Maximum torque required to turn the wheel along the vertical axis

[Urban Spoken]

TL;DR

An electric motor via rack and pinion connceted to a wheel standing on the ground. Find maximum torque requied to turn the wheel along the vertical axis. The connection point is offset from the horizontal axis of the wheel.

Hello!

I'm trying to design a system which will rotate a wheel utilizing electric motor. It is something pretty much similar as a car steering system made independently for each wheel.
Please look at the images attached to the post - I made top and front view diagrams.

I want to understand motor characteristics - max torque required to turn the wheel with the velocity v (m/s) along the wheel's vertical axis (specified on the fron view diagram), taking into account the mass of the wheel Y (kg), the maximum angle θ (degrees), the distance between the motor and wheel's center of mass L (m), and the motor and the point where steering rod connects to the wheel (l, m) [the distance between this point and the horizontal axis of the wheel is k, m]. The rack which pushes the wheel has mass of R kg and length of x meters.

Previously I solved a problem where a motor has to keep a load (mass M) [connected via lever (mass m, length L, center of mass L/2)] still on horizontal position. The maximum torque here is straightforward: mg x (L/2) + Mg x L

But I can't understand how to resolve the issue I specified above, even the starting point. Please lead me in the correct direction.

 

[Juanda]

The diagrams are neat. It makes tackling the problem much easier. There are too many times when people don't provide enough information to be helped properly.
I can't work on this now but I'll try it later.

Just a few comments on what you posted.

1. The turning velocity of the wheel I believe is not relevant. It's the forces/torques and accelerations that matter since that affects the inertia.
2. Why are you saying that $\sum\tau = 0$ in one of your diagrams? Are you saying the wheel is turning without acceleration?
3. Is the wheel rotating on its own axis (like rolling)? That could introduce more resistance since it'd be necessary to change the direction of the angular momentum of the wheel making things more complex.

 

[Baluncore]

There is a contact patch between the wheel and the ground. That patch will be elliptical, but let's assume it is circular. The area of the circle is the wheel load divided by the tire internal air pressure.

When you turn that wheel, the patch will rotate with friction against the ground.

The centre of wheel rotation is different between the two drawings. If the line of the steering king-pin, does not pass through the centre of the contact patch, there will be additional friction, as the inside and outside of the wheel will roll with different velocities.

 

[jack action]

Look for the self-aligning torque of a tire.

​

 

[Urban Spoken]

The diagrams are neat. It makes tackling the problem much easier. There are too many times when people don't provide enough information to be helped properly.
I can't work on this now but I'll try it later.

Just a few comments on what you posted.

1. The turning velocity of the wheel I believe is not relevant. It's the forces/torques and accelerations that matter since that affects the inertia.
2. Why are you saying that $\sum\tau = 0$ in one of your diagrams? Are you saying the wheel is turning without acceleration?
3. Is the wheel rotating on its own axis (like rolling)? That could introduce more resistance since it'd be necessary to change the direction of the angular momentum of the wheel making things more complex.

Thank you for responding so quickly

Answering the second question, I guess I made a mistake - I took it from the problem I solved for lifting a weight using a lever, which states that in a horizontal position, we must satisfy the rotational equilibrium in order to maintain the load steady. Furthermore, the formula indicates that the system has no net torque. In the actual problem it is not the case since starting to turn the wheel is the most difficult part. Maintaining it moving up to 45 degrees after that shouldn't be a problem.

Regarding the third question - the wheel rotates over the axis passing via control arms' joints. I've just edited the diagrams - I added the upper and lower control arm. I hope that the friction there (inside ball joints) is negligible.

 

[Urban Spoken]

There is a contact patch between the wheel and the ground. That patch will be elliptical, but let's assume it is circular. The area of the circle is the wheel load divided by the tire internal air pressure.

When you turn that wheel, the patch will rotate with friction against the ground.

The centre of wheel rotation is different between the two drawings. If the line of the steering king-pin, does not pass through the centre of the contact patch, there will be additional friction, as the inside and outside of the wheel will roll with different velocities.

Thank you for the insights!

Now I guess I see how to incorporate tire-ground friction force into the max torque calculation. It's also important to consider the worst-case friction scenarios, such as the friction coefficient between tires and asphalt or concrete.

Should I somehow include the mass of the rack to the calculation?

And sure, I made a mistage drawing the axis of rotation
I've just edited the drawings assuming that the wheel is fixed by lower and upper control arms. Let's assume that the axis of rotation is 90 degree to the horizontal line so it passess right through the control arms' ball joints. I indicated these on the plots. I suppose the friction force here must be tiny so I can ignore it. Am I right?

 

[Baluncore]

Should I somehow include the mass of the rack to the calculation?

You must include the mass of the wheel, rack and vehicle that are supported on that wheel.

Let's assume that the axis of rotation is 90 degree to the horizontal line so it passess right through the control arms' ball joints.

It is usual for the upper ball joint to be moved inwards, so the line that passes through the two ball joints, passes through the centre of the contact patch on the road, in the middle of the tire. That balances many forces on the steering rack from the road, such as when braking.

 

[Baluncore]

https://en.wikipedia.org/wiki/Kingpin_(automotive_part)#Kingpin_inclination
https://en.wikipedia.org/wiki/Double_wishbone_suspension#Short_long_arms_suspension

 

[Juanda]

The diagrams are neat. It makes tackling the problem much easier. There are too many times when people don't provide enough information to be helped properly.
I can't work on this now but I'll try it later.

That "later" took longer than I expected.

I'd say the key to solving your problem is solving this simplified diagram.
Note: I'm aware yours is a 3D mechanism because forces are not aligned in the front view but that just makes things more mathematically tedious. The overall concept is basically the same. Since the most relevant plane is the one shown from the top view (because the misalignment in the other one is very small) I'm simplifying the problem in this example to consider only the top view.

Do you have experience with the kind of diagram I shared? Would you know how to solve its kinematics?
From your OP I believe you're interested in its dynamics (forces causing movement) rather than its kinematics (just movement) but I think understanding kinematics first is a proper starting point.

You're interested in knowing the torque experienced at A due to the action of the motor at D. That torque at A must be greater than the friction torque introduced by the wheel with the ground (ignoring other friction sources and assuming the wheel is not rolling).

To find out the torque caused by the mechanism, imagine there's so much friction at A that the motor can't turn the wheel. It's making a force but there's enough friction to keep everything steady (at equilibrium).
Make a free-body diagram for all the bodies until you can find what's the resultant torque in A due to the axial load pushing the rod DC so that the mechanism is at equilibrium. Notice that the elements CB and BA can only carry axial forces because they're articulated at both ends, unlike the rod DC.
Your mechanism only has one degree of freedom so you'll have to solve that body diagram for a given variable. I'd recommend choosing either the rotation of the wheel (the angle defined by AB) or the location of the rack and pinion (the length defined by DC).

For the friction of the wheel with the floor due to the non-perfect contact patch. As @Baluncore said in #3, friction can be complicated even when assuming the contact patch is a circle due to the center of rotation. If you want to keep complications to a minimum and you can allow for some "good enough" approximation, to calculate the friction I'd assume a circular contact patch rotating around its center and use the clutch equation.

 

[Juanda]

Notice that the elements CB and BA can only carry axial forces because they're articulated at both ends, unlike the rod DC.

I need to correct my previous comment.

What I said is generally true. However, in this case, the double-articulated rod defined by AB will carry more than just the axial load because a torque is applied at point A due to friction.

Double-articulated rods can only carry axial loads when the only external forces acting on them are applied at their ends and there are no torques on them.