Racing topic - Adding weight requires more or less KE?

[zanick]

In a race car, coming off a slow turn and going into a long straight approaching a slow turn.
if two identically powered cars (say 400hp at the rear tires but one at 3000lbs and the other at 3500lbs) are racing, would one car require more capable brakes than the other?
I subscribe to the laws of the conservation of energy here, but it seems that , against common opinion, that the lighter car might reach higher speeds and be able to slow at a faster rate approaching the turn. even though the same KE might be reached by both , ones rate of KE change on the decel might be higher thus demanding more power dissipation capability of the braking system.
From many years of pro and club racing experience, I have not seen a lesser demand for lowering weight, and certainly have seen a greater demand for increasing power for the same weight.
There are some great simulators out there to plug in the values, but wanted to ping the list to see if I am overlooking something.
Generally, this is to combat the thought from many, that if they have a heavy car, (and the same hp as their competitor) that they need more braking capacity. I disagree and think there is a trade off as the lighter car hits higher speed and that contributes to keeping the KE reached and released on decel, near the same.

 

[mfb]

There are many free parameters, and the answer will depend on your choice which parameters you want to fix in which way.

 

[zanick]

There are many free parameters, and the answer will depend on your choice which parameters you want to fix in which way.

We should change the title to (" ... More or less rate of KE change") as we know the heavier car will always end up with more KE, but not as much as it's weight change would appear to indicate.

Well, let's start with the fact that a heavier car with the same HP as the lighter car, will accelerate to a slower top speed for given distance.. (say to the point at which both cars need to engage the brakes to slow to get to turn in speed for the next turn)
General parameters:
3000lbs car (1363.6 kg) vs 3500lb car (1666 kg)
both launch onto the straight at 45mph (20.1 m/s)
Heavy car reaches 114mph (50.9mps) before braking at the same spot at initial braking point,
light car reaches 120mph (53.6mps) before braking at the same spot at initial braking point.
note: i ran the cars through a simulator to get these speeds, complete with engine torque curves, drag values, grip and geometry equivilance.

I also made the assumption that the tires would give the same g's of deceleration for both cars. assuming that the 3000lb car could decelerate at 1g, was first thinking that the heavier car, with no changes to tires, would slow at 15% slower speed for the 15% higher mass.

Many think that the heavier car is intuitively harder on the brakes. in other words, does a race car that suddenly takes on a 500lb guy as a passenger, have to worry more or less about brakes?

since the simulation takes the lighter of the two cars to about 5 % higher speed,, it has about 5% less KE as well. since the heavy car will decel at 15% less of a rate (.85g vs 1g) the rate of decel being greater for the lighter car gives it about 3.-4% more KE to dissipate. per second.. thus being harder on the brakes, not easier.
some then say, well,if the heavier car slows to a lower speed, say 30mph (13.4m/s) vs 33mph, (14.7m/s) then it will dissipate even more KE. .. True, but the rate of KE dissipation goes down even further. just comparing .85g for heavier car's distance that both would cover in 4.3seconds vs 4 seconds ((that's the time comparison for both masses to slow from their respective speeds to the final speed in their respective g deceleration rate) it would go to 4.5 seconds . this actually lowers the power/second of the heavier car's deceleration KE dissipation. by the way, the distances that both cars stop at , is within 10ft of each other... (in this simulation both cars got to 1840ft before braking and then i calculated the distance they would slow to the same target speed based on their g value of deceleration.

Let me know if I've missed anything here

If what I have run in the above simulation is true, then the thing that i think can change everything is the assumption that the tires of both the heavy vs light car, have the same deceleration force to the ground. there are limits to the tires ability in the way of slip percentage do to the deformation of the rubber on the surface, giving the mu for friction coefficient. but I've heard that there is some maximum force that the tires can apply, and if we found that with the lighter car, then the heavier car couldn't create any more force at the tire,even with more mass.
and if that is true, then, what if the car is lightened to half its weight again... can it impose the same force of deceleration as it did when it weighed 3000lbs? if it weighted 6000lbs?

The same scenareo, with the KE values attached:

Two cars , CAR A (light car at 3000lbs) vs Car B (heavy car 500lbs heavier at 3500lbs). two cars with the same HP are running:

car A 120mph (53.64 m/s)
car B 114mph (50.9 m/s)

the simple part of this example is that we can just plug in the numbers if we agree that the heavy car will slow at 15% less deceleration rate. we will call this 1g vs .85g respectively . (need to confim if this is a real possibility, or does the deceleration force go up proportional to weight?) for the sake of argument, let's assume that the heavier car breaks at a lower rate due to tire limitations.

This means in 4 seconds the light car slows to 33mph (15m/s)
this also means in 4.3 seconds, the HEAVY car slows to the same speed

keeping it really simple. the KE at the start is:
1,991956 J Light
2,085,914 J Heavy

the KE at the end at the final same exact speed is:
153,405 J
178,977 J

this ends up with the lighter car dissipating a total of 1,838,551 for the light car
and 1,906,937 J for the Heavy car. heavy car dissipates more energy by 3.7%

BUT

Because the lighter car slows to the same speed in 4 seconds and the heavy car slows to this same speed in 4.3 seconds, the RATE OF KE DISSIPATION is HIGHER for the ligher car by about 3.7%

616HP/sec for the light car ( example : 1,838,551 J /4 sec /746watt =hp/sec)
594HP/sec for the heavy car ( example : 1,906,937 J /4.3 sec /746watt =hp/sec)

the rate of heat dissipation for the light car is HIGHER than the heavy car for its decel rate and its higher speed at the moment of braking (at the same spot on the track)

Therefore, this shows that adding weight, under these normal conditions, actually lessens the burden on the braking system.

 

[mfb]

Therefore, this shows that adding weight, under these normal conditions, actually lessens the burden on the braking system.

With arbitrary assumptions on the deceleration rate, sure. If you reduce the deceleration, you reduce power dissipated. That is not surprising.

 

[zanick]

We assumed 1 g decel rate, which is a pretty acceptable value. the rest is where we have questions. does the mu of the tire go down with increased weigtht. If we are at the limit with the lighter car, the tire might not be able to increase mu with the extra weight. (assuming we were at the limit with the original car.

is the terminology correct. Its been a long time since Physics classes!
"616HP/sec for the light car ( example : 1,838,551 J /4 sec /746watt =hp/sec)"
is it just 616hp ave, because the joules/sec is a watt-second/second, which is just watts, or divided by 746 for HP, right?

