# Slip -- should it be used with effective radius?

1. Jan 17, 2017

### Payam30

Hi,
I was thinking of the slip that is defined as following:
$$s = \frac{R\omega}{v_x} -1$$
The definition of the effective radius according to Chand and Sandu [1]
$$R_e = \left\{ \begin{array}{l l} R - R \left(1- \frac{1 -\frac{\delta}{R_g}}{\cos\theta} \right) & \text{if } \theta_r < \theta \leq \theta_f \\ & \\ R - R \left(1- \frac{1 -\frac{\delta}{R_g}}{\cos\theta_f} \right) e^{-\beta z (\theta -\theta_f}) & \text{if } \theta_f < \theta \leq \pi \\ &\\ R - R \left(1- \frac{1 -\frac{\delta}{R_g}}{\cos(2\pi + \theta_r)}\right) e^{\beta z'(\theta -(2\pi + \theta_r))} & \text{if } \pi < \theta \leq 2\pi + \theta_r \\ \end{array} \right.$$

The effective radius is depending of the variabel $$\theta$$ which is the angel defined on the contact patch. Reza N.Jazar [2] propose the unloaded radius and not the effective radius when calculating the slip. while some people go with effective radius. what are the arguments for using and not using the effective radius in calculating the slip?

[1] J. Y. Wong.Theory Of Ground Vehicles. John Wiley & Sons New York, 3rd edition, 2001
[2] Reza N.Jazar.Vehicle Dynamics Theory and Application 2th Edition. Springer New York
Heidelberg Dordrecht London.2014

2. Jan 17, 2017

### jack action

For zero slip, the velocity of the tire at the contact patch must be zero with respect to the ground. Therefore, the velocity of the wheel axle must be $v_x = \omega R$ where $R$ is the true distance between the ground and the center of the axle.

3. Jan 17, 2017

### Baluncore

The circumference of the belt of fabric inside the tyre that is bonded against the tread dictates the rolling circumference and so the rolling radius. The flat contact patch with a shorter radius, is balanced by a slightly greater radius where there is no contact. For that reason the height of the axle is not as relevant as the dimension of the tyre construction.

The thickness of the tread above the contact patch is included in the height of the axle above the pavement. This brings up the question of tread thickness and wear. Since the length of the tread is variable it is the rolling circumference of the internal fabric that counts. Tread wear does not change the rolling radius but it does change axle height.

4. Jan 18, 2017

### jack action

Sorry, but I cannot agree with that. Slip (especially as defined in the OP) cannot have anything to do with tire construction. What about a tire made out of solid rubber? Or any other material for that matter. Slip is a purely motion definition. It's what is in contact with the ground that counts, nothing else.

Slip relates to this rolling motion concept:

For pure rolling motion, slip is zero and $v_{@P}$ is also zero by definition. From that, $v_{CM} = R\omega$.

There is slip whenever $v_{CM} \neq R\omega$. If slip is $\infty$, then $v_{CM}$ = 0 (i.e. no axle translation) and $v_{@P} = - v_{@P'}$ (i.e pure rotation about the axle).

5. Jan 18, 2017

### Baluncore

What about a dreadnaught = endless railway wheel? The thickness of the shoe pads does not change the ground speed, but it does change the axle height. The slip takes place at the axle height while the wheel rolls at the rim radius.

As I see it there are different interpretations of wheel slip, by different analysts, with different assumptions about the contact patch and tread flexibility. It appears that in some interpretations, a contact patch, a thick tread or a tractor tyre lug makes zero slip unatainable.

I do not have the Chand and Sandu reference [1] in the OP, so I cannot see how they define theta on the contact patch.

6. Jan 18, 2017

### jack action

That is not a wheel, it is a mechanism including a wheel. Of course, in this case, the slip happens between the wheel and the rail, or as I called it earlier 'the ground'. Still, in this case, there could be slip between the rail and the real ground as well and it should be included in the overall slip of the mechanism.

That is like having a normal rim & tire configuration and having the rim slipping within the tire. The 'total' slip would be the combination of the rim-tire and tire-ground slips. But nobody would say that the rim diameter is a important characteristic for slip, as it is not supposed to do that.

You said it yourself:
They represent one single link, so they are not supposed to move with respect to each other.

7. Jan 18, 2017

### Baluncore

You misunderstand it. The dreadnaught shoe/pad/rail is bonded to the rim with fixed short wire ropes, better than, but with the same effect as the bead of a tyre is firmly seated on the rim without slip.

8. Jan 18, 2017

### jack action

I found another image of a Dreadnaught wheel:

I see better how the shoe/pad/rail are linked to the wheel. The shoes are loosely attached to the wheel. The shape of the hanger is only for proper positioning of the shoe for 'entrance' between ground & wheel (going forward or backward).

Imagine the wheel-side of the shoes are slippery as ice, but the ground-side are sticky as rubber. When the wheel sits on a shoe, there is enough play to permit slipping between the wheel and the shoe. But once the play is gone, the peg of the wheel will mechanically 'lock' with the hanger of the shoe and further slipping (if any) will be between the shoe and the ground. As I said earlier, slipping can occur at both levels, even simultaneously.

I still don't see how this relates to a tire where the belt is mechanically bonded to the tread. The way you made your statement, it suggests that if the treaded part of a tire was as thick as its belt radius for example, that it would have no effect on the speed of the axle. If this is what you meant, I can't see how I can agree with that.

