ASTRONOMY NOTES: “GEOGRAPHICAL ASTRONOMY”
Declination, RA independent of where you are on Earth;
The above pic is not perfect. It looks as if its zenith is at 30 deg. declination, 90-70 =20
which is a discrepancy. (90 – declination for northern horizon limit = your latitude)
This is because declination is referenced to Polaris (which is not seen on the south pole,
but still referenced to it)
and RA is referenced from the point where you are looking horizontally(in same plane) at
the sun if you suddenly moved to a place in orbit where Earth is on March 21.
This is not a time, works as a place, but is actually an orientation. It is a grid that moves
with the Earth that has its 0 hrs always pointed in the same direction.
RA referenced from vernal day position so autumnal night(autumnal night shown below)
is the same absolute direction.
0,6,12,18,24 hrs going counterclockwise(ccw)(24 going back to 0);
The grid keeps same orientation as earth moves around sun. The grid does not turn or
rotate.
R.A.of the SUN and DEC of the center of all of the parallel rays coming from the Sun
referenced from earth(imagine a laser pointer hitting the earth):
RA 0 6 12 18
DEC 0 23.5 0 -23.5
24
0
You can draw a sine wave: sun zenith can only occur
between -23.5 and +23.5 degrees latitude:
The pic above could have been drawn more exactly,
but hopefully you get the idea:
On the x axis is R.A. and the y axis is Dec.
Look at RA =6 hours; the declination
is below 30 deg.(it’s actually +23.5 deg);
declination = your latitude= +23.5 if sun at zenith; it must also be about 12 noon!
For the above pic., notice that this is referenced to the vernal equinox as well, looking
into the Sun, where you should place Sun between Earth and Aquarius in the pic.,
where Earth should be moving counterclockwise about the Sun!!! (summer on left,
winter on right); Notice precession(CW) is opposite direction of Earth’s
revolution(CCW) around Sun.
NOW: This pic below clearly shows the CCW revolution of Earth around Sun.
WHERE we are, more precisely:
STARS: For the most part:
RA stays put because referenced to the direction
of the vernal equinox (1 deg change over 100 years);
There are no stars between the earth and the sun.
All stars are very far away.
DEC referenced to Polaris, so circles on same
LATITUDE or DECLINATION line over the
course of the night (Polaris at zenith only at 90 degrees latitude);
Declination doesn’t change much for stars(change is only due to precession which
takes 25,000 years for a cycle), so for the astrologers out there, for one sign on the
zodiac(27.69 degree span), it takes 1923.08 years to see that age to be represented(13
signs now) (69.44 years is 1 degree of precession, so you only need to buy a new star
chart every 70 years or so;
furthermore,
because referenced to Polaris, all stars just seem to rotate around this “axis” from
the north star(now Polaris) to the southern pole star due to Earth’s rotation ;
however,
altitude changes a lot over the course of the night,
and azimuth changes a tiny bit over the course of the year
Shoot a laser at the globe at different latitudes:
p.s. this is earth from the sun’s perspective in the spring
DECLINATION = LATITUDE only where light hits;
parallel rays from sun hits “same line” as Earth spins
around on axis (think of shooting laser at 1 point on a
rotating ball; this follows circumference)
The lit side of earth is a hemisphere of non-uniform brightness.
Shown below is sun hitting directly (center ray) at -23.5 deg. Latitude.
Notice how south pole has no night. This is Dec. 21st, the winter solstice.
Sun is hitting -23.5 deg. (South) in above pic on Dec.21;
(Northern Hemisphere summer sun hits +23.5 latitude
as Earth rotates on June 21)
ALTITUDE AND AZIMUTH:
In the pic below , the Altitude (at azimuth = 180 degrees) is 
ALTITUDE GENERAL FORMULA:
for altitude (A), there are 2 possibilities:
*1) az=180 degrees(A from southern horizon), A= (90-L +-D) {the most popular}
2) az=0 deg (A from northern horizon), 180-A=(90-L +-D)
declination, D  -23.5<D<+23.5 deg
altitude, A 0<A<180 deg from south or north for stars(180 deg span)
Change in Altitude is limited to a 47 deg span for the sun over the course of a year, no
matter where you are on earth, because 23.5- -23.5=47 degrees.
***Important: There are 2 latitudes: zenith latitude and your latitude;
L= your latitude or observational latitude;
D = zenith latitude (-23.5<D<+23.5)
Altitude, A 66.5<A<113.5 deg from south for sun if you’re at the equator(L=0)
90-23.5=66.5 for winter
90+23.5=113.5 for summer
This is a 47 deg span of the sky over the course of a year
You can use same logic for other latitudes, but it’s easier to use the altitude formula.
