Sdo_intersection across the antimeridian

Question

Hello All,
I'm trying to create a function which splits geometries at the antimeridian, producing multipart geometries.  I'm approaching this using a sdo_intersection and a clipping geometry representing each of the hemispheres.  I'm confused about the behavior of sdo_geom.sdo_intersection though.  Why does the following not produce a box like 'POLYGON((-179.9 0, -170 0, -170 10, -179.9 10, -179.9 0))'
Thanks!
--john
  1  select sdo_util.to_wktgeometry(sdo_geom.sdo_intersection(
  2 MDSYS.SDO_GEOMETRY('POLYGON((170 0, -170 0, -170 10, 170 10, 170 0))', 8307),
  3 MDSYS.SDO_GEOMETRY(2003,8307,NULL, sdo_elem_info_array (1,1003,1), sdo_ordinate_array (-179.9, 89.9, -179.9, -89.9, -0.1, -89.9, -0.1, 89.9, -179.9, 89.9)),
  4 0.05
  5 )
  6* ) from dual

Answer 1

Hi John,
It would be helpful to mention the version of Oracle you are using.  But in any case the answer has less to do with the date line than with the size of your geodetic polygons. 
Here is the link for 11g with this caveat
http://docs.oracle.com/cd/B28359_01/appdev.111/b28400/sdo_cs_concepts.htm#SPATL624
The 12g blurb actually expands on things a bit
http://docs.oracle.com/cd/E16655_01/appdev.121/e17896/sdo_cs_concepts.htm#SPATL624
It's always nice to see the documentation getting a little better.
So you could break your polygons down into smaller pieces but I guess I wonder what you are up to with these globe spanning boxes.  Perhaps if you explained your use case other folks on the list could chime in with some suggestions.
Cheers,
Paul