Calculating required torque, if load is on a lazy suzan 2

[JedOs]

I am having trouble with calculating the required torque to rotate a disk (composed of 2 steel disks seperated by steel ball bearings) 90 degrees. There is also a 35 pound object on the disk. Also, the disk is to only move at 6 RPM. Everytime I calculate a required torque I get astonishingly high torque and watt requirements, for instance I just used some formulas for micromo.com and the formula told me I needed a motor that could output 10 Watts! and 144 lb-ins!

 

[ScottyUK]

You need a gearbox. To a first approximation, they behave like an electrical transformer: constant power throughput, input torque x input rpm = output torque x output rpm. Of course, that is a really simplistic view, but you get the idea. Friction and other losses screw up the analogy in the real world.


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I don't suffer from insanity. I enjoy it...

 

[Clyde38]

If you are concerned about the calculation of your required torque, why don't you just measure it? Something as simple as a fish scale attached to the outside radius of your device will give you the force required to rotate the device. You know the radius and you measure the force, so you will have torque (force X radius). Just a sanity check. For instance, to get 144 lb-ins of torque, if your device is 24 inches in diameter then it should take 12 pounds of force to rotate it. Likewise if it is 48 inches in diameter it should take 6 pounds of force to move it etc.

 

[aolalde]

You have two steps:
1- Torque (T) to accelerate (a) the load inertia (j) and reach certain speed (wf) in certain time (dt).
T = j*a were a = (wf-wi)/dt
2- Power requirement (HP) to keep the speed(rpm) under certain load torque(Tl), which could be due only to friction losses.
P = Tl*rpm/5252

 

[Barry1961]

10 watts does not seem out of bounds for a lazy suzan. The type of bearing and accel/decel will be the two factors. I think the acceleration would be in radians per second squared.

I see Brother has a 1/50hp, 300:1 that puts out 148 in-lb so the gear motor is available.

Not sure what you are concerned about. Maybe if you think of spinning the lazy suzan by a small peg in the center instead of on the circumference it might seem more reasonable. 35lbs will take a bit of force to stop and start.

Barry1961

 

[JedOs]

Hey guys thanks for helping me with this, I really appreciate it.
Aolaide I have seen the formula you posted before, but it keeps on eluding me: I am confused about what units to use in the formula.

The guys over at globo-motor sent me back an e-mail, I guess my 10 watts wasnt too off. This is what they said:

HP = torque (in-oz) x speed



HP = 35x16X6 (6”RADIUS) x 6 RPM

1,000,000



HP= .020 WATT = 746 X .02



Would need 14.92 watts out. Assuming 50% efficiency of the motor now = 22 watts in.

 

[JedOs]

Barry1961, I am checking out Brother site right now. Thanks for the recomendation.

 

[aolalde]

For the power in HP proposed formula, the torque is in LbxFt and speed in rpm

Assuming the torque required is; T = 144 (Lbxin) = 144/12 (Lb x FT) = 12 (Lb x Ft) at 6 rpm:

HP = 12*6/5252 =0.0137 HP

Watts = HP x 746 = 0.137 x 746 = 10.22 Watts

I guess you should follow the recomendation for manual messurement of the friction torque and yes, depending on the motor-reducer efficiency (EFF in per unit) the electric input could be doubled for this small size unit(EFF~0.5).

Watts in = 10.22/EFF

 

[electricpete]

If the motor is only intermittent duty, converting torque to horsepower may result in buying a motor that is too big.

A horsepower rating implies the motor is designed to continuously run at that loading level.

If intermittent, just specify the torque and the speed and the duty cycle and let your motor supplier work with that.


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