Worked CTS: Calculation, Plot and Command Brief

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. With 2°W variation, add west: 071°T becomes 073°M. With 1°E deviation, subtract east: 073°M becomes 072°C. State that variation must come from the chart for its position and date and that deviation must come from the vessel's current card for the intended heading.

At Yachtmaster standard, present the answer as a command plan: ‘Steer 072°C initially to make good 090°T; expected SOG 5.7 kn. This assumes 6 kn through the water, tide 180°T at 2 kn, negligible leeway, and the stated compass errors.’ Then identify the controls: cross-track limits, clearing marks, fix interval, depth comparison, and the point at which you will recalculate. The plotted result is never authority to continue towards danger when observed evidence disagrees.

Use an independent sense-check before briefing the course. Because the tide sets south, the water track must point north of east; 071°T satisfies that test, whereas 109°T would worsen the southerly displacement. Vector components also recombine correctly: 4 NM north through the water plus 4 NM south from tide gives zero north-south ground displacement.