Trying to calculate motor kW/reduction drive for lift of 100kg at 0.8m /sec

Hi
need to calculate 240v/315v motor size attached to reduction gearbox (reduction unknown) to lift a weight of 100kg on cable drum .8mtr/sec.
Drum dia not determined yet.
motor size not determined yet.
any help is greatly appreciated as this seems to be out of depth of many motor drum winch suppliers and myself.
thanks

Welcome to PF.

What is the application? How far do you have to lift it?

Single or 3 phase?

Step 1: calculate the power you need, then double it.

Step 2: pick a common motor rpm and "reasonable" drum size, then calculate the reduction. Adjust as seems appropriate.

Ignoring acceleration.
The change in PE per second = m·g·h = 100 * 9.8 * 0.8 = 784 watt ≈ 1 HP.

Finally, the gear reduction ##GR##, drum diameter ##d## (in meters), and motor ##RPM## are linked in the following way:
$$v = \frac{\pi}{60} \times d \times \frac{RPM}{GR}$$
Or
$$d = \frac{19.1 \times GR \times v}{RPM}$$
Where ##v## is the cable speed in meters per second. For example, given a motor of 1750 RPM and a gear reduction of 10:1, the drum diameter would be 0.087 meters or 8.7 cm to get a cable velocity of 0.8 m/s.

I was hoping OP would calculate them...