Conversion between horizontal and equatorial systems

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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.}

To convert between the horizontal and equatorial
coordinates for an object X,

we use a spherical triangle often
called *"The" Astronomical Triangle*: XPZ,

where Z
is the zenith, P is the North Celestial Pole, and X is the object.

The sides of the triangle:

PZ is
the observer's co-latitude = 90°-φ.

ZX
is the zenith distance of X = 90°-a.

PX is
the North Polar Distance of X = 90°-δ.

The angles of the triangle:

The
angle at P is H, the local Hour Angle of X.

The
angle at Z is 360°-A, where A is the azimuth of X.

The
angle at X is q, the **parallactic angle**.

We assume we know the observer’s latitude φ
and the Local Sidereal Time LST.

(LST may be obtained, if
necessary, from Greenwich Sidereal Time and observer’s
longitude.)

**To
convert from equatorial to horizontal coordinates**:

Given RA α and
declination δ, we have

Local
Hour Angle H = LST - RA, in hours;

convert H to degrees (multiply
by 15).

Given H and δ, we
require azimuth A and altitude a.

By the cosine rule:

cos(90°-a) = cos(90°-δ)
cos(90°-φ) + sin(90°-δ)
sin(90°-φ) cos(H)

which
simplifies to:**sin(a) = sin(δ)
sin(φ) + cos(δ)
cos(φ) cos(H)**

This gives us
the altitude a.

By the sine rule:

sin(360°-A)/sin(90°-δ)
= sin(H)/sin(90°-a)

which simplifies to:

- sin(A)/cos(δ)
= sin(H)/cos(a)*i.e. ***sin(A) = - sin(H) cos(δ)
/ cos(a)**

which gives us the
azimuth A.

Alternatively, use the cosine rule again:

cos(90°-δ)
= cos(90°-φ) cos(90°-a) +
sin(90°-φ) sin(90°-a)
cos(360°-A)

which simplifies to

sin(δ)
= sin(φ) sin(a) + cos(φ)
cos(a) cos(A)

Rearrange to find A: **cos(A) = { sin(δ)
- sin(φ) sin(a) } / { cos(φ)
cos(a) }**which again gives us
the azimuth A.

Here
are all the equations together:**H = t -
α****sin(a)
= sin(δ) sin(φ)
+ cos(δ) cos(φ)
cos(H)**

sin(A) = -
sin(H) cos(δ) / cos(a)

cos(A) =
{ sin(δ) - sin(φ)
sin(a) } / { cos(φ) cos(a) }

Now for the inverse problem:**to
convert from horizontal to equatorial coordinates:**

Given φ, a and A, what are α and δ?

Start by using the cosine rule to get δ,
as shown above: **sin(δ) =
sin(a)sin(φ) + cos(a) cos(φ)
cos(A)**

We can now use the sine rule to get H, using the same
formula as above: **sin(H) = - sin(A) cos(a) / cos(δ)
**

Or use the cosine rule instead:

sin(a) = sin(δ)sin(φ)
+ cos(δ) cos(φ)
cos(H)

and rearrange to find H: **cos(H) = { sin(a) - sin(δ)
sin(φ) } / cos(δ)
cos(φ)**

Having calculated H, ascertain the Local Sidereal
Time t.

Then the R.A. follows from

**α
= t – H **.

Here are all the equations together:
**sin(δ)
= sin(a)sin(φ) + cos(a) cos(φ) cos(A)
sin(H) = - sin(A) cos(a) / cos(δ)
cos(H) = { sin(a) - sin(δ) sin(φ)} / cos(δ) cos(φ)
α = t – H**

**Exercise:**

Prove that the celestial
equator cuts the horizon at azimuth 90° and 270°,

at any
latitude (except at the North and South Poles).

At what *angle *does the
celestial equator cut the horizon, at latitude φ
?

Click here for the answer.

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Sidereal time

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