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References

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  1. H. H. Higbie, “Bibliography of Natural Lighting,” Trans. I. E. S. 24, 315 (1929). 154 items.
  2. McAllister, Trans. I. E. S. 6, 713 (1911).
  3. H. H. Higbie, Lighting Calculations (John Wiley and Sons, 1934), p. 152.

1929 (1)

H. H. Higbie, “Bibliography of Natural Lighting,” Trans. I. E. S. 24, 315 (1929). 154 items.

1911 (1)

McAllister, Trans. I. E. S. 6, 713 (1911).

Higbie, H. H.

H. H. Higbie, “Bibliography of Natural Lighting,” Trans. I. E. S. 24, 315 (1929). 154 items.

H. H. Higbie, Lighting Calculations (John Wiley and Sons, 1934), p. 152.

McAllister,

McAllister, Trans. I. E. S. 6, 713 (1911).

Trans. I. E. S. (2)

H. H. Higbie, “Bibliography of Natural Lighting,” Trans. I. E. S. 24, 315 (1929). 154 items.

McAllister, Trans. I. E. S. 6, 713 (1911).

Other (1)

H. H. Higbie, Lighting Calculations (John Wiley and Sons, 1934), p. 152.

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Figures (8)

Fig. 1
Fig. 1

The luminous disk whose trace is MN illuminates the sphere with an intensity proportional to the spherical area concealed behind the disk.

Fig. 2
Fig. 2

The illumination received from MN is equal on planes AB and CD which are tangent to the sphere and parallel to the disk, respectively.

Fig. 3
Fig. 3

The projection of the light source ABCD toward the point Q is shown to cut a spherical area from the sphere equal to the area cut by double projection on the base plane FGHI.

Fig. 4
Fig. 4

The line H0, in a plane perpendicular to the base plane, projects equal arcs on the hemisphere and sphere.

Fig. 5
Fig. 5

This drawing is an engineering solution of an existing problem in applied illumination where the lower boundaries of the light sources are the broken line of the horizon, and hence outside the realm of solution by equations. The solution is for illumination on a vertical plane.

Fig. 6
Fig. 6

This is a photograph through window B of Fig. 5.

Fig. 7
Fig. 7

The windows of Fig. 5 are here investigated for illumination on a horizontal plane.

Fig. 8
Fig. 8

On the left of the sketch is shown the construction to determine the illumination received by a vertical plane from two lines of skylights, while on the right the same skylights are reversed in directions, as a matter of convenience, and investigated for illumination on a horizontal plane.

Equations (18)

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E 1 = 2 π R 2 ( 1 - cos 2 θ ) 4 π R 2 B = 1 - cos 2 θ 2 B ft.-c.
E 2 = 1 + cos 2 θ 2 B ft.-c.
E 1 + E 2 2 = B ft.-c. .
E Av = 1 2 ( 1 - cos 2 2 θ ) B ft.-c. ,
E Av = 1 2 B ft.-c. ,
F = x 2 + y 2 - r 2 2 y
R = ( F 2 + r 2 ) 1 2 .
E ( x , y ) = r 2 B r 2 + ( R + F ) 2 ft.-c. ,
E 1 = B sin 2 θ ft.-c. ,
sin 2 θ = r 2 r 2 + ( F + R ) 2 .
E 1 = π r 2 a 2 ( B π ) ft.-c. .
E = B a u / 4 π R 2 ft.-c. ,
a h = H h W h
a u = H u W u .
H u = H h cos θ sec θ = H h .
W u = W h cos θ .
a u = a h cos θ .
a p = a h cos θ ,