# Worm gear & motor selection calculation

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.

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.

• 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. .... Mar 28, 2015 at 9:46
• Bonus: Power - watts ~= torque in kg.m x RPM. Mar 28, 2015 at 9:46
• @RussellMcMahon can you give some appropriate calculation part Mar 30, 2015 at 3:48
• 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. Mar 30, 2015 at 5:11
• @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 Mar 31, 2015 at 4:41