John R Weiss wrote:

"Jim Woodward" ...
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?

Brian Whatcott wrote...

Yes, ignoring scaling constants:

Horse power = thrust X speed

It appears there are a few means of conversion...

My MinnKota EX42 is advertised to put out 42 lb thrust at 12V and 36 amps. What
is that in electrical power consumption, converted to HP?

It also pushes my 21' boat at about 2.5 - 3 knots at full thrust on flat water.
What is that in power output or work done?

I don't have my calculator and conversion constants handy, but at first glance
the electrical power draw appears considerably higher than mechanical power
output or work. OTOH, maybe it isn't pulling all 36 amps when pushing the boat
at max speed (I assume the prop RPM is electronically governed).

that works out (at 100% conversion efficiency) to .57 hp. figure the motor at about
85% efficiency (at best). and you are at just under 1/2 hp.


--
Beer, it's not just for breakfast anymore.......

What no-one has considered here is that powering calcs are usually done from
the other point of view:

My boat has X lbs of drag at 5 knots, how many horsepower do I need?

Factor in windage (the lee shore situation), foul hulls, appendage drag, prop
efficiency (an oxymoron if ever I saw one) and other such truck like Taylor
Wake Fraction, and you have a real problem on your hands.

Doing it in reverse? Priceless....

Steve

On 17 Sep 2003 02:52:45 GMT, ospam (Stephen Baker)
wrote:

What no-one has considered here is that powering calcs are usually done from
the other point of view:

My boat has X lbs of drag at 5 knots, how many horsepower do I need?

Factor in windage (the lee shore situation), foul hulls, appendage drag, prop
efficiency (an oxymoron if ever I saw one) and other such truck like Taylor
Wake Fraction, and you have a real problem on your hands.

Doing it in reverse? Priceless....

Steve


Once you seize the idea that if a vessel (of any sort) is not
accelerating wrt the water mass, then the total drag = total thrust.

Brian Whatcott Altus OK