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I am curious to find the braking distance for a car on a road.

In attempting to find this out, I found that the braking distance for a car (on a flat road) is $$ d = \frac{v^2}{2\mu g} $$ where $\mu$ is the coefficient of friction between the road and tires (CRF), $g$ is gravity, and $d$ is the distance traveled. However, this book mentioned that $\mu$ increases as velocity increases. Right now, I'm having trouble finding said coefficient of friction.

This transportation engineering site gave several examples of CRFs at various speeds on a rainy road, but I'd like to understand if they needed to empirically measure it.

(EDIT: Of course $\mu$ increases as $v$ does; there wouldn't be a table if it didn't.)

In short, my question is this: Is there a general method of calculating $\mu$, given that the car is on a 0% grade, wet, asphalt-covered road? Or must $\mu$ be measured/calculated using empirical data?

(Note: not really a homework question; if it was, I'd just guesstimate the answer based on the data from the second link. I just want to understand the data better, if possible, as opposed to just having the value.)

PeterJ
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éclairevoyant
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3 Answers3

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I think there is not one general method to calculate this just because its too complex. If there was one F1 engineers would have an easier job.

Nevertheless you can guess $\mu$'s dependence from the data, as you pointed out, if you plot it. I just did it here

http://genflux.chartle.net/embed?index=47502

and seems an exponential decay which, makes sense. As you increase the velocity your tires will lack of adherence due to less contact. So my first guess is that you can write

$$\mu = e^{-b |\vec v|} + a \quad ; \quad a > 0, \quad b>0 $$

where $a$ and $b$ will depend on the data to be fit (your tires and conditions).

stringparser
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Using the answer provided by fénix, I came up with a sufficiently good estimate of the data. CRF values

éclairevoyant
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  • Looking good ;) You did it by hand or using software to fit it? PS. There is a mistake on the key: shouldn't it be $e^{-0.30... v}$ instead of $v^{-0.30}$? – stringparser Nov 08 '11 at 20:43
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    I used this online tool to figure out the best approximation, and it seems that a power regression ($av^b$) seems to fit better than an exponential ($e^{av}+b$) one. Still, your guess helped me get to the answer, so thank you for that :) – éclairevoyant Nov 09 '11 at 04:15
  • I can not understand why increase in velocity will lead to lesser contact. Is it because of lift or something else? – user43794 Feb 29 '16 at 14:36
  • @user43794 This answer might be helpful in explaining why: http://physics.stackexchange.com/a/48536/5892 – éclairevoyant Feb 29 '16 at 20:08
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There is a formula in German theory booklet of driving or better to say driver's instruction:
Braking Distance in 1 second= car speed/10 x car speed/10

Example: When you are driving with 100 km/h and you see a barrier and deciding to stop (Reaction distance: speed/10 x 3) in 1 second, your braking distance will be 100 m and your stopping distance will be calculated in this way:

Reaction distance + braking distance, so in this example we will have:

30 m + 100 m = 130 m

Waffle's Crazy Peanut
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Arash
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  • I think that's more of an estimate than an actual formula. Besides, if it's in a driving instruction book, it would probably have to account for more than just one or two weather conditions. Moreover, there's no mention of $\mu$ in there (although it could be derived, of course, from something like $\mu = \frac{v^2}{2dg}$). – éclairevoyant May 25 '13 at 15:15