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

[William Adams]

On Sep 7, 2010, at 3:34 PM, Lantz Susan wrote:

> If I have followed this discussion correctly, the arrow's shaft is
> supported by two "centers," 26 inches apart, and the arrow shaft and the
> centers are on top of a lever, but at the end of the lever farthest from
> the fulcrum.(I.e., the arrow is not at the end of the lever that is
> pushed down.) Then a two-pound weight is applied at the fulcrum-end of
> the lever.

I believe that's it.

Here's a picture:

http://media.photobucket.com/image/how%20to%20build%20a%20spine%20tester/artcher1/spinetester1001.jpg

> 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.

Thanks!

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

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