I did a little calculation that shows me you might be on to something. The deflection of a shaft in a spine tester would vary as a a function of the span to the third power, L^3. Technically, you would have to vary both supports to keep the weight in the center of the span; however, applying the weight at small distances from the center won't have too much affect, try it if you can. I can calculate how much it is affected, but not tonight.
The calculation I did that leads me to think that you may have something is as follows:
K2*L1^3 = K1*L2^3
Were K1 and K2 are the respective stiffnesses of two shafts, and L1 and L2 are the two spans.
Suppose we have a 50# spine at 26 inches and want to find the span for a 60# spineto get the same deflection, we have:
60*(26^3) = 50*L2^3
L2^3 = (60/50)*(26^3)
L2=27.63 inches after doing the math.
This is pretty close to "add 5 pounds per inch" which would give us the answer as 28 inches.
In the end it is probably more complicated than this but this is pretty close to years of established methods.
This is not dynamic spine, I believe. To me dynamic spine has to do with the natural frequency of the arrow. Changing length, and point weight changes this natural frequency.
I have to stop here and get to bed, and we can leave natural frequency for another time.
Kevin