Quote:
Originally Posted by James Critchley
Interesting question.
A flat mechanical efficiency (single number such as 90%) is a Coulomb friction model. It represents the load dependent binding of the gears and shafts due to tolerances in the bearings and the relative sliding of gear tooth surfaces under load.
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By definition there is no load at free speed. At no load, there is no friction.
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So, going back to the original post (excerpt appended below), what I hear you saying is this (the highlighted parts):
Quote:
For example, suppose I bolt a CIM to a gearbox that has ratio 10:1 and efficiency 90%. Then:
- the free speed at the gearbox shaft should be slightly slower than 1/10th of the CIM's free speed, because of the friction in the gearbox... but how much slower? answer: no load, no friction, so the free speed at the gearbox output of the combo would be modeled as exactly 1/10th motor free speed (ignoring windage for simplicity)
- the free current should be slightly more than the CIM's free current (because of the torque load of the gearbox friction), but how much more? answer: no load, no friction, so the free current of the combo would be the same as the motor-alone free current
- the stall current should be exactly the same as the CIM stall current answer: yes
- the stall torque would be less than 10 times the CIM's stall torque, but how much less? answer: 10% less
- assume a linear behavior between free and stall points? answer: if you ignore windage for simplicity, then yes
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So:
N
combofree = N
motorfree/G
T
stallcombo = T
stallmotor*G*eff
... and connect those points with a line to get the speed vs torque for the combo.
The "no friction at no external load" assumption is (as previously stated) an approximation, and I suspect that approximation becomes increasingly less tenable as the gear ratio increases. Does anyone have free speed data for a Banebots motor with and without a 256:1 gearbox?