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A HRB=10:00, tomamos simultáneamente demora de aguja a Cabo Trafalgar 327º y demora de aguja a Barbate (FI (2) WR 7S 10/7M) 033º. Una vez situados, ponemos rumbo a pasar a 3 millas de Punta Paloma, a una velocidad de máquinas de 10 nudos. Se pide calcular la HRB cuando pasamos al través de Punta Gracia (Torre de Gracia), suponiendo un desvío de aguja de 1º (-) y una declinación magnética en la zona, en el momento de la navegación, de 4º (-).At 10:00 you take bearings Da Cape Trafalgar = 327º and Da Barbate = 033º (deviation –1º, variation –4º), then head at 10 knots to pass 3 miles off Punta Paloma. Find your time when abeam of Punta Gracia (Torre de Gracia):

Show the answer

Correct answer: d) HRB = 10:46.

Why it is correct

With Ct = −4º − 1º = −5º, true bearings are Dv(Trafalgar) = 322º and Dv(Barbate) = 028º, fixing the vessel at 36º 07.3′ N, 005º 58.4′ W. On the course tangent to 3 miles off Punta Paloma (Rv = 123º), abeam of Punta de Gracia (perpendicular bearing 033º) is reached after 7.7 miles (46 min at 10 knots ⇒ 10:46).

Why the others are not

  • a) Does not match the 46-minute run (7.7 miles at 10 knots) to the perpendicular from Punta de Gracia.
  • b) Does not match the 46-minute run (7.7 miles at 10 knots) to the perpendicular from Punta de Gracia.
  • c) Does not match the 46-minute run (7.7 miles at 10 knots) to the perpendicular from Punta de Gracia.

How to work it out in the exam

  1. Fix the start with Dv(Trafalgar) = 322º and Dv(Barbate) = 028º and draw the tangent 3′ off Punta Paloma.
  2. Drop the perpendicular to the track from Punta de Gracia and measure 7.7 miles (46 min at 10 knots ⇒ 10:46).

Where this question comes from

Try it: Navigation calculations

Pending instructor review. Machine translation, not reviewed.