The elevation on the tangent at station 29230.12 + 4.50 = 234.62 feet.The elevation on the tangent at station 30234.62 + 4.50 = 239.12 feet.+ 50 is+ 00 isIn this problem, to find the elevation on the tan-gent at any 50-foot station starting at the PVC, add4.50 to the elevation at the preceding station until youreach the PVI. At this point use a slightly differentmethod to calculate elevations because the curveslopes downward toward the PVT. Think of the eleva-tions as being divided into two groups—one grouprunning from the PVC to the PVI; the other grouprunning from the PVT to the PVI.Going downhill on a gradient of –7 percent fromthe PVI to the PVT, there will be a drop of 3.50 feetfor every 50 feet of horizontal distance. To find theelevations at stations between the PVI to the PVT inthis particular problem, subtract 3.50 from the eleva-tion at the preceding station. The elevation on thetangent at station 30 + 50 is239.12-3.50, or 235.62 feet.The elevation on the tangent at station 31 + 50 is235.62-3.50, or 232.12 feet.The elevation on the tangent at station 31 + 50 is232.12-3.50, or 228.62 feet.The elevation on the tangent at station 32+00 (PVT) is228.62-3.50, or 225.12 feet,The last subtraction provides a check on the work youhave finished. List the computed elevations under col-umn 2.STEP 4: Calculate (e), the middle vertical offsetat the PVI.First, find the (G), the algebraic difference of thegradients using the formulaG = g_{2}– g_{1}G = -7 –(+9)G = –16%The middle vertical offset (e) is calculated as follows:e = LG/8 = [(4)(–16) ]/8 = -8.00 feet.The negative sign indicates e is to be subtracted fromthe PVI.STEP 5: Compute the vertical offsets at each50-foot station, using the formula (x/l)^{2}e. To findthe vertical offset at any point on a vertical curve,first find the ratio x/l; then square it and multiplyby e; for example,x/l = 50/200 = 1/4.at station 28 + 50, the ratio ofTherefore, the vertical offset is(1/4)^{2 }e = (1/16) e.The vertical offset at station 28 + 50 equals(1/16)(–8) = –0.50 foot.Repeat this procedure to find the vertical offset ateach of the 50-foot stations. List the results undercolumns 3, 4, and 5.STEP 6: Compute the grade elevation at each ofthe 50-foot stations.When the curve is on a crest, the sign of the offsetwill be negative; therefore, subtract the vertical offset(the figure in column 5) from the elevation on thetangent (the figure in column 2); for example, thegrade elevation at station 29 + 50 is234.62 – 4.50 = 230.12 feet.Obtain the grade elevation at each of the stations in asimilar manner. Enter the results under column 6.Note: When the curve is in a dip, the sign will bepositive; therefore, you will add the vertical offset(the figure in column 5) to the elevation on the tangent(the figure in column 2).STEP 7: Find the turning point on the verticalcurve.When the curve is on a crest, the turning point isthe highest point on the curve. When the curve is in adip, the turning point is the lowest point on the curve.The turning point will be directly above or below thePVI only when both tangents have the same percent ofslope (ignoring the algebraic sign); otherwise, theturning point will be on the same side of the curve asthe tangent with the least percent of slope.The horizontal location of the turning point iseither measured from the PVC if the tangent with thelesser slope begins there or from the PVT if the tangentwith the lesser slope ends there. The horizontal loca-tion is found by the formula:Where:x_{t}= distance of turning point from PVC or PVT_{g}= lesser slope (ignoring signs)L = length of curve in stationsG = algebraic difference of slopes.11-17