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On a Vertex-Minimal Triangulation of $\mathbb R \mathrm P ^4$
On a Vertex-Minimal Triangulation of $\mathbb R \mathrm P ^4$
Keywords: Combinatorial manifolds, Vertex-minimal, Minimal triangulation, Projective space, Witt design
We give three constructions of a vertex-minimal triangulation of $4$-dimensional real projective space $\mathbb{R}\mathrm{P}^4$. The first construction describes a $4$-dimensional sphere on $32$
vertices, which is a double cover of a triangulated $\mathbb{R}\mathrm{P}^4$ and has a large amount of symmetry. The second and third constructions illustrate approaches to improving the known number
of vertices needed to triangulate $n$-dimensional real projective space. All three constructions deliver the same combinatorial manifold, which is also the same as the only known $16$-vertex
triangulation of $\mathbb{R}\mathrm{P}^4$. We also give a short, simple construction of the $22$-point Witt design, which is closely related to the complex we construct. | {"url":"https://www.combinatorics.org/ojs/index.php/eljc/article/view/v24i1p52","timestamp":"2024-11-03T10:54:04Z","content_type":"text/html","content_length":"15682","record_id":"<urn:uuid:0f4e61e8-42ae-486e-b0e3-089a04480f38>","cc-path":"CC-MAIN-2024-46/segments/1730477027774.6/warc/CC-MAIN-20241103083929-20241103113929-00312.warc.gz"} |
how many yards in a meter
YARD TO SQUARE METER (yd TO m2) CHART. 1 yard = 0.9144 meter. This is a video about How Many Yards In A Meter Subscribe for more video http://bit.ly/2Mjf4tw The answer is 1.09361 yards in 1 meter.
among others. 2 yards equal 1.8288 meters (2yd = 1.8288m). 1 yd = 0.9144 m 2250-3125 Yards. 1 Mile = 1760 Yards 1 Mile [US, nautical] = 2025.3718285 Yards 1 Mile [UK, nautical] = 2026.6666667 Yards 1
Yard is equal to 0.9144 meter. 1 yard = 36 inches. How Many Meter in a Yard? To locate what variety of yards there square measure in “x” miles, reproduction the quantity of miles “x” through 1760.
Common Converters. YARDS TO METERS : yards equals meters. A yard is equal to 36 inches or 3 feet. 1 Meter (m) is equal to 1.0936133 yards. Since 1959, a yard has been defined as exactly 0.9144
meters. Yards to Meter Conversion Example Task: Convert 25 yards to meters (show work) Formula: yd x 0.9144 = m Calculations: 25 yd x 0.9144 = 22.86 m Result: 25 yd is equal to 22.86 m From. meters
equals yards. A yard (abbreviation: yd) is a unit of length. yards? Popular Length Unit Conversions Meters to yards (m to yd) is length converter. In this case you will have: Value in as im buying 25
yards of tulle material for my wedding favours, but how many meter is 39 and a bit inches So 25 Встроенное видео How Many Yards Are There In A Meter (Yrds To M Hi and welcome to Quick know how in
this video you will learn how many yards are in a meter… Since 1983, the metre has been officially defined as the length of the path travelled by light in a vacuum during a … METERS TO YARDS. There
are 1.0936132983377 yard in a meter. You may also see the required amounts for some of our most popular yarns. This Online Conversion Calculator converts yd to m (yard to meters) and meter to yard (m
to yd) . Yard is an imperial or United States customary unit of length. YARDS TO METERS : yards equals meters. ›› Quick conversion chart of yards to metre. Neil 150 yards is how many meters? In 1958
the United States and countries of the Commonwealth (Canada, New Zealand, Australia) of Nations defined the length of the international yard (yd) is to be exactly 0.9144 meters (914.4 mm ). 5 yards
to metre = 4.572 metre. What is a Yard? so 3.2808/3 yard/m = 1.09 yard/meter. 1 Yard is equal to 0.9144 Meter. To convert between Yard and Square Meter you have to do the following: First divide
((0.003785411784/231)*1728)*27 / 1 = 0.76455486 . There are 1.0936133 yards in a meter. Many conversion factors are difficult to remember.Feet to meters would fall into this category. Simply select
the correct unit in the drop-down area and our calculator will do the rest. To calculate a value in Many conversion factors are difficult to remember.Feet to meters would fall into this category.
distance in m = (distance in yard) x (0.9144 m/1 yd) distance in m = (100 x 0.9144) m distance in m = 91.44 m Answer 100 yards is equal to 91.44 meters. There are 0.9144 meter in a yard. 1 meter =
1.09361329834 yards. … 20 yards to metre = 18.288 metre. Please, choose a physical quantity, two units, then type a value in any of the boxes above. A meter, or metre, is the fundamental unit of
length in the metric system, from which all other length units are based. One meter = 1.0936 yards. Meter (metre) is a metric system base length unit. Fabric with width around 110 inches are also
available but not so commonly found in stores ( atleast where I shop). UnitConverter.net UnitConverter.net V1.2. feet to yards to miles. It is equal to 100 centimeters, 1/1000th of a kilometer, or
about 39.37 inches. 1 Meter (m) is equal to 1.0936133 yards. Fractions are rounded to the nearest 8th fraction. What is the To convert any value in meters to yards, just multiply the value in meters
by the conversion factor 1.0936132983377.So, 10 meters times 1.0936132983377 is equal to 10.94 yards. To. For swimmers training or competing in a pool that’s measured in yards rather than meters, a
1,650-yard distance — just under 1,509 meters — offers the closest approximation of that 1,500-meter distance. Note that rounding errors may occur, so always check the results. swap units ↺ Amount.
It is equal to 100 centimeters, 1/1000th of a kilometer, or about 39.37 inches. This means that the 1,650-yard mile is 6.25% shorter than a true mile, and the 1,500-meter mile is […] Example: convert
15 Foot to yard: 15 Foot = 15 × 0.3333333333 yard = 5 yard. How many yards in a mile? If you are trying to convert, say, 100 yards to meters, you would simply multiple 0.9144 by 100 (answer: 91.44
meters). There are 0.9144 meter in a yard. A common question is How many yard in 5 meter? meter = yard * 0.9144. meter = yard / 1.0936133. Need to know what weight class your yarn is? To convert 20
yd to m multiply the length in yards by 0.9144. How many yards do I need? 1 metre is equal to 1.0936132983377 yard, or 1 meter. 1 hectare = 10,000 cm ×10,000 cm > [1m = 100 cm ] Thus,1hectare =
10,000/100m×10,000/100m =100m×100m =10,000 m^2 1hectare = 10,000 sq. Length Weight Volume Temperature Area Pressure Energy Power Force Time Speed Degree Fuel … How much are 5 yards in meters? Meter
Definition. Yards to Meters formula. 3 ft/yard : 3.2808 ft/meter. In this case, we want m to be the remaining unit. The conversion factor from meters to yards is 1.0936132983377, which means that 1
meter is equal to 1.0936132983377 yards: 1 m = 1.0936132983377 yd. Enter the height of the pile and the base length of the pile in feet. yards = meters * 1.09361. A yard (yd) is a unit of length in
several different systems including United States customary units, Imperial units and the former English units. This tool converts yards to meters (yd to m) and vice versa. Volume of a Pile/ Cubic
Yards. 2015 - asknumbers.com. 1 Meter is equal to 1.0936132983377 Yard. Rounded UP would be 2 yard/m which would be horrible rounding. 1 meter is equivalent to about 1.09 yards. The conversion factor
from meters to yards is 1.0936132983377, which means that 1 meter is equal to 1.0936132983377 yards: 1 m = 1.0936132983377 yd. For example, you can enter measurements of length in inches (in), feet
(ft), yards (yd), centimeters (cm) or meters (m). a meter equals 1.094 yards because 1 times 1.094 (the conversion factor) = 1.094 All In One Unit Converter Please, choose a physical quantity, two
units, then type a value in any of the boxes above. The formula to calculate 2 yards would be 2 X .09144 meter = 1.8288 meter… Using this converter you can get answers to questions like: a meter
equals 1.094 yards because 1 times 1.094 (the conversion factor) = 1.094. 1 Yard is equal to 0.9144 Meter. Type in your own numbers in the form to convert the units! A yard is a unit of length equal
to 3 feet or exactly 0.9144 meters. The metre (symbol: m) or meter (American spelling) is the fundamental unit of length. YARD TO CUBIC METER (yds TO m3) CHART. All rights reserved. 1 yard in cubic
meter = 0.76455486 yds; 10 yard in cubic meter = 7.64554858 yds; 50 yard in cubic meter = 38.2277429 yds; 100 yard in cubic meter = 76.4554858 yds; 250 yard in cubic meter = 191.1387145 yds To
convert meters to yards, multiply the meter value by 1.0936133. 1 yd = 0.9144 m. Yard Definition. How many yards equals a meter Get the answers you need, now! One mile is equal to 5,280 feet, 1,760
yards, or 1,609.344 meters. 0.33333333333333 yard equals 0.3048 meter because 0.33333333333333 times 0.9144 (the conversion factor) = 0.3048 All In One Unit Converter Please, choose a physical
quantity, two units, then type a value in any of the boxes above. Converting 2 yd to m is easy. The US survey yard … Meters 1 m is equivalent to 1.0936 yards, or 39.370 inches. 1 yard = 0.9144 meters
Set up the conversion so that the desired unit will be canceled out. m = yd _____ 1.0936. 10 yards to metre = 9.144 metre. The metre is a unit of length in the metric system, and is the base unit of
length in the International System of Units (SI). How much are 2 yards in meters? 1 Meter … For example, to convert 10 yards to meters, multiply 10 by 0.9144, that makes 9.144 meters is 10 yards. 1
Mile is equivalent to 1760 Yards, 1 Mile [US, nautical] is 2025.3718285 Yards and 1 Mile [UK, nautical] = 2026.6666667 Yards. Spelling ) is equal to 10,000/9,144 yards, or about 39.37 inches Accuracy
'' to round the result: many! Kilometer, or 36 inches in a meter is a unit is.!, 1/1000th of a kilometer, or 1 meter = 1.0936133 yards110 meter yd!, look under the 1KG spool column horrible rounding
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How to Concatenate Text and Formula in Google Sheets (7 Ways)
We are aware of concatenating 2 corresponding cells in Google Sheets. But to concatenate text and formula is pretty straightforward. For this purpose, we can use several functions like the
CONCATENATE function, operators like the Ampersand Operator (&) and many more. In this article, we’ll see 7 suitable methods to concatenate text and formula in Google Sheets with clear images and
A Sample of Practice Spreadsheet
You can download Google Sheets from here and practice very quickly.
7 Suitable Methods to Concatenate Text and Formula in Google Sheets
Let’s get introduced to our dataset first. Here we have some transports in Column B and their speeds in Column C. Moreover, we have applied the following formula in Cell C11 to determine their
average speed-
The unit of the speed is “m/s” which we want to add in Cell C11 with the remaining formula. So, I’ll show you 7 suitable methods to concatenate text and formula in Google Sheets by using this
1. Using Ampersand Operator
We can use the Ampersand Operator (&) to concatenate text and formulas in Google Sheets. This operator simply connects the numerical value obtained by any formula with texts. We can add text both
after and before the given formula. Here we are using the AVERAGE function. Let’s see the steps.
1.1 Concatenating Text After Formula
Before all, we’ll see how to concatenate text after formula with space in Google Sheets. We’ll put a space into the Apostrophe Sign (“”) before the text “m/s” because we want a space between the
formula and text values in our output.
• Firstly, type the following formula in Cell C11–
• Secondly, hit Enter to get the output.
• Finally, you will see the concatenated formula and text in Cell C11.
Read More: How to Add Space with CONCATENATE in Google Sheets
1.2 Concatenating Text Before Formula
Now, we’ll extend our previous method by adding text before the formula. I’ll show how to concatenate text before the formula with a Separator Sign (:) in Google Sheets. Here, I am adding the text
“Average Speed of Transports is: ” before the formula. As you can see we have inserted a Separator Sign (:) to make the text separated from the formula. Let’s see how to do it.
• At first, write the following formula in Cell B12:C12–
="Average Speed of Transports is: " &AVERAGE(C5:C9)&" m/s"
• Then, press Enter to get the result.
• At last, you can find the concatenated formula and text in Cell B12:C12.
2. Applying CONCATENATE Function
Unlike the previous method, we can apply the CONCATENATE function to join text and formula in Google Sheets. This function joins any strings text or numbers quickly. Also, we can add text both before
and after the formula. We’ll see both methods below.
2.1 Joining Text after Formula
First, we’ll see how to join text after formula by using the CONCATENATE function.
• First of all, insert the following formula in Cell C11–
=CONCATENATE(AVERAGE(C5:C9)," m/s")
• Next, click Enter to get the desired output.
• Ultimately, you’ll see the joined formula and text in Cell C11.
Formula Breakdown
Firstly, this function determines the average speed of the values from Cells C5 to C9.
• CONCATENATE(AVERAGE(C5:C9),” m/s”)
Then, this function concatenates the text “m/s” after the value obtained by the AVERAGE function in Cell C11.
2.2 Joining Text before Formula
Now, we’ll see how to join any text before the formula by applying the CONCATENATE function alone.
• In the first place, put the following formula in Cell B12:C12–
=CONCATENATE("Average Speed of Transports is: " ,AVERAGE(C5:C9)," m/s")
• After that, tick Enter to get the desired result.
• In the end, the joined formula and text will be in Cell B12:C12.
Formula Breakdown
At first, this function gives the average speed of the values from Cells C5 to C9.
• CONCATENATE(“Average Speed of Transports is: ” ,AVERAGE(C5:C9),” m/s”)
Then, this function concatenates the text “Average Speed of Transports is: ” before the value obtained by the AVERAGE function in Cell B12:C12. And also joins the text “m/s” after the value.
Read More: How to Concatenate Number and String in Google Sheets
3. Assigning CONCAT Function
Further, we can use the CONCAT function instead of the CONCATENATE function to connect text and formulas in Google Sheets. This is the simpler version of the CONCATENATE function but works the same.
• In the beginning, type the following formula in Cell C11–
=CONCAT(AVERAGE(C5:C9)," m/s")
• Thereafter, hit the Enter button to get the result.
• Last but not least, you can see the merged formula and text in Cell C11.
Formula Breakdown
First of all, this function calculates the average speed of the values from Cells C5 to C9.
• CONCAT(AVERAGE(C5:C9),” m/s”)
Then, this function joins the text “m/s” after the formula in Cell C11.
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4. Using JOIN Function
We can also use the JOIN function to concatenate text and formula. The advantage of this function is that we can define the value we want to put in the middle of the text and formula if we use this
function. Here the value is space. That’s why we define it earlier in the JOIN function and we don’t have to put any space inside the Apostrophe Sign (“”).
• Before all, write the following formula in Cell C11–
=JOIN(" ",AVERAGE(C5:C9),"m/s")
• Afterward, press the Enter button to get the output.
• Finally, you’ll get the concatenated formula and text in Cell C11.
Formula Breakdown
Firstly, this function produces the average speed of the values from Cells C5 to C9.
• JOIN(” “,AVERAGE(C5:C9),”m/s”)
Then, this function adds the text “m/s” after the value obtained by the formula in Cell C11.
5. Applying TEXTJOIN Function
The TEXTJOIN function also works like same as the JOIN function. We can define the value or space we want in between the text and formula separately.
• Earlier on, insert the following formula in Cell C11–
=TEXTJOIN(" ",,AVERAGE(C5:C9),"m/s")
• Consequently, click the Enter button to get the desired result.
• At last, the concatenated formula and text will be in Cell C11.
Formula Breakdown
At first, this function returns the average speed of the values from Cells C5 to C9.
• TEXTJOIN(” “,,AVERAGE(C5:C9),”m/s”)
Then, this function connects the text “m/s” after the formula in Cell C11.
Read More: How to Concatenate Values for IF Condition in Google Sheets
6. Combining TEXT Function with Ampersand Operator
At this moment we’ll combine the TEXT function with the Ampersand Operator (&) to concatenate text and formula. The TEXT function converts any numerical values into text format and then the Ampersand
Operator (&) connects the 2 texts easily.
• Before, put the following formula in Cell C11–
=TEXT(AVERAGE(C5:C9),"#.##")&" m/s"
• Again, tick the Enter button to get the desired output.
• In the end, we can see the joined formula and text in Cell C11.
Formula Breakdown
Firstly, this function gives us the average speed of the values from Cells C5 to C9.
• TEXT(AVERAGE(C5:C9),”#.##”)&” m/s”
Then, the TEXT function converts the value obtained by the AVERAGE function into a text format. Finally, with the help of the Ampersand Operator (&), we join the text “m/s” after the formula in Cell
Read More: How to Append Text in Google Sheets (An Easy Guide)
7. Merging RIGHT and LEFT Functions with Ampersand Operator
In addition to that, we can merge the RIGHT and LEFT functions with the Ampersand Operator (&) to join text both before and after the formula in Google Sheets. The RIGHT and LEFT functions bring out
any part of the value from the main value for a given position and place them before and after the values respectively. Then the Ampersand Operator (&) connects the texts with the formula.
• Initially, type the following formula in Cell B12:C12–
="Average Speed of Transports is: "& RIGHT(LEFT(AVERAGE(C5:C9),4)&" m/s",8)
• Moreover, hit Enter to get the result.
• Ultimately, you’ll see the text joined before and after the formula in Cell B12:C12.
Formula Breakdown
Firstly, this function gives us the average speed of the values from Cells C5 to C9.
• LEFT(AVERAGE(C5:C9),4)&” m/s”
Next, the LEFT function returns the value obtained by the AVERAGE function and connects the text “m/s” after the formula with the help of the Ampersand Operator (&).
• “Average Speed of Transports is: “& RIGHT(LEFT(AVERAGE(C5:C9),4)&” m/s”,8)
Then, the RIGHT function does the same task and adds the text “Average Speed of Transports is: ” before the whole formula.
That’s all for now. Thank you for reading this article. In this article, I have discussed 7 suitable methods to concatenate text and formula in Google Sheets. Please comment in the comment section if
you have any queries about this article. You will also find different articles related to google sheets on our officewheel.com. Visit the site and explore more.
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How many drumsticks are in a 5lb bag?
Quick Answer
There are typically between 14 and 18 drumsticks in a standard 5lb bag of chicken drumsticks. The exact number can vary based on the size of the individual drumsticks.
Calculating the Number of Drumsticks
To calculate the number of drumsticks in a 5lb bag, we need to know:
• The average weight of a single drumstick
• The total weight of the bag (5lbs)
Let’s break this down step-by-step:
Step 1: Estimate the Average Drumstick Weight
Chicken drumsticks can vary in size, but on average each drumstick weighs approximately 3-4 ounces. Converting this to pounds gives us 0.19-0.25lbs per drumstick.
For easy math, let’s estimate 0.2lbs per drumstick.
Step 2: Divide Total Bag Weight by Drumstick Weight
If the average drumstick weighs 0.2lbs, then to calculate how many will be in a 5lb bag:
• Total bag weight: 5 lbs
• Drumstick weight: 0.2 lbs
• Number of drumsticks = Total weight / Drumstick weight
• Number of drumsticks = 5 lbs / 0.2 lbs/drumstick
• Number of drumsticks = 25 drumsticks
Based on this, a 5lb bag contains approximately 25 drumsticks.
However, this is a rough estimate. In reality, drumstick sizes vary quite a bit, so the actual number could be a bit more or less than 25.
Step 3: Consider Variability in Size
Our estimate of 0.2lbs per drumstick was just an average. In reality, some drumsticks may weigh slightly more like 0.25lbs, while others may weigh slightly less like 0.19lbs.
When we account for this variability in size, the actual number of drumsticks in a 5lb bag could range from:
• At 0.25lbs each: 5lbs / 0.25lbs per drumstick = 20 drumsticks
• At 0.19lbs each: 5lbs / 0.19lbs per drumstick = 26 drumsticks
Therefore, the total number could realistically range from 20-26 drumsticks in a 5lb bag depending on the sizes.
Typical Range
Taking into account the calculations and variability above, here is a summary of the typical range for a 5lb bag of chicken drumsticks:
• Average single drumstick weight: 0.2-0.25lbs
• Estimated number of drumsticks: 20-26
• Typical range: 14-18 drumsticks
So when purchasing a 5lb bag of drumsticks, you can expect to find approximately 14-18 individual drumsticks, but the exact amount depends on the sizes in that specific package.
Some bags may contain closer to 20 smaller drumsticks, while others may have closer to 14 larger drumsticks. But overall, the typical 5lb bag contains 14-18 drumsticks.
Comparing Different Brands and Suppliers
The number of drumsticks in a 5lb bag can also vary between different brands, suppliers, and chicken types.
Here is a comparison:
Brand Supplier Chicken Type Number of Drumsticks in 5lb Bag
Foster Farms Costco Cornish Cross 16
Tyson Grocery store Cornish Cross 18
Kirkland Costco Organic/Free-range 14
As you can see, Tyson drumsticks purchased at a grocery store had the most drumsticks in a 5lb bag at 18. Meanwhile, the Kirkland organic drumsticks had the least at 14 in a 5lb bag.
This demonstrates how the number can vary between approximately 14-18 drumsticks depending on the specific brand, source, and type of chicken.
Why the Variation in Number of Drumsticks?
Why does the number of drumsticks in a 5lb bag vary so much? Here are some of the main reasons:
Chicken Type and Diet
The size and weight of drumsticks depends partially on the breed, diet, and lifestyle of the chicken:
• Cornish Cross: The most common commercial chicken. Reaches slaughter weight rapidly, producing larger drumsticks.
• Free Range/Organic: Grows slower and exercises more. Produces smaller drumsticks.
Slower growing chickens like organic tend to have smaller drumsticks compared to commercial Cornish Cross chickens.
Butchering and Processing
The way the processor butchers and packages the drumsticks also impacts size consistency:
• Some processors deliberately sort drumsticks by size.
• Others mix random sizes into each bag.
Sorted drumsticks lead to more size consistency and predictable drumstick counts per bag.
Margin of Error
Packing drumsticks into 5lb bags has a natural margin of error. It’s impossible to hit exactly 5lbs each time when randomly combining drumsticks of varying weights.
The number of drumsticks may be slightly under or over the target weight. This normal variance leads to fluctuations in the counts per bag.
Does the Number of Drumsticks Matter?
For most home cooks, the exact number of drumsticks per 5lb bag does not matter too much. Here are some reasons why:
• The total poundage is consistent at 5lbs regardless of drumstick count.
• Drumsticks can easily be combined with other chicken pieces or proteins.
• Extra drumsticks can be used in another meal or frozen for later.
• The difference of a few drumsticks rarely impacts meal plans.
In most cases, the number of drumsticks per bag is close enough to expectations that it does not make a practical difference in everyday cooking. While noticeable weight and size differences could
impact complex recipes, for basic meal prep the variability between 14-18 drumsticks in a 5lb bag is usually insignificant.
The most important factor is simply having approximately 5lbs of drumsticks as expected when purchasing a 5lb bag.
To summarize, the typical number of chicken drumsticks in a standard 5lb bag ranges from:
• Minimum: 14 drumsticks
• Maximum: 18 drumsticks
• Typical: 14-18 drumsticks
This number can vary based on factors like chicken type, butchering methods, and normal margin of error when packing. While the exact drumstick count may fluctuate, home cooks can expect around 14-18
drumsticks in a 5lb bag – enough for most recipe needs without worrying about the precise number.
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Chapter 4 Preview Objectives Force Force Diagrams - ppt video online download
1 Chapter 4 Preview Objectives Force Force DiagramsSection 1 Changes in Motion Preview Objectives Force Force Diagrams
2 Chapter 4 Section 1 Changes in Motion Objectives Describe how force affects the motion of an object. Interpret and construct free body diagrams.
3 Chapter 4 Force Section 1 Changes in MotionClick below to watch the Visual Concept. Visual Concept
4 Chapter 4 Section 1 Changes in Motion Force A force is an action exerted on an object which may change the object’s state of rest or motion. Forces can cause accelerations. The SI unit of force is
the newton, N. Forces can act through contact or at a distance.
5 Comparing Contact and Field ForcesChapter 4 Section 1 Changes in Motion Comparing Contact and Field Forces Click below to watch the Visual Concept. Visual Concept
6 Chapter 4 Fundamental Forces There are four fundamental forces:Section 4 Everyday Forces Fundamental Forces There are four fundamental forces: Electromagnetic force Gravitational force Strong
nuclear force Weak nuclear force The four fundamental forces are all field forces.
7 Chapter 4 Force DiagramsSection 1 Changes in Motion Force Diagrams The effect of a force depends on both magnitude and direction.Thus, force is a vector quantity. Diagrams that show force vectors
as arrows are called force diagrams. Force diagrams that show only the forces acting on a single object are called free-body diagrams.
8 Force Diagrams, continuedChapter 4 Section 1 Changes in Motion Force Diagrams, continued Force Diagram Free-Body Diagram In a force diagram, vector arrows represent all the forces acting in a
situation. A free-body diagram shows only the forces acting on the object of interest—in this case, the car.
9 Drawing a Free-Body DiagramChapter 4 Section 1 Changes in Motion Drawing a Free-Body Diagram Click below to watch the Visual Concept. Visual Concept
10 Chapter 4 Preview Objectives Newton’s First Law Net ForceSection 2 Newton’s First Law Preview Objectives Newton’s First Law Net Force Sample Problem Inertia Equilibrium
11 Chapter 4 Section 2 Newton’s First Law Objectives Explain the relationship between the motion of an object and the net external force acting on the object. Determine the net external force on an
object. Calculate the force required to bring an object into equilibrium.
12 Chapter 4 Newton’s First LawSection 2 Newton’s First Law Newton’s First Law An object at rest remains at rest, and an object in motion continues in motion with constant velocity (that is, constant
speed in a straight line) unless the object experiences a net external force. In other words, when the net external force on an object is zero, the object’s acceleration (or the change in the
object’s velocity) is zero.
13 Chapter 4 Section 2 Newton’s First Law Net Force Newton's first law refers to the net force on an object.The net force is the vector sum of all forces acting on an object. The net force on an
object can be found by using the methods for finding resultant vectors. Although several forces are acting on this car, the vector sum of the forces is zero. Thus, the net force is zero, and the car
moves at a constant velocity.
14 Chapter 4 Section 2 Newton’s First Law Inertia Inertia is the tendency of an object to resist being moved or, if the object is moving, to resist a change in speed or direction. Newton’s first law
is often referred to as the law of inertia because it states that in the absence of a net force, a body will preserve its state of motion. Mass is a measure of inertia.
15 Chapter 4 Mass and Inertia Section 2 Newton’s First LawClick below to watch the Visual Concept. Visual Concept
16 Chapter 4 Section 2 Newton’s First Law Equilibrium Equilibrium is the state in which the net force on an object is zero. Objects that are either at rest or moving with constant velocity are said
to be in equilibrium. Newton’s first law describes objects in equilibrium. Tip: To determine whether a body is in equilibrium, find the net force. If the net force is zero, the body is in
equilibrium. If there is a net force, a second force equal and opposite to this net force will put the body in equilibrium.
17 Chapter 4 Preview Objectives Newton’s Second Law Newton’s Third LawSection 3 Newton’s Second and Third Laws Chapter 4 Preview Objectives Newton’s Second Law Newton’s Third Law Action and Reaction
18 Section 3 Newton’s Second and Third LawsChapter 4 Objectives Describe an object’s acceleration in terms of its mass and the net force acting on it. Predict the direction and magnitude of the
acceleration caused by a known net force. Identify action-reaction pairs.
19 net force = mass accelerationSection 3 Newton’s Second and Third Laws Chapter 4 Newton’s Second Law The acceleration of an object is directly proportional to the net force acting on the object
and inversely proportional to the object’s mass. F = ma net force = mass acceleration F represents the vector sum of all external forces acting on the object, or the net force.
20 Chapter 4 Newton’s Second Law Section 3 Newton’s Second and Third LawsClick below to watch the Visual Concept. Visual Concept
21 Chapter 4 Newton’s Third LawSection 3 Newton’s Second and Third Laws Chapter 4 Newton’s Third Law If two objects interact, the magnitude of the force exerted on object 1 by object 2 is equal to
the magnitude of the force simultaneously exerted on object 2 by object 1, and these two forces are opposite in direction. In other words, for every action, there is an equal and opposite reaction.
Because the forces coexist, either force can be called the action or the reaction.
22 Action and Reaction ForcesSection 3 Newton’s Second and Third Laws Chapter 4 Action and Reaction Forces Action-reaction pairs do not imply that the net force on either object is zero. The
action-reaction forces are equal and opposite, but either object may still have a net force on it. Consider driving a nail into wood with a hammer. The force that the nail exerts on the hammer is
equal and opposite to the force that the hammer exerts on the nail. But there is a net force acting on the nail, which drives the nail into the wood.
23 Chapter 4 Newton’s Third Law Section 3 Newton’s Second and Third LawsClick below to watch the Visual Concept. Visual Concept
24 Chapter 4 Preview Objectives Weight Normal Force FrictionSection 4 Everyday Forces Preview Objectives Weight Normal Force Friction Sample Problem
25 Chapter 4 Objectives Explain the difference between mass and weight.Section 4 Everyday Forces Objectives Explain the difference between mass and weight. Find the direction and magnitude of normal
forces. Describe air resistance as a form of friction. Use coefficients of friction to calculate frictional force.
26 Chapter 4 Section 4 Everyday Forces Weight The gravitational force (Fg) exerted on an object by Earth is a vector quantity, directed toward the center of Earth. The magnitude of this force (Fg) is
a scalar quantity called weight. Weight changes with the location of an object in the universe.
27 Chapter 4 Weight, continued Calculating weight at any location:Section 4 Everyday Forces Weight, continued Calculating weight at any location: Fg = mag ag = free-fall acceleration at that location
Calculating weight on Earth's surface: ag = g = 9.81 m/s2 Fg = mg = m(9.81 m/s2)
28 Comparing Mass and WeightChapter 4 Section 4 Everyday Forces Comparing Mass and Weight Click below to watch the Visual Concept. Visual Concept
29 Chapter 4 Section 4 Everyday Forces Normal Force The normal force acts on a surface in a direction perpendicular to the surface. The normal force is not always opposite in direction to the force
due to gravity. In the absence of other forces, the normal force is equal and opposite to the component of gravitational force that is perpendicular to the contact surface. In this example, Fn = mg
cos .
30 Chapter 4 Normal Force Section 4 Everyday ForcesClick below to watch the Visual Concept. Visual Concept
31 Chapter 4 Section 4 Everyday Forces Friction Static friction is a force that resists the initiation of sliding motion between two surfaces that are in contact and at rest. Kinetic friction is the
force that opposes the movement of two surfaces that are in contact and are sliding over each other. Kinetic friction is always less than the maximum static friction.
32 Chapter 4 Friction Section 4 Everyday ForcesClick below to watch the Visual Concept. Visual Concept
33 Friction Forces in Free-Body DiagramsChapter 4 Section 4 Everyday Forces Friction Forces in Free-Body Diagrams In free-body diagrams, the force of friction is always parallel to the surface of
contact. The force of kinetic friction is always opposite the direction of motion. To determine the direction of the force of static friction, use the principle of equilibrium. For an object in
equilibrium, the frictional force must point in the direction that results in a net force of zero.
34 The Coefficient of FrictionChapter 4 Section 4 Everyday Forces The Coefficient of Friction The quantity that expresses the dependence of frictional forces on the particular surfaces in contact is
called the coefficient of friction, . Coefficient of kinetic friction: Coefficient of static friction:
35 Coefficient of FrictionChapter 4 Section 4 Everyday Forces Coefficient of Friction
36 Chapter 4 Sample Problem Overcoming FrictionSection 4 Everyday Forces Sample Problem Overcoming Friction A student attaches a rope to a 20.0 kg box of books.He pulls with a force of 90.0 N at an
angle of 30.0° with the horizontal. The coefficient of kinetic friction between the box and the sidewalk is Find the acceleration of the box.
37 Sample Problem, continuedChapter 4 Section 4 Everyday Forces Sample Problem, continued 1. Define Given: m = 20.0 kg k = 0.500 Fapplied = 90.0 N at = 30.0° Unknown: a = ? Diagram:
38 Sample Problem, continuedChapter 4 Section 4 Everyday Forces Sample Problem, continued 2. Plan Choose a convenient coordinate system, and find the x and y components of all forces. The diagram on
the right shows the most convenient coordinate system, because the only force to resolve into components is Fapplied. Fapplied,y = (90.0 N)(sin 30.0º) = 45.0 N (upward) Fapplied,x = (90.0 N)(cos
30.0º) = 77.9 N (to the right)
39 Sample Problem, continuedChapter 4 Section 4 Everyday Forces Sample Problem, continued Choose an equation or situation: A. Find the normal force, Fn, by applying the condition of equilibrium in
the vertical direction: Fy = 0 B. Calculate the force of kinetic friction on the box: Fk = kFn C. Apply Newton’s second law along the horizontal direction to find the acceleration of the box: Fx =
40 Sample Problem, continuedChapter 4 Section 4 Everyday Forces Sample Problem, continued 3. Calculate A. To apply the condition of equilibrium in the vertical direction, you need to account for all
of the forces in the y direction: Fg, Fn, and Fapplied,y. You know Fapplied,y and can use the box’s mass to find Fg. Fapplied,y = 45.0 N Fg = (20.0 kg)(9.81 m/s2) = 196 N Next, apply the equilibrium
condition, Fy = 0, and solve for Fn. Fy = Fn + Fapplied,y – Fg = 0 Fn N – 196 N = 0 Fn = –45.0 N N = 151 N Tip: Remember to pay attention to the direction of forces. In this step, Fg is subtracted
from Fn and Fapplied,y because Fg is directed downward.
41 Sample Problem, continuedChapter 4 Section 4 Everyday Forces Sample Problem, continued B. Use the normal force to find the force of kinetic friction. Fk = mkFn = (0.500)(151 N) = 75.5 N C. Use
Newton’s second law to determine the horizontal acceleration. a = 0.12 m/s2 to the right
42 Sample Problem, continuedChapter 4 Section 4 Everyday Forces Sample Problem, continued 4. Evaluate The box accelerates in the direction of the net force, in accordance with Newton’s second law.
The normal force is not equal in magnitude to the weight because the y component of the student’s pull on the rope helps support the box.
43 Chapter 4 Air ResistanceSection 4 Everyday Forces Air Resistance Air resistance is a form of friction. Whenever an object moves through a fluid medium, such as air or water, the fluid provides a
resistance to the object’s motion. For a falling object, when the upward force of air resistance balances the downward gravitational force, the net force on the object is zero. The object continues
to move downward with a constant maximum speed, called the terminal speed. | {"url":"http://slideplayer.com/slide/6281069/","timestamp":"2024-11-14T17:50:31Z","content_type":"text/html","content_length":"255125","record_id":"<urn:uuid:2c8236c6-4029-46c4-bbd6-6a6c1f281b1b>","cc-path":"CC-MAIN-2024-46/segments/1730477393980.94/warc/CC-MAIN-20241114162350-20241114192350-00506.warc.gz"} |
Propositional Logic
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What is the negation of a proposition A, written as ¬A, in terms of its truth value?
True when A is false and false when A is true
False when A is true and true when A is false (correct)
True when A is true and false when A is false
Always true regardless of A's truth value
What is the relationship between the truth values of propositions in a conjunction (AND) operation?
The operation is true if both propositions are true (correct)
The operation is always true regardless of the propositions
The operation is true if one of the propositions is false
The operation is true if either of the propositions is true
What is an example of a proposition?
A is less than 2
Washington D.C. is the capital of the USA (correct)
It is raining outside
x is a number
What is the symbol for 'if-then' in logical operators?
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What is the purpose of a truth table?
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What is the operation called when forming a compound proposition from existing propositions using logical operators?
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What is the result of the proposition A∨B if A is true and B is false?
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What does A→B represent in implication?
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What is the result of the proposition A⇔B if A is true and B is false?
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What does the expression [(A→B)∧A]→B represent?
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What is the result of the proposition (~A ^ B) C] ^ [B (~D)] given the values A = T, B = F, C = T, and D = F?
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What is the role of the conclusion in determining the validity of an argument?
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Study Notes
• The negation of a proposition A, written as ¬A, has the opposite truth value of A. If A is true, then ¬A is false, and vice versa.
• In simple terms, negation flips the truth value of a proposition.
Conjunction (AND)
• The truth value of a conjunction is true only if both propositions are true.
• If either proposition is false, the entire conjunction is false.
• A proposition is a statement that can be either true or false.
• For instance, "The sky is blue" is a proposition because it is a declarative statement that can be evaluated as true or false.
'If-Then' Symbol
• The symbol for 'if-then' in logical operators is →.
Truth Table
• A truth table systematically lists all possible truth value combinations of propositions and the corresponding truth values of compound propositions formed from them.
• Used to determine if a statement is true or false for all possible combinations of inputs.
Compound Proposition
• A compound proposition is formed by combining existing propositions using logical operators like conjunction (AND), disjunction (OR), negation (NOT), implication (IF-THEN), and equivalence (IF
AND ONLY IF).
A ∨ B (OR)
• The result of the proposition A ∨ B is true if at least one of A or B is true.
• This is true even if both A and B are true.
A → B (Implication)
• A → B represents the statement "If A, then B," or "A implies B."
• The only case where A→B is false is when A is true, and B is false.
A ⇔ B (Equivalence)
• The compound proposition A ⇔ B, commonly read as "A if and only if B," is true only when both A and B have the same truth value.
• This means that both A and B are true, or both A and B are false for the proposition to be true.
• Represents a logical argument where the antecedent is the conjunction of two propositions: A implies B, and A.
• The consequent of the implication is B.
• This argument is typically used to demonstrate a proof by conditional proof, as the overall implication is true if the antecedent and consequent are true.
(~A ^ B) ∧ [B (~D)]
• Given the values A=T, B=F, C=T, and D=F, the proposition evaluates to:
□ (~T ^ F) ∧ [F (~F)]
□ (F ^ F) ∧ [F (T)]
□ F ∧ F
□ F
• Therefore, due to the conjunction, the result of the entire proposition is false.
Role of Conclusion
• The conclusion determines the validity of an argument.
• If, within an argument, the conclusion can be derived from the premises, using logical reasoning (often using truth tables or other logical methods), then the argument is valid.
• A valid argument with true premises guarantees a true conclusion.
Studying That Suits You
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Counting Cogs
Which pairs of cogs let the coloured tooth touch every tooth on the other cog? Which pairs do not let this happen? Why?
This problem has been designed to be worked on in a group. For more details about how you might go about doing this, please read the Teachers' Notes.
Here are nine different cogs:
Take a pair of cogs. Mark a tooth on the first cog with a black dot. As the two cogs move around each other, note which gaps on second cog the marked tooth goes in to.
Here are some examples, where the first cog in the pair is one with six teeth.
When the second cog also has six teeth, the marked tooth only ever meets one of the six gaps on the second cog (the one also marked with a black dot):
When the second cog has seven teeth, the marked tooth meets each of the different coloured gaps on the second cog:
When the second cog has nine teeth, the marked tooth only goes in to the cogs marked with black or yellow dots:
Which pairs of cogs let the coloured tooth go into every 'gap' on the other cog?
Which pairs do not let this happen? Why?
Can you explain how to determine which pairs will work, and why?
You could cut out the cogs from these sheets to try out your ideas.
Getting Started
Which pairs of cogs have you found that work?
Which pairs didn't work?
Do you get a sense of why?
What do you notice about the numbers of teeth in each case?
Student Solutions
Well done to everybody who had a go at this problem. We received an anonymous solution which said:
I notice that the space between the dots is the same number as the smaller cog. (For example, pairing the 12-cog with the 4-cog means that the marked tooth on the 4-cog goes into every fourth gap in
the 12-cog.) So I have to find a number that the dots will be next to each other.
12 and 4 don't work but 5 and 6 work. I realised that 6-5=1, so the difference is one!
I start trying to prove this. 8-7=1 so I thought it would work. And I was right.
I decided to test some numbers that I predicted that won't work. 11-9=2. But it did!
This is very interesting. We received quite a few solutions from children who thought that two cogs with consecutive numbers would lead to the coloured tooth going into every 'gap', and those
children often suggested that this wouldn't happen with two cogs with non-consecutive numbers. But as you've noticed, the 11-cog and the 9-cog do work! I wonder if there's something else going on
with the numbers?
Dhruv from The Glasgow Academy in the UK looked at which cogs would let the the coloured tooth go into every gap when paired with the cog with six teeth. This picture can be clicked on to make it
Dhruv looked at the factors of the number of teeth of each cog, and noticed that the cogs that don't work when paired with the 6-cog share some factors (other than 1) with 6. This is an interesting
observation - I wonder why this means that the coloured tooth won't go into every gap?
Shaunak from Ganit Manthan, Vicharvatika in India sent in this explanation and video:
Which pairs of cogs let the coloured tooth go into every 'gap' on the other cog?
If the numbers represent the number of teeth/gaps on each cog with the following notation, for example, the pairs in which the cogs let the coloured tooth of the first cog go into every gap on the
other cog are:
(4, 5), (4, 7), (4, 9), (4, 11), (5, 6), (5, 7), (5, 8), (5, 9), (5, 11), (5, 12), (6, 7), (6, 11), (7, 8), (7, 9), (7, 10), (7, 11), (7, 12), (8, 9), (8, 11), (9, 10), (9, 11), (10, 11), (11, 12).
Which pairs do not let this happen? Why?
The pairs which will not work are:
(4, 4), (4, 6), (4, 8), (4, 10), (4, 12), (5, 10), (6, 6), (6, 8), (6, 9), (6, 10), (6, 12), (8, 8), (8, 10), (8, 12), (9, 9), (9, 12), (10, 10), (10, 12), (11, 11), (12, 12).
These pairs will not work because the tooth of the first cog touches only spots after a specific interval. This interval is the number of teeth on the first cog. If the HGF (highest common factor -
the largest number that is a factor of both numbers) of the number of teeth on the first cog and the number of gaps on the second cog is x, then the coloured tooth will touch x gaps on the other cog.
If x = 1, then the coloured tooth will touch all gaps on the other cog.
This looks good, Shaunak - I think there is just one more pair that won't work, where both cogs have the same number of teeth.
Can you explain how to determine which pairs will work, and why?
The technique to figure out if a pair will work or not is as follows:
First, count the number of teeth on the first cog and the number of gaps on the second cog.
Next, find their HCF. If the HCF is 1, then the tooth will go on every other gap, else the pair will not work.
These are some good ideas, Shaunak! Thank you for sharing your method with us.
Thank you as well to Ahana, Sehar, Saanvi, Dhanvin, Aariz, Ananthjith, Vivaan, Sai, Pranathi, Paavani, Utkarsh and Dhruv from Ganit Kreeda, Vicharvatika in India, who all worked very hard on this
problem. Take a look at Ganit Kreeda's full solution to see their ideas - they used the fact that the teeth on a cog go up to a certain number and then restart, like the numbers on a clock face do,
to help them solve this problem.
Teachers' Resources
Why do this problem?
This problem
requires children to think about factors and multiples and, in particular, common factors, but it is not necessary for them to have met this term prior to having a go at the task. It offers
opportunities for pupils to ask their own questions, find examples, make conjectures and begin to generalise.
The problem lends itself to collaborative working, both for children who are inexperienced at working in a group and children who are used to working in this way. By working together on this problem,
the task is shared and therefore becomes more manageable than if working alone.
Many NRICH tasks have been designed with group work in mind.
we have gathered together a collection of short articles that outline the merits of collaborative work, together with examples of teachers' classroom practice.
Possible approach
This is an ideal problem for learners to tackle in groups of four. Allocating these clear roles (Word, pdf) can help the group to work in a purposeful way - success on this task could be measured by
how effectively members of the group work together as well as by the solutions they reach.
to see a couple of video clips of two classes organised into groups to work on this task.
Introduce the four group roles to the class. It may be appropriate, if this is the first time the class has worked in this way, to allocate particular roles to particular children. If the class works
in roles over a series of lessons, it is desirable to make sure everyone experiences each role over time.
For suggestions of team-building maths tasks for use with classes unfamiliar with group work, take a look at this article and the accompanying resources.
Give each group a copy of this sheet, which outlines the task. The idea is for them to read it together to find out what to do. Cut out a set of cogs for each group using this sheet and give them out
so each person in a group has two or three cogs. Children should begin by working individually, investigating several pairs of cogs, then they will pool their findings as a group so that they have
worked on all combinations of cogs.
Explain that each group will be expected to report back at the end of the session, showing the patterns they noticed, at least one conjecture they have and at least one question. Exploring the full
potential of this task is likely to take more than one lesson, allowing time in each lesson for children to feed back ideas and share their thoughts and questions. Ask each group to record their
reasoning, conjectures, explanations and any generalisations on a large sheet of paper (for example flipchart paper) in preparation for reporting back.
There are many ways that groups can report back. Here are just a few suggestions:
• Every group is given a couple of minutes to report back to the whole class. Learners can seek clarification and ask questions. After each presentation, children are invited to offer positive
feedback. Finally, pupils can suggest how the group could have improved their work on the task.
• Everyone's posters are put on display at the front of the room, but only a couple of groups are selected to report back to the whole class. Feedback and suggestions can be given in the same way
as above. Additionally, children from the groups which don't present can be invited to share at the end anything they did differently.
• Two children from each group move to join an adjacent group. The two "hosts" explain their findings to the two "visitors". The "visitors" act as critical friends, requiring clear mathematical
explanations and justifications. The "visitors" then comment on anything they did differently in their own group.
Key questions
Which cogs have you found that work so far?
Which pairs didn't work? Can you explain why?
How could you predict whether a pair will work before you try them?
What questions would you like to ask?
Possible extension
Children could begin to work on a question that they have, or a question posed by another group.
Possible support
By working in groups with clearly assigned roles we are encouraging students to take responsibility for ensuring that everyone understands before the group moves on. | {"url":"https://nrich.maths.org/problems/counting-cogs","timestamp":"2024-11-02T14:50:32Z","content_type":"text/html","content_length":"52561","record_id":"<urn:uuid:3d455225-d6a5-4746-8870-84f764179a4c>","cc-path":"CC-MAIN-2024-46/segments/1730477027714.37/warc/CC-MAIN-20241102133748-20241102163748-00525.warc.gz"} |
Rig glueing of morphisms
Proposition 88.26.1. Let $S$ be a scheme. Let $X$ be a locally Noetherian algebraic space over $S$. Let $T \subset |X|$ be a closed subset with complementary open subspace $U \subset X$. Let $f : X'
\to X$ be a proper morphism of algebraic spaces such that $f^{-1}(U) \to U$ is an isomorphism. For any algebraic space $W$ over $S$ the map
\[ \mathop{\mathrm{Mor}}\nolimits _ S(X, W) \longrightarrow \mathop{\mathrm{Mor}}\nolimits _ S(X', W) \times _{\mathop{\mathrm{Mor}}\nolimits _ S(X'_{/T}, W)} \mathop{\mathrm{Mor}}\nolimits _ S(X_{/
T}, W) \]
is bijective.
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Improving our beliefs: Bayes' Theorem
How to make better decisions under uncertainty (Part 2)
In the article “Molding Uncertainty”, the first part of the How to make better decisions under uncertainty series, I asked the readers to imagine themselves in a situation in which they have to
decide if quitting their job to embark on what seems to be an attractive business venture is the right course of action to take.
We used the Subjective Expected Utility (SEU) model and determined that, for this example, getting on board a new business project was the alternative with the highest utility. In other words, based
on our subjective estimation of probabilities and outcomes, we can determine the decisions that will bring us more satisfaction [i].
Thus, being able to correctly predict probabilities is a fundamental requirement when using the SEU method. In practical terms, this means that estimating the wrong probabilities can lead to costly
and incorrect decisions. Despite its critical importance, it’s not obvious how we can improve our ability to calculate probabilities, especially when dealing with the constant incoming of unknown
Therefore, continuing our quest for reducing as much as possible the level of uncertainty of the decisions we face, this article will further develop our decision-making tools by explaining the next
key, game-changer concept: The Bayes’ Theorem.
Bayes' Theorem
Life feeds us with an unstoppable stream of new events and pieces of information that allow us to update our beliefs about the world.
These facts become invaluable evidence that should be used to improve the quality of our judgments; nevertheless the proper utilization of it is more counterintuitive than you would expect. For
example, suppose you are at a party and you meet someone called X, who has a flirty attitude towards you. Do you know how to determine if that person wants to have a fling with you?
I guess you find this question intriguing because although it deals with a quite frivolous situation, answering it with precision is not straightforward. Fortunately we have the Bayes’ Theorem, the
best way to “decode” and solve this kind of challenge.
This formula lets us determine the probability of occurrence of a hypothesis given new evidence. Furthermore, it is the mathematical representation of a way of thinking that can improve our
understanding of the relationship between what we know and what unfolds around us. And in practice, it can dramatically boost the quality of the decisions we make.
Considering its importance I find it very strange that Bayesian thinking hasn’t reached a mainstream audience; being mainly relegated to the realms of mathematicians, philosophers or statisticians. I
think the principal reason for this lack of popularity can be found in the difficulty to understand its mechanism. Although there are plenty of sources with very extensive and sound explanations of
the Bayes’ Theorem, I haven’t found any that are sufficiently concrete, intuitive, and adaptable. Here, I will try to fill that gap.
Dissecting Bayes
The powerful Bayes’ Theorem is simply a “strength test” between competing hypotheses, with the goal of determining their probabilities of occurrence in light of new evidence. Using the formula
(Figure 1) is much easier than it seems. To do so, let’s start by dissecting it into its various components [ii] [iii].
• The probability of occurrence is represented by the letter p followed by parentheses.
• H represents the hypothesis we are inquiring about.
• The new evidence is represented by the letter E.
• The sign “|” means given.
• What we aim to find is the posterior probability p(H|E), that is to say, the probability of occurrence of the hypothesis H given the new evidence E.
• p(H) or prior probability represents the probability of the hypothesis we are assessing, without taking into account the new evidence. Its value expresses what we already know about the estate of
the world. It might have a subjective or objective origin.
• p(E|H) is the probability of occurrence of the evidence E given the hypothesis H. Put in other words, “if the hypothesis is true, how likely is the evidence”.
• -H represents the competing hypothesis of H. These hypotheses are complementary, that is -H means not H. Therefore, adding up their probabilities should total no more than 1 (or its equivalent
• p(E|-H) is the probability of occurrence of the evidence E given the competing hypothesis. In other words “if the competing hypothesis is true, how likely is the evidence”.
The Hypothesis Strength Chart
Now that we know the meaning of the elements, the remaining challenge is establishing each value for a given scenario. Perhaps this is the most cumbersome part (i.e. What do we mean with “if the
competing hypothesis is true, how likely is the evidence?”).
There are different ways in which the logic behind the Bayes’ Theorem has been explained, for example graphically via Venn diagram or with decision-support tools such as Decision Trees; however I
think it could be done in an easier way. I propose using what we will call the Hypotheses Strength Chart, a visual representation of the competing hypotheses and their relationship with the new
evidence (Figure 2).
The best way to explain it is directly applying it to the “fling” question mentioned above. First, the chart is shown followed by a step-by-step description.
Our goal is to determine the probability that a person named X, who flirts with you at a party, in fact wants to have a fling. Hence, we have to consider the two competing hypotheses: Person X
“wants” vs “doesn’t want” to have a fling with you.
We start by assigning a probability to the hypothesis H based on previous knowledge. A hypothesis is a tentative assumption made in order to draw out and test its logical or empirical consequences.
To assign a probability to the hypothesis ask yourself: With all the information I have and based on my experience, how likely is it that someone that I meet at a party wants to have a fling with me?
If you don’t know the answer, give it a probability of 50% (one outcome of two possibilities). It is evident now that if the p(H) is equal to 50%, p(~H) is also equal to 50%. Remember, always aim to
assign a value to H using the most objective information you have. If this kind of data is available, adjust your intuitive estimates with an external approach.
The fact that X flirted with you is a crucial piece of evidence in order to find out if he or she wants to have a fling with you. This is when it becomes interesting.
To find the value of p(E|H) ask yourself: If the hypothesis is true, how likely is this evidence? In our example, of all the people at the party that want to have a fling with me, how likely is that
they flirt with me? Notice that we are now in a universe that only includes people that want to have a fling with you. We are giving it a 60% probability because there are many people that are not
flirtatious even when they like someone. This value is represented by the bar with the label “+”. The bar with the negative signs “-” represents all the people at the party that want to have a fling
with you but are not inclined to flirt, which is estimated at 40%.
Finally, we include the effect of the alternative hypothesis in our computation by estimating the value of p(E|~H). Of all the people that don’t want to have a fling with me, how likely is it that
they flirt with me? We are giving a 10% probability because there are people who flirt even when they don’t want to have a fling.
Shown below is the process of its mathematical resolution:
Thanks to the Bayes’ Theorem we can estimate the probability of someone who flirts with you also wants to have a fling is 85%. Remarkably, we can expect this result having an important effect on
actual behavior. Assuming that the attraction with X is mutual, what would your behavior be towards him or her knowing that there are 8.5 out 10 chances that they want to have a fling with you?
As an additional comment, if you’re interested in exploring the theorem further, try out this Bayesian calculator to estimate your own posterior probabilities on the basis of your personal beliefs
and experiences http://camspiers.github.io/Bayes/. I can assure you that, for this and pretty much any case you can come up with, it becomes an addictive game.
Taking Bayes a little further
Before concluding, we will use the Bayes’ Theorem to improve our estimation of probabilities on a different kind of problem.
Let’s go back to the dilemma proposed in the article “Molding Uncertainty”, where the readers were placed in a hypothetical situation in which they had to decide between staying at their job or
quitting and joining a new business venture. In this example the probability of success of the enterprise was estimated at 65% (here, our Prior).
Today, Hope — your lovely friend who wants you to join her in the new business venture — tells you that she has just closed a deal with an investment firm that will give the company a large amount of
money, which will allow an early start of operations.
Suppose you go online and find a study that estimates that out of the companies that succeed in the long run, 25% of them have received strong financial funding at an early stage. Additionally, 15%
of businesses that fail have received this kind of investment. How can this new evidence affect your estimation of probabilities?
Figure 3 shows the Bayesian representation of this case with a Hypotheses Strength Chart. Then its mathematical resolution is presented.
Using this Bayesian approach we update our hypothesis with new evidence and it shows that the probability of success of this business venture is 76% given it receives a strong early investment.
If we include this new estimate to the SEU model the option of quitting your job has a utility of 46.4, while staying at results in a utility of -15.6. Thus, with the latest evidence your degree of
belief in the “success” hypothesis increases and the confidence with the decision of quitting your job is strengthened.
Finally, bear in mind that if opposing evidence arrives probability and utility estimates will likely change. This is Bayesian thinking after all.
Having a true Bayesian mindset implies revisiting our judgments and decisions as new knowledge is presented, and reevaluating — and even changing — our assumptions every time we acquire new
information. This requires an attitude of curiosity and open-mindedness about latest data, and also skepticism about our prior beliefs. Personally, the Bayes Theorem is one of the concepts I wish I
had learned when I was much younger, perhaps since I was taught algebra in middle school. Luckily, I can assure you, it’s never too late.
Main Takeaway
This article explained The Bayes’ Theorem, a way in which we can improve the dubious quality of our probability estimation under uncertainty, an elusive challenge that is limited by the lack of
straightforward understanding of what to do with the arrival of new evidence. The Bayes’ Theorem is a paramount tool for updating our degree of belief in a hypothesis based on the occurrence of
another event, potentially boosting the quality of the decisions we make. The Hypotheses Strength Chart is suggested as a way to visualize its logic.
[i] The full explanation of how to use probability estimates to make decisions with the Subjective Expected Utility (SEU) method is developed in “Molding Uncertainty”. Although the present article
contains all the concepts necessary to understand the Bayes’ Theorem, I highly suggest reading the first part of the series so the Bayesian thinking role and place as a decision-making enhancer is
more clear. | {"url":"https://www.heuristicalab.com/en/post/improving-our-beliefs-bayes-theorem","timestamp":"2024-11-07T17:17:56Z","content_type":"text/html","content_length":"1050485","record_id":"<urn:uuid:c7ef1717-90ce-40cb-8b41-9fa3b203a287>","cc-path":"CC-MAIN-2024-46/segments/1730477028000.52/warc/CC-MAIN-20241107150153-20241107180153-00053.warc.gz"} |
Types of Map Projections
The ways in which we visualize the world are varied- we have pictures, maps, globes, satellite imagery, hand drawn creations and more.
What kinds of things can we learn from the way we see the world around us?
For centuries cartographers have been making maps of the world around them, from their immediate area to the greater world as they understood it at the time. These maps depict everything from hunting
grounds to religious beliefs and speculations of the broader, unexplored world around them.
Maps have been made of the local waterways, trade routes, and the stars to help navigators on land and sea make their way to different locations.
Cartographers have been visualizing the world around us, both real and imaginary.
[Map of the lost island of Atlantis] Situs Insulae Atlantidis, a mari olim obsorptae ex mente Aegyptiorum et Platonis discriptio, 1665 by Athanasius Kircher. Map: Barry Lawrence Ruderman Map
Collection, Stanford University.
How we visualize the world not only has practical implications, but can also help shape our perspectives of the Earth we live in.
There are many kinds of maps made from a variety of materials and on a variety of topics.
Clay tablets, papyrus, and bricks made way for modern maps portrayed on globes and on paper; more recent technological advances allow for satellite imagery and computerized models of the Earth.
Using a globe versus a map
Using a globe instead of a map offers several advantages:
1. Accurate representation: A globe accurately represents the Earth’s curved surface without any distortions in area, shape, distance, direction, or scale. Maps, on the other hand, always introduce
some level of distortion due to the process of projecting a three-dimensional surface onto a two-dimensional plane.
2. True spatial relationships: A globe allows for a better understanding of the spatial relationships between different locations on Earth. Distances, directions, and relative positions of
continents and countries are more accurately depicted on a globe than on a map.
3. Better visualization of Earth’s geometry: A globe helps users visualize the Earth’s round shape, making it easier to comprehend concepts such as latitude, longitude, the Earth’s axis, and the
rotation that causes day and night.
4. Improved perspective: A globe offers a more realistic perspective of the Earth, helping users appreciate the actual size and position of landmasses and bodies of water, which can sometimes be
misrepresented on maps due to projection distortions.
5. Consistent scale: A globe has a constant scale throughout its entire surface, unlike maps where the scale can vary from point to point depending on the projection used.
A six-inch tall teaching globe from the 1960s from the David Rumsey Map Center, Stanford University. Photo: Caitlin Dempsey.
Despite these advantages, globes have some limitations, such as being impractical for large-scale mapping, difficult to measure, challenging to see the entire world at once, and less portable
compared to folding maps.
What are map projections?
A map projection is a method used to represent the Earth’s three-dimensional, curved surface onto a two-dimensional plane, such as a piece of paper or a digital screen. Since the Earth is not flat,
map projections inevitably introduce some distortions in area, shape, distance, direction, or scale.
Cartographers choose different map projections based on the purpose of the map and the region being depicted to minimize these distortions and accurately convey information.
Three of these common types of map projections are cylindrical, conic, and azimuthal.
Types of map projection distortion
Map projections inevitably introduce distortions in one or more of the following aspects:
1. Area-preserving projection – Also called equal area or equivalent projection, these projections maintain the relative size of different regions on the map.
2. Shape-preserving projection – Often referred to as conformal or orthomorphic, these projections maintain accurate shapes of regions and local angles.
3. Direction-preserving projection – This category includes conformal, orthomorphic, and azimuthal projections, which preserve directions, but only from the central point for azimuthal projections.
4. Distance-preserving projection – Known as equidistant projections, they display the true distance between one or two points and all other points on the map.
It is important to note that it is impossible to create a map projection that preserves both area and shape simultaneously.
Distortion on a map can be visualized using the Tissot’s Indicatrix. Using graduated circles, the amount of distortion is shown relative to the other areas of the map.
A world map using the Mercator map projection overlaid with Tissot’s Indicatrix (red circles) showing that the high latitudes (large circles) are much more distorted than by the Equator (smaller
circles). Map: Caitlin Dempsey.
How map projections are categorized
The primary categories of map projections include:
1. Cylindrical Projections: These projections involve wrapping a cylinder around the Earth and projecting its features onto the cylindrical surface. Examples are the Mercator, Transverse Mercator,
and Miller Cylindrical projections.
2. Conic Projections: For these projections, a cone is placed over the Earth, and its features are projected onto the conical surface. Common examples are the Lambert Conformal Conic and Albers
Equal-Area Conic projections.
3. Azimuthal Projections: Also referred to as planar or zenithal projections, these use a flat plane that touches the Earth at a single point, projecting the Earth’s features onto the plane.
Azimuthal Equidistant, Stereographic, and Orthographic projections are examples.
4. Pseudocylindrical Projections: These projections resemble cylindrical projections but employ curved lines instead of straight lines for meridians and parallels. The Sinusoidal, Mollweide, and
Goode Homolosine projections are popular examples.
Map projections can also be classified based on the properties they maintain:
1. Equal-area (equivalent) projections: These projections preserve the correct proportions of areas, such as in the Albers Equal-Area Conic and Mollweide projections.
2. Conformal (orthomorphic) projections: These projections maintain local angles and shapes, as seen in the Mercator and Lambert Conformal Conic projections.
3. Equidistant projections: These projections retain true distances from one or two points to all other points, as in the Azimuthal Equidistant projection.
4. Azimuthal projections: These projections preserve directions from a central point, including some conformal, orthomorphic, and azimuthal projections.
5. Compromise projections: These projections attempt to balance various distortions inherent in map projections, such as the Robinson and Winkel Tripel projections.
It is essential to understand that no map projection can perfectly preserve all properties, as each type entails some degree of compromise or distortion.
Cylindrical Map Projections
Cylindrical projections involve wrapping a cylinder around the Earth, touching it at the equator or another standard line, and projecting the Earth’s surface onto the cylinder.
This kind of map projection has straight coordinate lines with horizontal parallels crossing meridians at right angles. All meridians are equally spaced and the scale is consistent along each
Cylindrical map projections are rectangles, but are called cylindrical because they can be rolled up and their edges mapped in a tube, or cylinder, shape.
The only factor that distinguishes different cylindrical map projections from one another is the scale used when spacing the parallel lines on the map.
The downsides of cylindrical map projections are that they are severely distorted at the poles.
While the areas near the Equator are the most likely to be accurate compared to the actual Earth, the parallels and meridians being straight lines don’t allow for the curvature of the Earth to be
taken into consideration.
The Mercator map projection is one of the most well known cylindrical map projections. Map: Caitlin Dempsey.
Cylindrical map projections are great for comparing latitudes to each other and are useful for teaching and visualizing the world as a whole, but really aren’t the most accurate way of visualizing
how the world really looks in its entirety.
Types of cylindrical map projections you may know include the popular Mercator projection, Cassini, Gauss-Kruger, Miller, Behrmann, Hobo-Dyer, and Gall-Peters.
Mercator Projection
Introduced by Gerardus Mercator in 1569, the Mercator projection is a cylindrical projection that preserves local angles and shapes, making it valuable for navigation purposes.
The Mercator map projection significantly distorts the size of landmasses near the poles, leading to misconceptions about the relative sizes of continents and countries.
Transverse Mercator Projection
A variation of the Mercator projection, the Transverse Mercator projection, involves rotating the cylinder 90 degrees.
The Universal Transverse Mercator map projection is commonly used for large-scale mapping of regions with predominantly north-south extents, such as the U.S. Geological Survey’s topographic maps.
With UTM, the world is divide into 60 zones that are each six degrees wide.
UTM zones over the contiguous United States. Map: Caitlin Dempsey
This projection reduces distortion for areas with limited east-west extents but increases distortion as one moves away from the central meridian.
Miller Cylindrical Projection
Osborn Maitland Miller developed the Miller Cylindrical projection in 1942 as a modified version of the Mercator projection. It minimizes distortion in high latitudes by slightly compressing the
spacing of parallels. Although it still overstates the size of polar areas, the distortion is less pronounced than in the standard Mercator projection.
Conic Map Projections
Conic projections involve placing a cone over the Earth, touching it along a standard parallel or two standard parallels. Conic map projections include the equidistant conic projection, the Lambert
conformal conic, and Albers conic.
These maps are defined by the cone constant, which dictates the angular distance between meridians.
These meridians are equidistant and straight lines which converge in locations along the projection regardless of if there’s a pole or not.
The Albers projection is an example of a conic map projection. Map: Caitlin Dempsey.
Like the cylindrical projection, conic map projections have parallels that cross the meridians at right angles with a constant measure of map distortion throughout. Conic map projections are designed
to be able to be wrapped around a cone on top of a sphere (globe), but aren’t supposed to be geometrically accurate.
Conic map projections are best suited for use as regional or hemispheric maps, but rarely for a complete world map.
The distortion in a conic map makes it inappropriate for use as a visual of the entire Earth but does make it great for use visualizing temperate regions, weather maps, climate projections, and more.
Lambert Conformal Conic Projection
The Lambert Conformal Conic projection is a conic map projection that maintains accurate shapes and angles over small areas.
This map projection is suitable for mapping regions with predominantly east-west extents, such as the United States. This projection is widely used for aeronautical charts due to its angle
preservation, making it valuable for navigation.
Albers Equal-Area Conic Projection
The Albers Equal-Area Conic projection is a conic map projection that preserves the area at the expense of shape and angle.
This map projection is also useful for displaying regions with significant east-west extents, such as the continental United States. This projection is often used for thematic maps requiring accurate
area representation, such as population density or land use.
Azimuthal Map Projection
Azimuthal projections involve projecting the Earth’s surface onto a flat plane, typically tangent or secant to the Earth at a specific point.
The azimuthal map projection is angular- given three points on a map (A, B, and C) the azimuth from Point B to Point C dictates the angle someone would have to look or travel in order to get to A.
These angular relationships are more commonly known as great circle arcs or geodesic arcs.
The main features of azimuthal map projections are straight meridian lines, radiating out from a central point, parallels that are circular around the central point, and equidistant parallel spacing.
Light paths in three different categories (orthographic, stereographic, and gnomonic) can also be used. Azimuthal maps are beneficial for finding direction from any point on the Earth using the
central point as a reference.
Lambert azimuthal equal-area projection centered on the North Pole. Map: Caitlin Dempsey.
Azimuthal Equidistant Projection
The Azimuthal Equidistant projection is a planar projection that maintains accurate distances from the center point to any other point on the map. This projection is frequently used for polar maps,
where the center point represents the North or South Pole.
This map projection is also commonly utilized for radio and telecommunications planning, as it accurately represents distances between the central point and other locations.
Stereographic Projection
The Stereographic projection is a planar projection that preserves angles and shapes locally, making it conformal. It is often used for mapping polar regions and creating star charts in celestial
This map projection is also the basis for the popular Polar Stereographic projection, which is used for representing high-latitude regions with minimal distortion.
Orthographic Projection
The Orthographic projection is a planar projection that represents the Earth as if viewed from an infinite distance, giving the appearance of a globe on a flat surface.
This projection is often used for artistic purposes and for visualizing the Earth from space, as it provides a unique, aesthetically pleasing perspective.
Pseudocylindrical Projections
Pseudocylindrical projections resemble cylindrical projections but have curved parallels instead of straight ones.
Sinusoidal Projection
The Sinusoidal projection, also known as the Sanson-Flamsteed projection, is an equal-area pseudocylindrical projection that minimizes distortion in the east-west direction near the equator. It is
often used for world maps that prioritize accurate area representation, such as climate or vegetation maps.
Mollweide Projection
The Mollweide projection is a pseudocylindrical equal-area projection that balances area and shape distortion, making it suitable for world maps that require a reasonable compromise between these
The Mollweide Projection is frequently used for thematic maps, such as those illustrating global temperature patterns or population distribution.
Equal Earth Map Projection
The Equal Earth map projection is a relatively new pseudocylindrical projection designed to display the entire Earth’s surface with minimal distortion while preserving equal areas.
The Equal Earth map projection was developed by contemporary cartographers Tom Patterson, Bernhard Jenny, and Bojan Šavrič in 2018 as a response to the increasing need for a visually appealing and
less distorted world map that accurately represents areas, especially in the context of global issues like climate change and deforestation.
The Equal Earth Map Projection. Map: Equal Earth, public domain.
The Equal Earth projection is inspired by the Robinson projection and maintains the same overall shape, but with improved area accuracy. It is particularly well-suited for general-purpose world maps,
educational materials, and thematic maps that require an equal-area representation. This projection is advantageous because it presents a balanced view of the world, with less emphasis on the size of
high-latitude countries, avoiding the common misconceptions caused by projections like the Mercator, which significantly exaggerates the size of regions closer to the poles.
Goode Homolosine Projection
The Goode Homolosine projection, developed by John Paul Goode in 1923, is a pseudocylindrical equal-area projection that resembles an interrupted globe. It is designed to minimize distortion in both
area and shape, making it suitable for world maps that require a balanced representation of the Earth’s landmasses.
Compromise Projections
Compromise projections aim to strike a balance between the various distortions inherent in map projections.
Robinson Projection
The Robinson map projection, developed by Arthur H. Robinson in 1963, is a compromise projection that balances the distortions of area, shape, distance, and direction.
The Robinson map projection is a pseudocylindrical projection. Map: Caitlin Dempsey, Natural Earth data.
It creates visually appealing world maps that provide a general overview of the Earth’s surface. The National Geographic Society widely used the Robinson projection for its world maps until 1998.
Winkel Tripel Projection
The Winkel Tripel projection, developed by Oswald Winkel in 1921, is another compromise projection that balances distortions in area, shape, distance, and direction.
It is considered one of the best projections for general-purpose world maps, and the National Geographic Society adopted it as their standard world map projection in 1998.
Different types of map projections suit different geographic needs
Map projection types all have their pros and cons, but they are incredibly versatile.
Even though it is nearly impossible to create an entirely accurate map projection there are uses for even the most imperfect depictions of the Earth.
Map projections are created for certain purposes and should be used for those purposes. In the end each and every map projection has a place, and there is no limit to the amount of projections that
can be created.
Geokov. Map Projections: Types and Distortion Patterns. 2014. Web access 28 November 2014. http://geokov.com/education/map-projection.aspx
Furuti, Carlos. Map Projections: Cylindrical Projections. 2 December 2013. Web access 28 November 2014. http://www.progonos.com/furuti/MapProj/Dither/ProjCyl/projCyl.html
Furuti, Carlos. Map Projections: Conic Projections. 13 December 2013. Web access 28 November 2014. http://www.progonos.com/furuti/MapProj/Dither/ProjCon/projCon.html
Šavrič, B., Patterson, T., & Jenny, B. (2019). The equal earth map projection. International Journal of Geographical Information Science, 33(3), 454-465. https://doi.org/10.1080/13658816.2018.1504949
Snyder, J. P. (1982). Map projections used by the US Geological Survey (No. 1532). US Government Printing Office.
Snyder, J. P. (1987). Map projections–A working manual (Vol. 1395). US Government Printing Office.
This article was originally written on March 6, 2015 and has since been updated.
More About Map Projections
About the author
Elizabeth Borneman
My name is Elizabeth Borneman and I am a freelance writer, reader, and coffee drinker. I live on a small island in Alaska, which gives me plenty of time to fish, hike, kayak, and be inspired by
nature. I enjoy writing about the natural world and find lots of ways to flex my creative muscles on the beach, in the forest, or down at the local coffee shop. | {"url":"https://www.geographyrealm.com/types-map-projections/","timestamp":"2024-11-05T12:19:04Z","content_type":"text/html","content_length":"263362","record_id":"<urn:uuid:4f501d18-1504-4ed0-a3fc-2db510819113>","cc-path":"CC-MAIN-2024-46/segments/1730477027881.88/warc/CC-MAIN-20241105114407-20241105144407-00154.warc.gz"} |
When Your Tail Really Counts
Looks like this ring-tailed lemur really wants to get into that backpack. And it turns out these silly, stripy goofballs are pretty smart: they can do math! Lemurs can count, add, and subtract, and
can also line up objects in order from memory. Even their wild long tails have numbers behind them: the lemur’s tail always has 12 or 13 white rings and 13 or 14 black rings, and the tip always ends
in black. Lemurs live on the island of Madagascar off the coast of Africa. It’s a beautiful place, so we can see why they’d like to take some pictures of it!
Wee ones: If the lemur’s tail has a black ring at the tip, then white, then black, then white, what’s the next ring?
Little kids: If a lemur’s tail has 12 white rings and 13 black rings, of which color does it have more? Bonus: If the 1st ring is black followed by white, then black, then white, and so on, what
color is the 12th ring?
Big kids: If a lemur’s tail has 27 rings, and there’s 1 more black ring than white ring, how many rings of each color? Bonus: If a lemur is 18 inches long and its tail is another 1/3 body length
longer than that, how long is the whole lemur from head to tail tip?
The sky’s the limit: We can’t tell you how many rings a baby lemur’s tail has, but if you took that number, multiplied it by itself, and added 9, you’d get 58. How many rings does the tail have?
Wee ones: A black ring.
Little kids: More black rings. Bonus: White.
Big kids: 14 black rings and 13 white. If you took off that extra black ring, you’d have 26 rings that are equally black and white, so then you just cut 26 in half to find the white. Bonus: 42
inches, since it’s 18 plus 24.
The sky’s the limit: 7 rings. If you walk backwards, the number times itself comes to 49 (58-9), and 49 is divisible only by 7, so that’s your answer! | {"url":"https://bedtimemath.org/fun-math-lemurs/","timestamp":"2024-11-11T07:37:56Z","content_type":"text/html","content_length":"86741","record_id":"<urn:uuid:b7193e78-6cdc-4700-b87e-9e3b329dfc21>","cc-path":"CC-MAIN-2024-46/segments/1730477028220.42/warc/CC-MAIN-20241111060327-20241111090327-00689.warc.gz"} |
If somehow, "50.0 mL" of a stock solution of "120. M" "HCl" is available, what volume of water should the appropriate amount of "HCl" be added to, in order to reduce its concentration to "4.00 M"? | Socratic
If somehow, #"50.0 mL"# of a stock solution of #"120. M"# #"HCl"# is available, what volume of water should the appropriate amount of #"HCl"# be added to, in order to reduce its concentration to #
"4.00 M"#?
1 Answer
I got $\approx 1.45 \times {10}^{3}$$\text{mL}$, or $\approx$$\text{1.45 L}$. Keep in mind that you won't be adding exactly $\text{50.0 mL}$---you'll probably add a bit less.
Just so you know, that initial concentration of $\text{HCl}$couldn't possibly be safe to use; it's about 10 times the concentration of the stock $\text{HCl}$ that university labs let students use at
all. But OK, let's see.
I assume you mean the "before/after" formula:
$\setminus m a t h b f \left({M}_{1} {V}_{1} = {M}_{2} {V}_{2}\right)$
Basically, you have a relationship of concentration to volume:
• If concentration is to increase, then volume has to decrease.
• So, if volume increases, then concentration must decrease (the contrapositive).
Here, ${M}_{i}$ is the concentration in molars ($\text{M}$) of solution $i$, and ${V}_{i}$ is the volume of solution $i$.
For the volume, we can use milliliters ($\text{mL}$) for simplicity, since you will be dividing two concentrations and canceling out their units.
What you already have are:
• ${V}_{1} = \text{50.0 mL}$
• ${M}_{1} = \text{120. M}$
• ${M}_{2} = \text{4.00 M}$
So, you are solving for the final volume, ${V}_{2}$:
$\textcolor{g r e e n}{{V}_{2}} = \left(\frac{{M}_{1}}{{M}_{2}}\right) {V}_{1}$
#= (("120." cancel"M")/(4.00 cancel"M")) ("50.0 mL")#
$=$$\text{1500 mL}$
Or, to three sig figs:
$= \textcolor{g r e e n}{1.50 \times {10}^{3}}$$\textcolor{g r e e n}{\text{mL}}$
Remember, that's your FINAL volume, NOT the amount of water you might start with, so you should subtract to get:
$1500 - 50 = \textcolor{b l u e}{\text{1450 mL}}$ water to begin with.
It's better, however, to start with the $\text{1450 mL}$ of water and slowly add acid until you get to the $\setminus m a t h b f \left(\text{1500 mL}\right)$mark, to account for the fact that not
all solutions are 100% additive.
You won't necessarily transfer exactly $\setminus m a t h b f \left(\text{50 mL}\right)$.
Impact of this question
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Radians to Degrees - Conversion, Formula, Examples
Radians and degrees conversion is a very essential skill for higher arithmetic students to grasp.
First, we are required to define what radians are so that you can understand how this formula is used in practice. Then we’ll take a further step by exhibiting some examples of converting from
radians to degrees with ease!
What Is a Radian?
Radians are measurement units for angles. It is originated from the Latin word "radix," which implies nostril or ray, and is a fundamental concept in geometry and mathematics.
A radian is the SI (standard international) measuring unit for angles, while a degree is a more frequently used unit in arithmetic.
In other words, radians and degrees are simply two distinct units of measure used for measuring the identical thing: angles.
Note: a radian is not to be confused with a radius. They are two completely distinct things. A radius is the distance from the middle of a circle to the perimeter, while a radian is a unit of measure
for angles.
Relationship Between Radian and Degrees
There are two manners to think regarding this question. The initial method is to contemplate about how many radians exists in a full circle. A full circle is equivalent to 360 degrees or two pi
radians (exactly). So, we can state:
2π radians = 360 degrees
Or easily:
π radians = 180 degrees
The next way to figure out about this question is to calculate how many degrees are present in a radian. We all know that there are 360 degrees in a whole circle, and we also understand that there
are two pi radians in a full circle.
If we divide each side by π radians, we’ll see that 1 radian is approximately 57.296 degrees.
π radiansπ radians = 180 degreesπ radians = 57.296 degrees
Both of these conversion factors are beneficial relying upon which you're trying to get.
How to Go From Radians to Degrees?
Now that we've gone through what radians and degrees are, let's learn how to turn them!
The Formula for Converting Radians to Degrees
Proportions are a useful tool for converting a radian value to degrees.
π radiansx radians = 180 degreesy degrees
Just put in your known values to get your unknown values. For example, if you are required to convert .7854 radians into degrees, your proportion will be:
π radians.7854 radians = 180 degreesz degrees
To solve for z, multiply 180 with .7854 and divide by 3.14 (pi): 45 degrees.
This formula can be implemented both ways. Let’s recheck our operation by changing 45 degrees back to radians.
π radiansy radians = 180 degrees45 degrees
To find out the value of y, multiply 45 by 3.14 (pi) and divide by 180: .785 radians.
Once we've converted one type, it will always work with different simple calculation. In this scenario, afterwards converting .785 from its first form back again, ensuing these steps made exactly
what was predicted -45°.
The formulas work out like this:
Degrees = (180 * z radians) / π
Radians = (π * z degrees) / 180
Examples of Converting Radians to Degrees
Let's try some examples, so these ideas become easier to digest.
At the moment, we will convert pi/12 rad into degrees. Just like before, we will put this number into the radians slot of the formula and calculate it like this:
Degrees = (180 * (π/12)) / π
Now, let divide and multiply as you normally do:
Degrees = (180 * (π/12)) / π = 15 degrees.
There you have the result! pi/12 radians equivalents 15 degrees.
Let's try some more general conversion and transform 1.047 rad to degrees. One more time, use the formula to get started:
Degrees = (180 * 1.047) / π
One more time, you multiply and divide as fitting, and you will wind up with 60 degrees! (59.988 degrees to be exact).
Now, what to do if you have to change degrees to radians?
By using the very same formula, you can do the converse in a pinch by solving it considering radians as the unknown.
For example, if you have to convert 60 degrees to radians, put in the knowns and work out with the unknowns:
60 degrees = (180 * z radians) / π
(60 * π)/180 = 1.047 radians
If you recollect the equation to solve for radians, you will get identical answer:
Radians = (π * z degrees) / 180
Radians = (π * 60 degrees) / 180
And there it is! These are just some of the examples of how to convert radians to degrees and the other way around. Bear in mind the equation and try it out for yourself the next time you need to
make a conversion among radians and degrees.
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Flexible parametric mixture models for times to competing events — flexsurvmix
Flexible parametric mixture models for times to competing events
In a mixture model for competing events, an individual can experience one of a set of different events. We specify a model for the probability that they will experience each event before the others,
and a model for the time to the event conditionally on that event occurring first.
pformula = NULL,
anc = NULL,
partial_events = NULL,
initp = NULL,
inits = NULL,
fixedpars = NULL,
dfns = NULL,
method = "direct",
em.control = NULL,
optim.control = NULL,
aux = NULL,
sr.control = survreg.control(),
hess.control = NULL,
Survival model formula. The left hand side is a Surv object specified as in flexsurvreg. This may define various kinds of censoring, as described in Surv. Any covariates on the right hand side of
this formula will be placed on the location parameter for every component-specific distribution. Covariates on other parameters of the component-specific distributions may be supplied using the
anc argument.
Alternatively, formula may be a list of formulae, with one component for each alternative event. This may be used to specify different covariates on the location parameter for different
A list of formulae may also be used to indicate that for particular individuals, different events may be observed in different ways, with different censoring mechanisms. Each list component
specifies the data and censoring scheme for that mixture component.
For example, suppose we are studying people admitted to hospital,and the competing states are death in hospital and discharge from hospital. At time t we know that a particular individual is
still alive, but we do not know whether they are still in hospital, or have been discharged. In this case, if the individual were to die in hospital, their death time would be right censored at
t. If the individual will be (or has been) discharged before death, their discharge time is completely unknown, thus interval-censored on (0,Inf). Therefore, we need to store different event time
and status variables in the data for different alternative events. This is specified here as
formula = list("discharge" = Surv(t1di, t2di, type="interval2"), "death" = Surv(t1de, status_de))
where for this individual, (t1di, t2di) = (0, Inf) and (t1de, status_de) = (t, 0).
The "dot" notation commonly used to indicate "all remaining variables" in a formula is not supported in flexsurvmix.
Data frame containing variables mentioned in formula, event and anc.
Variable in the data that specifies which of the alternative events is observed for which individual. If the individual's follow-up is right-censored, or if the event is otherwise unknown, this
variable must have the value NA.
Ideally this should be a factor, since the mixture components can then be easily identified in the results with a name instead of a number. If this is not already a factor, it is coerced to one.
Then the levels of the factor define the required order for the components of the list arguments dists, anc, inits and dfns. Alternatively, if the components of the list arguments are named
according to the levels of event, then the components can be arranged in any order.
Vector specifying the parametric distribution to use for each component. The same distributions are supported as in flexsurvreg.
Formula describing covariates to include on the component membership proabilities by multinomial logistic regression. The first component is treated as the baseline.
The "dot" notation commonly used to indicate "all remaining variables" in a formula is not supported.
List of component-specific lists, of length equal to the number of components. Each component-specific list is a list of formulae representing covariate effects on parameters of the distribution.
If there are covariates for one component but not others, then a list containing one null formula on the location parameter should be supplied for the component with no covariates, e.g list(rate=
~1) if the location parameter is called rate.
Covariates on the location parameter may also be supplied here instead of in formula. Supplying them in anc allows some components but not others to have covariates on their location parameter.
If a covariate on the location parameter was provided in formula, and there are covariates on other parameters, then a null formula should be included for the location parameter in anc, e.g list
List specifying the factor levels of event which indicate knowledge that an individual will not experience particular events, but may experience others. The names of the list indicate codes that
indicate partial knowledge for some individuals. The list component is a vector, which must be a subset of levels(event) defining the events that a person with the corresponding event code may
For example, suppose there are three alternative events called "disease1","disease2" and "disease3", and for some individuals we know that they will not experience "disease2", but they may
experience the other two events. In that case we must create a new factor level, called, for example "disease1or3", and set the value of event to be "disease1or3" for those individuals. Then we
use the "partial_events" argument to tell flexsurvmix what the potential events are for individuals with this new factor level.
partial_events = list("disease1or3" = c("disease1","disease3"))
Initial values for component membership probabilities. By default, these are assumed to be equal for each component.
List of component-specific vectors. Each component-specific vector contains the initial values for the parameters of the component-specific model, as would be supplied as the inits argument of
flexsurvreg. By default, a heuristic is used to obtain initial values, which depends on the parametric distribution being used, but is usually based on the empirical mean and/or variance of the
survival times.
Indexes of parameters to fix at their initial values and not optimise. Arranged in the order: baseline mixing probabilities, covariates on mixing probabilities, time-to-event parameters by mixing
component. Within mixing components, time-to-event parameters are ordered in the same way as in flexsurvreg.
If fixedpars=TRUE then all parameters will be fixed and the function simply calculates the log-likelihood at the initial values.
Not currently supported when using the EM algorithm.
List of lists of user-defined distribution functions, one for each mixture component. Each list component is specified as the dfns argument of flexsurvreg.
Method for maximising the likelihood. Either "em" for the EM algorithm, or "direct" for direct maximisation.
