# Car braking distance

1. Aug 13, 2014

### Feeh

I always found this an interesting question
How can you measure the braking distance of a car with a good reliability in normal conditions?

From classical physics, we can assume that the braking distance can be found by determining the work needed to dissipate the vehicle kinetic energy.

After some work, we can reach the following formula:
$$d=vt + \frac{v^2}{2gμ}$$
with the first part being the distance traveled before reaction time (which I will ignore)
assuming that μ=0.7, g=9.81m/s2 and v=100km/h
I've done some research and found 0.7 for μ
With those values, the braking distance is about 56 meters, far different from a (first reference car I've found) Nissan GTR (R35 - 2011) 32.75 meters braking distance.

Ok, if we find another value for μ such as 1.0 (hi-performance tire / hpwizard.com) which is close to a treadwear friction coef from a Dunlop 7010 tires (TW=240)
the braking distance become about 39 meters, which is more plausible but not the accuracy I want.

My first is: the classic physics formulas can return the approximated braking distance?
I mean, the formula does not take into account the drag force / the car weight distribution / the dive effect when the car brakes / maybe μ varies from tire temp, speed / rolling resistance / etc...

These are the values we can return or there are some others important values we need to take into account?

2. Aug 13, 2014

### jack action

Before blaming the accuracy of the equation, you might question the source of your data.

This braking distance table seems to be your reference.

First, it is said to be careful using the data, citing speedometer accuracy for one. We also notice that all cars are in the 36-44 m except for the Nissan GTR and that all tests do not seem to come necessarily from the same source.

So assuming this car was a «special» car with race-oriented tires, a friction coefficient of 1.2 could easily be possible. If we assume there was a speedometer inaccuracy, that would mean that a car with a CoF of 1.0 could have a true speed of 91 km/h instead of 100 km/h and that would give a 32.75 m braking distance. That speedometer error is more than possible according to this source:

Also, if you want to consider aerodynamic effect, you can use the more accurate equation for braking distance found on hpwizard.com. But, in most cases for 100-0 km/h, the effect is negligible.

As for a CoF of 0.7, it is very unrealistic with today's car tires unless you are on a wet surface. Though, it might be possible for truck tires (worst case scenario).

3. Aug 14, 2014

### Kozy

The problem is that simply taking the tyre friction as the limiting factor is so far wide of the mark that you're never going to get any kind of accuracy out of it.

You could have μ=1.5, but if you're only operating the back brakes you might only be using 30% of that available grip, so your calculation would be totally wrong.

Essentially, you need to figure out the total deployable brake force, that is how much force can be transmitted to the ground before one or more of the tires locks up. To know that, you need to know the bias which in turn means you know everything about the brake system, and I do mean everything. This ranges from the sizing of the brake components, to the specific pressure ratios for the proportioning valve. For most cars, this information is simply not available.

If you've got that all that, and can figure out exactly how much brake force the car can generate, you then need to figure out how much it can generate before the front tyres lock up. For this you must know the vehicles weight, weight distribution and CGz, as well as the μ of the tyres.

Then it's a simple calculation*, you simply solve the following:

Front Grip - Front Brake Force = 0

At that point, your braking is maximised for that set of parameters, your brake force can then be used in the original calculation for stopping distance and you will have an accurate figure.

*This is not a simple calculation in the slightest. I've done it, it ain't pretty.

4. Aug 14, 2014

### Staff: Mentor

Welcome to PF!
Typically when the question is asked and the test is done, the intent is that the tires are utilizing as close to 100% of the available grip as possible. While I agree that that's probably the largest source of error, it isn't going to be 70% off. Maybe 10 or 20%.

But while your OP said "measure" it appears you are looking to calculate. So you get to pick what you want to calculate. If you want the theoretical minimum braking distance, assume 100%.
The brakes on any car are significantly stronger than the tires in most performance regimes in most cars, so the assumption that they can utilize 100% of the braking force is generally valid.

Last edited: Aug 14, 2014
5. Aug 14, 2014

### Kozy

T'was but an exteme example to illustrate the weakness is using μ as the only input.

