vertical curves
TRANSCRIPT
![Page 1: Vertical Curves](https://reader033.vdocuments.us/reader033/viewer/2022061106/54482a97b1af9f66018b48b8/html5/thumbnails/1.jpg)
Vertical Curves
Chapter 25
![Page 2: Vertical Curves](https://reader033.vdocuments.us/reader033/viewer/2022061106/54482a97b1af9f66018b48b8/html5/thumbnails/2.jpg)
Profiles:Curve a: Crest Vertical Curve (concave downward)
Curve b: Sag Vertical Curve (concave upward)
Tangents: Constant Grade (Slope)
![Page 3: Vertical Curves](https://reader033.vdocuments.us/reader033/viewer/2022061106/54482a97b1af9f66018b48b8/html5/thumbnails/3.jpg)
Equal-Tangent Vertical Parabolic Curve:
![Page 4: Vertical Curves](https://reader033.vdocuments.us/reader033/viewer/2022061106/54482a97b1af9f66018b48b8/html5/thumbnails/4.jpg)
Terms:BVC: Beginning of Vertical Curve aka PVC
V: Vertex aka PVI
EVC: End of Vertical Curve aka PVT
g1: percent grade of back tangent
g2: percent grade of forward tangent
L: curve length (horizontal distance) in feet or stations
x: horizontal distance from any point on the curve to the BVC
r: rate of change of grade
![Page 5: Vertical Curves](https://reader033.vdocuments.us/reader033/viewer/2022061106/54482a97b1af9f66018b48b8/html5/thumbnails/5.jpg)
Equations:r = (g2 – g1)/L
where:
g2 & g1 - in percent (%)
L – in stations
and
Y = YBVC + g1x + (r/2)x2
where:
YBVC – elevation of the BVC in feet
![Page 6: Vertical Curves](https://reader033.vdocuments.us/reader033/viewer/2022061106/54482a97b1af9f66018b48b8/html5/thumbnails/6.jpg)
Example: Equal-Tangent Vertical CurveGiven the information show below, compute and tabulate the curve for stakeout at full 100’ stations.
![Page 7: Vertical Curves](https://reader033.vdocuments.us/reader033/viewer/2022061106/54482a97b1af9f66018b48b8/html5/thumbnails/7.jpg)
Solution:
L = STAEVC – STABVC
L = 4970 – 4370 = 600’
or 6 full stations
r = (g2 – g1) / L
r = (-2.4 – 3) / 6
r = -0.90 r/2 = -0.45 % per station
STABVC = STAVertex – L / 2 = 4670 – 600/2 = STABVC= STA 43 + 70
STAEVC = STAVertex + L / 2 = 4670 + 600/2 = STAEVC= STA 49 + 70
ElevBVC = Elevvertex – g1 (L/2) = 853.48 – 3.00 (3) = 844.48’
ElevEVC = Elevvertex – g2 (L/2) = 853.48 – 2.40 (3) = 846.28 ’
![Page 8: Vertical Curves](https://reader033.vdocuments.us/reader033/viewer/2022061106/54482a97b1af9f66018b48b8/html5/thumbnails/8.jpg)
Solution:(continued)
r/2 = -0.45 % per station
Elevx = ElevBVC + g1x + (r/2)x2
Elev 44 + 00 = 844.48 + 3.00(0.30) –0.45(0.30)2 = 845.34’
Elev 45 + 00 = 844.48 + 3.00(1.30) –0.45(1.30)2 = 847.62’
Elev 46 + 00 = 844.48 + 3.00(2.30) –0.45(2.30)2 = 849.00’
etc.
Elev 49 + 00 = 844.48 + 3.00(5.30) –0.45(5.30)2 = 847.74’
Elev 49 + 70 = 844.48 + 3.00(6.00) –0.45(6.00)2 = 846.28’ (CHECKS)
![Page 9: Vertical Curves](https://reader033.vdocuments.us/reader033/viewer/2022061106/54482a97b1af9f66018b48b8/html5/thumbnails/9.jpg)
Solution:(continued)
Stationx
(stations) g1x r/2 x2
Curve Elevation
43 + 70 BVC 0.0 0.00 0.00 844.4844 + 00 0.3 .90 -0.04 845.3445 + 00 1.3 3.90 -0.76 847.6246 + 00 2.3 6.90 -2.38 849.0047 + 00 3.3 9.90 -4.90 849.4848 + 00 4.3 12.90 -8.32 849.0649 + 00 5.3 15.90 -2.64 847.7449 + 70 EVC 6.0 18.00 -6.20 846.28
![Page 10: Vertical Curves](https://reader033.vdocuments.us/reader033/viewer/2022061106/54482a97b1af9f66018b48b8/html5/thumbnails/10.jpg)
High and Low Points on Vertical Curves
Sag Curves:
Low Point defines location of catch basin for drainage.
Crest Curves:
High Point defines limits of drainage area for roadways.
Also used to determine or set elevations based on minimum clearance requirements.
