Shorter Stopping Distance for ultralight vehicles?

In summary, the concept of ultralight vehicles focuses on reducing mass by 2 or 3 times to improve fuel efficiency. While there is mention of shorter stopping distances as a safety benefit, there is no elaboration on why this is so. The standard stopping distance derivation shows that stopping distance is independent of mass, but there may be other factors such as chassis, suspension, tires, and brake design that contribute to this claim. The article on ultralight vehicles discusses the safety benefits of lightweight design, but the exceptions are high-speed head-on collisions and side impacts with significantly heavier vehicles. Lighter vehicles may have advantages in terms of acceleration, brake ventilation, and use of softer tires, but there is no fundamental reason why they can stop
  • #36
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...
 
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  • #37
shamrock5585 said:
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.
 
  • #38
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...
 
  • #39
shamrock5585 said:
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 [itex]F_{\mathrm{max}} = \mu mg[/itex]. 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.
 
  • #40
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
 
  • #41
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)
 
  • #42
shamrock5585 said:
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.
 
  • #43
shamrock5585 said:
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.
 
  • #44
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
 
  • #45
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
 
  • #46
ank_gl said:
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.
 
  • #47
ank_gl said:
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.
 
  • #48
Hurkyl said:
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:frown:?? 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
 
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  • #49
ank_gl said:
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.
 
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  • #50
ank_gl said:
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.

ank_gl said:
Are you a mechanical engineer:frown:??

Yes.

ank_gl said:
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).

ank_gl said:
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?

ank_gl said:
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):

Mech_Engineer said:
But look at this next graph:

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

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" [Broken]
 
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  • #51
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.
 
  • #52
ank_gl said:
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:
[tex]F=ma[/tex]

Then we have
[tex]F=\mu \times mg[/tex]
so
[tex]\mu \times mg = ma[/tex]
[tex]\mu \times g = a[/tex]
so to a first order approximation, the acceleration is independent of the mass of the vehicle.
 
  • #53
NateTG said:
Newton gives us:
[tex]F=ma[/tex]

Then we have
[tex]F=\mu \times mg[/tex]
so
[tex]\mu \times mg = ma[/tex]
[tex]\mu \times g = a[/tex]
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, [tex]F=\mu \times mg[/tex] 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.
 
  • #54
mheslep said:
Arg. We're going backwards. For ABS or some kind of non-skidding braking scenario, [tex]F=\mu \times mg[/tex] 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.
 
  • #55
Stingray said:
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.
 
  • #56
mheslep said:
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 [itex]\mu_s mg[/itex]. 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).
 
  • #57
Stingray said:
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.
 
  • #58
mheslep said:
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.
 
  • #59
Stingray said:
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.
 
  • #60
You know you're saying that a heavier car stops faster than a lighter car...
 
  • #61
Mech_Engineer said:
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 ame set of tires) and has ABS. The Lamborghini puts down an average of 606 braking hp, versus the Saleen's "paltry" 370 braking hp.

So there you have it, a case where being heavier means a shorter stopping distance... :wink:

I suspect the engine location in these vehicles also plays a role. In most front-engine cars the braking is proportioned about 70%F, 30%R. By moving the C.O.G. closer to the rear of the car, the rear brakes can actually do something useful and decrease the cars overall chance of skidding.

BTW...some quick and dirty Dynamics

[tex]N_R=mg(\frac{a-uc}{l})[/itex]

[tex]N_F=mg-N_R[/itex]

