[Paddlewise] Efficiency Analysis.

At 11:38 PM -0700 7/26/02, Peter Chopelas wrote:
><snip>
>Consider it this way, in an ideal world, if the amount of drag you push
>against at the shaft is exactly equal to the amount of forward thrust you
>get out of it, you would have 100 percent hydrodynamic paddle efficiency.
>This is ignoring the efficiency of the human "machine" of course because we
>just want to compare energy input at the handle, to the forward thrust
>output.  This also ignores the energy it takes to raise and lower the paddle
>since we want to compare the hydrodynamic efficiency, not the mechanical
>efficiency.  A paddle that weighs the same, with the same inertia and
>stiffness, will have the same mechanical efficiency [and a paddle with zero
>weight, and infinite stiffness, is 100 percent mechanically efficient].
><snip>
>Since the efficiency we are looking for is the power-out (i.e. thrust)
>divided by the power-in (resistance at the paddle handle) we have to convert
>these forces (in lbs. for example) to units of power by multiplying them by
>the velocity of the paddle blade through the water.  So the equation will
>reduce to the following:
>
>   efficency=P-out/P-in = TxV/DxV = T/D    since the velocity cancels
>
>substituting the above relationships in we get the following:
>
>   Efficency= [S(Ct(rho)V^2)/2]/[S(Cd(rho)V^2)/2]

Thrust and drag are both forces. They are not power. The amount of 
force you apply to the paddle is exactly matched by the amount of 
force generated by the water on the paddle. It is impossible to push 
with 1 pound of force and have what you are pushing against push back 
at with 1/2 or 2 pounds of force. The forces must be equal and 
opposite. If they aren't equal there must be something wrong with 
your analysis.

In the above equation for efficiency you incorporate the velocity. 
This presumably converts the force to power. However, you drop it 
immediately, saying the velocities cancel. This leaves an equation 
where efficiency is the ratio of two opposing forces "T" and "D". 
That ratio is always going to be 1. If it is not, there would have to 
be some outside force acting on the paddle, not just the paddler and 
the water. While drag through the air would work, in your analysis 
there is no third force, only the force applied by the paddler 
(Thrust), and the force caused by the water (Drag).

Your equation depends on two opposing forces being different. That is 
a physical impossibility. I must assume that it is incorrect to 
cancel out the velocities. Since the power applied by the paddler is 
near the center of the paddle and the power applied by the water is 
near the end of the blades, I don't know why the velocities involved 
would be the same.
-- 
Nick Schade
Guillemot Kayaks
824 Thompson St
Glastonbury, CT 06033
(860) 659-8847
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