What a Difference an Inch Makes.

My first anxiety about writing this post is that either my Integral Calculus professor (Hi, Prof. Tsuchida!) or my Kinematics professor (Hi, Prof. Koskelo!) will read this and grade it.  I really need to get over these anxieties.

But this article isn’t about calculus or physics.  Well, maybe it is.  Sorta.  Indirectly.  It’s about archery and the way we teach it.

When we see our students anchoring incorrectly, and specifically when we see them do a short-draw, we often tell them, by way of emphasis, that for every inch they short draw, they lose a pound of thrust to the arrow.  But do they?  Had anybody ever actually measured this?  Well, probably.  However, I decided to do some experimenting myself.

Recurve bows – the type we use to teach western traditional archery at San Francisco Archers – are wonderful machine for storing energy supplied by human muscles and giving it all back in the blink of an eye, sending a 20 gram arrow a couple hundred yards down range.  Some minimal Google research indicates that they were invented independently by cultures all over the world, and are much older than the written records.  The engineering of the recurve bow is fascinating, in that the recurve part of it gives the arrow an additional little kick at the end of the process.  I’m not going to worry about that part, because I’m primarily concerned about the beginning of the process.

The claim that “one inch equals one pound of thrust” comes from the approximation that the average beginner bow requires about 25 pounds of tension to pull to 28 inches, which is the standard draw-length, so we estimate that the difference between 27 inches and 28 inches is approximately 28/25 of a pound.  This has always bothered me because the amount of thrust imparted to the arrow would be the force on the arrow INTEGRATED over the distance between 27 and 28 inches, which surely had to be more than a pound.  What is this value?  I had no idea.  I had to measure it.

Using my Martin Hatfield, which has an AMO weight of 40# at 28 inches, and using the fish scale at the SFA clubhouse, I went to work.  Caveat: The fish scale in the clubhouse isn’t calibrated, and the process was rife with human error.  It involved holding the bow at marked intervals while somebody else made a valiant attempt to read the shaky, unsteady needle.  Also, it’s not a beginner’s 25# bow.  I don’t own one, but I believe we can make some easy extrapolations.  I did my best to minimize experimental errors by taking three readings at each distance and averaging them, but we can still only consider the results to be an approximation.

Martin Response

Response of the Martin Hatfield bow. X-axis is draw distance in inches, the y-axis is tension in pounds.

After taking measurements, I used Wolfram Alpha to find the best fit curve.  Turns out the best fit is a quartic, demonstrating that either the response of wood is far more complex and interesting than I imagined, or the data are really funky.  I leave the judging up to you. However, even if we use the linear equation, the numbers do not differ much.

The math is pretty straight forward – we just take the equation for the curve and integrate.  Interestingly, if we ignore friction and other real life messiness, the total amount of thrust imparted to the arrow is shown the integration from 7.5 inches (brace height) to 28 inches of the equation for the curve, which comes out to 465 pounds.  This is how amazing the bow is!  Using just your own human muscles, you can impart nearly a quarter ton total thrust into an arrow with the mass of 20 grams.  And this bow is only half the pull weight of a typical war-bow from eight centuries past!  (Estimates of what constituted a “typical war-bow” will differ depending on your local expert.)

So what’s the difference in an inch?  Integrating from 27 to 28 inches, this comes out to about 40 pounds.  That’s how much total thrust you loose by short drawing by only one inch.

However, it’s not typical for a beginner to short draw by only one inch.  The most common anchoring error we see is the floating anchor, and that would be nearly impossible to measure.  The second most common anchoring error we see is to anchor with the wrist to the chin.  Using my own hand as a model, that’s a difference of about 4 inches in draw length.  If we integrate fro 24 to 28 inches on the curve, we get nearly 150 lbs of total thrust lost.

Granted, this does not take into account different sized people with different sized bows.  However, we can safely start saying to students who short draw, “For every inch, you lose about 40 pounds of thrust.”  A little more convincing, I think.

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Polygons of Infinite Perimeter but Finite Area

There are damn few people whom I know will be interested in this post, but I wanted to record the idea anyway.

I came across an online math puzzle which states:

It is possible to increase the area of a regular triangle by placing smaller regular triangles on the middle thirds of its three sides. By so doing, you obtain a six-pointed star. The process can continue indefinitely. At each step, a smaller regular triangle is placed on the middle third of all the line segmens on the perimeter of the figure obtained from the previous step. Sketching the shapes obtained for the first few steps of this process is an interesting way to spend a few moments.

The perhaps surprising result is that this process converges to a fractal-like figure of infinite perimeter but of finite area. Can you determine the area limit?

A more interesting question arises. Can some similar process converge to a fractal-like figure of infinite perimeter but of zero area?

Regarding the first question:  in solving this problem, I discovered that the resulting figure is called a Koch Snowflake.  I solved it thus:

Image

Image

Regarding Part Two of the question: Can some similar process converge to a fractal-like figure of infinite perimeter but of zero area?

My first thought was to use a variation on the Koch Snowflake and subtract area instead of adding it.  I should have known that this process wouldn’t have worked, but it fascinated me for a couple of hours.

Image.

Image

So, it sums out to 6/15.  Just over a third of the original area.  Still an infinite perimeter.

Finally I lost interest and went to the website where the question was originally posted looking for the answer to Part Two of this problem, and the author said that the method which I just described above would result in “zero area.”  I tried to create an account so that I could respond to his answer and point out that, although 6/15 was close to zero (relatively) it just was no cigar.  However, the system was balky and wouldn’t let me create an account.

Even though the method didn’t work, it made for same pretty pictures, which I’m happy to share here.  If anybody reading this has another suggestion, I’m happy to talk about it.