From: Guillemot@aol.com
Date: Mon, 20 Mar 1995 22:30:35 -0500
Message-Id: <950320223029_55797320@aol.com>
To: baidarka@imagelan.com
Subject: Re: Paddling...
> OK... let's say I rig up a contraption on my kayak that
> provides for a 5 lb weight to be suspended by a line and set
> of pulleys such that when the weight is released, it moves a
> 'clamp' back, in a straight line, toward the stern... I can
> attach anything to the clamp... a pencil, a
> pizza, a paddle, whatever...
> Are you saying that if I attach a pencil to my clamp, the
> kayak will move forward as far and as fast as it would if I
> used a paddle in the clamp?
Force = mass * acceleration. Weight refers to force not mass and is equal to
mass * [the acceleration of gravity (G)]. A mass weighing 5 lbs does not
always apply a 5 lbs force. It only weighs 5 lbs at 1 gravity. If it is in
free fall it weighs nothing but has the same mass. Lets use M = 5 kg for our
example because the units are easier. At 1 gravity (G) (about 10
meters/second^2) this mass will apply a force (weight) 50 Newtons.
For your experiment to apply 50 newtons of force it must not accelerate as if
falls. If it does accelerate at A m/sec^2 the force applied will be Fm= 5 kg
(G - A). Notice in free fall (A = G) no force is applied.
So your pencil provides less drag than a pizza at a given velocity (V). Drag
force (Fd) is approximately equal to K * V^2 where K is some drag
coefficient. K is dependant upon the form of the paddle. This is some complex
combination of area, shape, smoothness, and orientation, water viscosity,
etc, etc. When you first drop the mass, V=0, therefore Fd=0, therefore,
nothing restraining your falling mass, so it goes into freefall A=G.
Immediately V increases, causing Fd to increase. This creates an equal and
opposite force on your falling mass, reducing A. The forces are matched and
continue to match as the drag force increases until the point where Fd= K *
V^2 = 50 newtons. At this point the falling mass is no longer accelerating
(A=0). This is at it's terminal velocity where 5 kg * G = K * V^2. A smaller
paddle will have a smaller K, resulting in a higher terminal velocity.
_
/O\
| |
| |
(M) |
V| A| Fm| |
v v v |
|
______________|____________
|
\O______| -->
| Fd
<-
V
The point of all the above is that the force applied is always exactly the
same as the force pulling back. If you pull with 5 lbs of force the object
will push back with 5 lbs of force. Take two spring scales, hook them
together, and pull on either end. Obviously they will always read the same
thing.
So how about all those vortices? Remember conservation of momentum. Before
you let go of the mass the momentum was zero. When the mass reaches terminal
velocity the momentum must still be zero. Momentum (M) = mass * velocity. If
V= 1 m/sec, the momentum of the mass is 5 kg m/sec. Since the momentum of the
system must be 0 the momentum of the water must be -5 kg m/sec. So the water
sits there and spins in little vortices to create enough momentum.
> But you're also right about something else - there *is* a real problem with
> this force-only analysis. We also need to look at power and energy, and
> this is where smaller paddles show their inefficiency. In terms of work
> done (I'm referring to the physics definition of work here), each paddle
> stroke using the same force pulled through the same arc does the same
amount
> of work. Work = Force x distance. So if you make twice as many paddle
> strokes, but only result in the same average force on the kayak, you've
done
> twice as much work to do that!
Strokes are not the same as distance. The distance you need to use is how far
_you_ moved, not how far the paddle moved.
> Another way to look at it is the power situation. The power you expend in
> paddling is equal to the force x velocity -- so if you paddle faster,
you're
> expending more power. Power is also the energy/time, so if your average
> power expenditure is twice as high, your energy expenditure over that time
> will also be twice as high.
If you apply the same force to the same mass you will get the same
acceleration therefore the same velocity.... ad infinitum. The energy is the
same. Again, you can't look at the velocity of the paddle. You need to look
at the velocity of the kayak. The paddle is only the conduit through which
you apply the force. The force from the water is applied to the kayak via the
paddle. The paddle shaft serves the same function as the rope in the above
experiment
> Airplanes and helicopters paddle through the air quite successfully.
One advantage a paddler has over a airplane or helicopter, is we can remove
our paddle from the fluid we are pushing against. In other words we can keep
tossing a parachute back up with less force than it took to pull it down. An
airplane must claw its way up through a consistant medium. To do this it must
use lift. We can you drag. Lift neccessarily also produces drag.
Unfortunately it is not in a direction that is useful.
> Wait a minute - it's not useful to separate lift and drag in this manner.
> You don't have to move the paddle in either up-and-down or horizontally
> back; you can combine the two motions. If you do so appropriately, there
is
> no wasted force or energy. Think of it this way: lift and drag are
merely
> the two perpendicular components of the hydrodynamic force. The part of
> that force that is parallel to the direction of motion of the paddle is
> called drag, and the part that is perpendicular to the direction of motion
> is the lift. If you take that total hydrodynamic force vector, instead of
> worrying about whether you're using "lift" or "drag", you should be able to
> move the paddle so that the total force vector is horizontal, pulling the
> kayak forward - without any wasted force.
True, you cannot seperate the paddle stroke into seperate lift portions and
drag portions. You are probably always providing some lift and you are
definitely always providing drag. However, the only forces that are useful
are those in the direction of motion. Useful lifting force comes from that
component of paddle motion that is perpendicular to the kayak motion and
useful drag comes from that component of paddle motion that is parallel to
kayak motion. If there is any paddle motion perpendicular to the kayak motion
there is neccesarily drag in that direction as well. There is no possible way
to direct the vector of this drag backward. You can minimize the
non-backwards force but you can't eliminate it.
Looking at the vectors. Your paddle is on the left, the blade angle is
perpendicular to the kayak. You pull it backward towards the right. Drag is
parallel to motion, so the drag force is forward to the left. Lift is
produced perpendicular to the blade (not neccessarily perpendicular to drag)
or directly forward. You still have the drag force to the left. So you angle
the inner edge of the blade forward to create more lift. This vectors the
lift towards the right but increases the drag force towards the left. That's
no good. So you angle the inner edge of the blade back. This reduces the drag
but vectors the lift toward the left. Regardless of how you angle the blade
you will end up a component of wasted force to the left.
> Caveat: all of this ignores the biological aspects of power generation.
The body is the truly inefficient system in this whole equation. There will
be an optimum force and rate of motion for a body. The best combination of
force and rate of motion will be different for everyone. Remember our goal as
recreational paddlers is to get where we are going in the most comfortable
manner. Some people are comfortable pushing hard and fast, others like hard
and slow, some prefer soft and fast, while others would much rather stay on
the beach eating donuts. Your preferred mode of travel will determine what
style and thus what paddle is best for you.
Boy, this was a long response.