I've been reading about how the conical shape of train wheels helps trains round turns without a differential. For those who are unfamiliar with the idea, the conical shape allows the wheels to shift and slide across the tracks, thus effectively varying their radii and allowing them to cover different distances while rotating at the same angular velocity.

A cross-sectional view of the tracks and wheels generally looks something like: enter image description here

But what about a configuration like the following?

enter image description here

I read in an online article that wheels in the second configuration may more easily slip and derail from the tracks (assuming there are no flanges to prevent them from doing so). But I can't convince myself using physics why that might be.

Is one of these two configurations actually more reliable than the other?

  • 1
    $\begingroup$ You really should check out Feynman's video on a very related topic, under the "Fun to Imagine" videos. youtube.com/watch?v=y7h4OtFDnYE $\endgroup$
    – Aritra Das
    Aug 5, 2015 at 20:00
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    $\begingroup$ @AritraDas Ha ha, I actually linked that video in my original post! $\endgroup$
    – Rations
    Aug 5, 2015 at 20:36

6 Answers 6


Shift the upper configuration to the left a short distance at equilibrium. Result: the left wheel goes a little up, the right goes a little down, the train tilts clockwise, the center of mass is to the right of the centerline between the wheels, and therefore the center of mass provides a restorative force to push the train back to the right.

Shift the lower configuration to the left a short distance at equilibrium. The argument proceeds in reverse and the center of mass provides an anti-restorative force, pushing the train further to the left. Pain ensues.

You're trading $m \ddot x = - k x$ (harmonic oscillator) for $m \ddot x = k x$ (exponential diverger) and praying that the implicit drag forces keep the thing diverged only a small amount. That's a risky game, no doubt.

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    $\begingroup$ It's not the center of mass that provides a restorative force, it's that the relative radii of the wheels causes their linear velocities to be different. In the first case as the wheel shifts more to the right, the left wheel will cover less linear distance than the right wheel causing a left turning tendency (towards the center of the track). In the second case shifting the wheels to the right reduces the radius of the right wheel causing it to cover less linear distance. This causes a right turning tendency, which is away from the center. $\endgroup$
    – Kyle
    Jul 30, 2015 at 0:49
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    $\begingroup$ Also, I'm pretty sure the center-of-mass argument is backwards. It really doesn't matter whether the CoM moves left or right; what matters for stability is whether it moves up or down, and that's not so obvious from the geometry. (In particular, it depends on the curvature of the wheel surfaces, not just on their slope.) But, of course, the wheel radius effect described by @Kyle probably dominates this anyway. $\endgroup$ Jul 30, 2015 at 7:41
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    $\begingroup$ If the restorative force were being provided by the centre of mass, the system would restore itself by having the axle slide perpendicularly across the rails as the train's CoM returned to its lowest point. That would create huge amounts of mechanical wear on both the wheels and the track: that's not what happens. $\endgroup$ Jul 30, 2015 at 8:36
  • $\begingroup$ @DavidRicherby: Your assumption that the only way to balance out a force is sliding friction is 100% unconditionally false. You can also balance it out with static friction or centrifugal forces due to a track with a radius of curvature, for example. The axle will pick up a force $\pm M_{\text{train}}~g~\sin\theta$ in its axial direction, which may be "restorative" or not. $\endgroup$
    – CR Drost
    Jul 30, 2015 at 14:03
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    $\begingroup$ @ChrisDrost: If the axle were not rigid, the wheel assembly would have little reason to want to follow a curve. Because the axle is rigid, however, if the wheels have radii x and y (with x being larger) and are separated by distance s, the axle will want to rotate about an axis located at distance sy/(x-y) from the smaller wheel. The difference between the radii will tend to increase or decrease so as to make the aforementioned distance match the radius of the curve the wheel is following. $\endgroup$
    – supercat
    Jul 30, 2015 at 19:41

In both diagrams in the question, the left wheel has a smaller radius at the contact point than the right wheel. Because they're fixed to a common axle, in any given amount of time, the right wheel will travel a greater distance than the left, so the axle as a whole will rotate anti-clockwise (when viewed from above) about a vertical axis. As it does this, it will start to lie diagonally across the track, rather than perpendicular to the two rails.

Suppose we're using the wheel profile in the first diagram. As the axle gets farther from perpendicular, the right wheel moves ahead and drops down to have a lower-radius part in contact with the rail. This means that the axle is self-straightening since the more it steers, the more it tends to push itself back towards being perpendicular.

However, if we use the second wheel profile, as the axle gets farther from perpendicular, the right wheel climbs up the rail, causing the contact point to move to an area of higher radius. That means the right wheel goes even farther in one revolution, so it turns even more. That's completely unstable.

