Geometric Design Question

[John QPE]

Is that not a shift to the right? New PI of 572+12?

No, it is a shift/offset the left. Remember, you should be looking up-station.

I'm sorry. Would you mind explaining this? Or is this a green book concept you can point me too?

Thanks

 

[ptatohed]

Is that not a shift to the right? New PI of 572+12?

No, it is a shift/offset the left. Remember, you should be looking up-station.

I'm sorry. Would you mind explaining this? Or is this a green book concept you can point me too?

Thanks

[SIZE=10.5pt]No problem JQ. (I think) you are saying that the forward tangent is being shifted to the right simply because when you view the diagram (Post #17 above), it is being shifted to the right per the perspective of someone viewing the graphic. Is that a correct assumption? Well, right and left are relative. What if we turned the graphic upside down? Then, it would be a shift to the left, per the perspective of the viewer. Therefore, so that right and left stay relative regardless of the orientation of the graphic, we refer to right and left in relationship to the centerline (CL) of the alignment. And most often, this is when facing up-station. Let’s say you have a simple alignment running from west to east, with the stationing increasing to the east. If we say there is a street light at Station 5+00 @ 25’ Left, that means that when you stand at Sta 5+00 and face up-station (you’re facing east – you’re facing Station 6+00), the SL would be 25’ Left of CL, or 25’ to the north. If the Stationing increased from east to west, then a SL 25’ Left of CL would be 25’ to the south. So, in this case, for the horizontal curve we have been talking about, if you were standing at, say, the PT and looking up-station, you’d be facing “down” (relative to the orientation of the graphic) with your back to the PI and, thus, the newly offset forward tangent would be to your left. Picture yourself in a car driving along the roadway alignment, driving up-station (from lower station to higher station – from PC to PT). As you go around the curve, the newly offset forward tangent is to your left. I hope I explained this clearly and I hope this helps. [/SIZE]

 

[John QPE]

Totally clear. Thank you very much!

 

[rejectedbytexas]

ptaohed, I spent all :beat: these 19 months to get my answer. Hahaha.

1806'

1806.8' actually. So you bumped a 19 month old thread to provide an answer (with no solution) that has already been provided?