[OS X TeX] RE: OT: A favour --- formula names for beam

Lantz Susan lantzs at trine.edu
Thu Sep 9 18:14:58 EDT 2010


Your testing set-up looks quite different from what I'd envisioned. 

Your method of using the Pythagorean Theorem to determine the angle will work only as long as the deflection remains linear. (And from your photo, it's obvious that the deflection doesn't remain linear for long.)

If you strung a piece of string/wire/dental floss between the supports (to indicate the original position of the arrow before the load was applied), then applied the load, the equation for the angle between the arrow and the string/wire/dental floss is
        Angle = W*(3L^2 - 12x^2)/(48EI),
where W is the applied load (your weight), L is the distance between the supports (26 inches, I think you said), x is any distance along the arrow you choose, E is the modulus of elasticity of the arrow shaft, and I is the moment of inertia of the arrow shaft [I = pi*(diameter)^4/(64)]. 

This equation will give you the angle in radians. To convert to degrees, multiply by (180/pi).

Susan Lantz

-----Original Message-----

Date: Thu, 9 Sep 2010 09:01:42 -0400
From: William Adams <will.adams at frycomm.com>
Subject: Re: [OS X TeX] Re: OT: A favour --- formula names for beam
To: TeX on Mac OS X Mailing List <macosx-tex at email.esm.psu.edu>

Here's a picture:


> Since (if I've understood what you're asking) the lever is deflecting,
> not the arrow, the deflection equation isn't going to tell you anything
> about the arrow. It will, however, tell you about the lever.


I finally figured it out. I use (of course) the Pythagorean theorem to determine the angle the lever is rotated to based on the right triangle defined by it having moved down by a given deflection --- the endpoint of the lever is determined by its intersection w/ a circle of a radius equal to the length of that arm of the lever (or can be calculated using the known angle and the radius as the hypotenuse).

Turns out there're some strange values in the published spine charts (from 29 to 30 pounds the change increases .034", but adding one more pound increases the amount of the change to 0.035" &c. --- resistance should increase consistently and the amount of change should decrease as the arrow becomes stiffer, right? but this happens over a dozen times between 20 and 95# in the chart at http://www.rosecityarchery.com/AMOspine.html) and I made some naïve mistakes in drawing things up (in FreeHand), but I believe I should be able to draw this up using Asymptote presently.


William Adams
senior graphic designer
Fry Communications
Sphinx of black quartz, judge my vow.

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