Hi, I am new to this!

I am struggling to answer the following:

What size ( torque etc. ) 6 volt dc. Planetary gear motor would I need to lift
A 3kg. Load vertically.
I plan to couple the gear motor directly to a 12mm dia. Steel acme lead screw
With a 3mm pitch.
The load I will attach to a brass lead screw nut.
A stainless steel guide will stop the nut rotating and so give it vertical travel.

Any better guesses out there than mine ?

Thanks all.

It all depends on how quickly you want to be able to raise the load, ir stated another way - the speed of the motor and the gear arrangement. The power required equals the load (weight of the object you're lifting in newtons) times the velocity with which you want to raise it in meters/second. So if you're willing to use a gear set to convert the high speed RPM of the motor to a low speed RPM lifting gear, in theory you could get away with a very small motor. But using the mechasim that you describe with a simple 6mm radius gear you have:

Torque = 3Kg x 9.8 m/s^2 x 0.006m = 0.176 N-m
Power = Torque x radial velocity = Torque x RPM x (2 pi/60) = 0.018 x RPM watts

Form this if you know the RPM of the motor you can calculate power it must produce in watts.

Hi,
Thanks for your kind reply.
That has moved me a long way on.
I need to lift the 3 KG.load 150 MM. In 10 seconds

Power required = F x V = 3 Kg x 9.8m/s^2 x 0.015m/s = 0.44 watts. You should round up a bit for a safety factor to account for friction etc.

This does not take into the energy required to accelerate the load from 0m/s to 0.015 m/s, but assuming the acceleration is small (much less than 1 g) it's not significant.

That helps hugely.

I will now surf the net for a gear motor with these characteristics and small enough to fit into my contraption.

Many thanks again.