 

[zanick]

We assumed 1 g decel rate, which is a pretty acceptable value. the rest is where we have questions. does the mu of the tire go down with increased weigtht. If we are at the limit with the lighter car, the tire might not be able to increase mu with the extra weight. (assuming we were at the limit with the original car.

is the terminology correct. Its been a long time since Physics classes!
"616HP/sec for the light car ( example : 1,838,551 J /4 sec /746watt =hp/sec)"
is it just 616hp ave, because the joules/sec is a watt-second/second, which is just watts, or divided by 746 for HP, right?

Also...this being an average, the HP or power at the end of the deceleration will dissipate far less energy than the beginning of the decel. after all , acceleration = power/(mass*velocity), so as the speed goes up or down, the power goes up or down proportionately.

 

[CWatters]

Is the straight is long enough for both cars to reach the same maximum speed (dictated by engine power)?

 

[mfb]

does the mu of the tire go down with increased weigtht.

It depends on details of the cars.

the tire might not be able to increase mu with the extra weight.

It does not have to increase it, keeping it constant is sufficient to make your assumption of a lower acceleration invalid.

 

[cjl]

Is the straight is long enough for both cars to reach the same maximum speed (dictated by engine power)?

That would be relatively rare for most cars and racetracks in existence. If it were long enough though, the heavier car would definitely need better brakes (assuming that the cars were identical in every way except weight) - top speed is not significantly dependent on weight for cars, so they'd both be slowing from the same speed and the heavier car would be dissipating much more KE.

 

[zanick]

MFB: if the car is the same, can the mu go down with increased weight
if the mu doesn't increase...based on your second comment, your saying that the friction coefficient says the same to make the comment where II assume you mean, "deceleration " the same. so, if that mu was constant for cars , heavy and light, the force would go up with weight and the deceleration would be the same. In real life, we see the fastest pro cars , like prototypes running the same tires, decelerate MUCH faster than the heavier cars on the same tires. It seems that if the limit of deceleration was found with the 3000lb car, that same force might only be available for the heavier car. (i.e the mu would go down with extra weight... which would make my assumption of lower rates of deceleration, valid)

Cjt: as you explained to Watters, there is no way both cars can reach the same maximum speed , even if there was an infinitely long straight. the heavier car would have slightly more rolling resistance. But, as you said, this is not probable in a track condition. But in reality, if both cars could reach their terminal velocity, and then brake to a stop, only if they had the same rate of deceleration force, would the heavier car dissipate heat at a higher rate. if their weight determines their decel rate ( 15% more weight causes 15% less decel rate), then the faster rate of decel could cause a faster rate of KE dissipation The example is talking about two cars reaching the same speed... on a race track, they wouldn reach the same speed, so if the force available is the same regardless of weight, the rate of KE dissipation would depend on the decel rate and would be higher with the higher rate of deceleration of the lighter car.

 

[mfb]

MFB: if the car is the same, can the mu go down with increased weight

It can go up, it can stay the same, or it can go down. It depends on details of the cars.

Actual racing cars have an additional complication we did not consider so far: they use the air flow to create an additional downwards force. It increases the achievable friction, but not the mass, so acceleration (both positive and negative) can be faster.

the heavier car would have slightly more rolling resistance.

It can also have a more powerful motor. The additional weight has to be somewhere.

 

[cjl]

MFB: if the car is the same, can the mu go down with increased weight
if the mu doesn't increase...based on your second comment, your saying that the friction coefficient says the same to make the comment where II assume you mean, "deceleration " the same. so, if that mu was constant for cars , heavy and light, the force would go up with weight and the deceleration would be the same. In real life, we see the fastest pro cars , like prototypes running the same tires, decelerate MUCH faster than the heavier cars on the same tires. It seems that if the limit of deceleration was found with the 3000lb car, that same force might only be available for the heavier car. (i.e the mu would go down with extra weight... which would make my assumption of lower rates of deceleration, valid)

The fastest (and lightest) cars decelerate faster for a couple of reasons. First off, as mfb alluded to, there's downforce. This makes comparison of mu very difficult, since it makes normal force non-constant and not just dependent on car mass. For a given amount of downforce, a lighter car benefits more than a heavier car (proportionally), so even with the same friction coefficient, the lighter car will stop faster. Secondly, there's tire compound - a heavier car will tend to wear out its tires faster, requiring a harder rubber compound to last the duration of the race. A lighter car can run a softer compound and not wear it out as quickly, so this would promote higher coefficients of friction for lighter cars.

Cjt: as you explained to Watters, there is no way both cars can reach the same maximum speed , even if there was an infinitely long straight. the heavier car would have slightly more rolling resistance. But, as you said, this is not probable in a track condition. But in reality, if both cars could reach their terminal velocity, and then brake to a stop, only if they had the same rate of deceleration force, would the heavier car dissipate heat at a higher rate. if their weight determines their decel rate ( 15% more weight causes 15% less decel rate), then the faster rate of decel could cause a faster rate of KE dissipation The example is talking about two cars reaching the same speed... on a race track, they wouldn reach the same speed, so if the force available is the same regardless of weight, the rate of KE dissipation would depend on the decel rate and would be higher with the higher rate of deceleration of the lighter car.

Actually, that's exactly backwards from what I said. I said that the heavier car and lighter car would reach about the same top speed (given sufficient distance), since rolling resistance is a relatively insignificant portion of the overall drag at top speed. Air resistance is dramatically more important. Also, I said nothing about deceleration rates - I said that the heavier car would be dissipating more kinetic energy, which is also accurate. Also, if the braking force available were the same, the rate of KE dissipation would be the same for both cars, but the heavier car would need to apply the brakes earlier (due to the lower deceleration rate) and would put more total energy into the brakes.

 

[zanick]

MFB: let's not add too many additional factors. assume the down force is the same for both cars, assume they are the same car. if they are not the same car, when comparing the prototypes... let's say the braking analysis is done in an area where the downforce aid comparisons are negligible
as far as HP concerned... both cars are the same. one has more weight because of a penalty or carrying a passenger. power is the same. again, let's keep the variables constant.

 

[zanick]

The fastest (and lightest) cars decelerate faster for a couple of reasons. First off, as mfb alluded to, there's downforce. This makes comparison of mu very difficult, since it makes normal force non-constant and not just dependent on car mass. For a given amount of downforce, a lighter car benefits more than a heavier car (proportionally), so even with the same friction coefficient, the lighter car will stop faster. Secondly, there's tire compound - a heavier car will tend to wear out its tires faster, requiring a harder rubber compound to last the duration of the race. A lighter car can run a softer compound and not wear it out as quickly, so this would promote higher coefficients of friction for lighter cars.
Actually, that's exactly backwards from what I said. I said that the heavier car and lighter car would reach about the same top speed (given sufficient distance), since rolling resistance is a relatively insignificant portion of the overall drag at top speed. Air resistance is dramatically more important. Also, I said nothing about deceleration rates - I said that the heavier car would be dissipating more kinetic energy, which is also accurate. Also, if the braking force available were the same, the rate of KE dissipation would be the same for both cars, but the heavier car would need to apply the brakes earlier (due to the lower deceleration rate) and would put more total energy into the brakes.

