396 
Mr, Christie on the magnetism of 
Also if (p is the angle of deviation of the magnetic particle 
due to the mass alone of the iron, then 
— 3 sin. X cos. X 
tan«.= ^ 
F r 
+ 3 sin.*x — I 
( 4 ). 
Since when the rotation of the iron produces no effect, 
= ^ s= we must have in this case, 
3 sin. (il/ + . cos. A — sin. 4 - 3 sin. (4 — x) . cos. \ — sin. ■«)/ 
3 sin. (4 + ^) • sin. A — cos. 4 3 sin. (4 • sin. x -f cos. 4 ’ 
whence 
sin. . cos. >)/ = o. 
The value of which satisfy this equation are o°, 90°, 180®, 
270°; and since 0° and 180° would in all cases give 
these must be excluded; also ^(/=27o° would merely give 
the same value for q>' which = 90° would give for 4^, and 
vice versa ; consequently we have %j/ = 90°. - 
If now we reduce the deviations 4', 4, 4^, of the magnetic 
particle in the line of the dip, to the horizontal plane, and 
call the corresponding horizontal deviations 6, 0^, and the 
angle of the dip S ; then since tan. 4 = cos. S tan. 9, and 
= 90°, the equations ( 2 ) , ( 3 ) , ( 4 ) will become 
Tan. 9 ^ * zmR* 
zfp 
3 sin. 2 X + Yy ' (3 cos. 2 A + i) 
_ — (3 cos. 2 X — i) + . 3 sin. 2 X 
Fr F r 
, 3 sin. 2 X — ^-'(3 C 0 S. 2 X + I) 
Tan. 9 = — r • — — — 
cos.^ l^._( 3 cos. 2 X-i)-^. 3 sin. 2 X 
F r 
Fr 
Tan. 9 • 7—^ 
cos. 2 m 
3 sin. 2 X 
F r 
— (3COS.2X— 1) 
(5), 
( 6 ). 
(7), 
The angles 9^ 9, 9„ are the horizontal angles of deviation of 
the south end of the magnetic particle towards the west ; so 
