Russell McMahon wrote:

>> would like to know what HP the motors actually are,
>=20
>> 1.8m (6ft) / minute
>> 900kg capacity, pulling with load (presumably this means dragging ?)
>=20
>> How would I work backwards to get the HP of the motor?

> I'll assume the 900 kg is dead load capacity=20
> Work =3D force x distance.
> Power =3D work/time
>=20
> Force in Newton =3D kg x g

Possibly better written as (to avoid mixing units and symbols)=20

Force =3D m =D7 g =20
(or force in Newton =3D mass in kg times gravity acceleration in m/s^2)

> Work =3D 1.8m x 900 kg x 9.8 (=3Dg) =3D~~  16,000 Nm (or Watt-seconds)
> Power =3D 16,000/60s =3D 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, s=
o
something between 1/2 and 1 HP.

Gerhard "aren't SI units nice" Fiedler

--=20
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