- #1

Alex Paul

- 7

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I was doing some behavioural modelling of the torque transfer characteristics of a belt drive system from the driver pulley to the driven pulley. While doing the same, i have tried to see how the angular velocity is getting transferred as well. I would explain my point with the following equations.

G_omega= (Omega_driven)/(Omega_driver )

G_Torque= (Torque_driven)/(Torque_driver )

G_Torque= (I∗alpha_driven)/(I∗alpha_driver )

where I is the Moment of inertia and alpha is the angular acceleration

G_Torque= (I∗(d(omega_driven))/dt)/(I∗(d(omega_driver))/dt)

where omega is the angular velocity

In Laplace domain, with zero initial conditions,

G_Torque= (I∗s∗omega_driven)/(I∗s∗omega_driver )

After cancelling I and s terms,

G_Torque=G_omega.

So does this really mean that the angular velocity transfer from the driver to the driven pulley has the same behaviour as the torque transfer and the step response of both of them will look the same ?

I feel like cancelling the moment of inertia terms is valid only if there is no slip or flex in the belt (could anybody confirm?). But even if there is a slip in the belt (then different moment of inertias felt by the driver and driven pulleys), doesn't it mean that this is already captured by the omega differences at both sides and this model will again work for torque even in the case of torque ?

The reason for this is obviously due to the fact that anglular velocity can be measured easily unlike torque.

Much appreciated if somebody could point me in the right direction.

Thank you,

Alex