left(\prod _{{i}^{\prime }\in {V}_{j}\\left\{i\right\}}\mathrm{tanh}\left(\frac{1}{2}L\left({q}_{{i}^{\prime }j}\right)\right)\right)
L\left({q}_{ij}\right)=L\left({c}_{i}\right)+\sum _{j\text{'}\in {C}_{i}\\left\{j\right\}}L\left({r}_{{j}^{\prime }i}\right)
L\left({q}_{ij}\right)=L\left({c}_{i}\right)
L\left({Q}_{i}\right)=L\left({c}_{i}\right)+\sum _{{j}^{\prime }\in {C}_{i}}L\left({r}_{{j}^{\prime }i}\right)
L\left({Q}_{i}\right)
{c}_{i}
L\left({Q}_{i}\right)
{c}_{i}
L\left({Q}_{i}\right)
L\left({Q}_{i}\right)
{C}_{i}\\left\{j\right\}
{V}_{j}\\left\{i\right\}
{C}_{i}
{V}_{j}
H
H{c}^{T}=0
The function implements the layered BP algorithm based on the decoding algorithm presented in Section II.A of [3]. The decoding loop iterates over subsets of rows (layers) of the PCM.
For each row, m, in a layer and each bit index, j, the implementation updates the key components of the algorithm based on these equations.
L\left({q}_{mj}\right)=L\left({q}_{j}\right)-{R}_{mj}
\Psi \left(x\right)=\mathrm{log}\left(|\mathrm{tanh}\left(x/2\right)|\right)
{A}_{mj}=\sum _{n\in N\left(m\right)\\left\{j\right\}}\Psi \left(L\left({q}_{mn}\right)\right)
{s}_{mj}=\prod _{n\in N\left(m\right)\\left\{j\right\}}\mathrm{sgn}\left(L\left({q}_{mn}\right)\right)
{R}_{mj}=-{s}_{mj}\Psi \left({A}_{mj}\right)
L\left({q}_{j}\right)=L\left({q}_{mj}\right)+{R}_{mj}
L\left({q}_{mj}\right)
{R}_{mj}
{A}_{mj}={\mathrm{min}}_{n\in N\left(m\right)\\left\{j\right\}}\left(\alpha |L\left({q}_{mn}\right)|\right)
{A}_{mj}=\mathrm{max}\left({\mathrm{min}}_{n\in N\left(m\right)\\left\{j\right\}}\left(|L\left({q}_{mn}\right)|-\beta \right), 0\right)
[1] IEEE STD 802.11ac™-2013 (Amendment to IEEE Std 802.11-2012, as amended by IEEE Std 802.11ae™-2012, IEEE Std 802.11a™-2012, and IEEE Std 802.11ad™-2012). “Part 11: Wireless LAN Medium Access Control (MAC) and Physical Layer (PHY) Specifications. Amendment 4: Enhancements for Very High Throughput Operation in Bands below 6 GHz.” IEEE Standard for Information technology — Telecommunications and information exchange between systems. Local and metropolitan area networks — Specific requirements.
wlanVHTConfig | wlanVHTData | wlanVHTLTFDemodulate | wlanVHTLTFChannelEstimate | wlanVHTSIGARecover | wlanVHTSIGBRecover
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Create tunable gain surface for gain scheduling - MATLAB tunableSurface
K\left(n\left(\sigma \right)\right)=\gamma \left[{K}_{0}+{K}_{1}{F}_{1}\left(n\left(\sigma \right)\right)+\dots +{K}_{M}{F}_{M}\left(n\left(\sigma \right)\right)\right],
K\left(n\left(\sigma \right)\right)=\gamma \left[{K}_{0}+{K}_{1}{F}_{1}\left(n\left(\sigma \right)\right)+\dots +{K}_{M}{F}_{M}\left(n\left(\sigma \right)\right)\right].
K\left(t\right)={K}_{0}+{K}_{1}n\left(t\right)+{K}_{2}{\left(n\left(t\right)\right)}^{2}.
{K}_{0}
{K}_{1}
{K}_{2}
n\left(t\right)=\left(t-20\right)/20
{f}_{0}\left(t\right)=1
\left[{K}_{0},{K}_{1},{K}_{2}\right]
{K}_{0}
{K}_{1}
{K}_{2}
x=\frac{\alpha -7.5}{7.5},\phantom{\rule{1em}{0ex}}y=\frac{V-450}{150}.
K\left(\alpha ,V\right)={K}_{0}+{K}_{1}x+{K}_{2}y+{K}_{3}xy,
{K}_{0},...,{K}_{3}
\begin{array}{c}{F}_{1}\left(x,y\right)=x\\ {F}_{2}\left(x,y\right)=y\\ {F}_{3}\left(x,y\right)=xy.\end{array}
{\alpha }_{N}
{\beta }_{N}
K\left({\alpha }_{N},{\beta }_{N}\right)={K}_{0}+{K}_{1}{\alpha }_{N}+{K}_{2}{\beta }_{N}+{K}_{3}{\alpha }_{N}{\beta }_{N}.
\left(\alpha ,\beta \right)
\left(\alpha ,\beta \right)=\left\{\begin{array}{l}\left(-0.9,0.05\right)\\ \left(-1.5,0.6\right)\\ \left(-1.5,0.95\right)\\ \left(-2.5,0.5\right)\\ \left(-3.2,0.7\right)\\ \left(-3.9,0.3\right)\end{array}.
\left(\alpha ,\beta \right)
\left(\alpha ,\beta \right)
\left(\alpha ,\beta \right)
\left(\alpha ,\beta \right)
\alpha
\beta
{K}_{ij}\left(n\left(\sigma \right)\right)={K}_{i{j}_{0}}+{K}_{i{j}_{1}}{F}_{1}\left(n\left(\sigma \right)\right)+\dots +{K}_{i{j}_{M}}{F}_{M}\left(n\left(\sigma \right)\right).
G\left(\alpha ,V\right)={G}_{0}+{G}_{1}{\alpha }_{N}+{G}_{2}{V}_{N}+{G}_{3}{\alpha }_{N}{V}_{N}
{K}_{ij}\left(n\left(\sigma \right)\right)={K}_{i{j}_{0}}+{K}_{i{j}_{1}}{F}_{1}\left(n\left(\sigma \right)\right)+\dots +{K}_{i{j}_{M}}{F}_{M}\left(n\left(\sigma \right)\right).
K\left(\sigma \right)=\text{OutputScaling}\left[{K}_{0}+{K}_{1}{F}_{1}\left(n\left(\sigma \right)\right)+\cdots +{K}_{m}{F}_{m}\left(n\left(\sigma \right)\right)\right],
n\left(\sigma \right)=\frac{\sigma -\text{InputOffset}}{\text{InputScaling}}.
\sqrt{\sigma }
|
{x}_{0}
{x}_{0}
{x}_{0}
X\left(t\right)
\mathrm{dX}\left(t\right)=\mathrm{\mu }\left(X\left(t\right),t\right)\mathrm{dt}+\mathrm{\sigma }\left(X\left(t\right),t\right)\mathrm{dW}\left(t\right)
\mathrm{\mu }\left(X\left(t\right),t\right)
\mathrm{\sigma }\left(X\left(t\right),t\right)
W\left(t\right)
{x}_{0}
X
is an
{X}_{1}
{X}_{n}
{\mathrm{\mu }}_{1}
{\mathrm{\mu }}_{n}
{\mathrm{\sigma }}_{1}
{\mathrm{\sigma }}_{n}
be the corresponding drift and diffusion terms. The ItoProcess(X, Sigma) command will create an
Y
{\mathrm{dY}\left(t\right)}_{i}={\mathrm{\mu }}_{i}\left({Y\left(t\right)}_{i},t\right)+{\mathrm{\sigma }}_{i}\left({Y\left(t\right)}_{i},t\right){\mathrm{dW}\left(t\right)}_{i}
W\left(t\right)
is an
\mathrm{with}\left(\mathrm{Finance}\right):
Y≔\mathrm{ItoProcess}\left(1.0,\mathrm{\mu },\mathrm{\sigma },x,t\right)
\textcolor[rgb]{0,0,1}{Y}\textcolor[rgb]{0,0,1}{≔}\textcolor[rgb]{0,0,1}{\mathrm{_X0}}
\mathrm{Drift}\left(Y\left(t\right)\right)
\textcolor[rgb]{0,0,1}{\mathrm{\mu }}
\mathrm{Diffusion}\left(Y\left(t\right)\right)
\textcolor[rgb]{0,0,1}{\mathrm{\sigma }}
\mathrm{Drift}\left(\mathrm{exp}\left(Y\left(t\right)\right)\right)
\textcolor[rgb]{0,0,1}{\mathrm{\mu }}\textcolor[rgb]{0,0,1}{}{\textcolor[rgb]{0,0,1}{ⅇ}}^{\textcolor[rgb]{0,0,1}{\mathrm{_X0}}\textcolor[rgb]{0,0,1}{}\left(\textcolor[rgb]{0,0,1}{t}\right)}\textcolor[rgb]{0,0,1}{+}\frac{{\textcolor[rgb]{0,0,1}{\mathrm{\sigma }}}^{\textcolor[rgb]{0,0,1}{2}}\textcolor[rgb]{0,0,1}{}{\textcolor[rgb]{0,0,1}{ⅇ}}^{\textcolor[rgb]{0,0,1}{\mathrm{_X0}}\textcolor[rgb]{0,0,1}{}\left(\textcolor[rgb]{0,0,1}{t}\right)}}{\textcolor[rgb]{0,0,1}{2}}
\mathrm{Diffusion}\left(\mathrm{exp}\left(Y\left(t\right)\right)\right)
\textcolor[rgb]{0,0,1}{\mathrm{\sigma }}\textcolor[rgb]{0,0,1}{}{\textcolor[rgb]{0,0,1}{ⅇ}}^{\textcolor[rgb]{0,0,1}{\mathrm{_X0}}\textcolor[rgb]{0,0,1}{}\left(\textcolor[rgb]{0,0,1}{t}\right)}
\mathrm{\mu }≔0.1
\textcolor[rgb]{0,0,1}{\mathrm{\mu }}\textcolor[rgb]{0,0,1}{≔}\textcolor[rgb]{0,0,1}{0.1}
\mathrm{\sigma }≔0.5
\textcolor[rgb]{0,0,1}{\mathrm{\sigma }}\textcolor[rgb]{0,0,1}{≔}\textcolor[rgb]{0,0,1}{0.5}
\mathrm{PathPlot}\left(\mathrm{exp}\left(Y\left(t\right)\right),t=0..3,\mathrm{timesteps}=100,\mathrm{replications}=10\right)
\mathrm{\mu }≔'\mathrm{\mu }'
\textcolor[rgb]{0,0,1}{\mathrm{\mu }}\textcolor[rgb]{0,0,1}{≔}\textcolor[rgb]{0,0,1}{\mathrm{\mu }}
\mathrm{\sigma }≔'\mathrm{\sigma }'
\textcolor[rgb]{0,0,1}{\mathrm{\sigma }}\textcolor[rgb]{0,0,1}{≔}\textcolor[rgb]{0,0,1}{\mathrm{\sigma }}
\mathrm{X0}≔〈100.0,0.〉
\textcolor[rgb]{0,0,1}{\mathrm{X0}}\textcolor[rgb]{0,0,1}{≔}[\begin{array}{c}\textcolor[rgb]{0,0,1}{100.0}\\ \textcolor[rgb]{0,0,1}{0.}\end{array}]
\mathrm{Μ}≔〈\mathrm{\mu }X[1],\mathrm{\kappa }\left(\mathrm{\theta }-X[2]\right)〉
\textcolor[rgb]{0,0,1}{\mathrm{Μ}}\textcolor[rgb]{0,0,1}{≔}[\begin{array}{c}\textcolor[rgb]{0,0,1}{\mathrm{\mu }}\textcolor[rgb]{0,0,1}{}{\textcolor[rgb]{0,0,1}{X}}_{\textcolor[rgb]{0,0,1}{1}}\\ \textcolor[rgb]{0,0,1}{\mathrm{\kappa }}\textcolor[rgb]{0,0,1}{}\left(\textcolor[rgb]{0,0,1}{\mathrm{\theta }}\textcolor[rgb]{0,0,1}{-}{\textcolor[rgb]{0,0,1}{X}}_{\textcolor[rgb]{0,0,1}{2}}\right)\end{array}]
\mathrm{\Sigma }≔〈〈\mathrm{sqrt}\left(X[2]\right)X[1]|0.〉,〈0.|\mathrm{\sigma }X[2]〉〉
\textcolor[rgb]{0,0,1}{\mathrm{\Sigma }}\textcolor[rgb]{0,0,1}{≔}[\begin{array}{cc}\sqrt{{\textcolor[rgb]{0,0,1}{X}}_{\textcolor[rgb]{0,0,1}{2}}}\textcolor[rgb]{0,0,1}{}{\textcolor[rgb]{0,0,1}{X}}_{\textcolor[rgb]{0,0,1}{1}}& \textcolor[rgb]{0,0,1}{0.}\\ \textcolor[rgb]{0,0,1}{0.}& \textcolor[rgb]{0,0,1}{\mathrm{\sigma }}\textcolor[rgb]{0,0,1}{}{\textcolor[rgb]{0,0,1}{X}}_{\textcolor[rgb]{0,0,1}{2}}\end{array}]
S≔\mathrm{ItoProcess}\left(\mathrm{X0},\mathrm{Μ},\mathrm{\Sigma },X,t\right)
\textcolor[rgb]{0,0,1}{S}\textcolor[rgb]{0,0,1}{≔}\textcolor[rgb]{0,0,1}{\mathrm{_X2}}
\mathrm{Drift}\left(S\left(t\right)\right)
[\begin{array}{c}\textcolor[rgb]{0,0,1}{\mathrm{\mu }}\textcolor[rgb]{0,0,1}{}{\textcolor[rgb]{0,0,1}{\mathrm{_X2}}\textcolor[rgb]{0,0,1}{}\left(\textcolor[rgb]{0,0,1}{t}\right)}_{\textcolor[rgb]{0,0,1}{1}}\\ \textcolor[rgb]{0,0,1}{\mathrm{\kappa }}\textcolor[rgb]{0,0,1}{}\left(\textcolor[rgb]{0,0,1}{\mathrm{\theta }}\textcolor[rgb]{0,0,1}{-}{\textcolor[rgb]{0,0,1}{\mathrm{_X2}}\textcolor[rgb]{0,0,1}{}\left(\textcolor[rgb]{0,0,1}{t}\right)}_{\textcolor[rgb]{0,0,1}{2}}\right)\end{array}]
\mathrm{Diffusion}\left(S\left(t\right)\right)
[\begin{array}{cc}\sqrt{{\textcolor[rgb]{0,0,1}{\mathrm{_X2}}\textcolor[rgb]{0,0,1}{}\left(\textcolor[rgb]{0,0,1}{t}\right)}_{\textcolor[rgb]{0,0,1}{2}}}\textcolor[rgb]{0,0,1}{}{\textcolor[rgb]{0,0,1}{\mathrm{_X2}}\textcolor[rgb]{0,0,1}{}\left(\textcolor[rgb]{0,0,1}{t}\right)}_{\textcolor[rgb]{0,0,1}{1}}& \textcolor[rgb]{0,0,1}{0}\\ \textcolor[rgb]{0,0,1}{0}& \textcolor[rgb]{0,0,1}{\mathrm{\sigma }}\textcolor[rgb]{0,0,1}{}{\textcolor[rgb]{0,0,1}{\mathrm{_X2}}\textcolor[rgb]{0,0,1}{}\left(\textcolor[rgb]{0,0,1}{t}\right)}_{\textcolor[rgb]{0,0,1}{2}}\end{array}]
\mathrm{\mu }≔0.1
\textcolor[rgb]{0,0,1}{\mathrm{\mu }}\textcolor[rgb]{0,0,1}{≔}\textcolor[rgb]{0,0,1}{0.1}
\mathrm{\sigma }≔0.5
\textcolor[rgb]{0,0,1}{\mathrm{\sigma }}\textcolor[rgb]{0,0,1}{≔}\textcolor[rgb]{0,0,1}{0.5}
\mathrm{\kappa }≔1.0
\textcolor[rgb]{0,0,1}{\mathrm{\kappa }}\textcolor[rgb]{0,0,1}{≔}\textcolor[rgb]{0,0,1}{1.0}
\mathrm{\theta }≔0.4
\textcolor[rgb]{0,0,1}{\mathrm{\theta }}\textcolor[rgb]{0,0,1}{≔}\textcolor[rgb]{0,0,1}{0.4}
A≔\mathrm{SamplePath}\left(S\left(t\right),t=0..1,\mathrm{timesteps}=100,\mathrm{replications}=10\right)
\textcolor[rgb]{0,0,1}{A}\textcolor[rgb]{0,0,1}{≔}\begin{array}{c}[\begin{array}{cc}\textcolor[rgb]{0,0,1}{100.}& \textcolor[rgb]{0,0,1}{100.670084719786}\\ \textcolor[rgb]{0,0,1}{100.100000000000}& \textcolor[rgb]{0,0,1}{101.034139425728}\\ \textcolor[rgb]{0,0,1}{100.280880089808}& \textcolor[rgb]{0,0,1}{101.924198818577}\\ \textcolor[rgb]{0,0,1}{102.915077811759}& \textcolor[rgb]{0,0,1}{99.6518477031121}\\ \textcolor[rgb]{0,0,1}{103.858818858166}& \textcolor[rgb]{0,0,1}{100.628185358730}\\ \textcolor[rgb]{0,0,1}{104.476699657855}& \textcolor[rgb]{0,0,1}{98.7691518445139}\\ \textcolor[rgb]{0,0,1}{103.737362966326}& \textcolor[rgb]{0,0,1}{95.0859221941374}\\ \textcolor[rgb]{0,0,1}{102.574346549913}& \textcolor[rgb]{0,0,1}{94.1008617878134}\\ \textcolor[rgb]{0,0,1}{101.159282939668}& \textcolor[rgb]{0,0,1}{92.9644135833222}\\ \textcolor[rgb]{0,0,1}{100.709702216007}& \textcolor[rgb]{0,0,1}{93.6061768383076}\end{array}]\\ \hfill \textcolor[rgb]{0,0,1}{\text{slice of 10 × 2 × 101 Array}}\end{array}
\mathrm{PathPlot}\left(A,1,\mathrm{thickness}=3,\mathrm{markers}=\mathrm{false},\mathrm{color}=\mathrm{red}..\mathrm{blue},\mathrm{axes}=\mathrm{BOXED},\mathrm{gridlines}=\mathrm{true}\right)
\mathrm{PathPlot}\left(A,2,\mathrm{thickness}=3,\mathrm{markers}=\mathrm{false},\mathrm{color}=\mathrm{red}..\mathrm{blue},\mathrm{axes}=\mathrm{BOXED},\mathrm{gridlines}=\mathrm{true}\right)
\mathrm{ExpectedValue}\left(\mathrm{max}\left(S\left(1\right)[1]-100,0\right),\mathrm{timesteps}=100,\mathrm{replications}={10}^{4}\right)
[\textcolor[rgb]{0,0,1}{\mathrm{value}}\textcolor[rgb]{0,0,1}{=}\textcolor[rgb]{0,0,1}{21.41114565}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{\mathrm{standarderror}}\textcolor[rgb]{0,0,1}{=}\textcolor[rgb]{0,0,1}{0.3390630872}]
X≔\mathrm{GeometricBrownianMotion}\left(100.0,0.05,0.3,t\right)
\textcolor[rgb]{0,0,1}{X}\textcolor[rgb]{0,0,1}{≔}\textcolor[rgb]{0,0,1}{\mathrm{_X4}}
Y≔\mathrm{GeometricBrownianMotion}\left(100.0,0.07,0.2,t\right)
\textcolor[rgb]{0,0,1}{Y}\textcolor[rgb]{0,0,1}{≔}\textcolor[rgb]{0,0,1}{\mathrm{_X5}}
\mathrm{\Sigma }≔〈〈1|0.5〉,〈0.5|1〉〉
\textcolor[rgb]{0,0,1}{\mathrm{\Sigma }}\textcolor[rgb]{0,0,1}{≔}[\begin{array}{cc}\textcolor[rgb]{0,0,1}{1}& \textcolor[rgb]{0,0,1}{0.5}\\ \textcolor[rgb]{0,0,1}{0.5}& \textcolor[rgb]{0,0,1}{1}\end{array}]
Z≔\mathrm{ItoProcess}\left(〈X,Y〉,\mathrm{\Sigma }\right)
\textcolor[rgb]{0,0,1}{Z}\textcolor[rgb]{0,0,1}{≔}\textcolor[rgb]{0,0,1}{\mathrm{_X6}}
\mathrm{Drift}\left(Z\left(t\right)\right)
[\begin{array}{c}\textcolor[rgb]{0,0,1}{0.05}\textcolor[rgb]{0,0,1}{}{\textcolor[rgb]{0,0,1}{\mathrm{_X6}}\textcolor[rgb]{0,0,1}{}\left(\textcolor[rgb]{0,0,1}{t}\right)}_{\textcolor[rgb]{0,0,1}{1}}\\ \textcolor[rgb]{0,0,1}{0.07}\textcolor[rgb]{0,0,1}{}{\textcolor[rgb]{0,0,1}{\mathrm{_X6}}\textcolor[rgb]{0,0,1}{}\left(\textcolor[rgb]{0,0,1}{t}\right)}_{\textcolor[rgb]{0,0,1}{2}}\end{array}]
\mathrm{Diffusion}\left(Z\left(t\right)\right)
[\begin{array}{cc}\textcolor[rgb]{0,0,1}{0.3}\textcolor[rgb]{0,0,1}{}{\textcolor[rgb]{0,0,1}{\mathrm{_X6}}\textcolor[rgb]{0,0,1}{}\left(\textcolor[rgb]{0,0,1}{t}\right)}_{\textcolor[rgb]{0,0,1}{1}}& \textcolor[rgb]{0,0,1}{0.15}\textcolor[rgb]{0,0,1}{}{\textcolor[rgb]{0,0,1}{\mathrm{_X6}}\textcolor[rgb]{0,0,1}{}\left(\textcolor[rgb]{0,0,1}{t}\right)}_{\textcolor[rgb]{0,0,1}{1}}\\ \textcolor[rgb]{0,0,1}{0.10}\textcolor[rgb]{0,0,1}{}{\textcolor[rgb]{0,0,1}{\mathrm{_X6}}\textcolor[rgb]{0,0,1}{}\left(\textcolor[rgb]{0,0,1}{t}\right)}_{\textcolor[rgb]{0,0,1}{2}}& \textcolor[rgb]{0,0,1}{0.2}\textcolor[rgb]{0,0,1}{}{\textcolor[rgb]{0,0,1}{\mathrm{_X6}}\textcolor[rgb]{0,0,1}{}\left(\textcolor[rgb]{0,0,1}{t}\right)}_{\textcolor[rgb]{0,0,1}{2}}\end{array}]
\mathrm{ExpectedValue}\left(\mathrm{max}\left(X\left(1\right)-Y\left(1\right),0\right),\mathrm{timesteps}=100,\mathrm{replications}={10}^{4}\right)
[\textcolor[rgb]{0,0,1}{\mathrm{value}}\textcolor[rgb]{0,0,1}{=}\textcolor[rgb]{0,0,1}{14.32896059}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{\mathrm{standarderror}}\textcolor[rgb]{0,0,1}{=}\textcolor[rgb]{0,0,1}{0.2447103632}]
\mathrm{ExpectedValue}\left(\mathrm{max}\left(Z\left(1\right)[1]-Z\left(1\right)[2],0\right),\mathrm{timesteps}=100,\mathrm{replications}={10}^{4}\right)
[\textcolor[rgb]{0,0,1}{\mathrm{value}}\textcolor[rgb]{0,0,1}{=}\textcolor[rgb]{0,0,1}{8.103315185}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{\mathrm{standarderror}}\textcolor[rgb]{0,0,1}{=}\textcolor[rgb]{0,0,1}{0.1520913055}]
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UBC secondary parameters - OpenVZ Virtuozzo Containers Wiki
UBC secondary parameters
(Redirected from Dgramrcvbuf)
Secondary (dependant) UBC parameters are directly connected to the primary ones and can't be configured arbitrarily.
