I am looking for two mathematical formulas.....any help??????

Question

FORMULA 1: need to calculate the noonday sun angle for given latitude (X) and given date (Y)

FORMULA 2: How to calculate the sun angle in comparison to the equinox and solstice

Answer 1

Instruments for celestial navigation - tools that
allow determination of position from a reference
such as the sun or another star - have
developed from the time of the Ptolemy; the
sextant is the culmination of these tools and has
remained virtually unchanged since its invention
in the early 1700's. The sextant is still used today
because of its accuracy and perhaps because of
its mystique. Gyroscopes and radars are great,
but the sextant functions under all sorts of
conditions and never runs out of electricity. This
article discusses the principles of sextant use
and demonstrates how to determine latitude and
longitude from a noon day sun sighting.
(photo courtesy Defense Mapping Agency)
U.S. Navy mark 2 micrometer drum sextant: Light
follows the red line ABC, and then is reflected into
the eyepiece, D. A reflected image from A is
aligned with one observed directly through the
eyepiece from C. The altitude of the image (angle)
is given on the scale at the bottom of the sextant.
A sextant is an instrument designed to measure the angle of an object with respect to a reference point. In
this case, the angle measured is that of the sun over the horizon at local area noon. This angle, together with
the time of the measurement and the corresponding angles at some reference location, will tell you your
longitude and latitude.
To understand how this measurement works, assume that the earth is a sphere and that its orbit around the
sun is a circle. Also assume that the tilt of the earth with respect to the plane of its orbit is zero degrees
instead of 23 degrees. With these assumptions, one rotation of the earth will be exactly 24 hours and the
sun's altitude at a given time of day will be the same for each day of the year.
If you were to stand on the equator (0 degrees latitude) at your local noon, the angle of the sun to the horizon
would be 90 degrees (sun directly overhead). Local area noon (LAN) is defined as the time when the angle
the sun makes with the horizon is at a maximum (zenith). If you were at the north pole the sun would be on the
horizon all day, or an angle of zero degrees. In this example, your latitude is
LAT 90
H
o
where H
o
is the observed height of the sun at local area noon in degrees
(and minutes).
The longitude calculation is based on the time of your local area noon compared with noon at Greenwich.
Since the ideal earth rotates once in 24 hours, the sun appears to move across the sky with an angular
velocity of
=
.
360 deg
.
24 hr
15
deg
hr
or, alternatively, a time zone per hour. One time zone covers 15
degrees.
Page 2
Take the time difference of your LAN, expressed in hours Greenwich mean time (GMT), and 12.00 hours
GMT. At 12.00 hours GMT, the sun is directly over the Greenwich meridian or 0 degrees of longitude.
Multiply the time difference by the sun's angular velocity to obtain your longitude. Suppose we're in
California, and LAN is at 1 pm (PDT). 1pm Pacific daylight time equals 20:00 GMT.
T
gmt
20
∆TIME
T
gmt
12.00
Time zone:
=
∆TIME
8
LONG
.
∆TIME 15
West is positive, east is negative.
=
LONG
120
To convert more precise readings of angle, given in degrees and minutes, into a single number, we'll
need a function. Also, lets set up a function that converts more precise readings of time into a single
number.
degrees
,
deg min
deg
min
60
hours
,
,h m s
h
m
60
s
3600
Now, if the universe were really as simple as our model, we could just take the angular reading with the
sextant, determine the height of the sun, and figure out our latitude and longitude as described above.
There will be, however, a number of corrections to the reading having to do with the shape of the earth's
orbit, the angle of the earth with respect to the sun, how we take the readings, the accuracy of our
instruments, etc.
Corrections
Geometric errors
First, consider the earth's orbit. The earth's orbit is really elliptical, and its changing orbital velocity
causes the length of a day to be different than 24 hours. To correct for small changes in the length of
the day, the daily page of the Nautical Almanac is used to determine Meridian Passage and Greenwich
hour angle (GHA) of the Sun (with min./sec. correction from the increments and corrections pages).
GHA is the angle the sun makes with a vertical line passing through Greenwich at the approximate time
