GT4 and Brakes

  • Thread starter Scaff
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Better pedal modulation and control could be part of the answer, however I would ask if you could posts a specific example.
It was in an old Car and Driver I think. I'll look around for it.
Scaff
BTW - No offence is ever taken to a question, as you can clearly see from my replies in this entire thread.
I know. I just didn't want you to think I was mocking you or anything like that, especially after the recent revelation.
 
Wait a minute here... I must be misunderstanding something very basic here. Scaff, are you (and as it seems the motor industry) saying that it doesn´t actually matter how good your brakes are, it´s the tyre that matters???
An example: Take two F1 cars (of the same make), set up in the exact same way, but provided with different brakesystems (let´s say one has state of the art F1 brakes, and the other has the brakes from a ´70 Dodge Charger.), these two cars will have the same stoppingdistance, if they have the same kind of tyres? I must say I have a very hard time beliveing that!
Or are you (and the motor industry) saying that it´s pointless to have brakes that are better than the grip the tyre can provide (like having F1 brakes on a ´70 Charger with original tyres)? That is much more understandable in my ears!
 
Or are you (and the motor industry) saying that it´s pointless to have brakes that are better than the grip the tyre can provide (like having F1 brakes on a ´70 Charger with original tyres)? That is much more understandable in my ears!

That's exactly how I've always understood what Scaff says on brakes. The problem (or perhaps misunderstanding) when talking about brakes in GT4 is that on any given car, the stock brakes are almost always more than strong enough to overwhelm the grip of the tyres (after adjusting the balance controller if necessary) - so there's no point in upgrading to racing brakes.

At least, that's my take on it...
 
Wait a minute here... I must be misunderstanding something very basic here. Scaff, are you (and as it seems the motor industry) saying that it doesn´t actually matter how good your brakes are, it´s the tyre that matters???
An example: Take two F1 cars (of the same make), set up in the exact same way, but provided with different brakesystems (let´s say one has state of the art F1 brakes, and the other has the brakes from a ´70 Dodge Charger.), these two cars will have the same stoppingdistance, if they have the same kind of tyres? I must say I have a very hard time beliveing that!
Or are you (and the motor industry) saying that it´s pointless to have brakes that are better than the grip the tyre can provide (like having F1 brakes on a ´70 Charger with original tyres)? That is much more understandable in my ears!

The latter is the correct one (that I have highlighted in bold), as your stopping potential is always limited by the tyres.

However I've never said it pointless to have better brakes, just that better brakes will not reduce stopping distances (unless the original brake system is very poor).

The advantages of better brakes are many, they reduce brake fade, increase modulation and improve feel. However with this also comes problems, such as an increased need for cooling (and the drag caused by cooling ducts) and an increase in unsprung weight.

The first example you used of the two F1 cars would cause confusion because the Charger brakes would most likely not be able to produce sufficient force to reach the tyres limits. However if you put say the brakes from a modern performance car (say a 911) on an F1 car they may be able to stop the car in the same distance as the F1 brakes if they could use all the tyres available grip. What they certainly would not be able to do is repeat the process more than once (brake fade would certainly set in here).

Its this reason why the industry (both motor and motorsport) place such importance on the tyres, they are the ultimate limiting force for stopping distances. Once you reach a certain point larger, more powerful brakes will not stop you any quicker, but they will allow you to do it more often and with more control.

Which is why Stop-Tech (who make motorsport brake systems) say

Stop-Tech
“You can take this one to the bank. Regardless of your huge rotor diameter, brake pedal ratio, magic brake pad material, or number of pistons in your calipers, your maximum deceleration is limited every time by the tire to road interface. That is the point of this whole article. Your brakes do not stop your car. Your tires do stop the car. So while changes to different parts of the brake system may affect certain characteristics or traits of the system behavior, using stickier tires is ultimately the only sure-fire method of decreasing stopping distances.

Regards

Scaff
 
Hello All Brake-Nuts and Scaff!

After a long hiatus, just wanted to drop by and post a Hello, and great job with the summary!

I'm going to be linking all my new BMW Z4 forum friends to this discussion... I got in a bit of a discussion about improving brake performance on our cars...