1 kmemsize
2 tcpsndbuf
3 tcprcvbuf
4 othersockbuf
5 dgramrcvbuf
6 oomguarpages
7 privvmpages
8 System-wide limits
kmemsize[edit]
Size of unswappable memory in bytes, allocated by the operating system kernel.
It includes all the kernel internal data structures associated with the container's processes, except the network buffers discussed below. These data structures reside in the first gigabyte of the computer's RAM, so called “low memory”.
This parameter is related to the number of processes (numproc). Each process consumes certain amount of kernel memory — 24 kilobytes at minimum, 30–60 KB typically. Very large processes may consume much more than that.
It is important to have a certain safety gap between the barrier and the limit of the kmemsize parameter (for example, 10%, as in UBC configuration examples). Equal barrier and limit of the kmemsize parameter may lead to the situation where the kernel will need to kill container's applications to keep the kmemsize usage under the limit.
Kmemsize limits can't be set arbitrarily high. The total amount of kmemsize consumable by all containers in the system plus the socket buffer space (see below) is limited by the hardware resources of the system. This total limit is discussed in “low memory”.
tcpsndbuf[edit]
The total size of buffers used to send data over TCP network connections. These socket buffers reside in “low memory”.
Tcpsndbuf parameter depends on number of TCP sockets (numtcpsock) and should allow for some minimal amount of socket buffer memory for each socket, as discussed in UBC consistency check:
{\displaystyle tcpsndbuf_{lim}-tcpsndbuf_{bar}\geq 2.5KB\cdot numtcpsock{\rm {.}}}
If this restriction is not satisfied, some network connections may silently stall, being unable to transmit data.
Setting high values for tcpsndbuf parameter may, but doesn't necessarily, increase performance of network communications. Note that, unlike most other parameters, hitting tcpsndbuf limits and failed socket buffer allocations do not have strong negative effect on the applications, but just reduce performance of network communications.
If you use rtorrent in a container, a low value for tcpsndbuf may cause rtorrent to take unusual amount of cpu. In this case, you must put a higher value. Also watch the number of failcnt in /proc/user_beancounters.
Tcpsndbuf limits can't be set arbitrarily high. The total amount of tcpsndbuf consumable by all containers in the system plus the kmemsize and other socket buffers is limited by the hardware resources of the system. This total limit is discussed in “low memory”.
tcprcvbuf[edit]
The total size of buffers used to temporary store the data coming from TCP network connections. These socket buffers also reside in “low memory”.
Tcprcvbuf parameter depends on number of TCP sockets (numtcpsock) and should allow for some minimal amount of socket buffer memory for each socket, as discussed in UBC consistency check:
{\displaystyle tcprcvbuf_{lim}-tcprcvbuf_{bar}\geq 2.5KB\cdot numtcpsock{\rm {.}}}
If this restriction is not satisfied, some network connections may stall, being unable to receive data, and will be terminated after a couple of minutes.
Similarly to tcpsndbuf, setting high values for tcprcvbuf parameter may, but doesn't necessarily, increase performance of network communications. Hitting tcprcvbuf limits and failed socket buffer allocations do not have strong negative effect on the applications, but just reduce performance of network communications. However, staying above the barrier of tcprcvbuf parameter for a long time is less harmless than for tcpsndbuf. Long periods of exceeding the barrier may cause termination of some connections.
Tcprcvbuf limits can't be set arbitrarily high. The total amount of tcprcvbuf consumable by all containers in the system plus the kmemsize and other socket buffers is limited by the hardware resources of the system. This total limit is discussed in “low memory”.
othersockbuf[edit]
The total size of buffers used by local (UNIX-domain) connections between processes inside the system (such as connections to a local database server) and send buffers of UDP and other datagram protocols.
Othersockbuf parameter depends on number of non-TCP sockets (numothersock).
Othersockbuf configuration should satisfy
{\displaystyle othersockbuf_{lim}-othersockbuf_{bar}\geq 2.5KB\cdot numothersock.}
Increased limit for othersockbuf is necessary for high performance of communications through local (UNIX-domain) sockets. However, similarly to tcpsndbuf, hitting othersockbuf affects the communication performance only and does not affect the functionality.
Othersockbuf limits can't be set arbitrarily high. The total amount of othersockbuf consumable by all containers in the system plus the kmemsize and other socket buffers is limited by the hardware resources of the system. This total limit is discussed in “low memory”.
dgramrcvbuf[edit]
The total size of buffers used to temporary store the incoming packets of UDP and other datagram protocols.
Dgramrcvbuf parameters depend on number of non-TCP sockets (numothersock).
Dgramrcvbuf limits usually don't need to be high. Only if the containers needs to send and receive very large datagrams, the barriers for both othersockbuf and dgramrcvbuf parameters should be raised.
Hitting dgramrcvbuf means that some datagrams are dropped, which may or may not be important for application functionality. UDP is a protocol with not guaranteed delivery, so even if the buffers permit, the datagrams may be as well dropped later on any stage of the processing, and applications should be prepared for it.
Unlike other socket buffer parameters, for dgramrcvbuf the barrier should be set to the limit.
Dgramrcvbuf limits can't be set arbitrarily high. The total amount of dgramrcvbuf consumable by all containers in the system plus the kmemsize and other socket buffers is limited by the hardware resources of the system. This total limit is discussed in “low memory”.
oomguarpages[edit]
The guaranteed amount of memory for the case the memory is “over-booked” (out-of-memory kill guarantee).
Oomguarpages parameter is related to vmguarpages. If applications start to consume more memory than the computer has, the system faces an out-of-memory condition. In this case the operating system will start to kill container's processes to free some memory and prevent the total death of the system. Although it happens very rarely in typical system loads, killing processes in out-of-memory situations is a normal reaction of the system, and it is built into every Linux kernel[1].
Oomguarpages parameter accounts the total amount of memory and swap space used by the processes of a particular container. The barrier of the oomguarpages parameter is the out-of-memory guarantee.
If the current usage of memory and swap space (the value of oomguarpages) plus the amount of used kernel memory (kmemsize) and socket buffers is below the barrier, processes in this container are guaranteed not to be killed in out-of-memory situations. If the system is in out-of-memory situation and there are several containers with oomguarpages excess, applications in the container with the biggest excess will be killed first. The failcnt counter of oomguarpages parameter increases when a process in this container is killed because of out-of-memory situation.
If the administrator needs to make sure that some application won't be forcedly killed regardless of the application's behavior, setting the privvmpages limit to a value not greater than the oomguarpages guarantee significantly reduce the likelihood of the application being killed, and setting it to a half of the oomguarpages guarantee completely prevents it. Such configurations are not popular because they significantly reduce the utilization of the hardware.
The meaning of the limit for the oomguarpages parameter is unspecified in the current version.
The total out-of-memory guarantees given to the containers should not exceed the physical capacity of the computer, as discussed in UBC systemwide configuration#Memory and swap space. If guarantees are given for more than the system has, in out-of-memory situations applications in containers with guaranteed level of service and system daemons may be killed.
privvmpages[edit]
Memory allocation limit in pages (which are typically 4096 bytes in size).
Privvmpages parameter allows controlling the amount of memory allocated by applications.
The barrier and the limit of privvmpages parameter control the upper boundary of the total size of allocated memory. Note that this upper boundary doesn't guarantee that the container will be able to allocate that much memory, neither does it guarantee that other containers will be able to allocate their fair share of memory. The primary mechanism to control memory allocation is the vmguarpages guarantee.
Privvmpages parameter accounts allocated (but, possibly, not used yet) memory. The accounted value is an estimation how much memory will be really consumed when the container's applications start to use the allocated memory. Consumed memory is accounted into oomguarpages parameter.
Since the memory accounted into privvmpages may not be actually used, the sum of current privvmpages values for all containers may exceed the RAM and swap size of the computer.
There should be a safety gap between the barrier and the limit for privvmpages parameter to reduce the number of memory allocation failures that the application is unable to handle. This gap will be used for “high-priority” memory allocations, such as process stack expansion. Normal priority allocations will fail when the barrier of privvmpages is reached.
Total privvmpages should correlate with the physical resources of the computer. Also, it is important not to allow any container to allocate a significant portion of all system RAM to avoid serious service level degradation for other containers. Both these configuration requirements are discussed in UBC systemwide configuration#Allocated memory.
There's also an article describing how user pages accounting works.
System-wide limits[edit]
All secondary parameters are related to memory. Total limits on memory-related parameters must not exceed the physical resources of the computer. The restrictions on the configuration of memory-related parameters are listed in UBC systemwide configuration. Those restrictions are very important, because their violation may allow any container cause the whole system to hang.
↑ The possible reasons of out-of-memory situations are the excess of total vmguarpages guarantees the available physical resources or high memory consumption by system processes. Also, the kernel might allow some containers to allocate memory above their vmguarpages guarantees when the system had a lot of free memory, and later, when other containers claim their guarantees, the system will experience the memory shortage.
Retrieved from "https://wiki.openvz.org/index.php?title=UBC_secondary_parameters&oldid=11453#dgramrcvbuf"
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Circumference - Simple English Wikipedia, the free encyclopedia
In geometry, circumference is the distance around a closed curve; for example, a circle. It is a special kind of perimeter. The length of the circumference of a circle is often written as
{\displaystyle C}
,[1] with:[2][3]
{\displaystyle C=\pi d}
↑ Weisstein, Eric W. "Circle". mathworld.wolfram.com. Retrieved 2020-09-24.
The Simple English Wiktionary has a definition for: circumference.
Retrieved from "https://simple.wikipedia.org/w/index.php?title=Circumference&oldid=7982856"
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Home : Support : Online Help : Programming : Logic : Boolean : verify : table
verify a relation between the entries of two tables
verify(expr1, expr2, table)
verify(expr1, expr2, 'table'(ver))
anything, assumed to be of type table
verification for the table entries
The verify(expr1, expr2, table) and verify(expr1, expr2, 'table'(ver)) calling sequences return true if it can be determined that the two tables satisfy a relation entrywise, either by testing with equality or using the verification ver.
Since table look-up is done with Boolean comparisons, the indices of two tables are never verified. For example, the tables table([2.0 = 3]) and table([2. = 3]) will never be verified as being equal.
If the two tables have unequal indexing functions, false will be automatically returned. The one exception to this is when one table is symmetric and the other has no indexing function. In this case, the tables will be declared equal if the unindexed table is symmetric and equal to the index table.
The verification table is symmetric and a verification 'table'(ver) is symmetric if and only if the verification ver is symmetric.
Because table is a Maple function, it must be enclosed in single quotes to prevent evaluation.
If either expr1 or expr2 is not of type table, then false is returned.
The special verifications %NULL, seq, and &, can be used to verify an expression sequence in a table. See verify/exprseq.
A≔\mathrm{table}\left(\mathrm{symmetric},[\left(a,b\right)=3]\right)
\textcolor[rgb]{0,0,1}{A}\textcolor[rgb]{0,0,1}{≔}\textcolor[rgb]{0,0,1}{table}\textcolor[rgb]{0,0,1}{}\left(\textcolor[rgb]{0,0,1}{\mathrm{symmetric}}\textcolor[rgb]{0,0,1}{,}[\left(\textcolor[rgb]{0,0,1}{a}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{b}\right)\textcolor[rgb]{0,0,1}{=}\textcolor[rgb]{0,0,1}{3}]\right)
B≔\mathrm{table}\left([\left(a,b\right)=3]\right)
\textcolor[rgb]{0,0,1}{B}\textcolor[rgb]{0,0,1}{≔}\textcolor[rgb]{0,0,1}{table}\textcolor[rgb]{0,0,1}{}\left([\left(\textcolor[rgb]{0,0,1}{a}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{b}\right)\textcolor[rgb]{0,0,1}{=}\textcolor[rgb]{0,0,1}{3}]\right)
C≔\mathrm{table}\left([\left(a,b\right)=3,\left(b,a\right)=3]\right)
\textcolor[rgb]{0,0,1}{C}\textcolor[rgb]{0,0,1}{≔}\textcolor[rgb]{0,0,1}{table}\textcolor[rgb]{0,0,1}{}\left([\left(\textcolor[rgb]{0,0,1}{a}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{b}\right)\textcolor[rgb]{0,0,1}{=}\textcolor[rgb]{0,0,1}{3}\textcolor[rgb]{0,0,1}{,}\left(\textcolor[rgb]{0,0,1}{b}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{a}\right)\textcolor[rgb]{0,0,1}{=}\textcolor[rgb]{0,0,1}{3}]\right)
\mathrm{verify}\left(A,B,\mathrm{table}\right)
\textcolor[rgb]{0,0,1}{\mathrm{false}}
\mathrm{verify}\left(A,C,\mathrm{table}\right)
\textcolor[rgb]{0,0,1}{\mathrm{true}}
A≔\mathrm{table}\left([a=\frac{{x}^{2}-{y}^{2}}{{\left(x-y\right)}^{3}}]\right)
\textcolor[rgb]{0,0,1}{A}\textcolor[rgb]{0,0,1}{≔}\textcolor[rgb]{0,0,1}{table}\textcolor[rgb]{0,0,1}{}\left([\textcolor[rgb]{0,0,1}{a}\textcolor[rgb]{0,0,1}{=}\frac{{\textcolor[rgb]{0,0,1}{x}}^{\textcolor[rgb]{0,0,1}{2}}\textcolor[rgb]{0,0,1}{-}{\textcolor[rgb]{0,0,1}{y}}^{\textcolor[rgb]{0,0,1}{2}}}{{\left(\textcolor[rgb]{0,0,1}{x}\textcolor[rgb]{0,0,1}{-}\textcolor[rgb]{0,0,1}{y}\right)}^{\textcolor[rgb]{0,0,1}{3}}}]\right)
B≔\mathrm{table}\left([a=\frac{{x}^{2}-{y}^{2}}{{\left(x-y\right)}^{3}}]\right)
\textcolor[rgb]{0,0,1}{B}\textcolor[rgb]{0,0,1}{≔}\textcolor[rgb]{0,0,1}{table}\textcolor[rgb]{0,0,1}{}\left([\textcolor[rgb]{0,0,1}{a}\textcolor[rgb]{0,0,1}{=}\frac{{\textcolor[rgb]{0,0,1}{x}}^{\textcolor[rgb]{0,0,1}{2}}\textcolor[rgb]{0,0,1}{-}{\textcolor[rgb]{0,0,1}{y}}^{\textcolor[rgb]{0,0,1}{2}}}{{\left(\textcolor[rgb]{0,0,1}{x}\textcolor[rgb]{0,0,1}{-}\textcolor[rgb]{0,0,1}{y}\right)}^{\textcolor[rgb]{0,0,1}{3}}}]\right)
C≔\mathrm{table}\left([a=\frac{y+x}{{\left(-x+y\right)}^{2}}]\right)
\textcolor[rgb]{0,0,1}{C}\textcolor[rgb]{0,0,1}{≔}\textcolor[rgb]{0,0,1}{table}\textcolor[rgb]{0,0,1}{}\left([\textcolor[rgb]{0,0,1}{a}\textcolor[rgb]{0,0,1}{=}\frac{\textcolor[rgb]{0,0,1}{y}\textcolor[rgb]{0,0,1}{+}\textcolor[rgb]{0,0,1}{x}}{{\left(\textcolor[rgb]{0,0,1}{-}\textcolor[rgb]{0,0,1}{x}\textcolor[rgb]{0,0,1}{+}\textcolor[rgb]{0,0,1}{y}\right)}^{\textcolor[rgb]{0,0,1}{2}}}]\right)
\mathrm{evalb}\left(A=B\right)
\textcolor[rgb]{0,0,1}{\mathrm{false}}
\mathrm{verify}\left(A,B,'\mathrm{table}'\right)
\textcolor[rgb]{0,0,1}{\mathrm{true}}
\mathrm{verify}\left(A,C,'\mathrm{table}'\right)
\textcolor[rgb]{0,0,1}{\mathrm{false}}
\mathrm{verify}\left(A,C,'\mathrm{table}\left(\mathrm{normal}\right)'\right)
\textcolor[rgb]{0,0,1}{\mathrm{true}}
The verify/table command was updated in Maple 2015.
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Estimate Tezos performances - Nomadic Labs knowledge center
NFT transfer performances
At the time of writing (February 2022, Hangzhou):
a typical NFT transfer on Tezos consumes¹ 2,691 gas units per transaction,
the Tezos maximum block gas limit is 5,200,000 gas units,
To maximize throughput, we can use batching - grouping several transactions into a single block.
We performed experiments and measured how gas consumption varies depending on how many NFT transfers we group into a single batch of transactions, as presented in the following experimental values table:
We had to stop at 650, because we had reached the operation size limit imposed by the protocol. Thus, the largest batch we can use is 650. Now we estimate how many batches we can have in a block, considering that the max block gas limit is 5,200,000 gas units:
\text{Max batches} = \frac{5,200,000}{501,048} \approx 10
We now need to process the leftover gas:
\text{Leftover gas} = 5,200,000 - (10 \times 501,048) \approx 189,520
By taking this value and comparing it to our experimental values table we can see that we can add an extra batch of 240 NFT transfers approximately.
