 | Charles William Hackley - Trigonometry - 1851 - 536 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... | |
 | United States. Army. Corps of Engineers - Military engineering - 1873 - 350 pages
...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... | |
 | 1876 - 172 pages
...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... | |
 | Anthony Charles Cooke - 1879 - 438 pages
...68-198 Dip. n 8 33-4 9 4-54 9 34 11 43 13 31'78 16 34-2 19 8 21 as-r, 30 15'1 42 47 51 23-9 60 30'S — 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 t» tho spherical... | |
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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. 