Draw
the triangle KPX,
where P is the North Celestial Pole,
K is the
north pole of the ecliptic,
and X is the object in question.
Apply the cosine rule:
cos(90°-δ) = cos(90°-β) cos(ε) +
sin(90°-β) sin(ε) cos(90°-λ)
i.e. sin(δ) = sin(β) cos(ε) + cos(β) sin(ε) sin(λ)
Alternatively, apply the same rule to the other
corner, and get:
cos(90°-β) = cos(90°-δ) cos(ε) + sin(90-δ) sin(ε)
cos(90°+α)
i.e. sin(β) = sin(δ) cos(ε) - cos(δ)
sin(ε) sin(α)
Now try applying the sine rule to the same triangle,
sin(90°-β) / sin(90°+α) = sin(90°-δ) /
sin(90°-λ)
i.e. cos(λ) cos(β) = cos(α) cos(δ)
Grouping these three
relations together, we have:
sin(δ) = sin(β) cos(ε) + cos(β) sin(ε) sin(λ)
sin(β) = sin(δ) cos(ε) - cos(δ) sin(ε)
sin(α)
cos(λ) cos(β) = cos(α) cos(δ)
Exercise:
Aldebaran has Right Ascension
4h36m, declination +16°31'.
What are its ecliptic coordinates?
Click here for the answer.
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coordinates
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