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Partiendo de la situación conocida latitud: 39º 20' N y Longitud: 016º 45'W se navega a rumbo de aguja Ra=050º y a velocidad de máquina de 15 nudos, la corrección total para ese rumbo es CT= -3º24'. Determínese la situación estimada al cabo de 10 h.From known position 39º 20' N, 016º 45' W you sail on compass course Ra = 050º at 15 knots with total compass error CT = –3º 24'. Determine your dead-reckoning position after 10 hours:
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Correct answer: b) 41º 03.1' N 014º 22.4' W.
Why it is correct
With Rv = 050º − 3.4º = 046.6º and D = 15 × 10 = 150′, Δl = 150 × cos 46.6º = 103.1′ N ⇒ l′ = 41º 03.1′ N. Departure is A = 150 × sin 46.6º = 109.0′ E; with lm = 40º 11.5′ N, ΔL = 109.0 / cos 40.19º = 142.6′ E = 2º 22.6′ E ⇒ L′ = 014º 22.4′ W.
Why the others are not
- a) Does not correctly apply the plane-sailing DR formulas for Rv = 046.6º and D = 150 miles.
- c) Does not correctly apply the plane-sailing DR formulas for Rv = 046.6º and D = 150 miles.
- d) Does not correctly apply the plane-sailing DR formulas for Rv = 046.6º and D = 150 miles.
How to work it out in the exam
- Compute Rv = 046.6º, D = 150′ and Δl = 150 × cos 46.6º = +103.1′ ⇒ l′ = 41º 03.1′ N.
- Find A = 109.0′ E and ΔL = 109.0 / cos 40.19º = 2º 22.6′ E ⇒ L′ = 014º 22.4′ W.
Where this question comes from
- Ministerio de Transportes (Madrid) · 2022 · Convocatoria de abril · PER Test 3 · question 44 Official paper · Official answer key
- Ministerio de Transportes (Madrid) · 2022 · Convocatoria de abril · PER Test 4 · question 42 Official paper · Official answer key
Try it: Navigation calculations
Pending instructor review. Machine translation, not reviewed.