1st, again, let's keep downforce out of this as an added variable, so let's keep that out of it for simplicity sake. but regarding the tire compound, if you are using the tire to the limit and not beyond, ,the tires tend to wear out at the same rate. again, let's not add too many variables... we are talking same car , same tires. (same tire compound and size. just picture the one car picked up 500lbs of ballast weight and it was evenly distributed. so, mu of the tires does what with extra weight during a at limit decel?
intuitively, we know that 6000lb cars vs 3000llb cars don't decel at the same rate with the same tires from the same speed. theoretically, they should, but that's only if the mu stays the same, and I don't think the tires limit allows for this. they deform and have an optimal slip percentage. add more weight and this deformation goes over their limits, temps go up, rubber liquefies, etc etc.

2nd. yes, I agree. a long enough road, the KE added to both cars goes down with speed, most fighting the aero drag, but in the end, they both need to slow. yes, the heavier car slows starting with more KE and dissipates more KE, BUT, if the mu goes down with the heavier weight, then the KE dissipation rate is less for the heavier car. this is the KEY point. the heavier car puts more KE into the brakes, but over a longer period of time, thus lightning the thermal load on the brakes for an even higher total KE dissipation.

I think you already agreed to this at the beginning... if so, then we can discuss if the heavier car can have the same mu at the tires, and have greater deceleration forces as its weight goes up. I don't think this is possible. I guess that the force is kept near constant due to limits of the tire, and that means with weight going up, deceleration rates might go down proportionally... is this possible??

 

[cjl]

2nd. yes, I agree. a long enough road, the KE added to both cars goes down with speed, most fighting the aero drag, but in the end, they both need to slow. yes, the heavier car slows starting with more KE and dissipates more KE, BUT, if the mu goes down with the heavier weight, then the KE dissipation rate is less for the heavier car. this is the KEY point. the heavier car puts more KE into the brakes, but over a longer period of time, thus lightning the thermal load on the brakes for an even higher total KE dissipation.

Ignoring decel rates entirely for a moment (since there are so many variables here), the thermal load on the brakes is much more dependent on the total KE dissipated than the rate at which it is dissipated. Even if the heavier car is putting KE into the brakes at a lower rate, the brakes will still end up hotter because of the higher total energy put into them.

 

[zanick]

Ignoring decel rates entirely for a moment (since there are so many variables here), the thermal load on the brakes is much more dependent on the total KE dissipated than the rate at which it is dissipated. Even if the heavier car is putting KE into the brakes at a lower rate, the brakes will still end up hotter because of the higher total energy put into them.

I think you might want to think about that one a little longer. that would be true if the heat was stored in a "bucket". but, it has a dissipation rate capability. just like I could ride my brakes from the top of that mountain on Mars down to the surface and (assuming Earth atmosphere) and the brakes would reach some equilibrium temp and not change, even though in the end, I would be dissipating the amount of power of a small city by the end. however, I could slow in 4 seconds, as my example states and dissipate at a rate that is well over the dissipation rate of the components and they would get much hotter, and could go over the thermal limits of the pads dropping their mu values substantially. so , I have to disagree strongly with the dependence of KE rate vs total KE dissipated for the reasons I just mentioned.
again, look at my example at the limit of the tires being a constant force, the KE rate is 616hp vs 594hp (ave) with the light vs heavy car. over the same distance with the heavier car dissipating much more KE than the lighter car over all, but the rate is higher for the lighter car due to a greater rate of deceleration.

 

[cjl]

I think you might want to think about that one a little longer. that would be true if the heat was stored in a "bucket". but, it has a dissipation rate capability. just like I could ride my brakes from the top of that mountain on Mars down to the surface and (assuming Earth atmosphere) and the brakes would reach some equilibrium temp and not change, even though in the end, I would be dissipating the amount of power of a small city by the end. however, I could slow in 4 seconds, as my example states and dissipate at a rate that is well over the dissipation rate of the components and they would get much hotter, and could go over the thermal limits of the pads dropping their mu values substantially. so , I have to disagree strongly with the dependence of KE rate vs total KE dissipated for the reasons I just mentioned.
again, look at my example at the limit of the tires being a constant force, the KE rate is 616hp vs 594hp (ave) with the light vs heavy car. over the same distance with the heavier car dissipating much more KE than the lighter car over all, but the rate is higher for the lighter car due to a greater rate of deceleration.

Not only have I thought about it, I've also driven quite a few cars around racetracks, and I promise, even if they have less sticky tires, the heavier, more powerful ones that have to dissipate more total KE are the ones that tend to have problems with brake overheating. On a race track, there is basically no such thing as brakes that can dissipate heat anywhere near as fast as you can put heat into them during a braking event - you're really relying on the periods between brake events to keep the brakes at a reasonable temperature. During each braking event, the brake temperature spikes dramatically, then it cools until the next braking event.

(One caveat - I'm not tremendously familiar with super high end carbon ceramic brake setups, so this may not apply in those cases)

 

[zanick]

Not only have I thought about it, I've also driven quite a few cars around racetracks, and I promise, even if they have less sticky tires, the heavier, more powerful ones that have to dissipate more total KE are the ones that tend to have problems with brake overheating. On a race track, there is basically no such thing as brakes that can dissipate heat anywhere near as fast as you can put heat into them during a braking event - you're really relying on the periods between brake events to keep the brakes at a reasonable temperature. During each braking event, the brake temperature spikes dramatically, then it cools until the next braking event.

(One caveat - I'm not tremendously familiar with super high end carbon ceramic brake setups, so this may not apply in those cases)

Well, there is a huge difference between running around a track, and running at the limit with the consciousness to optimize brake efficiency.. Having raced at many levels for over 15 year and instructing, I can 100% assure you that the heavier car doesn't burden the brakes any more than the light cars when driven at the edge. when you drove, if you took longer to slow than the ligher car, you proved the case. less rate of KE dissipation if you took longer to slow with the heavy car (which 100% of students I've coached, do in the street cars vs their race cars on similar tires) Brakes are so capable these days, that if you know what you are doing, you never over heat them, as long as you have the right pads. if you don't have a performance brake pad, then its easy to get over the temp range to get an effective mu out of them... that just gives a ton of fade. those same brakes with the right pad, do the job without issue at the higher temp ranges. so, its not a matter of brakes being able to dissipate the heat, its what they do with the heat. the rotors glow red, heat goes into the calipers , hubs, bearings, wheels, and the temps skyrocket. during the braking event. as long as the pad can operate in those temp ranges, you are good because, thank god, that process only lasts for 4 seconds or so worst case. then, you have only quick high energy braking force for a few corners, or no braking at all while the brakes get fed 100 to 130mph air flow to get ready for that next big braking event.

