STEP 1: Prepare a table as shown in figure 11-18.In this figure, column 1 shows the stations; column 2,the elevations on tangent; column 3, the ratio of x/l;column 4, the ratio of (M)*; column 5, the verticaloffsets [(x/l)*(e)]; column 6, the grade elevations onthe curve; column 7, the first difference; and column8, the second difference.STEP 2: Compute the elevations and set thestations on the PVC and the PVT.Knowing both the gradients at the PVC and PVTand the elevation and station at the PVI, you cancompute the elevations and set the stations on the PVCand the PVT. The gradient (g_{1}) of the tangent at thePVC is given as +9 percent. This means a rise inelevation of 9 feet for every 100 feet of horizontaldistance. Since L is 400.00 feet and the curve issymmetrical, l_{1 }equals l_{2 }equals 200.00 feet; therefore,there will be a difference of 9 x 2, or 18, feet betweenthe elevation at the PVI and the elevation at the PVC.The elevation at the PVI in this problem is given as239.12 feet; therefore, the elevation at the PVC is239.12 – 18 = 221.12 feet.Calculate the elevation at the PVT in a similarmanner. The gradient (g_{2}) of the tangent at the PVT isgiven as –7 percent. This means a drop in elevation of7 feet for every 100 feet of horizontal distance. Sincel_{1 }equals l_{2 }equals 200 feet, there will be a differenceof 7 x 2, or 14, feet between the elevation at the PVIand the elevation at the PVT. The elevation at the PVItherefore is239.12 – 14 = 225,12 feet.In setting stations on a vertical curve, rememberthat the length of the curve (L) is always measured asa horizontal distance. The half-length of the curve isthe horizontal distance from the PVI to the PVC. Inthis problem, l_{1 }equals 200 feet. That is equivalent totwo 100-foot stations and may be expressed as 2 + 00.Thus the station at the PVC is30 + 00 minus 2 + 00, or 28 + 00.The station at the PVT is30 + 00 plus 2 + 00, or 32 + 00.List the stations under column 1.STEP 3: Calculate the elevations at each 50-footstation on the tangent.From Step 2, you know there is a 9-foot rise inelevation for every 100 feet of horizontal distancefrom the PVC to the PVI. Thus, for every 50 feet ofhorizontal distance, there will be a rise of 4.50 feetin elevation. The elevation on the tangent at station28 + 50 is221.12 + 4.50 = 225.62 feet.The elevation on the tangent at station 29 + 00 is225.62 + 4.50 = 230.12 feet.Figure 11-18.—Table of computations of elevations on a symmetrical vertical curve.11-16