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
Solution:
We need to perform the matrix operations in
the specific order:
(x)
(y)
(z) A,a = R3(-180o) R2(90 - lat.) (XYZ(h, d))
(y)
(z) A,a = R3(-180o) 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 o - lat.) =
(sin lat. 0 - cos lat.)
(0 1 0 )
(cos lat. 0. sin lat.)
For which we have:
(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 )
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
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:
(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 )
(-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
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’.
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