GT4 and Brakes

[Scaff]

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

I'm not specifically refering to any car (as I do not have the data to hand), but as a general rule, mid-engined cars do have less weight transfer to the front wheels under braking, combined with a (again general) lower COG then for braking purposes mid-engined cars are not ideal.

The very low central COG for mid-engined cars benefits cornering and acceleration more than braking. Its one of the reasons why mid-engined cars are generally faster on the limit, but harder to control.

One thing I would add is that while a Range Rover will stop (to zero) quicker than an Elise, on the track the Elise would be able to carry more speed into a corner than the Range Rover, and as such would not need to brake as much as the Range Rover.

This is one of the central points of my original thread, while weight will not affect your overall stopping distance, it will change the handling characteristics of the car, if you can carry more speed into a corner then you need to brake less.

For example if our Range Rover and Elise both approach a corner at 100mph, and the elise can take the corner at 60mph and the Range Rover take it at 45mph. Then the Elise will be able to brake later and less than the Range Rover, despite the fact that the Range Rover can brake to zero in less time than the Elise.

You must remember the reason I started this thread was to dismiss the myth that certain cars had been incorrectly modeled by PD, when people were just underetsimating the correct braking for the car and tyre. When reducing weight did not work, they dismiss the car as one PD 'messed' up, when in fact this would not help them.

In fact if you take a look at the text in GT4 for weight reduction, you will find that it says that reducing weight will benefit braking characteristics, not reduce braking distance.

Hope this helps to clarify my standpoint on this.

 

[ferret-turbo]

dgoai lol

 

[mudia]

I'm not specifically refering to any car (as I do not have the data to hand), but as a general rule, mid-engined cars do have less weight transfer to the front wheels under braking, combined with a (again general) lower COG then for braking purposes mid-engined cars are not ideal.

The very low central COG for mid-engined cars benefits cornering and acceleration more than braking. Its one of the reasons why mid-engined cars are generally faster on the limit, but harder to control.

One thing I would add is that while a Range Rover will stop (to zero) quicker than an Elise, on the track the Elise would be able to carry more speed into a corner than the Range Rover, and as such would not need to brake as much as the Range Rover.

This is one of the central points of my original thread, while weight will not affect your overall stopping distance, it will change the handling characteristics of the car, if you can carry more speed into a corner then you need to brake less.

For example if our Range Rover and Elise both approach a corner at 100mph, and the elise can take the corner at 60mph and the Range Rover take it at 45mph. Then the Elise will be able to brake later and less than the Range Rover, despite the fact that the Range Rover can brake to zero in less time than the Elise.

You must remember the reason I started this thread was to dismiss the myth that certain cars had been incorrectly modeled by PD, when people were just underetsimating the correct braking for the car and tyre. When reducing weight did not work, they dismiss the car as one PD 'messed' up, when in fact this would not help them.

In fact if you take a look at the text in GT4 for weight reduction, you will find that it says that reducing weight will benefit braking characteristics, not reduce braking distance.

Hope this helps to clarify my standpoint on this.

scaff ok i appreciate the points you are making..

but what do you say regard the other forces that are acting on a car to slow it down .. (read above)...

with the retardive force being F

we have F = x + y.v + z.v^2

where x are the frictional forces in bearing etc and resistance caused by rotation of tyres which are independant of speed

y are the frictional forces in bearing etc and resistance caused by rotation of tyres which are dependant of speed

and z is the drag effect on the car...

as you can see from that equation the force grows very quickly with speed... and it is also independant of weight... so whereas (in theory) there is and increase in frictional force of the tyre due to increase in wt which offsets the effects of increased weight.. the effects of drag and internal friction remain the same for too identically shaped cars.

if this force is constant ... then the lighter car must be acclereated (negatively) quicker by it .. this goes without saying... theefore a lighter car will decelerate quicker from highspeeds .. and as a result will have a much shorter braking distance than a heavier car...

the fact is an elise will outbrake a range rover from 100- 0 every time. and the gap will increase with speed....

I doubt that the difference that one observes with braking distance in such cars is down to drag alone .. but that is at least something that i can think of...

 

[Scaff]

scaff ok i appreciate the points you are making..

but what do you say regard the other forces that are acting on a car to slow it down .. (read above)...

with the retardive force being F

we have F = x + y.v + z.v^2

where x are the frictional forces in bearing etc and resistance caused by rotation of tyres which are independant of speed

y are the frictional forces in bearing etc and resistance caused by rotation of tyres which are dependant of speed

and z is the drag effect on the car...

as you can see from that equation the force grows very quickly with speed... and it is also independant of weight... so whereas (in theory) there is and increase in frictional force of the tyre due to increase in wt which offsets the effects of increased weight.. the effects of drag and internal friction remain the same for too identically shaped cars.

if this force is constant ... then the lighter car must be acclereated (negatively) quicker by it .. this goes without saying... theefore a lighter car will decelerate quicker from highspeeds .. and as a result will have a much shorter braking distance than a heavier car...

the fact is an elise will outbrake a range rover from 100- 0 every time. and the gap will increase with speed....

I doubt that the difference that one observes with braking distance in such cars is down to drag alone .. but that is at least something that i can think of...

Quite agree that drag has a place to play in it, particularly from very high speeds, this can be a major problem with drafting. Brake when in the draft and leave it too late and you will be in big trouble.

From speeds in excess of approx. 100 - 120mph (depends on the car) then drag will have an effect, but not a major one. Still in motorracing, you don't get big gains, you work for tenths of a second.

Cars with highdown force suffer a similar issue, the downforce at speed does effect braking distances (it acts in a manner completely different to weight) but lessens suddenly as the speed drops. I can remember an interview with Derek Bell describing braking at LeMans at the end of the Mulsanne, and how he said they could apply maximum braking force at first, as the downforce would help, but as the car slowed they had to back of the brakes or risk locking the wheels.

If you have a copy watch in the DVD In-Car 956, the braking force the ground effect 956 allows them to use from speed is amazing, but you can see him fighting with the car as it slows and the downforce reduces.

