The relation between ecliptic and equatorial coordinates

**{Note:** If your browser does not distinguish
between "a,b" and "a,b" (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°-d)
= cos(90°-b) cos(e)
+ sin(90°-b) sin(e)
cos(90°-l)*i.e.*
sin(d) = sin(b)
cos(e) + cos(b)
sin(e) sin(l)

Alternatively, apply the same rule to the other
corner, and get:

cos(90°-b)
= cos(90°-d)
cos(e) +
sin(90-d)
sin(e)
cos(90°+a)*i.e.*
sin(b) = sin(d)
cos(e) - cos(d)
sin(e) sin(a)

Now try applying the sine rule to the same triangle,

sin(90°-b) / sin(90°+a)
= sin(90°-d) / sin(90°-l)*i.e.*
cos(l) cos(b)
= cos(a) cos(d)

Grouping
these three relations together, we have:

sin(d)
= sin(b) cos(e)
+ cos(b) sin(e)
sin(l)

sin(b)
= sin(d) cos(e)
- cos(d) sin(e)
sin(a)

cos(l)
cos(b) = cos(a)
cos(d)

**Exercise:**

Aldebaran has Right Ascension
4h36m, declination +16°31'.

What are its ecliptic coordinates?

Click here for the answer.

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Ecliptic coordinates

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effect on timekeeping

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