5 ms·

Seventy versus One Hundred

65 points by ColinWright 3y ago

didgeoridoo 3y ago
Nerd snipe alert. Do not open this if you’re just getting into bed.
- cheschire 3y ago
  https://www.explainxkcd.com/wiki/index.php/356:_Nerd_Sniping https://www.explainxkcd.com/wiki/index.php/356:_Nerd_Sniping
dylan604 3y ago
"(keep scrolling ...) " ... "(was that far enough?) " oh man, i was hoping this was going to be one of those tricks of no matter what your window size was, this would always be just below the fold with CSS. i just got lucky for it to line up. after trying it again with a smaller window, it was just in the scroll. =( then again, looking at the site's layout, it does not really suggest heavy use of CSS. at. all.
dmoy 3y ago
TL;DR v^2 is bad for your health in a car crash (But probably read it because it's really short)
- dylan604 3y ago
  yeah, if we need a TL;DR version of that...well, boy, I just don't know
- thriftwy 3y ago
  But at least it is halvedz or it would be much worse!
- 082349872349872 3y ago
  Old (70s?) Citroens used to have a "braking distance" bar on the speedo that grew quadratically, attempting to keep this point in the drivers' minds?
krukah 3y ago
> Let's assume that the brakes are working at their limit, and as they do so, they are shedding energy at a maximum rate that doesn't change. The cars are identical, so they are both shedding energy at the same rate per unit distance. Interesting (and I think very reasonable) first-order assumption. Maybe to second-order, the calculation could assume the braking force is proportional to velocity, which I think is roughly true of friction generally, but is also harder to model.
- bell-cot 3y ago
  IANAE (...not an engineer), but "braking force is proportional to velocity" is not how brakes, or friction, work.
  - purpleflame1257 3y ago
    No, but it is how first-order atmospheric drag is modeled, which may be what confused the GP comment.
- Karellen 3y ago
  > > they are both shedding energy at the same rate per unit distance. > Interesting (and I think very reasonable) first-order assumption. Really? That triggered alarm bells in my head immediately. I mean, it might be true, but I'd need to do the math to figure it out one way or another. > the calculation could assume the braking force is proportional to velocity, which I think is roughly true of friction generally Again, not my intuition at all. I'd have gone with the braking force being constant at non-zero velocity. (And force is variable up to the limit of static friction when at zero velocity.)
  - calfuris 3y ago
    Your intuition about braking force matches the thing that triggered alarm bells in your head: work = force x distance, so if the force is approximately constant the kinetic energy dissipated will be approximately constant per unit distance.
lmm 3y ago
I assumed constant deceleration. Then u = sqrt(v^2 - 4as), a and s are constant so = sqrt(v^2-70^2) = approximately 70mph.
- FabHK 3y ago
  That's the same assumption as the (first) article makes.
  - lmm 3y ago
    It talks about shedding energy at a constant rate per distance. Which, sure, ends up giving the same result, but it's a much weirder starting assumption, IMO.
    - Ma8ee 3y ago
      I thought it was a quite natural assumption, considering the same brake friction force and that the work done by that is the force times the distance. (W = Fs).
tempestn 3y ago
Cars can generally decelerate at a roughly constant rate of acceleration, based on the friction between tires and ground. Given the cars are identical, you can therefore define braking acceleration (deceleration) a, which is equal to both 70/t1 and (100-v)/t2, where v is the final velocity we're solving for. We also know that the average velocity for each car equals the distance travelled over time elapsed, where the distance is equal for both cars, but the time isn't. So, 70/2 = d/t1 and (100+v)/2 = d/t2. Can then solve these equations for v ~= 71.4 The gotcha is that it's much easier to solve if you take both distance and time as constant for both cars, but that would be a mistake. Of course you could get even more meta with it. Since the faster car is going to reach the obstacle first and smash into it, it probably won't be in the same position when the slower car reaches it either!
- neerajsi 3y ago
  Yeah, I also got 71.4 according to the same logic, but I can buy the argument in the follow on article that the amount of energy dissipated per unit time is what's constant.
  - tempestn 3y ago
    Why? No production car is brake limited; slam on the brakes in any car and you'll go into (and stay in) ABS, meaning you're at the limit of tire grip, and therefore at roughly constant deceleration. Where would energy dissipation come into the picture? The energy is being dissipated by the brakes, not the tires, and they're not the limiting factor.
    - eschneider 3y ago
      Not true. Brakes need time to dump heat. With repeated hard braking it's not too hard to overwhelm the brakes on most cars.
      - tempestn 3y ago
        But the scenario we're talking about here isn't repeated hard braking.
Ntrails 3y ago