Now, its very easy to over heat brakes by dragging the brakes and using them in a preventative fashion. This is how most begnner drivers get into trouble, they are on the brakes for too long a period and they end up boiling brake fluid and operating the pads in the fade range , where the additional pressure to brake, only exacerbates the problem with further heating.

Now, add the weigiht, and that top speed goes down before the same braking point, the heavier car takes longer to slow down and slows to a slower speed as well. this creates Less KE dissipation rates, vs the light car that decelerates faster putting more thermal load on the braking system.

yes, you have More KE to deal with, adding the weight. lots of ways to over heat things with more heating potenital. but, if properly, and equally managed, there should be no difference with more weight vs lighter weight race cars, but in fact, if the heavier car can't keep the mu the same with the extra weight, it will have a lesser rate of KE dissipation . However, if you think a heavy vs light racer can slow at the same rate, then the heavy car will dissipate its KE at a higher rate, but not as much as its increase in weight would at first glance , would indicate.

 

[rcgldr]

Assuming the same size tires, then a heavier car increases the tire load, which decreases the coefficient of friction. Wiki article on this:

http://en.wikipedia.org/wiki/Tire_load_sensitivity

 

[cjl]

Well, there is a huge difference between running around a track, and running at the limit with the consciousness to optimize brake efficiency.. Having raced at many levels for over 15 year and instructing, I can 100% assure you that the heavier car doesn't burden the brakes any more than the light cars when driven at the edge.

Well, I don't know what to say other than that you're wrong. The heavy car will need to dissipate more KE, and will typically dissipate it at a higher rate, necessitating larger brakes, more brake cooling, and more brake pad area/thickness. Out of curiosity, what types of cars do you race/instruct in?

when you drove, if you took longer to slow than the ligher car, you proved the case. less rate of KE dissipation if you took longer to slow with the heavy car (which 100% of students I've coached, do in the street cars vs their race cars on similar tires) Brakes are so capable these days, that if you know what you are doing, you never over heat them, as long as you have the right pads. if you don't have a performance brake pad, then its easy to get over the temp range to get an effective mu out of them... that just gives a ton of fade. those same brakes with the right pad, do the job without issue at the higher temp ranges. so, its not a matter of brakes being able to dissipate the heat, its what they do with the heat. the rotors glow red, heat goes into the calipers , hubs, bearings, wheels, and the temps skyrocket. during the braking event. as long as the pad can operate in those temp ranges, you are good because, thank god, that process only lasts for 4 seconds or so worst case. then, you have only quick high energy braking force for a few corners, or no braking at all while the brakes get fed 100 to 130mph air flow to get ready for that next big braking event.

When I drive around a track, aside from a warm up and cool down lap, I'm threshold braking at every brake zone, and trailbraking into several of the corners. Believe me, I'm not taking it easy on the brakes, nor am I dragging them. Brakes on the vast majority of street cars these days do not have adequate airflow nor do they have sufficient rotor area to effectively dissipate the amount of heat that you can put into them on the track, and as a result, even with track pads, you will often boil the brake fluid unless you add additional brake cooling and/or go to an extremely high temperature fluid (I like Castrol SRF, personally). I agree with you about the pads too, but I figured that would go without saying. That having been said, the brakes hold onto a remarkable amount of heat long after the major braking event, and that residual heat is what can boil the fluid and cause braking issues. Last summer, I went on the track in a 2006 Porsche Cayman S that had been upgraded with larger than stock brake cooling scoops, and after a full lap (~2 miles on this track) of cooldown with minimal braking, the brake disks were still at a temperature of over 750F when I pulled into the paddock. This is still more than a hundred degrees F hotter than the boiling point of even high end brake fluid.

Now, its very easy to over heat brakes by dragging the brakes and using them in a preventative fashion. This is how most begnner drivers get into trouble, they are on the brakes for too long a period and they end up boiling brake fluid and operating the pads in the fade range , where the additional pressure to brake, only exacerbates the problem with further heating.

Sure, dragging the brakes will overheat them, but so will maximum energy threshold braking at the limit.

Now, add the weigiht, and that top speed goes down before the same braking point, the heavier car takes longer to slow down and slows to a slower speed as well. this creates Less KE dissipation rates, vs the light car that decelerates faster putting more thermal load on the braking system.

Add weight and the maximum braking force increases. This increases rate of KE dissipation. In addition, the heavier car (assuming the same horsepower) will have more total KE to dissipate. Higher KE dissipation rate plus more KE to dissipate means that the heavier car (all else equal) will always put a greater thermal load on the braking system.

yes, you have More KE to deal with, adding the weight. lots of ways to over heat things with more heating potenital. but, if properly, and equally managed, there should be no difference with more weight vs lighter weight race cars, but in fact, if the heavier car can't keep the mu the same with the extra weight, it will have a lesser rate of KE dissipation . However, if you think a heavy vs light racer can slow at the same rate, then the heavy car will dissipate its KE at a higher rate, but not as much as its increase in weight would at first glance , would indicate.

Yes, the friction coefficient (mu) decreases slightly with increased weight, but the total frictional force available increases. Rate of KE dissipation is equal to total frictional force multiplied by velocity, and the average rate of KE dissipation will be higher for the heavier car. In addition, the change in frictional coefficient due to speed won't be very large for reasonable weight ranges, and thus the heavier car will slow nearly as quickly as the lighter one.

 

[zanick]

Assuming the same size tires, then a heavier car increases the tire load, which decreases the coefficient of friction. Wiki article on this:

http://en.wikipedia.org/wiki/Tire_load_sensitivity

Ive got quite a few charts on degrading mu and traction based on adding weight to a tire. thanks for posting. the question is, "how much? " :)

 

[zanick]

I don't know how to insert comments.. so ill just use a >>>>>> and underline

Well, I don't know what to say other than that you're wrong. The heavy car will need to dissipate more KE, and will typically dissipate it at a higher rate, necessitating larger brakes, more brake cooling, and more brake pad area/thickness. Out of curiosity, what types of cars do you race/instruct in?

