Ok I was thinking aobut this- assuming tires are equal and all-

if every car had brakes strong enough to lock all 4 wheels at a given speed (say 60) all the way to stop (so heat fade isn't an issue), would every car have the same braking capabilites (stopping distance)? - lets not get into technicalities like who has the most aerodynamic drag and all. All 4 wheels can be locked so weight transfer isn't an issue. Assume all have equal sized wheels/tires.
ALSO!- Assume the perfect amount of force is given- IE, max force before lock up.

Before you scorn me and say "duh lightest car stops fastest" take into concideration that the lightest car will take the least amount of force to lock the wheels. And we all know maximum braking comes when neg torque from applied brakes = torque produced from wheel/ground contact/motion.

So the heavy car's brake system can apply a GREAT deal of force, but it has more mass to stop.

The lighter car has less momentuim, but the brakes can not apply much force before they lock up.

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now if the above is true- every vehicle would stop in the same distance, I argue this point.

What is we increased the strength of the braking system and threw on some bling bling duba dub 22' wheels

The braking force car be MUCH more before lock up (ie 22 chrome dumbass wheels are much harder to lock up) so now we have more braking force, almost equal linear mass. What happens now?

Sorry, my phys classes (ok class) has only covered things like a tire rolling down hill. Nothing when a solid body and a wheel - like a car- are mixed.

My intuition tells me as long as the brake can lock it up, all cars stop the same distance. I stumed some brake systems engineers with this question so its no so simple, clear cut or dry. And they are some SMART guys. Then again i dont think they cared much

 

I assume the force it takes to lock the wheels only take the wheel/disk into account, not the weight of the entire car.

 

umm, when you lock the wheels, only the wheels stop. not the whole car.

We are assuming "the perfect amount of force is given- IE, max force before lock up."

 

im not sure if i completely understand the question but here goes
1st question is wether the wheels are locked up...or if youre using an abs system that is able to apply maximum force to the brakes so the wheel is just before lock up...ill answer both questions as best i can.
if the wheels were locked up the car would stop in the same distance...without taking technicalities into account (contact patch, etc) the bigger tire would have more mass so it would add its inertia to the entire vehicle...but then the tire would have that much more weight pushing down on it, so it would respond with more force pushing back, thereby meaning that a car with a heavier weight will stop just as fast as a lighter car as long as the wheels are locked up...now on to the other part.

i sat here thinking for about an hour about angular momentum and all that then realized that the only force the tires will see is the stopping of the car...none of the angular momentum, thats just something the brakes have to overcome...so even with the heavy wheels it should still stop the same. its been awhile since ive been in physics so i might be wrong. it seems just like it would be adding more weight (static mass i guess). so the only reason you would need more brakes with bigger wheels is just because of that increased angular momentum.

 

IMO, your intuition isn't correct, but I think your final analysis of the SMART guys was correct.

My $.02

 

The amount of "force" available to stop the car depends on friction - Cf * Fn (coeffecient of friction * force normal). The actual stopping time is going to be based on dissipating the momentum of the car - Mc * Vc (mass of car * velocity of car).

So Lets say Cf * Fn = Cf * G * Mc (Gravity * Mass Car = Fn)

The stopping time is Momentum / Force, or

(Mc * Vc) / (Cf * G * Mc)

So Mc - Mass of car - cancels out, and stopping time simply depends on the initial velocity and your coefficient of friction (since gravity is effectively fixed for our frame of reference) - or

Vc / (Cf * G) = Tstop

So it really doesn't make any difference how heavy the car is - stopping distance just depends on how fast you are going when you apply the brakes and how "sticky" your tires are. Sure the havier car will "take" more torque to lock up the brakes, but it also has more momentum to dissipate.

Also you don't actually want to lockup the brakes - since the Cf for a sliding object is less than that for a static object (rolling wheel, since each contact patch is effectively static)

In the real world you can't lock up the tires all the time - try slamming on the brakes at 150 - so "better" brakes help. Also, though you may lock them up at 70 on the first stab what about the 10th, or 100th? Better brakes also have more fade resistance.


Chris

 

the wheels AREN'T LOCKED UP

perhaps i didn't word it propperly.


Most braking occurs RIGHT BEFORE lock up right? in this completly theoretical situation, every car has the perfect driver and the MAX force is applied BEFORE lockup-

 

Thank you Chris-
this was exactly the answer i was looking for, and i assume to applies when the tires arn't locked up either, i like iwas trying to get at.

I knew the "real world" part as well. Trust me, stopping a silverado around GVW on a downhill at 45mph is an issue

so assuming brakes are the only thing causeing the car to slow down and calipers are strong enough to lock the wheel. If the perfect about of braking force is applied (so we have maxium braking right before lockup), all vehicles would stop just as fast

 

The mistake here is in assuming that the coefficient of friction is independent of the normal force applied to the contact patch - it isn't. Since the tire depends on the interaction between the tread block and the road surface, and since this interaction is heavily dependent on the mechanical strength of the tread block, you can't expect to model this interaction using simple equations. At some point, the tread will give way, and all that high school physics goes out the window. The additional force not only tests the mechanical strength of the tread, but it also increases the thermal load and makes the problem even worse!

