In the course of other on-going research, Terry Smith reminded me of the February 11, 1888 Scientific American article written about the Bridgton and Saco. I had been remiss in reviewing my copy of this publication prior to giving my MARPM presentation this year, as it would have allowed me to provide better information on the use of superelevation on the railroad.

For those researching designs and information for their own layouts, superelevation is the practice of making the outer rail higher than the inner rail, within a curve, in order to decrease or eliminate the lateral (outward) pressure on the rails and to avoid the danger of derailment. This can be accomplished with a very simple mathematical equation, which I provide an example of below.

Within the presentation, I had commented that I would build my layout using 45 MPH as the railroad’s design speed, with this number being drawn from a comment Ernest Ward made within his biography “My First Sixty Years in Harrison, ME” (pg 16):

Fastest speed recorded, Number 5 with two passenger cars – three and one-fourth miles in four minutes.

This equates to 48 MPH, therefore the use of 45 MPH. What is missing from the statement is where this show of speed occurred. It is unlikely that this was an average speed on the line, or even over the majority of the line. It is more likely this was exhibited on a (relatively) straight portion of the line.

The extra bit of information I was reminded of when reviewing the Scientific American (Vol. 58) article was in relation to an actual exhibition of the railroad’s capabilities to visitors from Central and South America:

The visitors were disembarked at the beginning of the 16 deg. curve, and, despite their fears and misgivings when Mansfield, who chaperoned the party, told them the train should round that sharp arc at a speed of 25 miles an hour, the thing was done before their very eyes.

The equation to calculate superelevation, from my 1908 reference book on railroad engineering is:

e = (G’ * v^2) / (g * R)


  • e = superelevation, in feet;
  • G’ = horizontal distance between the rail head centers, in feet;
  • R = radius of curvature, in feet;
  • v = velocity of train, in feet per second;
  • g = acceleration due to gravity (32.174 ft/s^2)

The distance between the rails for a two footer is 24 inches, and the rail head width for 30 lb ASCE rail is 1 11/16 inches, per LB Foster. This sets G’ equal to 2.14 feet. The radius of curvature given in the Scientific America demonstration is 16 degrees, or 358 feet radius, using R = 5,730/D, a fairly accurate conversion equation. The velocity of the train George Mansfield had run by was 25 MPH, or 36.7 ft/s.

We mix all of these numbers up in the equation and get a superelevation, in feet:

.25 ft = [2.14 ft * (36.7 ft/s)^2] / (32.174 ft/s^2 * 358 ft)

The rails on this particular curve referenced likely had 3 inches of superelevation, resulting in a “lean” into the curve of approximately seven degrees. This would equate to 1/16 inch rail elevation for those of us modeling in O scale, and 0.034 inches for those practicing the HO arts. Not too shabby.

Within my MARPM presentation, I used scale radii (48″, 60″, 72″, etc.) converted to real world curvatures (a 72″ scale radius equals a real world 20 degree curve). In light of this “new” information, I will recalculate my curvature elevations using a 25 or 30 MPH design speed, rather than the “more modelgenic” numbers I had created using a 45 MPH design speed.