Shorter Stopping Distance for ultralight vehicles?

[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...

 

[Jupiter6]

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...

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

N_R=mg(\frac{a-uc}{l})[/itex]<br /> <br /> N_F=mg-N_R[/itex]&lt;br /&gt; &lt;br /&gt; c=height of COG from ground&lt;br /&gt; &lt;br /&gt; l=wheelbase&lt;br /&gt; &lt;br /&gt; a=distance between CoG and front axle

 

[viperblues450]

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)

 

[mheslep]

...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.

 

[viperblues450]

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.

 

[viperblues450]

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.

 

[mheslep]

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.

 

[Mech_Engineer]

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:

...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.

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-

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:

...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.

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.

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.

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.

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.

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.

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.

 

[mender]

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

 

[viperblues450]

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."

 

[viperblues450]

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.

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.

 

[mender]

viperblues450, the next issue is controlling the transients while the car is doing its thing on the track. The shocks have to smooth out the body motions in such a way as to minimize instantaneous loading of any of the tires, again because of the load to grip curve of the tires.

 

[viperblues450]

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:
...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.

Untrue. The name of the thread is "Shorter distances for ultralight vehicles?"
The answer is yes. And everything I've talked about is applicable to understanding why the answer is "yes."

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.

You are leaving out a very important factor, the frictional coefficient of the brake pads to the rotors. This is much more important over a single stop than the size or heat capacity of the rotors. This is also a reason why the Saleen could not stop as fast as the Lambo. Looking at the size of the rotors is simply not enough. The bite of the brake pad material is much more important (until the discs heat up and brakes fade).

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.

It is all about the relationship between the brake/tire/road system and the mass of the vehicle. The lighter car will be able to stop quicker assuming the maximum braking and tire grip is achieved with both vehicles.

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

As I explained, the first order approx is invalid in this application because the "m" term in the F=mu*m*g equation for normal force is not actually equal to m. See the tire grip chart level to understand this. I know that its a weird concept, but try understand why those simple equations don't work here.

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.

This is all true except for the part where you say " the extra momentum from any extra weight in the vehicle is offset by the fact that there is more available frictional force"
I don't mean to keep repeating, but the nonlinearity tire characteristics invalidates this statement. Does the additional force provide additional grip? Yes, but it does not fully cancel out the additional mass.

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..

Case unclosed, check the link. The lighter vehicle will stop quicker.

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.

ABS NEVER gives you the perfect stop, because it has to wait for tires to lock up before it engages, where a perfect brake w/o ABS could brake right on the limit the whole time.

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.

The Saleen most definately should be able to stop quicker than the Lambo. Its poor performance here is probably due to the driver (pro or not) being unable to give a perfect braking run. This is very hard to do, and even a pro driver might not be able to do it. Other factors that the Saleen might not have been able to stop well are the conditions of the tires, tread, tire temp, tire pressure, brake pad material (as I previously explained, probably the most important factor besides tires and weight). The feel of the brakes may have been numb, not allowing the driver to get a good feel. Your argument that the Saleen SHOULD not be able to stop faster than the Lambo is just wrong. This assumes that the Saleen and Lambo has similar weight distribution, wheelbase, and CG height. If the Lambo has significantly advantagous geometery, its possible that also helped it, however I doubt this is the case.
The Porsche does so well because it starts with more weight over the rear wheels, and transfers weight to a near 50/50 distribution under braking. Having a equal load on all tires provides the greatest overall grip and braking force.
Using your incorrect understanding of tire grip, the extra weight on the front tires of say the Lambo would completely cancel out the weight lost on the rear wheels. This is not the case, and it shows up in how rear engined Porsche consistantly have some of the shortest braking distances.

 

[mheslep]

Note that the grip vs normal force graph mender linked to was for one tire. That by itself is not helpful, as obviously the heavier vehicles can, and do, use tires with a higher coefficient of friction. So VP450, you are changing the scenario I posed above - vehicle A vs vehicle B, to the scenario of vehicle A vs vehicle A with more or less load. That's not useful for my problem.