of your LAN, and therefore gives your longitude, using the method shown above. From the almanac,
for the date of the particular sightings discussed in this article,
predicted time of LAN:
EstTimeLAN
hours
,
,
12 52 44
Date of sightings:
GHA for 1900Z (12 noon) on 4/18/93
is:
date
4 18 1993
GHAhr
degrees
,
105 11.4
Note: Z (or Zulu) is shorthand for GMT.
Page 3
In addition to inaccuracies in time
introduced by the elliptic orbit, the
angle of the reading must be
corrected for the declination angle.
Declination describes the tilt of the
earth's axis to the plane of its orbit.
During the summer solstice (in the
northern hemisphere), the sun is 23
degrees too high at LAN. During
the winter solstice, the sun is 23
degrees too low.
Demonstration of declination
To correct for the tilt, or declination, on any particular day, look up the correction factors on the daily page of
the Almanac.The declination, along with the d correction, is added to the zenith distance to calculate the
latitude. From the almanac, the Sun's declination, DecHr, with a d correction from the Increments and
Corrections table on the day of sighting is
DecHr
degrees
,
11 01.7
d
0.9
change of declination per hour
Later, when a corrected time is calculated, more precise declination corrections can be added
corresponding to the changes per minute and second.
The altitude we wish to determine, the solar altitude, is the angle from the center of the sun to the horizon,
with the sextant, or more precisely, the eye at the apex. Since it is hard to tell where the exact center of the
sun is, the image of either the lower or upper limb (side) of the sun is brought down to the image of the
horizon by moving the mirrors on the sextant. The image of the sun is positioned so that it's edge is just
tangent to the horizon line. This procedure then requires a correction for the solar (limb) altitude: the reading
will be off by the semi-diameter of the sun. The semi-diameter of the sun is approximately 15 minutes of arc.
The A2 table in the almanac contains the semi-diameter corrections for upper and lower limbs of the sun;
these corrections also have built-in factors for the normal refraction of the atmosphere. Since the A2 table
corrects for normal refraction, the table must be entered with the solar altitude along with the time of year. If
there are unusual conditions of air pressure and temperature then the A4 table contributes a small correction.
From the almanac, the A2 and A4 corrections for the lower limb of the sun on 4/18/93 are
A2
degrees ,0 15.8
A4
degrees ,0 0.0
There is one other correction for physical and geometrical
effects: the dip. If the eye is higher than sea-level, then
the visual horizon is below the actual horizon, which is the
tangent plane to the earth at the point you are located.
Page 4
The correction is approximately equal to
dip he
he
he is the eye height in feet and dip is in minutes of arc.
or you can look it up in the almanac. The dip correction is subtracted from the height of the sun given by
the sextant reading. In the present case, the eye height is
Height of eye in feet:
hE
8
=
dip hE
2.828
The value given in the almanac:
dip
0 2.7
units of (degrees minutes)
Instrument errors and corrections
The index correction removes zero error from the sextant. If the error is on the arc then IC is subtracted
from Hs. If its off the arc, then IC is added to Hs.
Index error of sextant
IE
degrees ,0 1.0
Before the start of the sun sightings, the index error of the sextant was adjusted to 0 degrees and 0.0
minutes of arc (see reference 3 or 4 for sextant adjustments). At the end of the sun sightings, the index
error was one minute on the arc . This was most likely due to the temperature change from when the
sextant was taken out of its box in preparation for the noon sun shots and the time the shots were
recorded.
Watch correction from WWV. If the watch is fast, WC is negative. If watch is slow, then WC is positive.
clock error
WC
hours
,
,0 0
1
1 sec fast
Time zone
Zd
8
Calculating Position from Sightings
The actual use of the sextant is an art. Practice in using the sextant is required before consistently
accurate fixes can be obtained; missing the true time of LAN by 4 seconds will throw you off by a mile.
Due to inexperience with the sextant, the author will first examine the raw data by hand. Data points
(sightings) that are obviously bad will be dropped. After this elimination, least squares curve fitting will be
used to determine the time of LAN and the solar altitude at LAN.
Below, the raw sextant data for the noon day sun shot is read from two .prn files. The Times matrix
