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Between latitudes 45° and 25° the spherical excess amounts to about 1" for an area of 75.5 square miles. Hence, if the area in square miles be known, a close approximation to the spherical excess will be had by dividing the area by 75.5.
A Collection of Tables and Formulć Useful in Surveying, Geodesy, and ... - Page 54
by Thomas Jefferson Lee - 1853 - 242 pages

## A Treatise on Trigonometry, Plane and Spherical: With Its Application to ...

Charles William Hackley - Trigonometry - 1851 - 372 pages
...being known, that of the others may be obtained by the solution of the plane triangle. Between the latitudes 45° and 25° the spherical excess amounts to about 1" for an area of 75*5 square miles. To find the spherical excess in seconds of space therefore divide the area in square...

## Professional Papers of the Corps of Engineers of the United States Army

...c s being = — I - - — Between latitudes 45° and 25° the spherical excess amounts to about i" for an area of 75.5 square miles. Hence, if the area...spherical excess will be had by dividing the area by 7S-Slog mean radius of the eanh in yards = 6.8427917 If the three angles of a triangle are assumed...