Keep up the good work, eh!
 
thanks Scaff for these facts!.:bowdown:
You've learnt me a whole lot, and I haven't read the other guides yet!:scared:
Sadly they don't have this stuff in physics class...(yet??)
 
thanks Scaff for these facts!.:bowdown:
You've learnt me a whole lot, and I haven't read the other guides yet!:scared:
Sadly they don't have this stuff in physics class...(yet??)

Glad to hear this thread is still of interest and use 👍 and thanks for the feedback, I hope that you find the guides just as enjoyable/interesting.

Thanks

Scaff
 
I may be a bit of topic. And I'm sorry for my bad English but there is something that bothers me a lot about GT4- brakes setups and i dont know
where else i can get this question answered. Why have even performance cars like the Toyota Minolta a braking force distribution of 3 to 3 thats somehow strange because this is the starting value when you built in the brake controler and its obviously not the setup is really used on cars like that. That makes adjusting the brakes difficult cause the braking performance on the minolta is quite good actually a lot better than it can be with this stupid default setup. Why had PD done such a thing its like with fully costum LSD it has always this standart setup of 10 40 20 and works even on the fastest cars. That is confusing to me. At least they called it the real driving simulator. In my opinion they missed something with brakes and default setups at least they should have been a bit more like the setups of the real cars. Thats pretty loosy when you think about the time and money wich was invested in the game.
 
The default values are only meant as a starting point for you to begin your own work from. Setups of real race cars are impossible to replicate as every driver uses different setups on different tracks, even in different weather conditions. In short races the setup can be very different to that of long races, let alone very different course profiles. On a slow twisty course a rear-biased setup can be worth gold whereas on a fast course with hard brakings it could be fatal.

To be honest, very few default settings work perfectly. Pretty much everything needs tweaking if one wants the car to drive really well, the stock LSD is good for FR cars of around 500bhp but drive anything else and you will find that fine tuning is needed.

They didn't save any money or time when doing the setups, the fact is just that no car will fit everybody. There would be many people complaining about bad default values even if the values were those of the real cars. One of the main points of this game is making the car drive like you want it to, and you have the chance of doing it.

- R -
 
I may be a bit of topic. And I'm sorry for my bad English but there is something that bothers me a lot about GT4- brakes setups and i dont know
where else i can get this question answered. Why have even performance cars like the Toyota Minolta a braking force distribution of 3 to 3 thats somehow strange because this is the starting value when you built in the brake controler and its obviously not the setup is really used on cars like that. That makes adjusting the brakes difficult cause the braking performance on the minolta is quite good actually a lot better than it can be with this stupid default setup. Why had PD done such a thing its like with fully costum LSD it has always this standart setup of 10 40 20 and works even on the fastest cars. That is confusing to me. At least they called it the real driving simulator. In my opinion they missed something with brakes and default setups at least they should have been a bit more like the setups of the real cars. Thats pretty loosy when you think about the time and money wich was invested in the game.

Its simply because no such thing as the 'right' set-up exists, take a look at every default value for any upgrade in GT4 (and the entire series for that matter) and everything has a standard set of starting values (if I recall correctly only spring rates vary from car to car, but even these increase as the suspension gets more race orientated).

What they do is put in place a set of default values simply because something has to be put in place, however you will need to work with these values, you driving style, any modifications you make to the car and the track you are racing at to get the 'right' value for you. Any 'default' value will be wrong from this point of view.

I suggest that you don't worry about the default values and simply get the car set-up for the track you need.

Regards

Scaff
 
Its simply because no such thing as the 'right' set-up exists, take a look at every default value for any upgrade in GT4 (and the entire series for that matter) and everything has a standard set of starting values (if I recall correctly only spring rates vary from car to car, but even these increase as the suspension gets more race orientated).
The Stabilizers vary from car to car also, as well as ride height. 👍
 
I suggest that you don't worry about the default values and simply get the car set-up for the track you need.

Regards

Scaff

yes it is same thing happening on aids setting , it's have two defaults setting ones 10/10/7 and the other 10/10/5

how can I fix aids setting (oversteer and understeer)?
where is the best track to try my setting ?

I really need to learn that ,who knows maybe it will make your race easier.
 
how can I fix aids setting (oversteer and understeer)?
By turning them off. Seriously, all aids you will ever need is the occasional "1" of TCS and even that isn't needed in cars with less than 800 bhp, not even in them if you have good throttle control.

