Geographic Information Systems Stack Exchange is a question and answer site for cartographers, geographers and GIS professionals. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I have a bunch of objects which have associated Eastings and Northings in format OSGB36

for example:

265397  869562
547297  179002

If I have a starting Easting and Northing, and wanted to find related objects within X km is it possible to compute a Start/End Northing and Start/End Easting range that I can search for programatically?

I am using C#.

I.e. searching for objects using pseudo-code :

Easting >= 265390 && Easting <= 265400 &&
Northing >= 869560 && Northing <= 869564

would this find nearby objects? What resolution do these coordinates operate to? I can operate in miles or km, just need to know the scale - if such a thing is possible.

Sorry this is new to me.

share|improve this question
Further to Polygeo's comment, could you also say what coordinate reference system those eastings and northings are in? If UTM, which UTM zone? – BradHards Apr 12 '14 at 23:11
Sorry for lack of detail - edited post as requested. – Phil Apr 12 '14 at 23:25
See… for help on precision of OSGB map refs. – Martin F Apr 12 '14 at 23:40
Thanks Martin - that's really helped. – Phil Apr 13 '14 at 0:12
up vote 3 down vote accepted

See National Grid Reference Converter for help on the precision of OSGB map references.

If you just use a search range to create a box surrounding your query point, as indicated by

(east_QP - range)  <= east  <= (east_QP + range)
(north_QP - range) <= north <= (north_QP + range)

a variation on your pseudo code, it is fast but biased.

If you "create a circle" around your query point

hypot ((east_QP - east), (north_QP - north)) <= range

where hypot() computes distance, it is a tiny bit slower but less biased as it cuts out the box corners.

share|improve this answer
Thanks Martin - very helpful – Phil Apr 14 '14 at 15:03

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.