I need someone to shed some light on how transforms are calculated when converting from a DWF version < 6 to DWF version >= 6. As an example lets look at the readable ascii DWF 00.42 file that is one of the examples of the WHIP specs. The units there are:
(Units inches
((56251.8 0 0 0)
(0 56251.8 0 0)
(0 0 1 0)
(926921 552481 0 1)))
When this file is opened and then saved as DWF version 6+ by ADR this is what I see in the descriptor.xml file:
<ePlot:Resources><ePlot:GraphicResource role="2d streaming graphics" mime="application/x-w2d" href="com.autodesk.dwf.ePlot_5EF7548A-AC2F-4362-BD4F-3B3E77D3E5F4\F6F64A53-C75B-4D8B-A597-080FFD02C934.w2d" title="dwf042" size="417" objectId="5EF7548B-AC2F-4362-BD4F-3B3E77D3E5F4" extents="0.5 0.6863570208188321 10.5 7.8136429791811679" transform="3.9549452635575516e-005 0 0 0 0 3.9549452635575516e-005 0 0 0 0 1 0 -41.324613997342269 -29.448506612668474 0 1"/></ePlot:Resources>
How are the initial units mapped to this transform, size and extents?