The aids use brakes and throttle to control the car. Even my logic says that if the brakes are engaged, the speed will be going down and that's not what we're trying to do. They will make the driving easier to a point, but also greatly reduce from the experience and in many cases actually make the car difficult to drive. We've heard of cars accelerating by themselves thanks to the ASM, that doesn't sound normal to me anymore. The biggest problem of the aids is that they allow a certain amount of slip and then cut in just as the driver is beginning to correct the line. As a result you get double correction and an about 30% power loss as the system begins to brake the slipping wheels. By driving yourself you would only have made that one correction and continued at full throttle.

Got the idea? ;)

- R -
 
Greycap is right just the other day I was driving the 1989 Toyota 88C-V in the Sears point (Infineon) Raceway 2h:45min endurance race and started off with
965 HP on Hard racing slicks wihtout any driver aids and that was okay I could handle it but it wore the tires out too fast. I liked the controllable oversteer and steering the car with the throttle but there are so many slower corners on that circuit that I had to constantly feather the throttle out of the slower 2nd and 3rd gear corners to avoid lighting up the rear tires every time. So instead I did the race with TCS on the 1 setting adn my tires lasted 4 laps longer and my lap times went down by 2 to 3 seconds per lap! So when you have a car with that much power you may need a little TCS on the slower circuits where you need a very fast race car, or you may not, just experiment hey if you get frustrated just hit the reset button on the PS!
 
Sorry, but you have either not read or not understood the 'Physics of racing' series; weight is not a critical factor in the overall stopping distance.

If you believe it is then explain why a Range Rover has a quicker stopping time than an Elise?

Losing weight has many, many advantages when it comes to cars, but braking is not one of them.

I have provided links to detailed explanations of why this occurs, either you do not believe the racer (and Physics Phd) who wrote them or you have an alternative theory.

The Following is from the Physics of racing and can be applied to braking and acceleration as they discuss maximum levels of grip available during straight line travel. The second you start looking at accel/decel during cornering it gets far more complex.

"Finally, by Newton's second law again, the acceleration of the tire due to the force applied to it is We can now combine the expressions above to discover a fascinating fact: 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, and it can take it under load as well as when lightly loaded. We hinted at this fact in Part 2, but the analysis above hopefully gives some deeper insight into it. We note that being constant is only approximately true, because changes slightly as tire load varies, but this is a second-order effect (covered in a later article)."



Nothing theoretical about it, almost any modern braking system (including drum brakes) can exert sufficent force to make the tyres lock. You are quite right about why better brakee setups are an advantage, as they allow the driver to exploit the system to its full, with better feel and modulation availiable, with low levels of fade.

I am not sure if anyone reads these threads anymore, but I would like to add my 2 cents...

Since I am a new poster to this forum, you might not see me as an "expert", especially since you work in the industry (so do I btw).

I think you missed the point of MR-GT.

The "friction" a tire creates is based upon a coefficient and the vertical "load" on that tire. Take a 2000 lb car. at 50/50 distribution, it puts 500 lbs on each tire. Now if each tire is a "low-grade" tire only capable of 1.0 G of force at this weight, this means each tire produces 500 lbs of stopping force, which summed up, produces a total of 2000 lbs, to slow the car at -1.0 G. This is the "coefficient" mentioned above.

Now in your world, quoting newton, this coefficient is static. The problem, as Mr-GT tried to explain, is that this coefficient is NOT static, and actually varies upon load.

Now let's say this -1.0g decel transfers 100 lbs onto each front tire. One now has the rears only loaded for 400 lbs, while the fronts each have 600 lbs. This still totals to 2000 lbs of the car, but the traction coefficient for the rears has (probably) improved slightly, while that for the front has decreased. Let's say the rears went to 1.1G and the fronts dropped to 0.9g. This puts a total of (400*1.1*2 = 880 rear + 600*0.9*2 = 1080) 1960 lbs total for a REDUCTION to -0.98G of stopping power.

Now with a heavier car, lets say a 2400 lb car, the tires START at 600 lbs each, so the BEST it could do is -0.9G as each tire becomes "overloaded".

These numbers are not very accurate and are just used to illustrate the fact that as tires deform under weight, their friction reduces, so one does not get a 1:1 ratio between added force and added friction. Physically, as weight is added, the tires generate a LESS than proportional amount of added grip.

