GT4 and Brakes

[Uncle Harry]

But doesnt adding mass to the car affect the acceleration properties of the trye in that the radius of the trye will decrease and rotation acceleration is dependent on radius?
Also the contact patch will be larger which will change the tryes grip properties.
In the real world if I want to stop my car when it is fully loaded as compared to lightly loaded It takes longer to stop when I apply equal brake pedal pressure.
Also adding weight changes the C of G and thus the weight transfer you talk about so the net result is that added weight will effect braking performance.
A fully loaded 40 tonne truck will never stop as quickly as my 1000 kg car.
Also a fully loaded wheel barrow takes a lot more effort to stop than a unloaded one. This example does have a air filled tire!
I agree fitting better tires is the first step because they can handle greater acceleration forces, but it is the brake system that adds the acceleration force (deceleration in this case) to the tire. If the brakes can not slow the tire down at its maximum rate then all is wasted.

 

[Crayola]

Everyone seems to misunderstand this: If 2 cars were exactly the same except one was heavier BUT still applied the same downward loading to the front wheels, then and only then would the lighter car stop considerably quicker, because F=ma. Everyone seems to be getting confused though, there are ALOT of forces to take into account in working out a cars deceleration. The most important is friction, because this is what is doing the majority of the work.

The thing with the friction however and I stress that this single fact is the most important thing you could learn in reference to braking is: F = μmg. I'll explain what this means:
F, the force due to friction is equal to mass multiplied by gravitational force multiplied by the coefficient of grip. This means that the cars friction is DIRECTLY proportional to its tire compund and weight transfer.
People are right in saying that the force required to stop the car is greater when its heavier, but as the mass increases the friction does too. Increasing the coefficient of grip or increasing the weight transfer will always result in better braking where the mass sort of cancels itself out.

 

[MR_GT]

The thing with the friction however and I stress that this single fact is the most important thing you could learn in reference to braking is: F = μmg. I'll explain what this means:
F, the force due to friction is equal to mass multiplied by gravitational force multiplied by the coefficient of grip. This means that the cars friction is DIRECTLY proportional to its tire compund and weight transfer.
People are right in saying that the force required to stop the car is greater when its heavier, but as the mass increases the friction does too. Increasing the coefficient of grip or increasing the weight transfer will always result in better braking where the mass sort of cancels itself out.

If so...a heavier car adding more mass, which adds more friction to the tires would make for faster cornering as well.

Tire grip controls stopping, but it controls cornering too.

 

[Crayola]

If so...a heavier car adding more mass, which adds more friction to the tires would make for faster cornering as well.

Tire grip controls stopping, but it controls cornering too.

It'd increase friction but the gravitational weight wouldnt increase unless there was a lot of weight transfer. Yes a heavier car would have more grip, but you need that grip to corner in a heavy car. My whole point was the mass isnt as important as the weight transfer. If a heavier car can reach the maximum potential of the tires without overloading them and a lighter car cant, then the heavier car may well turn in better, cornering overall depends much more on the suspension than braking does.

 

[lazydog]

If so...a heavier car adding more mass, which adds more friction to the tires would make for faster cornering as well.

No, because the centrifugal forces trying to "pull" the car to the outside would increase too.

hey i really dont mean to be rude.. but if you cannot tell.. that having a pillion on the back of your bike increases breaking distance ...

The only thing I know is that I have to pull harder on the brakes when someones riding with me. I never went off the bike and actually measured the braking distance with and with someone on the back, therefore I can not say for sure if there's a difference.

That's the reason why I didn't comment your other examples with trucks, forumla one cars with empty fuel tanks and so on: I don't know of these and have no exact data availabe to make comparisons, so unfortunately can not say anything about it.

Back to the "braking issue" Crayola gave a very good summary / another approach:

Everyone seems to misunderstand this: If 2 cars were exactly the same except one was heavier BUT still applied the same downward loading to the front wheels, then and only then would the lighter car stop considerably quicker, because F=ma. Everyone seems to be getting confused though, there are ALOT of forces to take into account in working out a cars deceleration. The most important is friction, because this is what is doing the majority of the work.

