70 
ON A SEXTIO TORSE. 
[652 
We have r 2 =l+p 2 ; and thence also, if tan a = 2 p, = ± 2 \J(r 2 — 1), that is, 
then 
cos a. = 
1 
V(4r 2 — 3) ’ 
2 s/(r 2 - 
± V(4r 2 - 3) 
z — V(4r 2 — 3) cos (20 + a), 
showing that for a given value of r (or section by the cylinder x 2 +- y 2 = r 2 ) the 
maximum and minimum values of £ are z = ± \/(4r 2 — 3). 
that is, 
or, finally, 
But proceeding to eliminate 0, we find 
r 2 cos 20 = (1 — p 2 ) cos 20 — 2p sin 20, 
r 2 sin 20 = 2p cos 20 + (1 — p 2 ) sin 2$ ; 
or multiplying these by 1 + 3p 2 and 2p 3 and adding 
r 2 {(1 + 3p 2 ) cos 20 + 2p 3 sin 20} = (1 + p 2 ) 2 (cos 20 — 2p sin 20), 
r 2 {(3r 2 — 2) cos 20 + 2 (r 2 — 1)® sin 20} = r*z; 
r 2 z = (3r 2 — 2) cos 20 + 2 (r 2 — 1)« sin 20, 
which is the equation of the surface in terms of the coordinates r, 0, z. 
Observing that (3r 2 — 2) 2 + 4 (r 2 — l) 3 = r 4 (4r 2 — 3), we may write 
r 2 V(4r- — 3) cos /3 = 3r 2 — 2, 
r 2 V(4r 2 — 3) sin /3 = 2 (r 2 — 1)-, 
and therefore also 
and the equation thus becomes 
3r 2 - 2 ’ 
z = V(4r 2 — 3) cos (20 + /3), 
where £ is the altitude belonging to the azimuth 0 in the cylindrical section, radius r. 
The maxima and minima altitudes are + \/(4r 2 — 3), and these correspond to the values 
0 = + |/3, ■|'7r±|-/3, 7r + |/3, f 7T + ; it is to be further noticed that when r = 1, we 
have /3 = 0, but as r increases and becomes ultimately infinite, /3 increases to r, 
that is, -}¿/3 increases from 0 to \tt. 
It may be noticed that the surface is a peculiar kind of deformation, obtained by 
giving proper rotations to the several cylindrical sections of the surface z = V(4r 2 — 3) cos 20; 
viz. in rectangular coordinates this is r 2 z = \J(^r 2 — 3) (x 2 — y 2 ), that is, 
(x- +- y 2 f z 1 — (4 (x 2 + y 2 ) — 3} (x 2 — y 2 ) 2 = 0. 
To obtain the equation in rectangular coordinates, we have 
{*** - v “ 16 ( r2_ i ) 3 =°> 
viz. this is 
r*z 2 - 2z (Sr 2 - 2) (x 2 - y 2 ) + (3r 2 - 2) 2 (l - - 16 (r 2 - l) 3 = 0,