So as in pure turning grip, braking power reduces as the total weight of the car increases. The reasons for the Land rover are prolly as the other guy stated.. better/bigger tires, maybe better ABS (for more consistent stops if the magazine took samples).. or some other factor.

I will agree that Tires/weight transfer play a big role, but TOTAL weight does too.. just not as much as one would think.
 
I am not sure if anyone reads these threads anymore, but I would like to add my 2 cents...

Since I am a new poster to this forum, you might not see me as an "expert", especially since you work in the industry (so do I btw).

I think you missed the point of MR-GT.
Thanks for the reply, it is more than welcome. It would however appear that you have read part of the thread or has scan read it (both of which given its size are understandable), which has lead you to a few in-accuracies regarding my views on this topic.


The "friction" a tire creates is based upon a coefficient and the vertical "load" on that tire. Take a 2000 lb car. at 50/50 distribution, it puts 500 lbs on each tire. Now if each tire is a "low-grade" tire only capable of 1.0 G of force at this weight, this means each tire produces 500 lbs of stopping force, which summed up, produces a total of 2000 lbs, to slow the car at -1.0 G. This is the "coefficient" mentioned above.
I'm fully aware of this and its discussed here at length and while I understand that you use the number for simplicity a tyre capable of 1g of grip would most certainly not be a low grade tyre at all. On road tarmac most standard mass market tyres would struggle to generate much above 0.8g.


Now in your world, quoting newton, this coefficient is static. The problem, as Mr-GT tried to explain, is that this coefficient is NOT static, and actually varies upon load.
If you take the time to read this thread fully you will see that point is not dismissed at all, however you yourself have hit upon the small but rather fundamental point in that it is not curb weight that is important here but load.


Now let's say this -1.0g decel transfers 100 lbs onto each front tire. One now has the rears only loaded for 400 lbs, while the fronts each have 600 lbs. This still totals to 2000 lbs of the car, but the traction coefficient for the rears has (probably) improved slightly, while that for the front has decreased. Let's say the rears went to 1.1G and the fronts dropped to 0.9g. This puts a total of (400*1.1*2 = 880 rear + 600*0.9*2 = 1080) 1960 lbs total for a REDUCTION to -0.98G of stopping power.

Now with a heavier car, lets say a 2400 lb car, the tires START at 600 lbs each, so the BEST it could do is -0.9G as each tire becomes "overloaded".
It all depends how you use the figures, now assume all the above is true, but the static weight distribution of the car is 40% front and 60% rear (classic mid-engined), and 100 lbs is moved forward under deceleration again (on each tyre), we now have our full 2000 lbs of braking force back.

Static = Front 40% (400lbs each tyre) and Rear 60% (600lbs each tyre)
Under Braking (100lbs transfer per tyre) = Front 500lbs * 1 * 2 = 1000lbs and Rear 500lbs * 1 * 2 = 1000lbs. Total = 2000lbs


Switch that around to a FWD layout and we get 60% front and 40% rear, resulting in 700lbs on each front tyre and 300 lbs on each rear tyre under braking and a reduction in braking force avaliable to use (and the numbers here would be less than the initial 50/50 static distribution).

Static = Front 60% (600lbs each tyre) and Rear 40% (400lbs each tyre)
Under Braking (100lbs transfer per tyre) = Front 700lbs * 0.8 * 2 = 1120lbs and Rear 300lbs * 1.2 * 2 = 720lbs. Total = 1840lbs



Now without a single change in the curb weight we have three very different sets of figures.


However we can also look at differing curb weights and get some interesting results.

Car A
Curb weight - 2400lbs
Static dist. - Front 40% (480lbs per wheel) Rear 60% (720lbs per wheel)

100lbs per tyre forward under braking gives

Under braking - Front 580lbs per wheel and Rear 620 lbs per wheel

Using 580lbs = 0.87g and 620lbs = 0.93g (consistent with the scale you are using)

Front = 580 * 0.87 * 2 = 1009lbs
Rear = 620 * 0.93 * 2 = 1153lbs

Total = 2162lbs



Car B
Curb Weight - 2000 lbs

Static dist. - Front 60% (600lbs per wheel) Rear 40% (400lbs per wheel)

100lbs per tyre forward under braking gives

Under braking - Front 700lbs per wheel and Rear 300lbs per wheel

Using roughly 700lbs = 0.8g and 300lbs = 1.2g (consistent with the scale you are using)

Front = 700 * 0.8 * 2 = 1120lbs
Rear = 300 * 1.2 * 2 = 720lbs

Total = 1840lbs


So here we have a lighter car exerting less braking force (considerably less) than the heavier one, because load distribution allows the heavier car to make better use of it.

