Question

25. A spiral curve with a length of 50 feet connects a tangent to a curve having a radius of 100 feet. Deter- mine the offset from the tangent to the point where the spiral connects to the curve and the length of throw from the tangent. 26. Given a spiral curve with a length of 150 feet, which connects a tangent to a curve having a radius of 200 feet, determine the offsets from the tangent to the midpoint of the spiral. 27. For a preliminary vertical alignment of a road- way, the straight tangent sections are established as follows: Station 0 + 00, tangent elevation = 1055.00 Station 23 + 00, tangent elevation = 1107.75 Station 40 + 00, tangent elevation = 1056.75 Station 65 +00, tangent elevation = 1156.00 Determine the gradient of each tangent and the ele- vation at 1000-ft intervals along the tangents. 28. For a preliminary vertical alignment of a road- way, the straight tangent sections are established as follows: Station 0 + 00, tangent elevation = 73.00 Station 9 + 50, tangent elevation = 49.25 Station 22 + 25, tangent elevation = 93.00 Station 29 + 00, tangent elevation = 105.00 Determine the gradient of each tangent and the ele- vation at 500-ft intervals along the tangents. 29. A vertical parabolic curve has its PVI at station 29 + 25.00 and elevation 87.52. The grade of the back tangent is 2.5 percent, the grade of the forward tangent is -4.5 percent, and the curve length is 550 ft. Set up a table showing curve eleva- tions at the PVC, at the PVT, and at half-station points along the curve. Compute the station and elevation of the curve turning point.

Answer 1

267 sols Criven, Length of spiral curve (Ls) = 15oft Lx= {. Lg = } (150) = 75 Ft Radius (R) = 200 Ft & offset distance (6x)= KLA و اما x3 - GREC =(75)3 6x 206X150 You = 2.344 ft distance from Tangent to mid point is 2.344st. .. offset 1 of spiral 27 Sor each station 2 The gradient between a 51107-75 - 1055 1 à 300 - 0 = 2.293:/. X 100 9. [1056-75-1107.45) _ x 100 = -3.7. 4000 – 2300 9. z NIDO - 3-97). 1156 – 1056.75 l 6500 – 4000 / curve L= Axe 2009 Length of

= (3.97-2-293) *20002 200*(6500-0) X 100 = 516 ft General equation for elevation is 9, 7 x ² + 9, X + BVC elevation - (703500823 ](109) + 0*22293400 + 1055 = 10 2664 It Y = 592-2)2+ 9,* + BVC elevation = 0.0397- +0.02293 X1000 + 1055 - 516 =1138.45 ft
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