Question

Question: An existing tunnel needs to be connected to a newly constructed bridge with sag and crest vertical curves. The sag curve begins at station 0+00 and elevation 100 ft. and the crest curve ends at station 12+00 and elevation 140 ft. Station 12 +00, PVT Bridge deck elevation 140 ft Tunnel floor elevation 100 ft WURUK Station 0 + 00, PVC, 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. Crest Vertical Curve K-values speed 80 115 11.1 30 US Customary Stopping Design Rate of vertical sight curvature, K distance (mi/h) (ft) Calculated Design 15 3. 03 20 6.1 7 25 155 200 18.5 35 250 29.0 43.1 44 45 360 60.1 425 83.7 84 113.5 114 60 570 150.6 151 645 192.8 193 70 246.9 247 75 820 311.6 80910 383. 7 384 40 305 495 65 730 312

Answer 1

Question 7. Since the station as otoo it is = 6 +40.0 of clear the Pvc is given that PVT = PVC e ophia ( Question & The vertical intersection of the crest Curve PVC = 9+ 2000 option © The elevation of the point of vertical fangent of the sog curve is Question a 100+ Anekdo .AL 200 [ 100 + (8591) (640) 200 = 127.4367 6 option