From: Guillemot@aol.com
Date: Tue, 21 Mar 1995 21:56:27 -0500
Message-Id: <950321215523_56927552@aol.com>
To: baidarka@imagelan.com
Subject: Re: Paddling...
Bob Myers:
B: You skipped quoting my argument about using the weight. I'll quote it
again
B: here:
B: > To get back to your dropping weight thought experiment, if we have to
drop
B: > the weight more often to get the same time-average force, we have to
raise
B: > the weight more often - and that takes more energy, without using any
more
B: > force.
B: Nick, from your argument about the weight dropping, you agree that the
B: weight has to drop faster to produce the same force with a pencil than
with
B: a paddle, right? That means the cycle time is shorter for each dropping
of
B: the weight*, and therefore the weight is dropped much more often with a
B: pencil than with a paddle. Each time the weight is lifted for the next
B: cycle, the same amount of energy is used to lift the weight. So the total
B: energy expenditure is equal to the number of times you lift the weight
times
B: the energy to lift it once. Since you're increasing the number of times
you
B: lift it, your energy expenditure has increased.
This is a good point and I can't see any holes in it. However I am still
uncomfortable with the results. There must be something I am missing. Another
experiment: paddle your kayak for 1 minute, once with a pencil, once with a
pizza, applying the same force each time. In both cases the work done on the
kayak is the same. I guess the question is did the paddler do any more work.
I'm not sure. He applied the same force for the same time both times. Does
the fact that the smaller paddle was moving faster really matter?
If there is some waste energy with the smaller paddle where did it go. I
don't buy the argument that it is in the vortices. Sure there is now angular
momentum where what we want to conserve is linear. But after we push on the
water we don't care what it does. It can start spinning and still conserve
angular and linear momentum. I think the I know how the angular is preserved.
The vortices show up as pairs on the surface. One turning CW, one CCW. The
linear momentum can thus become angular and still preserve total momentum.
Water is moving, but it has to. You need some reaction mass. Its like
shooting BBs or boulders out the back of a space ship. With BBs you need to
shoot them fast to get the same reaction as a boulder. Isn't a small paddle
like shooting BBs?
Furthermore, if the smaller paddle wastes energy by creating more vortices
you would expect to see more when using a smaller paddle. My purely
subjective opinion is that you don't. I find it much easier to create big
mussle shell swallowing whirlpools with a big paddle. Is this relevant?
What am I missing?
Oh,... ...I've got an idea!
Kinetic Energy (KE) = 1/2 mass * velocity squared.
Momentum = mass * velocity.
Momentum of the kayak is the same in both the large and small paddle case. We
will call it MOM. KE of the kayak is the same in both cases as well.
Momentum is conserved. The momentum of the kayak (MOM) equals the momentum of
the water.
Kinetic energy of the water in the small paddle case: KE(ws) = 1/2 M(ws) *
V(ws)^2,
Large paddle case: KE(wl) = 1/2 M(wl) * V(wl)^2.
You can substitute MOM into KE: KE = 1/2 MOM * V:
KE(ws) = 1/2 MOM * V(ws),
KE(wl) = 1/2 MOM * V(wl).
The only variable left that matters is the velocity of the water in each
case. The smaller paddle must move faster to create the same force so it
follows that the water moves faster V(ws)>V(wl).
Therefore: The kinetic energy of the water is higher in the case of the
smaller paddle. If you've added more KE to the water without increasing the
KE of the boat you have wasted energy.
OK, I get it. ....Now is there anything wrong with the above?
Now to the lift question.
> Lift is
> produced perpendicular to the blade (not neccessarily perpendicular to
drag)
B: Exactly what definition of lift are you using? Lift is by definition
B: perpendicular to drag. Lift and drag are merely orthogonal components of
B: the total hydrodynamic force. Drag is the component in the direction of
B: motion, and lift is the component perpendicular to the direction of
motion.
The lift can be broken into two components. I guess you are right "Lift" is
defined as perpendicular to drag, but the "effective" lift on a wing is
roughly perpendicular to the chord of the wing. The difference between the
two (Lift and effective lift) is "induced" drag, which is in the same
direction as the "form" drag, or the drag created by moving any form through
a fluid. Therefore, the drag can be viewed as the sum of the form drag, skin
friction, and induced drag, all acting in the same direction. The remaining
force created by a wing is the effective lift. The induced drag can be
eliminated, making the effective lift equal the Lift, but you can not
eliminate the form drag and skin friction. If you want lift you must have
drag and there is nothing you can do to make all the vectors add up to a
100% useful direction.
In light of the above KE solution I can see where by using both lift
propulsion and drag propulsion you may end up imparting less KE to the water
and end up with a more efficient paddle stroke.
Nick Schade