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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