c=height of COG from ground

l=wheelbase

a=distance between CoG and front axle
 
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  • #62
I stumbled across this thread when trying to learn more about braking. There have been some good things written here, but there has been a lot of total crap as well. Many posters know only enough to be dangerous. The purpose of me creating an account and a post is more for those in the future who (like me) will come across this post. I doubt I will change the mind of many previous posters, but hopefully I will bring some things up they haven’t thought about.
Mass – Mass has a significant affect on braking. It does not “cancel out” of equations involving the full braking system. Mass certainly doesn’t create an advantage as Mech Engineer said! Have you ever tried braking in a car loaded with more passengers than seats? Your braking distance obviously INCREASES, and significantly.
Tires - Why does the braking distance increase with mass? F=ma. An increase in mass yields a decrease in acceleration given the same force. But doesn’t an increase in the mass of the car increase the force the tire can put on the road? Yes. Does that mean they cancel out? NO.
There has been a fatally flawed assumption about this throughout the thread. Many posters have used the old high school physics equation for normal force, F=mu*m*g . This was a basic approximation for something like sliding a block across a desk. It cannot be used for something as complicated as tire compounds and road surfaces. The curve of load vs. traction is NON-linear for a tire. As you increase load on the tire, the grip increases, but less and less with additional load until the tire has reach the max grip and additional load does not increase grip. Because of the shape of the curve, when you take load off of a tire, the grip drops off greater.
This phenomenon is illustrated when cornering. When a car is turning, load is transferred from the inside wheels to the outside wheels. The additional grip on the outside tires is LESS than the grip lost by the inside tires, so the overall sum of all four tires deceases with more transfer. Cars with a lower center of gravity have less load transfer and more overall grip.
ABS – the antilock braking system tries to prevent tires from skidding because, as was mentioned in the thread, the static grip (tires rotating) is higher than dynamic (skidding). When brakes/wheels lock, the ABS system engages, it releases the brakes allowing the tire to roll, but usually allows the brakes to lock again (and you get the pulsing effect). If a driver was able to apply the brakes at the exact limit of the tires, they would stop shorter than with the ABS system (since the ABS is going over and under the limit). This is probably why the Saleen had a longer stopping distance than the Lambo, the Lambo driver could aggressively stomp on the brakes and hold them there, letting the ABS sorting out the rest. The Saleen driver probably did not have the confidence or skill to effectively brake at absolute limit of the tires. The fact remains that if the driver were able to, the Saleen should have stopped shorter. (assuming same brakes, same tires, and the Saleen with less weight)
 
  • #63
viperblues450 said:
...Mass – Mass has a significant affect on braking. It does not “cancel out” of equations involving the full braking system. Mass certainly doesn’t create an advantage as Mech Engineer said! Have you ever tried braking in a car loaded with more passengers than seats? Your braking distance obviously INCREASES, and significantly.
Tires - Why does the braking distance increase with mass? F=ma. An increase in mass yields a decrease in acceleration given the same force. But doesn’t an increase in the mass of the car increase the force the tire can put on the road? Yes. Does that mean they cancel out? NO.
...
As phrased here you are changing the domain of the problem a bit. You make the point that braking changes in the case of a vehicle overload (beyond the design parameters) so that, for instance, the suspension no longer optimally distributes the vehicle load during deceleration. The intent of my OP was to discover whether there is a pay off in braking distance for mass reduction in a given vehicle design, operating inside its design parameters. That is, does vehicle A, mass X have a stopping distance advantage over vehicle B, mass greater than X if both have similar but size appropriate braking systems and tires.
 
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  • #64
Yes absolutely it has an advantage to be lighter. The advantage from the additional weight on the tires is less than the disadvantage from slowing the additional mass.
There is no difference between inside and outside the design parameters, the performance is still governered by the same laws of physics. One passenger increases stopping distance by x, two passengers by y (not 2x), three passengers by z, and 40 passengers by even more. Even the weight of the original driver will slightly affect the distance (negligable in reality).

You are correct that adding additional weight like passengers that raise the center of gravity with increase load transfer from the rear to the front tires, and with that transfer, the overall tire grip will decrease (due to the tire characteristics I explained before).
This is not changing the domain of the problem though, because (in your example) vehicle B, with mass X+Y will have more load transfer than vehcile A with mass X. The suspension NEVER optimally distributes the load under braking because the optimal distribution would be an equal load on all tires. There is nothing inside or outside of design parameters that does this. Also, suspension does not distribute total dynamic load, when a car is accelerating, braking, or turning, the total load transfer is only a funtion of geometry (and total weight), not of the suspension components.
 
  • #65
What you said though makes me realize that the additional load transfer probably attributes more to the decrease in braking than the nonlinearity tire characteristics. The tires probably behave near the linear region in the longitudinal direction, the nonlinearity shows up much more in the lateral grip.
An increase in vehichle weight will increase the size of the brakes needed to max out the tire's grip. Assuming the brakes are never the limiting factor (not usually the case), and assuming the increase in weight will not raise the center or gravity (unlikely unless the weight is place very low), an incease in vehichle weight can cancel itself out.
 
  • #66
viperblues450 said:
Yes absolutely it has an advantage to be lighter. The advantage from the additional weight on the tires is less than the disadvantage from slowing the additional mass.
There is no difference between inside and outside the design parameters, the performance is still governered by the same laws of physics. One passenger increases stopping distance by x, two passengers by y (not 2x), three passengers by z, and 40 passengers by even more. Even the weight of the original driver will slightly affect the distance (negligable in reality).
So that we don't talk past each other here, can you state your point mathematically? As discussed up thread, the data from various vehicle stopping distances is mixed, it somewhat suggestive that lighter cars have and advantage but its by no means conclusive.

