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I am building a machine for a project that should slowly tip a screen forward about 95 degrees, hold for a couple of seconds, and then slowly return to vertical. The design I am working on uses two stepper motors connected to the screen with a belt drive. I can change the gear ratio of this drive to anything up to 2.8.

I need to select the smallest motor that will provide enough power for this application. I have the torque - speed charts for the motors, as well as the weight and moment of inertia of the screen. Is there a formula that I can use to calculate if the motor will be strong enough?

The screen will begin vertical, rotate forward about 95 degrees, and then return to vertical. The weight of the screen assembly is ~5.7 lbs. It will rotate about an axis ~2.75 inches below the center of mass. The screen assembly consists of a display surrounded on all sides by 1" t-slotted aluminum. The axis of rotation runs through the center of the bottom aluminum rail. There are two small stepper motors mounted just below this rail as well.The aluminum frame is 11.75" wide by 10.0625" tall. The motors and mounting plates extend an additional 1.66" down. Specs for the family of motors I plan to use can be found at automationdirect.com.

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  • $\begingroup$ What is a "screen" in this context? Are you talking about a display screen, such as an LCD monitor, or a metal mesh that's being use to separate items by size, or something else? Are there any forces acting on this screen other than gravity? $\endgroup$ – Dave Tweed Jul 1 '15 at 17:05
  • $\begingroup$ The third paragraph has been updated to include details about the screen assembly. $\endgroup$ – Kevin Jul 1 '15 at 17:26
  • $\begingroup$ a picture could clarify things even further. Do you know the reduced innertia of your belt drive? Also do you know the speed at which this tipping is to happen. $\endgroup$ – joojaa Jul 2 '15 at 7:37
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I'm going to assume from your use of the word "slowly" in the question that speed is not going to be an issue for you; therefore this is simply a question of torque. It sounds like the screen will be passing through the point which will generate the greatest amount of leverage, so lets calculate the torque at that point: 2.6 kg (~5.7 lbs) on the end of a 0.07 m (~2.75 inches) lever requires 1.79 Nm of torque, and through your maximum gear ratio of 2.8 this reduces to 0.64 Nm at the motor.

As we've not allowed for any friction or inefficiency in the system then we should ensure that the motor is capable of comfortably exceeding this figure. The third motor up in the range you specified (STP-MTR-17060(D)) may be sufficient at 0.81 Nm, but you may wish to go a little higher in the range to give some greater margin for friction. If you are using two motors then somewhere near the smallest from that range may be sufficient.

Once you have a candidate motor then you can use the torque speed curves to return to our speed assumption and calculate how quickly your screen will actually move (remembering to take into account the gear ratio) at the motors top speed. If this turns out to be unacceptably slow then you should immediately look at selecting a larger motor and perhaps adjusting the gear ratio.

If the top speed is close to what you would consider an acceptable average for the whole movement then you also need to factor in the effects of inertia. Extra torque above the holding torque is required to accelerate the inertias involved (your screen, gearing and motor) up to the top speed, so again you may wish to look at stepping up to a larger motor in the range.

Finally you will want to consider the voltage and current requirements of the motors, as achieving the torque we've calculated is reliant upon the necessary current being available.

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