ColorChange

Larry: When you are talking dynamic situations, the weight is critical. It is just that in this specific case, weight is insignificant in stopping distance. The equation Force = masss x acceleration applies. That is why the downforce is So beneficial, you get much higher (though nonlinear) tire force, but you get to use the same mass so the acceleration goes throught the roof. That is why these guys pull 5 g's in braking and almost 4 in cornering. The tire force is way up do to downforce but mass stays the same so the acceleration gets pegged.

Now in the example you give, if you increase the mass, the tire force doesn't change much (the point of this thread) but the acceleration suffers. So instead of 3.0 g's throught the turn, the driver can only pull 2.95, and these all add up to the 2 seconds you correctly identified when they drop the 20 or 30 kgs of fuel.

2BWise

I've quietly been following this thread and think that maybe its time I try and contribute some information that may help our understanding. A few equations have been thrown out there for Force and Energy but we're only looking at one half of each equation.

First: Force=mass * acceleration, which is then equal to the braking
Force=(cf)*N, where N is the normal force equal to mass * gravity
-> Force=mass *acceleration=(cf)*mass*gravity, mass is constant so it cancels. Then if you are on the threshold of braking force the cf will be at its max therefore they will be capable of stopping in the same distance. (I have not included drag in this equation since it is the same for identical cars)
I know that this is a very simple model, but it implies that mass has no impact.

Secondly: The energy of the car.
Through conservation of energy -> the change in energy = the change in KE + the change in PE + the change in internal energy + work done on the car = 0.

State 1: KE = .5*mass*velocity squared.
PE = zero, assumed to be zero because its in motion.
U (internal energy) = a constant
Work = Braking Force*distance
State 2: KE = .5*mass*velocity squared.
PE = zero, assumed to be zero because its in motion.
U (internal energy) = a constant with the same value or nearly the same value as U in ST 1

Change in Energy -> ST 1 - ST 2 = 0, so ST 1 = ST 2
(.5*mass*(V initial)^2) + 0 + U + (Braking Force*distance) = (.5*mass*(V final)^2 )+ 0 + U

U's will be very close so we assume their impact is neglible, so
(.5*mass*(V initial)^2)+(cf*mass*gravity*distance) = (.5*mass*(V final)^2 )

Again mass cancels, so if cf is maxed, and the velocities are the same for both vehicles, distance will be the same value for both.

CC you seem to understand this, please correct me if I'm wrong, but I think I've adequately shown that braking distance is not dependent on mass.

jetskied

My take is given the same car, tire and brake and if you can reduce the weight you will obtain some reduction in stopping distance. Is it measurable? Pobably not significant. Its a simple fact of F=M*A. With a little more weight there is more force to stop. Given every thing is the same as far as the mechanical parts more weight should take longer to stop. This can be negated if you place the weight in a stratigic location that will help get better balance from the car which can have a better contact patch for all four brakes and tire.

RJay

Totally agree. Of course once you exceed that threshold and lock them up, as you flat spot those tires, which will stop faster? That might seem like a silly thing to consider except given the initial premise of this thread I can imagine a pretty important on-track situation when that happens, like when you've lost it and you're both feet in.

There is an assumption here, which to my mind is as yet unproved, that the braking system can overwelm the tires grip at any arbitrary speed. On the rare occassions that I've track my 996, I can't recall ever going into abs above what I'd guess was 70 MPH (stock brakes and pads, 225/285 Dunlop SSRs, 3200#s), despite having dragged the car down from 130+ pressing for all I'm worth. With F1 cars with the grippiest tires, gobs of downforce and the most powerful brakes on the planet, I've never seen them lockup on initial or even intermediate braking, its always seems to be the result of a failure to reduce pedal pressure after the car has slowed significantly and downforce is lowered at which point the grip is exceeded. This suggests to me that even an F1 car at relatively low speeds by F1 standards (say above 100) can't use all the available grip of the tires for braking. I'd really be surprised if the Toureg could decel from 130-70 within 8 feet of an Elise. So yeah, perhaps under some threshold speed that a mathematically challenged ignoramus such as myself will never be able to calculate, weight is relatively unimportant, but when the speeds are higher, I suspect it plays an important role.

Rich Sandor

RJay, F1 braking depends a lot on how hard the driver can push the pedal in. ASAIK, they have no power assist for the brakes. I've read that they literally stand on the pedal into high speed braking zones.

Cupcar

Look at it this way, would a given Porsche stop in the same distance if ballast were added in a fashion that did not change the center of gravity or the weight distribution and all other things, tires, brakes, etc. are left unchanged.

Lets say we add 1000 pounds in this way to a 3000 pound Porsche and increase the weight to 4000 pounds, would the car stop in the same distance, a greater distance or a lesser distance from a given speed under a maximal deceleration attempt?

