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El 28 de Abril de 2024 el yate PUNTA PRIMA zarpa de las coordenadas GPS l: 38º20 `N L:000º 22`W a las coordenadas l: 38º 38’ N L:001º 21`E. Calcular analíticamente rumbo directo y distancia navegada.Sailing from GPS 38º 20' N, 000º 22' W to 38º 38' N, 001º 21' E on 28 April 2024, the yacht PUNTA PRIMA computes analytically the direct course and distance:
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Correct answer: d) R = 077.4º, Distance = 82.5 miles
Why it is correct
ΔLat = 38º 38′ N − 38º 20′ N = +18.0′ N. Crossing the prime meridian from W to E, the longitudes add: ΔLong = 22′ + 81′ = 103.0′ E. With mean latitude 38º 29′ N, departure is A = 103.0 × cos 38º 29′ = 80.63′ E; tan R = 80.63 / 18.0 ⇒ R = 077.4º and d = √(18.0² + 80.63²) = 82.6 ≈ 82.5 miles.
Why the others are not
- a) Subtracts the longitudes instead of adding them when crossing the Greenwich meridian from W to E.
- b) Subtracts the longitudes instead of adding them when crossing the Greenwich meridian from W to E.
- c) Subtracts the longitudes instead of adding them when crossing the Greenwich meridian from W to E.
How to work it out in the exam
- Find ΔLat = +18′ N and add the opposite-sign longitudes: ΔLong = 22′ + 81′ = 103′ E.
- Compute departure A = 103′ · cos 38º 29′ = 80.6′ E, tan R = A / ΔLat (R = 077.4º) and d = √(ΔLat² + A²) = 82.5 miles.
Where this question comes from
- Región de Murcia · 2025 · 1ª convocatoria · PY Tipo 1 · question 39 Official paper · Official answer key
Try it: Navigation calculations
Pending instructor review. Machine translation, not reviewed.