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'relating torque to rpm'
1999\04\28@213439 by PDRUNEN

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Hi smart people,

If I am using a fly wheel to count pulses on my motor, I know how many pulses
I get
per revolution so I know the speed of the motor - right?

I have an open argrument with an associate who says that he is really
measuring torque
and not speed,  So is there a relationship here?  I could understand if the
measurement was motor current, but torque in ft*lbs or N*m  related back to
rpm.

I guess you could say the faster the motor runs the less or more torque you
product as in being in a low gear at 2300 rpm going 50mph or higher gear,
1300 rpm going 50mph.


Thanks

1999\04\29@113536 by John Payson

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|If I am using a fly wheel to count pulses on my motor, I know how many pulses
|I get
|per revolution so I know the speed of the motor - right?

If you're using some sort of rotary encoder you'd be measuring rotational
speed.

|I have an open argrument with an associate who says that he is really
|measuring torque
|and not speed,  So is there a relationship here?  I could understand if the
|measurement was motor current, but torque in ft*lbs or N*m  related back to
|rpm.

For an "ideal" DC motor, torque is proportional to current; speed is
proportional to (voltage - (current * winding_resistance)).  In practice
both formulae work quite well; the main caveat is that in many motors the
commutator switching action isn't perfect which means the motor resistance
may not be constant.

|I guess you could say the faster the motor runs the less or more torque you
|product as in being in a low gear at 2300 rpm going 50mph or higher gear,
|1300 rpm going 50mph.

For a given supply voltage, the motor will run more slowly the more torque
it is required to provide.  A frictionless motor, with no load, would run
at an speed controlled by the input voltage and would need no current.  As
mechanical load is added to the motor (requiring it to supply torque) the
current increases, thus the (voltage-(current*winding_resistance)) goes
down.

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