Motor combination to drive multiple tyres

[Jay1298]

If I need 50,000 Nm of torque to rotate a wheel, and I am rotating it about its rim (like the London eye), would 5 10,000 Nm motors each connected to a set of tyres to rotate it (the motors are not connected to each other), or would these motors first need to be connected to each other and then drive a set of tyres or would I need a single 50,000 Nm motor connected to a set of tyres?

Thank you.

 

[BvU]

Hello Jay,

The total torque is simply the net (vector) sum of the torques applied, so it isn't necessary to connect the motors or have only a single one.

My first guess is that 10000 Nm is way too much to ask from a tyre, but you may have better info.

Good luck with your design !

 

[Jay1298]

The plan is to split the 10,000 between 4-5 tyres

Hello Jay,

The total torque is simply the net (vector) sum of the torques applied, so it isn't necessary to connect the motors or have only a single one.

My first guess is that 10000 Nm is way too much to ask from a tyre, but you may have better info.

Good luck with your design !

 

[BvU]

My reasoning: accelerating a 1000 kg car with 4 m/s² requires 1000 N per tyre.
With a radius of 0.25 m that is 250 Nm of torque per wheel

 

[Jay1298]

My reasoning: accelerating a 1000 kg car with 4 m/s² requires 1000 N per tyre.
With a radius of 0.25 m that is 250 Nm of torque per wheel

I understand but our acceleration is tiny, the wheel runs at 1 revolution per 24.17 minutes and an acceleration time of around 3 minutes, giving an angular acceleration of something like 10^-5 if I remember right.

 

[BvU]

But you said you needed 50000 Nm ?

 

[Jay1298]

I understand but our acceleration is tiny, the wheel runs at 1 revolution per 24.17 minutes and an acceleration time of around 3 minutes, giving an angular acceleration of something like 10^-5 if I remember right.

Yes we are going to have 5 10,000 Nm motors each connected to 4-5 tyres

 

[BvU]

Still don't understand: your final angular velocity is $\displaystyle{{2\pi\over 24.17 * 60 }\approx 0.0043}$ rad/s. If you reach that in 180 s, your angular acceleration is 2.4 10^-5 rad/s².

With 50 kNm you could handle a moment of inertia of 2.1 10⁹ kg$\cdot$m² which I guess is not far from the London eye !?

 

[Jay1298]

Still don't understand: your final angular velocity is $\displaystyle{{2\pi\over 24.17 * 60 }\approx 0.0043}$ rad/s. If you reach that in 180 s, your angular acceleration is 2.4 10^-5 rad/s².

With 50 kNm you could handle a moment of inertia of 2.1 10⁹ kg$\cdot$m² which I guess is not far from the London eye !?

My mistake with the numbers i didn't have access to exact numbers of our project when writing this thread so I used approximations, here are my workings:

The time to accelerate is 240 s
The angular velocity is 0.00433 rad/s
The inertia is 4.3*10⁹
Therefore the torque needed is 77500 Nm approx.

This is spread over 8 motors so for simplicity let's say 10,000 Nm per motor, each motor will drive 4 x 0.5m radius tyres.

 

[BvU]

https://www.patana.ac.th/parents/curriculum/Physics_K5/units/010304.html. With 1900 tonnes and 70 m radius London eye has $I =$ 9.3 10⁹ kg$\cdot$m².

Your thingy half that $I$ ? If so, I agree with your calculations. Still worry about 5000 N per tyre, but I suppose truck tyres manage it too.

 

[Jay1298]

https://www.patana.ac.th/parents/curriculum/Physics_K5/units/010304.html. With 1900 tonnes and 70 m radius London eye has $I =$ 9.3 10⁹ kg$\cdot$m².

Your thingy half that $I$ ? If so, I agree with your calculations. Still worry about 5000 N per tyre, but I suppose truck tyres manage it too.

Yes we are planning on using big tyres, and our radius is 50m

 

[CWatters]

Just a heads up...you can have problems with asymetric load sharing if you connect some types of motors in parallel.