>>>>>> racing world challenge GT, 15 years of American V8 supercars, ST2 NASA, ITE, GT2 SCCA, GT2 Porsche. etc etc. and coaching in cars from 3500lb aston martins and 400hp spec mustangs. I can tell you, flat out the heavier car gets to a slower top speed (in my example , 15% more weight, , gives only 7% more KE, at 5% less speed. don't confuse, total KE with KE rate of dissipation. 7% the higher KE , heavier car, if it takes 4.3 seconds vs 4 seconds for the ligher car, will dissipate more heat yes...( and heat is being shed by the braking system, as you know) but because the rate of dissipation is lower (lower decel rate) , the KE dissipation rate can be lower as well. in fact, if we assume my variables as traction staying constant, the heavier car will dissipate at a rate 3.5% less than the ligher car, even though it dissipated more heat. this is because it does it over a longer period of time... so the temp rise on the rotors and pads will be less.When I drive around a track, aside from a warm up and cool down lap, I'm threshold braking at every brake zone, and trailbraking into several of the corners. Believe me, I'm not taking it easy on the brakes, nor am I dragging them. Brakes on the vast majority of street cars these days do not have adequate airflow nor do they have sufficient rotor area to effectively dissipate the amount of heat that you can put into them on the track, and as a result, even with track pads, you will often boil the brake fluid unless you add additional brake cooling and/or go to an extremely high temperature fluid (I like Castrol SRF, personally). I agree with you about the pads too, but I figured that would go without saying. That having been said, the brakes hold onto a remarkable amount of heat long after the major braking event, and that residual heat is what can boil the fluid and cause braking issues. Last summer, I went on the track in a 2006 Porsche Cayman S that had been upgraded with larger than stock brake cooling scoops, and after a full lap (~2 miles on this track) of cooldown with minimal braking, the brake disks were still at a temperature of over 750F when I pulled into the paddock. This is still more than a hundred degrees F hotter than the boiling point of even high end brake fluid.

>>>>>>>ive coached a lot of racers that thought they were threshold braking, but ended up braking too early and too long, and would induce a lot more heat in the system. not saying you are doing that, just an observation. you sound like you know what you are doing out there. I've never boiled brake fluid and I have used the crappy stuff too (ATE ), but use the SRF now. (longer change cycles), because I am very prudent on the brakes and do a lot of trail braking. I always come into the pits with rotor temps in the 5-600 degree mark when the rotors during the session are well over 1200 degrees. (we use temp paint to see the results) the brake fluid doesn't boil, unless the calipers are seeing that heat. by braking technique, you can get the temps very high and if the release is long enough, with high speed and cooling the temp never reaches the fluid. I've instructd in a lot of 911s and caymans. I would have no problem putting one through its paces with the stock cooling at the speeds it can run under stock power. if you can give me a link to a video of you driving, it would be interesting to see if I see anything that would indicate over braking. usually, with lower powered , heavy stock cars, its near ipossible for me to hurt or over heat the brakes, unless they are not race pads and not equipped with at least ATE level of quality brake fluid.. for many years I raced a 3100lb 350hp car with only 11" rotors.. Now, with 3000bs and 450hp, the 13" are needed and used.
be nice to get away from the little fade I get on the most challenging turn approachs, but that proves that the brakes can dissipate the heat, as I never get boil, just fade. that's a KE dissipation deal. by the way, with passenger seat in and a 250lb guy riding, there is no braking issues... as my calculations show, I get to a little slower speed, and the rate of decel is a little less.Sure, dragging the brakes will overheat them, but so will maximum energy threshold braking at the limit.
>>>>>>this is true. tough to get to this point with most stock'ish porsches and other production vehicles. most all are equipped with 12" rotors and brembos these days. I've never seen a braking performance issue when the car is well driven., even on race rubber, which by the way, increases the demanss WELL OVER the extra weight. why? because the mu goes way up, again, well decreasing the RATE OF KE DISSIPATION. that's key here. not total KE start- ending KE = x KE...
Its x KE / time that is the killer here!Add weight and the maximum braking force increases. This increases rate of KE dissipation. In addition, the heavier car (assuming the same horsepower) will have more total KE to dissipate. Higher KE dissipation rate plus more KE to dissipate means that the heavier car (all else equal) will always put a greater thermal load on the braking system.

>>>>>>>>>>>Well, some true and some not so true... as far as braking forces, possible, , but by how much is the question. I assumed in my calculations, that with extra weight, the mu changed by the same percentage. 15% more weight caused 15% less decel rate , (e.g. 1g vs .85g) what if its 10% or 5% less decel for the 15% more weight... still this would then increase the force as you say, but by how much. THEN, don't forget, even though the heavier car has more KE before braking, its slow down will be at a lower rate. KE/second or HP will be less . this is what drives the temp up. not total KE dissipated. the longer the time, the lower the heat dissipation RATE.
you say the KE rate is higher Its not, if its decelerating slower . look at my example ..making one assumption that decel rates go down proportional to increase weight , shows clearly that the rate of KE dissipation is greater for the ligher car.. if this is too high of an estimation, what is the difference. example:
two cars, one 3500lb vs 3000lb, heavier one going 3-5% slower than light car at end of straight, what is the deceleration rate of each? the same?? doubtful. what is the difference in mu? is there a difference? :)Yes, the friction coefficient (mu) decreases slightly with increased weight, but the total frictional force available increases. Rate of KE dissipation is equal to total frictional force multiplied by velocity, and the average rate of KE dissipation will be higher for the heavier car. In addition, the change in frictional coefficient due to speed won't be very large for reasonable weight ranges, and thus the heavier car will slow nearly as quickly as the lighter one.
>>>>>>>>>>>>>>>>>>>>yes, the force x velocity is the power dissipated but both cars are going different speeds for different periods of time. that's why KE/time is the overall average of heat dissipation rate. But, if you think the heavy car by 15% , with 500lbs extra weight will slow as fast as the lighter car, I think that's a stretch. I run against guys (same driver) that has a mustang and a cup car2500lbs vs 3500lbs the cup car decelerates at much faster rate (sure certain designs make that possible, but we see other cars with huge decel advantages due to weight. I don't intuitively think that the heavier car will slow as well as the light car (same car, just with weight diff) do you really? if not the difference has to be in the changing mu and we know it changes . but by how much? :)

 

[rcgldr]

Ive got quite a few charts on degrading mu and traction based on adding weight to a tire. thanks for posting. the question is, "how much? "

Depends on the tire. Increasing the width of the tire reduces the load per unit contact area, and provides more area for cooling of the tire, but increases the weight of the tire, which affects suspension response. For race cars, tire sizes and vehicle weights are regulated, so I don't know if the tire sizes are ideal for each of those race classes.