![Page 11: Vertical Curves](https://reader033.vdocuments.us/reader033/viewer/2022061106/54482a97b1af9f66018b48b8/html5/thumbnails/11.jpg)
Equation for High or Low Point on a Vertical Curve:
y = yBVC + g1x + (r/2)x2
Set dy/dx = 0 and solve for x to locate turning point
0 = 0 + g1 + r x
Substitute (g2 – g1) / L for r
-g1 = x (g2 – g1) / L
-g1 L = x (g2 – g1)
x = (-g1 L) / (g2 – g1)
or
x = (g1 L) / (g1 – g2) = g1/r x – distance from BVC to HP or LP
![Page 12: Vertical Curves](https://reader033.vdocuments.us/reader033/viewer/2022061106/54482a97b1af9f66018b48b8/html5/thumbnails/12.jpg)
Example: High Point on a Crest Vertical Curve
From previous example:
g1 = + 3 %, g2 = - 2.4%, L = 600’ = 6 full stations, r/2 = - 0.45,
ElevBVC = 844.48’
x = (g1 L) / (g1 – g2)
x = (3)(6) / (3 + 2.4) = 3.3333 stations or 333.33’
HP STA = BVC STA + x
HP STA = 4370 + 333.33 = HP STA 47 + 03.33
ELEVHP = 844.48 + 3.00(3.3333) – 0.45(3.3333)2 = 849.48’
Check table to see if the computed elevation is reasonable!
![Page 13: Vertical Curves](https://reader033.vdocuments.us/reader033/viewer/2022061106/54482a97b1af9f66018b48b8/html5/thumbnails/13.jpg)
Unequal-Tangent Parabolic Curve
A grade g1of -2% intersects g2 of +1.6% at a vertex whose station and elevation are 87+00 and 743.24, respectively. A 400’ vertical curve is to be extended back from the vertex, and a 600’ vertical curve forward to closely fit ground conditions. Compute and tabulate the curve for stakeout at full stations.
![Page 14: Vertical Curves](https://reader033.vdocuments.us/reader033/viewer/2022061106/54482a97b1af9f66018b48b8/html5/thumbnails/14.jpg)
The CVC is defined as a point of compound vertical curvature. We can determine the station and elevation of points A and B by reducing this unequal tangent problem to two equal tangent problems. Point A is located 200’ from the BVC and Point B is located 300’ from the EVC. Knowing this we can compute the elevation of points A and B. Once A and B are known we can compute the grade from A to B thus allowing us to solve this problem as two equal tangent curves.Pt. A STA 85 + 00, Elev. = 743.24 + 2 (2) = 747.24’Pt. B STA 90 + 00, Elev. = 743.24 + 1.6 (3) = 748.04’
Solution:
![Page 15: Vertical Curves](https://reader033.vdocuments.us/reader033/viewer/2022061106/54482a97b1af9f66018b48b8/html5/thumbnails/15.jpg)
The grade between points A and B can now be calculated as:gA-B = 748.04 - 747.24 = +0.16%
5and the rate of curvature for the two equal tangent curves can be computed as:
and Therefore: r1/2 = +0.27 and r2/2 = +0.12
1
0.16 2.00.54
4r
Solution (continued):
1
0.16 2.000.54
4r
2
1.60 0.160.24
6r
![Page 16: Vertical Curves](https://reader033.vdocuments.us/reader033/viewer/2022061106/54482a97b1af9f66018b48b8/html5/thumbnails/16.jpg)
The station and elevations of the BVC, CVC and EVC are computed as:
BVC STA 83 + 00, Elev. 743.24 + 2 (4) = 751.24’EVC STA 93 + 00, Elev. 743.24 + 1.6 (6) = 752.84’CVC STA 87 + 00, Elev. 747.24 + 0.16 (2) = 747.56’
Please note that the CVC is the EVC for the first equal tangent curve and the BVC for the second equal tangent curve.
