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El día 29 de junio, al ser Hrb = 11:30, navegando a Ra= 208º, Ct= 2º (+), nos encontramos en la oposición Punta Europa – Punta Almina y, simultáneamente, tomamos marcación al Faro de Punta Carnero 88º por estribor. En esta posición damos Ra= 253º, Ct = 1º (+). La velocidad de la embarcación es 8 nudos. Se pide la situación estimada a Hrb = 13:00.On 29 June at 11:30 on Ra = 208º (Ct = +2º) along the Punta Europa–Punta Almina opposition with M = 88º Starboard to Punta Carnero, you alter to Ra = 253º (Ct = +1º) at 8 knots. Find your DR position at 13:00:

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Correct answer: c) l = 35º 58.7' N; L = 005º 33.4' W.

Why it is correct

At 11:30, Rv = 208º + 2º = 210º and Dv(Punta Carnero) = 210º + 88º = 298º, intersecting the Europa–Almina opposition at 36º 02.0′ N, 005º 19.2′ W. Running 1.5 h at 8 knots (12 miles) on the new Rv = 253º + 1º = 254º reaches l = 35º 58.7′ N, L = 005º 33.4′ W.

Why the others are not

  • a) Does not match the position after running 12 miles on Rv = 254º from the Europa–Almina opposition.
  • b) Does not match the position after running 12 miles on Rv = 254º from the Europa–Almina opposition.
  • d) Does not match the position after running 12 miles on Rv = 254º from the Europa–Almina opposition.

How to work it out in the exam

  1. Compute Dv(Carnero) = 210º + 88º = 298º and intersect with the Europa–Almina opposition (36º 02.0′ N, 005º 19.2′ W).
  2. Plot 12 miles (1.5 h at 8 knots) on Rv = 254º to 35º 58.7′ N, 005º 33.4′ W.

Where this question comes from

Try it: Navigation calculations

Pending instructor review. Machine translation, not reviewed.