Shorter Stopping Distance for ultralight vehicles?

[shamrock5585]

ok fools here's the math to prove it... you proved that your work from the friction and the momentum of the vehicle add up to be your total force and that will be independant of mass... but then we got good old Newton on my side... F=m*a now the force could end up being the same if your math is correct... but if you divide the equal force by a larger mass then your acceleration (negative so its deceleration) will be smaller therefore the care with less mass will have a faster deceleration!

 

[Mech_Engineer]

I don't know what it is you're not getting, but your "math" was ineffective to say the least. You lost me at adding work and momentum to get force...

 

[mheslep]

fair enough... seeing as almost all cars these days have abs, they're advertisement would be true. that is a pretty interesting concept tho that by locking the tires the stopping distance could be independant of mass. it is kind of a hard concept to grasp. i don't know if i buy it. ...

Yes, IIRC, when I learned to drive back sometime in Model T days and everyone just skidded on locked tires to a stop, they used to just declare in driving courses that all 4 wheel cars (regardless of weight) stopped from 60mph on average in something like 160ft (i think) on dry payment; some multiple of that on wet pavement, etc. You had to know it for the test.

 

[Stingray]

ok fools here's the math to prove it... you proved that your work from the friction and the momentum of the vehicle add up to be your total force and that will be independant of mass... but then we got good old Newton on my side... F=m*a now the force could end up being the same if your math is correct... but if you divide the equal force by a larger mass then your acceleration (negative so its deceleration) will be smaller therefore the care with less mass will have a faster deceleration!

Lose the attitude. You don't know what you're talking about.

To a first approximation, every car can brake at the capacity of its tires (at least if you don't keep doing it so much that everything overheats). The braking force those tires generate is roughly proportional to the vertical load on them. The total force is therefore proportional to the car's mass. Newton's second law then says that the maximum deceleration is independent of mass.

This is not the greatest approximation in real life, but it is pretty much correct that properly designed vehicles all stop in similar distances on similar tires. One problem in real life is distributing the braking force optimally to all tires. No fixed ratio between the front and rear axles will always be best. This is largely fixed nowadays with electronic brake distribution, but is not a trivial problem. There are also tradeoffs between stopping distance and stability. Locking up the rear tires and not the fronts will cause a car to spin, for example.

It is also not quite true that the maximum braking force a tire can produce is directly proportional to load. The effective friction coefficient actually decreases a little at higher loads. This is a relatively small effect, but very important in racing.

 

[shamrock5585]

The braking force those tires generate is roughly proportional to the vertical load on them. The total force is therefore proportional to the car's mass. Newton's second law then says that the maximum deceleration is independent of mass.

ok the original argument says that, from the math... your friction force working against you and your force due to momentum will end up being the same for small and large vehicles because in the equations mass cancels out... but that's just your force... so your force, which is the same for a mac truck and a small car, can be plugged into F=m*a right? so for a small car your force is divided by a smaller number... rendering your deceleration to be larger which means you will stop faster... i don't see a flaw in that logic but if I am wrong I am wrong

 

[shamrock5585]

ok i see the flaw in my math... need to reup on the work energy theories instead of always dealing with force... doesn't seem logical that they would stop at the same distance... in real life they wont... but theoretically they should...

 

[Stingray]

ok i see the flaw in my math... need to reup on the work energy theories instead of always dealing with force... doesn't seem logical that they would stop at the same distance... in real life they wont... but theoretically they should...

I don't know if you're really seeing your errors or not. This has nothing to do with the amount of work required to stop a car.

In any case, cars with very different weights do stop in similar distances in real life. The test posted above shows that. Just to drive home the point, I looked up stopping distances for a 2,000 lb Smart FourTwo and a 5,300 lb Lexus LS600h L. According to Road and Track, the heavier Lexus was actually a little better. Stopping distances from 60 mph differed by only 8%. There was a 6% difference from 80 mph. These stats shouldn't be taken too seriously, but there's clearly no significant problem for heavy cars.

