RYA Yachtmaster Tidal Streams & Secondary Ports 2

Question 1

A vessel is heading 060°T at 6 knots through water. The tidal stream is 090°T at 2 knots. Using a vector triangle, the course and speed over ground is approximately:

Accepted Answer

075°T at 6.3 knots (stream adds an easterly component, shifting track right and increasing speed)

Answer

The stream (090°T) pushes the vessel east while it heads northeast. The resultant course is to the right of 060°T and the speed increases. Vector addition gives approximately 075°T at 6.3 knots.

Question 2

An eddy forms on the lee side of a headland during a strong tidal stream. A passage plan should:

Accepted Answer

Consider using the counter-current in the eddy for upwind or uptide passages, hugging the inshore side

Answer

Experienced passage planning uses inshore counter-currents (eddies behind headlands) to gain a favourable stream when the main stream is adverse — a valuable offshore technique.

Question 3

The rule of thumb 'tide over a shoal' states:

Accepted Answer

A vessel should have enough tidal height to pass safely over a charted shoal with the required clearance at the worst state of tide expected

Answer

Passing over shoals requires calculating the minimum depth of water at the time of passage: charted depth + minimum height of tide must exceed draft plus required clearance.

Question 4

How does an anomalous (non-sinusoidal) tidal curve at some UK ports (e.g., Southampton) affect Admiralty method calculations?

Accepted Answer

Ports like Southampton have double high waters — the standard tidal curve does not apply and the port-specific curve must be used

Answer

Southampton and the Solent have double high waters (caused by tidal resonance in the English Channel). The standard Admiralty curve does not apply — the port-specific graphical prediction must be used.

Question 5

A skipper needs to anchor overnight. They calculate the anchor needs to be deployed when tide height = 3.2 m (HW). The range is 4.0 m. By low water, the tide will fall 4.0 m. What depth of water will remain at the anchor at LW if charted depth is 2.5 m?

Accepted Answer

1.7 m — charted depth + LW height = 2.5 + LW height; LW height = 3.2 − 4.0 = −0.8 m, but CD is zero minimum → actual depth = 2.5 − (4.0 − 3.2) = 2.5 − 0.8 = 1.7 m

Answer

At LW, the tide height = HW height − range = 3.2 − 4.0 = −0.8 m (below CD). Depth at LW = charted depth + LW height = 2.5 + (−0.8) = 1.7 m. The area dries by 0.8 m below CD.

RYA Yachtmaster Practice Test

RYA Yachtmaster Practice Test

RYA Yachtmaster Tidal Streams & Secondary Ports 2