Re: [Paddlewise] PaddleWise [wing theory]

As usual, Peter posts a bunch of meaningless* equations when we get to 
this subject.  For those of you who might be awed by such, don't be - 
Peter understands equations but not physics, and his gobbledy-gook is 
not accepted by anybody here with a real clue about physics or fluid 
dynamics.

(* physical meaningless because they don't apply to the physical 
conditions he's talking about)

I'll just point out for starters this:

>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
>
Note that the velocity cancels because he says it does - because he uses 
the same velocity in both places.  But of course the paddle handle does 
not move at the same velocity as the paddle blade, so this is complete 
nonsense.

This is the very briefest sample of the kind of physics mistakes 
permeating Peter's work, and I have no intention of going through all 
his work and pointing out all his mistakes.






Peter Chopelas wrote:

>Someone asked for the theory of paddles, and I had happened so save some
>theorectical equations from the thread that was much debated last year and I
>have post it below for the benefit of all.
>
>And much of the recent speculation of the uninformed is incorrect, fluid
>mechanics is very complex and NOT intuitively obvious.  You can not compare
>"pushing" against a fluid with pushing against a solid object.  By
>definision fluids only generate thrust when you create motion in the fluid,
>and the best thrust DOES NOT occur with the paddle blade with the highest
>drag (in fact such a paddle would have the worst efficiency).
>
>It works out that for low speed, long distance curising, a thinner and
>smoother blade (on both surfaces), with the higher the aspect ratio of the
>paddle, the better the efficiency.  This is why the high aspect ratio native
>style paddles are superior for long distance sea kayaking.  And the total
>area of the blade and the speed you pull the paddle through the water is
>irrelevant to efficiency.  For maxium thrust however, maximum blade area is
>desirable, as in WW kayaking or surf kayaking where short bursts of sudden
>acceleration are necessary, but this is not efficient for long distance low
>speed cruising.
>
>For those so enclinded, below is the "proof" that the higher the aspect
>ratio for a smooth blade, the more effiecnt the paddle design.
>
>
>+++++++++++++++++++++++++++++++++++++++++++++++++++++++++++
>
>Paddle Aspect Ratio
>
>Some time ago I was clearing out my garage to make room to build another
>kayak and ran across a very old box of my engineering textbooks and the
>rather lengthy thread about high aspect ratio paddles came to mind.  I spent
>a few evenings thumbing through them and I put together a few equations for
>you all.
>
>Therefore just for grins I have listed a mathematical proof that you
>technical types might find interesting.  If the rest of you just read my
>text you should be able to follow the idea of the math without having to do
>it, at the very least if you may find the conclusion very interesting.
>
>The aspect ratio (AR) of a surface is the span squared divided by the area
>of the surface: AR=b^2/s  where b=span s=surface area
>
>This form of the AR equation is used to accommodate all shapes, notice that
>for rectangular surfaces the AR simply becomes the length divided by the
>width (or chord length c) or  AR=b/c
>
>Lets define some terms so everyone can follow.  As you move a paddle blade
>(technically a "foil") through the water what you feel at the handle end is
>the drag which you are pulling against to push your kayak forward, the blade
>OTOH must generate "thrust" so you have something to push against at the
>handle end.  The total "drag" you feel at the handle is directly related to
>the "thrust" the blade experiences in the water.  I think this relationship
>was confusing everyone, keep this strait and the rest will make sense.
>
>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 out.
>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].
>
>Also consider that you can never get more thrust out than the drag you put
>in.  It would be nice if you could more out then you put into it, but that
>does not happen in this universe [nor in the Brozian Universe either].
>
>The drag is composed of two parts, parasitic drag and induced drag.  The
>parasitic drag is what you would feel if you just slide the blade through
>the water sideways without producing any hydrodynamic thrust.  It is
>composed of the skin friction drag, the interference drag of the sharp edges
>and irregularities, the drag of the volume of the blade displacing water as
>you pull it through it.  To minimize this you would want very smooth foil
>shapes, very thin, and shapes that would not cause turbulence.  This is also
>known as the "base" drag on any surface when it moves through a fluid even
>when it is not generating any thrust or lift.  For automobiles for example
