Basic wheel calculations

  • #1

Main Question or Discussion Point

Hi there, really trying to grasp these concepts for one of my engineering papers. This is basic stuff I realised now that I never really understood.

If I've got two shafts, two wheels per shaft meaning four wheels in total, each wheel is 60 mm in diameter, and the power consumed by the system in providing a constant horizontal velocity of 0.15 m/s is 200W, what is the torque in each wheel?

The wheels are rubber coated, and the wheels are acting upon a steel surface (take coefficient of friction to be 0.7).

This is how I've tried solved the problem:
Power = Torque x angular velocity
w = v / r = 0.15 / 0.030 = 5 rad/s

Torque = P / w = 200 / 5 = 40 Nm

Now here's where I get stuck; the concept of what I've calculated:
Now since there are 4 wheels (instead of just 1 wheel) AND two shafts (two wheels in each shaft), how do I interpret this torque of 40 Nm? Is the force per wheel just force = Power / velocity or is this force divided amongst the 4 wheels?

With the friction, since the system is moving at a constant speed of 0.15 m/s, will the thrust force equal the frictional forces? The force here doesn't equal the force calculated via the torque of 40 Nm, is this because of a transmission efficiency?

So to summarise:
If power consumed in moving all four wheels at a constant speed is 200W, is the torque of 40 Nm I calculated split between the four wheels equally?
Power = force x velocity, calculating the force from here doesn't equal the total frictional forces etc. etc.

Thanks in advance.

Answers and Replies

  • #2
jack action
Science Advisor
Insights Author
Gold Member
In theory, if the vehicle goes in straight line, the torque per wheel is the total torque divided by the number of wheels, so 10 N.m per wheel.

In practice, to keep it in a straight line, it is possible that the inner and outer wheels have different values, as long as the moment about the vertical axis the same between the left and right side. Usually this is due to some flexibility introduced in construction. For example, the axle bends under the weight, lifting the outer wheel. This shifts the normal force from the outer to the inner wheel. The normal force of the outer wheel could be not enough to support the 10 N.m input torque, so it's the inner wheel that will take the «left over» torque with its increased normal force. If your inner and outer wheels are relatively far apart, this weight transfer could also happen (Do a free body diagram).

If the moment is different between the left and right side, then the vehicle will turn (unless you introduce some other moment elsewhere, like another axle).

That is for a solid axle. If you have a differential, it will ensure you that the torque is always split 50/50 between the 2 sides, no matter the conditions. That is why if you lift one side off the ground, the vehicle don't move: no reaction torque on the wheel in the air, so you have also zero torque on the other side (The power is solely used to accelerate the lifted wheel). With a solid axle the input torque would completely transfer to the other side (40 N.m). But the torque between the inner and outer wheels can still be different.

The coefficient of friction specifies the maximum force that can be applied, thus it is possible to have a smaller friction force. The actual friction force reacts to the input torque.