1

I'm afraid I've a very basic question about interpreting UTM grid values:

Are the grid references in metres?

I have data from the Spanish catastral map labelled as UTM Zone 31 ETRS89, and want to calculate some point-to-point distances, by pythagoras, over small distances used in urban planning.

If so, then over tens of metres we should expect extremely small errors from treating the ground as flat and simply using pythagoras, in order of 10-10 metre? (Obviously other errors will dominate!)

Reasoning:

  • 2 x π x 40 metres / 40,000,000 metre circumference = 6 x 10-6 radian
  • 6 x 10-6 - sin(6 x 10-6) = 4 x 10-17 (very possibly beyond accuracy of calculator)
  • 4 x 10-17 x 6,300,000 metre radius = 2 x 10-10

This is what I calculated for a building:

372954.477 4316151.012
372961.247 4316155.121 dx   6.770 dy   4.110 =>  7.920 m
372971.827 4316161.551 dx  10.580 dy   6.430 => 12.380 m
372980.437 4316147.931 dx   8.610 dy -13.620 => 16.113 m
372963.327 4316137.601 dx -17.110 dy -10.329 => 19.986 m
372954.477 4316151.012 dx  -8.850 dy  13.410 => 16.067 m
  • beware that while UTM is indeed in meters, ETRS89 seems to be in degrees --> spatialreference.org/ref/epsg/4258/html nevertheless the units you pasted seem to be in meters (UTM). as for the error with an euclidean measurement --> math.stackexchange.com/questions/738529/… – luckyshonway Aug 23 '18 at 4:22
  • Yes they're in meters, but UTM uses transverse Mercator which maintains shapes (angles), not distances. There will be distance distortion due to 1) projection, 2) the projection plane is not at local elevation, 3) data accuracy, 4) etc. – mkennedy Aug 23 '18 at 16:27
1

@jonathanjo if I try to find requested by you "UTM zone 31 ETRS89" I will end up with its correspondent EPSG:25831, check here: epsg projection 25831 - etrs89 / utm zone 31n. If I go further to this source Coordinate Systems Worldwide, I can find the answer to your question:

Unit: metre

Accuracy 1.0 m (default)

  • No, that's the accuracy of the default transformation between ETRS89 and WGS84 that epsg.io has set. – mkennedy Aug 23 '18 at 16:24
0

Undestanding this was greatly helped by the excellent "Guide to coordinate systems in Great Britain" by the Ordnance Survey, which explains the concepts very clearly as well as the mathematics. Although focussed on OSGB36 as used in the UK, it also covers other systems including WGS84 and ETRS89.

The coordinates from the Spanish catastral maps are labelled "UTM Husa 31 ETRS89".

This means:

  • It's a UTM, a universal transverse Mercator coordinate system, which defines eastings and northings, in this case in metres ref
  • The latitude/longitudes are defined as ETRS89, the European Terrestrial Reference System 1989, which defines which lat/long is exactly where, taking into account tectonic shift etc
  • This is based on an GRS80: "geodetic reference system 1980" ref which defines the exact shape of the ellipsoid
  • It's zone 31 [husa = zone], a 6°-wide band in the northerm hemisphere, covering a thin piece of eastern Spain (including Baleares), France, eastern Britain, etc. The bands are obviously wider at the equator and thin at higher latitudes.
  • Conversion between the UTM and lat/long coordinates is done with a geodetic transformation.
    • Simple ones are not accurate; accurate ones are not simple
    • One common one is the Helmert transformation ref, more is OS Guide p35 but this is coarse (3m across UK)
    • Better are Gauss-Krüger ref and simplified derivations (millimetres) ref and (nanometres) ref

Considering the original question of Using pythagoras to work out the sizes of plots of land from catastral data, which has scale in the tens of metres, and maximum accuracy of 1mm.

  • Indeed the grid coordinates are in metres ref
  • Error due to slope of ground might be several percent (1 in 4 slope gives about 3% by 1 - cos atan 1/4)
  • Accuracy on the ground as measured by surveying equipment is less than 10-3 ref
  • Error due to height-above-geoid might be around 10-6 (based on London 35m, Madrid 600m above sea level, earth radius 6,300 km)
  • The error for the projection is zero because catastral map is already in UTM.
  • The grid can be considered flat and the error due to small angle approximation will be order of 10-10 (for 40 m: earth radius times second term of Maclaurin series for sine, over distance is 6.3 x 106 x ((6.28 x 10-6)3)/6) / 40

My thanks to commenters.

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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