I assume the limiting factor during non-emergency braking is energy absorption, i.e. you can only brake so hard before you overtemp the brakes and wreck your ability to stop at all. But in an emergency - say, you’re about to hit/kill a bus full of people - ISTM the conductor could instead just apply so much braking force that the wheels lock up, causing them to slide on the rails and achieve that rapid decel without cooking the brakes. Instead, you’d flat-spot a whole lot of steel wheels, and probably damage the rails as well. This would be an expensive/logistical mess, but perhaps it beats killing people.
So…are train brakes capable of enough braking force to lock up the wheels? And can they develop that braking force rapidly enough to cause lockup before the brakes themselves overtemp? If this is possible, is it simply not justified by the logistics/expense of it all, i.e. it’s not worth a $100M repair if it only saves five lives?
Just as in your car; once the wheels lock, braking force dramatically reduces.
Trains have abs or something similar. There is a point, just before the wheels lock up that maximum stopping force is in effect. The system will rapidly apply the brakes and release them when the wheels stop turning.
There is another point to consider with a emergency stop. A train can weigh 1000 tonnes or more, and all of that energy is transferred to the rails. An emergency stop of a goods train would probably require some expensive repairs to the track. A heavy train would normally have no chance of stopping for anything that the driver could see on the track.
I’m not sure how much of a factor this is, but a train is not a solid body either, it’s a bunch of cars linked together. If the engine stops faster than the loaded cars behind it, you’re going to have a much bigger mess than some flat-spot wheels and deformed track. The train has to stop not only as quickly as possible, but at the same rate all along the train or it will derail.
…down to a kinetic (i.e. sliding) friction coefficient of about 0.6.
All understood, but I believe (pending confirmation in this thread by a railhead) the brakes can only absorb energy at a limited rate without overtemping, which is why they need to take a mile to stop from high speed. Thus my suggestion of using sliding action between the wheels and rails instead, where galling/thermal damage won’t impede braking ability.
Sliding the wheels down the rail, heavy train or light train, we’re talking about steel-on-steel friction; either way you should get a similar coefficient of friction, and similar stopping distance, unless I’m misunderstanding something. Why are you claiming a heavy train would take much longer?
I suspect the problem is going to be that you are not going to get that 0.6 of friction for very long. As you note, the wheels will flat spot - what that tells you is that it isn’t steel-on-steel any longer, but rather steel on very hot failing steel, lubricated by tiny bits of steel. Steel starts to soften with temperature, and the COF is known to drop with temperature. The brakes themselves failed due to excessive temperature, there is little reasons to think that the contact point on the wheels won’t heat up and the COF drop away in a similar manner. The entire energy of the train is being dissipated in this patch, and rather than trying to get rid of the energy into the air via brake pads designed for purpose (and designed to maintain COF at elevated temperatures), you will be relying on somehow getting the energy out of the tiny contact patch with the rail. Such systems tend to come to a stable point at very high temperatures and low COF.
I’m not any sort of metallurgist, but I expect if a train locked its wheels to try to emergency stop that the wheels and rails would ablate, spraying molten metal and likely starting a fire on the train and nearby property. And if the train stopped without derailing, the what’s left of the wheels would be welded to the rails. With the destruction of property by fire, removing the remains of the train, re-laying track, and the months of traffic disruption, the bus would have to be full of pregnant nuns taking blind orphans to the candy factory for the lives to be worth costs.
The underlying question is: what’s the most effective way of removing kinetic energy from a train? I don’t know. Maybe have all the axles connected to generators that charge up huge capacitors?
The next question is of course the cost to develop/deploy/maintain such a system.
First off, the brakes are pneumatic. This means that when the engineer applies the brakes, the pressure has to build in the system, and not only that, but it can only build at the speed of sound, so if the train is a mile long, it will be about 5 seconds after the engineer applies the brakes before the brake line pressure changes in the caboose.
Next, the train axles stay centered between the tracks due to the bi-conic nature of the wheel/axle pairs. This only works if they are turning, and fails if they are skidding. That means that once the wheels are skidding, it is only the flanges that are keeping the train from derailing.
Now you combine these two effects. The engineer applies enough brake to skid the wheels, but the rear cars are still pushing because the brake pressure hasn’t had time to travel back there… Even on a straight track, the train will be trying to buckle, and pronouncedly so if it is on any sort of curve. Some of the buckling force will be upward, and if it is high enough, may lift some lightly loaded cars enough for them to derail.
Once one car derails, the whole shebang becomes very ugly.
So ideally the train never brakes hard enough to skid any of the wheels. Braking harder than that means that situation is such that derailing the train is preferable to continuing forward.
Hard braking also increases track maintenance issues. The ties (UK sleepers) gradually move in the ballast due to sideways and braking forces on the track, as well a diurnal temperature cycling causing the rails to expand and contract. Eventually they “walk” enough that the track must be realigned. Nowadays this is done by specialized machines. In earlier times this was done by “Gandy Dancers” …teams of men (usually AA) with crowbars, who would sing colorful cadence songs to synchronize their efforts and movements.
That tiny contact patch is moving down the rail as the train moves. A train stopping from 60 MPH in my proposed five seconds would cover about 220 feet. That’s about 4 standard boxcar lengths; most of the train’s wheels would be rolling over rail that had already been heated by other sliding wheels, but on average each wheel’s total sliding friction energy would be dispersed over (220 feet/16 wheels =) 14 linear feet of rail. Not a lot of distance I suppose, but it’s not quite like spinning the wheels in place.
OTOH, I found this, which supports your claim of reduced CoF. See figure 94 on page 83, which shows that molten steel at the sliding interface reduces the CoF to 0.15. This is still better than the commonly cited 0.1-G decel for a controlled stop of 60-0 in 30 seconds, but not by my hoped-for huge margin.
So you’re probably right: you only have CoF=0.6 for the first few milliseconds of sliding, after which it falls to a not-very-useful 0.15. So no major benefit to locking up the wheels, even in the absence of other complicating factors.
Electric trains commonly dump energy back into the grid during braking - but if your e-motors and wiring are sized for modest acceleration, then they may not be able to move enough current to achieve very rapid braking, either.
Diesel-electrics can dissipate braking energy in an on-board resistor bank. This is useful for descending long mountain grades without cooking the air brakes, but AIUI it’s limited to the locomotive(s) only, and so can only provide very limited braking compared to what you can achieve by using the air brakes on all of the train’s cars.
It doesn’t matter. Frictional force is proportional to normal force which is proportional to mass. Doubling the mass of the train therefore doubles the force of friction, which implies that the acceleration remains constant (since force equals mass times acceleration.)
In Germany, the Trains have sandboxes that can blow sand onto the tracks installed right before the Brakes Sandstreuer – Wikipedia This is to help with braking and not-locking-up in cases of too low a friction: in autumn, when wet leaves make the tracks very Slippery, in winter when ice or slush does the same, and during emergency brakings.
A braking train shouldn’t be breaking anything - but I was under the impression that all of the cars in a train were fitted with air brakes. As long as nobody breaks them, every car should be able to contribute to the braking effort.
Several answers have addressed the potential of brake failure from excessive heat. That has to be avoid so the train doesn’t end up with no brakes after some partial reduction in speed. Derailing could also result in far more loss of life than the people in the bus.
