The relation between ecliptic and equatorial coordinates

**{Note:** If your browser does not distinguish
between "a,b" and "α, β" (the Greek letters
"*alpha, beta*")
then I am afraid you will not be able to make much sense of the
equations on this page.}

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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