The altitude of the sun

It is worth considering two cases:

(a) the special case for the altitude at noon;

(b) the general case for the altitude at any time of the day.

(a) The altitude at noon (a’)

(7.2)

This is easily illustrated in geometrical terms, as in Figure 7.6, in which it can be seen that a very simple expression is derived from a consideration of the latitude of the place in question and the declination for the particular date:

A’ = 90 — (L — d)

Place on the surface of the earth

S

Fig. 7.6 The special case of the solar altitude at noon.

It should be bome in mind when considering Figure 7.6 that all parallels of latitude are parallel to the plane of the equator, and that, because the sun is so far away from the earth, all its rays parallel to one another when they strike the earth.

EXAMPLE 7.1

Calculate the maximum and minimum altitudes of the sun at noon, in London, given that the latitude of London is 51°N.

Answer

The greatest altitude occurs at noon on 21 June and the least at noon on 21 December. Hence, from equation (7.2), and knowing that the declinations are 23 j °N and S respectively in June and December, we can write

0′(max) = 9Oo-(51°-23i°) = 62^ a'(min) = 90°-(51°-(-23{°)) = 15|°

(b) The general case of the sun’s altitude (a)

Although a geometrical illustration is possible, it is in three dimensions and consequently, being difficult to illustrate, is not shown. The result, obtained from such geometrical considerations, is that

Sin a = sin d x sin L + cos d x cos L x cos h (7.3)

This equation degenerates, to give the special result obtained from equation (7.2), if a value of h = 0 is taken for noon.

EXAMPLE 7.2

Calculate the altitude of the sun in London on 21 June (a) at 13.00 h and (b) at 17.00 h (British Summer Time), making use of equation (7.3) in both cases.

Answer

(a)

D = 23^°N = +23^°

L = +51°

T = 0 (because sun time is British Summer Time minus 1 h) therefore

H = 0

Hence

Sin a = sin 23 ° x sin 51 ° + cos 23 0 x cos 51 ° x cos 0 = 0.399 x 0.777 + 0.917 x 0.629 x 1.0 = 0.887

Therefore

A = 62

(a) At 17.00 h, T = 16.00 h, hence h = 60° and

Sin a = sin 23x sin 51° + cos 23-j° x cos 51° x cos 60°

= 0.399 x 0.777 + 0.917 x 0.629 x 0.5 = 0.598

Therefore

A = 36|°

Note that using a more exact value of 51 °N for London gives a slightly different answer.

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