The thing with the friction however and I stress that this single fact is the most important thing you could learn in reference to braking is: F = μmg. I'll explain what this means:
F, the force due to friction is equal to mass multiplied by gravitational force multiplied by the coefficient of grip. This means that the cars friction is DIRECTLY proportional to its tire compund and weight transfer.
People are right in saying that the force required to stop the car is greater when its heavier, but as the mass increases the friction does too. Increasing the coefficient of grip or increasing the weight transfer will always result in better braking where the mass sort of cancels itself out.

Or, if we look at your initial formula F = m * a , then its the tires which limit the maximum possible a, but the stopping force F can be increased with increasing mass m, and that's what modern braking system always can do.

 

[Crayola]

That doesnt mean jack though, because the force is provided by the tires, not by the mass. To make this simpler you need to rearange the formula: f/,=a, by the way a is infact negative acceleration or acceleration. increasing the mass will not increase the force, it just means you need more force to get the same (negative) acceleration.

Edit: Infact I read your post again and I see what your saying but you're not quite right. You cant increase mass through a system, mass is the amount of stuff you have. If you brake hard the weight (force due to mass*gravity) will increase at the front and you will gain more grip which equats to more force and a higher rate of deceleration.

This is starting to get overly complicated, Ill try my best to make a quick summary:
to find a cars deceleration, we use the formula F/m=a. Where F is the force of friction being applied by the brakes. Now to find this we use another formula F=umw. Where U is the "grip multiplier" and w is weight. Since in one formula we are multiplying by the mass, and in the other we are deviding, any increase in mass cancels out. This can be proven mathematically. Therefore the 2 most important factors in stopping are the "grip multplier", this is the tire compound, temperature, pressure and everything and the weight transfer of the car.

 

[mudia]

Or, if we look at your initial formula F = m * a , then its the tires which limit the maximum possible a, but the stopping force F can be increased with increasing mass m, and that's what modern braking system always can do.

I dont get what you mean

f = m * a is the same as a = f/m ie a is dependant on 2 variable f and m ... so changing either of those will affect a.

now with regards to braking ... IN THEORY .. weight should not matter... this is because as the car gets hevier the f (ie the braking force produced by the tyre) increases .. becasue the retardive forece F = mu * mass where mu is coefficient of friction ... so theoreticaly as mass increases F increases... this is essentially what that article is saying.

however this is incomplete for the real world... this is why i keep asking if you believe two cars which are identical in everyway except one is many times the weight of the other ( assuming strong enough brakes) the formulas above do not hold exactly true... and so the heavier the car the longer it will take to stop. ...

i dont know the physics behind it ... maybe coefficent of friction is changed by weight and or tyres just behave totally different when supporting a diffenernt amount of mass.

as i said it can be seen all around us.... i dont know why you would choose to ignore your observations based on incomplete physics

 

[Crayola]

I dont get what you mean

f = m * a is the same as a = f/m ie a is dependant on 2 variable f and m ... so changing either of those will affect a.

now with regards to braking ... IN THEORY .. weight should not matter... this is because as the car gets hevier the f (ie the braking force produced by the tyre) increases .. becasue the retardive forece F = mu * mass where mu is coefficient of friction ... so theoreticaly as mass increases F increases... this is essentially what that article is saying.

however this is incomplete for the real world... this is why i keep asking if you believe two cars which are identical in everyway except one is many times the weight of the other ( assuming strong enough brakes) the formulas above do not hold exactly true... and so the heavier the car the longer it will take to stop. ...

i dont know the physics behind it ... maybe coefficent of friction is changed by weight and or tyres just behave totally different when supporting a diffenernt amount of mass.

as i said it can be seen all around us.... i dont know why you would choose to ignore your observations based on incomplete physics

What your saying is not quite complete, to get the same a , when increasing mass you need a larger force; however the force is automatically greater because of the friction. This is the whole reason mass doesnt effect stopping distances and it is reliant on the coefficient of grip and the weigh transfer.


