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• # Torque Calculations

Discussion in 'Calculations' started by salman22, Jan 8, 2020.

1. ### salman22New Member

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Simply I need to calculate the torque required to move this solar tracker in the attachment assuming that i know every part's weight. What is the criteria should i follow?

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3. ### ErichWell-Known Member

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Estimates of friction at all joints,
Rate of acceleration desired as it starts up.
What loads does wind have on the structure?

Hint start with a free body diagram...……..

4. ### jasonmeurNew Member

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Thank its Realy helpful...

5. ### shachmatNew Member

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When i selecting a motor I'm checking all the moments in the system @ each angle and literally build the movement profile.
After that its easy to understand what is the required "continues moment and what is the peak moment.
Mt=F*L for movement.
Mt= I[Kg*m^2] × alfa[rad/sec^2].for acceleration.

Here is a good guide for motor selecting:
https://www.machinedesign.com/motors-drives/article/21831643/how-to-pick-motors-for-linear-motion.

Try to think about direct drive instead of spour gear
Good luck

6. ### OilytrunkWell-Known Member

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7. ### JustcurioustwoWell-Known Member

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Attached is a drawing of a machine that produces 118,428 foot pounds of force traveling at a speed of three (3) feet per second.

How can I convert this to torque and then to potential electric output?

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8. ### JustcurioustwoWell-Known Member

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9. ### ErichWell-Known Member

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Units of force are pounds or Newtons.
Foot-lbs are units of Torque or energy, not force.
Clarify what you have.

10. ### s.weinbergWell-Known MemberEngineeringClicks Expert

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What he has is a thoroughly debunked basic, run-of-the-mill, perpetual motion machine.

11. ### JustcurioustwoWell-Known Member

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Thanks for your clarification.

Do you agree with the following.

To determine the SeaEngine's potential energy output I have to determine the value of two (2) things-

(1) energy it takes to compress 40,000 cubic feet of air down to 2,666.66 cubic feet in watts.

(2) energy output of 118,428 foot pound in watts.

If (watts-2)'s output is greater than (watts-1), then the SeaEngiine is producing useful work.
I believe my drawing proves that the SeaEngine is producing useful work.
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