49 
at the total solar Eclipse in 1882. 
Therefore the width (ra 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 3 + D 3 B 4 = A B s + B 3 B± sin i = 5 + 331 sin 34° 6' = 6 - 85 inches. 
Further, if the line B 4 B 3 be continued till it cuts the axis in 
B 3 , then B±B 3 is the generating line of the complete cone of 
which the outer mirror is a portion. 
Its length is obviously 
• •» 7 ^ 7 ,v = 1223 inches. 
sm34 6 
The width of the mirror is 3*31 inches. Therefore the annular 
band of silvered metal has an outer radius of 12 23 inches and an 
inner radius of 8*92 inches, and the sector to be removed from it 
has an amplitude of 
360° 
12-23-6-85 
12-23“ 
158-5°. 
Inner Mirror. Through B 2 draw 0 2 P 2 , and on it lay off, below 
the line BB 2 , the length B. 2 A 1 = BA=2 inches. From A x at 
distance A 1 B 2 describe a circular arc. Join A X B. This line cuts 
the arc in B x . Join B X B 2 . B X B 2 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 x =-r-^ == # = 1’5 ; 
AB 2 
.-. A, = 56° 19', 
whence i x = 61° 51', 
and m x = 2 . AB cos i x — 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 X be continued to cut the axis in B x , B 2 B X 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 2 
sin 61° 51' 
= 3‘40 inches. 
VOL. XI. PT. I. 
4 