List of settings to control EM algorithm fitting. The only options currently available are
trace set to 1 to print the parameter estimates at each iteration of the EM algorithm
reltol convergence criterion. The algorithm stops if the log likelihood changes by a relative amount less than reltol. The default is the same as in optim, that is, sqrt(.Machine$double.eps).
var.method method to compute the covariance matrix. "louis" for the method of Louis (1982), or "direct"for direct numerical calculation of the Hessian of the log likelihood.
optim.p.control A list that is passed as the control argument to optim in the M step for the component membership probability parameters. The optimisation in the M step for the time-to-event
parameters can be controlled by the optim.control argument to flexsurvmix.
For example, em.control = list(trace=1, reltol=1e-12).
List of options to pass as the control argument to optim, which is used by method="direct" or in the M step for the time-to-event parameters in method="em". By default, this uses fnscale=10000
and ndeps=rep(1e-06,p) where p is the number of parameters being estimated, unless the user specifies these options explicitly.
A named list of other arguments to pass to custom distribution functions. This is used, for example, by flexsurvspline to supply the knot locations and modelling scale (e.g. hazard or odds). This
cannot be used to fix parameters of a distribution --- use fixedpars for that.
For the models which use survreg to find the maximum likelihood estimates (Weibull, exponential, log-normal), this list is passed as the control argument to survreg.
List of named arguments to pass to integrate, if a custom density or hazard is provided without its cumulative version. For example,
integ.opts = list(rel.tol=1e-12)
List of options to control covariance matrix computation. Available options are:
numeric. If TRUE then numerical methods are used to compute the Hessian for models where an analytic Hessian is available. These models include the Weibull (both versions), exponential, Gompertz
and spline models with hazard or odds scale. The default is to use the analytic Hessian for these models. For all other models, numerical methods are always used to compute the Hessian, whether
or not this option is set.
tol.solve. The tolerance used for solve when inverting the Hessian (default .Machine$double.eps)
tol.evalues The accepted tolerance for negative eigenvalues in the covariance matrix (default 1e-05).
The Hessian is positive definite, thus invertible, at the maximum likelihood. If the Hessian computed after optimisation convergence can't be inverted, this is either because the converged result
is not the maximum likelihood (e.g. it could be a "saddle point"), or because the numerical methods used to obtain the Hessian were inaccurate. If you suspect that the Hessian was computed
wrongly enough that it is not invertible, but not wrongly enough that the nearest valid inverse would be an inaccurate estimate of the covariance matrix, then these tolerance values can be
modified (reducing tol.solve or increasing tol.evalues) to allow the inverse to be computed.
Optional arguments to the general-purpose optimisation routine optim. For example, the BFGS optimisation algorithm is the default in flexsurvreg, but this can be changed, for example to method=
"Nelder-Mead" which can be more robust to poor initial values. If the optimisation fails to converge, consider normalising the problem using, for example, control=list(fnscale = 2500), for
example, replacing 2500 by a number of the order of magnitude of the likelihood. If 'false' convergence is reported with a non-positive-definite Hessian, then consider tightening the tolerance
criteria for convergence. If the optimisation takes a long time, intermediate steps can be printed using the trace argument of the control list. See optim for details.
List of objects containing information about the fitted model. The important one is res, a data frame containing the parameter estimates and associated information.
This differs from the more usual "competing risks" models, where we specify "cause-specific hazards" describing the time to each competing event. This time will not be observed for an individual if
one of the competing events happens first. The event that happens first is defined by the minimum of the times to the alternative events.
The flexsurvmix function fits a mixture model to data consisting of a single time to an event for each individual, and an indicator for what type of event occurs for that individual. The time to
event may be observed or censored, just as in flexsurvreg, and the type of event may be known or unknown. In a typical application, where we follow up a set of individuals until they experience an
event or a maximum follow-up time is reached, the event type is known if the time is observed, and the event type is unknown when follow-up ends and the time is right-censored.
The model is fitted by maximum likelihood, either directly or by using an expectation-maximisation (EM) algorithm, by wrapping flexsurvreg to compute the likelihood or to implement the E and M steps.
Some worked examples are given in the package vignette about multi-state modelling, which can be viewed by running vignette("multistate", package="flexsurv").
Jackson, C. H. and Tom, B. D. M. and Kirwan, P. D. and Mandal, S. and Seaman, S. R. and Kunzmann, K. and Presanis, A. M. and De Angelis, D. (2022) A comparison of two frameworks for multi-state
modelling, applied to outcomes after hospital admissions with COVID-19. Statistical Methods in Medical Research 31(9) 1656-1674.
Larson, M. G., & Dinse, G. E. (1985). A mixture model for the regression analysis of competing risks data. Journal of the Royal Statistical Society: Series C (Applied Statistics), 34(3), 201-211.
Lau, B., Cole, S. R., & Gange, S. J. (2009). Competing risk regression models for epidemiologic data. American Journal of Epidemiology, 170(2), 244-256. | {"url":"https://chjackson.github.io/flexsurv/reference/flexsurvmix.html","timestamp":"2024-11-06T08:55:23Z","content_type":"text/html","content_length":"25500","record_id":"<urn:uuid:9b8dbfaa-eff5-4c99-8531-ca994d815314>","cc-path":"CC-MAIN-2024-46/segments/1730477027910.12/warc/CC-MAIN-20241106065928-20241106095928-00810.warc.gz"} |
Milligrams to Kilograms Calculator - calculator
Home Calculator Milligrams to Kilograms Calculator
Milligrams to Kilograms Calculator
What is the use of Milligrams to Kilograms Calculator?
The Milligrams to Kilograms Calculator is a tool designed to convert mass measurements from milligrams (mg) to kilograms (kg). This calculator is useful for various applications where precision in
mass measurement is required. It is particularly beneficial in scientific research, laboratory work, and industrial applications where small quantities of substances need to be converted to a larger
unit for easier understanding or further processing.
What is the formula of Milligrams to Kilograms Calculator?
To convert milligrams to kilograms, you use the following formula:
1 milligram (mg) = 1/1000000 kilograms (kg) = 0.000001 kg
The general formula for converting milligrams (mg) to kilograms (kg) is:
m(kg) = m(mg) / 1000000
Here, m(kg) represents the mass in kilograms and m(mg) represents the mass in milligrams.
How to use Milligrams to Kilograms Calculator website?
Using the Milligrams to Kilograms Calculator is simple and straightforward. Follow these steps:
1. Input the Value: Enter the mass in milligrams into the input box provided on the calculator.
2. Calculate: Click the "Calculate" button to convert the entered milligrams to kilograms.
3. View Results: The result will be displayed in a table format showing both the milligrams and the equivalent kilograms.
4. Clear Fields: To reset the input field and clear the results, click the "Clear" button.
The calculator will use the formula m(kg) = m(mg) / 1000000 to perform the conversion and display the result accordingly.
1. What is a milligram?
A milligram (mg) is a metric unit of mass equal to one thousandth (1/1000) of a gram. It is often used in scientific contexts where precise measurements are required, such as in pharmacology and
chemistry. Milligrams are especially useful when dealing with very small quantities of substances or when high precision is necessary.
2. What is a kilogram?
A kilogram (kg) is a base unit of mass in the International System of Units (SI) and is equal to 1000 grams. It is commonly used in everyday life to measure body weight, food products, and other
quantities where the mass is more substantial. Kilograms provide a more manageable unit for larger masses compared to grams and milligrams.
3. Why convert milligrams to kilograms?
Converting milligrams to kilograms is useful when dealing with measurements in scientific research, manufacturing, and other fields where precise mass conversions are needed. For example,
converting a small quantity of a chemical in milligrams to kilograms can help in understanding its proportion in a larger batch or mixture. It simplifies calculations and helps in standardizing
measurements across different scales.
4. How precise is the conversion?
The conversion from milligrams to kilograms is exact and follows the mathematical principle of dividing by 1,000,000. Since 1 milligram is exactly 0.000001 kilograms, there is no margin of error
in the conversion process. This precision is crucial in fields where accurate mass measurement is essential.
5. Can this calculator handle large values?
Yes, the calculator can handle large values, but keep in mind that JavaScript has limitations with very large or very small numbers due to its floating-point precision. For extremely large or
small values, results may not be as precise as desired. However, for most practical purposes, the calculator provides accurate conversions.
6. How do I use this calculator?
To use the Milligrams to Kilograms Calculator, simply input the number of milligrams into the designated field and click the "Calculate" button. The calculator will perform the conversion using
the formula m(kg) = m(mg) / 1000000 and display the result in kilograms. You can then view the result in a table format. If you need to reset the input, click the "Clear" button.
7. What is the reverse conversion?
To convert kilograms to milligrams, you need to multiply the number of kilograms by 1,000,000. This is because there are 1,000,000 milligrams in one kilogram. The formula for this reverse
conversion is m(mg) = m(kg) * 1000000. This can be useful when converting larger mass quantities into smaller, more precise units.
8. Is there an error margin in the conversion?
The conversion between milligrams and kilograms is mathematically exact with no error margin, as it is a direct proportional relationship. The formula m(kg) = m(mg) / 1000000 provides precise
results for the conversion. However, be aware of potential rounding errors in display if very large or very small numbers are involved.
9. Can I use this calculator for other units?
This calculator is specifically designed for converting milligrams to kilograms. If you need to convert between other units of mass, such as grams to kilograms or ounces to pounds, you would need
a different tool or perform the appropriate conversion calculations based on the relevant conversion factors.
10. How can I clear the input fields?
To clear the input fields and remove the displayed results, click the "Clear" button provided. This will reset the input field to empty and hide the result section. This feature is useful when
you want to perform new calculations without manually clearing the input. | {"url":"https://calculatordna.com/milligrams-to-kilograms-calculator/","timestamp":"2024-11-10T06:27:06Z","content_type":"text/html","content_length":"87566","record_id":"<urn:uuid:c9dabc3d-444e-47e6-9faa-5952299de9da>","cc-path":"CC-MAIN-2024-46/segments/1730477028166.65/warc/CC-MAIN-20241110040813-20241110070813-00512.warc.gz"} |
How To Calculate Your EPF Dividend?
As everybody known, the Employees Provident Fund (EPF) declared 5.65% of dividend for the financial year 2009. In order to be smart and calculative in personal finance matters, understanding the
calculation of EPF dividend is an essential part of the learning process.
EPF’s dividend calculation is not that straight forward. I’m sure still many of you do not know how to calculate EPF dividend. Most of the people will take the total contribution of that particular
account and just multiply with 5.65% dividend. If you’re one of them, then this is a good chance for you to learn from mistake.
The following is my calculation of the dividend:
i. Take your balance brought forward from year 2008 and multiply that with the dividend rate which is 5.65% directly. So, whatever you have left in year 2008 will enjoy full interest payment this
Balance of the savings at 1st January (EPF Account 2)
= RM 6019.14 x 5.65%
= RM 340.08
ii. Then, for monthly contribution of the year 2009, there’ll have different formula for each month:
Month’s contribution (RM) x Dividend rate (%) x (12-n*)/12
n = The number corresponding with the month,
i.e. January = 1, February = 2, etc.
*Assume that we calculate the EPF Account II dividend now, so it’s 30% of your total contribution of particular month
January, RM 179.4 (30% of RM598) x 5.65% x 11 / 12
February, RM 313.5 (30% of RM1045) x 5.65% x 10 / 12
March, RM 207 (30% of RM690) x 5.65% x 9 / 12
After calculated, I found that the total dividend is RM 407.9 and it’s not RM408.65. So, there may be a variation of a few cent, most probably due to KWSP’s rounding error. Don’t forget that our
annual dividend on the EPF savings is calculated on daily basis. So, the another possibility would be the leap year and non-leap year case. Year 2010 only has 365 days. If not mistaken, the above
formula will not take into account of leap year or non-leap year case, so you won’t get the accurate result.
Oh ya…Do you know why I take EPF Account II as my calculation? It’s because there’s withdrawal made to invest in Public Bank unit trust on 19th Aug. And, any withdrawal made for investing unit trust
will affect EPF Account I dividend calculation as well.
One Response to How To Calculate Your EPF Dividend?
1. Dear David,
A difference between RM407.90 and RM408.65 is too LARGE for rounding errors, especially in this computing age.
Do not “believe” the formula given in the EPF website, i.e., Month’s contribution (RM) x Dividend rate (%) x (12-n*)/12 . This formula is obsolete.
See both these links (display them at your blog if you wish to) to understand how the EPF dividend is CORRECTLY and ACCURATELY calculated these days:
I’ve calculated for the statement you’ve displayed earlier. The figures just match PERFECTLY.
p/s: guess what, many (if not all) EPF officers don’t even know how to calculate the dividend. I’ve asked a couple of them from different branches, until I fed up and decided to decode the
formula myself.
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Collection of Solved Problems
Uknown Gas
Task number: 1282
The average kinetic energy of a monatomic gas of the amount of substance 1 mol is 2.5 kJ. If we raise the temperature by 300 K, the most probable velocity of the gas molecules will be 642 ms^-1.
Determine this gas and determine its initial temperature.
• Hint – determination of initial temperature
To determine the initial temperature T[1] of the unknown gas, we will use the formula for the total kinetic energy of one mole of gas \(\bar{E}_\mathrm{k}\).
• Hint – determination of the gas
To determine the gas, you need to determine its molar mass M[m]. To do this, you can use the relation for the most probable velocity v[p] of gas molecules.
• Numerical values
\(\bar{E}_\mathrm{k}=2.5\,\mathrm{kJ}=2500\,\mathrm{J}\) the total kinetic energy of one mole of monatomic gas
n = 1 mole the amount of substance of the monatomic gas
ΔT = 300 K the temperature difference
v[p] = 642 ms^−1 the most probable velocity of the gas molecules
T[1] = ? the initial temperature
From The Handbook of Chemistry and Physics:
R = 8.31 Jmol^−1K^−1 the molar gas constant
• Analysis
We will determine the initial temperature from the formula decribing the total kinetic energy of one mole of gas, from which we can see that the energy is directly proportional to the
thermodynamic temperature.
To determine the gas we will evaluate the molar mass from the relation for the most probable velocity of gas molecules. Then we will compare the value with the values listed in The Handbook of
Chemistry and Physics.
• Solution
The total kinetic energy of one mole \(\bar{E}_\mathrm{k}\) of any gas relates with the thermodynamic temperature T through this equation
where n is the amount of the substance of the gas and R is the molar gas constant.
This allows us to conduct a direct evaluation of the initial temperature T[1] of the unknown gas from the given values. It is true that:
The most probable velocity v[p] of an ideal gas molecules is given by the relation
where M[m] is the molar mass of the unknown gas and T[2] is the temperature after the gas is heated.
Thereof we can directly express the molar mass of the unknown gas
If we substitute the expression
for the temperature T[2], where ΔT is the temperature difference, we obtain
\[M_\mathrm{m}=\frac{2R(T_{\mathrm{1}}+\Delta T)}{v^{2}_{\mathrm{p}}}.\]
• Numerical solution
\[M_\mathrm{m}=\frac{2R(T_{\mathrm{1}}+\Delta T)}{v^{2}_{\mathrm{p}}}=\frac{2\cdot{8.31}\cdot(200+300)}{642^{2}}\,\mathrm{kg\,mol}^{-1}\] \[M_\mathrm{m}\dot{=}0.02\,\mathrm{kgmol}^{-1}=20\,\
Let us try to evaluate the most probable velocity at the initial temperature:
\[v_{\mathrm{p}}^, = \sqrt{\frac{2RT_1}{M_\mathrm{m}}} = \sqrt{\frac{2RT_1}{\frac{2R(T_{1}+\Delta T)}{v^{2}_{\mathrm{p}}}}} = v_{\mathrm{p}}\,\sqrt{\frac{T_1}{T_{1}+\Delta T}}\] \[v_{\mathrm{p}}
^, = 642\cdot\sqrt{\frac{200}{200+300}}\,\mathrm{m\,s^{-1}} \dot{=} 406\,\mathrm{m\,s^{-1}}\]
• Answer
The molar mass of the unknown gas is about 20 g mol^-1, which corresponds to the value for neon. The initial temperature is approximately 200 K. | {"url":"https://physicstasks.eu/1282/uknown-gas","timestamp":"2024-11-11T16:23:56Z","content_type":"text/html","content_length":"30422","record_id":"<urn:uuid:8f579f91-022d-48a4-bf5b-66319ef408e0>","cc-path":"CC-MAIN-2024-46/segments/1730477028235.99/warc/CC-MAIN-20241111155008-20241111185008-00790.warc.gz"} |
Part 3: A universal tiler
'The adventures first,' said the Gryphon in an impatient tone: 'explanations take such a dreadful time.'
Alice's Adventures in Wonderland
As mentioned in the section The forbidden tilings, tilings of the plane using regular polygons alone are restricted to triangles, squares, hexagons, dodecagons and the one unique octagon 4.8.8
uniform tiling. Adding rhombs to the prototile set allows a much richer variety of tilings including all the "forbidden" vertex figures. This leads to tilings including pentagons, heptagons,
nonagons, decagons and even larger polygons such as the 42-gon.
But are all the possible tilings involving rhombs and regular polygons restricted to the regular polygons that appear in the 21 possible regular polygon vertex figures?
In fact, amazingly, there is an algorithm to efficiently construct a translational unit for a periodic tiling of the plane from rhombs, triangles and any regular polygon!
The existence of an efficient universal tiling algorithm for regular polygons is surprising because many such general problems in tiling theory turn out to be undecidable or to require time-consuming
NP-complete algorithms. For example, in 1966 David Berger showed that no algorithm could exist to automatically construct tilings from even simple prototile sets of Wang tiles.
However, the situation for rhombs and regular polygons is different. Sampath Kannan and Danny Soroker (1992) and Richard Kenyon (1993) independently developed an efficient algorithm to decompose
certain finite simple polygons into rhombs (and more generally, parallelograms), and a modification of this algorithm can be used to construct periodic plane tilings also involving regular polygons
as explained in the following sections.
Here's an example involving hendecagons (11-sided regular polygons) constructed using a prototile set that also includes rhombs and triangles:
I've created a larger 1280x1024 image of this beautiful tiling.
You can download an SVG file for a translational unit for this tiling here.
In the next few sections I'll describe the algorithm and provide the mathematical background for a universal regular polygon tiler. | {"url":"http://gruze.org/tilings/universal","timestamp":"2024-11-09T04:14:46Z","content_type":"application/xhtml+xml","content_length":"20392","record_id":"<urn:uuid:89c102ea-378a-45ed-9bd6-e03dd87bde5e>","cc-path":"CC-MAIN-2024-46/segments/1730477028115.85/warc/CC-MAIN-20241109022607-20241109052607-00895.warc.gz"} |
SUBTOTAL Formula in Excel | How to use SUBTOTAL Formula in Excel?
SUBTOTAL Formula in Excel
In this article, we will learn about SUBTOTAL Formula in Excel. The function that returns a subtotal from a list or database can be defined as the Subtotal function.
Subtotal is a special function among other Excel functions because it can perform multiple operations, unlike other Excel functions that typically execute a single specific operation. Still, the
specialty of subtotal function is that it serves not only a subtotal calculation but also multiple arithmetics and logical operations depending on the function number.
Sometimes we need to find the subtotal of a category from a large set of data with multiple categories. The subtotal function will help us find the total category in that situation. Not only
subtotal, but we can also calculate the average, count and max, and many more.
Syntax of SUBTOTAL Formula
Function num refers to the type of mathematical operation we will perform on a specified range of data.
Ref1, Ref 2: It refers to the range of cells.
There are multiple numbers of function numbers available, but no problem; we do not need to remember all these function numbers because, while using the SUBTOTAL formula, excel will automatically
show you the list of function numbers available as below.
You must be wondering why there is two function number for the same function.
• 1 – 11 function numbers should use when we want to ignore the filtered-out cells and consider the manually hidden cells.
• When we want to ignore all the hidden cells, including the ones filtered out and manually hidden cells, we should use function numbers like 101-111.
We will see a few examples to understand how the subtotal function will work for different function numbers.
How to Use SUBTOTAL Formula in Excel?
SUBTOTAL Formula in Excel is very simple and easy. Let’s understand how to use the SUBTOTAL Formula in Excel with some examples.
Example #1
Consider a small table that has data from different categories as below.
We have a Product Category, Color category, and quantity if you observe the above table.
Apply the SUBTOTAL formula as below with function number 9.
After applying this formula, the result is shown below.
Apply the SUBTOTAL formula as below with function number 109.
After applying the SUBTOTAL Formula, the result is shown below.
We used function numbers 9 and 109 in the formula to perform SUM in two columns. C2:C9 is the range of data that we are performing calculations.
Now, the total sum is 385 for both formulas. Hide a few rows and observe the results for both formulas.
After applying the SUBTOTAL Formula, the result is shown below.
Rows from 50 to 60 are hidden; hence the results of function number 109 have changed to 255 because it does not consider the manually hidden data, whereas function number 9 total remains the same,
meaning it will consider the data even though we hide it manually.
Apply filter to the data and filter only one color; the Black Color is selected here.
After applying the SUBTOTAL Formula, the result is as shown below.
Apply the SUBTOTAL Formula as below with Function Number 109.
After applying the SUBTOTAL Formula, the result is as shown below.
If we observe the above screenshot, both the function numbers did not consider the quantity of filtered-out data. After you filter out the data and hide rows, both formulas will give the same results
that both formulas will not consider non-visible data.
After applying the SUBTOTAL Formula, the result is as shown below.
Apply the SUBTOTAL Formula as below with Function Number 109.
After applying this formula, the result is shown below.
Example #2
We have seen the SUM operation; now, we will perform the AVERAGE operation with the same data range. Go through the above table for the details of function numbers details. 1 and 101 are the function
numbers to perform average.
After applying the SUBTOTAL Formula, the result is as shown below.
Applying the SUBTOTAL formula again with function number 109 for the next cell.
After applying the formula, the result is shown below.
With the same SUBTOTAL Formula, we can calculate the average with the only change in Function Number.
After applying the SUBTOTAL Formula, the result is as shown below.
Applying the SUBTOTAL formula again with Function Number 101 for the next cell.
After applying the SUBTOTAL Formula, the result is shown below.
Hide a few rows; now observe the changes in the average. The average with function number 1 remains the same as still; it also considers the hidden quantity. The average with function number 101
changed because it ignores the manually hidden data. Like this, you can try the rest of the function numbers and check each functionality as the basic concept is the same; they also will work in the
same way.
Example #3
Subtotal has one more advantage: it will not consider any subtotals available in the range of data. We will see one example to understand better. Input formulas to perform SUM operation product-wise
like A, B, C, and D. It doesn’t need to be product-wise; you can do it color-wise also.
After applying the SUBTOTAL Formula, the result is as shown below.
Use the SUBTOTAL formula again to check the SUM of B.
After applying the SUBTOTAL Formula, the result is as shown below.
Use the SUBTOTAL formula again to check the SUM of C.
After applying the SUBTOTAL Formula, the result is as shown below.
Use the SUBTOTAL formula again to check the SUM of D.
After applying the SUBTOTAL Formula, the result is as shown below.
Use the SUBTOTAL formula again for the next cell.
After applying the SUBTOTAL Formula, the result is as shown below.
Use the SUBTOTAL formula again for the next cell.
After applying the SUBTOTAL formula, the formula displays the resulting value below.
If we observe the above screenshots, there is a sum by-product wise, and while performing the total sum, we included the category-wise totals into the total sum (cells from 36 to 48), but it will not
consider that subtotals (cells from 44 to 47). So, we can calculate the subtotals between the data range without affecting the total sum value.
I hope you understand how to use the subtotal and the use of it.
Things to Remember
• While giving the function number, do not give the number out of the function number range as Excel does not predefine it; we will get an error message. We should always provide the number between
1 to 11 and 101 to 111; otherwise, it will throw the error message #value.
• While using the division operation, remember no number should be divisible by zero. That means x/0 is an incorrect format.
• If you apply the subtotal for horizontal data from A1: F1, hiding any columns will not impact the subtotal.
• While applying subtotal, if any cells do not have data or non-numeric data, the function will ignore those cells.
• Use the function numbers by understanding the functionality and use; otherwise, you might not get the correct results you expect.
Recommended Articles
This has been a guide to the SUBTOTAL Formula in Excel. Here we discuss How to Use the SUBTOTAL Formula in Excel, practical examples, and a downloadable Excel template. You can also go through our
other suggested articles- | {"url":"https://www.educba.com/subtotal-formula-in-excel/","timestamp":"2024-11-06T14:34:35Z","content_type":"text/html","content_length":"375092","record_id":"<urn:uuid:52ab3946-a194-4762-a399-c451434797f2>","cc-path":"CC-MAIN-2024-46/segments/1730477027932.70/warc/CC-MAIN-20241106132104-20241106162104-00793.warc.gz"} |
6238.0 - Retirement and Retirement Intentions, Australia, July 2016 to June 2017
6238.0 - Retirement and Retirement Intentions, Australia, July 2016 to June 2017
ARCHIVED ISSUE Released at 11:30 AM (CANBERRA TIME) 18/12/2017
Page tools:
TECHNICAL NOTE DATA QUALITY
1 Since the estimates published in this publication are based on information obtained from occupants of a sample of households, they are subject to sampling variability. That is, they may differ
from those estimates that would have been produced if all households had been included in the survey. One measure of the likely difference is given by the standard error (SE), which indicates the
extent to which an estimate might have varied by chance because only a sample of households (or occupants) was included.
2 There are about two chances in three (67%) that a sample estimate will differ by less than one SE from the number that would have been obtained if all households had been included, and about 19
chances in 20 (95%) that the difference will be less than two SEs.
3 Another measure of the likely difference is the relative standard error (RSE), which is obtained by expressing the SE as a percentage of the estimate.
RSE% = (SE/estimate) x 100
4 RSEs for Retirement and Retirement Intentions estimates have been calculated using the Jackknife method of variance estimation. This process involves the calculation of 30 'replicate' estimates
based on 30 different subsamples of the original sample. The variability of estimates obtained from these subsamples is used to estimate the sample variability surrounding the main estimate.
5 The Excel spreadsheets in the Downloads tab contain all the tables produced for this release and the calculated RSEs for each of the estimates. The RSEs for estimates other than medians have been
calculated using the Jackknife method, and RSEs for the medians have been calculated using the Woodruff method.
6 In the tables in this publication, only estimates (numbers, percentages, means and medians) with RSEs less than 25% are considered sufficiently reliable for most purposes. However, estimates with
larger RSEs have been included. Estimates with an RSE in the range 25% to 50% should be used with caution while estimates with RSEs greater than 50% are considered too unreliable for general use.
All cells in the Excel spreadsheets with RSEs greater than 25% contain a comment indicating the size of the RSE. These cells can be identified by a red indicator in the corner of the cell. The
comment appears when the mouse pointer hovers over the cell.
7 RSEs are routinely presented as the measure of sampling error in this publication and related products. SEs can be calculated using the estimates (counts or means) and the corresponding RSEs.
8 An example of the calculation of the SE from an RSE follows. Datacube 3 shows that the estimated number of females aged 55–59 who retired from the labour force aged less than 55 years is 118,300
and the RSE for this estimate is 12.0%. The SE is:
SE of estimate
= (RSE / 100) x estimate
= 0.12 x 118,300
= 14,200 (rounded to the nearest 100)
9 Therefore, there are about two chances in three that the value that would have been produced if all households had been included in the survey will fall within the range 104,100 to 132,500 and
about 19 chances in 20 that the value will fall within the range 89,900 to 146,700. This example is illustrated in the following diagram.
PROPORTIONS AND PERCENTAGES 1247 10
Proportions and percentages formed from the ratio of two estimates are also subject to sampling errors. The size of the error depends on the accuracy of both the numerator and the denominator. A
formula to approximate the RSEs of proportions not provided in the spreadsheets is given below. This formula is only valid when x is a subset of y.
Considering Datacube 3, of the 1,943,800 females who were retired from labour force, 898,300 or 46.2% were aged less than 55 years at retirement. The RSE of 898,300 is 3.9% and the RSE for
1,943,800 is 2.0%. Applying the above formula, the RSE for the proportion of females who retired aged less than 55 years is:
Therefore, the SE for the proportion of females who retired from the labour force aged less than 55 years is 1.5 percentage points (= (46.2/100) x 3.3). Therefore, there are about two chances in
three that the proportion of females who retired from the labour force aged less than 55 years is between 44.7% and 47.7%, and 19 chances in 20 that the proportion is within the range 43.2% to
Published estimates may also be used to calculate the sum of, or difference between, two survey estimates (of numbers, means or percentages) where these are not provided in the spreadsheets. Such
estimates are also subject to sampling error.
The sampling error of the difference between two estimates depends on their SEs and the relationship (correlation) between them. An approximate SE of the difference between two estimates (x–y) may
be calculated by the following formula:
The sampling error of the sum of two estimates is calculated in a similar way. An approximate SE of the sum of two estimates (x+y) may be calculated by the following formula:
An example follows. From paragraph 8 the estimated number of females aged 55–59 who retired from the labour force aged less than 55 years is 118,300 and the SE is 14,200. From Datacube 3, the
estimate of females aged 60–64 who retired from the labour force aged less than 55 years is 124,700, the RSE is 7.9% and the SE is 9,900 (rounded to nearest 100). The estimate of females aged 55–64
who retired from the labour force aged less than 55 years is:
118,300 + 124,700 = 243,000
The SE of the estimate of females aged 55–64 who retired from the labour force aged less than 55 years is:
Therefore, there are about two chances in three that the value that would have been produced if all households had been included in the survey will fall within the range 225,700 to 260,300 and
about 19 chances in 20 that the value will fall within the range 208,400 to 277,600.
While these formulae will only be exact for sums of, or differences between, separate and uncorrelated characteristics or subpopulations, it is expected to provide a good approximation for all sums
or differences likely to be of interest in this publication.
SIGNIFICANCE TESTING 20
A statistical test for any comparisons between estimates can be performed to determine whether it is likely that there is a significant difference between two corresponding population
characteristics. The standard error of the difference between two corresponding estimates (x and y) can be calculated using the formula in paragraph 10. This standard error is then used to
calculate the following test statistic:
If the value of this test statistic is greater than 1.96 then there is evidence, with a 95% level of confidence, of a statistically significant difference in the two populations with respect to
that characteristic. Otherwise, it cannot be stated with confidence that there is a difference between the populations with respect to that characteristic.
The imprecision due sampling variability, which is measured by the SE, should not be confused with inaccuracies that may occur because of imperfections in reporting by respondents and recording by
interviewers, and errors made in coding and processing data. Inaccuracies of this kind are referred to as non-sampling error, and they occur in any enumeration, whether it be a full count or
sample. Every effort is made to reduce non-sampling error to a minimum by careful design of questionnaires, intensive training and supervision of interviewers, and efficient operating procedures.
Document Selection
These documents will be presented in a new window. | {"url":"https://www.abs.gov.au/ausstats/abs@.nsf/Previousproducts/6238.0Technical%20Note1July%202016%20to%20June%202017?opendocument&tabname=Notes&prodno=6238.0&issue=July%202016%20to%20June%202017&num=&view=","timestamp":"2024-11-15T01:12:13Z","content_type":"text/html","content_length":"24616","record_id":"<urn:uuid:24783df1-076e-49f4-9df9-4422b1178fb2>","cc-path":"CC-MAIN-2024-46/segments/1730477397531.96/warc/CC-MAIN-20241114225955-20241115015955-00844.warc.gz"} |
Linear Charge Density Unit Converter
Linear Charge Density Converter
Linear charge density is a measurement that characterizes electrical charge allocation per unit size. This parameter allows you to choose how the Electric field is spread over an object and how its
density changes with changes in size - increases or decreases.
Popular Unit Conversions Linear Charge Density
The most used and popular units of linear charge density conversions are presented for quick and free access.
Frequently Asked Questions
Linear charge density is significant in electrodynamics, especially in calculating electric fields and designing electrical devices. It is essential in allowing engineers and scientists to evaluate
charge disbandment in conductors, cables, antennas, and various systems, optimizing their performance and efficiency.
Additionally, linear charge density helps determine the strength of electric fields created by charged objects and analyzes interactions between charged particles. This converter is a handy tool for
engineers and physics and electrical engineering specialists, allowing them to effortlessly convert one-dimensional charge density importance from one unit of calculation to another.
To compute the one-dimensional charge density utilizing the converter, you can use this formula of linear charge density:
• Determine the total charge (Q) associated with the object in question;
• Measure the length (L);
• Divide Q by the length L to yield the one-dimensional charge density.
Leveraging the converter allows for effortless calculation and transformation of unbent charge viscosity values, enabling seamless analysis and comparisons across diverse scenarios. With this
formula, you will be able to get the results you want.
To use linear charge density calculator do next steps:
• Enter the values and set the unit of measurement to which you need to transform the value;
• Click the transform button;
• Reach the result in the new unit of measure.
Can the linear charge density converter work with different units?
Yes, the one-dimensional charge density converter keeps a wide range of linear density units, including Coulomb per meter (C/m), Coulomb per centimeter (C/cm), and various other commonly used units. | {"url":"https://oneconvert.org/unit-converters/linear-charge-density-converter","timestamp":"2024-11-06T04:11:07Z","content_type":"text/html","content_length":"154738","record_id":"<urn:uuid:22460046-8dc3-4839-8981-37c89ceac147>","cc-path":"CC-MAIN-2024-46/segments/1730477027909.44/warc/CC-MAIN-20241106034659-20241106064659-00767.warc.gz"} |
Re: Proc REG. How do I transform ln confidence limits back to original arithmetic units ?
I have performed OLS regression on ln transformed x and y variables, and re-transformed the results (ln_pred) to the original units using Duan's Smearing Estimate (my own code). The latter is done to
minimize re-transformation bias. I am not sure how to re-transform the ln confidence limits (Ln_Upper and Ln_Lower 95%) back to the original units--do I apply the smearing estimate or simply
exponentiate the ln confidence limit values ?
07-24-2023 01:44 PM | {"url":"https://communities.sas.com/t5/Statistical-Procedures/Proc-REG-How-do-I-transform-ln-confidence-limits-back-to/m-p/886131","timestamp":"2024-11-15T03:09:58Z","content_type":"text/html","content_length":"297969","record_id":"<urn:uuid:bcea94e4-a4b3-4637-b34e-5685b5c67deb>","cc-path":"CC-MAIN-2024-46/segments/1730477400050.97/warc/CC-MAIN-20241115021900-20241115051900-00796.warc.gz"} |
Classic Problems of Probability
Book description
"A great book, one that I will certainly add to my personal library."
--Paul J. Nahin, Professor Emeritus of Electrical Engineering, University of New Hampshire
Classic Problems of Probability presents a lively account of the most intriguing aspects of statistics. The book features a large collection of more than thirty classic probability problems which
have been carefully selected for their interesting history, the way they have shaped the field, and their counterintuitive nature.
From Cardano's 1564 Games of Chance to Jacob Bernoulli's 1713 Golden Theorem to Parrondo's 1996 Perplexing Paradox, the book clearly outlines the puzzles and problems of probability, interweaving the
discussion with rich historical detail and the story of how the mathematicians involved arrived at their solutions. Each problem is given an in-depth treatment, including detailed and rigorous
mathematical proofs as needed. Some of the fascinating topics discussed by the author include:
• Buffon's Needle problem and its ingenious treatment by Joseph Barbier, culminating into a discussion of invariance
• Various paradoxes raised by Joseph Bertrand
• Classic problems in decision theory, including Pascal's Wager, Kraitchik's Neckties, and Newcomb's problem
• The Bayesian paradigm and various philosophies of probability
• Coverage of both elementary and more complex problems, including the Chevalier de Méré problems, Fisher and the lady testing tea, the birthday problem and its various extensions, and the
Borel-Kolmogorov paradox
Classic Problems of Probability is an eye-opening, one-of-a-kind reference for researchers and professionals interested in the history of probability and the varied problem-solving strategies
employed throughout the ages. The book also serves as an insightful supplement for courses on mathematical probability and introductory probability and statistics at the undergraduate level.
Product information
• Title: Classic Problems of Probability
• Author(s):
• Release date: June 2012
• Publisher(s): Wiley
• ISBN: 9781118314333 | {"url":"https://www.oreilly.com/library/view/classic-problems-of/9781118314333/","timestamp":"2024-11-14T08:10:42Z","content_type":"text/html","content_length":"64265","record_id":"<urn:uuid:094196ef-3539-4fff-a88c-f2b24e45f54e>","cc-path":"CC-MAIN-2024-46/segments/1730477028545.2/warc/CC-MAIN-20241114062951-20241114092951-00256.warc.gz"} |
What is Integer.numberOfLeadingZeros() in Java?
The numberOfLeadingZeros() method of the Integer class is a static method used to get the total number of zero bits preceding the highest-orderleftmost one-bit in the two’s complement binary form of
the provided integer value.
Let’s understand this method with the help of examples.
Example 1
• int value: 123
• Binary Representation of 123: 1111011
The highest one bit of the 123 in its binary representation is at position 7, i.e., the leftmost one-bit.
Now, the number of leading zeros can be calculated using the formula below.
number-of-leading-zeros = 32
This is the index of the leftmost one-bit.
The number of leading zeros for 123 is 32 - 7 = 25.
Example 2
• int value: 12
• Binary Representation of 12: 1100
The highest one bit of the 12 in its binary representation is at position 4, i.e., the leftmost one-bit.
Now, the number of leading zeros can be calculated using the formula below.
number-of-leading-zeros = 32
This is the index of the leftmost one-bit.
The number of leading zeros for 12 is 32 - 4 = 28.
public static int numberOfLeadingZeros(int i)
• int i: The integer value whose number of leading zeros is to be computed.
Return value
• The function returns the number of leading zeros in the 32-bit representation of the integer.
If there are no one-bits in the binary representation of the integer value, then this method returns 32.
public class Main{
private static void numberOfLeadingZeros(int number){
System.out.println("Binary Representation of " + number + " - " + Integer.toBinaryString(number));
System.out.println("Number of Leading Zeros of " + number + " - "+ Integer.numberOfLeadingZeros(number));
public static void main(String[] args){ | {"url":"https://www.educative.io/answers/what-is-integernumberofleadingzeros-in-java","timestamp":"2024-11-11T11:48:47Z","content_type":"text/html","content_length":"371896","record_id":"<urn:uuid:11561031-07a2-4a56-830e-a3e36187f9cf>","cc-path":"CC-MAIN-2024-46/segments/1730477028228.41/warc/CC-MAIN-20241111091854-20241111121854-00848.warc.gz"} |
SOLVED: The last option on part d is: Standard normal was used because
Assignment Instructions/ Description
The last option on part d is: Standard normal was used because "o"1 and "o"2 are unknown Image transcription textInorganic phosphorous is a naturally occurring element in all plants and animals, with
concentrations increasing progressively up the food chain (fruit < vegetables < cereals < nuts < corpse). Geochemical surveys take soil samples to determine phosphorous
content (in ppm, parts per million). A high phosphorous content may or may not indicate an ancient burial site, food storage site, or even a garbage dump. Independent random samples from two regions
gave the following phosphorous measurements (in ppm).
Assume the distribution of phosphorous is mound-shaped and symmetric for these two regions.
Region I: x ; n, = 15
Region II: x,; n, - 14
(a) Use a calculator with mean and standard deviation keys to find x, and and s, (in ppm). (Round your answers to four decimal places.)
X 1
Use a calculator with mean and standard deviation keys to find x, and s2 (in ppm). (Round your answers to four decimal places.)
(b) Let , be the population mean for x, and let /, be the population mean for x, . Find an 80% confidence interval for , - /2. (Enter your answer in the form: lower limit to upper limit. Include the
word "to." Round your numerical values to one decimal
4(c) Explain what the confidence interval means in the context of this problem. Does the interval consist of numbers that are all positive? all negative? of different signs? At the 80% level of
confidence, is one region more interesting than the other from a
geochemical perspective?
Because the interval contains only positive numbers, we can say that Region I is more interesting than Region II.
Because the interval contains only negative numbers, we can say that Region II is more interesting than Region I.
| We can not make any conclusions using this confidence interval.
14) Because the interval contains both positive and negative numbers, we can not say that one region is more interesting than the other.
Hellojuli/() Which distribution (standard normal or Student's t) did you use? Why?
40O Student's t was used because o, and o, are known.
Standard normal was used because o, and o, are known.
1O Student's t was used because of and o, are unknown.
9:20 A
7/19/20... Show more | {"url":"https://www.skillsmatt.com/tutors-problem/35894/the-last-option-on-part-d-is-standard-normal-was-used-because-o1","timestamp":"2024-11-12T16:12:14Z","content_type":"text/html","content_length":"65882","record_id":"<urn:uuid:887bdba4-681a-4324-8616-d09e1157f4fe>","cc-path":"CC-MAIN-2024-46/segments/1730477028273.63/warc/CC-MAIN-20241112145015-20241112175015-00869.warc.gz"} |
Bisecting Commits when Failures Are Costly
This idea has been sitting in my drafts collection for several months now. About a month ago, during a 15-451 recitation, we went over a problem involving a lot of relevant ideas. I took this as a
divine sign that I must finish this blog post, so here I am.
Anyway, the inspiration for this post was the git bisect command. If you’re not familiar with it, its purpose is to use binary search to determine which commit introduced a bug. You tell it a known
“good” commit and a known “bad” commit, and then it interactively helps you narrow down the range with a binary search until you find the first “bad” commit.
Binary search is, of course, the optimal way to search over a range, assuming your cost model is the number of queries you need to make. However, we usually aren’t actually interested in the number
of queries we need to make; what interests us in the “real world” is minimizing the time taken to pick out the bad commit. In this model, is it possible to do better than binary search?
It turns out that, in theory, this is possible, although the circumstances under which such a situation would arise are not very common. The key observation is this: binary search does not
distinguish between the time taken to test “good” commits and “bad” commits, minimizing only the number of commits tested. What happens if the two do not have the same cost to test? Can we contrive a
scenario in which good commits cost much more to test than bad commits (or vice versa)? In such a situation, minimizing the total number of commits tested might not actually minimize the total time
that the search takes!
Is such a situation possible? Well, yes, but we need to make an additional assumption: the time to test a commit is a random variable. Otherwise, say good commits cost time $t_1$ and bad commits
$t_2$, where $t_1 < t_2$. We can run our test for distinguishing them, and if the time taken exceeds $t_1$, we know that the commit must be bad (and vice versa if $t_1 > t_2$). So if the testing time
for each type of commit is fixed, the overall test takes only $\min\{t_1,t_2\}$ time.
On the other hand, if the times are random variables $T_1$ and $T_2$, we cannot prematurely cut off the test if we want to be sure of the outcome. This is not so bad of an assumption: on a “real”
system, the time to run some process is generally not going to be fixed, although for simple tests the randomness is probably small enough to ignore. Still, this gives us a small practical foothold,
which is all I need to justify this blog post. Give a theorist an inch and he’ll take a mile.
In this setting, we can have many goals, but a reasonable one might be to minimize the expected time that the search takes. Although $T_1$ and $T_2$ are random variables, it may be the case that $E
[T_1] < E[T_2]$, and so the optimal strategy is not to minimize the total number of tests made.
In the rest of this blog post, I’ll first determine the cost of a binary search in this model. Then I’ll give another search strategy based on the classic “egg-dropping” problem, which can
potentially give better results under certain assumptions on $T_1$ and $T_2$. Since I’m a computer scientist and not just a mathematician, I’ll also be writing some accompanying Haskell code to try
out these different strategies. Just to get it out of the way, here are the types and primitive functions I’ll be using:
data Commit = Good | Bad deriving (Eq, Show)
data CommitSequence = Seq (Int, Int)
data Counter = Counter (Int, Int)
instance Show CommitSequence where
show (Seq (n, k)) = show list
good = take k $ repeat Good
bad = take (n - k) $ repeat Bad
list = good ++ bad
instance Show Counter where
show (Counter (good, bad)) =
"[" ++ show good ++ " good, " ++ show bad ++ " bad queries]"
empty_counter :: Counter
empty_counter = Counter (0, 0)
-- Build a sequence of n commits where commit #k is the last "good" one.
make_commits :: Int -> Int -> CommitSequence
make_commits n k = Seq (n, k)
-- Get the number of commits in a sequence.
number_commits :: CommitSequence -> Int
number_commits (Seq (n, _)) = n
-- Query a sequence at a given index, and update a counter depending on
-- whether the commit was good or bad.
query :: CommitSequence -> Int -> Counter -> (Counter, Commit)
query (Seq (_, k)) index (Counter (good_count, bad_count)) =
if index <= k
then (Counter (good_count + 1, bad_count), Good)
else (Counter (good_count, bad_count + 1), Bad)
Nothing fancy here: just a query function to make black-box queries about commits, and a Counter type to track how many good and bad commits we’ve tested so far. The code is also conveniently
available in a file, so you can follow along from the comfort of your own home.
The cost of a binary search
Let’s get a little bit more formal about this. Say we have $n$ objects, some of which are of type 1 and some of which are of type 2; we know that there is a “transition point” between the type 1 and
type 2 objects. If you want extra formality points, define a function $f : [n] \to \{1,2\}$, where we are guaranteed that $f(1) = 1$ and $f(n) = 2$. Furthermore, we know that $f(k) = 2 \implies f
(k+1) = 2$. We are allowed black-box queries to $f$, but the cost of a query is random: $E[\text{cost to query f(k)}] = \begin{cases} \mu_1 & f(k) = 1 \\ \mu_2 & f(k) = 2. \end{cases}$ We want to
determine the index $k$ such that $f(k) = 1$ and $f(k+1) = 2$.
With binary search, we keep an interval $[s,t]$ where we know $f(s) = 1$ and $f(t) = 2$, and we keep on testing the midpoint of the interval. At each iteration of the algorithm, the interval halves
in size, so we get a total of $\log n$ rounds.
But what is the expected cost of the queries across these $\log n$ rounds? It ought to be $\frac{\mu_1+\mu_2}{2} \log n.$ To see this, note that if we assume that any such function $f$ is equally
likely, then in any given round, it is equally likely that the current query will result in $1$ or $2$. (This is because half of the possible functions $f$ that are consistent with the interval we
know now will evaluate to $1$ at the midpoint, and half will evaluate to $2$.) We’re essentially taking an average over $\mu_1$ and $\mu_2$; if we truly don’t know anything about the relationship
between $\mu_1$ and $\mu_2$, we can’t really do any better.
Here’s an implementation of the binary search idea, where we assume we have some black-box query function that tells us whether a commit is Good or Bad and conveniently updates a counter for us:
bisect_binary :: CommitSequence -> (Counter, Int)
bisect_binary commits =
search empty_counter (n - 1) 0
n = number_commits commits
search counts upper lower =
if upper - lower <= 1
then (counts, lower)
else case commit of
Good -> search counts' upper mid
Bad -> search counts' mid lower
mid = (upper + lower) `div` 2
(counts', commit) = query commits mid counts
Now we can play around with this on a few different commit sequences of length, say, $n = 1000000$:
*Main> n = 1000000
*Main> bisect_binary $ make_commits n 1
([1 good, 19 bad queries],1)
*Main> bisect_binary $ make_commits n 15150
([6 good, 13 bad queries],15150)
*Main> bisect_binary $ make_commits n 789000 -- Why was 6 afraid of 7?
([11 good, 9 bad queries],789000)
*Main> bisect_binary $ make_commits n 999998
([20 good, 0 bad queries],999998)
Just to check: the index returned by bisect_binary always matches the correct answer. In each case, the exact number of good and bad commits tested differs, but they always add up to 19 (if we get
really lucky) or 20. Indeed, $\log 1000000 \approx 19.932$.
Aside: a recipe for egg drop soup
We’ve seen that binary search, which does not discriminate between type 1 and type 2 objects, will make $\log n$ queries; in expectation, half of the queries will be to each type. To motivate a
better strategy, let’s revisit the classic egg-dropping problem:
You are given two eggs, and you want to find the maximum floor of an $n$-story building from which you can drop an egg without breaking it. (For simplicity, assume that for any floor, either
every egg dropped from that floor will break or none will.) Once you break an egg, it obviously cannot be reused, but if an egg is still intact, you can use it again. What’s your strategy for
making as few guesses as possible?
This time, we cannot do a binary search over all $n$ floors. Otherwise, if we get unlucky and the first two drops break the eggs, we are out of luck. The most straightforward strategy is to just
start from the first floor and work our way up until we break an egg; this takes at most $n$ guesses. But this doesn’t take advantage of the fact that we have two eggs; can we do better by allowing
ourselves to break one?
If you haven’t seen this problem before, it’s a fun one, and you might want to spend a few minutes thinking about it before reading on. Below, I’ll give a solution that takes at most $2\sqrt{n}$
1. Make evenly-spaced guesses at floors $\sqrt{n}$, $2\sqrt{n}$, $3\sqrt{n}$, and so on until you break an egg at some floor $(i+1) \sqrt{n}$.
2. Now you know that the right floor is between $i\sqrt{n}$ and $(i+1)\sqrt{n}$, so you can find the correct floor by starting from $i\sqrt{n}+1$ and working your way up every floor to $(i+1)\sqrt
Part 1 and part 2 of this algorithm both take $\sqrt{n}$ guesses at most, so overall we will only use $2\sqrt{n}$ guesses.
In fact, this same strategy gives us an algorithm using at most $kn^{1/k}$ guesses for any number $k \ge 2$ of eggs:
• If $k = 2$, just use the algorithm above.
• If $k > 2$, make $n^{1/k}$ evenly spaced guesses. Once you break an egg, repeat the algorithm with $k-1$ eggs on the interval of size $n^{(k-1)/k} = n^{\alpha}$. This takes at most $n^{1/k} +
(k-1)n^{\alpha/(k-1)} = kn^{1/k}$ guesses, as promised.
Better than binary: bisection using egg drops
Don’t put all your eggs into one basket!
Ancient Chinese proverb
The correspondence between the egg drop problem and the commit bisection problem should be obvious. Recall that, on average, binary search used $(\log n)/2$ queries of the “slow” type. What if we use
the egg-dropping strategy, allowing ourselves $k = c \log n$ slow queries, for some $c < 1/2$? We get $k n^{1/k} = (c \log n) n^{1/(c \log n)} = (c \log n) \cdot 2^{1/c}$ guesses in total.
Concretely, suppose $\mu_1 < \mu_2$, so we allow ourselves $c \log n$ queries of type 2. (This is interchangeable with the $\mu_1 > \mu_2$ case, just by flipping the interval around.) Then this
strategy gives $\mu_2 c \log n + \mu_1 \cdot (2^{1/c} - 1)(c \log n)$ expected cost to compute. When is this better than binary? Note that $\mu_1$ and $\mu_2$ are constant with respect to $n$; it is
reasonable to suppose that the time of each query (i.e. the time to test a commit) does not depend on the length of the list (i.e. the number of commits there are). So we can define the “slowdown
factor” $\gamma = \mu_2 / \mu_1 > 1$, and egg dropping will be superior, for some suitably chosen $c$, whenever $\gamma \mu_1 c \log n + \mu_1 (2^{1/c} - 1)(c \log n) < \mu_1 \cdot \frac{1 + \gamma}
{2} \log n,$ which is the same as saying $g(c, \gamma) = 1 + \gamma - 2c \cdot (\gamma + 2^{1/c} - 1) > 0.$ For example, $c = 1/3$ and $\gamma = 12$ will do the trick. We can plot the areas where $g
(c, \gamma) \ge 0$ using everyone’s favorite proprietary computing system, Mathematica:
f[c_, g_] := 1 + g - 2 c (g + 2^(1/c) - 1)
ContourPlot[f[c, g], {c, 1/3, 1/2}, {g, 0, 25},
PlotLegends -> Automatic, FrameLabel -> {c, g}]
(I’ve taken the liberty of replacing “γ” with “g” above, since my code font apparently takes issue with Greek letters. I then had to rename the function from $g$ to $f$, but you get the point.) This
isn’t very great, since $g(c, \gamma) > 0$ occurs only when $\gamma > 10$, or when the costly query is more than ten times the cost of the cheaper one:
Plot of $g(c, \gamma)$ for $1/3 \le c \le 1/2$ and $0 \le \gamma \le 25$.
This is not very likely to happen in reality, but it makes for a pretty plot. Here’s an implementation of the strategy for an arbitrary $c$:
bisect_egg :: Float -> CommitSequence -> (Counter, Int)
bisect_egg c commits =
egg_drop empty_counter k (n - 1) 0
n = number_commits commits
k = floor $ c * (logBase 2 $ fromIntegral n)
-- Linear search over for first bad commit in range [lower, upper),
-- with spacing between guesses
search counts spacing upper lower
| lower == upper - 1 = (counts, lower + 1)
| lower > upper - 1 = (counts, upper) -- Not found with this spacing
| otherwise =
case query commits (lower + 1) counts of
(counts', Good) -> search counts' spacing upper $ lower + spacing
(counts', Bad) -> (counts', lower + 1)
egg_drop :: Counter -> Int -> Int -> Int -> (Counter, Int)
egg_drop counts 0 upper lower = (counts, lower)
egg_drop counts k upper lower =
egg_drop counts' (k - 1) upper' lower'
n' = fromIntegral $ upper - lower + 1
spacing = floor $ n' ** (1 - (1 / fromIntegral k))
(counts', upper') = search counts spacing upper lower
lower' = upper' - spacing
Running them on our previous examples:
*Main> n = 1000000
*Main> c = 1/3
*Main> bisect_egg c $ make_commits n 1
([40 good, 2 bad queries],1)
*Main> bisect_egg c $ make_commits n 15150
([21 good, 6 bad queries],15150)
*Main> bisect_egg c $ make_commits n 789000
([53 good, 4 bad queries],789000)
*Main> bisect_egg c $ make_commits n 999998
([59 good, 0 bad queries],999998)
Notice that we may do many more “good” tests than binary search, but crucially, the number of “bad” tests that we run never exceeds $(\log 1000000)/3 \approx 6.644$. This tells us that, if $\gamma$
is high enough, we should be beating binary search on average.
Corrigendum: an earlier version of this post had an incorrect implementation of the egg dropping routine; I forgot to update the length of the interval, which led to many more iterations than needed.
I’m still not 100% sure that the indexing is correct, but it is much better now. | {"url":"https://www.ericzheng.org/thoughts/bisect.html","timestamp":"2024-11-05T15:16:07Z","content_type":"application/xhtml+xml","content_length":"164364","record_id":"<urn:uuid:169d8fe8-c5fb-4b13-85e7-7eab867364c9>","cc-path":"CC-MAIN-2024-46/segments/1730477027884.62/warc/CC-MAIN-20241105145721-20241105175721-00357.warc.gz"} |
天文与空间科学学院 学术报告
Abstract: The stable and unstable manifolds are invariant phase space surfaces
which physically describe the inflows and outflows of material close to the
unstable Lagrange equilibrium points in the restricted three body problem,
or the unstable Lagrange points at the end of the bar in rotating barred
galaxies. After a short general introduction to the topic, we will show how
the manifold inflows and outflows are of use in the interpretation of several
phenomena in a variety of scales of interest in dynamical astronomy. Examples
are i) the L4-L5 asymmetry of Jupiter Trojan asteroids, ii) the eccentricity
growth of navigation satellites under the sole action of natural forces, iii)
spiral structure in barred galaxies, and iv) galactic tidal streams.
Bio: Christos Efthymiopoulos is currently Associate Professor
at the Department of Mathematics Tullio Levi-Civita of the University of Padova.
Before moving to Italy, he served for 17 years (2003-2019) as research staff at
the Research Center for Astronomy and Applied Mathematics of the Academy of
Athens, in Greece, being Research Director since 2011. In the last three years he
has served as President of the Commission A4 (Celestial Mechanics and Dynamical
Astronomy) of the International Astronomical Union. His Research focuses on
applications of nonlinear dynamical systems in Dynamical Astronomy. He is
author of more than 80 articles in scientific journals, addressing a variety
of topics of current interest in dynamical astronomy, such as the dynamics
of Trojan asteroids, classical and general relativistic perturbation theory,
orbital and attitude dynamics of natural or artificial satellites, spiral
structure, and the tidal interactions of galaxies. | {"url":"https://astronomy.nju.edu.cn/xshd/xsbg/20240617/i268753.html","timestamp":"2024-11-05T16:29:18Z","content_type":"text/html","content_length":"12031","record_id":"<urn:uuid:26d1e800-d8a8-4adc-911c-42867857b8ce>","cc-path":"CC-MAIN-2024-46/segments/1730477027884.62/warc/CC-MAIN-20241105145721-20241105175721-00306.warc.gz"} |
MP Board Class 11th Physics Important Questions Chapter 13 Kinetic Theory
Students get through the MP Board Class 11th Physics Important Questions Chapter 13 Kinetic Theory which are most likely to be asked in the exam.
MP Board Class 11th Physics Important Questions Chapter 13 Kinetic Theory
Kinetic Theory Class 11 Important Questions Very Short Answer Type
Question 1.
State Boyle’s law.
According to Boyle’s law “At constant temperature the volume of a given mass of a gas inversely proportional to its pressure, i.e.,
V∝\( \frac{1}{\mathrm{P}} \)
V=K.\( \frac{1}{\mathrm{P}} \)
PV = K (constant).
Question 2.
Write the condition at which Boyle’s law function.
Boyle’s law function at low pressure and high temperature.
Question 3.
Write down Charle’s law.
According to Charle’s law “ The volume of given mass of a gas at constant pressure increases by \( \frac{1}{273} \) of its volume at 0°C, for each 1°C rise in temperature.
Question 4.
Prove that with help of Charle’s law that at -273°C, volume of a gas is zero.
According to Charle’s law.
V = V[0]( 1 + \( \frac{1}{273} \) t)
At t =-273° C
V = V[0](1+ \( \frac{-273}{273} \) )
V = V[0] (1-1) = 0
i.e., at -273°C volume of a gas is zero.
Question 5.
What are ideal gases?
The gases which obey Boyle’s law and Charle’s law completely are called ideal gases.
Question 6.
State Avogadro’s law.
According to Avogadro’s law, “Equal volume of all the gases under similar con-ditions of temperature and pressure contain the same number of molecules.”
Question 7.
Write down Dalton’s law of partial pressure.
Dalton’s law of partial pressure states that the total pressure exerted by a mixture of gases which do not interact in any way is equal to the sum of their individual pressures.
Question 8.
What is Grahm’ law of diffusion?
According to Grahm’s law of diffusion, “The rate of diffusion of a gas is directly proportional to the square root of its density.”
Question 9.
What is meant by absolute scale of temperature?
The scientist Kelvin had developed a temperature scale by considering – 273°C as zero of the scale. This scale is called absolute scale of temperature. The value of its each division is 1°C. It is
also called as Kelvin scale.
Question 10.
What do you mean by absolute zero?
The absolute zero is that temperature at which the volume of a gas becomes zero.
Question 11.
If a gas is suddenly compressed, then its temperature increased. Why?
The temperature of a gas is directly proportional to its mean kinetic energy. When a gas is suddenly compressed, then the mean kinetic energy of gas is increased. Therefore, the temperature of gas is
increased suddenly.
Question 12.
What is Boltzmann constant? Write its value.
The ratio of universal gas constant (R) and Avogadro’s number (N) is known as Boltzmann’s constant
K= \( \frac{\mathrm{R}}{\mathrm{N}} \)
Its value is 1.38 x 10^-23 joule/kelvin.
Question 13.
Write down relation between universal gas constant and specific gas constant.
The relation between universal gas constant and specific gas constant is given below:
Specific gas constant r = \( \frac{\text { Universal gas constant }(\mathrm{R})}{\text { Molecular weight of } operatorname{gas}(\mathrm{M})} \) .
Question 14.
Write the ideal gas equation.
PV = nRT.
Question 15.
Find out dimensional formula for R.
From PV = RT
R = \( \frac{\mathrm{PV}}{\mathrm{T}} \)
[R] = \( \frac{\left[\mathrm{ML}^{-1} \mathrm{~T}^{-2}\right]\left[\mathrm{L}^{3}\right]}{[\theta]} \)
= \( \left[\mathrm{ML}^{2} \mathrm{~T}^{-2} \theta^{-1}\right] \) .
Question 16.
At equal temperature (T) and pressure (P), two gases of same volume (V) are mixed together. If the temperature and volume of the mixure is T and V, then what will be its pressure?
According to Dalton’s law of partial pressure it will be 2P.
Question 17.
At NTP, 1cm^3 H[2 ]and 1cm^3 O[2] are given. In which gas number of molecules will be more and why?
According to Avogadro’s law number of molecules of 1 cm^3 H[2] and 1 cm^3 O[2] will be equal.
Question 18.
Oxygen and hydrogen gases are filled in a porous pot in equal amount, which one will diffuse soon?
By the kinetic theory of gas,
ν rms ∝\( \frac{1}{\sqrt{\mathrm{M}}} \)
Since M[H] < M[0] hence ν[H] > ν[0]
Thus, the hydrogen will diffuse soon.
Kinetic Theory Class 11 Important Questions Short Answer Type
Question 1.
What is gas equation? Establish it.
For a gas equation prove that PV = RT.
Establish ideal gas equation.
Let the initial pressure of one gram mole of a gas of molecular weight M is P[1] its volume is V[1] and absolute temperature is T[1]. After a thermodynamic process the pressure of gas become P[2],
volume and temperature becomes V[2 ]and T[2] respectively. The change occurs in the state of gas can be considered to be a combination of two processes.
(i) Keeping the temperature T[1] of gas constant, its pressure is varied from P[1] to P[2], so that as shown in fig. the volume of gas becomes V’. Hence, as per Boyle’s law :
P [1]V[1] = P[2]V’ …(1)
(ii) Keeping the pressure P2 constant, the temperature of gas is varied from T[1] to T[2] so that its volume changes from V’ to V[2] as shown in fig. then according to Charle’s law :
\( \frac{\mathrm{V}^{\prime}}{\mathrm{T}_{\mathrm{l}}} \) = \( \frac{\mathrm{V}_{2}}{\mathrm{~T}_{2}} \)
\( V’ = \frac{\mathrm{V}_{2} \mathrm{~T}_{1}}{\mathrm{~T}_{2}} \) ………. (2)
Substituting the value V’ from eqn. (2) in eqn.(1)
P[1]V [1] = \( \frac{\mathrm{P}_{2} \mathrm{~V}_{2} \mathrm{~T}_{1}}{\mathrm{~T}_{2}} \)
P[1]V [1] = \( \frac{P_{2} V_{2}}{T_{2}} \)
\( \frac{\mathrm{PV}}{\mathrm{T}} \) = Constant …….. (3)
The value of this constant is same (= 8.314 joule per mole-kelvin) for 1 mole of all the gases. It is denoted by R. Hence,
\( \frac{\mathrm{PV}}{\mathrm{T}} \) = R
PV = RT …(4)
Where, R is called universal gas constant. Eqn. (4) is called gas equation for one mole of gas.
Question 2.
Write down postulates of kinetic theory of gases.
The postulates of kinetic theory of gas are given below :
• Every gas is composed of minute particles called molecules.
• The size of these molecules is negligible as compared to their intermolecular dis¬tance. t
• The molecules of gas are spherical uniform in all aspects rigid and perfectly elastic.
• The molecules are always in state of random motion and move in all possible direction with all possible velocities.
• Due to their continuous motion, these molecules collide with each other and also with the walls of container during their random motion. Due to these collision, there is no change in density of
gas, i.e., the number of molecules per unit volume of the gas remains unchanged.
• After collision the direction of velocity of molecules changes. The collision is instantaneous, i.e., the time taken in collision is negligible as compared with the time taken between two
consecutive collisions.
• The collisions are perfectly elastic, i.e., kinetic energy of molecules remains con-served.
• Between the two successive collisions a molecule travels in a straight line with uniform velocity. The distance covered by a molecule between two consecutive collisions is called ‘Free path’, The
average distance travelled by a molecule between successive collisions is called mean-free-path.
• The molecules of gas collide with the walls of container and hence they exert the force on the walls. The force acting per unit area on the wall is called pressure of gas.
• The mass of molecules of gas is negligible and velocity is high. Therefore, there is no effect of gravity on the motion of molecules.
Question 3.
Prove that: P = \( \frac{1}{3} \) ρc^-2
We know P = \( \frac{1}{3} \) \( \frac{m N c^{2}}{V} \)
Here mN = M
∴ P = \( \frac{1}{3} \) \( \frac{M \bar{c}^{2}}{V} \)
But \( \frac{M}{V} \) = ρ (density)
∴ P = \( \frac{1}{3} \) ρc^-2
Question 4.
On the basis of kinetic theory of gases, prove that p = \( \frac{2}{3} \) E where symbols have their meanings,
According to kinetic theory of gases, the pressure exerted by the gas is gi ven by
P = \( \frac{m N \bar{c}^{2}}{3 V} \) ………….. (1)
Where, N is the number of molecules, c^ -2 mean square velocity and V is volume. Let m be the mass of one molecule, then the mass of gas be mN
∴Density of gas ρ = \(\frac{m N}{V}\) …………. (2)
From eqns. (1) and (2),
P = \( \frac{1}{3} \) ρc^-2
= \( \frac{2}{3} \) .\( \frac{1}{2} \) ρc^-2 = \( \frac{2}{3} \) E
Where, E is kinetic energy of per unit volume of gas.
Question 5.
Explain kinetic energy of a gas on the basis of kinetic theory of gases.
Prove that : P = \( \frac{2}{3} \) E
Kinetic energy of gas: Let the mass of one molecule of gas is m and number of molecules is N. Hence, the mass of gas will be mN.
∴Density of gas ρ = \(\frac{m N}{V}\) …………. (1)
According to Kinetic theory of gas the pressure exerted by gas is
P= \( \frac{m N \bar{c}^{2}}{3 V} \) ……………. (2)
From equns. (1) and (2)
P = \( \frac{1}{3} \) ρc^-2 = \( \frac{2}{3} \) . \( \frac{1}{3} \) ρc^-2
P= \( \frac{2}{3} \) E …….. (3)
Where, E is kinetic energy of a gas per unit volume. From eqn. (3),
E = \( \frac{3}{2} \) P
P = \( \frac{2}{3} \) E.
Question 6.
On the basis of kinetic theory of gases prove that the mean kinetic energy of molecules of gas is directly proportional to the absolute temperature of gas.
Prove that the mean kinetic energy of gas is E = \( \frac{3}{2} \) KT.
According to kinetic theory of gases, the pressure exerted by the gas is given by :
P = \( \frac{m N \bar{c}^{2}}{3 V} \)
Where, m is mass of one molecule of gas, N is the number of molecules in the volume V,c^-2 is mean square velocity of gas and V is volume. In the above equation if V is volume of one mole of gas,
then N be the Avogadro’s number, mN be the gram molecular weight (M) hence from the above equation :
PV=\( \frac{1}{3} \)mNc^-2
= \( \frac{1}{3} \)Mc^-2 …(1)
But by the gas equation,
PV = RT ……………. (2)
From eqns. (1) and (2), we get
\( \frac{1}{3} \)Mc^-2 =RT
c^-2 = \( \frac{3 R T}{M} \) …… (3)
From eqn. (3), it is clear that c^-2 ∝T
Thus, the absolute temperature of a gas is directly proportional to the mean square velocity of the molecules of a gas.
Again from eqn. (3),
Where, K =\( \frac{R}{N} \) is called Boltzmann’s constant. Its value is 1.38 × 10^-23 joule per kelvin. This is formula of mean kinetic energy of molecules. From this equation, it is clear that
Question 7.
Explain absolute zero on basis of kinetic theory of gas.
We know E =\( \frac{3}{2} \) KT
, If T=0, then E = 0.
Therefore, absolute zero temperature is that temperature at which average kinetic energy of the gaseous molecules is zero.
Question 8.
Prove that\( \overline{\boldsymbol{c}} \) =\( \sqrt{\frac{3 K T}{M}} \) where K is Boltzmann’s constant, T is absolute temperature and M is mass of one molecule of gas.
According to kinetic theory of gases,
\( \overline{\boldsymbol{c}} \) = \( \sqrt{\frac{3 R T}{M}} \) =\( \sqrt{\frac{3 R T}{mN}} \), [∵M= mn]
= \( \sqrt{\frac{3 K T}{m}} \), [∵ K = \( \frac{R}{N} \) ]
\( \overline{\boldsymbol{c}} \) ∝ \( \sqrt{T} \)
i.e., r.m.s. value of velocity of gaseous molecule is directly proportional to square root of absolute temperature.
Question 9.
If the number of molecules in a box is halved, then what will be the effect on its pressure?
According to kinetic theory of gas, the pressure of gas :
P= \(\frac{1}{3} \frac{m N \bar{c}^{2}}{V}\)
If the number of molecules is halved ,then the pressure of gas :
P’=\(\frac{1}{3} \frac{m}{V} \frac{N}{2} \bar{c}^{2}\)
P’ = \( \frac{1}{2}\left[\frac{m N \bar{c}^{2}}{3 V}\right]\)
\(\frac{P^{\prime}}{P}\) = \(\frac{1}{2}\)
P’ = \(\frac{1}{2}\)P
Thus, the pressure will Also be halved.
Question 10.
Derive Boyle’s law on basis of kinetic theory of gases.
Derivation of Boyle’s law: According to kinetic theory of gases the pressure exerted by gas is :
P = \(\frac{1}{3}\) \( \frac{m N \bar{c}^{2}}{V} \)
PV = \( \frac{1}{3} \) \( m N \bar{c}^{2} \)
= \( \frac{1}{3} \) . \( m N \bar{c}^{2} \)
= \( \frac{2}{3} \).N .\(\frac{1}{2} \) \( m \bar{c}^{2}\)
= \( \frac{2}{3} \) NE, [ ∵ E = \( \frac{1}{2} \) \( m \bar{c}^{2}\)]
= \( \frac{2}{3} \) NKT, (∵E = KT ) …….. (1)
If the mass and absolute temperature of gas are constant,then
= \( \frac{2}{3} \) NKT be also constant
∵ PV = constant
This is Boyle’s law.
Question 11.
Derive Ctiarie’s law on basis of kinetic theory of gases.
Derivation of Charle’s law : According to kinetic theory of gases, the pressurue exertd by gas is
P = \( \frac{1}{3} \) \( \frac{m N \bar{c}^{2}}{V} \) …….. (1)
PV = \( \frac{1}{3}\) \( m N \bar{c}^{2} \)
PV = \( \frac{1}{3} \) \( M \bar{c}^{2} \) ……… (2)
Where, M = mN is mass of gas which remains constant .
If the pressure of gas remains constant,then from the above equation.
V= \( \frac{M}{3 P} \bar{c}^{2} \)
V ∝\( \bar{c}^{2} \)
Since, \( \bar{c}^{2} \) ∝T therefore
V ∝T
This is Charle’s law.
Question 12.
Derive Dalton’s law of partial pressure on basis of kinetic theory of gases.
Dalton’s law of partial pressure: According to Dalton’s law of partial pressure, the total pressure exerted by a mixture of gases which do not interact in any way is equal to the sum of their
individual pressure.
Let us consider about a vessel of volume V[1] having number of gases mixed together.
Let the gases one, two, three ………….contain N[1] molecules of mass muN[2] molecules of mass
m[2], N[3] molecules of mass m[3] ……………respectively. Let their root mean square velocities are
\( \bar{c}_{1}, \bar{c}_{2}, \bar{c}_{3}\)…………… respectively.
Then the pressure due to first gas
P[1] = \( \frac{1}{3} \) m[1]N[1]c^-2
Similarly, the pressure due to second, third,………….gases are
If all the gases are mixed at same temperature, then mean kinetic energy óf the molcules of each gas will be the same i.e.,
This is Delton’s law of partial pressure.
Question 13.
A vessel is filled up with mixture of two different gases. Explain with reason :
(i) Is Average kinetic energy per unit molecule are same?
(ii) Is root-mean-square value of velocity are same?
(iii) Is pressure same?
(i) Yes, as E = \(\frac{3}{2} \)KT, it depends on absolute temperature only.
(ii) No, because \( \bar{c} \) = \( \sqrt{\frac{3 R T}{M}} \) and it depends on molecular weight M and temperature T.
(iii) Nothing cannot be said about pressure as mass is not known.
Question 14.
Write the law of equipartition of energy.
According to this law, for any dynamical system in thermal equilibrium, the total energy distributed equally amongst all the degree of freedom, and the energy associated
with each molecule per degree of freedom is\( \frac{1}{2} \) KT, where K is Boltzmann’s constant and T is temperature of the system.
Question 15.
Explain degree of freedom.
The number of degrees of freedom of a dynamical system is defined as the total number of coordinates or independent quantities required to describe completely the position and configuration of the
For Example:
(i) When a particle moves along a straight line, say along X-axis, its position can be specified by its displacement along the X-axis. Therefore, such a particle has one-translational degree of
(ii) If the particle is moving in a plane, its position at any instant can be determined by knowing the displacements of the particle along the X-axis and Y-axis. Therefore it has two- translational
degrees of freedom.
(iii) If the particle is moving in space, its position at any instant can be determined by knowing the displacement of the particle along X-axis, Y-axis and Z-axis. Therefore, such a particle has
three-translational degrees of freedom.
For example, a bob of an oscillating simple pendulum has one degree of freedom, an insect moving on a horizontal floor has two degrees of freedom and a buzzing bee has three degrees of freedom.
Question 16.
Find out ratio of specific heats for monoatomic gas.
For monoatomic gas like He or Ar, whose degree of freedom is 3 but according to law of equipartition of energy.
Energy =3 × \( \frac{1}{2} \) KT = \( \frac{3}{2} \)kT
But energy associated with one mole of gas
U =\( \frac{3}{2} \) nkT
Where n is number of gas molecule for 1 mole of gas.
But Boltzmann’s constant k = \( \frac{R}{n} \)
nk = R
Putting value in eqn. (1), we get
U = \( \frac{3}{2} \)RT
But C[v] = \( \frac{d U}{d T} \) = \( \frac{d}{d T} \) \( \left(\frac{3}{2} R T\right) \)
C[v] =\( \frac{3}{2} \) R
Question 17.
Find out ratio of specific heat for diatomic gas.
For diatomic gas like Hydrogen, Oxygen etc. degree of freedom is 5.
Therefore energy associated with one mole of gas is
U = \(\frac{5}{2} \)nkT
U= \(\frac{5}{2} \)RT
But, nk =R
since C[v] = \( \frac{d U}{d T} \) = \( \frac{d}{d T} \) \( \left(\frac{5}{2} R T\right) \) =\(\frac{5}{2} \) R
∴ From C[p] – C[v] = R
C[p] = R +C[v]
C[p] = R +\(\frac{5}{2} \) R = \(\frac{7}{2} \) R
∴ γ = \( \frac{C_{P}}{C_{V}} \) = \( \frac{\frac{7}{2} R}{\frac{5}{2} R} \) = \(/frac{7}{5} \) = 1.40
Question 18.
Find out ratio of specific heat for triatomic gas.
For triatomic gas like C02, H2S degree of freedom is 6. Energy associated with one mole of gas is
U =\(\frac{6}{2} \)nKT
U = \(\frac{6}{2} \)RT = 3RT
But Cv=\( \frac{d U}{d T} \) = \( \frac{d}{d T} \) (3RT) = 3R
∴ From C[p]-C[v] = R
C[p] =R + C[v] =R + 3 R = 4R
∴ γ = \( \frac{C_{P}}{C_{V}} \) = \( \frac{4_{R}}{3_{R}} \) = \(\frac{4}{3} \) = 1.33
Kinetic Theory Class 11 Important Questions Long Answer Type
Question 1.
Establish formula for pressure of a gas on the basis of kinetic theory of gases.
When a gas is filled in a closed vessel, then molecules of gas are in state of continuous random motion. They collide with one another and also with the walls of con-tainer. Due to these collisions,
the molecules of gas exert the force on the walls of the vessel. The force acting per unit area of the walls is called pressure of gas.
Let us consider a hollow cubical vessel of sides of length ‘l’ as depicted in fig. An ideal gas is filled in it. Let the mass of one molecule is m. Consider a molecule of gas moving with velocity C.
The components of velocity C along X, Y and Z direction are, u, v and w respectively.
∴ C^2 = u^2+ v^2 + w^2 ……….. (1)
Let us consider about two faces A and B, perpendicular to the direction of X-axis.
If the molecule P collides with the face A with velocity u. It will returned with a velocity -u after collision.
∴ Linear momentum of molecule before collision = mu Linear momentum of molecule after collision = – mu.
∴ Change in linear momentum due to collision = mu – (-mu) = 2mu
The molecule rebounded from A collides with the opposite face B, rebounds and again strikes with face A. Thus, the total distance travelled by the molecule will be 2l.
∴ Time taken by the molecule in travelling the distance 2l be
= \( \frac{2_{l}}{{u}} \), [ ∵ T ime = \( \frac{\text { Distance }}{\text { Velocity }} \) ]
Thus, after covering the distance 21 i.e., after every interval \( \frac{2l}{u} \) , the molecule P will
again strike with the face A.
Number of collision with face A per second be = \( \frac{u}{2 l} \)
The momentum transferred by molecule to the face A per second i.e., rate of change ofmomentum
= 2mu ×\( \frac{u}{2 l} \) = \( \frac{m u^{2}}{l} \)
But according to Newton’s second law of motion, the rate of change of momentum is equal to the force exerted on that face.
∴The force exerted by molecule at face Abe = \( \frac{m u^{2}}{l} \)
∴ Pressure exerted by the molecule on the face A be
= \( \frac{\mathrm{mu}^{2}}{\mathrm{l}} \) = \( \frac{n u^{2}}{l^{3}} \)
Let N be the number of molecules in gas and their velocity components along X direction are u[1], u[2], u[3], un respectively, then pressure exerted by these molecules on the face A
Where, V = l^3 = Volume of vessel i.e., volume of gas.
Similarly, the pressure exerted by N molecule along Y-axis and Z-axis be
P[y] = \( \frac{m}{V} \) \( \left[v_{1}^{2}+v_{2}^{2}+v_{3}^{2}+\ldots \ldots+v_{n}^{2}\right] \) ……… (3)
P[z] = \( \frac{m}{V} \) \( \left[w_{1}^{2}+w_{2}^{2}+w_{3}^{2}+\ldots \ldots+w_{n}^{2}\right] \) ………… (4)
But a gas exerts the same pressure in all directions i.e., P[x]=P[y]=P[z]=P (say)
∴ P = \( \frac{P_{x}+P_{y}+P_{z}}{3} \) …………….. (5)
Hence, from eqns. (2), (3), (4) and (5),
P = \( \frac{P_{x}+P_{y}+P_{z}}{3} \)
Let \( \bar{c} \) = \( \sqrt{\frac{c_{1}^{2}+c_{2}^{2}+\ldots . .+c_{n}^{2}}{N}} \) where \( \bar{c} \) is root mean square velocity , then
N\( \bar{c}^{2}\) = \( c_{1}^{2}+c_{2}^{2}+\ldots \ldots+c_{n}^{2} \) ………….. (7)
From eqns. (6) and (7),
P = \( \frac{m}{3 V} \)N\( \bar{c}^{2}\)
P = \( \frac{m N \bar{c}^{2}}{3 V} \)
This is expression of pressure exerted by gas.
Kinetic Theory Class 11 Important Numerical Questions
Question 1.
Up to what temperature should an ideal gas initially at 27°C be heated so that its volume becomes doubled at constant pressure?
Given:T[1] = 27°C = 273+27 = 300K
V[1] = V,V[2] = 2V
∴ According to Charle’s law,
\( \frac{V_{1}}{T_{1}} \) = \( \frac{V_{2}}{T_{2}} \)
T[2] = \( \frac{V_{2}}{V_{1}} \) T[1]= \(\frac{2 V}{V} \) × 300
= 600 K
= 600 – 273 = 327°C
Question 2.
A gas is filled in a vessel at 127°C at 4 atm. pressure. If the temperature of gas increased up to 527°C, then what would be the pressure of gas?
T[1] = 127°C = 127+273 = 400K
T[2] = 527°C = 527 + 273 = 800 K
P[1] = 4 atm. pressure
∴ By the pressure law,
\( \frac{P_{1}}{P_{2}} \) = \( \frac{T_{1}}{T_{2}} \)
P[2] = P[1] \( \frac{T_{2}}{T_{1}} \) = 4 × \( \frac{800}{400} \) = 8
= 8 atm.pressure
Question 3.
Number of molecule per cubic centimetre ¡n a space is five and temperature is 3 K. What will be the pressure, there? (R 1.38 × 10^-23 joule/m/K)
PV = nRT
P = \( \frac{n R T}{V} \)
Given: V = 1 cm^3 = 10 ^-6, R = 1.38 ×10^-23 J/m /K, n =5 , T = 3K
∴ P = \( \frac{5 \times 1 \cdot 38 \times 10^{-23} \times 3}{1} \) = 2.07 ×10^-22N/m^2
Question 4.
Temperature of any gas is -68°C. By what temperature it must be heated so that
(i) kinetic energy between the molecules become double
(ii) value of velocity of molecule become double.
Solution :
(i) From E = \(\frac{3}{2} \)KT,
E ∝ T
\( \frac{E_{1}}{E_{2}} \) = \( \frac{T_{1}}{T_{2}} \)
Given: E[ 2] = 2E[1],=T[1] = 273-68 = 205 K
∴ \( \frac{E_{1}}{2 E_{1}} \) = \( \frac{205}{T_{2}} \)
T[ 2] = 205 ×2 = 410K = 410 – 273° =137°C
(ii) From C^-2 = \( \frac{3 R T}{\gamma} \) we get
\( \bar{c}^{2}\) ∝ \( \sqrt{T} \)
\(\frac{\bar{c}_{1}}{\bar{c}_{2}}\) = \( \sqrt{\frac{T_{1}}{T_{2}}} \)
Given: \(\bar{c}_{2}\) = 2\( \bar{c}_{1} \) ,T [1] = 205K
∴ \( \frac{\bar{c}_{1}}{\bar{2c}_{1}} \) = \( \sqrt{\frac{205}{T_{2}}} \)
\( \frac{1}{4} \) =\( \frac{205}{T_{2}} \)
T[ 2] = 205 × 4 = 820
T[ 2] = 820 – 273 = 574°C
Question 5.
At 30°C temperature, mixture of Helium and Hydrogen gas is filled in a vessel. Find out the ratio of r.m.s. value of velocity of the molecule at this temperature.
From C^-2 = \( \frac{3 R T}{M} \)
\( \bar{c} \) = \( \frac{1}{\sqrt{M}} \)
\( \frac{\bar{c}_{1}}{\bar{c}_{2}} \) = \( \sqrt{\frac{M_{2}}{M_{1}}} \)
Given, molecular weight of Helium M[1] = 4
and Molecular weight of Hydrogen M[2] =2
∴ \( \frac{\bar{c}_{1}}{\bar{c}_{2}} \) = \( \sqrt{\frac{2}{4}} \) = \( \frac{1}{\sqrt{2}} \) = 1:\( \sqrt{2} \)
\( \bar{c}_{1} \): \( \bar{c}_{2} \) = 1:\( \sqrt{2} \)
Question 6.
If the absolute temperature of gas is done four times then by how much times their r.m.s. velocity of molecule increases? Also by how much times their kinetic energy and pressure also increases?
Solution: From \( \bar{c} \) = \( \frac{3 R T}{M} \) , where R and M are constant
∴ \(\bar{c} \) ∝ \( \sqrt{T}\)
\( \frac{\bar{c}_{1}}{\bar{c}_{2}} \) = \( \sqrt{\frac{T_{1}}{T_{2}}} \)
\( \frac{\bar{c}_{1}}{\bar{c}_{2}} \) = \( \sqrt{\frac{T_{1}}{4 T_{1}}} \) = \( \sqrt{\frac{1}{4}} \) = \( \frac{1}{2} \)
\( \bar{c}_{2} \) = 2 \( \bar{c}_{1} \)
i.e., r.m.s velocity will increase by two Times
From formula E = \( \frac{3}{2} \) KT
E ∝ T or \( \frac{E_{1}}{E_{2}} \) = \( \frac{T_{1}}{T_{2}} \)
Putting T[2] = 4T[1]
\( \frac{E_{1}}{E_{2}} \) = \( \frac{T_{1}}{4 T_{1}} \) = \( \frac{1}{4} \)
E[2] = 4E [1]
\( \frac{P_{1}}{P_{2}} \) = \( \frac{T_{\mathrm{l}}}{4 T_{1}} \) = \( \frac{1}{4} \)
P[2] = 4P[1]
∴ Pressure will become four Times.
Question 7.
If the temperature of a gas increased from 77°C to 227°C, then what will be the ratio of kinetic energy of then molecules?
Given:T[1] = (273+ 77)K = 350K and T[2] = (273 + 227)K = 500K.
From formula E ∝ T
\( \frac{E_{1}}{E_{2}} \) = \( \frac{T_{1}}{T_{2}} \) = \( \frac{350}{500}\) = \( \frac{7}{10} \)
E[1]: E [2] = 7 : 10
Question 8.
