Torque required to backdrive planetary gearbox
Torque required to backdrive planetary gearbox
(OP)
How do you calculate the amount of torque required to backdrive a gearbox and motor? I need to find this for a three-stage planetary with a servo motor attached. The system will be powered off and we have to manually move the instrument back to home. Thanks!
Norm
Norm





RE: Torque required to backdrive planetary gearbox
TygerDawg
RE: Torque required to backdrive planetary gearbox
As pointed out above, the backdrive torque is going to depend on alot of variables, primarily the combined reduction of all of the stages and the reflected residual torque or friction all the way back to and including the servomotor. Based on my experience trying to backdrive a two stage planetary, 64:1 total, you are in trouble. You may end up twisting the output shaft off before backdriving occurs.
--CCW
RE: Torque required to backdrive planetary gearbox
the reciprocal of the individual
gear efficencies times the gear
ratio times 1/20th the output
of the motor might be a good
approximation. Too bad you cannot
reverse the current in the motor.
You might want to put on a projected
moment arm and test the pull at
that radius to find out.
RE: Torque required to backdrive planetary gearbox
Irefl=Imotor*(n)^2 or the inertia of the motor x fwd gear ratio squared. For example if it takes 10 seconds and the acceleration is constant and the speed at the turning point is ramps up to 10 RPM, then the acc=10/10*2*pi/60=0.1 radian/sec^2.
Further, if the gear ratio is 100:1 and the motor inertia is .03 in-oz sec^2, then the torque due to motion for 10 seconds would be:
.03*100^2*0.1=30 in-oz , which is negligible.
The main torque would be to accomodate friction and churning. You could get a handle on this by running the system at the design speed and from its inefficiency determine the loss of power after subtracting IR losses.
As a simple example, suppose we have a DC motor at 24V putting out 1 amp of power under no load and having an armature resistance of 1 ohm. Then the loss of power at 1000RPM (consistent with the above example) attributed to the friction/churning is 24W- 1W= 23 W =23/746*550=17 ft-lb/sec. Since the input speed in our example is 1 RPM, equating the mechanical power to 17 ft lb/sec I get
17=T*1*2*pi/60 and
T=170 ft lb, which is quite a bit but not necessarily realistic.
RE: Torque required to backdrive planetary gearbox
Consider manually controlled 'jog' inputs to the servomotor, or a round 'safety' handwheel on an intermediate shaft.
Mike Halloran
NOT speaking for
DeAngelo Marine Exhaust Inc.
Ft. Lauderdale, FL, USA
RE: Torque required to backdrive planetary gearbox
Could you add a manual detent/pin clutch in the drive train?
Barry1961