For story purpose I am trying to design a mass transit system for a perfect city. Right ... maybe not so perfect, if you look carefully into the details, but actually the mass transit system is supposed to be highly efficient.

Nevertheless, the city is built from scratch with heavy inspiration from New Urbanism (mass transit, high buildings, mixed zoning, nice parks, very little cars). In RL the metro suburban rail system is able to waste even 45% of used energy for braking (and engineers hope to recover a small fraction for regenerative breaking). In RL there are also attempts to reduce wasted energy by working with slopes - after leaving the station the train goes down a bit to go up just before the next one, thus saving all this accelerating /decelerating stuff.

I want to go one step further - I want to position the stations clearly higher than the track, thus effectively no breaking would be needed. (Yeah, the metro would resemble a bit like a roller coaster, I know). The system would work perfectly, if all stations were on the same height, just a bit energy would be used to compensate for rolling and air friction. If the altitude difference would be minimal, then train would need to use energy only in this slightly uphill direction, while on the opposite, everything would be provided by gravity.

Is there any idea how to calculate (rule of thumb, whatever) what the maximum height difference between stations would be, under which such a system may indeed work?

The tech level is comparable to contemporary Earth.

(Clarification: I know that you have to include rolling friction, air friction, etc... The issue is just how to either find data to put into such a formula, or how to make a very rough adjustment based on some RL life example, to derive calculations, that would not make any engineer cry)

  • $\begingroup$ Don't forget to account for the length of the carriages. There should be a maximum possible curvature of the rails in order to assure that the carriages are not hitting each other or hitting the ground and that the wheels stay on the rails when the train bends in the vertical around the connections. $\endgroup$
    – Pahlavan
    Jul 20, 2017 at 9:33
  • 2
    $\begingroup$ Did you try to ask that question on physics.stackexchange.com? I believe people there may have knowledge required to help you. $\endgroup$
    – running.t
    Jul 20, 2017 at 9:50
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    $\begingroup$ You don't need stations to be at varying heights; you need the platforms to be at varying heights. The rest of it: concourse, ticket counters, food stalls etc. can be at whatever height the local terrain prescribes. The advantage: if you have separate tracks for either direction (you should), you can force trains to roll uphill when entering the station, and downhill when leaving. Since both platforms are independently designed, you won't have a problem with trains entering from one direction facing downhill if the other direction is pointing uphill. $\endgroup$
    – nzaman
    Jul 20, 2017 at 10:42
  • $\begingroup$ There might be some information you find useful here specifically check out in text note #7 $\endgroup$
    – bendl
    Jul 20, 2017 at 13:05

2 Answers 2


The resistance of the train is dependent on a lot of things. I've tried to bring all those things together as best as I can for you here. You can play with the value x (in radians) and the output is the maximum velocity the train will reach at that slope. The coefficients of rolling and air resistance were taken from here and some of the math borrowed from here. Solving for x lets you play with V to get the appropriate slope for a given speed.

Finally, the difference in height will be proportional to the loss in energy do to friction, which itself depends on the slope of the rail, the velocity of the train and the distance it travels between stations. You can calculate that here.

These are all rough calculations and I haven't practiced my physics in a long time. There's also a good chance that I used the wrong unit of air density so please take this answer with a grain of salt, and I'd definitely recommend checking in with our friends at physics.stackexchange.com

  • $\begingroup$ (still working to understand your answer) $\endgroup$
    – Shadow1024
    Jul 27, 2017 at 16:41
  • $\begingroup$ Let me know what part gives you trouble and I'll see if I can expand on it! $\endgroup$
    – bendl
    Jul 27, 2017 at 16:43
  • $\begingroup$ I had trouble to figure out which number is which, but it seems I got it. (at 4th trial) In the last equation you assume constant speed all over the route, or did some trick to work around the issue of changing air friction during acceleration and deceleration? $\endgroup$
    – Shadow1024
    Jul 27, 2017 at 17:24
  • $\begingroup$ There are a lot of cut corners in these formulas that would make a difference if you were actually designing something, but these should give a reasonable estimate. To the point of variable speed, adding this sort of thing in would require information about train acceleration and braking power as well as some pretty tough calculus to tie it all together. In the end I assumed that you're going to be traveling at one speed for the majority of your trip so the formula only reflects that one speed. $\endgroup$
    – bendl
    Jul 27, 2017 at 17:49

I can provide you with a number. For a train system you describe, the maximum gradient is about 4%. This means that the steepest track you can lay is a climb of 4m every 100m of horizontal travel. For comparison, freight trains can only handle an absolute maximum of 2%. Question not asked, but I normally use a radius of 300m for my sharpest curves. I don't take super-elevation (banking) in my designs.

  • $\begingroup$ Is the 4% only for climbing? I can imagine that too steep and the train might lose its frictional grip on the rails and so not be able to climb. But going down would be ok. If each track were a 1 way affair (as suggested by @nzaman in comment on OP) could the downslope be steeper? $\endgroup$
    – Willk
    Jul 20, 2017 at 12:14
  • $\begingroup$ @Will, Those numbers were pulled out from memory. After your comment, I quickly googled it, and I still stand by my statement. See en.wikipedia.org/wiki/… $\endgroup$ Jul 20, 2017 at 12:20
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    $\begingroup$ I disagree with this. If the train were trying to climb a 4% gradient, that would be a problem, but the question is about stopping at the top of it. The train would already have enough forward momentum at the beginning of the gradient to coast to the top if the system were designed correctly. $\endgroup$
    – bendl
    Jul 20, 2017 at 13:10
  • $\begingroup$ @bendl - what is for any reason train is not going at full speed just before the station? It would not be able to climb the hill and would be stuck on tracks outside the station. $\endgroup$
    – Alexander
    Jul 20, 2017 at 16:36
  • $\begingroup$ @Alexander I've thought about that, but I don't think that's within the scope of the question. That's a design problem that would have to be addressed, but not here. $\endgroup$
    – bendl
    Jul 20, 2017 at 16:41

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