Now while the initial coefficient of the heavier car is not 1.0g, it must be kept in mind that this figure only plays a part for a very sort amount of time and that full braking is not achieved instantly (you don't go from 0g to 1g instantly) but as soon as you start braking the load shifts. As a result the 'under braking' values are far more relivent than the initial ones (and if you go from accelerating to braking then the initial values are completely irrelivant).


The only time you can say that a lighter car will always make better use of its tyres is if every single other value is identical, and lets be quite honest that is never going to happen. In an earlier post I did mention and example of this, take a car and prep it for racing. So strip out the seats, interior and anything not needed and it will be lighter, add in a roll-cage (which will almost certainly be needed) and while this does push the weight up a bit the car is now much lighter overall. However far more than the weight has cahnged, we have totally changed the cog of this car and almost certainly changed its static weight distribution, all factors that will effect how much load is transfered under braking, cornering and acceleration.


These numbers are not very accurate and are just used to illustrate the fact that as tires deform under weight, their friction reduces, so one does not get a 1:1 ratio between added force and added friction. Physically, as weight is added, the tires generate a LESS than proportional amount of added grip.
Again I don't disagree, but its the load that causes this to occur not the weight itself (as the example of three differeing static distributions for a single curb weight illustrate).

Now while I have not taken this into account in the figures I have used above (that is a simple linear scale), if I had it would actually be more of a problem for the lighter FWD car than it would be for the heavier Mid-engined car, once again illustrating that how load is distributed is far more important than the weight itself.

Simply saying lighter = more braking force is overly simplistic and assumes that all other factors are exactly the same, as soon as any other variable changes the load transfer and distribution become far more important than the weight.


I will agree that Tires/weight transfer play a big role, but TOTAL weight does too.. just not as much as one would think.

Again I have said exactly this repeatedly throughout this entire thread, you have simple picked on one example in which I was 'trying' to get a single point across, If you have the time to read the whole thread I think you will find that our opinions differ very little.


Regards


Scaff
 
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Thanks for the reply, it is more than welcome. It would however appear that you have read part of the thread or has scan read it (both of which given its size are understandable), which has lead you to a few in-accuracies regarding my views on this topic.

I'm fully aware of this and its discussed here at length and while I understand that you use the number for simplicity a tyre capable of 1g of grip would most certainly not be a low grade tyre at all. On road tarmac most standard mass market tyres would struggle to generate much above 0.8g.

If you take the time to read this thread fully you will see that point is not dismissed at all, however you yourself have hit upon the small but rather fundamental point in that it is not curb weight that is important here but load.

I am glad to see that this is still alive and well :)

You are right in that I only read about 1/3 of the post 30 pages!! :) ) I will make an effort to try to read the entire post as it seems the concept I am talking about is rather well known around here.

It all depends how you use the figures, now assume all the above is true, but the static weight distribution of the car is 40% front and 60% rear (classic mid-engined), and 100 lbs is moved forward under deceleration again (on each tyre), we now have our full 2000 lbs of braking force back.

Static = Front 40% (400lbs each tyre) and Rear 60% (600lbs each tyre)
Under Braking (100lbs transfer per tyre) = Front 500lbs * 1 * 2 = 1000lbs and Rear 500lbs * 1 * 2 = 1000lbs. Total = 2000lbs

Switch that around to a FWD layout and we get 60% front and 40% rear, resulting in 700lbs on each front tyre and 300 lbs on each rear tyre under braking and a reduction in braking force avaliable to use (and the numbers here would be less than the initial 50/50 static distribution).

Static = Front 60% (600lbs each tyre) and Rear 40% (400lbs each tyre)
Under Braking (100lbs transfer per tyre) = Front 700lbs * 0.8 * 2 = 1120lbs and Rear 300lbs * 1.2 * 2 = 720lbs. Total = 1840lbs

Now without a single change in the curb weight we have three very different sets of figures.