That would make the total of NFT transfers per block equal to:
\text{Max NFT transfers per block} = (10 \times 650) + (1 \times 240) = 6,740
Since we have two blocks per minute, the number of transactions per second would be:
\frac{6,740}{30} \approx 224 \ \text{NFT transfer/second}
Standard Tez transfer performances
For standard tez transfers, we can process the performance of the Tezos blockchain with the following formula:
\frac{\text{max gas per block}}{\text{tx gas consumption} \times \text{time per block}} = \text{tx/second}
In that scenario, the gas consumption is 1,420 gas units, and we will keep the same values as in the previous calculation:
Tezos maximum block gas limit: 5,200,000 gas units
\frac{5,200,000}{1,420 \times 30} \approx 122\ \text{tx/second}
Performance should improve further with the planned Tezos upgrade to adopt the Tenderbake consensus algorithm ², and the Hangzhou amendment.
[1]: With the minimal contract implementing the FA2 standard.
[2]: https://blog.nomadic-labs.com/a-look-ahead-to-tenderbake.html
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Artin’s conjecture for abelian varieties
December 2016 Artin’s conjecture for abelian varieties
A
an abelian variety of dimension
r
\mathbb{Q}
{rank}_{\mathbb{Q}}A\ge g
g\ge 0
{a}_{1},\dots ,{a}_{g}\in A\left(\mathbb{Q}\right)
be linearly independent points. (So, in particular,
{a}_{1},\dots ,{a}_{g}
have infinite order, and if
g=0
\left\{{a}_{1},\dots ,{a}_{g}\right\}
is empty.) For
p
a rational prime of good reduction for
A
\overline{A}
be the reduction of
A
p
{\overline{a}}_{i}
i=1,\dots ,g
{a}_{i}
p
〈{\overline{a}}_{1},\dots ,{\overline{a}}_{g}〉
\overline{A}\left({\mathbb{F}}_{p}\right)
{\overline{a}}_{1},\dots ,{\overline{a}}_{g}
\mathbb{Q}\left(A\left[2\right]\right)=\mathbb{Q}
\mathbb{Q}\left(A\left[2\right],{2}^{-1}{a}_{1},\dots ,{2}^{-1}{a}_{g}\right)\ne \mathbb{Q}
. (Note that this particular assumption
\mathbb{Q}\left(A\left[2\right]\right)=\mathbb{Q}
forces the inequality
g\ge 1
, but we can take care of the case
g=0
, under the right assumptions, also.) Then in this article, in particular, we show that the number of primes
p
\frac{\overline{A}\left({\mathbb{F}}_{p}\right)}{〈{\overline{a}}_{1},\dots ,{\overline{a}}_{g}〉}
\left(2r-1\right)
cyclic components is infinite. This result is the right generalization of the classical Artin’s primitive root conjecture in the context of general abelian varieties; that is, this result is an unconditional proof of Artin’s conjecture for abelian varieties. Artin’s primitive root conjecture (1927) states that, for any integer
a\ne ±1
or a perfect square, there are infinitely many primes
p
a
is a primitive root
\left(modp\right)
. (This conjecture is not known for any specific
a
Cristian Virdol. "Artin’s conjecture for abelian varieties." Kyoto J. Math. 56 (4) 737 - 743, December 2016. https://doi.org/10.1215/21562261-3664896
Received: 25 December 2014; Revised: 17 September 2015; Accepted: 24 September 2015; Published: December 2016
Keywords: abelian varieties , Artin’s conjecture , primitive-cyclic points
Cristian Virdol "Artin’s conjecture for abelian varieties," Kyoto Journal of Mathematics, Kyoto J. Math. 56(4), 737-743, (December 2016)
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Star Knowpia
Stars condense from regions of space of higher matter density, yet those regions are less dense than within a vacuum chamber. These regions—known as molecular clouds—consist mostly of hydrogen, with about 23 to 28 percent helium and a few percent heavier elements. One example of such a star-forming region is the Orion Nebula.[54] Most stars form in groups of dozens to hundreds of thousands of stars.[55]Massive stars in these groups may powerfully illuminate those clouds, ionizing the hydrogen, and creating H II regions. Such feedback effects, from star formation, may ultimately disrupt the cloud and prevent further star formation.[56]
{\displaystyle \Delta {m}=m_{\mathrm {f} }-m_{\mathrm {b} }}
{\displaystyle 2.512^{\Delta {m}}=\Delta {L}}
{\displaystyle E=mc^{2}}
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Modeling of Economic Processes, Instability and Chaos
Modeling of Economic Processes, Instability and Chaos
A. F. Mozhaisky Military Space Academy, St. Petersburg, Russia
We consider a mathematical model describing the dynamics of the three components of the economic activity of enterprises―production, management and resource consumption. For the analysis of the dynamics, the methods and criteria formulated by the authors earlier were used; new results of a practical nature were obtained; in particular, the calculation of the spectrum of small perturbations of the system depending on the input resource and system parameters is given. It is shown that the calculated values of the criteria of instability and chaos are fully consistent with the calculation of the spectrum, indicated the stability intervals of the system, as well as areas of instability and chaos.
Mathematical Modeling of Economic Processes, Methods for Analyzing Dynamics, Instability, Chaos
Methods of mathematical modeling are widely used to describe the economic processes [1] - [6] .
Modeling of economic processes considers: 1) method of studying phenomena and processes in the economic system or its elements by building and studying their models; 2) transfer of the results obtained during the study of models to the original; 3) using models to determine or refine the characteristics; 4) building newly constructed objects in the economic system.
Forms of modeling are different and depend on models and scope. The idea of modeling is based on any method of scientific research of economic processes. There are many models that display various aspects of economic processes.
Modeling serves as a mean of studying the economy and the phenomena occurring in it, substantiating decisions made, forecasting, planning, managing economic processes and objects. A model of an economic object is usually supported by real statistical data, and the results of calculations carried out within the framework of the constructed model make it possible to build forecasts and carry out objective evaluations.
For successful modeling, justified simplifications in the model compared to the original are important; an analysis of fairly simple and representative dynamic models is necessary [1] - [6] .
This article is a continuation of the work of the authors on the study of the dynamics of economic processes on the example of the “production―management―resource” model [7] .
The considered model PMR (production, management, resource) refers to the dynamic models of enterprises. Along with sufficient simplicity, it takes into account the main factors of production: the dynamics of production, management and the balance of external and system-produced resources. The model consists of three balanced, non-linear, differential equations corresponding to the three components of the required economic object subsystems―production, management and resource.
To analyze the dynamics of the system, assessing stability and possible instabilities, the methods and criteria given by the authors in [8] [9] [10] are used.
2. Initial Relations
Equations of the PMR model in a dimensionless form
\begin{array}{l}{\partial }_{t}x=x\left(a-y+bz\right)\\ {\partial }_{t}y=y\left(x-c+dz\right)\\ {\partial }_{t}z=z\left(-e-dy-bx\right)-hx-ky+gxy+r\end{array}
\frac{\left(x,y,z\right)}{{x}_{0}+{y}_{0}+{z}_{0}}
the main dynamic values having the same resource
dimension (for example, monetary) are related to the sum of the initial values and corresponding to the dynamics of funds in the three subsystems of the object under study; x―production, y―control, z―resource.
t=\tau /{\tau }_{0}
―dimensionless time,
{\tau }_{0}
―a characteristic financial time scale (quarter, year).
In the first equation of the system (1) describing the dynamics of production, the components ax, bxz correspond to the profits from the production and input resource; the term xy is a decrease in management (for example, for training, professional development, etc.). In the second equation for controlling yx, dyz, the flow of funds from production and the input resource; cy―direct management costs.
The third equation is the balance of the internal and external r resource (r > 0―credit, r < 0―debt repayment).
The terms hx, ky are the direct loss of the salary resource in production and management; ez―resource costs for repayment of interest in lending or receiving interest on deposits, bxz―resource costs for raw materials for production, energy, etc. The terms gxy, dyz, respectively, are the resource profit due to production b and decrease on innovations in management and production automation.
The main task in the study of the dynamics is to identify areas of stable functioning of system (1), as well as areas and conditions for the occurrence of instabilities. This problem is solved by studying the spectrum of small perturbations of stationary states. The stationary state of the system―St (xs, ys, zs) are
\begin{array}{l}x\left(a-y+bz\right)=0\\ y\left(x-c+dz\right)=0\\ z\left(-e-dy-bx\right)-hx-ky+gxy+r=0\end{array}
zeros of the right-hand sides of system (1), i.e. solutions of Equation (2). In this case, there are five stationary solutions, respectively
{\text{St}}_{1}\left(0,0,r/e\right);{\text{St}}_{2}\left(0,{y}_{s},c/d\right);{\text{St}}_{3}\left({x}_{s},0,{z}_{s}\right);{\text{St}}_{4,5}\left({x}_{s},{y}_{s},{z}_{s}\right)
By linearizing system (1), we obtain a system of equations for perturbations
{\partial }_{t}\delta {x}_{\alpha }={e}_{\alpha \text{ }\beta }\delta {x}_{\beta },\text{ }\left\{\delta {x}_{\alpha }\right\}=\left\{\delta x,\delta y,\delta z\right\}
where the elements of the evolutionary matrix are equal
\begin{array}{l}\stackrel{^}{E}=\left\{{e}_{\alpha \beta }\right\};\text{\hspace{0.17em}}\text{\hspace{0.17em}}{e}_{11}=a-{y}_{s}+b{z}_{s}=0,\text{\hspace{0.17em}}\text{\hspace{0.17em}}{e}_{12}=-{x}_{s}=d{z}_{s}-c,\text{\hspace{0.17em}}\text{\hspace{0.17em}}{e}_{13}=b{x}_{s},\\ {e}_{21}={y}_{s}=a+b{z}_{s},\text{\hspace{0.17em}}\text{\hspace{0.17em}}{e}_{22}={x}_{s}-c+d{z}_{s}=0,\text{\hspace{0.17em}}\text{\hspace{0.17em}}{e}_{23}=d{y}_{s},\\ {e}_{31}=-b{z}_{s}-h+g{y}_{s}=-h+ga-bg\left(1-g\right),\\ {e}_{32}=-d{z}_{s}-k+g{x}_{s}=gc-k-d{z}_{s}\left(1+g\right),\\ {e}_{33}=-e-d{y}_{s}-b{x}_{s}=-e-da-bc.\end{array}
The condition of the solvability of the system (4) is the spectral equation (SE) equal to
\mathrm{det}\left(\lambda {\delta }_{\alpha \beta }-{e}_{\alpha \beta }\right)={\lambda }^{3}+s{\lambda }^{2}+p\lambda +q=0
where λ is the spectral parameter. The roots of this equation determine the dynamics of the system. Namely: if all the roots of SE have negative real parts, the state under study is stable with respect to small perturbations; if there is at least one root with a positive real part, the state is unstable; if there are roots in the spectrum with positive and negative real parts (saddle focus) and some of them are complex, then this is a special kind of instability―chaos (irregular chaotic oscillations).
div\stackrel{\to }{G}\left({\partial }_{t}x,{\partial }_{t}y,{\partial }_{t}z\right)=-0.7+0.7x-0.8y+0.9z
Expression (6)―the divergence of the vector of the right-hand sides of equations (1) indicates the areas of growth and reduction of the phase volume and is one of the conditions for stability-instability. Depending on the sign
div\stackrel{\to }{G}
, the dynamics is called active (
div\stackrel{\to }{G}>0
), neutral (
div\stackrel{\to }{G}=0
), dissipative (
div\stackrel{\to }{G}<0
). From (6) it follows that, depending on the magnitude of the resource and the steady state, the system may have different types of dynamics.
3. Spectral and Criterial Analysis of the Dynamics
In view of the complexity of the system (1), only a numerical analysis of the spectrum is possible, for which it is necessary to specify the numerical values of the dimensionless coefficients, as well as select several (one, two) control parameters, varying the values of which monitor the changes in the spectrum and thus the system dynamics. As the control parameter, we select the parameter of the external resource―r, the parameters of innovation in management and production―d, g. The numerical values of the remaining parameters in the system (1) are selected in accordance with the real values on modern production and taking into account the preliminary account. So, in view of the above, the values of the parameters are chosen equal
\left\{\begin{array}{l}a=0.3,\text{\hspace{0.17em}}b=0.3,\text{\hspace{0.17em}}c=0.8,d=\left(-0.2;0.4\right)\\ e=\left(-0.2;0.2\right),\text{\hspace{0.17em}}h=0.2,\text{\hspace{0.17em}}k=0.4,\text{\hspace{0.17em}}g=\left(0.2;0.4;0.8\right),\text{\hspace{0.17em}}r\in \left(-0.4;0.4\right)\end{array}\right\}
The spectral coefficients in (5) taking into account (7) take the form
For the analysis of SEs, we use both the direct calculation of the roots of the spectral equation and the spectral criteria of instability and chaos given by the authors in [8] [9] [10] . According to these criteria, the relations defining the boundaries of the regions of instability and chaos are of the form
q<0,\text{ }K1=q-sp\ge 0;\text{ }б\right)\text{ }p>0;\text{ }в\right)\text{ }K2=q\cdot K1>0
The first inequalities in (9) are instability conditions, the second is a sufficient condition for oscillations, the third is the saddle-like character of the spectrum, the realization of all three conditions reflects the presence of chaotic oscillations (chaos) in the system.
Of greatest practical interest are stationary states with nonzero values of quantities, i.e. St4, St5, for which, respectively, taking into account (2), (7), we obtain
\begin{array}{l}{x}_{4,5s}=c-d{z}_{4,5};\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{y}_{4,5s}=b\left(1+{z}_{4,5}\right);\\ {z}_{4.5s}=\frac{-B\pm \sqrt{{B}^{2}-4AD}}{2A},\text{\hspace{1em}}A=-bdg,\text{\hspace{1em}}D=acg-hc-ka+r\\ B=-da\left(1+g\right)-bc\left(1-g\right)+hd-e.\end{array}
The values of the roots of the SE (5) and the criteria q, K1, and K2 for these states, depending on the parameter r, are given in Tables 1-4, respectively.
In Tables 1-2, it is taken into account that at r > 0.18, the states 4.5 cease to be stationary. From Tables 1-2 it also follows that in the whole area of resource change, both with a shortage of credit and with its excess, the system is unstable, there are random fluctuations in it. Moreover, in the St4 state, they weakly decay, in the St5 state, damped oscillations are transformed into growing ones. In both
Table 1. Estimated values of roots and criteria for the state St4 (e = d = −0.2, g = 0.4).
Table 2. Calculated values of the roots and criteria for the state St5 (e = d = −0.2, g = 0.4).
Table 3. Calculated values of the roots and criteria for the state St4 (e = 0.2, d = −0.2, g = 0.4).
cases, the signs of the criteria correspond to the signs of the roots and the presence of chaos.
Tables 3-4 reflect the dynamics of the system with increased resource due to management.
It follows from them that the state St4 is stable in the entire range of the parameter r, i.e. management maintains state stability regardless of resource change. In contrast, the St5 state is more dynamic than the St4 state; in it areas of stability
r\in \left(-0.2;0.0\right)\cup \left(0.2;0.3\right)
alternate with areas of chaos
r<-0.2:r\sim 0.1
The tables imply a significant influence on the system dynamics of the parameter d.
Also in these cases, the signs of the criteria correspond to the signs of the roots and the presence of chaos.
Figures 1-4 show the graphs of the dependence of the conditions of instability and chaos on the parameters r, g, d demonstrating the influence of these parameters on the dynamics of the system.
Figure 1. r = −0.2, g = 0.2.
Figure 3. r = 0.2, g = 0.2.
In Figures 1-4 (F = K2 is the condition of chaos, x = 10d). The dependence of the condition of the existence of chaos on the parameter d is shown. From Figures 1-4 it follows that
1) The system is very sensitive to changes in the parameter d;
2) With a small innovation parameter (small value of g parameter), the growth of d parameter stabilizes the system (see Figures 1-3);
3) With an increase in innovation (a larger value of the parameter g), the parameter d destabilizes the system, windows of chaos appear (see Figure 2-4);
4) Significant negative values of the parameter d, i.e. a decrease in the contribution to control and a corresponding increase in accumulations leads to the initiation of chaotic oscillations―the “fever” system.
In general, analyzing the results of calculations, the following can be noted:
1) The proposed model provides ample opportunities to study the impact on the economic activity of enterprises of various factors of the production, management and resource sphere.
2) This model describes a very dynamic system, in which, along with the areas of stability, there are areas of instability and chaos, significantly dependent on the parameters of the system.
3) From the tables and graphs it can be seen that the increase in the financing of management stabilizes the processes in the system and that the system loses stability both with a shortage and with an excess of resources.
4) The criteria of instability and chaos proposed by the authors are fully consistent with the calculation of the spectrum, i.e. can be used to predict areas of stability as well as instability of dynamic systems.
Perevoznikov, E. and Lomteva, E. (2019) Modeling of Economic Processes, Instability and Chaos. Journal of Applied Mathematics and Physics, 7, 356-363. https://doi.org/10.4236/jamp.2019.72027
1. Mazhukin, V.I. and Flinta, M. (2004) Moskovskij gumanitarnyj universitet. Matematicheskoe modelirovanie v ehkonomike, 232 s.
2. Markin, M.P. (2007) Matematicheskie metody i modeli v ehkonomike. M. Vysshaya shkola, 422 s.
3. Hachatryan, N.K. (2008) Matematicheskoe modelirovanie ehkonomicheskih sistem. M., EHkzamen, 158 s.
4. Vihanskij, O.S. and Naumov, O.I. (2010) Menedzhment: chelovek, strategiya, organizaciya, process. M. Delo, 383 s.
5. Kabushkin, N.I. (2008) Osnovy menedzhmenta. M.: Novoe znanie, 336 s.
6. Klejner, G.B. (2010) Razvitie teorii ehkonomicheskih sistem i ee primenenie v operativnom i strategicheskom upravlenii, WP /2010/CEHMI RAN, 209 s.
7. Perevoznikov, E.N. and Skvorcov, G.E. (2018) Analiz dinamiki model’noy sistemy “Vestnik Sankt—Peterburgskogo gosudarstvennogo universiteta tekhnologii i dizayna. ser. Yestestvennyye i tekhnicheskiye nauki, 2, 41-48.
8. Perevoznikov, E.N. (2006) Metody analiza ustojchivosti neravnovesnyh system. Izv. Vuzov, Fizika, 10, 34-39.
9. Skvorcov, G.E. and Perevoznikov, E.N. (2016) Teoriya neustojchivosti i kriterii haosa. Mezhdunarodnyj nauchno-issledovatel’skij zhurnal, fiziko-matematicheskie nauki, ch.4.-C.98-101.
10. Perevoznikov, E.N. and Skvortsov, H.E. (2018) Criteria for Instability and Chaos in Nonlinear System. Journal of Applied Mathematics and Physics, 6, 382-388. https://doi.org/10.4236/jamp.2018.62036
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Systems of Multiple Equilibria - Course Hero
General Chemistry/Equilibria of Other Reaction Classes/Systems of Multiple Equilibria
If a solution contains many different solutes, the solution must also contain many different equilibria. The different solutes can affect each other's equilibria by changing their solubilities and causing something that is insoluble on its own to dissolve or by causing something that is soluble to precipitate out of solution at a concentration below its saturation point.
Both the common ion effect (observed when KI is added to a solution of AgI) and the example of the effect of the complex ion [Ag(NH3)2]+ can be considered systems of multiple equilibria. In each case, the ions formed by one equilibrium reaction caused the equilibrium of a second reaction to shift one way or another. In the first case, the addition of KI to a solution of AgI causes a precipitate to form, and in the second case, the introduction of NH3 causes the previously insoluble AgCl to dissolve into the solution. So the equilibrium of one reaction can affect the equilibrium of another. To put it another way, introducing additional equilibria to a system can change the concentrations of species involved.
Consider a sample of AgCl in an aqueous solution of 1.0 M NH3. In this system, there are two relevant equilibrium equations.
\begin{alignedat}{2}{\rm{AgCl}}(s) &\rightleftharpoons{\rm{Ag^+}}(aq)+{\rm{Cl^-}}(aq)\quad &&K_{\rm{sp}}=1.77\times10^{-10}\\{\rm{Ag^+}}(aq)+{2\rm{NH_3}}(aq) &\rightleftharpoons{\left[\rm{Ag(NH_3)_2}\right]^+}(aq) &&K_{\rm{f}}=1.6\times10^{7}\hspace{25pt}\end{alignedat}
AgCl is a white powder that is essentially insoluble in water (
K_{\rm {sp}}=1.77\times10^{-10}
). When NH3 is introduced into the solution, the solid AgCl dissolves because NH3 forms a complex ion with Ag+.
NH3 increases the molar solubility of AgCl, and the equilibrium constants can be used to calculate the new solubility. The first step is to calculate the overall equilibrium constant for the entire system by combining the equilibrium equations for AgCl into one overall equation:
{\rm{AgCl}}(s)+{2\rm{NH}_{3}}(aq)\rightleftharpoons \left[\rm{Ag}(\rm{NH}_{3})_{2}\right]^{+}\!(aq)+{\rm{Cl}^{-}}(aq),\hspace{15pt}K=?
