The Rev. R. Murphy on Atmospheric Refraction. 311 



Hence by section 3, 



sm z = . sin ( » + w). 



l+m \^ ^ I 



§ 5. — Now ^ being the true zenith distance, z + p + w is 

 the angle which the issuing ray makes with the spectator's 

 vertical, and this angle diminished by the refraction r is the 

 apparent zenth distance. Hence 



sm (z + p + oj —r) = sm (p + u). 



§ 6. — The right line drawn from the moon to the spectator 

 makes with that drawn to the centre of the earth the angle 

 p, and with the spectator's vertical z + p; hence 



sin {z + p) : sin p : : a : 1, 



or sin [z + p) = a sin p. 



Hence sin z cos p = (ja — cos z) sin p 



sin z 



tan p = 



a — cos s 



c a^ — 2a cos z+l 



sec^ p = - 1^ ^^^ g — 



cos^ p = 



(a — cos zf 

 {a — cos z) 



•2acos« + l 



sin^s 



sm'' p — —g — , . 



•^ a^— 2fl!cosz+l 



i2 



a — cos z 



Iheretore cosp = — —-5 — :r \ 



^ x^ (a^—2acosz+l) 



sm 



^'"^ - -/(a2-2acos2 + l)* 



7. — By section 5, 



sin (r + oj— r) . cos p + cos (z + w—r) sin p 



(sin w cos p + cos w sin js). 



1+7K 



Hence cos p {(1 + m) sin{z + a> — ?•)— asin coj 



= sin p {a cos w— (1 + m) cos (2 + w— ;•) J. 



Therefore, by section 6, 



sin 2 {a cos w— (1 + m) cos (r + w — r)] 



= (a — cos :;) {(1 + m)sm {x + w — ?-) — a sin w}. 



This would be the complete solution of the question, were 

 w known. 



§ 8. — The angle &> is a function of the parallax, and is evi- 



