1899 - 1900 .] Prof. Macfarlane on Hyperbolic Quaternions. 175 
completely justified. Let the two versors POA and AOQ (fig. 4) 
be denoted by e b P and e c y, the axes f3 and y being both perpen- 
dicular to the axis of revolution a, and of unit length. 
Then eW e c v = (S + Y)(S' + Y') 
= SS' + S'Y + SY' + YY' 
f= cosh b cosh c 4- cosh c sinh b’fi + cosh b sinh c*y 
+ sinh b sinh c'/3y. 
Now /3y = cos /3y + J - 1 sin /3y’/3y 
— cos /3y + V — 1 sin /3y‘a. 
Hence e & % c v = cosh b cosh c + sinh b sinh c cos /3y 
+ cosh c sinh b’/3 + cosh b sinh c’y + - Isinh b sinh c sin py'a. 
Hence cosh ePPe°y = cosh b cosh 6 + sinh b sinh c cos /3y 
and Sinh e b Pe°y = cosh c sinh b'/3 4- cosh b sinh c‘y 
+ v - 1 sinh b sinh c sin fiy'a. 
The first and second components of the directed sinh (denoted 
by Sinh) are perpendicular to the axis of revolution, hence their 
resultant cosh c sinh b’/3 + cosh b sinh c y is also perpendicular to the 
principal axis. Let it be represented by OY in the figure. The 
difficulty consists in finding the true direction of the third com- 
ponent J _ i sinh b sinh c sin /3y’a on account of the presence of 
s/ - 1. It will be found that n/ - 1 has here nothing to do with 
the direction ; and as the term is otherwise in the positive direc- 
tion of a, we represent it by OU in the figure. In the case of the 
sphere OU is drawn in the direction opposite to a. Let OR be 
the resultant of OU and OY ; it represents the directed Sinh both 
in magnitude and direction. 
The square of the length of OR is 
cosh 2 c sinh 2 b + cosh 2 b sinh 2 c + 2 cosh c cosh b sinh c sinh b cos /3y 
+ sinh 2 b sinh 2 c sin 2 /3y. 
Rut the square of the modulus of OR is the same with a nega- 
tive sign before the last term ; added to the square of cosh e^e c v it 
yields 1. 
The directed sinh OR is not normal to the plane POQ ; how is 
it related to that plane? If we draw OU'= - OU and find OR' 
