Eastern Railroad Discussion > Speed on curves

I hope some civil engineers can help with this. i know the degree of curvature is 5730ft divided by feet in radii of the curve. I'm trying to develop maximum safe train speeds for certain degrees of curves as measured by tenths of a mile on a map.

Assume well maintained mainline track superelevated to the proper height for safe speed of 4 and six axle passenger and freight equipment at 26 tons maximum axle load. (lighter than North American standards).

Radius of two, three, four and five tenths of a mile equals curvature degrees of 10.85, 7.23,5.43, and 4.34 degrees respectively. 10.85 degrees is sharper than PRR's Horseshoe Curve which is 9 something degrees and train speed is 25mph I think, maybe 30 for passenger.

Thank you.

I ain't no engineer, well a locomotive type that is, but I can comment just on some changes where I work on Caltrain which operates from San Francisco to Gilroy, CA. During the CTX construction project which rehabilitated the RR for the express service that was to come after the project, many of the curves were relaid with welded rail, new ties (wood & concrete) different anchors and some type of screws instead of spikes. Two curves in particular had their speeds increased, one being the curve exiting Tunnel #1 at MP 1.2 roughly, was increased from 25 to 40mph. Another was the curve at Sierra Point, which goes under US 101, it had it's speed increased from 60 to 70mph. Additionally other curves had increased superelevation, which also boosted their speeds overall on the RR to 79mph, a tremendous difference given the fact that the RR for many years was mostly 60mph & 70mph with some 75mph trackage closer to San Jose.

Isn't an civil engineering answer to your question, but I thought I'd throw this out your way for general information.

Frank


jo37 Wrote:
-------------------------------------------------------
> I hope some civil engineers can help with this. i
> know the degree of curvature is 5730ft divided by
> feet in radii of the curve. I'm trying to develop
> maximum safe train speeds for certain degrees of
> curves as measured by tenths of a mile on a map.
>
> Assume well maintained mainline track
> superelevated to the proper height for safe speed
> of 4 and six axle passenger and freight equipment
> at 26 tons maximum axle load. (lighter than North
> American standards).
>
> Radius of two, three, four and five tenths of a
> mile equals curvature degrees of 10.85, 7.23,5.43,
> and 4.34 degrees respectively. 10.85 degrees is
> sharper than PRR's Horseshoe Curve which is 9
> something degrees and train speed is 25mph I
> think, maybe 30 for passenger.
>
> Thank you.

Complete federal regulations regarding this may be found in Code of Federal Regulations CFR 49 Section 213.57, which may be downloaded from this site.

http://www.access.gpo.gov/nara/cfr/waisidx_98/49cfr213_98.html

MTMEngineer:

I did try the cfr49 website as you suggested and did find the pages for formulas. I'm no engineer so find them hard to use. There is reference to Table 1 Appendix A which is a table of speeds on cures based on degree and super-elevation.

However, I can't figure out how to get to it after 20 minutes trying. Probably my lack of knowledge on how to use websites.

I thought I had this info in my own resources but apparently not. I'm following alignments of certain rail lines using Google Earth and trying to figure if some very curvy alignments are really sharp or not. This is on rail lines that probably have maximum speeds of 80mph for freight and 50mph or less for freight.

I'll try the website again, meanwhile possibly someone else will answer the four examples.

To Frank, what you are probably seeing is restoration of superelevation for the passenger trains mentioned. On many rail lines superelevation was reduced, sometimes to almost none, when railroads began large installations of welded rail in the 60's and 70's. This also co-incided with removal of most if not all passenger trains. Lower freight train speed did not need the superelevation and where 7 and 8 inches (the maximium) elevations were used slow freight trains placed undue wear on the inside rail, and stopped freight cars would someties lean precariously on those curves.

You mention "safe" train speed. FRA probably has nothing to say about that-- their formula is just the traditional "comfortable for passengers" speed. Axle load and so on doesn't enter into it-- it's just geometry.

Just using geometry and physics, you can calculate the speed that just matches the superelevation, with the gravity-plus-centrifugal-force vector perpendicular to the plane of the rails. Right? Traditionally in the US, the "comfortable" speed for conventional equipment was the speed (calculated as above) based on a superelevation 3 inches greater than the actual.

For instance: a 5730-ft (1 degree) curve with 4 inches actual superelevation. The allowed-by-FRA speed is the square root of

4 inches, plus 3, (equals 7)

divided by 0.0007 (the necessary conversion factor) (result 10000)

divided by 1 degree (result 10000)

which comes out to 100 mph.

Thanks timz. I will try to use your example on sharper curves to see how it works, if I understand it properly.
The curves I think are common on the rail lines I am following are 4 to 8 degrees.