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Solar Geometry
The Earth’s daily rotation about the axis
through its two celestial poles (North and South) is perpendicular to the equator, but it is
not perpendicular to the plane of the Earth’s orbit. In fact, the measure of tilt or obliquity
of the Earth’s axis to a line perpendicular to the plane of its orbit is currently about 23.5°.
We call the plane parallel to the Earth’s celestial equator and through the center
of the sun the plane of the Sun.
The Earth passes alternately above and below
this plane making one complete elliptic cycle
every year.
Winter Solstice (z Dec 21)
Vernal Equinox (z March 21 )
SUN
P L A N E O F S U N
Autumnal Equinox (z Sept 23)
Summer Solstice (z June 21 )
Dennis Nilsson [CC BY 3.0 (https://creativecommons.org/licenses/by/3.0)]
Summer Solstice
On the occasion of the summer solstice, the Sun shines down most directly on the Tropic of Cancer in the
northern hemisphere, making an angle δ = +23.5° with the equatorial plane.
In general, the Sun declination angle , δ, is defined to be that angle made between a ray of the Sun, when
extended to the center of the earth, O, and the equatorial plane. We take δ to be positively oriented whenever the Sun’s rays reach O by passing through the Northern hemisphere.
On the day of the summer solstice, the sun is above the horizon for the longest period of time in the
northern hemisphere. Hence, it is the longest day for daylight there. Conversely, the Sun remains below
the horizon at all points within the Antarctic Circle on this day.
Winter Solstice
On the day of the winter solstice, the smallest portion of the northern hemisphere is exposed to the Sun
and the Sun is above the horizon for the shortest period of time there. In fact, the Sun remains below the horizon everywhere within the Arctic Circle on this day. The Sun shines down most directly on the tropic
of Capricorn in the southern hemisphere on the occasion of the winter solstice.
We call δ the Sun declination angle.
sin δ = 0.39795 ⋅ cos [ 0.98563 ⋅ ( N – 173 ) ] ( 1 )
where the argument of the cosine here is in degrees and N denotes the number of days since January 1.
Vernal and Autumnal Equinox
There are two occasions throughout the year when the center of the Earth lies in the plane of the Sun. Since the Earth’s North – South axis of rotation is perpendicular to this plane, it follows that on these two days every location on the Earth receives 12 hours of sunshine. These two events are known as the vernal and autumnal equinoxes.
Note that
- The Earth is above the plane of the Sun during its motion from the autumnal equinox to winter solstice to vernal equinox. Hence, δ < 0 during the fall and winter.
- The Earth is below the plane of the sun as it moves from vernal equinox to summer solstice and back to autumnal equinox (i.e. during spring and summer). So δ > 0 during these seasons.
The latitude of a location on the Earth is the angle between the line joining that location to the center of the earth and the equatorial plane. For example, Chicago, Illinois is on the circle of latitude 41.8° N. All locations at the same latitude experience the same geometric relationship with the sun. The great semicircles along the surface of the Earth joining the North to the South poles are called lines of longitude.
The location P on the globe has longitude λ ο^ East and latitude ϕ ο^ North.
Cities on the same circle of latitude
Cities on the same line of longitude
The Sun declination angle has measure: δ = 0° on the days of the vernal and the autumnal equinox.
The line of longitude through Greenwich, England is called the prime meridian.
Solar Noon
Solar Noon is defined to be that time of day at which the Sun’s rays are directed perpendicular to a given line of longitude. Thus, solar noon occurs at the same instant for all locations along any common line of longitude.
Solar Noon will occur one hour earlier for every 15 degrees of longitude to the east of a given line and one hour later for every 15 degrees west. (This is because it takes the Earth 24 hours to rotate 360°.)
Problem: Determine the elevation angles at which the sun’s rays reach the surface of the Earth in Chicago, IL at solar noon at the times of an equinox or a solstice.
- At the Vernal and the Autumnal Equinox
Thus, the most direct angle at which the Sun’s rays strike the ground in Chicago is at about 48° on the days of a vernal and autumnal equinox.
On the day of the winter solstice, the Sun’s rays strike the ground at an angle of about 25° at solar noon in Chicago, IL.
It is evident that, on any day of the year, the most direct angle of solar radiation at a given location occurs at solar noon. From our previous examples, it follows that for points in the Northern hemisphere , the angle of solar elevation at noon is given by:
α = 90° – ( ϕ – δ ) ( 2 )
The symmetrical formula for solar elevation at noon for points in the Southern hemisphere is given by:
α = 90° + ( ϕ – δ ) ( 3 )
We have seen that at solar noon, the range of angles at which the sun strikes the ground at Chicago, Illinois varies from a minimum of 24.7° at the winter solstice to a maximum of 71.7° at the time of the summer solstice.
A sun path diagram for Chicago appears below.