To illustrate, I prepared a naive family of curves extension for the graph mender linked. It shows how, for example, one could maintain the same coefficient of friction (slope) at say, 600, 1000, and 1400 lbs per tire in three different vehicles by selecting different tires in the design process.

 

[viperblues450]

Note that the grip vs normal force graph mender linked to was for one tire. That by itself is not helpful, as obviously the heavier vehicles can, and do, use tires with a higher coefficient of friction. So VP450, you are changing the scenario I posed above - vehicle A vs vehicle B, to the scenario of vehicle A vs vehicle A with more or less load. That's not useful for my problem.

To illustrate, I prepared a naive family of curves extension for the graph mender linked. It shows how, for example, one could maintain the same coefficient of friction (slope) at say, 600, 1000, and 1400 lbs per tire in three different vehicles by selecting different tires in the design process.

I'm really not understanding why you think its an unfair comparison. High coefficient of friction tires are not reserved for heavy cars! Any car can use a hard or soft (grippy) tire. In order to compare the effect of mass on braking, you have to use the same tires. You can ONLY compare vehicle A to vehicle A with more load.

If you are comparing car A to car B which have both been DESIGNED to have similar braking performance, then you are comparing nothing! Oh course they will have similar performance, that's how they are designed! The only thing you could say from that is that the heavier vehicle needed larger brakes, high friction pads, larger tires to have comparable performance. If both cars are designed with max braking, the lighter car sport better braking performance.

 

[viperblues450]

Note that the grip vs normal force graph mender linked to was for one tire. That by itself is not helpful, as obviously the heavier vehicles can, and do, use tires with a higher coefficient of friction. So VP450, you are changing the scenario I posed above - vehicle A vs vehicle B, to the scenario of vehicle A vs vehicle A with more or less load. That's not useful for my problem.

To illustrate, I prepared a naive family of curves extension for the graph mender linked. It shows how, for example, one could maintain the same coefficient of friction (slope) at say, 600, 1000, and 1400 lbs per tire in three different vehicles by selecting different tires in the design process.

Also, about your graph, choose the highest grip tire for lighter car and it will brake the best. You seem to be trying to change two variables at once and I'm not sure why.

 

[Stingray]

Viper,
I think that you're overstating the significance of these nonlinear tire effects. mheslep should probably have used an example where different tires "saturate" at different loads, but have the same initial friction coefficient. Plenty of examples like that exist in the real world, which explains why braking performance does not strongly correlate to weight in (say) street-legal sports cars.

So why not put the more linear tire on a small car? I'm guessing that it would probably be too big. There could be packaging or weight constraints as well as issues with the suspension geometry. Rolling resistance could be made much worse, or maybe other properties of the tire aren't very desirable.

 

[Mech_Engineer]

Untrue. The name of the thread is "Shorter distances for ultralight vehicles?"
The answer is yes. And everything I've talked about is applicable to understanding why the answer is "yes."

There are of course many cases of cars that can stop more quickly than other cars that are lighter than them, so your broadly-sweeping generalization is invalid.

You are leaving out a very important factor, the frictional coefficient of the brake pads to the rotors. This is much more important over a single stop than the size or heat capacity of the rotors.

The frictional coefficient between the brake pads and the brake rotor is unimportant, as long as the tires have enough braking power to lock up. Brake heat fade is beyond the scope of this thread.

As I explained, the first order approx is invalid in this application because the "m" term in the F=mu*m*g equation for normal force is not actually equal to m. See the tire grip chart level to understand this. I know that its a weird concept, but try understand why those simple equations don't work here.

I understand what you're arguing, and I agree that if there are frictional nonlinearities in a tire when braking they can be a factor in a car's overall stopping distance.

This is all true except for the part where you say " the extra momentum from any extra weight in the vehicle is offset by the fact that there is more available frictional force"
I don't mean to keep repeating, but the nonlinearity tire characteristics invalidates this statement. Does the additional force provide additional grip? Yes, but it does not fully cancel out the additional mass.

