372 



♦ KNOWLEDGE ♦ 



June 22, 1883. 



LAWS OF BRIGHTNESS. 



IV. 

 By Richard A. Pkoctob. 



BEFORE passing to the consideration of tlic total 

 brightness of globes when viewed so as to present 

 various phases, I propose to make a few remarks on a diffi- 

 culty which may suggest itself as respects what I said in 

 my last paper about the brightness of an illuminated 

 hemisphere viewed directly. It seems at first sight that 

 such a hemisplicre, liaving a larger surface than a disc of 

 equal diameter, ought to send to the observer at least as 

 much light, notwithstanding the obliquity of the illumina- 

 tion near the edge of the hemisphere. This reasoning really 

 amounts to the supposition that the relative faintness of 

 the illumination of a part near the edge of the hemisphere 

 is compensated by the greater area. Thus, if F E be a 

 small circular cylinder intersecting the surface of the hemi- 

 sphere in the ellipse M L (Fig. 8) the illumination being 



received in direction I P ; it seems as though the faint 

 illumination of L M should be made up by the greater area 

 of L M, compared with the area of the small circle D E in 

 the case of an illuminated disc as H K. As a matter of 

 fact, the large area L M receives the same amount of light 

 precisely as the small area D E ; but it sends it away in a 

 difierent manner. D E gives back most light in direction 

 P I ; but L M sends out most light in direction P N. As 

 seen from that direction the space L M appears as an oval, 

 having a much larger area than D E as seen (from the same 

 distance) in the direction I P. It is obvious, therefore, 

 that M L, viewed in the direction P I, in which it does not 

 give out its maximum amount of light, cannot look so bright 

 as D E (which receives exactly the same amount of light) 

 viewed in the same direction P I, in which it gives out its 

 maximum amount of light. 



Or we may xievr the matter in this way. The hemisphere 

 receives precisely the same amount of light as a disc of 

 equal diameter. But the light received from the disc grows 

 less and less down to nothhig as the disc is viewed at a 

 smaller and smaller angle from the right angle down to 

 zero (when the disc is viewed sideways). But the light 

 received from the hemisphere, though it also is reduced as 

 the illuminated hemisphere is viewed more obliquely, is 

 nevertheless not reduced to notJdng when the hemisphere 

 is viewed sideways, for then the hemisphere appears as a 

 half disc of light. The light is not reduced absolutely to 

 zero until the exactly opposite hemisphere is looked at 

 directly. Now, neither the disc nor the hemisphere can 

 give out more light than they receive, and they receive the 

 same quantity of light; hence, since the hemisphere gives 

 out light over a greater range as respects direction, this 

 must bo compensated by the circumstance that in that par- 

 ticular direction in which both the hemisphere and the disc 

 give out the maximum of light, the disc gives out more 

 light than the hemisphere. 



I have failed, after some trying, to find any method by 



which the total brilliancy of an illuminated sphere, when 

 presenting any given phase, can bo determined so that the 

 general reader can follow the demonstration. Perhaps 

 some of our mathematical contributors may be able to solve 

 the problem in a geometrical way. My analytical solution 

 is given below.* The result, which was long ago obtained 

 by Lambert (I do not know in what way), may be con- 

 veniently presented as follows : — 



Let A D B E (Fig. 9) be a smooth sphere illuminated 

 by a distant body, and let the length of the semicircular 

 arc A D B be taken to represent the total light received 

 from this sphere when A D B E is the illuminated hemi- 

 sphere. Then, when D G E is the terminator and D G E A 

 illuminated, the total light is represented liy the length 

 of the mixed line A H K G, obtained by describing the 

 semicircular arc F L G with G as radius, and drawing 

 thereto the tangent A H ; and when D G E is the ter- 

 minator, but D G E B illuminated, the total light is repre- 

 sented by the length of the line K B, diminished by the 

 length of the circular arc K G. It follows that when 

 D C E is the terminator, or the sphere " half full," the 



* Lot A D Q, Fig. 10, be the illnniinated portion of a smooth 

 disc, and let the arc D E (or the circular measure of the angle 

 DOE) = a. Let B be the centre of the ilhiminated hemisphere, 

 so that B D is an arc of 90°. Let K and S be two points in Q, 

 such that CS = ic, RS = Ji, and let two planes, at right angles to 



Fig. 10. 



O Q and passing through tlie points E and S, intersect the sphere 

 in F H, G K, so that G H is a narrow zone of the sphere ; and let 

 two pianos through O Q, inclined at angles 6 and 9 + B to O DQ, 

 rut the zone G H in L P and NM. Then the elementary area PN 

 is equal to rS 6 . S x approximately ; but it is reduced by fore- 

 shortening, so as to appear = rS 6 . 1 x cos C P approximately. And 

 if the illumination of a unit of area at C be I, the illumination at 

 P is equal to I cos B P. Hence tlie light received from the ele- 

 mentary area PN = L r . S 9 .S x cos B P. Cos C P approximately. 