Mathematically, assuming the coefficient of grip remains constant:

(1) F/m=a
(2) F=umw

(2)-->(1)
umw/m=a
uw=a
As you can see, the only way the mass relates to the stopping distance is because as mass increase the weight does too. But there is no direct connection unlike with the coefficient of grip.

 

[mudia]

What your saying is not quite complete, to get the same a , when increasing mass you need a larger force; however the force is automatically greater because of the friction. This is the whole reason mass doesnt effect stopping distances and it is reliant on the coefficient of grip and the weigh transfer.


Mathematically, assuming the coefficient of grip remains constant:

(1) F/m=a
(2) F=umw

(2)-->(1)
umw/m=a
uw=a
As you can see, the only way the mass relates to the stopping distance is because as mass increase the weight does too. But there is no direct connection unlike with the coefficient of grip.

CRAYOLA... first you have merely just repeated what i wrote... secondly ...
i am asking everyone not to make assumptions (ie mu staying the same).. and not to use theoretical mechanics ...

you are stating that stopping distance is independent of mass... but i ask you will a fully loaded truck which has loads of contact with the road.. be able to stop first than my r1 bike .. which has contact patch that is smaller than a credit card...

no actually i will rephrase... WHY will my bike with so much less contact with the road stop in a shorter distance... .. surely can see my point ... the truck has SO MUCH MORE contact with the road but cannot stop in the same distance...

surely there must be a reason that these simple formulae do not explain ...

the effect of mass on stopping distance is something we observe... to dismiss it because we do not understand the complete picture just does not make sense!!

 

[Crayola]

Its all about weight transfer, thats my whole point. How mass relates to stoppings distance is that the weight force is equal to the mass, multiplied by the g forces acting downward.

 

[mudia]

Its all about weight transfer, thats my whole point. How mass relates to stoppings distance is that the weight force is equal to the mass, multiplied by the g forces acting downward.

wait a second... my whole argument .. has been that the weight of a vehicle plays a part in the stopping distance of that vehicle... is with all things being equal a heavier car (with same wt distribution) will take longer to stop... This is in the real world ...

are you disagreeing with this ?

 

[Crayola]

A heavier car with identical transfer characteristics will stop in the same distance in a straight line. You seem to be confusing weight with mass though.

 

[lazydog]

wait a second... my whole argument .. has been that the weight of a vehicle plays a part in the stopping distance of that vehicle... is with all things being equal a heavier car (with same wt distribution) will take longer to stop... This is in the real world ...

are you disagreeing with this ?

Yes, because it's not generally true. Take a look at this example:

Stopping times from 60-0:

Range Rover Vogue 2.570 kg: 2,7s
Lotus Elise 905 kg: 3,0s

How can this be? The Rover weighs nearly 3 times as much!

That's what this whole thread says, and it is explained several times in detail: The maximum achievable deceleration is limited by the tires of a vehicle and is independent of its mass.

//edit: typos

 

[mudia]

A heavier car with identical transfer characteristics will stop in the same distance in a straight line. You seem to be confusing weight with mass though.

firstly i disagree with what you just wrote ... simply because this is not observed in real life... like i said i understand the maths behind what you are saying but i am suggesting that it is incomplete in the real world...

can you please explain to me why a fully loaded truck will not stop in the same distance as a caterham or radical ??? in fact if mass is not a factor .. the truck should theoretically stop in a MUCH SHORT distance since it has so much more contact with the road...

wt = m * g... from now on i will use mass (so as not to confuse)

 

[mudia]

Yes, because it's not generally true. Take a look at this example:

Stopping times from 60-0:

Range Rover Vogue 2.570 kg: 2,7s
Lotus Elise 905 kg: 3,0s

How can this be? The Rover weighs nearly 3 times as much!

That's what this whole thread says, and it is explained several times in detail: The maximum achievable deceleration is limited by the tires of a vehicle and is independent of its mass.

