Subject: [baidarka] Superiority of Native paddles
From: Peter A. Chopelas (pac@premier1.net)
Date: Fri Jun 15 2001 - 21:24:33 EDT
Hi all,
Since things have been a bit slow I thought I would give you all something
to chew on.
Last spring I started a thread about the efficiency of high aspect ratio
paddles (such as most native paddles compared to Euro style paddles) and
the subject also recently came up on the PaddleWise list. I thought I
would share some technical analysis with you to make it available and have
it in the archives.
A few weeks 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.
Since I want to find the efficacy of the paddle we must define what that
is, it is the USEFUL power output (power it takes to move the kayak forward
at cruising speed), divided by the total power input (from your efforts).
Or Efficiency = Power-out/power-in
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 is used to accommodate all shapes such as ellipse or round plan forms,
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 more 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 can confuse you if you do not keep this strait and the rest
will make sense. So for this analysis I have defined the following terms
thus:
Drag= total forces you feel at the handle
Thrust=forward component of force on blade.
Note we are only interested in the forward component of the thrust since,
for moving the boat forward, that is the only USEFUL part of the force
acting against the water, the rest is just wasted energy.
Also note that not all of the effort you put into the handle goes into
creating useful forward thrust, the degree to which your effort does not go
into creating this forward thrust is the loss in efficiency. If you could
covert your effort at the handle into 100 percent forward thrust you would
have 100 percent efficiency by this definition. You put power into the
handle to raise and lower it, over come it's inertia, create turbulence,
etc. but none of this creates useful forward thrust (though it must be done
to do so).
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 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 know 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 much
theoretical vigorous 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.
So the total drag on the 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 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. 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
Note that the higher the velocity of paddle through the water the higher
the drag at the handle, and the higher the trust too, in exactly the same
proportions. That is why the velocity cancels, it had nothing to do
efficiency as some have argued.
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 the area of the paddle is
irrelevant to the efficiency (more on this later). Clearly the size of the
blade is not part of the efficiency equation.
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 this
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
>From this you can clearly see that as the Aspect ratio increases, the
efficiency 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).
You can also see from this relationship that there is no component of the
drag that is useful, any drag on the blade reduces it's efficiency.
Intuitively if you just put 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 efficiency is
really about.
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 reaction on the hull. 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 cursing 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:
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 a really long, very high AR paddle 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 out is 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), the 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,
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 where
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 it 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 work
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 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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