The expression for K follows the usual rules of equilibrium constants. Writing out the expression for K, however, reveals something interesting:
K=\frac{\left[\left[\rm{Ag}(\rm{NH}_{3})_{2}\right]^{+}\right][\rm{Cl}^{-}]}{[\rm{NH}_{3}]^{2}}=\frac{\left[\left[\rm{Ag}(\rm{NH}_{3})_{2}\right]^{+}\right]}{[\rm{Ag}^{+}][\rm{NH}_{3}]^{2}}\times{[\rm{Ag}^{+}]}{[\rm{Cl}^{-}]}=K_{\rm{f}}\times K_{\rm{sp}}
K is simply equal to the product of Kf and Ksp. Recall that Kf is the formation constant and Ksp is the solubility constant. Both are calculated in a similar way, but Kf describes the formation of a complex ion and Ksp describes the formation of a saturated solution. Use the known values of Kf and Ksp to calculate the value of K. Note that extra digits are included since this is an intermediate calculation.
\begin{aligned}K&={K_{\rm{f}}} \times {K_{{\rm{sp}}}}\\& = \left( {1.6 \times {{10}^7}} \right)\left( {1.77 \times {{10}^{-10}}} \right)\\&= 2.832 \times {10^{-3}}\end{aligned}
Now let s equal the molar solubility of AgCl. Remember that the molar solubility is equal to the concentration of a solute at equilibrium. If s mol of AgCl/L dissolves, the equilibrium equation requires that the concentrations of [Ag(NH3)2]+ and Cl– are also s mol/L. Then calculate the final concentrations of all species.
Let s represent the number of moles of AgCl dissolved. The balanced equation then indicates that 2s moles of NH3 react, producing s moles of [Ag(NH3)2]+ and s moles of Cl-. The final concentrations can be determined by subtracting or adding.
Next, use the final concentrations and the value of the equilibrium constant, K, to solve for s:
\begin{aligned}K &= \frac{\left[\left[\rm{Ag(NH}_3\rm{)}_2\right]^+\right]\![\rm{Cl}^-]}{\left[\rm{NH}_3\right]^2}\\2.832\times10^{-3}&= \frac{{{s^2}}}{{{{(1.0-2s)}^2}}}\\\sqrt {2.832 \times {{10}^{-3}}} &= \frac{s}{{1.0-2s}}\\s &= \left( {1.0-2s} \right)\sqrt {2.832 \times {{10}^{-3}}} \\s&= 4.8 \times {10^{-2}}\end{aligned}
Remember that the variable s is defined to be the molar solubility of AgCl, so the molar solubility of solid AgCl in 1.0 M aqueous NH3 is
4.8\times10^{-2}\,\rm{M}
. Compare that to the molar solubility of AgCl without NH3, which is a simpler calculation that requires only Ksp. Once again let s represent the molar solubility so that
s=\rm{[Ag}^+\rm{]}=\rm{[Cl}^-\rm{]}
, and use the Ksp expression to solve for s:
\begin{aligned}{K_{{\rm{sp}}}} &= [ {{\rm{Ag}}^ +}][ {{\rm{Cl}}^-} ]\\&= s \cdot s\\&= 1.77 \times {10^{-10}}\\s &= \sqrt {1.77 \times {{10}^{-10}}} \\&= 1.33 \times {10^{-5}}\end{aligned}
Without NH3, the molar solubility of AgCl drops from
4.8 \times {10^{-2}}{\,\rm{M}}
1.33 \times {10^{-5}}{\,\rm{M}}
. So a 1.0 M NH3 solution increases the molar solubility of AgCl by three orders of magnitude. This increase in solubility occurs because the formation of the complex ion [Ag(NH3)2]+ reduces the concentration of [Ag+] and shifts the equilibrium of the reaction to the right.
{\rm{AgCl}}(s)\rightleftharpoons {\rm{Ag}^{+}}(aq)+{\rm{Cl}^{-}}(aq)
<Lewis Acids and Bases>Weak versus Strong Electrolytes
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Flux - Wikipedia @ WordDisk
This article needs attention from an expert in physics. The specific problem is: confusion between flux and flux density. (September 2016)
{\displaystyle \mathbf {F} }
{\displaystyle S}
{\displaystyle dS}
{\displaystyle \mathbf {F} (\mathbf {x} )}
{\displaystyle {\hat {\mathbf {n} }}(\mathbf {x} )}
{\displaystyle \mathbf {x} }
{\displaystyle dS}
{\displaystyle \mathbf {F} \cdot {\hat {\mathbf {n} }}\,dS}
This article uses material from the Wikipedia article Flux, and is written by contributors. Text is available under a CC BY-SA 4.0 International License; additional terms may apply. Images, videos and audio are available under their respective licenses.
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33 CFR § 183.33 - Maximum weight capacity: Inboard and inboard-outdrive boats. | CFR | US Law | LII / Legal Information Institute
33 CFR § 183.33 - Maximum weight capacity: Inboard and inboard-outdrive boats.
(a) The maximum weight capacity (W) marked on a boat that has one or more inboard or inboard-outdrive units for propulsion must not exceed the greater value of W obtained from either of the following formulas:
\begin{array}{c}W=\frac{\left(\text{maximum displacement}\right)}{5}-\frac{\text{boat weight}}{5}-\frac{4\left(\text{machinery weight}\right)}{5}\\ \mathrm{or}\\ W=\frac{\left(\text{maximum displacement}-\text{boat weight}\right)}{7}\end{array}
(3) “Machinery weight” is the combined weight of installed engines or motors, control equipment, drive units, and batteries.
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Biological Effects of Radiation - Course Hero
General Chemistry/Radioactivity/Biological Effects of Radiation
Radioisotopes, isotopes that can undergo radioactive decay at relatively fast rates, are found throughout the natural environment. Organisms are exposed to different kinds of radiation throughout their lives.
Radiation can be categorized as two main types: nonionizing radiation and ionizing radiation. Nonionizing radiation is radiation that does not have sufficient energy to create ions from the atoms or molecules it strikes. Nonionizing radiation cannot remove electrons from any molecule. Visible light is nonionizing radiation. Wavelengths of electromagnetic radiation longer than visible light, such as infrared and radio waves, are also nonionizing radiation. Nonionizing radiation does not have biological impact. Ionizing radiation, however, is radiation that has sufficient energy to create ions from the atoms or molecules it strikes. Alpha and beta particles, as well as gamma rays and X-rays, are all ionizing radiation.
Ionizing radiation has sufficient energy to create ions by breaking molecular bonds. Nonionizing radiation lacks this amount of energy. Radioisotopes emit ionizing radiation.
Matter can block ionizing radiation, and different types of ionizing radiation can move different distances in matter before they are blocked. Alpha particles have the least penetrating power—they can be blocked by a sheet of paper. This is because
\rm\alpha
particles have very large mass; they are helium nuclei. Alpha particles are mostly blocked by skin but are still dangerous if ingested or inhaled.
Beta particles have slightly more penetrating power. They have much less mass than alpha particles, making it easier for them to pass between molecules, such as those in a sheet of paper. About 5 mm of acrylic plastic or a thin sheet of aluminum, however, can stop a
\rm\beta
Gamma rays have the least mass—none at all. This makes it possible for them to pass through materials that can stop alpha and beta particles. About 5 cm of lead can stop gamma rays. Note that these figures represent the amount required to stop a single particle or photon. In practice, the amount of material required to block radiation depends largely on the quantity of radioisotope present or on the amount of radiation emitted.
Penetrating Power of Different Types of Radiation
Alpha particles are easily blocked. Beta particles can be blocked by thin sheets of metal or acrylic. Gamma rays can be blocked by lead about 5 cm thick.
The biological effects of ionizing radiation can be broken down into categories in different ways. One division is whether the effects are deterministic or stochastic. Deterministic effects depend on the amount of exposure, or dose, and do not occur below a particular threshold. Stochastic effects do not depend on the dose, and no threshold exists. Instead, the chance of the effect happening depends on the dose. For example, cataracts of the eye are a deterministic effect. Below 5.0 sieverts (Sv) of exposure, cataracts will not form. Radiation-induced cancer is an example of a stochastic effect. The likelihood of developing cancer increases with increased dose.
Another division between types of effects of radiation is whether the effects are acute or latent. Acute effects happen immediately and go away over time if exposure stops. Latent effects are defined as those that appear six months or more following exposure. Acute effects are called radiation sickness and include symptoms such as nausea and vomiting and reduced red blood cells. The severity of acute effects depends on the dose. Latent effects are further divided into somatic effects and genetic effects. Somatic effects occur in the individual who received the exposure, while genetic effects occur in the descendants of that individual.
Characteristics of Effects
Location of Effects
Deterministic effects Acute effects Somatic effects Skin damage
Damage to blood-forming organs
Latent effects Cataracts
Stochastic effects Cancer
Genetic effects Hereditary effects
Effects of ionizing radiation on humans can be deterministic (occurring only once an exposure threshold has been exceeded) or stochastic (occurring without a threshold). The timing can be acute (occurring soon after exposure) or latent (delayed; possibly occurring years later). Genetic effects are passed on to offspring, but somatic effects occur only in an individual.
It is impossible to completely protect against radiation exposure. However, limiting exposure to levels below which no harmful deterministic effects occur greatly reduces an individual's risk of developing any type of effect. In order to measure how much radiation a person (or other object) is exposed to, various instruments and devices have been invented.
The best-known instrument used to measure radiation is a Geiger counter, a device that measures ionizing radiation using a tube filled with inert gas and metal electrodes. This tube is called a Geiger-Müller tube and gives the counter its name. When ionizing radiation hits a Geiger-Müller tube, some of the gas in the tube is ionized. Ions have an electrical charge and move toward the electrodes of the tube. This motion changes the electric field inside the tube, which can be measured. Geiger counters are portable, inexpensive, and durable, making them quite popular.
A scintillation counter is an instrument that uses the photoelectric effect, the emission of electrons from a surface when light shines on it, to measure ionizing radiation. Liquid scintillation counters, in which beta particle emitters are immersed in a liquid before being placed in the counter, can precisely measure the quantities of radioisotope present in a sample. They can also be connected to storage or output devices, such as computers and printers, making them ideal for academic and industrial uses.
A dosimeter is a device that uses a metal chip or crystal to absorb radiation and then indicates the total absorbed amount later when subjected to light or heat respectively. Dosimeters measure exposure over time and are used to track exposure of people or objects with high exposure risk, such as X-ray technicians and nuclear power plant workers. Dosimeters are formed as a small badge that can be carried in a pocket, affixed to a lapel, or worn as a ring on a finger.
Because the reasons for measuring radioactivity vary, so too do the units. Radiation detectors often describe amounts of radiation in terms of counts per minute (cpm). The amount of radiation emitted by radioactive material, or the amount released into the environment by an accidental release of radiation, is often described using the unit curie (Ci) or the more modern SI unit becquerel (Bq). The amount of radiation absorbed by a person exposed to radiation is described by the unit radiation absorbed dose (rad) or the SI unit gray (Gy), where
1\;\rm{Gy} = 100\;\rm{rad}
. The biological risk of exposure from radiation is described by the unit roentgen equivalent man (rem) or the SI unit sievert (Sv).
<Radioactive Decay>Uses of Radioisotopes
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Terahertz Detection of Halogen Additive-Containing Plastics
Optics and Photonics Journal > Vol.10 No.12, December 2020
Terahertz Detection of Halogen Additive-Containing Plastics ()
Tadao Tanabe1*, Yuki Makino2, Akio Shiota2, Mariya Suzuki1, Ryo Tanuma2, Masafumi Miyajima3, Noriyuki Sato4, Yutaka Oyama2
1Department of Engineering and Design, Shibaura Institute of Technology, Tokyo, Japan.
2Department of Materials Science, Tohoku University, Sendai, Japan.
3Bosch Corporation, Shibuya, Tokyo, Japan.
4Industrial Technology Institute, Miyagi Prefectural Government, Sendai, Japan.
DOI: 10.4236/opj.2020.1012025 PDF HTML XML 209 Downloads 671 Views Citations
Terahertz (THz) waves are transparent with respect to non-polar substances, such as plastics. The refractive index at THz frequencies and the specific spectral features of plastic materials, such as polypropene (PP), polyethylene (PE), polyvinyl alcohol (PVC) and polyethylene terephthalate (PET) can be used to discriminate waste plastics into each material regardless of color. This is important since colored plastic cannot be identified by the present method of near-infrared (NIR) reflection spectroscopy due to absorption of NIR light by the coloring. In addition, the THz refractive index of acrylonitrile butadiene styrene (ABS) increases in proportion to the content of bromine (Br), which can be used to quantitatively evaluate the presence of halogen additives in waste plastic. We show that the non-contact THz identification of materials and additives in plastics can be an effective method for sorting plastic waste. Safe & efficient waste management is one of the most urgent social requirements for smart city innovation.
Terahertz Sorting, Plastic Waste, Halogen Additive
Tanabe, T. , Makino, Y. , Shiota, A. , Suzuki, M. , Tanuma, R. , Miyajima, M. , Sato, N. and Oyama, Y. (2020) Terahertz Detection of Halogen Additive-Containing Plastics. Optics and Photonics Journal, 10, 265-272. doi: 10.4236/opj.2020.1012025.
At present, the status of plastic waste recycling in Japan is that 56% of plastic waste generated in 2018 was treated by thermal recycling, 23% by material recycling and 4% by chemical recycling, with the remaining 16% untreated [1]. Large quantities of plastic waste are produced from packaging containers [2], electronic devices [3] and vehicles [4]. The separation of plastic wastes is usually performed based on their differences in specific gravity and in their near infrared reflectivity. In material and chemical recycling, plastic wastes are decomposed to raw materials, but it is difficult to discriminate each kind of material. Furthermore, commercial plastics contain bromine for fire resistance. Harmful chemicals such as brominated dioxins can be generated during the recycling process. Therefore, it is required to detect and separate halogen additive-containing plastics.
Plastic materials are frequently identified using near infrared reflectivity are then separated by an air jet. But colored plastics cannot be sorted in this way since near infrared light is absorbed by the color. Therefore, Terahertz (THz) evaluation is expected to be useful to identify plastic materials. THz light is an electromagnetic wave that is located between radio waves and infrared, with a frequency in the range 0.1 - 10 THz. THz light has the property of high permeability for non-polar substances, so it is available for non-destructive diagnosis [5] - [12]. Also, it can be expected to be available for use regardless of color. Besides, THz light has low quantum photon energy, so that it is safe for human tissues. Our group has created a database of THz permeability characteristics for industrial materials and successfully constructed non-destructive THz diagnosis of metals in building concrete [13] as well as for insulated electronic cable [14]. For the THz identification of plastic paste, this study investigates the THz refractive index and spectral features of single plastic materials and bromine additive-containing plastics. The THz refractive index and spectra were measured for polypropene (PP), polyethylene (PE), polyvinyl alcohol (PVC), colored polyethylene terephthalate (PET), and bromine (Br) additive-containing acrylonitrile butadiene styrene (ABS), which is used as THz spectral library for discrimination of waste plastic.
Figure 1 indicates a schematic diagram of experimental procedure between THz spectral library and THz applications for discrimination of waste plastic plastics and evaluation of additive. For THz spectral library, THz refractive index and
Figure 1. Schematic diagram of experimental procedure between THz Spectral library and THz applications for plastics.
spectrum of plastic can be measured by using THz time-domain spectroscopy (THz-TDS) and GaP THz Spectrometer, respectively. THz spectrum has information of reflectivity and absorbance, which are effective for practical THz application.
2.1. Plastic Samples
Plastic plates of PE, PP, PVC, PET and ABS were used in this study. Their thickness was 0.5 - 2 mm. ABS plates with various bromine contents were fabricated as a halogen additive-containing plastic. To compare with the additive composition, PE plates were also fabricated with various contents of carbon black. Colored PET plates were commercially prepared as black, white, and clear.
2.2. THz Time-Domain Spectroscopy (THz-TDS) [15] [16] [17]
The optical configuration of THz-TDS was as follows. A femto-second pulse from Ti:Sapphire laser was split into pump light and probe light by a beam splitter and the pump light is introduced to the THz generating device and probe light guided to the THz detection device, where each device is a low-temperature grown GaAs photoconductive antenna. The time delay of THz pulse through the plastic sample can be detected by sweeping the delay. By calculating the transmitted THz pulse shape and phase delay with Fourier transform it is possible to obtain the refractive index of sample in THz region. The refractive index is averaged in the frequency range of 0.2 to 0.7 THz.
2.3. GaP THz Spectrometer [18]
Phonon-polaritons in GaP crystals can afford THz wave generation in a widely tunable frequency range from 1 THz to 4 THz. We have realized a THz wave signal generator as well as THz spectrometer. The GaP signal generator uses Cr:Forsterite lasers for both the pump source and signal source.
2.4. Fabrication of Bromine Additive-Containing Plastics
Bromine additive-containing ABS samples were fabricated by the injection molding of ABS (Cevian V-320SF, Daicel) and flame-retardant ABS (FR-ABS, Cevian V-SER2, Daicel). The mixing rate of natural ABS and FR-ABS is shown in Table 1. In this injection molding process, the material is heated and then extruded into a mold by a screw feed. Each material was molded with a resin temperature of 210˚C and a mold temperature of 45˚C, respectively. The content
Table 1. Details of halogen additive-containing ABS plastics.
of halogen-based flame retardants and Antimony measured by X-ray fluorescence (XRF) is as shown in Table 1. Also, carbon black doped PE plates were fabricated with various ratios of HDPE (Novatec HD HJ260, Japan polyethylene Corporation) and HDPE/carbon black master batch composite (Japan polyethylene Corporation) using the same method. The master batch has a 20 magnitude black color.
3.1. Dependence of the THz Refractive Index on Plastic Materials
Figure 2 shows the dependency of the THz refractive index on the specific gravity of the plastic samples, where the specific gravity is evaluated for each sample by the Archimedes method with water or ethanol. It is found that a linear relationship exists between the specific gravity and the refractive index, where the slope is 0.45. This result suggests that non-contact THz evaluation can separate the plastic waste of PP, PE, ABS, PET and PVC into each material based on its refractive index. As a discriminating component of plastic material, this evaluation can be applied for micro-plastics using THz wavelength, but which is affected by Rayleigh scattering with particle size [19].
3.2. THz Spectra of Plastic Materials
THz spectra of plastic samples are shown in Fig. 3. Based on the difference between their THz spectral characteristics, plastic materials can be identified even for colored plastics. PP has the property of high permeability over a wide frequency range. PE has strong absorption band around 2.2 THz (73 cm−1), which is attributed to the B1u translational lattice vibration mode of PE [20]. Polarization THz measurements are also helpful for the non-destructive diagnosis of mechanical deformation in polymer [21]. PVC shows a higher transparency at lower frequencies and a lower transparency in higher frequency, respectively. PET samples have lower transmission characteristics even in the lower frequency region, regardless of color. As shown in Figure 3, the THz spectral features of PP, PE, PVC can be used to discriminate waste plastics into each material. An
Figure 2. Dependence of the THz refractive index on the specific gravity.
imaging system of 300 - 350 GHz has been realized for sorting black plastics [22].
3.3. Dependence of ABS THz Refractive Index on Br Content
Figure 4 shows the refractive index of bromine additive-containing ABS samples versus the bromine content. The refractive index is constant over the frequency range of 0.2 to 0.7 THz. The horizontal axis indicates the mass% of bromine as measured by XRF. It is found that the refractive index nABS increases linearly with the bromine content xBr, which enables us to evaluate the presence of halogen additives in waste plastic by the following equation:
{n}_{ABS}=0.00\text{48}{x}_{Br}+\text{1}.\text{634}
The detection limit considered from the stability of THz measurement is 0.5 mass% due to the signal/noise ratio of the measurement.
3.4. THz Refractive Index of PE with Various Contents of Carbon Black
Figure 5 shows the THz refractive index of PE samples with various contents of
Figure 3. THz spectra of plastic samples.
Figure 4. Dependence of THz refractive index on the Br content for ABS plastic.
Figure 5. THz refractive index of PE mixed with HDPE/carbon black master batch (MB).
carbon black, which were constant and averaged over the frequency range of 0.2 to 0.7 THz. The horizontal axis indicates the mass% of the HDPE/carbon black master batch (MB) from the fabrication process. The refractive index is almost constant below 10 mass % content of this MB. As shown in Figure 5, over the 10 mass %, the slope of refractive index nPE with MB content xMB is 0.00092, which is fifth smaller than that of nABS with xBr. These results indicate that bromine has a more sensitive influence on the THz refractive index compared to carbon black as an additive.
