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It is my understanding that every axle on a modern train has brakes.

According to this page https://www.engineeringtoolbox.com/friction-coefficients-d_778.html clean/dry steel on steel has a (static) friction coefficient of .5 to .8 and a tire on a dry road has 1.0 or so.

Trains weight more, BUT the normal force for placed on the wheels should go up in direct proportion. If the (max braking) force of friction is the coefficient times the normal force, then I would expect that the max braking force would also proportionally go up with more weight.

So it seems like the only thing different between a heavily loaded truck and train affecting the (max braking) force you could generate is the coefficient of friction. Because this is about half, I would expect the braking force to be half.

Kinetic energy is 1/2 * mass * (velocity)2, which gives you units of joules. As a unit, joules can be re-written as newtons of force through meters of distance. If we have half as many newtons (because half as much coefficient of friction for steel on steel vs tire on pavement) we could assume 2 times the number of meters to stop the same mass.

On top of that, it seems that increasing the mass wouldn't increase stopping distance (assuming the brakes and discs survive the increased energy dissipated): an increase in mass should proportionally increase the kinetic energy AND the maximum static friction (= max braking force), causing the two to cancel for a same stopping distance.

So why is it "common knowledge" that trains take miles to stop but not trucks? Assuming that the braking systems on a train car had the controls a truck does to stay in static friction (anti-lock braking) why can a train car not be expected to stop in about 2x the distance of a truck (in ideal conditions, of course)? And if that can be expected, why can't we expect a whole train of train cars to stop in about 2x the distance of a truck?

(And maybe it could be even less if the train can do eddy current braking to the rails where a truck has no such option.)

Edit to add a few assumptions:

  • I'm assuming from the same speed (when comparing train vs truck).
  • I'm assuming that the surface that we are stopping on is built to take it: the rails won't topple, splay out, widen gauge, or otherwise detach from the surface below, asphalt/concrete doesn't ripple up, any bridges the vehicles are passing over can withstand the horizontal loads imposed, etc.
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    $\begingroup$ "It seems that increasing the mass wouldn't increase stopping distance (assuming the brakes and discs survive the increased energy dissipated)." That's a pretty big assumption and can't be ignored. Also, remember the energy needs to be dissipated in the rails. If train cars weigh significantly more for their size than trucks that power dissipation could be a problem. There are trucks that carry train containers so I am not sure. $\endgroup$
    – DKNguyen
    Jan 23 at 19:32
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    $\begingroup$ Did you find out how hard you need to stop a train to derail it? Cars won't all be identical in slowing down and that might cause a pretty catastrophic chain reaction with a lot of cars. Large semi-trailers jack knife with just one joint and trains present far more links between cars where that can happen. $\endgroup$
    – DKNguyen
    Jan 23 at 19:32
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    $\begingroup$ @SolarMike OP isn't asking about that. OP is basically assuming (which may or may not be true, I don't know but it certainly feels true) that one train car takes some distance to stop which is not anywhere on the order of miles and is asking why when a bunch of cars are connected up it takes the entire train takes miles to stop when each car has its own brakes. If a lone train car still takes miles to stop then the question is based on a false premise. $\endgroup$
    – DKNguyen
    Jan 23 at 19:54
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    $\begingroup$ @SolarMike Part of OP's question above already does involve the independence between contact area and friction (ideally at least). But if indeed trains just have less braking power for their weight, then the question could be why. It wouldn't be surprising since the engine is a much smaller fraction of the weight in a train than in a truck. It would also beg the question why (and I wouldn't be surprised if the answer was as simple as derailing but more likely a huge locomotive would be needed where most of its power was dedicated to braking rather than moving the train). $\endgroup$
    – DKNguyen
    Jan 23 at 20:00
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    $\begingroup$ @Azendale You do not need to necessarily lift. You can topple and flip over the rail. $\endgroup$
    – DKNguyen
    Jan 23 at 20:43

5 Answers 5

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TL;DR: temperature dependence of the coefficient of friction

After doing some calculations IMHO the main factors IMHO affects the braking distance of trains compared to trucks is the increase of wheel temperatures during braking which leads to a corresponding drop of coefficient of friction and at a (significantly) lesser degree the reduced Aerodynamic drag per kg of the vehicle.

Increase of Wheel temperature

The wheels of a train at full braking locally develop high temperatures in the order of 200-400 degrees (see following link, table 10).

However, increasing the temperature beyond 300 $^o C$ decreases rapidly the friction coefficient. See image below that the coefficient of friction loses approximately 40% of its value from 300 to 400 $^o C$ (for that specific metal)

enter image description here

Figure: Friction temperature dependence (source research gate)

Given the the deceleration $a$ is $a= \frac{\mu \cdot m\cdot g}{m}= \mu \cdot g $, the maximum braking distance can be calculated as:

$$s=\frac{v_0^2}{2 \mu g}$$

Since the friction coefficient is smaller and drops further the train is expected to require more braking distance during emergency braking.


