El día 1 de febrero de 2023 a Hrb 04:45 nos encontramos en situación 35° 51,0´N; 006° 11,0´W navegando al rumbo de aguja 083°, el desvío es de 2° (+), la declinación magnética será de 1°W y nuestra velocidad de máquinas es de 4,5 nudos. Situados, entramos en una zona de corriente desconocida. A Hrb 07:05 tomamos una demora de aguja al Faro de Punta de Gracia de 035° y simultáneamente otra demora de aguja al Faro de Cabo Espartel de 150°. Determinar el rumbo e intensidad de la corriente.On 1 February 2023 at 04:45 in 35° 51.0' N, 006° 11.0' W steering Ra = 083° (dev +2°, dm 1° W) at 4.5 knots in an unknown current, at 07:05 Da Punta de Gracia = 035° and Da Spartel = 150°. Find the current's set and drift:

Correct answer: c) Rc = 330° / Ihc = 0.70 knots.

Why it is correct

With Ct = −1º + 2º = +1º, Rv = 084º. In 2h 20m (2.333 h) at 4.5 knots you run 10.5 miles along 084º to a DR at 35º 52.1′ N, 005º 58.1′ W. At 07:05, Dv(Gracia) = 036º and Dv(Spartel) = 151º fix the vessel at 35º 53.5′ N, 005º 59.1′ W. From DR to fix is 1.63′ towards 330º ⇒ Ihc = 1.63 / 2.333 = 0.70 knots.

Why the others are not

• a) Does not divide the 1.63-mile drift by the 2.333 hours elapsed or reverses the direction.
• b) Does not divide the 1.63-mile drift by the 2.333 hours elapsed or reverses the direction.
• d) Does not divide the 1.63-mile drift by the 2.333 hours elapsed or reverses the direction.

How to work it out in the exam

1. Compute the 07:05 DR position (10.5′ on Rv = 084º) and observed fix (Dv = 036º and 151º).
2. Measure from DR to fix: Rc = 330º and drift = 1.63′ in 2.333 h ⇒ Ihc = 0.70 knots.

Where this question comes from

• Ministerio de Transportes (Madrid) · 2023 · Convocatoria de abril · PY Test 1 · question 36 Official paper · Official answer key
• Ministerio de Transportes (Madrid) · 2023 · Convocatoria de abril · PY Test 2 · question 33 Official paper · Official answer key

Try it: Navigation calculations

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