 

[Grip]

That is correct, bigger calipers and discs will not stop you quicker.

If you do have time it is worth reading through the posts, however if time is short start with the first post and work your way throughthe rest when you have time.

In particular the StopTech white paper on the first post is worth a look.

Maybe, but most aftermarket brakes are more effective,

what brakes do is take kinetic energy (going foward) and turn that into Heat energy (from fricton)

So, if you put the same amount of pressure on the brake diske with a larger pad you create more friction, which creates heat and therefore slows you down (transforming energy).

 

[Scaff]

Maybe, but most aftermarket brakes are more effective,

what brakes do is take kinetic energy (going foward) and turn that into Heat energy (from fricton)

So, if you put the same amount of pressure on the brake diske with a larger pad you create more friction, which creates heat and therefore slows you down (transforming energy).

This has been covered before, bigger discs, pads, etc do not stop you quicker.

No maybe exists here; upgrading your braking components will give you better brake feel, reduce fade, etc; but they will not stop you quicker.

 

[Crayola]

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

I was actually think about this in reference to the wagon vs. sedan problem however I still dont think it explains the truck stopping much slower. I think there is quite a difference between truck tires and car tires too. Truck tires are designed to carry a much greater load and last alot longer, for this reason they are or a much harder compund than regular tires. Also, drag increases exponentially with speed, so below 100mph I dont think it would have any pronounced effect. I do agree that the theoretical physics doesn't work entirely in the real world but weight is still more important than mass in respect to braking.

 

[FIDO69]

You should ALL go buy Skip Barber’s book, “Going Faster” and read the chapter on tires. Especially you, Scaff.

What you say in this thread is mostly true, and for that you are to be congratulated. But you should have sprinkled a little sugar on what you say about weight not being significant to stopping distance because you are wrong, and your pedantic tone is about to be shoved back down your throat. (please to excuse)

***What you and your referenced materials miss is that the CF (coefficient of friction) is NOT a constant, it is a function of load applied***

Heinlein once wrote that people who don’t understand mathematics are just little apes. So here’s the math:

Relevant formula:

A1 = F1/m1
A2 - F2/m2
F’ = u * F’’,
F’’ = m * g

Where A1 is the deceleration, F1 is the force applied, and m1 is the mass of car 1. Ditto for A2, F2, m2. F’ is frictional force, u is the CF, and F’’ is the normal force on the object in question (in this case, downward). Let’s assume this is a car with only one wheel and one tire, in order to eliminate weight transfer as a factor (for the moment).

What you are saying is that A1 = A2 because F1 and F2 are proportional to m1 and m2. (i.e., A = u*m*g / m, therefore, A = u*g, a constant! All hail Galileo!) 👍 This is true as long as u is a constant.

But, u is NOT a constant. This is the assumption that practically every high school science teacher (and college professor) who doesn’t understand tire physics makes. (how did you make this mistake??) u, CF, is a function of many variables including slip angle, percent slip (static vs. dynamic friction, a whole ‘nother ball of wax), tire temperature, track temperature, tire material, track material, AND weight.


Yes, weight. And yes, while it is true that CF sometimes INCREASES as load increases. It only increases at very low normal forces and then decreases! And it doesn’t decrease linearly. I’ll have to scan in Skip Barber’s graph, but it looks basically like this:

Maximum static CF vs. Load (edited after looking at the book…)


Why is this? Well, tire manufacturers write heavy volumes about that subject, but essentially, it’s because nothing is unbreakable. Just think of a rubber eraser. You rub very lightly, and there is almost no friction. You push harder, and you get lots of friction mostly due to a larger contact patch (the rubber bites into the paper and the graphite markings are lifted up). You push really hard, and the eraser material breaks down, and you get slippage, shredded paper, and rolling on those wee little bits of broken off eraser bits.

(edit: this site has interesting pictures backing up the above statements: http://www.insideracingtechnology.com/tirebkexerpt1.htm)

So, imagine you have a car (a) with a mass 1000kg, CF of 1.2. It will have a frictional force of 12,000Newtons, and decelerate at 1.2g (12m/s^2). And, if you have a heavier car (b) 1500kg with CF of 1.1, it will have a larger frictional force, 16,500N, and decelerate slower at 1.1g. And, if you take a third car (c), one that is too heavy, mass 2000kg, the CF decreases to 0.9 (your tires are starting to fail). While the friction is larger, 18,000N (I hope your ’71 VW Bug’s brakes can handle this – good luck), the deceleration is less, 0.9g. (do the math) We could write a formula fitting the u function, but I don’t think that’s necessary – is it?

Now of course, the graph I drew here is just an idealized theoretical one (and doesn’t even take into account dynamic sliding CF (~30% less than static friction CF), which I would hazard to guess is also a function of normal force, amongst other factors), and RL tires are far more complex, but they ALL follow this basic curve.

You can get this info for different RL tires from the manufacturers (maybe), or you can look in Skip Barber’s book (how do you like my ad homonym appeals to authority? Eh? My Skip Barber can beat up your “professional physicist, amateur racer”). Weight transfer will not help you, because you transfer force to the front tires you remove it from the rear tires in proportion to the rate of deceleration, the distance of the CG from the front tires, and the CG’s height above the ground.

Ftransfer = a * m * h / L

Where Ftransfer is the force transferred, a is the rate of deceleration, m is the mass of the car, h is the height, and L is the length from wheel to CG. (see Skip, p. 204)

So for our two cars (m1 and m2), h and L are the same. Ftransfer proportional to a * m. If the two cars decelerate at the same a, then the more massive car would have a larger Ftransfer.

If m2 > m1, then
Ftransfer2 > Ftransfer1
Since a greater Ftransfer leads to a smaller CF, the larger car has a proportionally lower F’. And you can see the example cars a, b, and c apply.


To summarize: assuming all other things being equal (car type, tire type, brake power, aerodynamics, suspension, CG location, etc) a heavier car will decelerate LESS THAN a lighter car.

Size matters.





