B-Unequal-Tangent Parabolic (Vertical) Curve: Example: A grade g1 of -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. 83 +00 BVC (751.24) 84 +00 - 85 +00 A (747.24) ناوه شرق 40 1-86 +00 00+ 28- CVC (747.56) -2.00% Stations 00 +88 89 +00 +1.60% V (743.24) 00+ 06- 1-91 +00 B (748.04) Point A STA 85 +00: Elev. = 743.24+2 (2) = 747.24' Point B STA 90+00: Elev. = 743.24 +1.6 (3)=748.04' 00 + 26 وه ی 600 93 +00 EVC (752.84) ونا تسلیة شنانه و پوینتة نوت ده دوام وه Solution: 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.
B-Unequal-Tangent Parabolic (Vertical) Curve: Example: A grade g1 of -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. 83 +00 BVC (751.24) 84 +00 - 85 +00 A (747.24) ناوه شرق 40 1-86 +00 00+ 28- CVC (747.56) -2.00% Stations 00 +88 89 +00 +1.60% V (743.24) 00+ 06- 1-91 +00 B (748.04) Point A STA 85 +00: Elev. = 743.24+2 (2) = 747.24' Point B STA 90+00: Elev. = 743.24 +1.6 (3)=748.04' 00 + 26 وه ی 600 93 +00 EVC (752.84) ونا تسلیة شنانه و پوینتة نوت ده دوام وه Solution: 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.
Chapter2: Loads On Structures
Section: Chapter Questions
Problem 1P
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Transcribed Image Text:B- Unequal-Tangent Parabolic (Vertical) Curve:
Example:
A grade gl of -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.
Stations
00
00
00
00
EVC
(752.84)
BVC
(751.24)
A
CVC
(748.04)
(747.24)
(747.56)
-2.00%
400
+1.60%
V(743.24)
>と じ。
Solution:
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.
Point A STA 85+00: Elev. = 743.24+2 (2) = 747.24'
Point B STA 90 + 00: Elev. = 743.24+ 1.6 (3) = 748.04'
%3D
%3D
83+00
-85+ 00
89 + 00
1.
- 92 + 00
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