# Calculations for a DC motor

I want to open and close a window using a DC motor and a lead screw. My window is 20 inches long.

I know buying recommendations are outside the scope of this forum. I have little to no knowledge about torque and RPM. Could someone tell me how to calculate the required torque and RPM to move a window 20 inches in less than 5-10 s? The window is pretty easy to open does that mean less torque is ok? Also, are brush-less motors better in this case?

Edit: I am using a lead screw(500mm long), which has 2mm spacing and a diameter of 8mm. Furthermore, my window requires about 2kg's of force to move.

This is the type of window i have: link.(opens horizontal)

• Without information about the leadscrew you can't do this since it acts as a gearbox. You need to measure how much force it takes to open and close the window. "Easy to open" doesn't mean very much because you are using all the muscles in your arm and back to open the window which are probably huge compared to the motor you have in mind. May 12 '20 at 15:45
• @DKNguyen What information would you like? My lead screw is an 8mm diameter,screw length 500mm and spacing 2mm. Please let me know if you want more information. Also, about the force taken to open the window, I will try to measure the force needed. But due to lockdown i may not be able to buy a forcemeter. May 12 '20 at 16:15
• @DKNguyen I used a hanging weight scale and hung it on the window handle. moving from closed to open it took about 2kg of force max to get the window starting and afterwards it moved in about 1.3kgs of force. While closing the window took 1 kg throughout. May 12 '20 at 16:31
• That's good. Could you show us a photo of the window and leadscrew just so we have an idea of the setup? There are many ways types of windows so it would be more accurate than trying to make up in our mind what it is you have to work with. May 12 '20 at 17:18
• @DKNguyen I have added pics to my answer. May 12 '20 at 18:13

Waiting for more detailed information but if it seems you measured:

Force to overcome static friction: $$\F_{static} = ma = 2kg * 9.81m/s^2\$$ = 19.62N

Force to overcome dynamic friction: $$\F_{dynamic} = ma = 1.3kg * 9.81m/s^2 = 12.753N\$$

20 inches = 0.508m

If you are applying just enough force to overcome dynamic friction the entire time (static friction is only at the start so let's ignore that for now), the work done is:

$$\W = F \times d = 12.753N \times 0.508m = 6.478J\$$

If you want that work to be done to occur over 10 seconds then the power required is:

$$\P = \frac{W}{t} = \frac{6.478J}{10s} = 647.8mW\$$ of continuous power

But there is also the high force required to overcome static friction at the beginning plus the friction losses in the leadscrew. I'm not sure how efficient leadscrews are but if they are anything like worm gears it's really bad. So I'm just going to guess 10% efficient.

So you need a motor that is capable of $$\\frac{647.8mW}{10%} = 6.478W\$$ of continuous power

And since the peak force required is 1.538x larger than the continuous torque, let's increase the continuous power by that much for peak power for a 10W motor.

$$\P = \tau \times \omega = [N-m] \times [rad/s] \$$

So to convert the power into required torque and RPM you need information on your leadscrew and any potential gearboxes present. For the leadscrew, you need the gear reduction ratio or to measure it and figure it out.

NEEDS VERIFICATION

If no gearbox other than the lead-screw is present, then:

$$\[Leadscrew Reduction Ratio] = \frac{[Circumferential Displacement(m)]}{[Linear Displacement(m)]}=\frac{[Angular Displacement(rad)] \times [Leadscrew Radius (m)]}{[Linear Displacement(m)]}\$$

$$\ \tau_{motorPeak} = 19.62N \times [Leadscrew Radius (m)] \times \frac{1}{[Leadscrew Reduction Ratio]}\$$

$$\ \tau_{motorContinuous} = 12.753N \times [Leadscrew Radius (m)] \times \frac{1}{[Leadscrew Reduction Ratio]}\$$

$$\RPM_{motor} = \frac{0.508m}{10s} \times [Leadscrew Reduction Ratio]\$$

NOTE: I am a bit uncertain about the equation for the lead screw reduction ratio. I came up with it on the spot. There's also probably a way to figure it out based on the thread angle and leadscrew radius but I don't feel like trying to figure that out.

I am also uncertain about how the $$\[Leadscrew Radius (m)]^2\$$ pops out in the torque equation which I was not expecting. It may not be correct.

• I have not bought the moter as of yet. Would a moter with a gearbox be better or worse? May 12 '20 at 18:16
• What other information can i give? Also in your RPM moter and Tmoter formulae, What is "LeadscrewReductionRatio" sorry if this is a noob question, but i have no idea what that means and how or where to find it.Thanks! May 12 '20 at 18:19
• @abdullahmohsin If the reduction ratio on your leadscrew is not enough for the speeds available at the motor power you need then you need to get a motor with a gearbox to augment that reduction ratio. May 12 '20 at 18:19
• Not really any more information I can think to ask from you. Do you know what a lever does? Gears are just round levers. Your leadscrew is like a gearbox. Takes something moving fast (or moving over a large distance) with low torque and makes it move slower (or over a smaller distance) with higher torque. A motor might put out enough power for your task but it spins too fast and its torque is too low. Too much of its power is the form of speed and not torque, so you use a gearbox to change some of that speed to torque. May 12 '20 at 18:20
• I now understand what gearboxes are! But how can one calculate LeadscrewReductionRatio. I dont need it to be extremely accurate, just enough for me to know what type of moter i am looking for. If calculating the reduction ratio is difficult is there a rough rpm/torque you think should do fine. If not, How do i calculate it, to get a torque/rpm number May 12 '20 at 18:30