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Figures (12)

Fig. 1
Fig. 1

The solid-angle method of calculating illumination.

Fig. 2
Fig. 2

The two-dimensional problem.

Fig. 3
Fig. 3

Coordinate system for illumination calculation.

Fig. 4
Fig. 4

The rectangular source.

Fig. 5
Fig. 5

Flux density from a rectangular source, in terms of the angles β and γ subtended by the source.

Fig. 6
Fig. 6

Flux density from a rectangular window, in terms of window dimensions.

Fig. 7
Fig. 7

Helios distribution for an overcast sky.

Fig. 8
Fig. 8

Flux density on tilted surfaces outdoors.

Fig. 9
Fig. 9

Flux density from a rectangular window with overcast sky.

Fig. 10
Fig. 10

A troffer system.

Fig. 11
Fig. 11

Helios distribution in a semi-circular cylinder infinite length.

Fig. 12
Fig. 12

Average flux density on a horizontal plane. Cubical room, ρ1=reflection factor of walls, ρ2=reflection factor of ceiling.

Equations (19)

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H = π lim ω 0 D ω ,
d D = ( H / π ) d ω .
d D N = ( H / π ) d ω cos ψ = ( H / π ) d σ .
D N = 1 π S H d σ
D N = H π S d σ = H π σ .
D N = H 2 s .
H π [ cos ξ cos θ sin θ + sin ξ cos η sin 2 θ cos φ + sin ξ sin η sin 2 θ sin φ ] d θ d φ .
D N = D x sin ξ cos η + D y sin ξ sin η + D z cos ξ ,
D x = 1 π H sin 2 θ cos φ d θ d φ , D y = 1 π H sin 2 θ sin φ d θ d φ , D z = 1 π H cos θ sin θ d θ d φ . }
D x = H 2 π [ γ - γ 1 cos β ] , D y = H 2 π [ β - β 1 cos γ ] , D z = H 2 π [ β 1 sin γ + γ 1 sin β ] , }
tan β 1 = tan β cos γ , tan γ 1 = cos β tan γ .
H ( θ ) = H 0 + H 1 cos θ .
D N = D u 14 π { π ( 3 + 11 ρ ) + [ 11 π ( 1 - ρ ) - 8 γ ] cos γ + 8 sin γ } ,
D x = D u 14 π [ ( 3 γ + 4 ) - ( 3 γ 1 + 4 ) cos β - 4 cos γ ( 1 - cos β ) ] , D y = D u 14 π [ ( 3 β + 4 ζ ) - ( 3 β 1 + 4 sin γ sin β 1 ) cos γ ] , D z = D u 14 π [ 3 ( β 1 sin γ + γ 1 sin β ) + 4 ( sin β - cos 2 γ sin β 1 ) ] , }
div D = 0.
curl D = 0.
H ( s ) = H 0 ( s ) + ρ a b K ( s , t ) H ( t ) d t ,
K ( s , t ) = 1 / 4 π a 2 ;
K ( s , t ) = 1 4 sin | s - t 2 | .