Motor Sizing for Turret Unwinder

[brenmech]

I am trying to size the motor for a turret unwinder that I am building.

My design is basically an arm that pivots on A-frame supported pillow-block bearings. The arm supports a roll at both ends. My worst-case-scenario is when one roll is unloaded and I have an unbalanced load of a 3,000 lb roll at one end and nothing to counter-balance it at the other. I have to lift the roll from horizontal, up over the top, and back down to horizontal. The roll centre is located 53" from the pivot.

I know that this should be simple, but I can't for the life of me get the numbers correct. I've tried it a number of ways, but I just can't get a number that feels right.

I even checked with two well-known drive manufacturers, and one came back with 1/2 HP, the other 25 HP. No doubt one of them is wrong (perhaps both of them), but I just can't seem to figure it out.

Thanks in advance,
brenmech

 

[MintJulep]

Not enough information provided.

Power is the rate of doing work, hence time is needed. How quickly do you need to make the move?

3000 lb x 53 in / 12 in/ft = 13250 lb-ft of torque needed.

1 hp = 550 lb-ft/sec.

So, that 25 hp motor could make the move in 1 second. The 1/2 hp drive would need 50 seconds.

 

[brenmech]

Thank you MintJulep,

I forgot to mention how quickly I wanted to rotate my roll, and the fact that I need to accelerate the roll. I would like the roll to accelerate from a dead stop, travel 180 degrees and decelerate to a dead stop again, all in 2-4 seconds (I'm not too picky about the speed).

The formula I was looking at was T=I(ALPHA)

The 'I' component was giving me the trouble.

Thanks again,
brenmech

 

[tygerdawg]

total torque = static torque + dynamic torque

mass moment of inertia of a cylinder reference to its central axis is J = (1/2)*M*r^2 , M=mass of cyl., r=radius of cyl. Absolutely necessary to get your units right on this calculation.

You seem to be already aware of the dynamic torque (for accelerating the load). You must take into account the static & dynamic load of the arms, drives, drive components, etc.

It may be required to add to your torque equation gravity effects.



TygerDawg

 

[Barry1961]

Just as a rough number to start with. If you have a constant acceleration and a triangular move. This would be half the move.

Weight 3000lb
Mass 93 ft-lb-sec^2
JM 1802 ft-lb-sec^2
Move 1.57 rad(90 degree)
Time 2 sec
Vel .785 rad-sec
Accel .79 rad-sec^2

Torque = (JM*Vel)/Time=707 ft-lb

So if you add this to gravity, 13,250 ft-lb, you get 13,957 ft-lb @ 7.5 rpm which is about 19 horsepower. This is before friction, efficiency and safety.

Barry1961