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profiles - tanget from station along vertical curve

4 REPLIES 4
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Message 1 of 5
Anonymous
661 Views, 4 Replies

profiles - tanget from station along vertical curve

I am not too familiar with the various profile creation tools.  I googled a bit but couldn't find an answer to my question.

 

I have a profile that has a vertical curve.  I am curious is there a way to create a profile subentity (apart of the same profile) that is tangent to this vertical curve at a particular station.


I am asking this because I have a complex geometric situation. 

 

I have an arterial roadway that is being developed to go over a railroad, so the arterial roadway has a the typical sag/crest/sag vertical curve. 

Adjacent to this arterial is an expressway that is at a slightly higher elevation (not by much) that also goes over the railroad. 

I am trying to create the ramp profile that will connect these two roadways. However, due to geometric constraints, the ramp profile needs to connect to the arterial as soon as possible.  Since I already have the sag/crest/sag on the arterial, I'd like to create a profile that is meets curvature requirements but is tangential to the crest of the arterial. 

 

The challenges I am facing are two fold:  

 

1) I find that in order to meet my K value, my length places me on the crest of the vertical curve (hence why I am asking if I can shoot a tangent off a point along a vertical curve).

 

2) Since (a) the length of curve is a function of algebraic difference and (b) the grade changes along the vertical crest curve I get into a recursive (chicken/egg scenario).  Hence why I was thinking of working backwards (e.g. from the known vertical curve).  This is why I am asking to create a tangent off a station/point along a vertical curve.  I was curious, if there is a quick way to determine the grade at a particular point on the curve?

 

I'd welcome any thoughts and feedback on how to solve/approach this problem and how to better use the tools available in Civil 3D.

 

Thanks!

4 REPLIES 4
Message 2 of 5
Anonymous
in reply to: Anonymous

Have you tried the Free Vertical Parabola curve, this option will allow you to use ether a K value, radius or Pass-through point.

You could set the pass through t the end of your fixed vertices and then use the Profile View Editor to adjust the K value if it is not at the required value without moving the tangential connection point

Message 3 of 5
Anonymous
in reply to: Anonymous

Thank you for your reply.

 

I've attached two quick MS paint drawn graphics to help explain what I am trying to accomplish.

 

If you look at the VC.png attachment, I created the profile (the one with the crest).  The crest curve is a floating curve between the two fixed tangents.

I am trying to create new profile [read: not a sub-entity, new profile object] (e.g. the sag) that comes in and is tangent off crest vertical curve at a particular station (VC2.png might help understand).

 

I don't know if AutoCAD can share information between profiles (e.g. being able to create a new profile object that is in relation to another separate profile object). So I was thinking I could probably figure it out geometrically by creating a line that is tangent to a profile object at a particular station. So, I was wondering if there was a way to create a profile sub-entity (or even a regular line) that is tangent to a profile object at a particular station location.

 

I noticed that you said the free vertical parabolic profile tool works from the end of the fixed vertices.  I have two issues: My crest curve is a floating curve; and I need to shoot off in the middle of the profile at a particular station.

 

Thanks again for your thoughtful reply!

Message 4 of 5
lim.wendy
in reply to: Anonymous

Hi there,

 

I hope I understand you correctly. Otherwise, feel free to correct me.

I think you should be able to create a new profile that is tangent to an existing profile by adding a new entity before the existing vertical profile. Make a copy of the profile and choose only the station range which you require.

Not directly but this is something I can think of. Smiley Wink

See the following screencast:

 



Wendy Lim

Data Nerd | Community Advocate | AEC Industry


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Message 5 of 5
Joe-Bouza
in reply to: Anonymous

You kind of answered your own question.
You know the k value and you know wher this tangent starts.... so from k & l you know the required grade of the tangent and what station it begins

Joe Bouza
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