# Relationship between rolling resistance and velocity?

1. Nov 25, 2015

### electrodruid

I'm a games programmer working on a (simple, so far) physics engine for a driving game. Right now I'm interested in acceleration, and in particular trying to estimate the forces that go into it. I've got basic data for a bunch of cars (0-60 time, 0-100 time, top speed, mass) and a brute-force algorithm that tries out a bunch of combinations of values for tractive force, drag and rolling resistance to try to generate acceleration curves that match the data that I have for each car. I'm aware that the physics model I'm using is a good deal simpler than real life so I'm just looking for something that's broadly within the bounds of plausibility and provides a reasonable fit. Not total accuracy in the values, but something in the right ballpark.

I'm using this page, and the sources it cites, as my inspiration. And I'm only interested in acceleration in one dimension, so my calculations are all just using floats rather than vectors. It looks like this:

Code (Text):
float Ftraction = EngineForce;
float Fdrag = -Cdrag * v * v;
float Frr = -Crr * v;
float totalForce = Ftraction + Fdrag + Frr;
float a = totalForce / Mass;
v += a * dt;
I call this in a loop that tries different values for EngineForce and Cdrag, and for each combination calculates Crr such that (Cdrag * topSpeed * topSpeed) + (Crr * topSpeed) = EngineForce. I calculate the resulting velocity curve, and use the least squares method to find the combination that gets closest to 60mph at the 0-60 time, to 100 mph at the 0-100 time and to the top speed.

The problem I have is that the closest curve always ends up with Cdrag = 0, and just uses a high value of Crr to enforce the speed limit. Seems wrong to me, like driving a car in a vacuum but on a really sticky surface. I'd expected drag to be the main factor, with RR being of less importance. That page I'm working from (and its sources) are pretty wooly on the subject of the subject of rolling resistance: one paper says that for one particular car, RR = drag at "about 60mph", and from that concluded that RR rose linearly with velocity and should be "about" Cdrag * 30. But nowhere does anything state how the resistances were measured, whether the same is true for all cars, etc. Conversely, I find pages like this which state that RR has nothing to do with velocity, and is just derived from properties of the tyres and the vehicle's weight... But if that was the case, how does that even work? RR is not pushing backwards on a stationary car, is it? Velocity has got to be a factor in there somewhere, hasn't it?

What is the relationship between a car's velocity and the amount of rolling resistance acting upon it?

2. Nov 25, 2015

### CWatters

3. Nov 25, 2015

### electrodruid

The claim from that page that Fr = c W kind of makes sense, except I'm confused by the fact that there's no velocity in that equation. What's the rolling resistance for a stationary vehicle? Or one moving very slowly?

4. Nov 25, 2015

### CWatters

See that page again. Velocity is included in the Rolling Coefficient (c) which is used to calculate the Rolling Resistance.

5. Nov 25, 2015

### electrodruid

This?

I did miss that first time around. But... What IS that? What do all those "magic numbers" mean? Might they vary, or are they magical constants of the universe? How can that be adapted to m/s rather than km/h? I'm having trouble even wrapping my head around the order of operations.

EDIT: And again, in the section immediately beneath that way, the coefficient is just assumed to be 0.03 for that situation: no velocity involved or even mentioned in the calculation, rolling resistance expressed in Newtons regardless of the vehicle speed.

6. Nov 25, 2015

### CWatters

I believe that equation comes from testing real tyres.

Converting from Km/h to m/s is easy. 1km/h = 0.278m/s

Lets try plugging in some numbers..

c = 0.005 + 1/p (0.01 + 0.0095(v/100)2)
where
c = rolling coefficient
p = tyre pressure (bar).

1bar = 14.5psi. On my car the tyre pressure is 38psi or 2.62bar

v = velocity (km/h).

Pick a velocity. Lets say 20m/s = 72km/h

That gives
c = 0.005 + 1/p (0.01 + 0.0095(v/100)2)

= 0.005 + 1/2.62 (0.01 + 0.0095(72/100)2)
= 0.005 + 1/2.62 (0.01 + 0.0095*0.5184)
= 0.005 + 1/2.62 * 0.0149
=0.01 approx.

Which is close to the value they give in the table for "ordinary car tires on concrete". They used 0.03 in the next section which they say is typical for "car tires on tar or asphalt".

7. Nov 26, 2015

### electrodruid

Thanks for clearing up how to run the numbers through the formula, but I still don't feel like I'm getting to the root of my question. In fact, this is raising more questions than it answers...

• The equation doesn't seem to take the tyre composition into account (unless the main difference in tyre composition is the pressure the tyre is pumped to?). Whereas the table of fixed values just above does differentiate.
• Likewise, it seems only applicable to dry (asphalt?) roads, no other road surfaces.
• It's no use at all for other types of rolling resistance (e.g. steel train wheels)
• Even plugging 0km/h into the equation yields a coefficient of ~0.0088. For the 1500kg car in the following example, that would mean a force of ~129N. On a stationary vehicle! What exactly opposes that force to stop the car from spontaneously rolling backwards?
• What is the table of Rolling Friction Coefficients? Rolling Friction Coefficients at what velocity?
I'll try to phrase the question a different way. Say we've got a 1500kg car, on dry asphalt, on normal tyres, under earth gravity. But say this car is in some experimental test centre, inside a giant vacuum under a dome, so we don't have to consider drag. The car starts off stationary, and then starts its engine and drives forwards in a straight line. What happens to the rolling resistance as the car's velocity changes? And just to ensure we're being clear about all the forces at play, how does the effect on the car differ from the rolling resistnace on a train engine weighing 100,000kg, with steel wheels driving from a standing start along steel track in the same vacuum dome?

8. Nov 26, 2015

### CWatters

To get an exact value for a friction force you would have to write and solve equations for the forces between all the molecules in both the tyre, road surface and anything in between which clearly isn't practical. Friction and rolling resistance equations will always be the result of an attempt to model the real world rather than fully calculate it from first principles.

I can't explain why you get a non zero rolling resistance when stationary. I've not been able to find out where that equation on the Engineering toolbox site comes from. I did find a few other places that use the same equation but none I can find explain it's origin.

I'm sure numerous other models exist and I did find this which might be worth more research..

https://en.wikipedia.org/wiki/Low_rolling_resistance_tire

I guess you would have to look at SAE J2452 to find out the coefficients.

9. Nov 28, 2015

### 256bits

Why not a rolling resistance at 0 kph?

When the vehicle is stopped, the bottom of the tires are a still a little bit flat. A minimum force would be necessary to overcome that flatness.

10. Nov 28, 2015

### A.T.

Or the ground has a little dent, and you need a certain horizontal force component to get out of that dent.

11. Nov 28, 2015

### lightarrow

Or the tire is attached to the ground as with a glue and you have to overcame the binding energy
Ok, that was a joke...

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lightarrow

12. Nov 28, 2015

### A.T.

Why a joke? The stickiness of the surface would be lumped together with all the other effects, when rolling resistance is determined empirically.

13. Nov 28, 2015

### lightarrow

Well, for a car with normal tires on normal grounds, I presume the stickiness is negligible. In general it's not, and if this is the case, it's not even true that T < μN where T is the static friction between two bodies' plane surfaces, N the normal reaction between them and μ the friction coefficient; infact it could even be T > μN (just to remind that what it's written in books could be misleading if improperly generalized).

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lightarrow

14. Nov 29, 2015

### A.T.

Static friction needs to be overcome by a horizontal force to make the contact patches slip. I was talking about overcoming a adhesive normal force to make the wheel roll without slippage, which is part of rolling resistance. The horizontal adhesive force would be part of static friction.