Turntable torque requirement

[laengineer]

Hello,

I am designing material handling car with a rotate function. The car uses a 47" slewing ring bearing. The load being turned is approximately 70,000 lbs and needs to rotate 180 deg in 5 seconds. I am trying to determine the amount of torque needed to size a hydraulic motor that will be used to rotate the assembly. I've started with T=Ia but am not getting any numbers that make and sense. What am I missing here??? I've attached a jpeg of the setup. Thanks in advance,

[Kelly_Bramble]

For the dirty math (estimating), see http://www.engineersedge.com/motors/...e_equation.htm

Also, see: http://www.engineersedge.com/motors/motor_menu.shtml

Bottom of webpage..

[wilsonr]

Cool problem..

An object in motion will stay in motion unless acted on by some force.

There are two parts to this that you must consider. First torque has to be enough to

1) get the object from rest to some desired rotation velocity.
2) keep the object rotating at desired velocity by overcoming friction.

I think people often forget about that second part... Any way... You are on the right track but make sure your calculation the rotational inertia using the correct formula. In this case the axis of rotation is perpendicular to the axis of your wheel (you mention car, but I see a wheel in that pic my friend). The following formula for inertia should be used: (1/4)MR^2 + (1/12)ML^2, this is for a solid cylinder but if anything this calculation will give you a more conservative value (i.e. the torque to get it to accelerate will be more than you need).

Now the torque to overcome friction... well my friend that is a bit more complicated. I would try looking at the bearing specs to see how it relates the applied load to the bearings friction. Once you have that figured out your golden. Torque to over come friction will be equal to the friction in the ball bearings multiplied by the distance between that force and the center of the bearing.

This is all assuming the axis of the motor will be in the center of your bearing.

[Cragyon]

Check out the following:

Mass Moment of Inertia