74 Mr. Robertson on the Precession 
The point E of the equator, nearest the sun, is at the dis- 
tance of a quadrant from either extremity of the axis of 
libration. For, by hypothesis, SC is at right angles to the 
plane DCF, and therefore the axis of libration, which is in 
this plane, is at right angles to SC. Again, as PC the axis of 
the earth is at right angles to the plane of the equator, the 
axis of libration, which is also in the equator, is at right angles 
to CP. The axis of libration, therefore, being at right angles 
to CS, CP in the plane PELQ, is at right angles to CE in the 
same plane. 
18. Let ADBE (Fig. 10.) be an oblate spheroid of which 
AB is the transverse axis, DE the conjugate axis, and C the 
centre. Let AKMBLH be the equator of this spheroid, and 
consequently at right angles to ADBE the generating ellipse. 
Let the spheroid be cut through DCE by a plane DMEL at 
right angles to ADBE, and let MCL be its common section 
with the equator. Then (19. XI.) MCL is at right angles to 
ADBE, and therefore ACM is a right angle ; and as ACD is 
a right angle, AC is at right angles to the plane DMEL, and 
consequently at right angles to any plane parallel to it. Let 
the spheroid be cut by a plane parallel to DMEL, and let the 
common section of this plane with the spheroid be the ellipse 
FKGPI. Let this ellipse cut the plane ADBE in the straight 
line FrG, and the plane of the equator in KrH, the point r 
being the centre of this last formed ellipse, or that point in 
which it meets AB. Then, by a well known property of 
the spheroid, the ellipses DMEL, FKGH are similar, and the 
area of the first is to that of the other as CA* or its equal CM® 
to rK\ Put a — AC or CM, e = DC, x = Cr, and the force 
of each particle in the spheroid being as its distance from the 
