Does anyone know how to calculate the width of the ROW with double tracks?
This is my attempt:
https://tc.canada.ca/en/rail-transpo...ay-clearance#5
Based on 5.1a, 5.3b and 5.3b, I first get 1.435 m (centre to track, x2) + 3.96 m (2 main tracks) + 0.508 m (1 degree of curvature) + 0.635 m (superelevation) = 6.538 m.
Then, I've read from somewhere that the distance from the centre of the outer track to the boundary of ROW is 15.24 m, so the overall width (from the outer edge of the rail tie to the other side) should be 6.538 m - 0.7175 m + 15.24 m = (rounded up) 21.061 m. Is that correct?
Edit: Oops, I got the decimal points wrong. BRB redoing the calculation!
1.435 m (centre to outer frame of rail tie, x2) + 3.96 m (2 main tracks) + 0.0508 m/degree x 1 degree difference* + 0.0635 m/m x 2 m elevation difference = 5.5728 m
Then add 15.24 m - 0.7175 m** to it:
20.0953 m. Yay or nay?
* This is according to
https://trn.trains.com/railroads/ask...0radii%20lines.
** This is to avoid double-counting the distance from center to outer frame of the rail tie of the outer track.