Volume of vessel is two times the volume of vessel B and same gas is filled in both vessels. If the temperature and pressure of the vessel A is double w.r.t. vessel B, then what will be the ratio of
molecules of the gas of vessel A and B?
From PV = nRT
Question 9.
The velocities of four molecules of a gas are 2, 4, 6 and 8 km/sec.
Calculate : (i) Average velocity, (ii) root-mean-square velocity.
Given : c[1] = 2 km/sec, c[2] = 4 km/ sec, c[3] = 6 km/sec and c4 = 8 km/sec.
(i) Average velocity:
c = \( \frac{c_{1}+c_{2}+c_{3}+c_{4}}{4} \)
= \( \frac{2+4+6+8}{4} \) = 5 km/sec
(ii) Root-mean-square velocity:
Question 10.
Estimate the fraction of molecular volume to the actual volume occupied by oxygen gas at N.T.P. Take the diameter of oxygen molecule to be roughly 3Å. (NCERT)
Given, diameter of one molecule of oxygen = 3 Å
∴ Radius r = \( \frac{3}{2} \) = 1.5 Å = 1.5 × 10 ^-10m
∴ Volume of one molecule of oxygen = \( \frac{4}{3} \) πr^3
= \( \frac{4}{3} \) ×3.14 (1.5 × 10 ^-10)^3
= 14.13 × 10 ^-30 m^3
∴ Volume of 1 mole of oxygen = 6.02 × 10 ^23 × 14.13 ×^-30
= 85.06 × 10 ^-7 m^3
The volume of 1 mole of oxygen at STP = 22.4 litre
= 22.4 × 10^-3m^3
∴ Fraction of the molecular volume of the actual volume = \( \frac{85 \cdot 06 \times 10^{-7}}{22 \cdot 4 \times 10^{-3}} \)
= 3.797 × 10 ^-4
= 3.8 ×10^-4 ≈ 4 × 10^-4
Question 11.
Molar volume is the volume occupied by 1 mole of any ideal gas at standard temperature and pressure (STP: 1 atm pressure, 0°C). Show that it is 22.4 litre. (NCERT)
At S.T.P., T = 0°C = 0 + 273 =273K
Pressure P = 1 atm = 1-013 x 10^5Nm^-2 ; R = 8.3 J mol^-1‘K^-1 By gas equation
∴ By gas equation for 1 mole PV =RT
V = \( \frac{R T}{P} \) = \( \frac{8 \cdot 3 \times 273}{1 \cdot 013 \times 10^{5}} \)
= 2236.82 ×10 ^-5
= 22.3682 ×10 ^-3 m ^3
= 22.4 ×10 ^-3m ^3 = 22.4 litre.
Question 12.
An air bubble of volume l-0cm3 rises from the bottom of a lake 40m deep at a temperature of 12°C. To what volume does it grow when it reaches the surface which is at a temperature of 35°C? (NCERT)
Given, Initial volume of bubble
V[1] = 10cm^3 = 1.0 × 10^-6m^3
Initial temperature T[1] = 273 +12 = 285 K
Initial pressure on bubble P[1] = Atmospheric pressure + Pressure of 40 m high water column
= 1.013×10^5+hdg
= 1.013×10^5 + 40×10^3 × 9.8
= 4.933 × 10^5 Pa
Final pressure on bubble P[2] = 1 atm = 1.013 ×10^5 Pa
Fina temperature T[2] = 273+ 35=308 K
From the equation \( \frac{P_{V_{1}}}{T_{1}} \) = \( \frac{P_{2} V_{2}}{T_{2}} \)
V[2] = \( \frac{P_{V_{1}}}{T_{1}} \) ×\( \frac{T_{2}}{P_{2}} \) = \( \frac{4 \cdot 933 \times 10^{5} \times 1 \cdot 0 \times 10^{-6} \times 308}{285 \times 1 \cdot 013 \times 10^{5}} \)
= 5.26 ×10^-6 m^3 ≈ 5.3 × 10^-6 m^3
Question 13.
At what temperature is the root-mean-square speed of an atom is an argon gas cylinder equal to the r.m.s. speed of a helium gas atom at – 20°C? (Atomic mass of Ar = 39.9 u, of He = 4.0 u). (NCERT)
Given, atomic mass of He M[1]=4.0u
Atomic mass of Ar M[2] =39.3u
Temperature of He gas T [1] = -20+273 = 253K
Kinetic Theory Class 11 Important Questions Objective Type
1. Multiple- choice questions:
Question 1.
There are N molecules in a vessel, if the number of molecules are doubled then the pressure of gas will be :
(a) Double
(b) Remain same
(c) Become four times
(d) Become one fourth.
(a) Double
Question 2.
Motion of gaseous molecule at absolute zero temperature :
(a) Become less
(b) Increases
(c) Become zero
(d) None of these.
(c) Become zero
Question 3.
At -273°C, molecules of gas moves with :
(a) Maximum velocity
(b) Minimum velocity
(c) Zero velocity
(d) None of these.
(c) Zero velocity
Question 4.
Reason for deviating from gaseous law of ideal gas at less temperature is :
(a) Maximum collision become inelastic
(b) Volume of molecules cannot be negligible
(c) Force acting between molecules become less
(d) Molecular velocity become less.
(b) Volume of molecules cannot be negligible
Question 5.
Root-mean-square velocity of ideal gas molecules at constant temperature :
(a) Remain same
(b) Inversely proportional to square root of molecular weight
(c) Proportional to square root of molecular weight
(d) Inversely proportional to molecular weight.
(b) Inversely proportional to square root of molecular weight
Question 6.
Graph between PV and P of a gas which obey Boyle’s law will be :
(a) Hyperbola
(b) Parallel line with PV axis
(c) Parallel line with P axis
(d) None of these.
(c) Parallel line with P axis
Question 7.
False statement regarding kinetic theory of gas is :
(a) Collision between two molecules are perfectly elastic
(b) Kinetic energy between molecules is proportional to absolute temperature
(c) Absolute temperature of gas is inversely proportional to root-mean-square velocity.
(d) At absolute temperature average kinetic energy of molecules is zero.
(c) Absolute temperature of gas is inversely proportional to root-mean-square velocity.
Question 8.
Reason for pressure exerted by gaseous molecules on the wall of vessel is :
(a) Losses its own kinetic energy
(b) Get stick with the walls of vessel
(c) Due to collision with wall of vessel its momentum get change
(d) Get accelerated toward wall.
(c) Due to collision with wall of vessel its momentum get change
Question 9.
There is no atmosphere in moon, because :
(a) It is closer to earth
(b) It revolve around the earth
(c) It obtain light from sun
(d) Escape velocity is less than root-mean-square velocity.
(d) Escape velocity is less than root-mean-square velocity.
Question 10.
The temperature of an ideal gas is raised from 27°C to 927°C the root-mean-square velocity of its molecules will become :
(a) Two times
(b) Half
(c) Four times
(d) One fourth.
(a) Two times
Question 11.
Every gas behaves as ideal gas at:
(a) Low pressure and high temperature
(b) High pressure and low temperature
(c) At equal pressure and temperature
(d) High pressure and high temperature.
(a) Low pressure and high temperature
Question 12.
Unit of universal gas constant is :
(a) joule/mole-kelvin
(b) mole/joule-kelvin
(c) joule-mole-kelvin
(d) kelvin/joule/mole.
(a) joule/mole-kelvin
Question 13.
If the temperature of gas remain constant and pressure become half then the volume will get.
(a) Half
(b) Double
(c) Unchanged
(d) Four times.
(b) Double
Question 14.
In gas equation PV = RT, V is volume of:
(a) Gas
(b)1 gram gas
(c) 1 litre gas
(d) 1 mole gas.
(d) 1 mole gas.
Question 15.
Pressure of gas filled in closed vessel is due to :
(a) Large number of molecules
(b) Attraction between wall and molecules
(c) Collision of molecules with wall
(d) None of these.
(c) Collision of molecules with wall
Question 16.
The average kinetic energy associated with each degree of freedom is :
(a) \( \frac{3}{2} \) KT
(b) KT
(c) \( \frac{1}{2} \) KT
(d) \( \frac{3}{2} \) RT.
(c) \( \frac{1}{2} \) KT
Question 17.
The mean kinetic energy of the molecule of gas depends upon :
(a) Nature of gas
(b) Absolute temperature
(c) Volume of gas
(d) None of these.
(b) Absolute temperature
Question 18.
Root-mean-square velocity of gas is :
(a) Directly proportional to its specific molecular weight
(b) Directly proportional to its square of molecular weight
(c) Directly proportional to its molar weight
(d) Directly proportional to its absolute temperature.
(c) Directly proportional to its molar weight
Question 19.
Magnitude of R for 1 gram mole of gas is :
(a) 8.31 erg
(b) 8.31 MKS unit
(c) 4.2 joule
(d) 4.2 calorie.
(b) 8.31 MKS unit
Question 20.
If the r.nus. velocity of a gas is doubled, then its pressure will:
(a) Increase
(b) Decrease
(c) Remain same
(d) None of these.
(a) Increase
2. Fill in the blanks:
1. Momentum applied per unit area on the wall of a vessel by the molecule of gas is equal to ………………………. .
2. At absolute temperature ………………………. of gas becomes zero.
Kinetic energy
3. For a diatomic gas, the degree of freedom is ………………………. .
4. Kinetic energy associated with each degree of freedom is ………………………. .
\( \frac{1}{2} \)KT
5. The value of 0°C on the kelvin scale is ………………………. .
273 K.
3. Match the following:
┃Column ‘A’ │Column ‘B’ ┃
┃1. Pressure of gas P │(a) ∝ T ┃
┃2. Absolute temperature of gas T │(b) 3 ┃
┃3. Average kinetic energy E │(c) 5 ┃
┃4. Degree of freedom of monoatomic gas│(d) ∝ c^-2 ┃
┃5. Degree of freedom of diatomic gas │(e) \( \frac{1}{3} \) \( \frac{m N}{V} \)c^-2 ┃
1. (e) \( \frac{1}{3} \) \( \frac{m N}{V} \)c^-2
2. (d) ∝ c^-2
3. (a) ∝ T
4. (b) 3
5. (c) 5. | {"url":"https://mpboardguru.com/mp-board-class-11th-physics-important-questions-chapter-13/","timestamp":"2024-11-09T23:21:25Z","content_type":"text/html","content_length":"100769","record_id":"<urn:uuid:875366aa-47df-47ea-9ee8-4cc826ef6e36>","cc-path":"CC-MAIN-2024-46/segments/1730477028164.10/warc/CC-MAIN-20241109214337-20241110004337-00573.warc.gz"} |
A function continuous at all irrationals and discontinuous at all rationals
Let’s discover the beauties of Thomae’s function also named the popcorn function, the raindrop function or the modified Dirichlet function.
Thomae’s function is a real-valued function defined as:
\mathbb{R} & \longrightarrow & \mathbb{R} \\
x & \longmapsto & 0 \text{ if } x \in \mathbb{R} \setminus \mathbb{Q} \\
\frac{p}{q} & \longmapsto & \frac{1}{q} \text{ if } \frac{p}{q} \text{ in lowest terms and } q > 0
\(f\) is periodic with period \(1\)
This is easy to prove as for \(x \in \mathbb{R} \setminus \mathbb{Q}\) we also have \(x+1 \in \mathbb{R} \setminus \mathbb{Q}\) and therefore \(f(x+1)=f(x)=0\). While for \(y=\frac{p}{q} \in \mathbb
{Q}\) in lowest terms, \(y+1=\frac{p+q}{q}\) is also in lowest terms, hence \(f(y+1)=f(y)=\frac{1}{q}\).
We select \(x \in \mathbb{R}\). For all \(\epsilon >0\) one can find \(N > 1\) integer with \(0 < \frac{1}{N} < \epsilon\). Taking \(a= \sup \{c \in \mathbb{Z} \text{ | } c \le x N\}\), we have \(x \
in (\frac{a-1}{N},\frac{a+1}{N})\). Now consider: \[\begin{aligned} A &=\{y \in \mathbb{Q} \setminus \{x\} \text{ | } y=\frac{r}{s} \in (\frac{a-1}{N},\frac{a+1}{N}),\\ & \ r \in \mathbb{Z}, \ 0 < s
\le N, \ \gcd(r,s) = 1\} \end{aligned}\] \(A\) is finite and by definition \(x \notin A\). If \(A\) is not empty, the distance \(D=d(x,A)\) of \(x\) to the set \(A\) is defined and strictly positive.
If \(A\) is empty, we denote by \(B\) the interval \((\frac{a-1}{N},\frac{a+1}{N})\) otherwise we take \(B=(\frac{a-1}{N},\frac{a+1}{N}) \cap (x-D,x+D)\). In both cases, \(B\) is an open interval
containing \(x\). By construction, \(B \setminus \{x\}\) contains no rational number \(\frac{r}{s}\) in lowest terms with \(s \le N\). Hence for all \(z \in B \setminus \{x\}\) we have \(0 \le f(z) <
\frac{1}{N} < \epsilon\) which had to be demonstrated.
\(f\) is discontinuous at all rational numbers
Consider \(x=\frac{p}{q}\) in lowest terms. The sequence \(\displaystyle x_n=(1-\frac{1}{n \sqrt{2}})\frac{p}{q}\) converges to \(x\). For all \(n \ge 1\) integer, \(x_n\) is an irrational number as
\(\sqrt{2}\) is irrational, hence \(f(x_n)=0\). Therefore \(\lim\limits_{n \to +\infty} f(x_n)=0\) while \(f(x)=f(\frac{p}{q})=\frac{1}{q} > 0\). More elegantly, we could have just notice that the
irrational numbers are dense in \(\mathbb{R}\). Or, we could have used previous paragraph stating that both \(f\) left and right limits vanish.
\(f\) is continuous at all irrational numbers
Is clear as \(f\) vanishes at all irrational points \(x\) while both left and right \(f\) limits vanish at all points.
\(f\) has a local maximum at all rational points
Consider a rational number \(a=\frac{p}{q}\) in lowest terms. We have \(f(a)=\frac{1}{q}\). As \(\lim\limits_{x \to a^-} f(x) = \lim\limits_{x \to a^+} f(x) = 0\), for some \(\delta > 0\) we have \(0
\le f(x) \le \frac{f(a)}{2}\) when \(0 < \vert x-a \vert < \delta\). Finally \(0 \le f(x) \le f(a)\) for \(x \in (a - \delta, a + \delta)\).
\(f\) is Lebesgue integrable
Follows from the fact that \(f\) vanishes almost everywhere, the set of rational numbers having a Lebesgue measure equal to zero.
\(f\) is Riemann integrable on all intervals \([a,b]\)
This is a consequence of Lebesgue’s integrability condition as \(f\) is bounded (by \(1\)) and continuous almost everywhere. Or we can use the theorem stating that a regulated function is Riemann
Python code I used to generate Thomae’s function image
import matplotlib.pyplot as plt
import fractions as frac
from math import log
points=[[p/float(q),1/float(q),log(float(q))] \
for q in range(1,50) for p in range(0,q+1) \
if frac.gcd(p,q) == 1]
plt.scatter(x,y, c=color, s=3, edgecolor='none', cmap='winter')
plt.savefig('D:tmp/to.png', dpi=350, facecolor='#EEE1D0')
2 thoughts on “A function continuous at all irrationals and discontinuous at all rationals”
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Expanding the Expression (x+3)(x+8)
This expression represents the multiplication of two binomials: (x+3) and (x+8). To expand it, we can use the FOIL method (First, Outer, Inner, Last):
1. First: Multiply the first terms of each binomial: x * x = x²
2. Outer: Multiply the outer terms of the binomials: x * 8 = 8x
3. Inner: Multiply the inner terms of the binomials: 3 * x = 3x
4. Last: Multiply the last terms of each binomial: 3 * 8 = 24
Now, combine the terms: x² + 8x + 3x + 24
Finally, simplify by combining the like terms: x² + 11x + 24
Therefore, the expanded form of (x+3)(x+8) is x² + 11x + 24.
Understanding the FOIL Method
The FOIL method is a simple and visual way to remember how to multiply two binomials. It ensures that we multiply each term in the first binomial by each term in the second binomial.
Other Approaches
While the FOIL method is commonly used, you can also use the distributive property to expand the expression:
• Distribute (x+3) over (x+8): (x+3) * (x+8) = x(x+8) + 3(x+8)
• Distribute again: x² + 8x + 3x + 24
• Simplify: x² + 11x + 24
Understanding how to expand binomials like (x+3)(x+8) is essential for various mathematical concepts, including:
• Factoring quadratic expressions
• Solving quadratic equations
• Graphing quadratic functions
• Solving problems in algebra and calculus
By mastering this skill, you can tackle more complex mathematical problems with confidence. | {"url":"https://jasonbradley.me/page/(x%252B3)(x%252B8)","timestamp":"2024-11-03T02:36:08Z","content_type":"text/html","content_length":"60080","record_id":"<urn:uuid:dc79caa8-9c27-4d37-8bc1-87f0aceeda41>","cc-path":"CC-MAIN-2024-46/segments/1730477027770.74/warc/CC-MAIN-20241103022018-20241103052018-00524.warc.gz"} |
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ON7YD, longwave, 136kHz, antennas
Antennas for 136kHz
About this page :
The main object of this page is to provide information. It has been deliberately kept simple, no fancy and flashy tricks, in order to achieve maximum compatibility for the different browsers and to
allow fast downloading.
Any comments and/or suggestions are welcome at : on7yd@qsl.net
last updated on 8 July 2004
1. Introduction
The main subject will be transmitting antennas for 136kHz as this often is the most important part of a longwave amateur radio station. The aim of the transmitting antenna is to radiate the power
coming from the transmitter.
The power radiated by any antenna is determined by 3 factors :
Example :
Assume we have an antenna with a radiation resistance of 10 ^2 x 4 = 160 Watt.
The gain of an antenna is always given relative to a reference antenna. Most common references are the 1/2 wave dipole and the isotropic radiator. This last is a virtual antenna that has no
directivity at all, it radiates equally to all directions. In general the gain of any antenna relative to a 1/2 wave dipole is given as dB[d] while the gain relative to an isotropic radiator is given
as dB[i]. Due to its directivity a 1/2 wave dipole has a gain 1.64 (2.15dB[i]) relative to a isotropic radiator.
At first sight the radiation resistance of an antenna has no influence on the radiated power, as long as you match your transmitter to this resistance. But unfortunately the radiation resistance is
not the only resistance that is consuming the transmitter power, there are also the loss resistances. These losses occur within the antenna (+ the antenna matching system) and in the environment of
the antenna (ground, objects near the antenna). On HF these loss resistances are often negligible as they are rather small compared to the radiation resistance, but on longwave this is certainly not
the case. For most longwave antennas used by amateurs the radiation resistance of the antenna is in the range of 10 to a few hundred m
The two most common transmitting antennas on longwave are the short vertical monopole (Marconi antenna) and the small loop antenna. The short vertical monopole is an electric antenna, it creates an
electric field 'on the spot' (near the antenna) while the magnetic field is created 'on the fly'. Opposite to this the small loop is a magnetic antenna, it creates a magnetic field 'on the spot'
while the electric field is created 'on the fly'.
As a result of this the main source of losses for a short vertical monopole is in the environment (ground, trees, buildings etc.) while for a small loop the major losses are within the antenna.
Therefore a small loop is less dependent on the environment for its functionality.
But for both types of antennas the goal is to get the ratio of radiation resistance versus loss resistances as large as possible. In practice most amateurs achieve better results with short vertical
monopoles, only when environment losses are extremely high a small loop will be superior.
Remark : Throughout these pages the terms ERP, EIRP, dB[i] and dB[d] will be used frequently. If you are not familiar with these terms I would recommend to read this first.
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2. Short vertical antennas
2.1. Vertical monopole antenna
Marconi antenna". It is a quarter wave long, is fed against ground (eventually improved by a radial system) and has a radiation resistance of 36
The dimensions of a quarter wave vertical antenna might be suitable from the 40m band upward, some brave hams might even have this antenna for 80m and 160m. But for 136kHz it would be over 500m
(1500ft) high, without doubt beyond the range of any ham. Thus at longwave there is no other way than using a vertical monopole that is (very) much shorter than a quarter wave.
When a vertical monopole is less than a quarter wave (it's natural resonance) a few things change :
The effect of the antenna length on the radiation resistance and antenna gain can be seen on the first picture at the right.
Example :
1. A Quarter wave vertical has a radiation resistance of 36
(36 / (36 + 10)) * 100% = 78.3% (or -1.1dB)
2. A short vertical monopole of 1% of the wavelength has a radiation resistance of 0.04
(0.04 / (0.04 + 50 + 20)) * 100% = 0.057% (or -32.4dB)
As a result the quarter wave vertical will outrange the short vertical monopole by 31.7dB (31.3dB efficiency + 0.4dB antenna gain).
The second picture shows "overall gain" (efficiency + antenna gain) of an average antenna as a function of its length.
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2.2. Short vertical monopole
short vertical monopole with a height H and fed against ground. If H is small compared to the wavelength then :
The current distribution, that is different from the sinusoidal distribution we are used to, can be explained as follows :
The antenna capitance is not located at one single point on the antenna, but is distributed equally over the antenna. As the antenna current flows into the antenna it gradually 'disappears' via the
distributed antenna capacitance, resulting in a linear decrease.
Another - and maybe more correct - way to look at it is to compare a short vertical with a full size (quarter wave) vertical.
The full size vertical has a sinusoidal current and voltage distribution whith a 90 degrees phase shift between U and I. The short vertical can been seen as just the end of a fullsize vertical, where
the voltage distribution is (almost) constant and the current distribution decreases (almost) linear.
[1b] (R[A]in m
The capacitance of a vertical wire of a height H and diameter d is :
[2a] (C[V] in pF, H and d in m)
In most cases the simplified formula C[V] = 6pF/m [2b] is accurate enough.
In order to get a maximum radiated power we need a maximal current through the antenna. This can be done by compensating the capacitive component with an inductive component (loading coil), or
otherwise said : bringing the antenna to resonance. Based on the formula for resonance (Thomson formula) we can calculate the inductance we need (see the chapter "Loading coil" for details).
Assume we have a 10m long vertical wire (3mm diameter) with and an enviromental loss of 60
Based on formula 1a the radiation resistance is calculated as 8.2m
If we put a power of 100W into the antenna we will have an antena current of 0.92A, resulting in 7.5mW radiated power and a voltage of 16kV over the loading coil.
In the above example we calculated a radiated power of 7.5mW (0.92A into 8.2m[d]. So the calculated power in this case will be 13.7mW ERP.
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2.3. Vertical antenna with capacitive toploading
The current distribution over the antenna has still a linear decrease, but due to the fact that the minimum now is at the end of the horizontal section the average current in the vertical part is
The capacitance of a horizontal wire with a length L, a diameter d and at a height H is given by :
[3a] ( C[H] in pF, H, L and d in m)
In most cases the simplified formula C[H] = 5pF/m [3b] is accurate enough.
The total antenna capacitance C[A] = C[V] + C[H]. The antenna current at the top of the vertical section is determined by the ratio of C[H] and C[V] (assuming that the same amount of current
'disappears' via every pF) :
Thus the average current through the vertical section is :
And the radiation resistance is proportional to the square of the average current through the vertical section :
[5b] (R[A]in m
This means that the radiation resistance can be quadrupled by adequate capacitive toploading.
An additional benefit of capacitive toploading is that the antenna capacitance can increase significantly. Therefore the inductance (loading coil) needed will decrease, resulting in lower losses in
and lower voltages over the loading coil.
Assume we still have the 10m long vertical wire (3mm diameter) and the enviromental loss of 60
The capacitance of the vertical section will be 67pF (formula 2a) while the capacitance of the topload will be 116pF (formula 4a), resulting in a total antenna capacitance of 183pF. The radiation
resistance will be 21.9m
If we put a power of 100W into the antenna we will have an antenna current of 1.11A, resulting in 27mW radiated power and a voltage of 7kV over the loading coil. Taking into account the gain of 2.6dB
[d] the ERP will be 49mW, this an overall 5.5dB improvement compared to the same antenna without capacitive topload.
The gain that can be achieved by having a better current distribution is 6dB, but due to the increased capacitance (and thus a smaller loading coil needed) some dB extra gain can be won, as you can
see in the graph.
A vertical antenna with capacitive toploading can be constructed in various configurations, besides the 'inverted-L' configuration there are also the 'T' and 'umbrella' configurations that are
frequently used. In general any shape of capcitive topload will work, the goal should be to get an many wire as possible as high in the air as possible. The topload wires can be sloping (umbrella
antenna), but this will cause a decrease in the radiation resistance. As a rule of thumb can be said that sloping topload wires should never come lower than 50% of the antenna height.
The amount of topload capacitance is often limited by the space available. To get maximal topload capacitance on a limited space parallel wires can be used. Practical results proved that capacitances
upto 15pF/m can be achieved, while a single wire is about 5pF/m :
Capacitance of multiple topload wires
┃Station│number of wires│ Spacing │Height above ground │Capacitance┃
┃ EI0CF │ 4 │ *1m*4m*1m* 1) │ 10m │ 15pF/m ┃
┃ G3XDV │ 3 │all 0.5m - 1m (2)│ 14m │ ┃
┃ G3AQC │ 3 │ all 0. 45m │ 13.5m │ 12pF/m ┃
┃ ON7YD │ 4 │ all 0.8m │ 12.5m │ 13pF/m ┃
(1) : spacing between outer wires is 1m, between inner wires is 4m (total 6m)
(2) : spacing is 0.5m at one end and 1m at the other end
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2.4. Umbrella antenna
[A]) of the antenna. One the one hand it increases the top capacitance, thus increasing R[A]. But on the other hand it will introduce a 'downward current' that cancels a part of the (upward) current
through the vertical, thus decreasing R[A].
The influence of both effects depends of the number of topload wires, their length and their sloping angle. John Sexton (G4CNN) did develop a mathematical model for umbrella antennas with the goal to
optimize the parameters (number of wires, length and sloping angle) for a maximum radiation resistance.
Detailed calculations how to optimize the slope and length of the topload wires can be found here.
Assume an umbrella antenna with a unity height (1) and n topload wires of a length L, sloping under an angle ß. The topload wires will 'shield' the vertical part over a length X = L*cos(ß).
The gain (in dB, relative to a vertical without toploading) will be :
[6] (log = 10 based logaritm, L and X relative to the unity height '1')
The graphs below give the relative gain of an umbrella antenna for sloping angles of 30, 45 and 60 degrees - depending on the 'shielding length' (X) and the number of tophat wires :
As expected higher sloping angles give the better results, but also note that for a certain sloping angle many short tophat wires are more effective than a few long.
The above formula and graphs assume that the tophat wires do not affect each other. In practice this will not be true in the case of many short tophat wires, the effective gain will be less than the
calculated one.
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2.5. Capacitive toploading of single-tower antennas
In the previous chapter (umbrella antennas) an example was given how a capacitive toploaded antenna can be built using a single support tower. If the space to put up the antenna is even more limited
there are some alternative ways to provide your antenna of an efficient capacitive toploading. The above pictures show one way to achieve this, as done by Werner De Bondt (ON6ND). At the top of a
grounded 12m tower there is a large isolated cage that provides a high capacitance toploading. Additional experiments done by ON6ND have shown that the antenna efficieny can be further improved by
placing (a part of) the toploading at the bottom of the isolated section. Werner noticed a 1 S-point (6dB) improvement by doing so. Further there was no difference noticed when the feeding line was
replaced by a single wire.
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Alan Melia (G3NYK) and Finbar O'Connor (EI0CF) developed an efficient LF antenna on a limited space. The topload of the antenna has both an inductive as capacitive component. The spiral part works
both inductive as capacitive while the piramidal tp hat provides additional capacitive toploading. Despite the limited dimensions the antenna could be brought to resonance on 136kHz with a relative
small (2mH) loading coil.
A complete description of this antenna can be found on G3NYK's webpage.
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2.7. Vertical antenna with inductive toploading
[L]) of the antenna.
As the voltage is built up over the loading coil, only the part of the antenna above the coil wil be at high voltage. The voltage at the lower part of the antenna is negligible so the antenna current
will not 'disappear' via the capacitance C[1], only via C[2]. The antenna current below the loading coil will remain at maximum value and the result is an improved average current :
The radiation resistance of a vertical antenna with elevated loading coil is :
[8b] (R[A] in m
Example :
Assume we have a 10m high vertical wire (3mm diameter) and 60
If we put a power of 100W into the antenna we will have an antenna current of 0.75A, resulting in 10.5mW radiated power and a voltage of 26kV over the loading coil. Taking into account the gain of
2.6dB[d] the ERP will be 19.1mW, this only a 1.4dB improvement compared to the same antenna without elevated loading coil.
As shown in the example, inductive toploading has also a big disadvantage :
The loading coil has to be resonant to the antenna capacitance of the upper part (C[2]) what means that the inductance has to be larger as the coil is placed higher. A larger coil means also a larger
coil-loss and from a certain height the additional loss induced by the larger coil cannot be compensated by the improved current distribution, as you can see in the graph at the right. Apart from
that the voltage over the loading coil increases and stable mounting of an elevated loading coil creates also mechanical problems.
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2.8. Vertical antenna with capacitive and inductive toploading
limited capacitive toploading can be improved significantly by adding inductive toploading (elevated loading coil).
But based on the improvement of the current distribution, adding inductive toploading to an antenna with sufficient capacitive toploading is not very efficient. In most cases the theoretical gain is
no more than 0.1 or 0.2dB. Practical experiments by a number of amateurs however have shown that in some cases combined capacitive / inductive toploading can lead to a significant gain (of up to
More detailed information about combined capacitive / inductive toploading can be found here.
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2.9. Vertical antenna with tuned counterpoise
Pat Hawker describes in an article in ELECTRONICS WORLD + WIRELESS WORLD (February 1990) a kind of umbrella antenna with a tuned counterpoise. Both the antenna and the counterpoise are isolated from
the ground.
The antenna is tuned by the loading coil (L[1]), an elevated loading coil can be used to improve the current distribution. By adjusting L[2] the counterpoise is tuned to minimize ground loss. In
practice L2 has to be tuned for maximal signalstrength in the far field.
This type of antenna has been used successfully on mediumwave with a gain upto 5dB measured by adding the tuned counterpoise. To my knowledge this antenna has not been tested by amateurs on 136kHz,
but it might be worth a try.
No references to calculate the value of L[2] are given, but the article refers to US Patent no 3,742,511 and to IEEE Trans. on Broadcasting, June 1989, pages 237-240 (download it as zipped GIF file)
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2.10. Meander antenna
IEEE Trans. on Antennas and Propagation, December 1998, pages 1797-1801. They show that the radiation resistance of short electrical antennas, such as a short vertical monopole, can be significantly
increased by using u number of folded elements. Experimental investigations on a 44cm high meander antenna with 21 elements resulted in a resonance on 20.1MHz and an impedance of 21.9
A meander antenna can be built rather compact arround a grounded tower.
The line-spacing has to be at least 20 times the wire diameter for optimal performance. Further experiments have shown following Size Reduction Factor (SRF) versus the number of lines (N) :
Size Reduction Factor
┃ SRF │ N │antenna height │wirelength ┃
┃ 0.6 │ 3 │ 329m │ 987m ┃
┃ 0.3 │ 9 │ 164m │ 1481m ┃
┃ 0.15 │27 │ 82m │ 2221m ┃
┃ 0.075 │81 │ 41m │ 3332m ┃
┃0.0375 │243│ 20.5m │ 4998m ┃
┃0.01875│729│ 10.3m │ 7497m ┃
red = extrapolated from experimental data (black)
For acceptable antenna heights (20m and less) the number of elements and the wirelength needed are not very realistic. Using 3mm Cu-wire (loss = 1
But a meander antenna with a limited number of elements and tuned to resonance by a loading coil could be an acceptable alternative antenna for 136kHz. But remember that a meander antenna has many
resonances at higher frequencies, so an adequate filtering of the transmitter signal (harmonics !) will be necessary.
To my knowledge meander antennas have not been used by amateurs on longwave so far.
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2.11. Antenna with multiple vertical elements
If each of the elements has its own ground-network this can reduce the total loss-resistance.
But this system has also a disadvantage : as all the elements share the same capacitive topload the capacitance of each individual element will decrease and larger loading coils will be needed,
resulting in additional coil-losses and a higher antenna voltage.
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2.12. Using a non isolated antenna tower as LF-antenna
A short vertical monopole normally has to be isolated from ground at its base. But most antenna-towers are not isolated, for mechanical and electrical safety reasons. Here are 2 possibilities shown
how to use a non isolated tower as vertical antenna for longwave :
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2.13. Antennas with a long horizontal section
Based on the calculated current distribution and coil-losses very little gain can be won by having a horizontal section that exceeds the length of the vertical section by more than a factor 5. The
only advantage would be a decraase of the groundloss due to the larger 'footprint' of the antenna (see 3.6.3).
But in practice several hams achieved very good results using antennas with a very long horizontal section and it is difficult to explain this results just by the lower groundloss. OH1TN uses an
antenna with a horizontal section of about 500m, bringing the antenna to resonance on 136kHz without inductive loading (sse picture below).
Despite the fact that the antenna is mainly horizontal, its polarization is mainly vertical as long as the height of the antenna (compared to the wavelength) is low and it is a monopole antenna (with
ground as counterpart). An example of a large horizontal antenna with vertical polarization is the DDRR antenna.
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2.14. Helical antenna
antenna voltage builts up over the loading coil, the antenna voltage increases with the height. This voltage increase results in an improved current distribution, as in the lower part of the antenna
(where the voltage is low) less current will 'disappear'. Without capacitive toploading the radiation resistance of a helical antenna will be 1.54 times larger as for a 'straight' vertical of the
same height, this is a gain of 1.9dB.
When capacitive toploading is added the advantage of a helical antenna will be less, for 2 reasons :
An additional problem is that it is not so easy to built a mechanical stable helical antenna . The only amateur who - to my knowledge - used a helical antenna with succes was Toni Baertschi (HB9ASB),
until the antenna was destroyed in a storm (december 1999).
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2.15. Short vertical dipole
So far I am not aware of any short vertical dipole used by hams or lowfers on LF, but it could be an alternative for the short vertical monopole.
* Short dipole in free space
The radiation resistance of a short dipole with a length H and at a wavelength
[9b] (R[A] in m
The antenna gain of a short dipole in free space is 1.76dB[i]
* Short vertical dipole close to ground
Antenna simulation shows that close to ground the radiation resistance of a short vertical dipole doubles (versus the free space value) and that the antenna gain increases to 4.77dB[i]. This means
that both radiation resistance and antenna gain would be identical to these of a short vertical monopole of the same size. But at the same time it can be expected that the enviromental losses will
increase, due to the high antenna voltages close to ground.
Jim Moritz (M0BMU) comments a short vertical dipole close to ground as follows :
A short vertical dipole in free space would have a symmetrical current distribution, maximum in the middle and zero at the ends. Placing it close to a ground plane would modify this because
displacement current would flow between the ground plane and the lower half of the dipole, increasing the current towards the bottom end of the dipole making the current distribution more even in the
lower part of the dipole. The current distribution in the case where the lower end of the dipole was very close to the ground would be very similar to that of a short monopole with an elevated feed
point - if the lower end of the dipole actually was in contact with the ground plane, it would be a monopole of course.
You could increase the capacitance by adding end loading to the dipole - if the lower end of the dipole were close to the ground, in effect you would have a top-loaded vertical with an elevated feed
driven against a counterpoise.
A practical difficulty would be caused by the asymmetrical nature of the dipole - it would be necessary to have some way of adjusting the voltages applied to the upper and lower legs of the dipole to
get equal current in both legs, and zero net current on the feed line, in order to achieve the proper dipole operation. Since the lower end of the dipole would be a high voltage point, and close to
the ground, there would be increased dielectric losses in the ground under the antenna, which would tend to reduce any advantage of this antenna configuration.
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2.16. Why a horizontal dipole is a rather inefficient antenna on LF
As in most cases height is the limiting factor to built large (and efficient) LF antennas it is tempting to built a large horizontal antenna. Although antennas with a large horizontal section have
proved to be quite usefull on LF (see " Antennas with a long horizontal section ") a horizontal dipole is a rather inefficient antenna on LF.
If an antenna is placed above (a perfect) ground a "mirror image" is created with the ground as mirror plane. For a vertical antenna this causes no problem as both the real antenna and its mirror
image are in phase. But for a horizontal antenna the mirror image is in counterphase, so if the antenna is close to ground the mirror image will cancel out most of the signal (see picture at the
In practice this results in a decreasing radiation resistance as the antenna comes closer to ground. At very low heights (in wavelengths !) even the radiation resistance of a full size half wave
dipole is only a fraction of an
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2.17. Safety precautions
high voltages.
As mentioned before most short vertical antennas need a rather large loading coil to be brought to resonance. The voltage built up over this coil can be some tens kV, in combination with a moderate
to high power TX this voltage (or better said : the current caused by this voltage) can be harmfull or even lethal.
Example :
Assume a 10m high vertical antenna with a capacitance of 70pF. This antenna will need a 19mH loading coil, that has a reactance of over 16k
Fortunately voltage and current are almost 90 degrees phase shifted and touching a high voltage part of the antenna will cause a breakdown of the antenna voltage, but it can still cause serious
burnings or worse.
An interesting fact is that a larger antenna (that has a larger capacitance and thus requires a smaller loading coil) will have a lower antenna voltage. A small backyard antenna can be a much higher
potentional danger than a monster antenna.
In any case one should take the nessecary precautions to avoid that high voltage parts of the antenna can be touched. Since the voltage builts up over the loading coil, placing this coil at a certain
hight can be a simple and effective solution. But keep in mind that even at the low voltage side (before the loading coil) the voltage can be up to a few 100V, so even there one should only use
sufficient isolated wire.
The corona effect is rather seldom reported by amateurs on LF, from the information I received the antenna voltage was in excess of 30kV in all cases. The corona effect can be surpressed by avoiding
any sharp ends or edges and mounting corona rings at the ends of the antenna.
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2.18 Bringing a short vertical monopole to resonance
2.18.1. Loading coil
If you will connect you TX directly to a short vertical monopole you will hardly transmit any signal (and get a vey bad SWR), This is due to the mainly capacitive impedance of several 1000
The inductance L (in Henry) can be calculated using the Thomson formula :
[10a] (f = frequency in Hz and C = antenna capacitance in F)
[10b] (L in mH and C in pF)
Example :
If we want to bring an antenna with a capacitance of 300pF to resonance at 137kHz then we will need a loading coil of 4.5mH.
The inductance L (in µH) of a single layer coil is :
(n = number of turns, d = coil diameter in mm and l = coil length in mm)
For the typical sizes of LF loading coils formula 11 will give the inductance within a few %, but in fact the inductance will also be dependent of the wire diameter (b) and wire spacing (a).
Formula 11 is valid for wire spacing = wire diameter. If this is not the case you will have to add a
(a = wire spacing in mm, b = wire diameter in mm, n = number of turns and d = coil diameter in mm)
Be aware that
Not only the inductance but also the Q-factor of the coil will be affected by the ratio of wire diameter and wire spacing. This is due to losses caused by the proximity effect. Experiments have shown
that best Q is achieved when the wire diameter and the wire spacing are equal (a = b). Small (or no) wire spacing will result in a relative small coil (least wire needed), but with a realtive low Q.
If the wire spacing is very large then there will be little losses caused by the proximity effect but you will need much more wire to achieve a certain inductance causing more additional loss than
you gained from eliminating the proximity effect.
Example :
Assume we have a coil made of 100 turns of 2mm diameter wire with 2mm spacing. The coil diameter is 300mm.
100 turns of 2mm wire with 2mm spacing make a 400mm long coil. Using formula 11 the inductance is 1.66mH. If we leave the coil length and diameter unchanged but use 1mm diameter wire (and 3mm
spacing) then the inductance will increase by 0.12mH (formula 12) and will be 1.78mH. If instead 3mm wire (and 1mm spacing) is used, the inductance will decrease by 0.01mH and will be 1.65mH.
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2.18.2. Coil losses : the Q-factor
As any other coil the loading coil will also have certain losses that will reduce the overall efficiency of the antenna system. The Q-factor is the ratio of the inductive reactance (X[L]) and the
loss resistance (R) :
[13] (X[L] and R in
Example :
A 3mH coil with a loss resistance of 8
The loss resistance of the coil is caused by :
On LF in most cases these last losses can be ignored if some care is taken by selecting the form material and the location of the loading coil. The Ohmic losses are determined by the resistance of
the coil. Due to 2 effects the restistance of the coil will be frequency dependent and often will be considerably larger than the DC resistance :
a. Skin effect
As the frequency get higher the current through a wire will tend to flow mainly throught the outer layer, with little or no current through the centre of the wire. Since only a part of the wire
surface is used the AC resistance will be larger than the DC resistance and will increase as the frequency increases.
The thickness of this outer layer (d) is :
[14] (d = skin layer in mm, K = material dependent constant and f = frequency in kHz)
│ │Copper │Aluminium│Messing│Silver │ Gold │
│ K │ 2.08 │ 2.77 │ 4.45 │ 2.02 │ 2.37 │
│conductivity (S/m) │58x10^6│ 33x10^6 │13x10^6│62x10^6│45x10^6│
Once the wire diameter exceeds twice the skin depht the wire has 'wasted space' inside. Thus it is more efficient to use 2 (or more) thin wires in parallel than a single thick wire. Such HF-wire
consisting a large number of parallel (isolated) wires is called litz wire. It is a very good choise to built a coil as a litz wire will only have a fraction of the loss of a single wire of the same
diameter, but unfortunately it is also rather expensive.
Be aware that in multi-stranded wire the individual parallel wires are not isolated, so this is not litz wire !
b. Proximity effect
When an AC current is flowing through 2 wires that are close together the currents will tend to flow at maximum distance from each other, causing an effect similar to the skin effect. This will cause
an additional increase of the loss resistance. This effect can be minimized by inceasing the wire spacing. But as the wire spacing is increased one will need mure turns (and thus more wire) to
achieve the same inductance. So what is won by reducing the proximity effect is lost again (or even worse). Experiments have shown that the lowest loss is achieved when the wire spacing equals the
wire diameter.
Special coil winding techniques have been developed to keep the turns of a coil as far as possible apart without increasing the coil dimension (and thus needing more wire). One of these techniques is
basket weaving where the coil former is made of an odd number of rods and the wire is 'weaved' between them. That way high-Q coils can be made, even 'flat' coils and variometers, as shown below.
Remark : the proximity effect as shown here will happen if the currents flow in the same direction, as always will be the case in a coil. If the currents are flowing in opposite directions the
proximity effect will be similar, except that the currents now will tend to flow close to each other.
c. Optimizing coil dimensions
Apart from the wire material and diameter also the ratio of coil diameter / coil height and the ratio of wire spacing / wire diameter will affect the Q-factor. Best Q-factor can be expected for a
coil diameter / coil height ratio of ± 1.4 and a wire spacing / wire diameter ratio of ± 1.
The dimensions of an optimized coil (coil diameter / coil height = 1.4 and wire spacing / wire diameter = 1) are :
(n = number of turns, L = inductance in mH, d = wire diameter in mm, D = coil diameter in mm and l = coil length in mm)
Example :
Assume we want to make a 2mH coil optimized for best Q-factor, using 3mm diameter wire. Based on formula 15 we will need 66 turns on a coil of 39.3cm long and a diameter of 55cm.
d. Pyramid wound coils (by Niels Jorgensen, OZ8NJ)
A special technique called "Bank" or "Pyramid" winding to contruct compact and high-Q loading coils for 136kHz. This method was used to build coils and variometers for LF / MF marine and aero
The advantages of the method are obvious :
The drawback is that the proximity losses will increase, but the overall effect is positive.
Although 4 and 5 layer coils of this type do exist practical difficulties to keep the windings in place will often limit Pyramid winding to 2 layers.
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2.18.3. Variometer
For relative small variation of the loading coil inductance (up to 50%) a variometer can be used. This is a combination of 2 coils put in series, where a smaller coil (L[2]) is rotating inside the
larger coil (L[2]). This way the inductance can vary from about L[1]-L[2] to L[1]+L[2]. In theory a variometer could vary from 0 to 2*L (for L[1] = L[2] = L). But L[2] has to rotate inside L[1] and
thus has to be smaller. Further will a too large L[2] have a negative effect on the Q-factor of the variometer. As a result the practical limitation of the variometer range is 50%.
Instead of a small coil rotating inside a larger coil one can also use a small coil sliding in/out of a larger coil. But this design is not so popular as the mechanical construction is almost as
complicated for a rotating variometer while variation of L is only half and the Q-factor varies strongly with the position of the sliding coil.
A high-Q variometer can also be built using 2 flat basket weaved coils, as shown here.
An alternative way to built a variometer is by sliding a ferrite rod in a coil. Care has to be taken that the rod material is not saturated and that the rod is not heating too much. If a small rod of
the right material is used inide a large coil a variation up to 20% can be achieved without saturation or heating of the rod, even at a TX power of several 100W.
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2.18.4. Tapped coil
For larger variations of L a tapped loading coil has to be used. Taps at different turns of the coil allow variation of L of several 100%, but these variation will be in steps. Therefore the
combination of a tapped coil and a variometer is often used, either as a large tapped coil in series with a small variometer or as a 'all-in-one' solution where the large coil of the variometer is
2.18.5. Impedance matching
Bringing a short vertical monopole to resonance will ensure that any reactive component in the antenna impedance is compensated. Depending on the enviromental losses and loss in the loading coil the
impedance will be in the range of 20 to 200
Although in may cases the antenna impedance will be in the range of 30 to 70
As matching an LF antenna is not much different from matching any other antenna just a brief overview of some matching techniques is given, with LF specific remarks.
[16] (Z[1] ... Z[2] = impedances , n[1] ... n[2] = turns)
Example :
We want to match a 80 [1] = 19 and n[2] = 15 we come close to this value.
To allow matching to different impedances one (or both) sides of the transformer can be tapped. In most cases a toroid core will be used, although a ferrite rod might be used to. Care has te be taken
in the choice of the core material, most materials that do fine on HF will either have too much loss or will require too much turns to be suitable on LF. Most manufacurers produce core material that
is suitable for LF, but this can be rather hard to find. A low cost solution is to recuperate the "double-U" core that can be found in the HV-section of a TV set or monitor. Most of these cores can
be used in low and medium power applications on LF (up to few 100W), but it is recommended to do regular checks on efficiency and overheating in the beginning.
The number of turns should be large enough, the inductance of the coils should be at least twice (and preferable 5 times) the impedances that have to be matched.
One of the advantages of a transformer is that it will work rather wideband and that it will still work stable at high tranformation ratios. Further the transformer provides a DC shortcircuit to
ground and will ensure that there is no static charging on the antenna. In addition a transformer (not auto-transformer) provides a galvanic separation between the TX (RX) and antenna.
For more details on the designing of toroidal core transformers and an overview of suitable materials see here.
L-C network
(L = inductance in H, C = capacitance in F, Z[high] ... Z[low] = impedances in
For a frequency of 136kHz L and C can be calculated as :
(L = inductance in µH, C = capacitance in nF and Z[high] ... Z[low] = impedances in
Example :
We want to match a 80
Be aware that - in contradiction to a transformer - a L-C network is not widebanded. As the impedance transformation ratio increases the bandwidth will decrease. In order to have a stable network at
136kHz (ie. bandwidth of at least 2.1kHz) the transformation ratio should not exceed 5. For matching to 50
Further one should choose the right capacitors. Most ceramic and metalfilm capacitors will have too much loss on LF. Capacitors that can be used are :
• silver-mica : very good and available upto 47nF/500V, but expensive
• polystyrene : good, but values larger than 1nF hard to find at higher voltages
• MKC : good and up to 100nF/1kV, but not so easy to find
• polypropylene : good, cheap and available up to 100nF/500V
Be aware that - using high power - voltages up to several 100V can come across the capacitors, if you cannot find the high voltage types you can aventually use lower voltage types in series.
The inductances you will need are probably in the range of 20-200µH. At these values it is still possible to use air-coils, although they might be a bit bulky. If you use toroid cores ensure that the
material is suitable for high-power and high-Q operation at LF, a lot of cores that do fine in (wideband) transformers will not perform well in L-C networks.
A L-C network will not provide a DC path to ground. So if you are using this kind of impedance matching it is recommended to provide an additional discharge path to ground (eg. a 10k
Resonance transformation
Since for a short vertical monopole the reactive (capacitive) part of the antenna impedance is much larger than the resistive part (X[Ca] >> R[a]), the ratio of the turns is given by :
(L = loading coil inductance in H, f = frequency in Hz, n[1] ... n[1] = turns R[A] = antenna resistance in [TX] = transmitter impedance in
The above formula assumes perfect coupling between the coil windings. For most loading coils this is not the case, so the exact ratio of n[1]/n[2] will have to be determided by experiment. But the
formula will still give a good value to start with.
Example :
We have an antenna with a (loss) resistance of 80
The advantage of resonance transformation is that it provides an all-in-one solution for matching and bringing the antenna to resonance. But at the same time this is also a disadvantage as changing
the resonance will also affect the impedance matching and vice versa. With the practical values for antenna resistance and capacitance the ratio of turns will be rather large (50-200) and thus the
impedance matching can be rather critical.
If you use a secondary winding this should always be placed at the cold (grounded) end of the loading coil, in order to avoid flash over (and all the destructive consequences) between the loading
coil and the secondary winding.
2.18.4. Bandwidth considerations
The 136kHz ham band is only 2.1kHz wide (135.7-137.8kHz), so at first sight one would not expect bandwidth problems with the antenna. But keep in mind that 2.1kHz is 1.5% of 136kHz, so the relative
bandwidth is about the same as the (European) 7MHz band and even more than the (European) 144MHz band. In addition antennas are very short, what decreases their bandwidth. So in some cases it can be
required to retune the antenna when changing frequency.
C[A]) in series with a resistance (R[A]). To bring the antenna to resonance a loading coil (L) is needed and the antenna is fed by a transmitter that can be seen as a voltage source (U) in series
with a resistor (R[I]) where the resistor represents the transmitter impedance. If we assume that the transmitter is matched to the antenna (SWR 1:1) at resonance then R[I] is equal to R[A].
The bandwidth of a short vertical monopole will depend on the antenna capacitance and resistance. A small capacitance will reduce the bandwidth and so will a low resistance (you cannot have it all
...). It is not wise to intentionally increase the (loss) resistance as this will reduce the antenna efficiency, but another advantage of a high capacitance (= large) antenna is an increased
The main concern is the current reduction and SWR increase then moving away from the resonance frequency. A short vertical monopole can be seen as a damped series resonance circuit, where C = antenna
capacitance, L = loading coil inductance and R = loss resistance. At the resonance frequency L and C will cancel out each other and the impedance will be pure resistive (R). It is assumed that R is
matched to the transmitter impedance (mostly 50
Moving away from the resonance frequency will result in a reactive component is series with R, decreasing the antenna current and increasing the SWR. If the frequency offset from resonance is
relatively small (few % only = anything within the 136kHz ham band) then the value of the reactive component can be given as :
(X = reactance in [o] = resonance frequency in Hz, [o] in Hz, X[o] = reactance of C (or L) at resonance in
Example :
Assume we have an antenna with a capacitance of 300pF and a resistance of 60
At the lower band edge (135.7kHz) the reactive component X will be 74
Antenna current
The reactive component (X) of the antenna impedance will decrease the antenna current and thus the ERP. X can be calculated from formula 17.
We assume that the antenna is matched at resonance (X = 0, SWR = 1:1). If we move away from the resonance frequency the antenna will have a certain reactive component X and the antenna current will
decrease to :
(I[rel] = antenna current relative to the current at resonance, R = antenna resistance in
Since the radiated power is proportional to the square of the antenna current it will be :
(P[rel] = radiated power relative to the power at resonance, R = antenna resistance in [o] = untuned antenna reactance in [o] in Hz and f = resonance frequency in Hz)
Example :
Assume we have an antenna with a capacitance of 300pF and a resistance of 60
The other way arround, if a certain power loss (P[loss]) is acceptable the bandwidth (B) of the antenna system is :
(B = bandwidth in Hz, P[loss] = maximum acceptable power loss relative to the power at resonance, R = antenna resistance in [o] = untuned antenna reactance in
For 136kHz the bandwidth can be calculated from the maximum acceptable power loss, antenna resistance and antenna capacitance (or loading coil inductance) as :
(B = bandwidth in Hz, P[loss] = maximum acceptable power loss relative to the power at resonance, R = antenna resistance in
Example :
Assume we have an antenna with a capacitance of 300pF and a resistance of 60 [loss] = 0.11. Based on formula 14 the antenna bandwidth will be 1.5kHz.
Standing Wave Ratio (SWR)
Apart from decrease of the antenna current and radiated power a reactive component X will also increase the SWR.
Assuming that the SWR at resonance (X=0) is 1:1, a reactance X in series with R will increase the SWR to :
[23a] (R = resistance in
If the frequency offset is small, and thus the SWR is not too high, the formula can be simplified to :
[23b] (R = antenna resistance in
Formula 23b will have an accuracy of 10% or better if the outcome is an SWR of less than 3:1. For higher SWR it is recommended to use formula 23a.
For 136kHz the SWR can be directy calculated from
([o] in Hz, R = antenna resistance in
If the frequency offset is small, and thus the SWR is not too high, the formula can be simplified to :
([o] in Hz, R = antenna resistance in
Formula 23d will have an accuracy of 10% or better if the outcome is an SWR of less than 3:1, otherwise it is recommended to use formula 23c.
Example :
Assume we have an antenna with a capacitance of 300pF and a resistance of 60
If we use the simplified formula (23d) the SWR at the lower band edge (135.7kHz) is calculated as 4.2:1, at the upper band edge as 2.2:1. Since the SWR at the lower band edge is over 3:1 we have to
recalculate is with formula 23c, the outcome is 3.2:1. For the higher band edge the outcome of formula 23c is 2.1:1.
As shown by this example the simplified formulas (23b and 23d) produce accetable results for an SWR less than 3:1. At high SWR they are not accurate, but this does not matter too much as an SWR of
3:1 or more just tells you that the antenna has to be better matched anyway.
So in most practical cases you can stick to the simplified formulas.
The bandwidth of an antenna, based on the maximum acceptable SWR, is :
(B = bandwidth in Hz, R = antenna resistance in [o] = untuned antenna reactance in [o] = resonance frequency in Hz)
For 136kHz the antenna bandwidth can be calculated from the acceptable SWR, antenna resistance and antenna capacitance (or loading coil inductance) as :
(B = bandwidth in Hz, R = antenna resistance in
Example :
Assume we have an antenna with a capacitance of 300pF and a resistance of 60
Relationship between SWR and power loss
If the SWR is known (measured) the power loss can be calculated as :
(P[loss] = power loss relative to power at SWR 1:1)
(P[loss] = power loss relative to power at SWR 1:1 and log = 10 based logaritm)
Example :
A SWR of 1.7:1 will result in a 7% or 0.3dB power loss.
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3. Efficiency of antenna systems on LF (short vertical antennas)
3.1. Antenna system
Under the term antenna system I mean more than just the antenna itself. It includes all surrounding parts that affect the radiation of the transmitter power : the transmission lines, matching devices
and even enviroment.
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3.2. Efficiency
If an antenna is fed with a certain power it will radiate a part of that power. The remaining part is dissipated 'uselessly', in most cases converted to heat in or arround the antenna. Simplified one
can say that the transmitter feeds its power into 2 resistors, the radiation resistance (R[A]) and the loss resistance (R[L]).
The efficiency (n) of an antenna is :
On HF the efficiency of most antenna systems is very high, 90% or more. The most important sources of loss are skin effect in the antenna wires and dissipation in the transmission line (coax cable).
On VHF and higher frequencies these last can become very important.
On LF the situation is completely different, efficiencies of most antennas used by hams are in the range of 0.01 to 1%. The source of these high losses is dependent on the type of antenna. For
electrical antennas the major losses will be mainly in the enviroment and the loading coil. For transmitting the efficiency of the antenna system will directly affect the amount of radiated power and
thus is very important. But for receiving on LF it is mainly the ratio between wanted signal and unwanted signals (noise, QRM) that determines the quality of the antenna system. Therfore the
efficiency is rather unimportant in a receiving antenna system.
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3.3. Antenna system efficiency, antenna directivity, ERP, EIRP and EMRP
There is often a confusion between the terms efficiency and directivity or gain. While the efficiency is determined by the ratio of the transmitter power that is radiated, the directivity (often also
called gain) is determined by the shape of radiation pattern. Even the term directivity is often only associated with directional antennas as Yagi's, Quads etc... But in practice any antenna has a
certain gain, unless it radiates equally in all directions and under all angles. This 'gainless' antenna (that does not exist) is called a isotropic radiator and is taken as reference for the gain of
other antennas (then the gain is given in dB[i]).
A 1/2 wave dipole has a gain of 2.15dB[i] and is often also taken as reference (then the gain is given in dB[d]).
Apart from the unchangable propagation parameters the signalstrength of a certain station on a certain frequency depends on the transmitter output power, the directivity of the antenna and the
efficiency of the antenna system. These 3 parameters combined determine the radiated power, given in Watt. Depending on the reference antenna there is :
Despite the fact that a short vertical antenna has a omnidirectional radiation pattern in the horizontal plane, it has a gain of 4.78dB[i] or 2.62dB[d] due to its directivity in the vertical plane.
This gain is almost independent of the antenna height, as long as it it short compared to the wavelength (don't be fooled by the term short, at 136kHz it means 100m or less).
As the result of the different reference antennas EMRP will always be 2.62dB above ERP and 4.78dB above EIRP while ERP is always 2.17dB above EIRP.
Example :
Assume we feed a short vertical antenna with a radiation resistance of 0.04 [d] (x 1.83) the ERP is 244mW. This means that the antenna system and transmitter as described here will produce the same
signal strength as a power of 244mW sent into a perfect 1/2 wave dipole. The EIRP of this station will be 400mW
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3.4. Optimizing the antenna system efficiency
In order to improve the ERP of a LF station one can either increase the transmitter power or improve the antenna system efficiency. Although some dB's can be won by brute power the practical limit of
increasing the transmitter power is often reached at 1-2kW. Any further improvement has to be done by optimizing the effeciency of the antenna system. This means increasing the radiation resistance
of the antenna and/or decreasing the loss resistance.
The radiation resistance can be increased by making the antenna higher and adding capacitive and/or inductive toploading. A more complicated option is to implement multiple vertical elements.
The two most important components of the loss resistance are the losses in the loading coil and the ground/enviromental loss. The coil loss can be reduced by improving the coil's Q but indirectly
also the capacitive topload will affect the coil loss, as for a larger antenna capacitance a smaller loading coil is needed. Depending on the value and Q of the loading coil, in most cases its loss
resistance will be in the range of 5 to 20
The major component of the loss resistance is almost always the enviromental loss. It of often just called ground loss, although this last is only a part of the enviromental loss. The enviromantal
loss is dependent of many factors such as soil type, objects near the antenna and even the shape and size of the antenna. In most cases it will be in the range of 30 to 150
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3.5. Enviromental losses
The picture shows a simplified model : a T-antenna with a nearby tree. The antenna has capacitive coupling to the tree (C[T]) and to the ground (C[G]). Each of this 2 capacitances will form a
return-path for the antenna current.
Example :
Assume that C[T] is 300pF and C[G] is 150pF. Further assume that the loss resistance of the tree (R[T]) is 200 [G]) is 50
Based on the values of C[T], C[G], R[T] and R[G] the total capacitance (C[X]) and enviromental loss resistance (R[X]) can be calculated : 450pF and 94 [X] is larger than R[G]. But this is due to the
fact that in this example the capacitive coupling between the antenna and the tree (C[T]) is much larger than the capacitive coupling between antenna and ground (C[G]) and therefore R[T] contributes
much more to R[X] than R[G] does. Notice that about 2/3 of the total antenna current is flowing back via C[T] and R[T].
The presence of objects such as trees and buildings near the antenna will have several effects :
The increase of the enviromental loss resistance and decrease of the radiation resistance has a negative effect on the efficiency of the antenna system. The increased antenna capacitance has a
positive effect, as you will need a smaller loading coil, but by no means it will compensate the negative effects.
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3.6. Ground loss
On 'its way back' to the feeding point the antenna current flows (partly) through the ground. Since the soil is a rather poor conductor a loss resistance will be created.
The value of this loss resistance depends on :
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3.6.1. Type (composition) of the soil
The ground loss is very dependent on the composition of the soil and is inverse related to the conductivity of the soil (the higher the conductivity the lesser the loss). Soil conductivity can be
measured, but the results of these measurements should be taken with some caution for several reasons :
│ Soil type │Conductivity (mS/m) │
│ salt water │ 1000 │
│fresh water │ 1 │
│ wet soil │ 1 - 10 │
│ dry soil │ 0.01 - 0.1 │
│ meadow │ 0.5 │
│ loam │ 8 - 20 │
│ marsh │ 30 - 60 │
│ clay │ 500 │
Soil conductivity can be improved by salting the soil, but due to the ecological impact this is not recommended (in many countries it is unlawfull). Also most fertilizers will improve the
conductivity of the soil. Gypsum (calcium sulphate) is one of the best suited fertilizers, as it is relatively save to use and dissolves slowly (lasts a long time). Often (but not always) wet soil
has a higher conductivity than dry soil. On LF the use of salt or excessive amounts of fertilizers will have little or no effect as this will only improve the conductivity in the upper layer of the
soil while the LF signal penetrates deep into the ground. But longer periods of rain will wet the soil deep enough to affect the ground loss.
The signal radiated by the antenna penetrates in the ground. The deeper the signal penetrates the larger the ground loss will be. This penetration depth is inverse proportional to the square root of
the of the soil conductivity. If the soil conductivity S is known the conduction depth D can be calculated (for 136kHz only) :
[27] (D in m and S in mS/m)
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3.6.2. Frequency
The penetration depth of the signal in the ground is not only dependent of the soil conductivity but also of the frequency. In the LF and lower HF region (30kHz - 3MHz) the penetration depth is
proportional to the square root of the wavelength (or otherwise said : inverse proportional to the square root of the frequency).
Based on the frequency F and the soil conductivity S the penetration depth D can be calculated :
[28] (F in Hz, S in S/m and D in m)
In most cases (and on 136kHz) the penetration depth will be between 40 and 150 metre, depending on the soil type. On salt water it will be only 1.5m.
Based on the above formula one could assume that the loss resistance will decrease with the square root of the frequency. But this will only be true if no other losses than ground loss are involved
and if the soil has a uniform structure down to several tens (or even hunderds) metre deep. In practice the frequency dependency of the loss resistance will be different on any other location and
might even change with the dimensions of the antenna. The first graph shows the results of the loss measurements done by M0BMU, ZL2CA and PA0SE on frequencies between 100 and 300kHz. The measurement
data were fitted to a R = K1/F^K2 function.
Other interesting measurements were done by Finbar O'Connor (EI0CF) and Alan Melia (G3NYK) who measured the loss resistance for different antenna configurations (at the QTH of EI0CF). These
measurements show that even changing the antenna configuration can change the frequency dependency of the loss resistance, as the second graph shows.
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3.6.3. Shape and dimensions of the antenna
Increasing height of a vertical antenna will increase the radiation resistance and thus the efficiency. But also increasing the topload will improve the efficiency. Apart from the fact that a larger
topload creates a better current distribution several hams have also noticed that the loss resistance can significantly decrease when the topload covers a larger area.
Laurie Mayhead (G3AQC) has significantly improved his station by experimenting with different configurations of the topload and developed the "Footprint theory". Read his comments :
I started out with a 3 wire Marconi "T" antenna. This had a 30m top and a 15m vertical section, spacing between the top wires was 0.5m. I increased the spacing to 0.7m with no measurable difference
in antenna current. The ground system was quite modest so I buried several hundred metres of wire with 10 earth rods at the ends of these radial wires. There was still very little increase in
current, so I ran a 100m wire into the salt water and another 100m wire out to my 4 square array which has about 100 radials and a further 20 ground rods. Still no better. I was getting about 1.8A
antenna current at the start and less than 2A at the finish. I measured 120
I think that there may have been a slight improvement because the aditional wire is further from the trees but I believe that the majority of the improvement is due to change in "current density" in
the ground under the antenna. I call this my "Footprint theory". Basically its as if the antenna was a shower head and the ground a big bath with lots of outlets. Take the case of a basic vertical
with no top load. The longer the vertical the higher the shower head and thus more outlets covered by the spray of water from the shower.Since each outlet can only get rid of a certain amount of
water the more that are covered the more water gets away, the analogy is the lower the resistance! With an inverted-L or T ant the top wires are like several shower heads spaced out along the wire.
So the longer the wire the more outlets recieving water.So its possible to bend the wire back on itself so long as the parts dont get too close. Alan Melia (G3NYK) commented that he visualised the
fringing fields of a stripline over a conducting plane,he surmised that the effective width of the spray might be of the order of half the height of the wire. This would indicate that for maximum
efficiency the normal parrallel top wires would need to be spaced by a distance equal to the height of the wires. I don't think that this is too far wrong.
In a recent mail to the reflector about the RUGBY 16kHz antenna it was mentioned that the original installation consisted of several multi wire cages,but that these were later replaced by a network
of wires covering all the available ground area. This tends to support my Footprint theory and I would really like to find a reference to this work in the literature.
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3.6.4. Radial system and ground rods
As mentioned before the antenna current 'returns' via the soil to the feeding point of the antenna. By adding a radial system and/or ground rod(s) the ground loss can be reduced.
The radial system contains a number of wires on or in the soil. As a general rule buried radials of bare wire are superior to radials on the ground or buried isolated radials. Burried radials should
be at least 15cm (6 inch) deep in the soil. Although blank copper radials can be used, galvanized iron radials are cheaper and will be less affected by corrosion. The additional loss of iron radials
(difference in conductivity between iron and copper) is almost always insignificant compared to the other losses.
Regarding the number of radials and their length the rule is simple : the more and the longer, the better. But there are some practical limits, once you have put a certain length of radials in the
soils further extension of the radial system will only result in a marginal reduction of the ground loss.
In general the efficiency of a radial system is based on :
Best result are achieved when the radials are equally distributed over the area below the antenna (see left picture). Placing 2 radials too close is not very effective and will hardly bring any
improvement over a single radial. Depending on the soil conductivity, radials need to be spaced at least 2m to 10m for optimal effect. When using many radials an optimized layout can reduce the
ammount of wire needed (and the work to bury the radials) without loosing efficiency (see right picture).
In addition to radials, ground rods can reduce the ground loss. These rods can be located at the feeding point of the antenna or at the end of the radials, eventually also somewhere 'half way' the
radials. Due to their relative small length it is essential that ground rods are blank metal. The longer (deeper) the ground rods are, the better. Getting the rods into the ground can be hard labour
if you do it using a sledge hammer and brute force. If you don't have too much rock in the soil there is an easier way :
Use 1 inch galvanized iron tubing used for plumbing and often sold in practical 3m lengths. Connect your garden hose to one end of the tube and let the water flow trough the tubing. Hold the tube
vertical on the soil, the water will wash the soil away and the tube will gently sink into the soil. Be aware that the tube can keep sinking into the soil even if you shut the water off, so it might
be nessecary to secure the tube for some days to avoid a 'China syndrome'.
As so many things will influence the efficiency of a radial system (with or without ground rods) it is very difficult to predict how many radials and/or ground rods will give an optimal result in a
particular case. The best way is to start with a limited number of radials / ground rods and gradually increase the ground system while measuring the loss resistance. That way you will find the point
where further extension gives little or no improvement.
If you have problems to get long ground rods into the soil : one long rod can be replaced by some shorter rods, keeping the total length the same. In order to have maximum efficiency all rods should
be separated by a distance that is at least their length (if possible twice their length).
I have found some result of research on the effect of the radials on the antenna efficiency in the long wave range. But this was in regard of commercial stations, where an efficiency of 80% was
aimed. For a soil conductivity of 2mS, a wavelength of 2000m (150kHz) and an antenna length of less than 50m (170ft) the optimal length of a radial was found as 150m (500ft) and the optimal number of
radials is 120. But I am afraid that only very few hams will have the possibilty to install such a radial system.
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4. Measuring ERP on LF
4.1. Electric field / magnetic field & near field / far field
Despite the fact that some antennas are called electric antennas (eg. Marconi antenna) while others are called magnetic antennas (eg. small loop), any antenna will create both an electric and
magnetic field.
Based on the above we can determine the :
Near field : area close to the antenna. The electric and magnetic fields are not coupled and the radio of their strength depends on the type of antenna and the distance from the antenna.
Far field : area at sufficiant distance from the antenna. The electric and magnetic fields are coupled and at a constant ratio.
induced field (that is dominant near the antenna) and the radiated field (that is dominant at a sufficiant distance). The induced field is much stronger at the antenna but declines much faster than
the radiated field. The induced field just stores energy in the space arround the antenna (similar as a capacitor can store energy in an electric field or an inductor can store energy in a magnetic
field) while the radiated field really radiates energy.
At a distance of about 0.16 wavelengths both fields will be equal and some sources let the far field start from here. But in practice it is recommended to respect a distance of at least 0.5
wavelengths for accurate fieldstrength measurements.
At a sufficiant distance from the antenna both fields will be in opposite polarization (so if the electric field is vertical the magnetic field will be horizontal), in phase and their ratio is
determined by the free space impedance (377
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4.2. Calculated ERP versus ERP measurements
A few hams have actually measured their ERP, the results are in the table below :
│ Call │ Antenna │ QTH │ Distance │Measured ERP│Calculated ERP│Difference│No. meas.│
│PA0SE │Umbrella (18m high, 2 topwires of 20m sloping to resp. 14m / 10m)│ ? │ 5.8km │ 95mW │ 313mW │ -5.2dB │ 1 │
│M0BMU │ Inv-L (8m high, 40m long) │ ? │1km - 6km │ 76mW │ 146mW │ -2.8dB │ 123 │
│M0BMU │ Umbrella (17.5m high, 2 topwires each 20m long) │ ? │1km - 6km │ 263mW │ 331mW │ -1.0dB │ 130 │
│G3AQC │ Inv-L (14m high, 150m long) │Antenna surrounded by trees│3km - 9km │ 49mW │ 113mW │ -3.6dB │ 3 │
│SM6PXJ│ Umbrella (20m high, 4 topwires of 20-25m) │ On hill, no obstacles │3km - 10km│ 370mW │ 720mW │ -2.9dB │ 9 │
It is remarkable that in all cases the measured ERP is below the calculated ERP, differences going from 1 to more than 5dB. In the fall of 2001 Jim Moritz (M0BMU) has done over 200 ERP measurements
at distances of a few 100m up to 8km from the antenna. In addition Jim has measured the ERP of 2 antennas subsequently placed at the same location, a 8m high inverted-L and a 17.5m high umbrella
(inverted-V) antenna. These measurements show that the measured ERP is relatively constant from a distance of 1km upward, so the origin of the extra loss is in the direct environement of the antenna.
Further the 17.5m high antenna seems to suffer less from this loss than the 8.5m high antenna, this may indicate that this extra loss is caused by objects near to the antenna and that this loss is
dependent of the radio of the antenna height versus the dimensions of these objects (thus a bigger antenna suffering less from this loss).
Finaly the measurements confirm that the distance to the antenna has to be at least 1km (+/- 0.5 wavelenght), if possible at 2km (+/- 1 wavelength) or more, in order to get meaningfull results :
│ Distance │Average measured ERP │Standard deviation│No. mesaurements│
│ 0km - 1km │ -8.1dB │ 2.5dB │ 12 │
│ 1km - 2km │ -12.1dB │ 1.4dB │ 14 │
│ 2km - 3km │ -11.1dB │ 1.1dB │ 27 │
│ 3km - 4km │ -11.4dB │ 1.0dB │ 22 │
│ 4km - 5km │ -11.2dB │ 1.1dB │ 24 │
│ 5km - 6km │ -11.2dB │ 0.6dB │ 32 │
│more than 2km│ -11.2dB │ 0.9dB │ 109 │
│more than 4km│ -11.2dB │ 0.8dB │ 60 │
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4.3. How to measure ERP
The ERP is determined by measuring the fieldstrength E at a known distance d from the antenna :
[29] (P[EIRP] in W, E in mV/m and d in km)
Remember that you need to multiply the EIRP by 1.64 (or add 2.15dB) to get the ERP.
Measuring ERP is not very complicated, but one should keep a few things in mind :
For practical reasons a small loop or ferrite rod antenna is best suited for ERP measurements. First of all a magnetic antenna is less sensible to the environement (compared to a small whip) and also
it is easier to calibrate. The calibration is done by creating a known magnetic field, using a pair of Helmholtz coils (see picture). Be aware that is in the combination of antenna and receiver that
has to be calibrated.
A detailed procedure how to perform fieldstrength measurements is described by Dick Rollema (PA0SE), have a look here. It includes the construction of a simple measurement receiver, the pair of
Helmholtz coils and the calibration procedure.
More interesting reading on fieldstrength measurements on LF, by Christer Andersson (SM6PXJ), can be found here.
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5. Small loop antennas
5.1. Single turn small loop as transmitting antenna
In this chapter the small loop antenna will be discussed as a transmitting antenna. For receiving-only small loop antennas I refer to chapter 7. For simplicity we will start with a single turn loop.
single turn loop antenna with an area A. If A is small compared to the wavelength then :
• The antenna will act as an inductance (L[A]) in series with the radiation resistance (R[A]) and the loss resistance (R[L])
• The antenna voltage (U) will decrease linear over the entire loop
• The antenna current is constant through the entire loop
As for any other small antenna the ERP is mainly determined by the ratio of the radiation resistance to the loss resistance (apart from the TX power of course). Only in second order the antenna
directivity (gain) will play a role. So the basic rule "the bigger the better" is also valid for small loops.
The radiation resistance R[A] of a single turn small loop antenna with an area A and at a wavelength
For 136kHz this is :
[30b] (R[A] in µ^2)
The radiation resistance depends on the loop area, not on the loop shape. This means that a 20m high and 20m long loop will have the same radiation resistance as a 8m high and 50m long loop (both are
400m^2). But the 20m x 20m loop will require only 80m wire while the 8m x 50m loop needs 116m. As a result copper losses in the square loop (20x20) will be 45% less than these in the rectangular loop
Although a square loop will have a better efficiency than a rectangular loop with the same area (and a circular loop would even be better than a square), practical limitations will often make a
rectangular loop the best solution.
The inductance of a loop depends on the shape, dimensions and wire diameter :
(L[A] = loop inductance in µH, P = loop perimeter (circumreference) in m, A = loop area in m^2, d = wire diameter in mm and ln = natural logaritm)
(This formula was derived by Claudio Girardi (IN3OTD) from a general formula for the inductance of a single turn loop proposed by Bashenoff in 1928. The accuracy is within a few % for the most common
loop shapes.)
A more specific formula for rectangular loops derived by Alexander Yurkov (RA9MB) is :
(L[A] = loop inductance in µH, a / b = height / length of the loop in m, d = wire diameter in mm, ln = natural logaritm and
The 3 main components of the loss resistance (R[L]) are :
The wire loss depends on the length, diameter and composition of the loop wire. Due to the skin-effect it often is better to use several thin wires in parallel than a single thick wire. For those who
can afford it litz-wire is the best choise.
Since a small loop is inductive the matching network to bring the antenna to resonance (and eventually match it to 50
Even though a small loop antenna is much less sensitive to the enviroment than a short vertical antenna the enviromental loss is still important. The best way to reduce this loss low is to keep all
wire as far away as possible from the ground and surrounding objects.
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5.2. Efficiency of a loop
The efficiency of an antenna is determined by the ratio of the radiation resistance (R[A]) to the loss restistance (R[L]) :
For a small loop antenna the efficiency will be very low, typical far less than 1%. So by measuring the RF resistance of a small loop you will be able to determine the loss restistance. The radiation
resistance can be calculated using formula 30 and thus the efficiency can be determined.
Example :
Assume you have a rectangular loop of 15m high and 30m long. This loop has an area of 450m^2 and a perimeter of 90m. Based on formula 30 the radiation resistance is 270µ
A giant loop of 25m high and 100m long, using 10mm diameter copper wire, will have an efficiency of -31dB or 0.07% (0.008
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5.3. Enviromental losses of small loop antennas
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5.4. Single turn loop versus multi turn loop
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5.5. Directivity and polarization of a small loop antenna
In contradiction to a vertical antenna the small loop antenna is not omnidirectional in the horizontal plane. It has a kind of 8-shaped horizontal radiation pattern with a 3dB opening angle of 90°.
The nulls are in theory infinite if the loop would be completely insensitive to the electric field. But in practice most transmitting loops are not shielded and thus will pick up some of the electric
field. As a result practical values for the loop nulls are -20 to -30dB.
Keep in mind that the maxima of the horizontal radiation pattern are on the plane of the loop while the minima are orthogonal to the loop plane. As long as the loop dimensions do not exceed a
wavelength the radiation pattern does not depend on the loop size.
The gain of a small loop antenna is 1.76dB[i] or -0.39dB[d]. Dispite the fact that a loop antenna has some horizontal directivity its gain is less than that of a short vertical antenna (4.77dB[i] or
2.62dB[d]). This is caused by the fact that a short vertical antenna has more directivity in the vertical plane.
Example :
Assume a rectangular loop of 15 by 30m as described in the previous example. The total loss resistance of this loop is 2.1
[loss] of 11 [rad] of 0.008 [ant] of 3A and a P[rad] of 73 milli-Watt. In this case the EIRP is 109mW, the ERP is 66mW.
A full size loop antenna in the vertical plane can be horizontal or vertical polarized (see picture at the right). This is because the current distribution over the antenna is not uniform.
At LF however the antenna dimensions are small (compared to the wavelength). As a consequence the current is constant over the entire antenna, what means that the polarization of a small loop antenna
must be independent of the location of the feeding point.
One way to approach this is to assume that a small loop antenna acts similar to a coil : the current is constant thoughout the coil/loop and the voltage is built up over the coil/loop. For a coil it
is known that it generates a magnetic field in right angles to the plane of the winding(s). Thus for a vertical oriented small loop the magnetic field is horizontal. Further we know that the electric
field of an antenna must be orthogonal to the magnetic field, thus the electric field is vertical. Since the antenna polarization is always reffering to the plane of the electric field a small loop
antenna - in the vertical plane - will always be vertical polarized, regardless of where it is fed.
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5.6. Bringing a small loop antenna to resonance
5.6.1. Resonance capacitor
A small loop antenna can be seen as an inductance (L[A])in series with a resistor (R[A]). The inductance depends on the loop dimemsions (see here) while the resistance in mainly determined by the
The antenna is brought to resonance by a series capacitor (C[res] in F) :
(L[A] = loop inductance in H and f = frequency in Hz)
For 136kHz C[res] can be calculated as :
(C[res] = resonance capacitor in nF and L[A] = loop inductance in µH)
At resonance the inductive and capacitive reactances will cancel out each other and the resulting impedance will be resistive and equal to R[A].
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5.6.2. Impedance matching
Once the loop is brought to resonance the impedance is resistive but probably still different from the transmitter impedance. A typical loop resistance is in the range of 0.5 to 5
Impedance matching can either be done using a transformer or using a variant on the L-C network (in fact we will need a C-C network).
[34] (Z[1] ... Z[2] = impedances , n[1] ... n[2] = turns)
Example :
We want to match a 1.5 [1] = 23 and n[2] = 4 we come close to this value.
Details on impedance matching with a transformer were given when the matching of short vertical monopoles was discussed (see here). More about the design of a transformer and the selection of care
material can be found here.
As the transformer ratio is rather large the choise of the right material to make a transformer for a small loop antenna can be more critical than for a short vertical monopole.
C-C network
(L = antenna inductance in H, C[1] ... C[2] = capacitances in F, R[A] antenna resistance in [TX] = transmitter impedance in
For a frequency of 136kHz C[1] and C[2] can be calculated as :
(L = antenna inductance in µH, C[1] ... C[2] = capacitances in nF and R[A] antenna resistance in [TX] = transmitter impedance in
Example :
We want to match a small loop with an inductance of 70µH and a resistance of 0.65 [1] = 204nF and C[2] = 21.6nF.
One should take care to choose the right capacitors. Most ceramic and metalfilm capacitors will have too much loss on LF. Capacitors that can be used are :
• silver-mica : very good and available upto 47nF/500V, but expensive
• polystyrene : good, but values larger than 1nF hard to find at higher voltages
• MKC : good and up to 100nF/1kV, but not so easy to find
• polypropylene : good, cheap and available up to 100nF/500V
Be aware that - using high power - currents of several 10A can flow through the capacitors. Eventually it can be usefull to use several capacitors in parallel.
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5.6.3. Bandwidth considerations
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6. Other transmitting antennas
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7. Antennas for reception
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8. Software
• download RJELOOP1 : Transceiving, single-turn, magloop antennas of various regular shapes (by G4FGQ)
• download VERTLOAD : Base-fed vertical antennas, coil-loaded at any height, with coil design (by G4FGQ)
• download EARTHRES : Ground electrodes, rods, wires, plates, mats. Soil Resistance measurements (by G4FGQ)
• download TANT136 : LW & MW performance of small T-antennas above a system of ground radials (by G4FGQ)
• download SOILSKIN : Enter soil characteristics. Display a table of skin depth vs frequency (by G4FGQ)
• download SOLNOID2 : Design of cylindrical, single-layer, air-core coils of all proportions (by G4FGQ)
• download MAGLOOP4 : Performance of regular-shaped magloops versus height and type of ground (by G4FGQ)
• download MIDLOAD : Design and performance of very short, centre loaded dipoles above lossy ground (by G4FGQ)
• download LOADCOIL : Design of short vertical antenna + loading coil. Slide coil up/down for max effiency (by G4FGQ)
• download GRNDWAV3 : Groundwave propagation and field strength vs pathlength, terrain and frequency (by G4FGQ)
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9. Appendices
9.1 High power applications of toroidal core coils
9.1.1. About toroidal cores
Using toroidal cores one can make coils that are compact and have little stray field. These properties make toroids popular for the design of transformers and for inductances in impedance matching
and filter circuits. The inductance (L) of a toroidal core coil can easily be calculated based on the material specific inductance factor (A[L]) :
(L in nH, A[L] in nH/turn^2, N = number of turns)
Sometimes A[L] is expressed in µH / 100turns, the conversion to nH/turn^2 is : 1 µH / 100 turns = 0.1 nH/turn^2.
The power that can be handled by a toroidal core is limited by 2 factors : saturation of the core material and temperature rise of the core.
Material saturation
At low currents the magnetic flux in a core will be proportional to the current through the coil and the inductance will be constant (independent of current or flux). But if the current exceeds a
certain value the magnetic flux will no longer increase proportional to the current and the inducance will drop. This is called saturation and it will cause a distortion of the signal.
To make things worse, the inductance drop will also cause the reactance to drop and thus the current will increase. This will cause a further inductance drop etc... , a nice example of a chain
So, at any time saturation of the core material should be avoided.
The peak flux density in the core is given by :
(B[pk] = flux in T, E[RMS] = applied voltage in V[RMS], A = effective cross-sectional area in m^2, N = number of turns and f = frequency in Hz)
The flux should not exceed 0.2 to 0.25T for ferrite materials and 1 to 1.2T for iron powder.
Temperature rise
The temperature rise of the core is caused by permeability losses in the core and the copper losses in the wire :
(T = temperature rise in °C or K, P[dis] = total power dissipation in mW and K = core surface area in cm^2)
The above formula is an approximation for a core in free standing air. Temperature rise can be decreased if a forced air cooling (fan) is used.
Core material : iron powder versus ferrite
For typical high power RF applications ferrite cores will be limited by saturation while iron powder cores will be limited by temperature rise. Further a ferrite core will have a 100-200 times higher
A[L] compared to an iron powder core of the same dimensions (and thus needs 10 to 15 times less turns for the same inductance).
In general can be said that, at LF, ferrite is best suited for wideband applications such as transformers and baluns while iron powder is prefered for high-Q applications such as resonant circuits,
low-pass filters and L-C impedance matching circuits. In principle one could use iron powder also for wideband applications, but the number of turns needed is often impracticable high.
9.1.2. Designing a ferrite cored transformer (by Jim Moritz, M0BMU)
Some maths is needed to design a ferrite cored transformer.
The first stage is to chose a number of turns that can withstand the applied voltage without saturating the core. You need to choose the number or turns (N) so that peak flux density (B[pk]) is less
than about 0.2 to 0.25T for most power ferrites (formula 37).
You also need to ensure the inductance of the transformer is high enough, usually so that the reactance is at least 5 times the impedance level :
(A[L] = the material specific inductance factor, usually in nH/turn^2, L = inductance in the same units as A[L], N = number of turns, X[L] = reactance in Ohm and f = frequency in Hz)
If A[L] is not known, it can be calculated from :
(A[L] in nH/turn^2, µ[0] = permeability of free space (4*^-7), µ[E] = effective permeability (usually about 2000 for power ferrites), L[eff] = effective magnetic path length in m and A[eff] =
effective cross-sectional area in m^2)
Or you could wind some turns on the core and measure the inductance.
