5

I want to convert a square polygon that is close to the edge of UTM zones 15N and 16N. The polygon's coordinates are in UTM 16N, but it is spatially positioned in the UTM15 area.

My question is that if the square will still be square (angles between lines will be preserved) after the coordinate transformation from UTM16N to UTM15N.

  • Can't you reproject your data from Utm 16 to Utm 15 and review the results? – klewis Apr 22 '16 at 15:52
  • Yes, and they kind of look square in both projections, but I would like to know from theoretical point of view to be precise. – Blind squirrel Apr 22 '16 at 17:13
2

The transverse Mercator Projection is conformal and preserves angles. So the square should still be a square. Area will change though.

  • The square will not remain square--but the angles will remain at 90 degrees, according to the (correct) argument in the answers by csd and nickves. – whuber Apr 23 '16 at 0:56
  • But the questions was if angles will be preserved. But what other is it when it has still 90 degree angles but a square? Rotated of course. Or do i really have a knot in my brain... Did create one and measured it also. – Matte Apr 23 '16 at 5:05
  • A "square," by definition, has equal sides as well as 90 degree angles. Figures with four (geodesic) sides and four 90 degree angles are called rectangles. Although these don't actually exist on the sphere, close approximations of them do, so it's worthwhile respecting these definitions. – whuber Apr 23 '16 at 16:20
  • Your right, didn´t thought of the small difference in length distortion for x and y axis. Will keep this in mind, thank you. – Matte Apr 24 '16 at 17:10
  • 1
    The important thing here is to recognize that these changes of projection will change the lengths of the sides of figures. They will also tend to change them disproportionately: almost all the change occurs in the east-west direction. As Matte points out, those length changes for a figure straddling two UTM zones will be tiny. That gets back to the more fundamental point: since they're tiny, why bother? – whuber Apr 25 '16 at 12:55
1

There's a reason UTM projections don't extend to infinitive and engulf the whole globe.

As you move further away from the UTM's central meridian the distortion becomes more and more apparent. In the case of a transverse Mercator projection the distortion, as its a conformal projection, is applied equally both in the X and Y axis, therefore your angles comprising your shape will not change, retaining the shape.

The area and the orientation of your shape will change.

If you have shapes that cross over different UTM zones, consider using another projection that engulf your study area

  • 1
    This argument is irrelevant, because there's still little distortion at UTM zone boundaries. In fact, UTM zones are designed to work outside their margins. I forget the size of the allowance, but it's around 1/2 to 1 degree at the Equator. At higher latitudes one can move even further east or west before encountering any distortion that's greater than the maximum encountered within any single UTM zone. – whuber Apr 23 '16 at 0:59
  • Looking back at one of your answers (gis.stackexchange.com/questions/31701/…) you demonstrated that the distortion can be acceptable even outside the boundaries. – nickves Apr 23 '16 at 9:55
1

Yes, they will be square in both projections! Breaking it down:

UTM stands for Universal Transverse Mercator. From the Wikipedia article:

The Universal Transverse Mercator (UTM) conformal projection ...

i.e. UTM is "conformal." As for what "conformal" means, again from Wikipedia:

Conformal, or orthomorphic, map projections preserve angles locally ...

So the angles at the corners of your square will be 90 degrees in any UTM zone (and on the earth's surface).

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.