# Equivalence of torque & angular velocity transfer function

• Alex Paul
In summary, Alex's behavioural model showed that the angular velocity is transferred in a similar way to the torque. The step response of both models will look the same.

#### Alex Paul

Hello all,

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

Hello Alex,

If I draw a simple driving wheel 1, a belt and a driven wheel 2, then from the fixed length of the belt I must conclude that ##\omega_1 r_1 = \omega_2 r_2## and ##\alpha_1 r_1 = \alpha_2 r_2##

Newton 3 says action = minus reaction
In other words your | Gtorque | = 1 ?

And your Gomega is definitely equal to rdriver/rdriven

Alex Paul
BvU said:
Hello Alex,

If I draw a simple driving wheel 1, a belt and a driven wheel 2, then from the fixed length of the belt I must conclude that ##\omega_1 r_1 = \omega_2 r_2## and ##\alpha_1 r_1 = \alpha_2 r_2##

Newton 3 says action = minus reaction
In other words your | Gtorque | = 1 ?

And your Gomega is definitely equal to rdriver/rdriven

BvU said:
Hello Alex,

If I draw a simple driving wheel 1, a belt and a driven wheel 2, then from the fixed length of the belt I must conclude that ##\omega_1 r_1 = \omega_2 r_2## and ##\alpha_1 r_1 = \alpha_2 r_2##

Newton 3 says action = minus reaction
In other words your | Gtorque | = 1 ?

And your Gomega is definitely equal to rdriver/rdriven

Hello BvU,

I understand your point. In static cases, with equal pulley sizes and no loss,
G_torque will be 1 since the entire torque is getting transferred.
I made a mistake of not mentioning the fact that i was looking at the dynamic or transient cases.
For example, if i make a jump in torque (a step for example) -
Then at the driver pulley side,
the angular velocity looks like an underdamped system which oscillates a bit before coming to a steady increase (constant torque --> steady increase of RPM).
The angular velocity at the driven side also looks similar with some small delay or may be reduced oscillations due to the flex or small slip of the belt.
By analysing this, i will be able to define a system (for example using an input output system identification). And now, if i plot the step response of this model identified using the angular velocity, will it look like the step response of the system which was identified using torques measured at both the pulleys.

With the equations i posted originally, i was trying to show that both these input output models look the same.
Apologies if i made the question confusing. Its a bit difficult to explain i think.

Alex

The dynamic case as you describe it adds a lot of detail, with corresponding parameters. You need to think about how you want to describe the slip, for example. That brings in details of the belt itself and the interaction with the wheels. And prettty soon you'll have to switch to numeric simulation because the usual analytic tools will fall short.

Make a list of equations you think you need. Consider which effects are first order and which are refinements you can do without. It's hard enough even with the first non-idealities brought in.

Alex Paul
For a Pulley and Belt the torque and speed ratios remain the same during a change in motor speed, only the acceleration rate of the entire drive system is affected by the inertia of the of the system components.
This assumes there are no belt to pulley slips in the drive.

Alex Paul
BvU said:
The dynamic case as you describe it adds a lot of detail, with corresponding parameters. You need to think about how you want to describe the slip, for example. That brings in details of the belt itself and the interaction with the wheels. And prettty soon you'll have to switch to numeric simulation because the usual analytic tools will fall short.

Make a list of equations you think you need. Consider which effects are first order and which are refinements you can do without. It's hard enough even with the first non-idealities brought in.

Thank you for the reply. I will try working on it.
The thing is i know my torque inputs and i was planning to construct a model with the input-output angular velocities and then use this model to estimate the output torque. (After scaling to include the fact that if torque is doubled at the driven pulley then omega is halved).
Regards,
Alex

JBA said:
For a Pulley and Belt the torque and speed ratios remain the same during a change in motor speed, only the acceleration rate of the entire drive system is affected by the inertia of the of the system components.
This assumes there are no belt to pulley slips in the drive.

Hello JBA,

Regards, Alex

Alex Paul said:
i know my torque inputs
That's good. We do not!
We don't know much about the system you are taking under consideration, and it's hard to come with specific comments in a general discussion...
Alex Paul said:
to include the fact that if torque is doubled at the driven pulley then omega is halved
That is so strange to read that I really have to ask for clarification ! What torque ?
Perhaps you can give a more extensive description of what you are talking about and include the arguments why you want to introduce so much non-ideal behaviour ?

BvU said:
We don't know much about the system you are taking under consideration
To explain in detail, I am trying to analyze a system where torque from an electric motor is transferred to the crank shaft of a vehicle. From the motor control unit, i know what is the torque its producing . During steady state conditions, the torque at the crankshaft will be the same as the torque from the motor. When there is a change in torque, during the transient conditions, there will be a difference in the torque "felt" at the crank shaft before reaching the steady state. The only thing i can measure at the crank shaft level is the RPM.
By observing the RPMs at the motor and crank shaft during transient conditions, i was able to model this behaviour. i just wanted to know if this model holds true for the troque behaviour as well.
In short, i make a model with the RPMs at both side. I feed my torque input to the same model and see how the torque changes. in short a step change in RPM and a step change in torque should have the same behaviour (?).
BvU said:
That is so strange to read that I really have to ask for clarification ! What torque ?
Now regarding this, i was talking about the fact that incase the pulley diameters were different which brings a gearing effect, a model made with omega will have to be scaled with the square of inverse of the omega amplification factor to get the torque amplification.
Thank you once again for taking your time for this.

Alex Paul said:
torque from an electric motor is transferred to the crank shaft of a vehicle
What is the ratio of the inertia of the gearing system wrt the inertia of the vehicle ? I.e. is it really worth while trying to model transient behaviour with a detailed model of the belt drive system ?

Alex Paul
BvU said:
is it really worth while trying to model transient behaviour with a detailed model of the belt drive system ?

That is a good point. Even i have felt like it is bit of an overkill.
BvU said:
What is the ratio of the inertia of the gearing system wrt the inertia of the vehicle

The moment of inertia values are not really known for the vehicle with the drivetrain connected. i was trying to avoid this by using what we have - the RPMs, to find out what happens to the torque at the crank shaft level. The final aim will be to filter the torque requests (by rate limiting) sent to the motor controller so that the system stays far from resonant regions and the drivability of the vehicle is also improved by not over filtering the request.

JBA said:
For a Pulley and Belt the torque and speed ratios remain the same during a change in motor speed, only the acceleration rate of the entire drive system is affected by the inertia of the of the system components.
This assumes there are no belt to pulley slips in the drive.
With no slip, its true 100%. But when there is a slip, then that means the increase in speed from the driver pulley is not transferred to the driven pulley and hence its logical to assume that no torque was also transferred to the driven pulley because with torque, there should be acceleration and increase in omega at the driven end. So the rate of change of omega ( which is the angular acceleration) is proportional to the torque transferred. In my head this is what gave me the idea that the input output model from omega could may work for torque as well and this is what i meant by this
Alex Paul said:
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 ?

Hope i explained it.