I have an existing inlet ('catchbasin' for us Canadian folks) that I need to keep in it's current place. I want to design a new sag vertical curve with this inlet at the low point of the curve...ie. the grade of the existing road will be reconstructed on both sides of the inlet.
Can I draw a vertical curve, specifying the low point as a parameter?
Thanks for the help.
Civil 3D 2010
in the profile tool bar there a vertical curve through a point.
Maybe someone else on this board more knowledgeable than me has figured out a way, but I've had to do similar elevation matching on vertical curves. I've drawn in a profile vertical curve that I wanted, and moved the curve by the low point to my target elevation, then building the rest of my profile that way.
I also found that using the dynamic nature of Civil 3D and the grip-edits, I can get my vertical curves to hit the elevation targets is nice, quick, and easy. I use a polyline drawn onto the profile grid to point to the elevation I need to hit, then 'tweak' my profile with the curve in it until it hits the target. You can shift the PI's to adjust grade in-and-outs, and grip-edit the length of the vertical curve to shift the low point around.
The only other way I can think of is to break out the scratch pad or engineering paper, and mathematcially work out the curve solution on paper.
I have to agree with ccoles here, this is one example where using the graphical editing tools in Civil3D will be faster than calculating it. I would suggest that you place some sort of mark on your profile at the appropriate location - you can use the transparent commands to make this step easier. Then add profile labels which include some sort of mark at the low point of the sag curves. My version draws a green line from the LPt to the middle of the dimension line.
Once you have that it's fairly easy to adjust your curve until it matches the marked point.
Right, I should have thought about that.
Obviously, (to those who think before they post), the Vc and its lowpoint are a function of the leading and trailing tangents and the rate of curvature K. In lieu of all the hand calculating required to determine the tangent angle and all, I agree with the graphical solution of dragging to a target.