Subject: RE: [baidarka] Superiority of Native paddles
From: Mr. Softeee (mr_softeee@hotmail.com)
Date: Wed Jun 20 2001 - 08:28:06 EDT
My question is if these paddles are superior paddles, How come Olympic kayak
racers don't use them? With all the research that is spent on improving
equipment why haven't they realized, according to you guys, that the native
paddle is superior????? Why don't surf ski racers use them for the long
grueling races in the Pacific??? I tell you I swear by my Werner Carbon
fiber feathered paddle.
sCOTT
>From: "Peter A. Chopelas" <pac@premier1.net>
>To: baidarka list <baidarka@lists.intelenet.net>
>Subject: RE: [baidarka] Superiority of Native paddles
>Date: Tue, 19 Jun 2001 20:42:20 -0700
>
>On Monday, June 18, 2001 6:14 PM, Michael Daly [SMTP:michaeldaly@home.com]
>wrote:
>
> >
> > Derivation - not proof.
>
>Fine, call it what you want. It is still within normal usage to "derive" a
>proof, it is a mathematical proof if you like, vs. an experimental proof.
> which I am confident will show the same thing and help convince more
>people, though you will no doubt still have some people STILL argue even
>against experimental proof.
>
> >
> > The chord is measured streamwise. Below you use the width of the
>paddle,
> > which in typical usage is normal to the flow. Why are you not using the
>thickness
> > of the blade?
>
>Typical usage is not normal to the flow, try it. You get flutter, vortex
>streets, turbulence, loss of thrust.
>
>But more importantly, by definition the aspect ratio HAS to be a function
>of the area of the blade, you do not affect the area with different
>thickness. The thrust is not affected by thickness either, so how could it
>affect the efficiency?
>
>
> >
> > If you use the thickness, you'll find the difference in aspect ratios
>between Euro
> > and native paddles not so great because if the considerable mean
>thickness of
> > the native paddles.
>
>Thickness not related to AR, and not related to thrust you get from the
>paddle.
>
>
> >
> > > So this relationship between the
> > > induced drag and the AR is just a natural phenomenon that is both
>observed
> > > and mathematically valid as well.
> >
> > Assuming you use the definition of AR for which the equation is derived.
>
>There is no other definition for the AR. If you do not want to take my
>word for it go look it up for yourself because I can not prove a negative.
>
> >
> > > Since the efficiency we are looking for is the power-out (i.e. forward
> > > thrust) divided by the power-in (resistance at the paddle handle) we
>have
> > > to covert these forces (in lbs. for example) to units of power by
> > > multiplying them by the velocity of the paddle through the water. so
>the
> > > equation will reduce to the following:
> > >
> > > efficency=P-out/P-in = TxV/DxV = T/D since the velocity cancels
> > >
> >
> > There are several problems with this assumption.
>
>This is NOT an assumption! It is my definition of efficiency, it is what I
>want to compare.
>
>
> >
> > One - I haven't got any paddles with handles.
>
>Fine, call it a shaft if you like. I presume you do not have any shovels,
>hammers, rakes or brooms at your house with handles either, they must all
>have shafts, right?
>
>Mine all use a shaft to transmit
> > the force to the blade and the forces in the shaft are derived from a
>moment
> > produced by my torso's muscles (I don't change the angle of my elbows
>much
> > and thus don't use arm muscles for generating thrust). This puts the
>paddle
> > blade on a roughly circular path. The path is about an eighth of a
>circle,
> > possibly more. I'll go along with your assumption of multiplication of
>velocity
> > for small angles, but an eighth of a circle isn't a small angle.
>
>Not relevant. I want to measure Useful-Power-out/total-power-in. You are
>getting stuck on details. think big picture, not details.
>Besides, did not you just say it is typically used normal to flow? Now you
>say there is a change in angle?
>
> >
> > Two - A long native blade can easily be shown to have a tip velocity
>about
> > twice the velocity where the blade is at the surface; perhaps more.
> Again,
> > I think that a multiplication is an oversimplification. Think
>convolution.
> >
>
>So what? This kind of detail is important when trying to optimize shapes
>at each cross section. This is why a propeller on an airplane for example
>has a slight amount of twist in it. In fact I suspect a slight amount of
>twist would benefit a high AR paddle too and am considering trying it. But
>it is not relevant to comparing ARs: if you make two identical paddles
>with the same area, shape etc. except with different AR, the higher AR will
>be more efficient. With twist, or without twist. You are comparing AR,
>not twist vs. no twist.
>
>
> > Three - Why is the "handle" velocity the same as the blade velocity?
> Given
> > the mechanics of paddling, I'm having a hard time mapping your "handle"
> > to a real paddle in such a way that I can see the equality.
> >
>
>Has nothing to do with handle velocity. The faster you move the blade
>through the water the more thrust you produce, and the more drag you
>produce in exactly the same proportion. This is based on the blade speed
>through the water only, since both are the same velocity, both cancel. The
>handle (or shaft if you wish) velocity (and the force you feel on it) will
>very depending on the length, but that does not change the POWER you have
>to add to it, it will be exactly the same. Since power=V x F... Long
>shaft, high velocity-low force; short shaft, low velocity-high force. The
>power input will be the same.