I mean this also misses the fact that if they _see_ the obstacle at the exact same distance + time (a sensible assumption to compare) by the time they hit the breaks a significant difference in distance has already been generated. Stopping distances are part of the theory test in the UK, albeit with nonsense values. The frustrating part for me is how they completely ignore differences due to vehicles and gloss over road conditions.
- HarHarVeryFunny 3y ago
  Even tire brand/model (and wear) can have a significant effect on braking distance.
  - seabass-labrax 3y ago
    Do you know of any independent tests of various brands? I'm sure motor racing teams and other professional driving setups know, but as an ordinary car owner where would one objectively find out whether Firestone 123/45R56 or Dunlop 123/45R56 is nominally better? Even Which magazine decides it's too much effort, and just reviews them by the aesthetic value of their tread!
    - hansvm 3y ago
      - Brands only matter a little. Most brands have a variety of price/performance offerings. - For most drivers, more tread is better, and more channels (and a variety of sizes thereof) are better. CA roads don't shed water very well for example, so in heavy rains you need the tire to get the water out from between you and the road. - Assume tire manufacturers are pareto optimal. Grippier, safer tires have lower MPG and expected lifetimes. The latter is usually somewhat advertised. Buy the 30k tires instead of the 80k mile options. - If you're not driving in cold (<15F or so) weather, don't get all-season tires. If you are driving in cold weather, get snow tires (better, Nokian studded options don't tear up the road and offer excellent grip if they're legal in your locale). If you are driving in cold weather and not springing for snow tires, do get the all-season option. - Properly inflated tires matter a lot more than you might think, making cheap tires outperform good options. Checking (or asking a mechanic to check for free) after any major temperature swings or every 3k miles or so is a good idea. - Several tire sizes fit most rims. The wider options perform a bit better (total tire grip scales sub-linearly with road pressure, so distributing the weight over more area slightly improves grip). You won't pick up more than 3-5% with this parameter, but all-else equal (and if you don't mind worse fuel economy) it's good to know. The best tires vary from year to year, and there's an inherent tradeoff between the different driving conditions you'll subject them to and how many sets of tires you want to keep around (most consumers want 1 set, most northern consumers at least recognize that 2 sets isn't a bad tradeoff if you have the space for as much of a grip improvement as proper snow tires offer). If you have a competent mechanic you trust, just tell them how you drive and ask what they would do for their car. They test drive a dozen cars a day and have a pretty good feel for what works well enough and gives you a good bang for your buck. Actually answering your question now, AutoBild, Consumer Reports, Tire Rack, MotorTrend, and a number of other organizations test at least a few models (some of those test lots of models) of tires each year. Actually ranking the best from second-best is hard and maybe prone to bribes. Generally though, they provide objective measurements, and those measurements correlate pretty well with how the tire will handle when placed on your car. If a few choices are close, choosing based on what you can easily find locally or price or shipping or other factors is probably a good idea.
HarHarVeryFunny 3y ago
I don't know how valid the "brakes shedding energy at same rate per unit distance" is, but other sources confirm how massively braking distance increases with speed. For example, from here: https://www.automotive-fleet.com/driver-care/239402/driver-care-know-your-stopping-distance https://www.automotive-fleet.com/driver-care/239402/driver-c... At 50 mph braking distance is 125 ft. At 60 mph it's 245 ft At 80 mph it's 439 ft Now, when you consider the emergency circumstances where you might be flooring the brakes, this is not good ... e.g. A deer or child runs out in front of you, so close that emergency braking is called for .. how far away are they going to be from you ? Or, say the car in front of you on highway slams on their brakes, and they were travelling at 50mph, with you approaching at 60mph. Unless you are at least 120 ft behind them (vs the typical highway following distance of a few car lengths), you are going to be in a bad head-on collision.
- zamadatix 3y ago
  > At 50 mph braking distance is 125 ft. > At 60 mph it's 245 ft > At 80 mph it's 439 ft It looks like you're jumping between columns and sometimes transposing rows here. 125' @ 50 mph is from the raw braking distance column, 245' is from the same column but for 70 mph rather than 60 mph, and 439' @ 80 mph is from the overall stopping distance column. I'm not pointing this out just to nit, rather because: > say the car in front of you on highway slams on their brakes, and they were travelling at 50mph, with you approaching at 60mph. Unless you are at least 120 ft behind them (vs the typical highway following distance of a few car lengths), you are going to be in a bad head-on collision. You've concluded nearly twice the stopping footage as the table actually says and, as is more directly said in the last column, it's really roughly a 4 car braking length difference between the two speeds (including reaction time).
