Determination of the Earth's Horizontal Magnetic Force. 405 



division 30 or 40 cm. on the bar. From the centre of the upper edge 

 of the frame a pillar extends vertically upwards, so as to fit into the 

 hollow shank which forms part of the appendages of the collimator 

 magnet. The vertical plane perpendicular to the bar through the 

 fiducial mark should contain the centre of the vertical pillar and also 

 that of the magnet. Supposing the frame to have fiducial marks on 

 both sides, and that these agree, one can test the existence of the first 

 possible cause of asymmetry by turning the frame end for end, keep- 

 ing the direction of the magnet unchanged ; the azimuth of the mirror 

 magnet will serve as a criterion. This kind of asymmetry has not been 

 detected in the Kew unifilar ; and if existent its influence would be 

 nearly eliminated in a large number of observations in which either 

 position of the sliding frame occurred promiscuously. The existence 

 of the second species of asymmetry is less easily tested, but is more 

 probable a priori. Accordingly my second assumption is that during 

 the deflection experiments the centre of the magnet is at a distance y 

 from the vertical plane which passes through the fiducial mark, and is 

 perpendicular to the magnet's length. Here y is counted positive 

 when the magnet's centre and north pole are on the same side of the 

 fiducial mark. 



§ 41. On these two hypotheses it is obvious from fig. 1, that when 

 the apparent distance between the centres of the magnets is r, the true 

 distances are as follows : — 



Position. 1. 2. 3. 4. 



True distance r + y - z r — y- z r + y + z r - y -f z. 



Now suppose that u and u + 8u are the deflection angles of the 

 mirror magnet answering to the true distances r and r + oY. Then, 

 according to the first approximation formula, 



(r + 6V) 3 sin (u + 8u) = r 2 sin u ; 



whence, neglecting squares and products of 8u and 8r, we have 



Sit = -3r~ 1 tan«8r (10). 



In what follows I employ u when r — 30 cm., and u' when r = 40 

 cm., and distinguish the several positions of fig. 1 by suffixes. 

 Eeferring to the scheme of values of 8r given above, we find — 



Position 1. Position 2. 



= - To (V ~ z ) tan S% 2 = to (y + Z) tan 7I > 



tu\ = - (y - z) tan u'. 8u'. 2 = ^ (jj + z) tan v! . 



Position 3. Position 4. 



s % = - T V(y+'-) tan? '> ^4 = A (y-z)tanw, 



6V 3 = - - 4 % (y + z) tan u. Sw' 4 = fa (y - z) tan u'l 



