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

I have no experience with Electric Propulsion, but have seen some
interesting technical info in the Vetus catalog. They also describe
their system at:

http://www.vetus.com/frame-cat.htm

then select the Electric Propulsion page.

MW

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

Seems to me I once calculated 32 lbs is about 1/4 hp ... if I find the
equations, I'll post ... but don't hold your breath :-)

"Paul Squire" wrote in message
...
Is there a simple equation between lbs thrust (as specified in electric
outboard motors) and Horsepower?

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

"Stephen Baker" wrote :
Don McPh says:

snip technicalities

Don, The OP asked for a "simple formula".......

;-)

Actually, I asked if a simple formula existed. I had a sneeking suspicion
that the answer would be no. I often recall the words of a colleague I was
working on a bank's data model with almost 20 years ago: "For every
complex, intricate & detailed problem there is a solution that is simple,
concise, to the point, and wrong."

I am pleased with the technicalities in the responses to my question as I am
always keen to learn. And I have now learned that thrust can only be
compared to HP if distance & time are considered.

I haven't quite reached full understanding of the responses and will think
on them more. In the meantime, I'd be grateful if someone who did have a
good understanding would be kind enough to do the calculations for the
particular case I have in mind.

The boat is a Monarch 17 foot trailer sailer. In cruising trim it displaces
2500lb on a waterline of 15'6"(L) x 5.4"(B) x 1'0"(D). On flat water 2HP is
adequate, 5HP would be normal, but if I ever expected to find myself trying
to motor off a lee shore into 30 knots I think I'd want 8hp. Lets use 5hp
for this exercise which equates to 3250 watts. Unfortunately makers of
electric trolling motors rarely specify wattage, preferring to specify lbs
of thrust. Trolling speed is 3-5 knots which is also the speed that the
Monarch would motor at so we're lucky there.

My gut feel is that we'd be looking at an electric trolling motor designed
for a 25-30' inboard launch. This is larger than the Monarch but the
trolling motor is not expected to push the vessel off a lee shore - the main
motor would be called in to play for that. I have no science to back up
this guess and do not know what thrust the manufacturers would recommend for
this sized launch but it feels about right.

So, If I've understood the technicalities, the variables necessary to
calculate the thrust required a

5 Hp at 4 knots. = ? lbs thrust.

Paul Squire.

"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).