Sky luminous intensity models

To calculate the daylight distribution in a room, models of the luminous intensity distribution of the sky are required. From the luminous intensity distribution, the effective surface and the solid angle, the light flux onto any recipient surfaces, for example windows, can then be calculated.

As the simplest model, the illuminance of the sky is set as constant, corresponding to an isotropic sky model. The density of light is calculated using Equation (8.14):

d Ф r

E =——- – = J L cosQ d

dAr J * r 2


The solid angle d02 is selected using Equation (8.9) as a spherical cone with half opening angle в, with в = 0 corresponding to the surface-normal of a horizontal recipient surface. With a horizontal recipient surface, the opening angle is equal to the angle of incidence вг and the following density of light results:

Подпись: О Figure 8.11: Projection of the sky dome with constant luminance onto a horizontal recipient surface.

The sky luminance can thus be calculated from the horizontal irradiance, if the radiating power is converted via the photometric radiation equivalent into a density of light.

Example 8.5

Calculation of the isotropic luminance of the sky for a horizontal irradiance of the overcast sky of 100 W/m2 and a photometric radiation equivalent of 115 lm/W.

image1038 image1039

The illuminance on the horizontal surface is 11 500 lm/m2. The illuminance then produces a sky luminance of

Подпись: La Подпись: —Lz (1 + 2sina 3 zV Подпись: (8.19)

Apart from the isotropic luminance distribution, standardised distributions of the international lighting engineering commission (CIE) for a clear and an overcast sky are used. The overcast, so-called Moon and Spencer sky, is characterised by a rise in the luminance La with the elevation angle a, with the zenith luminance Lz assuming three times the value of the horizon luminance,

and the zenith luminance with an overcast sky is a function of the sun height angle as:

Подпись: (8.20)Lz = 7 n(300 + 21000sinaj)

For a horizontal surface, the conical solid angle can be used again and the illuminance can be represented as a function of the angle of incidence 6r on the horizontal (which corresponds to the zenith angle) instead of the elevation angle a.

n/2( 1

Erh = Jl — Lz (1 + 2cos9r) cos9rd(2n(1 – cos9r))

image1044 Подпись: (8.21)

0 v 3


( 1 3 о

П / 2 Л


7 4)

+ 2

—cos3 9


3 r V.

0 ,

= 3


V 3 J

sin2 9r

= L 7 n

z 9

For vertical surfaces, the solid angle must be selected in such a way that the angles of incidence on the vertical, which change with the azimuth, can be taken into account. For example, two-dimensional elements on a ring zone of the sky hemisphere can be used, indicated as a function of the zenith and azimuth angles.

Подпись: r sin 9zd9z rdy Подпись: sin9zd9z dy Подпись: (8.22)

In spherical polar coordinates, a two-dimensional surface in the sky hemisphere with radius r is given by the product of the sides r d9 on the meridian and r sin9 dy on the parallel circle, with 9 corresponding to the zenith angle 9z. The solid angle dQ. is thus:

and the solid angle of a ring zone

Подпись: = 3 LzПодпись: 10 ——sin (20 2 z 4 v z Подпись: f n2 і .4 = Lzn f 1 4 4 + 2-2 + V4 3 J SSL 9п у Подпись: = 3 LzПодпись: 4 f1 4 3sin (0,) S3 ) n/2 in /2 Sin d—d2 d' 2 ' 1/3 J Подпись: (8.25)2n 02

Qz = J J sin0d0dy = 2n(cos02 – cos01) (8.23)

Подпись: Figure 8.12: Solid angle element of a ring zone on a sphere.

Y=0 0

image1056 Подпись: (8.24)

The angle of incidence 0r between the solid angle element of the sky with the coordinates zenith angle 0z and azimuth у and the vertical surface normal (angle of inclination в = 90°) is calculated with the known sun-position equations. Since the luminance does not depend on the azimuth, the vertical surface can be arbitrarily oriented in azimuth direction. For simplification, the surface azimuth ys = 0 is selected.

Ev = J L(0z )cos0rdQ = J—’П/2 J^[2] [3]V-3(1 + 2cos0) sin(^z cosiK,^3 sm. ezd9zdy

Подпись: dQ Подпись: dQ

The integration over the solid angle is carried out with the integration limits of the zenith angle from 0 to n/2 and of the azimuth of —n/2 to +n/2, since only half the sky is seen from the vertical surface.

Подпись: Erv Eh image1061 Подпись: (8.26)

In contrast to the isotropic sky model, in which from a vertical surface exactly half of the horizontal illuminance is seen, in the Moon and Spencer sky model only 40% of the horizontal value is obtained.

The solid angle element of the spherical ring zone dQ = sin0zd0zdy can (as an alternative to the spherical cone) also be used to calculate the horizontal illuminance, with of course the same results being obtained as with the conical solid angle. This sphere ring zone solid angle is used to calculate daylight coefficients, since in this way integration can take place over limited azimuth ranges of windows.


Updated: August 21, 2015 — 7:12 am