• Squealing Flanges

  • General discussion about railroad operations, related facilities, maps, and other resources.
General discussion about railroad operations, related facilities, maps, and other resources.

Moderator: Robert Paniagua

  by benltrain
I am doing research into how trains take curves, including tilting trains, superelevation of tracks, in-car forces taking different types and degrees of curves, speeds possible for different degrees of curves, etc. This especially includes the *simple* physics of how this works (I don't want an Einsteinian theory on how to build a nuclear bomb).

If anybody can give some info to me on this, or even better point to previous threads, websites, other sources, that would be very greatly appreciated.


  by 3rdrail
When I get the time, I'll look through some of my accident reconstruction formulas for a simple one which may demonstrate the theory. Cars, trucks, trolleys, trains, giraffes - they're all the same. They're all at the mercy of Sir Isaac Newton's Laws of Physics. One rule that comes to mind - "For every action, there is an equal but opposite reaction". (One question which has always mystified me is what amount of a trains steel wheel actually comes into contact with the steel rail ? (not counting the flange). Is it the same amount for different size wheels ?) I'll write you again in the next few days. Paul Joyce.

  by drayrr
The allowable speed through a curve is a function of the degree of curve (sharpness) plus the amount of superelevation in the curve.

For railroads superelevation (normally expressed in inches) consists of the actual amount of superelevation plus an allowed amount of additional value (called the unbalanced elevation) that is not actual superelevation but rather an amount of cant (superelevation) deficiency. The Federal Railroad Administration (FRA) currently allows 3" of ubalanced elevation for most passenger equipment and up to 4" in some cases. I believe that FRA allows the Acela equipment to operate at up to 7" of unbalanced elevation.

So, Ee or equilibrium elevation is the amount of superelevation in inches that would be necessary for a train on a given degree of curve at a given speed to be in equilibrium, that is there would be no outward or centrifugul force, all of the enegry of the angular acceleration would be directed perpedicular into the track.

Ea is the actual superelevtion in inches of the low rail to the high rail which most railroads limit to 6" and not more than 4" where freight trains operate.

Eu is unbalanced elevation or really a deficiency in elevation. Most railroads use the FRA maximum allowed of 3". That results in not balancing the centrifugul force so that someone riding in the train would feel an outward pull and the car would tend to be rolled slightly on its bolsters.

The formulae for speed/superelevation/degree of curve are as follows:
Ee = Ea + Eu

Ee = 0.0007 * D * Vsquared

Where Ee is the amount of superelevation in inches for equilibrium, D is the degree of curve and V is the speed in miles per hour.


V = square root of Ea + Eu / 0.0007D

In the above, Ea plus Eu will normally not exceed 9" ( 6" max actual elevation plus the 3" Eu)

Railroads use degree of curve rather than radius in feet. Railroads also use chord definintion rather than arc (They measure the length of the curve and deflections by 100 foot long chords (straight lines) rather than around the curve using pi.

So that R = 50/sinD/2

Where R is the radius in feet and D is the degree of curve.

A one degree curve has a radius of just under 5,730 feet, a two degree curve half of that and so on.

Hope this helps, and sorry to be so technical but that is about as simple as I can explain it. Just remember that Eu is basicaly an allowed amount of "cheating" to get additional speed on given curve.

  by 3rdrail
That was a great professional explanation by Drayrr. I won't attempt at that level of sophistication. I am not an engineer. Very simply put, a number of Newton's Laws of Physics are at play here. A train taking a curve on rail is pushing against the rail while the rail is pushing back at the same precise amount of energy. The center of mass of the train (or car) is experiencing centrifugal force as it does so. The centrifugal force may act as an unbalanced external force upon the train, causing it to push against the rail in excess of the rail's ability to push back. Remember, forces must equal out, so at the point that the centrifugal force against the rail cannot be matched by the rail, the body at motion (train) seeks to equalize its outward or centrifugal force, and thus, there is a derailment. Superelevation re-directs the centrifugal force with centripetal force (opposite direction). Drayy - what is your take on my round wheel on a straight rail question ? I believe that a round surface coming into contact with a straight surface must come into contact almost at a microscopic level if we are dealing with true straight and round. No ? Paul Joyce.

  by benltrain
Thank you all very much, this is very helpful.

  by drayrr

Wheel/Rail interface is not my area of expertise, but sufficeth to say it is a complicated subject only recently being understood to any great degree.

