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.

Design the curve, of An equal tangent vertical curve is to be constructed between grades of -2.0% (initial) and 1.0% (final). The PVI is at station 11 + 000.000 and at elevation 420 m. Due to a street crossing the roadway, the elevation of the roadway at station 11 + 071.000 must be at 421.5 m.  GIVEN DIMENSIONS ON THE ATTACHED PICTURE

solve using grade diagram

3. A grade g1 of +3.25% intersects grade g2 of -2.75% at a vertex whose station and elevation are STA 257+17.65 and 3,820.14 ft, respectively. An equal-tangent parabolic curve 900 ft long has been selected to join the two tangents. a. Compute the station and elevation of PVC. b. Compute the station and elevation of the first full station inside the curve. c. Identify the station and elevation of the highest point on the curve.