O.K., here's an excerpt that explains why ke goes up with arrow mass. It's not my opinion, its archery mechanics. Look it up for yourselves. Bigcountry, you might want to pay close attention, there will be a quiz at the end of the class to see if you have grasped the construct.

Note the expression m / (m + v) used in the determination of bow efficiency. Note that mass and virtual mass are summed in the denominator thus forcing efficiency and of course ke to increase as mass increases.

You candisagree all you want but don't argue with me. I'mmerely giving you the established physics regarding this. Argue with the scientists and engineers that havedeveloped the knowledge over the last 400 years or so.

O.K. here goes....


BOW EFFICIENCY AND THE CONCEPT OF VIRTUAL MASS
Defining Virtual Mass[align=left]Starting with a stationary bow at the brace height position, as you draw it back you store energy in the bow. The area under the force draw curve equals the total energy stored (Et). When the arrow is released the bow ends up stationary again at the bracing height. The total energy stored has gone somewhere. Most of this energy ends up where you want it as arrow linear kinetic energy (Ea) but the rest (Ew)is wasted.[/align]The energy equation for the bow is thus:
Et = Ea + Ew
The energy efficiency of the bow (F) is defined as the ratio of the arrow energy to the total stored energy. i.e.
F = Ea/Et
Supposing the arrow leaves the bow with a particular speed (S) then you can write the total stored energy in the bow as being equal to some imaginary mass (M) travelling at the same speed as the arrow. i.e.
Et = MS2/2
The value "M" includes the mass of the arrow (m) with the remaining mass (v) called the Virtual Mass. i.e.
M = m + v
This gives you that Et = MS2/2 = (m + v)S2/2 = mS2/2 + vS2/2
mS2/2 is just the kinetic energy of the arrow and vS2/2 is the wasted energy Ew in tems of the kinetic energy of an imaginary Virtual Mass.

[/align]The bow energy efficiency F = Ea/Et thus becomes:
F = (mS2/2)/((m + v)S2/2) = m/(m + v)
The bow energy efficiency can be defined in terms of the arrow mass and the value of the virtual mass. What makes this useful is that it is found by experiment that for a given bow the value of the virtual mass is a constant over a sensible range of arrow mass.

Note that this expression for bow energy efficiency and the value of v being constant indicates that the heavier the arrow then the more energy efficient the bow becomes. We'll look at the practical effects of this later on.

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This is a pretty abridged version. If anyone has a question or needs further explanation I'd be glad to help if I can but I'm not going to argue about this. The physics of archery has been well understood for the last 80 years or so, with vortex shedding being one of the last areas to explore. It can be discovered and understood by most everybody but debating fundamental concepts that have been well understood and established for many years is a waste of time.

Have a ball!

Oh yea, I should have added that the above is ALWAYS true. If the bow is capable of launching the spectrum of arrow masses you want to test, the efficiency will ALWAYS increase as the mass increases. The equation F = (mS2/2)/((m + v)S2/2) = m/(m + v) will always apply.