As shown in the figure below, a horizontal alignment consists of two tangent lines smoothened by a horizontal curve. The station of PC is at sta. 20+30.00 and the station of PT is at sta. 24+00.00. The superelevation runoff length is 210 feet. Calculate the distance from the station PC to the station where the full superelevation starts to be lowered. (Round to the nearest foot and fill in the blank with a number only.) PC Sta 20+30.00 R R +2% PT Rotation axis -2%
As shown in the figure below, a horizontal alignment consists of two tangent lines smoothened by a horizontal curve. The station of PC is at sta. 20+30.00 and the station of PT is at sta. 24+00.00. The superelevation runoff length is 210 feet. Calculate the distance from the station PC to the station where the full superelevation starts to be lowered. (Round to the nearest foot and fill in the blank with a number only.) PC Sta 20+30.00 R R +2% PT Rotation axis -2%
Chapter2: Loads On Structures
Section: Chapter Questions
Problem 1P
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Transcribed Image Text:Question 16
As shown in the figure below, a horizontal alignment consists of two tangent lines
smoothened by a horizontal curve. The station of PC is at sta. 20+30.00 and the station of
PT is at sta. 24+00.00. The superelevation runoff length is 210 feet.
Calculate the distance from the station PC to the station where the full
superelevation starts to be lowered.
(Round to the nearest foot and fill in the blank with a number only.)
PC
Sta 20+ 30.00
R
R
+2%
PT
Rotation axis
-2%
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