You are correct that adding additional weight like passengers that raise the center of gravity with increase load transfer from the rear to the front tires, and with that transfer, the overall tire grip will decrease (due to the tire characteristics I explained before).
This is not changing the domain of the problem though, because (in your example) vehicle B, with mass X+Y will have more load transfer than vehcile A with mass X. The suspension NEVER optimally distributes the load under braking because the optimal distribution would be an equal load on all tires. ...
I think this confuses optimal with perfect. For instance, the CoG change could be nearly eliminated with a perfectly rigid carriage, but that discards other vehicle desirable characteristics.
 
  • #67
Viperblues, I have to reply to this post just because you are completely misinterpreting what has been argued over in the past 4 pages. My argument can be summed up as such:

Mech_Engineer said:
...There isn't any fundamental reason an ultra-light car can stop faster than a heavy one, as long as the brakes and tires on each car are sized appropriately.

It could be argued that it is easier and cheaper to make a light car stop quickly, but that's about it (and it's easier and cheaper to do most anything performance-based in a lightweight car).

What you're trying to argue is not what this thread is about, period.

viperblues450 said:
Mass – Mass has a significant affect on braking. It does not “cancel out” of equations involving the full braking system. Mass certainly doesn’t create an advantage as Mech Engineer said! Have you ever tried braking in a car loaded with more passengers than seats? Your braking distance obviously INCREASES, and significantly.

Your argument is not addressing the fundamental issue that is being argued in this thread. The original poster asked a very simple question-

mheslep said:
In several discussions of these [ultra-light] vehicles I have seen and heard mention of the supposed additional safety benefit of shorter stopping distances, but I have not found any elaboration on why this is so, implying I fear that I missing something obvious.

The answer is of course that it takes more than mass to determine how quickly a vehicle can stop. The primary factors that will determine how quickly a vehicle can stop are the friction between the road surface and the tire (tire compound) and the power dissipation capacity of the brakes. This has been repeated over and over for 5 pages now.

Adding more mass to a vehicle without changing its braking capacity will of course make it stop more slowly, and that topic has also already been covered in this very thread by me; in the very first page:

Mech_Engineer said:
...what this proves is that increasing the weight while keeping the same brakes means the vehicle will take longer to stop. This is because brakes have an associated "power rating," which can be thought of in terms of horsepower or watts.

Since the brakes at maximum clamping force can only convert a specific amount of kinetic energy per second to heat, having more weight means more kinetic energy which in turn means it takes longer to convert all of the kinetic energy to heat.

That's not what's being argued here. Given two cars that have been designed by two different manufacturers, the lighter one will not automatically be able to stop more quickly than the heavier one. Heavier cars tend to have heavier capacity brakes, and as such they will tend to be able to stop as quickly as lighter cars. This is especially true in sports cars, which I covered in extreme detail.

viperblues450 said:
Tires - Why does the braking distance increase with mass? F=ma. An increase in mass yields a decrease in acceleration given the same force. But doesn’t an increase in the mass of the car increase the force the tire can put on the road? Yes. Does that mean they cancel out? NO.

Actually, it does to a first-order approximation, and the VERY simplified math was presented on page 4 by NateTG.

NateTG said:
Newton gives us:
[tex]F=ma[/tex]

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

The dynamics of vehicle braking are indeed quite complex, but your vehement argument is completely missing the point of this thread. Ironically, your argument assumes that a car with more people in it will ALWAYS stop slower than one with less people in it, which isn't true either.

If a car is carrying 4 people, and the brakes were sized appropriately during the vehicle's design phase to take this extra weight into account (read- the tires can still lock up on a full stop and the ABS system engages), the car will stop very close to as quickly as the same car with only one person in it. Any difference in stopping distance will not have to do with the increased weight, it will instead probably be due to minute shifts in the vehicle's center of gravity or weight distribution. If we assume the vehicle's brakes can always lock the tires (properly sized brakes for the vehicles estimate operating weight, the vehicle is not overloaded), the extra momentum from any extra weight in the vehicle is offset by the fact that there is more available frictional force available to decelerate that extra weight as well.

viperblues450 said:
There has been a fatally flawed assumption about this throughout the thread. Many posters have used the old high school physics equation for normal force, F=mu*m*g . This was a basic approximation for something like sliding a block across a desk. It cannot be used for something as complicated as tire compounds and road surfaces. The curve of load vs. traction is NON-linear for a tire. As you increase load on the tire, the grip increases, but less and less with additional load until the tire has reach the max grip and additional load does not increase grip. Because of the shape of the curve, when you take load off of a tire, the grip drops off greater.

