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A HRB = 10h 20m nos encontramos en situación 35º 55,0´ N, 005º 50,0´ W y damos rumbo a Barbate (faro de tierra), teniendo en cuenta que nos afecta una corriente de Rc = 110º e Intensidad horaria 2 millas. Calcular el rumbo de aguja y la velocidad del buque para llegar a Barbate a HRB = 13h 08m. Declinación magnética = 1,5º NE, desvío de la aguja = –11,5º (menos).At HRB 10h 20m in position 35º 55.0' N, 005º 50.0' W you head for Barbate landfall light in a current setting Rc = 110º at Ihc = 2 knots (variation 1.5º NE, deviation –11.5º) to arrive at HRB 13h 08m. Find Ra and Vb:
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Correct answer: b) Ra = 342°, Vb = 7.2
Why it is correct
From 35º 55.0′ N, 005º 50.0′ W to Barbate (36º 11.3′ N, 005º 55.3′ W) is 16.8 miles along Ref = 345º. To arrive in 2h 48m (2.8 h, from 10h 20m to 13h 08m) effective speed is Vef = 16.8 / 2.8 = 6.0 knots. Offsetting the current (Rc = 110º, Ihc = 2 knots) in the vector triangle gives Rv = 332º and Vb = 7.2 knots. With Ct = +1.5º − 11.5º = −10º, Ra = 332º + 10º = 342º.
Why the others are not
- a) Confuses effective speed (6.0 knots) with ship speed through the water (7.2 knots) or misapplies Ct = −10º.
- c) Confuses effective speed (6.0 knots) with ship speed through the water (7.2 knots) or misapplies Ct = −10º.
- d) Confuses effective speed (6.0 knots) with ship speed through the water (7.2 knots) or misapplies Ct = −10º.
How to work it out in the exam
- Compute Vef = 16.8 / 2.8 = 6.0 knots along Ref = 345º and solve the triangle with Rc = 110º (2 kn) ⇒ Rv = 332º and Vb = 7.2 knots.
- With Ct = +1.5º − 11.5º = −10º, compute Ra = 332º + 10º = 342º.
Where this question comes from
- Junta de Andalucía · 2025 · 2ª convocatoria · PY Módulo de navegación · question 17 Official paper · Official answer key
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Pending instructor review. Machine translation, not reviewed.