Positional Astronomy:
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.}

diagramDraw 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 +1631'.
What are its ecliptic coordinates?

Click here for the answer.



Previous section: Ecliptic coordinates
Next section: The Sun's motion, and its effect on timekeeping
Return to index