In Calculating View Angle? the accepted answer contains the following calculations:
Given two points (x,y,z) and (x',y',z') in an earth-centered coordinate system, the vector from the first to the second is (dx,dy,dz) = (x'-x, y'-y, z'-z)
And then provides an example:
The XYZ coordinates of the airplanes are (x,y,z) = (1285410, -4797210, 3994830) and (x',y',z') = (1202990, -4824940, 3999870), respectively (in the ITRF00 datum, which uses the GRS80 ellipsoid). The pilot's view vector therefore is (dx,dy,dz) = (-82404.5, -27735.3, 5034.56).
What am I missing such that:
1202990 - 1285410 = -82404.5 (vice -82420)
-4824940 - -4797210 = -27735.3 (vice -27730)
3999870 - 3994830 = 5034.56 (vice 5040) ?