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Día 17 de Abril de 2021. Navegan por la ortodrómica al rumbo inicial con viento y mar por la aleta de babor y a hora civil de Greenwich (HcG)= 17:30 la situación estimada es 27º 17,6´N – 073º 26,4´W y habiendo obtenido previamente del Almanaque Náutico el horario en Greenwich del Sol (HGO)= 82º 37,8´, calcula el valor del ángulo en el Polo con el Sol ( P ) en ese instante:On 17 April 2021 at Greenwich civil time (HcG) = 17:30, in DR position 27º 17.6' N, 073º 26.4' W, given the Sun's Greenwich hour angle (hGO) = 82º 37.8', calculate the meridian angle (P) to the Sun:
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Correct answer: a) P = 9º 11.4' (W)
Why it is correct
In west longitude, the local hour angle is hL = hG − L = 82º 37.8′ − 73º 26.4′ = 9º 11.4′. Since hL is less than 180º, the Sun has passed the upper meridian and the meridian angle is P = 9º 11.4′ W.
Why the others are not
- b) Labels it East when hG > L(W) means the body is already west of the meridian.
- c) Meridian angle can only be East or West, never North.
- d) Meridian angle can only be East or West, never South.
How to work it out in the exam
- hL = hG − L(W); if 0º < hL < 180º, the meridian angle P is West.
Where this question comes from
- Región de Murcia · 2021 · 1ª convocatoria · CY Tipo 1 · question 14 Official paper · Official answer key
Pending instructor review. Machine translation, not reviewed.