Meridian distance, DMD and area calculation

Meridian distance (M.D.)

The meridian distance of a line is the perpendicular distance of the middle point of the line from the reference meridian (a N-S line). It is positive east of the meridian. For convenience, the reference meridian is taken through the most westerly station of the traverse, so that all meridian distances are positive.

Double meridian distance (D.M.D.)

The D.M.D. of a line is twice its meridian distance. It is used because it avoids fractions and halves. For the first line AB, with the reference meridian through A, M.D. = ½ × departure of AB, so D.M.D. = departure of AB.

Calculation of D.M.D.

The latitudes and departures are first adjusted (so that $\Sigma L = 0$ and $\Sigma D = 0$). Then:

1. D.M.D. of the first line = its departure.
2. D.M.D. of any line = D.M.D. of the previous line + departure of the previous line + departure of the line itself (with signs; east +, west -).
3. D.M.D. of the last line = its departure with the opposite sign (a check on the work).

   reference
   meridian        B
      |          /
      |     m  /        M.D. = distance of the
      |<-----*/          mid-point m of AB from
      |    /             the meridian
      |  A

Area of a traverse by the D.M.D. method

The area between each line and the meridian is a trapezium; its area equals the meridian distance of the line (the mean width) multiplied by its latitude, that is, half of D.M.D. × latitude. The sum of these areas, with signs, gives the area of the traverse:

$$\text{Area} = \frac{1}{2}\sum \left(\text{D.M.D.} \times \text{Latitude}\right)$$

(The sum is taken with the signs of the products, north latitude + and south -; its numerical value is the area.)

Example

Consider the closed traverse with stations A (E 0, N 0), B (E 80, N 30), C (E 100, N 100), D (E 20, N 90). The most westerly station is A, so the reference meridian passes through A.

The D.M.D. of the last line DA is 20, equal to the departure of DA (-20) with the sign reversed, which checks the work.

$$\text{Area} = \frac{1}{2}\left|12000\right| = 6000.0\ \text{m}^2$$

This agrees with the area from the coordinates, 6000.0 m².