We've all probably seen this question asked roughly 1000 times, so in the spirit of sharing I'm going to divulge my method (read secret) in answering this question.
All bows, experience the highest stress at the limb surfaces (i.e. back and belly).
Therefore:
(Assumption #1) most if not all failures are due to overstressing the absolute outer surface first.
(assumption #2) pretend for a minute that all the energy is stored at the surface (back and belly).
(Assumption #3) All of the limb does an equal amount of work or the proportion of the limb which does the work is the same between the two designs being compared.
If these assumptions are true then limb surface area is proportional to maximum energy storage at failure of the bow (basically).
So what does it all mean?
It means if you've built or know of a hickory board bow thats 64" long with 8" handle (i.e. 56" of bending limb) and the limb is 2" wide at the fades and 1" wide at the tips (for easy math), then the limb surface area is 56in*((2+1)/2))in = 56in*1.5in = 84in^2.
Lets say that bow draws 50# @ 28" with an 8" brace height and stores approximately 50#/2 * (28-8)in = 1000#*in of energy. so for every 100#*in of energy you need 8.4in^2 of limb surface area.
So then you ask the question like this:
How wide does a 60" hickory bow with a 10" handle section have to be to draw the same weight at 26"?
50#/2 * (26-8)in = 900#*in of energy (approximately); therefore 9*8.4in^2 of limb area = 75.6in^2 of limb area.
Lets then assume the tips will also be 1" wide ... how wide do the fades need to be?
60"-10" = 50" bending limb,
75.6in^2/50in = 1.5inch. so the average limb width needs to be 1.5" approximately. For a pyrimidal limb that would be 2" fades tapering to 1" tips. or nearly 3" wide fades tapering to very thin (say 1/4") tips.
So thats basically it.
Ao another simple example If you cut limb length in half, you have to double the width assuming the same profile and so on an so forth.
Making it simpler, if you wanna build yourself another bow at the same draw length and draw weight as a bow you already have of the same material, and the tip width and limb profile is the same then you can simplify this all to:
Limb length 1 / limb length 2 = limb width 2 / limb width 1.
or (limblength1/limblength2)*limbwidth1 = limbwidth2
so for two bows with the same 8" handle, the first 68" long and 1" wide, and the second 58" long and unknown width:
(60"/50")*1" = limbwidth 2 = 1.2"
so the shorter bow needs to be 1.2" wide.
Hope that makes sense and someone can make use of it. This is the method I use all the time, and I successfully adjusted the width for a 58" bows based on a 68" bow just this month. It works.
There's already a mass formula so a part of me dreams this will become the surface area model or something like that.
Cheers,
Ben