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A HRB= 1030, navegando al rumbo de aguja = 140º (variación local= 2º W y desvío de aguja= -4º) y velocidad del yate= 8’ (sin viento y sin corriente), toman demora de aguja al faro de Trafalgar= 026º y continúan navegando en las mismas condiciones A HRB= 1115 toman otra demora de aguja a Cº Trafalgar= 340º. Calculan la situación a HRN= 1115At HRB 1030 steering Ra = 140º (variation 2º W, deviation –4º) at 8 knots without wind or current, you take Da Trafalgar = 026º; at HRB 1115 you take Da Trafalgar = 340º. Calculate your position at 1115:
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Correct answer: d) 36º 04.4' N – 005º 58.0' W.
Why it is correct
With Ct = −2º − 4º = −6º, true course is Rv = 140º − 6º = 134º, the first true bearing to Trafalgar is Dv₁ = 026º − 6º = 020º and the second is Dv₂ = 340º − 6º = 334º. In 45 minutes (10:30 to 11:15) at 8 knots you run 6 miles along 134º; advancing Dv₁ by 6 miles along 134º to intersect Dv₂ gives 36º 04.4′ N, 005º 58.0′ W.
Why the others are not
- a) Does not advance the bearing Dv₁ = 020º by 6 miles along Rv = 134º to intersect Dv₂ = 334º.
- b) Does not advance the bearing Dv₁ = 020º by 6 miles along Rv = 134º to intersect Dv₂ = 334º.
- c) Does not advance the bearing Dv₁ = 020º by 6 miles along Rv = 134º to intersect Dv₂ = 334º.
How to work it out in the exam
- Compute Ct = −6º, Rv = 134º, Dv₁ = 020º, Dv₂ = 334º and distance run d = 6 miles in 45 min.
- Advance the first bearing line 6 miles along 134º to intersect the second: 36º 04.4′ N, 005º 58.0′ W.
Where this question comes from
- Región de Murcia · 2021 · 3ª convocatoria · PY Tipo 1 · question 33 Official paper · Official answer key
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Pending instructor review. Machine translation, not reviewed.