I am trying to build a worm gear based on a single axis solar tracker for a worm gear box. I need some guidance choosing a worm gear & drive motor for the application. These are the steps I have explored so far - if I am wrong could someone please give links to material which gives a good explanation so I can learn and understand it better.

Motor planning to choose; Worm gear selection

General equations for worm gear & motor:

1) $Efficiency = \frac{\text{output Power of worm gear}}{\text{input Power of worm gear}} = \frac{P_o}{P_\text{Input}}$

2) $\text{Output power} = P_o = 2*\pi*\text{Output speed} * \text{Output torque}/60$

  • Output power in kW
  • Output speed in rev/min
  • Output torque in Nm

3) $\text{Input power} = P_\text{Input} = 2*\pi*\text{Input speed} * \text{Input torque}/60$

4) $\text{Net Torque}= T_\text{output} + T_\text{Input} + T_\text{holding} = 0$

In order to drive a 250 kW plant I have assumed 15 kW per row, this gives me 16 rows with 1 column.

The worm gear has to drive 16 rows with coupling with single column item.

Each panel weighs 20 kg and there are 60 panel per row which gives 1200 kg. Each row has a structural weight around 700 kg.

This gives a net weight of 1900 kg per row.

So, the load on the worm gear would be 1900*16 = 30400 kg.

Based on my calculations above, how can I calculate the exact motor & drive motor required. I need some specific relation here.

  • 2
    $\begingroup$ Is this homework? | The assumption that you have to provide a force equal to the total mass is a bad one. Practical systems may be balanced around or near a centre of mass and/or counterweighted and/or could use springs or pneumatics if absolutely essential (which it should not be. | Using a single motor with mechanical coupling makes no sense in modern practice. A motor per panel or set of panels is far far more likely for reasons or reliability cost mass danger and more. i| Basic formula in all suh is power = force x distance per time and work or energy = force x distance. .... $\endgroup$ Mar 28, 2015 at 9:46
  • $\begingroup$ Bonus: Power - watts ~= torque in kg.m x RPM. $\endgroup$ Mar 28, 2015 at 9:46
  • $\begingroup$ @RussellMcMahon can you give some appropriate calculation part $\endgroup$
    – user50949
    Mar 30, 2015 at 3:48
  • $\begingroup$ You have to do your part. I asked i this is "homework" (or similar). You still get answers but they differ in approach. | You need to show you are thinking and understanding and show what calculations you have done. Your question re "giving some appropriate calculations part" does not indicate that you have taken ANY notice of what I said above. I made suggestions on balancing, on using multiple motors and on a formula for power from RPM and torque. YOU need to provide some more input now. | Is this an assignment or homework. If so, when is it due. $\endgroup$ Mar 30, 2015 at 5:11
  • $\begingroup$ @RussellMcMahon : I am just student & doing internship . Can you explain me with example.What are input you need from my side.Basic calculation i given here.Other involved formula if known need to mention. question is whether i am on right track of calculation or not. If not suggest with relevant link $\endgroup$
    – user50949
    Mar 31, 2015 at 4:41

2 Answers 2


As mentioned in the comments, the weight of the solar panels is not the issue as the weight can be balanced such that it contributes negligible torque to the system. Also mentioned in the comments is that the driving mechanism should be for each panel as this simplifies the mechanics and will increase the reliability and ease of maintenance of the system. So the real question is "What driving and holding torque do I need for an individual panel?" To answer this you need to know the wind load. How much torque would high winds for the area produce on panels of your size?

Unfortunately, this is not a particularly easy calculation. I suggest opening a new question to ask about this (preferably after looking up the wind-speed so you can give the relevant info, including panel size) as it differs significantly from the current question.

As for your concerns about power, most motor and worm gear applications will be used in much much higher speed systems, and thus will easily be able to handle the power requirements if they can handle the torque. As such, once you find the wind torque, you can size your worm gear(s) for that torque and then the motor for the input torque.


There are many companies that sell gearmotors, and their specifications will give output torque for input power. A few gearmotors link are below, and you can compare their ratings to your calculations:

http://www.baldor.com/brands/baldor-reliance/products/motors/gearmotors/ac-gearmotors http://www.mcmaster.com/#gearmotors/=10jct48 http://www.globalindustrial.com/p/motors/Motors-Gearmotors/dc-right-angle-motors/gearmotor-dc-right-angle-90rpm-cm34d17nz35a?infoParam.campaignId=T9F&gclid=Cj0KEQiAzai0BRCs2Yydo8yptuIBEiQAN3_lFq10KrIML5XDwgHC3HtQlVn68AaBDrbCJwKGaV-TxBYaArPz8P8HAQ


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.