Trucks have problems for two reasons. The most important is that their weight distribution can change drastically. This makes it very difficult to design a braking system that will always work well. Something that might be optimal for a fully loaded truck would cause an empty one to be completely unstable. Trucks also tend to come with tires that are optimized more for large loads, slow wear, and low rolling resistance. This usually has a negative impact on their maximum braking or cornering performance. Large heavily loaded trucks might also get far into the nonlinear friction regime of the tires I mentioned before. I'm not sure.

 

[shamrock5585]

ok this has a lot to do with work, i don't know why you say that... work energy is directly related to stopping the car and is the basis for the equation that proves that stopping distance is not dependant upon mass. I don't care about two small cars traveling at the same speed... an 8% difference is a big difference and these cars are completely different... my whole thing was that a mac truck and a motorcycle will not stop at the same distance... theoretically they would, but if you stop a mac truck that fast the axles will snap and your cargo will role right over you...

 

[Stingray]

ok this has a lot to do with work, i don't know why you say that... work energy is directly related to stopping the car and is the basis for the equation that proves that stopping distance is not dependant upon mass.

No. That comes from saying that frictional processes tend to satisfy F_{\mathrm{max}} = \mu mg. All that's being used is the maximum force a tire can generate with a given weight on top of it. Talking about the work done is a pointless complication.

I don't care about two small cars traveling at the same speed... an 8% difference is a big difference and these cars are completely different...

An 8% difference between two cars with a 170% difference in mass is not much; especially when the heavier car stops faster. Braking performance isn't that repeatable anyway. 5-10%variations could probably be expected in the same car on different (dry) days. If you need another example, most passenger cars tend to brake a little better than motorcycles.

my whole thing was that a mac truck and a motorcycle will not stop at the same distance... theoretically they would, but if you stop a mac truck that fast the axles will snap and your cargo will role right over you...

I doubt that trucks are so fragile that their axles would snap trying to stop that quickly. The trailer may become unstable and snap around. I don't know. Your example introduces additional problems that have very little to do with a vehicle's mass. If you just focus in the point of this discussion, mass is found to be only a minor factor in one-time stopping performance.

 

[shamrock5585]

hahaha... dude you realize that just because the stopping distance is the same does not mean that the energy used to stop the vehicle is the same... it takes a lot of energy to bring a truck to a stop and that amount of energy is due to the mass. your telling me it has nothing to do with work when in the opening discussion he presents an equation directly showing work due to friction... and... if you stop a truck as fast as a motorcycle you would definately be f ucked

 

[shamrock5585]

one thing to keep in mind is that anti lock brakes void pretty much all of this because then all you are paying attention to is the friction from the brakes applied to the wheels themselves and that friction has nothing to do with the mass. So then you are dealing with energy put into the brakes vs. the momentum of the vehicle (which has to do with the mass)

 

[mheslep]

one thing to keep in mind is that anti lock brakes void pretty much all of this because then all you are paying attention to is the friction from the brakes applied to the wheels themselves and that friction has nothing to do with the mass. So then you are dealing with energy put into the brakes vs. the momentum of the vehicle (which has to do with the mass)

No, air brakes can stop rolling or lock the wheels. For the latter you are back to a mass x tire-road friction x g stop again.

 

[Mech_Engineer]

one thing to keep in mind is that anti lock brakes void pretty much all of this because then all you are paying attention to is the friction from the brakes applied to the wheels themselves and that friction has nothing to do with the mass. So then you are dealing with energy put into the brakes vs. the momentum of the vehicle (which has to do with the mass)

No, you have that completely backwards. If the brakes are strong enough to lock up the tires on a full power application (necessitating the use of the ABS system) then the limiting factor for the stopping distance of the vehicle is the coefficient of friction between the road and the tires.

The case you describe is only applicable if the brakes are not strong enough to lock up the tires, e.g. they are undersized. In this case, the stopping ability of the car will be dictated by the maximum power dissipation capacity of the brakes, an ABS system does not even get used since the tires do not lock up.

 

[ank_gl]

you guys, you confuse a lot.
A brake is used because a skidding surface creates a lot more friction than the pure rolling(a car wheel has something close to a pure rolling, an engine is just balancing the friction just like jet engines balances only drag for aircrafts). A brake locks up the tire which starts to skid, hence causing more friction, hence causing more motion opposing force(friction always opposes motion).

