In this question I got everything correct up to where I have to find elevation of PVT. (Not sure how the textbook got 991.0 ft) and I also don’t understand why the textbook found “a” by not putting G1 and G2 in percentages? Handwritten work only please essages File Edit View Conversations Window HelpKTransportation engineering textbook.pdf88♫curve will occur when the first derivative of the parabolic function (Eq. 3.1) is zero becausethe initial and final grades are opposite in sign. When initial and final grades are noopposite in sign, the low (or high) point on the curve will not be where the first derivative iszero because the slope along the curve will never be zero. For example, a sag curve with aninitial grade of -2.0% and a final grade of -2.0% will have its lowest elevation at the PVTand the first derivative of Eq. 3.1 will not be zero at any point along the curve. However, inour example problem the derivative will be equal to zero at some point, so the low pointwill occur whenFrom Eq. 3.3 we haveSubstituting for a and b givesb=G₁ = -3.5with G₁ in percent. From Eq. 3.6 (with L in stations and G₁ and G₂ in percent),0.5-(-3.5)2(6)IOB RADIOa=1.Kdydx7.281= 2ax+b=0This gives the stationing of the low point at 175 + 25 (5 + 25 stations from the PVC). Forthe elevation of the lowest point on the vertical curve, the values of a, b, c (elevation of thePVC), and x are substituted into Eq. 3.1, givingdy-= 2(0.33333)x+ (-3.5) = 0dxx = 5.25 stationsNote that the preceding equations can also be solved with grades expressed as the decimalequivalent of percent (for example, 0.02 ft/ft for 2%) if x is expressed in feet instead ofstations Care must be taken not to mix unite A dimensional analysis of Ea 31 must ensure-= 0.33333y = 0.33333(5.25)²+(-3.5) (5.25)+1000= 990.81 ft125Subscribed ✓MAR25Like CtvG a 600 ft equal tangent sag vertical curve has the PVC at Station 170+00 and elevation1000 ft. The initial grade is -3.5% and the final grade is +0.5%. Determine the stationingAnd elevation of the PVI, the PVT, and the lowest point on the curve.L=600 ftT=300 ftOptions-3.5%PVC = PVT-L17000 = PVT -60017600 = PVT176 +00= PVTPVC170+00PUI = PVC + T54°FCloudyElevation1000= 17,000 + 300- 17300PVI = 173+00-y = 2ax + bO SearchPVI173+00Elevation =a989.5'Elevation PUI = E pvc + 9²/12= 1000 + (-0.035) (600)=989.5'G₂+0.5%PVT1776+00elevation = 974.5¹Elevation PVT = EPI-9²/2/2b== 989.5 - (+0.05)(b)= 974.5'a== G₂-G₁₂₁ = 0.05-(-0.035)2L2(600)= 7.0833•10-50.0354:54 PM

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Accepted Answer

Data given
G1 = -3.5 %
G2=  +0.5 %
L = 600 ft
PVC, station = 170+00
        Elevation = 1000 ft…

a 600 ft equal tangent sag vertical curve has the PVC at Station 170+00 and elevation 1000 ft. The initial grade is -3.5% and the final grade is +0.5%. Determine the stationing And elevation of the PVI, the PVT, and the lowest point on the curve.