On Shooting Stars. 99 
Let A be one, and B the other place of observation upon the 
earth, whose center is C, pole P, equator NQM, and X the observ- 
ed point. From A,B,X drop perpendiculars to the plane of the 
equator, meeting itin a,b,r. Put A’=the Right Ascension of mid- 
heaven for B, B’ its geographical latitude, R’ its distance from the 
center of the earth; and let A”, B’, R” represent the same for the 
point A, then is Ch=R’ cos. B’, Ca=R” cos. B”, and aCh=A” — 
A’. Call the Right Ascension of the meteor as seen from B=a’, 
its Declination=0’, and let a”, b’ represent the same for the point A, 
then is xbL=a’ — A’, xaK=a—A”; and if x represent the 
Right Ascension of midheaven for the point U where the meteor — 
stood in the zenith, thenis tCa=x - A”, tCh=x—A’. Wethen 
Cb. sine xbL Ca. sin. raK , 
shiv Casein. Cab 1 en, 
R’ cos. B’. sin. (a’— A’) R” cos. BY”. sin. (a”—A”) 
Com ; Pe aE Say 2 a 
sin. (a’— <2) sin. (a” — x) 
whence we easily obtain | 
R’cos. B’sin. (a’ — A’)sin. a” —R“cos. BYsin. (a’’— A”’)sin.a” 
*®=R’cos. B’sin. (a’ — A’)cos. a” — R”cos. B’sin.(a’’ — A”)cos.a’ 
Hence x — A’, the difference of longitude between the place of ob- 
servation B, and the place where the meteor stood in the zenith, is 
given. 
tang 
