<div class="gmail_quote">On Fri, Nov 21, 2008 at 6:29 AM, J JIH <span dir="ltr"><<a href="mailto:jus168jih@gmail.com">jus168jih@gmail.com</a>></span> wrote:<br><blockquote class="gmail_quote" style="margin:0 0 0 .8ex;border-left:1px #ccc solid;padding-left:1ex;">
Neither am I your lawyer, but your points are very interesting and<br>
potentially encouraging. Once the Wikimedia Foundation is able to<br>
accept your points, recent pictures of very old coins may also become<br>
acceptable on Commons, similar to mere copies of very old<br>
2-dimensional works.<br>
<br>
Jusjih<br></blockquote><div><br></div><div>Lupo points out a flaw in the argument on my talk page: <a href="http://commons.wikimedia.org/w/index.php?title=User_talk:Howcheng&diff=16179935&oldid=15133401">http://commons.wikimedia.org/w/index.php?title=User_talk:Howcheng&diff=16179935&oldid=15133401</a></div>
<div><br></div><div>"Patry argues a fine point whether photos of 3D objects are derivative
works. However, please note that even if such photos are not
derivatives, they are still <i>copies</i> of the 3D object. <a href="http://en.wikisource.org/wiki/United_States_Code/Title_17/Chapter_1/Section_101" class="external text" title="http://en.wikisource.org/wiki/United_States_Code/Title_17/Chapter_1/Section_101" rel="nofollow">17 USC 101</a>
defines copies as "material objects, other than phonorecords, in which
a work is fixed by any method now known or later developed, and from
which the work can be perceived, reproduced, or otherwise communicated,
either directly or with the aid of a machine or device." <a href="http://en.wikisource.org/wiki/United_States_Code/Title_17/Chapter_1/Sections_105_and_106" class="external text" title="http://en.wikisource.org/wiki/United_States_Code/Title_17/Chapter_1/Sections_105_and_106" rel="nofollow">17 USC 106(1)</a>
gives the copyright owner of the 3D object the exclusive right to "to
reproduce the copyrighted work in copies or phonorecords". Hence the
photographer would still need to get the consent of the copyright owner
of the 3D object to even make a photo. Patry acknowledges this himself,
read the comments on the blog article you linked, in particular <a href="http://williampatry.blogspot.com/2008/02/photographs-and-derivative-works.html?showComment=1202223060000#c8671324199605658790" class="external text" title="http://williampatry.blogspot.com/2008/02/photographs-and-derivative-works.html?showComment=1202223060000#c8671324199605658790" rel="nofollow">this comment</a> <<a href="http://williampatry.blogspot.com/2008/02/photographs-and-derivative-works.html?showComment=1202223060000#c8671324199605658790">http://williampatry.blogspot.com/2008/02/photographs-and-derivative-works.html?showComment=1202223060000#c8671324199605658790</a>>: "yes I think a picture of a copyrighted object is a reproduction of that object". Also <a href="http://williampatry.blogspot.com/2008/03/photographs-are-not-derivative-works.html?showComment=1206038640000#c6248389526681202947" class="external text" title="http://williampatry.blogspot.com/2008/03/photographs-are-not-derivative-works.html?showComment=1206038640000#c6248389526681202947" rel="nofollow">this comment</a> <<a href="http://williampatry.blogspot.com/2008/03/photographs-are-not-derivative-works.html?showComment=1206038640000#c6248389526681202947">http://williampatry.blogspot.com/2008/03/photographs-are-not-derivative-works.html?showComment=1206038640000#c6248389526681202947</a>> from a follow-up blog post on this issue: "even though a photograph
isn't a derivative work of the object photographed doesn't mean there
might not be violation of the reproduction right. If I take a photo of
a copyrighted work of art and sell copies, I am violating the
reproduction art even though my photo is not a derivative work of the
art work." So, it doesn't matter for us: for photos of copyrighted
artwork, we need the consent of the rights holder of the artwork shown
in the photo (plus that of the photographer, if he isn't identical to
the uploader)"</div><div><br></div><div>But hopefully Mike can weigh in on this too.</div><div><br></div><div>-h</div></div>