Calculating required torque, if load is on a lazy suzan
Calculating required torque, if load is on a lazy suzan
(OP)
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!





RE: Calculating required torque, if load is on a lazy suzan
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I don't suffer from insanity. I enjoy it...
RE: Calculating required torque, if load is on a lazy suzan
RE: Calculating required torque, if load is on a lazy suzan
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
RE: Calculating required torque, if load is on a lazy suzan
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
RE: Calculating required torque, if load is on a lazy suzan
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.
RE: Calculating required torque, if load is on a lazy suzan
RE: Calculating required torque, if load is on a lazy suzan
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
RE: Calculating required torque, if load is on a lazy suzan
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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