Actually Braking force = μ*normal reaction(=weight of vehicle), μ heavily depends on braking system(heavier braking => more skidding => more μ)
But assuming braking force is same for both vehicles,
F = Mass * Acc
So a lighter vehicle observes a greater acceleration.

Also
v^2 - u^2 = 2*Acc*Distance
Assuming both vehicles applied brakes at same initial velocity,
Distance = Constant / Acc

Larger the acceleration(retardation actually), smaller the distance for stopping.

Assumption:
1. Both vehicles have same braking force(depending heavily on brake assembly and surface contact between tire and road).
2. Both start braking from same initial velocity

Result:
Lighter car stops earlier.

Advantage:
I don't really see any advantage, distance for stopping depends heavily on braking system and tire for reasons stated above. Only advantage with a lighter body is that force needed to reach a velocity is smaller => smaller power unit

 

[ank_gl]

of course the parameters are too many and too difficult to control, so it won't be appropriate to reach a conclusion like this and say vaguely that a lighter vehicle ll stop earlier

 

[Hurkyl]

you guys, you confuse a lot.
A brake is used because a skidding surface creates a lot more friction than the pure rolling(a car wheel has something close to a pure rolling,

Brakes, when functioning properly, use neither kinetic friction nor rolling friction: they use the static friction between the wheel and the road.

 

[Mech_Engineer]

you guys, you confuse a lot.
A brake is used because a skidding surface creates a lot more friction than the pure rolling(a car wheel has something close to a pure rolling, an engine is just balancing the friction just like jet engines balances only drag for aircrafts). A brake locks up the tire which starts to skid, hence causing more friction, hence causing more motion opposing force(friction always opposes motion).

This is just flat-out wrong. Static friction between a tire and the road (tire is rolling) is greater than dynamic friction (e.g. the tire is skidding). This is why ABS-equipped cars stop faster than ones without ABS.

This can be seen in braking and acceleration tests everywhere. Cars that don't lock up their wheels stop faster than ones that do, and cars that don't spin their tires excessively on acceleration (traction control) accelerate faster than ones that do burnouts. All proof that static friction is greater than dynamic.

 

[ank_gl]

Brakes, when functioning properly, use neither kinetic friction nor rolling friction: they use the static friction between the wheel and the road.

I meant that only, static friction is much more than rolling friction, i just missed the term "static"

This is just flat-out wrong. Static friction between a tire and the road (tire is rolling) is greater than dynamic friction (e.g. the tire is skidding).

Are you a mechanical engineer?? When tire rolls(assuming true rolling), there is no relative motions, there exists a point contact, so friction(call it anything static, kinetic or rolling, just anything) is less. When tire skids(tire locks up), there IS a relative motion, friction is more. A practical case is like a 95% rolling & 5% skidding(because a point contacts flats out to a surface contact).
No offense, but you should rather try to predict results from basic physics, not making laws from applications.

This is why ABS-equipped cars stop faster than ones without ABS.

Is it really so?? I always thought ABS equipped car stops slower because of intermittent braking action. Sorry, i don't like driving that much, so i don't know. Can you show some test results

 

[Hurkyl]

I meant that only, static friction is much more than rolling friction, i just missed the term "static"

You weren't talking about static friction: you were talking about kinetic friction (two surfaces actually sliding past each other).

Static friction is generally much greater than kinetic friction, and it is the principle upon how brakes (and accelerators!) act: they apply a torque to the wheel. Since the bottom of the wheel is stationary relative to the road, we get a static frictional force applied to the bottom of the wheel that acts to oppose the torque. This force is what causes you to decelerate/accelerate.

If the wheels start skidding (during braking) or start burning rubber (during acceleration), then you are now experiencing kinetic friction, which means you cannot brake or accelerate effectively.

 

[Mech_Engineer]

I meant that only, static friction is much more than rolling friction, i just missed the term "static"

Rolling friction is a negligible effect on a car that is braking.

Are you a mechanical engineer??

Yes.

