I am creating some simple grading objects from feature lines. I am targeting my existing ground surface with a 2:1 fill slope. My resulting surface shows me 5' contours, whcih I think should all be spaced out at constant 10' offsets, but they are not. When I measure the offsets I get different vaues ranging on 10.2' to 9.7' (depeding on the section of the feature line I am looking at). Anyone know a setting to make the slope build properly with a true 2:1 offset between contours? What good does this grading tool do if you can't get a properlty constructed slope?
Is there any way to maintain a constant resultant even when the slope of the base feature line varies? I'm trying to maintain a maximum 4:1 slope on a road.
If you are using a corridor to do the modeling, you can create a subassembly using the subassembly composer that will daylight with a resultant slope set to a certain value. Check the AU archives for a class by Kati Mercier.
I'd like to bump this post to see how others are dealing with this when grading with feature lines?
When grading a 60' tall slope with a road on it, the perpendicular projection from the road cause the toe of the slope to be off by several feet. By code, the slope needs to be 2:1 perpendicular to the contours.
Any thoughts without using corridors?
Two Grading. First grading off the feature line to the nearest contour elevation. second from contour FL to daylight. maybe?
Thanks Joe I'll give it a shot.
Steve, I'm not getting the procedure you are doing. Could you elaborate a bit more?
First off, let me explain why feature line gradings are not always actual slopes. A feature line grading will grade from each elevation point to the daylight at the desired slope. What the gradings do not do is account for the sloped along the feature line from one elevation point to the next. If you enter a 2:1 slope and there is a 50% grade between adjacent elevation points then the resultant slope between these two points is actually 2.82 (by pathagorean's theorem 2^2 + 2^2 = sq. rt. of 8 or 2.82:1.
If the feature line is level then then the resultant slope is the same as the entered slope. What you need to do is create a level feature line.
To do this I will take the lowest or highest point (lowest if grading up and highest if grading down). If the feature line ranges from 123.5' to 162.4, I will use the 'grade to elevation' to grade to elevation 125' (or maybe 130'). This will create a grading to elevation 125'. Areas below 125 will grade up to 125 and areas above 125' will grade down to 125. You only need the portion grading up to 125'.
Explode the new feature line and break the 3d polyline at the point where it intersects the original feature line. Create a feature line from this 3d polyline (it will be at elevation 125). Break the original feature line at the same location (elevation 125) and combine the two. What you will end up with is a new feature line with an elevation ranging from 125 to 162.4. From this new feature line I grade to elevation 130 and repeat the process.
After several iterations, you end up with a level feature line. From that you can accurately daylight to a surface using the entered slope. Hope that helps. It really is not as difficult or time consuming as it sounds.
If there is a better way to find a max slope I would love to learn it.
Steve.
Steve,
I guess with your solution you could actually just use c^=a^2+b^2 on the average run of a slope and project a long the entire run all at once. AKA @ 10% and wanting a 2:1
b=(50%^2-10%^2)^0.5 = 48.9898% (0.489898), which 1/answer = 2.0412 slope projection in order to get contours to be perpendicular at 2:1.
Now the question is: can this be done in a grading group automatically? It doesn't appear the grading group can't check the slope it is on then do the math
This seems all too easy to program and it is curious to see if the function is already in there... thoughts?
@EngineeringSam wrote:
I'd like to bump this post to see how others are dealing with this when grading with feature lines?
When grading a 60' tall slope with a road on it, the perpendicular projection from the road cause the toe of the slope to be off by several feet. By code, the slope needs to be 2:1 perpendicular to the contours.
Any thoughts without using corridors?
Why not use a corridor? You can create a corridor from a featureline (don't need an alignment/profile) in 2017.
@steve.mccoy01 wrote:
That is great. Is this available in 2016?
Nope. You'll need to upgrade to 2017 to be able to use featurelines as a baselines in corridors.