Hello everyone,

I would like to know how to calculate the the smallest/ least powerful electric motor I would need that has enough torque to move an object that requires 8.9 - 22.2 Newtons of Force at at distance of 105mm?

Hello everyone,

I would like to know how to calculate the the smallest/ least powerful electric motor I would need that has enough torque to move an object that requires 8.9 - 22.2 Newtons of Force at at distance of 105mm?

Well, thats 2.331 Newton-meters, how far and how fast?

I want it to move a distance of 0.105m. I never really considered speed but lets I want it to cover that distance in 1 second?

2.3 Joules in 1 second = 2.3 watts.

You may be confusing work done (Joules) with torque.

Can you describe more clearly what you want to do?

Power = Force x Velocity

= Force x Displacement/ (Time in which you want to have the load displaced)

= 22.2 (N) x 0.105 (m)/ time

= 2.331 Watt (When you want to have this displaced in one second)

Depending on the efficiency of the motor you'll have to select the motor suitable to deliver the output required.

I'm just trying to close a window using a small motor. Because of all the uncertainties such as the wind outside and friction of the window mechanism, I measure the force required to pull in a window using a spring scale connected to the window handle. I read 8.9N (when i close it slowly) and 22.2N (if I close it fast and tight.) Because this isn't a lot of force I assumed that I would only need a small motor to do so. So basically I am just trying to find out the smallest electric motor I would need to close the window securely (but not so fast that it slams shut) but also understand the reasoning behind it/ equations. Hope this helps clear things up!

Power = speed x force = angular velocity x torque

2.3Nm is a large heavy motor, torque depends on motor volume, *unless* you employ gearing or

some sort to trade speed for torque (which you obviously should here).

If you are prepared to wait minutes and have enough gearing a very small motor would

be enough. To work in 1 second to move 0.1m at 0.1m radius is about 1 rad/s (10 rpm).

Say you have a small 5W motor at about 4000rpm, then 400:1 reduction gear box, that would

probably in the right area - lots of reduction stages in the gear box will waste a lot of your power,

so 30 to 40% efficiency is likely.

Plenty of small/medium gearmotors available on eBay these days which is helpful.

As an exercise for the reader can you work out the nominal torque of a 5W 4000rpm motor?

Thanks for the replies!

T = P / 2 π n

= (5 W) / 2 π (4000 rpm) / (60)

= 0.012 Nm

Is this correct?

Yup - see its pretty easy to do these mechanics calculations, and I strongly recommend sticking

to SI units whenever doing so as there are no random constants to remember (except 2 pi).

The other thing I didn't mention is that for intermittent use you can usually over-drive a motor

quite a bit, since the continuous rating is mainly about avoiding cooking the motor. So a 2W or

even 1W motor might be plenty here - only operating occassionally for a second or so allows

cooling-down time for the windings.

Good motor datasheets show the maximum peak drive current as well as continuous rating.

Please correct me if I'm wrong but would another method also be to calculate the torque required to pull the window shut with:

T = F * r * sin(theta)

where the r = 1m (lenght of the window from its pivot)

F = 8.9 - 22.2N

theta = assuming the angle it is always being pulled on is 45 degrees

= 6.3 - 15.7Nm of torque

Then purchase of the DC gearmotor based on its rated torque and load speed?

For example if the motor was rated at 20Nm at 46rpm (close enough to 1 second) like the one below.

https://www.precisionmicrodrives.com/dc-gearmotor/12mm-dc-gearmotor-22mm-type

Completely the wrong motor by a long long way.

It is rated at 0.02Nm at 46rpm, or 0.035Nm at maximum power point. Maximum power is

0.15W.

ah sorry I read it as Nm not mNm. Ok I've realised what I've done now and my torque value is way off.

T = F r = 22.9N x 0.105m = 2.4045Nm

P = wt = 22.9N x 0.0525m/s = 1.20225W (Now decided I want it to close the window in 2 seconds and 105mm was the distance it needed to move)

do these values make a lot more sense?

Yes

BTW that tiny motor is very interesting as it uses a harmonic gearset, giving 120:1 reduction in

one set of planet gears. You don't see cheap harmonic gears very often and they are great

for large reduction in one step (less friction). It would be a great choice for a tiny robot.

OK so last question.

Why for the distance am I using the distance the window needs to travel to shut which is 0.105m and not the radius of the lever arm? Lets say I attach one side of a wire onto the window handle and I pull using a wire reel that is fitted around the shaft of the motor. Surely I use radius from the axis of the motor to where the wire is connected to around the reel?

so instead of

T = F r = 22.9N x 0.105m = 2.4045Nm

it would be

T = F r = 22.9N x 0.0167m = 0.38243Nm

(If I want to close the window with 1 revolution then I know the circumference of the reel is 105mm. So I got the radius from that which is 16.71126902mm)

Yes, that all figures. Or use a leadscrew arrangement for greater gearing ratio.