Mr. W. Brennan-1. 



QR, and the angular distance of an element in the zone QR from 

 SQ. 



Denote by I, the chemical action exerted by a circular area x of 

 the sphere, on the plane at right angles to the Sun. 



The area of an element will be rf0 dO sin 0, the intensity of the 

 chemical action will be i cosec 0. 



The angle between the normal to the element considered and that 

 to the plane AYX is 0. 



. ' . d . I, = d0 d$ sin x t a cosec x cos 



d<j> . de . COB = 2jn'sin0. 

 o 



Or, for the whole hemisphere, of which the Sun is the pole, 



from which, to get our desired result, we have to subtract the chemical 

 action I, of the gore XYH. 



nRSH p 



cos 0.d0.d<j> = 2i a RSH cos . dO. 

 -RSH - a 



(cos RSH = tan SH cot SR = tan * cot 0). 

 ..I = 2i a cos" 1 (tan * cot 0) cos dO 



J a. 



w 



= 2r. [Lim T (cos'i (tan cot 0) sin 0} - [* , ^""^ "I 

 L J a v/(l-sec 2 cos 2 (?)J 



_ 2 . Tw-_ f 1 tan a. . dO "| 



" a L^~| v/(l-sec 2 cos 2 e)J' 



Whence, subtracting this from equation (Q), 



This expression cannot be integrated in finite terms, but, by using : 

 formula of reduction in series, it gives 



which is the formula I have used in numerical computations. IH~^ 

 is the numerical value in the column headed "Sky alone" in 



