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can someone please help with this maths calculation for torque?

Thanks for looking.

I have a brushless motor that has:

a nominal voltage of 24V
rated torque of 0.25N.m (what does this mean?)
Torque constant of 0.0355N.m/A
Peak torque of 0.75N.m

Unfortunately for me these figures don't mean a lot to me! But I am hoping someone can help me to calculate the torque when the motor is at 26v and has a current of 3.5A

If you are able to help, please can you explain the method of your calculation so that I can use the method to calculate the torque output at various voltages and current. I have tried but I don't get a meaningful answer so I thought I turned to the forum for help and advice.

Thanks very much.
 
It says: "Torque constant of 0.0355N.m/A"

Which means for each Amp you get 0.0355 Nm of torque.

So, at 3.5A you get 3.5 x 0.0355 = 0.1245 Nm.

Which is about half the rated torque.

Bob
 
Torque here has the units Newton.metres.
At 1 metre radius the tangential force would be 0.25 Newtons when the motor is drawing its rated (normal long-term) current.
 
Most motor come with a torque chart and usually show continuous torque and peak torque.
Continuous torque is usually maximum from 0 rpm and stays fairly flay up to the rated rpm.
Peak torque (current) should only be experienced for a brief time, otherwise motor burn out can ensue.
Your motor would be capable of operating a continuous force equal to .25Nm, you would need to consult the motor spec sheet to find out the torque at 3.5 amps, the alternative is to find it empirically by fitting a pulley of a known diameter and use such as a cord and spring scale to measure the force or torque when the motor is conducting 3.5 amps, and calculate the torque via the radius of the pulley and the reading on the scale..
figure-1-torque-curves-of-a-synchronous-servo-motor.gif

M.
 
Hi All

Thanks for helping.

I wasn't sure if there would be any effect due to the voltage (26V) being higher than the rated nominal voltage of 24V. So this makes no difference does it?

Thanks
 
2v over the rated voltage is trivial and should not cause any problem, RPM is rated to voltage, so if it is exceeded by a greater amount it could cause overspeeding.
M.
 
Hi All

Thank you for helping with this. It was the 2V over the rated voltage which I thought would add a problem to the maths but thanks for confirming that this is not an issue.
 
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