You have a point there. However I'm curious, all of these charts are measuring "lateral grip" in the tires, but do they also apply to acceleration and braking? Are these charts basically taking into account tire side wall roll-over and tread deformation in lateral cornering, which are not present when braking?

ABS NEVER gives you the perfect stop, because it has to wait for tires to lock up before it engages, where a perfect brake w/o ABS could brake right on the limit the whole time.

If you do a perfect brake right on the limit, the tires will never actually lock up and therefore ABS would never even come into play. Regradless, ABS equipped vehicles are able to stop very quickly regardless of driver skill, and I'm seriously doubtful ANY driver could stop a car "perfectly" without ABS.

The Saleen most definately should be able to stop quicker than the Lambo.

"Should" and "did" are miles apart in this case, sorry. It's probable the Saleen would have stopped slightly more quickly than the Lamborghini if it had ABS brakes, but it's likely the difference would have been small (less than 5%) while their weight difference is far larger than that.

Its poor performance here is probably due to the driver (pro or not) being unable to give a perfect braking run. This is very hard to do, and even a pro driver might not be able to do it.

In the article, the driver says that he was unable to properly modulate the brakes in the Saleen to get a braking run as an ABS equipped vehicle. It is my suspicion that given two identical cars, one with ABS and one without, the ABS equipped vehicle will stop more quickly 99 times out of 100.

Other factors that the Saleen might not have been able to stop well are the conditions of the tires,

They were the same exact tires as on the Lamborghini. Same brand, model, size, and wheel size. They were installed that day for that series of tests, same as the Lamborghini. All of the cars had the same brand and model of tire installed that day to ensure Tires were not a variable in the test (other than the car's required sizing).

tread,

AFAIK, all the tires on all of the cars were installed brand new for that series of tests so that all of the cars had a level playing field in that respect.

brake pad material (as I previously explained, probably the most important factor besides tires and weight).

Brake pad material doesn't matter, as long as the brakes are stong enough to lock up the tires. The brakes were clearly fine because the driver had to try and modulate the pedal because it way very easy to lock up the tires.

Your argument that the Saleen SHOULD not be able to stop faster than the Lambo is just wrong. This assumes that the Saleen and Lambo has similar weight distribution, wheelbase, and CG height. If the Lambo has significantly advantagous geometery, its possible that also helped it, however I doubt this is the case.

Look, the point of this thread is that just because a car is lighter doesn't mean it can stop more quickly. The Saleen is much lighter than the Lamborghini, yet it takes longer to stop. This is not a good comparison because the Saleen doesn't have ABS, but it proves the point of the thread.

We obviously agree on the core topic in this thread- it takes a lot more than just the weight of the vehicle to determine how quickly it will stop. Saying that a superlight car is fundamentally able to stop faster than a "standard" car is a shady hedge-bet. Now saying a well tuned sports car will stop faster than a minivan, that's obviously a safer bet.

 

[viperblues450]

Yeah Stingray I agree with you. I may be overstating the nonlinearity, but I wanted to bring it up because I wanted future readers (like myself) to know that tire characteristics are more complex than the simple physics equations that were being used in the thread.
In reality you're right, there is not a huge advantage in braking to decrease the mass because the braking components can usually make up most of the difference--but not all.
I am an armature racer and am used to competition where the "small" difference becomes a "large" competitive edge. Things can get down into such detail that adding larger brakes increases the rotational inertia and works against braking.
Also ultralight cars are not often outfitted with max performance brakes and tires. In a quick search I did find that found a Ferrari Enzo stopping 70mph-0 in 151ft
http://www.supercarstats.com/cars/ferrari/enzo_ferrari/
Where a $10k Locost 7 kit car (1297lbs) can brake from 70-0 in 141ft (no ABS I think)
http://www.caranddriver.com/reviews..._winterhalter_s_locost_sports_car+page-5.html
That Seven didn't even have super tires like the Ferrari would have had. I know these we separate tests, but I'm trying to show that even all the money and technology Ferrari can throw at their ultimate supercar cannot match simply "adding lightness." The Ferrari is almost there, but not quite, and that's all I've been trying to say. Shorter stopping distance for ultralight vehicles? Yes.

