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'[EE] HP of winch motor'
2005\07\28@211622 by Jinx

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I've seen 12V boat winches for sale locally at NZ$99 and would
like to know what HP the motors actually are, as I've another use
for them if they're up to it. Seems a very good price compared
with bare motors I've heard about for NZ$200+

The brand is Super Gear, PLU No. 0921, Code No. 21016, made
in China, Google brings up nowt. I estimate the weight is 9-10kg
which indicates a fairly big motor

On the box (which the store won't let me open, and the motor
and its label are likely to be enclosed anyway), is

1.8m (6ft) / minute
2700kg capacity, rolling with load
900kg capacity, pulling with load (presumably this means dragging ?)

How would I work backwards to get the HP of the motor? At a
guess the gear train would include a worm ? which I'm led to
believe isn't the most efficient power transfer so that would need
to be factored in. No idea what the reduction is though. Maybe
someone familiar with winches can suggest a common ratio ?

TIA

===============================================
If you aren't part of the solution, you're part of the precipitate

2005\07\28@214840 by Richard Prosser

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At a Very rough guess (and probably no more accurate than looking at
the size of the thing).
If it'll drag 900kg at a rate of 1.8m/min.

If the coeff of friction is 0.7 (very wide guess #1 )
then the actual force being applied is 900 * 0.7 * 9.8 = 6174 Newtons
If it is dragging at the rate of 1.8m/min then this is 0.03m/sec
So the Power is 185watts.
Seems reasonable. - at 13.6V this is 13.6Amps, not allowing for gear
train losses.

I'd guess that gear losses are likely to be in the 70% area (very wide
guess # 2)

So the actual motor power is likely to be around 265 Watts. And the
current at 13.6V ~ 20Amps.

But I could be out by a large margin.

Any idea what size connector or cable is used?

Richard P

On 29/07/05, Jinx <spam_OUTjoecolquittTakeThisOuTspamclear.net.nz> wrote:
{Quote hidden}

> -

2005\07\28@221753 by Russell McMahon

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> I've seen 12V boat winches for sale locally at NZ$99 and would
> like to know what HP the motors actually are,

> 1.8m (6ft) / minute
> 900kg capacity, pulling with load (presumably this means dragging ?)

> How would I work backwards to get the HP of the motor?

I'll assume the 900 kg is dead load capacity - that may not be what
they mean. If not it'll be lower and following figures will be reduced
accordingly.

Work = force x distance.
Power = work/time

Force in Newton = kg x g

Work = 1.8m x 900 kg x 9.8 (=g) =~~  16,000 Nm (or Watt-seconds)

Power = 16,000/60s = 267 Watt

About 1/3 HP.

Expected current drain at 12 V would be > 267/12 = 22 amps at above
load. Probably 25 - 30 amps, maybe more.



       R "E&OE" M



2005\07\28@222125 by Jinx

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> So the actual motor power is likely to be around 265 Watts. And the
> current at 13.6V ~ 20Amps.
>
> But I could be out by a large margin.
>
> Any idea what size connector or cable is used?

None at all sorry. I had 20A, 1/3rd HP in mind for some reason.
They're at Supercheap Auto BTW (who have some absolute
bargains from time to time so I'm not assuming it's rubbish because
of the price)

http://www.supercheapauto.co.nz/www/index.cfm


2005\07\29@081251 by Gerhard Fiedler

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Russell McMahon wrote:

>> would like to know what HP the motors actually are,
>
>> 1.8m (6ft) / minute
>> 900kg capacity, pulling with load (presumably this means dragging ?)
>
>> How would I work backwards to get the HP of the motor?

> I'll assume the 900 kg is dead load capacity
> Work = force x distance.
> Power = work/time
>
> Force in Newton = kg x g

Possibly better written as (to avoid mixing units and symbols)
Force = m × g  (or force in Newton = mass in kg times gravity acceleration in m/s^2)

> Work = 1.8m x 900 kg x 9.8 (=g) =~~  16,000 Nm (or Watt-seconds)
> Power = 16,000/60s = 267 Watt
> About 1/3 HP.

Since the gears have an efficiency <1 (I've found ratios between 30% and
70% for smaller motors), the motor power will be bigger by the inverse, so
something between 1/2 and 1 HP.

Gerhard "aren't SI units nice" Fiedler

2005\07\29@085511 by Russell McMahon

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>> Force in Newton = kg x g
>
> Possibly better written as (to avoid mixing units and symbols)
>
> Force = m × g
> (or force in Newton = mass in kg times gravity acceleration in
> m/s^2)

SI units are indeed nice. But sometimes it MAY help people's brains to deviate very slightly. But often enough doing so makes things worse :-).
In this case as the "force" being winched was expressed in kg I started with that as a force rather than calling it a mass.

We are of course very accustomed to the mongrel unit "kilograms force" (except for Amercians) as we insist on expressing "weights" in kilogram rather than Newton. This would be eg the equivalent of asking for a 1/32 slug of tomatoes.

If one was were to start talking about cgs units for 'weight' one could dyne out on the results for a considerable period.

groan


           RM

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