https://bugzilla.wikimedia.org/show_bug.cgi?id=55817
--- Comment #2 from Laddo <[email protected]> --- Indeed the "linear precision" is a function of the latitude at that point. Here are the details of the calculation, using an ellipsoid rather than a sphere for more precision. Given Req = Radius of Globe at equator (Earth=63781370m) Rpo = Radius of Globe at pole (Earth=63567523m) lat = (angular) latitude of point Pan = angular precision (in degrees) Sought value Pln = linear precision Compute Tlat = eccentric anomaly = arctan( (Req/Rpo) * tan(lat) ) Rlat = Radius of Globe at latitude of point = Req * cos(Tlat) Plat = Perimeter of Globe at latitude of point = 2*π*Rlat Pln = Plat * (Pan/360) Thus linear precision is Pln = (Pan/180)*(π*Req*cos(Tlat)) Ex: lat 0° 30° 50° 70° 85° Tlat 0 30.08 50.09 70.06 85.016 ±0.1° ±11.1km ±9.63km ±7.14km ±3.80km ±967m ±1' ±1.85km ±1.61km ±1.19km ±633m ±161m ±1" ±30.9m ±26.8m ±19.8m ±10.5m ±2.69m -- You are receiving this mail because: You are on the CC list for the bug. _______________________________________________ Wikibugs-l mailing list [email protected] https://lists.wikimedia.org/mailman/listinfo/wikibugs-l
