286 Mr. L. Fletcher on the Dilatation of Crystals 
Peop. IV. If two planes at right angles to the plane of sym- 
metry of an Oblique crystal be permanently atropic, no other 
planes in this zone retain their mutual inclination at all tempe- 
ratures (fig. 3). 
As before, let a, a be the planes which are permanently 
atropic, and let P, Q be planes isotropic for one pair of tempe- 
ratures; whence Pl^P'Q' and PP = QQ'. From the last 
corollary aP = QV if p, q, S be the points of bisection of the 
arcs PP 7 , QQ', act respectively, 
ap = qcc, pq = ?Q = F'Q', and ;?S=S?= ^- 
Whence, if the angle between the isotropic planes P,Qbe given, 
the positions of p, q, the middle points of the arcs of displace- 
ment PP", QQ', are absolutely definite. 
But if the temperature of the crystal be again changed and 
the poles P ; Q' take up new positions P ;/ Q", it is seen that the 
middle points of the arcs PP" and QQ" do not now coincide 
with p q; in other words, the planes P Q no longer retain their 
mutual inclination. 
Pkop. V. The following method of calculating the displace- 
ment of a given pole is sometimes more convenient than the 
method indicated in Prop. I. 
Let a, a be the poles of any pair of isotropic planes, and 
d, d' the positions of a third pole at the two temperatures : if 
P, P' be the corresponding positions of a fourth pole, we have, 
as before, 
[aP<fe]=j>TW], 
sin Pa sin da _ sin Vol sin d'al 
sin Pa sin da ~~ sin PV sin d'a! 
or 
sin Pa sin P V _ sin da sin d'a! . 
sin Pa sin PV sin da. sin d'a! 
and is therefore, so long as we keep to the same pair of tem- 
peratures, constant for all positions of P. 
Let 
PV-Pa = PV-Pa = 6>; 
then 
whence 
sin Pa sin PV __ 1 + tan 6 cot Pa # 
sin Pa sin PV ~ 1 + tan 6 cot Pa' 
1 + tan 6 cot Pa 
l+tan<9cotPa 1 + ^ 
where e is some small quantity dependent upon the tempera- 
tures, but independent of P. 