 

[cjl]

I don't know how to insert comments.. so ill just use a >>>>>> and underline

>>>>>> racing world challenge GT, 15 years of American V8 supercars, ST2 NASA, ITE, GT2 SCCA, GT2 Porsche. etc etc. and coaching in cars from 3500lb aston martins and 400hp spec mustangs. I can tell you, flat out the heavier car gets to a slower top speed (in my example , 15% more weight, , gives only 7% more KE, at 5% less speed. don't confuse, total KE with KE rate of dissipation. 7% the higher KE , heavier car, if it takes 4.3 seconds vs 4 seconds for the ligher car, will dissipate more heat yes...( and heat is being shed by the braking system, as you know) but because the rate of dissipation is lower (lower decel rate) , the KE dissipation rate can be lower as well. in fact, if we assume my variables as traction staying constant, the heavier car will dissipate at a rate 3.5% less than the ligher car, even though it dissipated more heat. this is because it does it over a longer period of time... so the temp rise on the rotors and pads will be less.

OK, so it sounds like you've done most of your stuff in either very sporty cars or race cars. I agree that it's awfully hard to overheat the brakes in those cases - the ones that I've seen experience real difficulty with braking when driven well tend to be the cheaper sports cars, which come with substantially worse braking capability (370Z and base WRX come to mind). As for your math here though, your traction definitely goes up with increased weight - I'd actually expect the heavier vehicle in your example here to be braking later, and spending less time braking compared to the lighter vehicle, even though it had more energy to dissipate. It can brake later because its speed is lower, and tire friction coefficient isn't going to vary that much with a 15% difference in load.

>>>>>>>ive coached a lot of racers that thought they were threshold braking, but ended up braking too early and too long, and would induce a lot more heat in the system. not saying you are doing that, just an observation. you sound like you know what you are doing out there. I've never boiled brake fluid and I have used the crappy stuff too (ATE ), but use the SRF now. (longer change cycles), because I am very prudent on the brakes and do a lot of trail braking. I always come into the pits with rotor temps in the 5-600 degree mark when the rotors during the session are well over 1200 degrees. (we use temp paint to see the results) the brake fluid doesn't boil, unless the calipers are seeing that heat. by braking technique, you can get the temps very high and if the release is long enough, with high speed and cooling the temp never reaches the fluid. I've instructd in a lot of 911s and caymans. I would have no problem putting one through its paces with the stock cooling at the speeds it can run under stock power. if you can give me a link to a video of you driving, it would be interesting to see if I see anything that would indicate over braking. usually, with lower powered , heavy stock cars, its near ipossible for me to hurt or over heat the brakes, unless they are not race pads and not equipped with at least ATE level of quality brake fluid.. for many years I raced a 3100lb 350hp car with only 11" rotors.. Now, with 3000bs and 450hp, the 13" are needed and used.
be nice to get away from the little fade I get on the most challenging turn approachs, but that proves that the brakes can dissipate the heat, as I never get boil, just fade. that's a KE dissipation deal. by the way, with passenger seat in and a 250lb guy riding, there is no braking issues... as my calculations show, I get to a little slower speed, and the rate of decel is a little less.

I actually think we mostly agree here. Everything you say matches my experience pretty well. Sadly, I don't have any good videos of me driving at the moment - I really need to get a good traqmate setup or something similar sometime soon, since it would be really interesting. A friend of mine has some video (with data) from his racecar from earlier this summer that I drove several sessions of a race in, but I don't know where that's stored at the moment. If I get my hands on it, I may post it here, though that was a car that definitely didn't need much in the way of brakes (it was a 12 hour World Racing League enduro race, and we were in the GP4 class, which is a pretty slow category. Our car had all of about 130hp, but we did pretty well because our reliability was good).

>>>>>>this is true. tough to get to this point with most stock'ish porsches and other production vehicles. most all are equipped with 12" rotors and brembos these days. I've never seen a braking performance issue when the car is well driven., even on race rubber, which by the way, increases the demanss WELL OVER the extra weight. why? because the mu goes way up, again, well decreasing the RATE OF KE DISSIPATION. that's key here. not total KE start- ending KE = x KE...
Its x KE / time that is the killer here!

Again, we don't disagree much here. I will add that stock-ish Porsches tend to have excellent brakes compared to a lot of cheaper sports cars though, and they tend to be relatively light and low powered for the price as well, all of which contributes to them performing well on stock brakes. Porsches also have the benefit of having a rearward weight bias, which substantially evens out the braking energy amongst the wheels. Most cars will dump the vast majority of the braking KE into the front brakes, but a rear-engined car like a 911 will spread it out much more evenly between front and rear, helping a lot with temperature and wear. As I said above, the vehicles that I've noticed having the most trouble tend to be cheaper sports cars, not high end ones like Corvettes, Porsches, or the like, which tend to do just fine with decent fluid and race pads (I always liked the Pagid RS-19s personally, though with my new car I'm tempted to try out the RS-29).

(I will also point out that race rubber doesn't just increase mu, it also increases total KE, since you carry more speed onto the start of the straight and can brake later at the end of the straight, increasing peak speed)

>>>>>>>>>>>Well, some true and some not so true... as far as braking forces, possible, , but by how much is the question. I assumed in my calculations, that with extra weight, the mu changed by the same percentage. 15% more weight caused 15% less decel rate , (e.g. 1g vs .85g) what if its 10% or 5% less decel for the 15% more weight... still this would then increase the force as you say, but by how much. THEN, don't forget, even though the heavier car has more KE before braking, its slow down will be at a lower rate. KE/second or HP will be less . this is what drives the temp up. not total KE dissipated. the longer the time, the lower the heat dissipation RATE.
you say the KE rate is higher Its not, if its decelerating slower . look at my example ..making one assumption that decel rates go down proportional to increase weight , shows clearly that the rate of KE dissipation is greater for the ligher car.. if this is too high of an estimation, what is the difference. example:
two cars, one 3500lb vs 3000lb, heavier one going 3-5% slower than light car at end of straight, what is the deceleration rate of each? the same?? doubtful. what is the difference in mu? is there a difference? :)

That's a faulty assumption I think - if mu changed by the same percentage as extra normal force, downforce would be pointless, since the additional load on the tires would not increase the available grip. From the data I can find, typical tire load sensitivity is more along the lines of a 10% loss in peak frictional coefficient with a doubling of load. I suspect if you use this kind of data, you'll find that the heavier car has both higher KE and higher KE loss rate, rather than the lower loss rate you calculate here. Also, KE and KE rate both have a substantial impact on temperature, depending on the details of the braking system. A system with a high brake component mass and little airflow will heat up almost entirely based on total KE dissipated, while a system with low brake component mass and high airflow will heat up more based on KE rate (for extreme examples of those, compare the braking system on jet aircraft with the braking system on formula 1 racecars - the aircraft brakes heat up almost purely based on total energy, while the F1 brakes heat up almost entirely based on rate of energy dissipation).