In the second diagram, the only way the axle can return to running straight is for the whole axle to rotate clockwise about a horizontal axis, with the wheels sliding perpendicularly across the rails. That would cause a huge amount of wear to both the wheels and the track, assuming the thing didn't just derail.

There's a Numberphile video demonstrating all of this with a few taped-together espresso cups.

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    $\begingroup$ This is IMHO the right answer. I don't think the center of mass (mentioned in other answers) has anything to do with it. $\endgroup$
    – supercat
    Jul 30, 2015 at 19:50

The contact with the rail creates a kinematic center of rotation where the reaction forces meet. The rail car will tend to rotate about this center as a result of side loads.

  • If the center is above the center of mass, the rail car acts like a hanging pendulum. A small deflection will cause a restoring torque opposing the swing.

  • If the center is below the center of mass, the rail car acts like an inverted pendulum. A small deflection will cause a positive feedback amplifying the swing.


As a side effect the rail car will turn away from the turn instead of into the turn when the cone is the other way around.


To see why the first configuration is used rather than the second, perform the following experiment:

Hold a bowl in your hand and place a small ball inside. Move the bowl in circles at various speeds and observe the behavior of the ball.

Now turn the bowl over and balance the ball on top. Again, move the bowl around and observe the ball.

Which is more stable?

In both cases, there is a low center of mass, the motion of which is constrained by the geometry of the parts. In the first of each case - your first diagram of train wheels, or the upright bowl - the geometry pits the rotations caused by centrifugal forces against gravity; as the train wheels slide off-center, the center of mass is lifted up, and when gravity pulls it back down it centers itself again. In the second of each case, a lateral movement causes the center of mass to slide "downhill," and gravitational acceleration exacerbates the problem rather than correcting it.

A train with the first configuration of wheels keeps itself on the track naturally; the second configuration would require an expensive and challenging auto-ballast system to stay upright even when stationary.


Previous answers are great and explain the dynamics very well. I'd like to point out that this can be explained just as easily in a static situation.

Imagine the weight the shaft has to carry. You don't even have to imagine a curve to note that the weight will automagically center (and lower the center of gravity of) the train in the first image.

In the bottom image the flex of the shaft will cause the wheels to straighten out (thus raising the center of gravity).

As previously mentioned, even if you could lower the center enough not to cause instability, the wear and tear on both wheels and tracks due to constant "climbing" would be reason enough not to proceed...

That and the typical "This is how we've always done it. Don't think, just do." ;)

RE: dynamic situation: Now as the train rounds the curve and the centripedal (centrifugal?) force tries to topple the car, the conical wheels actually raise one side of the car while lowering the other thus moving the center of gravity closer to the pivot point. Also note that the shaft will flex more due to weight distribution on the inner wheel than on the outter. This will increase the angle of contact on the inner and decrease the outter counteracting the Newton's first law. Win, win, win.


I can see a difference by thinking about the flanges...

Given that

1) the flanges are on the inside of the wheels

2) the right handside of the diagrams is the outer part of the track where the flange will press against the rail...


enter image description here


enter image description here

where the red lines indicate the plane of the flange.... in the upper case the flange neatly pushes against the edge of the rail and will not easily jump past it, but in the lower case the flange looks like it could slide up the outwardly angled rail and cause the train to derail.

Edit after interesting comment - if the flanges were on the outside of the wheels then the lower case would be potentially as good as the upper case with the flanges on the inside. I am not sure how easy it would be to make points etc. work with flanges on the outside...

...the main advantage I can see with inner flanges is that when building the railway lines a suitable length rod between the rails can check the gap between them that might be easier than having some large calipers with jaws to check the distance between the outer edges of the rails. There may be other challenges as suggested by JonCuster

So as noted above this answer assumes that the wheel flanges are inside the wheels and not outside.

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    $\begingroup$ Wouldn't it make more sense to put the flanges in the outside in the second case? $\endgroup$ Jul 29, 2015 at 21:22
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    $\begingroup$ And you really want those flanges on the inside of the rails or engineering gets really difficult... $\endgroup$
    – Jon Custer
    Jul 29, 2015 at 21:22
  • $\begingroup$ I agree with Harry. If we did bring flanges into the picture, they would be installed on the outer part of the wheels in the second configuration. $\endgroup$
    – Rations
    Jul 29, 2015 at 22:50
  • $\begingroup$ @HarryWilson good point - the answer I gave is on the assumption that the flanges are inside the rails, which is the way I have seen every single railway - (except for one Paris metro line that uses rubber wheels or something similar) $\endgroup$
    – tom
    Jul 29, 2015 at 22:56
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    $\begingroup$ @ChrisH - but as we are talking about derailing in the this question - say the train approaches a corner too quickly - the flanges (on inside or outside) need to contact to provide extra centrapetal acceleration to keep the train on the tracks as it goes around the corner.... $\endgroup$
    – tom
    Aug 3, 2015 at 12:17

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