Situados el 13 de marzo de 2023 a Hrb 14:15 en posición 36° 13,0´N; 005° 13,0´W nos encontramos en zona de corriente de dirección N20W e Ih = 2,4 nudos, y ha comenzado a soplar un viento en la zona de componente Sur que nos abate 2°. Nuestro rumbo de aguja es de 215°, el desvío es de 1° (+) y la declinación magnética es 1°W. Calcular nuestra velocidad de máquinas más aproximada para estar en el Faro de Punta Europa a Hrb 14:45.On 13 March 2023 at 14:15 in 36° 13.0' N, 005° 13.0' W in a current setting N20W at 2.4 knots with a South wind causing 2° leeway (Ra = 215°, dev +1°, dm 1° W), find the engine speed needed to reach Punta Europa at 14:45:

Correct answer: a) 18.5 knots.

Why it is correct

From 36º 13.0′ N, 005º 13.0′ W to Punta Europa lighthouse (36º 06.5′ N, 005º 20.7′ W) is a direct distance of 9.0 miles (Ref = 223º). Arriving in 30 minutes (14:15 to 14:45) requires Vef = 18.0 knots; offsetting the opposing component of the current (Rc = 340º, Ihc = 2.4 knots) along Rs = 217º yields engine speed Vm ≈ 18.5 knots.

Why the others are not

• b) Fails to account for covering the 9 miles in half an hour (Vef = 18 knots) against the current.
• c) Fails to account for covering the 9 miles in half an hour (Vef = 18 knots) against the current.
• d) Fails to account for covering the 9 miles in half an hour (Vef = 18 knots) against the current.

How to work it out in the exam

1. Measure the distance from 36º 13.0′ N, 005º 13.0′ W to Punta Europa: 9.0 miles in 0.5 h ⇒ Vef = 18.0 knots.
2. Solve the vector triangle with Rc = 340º and Ihc = 2.4 knots: Vm = 18.5 knots.

Where this question comes from

• Ministerio de Transportes (Madrid) · 2023 · Convocatoria de abril · PY Test 1 · question 31 Official paper · Official answer key
• Ministerio de Transportes (Madrid) · 2023 · Convocatoria de abril · PY Test 2 · question 34 Official paper · Official answer key

Try it: Navigation calculations

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