THz properties of single plastic materials and bromine additive-containing plastics were investigated in order to develop a THz identification technique for plastic waste. Refractive indexes were measured for PE, PP, PVC, PET and ABS in the frequency region of 0.2 - 0.7 THz, for which a linear relationship was confirmed between the specific gravity and the refractive indexes. The THz spectra of these plastic samples show distinguishable properties. The refractive index of ABS plastic is also increased as a function of Br content. Based on the different properties of plastic samples, THz radiation can be used successful to distinguish between plastic materials and identify halogen additive-containing materials within waste plastics.
This research was supported by JST (Japan Science and Technology Agency)-SCORE program in 2019.
[1] Plastic Waste Management Institute, Report 2018.
http://www.pwmi.or.jp
[2] Sobhani, Z., et al. (2020) Microplastics Generated When Opening Plastic Packaging. Scientific Reports, 10, Article No. 4841.
[3] Choi, Y., Choi, H. and Rhee, S. (2020) Estimation on Hazardous Characteristics of the Components from Linear Type of End-of-Life Light-Emitting Diode Lamps. Journal of Material Cycles and Waste Management, 22, 307-314.
[4] Yu, J., Wang, H.Y., Sosuke, T.K., Serrona, K.R.B., Fan, G. and Erdenedalai, B. (2017) Environmental Impacts of Road Vehicles: Past, Present and Future. The Royal Society of Chemistry, London, 174-213.
[5] Hu, B.B. and Nuss, M.C. (1995) Imaging with Terahertz Waves. Optics Letters, 20, 1716-1719.
[6] Koch, M., Hunsche, S., Schumacher, P., Nuss, M.C., Feldmann, J. and Fromm, J. (1998) THz-Imaging: A New Method for Density Mapping of Wood. Wood Science and Technology, 32, 421-427.
[7] Mittleman, D.M., Gupta, M., Neelamani, R., Baraniuk, R.G., Rudd, J.V. and Koch, M. (1999) Recent Advances in Terahertz Imaging. Applied Physics B, 68, 1085-1094.
[8] Ferguson, B. and Zhang, X.C. (2002) Materials for Terahertz Science and Technology. Nature Materials, 1, 26-33.
[9] Kawase, K., Ogawa, Y., Watanabe, Y. and Inoue, H. (2003) Non-Destructive Terahertz Imaging of Illicit Drugs Using Spectral Fingerprints. Optics Express, 11, 2549-2554.
[10] Nishizawa, J. (2004) Development of THz Wave Oscillation and Its Application to Molecular Sciences. Proceedings of the Japan Academy Ser. B, 80, 74-81.
[11] Tonouchi, M. (2007) Cutting-Edge Terahertz Technology. Nature Photonics, 1, 97-105.
[12] Naftaly, M., Vieweg, N. and Deninger, A. (2019) Industrial Applications of Terahertz Sensing: State of Play. Sensors, 19, 4203.
[13] Oyama, Y., Zhen, L., Tanabe, T. and Kagaya, M. (2009) Sub-Terahertz Imaging of Defects in Building Blocks. NDT&E International, 42, 28-33.
[14] Takahashi, S., Hamano, T., Nakajima, K., Tanabe, T. and Oyama, Y. (2014) Observation of Damage in Insulated Copper Cables by THz Imaging. NDT&E International, 61, 75-79.
[15] Miles, R.E., Harrison, P. and Lippens, D. (2001) Terahertz Sources and Systems. Springer, Dordrecht.
[17] Mittleman, D. (2003) Sensing with Terahertz Radiation. Spriger, Berlin.
[18] Suto, K., Sasaki, T., Tanabe, T., Saito, K., Nishizawa, J. and Ito, M. (2015) GaP Terahertz Wave Generator and THz Spectrometer Using Cr:Forsterite Lasers. Review of Scientific Instruments, 76, Article ID: 123109.
[19] Lord Rayleigh, F.R.S. (1879) XXXI. Investigations in Optics, with Special Reference to the Spectroscope. The London, Edinburgh, and Dublin Philosophical Magazine and Journal of Science Series, 5, 261-274.
[20] Tasumi, M. and Shimanouchi, T. (1965) Crystal Vibrations and Intermolecular Forces of Polymethylene Crystals. Journal Chemical Physics, 43, 1245.
[21] Tanabe, T., Watanabe, K., Oyama, Y. and Seo, K. (2010) Polarization Sensitive THz Absorption Spectroscopy for the Evaluation of Uniaxially Deformed Ultra-High Molecular Weight Polyethylene. NDT&E International, 43, 329-333.
[22] Nüßler, D., Pohl, N., Küls, J., Hein, K. and Stein, D. (2014) THz Imaging for Recycling of Black Plastics. GeMiC 2014 German Microwave Conference, Aachen, 1-4.
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949.2 Speed Limit Guidelines - Engineering_Policy_Guide
Revision as of 12:15, 24 November 2010 by Smithk (talk | contribs) (→949.2.2.5 Adjacent Development: minor clarification)
Traffic Pamphlet
The following guidelines promote consistency of speed limit postings on the state of Missouri’s highway system. In all cases, engineering judgment is necessary to determine the most appropriate speed limit at a particular location.
1 949.2.1 Maximum Speed Limits
1.1 949.2.1.1 Definitions (per MRS 304.010)
1.2 949.2.1.2 Discrepancies
1.3 949.2.1.3 Length of Section
2 949.2.2 Prevailing Speed Determination
2.1 942.2.2.1 Fatality and Disabling Injury Crash Rate
2.1.1 Table 949.2.2.1 Prevailing Speed Reduction for Severe Crash Rates
2.2 942.2.2.2 Total Crash Rate
2.2.1 Table 949.2.2.2 Prevailing Speed Reduction for Total Crash Rates
2.3 949.2.2.3 Pedestrian Traffic
2.4 949.2.2.4 Parking
2.5 949.2.2.5 Adjacent Development
2.6 949.2.2.6 Procedures for Obtaining Prevailing Speed Data
3 949.2.3 Additional Guidelines for Establishing Speed Limits
949.2.1 Maximum Speed Limits
Maximum speed limits in Missouri are governed by the Missouri Revised Statutes, Section 304.010. The following table summarizes maximum speed limits as established by the statutes:
70 mph Interstate and Freeways
65 mph Expressways
55 mph State Lettered Routes 1
60 mph All other roadways, not urbanized
60 mph Interstate, Freeways and Expressways
1 May be up to 60 mph based on engineering judgement
At no time shall any Missouri road be posted above 70 mph.
“…the Highways and Transportation Commission may set a speed limit higher or lower than the uniform maximum speed limit…if a higher or lower speed limit is recommended by the Department of Transportation.”
949.2.1.1 Definitions (per MRS 304.010)
Freeway: A limited access divided highway of at least ten miles in length with four or more lanes that is not part of the federal interstate system of highways and does not have any crossovers or accesses from streets, roads or other highways at the same grade level as such divided highway within such ten miles of divided highway;
Expressway: A divided highway of at least 10 miles long with 4 or more lanes that is not part of the federal interstate system of highways and has crossovers or accesses from streets, roads or other highways at the same grade level as such divided highway;
949.2.1.2 Discrepancies
Before raising a posted speed limit above the maximums in the state statute, the district should notify the State Traffic Engineer.
Districts should coordinate across district borders to ensure consistent posted speed limits on inter-district routes in accordance with the speed limit guidelines.
949.2.1.3 Length of Section
Any speed limit section should have a logical beginning and ending point. Examples include city limits, roadway type changes, etc. Additionally, speed limits in unincorporated or “non-community” areas should stay consistent for a minimum of 2 miles.
949.2.2 Prevailing Speed Determination
Prevailing speed determination is the starting point to setting speed limits in incorporated and unincorporated areas, pursuant to the aforementioned state statutes. The prevailing speed of free flowing traffic shall be determined using one or more of the three criteria below:
85th percentile speed;
Upper limit of the 10 mph pace;
Average test run speed.
The selected speed limit (in 5 mph increments) should not exceed the established prevailing speed by more than 3 mph.
The following factors may allow for reduction of the prevailing speed. The prevailing speed shall not be reduced below the 50th percentile (average) speed using these factors.
1. Fatality and Disabling Injury Crash Rate
2. Total Crash Rate
3. Pedestrian Traffic
4. Parking
5. Adjacent Development
6. Physical Roadway Conditions
942.2.2.1 Fatality and Disabling Injury Crash Rate
If either the fatal or disabling injury crash rates, based on reportable accidents within the proposed area, are 50% higher than the statewide average crash rate for the same roadway type, prevailing speed may be reduced by 5%.
If the fatal or disabling crash rate is more than twice the statewide rate for the roadway type, the prevailing speed may be reduced by 10%.
Table 949.2.2.1 Prevailing Speed Reduction for Severe Crash Rates
Prevailing Speed Reduction
Fatal or Disabling Injury Crash Rate > 1.5 SWAR 5%
Fatal or Disabling Injury Crash Rate > 2.0 SWAR 10%
942.2.2.2 Total Crash Rate
If the overall crash rate, based on reportable accidents within the proposed area, is 50% higher than the statewide average crash rate for the same roadway type, prevailing speed may be reduced by 5%.
If the crash rate is more than twice the statewide rate for the roadway type, the prevailing speed may be reduced by 10%.
Table 949.2.2.2 Prevailing Speed Reduction for Total Crash Rates
Total Crash Rate > 1.5 SWAR 5%
Total Crash Rate > 2.0 SWAR 10%
949.2.2.3 Pedestrian Traffic
Where sidewalks are not provided adjacent to the route and a total pedestrian traffic along the route exceeds 10 per hour for 3 hours of any 8-hour period, the prevailing speed may be reduced by 5 percent. Pedestrians crossing the route may only be counted if the point of crossing is not protected by a traffic control device.
949.2.2.4 Parking
Where parking is permitted adjacent to the traffic lane, the prevailing speed may be reduced by 5 percent.
949.2.2.5 Adjacent Development
Where significant adjacent development includes residential and commercial access points, the prevailing speed may be reduced to account for driveway conflicts. If necessary, this affect on the prevailing speed may be determined as outlined below.
The effect of driveway entrances can be determined by using a Driveway Conflict Number. For this purpose, a private or field entrance should have a Driveway Conflict Number of 1. Minor commercial entrances should have a number of 5. Major commercial entrances, shopping centers or industrial plant entrances that generate relatively high volumes of traffic, and public streets should have a number of 10. If the total Driveway Conflict Number exceeds a rate of 40 per mile through the proposed zone, the prevailing speed may be reduced by 5 percent. If the total number exceeds a rate of 60 per mile, the prevailing speed may be reduced by 10 percent. However, before a reduction is made due to the Driveway Conflict Number, the Poisson Curve should be used to determine if the accident reduction is statistically significant.
a) Consult Fig. 949.2.2.5, using the current number of accidents for the last year.
b) Determine the graph’s percent value from the Poisson Curve.
c) Compute accident rate for last year.
{\displaystyle {\mbox{Accident Rate}}={\frac {\mbox{no. of crashes x 100,000,000}}{\mbox{365 x ADT x length in miles}}}}
d) Obtain Statewide Accident Rate (SWAR). The accident data is obtained from the Transportation Management System (TMS).
e) Calculate percent reduction between accident rate and statewide accident rate by:
{\displaystyle {\mbox{Percent Reduction}}={100}{\frac {\mbox{AR-SWAR}}{\mbox{AR}}}}
f) Compare % reduction with value obtained in step b.
g) If % reduction equals or exceeds value obtained in step b, the Driveway Conflict Number may be used as the action would be statistically significant.
8.0-mile section of roadway, AADT = 3,100, 30 accidents within last year.
a & b). Value obtained from Fig. 949.2.2 = 30 %
{\displaystyle {\mbox{AR}}={\mbox{331.4}}={\frac {\mbox{30(100,000,000)}}{\mbox{(365)(3,100)(8.0)}}}}
d). SWAR = 242.04
{\displaystyle {\mbox{26.9 percent reduction}}={100}{\frac {\mbox{331.4-242.04}}{\mbox{331.4}}}}
f). Compare % reduction with value.
Since 26.9 is less than 30, the Driveway Conflict Number cannot be considered.
949.2.2.6 Procedures for Obtaining Prevailing Speed Data
Spot speed studies can be made with any speed measurement device such as radar, count cards and inductive loops that determine the 85th percentile speed and the upper limit of the 10 mph pace. The 85th percentile speed is defined as the speed at or below which 85 percent of the vehicles are traveling. The 10 mph pace is defined as the 10 mph range containing the most vehicles.
Spot speed studies should be made as close to the center of the proposed zone as is practical. If the zone is lengthier than one mile, studies should be made at a minimum of two locations. If the difference in data between the two locations is minor, the higher value should be used. If the difference in data is over 5 mph, consideration should be given to designating separate zones.
Exercise care to collect the data in a manner and at times that indicate normal conditions. Normal conditions will be assumed to prevail under good weather conditions on dry pavement, following morning rush hours and prior to the evening rush hours, on any day of the week except Saturdays, Sundays or holidays. Observations should not be made immediately following a significant event or during a period of greater than normal police enforcement. Every effort should be made to disguise or conceal the fact that speeds are being reported.
Speeds should be observed for at least 100 passenger cars and pick-up trucks. Trucks over 4 tons will not be included in the data for determining a revised limit.
Average test run speeds shall be determined on the basis of at least two runs in each direction over the length of the proposed zone. The prime consideration in these test runs is to determine a maximum permissible speed. Speeds are to be recorded at 0.1 mile intervals. While making the test run, the driver will try to “float” in the traffic stream, passing as many vehicles as pass the test car.
949.2.3 Additional Guidelines for Establishing Speed Limits
Normally, isolated curves and turns, areas of restricted sight distance, no passing zones, etc. will not be considered as areas for lowering speed limits. The accident record on such highways will be the principle criterion. Critical speed signs shall be used in connection with the appropriate warning signs at these locations.
Except on divided highways, different speeds for traffic in opposite directions cannot be justified.
Speed limits shall begin at a point on or as near in advanced of the point as possible where the speed limit is warranted and shall end at the point or as near the point as possible where the speed limit is not warranted.
Retrieved from "https://epg.modot.org/index.php?title=949.2_Speed_Limit_Guidelines&oldid=24682"
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Diameter Knowpia
Circle with circumference (C) in black, diameter (D) in cyan, radius (R) in red, and centre or origin (O) in green.
In more modern usage, the length
{\displaystyle d}
of a diameter is also called the diameter. In this sense one speaks of the diameter rather than a diameter (which refers to the line segment itself), because all diameters of a circle or sphere have the same length, this being twice the radius
{\displaystyle r.}
{\displaystyle d=2r\qquad {\text{ or equivalently }}\qquad r={\frac {d}{2}}.}
For a convex shape in the plane, the diameter is defined to be the largest distance that can be formed between two opposite parallel lines tangent to its boundary, and the width is often defined to be the smallest such distance. Both quantities can be calculated efficiently using rotating calipers.[1] For a curve of constant width such as the Reuleaux triangle, the width and diameter are the same because all such pairs of parallel tangent lines have the same distance.
For an ellipse, the standard terminology is different. A diameter of an ellipse is any chord passing through the centre of the ellipse.[2] For example, conjugate diameters have the property that a tangent line to the ellipse at the endpoint of one diameter is parallel to the conjugate diameter. The longest diameter is called the major axis.
The word "diameter" is derived from Ancient Greek: διάμετρος (diametros), "diameter of a circle", from διά (dia), "across, through" and μέτρον (metron), "measure".[3] It is often abbreviated
{\displaystyle {\text{DIA}},{\text{dia}},d,}
{\displaystyle \varnothing .}
The definitions given above are only valid for circles, spheres and convex shapes. However, they are special cases of a more general definition that is valid for any kind of
{\displaystyle n}
-dimensional (convex or non-convex) object, such as a hypercube or a set of scattered points. The diameter or metric diameter of a subset of a metric space is the least upper bound of the set of all distances between pairs of points in the subset. Explicitly, if
{\displaystyle S}
is the subset and if
{\displaystyle \rho }
is the metric, the diameter is
{\displaystyle \operatorname {diam} (S)=\sup _{x,y\in S}\rho (x,y).}
{\displaystyle \rho }
is viewed here as having codomain
{\displaystyle \mathbb {R} }
(the set of all real numbers), this implies that the diameter of the empty set (the case
{\displaystyle S=\varnothing }
{\displaystyle -\infty }
(negative infinity). Some authors prefer to treat the empty set as a special case, assigning it a diameter of
{\displaystyle 0,}
[4] which corresponds to taking the codomain of
{\displaystyle d}
to be the set of nonnegative reals.
For any solid object or set of scattered points i{\displaystyle n}
-dimensional Euclidean space, the diameter of the object or set is the same as the diameter of its convex hull. In medical parlance concerning a lesion or in geology concerning a rock, the diameter of an object is the least upper bound of the set of all distances between pairs of points in the object.
In planar geometry, a diameter of a conic section is typically defined as any chord which passes through the conic's centre; such diameters are not necessarily of uniform length, except in the case of the circle, which has eccentricity
{\displaystyle e=0.}
Sign U+2205 ∅ EMPTY SET from an AutoCAD drawing in dim.shx font with an angle 16°. This font does not contain U+2300 ⌀ DIAMETER SIGN.
The symbol or variable for diameter, ⌀, is sometimes used in technical drawings or specifications as a prefix or suffix for a number (e.g. "⌀ 55 mm"), indicating that it represents diameter. For example, photographic filter thread sizes are often denoted in this way.
In German, the diameter symbol (German Durchmesserzeichen) is also used as an average symbol (Durchschnittszeichen).
It is similar in size and design to ø, the Latin small letter o with stroke. The diameter symbol ⌀ is distinct from the empty set symbol ∅, from an (italic) uppercase phi Φ, and from the Nordic vowel Ø (Latin capital letter O with stroke).[5] See also slashed zero.
The symbol has a Unicode code point at U+2300 ⌀ DIAMETER SIGN, in the Miscellaneous Technical set. On an Apple Macintosh, the diameter symbol can be entered via the character palette (this is opened by pressing ⌥ Opt⌘ CmdT in most applications), where it can be found in the Technical Symbols category. In Unix/Linux/ChromeOS systems, it is generated using Ctrl+⇧ Shift+U 2300space. It can be obtained in Unix-like operating systems using a Compose key by pressing, in sequence, Composedi.[6] In Windows, it can be entered in most programs with Alt code 8960.
The character will sometimes not display correctly, however, since many fonts do not include it. In many situations, the Nordic letter ø at Unicode U+00F8 ø LATIN SMALL LETTER O WITH STROKE (ø) is an acceptable substitute. It can be entered on a Macintosh by pressing ⌥ OptO (the letter o, not the number 0). In Unix/Linux/ChromeOS systems, it is generated using Ctrl+⇧ Shift+U F8space or Composeo/. AutoCAD uses U+2205 ∅ EMPTY SET available as a shortcut string %%c.
In Microsoft Word, the diameter symbol can be acquired by typing 2300 and then pressing Alt+X.
In LaTeX, the diameter symbol can be obtained with the command \diameter from the "wasysym" package.[7]
Angular diameter – How large a sphere or circle appears
Conjugate diameters – Perpendicular diameters of a conic section
Diameter (group theory), a concept in group theory
Semidiameter – Term in geometry; half of a shape's diameter
The diameters of a screwthread
^ Toussaint, Godfried T. (1983). "Solving geometric problems with the rotating calipers". Proc. MELECON '83, Athens. CiteSeerX 10.1.1.155.5671. {{cite journal}}: Cite journal requires |journal= (help)
^ Bogomolny, Alexander. "Conjugate Diameters in Ellipse". www.cut-the-knot.org.
^ "diameter - Origin and meaning of diameter by Online Etymology Dictionary". www.etymonline.com.
^ "Re: diameter of an empty set". at.yorku.ca.
^ Korpela, Jukka K. (2006), Unicode Explained, O'Reilly Media, Inc., pp. 23–24, ISBN 978-0-596-10121-3 .
^ Monniaux, David. "UTF-8 (Unicode) compose sequence". Retrieved 2018-07-13.
^ "wasysym – LaTeX support for the wasy fonts". Comprehensive TeX Archive Network. Retrieved 2022-03-11.
Look up diameter in Wiktionary, the free dictionary.
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2022 Hausdorff dimension and projections related to intersections
Pertti Mattila1
1Department of Mathematics and Statistics, P.O. Box 68, FI-00014 University of Helsinki, Finland
{S}_{g}\left(x,y\right)=x-g\left(y\right)
x,y\in {ℝ}^{n}
g\in O\left(n\right)
, we investigate the Lebesgue measure and Hausdorff dimension of
{S}_{g}\left(A\right)
given the dimension of
A
, both for general Borel subsets of
{ℝ}^{2n}
and for product sets.
