Calculating Torque from motor specs

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To calculate torque from motor specifications, first determine the output power using the formula P(output) = Amps x Volts x efficiency (assumed at 80%). For the given motor specs, the output power at 1.5V and 0.075A is calculated to be 0.09 Watts. Torque can then be calculated using the equation T = (P*60)/(2*pi*RPM), resulting in a torque of approximately 0.0762 N*m at 3300 RPM. The discussion also highlights the importance of converting RPM to radians per second for accurate calculations, and it concludes with an inquiry about whether this torque is sufficient to lift a 5-pound weight.
Leinad
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Greetings folks. I am unsure how to calculate the torque produced by a motor if the known voltage input and amps are known. Here are the specs:

3300 RPM @ 1.5Vdc @ 0.075 Amps. 6900 RPM @ 3VDC @ 0.095 Amps. 0.93" Diameter x 1.5" Long body. Operating Range 1.5 to 4.5 VDC. Shaft: 0.07" Diameter x 0.28" Long.

What is(are) the equation(s) to calculate torque with the known variables given?
 
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From the current and voltage you can calculate the input power assume an efficiancy (80% ?) you know the speed so you can calculate the torque.
 
I do not know me what good power or speed do me. What equation would I use to calculate torque? I do not know the relationships. It has been quite some time since a physics course.
 
To get power output (watts), multiply current x volts x 80%
To get torque in Newton-meters, multiply power by 60, and divide by 2 pi RPM = 6.28 RPM.
Be sure you are using full-load volts, amps, and RPM

Bob S
 
So, going along with what has been provided...

P(output) = Amps x Volts x 0.8
P(output) = 0.075 Amps * 1.5 volts * 0.8 = 0.09 Watts
Where P = Power

T = (P*60)/(2*pi*RPM)
T = (0.09 Watts * 60) / (6.28*3300 RPM * (1 min / 60 sec)) = 5.4 / 69.08 = 0.0762 N*M

As for the information you gave me, when dividing by RPM, if a Watt is (1 Joule / second), doesn't Rotations Per Minute (RPM) need to be converted to Rotations Per Second? That would account for (1 min / 60 sec).

I am not too familiar working with torque values, so would you say this could turn a wheel to move 5lb?
 
Leinad said:
So, going along with what has been provided...

P(output) = Amps x Volts x 0.8
P(output) = 0.075 Amps * 1.5 volts * 0.8 = 0.09 Watts
Where P = Power

T = (P*60)/(2*pi*RPM)
T = (0.09 Watts * 60) / (6.28*3300 RPM * (1 min / 60 sec)) = 5.4 / 69.08 = 0.0762 N*M

As for the information you gave me, when dividing by RPM, if a Watt is (1 Joule / second), doesn't Rotations Per Minute (RPM) need to be converted to Rotations Per Second? That would account for (1 min / 60 sec).
Torque (Newton-meters) = Power (watts) divided by angular velocity (radians per second)

radians per second = 2 pi RPM/60

I am not too familiar working with torque values, so would you say this could turn a wheel to move 5lb?
to lift 5 pounds with torque = 0.0762 N, 5 pounds= ~22 Newtons, so arm length r = 0.0762 N-m/22 N = 0.00346 meters
 
For simple comparison, I think the same thought process can be followed as a block slides down a hill, - for block down hill, simple starting PE of mgh to final max KE 0.5mv^2 - comparing PE1 to max KE2 would result in finding the work friction did through the process. efficiency is just 100*KE2/PE1. If a mousetrap car travels along a flat surface, a starting PE of 0.5 k th^2 can be measured and maximum velocity of the car can also be measured. If energy efficiency is defined by...

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