 

[jimgram]

I haven' seen any posts that relate net contact force (I.E. vehicle weight) to the maximum lateral force (braking) that can be applied without skidding. The standard fomula is Net (braking) force=Cf(coeff. of friction)X Vehicle weight. A linear relationship. So, if all other factors remained equal (e.g. contact area, surface, etc.) a heavy vehicle and a light vehicle will stop in the same distance.

 

[viperblues450]

There are of course many cases of cars that can stop more quickly than other cars that are lighter than them, so your broadly-sweeping generalization is invalid.

Broadly sweeping generalization in engineering discussions assume other variables constant.

The frictional coefficient between the brake pads and the brake rotor is unimportant, as long as the tires have enough braking power to lock up. Brake heat fade is beyond the scope of this thread.
Brake pad material doesn't matter, as long as the brakes are stong enough to lock up the tires. The brakes were clearly fine because the driver had to try and modulate the pedal because it way very easy to lock up the tires.

Well its very important as different compounds take different about of time to heat up to the temperature where they have the coefficient of friction to lock the wheels. The time that takes to heat up is short, but important if your comparing a brake test.
Formula 1 car brakes and tires can performance an INCREDIBLE 5 g's of braking! However, they have like a .5 second delay where there is almost no braking force from the "pads." It's interesting and makes the driver apply the brakes before they need them.

I understand what you're arguing, and I agree that if there are frictional nonlinearities in a tire when braking they can be a factor in a car's overall stopping distance.

You have a point there. However I'm curious, all of these charts are measuring "lateral grip" in the tires, but do they also apply to acceleration and braking? Are these charts basically taking into account tire side wall roll-over and tread deformation in lateral cornering, which are not present when braking?

I am also very curious about this question. I will consult my automobile dynamics textbook or do further research. If you find anything, please post.

If you do a perfect brake right on the limit, the tires will never actually lock up and therefore ABS would never even come into play. Regradless, ABS equipped vehicles are able to stop very quickly regardless of driver skill, and I'm seriously doubtful ANY driver could stop a car "perfectly" without ABS.

Thats true. What I learned in my automobile dynamics class was that tires actually provide the most grip with about 10% slip (both acceleration and braking). I'm not sure why that is, but if that's true, it would mean that an optimal braking performance would have the tires rotating about 90% of the car's speed.

"Should" and "did" are miles apart in this case, sorry. It's probable the Saleen would have stopped slightly more quickly than the Lamborghini if it had ABS brakes, but it's likely the difference would have been small (less than 5%) while their weight difference is far larger than that.

In the article, the driver says that he was unable to properly modulate the brakes in the Saleen to get a braking run as an ABS equipped vehicle. It is my suspicion that given two identical cars, one with ABS and one without, the ABS equipped vehicle will stop more quickly 99 times out of 100.

I agree, I think the limiting factor here is tires, and if the Saleen had ABS, I could see it only stopping about 5% shorter. Which was my point, less weight helps, not hurts or cancels out, even though it might be only 5-10% when both cars have high performance braking systems.

They were the same exact tires as on the Lamborghini. Same brand, model, size, and wheel size. They were installed that day for that series of tests, same as the Lamborghini. All of the cars had the same brand and model of tire installed that day to ensure Tires were not a variable in the test (other than the car's required sizing).

AFAIK, all the tires on all of the cars were installed brand new for that series of tests so that all of the cars had a level playing field in that respect.

Good to know.

Look, the point of this thread is that just because a car is lighter doesn't mean it can stop more quickly. The Saleen is much lighter than the Lamborghini, yet it takes longer to stop. This is not a good comparison because the Saleen doesn't have ABS, but it proves the point of the thread.

I think we are agreeing on most things now, but then our conclusions are opposite. I still retain that a car being lighter will stop more quickly. Since the Saleen doesn't have ABS, it does not prove the point you are making. It is only making the case for ABS. Look at the Ferrari Enzo and Locost 7 Kit car I posted.

