Torque Calculation

I'm just an all rounder amerture, but I reckon multiply by 90.

T figure sems really big already, what sort of motor is it ?

You could compare similar physical sized motors and see what torque others have, or look up specs on the motor from its manufacturers plate.

Output Torque = 15X9550/20 = 7162 Nm( Not cosidering eficiency of the gear unit)

Sir,

Kindly generallised the above formula.

MOHD. SAEED KHAN

Ordinarily, the output torque is equal to the input torque times the gear ratio. So in this case, the first estimate would be 1460 x 90 = 131,400 Nm. In a 90:1 gearbox, however there will be losses due to friction, but even at that high ratio, the gearbox might still be 90% efficient. The gearbox vendor can tell you the efficiency under any running condition.

• HOWEVER:

I'd expect input torque to be more like 80 Nm for a 15kW motor operating at 1800 rpm -- so something seems wrong here. I'd agree with answer number 2, which provides the amount of torque on a shaft rotating at 20 rpm with 15Kw.

I agree with the colleagues. It's just to clarify to you know that, in SI units, the relation between the power and torque is

Power[W] = Torque[N.m] x Angular speed[rad/s]

Note that 1 RPM = 2.pi rad/min = 0.03333.pi rad/s

You are suposed to have 80 Nm in the motor and 7200 Nm in the output shaft.

In addition, as it is a resoanable big issue, check the gear efficiency. The efficiency of a gear reduction can drop to less than 80% depending upon the constructive quality and general conditions...

Quite. Worm gears are the least efficient of all gear configurations.

I am shocked by your tagline. In fact, studies have shown that 86% of all statistics are made up on the spot!

Sorry. Consider it changed...

50 Hz motors run at 1450 RPM or 2800 RPM (4 poles / 2 poles). If output speed is 20 RPM the gearbox ratio is 72.5. If ratio is 90:1, output speed must be 16 RPM. Please reconsider calculations.

Using the formula (120 x freq) / # poles, you get 1500 or 3000 for standard speeds. Perhaps, in this motor, slip is 3%, and the 1460 is actually an rpm rather than a torque. (Maybe the nameplate reads rpm at max torque: 1460, and the person reading it saw only the torque: 1460 part) Who knows? No matter how one does the calculations, something does not fit. Maybe the 20 rpm was hand clocked -- or even done by counting seconds: one one thousand, two one thousand... ? If 20rpm is real, and 90:1 is real, then the motor rpm is 1800, and the motor is running at 60 Hz (perhaps driven by a variable frequency drive, or in a country where 60Hz is the line freq.) Puzzling.

Your figures for the Inpit Power and Input Torque are not consistent. The formula that connects these quantities is;

Power (W) = 2xpi x Torque (N.m) x Rotational speed (rps)

The value of Rotational speed is 20 x 90 /60 = 30 rps

Therefore Power (W) = 2 x pi x 1460 x 30 = 275 kW !!!!

Indeed, it agrees with the above comments that the motor should be running at 1800 RPM and, if the motor has indeed 15KW,

output shaft speed 20 rpm -> input shaft speed = 1800 RPM (20 x 90)

Normally it is obtained with 60 Hz...

15000 [W] = Torque[Nm] x 2.pi/60 x 1800 resulting a torque in the motor of 80 Nm.

output torque would be then 80 x 90 = 7200 Nm

and the output power calculation is:

Power [W] = 7200 [Nm] x 2.pi.20/60 [rad/s] = aprox. 15079 [W] (a lot of rounding values here)

One of the initial values ( input speed, input power, output speed, or gear reduction) is wrong. AND: take in count the gearbox efficiency.

As various posters have said, there are some inconsistencies here. 20rpm x gearbox ratio 90 gives 1800rpm, but it says 50Hz.

1460Nm @ 20rpm gives 3.1kW, so at say 80% gearbox efficiency 3.8kW input. So this could be the situation, the kit running light.

Also 100mm dia. shaft has torsional Z-value 0.5*pi*50³ mm³ so stress at 1460Nm = 7.4MPa, again on low side, but torque 90 x 1460Nm gives stress 669MPa which is far too high. If it is running light, at design condition stress is ~ 7.4*15/3.8 = 29MPa, which sounds reasonable.