Distance Between Topo/Contour Lines



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    Kurt Fanderclai

    If I'm understanding what youre asking -- sometimes it helps to think of a percent grade or slope as "something over 100".

    If they had said it's a 55.5% grade -- that's 55.5/100, which is the same thing as 10/18.  From knowing that, you'd plug in numbers exactly the same way as your example.    55.5/10 x 8 = 4.44.........so 824.44 

    It also helps to sketch a right triangle, and fill in the knowns -- a little trig keeps things straight.

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    Benjamin Marcionek

    Thanks Kurt! That's helpful. Pluralsight also suggested the following formulas.

    S = DE / L        and        S x L = DE

    Where S=Slope, L=Total Length, DE=Depth (i.e. elevation change).

    So for the above example.

    S = 10' / 18'  =  0.555 (or 55.5% slope)

    0.555 x 8' = DE  = 4.44'

    So the elevation at the point is 820' + 4.44' = 824.44'

    If the slope and elevation change are given and we needed to find the distance between the contours we could use the first equation and solve for the length. So  0.555 = 10' / L  =  18'





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