What factors affect car braking distance?

[electrodruid]

I'm a game programmer trying to write code to simulate a variety of cars braking at different speeds and on surfaces with different coefficients of friction. For now, consider that I'm interested in a fairly wide range of cars (people carriers, hatchbacks, saloons, sports cars etc) but that they're all equipped with the same decent-quality tyres and are driving in a straight line on dry asphalt. I'm not considering thinking time, just the actual braking time/distance.

So, the Wikipedia article on braking distance tells me that the distance is calculated like this:

.

The mass of the vehicle isn't a factor because it cancels out of the equations that lead to this one, gravity is constant, mu is (in my example) constant, so really everything is relative to v^2 and nothing else. So, at 60mph (26.8224 m/s), with mu = 0.7 and g = 9.8, the stopping distance is always going to be ~52.43 metres, right? Regardless of whether the vehicle is a Land Rover or a Ferrari.

Thing is, I've got a copy of Autocar magazine here, which features road tests for a whole bunch of different vehicles, including 60-0 braking times. I know I'm comparing braking distance to braking speed now, but their road tests say that a Ferrari F12 stops in 2.2 seconds, but a Land Rover Defender takes 3.5 seconds. Even comparing more similar vehicles like the Peugeot 3008 and the Peugeot 5008 yield braking times of 2.1 and 3.1 seconds respectively.

So, my questions:

1 - What factors could be causing this difference in braking time?
2 - How might those factors be included in an equation to calculate braking distance/braking time?
3 - Given a vehicle's braking time at a given velocity, is there a reasonable way to calculate its braking distance? Or vice versa?

 

[mfb]

The mass of the vehicle isn't a factor because it cancels out of the equations that lead to this one, gravity is constant, mu is (in my example) constant, so really everything is relative to v^2 and nothing else. So, at 60mph (26.8224 m/s), with mu = 0.7 and g = 9.8, the stopping distance is always going to be ~52.43 metres, right? Regardless of whether the vehicle is a Land Rover or a Ferrari.

With the assumptions you made: right.

For cars, usually contact between tyres and street is the limiting factor. This can be different for larger, more massive vehicles, if the brakes are not scaled up in the same way.

3 - Given a vehicle's braking time at a given velocity, is there a reasonable way to calculate its braking distance? Or vice versa?

With constant acceleration, braking distance is just 1/2 * initial speed * braking time. Or, more general, but still with constant acceleration, distance traveled = 1/2 (initial speed + final speed) * time, where the final speed is zero for the braking distance.

 

[electrodruid]

For cars, usually contact between tyres and street is the limiting factor. This can be different for larger, more massive vehicles, if the brakes are not scaled up in the same way.

Can you elaborate a bit about why this is the case? Is it because the fastest way to brake is to have the brakes apply as much force as possible without locking the wheels? A bigger contact area between the tyres and the street means more static friction, so you can apply the brakes more heavily without locking the wheels than you can on a car with a smaller contact area with the road?

With constant acceleration, braking distance is just 1/2 * initial speed * braking time. Or, more general, but still with constant acceleration, distance traveled = 1/2 (initial speed + final speed) * time, where the final speed is zero for the braking distance.

Okay, that's useful, thanks. Is it reasonable to assume that deceleration is constant, then? I found this page which talks about constant deceleration, where the amount of deceleration is apparently largely dependent on driver skill rather than any inherent property of the vehicle, but I wasn't sure if that was an oversimplification (their stopping distance calculator outputs weird results, which makes me a bit cautious about the rest of the information on the page)

 

[mfb]

Is it because the fastest way to brake is to have the brakes apply as much force as possible without locking the wheels?

Right.

A bigger contact area between the tyres and the street means more static friction

It does not. Unless you make your tyres out of glue that sticks even without a normal force. The contact area does not matter.

Is it reasonable to assume that deceleration is constant, then?

It is a reasonable approximation.

If you try to brake too hard, the tyres start slipping and braking gets a bit less effective (sliding friction instead of static friction). It also makes the car hard to control, so most cars have electronics preventing you from doing that.

 

[rcgldr]

One issue is:

http://en.wikipedia.org/wiki/Tire_load_sensitivity

 

[cjl]

A huge part of it is that high performance sports cars tend to be equipped with sticky, high performance tires, decreasing their stopping distance. Equip the Land Rover and the Ferrari with the same tires, and a large proportion of that difference will go away. There are also some effects of tire size and pressure, as well as load (as shown in the wikipedia link from rcqldr above), but these are smaller effects than the tire compound. The suspension setup also matters a bit due to the load shift that occurs as the vehicle starts braking, but again, this is a minor effect compared to tire stickiness. Put the Ferrari F12 on knobbly, off-road capable tires or hard, high-mileage low rolling resistance economy tires, and you'll see that 2.2 second stopping time increase rather dramatically. Alternatively, put Formula 1 racing slicks on the Land Rover and it'll stop on a dime.

For realistic street tires, the coefficient of friction can vary by nearly a factor of 2, from about 0.6-0.7 on the low end up to 1.1 or so for extremely sticky high performance summer tires.

 

[electrodruid]

Thanks, this is all really helpful. So if I'm reading the Tyre Load Sensitivity issue correctly, even if you did put the same tyres on the Land Rover and the Ferrari, the Land Rover would still have the bigger stopping distance because its additional weight would have a detrimental effect on the coefficient of friction? (Although there would be much less difference in braking speed/distance compared to a test where the Ferrari has racing tyres and the Land Rover has off-road ones). And the example with the two similar Peugeots can most likely be attributed to slightly different tyres, or differences in mass or suspension. Right?

One last thing I want to get fixed in my head is this idea of the deceleration being constant. I'd just like to make sure I properly understand why that would be the case. My logic goes something like this:

- For a given vehicle, assume the brakes, tyres, etc are in good condition...
- Assume the driver is a skilled racing driver, and is trying to brake as quickly as possible
- The brakes are capable of applying enough force to lock the wheels, but the optimal amount of braking is slightly less than that, so that the wheels keep rolling rather than lock.
- The amount of deceleration required to make the wheels lock is determined by the static friction between the tyres and the road.
- The static friction remains the same (or very nearly the same) regardless of how fast the car is travelling.

Have I got that right?

 

[mfb]

Thanks, this is all really helpful. So if I'm reading the Tyre Load Sensitivity issue correctly, even if you did put the same tyres on the Land Rover and the Ferrari, the Land Rover would still have the bigger stopping distance because its additional weight would have a detrimental effect on the coefficient of friction? (Although there would be much less difference in braking speed/distance compared to a test where the Ferrari has racing tyres and the Land Rover has off-road ones). And the example with the two similar Peugeots can most likely be attributed to slightly different tyres, or differences in mass or suspension. Right?

Right.

One last thing I want to get fixed in my head is this idea of the deceleration being constant. I'd just like to make sure I properly understand why that would be the case. My logic goes something like this:

- For a given vehicle, assume the brakes, tyres, etc are in good condition...
- Assume the driver is a skilled racing driver, and is trying to brake as quickly as possible
- The brakes are capable of applying enough force to lock the wheels, but the optimal amount of braking is slightly less than that, so that the wheels keep rolling rather than lock.
- The amount of deceleration required to make the wheels lock is determined by the static friction between the tyres and the road.
- The static friction remains the same (or very nearly the same) regardless of how fast the car is travelling.

Right. With the usual caveats that all those things are often just approximations. At high speeds, air resistance will help braking a bit, for example.