Torque and Size?

SB316

Hey, I have been thinking of a new robot design. Although, like all of my designs, I am not ready to make it. I like to figure out what I would do beforehand. And I came up with this question. How does one convert robot size into motor torque? For instance, if I knew I wanted to build a 5 lb robot, how would I decide that I need X foot-pounds of torque?

Thanks,

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SB 3:16

Beau Schwabe

SB316,

Excellent question... See if this link helps... www.tutorvista.com/content/physics/physics-iii/motion-laws/friction-angle.php

Basically you want to figure the amount of work required for a given weight, incline and distance from point A to point B on the incline (note: the incline could be Zero or a flat surface)... based on the speed and torque availability of your motor selection you can factor in how quickly you want to apply the required amount of work when going from point A to point B.


Edit:
When thinking about the foot-pounds, this is the amount of force you have at a 1 foot radius from your motor axle ... Suppose you have a motor that has 1ft pound of torque, if your wheel diameter is only 3 inches, you have a 'force' where the wheel meets the road of 8 pounds. This would be the counter acting force needed in the URL above to move the robot up an incline.

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Beau Schwabe

IC Layout Engineer
Parallax, Inc.

Post Edited (Beau Schwabe (Parallax)) : 2/6/2010 5:43:29 AM GMT

SB316

Thanks so much,
So, you're saying that if I had an 8 pound robot with three inch wheels, I could move that robot with a 1 foot-pound motor?

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SB 3:16

Beau Schwabe

SB316,

Essentially, yes ... with that example, the 8 pound robot would be able to lift itself. On a level surface, the 8 pound force at the wheels would only need to overcome "static friction" as you increase the inclination it becomes a trade off with speed... the steeper the inclination, obviously the slower the speed, so you would want to calculate in some 'torque overhead' so that you can maintain an acceptable velocity at the inclination angle you decide to be within your specs.


Edit:
Now, I'm not 100% sure, but I think that if you take the inverse of 1 minus the SIN of the inclination angle for example 30 Deg ... 1/(1-sin(30)) = 2 ... this means that in order to maintain the same speed at 30 Deg that you have on a flat surface (0 Deg) that the robot on the inclination of 30 Deg would need twice the torque in order to keep up. In other words... two robots each weighing 8 pounds would be able to travel at the same velocity if the one on the 30 Deg incline had a 2 ft pound torque motor and the one on the flat surface had a 1 ft pound torque motor.


1/(1-sin(30)) = 2

1/(1-sin(45)) = 3.41

1/(1-sin(60)) = 7.46

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Beau Schwabe

IC Layout Engineer
Parallax, Inc.

Post Edited (Beau Schwabe (Parallax)) : 2/6/2010 5:35:23 PM GMT

erco

Too many variables to say. Rolling resistance (tires, wheels, carpet), motor & geartrain losses (# of gears, grease) & miscellaneous factors (alignment, poor build) eat up much more power than most people realize. Typically, you'll have HALF or less the output power (speed & torque) that you calculate. Harsh realities of the real world.

As Beau states, run the math and calculate the loads. From my experience, DOUBLE the expected power requirements just to operate marginally in the real world.

And do lots of bench testing before you begin. Nothing takes the place of real testing.

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·"If you build it, they will come."

SB316

Thanks so much, I will be sure to do what you both are saying.

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SB 3:16