Aerodynamic drag

Aerodynamic drag does not play a significant role according to the calculation. However since I did the calculation I am adding it for reference. The characteristic I considered were.

train Truck
image <span class=$ m^2$" /> <span class=$2 v_0$" />
Frontal Area $12 m^2$ $10.5 m^2$
$C_d$ $1.8$ $0.8$
Weight Engine: 200000 kg, Passenger wagon:17000 kg $15000 kg$
coefficient of friction 0.5 1

Based on the above characteristics assuming a velocity of 30 m/s (108 kph = 67 mph)

train Truck
Aerodynamic drag ~12kN ~3 kN
Aerodynamic drag per kg mass $0.0541 \frac{m}{s^2}= 0.0054g$ $0.3 \frac{m}{s^2}= 0.03g$

So the aerodynamic drag per kg is approximately 6 times times more for a truck compared to a train (with one passenger cart), although its is still very small compared to the overall acceleration.

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  • $\begingroup$ Ah, sorry, I realize I didn't specify this in the question: I'm assuming 2x the stopping distance for the train over the truck, with them starting out at the same velocity. $\endgroup$
    – Azendale
    Jan 23 at 19:44
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    $\begingroup$ @Azendale Yeah you're assuming one train car can stop over the length of only a few times its length right? I am not sure if that is true though, but it certainly doesn't feel like one train car should take miles to stop but I don't know for sure. And you're wondering why it doesn't stay that way even as you connect up more train cars when each car has its own brakes? $\endgroup$
    – DKNguyen
    Jan 23 at 19:45
  • $\begingroup$ @DKNguyen, My assumption is: if you have truck going down the highway next to a train, neck and neck (front of the truck and train equal). Truck and train are going the same speed. There is a painted white line perpendicular across the highway and track. As soon as the front of the truck or train touches the line, they try to stop as fast as they can. I think the front of the train should end up 2x as far past the line as the front of the truck. (Regardless of where the tail of each is.) $\endgroup$
    – Azendale
    Jan 23 at 19:50
  • $\begingroup$ @DKNguyen For clarity: Yes, you summed up what I was thinking in your comment on this answer. $\endgroup$
    – Azendale
    Jan 23 at 20:27
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    $\begingroup$ @Azendale, I performed some calculations (after clarifying the velocity assumption). They resulted in a major update. $\endgroup$
    – NMech
    Jan 24 at 9:46
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Train brakes don't know how much load is in the car. They are applied the same whether the car is loaded or empty. If you had a bunch of empty cars, then a bunch of full cars, there would be a massive compression of the full cars pushing on the empty ones during braking. So braking is done very slowly on freight trains to manage coupler loads. On passenger trains, the weight variation is much less of a problem, and the trains are shorter and can distribute the weight, so they are able to safely stop much more rapidly.

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    $\begingroup$ It might be good to augment this answer by observing that tension and compression forces on couplers need to be kept within certain limits. If one had a bunch of lightweight cars in the middle of a train, and the brakes at the front were too much more effective than those in the back, the excessively strong brakes at the front could cause a derailment. $\endgroup$
    – supercat
    Jan 24 at 22:40
  • $\begingroup$ So if we were adapting the idea of ABS to trains, then it should probably be adjusted to also account for a car being pushed -- when the push force from the car behind hits a threshold (which would depend on loading of the current car, realistically) the brakes for that car should release a little, just like it would if skidding had started. $\endgroup$
    – Azendale
    Jan 25 at 5:57
  • $\begingroup$ Do you have any opinion about the classic movie scenario where the emergency brakes on the train are slammed as it hurls toward something like a broken bridge? Because it's certainly understandable that trains would brake more gently over a longer distance if they could, but I'm under the impression they are unable to brake faster even if they really, really needed to. Of course, that impression could just be wrong. $\endgroup$
    – DKNguyen
    Jan 25 at 19:09
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From the moment the wheel of a moving object is fully locked, due to momentum, the object will continue to skid forward until the inertia force is exhausted by the friction (inertia force = shear friction), which can be expressed as below:

$F + \mu N = 0$

$ma + \mu mg = 0$ -----> $a = -\mu g$

From kinematic equation, the stopping distance can be obtained,

$s = \dfrac{v_f^2 - v_0^2}{2a}$, $v_f = 0$ and $a = -\mu g$

$s = \dfrac{v_0^2}{2\mu g}$

For comparison, assume the train and the truck both have the same speed when the brake is fully engaged (the wheel is locked), and ignoring all other factors, then

$s_{truck} = \dfrac {v_0^2}{2g}$, assume $\mu_{pave} = 1.0$

$s_{train} = \dfrac{v_0^2}{g}$, assume $\mu_{track} = 0.5$

The train would require a stopping distance that is double the length the truck would require.

