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