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A HRB= 1124 toman demora verdadera y distancia radar del faro de Punta de Gracia N49ºE y 10,8’ millas y se sitúan. En este momento se les para el motor con viento en calma y quedan a merced de la corriente general de la zona que tira al rumbo de corriente= 112º y velocidad de corriente= 2 nudos. Dos horas más tarde (a HRB=1324), vuelve a funcionar el motor y reanudan la navegación y calculan la posición en la carta tras la deriva de 2 horas:At HRB 1124 you fix your position with Dv Punta de Gracia = N49ºE at 10.8 miles, then lose engine power in calm wind and drift for 2 hours in a current setting 112º at 2 knots. Calculate your position at HRB 1324:

Show the answer

Correct answer: a) 35º 56.8' N - 005º 53.8' W

Why it is correct

At 11:24, with Dv(Punta de Gracia) = 049º (reciprocal 229º) and distance 10.8 miles, the fix is 35º 58.3′ N, 005º 58.6′ W. Drifting without engine or wind for 2 hours in a current setting Rc = 112º at 2 knots moves the vessel 4 miles along 112º to 35º 56.8′ N, 005º 53.8′ W.

Why the others are not

  • b) Does not lay off the 4 miles of drift along 112º from the initial fix 10.8′ on 229º from Punta de Gracia.
  • c) Does not lay off the 4 miles of drift along 112º from the initial fix 10.8′ on 229º from Punta de Gracia.
  • d) Does not lay off the 4 miles of drift along 112º from the initial fix 10.8′ on 229º from Punta de Gracia.

How to work it out in the exam

  1. Plot the 11:24 fix on 229º true at 10.8 miles from Punta de Gracia (35º 58.3′ N, 005º 58.6′ W).
  2. Lay off 4 miles (2 h at 2 knots) along Rc = 112º: 35º 56.8′ N, 005º 53.8′ W.

Where this question comes from

Try it: Navigation calculations

Pending instructor review. Machine translation, not reviewed.