Case closed.

yeah right

(please note, I'm sorry for the flame, but you've got me on tilt. Also, you are absolutely correct in saying that the tires are the MOST IMPORTANT factor in braking. You are just plain wrong about the idea that weight is not relevant. And in this case, every 5yr old kid knows it, and is actually right. Tho, he thinks it for the wrong reason... but let's not get into that...)

 

[ferret-turbo]

god knows about all these formulae
what I do know is if I drive a truck without load it stops quicker then when it is fully laiden, at least if driving for any length of time
what you are saying may be true but it's the brake fade that causes the problems !!, after slowing down from fairly high speeds you wil get brake fade very quickly on most cars, (trackday for example) even 'sport' versions, same with if you have extra weight on a truck!

not doubting anything you're saying, just think it's misleading, especially since the use of the brakes in GT4 are for racing ! (slowing down from high speeds with very short distances in between)

 

[TruenoAE86]

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

Okay I think I can fill in some gaps in the physics.

Contact area has very little to do with the amount of friction a tire can produce.

Like other people have mentioned friction = weight * friction coefficient. As you can see contact area is not included in this equation. This means that a skinny tire will provide the exact same traction as a wider tire or 2 or 10 tires given that it is made of the same material and supports the same weight (including the weight of the tire(s)) . Now you may be asking your self “But why do race cars use wider tires then?” . The answer is simple a wider tire spreads the supported weight out over a more area meaning less weight on every square inch of the tire; this leads to less tire wear. With less tire wear this allows the use of softer stickier tires with higher friction coefficients which leads to more friction/grip.

This leads to a second point; some one asked why do companies spend millions of dollars on lighter materials, well again a lighter car will induce less pressure on tires and again will allow for softer sticker tires with out increased tire wear.

Again as people have mentioned Force = Mass * acceleration. As mass increases the force must also increase in order to keep the same acceleration. Take 2 identical cars with 2 identical braking systems. Lets make 1 car heavier then the other. The heavier car will require more force to stop in the same distance. However since the tires on the heavier car are supporting more weight they will be able to provide more traction or force, this added force just happens to be the exact amount of force needed to slow the heavier car at the same rate as the slighter car. The difference is that the brakes have to work harder in the heavier car to apply this force to the tires. Since the maximum force the brakes can apply has nothing to do with the weight of the car this can remain constant for our purposes. If we increase the weight of the car further, to the point that the added force needed to slow the car down at the same rate exceeds the maximum force of the brakes, then the car will not longer have the same braking distance as the lighter car. This is why a 40 tone truck can not slow down as fast as a light car, not to mention the tires probably have very low friction coefficient since they have to support so much weight. So it is harder to slow a heavier object but as long as the brakes are strong enough they will slow at the same rate.

This is getting pretty long so I’ll finish explaining my ideas another day. I still don’t really know why a land rover slows faster then a lotus but I have a few ideas.

Also I did not have time to read every post so some of this might already have been covered but I think it addresses some recent questions.


Edit: While CF does not directly depend on surface area at all ( If you don’t agree, google it) , It does depend on load. And since CF will generally decrease with load then 4 tires supporting 25% of a given load will have more traction then 2 tires supporting 50% each. This also applies to wider tires since there will be less pressure (psi, Nm^2) applied on the contact area, and therefore they will have a higher CF. What I said in my original post is still valid to a certain extent; this simply adds to it to make it more correct.

 

[TruenoAE86]

***What you and your referenced materials miss is that the CF (coefficient of friction) is NOT a constant, it is a function of load applied***

Ah cool, I didn't know that. This explains some, not the land rover though, i haven't read the entire post yet maybe the rest will.

 

[JOHN WAYNE]

some cars just dont wanna stop 🤬 🤬 🤬

 

[Crayola]

Surface area i part of the coefficeint of friction, if you believe that its not you're stupid. Why do you think drag cars have such fat tires? because it looks cool? You said its too do with tire wear, its not, you get more grip if there is more contact area. Its not possible for it to be any different. Think of it this way, you have tire a with 'a' certain amount of grip now, according to you if you had 2 tires, you would still have the same amount of grip, thats not true you'd obviously have double.

You make a good point but its been pointed out before. I distinctly remember saying "up to the maximum possible grip" Ie, once you have to much weight, the tire simply wont support any more. Although what you're saying is true your physics arent. We've already worked out that the strength of the brakes isnt that important. Also, the extra friction in the heavier car doesnt "just happen" to be that required, its a mathematical certainty.

The reason the range rover stops quicker is simple: It has bigger tires, and more weight at the front, this means it has a lot more braking force than the elise whos engine is at the back. We can extrapolate from the results that the range rover must produce more frictional force in proportion to its mass than the elise. However at speed this would be quite different, the elise would be able to transfer more weight and the range rover would most likely overload its brakes.

 

[Crayola]

Ah cool, I didn't know that. This explains some, not the land rover though, i haven't read the entire post yet maybe the rest will.

This isnt true, coefficient of friction has nothing to do with the weight. It is variable but that doesnt have to do with load applied. The weight (load) comes into the equation in the next stage where F=cf*w*m.

 

[FIDO69]

Nice try Crayola, but your assertion is trumped by my evidence:

Going Faster! Mastering the Art of Race Driving, p. 193, Carl Lopez, Skip Barber Racing School.


I'll highlight the key points: "A tire's coefficient of friction is a representation of its ability to convert download into traction. The CF declines with increases in download"

"This phenomenon... is one of the foundations behind the fact that lighter cars can attain higher cornering, braking and acceleration forces than cars with greater mass." (he mixes "force" and "acceleration" here, but I'll forgive him).

"The greater the load transferred to the outside, the more the CF of the loaded tires will suffer. Less load transfer means a higher overall CF for all four tires and, consequently, greater traction and cornering speed" (same is true for braking.... but replace slip angle with % slip).

In fact, I screwed up my plot and hurt my own argument. The CF does not increase with added force at all. It just decreases. (although I'm pretty sure athat what I wrote originally is true if the normal force is REALLY low... I'm sure that Lopez just ignored that really trivial case for cars).

Sorry. BUY THE BOOK. You can get it in most big bookstores.

