Is there a simple equation between lbs thrust (as specified in electric
outboard motors) and Horsepower?

Paul Squire says:

Is there a simple equation between lbs thrust (as specified in electric
outboard motors) and Horsepower?

Yes, but it involves speed as a variable.
Be very careful with manufacturer's figures for thrust, as it is usually taken
at very low (if any) speed, where the thrust is highest.

Steve

On Tue, 16 Sep 2003 22:51:57 +1200, "Paul Squire"
wrote:

Is there a simple equation between lbs thrust (as specified in electric
outboard motors) and Horsepower?


Yes, ignoring scaling constants:

Horse power = thrust X speed

Brian Whatcott Altus OK

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
Brian Whatcott wrote in message . ..
On Tue, 16 Sep 2003 22:51:57 +1200, "Paul Squire"
wrote:

Is there a simple equation between lbs thrust (as specified in electric
outboard motors) and Horsepower?


Yes, ignoring scaling constants:

Horse power = thrust X speed

Brian Whatcott Altus OK

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

SI may be a nuisance to some folks - but it shines here, because there
is NO scaling constant in this relation:
power (watts) = force (newtons) X speed (meter/sec)

Now let's put in all the scaling constants to US customary units
(Can't say British, or Imperial, any more...)

1 watt X 1HP/746watts =
1 newton X 1kg/9.81newtons X 2.2 lb/kg X
1 meter/sec X 39.37 inch/meter X 1ft/12 inches X
1 mile/5280ft / (1min/60 sec X 1hr/60 min)

And this boils down to
0.00134 HP = .2243 lb X 2.237 mph
or even more succinctly:
1HP = 374 lb X mph
(Your solution gives 6.25 X 60 = 375 lb X mph - close enough)

As James Watt defined the HP as a horse capable of exerting a force of
550 pounds at a speed of one foot per second (which is about 50% more
than a good horse can do for a shift, without then falling down dead)
your figure of 5.5 lbs at 100 ft/sec is *exact*

Brian Whatcott Altus OK

On 16 Sep 2003 11:22:30 -0700, (Jim Woodward)
wrote:

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
Brian Whatcott wrote in message . ..
On Tue, 16 Sep 2003 22:51:57 +1200, "Paul Squire"
wrote:

Is there a simple equation between lbs thrust (as specified in electric
outboard motors) and Horsepower?


Yes, ignoring scaling constants:

Horse power = thrust X speed

Brian Whatcott Altus OK

"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).

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 Tue, 16 Sep 2003 17:56:16 -0700, "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).
//
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).

12 V X 36 A = 432 W
432 W X 1HP/746W = 0.58 HP

IF notice IF 0.58 HP is the electrical rate for 2.5 kt ( = 2.5kt X
1NM/hr / 6080ft/hr X 1hr/60min) = 253.3 ft/min
\so the required HP at 100% efficiency is 253.3/550 HP = 0.46HP

So the efficiency might be 0.46/0.58 (power out/power in) = 79%
which is rather high.....

Alternatively,

42 lb thrust from 0.58 HP would give at most 7.6 ft/min
which is 0.075 kt. Hence, unsurprizingly, the maker is quoting
quasi-stationary testing.

Brian Whatcott Altus OK