1835.] Observations 071 the value of the cipher, Sfc, 375 



-it k here required to find the Moon's declination at the 

 moment of her rising without knowing when this hap-^ 

 pens ; if we adhere to direct calculation this problem 

 would remain unsolved, but by trial and approximation 

 we may gradually arrive at the truth. 



By inspecting an ephemeris of the Moon^we can easily 

 guess the declination of the Moon within a degree^ from 

 v/hich we can find the semidiurnal arc by the trigono- 

 metrical expression 



COS. hour angle — — cot Pol. Dis. x cot. lat. 

 the Moon in one hour moves from the meridian of any 

 place by (1 0 9.86 — ?n) therefore x- the time of describ- 

 ing this quantity will be 



^ _— sem. di. arc 



T — X ~ T' the approximate time of rising 

 If it wore required to ascertain the time of Moon ris- 

 ing to 5 or 6 minutes only, this first approximation would 

 be sufficient, but our object being to enter into the de- 

 tails of the calculation we will proceed. 

 . . . C Let a represent the A. R. of the Moon. 



At approxi-S , the sidereal time, 

 mate rism^. ^ jj ^^^^ ^^^^^ ^^^^ 



A the Moon's hourly motion in polar 



distance. 

 p the horizontal Paralax. 

 r the refraction in the horizon, 

 we have 



= S' -h T' + acc. of 



When the Moon's apparent zenith distance is 90° the 

 true zenith distance (which we will call z) == 90^ — • r 

 4- ^ sin (90 — ?•) and to compute 0 the hour angle of 

 the moon corresponding to this zenith distance, we have 

 in a spherical triangle, given the three sides z the zenith 

 distance, p the polar distance^ and c the co -latitude of 