Another example: When a tire is freely spinning on its axle (burnout, $v_{CM}$ = 0), the slipping obviously happens at ground level, and there is no slipping between the belt and the tread. Why would I use a smaller length than ground-axle distance for reference?

9. Jan 18, 2017

### Baluncore

Not quite. The detail is important but rarely reported.
Text and pictures at; http://pete-n-pam.com/main/061_080/page063.htm
The two tensioned cables for each shoe prevented rim slip on the top of the shoe. The looped cable symmetrically balanced the cable tension against the shoe, while still allowing azimuth rotation when cornering. Any slip with the Dreadnaught wheel must occur between the shoe and the ground.

The many wire ropes on a Dreadnaught wheel lie tight against the rim. Together they are the direct equivalent to the bonded belt in a tyre. The Dreadnaught wheel is an example of a thick tread that has no contact patch flattening, but that changes the height of the axle, not the rolling radius or circumference of the belt. It demonstrates that axle to ground height is applicable only to the idealised mathematical world of truely round wheels without tread.

You should use the rolling radius, based on the length of the belt in the tyre that slides past a point on the ground with each revolution of the rim. That may be bigger or smaller than axle height depending on the relative magnitude of thread thickness and contact patch flattening.

Hopefully, examining the extreme endpoints of wheel construction will identify the alternative possible definitions of wheel slip and the conditions where each might be applicable.

10. Jan 19, 2017

### jack action

I must say I've been looking at this and it raises more questions than answers to me.

I don't argue with the principal behind that wheel, but I can't wrap my head about what "tread" would mean.

You seem to apply the concept of that wheel to a tire. But what about a wooden wheel with a metal ring (or "tread") bonded to it? You should threat the thickness of the ring as part of the radius of the wheel, right? But if that ring is made of rubber, you don't? Where do we draw the line? Is there a type of material that will set the end of the effective wheel radius at half the thickness of the tread?

I don't grasp the meaning of that sentence. Is "the length of the belt" the circumference of the steel belt?

Based on what you said, I may understand 'smaller', but how can the effective wheel radius be 'bigger' than the axle height?

11. Jan 19, 2017

### Baluncore

That is good. There is probably not one case, so there will be more than one answer.

I still don't know how the OP defines the theta parameter for the contact patch.

The Dreadnaught shoes are equivalent to the blocks of tread bonded to the belt of a tyre. The rolling radius is fixed by the steel rim.

A wooden wheel with a steel tyre does not change radius significantly where it contacts the pavement. It remains circular. A rubber tyre on the other hand has flexible sidewalls so the contact patch increases area and reduces axle height as the load increases. But the length of the belt in the rubber tyre does not change significantly as the belt and tyre develop a variable radius, (in polar coordinates).

Yes. That fabric belt is bonded through the sidewalls to the rim. It sets the effective circumference of the tyre as it rolls along the pavement. A wheel driven speedometer on a vehicle does not change it's calibration as the tread wears and axle height becomes lower. That is because the belt sets the rolling circumference of a rubber tyre on the pavement. Likewise, a deformed loaded tyre does no change speedometer calibration.

An over-loaded pneumatic tyre can flatten at the contact patch, to place the rim very close to the pavement, risking damage to the folded sidewall. The belt has not changed length, so the circumference and the effective rolling radius will remain constant. The effective wheel radius, the rolling radius, is then significantly greater than the axle height.

12. Jan 19, 2017

### jack action

That is true, I never thought about it before. Speedometer reading shouldn't vary whether the tire is under-inflated or over-inflated. Or loaded or unloaded, for that matter.

So what is the wheel radius of a tracked vehicle? The length of the track divided by 2$\pi$ ?

13. Jan 20, 2017

### Baluncore

It is clearly not track length / 2Pi.

Tracks normally have a sprocket drive that should not slip. I think I can see a central line of lugs in the picture that run in the central groove of the drive wheels. The track then becomes the equivalent of Dreadnaught shoes, or of the block tread on a pneumatic tyre.

To be accurate we should count the number of teeth on the sprocket and measure the pitch of the rollers on the track. Rolling circumference = Number of teeth * Pitch. Divide that by TwoPi to get the effective rolling radius.

In the case of the John Deere pictured, there are teeth moulded onto the inside of the rubber track, but I cannot see where the sprocket might be hidden. It does appear that the track is driven as per normal, at the rear.

14. Jan 20, 2017

### jack action

Nevermind, After writing that last post I got to think more about it and realized my mistake I turned off my computer just after writing that post yesterday, but I'm correcting it today.

The reason why the tire radius doesn't change is because no matter the shape of tire belt, whenever the wheel makes one revolution, the belt travels its entire length and comes back to its original starting point. Thus the traveled distance per revolution by the tire is always the same. It's the flexing side walls that permits such characteristic, a feature that tracked vehicles don't have. So you have to treat them solely as belts or chains.

For the John Deere, I think it relies on friction (rubber on rubber), like a belt. The "teeth" you see are only there to take side force and keep the track aligned with the wheels. This picture is more clear:

15. Jan 20, 2017

### Baluncore

That is the fundamental principle behind the rolling radius of pneumatic tyres. I believe it is the radius that should be used when computing slip.

That second picture is actually a different system. It does not seem to have the independent rocking track assembly and split wheels. Maybe it is friction drive, with a hydraulic track tensioner to prevent track drive slip.