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A Hrb=11:00, situados 5 millas al NW verdadero del Faro de Cabo Espartel, encontrándonos en zona de corriente de Rc= 045º e intensidad horaria de la corriente Ihc= 3 nudos, navegamos al Ra= 060º con velocidad de máquina de 12 nudos. La declinación magnética es 2º(W) y el desvío 2º(E). Determinar el rumbo efectivo y la velocidad efectiva.At 11:00 ship’s time, 5 miles NW true of Cape Spartel in a current setting Rc = 045º at Ihc = 3 knots, you steer Ra = 060º at 12 knots engine speed (dm = 2º W, deviation = 2º E). What are the course and speed made good?
Show the answer
Correct answer: d) Ref = 057º; Vef = 14.8 knots.
Why it is correct
Total compass error is Ct = (−2º) + (+2º) = 0º, so true heading is Rv = 060º at 12 knots. Adding the current vector (Rc = 045º, Ihc = 3 knots) gives course made good Ref = 057º and speed made good Vef = 14.8 knots.
Why the others are not
- a) Does not match the vector sum of the ship vector (060º, 12 kn) and the current (045º, 3 kn).
- b) Does not match the vector sum of the ship vector (060º, 12 kn) and the current (045º, 3 kn).
- c) Does not match the vector sum of the ship vector (060º, 12 kn) and the current (045º, 3 kn).
How to work it out in the exam
- Find total compass error Ct = dm + Δ = 0º, so Rv = 060º.
- Add the ship vector (060º, 12 knots) and the current vector (045º, 3 knots).
Where this question comes from
- Ministerio de Transportes (Madrid) · 2025 · Convocatoria de noviembre · PY Test 1 · question 38 Official paper · Official answer key
- Ministerio de Transportes (Madrid) · 2025 · Convocatoria de noviembre · PY Test 2 · question 32 Official paper · Official answer key
Try it: Navigation calculations
Pending instructor review. Machine translation, not reviewed.