  - HarHarVeryFunny 3y ago
    Umm.. yeah. Not sure how I botched up reading that quite so badly! I think the point still stands though - by definition in an emergency situation there is not going to be much distance between you and that deer/whatever, and an extra 10mph can make all the difference!
- 3y ago
beefman 3y ago
Ignoring the obstacle: * It may be reasonable to assume the brakes produce the same force on both cars. With both cars weighing the same, this produces the same acceleration. Then the speeds decrease at the same rate, which means the faster car is going 100 - 70 = 30 kph when the slower car stops. * The faster car must dissipate more energy to do this. We may assume that braking is limited by heat dissipation and that the brakes shed energy at the same rate. Then the answer is sqrt(100^2 - 70^2) ~ 71 kph. Adding the obstacle: * As noted in the text, brakes don't shed energy at the "same rate per unit distance". So the faster car hits the obstacle first, at greater speeds than above in each case. We aren't given the information required to calculate the stopping time or distance of the slower car, so the problem is underspecified.
aabajian 3y ago
Note that you don't have to assume constant deceleration. You can just assume that the kinetic energy lost by the first car is the same as the second car. I think this is the more reasonable assumption given we know nothing about the breaks or drivers. The first car loses kinetic energy equal to 1/2 * m * 70^2 while the second car starts with 1/2 * m * 100^2. Assuming the second car loses the same energy as the first car while breaking, the formula is: 1/2 * m * 100^2 - 1/2 * m * 70^2 = 1/2 * m * v^2 or v = sqrt(100^2 - 70^2) = 71.4 mph
thsksbd 3y ago
This is at best a first order model. But the assumptions are wrong. It illustrates an important principle, the quadratic dependance of KE. It encourages the argument to lower speed limits. All good things. But the vehicle dynamics are wrong. Vehicle stopping force depends on tire traction not the brakes [1]. You'd think it's a wash but tire traction depends (non linearly) on the normal force which is vehicle weight plus aerodynamics. At 100 kph, aero matters. Aero can go either way. A ferrari or a rice racer with a massive wing will get decent down force and have improved braking performance lowering final impact speed. A van will probably have lift and therefore worse traction. Finally there are drag effects, not at all negligible at 100kph. [1] obviously; brakes have hydraulic calipers. As long as they're at operating temperature, they're fine. Brakes are sized for far more abuse than a single emergency slam. If your brakes can skid the tires, the brakes are not the limiting factor.
- shagie 3y ago
  > A ferrari or a rice racer with a massive wing will get decent down force and have improved braking performance lowering final impact speed. A memory of a groundbreaking car in the history of racing there... A video on the the Chaparral 2J - which has some interesting things with its downforce. https://youtu.be/WDaX3gFop7I https://youtu.be/WDaX3gFop7I From Wikipedia - https://en.wikipedia.org/wiki/Chaparral_2J https://en.wikipedia.org/wiki/Chaparral_2J > The skirting produced a zone within which the fans could create a vacuum producing downforce on the order of 1.25 to 1.50 g when the car was fully loaded (fuel, oil, coolant). Tremendous gripping power and greater maneuverability at all speeds were produced. Since it created the same levels of low pressure under the car at all speeds, downforce did not decrease at lower speeds. With other aerodynamic devices, downforce decreased as the car slowed or reached too much of a slip angle.
  - thsksbd 3y ago
    The fan car! Classic. I was thinking more of a road going Ferrari. But, wing, fan or whatever, the principle is the same: as soon as the normal force on the tires is anything other than the weight of car, the quadratic model brakes down.
- 3y ago
0x1ceb00da 3y ago
The author makes assumptions (rate of loss of energy per unit distance is same) but doesn't justify them.
jefftk 3y ago
One way to approximate thisb intuitively: braking distance is approximately quadratic in speed. 100^2/70^2 is ~2, so the faster car will need about twice as far to stop. Since we know this car slows down from 70 to 0 in the given distance, then that's also the distance to slow from 100 to 70.
at_a_remove 3y ago
I have a wee degree in physics and I would say "Run an experiment." There's too many unknowns in this, to me. Do brakex primarily shed speed, kinetic energy, or momentum? Is there a secondary effect and if so, how large? Is any of this linear? There's friction here and that can be notoriously non-linear. Would this change for disc brakes versus the other kind? I know this varies for older versus newer vehicles. What about the tendency of the car body to lurch forward and tilt? Does that affect things?

