The elevation on the tangent at station 29 230.12 + 4.50 = 234.62 feet. The elevation on the tangent at station 30 234.62 + 4.50 = 239.12 feet. + 50 is + 00 is In this problem, to find the elevation on the tan- gent at any 50-foot station starting at the PVC, add 4.50 to the elevation at the preceding station until you reach the PVI. At this point use a slightly different method  to  calculate  elevations  because  the  curve slopes  downward  toward  the  PVT. Think of the eleva- tions  as  being  divided  into  two  groups—one  group running from the  PVC to the PVI; the  other  group running from the  PVT to the PVI. Going downhill on a gradient of –7 percent from the PVI to the PVT, there will be a drop of 3.50 feet for every 50 feet of horizontal distance. To find the elevations at stations between the PVI to the PVT in this  particular  problem,  subtract  3.50  from  the  eleva- tion at the preceding station. The elevation on the tangent at station 30 + 50 is 239.12-3.50, or 235.62 feet. The elevation on the tangent at station 31 + 50 is 235.62-3.50,  or  232.12  feet. The elevation on the tangent at station 31 + 50 is 232.12-3.50,  or  228.62  feet. The elevation on the tangent at station 32+00 (PVT) is 228.62-3.50, or 225.12 feet, The last subtraction provides a check on the work you have finished. List the computed elevations under col- umn 2. STEP 4: Calculate (e), the middle vertical offset at the PVI. First, find the (G), the algebraic difference of the gradients using the formula G = g₂–  g₁ 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 from the PVI. STEP  5:  Compute  the  vertical  offsets  at  each 50-foot  station,  using  the  formula  (x/l)²e. To  find the vertical offset at any point on a vertical curve, first find the ratio  x/l; then square it and multiply by e; for   example, x/l = 50/200 = 1/4. at  station  28  +  50,  the  ratio  of Therefore, 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 at each  of  the  50-foot  stations.  List  the  results  under columns 3, 4, and 5. STEP 6: Compute the grade elevation at each of the  50-foot  stations. When the curve is on a crest, the sign of the offset will  be  negative;  therefore,  subtract  the  vertical  offset (the  figure  in  column  5)  from  the  elevation  on  the tangent  (the  figure  in  column  2);  for  example,  the grade elevation at station 29 + 50 is 234.62 – 4.50 = 230.12 feet. Obtain the grade elevation at each of the stations in a similar manner. Enter the results under column 6. Note: When the curve is in a dip, the sign will be positive;  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 vertical curve. When the curve is on a crest, the turning point is the highest point on the curve. When the curve is in a dip, the turning point is the lowest point on the curve. The turning point will be directly above or below the PVI only when both tangents have the same percent of slope  (ignoring  the  algebraic  sign);  otherwise,  the turning point will be on the same side of the curve as the tangent with the least percent of slope. The  horizontal  location  of  the  turning  point  is either measured from the  PVC if the tangent with the lesser slope begins there or from the  PVT if the tangent with  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 stations G = algebraic difference of slopes. 11-17