0

I'm trying to define the SRS EPSG::3857 inside Terrasolid software. I ran into a problem because the definition of the reference ellipsoid is given with a major axis and inverse flattening, not via both axis. Since flattening of a sphere is 0, inverse of a zero is infinite and I can't input that. Doesn't work with a 0, 1 or some very high value either. Any thoughts on how to define it, and if its even possible?

EDIT:

The issue is, a sphere (reference body for EPSG::3857) is a kind of ellipsoid that has no flattening, hence you cannot define 1/flattening. I'm looking for a workaround in this particular software, or just a confirmation that its in fact impossible to do at this point, since I cannot define datum that involves a sphere as a reference body.

Here's a screenshot of the datum input screen: Screenshot

  • Could you please be more specific about the way(s) in which the software "doesn't work"? What precisely do you do and how does it exhibit failure in response? – whuber Dec 27 '12 at 16:03
1

The EPSG has defined the code 3857 here

official values are:

Semi major axis (a)     6378137     metre

Inverse flattening  298.257223563   unity
| improve this answer | |
  • I feel so stupid, reference ellipsode is not a sphere, duuuuuh :) Thanks for pointing out the obvious – U2ros Dec 28 '12 at 9:54
  • I have to correct myself again. It is in fact using a sphere, not wgs84 ellipsoid. The definition in the link you sent is wrong, theres nothing more to it. Also, i Terrasolid doesn't support it, i ve done transformation of lidar data via FME, where you can define a projection with a sphere – U2ros Dec 28 '12 at 12:27
  • The EPSG definition is correct. The projection algorithm uses the semimajor axis value for the radius of the sphere, and ignores the flattening (which is what Google and Bing/Microsoft do too). – mkennedy Dec 28 '12 at 17:30
  • So the fault is in the projection, not in the ellipsoid. Here is some further reading: alastaira.wordpress.com/2011/01/23/… – AndreJ Dec 28 '12 at 19:40

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.