Sid,
Agreed.
Did some calculations with a model of a linear bow, 57#@ 28 inches (19 inch power stroke).
This gives a constant gain of 3#/inch.
Say, draw to 27 1/2 instead of 28.
First case: STORED KE = ((57/2) *19)/12 = 45.125 foot pounds.
Second case: Stored Energy = ((55.5/2) * 18.5)/12 = 42.78 foot pounds
Scenario 2 gives 97% stored energy of scenario 1.
When extrapolated into actual arrow drop, for every 10 inches of drop in scenario 2, the arrow drop in scenario 1 would be 9.4 inches (.97*.97) * 10.
In the case of your bows, where the draw weight drops off significantly as the draw length is approached, the difference would be much less significant.
To me, you can't have a bow that is "too smooth"
Agreed.
Did some calculations with a model of a linear bow, 57#@ 28 inches (19 inch power stroke).
This gives a constant gain of 3#/inch.
Say, draw to 27 1/2 instead of 28.
First case: STORED KE = ((57/2) *19)/12 = 45.125 foot pounds.
Second case: Stored Energy = ((55.5/2) * 18.5)/12 = 42.78 foot pounds
Scenario 2 gives 97% stored energy of scenario 1.
When extrapolated into actual arrow drop, for every 10 inches of drop in scenario 2, the arrow drop in scenario 1 would be 9.4 inches (.97*.97) * 10.
In the case of your bows, where the draw weight drops off significantly as the draw length is approached, the difference would be much less significant.
To me, you can't have a bow that is "too smooth"
