The basic principle involves relating the Cartesian coordinates (rectilinear) of a point on the celestial sphere (diagram) to the curvilinear coordinates measured in the primary and secondary reference planes. One has then, for example:

(x)

(y)

(z) u,v =

(cos v .....cos u)

(cos v .....sin u)

(sin v..............)

After conversion the curvilinear coordinates may be calculated according to:

u = arctan (y/x) and v = arcsin (z)

Now, consider conventional orthogonal matrices of 3 x 3 dimensions, given as functions: R1(Θ), R2(Θ) and R3(Θ), to rotate the general system by the angle Θ about axes x, y and z, respectively. Thus we obtain:

R1(Θ) =

(1..........0................0)

(0.....cos(Θ)..... sin(Θ))

(0......-sin(Θ)....cos(Θ))

R2(Θ) =

(cos (Θ)......0........- sin(Θ))

(0................1...............0.. )

(sin(Θ)........0.......cos(Θ) )

R3(Θ) =

(cos(Θ)..........sin(Θ)..........0)

(-sin (Θ)......cos(Θ)...........0)

(0 ..................0..................1)

We now apply this method to the following generic problem:

You are located in

^{ o}56’). If your latitude is 25.

^{ o}75 north, find Saturn’s position in terms of its altitude and azimuth

Solution:

We need to perform the matrix operations in
the specific order:

**(x)**

(y)

(z) A,a=

(y)

(z) A,a

**R3(-180**

^{o}) R2(90 - lat.) (XYZ(h,**d**

**))**

Where:

R3(180) =

(cos(180)..........sin(180)..........0)

(-sin (180)...... cos(180)...........0)

(0 ..................0..................1)

(cos(180)..........sin(180)..........0)

(-sin (180)...... cos(180)...........0)

(0 ..................0..................1)

Therefore: R3(-180) =

(-1 0 0)

(0 -1 0)

(0 0 1)

(-1 0 0)

(0 -1 0)

(0 0 1)

And: R2(90

(sin lat. 0 - cos lat.)

(0 1 0 )

(cos lat. 0. sin lat.)

For which we have:

^{ o}- lat.) =(sin lat. 0 - cos lat.)

(0 1 0 )

(cos lat. 0. sin lat.)

For which we have:

sin (lat.) = sin (25.

cos (lat.) = cos (25.

Thence, R2(90

(0.434 0 -0.900 )

(0 1 0 )

(0.900 0 0.434 )

^{ o}75)= 0.434cos (lat.) = cos (25.

^{ o}75) = 0.900Thence, R2(90

^{ o}- lat.) =(0.434 0 -0.900 )

(0 1 0 )

(0.900 0 0.434 )

**:**

*Finally*(x)

(y)

(z) h, d =

(cos d cos h)

(cos d sin h)

(sin d - )

where:

sin (d) = sin (- 7.

cos (d) = cos (- 7.

cos h = cos (-52

^{ o}93) = -0.138cos (d) = cos (- 7.

^{ o}93) = 0.990cos h = cos (-52

^{ o}.5) = 0.608
sin h = sin (-52

^{ o}.5) = -0.793
(Rem: h =
9 h 13 m – 13 h 44 m = - 3h 31m. There
are 15 degrees/ hr.

So: -3 h 31 m » -52

^{ o}.5 )

Assembling the foregoing into the applicable matrix:

(0.990
0.608)

(0.990 -0.793)

(-0.138 .......... ) =

(0.601)

(-0.785)

(-0.138)

Whence:

(0.990 -0.793)

(-0.138 .......... ) =

(0.601)

(-0.785)

(-0.138)

Whence:

R3(-180

(-0.385)

(0.785 )

(0.481 )

^{ o}) R2(90^{ o}- lat.) (XYZ(h, d)) =(-0.385)

(0.785 )

(0.481 )

The last element in the column yields the

a = arc sin(0.481) and a = 28.

*altitude*, so:a = arc sin(0.481) and a = 28.

^{ o }75Meanwhile, the azimuth A =

arc tan (y/x) = arc tan (0.785/ -0.385) = -2.03

Therefore: A = arc tan(-2.03) = -63.

^{ o}8

And, since its' negative, we

*must subtract*from 360 degrees:

A= 360

^{ o}- 63.

^{ o}8 = 296.

^{ o}2

Further practice problem:

Apply the

*matrix method*for the same location in the example problem and for the same sidereal time – but applied to the case of the planet Mars which is also visible at the same local time but at: RA = 10h 28m, and d = +12^{ o}51’.
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