Motor Sizing

Main Question or Discussion Point

Hello,

I need to size a motor for the following conditions,and I have managed to completly confuse myself:

Motor will turn a hollow cylinder
Cylinder mass = 10000kg
acceleration = 0.25m/sec/sec
Cylinder speed = 19 rpm

I've been calulating the torque by:

T = (I * a) where
I = moment of inertia
a = acceleration

and the horsepower by

HP = (n * T) / 5252 where
n = cylinder RPM
T = torque

I can do all of this fine and the answer come out to around 2hp. However, I am confused about how adding a gearbox to this arrangement will affect the HP needed. I don't see anything in any of these formulas that take into account gear ratio. It seesm to me that if I were to add a significant gear reduction, I should be able to use a much smaller motor.

Where am I going wrong?

Thanks,

Eric

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AlephZero
Homework Helper
If you use a gearbox, you will get a different torque and a different RPM for the motor. It's a lot easier to make or buy a 2HP motor that runs at 1900 RPM than one that runs at 19 RPM, and the 1900 RPM motor only needs to produce 1/100 of the torque compared with the 19 RPM motor.

But if the acceleration of the cylinder doesn't change, the power required to accelerate it doesn't change, so you still need a 2HP motor whatever speed it is designed to run at.

Because using a gearbox will increase the torque and decrease the speed (or vice-versa) proportionally, the power output required from the motor would be equal (ignoring the slight amount of friction in the gearbox).

You have to take into account inertia ratio hollow cylinder and motor armature inertia it should be less than 3 , if other wise you have to go for higher power motor to reduce the shock
Let Imr be the moment of inertia reflected at motor shaft, Ic be the moment of Inertia of cylinder, N1 be motor speed and N2 be the cylinder speed

then Imr = Ic x (N2)^2/(N1)^2 it is the moment of Inertia reflected at the motor shaft and it should less than 3

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