Alignment of Vertical Curves

An existing tunnel needs to be connected to a newly constructed bridge
with sag and crest vertical curves. The profile view of the tunnel and bridge
is shown. Develop a vertical alignment to connect the tunnel and bridge by
determining the highest possible common design speed for the sag and
crest (equal-tangent) vertical curves needed.
Compute the stationing and elevations of PVC, PVI, and PVT curve points.
0+366
112 m
100 m

PVIc
PVTc(0+366)@112
g2c
Let L = total length of Curve
Ls = Length of Sag Curve
Lc = Length of Crest Curve
L = Ls + Lc
366 = Ls + Lc (eq. 1)
PVCc
PVTs
Ls = Ks As & Lc = Kc Ac
As = |g2s-g1s| = |g2s-0|=g2s
Ac = |g2c-g1c| = |0-g1c|=g1c
g1s
PVCs (0+000)@100
Ls
PVIs
L = 366
Lc
From figure, g2s = g1c
Thus, As = Ac = A
366 = Ls + Lc
366 = Ks A + Kc A

366 = Ks A + Kc A
366 = A (Ks + Kc) equation 2
PVIc
PVTc(0+366)@112
g2c
12 = Yfs + Yfc
12 = +
12 =
g2s
PVCc
g1c
PVTs
Yfc
Yfs
12
12 =
A = 6.5574%
A in eq 2
g1s
PVCs (0+000)@100
PVIs
366 = A (Ks + Kc)
366 = 6.5574(Ks + Kc)
L = 366
(Ks + Kc) = 55.815

(Ks + Kc) = 56
Kc = 26 m
Ks = 30 m

PVIc
PVTc(0+366)@112
g2c
g2s
PVCc
g1c
PVTs
PVIs = 0+098 @ 100
PVTs/PVCc = 0+196 @
g1s
PVCs (0+000)@100
Ls
PVIs
Lc
L = 366
STATION
ELEVATION (m)
Kc = 26 m
Ks = 30 m
A =6.5574
Ls = Ks A
Ls = 30 (6.5574)
Ls = 196.72 m =196 m
Lc = Kc A
Lc = 26 (6.5574)
Lc = 170.49 m = 170 m
PVCs 0+000 100
PVIs ? ?
PVCc/ PVTs ? ?
PVIc ? ?
PVTc 0+366 112
Highest Design speed = 80 kph