(No, I don't own the stock, nor did I go to his school -- I went to Bob Bondurant's school)

 

[Scaff]

You should ALL go buy Skip Barber’s book, “Going Faster” and read the chapter on tires. Especially you, Scaff.

What you say in this thread is mostly true, and for that you are to be congratulated. But you should have sprinkled a little sugar on what you say about weight not being significant to stopping distance because you are wrong, and your pedantic tone is about to be shoved back down your throat. (please to excuse)

***What you and your referenced materials miss is that the CF (coefficient of friction) is NOT a constant, it is a function of load applied***

Heinlein once wrote that people who don’t understand mathematics are just little apes. So here’s the math:

Relevant formula:

A1 = F1/m1
A2 - F2/m2
F’ = u * F’’,
F’’ = m * g

Where A1 is the deceleration, F1 is the force applied, and m1 is the mass of car 1. Ditto for A2, F2, m2. F’ is frictional force, u is the CF, and F’’ is the normal force on the object in question (in this case, downward). Let’s assume this is a car with only one wheel and one tire, in order to eliminate weight transfer as a factor (for the moment).

What you are saying is that A1 = A2 because F1 and F2 are proportional to m1 and m2. (i.e., A = u*m*g / m, therefore, A = u*g, a constant! All hail Galileo!) 👍 This is true as long as u is a constant.

But, u is NOT a constant. This is the assumption that practically every high school science teacher (and college professor) who doesn’t understand tire physics makes. (how did you make this mistake??) u, CF, is a function of many variables including slip angle, percent slip (static vs. dynamic friction, a whole ‘nother ball of wax), tire temperature, track temperature, tire material, track material, AND weight.


Yes, weight. And yes, while it is true that CF sometimes INCREASES as load increases. It only increases at very low normal forces and then decreases! And it doesn’t decrease linearly. I’ll have to scan in Skip Barber’s graph, but it looks basically like this:

Maximum static CF vs. Load (edited after looking at the book…)


Why is this? Well, tire manufacturers write heavy volumes about that subject, but essentially, it’s because nothing is unbreakable. Just think of a rubber eraser. You rub very lightly, and there is almost no friction. You push harder, and you get lots of friction mostly due to a larger contact patch (the rubber bites into the paper and the graphite markings are lifted up). You push really hard, and the eraser material breaks down, and you get slippage, shredded paper, and rolling on those wee little bits of broken off eraser bits.

(edit: this site has interesting pictures backing up the above statements: http://www.insideracingtechnology.com/tirebkexerpt1.htm)

So, imagine you have a car (a) with a mass 1000kg, CF of 1.2. It will have a frictional force of 12,000Newtons, and decelerate at 1.2g (12m/s^2). And, if you have a heavier car (b) 1500kg with CF of 1.1, it will have a larger frictional force, 16,500N, and decelerate slower at 1.1g. And, if you take a third car (c), one that is too heavy, mass 2000kg, the CF decreases to 0.9 (your tires are starting to fail). While the friction is larger, 18,000N (I hope your ’71 VW Bug’s brakes can handle this – good luck), the deceleration is less, 0.9g. (do the math) We could write a formula fitting the u function, but I don’t think that’s necessary – is it?

Now of course, the graph I drew here is just an idealized theoretical one (and doesn’t even take into account dynamic sliding CF (~30% less than static friction CF), which I would hazard to guess is also a function of normal force, amongst other factors), and RL tires are far more complex, but they ALL follow this basic curve.

You can get this info for different RL tires from the manufacturers (maybe), or you can look in Skip Barber’s book (how do you like my ad homonym appeals to authority? Eh? My Skip Barber can beat up your “professional physicist, amateur racer”). Weight transfer will not help you, because you transfer force to the front tires you remove it from the rear tires in proportion to the rate of deceleration, the distance of the CG from the front tires, and the CG’s height above the ground.

Ftransfer = a * m * h / L

Where Ftransfer is the force transferred, a is the rate of deceleration, m is the mass of the car, h is the height, and L is the length from wheel to CG. (see Skip, p. 204)

So for our two cars (m1 and m2), h and L are the same. Ftransfer proportional to a * m. If the two cars decelerate at the same a, then the more massive car would have a larger Ftransfer.

If m2 > m1, then
Ftransfer2 > Ftransfer1
Since a greater Ftransfer leads to a smaller CF, the larger car has a proportionally lower F’. And you can see the example cars a, b, and c apply.


To summarize: assuming all other things being equal (car type, tire type, brake power, aerodynamics, suspension, CG location, etc) a heavier car will decelerate LESS THAN a lighter car.

Size matters.





Case closed.

yeah right

(please note, I'm sorry for the flame, but you've got me on tilt. Also, you are absolutely correct in saying that the tires are the MOST IMPORTANT factor in braking. You are just plain wrong about the idea that weight is not relevant. And in this case, every 5yr old kid knows it, and is actually right. Tho, he thinks it for the wrong reason... but let's not get into that...)

Thank you for you reply, and the information from Skip Barber's book.

If you do not mind I am going to look at your post in a number of sections, as to why I do not disagree with you and in particular Skip Barber.

Firstly your formula, which I have no problem with at all. You seem to have missed the point of my reasons for starting this thread, which was not to argue braking mechanics or physics. As such I have tried to keep things as straightforward as possiable.

Secondly, the exert from Skip Barber's book, some of which you apear to have edited out of your post, but in particular the graph. CF vs Load, you say that as load increases then CF increases until tyre slip occurs and then CF decreases.

Quite agree, and what is the principal cause of load changes.

Well the weight of the car at rest and its Front:Rear and Left:Right weight distribution will (on a flat level surface) will determine the initial load each tyre has to carry. I'm sure you would agree with this, but we are not talking about a car at rest, but one decelerating. When moving (and in this case braking) the load each tyre must carry varies through weight transfer.

How this weight transfer occurs is down to the cars set-up, wheel base, height, COG, etc; and I am more than happy to agree (and have never said otherwise) than generally a heavier car will have a higher COG and will therefore transfer more of its load forward under braking, and that this will incraese the load on the front tyres to the point that they lose grip.

