Jim:

You've described the conversion of force into power units - and it is all
OK. You'll note from your equation that thrust-per-hp is inversely
proportional to speed. You can generate more thrust at low speed. But if you
approach zero speed, you get infinite thrust - which, of course, means we're
missing something.

When considering the relationship between a propulsor's developed thrust and
its absorbed power (from an engine or motor), you have to introduce a number
of different efficiencies. In somewhat simple terms, from engine/motor power
to applied thrust, you pass through mechanical energy losses (mostly
friction and heat in shafting and transmission), propulsor energy losses
(such as friction and non-useful rotation of some of the water), and thrust
application losses (where the propulsor's suction side actually creates a
detrimental "suction" on the hull).

So, your thrust needs to be reduced to account for these losses - even for
small electric trolling motors and propellers. You'll need to include the
efficiency, and these overall system efficiencies also change with speed. At
running speeds, a good typical system efficiency is 65%. As speed approaches
zero, however, efficiencies also approach zero to keep the original
relationship from getting out of hand.

Regards,

Don

Donald M. MacPherson
VP Technical Director
HydroComp, Inc.
email: dm~AT~hydrocompinc~DOT~com



"Jim Woodward" wrote in message
m...
Since one HP is 550 ft-lbs per second, that suggests that one hp is
5.5 pounds thrust at 100 feet per second or 6.25 pounds at 60mph
(statute miles, here).

That feel OK, but my college days are in the distant past -- any
comments?

Jim Woodward
www.mvfintry.com