Register to reply

Calculating Torque from motor specs

by Leinad
Tags: motor, specs, torque
Share this thread:
Leinad
#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
Physicists discuss quantum pigeonhole principle
First in-situ images of void collapse in explosives
The first supercomputer simulations of 'spin?orbit' forces between neutrons and protons in an atomic nucleus
Jobrag
#2
Nov6-09, 11:58 AM
P: 475
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
#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
#4
Nov6-09, 01:12 PM
P: 4,663
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
#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
#6
Nov6-09, 05:07 PM
P: 4,663
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