Pertti Mattila. "Hausdorff dimension and projections related to intersections." Publ. Mat. 66 (1) 305 - 323, 2022. https://doi.org/10.5565/PUBLMAT6612212
Received: 26 May 2020; Accepted: 22 September 2020; Published: 2022
Keywords: Fourier transform , Hausdorff dimension , intersection , projection
Pertti Mattila "Hausdorff dimension and projections related to intersections," Publicacions Matemàtiques, Publ. Mat. 66(1), 305-323, (2022)
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(Redirected from Joseph Louis Lagrange)
{\displaystyle y=a\sin(mx)\sin(nt)\,}
{\displaystyle {\textbf {F}}={\frac {d}{dt}}(m{\textbf {v}})}
{\displaystyle ax^{n}}
{\displaystyle b=0}
{\displaystyle {\frac {d}{dt}}{\frac {\partial T}{\partial {\dot {x}}}}-{\frac {\partial T}{\partial x}}+{\frac {\partial V}{\partial x}}=0,}
{\displaystyle a^{p-1}-1\equiv 0{\pmod {p}}}
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Fuguang Ding, Jing Wu, Yuanhui Wang, "Stabilization of an Underactuated Surface Vessel Based on Adaptive Sliding Mode and Backstepping Control", Mathematical Problems in Engineering, vol. 2013, Article ID 324954, 5 pages, 2013. https://doi.org/10.1155/2013/324954
Fuguang Ding,1 Jing Wu,1 and Yuanhui Wang1
The paper studied controlling problem of an underactuated surface vessel with unknown interferences. It proved that the control problem of underactuated surface vessel can be transformed into the stabilization analysis of two small subsystems. This controller was designed by backstepping method and adaptive sliding mode, was suitable for solving the problem of the control of higher systems, can keep the system global asymptotic stability, and can inhibit unknown interference, and boundary layer can weaken the buffeting generated by sliding mode. The unknown interference was estimated by adaptive function. Finally, the simulation results are given to demonstrate the effectiveness of the proposed control laws.
With the rapid development of the modern shipping industry, the study about the controller of the underactuated surface vessel has received extensive attention. Since stabilization control problem of the underactuated surface vessel belongs to the field of nonlinear control, nonlinear control theory is the basic foundation of the underactuated stabilization control theory [1]. Underactuated ship system is a nonholonomic system, and the nonholonomic system which does not meet the sufficient condition of Brockett system linear approximation is not controllable; so, stabilization control of underactuated systems cannot be continuous and smooth [2]. Similarly, it cannot use output feedback or smooth state feedback on a linear system, and the linear control method is not satisfactory to the stabilization control of underactuated system.
So far, underactuated ship stabilization control problem is more studied without interferences, just like [3–7] for the case of noninterference. But for a bounded environment disturbance (or unknown disturbances) and the uncertainty of the ship systems, underactuated stabilization controller is few, except for [8], but from the research results of them shows that the antidisturbance (or uncertainty) ability of the designed controller is less than ideal. Backstepping and sliding mode control method is the combination of backstepping and sliding mode control and takes full advantages of their own characteristics in [9]. Boundedness and convergence can keep the equilibrium point; sliding mode control system can match perturbation and disturbance, and when using backstepping under less restrictive conditions, it can ensure that the closed-loop system has strong robustness. The controller exist sliding mode; so, it will cause system chattering and make control input change severely, and it cannot meet the requirements of the actual system, but boundary layer method can solve it. Adaptive function is used to cope with approximation of uncertain functions. The paper studies underactuated surface vessel, which under the conditions of bounded variable environmental interference effect. A new adaptive sliding mode and backstepping control method is designed to deal with the underactuated ship stabilization problem, and according to the experimental simulation, the designed controller is effective.
2. Model of an Underactuated Surface Vessel
We consider an underactuated surface vessel with simplified dynamics off-diagonal terms of the linear and nonlinear damping matrices which are ignored. Based on [9], the dynamics and kinematics of an underactuated surface vessel are described as follows: where denote position and orientation in inertial frame, denote linear velocities in surge sway and angular velocity in yaw, are the constants of inertia matrix, and are the constants of damping matrix, and we consider the case that the surface vessel has no side thruster; so, the control input can be described as , and are the environmental disturbances in the body-fixed frame.
3.1. Coordinate and Feedback Transformation
In order to design the controller conveniently, the underactuated ships model will be converted to a suitable form. Based on [10], the coordinate and input transformations are as follows:
By using (2), we can transform the differential equation as follows: where , and .
Remark 1. Because state transformation (2) is a global diffeomorphism transformation, so for is convergent by exponential form, and thus we imply that .
Proposition 2. , is globally exponential stable if , is globally exponential stable.
Proof. The similar proof has been given in [1].
According to Proposition 2, we know that in order to make the stabilization of underactuated surface vessels from the initial position to the origin, the system only needs to make transformation (4). It can be seen as strict feedback nonlinear subsystem, and the cascade system consists of two components which are shown in Figure 1 as follows:
Block diagram of the system of (4).
Figure 1 shows that the controller of underactuated surface vessels can be transformed into two single-input and single-output systems. So, we only need to design the controller to satisfy the fact that and can be controlled in three directions of the ship.
3.2.1. Design the Control Law of System
Firstly, we consider the subsystem and define Lyapunov function , and the time derivative of is as follows:
We take as a virtual control input of subsystem and define . In order to ensure that the subsystem is asymptotically stable when , we choose , where , and . Then, we can get equations as follows:
Secondly, we consider the whole system
We can assume the uncertain parameters exists the upper bound and choose a linear sliding mode switch function
Define the Lyapunov function ; the time derivative of becomes
Choose the control law where is the sign function, , and is the upper bound of uncertain parameter . Combining (12) into (11), we can obtain
Finally, because the upper bound of the general object is difficult to predict, and in order to determine the uncertainties of the upper bound, we use the adaptive algorithm to estimate it. is the upper bound of uncertain parameter . And its adaptive rate is .
Firstly, consider the subsystem , and define Lyapunov function . We see as a virtual control input of subsystem and define . The time derivative of is as follows:
Assuming that lateral disturbance force decrease with time, and . In this paper, we can make ; so, is asymptotically convergent to zero when (or ). In order to make asymptotically converging to zero, we can adjust the (and ) such that the convergence rate of is greater than the convergence rate of (or ).
Define . In order to ensure that the subsystem is asymptotically stable when , we choose , where . Then, we can get equations as follows:
Secondly, consider the whole system
We assume the uncertain parameters exists the upper bound. Choose a linear sliding mode switch function . Define the Lyapunov function ; the time derivative of is as follows:
Choose the control law where , and is the upper bound of uncertain parameter .
Combining (18) into (17), we can obtain
Finally, we use the adaptive algorithm to estimate . And its adaptive rate is .
Firstly, we should analyze the stability of the system , defined as follows:
From (21), choosing the adaptive value of , and , we can make ; then, . The wide range of Lyapunov asymptotic stability theorem shows that the system is stable.
Secondly, system and system use the same method; so, the stability of the system can refer to system .
Proposition 3. The system shown in (1) can be controlled by the input of (12) and (18), and a wide range of asymptotically stabilization can be obtained by (13) and (19).
Proof. The certification process has been given in the design steps.
From field of the theory, the motion choosing sliding mode is that parameter variations and external interference have nothing to do with the systems; so, the robustness of the system using sliding mode controller is better than the general control system. But in fact, sliding mode control can cause the system chattering for the reason of noncontinuous switch; so, the boundary layer method will be introduced to weaken this vibration. Saturation function is used to replace the ideal relay functions in the appropriate boundary layer; in other words, is used instead of as follows: where is the thickness of boundary layer; is a constant. And ; .
In order to verify that the controller is able to calm from the initial point to the origin, consider an underactuated surface vessel with the model parameters in [11], and assume the initial condition to be as follows: , , , , , , and .
Take the disturbances as follows: , and , .
In order to make ship at low speed operation, the longitudinal limited value of thrust is , and torque limit value is . The parameters of the control law are designed with the following: , , , , , , , and . Simulation results are shown as follows.
From Figure 2, we can see the trajectory of the system to be better when there exist disturbances. Designed controller can make the underactuated surface vessel stabilization. From Figures 3 and 4, although the response time is short, designed controller can make all the states of the system stable to the origin. From Figure 5, because of introduction of the boundary layer approach to weaken the chattering phenomenon of the control force and moment, there is no change when the system is balanced, and it can save the power as well.
Curve of the ship’s trajectory.
Curve of ship position and heading angle variables.
Curve of ship speed.
Curve of the longitudinal force and yawing moment.
In this paper, underactuated surface vessels use the combination of adaptive sliding mode and backstepping to design the controller with the consideration of environmental disturbance and use the boundary layer method to solve the chattering problem. From the analysis of stabilization, the system is state global stabilization to the origin. The designed controller overcomes the problems of anti-to perturbation poor difficulties of the other controllers [8]; so, the designed controller improves the robustness of the system.
This work is supported by the National Natural Science Foundation of China (51209056) and the Fundamental Research Funds for the Central Universities (HEUCF 100420 and HEUCF 110430).
K. D. Do and J. Pan, Control of Ships and Underwater Vehicles: Design for Underactuated and Nonlinear Matine Systems, Springer, London, UK, lst edition, 2009.
W. Dong and Y. Guo, “Global time-varying stabilization of underactuated surface vessel,” IEEE Transactions on Automatic Control, vol. 50, no. 6, pp. 859–864, 2005. View at: Publisher Site | Google Scholar | MathSciNet
Y. Liu, C. Guo, and R. Zhou, “Robust feedback stabilization control of an underactuated surface vessel,” in Proceedings of WRI World Congress on Computer Science and Information Engineering (CSIE '09), pp. 46–50, March-April 2009. View at: Publisher Site | Google Scholar
L. C. McNinch, H. Ashrafiuon, and K. R. Muske, “Sliding mode setpoint control of an underactuated surface vessel: simulation and experiment,” in Proceedings of American Control Conference (ACC '10), pp. 5212–5217, July 2010. View at: Google Scholar
J. Cheng, J. Yi, and D. Zhao, “Stabilization of an underactuated surface vessel via discontinuous control,” in Proceedings of the American Control Conference (ACC '07), pp. 206–211, New York, NY, USA, July 2007. View at: Publisher Site | Google Scholar
Y. Yan, S. Yu, J. Du, and K. Zheng, “Observer based finite-time switched controller of an underactuated surface vessel,” in Proceedings of the 3rd International Symposium on Systems and Control in Aeronautics and Astronautics (ISSCAA '10), pp. 1121–1125, June 2010. View at: Publisher Site | Google Scholar
Y. Liu, C. Guo, and R. Zhou, “Asymptotic stabilization control of an underactuated surface vessel with optimization based on genetic algorithm,” in Proceedings of the 2nd International Symposium on Intelligent Information Technology Application (IITA '08), pp. 622–626, December 2008. View at: Publisher Site | Google Scholar
N. Djeghali, M. Ghanes, and S. Djennoune, “Bachstepping fault tolerant control based on second order sliding mode observer: application to induction motors,” in Proceedings of Decision and Control and European Control Conference, pp. 4598–4603, 2011. View at: Google Scholar
B.-l. Ma, “Global
\kappa
-exponential asymptotic stabilization of underactuated surface vessels,” Systems & Control Letters, vol. 58, no. 3, pp. 194–201, 2009. View at: Publisher Site | Google Scholar | MathSciNet
C. Guo, Y. Liu, and S. Zhang, “Global sliding mode based trajectory tracking control for underactuated surface vessels with uncertain dynamics,” in Proceedings of the 31st Chinese Control Conference, pp. 4456–4461, 2012. View at: Google Scholar
Copyright © 2013 Fuguang Ding et al. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.
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Discrete Poisson equation - Wikipedia
In mathematics, the discrete Poisson equation is the finite difference analog of the Poisson equation. In it, the discrete Laplace operator takes the place of the Laplace operator. The discrete Poisson equation is frequently used in numerical analysis as a stand-in for the continuous Poisson equation, although it is also studied in its own right as a topic in discrete mathematics.
1 On a two-dimensional rectangular grid
On a two-dimensional rectangular grid[edit]
Using the finite difference numerical method to discretize the 2-dimensional Poisson equation (assuming a uniform spatial discretization,
{\displaystyle \Delta x=\Delta y}
) on an m × n grid gives the following formula:[1]
{\displaystyle ({\nabla }^{2}u)_{ij}={\frac {1}{\Delta x^{2}}}(u_{i+1,j}+u_{i-1,j}+u_{i,j+1}+u_{i,j-1}-4u_{ij})=g_{ij}}
{\displaystyle 2\leq i\leq m-1}
{\displaystyle 2\leq j\leq n-1}
. The preferred arrangement of the solution vector is to use natural ordering which, prior to removing boundary elements, would look like:
{\displaystyle \mathbf {u} ={\begin{bmatrix}u_{11},u_{21},\ldots ,u_{m1},u_{12},u_{22},\ldots ,u_{m2},\ldots ,u_{mn}\end{bmatrix}}^{\mathsf {T}}}
This will result in an mn × mn linear system:
{\displaystyle A\mathbf {u} =\mathbf {b} }
{\displaystyle A={\begin{bmatrix}~D&-I&~0&~0&~0&\cdots &~0\\-I&~D&-I&~0&~0&\cdots &~0\\~0&-I&~D&-I&~0&\cdots &~0\\\vdots &\ddots &\ddots &\ddots &\ddots &\ddots &\vdots \\~0&\cdots &~0&-I&~D&-I&~0\\~0&\cdots &\cdots &~0&-I&~D&-I\\~0&\cdots &\cdots &\cdots &~0&-I&~D\end{bmatrix}},}
{\displaystyle I}
is the m × m identity matrix, and
{\displaystyle D}
, also m × m, is given by:[2]
{\displaystyle D={\begin{bmatrix}~4&-1&~0&~0&~0&\cdots &~0\\-1&~4&-1&~0&~0&\cdots &~0\\~0&-1&~4&-1&~0&\cdots &~0\\\vdots &\ddots &\ddots &\ddots &\ddots &\ddots &\vdots \\~0&\cdots &~0&-1&~4&-1&~0\\~0&\cdots &\cdots &~0&-1&~4&-1\\~0&\cdots &\cdots &\cdots &~0&-1&~4\end{bmatrix}},}
{\displaystyle \mathbf {b} }
{\displaystyle \mathbf {b} =-\Delta x^{2}{\begin{bmatrix}g_{11},g_{21},\ldots ,g_{m1},g_{12},g_{22},\ldots ,g_{m2},\ldots ,g_{mn}\end{bmatrix}}^{\mathsf {T}}.}
{\displaystyle u_{ij}}
equation, the columns of
{\displaystyle D}
correspond to a block of
{\displaystyle m}
components in
{\displaystyle u}
{\displaystyle {\begin{bmatrix}u_{1j},&u_{2j},&\ldots ,&u_{i-1,j},&u_{ij},&u_{i+1,j},&\ldots ,&u_{mj}\end{bmatrix}}^{\mathsf {T}}}
while the columns of
{\displaystyle I}
{\displaystyle D}
each correspond to other blocks of
{\displaystyle m}
components within
{\displaystyle u}
{\displaystyle {\begin{bmatrix}u_{1,j-1},&u_{2,j-1},&\ldots ,&u_{i-1,j-1},&u_{i,j-1},&u_{i+1,j-1},&\ldots ,&u_{m,j-1}\end{bmatrix}}^{\mathsf {T}}}
{\displaystyle {\begin{bmatrix}u_{1,j+1},&u_{2,j+1},&\ldots ,&u_{i-1,j+1},&u_{i,j+1},&u_{i+1,j+1},&\ldots ,&u_{m,j+1}\end{bmatrix}}^{\mathsf {T}}}
From the above, it can be inferred that there are
{\displaystyle n}
block columns of
{\displaystyle m}
{\displaystyle A}
. It is important to note that prescribed values of
{\displaystyle u}
(usually lying on the boundary) would have their corresponding elements removed from
{\displaystyle I}
{\displaystyle D}
. For the common case that all the nodes on the boundary are set, we have
{\displaystyle 2\leq i\leq m-1}
{\displaystyle 2\leq j\leq n-1}
, and the system would have the dimensions (m − 2)(n − 2) × (m− 2)(n − 2), where
{\displaystyle D}
{\displaystyle I}
would have dimensions (m − 2) × (m − 2).
For a 5×5 (
{\displaystyle m=5}
{\displaystyle n=5}
) grid with all the boundary nodes prescribed, the system would look like:
{\displaystyle {\begin{bmatrix}U\end{bmatrix}}={\begin{bmatrix}u_{22},u_{32},u_{42},u_{23},u_{33},u_{43},u_{24},u_{34},u_{44}\end{bmatrix}}^{\mathsf {T}}}
{\displaystyle A=\left[{\begin{array}{ccc|ccc|ccc}~4&-1&~0&-1&~0&~0&~0&~0&~0\\-1&~4&-1&~0&-1&~0&~0&~0&~0\\~0&-1&~4&~0&~0&-1&~0&~0&~0\\\hline -1&~0&~0&~4&-1&~0&-1&~0&~0\\~0&-1&~0&-1&~4&-1&~0&-1&~0\\~0&~0&-1&~0&-1&~4&~0&~0&-1\\\hline ~0&~0&~0&-1&~0&~0&~4&-1&~0\\~0&~0&~0&~0&-1&~0&-1&~4&-1\\~0&~0&~0&~0&~0&-1&~0&-1&~4\end{array}}\right]}
{\displaystyle \mathbf {b} =\left[{\begin{array}{l}-\Delta x^{2}g_{22}+u_{12}+u_{21}\\-\Delta x^{2}g_{32}+u_{31}~~~~~~~~\\-\Delta x^{2}g_{42}+u_{52}+u_{41}\\-\Delta x^{2}g_{23}+u_{13}~~~~~~~~\\-\Delta x^{2}g_{33}~~~~~~~~~~~~~~~~\\-\Delta x^{2}g_{43}+u_{53}~~~~~~~~\\-\Delta x^{2}g_{24}+u_{14}+u_{25}\\-\Delta x^{2}g_{34}+u_{35}~~~~~~~~\\-\Delta x^{2}g_{44}+u_{54}+u_{45}\end{array}}\right].}
As can be seen, the boundary
{\displaystyle u}
's are brought to the right-hand-side of the equation.[3] The entire system is 9 × 9 while
{\displaystyle D}
{\displaystyle I}
are 3 × 3 and given by:
{\displaystyle D={\begin{bmatrix}~4&-1&~0\\-1&~4&-1\\~0&-1&~4\\\end{bmatrix}}}
{\displaystyle -I={\begin{bmatrix}-1&~0&~0\\~0&-1&~0\\~0&~0&-1\end{bmatrix}}.}
Methods of solution[edit]
{\displaystyle {\begin{bmatrix}A\end{bmatrix}}}
is block tridiagonal and sparse, many methods of solution have been developed to optimally solve this linear system for
{\displaystyle {\begin{bmatrix}U\end{bmatrix}}}
. Among the methods are a generalized Thomas algorithm with a resulting computational complexity of
{\displaystyle O(n^{2.5})}
, cyclic reduction, successive overrelaxation that has a complexity of
{\displaystyle O(n^{1.5})}
, and Fast Fourier transforms which is
{\displaystyle O(n\log(n))}
. An optimal
{\displaystyle O(n)}
solution can also be computed using multigrid methods. [4]
Poisson convergence of various iterative methods with infinity norms of residuals against iteration count and computer time.
In computational fluid dynamics, for the solution of an incompressible flow problem, the incompressibility condition acts as a constraint for the pressure. There is no explicit form available for pressure in this case due to a strong coupling of the velocity and pressure fields. In this condition, by taking the divergence of all terms in the momentum equation, one obtains the pressure poisson equation.
For an incompressible flow this constraint is given by:
{\displaystyle {\frac {\partial v_{x}}{\partial x}}+{\frac {\partial v_{y}}{\partial y}}+{\frac {\partial v_{z}}{\partial z}}=0}
{\displaystyle v_{x}}
is the velocity in the
{\displaystyle x}
{\displaystyle v_{y}}
is velocity in
{\displaystyle y}
{\displaystyle v_{z}}
{\displaystyle z}
direction. Taking divergence of the momentum equation and using the incompressibility constraint, the pressure Poisson equation is formed given by:
{\displaystyle \nabla ^{2}p=f(\nu ,V)}
{\displaystyle \nu }
is the kinematic viscosity of the fluid and
{\displaystyle V}
is the velocity vector.[5]
The discrete Poisson's equation arises in the theory of Markov chains. It appears as the relative value function for the dynamic programming equation in a Markov decision process, and as the control variate for application in simulation variance reduction.[6][7][8]
^ Hoffman, Joe (2001), "Chapter 9. Elliptic partial differential equations", Numerical Methods for Engineers and Scientists (2nd ed.), McGraw–Hill, ISBN 0-8247-0443-6 .
