565 
figure of the earth. 
The irregularities in the lengths at the places nearest to the 
equator are considerable ; but the number of these places is 
so great, that the errors will probably destroy each other. 
Assuming the length of the seconds’ pendulum = M + 
N cos 2 lat -j- P cos 4 lat. the errors are 
— 39,02074 + M + N x ,99989 + P x ,99959 
— 39,01214 + M + N x ,99610 -f Px ,98445 
— 39,02410 + M 4- N x ,96194 + P x ,85063 
— 39,01997 + M + N x ,95640 4- P x ,82938 
— 39,01884 4- M 4- N x ,93171 4 -Px ,73611 
— 39,02425 4- M 4- N x ,89895 4 “ P x ,61626 
— 39,03510 4- M 4- N x ,81034 + P x ,31330 
— 39,10168 4- M 4- N x ,14910 — P x ,95552 
— 39,13929 + M — N x ,22559 — P x ,89821 
— 39,i 7456 4“ M — N x ,59992 — P x ,28024 
— 39,19519 4 - M — N x ,78082 4- P x ,21936 
— 39,20335 4 -M-Nx ,85784 4- Px ,47184 
— 39,21469 4“ M — N x ,93767 4“ p x ,75849 
To determine M, N, and P, by the method of least squares, 
we must form three equations, by making equal to nothing 
the sum of these errors — -first, when each is multiplied by 
the coefficient of M in that line, then when each is multi- 
plied by the coefficient of N, and finally when each is multi- 
plied by the coefficient of P. Thus we find 
0= — 508,18390 + M x 13 + N x 3,30259 4- P x 4,64544 
o = — 128,29476 4 - Mx 3,30259 4- Nx 8, 82265 4 - Px 4, 02629 
0= — 181,31213 4" Mx 4,64544 4“ ^x 4,02629 4- P X 7,04391 
By the solution of these equations M = 39,11647: N = 
— ,10146 ; P = ,00106 ; and the length of the seconds’ pen- 
dulum in inches = 39,11647 — ,10146 x cos 2 lat 4“ ,00106 
MDCCCXXVI. 4 D 
