# Math Question **URL:** \ **Category:** Factual Questions\ **Created:** [February 7, 2016, 4:04pm UTC](https://boards.straightdope.com/t/math-question/745432 "2016-02-07T16:04:41Z")\ **Posts on this page:** 20\ **Page:** 1
**Author:** ![HoneyBadgerDC](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/honeybadgerdc/32/1033_2.png) [@HoneyBadgerDC](https://boards.straightdope.com/u/HoneyBadgerDC)\ **Post date:** [February 7, 2016, 4:04pm UTC](https://boards.straightdope.com/t/math-question/745432/1 "2016-02-07T16:04:41Z")
I will do my best posing this question. I hope you can understand it. When measuring the stored energy on an archery bow we normaly draw the bow back and take a force reading every 1 " increment, we add those up usually at 28" and then divide that number by 12 to convert from inch pounds of SE to Ft pounds of SE. My question is, if I were to take those same measurements in 1/10 of an inch increments I get a lower number, 1/100th increments even lower but not by much 1/1000 still lower. Is there a way or a formula I can use to still take my measurements at 1" icrements but average it out as if I took those same measurements in 1/1000 of an inch increments?? ---
**Author:** ![Xema](https://avatars.discourse-cdn.com/v4/letter/x/9de053/32.png) [@Xema](https://boards.straightdope.com/u/Xema)\ **Post date:** [February 7, 2016, 4:26pm UTC](https://boards.straightdope.com/t/math-question/745432/2 "2016-02-07T16:26:10Z")
If I understand the problem correctly, one way to do this would be: Take a bunch of force/distance readings (e.g. at every inch of pull) Find a formula\* (quadratic? cubic?) that best fits these points Integrate the formula to find the area under the curve \*Here’s a wiki link on [curve-fitting](https://en.wikipedia.org/wiki/Curve_fitting) ---
**Author:** ![HoneyBadgerDC](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/honeybadgerdc/32/1033_2.png) [@HoneyBadgerDC](https://boards.straightdope.com/u/HoneyBadgerDC)\ **Post date:** [February 7, 2016, 4:48pm UTC](https://boards.straightdope.com/t/math-question/745432/3 "2016-02-07T16:48:54Z")
You may have answered my question but my math skills may not be up to working it out. A force curve does accurately represent the power under the curve but finding the exact number under the curve is what I am struggling with. Typically a curve is not linear overall but straight enough between 1" increments to give an accurate depiction. ---
**Author:** ![HoneyBadgerDC](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/honeybadgerdc/32/1033_2.png) [@HoneyBadgerDC](https://boards.straightdope.com/u/HoneyBadgerDC)\ **Post date:** [February 7, 2016, 4:53pm UTC](https://boards.straightdope.com/t/math-question/745432/4 "2016-02-07T16:53:00Z")
I wonder if a program exists that could fill the bottom of the curve with a virtual liquid that could be measured?? ---
**Author:** ![John\_Mace](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/john_mace/32/185_2.png) [@John\_Mace](https://boards.straightdope.com/u/John_Mace)\ **Post date:** [February 7, 2016, 4:56pm UTC](https://boards.straightdope.com/t/math-question/745432/5 "2016-02-07T16:56:49Z")
> [@HoneyBadgerDC](#): > > I wonder if a program exists that could fill the bottom of the curve with a virtual liquid that could be measured?? Yes, very simple. Just square off the top of each increment, and you have a rectangle below and triangle above. Integral calculus is just a way of making those increments arbitrarily small. You could improve the accuracy by squaring each increment off both possible ways and then averaging the result. ---
**Author:** ![HoneyBadgerDC](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/honeybadgerdc/32/1033_2.png) [@HoneyBadgerDC](https://boards.straightdope.com/u/HoneyBadgerDC)\ **Post date:** [February 7, 2016, 5:02pm UTC](https://boards.straightdope.com/t/math-question/745432/6 "2016-02-07T17:02:10Z")
> [@John\_Mace](#): > > Yes, very simple. Just square off the top of each increment, and you have a rectangle below and triangle above. Integral calculus is just a way of making those increments arbitrarily small. > > You could improve the accuracy by squaring each increment off both possible ways and then averaging the result. Your right, that would work!! ---