//edit: typos

lazydog ... there might be several reasons.. such as the range has more contact with the road.. and has a superior braking system which plays a bigger part at slow speeds .. the tyres might even be better .. who knows!

but you can be sure .. that if you stopped both cars from 100 mph .. the effect of mass will be very pronounced.. and if you stop it from 200 mph .. even more so ... and the elise will stop in a shorter distance !!!!

further to that ..lets not confuse stopping times and stopping distances please... because a car with superior stopping at high speeds might take longer to stop than one with superior stopping at low speeds.... but still stop in a shorter distance

 

[MR_GT]

Range Rover and Elise have different tires so you can't really prove the theory. Also, the comparison is made from 60-0 which is not fast at all...which means the load on the tires is minimal during braking, so the Elise probably didn't even get full load on the tires.

If weight has nothing to do with braking distance, than it has nothing to do with cornering, or acceleration. Racing teams are wasting a lot of money on exotic light weight materials.

 

[mudia]

Range Rover and Elise have different tires so you can't really prove the theory. Also, the comparison is made from 60-0 which is not fast at all...which means the load on the tires is minimal during braking, so the Elise probably didn't even get full load on the tires.

If weight has nothing to do with braking distance, than it has nothing to do with cornering, or acceleration. Racing teams are wasting a lot of money on exotic light weight materials.

well said MR_ GT.. however it is important to note that the thoery behind what these guys have been saying is not wrong... it just seems to be incomplete !

a good point of comparison ... is if you have 2 cars one a salon .. and the other a station wagon (estate) of the same model... you will see that they have similar 30 - 0 and 60-0 ... but stop them from 120-0 and there is a huge differece... i appreciate that weight shift plays a part .. however it should not cause such a big difference...

it would be interesting if someone could explain this phenomenon

 

[Crayola]

firstly i disagree with what you just wrote ... simply because this is not observed in real life... like i said i understand the maths behind what you are saying but i am suggesting that it is incomplete in the real world...

can you please explain to me why a fully loaded truck will not stop in the same distance as a caterham or radical ??? in fact if mass is not a factor .. the truck should theoretically stop in a MUCH SHORT distance since it has so much more contact with the road...

wt = m * g... from now on i will use mass (so as not to confuse)

It is observed in rela life though, A truck simply doesnt transfer its weight effectivly enough, and doenst have tires that are as grippy as racing tires. Its independant of the mass, not the weight loading, and a truck would surely overload the tires because its momentum is huge, thats another problem altogether, trucks brakes arent as good as cars for a start. Up to a point the mass of a car wont make a differnce.

 

[Crayola]

Range Rover and Elise have different tires so you can't really prove the theory. Also, the comparison is made from 60-0 which is not fast at all...which means the load on the tires is minimal during braking, so the Elise probably didn't even get full load on the tires.

If weight has nothing to do with braking distance, than it has nothing to do with cornering, or acceleration. Racing teams are wasting a lot of money on exotic light weight materials.

You guys are confusing weight and mass, it has all to do with weight. Not so much to do with mass. In a striaght line (cornering is totally different, assume the forces are symetrical) up to a maximum loading of the tires, the heavier car will brake at the same speed as the lighter one.

 

[Crayola]

well said MR_ GT.. however it is important to note that the thoery behind what these guys have been saying is not wrong... it just seems to be incomplete !

a good point of comparison ... is if you have 2 cars one a salon .. and the other a station wagon (estate) of the same model... you will see that they have similar 30 - 0 and 60-0 ... but stop them from 120-0 and there is a huge differece... i appreciate that weight shift plays a part .. however it should not cause such a big difference...

it would be interesting if someone could explain this phenomenon

There difference in mass in not as great as the difference between the weight balance. In a station wagon more of the weight is at the back usually which is not good for braking. Also station wagons dont usually have the same sports packages and such as regular sedans. If you tested the audi S4 and RS4 im sure their braking distances wouldn't be that great at all.