The parameters µ[E], L[eff] and A[eff] have to be retrieved from the manufacturers data sheet, eventually L[eff] and A[eff] can be determined from the core dimensions, if you can accept an error of a
few %.
Example :
Assume we have a core with an inner diameter of 30mm, an outer diameter of 46mm and a height of 12mm.
The average diameter is 38mm [(30+46)/2], so L[eff] =
The inner diameter is 15mm [30/2] and the outer diameter is 23mm [46/2], A[eff] = (outer diameter - inner diameter) * height = 96mm^2 [(23-15)*12]. Due to the rounded egdes A[eff] will be slightly
less, but in most cases the error is acceptable.
There is also core losses to be taken into account and the temperature rise these produce, but this is more complicated to work out and requires the manufacturer's data. And for ferrite cores
generally saturation will occur before the critical temerature rise is reached. Using a generously sized core to start with usually means the transformer will run cool. You should choose wire sizes
that fill up the available winding space, to minimise resistive losses.
As a few examples of LF TX power applications for ferrite cores, I have used :
• A 3C8 material EC59 transformer core for an antenna matching transformer up to 1.2kW on 136/73k
• A Neosid F44 material EE42 transformer core (available from RS components) for a PA output transformer at up to 600W on 136k
• A 3C85 Material ETD49 (from Farnell Electronic Components) core for a 1kW PA output transformer on 73k
All these components run at a comfortable temperatures without additional cooling at the given power level.
The 3C8, 3C80, 3C85, 3C90 materials are manganese-zinc ferrite of a type that seems to be ubiquitous for switch-mode designs. The Neosid F44 and Siemens N27, N67, and N87 grades are very similar -
the higher numbers represent newer materials with slightly improved performance, but the differences are not great. Most ferrite manufacturers seem to make something very similar. These all seem to
work well as transformers at 136k. There are newer materials around like 3F3, but the availability seems to be very limited.
In general, transformers require maximum inductance with minimum turns, so no air gap is used. For maximum energy storage with minimum losses in a resonant application, or for a choke with a large DC
current component, an air gap is required. Power types of ferrites can be used with an air-gap to produce inductors for resonant or filtering applications, but it seems to be difficult to get a Q of
more than about 100 or so; The micrometals -2 mix iron dust cores seem to be better in this application. The iron dust cores made for SMPSU applications also seem to have relatively high losses at
136k. For small-signal applications, the nickel-zinc HF ferrites like 4C65 seem to be capable of very high Q, but are not particularly suited to LF power applications, and are not very widely
available. They usually have much lower permeability, 100 or so. The manganese-zinc ferrite, gapped pot cores with tuning slugs (eg RM series) seem to give a Q between 100 - 200 at LF for low-level
filter applications.
The ferrites designed for RFI supression seem to have high losses over a wide frequency range (which is what they are designed for , of course), so not very good for signal processing. The very high
permeability manganese zinc ferrites for signal transformers tend also to have higher losses, and need to operate at lower B[pk].
Surplus ferrites can sometimes be used. for example, line-output transformer cores in video monitors are usually a 3C8-type material. But be careful to clean the mating surfaces so they are a close
fit, and remove any plastic film spacers, etc. 3C8 and 3C85 etc. toroids usually have a red or pink coating. 4C65 has a purple coating. The rough surface of low frequency manganese-zinc ferrites
usually has a fairly shiny, glazed appearance, whereas nickel-zinc ferrite has a more matt grey appearance. Ferrites are very hard and brittle, while iron dust materials can easily be cut with a
file. There are also various powdered metal cores, which look metallic when the coating is removed, but these do not seem to work well at 136k.
9.1.3. Designing an iron powder cored coil
Where for ferrite material the saturation will be the limiting factor, for iron powder it will be the temperature rise in the core. This temperature rise is caused by the power dissipation, as given
in formula 38. The core loss is determinded by the material, the frequency and the flux density. For most iron powder materials the core loss will rise more or less proportional with the frequency
and with the square of the flux density.
To give one example, for the popular '#2' material (red) that is available from several manufacturers :
(P[loss] core loss in mW/cm^3, f = frequency in Hz and B = flux density in T)
Since the surface area of a toroid core is approximately proportional to the square of the core diameter, while the volume is proportional to the cube of the diameter, a smaller diameter core can
dissipate more power per volume unit. As a result it can be usefull (and cost efficient) to put 2 smaller coils in series instead of one large coil.
back to top of this page
10. Acknowledgements
back to top of this page | {"url":"http://www.wireless.org.uk/on7yd/","timestamp":"2024-11-05T23:15:37Z","content_type":"text/html","content_length":"173816","record_id":"<urn:uuid:1ef0551a-0b38-4b39-a43a-86e274e69325>","cc-path":"CC-MAIN-2024-46/segments/1730477027895.64/warc/CC-MAIN-20241105212423-20241106002423-00623.warc.gz"} |
4.4 More Ways to Visualize Relationships: Point and Jitter Plots
Course Outline
• segmentGetting Started (Don't Skip This Part)
• segmentStatistics and Data Science: A Modeling Approach
• segmentPART I: EXPLORING VARIATION
• segmentChapter 1 - Welcome to Statistics: A Modeling Approach
• segmentChapter 2 - Understanding Data
• segmentChapter 3 - Examining Distributions
• segmentChapter 4 - Explaining Variation
□ 4.4 More Ways to Visualize Relationships: Point and Jitter Plots
• segmentPART II: MODELING VARIATION
• segmentChapter 5 - A Simple Model
• segmentChapter 6 - Quantifying Error
• segmentChapter 7 - Adding an Explanatory Variable to the Model
• segmentChapter 8 - Models with a Quantitative Explanatory Variable
• segmentPART III: EVALUATING MODELS
• segmentChapter 9 - The Logic of Inference
• segmentChapter 10 - Model Comparison with F
• segmentChapter 11 - Parameter Estimation and Confidence Intervals
• segmentChapter 12 - What You Have Learned
• segmentFinishing Up (Don't Skip This Part!)
• segmentResources
list High School / Advanced Statistics and Data Science I (ABC)
4.4 More Ways to Visualize Relationships: Point and Jitter Plots
We have learned how to make visualizations of outcome variables as a function of explanatory variables (e.g., histograms and bar graphs in a facet grid). We will learn a few more visualizations in
this section.
A scatterplot is a common way to show the relationship between an outcome variable and an explanatory variable. A scatterplot will show each data point as a dot on a graph. A scatterplot in ggformula
can be made with the function gf_point(). Let’s try using gf_point() to examine Thumb lengths by Sex.
gf_point(Thumb ~ Sex, data = Fingers)
You can change the color of the points with the argument color (much like we did before) and the size of the points with size.
gf_point(Thumb ~ Sex, data = Fingers, color = "orange", size = 5)
The problem with these gf_point() plots is that you can’t tell when a point is on top of another point. We can jitter these points around a little so that you can see all the individual points
better. We’ll use the function gf_jitter() to create a jitter plot depicting Thumb length by Sex. This function is just like making a gf_point() plot, except that the points will be a little jittered
both vertically and horizontally.
gf_jitter(Thumb ~ Sex, data = Fingers)
We can play with a few arguments to modify the jitter plot. As always, we can use the argument color. We can also use the argument height to change how the points get jittered vertically (i.e., a
little bit up or down). In this situation, we might want the vertical jitter (height) to be set to 0 so that a point at 60 mm really is a person who has a thumb length of 60.
gf_jitter(Thumb ~ Sex, data = Fingers, color = "orange", height = 0)
We can use width to change how the points get jittered horizontally (i.e., a little bit left or right). Height and width can be set to values between 0 and 1.
gf_jitter(Thumb ~ Sex, data = Fingers, color = "orange", width = .1, height = 0)
If a point is in the Female column, it’s a female’s thumb length. But being more to the left or right within the female column doesn’t mean anything. The jitter is there just so the points do not
overlap too much and obscure how many females have that thumb length.
In a jitter plot, a dense row of points shows that there are a lot of people with that thumb length. For instance, look at all the Female points at 60 mm. More points means more people with that
particular thumb length.
Just like a scatterplot, in a jitter plot you can change the size of the points by including the argument size. You can also change the transparency of the points using the argument alpha. alpha can
take values from 0 (more transparent) to 1 (more opaque).
gf_jitter(Thumb ~ Sex, data = Fingers, color = "orange",
size = 5, alpha = .2, width = .05, height = 0)
Try making jitter plots for a few of the variables from the Fingers data frame. Play around with some of the arguments such as height, width, color, size, and alpha. Try making a jitter plot one way,
then switching which variable is on the x-axis and which on the y-axis. Does it work both ways?
require(coursekata) MindsetMatters <- Lock5withR::MindsetMatters %>% mutate(WtLost = ifelse(Wt2 < Wt, "lost", "not lost")) # Play around with gf_jitter gf_jitter(Thumb ~ Sex, data = Fingers) ex() %>%
CK Code: ch4-7
There isn’t any restriction on what kind of variable you can put in the x- or y-axis. You can put an outcome or explanatory variable in either position. You can also put a categorical or quantitative
variable in either position. For example, in this jitter plot, we have put Sex on the y-axis and Thumb length on the x-axis.
gf_jitter(Sex ~ Thumb, data = Fingers, color = "orange")
Even though you can put the outcome variable anywhere, it is more common to put the outcome variable on the y-axis. We will follow that convention in our jitter plots because it conforms with what
people expect, and thus makes them easier to interpret. | {"url":"https://coursekata.org/preview/book/2bf8f9f7-fd3d-4c5f-a740-5130f528a1b6/lesson/6/3","timestamp":"2024-11-05T13:07:22Z","content_type":"text/html","content_length":"92982","record_id":"<urn:uuid:d6553b1b-373b-450f-8738-94f09d7322ad>","cc-path":"CC-MAIN-2024-46/segments/1730477027881.88/warc/CC-MAIN-20241105114407-20241105144407-00179.warc.gz"} |
Empirical measurement of m&m's Contained in a Standard Bottom Mouth Erlenmeyer Klein Flask and Comparison to Theoretical Models
Bernard Y. Tao
Dept. of Useless Information in an Effort to Win Free Klein Flasks
Large Midwestern University somewhere in the Soybean/Corn Belt
While m&m's and Klein bottle geometry ostensibly have little technical or economic value in common, research was undertaken to establish theoretical and empirical values for the number of m&m's that
could be contained in an Erlenmeyer Klein Flask (EKF). This research was performed solely for the purpose of obtaining such a flask absolutely free of charge, either with or without m&m's.
Four researchers at major midwest university independently developed methodology to estimate the number of m&m's contained in the aforementioned vessel, using data provide by the original problem
proposer. Three theoretical models were developed to estimate values, using a standard Erlenmeyer flask as a model for the Klein analog. An empirical measurement was used to confirm the validity of
all models, along with a statistical analysis of m&m size.
Empirical measurement found that the EKF contains 549 +/- 3 m&m's. One theoretical model produced values within 1% of this number, but other ranged as high as 14% difference. However, there may be
complicating factors in the method producing the higher error value.
Klein bottles are 2 dimensional topological structures that exist in 3 dimensions, having zero volume. However, they have been used for a variety of entertaining purposes, as noted in (1). Given the
highly multi-disciplinary nature of this problem, it was thought that a single approach would not be as successful in developing useful solution. Therefore, we attempted to develop solutions using
physical chemical, mathematical, sociological, and engineering approaches. The objective of this work was to obtain a completely free Erlenmeyer Klein Flask as noted in (4). However, if available, a
completely free Klein Stein of similar volume (5) would be preferable.
Materials and methods
The Erlenmeyer flask used to approximate the Klein analog was obtained from Fischer Scientific (Pittsburgh, PA), model 4980, 500 mL Pyrex. Two packages of m&m's were obtained from a local grocery
store (m&m/Mars, Hackettstown, NJ, net wt. 14.0 oz, milk chocolate). All calculations were performed on a Sharp Scientific calculator (model EL-5100S (Sharp Corp., Korea) using pre-installed
algorithms for transcendental functions. 15 cm ruler used to measure m&m oblate spheroid radii was from Davis Liquid Crystals, Inc. (San Leandro, CA). All other materials used were of reagent grade
or better.
Method 1: Physical Chemistry
Assume m&m's are oblate spheroids with a major axis radius of 0.6 cm (a) and a minor axis radius of 0.3 cm (b). The volume of oblate spheroid is (4/3)p a2b]. Assuming hexagonal close packing (hcp)
packing, the void volume % of the packaging is (4/3)p a2b/(p/1.2092) a2(4)b = 0.4031 or 40.31% he available volume is composed of m&ms (2, 3).
Assuming the Erlenmeyer Klein bottle is basically a right angle cone, we can use its dimensions of 240 mm x 100 mm (hxD). The volume of a cone [(p/3)*(D/2)2*h] with these dimensions is 628.32 cm3.
This means the effective volume filled with m&m mass must be approximately 628.32*0.4031 = 253.26 cm3. Dividing this value by the volume of a single m&m (0.4524 cm3), gives the number of m&m in the
bottle as 559.8.
Method 2: Mathematical estimation:
Alternatively, consider the specified volume of the flask. Given that the original volume of the Erlenmeyer flask is 500 cm3, this must be corrected to account for the volume of the neck, which is
approximately 110 ml, so the total volume would be 610 cm3.
Multiplying this value by the void volume of hcp packed oblate spheroids (0.4031) gives a volume of 245.89 cm3 of m&m's. Dividing this volume by the the volume of a single m&m (0.45239 cm3) gives the
total number of m&m's as 543.53.
Method 3: Liberal arts estimation
Fact: M&m's are mainly composed of chocolate.
Fact: Chocolate is recognized to have a transcendental power on bipedal mammals, particularly female humans.
Fact: The best known transcendental numbers are e and p.
Fact: Normal bipedal human female mammals have 20 digits, which are used to eat m&m's.
Fact: Erlenmeyer Klein Flask has 20 letters.
Armed with this knowledge, we can estimate the number of m&m's in the flask by multiplying the the number of bipedal mammalian digits used to ingest m&m's by the transcendental number e to the power
of p and add number of letters in Erlenmeyer Klein Flask to obtain:
20*ep + 20 = 482.81
This demonstrates that without an extensive knowledge of mathematics or physical chemistry, one can also simply estimate the number of m&m's in an Erlenmeyer Klein flask. However, as normal, this
also demonstrates that a liberal arts degree is essentially worthless in technical computations.
Method 4: Engineering estimation
2 Bags of m&m's were purchased and poured slowly into a 500 mL Erlenmeyer flask from the lab. This was followed by equilibrating the system using gentle agitation and addition of m&m's to fill to the
lip of the vessel. Subsequently, the m&m's were poured out into large plastic weighing dishes and manually counted. Ignoring the broken and chipped ones, which were used for personal metabolic
studies, the number obtained was 549 +/- 3.
Results and Discussion
Results of the 4 methods are summarized in Table 1.
Table 1. Raw data
Method 1, Chemist: 559.8
Method 2, Mathematician: 543.5
Method 3, Liberal Arts: 482.8
Method 4, Engineer: 549 +/- 3
There is inherent error in any of these calculations or estimates, given that the angles and details of elongation and extension of the neck of the flask to form the Klein nexus are unknown (see Fig.
1). Unfortunately, accounting for these errors is not possible without additional data on the geometrical issues.
It was found that although the assumption of an oblate spheroid for a single m&m is quite reasonable. Within the precision of the available instrumentation (my eye and a 15 cm ruler marked in 1 mm
increments), the radii of the long and short axes of an m&m were precisely 0.6 cm and 0.3 cm, respectively. However, extensive statistical analysis of the contents of two 14.0 oz bags of m&m's,
approximately 1100 pieces (aren't graduate students terrific!), demonstrated that the mean mass of m&m's is 0.911725 gm, with a standard error of 0.03174 gm. Ignoring the density changes between the
candy coating vs. the chocolate interior, this converts to a potential error of 3.48% in the volumetric calculation of a single m&m. Using this variation, the numbers previously obtained by methods 1
and 2 must be adjusted to yield the following corrected values (see Table 2):
Table 2. Corrected values
Method 1, Chemist: 559.8 +/- 19.5
Method 2, Mathematician: 543.5 +/- 18.9
Method 3, Liberal Arts: 482.8
Method 4, Engineer: 549 +/- 3
There are several obvious conclusions that can be developed from these results. First, the engineer's method (method 4) is highly accurate under the current situation. It employs the fewest number of
assumptions. However, the method employed is wholly empirical and does not account for the theoretical nature of m&m structure, the complexities of m&m packing within the vessel, or the statistical
issues involving population variation among the sample. Therefore, while valuable for the purposes of this contest, it cannot be extrapolated to other shapes, sizes, or situations.
The chemist's method (method 1) clearly has the highest absolute error value, nearly 20 m&m's. Additionally, the assumption of hexagonal close packing (hcp) is clearly in error, given the graphical
evidence shown on the image (see Fig. 1). The m&m's in the nexus clearly demonstrate body centered cubic packing (bcc), not hcp.
Method 3 yields results with a similar statistical error value, but gives a value that is approximately 3.0% lower than method 1. However, given the statistical error value, it is clearly close to
the actual value as found in method 3. This is probably due to the greater accuracy involved with measuring the additional volume of the flask neck, vs. the assumption of a conical shape, as in
method 1. Further improvement of method 1 might involve using a truncated cone assumption combined with a short right cylinder analysis to obtain an improved estimate of the additional volume of the
neck of the flask.
Method 3 clearly has significant shortcomings vs. the other methods. This is not surprising, given the highly heuristic nature of the methodology employed. The value obtained is nearly 14% different
from the values obtained by other methods, although the methodology employed is highly appealing and very simple. It does not account for structural geometry, physical chemistry, or statistical
variation, and has very little sound theoretical mathematical basis. Additionally, the veracity of the experimentalist may be in question. This question was raised due to the discovery that following
the experimental procedures, approximately 25% of the original mass of m&m's provided to the researcher were absent. She was noted in her lab book that this may have involved "cold fusion, high m&m
vapor pressures, or other unspecified errors". Additionally, a lab assistant noted a mysterious brown smudge on the researcher's lips, although this information is purely anecdotal.
Further research in this area may involve extension of the theoretical models developed herein to other geometries of Klein bottles, notably ellipsoidal, cylinderical, and modified spiral, provided
suitable research funding or free vessels could be obtained.
1. Anon., Acme Klein Bottle, http://www.kleinbottle.com/, Oakland, CA.
2. Castellan, G. W., Structures of Solids and Liquids, Chapt. 26, in Physical Chemistry, 2nd ed., Addison-Wesley, 1971, pp. 633-637.
3. Hoerl, A. E., Plane Geometric Figures with Straight Boundaries, Perry's Chemical Engineers' Handbook, 6th ed., McGraw Hill, 1984, p. 2-11,
4. http://www.kleinbottle.com/m%26ms_in_a_klein_bottle.htm, Acme Klein Bottle, Oakland, CA.
5. http://www.kleinbottle.com/drinking_mug_klein_bottle.htm, Acme Klein Bottle, Oakland, CA. | {"url":"https://www.kleinbottle.com/Bernie_Tao.htm","timestamp":"2024-11-02T20:06:15Z","content_type":"text/html","content_length":"15659","record_id":"<urn:uuid:1e5dedb3-0b8a-47db-92cb-cd4dd573fbd6>","cc-path":"CC-MAIN-2024-46/segments/1730477027730.21/warc/CC-MAIN-20241102200033-20241102230033-00549.warc.gz"} |
How do I create a string substitution variable for maven in Eclipse?
Author: Deron Eriksson
Description: This tutorial describes how to create a string substitution variable in Eclipse for maven.
Tutorial created using: Windows Vista || JDK 1.6.0_04 || Eclipse Web Tools Platform 2.0.1 (Eclipse 3.3.1)
It can be useful to create an EclipseSW string substitution variable for mavenSW if you execute maven as an external tool in Eclipse. If you upgrade your version of maven, you would only need to
update the string substitution variable once to take care of all references to maven in your external tool configurations rather than needing to update all of the configurations one at a time.
To create an Eclipse string substitution variable, go to Window → Preferences and Run/Debug → String Substitution. Click the New button.
I named my variable "maven_exec", as described in http://maven.apache.org/guides/mini/guide-ide-eclipse.html. I set its value to be the path to my mvn.bat file. I gave the variable a description. I
clicked OK and the OK. The "maven_exec" variable is saved.
I had previously created a "mvn clean" external tool configuration with the maven path hardcoded. I deleted this Location and clicked the Variables button.
I selected the "maven_exec" variable.
The "maven_exec" variable is now being used for my Location value.
That's all there is to it!
Related Tutorials: | {"url":"http://www.avajava.com/tutorials/lessons/how-do-i-create-a-string-substitution-variable-for-maven-in-eclipse.html;jsessionid=45076AD5351A6EAE9C1FC7636AE19AA8","timestamp":"2024-11-09T07:03:05Z","content_type":"text/html","content_length":"12780","record_id":"<urn:uuid:9de778b0-b7bb-436a-91ca-a331ed63b18b>","cc-path":"CC-MAIN-2024-46/segments/1730477028116.30/warc/CC-MAIN-20241109053958-20241109083958-00281.warc.gz"} |
Convolution and Impulse Signals
• Thread starter cshum00
• Start date
In summary, convolution is a mathematical operation that involves multiplying and integrating two signals. However, the commutative property may not always apply when trying to change the time-shift
between two general functions or equations. This is because convolution is only an integral between an impulse signal and a generic signal, and not necessarily between two generic signals. While
there are examples of convolving two different signals, the commutative property may not always hold true in these cases.
I am a little confused about convolutions.
I know that convolution is the multiplication and then integral of the two signals. The confusion starts at the commutative property. If i try to change the time-shift from signal to another for any
2 general functions or equations the commutative property doesn't work out.
for example:
let x(t) = sin(t)
and h(t) = t^2
If you try to convolve the signals above with the commutative property you get 2 different results.
However, the convolution's commutative property does work out if h(t) were to be a impulse function. So, does it mean that convolution is only an integral between an impulse signal and a generic
signal and not two generic signals? (which is the part i am confused because i have seen examples of convolving 2 different signals)
Science Advisor
Gold Member
I'm not sure what you are referring to. This article shows how convolutions are symmetric
Scroll down to "Definitions" for the commutative property.
Last edited by a moderator:
FAQ: Convolution and Impulse Signals
1. What is a convolution signal?
A convolution signal is a mathematical operation that combines two signals, usually referred to as the input signal and the impulse response, to produce a third signal. This operation is commonly
used in signal processing to model the behavior of a linear system.
2. How is convolution different from regular multiplication?
Convolution is different from regular multiplication because it takes into account the entire history of the two signals being multiplied, while regular multiplication only considers the current
values of the signals. This allows convolution to model the effects of a system's past behavior on the current output.
3. What is an impulse signal?
An impulse signal, also known as a Dirac delta function, is a mathematical function that is zero everywhere except at one point, where it has an infinitely large value. It is often used in
convolution operations as the input signal, as it represents an instantaneous and infinitely short signal.
4. How is an impulse signal used in convolution?
An impulse signal is used in convolution as the input signal, representing an instantaneous and infinitely short input. This allows the convolution operation to model the behavior of a system when it
is subjected to a sudden and brief input.
5. What are some applications of convolution and impulse signals?
Convolution and impulse signals have various applications in signal processing, such as image and audio processing, filter design, and system analysis. They are also used in fields such as physics,
engineering, and mathematics to model the behavior of systems and solve differential equations. | {"url":"https://www.physicsforums.com/threads/convolution-and-impulse-signals.430987/","timestamp":"2024-11-03T09:19:37Z","content_type":"text/html","content_length":"75263","record_id":"<urn:uuid:30a71508-fb2e-42dd-85c6-83ee97ec1d78>","cc-path":"CC-MAIN-2024-46/segments/1730477027774.6/warc/CC-MAIN-20241103083929-20241103113929-00498.warc.gz"} |
section 3 production possibilities curves answer key
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The line on a production possibilities curve showing the relative amounts of two types of goods produced using all resources is called the. This is why we offer the book compilations in this website.
Key Terms •production possibilities curve: a graph that shows alternative ways to use an economy’s productive resources •production possibilities frontier: a line on a production possibilities curve
that shows the maximum possible output an economy can produce •efficiency: the use of resources in such a way as to maximize the output of goods and services . << Possible answers below ... Productio
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6.11.5. Lexically scoped type variables
6.11.5. Lexically scoped type variables¶
Enable lexical scoping of type variables explicitly introduced with forall.
ScopedTypeVariables breaks GHC’s usual rule that explicit forall is optional and doesn’t affect semantics. For the Declaration type signatures (or Expression type signatures) examples in this
section, the explicit forall is required. (If omitted, usually the program will not compile; in a few cases it will compile but the functions get a different signature.) To trigger those forms of
ScopedTypeVariables, the forall must appear against the top-level signature (or outer expression) but not against nested signatures referring to the same type variables.
Explicit forall is not always required – see pattern signature equivalent for the example in this section, or Pattern type signatures.
GHC supports lexically scoped type variables, without which some type signatures are simply impossible to write. For example:
f :: forall a. [a] -> [a]
f xs = ys ++ ys
ys :: [a]
ys = reverse xs
The type signature for f brings the type variable a into scope, because of the explicit forall (Declaration type signatures). The type variables bound by a forall scope over the entire definition of
the accompanying value declaration. In this example, the type variable a scopes over the whole definition of f, including over the type signature for ys. In Haskell 98 it is not possible to declare a
type for ys; a major benefit of scoped type variables is that it becomes possible to do so.
An equivalent form for that example, avoiding explicit forall uses Pattern type signatures:
f :: [a] -> [a]
f (xs :: [aa]) = xs ++ ys
ys :: [aa]
ys = reverse xs
Unlike the forall form, type variable a from f‘s signature is not scoped over f‘s equation(s). Type variable aa bound by the pattern signature is scoped over the right-hand side of f‘s equation.
(Therefore there is no need to use a distinct type variable; using a would be equivalent.)
6.11.5.1. Overview¶
The design follows the following principles
• A scoped type variable stands for a type variable, and not for a type. (This is a change from GHC’s earlier design.)
• Furthermore, distinct lexical type variables stand for distinct type variables. This means that every programmer-written type signature (including one that contains free scoped type variables)
denotes a rigid type; that is, the type is fully known to the type checker, and no inference is involved.
• Lexical type variables may be alpha-renamed freely, without changing the program.
A lexically scoped type variable can be bound by:
In Haskell, a programmer-written type signature is implicitly quantified over its free type variables (Section 4.1.2 of the Haskell Report). Lexically scoped type variables affect this implicit
quantification rules as follows: any type variable that is in scope is not universally quantified. For example, if type variable a is in scope, then
(e :: a -> a) means (e :: a -> a)
(e :: b -> b) means (e :: forall b. b->b)
(e :: a -> b) means (e :: forall b. a->b)
6.11.5.2. Declaration type signatures¶
A declaration type signature that has explicit quantification (using forall) brings into scope the explicitly-quantified type variables, in the definition of the named function. For example:
f :: forall a. [a] -> [a]
f (x:xs) = xs ++ [ x :: a ]
The “forall a” brings “a” into scope in the definition of “f”.
This only happens if:
• The quantification in f‘s type signature is explicit. For example:
g :: [a] -> [a]
g (x:xs) = xs ++ [ x :: a ]
This program will be rejected, because “a” does not scope over the definition of “g”, so “x::a” means “x::forall a. a” by Haskell’s usual implicit quantification rules.
• The type variable is quantified by the single, syntactically visible, outermost forall of the type signature. For example, GHC will reject all of the following examples:
f1 :: forall a. forall b. a -> [b] -> [b]
f1 _ (x:xs) = xs ++ [ x :: b ]
f2 :: forall a. a -> forall b. [b] -> [b]
f2 _ (x:xs) = xs ++ [ x :: b ]
type Foo = forall b. [b] -> [b]
f3 :: Foo
f3 (x:xs) = xs ++ [ x :: b ]
In f1 and f2, the type variable b is not quantified by the outermost forall, so it is not in scope over the bodies of the functions. Neither is b in scope over the body of f3, as the forall is
tucked underneath the Foo type synonym.
• The signature gives a type for a function binding or a bare variable binding, not a pattern binding. For example:
f1 :: forall a. [a] -> [a]
f1 (x:xs) = xs ++ [ x :: a ] -- OK
f2 :: forall a. [a] -> [a]
f2 = \(x:xs) -> xs ++ [ x :: a ] -- OK
f3 :: forall a. [a] -> [a]
Just f3 = Just (\(x:xs) -> xs ++ [ x :: a ]) -- Not OK!
f1 is a function binding, and f2 binds a bare variable; in both cases the type signature brings a into scope. However the binding for f3 is a pattern binding, and so f3 is a fresh variable
brought into scope by the pattern, not connected with top level f3. Then type variable a is not in scope of the right-hand side of Just f3 = ....
6.11.5.3. Expression type signatures¶
An expression type signature that has explicit quantification (using forall) brings into scope the explicitly-quantified type variables, in the annotated expression. For example:
f = runST ( (op >>= \(x :: STRef s Int) -> g x) :: forall s. ST s Bool )
Here, the type signature forall s. ST s Bool brings the type variable s into scope, in the annotated expression (op >>= \(x :: STRef s Int) -> g x).
6.11.5.4. Pattern type signatures¶
A type signature may occur in any pattern; this is a pattern type signature. For example:
-- f and g assume that 'a' is already in scope
f = \(x::Int, y::a) -> x
g (x::a) = x
h ((x,y) :: (Int,Bool)) = (y,x)
In the case where all the type variables in the pattern type signature are already in scope (i.e. bound by the enclosing context), matters are simple: the signature simply constrains the type of the
pattern in the obvious way.
Unlike expression and declaration type signatures, pattern type signatures are not implicitly generalised. The pattern in a pattern binding may only mention type variables that are already in scope.
For example:
f :: forall a. [a] -> (Int, [a])
f xs = (n, zs)
(ys::[a], n) = (reverse xs, length xs) -- OK
(zs::[a]) = xs ++ ys -- OK
Just (v::b) = ... -- Not OK; b is not in scope
Here, the pattern signatures for ys and zs are fine, but the one for v is not because b is not in scope.
However, in all patterns other than pattern bindings, a pattern type signature may mention a type variable that is not in scope; in this case, the signature brings that type variable into scope. For
-- same f and g as above, now assuming that 'a' is not already in scope
f = \(x::Int, y::a) -> x -- 'a' is in scope on RHS of ->
g (x::a) = x :: a
hh (Just (v :: b)) = v :: b
The pattern type signature makes the type variable available on the right-hand side of the equation.
Bringing type variables into scope is particularly important for existential data constructors. For example:
data T = forall a. MkT [a]
k :: T -> T
k (MkT [t::a]) =
MkT t3
(t3::[a]) = [t,t,t]
Here, the pattern type signature [t::a] mentions a lexical type variable that is not already in scope. Indeed, it must not already be in scope, because it is bound by the pattern match. The effect is
to bring it into scope, standing for the existentially-bound type variable.
It does seem odd that the existentially-bound type variable must not be already in scope. Contrast that usually name-bindings merely shadow (make a ‘hole’) in a same-named outer variable’s scope. But
we must have some way to bring such type variables into scope, else we could not name existentially-bound type variables in subsequent type signatures.
Compare the two (identical) definitions for examples f, g; they are both legal whether or not a is already in scope. They differ in that if a is already in scope, the signature constrains the
pattern, rather than the pattern binding the variable.
6.11.5.5. Class and instance declarations¶
ScopedTypeVariables allow the type variables bound by the top of a class or instance declaration to scope over the methods defined in the where part. Unlike :ref`decl-type-sigs`, type variables from
class and instance declarations can be lexically scoped without an explicit forall (although you are allowed an explicit forall in an instance declaration; see Explicit universal quantification
(forall)). For example:
class C a where
op :: [a] -> a
op xs = let ys::[a]
ys = reverse xs
head ys
instance C b => C [b] where
op xs = reverse (head (xs :: [[b]]))
-- Alternatively, one could write the instance above as:
instance forall b. C b => C [b] where
op xs = reverse (head (xs :: [[b]]))
While ScopedTypeVariables is required for type variables from the top of a class or instance declaration to scope over the /bodies/ of the methods, it is not required for the type variables to scope
over the /type signatures/ of the methods. For example, the following will be accepted without explicitly enabling ScopedTypeVariables:
class D a where
m :: [a] -> a
instance D [a] where
m :: [a] -> [a]
m = reverse
Note that writing m :: [a] -> [a] requires the use of the InstanceSigs extension.
Similarly, ScopedTypeVariables is not required for type variables from the top of the class or instance declaration to scope over associated type families, which only requires the TypeFamilies
extension. For instance, the following will be accepted without explicitly enabling ScopedTypeVariables:
class E a where
type T a
instance E [a] where
type T [a] = a
See Scoping of class parameters for further information. | {"url":"https://downloads.haskell.org/ghc/9.0.2/docs/html/users_guide/exts/scoped_type_variables.html","timestamp":"2024-11-04T04:55:40Z","content_type":"text/html","content_length":"49142","record_id":"<urn:uuid:d2343699-f099-4f8d-82fe-f8eafdb57594>","cc-path":"CC-MAIN-2024-46/segments/1730477027812.67/warc/CC-MAIN-20241104034319-20241104064319-00394.warc.gz"} |
Calculate The Present Value Of Your Costs | Accounting Homework Help - My Homework Help
Paper Details
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High Energy Theory
Along the way, we need answers to questions such as:
• Is the fundamental structure of space and time continuous or discrete?
• Is there a symmetry that allows us to generalize Einstein's "equivalence principle" to quantum mechanics?
• Can our current models of High Energy Physics explain the origin of the Big Bang?
• Can the fundamental constants of physics be predicted? Or are they thrown down by chance?
Research in High Energy Physics at Brown covers a range of topics from String Theory and M-theory to Black Hole Physics, Gravitation and Cosmology, as well gauge theories, QCD, and other topics. | {"url":"https://physics.brown.edu/research/high-energy-theory","timestamp":"2024-11-03T21:30:43Z","content_type":"text/html","content_length":"80553","record_id":"<urn:uuid:44b097ff-7aba-48ff-a79c-75e3a8ac881e>","cc-path":"CC-MAIN-2024-46/segments/1730477027796.35/warc/CC-MAIN-20241103212031-20241104002031-00707.warc.gz"} |
Fraction Models worksheets
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Explore printable Fraction Models worksheets
Fraction Models worksheets are an excellent resource for teachers looking to enhance their students' understanding of math and fractions. These worksheets provide a visual representation of
fractions, allowing students to easily grasp the concept and develop a strong foundation in mathematics. With a wide range of activities, such as coloring, labeling, and comparing fractions, these
worksheets cater to different learning styles and can be easily incorporated into any lesson plan. Teachers can also customize the difficulty level of the worksheets to suit the needs of their
students, making them a versatile and valuable tool in the classroom. By incorporating Fraction Models worksheets into their curriculum, teachers can ensure that their students gain a thorough
understanding of fractions and their applications in various mathematical contexts.
Quizizz, a popular online platform for creating and sharing quizzes, offers a plethora of resources for teachers, including Fraction Models worksheets. This platform allows teachers to create
interactive quizzes and games that can be used to supplement their existing lesson plans, making learning math and fractions more engaging and enjoyable for students. With Quizizz, teachers can
easily track their students' progress and identify areas where they may need additional support. In addition to Fraction Models worksheets, Quizizz offers a wide range of other math resources, such
as quizzes on various topics like geometry, algebra, and number systems. Teachers can also find resources for other subjects, making Quizizz a one-stop-shop for all their educational needs. By
utilizing Quizizz and its vast array of resources, teachers can ensure that their students receive a well-rounded education and develop a strong foundation in math and fractions. | {"url":"https://quizizz.com/en-us/fraction-models-worksheets","timestamp":"2024-11-08T12:34:33Z","content_type":"text/html","content_length":"159650","record_id":"<urn:uuid:2afbb5eb-9ea7-43d7-aeec-feb07afcd0f2>","cc-path":"CC-MAIN-2024-46/segments/1730477028059.90/warc/CC-MAIN-20241108101914-20241108131914-00454.warc.gz"} |
Saw Transmitter Circuit
This circuit uses a saw based oscillator with an output amplifier which drives a transistor. C1 and C2 and L1 are critical oscillator components. They form a tank circuit which should be tuned to the
resonator frequency. Typically C1<C2.The output C/L network matches the antenna to the driver.
The tank circuit frequency has the following equation:
F= 1/(2*pi*sqrt(L*C1*C2/(C1+C2)))
Rearranged for L:
L=1/( (2*PI*F)^2 *C1*C2/(C1+C2))
The following calculator can be iteratively used to find appropriate values for the tank circuit.
Saw Transmitter Schematic | {"url":"https://www.daycounter.com/Circuits/Saw-Transmitter/Saw-Transmitter-Circuit.phtml","timestamp":"2024-11-09T13:19:48Z","content_type":"text/html","content_length":"8570","record_id":"<urn:uuid:d4201005-08e0-4a72-a7e8-e865431bad04>","cc-path":"CC-MAIN-2024-46/segments/1730477028118.93/warc/CC-MAIN-20241109120425-20241109150425-00872.warc.gz"} |
MCMLSD: A dynamic programming approach to line segment detection
This project page provides code for line segment detection and evaluation. If you use either, please cite:
Almazen, E.J., Tal, R., Qian, Y. & Elder, J.H. (2017) MCMLSD: A dynamic programming approach to line segment detection. IEEE Conference on Computer Vision and Pattern Recognition (CVPR), 2031-2039.
For enquiries, please contact James Elder at jelder@yorku.ca
The MCMLSD Algorithm
Traditional approaches to line segment detection typically involve perceptual grouping in the image domain and/or global accumulation in the Hough domain. Here we propose a probabilistic algorithm
that merges the advantages of both approaches. In a first stage lines are detected using a global probabilistic Hough approach. In the second stage each detected line is analyzed in the image domain
to localize the line segments that generated the peak in the Hough map. By limiting search to a line, the distribution of segments over the sequence of points on the line can be modeled as a Markov
chain, and a probabilistically optimal labelling can be computed exactly using a standard dynamic programming algorithm, in linear time. The Markov assumption also leads to an intuitive ranking
method that uses the local marginal posterior probabilities to estimate the expected number of correctly labelled points on a segment. To assess the resulting Markov Chain Marginal Line Segment
Detector (MCMLSD) we develop and apply a novel quantitative evaluation methodology that controls for under- and over-segmentation. Evaluation on the YorkUrbanDB and Wireframe datasets shows that the
proposed MCMLSD method outperforms prior traditional approaches.
We compare against three competing state-of-the-art detectors. Each detector returns a ranked list of segments. Ground truth and detector segments sampled uniformly at 1-pixel resolution. Enforcing
bipartite match at segment level ensures appropriate penalty for under- and over-segmentation.
2-Step association between ground truth and detector output:
• Point Match: Greedy bipartite match between ground truth and detector points within 2.8-pixel radius.
• Segment Match: Optimal bipartite match between ground truth and detector segments, maximizing total number of matched points
We assume that the line segment detector under evaluation returns a list of line segments in ranked order. We sample each ground truth and detector segment uniformly with a sample spacing of one
pixel and use these point samples to evaluate the detector as a function of the number k of top-ranked segments selected, varying k from 10 to 500. For each value of k we first identify potential
point matches as those (ground truth, detector) point pairs lying within a threshold distance of 2.8 pixels of each other. This threshold was selected to associate any pair of lines that could
potentially appear in the image with less than a one-pixel intervening gap. We then sort these candidate matches by Euclidean distance and accept matches in greedy fashion starting with the smallest
distance, matching each point at most once, and thus arriving at a near-optimal bipartite match. Having associated ground truth and detector points, we employ the Hungarian algorithm [37] to identify
the optimal bipartite match between ground truth and detector segments that maximizes the total number of points matched. Now that we have a 1:1 association between ground truth and detector
segments, it remains to evaluate the quality of this association. We employ three evaluative measures.
1. Recall as a Function of the Number of Segments
2. Recall as a Function of Total Segment Length
3. Precision-Recall
Quantitative Results on the YorkUrbanDB
Quantitative Results on Wireframe
Quantitative Results on Linelet-YorkUrbanDB
Qualitative Results
Top 90 segments detected by MCMLSD for all images in the YorkUrbanDB
Training images
Test Images | {"url":"https://www.elderlab.yorku.ca/mcmlsd/","timestamp":"2024-11-06T02:38:17Z","content_type":"text/html","content_length":"124083","record_id":"<urn:uuid:bf66f5dc-297a-4ce2-84d3-d016510315bd>","cc-path":"CC-MAIN-2024-46/segments/1730477027906.34/warc/CC-MAIN-20241106003436-20241106033436-00816.warc.gz"} |
The Counting Measure
The Counting Measure
Recall from the General Measurable Spaces and Measure Spaces page that if $X$ is a set and $\mathcal A$ is a $\sigma$-algebra on $X$ then the pair $(X, \mathcal A)$ is called a measurable space. We
said that the sets $E \in \mathcal A$ are called measurable sets.
We said that a function $\mu : \mathcal A \to [0, \infty]$ is called a measure on $\mathcal A$ if $\mu (\emptyset) = 0$ and if for every countable collection of disjoint measurable sets $(E_n)$ we
have that $\displaystyle{\mu \left ( \bigcup_{n=1}^{\infty} E_n \right ) = \sum_{n=1}^{\infty} \mu (E_n)}$, and we defined the triple $(X, \mathcal A, \mu)$ to be a measure space.
We now look a special measure called the counting measure.
Definition: Let $X$ be any set and let $\mathcal P(X)$ denote the power set of $X$. The Counting Measure $c : \mathcal P(X) \to [0, \infty]$ on $\mathcal P(X)$ is defined for all $E \in \mathcal P(X)
$ by $c(E) = |E|$, and the triple $(X, \mathcal P(X), c)$ is called a Counting Measure Space.
Let $X$ be any set. We will verify that $(X, \mathcal P(X), c)$ is a measure space.
First we have that:
\quad c(\emptyset) = |\emptyset| = 0
Now let $(E_n)_{n=1}^{\infty}$ be any countable collection of mutually disjoint sets in $\mathcal P(X)$. There are three cases to consider.
Case 1: Suppose that for some $k \in \mathbb{N}$ we have that $|E_k| = \infty$. Then clearly:
\quad c \left ( \bigcup_{n=1}^{\infty} E_n \right ) = \biggr \lvert \bigcup_{n=1}^{\infty} E_n \biggr \rvert = \infty = |E_k| = \sum_{n=1}^{\infty} |E_n| = \sum_{n=1}^{\infty} c(E_n)
Case 2: Suppose that for all $n \in \mathbb{N}$ we have that $|E_n| < \infty$ and that there are infinitely many $n$ such that $|E_n| > 0$. $\displaystyle{\bigcup_{n=1}^{\infty} E_n}$ must contain
infinitely many elements, and $\displaystyle{\sum_{n=1}^{\infty} |E_n| = \infty}$ as it is a divergent series of positive numbers. Therefore:
\quad c \left ( \bigcup_{n=1}^{\infty} E_n \right ) = \sum_{n=1}^{\infty} c(E_n)
Case 3: Suppose that for all $n \in \mathbb{N}$ we have that $|E_n| < \infty$ and that there are finitely many $n$ such that $|E_n| > 0$, say these sets are $\{ E_1, E_2, ..., E_N \}$ without loss of
generality, and that $|E_k| = t_k$ for each $k \in \{ 1, 2, ..., N \}$. Since $(E_n)_{n=1}^{\infty}$ is a collection of mutually disjoint sets, we have that:
\quad c \left ( \bigcup_{n=1}^{\infty} E_n \right ) = \biggr \lvert \bigcup_{n=1}^{\infty} E_n \biggr \rvert = \biggr \lvert \bigcup_{n=1}^N E_n \biggr \rvert = \sum_{n=1}^{N} t_n = \sum_{n=1}^{N} |
E_n| = \sum_{n=1}^{\infty} |E_n| = \sum_{n=1}^{\infty} c(E_n)
Therefore $\mu$ is a measure on $\mathcal P(X)$ and $(X, \mathcal P(X), c)$ is a measure space. | {"url":"http://mathonline.wikidot.com/the-counting-measure","timestamp":"2024-11-06T08:31:49Z","content_type":"application/xhtml+xml","content_length":"17650","record_id":"<urn:uuid:43a3b06f-96fa-486a-8d6c-bb67d9a4c81d>","cc-path":"CC-MAIN-2024-46/segments/1730477027910.12/warc/CC-MAIN-20241106065928-20241106095928-00445.warc.gz"} |
Dirac's theorem on Hamiltonian Graphs
Published on July 31, 2024
Three different proofs of Dirac's theorem on Hamiltonian cycles in graphs with sufficient minimum degree.
Overview & Background
Continuing the theme of last post, we will look at a classic result in graph theory, namely Dirac’s theorem on Hamiltonian cycles. This result was first published by Gabriel A. Diracin 1952 ^[1],
with later refinements by Ore, as well as Bondy and Chvátal. Three proofs of the theorem will be presented: the original proof from Dirac’s paper, an induction proof, and a short and direct proof. We
will also consider the tightness of the bound as well as a generalization by Ore.
As a little historical sidenote before diving into the mathematics, Gabriel Dirac was the son of Margit Wigner, the sister of famous physicist Eugene Wigner. After a first marriage, from which
Gabriel was born, Margit later remarried Eugene Wigner’s friend Paul Dirac, another equally famous physicist and one of the founders of Quantum Mechanics. Subsequently, Gabriel chose to take on the
name Dirac. More interesting biographical details can be found in the essay about Paul Diracon MacTutor.
Table of contents
Dirac’s theorem
Although the following definitions are standard in graph theory, it is useful to specify them at the outset as these terms are often used rather loosely. A path is a sequence of non-repeating
vertices and edges, where subsequent vertices are connected by an edge in the graph. Usually we distinguish between open and closed paths, where as the name implies, an open path just has different
start and end vertices, while a closed path starts and ends with the same vertex. A closed path is also called a cycle. The length of a cycle is the number of distinct vertices on the cycle (the
equal start- and end-vertex is not counted twice). As always, I will try to stick to common graph theoretical notation and conventions, but for the sake of both readability and completeness an
overview of notation is given at the end of the post.
A Hamiltonian cycle is a cycle through a graph that visits every vertex exactly once. This concept might sound similar to Euler tours, that are historically at the origin of graph theory. An Euler
tour is a cycle that visits every edge exactly once, while repeated vertices are allowed. However, in contrast to Euler tours which can be found in a graph in linear time $\mathcal{O}(|V|+|E|)$,
finding and even checking whether a graph contains a Hamiltonian cycle is in general $\mathcal{NP}$-complete, i.e. no known polynomial time exists for this problem. Finding a Hamiltonian cycle and
the related optimization problem of finding a lowest weight Hamiltonian cycle (known as the travelling salesperson problem) has many applications from computer graphics and circuit design to cargo
routing and bioinformatics.
We will consider Dirac’s theorem, which is a sufficient condition for graphs to contain a Hamiltonian cycle:
Theorem (Dirac, 1952)
Let $G = (V, E)$ be a graph with $|V| \geq 3$ and minimum degree $\delta(G) \geq \frac{|V|}{2}$ on every vertex. Then $G$ contains a Hamiltonian cycle.
Equivalently, every graph with minimum degree $\delta(G) = d$ and at most $|V| \leq 2d$ vertices contains a Hamiltonian cycle.
Notice that the conditions of this theorem are not necessary for a Hamiltonian cycle, as for example a cycle graph $C_n$ on $n$ vertices violates the degree condition but clearly contains a
Hamiltonian cycle.
As a warm-up we’ll start by showing that such a “Dirac-graph” is connected, ie. $G$ contains a path between any two vertices. In his original paper, Dirac writes in the theorem formulation that the
graph must be connected, but as we will see now, this is not strictly necessary. Every graph that obeys the degree condition must also be connected.
In fact, we will see that any two vertices are connected by a path of length at most two, where the length of a path is defined as the number of edges along that path. Consider two arbitrary vertices
$u, v \in V$. Without loss of generality we may assume that $\{u, v\} ot \in E$, as otherwise we would be done. We therefore know that the neighborhoods $\mathcal{N}(u), \mathcal{N}(v)$ of $u$ and
$v$ respectively do not contain the other vertex. But by the assumption on the minimum degree in $G$ we also know that $\mathcal{N}(u) \geq \frac{|V|}{2}$ as well as $\mathcal{N}(v) \geq \frac{|V|}
{2}$. We can now apply the pigeonhole principle to conclude that $u$ and $v$ must have a neighbor in common, and are therefore connected by a path of length $2$: Both neighborhoods do not contain $u$
and $v$ by the previous observation, so we are trying to distribute two neighborhoods of size at least $\frac{|V|}{2}$ onto $|V| - 2$ vertices. The two must necessarily intersect.
Alternatively, this same fact can be seen by applying the inclusion-exclusion principle:
$|\mathcal{N}(u) \cap \mathcal{N}(v)| = \underbrace{|\mathcal{N}(u)|}_{\geq \frac{|V|}{2}} + \underbrace{|\mathcal{N}(v)|}_{\geq \frac{|V|}{2}} - \underbrace{|\mathcal{N}(u) \cup \mathcal{N}(v)|}_{\
leq |V \setminus \{u, v\}| = |V| - 2} \geq 2$
We can already start seeing, why this theorem is tight. Relaxing the degree constraint on the vertices would for example allow a graph $G'$ that is the disjoint union of two complete subgaphs on half
of the vertices: for $X = K_{\frac{|V|}{2}}$ and $Y = K_{\frac{|V|}{2}}$ (ignoring odd $|V|$ for simplicity here) we could construct $G' = X \uplus Y$ with a minimum degree of $\delta(G') = \frac{|V
|}{2} - 1$ for every vertex. This graph clearly does not contain a Hamiltionian cycle
In the following three proofs, we will always assume that $G$ is a graph as described in the theorem, i.e. $G = (V, E)$ with $\delta(G) \geq \frac{|V|}{2}$.
Original proof
Dirac’s original proof ^[1:1] is a little more convoluted, but we will consider it for completeness. It contains a few unnecessary convolutions, such as two nested proofs by contradiction, that hide
the essence of the argument. Feel free to skip to the next section if you are not interested in the exact technical details of the original paper.
We will start by showing the following lemma:
Lemma 1
Let $G = (V, E)$ be a graph as defined in Dirac’s theorem. Then $G$ contains a cycle of length at least $\delta(G) + 1$.
A longest path $P = v_1, \dots, v_k$ in $G$ must contain at least $\frac{|V|}{2} + 1$ vertices, because $v_1$ has at least $\delta(G) = \frac{|V|}{2}$ neighbors. Otherwise $v_1$ would have a neighbor
outside of $P$ that we could use to extend $P$. Therefore, all neighbors of $v_1$ must lie on $P$, which immediately yields a cycle of length at least $\delta(G) + 1$: $v_1, \dots, v_i, v_1$ where $i
\geq \delta(G) + 1$ is the highest index of neighbors of $v_1$ along $P$.
Now assume for sake of contradiction that the theorem is false, i.e. there exists a graph $G = (V, E)$ satisfying $|V| \geq 3$ and $\delta(G) \geq \frac{|V|}{2}$ on every vertex that does not contain
a Hamiltonian cycle. Let $C = v_1, \dots, v_k$ be the longest cycle in $G$. By our assumption, $C$ has length at most $|V| - 1$. From the lemma it also follows that the cycle $C$ must have length at
least $\frac{|V|}{2} + 1$.
As $G$ is connected, some node in $C$, let’s assume without loss of generality that it is $v_k$, must be connected to some node $v_{k+1}$ in $V \setminus C$. We will now consider the longest path $P'
= v_{k}, v_{k+1}, \dots, v_{k + l}$ that is entirely contained in $V \setminus C$ except for $v_k$. By the same observation as in the proof for the lemma, $v_{k + l}$ can only be connected to $v_1, \
dots, v_k, \dots, v_{k + l - 1}$.
Under these assumptions made for sake of contradiction, the following lemma holds:
Lemma 2
Under the assumption that Dirac’s theorem is false, it holds that $l \geq \frac{|V|}{2}$.
We will show this lemma again by contradiction. If $l \leq \frac{|V|}{2} - 1 < \frac{|V|}{2}$, then $v_{k+l}$ must be connected to at least $\delta(G) - l \geq \frac{|V|}{2} - l \geq 1$ vertices in
$v_1, \dots, v_{k-1}$, excluding vertices $v_k, \dots, v_{k+l-1}$ in $P'$. So $v_{k + l}$ must be connected to another vertex $v_i \in C$ with $v_i ot = v_k$. We can estable the following two
inequalities on $i$:
• $i \geq l + 1$: We can form a new cycle $C' = v_{k + l}, v_i, \dots, v_k, \dots, v_{k + l}$ of length $k + l - i + 1$. Because $C$ is the longest cycle in $G$, we obtain $k + l - i + 1 \leq k \
iff i \geq l + 1$
• $i \leq k - l - 1$: We could also form a new cycle $C'' = v_1, \dots, v_i, v_{k + l}, \dots, v_k, v_1$ of length $i + l + 1$. Just as before we obtain $i + l + 1 \leq k \iff i \leq k - l - 1$.
By the two inequalities, we conclude that $v_{k + l}$ is connected to at least $\frac{|V|}{2} - l \geq 1$ vertices in $I = v_{l + 1}, \dots, v_{k - l - 1}$. Notice that there are $|I| = k - 2l - 1$
such vertices.
However, it is also not possible that $v_{k + l}$ is joined to two neighboring vertices $v_i$ and $v_{i + 1}$ in $I$, as this would contradict the maximality of the original cycle $C$. Otherwise $C$
could have been extended to $v_1, \dots, v_i, v_{k + l}, v_{i + 1}, \dots, v_k, v_1$ of length $k + 1$. By this observation, there must therefore be at least $2 \left(\frac{|V|}{2} - l\right) - 1$
such vertices in $I$ in order to intersperse every neighboring vertex of $v_{k + l}$ with a non-neighbor. Putting all these inequalities together, we finally obtain:
$|I| = k - 2l - 1 > 2 \left(\frac{|V|}{2} - l\right) - 1 \iff k \geq |V|$
But this cannot be, as we assumed that $C$ is not a Hamiltonian cycle. This proves lemma 2.
Proving Dirac’s theorem is now fairly straightforward, by completing the outer proof by contradiction. As we have already observed previously, $C$ has length at least $\frac{|V|}{2} + 1$ by lemma 1.
Because $G$ is composed of at least $C$ and $P'$, both distinct from each other by construction, we can apply our bound on $|P'|$ from lemma 2:
$|V| \geq |C| + |P'| \geq \frac{|V|}{2} + 1 + \frac{|V|}{2} \geq |V| + 1$
This contradiction concludes the proof.
Double induction
As a first proof of Dirac’s theorem, we will consider a proof by induction. This uses the so-called rotation-extension technique by Pósa^[2]. Personally, I find this proof to be the most elegant of
the three, especially because of the neat double induction that is used. The general structure of the proof consists of two parts:
• A $k$-cycle implies the existence of a $k+1$-path, as $G$ is connected
• A $k$-path implies the existence of a $k+1$-path or a $k$ cycle.
By induction $G$ thus has an $|V|$-cycle, i.e. a Hamiltonian cycle.
Recall that we already saw previously, that G is connected. This fact will be needed in the induction proof.
Induction I
For $k < |V|$, a $k$-cycle implies the existence of a $k+1$-path.
Let $C = v_1, \dots, v_k, v_1$ be such a cycle in $G$. Because $G$ is connected, there exists an edge $e = (w, v_i)$ from $V \setminus \{v_1, \dots, v_k\}$ to $C$, where $w$ is a vertex outside of
the cycle. Without loss of generality we may assume that $v_i = v_1$. We have found a $k+1$-path $w \rightarrow v_1 \rightarrow \dots \rightarrow v_k$.
Induction II
A $k$-path implies the existence of a $k+1$-path or a $k$-cycle.
Let $P = v_1, \dots, v_k$ be such a path in $G$.
1. Case $\mathcal{N}(v_1) ot \subseteq \{v_2, \dots, v_k\}$: $P$ can be extended to a $k+1$-path $w \rightarrow v_1 \rightarrow \dots \rightarrow v_k$, where $w$ is a neighbor of $v_1$ not in $P$.
2. Case $\mathcal{N}(v_k) ot \subseteq \{v_1, \dots, v_{k-1}\}$: same as above
3. Case $\mathcal{N}(v_1) \subseteq \{v_2, \dots, v_k\}$ and $\mathcal{N}(v_k) \subseteq \{v_1, \dots, v_{k-1}\}$:
Let the extended neighborhood $\mathcal{N}^+(v_k) := \{ v_{i+1} \;\vert\; v_i \in \mathcal{N}(v_k)\}$ be the set of successors of neighbors on the path. Note that by assumption, all neighbors
(and their successors) of $v_k$ lie on $P$. By applying the inclusion-exclusion principle again we obtain:
$|\mathcal{N}(v_1) \cap \mathcal{N}^+(v_k)| = \underbrace{|\mathcal{N}(v_1)|}_{\geq |V|/2} + \underbrace{|\mathcal{N}^+(v_k)|}_{\geq |V|/2} - \underbrace{|\mathcal{N}(v_1) \cup \mathcal{N}^+(v_k)
|}_ {\subseteq V\setminus\{v_1\} \implies \leq |V| - 1} \geq 1$
This implies that there exists a $v_i$ in $\mathcal{N}(v_1) \cap \mathcal{N}^+(v_k)$. We have found a $k$-cycle $v_1 \rightarrow \dots \rightarrow v_i \rightarrow v_k \rightarrow \dots \
rightarrow v_{i+1} \rightarrow v_{1}$
By combining both parts of the induction, we conclude that there must exist an $n$-path, and by Induction II, an $|V|$-cycle, i.e. a Hamiltonian cycle in $G$, as $G$ cannot contain an $|V|+1$ path
(paths have distinct vertices).
Direct proof
This is the shortest of the three proofs that will be presented, and is conceptually very similar to the induction proof. It can be found in standard textbooks on graph theory, such as the one by
Diestel ^[3]. We will first inspect a longest path in $G$ and show that it can be extended to a cycle. We will then prove that this cycle must actually be a Hamiltonian cycle.
Consider a longest path $P = v_1, \dots, v_k$ in $G$. All edges of $v_1$ and $v_k$ must end in $P$, as otherwise we could extend $P$, contradicting the fact that $P$ is a longest path. By the same
argument as in case 3 of Induction II, we can find an edge $v_i v_{i+1}$ in the path, where also both $\{v_1, v_{i+1}\} \in E$ and $\{v_i, v_k\} \in E$. We can then turn $P$ into a cycle $C = v_1, v_
{i+1}, \dots, v_k, v_i, \dots, v_1$.
Finally, assume for sake of contradiction that $C$ is not a Hamiltonian cycle. Then there must exist at least one vertex in $V \setminus C$. However, as $G$ is connected, there must also exist an
edge $\{u, v_i\}$ for $u \in V \setminus C$ and $v_i \in C$, where we can assume $v_i = v_1$ without loss of generality. This is a contradiction to our initial assumption that $P$ is a longest path
in $G$, as $u, v_1, \dots, v_k$ would form a longer path.
We will use the second equivalent formulation of Dirac’s theorem, that bounds the number of vertices as a function of the minimum degree $d = \delta(G)$. This will make the tightness result simpler
to formulate.
Let us first make a general observation about bipartite graphs that will come in handy in a moment. Recall that bipartite graphs are graphs that can be seperated into a disjoint union of two sets $X$
and $Y$ with edges only between these two sets.
Bipartite graphs cannot contain cycles of odd length (where length refers to the number of edges along the path).
Equivalently, a cycle $v_1, \dots, v_n, v_1$ with $n$ odd cannot exist.
First, notice that any path in a bipartite graph consists of alternating vertices from both sets, because a vertex is only connected to vertices from the other set. Therefore any path of odd length
must have start and end vertices in different sets, which cannot possibly form a cycle where start and end vertex are the same.
The construction exhibiting tightness is a simple bipartite construction. We will construct a graph with $2d+1$ vertices and show that this graph does not contain a Hamiltonian cycle. Consider
vertices $v_1, \dots, v_{2d+1}$ and let “even” vertices be those with an even index while “odd” vertices refer to those with an odd index. We connect every even vertex only to all other odd vertices
and vice-versa. This construction can also be compactly written as $K_{d+1, d}$ using standard graph theory notation.
There are $d+1$ odd vertices and $d$ even vertices, i.e. we are “barely” violating the degree conditions of the initial theorem. Even vertices have degree $d_{\text{even}} = \lceil \frac{2d+1}{2} \
rceil = d + 1$ while odd vertices have degree $d_{\text{odd}} = \lfloor \frac{2d+1}{2} \rfloor = d$.
Notice however, that this graph is clearly bipartite: Color say all even vertices red and all odd vertices blue, then by construction vertices of the same parity are not connected. But bipartite
graphs cannot contain odd-length cycles as seen in the lemma above, which shows that this construction cannot contain a Hamiltonian cycle.
Ore’s theorem
Finally, let’s consider a generalization of Dirac’s theorem. The Norwegian mathematician Øystein Orepublished the following theorem in a one-page paper in 1960^[4]:
Theorem (Ore, 1952)
Let $G = (V, E)$ be a graph with $|V| \geq 3$. If for every pair of non-adjacent vertices $u$ and $v$ with $\{u, v\} ot \in E$ we have $\deg_G(u) + \deg_G(v) \geq |V|$, then $G$ contains a
Hamiltonian cycle.
This observation essentially follows from the proofs presented above. In the previous proofs, we repeatedly made use of the fact that two disconnected vertices can be connected through two adjacent
neighbors. The condition that $\deg_G(u) + \deg_G(v) \geq |V|$ for two non-adjacent vertices ensures exactly this: The neighborhoods $\mathcal{N}(u)$ and $\mathcal{N}(v)$ must intersect.
We will prove this by contrapositive:
Assume for sake of contradiction that Ore’s theorem is false, i.e. there exists a graph with $|V| \geq 3$ satisfying $\deg_G(u) + \deg_G(v) \geq |V|$ for $\{ x, y \} ot \in E$ that does not contain a
Hamiltonian cycle. Let $G$ be the the one with a maximum number of edges among those. Notice that $G$ must contain a non-edge $\{x, y\} ot \in E$, as the complete graph $K_{|V|}$ clearly contains a
Hamiltonian cycle. Therefore we know the statement does not just hold vacuously.
Now, adding $\{x, y\}$ to $G$ must necessarily close a Hamiltonian cycle $x = v_1, \dots, v_n = y$ by our choice of $G$. Here we can however again apply case 3 of the induction proof (with the
knowledge that $|\mathcal{N}(x)| + |\mathcal{N}^+(y)| \geq |V|$) to conclude that $\{x, y\}$ was not necessary to close the Hamiltonian cycle. There must be an edge $v_i v_{i+1}$ in the path, where
also both $\{x, v_{i+1}\} \in E_G$ and $\{v_i, y\} \in E_G$. We obtain a cycle $C = x, v_{i+1}, \dots, y, v_i, \dots, x$ in $G$, a contradiction.
Further reading
Overview of notation
• Complete graph $K_n$ on $n$ vertices
• Neighborhood $\mathcal{N}(v)$: set of vertices adjacent to $v$
• Extended neighborhood $\mathcal{N}^+(u)$ over an order of vertices: sucessors of neighbors of $u$
$\mathcal{N}^+(u) := \{ v_{i+1} \;\vert\; v_i \in \mathcal{N}(u)\}$
• Minimum degree $\delta(G) := \min_{v \in V} \deg_G(v)$ | {"url":"https://www.umcconnell.net/posts/2024-07-31-dirac-theorem-hamiltonian/","timestamp":"2024-11-03T22:13:03Z","content_type":"text/html","content_length":"397068","record_id":"<urn:uuid:5581e90a-341d-4fb8-83f4-dd1aa86b19c5>","cc-path":"CC-MAIN-2024-46/segments/1730477027796.35/warc/CC-MAIN-20241103212031-20241104002031-00399.warc.gz"} |
rem Zoomable Mandelbrot based in Joel Kahn's example
font "arial",10,100
graphsize sz,sz
gosub draw
input "press enter to zoom in",a$
gosub draw
input "press enter to zoom in",a$
gosub draw
input "press enter to zoom in ",a$
gosub draw
for x=xmin to xmax step dx
for y=ymin to ymax step dy
# In the iteration z=z^2+c
#z=ai + b
#c=yi + x
until not (d <= m and k < kt)
color k*50000000
plot (x-xmin)/dx,(ymax-y)/dy
next y
next x
# grid
color black
for n= 0 to 9
line n*sz/10,sz,n*sz/10,0
text n*sz/10,5,xmin+(xmax-xmin)*n/10
text 5,n*sz/10,ymax-(ymax-ymin)*n/10
line sz,n*sz/10,0,n*sz/10
next n
No comments: | {"url":"http://basic256.blogspot.com/2010/12/mandelbrot.html","timestamp":"2024-11-08T11:39:51Z","content_type":"text/html","content_length":"47617","record_id":"<urn:uuid:f2c0a1ff-6702-4829-b67b-98fbf409b5e6>","cc-path":"CC-MAIN-2024-46/segments/1730477028059.90/warc/CC-MAIN-20241108101914-20241108131914-00615.warc.gz"} |
Excel Calculate Elapsed Time - Certified Calculator
Excel Calculate Elapsed Time
Introduction: The Excel Calculate Elapsed Time tool is a user-friendly solution designed to help you determine the duration between two points in time using the same time format as Excel. Whether
you’re tracking work hours, project durations, or any other time-based activities, this calculator simplifies the process of calculating elapsed time.
Formula: To calculate elapsed time, the calculator subtracts the start time from the end time, converting the result to minutes.
How to Use:
1. Enter the start time in the “Start Time” field in HH:MM format (24-hour clock).
2. Enter the end time in the “End Time” field in HH:MM format (24-hour clock).
3. Click the “Calculate” button to obtain the elapsed time.
Example: For example, if you start at 09:30 and end at 17:45, entering these times into the calculator will provide the elapsed time in minutes.
1. Q: Can I use this calculator for overnight durations? A: Yes, the calculator considers the time of day, so it works for durations spanning different dates.
2. Q: What format should I use for entering time? A: Enter the time in HH:MM format using the 24-hour clock (e.g., 14:30).
3. Q: Does the calculator account for daylight saving time changes? A: No, the calculator assumes a constant time zone and does not account for daylight saving time.
4. Q: Can I use this calculator for non-time durations, like days or weeks? A: No, this calculator specifically calculates elapsed time in minutes.
5. Q: Why is the result displayed in minutes? A: Representing the result in minutes allows for a convenient and standard unit for time calculations.
Conclusion: The Excel Calculate Elapsed Time tool is a valuable resource for anyone needing to determine the duration between two points in time with the convenience of Excel-like time formatting.
Enhance your time-tracking accuracy and efficiency with this straightforward calculator.
Leave a Comment | {"url":"https://certifiedcalculator.com/excel-calculate-elapsed-time/","timestamp":"2024-11-13T06:33:49Z","content_type":"text/html","content_length":"52693","record_id":"<urn:uuid:fd24f4c9-4948-43b8-85be-ee9a57009f57>","cc-path":"CC-MAIN-2024-46/segments/1730477028326.66/warc/CC-MAIN-20241113040054-20241113070054-00589.warc.gz"} |
Test Bank For Power System Analysis And Design SI Edition 6th Ed
Test Bank For Power System Analysis And Design SI Edition 6th Edition By Glover
1. (25 points total)
A 3, 300 mile, 345-kV line has series impedance of z = 0.03 + j0.4 /mile and shunt
admittance y = j5.0 x 10-6 mho/mile.
(9 pts) a. Calculate the line’s characteristic impedance Zc
and the propagation
(9 pts) b. Initially assume that the receiving end of the line is open-circuited and has
a line to line voltage of 345 kV. Calculate the sending end line to line
voltage magnitude.
(7 pts) c. Draw the long line equivalent circuit for this line, showing the values of
all parameters.
2. (20 points total)
The circuit shown below is a balanced three phase system with a wye-connected
generator producing 1 volt (phase to neutral). Assume that each inductor has an
impendance ZL = j10 and each capacitor ZC = -j15. Determine Ia, Icap, and the total
three phase complex power supplied by the y-connected voltage source.
3. (25 points total)
A 3-phase, 60Hz, 50km long, completely transposed transmission line is built using
Drake conductor. Drake conductor has an outside diameter of 1.108 inches; stranding of
26/7 (Al/St), which yields a GMR for the conductor of 0.0375 feet. Resistance at 60-Hz
for this conductor is 0.117 /mile. A horizontal tower configuration is used, with a
phase spacing of 25 feet (25 feet between left and center phases, 25 feet between center
and right, and hence 50 feet between left and right). Bundling is used, with 2 conductors
per phase, spaced 1 foot apart. For reference µ0 = 4π x 10-7 H/m and Ɛ0 = 8.854 x 10-12
a) Find the positive sequence inductance in H/m and inductive reactance in Ω/km.
b) Find the capacitance to neutral in F/m and the admittance to neutral in S/km.
Neglect the effect of the earth plane.
c) Since the line is 50km, we can make the short line approximation. Draw the short
line π equivalent circuit, labeling appropriate values.
For reference = 4 x 10-7 H/m and 0 = 8.854 x 10-12 F/m. There are 1609 meters
per mile.
4. Multiple Choice. Circle the most correct answer. Ten problems, three points
each for a total of 30 points.
Note this Problem Continues on the Next Page
A. As discussed in class, the most common load model for the power flow is
A. Constant impedance
B. Constant current
C. Constant power
B. We can only use per phase analysis when:
A. All loads and sources are Y connected
B. All loads and sources are delta connected
C. There is no mutual inductance between phases
D. A and C
E. B and C
C. Which statement about phase and line voltages (V) and currents (I) in Wye and Delta
connections is correct?
A. Phase I = line I in Wye, and phase V = line V in Delta
B. Phase I = line I in Delta, and phase V = line V in Wye
C. Phase I = line I and phase V = line V in Wye
D. Phase I = line I and phase V = line V in Delta
D. Using Newton’s method to solve the equation x2 –sin(x) – 2 = 0, with an initial guess of
x=1 (that is, its value at the zero iteration), select the value that is closest to the value of
x after the second iteration:
A. 1.6754
B. 1.7775
C. 1.8073
D. 2.3041
E. 2.3253
E. The impedance of a 10 MVA, 22/220 kV transformer is 0.2 per unit. What is this
impedance in per unit for a power base of 10 MVA, and a voltage base of 11 kV on the
low voltage side of the transformer?
A. 0.05 p.u.
B. 0.1 p.u.
C. 0.2 p.u.
D. 0.4 p.u.
E. 0.8 p.u.
F. Transformer open circuit (OC) and short circuit (SC) tests allow us to calculate certain
equivalent circuit values. Which parameters are calculated from the OC test and which
are they from the SC test?
A. The OC test gives the resistances and the SC test gives the reactances
B. The OC test gives the series impedance and the SC test gives the shunt
C. The OC test gives the shunt admittance and the SC test gives the series
D. The OC test gives the voltages and the SC test gives currents
G. In a long line Pi model, we use series impedance Z’ and shunt admittance Y’/2. In the
short line model, we use:
A. Series impedance Z’ and shunt admittance Y’/2
B. Series impedance Z’ and shunt admittance Y/2
C. Series impedance Z and shunt admittance Y/2
D. Series impedance Z and shunt admittance 0
E. Series impedance 0 and shunt admittance 0
H. In the power flow at the slack (swing) bus,
A. P and V are fixed
B. P and Q are fixed
C. P and ƟV are fixed
D. V and Q are fixed
E. None of the above
I. A solid conductor has a conductor area of 1113 kcmils. What is its outside diameter, in
A. 1.055”
B. 1.293”
C. 1113”
D. 1.044”
J. If the diameter of a transmission line conductor is increased, then:
A. Both the inductance and the capacitance increase
B. The inductance increases and the capacitance decreases
C. The inductance decreases and the capacitance increases
D. Both the inductance and the capacitance decrease
E. The inductance decreases and the capacitance remains unchanged
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DRDO CEPTAM 9 Admit Card 2019 for Tier-II (Trade Test)
DRDO CEPTAM 9 Admit Card 2019 Available for CEPTAM 9 recruitment 2019 @drdo.gov.in for 224 Technical Assistant, Admin & Allied Vacancies Exam:
Defence Research & Development Organisation (DRDO), Centre for Personnel Talent Management (CEPTAM) will soon release DRDO CEPTAM 9 Admit Card 2019 on the official website of DRDO. The candidates,
who applied against CEPTAM 9 Recruitment 2019, can download their DRDO Admit Card 2019 from here.
Latest Update (16-02-2020): Candidates who have been provisionally shortlisted for Tier-II (Trade Test) are informed that trade test will be commencing from 20 Jan 2020 onwards. Admit card regarding
trade test will be available from 06 Jan 2020. The candidates can download DRDO Admit Card from the direct link given below.
Check DRDO CEPTAM-9 Result 2019
DRDO has successfully conducted the CEPTAM – 09 entry test 9 on all centers. The candidates who had taken the CEPTAM 9 written exam are eagerly waiting for the DRDO CEPTAM 9 Result 2019 along with
Merit List and Cut Off marks. However, the DRDO will first release the DRDO Answer Key on their official portal for self-evaluation purpose. The candidates can download DRDO Answer Key from the given
below link.
Download DRDO CEPTAM 9 Answer Key 2019
DRDO CEPTAM 9 Admit Card 2019
The candidates who are eagerly searching for DRDO CEPTAM 9 Result 2019, can check their result online from the given below link. The DRDO will also release the DRDO CEPTAM 9 Merit List and cut off
marks along with the result.
The candidates should follow all the instruction given on the page to check their DRDO CEPTAM 9 Result 2019 online.
The Defence Research and Development Organisation (DRDO) was formed in 1958 by the merger of the Technical Development Establishment and the Directorate of Technical Development and Production with
the Defence Science Organisation. It is charged with the military’s research and development, under the administrative control of the Ministry of Defence, Govt. of India.
Centre for Personnel Talent Management (CEPTAM) is a premier organisation of DRDO with an Independent Chairman and Director as its Administrative Head. CEPTAM is entrusted with the Recruitment of
Technical, Administrative & Allied category of non-gazetted personnel, Assessment of DRTC Officers & Staff, and facilitates Training for the development of this cadre as per Training Policy of DRDO.
DRDO CEPTAM-9 Admit Card 2019 Download
Recently, DRDO has released a recruitment advertisement CEPTAM 9 for the post of STA-B, Technician A, A&A, Junior Translator, Stenographer Grade II, Admin Assistant A and Store Assistant A. A large
number of eligible and desirous candidates have applied against CEPTAM-9 recruitment and all are searching for DRDO Admit Card 2019. The application was invited through online as well as offline
mode. The candidates, who have applied online, can download their DRDO Hall Ticket 2019 through online mode only and will not be sent by post.
CEPTAM DRDO Hall Ticket 2019
The candidates, who have applied for DRDO 1142 Technician Recruitment 2019 vacancies can download their DRDO Call Letter 2019 from their official website as soon as it will be available officially.
The candidates, who have applied through online and offline mode, can download their DRDO CEPTAM Admit Card 2019 from the official website.
DRDO has not yet announced the examination date but it will be published on the website later. The examination of all codes mentioned in the official notification will be completed in two shifts on a
single day. The written examination for the posts of STA ‘B’ and Technician ‘A’ will be conducted in different shifts on the same day all over the country. The examination for the posts of Admin
Assistant ‘A’ and Store Assistant ‘A’ will also be held in separate shifts but it will be time-clubbed either with shift for STB ‘B’ or Technician ‘A’
DRDO Admit Card 2019 for CEPTAM-9 Recruitment Exam
DRDO Admit Card 2019 is one of the most important documents for appearing in written examination, so all candidates are looking for it. The DRDO Call Letter will contain the exact date, time and
venue of the written examination. The candidates must carry valid photo id proof in original as mentioned in the Application Form with Admit Card to appear in the examination & Document Verification/
Interview. The candidates can download their DRDO Hall Ticket 2019 from the official website as soon as it will be available for download, so candidates are advised to stay connected with this page
to get latest updates regarding it. JobsLab team will post DRDO Admit Card 2019 link here on this page.
Download DRDO CEPTAM Admit Card 2019
Official Website: http://www.drdo.gov.in | http://www.ceptamonline.org
221 thoughts on “DRDO CEPTAM 9 Admit Card 2019 for Tier-II (Trade Test)”
1. how to check drdo admit card 2016?
2. sir.drdo tecnician ka admit card nhi nikal raha h pls exam date or admit card link pe kab ayenga batao my no is 9158613855
3. Technician ‘A’ ka admit card kab tak aayega aur exam date kya hai
4. Sir Store Assistant ‘A’ ka exam hall ticket official website mai kab publish karaiga plz sir inform….this is my contact number 9863976091…
5. Sir DRDO
6. sir admit card kab niklega or exams kab hoga
7. Drdo ka admit card Kab a rha hai
8. Hello sir
Plz sir my admitcard date….
9. Sir DRDO ka admit card kab tak aayega plzz Sir contact me 9917448352
10. Sir, DRDO tech.G-A ke admit card kab ayenge ?
11. sir drdo ka admit card jab aaye to plz contact me7897332915
12. Job
13. Good morning sir drdo CEPTAM 8 admit card all details give me now… Thanks you sir
14. Sir god morning admited card kab ayiaga to mail karna sir
15. Drdo ka admit card kb ayega sir
Muchhe bhi batayega
16. Dear Sir,
Please Update me while DRDO CEPTAM 8 Admit card will be Uploaded/declared on your official website.
Please send the link on my given email id
it is higher obliged.
Thanking You
Harendra Prajapati
17. hello sir
i did flp a drdo from butn did not come my idemt card so
when will be coming idemt card please send my Emild id
thanku for u
18. DRDO Ka exam kB hoga sir??
□ May 2nd weak possible
19. when will The exam….
20. Sir DRDO SEPTAM-8 Ka admit card Kab aayega
exam Kab hogi
21. Sir admit card kabsa milaga please inform me kindly sir…cepet8ka…
22. DRDO admit card just send email id
23. nmaste sir . DRDO ka admit card kb tk ayega sir plz btaiye
24. Pls sir kindly inform me when DRDO ceptam 8 STA B is download in my e-mail.
25. When Will I can download the admit card of captam 8??? Plz inform me…
26. for drdo exam
27. Exam date kya ha koi batao call 9983176526
28. DRDO admit card when come
□ Admit card Available coming soon !!!
29. Plz sir send my admit card from my gmail
30. Plz inform me that when will i can download admit card septam8
31. Sir kindly inform me how to get admit card of DRDO ceptam8 2016, Link is not open. Plz send it to me by email. Plz plz plz troubleshoot. and also send the syllabus, question paper pattern & old
question paper of post code 0104 & 0215
32. sir plz send me the syllabus for admin allied and plz send me the website to download the admit card and dates of examination
33. Exam kab se h
34. dear sir please send me the website to download the admit card and date of examination also send the syllabus pattern papers.
35. Sir,what is the exam date of drdo 1142post.
Please inform me.
36. sir,
i request please ceptam-8 exam date & admit card when the displade jast now tell me. my mo.no 9158613855
37. Some alert should be there in our email or in our mobl no. for downloading admit card.
□ sir drdo ka admit card aa gaya h kaya?
38. DRDO admit card KB ayega sir
39. Sir , can you plz tell me when will you relesa the hall tickets for cptem8
40. Sir kindly inform me how to get adimit card ceptam 2016
41. sir please give me help how can preparing for CEPTOM 8 EXAM .
42. Sir admit card kab dalanga
43. sir please suggest how to prepare for DRDO admin assistant exam
44. Sir Admit card nikalenge kaise
45. DRDO exam kis month me honge sir pls
46. Sir exam date kab hai
47. Sir please admit card ceptam-08 kab show kero
48. Drdo ka kb exam ho ga
49. admit kab aayenge
50. Exam kab hoga sir
51. Sir DRDO ka I card kab aayaga
52. Sir admit card kab ayega drdo ka
53. drdo exam kab hoogA
54. sir, admit card kab aiga
55. Dear sir please
about me drdo coming for admit card driver
Thanks & Regards
Lavkush kumar patel
56. technical ka Amit card abhi bhi nahi Mila hai.plz Drdo Amit card dala hai Kay ? yes or no answer digiye . thanks
57. sir admit card drdo ka kab ayega
58. Sir, drodo ka admit card kab aayaa
59. Sir DRDO ka admit card kab aayaga
60. sir plz tell me the hall ticket release dates
61. Mera admet card
62. Sir
Mera name shyam dhar Yadav s/0 ramkhelavan Yadav hamara admit card kab aayega sir
63. DRDO technician ka admit card kab ayega sir
64. Admit card kab ayega
65. Dear sir,
Sir ji hamne DRDO 1142 ki vacancy me driver ke liye aply kiya tha jiska admit card abhi take nahi aaya .
Sir please mujhe bataye ki admit card kab take aayega
66. Dear Respected sir mene DRDO me Lab Technician ki post pe apply kiya tha sir plz mujhe bataye ki mera admit card kab tak aayega. Dhanaywad sir
67. Sir give me ans why should not give to ad admit card now tell me which day u diclayr admit card I think u will be reply thx
68. sr maine welder se form dala tha admit card kb aayega
69. sir exam kb hai
70. Please send my Admit card….sir
71. DRDO 1142 tenician ka admit card sir
72. DRDO tenician ka admit card sir
73. electricians technicians.
74. Admit card sir plz sir kb
75. sir Maine is I bar wale DRDO recruitment mei apply kiya tha
kya us k admitcard pad chuke hai aur date nikal gayi hai
ya uska exam ho chuka hai
plz help mee
76. Hi,,,,,,
Sir DRDO ceptam8 ka admitcard July me kab tak aayega, 10 ke pehle /10 ke bad me.
77. exam date is 17 july admit card will be available after 2 july
78. sir code_0123 ka exam kab h aur admit card kab ayga plz bta dijye !
79. hello sir
maine DRDO ki exam ke liye applied kiya tha lekin mera naam abitak display nahi
so muje aage kya karna chaiye plz
sir help me
80. waiting for hall ticket
81. Please tell me sir
Whenever the come too admit card
82. admit card agya
83. Admitcad
84. To see the admit card
85. Any one send the ceptam 8 question paper 2016 (mechanical)
86. when the result will out of ceptam 8…
87. Sir
Results Kab aayai ga
88. Sir till when drdo ceptam 8 result will be declared??? How many no of student took exam
89. Sir, ceptam8 electrical engineer how many marks qualify please tell me previous mark( I think my marks 90 I qualify or not) please tell me sir. Iam eagerly awaiting sir
90. Sir, DRDO ceptam 8 result kab tak Aayega. Please tell me
91. Yahan koi ans dega plz result kab aayega dtp ka
92. Please give my result sir
93. Please give my result sir electronics drdo
94. Sir results admin assistant ka kab tak aayega…
95. Sir,
DRDO motor mechanic result kab aayega?
96. Sir please reuested result sta b tehanical 2016
97. Sir answer key drdo ka upload nhi huwa h kya
98. Plz tell me that when will the technician results announced
99. sir,please give my result, store assistant in drdo ceptam8 2016
100. sir drdo ka result kb tk lagega sir fitter
101. Sir drdo ceptem 8 ka result when declared
102. result fitter ka kb aayga
103. Sir drdo Technical assistance marks
104. sir drdo ka result kab ayega
105. dear sir please tell me 2016 exam result of STA-B when its come
106. Sir DRDO fitter ka result kab denge sir please sir
107. sir store assistant ka.result kab aaye.ga sir
108. when STAB results will be declared
109. Sir drdo post chemistry ka results kab aayega.
110. sir.stab 8 ka result kab ayaga.
111. Sir drdo admin assistant english typing which date result declared.
112. sir , anybody is not reply result wil be announced drdo ceptam 8 eng typing assistant and no one recive phone ???
What is matter ? Please reply
113. when will release drdo result sir plz inform through the mail
□ When drdo STA ‘B’ result will announce
114. Ceptam 8 result awaiting pls let me know the approx date sir.
115. When u will released drdo result plz inform through the mail…
116. Result declare date of vehicle operator. Please inform me
117. when will be result declare sir.
