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En el Estrecho de Gibraltar, navegando con Rumbo de aguja, Ra=270º, velocidad, v=12’, desvío=-2º y declinación magnética, dm=3ºNW.  A Hrb=1000 se toma demora de aguja del Faro de la isla de Tarifa, Da=065º.  Se sigue navegando en las mismas condiciones y a Hrb=1030, se toma demora de aguja del Faro de Punta de Gracia, Da=345º. - Se pide situación a Hrb=1030In the Strait of Gibraltar on Ra = 270º at v = 12 knots (deviation –2º, variation 3º NW), you take Da Isla de Tarifa = 065º at Hrb 1000 and Da Punta de Gracia = 345º at Hrb 1030. Find your position at Hrb 1030:

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Correct answer: c) l = 35º 58.8' N – L = 005º 45.5' W

Why it is correct

With Ct = −3º − 2º = −5º, true course is Rv = 270º − 5º = 265º, the first true bearing to Tarifa is Dv₁ = 065º − 5º = 060º and the second to Punta de Gracia is Dv₂ = 345º − 5º = 340º. In 30 minutes (10:00 to 10:30) at 12 knots you run 6 miles along 265º; advancing Dv₁ by 6 miles along 265º to intersect Dv₂ gives l = 35º 58.8′ N, L = 005º 45.5′ W.

Why the others are not

  • a) Does not advance the Tarifa bearing 6 miles along Rv = 265º to intersect the Punta de Gracia bearing.
  • b) Does not advance the Tarifa bearing 6 miles along Rv = 265º to intersect the Punta de Gracia bearing.
  • d) Does not advance the Tarifa bearing 6 miles along Rv = 265º to intersect the Punta de Gracia bearing.

How to work it out in the exam

  1. Compute Ct = −5º, Rv = 265º, Dv(Tarifa) = 060º and Dv(Gracia) = 340º.
  2. Advance the Tarifa bearing 6 miles along 265º and intersect with Gracia: 35º 58.8′ N, 005º 45.5′ W.

Where this question comes from

Try it: Navigation calculations

Pending instructor review. Machine translation, not reviewed.