When tire rolls(assuming true rolling), there is no relative motions, there exists a point contact, so friction(call it anything static, kinetic or rolling, just anything) is less. When tire skids(tire locks up), there IS a relative motion, friction is more.

You clearly have a poor understanding of which constants or physical quantities apply to a decelerating vehicle, and you're not defining your frictional constants properly. You can't call them anything you want because each term has a specific meaning. These are the definitions:

Rolling Friction- http://en.wikipedia.org/wiki/Static_friction#Kinetic_friction
Drag coefficient used as a descriptive quantity to approximate the rolling drag on a rotating object due to surface imperfections; it has nothing to do with any braking force that is being applied. This is not a useful quantity in determining how quickly a car will stop since it tends to be small WRT other frictional quantities.

Static Friction- http://en.wikipedia.org/wiki/Static_friction#Static_friction
Frictional constant used to describe the friction between the tire and the road when it is NOT skidding. It is applicable because although the tire is rotating, the tread on the contact patch is stationary WRT the ground it is touching. This coefficient can be used to approximate the maximum acceleration force available for the car, since it is generally larger than Kinetic friction.

Kinetic Friction- http://en.wikipedia.org/wiki/Static_friction#Kinetic_friction
Frictional constant used to describe friction between two bodies that are moving WRT each other. In the case of a tire, it has to be skidding for Kinetic friction to apply. Kinetic friction is generally smaller than static friction. In applications on vehicles, kinetic frictional force decreases as velocity increases; so the faster a tire is spinning the less force it imparts towards accelerating the car (this may have to do with the tire's temperature increasing).

No offense, but you should rather try to predict results from basic physics, not making laws from applications.

I have been doing just that for four pages now. Did you read any of it?

Is it really so?? I always thought ABS equipped car stops slower because of intermittent braking action. Sorry, i don't like driving that much, so i don't know. Can you show some test results

Yes. I already posted a result that shows exactly that (emphasis added):

But look at this next graph:

http://www.roadandtrack.com/assets/image/7162003125016.gif
http://www.roadandtrack.com/article.asp?section_id=3&article_id=663&page_number=9

In the "exotic" class, the Saleen S7 is pitted against the Lamborghini Murcielago. The S7 weighs in at 3050 lb, a full 1140 lbs lighter than the Lamborghini. Yet, the Lamborghini stops 70 feet shorter and 0.8 seconds faster from 100mph than the S7. Why?

Both cars have the exact same tires fitted (Pirelli P Zero Rosso's, 245/ 35ZR-18 front and 335/ 30ZR-18 rear), so the answer has to be a combination of more traction available to the Lamborghini because it weighs more, and the fact that the Saleen does not have ABS. The Saleen should have more braking power available, since it has 1" larger discs in the front and 0.8" larger dics in the rear, but its traction is limited by its lighter weight, and its lack of ABS causes the tires to lock up easily...

The effects of no ABS can be seen in the graph, where the Lamborghini's braking curve is completely linear all the way to from 100 to 0 mph, while the Saleen's fluctuates wildly since the driver has to modulate the pedal to try and make up for the lack of ABS. Even though the Saleen was much faster to 100 mph, it ironically loses the 0-100-0 because the Lamborghini is HEAVIER (more traction available from the same set of tires) and has ABS. The Lamborghini puts down an average of 606 braking hp, versus the Saleen's "paltry" 370 braking hp.

https://www.physicsforums.com/showpost.php?p=1746924&postcount=15

[NOTE- Of all the cars tested, the Saleen S7 was the slowest to stop from 100mph, despite having one of the lightest weights. It was also the only car not equipped with ABS.]

The links to the article I reference changed, so here is the updated link:

http://www.roadandtrack.com/article.asp?section_id=3&article_id=663&page_number=1"

 

[ank_gl]

short on time, all i am saying is the weight of car is not the only deciding factor, the braking capability is too. How fast a tire locks and how heavy it does, determines 'mu'. So
force = 'mu' * weight
Its a combination of both.

 

[NateTG]

short on time, all i am saying is the weight of car is not the only deciding factor, the braking capability is too. How fast a tire locks and how heavy it does, determines 'mu'. So
force = 'mu' * weight
Its a combination of both.