65 points by ColinWright 3y ago

didgeoridoo 3y ago

Nerd snipe alert. Do not open this if you’re just getting into bed.

cheschire 3y ago
https://www.explainxkcd.com/wiki/index.php/356:_Nerd_Sniping https://www.explainxkcd.com/wiki/index.php/356:_Nerd_Sniping

dylan604 3y ago

"(keep scrolling ...) " ... "(was that far enough?) " oh man, i was hoping this was going to be one of those tricks of no matter what your window size was, this would always be just below the fold with CSS. i just got lucky for it to line up. after trying it again with a smaller window, it was just in the scroll. =( then again, looking at the site's layout, it does not really suggest heavy use of CSS. at. all.

dmoy 3y ago

TL;DR v^2 is bad for your health in a car crash (But probably read it because it's really short)

dylan604 3y ago
yeah, if we need a TL;DR version of that...well, boy, I just don't know
thriftwy 3y ago
But at least it is halvedz or it would be much worse!
082349872349872 3y ago
Old (70s?) Citroens used to have a "braking distance" bar on the speedo that grew quadratically, attempting to keep this point in the drivers' minds?

krukah 3y ago

> Let's assume that the brakes are working at their limit, and as they do so, they are shedding energy at a maximum rate that doesn't change. The cars are identical, so they are both shedding energy at the same rate per unit distance. Interesting (and I think very reasonable) first-order assumption. Maybe to second-order, the calculation could assume the braking force is proportional to velocity, which I think is roughly true of friction generally, but is also harder to model.

