60 W. A. Norton on the Dimensions of Donati’s Comet. 
tang 6=0:018965 (arc° tang aan and ¢=interval from Sept. 
7 Hig to Oct. 5¢-0776=54-325. The calculation gives 6=16° 
60,568,700 _=59 “ 7 a. ae <1 le 
"F101 96 cus 18° TBO ee ee Flee UA 
With these values of > a 7 we obtain, 
; R. Asc.=214° 1’ 10"; Dec.=22° 23’ 0” N. 
As another example we will determine the position on Oct. 
54-0776, of a particle which left the sphere of influence of the 
nucleus on Sept. 244-360, and — emitted from the nucleus on 
its preceding side, and under an angle to the radius-vector of 
19°. is is the average limiting angle of emission, for the outer 
envelope, on Sept. 24th and Oct. 2d. We have W=1-218. 
By equs. (10) and (11), 0,=16° 50’; r,=56,209,000™; Vi= 
84m-785; @=81° 85’. By equs. (7) and (8), (aking k= 141-218 
=2°213, ‘and reducing from distance 55,000,000™ to 56,209,000"), 
== Oar2 oy Thus initial velocity =34m- 785-+-0m- ‘254=35™-039. 
mi =-=1- 676085; m/’ sin?6=1-64009; e=/m! sin*6(m'+2)+1 
=2°65125. By equ. (15) ®=5° 15’ 80”. 
94: tang @\., 
Equ. (16) becomes, 7=24'5174 tang (are® are whence 
T=1*-7061. 
Interval from Sept. 244-360 to Oct. 54-:0776=104-7176. t= 
10¢-7176—14-7061=94-0115. 
Equ. (17) becomes, tang 9=0-0231238 (are® tang aa )i 
whence 6=25° 0’ 55”. 
rie 3 (m' sin?8. R) == 79646743; and by equ. (18), 
*=65 736,500 
4=0,—®= Ae 50’—5° 15’ 80"=11° 34’ 30"; y= 0—4 =25° 
0! 55" ~11° 34’ 30"=18° 26! 25”. 
These values of r and 7 give 
R. Asce.=219° 14’ 40”; Dec.=29° 4’ 20” N. 
RESULTS. 
The positions of particles have all been determined for Oct. 
540776. The several dates at which they are supposed to have 
left the sphere of ee =" the nucleus, are Oct. 24-6048, 
Sept. 291-752 (Per. pam ), Sept. 264-9003, Sept. 244360, Sept. 
194-049, and Sept. 134 834. The several values attributed to the 
intensity of the sun’s effective force are as follows; for the ae 
sive force, 0, 0-455, 1:218, 1°82, and 2°73; for the attractive force, 
0, 0°308 0-455. 