Now the following is the area that we seem to disagree on, while weight plays a fundimental part in the load that is transfered, it is the actual load that is transfered that is the critical factor.

You use an example of two cars identical in all areas except weight and state that the lighter car will decelerate more, the difference will be small, but I would not disagree with you, because all things are equal.

However if you take a heavier car and remove weight from it, the way that the remaining weight is distributed at rest, and the cars COG are very, very unlikely to be the same. A car that is striped of weight is an unconsidered manner may result in the balance being moved to far in one direction and even raise the COG.

The latter is actual very common in club racing, where the seats, carpets, spare wheel and trim are the first to go, combined with the fitting of a roll cage, this will almost always raise the COG.

This can cause problems as even if the car is lighter, it's raised COG and changed distribution may actually cause more load to be applied to a certain tyre and the grip level exceeded sooner.

In my very first post I said.

the weight of a car does not significantly affect its straight line braking performance.

What can affect a cars straight line braking performance is weight distribution and transfer, as discussed in the StopTech white paper.

Now lets look at this, I did not say that weight was not a factor, I said it is not a significant factor. I am not dismissing it, I am saying that how the weight is distributed as load under braking (or accelerating or cornering) is more important than the weight itself.

The extract you provide by Skip Barber (and the graph) support this, as does the StopTech white paper, as does The Physics o Racing in later chapters, as does Russ Bentley's Book Speed Secrets.

It seems to me that you may have scan-read this thread (and who can blame you) and miss-interprated me, or maybe at times I have not always been as clear as I could have been.

THis principal of how the load is distributed being more critical than the total mass i sone of the reasons that a Range Rover can stop quicker than an Elise. Tyres (as we agree) are the most critical aspect here, but how the two vehicles distribute load under braking also plays a part.

In my original post I said that the single most important factor in stopping a car are the tyres (which we agree on), they are the only part of the car in contact with the road and provide (as you have explained at the start of your post) the initial grip.

The second factor is how the weight of the vehicle is then applied as load to the tyres (because this causes the existing grip level to vary - generally by a small amount). How this load is distributed is down to the weight distribution and transfer characteristics of the car, these factors determine how great the load taken by each tyre is.

Now weight does of course affect a cars transfer and distribution characteristics, but so does the wheelbase vs overall size, COG, suspension, unsprung mass, tyre pressure, gradient, etc.

The StopTech white paper explains this (in relation to Brake Bias) very well.

I closing I would just like to say that I apologise if you have found my tone pedantic or offensive in any way, but would add that my posts were not directed to you and I'm sure that existing members of GTp do not need you to take offence on their behalf.

Also for one who appears to dislike 'pedantic' tones you should look more closly at how you come across in text yourself, I personally do not care to be refered to as an ape, even if you do resort to the third person to do so. Please do not deny this, as if this was not your intention then I fail to see the relevence of this

Heinlein once wrote that people who don’t understand mathematics are just little apes. So here’s the math:

The above is close to a breach of the sites TOS, this is not a discussion about Heinlein or apes and simply saying So Here's the math would have been sufficent.

As a new member you will also find that as a point of clarity most members do not use the edit function to change sections of a post, but to clear up spelling and gramatical errors. Whole sections of your post changed from last night to this morning (UK time), it is generaly considered more appropriate to put up a new post.

Meanwhile may I take this opportunity to welcome you to GTP.

 

[FIDO69]

Also for one who appears to dislike 'pedantic' tones you should look more closly at how you come across in text yourself, I personally do not care to be refered to as an ape, even if you do resort to the third person to do so. Please do not deny this, as if this was not your intention then I fail to see the relevence of this



The above is close to a breach of the sites TOS, this is not a discussion about Heinlein or apes and simply saying So Here's the math would have been sufficent.

Meanwhile may I take this opportunity to welcome you to GTP.

Thank you for the welcome.

I am embarassed that I wrote in such a flame worthy manner. But I did not mean to call YOU an ape. And you will believe me, I hope, because I meant to (subtley) call some of your detractors apes -- those other posters who do not make any attempt at understanding the underlining math. You clearly understand the math. 'nuff said.

Sorry about the edits... I had been posting over at GranTurismo.com (until they croaked), but I'm still a relative newbie. I wasn't sure if I should edit the post, or repeat large chunks of it with the new graph... I'll do the latter from now on.

 

[FIDO69]

Hmm... I'd like to try to contact Skip Barber about that plot... I wonder how they produced it? Was it done on real tires, on a block of rubber, what? Was it a dynamic or static test?

I don't doubt it's essential truth, but just it's veracity for the following (back of the envelope) calculation...

Question I'd like answer:

If you reduce your car's weight by x% (and don't change the COG -- good points Scaff, all the minor weight reduction I've done on my RL car has raised the COG...), how much can you expect to improve your braking?

Well, my previous posts showed that
a = u*g
(deceleration is a function of coefficient of friction)

And Skip Barber's plot shows that u is a function of weight. I'll linearize around the two points (200lbs, 1.75) and (400lbs, 1.4). This gives me a formula for u (in the neighborhood of 300lbs).

u = 1.4 + 0.35/200*(400 - w)
= 1.4 + 0.00175 * (400 - w)

where u is CF and w is the weight of the car.

So to find the sensitivity of u WRT w, you take the derivative of the function... which is just the slope of the line, or -0.00175

du/dw = -0.00175

Right?

Of course, this has the units of -0.00175 g/lbs

So, if our example 300lb car were to change by 30lbs, the g's would change by 0.0525 which in percent is 3.33% (@300lbs, the plot shows 1.575g) (g's and u are essentially interchangeable).

Ah ha! (Roughly) changing the weight of a car with sticky rubber (as these high CF tires show) by 10% decreases breaking distance by 3.3%!!! Not insignificant, but not as much as most would expect either!

Interesting... (please check my math, I'm sneaking at work, and I'm hungry!)