Currently, railroads are operating cars weighing up to 315,000 lbs on four axles (eight wheels) resulting in wheel loads of just under 40,000 lbs. It is known that these wheel loads can make even head hardened rail plastic (a state where the steel "flows" away from the load). Since this is not a healthy state for rail it is necessary to try and maintain a sufficient contact area between the wheel tread and the head of the rail. Ideally, when the rail head and wheel treads are properly contoured, there are two points of contact between the rail and wheel, one near the center of the rail and the second more towards the gauge side, just short of the flange. In that case the effective contact area is probably in the range of close to one squre inch and at that load, the rail is not plastic. When the point of contact is reduced to a lower surface area due to too narrow a contact band, then the wheel loads increase to somewhere in the range of 55-60,000 pounds per square inch or more and the rail becomes plastic, quickly deforming to a shape that exacebates the situation. The same situation occurs to wheels themselves with freight cars experiencing hollow treads (center worn).

That is why the Class 1 railroads (and many transit authorites) are spending more and more on profile rail grinding to try and maintain the proper wheel rail contact.

There is a trade organization that puts out a publication called Wheel-Rail Interface Journal that discusses this whole subject in great detail. Can't remember offhand the exact name of this group. They had an annual meeting in Boston a few years ago.

  by 3rdrail
Thanks for your input Drayrr. I can't understand, however, how the point of wheel tread/rail contact can be one square inch if the rail or wheel is not plasticised. It would seem to me, that an area of wheel interface with a rail would be smaller than the area of touching a wall with a very thin tapered blade which came to an extremely fine point (microscopic), as the realities of geometry do not allow true straight and true round to interface at more than an infinitesimal amount at the point of highest contact. Modern rail is curved, so this even tends to disallow perpendicular, as well as lateral abutting, unless the wheel profile exactly contours to the rail. Which brings about some very interesting ideas relative to how strong rail and wheel construction has to be in order to not deflect under these loads. The standard grooved-rail and grooved-wheel with mucho miles on it is probably an attempt by which this coupling attempts to enlargen this small contact area, and in doing so, perhaps increases contact perpendicular to the wheel only. Flat spots would probably enlarge the contact area in both directions. I guess I know why trolleys blow sand during winter on rail and why wheels spin on starts.(Think of this the next time you're doing 125 on Acela !) Thanks again.

  by gcarp1
When Drayrr says Plastic deformation, he means deformation which is not reversible. By that I mean it tends to stay deformed. Before reaching that point, most all materials have ELASTIC deformation which IS reversible. With elastic deformation, the material will flex under pressure and then return to its original shape (both the wheel and the rail).


  by 3rdrail
An interesting principle. It would seem that manufacturers would attempt (as they do) to reduce such elasticism as much as possible (ie. w/ magnesium steel rail, etc.), as elasticism=more friction=more energy required to move the train. Conversely, I suppose, to keep parts subject to temporary deformity, allows for greater tolerances (???)

  by steemtrayn
With 12 tons concentrated on an area the size of a dime, one would expect SOME flexing.
  by obienick
I'm not sure if this is the right forum, but is there a certain degree of curvature responsible for causing flanges to squeal?
Try using the "search" feature. We've done this one, twice already. The flanges squeel, due to the wheel being forced to rotate att a speed unequal for the opposite wheel, for the length of track it's being operated on. The tread is "scrubbing" as well as the flange attempting to climb, the outside rail, causing the ringing/squeeling sound.It's been answered dynamically, elsewhere. Regards...
  by UPRR engineer
The merger of these two posts didnt go like i wished, but theres some more info for you dude.