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Situados el 13 de marzo de 2025 a Hrb= 08:30, cuatro millas al Norte verdadero del Faro de Punta Almina, nos encontramos en zona de corriente de Rc=N45°E e Ih=2 nudos, navegando al rumbo de aguja Ra=275°. El desvío es 3° E, la declinación magnética es de 1°W y nuestra velocidad de máquinas es de 6 nudos. Determinar el rumbo y velocidad efectivos a Hrb= 09:30, suponiendo que las condiciones de navegación se mantienen estables desde Hrb= 08:30.On 13 March 2025 at 08:30 ship’s time, 4 miles N true of Punta Almina in a current setting Rc = N45ºE at Ih = 2 knots, you steer Ra = 275º (deviation 3º E, variation 1º W, speed 6 knots). What are the course and speed made good?
Show the answer
Correct answer: a) Ref = 296º; Vef = 5.0 knots.
Why it is correct
Total compass error is Ct = (−1º) + (+3º) = +2º, giving true heading Rv = 275º + 2º = 277º at 6 knots. Adding the current vector (045º, 2 knots) to the vessel vector (277º, 6 knots) gives Ref ≈ 296º and Vef = 5.0 knots.
Why the others are not
- b) Does not match the vector composition of Rv = 277º (6 knots) and Rc = 045º (2 knots).
- c) Does not match the vector composition of Rv = 277º (6 knots) and Rc = 045º (2 knots).
- d) Does not match the vector composition of Rv = 277º (6 knots) and Rc = 045º (2 knots).
How to work it out in the exam
- Compute Ct = −1º + 3º = +2º and Rv = Ra + Ct = 277º.
- Combine the ship vector (277º, 6 knots) with the current vector (045º, 2 knots).
Where this question comes from
- Ministerio de Transportes (Madrid) · 2025 · Convocatoria de abril · PY Test 1 · question 35 Official paper · Official answer key
- Ministerio de Transportes (Madrid) · 2025 · Convocatoria de abril · PY Test 2 · question 34 Official paper · Official answer key
Try it: Navigation calculations
Pending instructor review. Machine translation, not reviewed.