Seventy versus One Hundred

https://www.solipsys.co.uk/new/SeventyVersusOneHundred.html?wk27hn

by ColinWright • 3 years ago

65 46 3 years ago

solipsys.co.uk

didgeoridoo• 3 years ago

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

cheschire• 3 years ago

https://www.explainxkcd.com/wiki/index.php/356:_Nerd_Sniping

dylan6042• 3 years 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• 3 years ago

TL;DR v^2 is bad for your health in a car crash

(But probably read it because it's really short)

dylan6042• 3 years ago

yeah, if we need a TL;DR version of that...well, boy, I just don't know

thriftwy• 3 years ago

But at least it is halvedz or it would be much worse!

082349872349872• 3 years 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• 3 years 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• 3 years ago

IANAE (...not an engineer), but "braking force is proportional to velocity" is not how brakes, or friction, work.

purpleflame1257• 3 years ago

No, but it is how first-order atmospheric drag is modeled, which may be what confused the GP comment.

Karellen• 3 years 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• 3 years 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.

lmm2• 3 years 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• 3 years ago

That's the same assumption as the (first) article makes.

lmm2• 3 years 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.

tempestn5• 3 years 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• 3 years 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.

tempestn5• 3 years 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.

Ntrails2• 3 years 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.

HarHarVeryFunny4• 3 years ago

Even tire brand/model (and wear) can have a significant effect on braking distance.

seabass-labrax• 3 years 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!

HarHarVeryFunny4• 3 years 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-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• 3 years 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).

HarHarVeryFunny4• 3 years 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!

adewinter• 3 years ago

Kinetic energy scales with velocity squared (0.5mv^2), so this result makes sense. Brakes dissipate constant energy, but the amount of energy it needs to disapate is going up much faster.

margalabargala• 3 years ago

I don't know how valid the "brakes shedding energy at same rate per unit distance" is

From the follow-up post, the author gives a justification for that.

the force applied by the brake pads on the brake disc is going to be roughly constant, and so the work done, which in this case is the energy shed, is the force times the distance over which the force is applied.

There will of course be confounding factors like temperature, etc. But it's probably pretty close.

beefman• 3 years 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• 3 years 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

thsksbd2• 3 years 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• 3 years 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

From Wikipedia - 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.

thsksbd2• 3 years 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.

andromaton• 3 years ago

This is all correct. I must be getting older or wiser, but to me an increasing number of explainers and even journal articles look like "oh, that's how you ride a bike! I will write an article on velodrome design."

0x1ceb00da• 3 years ago

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

jefftk• 3 years 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• 3 years 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?

Seventy versus One Hundred

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