bell-cot 3y ago
IANAE (...not an engineer), but "braking force is proportional to velocity" is not how brakes, or friction, work.
- purpleflame1257 3y ago
  No, but it is how first-order atmospheric drag is modeled, which may be what confused the GP comment.
Karellen 3y ago
> > they are both shedding energy at the same rate per unit distance. > Interesting (and I think very reasonable) first-order assumption. Really? That triggered alarm bells in my head immediately. I mean, it might be true, but I'd need to do the math to figure it out one way or another. > the calculation could assume the braking force is proportional to velocity, which I think is roughly true of friction generally Again, not my intuition at all. I'd have gone with the braking force being constant at non-zero velocity. (And force is variable up to the limit of static friction when at zero velocity.)
- calfuris 3y ago
  Your intuition about braking force matches the thing that triggered alarm bells in your head: work = force x distance, so if the force is approximately constant the kinetic energy dissipated will be approximately constant per unit distance.

lmm 3y ago

I assumed constant deceleration. Then u = sqrt(v^2 - 4as), a and s are constant so = sqrt(v^2-70^2) = approximately 70mph.

FabHK 3y ago
That's the same assumption as the (first) article makes.
- lmm 3y ago
  It talks about shedding energy at a constant rate per distance. Which, sure, ends up giving the same result, but it's a much weirder starting assumption, IMO.
  - Ma8ee 3y ago
    I thought it was a quite natural assumption, considering the same brake friction force and that the work done by that is the force times the distance. (W = Fs).

tempestn 3y ago

Cars can generally decelerate at a roughly constant rate of acceleration, based on the friction between tires and ground. Given the cars are identical, you can therefore define braking acceleration (deceleration) a, which is equal to both 70/t1 and (100-v)/t2, where v is the final velocity we're solving for. We also know that the average velocity for each car equals the distance travelled over time elapsed, where the distance is equal for both cars, but the time isn't. So, 70/2 = d/t1 and (100+v)/2 = d/t2. Can then solve these equations for v ~= 71.4 The gotcha is that it's much easier to solve if you take both distance and time as constant for both cars, but that would be a mistake. Of course you could get even more meta with it. Since the faster car is going to reach the obstacle first and smash into it, it probably won't be in the same position when the slower car reaches it either!

neerajsi 3y ago
Yeah, I also got 71.4 according to the same logic, but I can buy the argument in the follow on article that the amount of energy dissipated per unit time is what's constant.
- tempestn 3y ago
  Why? No production car is brake limited; slam on the brakes in any car and you'll go into (and stay in) ABS, meaning you're at the limit of tire grip, and therefore at roughly constant deceleration. Where would energy dissipation come into the picture? The energy is being dissipated by the brakes, not the tires, and they're not the limiting factor.
  - eschneider 3y ago
    Not true. Brakes need time to dump heat. With repeated hard braking it's not too hard to overwhelm the brakes on most cars.
    - tempestn 3y ago
      But the scenario we're talking about here isn't repeated hard braking.

Ntrails 3y ago

I mean this also misses the fact that if they _see_ the obstacle at the exact same distance + time (a sensible assumption to compare) by the time they hit the breaks a significant difference in distance has already been generated. Stopping distances are part of the theory test in the UK, albeit with nonsense values. The frustrating part for me is how they completely ignore differences due to vehicles and gloss over road conditions.