contains times of the sun shots in hours, minutes, and seconds. The Angles matrix contains the solar
altitude in degrees and minutes. There is a one to one correspondence between the entries in the
Times matrix and the Angles matrix. The data from the first sun shot is place in the first row of the
Times and Angles matrices. To calculate your own sighting, replace the contents of Times and
Angles with the data from your sun shots. Mathcad will reduce them for you. You will also need an
almanac to look up new correction factors.
Times
READPRN times
Angles
READPRN angles
Page 5
The times are entered in (hr min sec) format, while the angles are in (deg min). The functions hours
and degrees defined above can convert this data into a single number. To convert back at the end of
the calculation, the following functions are defined.
ghrs h
floor h
gdeg ang
floor ang
gmin h
floor
.
h
ghrs h
60
gamin ang
.
ang
gdeg ang
60
gsec h
.
.
h
ghrs h
60
gmin h
60
The function dtim is used to convert time in hours to time in hours, minutes and seconds
for display purposes. The function ddeg is used to convert angles in degrees to angles in degrees and
minutes for display purposes.
dtim hours
ghrs hours
gmin hours
gsec hours
ddeg degs
floor degs
gamin degs
Define a range variable i to enumerate all the sun shots. Since all the columns in the Times matrix
and Angles matrix are of the same length, we can use the length of the first column in Times to
calculate the size of the range.
i
..
0 last
< >
Times
0
Compute the vectors hrs and deg. hrs contains the time of the sun shots in hours and
deg contains the corresponding altitudes in degrees.
hrs
i
hours
,
,
Times
,i 0
Times
,i 1
Times
,i 2
deg
i
degrees
,
Angles
,i 0
Angles
,i 1
Now plot the raw sextant altitudes against the times of the shots and see what it looks like.
12.6
12.8
13
13.2
66.5
66.65
66.8
66.95
67.1
deg
i
hrs
i
You should be able to see the trend
toward a curve with a maximum
around 12:30 or 1:00 in the
afternoon. A closer guess can be
determined by looking at the actual
data. Now is also a good opportunity
to remove bad sightings from the
calculation.
Angle of sun vs. time of day
Page 6
Examine the data and determine when local area noon occurs.
i
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
hrs
i
12.656
12.673
12.687
12.702
12.718
12.742
12.761
12.778
12.79
12.814
12.831
12.847
12.861
12.881
12.893
12.91
12.926
12.939
12.957
12.975
12.99
13.007
13.061
13.079
13.095
13.11
13.126
13.138
13.151
13.163
deg
i
66.723
66.98
66.747
66.887
66.985
66.837
66.783
67
66.897
66.867
66.863
66.907
66.857
66.867
67.003
66.893
66.9
66.91
66.823
66.793
66.78
66.817
66.78
66.687
66.693
66.593
66.54
66.613
66.52
66.51
From a study of the raw data we can conclude
that the sightings at i=1,4,7,14 are most likely
in error, because the angles don't fall smoothly in
sequence with the other sightings. These values will be
ignored. A second range, j, will select only the valid sun
shots.
My initial guess at noon is the set of values for
i = 12:
=
dtim hrs
12
12
51
40
=
deg
12
66.857
j
0
2
3
5
6
8
9
10
11
12
13
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
=
ddeg deg
12
66
51.4
The HRS and DEG vectors contain the
data from the sun shots selected byj.
max
last j
=
max
25
ii
..
0 max
HRS
ii
hrs
j
ii
DEG
ii
deg
j
ii
The least squares method can be used to calculate the best fitting curve through the remaining data
points. Since the earth revolves once every 24 hours, approximately, a periodic curve should fit the sun
sighting data very closely. Since we are only modeling a short section of the periodic, a second order fit
should be sufficient.
Use a version of a least squares fit, assuming a function of the form y(x) = b
0
+ b
1
x + b
2
x
2
; matrix operations
can be used to find the fit coefficients for the sun sighting data.
Page 7
Quadratic fit using matrix operations
Create the second variable column (x
2
) and construct the X matrix:
x2
HRS
2
X
,
ii 0
1
< >
X
1
HRS
< >
X
2
x2
b
.
.
T
X X
1
.
T
X DEG
=
b
588.922
102.034
3.969
Fitted curve:
quad x
b
0
.
b
1
x
.
b
2
x
2
Compute mean squared error: this is a check of how well the curve fits the data.
SSE
Q
DEG
quad HRS
2
MSE
Q
SSE
Q
max
2
=
MSE
Q
1.568 10
3
Page 8
ii
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
HRS
ii
12.656
12.687
12.702
12.742
12.761
12.79
12.814
12.831
12.847
12.861
12.881
12.91
12.926
12.939
12.957
12.975
12.99
13.007
13.061
13.079
13.095
13.11
13.126
13.138
13.151
13.163
DEG
ii
66.723
66.747
66.887
66.837
66.783
66.897
66.867
66.863
66.907
66.857
66.867
66.893
66.9
66.91
66.823
66.793
66.78
66.817
66.78
66.687
66.693
66.593