Newton gives us:
F=ma

Then we have
F=\mu \times mg
so
\mu \times mg = ma
\mu \times g = a
so to a first order approximation, the acceleration is independent of the mass of the vehicle.

 

[mheslep]

Newton gives us:
F=ma

Then we have
F=\mu \times mg
so
\mu \times mg = ma
\mu \times g = a
so to a first order approximation, the acceleration is independent of the mass of the vehicle.

Arg. We're going backwards. For ABS or some kind of non-skidding braking scenario, F=\mu \times mg where F is the tire/road force, is not the applicable equation. The thread clearly establishes for this scenario that stopping distance and vehicle deceleration clearly are dependent on mass in addition to the wheel-brake force.

 

[Stingray]

Arg. We're going backwards. For ABS or some kind of non-skidding braking scenario, F=\mu \times mg where F is the tire/road force, is not the applicable equation. The thread clearly establishes for this scenario that stopping distance and vehicle deceleration clearly are dependent on mass in addition to the wheel-brake force.

The same arguments work whether the tires are locked or operated at their optimum slip ratio (via ABS or a careful foot). The only difference is that the effective friction coefficients are different in the two cases.

 

[mheslep]

The same arguments work whether the tires are locked or operated at their optimum slip ratio (via ABS or a careful foot). The only difference is that the effective friction coefficients are different in the two cases.

The relevant difference is where the work is done: locked tires - work is done on the tires/road surface involving vehicle mass, optimum slip - work is done mostly on the brake pads/disks/wheels and does not involve vehicle mass.

 

[Stingray]

The relevant difference is where the work is done: locked tires - work is done on the tires/road surface involving vehicle mass, optimum slip - work is done mostly on the brake pads/disks/wheels and does not involve vehicle mass.

It doesn't matter which components have the most heat transferred to them. ABS (roughly speaking) tries to keep the tires operating at a point where they generate the most longitudinal force. That force is approximately \mu_s mg. That's all that's important. This obviously depends on the driver applying enough force on the brake pedal and the various components translating that force to the calipers. Any production car in reasonable working order will be able to reach this limit at least for one stop from highway speeds (repeated tries will eventually overheat things).

 

[mheslep]

It doesn't matter which components have the most heat transferred to them...

How do you conclude that? Heat build up drives the braking power limitations. If one wants more braking power you need more thermal mass in the brakes.

In any case, the topic is degree to which vehicle mass may / may not impact stopping distance: mass is a factor in the derivation of stopping distance in ABS / careful foot vehicles.

 

[Stingray]

How do you conclude that? Heat build up drives the braking power limitations. If one wants more braking power you need more thermal mass in the brakes.

In any case, the topic is degree to which vehicle mass may / may not impact stopping distance: mass is a factor in the derivation of stopping distance in ABS / careful foot vehicles.

The point that has been repeated multiple times is that braking power is not affected by thermal issues for normal vehicles under normal conditions. If you just want to know the shortest distance for a single stop, tires are always the limiting factor.

Overheating is only an issue when considering multiple hard stops from high speeds in a short amount of time. That's what happens when racing or otherwise driving in a very illegal manner on winding roads. I've managed to overheat the brake pads in a couple of cars, but it is honestly very hard to do. It is not what has been discussed here so far, and is not relevant for most road vehicles.

I'm really getting tired of repeating myself in this thread (especially since other people have been saying the same thing). To lowest order, mass is not a factor in a normal ABS stop. If you want to look at it in terms of work, that's proportional to force, which is in turn proportional to mass. The kinetic energy of the car is also proportional to mass, so it cancels out. That's the same for ABS or skidding stops.

 

[Mech_Engineer]

I'm really getting tired of repeating myself in this thread (especially since other people have been saying the same thing). To lowest order, mass is not a factor in a normal ABS stop. If you want to look at it in terms of work, that's proportional to force, which is in turn proportional to mass. The kinetic energy of the car is also proportional to mass, so it cancels out. That's the same for ABS or skidding stops.

Exactly. Nothing new has been said in this thread in 2 pages.

 

[mtworkowski@o]

You know you're saying that a heavier car stops faster than a lighter car...