We left off last time with the reader invited to use the matrix format for coordinate transformation:

http://brane-space.blogspot.com/2011/03/spherical-astronomy-matrix-methods.html

to solve the Mercury position problem for horizontal coordinates:

http://brane-space.blogspot.com/2011/03/more-spherical-astronomy.html

To briefly recap we set out the order of transformation thus:

(x)

(y)

(z) A,a = R3(-180 deg) R2(90 - lat.) (XYZ(h, decl.))

where : (XYZ(h, decl.)) =

(x)

(y)

(z) h,decl.

And we found:

R3(Θ) =

(cos(Θ)..........sin(Θ)..........0)

(-sin (Θ)......cos(Θ)...........0)

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

= R3(-180) =

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

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

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

At the end, I noted that obtaining R2(Θ) = R2(90 - lat.) is just as easy, if one recalls the basic trig identity:

cos (90 - φ) = sin (φ)

Applying this:

R2(90 - lat.) =

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

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

(cos lat. ......0.........sin lat.)

And for which we have:

sin (lat.) = sin 40 = 0.6427

cos (lat.) = cos 40 = 0.7660

Thence, R2(90 - lat.) =

(0.6427.........0.....-0.7660)

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

(0.7660 .......0.......0.6427)

Meanwhile:

(x)

(y)

(z) h,decl. =

(cos decl. .......cos h)

(cos decl. ......sin h)

(sin decl...............)

where:

sin (decl.) = sin 12.45 = 0.2155

cos (decl.) = cos 12.45 = 0.9764

cos h = cos 90 = 0, and sin h = sin 90 = 1

Therefore:

(cos decl. .......cos h)

(cos decl. ......sin h)

(sin decl...............)

=

(0)

(0.9764)

(0.2155)

So we're now set to perform the matrix multiplication:

R3(-180 deg) R2(90 - lat.) (XYZ(h, decl.)) =

(0.165)

(-0.976)

(0.139)

Thereby the rectilinear coordinates can easily be referenced to curvilinear ones in the horizon system - noting the first and easiest element to find is the altitude, so

a = arc sin(0.139) and a = 8 degrees

Meanwhile, the azimuth A =

arc tan (y/x) = arc tan (-0.976/ 0.165) = -5.91

Therefore: A = arc tan(-5.91) = -80.4 deg

And, since its' negative, we must subtract from 360 degrees:

A= 360 - 80.4 = 279.5 deg

which yields the same result we obtained using the spherical trig triangle!

(Note: the final matrix multiplication to obtain the final column or solution (a, A) parameters is shown in the sketch)

http://brane-space.blogspot.com/2011/03/spherical-astronomy-matrix-methods.html

to solve the Mercury position problem for horizontal coordinates:

http://brane-space.blogspot.com/2011/03/more-spherical-astronomy.html

To briefly recap we set out the order of transformation thus:

(x)

(y)

(z) A,a = R3(-180 deg) R2(90 - lat.) (XYZ(h, decl.))

where : (XYZ(h, decl.)) =

(x)

(y)

(z) h,decl.

And we found:

R3(Θ) =

(cos(Θ)..........sin(Θ)..........0)

(-sin (Θ)......cos(Θ)...........0)

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

= R3(-180) =

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

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

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

At the end, I noted that obtaining R2(Θ) = R2(90 - lat.) is just as easy, if one recalls the basic trig identity:

cos (90 - φ) = sin (φ)

Applying this:

R2(90 - lat.) =

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

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

(cos lat. ......0.........sin lat.)

And for which we have:

sin (lat.) = sin 40 = 0.6427

cos (lat.) = cos 40 = 0.7660

Thence, R2(90 - lat.) =

(0.6427.........0.....-0.7660)

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

(0.7660 .......0.......0.6427)

Meanwhile:

(x)

(y)

(z) h,decl. =

(cos decl. .......cos h)

(cos decl. ......sin h)

(sin decl...............)

where:

sin (decl.) = sin 12.45 = 0.2155

cos (decl.) = cos 12.45 = 0.9764

cos h = cos 90 = 0, and sin h = sin 90 = 1

Therefore:

(cos decl. .......cos h)

(cos decl. ......sin h)

(sin decl...............)

=

(0)

(0.9764)

(0.2155)

So we're now set to perform the matrix multiplication:

R3(-180 deg) R2(90 - lat.) (XYZ(h, decl.)) =

(0.165)

(-0.976)

(0.139)

Thereby the rectilinear coordinates can easily be referenced to curvilinear ones in the horizon system - noting the first and easiest element to find is the altitude, so

a = arc sin(0.139) and a = 8 degrees

Meanwhile, the azimuth A =

arc tan (y/x) = arc tan (-0.976/ 0.165) = -5.91

Therefore: A = arc tan(-5.91) = -80.4 deg

And, since its' negative, we must subtract from 360 degrees:

A= 360 - 80.4 = 279.5 deg

which yields the same result we obtained using the spherical trig triangle!

(Note: the final matrix multiplication to obtain the final column or solution (a, A) parameters is shown in the sketch)

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