However we can also look at differing curb weights and get some interesting results.

Car A
Curb weight - 2400lbs
Static dist. - Front 40% (480lbs per wheel) Rear 60% (720lbs per wheel)

100lbs per tyre forward under braking gives

Under braking - Front 580lbs per wheel and Rear 620 lbs per wheel

Using 580lbs = 0.87g and 620lbs = 0.93g (consistent with the scale you are using)

Front = 580 * 0.87 * 2 = 1009lbs
Rear = 620 * 0.93 * 2 = 1153lbs

Total = 2162lbs

Car B
Curb Weight - 2000 lbs

Static dist. - Front 60% (600lbs per wheel) Rear 40% (400lbs per wheel)

100lbs per tyre forward under braking gives

Under braking - Front 700lbs per wheel and Rear 300lbs per wheel

Using roughly 700lbs = 0.8g and 300lbs = 1.2g (consistent with the scale you are using)

Front = 700 * 0.8 * 2 = 1120lbs
Rear = 300 * 1.2 * 2 = 720lbs

Total = 1840lbs

So here we have a lighter car exerting less braking force (considerably less) than the heavier one, because load distribution allows the heavier car to make better use of it.

I still think a piece here is missing.

While the 2400 lb car can exert a higher force, it takes more to stop it. Using your figures we see:
Car A
2162 lbs braking force/ 2400 lbs static weight = -0.9008333 G Decel

Car B
1840 lbs braking force/ 2000 lbs static weight = -0.92 G Decel

So even in your example.. the lighter car stops quicker..

Now while the initial coefficient of the heavier car is not 1.0g, it must be kept in mind that this figure only plays a part for a very sort amount of time and that full braking is not achieved instantly (you don't go from 0g to 1g instantly) but as soon as you start braking the load shifts. As a result the 'under braking' values are far more relivent than the initial ones (and if you go from accelerating to braking then the initial values are completely irrelivant).

The only time you can say that a lighter car will always make better use of its tyres is if every single other value is identical, and lets be quite honest that is never going to happen. In an earlier post I did mention and example of this, take a car and prep it for racing. So strip out the seats, interior and anything not needed and it will be lighter, add in a roll-cage (which will almost certainly be needed) and while this does push the weight up a bit the car is now much lighter overall. However far more than the weight has changed, we have totally changed the cog of this car and almost certainly changed its static weight distribution, all factors that will effect how much load is transfered under braking, cornering and acceleration.

Again I don't disagree, but its the load that causes this to occur not the weight itself (as the example of three differeing static distributions for a single curb weight illustrate).

Now while I have not taken this into account in the figures I have used above (that is a simple linear scale), if I had it would actually be more of a problem for the lighter FWD car than it would be for the heavier Mid-engined car, once again illustrating that how load is distributed is far more important than the weight itself.

Simply saying lighter = more braking force is overly simplistic and assumes that all other factors are exactly the same, as soon as any other variable changes the load transfer and distribution become far more important than the weight.

Again I have said exactly this repeatedly throughout this entire thread, you have simple picked on one example in which I was 'trying' to get a single point across, If you have the time to read the whole thread I think you will find that our opinions differ very little.

Regards

Scaff

I am sure there are cases (like one where the CG is high enough to cause significant transfer and the "heavy" read-biased car balances out to 50/50 where the "light" front bias car puts about 80-90% of its weight on 2 tires) where a heavier car will outbrake a lighter car given the same tires and braking systems (one that is "good enough" to lockup all tires), but I would say that this is not "normal".

the heavier car starts out at a disadvantage, and it takes a good deal of weight distribution "sloshing" to overcome this disadvantage. Given the "perfect" circumstances where both cars are 50/50 under braking, the lighter car generates -1.0g of braking force while the heavier car generates -0.9g. This *can* be overcome by favorable/unfavorable weight distribution, but it is still a hill to overcome.

-----------

I will come back to this again when I finish reading the entire thread. I am sure you understand these concepts, but others were not quite grasping them. In particular, the case where one thought weight matter because it was easier to push his remote with a heavy object instead of a lighter. The same is true for cars in that it takes more force to stop a heavier car than a lighter one. The point you are making is that most brake systems are strong enough to stop all but the heaviest of cars, thus it was the tires that is the "weak link". I mostly agree with this (though I dunno if modern braking systems can generate the 3.5-4.0 G of force an F1 car can utilize braking at 200 mph), but I was trying to add the subtleties of tire deformation under load starts to question some of the assumptions that were being made.