^ Golub, Gene H. and C. F. Van Loan, Matrix Computations, 3rd Ed., The Johns Hopkins University Press, Baltimore, 1996, pages 177–180.
^ Cheny, Ward and David Kincaid, Numerical Mathematics and Computing 2nd Ed., Brooks/Cole Publishing Company, Pacific Grove, 1985, pages 443–448.
^ CS267: Notes for Lectures 15 and 16, Mar 5 and 7, 1996, https://people.eecs.berkeley.edu/~demmel/cs267/lecture24/lecture24.html
^ Fletcher, Clive A. J., Computational Techniques for Fluid Dynamics: Vol I, 2nd Ed., Springer-Verlag, Berlin, 1991, page 334–339.
^ S. P. Meyn and R.L. Tweedie, 2005. Markov Chains and Stochastic Stability. Second edition to appear, Cambridge University Press, 2009.
^ S. P. Meyn, 2007. Control Techniques for Complex Networks, Cambridge University Press, 2007.
^ Asmussen, Søren, Glynn, Peter W., 2007. "Stochastic Simulation: Algorithms and Analysis". Springer. Series: Stochastic Modelling and Applied Probability, Vol. 57, 2007.
Hoffman, Joe D., Numerical Methods for Engineers and Scientists, 4th Ed., McGraw–Hill Inc., New York, 1992.
Sweet, Roland A., SIAM Journal on Numerical Analysis, Vol. 11, No. 3 , June 1974, 506–520.
Press, WH; Teukolsky, SA; Vetterling, WT; Flannery, BP (2007). "Section 20.4. Fourier and Cyclic Reduction Methods". Numerical Recipes: The Art of Scientific Computing (3rd ed.). New York: Cambridge University Press. ISBN 978-0-521-88068-8.
Retrieved from "https://en.wikipedia.org/w/index.php?title=Discrete_Poisson_equation&oldid=1073964553"
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821.11 Calibration - Engineering_Policy_Guide
Definition: Calibration ensures that your equipment delivers the correct amount of herbicide uniformly over the target area.
When: The calibration of spray equipment is to be done prior to the spray season, after any alterations of spray equipment. Equipment calibration is to be routinely checked.
1. Proper calibration is an essential part of any herbicide application dependent on:
a. Constant volume
b. Constant speed
c. Constant pressure
d. Proper application rate (ex. ounces per acre)
2. The speed, volume, and pressure that the sprayer is calibrated must be the same as those during applications
3. Accurate rates are critical for successful results
Important information about an herbicide application is gained by calibrating a spray system.
How much volume is released on a given area or in a given time.
1. How much product to add to tank for area to be treated.
2. How many tanks it will take to complete a large area.
3. How much to mix for small areas.
4. Applications can be done without left over mixed product.
1. Acres treated
2. Amount applied
A. Check spray system. Use only water.
Always check if equipment is working properly. Use water only in the tank to test operation of all booms and handguns.
1. Check for leaks, cracks in hoses, proper pump operation, etc.
2. Check spray bar or boom for proper nozzle spacing and operation.
3. Perform a test application with water on your lot at your desired speed and pressure.
4. Make a visual observation of the spray pattern to see if it dries evenly. If not make needed adjustments before beginning actual applications.
5. Recheck pattern after adjustments are made.
Make sure all nozzles are putting out the same amount. Operate sprayer at desired pressure (ex. 30 psi). Catch water from each nozzle for the same amount of time (ex. 30 seconds). Add all nozzles together and divide by the number of nozzles to get average output. All nozzles should operate with no more than 5% difference.
B. Calibration methods. Use only water.
1. Mark a distance (ex. 500 ft.). Fill tank, set operating pressure (ex. 30 psi), drive marked area at desired speed (ex. 8 mph).
Hand Sprayers – Mark area (ex. 20 ft. x 20 ft.), spray area as if actually applying product
2. Using a stopwatch, track the time taken to travel marked area. Operate boom at desired pressure and catch liquid from nozzles for same time it took to travel marked area.
a. Hand Sprayers – Measure amount of water needed to refill sprayer
Tip: Hand sprayers can be calibrated using the same method on a smaller area
Output from one nozzle can be caught and multiplied by the number of nozzles. This method is slightly less accurate.
Divide the number of ounces caught by 128 to get the gallons used.
Calculate gallons sprayed as follows:
{\displaystyle {\mbox{Gallons per Acre}}={\frac {\mbox{(Gallons Used)}}{\mbox{(Test Area in Acres)}}}}
The speed, pressure and volume used during calibration must be used during the application.
A. Multiple nozzle spray bars - Even distribution.
1. Time water is caught________seconds.
2. Nozzle 1 ___________oz. Nozzle 7 __________oz.
Nozzle 2 ___________oz. Nozzle 8 __________oz.
Nozzle 4 ___________oz. Nozzle 10 _________oz.
3. Total ___________oz.
4. Divide total oz. by # of nozzles.
___________oz. / ________nozzles = ___________Avg. output
5. Figure 5% error of operating range. Any nozzle not performing within this range (+ or -) should be checked and replaced if needed.
__________Avg. output x .05 = __________ 5% error.
1. Desired speed ____________mph
a. Truck#_____________
b. Gear ____________ High / Low (circle)
c. RPM_____________
2. Operating pressure___________psi
3. Test distance____________ft.
4. Time to travel test distance ___________seconds
5. Catch water for same time took to travel test distance at desired pressure.
__________ounces used divided by 128 oz. = _________ gallons used.
C. Gallons per acre (GPA)
1. Test Area ________ = (Test Distance________ x Pattern Width) / Acre (43,560 sq. ft.)
2. Gallons Per Acre(GPA) ____________= (Gallons Used) / (Test Area)
D. Convert miles to acres. Multiply miles to spray by the number of acres in a mile at desired width.
1. _________ miles x 2 (both sides) = _____________ miles
2. _________miles x _________ acres/mile = ___________ acres to spray
Table 821.11.1 Area Covered to Acres per Mile
Acres/Mile
E. Calculate gallons needed to complete spray operation.
_________GPA x __________acres = ____________ gallons needed
F. Calculate amount of herbicide to add to tank or needed for whole job.
________acres x ________rate per acre = ___________product needed
G. Calculate acres per tank.
_____(Tank capacity gallons)/ ____ (gallons per acre) = ____ acres/tank
Mix only amount needed to complete job. Avoid the need to store unused mixtures in spray tank.
H. Conversions
128 oz. = 4 quarts = 1 gallon
32 oz. = 1 quart
16 oz. = 1 pint
43,560 sq.ft. = 1 acre
5,280 ft. = 1 mile
A. Read the label.
B. Any inconsistency in the application rates warrants a re-calibration of equipment.
Backpack Sprayer Calibration:
The following step-by-step method of calibrating a backpack or hand-gun sprayer involves very little math or formulas. It is based on the following principle:
One gallon = 128 fluid ounces and your calibration area to be sprayed is 1/128th of an acre, thus fluid ounces collected = gallons per acre
1. Clean sprayer and nozzle thoroughly. Then, fill the spray tank with clean water. Spray with water only to check to see that the nozzle forms a uniform spray pattern. If the pattern is uneven, check to make sure the nozzle is clean and replace if needed. Adjustable nozzles should be set and marked to permit repeated use of the selected spray pattern. If necessary, add a marker dye to the water to more easily see your spray pattern.
2. Measure an area 18.5 ft. by 18.5 ft., which is 1/128th of an acre. If possible, this should be done in the field on which you will be spraying.
3. Time the number of seconds it takes to spray the measured area uniformily with water using gentle side-to-side sweeping motion with the spray wand similar to painting a home or automobile. Record the number of seconds required to spray the area. During application be sure to maintain a constant sprayer pressure and cover the entire area. Repeat this step and take the average of the two times.
4. Spray into a container for the average time calculated in Step 3. Be sure to maintain constant sprayer pressure while you spray into the container.
5. Measure the number of fluid ounces of water in the bucket. The number of fluid ounces collected from the bucket is equal to the number of gallons of water per acre the sprayer is delivering. Volume sprayed in fluid ounces = gallons of water per acre.
6. Add the proper amount of herbicide to the tank based on label directions.
Table 821.11.2 Rates per Acre
DMA 4 IVM 2 qt. Selective and Aquatic Label Fumes Drift Off R/W
Aquatic Glyphosate 1 - 2% solution Safe, Versatile and Aquatic Non-Selective do not add surfactant is spraying over water
Krenite S 6 qt. Safe, Selective Rate and Time Critical
Plateau or Panoramic 3-5 oz. Safe, Selective -
Outrider 1 oz. Low Rate Johnsongrass control
Escort or generic ¼ - 1/3 oz. Low Rate -
Vastlan 2 qt. Low odor -
Total Veg Control – Custom Mix 2 qt. Low Rate season long control Non selective – Total Vegetation Control
Perspective 3 – 4.75 oz. Dry, broadleaf control Great for fall applications
Milestone 5-7 oz. Broadleaf control Use in PGR mixes or stand alone
Method 2 – 4 oz. for general weed control and 8 oz. for stands of target species Broadleaf Control Use in PGR mixes or stand alone
Basal Bark oil RTU Pre-mixed and ready to use Dormant season and cut stump treatments
Retrieved from "https://epg.modot.org/index.php?title=821.11_Calibration&oldid=38971"
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Effective dose (radiation) - Wikipedia
Effective dose is a dose quantity in the International Commission on Radiological Protection (ICRP) system of radiological protection.[1]
It is the tissue-weighted sum of the equivalent doses in all specified tissues and organs of the human body and represents the stochastic health risk to the whole body, which is the probability of cancer induction and genetic effects, of low levels of ionizing radiation.[2][3] It takes into account the type of radiation and the nature of each organ or tissue being irradiated, and enables summation of organ doses due to varying levels and types of radiation, both internal and external, to produce an overall calculated effective dose.
The SI unit for effective dose is the sievert (Sv) which represents a 5.5% chance of developing cancer.[4] The effective dose is not intended as a measure of deterministic health effects, which is the severity of acute tissue damage that is certain to happen, that is measured by the quantity absorbed dose.[5]
The concept of effective dose was developed by Wolfgang Jacobi and published in 1975, and was so convincing that the ICRP incorporated it into their 1977 general recommendations (publication 26) as "effective dose equivalent".[6] The name "effective dose" replaced the name "effective dose equivalent" in 1991.[7] Since 1977 it has been the central quantity for dose limitation in the ICRP international system of radiological protection.[1]
1.1 Usage for external dose
1.2 Usage for internal dose
2 Calculation of effective dose
2.1 Use of tissue weighting factor WT
3 By medical imaging type
5 Regulatory nomenclature
5.2 US effective dose equivalent
7 Future use of Effective Dose
According to the ICRP, the main uses of effective dose are the prospective dose assessment for planning and optimisation in radiological protection, and demonstration of compliance with dose limits for regulatory purposes. The effective dose is thus a central dose quantity for regulatory purposes.[8]
The ICRP also says that effective dose has made a significant contribution to radiological protection as it has enabled doses to be summed from whole and partial body exposure from external radiation of various types and from intakes of radionuclides.[9]
Usage for external dose[edit]
The calculation of effective dose is required for partial or non-uniform irradiation of the human body because equivalent dose does not consider the tissue irradiated, but only the radiation type. Various body tissues react to ionising radiation in different ways, so the ICRP has assigned sensitivity factors to specified tissues and organs so that the effect of partial irradiation can be calculated if the irradiated regions are known.[10] A radiation field irradiating only a portion of the body will carry lower risk than if the same field irradiated the whole body. To take this into account, the effective doses to the component parts of the body which have been irradiated are calculated and summed. This becomes the effective dose for the whole body, dose quantity E. It is a "protection" dose quantity which can be calculated, but cannot be measured in practice.
An effective dose will carry the same effective risk to the whole body regardless of where it was applied, and it will carry the same effective risk as the same amount of equivalent dose applied uniformly to the whole body.
Usage for internal dose[edit]
Effective dose can be calculated for committed dose which is the internal dose resulting from inhaling, ingesting, or injecting radioactive materials.
The dose quantity used is:
Committed effective dose, E(t) is the sum of the products of the committed organ or tissue equivalent doses and the appropriate tissue weighting factors WT, where t is the integration time in years following the intake. The commitment period is taken to be 50 years for adults, and to age 70 years for children.[11]
Calculation of effective dose[edit]
Graphic showing relationships of protection dose quantities in SI units
Ionizing radiation deposits energy in the matter being irradiated. The quantity used to express this is the absorbed dose, a physical dose quantity that is dependent on the level of incident radiation and the absorption properties of the irradiated object. Absorbed dose is a physical quantity, and is not a satisfactory indicator of biological effect, so to allow consideration of the stochastic radiological risk, the dose quantities equivalent dose and effective dose were devised by the International Commission on Radiation Units and Measurements (ICRU) and the ICRP to calculate the biological effect of an absorbed dose.
To obtain an effective dose, the calculated absorbed organ dose DT is first corrected for the radiation type using factor WR to give a weighted average of the equivalent dose quantity HT received in irradiated body tissues, and the result is further corrected for the tissues or organs being irradiated using factor WT, to produce the effective dose quantity E.
The sum of effective doses to all organs and tissues of the body represents the effective dose for the whole body. If only part of the body is irradiated, then only those regions are used to calculate the effective dose. The tissue weighting factors summate to 1.0, so that if an entire body is radiated with uniformly penetrating external radiation, the effective dose for the entire body is equal to the equivalent dose for the entire body.
Use of tissue weighting factor WT[edit]
The ICRP tissue weighting factors are given in the accompanying table, and the equations used to calculate from either absorbed dose or equivalent dose are also given.
Some tissues like bone marrow are particularly sensitive to radiation, so they are given a weighting factor that is disproportionately large relative to the fraction of body mass they represent. Other tissues like the hard bone surface are particularly insensitive to radiation and are assigned a disproportionally low weighting factor.
Weighting factors for different tissues[12]
Calculating from the equivalent dose:
{\displaystyle E=\sum _{T}W_{T}\cdot H_{T}=\sum _{T}W_{T}\sum _{R}W_{R}\cdot {\bar {D}}_{T,R}}
Calculating from the absorbed dose:
{\displaystyle E=\sum _{T}W_{T}\sum _{R}W_{R}\cdot {\frac {\int _{T}D_{R}(x,y,z)\rho (x,y,z)dV}{\int _{T}\rho (x,y,z)dV}}}
{\displaystyle E}
is the effective dose to the entire organism
{\displaystyle H_{T}}
is the equivalent dose absorbed by tissue T
{\displaystyle W_{T}}
is the tissue weighting factor defined by regulation
{\displaystyle W_{R}}
is the radiation weighting factor defined by regulation
{\displaystyle {\bar {D}}_{T,R}}
is the mass-averaged absorbed dose in tissue T by radiation type R
{\displaystyle D_{R}(x,y,z)}
is the absorbed dose from radiation type R as a function of location
{\displaystyle \rho (x,y,z)}
is the density as a function of location
{\displaystyle V}
{\displaystyle T}
is the tissue or organ of interest
The ICRP tissue weighting factors are chosen to represent the fraction of health risk, or biological effect, which is attributable to the specific tissue named. These weighting factors have been revised twice, as shown in the chart above.
The United States Nuclear Regulatory Commission still uses the ICRP's 1977 tissue weighting factors in their regulations, despite the ICRP's later revised recommendations.[15]
By medical imaging type[edit]
Ionizing radiation is generally harmful and potentially lethal to living things but can have health benefits in radiation therapy for the treatment of cancer and thyrotoxicosis. Its most common impact is the induction of cancer with a latent period of years or decades after exposure. High doses can cause visually dramatic radiation burns, and/or rapid fatality through acute radiation syndrome. Controlled doses are used for medical imaging and radiotherapy.
Regulatory nomenclature[edit]
The UK Ionising Radiations Regulations 1999 defines its usage of the term effective dose; "Any reference to an effective dose means the sum of the effective dose to the whole body from external radiation and the committed effective dose from internal radiation."[19]
US effective dose equivalent[edit]
The US Nuclear Regulatory Commission has retained in the US regulation system the older term effective dose equivalent to refer to a similar quantity to the ICRP effective dose. The NRC's total effective dose equivalent (TEDE) is the sum of external effective dose with internal committed dose; in other words all sources of dose.
In the US, cumulative equivalent dose due to external whole-body exposure is normally reported to nuclear energy workers in regular dosimetry reports.
deep-dose equivalent, (DDE) which is properly a whole-body equivalent dose
shallow dose equivalent, (SDE) which is actually the effective dose to the skin
The concept of effective dose was introduced in 1975 by Wolfgang Jacobi (1928–2015) in his publication "The concept of an effective dose: a proposal for the combination of organ doses".[6][20] It was quickly included in 1977 as “effective dose equivalent” into Publication 26 by the ICRP. In 1991, ICRP publication 60 shortened the name to "effective dose."[21] This quantity is sometimes incorrectly referred to as the "dose equivalent" because of the earlier name, and that misnomer in turn causes confusion with equivalent dose. The tissue weighting factors were revised in 1990 and 2007 due to new data.
Future use of Effective Dose[edit]
At the ICRP 3rd International Symposium on the System of Radiological Protection in October 2015, ICRP Task Group 79 reported on the "Use of Effective Dose as a Risk-related Radiological Protection Quantity".
This included a proposal to discontinue use of equivalent dose as a separate protection quantity. This would avoid confusion between equivalent dose, effective dose and dose equivalent, and to use absorbed dose in Gy as a more appropriate quantity for limiting deterministic effects to the eye lens, skin, hands & feet.[22]
It was also proposed that effective dose could be used as a rough indicator of possible risk from medical examinations. These proposals will need to go through the following stages:
Discussion within ICRP Committees
Revision of report by Task Group
Reconsideration by Committees and Main Commission
Deep-dose equivalent
^ a b ICRP publication, 103 para 103
^ ICRP publication 103, glossary
^ ICRP publication 103, para 104 and 105
^ ICRP publication 103
^ a b Journal of Radiological protection Vol.35 No.3 2015. "Obituary - Wolfgang Jacobi 1928 - 2015."
^ ICRP publication 103 executive summary para 101
^ ICRP publication 103 executive summary para j
^ ICRP publication 103 para 101
^ ICRP publication 103, para 22 & glossary
^ Vennart, J. (1991). "1990 Recommendations of the International Commission on Radiological Protection". Annals of the ICRP. ICRP publication 60. 21 (1–3): 8. Bibcode:1991JRP....11..199V. doi:10.1016/0146-6453(91)90066-P. ISBN 978-0-08-041144-6.
^ 10 CFR 20.1003. US Nuclear Regulatory Commission. 2009. Retrieved 25 November 2012.
- "Radiation Dose in X-Ray and CT Exams". RadiologyInfo.org by Radiological Society of North America. Retrieved 2017-10-23.
^ The UK Ionising Radiations Regulations 1999
^ Jacobi W (1975). "The concept of effective dose - A proposal for the combination of organ doses". Radiat. Environ. Biophys. 12 (2): 101–109. doi:10.1007/BF01328971. PMID 1178826. S2CID 44791936.
^ "Use of Effective Dose", John Harrison. 3rd International Symposium on the System of Radiological Protection, October 2015, Seoul.
