56 On a solar Calorimeter used in Egypt 



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

 ! is evidently equal to CBB 1 and to CAA l ; therefore 



and these are known from the coordinates of the given point 

 /?u therefore A^ is known and the values of t\ and ;//j follow 

 as above. Through B z draw B^D^ at right angles to A l B l 

 and cutting it at D. Then 



and 



Then tan 



~ACAB-m 1 cos ^ ' 

 whence /3 is found. 



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

 tan A 3 = 



CB l + 7/2, sin z'i + m. 2 sin i. 2 



A C AB mi cos t\ m. 2 cos z" 2 ' 



whence A 3 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 = 3O millimetres, BC = 20 millimetres, 

 and CJ5 1 = 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 always the quotient - of the coordinates of the point 

 A, and that the values of y change sign at the origin. Thus 

 for AI, jj = 50 30 = 20; 

 for A z , jj/ 2 = 50- 11-25-30 = 875 ; 

 for A 3 , 7 3 = 50- 11-25 - 24-66- 30 = - 15-91 ; 

 and so on. 



