Elegant, but you are going to waste too much time doing it this way. If you realize the area under the diagram of grades is the elevation, then: 1) make a diagram of the grades. Horizontal axis will go from 1812 ft, to 2512 ft [given]. Vertical axis from 0.02 (2%) to -0.04 (-4%) [given]. Draw a line from 0.02 to -0.04. Do you visualize the triangles?, good. Then: 2) Calculate the horizontal distance to the summit of the parabole (which is where the grade line crosses the horizontal axis). Doing a simple triangle equivalency: (0.02)/x = (0.04+0.02)/700 ft, x = 233.33 ft, so that station is also 1812’ + 233.33’ = 2045.33’. 3) Calculate the elevation at PC using EL PVI: 439.20’ - 0.02*(700’/2) = 432.20’. 4) Calculate the grade at station 21+00 using another triangle equivalency: -g/(2100’ - 2045.33’) = (0.04+0.02)/700’ => g = -0.004686. 5) Calculate the elevation at the summit adding the first area of the triangle: y = 432.20’ + (1/2)*0.02*233.33’ = 434.53’. 6) Calculate the elevation at station 21+00 adding the next area of the triangle to the elevation of the summit: y = 434.53’ + (1/2)*(2100’ - 2045.33’)*(-0.0046857) = 434.40’.
Here's an organized set of playlists that might be helpful: sites.google.com/ncsu.edu/daniel-findley/educational-resources Specifically, I think these playlists might be helpful for you: ruclips.net/video/0E6E3WTSiEg/видео.html ruclips.net/video/ELzTygJbZYE/видео.html ruclips.net/video/qZVZyf_9r2k/видео.html
one of the best vertical curve primers on yt. loved the pointers at the end!
ncees hb 1.1, page 278.
Elegant, but you are going to waste too much time doing it this way. If you realize the area under the diagram of grades is the elevation, then:
1) make a diagram of the grades. Horizontal axis will go from 1812 ft, to 2512 ft [given]. Vertical axis from 0.02 (2%) to -0.04 (-4%) [given]. Draw a line from 0.02 to -0.04. Do you visualize the triangles?, good. Then:
2) Calculate the horizontal distance to the summit of the parabole (which is where the grade line crosses the horizontal axis). Doing a simple triangle equivalency: (0.02)/x = (0.04+0.02)/700 ft, x = 233.33 ft, so that station is also 1812’ + 233.33’ = 2045.33’.
3) Calculate the elevation at PC using EL PVI: 439.20’ - 0.02*(700’/2) = 432.20’.
4) Calculate the grade at station 21+00 using another triangle equivalency: -g/(2100’ - 2045.33’) = (0.04+0.02)/700’ => g = -0.004686.
5) Calculate the elevation at the summit adding the first area of the triangle: y = 432.20’ + (1/2)*0.02*233.33’ = 434.53’.
6) Calculate the elevation at station 21+00 adding the next area of the triangle to the elevation of the summit: y = 434.53’ + (1/2)*(2100’ - 2045.33’)*(-0.0046857) = 434.40’.
Thanks for the suggestion! I'm going to try working it out this way too.
Any more vertical and horizontal curve problems for PE ?
Here's an organized set of playlists that might be helpful: sites.google.com/ncsu.edu/daniel-findley/educational-resources
Specifically, I think these playlists might be helpful for you:
ruclips.net/video/0E6E3WTSiEg/видео.html
ruclips.net/video/ELzTygJbZYE/видео.html
ruclips.net/video/qZVZyf_9r2k/видео.html
@@FindleyDaniel thank you.