HarHarVeryFunny 3y ago
Even tire brand/model (and wear) can have a significant effect on braking distance.
- seabass-labrax 3y ago
  Do you know of any independent tests of various brands? I'm sure motor racing teams and other professional driving setups know, but as an ordinary car owner where would one objectively find out whether Firestone 123/45R56 or Dunlop 123/45R56 is nominally better? Even Which magazine decides it's too much effort, and just reviews them by the aesthetic value of their tread!
  - hansvm 3y ago
    - Brands only matter a little. Most brands have a variety of price/performance offerings. - For most drivers, more tread is better, and more channels (and a variety of sizes thereof) are better. CA roads don't shed water very well for example, so in heavy rains you need the tire to get the water out from between you and the road. - Assume tire manufacturers are pareto optimal. Grippier, safer tires have lower MPG and expected lifetimes. The latter is usually somewhat advertised. Buy the 30k tires instead of the 80k mile options. - If you're not driving in cold (<15F or so) weather, don't get all-season tires. If you are driving in cold weather, get snow tires (better, Nokian studded options don't tear up the road and offer excellent grip if they're legal in your locale). If you are driving in cold weather and not springing for snow tires, do get the all-season option. - Properly inflated tires matter a lot more than you might think, making cheap tires outperform good options. Checking (or asking a mechanic to check for free) after any major temperature swings or every 3k miles or so is a good idea. - Several tire sizes fit most rims. The wider options perform a bit better (total tire grip scales sub-linearly with road pressure, so distributing the weight over more area slightly improves grip). You won't pick up more than 3-5% with this parameter, but all-else equal (and if you don't mind worse fuel economy) it's good to know. The best tires vary from year to year, and there's an inherent tradeoff between the different driving conditions you'll subject them to and how many sets of tires you want to keep around (most consumers want 1 set, most northern consumers at least recognize that 2 sets isn't a bad tradeoff if you have the space for as much of a grip improvement as proper snow tires offer). If you have a competent mechanic you trust, just tell them how you drive and ask what they would do for their car. They test drive a dozen cars a day and have a pretty good feel for what works well enough and gives you a good bang for your buck. Actually answering your question now, AutoBild, Consumer Reports, Tire Rack, MotorTrend, and a number of other organizations test at least a few models (some of those test lots of models) of tires each year. Actually ranking the best from second-best is hard and maybe prone to bribes. Generally though, they provide objective measurements, and those measurements correlate pretty well with how the tire will handle when placed on your car. If a few choices are close, choosing based on what you can easily find locally or price or shipping or other factors is probably a good idea.

HarHarVeryFunny 3y ago

I don't know how valid the "brakes shedding energy at same rate per unit distance" is, but other sources confirm how massively braking distance increases with speed. For example, from here: https://www.automotive-fleet.com/driver-care/239402/driver-care-know-your-stopping-distance https://www.automotive-fleet.com/driver-care/239402/driver-c... At 50 mph braking distance is 125 ft. At 60 mph it's 245 ft At 80 mph it's 439 ft Now, when you consider the emergency circumstances where you might be flooring the brakes, this is not good ... e.g. A deer or child runs out in front of you, so close that emergency braking is called for .. how far away are they going to be from you ? Or, say the car in front of you on highway slams on their brakes, and they were travelling at 50mph, with you approaching at 60mph. Unless you are at least 120 ft behind them (vs the typical highway following distance of a few car lengths), you are going to be in a bad head-on collision.

zamadatix 3y ago
> At 50 mph braking distance is 125 ft. > At 60 mph it's 245 ft > At 80 mph it's 439 ft It looks like you're jumping between columns and sometimes transposing rows here. 125' @ 50 mph is from the raw braking distance column, 245' is from the same column but for 70 mph rather than 60 mph, and 439' @ 80 mph is from the overall stopping distance column. I'm not pointing this out just to nit, rather because: > say the car in front of you on highway slams on their brakes, and they were travelling at 50mph, with you approaching at 60mph. Unless you are at least 120 ft behind them (vs the typical highway following distance of a few car lengths), you are going to be in a bad head-on collision. You've concluded nearly twice the stopping footage as the table actually says and, as is more directly said in the last column, it's really roughly a 4 car braking length difference between the two speeds (including reaction time).
- HarHarVeryFunny 3y ago
  Umm.. yeah. Not sure how I botched up reading that quite so badly! I think the point still stands though - by definition in an emergency situation there is not going to be much distance between you and that deer/whatever, and an extra 10mph can make all the difference!
3y ago

beefman 3y ago

Ignoring the obstacle: * It may be reasonable to assume the brakes produce the same force on both cars. With both cars weighing the same, this produces the same acceleration. Then the speeds decrease at the same rate, which means the faster car is going 100 - 70 = 30 kph when the slower car stops. * The faster car must dissipate more energy to do this. We may assume that braking is limited by heat dissipation and that the brakes shed energy at the same rate. Then the answer is sqrt(100^2 - 70^2) ~ 71 kph. Adding the obstacle: * As noted in the text, brakes don't shed energy at the "same rate per unit distance". So the faster car hits the obstacle first, at greater speeds than above in each case. We aren't given the information required to calculate the stopping time or distance of the slower car, so the problem is underspecified.

