# User:Peter Mercator/Draft for Legendre

## Legendre's theorem (geodesy)

Let ABC be a spherical triangle on the unit sphere with small sides a, b, c. Let A'B'C be the planar triangle with sides of the same length. Legendre's theorem states that the angles of the spherical triangle exceed the corresponding angles of the planar triangle by approximately one third of the spherical excess. The theorem was very important in simplifying the heavy numerical work in calculating the results of traditional (pre-GPS and pre-computer) geodetic surveys from about 1800 until the middle of the twentieth century.

The theorem was stated by Legendre (1787) who provided a proof (1798) in a supplement to the report of the measurement of the French meridional arc used in the definition of the metre (Delambre 1798). Legendre does not claim that he was the originator of the theorem despite the attribution to him. Tropfke (1903) maintains that the method was in common use by surveyors at the time and may have been used as early as 1740 by La Condamine for the calculation of the Peruvian meridional arc.

Girard's theorem states that the spherical excess of a triangle, E, is equal to its area, Δ, and therefore Legendre's theorem may be written as

\begin{align} A-A'\;\approx\; B-B'\;\approx\; C-C'\;\approx\;\frac13 E\;=\; \frac13\Delta,\qquad a,\;b,\;c\,\ll\, 1. \end{align}

The excess, or area, of small triangles is very small. For example consider an equilateral spherical triangle with sides of 60km on a spherical Earth of radius 6371km; the side corresponds to an angular distance of 60/6371=.0094, or approximately 10-2 radians (subtending an angle of 0.57° at the centre). The area of such a small triangle is well approximated by that of a planar equilateral triangle with the same sides: 12a2sin(π/3)=0.0000433 radians corresponding to 8.9″.

When the sides of the triangles exceed 180km, for which the excess is about 80″, the relations between the areas and the differences of the angles must be corrected by terms of fourth order in the sides, amounting to no more than 0.01″:

\begin{align} \Delta &=\Delta'\left( 1+\frac{a^2+b^2+c^2}{24} \right),\\ A&=A'+\frac{\Delta}{3} +\frac {\Delta}{180} \left({-2a^2+b^2+c^2} \right),\\ B&=B'+\frac{\Delta}{3} +\frac {\Delta}{180} \left({\quad a^2-2b^2+c^2} \right),\\ C&=C'+\frac{\Delta}{3} +\frac {\Delta}{180} \left({\quad a^2+b^2-2c^2} \right). \end{align}

This result was proved by Buzengeiger (1818)—an extended proof may be found in Osborne (2013) (Appendix D). Other results are surveyed by Nádeník (2004).

The theorem may be be extended to the ellipsoid if a, b, c are calculated by dividing the true lengths by the square root of the product of the principal radii of curvature (see Osborne (2013) Chapter 5) at the median latitude of the vertices (in place of a spherical radius). Gauss (1828, Art. 26–28) provided more exact formulae.

Example

Legendre's theorem may be used to correct observational errors as well as to facilitate the calculations of triangles or networks of triangles (Clarke 1880, Ch.9). Suppose that angle observations are made at three locations A, B, C and further suppose that the baseline BC=a=38,386.67m is a known distance. The measured angles, which will have errors, are

A0=50°01'59.15″,    B0=86°03'08.44″,    C0=43°54'55.07″.

The observed spherical excess calculated from these angles is 2.66″. But we can also calculate the excess as the area of the planar triangle using Δ=Δ' if we neglect the fourth order terms. A first approximation of the area is obtained by using the raw observed angles to calculate

$\Delta'=\frac{1}{2}bc\sin A_0=\frac{a^2\sin B_0 \sin C_0}{2\sin(B_0+C_0)}.$.

The second equality follows by applying the planar sine rule twice and setting A≈π-B-C. The corrections to this approximation will be small. For the given values of a, B and C the area, and excess, work out as 3.38″. This contradiction in values for the spherical excess indicates errors in the observed values which must be increased by 0.24″ each to compensate. The new values are

A0=50°01'59.39″,    B0=86°03'08.68″,    C0=43°54'55.31″.

These values are then used for a more accurate calculation. Note that the angles reduced by one third of the excess, 1.13″ are

A1=50°01'58.26″,    B1=86°03'07.56″,    C1=43°54'54.18″

and the sum of these angles is exactly 180°, to the second decimal place of seconds, as it must be for a planar triangle. The final step is to calculate the sides of the planar triangle (with the planar sine rule) using a and these angles. The result is b=49,967.30m and c=34,739.31m. This level of accuracy is not improved by using the fourth order expressions.