566 
Mr. Airy on the 
x cos 4 lat = 39,01606 -j- ,20292 x sin 2 lat —,00844 x sin 2 
lat . cos 9 lat . = 39,01606 1 1 + ,005201 x sin a lat — ,000216 
x sin 2 lat . cos 2 lat^ ; and gravity is proportional to this. 
Comparing the coefficients of sin 2 lat and sin a lat . cos* lat 
with those in the expression of Art. 19, we have the follow- 
ing equations 
Y™ e -S*e*-"^me-|- y A = >005201 
— me — - + 3 A =,000216 
2 2 8 7 
From the length of the equatorial pendulum, supposing the 
equatorial radius of the earth = 3486908 fathoms, we find 
m = ,003464. And in the terms of the second order instead 
of e we may put A., which is certainly not far from the truth. 
Making these substitutions, the equations become 
,008657 — e + y A = ,005201 
,000024 -|- 3 A =,000216 
From these, e = ,003474 = ~^Ts ’ an< ^ A = 000064. The 
sign of A shows that the earth is less protuberant at the 
latitude of 45 0 than an ellipsoid of the same polar and equa- 
torial radii. For the elevation of the spheroid above the 
ellipsoid = a A . ^ . fx 2 — 1, where f* = sine of latitude ; 
making the latitude = 45°, jx— and the elevation of the 
spheroid is — A : as A here is positive, this elevation is ne- 
gative, or the spheroid at that latitude is depressed below the 
ellipsoid. This I should be inclined to expect : for though I 
have not been able to solve the differential equation in A , 
even in the cases in which the differential equation for e can 