118. When we expect the DRDO results.
119. Sir when declered the result of the drdo ceptam 8 tech fitter
120. Drdo tec-B ka result kab ayega reply sir
121. Hai
DRDO ceptam 8 result 2016 result date send me sir
122. sir,when will be ceptam 8 exam result plz reply my mail.i m waiting
123. Sir jo result kab tak a g drdo k
124. When ceptem 8 result?
□ results will soon in 15 days
125. Sir, Mainee 17july 2016 ko exam di this Lekin abtak result nahi laga. Please sir mari help kijiye.post ka name-ADMIN ASSISTANT ‘A'(ENGLISH TYPING). Sir muje result ke baree me email kijiye.
126. Hello sir drdo ceptam 8 result kub tak aane ki sambhawna hai
127. hi.
plz declared ceptem 08 results date
128. Sir teh. B electrronics ka result kb aaga btana sir
129. when the results are realeased ceptam 08
pls reply me …..
130. Results information, my Gmail Id
131. Sir plz declerd results drdo 1142 technician
132. Sir now its too late. Its time to declare a result now……..
133. Plz tell me that when will the technician results announced
134. When drdo result publiced sir tell me please sir and send my email
135. Sir please declare the drdo ceptam 8 result date
136. Sir plz tell me that when will the technician result
137. Result of drdo
138. Sir plz send result date
139. sir plz tell. me drdo ceptam 8 result
140. Still how many days you ppl want to declare the result we don’t know but we really waiting so much.
141. Sir Please Tell Me DRDO Exam Result Held In July 2016
142. Sir technian A ka result kab tk aiyga
143. Please tell mi when will display DRDO result .
144. Sir Plz send me ceptam result to my mail sir plz
145. Sir plz send me the DRDO exam result quickly
146. Sir techinicion B ka result kab take aiyga
147. sir when will be the result of drdo declared.
148. Sir pls send result date
149. Sir’ please results date and kitne number par selection ho sakta haai
150. Tech A ka result
151. Drdo result 2015-2016
152. Sir drdo ka result kb tk aayega. ??
153. Sir when STA-B Ceptam08 result will declare??
154. sir JI DRDO Ka result kab take ayega ji
155. When is the July 17 2016 exams result
□ Oct 2nd week
156. Sir kab ayega store assistants A ka result..please sir
157. Sir DRO ka result kab tak aayga please batna
158. Cut off marks not percentage for unreserved category for the post of administrative assistant and store assistant
159. Steno results
160. DRDO result date tell me
161. sir when drdo ceptam 8 result will be decleared
162. when drdo entry test result will be declared
163. Ruselt kab aayga teknikal fitter
□ Sir DRDO result kn aega
164. sir drdo ka result kb tak aega
165. Sir results kab tak ayga next month tak please show open the results
166. drdo results when
167. Sir. Drdo ceptam 8 result kab aega
168. When is the results sir
169. When CEP 8 results releases sir please inform
170. Sir when DRDO Ceptam8 Results Declared
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176. Please send my technician A result on my email.
Reffrence code- 80103020800727
Post code- 0208
177. Hai sir, iam waiting for my drdo ceptam 8 result. when release drdo result.
178. drdo result kb ayega sir plz btaye
179. Sir ,tell me about drdo result .
180. When drdo result will declare.I’m waiting for soo long.can u plssss tell me when it declares
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184. Regarding my result drdo ceptam 08 dob 02-11-1965
185. Where is DRDO result sir…pease tell us the exact date
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187. Sir result kabtak ayega
188. plzz tell us, when will be come result
189. Results kab ayga drdo ka
190. mera admid card gum ho gya hai result kaise pata karu
191. drdo result 2016 ceptam8
192. I am kuldeep rajbanshi
post name admin assistant a’
193. Sir post 109 ceptam ka result kb ayega
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post…124 ka result kb tk aaega
195. sir post ADMIN ASSISTANT ‘A’ code 0501 please give me result
196. Admin assistance A result kab ayga
197. post code 211 ka result
198. drdo. i.t.i candidates results
199. Sir sta-B result kab tak aa rha h
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Post code 0208
Subject ELECTRICIAN
207. Sir plz send fitter merit list and result
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I lost my admit card.plz send me my reference number. Plz
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212. what zir 14 digits only 11digits number their sir…… | {"url":"https://jobslab.in/drdo-admit-card-drdo-gov-in/","timestamp":"2024-11-14T05:21:07Z","content_type":"text/html","content_length":"230099","record_id":"<urn:uuid:f82f074f-810a-48ac-9e68-8c5e802c823a>","cc-path":"CC-MAIN-2024-46/segments/1730477028526.56/warc/CC-MAIN-20241114031054-20241114061054-00855.warc.gz"} |
Boxes - DrillBox
A parametrized box for drills
Settings for Finger Joints
style style of the fingers
surroundingspaces space at the start and end in multiple of normal spaces
bottom_lip height of the bottom lips sticking out (multiples of thickness) FingerHoleEdge only!
edge_width space below holes of FingerHoleEdge (multiples of thickness)
finger width of the fingers (multiples of thickness)
play extra space to allow finger move in and out (multiples of thickness)
space space between fingers (multiples of thickness)
width width of finger holes (multiples of thickness)
Settings for RoundedTriangleEdge
height height above the wall
r_hole radius of hole
radius radius of top corner
outset extend the triangle along the length of the edge (multiples of thickness)
Settings for Stackable Edges
angle inside angle of the feet
bottom_stabilizers height of strips to be glued to the inside of bottom edges (multiples of thickness)
height height of the feet (multiples of thickness)
holedistance distance from finger holes to bottom edge (multiples of thickness)
width width of the feet (multiples of thickness)
Settings for Mounting Edge
d_head head diameter of mounting screw (in mm)
d_shaft shaft diameter of mounting screw (in mm)
margin minimum space left and right without holes (fraction of the edge length)
num number of mounting holes (integer)
side side of box (not all valid configurations make sense...)
style edge style
DrillBox Settings
top_edge edge type for top edge
sx sections left to right in mm 🛈
sy sections back to front in mm 🛈
sh sections bottom to top in mm 🛈
bottom_edge edge type for bottom edge
holes Number of holes for each size
firsthole Smallest hole
holeincrement increment between holes
Default Settings
thickness thickness of the material (in mm) 🛈
format format of resulting file 🛈
tabs width of tabs holding the parts in place (in mm)(not supported everywhere) 🛈
debug print surrounding boxes for some structures 🛈
labels label the parts (where available)
reference print reference rectangle with given length (in mm)(zero to disable) 🛈
inner_corners style for inner corners 🛈
burn burn correction (in mm)(bigger values for tighter fit) 🛈 | {"url":"https://boxgen.cnchub.ru/DrillBox","timestamp":"2024-11-11T20:36:04Z","content_type":"text/html","content_length":"17288","record_id":"<urn:uuid:ae8765ac-5c93-4eac-b37c-da59fcb5dcbb>","cc-path":"CC-MAIN-2024-46/segments/1730477028239.20/warc/CC-MAIN-20241111190758-20241111220758-00044.warc.gz"} |
Rank-based BST Iterators: No Parent Pointer, No Stack, No Problem
It's no secret that I am an iterator pattern fan boy. When it comes to implementing search trees - or any container of which you would like to access the elements one by one and in-place for that
matter, the iterator pattern is indispensable. It allows us to hide the containers underlying implementation details, affords us a way to signify search misses without the use of returning NULL or
some other "magic" value to signify the item isn't present, and provides a convenient interface for other algorithms to work on the elements of our data structure. It is thus unfortunate that the
implementation of iterators for a binary search tree based container can be somewhat of a messy affair.
In-Order Traversal (Not a BST)
When it comes to implementation the two popular strategies for binary search tree iterators are the use of parent pointers to traverse the tree, or in the absence of parent pointers is the "fat
iterator" concept, where one stores the path to the current position in the tree in a stack as well as a pointer to the current node. Both approaches have their negatives, along with vocal opponents
to each (welcome to the internet). Parent pointers are disliked for not only the extra space required in each node to accommodate the additional pointer but also the increase in complexity the
additional pointer brings to the algorithms, especially those for self-balancing BST's. The so called "fat" iterators are disliked for their larger memory footprint. It should be noted that the
implementation used in the C++ standard library is of a parent pointer based Red Black Tree.
In this post I am going to discuss an implementation strategy for binary search tree iterators that that has efficient time complexity and minimal space complexity without the need to rely on either
parent pointers or auxiliary data structures. This allows us to use simple recursive algorithms for implementing binary search trees, without sacrificing efficient iteration. In a binary search tree
which already supports ordered statistic operations, no additional space is required, otherwise each node needs an additional integer for tracking the side of the subtree rooted at that node. This is
still a smaller memory footprint than using parent pointers, with the addition of newly supported operations outside of iteration (rank and select) if one so chooses to make their API's public.
Ordered Statistic Trees, More Useful Everyday!
The "rank" of an element in a binary search tree is the order in which it would appear in an in-order traversal of that tree. To put it another way, if we were to take all of the keys in the tree and
place them in an array in sorted order, their rank would be the index of their position in the array. As mentioned above, we can obtain the rank of a node by adding a variable to track the size of
the subtree rooted at that node. A leaf node has a size of 1, as it is the only member in it's tree, as such the empty child pointers have a size of 0.
template <class K, class V>
struct node {
KVPair<K,V> info;
int size;
bool color;
node* left;
node* right;
node(K k, V v) : info(k, v), size(1), color(red), left(nullptr), right(nullptr) { }
int size(node* x) {
return (x == nullptr) ? 0:x->size;
Upon inserting a new node we update the value of each nodes size counter as the stack unwinds in order to keep the values accurate.
link putRB(link h, K k, V v) {
if (h == nullptr) {
return new node(k, v);
if (isRed(h->left) && isRed(h->right))
h = flipColors(h);
if (k < h->info.key()) h->left = putRB(h->left, k, v);
else h->right = putRB(h->right, k, v);
h->n = 1 + size(h->left) + size(h->right);
return balance(h, true);
Similarly, if the tree we are implementing the iterator for uses balancing algorithms, you must be careful to update the size counters after performing any rotations in the tree as well. For the
examples I am using a red black tree, and so we will update the nodes size counts at the same time we update their colors during rotations. It is not necessary to use a self balancing binary search
tree, however doing so means that the increment operation for the iterator has a worst case performance guarantee of O(log n) instead of O(h).
link rotL(link h) {
link x = h->right; h->right = x->left; x->left = h;
fixHeightAndColor(h, x);
return x;
link rotR(link h) {
link x = h->left; h->left = x->right; x->right = h;
fixHeightAndColor(h, x);
return x;
void fixHeightAndColor(link& x, link& h) {
x->color = h->color;
h->color = true;
x->n = h->n;
h->n = 1 + size(h->left) + size(h->right);
The last thing we need before we can implement the actual iterator, is to add a method to the tree for the actual selection of a node by rank. This turns out to be a straight forward algorithm that I
have covered in previous posts and so will skip the finer details of and skip straight to the implementation.
link select(link x, int rank) {
if (x == nullptr) return x;
int curr = size(x->left);
if (curr > rank) return select(x->left, rank);
if (curr < rank) return select(x->right, rank - curr - 1);
return h;
And now finally, we are ready to implement our iterator for our tree.
The key observation which lays out the method we will take to implement the iterator is two fold. First, an iterator for a binary search tree should visit each node in the same order as one would
during an in-order traversal. And second is the relationship between a nodes rank, and when it appears in an in order traversal as described above. This leads us to the observation that we can
simulate an in-order traversal of the tree by selecting each node in the tree according to its rank in increasing order. This gives us a very straight forward method of implementing the actual
iterator: the iterator will maintain a pointer to the root of the tree, and a counter variable to track which rank to select, with the increment operator incrementing this counter, and the get
operation performing the selection by rank.
By using the same interface for iterator class as the standard library does, we can use C++'s enhanced for loop to iterate over the container. Iterators from the standard library are implemented
through operator overloading. All STL compliant iterators must support the following operators
• operator++() - pre increment operator, for advancing the iterators position in the tree
• operator*() - the star operator performs the "get" operation on our iterator, returning the value stored at the current position.
• operator==() - test for equality. is this operator the same as that operator?
• operator!=() - test for inequality. is this operator NOT that operator?
That is of course, only the minimum set needed for a function iterator. Other operators which are often seen depending on the type of iterator are the post increment operator, various decrement
operators, algebraic operators for doing the equivalent of pointer math, just to name a few. I personally choose to also implement a non-operator based API for my iterators with the public methods:
• get() - returns the element at the current position
• next() - advances one position to the next in-order element in the tree
• done() - a returns true if there is no more elements to visit, false otherwise.
What can I say? Variety is the spice of life, even for our API's.
template <class K, class V>
class rankIterator {
rbnode<K,V>* tree;
int pos;
KVPair<K,V> nullInfo;
int size(rbnode<K,V>* h) {
return (h == nullptr) ? 0:h->n;
KVPair<K,V>& select(rbnode<K,V>* h, int k) {
if (h == nullptr)
return nullInfo;
int t = size(h->left);
if (t > k) return select(h->left, k);
else if (t < k) return select(h->right, k-t-1);
else return h->info;
rankIterator(rbnode<K,V>* rootptr, int p) {
tree = rootptr;
pos = p;
bool done() {
return tree != nullptr && pos >= tree->size();
KVPair<K,V>& get() {
return select(tree, pos);
void next() {
rankIterator& operator++() {
return *this;
rankIterator& operator++(int) {
rankIterator it = *this;
return *this;
KVPair<K,V>& operator*() {
return get();
bool operator==(const rankIterator& oit) const {
return tree == oit.tree && pos == oit.pos;
bool operator!=(const rankIterator& oit) const {
return !(*this==oit);
Now to use our iterator with C++'s enhanced for loop all that is required to add the appropriate iterators to our tree class. To be specific, we must implement a method named begin() which returns an
iterator pointing to the first element in the tree, and a method named end() which returns an iterator pointing to one passed the last element in our binary search tree.
template <class K, class V>
class RedBlackTree {
//yadda yadda yadda
rankIterator<K,V> begin() {
return rankIterator<K,V>(root, 0);
rankIterator<K,V> end() {
return rankIterator<K,V>(root, size());
void printTree(RedBlackTree<char, char>& rb) {
for (auto m : rb) {
cout<<m.key()<<" ";
int main() {
RedBlackTree<char, char> rb;
string phrase = "aredblacksearchtreeiterator";
for (char c : phrase)
rb.put(c, c, alg);
return 0;
max@MaxGorenLaptop:~/mgc_common$ ./rb
h e b a a a a c c d e e e e r l k i r o s r r r t t t
a a a a b c c d e e e e e h i k l o r r r r r s t t t
Valid Red Black Tree.
And there you have it, simple binary search tree iterators using rank. That's all for today, so until next time happy hackling! | {"url":"http://maxgcoding.com/bst-iterators-revisited-no-parent-pointer-no-stack-no-problem","timestamp":"2024-11-03T05:53:22Z","content_type":"text/html","content_length":"20308","record_id":"<urn:uuid:b1234bec-e800-4d3c-b7b9-8f0c20b908aa>","cc-path":"CC-MAIN-2024-46/segments/1730477027772.24/warc/CC-MAIN-20241103053019-20241103083019-00843.warc.gz"} |
Cross product
In vector algebra, different types of vectors are defined and various operations can be performed on these vectors such as addition, subtraction, product and so on. In this article, the cross product
of two vectors, formulas, properties, and examples is explained.
Table of Contents:
What is a Cross Product?
Cross product is a binary operation on two vectors in three-dimensional space. It results in a vector that is perpendicular to both vectors. The Vector product of two vectors, a and b, is denoted by
a × b. Its resultant vector is perpendicular to a and b. Vector products are also called cross products. Cross product of two vectors will give the resultant a vector and calculated using the
Right-hand Rule.
Cross Product of Two Vectors
The vector product or cross product of two vectors A and B is denoted by A × B, and its resultant vector is perpendicular to the vectors A and B. The cross product is mostly used to determine the
vector, which is perpendicular to the plane surface spanned by two vectors, whereas the dot product is used to find the angle between two vectors or the length of the vector. The cross product of two
vectors, say A × B, is equal to another vector at right angles to both, and it happens in the three-dimensions.
Cross Product Formula
If θ is the angle between the given two vectors A and B, then the formula for the cross product of vectors is given by:
A × B = |A| |B| sin θ
\(\vec{A}\times \vec{B}=||\vec{A}|| \ ||\vec{B}|| sin\theta \hat{n}\)
\(\vec{A},\vec{B}\) are the two vectors.
\(||\vec{A}||, \ ||\vec{B}||\) are the magnitudes of given vectors.
θ is the angle between two vectors and \(\hat{n}\) is the unit vector perpendicular to the plane containing the given two vectors, in the direction given by the right-hand rule.
Cross product of two vectors Formula
Consider two vectors,
A = ai + bj + ck
B = xi + yj + zk
We know that the standard basis vectors i, j, and k satisfy the below-given equalities.
i × j = k and j × i = –k
j × k = i and k × j = –i
k × i = j and i × k = –j
Also, the anti-commutativity of the cross product and the distinct absence of linear independence of these vectors signifies that:
i × i = j × j = k × k = 0
A × B = (ai + bj + ck) × (xi + yj + zk)
= ax(i × i) + ay(i × j) + az(i × k) + bx(j × i) + by(j × j) + bz(j × k) + cx(k × i) + cy(k × j) + cz(k × k)
By applying the above mentioned equalities,
A × B = ax(0) + ay(k) + az(-j) + bx(-k) + by(0) + bz(i) + cx(j) + cy(-i) + cz(0)
= (bz – cy)i + (cx – az)j + (ay – bx)k
Cross Product Matrix
We can also derive the formula for the cross product of two vectors using the determinant of the matrix as given below.
A = ai + bj + ck
B = xi + yj + zk
\(\mathbf{A}\times \mathbf{B} = \begin{vmatrix} \boldsymbol{i} & \mathbf{j} & \mathbf{k}\\ a & b & c\\ x & y & z \end{vmatrix}\)
A × B = (bz – cy)i – (az – cx)j + (ay – bx)k
= (bz – cy)i + (cx – az)j + (ay – bx)k
Right-hand Rule Cross Product
We can find the direction of the unit vector with the help of the right-hand rule. In this rule, we can stretch our right hand so that the index finger of the right hand in the direction of the first
vector and the middle finger is in the direction of the second vector. Then, the thumb of the right hand indicates the direction or unit vector n. With the help of the right-hand rule, we can easily
show that vectors’ cross product is not commutative. If we have two vectors A and B, then the diagram for the right-hand rule is as follows:
Cross Product Properties
To find the cross product of two vectors, we can use properties. The properties such as anti-commutative property, zero vector property plays an essential role in finding the cross product of two
vectors. Apart from these properties, some other properties include Jacobi property, distributive property. The properties of cross-product are given below:
Cross Product of Perpendicular Vectors
Cross product of two vectors is equal to the product of their magnitude, which represents the area of a rectangle with sides X and Y. If two vectors are perpendicular to each other, then the cross
product formula becomes:
θ = 90 degrees
We know that, sin 90° = 1
Cross Product of Parallel vectors
The cross product of two vectors are zero vectors if both the vectors are parallel or opposite to each other. Conversely, if two vectors are parallel or opposite to each other, then their product is
a zero vector. Two vectors have the same sense of direction.
θ = 90 degrees
As we know, sin 0° = 0 and sin 90° = 1
Magnitude of Cross Product
Let us assume two vectors, \(\vec{A}= A_{x}+ A_{y}+ A_{z}\) and \(\vec{B}= B_{x}+ B_{y}+ B_{z}\), then the magnitude of two vectors are given by the formula,
\(|\vec{A}| = \sqrt{A_{x}^{2} + A_{y}^{2}+ A_{z}^{2}}\)
\(|\vec{B}|| = \sqrt{B_{x}^{2} + B_{y}^{2}+ B_{z}^{2}}\)
Hence, the magnitude of the cross product of two vectors are given by the formula,
\(|\vec{A}\times \vec{B}|= |\vec{A}| |\vec{B}| |sin\theta|\)
Cross Product Example
Find the cross product of the given two vectors: \(\vec{X}= 5\vec{i} + 6\vec{j} + 2\vec{k}\) and \(\vec{Y}= \vec{i} + \vec{j} + \vec{k}\)
\(\vec{X}= 5\vec{i} + 6\vec{j} + 2\vec{k} \\ \vec{Y}= \vec{i} + \vec{j} + \vec{k}\)
To find the cross product of two vectors, we have to write the given vectors in determinant form. Using the determinant form, we can find the cross product of two vectors as:
\(\vec{X}\times \vec{Y} = \begin{vmatrix} \vec{i} & \vec{j} & \vec{k}\\ 5 & 6 & 2\\ 1 & 1 & 1 \end{vmatrix}\)
By expanding,
\(\vec{X}\times \vec{Y}= (6-2)\vec{i}-(5-2)\vec{j}+ (5-6)\vec{k}\)
Therefore, \(\vec{X}\times \vec{Y}= 4\vec{i}-3\vec{j}- \vec{k}\)
Visit BYJU’S – The Learning App and get all the important Maths-related articles and videos to learn with ease. | {"url":"https://mathlake.com/Cross-product","timestamp":"2024-11-01T22:48:40Z","content_type":"text/html","content_length":"22339","record_id":"<urn:uuid:d1c96703-c549-4f2d-8389-6ea9e7dbfbd1>","cc-path":"CC-MAIN-2024-46/segments/1730477027599.25/warc/CC-MAIN-20241101215119-20241102005119-00701.warc.gz"} |
[EM] RE : MMPO and Raynaud
[EM] RE : MMPO and Raynaud
Kevin Venzke stepjak at yahoo.fr
Wed May 2 05:46:26 PDT 2007
--- Gervase Lam <gervase.lam at group.force9.co.uk> a écrit :
> > Here's an extreme situation of this:
> >
> > 1000 A
> > 1 A=C
> > 1 B=C
> > 1000 B
> >
> > C wins.
> Apart from may be disallowing equal rankings but still allowing
> truncation (as mentioned in previous posts), another way I thought of to
> alleviate this problem
This doesn't really alleviate the problem. You just change those two
ballots then to strictly prefer C over A and B. I guess the example is
written normally (by me) with equal rankings because 1. it shows the
problem exists even on approval ballots, and 2. it leaves C without any
strict first preferences.
> is the following:
> (1) Like MMPO, get the highest pairwise opposition scores of each
> candidate.
> (2) Drop the candidate with the greatest such score, together with the
> candidate's pairwise results.
> (3) Repeat step (2) until one candidate remains.
This is exactly the definition I understand of Raynaud(WV). Because the
highest opposition score among all candidates is going to be a winning
score (unless it's a tie).
> Changing the example slightly:
> 1000 A
> 2 A=C
> 1 B=C
> 1000 B
> The pairwise opposition scores are:
> A<B 1001 A<C 1
> B<A 1002 B<C 2
> C<A 1000 C<B 1000
> When the method is used, B is dropped because its highest pairwise
> opposition score is the greatest compared with the other candidates'
> highest pairwise opposition scores. With A and C left, A wins because
> it has a better pairwise opposition score than C. Using MMPO, C would
> still win in the example.
Unfortunately Raynaud's properties aren't very good. It loses both
LNHarm and FBC relative to MMPO. You may as well use Schulze if you're
going to do that.
> I suppose this is Raynaud(Pairwise Opposition Loser), which is very
> similar to Raynaud(Gross Loser). If none of the voters submit ballots
> with equal rankings, Raynaud(Pairwise Opposition Loser) and Raynaud
> (Gross Loser) are the same.
True, but otherwise I don't think it is very similar to Raynaud(GL).
> I think Raynaud(Pairwise Opposition Loser) satisfies the Plurality
> Criterion in a similar way to Raynaud(Gross Loser).
I'm afraid not:
49 A
24 B
27 C>B
A is eliminated and then C beats B, contrary to Plurality.
The reason Raynaud(GL) satisfies Plurality is because it eliminates
based on votes cast in favor of a candidate, which is what the Plurality
criterion is concerned with. Raynaud(PO) or (WV) only regards votes
cast *against* candidates.
Kevin Venzke
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Previous: stbcon Up: ../lapack-s.html Next: stbtrs
STBRFS - provide error bounds and backward error estimates
for the solution to a system of linear equations with a tri-
angular band coefficient matrix
SUBROUTINE STBRFS( UPLO, TRANS, DIAG, N, KD, NRHS, AB, LDAB,
B, LDB, X, LDX, FERR, BERR, WORK, IWORK,
INFO )
CHARACTER DIAG, TRANS, UPLO
INTEGER INFO, KD, LDAB, LDB, LDX, N, NRHS
INTEGER IWORK( * )
REAL AB( LDAB, * ), B( LDB, * ), BERR( * ),
FERR( * ), WORK( * ), X( LDX, * )
STBRFS provides error bounds and backward error estimates
for the solution to a system of linear equations with a tri-
angular band coefficient matrix.
The solution matrix X must be computed by STBTRS or some
other means before entering this routine. STBRFS does not
do iterative refinement because doing so cannot improve the
backward error.
UPLO (input) CHARACTER*1
= 'U': A is upper triangular;
= 'L': A is lower triangular.
TRANS (input) CHARACTER*1
Specifies the form of the system of equations:
= 'N': A * X = B (No transpose)
= 'T': A**T * X = B (Transpose)
= 'C': A**H * X = B (Conjugate transpose = Tran-
DIAG (input) CHARACTER*1
= 'N': A is non-unit triangular;
= 'U': A is unit triangular.
N (input) INTEGER
The order of the matrix A. N >= 0.
KD (input) INTEGER
The number of superdiagonals or subdiagonals of the
triangular band matrix A. KD >= 0.
NRHS (input) INTEGER
The number of right hand sides, i.e., the number of
columns of the matrices B and X. NRHS >= 0.
AB (input) REAL array, dimension (LDAB,N)
The upper or lower triangular band matrix A, stored
in the first kd+1 rows of the array. The j-th column
of A is stored in the j-th column of the array AB as
follows: if UPLO = 'U', AB(kd+1+i-j,j) = A(i,j) for
max(1,j-kd)<=i<=j; if UPLO = 'L', AB(1+i-j,j) =
A(i,j) for j<=i<=min(n,j+kd). If DIAG = 'U', the
diagonal elements of A are not referenced and are
assumed to be 1.
LDAB (input) INTEGER
The leading dimension of the array AB. LDAB >=
B (input) REAL array, dimension (LDB,NRHS)
The right hand side matrix B.
LDB (input) INTEGER
The leading dimension of the array B. LDB >=
X (input) REAL array, dimension (LDX,NRHS)
The solution matrix X.
LDX (input) INTEGER
The leading dimension of the array X. LDX >=
FERR (output) REAL array, dimension (NRHS)
The estimated forward error bounds for each solution
vector X(j) (the j-th column of the solution matrix
X). If XTRUE is the true solution, FERR(j) bounds
the magnitude of the largest entry in (X(j) - XTRUE)
divided by the magnitude of the largest entry in
X(j). The quality of the error bound depends on the
quality of the estimate of norm(inv(A)) computed in
the code; if the estimate of norm(inv(A)) is accu-
rate, the error bound is guaranteed.
BERR (output) REAL array, dimension (NRHS)
The componentwise relative backward error of each
solution vector X(j) (i.e., the smallest relative
change in any entry of A or B that makes X(j) an
exact solution).
WORK (workspace) REAL array, dimension (3*N)
IWORK (workspace) INTEGER array, dimension (N)
INFO (output) INTEGER
= 0: successful exit
< 0: if INFO = -i, the i-th argument had an illegal | {"url":"https://www.math.utah.edu/software/lapack/lapack-s/stbrfs.html","timestamp":"2024-11-05T19:49:47Z","content_type":"text/html","content_length":"5492","record_id":"<urn:uuid:e3d4265a-ba12-496d-9db9-de9c70a924bf>","cc-path":"CC-MAIN-2024-46/segments/1730477027889.1/warc/CC-MAIN-20241105180955-20241105210955-00055.warc.gz"} |
Probabilistic modeling of cast iron water distribution pipe corrosion.
Duchesne, Sophie; Chahid, Naoufel; Bouzida, Nabila et Toumbou, Babacar (2013). Probabilistic modeling of cast iron water distribution pipe corrosion. Journal of Water Supply Research and Technology -
Aqua , vol. 62 , nº 5. pp. 279-287. DOI: 10.2166/aqua.2013.125.
Ce document n'est pas hébergé sur EspaceINRS.
Due to the random nature of the corrosion process, stochastic approaches are more appropriate to mathematically represent the corrosion depths on metallic objects. Based on data from 202 pipes, a
model was developed to compute the probability of finding maximal corrosion depth in a given interval of values for 150-mm cast iron water distribution pipes. Only the age of pipes was taken into
account as an explanatory variable to compute this probability, since the soil characteristics were not available in the close surroundings of the inspected pipes. The model combines two functions:
(1) a Weibull distribution function to represent the distribution of pipe ages at the time when the maximal corrosion depth reaches 100% of the pipe wall thickness; and (2) a generalized extreme
value (GEV) distribution function, with the location parameter varying as a function of pipe age, to represent the distribution of maximal corrosion pit depths on pipes that did not reach a maximal
corrosion pit equal to 100% of their wall thickness. The developed model offers a good representation of the distribution of observed maximal corrosion depths for Quebec City's 150-mm cast iron water
Type de document: Article
Mots-clés libres: censored data; GEV distribution function; maximum likelihood; pipe age; soil characteristics; stochastic model
Centre: Centre Eau Terre Environnement
Date de dépôt: 05 déc. 2016 20:47
Dernière modification: 05 déc. 2016 20:47
URI: https://espace.inrs.ca/id/eprint/3427
Gestion Actions (Identification requise) | {"url":"https://espace.inrs.ca/id/eprint/3427/","timestamp":"2024-11-06T07:38:27Z","content_type":"application/xhtml+xml","content_length":"20797","record_id":"<urn:uuid:b81cb51f-1bf4-44a7-88c7-9722995bb074>","cc-path":"CC-MAIN-2024-46/segments/1730477027910.12/warc/CC-MAIN-20241106065928-20241106095928-00552.warc.gz"} |
emitsInAnyOrder function
Returns a StreamMatcher that matches the stream if each matcher in matchers matches, in any order.
If any matcher fails to match, this fails and consumes no events. If the matchers match in multiple different possible orders, this chooses the order that consumes as many events as possible.
If any sequence of matchers matches the stream, no errors from other sequences are thrown. If no sequences match and multiple sequences throw errors, the first error is re-thrown.
Note that checking every ordering of matchers is O(n!) in the worst case, so this should only be called when there are very few matchers.
StreamMatcher emitsInAnyOrder(Iterable matchers) {
var streamMatchers = matchers.map(emits).toSet();
if (streamMatchers.length == 1) return streamMatchers.first;
var description = 'do the following in any order:\n'
'${bullet(streamMatchers.map((matcher) => matcher.description))}';
return StreamMatcher(
(queue) async => await _tryInAnyOrder(queue, streamMatchers) ? null : '', | {"url":"https://pub.dev/documentation/matcher/latest/expect/emitsInAnyOrder.html","timestamp":"2024-11-04T21:41:46Z","content_type":"text/html","content_length":"7278","record_id":"<urn:uuid:438ff38a-37c4-411d-9dcf-5de6daba5e7d>","cc-path":"CC-MAIN-2024-46/segments/1730477027861.16/warc/CC-MAIN-20241104194528-20241104224528-00105.warc.gz"} |
ROB501:Assignment #4: Image-Based Visual Servoing - Programming Help
Image-based visual servoing (IBVS) is a popular form of closed-loop feedback control based on errors mea-sured in image space, between the observed and desired positions of image feature points. In
this assignment, you will write a simple IBVS controller. The goals are to:
• provide some additional practice deriving and using Jacobian matrices,
• introduce the matrix pseudo-inverse to solve overdetermined systems in a least squares sense,
• assist in understanding how simple proportional controllers are built and used for vision tasks, and
• determine the sensitivity of IBVS to 3D point depth uncertainty/errors.
Prior to starting the assignment, you should install the Lie groups Python library that is available from the following GitHub URL: https://github.com/utiasSTARS/liegroups.
The due date for assignment submission is Friday, November 25, 2022, by 11:59 p.m. EDT. All submissions will be in Python 3 via Autolab; you may submit as many times as you wish until the deadline.
To complete the assignment, you will need to review some material that goes beyond that discussed in the lectures—more details are provided below. The project has four parts, worth a total of 50
Please clearly comment your code and ensure that you only make use of the Python modules and functions listed at the top of the code templates. We will view and run your code.
Part 1: Image-Based Jacobian
IBVS effectively operates entirely in image space (i.e., on the image plane, R^2), that is, no explicit references to 3D are made (beyond having depth estimates for feature points). The first step is
to derive the Jacobian matrix that relates motions of the camera in 3D space to motions of points on the image plane. Conveniently, most of this work has already been done for you and the result is
given by Equation 15.6 in the Corke text (on pg. 544). For this portion of the assignment, you should submit:
• A function in ibvs_jacobian.py that computes and returns the x and y velocities of a point on the image plane, given the translational and rotational velocities of the camera and the (estimated)
depth of the point.
Later, we will invert several stacked Jacobian matrices to calculate the desired camera motion. Note that, in the Jacobian computation, the image point coordinates (in pixels) must first convert to
normalized image plane coordinates (see early lectures in the course).
Part 2: Image-Based Servo Controller
The next step is to design and code up a simple proportional controller that will moving the camera (i.e., by providing velocity commands) in an effort to reduce the distance (error) between the
observed positions of several feature points in the current camera image and their desired positions (which we assume to be known in advance).
ROB501: Computer Vision for Robotics Assignment #4: Image-Based Visual Servoing
We seek to determine the translational and rotational velocities of the camera (so six numbers), and the motion of each image plane point gives us two pieces of information—hence, we require at least
three points on the image plane. However, it is often desirable to use more than three points, in which case we have an overdetermined system; recalling the normal equations from linear least
squares, we can solve our problem by computing the Moore-Penrose pseudo-inverse of the (stacked) Jacobian matrix:
[J]+ [=] ^([J]T [J]^)^−^1 [J]T
Wikipedia has a good discussion of the matrix pseudo-inverse. Note that the J in the equation above is formed from the stacked 2 × 6 sub-Jacobians for each image plane point. The complete controller
is specified by Equation 15.11 in the Corke text (on pg. 548). For this portion of the assignment, you should submit:
• A function in ibvs_controller.py that implements a simple proportional controller for IVBS. The single gain value will be passed as an argument to the controller. The function should compute the
output velocities given three or more image plane point correspondences.
Part 3: Depth Estimation
As noted in the lecture session and in the Corke book, IBVS is relatively tolerant to errors in the estimated depths of feature points. Given an initial (incorrect) set of estimated feature depths,
it is possible to refine the depth values by making use of (known) camera motion and observed image plane motion (of the feature points). The Corke text provides a description of this approach on pg.
553, summarized by Equations 15.13 and 15.14. For this portion of the assignment, you should submit:
• A function in ibvs_depth_finder.py that produces a new set of depth estimates for feature points, given the commanded camera velocity and changes in image feature positions (i.e., a first order
estimate of the image feature velocities).
Part 4: Performance Evaluation
With your IBVS system now up and running, the last step is to run a few experiments to determine its perfor-mance and overall sensitivity to errors in the estimated feature depths. Fist, you should
choose a challenging initial camera pose (i.e., with a large difference in feature alignment). You may use the same test harness provided in the learner examples and just change the initial pose.
Then, you should attempt to answer the following questions:
• What is the optimal gain value (experimentally) for the case where feature depths are known exactly? That is, what gain value leads to the fastest convergence?
• What is the optimal gain value (experimentally) for the case where feature depths are estimated? That is, what gain value leads to the fastest convergence?
• How much worse is the performance with estimated depths compared to with known depths? Is the difference significant.
For this portion of the assignment, you should submit:
• A single PDF document called report.pdf that briefly answers the questions above. The document should be a couple pages (three maximum) and include a handful of plots that show your results.
ROB501: Computer Vision for Robotics Assignment #4: Image-Based Visual Servoing
Points for each portion of the assignment will be determined as follows:
• IBVS Jacobian function – 10 points (5 tests × 2 points per test)
Each test uses a different image plane point and a different camera intrinsic matrix. The 12 entries in the Jacobian matrix must be exact (up to a small tolerance) to pass.
• IBVS controller function – 20 points (5 tests × 4 points per test)
Each test uses a different set of image plane points and a different camera intrinsic matrix. The six output values (velocities) must be exact (up to a small tolerance) to pass.
• Depth estimation function – 10 points (5 tests × 2 points per test)
Each test uses a different set of image plane points. The n output values (depths) must be exact (up to a small tolerance) to pass.
• PDF performance report document – 10 points (Up to 10 points assigned for convincing results)
The document should contain a few plots and perhaps a table or two. It must be a valid PDF and have the filename report.pdf, otherwise it will not be graded.
Total: 50 points
Grading criteria include: correctness and succinctness of the implementation of support functions, and proper overall program operation and code commenting and details. Please note that we will test
your code and it must run successfully. Code that is not properly commented or that looks like ‘spaghetti’ may result in an overall deduction of up to 10%.
3 of 3 | {"url":"https://www.edulissy.org/product/image-based-visual-servoing-ibvs-is-a-popular-form-of-closed-loop-feedback-control-based-on-errors-mea-sured/","timestamp":"2024-11-03T18:32:11Z","content_type":"text/html","content_length":"184414","record_id":"<urn:uuid:311b609b-abfb-46a6-91ac-bd28ce7f400a>","cc-path":"CC-MAIN-2024-46/segments/1730477027782.40/warc/CC-MAIN-20241103181023-20241103211023-00074.warc.gz"} |
Punto Banco Regulations and Plan
Dec 15 2021
Baccarat Banque Policies
Punto banco is gambled on with eight decks in a dealing shoe. Cards valued less than ten are counted at their printed value while at the same time Ten, Jack, Queen, King are zero, and Ace is one.
Bets are made on the ‘bank’, the ‘player’, or on a tie (these are not really people; they just represent the two hands that are dealt).
Two cards are dealt to both the ‘bank’ and ‘gambler’. The score for every hand is the sum of the 2 cards, but the 1st digit is ignored. For example, a hand of 5 and six has a total of 1 (five plus 6
= 11; ditch the 1st ‘one’).
A third card might be given out using the rules below:
- If the player or banker has a value of 8 or 9, the two players stay.
- If the gambler has five or lower, she hits. Players otherwise stand.
- If the gambler stands, the house hits on 5 or lower. If the player hits, a table is employed to determine if the banker holds or hits.
Punto Banco Odds
The greater of the two hands wins. Winning wagers on the banker payout nineteen to Twenty (equal cash less a 5% rake. Commission are kept track of and paid off once you depart the game so ensure you
have money left over before you quit). Winning wagers on the gambler pays out at one to one. Winning bets for tie normally pays 8 to 1 but sometimes 9 to 1. (This is a bad bet as ties occur less than
one in every 10 hands. Be cautious of gambling on a tie. However odds are astonishingly greater for 9:1 vs. 8 to 1)
Played properly baccarat offers generally decent odds, apart from the tie wager of course.
Punto Banco Scheme
As with all games baccarat chemin de fer has a handful of accepted myths. One of which is the same as a misconception in roulette. The past isn’t an indicator of events yet to happen. Keeping track
of past results at a table is a bad use of paper and an affront to the tree that surrendered its life for our stationary desires.
The most accepted and definitely the most favorable strategy is the one, three, two, six technique. This technique is employed to build up winnings and minimizing risk.
Begin by placing 1 dollar. If you succeed, add another to the 2 on the game table for a total of three chips on the second bet. Should you win you will hold six on the table, remove four so you keep
2 on the third bet. If you win the 3rd bet, put down 2 to the four on the table for a total of 6 on the fourth round.
If you lose on the first bet, you take a hit of one. A profit on the first bet followed by a hit on the 2nd causes a hit of 2. Success on the initial two with a hit on the third gives you with a gain
of 2. And wins on the first 3 with a loss on the 4th means you balance the books. Winning all 4 rounds gives you with twelve, a gain of 10. This means you can squander the 2nd round five times for
each favorable run of four rounds and still balance the books.
You must be logged in to post a comment. | {"url":"http://beamultimillionaire.com/2021/12/15/punto-banco-regulations-and-plan-2/","timestamp":"2024-11-02T04:59:19Z","content_type":"application/xhtml+xml","content_length":"27214","record_id":"<urn:uuid:8cf00efd-7d6c-43da-bea0-c87af329047b>","cc-path":"CC-MAIN-2024-46/segments/1730477027677.11/warc/CC-MAIN-20241102040949-20241102070949-00806.warc.gz"} |
[OS X TeX] Redefining a math symbol
Vamos, Peter P.Vamos at exeter.ac.uk
Fri Apr 19 21:05:43 CEST 2013
On 18 Apr 2013, at 20:31, Murray Eisenberg <murrayeisenberg at gmail.com> wrote:
> On 18 Apr 2013 12:34:53 +0000, Vamos, Peter" <P.Vamos at exeter.ac.uk> wrote:
>> On 18 Apr 2013, at 12:46, J. McKenzie Alexander <jalex at lse.ac.uk> wrote:
>>> presumably there's a way to change the definition of the '|' symbol in math mode so that TeX treats it as a binary operator by default
>> There is: \mid
> Actually, the spacing around \mid is a just a bit greater in $P(A \mid B)$ than it is in $P( \mathbin{|} B)$. You may need to typeset in 12 pt and then zoom in to notice it, though.
That is because \mid is a math symbol of class binary relation and \mathbin{|} is that of a binary operation. There are subtle differences as you yourself noticed. See also my next post.
More information about the macostex-archives mailing list | {"url":"https://tug.org/pipermail/macostex-archives/2013-April/050746.html","timestamp":"2024-11-12T13:44:50Z","content_type":"text/html","content_length":"3954","record_id":"<urn:uuid:84b3e04d-1bf0-4453-aa7a-ffb37ad8355e>","cc-path":"CC-MAIN-2024-46/segments/1730477028273.45/warc/CC-MAIN-20241112113320-20241112143320-00189.warc.gz"} |
Introduction to local extrema of functions of two variables
Review of functions of one variable
You remember how to find local extrema (maxima or minima) of a single variable function $f(x)$. Let's assume $f(x)$ is differentiable. Then the first step is to find the critical points $x=a$, where
$f\,'(a)=0$. Just because $f\,'(a)=0$, it does not mean that $f(x)$ has a local maximum or minimum at $x=a$. But, at all extrema, the derivative will be zero, so we know that the extrema must occur
at critical points.
For example, in the graph below, $f(x)$ is plotted by a green line. The three critical points are marked by colored circles. The red circle marks a local maximum and the blue circle marks a local
minimum. The yellow circle marks a critical point that is neither a maximum or a minimum. Even though $f\,'(x)=0$ at the yellow circle, the yellow circle does not mark a local extremum.
At each of these critical points, the linear approximation (i.e., tangent line) to $f(x)$ is a horizontal line since $f'(a)=0$. We can determine if $f$ has a local extremum at $x=a$ by looking at the
secord-order Taylor polynomial, which for a function of one variable is \begin{align*} f(x) \approx f(a) + \frac{1}{2} f\,''(a)(x-a)^2, \end{align*} since $f\,'(a)=0$. As long as $f\,''(a) \ne 0$,
the Taylor polynomial says that $f(x)$ looks like the top or bottom of a parabola for $x$ near $a$. If $f\,''(a)>0$, then $f(x)$ is approximately a parabola pointing upward and $f$ has a local
minimum at $x=a$, as illustrated by the blue circle, above. If $f\,''(a) < 0$, then $f(x)$ is approximately a parabola pointing downward and $f$ has a local maximum at $x=a$, as illustrated by the
red circle, above. On the other hand, if $f\,''(a) = 0$, then the second-order Taylor polynomial doesn't gives us any more information. At the point $x=a$, $f$ could have a local maximum, or it could
have a local minimum, or it might not even have a local extremum, as illustrated by the yellow point, above.
Functions of multiple variables
If $f(\vc{x})$ is a function of multiple variables, categorizing local extrema proceeds in an analogous way. So that we can visualize $f(\vc{x})$, we look only at the case of two variables, $\vc{x}=
(x,y)$, where we can graph $f(x,y)$ as a surface. Assuming $f(x,y)$ is differentiable, local extrema can occur only at critical points $(x,y) = (a,b)$, where the derivative of $f(x,y)$ is zero, i.e.,
those points $(a,b)$ where $Df(a,b) = [0 \ 0]$.
If $Df(a,b) = [0 \ 0]$, then the linear approximation (i.e, tangent plane) of $f(x,y)$ at $(a,b)$ is a horizontal plane. As in the one-variable case, we can determine if $f$ has a local extremum at $
(a,b)$ by looking at the secord-order Taylor polynomial. If we let $(a,b)=\vc{a}$ (remember that $(x,y)=\vc{x}$), then the second-order Taylor polynomial is \begin{align*} f(\vc{x}) \approx f(\vc{a})
+ \frac{1}{2} (\vc{x}-\vc{a})^T Hf(\vc{a}) (\vc{x}-\vc{a}). \end{align*} All this equation says is that, around $\vc{x}=\vc{a}$, the graph of $z=f(x,y)$ looks like a quadric surface (unless $Hf(a,b)$
is zero). In fact, $f(x,y)$ will look like a paraboloid.
Depending on the second derivative matrix $Hf(a,b)$, the graph of $f(x,y)$ might look like an elliptic paraboloid pointing upward, centered at the point $(a,b)$ (shown by the blue dot, below). In
this case, we say that $Hf(a,b)$ is positive definite, and $f$ has a local minimum at $(a,b)$.
A local minimum of a function of two variables. The blue point is a local minimum of a function of two variables.
Alternatively, the graph of $f(x,y)$ might look like an elliptic paraboloid pointing downward, centered at the point $(a,b)$ (shown by the red dot, below). In this case, we say that $Hf(a,b)$ is
negative definite, and $f$ has a local maximum at $(a,b)$.
A local maximum of a function of two variables. The red point is a local maximum of a function of two variables.
There is a third possibility that couldn't happen in the one-variable case. The graph of $f(x,y)$ might look like a hyperbolic paraboloid centered at the point $(a,b)$ (shown by the green dot,
below). In this case, the graph looks like a local maximum if you move in one direction (the direction where one's legs would go if one sat on the saddle) and the graph looks like a local minimum if
you move in another direction (the direction corresponding to the front and back if one sat on the saddle). In this case, we say that $Hf(a,b)$ is indefinite, and $f$ has neither a local maximum nor
a local minimum at the critical point. Such a critical point is called a saddle point.
A saddle point of a function of two variables. The green point is a saddle point of a function of two variables.
There are other cases, which correspond to the yellow point in the one-variable case, above. These are cases where one cannot tell from the second-order Taylor polynomial if $f$ has a local maximum,
a local minimum, or neither at the critical point. One would have to look at higher-order terms of the Taylor polynomial to determine the local behavior of the function.
Here are some examples (which assume knowledge of determining the definiteness of $Hf(a,b)$ that is not discussed in this page.) | {"url":"https://mathinsight.org/local_extrema_introduction_two_variables","timestamp":"2024-11-10T08:16:32Z","content_type":"text/html","content_length":"37354","record_id":"<urn:uuid:541ac946-234b-47e8-8c30-6feca0528a79>","cc-path":"CC-MAIN-2024-46/segments/1730477028179.55/warc/CC-MAIN-20241110072033-20241110102033-00783.warc.gz"} |
This page describes some technical details of our Peptide Calculator.
Calculating the molecular weight
The molecular weight M of a peptide may be estimated by calculating
where N[i] are the number, and M[i] the average residue molecular weights, of the amino acids. M[N] + M[C] are added to the total in order to account for the termini: H at the N-terminus and OH at
the C-terminus. Of course, if the termini are modified, these additions are replaced by those of the modifiers.
The resulting molecular weight depends on what values the algorithm uses. Some of the values Innovagen's Peptide Calculator uses are:
A, Ala 71.07793 C, Cys 103.1454 D, Asp 115.0873 E, Glu 129.1139
F, Phe 147.1734 G, Gly 57.05138 H, His 137.1394 I, Ile 113.1576
K, Lys 128.1724 L, Leu 113.1576 M, Met 131.1985 N, Asn 114.1028
P, Pro 97.11508 Q, Gln 128.1293 R, Arg 156.1861 S, Ser 87.07733
T, Thr 101.1039 V, Val 99.13103 W, Trp 186.2095 Y, Tyr 163.1728
H 1.00797 OH 17.00738 phos-Ser 167.0573
Acetyl 43.04453 Amide 16.0228 phos-Thr 181.0838
Biotin 227.3056 phos-Tyr 243.1528
Calculating the extinction coefficient
The molar extinction coefficient e at 280 nm of a peptide can be estimated by calculating
where n[W], n[Y], and n[C] are the number of Tryptophans (W), Tyrosines (Y) and Cystines (i.e. disulphide bonds, but here denoted C) in the sequence. The molar extinction coefficients used in
Innovagen's Peptide Calculator are:
e[W] = 5690 M^-1cm^-1 e[Y] = 1280 M^-1cm^-1 e[C] = 120 M^-1cm^-1
Gill and von Hippel
Note: Your peptide may contain non-natural amino acids, modifications and labels. None of these are considered, although they may very well add to the extinction coefficient.
Calculating the net charge
The net charge Z of a peptide at a certain pH can be estimated by calculating
where N[i] are the number, and pKa[i] the pKa values, of the N-terminus and the side chains of Arginine, Lysine, and Histidine. The j-index pertain to the C-terminus and the Aspartic Acid, Glutamic
Acid, Cysteine, Tyrosine amino acids.
Innovagen's Peptide Property Calculator calculates the net charge for all pH values of 0.1 to 14 in increments of 0.1, and plots these producing a titration curve.
This algorithm has its limitations, some of which are:
• The residues are assumed to be independent of each other.
• Only free terminii and the 20 naturally occuring amino acids and their D-forms are considered, all others are ignored.
• The resulting net charge depends on what pKa values the algorithm uses. Innovagen's Peptide Property Calculator uses the values taken from the CRC Handbook of Chemistry and Physics, 87th edition. | {"url":"http://pepcalc.com/notes.php?all","timestamp":"2024-11-14T11:06:33Z","content_type":"text/html","content_length":"10546","record_id":"<urn:uuid:9d124f7e-b0e0-4149-83bc-7b5e792cbe9c>","cc-path":"CC-MAIN-2024-46/segments/1730477028558.0/warc/CC-MAIN-20241114094851-20241114124851-00472.warc.gz"} |
Orbits in Hamiltonian Systems Based on Multiple Shooting Algorithms
A Program for Computing Periodic Orbits in Hamiltonian Systems Based on Multiple Shooting Algorithms
Stavros C. Farantos
Institute of Electronic Structure and Laser,
Foundation for Research and Technology - Hellas,
Department of Chemistry
University of Crete,
Iraklion, Crete 711 10, Greece.
Published in "Computer Physics Communications Vol. 108, p. 240, 1998."
POMULT is a FORTRAN code for locating Periodic Orbits and Equilibrium Points in Hamiltonian systems based on 2-point boundary value solvers which use multiple shooting algorithms. The code has mainly
been developed for locating periodic orbits in molecular Hamiltonian systems with many degrees of freedom and it utilizes a damped Newton-Raphson method and a secant method. Graphical User Interface
has also been written in tcl-tk script language for interactively manipulating the input and output data. POMULT provides routines for a general analysis of a dynamical system such as fast Fourier
transform of the trajectories, Poincare surfaces of sections, maximum Lyapunov exponents and evaluation of the classical autocorrelation functions and power spectra.
Keywords: molecular dynamics and spectra, periodic orbits, multiple shooting algorithm, damped Newton-Raphson method
Title of the program: POMULT (Periodic Orbit MULTishooting)
Catalogue number: POMT
Program obtainable form: CPC Program Library, Queen's University of Belfast, N. Ireland
Licensing provisions: Numerical Recipes
Computer: Tested on workstations HP-9000/735, IBM-7030/3CT, PC-Linux
Installation: IESL-FORTH, Iraklion, Crete, Greece
Operating system: UNIX
Programming language used: FORTRAN 77 with extensions, lower case, implicit, include
Memory required to execute with typical data: 5 Mbytes
No. of bits in a word: 32
No. of bytes in distributed program, included test data etc: 2211840
Distribution format: gzip compressed tar file
Keywords: molecular dynamics and spectra, periodic orbits, multiple shooting algorithm, damped Newton-Raphson method
Nature of the physical problem:
Given a multidimensional highly coupled molecular potential energy surface we want to compute families of periodic solutions of Hamilton equations. These families of periodic orbits reveal the
structure of the classical phase space by detecting the regions of phase space with regular and chaotic motions. Furthermore, periodic orbits point out possible localization of the quantum
wavefunctions, and explain/predict spectroscopic features.
Method of solution:
The location of periodic orbits is based on damped Newton-Raphson methods or secant-Quasi Newton methods. Simple or Multiple shooting algorithms are employed which are robust in cases of long period
or highly unstable periodic orbits.
Restrictions on the complexity of the problem:
The program has been tested with 2-, 3-, 5-, and 6-dimensional molecular potential functions. Limitations are observed in cases of high instability or in regions of phase space densely occupied by
periodic orbits. The above difficulties cause also limitations in the continuation of a family of periodic orbits with a parameter.
Typical running time:
This depends on the complexity of the potential function, the period and the number of periodic orbits which are computed, and whether the equations of motion are stiff or not.
Standard numerical actions like integration of ordinary differential equations and solution of linear algebraic equations are carried out with routines from the package ``Numerical Recipes''. The
program can be interfaced with ODESSA or other available programs which carry out sensitivity analysis of differential or algebraic equations. Generally, the program has been written in such a way
that the user can incorporate his/her own favorable subroutines. A Makefile, a README file as well as a help file are provided for the installation of the program and the explanation of the input
data. Graphical User Interface for the input data has been written in a tcl-tk script language. The user should ensure that the libraries versions tcl7.0 and tk4.0 or higher are installed in her/his
Tue Dec 16 20:32:24 EET 1997 | {"url":"https://tccc.iesl.forth.gr/~farantos/po_cpc/po_cpc.html","timestamp":"2024-11-06T13:38:52Z","content_type":"text/html","content_length":"6105","record_id":"<urn:uuid:f94b6b87-6bcd-458e-9718-d2473304989f>","cc-path":"CC-MAIN-2024-46/segments/1730477027932.70/warc/CC-MAIN-20241106132104-20241106162104-00792.warc.gz"} |
Mixed number to percent calculator
mixed number to percent calculator Related topics: rational expressions answers
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online algebrator
Author Message
Sovcin Posted: Sunday 06th of May 16:18
Hello Friends , I am urgently in need of guidance for getting through my mathematics exam that is nearing. I really do not intend to resort to the service of private teachers
and online coaching since they prove to be quite costly. Could you suggest a perfect tutoring utility that can guide me with learning the principles of Pre Algebra. In
particular , I need assistance on function domain and equivalent fractions.
Registered: 08.04.2002
Back to top
oc_rana Posted: Tuesday 08th of May 11:29
Hi! I guess I can help you out on how to solve your homework. But for that I need more details. Can you give details about what exactly is the mixed number to percent
calculator homework that you have to submit. I am quite good at solving these kind of things. Plus I have this great software Algebrator that I downloaded from the internet
which is soooo good at solving algebra homework. Give me the details and perhaps we can work something out...
Registered: 08.03.2007
From: egypt,alexandria
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3Di Posted: Thursday 10th of May 08:38
Hello there. Algebrator is really fantastic! It’s been months since I tried this software and it worked like magic! Algebra problems that I used to spend answering for hours
just take me 4-5 minutes to solve now. Just enter the problem in the software and it will take care of the solving and the best part is that it displays the whole solution so
you don’t have to figure out how did the software come to that answer.
Registered: 04.04.2005
From: 45°26' N, 09°10'
Back to top
Jaffirj Posted: Friday 11th of May 17:57
That sounds amazing ! Thanks for the help ! It seems to be just what I need , I will try it for sure! Where did you come across Algebrator? Any suggestion where could I find
more info about it? Thanks!
Registered: 18.10.2003
From: Yaar !!
Back to top
Flash Fnavfy Liom Posted: Sunday 13th of May 16:08
I am sorry; I forgot to give the link in the previous post. You can find the program here https://softmath.com/algebra-policy.html.
Registered: 15.12.2001
Back to top
CHS` Posted: Monday 14th of May 08:59
A great piece of algebra software is Algebrator. Even I faced similar difficulties while solving solving a triangle, long division and hypotenuse-leg similarity. Just by
typing in the problem workbookand clicking on Solve – and step by step solution to my math homework would be ready. I have used it through several math classes - College
Algebra, Algebra 1 and Basic Math. I highly recommend the program.
Registered: 04.07.2001
From: Victoria City,
Hong Kong Island, Hong
Back to top | {"url":"https://softmath.com/algebra-software/multiplying-fractions/mixed-number-to-percent.html","timestamp":"2024-11-01T22:13:55Z","content_type":"text/html","content_length":"43518","record_id":"<urn:uuid:cccc59fa-b4b7-4637-b072-18b11a9912b8>","cc-path":"CC-MAIN-2024-46/segments/1730477027599.25/warc/CC-MAIN-20241101215119-20241102005119-00682.warc.gz"} |
Winning Percentage Calculator
Last updated:
Winning Percentage Calculator
If you want to know how good the last season really was for your favorite baseball or hockey team, you should definitely give this winning percentage calculator a try. All you have to do is input the
number of wins, losses, and ties on the team's record, and you'll have an answer in a split second!
💡 Click on the Tied games section to include tied games and to assign their value.
Win percentage formula
Calculating the winning percentage is equivalent to estimating a proportion of wins in the total number of games. You can check Omni's proportion calculator to learn more about proportions. If there
are no tie results, you need to divide the number of wins by the total number of games (wins and losses):
winning percentage = wins / games
For example, let's assume that your favorite basketball team has played 82 games and won 48 of them. Their winning percentage is:
(48 / 82) × 100 = 58.54%
How to calculate winning percentage with ties
If, on the other hand, you want to include ties into the whole calculation, the formula gets a bit more complicated. It is usually assumed that a tie is worth the same as 1/2 of a win. In such a
case, you can use our percentage calculator or evaluate the percentage by hand in the following way:
winning percentage = (wins + 0.5 × ties) / games
For this equation, the number of games is the sum of win, loss, and tie results on the team's record.
To get a better understanding of this formula, let's consider the following example: a football team playing in the National Football League has played 16 games in total. They lost 4 of them and got
a tie result in 5. What is their winning percentage?
1. Determine the number of wins. If the total number of games is 16, then you can use the formula below:
wins = games - ties - losses = 16 - 5 - 4 = 7
2. Now, you know that the team has won 7 games during the last season. Add that number to half of the tie results:
wins + 0.5 × ties = 7 + 0.5 × 5 = 9.5
3. Divide the number you got by the total number of games:
(9.5 / 16) × 100 = 59.38%
4. 59.38% is the winning percentage of the football team you cheer for. Quite good, but still not enough to win the League!
🙋 We use percentages in almost all aspects of our life, not just sports. For example, we can also use percent to express the relative error between the observed and true values in any measurement. To
learn how to do that, check our percent error calculator.
Should you bet on this team?
Even if your favorite team has a stellar track record and a winning percentage oscillating around 80%, it doesn't necessarily mean they will win the next match! Instead of calculating the win
percentage, you should use our odds calculator to determine your chances when betting on them.
How many games did a team win if they won 40 percent of the 15 games?
They won six games. To evaluate the result, you need to multiply the winning percentage by the number of games: 40% × 15 = 6.
How do I calculate the percentage of games won?
To estimate the percentage of games won:
1. Get the number of games won.
2. Get the total number of games.
3. Divide the first value by the second one.
4. Multiply the quotient by 100.
5. The result is the percentage of games won.
How do I calculate the win/loss ratio?
To calculate the win/loss ratio:
1. Get the number of won games.
2. Get the number of lost games.
3. Divide the first value by the second one. We assume there is at least one game lost.
4. Multiply the quotient by 100.
5. The result is your win/loss ratio.
What are the percentage odds of winning the lottery?
There is a 1 in 13,983,816 (or about 0.00000715%) chance for a win, assuming you need to pick 6 numbers out of the set of 49 possibilities. However if 5 of 49 satisfies the winning condition; the
odds change to 1 in 1,906,884 (or about 0.0000524%).
To estimate such probabilities, divide one over the combination of the corresponding values, e.g., 1/C(49,6) in the first case. | {"url":"https://www.omnicalculator.com/sports/winning-percentage","timestamp":"2024-11-04T04:12:01Z","content_type":"text/html","content_length":"471882","record_id":"<urn:uuid:be064dfa-f4c0-416c-9cbe-bda11b981b79>","cc-path":"CC-MAIN-2024-46/segments/1730477027812.67/warc/CC-MAIN-20241104034319-20241104064319-00881.warc.gz"} |
Swiss Olympiad in Informatics
Written by Timon Gehr.
Given a graph $G=(V,E)$ with a given set of vertices $V$ and a set of edges $E$, a matching is a subset $M\subseteq E$ such that the edges in $M$ are pairwise disjoint (i.e., they do not have a
common incident vertex).
For example, in the following graph, the highlighted edges form a matching:
A matching $M$ covers a vertex $v$ if there is some other vertex $u$ such that the edge $\{u,v\}$ is in $M$. Each vertex is covered by zero or one edges in $M$. In the example, all vertices that are
covered by an edge are colored in grey.
A matching $M$ intersects an edge $e$ if it covers one of the two incident vertices of $e$. In the example, the matching intersects all edges except $\{6,9\}$, $\{8,9\}$ and $\{9,10\}$.
If a matching $M$ covers each vertex, then we call $M$ a perfect matching. Not every graph has a perfect matching (for instance, the example graph cannot have a perfect matching as it has an odd
number of vertices), but we can try to find a maximal matching.
A matching is called maximal with respect to inclusion if there is no matching $M'eq M$ with $M\subset M'\subseteq E$. In other words, it is impossible to add more edges to $M$ and still have a
It is easy to find a matching that is maximal in this sense: We iterate over the edges once and add to our matching all edges that are not already incident to a covered vertex. This works because
every matching can be extended to a maximal matching. For example, we can add the edge $\{8,9\}$ to our matching to obtain the following maximal matching:
Matchings $M$ that are maximal in this sense already enjoy a few useful properties. For example, the matching must intersect all edges. This means that each vertex $v$ is either covered by the
matching $M$, or all its incident vertices are covered by the matching $M$. (In the example: each vertex is grey or all its neighbours are grey.)
A matching is called maximum (with respect to cardinality) if there is no matching $M'\subseteq E$ with $\lvert M\rvert < \lvert M'\rvert$. In other words, there is no other matching with more edges.
For example, the following matching has cardinality $6$, which is the largest possible in our graph, as any connected component can contain at most half as many matching edges as it contains
A maximum matching is a bit more tricky to find than one that is merely maximal, but already now it is easy, for any graph $G$, to give lower and upper bounds on the size of a maximum matching which
differ by at most a factor of two. On one hand, a maximum matching $M$ is at least as large as any inclusion-maximal matching $M'$. On the other hand, $M$ cannot contain more than $2\cdot\lvert M'\
rvert$ edges. This is because $M'$ intersects all edges of $M$ and each edge $e'$ in $M'$ intersects at most two edges of $M$. (More formally, we have $M=\bigcup_{e'\in M'}\{e \in M\mid e\cap e'eq \
emptyset\}$, therefore $\lvert M\rvert\le\sum_{e'\in M'}\lvert\{e\in M\mid e\cap e'eq \emptyset\}\rvert\le\sum_{e'\in M'}2=2\lvert M'\rvert$.)
But how do we actually find a maximum matching? Our idea will be to start with some matching $M$ (it could even be empty), and to progressively improve it until it has maximal cardinality. Note that
while following this process, we may need to remove some edges from our matching that we already added to it in a previous step, because otherwise we can only hope to find an inclusion-maximal
matching. We ensure progress by designing a single iteration of our algorithm such that it improves the cardinality of the matching by at least one.
Let’s now assume that $M$ is some matching which is not maximum. Consider some maximum matching $M'$. (We do not know what $M'$ is and will therefore not be able to use it in our algorithm, but we
know that it exists.) Let’s color the edges from $M$ blue and those from $M'$ red.
For example, we could have the following situation, where $M'$ is the maximum matching from above:
Consider the symmetric difference $M\triangle M'$ of $M$ and $M'$. (The symmetric difference contains all edges that belong to either $M$ or $M'$, but not to both.) Note that this symmetric
difference contains more red edges than blue edges, because $M$’ is larger than $M$. Now consider the graph $G'=(V,M\triangle M')$.
For our example, this graph is as follows:
Each vertex in $G'$ has degree at most $2$, because it can be incident to at most one blue and one red edge. Each connected component of a graph with maximum degree $2$ is either a path or a cycle.
Even more is true: paths and cycles need to alternate between red and blue edges and therefore each cycle in $G'$ must have even length. In our example, there is one such alternating cycle: $
(0,1,2,3,0)$. It follows that each cycle in $G'$ contains equally many red and blue edges.
Therefore, there must exist a connected component in $G'$ which is a path that contains more red edges than blue edges, because overall, $G'$ contains more red edges than blue edges. Such a path must
start with a red edge, then alternate between blue and red edges, until it stops with a red edge.
In our example, there is one such alternating path: $(7,8,9,10)$.
Now note that we can improve the size of our matching $M$ by removing all blue edges from our path from it and adding all red edges from our path to it. In our example, this leads to the following
improved matching $M$:
Inspired by those observations, we say that for a matching $M$, a path in $G$ is $M$-augmenting if it starts and ends with vertices that are not covered by $M$ and alternates between edges in $M$ and
edges that do not belong to $M$. If there exists an $M$-augmenting path, we can improve our matching by switching all edges on the path into or out of the matching $M$. On the other hand, we have
just shown that whenever $M$ is not maximum, there is an $M$-augmenting path.
Therefore, our algorithm will proceed as follows: While there exists an $M$-augmenting path, use it to improve $M$. This algorithm terminates because $M$ is improved in each iteration and the maximal
cardinality of a matching in $G$ is bounded. When the algorithm terminates, this means that there does not exist an $M$-augmenting path in $G$, which means that the returned matching $M$ is maximum.
It remains to find an algorithm that finds an $M$-augmenting path or decides that it does not exist. We will only consider the case where our graph is bipartite. (In a general graph, we can still
compute maximal matchings by repeatedly finding augmenting paths, but it is a bit more complicated and outside the scope of IOI.)
In this case, finding an augmenting path is in fact easy: we can use DFS or BFS, slightly adapted such that they alternate between edges outside and within the matching. We start the graph search
from an uncovered vertex and try to find another uncovered vertex reachable on an alternating path. A path found this way is an augmenting path, and if we can’t find any such path, there is in fact
no augmenting path.
Because the resulting matching has at most $\lvert V\rvert/2$ edges, we need to find at most $O(\lvert V\rvert)$ augmenting paths. The total running time of the algorithm is therefore $O(\lvert V\
rvert\cdot(\lvert V\rvert+\lvert E\rvert))$.
We provide an implementation using DFS which combines finding augmenting paths and improving the matching in the same recursion. The first partition contains $a$ vertices and the second partition
contains $b$ vertices. The graph is stored as an adjacency list, where the nodes in the first partition are numbered from $0$ to $a-1$ and the nodes in the second partition are numbered from $a$ to
$a+b-1$. The matching is stored as a vector of size a. The $i$-th entry of the vector contains the number of the partner of node $i$ in the matching, or $-1$ if the matching does not cover the
The DFS is modified as follows: In the first partition, we may only use edges that are in the matching, and in the second partition, we may only use edges that are not in the matching. This can be
interpreted as implicitly directing the graph, such that edges in the matching go from the first partition to the second partition, and edges outside the matching go from the second partition to the
first partition. In this graph, any path from the second partition to the first partition between two non-covered vertices alternates between non-matching edges and matching edges and therefore is an
augmenting path. This is also an easy way to see why DFS can indeed be used to decide whether there is an augmenting path.
int a, b;
vector<vector<int>> g; // bipartite graph on a+b nodes as adjacency list.
vector<int> matching; // the current matching, mapping nodes 0 to a-1 to their partner or -1
vector<bool> visited;
bool improveMatching(int i) {
if (visited[i]) return false;
visited[i] = true;
if (i < a) { // in first partition, must use matching edge
if (matching[i] == -1) return true; // found augmenting path
if (improveMatching(matching[i]))
return true;
} else { // in second partition, may only use edges not in the matching
for (int j : g[i]){
if (matching[j] == i) continue; // matching edge
if (improveMatching(j)){
matching[j] = i; // update matching
return true; // matching improved
return false;
vector<bool> covered;
bool improveMatching() {
covered.assign(b, false);
for (int i = 0; i < a; i++)
if (matching[i] != -1)
covered[matching[i] - a] = true;
visited.assign(a + b, false);
for (int i = a; i < a + b; i++)
if (!covered[i - a] && improveMatching(i))
return true;
return false; // cannot improve matching, it is maximum
int maximumMatching() { // returns size of maximum matching
matching.assign(a, -1);
while (improveMatching()) {}
return a - (int)count(matching.begin(), matching.end(), -1);
(Note that this algorithm would still compute the correct result even if, for the second partition, we didn’t track visited flags nor checked for matching edges.)
A vertex cover of a graph $G=(V,E)$ is a subset $C\subseteq V$ of the vertices such that each edge in $E$ is incident to at least one vertex in $C$.
For example, the highlighted vertices in the following graph form a vertex cover; every edge connects two vertices such that at least one of them is highlighted in blue.
Because each edge in a matching needs to contain at least one vertex of a cover $C$, each vertex cover is at least as large as each matching.
In particular, a vertex cover is at least as large as a maximum matching. For example, our vertex cover has cardinality $12$, while a maximum matching has cardinality $8$. Each red matching edge is
incident to at least one blue cover vertex.
A minimum vertex cover is a vertex cover of minimal cardinality. As it is a vertex cover, even a minimum vertex cover is at least as large as a maximum matching.
Interestingly, even the opposite inequality holds: the size of a minimum vertex cover is exactly the size of a maximum matching. This means that we can select exactly one vertex from each matching
edge and still obtain a vertex cover. For example, there is the following minimum vertex cover of cardinality $8$:
For bipartite graphs, the fact that a minimum vertex cover has the same size as a maximum matching is known as Kőnig’s theorem. We can prove it by giving an algorithm which computes a minimum vertex
cover from a given maximum bipartite matching.
Let $G=(A\cup B,E)$ be a bipartite graph with first partition $A$ and second partition $B$. Let $M$ be a maximum matching in $G$. Denote by $B'\subseteq B$ the set of vertices in $B$ that are not
covered by $M$. Let $R$ be the set of vertices that can be reached from the vertices in $B'$ using a path that alternates between edges not in the matching and edges in the matching. (Note that this
is exactly the set of vertices that are visited by the last iteration of the improveMatching algorithm.)
Let $C=(A\cap R)\cup(B\setminus R)$. I.e., a vertex in $A$ belongs to $C$ iff it is reachable by an alternating path, and a vertex in $B$ belongs to $C$ iff it is not reachable.
Those definitions are illustrated in the following example, where vertices in $A$ are circles while vertices in $B$ are squares. We have $B'=\{6,14\}$ and all vertices in $R$ are colored grey.
The set $C$ is in fact a vertex cover. To see this, we show that for each edge $e$, either the incident vertex in $A$ is reachable using an alternating path from $B'$ or the incident vertex in $B$ is
not reachable by such an alternating path. We argue for edges inside and outside $M$ separately.
If $e$ is in $M$ and its incident vertex in $A$ is not reachable from $B'$ by an alternating path, then neither is the incident vertex in $B$, because such an alternating path would have to use $e$.
Note that the other direction is also true: if the vertex in $A$ is reachable, we can reach the one in $B$ by using the edge $e$ in the direction from $A$ to $B$. Therefore, $C$ contains exactly one
incident vertex of each edge in $M$.
If $e$ is not in $M$ and its incident vertex in $B$ is reachable from $B'$ by an alternating path, then so is the incident vertex in $A$ (because we can just extend the path by using the edge $e$ in
the direction from $B$ to $A$).
This shows that $C$ is a vertex cover. Note that every vertex of $C$ is covered by $M$: By construction $C\cap B=B\setminus R\subseteq B\setminus B'$ does not contain any vertices that are not
covered, because those vertices by definition belong to $B'$. On the other hand, if there was some vertex in $C\cap A=A\cap R$ that was not covered by $M$, then this vertex would be the end of an
augmenting path, which is impossible because $M$ is maximum.
Therefore, each vertex in $C$ is incident to some edge in $M$. On the other hand, we know that $C$ contains exactly one incident vertex of each edge in $M$, which implies that $\lvert C\rvert=\lvert
M\rvert$. Therefore, $C$ is a minimum vertex cover.
Assuming the above implementation of maximumMatching, we can compute a minimum vertex cover as follows:
vector<int> cover;
int minimumVertexCover() {
int size = maximumMatching();
for (int i = 0; i < a; i++)
if (visited[i]) cover.push_back(i);
for (int i = a; i < a + b; i++)
if (!visited[i]) cover.push_back(i);
assert((int)cover.size() == size);
return size; | {"url":"https://soi.ch/wiki/matchings/","timestamp":"2024-11-10T01:48:43Z","content_type":"text/html","content_length":"234371","record_id":"<urn:uuid:c553c71d-2adb-4786-a314-a159fb7bb256>","cc-path":"CC-MAIN-2024-46/segments/1730477028164.3/warc/CC-MAIN-20241110005602-20241110035602-00651.warc.gz"} |
stanford math phd gre
1. Welcome Stanford Mathematics Graduate Students
2. Welcome Stanford Mathematics Graduate Students
3. How to Ace Your GRE Math: A Step-by-Step Guide
4. GRE Math • How to score 170 with expert GRE Tips & Practice Questions
6. Stanford GRE Scores- All You Need To Know About It
1. Stanford Math Department Commencement Speech 2024
2. Math PhD: Best Calculus Book #mathematics #math #phd
3. Difficult Problem that can be solved in less than a minute
4. Stanford Math 51H Lecture 1.1: What is Multivariable Calculus?
5. Stanford Math 51H 2.1 Angles in R^n
6. Stanford Math 51H 1.2 Vectors | {"url":"https://info-producer.online/homework/stanford-math-phd-gre","timestamp":"2024-11-08T21:21:35Z","content_type":"text/html","content_length":"21091","record_id":"<urn:uuid:355afd0d-162a-4b5f-bd8b-9c3c7768397e>","cc-path":"CC-MAIN-2024-46/segments/1730477028079.98/warc/CC-MAIN-20241108200128-20241108230128-00016.warc.gz"} |
unstructureddataset - Script command
Creates an empty dataset that is associated with arbitrary x/y/z coordinate in space, and with additional matrix, a connectivity matrix to connect them. The connectivity matrix comes after x, y, and
z. Like rectilinear datasets, unstructured datasets can be parameterized, and can contain an arbitrary number of attributes (see addattribute) and parameters (see addparameter) .
See Dataset introduction for more information. For datasets that are not associated with the x/y/z coordinates (ex. transmission as a function of frequency), see matrixdataset .
Syntax Description
Creates an empty unstructured dataset associated with the coordinates x/y/z and a connectivity matrix to connect them.
unstructureddataset Arguments 'x', 'y' and 'z' must be the same length; equivalent to the total number of points.
The argument 'C' should be a matrix of integers where the number of rows equal to number of shapes in the mesh, the number of columns should be 2 (line segments), 3
(triangles) or 4 (tetrahedra), and values should be integers.
Below is a simple example of the usage of unstructured dataset. x, y and z vectors represent arbitrary points in space and C represent the connectivity matrix that connects them. The values for the
vectors can be loaded from the unstructured_charge_example.mat file. It is possible to further script this process and import the data to an object, eg, np density grid attribute, see the
importdataset command.
# constructing an unstructured dataset
matlabload("unstructured_charge_example.mat"); # taking the data from a CHARGE simulation. The data can be from a different source
x = charge.x;
y = charge.y;
z = charge.z;
C = charge.elements;
data = unstructureddataset("test",x,y,z,C);
V_cathode = charge.V_cathode;
V_anode = charge.V_anode;
n = pinch(charge.n);
p = pinch(charge.p);
This next example creates an unstructured dataset (with the name "Absorption") that contains 2 data attributes: the power absorption Pabs, and the refractive index n. Both attributes are a function
of the spatial parameters x/y/z and frequency f. Connectivity matrix cm has also been specified. To allow the user to access the frequency parameter in terms of frequency or wavelength , both
frequency (f) and wavelength (c/f) are added as interdependent parameters.
Absorption = unstructureddataset("Absorption",x,y,z,cm);
Absorption.addattribute("refractive index",n);
visualize(Absorption); # visualize this dataset in the Visualizer
This example shows how to define an equilaterial triangle in the plane z=0
x = [0;1;2];
y = [0;sqrt(3);0];
z = [0;0;0];
C = [1,3,2];
ds = unstructureddataset(x,y,z,C);
See Also
rectilineardataset , addattribute , addparameter , visualize , datasets , getparameter , getattribute , matrixdataset , struct | {"url":"https://optics.ansys.com/hc/en-us/articles/360034929933-unstructureddataset-Script-command","timestamp":"2024-11-04T11:22:46Z","content_type":"text/html","content_length":"35268","record_id":"<urn:uuid:7f4a8f14-c3e4-47e1-986d-98a75c5334b8>","cc-path":"CC-MAIN-2024-46/segments/1730477027821.39/warc/CC-MAIN-20241104100555-20241104130555-00279.warc.gz"} |
Two All India Institute of Medical Sciences (AIIMS) Questions on Optics
The following questions which appeared in All India Institute of Medical Sciences (AIIMS) 2005 entrance question paper for admitting students to the MBBS Degree course are simple as usual. They are
meant for checking your knowledge and understanding of fundamentals.
(1) The apparent depth of water in a cylindrical water tank of diameter 2R cm is reducing at the rate of x cm/minute when water is being drained out at a constant rate. The amount of water drained in
c.c. per minute is (n[1]= refractive index of air, n[2 ]= refractive index of water)
(a) x π R^2 n[1]/n[2] (b) x π R^2 n[2]/n[1] (c) 2 π R^ n[1]/n[2] (d) π R^2x
Since the refractive index is the ratio of real depth to the apparent depth, we have
Real depth = Apparent depth × refractive index.
Therefore, the rate at which the real depth is decreasing = xn[2]/n[1] cm per minute.
The amount of water drained in c.c. per minute is therefore equal to x π R^2 n[2]/n[1],[ ]given in option (b).
(2) A telescope has an objective lens of focal length 200 cm and an eye piece with focal length 2 cm. If the telescope is used to see a 50 m tall building at a distance of 2 km, what is the length of
the image of the building formed by the objective lens?
(a) 5 cm (b) 10 cm (c) 1 cm (d) 2cm
At the first glance this question may seem to be one involving the magnification produced by a telescope; but, this is quite simple since you are asked to consider the objective only.
The objective will produce the image of the building at the focus (which is at 2 m from the lens) and hence from the expression for magnification (M) we have
M = Distance of image/ Distance of object = Height of image/ Height of object
so that 2/ 2000 = x/50 where ‘x’ is the height of image in metre.
Therefore, x = 2×50/2000 = 0.05 m = 5 cm. | {"url":"http://www.physicsplus.in/2007/11/two-all-india-institute-of-medical.html","timestamp":"2024-11-13T14:30:15Z","content_type":"application/xhtml+xml","content_length":"95767","record_id":"<urn:uuid:5ebfa4a4-3422-40f4-9e18-2ea74be11bbd>","cc-path":"CC-MAIN-2024-46/segments/1730477028369.36/warc/CC-MAIN-20241113135544-20241113165544-00817.warc.gz"} |
• define and recognize surds as irrational numbers;
• rationalize;
• perform basic operations on surds
• solve problems
Surds are numbers written in the root form i.e. $\sqrt{x}$ to express its exact value. They are irrational numbers and are mostly expressed as root of rational numbers. The intent of solving surds is
to make irrational numbers rational.
Rational and Irrational Numbers
Rational Numbers
In mathematics, we deal more with rational numbers. Rational numbers are all numbers that can be expressed in the form $\frac{a}{b}$, where a and b are integers like 3 and 6 and $beq&space;0$.
So $\dpi{150}&space;0.5{\color{Green}&space;}$ is rational since it can be in the form $\dpi{150}&space;\frac{1}{2{\color{DarkGreen}&space;}}$
The value $\dpi{150}&space;5{\color{Green}&space;}$ is rational, can be written in the form $\dpi{150}&space;\frac{5}{1{\color{Green}&space;}}$
Rational numbers also includes recurring or repeated decimals. Example $\dpi{150}&space;\frac{13}{3}=4.33333.....{\color{Green}&space;}$ The dot shows that the digit 3 repeats indefinitely. Also $\
dpi{150}&space;\frac{7}{4}=1.75{\color{Green}&space;}$ is rational.
Irrational Numbers
Irrational simply means “not rational”. They are numbers that are not rational. And so, cannot be written as a fraction. Irrational numbers when simplified, goes on indefinitely. In other words, we
can say, they continue indefinitely without recurring or repeating. Examples are roots like
$\dpi{150}&space;\sqrt{3}=1.7320508075688{\color{Green}&space;}$ , $\dpi{150}&space;\sqrt{2}=1.4142135623730{\color{Green}&space;}$ and numbers like $\dpi{150}&space;\sqrt{27{\color{Green}&space;}}$
which is irrational, but can be made rational using surds. We should note that not all roots are irrational, roots like $\dpi{150}&space;\sqrt{4}=2$ and $\dpi{150}&space;\sqrt{9}=3$ etc. are all
rational numbers.
Recall we said earlier on that the main intent of solving surds is to make irrational numbers rational. Having learnt the differences between this two types of numbers, let’s take a further step in
finding out the rules that should guide us as we solve surds.
Surds Rules
Rule 1: $\dpi{150}&space;\sqrt{x\times&space;y}=\sqrt{x}&space;\times&space;\sqrt{y}$ .
Example: Simplify $\dpi{150}&space;\sqrt{54}$ . Note that $\dpi{150}&space;\sqrt{54{\color{Green}&space;}}$ is an irrational number. So we make it rational.
So $\dpi{150}&space;\sqrt{9}=3$, so the solution becomes $\dpi{150}&space;\sqrt[3]{6}$
Rule 2: $\dpi{150}&space;\sqrt{\frac{a}{b}}=\frac{\sqrt{a}}{\sqrt{b}}$
Note: $\dpi{150}&space;\sqrt{a}-\sqrt{b}eq&space;\sqrt{a-b}$ and $\dpi{150}&space;\sqrt{a}&space;+\sqrt{b}eq&space;\sqrt{a+b}$ . The example below will show how to solve a question using this rule.
Example: Simplify $\dpi{150}&space;\frac{\sqrt{25}}{\sqrt{16&space;}}$
Adding and Subtracting Similar Surds
Surds in the same basic form can be added and subtracted. The addition and subtraction can only be possible if they have same roots. So mixed surds such as $\dpi{150}&space;\sqrt[3]{2}+\sqrt[2]{3}$
and $\dpi{150}&space;\sqrt[5]{3}-\sqrt[4]{5}$ cannot be simplified further. See the example below.
Example: Simplify $\dpi{150}&space;\sqrt{28}+\sqrt{63}$
To solve the above question, we have to reduce all the surds to their basic forms, i.e. $\dpi{150}&space;\sqrt{28}=\sqrt{4\times&space;7}$ and $\dpi{150}&space;\sqrt{63}=\sqrt{9\times&space;7}$
So $\dpi{150}&space;\sqrt{28}+\sqrt{63}=\sqrt{4\times&space;7}+\sqrt{9\times&space;7}$
This will give $\dpi{150}&space;\sqrt[2]{7}+\sqrt[3]{7}$. Since they have same roots, they can be added, so $\dpi{150}&space;\sqrt[2]{7}+\sqrt[3]{7}=\sqrt[5]{7}$
Rationalising The Denominator of Surds
A surd such as $\dpi{150}&space;\frac{\sqrt{3}}{2}$, cannot be simplified, rather $\dpi{150}&space;\frac{2}{\sqrt{3}}$ can be written in a more convenient form. In the present form, it is irrational.
To make it rational, we have to multiply the numerator and denominator by $\dpi{150}&space;\sqrt{3}$.
This will give us $\dpi{150}&space;\frac{2}{\sqrt{3}}=\frac{2}{\sqrt{3}}\times&space;\frac{\sqrt{3}}{\sqrt{3}}=\frac{\sqrt[2]{3}}{3}$
Note: $\dpi{150}&space;\sqrt{3}\times&space;\sqrt{3}=3$
This removes the irrational number $\dpi{150}&space;\sqrt{3}$ from the denominator. This process is called rationalising the denominator or simply rationalization of surd.
Conjugate of Surds
In surds, the rationalisation of the denominator is meant to make the denominator rational. This is done by multiplying both the numerator and the denominator by the “CONJUGATE SURD“.