M.A. Boyd. "The Confusing World of Radiation Dosimetry - 9444" (PDF). US Environmental Protection Agency. Archived from the original (PDF) on 2016-12-21. Retrieved 2014-05-26. – an account of chronological differences between USA and ICRP dosimetry systems
Retrieved from "https://en.wikipedia.org/w/index.php?title=Effective_dose_(radiation)&oldid=1040794104"
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How Institutional Distance and International Experience Affect the Success or Failure of Foreign Direct Investment by Chinese Enterprises?
\begin{array}{l}Logit\left(su{c}_{ij,t}\right)=\alpha +{\beta }_{1}Dist_inst{i}_{ij,t}+{\beta }_{2}Dist_insti{P}_{ij,t}+{\beta }_{3}Dist_insti{N}_{ij,t}\\ \text{}+{\beta }_{4}EX{P}_{ij,t}+\eta {Х}_{ij,t}+{\epsilon }_{ij,t}\end{array}
\begin{array}{l}Logit\left(su{c}_{ij,t}\right)=\alpha +{\beta }_{1}Dist_inst{i}_{ij,t}+{\beta }_{2}Dist_insti{P}_{ij,t}+{\beta }_{3}Dist_insti{N}_{ij,t}\\ \text{}+{\beta }_{4}EX{P}_{ij,t}+{\beta }_{5}Dist_inst{i}_{ij,t}\ast EX{P}_{ij,t}+{\beta }_{6}Dist_insti{P}_{ij,t}\ast EX{P}_{ij,t}\\ \text{}+{\beta }_{7}Dist_insti{N}_{ij,t}\ast EX{P}_{ij,t}+\eta {Х}_{ij,t}+{\epsilon }_{ij,t}\end{array}
su{c}_{ij,t}
i
j
t
su{c}_{ij,t}\text{=}1
su{c}_{ij,t}=0
Dist_insti{P}_{ij,t}
Dist_insti{N}_{ij,t}
Dist_inst{i}_{ij,t}
EX{P}_{ij,t}
{Х}_{ij,t}
{\epsilon }_{ij,t}
su{c}_{ij,t}
i
j
t
su{c}_{ij,t}=1
su{c}_{ij,t}=0
{\text{gendist}}_{i,j}={\sum }_{i}{\sum }_{j}{s}_{i,p}\cdot {s}_{j,q}\cdot {d}_{pq}
{\text{gendist}}_{i,j}
{s}_{i,p}
{s}_{j,q}
{d}_{pq}
{\text{Rgendist}}_{i,j}=\left[{\text{gendist}}_{i,USA}-{\text{gendist}}_{j,USA}\right]
{\text{Rgendist}}_{i,j}
i
j
\text{Rgendist}_\text{nei}_\text{dominan}
\text{res_instidt}
Yang, J. (2019) How Institutional Distance and International Experience Affect the Success or Failure of Foreign Direct Investment by Chinese Enterprises? American Journal of Industrial and Business Management, 9, 512-535. https://doi.org/10.4236/ajibm.2019.93035
1. North, D.C. (1990) Institutions, Institutional Change and Economic Performance. Cambridge University Press, Cambridge. https://doi.org/10.1017/CBO9780511808678
2. Blonigen, B.A.A. (2005) Review of the Empirical Literature on FDI Determinants. Atlantic Economic Journal, 33, 383-403. https://doi.org/10.1007/s11293-005-2868-9
3. Daniele, V. and Marani, U. (2008) Do Institutions Matter for FDI? A Comparative Analysis for the MENA Countries. MPRA Paper No. 2426.
4. Yan, C.L. and Zou, C. (2013) The Host Country’s System Quality and Institutional Distance and the Location of China’s Foreign Direct Investment. Contemporary Finance and Economics, No. 7, 100-110.
5. Chen, C.S., Liu, X.D. and Yi, C.J. (2017) The Influence of Overseas Chinese Business Network and Host Country System Environment on China’s OFDI—The Perspective Based on “One Belt and One Road”. Journal of Fujian Normal University (Philosophy and Social Sciences Edition), No. 1, 79-86.
6. Buckley, P.J., et al. (2007) The Determinants of Chinese Out-Ward Foreign Direct Investment. Journal of International Business Studies, 38, 499-518. https://doi.org/10.1057/palgrave.jibs.8400277
7. Kolstad, I. and Wiig, A. (2012) What Determines Chinese Out-Ward FDI? Journal of World Business, 47, 26-34. https://doi.org/10.1016/j.jwb.2010.10.017
8. Li, X.M. and Li, C.M. (2016) Institutional Risks of Countries along the “Belt and Road” and the Economic Logic of Chinese Enterprises to “Go Out”. Contemporary Economic Management, No. 38, 8-14.
9. Kostova, T. (1999) Transnational Transfer of Strategic Organizational Practices: A Contextual Perspective. Academy of Management Review, 24, 308-324. https://doi.org/10.5465/amr.1999.1893938
10. Habib, M. and Zurawick, L. (2002) Corruption and Foreign Direct Investment. Journal of International business Studies, 33, 291-307. https://doi.org/10.1057/palgrave.jibs.8491017
11. Bénassy-Quéré, A., Coupet, M. and Mayer, T. (2007) Institutional Determinants of Foreign Direct Investment. World Economy, 29, 38-53. https://doi.org/10.1111/j.1467-9701.2007.01022.x
12. Yan, D.Y. (2011) Research on the Institutional Distance, International Experience and the Success or Failure of Chinese Enterprises’ Overseas Merger and Acquisition. Nankai Economic Research, No. 5, 75-97.
13. Guo, S.W. and Huang, H.M. (2010) The Influence of Institutional Distance on China’s External FDI—An Empirical Study Based on Dynamic Panel Model. International Economics and Trade Research, No. 11, 21-26.
14. Du, J. and Song, Y.G. (2014) Institutional Distance, Factor Endowment and Chinese OFDI Location Choice Preference—An Empirical Study Based on Dynamic Panel Data Model. World Economy Study, No. 12, 47-52.
15. Jiang, G.H. and Jiang, D.C. (2012) Is the Host Country System Important—For China’s Investment in Developing Countries? Management World, No. 11, 45-56.
16. Xu, D. and Shenkar, O. (2002) Institutional Distance and the Multinational Enterprise. The Academy of Management Review, 27, 608-618. https://doi.org/10.2307/4134406
17. Li, X. and Wan, N. (2016) The Legitimacy Threshold of Multinational Enterprises: The Perspective of Institutional Distance. Management World, No. 5, 184-185.
18. Hymer, S.H. (1976) The International Operations of National Firms: A Study of Direct Foreign Investment. Doctoral Dissertation, Massachusetts Institute of Technology, Cambridge.
19. Sepulveda, M., Angel, J. and Rodriguez, D. (2014) Geographical and Industrial Spillovers in Entry Decisions across Export Markets. MPRA Paper No. 53249.
20. Delios, A. and Henisz, W.J. (2003) Political Hazards, Experience, and Sequential Entry Strategies: The International Expansion of Japanese Firms, 1980-1998. Strategic Management Journal, 24, 1153-1164. https://doi.org/10.1002/smj.355
21. Pogrebnyakov, N. and Mait-land, C.F. (2001) Institutional Distance and the Internationalization Process: The Case of Mobile Operators. Journal of International Management, 17, 68-82. https://doi.org/10.1016/j.intman.2010.12.003
22. Hernandez, V. and Nieto, M.J. (2015) The Effect of the Magnitude and Direction of Institutional Distance on the Choice of International Entry Modes. Journal of World Business, 50, 122-132. https://doi.org/10.1016/j.jwb.2014.02.002
23. Xu, D., et al. (2004) The Effect of Regulative and Normative Distance on MNE Ownership and Expatriate Strategies. Mir Management International Review, 44, 285-307.
24. Yang, L., Liu, X. and Zhang, J. (2016) How Bilateral Political Relations Affect Foreign Direct Investment—Based on the Perspective of Binary Margin and Investment Success and Failure. China Industrial Economy, No. 11, 56-72.
25. Huang, X., Shu, Y. and Yu, M. (2013) Institutional Distance and Cross-Border Income Gap. Economic Research, No. 9, 4-16.
26. Cavalli-Sforza, L.L., Menozzi, P. and Piazza, A. (1994) The History and Geography of Human Genes. Princeton University Press, Princeton.
27. Spolaore, E. and Wacziarg, R. (2009) The Diffusion of Development. Quarterly Journal of Economics, 124, 469-529. https://doi.org/10.1162/qjec.2009.124.2.469
28. Hilbe, J.M. (2011) Negative Binomial Regression. 2nd Edition, Cambridge University Press, Cambridge. https://doi.org/10.1017/CBO9780511973420
29. Zhou, H. and Zheng, Y. (2015) The Impact of Environmental Regulation on Industrial Transfer—Evidence from the Location of New Manufacturing Enterprises. Southern Economy, No. 4, 12-26.
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Long-Run Aggregate Supply - Course Hero
Macroeconomics/Aggregate Demand and Aggregate Supply/Long-Run Aggregate Supply
What Is Long-Run Aggregate Supply?
The long-run aggregate supply (LRAS) curve is a description of the relationship between the price level and the quantity of output produced in an economy over the long run.
The aggregate supply curve shows the relationship between price level and output. In the short run, this curve reflects the view that firms will respond to an increase in price by increasing their output. However, this action relies on the premise that the price at which consumers are buying is increasing while the input costs for production are remaining the same. While this may be true in the short run, this assumption does not hold in the long run. Over time, the input costs also increase, making it no longer advantageous to increase output in response to an increase in price. As a result, the long-run aggregate supply curve (LRAS) shows that output does not depend on price level, and the linear expression of output versus price is a vertical line, determined only by production costs. The main rationale is that there is potential output based on capacity and that the economy always returns to this level of potential output. Potential output or GDP occurs when the economy is at the natural rate of unemployment, the amount of unemployment that occurs when the economy is producing at potential output. Any remaining unemployment is considered to be frictional, structural, or voluntary. It is also assumed that all the material and capital resources in an economy are fully available to be utilized for the production of goods and services.
At full employment of all resources, the LRAS is a vertical line that intersects the aggregate demand (AD) curve, determining an equilibrium price level. The economy may shift away from this equilibrium point in the short run, but in the long run it will always come back to the same vertical LRAS, even though it may return at a different price level. For example, if there is an increase in AD, both price level and output increase in the short term, and the economy will move to the right along the SRAS. Over time, the economy will adjust and output will come back down. However, the price increase will remain, and the economy will return to a higher point on the LRAS corresponding to an increase in price level. Similarly, if AD decreases, both price level and output decrease, and the economy will move to a lower point on the LRAS.
Over the long run, GDP will remain at the same point (
Y^{\ast}
) at varying price levels, resulting in a long-run aggregate supply curve (LRAS) that is a vertical line.
The LRAS is described by
Y=Y^\ast
Y^{\ast}
is the natural level of output. This value is not affected by price levels.
The long-run aggregate supply curve can shift when the productive potential of an economy changes because of factors such as increased resources or technological development, but not because of changes in price level.
The long-run aggregate supply (LRAS) curve is a vertical line on a graph of output versus price level, indicating that in the long run, there is a potential level of output from an economy that is independent of price. The LRAS curve can be compared to the production possibilities frontier (PPF) model. The PPF model indicates various combinations of two products that an economy can produce in the most efficient manner, given a fixed supply of resources and a given state of technology. Points along the PPF represent the efficient use of the economy's resources to produce various combinations of the two goods. A change in resources used to produce one or both of the goods represented in the model or a change in the state of technology will shift the PPF. The LRAS curve explains very much the same idea. An improvement in the level of labor, capital, technology, or natural resources will cause the LRAS curve to shift. A shift in the LRAS curve indicates that this natural level of output has either increased or decreased, so the productive potential of the economy has changed. A shift in the PPF also illustrates changes in the economy's output potential. A shift of the LRAS curve to the left indicates a reduction in productive potential, and a shift to the right indicates an increase in productive potential.
In order to change the productive potential of an economy, something fundamental needs to alter its potential level of output, such as changing the quality and quantity of available resources. Resources could refer to labor, capital, technology, or natural resources. If, for instance, there is a growth in population that increases the amount of available labor in an economy, the economy's potential level of output level will increase and the LRAS curve will shift to the right. On the other hand, if there is less available labor, output will decrease and the LRAS curve will shift to the left. Technological development, such as better machinery, is another key factor for increasing production because it can enable labor to be more productive or increase the ability to make use of natural resources. During the Industrial Revolution (the development of mechanized industry over the 18th and 19th centuries), inventions such as the sewing machine and cotton gin increased production greatly. These technologies increased output and shifted the LRAS curve to the right.
As potential output increases (from Y1 to Y2),the long-run aggregate supply curve (LRAS) shifts to the right.
If governments want to increase or decrease their output to shift the LRAS curve, they can implement certain policies. The supply of labor can be controlled through work incentives and regulating immigration and migration. Infrastructure investments can also affect labor availability by improving workers' mobility. Additionally, innovation and technological development can be promoted through capital investment and favorable tax policies for businesses, enabling the natural output of the economy to increase.
<Long-Run versus Short-Run Macroeconomic Analysis>Long-Run Macroeconomic Equilibrium
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Pound_(force) Knowpia
The pound of force or pound-force (symbol: lbf,[1] sometimes lbf,[2]) is a unit of force used in some systems of measurement, including English Engineering units[a] and the foot–pound–second system.[3]
English Engineering units, British Gravitational System
Absolute English System
The pound-force is equal to the gravitational force exerted on a mass of one avoirdupois pound on the surface of Earth. Since the 18th century, the unit has been used in low-precision measurements, for which small changes in Earth's gravity (which varies from equator to pole by up to half a percent) can safely be neglected.[4]
Product of avoirdupois pound and standard gravityEdit
The pound-force is the product of one avoirdupois pound (exactly 0.45359237 kg) and the standard acceleration due to gravity, 9.80665 m/s2 (32.174049 ft/s2).[5][6][7]
The standard values of acceleration of the standard gravitational field (gn) and the international avoirdupois pound (lb) result in a pound-force equal to 4.4482216152605 N.[b]
{\displaystyle {\begin{aligned}1\,{\text{lbf}}&=1\,{\text{lb}}\times g_{\text{n}}\\&=1\,{\text{lb}}\times 9.80665\,{\tfrac {\text{m}}{{\text{s}}^{2}}}/0.3048\,{\tfrac {\text{m}}{\text{ft}}}\\&\approx 1\,{\text{lb}}\times 32.174049\,\mathrm {\tfrac {ft}{s^{2}}} \\&\approx 32.174049\,\mathrm {\tfrac {ft{\cdot }lb}{s^{2}}} \\1\,{\text{lbf}}&=1\,{\text{lb}}\times 0.45359237\,{\tfrac {\text{kg}}{\text{lb}}}\times g_{\text{n}}\\&=0.45359237\,{\text{kg}}\times 9.80665\,{\tfrac {\text{m}}{{\text{s}}^{2}}}\\&=4.4482216152605\,{\text{N}}\end{aligned}}}
{\displaystyle {\begin{aligned}1\,{\text{lbf}}&=1\,{\text{slug}}\times 1\,{\tfrac {\text{ft}}{{\text{s}}^{2}}}\\&=1\,{\tfrac {{\text{slug}}\cdot {\text{ft}}}{{\text{s}}^{2}}}\end{aligned}}}
Foot–pound–second (FPS) systems of unitsEdit
In some contexts, the term "pound" is used almost exclusively to refer to the unit of force and not the unit of mass. In those applications, the preferred unit of mass is the slug, i.e. lbf⋅s2/ft. In other contexts, the unit "pound" refers to a unit of mass. The international standard symbol for the pound as a unit of mass is lb.[8]
In the "engineering" systems (middle column), the weight of the mass unit (pound-mass) on Earth's surface is approximately equal to the force unit (pound-force). This is convenient because one pound mass exerts one pound force due to gravity. Note, however, unlike the other systems the force unit is not equal to the mass unit multiplied by the acceleration unit[11]—the use of Newton's second law, F = m ⋅ a, requires another factor, gc, usually taken to be 32.174049 (lb⋅ft)/(lbf⋅s2). "Absolute" systems are coherent systems of units: by using the slug as the unit of mass, the "gravitational" FPS system (left column) avoids the need for such a constant. The SI is an "absolute" metric system with kilogram and meter as base units.
Pound of thrustEdit
The term pound of thrust is an alternative name for pound-force in specific contexts. It is frequently seen in US sources on jet engines and rocketry, some of which continue to use the FPS notation. For example, the thrust produced by each of the Space Shuttle's two Solid Rocket Boosters was 3,300,000 pounds-force (14.7 MN), together 6,600,000 pounds-force (29.36 MN).[12][13]
Mass versus weight for the difference between the two physical properties
Pounds per square inch, a unit of pressure
^ Despite its name, this system is based on United States customary units and is only used in the US.
^ The international avoirdupois pound is defined to be exactly 0.45359237 kg.
^ Fletcher, Leroy S.; Shoup, Terry E. (1978), Introduction to Engineering, Prentice-Hall, ISBN 978-0135018583, LCCN 77024142, archived from the original on 2019-12-06, retrieved 2017-08-03. : 257
^ "Mass and Weight". engineeringtoolbox.com. Archived from the original on 2010-08-18. Retrieved 2010-08-03.
^ In 1901 the third CGPM Archived 2012-02-07 at the Wayback Machine declared (second resolution) that:
^ Barry N. Taylor, Guide for the Use of the International System of Units (SI), 1995, NIST Special Publication 811, Appendix B note 24
^ "Space Launchers - Space Shuttle". www.braeunig.us. Archived from the original on 6 April 2018. Retrieved 16 February 2018. Thrust: combined thrust 29.36 MN SL (maximum thrust at launch reducing by 1/3 after 50 s)
^ Richard Martin (12 January 2001). "From Russia, With 1 Million Pounds of Thrust". wired.com. Archived from the original on 25 September 2019. Retrieved 25 November 2019.
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Draw the graphs of following equations on the same graph sheet: x-y=0, x+y=0, y+5=0 Also, find the area enclosed between - Maths - Linear Equations in Two Variables - 7316401 | Meritnation.com
ABC is the required area:
area of triangle ABC = 1/2 * BC * height
=\frac{1}{2}*\left(5-\left(-5\right)\right)*5\phantom{\rule{0ex}{0ex}}=\frac{1}{2}*10*5\phantom{\rule{0ex}{0ex}}=25 square units
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Restore potion - OSRS Wiki
(Redirected from Restore potion(2))
The RuneScape Wiki also has an article on: rsw:Restore potionThe RuneScape Classic Wiki also has an article on: classicrsw:Stat restoration Potion
Restore potion(1)Restore potion(2)Restore potion(3)Restore potion(4)? (edit)1 dose2 dose3 dose4 dose? (edit)? (edit) ? (edit)File:Restore potion(1).pngFile:Restore potion(2).pngFile:Restore potion(3).pngFile:Restore potion(4).png? (edit)27 February 2002 (Update)27 February 2002 (Update)27 February 2002 (Update)29 March 2004 (Update)? (edit)YesNoYes? (edit)? (edit)NoNoYes? (edit)? (edit)Drink, EmptyDrink, EmptyDrink, EmptyDrink, Empty? (edit)Drop? (edit)1 dose of restore potion.2 doses of restore potion.3 doses of restore potion.4 doses of restore potion.? (edit)446688110? (edit)44 coins66 coins88 coins110 coinstrue? (edit)26 coins39 coins52 coins66 coins? (edit)17 coins26 coins35 coins44 coins? (edit)26395266? (edit)0.0200.0250.0300.035? (edit)0.02 kg0.025 kg0.03 kg0.035 kg? (edit)trueRestore potion(1)Restore potion(2)Restore potion(3)Restore potion(4)25517110825 coins (info)51 coins (info)71 coins (info)108 coins (info)infobox-cell-shown2,0002000243310,12318,989
SMW Subobject for (3)High Alchemy value: 52Buy limit: 2000Examine: 3 doses of restore potion.Is members only: trueIs variant of: Restore potionImage: File:Restore potion(3).pngUses infobox: ItemWeight: 0.030Value: 88Version anchor: 3 doseRelease date: 27 February 2002Item ID: 127
SMW Subobject for (4)High Alchemy value: 66Buy limit: 2000Examine: 4 doses of restore potion.Is members only: trueIs variant of: Restore potionImage: File:Restore potion(4).pngUses infobox: ItemWeight: 0.035Value: 110Version anchor: 4 doseRelease date: 29 March 2004Item ID: 2430
SMW Subobject for (1)High Alchemy value: 26Buy limit: 2000Examine: 1 dose of restore potion.Is members only: trueIs variant of: Restore potionImage: File:Restore potion(1).pngUses infobox: ItemWeight: 0.020Value: 44Version anchor: 1 doseRelease date: 27 February 2002Item ID: 131
A restore potion is a potion made by using red spiders' eggs on a harralander potion (unf), requiring 22 Herblore, yielding a restore potion(3) and 62.5 Herblore experience. A dose of restore potion restores levels of temporarily reduced stats in Attack, Strength, Defence, Ranged, and Magic equal to 10 + 30% of the related current level, rounded down, for each skill. It does not restore Prayer points or any other skill that is not listed like a super restore.
After partial completion of the In Aid of the Myreque quest, garlic can be added to a restore potion(3), requiring 22 Herblore, yielding a Guthix balance (unf) and 25 Herblore experience.
After Barbarian Herblore Training, roe or caviar can be added to a restore potion(2), requiring 24 Herblore, yielding a Restore mix and 21 Herblore experience. A restore mix has 2 doses, restores the listed stats equal to 10 + 30% of the related current level, rounded down, for each skill, and heals 3 Hitpoints.
A restore potion(4) is a reward from Jekyll and Hyde random events if a harralander is given to Dr Jekyll.
5.1 2 dose
When Attack, Strength, Defence, Ranged, or Magic are temporarily reduced, the amount of levels restored is calculated with:
{\displaystyle \lfloor SkillLevel\times {\frac {3}{10}}\rfloor +10}
for each of those skills up to the player's current level of that skill.
1–3 +10 34-36 +20 67-69 +30
10–13 +13 44-46 +23 77-79 +33
Harralander potion (unf) 1 1,042
Red spiders' eggs 1 500
Restore potion(3) 1 71
1 × Restore potion(1)
Restore potion(1) 25
2 dose[edit | edit source]
Jekyll and Hyde N/A 1 Always
Retrieved from ‘https://oldschool.runescape.wiki/w/Restore_potion?oldid=14284460#2_dose’
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