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  • $\begingroup$ "Assuming that the braking systems on a train car had the controls a truck does to stay in static friction (anti-lock braking)" -- I'm specifically assuming a control system that doesn't lock the brakes. IE, if they do lock, it releases the brake to go back to static friction. I'll be honest, I didn't exactly follow the first equation. I'm understanding 's' to be distance traveled beyond where the brakes were applied. I'm not sure I understand what you are saying/implying in the first half of your answer. The second half seem to fit with what I seemed to come up with. $\endgroup$
    – Azendale
    Jan 23 at 23:50
  • $\begingroup$ I was trying to simplify the situations and make the comparison easier. In reality, there are many factors involved besides the brak systems are never the same for a truck and a train. But the simple fact is that there is energy that needs to be dissipated after applying the brake. Without shear friction, the object will continue to slide endlessly and requiring another type of resintant force to comsume the energy. $\endgroup$
    – r13
    Jan 24 at 0:19
  • $\begingroup$ ah OK. There's two places that there's friction at play. I'm just focusing on how much traction is available on the wheels. I assumed that there was shear friction in the braking system, but that it was designed large enough to dissipate the required energy at the loads and speeds the vehicle would be rated/certified for. $\endgroup$
    – Azendale
    Jan 24 at 0:34
  • $\begingroup$ Yes, there are two shear frictions in play - the shear friction at the brake-wheel interface stops the rotation of the wheel due to mechanical power, it is out of the picture once the brake is applied; the car slides (a linear motion) due to momentum and inertia force after the wheel stops to rotate, thus generates the shear friction between the wheel and the pavement/track. Actually, there is another type of shear friction, which is responsible for moving the vehicle - traction, but it is outside of the scope of your current question. $\endgroup$
    – r13
    Jan 24 at 1:03
  • $\begingroup$ No! Not sliding! The question is specifically under the presumption of a control system that doesn't allow sliding/lockup. Wheels stay turning in sync with the surface until the vehicle is at a stop. We want dynamic friction in the brakes, but static (rolling) friction between the wheels and surface to maximize braking ("negative" acceleration) force, since you will have a higher friction coefficient in static friction with steel on steel. If you break out of static wheel friction into a slide, the control system would remove the normal force on the brake pads...that's what ABS does. $\endgroup$
    – Azendale
    Jan 24 at 3:39
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I see nobody came up highlighting something interesting: the wheels material. This won't be an exaustive answer but will point out something not trivial that must also be accounted for in such analysis.

Truck wheels are made of rubber, an interesting proprerties of polymers (rubber are polymeric blends at the end of the day) is that they have an energy histeresys cicle. A consequence of this is that we can tune the material to have a defined energetic behaviour within given stress frequency ranges. Think it like this, wouldn't be great to have a wheel that dissipate as less energy as possible when spinning fast but still able to dissipate energy when spinning at lower frequencies (like when braking)? Turns out this is not only possible but is something common rubber tires rely on. (for further info you can dig polymers' complex modulus, the imaginary part of such value is called also Loss Modulus and it is related to the viscoelastic nature of polymers, meaning their frequency-domain response)

This is obviously not the case of steel made rail wheels. Steel is not viscoelastic and it is not capable of the same energy dissipation thus brakes must account for it all.

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  • $\begingroup$ At least one of the answers uses values based on the different materials. Did you miss that? $\endgroup$
    – Solar Mike
    Jan 23 at 22:24
  • $\begingroup$ Furthermore the OP does mention the coefficients of friction in his question. $\endgroup$
    – DKNguyen
    Jan 24 at 1:51
  • $\begingroup$ What % of the overall dissipated energy do you estimate this accounts for in the braking system on a truck? Is that difference something that could not be overcome by slightly beefing up the brakes on a train? $\endgroup$
    – Azendale
    Jan 24 at 3:41
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If a train were designed to only travel in straight lines, it could be made to stop much more quickly and effectively than is possible with today's trains. Unfortunately, it is necessary to limit braking force when a train is on or near a curve for a number of reasons, most notably:

  1. If braking force in the front of a train is too much greater than in the rear, the train is likely to buckle and derail. If the braking force in the rear is too much greater than in the front, the train will be pulled off the inside of the curve.

  2. Tracks are designed primarily to resist vertical gravity loads. While adding calipers to bogeys might greatly increase the braking force a train could achieve, doing so might pull the rails away from the ties/sleepers.

If trains had separate mechanisms for an emergency stop triggered by an equipment malfunction, versus an emergency stop of a perfectly functional train, necessitated by outside events like a track obstruction, and if the latter could be constructed to distribute the maximum possible braking force safely throughout a train, stopping distances for the latter kinds of stops could be greatly reduced. Such a design, however, would increase cost and complexity, add many new points of failure, and complicate the notion of "fail safe". If a train is being operated properly, a braking system failure that triggers passive activation of all brakes would not create a dangerous condition. If a train were in a situation that required stopping faster than would be possible via passive braking, however, a failure that put all brakes into "passive stop" mode would extend the stopping distance, thus possibly creating a dangerous condition where none would otherwise have existed. Consequently, having a system trigger passive brakes if anything goes wrong would cease to be a "fail safe".

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