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Al ser HRB = 19:00 nos encontramos al noroeste (NW) verdadero del faro de Punta Alcázar y a una distancia de 5 millas. Navegamos al rumbo de aguja = 045º con velocidad del buque 8 nudos, con una corriente de Rc = Sur (S) e intensidad horaria = 3 millas. Declinación magnética = 4º NW, desvío de la aguja = +8º (más). Calcular el rumbo efectivo que hace el barco como consecuencia de la corriente.At 19:00 ship’s time, 5 miles NW true of Punta Alcázar, you steer compass course Ra = 045º at 8 knots in a current setting South (180º) at 3 knots (variation = 4º NW, deviation = +8º). What is the course made good (Ref)?
Show the answer
Correct answer: b) 070º
Why it is correct
Total compass error is Ct = (−4º) + (+8º) = +4º, so true heading is Rv = 045º + 4º = 049º (speed 8 knots). Vector-adding the south-going current (Rc = 180º, Ihc = 3 knots) gives a course made good Ref = 069.6º ≈ 070º.
Why the others are not
- a) Does not match the vector resultant of Rv = 049º (8 knots) and Rc = 180º (3 knots).
- c) Does not match the vector resultant of Rv = 049º (8 knots) and Rc = 180º (3 knots).
- d) Does not match the vector resultant of Rv = 049º (8 knots) and Rc = 180º (3 knots).
How to work it out in the exam
- Find Ct = −4º + 8º = +4º and true heading Rv = 045º + 4º = 049º.
- Vector-add the ship’s run (049º, 8 knots) and the current (180º, 3 knots) to get Ref = 070º.
Where this question comes from
- Junta de Andalucía · 2026 · 1ª convocatoria · PY Módulo de navegación · question 15 Official paper · Official answer key
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Pending instructor review. Machine translation, not reviewed.