(edit: I overstated my case, I said "decreases breaking distance by 3.3%" when I should have said, "increases deceleration by 3.3%". As the following post cleans up.)

 

[FIDO69]

Hmm, I forgot that perhaps we are more interested in braking DISTANCE, not deceleration rate.

The formula for braking distance(x) from an initial velocity (v) at constant deceleration (a) is derived below:

and
v = a * t

so, stopping time is

t = v/a

plugging into the first equation gives us:

x = v^2 / 2a

And now, we get to some interesting math. To get the sensitivity of x with respect to a, we take the derivative of x WRT a or

dx/da = -v^2/2a^2

In english, this says that change in stopping distance is negatively proportional to the inverse square of deceleration.

So (from my previous example of the 300lb car) a 3.3% deceleration improvement is good for a 6.8% improvement in stopping distance. Or, a 10% reduction in weight is worth 6.8% improvement in stopping distance. Not bad... 👍

Again, check my math.

 

[FIDO69]

Oops, I goofed again.

While it's true that

dx/da = -v^2/2a^2

this is not necessary, and my conclusion is not true.

dx/da is the slope of the x versus a curve, at a particular value of a. But that doesn't matter, because I'm interested in change in x versus change in a. And for that I only need to see the previous formula.

x = v^2 / 2a

This says that x is inversely proportional to a. And thus if 'a' doubles, then x is halved. And if 'a' is halved, x doubles. Therefore, if 'a' increases/decreases by 3.3%, x will decrease/increase by 3.3%.

Conclusion: using the Skip Barber plot, a 10% reduction in weight will result in a 3.3% change in stopping distance.

(My original conclusion...)

 

[Scaff]

Oops, I goofed again.

While it's true that

dx/da = -v^2/2a^2

this is not necessary, and my conclusion is not true.

dx/da is the slope of the x versus a curve, at a particular value of a. But that doesn't matter, because I'm interested in change in x versus change in a. And for that I only need to see the previous formula.

x = v^2 / 2a

This says that x is inversely proportional to a. And thus if 'a' doubles, then x is halved. And if 'a' is halved, x doubles. Therefore, if 'a' increases/decreases by 3.3%, x will decrease/increase by 3.3%.

Conclusion: using the Skip Barber plot, a 10% reduction in weight will result in a 3.3% change in stopping distance.

(My original conclusion...)

Fido69

Sorry for the delay in getting back, have been at a friends for the last couple of days.

However it was time well spent as his brother was there who is a design engineer for a prestige British car maker (he asked me not to say who and I have to respect that).

I got talking to him on this subject (and GT4 in general) and although it is not his direct field, he was able to confirm that what we have been discusssing is pretty much spot on.

Apparently most load vs Cf graphs are at best theoretical models (as you guessed) and are fine for theoretical modeling, he also said the same with the formula we have been looking at.

He did throw in a few 'buts', which is no big surprise, and they are as follows.

1. The majority of Load vs Cf graphs do not take into account the fact that tyres are not solid lumps of rubber (as you guessed).

2. When looking at weight transfer & load and it effect on grip the formula and models we have been looking at do not take into account the following:

a. Tyres are not solid; the composition, size (in particular the sidewall) and pressure will all effect how load is transfered to the contact patch and how it effects it.

b. Suspension plays a large part in how load from weight transfer is then transfered to the tyre

3. Braking is done gradually, you do not reach maximum threshold braking immediatly, nor do you stop braking immediatly. These are gradually applied force (even if the application is rapid) and as such the braking technique of the driver will have an effect on how the load is transfered.

I did ask him if he could let me know any way of modeling at least some of this, at which point he looked at me as if I was mad! Apparently the sheer number of variables involved are so great that you need very high level computer models to get even a semi-accurate model, and even then the final setup would be determined by drive evaluation.

The one thing he did say was that the above three points will almost always result in a loss in the benefit from the transfer of load, the energy used by the susspension components and the tyres will reduce the transfered forces (which sounds logical) its just very difficult to predict how.

He summerised this as follows.

To reduce stopping distances always start with the tyre, its the single most predictable element.

Weight reduction may reduce overall stopping distances, but this depends on how the load is transfered more than the weight reduction itself (and can sometimes increase the stopping distance if the result unbalances the car).

Also the results are very hard to model and generally small, which backs up your calculation above, a 10% reduction in weight (which is large) and very sticky tyres saw a 3.3% reduction in stopping distance and thats without taking into account suspension, tyres, etc.

Going back to the original reason I put this thread to gether we then got out GT4 and took one of the cars people have been complaining about and carried out a few very rough and ready tests.

Ford GT braking from 100mph at the test course. Braking started at the 300 metre mark. No driving aids were used, manual gears and a DFP.

Starting with the standard S2 tyres the car stopped at the start of the 400 metre text on the road.

Fitting Racing medium tyres saw the car stopping approx two and a half car lengthes from the 400 metre text (its hard to be accurate).

Refitting the S2 and then adding weight reduction one stage at a time the following happened.

Lvl 1 - no detectable difference, and any possiable reduction was so small that it could also be attributed to the driver (me).

Lvl 2 - same as above, but the car felt a lot more 'twitchy'

Lvl 3 - slight reduction in stopping distance appeared to be present, however the car was now a lot harder to control under hard braking. We did fit a brake balance controler and by increasing the front bias slightly were able to settle the car and a very small reduction in stopping distance was observed.

The above does show that tyres are by far and away the most cost effective route to stopping distance reduction (IRL and GT4), and while reducing the weight and the resultant impact in weight transfer and load you can reduce stopping distances, the results are slight and you may have to spend more money on the general set-up of the car.

Now I fully admit that the Ford GT tests we carried out are far from definative or controled, they are an interesting back-up to what we have been discussing and I hope a help to people looking to get more out of GT4.

I have attached the following to help people remember the level of forces at play due to weight transfer under hard braking, god bless the 240z and suspension thats as old as I am.