We obviously agree on the core topic in this thread- it takes a lot more than just the weight of the vehicle to determine how quickly it will stop. Saying that a superlight car is fundamentally able to stop faster than a "standard" car is a shady hedge-bet. Now saying a well tuned sports car will stop faster than a minivan, that's obviously a safer bet.

I will agree that the difference is not large, but there is an fundamental advantage to a lighter car with braking performance. And yes, it is complicated.

 

[viperblues450]

jimgram what we have been saying here that it is not a linear relationship, and that simple formula doesn't hold up.
Think of the extreme case of a car braking so hard the rear wheels are just about leaving the ground. The front wheels do not have twice the grip to make up the difference. The front wheels have more with with the increased weight, but they do not have enough extra grip to make up for the rears (which have zero). The net grip of the 4 tires is less than if they all had an equal share.
Also, I think you meant longitudinal grip for braking.

 

[mender]

To clarify where I'm coming from, I have a fairly extensive practical background to go with my book learning. I don't usually deal in formulas and equations; I mostly go by lap times instead. I have been reading about chassis dynamics for the last twenty years and have been working on street and race cars for about thirty. Not to brag, just to let you know what my background is. Some things I won't be able to point to a specific page in a book for.

The slip angle that the tire assumes at maximum grip is about the same in all directions. The tire tread needs to be held flat to the surface to work effectively; when that is achieved, the suspension of the car can be tuned to either maximize the car's performance or the driver's. Note that these are not always the same! Getting the most out of the software (driver) is the more important of the two. You want the driver driving the car, not the car driving the driver.

The tire patch is essentially an oval that twists and moves around a bit in response to the forces placed upon it.

"Note that the grip vs normal force graph mender linked to was for one tire."

This is true: however this was only to show the trend. All tires will show this trend, with individual characteristics determining the exact curve. That's why there are different brands and construction techniques, to offer a tire that can be better matched to a car's needs and the driver's wants.

The Saleen very likely suffered from rear wheel lock-up. By adjusting the brake balance it would have outbraked the Lambo with the same tires just because of the difference in weight. Straight forward. A pro driver cannot get maximum braking on a car that is locking up the rear brakes without flatspotting the tires and introducing sliding friction instead of rolling friction. It has nothing to do with the driver's ability.

This is a link to one of the best books that I have on the subject. It has more than enough material to satisfy most engineers and is also very helpful for others:

http://www.powells.com/cgi-bin/biblio?inkey=4-0973432004-0

 

[Mech_Engineer]

Broadly sweeping generalization in engineering discussions assume other variables constant.

Well, the problem with holding all other variables constant is that would mean you are taking away or adding weight to the same car, not comparing two completely different cars.

Think of it as a scientific hypothesis. You state your hypothesis "if a car is lighter, it will stop faster"; that hypothesis is easily disproven because there are many examples to the contrary, so obviously the hypothesis is too generalized to be true. Perhaps if the hypotheis were "given two identical cars in every respect save weight, the lighter one will stop faster" it would be closer to the truth.

I guess I'll just leave it at in general, its a bad idea to make generalizations

Well its very important as different compounds take different about of time to heat up to the temperature where they have the coefficient of friction to lock the wheels. The time that takes to heat up is short, but important if your comparing a brake test.

I seriously doubt that would have any measurable effect when compared to all of the other far more significant variables in the test.

Formula 1 car brakes and tires can performance an INCREDIBLE 5 g's of braking! However, they have like a .5 second delay where there is almost no braking force from the "pads." It's interesting and makes the driver apply the brakes before they need them.

I'm willing to bet Formula cars only achieve those kind of decelerations at high speeds where aerodynamic dorwnforce is very large, so they have more downforce on the tires and hence more frictional force available. The same is true for cornering, where they can only achieve maximum cornering ability once they have reached sufficient speed.

Thats true. What I learned in my automobile dynamics class was that tires actually provide the most grip with about 10% slip (both acceleration and braking). I'm not sure why that is, but if that's true, it would mean that an optimal braking performance would have the tires rotating about 90% of the car's speed.

I think that's probably true from what I've seen in high performance speed stops.

