at the total solar Eclipse in 1882. 49 



Therefore the width (m 3 ) of the outer mirror is 3-31 inches, and it 

 is inclined to the axis at an angle of 34° 6'. 



The radius of the upper rim of the outer mirror is 



AB S + A#4 = AB 3 + B 3 B, sin i = 5 + 331 sin 34° 6' = 6'85 inches. 



Further, if the line B^B^ be continued till it cuts the axis in 

 B 3 , then B 4 B 3 is the generating line of the complete cone of 

 which the outer mirror is a portion. 



Its length is obviously 

 6-85 



sin 34° 6 



-, = 12*23 inches. 



The width of the mirror is 3*31 inches. Therefore the annular 

 band of silvered metal has an outer radius of 1223 inches and an 

 inner radius of 8*92 inches, and the sector to be removed from it 

 has an amplitude of 



Qgn0 12-23 -6-8 5 



360 ~ 12-23 =1 ° 85 - 



Inner Mirror. Through B 2 draw 2 P 2 , and on it lay off, below 

 the line BB,, the length B,A 1 =BA=2 inches. From A 1 at 

 distance A X B, describe a circular arc. Join A X B. This line cuts 

 the arc in B x . Join B 1 B.,. BjB, is evidently the line of section 

 of the inner mirror. The demonstration is the same as in the case 

 of the outer mirror. 



Also tan A 1 = -^ = % = 1*5 ; 



AB - 



.-. ^ = 56° 19', 



whence i x = 61° 51', 



and m 1 = 2 . AB cos i, = 1'88 inches. 



Therefore the inner mirror has a width of 1-88 inches and is 

 inclined to the axis at an angle of 61° 51'. 



Further, if B 2 B 1 be continued to cut the axis in BJ, B. 2 B^ is 

 the length of the generating line of the complete cone of which 

 the mirror forms a part, and it is the greater radius of the annular 

 disc of silvered metal out of which the band is to be cut. Its 

 length is 



BB, 



sin 61° 51' 



VOL. XI. PT. I. 



3 '40 inches. 