>>>>>>>>>>>>>>>>>>>>yes, the force x velocity is the power dissipated but both cars are going different speeds for different periods of time. that's why KE/time is the overall average of heat dissipation rate. But, if you think the heavy car by 15% , with 500lbs extra weight will slow as fast as the lighter car, I think that's a stretch. I run against guys (same driver) that has a mustang and a cup car2500lbs vs 3500lbs the cup car decelerates at much faster rate (sure certain designs make that possible, but we see other cars with huge decel advantages due to weight. I don't intuitively think that the heavier car will slow as well as the light car (same car, just with weight diff) do you really? if not the difference has to be in the changing mu and we know it changes . but by how much? :)

Well, I'd expect the cup car to decelerate substantially faster (assuming you mean a Porsche GT3 cup car) - it doesn't take as long for the suspension to settle under braking, it has more aero drag (which is pretty much free braking force), it has more downforce (which increases grip significantly even with the same tire compound), and the difference in tire load is much more significant than just the difference in weight between the cars. The rear weight bias and low CG of the Porsche will load the tires much more evenly than the higher CG and front weight bias of the Mustang. I don't know the exact numbers for a 911 Cup, but here's an example:

Say the Porsche has a static weight distribution of 63% rear and 37% front (this is probably at least in the right ballpark), and the CG is around 14 inches off the ground (also at least in the right ballpark). With a wheelbase of 97", this means that the CG is about 36" in front of the rear axle. Now, let's say it is decelerating under heavy braking at a rate of 1.5G. Now, the front tires are experiencing 65% of the weight of the car, while the rears are experiencing 35%, so the front tires are under a load of about 1625lb, while the rears are under a load of about 875lb.

Now let's look at the Mustang. A 2015 Mustang GT has its CG 20 inches off the ground, with a F/R weight distribution of 53.8/46.2%. It has a wheelbase of 107.1", with the CG located 57.6" in front of the rear axle. Under 1.5G deceleration, it will load the front tires with nearly 82% of the car's weight, while the rear tires will have only 18%. Out of the car's 3800lb weight, 3100lb will be on the front axle, with only 700lb on the rear. Even though the car only weighs 50% more than the 911 cup, the front tires are under 90% higher load at this level of braking. As I mentioned earlier, the tire load sensitivity figures I can find indicate about a 10% reduction in available frictional coefficient with a doubling of load, so this will decrease the Mustang's braking deceleration by nearly 10% compared to the Porsche, even ignoring downforce. This also demonstrates why 911s are very good at braking - that rear-biased weight distribution combined with a low CG means that they load the front tires substantially less under braking than most cars of similar weight, giving them more grip and spreading the KE dissipation more evenly between the front and rear than most cars.

 

[zanick]

Depends on the tire. Increasing the width of the tire reduces the load per unit contact area, and provides more area for cooling of the tire, but increases the weight of the tire, which affects suspension response. For race cars, tire sizes and vehicle weights are regulated, so I don't know if the tire sizes are ideal for each of those race classes.

Im talkng of the experience where both classes of cars are using the same exact tire. the ligher car. (good example, because it mimicks the near 500lb difference in my example) Daytona prototypes and GT3 cup cars in Grand Am racing.

 

[zanick]

OK, so it sounds like you've done most of your stuff in either very sporty cars or race cars. I agree that it's awfully hard to overheat the brakes in those cases - the ones that I've seen experience real difficulty with braking when driven well tend to be the cheaper sports cars, which come with substantially worse braking capability (370Z and base WRX come to mind). As for your math here though, your traction definitely goes up with increased weight - I'd actually expect the heavier vehicle in your example here to be braking later, and spending less time braking compared to the lighter vehicle, even though it had more energy to dissipate. It can brake later because its speed is lower, and tire friction coefficient isn't going to vary that much with a 15% difference in load.
>>>>>>>>I agree. IN fact, we are agreeing more and more through this discussion. good stuff... makes you think though!
I've done the deceleration math.. actually, the deceleration rate and slower speed , don't equal a point where the heavier car can brake any earlier, due to the slower speed it needs to reach to make the turn in point. Ill post my latest calculations. both cars, braking at the same point. Mu only changing by 8% based on some curves I've seen of changing Mu (coefficient of friction) based on tire loading.I actually think we mostly agree here. Everything you say matches my experience pretty well. Sadly, I don't have any good videos of me driving at the moment - I really need to get a good traqmate setup or something similar sometime soon, since it would be really interesting. A friend of mine has some video (with data) from his racecar from earlier this summer that I drove several sessions of a race in, but I don't know where that's stored at the moment. If I get my hands on it, I may post it here, though that was a car that definitely didn't need much in the way of brakes (it was a 12 hour World Racing League enduro race, and we were in the GP4 class, which is a pretty slow category. Our car had all of about 130hp, but we did pretty well because our reliability was good).

>>>>>>>>>love to see them.. even in a very light car with a good hp/weight ratio. Probably lots of action! :)Again, we don't disagree much here. I will add that stock-ish Porsches tend to have excellent brakes compared to a lot of cheaper sports cars though, and they tend to be relatively light and low powered for the price as well, all of which contributes to them performing well on stock brakes. Porsches also have the benefit of having a rearward weight bias, which substantially evens out the braking energy amongst the wheels. Most cars will dump the vast majority of the braking KE into the front brakes, but a rear-engined car like a 911 will spread it out much more evenly between front and rear, helping a lot with temperature and wear. As I said above, the vehicles that I've noticed having the most trouble tend to be cheaper sports cars, not high end ones like Corvettes, Porsches, or the like, which tend to do just fine with decent fluid and race pads (I always liked the Pagid RS-19s personally, though with my new car I'm tempted to try out the RS-29).

