That first reply I made was too late at night. After posting I checked bdesmond's reference – looks like its on the right
track. Weight AND speed of the
revolve determine the horsepower. That's what I was alluding to at the end of that first reply. I also think that the solution is beyond a few posts at this site. And I now apologize in
advance for what will be a lengthy reply…
I'd look into a theater reference text about how to make and
power a
revolve. The reference I had shows a motorized drum and cable. And I suppose you can have a direct drive too. But motorized rotating equipment and moving cables are dangerous things – you may just want to make the
revolve manually operated/pushed. Also, the noise of the motor and
idler pulleys and gear reduction need to be considered – I'm not altogether sure that you want the motor on the
stage. And then once you get this mass moving, you have to stop it.
Be that as it may, I tracked down a couple equations that may help:
First, the
power equation, torque times angular speed (rad/minute). Then, convert ft-lb/minute to horsepower.
Rotation speed is a given. (Can't help you as to what it should be – there are
safety issues and noise issues, but it would have to be as slow as
practical, but it may be a function of your drive.) But there's that torque variable. Torque is inertia momentum times angular acceleration. (That's from the old physics
book I dug up – Resnick and Halliday.) But if the speed is constant then acceleration is zero – and that obviously isn’t the answer.
So I searched for horizontal torque and came across an equation at flo-tork.com. They have an equation for the friction torque of a horizontal table. The equation is Weight x coefficient of friction x bearing radius. Now assuming that the trolley wheels are on the outer
edge, that would be bearing radius. (Okay, there may be additional interior trolley wheels, but we'll be conservative.) The weight would be the weight of the
revolve, the wheels, the set, the
props, any thing permanently on the
revolve. Then I'd add a
safety factor – maybe even 100% more or 750 lb minimum to account for unknowns and stuff like people taking a ride or major design changes. The coefficient of friction (unitless) appears to be a
bit of a "fudge factor". It may be high at the start, but very low once it gets moving. I found a conservative reference of 0.8 to start and maybe as low as 0.015 once it gets rolling. But I'd use the higher number.
Now, an example. 20 feet diameter
revolve (Radius = 10 feet). Weight of
Revolve. Assuming typical
platform construction applies, then by my own experience, a 4 x 4 section (16 square feet) weighs about 90 lbs. The
revolve, at 20 feet diameter, has an area of 314 square feet. So the weight is about 314 x 90 /16 = 1,800 lb (!!!). Assume 500 lb for set/wall, so the total is 2,300 lb.
Safety factor of 2, so 4,600 lb. Finally, assume a relatively slow rotation speed of 1 rpm.
Friction Torque = 4,600 lb x 0.8 x 10 feet = 36,800 ft-lb
Power is Torque x rotation speed (radians/time) = 36,800 ft-lb x 2 x pi/minute x 1 minute/60 seconds = 3,863 ft-lb/sec (Note - 1 rpm = 2 x pi rad/min)
550 ft-lb/sec = 1 hp
So 3,853 ft-lb/sec x 1 hp/550 ft-lb/sec = 7 hp.
There are probably other minor inefficiencies in the
system, and the 7.5 hp motor probably won't be enough, under these conditions and assumptions.
Now, based on the above, you really need to either go to the library and/or
book store to get some better design references; or (as Supercow suggested)
track down a professional or someone who has designed these. And again, consider
safety and noise.
Joe