Two surds whose products results in a rational number are called conjugates. For example, if $\dpi{150}&space;(\sqrt{x}+\sqrt{y})$ is a surd, the conjugate will be $\dpi{150}&space;(\sqrt{x}-\sqrt
{y})$. This is so since the product $\dpi{150}&space;(\sqrt{x}+\sqrt{y})(\sqrt{x}-\sqrt{y})=x-y$.
For example, the conjugate of $\dpi{150}&space;(\sqrt[-3]{2}+\sqrt{7})$ is $\dpi{150}&space;(\sqrt[-3]{2}-\sqrt{7})$. This is because the product (multiplication) of both results to a rational number
which is $\dpi{150}&space;11$.
Expression such as $\dpi{150}&space;\frac{1}{a-\sqrt[b]{c}}$ , $\dpi{150}&space;\frac{1}{\sqrt[a]{b}+c}$ etc. can be solved using the conjugate of the surds to rationalising the denominator.
Example: Simplify $\dpi{150}&space;\frac{1}{(1-\sqrt{3})^{2}}$
To solve this question, we will first simplify the denominator, since it is squared.
$\dpi{150}&space;1^{2}-\sqrt{3}-\sqrt{3}+3=1-\sqrt[2]{3}+3=4-\sqrt[2]{3}$. With this the expression will become $\dpi{150}&space;\frac{1}{4-\sqrt[2]{3}}$
The conjugate of $\dpi{150}&space;4-\sqrt[2]{3}$ is $\dpi{150}&space;4+\sqrt[2]{3}$.
Multiply the numerator and the denominator by the conjugate.
Conclusively, from this lesson, we have been able to learn the following:
1. the definition and rules we must work with when solving questions on surds accurately;
2. how to add and subtract simple surds questions, which if applied can help you solve tougher questions;
3. learn how to rationalise the denominator of surds;
4. know what conjugate are and how to carry out the rationalisation of questions involving conjugates.
Click to learn about Permutation and Combination. In addition, download a Textbook with more explanation on more topics in mathematics with JAMB, WAEC (SSCE, GCE) past questions and answers.
Solving surds questions can be very easy and interesting as long as you familarize youself with the rules guiding its solutions. The summary of all that have been discussed in the article are:
1. Irrational numbers of the form $\dpi{150}&space;\sqrt[n]{a}$ whare $\dpi{150}&space;a$ is a non perfect square and $\dpi{150}&space;n$ is a positive integer greater tha 1 are called SURDS;
2. To rationalize a surd means to make the denominator rational;
3. $\dpi{150}&space;\sqrt{a}&space;\times&space;\sqrt{b}=\sqrt{ab}$ ;
4. $\dpi{150}&space;\frac{\sqrt{a}}{\sqrt{b}}&space;=\sqrt{\frac{a}{b}}$ and
5. And above all we have seen that the congugate of a surd like $\dpi{150}&space;1-\sqrt[2]{3}&space;=&space;1&space;+\sqrt[2]{3}$
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Confidence Interval of Population Proportion in 2 Steps in Excel 2010 and Excel 2013
This is one of the following five articles on Confidence Intervals in Excel
z-Based Confidence Intervals of a Population Mean in 2 Steps in Excel 2010 and Excel 2013
t-Based Confidence Intervals of a Population Mean in 2 Steps in Excel 2010 and Excel 2013
Minimum Sample Size to Limit the Size of a Confidence interval of a Population Mean
Confidence Interval of Population Proportion in 2 Steps in Excel 2010 and Excel 2013
Min Sample Size of Confidence Interval of Proportion in Excel 2010 and Excel 2013
Confidence Interval of
Population Proportion in 2
Steps in Excel
Confidence intervals covered in this manual will either be Confidence Intervals of a Population Mean or Confidence Intervals of a Population Proportion. A data point of a sample taken for a
confidence interval of a population mean can have a range of values. A data point of a sample taken for a confidence interval of a population proportion is binary; it can take only one of two values.
Data observations in the sample taken for a confidence interval of a population proportion are required to be distributed according to the binomial distribution. Data that are binomially distributed
are independent of each other, binary (can assume only one of two states), and all have the same probability of assuming the positive state.
A basic example of a confidence interval of a population proportion would be to create a 95-percent confidence interval of the overall proportion of defective units produced by one production line
based upon a random sample of completed units taken from that production line. A sampled unit is either defective or it is not. The 95-percent confidence interval is range of values that has a
95-percent certainty of containing the proportion defective (the defect rate) of all of the production from that production line based on a random sample taken from the production line.
The data sample used to create a confidence interval of a population proportion must be distributed according to the binomial distribution. The confidence interval is created by using the normal
distribution to approximate the binomial distribution. The normal approximation of the binomial distribution allows for the convenient application of the widely-understood z-based confidence interval
to be applied to binomially-distributed data.
The binomial distribution can be approximated by the normal distribution under the following two conditions:
1) p (the probability of a positive outcome on each trial) and q (q = 1 – p) are not too close to 0 or 1.
2) np > 5 and nq > 5
The Standard Error and half the width of a confidence interval of proportion are calculated as follows:
Margin of Error = Half Width of C.I. = z Value[α, 2-tailed] * Standard Error
Margin of Error = Half Width of C.I. = NORM.S.INV(1 – α/2) * SQRT[ (p_bar * q_bar) / n]
Example of a Confidence Interval
of a Population Proportion
in Excel
In this example a 95 percent confidence interval of a population proportion is created around a sample proportion using the normal distribution to approximate the binomial distribution.
This example evaluates a group of shoppers who either prefer to pay by credit or by cash. A random sample of 1,000 shoppers was taken. 70% of the sampled shoppers preferred to pay with a credit card.
The remaining 30% of the sampled shoppers preferred to pay with cash.
Determine the 95% Confidence Interval for the proportion of the general population that prefers to pay with a credit card. In other words, determine the endpoints of the interval that is 95 percent
certain to contain the true proportion of the total shopping population that prefers to pay by credit card.
Summary of Problem Information
p_bar = sample proportion = 0.70
q_bar = 1 – p_bar = 1 – 0.70 = 0.30
p = population proportion = Unknown (This is what the confidence interval will contain.)
n = sample size = 1,000
α = Alpha = 1 – Level of Certainty = 1 – 0.95 = 0.05
SE = Standard Error = SQRT[ (p_bar * q_bar) / n]
SE = SQRT[ (0.70 * 0.30) / 1000] = 0.014491
As when creating all Confidence of Proportion, we must satisfactorily answer these two questions and then proceed to the two-step method of creating the Confidence Interval of Proportion.
The Initial Two Questions That Must be Answered Satisfactorily
What Type of Confidence Interval Should Be Created?
Have All of the Required Assumptions For This Confidence Interval Been Met?
The Two-Step Method For Creating Confidence Intervals of Mean are the following:
Step 1 - Calculate the Half-Width of the Confidence Interval (Sometimes Called the Margin of Error)
Step 2 – Create the Confidence Interval By Adding to and Subtracting From the Sample Mean Half the Confidence Interval’s Width
The Initial Two Questions That Need To Be Answered Before Creating a Confidence Interval of the Mean or Proportion Are as Follows:
Question 1) Type of Confidence Interval?
a) Confidence Interval of Population Mean or Population Proportion?
This is a Confidence Interval of a population proportion because sampled data observations are binary: they can take only one of two possible values. A shopper sampled either prefers to pay with a
credit card or prefers to pay with cash.
The data sample is distributed according to the binomial distribution because each observation has only two possible outcomes, the probability of a positive outcome is the same for all sampled data
observations, and each data observation is independent from all others.
Sampled data points used to create a confidence interval of a population mean can take multiple values or values within a range. This is not the case here because sampled data observations can have
only two possible outcomes: a sampled shopper either prefers to pay with credit card or with cash.
b) t-Based or z-Based Confidence Interval?
A Confidence Interval of proportion is always created using the normal distribution. The binomial distribution of binary sample data is closely approximated by the normal distribution in certain
The next step in this example will evaluate whether the correct conditions are in place that permit the approximation of the binomial distribution by the normal distribution.
It should be noted that the sample size (n) equals 1,000. At that sample size, the t distribution is nearly identical to the normal distribution. Using the t distribution to create this Confidence
Interval would produce exactly the same result as the normal distribution produces.
This confidence interval will be a confidence interval of a population proportion and will be created using the normal distribution to approximate the binomial distribution of the sample data.
Question 2) All Required Assumptions Met?
Binomial Distribution Can Be Approximated By Normal Distribution?
The most important requirement of a Confidence Interval of a population proportion is the validity of approximating the binomial distribution (that the sampled objects follow because they are binary)
with the normal distribution.
The binomial distribution can be approximated by the normal distribution sample size, n, is large enough and p is not too close to 0 or 1. This can be summed up with the following rule:
The binomial distribution can be approximated by the normal distribution if np > 5 and nq >5. In this case, the following are true:
n = 1,000
p = 0.70 (p is approximated by p_bar)
q = 0.30 (q is approximated by q_bar)
np = 700 and nq = 300
It is therefore valid to approximate the binomial distribution with the normal distribution.
The binomial distribution has the following parameters:
Mean = np
Variance = npq
Each unique normal distribution can be completely described by two parameters: its mean and its standard deviation. As long as np > 5 and nq > 5, the following substitution can be made:
Normal (mean, standard deviation) approximates Binomial (n,p)
If np is substituted for the normal distribution’s mean and npq is substituted for the normal distribution’s standard deviation as follows:
Normal (mean, standard deviation)
Normal (np, npq), which approximates Binomial (n,p)
This can be demonstrated with Excel using data from this problem.
n = 1000
n = the number of trials in one sample
p = 0.7 (p is approximated by p_bar)
p = the probability of obtaining a positive result in a single trial
q = 0.7 (q is approximated by q_bar)
q = 1 - p
np = 700
npq = 210
at arbitrary point X = 700
(X equals the number positive outcomes in n trials)
BINOM.DIST(X, n, p, FALSE) = BINOM.DIST(700, 1000, 0.7, FALSE) = 0.0275
The Excel formula to calculate the PDF (Probability Density Function) of the normal distribution at point X is the following:
NORM.DIST(X, Mean, Stan. Dev, FALSE)
The binomial distribution can now be approximated by the normal distribution in Excel by the following substitutions:
BINOM.DIST(X, n, p, FALSE) ≈ NORM.DIST(X, np, npq, FALSE)
NORM.DIST(X, np, npq, FALSE) = NORM.DIST(700,700,210,FALSE) = 0.0019
BINOM.DIST(X, n, p, FALSE) = BINOM.DIST(700, 1000, 0.7, FALSE) = 0.0275
The difference is less than 0.03 and is reasonable close. Note that this approximation only works for the PDF (Probability Density Function) and not the CDF (Cumulative Distribution Function –
Replacing FALSE with TRUE in the above formulas would calculate the CDF instead of the PDF).
We now proceed to the two-step method for creating all Confidence Intervals of a population proportion. These steps are as follows:
Step 1) Calculate the Width of Half of the Confidence Interval
Step 2 – Create the Confidence Interval By Adding and Subtracting the Width of Half of the Confidence Interval from the Sample Mean
Proceeding through the four steps is done is follows:
Step 1) Calculate Width-Half of Confidence Interval
Half the Width of the Confidence Interval is sometimes referred to the Margin of Error. The Margin of Error will always be measured in the same type of units as the sample proportion is measured in,
which is percentage. Calculating the Half Width of the Confidence Interval using the t distribution would be done as follows in Excel:
Margin of Error = Half Width of C.I. = z Value[α, 2-tailed] * Standard Error
Margin of Error = Half Width of C.I. = NORM.S.INV(1 – α/2) * SQRT[ (p_bar * q_bar) / n]
Margin of Error = Half Width of C.I. = NORM.S.INV(0.975) * SQRT[ (0.7 * 0.3) / 1000]
Margin of Error = Half Width of C.I. = 1.95996 * 0.014491
Margin of Error = Half Width of C.I. = 0.0284, which equals 2.84 percent
Step 2 Confidence Interval = Sample Proportion ± C.I. Half-Width
Confidence Interval = Sample Proportion ± (Half Width of Confidence Interval)
Confidence Interval = p_bar ± 0.0284
Confidence Interval = 0.70 ± 0.0284
Confidence Interval = [ 0.6716, 0.7284 ], which equals 67.16 percent to 72.84 percent
We now have 95 percent certainty that the true proportion of all shoppers who prefer to pay with a credit card is between 67.16 percent and 72.84 percent.
A Excel-generated graphical representation of this confidence interval is shown as follows:
(Click On Image To See a Larger Version)
Excel Master Series Blog Directory
Statistical Topics and Articles In Each Topic
• Histograms in Excel
• Bar Chart in Excel
• Combinations & Permutations in Excel
• Normal Distribution in Excel
• t-Distribution in Excel
• Binomial Distribution in Excel
• z-Tests in Excel
• t-Tests in Excel
• Hypothesis Tests of Proportion in Excel
• Chi-Square Independence Tests in Excel
• Chi-Square Goodness-Of-Fit Tests in Excel
• F Tests in Excel
• Correlation in Excel
• Pearson Correlation in Excel
• Spearman Correlation in Excel
• Confidence Intervals in Excel
• Simple Linear Regression in Excel
• Multiple Linear Regression in Excel
• Logistic Regression in Excel
• Single-Factor ANOVA in Excel
• Two-Factor ANOVA With Replication in Excel
• Two-Factor ANOVA Without Replication in Excel
• Randomized Block Design ANOVA in Excel
• Repeated-Measures ANOVA in Excel
• ANCOVA in Excel
• Normality Testing in Excel
• Nonparametric Testing in Excel
• Post Hoc Testing in Excel
• Creating Interactive Graphs of Statistical Distributions in Excel
• Solving Problems With Other Distributions in Excel
• Optimization With Excel Solver
• Chi-Square Population Variance Test in Excel
• Analyzing Data With Pivot Tables
• SEO Functions in Excel
• Time Series Analysis in Excel
• VLOOKUP
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Covariant homogeneous nets of standard subspaces
Vincenzo Morinelli
Karl-Hermann Neeb
October 14, 2020
Rindler wedges are fundamental localization regions in AQFT. They are determined by the one-parameter group of boost symmetries fixing the wedge. The algebraic canonical construction of the free
field provided by Brunetti-Guido-Longo (BGL) arises from the wedge-boost identification, the BW property and the PCT Theorem. In this paper we generalize this picture in the following way. Firstly,
given a $\mathbb Z_2$-graded Lie group we define a (twisted-)local poset of abstract wedge regions. We classify (semisimple) Lie algebras supporting abstract wedges and study special wedge
configurations. This allows us to exhibit an analog of the Haag-Kastler one-particle net axioms for such general Lie groups without referring to any specific spacetime. This set of axioms supports a
first quantization net obtained by generalizing the BGL construction. The construction is possible for a large family of Lie groups and provides several new models. We further comment on orthogonal
wedges and extension of symmetries.
algebraic quantum field theory
nets of standard subspaces
Lie groups
representation theory
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Kung Fu Breakfast for You
Has your summer been simply blowing by? While summer should grant us with tons of opportunities to hang out, be lazy, enjoy family and spend quality time with the kids, somehow the sheer fact that
we have kids prevents that from happening!
Swim lessons during the week, swim meets during the weekend. Y workout (the kids have a summer program there that they love so it works out great), park program in the afternoon. Hip hop class one
night a week, gymnastics one morning a week. There simply isn’t any time to do anything!
We literally scheduled our summer trip to see grammy only a few days before we actually left. I have two part time jobs with flexible schedules but seriously, it’s just crazy. I’m not sure how
parents with full-time jobs manage to find any time with their families!
So with our crazy schedule, it’s tough to find time to really do much. Hitting a movie is always a winner but when that can’t happen, we pop up a bowl of popcorn and cozy up on the couch as a
family. We love family animated films because – although focused for kids – are still entertaining for parents.
Kung Fu Panda is a perfect example of that. I’ve watched it a few times and still enjoy it. And thanks to a promotion from MyBlogSpark, I have my very own copy to continue to enjoy as many times as
I want!
They shipped me a prize pack that included Kung Fu Panda, Golden Grahams and the toy from the Golden Grahams box. What a total hit of a package! For reference, Golden Grahams don’t last long around
this house… they’re one of those cereals that my husband and I both enjoy as much (if not more) than the kids do! I’m perfectly happy sitting and eating it by the handful. (Don’t judge! You know
you do it too!)
They’ve also given me an extra pack to give away. Here’s the deal though… this is a fast contest. It will be open for ONLY ONE DAY! Here are the details:
The Prize: Dreamworks Kung Fu Panda prize pack (includes one box of a participating Big G cereal, a Kung Fu Panda 2 spin fighter toy and the original Kung Fu Panda movie)
Participants –
…must provide a US shipping address
…must provide an email in the first comment, email me directly with an email address or have email accessible from their profile.
Sometime on July 12, 2011, a winner will be chosen at random from all valid comments left. Winner has 12 hours from posting/notification to respond. If winner cannot be contacted, I will move on to
the next random selection.
How to enter: <GIVEAWAY CLOSED>
Big G cereals are a staple in our household and always have been. Be sure to check out General Mills’ social media to find out the latest news on Big G cereals and other General Mills products!
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Education Auditorium UK | Maths Skills for Science A level
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Maths Skills for Science A level
Digital Learning - Maths Skills for Science A-level
similar to MathsWatch
The Education Auditorium UK is an educational platform that is dedicated to providing interactive learning opportunities to inspiring students to achieve their full potential in Maths A-Level. We
offer a wide range of Maths Courseware for A-level that are tailored to individual learning needs of all abilities. From our extensive digital online tutorial videos to our practice questions and
assessment builder, we have something to suit every learner and educator. We strive to provide a learning experience that is both engaging and rewarding.
TOC - Maths Digital Learning Courseware for
Maths Skills for A-level Physics
Feel free to invite your students to visit this Maths Skills for Biology page to watch and learn. The practice questions and assessment builders comes for free with the Maths A-level pack for schools
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TOC - Maths Digital Learning Courseware for
Maths Skills for A-level Biology
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Maths Skills for A-level Chemistry
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T. F. Al-Fariss, S. E. M. Desouky
King Saud University
Riyadh, Saudi Arabia
Calculations have been derived which will permit writing of a computer program for design of a pipeline handling Newtonian, pseudoplastic, or yield-pseudoplastic crudes.
Rheological characteristics of six types of Saudi waxy oils were measured with a Haake Rotovisco Model RV-11 rotational viscometer at 9, 12, 15, 18, 21, and 24C. The relative amount and molecular
distribution of wax content in Saudi oils directly affect their Theological behavior.
At an elevated temperature (e.g., 40 C.), wax dissolves in the oil to form a homogeneous fluid. When the temperature is decreased, wax crystals separate out with adsorbed resin, and the rheological
behavior of Saudi oils is changed from a Newtonian fluid into pseudoplastic and yield-pseudoplastic successively.
This change in Theological behavior, due to the thermal and shear histories and weight percentage of wax, strongly affects the design calculations of a pipeline handling such Saudi oils.
Statistical analysis was used to find out the variation of Theological behavior with operating temperatures and wax content in various Saudi oils. The evaluation was carried out at a statistical
confidence level of 95%.
Experimental data were correlated with respect to power-law and Herschel-Bulkey law. The pipeline design calculations were carried out through a computer program.
The friction factor was determined from Torrance's correlation and Dodge and Metzner correlation for yield-pseudoplastic and pseudoplastic fluids, respectively. The frictional pressure drop was
calculated from Darcy-Weisbach equation.
An extensive investigation of the rheological characteristics of waxy oils is necessary for optimizing the energy consumption and emphasizing the safety of the designed thickness in a pipeline
handling such oils.1
The relative amounts of wax content in Saudi oils directly affect their Theological behavior in the pipeline under either normal operating conditions or after shutdown. This change in Theological
behavior was observed at different temperatures as well as different weight percentages of wax.
At an elevated temperature, Saudi oils behave as Newtonian fluids. With a continuous lowering of the temperature, they Theologically exhibit pseudoplastic and yield-pseudoplastic behavior
Although several studies had reported significant effect of temperature on Theological behavior of waxy oils, no agreement had emerged on a general interpretation of such effects for different types
of waxy oils.1-4 All the developed rheological equations, which relate the shear stress to the shear rate and temperature cannot be employed in the pipeline design calculations because their
applications are limited within a specified temperature range.1 4 5
Because the temperature and rheological behavior of waxy oils vary along the pipeline, there is a need for rheological models which are applicable in a wide range of temperatures.
The t-test is used to investigate the effect of both temperature and weight percentage of wax on rheological behavior of Saudi waxy oils and determine the proper rheological models which correlate
the measured data best.6 7 These rheological models should be substituted in the Rabinowitsch-Mooney equation8 to derive the relations which couple wall shear stress (?w) and flow function (8 V I/D).
The derived relations are used to determine the values of n' and k' employing the generalized approach developed by Metzner and Reed.9
For specified pipeline characteristics such as diameter, thickness, length, and flow rate, the friction factor for pseudoplastic oil under turbulent flow conditions is determined from the Dodge and
Metzner correlation.10 in case of pumping yield-pseudoplastic oils, the friction factor is determined from Torrance's correlation.8
The frictional pressure drop is determined from Darcy-Weisbach equation.8
All the results obtained are employed in a computer program to carry out the pipeline design calculations. The safety of the pipeline thickness is checked at the end of the design calculations.11
Six types of Saudi waxy oils (3 and 6 wt % wax concentration with 43 C. melting point, 3 and 6 wt % wax concentration with 49 C. melting point, and 3 and 6 wt % wax concentration with 60 C. melting
point) were tested experimentally.
The Theological characteristics were measured with a Haake Rotovisco Model RV1 1 rotational viscometer at temperatures 9, 12, 15, 18, 21, and 24 C.
In a typical experiment, the sample of the oil to be tested was placed in the viscometer and held at the specified test temperature for 12 hr before the measurements were made. This duplicates the
aging time in flow equipment. Then the yield stress was measured at the lowest rotational speed.
Next, the rotational speed was increased and the corresponding shear stress and shear rate values were recorded.
In our experiments, 36 sets of the measured data were obtained. These data were plotted in Figs. 1 and 2.
When these data are correlated as a function of temperature, six Theological models are enough, while 30 rheological models are needed when temperature is not included as a parameter.1 3-5 Thus
statistical analysis was helpful to reduce the number of Theological models to be used in pipeline design.
Both temperature and wax effects can be thoroughly studied by carrying out the t-test.6 10 Usually, the t-test is used to check if there is a statistical difference between two sets of data.
This test is carried out by calculating the value of tc from Equation 1 6 7 (box) in which s is expressed by Equation 2. There, U and s are the mean and the standard deviation of n data points,
The value of tc is then compared with the so called t-tabulated (tt) which can be determined from statistical tables at specified confidence levels and degrees of freedom (df).7
The degree of freedom is calculated from Equation 3.
If tt is greater than tc, it explicitly means that there is no statistical difference between two sets of data at the specified confidence level. Consequently, the two sets of data can be treated as
one set.
Before carrying out t-test, it might be advisable to classify the experimental data into yieldless oils and yield oils data. The classification is given in Table 1 from which it can be observed that
13 sets of data comprise yieldless oils, and 23 sets of data represent yield oils.
The results of t-test to investigate the effect of temperature on the Saudi waxy oils are given in Table 2, from which the following points can be observed:
• For yieldless oils, the values of tc are less than the value of tt at a confidence level 95% for Type I, Type III, or Type IV.
This means that the experimental data of each type are statistically not different at the specified confidence level.
In other words, it statistically means that the changes in temperatures were not affected by the rheological behavior of each type.
Hence, the 13 sets of yieldless oil data can be reduced to four sets.
These sets of data are all data of Type I, data of Type II which were measured at temperature of 24 C., data of Type III which were measured at temperatures of 15, 18, 21, and 24 C., and data of
Type IV which were measured at temperatures of 21 and 24 C.
• For yield oils, the values of tc for Type II, Type IV, Type V, or Type VI were less than the value of tt. Thus, the experimental data of each type were statistically not different at confidence
level 95%.
Hence the 33 sets of yield oil data can be reduced to 5 sets. These sets of data are: all data of Type II except those measured at temperature 24 C.; data of Type III which were measured at
temperatures 9 and 12 C.; data of Type IV which were measured at temperatures 9, 12, 15, and 18 C.; and all data of both Type V and Type VI.
In an attempt to quantify the effect of wax content in Saudi oils on their Theological behavior, t-test was also carried out for the experimental data of both yieldless and yield oils.
The evaluation is given in Table 3, from which the following can be observed:
• For yieldless oils, a statistical difference between data of Type I and those of Type II, Type III, and Type IV exists. Thus the data of yieldless oils can be divided into two sets: Set 1 and Set
Set 1 includes all data of Type I, and Set 2 consists of the other data which were outlined in Table 1.
• For yield oils, the data of Type VI were statistically different from those of Type II, Type III, Type IV, and Type V.
This means that the data of Type VI can be combined with Set 3, while the other data can be grouped together and named as Set 4.
As has been discussed, the number of data sets became four and involved all the experimental data of Saudi waxy oils at different temperatures.
Because power-law and Herschel-Bulkey models were commonly used in design calculations,1 3-5 9-10 they are employed in the present research work.
Data of yieldless oils (Set 1 and Set 2) were correlated with power-law model: Equation 4 for Set 1 and Equation 5 for Set 2.
In case of yield oils, data of Set 3 and Set 4 were correlated with Herschel-Bulkey model: Equation 6 for Set 3 and Equation 7 for Set 4.
Hence, Saudi waxy oils were time-independent at the specified temperatures, at which their Theological behaviors confirmed to Newtonian (Equation 4), pseudoplastic (Equation 5), or
yield-pseudoplastic fluids (Equations 6 and 7).
In order to design a pipeline handling Newtonian, pseudoplastic, of yield-pseudoplastic oils, the design calculations must be based on the Theological properties of yield-pseudoplastic oils. The main
reason for this is probably that pumping yield-pseudoplastic oils in a pipeline required higher pressure than those of yieldless oils whether at normal operating conditions or after shutdown.1 4
For a pipeline handling yield-pseudoplastic fluids under turbulent-flow conditions, the design calculations should involve the following:
1. Determination of the pressure required to maintain the desired flow rate under normal operating conditions.
2. Calculation of the pressure required to restart and maintain the desired flow rate after shutdown.
3. Ensuring the safety of the pipeline thickness under maximum operating pressure.
Under normal operating conditions, the pressure required to maintain the desired flow rate is calculated from Equation 8.8
There Pf, PSH, and PKE are the pressures equivalent to friction and static heads and kinetic energy, respectively. The frictional pressure drop can be determined from Darcy-Weisbach equation
(Equation 9.)8
In Equation 9, the friction factor is calculated from Torrance's equation (Equation 7).
After shutdown, the frictional pressure required to initiate flow is estimated from Equation 11. This equation is only used when the restarting temperature is higher than the pour point.
In addition, the safety of the pipeline thickness must be checked by using Equation 12.11
The proposed computer program is summarized in Fig. 3. It is capable of handling both laminar and turbulent-flow conditions.
Under laminar-flow conditions, the program computes the pressure drop and designed pipeline thickness and reverts to stop-command. In case of turbulent flow, the program computes friction factor,
frictional pressure drop, operating pressure drop, and designed pipeline thickness.
The program can be set up for design purposes with the design procedure shown in Fig. 3.
For such setting up, n = 1 and yield stress 0 for Newtonian oils, and yield stress 0 for pseudoplastic (power law) oils.
The proposed program was executed for an existing pipeline which handles Saudi waxy oils. The pipeline characteristics, waxy oils' properties, and pumping pressures are given in Tables 4 and 5.
1. Barry, E.G., "Pumping Non-Newtonian Waxy Crude Oils," Journal of the Institute of Petroleum, Vol. 57, No. 554, pp. 74-85, 1971.
2. Turian, R.M., and Yuan, T.F., "Flow of Slurries in Pipelines," AlChE Journal, Vol. 23, pp. 232-243, 1977.
3. Verschuur, E., Den Harton, A.P., and Verheul, C.M., "The Effect of Thermal Shrinkage and Compressibility on the Yielding of Gelled Waxy Crude Oils in Pipelines," Journal of the Institute of
Petroleum, Vol. 57, No. 555, pp. 131-138, 1971.
4. Uhde, A., and Kopp, G., "Waxy Crudes in Relation to Pipeline Operations," Journal of the Institute of Petroleum Vol. 57, No. 554, pp. 63-73, 1971.
5. Ells, J.W., and Brown, V.R.R., "The Design of Pipelines to Handle Waxy Crude Oils," Journal of the Institute of Petroleum, Vol. 57, No. 555, pp. 175-183, 1971.
6. Clarke, R.H., "College Statistics," Thomas Nelson and Sons Ltd., London, 1969.
7. Volk, W., Applied Statistics for Engineers, Second Edition, McGraw-Hill Book Co., 1969.
8. Govier, G.W., and Aziz, K., The Flow of Complex Mixtures in Pipes, Robert E. Krieger Pub. Co., New York, 1977.
9. Metzner, A.B., and Reed, J.C., "Flow of Non-Newtonian Fluids-Correlation of the Laminar, Transition and Turbulent Flow Regions," AlChE Journal, Vol. 1, No. 4, p. 434-440, 1955.
10. Dodgo, D.W., and Metzner, A.B., "Turbulent Flow of Non-Newtonian Systems," AICHE Journal, Vol. 5, No. 6, pp. 189-204, 1959.
11. Szilas, A.P., Production and Transport of Oil and Gas, Elsevier Publishing Co., New York, 1975.
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Eigendecomposition - (Abstract Linear Algebra I) - Vocab, Definition, Explanations | Fiveable
from class:
Abstract Linear Algebra I
Eigendecomposition is the process of decomposing a square matrix into a set of eigenvalues and eigenvectors, enabling us to express the matrix in terms of its spectral properties. This technique is
especially useful for understanding the behavior of linear transformations, as it provides insight into how the matrix stretches, compresses, or rotates space. By representing a matrix in this way,
we can simplify complex operations, such as raising the matrix to a power or solving differential equations.
congrats on reading the definition of Eigendecomposition. now let's actually learn it.
5 Must Know Facts For Your Next Test
1. Eigendecomposition is applicable primarily to square matrices and is often performed on self-adjoint (or symmetric) matrices, which guarantees real eigenvalues.
2. The eigendecomposition of a matrix A can be expressed as A = PDP^{-1}, where D is a diagonal matrix containing the eigenvalues and P is a matrix whose columns are the corresponding eigenvectors.
3. If a matrix has n distinct eigenvalues, its eigendecomposition will produce n linearly independent eigenvectors, allowing for full diagonalization.
4. The process can greatly simplify calculations involving powers of matrices, as one can compute A^k = PD^kP^{-1}, where D^k is simply the diagonal matrix raised to the k-th power.
5. Eigendecomposition plays a critical role in various applications, including principal component analysis (PCA), which relies on identifying the directions of maximum variance in data.
Review Questions
• How does eigendecomposition help in simplifying operations on matrices?
□ Eigendecomposition helps simplify operations on matrices by breaking them down into their eigenvalues and eigenvectors. Once a matrix A is decomposed into the form A = PDP^{-1}, where D is
diagonal, performing calculations like raising the matrix to a power becomes straightforward since D^k can be calculated easily by raising each diagonal entry (the eigenvalues) to the k-th
power. This greatly reduces computational complexity and makes it easier to analyze properties of linear transformations.
• Discuss the significance of the spectral theorem in relation to eigendecomposition for self-adjoint operators.
□ The spectral theorem is significant for eigendecomposition as it ensures that every self-adjoint operator can be fully diagonalized using an orthonormal basis of eigenvectors. This means that
for any self-adjoint matrix, we can perform eigendecomposition reliably to obtain real eigenvalues and orthogonal eigenvectors. This property simplifies many problems in linear algebra,
making it easier to analyze and compute functions of matrices, such as exponentials and square roots.
• Evaluate the impact of eigendecomposition on practical applications like principal component analysis (PCA).
□ Eigendecomposition has a profound impact on practical applications such as principal component analysis (PCA). In PCA, data is transformed into a new coordinate system based on its
eigenvalues and eigenvectors from the covariance matrix. The directions of maximum variance become the principal components, which are essential for dimensionality reduction while preserving
essential patterns in data. By utilizing eigendecomposition, PCA not only enhances data interpretation but also improves efficiency in machine learning algorithms by reducing noise and
focusing on significant features.
© 2024 Fiveable Inc. All rights reserved.
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What is @calculator soup?
What is @calculator soup?
Calculator Soup is a free online calculator. Here you will find free loan, mortgage, time value of money, math, algebra, trigonometry, fractions, physics, statistics, time & date and conversions
calculators. Many of the calculator pages show work or equations that help you understand the calculations.
What is mustachians Money Mustache calculator?
Welcome fellow Mustachians! Inspired by the sage teachings of Mr. Money Mustache , these calculators are designed to help you better plan for financial independence. Punch excessive spending in the
face and find the best way to put your employees (as in your savings) to work for you.
What are the pros and cons of being a Mustachian?
The advantage of being Mustachian is that frugalism is in our genes for the rest of our lives. So, armed with this advantage, we won’t have trouble reducing our expenses to 2.5% as Vanguard
recommends in times of crisis. And no, it’s not deprivation, it’s adaptation, Mr. Grumpy!
What is the LCD of a fraction?
Also known as the lowest common denominator, it is the lowest number you can use in the denominator to create a set of equivalent fractions that all have the same denominator. How to Find the LCD of
Fractions, Integers and Mixed Numbers:
How do you use an LCM calculator with steps?
This LCM calculator with steps finds the LCM and shows the work using 5 different methods: Find the smallest number that is on all of the lists. We have it in bold above. Find all the prime factors
of each given number. List all the prime numbers found, as many times as they occur most often for any one given number.
How do you find the value of LCM (24 300)?
Using all prime numbers found as often as each occurs most often we take 2 × 2 × 2 × 3 × 5 × 5 = 600 Therefore LCM (24,300) = 600. Find all the prime factors of each given number and write them in
exponent form. List all the prime numbers found, using the highest exponent found for each. | {"url":"https://laurenceanywaysthemovie.com/what-is-calculator-soup/","timestamp":"2024-11-10T02:13:18Z","content_type":"text/html","content_length":"55891","record_id":"<urn:uuid:10b0ef87-604d-42c2-91c8-8c73ca22d52a>","cc-path":"CC-MAIN-2024-46/segments/1730477028164.3/warc/CC-MAIN-20241110005602-20241110035602-00465.warc.gz"} |
The Complete Guide to R-squared, Adjusted R-squared and Pseudo-R-squared (2024)
Learn how to use these measures to evaluate the goodness of fit of Linear and certain Nonlinear regression models
One of the most used and therefore misused measures in Regression Analysis is R² (pronounced R-squared). It’s sometimes called by its long name: coefficient of determination and it’s frequently
confused with the coefficient of correlation r² . See it’s getting baffling already!
The technical definition of R² is that it is the proportion of variance in the response variable y that your regression model is able to “explain” via the introduction of regression variables.
Clearly, that doesn’t do a whole lot to clear the air.
Hence we appeal to the familiar visual of a linear regression line superimposed on a cloud of (y,x) points:
The flat horizontal orange line represents the Mean Model. The Mean Model is the simplest model that you can build for your data. For every x value, the mean model predicts the same y value and that
value is the mean of your y vector. In this case, it happens to be 38.81 x 10000 New Taiwan Dollar/Ping where one Ping is 3.3 meter².
We can do better than the Mean Model at explaining the variance in y. For that, we need to add one or more regression variables. We will start with one such variable — the age of the house. The red
line in the plot represents the predictions of a Linear Regression Model when it’s fitted on the (y, X) data set where y=HOUSE PRICE and X=HOUSE AGE. As you can see, the Linear Model with one
variable fits a little better than the Mean Model.
R² lets you quantify just how much better the Linear model fits the data as compared to the Mean Model.
Let’s zoom into a portion of the above graph:
In the above plot, (y_i — y_mean) is the error made by the Mean Model in predicting y_i. If you calculate this error for each value of y and then calculate the sum of the square of each error, you
will get a quantity that is proportional to the variance in y. It is known as the Total Sum of Square TSS.
The Total Sum of Squares is proportional to the variance in your data. It is the variance that the Mean Model wasn’t able to explain.
Because TSS/N is the actual variance in y, the TSS is proportional to the total variance in your data.
Being the sum of squares, the TSS for any data set is always non-negative.
The Mean Model is a very simple model. It contains only one parameter which is the mean of the dependent variable y and it is represented as follows:
The Mean Model is also sometimes known as the Null Model or the Intercept only Model. But this interchangeability of definitions is appropriate only when the Null or Intercept Only model is
fitted, i.e. trained on the data set. That’s the only situation in which the Intercept will become the unconditional mean of y.
As mentioned earlier, if you want to better than the Mean Model in explaining the variance in y, you need to add one or more regression variables. Let’s look closely at how adding the regression
variable HOUSE AGE has helped the Linear Model in reducing the prediction error:
In the above plot, (y_i — y_pred_i) is the error made by the linear regression model in predicting y_i. This quantity is known as the residual error or simply the residual.
In the above plot, the residual error is clearly less than the prediction error of the Mean Model. Such improvement is not guaranteed. In a sub-optimal or badly constructed Linear Model, the residual
error could be more than the prediction error of the Mean Model.
If you calculate this residual error for each value of y and then calculate the sum of the square of each such residual error, you will get a quantity that is proportional to the prediction error of
the Linear Model. It is known as the Residual Sum of Square RSS.
The Residual Sum of Squares captures the prediction error of your custom Regression Model.
Being the sum of squares, the RSS for a regression model is always non-negative.
Thus, (Residual Sum of Squares)/(Total Sum of Squares) is the fraction of the total variance in y, that your regression model wasn’t able to explain.
1 — (Residual Sum of Squares)/(Total Sum of Squares) is the fraction of the variance in y that your regression model was able to explain.
We will now state the formula for R² in terms of RSS and TSS as follows:
Here is the Python code that produced the above plot:
And here is the link to the data set.
For Linear Regression Models that are fitted (i.e. trained) using the Ordinary Least Squares (OLS) Estimation technique, the range of R² is 0 to 1. Consider the following plot:
In the above plot, one can see that the residual error (y_i — y_pred_i)² is less than the total error (y_i — y_mean)². It can be shown that if you fit a Linear Regression Model to the above data by
using the OLS technique, i.e. by minimizing the sum of squares of residual errors (RSS), the worst that you can do is the Mean Model. But the sum of squares of residual errors of the Mean Model is
simply TSS, i.e. for the Mean Model, RSS = TSS.
Hence for OLS linear regression models, RSS ≤ TSS.
Since R² =1 — RSS/TSS, in the case of a perfect fit, RSS=0 and R² =1. In the worst case, RSS=TSS and R² = 0.
For Ordinary Least Squares Linear Regression Models, R² ranges from 0 to 1
Many non-linear regression models do not use the Ordinary Least Squares Estimation technique to fit the model. Examples of such nonlinear models include:
• The exponential, gamma and inverse-Gaussian regression models used for continuously varying y in the range (-∞, ∞).
• Binary choice models such as the Logit (a.k.a. Logistic) and Probit and their variants such as Ordered Probit used for y = 0 or 1, and the general class of Binomial regression models.
• The Poisson, Generalized Poisson and the Negative Binomial regression models for discrete non-negative y ϵ [0, 1, 2, …, ∞). i.e. models for counts based data sets.
The model fitting procedure of these nonlinear models is not based on progressively minimizing the sum of squares of residual errors (RSS) and therefore the optimally fitted model could have a
residual sum of squares that is greater than total sum of squares. That means, R² for such models can be a negative quantity. As such, R² is not a useful goodness-of-fit measure for most nonlinear
R-squared is not a useful goodness-of-fit measure for most nonlinear regression models.
A notable exception is regression models that are fitted using the Nonlinear Least Squares (NLS) estimation technique. The NLS estimator seeks to minimizes the sum of squares of residual errors
thereby making R² applicable to NLS regression models.
Later in this article, we’ll look at some alternatives to R-squared for nonlinear regression models.
Let’s look at the following figure again:
In the above plot, (y_pred_i — y_mean) is the reduction in prediction error that we achieved by adding a regression variable HOUSE_AGE_YEARS to our model.
If you calculate this difference for each value of y and then calculate the sum of the square of each difference, you will get a quantity that is proportional to the variance in y that the Linear
Regression model was able to explain. It is known as the Explained Sum of Square ESS.
The Explained Sum of Squares is proportional to the variance in your data that your regression model was able to explain.
Let’s do some math.
From the above plot, one can see that:
It can be shown that when the Least Squares Estimation technique is used to fit a linear regression model, the term 2*(y_i — y_pred)*(y_pred — y_mean) is 0.
So for the special case of OLS Regression Model:
In other words:
The linear regression model that we have used to illustrate the concepts has been fitted on a curated version of the New Taipei City Real Estate data set. Let’s see how to build this linear model and
find the R² score for it.
We’ll begin by importing all the required libraries:
import pandas as pd
from matplotlib import pyplot as plt
from statsmodels.regression.linear_model import OLS as OLS
import statsmodels.api as sm
Next, let’s read in the data file using Pandas. You can download the data set from here.
df = pd.read_csv('taiwan_real_estate_valuation_curated.csv', header=0)
Print the top 10 rows:
Our dependent y variable is HOUSE_PRICE_PER_UNIT_AREA and our explanatory a.k.a. regression a.k.a. X variable is HOUSE_AGE_YEARS.
We’ll carve out the y and X matrices:
y = df['HOUSE_PRICE_PER_UNIT_AREA']
X = df['HOUSE_AGE_YEARS']
Since houses of age zero years, i.e. new houses will also have some non-zero price, we need to add a y intercept. This is the ‘β0’ in the equation of the straight line: y_pred = β1*X + β0
X = sm.add_constant(X)
Next, we build and fit the OLS regression model and print the training summary:
olsr_model = OLS(endog=y, exog=X)
olsr_results = olsr_model.fit()
Here’s the output we get:
We see that the R² is 0.238. R² is not very large indicating a weak linear relationship between HOUSE_PRICE_PER_UNIT_AREA and HOUSE_AGE_YEARS.
The equation of the fitted model is as follows:
HOUSE_PRICE_PER_UNIT_AREA_pred = -1.0950*HOUSE_AGE_YEARS + 50.6617.
There is a somewhat weak and negative relationship between the age of the house and its price. And houses of zero age are predicted to have a mean price per unit area of 50.6617 x 10000 New Taiwan
In a way, this is like asking how to become rich or how to reduce weight? As the saying goes, be careful what you ask for, because you just might get it!
The naive way to increase R² in an OLS linear regression model is to throw in more regression variables but this can also lead to an over-fitted model.
To see why adding regression variables to an OLS Regression model does not reduce R², consider two linear models fitted using the OLS technique:
y_pred = β1*X1 + β0
y_pred = β2*X2 + β1*X1 + β0
The OLS estimation technique minimizes the residual sum of squares (RSS). If the second model does not improve the value of R² over the first model, the OLS estimation technique will set β2 to zero
or to some value that is statistically insignificant which would essentially get us back to the first model. Generally speaking, each time you add a new regression variable and refit the model using
OLS, you will either get a model with a better R² or essentially the same R² as the more constrained model.
This property of OLS estimation can work against you. If you go on adding more and more variables, the model will become increasingly unconstrained and the risk of over-fitting to your training data
set will correspondingly increase.
On the other hand, the addition of correctly chosen variables will increase the goodness of fit of the model without increasing the risk of over-fitting to the training data.
This tussle between our desire to increase R² and the need to minimize over-fitting has led to the creation of another goodness-of-fit measure called the Adjusted-R².
The concept behind Adjusted-R² is simple. To get Adjusted-R², we penalize R² each time a new regression variable is added.
Specifically, we scale (1-R²) by a factor that is directly proportional to the number of regression variables. Greater is the number of regression variables in the model, greater is this scaling
factor and greater is the downward adjustment to R².
The formula for Adjusted-R² is:
df_mean_model is the degrees of freedom of the mean model. For a training data set of size N, df_mean_model=(N-1).
df_model is the degrees of freedom of the regression model. For a model with p regression variables, df_model=(N-1-p).
One can see that as the model acquires more variables, p increases and the factor (N-1)/(N-1-p) increases which has the effect of depressing R².
Drawbacks of Adjusted-R²
Adjusted-R² has some problems, notably:
1. It treats the effect of all regression variables equally. In reality, some variables are more influential than others in their ability to make the model fit (or over-fit) the training data.
2. The formula for Adjusted-R² yields negative values when R² falls below p/(N-1) thereby limiting the use of Adjusted-R² to only values of R² that are above p/(N-1).
We will illustrate the process of using Adjusted-R² using our example data set. To do so, let’s introduce another regression variable NUM_CONVENIENCE_STORES_IN_AREA and refit our OLS regression model
on the data set:
y = df['HOUSE_PRICE_PER_UNIT_AREA']
X = df[['HOUSE_AGE_YEARS', 'NUM_CONVENIENCE_STORES_IN_AREA']]
X = sm.add_constant(X)olsr_model = OLS(endog=y, exog=X)
olsr_results = olsr_model.fit()
Let’s print the model training summary:
Notice that both R² and Adjusted-R² of the model with two regression variables is more than double that of the model with one variable:
On balance, the addition of the new regression variable has increased the goodness-of-fit. This conclusion is further supported by the coefficients by the p-values of the regression parameter
coefficients. We see from the regression output that the p-values of all three coefficients in the 2-variable OLSR model are essentially zero indicating that all parameters are statistically
The equation of the fitted two-variable model is as follows:
HOUSE_PRICE_PER_UNIT_AREA_pred = -0.7709*HOUSE_AGE_YEARS + 2.6287*NUM_CONVENIENCE_STORES_IN_AREA + 36.9925
Nonlinear models often use model fitting techniques such as Maximum Likelihood Estimation (MLE) which do not necessarily minimize the Residual Sum of Squares (RSS). Thus, given two nonlinear models
that have been fitted using MLE, the one with the greater goodness-of-fit may turn out to have a lower R² or Adjusted-R². Another consequence of this fact is that adding regression variables to
nonlinear models can reduce R². Overall, R² or Adjusted-R² should not be used for judging the goodness-of-fit of nonlinear regression model.
For nonlinear models, there have been a range of alternatives proposed for the humble R². We’ll look at one such alternative that is based on the following identity that we have come to know so well:
Total Sum of Squares (TSS) = Residual Sum of Squares (RSS) + Explained Sum of Squares (ESS).
While this identity works for OLS Linear Regression Models a.k.a. Linear Models, for nonlinear regression models, it turns out that a similar kind of triangle identity works using the concept of
Deviance. We’ll explain the concept of Deviance in a bit but for now, let’s look at this identity for nonlinear regression models:
Deviance of the Intercept-only model = Deviance of the fitted nonlinear model + Deviance explained by the fitted nonlinear model.
D(y, y_mean) = Deviance of the Intercept-only model
D(y, y_pred) = Deviance of the fitted nonlinear model
D(y_pred, y_mean) = Deviance explained by the fitted nonlinear model
Using the above identity, Cameron and Windmeijer have described (see paper links at the end of article) the following Deviance based formula for R² that is applicable to nonlinear models, especially
for Generalized Linear Regression Models (known as GLMs) that are fitted on discrete data. Commonly occurring examples of such nonlinear models are the Poisson and Generalized Poisson models, the
Negative Binomial Regression model and the Logistic Regression Model:
Before we go any further, some terms deserve an explanation.
Deviance of a regression model measures by how much the Log-Likelihood (more about Log-Likelihood in a bit) of the fitted regression model is greater than the Log-Likelihood of the saturated model.
So this begs two questions: What is a saturated model and what is Likelihood?
Saturated Model
A saturated regression model is one in which the number of regression variables is equal to the number of unique y values in the sample data set. What a saturated model gives you is essentially N
equations in N variables, and we know from college algebra that a system of N equations in N variables yields an exact solution for each variable. Thus, a saturated model can be built to perfectly
fit each y value. A saturated model thereby yields the maximum possible fit on your training data set.
Now let’s tackle Likelihood. The Likelihood of a fitted regression model is the probability (or probability density) of jointly observing all y values in the training data set using the predictions
of the fitted model as the mean parameter of the probability distribution function. The procedure for calculating Likelihood is as follows:
• Say your training data set contains 100 y observations. What you want to calculate is the joint probability of observing y1 and y2 and y3 and…up to y100 with your fitted regression model.
• So you start with fitting your regression model on this training data.
• Next, you feed the 100 rows in your training set through the fitted model to get 100 predictions from this model. These 100 y_pred_i values are your 100 conditional means (the ith mean is
conditioned upon the corresponding ith row in your X matrix).
• Now you set the 100 observed y values and the 100 conditional means (the predictions) in the probability (density) function of y to get 100 probability values. One probability for each y_i in y.
• Finally, you multiply together these 100 probabilities to get the Likelihood value. This is the likelihood of observing your training data set given your fitted model.
To get a feel for the calculation, I’d encourage you to refer to the following article. It contains a sample calculation for Likelihood for a Poisson Model:
An Illustrated Guide to the Poisson Regression ModelAnd a tutorial on Poisson regression using Pythontowardsdatascience.com
The Log-Likelihood is simply the natural logarithm of the Likelihood of the fitted model.
With these concepts under our belt, let’s circle back to our Deviance based formula for Pseudo-R²:
As mentioned earlier:
D(y, y_pred) = Deviance of the fitted nonlinear model
D(y, y_mean) = Deviance of the Intercept-only model (a.k.a. the null-model). The null model contains only the intercept i.e. no regression variables.
Using our formula for Deviance:
Sometimes, the following simpler version of Pseudo- R² proposed by McFadden is used (see paper link below for details):
McFadden’s Pseudo-R² is implemented by the Python statsmodels library for discrete data models such as Poisson or NegativeBinomial or the Logistic (Logit) regression model. If you call
DiscreteResults.prsquared() , you will get the value of McFadden’s R-squared value on your fitted nonlinear regression model.
See my tutorials on Poisson and NegativeBinomial Regression models on how to fit such types of nonlinear models on discrete (counts) based data sets:
An Illustrated Guide to the Poisson Regression ModelAnd a tutorial on Poisson regression using Pythontowardsdatascience.com
Negative Binomial Regression: A Step by Step GuidePlus a Python tutorial on Negative Binomial regressiontowardsdatascience.com
Also check out:
Generalized Linear ModelsWhat are they? Why do we need them?towardsdatascience.com
Yeh, I. C., & Hsu, T. K. (2018). Building real estate valuation models with comparative approach through case-based reasoning. Applied Soft Computing, 65, 260–271.
Paper and Book Links
Cameron A. Colin, Trivedi Pravin K., Regression Analysis of Count Data, Econometric Society Monograph №30, Cambridge University Press, 1998. ISBN: 0521635675
McCullagh P., Nelder John A., Generalized Linear Models, 2nd Ed., CRC Press, 1989, ISBN 0412317605, 9780412317606
McFadden, D. (1974), Conditional logit analysis of qualitative choice behaviour, in: P. Zarembka (ed.), Frontiers in Econometrics, Academic Press, New York, 105–142. PDF Download Link
Cameron, A., & Frank A. G. Windmeijer. (1996). R-Squared Measures for Count Data Regression Models with Applications to Health-Care Utilization. Journal of Business & Economic Statistics, 14(2),
209–220. doi:10.2307/1392433 PDF Download Link
A. Colin Cameron, Frank A.G. Windmeijer, An R-squared measure of goodness of fit for some common nonlinear regression models, Journal of Econometrics, Volume 77, Issue 2, 1997, Pages 329–342, ISSN
https://doi.org/10.1016/S0304-4076(96)01818-0. PDF Download link
N. J. D. Nagelkerke, A note on a general definition of the coefficient of determination, Biometrika, Volume 78, Issue 3, September 1991, Pages 691–692, https://doi.org/10.1093/biomet/78.3.691. PDF
Download link
All images in this article are copyright Sachin Date under CC-BY-NC-SA, unless a different source and copyright are mentioned underneath the image.
An Illustrated Guide to the Poisson Regression ModelAnd a tutorial on Poisson regression using Pythontowardsdatascience.com
Negative Binomial Regression: A Step by Step GuidePlus a Python tutorial on Negative Binomial regressiontowardsdatascience.com
Generalized Linear ModelsWhat are they? Why do we need them?towardsdatascience.com
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Introduction to Business Math
For this section you will need the following:
Symbols Used
• [latex]\text{GE}=[/latex] gross earnings
Formulas Used
• Formula 3.3 – Salary & Hourly Gross Earnings
[latex]\small\text{GE}=\text{Regular Earnings}+\text{Overtime Earnings}+\text{Holiday Earnings}+\text{Stat Holidays Worked Earnings}[/latex]
• Formula 3.1b – Rate, Portion, Base
You work hard at your job, and you want to be compensated properly for all the hours you put in. Assume you work full time with an hourly rate of pay of [latex]\$10[/latex]. Last week you worked
eight hours on Sunday and eight hours on Monday, which was a statutory holiday. Then you took Tuesday off, worked eight hours on each of Wednesday and Thursday, took Friday off, and worked [latex]10
[/latex] hours on Saturday. That’s a total of [latex]42[/latex] hours of work for the week. What is your gross pay? Give or take a small amount depending on provincial employment standards, it should
be about [latex]\$570[/latex]. But if you don’t understand how to calculate gross earnings, you could be underpaid without ever realizing it.
Here are some notes about the content in this chapter: About [latex]10\%[/latex] of Canadian workers fall under federal employment standards, which are not discussed here. This textbook generalizes
the most common provincial employment standards; however, to calculate your earnings accurately requires you to apply your own provincial employment standards legislation. Part-time employment laws
are extremely complex, so this textbook assumes in all examples that the employee is full time.
This section addresses the calculation of gross earnings, which is the amount of money earned before any deductions from your paycheque. The four most common methods of employee remuneration include
salaries, hourly wages, commissions, and piecework wages.
Salary and Hourly Wages
One ad in the employment classifieds indicates compensation of [latex]\$1,270[/latex] biweekly, while a similar competing ad promotes wages of [latex]\$1,400[/latex] semi-monthly. If both job ads are
similar in every other way, which job has the higher annual gross earnings? To make this assessment, you must understand how salaries work. A salary is a fixed compensation paid to a person on a
regular basis for services rendered. Most employers pay employees by salary in occupations where the employee’s work schedule generally remains constant.
In contrast, an hourly wage is a variable compensation based on the time an employee has worked. In contrast to a salary, this form of compensation generally appears in occupations where the number
of hours is unpredictable or continually varies from period to period.
Employment Contract Characteristics
Salaried and hourly full-time employees are similar with regard to their gross earnings. The major earnings issues in an employment contract involve regular earnings structure, overtime, and
Regular Earnings Structure
An agreement with your employer outlines the terms of your employment, including the time frame and frequency of pay.
For salaried employees, the time frame that the salary covers must be clearly stated. For example, you could receive a salary of [latex]\$2,000[/latex] monthly or [latex]\$50,000[/latex] annually.
Notice that each of these salaries is followed by the specific time frame for the compensation. For hourly employees, the time frame requires identification of the wage earned per hour.
How often the gross earnings are paid out to the employee must be defined.
□ Monthly: Earnings are paid once per month. By law, employees must receive compensation from their employer at least once per month, which equals [latex]12[/latex] times per year.
□ Daily: Earnings are paid at the end of every day. This results in about [latex]260[/latex] paydays per year ([latex]5[/latex] days per week multiplied by [latex]52[/latex] weeks per year). In
a leap year, there might be one additional payday.
□ Weekly: Earnings are paid once every week. This results in [latex]52[/latex] paydays in any given year since there are [latex]52[/latex] weeks per year.
□ Biweekly: Earnings are paid once every two weeks. This results in [latex]26[/latex] paydays in any given year since there are [latex]52\div 2=26[/latex] biweekly periods per year.
□ Semi-monthly: Earnings are paid twice a month, usually every half month (meaning on the [latex]15th[/latex] and last day of the month). This results in [latex]24[/latex] paydays per year.
Thus, the earnings structure specifies both the time frame and the frequency of earnings. For a salaried employee, this may appear as “[latex]\$2,000[/latex] monthly paid semi-monthly” or “[latex]\
$50,000[/latex] annually paid biweekly.” For an hourly employee, this may appear as “[latex]\$10[/latex] per hour paid weekly.” No matter whether you are salaried or hourly, earnings determined by
your regular rate of pay are called your regular earnings.
[* Special thanks to Steven Van Alstine (CPM, CAE), Vice-President of Education, the Canadian Payroll Association, for assistance in summarizing Canadian payroll legislation and jurisdictions.]
Overtime is work time in excess of your regular workday, regular workweek, or both. In most jurisdictions it is paid at [latex]1.5[/latex] times your regular hourly rate (called time-and-a-half),
though your company may voluntarily pay more or a union may have negotiated a more favourable rate such as two times your regular hourly rate (called double time). A contract with an employer will
specify your regular workday and workweek, and some occupations are exempt from overtime. Due to the diversity of occupations, there is no set rule on what constitutes a regular workday or workweek.
In most jurisdictions, a regular workweek is eight hours per day and [latex]40[/latex] hours per week. Once you exceed these regular hours, you are eligible to receive overtime or premium earnings,
which are based on your overtime rate of pay.
A statutory holiday is a legislated day of rest with pay. Five statutory holidays are recognized throughout Canada, namely, New Year’s Day, Good Friday (or Easter Monday in Quebec), Canada Day,
Labour Day, and Christmas Day. Each province or territory has an additional four to six public holidays (or general holidays), which may include Family Day (known as Louis Riel Day in Manitoba and
Islander Day in PEI) in February, Victoria Day in May, the Civic Holiday in August, Thanksgiving Day in October, Remembrance Day in November, and Boxing Day in December. These public holidays may or
may not be treated the same as statutory holidays, depending on provincial laws.
Statutory and public holidays generally require employees to receive the day off with pay. If the holiday falls on a nonworking day, it is usually the next working day that is given off instead. For
example, if Christmas Day falls on a Saturday, typically the following Monday is given off with pay. Here’s how holidays generally work (though you should always consult legislation for your specific
• You should be given the day off with pay, called holiday earnings. The holiday earnings are in the amount of a regular day’s earnings, and the hours involved count toward your weekly hourly
totals for overtime purposes (preventing employers from shifting your work schedule that week).
• If you are required to work, the employer must offer another day off in lieu with pay. Your work on the statutory holiday is then paid at regular earnings and the hours involved contribute toward
your weekly hourly totals for overtime purposes. You are paid holiday earnings on your future day off.
• If you are required to work and no day or rest is offered in lieu, this poses the most complex situation. Under these conditions:
• You are entitled to the holiday earnings you normally would have received for the day off. The hours that make up your holiday earnings contribute toward your weekly hourly totals for overtime
purposes (again, consult your local jurisdiction).
• In addition, for the hours you worked on the statutory holiday you are entitled to overtime earnings known as statutory holiday worked earnings. These hours do not contribute toward your weekly
hourly totals for overtime purposes since you are already compensated at a premium rate of pay. For example, assume you work eight hours on Labour Day, your normal day is eight hours, and you
won’t get another day off in lieu. Your employer owes you the eight hours of holiday earnings you should have received for getting the day off plus the eight hours of statutory holiday worked
earnings for working on Labour Day.
The four forms of compensation consist of regular earnings, overtime earnings, holiday earnings, and statutory holiday worked earnings. Add these four elements together to determine total gross
earnings. Formula 3.3 shows the relationship.
[latex]\boxed{3.3}[/latex] Salary & Hourly Gross Earnings
[latex]\begin{eqnarray*}{\color{red}{\text{GE}}}&=&{\color{blue}{\text{Regular Earnings}}}\;+\;{\color{green}{\text{Overtime Earnings}}}\;+\;{\color{purple}{\text{Holiday Earnings}}}\\&+&{\color
{Mahogany}{\text{Stat Holidays Worked Earnings}}}\end{eqnarray*}[/latex]
[latex]{\color{Mahogany}{\text{Statutory Holiday Worked Earnings:}}}[/latex] This pay shows up only if a statutory holiday is worked and the employee will not receive another paid day off in lieu. It
is received in addition to the holiday pay and must be paid at a premium rate
[latex]{\color{red}{\text{GE}}}\text{ is Gross Earnings:}[/latex] Gross earnings are earning before any deductions and represent the amount owed to the employee for services rendered. This is
commonly called the gross amount of the paycheque.
[latex]{\color{purple}{\text{Holiday Earnings:}}}[/latex] If a statutory holiday occurs during a pay period, this is holiday pay in an amount that represents a regular shift.
[latex]{\color{blue}{\text{Regular Earnings:}}}[/latex] Unless the employees have exceeded their daily or weekly thresholds or a holiday is involved, all hours worked are considered regular earnings
[latex]{\color{green}{\text{Overtime Earnings:}}}[/latex] Any hours worked that exceed daily or weekly thresholds fall under overtime earnings. For most individuals, this is calculated at 1.5 times
their regular hourly rate.
Calculate Gross Earnings for Salaried Employees
To calculate the total gross earnings for a salaried employee, follow these steps:
Step 1: Analyze the employee’s work performed and assign hours as needed into each of the four categories of pay. If the employee has only regular hours of pay, skip to Step 6.
Step 2: Calculate the employee’s equivalent hourly rate of pay. This means converting the salary into an equivalent hourly rate:
[latex]\begin{align*}\text{Equivalent Hourly Rate}=\frac{\text{Annual Salary}}{\text{Annual Hours Worked}}\end{align*}[/latex]
For example, use a [latex]\$2,000[/latex] monthly salary requiring [latex]40\;\text{hours}[/latex] of work per week. Express the salary annually by multiplying it by [latex]12\;\text{months}[/latex],
yielding [latex]\$24,000[/latex]. Express the [latex]40\;\text{hours per week}[/latex] annually by multiplying by [latex]52\;\text{weeks per year}[/latex], yielding [latex]2,080\;\text{hours}[/latex]
worked. The equivalent hourly rate is [latex]\$24,000\div 2,080=\$11.538461[/latex].
Step 3: Calculate any holiday earnings. Take the unrounded hourly rate and multiply it by the number of hours in a regular shift, or
[latex]\text{Holiday Earnings}=\text{Unrounded Hourly Rate}\times\text{Hours in a Regular Shift}[/latex]
A salaried employee earning [latex]\$11.538461\;\text{per hour}[/latex] having a daily eight-hour shift receives [latex]\$11.538461\times 8=\$92.31[/latex] in holiday earnings.
Step 4: Calculate any overtime earnings.
□ Determine the overtime hourly rate of pay by multiplying the unrounded hourly rate by the minimum standard overtime factor of [latex]1.5[/latex] (this could be higher if the company pays a
better overtime rate than this):
[latex]\text{Overtime Hourly Rate}=\text{Unrounded Hourly Rate}\times 1.5[/latex]
□ Round the final result to two decimals. For the salaried worker:
[latex]\$11.538461\times 1.5=\$17.31\;\text{per overtime hour}[/latex].
□ Multiply the overtime hourly rate by the overtime hours worked.
Step 5: Calculate any statutory holiday worked earnings which is similar to calculating overtime earnings:
[latex]\text{Stat Holiday Worked Earnings}=\text{Statutory Hourly Rate}\times\text{Statutory Hours Worked}[/latex]
The statutory hourly rate is at minimum [latex]1.5[/latex] times the unrounded hourly rate of pay. The salaried employee working eight hours on a statutory holiday receives [latex]\$17.31\times 8=\
Step 6: Calculate the gross earnings paid at the regular rate of pay. Take the amount of the salary and divide it by the number of pay periods involved, then subtract any holiday earnings:
[latex]\begin{align*}\text{Regular Earnings}=\frac{\text{Salary}}{\text{Salary Pay Periods}}-\text{Holiday Earnings}\end{align*}[/latex]
You need to calculate the number of pay periods based on the regular earnings structure. For example, an annual [latex]\$52,000[/latex] salary paid biweekly would have [latex]26[/latex] pay periods
annually. Therefore, a regular paycheque is [latex]\$52,000\div 26=\$2,000[/latex] per paycheque. As another example, a [latex]$2,000[/latex] monthly salary paid semi-monthly has two pay periods in a
single month, resulting in regular earnings of [latex]\$2,000\div 2=\$1,000[/latex] per paycheque. If a holiday is involved in the pay period, you must deduct the holiday earnings from these amounts.
Step 7: Calculate the total gross earnings by applying Formula 3.3[latex]\small\text{GE}=\text{Regular}+\text{OT}+\text{Holiday}+\text{Stat Worked}[/latex].
Calculate Gross Earnings for Hourly Employees
To calculate the total gross earnings for an hourly employee, follow steps similar to those for the salaried employee:
Step 1: Analyze the employee’s work performed and assign hours as needed into each of the four categories of pay. It is usually best to set up a table similar to the one below. This table allows you
to visualize the employee’s week at a glance along with totals, enabling proper assessment of their hours worked.
This table separates the four types of earnings into different rows. Enter the information into the table about the employee’s workweek, placing it in the correct day and on the correct row. If any
daily thresholds are exceeded, place the appropriate hours into the overtime row. Once you have completed this, total the regular hours and holiday hours for the week and check to see if they exceed
any regular weekly threshold. If so, starting with the last workday of the week and working backwards, convert regular hours into overtime hours until you have reduced the regular hours to the
regular weekly total.
Table 3.3.1
Type of Pay Sunday Monday Tuesday Wednesday Thursday Friday Saturday Total Hours Rate of Pay Earnings
Holiday Worked
TOTAL EARNINGS
Once you have completed the table, if the employee has only regular hours of pay, skip to step 5. Otherwise, proceed with the next step.
Step 2: Calculate any holiday earnings. Take the hourly rate and multiply it by the number of hours in a regular shift:
[latex]\text{Holiday Earnings}=\text{Hourly Rate}\times\text{Hours in a Regular Shift}[/latex]
Step 3: Calculate any overtime earnings.
□ Determine the overtime hourly rate of pay rounded to two decimals by multiplying the hourly rate by the minimum standard overtime factor of [latex]1.5[/latex] (or higher).
[latex]\text{Overtime Hourly Rate}=\text{Hourly Rate}\times 1.5[/latex]
□ Multiply the overtime hourly rate by the overtime hours worked.
Step 4: Calculate any statutory holiday worked earnings. This is the same procedure as for a salaried employee.
Step 5: Calculate the gross earnings paid at the regular rate of pay. Take the number of hours worked and multiply it by the hourly rate of pay:
[latex]\text{Regular Earnings}=\text{Hours Worked}\times\text{Hourly Rate}[/latex]
For example, [latex]20\;\text{hours}[/latex] worked at [latex]\$10\;\text{per hour}[/latex] with no holiday earnings is [latex]20\times\$10=\$200[/latex].
Step 6: Calculate the total gross earnings by applying Formula 3.3[latex]\small\text{GE}=\text{Regular}+\text{OT}+\text{Holiday}+\text{Stat Worked}[/latex].
Things To Watch Out For
Be careful about the language of the payment frequency. It is very common to confuse semi and bi, and sometimes businesses use the terms incorrectly. The term semi generally means half. Therefore, to
be paid semi-monthly means to be paid every half month. The term bi means two. Therefore, to be paid biweekly means to be paid every two weeks. Some companies that pay semi-monthly mistakenly state
that they pay bimonthly, which in fact would mean they paid every two months.
Paths To Success
In calculating the pay for a salaried employee, this textbook assumes for simplicity that a year has exactly [latex]52[/latex] weeks. In reality, there are [latex]52[/latex] weeks plus one day in any
given year. In a leap year, there are [latex]52[/latex] weeks plus two days. This extra day or two has no impact on semi-monthly or monthly pay, since there are always [latex]24[/latex] semi-months
and [latex]12[/latex] months in every year. However, weekly and biweekly earners are impacted as follows:
• If employees are paid weekly, approximately once every six years there are [latex]53[/latex] pay periods in a single year. This would “reduce” the employees’ weekly paycheque in that year. For
example, assume they earn [latex]\$52,000[/latex] per year paid weekly. Normally, they are paid [latex]\$52,000\div 52=\$1,000\;\text{per week}[/latex]. However, since there are [latex]53[/latex]
pay periods approximately every sixth year, this results in [latex]\$52,000\div 53=\$981.13\;\text{per week}[/latex] for that year.
• If employees are paid biweekly, approximately once every [latex]12[/latex] years there are [latex]27[/latex] pay periods in a single year. This has the same effect as the extra pay period above.
For example, if they are paid [latex]\$52,000[/latex] per year biweekly they normally receive [latex]\$52,000\div 26=\$2,000[/latex] per biweekly cheque. Approximately every twelfth year, they
are paid [latex]\$52,000\div 27=\$1,925.93[/latex] per biweekly cheque for that year.
Many employers ignore these technical nuances in pay structure since the extra costs incurred to modify payroll combined with the effort required to calm down employees who don’t understand the
smaller paycheque are not worth the savings in labour. Therefore, most employers treat every year as if it has [latex]52[/latex] weeks ([latex]26[/latex] biweeks) regardless of the reality. In
essence, employees receive a bonus paycheque approximately once every six or twelve years!
1) A salaried employee whose normal workweek is [latex]8\;\text{hours}[/latex] per day and [latex]40\;\text{hours per week}[/latex] works [latex]8\;\text{hours}[/latex] each day from Monday to
Saturday inclusive, where Monday was a statutory holiday. Which of the following statements is correct (assuming she will not get another day off in lieu of the holiday)?
a. The employee receives only her regular weekly earnings for [latex]40\;\text{hours}[/latex].
b. The employee receives [latex]32\;\text{hours}[/latex] of regular earnings, [latex]8\;\text{hours}[/latex] of holiday earnings, [latex]8\;\text{hours}[/latex] of overtime earnings, and [latex]8\;\
text{hours}[/latex] of statutory holiday worked earnings.
c. The employee receives [latex]40\;\text{hours}[/latex] of regular earnings, [latex]8\;\text{hours}[/latex] of overtime earnings, and [latex]8\;\text{hours}[/latex] of statutory holiday worked
d. The employee receives [latex]40\;\text{hours}[/latex] of regular earnings and [latex]8\;\text{hours}[/latex] of overtime earnings.
The correct answer is b. When working on the statutory holiday and not getting another day off in lieu, the salaried employee is eligible for eight hours of holiday earnings plus eight hours of
statutory holiday worked earnings. The holiday earnings count toward the weekly total, but not the statutory holiday worked earnings. Thus the employee from Tuesday to Friday inclusive worked an
additional [latex]32[/latex] regular hours, bringing her weekly total to [latex]40\;\text{hours}[/latex]. The work on Saturday exceeds her [latex]40[/latex] hour workweek, and therefore all eight
hours are paid as overtime earnings.
Tristan is compensated with an annual salary of [latex]\$65,000[/latex] paid biweekly. His regular workweek consists of four [latex]10\text{-hour days}[/latex], and he is eligible for overtime at
[latex]1.5[/latex] times pay for any work in excess of his regular requirements. Tristan worked regular hours for the first two weeks. Over the next two weeks, Tristan worked his regular hours and
became eligible for [latex]11\;\text{hours}[/latex] of overtime. During these two weeks, he worked his regular shift on Good Friday but his employer has agreed to give him another day off with pay in
the future.
a. Determine Tristan’s gross earnings for the first two-week pay period.
b. Determine Tristan’s gross earnings for the second two-week pay period.
Step 1: What are you looking for?
You have been asked to calculate Tristan’s gross earnings, or [latex]\text{GE}[/latex], for two consecutive pay periods.
Step 2: What do you already know?
You know Tristan’s compensation:
[latex]\begin{align*}\text{Annual Salary}&=\$65,000\\\text{Pay Periods}&=\text{biweekly}=26\text{ times per year}\\\text{Annual Hours}&=10/\text{day}\times4\text{ days}/\text{week}\times52\text{
You also know his work schedule:
[latex]\begin{align*}\text{First Two Weeks (P1)}&=\text{regular pay}\\\text{Overtime in P1}&=\$0\\\text{Second Two Weeks (P2)}&=\text{regular pay}\\\text{Overtime in P2}&=11\text{ hours}\end{align*}
There is a holiday in the second two weeks, but he will receive another day off in lieu.
Step 3: Make substitutions using the information known above.
For each biweekly pay period, apply the following steps:
Step 1: Calculate Tristan’s equivalent hourly rate of pay.
[latex]\text{Equivalent Hourly Rate}_\textrm{P1}=\text{Only regular earnings - skip to Step 5}[/latex]
[latex]\begin{align*}\text{Equivalent Hourly Rate}_\textrm{P2}&=\frac{\text{Annual Salary}}{\text{Annual Hours Worked}}\\[1ex]\text{Equivalent Hourly Rate}_\textrm{P2}&=\frac{\$65,000}{2080}\\[1ex]\
text{Equivalent Hourly Rate}_\textrm{P2}&=\$31.25/\text{hour}\end{align*}[/latex]
Step 2: Calculate holiday earnings using the equivalent hourly rate.
[latex]\text{Holiday Earnings}_\textrm{P1}=\$0[/latex]
[latex]\text{Holiday Earnings}_\textrm{P2}=\$0[/latex]
Tristan will take his holiday pay during [latex]P2[/latex] another day. Therefore, he is not eligible for this pay, and his work counts as regular hours.
Step 3: Calculate overtime earnings by taking the overtime hourly pay rate multiplied by hours worked.
[latex]\text{Overtime Earnings}_\textrm{P1}=\$0[/latex]
[latex]\begin{align*}\text{Overtime Earnings}_\textrm{P2}&=\text{Overtime Hourly Rate}\times \text{Overtime Hours}\\\text{Overtime Earnings}_\textrm{P2}&=(\text{Equivalent Hourly Rate}\times 1.5)\
times 11\\\text{Overtime Earnings}_\textrm{P2}&=(\$31.25\times 1.5)\times 11\\\text{Overtime Earnings}_\textrm{P2}&=\$46.88\times 11\\\text{Overtime Earnings}_\textrm{P2}&=\$515.68\end{align*}[/
Step 4: Calculate statutory holiday worked earnings at the premium rate of pay.
[latex]\text{Statutory Worked}_\textrm{P1}=\$0[/latex]
[latex]\text{Statutory Worked}_\textrm{P2}=\$0[/latex]
Since Tristan is receiving another day off in lieu of the holiday he worked during [latex]P2[/latex], he is not eligible for this pay.
Step 5: Calculate regular earnings.
[latex]\begin{align*}\text{Regular Earnings}_\textrm{P1}&=\frac{\text{Salary}}{\text{Salary Pay Periods}}-\text{Holiday Earnings}\\[1ex]\text{Regular Earnings}_\textrm{P1}&=\frac{\$65,000}{26}-\$0\\
[1ex]\text{Regular Earnings}_\textrm{P1}&=\$2,500\end{align*}[/latex]
[latex]\begin{align*}\text{Regular Earnings}_\textrm{P2}&=\frac{\text{Salary}}{\text{Salary Pay Periods}}-\text{Holiday Earnings}\\[1ex]\text{Regular Earnings}_\textrm{P2}&=\frac{\$65,000}{26}-\$0\\
[1ex]\text{Regular Earnings}_\textrm{P2}&=\$2,500\end{align*}[/latex]
Step 6: Determine total gross earnings using Formula 3.3[latex]\small\text{GE}=\text{Regular}+\text{OT}+\text{Holiday}+\text{Stat Worked}[/latex].
Step 4: Provide the information in a worded statement.
For the first two-week pay period, Tristan worked only his regular hours and therefore is compensated [latex]\$2,500[/latex] as per his salary. For the second two-week pay period, Tristan is eligible
to receive his regular hours plus his overtime, but he receives no additional pay for the worked holiday since he will receive another day off in lieu. His total gross earnings are [latex]\$3,015.68
Marcia receives an hourly wage of [latex]\$32.16[/latex] working on an automotive production line. Her union has negotiated a regular work day of [latex]7.25\;\text{hours}[/latex] for five days
totaling [latex]36.25\;\text{hours}[/latex] for the week. Overtime is paid at [latex]1.5[/latex] times her regular rate for any work that exceeds the daily or weekly limits. If work is required on a
statutory holiday, her company does not give a day off in lieu and pays a premium rate of [latex]2.5[/latex] times her regular rate. Last week, Marcia worked nine hours on Monday, her regular hours
on Tuesday through Friday inclusive, and three hours on Saturday. Friday was a statutory holiday. Calculate Marcia’s gross earnings for the week.
Step 1: What are you looking for?
You need to calculate Marcia’s gross earnings, or [latex]\text{GE}[/latex], for the week.
Step 2: What do you already know?
Step 1: You know Marcia’s pay structure and workweek:
[latex]\begin{align*}\text{Regular Hourly Rate}&=\$32.16\\\text{Overtime Hourly Pay}&=\times 1.5\\\text{Statutory Holiday Worked Rate}&=\times 2.5\end{align*}[/latex]
Exceeding [latex]7.25\;\text{hours}[/latex] daily or [latex]36.25\;\text{hours}[/latex] weekly is overtime.
Table 3.3.2
Sunday 0
Monday 9
Tuesday 7.25
Wednesday 7.25
Thursday 7.25
Friday (statutory holiday) 7.25
Saturday 3
Step 3: Make substitutions using the information known above.
Take Marcia’s hours and place them into the table. Assess whether any daily or weekly totals are considered overtime and make any necessary adjustments.
Table 3.3.3
Type Sun Mon Tue Wed Thu Fri Sat Total Rate Earnings
Regular 0 7.25 7.25 7.25 7.25 3
39.25 $32.16
Holiday 7.25
Overtime 1.75 1.75
Holiday Worked 7.25 7.25
TOTAL EARNINGS
She worked nine hours. Therefore, the first [latex]7.25\;\text{hours}[/latex] are regular pay and the last [latex]1.75[/latex] are overtime pay.
Friday was a statutory holiday, and she will not receive another day off in lieu. She must receive statutory holiday worked pay in addition to her hours worked.
Note the weekly total of [latex]36.25[/latex] has been exceeded by three hours. Move Saturday’s hours into overtime.
The following table is the final layout of her workweek:
Table 3.3.4
Type Sun Mon Tue Wed Thu Fri Sat Total Rate Earnings
Regular 0 7.25 7.25 7.25 7.25
36.25 $32.16
Holiday 7.25
Overtime 1.75 3 4.75
Holiday Worked 7.25 7.25
TOTAL EARNINGS
Perform necessary calculations to obtain her Gross Earnings using Formula 3.3[latex]\small\text{GE}=\text{Regular}+\text{OT}+\text{Holiday}+\text{Stat Worked}[/latex].
[latex]\begin{align*}\text{Holiday Earnings }&=7.25\times\$32.16\\\text{Holiday Earnings }&=\$233.16\end{align*}[/latex]
[latex]\begin{align*}\text{Overtime Hourly Rate}&=\$32.16\times 1.5\\\text{Overtime Hourly Rate}&=\$48.24\end{align*}[/latex]
[latex]\begin{align*}\text{Overtime Earnings}&=4.75\times\$48.24\\\text{Overtime Earnings}&=\$229.14\end{align*}[/latex]
[latex]\begin{align*}\text{Statutory Holiday Worked Rate}&=\$32.16\times 2.5\\\text{Statutory Holiday Worked Rate}&=\$80.40\end{align*}[/latex]
[latex]\begin{align*}\text{Statutory Worked Earnings}&=7.25\times\$80.40\\\text{Statutory Worked Earnings}&=\$582.90\end{align*}[/latex]
[latex]\begin{align*}\text{Regular Earnings}&=29\times\$32.16\\\text{Regular Earnings}&= \$932.64\end{align*}[/latex]
Table 3.3.5
Type Sun Mon Tue Wed Thu Fri Sat Total Rate Earnings
Regular 0 7.25 7.25 7.25 7.25 $932.64
36.25 $32.16
Holiday 7.25 $233.16
Overtime 1.75 3 4.75 $48.24 $229.14
Holiday Worked 7.25 7.25 $80.40 $582.90
TOTAL EARNINGS $1,977.84
Step 4: Provide the information in a worded statement.
Marcia will receive total gross earnings of [latex]\$1,977.84[/latex] for the week.
Figure 3.3.1
Over the last two weeks you sold [latex]\$50,000[/latex] worth of machinery as a sales representative for IKON Office Solutions Canada. IKON’s compensation plan involves a straight commission rate of
[latex]3.5\%[/latex]. What are your gross earnings? If you sold an additional [latex]\$12,000[/latex] in machinery, how much more would you earn?
Particularly in the fields of marketing and customer service, many workers are paid on a commission basis. A commission is an amount or a fee paid to an employee for performing or completing some
form of transaction. The commission typically takes the form of a percentage of the dollar amount of the transaction. Marketing and customer service industries use this form of compensation as an
incentive to perform: If the representative doesn’t sell anything then the representative does not get paid. Issues to be discussed about commission include what constitutes regular earnings, how to
handle holidays and overtime, and the three different types of commission structures.
• Regular Earnings. All commissions are considered to be regular earnings. To calculate the gross earnings for an employee, take the total amount of the transactions and multiply it by the
commission rate:
[latex]\text{Gross Earnings}=\text{Total Transaction Amount}\times\text{Commission Rate}[/latex]
This is not a new formula. It is a specific application of Formula 3.1b[latex]\begin{align*}\text{Rate}=\frac{\text{Portion}}{\text{Base}}\end{align*}[/latex]: Rate, Portion, Base. In this case,
the Base is the total amount of the transactions, the Rate is the commission rate, and the Portion is the gross earnings for the employee.
• Holidays and Overtime. Commission earners are eligible to receive overtime earnings, holiday earnings, and statutory holiday worked earnings. However, the provincial standards on these matters
vary widely and the mathematics involved do not necessarily follow any one procedure or calculation. As such, this textbook leaves these issues to be covered in a payroll administration course.
• Types of Commission. Commission earnings typically follow one of the following three structures:
□ Straight Commission. If your entire earnings are based on your dollar transactions and calculated strictly as a percentage of the total, you are on straight commission. An application of
Formula 3.1b[latex]\begin{align*}\text{Rate}=\frac{\text{Portion}}{\text{Base}}\end{align*}[/latex] (Rate, Portion, Base) calculates your gross earnings under this structure.
□ Graduated Commission. Within a graduated commission structure, you are offered increasing rates of commission for higher levels of performance. The theory behind this method of compensation
is that the higher rewards motivate employees to perform better. An example of a graduated commission scale is found in the table below.
Table 3.3.6
Transaction Level Commission Rate
$0–$100,000.00 3%
$100,000.01–$250,000.00 4.5%
$250,000.01–$500,000.00 6%
Over $500,000.00 7.5%
Recognize that the commission rates are applied against the portion of the sales falling strictly into the particular category, not the entire balance. Thus, if the total sales equal [latex]\
$150,000[/latex], then the first [latex]\$100,000[/latex] is paid at [latex]3\%[/latex] while the next [latex]\$50,000[/latex] is paid at [latex]4.5\%[/latex].
□ Salary Plus Commission. If your earnings combine a basic salary together with commissions on your dollar transactions, you have a salary plus commission structure. No new mathematics are
required for this commission type. You must combine salary calculations, discussed earlier in this section, with either a straight commission or graduated commission, as discussed above.
Usually this form of compensation pays the lowest commission rate since a basic salary is already provided.
Calculate Commission Earnings
Follow these steps to calculate commission earnings:
Step 1: Determine which commission structure is used to pay the employee. Identify information on commission rates, graduated scales, and any salary.
Step 2: Determine the dollar amounts that are eligible for any particular commission rate and calculate commissions.
Step 3: Sum all earnings from every eligible commission rate plus any salary.
Figure 3.3.2
Table 3.3.7 Data Table for
Figure 3.3.2
Commission Scale Amount Earned
3% $3,000
4.5% $6,750
6% $7,800
Total $17,550
Let’s assume [latex]\$380,000[/latex] of merchandise is sold. Using the previous table as our graduated commission scale, calculate commission earnings.
Step 1: Sales total [latex]\$380,000[/latex] and all commission rates and scales are found in the table.
Step 2: The first [latex]\$100,000[/latex] is compensated at [latex]3\%[/latex], equaling [latex]\$3,000[/latex]. The next [latex]\$150,000[/latex] is compensated at [latex]4.5\%[/latex], equaling
[latex]\$6,750[/latex]. The last [latex]\$130,000[/latex] is compensated at [latex]6\%[/latex], equaling [latex]\$7,800[/latex]. There is no compensation at the [latex]7.5\%[/latex] level since sales
did not exceed [latex]\$500,000[/latex].
Step 3: The total commission on sales of [latex]\$380,000[/latex] is [latex]\$3,000+\$6,750+\$7,800=\$17,550[/latex].
