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Have you ever noticed how, on a highway in conditions of heavy traffic, whenever you have to brake for any reason, the braking turns out to be sudden and hard?

The same does not seem to happen on slow and/or less busy roads. Can you explain the difference?

P.S: strictly speaking, this is not a physics problem but in the theory of queues, but the queue tag does not exist.

Hint:

Try to imagine under which (physical) conditions a smooth section of traffic is brought to a sudden stop rather than a mild slow-down.

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    $\begingroup$ This is just plain wrong, you are driving badly. If you maintain proper distance and stay aware of what's happening up ahead then you will not have to brake sharply under normal conditions - only in exceptional circumstances for example if someone pulls in front of you and slows down. $\endgroup$ – jhabbott Feb 10 '16 at 14:07
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    $\begingroup$ @jhabbott is correct. If you are frequently having to do this, you are simply not driving very well $\endgroup$ – Kevin Feb 10 '16 at 14:50
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    $\begingroup$ This question needs more elaboration about its constraints. It seems to assume that all drivers have not left sufficient following distance, but then claims that the solution applies in all circumstances. And the author is rejecting what seem to be perfectly valid answers because they are qualitative rather than quantitative even though that's not stated or made clear what qualifies in the question. $\endgroup$ – Shufflepants Feb 10 '16 at 17:00
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    $\begingroup$ In heavy traffic conditions on a highway, I find myself rarely ever braking hard. I tend to maintain a safe distance between my car and the car in front of me, and also look ahead past that car to see what the car in front of him is doing. If I cannot see past the car (because it is a large truck, for example), I leave even more room between me and the car in front of me. It is only when other drivers do dangerous things during heavy traffic (i.e. pulling into the lane right in front of me) that I would ever have to brake hard. You need to reassess the way that you drive. $\endgroup$ – Ian MacDonald Feb 10 '16 at 17:44
  • $\begingroup$ Abrupt braking is not a necessary feature of daily interstate highway choke points near me. Gentle braking is sufficient for a change from 4 lanes to 3 for instance. The only time i was ever in a car crash was on a neighborhood street. What exactly is the puzzle? $\endgroup$ – user662852 Feb 11 '16 at 5:23
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I have no illustration, but this happens due to the "concertina effect".

Example:

  • Let's say we have 5 cars driving in a straight line, one behind the other, with driver A at front and driver B behind him, and so on with driver E at the back.
  • Let's assume they are traveling at a moderate pace, all at the same speed, spaced two car-lengths from each other.

If driver A were to start braking normally (uniformly), driver B would take some time (some milliseconds) to respond to the brake lights of driver A, depending on his reaction speed. This already forces driver B to brake harder and less uniformly than driver A, as driver B has less distance than driver A to brake in. Now, this transpires with every following vehicle, until driver E experiences a sudden brake, much harsher than driver A would have braked, and therefor compensates by applying an even harder brake (not to mention his reaction time as well). This whole scenario of course neglects other factors which are apparent due to human error (driver not concentrating, driver riding very close to driver in front of him, incorrect handling of brakes, etc.). Therefor, due to the concertina effect, this happens.

The reason why the concertina effect is not as apparent in slower traffic is because there is more time to react. It therefor becomes negligent, relative to driver and brake-mechanics.

An interesting thought: if all cars at a red light would start to drive at the same time as the light turns green, the wait time would be the sum of red-light and braking time, with no added queuing (traffic) time. This might some day be the norm with interconnected vehicle communication.

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  • $\begingroup$ This is so qualitative I have troubles accepting it as a full solution. $\endgroup$ – MariusMatutiae Feb 10 '16 at 14:22
  • $\begingroup$ The question states "whenever you have to brake for any reason" but this answer explains hard braking only when the reason is that the car in front of you slows down. $\endgroup$ – JiK Feb 10 '16 at 14:31
  • $\begingroup$ This is better known as the accordion effect. $\endgroup$ – isanae Feb 10 '16 at 18:24
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    $\begingroup$ @MariusMatutiae if you wanted a precise, quantitative solution you should have done a better job explaining what criteria a solution needs to satisfy in order to be accepted. $\endgroup$ – user17947 Feb 10 '16 at 19:25
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    $\begingroup$ @isanae: A concertina is a type of accordion, so they're basically synonymous. $\endgroup$ – Darrel Hoffman Feb 10 '16 at 21:05
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Some students at Caltech did a summer research project on this about 30 years ago. Unfortunately, I cant' find any links to it; I'm not sure the results were ever formally published.

They spent the summer analyzing traffic on the Santa Monica Freeway, particularly during rush hour. What they found is that traffic will tend to run smoothly, but sometimes there will be a sudden stop for no apparent reason. All the cars slow to a stop, then continue. This creates a ripple of slowed cars that moves down the highway (against the flow of traffic).

Further study found that this effect is most pronounced when traffic density is highest.

The model they came up with is that when one driver slows briefly, the driver behind them will react, also slowing. The amount of reaction by the following driver depends on the distance between the cars (and the velocity). If the car in front is far enough away, then the car behind can ignore the glitch and the effect will not propagate. If the cars are closer together, then the following driver will react by slowing more than the car in front.

So this leads to a specific threshold density (which is a nearly-linear function of velocity) where if the density is below the threshold, then glitches in driving behavior will tend to fade out, while above the density, glitches will grow. If the density is high enough, a simple glitch can grow into a full traffic snarl where everybody stops for no reason.

Highways are really the only places where this sort of phenomenon can build up.

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  • $\begingroup$ I agree with all of this, (+1, in fact). Yet the condition discussed above can be made quantitative... $\endgroup$ – MariusMatutiae Feb 10 '16 at 15:17
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Not sure what you're going for, but here's my guess:

When driving on the highway, you're driving fast and expect to be driving fast with no interruptions. When something does happen to cause you to have to brake, it's a surprise, and because of your high speed, you react harder and brake faster. When driving on a slow neighborhood road, you're driving slowly and expect to be stopping/slowing more often, so you're ready.

Driving fast takes a harder deceleration to slow in a reasonable amount of time than while driving slowly

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The system is prone, under certain conditions, to an instability. Let the first car in a long line of cars, pairwise separated by a distance d, see a very remote obstacle, at distance D. The driver starts to break immediately, as he tries to avoid brusque brakings. His braking space X_1 is then D:

X_1 = D

The braking space of the next car is not D, it is instead

X_2 = D + d - v T

where d is the (mean) distance between cars, v is the car speed. and T is an average reaction time. Under certain circumstances,

X_2 < X_1

or,

d -v T < 0.

The braking space available to the n-th car is,by induction,

X_n = D + (n-1)(d - v T).

If d-vT < 0, it can occur that X_n <= 0; since d-vT does not depend on n, this will occur for a finite n. This n-th driver will be obliged to brake brusquely.

The condition d- v T < 0 is most easily satisfied when d is small (heavy traffic), or v is large (highways, as opposed to slower roads).

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    $\begingroup$ This is much less clear an answer than @nine9's above $\endgroup$ – question_asker Feb 10 '16 at 17:19

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