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Navegando al rumbo de aguja Ra=041º, corrección total Ct= -6º y velocidad v=12 nudos. A Hrb=0900 se toma demora de aguja al Faro de cabo Espartel Da= 076º. A Hrb=0930, se toma demora de aguja al Faro de cabo Espartel Da=196º. - Se pide situación a Hrb=0930Steering Ra = 041º with Ct = –6º at speed v = 12 knots, you take Da Cape Spartel = 076º at Hrb 0900 and Da Cape Spartel = 196º at Hrb 0930. Find your position at Hrb 0930:
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Correct answer: d) l = 35º 51.5' N – L = 005º 54.5' W
Why it is correct
With Ct = −6º, true course is Rv = 041º − 6º = 035º, the first true bearing to Spartel is Dv₁ = 076º − 6º = 070º and the second is Dv₂ = 196º − 6º = 190º. In 30 minutes (09:00 to 09:30) at 12 knots you run 6 miles along 035º; advancing Dv₁ by 6 miles along 035º to intersect Dv₂ gives l = 35º 51.5′ N, L = 005º 54.5′ W.
Why the others are not
- a) Does not advance the bearing Dv₁ = 070º by 6 miles along Rv = 035º to intersect Dv₂ = 190º.
- b) Does not advance the bearing Dv₁ = 070º by 6 miles along Rv = 035º to intersect Dv₂ = 190º.
- c) Does not advance the bearing Dv₁ = 070º by 6 miles along Rv = 035º to intersect Dv₂ = 190º.
How to work it out in the exam
- Compute Rv = 035º, Dv₁ = 070º, Dv₂ = 190º and distance run d = 6 miles in 30 min.
- Advance the first bearing line 6 miles along 035º to intersect the second: 35º 51.5′ N, 005º 54.5′ W.
Where this question comes from
- Región de Murcia · 2026 · 2ª convocatoria · PY Tipo 1 · question 33 Official paper · Official answer key
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Pending instructor review. Machine translation, not reviewed.