Heavier vehicles can also cause greater distorsion of the tire in the area of the contact patch, which causes greater inequalities in the distribution of the weight of the vehicle over the surface of the patch, which doesn't allow the contact patch to work as well as it could. This also adds to the thermal load as described above.

Additionally, you've got to consider that the tire is rolling, and that you've got an interesting situation where the tread comes down and "slaps" the pavement. This can actually set up a standing wave on the surface of the tread, and what you'd see is that you don't get consistant contact with the road at the leading edge of the patch. Heavier vehicles make the problem worse through the tire distorsion issue mentioned in the previous point.

No flames intended here, folks - just trying to demonstrate that this is yet another area where you need to take your physics to a deeper level than you're likely to in even a college classroom. There's a lot of non-linearities involved here, the kinimatics of a rotating tire are not as simple as the average 2-D physics problem, and the resulting set of equations quickly become complicated enough to justify a rather complex computer model.

 

Eric: I agree 100% - but at the same time you can take almost any problem, question and increase the precision/complexity of the model to the point where you can't easily get anything usefull out of it (without much more precise knowledge of the situation, etc.) - I definitely should have put a disclaimer on the "conclusion" quoted above.

I think within the context of the question tread/tire strength shouldn't be an issue though (or at least the context I assumed - something like a 2800lb car vs a 3800lb car both on good fresh tires, etc) - and though it would definitely be with repeated stops/breaking, I would submit at that point rotor/brake heat dissipation becomes a bigger issue with regards to performance. Definitely though during a long roadrace, etc. it is an issue.

I was not aware of a few of the other factors you pointed out - since you seem to have experience with this type of issue, can you give us a general idea of what type of real world difference we would see in a 2800 vs 3800lb vehicle due to the additional factors you mentioned (distortion of contact patch due to additonal weight, distortion for the tire slap,etc) - are we talking a 1% reduction in friction, 10%, etc?


Thanks, and good points!

Chris Bennight

 

Eric-
EVERY basic "problem" can be taken through the depths of physics hell

Thats why i wrote "competely theoretocal"
This situation is almost impossible to come about in th real world. My silverado doesn't have enough braking capabilities to lock the wheels form 30mph completely unloaded

 

Real world? You're probably talking in the 5-10% range, if you can truely set up an apples-to-apples comparision. It really depends on the tires, road surface, temperature, speeds, etc; all of these dramatically affect the tire's response to changes in "unit loading" (the ratio of contact patch to vehicle weight, which is highly dynamic).

My Impala SS can do a "real world" 60-0 stop in 117 ft, per my G-Tech (Motor Trend got a 120 ft 60-0 stop back in '94); I suspect than an optimal road surface would yield another 2 or 3 ft reduction. How? Wide tires with a large diameter (the larger the diameter, the less distortion of the contact patch for a given pressure and loading), proper biasing, and a killer stock ABS system. My friend's '96 Caprice (with an Impala rear axles so he's got discs all around) takes another 18 ft to stop on his 235/70-15 Michelins, so you can see that tires are a pretty big factor.

I don't mean to bust anyone for constructing theoretical arguments because they're a great place to start, but one has to incorporate a sufficient level of complexity into their model to ensure that it has some nominal amount of correlation to real-world results.

 

And the contact patch distortion, etc. issues would also explain the differences in traction/braking you see with wider tires (which you shouldn't see if Cf was constant). 5-10% is definitely significant enough to account for these differences - thanks for the information!


Chris

 

Did I say the whole car stopped when you lock the wheels???

 

You got it. If my buddy and I ran the same tire pressure in our cars of identical weight (we do, at least within a couple PSI or so), the contact patch would be the same size even though I've got much wider tires. However, due to the difference in width, the shape of the patch is different, and the skinnier tires on my friend's car require more carcass distortion/deflection to put down the same amount of rubber. This is one reason why wider tires will generate more traction than skinny ones for a given vehicle weight and tire pressure.

All of this just barely scratches the surface of the traction issue, and after looking back on the topic I think we've gone a bit off-course. From a brake standpoint, generating maximum stopping force actually isn't terribly difficult - if you've got enough force to lock the tires, the thermal capacity to avoid overheating during the particular length and number of stops that you'll be making, and the proportioning to make sure that you don't lock one up prematurely, stopping distance becomes a tire issue and not a brake issue. Generating stopping force isn't a problem, but generating it without a great deal of noise, dust, and wear and in a controlled manner is. Thermal capacity isn't difficult to acheive until you throw in cost and weight targets; oh, and we'd like to be able to stick these brakes behind tiny steel wheels on the base model, too. Proportioning appears to be a real SOB, at least in my (extremely limited) brake design experience. Anti-lock helps to cover up warts in your proportioning scheme, at least to a certain extent.