284 On a Method of fixing a Transit Instrument 
explain the use and application of the table, and the mode of 
operating in such cases. 
On July 1, 1819, I placed my transit instrument nearly in the 
meridian; and in order to ascertain how much it deviated from 
the true meridian I observed the two stars y Lyr@ and + Sagit- 
tarii. The passage of the former was observed at 18>. 52’, 37”,3, 
and of the latter at 18". 56’. 4”,5 sidereal time. The apparent 
right ascensions of those stars, on that day, were 18%. 52’. 9”,8 
and 18>. 55’, 39”,7 respectively: and their declinations were 
32°. 27’ north, and 27°, 55’ south. Consequently the operation 
will stand. thus 
An's 18")'52., 9°58 T°=18), 52/.37",3 
A218, 55. 39,7 T= 18, 56. 959 
adR=-— 3. 29,9 aT =—: 3.932,6 
whence (dT—d R)=—2",7. This value, being negative, shows 
that the deviation is to the west: and in order to determine the 
quantity of the deviation, we must take the sum of the declina- 
tions (or the difference of the polar distances) of the two stars ; 
‘which in this case is equal to 60°. 22’; or, for the sake of round 
numbers, equal to 60° : and the declination of N (or the northern 
star) is about 32°. Consequently against the number 60 and 
under the column headed 32° we shall find 1:39; which being 
multiplied by —2",7 will give —3”’,75 for the deviation of the 
instrument in ¢ime: and this multiplied by 15 will give —56’,3 
for the deviation in arc westerly. 
- Again, on Jan. 1, 1820, having reason to suspect that the 
transit instrument (from some motion which had been given to 
it) deviated from the plane of the meridian, J observed the 
passage of « Canis majoris, and of Castor: the former at 
6, 52’. 45”,6, and the latter at 7°. 24’. 28’,4. The apparent 
right ascension of those stars on that day was 6°. 51’. 34”,3 and 
7", 23’. 7",3 respectively; and their declinations were 28°. 44’ 
south and 32°. 16’ north. Consequently the operation will 
stand thus 
ARP ss PB. 28.0 7" 5a T= 7», 24’. 28",4 
AM= 6. 51. 34,3 T?= 6. 52, 45,6 
os 
| d@R=+ 31. 33,0 dT= + 31. 42,8 
whence (d T—dR) = + 9,8. The sum of the declinations 
(or difference of polar distances) being in this case 61°, we shall 
find that the value to be adopted is 1°36; which being multi- 
plied by +9",8 will give + 13”,33 for the value of A in time, or 
(multiplying this by 15) 4-3’. 20” for the value of A in are. 
And this quantity being positive shows that the deviation was to 
the east. te 
