# Car's braking distance formula

Hello everyone !
I'm working on my highway code and my book give me the approximate formula and the real formula of the braking distance. Here's the real formula according to the book :

$$Bd = \frac{V^2}{254 \times f}$$

With :
V : the velocity

But I have on question. Where does this 254 come from ?

Regards

Hugo

Doc Al
Mentor
That factor is a units conversion when V is measured in km/h instead of the usual m/s.

That factor is a units conversion when V is measured in km/h instead of the usual m/s.

It's strange, usually we multiply (or divide) by 3.6 when we want to convert m/s in km/h (km/h in m/s).

Doc Al
Mentor
It's strange, usually we multiply (or divide) by 3.6 when we want to convert m/s in km/h (km/h in m/s).
Right. That 254 is more than just a units conversion, it also contains part of the standard formula: $$\frac{V^2}{2 g \times f}$$

Right. That 254 is more than just a units conversion, it also contains part of the standard formula: $$\frac{V^2}{2 g \times f}$$
Sorry but I don't find the right calculus. 2*9.8 = 19.6 So I don't understand how you find this 254.

Doc Al
Mentor
Sorry but I don't find the right calculus. 2*9.8 = 19.6 So I don't understand how you find this 254.
2*9.8 takes care of the 2g factor. Now include the conversion from (km/h)^2 to (m/s)^2. That requires multiplying by (1/3.6)^2, which gives you a constant of 2*9.8*3.6^2 = 254 in the denominator.

2*9.8 takes care of the 2g factor. Now include the conversion from (km/h)^2 to (m/s)^2. That requires multiplying by (1/3.6)^2, which gives you a constant of 2*9.8*3.6^2 = 254 in the denominator.

AH I see. Highway code should tells us the unit of V. Thank you ! :)