# How to calculate the required torque for an azimuth/elevation antenna rotator?

I want to build an antenna rotator for a 1M dish. How do I calculate the required torque for the motor of the elevation axis?

Here's a sketch of the problem:

If the "bottom" of the dish is the pivot point, what would be the required torque to rotate the dish around this pivot point? Obviously the motor will have a series of reductions to reach the required torque and speed. The last stage a worm gear with a turn ratio enough to prevent rotation from the antenna weight when no power is being applied.

• Do you mean you want to swivel to right and left or up and down and how much is the maximum anticipated wind speed? the dashed line is what? a rooftop? Commented Mar 16, 2020 at 23:28
• The dashed line is the ground, or a rooftop. Either way it will be a concrete slab. I want it to swivel up and down (also right and left but that's a different problem). Winds here can be as high as 250km/h on a good storm, but usually not more than 40-60 on windy days. The dish is a grid, so the wind load is greatly reduced. The final reduction is a 9:1 worm gear used for outdoor roller curtains and retractable awnings so it should be pretty sturdy. If the motor can't spin during high winds it's not an issue as I don't expect to operate this during storms.
– hjf
Commented Mar 17, 2020 at 0:59
• You can add a counter balance to reduce the necessary torque. You also need to consider wind if you are outside. Commented Dec 12, 2020 at 3:24

Let's call the dish mass m and its depth H, and assume by just eyeballing the parabola's CG axis is at H*0.45. and say we want the torque $$\tau \$$ to deflect the dish from horizontal to 90 degrees up vertical.
$$\tau_{max }= m *0.45H/9*(1/0.9)=0.055mH$$