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Since there are brakes on every car, shouldn't the stopping time of the entire train be the same as the stopping time of a single car?

Is this because the cars themselves take a long time to stop, or because train braking is less efficient than typical braking, or another reason?

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  • $\begingroup$ Why do you think it isn't so? $\endgroup$
    – Steeven
    Jun 17 '16 at 15:23
  • $\begingroup$ Steeven beat me to it. What makes you think this isn't true? $\endgroup$
    – Gert
    Jun 17 '16 at 15:25
  • $\begingroup$ Long trains take longer time to stop than a single car because the braking effort is not synchronized. This was especially the case with brakes using pneumatic negative pressure (Westinghouse) because the pneumatic pulse took a relatively long time to propagate, it is less of a problem when the brakes are electrically controlled. $\endgroup$
    – hyportnex
    Jun 17 '16 at 15:31
  • $\begingroup$ @hyportnex Interesting. I guess you really really really want to start braking on the last car and propagate the braking action forward. $\endgroup$
    – garyp
    Jun 17 '16 at 15:59
  • $\begingroup$ @garyp there were all kinds of schemes to synchronize the cars, but none could really do it mechanically for very long freight trains because those could be miles long. The speed of the pressure pulse is 340m/sec so to get to the end it might take more than 10 seconds for a very long train. Passenger trains are always shorter and electric brake control came much earlier by wiring up a separate cable from the loco to the end. Now imagine the logistics of wiring up a long freight train. Radio link was also considered to control the brakes but powering them was the problem. $\endgroup$
    – hyportnex
    Jun 17 '16 at 17:36
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Suppose you have a vehicle with a mass $m$ moving at some speed $v$ and the friction coefficient between the wheels and the road is $\mu$. The maximum deceleration is then:

$$ a_\text{max} = \frac{F}{m} = \frac{mg\mu}{m} = g\mu $$

The time taken to stop will be:

$$ t = \frac{v}{a} = \frac{v}{g\mu} $$

And the distance taken to stop will be:

$$ s = \frac{v^2}{2a} = \frac{v^2}{2g\mu} $$

So the stopping time and distance are inversely proportional to the friction coefficient $\mu$.

The maximum braking force is determined by the friction coefficient between the wheels and the surface. In the case of cars even a basic family car will have a tyre-road friction coefficient of around 0.75, and for a sports card it will be nearer 1.

For trains the wheels and the rail are both steel, and the steel-steel friction coefficient is around 0.25. So the stopping time and distance will, at best, be three to four times greater than a car. In practice locking the wheels of a train causes damage to the wheels and rail that is very expensive to fix, so train brakes are designed to provide only about 75% of the maximum possible braking effort.

Finally we should note that a train under braking is a potentially unstable object. If the braking efficiency varies along the train it could concertina and derail. For this reason long trains will not use the maximum braking possible unless there's a very good reason.

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  • $\begingroup$ Do you have a reference for the 75%? $\endgroup$
    – Ryan Unger
    Jun 17 '16 at 15:56
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    $\begingroup$ This cannot be a reference, but is helpful. $\endgroup$
    – lucas
    Jun 17 '16 at 16:47
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    $\begingroup$ There are break shoe arrangements for trains that work just as well as breaks on cars. I have been on streetcars where the driver had to use them to save a pedestrian and I got almost thrown out of my seat. Now scale this up to a train running at 100mph with hundreds of people inside. Why in the world would one want to break that train so hard that passengers would get injured or die? It's not a technical problem. It's simply not useful to break a train hard. $\endgroup$
    – CuriousOne
    Jun 17 '16 at 18:42
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    $\begingroup$ Trains do not accelerate as fast as autos either; they are not likely to tolerate well the internal strains of deceleration at the panic-stop rate we can achieve in an auto on dry pavement, rubber against asphalt. $\endgroup$
    – Whit3rd
    Jun 18 '16 at 2:56
  • $\begingroup$ The above is true if braking is applied to all the wheels. But in many cases the distribution of braking power has to match the distribution of weight over the wheels. $\endgroup$ Mar 28 '17 at 19:16

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