Solution (continued):
![Page 17: Vertical Curves](https://reader033.vdocuments.us/reader033/viewer/2022061106/54482a97b1af9f66018b48b8/html5/thumbnails/17.jpg)
STATION x g1x (r/2)x2Curve Elevation
BVC 83 + 00 0 0 0 751.24'84 + 00 1 -2.0085 + 00 286 + 00 3
CVC 87 + 00 4 747.56'88 + 00 1 0.1689 + 00 290 + 00 391 + 00 492 + 00 5
EVC 93 + 00 6
g1x = -2 (1) = -2.00
g2x = .16(1) = 0.16
Computation of values for g1x and g2x
![Page 18: Vertical Curves](https://reader033.vdocuments.us/reader033/viewer/2022061106/54482a97b1af9f66018b48b8/html5/thumbnails/18.jpg)
STATION x g1x (r/2)x2Curve Elevation
BVC 83 + 00 0 0 0 751.24'84 + 00 1 -2.00 0.2785 + 00 2 -4.0086 + 00 3 -6.00
CVC 87 + 00 4 -8.00 747.56'88 + 00 1 0.16 0.1289 + 00 2 0.3290 + 00 3 0.4891 + 00 4 0.6492 + 00 5 0.80
EVC 93 + 00 6 0.96
(r1/2)x2 = (0.27)(1)2 = 0.27
(r2/2)x2 = (0.12)(1)2 = 0.12
Computation of values for (r1/2)x2 and (r2/2)x2
![Page 19: Vertical Curves](https://reader033.vdocuments.us/reader033/viewer/2022061106/54482a97b1af9f66018b48b8/html5/thumbnails/19.jpg)
STATION x g1x (r/2)x2Curve Elevation
BVC 83 + 00 0 0 0 751.24'84 + 00 1 -2.00 0.2785 + 00 2 -4.00 1.0886 + 00 3 -6.00 2.43
CVC 87 + 00 4 -8.00 4.32 747.56'88 + 00 1 0.16 0.1289 + 00 2 0.32 0.4890 + 00 3 0.48 1.0891 + 00 4 0.64 1.9292 + 00 5 0.80 3.00
EVC 93 + 00 6 0.96 4.32
Y1 = 751.24 - 2.00 + 0.27 = 749.51'
Y2 = 747.56 + 0.16 + 0.12 = 747.84'
Elevation Computations for both Vertical Curves
![Page 20: Vertical Curves](https://reader033.vdocuments.us/reader033/viewer/2022061106/54482a97b1af9f66018b48b8/html5/thumbnails/20.jpg)
STATION x g1x (r/2)x2Curve Elevation
BVC 83 + 00 0 0 0 751.24'84 + 00 1 -2.00 0.27 749.51'85 + 00 2 -4.00 1.08 748.32'86 + 00 3 -6.00 2.43 747.67'
CVC 87 + 00 4 -8.00 4.32 747.56'88 + 00 1 0.16 0.12 747.84'89 + 00 2 0.32 0.48 748.36'90 + 00 3 0.48 1.08 749.12'91 + 00 4 0.64 1.92 750.1292 + 00 5 0.80 3.00 751.36'
EVC 93 + 00 6 0.96 4.32 752.84'
Computed Elevations for Stakeout at Full Stations
(OK)
![Page 21: Vertical Curves](https://reader033.vdocuments.us/reader033/viewer/2022061106/54482a97b1af9f66018b48b8/html5/thumbnails/21.jpg)
Designing a Curve to Pass Through a Fixed Point
Design a equal-tangent vertical curve to meet a railroad crossing which exists at STA 53 + 50 and elevation 1271.20’. The back grade of -4% meets the forward grade of +3.8% at PVI STA 52 + 00 with elevation 1261.50.
![Page 22: Vertical Curves](https://reader033.vdocuments.us/reader033/viewer/2022061106/54482a97b1af9f66018b48b8/html5/thumbnails/22.jpg)
Solution:
21
2 1
1
22
(5350 5200) 150 ' 1.52 2 2
2
1261.50 4.002
4.00 4.00 1.52
3.80 4.00
3.80 4.001.5
2 2 2
BVC
BVC
L L Lx stations
ry y g x x
g gr
LL
Y
Lg x x
rL
r Lx
L
![Page 23: Vertical Curves](https://reader033.vdocuments.us/reader033/viewer/2022061106/54482a97b1af9f66018b48b8/html5/thumbnails/23.jpg)
Solution (continued):2
2
2
3.80 4.001271.20 1261.50 4.00 4.00 1.5 1.5
2 2 2 2
0.975 9.85 8.775 0
4
20.975
9.85
8.775
9.1152 911.52 '
L L L
L
L L
b b acx
aa
b
c
L stations
Check by substituting x = [(9.1152/2)+1.5] stations into the elevation equation to see if it matches a value of 1271.20’
![Page 24: Vertical Curves](https://reader033.vdocuments.us/reader033/viewer/2022061106/54482a97b1af9f66018b48b8/html5/thumbnails/24.jpg)
Sight Distance
Defined as “the distance required, for a given design speed to safely stop a vehicle thus avoiding a collision with an unexpected stationary object in the roadway ahead” by AASHTO (American Association of State Highway and Transportation Officials)Types
Stopping Sight DistancePassing Sight DistanceDecision Sight DistanceHorizontal Sight Distance
![Page 25: Vertical Curves](https://reader033.vdocuments.us/reader033/viewer/2022061106/54482a97b1af9f66018b48b8/html5/thumbnails/25.jpg)
Sight Distance Equations For Crest Curves For Sag Curves
21 2
1 2
2
1 2
1 2
2
22
S L
S g gL
h h
S L
h hL S
g g
22 1
1 2
4 3.5
4 3.52
S L
S g gL
SS L
SL S
g g
h1: height of the driver’s eye above the roadway
h2: height of an object sighted on the roadway
AASHTO recommendations: h1 = 3.5 ft, h2 = 0.50 ft (stopping), h2 = 4.25 ft (passing)
Lengths of sag vertical curves are based upon headlight criteria for nighttime driving conditions.