# How to calculate motor torque required?

Discussion in 'The Projects Forum' started by raviprakash.hegde, Jan 8, 2015.

1. ### raviprakash.hegde Thread Starter New Member

Nov 14, 2011
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I need to move 10 kg of payload from ground to 15 meters vertical height in 25 seconds(approx) on a metal pole. How to calculate motor torque required for this requirement?

12V, 100W DC Motor speed is 2500rpm, gear ratio 50:1, hence at output shaft speed is 50 rpm.
Wheel diameter is 10cm.

I know that, Torque = Force x length of lever arm.
But, when wheel is directly connected to motor shaft, what to put in 'length of lever arm' ?

How to calculate motor torque required for this requirement?

2. ### GopherT AAC Fanatic!

Nov 23, 2012
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12 m/second squared as acceleration of gravity plus vertical acceleration of mass * 10 kg = 120 N.

120 N * 0.05 meter (Wheel radius) = 6N*m of torque. This is not exact, but just directional. It depends on how much control (excess power) you need to accelerate and decelerate the mass.

Now, you can reduce the 6Nm with your 50:1 gear box. To 0.12Nm

raviprakash.hegde likes this.

Jul 18, 2013
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For slow accelerating loads or a high degree of reduction present, you can usually use the break-away torque as a guide, this is done by using a torque wrench or the pulley/string/fish scale method.
Your output torque will increase by motor torque x rate of reduction.
Max.

4. ### Alec_t AAC Fanatic!

Sep 17, 2013
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If the wheel is running directly on the pole it will travel π x diam x 50 cm/minute = ~ 15 metres/min. That's only half the speed you need .

5. ### GopherT AAC Fanatic!

Nov 23, 2012
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Righty right! 6.54 Meters per 25 seconds at full speed! Good catch Alec!

6. ### raviprakash.hegde Thread Starter New Member

Nov 14, 2011
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Thanks GopherT.
I did not clearly understand the 1st line. '12 m/second squared as acceleration of gravity plus vertical acceleration of mass * 10 kg = 120 N.'.
And, how did you come to calculation of 12m/sec ?

7. ### GopherT AAC Fanatic!

Nov 23, 2012
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For torque, you need force (as Newtons) instead of mass (kilograms), (remember, Force = mass * Acceleration). To keep the gearbox shaft in position, you need at least to match the force of gravity on your 10kg load so 9.81m/sec squared. Then you have to accelerate that load to move it up 15 meters in 25 seconds. I selected 2 m/second just because I wanted something sufficiently above overcoming gravity. 9.81 + 2 = 12 m/second squared as your acceleration. You can select what you want. Too low, and you risk that tolerance ranges, frictional forces and other losses might cause you not to be able to move your mass at all. Too much is not a big deal unless you run into issues with too much jerk in start and stop.

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8. ### shortbus AAC Fanatic!

Sep 30, 2009
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raviprakash - this website may help you, a lot of engineering formulas and information on about everything engineering --- http://www.engineersedge.com/

9. ### raviprakash.hegde Thread Starter New Member

Nov 14, 2011
26
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Thanks GopherT. I got it.

Nov 14, 2011
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11. ### raviprakash.hegde Thread Starter New Member

Nov 14, 2011
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Can you please suggest which would be right motor? Motor1 or Motor2 ? or not both ?

12. ### GopherT AAC Fanatic!

Nov 23, 2012
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Sorry but neither motor will meet your needs. You will need a torque at the output shaft for the 10cm diameter wheel to be 60 kg cm or more.

I looked and I didn't immediately find a motor with this torque range as a robotics style motor. Most likely a motor like an automotive windshield wiper will have that range of torque. This will be a big motor!

I assume this is for a robotics competition. Is that correct? Could you post the full rules (or a link) and let us know if the 10cm wheel (pulley) is required to connect to the rope that you are climbing up (if that is what you are doing).

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13. ### raviprakash.hegde Thread Starter New Member

Nov 14, 2011
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No GopherT, this is not for any robotics competition. This is to find a tech solution to a serious agriculture problem.
I am trying to build a robot to climb up the tree like this to spray medicine. Trees are of max.50feet height.
Currently being done manually, a skilled labor climbing slippery tree in heavily raining season, a seriously risky business.
Due to acute shortage of skilled labor, day by day problem is increasing manifold.
I am trying to find an engineered solution for the same.
I chose design similar to rope climbing robot. Instead of rope, there will be tree. Thankfully, trees are also round safe with even size/surface.
Two motor powered wheels should grip the tree and climb up with a DC diaphragm pressure pump and a 2L bottle with liquid in it to spray.

14. ### GopherT AAC Fanatic!

Nov 23, 2012
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I was trying to understand if you had certain limitations because there is a robotics competition that is similar.

I would find a motor similar to a windshield wiper motor. Use a smaller wheel if needed, 10 cm could be reduced and increase climbing torque.

15. ### raviprakash.hegde Thread Starter New Member

Nov 14, 2011
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What would be the torque of the windshield wiper motor? Will it produce 60kgcm? And, would it be possible to achieve speed of 15m / 25sec ?

16. ### Bernard AAC Fanatic!

Aug 7, 2008
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60 kg-cm sounds about right. 3 of the # 1 motors execpt use 30 RPM is still only 75 kg-cm . Forget 25 sec limit.

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