Dr. Brewster on the laws of polarisation, &c. 239 
4/ e=3 180° — 2 » + w = 2 ^4 « 
When A and C are both positive, or both negative, and 
C -7 A 
a = Sin. 2 CE , and 6 = Sin. 2 EA x Sin. 4 GC 
When A and B are both positive, or both negative, and 
A f B 
a — Sin. 2 AE , and b — Sin. 2 BE x Cos. 4 GC 
When B and C are the one positive, and the other ne- 
gative, 
a = Sin. 2 EC x Cos. 2 GC, and 
6 = Sin. 2 BE x Sin. 4 GC. 
The tints produced separately by each axis being thus de- 
termined, the tint resulting from their joint action will be 
found to be the diagonal of a parallelogram, whose sides are 
a,b and whose angle isv|/. In order to find this diagonal, 
We have Tang, f 2 and 
£ i 4 = Greater angle at the base ; 
xt nn cl Sin. 4/ 
Hence I = ^ — tft-t-tt- 
Sin. (f + i 40 
When a — b then T == 2 a (Cos. w + l ) 
When a = b, and the axes equal, then tt = & and 
T — (Cos. 2 tt) or T = 2 a (Cos. 2 ») and since <p = 6 
T = 2 Sin . 9 <p ( Cos. 2 7 r) 
When 4 = 90° then T = Vf + b z 
When 4 = 180° T = a — b 
When 4 = 360® T == a -f- 6 
This general law of the tints may be expressed in the 
following manner : the tint produced at any point of the sphere 
by the joint action of two axes, is equal to the diagonal of a 
parallelogram , whose sides represent the tints produced by each 
