الصلة بين الإحداثيات البروجية و الاستوائية

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الصلة بين الاحداثيات البروجية و الاستوائية

{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)

تمرين:

Aldebaran has Right Ascension 4h36m, declination +16°31'.
What are its ecliptic coordinates?

الحل:

First use
sin(β) = sin(δ) cos(ε) - cos(δ) sin(ε) sin(α)
where α = 4h36m = 69.00°, δ = 16.52°, ε = 23.43°
This gives β = -5.45°.

Now use
cos(λ) cos(β) = cos(α) cos(δ)
to obtain λ = 69.81°.

So the ecliptic co-ordinates of Aldebaran are
λ = 69.81°, β = -5.45°.