Calculating Torque from motor specs


by Leinad
Tags: motor, specs, torque
Leinad
Leinad is offline
#1
Nov6-09, 11:19 AM
P: 12
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?
Phys.Org News Partner Physics news on Phys.org
Information storage for the next generation of plastic computers
Scientists capture ultrafast snapshots of light-driven superconductivity
Progress in the fight against quantum dissipation
Jobrag
Jobrag is offline
#2
Nov6-09, 11:58 AM
P: 459
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.
Leinad
Leinad is offline
#3
Nov6-09, 01:00 PM
P: 12
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.

Bob S
Bob S is offline
#4
Nov6-09, 01:12 PM
P: 4,664

Calculating Torque from motor specs


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
Leinad
Leinad is offline
#5
Nov6-09, 02:35 PM
P: 12
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?
Bob S
Bob S is offline
#6
Nov6-09, 05:07 PM
P: 4,664
Quote Quote by Leinad View Post
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


Register to reply

Related Discussions
Calculating Motor Engine torque/power using accelerometer data Classical Physics 7
Motor Torque Requirements (Hub Motor) Mechanical Engineering 0
PWM/DC motor torque Mechanical Engineering 2
PWM and DC motor torque Electrical Engineering 0
thread torque specs Mechanical Engineering 3