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 =

(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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