74 TECHNICAL SURVEY 
south only. 
The azimuth of the sun may be calculated from 
the formula: 
sin HA 
cos tan 6 — sin @cos HA © 
tan B = — (1) 
B = bearing of the sun. 
The bearing is east or west of south when ¢-6 is 
positive. The bearing is east or west of north when 
#-6 is negative. The bearing is east in the morning 
(6 will be negative), and west in the afternoon (6 
will be positive). 
HA = hour angle of the sun. 
During the morning hours when the hour angle is 
greater than 12 hours, its value should be subtracted 
from 24 hours for use in the formula. 
= latitude of the place of observation. 
6 = declination of the sun at the time of observation. 
The signs of ® and 6 are important and each is 
positive when north of the equator and negative 
when south. 
The hour angle HA is the local apparent time 
[LAT] minus 12 hours. To convert the observed time 
‘into LAT the civil time at Greenwich [GCT] must 
be found and combined with the equation of time 
to correct for the apparent irregular motion of the 
sun. This gives Greenwich Apparent Time [GAT] 
which is converted to LAT by allowing for the lon- 
gitude. The equation of time and the declination of 
the sun are plotted in Figure 1 for 1945. The annual 
change is small, and these curves may be used for 
radar ‘work without regard to the year. Standard 
time meridians are every 15° east or west of 
DECLINATION>IN DEGREES 
EQUATION OF TIME IN MINUTES 
Greenwich, each zone corresponding to 1 hour. Care 
should be used to take daylight saving, or other 
changes from standard, into account correctly. 
Example 1. It is desired to compute the azimuth 
of the sun. 
Given: Date March 16, 
Time 1345 hours PWT 
Latitude 40° North 
Longitude 118° West 
Solution: 
The hour angle will be determined first: 
Observed time PWT 13" 45" 
Zone difference ab) fe 
Greenwich civil time 20" 45" 
Equation of time (Figure 1) — g™ 
Greenwich apparent time 20” =36” 
Longitude difference for 118° W Oe 
Local apparent time 12? 44™ 
LAT — 12 = HA — 12? 
Hour angle of sun + 0® 44m 
HA in arc (4” = 1°) + 11° 
Latitude ® + 40° 
Declination of sun 6 (Figure 1) — 2° 
Substituting in equation (1): 
° sin 11° 
tan B= ~ cos 40° tan (—2°) — sin 40° cos 11° 
bis 0.19 
"0.766 X (—0.0349) — 0.643 x 0.982 
= 0.29, 
B = 16°10’. 
Since ® — 6 is positive, 8 is the bearing from the south. 
The bearing is west of south since B is positive (p.m.). The 
azimuth of the sun is 
180° + 16°10’ = 196°10’. 
A quicker solution may be obtained from a book 
7 17 27,7 I7 27 
{il 2 3i 1020.2 1222 1 I 2) 4 Mt 21 31 10203010 20309 1929 8 16 268 8 18 28 2 172 
JAN FEB MAR APR MAY JUN JUL AUG SEP oct NOV DEC 
FicureE 1. Sun data from nautical almanac, 1945. 