**Author:** ![Thudlow\_Boink](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/thudlow_boink/32/320_2.png) [@Thudlow\_Boink](https://boards.straightdope.com/u/Thudlow_Boink)\ **Post date:** [February 7, 2016, 5:04pm UTC](https://boards.straightdope.com/t/math-question/745432/7 "2016-02-07T17:04:39Z")
> [@HoneyBadgerDC](#): > > When measuring the stored energy on an archery bow we normaly draw the bow back and take a force reading every 1 " increment, we add those up usually at 28" and then divide that number by 12 to convert from inch pounds of SE to Ft pounds of SE. > > My question is, if I were to take those same measurements in 1/10 of an inch increments I get a lower number, 1/100th increments even lower but not by much 1/1000 still lower. > > Is there a way or a formula I can use to still take my measurements at 1" icrements but average it out as if I took those same measurements in 1/1000 of an inch increments?? The process you describe screams “definite integral” to me: namely, the limit as the size of the increments approaches 0 would be a definite integral. This suggests to me that a formula for approximating definite integrals, like [Simpson’s Rule](http://www.mathwords.com/s/simpsons_rule.htm), might be the best way to do what you want, using your measurements at each increment as the f(x)'s in the formula. ---
**Author:** ![HoneyBadgerDC](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/honeybadgerdc/32/1033_2.png) [@HoneyBadgerDC](https://boards.straightdope.com/u/HoneyBadgerDC)\ **Post date:** [February 7, 2016, 5:13pm UTC](https://boards.straightdope.com/t/math-question/745432/8 "2016-02-07T17:13:22Z")
> [@Thudlow\_Boink](#): > > The process you describe screams “definite integral” to me: namely, the limit as the size of the increments approaches 0 would be a definite integral. > > This suggests to me that a formula for approximating definite integrals, like [Simpson’s Rule](http://www.mathwords.com/s/simpsons_rule.htm), might be the best way to do what you want, using your measurements at each increment as the f(x)'s in the formula. That is most definetly the process I need to use. I never made it to algebra 1 in school so I will have to study that a bit to see if I can figure out how to do it. ---
**Author:** ![Blue\_Blistering\_Barnacle](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/blue_blistering_barnacle/32/3386_2.png) [@Blue\_Blistering\_Barnacle](https://boards.straightdope.com/u/Blue_Blistering_Barnacle)\ **Post date:** [February 7, 2016, 5:38pm UTC](https://boards.straightdope.com/t/math-question/745432/9 "2016-02-07T17:38:34Z")
If the math is too hard… My dad was an engineer. He said he would sometimes do this by graphing the points and fitting a curve by hand. Then he would cut out the portion of the curve of interest and weigh that piece of graph paper. Then he would cut out and weigh (one or some multiple of) a unit square and use this to estimate the definite integral. ---
**Author:** ![HoneyBadgerDC](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/honeybadgerdc/32/1033_2.png) [@HoneyBadgerDC](https://boards.straightdope.com/u/HoneyBadgerDC)\ **Post date:** [February 7, 2016, 5:46pm UTC](https://boards.straightdope.com/t/math-question/745432/10 "2016-02-07T17:46:15Z")
> [@Blue\_Blistering\_Barnacle](#): > > If the math is too hard… > > My dad was an engineer. > > He said he would sometimes do this by graphing the points and fitting a curve by hand. Then he would cut out the portion of the curve of interest and weigh that piece of graph paper. Then he would cut out and weigh (one or some multiple of) a unit square and use this to estimate the definite integral. This was actually my first consideration, the finest weight measures I can take are grains, paper is so light that I don’t believe I could get much accuracy. ---
**Author:** ![John\_Mace](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/john_mace/32/185_2.png) [@John\_Mace](https://boards.straightdope.com/u/John_Mace)\ **Post date:** [February 7, 2016, 5:48pm UTC](https://boards.straightdope.com/t/math-question/745432/11 "2016-02-07T17:48:23Z")
> [@HoneyBadgerDC](#): > > This was actually my first consideration, the finest weight measures I can take are grains, paper is so light that I don’t believe I could get much accuracy. I’d graph it on a piece of plywood and cut it with a jigsaw. But it’s a very simple program to write if you input the data points. Very simple. ---
