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Siendo Hrb=11:00 obtenemos simultáneamente desde nuestra embarcación demora de aguja del faro de Isla de Tarifa 318º y demora de aguja al faro de Punta Carnero 022º. ¿Cuál será nuestra situación si tenemos una corrección total (CT) = -6º?At 11:00 ship’s time you simultaneously obtain compass bearings Da = 318º to Isla de Tarifa lighthouse and Da = 022º to Punta Carnero lighthouse with total compass error CT = –6º. Find your position:
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Correct answer: d) l = 35º 54.7' N; L = 005º 29.0' W.
Why it is correct
Adding total compass error CT = −6º gives true bearings Dv(Tarifa) = 318º − 6º = 312º and Dv(Punta Carnero) = 022º − 6º = 016º. Plotting their reciprocals (132º and 196º) from each light intersects at l = 35º 54.7′ N, L = 005º 29.0′ W.
Why the others are not
- a) Does not match the charted intersection of Dv(Tarifa) = 312º and Dv(Carnero) = 016º.
- b) Does not match the charted intersection of Dv(Tarifa) = 312º and Dv(Carnero) = 016º.
- c) Does not match the charted intersection of Dv(Tarifa) = 312º and Dv(Carnero) = 016º.
How to work it out in the exam
- Compute Dv(Tarifa) = 318º − 6º = 312º and Dv(Punta Carnero) = 022º − 6º = 016º.
- Plot their reciprocals from both lights: they intersect at 35º 54.7′ N, 005º 29.0′ W.
Where this question comes from
- Ministerio de Transportes (Madrid) · 2021 · Convocatoria de abril · PER Test 2 · question 42 Official paper · Official answer key
- Ministerio de Transportes (Madrid) · 2021 · Convocatoria de abril · PER Test 4 · question 43 Official paper · Official answer key
Try it: Navigation calculations
Pending instructor review. Machine translation, not reviewed.