Stopping distance increased by how much? with a mass increase

1. Apr 19, 2013

stephenn

Hello have a question and am interested to find a simple equation to include mass as a variable regarding stopping distances of a car

In the UK highway code..... a braking distance = 14m for a car at 30 mph (48 kph)

my questions are..... if you load the car up fully (increase mass)..... then that distance will increase????
Q1: by how much ?
Q2: what's a simple equation ?

NB
1. [The car and the road conditions are the same for each stop]
2. [the original car is say 1000kg]
3. [the fully loaded car is say 1500kg]

thanks for time / thoughts
stephenn

2. Apr 19, 2013

Staff: Mentor

If the brakes are good enough, the distance should not depend on the mass (significantly). Friction between tires and road increases in the same rate as the required force increases.

3. Apr 19, 2013

sophiecentaur

To get a worthwhile answer to this question you need to specify the conditions better. If the limiting factor is the brakes themselves and we assume that contact with the road is not a problem then the brakes can produce a certain maximum braking force (F), irrespective of the load.
So the brakes will dissipate the Kinetic Energy of the vehicle in a distance d where KE = Fd
KE is proportional to Mass so that means the braking distance will be proportional to the mass of the vehicle.
This will be over a certain limited range of conditions, imposed by factors like brake pad temperature (if the load is much bigger than the brakes are designed for, they will need to dissipate more energy than expected and they can fail by over-heating.
If the limiting factor is contact with the road then, 'ideally' the braking force will be proportional to the vehicle mass - hence the braking distance will be constant. However, tyres on roads don't behave nicely, like laboratory equipment, so you can't rely on this. Very light loads may produce aquaplaning on a wet road where a heavy load, at the same speed, might squish the water away and the tyres would grip better.
It's a big area of study; talk to Formula One race teams.

4. Apr 20, 2013

stephenn

very interesting..... I've heard a story sample regarding friction:
if a full size ship and a small replica (same materials) were placed on a slipway..... then the slip way tilted sideways...... which one would start to slide first?
The answer was said to be.... at the same time.... because it was a matter of the co-efficiciency of friction (same for both).... and mass made no difference
....most interesting

anyway back to the car:
I'll make it more practical:

* If I got into a mini.... set off west at 30mph..... then EMERGENCY STOP..... I could stop the car in around 6 car lengths (car and road conditions = being all good)
* Then if I loaded it up with four more passengers, the dog, camping gear, food food food etc.... then did as said above.....
Would the car take longer to get stopped.... Yes/No?

5. Apr 20, 2013

stephenn

very interesting.... I understand the principle of aerodynamic downforces (on F1 especially) would increase contact forces of car on the planet surfaces.... and so as you mention would displace water whereas a lighter massed or downforced car may aquaplane.
Thanks for thoughts.

Anyway, back to the car question.... I've gone for a simplistic angle on my original car question.....
as here:

* If I got into a mini.... set off west at 30mph..... then EMERGENCY STOP..... I could stop the car in around 6 car lengths (car and road conditions = being all good)
* Then if I loaded it up with four more passengers, the dog, camping gear, food food food etc.... then did as said above.....
Would the car take longer to get stopped.... Yes/No?

cheers

6. Apr 20, 2013

sophiecentaur

Longer if you don't skid. Definitely longer with ABS.

7. Apr 20, 2013

rcgldr

Last edited: Apr 20, 2013
8. Apr 21, 2013

stephenn

for a fully loaded small car of 1100 kg
with a speed: 30mph 48kmh (13.33m/s)
and a uk highway code breaking distance of 14m

I have found and use these two equations to give:
1. acceleration (decelleration)
2. brake force
..... assuming brake force to be near to a maximum..... the 2nd equation shows the relation of mass and acceleration.... being more mass = less aceleration (for the same brake force)

1. v2 = u2 + 2as

0 = 13.332 + 28a
0 = 177.69 + 28a
28a = -177.69
a = -177.69 / 28
a = -6.35 m/s2

2. F = ma

F = 1100 x -6.35
F = -6985 N

working out the extra distances that would accompany a decreased acelleration is next..... need a good equation

9. Apr 22, 2013

sophiecentaur

I would hesitate to use Highway Code figures for your sources. They will have been chosen from statistical results of practical 'experiments'. They are are an excellent rule of thumb because the last thing you want to do is to be doing calculations at the time you actually want to apply your brakes and they are, of course, pessimistic / optimistic (see also the caveats attached to the figures).

I am reluctant to get involved with figures before the actual equations have been established (chicken and egg situation I realise) but you have to decide whether the braking is limited by the brakes or by the tyres in any particular case.
If it's down to tyres and road then it's anyone's guess and you have to rely on a massive number of measured results. If it's down to the brakes themselves then you can be a bit more theoretical about it (again, within practical limits). Just considering the brakes and, assuming they can produce a certain braking force then the car will be brought to a halt when the work done Fd is equal to the KE (mv2/2.
Fd = mv2/2
tells you that, d is proportional to m for any given speed of vehicle and available brake force - but I have written this before.
You will find it very hard to get proper figures about the performance of actual tyres from the manufacturers. Just like the Pharma companies, they will provide a combination of measurements and results that make their particular product look best for mileage, cornering, wet road performance etc. etc.. So, apart from talking to an F1 team, I think you are unlikely to come up with anything like a reliable theory about the tyre / road situation. (Of course, F1 tyres are nothing like road tyres, in any case - except being made of rubber.)

