Jim, I believe what you're referring to is that the second moment of area of a rectangular cross section is proportional to the cube of its thickness. This second moment of area is the letter 'I' in the attached beam bending equation. Thus, for a given force applied to the end of the beam (e.g. string pulling on bow limb tip), the resultant tip deflection is inversely proportional to the cube of the beam thickness. For a given force, doubling the thickness cuts the deflection to 1/8th; halving the thickness multiplies the deflection by 8.
The maximum stress in the beam does not follow this cubic relationship, however, and is in fact approximately linear, as Steve suggests. For a linear material -- which wood essentially is as long as its not taking set -- the stress is proportional to the strain. If you take a strip of material of thickness T and bend it around a radius R, the resultant maximum strain S (i.e. elongation and compression at the outer and inner surface) is approximately:
S = T / (2*R+T)
Now, since stress and strain are proportional, the ratio of stresses for two different bent strips will be equal to the ratio of strains.
So say I have two strips of material: my bow of expected thickness T1 and my experimental backing strip of thickness T2. From experience I know my bow will experience maximum bend radius R1. The question is: what radius must I bend my experimental backing strip on its own to feel confident it can survive the stresses it will experience as a backing in the bent bow? Setting the maximum strains equal, we need to solve for R2 in the equation:
T1 / (2*R1+T1) = T2 / (2*R2+T2)
The result is R2 = T2/T1 * R1
In other words, the necessary bend radius is indeed linearly proportional to the ratio of the thicknesses.
So if I expect my bow limb to be 1/2" thick and bend to a 36 inch radius, then a 1/8" backing strip will need to survive a bend radius of:
R2 = (1/8")/(1/2") * 36" = 9"
This analysis of course depends on the backing material being the same stiffness as the rest of the bow, which for hickory-backed maple or bamboo-backed ipe is not an unreasonable assumption. If the backing is stiffer than the belly wood, then the neutral plane will shift toward the back, subjecting the backing to less elongation and the belly to more compression. So if your experimental backing is stiffer than your belly wood, it doesn't need to survive as tight of a test bend. If your belly wood is stiffer than your backing, then your backing will need to survive a tighter test bend.
