Ah ok... this looks good - the only problem is that I don't know the centre, only the top left corner of the square. I guess I can work this out.. My maths isnt too good though:
a square of 20m x 20m with its top-left at e.g. -5.2137, 122.870211 DD: so, I need to locate its centre, namely -5.2137 DD + 10m, 122.870211 DD + 10m: I could say that 10m is 10 * 0.00001, but this introduces large inaccuracy (well, about a metre...)... am I being too pedantic here? is 0.00001 DD really the best estimate for 1m? If that is ok, then I guess my new centre is at -5.21380, 122.870311 (correct?)... thanks for the help so far! -S On May 2, 1:39 pm, Rossko <[email protected]> wrote: > > I was wondering if it is possible to do the opposite > > (i.e. give a GLatLon a distance to in meters and it return the new > > GLatLon) somehow? > > Only if you have a bit more information ... like a bearing. > If you want to make a square that's easy enough! > See - > > http://groups.google.com/group/Google-Maps-API/browse_thread/thread/5... --~--~---------~--~----~------------~-------~--~----~ You received this message because you are subscribed to the Google Groups "Google Maps API" group. To post to this group, send email to [email protected] To unsubscribe from this group, send email to [email protected] For more options, visit this group at http://groups.google.com/group/Google-Maps-API?hl=en -~----------~----~----~----~------~----~------~--~---