**Author:** ![Blue\_Blistering\_Barnacle](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/blue_blistering_barnacle/32/3386_2.png) [@Blue\_Blistering\_Barnacle](https://boards.straightdope.com/u/Blue_Blistering_Barnacle)\ **Post date:** [February 7, 2016, 5:51pm UTC](https://boards.straightdope.com/t/math-question/745432/12 "2016-02-07T17:51:33Z")
> [@HoneyBadgerDC](#): > > This was actually my first consideration, the finest weight measures I can take are grains, paper is so light that I don’t believe I could get much accuracy. Yes, he described the process to me, but I never saw him do it. I think he said he used a laboratory balance. Looking at the page for simpson’s rule, it should be easy to do that in excel or numbers. Easier than graph paper, I bet. ---
**Author:** ![HoneyBadgerDC](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/honeybadgerdc/32/1033_2.png) [@HoneyBadgerDC](https://boards.straightdope.com/u/HoneyBadgerDC)\ **Post date:** [February 7, 2016, 5:57pm UTC](https://boards.straightdope.com/t/math-question/745432/13 "2016-02-07T17:57:03Z")
> [@John\_Mace](#): > > I’d graph it on a piece of plywood and cut it with a jigsaw. > > But it’s a very simple program to write if you input the data points. Very simple. I am going to check with an engineer buddy of mine. I bet he allready has the prorgam on excel. Thats probably just how he does it. ---
**Author:** ![Blue\_Blistering\_Barnacle](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/blue_blistering_barnacle/32/3386_2.png) [@Blue\_Blistering\_Barnacle](https://boards.straightdope.com/u/Blue_Blistering_Barnacle)\ **Post date:** [February 7, 2016, 8:06pm UTC](https://boards.straightdope.com/t/math-question/745432/14 "2016-02-07T20:06:42Z")
> [@Thudlow\_Boink](#): > > The process you describe screams “definite integral” to me: namely, the limit as the size of the increments approaches 0 would be a definite integral. > > This suggests to me that a formula for approximating definite integrals, like [Simpson’s Rule](http://www.mathwords.com/s/simpsons_rule.htm), might be the best way to do what you want, using your measurements at each increment as the f(x)'s in the formula. If you have measured data points (assume one measurement for each point), should you smooth or curve fit the data first? Looking at your link for Simpson’s rule, it seems to “smooth” a bit anyway. ---
**Author:** ![Isilder](https://avatars.discourse-cdn.com/v4/letter/i/cc9497/32.png) [@Isilder](https://boards.straightdope.com/u/Isilder)\ **Post date:** [February 8, 2016, 8:09am UTC](https://boards.straightdope.com/t/math-question/745432/15 "2016-02-08T08:09:07Z")
> [@HoneyBadgerDC](#): > > My question is, if I were to take those same measurements in 1/10 of an inch increments I get a lower number, 1/100th increments even lower but not by much 1/1000 still lower. … because the shape is roughly the same everytime, its quite close to a fixed percentage. ---
**Author:** ![Chronos](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/chronos/32/134_2.png) [@Chronos](https://boards.straightdope.com/u/Chronos)\ **Post date:** [February 8, 2016, 8:59pm UTC](https://boards.straightdope.com/t/math-question/745432/16 "2016-02-08T20:59:11Z")
Simpson’s Rule is equivalent to fitting pieces of the graph with parabolas, and then calculating the exact area under those parabolas. While most curves are not actually parabolas, they’re usually locally very close to it, so Simpson’s Rule in practice turns out to usually be a good enough approximation. ---
**Author:** ![newme](https://avatars.discourse-cdn.com/v4/letter/n/e0b2c6/32.png) [@newme](https://boards.straightdope.com/u/newme)\ **Post date:** [February 8, 2016, 9:48pm UTC](https://boards.straightdope.com/t/math-question/745432/17 "2016-02-08T21:48:48Z")
This page links to a spreadsheet which will do the calculations for you. Use at your own risk. I have no affiliation with this site: > **[How to make and read a force/draw curve (part 2)](http://buildyourownbow.com/how-to-make-and-read-a-forcedraw-curve-part-2/)** > > How to MAKE a force/draw curve Welcome to part two of the series! In part one, we learned how to build a bow weighing rack and measure the draw weights of our bow at each one inch increment of the … ---
