Re: [Paddlewise] PaddleWise [wing theory]

From: Nick Schade <nick_at_guillemot-kayaks.com>
Date: Wed, 11 Jun 2003 09:08:02 -0400
Bob said it well. This analysis doesn't pass even the most casual 
sanity check. Measurements of conventional "euro"  paddles show an 
efficiency of about 74% and for wing paddles of 89%. If a high aspect 
ratio paddle is twice as efficient this would mean a Greenland style 
paddle is 148% to 178% efficient. In other words the high AR paddle 
would actually be generating energy. Cool!

As Bob pointed out, the velocities aren't the same thus shouldn't 
cancel. And the final equation is a ratio of forces, essentially force 
in and force out as best as I can follow. The idea that the drag force 
at the paddle blade is different in anything but sign from the force 
being applied by the paddler's hands is self evident nonsense. For 
forces to balance this would require that there be some other force 
acting on the paddle besides the hands and the water, the analysis 
gives no indication of where the extra force goes. If this mystery 
force is the wind, I would expect to see some factor for the wind 
included in the calculation.

There are obviously some faulty assumptions in the analysis. The 
equations do not agree with available empirical data nor are they 
physically logical. I don't know what the faulty assumptions are, but 
the canceling of the velocity is a good place to start.

On Monday, June 9, 2003, at 07:30  PM, Peter Chopelas wrote:
>
>   Cd(total drag)=Cdo+Cdi
>
>   Ct=SQRT[3(pi)AR(Cdo)]
>
>   Thrust=S(Ct(rho)V^2)/2
>
>   Total drag=S(Cd(rho)V^2)/2
>
>   efficency=P-out/P-in = TxV/DxV = T/D    since the velocity cancels
>
>   Efficency= [S(Ct(rho)V^2)/2]/[S(Cd(rho)V^2)/2]
>
>   Eff. = Ct/(Cdo+Cdi)
>
>    = Ct/(Cdo+Ct^2/(pi)AR)
>
>   1/Eff. = Cdo/Ct + Ct2/(pi)AR(Ct) = Cdo/Ct + Ct/(pi) AR
>
>   1/Eff. = K1 + K2/AR
>
> a typical native style blade 3"x36" an AR=12:1
>
> hypothetical very high AR paddle of 2"x54"...AR=27
>
> the following sustained power at the handle would be required from the
> paddler to maintain the same thrust output:
>
> AR=3:1  P=0.4 hp
> AR=12:1 P=0.22 hp
> AR=27:  P=0.15 hp
>
>
> Useful Power-out:  5 lb drag at 5 knots = 5 LB x 5 x 1.688 
> feet/sec/knot =
> 42.2 foot-lb/min = 0.001279 hp
>
> efficiency of converting power-in to forward movement:  0.001279/0.025 
> =
> 0.051 or about 5 percent
Nick Schade

Guillemot Kayaks
824 Thompson St
Glastonbury, CT 06033
USA
Ph/Fx: (860) 659-8847
http://www.guillemot-kayaks.com/
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Received on Wed Jun 11 2003 - 06:29:23 PDT

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