vertical distance from your eye level to the ground is 5.7feet. ‘Then with the hand level at your eye and with youstanding on station A, the HI is112.5 + 5.7= 118.2 feet.If a level rod is set up anywhere on the 112.0-footcontour, the reading you would get from station A wouldbe118.2 – 112.0= 6.2 feet.Therefore, to determine the point where the 112.0 footcontour crosses AB, you only need to have the rodmanback out from point A along AB until he comes to thepoint where you read 6.2 feet on the rod. You candetermine the point where the 112.0-foot contourcrosses AD in the same reamer as AB. You can measurethe distance from A to each point and then record thedistance from A to the 112.0-foot contour on AB and AD.When all of the contours have been located on ABand AD, you can shift to station C and carry out the sameprocedure to locate the contours along BC and CD. Youhave now located all the points where contours at al-foot interval intersect the traverse lines. If the slope ofthe ground is uniform (as it is presumed to be in fig.8-13), you can plot the contour lines by simply drawinglines between points of equal elevation, as shown in thatfigure. If there were irregularities in the slope, youwould send the rodman out along one or more lines laidacross the irregular ground, locating the contours onthese lines as you located them on the traverse lines.Grid Coordinate SystemIn the grid coordinate system, the area is laid out insquares of convenient size, and the elevation of eachcomer point is determined. While this method lendsitself to use on relatively level ground, ridge or valleylines must be located by spot elevations taken along thelines. The locations of the desired contours are thendetermined on the ridge and valley lines and on the sidesof the squares by interpolation. This gives a series ofpoints through which the contour lines may be drawnFigure 8-14 illustrates this method. Assume that thesquares here measure 200.0 feet on each side. Points a,b, and c are points on a ridge line, also 200.0 feet apart.You need to locate and draw the 260.0-foot contour line.By inspection, you can see that the 260.0-foot contourmust cross AD since the elevation of A is 255.2 feet andthe elevation of D is 263.3 feet. However, at what pointdoes the 260.0-foot contour cross AD? This can bedetermined by using a proportional equation as follows.Figure 8-14.-Grid system of ground pointsAssume that the slope from A to D is uniform. Thedifference in elevation is 8.1 feet (263.3 – 255.2) for200.0 feet. The difference in elevation between 255.2and 260.0 feet (elevation of the desired contour) is 4.8feet. The distance from A to the point wherethe 260.0-foot contour crosses AD is the value of xin the proportional equation: 8.1:200 = 4.8:x orx = 118.5 feet. Lay off 118.5 feet from A on AD andmake a mark.In the same manner, you locate and mark the pointswhere the 260.0-foot contour crosses BE,EF, EH, andGH. The 260.0-foot contour crosses the ridge,obviously, between point b (elevation 266.1 feet) andpoint c (elevation 258.3 feet). The distance between band c is again 200.0 feet. Therefore, you obtain thelocation of the point of crossing by the same procedurejust described.You now have six plotted points: one on the ridgeline between b and c and the others on AD, BE, EF, EH,and GH. A line sketched by hand through these pointsis the 260.0-foot contour line. Note that the line is, ineffect, the line that would be formed by a horizontalplane that passed through the ridge at an elevation of260.0 feet. Note, too, that a contour line changesdirection at a ridge summit.Control PointsThis explanation illustrates the fact that any contourline may be located by interpolation on a uniform slopebetween two points of known elevation a knowndistance apart. We, also, demonstrated how a ridge lineis located in the same manner.If you locate and plot all the important irregularities8-14in an area (ridges, valleys, and any other points where

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