[HN Gopher] Seventy versus One Hundred
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Seventy versus One Hundred
Author : ColinWright
Score : 44 points
Date : 2023-12-27 11:14 UTC (1 days ago)
(HTM) web link (www.solipsys.co.uk)
(TXT) w3m dump (www.solipsys.co.uk)
| didgeoridoo wrote:
| Nerd snipe alert. Do not open this if you're just getting into
| bed.
| cheschire wrote:
| https://www.explainxkcd.com/wiki/index.php/356:_Nerd_Sniping
| dylan604 wrote:
| "(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 wrote:
| TL;DR v^2 is bad for your health in a car crash
|
| (But probably read it because it's really short)
| dylan604 wrote:
| yeah, if we need a TL;DR version of that...well, boy, I just
| don't know
| krukah wrote:
| > 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 wrote:
| IANAE (...not an engineer), but "braking force is proportional
| to velocity" is not how brakes, or friction, work.
| purpleflame1257 wrote:
| No, but it is how first-order atmospheric drag is modeled,
| which may be what confused the GP comment.
| Karellen wrote:
| > > 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.)
| lmm wrote:
| I assumed constant deceleration. Then u = sqrt(v^2 - 4as), a and
| s are constant so = sqrt(v^2-70^2) = approximately 70mph.
| FabHK wrote:
| That's the same assumption as the (first) article makes.
| lmm wrote:
| 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 wrote:
| 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 wrote:
| 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 wrote:
| 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 wrote:
| 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 wrote:
| 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 wrote:
| But the scenario we're talking about here isn't repeated
| hard braking.
| klodolph wrote:
| There is a different, clever way to analyze the problem
| thinking about energy and work.
|
| Work is by definition (work) = (force) x (distance), or the
| same thing expressed as an integral.
|
| Distance is the same, that's a parameter of the problem. If
| we say that the brakes apply constant deceleration, then
| that implies that the force is the same, between both cars.
| Then the amount of energy shed by both cars is the same
| before they reach the barrier. (We don't need to use the
| integral definition for work here, because force is a
| constant.)
|
| This explains why the people who assumed that both cars
| dump the same amount of energy arrived at the same result,
| even though some of the people in this thread appear to be
| using faulty reasoning.
| tempestn wrote:
| Ah, of course, you're right. Thanks for pointing that
| out! I didn't really think about it at the time, but the
| equation you end up with if you use the acceleration
| approach results in a v^2, which makes sense when you
| consider that the equal energy dissipation approach is
| equivalent.
| Ntrails wrote:
| 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 wrote:
| Even tire brand/model (and wear) can have a significant effect
| on braking distance.
| HarHarVeryFunny wrote:
| 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 wrote:
| > 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).
| adewinter wrote:
| Kinetic energy scales with velocity squared (0.5 _m_ v^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 wrote:
| > 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 wrote:
| 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.
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(page generated 2023-12-28 23:00 UTC)