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A la HRB= 11:15 estamos situados en l=36º 11´N y L=005º 11´W navegando con rumbo de aguja=190º, viento del SE que nos abate 3º y una corriente de intensidad horaria de 3 nudos y rumbo de la corriente” SW. Calcular el Rumbo efectivo y la Velocidad efectiva, si la Corrección Total es de 2,5º (-) y la velocidad de máquinas es de 10 nudos.At 11:15 ship’s time in 36º 11' N, 005º 11' W, you steer Ra = 190º at 10 knots (Ct = −2.5º) with a SE wind giving 3º leeway and a SW current of 3 knots. What are the course and speed made good?
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Correct answer: b) Ref = 199º; Vef = 12.5 knots
Why it is correct
Rv = Ra + Ct = 190º − 2.5º = 187.5º. The SE wind blows on the port side, pushing the vessel 3º to starboard: Rs = 187.5º + 3º = 190.5º (speed 10 knots). Adding the SW current (Rc = 225º, Ihc = 3 knots) yields Ref = 198.3º ≈ 199º and Vef = 12.5 knots.
Why the others are not
- a) Does not match the vector sum of Rs = 190.5º (10 knots) and Rc = 225º (3 knots).
- c) Does not match the vector sum of Rs = 190.5º (10 knots) and Rc = 225º (3 knots).
- d) Does not match the vector sum of Rs = 190.5º (10 knots) and Rc = 225º (3 knots).
How to work it out in the exam
- Convert Ra = 190º to Rv = 187.5º (Ct = −2.5º) and add starboard leeway (+3º) to get Rs = 190.5º.
- Vector-add (190.5º, 10 knots) and the SW current (225º, 3 knots).
Where this question comes from
- Ministerio de Transportes (Madrid) · 2021 · Convocatoria de julio · PY Test 1 · question 38 Official paper · Official answer key
- Ministerio de Transportes (Madrid) · 2021 · Convocatoria de julio · PY Test 2 · question 39 Official paper · Official answer key
Try it: Navigation calculations
Pending instructor review. Machine translation, not reviewed.