(I will also point out that race rubber doesn't just increase mu, it also increases total KE, since you carry more speed onto the start of the straight and can brake later at the end of the straight, increasing peak speed)
>>>>>>>>>love the comparisons of the 911 and the front engine cars. (im running a front engine car so i know those differences) yep, 911s can have double on the rear brake forces and the fronts have 15% less % to match the weight transfer and distrib of a front engine racer with near 50/50 weight. (another discussion)
as a note... I've got a lot of experience with the RS14s (RS15s are newer and better) and have best mu characteristics and ability to deal with heat. go to the enduro pad, which work well for endurance cars, last a long time, don't have the performance . but last a long time. (i.e. pagid black vs pagid yellows)That's a faulty assumption I think - if mu changed by the same percentage as extra normal force, downforce would be pointless, since the additional load on the tires would not increase the available grip. From the data I can find, typical tire load sensitivity is more along the lines of a 10% loss in peak frictional coefficient with a doubling of load. I suspect if you use this kind of data, you'll find that the heavier car has both higher KE and higher KE loss rate, rather than the lower loss rate you calculate here. Also, KE and KE rate both have a substantial impact on temperature, depending on the details of the braking system. A system with a high brake component mass and little airflow will heat up almost entirely based on total KE dissipated, while a system with low brake component mass and high airflow will heat up more based on KE rate (for extreme examples of those, compare the braking system on jet aircraft with the braking system on formula 1 racecars - the aircraft brakes heat up almost purely based on total energy, while the F1 brakes heat up almost entirely based on rate of energy dissipation).
>>>>>>>Thatsa good point. i think I've said before , 15% weight increase, 15% loss in mu.. after digging a little deeper, its probalby more like 15% increase, 8% loss in grip. in any case, the extra weight adds to the extra force. yes, the heavier car has the increased KE , but not the higher rate, due to the lower rate of decel (remember, if we agree 8% lower mu based on 15% increase rate, then the decel rate would go from 1g to .92 g )
yep, correct on why F1 brakes heat up so much... its all rate of heat dissipation! Well, I'd expect the cup car to decelerate substantially faster (assuming you mean a Porsche GT3 cup car) - it doesn't take as long for the suspension to settle under braking, it has more aero drag (which is pretty much free braking force), it has more downforce (which increases grip significantly even with the same tire compound), and the difference in tire load is much more significant than just the difference in weight between the cars. The rear weight bias and low CG of the Porsche will load the tires much more evenly than the higher CG and front weight bias of the Mustang. I don't know the exact numbers for a 911 Cup, but here's an example:

Say the Porsche has a static weight distribution of 63% rear and 37% front (this is probably at least in the right ballpark), and the CG is around 14 inches off the ground (also at least in the right ballpark). With a wheelbase of 97", this means that the CG is about 36" in front of the rear axle. Now, let's say it is decelerating under heavy braking at a rate of 1.5G. Now, the front tires are experiencing 65% of the weight of the car, while the rears are experiencing 35%, so the front tires are under a load of about 1625lb, while the rears are under a load of about 875lb.

Now let's look at the Mustang. A 2015 Mustang GT has its CG 20 inches off the ground, with a F/R weight distribution of 53.8/46.2%. It has a wheelbase of 107.1", with the CG located 57.6" in front of the rear axle. Under 1.5G deceleration, it will load the front tires with nearly 82% of the car's weight, while the rear tires will have only 18%. Out of the car's 3800lb weight, 3100lb will be on the front axle, with only 700lb on the rear. Even though the car only weighs 50% more than the 911 cup, the front tires are under 90% higher load at this level of braking. As I mentioned earlier, the tire load sensitivity figures I can find indicate about a 10% reduction in available frictional coefficient with a doubling of load, so this will decrease the Mustang's braking deceleration by nearly 10% compared to the Porsche, even ignoring downforce. This also demonstrates why 911s are very good at braking - that rear-biased weight distribution combined with a low CG means that they load the front tires substantially less under braking than most cars of similar weight, giving them more grip and spreading the KE dissipation more evenly between the front and rear than most cars.

>>>>>>>>>>Ill have to look into your example above... and illl post back my latest numbers

 

[zanick]

If we agree that the mu (coefficient of friction ) of the tires go down and i agree as well its to a level of 92% of the lighter car.
the thermal load is going to be near the same on the track by adding 15% more weight. the top speeds before braking are a result of a very good simulator with many factors equal to what we might see on our club racers... using actual weight, hp/torque curves, aero drag, car geometry, tire size etc etc. Here are my calculations:

Two cars , CAR A (light car at 3000lbs) vs Car B (heavy car 500lbs heavier at 3500lbs). They get the point that of the exact same initial braking point, the two cars with the same 400HP are running:

car A 120mph (53.64 m/s)
car B 114mph (50.9 m/s)

the simple part of this example is that we can just plug in the numbers if we agree that the heavy car will slow at 8% less deceleration rate. we will call this 1g vs .92g respectively. this is because of the changing of the mu for the tires as weight is added to them. (weight on the tire, is the total weight/4 plus weight transfer weight/2 based on deceleration rate)

This means in 4 seconds the light car slows to 33mph (15m/s)
this also means in 4 seconds, the HEAVY car slows to the same speed

keeping it really simple. the KE at the start is:
1,991956 J Light
2,085,914 J Heavy

the KE at the end at the final same exact speed is:
153,405 J light
178,977 J heavy

this ends up with the lighter car dissipating a total of 1,838,551 for the light car
and 1,906,937 J for the Heavy car. heavy car dissipates more energy by 3.7%
not anywhere near the increase that the 15% increase of weight would first indicate

So,

Because the lighter car slows to the same speed in 4 seconds (3.94secs) and the heavy car slows to this same speed in 4 seconds too the RATE OF KE DISSIPATION is HIGHER for the heavy car by about 2%. no surprise here

625HP/sec for the light car ( example : 1,838,551 J /4 sec (3.94actual) /746watt =hp/sec)
639HP/sec for the heavy car ( example : 1,906,937 J /4 sec /746watt =hp/sec)
(only about 15hp/sec greater dissipation rate)BUT, since we know the turn in point will happen at a slightly lower speed for the heavy car (13.5ms or 30mph), the result is: . Because now we are slowing at the same rate of .92g but for 4.15 seconds. NET NET 626hp ave dissipation rate... the heavier car has the same of KE dissipation as the light car

The heavy car:
2,085,914 J at 50.64/s (114mph)
144,971 J at 13.5m/s (30mph)
Total KE dissipated of 1,940,942. / 4.15 seconds = 626hp ave dissipation

The rate of heat dissipation for the HEAVIER CAR is the same vs the light car for its decel rate and its lower top speed at the moment of braking (at the same spot on the track) and slightly lower target speed upon brake release for turn in.

the interesting thing here is also, with both cars activating their brakes at the same point, they end up at the same spot before turn in, 320f vs 316ft, if you use the decel rates of 1g vs .92g .

Therefore, this shows that adding weight, under these normal conditions, actually doesn't change the burden on the braking system.acceleration calculator used to determine the time intervals to match the g rate of deceleration and starting and final speeds
http://www.smartconversion.com/unit_...alculator.aspx