7.2 Coordinate Geometry, Latitudes, and Departures

Resolving a Course into Components

A course defined by a distance and an azimuth resolves into two rectangular components. The latitude is the north-south projection and the departure is the east-west projection. With azimuths measured clockwise from north:

• Latitude = distance x cos(azimuth)
• Departure = distance x sin(azimuth)

The cosine carries the north-south sign and the sine carries the east-west sign automatically when you feed the full azimuth (0 to 360 deg) into the calculator. If you work from quadrant bearings instead, use the bearing angle with cos and sin and assign signs by quadrant: N is +, S is -, E is +, W is -.

Worked example. A course runs 250.00 ft at azimuth 120 deg 00 min (a SE line). Latitude = 250 x cos(120 deg) = 250 x (-0.5000) = -125.00 ft (south). Departure = 250 x sin(120 deg) = 250 x (+0.8660) = +216.51 ft (east). The negative latitude and positive departure correctly place the line in the southeast quadrant.

Accumulating Coordinates

Starting from a known point, each course adds its components to the running totals:

• N(next) = N(prev) + latitude
• E(next) = E(prev) + departure

This is the heart of COGO. A well-organized table has one row per course with distance, azimuth, latitude, departure, ending northing, and ending easting. Carrying that table cleanly prevents almost every traverse error you will meet on the exam.

A Short COGO Table

Start at point A with coordinates N 5000.00, E 5000.00. Two courses follow:

Latitude A-B = 300 cos 45 deg = +212.13; departure A-B = 300 sin 45 deg = +212.13. Add to A to get B at (5212.13, 5212.13). Repeat for B-C to reach C at (5087.13, 5428.64). Every coordinate is now available for inversing, intersection, or area.

The Inverse: Coordinates Back to Direction and Distance

The inverse is the reverse operation, used constantly in COGO. Given two coordinate pairs, recover the line:

• latitude = N2 - N1, departure = E2 - E1
• distance = sqrt(latitude^2 + departure^2)
• azimuth = atan2(departure, latitude), placed in the correct quadrant by the signs of the two components

Example. From A (5000, 5000) to C (5087.13, 5428.64): latitude = +87.13, departure = +428.64. Distance = sqrt(87.13^2 + 428.64^2) = 437.41 ft. Bearing angle = arctan(428.64 / 87.13) = 78.51 deg; both components positive means NE quadrant, so the bearing is N 78 deg 31 min E.

Area by Coordinates

With balanced coordinates, area follows from the coordinate cross-product (shoelace) formula:

Area = 0.5 x | sum of [ N_i x (E_{i+1} - E_{i-1}) ] |

The double meridian distance (DMD) and double parallel distance (DPD) methods give the same area from balanced latitudes and departures and are detailed where area is computed in Section 7.3. Whichever method you use, the latitudes and departures must close (sum to zero) before the area is meaningful.

COGO Intersection and Resection

Beyond inversing, the FS tests two coordinate-geometry constructions that locate a new point from existing control.

Intersection finds the point where two lines meet. In a bearing-bearing intersection you know two control points and the direction of a line through each; the new point lies where those two rays cross. In a distance-distance intersection you know a distance from each of two control points and solve the two circles for their crossing (there are two solutions, so you keep the one on the correct side). A bearing-distance intersection mixes the two.

These reduce to solving two simultaneous equations in the unknown northing and easting, and the FS handbook gives the closed-form expressions; the practical skill is feeding the right azimuths and distances into them.

Resection is the inverse situation: the occupied point is unknown and you observe angles or directions to three or more points of known coordinates, then solve for your own position. The classic three-point resection has a weak geometry warning, the danger circle: if the occupied point falls near the circle passing through the three control points, the solution becomes unstable. Choose control that spreads the angles well.

Putting the Pieces Together

Every boundary, route, and control problem on the FS is some combination of these five moves. Build the coordinate table first, keep the signs honest, and each operation becomes a short, checkable calculation rather than a guess. The exam rewards candidates who can name which operation a word problem is really asking for, then apply the correct formula without conflating a forward computation with an inverse.