Message-Id: <9503220021.AA13421@st.unocal.com>
From: stssram@st.unocal.com (Bob Myers)
Date: Tue, 21 Mar 1995 16:21:53 -0800
In-Reply-To: stssram@st.unocal.com (Bob Myers)
To: baidarka@imagelan.com
Subject: Re: Inefficiency of paddle slippage
On Mar 21, 15:41, Bob Myers wrote:
} Subject: Inefficiency of paddle slippage
> It occurs to me that there really is a simple way of showing that paddle
> slippage causes energy transfer inefficiency.
>
> As a kayak moves through the water at a constant speed V, there is a
> constant drag D on it. The power lost to drag, and therefore required to
> keep the kayak at constant speed, is D*V. Again at constant speed, the
> paddle must exert an equal and opposite force F=D to keep the kayak from
> decelerating. If the paddle *does not slip*, it moves at the same speed as
> the water, and therefore the work done/time is F*V = D*V. No losses. But
> if the paddle slips through the water, it moves faster than the water (frame
> of reference of the paddler, not a stationary observer), so the work
> done/time is D*(V+Vs), where Vs is the slip velocity (how fast the paddle
> slips through the water). You've lost D*Vs in power due to paddle slippage.
>
I should probably wait before posting on this until I've thought it all the
way through. Oh well. I hope this is the last posting on paddle slippage
from me.
I know where the extra energy is going - it's moving the water backwards.
Nick said earlier that you need to move the water to get momentum transfer,
and he's right, and George mentioned that you can move the planet or move
the water, and it doesn't make any difference for propulsion force - and
he's right. But - how you do that makes a big difference for energy
efficiency. It's much more efficient to move larger masses at slower
velocity than smaller masses at higher velocity.
Momentum = mass * velocity = mv
Kinetic energy = 1/2 * mass * velocity^2 = 1/2 mv^2
Momentum transfer to the water is constant at constant velocity, since it's
equal to the drag on the kayak * the time elapsed. So if you move half as
much water, you need to move it twice as fast. If you're moving the water
twice as fast, that water has 4 times the kinetic energy, though, due to the
v^2 term. So you want to move as much water as possible at as slow a
velocity as possible to maximize the energy efficiency.
I think this is the same reason modern "jet" aircraft use large fans
(turbofan engines) for the bulk of the propulsive force - the fan moves much
more air than the jet itself does, so it's much more efficient.
-- Bob Myers Unocal Tech. & Ops. I. S. Support Internet: Bob.Myers@st.unocal.com P. O. Box 68076 Phone: [714] 693-6951 Anaheim, California 92817-8076