SUN EARTH GEOMETRY

From SUN EARTH GEOMETRY figure, the following conclusions are

derived:
1. The earth’s orbit around the sun is elliptical with a mean
centre to centre distance from the sun is approximately 9.3 ×
106 miles (1.5 × 108 Km).
2. While the earth makes its daily rotation (24 hrs) and yearly
revolution (365 days), the sun also rotates on its axis
approximately once every month.
3. The earth’s axis of rotation (the polar axis) is always inclined
at an angle of 23.5° from the ecliptic axis.
4. This distance from the sun to the earth varies ±1.7% over the
average distance.
5. The sun is 109 times larger in diameter than the earth.
6. The sun appears to move across the sky in an arc from east to
west, owing to the rotation of the earth around its north-south
axis.
7. Viewing the sun from the average miles, it subtends an arc of
0.53° (32 min).
LATITUDE AND LONGITUDE
Numericals
Example 2.1

Calculate hour angle when it is 3 h after solar noon.

Solution Solar time = 12 + 3 = 15:00

Therefore, hour angle (w) = 15 × (ts – 12) = 15 × (15 – 12) = 45°

Example 2.2

Calculate hour angle when it is 2 h 20 min before solar noon.

Solution Solar time = –1/4 × (tm) = –1/4 × 140 = –35°

# Equation of Time
This is the difference between the local apparent solar time and the local mean solar time. The
actual equation of time (EOT), which is mathematically defined as apparent solar time minus
mean solar time, varies slightly from year to year due to variations in the earth’s eccentricity and
obliquity and in the time of the solstices and equinoxes. However, for a century, either side of the
year 2000, it may be approximated (to an accuracy of better than 1%) by the formula:

(2.3a)

Where,

(2.4)

Another formula equation for EOT can be approximated as

(2.3b)

Where,

n is the total number of days of the year (e.g., n = 1 on Jan 1 and n = 33 on Feb 2)

Important to remember: During leap year, February month will have 29 days.
The expressions for incidence angle (θ) can be further simplified as given below.

For horizontal surface, slope or tilt angle β = 0° and the angle of incidence θ becomes zenith
angle θZ of the sun. Therefore,

(2.13)

Numerical

Example 2.3

Calculate zenith angle of the sun at Lucknow (26.750 N) at 9:30 am on February 16, 2012.

Solution Total No. of days counted from January 1, 2012, till February 16, 2012, n = 47

From Eq. (2.8), the declination angle is given by

From Eq. (2.2), hour angle is given by ω = 1/4 × tm

Where tm = 12:00 – 9:30 = 150 min; therefore, ω = 1/4 × 150 = 37.5 (since time is before solar
noon, negative sign will be taken)
Therefore, θZ = cos–1(0.589) = 53.914

With vertical surface β = 90°, and then

(2.14)

Horizontal surface β = 0,

(2.14a)

The angle θ in this case is the zenith angle θz. The complement of zenith angle is called the solar
altitude angle.

For surface facing due south γ = 0°,

(2.14b)

And for vertical surface facing due south β = 90°, γ= 0°

The solar azimuth angle γs is the angular displacement from south of the projection of the
beam radiation on the horizontal plane.

Therefore, the solar azimuth γs can be written as

(2.15)

Equation (2.15) can be solved for the sunset hour angle ωss when θz = 90°.

Therefore,

(2.16)
# Sunrise, Sunset, and Day Length Equations
The sunrise equation can be used to derive the time of sunrise and sunset for any solar
declination and latitude in terms of local solar time (LST) when sunrise and sunset actually
occur.

Where

Since the hour angle at local solar noon is zero, with each 15° of longitude equivalent to 1 h, the
sunrise and sunset from local solar noon is derived as follows:

(2.17)

Therefore, daylight hour is given by 2TH.

Numerical

Example 2.4

Find the solar altitude angle at 2 h after local solar noon on 1 June 2012 for a city, which is
located at 26.75° N latitude. Moreover, find the sunrise and sunset hours and the day length.

Solution The declination on June 1 (n = 153) is

The hour angle at 2 h (120 min) after local solar noon is obtained by Eq. (2.2) as,

The solar altitude angle is calculated as follows:

Since solar altitude angle θZ = 90° – α

=
13.58 hrs

Therefore, the sun rises at 12:00 – 13.58/2 = 05:13 am

Sunset time = 12:00 + 13.58/2 = 06:47 pm

However, if the day under consideration lies between March 21 and September 22, the hour
angle at sunrise or sunset (ωst) would be smaller in magnitude than the value given its equation
and would be obtained by following equation:

(2.20)

Numerical
Example 2.5

Calculate the hour angle at sunrise and sunset on June 21 and December 21 for a surface inclined
at an angle of 10° and facing due south (γ = 0°). The surface is located in Mumbai (19°07 ′ N,
72°51′ E).

Solution For December 21, using Eq. (2.20),

For December 21, using Eq. (2.18),

2.3.10 Solar Time

It is the time based on the angular motion of the sun across the sky. With solar noon, the sun
crosses the meridian of the observer. Solar time does not coincide with the local clock time. It is,
therefore, necessary to convert standard time to solar time by applying the following correction.
It is the time based on the 24-h clock, with 12:00 as the time that the sun is exactly due south.
The concept of solar time is used in predicting the direction of sunrays relative to a point on the
earth. Solar time is location (longitude) dependent and is generally different from local clock
time, which is defined by politically defined time zones and other approximations. Solar time is
used extensively to define the rotation of the earth relative to the sun. The time used for
calculating the hour angle (ω) is the local apparent time. This LST can be obtained from the
standard time (ST) by making two corrections as given below:

1. First correction arises because of the difference between the longitude of a location and
the meridian on which the standard time is based.

2. It has a correction of 4 min for every degree difference in longitude. The factor of 4
comes from the fact that earth rotates 1° for every 4 min.

3. The second correction is called the EOT.

It is due to the fact that the earth’s orbit and rate of rotation are subject to small fluctuations.

Therefore, Time Correction Factor (T C) in minutes can be obtained as

Where TC is in minutes

(2.21)

In India, standard time is based on 82.5°E. Twelve noon LST is defined as when the sun is the
highest in sky.

LSTM = Local standard meridian time zone = 82.5°E (in India)

LLOL = Longitude of location in degrees

EOT = EOT in minutes

Numerical
Example 2.6

For a city located at 80.50 longitudes, calculate the solar time on March 15, 2011, at 10:30 am
Indian Standard Time.

Solution The standard meridian for IST zone is 82.50 E.

Total Number of days on March 15 counted from January 1, 2011, n = 74

B = 360 (n – 81)/365

B = 360 (74 – 81)/365 = –6.9 (in degrees)

EOT = 9.87 sin 2B – 7.53 cos B – 1.5 sin B

From Eq. (2.8), LST

Given that, LSTM = 82.5°E; LLOL = 80.5°; standard time = 10:30 am

Therefore, LST = 10:30 – 4 × (82.5 – 80.5) + (–9.65)