at the total solar Eclipse in 1882. 53 



To find the succeeding values of A let us take A 1 and A 2 . A 1 is 

 evidently equal to CBB 1 and to GAA X ; therefore 



, , _B X G 

 tan A. x — Yvn > 



and these are known from the coordinates of the given point B x , 

 therefore A x is known and the values of i x and m x follow as above. 

 Through B 2 draw B 2 D X at right angles to A X B X and cutting it at D x . 



Then D x B. 2 = m x sin i x , 



and D 1 B 1 = m 1 cos i x . 



GB X + D X B 2 



Then tan A, 



AG-{D X B X + A 2 B 2 ) 

 CB X + m x sin i x 



AG - AB -m 1 cosi 1 y 

 whence A 2 is found. 



The values of i 2 and m 2 follow as before, and we have 



. GB X + m 1 sin i x + m 2 sin i 2 



tan xLu — 



AG — AB — m 1 cos i x — m 2 cos i 2 



whence A s is found ; and by thus proceeding step by step the 

 elements of all the mirrors in the series are easily obtained. 



The diagram (Fig. 6) was actually constructed on the following 

 numerical data : — 



AB= 30 millimetres, BG = 20 millimetres, 



and CB X = 25 millimetres ; 



and the values of the different elements as calculated are collected 

 in the following table. It is to be remembered that tan A is 



CO 



always the quotient - of the coordinates of the point A, and that 



the values of y change sign at the origin. Thus for A 1} 



y x = 50 - 30 = 20 ; 



for A 2 , y 2 = 50 - 11-25 - 30 = 8-75 ; 



for A 3 , y s = 50 - 1125 - 2466 - 30 = - 15-91 ; 



and so on. 



In order to make the table complete the specifications of the 

 metal bands for the. mirrors are added. 



