Author Topic: How does dividing 1 Nm over a small wheel work?  (Read 792 times)

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Eithman

• Jr. Member
• Posts: 24
How does dividing 1 Nm over a small wheel work?
« on: May 03, 2012, 07:48:24 PM »
I found a servo that provides a little over 1 Nm of torque. Since a Nm = 0.74 Foot pounds, if I applied 1 Nm to a wheel that had a 10cm circumference, 100cm/10cm = 10,  And changed foot pounds to meter pounds .74 FP * .305 = .226 would each rotation of the 10cm wheel be pulling with 2.26  lbs of force? .226 * 10 = 2.26 lbs? Disregarding efficiency. Is this correct, if so could you explain where I went wrong?

jkerns

• Robot Overlord
• Posts: 214
Re: How does dividing 1 Nm over a small wheel work?
« Reply #1 on: May 03, 2012, 08:37:08 PM »
Torque = force * distance

Force = Torque / Distance.

The distance is the radius of the wheel, not the circumference. You use the circumference to convert revolutions to distance traveled, but think of torque as a force applied to a lever with a length that is the radius of the wheel.
I get paid to play with robots - can't beat that with a stick.

http://www.ltu.edu/engineering/mechanical/robotics_engineering.asp

Eithman

• Jr. Member
• Posts: 24
Re: How does dividing 1 Nm over a small wheel work?
« Reply #2 on: May 03, 2012, 10:10:36 PM »
Ok, so if the radius of the wheel was 10 cm and everything else was the same, would the answer be the same as before?

jkerns

• Robot Overlord
• Posts: 214
Re: How does dividing 1 Nm over a small wheel work?
« Reply #3 on: May 04, 2012, 09:35:54 AM »
10 cm radius = .33 feet.  (20 cm diameter)

.74 ft-lb / .33 ft = 2.2 pounds of tractive force.

« Last Edit: May 04, 2012, 09:38:00 AM by jkerns »
I get paid to play with robots - can't beat that with a stick.

http://www.ltu.edu/engineering/mechanical/robotics_engineering.asp

• Jr. Member
• Posts: 24