 

[TruenoAE86]

Surface area i part of the coefficeint of friction, if you believe that its not you're stupid. Why do you think drag cars have such fat tires? because it looks cool? You said its too do with tire wear, its not, you get more grip if there is more contact area. Its not possible for it to be any different. Think of it this way, you have tire a with 'a' certain amount of grip now, according to you if you had 2 tires, you would still have the same amount of grip, thats not true you'd obviously have double.

You make a good point but its been pointed out before. I distinctly remember saying "up to the maximum possible grip" Ie, once you have to much weight, the tire simply wont support any more. Although what you're saying is true your physics arent. We've already worked out that the strength of the brakes isnt that important. Also, the extra friction in the heavier car doesnt "just happen" to be that required, its a mathematical certainty.

The reason the range rover stops quicker is simple: It has bigger tires, and more weight at the front, this means it has a lot more braking force than the elise whos engine is at the back. We can extrapolate from the results that the range rover must produce more frictional force in proportion to its mass than the elise. However at speed this would be quite different, the elise would be able to transfer more weight and the range rover would most likely overload its brakes.

Coefficients of Friction

Static Dynamic
Steel on Steel 0.74 0.57


Aluminum on Steel 0.61 0.47


Copper on Steel 0.53 0.36


Rubber on Concrete 1.0 0.8


Wood on Wood 0.25-0.5 0.2


Glass on Glass 0.94 0.4


Waxed wood on Wet snow 0.14 0.1


Waxed wood on Dry snow - 0.04


Metal on Metal (lubricated) 0.15 0.06


Ice on Ice 0.1 0.03


Teflon on Teflon 0.04 0.04


Synovial joints in humans 0.01 0.003

http://www.physlink.com/Reference/FrictionCoefficients.cfm

CF has nothing to do with surface area; it has everything to do with the materials that are in contact. As you can see in a chart like this surface area does not come into play, other wise you would have different CF for 1 square meter, and 1 square foot if you don’t believe me you are NOT stupid, just look it up you will notice you will not find “metric” or “imperial” CF nor will you find different CF for 1 ft^2 or 2 ft^2.

Also I apologies that I did not show mathematically that added friction due to weight is directly proportional to added forces due to acceleration on a 1 to 1 basis, I just figured everyone would understand what I meant with out going onto details.

“once you have to much weight, the tire simply wont support any more”

I’m not sure what you are getting at here. If the tire isn’t supporting the weight, what is? Are you saying the wheel actually collapses? I can only assume that the brakes would overload long before this would happen.

On a further note the last few post have been very interesting and informative.

 

[63AvantiR3]

(edited)If you have access to an area that you can test this away fom public roads (an old air-field for example) then try applying maximum braking force as smoothly and rapidly as you can. You should get the ABS to kick in. Once you can get that to occur, then try braking in the same way, but modulate the pedal to apply force up to (but not beyond) the point the ABS kicks in. You should find that the overall stopping distance is shorter (if its dry).

Now as I said above do not try this on public roads, and be aware that it will give your pads and discs a hammering.

*Evil Grimace* I got to do this on public roads in my Rabbit GTi. I just recently upgraded my stock 238mm rotors and calipers to blank (not slotted or drilled, that crap isn't necessary) 256mm Brembo rotors and other 1986-1988 16V Scirocco brake components. Since I replaced the pads and the rotors I got to bed-in my pads at 2 AM on an expressway. I don't have ABS, as the car is an '83, so I got to threshold brake from 30 MPH and 50 MPH. Boy was that fun. My next "whoa" upgrade is to change the stock 180mm rear drums to either 200mm Quantum drums or rear discs from a donor Scirocco or Jetta.

I have 195/60-14 Falken Azenis tires, had them through the end of my last braking system, so I can compare equally. I will say that now the car is easier to make it stop in shorter distance. I also upgraded to stainless steel braided brake lines so less energy my foot puts down is 'wasted' to expanding stock rubber lines.

Before I had all of these new braking goodies, sure my car could lock the tires, but it took a lot more pedal travel. Now I am more comfortable and confident in how my brakes slow my car down. Sure, I can lock the tires easier now than I used to, but I have learned where my pedal modulates and where "impending lockup" happens.

I also drive at track days, and can't wait to drive Sears Point with better braking capabilities. These rotors will dissipate heat a lot better and with some track pads that better resist fade due to heat, I should have a lot of fun.

 

[Crayola]

Also I apologies that I did not show mathematically that added friction due to weight is directly proportional to added forces due to acceleration on a 1 to 1 basis, I just figured everyone would understand what I meant with out going onto details.

“once you have to much weight, the tire simply wont support any more”

I’m not sure what you are getting at here. If the tire isn’t supporting the weight, what is? Are you saying the wheel actually collapses? I can only assume that the brakes would overload long before this would happen.

On a further note the last few post have been very interesting and informative.

Thats really interseting, thanks for correcting me. Also, what I meant was, there is a maximum amount of grip the tire can attain before it starts to slip.

Edit: What I still dont understand is why the coefficient of friction depends on weight even though overall friction is F = cf*M*G

Edit 2: Skipped ahead in my physics and engineering textbooks, I understand why now. From what I've gathered:
the coefficient of static frictio = force of limiting friction/normal forces between surfaces. or Us=Fr/N

 

[FIDO69]

The one thing he did say was that the above three points will almost always result in a loss in the benefit from the transfer of load, the energy used by the susspension components and the tyres will reduce the transfered forces (which sounds logical) its just very difficult to predict how.

He summerised this as follows.

To reduce stopping distances always start with the tyre, its the single most predictable element.

Weight reduction may reduce overall stopping distances, but this depends on how the load is transfered more than the weight reduction itself (and can sometimes increase the stopping distance if the result unbalances the car).

Nice pic! 👍

Very interesting dinner conversation you guys must have had. Did S.O.'s get peeved?

I'm not sure what you are saying in the first paragraph I quoted (above). Are you saying that there is a benefit to transferring load?

I also did a GT4 test using a Nissan (?) Option Z (a prize car). I went to the test ring and recorded stops from 200mph before and after weight reduction and with and without 200kg of ballast. I looked at the analyzer results and all stopping distances were equal to within 0.1mi (I'm not sure what the units they are using in the Analyzer... it definitely isn't miles...).