While your argument is interesting (I'd be interested to see a curve that documents the phenomenea you are mentioning) the fact is that we are talking about basic, first order braking with respect to vehicle weight. The point is, an ultra-light vehicle cannot necessarily stop faster than a standard weight vehicle; there are of course MANY more things that have to be taken into account. Case closed.

viperblues450 said:
If a driver was able to apply the brakes at the exact limit of the tires, they would stop shorter than with the ABS system (since the ABS is going over and under the limit). This is probably why the Saleen had a longer stopping distance than the Lambo, the Lambo driver could aggressively stomp on the brakes and hold them there, letting the ABS sorting out the rest. The Saleen driver probably did not have the confidence or skill to effectively brake at absolute limit of the tires.

Sure, it's in theory possible that given the perfect driver and perfect conditions, ABS might not be necessary for a perfect stop. But the fact is they were using professional drivers in the test and they couldn't get the Saleen to stop anywhere close to as quickly as say the Porsche, even though it was lighter.

viperblues450 said:
The fact remains that if the driver were able to, the Saleen should have stopped shorter. (assuming same brakes, same tires, and the Saleen with less weight)

No, since the Lamborghini and the Saleen had the exact same tires, it's likely that at best the Saleen would have been able to stop as fast as the Lamborghini, but not faster. Slight differences in their stopping distances would have been due to differences in f/r weight distribution and center of gravity, but not overall weight.
 
  • #68
Here is where the problem is:

"Of course I reached for the standard stopping distance derivation: the kinetic energy of the vehicle and the work done by friction are both linearly related to mass, so that stopping distance is independent of mass as shown here:"

As viperblues450 and I think Stingray stated, the load to grip graph of a tire is not linear and shows that the higher the load the lower the traction of the tire for that weight. In other words, if everything else is a constant, and the only thing changed is weight, the vehicle will stop faster - every time. This is not a minor effect. It pretty well overrides everything else.

This concept works exactly the same way for weight transfer while cornering and accelerating as well.

The braking line on the Saleen shows braking instability, i.e., the driver had to modulate the brakes to maintain control or avoid tire lock-up. With a bit of time and some adjustments that could be cured. The Lamborghini actually has less grip for its weight but manages it better for various reasons and as a result the driver is able to stop more quickly. The Porsche has the best dynamic weight balance during braking because of the rear weight bias and should post the best braking rate if all the cars had the same tire size to weight ratio.

If the tire size to weight ratio weren't so important, race cars wouldn't use wide tires - but that might be another thread.

Here's a graph:

http://buildafastercar.com/node/10
 
Last edited:
  • #69
I will answer some of the other questions by Mech Engr and mheslep soon, but in the mean time you both need to open the wonderful link posted by mender. It explains many of the things I was trying to say, but better than I could explain them. The summary points are:

"*Peak grip exists when all four tires are evenly loaded.
* Reducing weight transfer (via a wider track or lower center of gravity) can increase the
mechanical grip of your tires.
* A lighter car will have more total grip than a heavier car when on the same set of tires."
 
  • #70
mheslep said:
So that we don't talk past each other here, can you state your point mathematically? As discussed up thread, the data from various vehicle stopping distances is mixed, it somewhat suggestive that lighter cars have and advantage but its by no means conclusive.
I think it would do a lot better to look at the tire characteristic curves in the link. Keep in mind that these are for lateral grip, not longitundinal grip, however I think the basic characteristics are the same. The idea is that what you gain with mass to additional force on the tires cannot be fully translated to additional grip. Therefor it is to your advantage to "add lightness" as a great auto legend once said.


mheslep said:
I think this confuses optimal with perfect. For instance, the CoG change could be nearly eliminated with a perfectly rigid carriage, but that discards other vehicle desirable characteristics.
Yes you're right, it will never be perfect, but optimal is hard to use here as well, because the optimal set up for braking will not be optimal for other the rest of the vehicle. Since usually we are trying to optimize the vehicle, the set-up is never optimal for braking.

You are also right that an perfectly rigid chassis will not allow the CoG to move. I just want to make sure you everyone understand this will NOT effect the weight transfer of the vehicle under braking. If two cars are identical, one has very soft suspension and the other very stiff. The soft suspensioned car will "dive" heavily under braking, while the stiff suspensioned car will remain flat. BOTH cars will still have the SAME weight transfer from the rear tires to the front. The movement of the cars is a RESULT of weight transfer, not the cause of it. The weight transfer is only a funtion of the wheelbase and CG height, and deceleration.
It is true that the car diving will very slight change the center of gravity height, and may make a very small impact on the weight transfer, but this effect is almost alway negligible.
 

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