Module 10 - Chartwork
Worked CTS: Complete Two-Hour Calculation
Assume a required ground track of 090°T for two hours, a boat speed of 6 kn through the water, and a steady tidal stream setting 180°T at 2 kn. Ignore leeway for this training example. The boat travels 12 NM through the water, while the tide carries it 4 NM south. The water vector must therefore contain 4 NM of northing to cancel the tide. Its eastward component is √(12² − 4²) = 11.31 NM, so the true course to steer is arctan(11.31 ÷ 4) = 70.5°T, plotted and rounded to 071°T.
The resulting ground distance is 11.31 NM east in two hours, so the planned speed over ground is 11.31 ÷ 2 = 5.66 kn. Now convert the plotted true course using the variation for the charted place and date and the deviation for the intended heading. With 2°W variation, add west: 071°T becomes 073°M. With 1°E deviation, subtract east: 073°M becomes 072°C. The planned compass course is therefore 072°C.
A chart construction will normally be read to the precision supported by the plot, so do not imply false decimal accuracy. Label the time, boat speed, tide source, variation year and correction, deviation, and the no-leeway assumption. Under way, compare the observed track and fixes with the plan and revise it when the assumptions change.
- Two hours at 6 kn gives a 12 NM water vector
- A 180°T stream at 2 kn for two hours gives 4 NM south
- Cancel that drift with 4 NM northing: CTS 70.5°T, rounded 071°T
- Ground distance 11.31 NM gives a planned SOG of 5.66 kn
- 071°T with 2°W variation and 1°E deviation becomes 072°C
- State assumptions and monitor the actual track rather than treating CTS as certainty
Continue studying Chartwork
This public lesson is part of Revision Module 10. Full revision access adds the guided Learn, Practise and Test flow, flashcards and revision progress.