Message-Id: <9503212341.AA13158@st.unocal.com>
From: stssram@st.unocal.com (Bob Myers)
Date: Tue, 21 Mar 1995 15:41:17 -0800
In-Reply-To: kork@imagelan.com (Kirk Olsen)
To: baidarka@imagelan.com
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.
QED
-- 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