Interpolation of Contours

Interpolation of the contours is the process of spacing the contours proportionately betn the plotted ground by indirect methods. The methods of interpolation are based on the assumptions that the slope of ground betn 2 pts is uniform.

The chief methods of interpolation are:

By Estimation

This method is very damn rough & is taken for small scale work only. The positions of contour pts betn the guide pts are located by estimation.

By Arithmetic Calculations

The method, though precise, is time consuming. The positions of contour pts betn the guide pts are located by arithmetic calculations.

For eg, Let A, B, D & C be the pts plotted on the map, having elevations of 607.4, 617.3, 612.5 & 604.3 ft resp’ly. Let AB = BD = CD = CA = 1 inch on the plan & let it be reqd to locate the position of 605, 610 & 615 ft contours on these lines. The vertical diff in elevation betn A & B is (617.3 – 607.4) = 9.9 ft. Hence, the dist’s of the contour pts from A will be:

Dist of 610 ft contour pt = 1/9.9 x 2.6 = 0.26″ (approx)

Dist of 615 ft contour pt = 1/9.9 x 7.6 = 0.76″ (approx)

These 2 contour pts may be located on AB. Similarly, the position of the contour pts on the lines AC, CD, BD & also on AD & BC may be located. Contour lines may then be drawn through appropriate contour pts.

By Graphical Method

In this method, the interpolation is done with the help of a tracing paper or a tracing cloth. There are 2 methods:

• 1st method

On a piece of tracing cloth, several lines are drawn llel to each other, say at interval of 0.2m. It reqd, each 5th line may be made heavier to represent each mtr interval. Let the bottom line of the diag, so prepared on the tracing cloth, represent an elevation of 99m & let it be reqd to interpolate contours of 99.5, 100 & 100.5 m values betn 2 pts A & B having elevations of 99.2 & 100.7 resp’ly. Keep the tracing cloth on the line insuch a way that pt A lie on a llel showing an elevation of 99.2 mtr. Now rotate the tracing cloth on drawing in such a way that pt B may lie on a llel showing 100.7 mtr. The pts at which the llel’s showing 99.5 (pt x), 100.0 (pt y) & 100.5 (pt z) may now be picked through the respective positions of the contour pt on the line AB.

• 2nd Method

A line XY of any convenient length is taken on a tracing cloth & ÷ed into several parts, each showing any particular interval, say 0.2m. On a line ⟂er to XY at its mid pt,  a pole O is chosen & radial lines are drawn joining the pole O & the division on the line XY. Let the bottom radial line represent an elevation of 97.0. If reqd, each 5th radial line showing 1 mtr interval may be made dark. Let it be reqd to interpolate contours of 98, 99, 100 & 101 mtr elevation of 97.6 & 101.8 mtr. Arrange the tracing cloth on the line AB in such way that the pt A & B lie at the same time on radial lines showing 97.6 & 101.8 mtr resp’ly. The pts at which radial lines at 98, 99, 100 & 101 may then be picked through.