Books Books 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 Formulæ 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... ## A Text Book on Surveying, Projections, and Portable Instruments.

1876
...of earth, logarithm of mean radius of earth in yards is 6.8427917 Between latitudes of 25° and 45° the spherical excess amounts to about 1" for an area...spherical excess will be had by dividing the area by 75.5 miles. If the three angles are assumed to have been equally well measured, the previous determination... 