# Draw the circle of illumination

Draw the circle of illumination

On the Figure 5.1,::c6iii’plete the following items for each diagram. 1. Extend the suns rays so that they are touching the earth’s surface. 2. Label the rays with the appropriate term: VR (for vertical ray), OR (for

oblique ray), or TR (for tangent ray) 3. Put an ‘X’ on the subsolar point 4. Draw the following latitudes as accurately as possible andJab~l them:

a) Equator – -~– — – —-· –

b) Tropic of Cancer c) Tropic of Capricorn d) Arctic and Antarctic Circle

Draw the circle of illumination and lightly shade the portion of earth that is experiencing night

s u n

R a y s —

June solstice December solstice

— s u n

R a y

— •

March equinox September equinox

Figure 5.1 Earth-Sun relationships

What is_the latitude (in degrees) of the subsolar point for each season? a) June Solstice. _ b) Dec. Solstice _ _ c) March Equinox. d) Sept Equinox _ _

“‘ c: E 0 …. ~ Q) al c.. 0 UJ UJ ::::: c: al – :=

Nvtl’?tl.~e_ ·. 7 Using the data in figure 5.3~plot and the insolation data for each month for

the following locations on the graph provided: Label each line on the graph with their corresponding letter below (a,b,c,d, ore). a) North Pole (already started for you) b) New York (already started for you) c) Equator d) Tropic of Capricorn e) South Pole

~ . Why does the North Pole exceed the amount of insolation received by the equator during the June Solstice? _

I

600 .–:–,—,—.-,–.-l.–,—-,—,.-r—-r:l-r-i—-i-r—r-,–:-Tl-r—r–r—i-r-T1:l600 l I ; l

5ool_l–L–l—1—–+-~~:1—J.:-1–t-,-+–+~1+-c–+-+–f-1-+14:+–+–+—t-t-r1n: 500 I I I l ! _l I

I I I l I I I

~ I / I ~: I I I

1~ –r1 1 I,’ 1 200 L_J__j_,. …….. 4 ·—-+–W~~”–if–f–H-, ~+-+-+-+-+1’t—t—t—t-llt1: 250 v : ~ I L 1 I I I : l t I fI j I ‘- I l ll_ 1 I I Ir I I

: p : I -+ !7 ! : :

ol…LJ-1-1-1.J1’1-L…J-1__.l_~·~~:’·L_LJ-1_:_l.-:-1-~.;.L.L~:–“-:~-‘-;:;~O Jan Feb Mar Apr May Jun Jul Aug Sep Oct Nov Dec

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In addition, the greater length of travel through the atmosphere at higher latitudes reduces solar iritensity due to increased reflection, absorption, and scattering within the atmosphere. This energy receipt also varies daily, sea- sonally, and annually as Sun angle and daylength vary-less toward the equator, more toward the poles.

A useful altitude at which to characterize insolation is the top of the atmosphere (480 km, 300 mi). The graph in Figure 5 .3 plots the daily variation in insolation for selected latitudes. Latitudes are marked along iheleft side in 10° intervals. The months of the year are marked and labeled across the top and bottom. The dashed curved.Tine shows the declination of the Sun (the latitude of the sub solar point) throughout the year. · ·

To read the graph, select a latitude line and follow it from left to right through the months of the year. The Cwly insolation totals are read from the curved lines, given in watts per square meter per day (W/m2 per day). For ei<:am- ple, at the top of the atmosphere above 30° N latitude the insolation values (in W/m2 per day) are as follows (re~d­ ings are approximate-extrapolate when necessary): 240 on January 1; 250 on January 15, and 325-en_Eebruary I.

Figure 5.3

0 ~ Arctic Circle :E 1ii …J

“”‘ ‘§ New York, z Rome

Tropic of Cancer

Equator

Months of the Year

Total daily insolation received at the top of the atmosphere charted in watts per square meter by latitude and month. (1 watt/m2 = 2.064 cal/cm2/day.) Adapted from the Smithsonian Institution Press from Smithsonian Miscellaneous Collections: Smithsonian Meteorological Tables, vol. 114, 6th Edition. Robert.List, ed. Smithsonian Institution, Washington, DC; 1984, p. 419, Table 134.

Lab Exercise 5 • 79

3

Find January 17th on the analemma (marked .with a dot). It lies on the line labeled 9 minutes, on the “sun slow” side. This means that the sun will reach its

“/ q~ h /I…,,__ 5” 1.1..’ use the analemma: to find:

“noon” zenith (highest point in the sky on that day) at 12:09 P.M., 9 minutes “late.”

a) the subsolar point on November 10 (marked) _ ~[~1~7°s~o’-‘u=th=-l=a=ti”-‘tu”‘d”‘”e]._ __ _

“b) the subsolarpoint on May 11————————-

c) the subsolarpointtoday _______________ _

~ d) the date(s) when the declination is 9° S · · ——————— e) the date(s) when the declination is 21° N ________ _

IO At what clock time does the Sun actually~ on:

October 13? (13 minute~timel March8? ____ _

May20? ___ _

Today? __ ~————

As we have noted, the noon Sun angle changes at any given latitude throughout the year. Using the analemma to de- termine the subsolar point (Sun’s declination), you can calculate the altihide_ofthe noon Sun (Sun L for any loca- tion by using the following formula:

L = 90° – (arc distance between your latitude and subs~la’r point)

EXAMPLES:

You are at 30° Non December 21. You know that on the December solstice the subsolar point of the noon Sun is ·23°30′ S. Using the above formul;<: – . L = 90° – (arc distance betwe’F your latitude and subsolar point) L = 90° – (30° N ++23°30′ s)(the symbol +.+is to indicate +.+arc distance between”)

L = 90° – 53° 30′ (Note: change 90° to 89°60′ in order to correctly calculate this.

Remember that each whole degree is equal to 60′.)

L = 36° 30′ above the southern horizon from your location

You determined that on December 21, if you were stru+ding at 30° north latitude, you observe the Sun’s noon altitude at 36°30′ above the southern horizon (since you aie north of the subsolar point on this day). ‘ · ,.;· . ‘

If you are at 40° Non April 20, you determine from the analemma that the subsolarpoint is 10° N; therefore:

L = 90° – (arc distance between your latitude and subsolar point)

L = 90° – ( 40° N ++ 10° N)

L=90°-30° L = 60° above the southern horizon*

*Note: You would observe the same Sun altitude on August 25 when the subsolar point is once again at 10° N. If you end up with a negative value for our Sun altitude, then the Sun is below the horizon and will not be visible on that day-as someone might experience north of the Arctic Circle in the Northern Hemisphere winter.
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