>all the aerodynamic drag on it can be considered parasitic since lift is not
>desired nor necessary (though most car shapes do generate lift, but it is
>undesirable since it reduces the traction of the tires, and causes induced
>drag on the body).
>
>The induced drag is the drag caused by the lift or thrust you are
>generating, IOW it is the "cost" of creating the thrust.  This has been
>experimentally determined (and verified many times) over a hundred years ago
>to act according to the following equation:
>
>   Coefficient of induced drag is: Cdi= Ct^2/(pi)AR
>
>Where Ct is the Coefficient of "thrust", pi is the mathematical constant
>3.1415927.  Coefficients are always used to normalize the factors from the
>variables that affect lift, drag, thrust, etc. for different flow
>conditions, more on this below. Notice that as the AR gets larger, the
>coefficient of induced drag gets smaller.  With finite shapes there is no
>way around induced drag, if you generate lift or thrust in a fluid, you must
>overcome this extra drag to get it (actually indirectly related to Newton's
>laws of action/reaction; when you get thrust, there is a drag reaction).
>And like many things in physics, it was this relationship between these
>variables that was first experimentally observed, and then later with
>vigorous theoretical mathematics "proven" to be valid (even though the
>experimentalists had already proven it).  So this relationship between the
>induced drag and the AR is just a natural phenomenon that is both observed
>and mathematically valid as well.  It is just a fact of nature that this
>relationship occurs and I can not "prove" it with mere words, you will have
>to go and prove it for yourself, or study the textbooks I have listed below.
>
>So the total drag on any surface creating thrust (or lift) like a paddle in
>the water is a combination of the base parasitic drag, and the induced drag
>expressed like this:
>
>  Cd(total drag)=Cdo+Cdi
>
>Also the expression for induced drag could be rearranged to express the
>relationship to the coefficient of thrust:
>
>  Ct=SQRT[3(pi)AR(Cdo)]
>
>Again as the AR goes up, the Thrust goes up too.  The total thrust force on
>the blade will be dependant on this coefficient of thrust, times the kinetic
>energy of the mass flow RATE, and the size of the paddle, or surface area=s.
>Kinetic energy is half of the mass times the velocity squared.  So thrust is
>expressed as follows:
>
>  Thrust=S(Ct(rho)V^2)/2
>
>the Greek letter rho is the mass density of the fluid, for sea water it is
>about 64 LBS/cubic foot divided by one G or 32.2 Feet/sec^2
>
>The total drag is expressed similarly:
>
>  Total drag=S(Cd(rho)V^2)/2
>
>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]
>
>This reduces to: Eff.=Ct/Cd since everything else cancels. Notice that the
>density cancels so the temperature of the water does not matter, the loss in
>thrust is exactly balanced by the lost in drag.  And also notice the area of
>the blade cancels, this is why I had wrote earlier that the area is
>irrelevant to the efficiency (more on this later).  Clearly the size of the
>blade is not part of the efficiency equation, nor is the velocity you pull
>the paddle through the water.
>
>Substituting these to determine the relationship between AR and Efficiency
>we get:
>
>  Eff. = Ct/(Cdo+Cdi)
>
>   = Ct/(Cdo+Ct^2/(pi)AR)
>
> to clarify this relationship further take the inverse so we can break the
>equation up:
>
>  1/Eff. = Cdo/Ct + Ct2/(pi)AR(Ct) = Cdo/Ct + Ct/(pi) AR
>
>To simplify further lets call the roughly constant ratios of Cdo/Ct and
>Ct/(pi) as constants K1 and K2 we get:
>
>  1/Eff. = K1 + K2/AR
>
>increases directly.  The base drag Cdo is going to be the same for similar
>designed blades of the same surface area even if the AR is different.  And
>the Coefficient of thrust is the same since we are producing the same thrust
>with each different AR to push the kayak at the same speed (even if the
>effort at the handle is different).  Notice too that even at infinite aspect
>ratio [where K2/AR=zero], the best efficiency you can get would be 1/K1.
>Which means the higher Ct/Cdo, the better.  Which in turn means, the more
>thrust, and least drag you can get out of the paddle, the better the paddle'
>s efficiency.
>
>You can see from this relationship that there is no component of the drag
>that is useful, any drag on the blade reduces its efficiency.  Intuitively
>this is easy to grasp if you just imagine putting a large drag device on the
>end of a paddle shaft instead of a blade, say something like a giant pine
>cone, it would not work as good as a smooth foil shape.  Although you can
>use it go forward, the equation above, and common sense, tells you that it
>would be a lot of effort for the amount of forward progress you would make.
>
>You can see clearly from this relationship that if you want to increase