aabajian 3y ago

Note that you don't have to assume constant deceleration. You can just assume that the kinetic energy lost by the first car is the same as the second car. I think this is the more reasonable assumption given we know nothing about the breaks or drivers. The first car loses kinetic energy equal to 1/2 * m * 70^2 while the second car starts with 1/2 * m * 100^2. Assuming the second car loses the same energy as the first car while breaking, the formula is: 1/2 * m * 100^2 - 1/2 * m * 70^2 = 1/2 * m * v^2 or v = sqrt(100^2 - 70^2) = 71.4 mph

thsksbd 3y ago

This is at best a first order model. But the assumptions are wrong. It illustrates an important principle, the quadratic dependance of KE. It encourages the argument to lower speed limits. All good things. But the vehicle dynamics are wrong. Vehicle stopping force depends on tire traction not the brakes [1]. You'd think it's a wash but tire traction depends (non linearly) on the normal force which is vehicle weight plus aerodynamics. At 100 kph, aero matters. Aero can go either way. A ferrari or a rice racer with a massive wing will get decent down force and have improved braking performance lowering final impact speed. A van will probably have lift and therefore worse traction. Finally there are drag effects, not at all negligible at 100kph. [1] obviously; brakes have hydraulic calipers. As long as they're at operating temperature, they're fine. Brakes are sized for far more abuse than a single emergency slam. If your brakes can skid the tires, the brakes are not the limiting factor.

shagie 3y ago
> A ferrari or a rice racer with a massive wing will get decent down force and have improved braking performance lowering final impact speed. A memory of a groundbreaking car in the history of racing there... A video on the the Chaparral 2J - which has some interesting things with its downforce. https://youtu.be/WDaX3gFop7I https://youtu.be/WDaX3gFop7I From Wikipedia - https://en.wikipedia.org/wiki/Chaparral_2J https://en.wikipedia.org/wiki/Chaparral_2J > The skirting produced a zone within which the fans could create a vacuum producing downforce on the order of 1.25 to 1.50 g when the car was fully loaded (fuel, oil, coolant). Tremendous gripping power and greater maneuverability at all speeds were produced. Since it created the same levels of low pressure under the car at all speeds, downforce did not decrease at lower speeds. With other aerodynamic devices, downforce decreased as the car slowed or reached too much of a slip angle.
- thsksbd 3y ago
  The fan car! Classic. I was thinking more of a road going Ferrari. But, wing, fan or whatever, the principle is the same: as soon as the normal force on the tires is anything other than the weight of car, the quadratic model brakes down.
3y ago

0x1ceb00da 3y ago

The author makes assumptions (rate of loss of energy per unit distance is same) but doesn't justify them.

jefftk 3y ago

One way to approximate thisb intuitively: braking distance is approximately quadratic in speed. 100^2/70^2 is ~2, so the faster car will need about twice as far to stop. Since we know this car slows down from 70 to 0 in the given distance, then that's also the distance to slow from 100 to 70.

at_a_remove 3y ago

I have a wee degree in physics and I would say "Run an experiment." There's too many unknowns in this, to me. Do brakex primarily shed speed, kinetic energy, or momentum? Is there a secondary effect and if so, how large? Is any of this linear? There's friction here and that can be notoriously non-linear. Would this change for disc brakes versus the other kind? I know this varies for older versus newer vehicles. What about the tendency of the car body to lurch forward and tilt? Does that affect things?