 

[mudia]

You guys are confusing weight and mass, it has all to do with weight. Not so much to do with mass. In a striaght line (cornering is totally different, assume the forces are symetrical) up to a maximum loading of the tires, the heavier car will brake at the same speed as the lighter one.

thanks but i know the difference between wt and m

 

[mudia]

It is observed in rela life though, A truck simply doesnt transfer its weight effectivly enough, and doenst have tires that are as grippy as racing tires. Its independant of the mass, not the weight loading, and a truck would surely overload the tires because its momentum is huge, thats another problem altogether, trucks brakes arent as good as cars for a start. Up to a point the mass of a car wont make a differnce.

firstly the brakes on truck a very good.. and are sufficient to lock the wheels.. therefore giving maximum braking....(actually max braking is achieved when the wheel is spinning about 5% less than cars speed)

secondly are you trying to say that the tyres on a caterham (have you very driven one ?? or looked at thme) are so much stickier than a trucks tyres that they would overcome such a defcit in terms of contact area....

i am asking if a truck has 10 x the contact area... ok lets say 8 due to better tyres on the car... shouldnt it stop 8 x as quickly... but why is it the car stops SOOOOOOOOO much faster.. when both are braking as best they can... you cant tell me that that is simply down to wt transfer

 

[mudia]

There difference in mass in not as great as the difference between the weight balance. In a station wagon more of the weight is at the back usually which is not good for braking. Also station wagons dont usually have the same sports packages and such as regular sedans. If you tested the audi S4 and RS4 im sure their braking distances wouldn't be that great at all.

the two cars in question were AMG E-Classes.. same kit on both.. and i appreciate the shift in mass to the back is a hinderence.. however my point is the cars had similar distances from 60 -0 30-0 but were far apart from 120-0 .. which suggests that comparing a range to an elise 0 does not really say much because there is not much difference between cars at low speed

while we are on this subject ... with regards to wt transfer .. are you trying to say that a range rover has a better wt transfer for braking than does an elise...?? I think not... yet the range stopped quicker... why is that? well obviously there must be other factors than mass of the car and wt transfer that slow the car down...

 

[lazydog]

If weight has nothing to do with braking distance, than it has nothing to do with cornering, or acceleration.

Noone said "weight has nothing to do with braking distance".

Please read and understand the technical details given, especially at the beginning of this thread and in this article: http://www.miata.net/sport/Physics/07-Circle.html


Instead we say: The tires of a vehicle give the upper limit for the maximum possible deceleration, and this limit is independent of total vehicle weight. (There are second order effects like weight transfer, temperature and so on, but let's put these aside for a moment).

To quote from the article:

The maximum acceleration a tire can take is , a constant, independent of the mass of the car! While the maximum force a tire can take depends very much on the current vertical load or weight on the tire, the acceleration of that tire does not depend on the current weight. If a tire can take one before sliding, it can take it on a lightweight car as well as on a heavy car[..]

This means, for instance, if you are not satisfied with the braking distance of your modern sports car, it would NOT help to

a) equip "stronger" brakes
b) reduce vehicle weight

What you could do is

* get other (better, gripier) tires
* try to increase braking performance by looking at other things, like weight transfer, center of gravity and so on.

"High performance" brakes do not reduce braking distance, instead they give better "feel" when braking, reduce fade and operate better at higher temperatures such when making heavy use of the brakes (i.e. driving curved downhill streets for long distances).

 

[mudia]

Noone said "weight has nothing to do with braking distance".

Please read and understand the technical details given, especially at the beginning of this thread and in this article: http://www.miata.net/sport/Physics/07-Circle.html


Instead we say: The tires of a vehicle give the upper limit for the maximum possible deceleration, and this limit is independent of total vehicle weight. (There are second order effects like weight transfer, temperature and so on, but let's put these aside for a moment).

To quote from the article:



This means, for instance, if you are not satisfied with the braking distance of your modern sports car, it would NOT help to

a) equip "stronger" brakes
b) reduce vehicle weight

What you could do is

* get other (better, gripier) tires
* try to increase braking performance by looking at other things, like weight transfer, center of gravity and so on.

