To be sure that it will work before you start putting parts together you need to do the math, and to do the math you need numbers. You also need to know what resources you have to create the interface between the motor and the chain.

The numbers you need are the weight you are lifting, the motor’s specs, the distance you need to lift it and assuming the use of a sprocket to engage the chain its effective diameter. Eventually you’ll need to know what is available to make the connection and gear reduction between the motor and sprocket.

The weight is straight forward, for the motor you need to know a couple of the specs, peak power (mechanical) output in Watts, the free speed rpm and stall torque.

Now that you know those numbers you can do the math. This page Power has the formulas needed to get started, specifically this section"

Example - Electric Winch lifting a Mass

An electric winch is lifting a mass of 100 kg 10m above the ground. The electric motor in the winch is 500 W. The force (weight) acting on the mass due to the acceleration of gravity can be calculated as

F = m g

= (100 kg) (9.81 m/s2)

= 981 N

= 0.98 kN

The work done by the winch can be calculated with (1b) as

W = F s

```
= (981 N) (10 m)
= 9810 Nm (J)
```

The time required for the winch with the actual motor to lift the mass can be calculated by modifying (1)

dt = W / P

```
= (9810 J) / (500 W)
= 19.6 s
```

However that is done in the magical world of physics where friction can be turned off at will. I’m going to assume that you will not be doing this in a lab where friction can be turned off and on at the teacher’s will.

In the real world there are several points of friction between the motor’s output and the point where the actual work is being done. We must account for this, the problem being that we just have to guess at the friction of an unknown system when we are at this stage. So run the above calculations assuming that 20% of that power is not available to do the work due to the friction of the gears, and bearings. Substituting 400 W into the equation and we find that it would take 24.5 sec.

Now the key is to force the motor to operate at its peak power.

The nature of a basic DC motor is such that it produces peak torque at 0 rpm and it falls in a straight line to 0 torque at the peak rpm also known as free speed. The rpm that the motor will operate at is dependent on the torque that is applied to it. A great example is a cordless screwdriver putting a screw into wood. As the screw goes deeper and deeper the drill turns slower and slower as the load due to friction increases. Peak power occurs at 1/2 the free speed rpm and it just so happens that the torque at that rpm is 1/2 the stall torque.

So how do we force the motor to operate at peak power? By ensuring it sees a load on its output shaft equal to 1/2 the stall torque. Since torque can be multiplied we just have to divide the torque on the shaft that is doing the lifting by 1/2 stall torque. The resulting number will be the gear ratio needed to force the motor to operate where we want it to. Of course that is assuming we are in a lab with the anti-friction device switched on. In the real world we will have to account for the unknown friction in the system.

Another consideration it trying to maximize the speed at which the work is done is what happens to the chain as you lift it. If for example if it is simply looped over the sprocket and is allowed to fall back down uninhibited the actual weight being lifted will drop as the length of the chain gets shorter on the side you are lifting and longer on the side that is falling and once the two sides are equal in length you won’t be “lifting” anymore even though you are still moving the chain.