Question A crest vertical curve and a horizontal curve on the same highway have the same design speed. The equal-tangent vertical curve connects a +3% initial grade with a +1% final grade and has a PVC at 101 + 78 and a PVT at 106 + 72. The horizontal curve has a PI at 150 + 10 and a central angle of 75 degrees. If the superelevation of the horizontal curve is 8% and the road has two 12-ft lanes, what is the stationing of the PT? A crest vertical curve and a horizontal curve on the same highway have the same design speed. The equal-tangent vertical curve connects a +3% initial grade with a +1% final grade and has a PVC at 101 + 78 and a PVT at 106 + 72. The horizontal curve has a PI at 150 + 10 and a central angle of 75 degrees. If the superelevation of the horizontal curve is 8% and the road has two 12-ft lanes, what is the stationing of the PT?
A crest vertical curve and a horizontal curve on the same highway have the same design speed. The equal-tangent vertical curve connects a +3% initial grade with a +1% final grade and has a PVC at 101 + 78 and a PVT at 106 + 72. The horizontal curve has a PI at 150 + 10 and a central angle of 75 degrees. If the superelevation of the horizontal curve is 8% and the road has two 12-ft lanes, what is the stationing of the PT? A crest vertical curve and a horizontal curve on the same highway have the same design speed. The equal-tangent vertical curve connects a +3% initial grade with a +1% final grade and has a PVC at 101 + 78 and a PVT at 106 + 72. The horizontal curve has a PI at 150 + 10 and a central angle of 75 degrees. If the superelevation of the horizontal curve is 8% and the road has two 12-ft lanes, what is the stationing of the PT?
A crest vertical curve and a horizontal curve on The same highway have. The same dceign speed. The equal tangent vertical curve connecis a+3% intial grade with a+1.1. tinal gradelength of the verticas curve{:[l=10672-10178],[l=494 tt]:}S > 1{:[l=2s-(200(sqrt(h_(1))+sqrt(h_(2)))^(2))/(A)],[l=2s-(200*(sqrt3.5+sqrt2)^(2))/(2)],[494=2s-10.79*15],[s=786.575 tt],[S=1.47 vt+1.075v^(2)],[786.575=1.47 v(2.5)+1.075v^(2)],[v=73.384mph]:}claign speed of the corve{:[R=(v^(2))/(15(e+1))=((73)^(2))/(15(0.08+0.10))],[R=1973.70 pt]:}length of the horizonial curvel=(xR Delta)/(180)=(pi(1973-70) ... See the full answer