I am sure that after reading all of the thread this point will probably already been made and taken into account :)
 
I still think a piece here is missing.

While the 2400 lb car can exert a higher force, it takes more to stop it. Using your figures we see:
Car A
2162 lbs braking force/ 2400 lbs static weight = -0.9008333 G Decel

Car B
1840 lbs braking force/ 2000 lbs static weight = -0.92 G Decel

So even in your example.. the lighter car stops quicker..
Not quite.

The above would be true is weight were the final part of braking, if F = M * A was the final factor in braking. However its not the case, as the following two links cover....

http://www.stoptech.com/tech_info/The Physics of Braking Systems.pdf

http://www.teamscr.com/grmbrakes.html

http://www.stoptech.com/tech_info/wp_brakebiasandperformance.shtml


.....the figure you use here are vital in specing the braking system but don't ultimately determine stopping distances.


Scratch that - going mad and posting while busy is not a good idea.

I am sure there are cases (like one where the CG is high enough to cause significant transfer and the "heavy" read-biased car balances out to 50/50 where the "light" front bias car puts about 80-90% of its weight on 2 tires) where a heavier car will outbrake a lighter car given the same tires and braking systems (one that is "good enough" to lockup all tires), but I would say that this is not "normal".

the heavier car starts out at a disadvantage, and it takes a good deal of weight distribution "sloshing" to overcome this disadvantage. Given the "perfect" circumstances where both cars are 50/50 under braking, the lighter car generates -1.0g of braking force while the heavier car generates -0.9g. This *can* be overcome by favorable/unfavorable weight distribution, but it is still a hill to overcome.
The main point I'm simply trying to make is that talking about weight alone is and can be very misleading, which is why every technical paper on vehicle dynamics, every braking company white paper or tech paper pretty much ignores it and focuses on the transfer of load.

Weight distribution and load transfer are far more important that the actual weight and by focusing on that area (weight distribution) we get both a more accurate picture and a better understanding of what is occurring.



I will come back to this again when I finish reading the entire thread. I am sure you understand these concepts, but others were not quite grasping them. In particular, the case where one thought weight matter because it was easier to push his remote with a heavy object instead of a lighter. The same is true for cars in that it takes more force to stop a heavier car than a lighter one. The point you are making is that most brake systems are strong enough to stop all but the heaviest of cars, thus it was the tires that is the "weak link". I mostly agree with this (though I dunno if modern braking systems can generate the 3.5-4.0 G of force an F1 car can utilize braking at 200 mph), but I was trying to add the subtleties of tire deformation under load starts to question some of the assumptions that were being made.

I am sure that after reading all of the thread this point will probably already been made and taken into account :)
I hope you enjoy it....👍

Please do keep in mind that in a lot of cases simplified examples are used to try and get across what is a very complex subject in the most straightforward manner.


Regards


Scaff
 
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Not quite.

The above would be true is weight were the final part of braking, if F = M * A was the final factor in braking. However its not the case, as the following two links cover....

http://www.stoptech.com/tech_info/The Physics of Braking Systems.pdf

http://www.teamscr.com/grmbrakes.html

http://www.stoptech.com/tech_info/wp_brakebiasandperformance.shtml


.....the figure you use here are vital in specing the braking system but don't ultimately determine stopping distances.


Scratch that - going mad and posting while busy is not a good idea.

yup.. was just about to go into how those articles actually show that F=MA IS what determines how fast one stops... Furthermore, the StopTech PDF not only gives the formulas to walk through what we have done (summed up the corner forces and divided them by static weight) it even hints at what I am talking about:

"Note that in this analysis it is assumed that Upeakf and Upeakr are independent of deceleration, when in practice they are sensitive to the loaded changes brought about by the weight transfer phenomenon. Consequently, as weight is transferred the total vehicle deceleration capability is diminished by a small amount."

The main point I'm simply trying to make is that talking about weight alone is and can be very misleading, which is why every technical paper on vehicle dynamics, every braking company white paper or tech paper pretty much ignores it and focuses on the transfer of load.

