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Intresting braking/speeding facts

chriswt

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Picked up Classics Momthly a week or so ago and came across a very intresting article comparing a Ford Anglia with a new S-Class.

As a side note to the article was the science behind braking distance and how that cars don't slow at a constant rate.

It stated that if you were travelling at 30mph and came to a stop just behind a staionary car in front and then imaged that you'd been travelling at 2mph more (32mph) you would have actually hit the car at 11mph. If you'd been doing 40mph you would have hit the car at 26mph.

...and this is the intresting one

A police Omega was driven in test conditions to 70mph and then braked to a stop. The point at which the Omega came to a stop was marked.

The Omega was then driven to 100mph and went over the marked spot at 71mph.

I know its all basic logic but when facts like these are put in simple terms like this it makes you think when you're next batting along on the motorway.

Sorry to be so dull, it's meant as a speeding lecture but I found it intresting. :D
 
Actually, it's not basic logic for many people who don't understand the relation between kinetic energy and velocity, so it's well worth pointing it out.
 
And the potential energy should you actually hit someone at the differing speeds!!
 
I've seen this quoted before somewhere.

If what they are saying is that stopping from 70mph can be done in less distance than slowing from 100-70mph then I'd like to see this proved under scientific conditions using a modern vehicle before I am convinced.
 
Kinetic energy = 1/2mv2 so at 100mph the kinetic energy is exactly twice that at 70.7mph, so if the braking is limited by the grip of the tyre on the road and that is constant the maths does work.
 
The emergency braking is governed by tyre / road friction - the brake itself doesn't have much to do with it.

Assuming a co-efficient of friction of 0.7 (not too far different from typical road conditions)

>> u=70*1609.3/3600;
>> s=(u^2)/(2*0.7*9.81)

s =

71.2965

Gives a braking distance of 71.3 metres from 70 (this compares with the 75m quoted in the highway code)

>> u_100=100*1609.3/3600;
>> v=sqrt((u_100^2)-2*0.7*9.81*s)*3600/1609.3

v =

71.4143

This confirms the test data - the car braking from 100mph would be travelling at over 71mph at the point where the car travelling at 70 had stopped.
 
The emergency braking is governed by tyre / road friction - the brake itself doesn't have much to do with it.

Assuming a co-efficient of friction of 0.7 (not too far different from typical road conditions)

>> u=70*1609.3/3600;
>> s=(u^2)/(2*0.7*9.81)

s =

71.2965

Gives a braking distance of 71.3 metres from 70 (this compares with the 75m quoted in the highway code)

>> u_100=100*1609.3/3600;
>> v=sqrt((u_100^2)-2*0.7*9.81*s)*3600/1609.3

v =

71.4143

This confirms the test data - the car braking from 100mph would be travelling at over 71mph at the point where the car travelling at 70 had stopped.

:dk::dk::dk:
 
One way to re-phrase my post would be to say that there are some things which need to be physically tested before you can understand what's going on - emergency braking of road cars isn't one of them - you don't need to waste your tyres, petrol, and brake linings where a quick calc will tell you what's important.
 

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