>efficiency you want maximum thrust at the paddle blade, with minimum effort
>(or drag) at the handle end where you are holding it.  Aha!  Now that makes
>sense, max thrust for the least effort!  And that is what we want to
>consider when we discus the efficiency of the paddle.
>
>Also keep in mind this is about efficiency and not total available thrust.
>Making the paddle as big as possible would provide the max thrust, this is
>from the momentum theory and Newton's laws of motion, the more and faster
>you accelerate a mass of water, the bigger the thrust reaction.  So for
>racing, rapid accelerations, or rapid control movements like you need in WW
>or surf kayaking, the biggest blade you can handle would be best, and aspect
>ratio is not as important.
>
>But for low speed cruising to minimize energy expenditure over long periods
>of time, the high AR paddle is king.  Plugging in a few numbers for a
>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 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
>
>Wow!  This makes me want to make one of those really long, very high AR
>paddles just to try it out.  It would have to be 10.5 feet long!  There may
>be a practical limit to AR, just like an aircraft wing or a propeller, but
>it would be worth a try.
>
>Something else I just worked out is that the power requirements for a
>typical paddle/paddler might be something like this:
>
>Power in at the handle:  824 ft-lb/min or about 0.025 hp (typical
>sustainable output for human arms in reasonably good condition)
>
>Though you could measure this indirectly with an actual person paddling by
>measuring oxygen uptake with a portable device (you would have to assume a
>oxygen uptake rate to power output ratio but that could also be
>experimentally determined on a human dynamometer).  Though O-2 uptake would
>give you a simple way to compare one paddle to another.
>
>
>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.
>
>how much a new paddle design could help this is unknown, but the equations
>indicate a good design could as much as double the efficiency.  One thing
>for sure, even with these assumed numbers (which are within reasonable range
>based on my experience doing testing on Olympic athletes), there is lots of
>room for improvement in paddle design.
>
>And certainly there are many other aspects that affect efficiency than just
>AR of the paddle:  the foil shape, the stroke mechanics, the surface finish,
>perhaps shaft/blade stiffness, etc.
>
>Needless to say, it appears that the design of the hull (which everyone
>always focuses on) is only one part of the total picture, and perhaps I
>suspect not even the majority part.  It is the paddles!!!
>
>Also unknown by me from this analysis is the effect of the surface wake that
>the paddle creates as it is moved through the water.  Though I suspect the
>surface wake does not produce any thrust but just drag.  This means the most
>efficient way to paddle would be to try to minimize the surface wake of the
>paddle as much as possible.  That would be from a fairly thin smooth blade,
>slicing downward through the water as you pull it back; it would make the
>least surface wake.
>
>Note that there is a native technique which I have found comfortable.  You
>start with the paddle at almost level and about a 45 deg angle across the
>front of your hull, reaching horizontally out, and then slicing downward as
>you pull it back, ending with the paddle nearly vertical at the end of the
>power stroke by your side, but ready to move it back across in front of you
>to start the next stroke.  Also there is a racing stroke where the racers
>push the blade out sideways from the gunwale in the water as they pull back.
>Both of these strokes would keep a surface wake from the paddle minimum
>since you are moving the blade through the water with the small dimension of
>the blade, not in the direction of the full width of it.
>
>Pulling a fat Euro paddle strait back through the water, making lots of
>vortexes and a large surface wake, would be the least efficient way to
>paddle (though you will accelerate just fine in a short burst, it just will
>not be efficient).  But regardless of the effects of a surface wake, you can
>not escape the importance of the AR.
>
>>From this you can see that smooth foil shaped, high aspect ratio blades are
>best for conserving energy over long hauls.  It will not work well for
>maximum accelerations, nor for quick manuvering strokes, and likely very
>poorly for a C-C type roll.  In fact for all of these needs you have to use
>very different techniques with a high AR paddle.
>
>Most of this I have pulled from "Theory of Flight" by Von Mises, and from
>"Simplified Aircraft Performance and Design" by Donald Crawford, and my
>class notes and a few other books if you are so inclined to verify the
>equations.
>
>There will be a quiz next week.
>
>Peter Chopelas
>
>
>
>
>
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