66.54
66.613
66.52
66.51
quad HRS
ii
66.731
66.775
66.795
66.837
66.853
66.871
66.881
66.885
66.887
66.887
66.885
66.876
66.868
66.859
66.846
66.83
66.815
66.796
66.719
66.687
66.659
66.629
66.596
66.569
66.54
66.51
Now graph the data points and the least squares curve
against the time of day. The x axis is the time of day in
hours, and the y axis is the solar altitude in degrees.
12.6
12.8
13
13.2
66.5
66.6
66.7
66.8
66.9
67
DEG
ii
quad HRS
ii
HRS
ii
You can see from both the graph and the displayed data that
the fit is pretty good, and will provide a good estimate of the
actual LAN time and uncorrected altitude. Note that, in
Mathcad PLUS 5.0, the genfit function could have been used
to fit these points.
Now that we have fit quad(x) to the sighting data, mathematically find the time of Local Area
Noon and the solar altitude at LAN by taking the derivative of the function, and setting it equal to zero.
d
d x
b
0
.
b
1
x
.
b
2
x
2
0
has solution(s)
LANtime
.
1
2
b
1
b
2
=
LANtime
12.855
Hours
The time of LAN is displayed below in more familiar notation:
Fitted sextant reading at LAN is
=
dtim LANtime
12
51
17.163
The altitude at LAN is then
LANalt
quad LANtime
=
ddeg LANalt
66
53.25
Page 9
Reduction of noon-day sun shots to actual position
Date of sun shot:
=
date
4
18
1993
Correct the altitude
Correct the time
Hs is the sextant height in degrees for the
lower limb of the sun, as shown above.
Tw is the LAN computed by the least
squares fit of the sighting data.
Hs
LANalt
Tw
LANtime
=
dtim Tw
12
51
17.163
Correct the altitude reading for index error on
the sextant:
Correct the time of LAN for watch error:
Hs
Hs
IE
Local time is
Tl
Tw
WC
=
dtim Tl
12
51
16.163
The dip is always subtracted from Hs to
obtain Ha.
compare:
=
dtim EstTimeLAN
12
52
44
DC
degrees
,
dip
,0 0
dip
,0 1
Tdl is the daylight savings time correction. If sightings
are taking during daylight savings time, add -1, if not
then add 0.
Ha
Hs
DC
Adjusted height is:
Tdl
1
Tz
Tl
Tdl
=
ddeg Ha
66
49.55
Zone descriptor correction includes the time
zone. If you are at a west longitude,
then Zd is positive
Combine the correction factors A2 and A4
into the height reading to get the observed
height, H
o
.
=
Zd
8
hours
Ho
Ha
A2
A4
=
ddeg Ho
67
5.35
LAN in Greenwich main time is
GMT
Tz
Zd
The zenith distance:
Adjusted time is
Zdist
90
Ho
=
dtim GMT
19
51
16.163
=
ddeg Zdist
22
54.65
Page 10
Using the final adjusted time, find the additional declination correction for the minutes and seconds. The
increments and corrections table in the almanac for 51 min and 16 sec gives
IandC
degrees
,
12 49.0
dCorr
degrees ,0 0.8
change for 51 min.
DecT
DecHr
dCorr
Total GHA is
GHAt
GHAhr
IandC
North declination is added to the zenith distance to
calculate the latitude.
Longitude in degrees and minutes:
LAT
Zdist
DecT
LONG
if
,
,
<
GHAt 180 GHAt GHAt
360
=
ddeg LAT
33
57.15
North
=
LONG
118.007
=
ddeg LONG
118
0.4
West is positive
The sun sighting data used in this paper was from the author's practice session on the beach at
Playa del Rey, California. The deduced reckoning (DR) position was determined from the 18744
nautical chart, so we can compare the sun sight fix to the known DR position and determine how
close we came.
Enter your DR position guess:
North Latitude:
DRLat
degrees
,
33 57.4
West Longitude:
DRlong
degrees
,
118 27.1
Determined North Latitude:
Known North Latitude:
=
ddeg LAT
33
57.15
=
ddeg DRLat
33
57.4
Find the error:
∆Lat
LAT
DRLat
=
ddeg ∆Lat
0
0.25
minutes of arc
Since a latitude difference of 1 minute is equal to 1 nautical mile, we are off by .25 mile in latitude. Not
bad at all.
E
LAT
.25
mile
Determined West Longitude:
Known West Longitude:
=
ddeg LONG
118
0.4
=
ddeg DRlong
118
27.1
Find the error:
∆Lon
GHAt
DRlong
=
ddeg ∆Lon
0
26.7
minutes of arc
Page 11
It's a little harder to calculate distance for a change of longitude. With latitude you are on a great circle
and a change of one minute equals one nautical mile. when dealing with changes of longitude, the
parallel isn't a great circle. Its circumference varies as the cosine latitude times one nautical mile per
minute of longitude.
E
LON
.
.
cos
.
.
2 π
360
DRLat 60 ∆Lon
=
E
LON
22.147
miles
Remember it is much easier to get accurate latitude than longitude. An error of four seconds in the
time of LAN will introduce error of one mile. We only missed the time of LAN by 1 minute and 28
seconds, as shown below.
=
dtim hours
,
,
11 52 44
Tz
0
1
27.837
=
E
LAT
2
E
LON
2
22.148
Miles

Answer 2

Pl try Astrologers guide book.

Answer 3

I don't have the formula, but you can find the values from this.

Answer 4

try google