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Main Boards => The Bowyer's Bench => Topic started by: OkKeith on January 05, 2009, 05:12:00 PM

Title: Formula for lam bow....
Post by: OkKeith on January 05, 2009, 05:12:00 PM

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

Title: Re: Formula for lam bow....
Post by: Greg Szalewski on January 05, 2009, 07:03:00 PM

http://www.binghamprojects.com/drawweight.htm

Title: Re: Formula for lam bow....
Post by: kennym on January 05, 2009, 08:09:00 PM

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

Title: Re: Formula for lam bow....
Post by: OkKeith on January 06, 2009, 09:09:00 AM

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