We work out the solution from last time, for the problem:

The sidereal time (ST) is given as: 14 h 15 m

Next, we change the Right Ascension (12h 47m) to hour angle using:

h = ST - RA

So: h = 14 h 15 m – 12h 47 m = 1 h 28 m,

This is then converted into degrees, using the fact that there are 15 degrees/ hr.

So: 1h 28m ~ 22.5 degrees (e.g. ~3/ 2h x 15 deg/ h = 22.5 deg)

We then find the zenith distance, z, using an astronomical triangle using sides with hour angle, declination and latitude. This implies the spherical version of the law of cosines, and by analogy with the previous solution we gave, e.g

.http://brane-space.blogspot.com/2011/03/introducing-practical-astronomy.html

we have:

cos z = sin (decl.) sin (Lat) + cos (decl.) cos h cos (Lat)

We note the following respective values:

sin (decl.) = sin (-2.16 deg) = -0.0376

sin (Lat) = sin (37 deg 46 ½' ) = sin 37.77 deg = 0.6124

sin h = sin 22.5 = 0.3826

cos (decl.) = cos (-2.16 deg) = 0.9992

cos h = cos (22.5) = 0.9238

cos(Lat) = cos 37.77 = 0.7904

so, effectively:

cos z = sin (-2.16) sin (37.77) + cos (-2.16) (cos 22.5) (cos 37.77)

cos z = (-0.0376)(0.6124) + (0.9992)(0.9238)(0.7904)

cos z = -0.023 + 0. 7295 = 0.7065

z = arc cos (0.7065) = 45. 0 deg

Then,

a (altitude) = 90 deg - z = 90 deg - 45 deg = 45 degrees

Which shows that we need to look about halfway up above the horizon, toward the zenith.

But which direction?

This requires the azimuth, A:

The appropriate astronomic triangle yields the following equation for A, azimuth:

tan A = -cos(decl.) sin h / [sin(decl.)cos(Lat) - cos(decl) cos h sin (Lat)]

Take the numerator first and compute it:

-cos (-2.16) (sin 22.5) = -(0.9992) (0.3826) = -0.3822

Then the denominator:

sin(-2.16)cos(37.77) - cos(-2.16) cos 22.5 sin (37.77)

= (-0.0376)(0.7904)) - (0.9992)(0.9238)(.6124)

= -0.0297 - 0.5652 = -0.5949

so that:

tan (A) = (-0.3822)/ ( -0.5949)

tan (A)= 0.6424

A = arc tan (0.6424) = 32.7 degrees

This is positive, so this angle is added to the azimuth for the south point of the horizon (180 deg) to get:

Az (Saturn) = 180 degrees + 32. 7 degrees = 212.7 degrees, or very near to the actual value (see star map for time and date) of 212.6 degrees (or

(Click on the sky diagram to enlarge- note the azimuth degrees marked on periphery, i.e. below the horizon)

Of course, one can note from these horizon coordinates, using separate computations, that the date in question is May 16, 2011.

*You are located in San Francisco, and the sidereal time at 11.45 p.m. local time for your location on this date, is 14 h 15 m, approximately. Saturn is visible and is at 12h 47 m Right Ascension, and Declination (- 2 deg 10m).**If your latitude is 37 deg 46 ½' north, find where Saturn will be visible in terms of its altitude and azimuth.*The sidereal time (ST) is given as: 14 h 15 m

Next, we change the Right Ascension (12h 47m) to hour angle using:

h = ST - RA

So: h = 14 h 15 m – 12h 47 m = 1 h 28 m,

This is then converted into degrees, using the fact that there are 15 degrees/ hr.

So: 1h 28m ~ 22.5 degrees (e.g. ~3/ 2h x 15 deg/ h = 22.5 deg)

We then find the zenith distance, z, using an astronomical triangle using sides with hour angle, declination and latitude. This implies the spherical version of the law of cosines, and by analogy with the previous solution we gave, e.g

.http://brane-space.blogspot.com/2011/03/introducing-practical-astronomy.html

we have:

cos z = sin (decl.) sin (Lat) + cos (decl.) cos h cos (Lat)

We note the following respective values:

sin (decl.) = sin (-2.16 deg) = -0.0376

sin (Lat) = sin (37 deg 46 ½' ) = sin 37.77 deg = 0.6124

sin h = sin 22.5 = 0.3826

cos (decl.) = cos (-2.16 deg) = 0.9992

cos h = cos (22.5) = 0.9238

cos(Lat) = cos 37.77 = 0.7904

so, effectively:

cos z = sin (-2.16) sin (37.77) + cos (-2.16) (cos 22.5) (cos 37.77)

cos z = (-0.0376)(0.6124) + (0.9992)(0.9238)(0.7904)

cos z = -0.023 + 0. 7295 = 0.7065

z = arc cos (0.7065) = 45. 0 deg

Then,

a (altitude) = 90 deg - z = 90 deg - 45 deg = 45 degrees

Which shows that we need to look about halfway up above the horizon, toward the zenith.

But which direction?

This requires the azimuth, A:

The appropriate astronomic triangle yields the following equation for A, azimuth:

tan A = -cos(decl.) sin h / [sin(decl.)cos(Lat) - cos(decl) cos h sin (Lat)]

Take the numerator first and compute it:

-cos (-2.16) (sin 22.5) = -(0.9992) (0.3826) = -0.3822

Then the denominator:

sin(-2.16)cos(37.77) - cos(-2.16) cos 22.5 sin (37.77)

= (-0.0376)(0.7904)) - (0.9992)(0.9238)(.6124)

= -0.0297 - 0.5652 = -0.5949

so that:

tan (A) = (-0.3822)/ ( -0.5949)

tan (A)= 0.6424

A = arc tan (0.6424) = 32.7 degrees

This is positive, so this angle is added to the azimuth for the south point of the horizon (180 deg) to get:

Az (Saturn) = 180 degrees + 32. 7 degrees = 212.7 degrees, or very near to the actual value (see star map for time and date) of 212.6 degrees (or

*212 deg 36 mins*., i.e. 32 degrees and 36 mins. west of the south point of the horizon)(Click on the sky diagram to enlarge- note the azimuth degrees marked on periphery, i.e. below the horizon)

Of course, one can note from these horizon coordinates, using separate computations, that the date in question is May 16, 2011.

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