!88 
MR. LOUIS VESSOT KING ON THE SCATTERING AND 
where 
<t.(C, i.) = (1-I!-““'>*){1-/(C)} 
+ cos^[B(C)-B{C(l+seo^)}+c-''“‘{B(C)-B[-C(sec^,-l)]}].. (56) 
We notice that for <p = 0, sec 0 = 1, 
4.(C,0) = (l-e-<=){l-/(C)}+B(C)-B(2C) + <;-={B(G)-y}, . . (57) 
where y here stands for Euler’s constant. 
For 0 = hir, sec <p = co (56) reduces to 
<I>(C,ix)= 1-/(0.(58) 
From (48) and (55) the expression for the intensity at the earth’s surface of 
radiation from that portion of sky which is in any direction 0 with the vertical and 
azimuth measured from a vertical plane through the sun is, for ^'’>0. 
T (f, t) = ^ 
S 
Csec 0 e G {C (sec sec 0)}( e )<h(C, 0) 
(59) 
while for ^<0 we have 
T (^„ i) = s 
Csec0e ‘^"‘'®^G{C(sec 0—seci"’)}+|-^( e ) <h (C, 0) 
\^ 2 / 
(60) 
From equations (3) and (7) we notice that 
Mo(^^) = f (1 + COS^ 0)/Zo = f ( 1 +COS^0)/Co/47r, 
I that the factor 
S/xo (^)/Ko = S (47r)“^ f (1 + cos^ d) cfC. . 
From the polar diagram fig. 1 we see that 
cos 6 = cos 0 cos ^ + sin <p sin ^ cos \p-. 
(61) 
(62) 
The first terms in equations (59) and (60) give the contribution to the intensity of 
sky light due to the sun’s radiation which has been once scattered by the atmosphere. 
If C is small and if the attenuation is due to scattering only, (c = C), the first 
term of (6O) gives for the scattered radiation coming from a direction 0 the value 
(47r) f (1 + cos^ 0 ) c sec 0 . 
(63) 
which agrees with the value obtained by Kelvin.* 
The first terms in (59) and (60) taken as they stand with C = c take into account 
the fact that both the incident and scattered radiations suffer attenuation ; this effect 
* Kkia'IN, loc. Hi. equation (19), p. 31.3. 
