# Shorter Stopping Distance for ultralight vehicles?

by mheslep
Tags: distance, stopping, ultralight, vehicles
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 Quote by Stingray 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.
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 Quote by mheslep 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).
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 Quote by Stingray 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.
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 Quote by mheslep 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.
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 Quote by Stingray 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.
 P: 207 You know you're saying that a heavier car stops faster than a lighter car.....
P: 120
 Quote by Mech_Engineer 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] [tex]N_F=mg-N_R[/itex] c=height of COG from ground l=wheelbase a=distance between CoG and front axle  P: 13 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) PF Patron P: 2,949  Quote by viperblues450 ...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.  P: 13 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.  P: 13 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. PF Patron P: 2,949  Quote by 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). 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. PF Patron Sci Advisor P: 2,207 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:  Quote by Mech_Engineer ...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.  Quote by viperblues450 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-  Quote by mheslep 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:  Quote by Mech_Engineer ...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.  Quote by viperblues450 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.  Quote by NateTG Newton gives us: [tex]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.

 Quote by viperblues450 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.

 Quote by viperblues450 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.

 Quote by viperblues450 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.
 P: 563 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
 P: 13 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."
P: 13
 Quote by mheslep 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.

 Quote by mheslep 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.
 P: 563 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.
P: 13
 Quote by 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.
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."

 Quote by Mech_Engineer 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).

 Quote by Mech_Engineer 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.

 Quote by Mech_Engineer 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 dont work here.

 Quote by Mech_Engineer 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.

 Quote by Mech_Engineer 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.

 Quote by Mech_Engineer 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.

 Quote by Mech_Engineer 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 definetly 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.

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