Siendo el rumbo de la corriente=200º, la intensidad horaria de la corriente 3,5 nudos y estando situados en la oposición del Faro de Punta Paloma y el Faro de Punta Malabata y a 6,4 millas del Faro de Punta Paloma, ponemos rumbo para pasar a 7,1´ del Faro de Cabo Espartel. Hallar el rumbo de aguja y la velocidad efectiva, si la corrección total=3,5º(+) y la velocidad de máquinas=11,2 nudos.In a current setting 200º at 3.5 knots, from the Punta Paloma–Punta Malabata opposition at 6.4 miles from Punta Paloma you head to pass 7.1' off Cape Spartel (Ct = +3.5º, engine speed 11.2 knots). Find Ra and Vef:

Correct answer: d) Rºa = 265º Vef = 12.9'

Why it is correct

At 6.4 miles from Punta Paloma on the Punta Malabata opposition we are at 35º 57.6′ N, 005º 43.9′ W. The tangent to pass 7.1 miles north of Cape Spartel gives effective course Ref = 253º. Solving the current triangle (Rc = 200º, Ihc = 3.5 knots, Vm = 11.2 knots) yields Rv = 268.5º and Vef = 12.9 knots. With Ct = +3.5º, Ra = 268.5º − 3.5º = 265º.

Why the others are not

• a) Does not properly solve the vector triangle for Ref = 253º with Rc = 200º (3.5 knots).
• b) Does not properly solve the vector triangle for Ref = 253º with Rc = 200º (3.5 knots).
• c) Does not properly solve the vector triangle for Ref = 253º with Rc = 200º (3.5 knots).

How to work it out in the exam

1. Plot the start 6.4′ from Paloma on the Malabata opposition and draw the tangent 7.1′ off Spartel: Ref = 253º.
2. Solve the triangle with Rc = 200º (3.5 kn) and Vm = 11.2 kn: Rv = 268.5º (Ra = 265º) and Vef = 12.9 knots.

Where this question comes from

• Ministerio de Transportes (Madrid) · 2020 · Convocatoria de diciembre · PY Test 1 · question 40 Official paper · Official answer key

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