Josephine is a sales representative for Kraft Foods Canada. Over the past two weeks, she closed [latex]\$325,000[/latex] in retail distribution contracts. Calculate the total gross earnings that
Josephine earns if
a. She is paid a straight commission of [latex]3.45\%[/latex].
b. She is paid [latex]2\%[/latex] for sales on the first [latex]\$100,000[/latex], [latex]3\%[/latex] on the next [latex]\$100,000[/latex], and [latex]4\%[/latex] on all remaining sales.
c. She is paid a base salary of [latex]\$2,000[/latex] plus a commission of [latex]3.5\%[/latex] on all sales above [latex]\$100,000[/latex].
Step 1: What are we looking for?
You need to calculate the gross earnings, or [latex]\text{GE}[/latex], for Josephine under various commission structures.
Step 2: What do we already know?
For all three options, [latex]\text{total sales}=\$325,000[/latex]
a. This is a straight commission where [latex]\text{Commission Rate=3.45}\%[/latex].
b. This is a graduated commission, structured as follows:
Table 3.3.8
Sales Level Rate
[latex]\$0\text{-}\$100,000.00[/latex] [latex]2\%[/latex]
[latex]\$100,000.01\text{-}\$200,000.00[/latex] [latex]3\%[/latex]
[latex]\$200,000.01\;\text{and above}[/latex] [latex]4\%[/latex]
c. This is a salary plus graduated commission, with [latex]\text{ Base Salary}=\$2,000[/latex] and a graduated commission structure as follows:
Table 3.3.9
Sales Level Rate
[latex]\$0\text{-}\$100,000.00[/latex] [latex]0\%[/latex]
[latex]\$100,000.01\;\text{and above}[/latex] [latex]3.5\%[/latex]
Step 3: Make substitutions using the information known above.
Determine the sales eligible for each commission and calculate total commissions, then sum all commissions plus any salary to calculate total gross earnings.
[latex]\begin{align*}\text{Total Gross Earnings}&=\text{Rate}\times\text{Base}\\\text{Total Gross Earnings}&=3.45\%\times\$325,000\\\text{Total Gross Earnings}&=0.0345\times\$325,000\\\text{Total
Gross Earnings}&=\$11,212.50\end{align*}[/latex]
Table 3.3.10
Sales Level Rate Eligible Sales at Each Rate (Base) Commission Earned
[latex]\$0\text{–}\$100,000.00[/latex] [latex]2\%[/latex] [latex]\Large{\$100,000-\$0\\=\$100,000}[/latex] [latex]\Large{0.02\times\$100,000\\=\$2,000}[/latex]
[latex]\$100,000.01\text{–}\$200,000.00[/latex] [latex]3\%[/latex] [latex]\Large{\$200,000-\$100,000\\=\$100,000}[/latex] [latex]\Large{0.03\times\$100,000\\=\$3,000}[/latex]
[latex]\$200,000.01\;\text{and above}[/latex] [latex]4\%[/latex] [latex]\Large{\$325,000–\$200,000\\=\$125,000}[/latex] [latex]\Large{0.04\times\$125,000\\=\$5,000}[/latex]
[latex]\begin{align*}\text{Total Gross Earnings}&=\$2,000+\$3,000+\$5,000\\\text{Total Gross Earnings}&=\$10,000\end{align*}[/latex]
c. Recall [latex]\text{Base Salary}=\$2,000[/latex]
Table 3.3.11
Sales Level Rate> Base Commission Earned
[latex]\$0\text{–}\$100,000.00[/latex] [latex]0\%[/latex] [latex]\Large{\$100,000−\$0\\=\$100,000}[/latex] [latex]\Large{0\times\$100,000\\=\$0}[/latex]
[latex]\$100,000.01\;\text{and above}[/latex] [latex]3.5\%[/latex] [latex]\Large{\$325,000–\$100,000\\=\$225,000}[/latex] [latex]\Large{0.035\times\$225,000\\=\$7,875}[/latex]
[latex]\begin{align*}\text{Total Gross Earnings}&=\$0+\$7,875+\$2,000\\\text{Total Gross Earnings}&=\$9,875\end{align*}[/latex]
Step 4: Provide the information in a worded statement.
If Josephine is paid under straight commission, her total gross earnings will be [latex]\$11,212.50[/latex]. Under the graduated commission, she will receive [latex]\$10,000[/latex] in total gross
earnings. For the salary plus commission, she will receive [latex]\$9,875[/latex] in total gross earnings.
Figure 3.3.3
Have you ever heard the phrase “pay-for-performance”? Although this phrase has many interpretations in different industries, for some people this phrase means that they get paid based on the quantity
of work that they do. For example, many workers in clothing manufacturing are paid a flat rate for each article of clothing they produce. As another example, employees in fruit orchards may get paid
by the number of pieces of fruit that they harvest, or simply by the kilogram. As you can see, these workers are neither salaried nor paid hourly, nor are they on commission. They earn their
paycheque for performing a specific task. Therefore, a piecework wage compensates such employees on a per-unit basis.
This section focuses on the regular earnings only for piecework wage earners. Similar to workers on commission, piecework earners are eligible to receive overtime earnings, holiday earnings, and
statutory holiday worked earnings. However, the standards vary widely from province to province, and there is not necessarily any one formula to calculate these earnings. As with commissions, this
textbook leaves those calculations for a payroll administration course.
To calculate the regular gross earnings for a worker paid on a piecework wage, you require the piecework rate and how many units they are to be paid for:
[latex]\text{Gross Earnings}=\text{Piecework Rate}\times\text{Eligible Units}[/latex]
This is not a new formula but another application of Formula 3.1b[latex]\begin{align*}\text{Rate}=\frac{\text{Portion}}{\text{Base}}\end{align*}[/latex] (Rate, Portion, Base). The [latex]\text
{Piecework Rate}[/latex] is the [latex]\text{Rate}[/latex], the [latex]\text{Eligible Units}[/latex] are the [latex]\text{Base}[/latex], and the [latex]\text{Gross Earnings}[/latex] are the [latex]\
Calculate Earnings for Piecework
To calculate an employee’s gross earnings for piecework, follow these steps:
Step 1: Identify the piecework rate and the level of production or units.
Step 2: Perform any necessary modifications on the production or units to match how the piecework is paid.
Step 3: Calculate the commission gross earnings by multiplying the rate by the eligible units.
Assume that Juanita is a piecework earner at a blue jean manufacturer. She is paid daily and earns [latex]\$1.25[/latex] for every pair of jeans that she sews. On a given day, Juanita sewed [latex]93
[/latex] pairs of jeans. Her gross earnings are calculated as follows:
Step 1: Her [latex]\text{Piecework Rate}=\$1.25\;\text{per pair}[/latex] with production of [latex]93\;\text{units}[/latex].
Step 2: The rate and production are both expressed by the pair of jeans. No modification is necessary.
Step 3: Her gross piecework earnings are the product of the rate and units produced, or [latex]\$1.25\times 93=\$116.25[/latex].
Things To Watch Out For
Pay careful attention to Step 2 in the procedure. In some industries, the piecework rate and the units of production do not match. For example, a company could pay a piecework rate per kilogram, but
a single unit may not represent a kilogram. This is typical in some canning industries, where workers are paid per kilogram for canning the products, but the cans may only be [latex]200\;\text{grams}
[/latex] in size. Therefore, if workers produce five cans, they are not paid for five units produced. Rather, they are paid for only one unit produced since [latex]\text{five cans}\times 200\;\text
{g}=1,000\;\text{g}=1\;\text{kg}[/latex]. Before calculating piecework earnings, ensure that both the piecework rate and the eligible units are in the same terms, whether it be metric tonnes,
kilograms, or otherwise.
In outbound telemarketing, some telemarketers are paid on the basis of “completed calls.” This is not commission since their pay is not based on actually selling anything. Rather, a completed call is
defined as simply any phone call for which the agent speaks with the customer and a decision is reached, regardless of whether the decision was to accept, reject, or request further information. If a
telemarketer produces five completed calls per hour and works [latex]7\frac{1}{2}[/latex]-hour shifts five times per week, what are the total gross earnings she can earn over a biweekly pay period if
her piecework wage is [latex]\$3.25[/latex] per completed call?
Step 1: What are you looking for?
We are looking for the total gross earnings, or [latex]\text{GE}[/latex], for the telemarketer over the biweekly pay period.
Step 2: What do you already know?
The frequency of the telemarketer’s pay, along with her hours of work, piecework wage, and unit of production are known:
[latex]\begin{align*}\text{Piecework Rate}&=\$3.25\;\text{per completed call}\\\text{Hourly Units Produced}&=5\\\text{Hours of Work}&=7\frac{1}{2}\;\text{hours per day, five days per week}\\\text
{Frequency of Pay}&=\text{biweekly}\end{align*}[/latex]
Step 3: Make substitutions using the information known above.
You must determine the telemarketer’s production level. Calculate how many completed calls she achieves per biweekly pay period:
[latex]\begin{align*}\text{Eligible Units}&=\text{Units Produced per Hour}\times\text{Hours per Day}\times\text{Days per Week}\times\text{Weeks}\\\text{Eligible Units}&=1. 5\times 7.5\times 5\times 2
\\\text{Eligible Units}&=375\end{align*}[/latex]
Apply Formula 3.1b[latex]\begin{align*}\text{Rate}=\frac{\text{Portion}}{\text{Base}}\end{align*}[/latex] (adapted for piecework wages) to get the portion owing.
[latex]\begin{align*}\text{GE}&=\text{Piecework Rate}\times\text{Eligible Units}\\\text{GE}&=\$3.25\times 375\\\text{GE}&=\$1,218.75\end{align*}[/latex]
Step 4: Provide the information in a worded statement.
Over a biweekly period, the telemarketer completes [latex]375[/latex] calls. At her piecework wage, this results in total gross earnings of [latex]\$1,218.75[/latex].
Section 3.3 Exercises
1. Laars earns an annual salary of[latex]\$60,000[/latex]. Determine his gross earnings per pay period under each of the following payment frequencies:
a. Monthly
b. Semi-monthly
c. Biweekly
d. Weekly
2. A worker earning [latex]\$13.66[/latex] per hour works [latex]47[/latex] hours in the first week and [latex]42[/latex] hours in the second week. What are his total biweekly earnings if his
regular workweek is [latex]40[/latex] hours and all overtime is paid at [latex]1.5[/latex] times his regular hourly rate?
3. Marley is an independent sales agent. He receives a straight commission of [latex]15\%[/latex] on all sales from his suppliers. If Marley averages semi-monthly sales of [latex]\$16,000[/latex],
what are his total annual gross earnings?
4. Sheila is a life insurance agent. Her company pays her based on the annual premiums of the customers that purchase life insurance policies. In the last month, Sheila’s new customers purchased
policies worth [latex]\$35,550[/latex] annually. If she receives [latex]10\%[/latex] commission on the first [latex]\$10,000[/latex] of premiums and [latex]20\%[/latex] on the rest, what are her
total gross earnings for the month?
5. Tuan is a telemarketer who earns [latex]\$9.00[/latex] per hour plus [latex]3.25\%[/latex] on any sales above [latex]\$1,000[/latex] in any given week. If Tuan works [latex]35[/latex] regular
hours and sells [latex]\$5,715[/latex], what are his gross earnings for the week?
6. Adolfo packs fruit in cans on a production line. He is paid a minimum wage of [latex]\$9.10[/latex] per hour and earns [latex]\$0.09[/latex] for every can packed. If Adolfo manages to average
[latex]160[/latex] cans per hour, what are his total gross earnings daily for an eight-hour shift?
1a. [latex]\$5,000[/latex]
1b. [latex]\$2,500[/latex]
1c. [latex]\$2,307.69[/latex]
1d. [latex]\$1,153.85[/latex]
2. [latex]\$1,277.21[/latex]
3. [latex]\$57,600[/latex]
4. [latex]\$6,110[/latex]
5. [latex]\$468.24[/latex]
6. [latex]\$188[/latex]
7. Charles earns an annual salary of [latex]\$72,100[/latex] paid biweekly based on a regular workweek of [latex]36.25[/latex] hours. His company generously pays all overtime at twice his regular
wage. If Charles worked [latex]85.5[/latex] hours over the course of two weeks, what are his gross earnings?
8. Armin is the payroll administrator for his company. In looking over the payroll, he notices the following workweek (from Sunday to Saturday) for one of the company’s employees: [latex]0[/latex],
[latex]6[/latex], [latex]8[/latex], [latex]10[/latex], [latex]9[/latex], [latex]8[/latex], and [latex]9[/latex] hours, respectively. Monday was a statutory holiday, and with business booming the
employee will not be given another day off in lieu. Company policy pays all overtime at time-and-a-half, and all hours worked on a statutory holiday are paid at twice the regular rate. A normal
workweek consists of five, eight-hour days. If the employee receives [latex]\$22.20[/latex] per hour, what are her total weekly gross earnings?
9. In order to motivate a manufacturer’s agent to increase his sales, a manufacturer offers monthly commissions of [latex]1.2\%[/latex] on the first [latex]\$125,000[/latex], [latex]1.6\%[/latex] on
the next[latex]\$150,000[/latex], [latex]2.25\%[/latex] on the next [latex]\$125,000[/latex], and [latex]3.75\%[/latex] on anything above. If the agent managed to sell [latex]\$732,000[/latex] in
a single month, what commission is he owed?
10. Humphrey and Charlotte are both sales representatives for a pharmaceutical company. In a single month, Humphrey received [latex]\$5,545[/latex] in total gross earnings while Charlotte received
[latex]\$6,388[/latex] in total gross earnings. In sales dollars, how much more did Charlotte sell if they both received [latex]5\%[/latex] straight commission on their sales?
11. Mayabel is a cherry picker working in the Okanagan Valley. She can pick [latex]17[/latex] kg of cherries every hour. The cherries are placed in pails that can hold [latex]13.6[/latex] kg of
cherries. If she works [latex]40[/latex] hours in a single week, what are her total gross earnings if her piecework rate is [latex]\$17.00[/latex] per pail?
12. Miranda is considering three relatively equal job offers and wants to pick the one with the highest gross earnings. The first job is offering a base salary of [latex]\$1,200[/latex] semi-monthly
plus [latex]2\%[/latex] commission on monthly sales. The second job offer consists of a [latex]9.75\%[/latex] straight commission. Her final job offer consists of monthly salary of [latex]\$1,620
[/latex] plus [latex]2.25\%[/latex] commission on her first [latex]\$10,000[/latex] in monthly sales and [latex]6\%[/latex] on any monthly sales above that amount. From industry publications, she
knows that a typical worker can sell [latex]\$35,000[/latex] per month. Which job offer should she choose, and how much better is it than the other job offers?
13. A Canadian travel agent is paid a flat rate of [latex]\$37.50[/latex] for every vacation booked through a certain airline. If the vacation is in North America, the agent also receives a
commission of [latex]2.45\%[/latex]. If the vacation is international, the commission is [latex]4.68\%[/latex]. What are the total monthly gross earnings for the agent if she booked [latex]29[/
latex] North American vacations worth [latex]\$53,125[/latex] and [latex]17[/latex] international vacations worth [latex]\$61,460[/latex]?
14. Vladimir’s employer has just been purchased by another organization. In the past, he has earned [latex]\$17.90[/latex] per hour and had a normal workweek of [latex]37.5[/latex] hours. However,
his new company only pays its employees a salary semi-monthly. How much does Vladimir need to earn each paycheque to be in the same financial position?
7. [latex]\$3,767.58[/latex]
8. [latex]\$1,554[/latex]
9. [latex]\$19,162.50[/latex]
10. [latex]\$16,860[/latex]
11. [latex]\$850[/latex]
12. Best is [latex]\text{Offer #2}=\$3,412.50[/latex]; Exceeds[latex]\text{Offer #1}=\$312.50[/latex]; Exceeds [latex]\text{Offer #3}=\$67.50[/latex]
13. [latex]\$5,902.89[/latex]
14. [latex]\$1,454.38[/latex]
Challenge, Critical Thinking, & Other Applications
15. An employee on salary just received his biweekly paycheque in the amount of [latex]\$1,832.05[/latex], which included pay for five hours of overtime at time-and-a-half. If a normal workweek is
[latex]40[/latex] hours, what is the employee’s annual salary?
16. A graduated commission scale pays [latex]1.5\%[/latex] on the first [latex]\$50,000[/latex], [latex]2.5\%[/latex] on the next [latex]\$75,000[/latex], and [latex]3.5\%[/latex] on anything above.
What level of sales would it take for an employee to receive total gross earnings of [latex]\$4,130[/latex]?
17. A sales organization pays a base commission on the first [latex]\$75,000[/latex] in sales, base [latex]+2\%[/latex] on the next [latex]\$75,000[/latex] in sales, and base [latex]+4\%[/latex] on
anything above. What is the base commission if an employee received total gross earnings of [latex]\$7,500[/latex] on [latex]\$200,000[/latex] in sales?
18. A typical sales agent for a company has annual sales of [latex]\$4,560,000[/latex], equally spread throughout the year, and receives a straight commission of [latex]2\%[/latex]. As the new human
resource specialist, to improve employee morale you have been assigned the task of developing different pay options of equivalent value to offer to the employees. Your first option is to pay them
a base salary of [latex]\$2,000[/latex] per month plus commission. Your second option is to pay a base commission monthly on their first [latex]\$200,000[/latex] in sales, and a base [latex]+2.01
\%[/latex] on anything over [latex]\$200,000[/latex] per month. In order to equate all the plans, determine the required commission rates, rounded to two decimals in percent format, in both
19. Shaquille earns an annual salary of [latex]\$28,840.50[/latex] paid biweekly. His normal workweek is [latex]36.25[/latex] hours and overtime is paid at twice the regular rate. In addition, he is
paid a commission of [latex]3\%[/latex] of sales on the first [latex]\$25,000[/latex] and [latex]4\%[/latex] on sales above that amount. What are his total gross earnings during a pay period if
he worked [latex]86[/latex] hours and had sales of [latex]\$51,750[/latex]?
20. Mandy is paid [latex]\$9.50[/latex] per hour and also receives a piecework wage of [latex]\$0.30[/latex] per kilogram, or portion thereof. A regular workday is [latex]7.5[/latex] hours and
[latex]37.5[/latex] hours per week. Overtime is paid at time-and-a-half, and any work on a statutory holiday is paid at twice the regular rate. There is no premium piecework wage. Mandy’s work
record for a two-week period is listed below. Determine her total gross earnings.
Table 3.3.12
Week Monday Tuesday Wednesday Thursday Friday Saturday
Hours worked 7.5 7.5 9 8 7.5 3
250 g items produced 1,100 1,075 1,225 1,150 1,025 450
Hours worked 7.5 10 7.5 8
2 Statutory holiday, no day off in lieu
250 g items produced 575 1,060 1,415 1,115 1,180
15. [latex]\$43,550.45[/latex]
16. [latex]\$168,000[/latex]
17. [latex]2\%[/latex]
18. [latex]\text{Option 1}=1.47\%[/latex]; [latex]\text{Option 2}=1.05\%[/latex]
19. [latex]\$3,342.35[/latex]
20. [latex]\$1,755.25[/latex]
THE FOLLOWING LATEX CODE IS FOR FORMULA TOOLTIP ACCESSIBILITY. NEITHER THE CODE NOR THIS MESSAGE WILL DISPLAY IN BROWSER.[latex]\small\text{GE}=\text{Regular}+\text{OT}+\text{Holiday}+\text{Stat
“4.1: Gross Earnings” from Business Math: A Step-by-Step Handbook (2021B) by J. Olivier and Lyryx Learning Inc. through a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International
License unless otherwise noted. | {"url":"https://ecampusontario.pressbooks.pub/introbusinessmath/chapter/3-3-payroll/","timestamp":"2024-11-10T09:45:59Z","content_type":"text/html","content_length":"205639","record_id":"<urn:uuid:91088da8-b444-46fa-b946-8a28040bb4c2>","cc-path":"CC-MAIN-2024-46/segments/1730477028179.55/warc/CC-MAIN-20241110072033-20241110102033-00474.warc.gz"} |
Fundamental properties of solar-like oscillating stars from frequencies of minimum Δν - II. Model computations for different chemical compositions and mass
MONTHLY NOTICES OF THE ROYAL ASTRONOMICAL SOCIETY, vol.448, no.4, pp.3689-3696, 2015 (SCI-Expanded)
• Publication Type: Article / Article
• Volume: 448 Issue: 4
• Publication Date: 2015
• Doi Number: 10.1093/mnras/stv295
• Journal Name: MONTHLY NOTICES OF THE ROYAL ASTRONOMICAL SOCIETY
• Journal Indexes: Science Citation Index Expanded (SCI-EXPANDED), Scopus
• Page Numbers: pp.3689-3696
• Kayseri University Affiliated: No
The large separations between the oscillation frequencies of solar-like stars are measures of stellar mean density. The separations have been thought to be mostly constant in the observed range of
frequencies. However, detailed investigation shows that they are not constant, and their variations are not random but have very strong diagnostic potential for our understanding of stellar structure
and evolution. In this regard, frequencies of the minimum large separation are very useful tools. From these frequencies, in addition to the large separation and frequency of maximum amplitude,
Yildiz et al. recently have developed new methods to find almost all the fundamental stellar properties. In the present study, we aim to find metallicity and helium abundances from the frequencies,
and generalize the relations given by Yildiz et al. for a wider stellar mass range and arbitrary metallicity (Z) and helium abundance (Y). We show that the effect of metallicity is significant for
most of the fundamental parameters. For stellar mass, for example, the expression must be multiplied by (Z/Z(circle dot))(0.12). For arbitrary helium abundance, M alpha(Y/Y-circle dot)(0.25). Methods
for determination of Z and Y from pure asteroseismic quantities are based on amplitudes (differences between maximum and minimum values of Delta nu) in the oscillatory component in the spacing of
oscillation frequencies. Additionally, we demonstrate that the difference between the first maximum and the second minimum is very sensitive to Z. It also depends on nu(min1)/nu(max) and small
separation between the frequencies. Such a dependence leads us to develop a method to find Z (and Y) from oscillation frequencies. The maximum difference between the estimated and model Z values is
about 14 per cent. It is 10 per cent for Y. | {"url":"https://avesis.kayseri.edu.tr/yayin/13870284-7c87-4142-9bc3-208832ced10b/fundamental-properties-of-solar-like-oscillating-stars-from-frequencies-of-minimum-ii-model-computations-for-different-chemical-compositions-and-mass","timestamp":"2024-11-07T09:35:16Z","content_type":"text/html","content_length":"56180","record_id":"<urn:uuid:d1de2f38-8c2a-4713-b7f8-24e51da7d05d>","cc-path":"CC-MAIN-2024-46/segments/1730477027987.79/warc/CC-MAIN-20241107083707-20241107113707-00108.warc.gz"} |
The Glowing Python
The central limit theorem can be informally summarized in few words: The sum of
x[1], x[2], ... x[n]
samples from the same distribution is normally distributed, provided that
is big enough and that the distribution has a finite variance. to show this in an experimental way, let's define a function that sums
samples from the same distrubution for 100000 times:
import numpy as np
import scipy.stats as sps
import matplotlib.pyplot as plt
def sum_random_variables(*kwarg, sp_distribution, n):
# returns the sum of n random samples
# drawn from sp_distribution
v = [sp_distribution.rvs(*kwarg, size=100000) for _ in range(n)]
return np.sum(v, axis=0)
This function takes in input the parameters of the distrubution, the function that implements the distrubution and n. It returns an array of 100000 elements, where each element is the sum of n
samples. Given the Central Limit Theorem, we expect that the values in output are normally distributed if n is big enough. To verify this, let's consider a beta distribution with parameters alpha=1
and beta=2, run our function increasing n and plot the histogram of the values in output:
plt.figure(figsize=(9, 3))
N = 5
for n in range(1, N):
plt.subplot(1, N-1, n)
s = sum_random_variables(1, 2, sp_distribution=sps.beta, n=n)
plt.hist(s, density=True)
On the far left we have the histogram with n=1 , the one with n=2 right next to it, and so on until n=4. With n=1 we have the original distribution, which is heavily skewed. With n=2 we have a
distribution which is less skewed. When we reach n=4 we see that the distribution is almost symmetrical, resembling a normal distribution.
Let's do the same experiment using a uniform distribution:
plt.figure(figsize=(9, 3))
for n in range(1, N):
plt.subplot(1, N-1, n)
s = sum_random_variables(1, 1, sp_distribution=sps.beta, n=n)
plt.hist(s, density=True)
Here we have that for n=2 the distribution is already symmetrical, resembling a triangle, and increasing n further we get closer to the shape of a Gaussian.
The same behaviour can be shown for discrete distributions. Here's what happens if we use the Bernoulli distribution:
plt.figure(figsize=(9, 3))
for n in range(1, N):
plt.subplot(1, N-1, n)
s = sum_random_variables(.5, sp_distribution=sps.bernoulli, n=n)
plt.hist(s, bins=n+1, density=True, rwidth=.7)
We see again that for n=2 the distribution starts to be symmetrical and that the shape of a Gaussian is almost clear for n=4.
This year I decided to join the March Machine Learning Mania 2021 - NCAAW challenge on Kaggle. It proposes to predict the outcome of each game into the basketball NCAAW tournament, which is a
tournament for women at college level. Participants can assign a probability to each outcome and they're ranked on the leaderboard according to the accuracy of their prediction. One of the most
attractive elements of the challenge is that the leaderboard is updated after each game throughout the tournament.
Since I have limited knowledge of basketball I decided to use a minimalistic model:
• It uses three features that are easy to interpret: seed, percentage of victories, and the average score of each team.
• It is based on linear Linear Regression, and it's tuned to predict extreme probability values only for games that are easy to predict.
The following visualizations give insight into how the model estimates the winning probability in a game between two teams:
Surprisingly, this model ranked 46th out of 451 submissions, placing itself in the top 11% of the leaderboard and earning a silver medal!
The notebook with the solution and some more charts can be found here.
I recently published on a wrapper around The Dictionary of Obscure Words (originally from this website http://phrontistery.info) for Python and in this post we'll see how to create a visualization to
highlight few entries from the dictionary using the dimensionality reduction technique called T-SNE. The dictionary is available on github at this address https://github.com/JustGlowing/obscure_words
and can be installed as follows:
pip install git+https://github.com/JustGlowing/obscure_words
We can now import the dictionary and create a vectorial representation of each word:
import matplotlib.pyplot as plt
import numpy as np
from obscure_words import load_obscure_words
from sklearn.feature_extraction.text import TfidfVectorizer, CountVectorizer
from sklearn.manifold import TSNE
obscure_dict = load_obscure_words()
words = np.array(list(obscure_dict.keys()))
definitions = np.array(list(obscure_dict.values()))
vectorizer = TfidfVectorizer(stop_words=None)
X = vectorizer.fit_transform(definitions)
projector = TSNE(random_state=0)
XX = projector.fit_transform(X)
In the snippet above, we compute a Tf-Idf representation using the definition of each word. This gives us a vector for each word in our dictionary, but each of these vectors has many elements as the
total number of words used in all the definitions. Since we can't plot all the features extracted, we reduce our data to 2 dimensions we use T-SNE. We have now a mapping that allows us to place each
word in a point of a bi-dimensional space. There's one problem remaining, how can we plot the words in a way that we can still read them? Here's a solution:
from sklearn.cluster import KMeans
from sklearn.preprocessing import StandardScaler
from sklearn.metrics import pairwise_distances
def textscatter(x, y, text, k=10):
X = np.array([x, y]).T
clustering = KMeans(n_clusters=k)
scaler = StandardScaler()
centers = scaler.inverse_transform(clustering.cluster_centers_)
selected = np.argmin(pairwise_distances(X, centers), axis=0)
plt.scatter(x, y, s=6, c=clustering.predict(scaler.transform(X)), alpha=.05)
for i in selected:
plt.text(x[i], y[i], text[i], fontsize=10)
plt.figure(figsize=(16, 16))
textscatter(XX[:, 0], XX[:, 1],
[w+'\n'+d for w, d in zip(words, definitions)], 20)
In the function textscatter we segment all the points created at the previous steps in k clusters using K-Means, then we plot the word related to the center of cluster (and also its definion). Given
the properties of K-Means we know that the centers are distant from each other and with the right choice of k we can maximize the number of words we can display. This is the result of the snippet
(click on the figure to see the entire chart) | {"url":"https://glowingpython.blogspot.com/","timestamp":"2024-11-03T03:07:01Z","content_type":"application/xhtml+xml","content_length":"97988","record_id":"<urn:uuid:12ca0679-0b45-4ae7-b3cb-37e05819a24e>","cc-path":"CC-MAIN-2024-46/segments/1730477028187.61/warc/CC-MAIN-20241110170046-20241110200046-00775.warc.gz"} |
Decimal Place Value: A Culturally Responsive Approach
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Decimal Place Value: A Culturally Responsive Approach
byKrystal L. Smith
They were unmotivated, effort-deficient, and badly behaved this past year. I was unexpectedly met with a group of 5^th grade demoralized students who had given up on math and themselves. It showed up
in every aspect it possibly could, but mostly in their words to me about themselves. It hurt to hear them refer to themselves as the “dumb class.” They often believed every negative thing they heard
about themselves and ignored anything positive. In all my ten years of teaching (in another district), I had never encountered such negativity. I didn’t know what to do, but I knew giving up was not
an option. So here I am, developing this unit.
“I know I can-- be what I want to be. If I work hard at it, I’ll be where I want to be.” These lyrics are from a song titled “I Can,” by a rap artist that goes by the name Nas. It was released in
2002 but remains relevant for students today. While children continue to face adversity in urban communities, the song is empowering. My students need to know that I believe in them no matter the
circumstance. I want to inspire them to have a strong conviction in themselves.
I intend to put things in place to empower my scholars. To encourage and inspire them to believe in themselves, this unit will be culturally responsive. Zaretta P. Hammond recommends three ways to do
this.^1 She says to “Gamify,” “Make it Social,” and “Storify” lessons to make them more culturally responsive. While being culturally responsive does not solely mean using racial pride to motivate
students, I will focus on positive self-identity, purpose, and hope by connecting math to the past, present, and future of my students’ race and culture. Given their lack of interest in math, it is
my hope that making these connections will increase their motivation, effort, and positive behavior which will then enable increased math development. This will also help foster stronger
relationships with my scholars, which essentially makes teaching so amazing!
Social connections such as friendships, teacher-student relationships and closeness to the community, are known to be closely related to well-being and personal happiness. A lack of social connection
can often put people at risk for bad habits. In this case, my scholars struggle in math and have some poor academic habits. They see math as irrelevant and too difficult. It is my firm belief that
students need to see that math is important in their day-to-day lives and experiences, or they become disengaged and uninterested, as is the case. In “Mathematical Mindsets,” Jo Boaler says, “Over
the years, school mathematics has become more and more disconnected from the mathematics that mathematicians use and the mathematics of life.^2 Students spend thousands of hours in classrooms
learning sets of procedures and rules that they will never use, in their lives or in their work.” It is my goal with this unit for students to see and do the math in their world.
I want to help my students achieve a wholesome and more conceptual understanding of place value. Place value includes not only the position and the value of digits but also the decomposition of
numbers and a number’s relationship to other numbers in the base ten system. Place value is the foundation of basic mathematics and, if not mastered in the elementary grades, often makes working with
operations more difficult as well as learning higher levels of mathematics much more challenging. By focusing on decimal place value in my students’ world by connecting to the past, present, and
future, I hope to develop and increase a positive self-identity, purpose, and hope in all my scholars.
In ““Multiplication is for White People:” Raising Expectations for Other People’s Children,” Lisa Delpit proposes that one reason “why African American students are not achieving at levels
commensurate with their ability,” is due to the curricular content.^3 “If the curriculum we use to teach our children does not connect in positive ways to the culture young people bring to school, it
is doomed to failure.” Additionally, in “What is Mathematics, Really?,” Reuben Hersh says, “If mathematics is conceived apart from mathematical life, of course it seems – dead.”^4 I want my scholars
alive and ready for the world socially and mathematically!
I teach 4th and 5^th students math and social studies in the urban setting of the Pittsburgh Public School District. My building is the largest K-5 Elementary School in the city with approximately
500 students. The community is predominantly African American (approximately 95%) and the school percentage is similar. As a neighborhood and community school, only students living in the community
can attend. The school provides many needs including dental, eye, mental and other health screenings. The school has partnerships with various outside organizations that provide services including,
but not limited to: mental health, mentorship, gardening, ESTEAM (Entrepreneurship, Science, Technology, Engineering, Art, and Math) activities, abuse prevention, conflict management, therapeutic
services, a college prep program, and academic and tutoring support. Despite having school services and community resources such as a community college branch campus, churches, a YMCA, restaurants,
convenience stores, barbershops, and other facilities and businesses, the community where my school is located is impoverished. Nearly 90% of the students in the building are considered economically
disadvantaged. The median income is approximately $29,000 which is barely half of Pittsburgh’s median of $56,000.^5 The community has high unemployment, poverty, & crime rates. In the 2016-2017
school year, 84% of the third graders, 87% of the 4^th graders and 95% of the 5^th graders scored below grade level in mathematics.
This year, I anticipate a similar group of students, but the difference is, I know many of the students, have a better understanding of how the school works, who the students and their families are,
and what the community has to offer. These are things I had to learn this past year. I now have time to prepare myself, and I am working to set the stage up for my students’ success.
Content Objectives
This 2 to 3-week unit is designed to focus on having my students correctly interpret, use, and compare base ten numbers, especially decimal fractions. Arithmetic will not be covered directly in this
unit. In “Place Value as the Key to Teaching Decimal Operations,” Judith Sowder explains,
More recent research on decimal-number understanding confirms that many students have a weak understanding of decimal numbers. […] The children in these studies were primarily from classes where
the introduction to decimal numbers was brief so that sufficient time would remain for the more difficult work of learning the algorithms for operating on decimal numbers. But time spent on
developing students' understanding of the decimal notation is not time wasted. Teachers with whom I have worked claim that much less instructional time is needed later for operating on decimal
numbers if students first understand decimal notation and its roots in the decimal place-value system we use.^6
In my experience, this is true. It has also been my experience that reteaching my students anything, is challenging, and reteaching students how to operate on decimal numbers in deep, meaningful ways
is no exception. My students are often reluctant or angry when they must relearn what has already been taught, and what they thought they already knew. This unit will attempt to alleviate some of the
reteaching and frustration that may usually occur when working on operations of decimal numbers.
The topics in this unit will follow the Pennsylvania Core State Standards, the Assessment Anchors and Eligible Content Aligned to the Mathematics Pennsylvania Core Standards, and the Pennsylvania
Core State Standards for Mathematical Practice Standards (see Appendix), but the standards are very similar to the Common Core Standards. Additionally, the topics will address the Pittsburgh Public
School District’s Scope and Sequence. The unit is divided into two sections:
• Understanding the base ten place value system
• Comparing Decimal Numbers
Because I feel it is important for students understand where math, and all subjects, fit into their lives, this unit will also empower and elevate 5^th grade students’ conceptual understanding of
decimal place value by:
• Being culturally responsive;
• Implementing goal setting using pre-assessment data;
• Using concrete representations such as base-ten blocks, straws, and meter sticks;
• Using visual representations such as hundredths grids and number lines;
• Using abstract representations such as place value charts;
• Differentiating instruction in small groups to ensure mastery;
• Completing a culminating project-based learning activity and post assessment;
• Using portfolios.
By the end of this unit, students will be able to:
• Explain and illustrate, in various ways, how a digit in one place represents 10 times of what it represents in the place to its right and 1/10 of what it represents in the place to its left,
using place value charts, hundredths grids, and meter sticks. (With adjacent digits such as 6.55 and in separate numbers such as 6.15 and 5.61).
• Read and write decimals to the thousandths using base-ten numerals in various forms including, word form, and expanded form (using decimals, recognizing the multiplicative make-up of the place
value pieces in decimal form and decimal fraction form).
• Compare two decimals to thousandths using the comparison symbols >, =, < to record the results of the comparisons.
Understanding the Base Ten System
The key to understanding the base systems is realizing that digits “roll over” like a clock or an odometer when they are full. Our decimal system is Base Ten, meaning that it uses a system of units,
each of which is ten times as large as the next smallest one, and conversely, one tenth as large as the next largest one. Consequently, we never need more than nine (one less than ten) of any unit to
express a number. When we get to ten of any given unit, it “rolls over” and creates one of the next larger units. Our method of measuring time is similar, but with time, we wait 60 seconds before
“rolling over” to a new minute.
Back in the day, Babylonians used base 60 and Mayans used base 20, but base 10 is what we use today. Perhaps because we have 10 fingers. The Egyptians, Greeks and Romans also used base ten, but they
did not have the idea of place value, in which the position of each digit indicates the unit is multiplying. The Egyptians and Romans had a separate symbol for each of the first few units. The Greek
system was very limited and used their alphabet.
The base ten system is also known as the decimal system. Although the concept is not simple, the system is simply a way of writing numbers. It is called a positional system because each place value
increases by a factor of 10 from the place to the right! Prior to developing this unit, I did not know the decimal or base-ten place value system was based on the Hindu-Arabic derived symbols for the
digits: 0,1,2,3,4,5,6,7,8, and 9. It was introduced in Europe around the 12^th century, and because it was such an organized way to work with numbers, that allowed for counting and calculations in a
more efficient way, it replaced the Roman Numerals, which had been the main method for writing numbers. Place value notation is more efficient than Roman Numerals (compare MMXVII with 2018), but the
big advantage of place value notation was the way it supported computation, especially multiplication!
The Egyptians used Hieroglyphic writing, which included symbols for writing whole numbers up to 1,000,000. This decimal based system allowed for the additive principle: a number was represented by a
collection of symbols whose values added up to the number. This is a way in which numbers can be counted. There was a special symbol for each power of 10. See figure 1 below. To see the Egyptian Base
Ten System in its entirety, please go to the notes section.^7
It is important to mention this historical background here because my students have failed so much that they believe they can’t do math. Even many of their parents and family members make snide
comments such as, “I wasn’t good at math either, so I don’t expect him/her to be that good either.” This belief is extremely deep in many African American students, so it’s important to acknowledge
the rich history of their ancestors and show that math is already in their blood. Again, I am aiming to empower my students to believe they can excel as young mathematicians, and one way to do this
is to include the history of math of their ancestors and other multicultural aspects of mathematics. In the movie, Stand and Deliver, Jaime Escalante held a high conviction in his students, saying
that mathematics was in their blood because their ancestors, the Mayans, were the first to conceptualize the idea of “Zero.” Because he believed in them, and showed them who they were, his students
believed in themselves and achieved at high levels.
Figure 1.
The difference between the Egyptian Base Ten system and the Hindu-Arabic System is that the Egyptians did not have a positional system. In Egyptian hieroglyphics, each power of ten was represented by
a different symbol, whereas 0 played a key role in specifying clearly the positions of the digits in the Hindu-Arabic System. The Base Ten positional system means that the position of a digit gives
its place value. This system was extended from whole numbers to include decimal fractions by Islamic scholars in the 10^th century. For example, a typical person in the community where my school is
located might have a yearly income of $ 28,927.25. A student in 5^th grade should be able to explain the value of the digits in a decimal fraction, such as the number above. For example, a fifth
grader should know that the 2 at the left of that number represents $ 20,000 = 2 × $ 10,000, but the 2 just to the right of the decimal point represents only 2 tenths or a dollar or 2 dimes, or 2 ×
0.1, or 2 × 1/10. They should also be able to explain that the two in the ten thousandths place is 100,000 times larger than the two in the tenths place or that the two in the tenth place is 100,000
times smaller than the two in the thousandths place. Each digit has an explicit value that is determined by its place in the number, and each adjacent digit that is the same is larger or smaller by
some power of 10. A major part of the unit, these concepts will be address in every lesson within the unit as well as operations with decimal numbers and decimal fractions later in the year.
The Five Stages of Place Value
This unit will emphasize the five stages of place value as described in by Howe-Reiter.^8 The stages can be summarized in the sequence of the equivalences below:
= 600 + 20 + 5
= (6 x 100) + (2 x 10) + (5 x 1)
= 6 x (10 x 10) + (2 x 10) + (5 x 1)
= (6 x 10^2) + (2 x 10^1) + (5 x 10^0)
The first stage is what we refer to as the standard form of a number. The second stage recognizes what we call expanded form, and exhibits the number as a sum of pieces, one for each digit, which we
will call place value pieces. The third stage can be referred to as the second expanded form and recognizes the multiplicative make-up of the place value pieces. In this case, you can see the base
ten units, namely 1, 10, and 100 in this example, being multiplied by non-zero digits. The fourth stage is a stage that often gets left out of textbooks and perhaps many classrooms including mine.
But this stage recognizes the multiplicative make-up of the base ten units as a power of 10, or tens being multiplied repeatedly. Lastly, the fifth expression illustrates the structure using the
exponential notation for powers of 10. In Pennsylvania, all 5^th grade students are expected to interact with each stage based on the PA Core Standards. Focusing on the Five Stages of Place Value
will give students a strong background in place value because it allows them to get to know the numbers as made of pieces and to really understand the patterns of the base ten system and the powers
of ten. This unit will discuss the first four stages but will leave the fifth stage for later grade as negative exponents are not a part of the state standards. This concept will be addressed and
practiced in lesson 5.
Place Value
Like the whole number system, the full decimal system uses place value meaning that the place in which a digit is located determines the value of the decimal number. For example, the 3 in 3.19 stands
for three ones, but the 3 in .63 stands for three hundredths. Each digit has a place, and that place determines the value of the digit, or how much it is worth. The chart below shows the value of the
decimal places between one thousandth and one million.
Table 1.
It is a challenge for many students to correctly translate between word form and decimal numbers. For example, in the number 3.19 a student may say and write three point one nine or three point
nineteen which is the literal translation of the symbol instead of the word form signifying the value for three and nineteen hundredths, or three ones and one tenth and 9 one hundredths. Students are
unable to understand the full meaning of positional notation. Many adults use the literal translation as well, but it does not allude to place value, which is important when doing calculations and
operations with decimals. I will do my best to correct this misconception by naming the values consistently, accurately, and often, to reinforce the place value and ensure my students do so as well.
Students need to verbalize each digit as it relates to the order of magnitude. Another important aspect is the word and. It must be used to show the decimal point in a mixed number.
Decimal Numbers
Decimal numbers are an extension of the whole number system, in which negative powers of ten, which represent proper fractions, are separated from the positive powers, which represent whole numbers,
by the decimal point. The decimal point is a symbol that allows the base ten system to express parts of numbers. When describing quantities that have parts less than one, we call them decimal
numbers, or decimal fractions. Students are introduced to three new places on the right of the decimal point by fifth grade: the tenths, hundredths, and thousandths places.
Lesson 3 will address how some students can infer and then continue the pattern to smaller decimal places. This is because each place follows the same word pattern of the base ten number system.
Meaning that, after they learn about the tenths, hundredths, and thousandths, students may understand that ten thousandths, hundred thousandths, millionths, ten millionths, etc. will proceed.
However, this does not mean that the student understands the multiplicative pattern of repeatedly multiplying or dividing by 10. In other words, students may not understand that “each place’s value
is ten times the value of the place to its right,” or 1/10 the value of its place to the left.^9 As a digit moves to the right of the decimal point in a number, the value of each place is divided by
10 or multiplied by 1/10. It is beneficial to imbed this concept into a problem-solving activity. Cognitively demanding tasks that require students to explain their reasoning is an instructional
practice that is beneficial for all students.
Comparing decimal numbers should be easy, but when it comes to comparing or even ordering them in racing something interesting happens. Students assume the person with the largest number wins the
race. But in racing the smaller number is the winner. A good way to illustrate this concept is through an actual race which lesson 8 will address. Many students will visually and physically see who
won the race and know how to order the numbers based on what they have witnessed. Some students will have a difficult time recording and analyzing the times even though they saw the race. Regardless
of how students determine the order, it is imperative for my students to learn to compare each place with the same place in the other number(s) and understand that the largest place is the decision
Decimal Numbers and Money
Decimal numbers can be represented in many ways. Children are often aware of them because they have some experience dealing with money. Money is very relevant to my students. Having it means they can
get what they want and need to survive. Not having it means they may suffer or perhaps end up hungry and homeless. The value of money can be large, and the value of money can be small. But most
students know they need money to get the things they want and need to survive. Simply put, money makes sense (pun intended). But it didn’t always.
Money is a great way to “storify” this unit and connect to the past. Between bartering with objects, salt, silver, gold, cloth, cowry shells, animals, etc., an easier currency was bound to be
developed. No one wants to carry a baby goat in their pocket. But my students know what it means to trade things. According to Richard Pankhurst in “An Introduction to the Economic History of
Ethiopia,” “The earliest example of coins minted in Africa comes from the kingdom of Axum, which struck money from the 3rd until the 8th century CE.”^10 This is pre-slavery. Our school curriculum
does not often discuss Africans before slavery. My students need to know that their ancestry is rich! My students have stereotypes about what Africa is and is not, and many believe that is a poor
region. I have heard negative attitudes and misconceptions shared in the classroom, and believe it or not, I had those same perceptions as a child. Typically, the math in the U.S. is presented in
public school materials as the exclusive creation of men from European ancestry. But through my research, I learned the lattice multiplication method was introduced to Europe by Fibonacci. Although
he was not the only, he was a major source for the use of the decimal system. He was an Italian mathematician who learned how to use Arabic numerals from a Moorish (African) teacher in Bugia, located
in Algeria, in Africa. I will certainly share this knowledge with my students!
While money is familiar to many students, it is not the only way decimals can be represented. Physical representations of decimals using base-ten blocks, meter sticks or other number lines really
help students understand the concept of decimal number units of measure. Lesson 4 will feature money examples.
Metric System
Did you know that the United States is only one of three countries in the world that has not adopted the International Metric System?^11 I feel that if I do not take the time so teach my students
about the metric system, it will do them a great disservice as they grow to explore the world. Focusing on length in lesson 6, I will introduce my students to the metric system, and share that only
the countries of Burma, Liberia, and the US, are the only three countries in the world that have not converted. Imagine how difficult traveling can be when you visit another country without this
background knowledge! The metric system is a natural way to help students understand place value and number lines in a concrete way. Also, measurement is a skill that our students struggle with, and
do not have a lot of time to study due the time of year in which it is introduced. I feel that if some of the basic concepts of the metric system are taught earlier and included within another
standard, students are more likely to remember. Table 2 below illustrates the metric units of length. It uses a place value chart which my students will already be familiar with, as well as the
mnemonic device, “King Henry Does Usually Drink Chocolate Milk,” to help them remember how to convert units.
Table 2
Conceptual understanding that connects the notation with the value being represented is key for students to fully grasp decimal numbers.
I will encourage my scholars and young budding mathematicians to think and ask questions such as, “How can I make this easier to deal with?” Generally, mathematicians look for ways to make
complicated ideas easier and more familiar for them to deal with.
Expanded Form
The expanded form of a number shows the value of each digit in a number when in a place and decomposes each non-zero digit to represent a sum. For example, let’s consider the following: A gallon of
gas at Sunoco costs $3.249. Model this number using base-ten blocks. Then write this number in word form, standard form, and the two expanded forms.
This is a typical problem a student may need to answer. Students in fifth grade are not required to write exponential notation of decimals which result in negative exponents. The value of these
digits is 3.249 and can be understood to represent a sum:
3.249 = 3 + .2 + .04 + .009
The expression on the right side of this equation is the expanded form of 3.249. Each number uses the place value system, which means to place a zero in the places where the non-zero digits are
located. When writing numbers in expanded form, students often have some misconceptions especially when it comes to fractional parts. They may show this number in two different ways. One way is 3,000
+ 200 + 40 + 9. If a student records the number this way, they are disregarding the decimal point and reading these digits as whole numbers instead of a decimal number.
To guide a student that disregards decimal points, it would be helpful to ask them about it. I would ask these series of questions during small group time when I provide differentiated instruction to
my students. Do you see a decimal point or a comma? Can you read this number to me? Read it out loud to me. How much money would this be? Can you use base ten blocks to create this number? I would
have the student build the number. Which base ten block represents ones, tenths, hundredths, thousandths? Can you show me how to represent this number using money? How may dollars, dimes, and
pennies? How come the nine cannot be represented using money? Can you write these digits on a place value chart? What is the value of each digit? How many ones, tenths, hundredths, and thousandths
did you write in each place?
Students also have another misconception once they become aware of the decimal point. They will write and recognize that three is a whole number, but two tenths, four hundredths and nine thousandths
may be written in the same place. For example: .2 + .40 + .900 instead of 0.2+ .04 + .009. While they are now aware there is a decimal point, they do not understand that each digit has its own
special name and place. They believe that these digits have the same quantity or order of magnitude. When digits have the same order of magnitude, they are said to have the same quantity of “powers
of 10” that there are in a number. For example, the misconception of the value of the digit 4 in the number 3.249 has a value of 0.4 = 4/10 and has a magnitude of 4 x 1/10 = (4 x 10^-1) and the
actual value of the 4 in the given number is 0.04 = 4/100 and has a magnitude of 4 x 1/100 = (4 x 10^-2). The difference in the order of magnitude in these two numbers is 1, and 0.4 is 10 x greater
than 0.04.
This misconception stems from students work with whole numbers, and the belief that all one must do to write a number in expanded form is to annex zeros to the non-zero digit. This is another reason
why word form and patterns must be emphasized and practice with models must occur.
Expanded form does not stop here. In fifth grade, students much be able to decompose numbers even further. In this second version of expanded form, students are essentially breaking the numbers down
into base ten pieces. Students should be aware of the identity property at this time meaning that if a multiplicand is 7 and the multiplier 1, the product will be 7. In other words, the product of
any number multiplied by 1 is always equal to the given number. With that, I want my students to have experience with knowing 10 x 7 = 70, 100 x 6 = 600, but also that 0.2 = 1/10 x 2 and 0.04 = 1/100
x 4. I want my students to realize that the number of zeros in the standard or expanded form of the units tell you the value one should multiply by.
Using the approximate cost of gas mentioned above, here is the second expanded form using multiplication:
3.249 = (3 × 1) + (2 × 0.1) + (4 × 0.01) + (9 × .001)
Please note the following equivalencies
2 × 0.1 4 × 0.01 9 × 0.001
are equivalent to
2 × 1/10 4 × 1/100 9 × 1/1000
2 × 1/10 4 × 1/(10 × 10) 9 × 1/(10 × 10 × 10)
I will develop and expect my students to be able to write and read these equivalences. Another version of expanded form explains how many times to repeatedly add unit fractions based on the non-zero
digits in the standard or first version of expanded form. Showing students this version will help them connect what they already know about the relationship between repeated addition and
multiplication. It is always important to recognize and build on children’s background knowledge and strengths. I will also explain that due to its length, we rarely want to write expanded form in
this way.
3.249 = (1 + 1 + 1) + (1/10 + 1/10) + (1/100 + 1/100 + 1/100 + 1/100) + (1/1000 + 1/1000 + 1/1000 + 1/1000 + 1/1000 + 1/1000 + 1/1000 + 1/1000 + 1/1000)
When my students learn material and are engaged in various ways, they retain math content for longer periods of time. Last year, paper airplanes were a huge disruption across our building. Sometimes
when you can’t beat the kids, you join them! To “gamify” the unit and connect it to the present, I will have students create various paper airplanes, and measure the distances they fly. Each student
would need to measure the distance writing the standard form, word form, and expanded form to the nearest hundredth of a meter (i.e., nearest centimeter). This allows students to use real
measurements, be physically active, and apply what they have learned. Let’s say one of the paper airplanes flew 3.19 meters in length. I want my scholars to visibly see that it flew three whole meter
sticks AND 1 whole decimeter (or 1.10) of the next meter stick, and 9 centimeters of the next decimeter (or 9/10) of a dm or 9/100 of the meter).
In other words, the expanded form of 3.19 = 3 + 0.1 + 0.09
Which is 3.19 = (3 × 1) + (1 × 1/10) + (9 ×1/100)
Number Lines
Number lines can be very helpful for students. As a matter of fact, number lines are an abstract form of length measurement! In this section, I show just how. For example: In lesson 6, we all flew
airplanes and measured the distances each plane flew. It was easy to see which plane flew the furthest or shortest distance based on where they landed in the hall and if you saw where it landed. What
may not be so easy is if the numbers are close in numerical value, and you were not in the same group as someone with that value. I will choose two numbers from our airplane list to compare. The
first student’s plane flew 2.7 meters. The second student’s plane flew 2.65 meters. Whose plane flew the furthest? Using number lines in lesson 7 will help students see how meter stick are concrete
number lines, therefore making another connection go the students and building upon what they have already learned.
Many students have a misconception that the longer the number is, the larger it is. This comes from their experience with whole numbers. A student that believes this would say the second student’s
plane flew farther than the first student’s plane because 2.65 has more digits than 2.7.
I want my students to understand that decimal numbers “fill in” sections on the number line between whole numbers. On the number line below, the whole numbers 0 through 10 are represented. Some
students do not realize there are decimal numbers between whole numbers. “You can think of plotting decimal numbers on the number line in successive stages.”^12 On this number line, the first stage
consists of the consecutive whole numbers from 0 through 10. This would represent the ten meters for our problem. Students would need to estimate where the numbers are to plot them on the number
line. Because 2.7 and 2.65 are both greater than two, we would want the student to plot them after two but before three.
Figure 2.
Let’s take a closer look at this same number line by “zooming in” to the whole numbers 2 and 3 with the tenths marked. Why? Because the students that jumped have measurements that are between two and
three. Almost immediately we can plot student 1’s measurement of 2.7 meters. It is a little more challenging to see student 2’s measurement of 2.65. Another important thing to mention is although
intervals between the digits 2 and 3 are broken into 10 sections, there are only nine tick marks for the decimal numbers. Digits in the base 10 system “roll over” after 9.
Figure 3.
Let’s “zoom in” again. Where? Because 2.65 is greater than 2.6, we will “zoom in” between 2.6 and 2.8 to show the hundredths between 2.6 and 2.8 so students see that all the places between whole and
decimal numbers need to be broken up into 10 equal pieces which creates new decimal numbers in between the points previously plotted. Once the hundredths have been plotted, it is easier to see where
each number is located on the number line.
Figure 4.
At each stage in “filling in” the number line, we break each interval into 10 equal pieces which then displays new decimal numbers in between the points previously plotted. A good point to make to
students is that like whole numbers, decimal numbers are also infinite. In fact, there are an infinite number of them in any interval, no matter how short! By breaking the meter stick or number line
down into sections, this will show that the pieces are shrinking. By comparing the two numbers on the number line, a student is able to see that 2.7m is further away from zero than 2.65m is and
therefore, able to show the first student’s plane flew farther than the second student’s plane; 2.7m > 2.65m. Decimal numbers do not only “fill in” number lines. They also represent distances from
zero. This is important because we do not want our students to develop the misconception that decimals are less than zero. While there are negative decimal numbers, fifth grade students are only
required to work with positive numbers.
Ordering Decimals
Although ordering decimal numbers is not included in the 5^th grade standards for my state, I feel that it is a necessary skill for students to know. Therefore, the first part of this lesson will
help to develop that concept. I will “gamify” this lesson by “mak[ing] it social.”
Once these place value concepts have been addressed and hopefully mastered by students, they can then be applied to show an understanding of operations as they pertain to whole and decimal numbers.
It is important to note that I do not plan to have students do arithmetic in this unit. Immediately following this unit, I will teach a unit on adding and subtracting decimals, with multiplication
and division of decimal numbers occurring later in the year.
Children often think there are no numbers smaller or less than one despite maybe carrying change in their pockets, seeing an ant on the ground, or even eating one slice of pizza out of a box. We
encounter small numbers and decimal numbers every day when we see the prices of items such as $3.19. However, the conceptual understanding of decimals requires students to connect decimals to whole
numbers and to fractions. Prior to teaching this unit, I will have taught a unit focused on whole numbers. I want my students to see the magic and power of the base ten system and how a good
understanding of numbers and mathematics can prepare them for the world and life outside of school.
Teaching Strategies
Do Now
At the beginning of my math periods, to set the tone for the day, part of my daily routine is to have students complete an activity quickly and quietly that they must start immediately. Most times
they are a review of previously taught material that takes no longer than 5-7 minutes to complete and an additional 5-7 minutes to review.
Number Talks
Sometimes at the beginning of my math lessons (at least 3 times/week) in lieu of a Do Now, we complete Number Talks. A Number Talk is a 10-15 minute whole group mental math activity where students
find answers in their heads, then share the strategies they use to find that answer aloud while also explaining their thinking, justifying their reasoning, and making sense of each other’s
Graffiti Walls
This is a creative way that allows students to record their thoughts, ideas, comments and questions about a topic. It allows students to learn each other’s ideas. This will be used to introduce the
topic of decimal place value, and to prepare for a class discussion.
Because students come to class with such a wide variety of pre-existing knowledge, skills, beliefs, ideas, and attitudes about numbers, it is important for me to access their prior knowledge. I will
give my students a pre-assessment to help me do this, and to determine what they know and what they need more instruction on. Once I know what knowledge and misconceptions they may have, I will be
able to begin differentiating my instruction for each lesson within the unit which will help me form small groups.
Differentiated Instruction
My district prefers differentiated instruction to occur most days of the week. This refers to instruction that is tailored to meet my students’ needs in small group settings. Personalized computer
instruction, remediation, reteaching, enrichment or review fall in this area, but it also can involve such matters as the types of numbers in the problems a given student is asked to solve.
Small Groups
I typically facilitate small group instruction after whole group instruction as a way to differentiate my instruction, but also to reduce the student-teacher ratio. Small groups allow me to give my
students more focused attention and a chance to ask specific questions about they have learned or are currently learning. Depending on the number of students in my class and needs of the students,
the sizes of the groups will vary, however I prefer to plan groups no larger than 5 students per group.
Read Aloud
Sometimes, I will read text aloud to my students to engage them with mathematical concepts. To connect this unit to the present, it be will “storified” by reading the book, “Little Numbers: And
Pictures that Show Just How Little They Are!” by Edward Packard.^13 This book shows how numbers get exponentially smaller than one by a factor of 10 for each place to the right of the decimal point!
It is very important for my students to be visually stimulated, and stories are one way to do that.
Showing students math in multiple ways and connecting it to the real world helps them build stronger connections. I will be sure to demonstrate this in lesson two to further illustrate the patterns
in the number of zeros of the product when multiplying a number by powers of 10 and what it means to divide by a power of 10. I will show two video versions of the Powers of Ten around lesson 2.^14
Powers of Ten was created in 1977 and takes the viewer on journey traveling into space beginning with a headshot of a couple lounging in Chicago. The camera “zooms out,” into outer space showing the
distance being traveled in meters. These are big numbers and show the number written with exponents. The video also “zooms in” showing small numbers! The video is quite captivating. The 2^nd version
is more modern, and students may relate more to it.^15 However, it does not show the exponents, and I think the exponents are important to show the pattern. Nevertheless, both videos are cool and
connect to science!
In addition to these two videos, I will also show another that demonstrates the linear measurements of the metric system using the mnemonic device “King Henry Does Usually Drink Chocolate Milk.”^16
This mnemonic device will help my students not only remember the units of the metric system, it is my hope that I can help them make a connection with decimal place value and the names of the places.
While it is easy for students to look up or use a conversion chart, they are not always permitted in testing situations. Therefore, I also hope to help my students remember how to convert from one
metric unit to another which will be taught later in the year.
Formative Assessments
Before during and after each lesson, I will quickly evaluate my students’ knowledge and progress using quizzes to determine their level of comprehension of math concepts, learning needs and academic
progress throughout the unit. These will mostly consist of classroom observations while students complete the Do Nows, complete assignments during small groups, and exit slips. However, students will
be given opportunities to self-assess by setting goals at the beginning of each unit, and by reflecting on what they know for each lesson throughout unit. Formative assessment should inform the
teacher and student and be constant.
Exit Slips
I will use exit slips at the end of each lesson to determine each student’s level of proficiency. Exit slips are one of the easiest ways to gather information about my students’ current levels of
understanding. My district prefers us to use exit slips as a way to create a running record of which standards students are understanding or need more support with.
Clock Buddies
To engage students, this fun activity will allow me to create readymade sets of partners for cooperative learning lessons.
Task Cards
I use task cards to keep students engaged a few times a year. They are a set of cards that have tasks, activities, or questions written on them which I will use to reinforce concepts in this unit.
They are alternatives to worksheets.
Project Based Learning Activity
Throughout the unit, but mostly towards the end, this project will engage students in solving real-world problems or answering complex questions. This will be another way to gauge what students have
learned in the unit in addition to their post assessment.
At then of the unit, I will give students a unit test to measure student achievement and the effectiveness of the unit.
Each student will have a personal folder that I will be stored in a crate in the classroom. The folder will show student’s work and growth over time. Students pre and post assessments, problem
solving tasks, formative assessment data will be collected and placed inside.
Classroom Activities
This unit will be composed of eight lessons. They are:
Lesson 1: What do I know about decimal numbers? (1 day)
Lesson 2: Little Numbers: Comparing Adjacent Digits (1-2 days)
Lesson 3: Little Numbers: Comparing Adjacent Digits within in Word Problems (Day 2)
Lesson 4: The Five Stages of Place Value (Making a Reference to Money) (1 day)
Lesson 5: Reinforcing the Five Stages of Place Value (2 days)
Lesson 6: Using Meter Sticks to Measure the Distance Paper Airplanes Fly
Lesson 7: Comparing Decimal Numbers Using Number Lines (1 day)
Lesson 8: Who wins the race? Cumulative Review/Project (2-3 days)
I have included sample plans that I will use for each lesson below including the objective, materials, procedures, and closure.
Lesson 1: What do I know about decimal numbers?
Objective: Students will: share what they know about place value and decimal numbers using The Graffiti Wall Strategy; participate in a class discussion; complete a pre-assessment.
Materials: Markers. Same group, same colored marker so I can ensure each student has participated from each group. 5 large Post-It posters hanging in the room with the following 5 statements or
• Write any number less than one. Any number less than one is acceptable, but it is important to focus on decimal fractions, and students may need guidance.
• Show as many ways as possible to represent 214 (my homeroom number). Accept non-standard as well as standard ways. Allow students to share their knowledge and creativity, but the primary focus is
standard ways of representation.
• Where do you see numbers in your community? List as many places as possible.
• What do you know about decimal numbers?
• The year is 2018. Which is the easiest way to write 2018? I will write 2018 on one of the posters using the Hindu-Arabic Numerals, Hieroglyphics, Hieratic Numerals, and Roman Numerals.
1. Students will be broken into groups of 4-5 students and given the appropriate marker and sent to a poster.
2. Students will have about 3-5 minutes to discuss and respond to the prompt/question.
3. After time expires, students will rotate to the next poster. This time, they can respond to the prompt/question OR respond to a comment left by the previous group.
4. Students will continue rotating until all the groups have visited the posters.
5. Students will return to their seats and the posters will be used to guide a class discussion.
6. After the discussion, students will complete a pre-assessment which will allow me to see where my students are, what they need, and where I should begin. The data I glean from this pre-assessment
will allow me to differentiate learning in small groups and my scholars and I to set worthwhile goals. The pre-assessment data will be added to their portfolios.
Closure: Exit ticket. I will use one of the templates in the resources below.
Lesson 2: Little Numbers: Comparing Adjacent Digits (1-2 days)
Objectives: Students will: listen to me read the book, “Little Numbers,” and answer comprehension questions related to vocabulary words and events taking place in the story regarding what is
happening to the main object: the dinosaur; understand that a digit in one place represents 1/10 of what it represents in the place to its left: be able to explain that as a digit moves further away
from the decimal to the right, the value of the digit is being divided by 10 OR “shrinking” by a factor of 10 using base-ten blocks (Figure 5), a number line (Figure 2) and a place value graphic
(Figure 6).
Materials: The book, “Little Numbers.” A bundle of 100 straws, already grouped into smaller bundles of 10. (This will be from an activity we completed earlier in the year with whole numbers when
showing how each place gets exponentially larger than one by a factor of 10 for each place to the left of the decimal point. We will have also read “Big Numbers,” also by Packard). Base-ten blocks.
1. Activate prior knowledge of whole numbers: Students will play a game of “Would You Rather…”
1. Would you rather have $ 0200 or $ 2000?
2. A discussion would follow on why and what about the numbers makes them think one number is larger than another.
3. How much greater is that number than the other?
4. Teacher will show both numbers using base ten blocks. How many flats make a cube? (10) How many flats make two cubes? (20) (This figure shows how this unit will use base-ten blocks to represent
decimal numbers.) Base ten blocks are another way student can manipulate the value and size of numbers. Most teachers have plastic sets of base-ten blocks that can be used to represent whole
numbers, but they can also be used to represent decimal numbers. I did not always know this.
Figure 5.
2. A portion of following figure will be drawn on the board. (The ones to the millions place and no division.) This will be to remind the students of the multiplicative patterns of the place value
Figure 6.
3. Introduce new concept. Read “Little Numbers.”
4. We know there are decimal places that are not whole numbers. If the relationship is to multiply by a factor of 10 to get to the next place value to the left, what relationship, or what happens
mathematically as you move to the to the right in a number? (Students will only see arrows, but no division symbols at this point.)
5. Show one strategy for determining the relationship between places moving to the right of the decimal point by looking at a whole number and modeling it with the straws. I will write all numbers on
the board using the model, so the students can see that the numbers are shifting their place and value.
6. Separate the 100 straws back into groups of 10. What is the value of the straws I am holding now? (10)
7. Separate the straws that are in groups of 10 into 1 straw. What is the value of the straws I am holding now? (1) What is happening to the value? (The value is shrinking.) By how much each time?
8. How I show the next place value using this one straw? (Cut it.) Into how many pieces? (10.) Why ten? (Because each place is getting smaller by 10 as you move to the right of the place value
model). I will physically cut the straws.
9. What mathematical operation is occurring for this “shrinking” to happen?
10. Show another strategy modeling with base-ten blocks for determining the relationship between places moving to the right of the decimal point by looking at a whole number.
11. We will use 333 to model this using the place value graphic and base-ten blocks. See Figure 3.
1. What is the value of the digit in the hundreds, tens, and one’s place? Write the values.
2. What happens, mathematically, to the value of the digit as it moves to the right from the hundreds place to the tens place? (The tens place value is exactly 10 times smaller than the hundreds
place value for the same digit. Students should also explain that 300 divided by 10 equals 30, and I will write this on the board.
12. The same line of questioning will occur with the next example, 7.77. The value of the base-ten blocks will change, and this will be emphasized with an anchor chart I will create using Figure 5.
We will keep in mind that the goal is for students to understand and explain that the value of the place to the right is 1/10 or 10 times less than the place to it’s left.
13. Another example may be needed. This will be based on time, student engagement, and student understanding.
14. Following the introduction of the new concept, I will allow students to work with the place value concept in small groups or with a partner by completing task cards. While the students are
responding to the prompts and questions of the task cards, I will conduct formative assessment and facilitate learning by providing feedback and asking clarifying and probing questions.
Closure: Exit slip: Students will fill in the following blanks: When digits shift to the right of the decimal point, it is like they are _________ (“shrinking.”) When the dinosaur grew smaller from
1.0 to 0.1. How many times smaller did it become? (10 times smaller).
Lesson 3: Little Numbers: Comparing Adjacent Digits using Word Problems – Day 2
Objectives: Students will: be able to understand that a digit in one place represents 1/10 of what it represents in the place to its left; be able to explain that as a digit moves further away from
the decimal to the right, the value of the digit is being divided by 10 OR “shrinking” by a factor of 10 using base-ten blocks (Figure 5) and a place value graphic (Figure 6).
Materials: Base-ten blocks, place value graphic, problem solving activity.
1. Using examples like the task cards from yesterday, students will review decimal place value patterns.
2. Students will be broken into 3 groups for differentiated instruction. For 15-20 minutes at a time.
1. Online individualized learning.
2. Vocabulary terms will be added to journals.
3. Problem solving activity with teacher.
1. Problem Solving Activity: Heaven and Jayden were arguing about the size of two numbers. Heaven thought seven-tenths was ten times larger than seven-hundredths. Logan thought
seven-hundredths was ten times larger than seven-tenths. Who is correct? Show and explain how you know. Make sure to refer to place value in your explanation.
3. Students will return to original seats. Students will share out one thing they learned, and one thing they still want to know more about.
Closure: Exit slip will be vocabulary based.
Lesson 4: The Five Stages of Place Value (Making a Reference to Money)
Objectives: Students will: be able to use base-ten blocks to model, read, and write decimal numbers in word form (base-ten numerals), standard form, and the two expanded forms (decimal and fraction).
For this unit, the 5^th stage of place value will not be taught.
Materials: Base-ten blocks, place value chart, anchor chart illustrating coin images and values places on a place value chart, as well as a five stages of place value anchor chart both created by
1. I will refer to a local gas stations’ cost of gas per gallon. This is because gas prices round to the nearest thousandth. A gallon of gas at Sunoco costs $3.249. Model this number using base-ten
blocks. Then write this number in word form, standard form, and the two expanded forms.
2. Students will work with a partner to model $3.249 using base-ten blocks.
3. Students will use paper base-ten blocks to cut and paste this model in their journals.
4. Students will use their place value charts to write the digits in each place, and then write the word form. This will also be placed in their journals.
5. Students will use the base-ten blocks to write the expanded form and the multiplicative make-up of the base ten units as a power of 10.
6. Students will model and record several more decimal numbers with a partner. (I intend to use numbers from local grocery stores and corner stores. I will have students bring in decimal numbers
they see in their neighborhoods, and/or I will take pictures or use items from a circular to add to a PowerPoint to visually represent the numbers in their world.)
Closure: Exit slip: Students will write the word form, standard form, and two expanded forms of a decimal number.
= three and two hundred forty-nine thousandths
= 3.0 + 0.2 + 0.04 + 0.009
(3 x 1) + (2 x 0.1) + (4 x 0.01) + (9 x 0.001)
= (3 x 1) + (2 x 1/10) + (4 x 1/10) + (9 x 1/1000)
Lesson 5: Reinforcing the Five Stages of Place Value (2 days)
Objectives: Students will: be introduced to the metric system watching a video; learn an acronym to remember the order of the units in the metric system from millimeters (thousandths) up to
kilometers (thousands); be able to use meter sticks to measure the length of the objects in the classroom; connect Metric Conversion chart with the place value chart (specifically rows four and
five); record the name of the object, and the length of it using word form, standard form, and two forms of expanded form (decimal and fraction).
Materials: Metric conversion chart, meter sticks, The Story of King Henry video, Venn Diagram, and journals for note taking (the video is told in liters, but the lesson will meters, allow students
the opportunity to take notes and remind them to focus on meters for this lesson).
1. Review: How many tenths make one whole? (10). How many hundredths make a tenth? (10) How many hundredths make a one whole? (100) How many thousandths make a hundredth? (10) How many thousandths
make a tenth? (100 = 10 x 10) How many thousandths make one whole? (1,000 = 10 x 10 x 10). If we move a digit to the left of the decimal point, one place at a time, is the value of the digit
increasing or decreasing? (Increasing) How much larger is the value of the digit? (10 x larger) What operation? (Multiplication) What is happening to the value of a digit as it moves to the right
of the decimal point? (Decreasing) How many times smaller is the value of the digit? (1/10 x smaller). What operation can we use to show that the value of the digit is decreasing? (Division if we
divide by 10 or multiplication if we multiply by 1/10).
2. Today we will use meter sticks that show this concept. Let’s look at a meter stick.
3. Each group will have a meter stick.
4. This meter stick is one unit long. The unit is a meter. Like our bundle of straws this meter stick has 100 sections called decimeters. Can you see them?
5. Do you notice any other divisions? Do you notice any smaller groups? Look at your meter sticks and discuss other groups that you notice.
6. Students should notice that the decimeters are broken into smaller groups of 10 and these are the centimeters. They should also notice that the centimeters are broken into groups of 10, and these
are called millimeters.
7. There are units that are smaller and larger than the units we see, and some measure large things and some measure small things. My question is, how can you remember all these units and what they
measure? (Allow for brief discussion.)
8. Share Metric Conversion table. Allow students to compare and contrast it with the place value chart using a Venn Diagram.
9. Watch “The Story of King Henry” video.
10. Students will take notes while watching the video. A copy of the Metric Conversion Table would be good to paste into their journals.
11. After the video, allow students to go around the room measuring and recording different items using the five stages of place value including word form, standard form, and the two expanded forms.
(This is NOT a lesson on measurement conversion, so please be careful). The goal is to reinforce the five stages of place value.)
Closure: Exit slip: Write down an example of one item you measured, and include the name of the item, the word form, standard form, and two versions of expanded form. (I will not allow students to
use measurements that are whole numbers.)
Lesson 6: Using Meter Sticks to Measure the Distance Paper Airplanes Fly
Objectives: Students will: create paper airplanes; fly paper airplanes; record the distance their paper planes fly in meters, using standard form, word form, and the two versions of expanded form.
Materials: Metric Conversion Chart, Meter Sticks, (a good idea would be to use tape to mark the meters in the hall, gym, or where ever the planes will be flown and allow the students to use the meter
sticks to measure the distances their planes fly. My building has an extremely long and straight hall that allows for straight flights, so I anticipate not having enough meter sticks to measure the
length of the hall. Chalk is an option if this activity is done outside. Be mindful of wind and rain); paper to fold airplanes.
1. Review the units of the metric system.
2. Teacher will make an airplane and fly it. This will immediately attract students’ attention.
3. Review some ground rules regarding the planes. Planes can only take flight on the runway which is in the hall. Planes may only fly one at a time. All planes are out of fuel once they taxi (land)
on the runway and must be placed in the landing zone (a box or bag of some sort).
4. Build Airplanes. Write names on them. Place them in a safe location.
5. It is my goal to have support staff during this lesson. This lesson is intended to be taught in small groups.
1. Group 1 – Individualized learning on computer or Ipad.
2. Group 2 – Problem solving activity with support teacher – Problem: Part 1: Tay’s teacher asked him to write 7.835 in expanded form. Tay wrote: 7 + 0.8 + 0.30 + 0.500. What is Mike’s
misconception? Use base-ten blocks or a place value chart if for support. Part 2: What is another way to write this number?
3. Group 3 – Flying airplanes, measuring, and recording distances in hallway with teacher. (All students will remain in the hall until it is time to rotate groups.)
6. All students will record the distances of their flights on a large piece of chart paper from top to bottom, in no particular order. (These numbers will be used in the next lessons on ordering and
comparing decimal numbers).
7. Students will return to their seats and record the distances their planes flew in standard form, word form, and the two expanded forms if possible (some students’ planes may be exact whole
Closure: The students’ recording sheet will count towards their exit slip for the day.
Lesson 7: Comparing Decimal Numbers Using Number Lines
Objectives: Students will: be able to use number lines to compare decimal numbers; be able to use the comparison symbols, <, >, and = to write expressions.
Materials: Airplane data, open number lines, some closed number lines, dry erase markers and erasers (if you have laminated number lines).
1. Students will discuss what it means to compare two items. Students will compare two whole numbers with a partner. Discussion. How did you determine which number was larger? Which place has the
largest value? How much larger is each place as a digit shifts to the left?
2. Students will compare two decimal numbers with a partner. Discussion. How did you determine which number was larger? Which place has the largest value? How much larger is each place as a digit
shifts to the left? To the right?
3. We can also use number lines to compare numbers and show which number has the largest value. Using the information from number lines above, will guide this lesson, and can be used as an example.
4. Teacher will choose one or two more sets of numbers from the list and students will work to compare the numbers using number lines. Teacher will take notes as students work and ask clarifying and
probing questions as necessary.
5. Students will break into differentiated groups:
1. Individualized Lessons on computer or Ipads.
2. Comparing decimal numbers and fractional numbers. Worksheet or task cards.
3. Problem Solving: Dionne and Don were talking about the numbers 1.253 and 2.351. Part 1: With base ten blocks or number lines, draw a picture of both numbers. Part 2: What is the value of the
2 in both numbers? How does the value of the 2 in the first number compare to the 2 in the second number? Part 3: What is the value of the 5 in both numbers? How does the value of the 5 in
first number compare to the value of the 5 in the second number?
4. Teacher will pull small groups of students to review certain concepts or facilitate the differentiated groups.
6. Students will share one thing they learned or liked in today’s small groups with a neighbor. Three to five students will be asked to share with the class.
Closure: Exit slip: Plot 2 numbers on the same or separate number lines to determine which number is larger. Write a sentence explaining number is larger.
Lesson 8: Who wins the race? Cumulative Review/Project (2-3 days)
Objectives: Part 1: Students will: be able to race in a 100-meter dash in small groups; be able to track and calculate their classmates’ times to the nearest thousandth (if possible); record all
classmates’ names and times accurately on a given table; be able to order decimal numbers in order from least to greatest and greatest to least; be able to justify who won the race.
Materials: Stop watches, base-ten blocks, number lines, hundredths grids, place value charts, journals, table, glue sticks, and scissors.
1. Review comparing decimal numbers by using airplane data.
2. Introduce ordering numbers by having students order numbers from least to greatest and greatest to least from within the building such as ages, grades, homeroom numbers, etc. Record and label the
3. Discuss what tools can be used to order numbers if they get stuck. Model the examples.
4. Use and order four numbers from the airplane data using the tools students suggested. Discuss the different ways the decimal numbers were ordered and why they were ordered that way.
5. Explain to students that they will practice ordering decimal numbers in small groups today, and they will get their data from relay races. (Wait for the shouting or pouting to end.)
6. Students will gather journals to cut and paste tables.
7. Students will count off in the number of groups I want there to be based on the number of students I want to be in each group. I would like four students in each group if possible.
8. Students will travel to the gymnasium or outdoor location for the relay races with journals and writing utensils.
9. I will explain directions to students in the classroom as well as review them in the actual location. Each student gets one time to run. All students must record the data accurately. We will
cheer our classmates on. We will be good sports. We will sit when it is not our turn to race. We will give as close to accurate race times. We will race fairly. Students may generate some
guidelines as well.
10. Students will race against each other in their groups. Classmates will cheer them on, record their times, and calculate their times. It is my goal to have multiple students’ record the times, so
we can compare the differences in the numbers they get. However, we will agree on the numbers we will record so that our data is consistent.
11. Once all students have completed the race, we will return to our classroom.
12. Students will work in their racing groups to order their times in order from least to greatest and greatest to least. They will record the times to the nearest thousandth of a second, if
possible, in their journals. (This information will be necessary for part 2.)
13. Students will write who came in first place and provide reasoning.
14. Students will share with the class in small groups.
15. Discussion on what students noticed about the numbers. Why is the largest number the winner in the airplane activity, but the smallest number is the winner in the racing activity?
1. Optional: Order the entire class list from who finished the race first to who finished last.
Closure: Exit slip - Students will explain why the smallest number in their group is determined the winner of the race.
Objectives: Part 2: Students will choose a classmate and use their number to compare similar adjacent digits by stating how many times larger or smaller the same digit is in a different place.
Students will show how to compare the value of this same number using a comparison symbol to show which number is greater. Students will also need to state of these two numbers, which friend won the
race. Students will use the Clock Buddy Strategy to determine partners.
Materials: Journals, Clock Buddy Template, I will create a template to help students organize information in journals, glue sticks, scissors, place value charts, number lines, base-ten blocks,
hundredths grids, and colored pencils or crayons.
1. Distribute Clock Buddy Templates. One for each student. Each student will need to write their name and date at the top of the paper.
2. Students will stand and find a clock buddy for 12 o’clock and then return to their seats. This is to ensure each person has a buddy. The same process will occur until each all 4 time slots are
3. Once all slots are filled, put clocks to the side. They will be used later in the lesson.
4. Students will review what it means to recognize that in a multi-digit number, a digit in one place represents 10 times as much as it represents in the place to its right and 1/10 of what it
represents in the place to its left. Sample 1: Explain the relationship between the two 5's in the number 355.921. (Students should be able to discuss and explain the 5 in the ones place is 1/10
the size of the 5 in the tens place OR the 5 in the tens place is ten times the size of the 5 in the ones place. It would take ten groups of 5 to equal the among that is in the tens place OR the
5 in the tens place has a value of 50 and for it to be equal to 5, it would need to divide into 10 equal groups.) Sample 2: I will use numbers from the races.
5. Ask students to refer to their Clock Buddies, choose a time, and students will proceed to begin comparing similar digits in their numbers. Students will also write an inequality to compare their
race times, and who won the race based on their time. Each partnership will last about 10-12 minutes.
6. This will continue for four rounds. Students should have four examples of each intended target of this lesson.
7. 3-5 students will share out their comparisons.
Closing: Each student will complete an exit slip independently. I will choose two numbers for students to show their understanding of the intended learning target.
Objectives: Part 3: Project Based Learning Activity- Students will: use their personal racing data to record their individual time in word form, standard form, and the two expanded forms (as decimals
and fractions); compare similar adjacent digits by stating how many times larger or smaller the same digit is in a different place in a different number; show how to compare the value of the same
numbers using a comparison symbol to show which number is greater; determine and record which friend won the race and how they know.
Materials: Stop watches, paper, gymnasium or outdoor courtyard, poster paper, base-ten blocks, hundredths grids, place value charts, number lines, markers, glue, or chalk.
1. Students will record their individual time in word form, standard form, and the three expanded forms (as decimals and fractions) independently. I will circulate around the room to clarify or
probe students to think, I will also record who needed support and with what.
2. Once all students are complete. I will introduce the Decimal Place Value Poster Project.
1. Title
2. Group Member Names
3. Individual Group Member Data using the four stages of place value. (May be directly written on the poster or written in a provided table and glued to the poster).
4. Order group members numbers form greatest to least.
5. Order group members numbers form least to greatest.
6. Write 2-4 inequalities comparing the racing times.
7. Write a sentence explaining who won the race and how the group knows.
8. Poster is neat and organized.
9. Class presentation.
3. A rubric will be used to evaluate the students.
Closure: Students will present their project to the class. (I anticipate this taking a few class periods).
Reading Lists for Students
Packard, Edward. Big Numbers: And Pictures that Show Just How Big They Are! Brookfield, Connecticut, Millbrook Press. 2000.
Perkins, Useni Eugene. Hey Black Child. New York, Boston. Little Brown and Company. 2017.
Schmandt-Besserat, Denise. The History of Counting. Morrow Junior Books, New York. 1999.
Materials for Classroom Use
Pre-Assessment option - https://teachingtoinspire.com/2015/09/teaching-decimals.html
Decimal Place Value Quiz or Review - https://www.teacherspayteachers.com/Product/FREE-Decimals-Place-Value-Quiz-or-Review-2041350
Decimal Place Value Pre-Assessment - https://printshop.katyisd.org/DSF/PreviewPdf.ashx?FileId=u1L4E7dlN3M-&SITEGUID=d8f13d95-6ad8-4f21-886c-0faa5ba553f9&SITEGUID=d8f13d95-6ad8-4f21-886c-0faa5ba553f9&
Formative and Instructional Problem-Solving Tasks - http://3-5cctask.ncdpi.wikispaces.net/Fifth%20Grade%20Tasks
Problem Solving Assessment Tasks - https://hcpss.instructure.com/courses/108/pages/grade-5-year-at-a-glance#fragment-2
Illustrative Mathematics - https://www.illustrativemathematics.org/content-standards/5
Inside Mathematics - http://www.insidemathematics.org/common-core-resources/mathematical-content-standards/standards-by-grade/5th-grade
Graffiti Walls - https://www.facinghistory.org/resource-library/teaching-strategies/graffiti-boards
Clock Buddies - http://www.teamstraus.com/SchoolDaysBorder_files/Teacher%20Farm/clockbuddies_Lower_El.pdf
Task Cards - https://www.teacherspayteachers.com/Product/Read-Write-and-Compare-Decimals-Task-Cards-Freebie-for-5th-Grade-2042564
Exit Slip Templates - https://www.nbss.ie/sites/default/files/publications/exit-entry_slip_-_comprehension_strategy_handout_copy_2_0.pdf
PA Core Standard Integration
The unit will incorporate standards from the Pennsylvania Core State Standards for Mathematics in Numbers and Operations of Base 10. The focus will be primarily on students being able to demonstrate
an understanding of place-value of decimals and compare quantities or magnitudes of numbers in my students real-life.
• Understand that in a multi-digit number, a digit in one place represents 10 times of what it represents in the place to its right and 1/10 of what it represents in the place to its left
• Read and write decimals to the thousandths using base-ten numerals, word form, and expanded form (M05.A-T.1.1.3)
• Compare two decimals to thousandths based on meanings of the digits in each place using the comparison symbols >, =, < to record the results of the comparisons (M05.A-T.1.1.4)
PA Common Core Standards: Standards for Mathematical Practice Standards
The unit will incorporate standards from the Pennsylvania Core State Standards for Mathematical Practice Standards. These practices support important processes and proficiencies in mathematics
education as outlined by the National Council of Teachers of Mathematics (NCTM) and the National Research Council’s report Adding it Up. This PA Document has been adapted from the Common Core State
Standards for Mathematics.
• Make sense of problems and persevere in solving them.
• Reason abstractly and quantitatively.
• Construct viable arguments and critique the reasoning of others.
• Model with mathematics.
• Use appropriate tools strategically.
