In the past instalments we examined how to obtain horizontal coordinates for celestial objects by using the so-called "astronmical triangle". (see image). There is, however, an alternative method which makes use of matrices - mainly of the 3 x 3 variety. In this blog we explore this other method, and show how it can be used.
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)
Now, I invite readers to use the example of the Mercury problem:
http://brane-space.blogspot.com/2011/03/more-spherical-astronomy.html
to apply the matrix methods.
To arrive at the same values (A, a) we will require:
R3(Θ) = R3(-180 deg)
R2(Θ) = R2(90 - lat.)
so that:
(x)
(y)
(z) A,a = R3(-180 deg) R2(90 - lat.) (XYZ(h, decl.))
where : (XYZ(h, decl.)) =
(x)
(y)
(z) h,decl.
To get you started, we have: R3(-180 deg) =
(cos 180........-sin 180................0)
(sin 180........cos 180................0)
(0......................0......................1)
Since we saw: R3(Θ) =
(cos(Θ)..........sin(Θ)..........0)
(-sin (Θ)......cos(Θ)...........0)
(0 ..................0..................1)
Therefore: R3(-180) =
(-1.....0.......0)
(0.......-1.....0)
(0.......0.......1)
Obtaining R2(Θ) = R2(90 - lat.) is just as easy, if one recalls the basic trig identity:
cos (90 - φ) = sin (φ)
Good luck!
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)
Now, I invite readers to use the example of the Mercury problem:
http://brane-space.blogspot.com/2011/03/more-spherical-astronomy.html
to apply the matrix methods.
To arrive at the same values (A, a) we will require:
R3(Θ) = R3(-180 deg)
R2(Θ) = R2(90 - lat.)
so that:
(x)
(y)
(z) A,a = R3(-180 deg) R2(90 - lat.) (XYZ(h, decl.))
where : (XYZ(h, decl.)) =
(x)
(y)
(z) h,decl.
To get you started, we have: R3(-180 deg) =
(cos 180........-sin 180................0)
(sin 180........cos 180................0)
(0......................0......................1)
Since we saw: R3(Θ) =
(cos(Θ)..........sin(Θ)..........0)
(-sin (Θ)......cos(Θ)...........0)
(0 ..................0..................1)
Therefore: R3(-180) =
(-1.....0.......0)
(0.......-1.....0)
(0.......0.......1)
Obtaining R2(Θ) = R2(90 - lat.) is just as easy, if one recalls the basic trig identity:
cos (90 - φ) = sin (φ)
Good luck!
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