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A Hrb 09:30, navegando al rumbo de aguja 238º, nos encontramos al sur verdadero de Punta Paloma, al tiempo que marcamos el Faro de Punta Cires a 126º por babor, siendo la corrección total= 3º(-). En ese momento, ponemos rumbo hacia un punto equidistante entre los dos faros de la oposición Cabo Espartel – Cabo Trafalgar, siendo la nueva corrección total de 1º(+). A las 12:00 horas, antes de llegar al mencionado punto, cambiamos el rumbo para dirigirnos a un nuevo punto situado a 3 millas náuticas al SSW (202,5º) del Cabo de Trafalgar, siendo la nueva corrección total de 2º(+). La velocidad del buque durante todo el ejercicio es de 5 nudos. Calcular el rumbo de aguja al punto situado a 3 millas al SSW del Cabo Trafalgar y la hora de llegada a ese punto.At 09:30 on Ra = 238º (Ct = –3º) due South of Punta Paloma with Punta Cires bearing 126º Port, you head at 5 knots towards the midpoint of the Spartel–Trafalgar opposition (Ct = +1º) until 12:00, then head for 3' SSW of Trafalgar (Ct = +2º). Find final Ra and ETA:

Show the answer

Correct answer: d) Ra = 333.5º ; Hrb: 13:59.

Why it is correct

At 09:30, Rv = 238º − 3º = 235º and Dv(Cires) = 235º − 126º = 109º; on the meridian of Punta Paloma (005º 43.2′ W) this fixes the vessel at 35º 58.0′ N, 005º 43.2′ W. Running 2.5 h at 5 knots (12.5 miles) towards the Spartel–Trafalgar midpoint reaches the 12:00 position from which the point 3′ SSW (202.5º) of Trafalgar bears Rv = 335.5º (Ra = 335.5º − 2º = 333.5º) at 9.9 miles (1h 59m ⇒ 13:59).

Why the others are not

  • a) Does not match final compass course Ra = 333.5º and the 9.9-mile (1h 59m) second leg.
  • b) Does not match final compass course Ra = 333.5º and the 9.9-mile (1h 59m) second leg.
  • c) Does not match final compass course Ra = 333.5º and the 9.9-mile (1h 59m) second leg.

How to work it out in the exam

  1. Fix the start due S of Paloma with Dv(Cires) = 109º and run 12.5 miles towards the Spartel–Trafalgar midpoint (12:00).
  2. From the 12:00 position measure to 3′ along 202.5º from Trafalgar: Rv = 335.5º (Ra = 333.5º) and d = 9.9′ (13:59).

Where this question comes from

Try it: Navigation calculations

Pending instructor review. Machine translation, not reviewed.