# Solar angles

• Sun’s position

The sun’s position can be determined by two angles:

Solar altitude, (3 Solar azimuth from the south, \$ Sin (3 = cos L cos 8 cos H + sin L sin 8 cos c[) = (sin (3 sin L — sin 8)/(cos (3 cos L) Where L = local latitude decime’.ion

8 = solar declination (TTable 6-1, ATable 8)

H = local solar time expressed as the hour angle

Example 7-2

Continue from example 7-1. Latitude of the PSU classroom is 41° N. Find the sun’s position. Solution

L =41°

H =11:34 am — 12:00 (true south) = — 26 minutes = — 26/(60 x 24) x 360° = — 6.5°

8 = -10.8° (A Table 6-1, T Table 8)

Sin (3 = cos L cos 8 cos H + sin L sin 8

= cos 41° cos (- 10.8°) cos (- 6.5°) + sin 41° sin (- 10.8°)

= 0.6136 (3 =37.85°

Cos c[) = (sin (3 sin L — sin 8)/(cos (3 cos L)

= (sin 37.85° sin 41° — sin (- 10.8°))/(cos 37.85° cos 41°)

= 0.9899 <|> = 8.1°

• Incident angle

The incident angle 0 of a surface:

Cos 0 = cos (3 cos y sin Z + sin (3 cos Z where 0 = angle between the sun’s rays and the normal to the surface y = <> —|/

j/ = surface azimuth defined as:

Orientation N NE E SE S SW W NW

j/ 180° -135° -90° -45° 0° 45° 90° 135°

X = tilt angle of surface (angle between the normal to the surface and the normal to the horizontal surface)

X = 0° for a horizontal surface and 90° for a vertical surface

Example 7-3

Continue from Example 7-2. Find the incident angle of a vertical window facing east.

Solution

|/= -90°

X = 90°

Y= 0 — j/= 8.1° — (-90°) = 98.1°

Cos 0 = cos Я cos y sin X + sin Я cos X

= cos 37.85° cos 98.1° sin 90° + sin 37.85° cos 90°

= -0.1113

0 = 96.39° — No solar on the surface