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bnsftonyau

New Member
Hi, how do you work out the ratio of a climbing track? I need to climb 10 inches and need to know how to work out the percentage and ratio needed. This is N scale
cheers Tony
 
Here is what I was told:

2% grade max... That is 2 inches in 100 inches or 1 inch in 50 inches.

% = height/length of grade
or
length of grade = height/% Percent as a decimal, 2% = .02 in this case.
However I think people commonly use 3% and even 4%. On my N scale layout I am staying with the 2% number. Likely you will get other opinions and comments.
 
Hi Tony -

The thread recommended in the previous post is a good one with lots of information.

To address your specific question, climbing 10" on a model railroad layout at a reasonable grade is going to require a long run of track.

For example, to climb 10" at a 2% grade will require:
10" / 0.02 = 500" = 41 feet, 8 inches of track

If you increase the grade to 3%, you can shorten the run:
10" / 0.03 = 333" = 27 feet, 9 inches

And if you go to 4% grade, which is considered to be rather steep:
10" / 0.04 = 250" = 20 feet, 10 inches

The general rule of thumb is that a 2% to 3% grade is about the maximum for a model railroad main line. Beyond that, you'll be restricted to short trains or multiple engines to pull a typical train up such a steep grade. The runs calculated above will bring you to the top of a 10" rise. You will need an equal length of track you get back down to the base level.

So if you want to climb 10", I hope you have a large layout (or a helix).

- Jeff
 
Nobody has mentioned the easement into the grade. You don't go from flat to 2% grade with a sharp bend in the track. If you are going to constrain yourself to a 2% grade, your actual runout of track will be more than the basic calculation of 2% at 10". You have to take into consideration the wheel base of the locomotives you are running, the length of cars and your coupler tolerance.
 
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