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## Shorter Stopping Distance for ultralight vehicles?

 Quote by 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.
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.

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 Quote 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.
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.

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 Quote by mheslep 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.

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

 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  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) Recognitions: Gold Member  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.  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.  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. Recognitions: Gold Member  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. Blog Entries: 2 Recognitions: Gold Member Science Advisor 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.