## Abstract

The solution to a problem in asymmetric reflector design for illumination is described. Working backward from a required rectangular and nonuniform light distribution pattern, the reflector geometry is arrived at by a combination of calculus and geometric construction. The central curve is computed from the general reflector equation lnR = tan (θd/2) , the integration being performed on an approximate matching function or graphically. The off-center contours are then obtained by means of an original drafting-board construction which is described, and the complete reflector shape is thus defined.

© 1966 Optical Society of America

### Cited By

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### Figures (9)

Fig. 1

Diagram of desired illumination coverage from street lighting unit.

Fig. 2

Ideal vertical candlepower distribution curve for street lighting unit.

Fig. 3

Reflector diagram.

Fig. 4

Candlepower in percent of luminaire maximum. A, bare lamp candlepower; B, luminaire candlepower; C, luminaire candlepower less bare lamp candlepower (BA); D, reflector candlepower (luminaire candlepower with glassware removed less bare lamp candlepower).

Fig. 5

Ray diagram of central reflector contour.

Fig. 6

Bare reflector candlepower vs vertical angle or reflected light.

Fig. 7

Reflector radius vs reflector angle in central contour plotted in (left) rectangular coordinates and (right) polar coordinates.

Fig. 8

Diagram of incident and reflected light rays striking rotating mirror.

Fig. 9

Geometric construction for determining off-center reflector contours.

### Equations (35)

$R s = O X O L .$
$C : n k C 0 = d θ : d α$
$C = n k C 0 d θ d α .$
$tan β = d R R d θ$
$θ = α + 2 β$
$β = θ - α 2 .$
$tan θ - α 2 = d R R d θ .$
$k = reflected flux incident flux = ∫ 0 α 1 C d α ∫ 0 θ 1 n C 0 d θ$
$n k ∫ 0 θ 1 C 0 d θ = ∫ 0 α 1 C d θ .$
$n k C 0 θ = ∫ 0 α 1 C d α .$
$C = A α 2 + B - C ( D - α ) - 1 / 2 .$
$C = 102.8 α 2 + 105.3 - 73.0 ( 1.4212 - α ) - 1 / 2 .$
$C d α = n k C 0 d θ .$
$∫ C d α = n k C 0 ∫ d θ = n k C 0 θ .$
$n K C 0 θ n K C 0 θ 1 = ∫ C d α ∫ 0 α 1 C d α = θ θ 1$
$θ = θ 1 ∫ 0 α 1 C d α C d α ,$
$θ = T ∫ C d α ,$
$T = θ 1 ∫ 0 α 1 C d α .$
$d R R = tan T ∫ c d α - α 2 d θ ,$
$d R R = tan ( T 2 ∫ c d α - α 2 ) d θ ,$
$d θ = d ( T ∫ c d α ) = 2 d [ T 2 ∫ c d α - α 2 + α 2 ] ,$
$d θ = 2 d [ T 2 ∫ c d α - α 2 ] + d α ;$
$d R R = 2 tan ( T 2 ∫ C d α - α 2 ) d ( T 2 ∫ c d α - α 2 ) + tan ( T 2 ∫ c d α - α 2 ) d α .$
$ln R + K 1 = - 2 ln cos ( T 2 ∫ c d α - α 2 ) + ∫ tan ( T 2 ∫ c d α - α 2 ) d α .$
$ln R R 0 = - ln cos 2 ( T 2 ∫ c d α - α 2 ) + ∫ tan ( T 2 ∫ c d α - α 2 ) d α$
$R R 0 = ln - 1 ∫ tan ( T 2 ∫ c d α - α 2 ) d α cos 2 ( T 2 ∫ c d α - α 2 ) .$
$θ = T ∫ [ A α 2 + B - C ( D - α ) - 1 / 2 ] d α ,$
$θ = T [ A α 3 3 + B α - 2 C D 1 / 2 + 2 C ( D - α ) 1 / 2 .$
$R R 0 = ln - 1 ∫ tan T 2 [ A α 3 3 + ( B - 1 T ) α - 2 C D 1 / 2 + 2 C ( D - α ) 1 / 2 ] d α cos 2 T 2 [ A α 3 3 + ( B - 1 T ) α - 2 C D 1 / 2 + 2 C ( D - α ) 1 / 2 ]$
$C = A α 2 + B - C ( D - α ) - 1 / 2 .$
$44 = B - C D - 1 / 2 ,$
$101 = 1.13 2 A + B - C ( D - 1.13 ) - 1 / 2 ,$
$0 = 2.26 A - c 2 ( D - 1.13 ) - 3 / 2 ,$
$0 = 1.36 2 A + B - C ( D - 1.36 ) - 1 / 2 .$
$C = 102.8 α 2 + 105.25 - 73.0 ( 1.4212 - α ) - ½ .$