Al ser HRB = 12h 12m nos encontramos al Sur verdadero del faro de Punta Europa y al Este verdadero del faro de Punta Carnero. Situados, damos rumbo al Puerto de Ceuta (luz verde de la bocana de entrada) con velocidad del buque 6 nudos, teniendo en cuenta que tenemos corriente de Rc = SW (Suroeste) e Intensidad horaria = 2 millas. Calcular el rumbo de aguja y la HRB de llegada a Ceuta. Declinación magnética = 3º NE y desvío de la aguja = –9º (menos).At HRB 12h 12m when due South true of Punta Europa and due East true of Punta Carnero, you head for Ceuta green entrance light at 6 knots in a SW current of Ihc = 2 knots (variation 3º NE, deviation –9º). Find Ra and arrival HRB:

Correct answer: b) Ra = 161º, HRB = 13h 47m

Why it is correct

The start (S of Europa, E of Carnero) is 36º 04.3′ N, 005º 20.7′ W; to Ceuta green light (35º 53.8′ N, 005º 18.6′ W) is 10.6 miles along Ref = 171º. With a SW current (225º, 2 knots) and Vb = 6 knots, the current triangle gives Rv = 155º and Vef = 6.7 knots. With Ct = +3º − 9º = −6º, Ra = 155º + 6º = 161º. Time is 10.6 / 6.7 = 1h 35m ⇒ arrival at 12h 12m + 1h 35m = 13h 47m.

Why the others are not

• a) Does not add the 6º negative compass error or computes arrival time with 6 knots instead of Vef = 6.7 knots.
• c) Does not add the 6º negative compass error or computes arrival time with 6 knots instead of Vef = 6.7 knots.
• d) Does not add the 6º negative compass error or computes arrival time with 6 knots instead of Vef = 6.7 knots.

How to work it out in the exam

1. Plot Ref = 171º (10.6′) to Ceuta and solve the triangle with Rc = 225º (2 kn) and Vb = 6 kn ⇒ Rv = 155º and Vef = 6.7 kn.
2. Convert to Ra = 155º + 6º = 161º and add t = 1h 35m to 12h 12m: 13h 47m.

Where this question comes from

• Junta de Andalucía · 2025 · 2ª convocatoria · PY Módulo de navegación · question 13 Official paper · Official answer key

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