500 Lieut. Col. Lambton's abstract of the results deduced 
Since the degree on the meridian, and the degree perpen- 
dicular to the meridian, are equal at the pole, we shall have 
by equations 7, and 1 3 , p ( ) = m ( ) ; 
where p and m are in the same latitude l. Now 
m ( ~ y + sL ~i ) = m (i + 3«) . ( 1 + 3C. Sin.’/) _ ' = 
~m (i 4 %e. Cos. 9 /) nearly ; and therefore p | — LLl_J == 
—P( l 4 ” e) . ( i 4 - e ’ Sin .*/)”" 1 = p (l 4 e. Cos. 8 /) nearly. 
Hence m(i 4 S e - Sin . 2 /) = p (l 4 " e * Cos.®/) which reduced 
• • • (14) 
° (3 m — p) Cos. 2 / 
Since m { i 4 * S e • Co s. “/) = /> ( i 4 *• Cos. 2 /), we get 
m : p : : i 4 ~ e Cos . 2 / : i 4* 3 e Cos.® I . . ( 15) 
and when / =s 0, and its Cos. equal 1 , then m : p : : 1 4 e : 
1 + 3 * .... (1 6) 
If we make use of the degrees of longitude, then let d‘ and 
d represent their measures in latitudes 'l and / ; and their 
respective radii of curvature at 'l and / will be expressed by 
rns and . .yr?r- T . - A. .and there- 
3 (Cos* 7+ (i — 
fore d’:d : : 
that is d! : d: 
2^) Sin. 2 7 )| 2 (Cos. 1 / -f (1 — ze). Sin. 2 /)§ : 
Cos. 7 Cos. I 
(Cos . 1 7 4 (i — ze) Sin. 2 , Z) 4 » (Cos. 1 / + (i — ze) Sin .*/)£ * 
Cos. */ Cos. / ,1 . • . 
: 7- — c . ■■ —7, 7 ; 7- - g . ■ ; that is d : d : ; 
(i ■— 2t?. Sin. 1 7)4 ’ (1 — ze- Sin, 2 /)y ’ 
Cos. 7 (1 — 2£. Sin. 2 7 )“ 1 ; Cos. / (1 — - 2e. Sin. 2 /)““£ ; that is 
d ' : d : : Cos. 7 (i+t. Sin. 2 7 ) : Cos. / (1 4 Sin. 2 /) . . (17) 
d! Cos. / — d. Cos. 7 . . 
d. Cos. 7 . Sin . 2 7 — d‘ . Cos. /. Sin. 2 / ( * ® ) 
and this reduced gives e 
since d' : d : : Cos. 7 ( 1 4 f • Sin . 9 7 ): Cos. / ( 1 4 e - Sin. 2 /) ; 
if d be at the equator where Sin. I vanishes ; then 
d : d r : : 1 : Cos. 7 (1 4 e • Sin . 2 7 ) .... ( lg) 
