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Tue: 12:30 PM to 2:30 PM
Wed: 9:00 AM to 10:30 AM & 12:00 PM to 2:00 PM
Thr: 9:00 AM to 10:30 AM
Course Syllabus can be found at:
http://www.wx4sno.com/portfolio/BSU/Fall_2011/
This lecture will be posted AFTER class at:
http://www.wx4sno.com/portfolio/BSU/Fall_2011/Lectures/
Lesson 11
Solar Angle
Hess, McKnight’s Physical Geography, 10 ed.
17-20 pp.
Sun Declination
Recall from last week, the vertical rays (direct
rays) are those from the sun that strike the
Earth at a 90° angle.
 The location of these vertical rays changes
throughout the year

◦ Located at the Tropic of Cancer during the summer
solstice
◦ Located at the equator during the equinoxes.

The latitude at any given time of the year where
the sun’s vertical rays strike the Earth is known
as the sun declination.
Sun Declination, cont.

The sun’s declination can be plotted on a
graph, known as an analemma (next slide)
◦ The sun’s declination is plotted along the
vertical axis
◦ The days of the year are along the analemma
itself

See page 52 of your lab manual for an
example
Solar Altitude

Solar altitude is the elevation of the sun in the
sky at noon local time
◦ i.e. the angle of the noon sun above the horizon

Can be calculated mathematically:
𝑆𝐴 = 90° − 𝐴𝐷
where SA is the solar altitude and AD is the arc distance
Arc distance (AD) is actually the difference
between the latitude your at and the declination
of the sun at that time of year.
 Let’s look at an example…

Solar Altitude Example

Suppose it’s noon and we’re outside on campus
during a snowstorm on January 12th. The clouds
and snow showers are blocking the sun from
being visible. Even though we can’t see the sun,
what is the solar altitude (sun’s elevation above
the horizon)?
Solar Altitude Example, cont.
The latitude of Muncie is 40° 11’ 36” N.
 Looking at our analemma, we find that on
January 12th, the sun’s declination is 21.5°
S.
 The arc distance (AD) can be found by
subtracting the sun’s declination from our
latitude:

40° ─ −21.5° = 61.5° = AD
◦ The reason there is a negative in front of the
21.5° is because the sun’s declination was “S”
and in the southern hemisphere.
Solar Altitude Example, cont.

Now that we know AD, plug into the equation
we were given:
𝑆𝐴 = 90° − 𝐴𝐷
𝑆𝐴 = 90° − 61.5°
𝑆𝐴 = 28.5°
 Therefore, if we could see the sun through the
clouds and snow, it would only be 28.5° above the
horizon.
◦ For additional examples, see page 51 and 53 in your lab
manual.
Tangent Rays and Daylight/Darkness
If we know the latitude of declination for the
sun (where it’s direct rays strike the Earth; from
the analemma), then we can easily find the
latitude of the tangent rays (those rays that
skim past the earth).
 Simply use this equation:
𝑇𝑅 = 90° − 𝐷𝑒𝑐

Where TR is the latitude of the tangent rays and Dec is the sun’s
declination
Tangent Rays and Daylight/Darkness, cont.

From our previous example, we found that the
declination of the sun on January 12th is 21.5° S.
So, plugging that in:
𝑇𝑅 = 90° − 𝐷𝑒𝑐
𝑇𝑅 = 90° − 21.5°
𝑇𝑅 = 68.5°

So, we know that those latitudes above 68.5° N
and 68.5° S receive either 24 hours of sunlight
or 24 hours of darkness…but which is which?
Tangent Rays and Daylight/Darkness, cont.

For the Northern Hemisphere, remember this:
◦ If it (the day) is between March 20 and September
22, then those areas north of the latitude of the
tangent rays are experiencing 24 hours of daylight.
◦ If the day is between September 23 and March 21,
then those areas north of the latitude of the tangent
rays are experiencing 24 hours of darkness.
Lesson 12
Insolation
Hess, McKnight’s Physical Geography, 10 ed.
pp. 70, 80-84, and Fig. 4.17 on p. 78
Insolation
From lesson 11 we now know that the sun’s
direct rays strike the earth at different locations
throughout the year.
 These differences give us our seasons and
influence the amount of average daily
insolation (incoming solar radiation)

◦ Average daily insolation is the rate or intensity of the
sun’s radiation that strikes the surface over a 24-hour
period
 Measured in watts per square meter (W· m-2)
 The average insolation hitting the Earth’s upper atmosphere is
~ 1372 W· m-2. This is known as the solar constant
Insolation, cont.

However, the amount of insolation hitting the
surface of the Earth varies widely due to three
factors:
1. The angle of incidence
2. Day length
3. Atmospheric obstructions

We will discuss each of these next…
Angle of Incidence

Angle of incidence: the angle at which the Sun’s rays
strike the surface of the Earth (solar altitude)
◦ This can be directly related to the intensity of radiation that
reaches the surface.

Areas that have a high angle of incidence have a
given amount of radiation concentrated on a
small area
◦ Therefore, radiation is higher in intensity

While areas with a low angle of incidence have
that same amount of radiation concentrated on
a larger area
◦ Lower intensity radiation
Angle of Incidence, cont.
Angle of Incidence, cont.
Length of Day
We all know that the length of daylight
influences how much solar radiation is received
(e.g. longer days generally mean warmer days)
 Even if it is cloudy, longer days generally mean a
significant increase in solar radiation received
 Take a look at Fig. 3 and Fig. 4 on pages 58 and
59. These provide the hours of daylight and
daily insolation, respectively, for location at the
equator, 45° N and 90 N°.

Atmospheric Obstruction

The amount of atmosphere that radiation has
to travel through affects the total amount
received.
◦ e.g. If the angle of incidence is low, then solar
radiation has to travel through more atmosphere,
thereby reducing the amount received when it finally
reaches the surface
Water droplets (clouds) and other atmospheric
particulates also affect the amount received.
 The percentage of solar radiation reaching
Earth’s surface through the atmosphere is listed
in Fig. 5