"High performance" brakes do not reduce braking distance, instead they give better "feel" when braking, reduce fade and operate better at higher temperatures such when making heavy use of the brakes (i.e. driving curved downhill streets for long distances).

i did not say that brakes are what determin stopping distance .. and i have agreed that the stopping power you get is from your tire... you guys repeatedly quote theoretical physics.. to make your point.. i am trying to say that IN THE REAL WORLD there is an effect on stopping distance that comes about as result of increased mass... this may be due to tyres actually behaving differently when supporting more wt... or problems with heat in the tyres.. i dont know.. however it is something that you see.. that the physics that you have shown us does not explain... because it is obvioulsy not considering the characteristics of the tyre when it has to deal with reducing a certain amount of speed.. or support a certain amount of weight...


and so i will disagree with you ... if you have a chance of reducing the mass of a car .. it will improve your braking... this effect will be seen the greater the speed you are stopping from.....

NB . i am not saying that this is a linear relationship.. ie halving the mass wont stop the car in half the distance... but it will improve your braking distance

 

[lazydog]

and so i will disagree with you ... if you have a chance of reducing the mass of a car .. it will improve your braking...

Yes, it can improve braking performance, yet not because of the reduced mass, but because if you reduce weight, usually some second order effect kicks in, like the center of gravity lowers, weight transfer changes and so on.

 

[Polyphony001]

Speaking of braking, have you used the lemans quattro? Ouch, where are the brakes?

 

[mudia]

OK I just thought of the explanation for what I am saying

has anyone also thought about the other force that is acting to slow a car down...

namely DRAG.... perhaps this explains the difference in breaking for lighter cars ... since whereas friction in tyres is dependent on wt ... drag is not .. so if you have two identical cars with the exception that one is comprised of materials that are 10 x as dense as the other....

if we foget about the brakes for a second .. and we assume both cars are doing 200mph .. and then they both just come off the throttle and disengage the clutch ... wont the lighter car slow down faster ??? sure it will...

so there IS an added benifit of having a lighter car for reducing stopping distances

lazyboy tell me what you think of this

 

[Scaff]

Range Rover and Elise have different tires so you can't really prove the theory. Also, the comparison is made from 60-0 which is not fast at all...which means the load on the tires is minimal during braking, so the Elise probably didn't even get full load on the tires.

If weight has nothing to do with braking distance, than it has nothing to do with cornering, or acceleration. Racing teams are wasting a lot of money on exotic light weight materials.

Any modern braking system is more than capable of almost immediatly applying full braking force, if the driver wishes. The quoted figures and shortest stoping times, so would be produced using maximum threshold braking, or the highest level of load the brakes can transfer to the tyres without slip. Transfer full braking load to tyres and they will slip.

You second paragraph demonstrates that you clearly have not read the material supply fully. While weight does not affect stopping distances (weight transfer does as a second order effect), it will affect acceleration if you lower a cars weight as you have just changed the cars power to weight ratio. Secondly the physics involved in cornering (and braking while cornering) are very different and covered by transient forces (see the physics of racing).

This is why I have stuck to discussing straight line braking only, once cornering values are introduced the forces involve decome far more complex.

 

[mudia]

Any modern braking system is more than capable of almost immediatly applying full braking force, if the driver wishes. The quoted figures and shortest stoping times, so would be produced using maximum threshold braking, or the highest level of load the brakes can transfer to the tyres without slip. Transfer full braking load to tyres and they will slip.

You second paragraph demonstrates that you clearly have not read the material supply fully. While weight does not affect stopping distances (weight transfer does as a second order effect), it will affect acceleration if you lower a cars weight as you have just changed the cars power to weight ratio. Secondly the physics involved in cornering (and braking while cornering) are very different and covered by transient forces (see the physics of racing).

This is why I have stuck to discussing straight line braking only, once cornering values are introduced the forces involve decome far more complex.

scaff are you saying that the wt transfer in range is better than an elise for braking purposes??