**Author:** ![Oukile](https://avatars.discourse-cdn.com/v4/letter/o/13edae/32.png) [@Oukile](https://boards.straightdope.com/u/Oukile)\ **Post date:** [February 8, 2016, 10:36pm UTC](https://boards.straightdope.com/t/math-question/745432/18 "2016-02-08T22:36:41Z")
Matlab (a scientific calculation programming language) would do that in a second. There is a free equivalent of Matlab called Octave that you can download or use online (for instance here: [Online Octave Terminal - octaveterm](http://www.tutorialspoint.com/octave_terminal_online.php) ) It might be a bit hard for you to use as it is not as friendly to use as excel… but here is an example. Give it a try if you like and see by yourself. Type the following lines one by one in the octave terminal: x = [1 2 3 4 5 6 7 8] ; y = [2 9 4 6 3 2 4 7] ; xx = 1:0.01:8 ; yy = interp1(x,y,xx,‘spline’); sum(yy)\*0.01 The first two lines are your input: x (from 1 to 8 by increments of 1) is how much you drew your bow and y the force you read. The third line defines a new set of length xx with the same range (from 1 to 8, this is important) but smaller increments (1/100 here). The fourth line calls an interpolation function to compute the corresponding yy values. The final line sums the values of yy and multiplies by the increment (1/100, you have to do that). When you type that line, you will see “ans = 34.18” appear. This is your integral. If you download a desktop version, you will be able to plot the original and interpolated values by typing: plot(xx,yy,‘k’); hold on; plot(x,y,‘or’); hold off ---
**Author:** ![HoneyBadgerDC](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/honeybadgerdc/32/1033_2.png) [@HoneyBadgerDC](https://boards.straightdope.com/u/HoneyBadgerDC)\ **Post date:** [February 8, 2016, 11:03pm UTC](https://boards.straightdope.com/t/math-question/745432/19 "2016-02-08T23:03:35Z")
> [@Oukile](#): > > Matlab (a scientific calculation programming language) would do that in a second. There is a free equivalent of Matlab called Octave that you can download or use online (for instance here: [Online Octave Terminal | Online Octave Compiler](http://www.tutorialspoint.com/octave_terminal_online.php) ) > > It might be a bit hard for you to use as it is not as friendly to use as excel… but here is an example. Give it a try if you like and see by yourself. Type the following lines one by one in the octave terminal: > > x = [1 2 3 4 5 6 7 8] ; > y = [2 9 4 6 3 2 4 7] ; > xx = 1:0.01:8 ; > yy = interp1(x,y,xx,‘spline’); > sum(yy)\*0.01 > > The first two lines are your input: x (from 1 to 8 by increments of 1) is how much you drew your bow and y the force you read. > The third line defines a new set of length xx with the same range (from 1 to 8, this is important) but smaller increments (1/100 here). > The fourth line calls an interpolation function to compute the corresponding yy values. > The final line sums the values of yy and multiplies by the increment (1/100, you have to do that). When you type that line, you will see “ans = 34.18” appear. This is your integral. > > If you download a desktop version, you will be able to plot the original and interpolated values by typing: > > plot(xx,yy,‘k’); hold on; plot(x,y,‘or’); hold off Thats good info, It might be over my head but I will see if I can work through it. Might be a few days before I report back. ---
**Author:** ![Blue\_Blistering\_Barnacle](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/blue_blistering_barnacle/32/3386_2.png) [@Blue\_Blistering\_Barnacle](https://boards.straightdope.com/u/Blue_Blistering_Barnacle)\ **Post date:** [February 10, 2016, 12:13am UTC](https://boards.straightdope.com/t/math-question/745432/20 "2016-02-10T00:13:23Z")
For Simpson’s rule, the pattern for the multipliers of the function values in the summation is: 1-4-2-4-2…2-4-1 What I’d there are only 3 points? Would it be 1-4-1 or 1-2-1?