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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: enter image description here

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.

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  • $\begingroup$ 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? $\endgroup$
    – kamran
    Mar 16, 2020 at 23:28
  • $\begingroup$ 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. $\endgroup$
    – hjf
    Mar 17, 2020 at 0:59
  • $\begingroup$ You can add a counter balance to reduce the necessary torque. You also need to consider wind if you are outside. $\endgroup$
    – Eric S
    Dec 12, 2020 at 3:24

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we just do it for gravity and assume 90% efficiency for the gear, or if they have a datasheet look that up.

Regardless of the dish being perforated in certain angles it could act as a wing and create flutter, lift, destructive vibration. So we need to use a light truss to stiffen the dish.

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$

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