# [Proj] X,Y UTM distance calculation (compare with latlong WGS84)

José Alberto Gonçalves jagoncal at fc.up.pt
Wed Sep 13 09:31:52 EDT 2006

```You can obtain the distance deformation using proj with option -V. It
gives the meridian and parallel scale factors, which are the same in the
case of UTM, since it is a conformal projection.

The deformation increases with the distance to the central meridian. For
a point on the equator, on the border of a standard UTM zone (i.e. 3
degrees away from the central meridian) you have k=1.00098, which
corresponds approx. to 1 metre per kilometer. For a point 4 degrees away
it gets to 1.0021, i.e., approx. 2 meters per kilometer.

If you need to apply the scale factor to a long distance you must take
into account that it is an integration problem. Anyway it is probably
enough to take the scale fator at the midpoint of the line.

Regards

Jose' Gonçalves

oon at informatika.org wrote:

>
>     Dear All,
>
> I want to ask a question about distance calculation of two points in
> X,Y (UTM).
> Is it ok if I get the distance of two points in X,Y (UTM) only with
> distance
> calculation of cartesian coordinate (pythagoras method)?
> The error getting larger (or getting huge) if I calculate the distance
> with the
> points crossing the UTM zone. (I have tried for point with (1)
> different zone,
> I don't know what will be happened if the points over more than 1 zone??)
>
> Is it any other method to get distance calculation of cartesian
> coordinates X,Y (UTM), so the result of the calculation have accuracy
> like distance calculation
> of WGS84?
> (It's ok if the error about hundred meters)
>