To me this says two things: 1. there is no modelling of variable CF due to load (if there was, there would be some change) 2. there is no change to the COG or suspension geometry due to the weight change.

Well, what do you expect? It's just a game that sells for $40 bucks, not a multimillion dollar racing team simulator or PhD research project.


Now there are two more things I want to test: does greater downforce improve braking distance? And do Racing Brakes (in GT4) improve braking distance? And if so, why??

 

[FIDO69]

Edit: What I still dont understand is why the coefficient of friction depends on weight even though overall friction is F = cf*M*G

Edit 2: Skipped ahead in my physics and engineering textbooks, I understand why now. From what I've gathered:
the coefficient of static frictio = force of limiting friction/normal forces between surfaces. or Us=Fr/N

Hi Crayola,

My short answer is, don't believe everything you read in physics textbooks.

The longer answer is partially covered by your second edit. cf IS constant if you measure it in a very in a particular, controlled, 'scientific' way. However, if you measure it some other way, lo and behold! It's... changed! (oh don't worry, just ignore that "spurious" answer, son!).

People do tests and ignore the results all the time - if they don't like the results (i.e., they don't fit what the experimenter wanted to show). Trust me, I know. I work for NASA...

cf is a function of many variables (T, age, radiation exposure, load, slip angle, percent slip, static or sliding, and the list goes on), don't let one simplified formula in a book tell you otherwise.

FIDO


(BTW, I pretty much assume that there are NO constants in this world, unless... well, I don't know, Einstein says so??) [shrug]

 

[Scaff]

Nice pic! 👍
Very interesting dinner conversation you guys must have had.

One of the best, up till about 3am.

Did S.O.'s get peeved?

Oh yes and we did get called ' a bit sad', I guess I will be paying for that for a while.

I'm not sure what you are saying in the first paragraph I quoted (above). Are you saying that there is a benefit to transferring load?

No, just saying that the results can be very unexpected as its difficult to tell in advance whtthe effects will be.

I also did a GT4 test using a Nissan (?) Option Z (a prize car). I went to the test ring and recorded stops from 200mph before and after weight reduction and with and without 200kg of ballast. I looked at the analyzer results and all stopping distances were equal to within 0.1mi (I'm not sure what the units they are using in the Analyzer... it definitely isn't miles...).

To me this says two things: 1. there is no modelling of variable CF due to load (if there was, there would be some change) 2. there is no change to the COG or suspension geometry due to the weight change.

Well, what do you expect? It's just a game that sells for $40 bucks, not a multimillion dollar racing team simulator or PhD research project.

Hmm this one got me interested, so I just carried out a GT4 test myself. I went for something with a definate weight distribution issue to begin with. The Ruf Yellowbird, with and without 40kgs of ballast, but pushed it as far to the front as I could.

The run with ballast had a better initial braking response, but was quicker to lock up the front tyres. The overall stopping distance was slightly longer, but not by a great deal.

Going to run this for a few more cars before I make my mind up, from this initial run I think something may be in place, but given your results it could be on a very basic level.

And as you say this is running on a PS2 and for the price I think what we have got is excellent, it certainly got this topic up and running. My friends brother was very impressed with GT4, well in regard to the tracks (Nurburgring got a big 👍 ), vehicles and physics; he did however describe the AI as ' too stupid for words', which just about sums that side of things up.

Now there are two more things I want to test: does greater downforce improve braking distance? And do Racing Brakes (in GT4) improve braking distance? And if so, why??

Downforce I have not got a chance to play around with yet, add this one to the list.

However I have tested with and without Racing Brakes and can see no difference with the braking distance (have done this with quite a few cars), it does seem to improve your ability to modulate the brakes, its a very slight effect and hard to put into words, but particularly on dirt/snow/ice surfaces it seems a bit easier to control your braking. However, I have only felt this when using the DFP, my thumbs are just not sensitive enough on the DS2.

Just to close one of my best pictures of weight transfer in GT4, amazing what you can do with a Fiat.

 

[63AvantiR3]

Just to close one of my best pictures of weight transfer in GT4, amazing what you can do with a Fiat.

OH that's HOT! So P.D. *did* take the time to factor in lifting the inside rear wheel off the ground. Volkswagen Rabbits (known as the Golf I in Europe and everywhere else in the world) are known for doing this. Race prepped ones anyway. The suspension in my daily driven Rabbit isn't stock at all and I lift the inside rear wheel frequently. People give me these weird looks because to them it seems like I'm about to roll on my side. Let's hear it for tie-bars, racing springs and tight shocks on a street driven car.

I can't wait to photograph the '76 Golf/Rabbit 'German Saluting' at Sears Point. I loved my track day there back in August; can't wait to go again.

 

[63AvantiR3]

All right Scaff. How did you get the little Fiat to lift its inside rear wheel under braking/cornering? The VW Rabbit (1976 Golf I in the game) is known for lifting its inside rear wheel and I've taken it to tracks, tried different settings, and I still can't get it to happen.

Should I do something like alter the suspension settings? I've tried racing suspension tweaks, stock suspension, semi-racing suspension and that darn wheel won't lift. If P.D. didn't program the chassis to do it in the game, I guess I just can't do it, but I hope they were smart enough to put it in.

 

[Scaff]

All right Scaff. How did you get the little Fiat to lift its inside rear wheel under braking/cornering? The VW Rabbit (1976 Golf I in the game) is known for lifting its inside rear wheel and I've taken it to tracks, tried different settings, and I still can't get it to happen.

Should I do something like alter the suspension settings? I've tried racing suspension tweaks, stock suspension, semi-racing suspension and that darn wheel won't lift. If P.D. didn't program the chassis to do it in the game, I guess I just can't do it, but I hope they were smart enough to put it in.

The car has not been modified in any way (well I gave it an oil change), just took it around the Paris Opera track in Photo Drive.

I was throwing the 500 into corners and then lifting off the throttle, just to get a tight line as its so slow you don't need to brake before th corner.

I did not realise the effect until I watched the replay and got the shots.