Spherical Astronomy Revisited (2) - Matrix Methods

In spherical astronomy one can also use matrix methods to obtain horizontal coordinates for celestial objects by using the so-called "astronmical triangle". (see image). 

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 Miami, Florida and the sidereal time = 9 h 13 m at 9.30 p.m.  local time for this date,  approximately. Saturn is visible and is at 13 h 44 m Right Ascension, and at a Declination of (- 7 ^o 56’).  If your latitude is 25. ^o75 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 = 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)

Therefore: R3(-180) =

(-1       0      0)
(0       -1      0)
(0        0      1) 

And:  R2(90 ^o - lat.) =

(sin lat.       0         - cos lat.)
(0                 1                   0   )
(cos lat.      0.           sin lat.)

For which we have:

sin (lat.) = sin (25. ^o75)= 0.434 

cos (lat.) = cos (25. ^o75) = 0.900


Thence, 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. ^o93) = -0.138

cos (d) = cos (- 7. ^o93) = 0.990

cos 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:

R3(-180 ^o) R2(90 ^o - lat.) (XYZ(h, d)) =

(-0.385)
(0.785 )
(0.481 )

The last element in the column yields the altitude, so:

a = arc sin(0.481) and a = 28. ^o  75

Meanwhile, 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’.