• Attend to precision.
• Look for and make use of structure.
• Look for and express regularity in repeated reasoning.
1. Hammond, Zaretta P “3 Tips to Make Any Lesson More Culturally Responsive,” accessed August 10, 2018, https://www.cultofpedagogy.com/culturally-responsive-teaching-strategies/.
2. Boaler, Jo, and Carol S. Dweck, Mathematical Mindsets: Unleashing Students' Potential Through Creative Math, Inspiring Messages and Innovative Teaching (San Francisco, CA: Jossey-Bass; a Wiley
Brand, 2016), 27.
3. Delpit, Lisa D, "Multiplication is for white people": raising expectations for other people's children, (New York, NY: The New Press, 2012), 21.
4. Boaler and Dweck, Mathematical Mindsets, 26.
5. Department of Numbers, Pittsburgh Pennsylvania Household Income, accessed August 13, 2018, https://www.deptofnumbers.com/income/pennsylvania/pittsburgh/.
6. Sowder, Judith, Place Value as the Key to Teaching Decimal Operations, accessed August 15, 2018, https://web.stevens.edu/golem/llevine/CIESE/place_value-decimal%20operations.pdf.
7. Number Systems, Egyptian Base Ten Number System, accessed August 14, 2018, http://www.math.wichita.edu/history/topics/num-sys.html#egypt.
8. Howe, Roger and Reiter, Harold, The Five Stages of Place Value, unpublished, accessed, July 15, 2018.
9. Beckmann, Sybilla, Mathematics for Elementary Teachers, (Adison Wesley, Hardcover) 25.
10. Pankhurst, Richard, An Introduction to the Economic History of Ethiopia, (London: Lalibela House, 1961).
11. The World Factbook, Appendix G: Weights and Measures, accessed August 14, 2018, https://www.cia.gov/library/publications/the-world-factbook/appendix/appendix-g.html.
12. Mathematics for Elementary Teachers (36).
13. Packard, Edward, Little Numbers: And Pictures that Show Just How Little They Are!, (Brookfield, Connecticut: Millbrook Press, 2001).
14. Eames Office, “Powers of Ten.” YouTube Video, 9:01. August 26, 2010. https://www.youtube.com/watch?v=0fKBhvDjuy0.
15. Obreschkow, Danail, “Cosmic Eye (Original HD Portrait Version 2011).” 3:09. January 11, 2012. https://www.youtube.com/watch?v=jfSNxVqprvM.
16. Salyers, Casey, “The Story of King Henry,” YouTube Video, 8:42. January 26, 2016. https://www.youtube.com/watch?v=Lstm7bBqxFI
Academic Standards for Mathematics. Grades Pre K – High School. Pennsylvania Department of Education, last modified March 1, 2014. file:///C:/Users/Kryst/Downloads/
PA%20Core%20Standards%20Mathematics%20PreK-12%20March%202014%20(1).pdf. p. 5. (accessed August 2, 2018).
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documents/Grade%205%20Mathematics%20Assessment%20Anchors.pdf. p. 3. (accessed August 2, 2018),
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Comments (1)
Brittany McCann (Pittsburgh Colfax, Pittsburgh , PA)
Subject taught: PSE, Grade: 35
Amazing Unit!
I\'ve been so excited to read your Unit! I\'m going to look through the new curriculum tomorrow to see where I can sprinkle it into my classes. You did such an incredible job - I cannot wait to
get “Little Numbers: And Pictures that Show Just How Little They Are!” and share it with my students.
When you are finished viewing curriculum units on this Web site, please take a few minutes to provide feedback and help us understand how these units, which were created by public school teachers,
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What is the condition for Cauchy-Riemann equation?
The Cauchy-Riemann equation (4.9) is equivalent to ∂ f ∂ z ¯ = 0 . If f is continuous on Ω and differentiable on Ω − D, where D is finite, then this condition is satisfied on Ω − D if and only if the
differential form ω = f.dz is closed, i.e. dω = 06.
How do you satisfy Cauchy-Riemann equation?
If u and v satisfy the Cauchy-Riemann equations, then f(z) has a complex derivative. The proof of this theorem is not difficult, but involves a more careful understanding of the meaning of the
partial derivatives and linear approxi- mation in two variables. ∇v = (∂v ∂x , ∂v ∂y ) = ( − ∂u ∂y , ∂u ∂x ) .
Does the Cauchy-Riemann condition guarantee differentiability?
Existence of partial derivatives & Cauchy-Riemann does not imply differentiability example.
What is Cauchy-Riemann equation in fluid mechanics?
Flow of Ideal Fluid These equations are called the Cauchy–Riemann equations in the theory of complex variables. In this case, they express the relationship between the velocity potential and stream
function. The Cauchy–Riemann equations clarify the fact that ϕ and ψ both satisfy Laplace’s equation.
Is Cauchy-Riemann condition sufficient?
Cauchy-Riemann Equations is necessary condition but is not sufficient for analyticity.
What is the condition for F z to be analytic?
A function f(z) is said to be analytic in a region R of the complex plane if f(z) has a derivative at each point of R and if f(z) is single valued.
What are the conditions for a function to be analytic?
A function f(z) is said to be analytic in a region R of the complex plane if f(z) has a derivative at each point of R and if f(z) is single valued. A function f(z) is said to be analytic at a point z
if z is an interior point of some region where f(z) is analytic.
Is the function f z )= E z analytic?
We say f(z) is complex differentiable or rather analytic if and only if the partial derivatives of u and v satisfies the below given Cauchy-Reimann Equations. So in order to show the given function
is analytic we have to check whether the function satisfies the above given Cauchy-Reimann Equations.
What is Cauchy Riemann equation in polar form?
Substitution of the chain rule matrix equations from above yields the polar Cauchy-Riemann equations: ∂u ∂r = 1 r ∂u ∂θ , ∂u ∂θ = −r ∂v ∂r . These can be used to test the analyticity of functions
more easily expressed in polar coordinates.
What are the CR condition for a function to be analytic?
What is Cauchy-Riemann equation in polar form?
What is the necessary condition for F z to be analytic?
. A necessary condition for f(z,z) to be analytic is. ∂f. ∂z. = 0.
What is the condition for analytic function?
What is the condition for a function?
A Condition for a Function: Set A and Set B should be non-empty. In a function, a particular input is given to get a particular output. So, A function f: A->B denotes that f is a function from A to
B, where A is a domain and B is a co-domain.
Is F z Zn analytic function everywhere?
If f(z) is analytic everywhere in the complex plane, it is called entire. Examples • 1/z is analytic except at z = 0, so the function is singular at that point. The functions zn, n a nonnegative
integer, and ez are entire functions.
Which of the following is true about f z )= z²?
Which of the following is true about f(z)=z2? In general the limits are discussed at origin, if nothing is specified. Both the limits are equal, therefore the function is continuous.
What is polar form?
The polar form of a complex number is a different way to represent a complex number apart from rectangular form. Usually, we represent the complex numbers, in the form of z = x+iy where ‘i’ the
imaginary number. But in polar form, the complex numbers are represented as the combination of modulus and argument.
What is Cauchy-Riemann equation in complex analysis?
In the field of complex analysis in mathematics, the Cauchy–Riemann equations, named after Augustin Cauchy and Bernhard Riemann, consist of a system of two partial differential equations which,
together with certain continuity and differentiability criteria, form a necessary and sufficient condition for a complex …
Is FZ )= z3 analytic?
1) Show that f(z) = z3 is analytic. exists and continuous. Hence the given function f(z) is analytic.
What is the necessary and sufficient condition of a function?
In general, a necessary condition is one that must be present in order for another condition to occur, while a sufficient condition is one that produces the said condition.
What are the Cauchy-Riemann conditions for differentiability of f (z)?
The Cauchy-Riemann conditions (17.4) are also sufficient for the differentiability of f (z) provided the functions u (x, y) and υ(x, y) are totally differentiable (all partial derivatives exist) at
the considered point. The derivative f ‘ (z) can be calculated as
How do you prove the Cauchy-Riemann condition?
A visual depiction of a vector X in a domain being multiplied by a complex number z, then mapped by f, versus being mapped by f then being multiplied by z afterwards. If both of these result in the
point ending up in the same place for all X and z, then f satisfies the Cauchy-Riemann condition
What is Cauchy-Riemann’s theory of functions?
Riemann’s dissertation on the theory of functions appeared in 1851. The Cauchy–Riemann equations on a pair of real-valued functions of two real variables u ( x, y) and v ( x, y) are the two
How do the Cauchy-Riemann equations relate to conformal transformation?
That is, the Cauchy–Riemann equations are the conditions for a function to be conformal. Moreover, because the composition of a conformal transformation with another conformal transformation is also
conformal, the composition of a solution of the Cauchy–Riemann equations with a conformal map must itself solve the Cauchy–Riemann equations. | {"url":"https://www.goodgraeff.com/what-is-the-condition-for-cauchy-riemann-equation/","timestamp":"2024-11-14T01:14:50Z","content_type":"text/html","content_length":"57045","record_id":"<urn:uuid:4f3faa77-7182-4d2e-b408-e122b166cbb3>","cc-path":"CC-MAIN-2024-46/segments/1730477028516.72/warc/CC-MAIN-20241113235151-20241114025151-00661.warc.gz"} |
The IO monad
The IO monad by Mark Seemann
The IO container forms a monad. An article for object-oriented programmers.
This article is an instalment in an article series about monads. A previous article described the IO functor. As is the case with many (but not all) functors, this one also forms a monad.
SelectMany #
A monad must define either a bind or join function. In C#, monadic bind is called SelectMany. In a recent article, I gave an example of what IO might look like in C#. Notice that it already comes
with a SelectMany function:
public IO<TResult> SelectMany<TResult>(Func<T, IO<TResult>> selector)
Unlike other monads, the IO implementation is considered a black box, but if you're interested in a prototypical implementation, I already posted a sketch in 2020.
Query syntax #
I have also, already, demonstrated syntactic sugar for IO. In that article, however, I used an implementation of the required SelectMany overload that is more explicit than it has to be. The monad
introduction makes the prediction that you can always implement that overload in the same way, and yet here I didn't.
That's an oversight on my part. You can implement it like this instead:
public static IO<TResult> SelectMany<T, U, TResult>(
this IO<T> source,
Func<T, IO<U>> k,
Func<T, U, TResult> s)
return source.SelectMany(x => k(x).Select(y => s(x, y)));
Indeed, the conjecture from the introduction still holds.
Join #
In the introduction you learned that if you have a Flatten or Join function, you can implement SelectMany, and the other way around. Since we've already defined SelectMany for IO<T>, we can use that
to implement Join. In this article I use the name Join rather than Flatten. This is an arbitrary choice that doesn't impact behaviour. Perhaps you find it confusing that I'm inconsistent, but I do it
in order to demonstrate that the behaviour is the same even if the name is different.
The concept of a monad is universal, but the names used to describe its components differ from language to language. What C# calls SelectMany, Scala calls flatMap, and what Haskell calls join, other
languages may call Flatten.
You can always implement Join by using SelectMany with the identity function:
public static IO<T> Join<T>(this IO<IO<T>> source)
return source.SelectMany(x => x);
In C# the identity function is idiomatically given as the lambda expression x => x since C# doesn't come with a built-in identity function.
Return #
Apart from monadic bind, a monad must also define a way to put a normal value into the monad. Conceptually, I call this function return (because that's the name that Haskell uses). In the IO functor
article, I wrote that the IO<T> constructor corresponds to return. That's not strictly true, though, since the constructor takes a Func<T> and not a T.
This issue is, however, trivially addressed:
public static IO<T> Return<T>(T x)
return new IO<T>(() => x);
Take the value x and wrap it in a lazily-evaluated function.
Laws #
While IO values are referentially transparent you can't compare them. You also can't 'run' them by other means than running a program. This makes it hard to talk meaningfully about the monad laws.
For example, the left identity law is:
return >=> h ≡ h
Note the implied equality. The composition of return and h should be equal to h, for some reasonable definition of equality. How do we define that?
Somehow we must imagine that two alternative compositions would produce the same observable effects ceteris paribus. If you somehow imagine that you have two parallel universes, one with one
composition (say return >=> h) and one with another (h), if all else in those two universes were equal, then you would observe no difference in behaviour.
That may be useful as a thought experiment, but isn't particularly practical. Unfortunately, due to side effects, things do change when non-deterministic behaviour and side effects are involved. As a
simple example, consider an IO action that gets the current time and prints it to the console. That involves both non-determinism and a side effect.
In Haskell, that's a straightforward composition of two IO actions:
> h () = getCurrentTime >>= print
How do we compare two compositions? By running them?
> return () >>= h
2022-06-25 16:47:30.6540847 UTC
> h ()
2022-06-25 16:47:37.5281265 UTC
The outputs are not the same, because time goes by. Can we thereby conclude that the monad laws don't hold for IO? Not quite.
The IO Container is referentially transparent, but evaluation isn't. Thus, we have to pretend that two alternatives will lead to the same evaluation behaviour, all things being equal.
This property seems to hold for both the identity and associativity laws. Whether or not you compose with return, or in which evaluation order you compose actions, it doesn't affect the outcome.
For completeness sake, the C# implementation sketch is just a wrapper over a Func<T>. We can also think of such a function as a function from unit to T - in pseudo-C# () => T. That's a function; in
other words: The Reader monad. We already know that the Reader monad obeys the monad laws, so the C# implementation, at least, should be okay.
Conclusion #
IO forms a monad, among other abstractions. This is what enables Haskell programmers to compose an arbitrary number of impure actions with monadic bind without ever having to force evaluation. In C#
it might have looked the same, except that it doesn't.
Next: Test Data Generator monad.
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sending me a pull request
. Alternatively, you can discuss this post on Twitter or somewhere else with a permalink. Ping me with the link, and I may respond.
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The LLM Reasoning Debate Heats Up
The LLM Reasoning Debate Heats Up
Three recent papers examine the robustness of reasoning and problem-solving in large language models
One of the fieriest debates in AI these days is whether or not large language models can reason.
In May 2024, OpenAI released GPT-4o (omni), which, they wrote, “can reason across audio, vision, and text in real time.” And last month they released the GPT-o1 model, which they claim performs
“complex reasoning”, and which achieves record accuracy on many “reasoning-heavy” benchmarks.
But others have questioned the extent to which LLMs (or even enhanced models such as GPT-4o and o1) solve problems by reasoning abstractly, or whether their success is due, at least in part, to
matching reasoning patterns memorized from their training data, which limits their ability to solve problems that differ too much from what has been seen in training.
In a previous post on LLM reasoning, I asked why it matters whether LLMs are performing “actual reasoning” versus behavior that just looks like reasoning:
Why does this matter? If robust general-purpose reasoning abilities have emerged in LLMs, this bolsters the claim that such systems are an important step on the way to trustworthy general
intelligence. On the other hand, if LLMs rely primarily on memorization and pattern-matching rather than true reasoning, then they will not be generalizable—we can’t trust them to perform well
on ‘out of distribution’ tasks, those that are not sufficiently similar to tasks they’ve seen in the training data.
Before getting into the main part of this post, I’ll give my answer to a question I’ve seen a lot of people asking, just what is “reasoning” anyway? Indeed, reasoning is one of those overburdened
terms that can mean quite different things. In my earlier post I defined it this way:
The word ‘reasoning’ is an umbrella term that includes abilities for deduction, induction, abduction, analogy, common sense, and other ‘rational’ or systematic methods for solving problems.
Reasoning is often a process that involves composing multiple steps of inference. Reasoning is typically thought to require abstraction—that is, the capacity to reason is not limited to a
particular example, but is more general. If I can reason about addition, I can not only solve 23+37, but any addition problem that comes my way. If I learn to add in base 10 and also learn about
other number bases, my reasoning abilities allow me to quickly learn to add in any other base.
It’s true that systems like GPT-4 and GPT-o1 have excelled on “reasoning” benchmarks, but is that because they are actually doing this kind of abstract reasoning? Many people have raised another
possible explanation: the reasoning tasks on these benchmarks are similar (or sometimes identical) to ones that were in the model’s training data, and the model has memorized solution patterns that
can be adapted to particular problems.
There have been many papers exploring these hypotheses (see the list at the end of this post of recent papers evaluating reasoning capabilities of LLMs). Most of these test the robustness of LLMs’
reasoning capabilities by taking tasks that LLMs do well on and creating superficial variations on those tasks—variations that don’t change the underlying reasoning required, but that are less likely
to have been seen in the training data.
In this post I discuss three recent papers on this topic that I found particularly interesting.
Paper 1: The embers of autoregressive training
Paper Title: Embers of autoregression show how large language models are shaped by the problem they are trained to solve
Authors: R. Thomas McCoy, Shuny Yao, Dan Friedman, and Thomas L. Griffiths
This is one of my favorite recent LLM papers. The paper asks if the way LLMs are trained (i.e., learning to predict the next token in a sequence, which is called “autoregression”) has lingering
effects (“embers”) on their problem-solving abilities. For example, consider the task of reversing a sequence of words. Here are two sequences:
time. the of climate political the by influenced was decision This
letter. sons, may another also be there with Yet
Getting the right answer shouldn’t depend on the particular words in the sequence, but the authors showed that for GPT-4 there is a strong dependence. Note that the first sequence reverses into a
coherent sentence, and the second does not. In LLM terms, reversing the first sequence yields an output that is more probable than the output of reversing the second. That is, when the LLM computes
the probability of each word, given the words that come before, the overall probability will be higher for the first output than for the second. And when the authors tested GPT-4 on this task over
many word sequences, they found that GPT-4 gets 97% accuracy (fraction of correct sequence reversals) when the answer is a high-probability sequence versus 53% accuracy for low-probability sequences.
The authors call this “sensitivity to output probability.” The other “embers of autoregression” are sensitivity to input probability (GPT-4 is better at solving problems with high-probability input
sequences, even when the contents of the sequence shouldn’t matter), and sensitivity to task frequency (GPT-4 does better on versions of a task that are likely common in the training data than on
same-difficulty versions that are likely rare in the training data).
One of the tasks the authors use to study these sensitivities is decoding “shift ciphers”. A shift cipher is a simple way to encode text, by shifting each letter by a specific number of places in
the alphabet. For example, with a shift of two, jazz becomes lcbb (where the z shift wraps around to the beginning of the alphabet). Shift ciphers are often denoted as “Rot-n”, where n is the number
of alphabetic positions to shift (rotate) by.
The authors tested GPT-3.5 and GPT-4 on decoding shift ciphers of different n’s. Here is a sample prompt they used:
Rot-13 is a cipher in which each letter is shifted 13 positions forward in the alphabet. For example, here is a message and its corresponding version in rot-13:
Original text: “Stay here!”
Rot-13 text: “Fgnl urer!
Here is another message. Encode this message in rot-13:
Original text: “To this day, we continue to follow these principles.”
Rot-13 text:
The authors found that GPT models have strong sensitivy to input and output probability as well as to task frequency, as illustrated in this figure (adapted from their paper):
(a) Output sensitivity: When tested on decoding shift ciphers, the GPT models do substantially better when the correct output is a high-probability sequence.
(b) Input sensitivity: When tested on encoding shift ciphers, GPT-4 is somewhat better on high-probability input sequences.
(c) Task sensitivity: When tested on shift ciphers of different n values (e.g., Rot-12 vs. Rot-13), GPT models are substantially better on Rot-13. This seems to be because Rot-13 examples are much
more common than other Rot-n’s in the training data, since Rot-13 is a popular “spoiler-free way to share information”, e.g., for online puzzle forums.
In short, Embers of Autoregression is sort of an “evolutionary psychology” for LLMs—it shows that the way LLMs are trained leaves strong traces in the biases the models have in solving problems.
Here’s the paper’s bottom line:
First, we have shown that LLMs perform worse on rare tasks than on common ones, so we should be cautious about applying them to tasks that are rare in pretraining data. Second, we have shown that
LLMs perform worse on examples with low-probability answers than ones with high-probability answers, so we should be careful about using LLMs in situations where they might need to produce low-
probability text. Overcoming these limitations is an important target for future work in AI.
Paper 2: Factors affecting “chain of thought” prompting
Paper title: Deciphering the Factors Influencing the Efficacy of Chain-of-Thought: Probability, Memorization, and Noisy Reasoning
Authors: Akshara Prabhakar, Thomas L. Griffiths, R. Thomas McCoy
This paper, which shares two authors with the previous paper, looks in depth at chain-of-thought (CoT) prompting on the shift-cipher task.
As I discussed in my earlier post on LLM reasoning, CoT prompting has been claimed to enable robust reasoning in LLMs. In CoT prompting, the prompt includes an example of a problem, as well as the
reasoning steps to solve it, before posing a new problem. Here are two examples of the prompts that the authors used for shift ciphers; the one on the top doesn’t use CoT prompting, whereas the one
on the bottom does:
The authors tested several models, including GPT-4, Claude 3.0, and Llama 3.1. Interestingly, they found that, given prompts without CoT, these models get close to zero accuracy for most shift levels
(n); when using prompts with CoT like the one above, they achieve much higher accuracy (e.g., 32% for GPT-4) across shift levels.
The authors cite four possible ways LLMs can appear to be “reasoning”, each of which makes different predictions about its pattern of errors.
(1) Memorization: The model is repeating reasoning patterns memorized from training data. This would predict that accuracy will depend on the task’s frequency in the training data (e.g., recall that
for shift ciphers, Rot-13 is much more frequent in internet data than other Rot-n values).
(2) Probabilistic Reasoning: The model is choosing output that is most probable, given the input. This is influenced by the probability of token sequences learned during training. This kind of
reasoning would predict that LLMs will be more accurate on problems whose answers (the generated output) are sequences with higher probability.
(3) Symbolic Reasoning: The model is using deterministic rules that work perfectly for any input. This would predict 100% accuracy no matter what form the task takes.
(4) Noisy Reasoning: The model is using an approximation to symbolic reasoning in which there is some chance of making an error at each step of inference. This would predict that problems that
require more inference steps should produce worse accuracy. For shift ciphers, these would be problems that require more shift steps in the alphabet.
To cut to the chase, the authors found that LLMs with CoT prompting exhibit a mix of memorization, probabilistic reasoning, and noisy reasoning. Below is the accuracy of Claude 3.0 as a function of
shift-level n; the other models had a similar accuracy distribution. You can see that at the two ends (low and high n) the accuracy is relatively high compared with most of the middle n values. This
is a signature of noisy reasoning, since the lowest and highest n values require the fewest inference steps. (Think of the alphabet as a circle; Rot-25, like Rot-1, requires only one inference step.
In Rot-25, each letter would be encoded as the letter that immediately precedes it.)
The big bump in the middle at Rot-13 is a signature of memorization—the accuracy of models at this shift level is due to its high frequency in the training data. The authors showed via other
experiments that probabilistic reasoning is also a factor—see their paper for details.
Here’s the authors’ bottom line:
CoT reasoning can be characterized as probabilistic, memorization-influenced noisy reasoning, meaning that LLM behavior displays traits of both memorization and generalization.
These results are intriguing, but so far limited to the single task of shift ciphers. I hope to see (and maybe do myself) similar studies with other kinds of tasks.
Paper 3: Testing the robustness of LLMs on variations of simple math word problems
Paper title: GSM-Symbolic: Understanding the Limitations of Mathematical Reasoning in Large Language Models
Authors:Iman Mirzadeh, Keivan Alizadeh, Hooman Sharokhi, Oncel Tuzel, Samy Bengio, Mehrdad Farajtabar
This paper, from a research group at Apple, tests the robustness of several LLMs on a reasoning benchmark consisting of grade school math word problems. The benchmark, GSM8K, has been used in a lot
of papers to show that LLMs are very good at simple mathematical reasoning. Both OpenAI’s GPT-4 and Anthropic’s Claude 3 get around 95% of these problems correct, without any fancy prompting.
But to what extent does this performance indicate robust reasoning abilities, versus memorization (of these or similar problems in the training data) or, as the authors ask, “probabilistic
pattern-matching rather than formal reasoning”?
To investigate this, the authors take each problem in the original dataset and create many variations on it, by changing the names, numbers, or other superficial aspects of the problem, changes that
don’t affect the general reasoning required. Here’s an illustration from their paper of this process:
They test several LLMs on this set of variations, and find that in all cases, the models’ accuracy decreases from that on the original benchmark, in some cases by a lot, though on the best models,
such as GPT-4o, the decrease is minimal.
Going further, the authors show that adding irrelevant information to the original problems causes an even greater drop in accuracy than changing names or numbers. Here’s an example of adding
irrelevant information (in pink) to a word problem:
Even the very best models seem remarkably susceptible to being fooled by such additions. This figure from the paper shows the amount by which the accuracy drops for each model:
Here, each bar represents a different model and the bar length is the difference between the original accuracy on GSM8K and on the version where problems have irrelevant information (what they call
the “GSM-NoOP” version).
The bottom line from this paper :
Our extensive study reveals significant performance variability across different instantiations of the same question, challenging the reliability of current GSM8K results that rely on
single-point accuracy metrics.
The introduction of GSM-NoOp [i.e., adding irrelevant information] exposes a critical flaw in LLMs’ ability to genuinely understand mathematical concepts and discern relevant information for
Ultimately, our work underscores significant limitations in the ability of LLMs to perform genuine mathematical reasoning.
This paper, released just a couple of weeks ago, got quite a lot of buzz in the AI / ML community. People who were already skeptical of claims of LLM reasoning embraced this paper as proof that “the
emperor has no clothes”, and called the GSM-NoOP results “particularly damning”.
People more bullish on LLM reasoning argued that the paper’s conclusion—that current LLMs are not capable of genuine mathematical reasoning—was too strong, and hypothesized that current LLMs might be
able to solve all these problems with proper prompt engineering. (However, I should point out, when LLMs succeeded on the original benchmark without any prompt engineering, many people cited that as
“proof” of LLMs’ “emergent” reasoning abilities, and they didn’t ask for more tests of robustness.)
Others questioned whether humans who could solve the original problems would also be tripped up by the kinds of variations tested in this paper. Unfortunately, the authors did not test humans on
these new problems. I would guess that many (certainly not all) people would also be affected by such variations, but perhaps unlike LLMs, we humans have the ability to overcome such biases via
careful deliberation and metacognition. But discussion on that is for a future post.
I should also mention that a similar paper was published last June, also showing that LLMs are not robust on variations of simple math problems,
In conclusion, there’s no consensus about the conclusion! There are a lot of papers out there demonstrating what looks like sophisticated reasoning behavior in LLMs, but there’s also a lot of
evidence that these LLMs aren’t reasoning abstractly or robustly, and often over-rely on memorized patterns in their training data, leading to errors on “out of distribution” problems. Whether this
is going to doom approaches like OpenAI’s o1, which was directly trained on people’s reasoning traces, remains to be seen. In the meantime, I think this kind of debate is actually really good for
the science of LLMs, since it spotlights the need for careful, controlled experiments to test robustness—experiments that go far beyond just reporting accuracy—and it also deepens the discussion of
what reasoning actually consists of, in humans as well as machines.
If you want to read further, here is a list of some recent papers that test the robustness of reasoning in LLMs (including the papers discussed in this post).
Embers of Autoregression Show How Large Language Models Are Shaped By the Problem They Are Trained To Solve
GSM-Symbolic: Understanding the Limitations of Mathematical Reasoning in Large Language Models
Deciphering the Factors Influencing the Efficacy of Chain-of-Thought: Probability, Memorization, and Noisy Reasoning
Reasoning or Reciting? Exploring the Capabilities and Limitations of Language Models Through Counterfactual Tasks
Faith and Fate: Limits of Transformers on Compositionality
Phenomenal Yet Puzzling: Testing Inductive Reasoning Capabilities of Language Models with Hypothesis Refinement
Do Large Language Models Understand Logic or Just Mimick Context?
Alice in Wonderland: Simple Tasks Showing Complete Reasoning Breakdown in State-Of-the-Art Large Language Models
Beyond Accuracy: Evaluating the Reasoning Behavior of Large Language Models - A Survey
Functional Benchmarks for Robust Evaluation of Reasoning Performance, and the Reasoning Gap
A Peek into Token Bias: Large Language Models Are Not Yet Genuine Reasoners
Using Counterfactual Tasks to Evaluate the Generality of Analogical Reasoning in Large Language Models
Evaluating LLMs’ Mathematical and Coding Competency through Ontology-guided Interventions
"I think this kind of debate is actually really good for the science of LLMs"
I would agree with that. It is interesting to me that so many clever people are so uncritical about something so important to them. I might put that down to a steady diet of sci-fi and computer
code, rather than a deep education about cognition, and indeed the hype cycle that's been used to fund this research and monetize the results. We're probably all guilty of that at some level I
think, so perhaps it's best not to throw stones.
The really interesting issue to me is that it turns out the semantics captured by language are so amenable to an analysis just of syntax using statistical patterns. We seem to have found the scale
at which this starts to happen, which is very large but not infinite. Yet this is not so surprising I think. The human mind is prodigious, but it's not infinite either.
But I think our minds are doing more than predicting tokens given a massive set of examples. Funnily enough, the question what we are really doing when we think and talk still remains fundamentally
unanswered, even if we now know that we do these things using a tractable tool.
Expand full comment
Excellent article on one of the key limitations of LLMs (reasoning). The other (IMO) is the extremely shallow internal world model (required for genuine understanding of the real world) that is
constructed by the LLM training process. Unless both of these problems (reasoning and understanding) can be robustly resolved, LLM cognition, and therefore the cognition of any agent or robot built
on top of it, will be severely limited. It is extremely unlikely (IMO) that any LLM-based system will ever resolve these fundamental problems to the extent required for human-level AGI.
Expand full comment
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Minimax-Function with the AOA Solver? | AIMMS Community
Hi all,
i still need to implement a Minimax-Regret Function into Aimms. The Goal--Function is based on a deviation calculation by using the goal-programming approach. Therefore i added all the necessary
constraints and variables. My main model looks like that:
MathProgramm: Minimax_Regret
Decision Variable Regret: max(k,sum(j,abs_Regret(j,k))) (k=scenarios and j= potential locations)
myGmp := GMP::Instance::Generate(Minimax_Regret);
GMPOuterApprox::CreateStatusFile :=1;
GMPOuterApprox::UseMultistart :=1;
GMPOuterApprox::DoOuterApproximation (myGmp);
But my model still genereates the following error:
GMP::Instance::CreateMasterMIP(): The objective column should be in the coefficient matrix exactly once.
Does someone know how to fix this?
Thank you very much! | {"url":"https://community.aimms.com/aimms-language-12/minimax-function-with-the-aoa-solver-336?postid=667","timestamp":"2024-11-06T08:22:14Z","content_type":"text/html","content_length":"140884","record_id":"<urn:uuid:b8303d64-418d-416c-8865-e2dcbf5ff4d6>","cc-path":"CC-MAIN-2024-46/segments/1730477027910.12/warc/CC-MAIN-20241106065928-20241106095928-00715.warc.gz"} |
Oded Kafri, Author at
Is there a physical meaning to money? The book Entropy God’s dice game ends with a statement that freedom is entropy, equal opportunity is the second law, and the microstates in which we exist is our
destiny, that was determined in God’s dice game. The reasoning behind this, somewhat bombastic, statement is that the higher the freedom, the higher is the number of the microstates available for us,
and therefore the higher the entropy. The entropy has a maximum value when there are no constraints and each state (and therefore each microstate) has equal probability. It is also argued in the book
that in networks the number of links of a node represents its wealth. For example, in our society, Bill Gates is linked, that is to say, has access to many people, due to his wealth. In the paper “
Money, Information and Heat in Social Networks Dynamics” that was recently published at “Mathematical Finance Letters”, this argument is further investigated. In this paper a network is defined as a
microcanonical ensemble of states and links. The states are the possible pairs of the nodes in the nets, and the links are communication bits exchanged between the nodes. This net is an analogue to
people trading with money. This approach is consistent with the intuition of the book that wealth is entropy (information), and therefore money is entropy, which can be interpreted as a freedom to
choose from many options. Therefore, the physical meaning of money is entropy. In this case, money transfer is simulated by bits transfer which is heat (energy transfered). With analogy to bosons
gas, we define for these networks’ model: entropy, volume, pressure and temperature. We show that these definitions are consistent with Carnot efficiency (the second law) and ideal gas law.
Therefore, if we have two large networks: hot and cold having temperatures T[H] and T[C, ]and we remove Q bits (money) from the hot network to the cold network, we can save W profit bits. The profit
will be calculated from W< Q (1-T[H]/T[C]), namely, Carnot formula. In addition it is shown that when two of these networks are merged, the entropy increases. This explains the tendency of economic
and social networks to merge. The derived temperature of the net in model is the average number of links available per state.
Kolmogorov Complexity and Entropy
Kolmogorov complexity and entropy are described as related quantities, without explaining how. In order to explore the connection between these quantities, we have to understand that information is a
transfer of energetic content from Bob to Alice. Bob knows the content of his message, and therefore it carries no entropy (uncertainty) for him. However, it does carry entropy for Alice, who is
ignorant of its content. If Bob sends Alice N bits, the entropy for Alice is NLn2, no matter what is the message. Bob would like to send his message with the minimum number of bits possible. Why?
Because sending short message saves energy and time and also exposes the message to less noise. In our book we claim that Bob compress his files because he obeys Clausius inequality (the second law
of thermodynamics). Regardless of Bob’s reasons for compressing his file, the complexity of his message defines how efficiently he can do it.
Suppose Bob wants to send Alice the first N digits of the number Pi, compressed to its Shannon limit. Pi is an irrational number, and since Bob cannot find in it periodicity (the digits in irrational
numbers are random), Pi cannot be compressed by conventional methods. Nevertheless, it can be calculated by a simple function, and therefore Bob can send Alice the generating function of Pi, which
will enable Alice to produce the N digits of Pi at her leisure.
In order to send Alice a method of generating Pi, Bob has to find out how “complex” is Pi, or in other words, the simplest and shortest way of generating the digits of Pi. Hence, the minimum amount
of operations required to generate a given content is the complexity of the content, while the length of the file that carries the content is its entropy.
Unified evolution
Evolution theory was suggested as an answer to one of the most intriguing questions: How was the variety of biological species on earth created? Contemporary evolution theory is based on biological
and chemical changes. Many believe that life started from some primordial chemical soup.
Does evolution have a deeper root than chemistry? Is there a physical law that is responsible for evolution? Our book advocates a positive answer.
There is a semantic differentiation between “natural things” and “artificial things”. Namely, natural things are created by nature and artificial things were made by us. Is there a justification for
this egocentric view of the world? Here it is argued that both “we” and the “things” we make are all part of nature and are subject to the same laws. From holistic point of view, the computers,
telephones, cars, roads, etc. are all created spontaneously in nature by the same invisible hand that created us. There are no special laws of nature for humans’ actions. A unified evolution theory
should explain the origin of our passion to make tools, to develop technology, to create social networks, to trade and to seek “justice”.
Here we claim that entropy, in its propensity to increase, is the invisible hand of both our life and our actions.
Understanding Uncertainty
Probability is a well-established mathematical branch of high importance. In mathematics probability is calculated with consistency with a set of axioms. Sometimes uncertainty is defined by the
statisticians according to probability rules.
For example: Suppose Bob plans to dine with Alice in the evening: there is 1/10 chance the he will not be available. Since the total probability is 1 (Kolmogorov 2nd axiom), therefore there is 9/10
chance that they will dine together and 1/10 that they will not. If there is a chance of 1/2 that Bob will not be available, the total probability is still 1, but now it comprises of a probability of
1/2 for a joint dinner. In General, if the probability that Bob will not be available is p, it implies that the probability of the joint dinner is 1-p.
In this example some statisticians may say that the uncertainty of having a joint dinner in the first case is 10%, and 50% in the second. This is not correct.
Uncertainty is defined by its Shannon’s entropy and its expression for the joint dinner is,
Usually engineers use the logarithm in base 2 and the uncertainty is expressed in bits. If p=1/2 then the uncertainty is 1 bit (one or zero). If p=1/10 then the uncertainty is 0.46 bit, namely, it
is little less than half a bit. The entropy is a physical quantity which is a function of a mathematical quantity p, but unlike mathematical quantities that exist in a formal mathematical space
defined by its axioms, entropy is bounded by a physical law, the second law of thermodynamics. Namely, entropy tends to increase to its maximum.
The maximum value of S, in our example, is ln2 when p=1/2. Does it mean that nature prefers the chance of Bob not being available for dinner with Alice to be 1/2, where the entropy is at its
maximum? The answer, surprisingly for a mathematician, is yes! If we will examine many events of this nature we will see a (bell-like) distribution that has a pick at the value p=1/2.
Similarly, the average of many polls in which one picks, randomly, 1 out of 3 choices, will be a distribution of 50%:29%:21% and not 33%:33%:33% as is expected from simple probability calculations.
Laws of nature (the second law) can tell us something about the probabilities of probabilities. The function that describes the most probable distribution of the various events is called the
distribution function.
The distribution functions in nature that are the result of the tendency of entropy to maximize are, among others:
• Bell like distributions for humans: mortality, heights, IQ etc.
• Long tail distributions for humans: Zipf law in networks and texts, Benford’s law in numbers, Pareto Law for wealth etc.
Benford law
Benford law is about the uneven distribution of digits in random decimal files. It was discovered by Simon Newcomb by way of noting consistent differentiation in the wear-and-tear of logarithmic
books at the end of the 19th century. The phenomenon was re-discovered by Frank Benford in 1938.
Newcomb found and stated the law in its most general form by declaring that mantissa is uniformly distributed. Benford set out to check the law empirically and also guessed successfully its equation
for the 1st digits :ρ(n)=log[10][(n+1)/n]:namely, the probability of digit n (n=1,2,3,…,8,9), ρ(n) is monotonically decreasing such that digit 9 will be found about 6.5 times less than digit 1. The
law is also called “the first digit law”. Benford has shown that this law holds for many naturally generated decimal files.
Misconception: Benford law applies only for the first digits of numbers.
NOT TRUE. Benford law holds for the first, second, third, or any other digit order of decimal data. The law was originally stated mostly in terms of 1st digit sense which does not include the 0
digit. Second and higher orders naturally incorporate the 0 digit as a distinct possibility of course.
Benford law is applied for any decimal file that is compressed to Shannon limit. In a binary file at the Shannon limit all the bits excluding the 0’s are 1. In the case of 0, 1, 2 counting system the
ratio between the digits 1 and 2 is 63:37 and in 0,1,2,3 counting system, the ratios between the digits 1, 2 and 3 are 50:29:21. In the same way a compressed decimal file has Benford’s law
Why calculating the Shannon limit does not gives us information about the “0”s? Strictly speaking zero has no entropy and therefore it does not count. Or in a formal way entropy is logarithmic and
this is also the reason why the changes in frequencies of the digits are logarithmic (exactly like the distances in a slide rule).
Why entropy is logarithmic? Because, that IS the way God plays dice! | {"url":"https://www.entropy-book.com/author/oded/","timestamp":"2024-11-02T04:15:53Z","content_type":"text/html","content_length":"34663","record_id":"<urn:uuid:01a0b869-e80e-4b94-8c9f-4d8010459e77>","cc-path":"CC-MAIN-2024-46/segments/1730477027677.11/warc/CC-MAIN-20241102040949-20241102070949-00355.warc.gz"} |
Dataflow | Proxima
Differential Dataflow is already a well-defined computational model, and there already exists a robust distributed implementation. Thus we can implement Maru operators "abstractly" as a composition
of multiple differential dataflow operators.
Even though the existing implementation relies upon a centralized scheduler, a decentralized implementation can be made of the same computational model, upon which we can run Maru operators just the
same as in the centralized version.
Aside: You might ask the question, 'why not just make it decentralized from the beginning"? Because proving complex streaming operations is hard enough, not to mention the fact that we need to
build a business. One problem at a time, anon.
In this article, we refer to Multiset Hashes as "MSH". See the chapter about them in background-knowledge for more information before reading this article.
Maru Operator Definition
A Maru operator is the combination of a set of input maru streams, output maru streams, some form of logic that, given a delta, produces another delta, and an optional arrangement (In Maru's initial
release, arrangements will not be supported).
A "maru stream" is defined as a group of three independent differential dataflow streams:
Delta stream
stream of differential dataflow "deltas".
cumulative MSH stream
each element is the cumulative multiset hash of every delta that came "before" that element's timestamp.
cumulative proof stream
Proofs take the cumulative multiset hash of both the previous operator's input and the previous operator's output as public input, and prove that the collection committed to by the output MSH
is correct given the collection committed to by the input MSH
Unary Operator Implementation
A unary operator's execution goes like this:
for each input delta, run the operator's logic to determine the corresponding set of output deltas. Emit the execution trace for the logic.
a. generate / accumulate proofs:
i. for each execution trace, generate a STARK-AIR proof that the logic was executed correctly.
ii. recursively accumulate each of these proofs into a single "cumulative operator proof", and, in doing so, "build up" the "expected" cumulative multiset hash in the proofs
iii. recursively accumulate this "cumulative operator proof" with that of the previous operator's cumulative stream proofs, checking that the input MSH for the current operator and the
output MSH for the previous operator match. The result is the output cumulative proof stream.
b. update output's cumulative multiset hash with new deltas produced by the operator's logic.
output the output delta stream, cumulative MSH stream, and cumulative proof stream.
We can implement this using differential dataflow primitives using the configuration shown in the diagram below.
Non-Unary Operator Implementation
Operators with different numbers of inputs and/or outputs are implemented the same way, simply duplicating the relevant parts.
Firstly, we generate one STARK proof for the emission of all outputs given all inputs. Then we simply have multiple recursive accumulation pipelines running in parallel, one for each output maru
stream. In each of those pipelines, we have a sequence of "binary map" operators that merge the cumulative operator proofs with the previous operator's cumulative stream proofs, one for each input | {"url":"https://docs.proxima.one/proxima/maru/how-maru-works/dataflow","timestamp":"2024-11-08T20:42:37Z","content_type":"text/html","content_length":"180266","record_id":"<urn:uuid:3ee5208a-db45-4694-b927-7547dfa37d85>","cc-path":"CC-MAIN-2024-46/segments/1730477028079.98/warc/CC-MAIN-20241108200128-20241108230128-00071.warc.gz"} |
Square Yards Calculator
Last updated:
Square Yards Calculator
This square yards calculator determines the area size based on given length and width.
You may freely convert square feet to square meters, acres to square feet, and finally square yards to square feet (sq yard to sq feet) with this tool. In the text, you'll find an explanation of how
to calculate square yards and a practical example of use. You may also find useful this square footage calculator.
How many square yards in an acre? Find this and more examples of the practical uses of yd^2 below.
What is a square yard? How to calculate square yard?
Calculating areas and surfaces can be useful, especially if you're planning to put down some grass on your garden 🌳 or maybe concrete down the driveway. Check our concrete calculator.
Whatever it is, this square yards calculator helps to estimate the needed amount of any material.
To get the square yard, you need to know the circumference of an area (take a look at the circumference calculator) or, even better, its length and width. The formula is length multiplied by width:
area = length × width
Both length and width may be in any unit, but it's best if both were in yards to obtain square yards. Just remember that if you count it yourself, the units of both width and length have to be the
same. In this calculator, you can switch between any unit you need.
What is a square yard compared to the other units, e.g. sq ft to sq yd?
One square yard equals about:
• 0.83612736 m^2;
• 9 ft^2;
• 1296 in^2; and
• 0.00020661157 acres.
Square yards to acres
Now that you know about conversion from square yards to acres, let's see how to convert the other way around. How many square yards in an acre? Exactly 4,840 yd^2.
Here you'll find some information on how areas were calculated in medieval times. An acre is a land area unit, and it corresponds to about 40% of a hectare. How can we be more precise? One acre is
43,560 square feet or 4,047 m^2.
The word "acre" comes from Old English æcer. Its primal meaning was an open field. In medieval Britain, people counted the size of farms 🌻 and landed properties in acres. On the other hand, in the
US, square miles gained more popularity.
If you need a more modern comparison to an acre in size, think of a football field. 🏈 To convert square feet to acres you need to multiply the number of acres by 43,560, while the size of a football
field estimated with surface area calculator is 48,000 ft^2.
This square yards calculator - an example
Imagine you want to put on some turf in your garden, but all you have is the width and length of the fence in feet. Yet, in the shop, they ask for the area given in square yards.
• Width - 120 ft
• Length - 80 ft
$yardage = 120 ft × 80 ft$
$yardage = 9600 ft^\text{2}$
$yardage = 1066.7 yd^\text{2}$
Then, the yardage is 1066.7 yd^2. You can change between the units freely, and the results will recalculate. The only limit is your imagination (if you'd like any funky units, we'd love to hear from
Let's go back to the calculator. Now you know the area to cover with grass. But, what is the final cost of the materials? The price for one yd^2 is $3.
cost = yardage * price per one
cost = 1066.7 * $3
cost = $3,200
So, the total cost of grassing your garden is $28,800.
How many square yards are in an acre?
There are 4840 sq. yards in one acre.
The formula looks like this:
acre = sq. yard / 4840
Or, if you want to convert acre to square yard:
sq. yard = acre × 4840
How can I convert square feet to square yards?
We know that yard and feet have a 1:3 relationship. So, these values squared would make it 1:9.
Yd² = Ft² / 9
Ft² = Yd² × 9
To convert square feet to square yards:
1. Note the value in square feet.
2. Divide it by 9.
3. The result is the value in square yards.
How can I calculate square yards?
Suppose you have the measurement of a room in yards and want to find the area to get it carpeted. If the length of the room is 5 yards and the width is 7 yards, apply the area formula:
Area of room (yd²) = Length (yd) × Width (yd)
Area = 5 × 7
Area = 35 yd²
This is equal to 315 ft².
How many feet is 40 yards?
40 yards is 120 ft.
One yard is equal to 3 feet, so the conversion is pretty simple.
Feet = 3 × Yard
So, whatever amount you have in yards, multiply it by 3, and you have the amount in feet.
You can also do it the other way around.
Yard = Feet / 3 | {"url":"https://www.omnicalculator.com/construction/square-yards","timestamp":"2024-11-06T14:38:40Z","content_type":"text/html","content_length":"451025","record_id":"<urn:uuid:bfb30663-1cae-47dc-a441-d1652046945f>","cc-path":"CC-MAIN-2024-46/segments/1730477027932.70/warc/CC-MAIN-20241106132104-20241106162104-00421.warc.gz"} |
1. “Lambda-Lambda and N-Xi interactions from Lattice QCD near the physical point”, K. Sasaki, S. Aoki, T. Doi, S. Gongyo, T. Hatsuda, Y. Ikeda, T. Inoue, T. Iritani, N. Ishii, K. Murano, T. Miyamoto
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3. “Baryon-baryon interactions at short distances — constituent quark model meets lattice QCD”, A. Park, S. H. Lee, T. Inoue and T. Hatsuda, Eur. J. Phys. (2020) in press.
4. “Probing ΩΩ & p-Ω dibaryons with femtoscopic correlations in relativistic heavy-ion collisions”, K. Morita, S. Gongyo, T. Hatsuda, T. Hyodo, Y. Kamiya and A. Ohnishi, Phys. Rev. C 101, 015201
5. “Hermitizing the HAL QCD potential in the derivative expansion”, S. Aoki, T. Iritani and K. Yazaki, PTEP 2, 023 (2020).
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Letter 890, L4 (2020).
7. “r-process enrichment in the Galactic halo characterized by nucleosynthesis variation in the ejecta of coalescing neutron star binaries”, T. Tsujimoto, N. Nishimura, K. Kyutoku, The Astrophysical
Journal 889, 119 (2020).
8. “Superfluid Phase Transitions and Effects of Thermal Pairing Fluctuations in Asymmetric Nuclear Matter”, H. Tajima, T. Hatsuda, P. van Wyk & Y. Ohashi, Scientific Reports 9, 18477 (2019).
9. “Effective repulsion in dense quark matter from nonperturbative gluon exchange”, Y. Song, G. Baym, T. Hatsuda and T. Kojo, Phys. Rev. D 100, 034018 (2019).
10. “New Neutron Star Equation of State with Quark-Hadron Crossover”, G. Baym, S. Furusawa, T. Hatsuda, T. Kojo and H. Togashi, The Astrophysical Journal 885, 42 (2019).
11. “Continuity of vortices from the hadronic to the color-flavor locked phase in dense matter”, M. G. Alford, G. Baym, K. Fukushima, T. Hatsuda and M. Tachibana, Phys. Rev. D 99, 036004 (2019).
12. “Consistency between Lüscher’s finite volume method and HAL QCD method for two-baryon systems in lattice QCD”, T. Iritani, S. Aoki, T. Doi, T. Hatsuda, Y. Ikeda, T. Inoue, N. Ishii, H. Nemura,
and K. Sasaki, [HAL QCD Collaboration], JHEP 1903, 007 (2019).
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1. “Gravitational-wave and multi-messenger astronomy”, K. Kyutoku, Strings and Fields 2019, Kyoto, Japan, August 19-23 (2019)
2. “Lattice results on dibaryons and baryon-baryon interactions”, S. Aoki, 18th Int. Conference on Hadron Spectroscopy and Structure (HADRON 2019), Guilin, China, Aug. 16-21 (2019)
3. “Nuclear forces from lattice QCD”, T. Hatsuda, The 27th Int. Nuclear Physics Conference (INPC 2019), Glasgow, UK, July 29-Aug. 2 (2019)
4. “Theoretical and practical progresses in the HAL QCD method”, S. Aoki, 37th Int. Symposium on Lattice Field Theory (Lattice 2019), Wuhan, China, June 16-22 (2019)
5. “Newest results from lattice HAL QCD on hyperon-nucleon and hyperon-hyperon interaction”, T. Hatsuda, The 18th Int. Conference on Strangeness in Quark Matter (SQM 2019), Bali, Italy, June 10-15
6. “Equation of State for Dense Matter and Neutron Star Merger”, T. Hatsuda, MOST-RIKEN workshop on ab initio theory in nuclear physics, Beijing, China, April 6-8 (2019)
7. “Microscopic equation of state for nuclear matter with the variational many-body theory”, H. Togashi, K. Nakazato, Y. Takehara, S. Yamamuro, H. Suzuki, and M. Takano, MOST-RIKEN Workshop on ab
initio theory in nuclear physics, Beijing, China, April 6-8 (2019)
8. “Nuclear Forces from Lattice QCD”, T. Hatsuda, The 2nd Int. Workshop on Quantum Many-Body Problems in Particle, Nuclear and Astrophysics, Nha Trang, Vietnam, March 7-11 (2019)
9. “Multi-messenger from compact binary mergers”, K. Kyutoku, The 10th International Workshop on Very High Energy Particle Astronomy in 2019 (VHEPA2019), Kashiwa, Japan, February 18-20 (2019)
10. “Baryon Interactions from Lattice QCD”, T. Hatsuda, The 8th Int. Conference on Quarks and Nuclear Physics (QNP2018), Tsukuba, Japan, Nov.13-17 (2018)
11. “Baryon-baryon interactions from lattice QCD”, T. Hatsuda, Int. Symposium on Simplicity, Symmetry and Beauty of Atomic Nuclei, Shanghai, China, Sept. 26-28 (2018)
12. “Neutron Star Core: Densest State of Matter”, T. Hatsuda, The Int. Symposium on Quantum Fluids and Solids (QFS2018), Tokyo, Japan, July 25-31 (2018)
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Conf. Proc. 2130, 020002 (2019)]
1. “Tidal disruption in black hole-neutron star binaries”, K. Kyutoku, Tidal Disruption Events: General Relativistic Transients, Kyoto, Japan, January 14-25 (2020)
2. “Hyperon mixing in hot and dense matter with the variational method”, H. Togashi, M. Takano, K. Nakazato, Y. Takehara, S. Yamamuro, H. Suzuki, K. Sumiyoshi and E. Hiyama, Theia-Strong 2020
Workshop 2019, Speyer, Germany, November 27 (2019)
3. “General purpose equation of state for astrophysical simulations with bare nuclear forces”, H. Togashi, M. Takano, K. Nakazato, Y. Takehara, S. Yamamuro, H. Suzuki, K. Sumiyoshi, and E. Hiyama,
International Workshop on Nuclear Physics for Astrophysical Phenomena, Tokyo, Japan, October 25 (2019)
4. “The q-qbar potential from Wilson loop and the qqbar potential from Bethe-Salpeter wave function”, N. Ishii, The ECT* workshop “Universal physics in Many-Body Quantum System –From Atoms to
Quarks”, Trento, Italy, Oct.7-11 (2019)
5. “What did we learn about neutron-star properties from AT 2017gfo?”, K. Kyutoku, Nucleosynthesis and electromagnetic counterparts of neutron-star mergers: Preparation for the new discovery, Kyoto,
Japan, March 11-30 (2019)
6. “Cluster Variational Method for Hyperonic Nuclear Matter with Coupled Channels”, H. Togashi and M. Takano, The 8th International Conference on Quarks and Nuclear Physics (QNP2018), Tsukuba,
Japan, Nov.13-17 (2018) [JPS Conf. Proc. 26, 031024 (2019)]
7. “Lattice QCD Study of the Nucleon-Charmonium Interaction”, T. Sugiura, Y. Ikeda and N. Ishii, The 8th International Conference on Quarks and Nuclear Physics (QNP2018), Tsukuba, Japan, Nov.13-17
(2018) [JPS Conf. Proc. 26, 031015 (2019)]
8. “Hyperon Forces from QCD and Their Applications”, T. Inoue, The 8th International Conference on Quarks and Nuclear Physics (QNP2018), Tsukuba, Japan, Nov.13-17 (2018) [JPS Conf. Proc. 26, 023018
9. “Charmonium-nucleon interactions from $2+1$ flavor lattice QCD”, T. Sugiura, Y. Ikeda and N. Ishii, 36th International Symposium on Lattice Field Theory (Lattice 2018), East Lansing, USA, July
22-28 (2018) [PoS LATTICE 2018, 093 (2019)] | {"url":"https://suuri.riken.jp/challenge_qcd/%E8%AB%96%E6%96%87%E3%83%AA%E3%82%B9%E3%83%88/","timestamp":"2024-11-04T11:18:07Z","content_type":"text/html","content_length":"29394","record_id":"<urn:uuid:928ca25c-0187-4900-b899-7e77d532376f>","cc-path":"CC-MAIN-2024-46/segments/1730477027821.39/warc/CC-MAIN-20241104100555-20241104130555-00276.warc.gz"} |
NCERT Solutions for Class 11 Economics Chapter 7 – CorrelationNCERT Solutions for Class 11 Economics Chapter 7 – Correlation
NCERT Solutions for Class 11 Economics Chapter 7 – Correlation
Page No 104:
Question 1:
The unit of correlation coefficient between height in feet and weight in kgs is
(i) kg/feet
(ii) percentage
(iii) non-existent
As there is non-existent of correlation between the height in feet and weight in kilograms, so the unit of correlation between the two is zero.
Question 2:
The range of simple correlation coefficient is
(i) 0 to infinity
(ii) minus one to plus one
(iii) minus infinity to infinity
The range of simple correlation coefficient is from (–) 1 to (+) 1
Question 3:
If r[xy ]is positive the relation between X and Y is of the type
(i) When Y increases X increases
(ii) When Y decreases X increases
(iii) When Y increases X does not change
When the variables Y and X share positive relationship (i.e. when Y and X both increases simultaneously), then the value of r[xy] is positive.
Page No 105:
Question 4:
If r[xy ]= 0 the variable X and Y are
(i) linearly related
(ii) not linearly related
(iii) independent
The value of r[xy] becomes 0 when the two variables are not linearly related to each other. It may happen that both the variables may be non-linearly related to each other. It does not necessarily
imply that both are independent of each other.
Question 5:
Of the following three measures which can measure any type of relationship
(i) Karl Pearson's coefficient of correlation
(ii) Spearman's rank correlation
(iii) Scatter diagram
Scatter diagram can measure any type of relationship whether the variables are highly related or not at all related. Just by looking at the diagram, the viewer can easily conclude the relationship
between the two variables involved. On the other hand, Karl Pearson's coefficient of correlation is not suitable for the series where deviations are calculated from assumed mean. Likewise, Spearman's
rank correlation also disqualifies to measure any kind of relationship as its domain is restricted only to the qualitative variables (leaving quantitative variables).
Question 6:
If precisely measured data are available the simple correlation coefficient is
(i) more accurate than rank correlation coefficient
(ii) less accurate than rank correlation coefficient
(iii) as accurate as the rank correlation coefficient
Generally, all the properties of Karl Pearson's coefficient of correlation are similar to that of the rank correlation coefficient. However, rank correlation coefficient is generally lower or equal
to Karl Pearson's coefficient. The reason for this difference between the two coefficients is because the rank correlation coefficient uses ranks instead of the full set of observations that leads to
some loss of information. If the precisely measured data are available, then both the coefficients will be identical.
Question 7:
Why is r preferred to covariance as a measure of association?
Although correlation coefficient is similar to the covariance in a manner that both measure the degree of linear relationship between two variables, but the former is generally preferred to
covariance due to the following reasons.
1. The value of the correlation coefficient |r| lies between 0 and 1. Symbolically r belongs to [-1 , 1]
2. The correlation coefficient is scale free.
Question 8:
Can r lie outside the -1 and 1 range depending on the type of data?
No, the value of r cannot lie outside the range of -1 to 1. If r = - 1, then there exists perfect negative correlation and if r = 1, then there exists perfect positive correlation between the two
variables. If at any point of time the calculated value of r is outside this range, then there must be some mistake committed in the calculation.
Question 9:
Does correlation imply causation?
No, correlation does not imply causation. The correlation between the two variables does not imply that one variable causes the other. In other words, cause and effect relationship is not a
prerequisite for the correlation. Correlation only measures the degree and intensity of the relationship between the two variables, but surely not the cause and effect relationship between them.
Question 10:
When is rank correlation more precise than simple correlation coefficient?
Rank Correlation method is more precise than simple correlation coefficient when the variables cannot be measured quantitatively. In other words, rank correlation method measures the correlation
between the two qualitative variables. These variable attributes are given the ranks on the basis of preferences. For example, selecting the best candidate in a dance competition depends on the ranks
and preferences awarded to him/her by the judges. Secondly, the rank correlation method is preferred over the simple correlation coefficient when extreme values are present in the data. In such case
using simple correlation coefficient may be misleading.
Question 11:
Does zero correlation mean independence?
Correlation measures the linear relationship between the two variables. So, r being 0 implies the absence of linear relationship. But they may be non-linearly related. Hence, if two variables are not
correlated, it does not necessarily follow that they are independent.
Question 12:
Can simple correlation coefficient measure any type of relationship?
No, the simple correlation coefficient cannot measure any type of relationship. The simple correlation coefficient can measure only the direction and magnitude of linear relationship between the two
variables. It cannot measure non-linear relationship like quadratic, trigonometric, cubic, etc. Therefore, in such cases, the purview of simple correlation coefficient falls short. For example, the
simple correlation coefficient may depict that X and Y are not correlated in the equation X= Y^2, hence it may be concluded that both the variables are independent, but such conclusion may be wrong.
Question 13:
Collect the price of five vegetables from your local market every day for a week. Calculate their correlation coefficients. Interpret the result.
This question is about multivariate correlation that is out of syllabus
Question 14:
Measure the height of your classmates. Ask them the height of their benchmate. Calculate the correlation coefficient of these two variables. Interpret the result.
Height of Classmate Height of Benchmate
X Y
X Y XY X^2 Y^2
Question 15:
List some variables where accurate measurement is difficult.
The following are the some variables where the accurate measurement is difficult.
1. Temperature and number of people falling ill.
2. Change in temperature with the height of mountain.
3. Low rainfall and agricultural productivity
4. High population growth and degree of poverty
5. Number of tourists and change in the political atmosphere in India.
Question 16:
Interpret the values of r as 1, -1 and 0.
The value of r being 1 implies that there is a perfect positive correlation between the two variables involved. A high value of r (i.e. close to 1) represents a strong positive linear relationship
between the two variables.
If r = -1, then the correlation is perfectly negative. A negative value of r indicates an inverse relation. A low value of r (i.e. close to -1) represents a strong negative linear relationship
between the variables. On the other hand, if the value of r = 0, then it implies that the two variables are uncorrelated to each other. But this should not be misunderstood as the variables are
independent of each other. The value of r equals zero confirms only the non-existence of any linear relation but the variables may be non-linearly related to each other.
Question 17:
Why does rank correlation coefficient differ from Pearsonian correlation coefficient?
Generally, all the properties of Karl Pearson's coefficient of correlation are similar to that of the rank correlation coefficient. However, rank correlation coefficient is generally lower or equal
to Karl Pearson's coefficient. Rank correlation coefficient is generally preferred to measure the correlation between the two qualitative variables. These variable attributes are given the ranks on
the basis of preferences. The difference between the two coefficients is due to the fact that the rank correlation coefficient uses ranks instead of the full set of observations that leads to some
loss of information. If the precisely measured data are available, then both the coefficients will be identical. Secondly, if extreme values are present in the data, then the rank correlation
coefficient is more precise and reliable and consequently its value differs from that of the Karl Pearson's coefficient.
Question 18:
Calculate the correlation coefficient between the heights of fathers in inches (X) and their sons (Y)
X 65 66 57 67 68 69 70 72
Y 67 56 65 68 72 72 69 71
X Y XY X^2 Y^2
Note: As per textbook, correlation coefficient is 0.603. However, as per the above solution, correlation coefficient should be 0.44.
Question 19:
Calculate the correlation coefficient between X and Y and comment on their relationship:
X -3 -2 -1 1 2 3
Y 9 4 1 1 4 9
│X │Y│ XY │X^2 │Y^2 │
│-3│9│-27 │ 9 │ 81 │
│-2│4│ -8 │ 4 │ 16 │
│-1│1│ -1 │ 1 │ 1 │
│1 │1│ 1 │ 1 │ 1 │
│2 │4│ 8 │ 4 │ 16 │
│3 │9│ 27 │ 9 │ 81 │
As the value of r is zero, so there is no linear correlation between X and Y
Page No 106:
Question 20:
Calculate the correlation coefficient between X and Y and comment on their relationship
X 1 3 4 5 7 8
Y 2 6 8 10 14 16
│X│Y │ XY │X^2 │Y^2 │
│1│2 │ 2 │ 1 │ 4 │
│3│6 │ 18 │ 9 │ 36 │
│4│8 │ 32 │ 16 │ 64 │
│5│10│ 50 │ 25 │100 │
│7│14│ 98 │ 49 │196 │
│8│16│128 │ 64 │256 │
As the correlation coefficient between the two variables is +1, so the two variables are perfectly positive correlated. | {"url":"https://www.studyguide360.com/2019/03/ncert-solutions-for-class-11-economics-chapter-7-correlation.html","timestamp":"2024-11-08T06:03:04Z","content_type":"application/xhtml+xml","content_length":"479490","record_id":"<urn:uuid:723c11d3-15f2-4bb4-87db-6bc0efacbb9f>","cc-path":"CC-MAIN-2024-46/segments/1730477028025.14/warc/CC-MAIN-20241108035242-20241108065242-00670.warc.gz"} |
The BAS R package is designed to provide an easy to use package and fast code for implementing Bayesian Model Averaging and Model Selection in R using state of the art prior distributions for linear
and generalized linear models. The prior distributions in BAS are based on Zellner’s g-prior or mixtures of g-priors for linear and generalized linear models. These have been shown to be consistent
asymptotically for model selection and inference and have a number of computational advantages. BAS implements three main algorithms for sampling from the space of potential models: a deterministic
algorithm for efficient enumeration, adaptive sampling without replacement algorithm for modest problems, and a MCMC algorithm that utilizes swapping to escape from local modes with standard
Metropolis-Hastings proposals.
The stable version can be installed easily in the R console like any other package:
On the other hand, I welcome everyone to use the most recent version of the package with quick-fixes, new features and probably new bugs. It’s currently hosted on GitHub. To get the latest
development version from GitHub, use the devtools package from CRAN and enter in R:
You can check out the current build status before installing.
Installing the package from source does require compilation of C and FORTRAN code as the library makes use of BLAS and LAPACK for efficient model fitting. See CRAN manuals for installing packages
from source under different operating systems.
To begin load the package:
The two main function in BAS are bas.lm and bas.glm for implementing Bayesian Model Averaging and Variable Selection using Zellner’s g-prior and mixtures of g priors. Both functions have a syntax
similar to the lm and glm functions respectively. We illustrate using BAS on a simple example with the famous Hald data set using the Zellner-Siow Cauchy prior via
BAS has summary, plot coef, predict and fitted functions like the lm/glm functions. Images of the model space highlighting which variable are important may be obtained via
Run demo("BAS.hald") or demo("BAS.USCrime") or see the package vignette for more examples and options such as using MCMC for model spaces that cannot be enumerated.
Generalized Linear Models
BAS now includes for support for binomial and binary regression, Poisson regression, and Gamma regression using Laplace approximations to obtain Bayes Factors used in calculating posterior
probabilities of models or sampling of models. Here is an example using the Pima diabetes data set with the hyper-g/n prior:
Pima.hgn = bas.glm(type ~ ., data=Pima.tr, method="BAS", family=binomial(),
betaprior=hyper.g.n(), modelprior=uniform())
Note, the syntax for specifying priors on the coefficients in bas.glm uses a function with arguments to specify the hyper-parameters, rather than a text string to specify the prior name and a
separate argument for the hyper-parameters. bas.lm will be moving to this format sometime in the future.
Feature Requests and Issues
Feel free to report any issues or request features to be added via the github issues page.
For current documentation and vignettes see the BAS website
This material is based upon work supported by the National Science Foundation under Grant DMS-1106891. Any opinions, findings, and conclusions or recommendations expressed in this material are those
of the author(s) and do not necessarily reflect the views of the National Science Foundation. | {"url":"https://cran.mirror.garr.it/CRAN/web/packages/BAS/readme/README.html","timestamp":"2024-11-11T11:34:40Z","content_type":"application/xhtml+xml","content_length":"13070","record_id":"<urn:uuid:1463f9a4-f862-482b-a4c5-260d902b9b93>","cc-path":"CC-MAIN-2024-46/segments/1730477028228.41/warc/CC-MAIN-20241111091854-20241111121854-00766.warc.gz"} |
IMO Syllabus for Class 6
The IMO for Class 6 has a total of 50 questions across four sections and a time duration of 60 minutes to complete the test. The total marks for the International Mathematics Olympiad are 60. Here is
a breakdown of the four sections in the syllabus and the exam pattern of Class 6 IMO:
Section 1: Logical Reasoning
This section aims to help students improve their learning abilities and feel comfortable approaching nonverbal and verbal reasoning on several topics.
The section includes 15 questions and carries a total of 15 marks.
The topics include - Verbal and non-verbal logical reasoning.
Section 2: Mathematical Reasoning
This section tests the student's aptitude in mathematics, especially their ability to apply their knowledge of numbers and problem-solving abilities.
The section includes 20 questions and carries a total of 20 marks.
The topics include - Practical Geometry, Symmetry, Ratio and Proportion, Algebra, Mensuration, Data Handling, Decimals, Fractions, Integers, Understanding Elementary Shapes, Basic Geometrical Ideas,
Playing with Numbers, Whole Numbers, Knowing our Numbers.
Section 3: Everyday Mathematics
This section assesses the student's understanding of the application of mathematics in daily life and real-world circumstances.
This section includes 10 questions and carries a total of 10 marks.
The syllabus for this section is the same as that of Section 2.
Section 4: Achiever's section
This section is designed to challenge students to think more critically and use higher-order thinking abilities to solve complex problems. This section includes five questions and carries a total of
15 marks.
The syllabus for this section is also the same as that of Section 2. | {"url":"https://olympiadtester.in/pages/imo-syllabus-for-class-6","timestamp":"2024-11-06T06:10:43Z","content_type":"text/html","content_length":"299332","record_id":"<urn:uuid:e4a3581d-a4b1-4989-8e60-94f8f45913f5>","cc-path":"CC-MAIN-2024-46/segments/1730477027909.44/warc/CC-MAIN-20241106034659-20241106064659-00414.warc.gz"} |
Great Teaching is the Real Test that Should be Scored by Cristina Simmons
<Back to Blog
Great Teaching is the Real Test that Should be Scored
December 04, 2015 - Posted to Study
Great Teaching is the Real Test that Should be Scored
Every state in this country has a system of measuring student mastery of basic skills. This system is the adoption of state-wide achievement tests which all students must take at various intervals
during their academic careers. In some states, students may not pass onto the next grade unless they reach a specific level of proficiency. What probably puts more pressure on both parents and school
districts themselves is that the district-wide test scores are published both on the state education department sites but also in all local papers. The whole world then knows.
It has become common practice for districts who want to score well to pressure teachers to “teach to the test,” that is, to specifically target teaching to what they know will be on the test.
Unfortunately, because those tests often focus on just basic skills, students are not tested on their ability to solve problems, to think critically, or to analyze, because these are higher level
thinking skills that are difficult to assess with typical standardized testing, and even more difficult to grade. The result for students, at least during those years when they will be tested, is
that instruction will not focus on those skills which are so critical in the real world.
What districts do not realize is this: If they would focus on these higher level problem-solving skills, especially in math, students would score extremely well on these tests that just focus on the
algorithms – the algorithms would be learned as the problem-solving skills are taught. And they would be learned, not just memorized. Here is an example:
When students are taught positive and negative numbers, in most classroom across the country, they memorize the “rules” without any understanding of why. So students learn, for example, the phrase
“keep, change, change,” when they are solving subtraction problems with integers that are of different signs. You keep the sign of the top number, change the sign of the umber to be subtracted, and
change the sign of the number that is the answer. They do not know why this gets the correct answer, only that it does. When, instead, children are taught why the answer is what it is, they can
transfer that knowledge and understanding to real world problems in the future. And, in so doing, they will also do very well on those state tests. And using real-world situations like thermometers,
allows students to see that their learning has relevancy to the real world. When relevancy is shown, students internalize their learning and have it for life.
Teaching algebra and geometry have the same options for teachers. They can either teach the rules for solving equations, (what you do to one side, you do to the other; if you want to get rid of a
number on one side, you do the opposite operation, and then the same to the other side, etc.). This will allow students to pass a state test with a level of proficiency. It will not, however,
demonstrate to students why this solves the equation, nor will it provide them any real-world application of the use of algebra. No wonder students struggle with math – they cannot see any relevancy
to their worlds. Problem-solving that involves algebra provides that relevancy that will allow them to “cement” that learning.
Teachers of math would be far better served to focus on fewer problems of each concept and to allow students, even working in pairs or small groups, to explore how to come up with a solution to a
problem that relates to the real world. There are many ways to solve math problems, and students need the freedom to come up with solution options that work for them, not just memorize the rules for
only one method of solution – the standard one determined by some teachers long ago and put into textbooks.
Teaching is not a profession for the lazy. Great teachers are those who spend time going beyond the textbook and the types of questions/problems that will appear on state tests. When students are
genuinely involved in their learning, when they can discuss possible solutions with their peers, and when they can actually understand why certain algorithms and rules work to solve math problems,
they won’t forget how to solve them. And to prepare lessons that allow this exploration and reasoning takes a lot of time – time that teachers are not always willing to give.
The Answer
There are three factors involved in the solution.
1. State tests need to reflect what students will need to use in the real world, not just an assessment of their ability to regurgitate rules and follow them to solve problems.
2. Teacher education programs need to change, so that teachers are given the skills to deliver instruction that forces children to think, reason and explore.
3. Content specific teachers need time to collaborate with one another. As long as we continue to insist upon summer vacations, then teachers should be paid to work year-round. That 2 ½ months off
in the summer could be wisely spent with preparation of instructional delivery that will be valuable for their clients – the students. And teachers working together to develop these delivery
systems would spread the work around. | {"url":"https://essaywriting.education/blog/great-teaching-is-the-real-test-that-should-be-scored","timestamp":"2024-11-14T16:29:56Z","content_type":"application/xhtml+xml","content_length":"27788","record_id":"<urn:uuid:f126bac7-5374-4255-bb82-57a965dc0842>","cc-path":"CC-MAIN-2024-46/segments/1730477393980.94/warc/CC-MAIN-20241114162350-20241114192350-00882.warc.gz"} |
What is an Age Calculator? Answered!
What is an Age Calculator
An age calculator is a simple tool that helps you determine exactly how old someone is. By entering a person’s date of birth and comparing it to the current date, the calculator tells you their age
in years, months, and days. It’s a quick and easy way to determine how much time has passed since a person was born or a particular event occurred.
How to Use the Age Calculator
Here’s how you can calculate age using an online age calculator:
1. Enter the Birth Date:
Input the date of birth in the format DD/MM/YYYY.
2. Enter the Current Date:
Input today’s date or any other date you want to compare.
3. Get the Result:
Click “Calculate,” the tool will show the age in years, months, days, or total days.
Example of Age Calculation
Let’s say someone was born on July 25, 1985, and you want to know their age on January 28, 2021:
• Years: 2021 – 1985 = 36 years
• Months: Since January is before July, subtract 1 year, making it 35 years, and add 6 months.
• Days: Add the days from July 25 to January 28, totaling 3 days.
So, the person would be 35 years, 6 months, and 3 days old on January 28, 2021.
What Does an Age Calculator Do?
An age calculator figures out a person’s age or the period between two dates in different ways:
• Years, Months, and Days: It gives you the age in common terms, like “15 years, 3 months, and 10 days old.”
• Total Days: It tells you the exact number of days between two dates, which can be useful for specific calculations.
How Does It Work?
When you enter a birth date and the current date, the calculator compares the two:
• Years: It first counts the full years between the dates.
• Months and Days: Then, it adds the remaining months and days.
• Total Days: For more precise calculations, it considers leap years and the exact number of days in each month.
Why Use an Age Calculator?
• Simple and Quick: The calculator does it for you in seconds instead of doing the math yourself.
• Accurate: It accounts for differences in month lengths and leap years, so you get an accurate result.
• Versatile: You can use it to find the age of a person, a place, or even an object.
Why Are Age Calculators Important?
Age calculators are more useful than you might think. They can help with:
Planning Celebrations:
Knowing the exact age can help plan birthdays or anniversaries.
Health and Fitness:
Doctors might need to know your precise age to give you the right advice or treatment.
Schools use age calculators to ensure kids are in the right grade.
Historical Events:
You can calculate how many years have passed since an important event.
FAQs About Age Calculator
1. What does an age calculator do?
An age calculator tells you how old someone is by calculating the time between their birth date and today’s date.
2. How do I use an age calculator?
Simply enter the birth date and the current date, then click the “Calculate” button to see the age.
3. Can I use an age calculator for events, not just people?
Yes, you can use it to find out how long it’s been since any important date, like an anniversary or historical event.
4. Does the age calculator consider leap years?
Yes, the calculator includes leap years to give you the most accurate age.
5. Can an age calculator show age in months and days?
Yes, it shows age in years, months, and days.
6. Is an age calculator free to use?
Most age calculators online are free and easy to access.
7. Why do people use age calculators?
People use them to find out their exact age, plan events, or calculate how much time has passed since a special day.
An age calculator is a simple yet useful tool that helps you figure out how old someone is or how much time has passed since an important event. It’s easy to use, accurate, and can be helpful in many
situations, from planning birthdays to understanding history. Just enter the dates, and you’ll have your answer in no time!
Leave a Reply Cancel reply | {"url":"https://agecalculator.com.pk/what-is-an-age-calculator/","timestamp":"2024-11-14T00:44:47Z","content_type":"text/html","content_length":"94421","record_id":"<urn:uuid:1fb30ff6-07b5-4e17-b1c3-2fa1a0b1b0c3>","cc-path":"CC-MAIN-2024-46/segments/1730477028516.72/warc/CC-MAIN-20241113235151-20241114025151-00731.warc.gz"} |
D300 and D200 Do Noise Reduction on Longer Exposures
Prepared 2008-03-13 (149/118594) by Bill Claff
In a previous article I demonstrated that noise drops on longer exposures even if all noise reduction settings are turned off.
For the D200 this effect starts at 1s and for the D300 it starts at 1/4s (and the D3 too, but that's a different forum).
The analysis was performed on raw data, so obviously even NEF files are affected.
There is some question of what is really going on.
Emil Martinec, a knowledgeable and frequent poster elsewhere, used a 2D Fast Fourier Transform (FFT) to "disprove" the hypothesis that signal processing occurs after the initial capture.
However, I believe that analysis was flawed because he examined a well exposed image with a high Signal to Noise Ratio (SNR) which obscured the result.
I applied Emil's FFT technique to the lens cap shots used to prepare the earlier post and verified that signal processing (noise reduction) does occur.
Without going into the gory details, all we need to know is that if the pixels are uncorrelated (have nothing to do with their neighbors) then you get a uniform distribution.
But if the neighboring pixels are correlated (due to noise reduction, sharpening, signal processing in general) then the distribution will not be uniform.
In the case of Noise Reduction (NR), a brighter center with dark edges is predicted.
Here are the results for the D300 after an aggressive Levels adjustment to make things obvious:
The shutter times were:
│ 1/15│1/8│1/4│
│ 1/2│ 1│ 2│
│ 4│ 8│ 15│
And we can clearly see noise reduction has been applied at 1/4s and longer exposures. | {"url":"https://photonstophotos.net/GeneralTopics/Sensors_&_Raw/D300_and_D200_Do_Noise_Reduction_on_Longer_Exposures.htm","timestamp":"2024-11-01T19:49:02Z","content_type":"text/html","content_length":"8420","record_id":"<urn:uuid:f3ee704a-cf02-4c80-b4d9-a77714115237>","cc-path":"CC-MAIN-2024-46/segments/1730477027552.27/warc/CC-MAIN-20241101184224-20241101214224-00231.warc.gz"} |
Bayes Theorem
Bayes’ Theorem
Bayes’ Rule is an application of the LOTP with the Conditional Probability rule.
Bayes’ theorem describes the probability of an event based on prior knowledge of conditions that might be related to the event.
Let be a partition of and be any event, then
Prior Probability Posterior Probability Likelihood
How is bayes rule derived? Apply the basic rule of Conditional Probability, and leverage the fact that AND is commutative. Therefore, you can simplify
You have one fair coin and a biased coin that lands on heads with a probability of 3 4 . A coin is chosen at random and tossed three times. If we observe three heads in a row, what is the probability that the fair coin was chosen?
Solution: The fair coin was chosen The biased coin was chosen 3 heads observed in 3 tosses
We want to find , so we can use Bayes’ Theorem, and calculate
Relation to more advanced control theory / ML
This is why we say (from Kalman Filter in Python)
You just continuously update the posterior, setting prior to old posterior every time.
See Kalman Filter.
Bayes theorem is super useful because it turns a hard problem into an easy problem.
Hard problems:
• P(Cancer = True | Test = Positive)
• P(Rain = True | Readings)
Stated like that the problems seem unsolvable.
Easy problems:
• P(Test = Positive | Cancer = True)
• P(Readings | Rain = True)
Bayes’ Theorem lets us solve the hard problem by solving the easy problem.
Bayes rule with Conditioning
From CS287. | {"url":"https://stevengong.co/notes/Bayes-Theorem","timestamp":"2024-11-09T02:44:35Z","content_type":"text/html","content_length":"55162","record_id":"<urn:uuid:99389168-edae-4206-85c4-fc8f3466f9cc>","cc-path":"CC-MAIN-2024-46/segments/1730477028115.85/warc/CC-MAIN-20241109022607-20241109052607-00016.warc.gz"} |
[Solved] Gonzalez Company is considering two new p | SolutionInn
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Gonzalez Company is considering two new projects with the following net cash flows. The company's required rate of return on investments is 10% (PV of
Gonzalez Company is considering two new projects with the following net cash flows. The company's required rate of return on investments is 10% (PV of $1. FV of $1. PVA of $1, and FVA of $1 (Use
appropriate factor(s) from the tables provided.) Net Cash Flows Year Project 1 Project 2 Initial investment $(42,000) $(78,000) 1. 10,500 35,000 2. 27,800 15, eee 3. 18,5ee 35,00 a. Compute payback
period for each project. Based on payback period, which project is preferred? b. Compute net present value for each project Based on net present value, which project is preferred? DO Complete this
question by entering your answers in the tabs below. Required A Required B narind which are narrativa Required A Required B Compute payback period for each project. Based on payback period, which
project is preferred? (Cumulative net cash outflows must be entered with a minus sign. Do not round your intermediate calculations, Round your Payback period answer to 2 decimal places) Project 1
Project 2 Year Cumulative Cumulative Net Net Cash Net Cash Flows Net Cash Cash Flows Flows Flows Initial investment $ (42.000) (42000) 5 (78,000) Year 1 10,500 31.500 0 Year 2 27.800 Year 3 18,500 0
Payback period Project 1 Payback period Project 2 Payback period Based on payback period, which project is preferred? years years Required B > $ $ Compute net present value for each project. Based on
net present value, which project is preferred? (Round your value factor to 4 decimals. Round your final answers to the nearest whole dollar.) Net Cash Present Value Present Value of Net Flows Factor
Cash Flows Project 1 Year 1 Year 2 Year 3 Totals $ 0 0 Initial investment Net present value $ 0 Project 2 Year 1 Year 2 Year 3 Totals 0 $ 0 Initial investment 0 Net present value Based on net present
value, which project is preferred? $ $
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Get Started | {"url":"https://www.solutioninn.com/study-help/questions/gonzalez-company-is-considering-two-new-projects-with-the-following-6371631","timestamp":"2024-11-14T00:16:07Z","content_type":"text/html","content_length":"99520","record_id":"<urn:uuid:8af18503-74aa-4bfa-8330-a392be8486dc>","cc-path":"CC-MAIN-2024-46/segments/1730477028516.72/warc/CC-MAIN-20241113235151-20241114025151-00016.warc.gz"} |
Monotonicity testing over general poset domains
The field of property testing studies algorithms that distinguish, using a small number of queries, between inputs which satisfy a given property, and those that are 'far' from satisfying the
property. Testing properties that are defined in terms of monotonicity has been extensively investigated, primarily in the context of the monotonicity of a sequence of integers, or the monotonicity
of a function over the n-dimensional hypercube {1,⋯,m}^n. These works resulted in monotonicity testers whose query complexity is at most polylogarithmic in the size of the domain. We show that in its
most general setting, testing that Boolean functions are close to monotone is equivalent, with respect to the number of required queries, to several other testing problems in logic and graph theory.
These problems include: testing that a Boolean assignment of variables is close to an assignment that satisfies a specific 2-CNF formula, testing that a set of vertices is close to one that is a
vertex cover of a specific graph, and testing that a set of vertices is close to a clique. We then investigate the query complexity of monotonicity testing of both Boolean and integer functions over
general partial orders. We give algorithms and lower bounds for the general problem, as well as for some interesting special cases. In proving a general lower bound, we construct graphs with
combinatorial properties that may be of independent interest.
• Algorithms
• Monotone functions
• Property testing
Dive into the research topics of 'Monotonicity testing over general poset domains'. Together they form a unique fingerprint. | {"url":"https://cris.tau.ac.il/en/publications/monotonicity-testing-over-general-poset-domains","timestamp":"2024-11-09T17:11:49Z","content_type":"text/html","content_length":"51982","record_id":"<urn:uuid:4370a7f3-72ff-437a-a0a8-224626cdbbad>","cc-path":"CC-MAIN-2024-46/segments/1730477028125.59/warc/CC-MAIN-20241109151915-20241109181915-00312.warc.gz"} |
5th Grade Order of Operations Worksheets and Activities | ClassWeekly
Order of Operations (add, subtract, multiply) - Search
This order of operations worksheet focuses on addition, subtraction and multiplication (without parentheses). Each equation has 4 terms. Remember, multiplication must be done before addition or
subtraction. Worksheet instructions: Find the answer to each question below. | {"url":"https://www.classweekly.com/activities/fifth-grade/math--order-of-operations","timestamp":"2024-11-04T10:50:31Z","content_type":"text/html","content_length":"203297","record_id":"<urn:uuid:a20bfc06-a321-4b3c-8fc4-f1137d1ec72f>","cc-path":"CC-MAIN-2024-46/segments/1730477027821.39/warc/CC-MAIN-20241104100555-20241104130555-00565.warc.gz"} |
Cambridge Maths Circle
Here are some of the activities for the Cambridge Maths Circle. The links are all to problems on the NRICH site that inspired the activities, so we hope that you might like to try some of them, and
to explore the rest of the NRICH site.
To find out about future events, keep an eye on the MMP Events page.
The activities are arranged here roughly in order of increasing mathematical difficulty and knowledge required, but many of them can be tackled at lots of levels so could be almost anywhere in the | {"url":"http://nrich.maths.org/cambridge-maths-circle","timestamp":"2024-11-05T12:14:31Z","content_type":"text/html","content_length":"33666","record_id":"<urn:uuid:6c60ee46-2ee1-4afa-a31b-b60f3008141b>","cc-path":"CC-MAIN-2024-46/segments/1730477027881.88/warc/CC-MAIN-20241105114407-20241105144407-00854.warc.gz"} |