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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"
 

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Perhaps my wording wasn't as good as it should be.

That is aexactly what I meant. Same peak draw weight but rate of climb drops off, therefore difference in draw length (at or near peak weight) is less significant in determining the stored KE ..
 
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