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A Hrb= 15:00, situados 5 millas al SE verdadero del Faro de Punta Europa, con velocidad de máquinas Vm= 8 nudos, entramos en zona de corriente con rumbo de la corriente Rc= N39E e intensidad horaria de la corriente Ihc= 2 nudos. Determinar el rumbo de aguja para pasar a 4 millas del Faro de Punta Cires y la velocidad efectiva de la embarcación. La declinación magnética es dm= 3º(W) y el desvío 2º(E).At 15:00, 5 miles SE true of Punta Europa at Vm = 8 knots in a current Rc = N39E at Ihc = 2 knots (variation 3º W, deviation 2º E), find the compass course to pass 4 miles off Punta Cires and your speed made good:
Show the answer
Correct answer: d) Ra = 242º; Vef = 6.2 knots.
Why it is correct
From 5′ SE of Punta Europa (36º 03.0′ N, 005º 16.3′ W), the tangent to pass 4 miles north of Punta Cires is Ref = 248º. Solving the current triangle (Rc = 039º, Ihc = 2 knots, Vm = 8 knots) gives Rv = 241º and Vef = 6.2 knots. With Ct = −3º + 2º = −1º, Ra = 241º − (−1º) = 242º.
Why the others are not
- a) Does not properly solve the current triangle for Ref = 248º with Ct = −1º.
- b) Does not properly solve the current triangle for Ref = 248º with Ct = −1º.
- c) Does not properly solve the current triangle for Ref = 248º with Ct = −1º.
How to work it out in the exam
- Plot 5′ SE of Punta Europa and draw the tangent 4′ off Punta Cires: Ref = 248º.
- Lay off the current vector (039º, 2′) and close with Vm = 8′: Rv = 241º (Ra = 242º) and Vef = 6.2 knots.
Where this question comes from
- Ministerio de Transportes (Madrid) · 2026 · Convocatoria de abril · PY Test 1 · question 38 Official paper · Official answer key
- Ministerio de Transportes (Madrid) · 2026 · Convocatoria de abril · PY Test 2 · question 35 Official paper · Official answer key
Try it: Navigation calculations
Pending instructor review. Machine translation, not reviewed.