Module 10 - Chartwork

Worked CTS: Complete Two-Hour Calculation

A fictional yacht must make good 090°T for two hours at 6 kn through the water. A steady stream sets 180°T at 2 kn. Leeway is ignored for this training example. The boat-distance arc is 2 × 6 = 12 M; tidal drift is 2 × 2 = 4 M south.

To finish on the eastbound ground track, the 12 M water vector needs 4 M of northing. Its eastward component is √(12² − 4²) = √128 = 11.31 M. The true water-track bearing is arctan(11.31 ÷ 4) = 70.5°T, read from the chart to appropriate precision and rounded here to 071°T. Ground distance is 11.31 M in two hours, so predicted SOG is 5.66 kn.

Now apply only the corrections specified. With no leeway in the example, true heading remains 071°T. With 2°W variation, add west to obtain 073°M. With 1°E deviation for the intended heading, subtract east to obtain 072°C. Real navigation uses the applicable place, date and deviation information, then monitors the actual track and position rather than treating the calculation as certainty.

Two hours at 6 kn gives a 12 M boat vector
Two hours of 2 kn set south gives 4 M tidal drift
The required true water track is about 071°T
Expected ground distance is 11.31 M and SOG 5.66 kn
071°T with 2°W variation and 1°E deviation becomes 072°C

Stage	Working	Result
Boat distance	6 kn × 2 h	12 M
Tidal drift	2 kn × 2 h south	4 M south
Water vector	√(12² − 4²)	11.31 M east; 071°T
Ground speed	11.31 M ÷ 2 h	5.66 kn
Compass	071°T + 2°W − 1°E	072°C

Optional quick check

Section 7 of 13

After obtaining 071°T, variation is 2°W and deviation is 1°E. What compass heading follows?

Sources and factual review

Reviewed 20 August 2026 - Compass Revision editorial review.

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