In the above diagram, the azimuth and solar elevation angles are the coordinates of an observer's local horizon system in Chicago, Illinois.
You may obtain a sun path diagram for any specified location on Earth at any specified time of the year by entering the relevant information at the University of Oregon’s Solar Radiation Monitoring Laboratory website: http://solardat.uoregon.edu/SunChartProgram.html.
It follows that at local noon, the hour angle is zero: ( 0 < t < 24, then t = 12 at local noon at six and one-quarter hours after local noon
Upon solving (4) for o , setting th we find that
Exceptional cases
Note that cases (a) and (b ) only relate to latitudes beyond the Antarctic Circle.
Angle of Solar Elevation for
It can be shown that the angle of solar elevation angle, , on a given date, the current hour angle (i.e. the equation: = sin –1^ ( sin
Azimuth Equation
One equation which relates the Sun’s azimuth angle, A, at a given location, current hour angle at the observer’s
(i) Sunris e occurs a
and
(ii) S unset occurs at
Diagram from: http://www.powerfromthesun.net/book.html
It follows that at local noon, the hour angle is zero: = 0°. Using a 24 hour time scale for the day, noon, t = 6.5 at five and one-half hours before local noon hours after local noon) , we have that at time to , the measure of the hour angle is
= 15 ( to – 12 )° (5)
, setting that result equal to the right si de of (5), and then solving
Exceptional cases: (a) If (tan tan > 1, then no sunset that day. (b) If (tan tan < –1, then no sunrise that day.
) only relate to latitudes beyond +65.5°, i.e. above the Arctic Circle or below
for Times of Day Other Than Local Noon
angle of solar elevation, , may be expressed in terms of the Sun’s declination , on a given date, the current hour angle (i.e. the time of day) , and an observer’s latitude
( sin sin + cos cos cos ). (7)
which relates the Sun’s azimuth angle, A, at a given location, its angle of elevation the observer’s latitude, and the Sun’s declination angle, , on this date
cos cos A = sin cos – cos
Solving for the azimuth, A, we find:
A” =
^
cos
cos 1 sin cos cos cos
where if < 0 then A = A
Otherwise, > 0 and A = 360
e occurs at (^) o
o 15
cos ^1 (tan tan ) hours before local noon
unset occurs at (^) o
o 15
cos ^1 (tan tan ) hours after local noon.
sing a 24 hour time scale for the day, hours before local noon and t = 18. the measure of the hour angle is:
de of (5), and then solving it for to ,
i.e. above the Arctic Circle or below
, may be expressed in terms of the Sun’s declination , and an observer’s latitude, via
its angle of elevation , , the on this date is:
cos sin (8)
Solving for the azimuth, A, we find:
cos sin (9) then A = A”.
0 and A = 360 ° – A”.
hours before local noon
hours after local noon.
A knowledge of the angle of solar elevation, α, and the azimuth, Α, allows us to calculate the length and the location of a simple shadow.
Example Describe the shadow due to a telephone post that is h ft high at the time of day when the solar elevation is α = αο and the azimuth is Α = Αο.
In the figure, we observe that this shadow has length | OP | where in the right triangle, △ OPQ , we find that tan αο = | OP |
h
⇒ | OP | =
tan (^) o
h α
ft.
Note that the tip of the shadow is located at the point P = ( x o , y o ) that is x o ft East and y o ft North of the base of the post.
By analyzing the right triangle, △ ORP , we find that
sin( Αο – 180° ) = | OP |
x (^) o (^) ⇒ (^) x o = |^ OP^ |^ ⋅^ sin(^ Αο – 180° ) ft.
Thus, x o = tan αo
h ⋅ sin( Αο – 180° ) = – tanα
sin o
h Α ο ft,
and
cos( Αο – 180° ) = | |
o OP
y (^) ⇒ y o = |^ OP^ |^ ⋅^ cos(^ Αο – 180° ) ft.
So y o = tan αo
h ⋅ cos( Αο – 180° ) = – tanα
cos o
h Α ο ft.
Therefore, the shadow’s vertex is at
P = ⎟⎟
⎠
⎞ ⎜⎜⎝
⎛ (^) Α −
Α − tanα
cos , tan α
sin o
ο o
h (^) ο h.
- Shadow Position on Autumnal Equinox 2 Hours After Solar Noon
= 0 = 41.8 = 30 = 40.2 = 220.9 ' = – 180^ = 40.9 Shadow Length = 28.4 ft y = 21.5 ft northward x = 18.6 ft eastward
- Shadow Position on Autumnal Equinox 4 Hours After Solar Noon
= 0 = 41.8 = 60 = 21.9 = 249.0 ' = – 180^ = 69.0 Shadow Length = 59.7 ft y = 21.5 ft northward x = 55.8 ft eastward
For much more detailed information regarding all of the
concepts and formulas discussed in this article, see the
website: http://www.powerfromthesun.net/book.html
