Re: [Paddlewise] paddle sizing heresy

From: David Carlson <qajaq_at_nwlink.com>
Date: Fri, 11 May 2001 17:11:50 -0700
This thread has finally lead me to attempt some very simplified calculations:

These calculations are made assuming that for both the boat and paddle, force is
proportional to the square of the velocity through the water (Fp=A*Vp^2 and
Fb=B*Vb^2 ) and also assuming that we can learn something from calculations
involving the very great simplification that the velocity (and resistance force)
of the boat and paddle remain constant.  This last is impossible, since it would
involve having no time between strokes, but here are the results for what they
are worth.  If people are interested in the derivation of the equations, I can
supply them.

First, if X is the length of the stroke and Y is the distance the paddle slips
through the water during a single stroke, Y=X(sqrt(B)/(sqrt(B)+sqrt(A)).  That
is, for a given stroke length the slippage depends only on the resistance
coefficients of the paddle and the boat, so of course, James is right.  No
matter how fast you paddle, for a given boat/paddle combination (neglecting boat
acceleration) the distance the paddle slips through the water with each stroke
remains the same.

This leads to a simple efficiency calculation, using Power out/ Power in as the
definition of efficiency and the constants defined above for boat and paddle
resistance:

    Efficiency = sqrt(A)/(sqrt(A)+sqrt(B))

It would be interesting to have some ballpark numbers for A and B.  From the
equations, the efficiency is only slightly greater than 80% when the paddle
coefficient A is 20 times greater than the boat coefficient B.  The efficiency
increases pretty slowly with increased paddle resistance after that.  To get 10%
slippage, A/B=80.

Note that for the 80% paddle/boat combination described above, the paddle would
slip through the water about 20% of the stroke length.

How reasonable are the equations?  I really don't know.  I think there was a
post that gave some resistance figures for paddles, but I have deleted it.  I
looked at one set of resistance curves in Sea Kayaker Mag and for those curves
at speeds less than 4-4.5 kn the resistance coefficient was fairly constant and
close to .22 lb/kn, giving a resistance force of about 5.2 lb at 4.5 kn.  That
means that for our 80+% paddle A = 4.4 lb/kn, the velocity of the paddle
relative to the water would be 1 kn and it's velocity relative to the boat would
be about 5.5 kn.  Since in reality the a paddle is on the water only part of the
time, either the paddle velocity through the water or A would have to increase.

Anyway, the relatively small effect of increased area (resistance coefficient)
of the paddle after a certain point was consistent with my experience switching
from a big Harmony paddle to a smaller Kawai.  If anyone knows some reasonable
figures for paddle resistance coefficients, it would be interesting to see what
kind of efficiencies can be expected.

Regards,

Dave Carlson

James Tibensky wrote:

> Maybe I'm missing something with my silly examples, but here goes again.
>
> Sit alongside the water, not in a boat but on solid ground, and make a
> paddling motion with a toothpick. The toothpick slips easily through the
> water with no noticeable resistance.  Try it with a popsicle stick.  Then a
> ping pong paddle.  Then an Inuit style blade.  They all slip through the
> water, as will whatever paddle you're using.  The bigger the surface area of
> the "blade" [I sholdn't call a toothpick a blade], the less slip and the
> more resistance.  But they all slip, for Pete's sake.  If a blade didn't
> slip in the water when paddling a kayak, there wouldn't be any cavitation,
> would there??  And there would be no need to dig deep.
>
> And we would go just as fast in concrete as we do in H2O.
>
> I think.
>
> Jim Tibensky
>
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Received on Fri May 11 2001 - 17:12:39 PDT

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