I see you have used the usual suvat formulae for your calculations and I assume your arithmetic is right. Putting it the way I have, above, cuts out one equation by just considering Energy but there are many ways of skinning a cat and the distance still will come out as being proportional to mass. (I would advise you to stick to algebra, rather than figures until right at the end.)

10. Apr 25, 2013

mrspeedybob

Stopping distance will be almost the same regardless of mass until you exceed the capabilities of the brakes and/or tires. After that there is no simple answer because it depends greatly on the specific design of the vehicle.

11. Apr 25, 2013

sophiecentaur

That's the point, isn't it? The 'capabilities' of the brakes depend on the force from your foot. It has to be assumed for this thought experiment that you are applying the (same) maximum force with your foot - it's an EMERGENCY So the braking distance will be proportional to the mass of the vehicle (as long as the tyres don't lose grip).

12. Apr 25, 2013

rcgldr

Using the wiki article on tire load sensitivity formula range, and choosing maximum braking force = maximum vertical force ^ (0.9), if you double the weight, maximum braking force increases by 2^.9 = 1.866, so coefficient of static friction has been reduced from 1.000 to 0.933 (1.866/2) when the weight is doubled.

The related formula for distance based on coefficient of static friction f:

a = g f (where g = gravitational acceleration constant)
d = 1/2 v^2 / a
d = 1/2 v^2 / (g f)

so if mass is doubled, then the braking distance increases by (1/0.933) = 1.0718 or 7.18 %.

That was using the lowest difference sensitivity factor. If it was the other extreme .7, then doubling the mass would increase braking distance by 1.231 or 23.1%.

Last edited: Apr 25, 2013
13. Apr 25, 2013

mrspeedybob

Lets say the co-efficient of friction of my tires is 1, just to make the math easy...
If my foot and my brakes are capable of supplying 3000 kg of stopping force and my car+me weighs 1500 kg then I can only apply 50% of my maximum stopping force before starting to slide.

If my car is loaded to 3000 kg then I should apply 100%.

So from 1500 to 3000 kg I am not applying the same force with my foot.

It also depends on the brake systems capacity to reject heat. As automotive brakes heat up they reach a temperature of maximum efficiency. As they rise past that temperature the coefficient of friction between the pads and rotors declines. This is called brake fade. In severe circumstances it can reduce the brakes maximum stopping force to a small fraction of it's normal maximum. So if my brakes from the example above are severely overheated I may only have 500 kg of available stopping force. How quickly the brakes heat up, how hot they can get before starting to fade, and how severely they fade are all factors that are going to depend on the specifics of the brake design.

14. Apr 25, 2013

rcgldr

I'm not aware of any car that can overheat the brakes from a single panic stop, assuming the weight increase is not absurdly large.

15. Apr 26, 2013

sophiecentaur

I think the Morris 1000 was the last one. They were tiny drum brakes all round, of course, which could fade at the drop of a hot.

16. Apr 26, 2013

mrspeedybob

Such as when a certain individual realizes that the moving truck they rented is too small and so they pack 1500 kg of stuff into a 500 kg trailer with no trailer brakes and hitch it to a vehicle rated to tow 250 kg, then procede to drive it 600 km up and down the mountains of West Virginia.

Not that I have any such personal experinece...

17. Apr 26, 2013

sophiecentaur

So you are saying that you have to maintain the force between tyres and road to a value below slipping value and that this is determined by the load on the tyres. That makes sense and that also applies to the condition for wheel spin - less likely when the wheel is loaded more. It also shows how ABS gives the best braking, even by backing off the force from the brakes.

But this only applies when the car's brakes are 'good'. If they are not capable of skidding the wheels (with very heavy loads, for instance) the braking distance is governed only by the force of your foot and would be proportional to mass.

18. Apr 26, 2013

rcgldr

But that is a combination of vehicles, not a car that's just heavily loaded. Say there's no increase in weight of the car, so that the braking force remains the same, and that the trailer with no brakes weighs as much as the car. In that case, same force, twice the mass, deceleration is halved, and the distance doubles (d_with_trailer = 1/2 v^2 / (1/2 a)).

19. Apr 26, 2013

mrspeedybob

True. I chose a bad example. Let me try again. I'll use the same 1500 kg car and 3000 kg brake system from my previous post.

Suppose I'm stopping the vehicle from 50 kph. about 145 Kj heat will be generated. Loading the same vehicle to 3000 kg and making the same stop will generate twice as much heat. If neither one of these scenerios make the brakes hot enough to fade, the stopping distance is the same (neglecting the effect of tire loading)

Now suppose I make 2 more stops from 200 kph. 16 times more heat will be generated. For the 1500 kg stop that may not be an issue, but it may cause a brake fade issue for the 3000 kg stop, especially since this is already using 100% of my braking capacity. In this scenerio the weight will significantly affect the stopping distance.

20. Apr 27, 2013

sophiecentaur

We live in a non-linear world. The best approach of Physics is often to identify the linear regimes and analyse them first. Your average brain doesn't usually work beyond that, without the help of Maths (algebra). We need to be sure that this thread has dealt with the linear bit before adding an unlimited amount of complication. Shifting the goal posts is fine as long as we take everyone with us.
Have we established the facts about the most basic situation, here, where brakes and tyres behave ideally?