I would say, since the coefficient of friction (grip) of a tire goes down as the load on the tire goes up, it would take a longer distance to stop at 4000 pounds than 3000 pounds.

Rich Sandor

If you push harder on an eraser, is there more or less friction?

ColorChange

RJay: OK, I can agree with you! Weight is insignificant as long as it's not a huge amount, and the stopping distance doesn't change as long as yyou don't overheat the brake system (fast starting speed).

CupCar: Please reread my post on page 3, and try to get the whole point of this thread. Weight does not have a significant effect on stopping distance from 60-0 in a normal car. In fact, adding even 1,000 lbs to a Porsche will NOT cause the car to stop in a much longer distance.

Rich: The harder you push, the more friction. This is always the case. The whole point of this thread is the increase in friction force does not go up liniearly with the normal (pushing) force. It is close for a long range, but not for all the range.

RJay

BTW,Car and Driver lists the following 70 to 0 results:

Elise: 160 ft.
Toureg V6: 189 ft.

Interestingly it also lists:
Cayenne (Presumably V6 as its listed as an S): 175 ft.
Cayenne Turbo: 170 ft.

Boxster: 153 ft
997: 150 ft

So from the data we can of course infer that as an SUV or a sports car gets heavier, it stops in a shorter distance, but sports cars, because they are so sporty, will stop in shorter distances than SUVs.

centerpunch

I think "coefficient of friction" is always a number less than 1.0, and the interaction between tires and pavement is more complex than simple friction (which is proven by some vehicle's abilities to exceed 1.0g cornering, even at low speeds where aerodynamic downforce is not a factor.....)

Larry Herman

That's the factor being left out of the equation. You cannot measure the grip of a tire by simple CF. It is much more complicated than that. You have some static friction where the tire is in firm contact with the road, some sliding friction where the tire is moving around a little, and some mechanical tension where the rubber is actually impressed into the surface of the road, clinging to all the edges and imperfections. That's why the physical models do not work, because they do not encompass all three, which probably can exist within the tire patch at the same time.

macnewma

The myth that the coefficient of friction of a tire cannot exceed 1.0g was perpetuated by the tire engineers of the 60s because they could not seem to build a tire that would generate more than 1.0g of grip. That has since been exceeded greatly. Stick Auto-x Hoosiers will grip far in excess of 1.0g. I have heard them estimated around 1.6-1.7g.

I agree that this makes it difficult to assign an absolute number to a tire/car for cF, but that only means they will have difficulty giving us an absolute stopping distance from an equation.

I think it has been determined that the cF and the speed of the car determine its stopping distance. What has yet to be determined is the extent to which cF is affected by increasing weight. (or maybe I missed it as this thread is getting long)

If someone felt so inclined, they could take a car like a 951 that is pretty well balanced front to back and take it for two runs with cold tires and brakes. One with about 400-500 lbs placed near the CG of the car and one without. Make a threshold braking run from 80 mph and check the distances. We should then be able to estimate the affect weight has on cF.

Anyone know a good way to secure 500lbs to the center console of a 951?

I think we have gotten to the point where we have determined that weight affects cF, but we can't agree to the extent.

Cupcar

A nice read is Paul Haney's book The Racing and High Performance Tire.

The Coefficient of Friction is the ratio of the normal force and the lateral force on a contact patch. If a tire had a normal (vertical force) of 100 pounds and supported a lateral force of 100 pounds before sliding then the coefficient of friction would be one.

The equation for Friction Force= Coefficient of Friction X Normal Force.

Rubber is wierd stuff it is not a simple friction material that responds to load as a linear model. Rubber drops it's "coefficient of friction" as the surface loading- pounds per square inch on the contact patch increases. Rubber looses relative grip as it is loaded. This is why sway bar tuning works for example as load is shifted across an axle pair of tires.

Another way to look at it is for each increment of vertical force the tire can support a relatively lesser increment of horizontal force can be supported before the tire slides.

This is also one reason why wings are such an effective device to improve cornering, they provide vertical load with essentially no increase in horizontal (cornering or braking) load. This is why weight is so important in a wing equipped car like an F1 car.

When you push an eraser harder on a table the Total Friction Force is increased through increased Normal Force, however if you looked carefully at the situation and measured the relative increase in lateral force as you pushed the eraser versus the increased normal force you applied you would find that the actual coefficient of friction dropped. The total goes up, the ratio goes down.

This would be the same with the heavier car, all other things being equal in the chassis, the heavier car has to take longer to stop. This would not be the case if rubber were a simple friction material and that is not to say that a heavier car could not be made to stop as fast as the lighter car given larger tires.

Larry Herman

I'm half way there

That is what I was trying to remember. That was the essence of the talk that Don *** gave. He went on about how this affects suspension tuning etc.

Cupcar

It is the essense of suspension tuning as I understand it.