>
>
> > > But for low speed cursing
> >
> > I may paddle slow, but I curse fast! (sorry, couldn't resist!)
>
>I think all of us curse our paddles pretty regularly too, though I have
>observed this seems to have little affect on efficiency. But I'll keep
>trying it to see if I can get an effect. Research on this continues.
> >
> > > typical kayak at 5 knots (about 5 LB of drag, therefore 5 Lbs. of
>thrust at
> > > five knots need to be generated) I came up with the following for
>different
> > > paddle AR but with the same blade area:
> > >
> > > for typical Euro touring blade it has a blade approx. 6"x18" and AR=
>3:1
> > >
> > > 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:
> > >
> > > AR=3:1 P=0.4 hp
> > > AR=12:1 P=0.22 hp
> > > AR=27: P=0.15 hp
> >
> > Compare this to Matt Broze's figures in Sea Kayaker: 0.09 hp at 5
>knots.
> > (August 1998, page 43)
> >
>
>That is for Useful-power-out, the number above is for Total-Power-in.
> Though I just pulled numbers from a typical kayak and made guesses. It
>would be interesting to actually attempt to measure power-out vs. power-in.
>
>
> > >
> > > 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
> > >
> > > which is about what I suspected. The rest of your input goes to other
> > > things like heating water, lifting the weight of the paddle, etc.
> >
> > I can think of few things, other than those powered by infernal
>combustion
> > engines, which would survive in the real world with such pitiful
>efficiencies.
>
>Not true: FACT most automobile engines are doing about 24 to 30 percent
>efficiency at the flywheel (I have measure as high as 36 percent but it
>would not have been very driveable). If you include the whole car all the
>friction of the drive train and tires it is about 12 to 14 percent. I have
>measured it both ways (though since it is impractical to have a car without
>a drive train and rubber tires it is part of the power you need to drive a
>car). Most "natural" processes are not that efficient at all, most of the
>calories you consume for example are to just sustain life, not do useful
>work (so "sleeping" for example would have zero efficiency). Also keep in
>mind I am only measuring the small faction of the total energy input that
>goes to forward thrust only. Steering and manuvering inputs would be
>counted as a loss, as is the power required to raise and lower the paddle,
>but are still very necessary to move a kayak forward.
>
>
> > Now try to convince me that I can
> > do that with such poor efficiency.
>
>How do you want to measure it? How much drag is there on you when walking?
> Only a few ounces I suspect, but lets be generous say 5 ounces at 3 MPH
>(4.4 FPS), power out then is 0.3125 LB x 4.4 FPS/550 ft-LB/sec/hp = 0.0025
>hp. Walking takes about 0.2 hp of caloric input. Efficiency is: (drum
>roll please) .0025/0.2x100 = 1.25 percent Ta-DAH! Worse than you thought
>I bet. Don't take my word for it, go find out your own numbers, you will
>not see a big difference. 5% for a paddle looks pretty good actually.
>
> If that were true, arm-powered bikes would
> > whomp the leg-powered variety any day.
>
>I do not know what you are talking about (and apparently neither do you).
> We are not talking about total power out, which would be higher for the
>legs because the muscle's cross sectional area is much bigger on the leg
>than the arms. Again we want to know power-out vs. power-in, a ratio or
>percentage. Total power out tells you nothing about efficiency. For an
>arm powered bike or a leg powered bike the ratio would be about the same
>because the mechanism is about the same.
>
> > Also note that John Winters has
> > published, in The Shape Of The Canoe, a graph of his experimental
>results
>that
> > show that efficiency _decreases_ with aspect ratio. You'll have to
>explain why
> > your hypothesis is more valid than his experiments with real paddles.
> >
>
>I do not know how he measured this, likely he is not measuring the same
>thing. I want USEFUL-power-out vs. total-power-in at a nominal cruise
>speed. He might just be measuring total thrust vs. AR, or measuring it at
>max. available thrust, these would give you very different results: you
>would not use a drag racing engine (max thrust) to try and get max economy
>in you Honda Civic. It just depends under what conditions you are trying
>to optimize. An unqualified statement like that is meaningless.
>
>Also there are a lot of other things that affect output: stroke mechanics,
>crossectional shape of the blade. twist, angle, etc. The most efficient
>angle for a low AR paddle for example would be different than for a high AR
>paddle. It is difficult to make comparisons if you have not done extensive
>parametric studies to know what is affected when you make changes so you
>can normalize for them. I do not know for sure what is affected either,
>though I have a pretty good idea what to watch for. I have done a lot of
>experimenting on very expensive test equipment to know there are no clear
>answers, the equations are just a starting point so you know what direction
>to try. You often discover other effects along the way that you previously
>ignored. No matter how good your analysis you still must test. I know of
>no development program that does not use at least one third of their
>development budget in testing (usually much more).
>
>My own observation is that the high aspect ratio paddle is more efficient,
>and the equations point in that direction too.
>
>Peter
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