This method can be adopted in two ways: (a) cross-sections, and (b) squares.
(a) Using Cross Sections
In this method, a bottom line , center line or profile line is taken into account . Cross sections are taken perpendicular to the present line at regular intervals (say 50 m. 100 m, etc.). After this, points are marked along the cross sections at a daily intervals (say, 5 m. 10 m, etc). A temporary bench-mark is about up near the location . Staff readings are taken along the bottom line and therefore the cross sections. The readings are entered within the level book: the bottom line and therefore the cross sections should even be mentioned. The RL of each points calculated. Then the bottom line and cross sections are plotted to an appropriate scale. Subsequently, the RLs of the respective points are noted on the map, after which the specified contour is drawn by interpolation.
This method is suitable for route survey, when cross sections are taken transverse to the longitudinal section.
(b) Using Squares
In this method, the area is divided into a number of squares. The size of those squares depends upon the character and extent of the bottom . Generally, they have a sides varying from 5 to 20 m. The corners of the squares are numbered serially, as 1, 2, 3….. A temporary bench-mark is set up near the site, and the level is set up at a suitable position. The staff readings on the corners of the squares are taken and noted within the level book maintaining the sequence of the serial numbers of the corners. The RLs of all the corners are calculated. The skeletons of the squares are plotted to an appropriate scale. The respective RLs are noted on the corners, after which the contour lines are drawn by interpolation.
Arithmetical Method
of Interpolation of Contour
The process
of locating the contours proportionately between the plotted points is termed
interpolation. Let A and B be two corners
of the squares. The RL of A is 98.75 m, and that of B 100.75 m. The horizontal
distance between A and B is 10 m.
Horizontal distance between A and B = 10 m
Vertical difference between A and B = 100.75 – 98.75 = 2 m
Let a
contour of 99.00 m be required. Then,
Difference
of level between A and 99.00 m contour = 99.00 - 98.75 = 0.25 m
Now, distance
of 99.00 m contour line from A = (10/2) x 0.25 = 1.25 m
This
calculated distance is plotted to the same scale in which the skeleton was plotted,
to obtain a point of RL of 99.00 m.
Similarly, the other points can be located.
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