| M = ∫φ2φ1 ρdφ = α ∫φ2φ1 | (1 − e²) dφ |
| (1 − e² sin²φ)3/2 |
where a²e² = a² − b²; instead of using the eccentricity e, put the ratio of the axes b : a = 1 − n : 1 + n, then
| M = ∫φ2φ1 | b (1 + n) (1 − n²) dφ | . |
| (1 + 2n cos2φ + n²)3/2 |
This, after integration, gives
| M/b = (1 + n + | 5 | n² + | 5 | n³)α0 − (3n + 3n² + | 21 | n³)α1 + ( | 15 | n² + | 15 | n³)α2 − ( | 35 | n³)α3, |
| 4 | 4 | 8 | 8 | 8 | 24 |
where
| α0 = φ2 − φ1 α1 = sin (φ2 − φ1) cos (φ2 + φ1) α2 = sin 2(φ2 − φ1) cos 2(φ2 + φ1) α3 = sin 3(φ2 − φ1) cos 3(φ2 + φ1). |
The part of M which depends on n³ is very small; in fact, if we calculate it for one of the longest arcs measured, the Russian arc, it amounts to only an inch and a half, therefore we omit this term, and put for M/b the value
| (1 + n + | 5 | n²)α0 − (3n + 3n²)α1 + ( | 15 | n²)α2. |
| 4 | 8 |
Now, if we suppose the observed latitudes to be affected with errors, and that the true latitudes are φ1 + x1, φ2 + x2; and if further we suppose that n1 + dn is the true value of a − b : a + b, and that n1 itself is merely a very approximate numerical value, we get, on making these substitutions and neglecting the influence of the corrections x on the position of the arc in latitude, i.e. on φ1 + φ2,