He said measure, but the question itself appeared to be centered around calculating an accurate figure. Also, you cannot assume 100% if you want accuracy, it's not representative of real life conditions.

Very few cars use 100% of the grip available, it would be a litigation nightmare if they did. All cars come with front bias, and as such most are only using between 70-90% of the available grip.

I'll admit, a technological masterpiece such as the GTR will probably be using 99.5% of the available grip by monitoring each wheel and keeping it right on the limit, so in this particular example, it should be pretty accurate, but it's certainly well wide of the mark on older vehicles.

Last edited: Aug 14, 2014
6. Aug 14, 2014

### Staff: Mentor

Sorry, I got my posters mixed-up and thought you were the OP.

7. Aug 14, 2014

### Kozy

Ah OK. :)

As an example of what I mean though (as I've actually done the aforementioned 'simple calculation'), all the required information, including prop valve pressure ratios, is available for the MX5/Miata.

The standard brake setup when used with a μ=0.9 tyre has a significant front bias, and the stopping distance is in fact limited to 0.8G. As such, the standard brake distance is calculated at 47m from 60mph. Adjust the bias to achieve 100% usage of the available grip, and the stopping distance drops to 39m.

In both instances, the grip is the same at 0.9, but there's a big difference in the accuracy of the result.

Last edited: Aug 14, 2014
8. Aug 14, 2014

### rcgldr

Some antilock braking systems control each wheel independently, so there's no bias other than what is needed to deal with the weight shift during braking.

Braking distance for 60 mph to 0 mph can be under 100 feet for some sports cars with high performance tires, but typically it ranges 120 feet to 140 feet which correspond to .85 g to 1.0 g deceleration with a good driver. The .75 figure used in some tables is a conservative value, perhaps taking into account tires and/or roads in poor conditions.

9. Aug 14, 2014

### Kozy

Yes I've got EBD in my Accord, and very good it is too.

It actually has a set up designed to make use of it too, in that the rear brakes are oversized, meaning the EBD controls rear pressure from relatively low levels of braking.

A lot of cars still use a standard mechanical prop valve arrangement even with the introduction of EBD, as such the rear brakes are still relatively weak and the EBD control does not kick in until pretty high levels of brake pressure are applied. Many are actually so weak that even with no pressure reduction, they still lock front first!

10. Aug 14, 2014

### jack action

The interest in using the kind of basic equation shown in the OP is that it represents the shortest braking distance possible. Of course, the brake system can be «under-designed», intentionally or not, and get longer braking distances. But with well designed brake systems (and of course in good working condition), you are not that far off, because it would be insane to not want to use the maximum friction force you can get from your tires.

The kind of equation in the OP is to braking performance what the Carnot cycle is to engine efficiency (which is a very simple equation as well): If someone claims that his engine gets a higher efficiency than the one from the Carnot cycle, then you know someone is lying or a mistake has been done somewhere.

That is the reason why someone should question the presented 32.75 m distance. It had some very special tires or someone wasn't truly driving 100 km/h when the brakes were applied.

11. Aug 14, 2014

### Kozy

When you put it like that, yes it makes a lot of sense!

I actually don't doubt the stopping distance of the GTR, but I expect it will be down to tyres with a μ≥1.

12. Aug 14, 2014

### olivermsun

I wouldn't doubt the number much at all. They are very much in line with published instrumented tests, which typically use accelerometer + GP, for the GT-R and other similar-class vehicles.

The tires are very special—they custom-designed for the GT-R, which is a \$100k super car that has achieved some of the fastest lap times at the Nürburgring ever recorded for a production car.

13. Aug 14, 2014

### jack action

A little search and I found the tires in question. Very special indeed, with a treadwear of 140 (not 240), which gives out an estimated friction coefficient of 1.07 ±10%. With a possible 1.177 friction coefficient, you get a 33.4 m braking distance with the OP equation, a merely 2% error over the 32.75 m.

14. Aug 14, 2014

### rcgldr

That's 107.45 feet. Link to to article:

22_cars_that_stop_in_less_than_100_feet

Some sports cars are sold with track oriented tires as their primary or optional tire. Take a look at the tires listed in the article about the cars that stop in less than 100 feet.