- #1
CraigJ
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Question: What parameters are needed to have no slippage between 2 small drive wheels and the bowl?
Startup of the system is obviously the moment where there is the greatest chance of slippage.
As in the illustration above this is a possible layout of a device for rotating a large bowl weighing 746 Lbs. there is an idler assembly and drive assembly, each with the same small wheels. This entire assembly will tilt up to around 50 degrees from horizontal. Assuming the 4 wheels take the force of the bowl resting on them at that angle they should experience about 476.5 Lbs of normal force cumulatively. Divided by 4 that leaves each wheel experiencing 476/4=119 Lbs each.
-Currently I've chosen a 1hp motor; rpm=1800, full load torque=2.99LbFt, locked rotor torque=13.17LbFt, DOL (starts at full speed)
-Gearbox is a 59.79:1 reducer.
-Wheels are 8" outer diameter, 2" wide, polyurethane outer, weighing around 10Lbs each.
-Large bowl is SS, 746 Lbs, 35" OD, rotates on a bottom shaft with bearings
*The possible variables in this scenario are wheel widths, wheel hardness variability (for increasing frictional forces), the rate at which the motor starts up (add a soft start feature), With these numbers the 8" diameter wheels should experience 30.1 RPM with a Locked rotor (startup) torque of 787 FtLb from the motor starting up. The bowl final RPM (if there is no slippage) is 6.88 RPM.
Using hoop moment of inertia (I=mr^2), the bowl I=1586.55 LbFt^2
The force of the bowl resisting motion I would think has to be smaller than the force being exerted by the 2 drive wheels, in the instant after the motor starts up. Does anyone have any ideas on where i should go next as far as figuring this thing out?
Your input is appreciated. This is a really neat problem and its driving me crazy right now haha.
-Craig