How to properly determine "size" motors, especially surplus motors?

[csilvest]

This might be a bit off topic here; if so, please suggest where I might ask.

I'm looking for a tutorial on how "size" a motor for a given task.  For example, it I want to pull a rope with 20kg/40lbs a distance of 3m/10' in 1 second.  I want to use a 12VDC motor so I can use an automotive battery for power, turning a rope pulley between 50mm/2" - 200mm/8" diameter.  I really want to do direct drive, so no belts or chains, but gearmotors would be OK, even offset or right-angle.  I know the RPM required for a given rope pulley.  But I don't know how much force (or power?) I need from the motor to turn this rope pulley.  I think this is a failry simply calculation at the first order approximation; that would be close enough for me.

Also, the reverse of this.  In other words, I have a motor that works, but I don't have a part# or data sheet.  It's easy to get the no-load RPM, but I want to know maximum output torque and current at max. torque.  I think the process could be as simple as somehow locking the shaft to a force-measuring device, and an ammeter to measure the current at max torque.  But is this valid?  Will it be close enough to be in the ballpark t start experimenting?

Thanks for listening!

[beanflying]

You need to look at your idea in a lot more detail. You don't mention if you are wanting to lift your load against gravity or slide it (and on what)? Your load and the motor will need time to accelerate and decelerate, assuming you require the load to soft stop without as hard point? Before knowing or having an idea on these factors you cannot get a peak speed and torque required if you want to meet your time constraint.

Is this a project for home? If so the quick and dirty option is a drum on the end of a small car starter motor and matching solenoid from a wreckers and possibly some limit switches to limit the motion. For short bursts locked rotor current is a few hundred amps @ 12V so plenty of torque to accelerate your load. Running them at a higher duty cycles will require adding bearings and potentially some cooling. I would still recommend an outboard bearing on your drum with one of these based on personal experience.

If it is for a commercial option then add some more information so a better idea on a motor size can be given.

[soldar]

This is more complex than you might think at first sight. If you are selecting a motor for a given task you need to know the power but also a few other characteristics like type of motor, number of phases, speed or speeds, torque curves for different loads, etc. Do you need a motor that can start under load? Or will it only start with no load? Then you search through catalogues of specs and choose the one that best suits your needs ... or the one you can afford.

People with some experience will just eyeball things. Once you have used several lathes with different power motors you get a pretty good idea of what a motor can do.

The other way around. You have a motor and want to know the specs. You will need to have some gear and measuring instruments. If you are just playing at home this is out of your league.

[Doctorandus_P]

It starts with some simple ballpark calculations.

First start with decent units. SI force is measured in Newton, distance in meters, etc.
If you need a force of 1N to move something with a (constant) speed of 1m/s, then you need 1W of power to do that.
So 20kg (approx 200N, no need to get accurate here) and 3m/s needs a raw power of 600W.

Your motor needs a gearbox, drum and / or some pulley system and there are always losses.
There are also caveats. With some motors the power rating is on the output shaft, while on other motors it is the electrical input power.

Considering this, then for your application a motor of around 1kW seems about right. This also leaves some room for not being able to use the maximum power of the motor, for example for electronic speed controll. Genreally (regardless of motor type !!!). The torque a motor can deliver is independent of the speed the motor runs. and becasue a rotation of 1 radian/s with an arm of 1m is also 1Watt of power, the availe motor power halves, if the motor runs at half speed.

Sometimes people play dirty. Powerfull motors are expensive, and most electrical motors can be overloaded for short periods of time.
If you only need short bursts of power (short enough that the motor does not get too hot, and has time to cool inbetween) your goal might be achievable with a 300W or maybe even a 200W motor, but that is really stretching the limits.


About gear ratio and speed.
Assume you have a 12V DC motor with a reasonable speed of 6000rpm under load (about 8000 rpm idle).
6000rpm is 100 revolutions per second.
If you put a drum on it with a radius of 1m then you wil have lifted your load 2 * pi * 100 = 630 meters.
Or, if you use such a big drum, you will need a gearbox of 630 / 3 = 110:1.
For a drum with a more sensible radius of 50mm (10 cm diameter). your drum must spin 20x faster to winch up the same amount of rope in the same time, and the gear radio would be around 6:1.

With a bit of logical thinking (winding rope around a drum) it's easy to calculate those speeds etc, and you will also get more insight in the math behind it and how the formulas work.

[rstofer]

If you expect the motor to be able to suspend the load (assuming a lift operation), you will need to use a worm gear drive.  Even then you may want to add a brake.

https://www.motioncontroltips.com/5-design-considerations-gearmotor-application/

Gear boxes are far from 100% efficient.  If you put 5 HP in, you don't get 5 HP out.  Very likely you lose 1 HP in the gearbox (assuming 80% efficiency).

https://www.machinedesign.com/archive/second-look-gearbox-efficiencies

More to the point:

http://www.epi-eng.com/piston_engine_technology/power_and_torque.htm