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From: Jeffrey Steck <jeffrey_steck_at_hotmail.com>
subject: [Paddlewise] Newtonian Physics
Date: Sun, 28 Jul 2002 11:58:20 -0400
To paraphrase a famous joke about physicists,
"Assume the paddle is a sphere . . ."

Seriously, the physics does not need to be all that complicated
to provide some insight. A reasonable question to ask is how
much power does the paddler have to put out for a given
boat speed.  The equation for power is simply:

POWER = FORCE * VELOCITY

The FORCE is just the resistance of the boat at that
speed, becasue of the constant speed assumption.
You can look up the resistance numbers in Sea Kayaker. (I am
ignoring the effects of "coasting" between strokes.)

The VELOCITY is the velocity of the paddle relative to the
paddler, which is the velocity of the boat through the
water PLUS the velociy of the paddle through the water, i.e.
the speed of paddle slippage.

Thus, the power that a paddler has to put in to keep
a constant boat speed is equal to:

     BOAT_RESISTANCE * (BOAT_SPEED + PADDLE_SLIPPAGE_SPEED)

In other words, all other things being equal, greater slippage
means more power output by the paddler with no speed advantage.
Note that this has nothing to do with paddle shape, except in
how much slippage the shape allows.  A giant pine cone would
work fine as a paddle, except that it would be hard to pull it
out of the water after every stroke.

Peter Chopelas wrote:
>  efficiency of converting power-in to forward movement:   0.001279/0.025 = 
>0.051 or about 5 percent

Efficiency of 5% would mean that for a boat speed of 3knots,
the paddle is slipping backwards at 57 knots, something none
of us have experienced.

Jeff Steck

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From: Peter Chopelas <pac_at_premier1.net>
subject: Re: [Paddlewise] Newtonian Physics
Date: Tue, 30 Jul 2002 19:20:48 -0700
Mike wrote:

>Let's work some numbers. Say we want to go 4 mph (5.87 fps), and say
>Matt's table says that takes a force of 8 lbs (It's a heavy boat).

>Say our big fat WW paddle moves through the water twice as fast as the
>boat to go 4mph:

        8lbs * (5.87fps + 11.74 fps) = 140 ft-lb/sec = 0.256 HP.

>  Now let's try it with a narrow Greenland paddle that requires paddling
>four times as fast as the boat:

>        8 lbs * (5.87fps + 23.48 fps) = 234.8 ft-lb/sec = 0.427 HP

>But that doesn't make sense...<snip> Ergo there's an error in the model.


Mike,

Your model is wrong because you are comparing the wrong forces.  In both
cases you used the resistance of the drag of the boat, 8 pounds, and then
multiplied that by an assumed paddle speed.  You have to multiply the hull
drag by the HULL speed and you get the power OUT-put.  Multiplying the hull
drag by the assumed paddle give you nothing useful because they are not
related.  The output it is the same with both paddles when you are pushing
the same hull at the same speed.

What you want to compare is the force at the paddle handle times the paddle
speed for each paddle type, which would be the power input.  There would
likely be a difference in the power input for each paddle, the one with the
lowest input (to maintain the same hull speed) would be the most efficient.
You do not have enough information to determine power input.

I think an easier way to get a comparison would be to measure air intake
(there are simple breath masks that can do this) it takes to sustain the
same speed with different paddles.  Another way is measure heart rate at the
same speed (there are sports watches that will do this), though you would
have to paddle for a bit to make sure your heart rate has stabilized.

Both of these measures will not give you power input, but you will have a
direct comparison.  And it would include the "human factors" in the test, so
the best cadence, best biomechanices, etc. will be built into the test.

Peter

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