Weight distribution and load transfer are far more important that the actual weight and by focusing on that area (weight distribution) we get both a more accurate picture and a better understanding of what is occurring.

I hope you enjoy it....👍

Please do keep in mind that in a lot of cases simplified examples are used to try and get across what is a very complex subject in the most straightforward manner.

Regards

Scaff

I have also found the part where this was brought up before by Scant on page 13, and he DID do a rather good job explaining things :) (especially the part where a loaded truck takes longer to stop than an unladen one)

He also tried to clear up the bogus response to "wide tires" that a High-school physics teacher (B.S.) "answered". While theoretically the teacher is correct in that the coefficient of friction shouldn't depend on area (pressure), tires, being deformable and all, DO change characteristics based upon pressure.

The elastic phenomenon that causes tires to lose grip when they are deformed under high load, is the same as what happens when a smaller tire has a larger average psi than a "large" tire. Pressure = load.. decreasing the pressure is the same as lightening the load, and is the better reason that larger tires give more grip over the "higher chance of making contact with the road" excuse (though there IS merit that a wide tire maintains it's contact patch better under camber change than a smaller one, so in part, it is true).
 
yup.. was just about to go into how those articles actually show that F=MA IS what determines how fast one stops... Furthermore, the StopTech PDF not only gives the formulas to walk through what we have done (summed up the corner forces and divided them by static weight) it even hints at what I am talking about:

"Note that in this analysis it is assumed that Upeakf and Upeakr are independent of deceleration, when in practice they are sensitive to the loaded changes brought about by the weight transfer phenomenon. Consequently, as weight is transferred the total vehicle deceleration capability is diminished by a small amount."

That's what happens when I try and post while getting my youngest daughter ready for Cub Scouts. Family and things about physics rarely go hand in hand.




I have also found the part where this was brought up before by Scant on page 13, and he DID do a rather good job explaining things :) (especially the part where a loaded truck takes longer to stop than an unladen one)

He also tried to clear up the bogus response to "wide tires" that a High-school physics teacher (B.S.) "answered". While theoretically the teacher is correct in that the coefficient of friction shouldn't depend on area (pressure), tires, being deformable and all, DO change characteristics based upon pressure.

The elastic phenomenon that causes tires to lose grip when they are deformed under high load, is the same as what happens when a smaller tire has a larger average psi than a "large" tire. Pressure = load.. decreasing the pressure is the same as lightening the load, and is the better reason that larger tires give more grip over the "higher chance of making contact with the road" excuse (though there IS merit that a wide tire maintains it's contact patch better under camber change than a smaller one, so in part, it is true).

If I recall correctly I believe I put a link to that post of Skants in the first post, it is a great summary.

Having had a chance to read more of the whole thread I hope you can see that views are not that different. Its not that weight doesn't play a role, more that a lot of other factors are just as, if not more significant in the braking process.

👍

Scaff
 
At 30 pages, I'm sure this has been discussed.

Can we talk about when to brake and full throttle at the same time? I noticed some people setting incredible GT4 lap times on youtube doing this.

Wikipedia has this to say:
One common race situation that requires left-foot braking is when a racer is cornering under power. If the driver doesn't want to lift off the throttle, potentially causing trailing-throttle oversteer, left-foot braking can induce a mild oversteer situation, and help the car "tuck", or turn-in better. Mild left-foot braking can also help reduce understeer.

Surely it's not wise to do this for every sweeping turn in every car. So what decides when this is the proper technique?
 
I'm still kinda new here, but anyway here's my 2 cents.

We shouldn't get too hung up on the maths and theory of braking (especially since we don't know for sure that the BBC is raw braking power to each axle- maybe Polyphony have already factored in static weight distribution for example).

I used to tune brakes by taking them to the test track and finding the setting that gave minimum stopping distance (250-100km/h). Most of the time this resulted in 24 rear brake strength. I have since found that this approach causes strange corner entry behavior, even though the rears are below the ABS threshold.

So these days I tune the BBC mainly on how it feels during braking and turn-in. Stopping distances into tight hairpins are longer, but it is well worth it for the improved balance in every other corner. I reckon this is proof that there are more important things that raw stopping power. Let's say you work out that a front heavy car will stop best with 80% front brake bias. I don't think this is much use if it sets the car up for plough understeer throughout the corner.
</rant>
 
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