I hope this isn't a daft question, but I'm not a lawyer, and would like to
check my understanding of the POV-Ray source license. I also hope it's in
the right group; it is relevant to MegaPOV, and to some unofficial
modifications of POV. I will just step through the matter to the point of
confusion and I
would be grateful if someone could tell me what I've got wrong:

I have read POV-Ray 3.6.1's source license (maybe not understood it). I
would like to experiment with adding features to the source (e.g.
additional pattern types, that sort of thing). I think this is clearly
permitted under clause 1.1 and 1.3 - I'd be making a "modified version" as
defined in the license, for my own personal experimentation, education and
perhaps submission back to POV or release as a "custom version" in the
unlikely event that what I end up with is of any use to anyone (doubtful).
No problem so far.

Now, at present I mainly use MegaPOV 1.2.1 instead of the official 3.6.1
release, since MegaPOV currently has several additional features I find
useful and at least one bug fix I find almost essential. MegaPOV obviously
releases its source as required by the source license. MegaPOV is clearly a
"custom version" of POVRay, as defined in the source license.

Here is the bit that I am confused about. I would like to modify the source
of MegaPOV rather than that of the official version, since I don't wish to
add my single new feature (say) while simultaneously losing all the MegaPOV
features I've gotten used to in my scenes. Now, the following part of clause
1.3 leaves me unsure as to whether I can produce modified versions based on
MegaPOV or not:

"[...] all Modified Versions that you create must, in substance, be
modifications of the Licensed Version."

where the Licensed Version is defined at the top of the license file as 3.6.
It seems to me that "in substance" could be read either way. Which way is
the right way; is it possible to modify MegaPOV while remaining with the
spirit of the license, or must I only modify the official source?

Thanks,

Tom

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