A= (90-L +D)
At the N pole(L=90), sun goes from A=90-90+23.5= +23.5 from south in summer(light
without setting)
At the N pole(L=90), sun goes from A=90-90+ -23.5=-23.5 from south in winter(dark
because under horizon)
Again, this is less than a 47 deg span of the sky. It is only a 23.5 deg span of the sky
because the sun is invisible below the horizon.
The arctic circle is at 66.5 deg north latitude.
A= 90-66.5+23.5=47 deg from south in summer
A=90-66.5-23.5=0 deg from south in winter; This is right on the horizon!
This is a 47 deg span of the sky
********Givens in bold:*******************
Answer is in italics and underlined:
i.e.
sun is at A = 106.5 degrees(from northern horizon),
and it is the northern hemisphere on June 21st
(Hint: D=23.5 degrees)
180-106.5=(90-L+23.5)
first, bring everything on left to right side and switch signs;
then, bring -L to the left side and switch signs to get:
L= (90+23.5) + (106.5-180)
= 90 + (130-180)
= 90 - 50 = 40 degrees
So, your latitude is 40 degrees and the zenith of the sun
is at 23.5 degrees
----------------------------------------------------------------------------------------If you’re on equator(L=0), and sun at zenith there(A=90 from north or south),
180-90 = 90-0 + D…….or 90=90-0 + D
D = 0 degrees on either equinox
So, your latitude is 0 degrees and the zenith
of the sun is at 0 degrees
1)If it is an equinox 
{equinox(both of them) means sun is highest in sky at 0 deg latitude (equator);
If it is a solstice, sun is highest in sky at 23.5 deg for summer
or -23.5 deg for winter}
D=0 degrees
{because declination is equal to latitude at location where sun at zenith)
A = 90-L + 0
Now,
2)let’s say, you are on north pole; (L=90)
3) so A = 0 degrees from the south
EXPLAINATION
So, your latitude is +90 degrees and the zenith of the sun is at 0 degrees from south
So, sun is parallel to the horizon;
Sun stays at 0 degrees from south for altitude
----------------------------------------------------------------------------------If D=0 still and L=40, A = 130 degrees from the north(A=50 degrees from the south)
So, your latitude is +40 degrees and the zenith of the sun is at 0 degrees(because equinox)
A=90-L+D
50=90-40+D…..D=0 deg!!!
If L=0 (on equator), and A =106.5 from south
106.5 = 90-0 + D
D= +16.5 degrees: this means a zenith is
occurring at +16.5 degrees(it’s between spring and summer
in the northern hemisphere)
So, your latitude is 0 degrees and the zenith of the sun is at +16.5 degrees;
There’s a zenith of the sun north of your location on the equator at +16.5 degrees
If L=0 (on equator), and A =106.5 from north
180-106.5 = 90-0 + D
D= -16.5 degrees: this means a zenith is
occurring at -16.5 degrees(it’s between fall and winter
in the northern hemisphere)
So, your latitude is 0 degrees and the zenith of the sun is at -16.5 degrees;
There’s a zenith of the sun south of your location on the equator at -16.5 degrees
____________________________________________________________________
For the sun, use logic. A=90-L +D from the southern horizon…..so:
The maximum altitude from the southern horizon
if you’re at the equator is A=90-0+23.5= 113.5 deg for summer
A=90+L -D
The maximum altitude from the northern horizon
at the equator is 90+0 - -23.5= 113.5 deg for winter or 180-113.5=66.5 deg from south.
Find azimuth first, then altitude
73.5 N Didson, Russia and Alaska
50 N Vancouver
40 N N.Y.C.
34 N L.A.
23.5 Dhaka, Bangladesh
24 N Key West and Hawaii (tropic of cancer)
At 50 N, A(from northern horizon)=116.5 on june 21, =140 at equinoxes, and 163.5 on dec 21
At 23.5 N, A(from northern horizon)=90 on june 21, =113.5 at equinoxes, and 137 on dec 21
At 0 N, A(from northern horizon)=66.5 on june 21, =90 at equinoxes, and 113.5 on dec 21
Note how the sun is highest in sky at A=90; don’t get confused; 90 is reference
For stars, it’s the same. Let’s look at Polaris. On the N. Pole, Polaris is at A=90 degrees all the time;
Polaris has D=+90 at L=+90: A=90-L +D from the southern horizon; A=90-90+90=90 deg from south
=180-90 =90 deg from the north
In L.A. at 34 deg N latitude, Polaris is 34 degrees from the North all the time;
Polaris has D=+90(because zenith there) but L=+34(observational latitude): A=90-L +D from the
southern horizon; A=90-34+90=146 deg from south =180-146=34 deg from the north
sinA=sinDsinL+ cosDcosLcosH
H is hour angle (15 degrees/hour) based from 0 to 360 degrees where
noon is 0 or 360 degrees,
midnite is 180 degrees 15(12)=180