I think we are agreeing on most things now, but then our conclusions are opposite. I still retain that a car being lighter will stop more quickly. Since the Saleen doesn't have ABS, it does not prove the point you are making. It is only making the case for ABS. Look at the Ferrari Enzo and Locost 7 Kit car I posted.

The difference is significant, but the problem is this ends up being an apples to oranges test. The tests were not done at the same time in the same location. In addition the Enzo probably had street legal touring tires fitted, while it's likely the kit race car had racing stickies, and the race car may have even been tested on a "sticky" race track... The only way to get a really definitive comparison is if they both have the same tires, and are tested on the same surface at the same time.

 

[viperblues450]

Well, the problem with holding all other variables constant is that would mean you are taking away or adding weight to the same car, not comparing two completely different cars.

Think of it as a scientific hypothesis. You state your hypothesis "if a car is lighter, it will stop faster"; that hypothesis is easily disproven because there are many examples to the contrary, so obviously the hypothesis is too generalized to be true. Perhaps if the hypotheis were "given two identical cars in every respect save weight, the lighter one will stop faster" it would be closer to the truth.

I think the latter hypothesis is the one that everyone has been referring to throughout the thread. I hope anyways because you are correct, the first one means nothing.

I seriously doubt that would have any measurable effect when compared to all of the other far more significant variables in the test.

Its true that there probably wasn't much difference between the Saleen and Lambo due to this effect, but this effect is still significant. Most race pads/rotors provide almost no bite until they are warm, requiring several easy stops to get them up to temperature. This is a consideration when racers are selecting pads depending on what type of racing they're doing. If the testers did the run on cold brakes, they would have gotten worse results. I'm sure R&T knows what they're doing though.

I'm willing to bet Formula cars only achieve those kind of decelerations at high speeds where aerodynamic dorwnforce is very large, so they have more downforce on the tires and hence more frictional force available. The same is true for cornering, where they can only achieve maximum cornering ability once they have reached sufficient speed.

Yes that's true, they require the downforce, but they still have amazing brake technology and tire technology to provide the braking and grip to the road. Think about that, that's more than 4X the hardest braking you could ever get from a street car.

The difference is significant, but the problem is this ends up being an apples to oranges test. The tests were not done at the same time in the same location. In addition the Enzo probably had street legal touring tires fitted, while it's likely the kit race car had racing stickies, and the race car may have even been tested on a "sticky" race track... The only way to get a really definitive comparison is if they both have the same tires, and are tested on the same surface at the same time.

From the pictures and several other tests on the 7 kit car, they just use street rubber, in the same league as the Ferrari tires. I know you can't compare these numbers exactly, but I think it shows the idea that low tech and lightweight can out perform or at least keep up with one of the best road cars ever. And for what its worth the 151ft of the Ferrari Enzo was also from a Car&Driver test I think.

 

[viperblues450]

mender thanks for the great post, its great to hear someone with a lot of experience confirm what's being said here.
Good thought about the Saleen too, if the bias isn't dialed it, there is nothing the driver can do.

So are you saying that the lateral tire curve is about the same shape as the longitudinal?

 

[mender]

I'll quote from that book:

"A tire's coefficient of friction, when considering longitudinal forces, is measured as percentage shear rather than shear angle. The "Percentage shear vs Coefficient of Friction" graph is much the same as the "Shear Angle vs Coefficient of Friction" graph. However, the longitudinal graph generally reaches its peak faster. It has a small bubble at the peak and a short flat area before it quickly drops off."

Most of my racing and vehicle dynamics books start off with the tire performance curve because almost all the suspension design and tuning revolves around that concept.

If you want to read some more from Warren's book, look on this page:

http://www.rowleyrace.com/publications.htm

No, I am not selling it but I consider it one of the best books out there and strongly recommend it despite the price.Here's a site that shows the lateral and longitudinal traction combined into another important concept:

http://www.teamassociated.com/racerhub/techhelp/marc/Handling.2.html

The traction circle helps drivers understand what is being demanded of them. I have several data acquisition programs that will map each part separately or together as needed. I use them when allowed (not in NASCAR) to "tune" the driver.