Formula for lam bow....

Started by OkKeith, January 05, 2009, 05:12:00 PM

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OkKeith

Hey Guys,

I am working up to trying to make a fiberglass laminated bow for myself. I guess I have yet to make a big enough mess making self and backed bows, now I want to try this.

I have read quite a bit on how to do it, logged several pointers and tips and gone through a lot of helpful hint type information. Here is what I can not find; any formulas leading to a bow of prescribed weight and dimension.

I basically want a straight profile longbow (maybe set back just a little at the handle) of about 45lbs at my 28 inch draw length somewhere in the neighborhood of 66-68 inches.

I am trying to find some generic tables that give info like; Use 2 parallel lams of 0.0X thickness and one tapered lam (of a given taper) along with 0.Y thickness glass, and if the bow is 1 and 1/2 inches wide it will be 45 pounds.

I realize there are a lot more factors involved; I just need some stats to get me in the ballpark.

What do all ya'll think?

OkKeith
In a moment of decision the best thing you can do is the right thing. The worst thing you can do is nothing.
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Greg Szalewski

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kennym

Greg gave you a good link there,I think what you want is the 68" lb. I think it uses a .002 limb taper,and an 18" riser.
Good luck ,if you can make a selfbow,a glass bow should be a breeze!LOL   JMHO
Stay sharp, Kenny.

   https://www.kennysarchery.com/

OkKeith

Thanks for all the help guys!

I'm not sure how I managed to miss that particular resource on the Bingham's site.

If I'm not totallty embarassed by my first trial run, I will of course post pictures.

Thanks again,
OkKeith
In a moment of decision the best thing you can do is the right thing. The worst thing you can do is nothing.
Theodore Roosevelt

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