--- title: "Meridian distance, DMD and area calculation" subject: "Surveying I" chapter: 9 key: "semester_3/surveying_i:meridian-distance-and-dmd-area" source: https://www.bctengineer.com/q/semester_3/surveying_i/meridian-distance-and-dmd-area site: BCT Engineering --- # Meridian distance, DMD and area calculation Surveying I · Chapter 9: Computation of Area and Volume - Asked in: 2069 Chaitra, 4 marks; 2068 Chaitra, 8 marks (asked 2 times) - Video: [Double meridian distance(DMD)method |Surveying-1 |Computation of area and volume #surveying ](https://www.youtube.com/watch?v=eG_EESLXyiY) — The civil classroom - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_3/surveying_i/meridian-distance-and-dmd-area - Whole chapter: https://www.bctengineer.com/md/semester_3/surveying_i/chapter-9 ## Question What are meridian distance and double meridian distance? Explain how they are calculated and how the area of a traverse is calculated by the double meridian distance method. ## Short answer Meridian distance is the perpendicular distance of a line’s midpoint from the reference meridian. Double meridian distance (DMD) = 2×MD; for the first line DMD = its departure, and for subsequent lines DMD = previous DMD + previous departure + line’s departure. Traverse area = ½ Σ (DMD × latitude), summing signed products. ## Answer #### 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). ```text 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. | Line | Latitude $L$ (m) | Departure $D$ (m) | DMD (m) | DMD × $L$ (m²) | |---|---|---|---|---| | AB | +30 | +80 | 80 | 2400 | | BC | +70 | +20 | 180 | 12600 | | CD | -10 | -80 | 120 | -1200 | | DA | -90 | -20 | 20 | -1800 | | **Sum** | 0 | 0 | | **12000** | 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. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What is the meridian distance of a line? A. Its length B. Its bearing C. Its latitude D. Perpendicular distance of its mid-point from the reference meridian Answer: D. M.D. is the perpendicular distance of the middle point from the reference meridian. (Key point: Meridian distance) 2. What are the D.M.D. of the first line and the last line? A. First: its departure; last: its departure with opposite sign B. First: zero; last: its length C. First: its latitude; last: its latitude D. Both equal to their lengths Answer: A. The first line DMD equals its departure; the last equals its departure with opposite sign, which is a check. (Key point: DMD of first and last lines) 3. How is the D.M.D. of a line found from the previous line? A. Previous DMD - own latitude B. Previous DMD + previous departure + own departure C. Previous DMD + previous latitude D. Twice previous departure Answer: B. DMD = DMD of previous line + departure of previous line + departure of the line, with signs. (Key point: DMD of any line) 4. What is the area of the traverse by the DMD method? A. Sum of (DMD x departure) B. Sum of latitudes C. Half the sum of (DMD x latitude) D. DMD x length Answer: C. Area = 1/2 sum(DMD x latitude), using signs; the numerical value is the area. 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