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	<title>Comments on: Matrix transforms based on an arbitrary registration point</title>
	<atom:link href="http://keith-hair.net/blog/2009/01/24/matrix-transforms-based-on-an-arbitrary-registration-point/feed/" rel="self" type="application/rss+xml" />
	<link>http://keith-hair.net/blog/2009/01/24/matrix-transforms-based-on-an-arbitrary-registration-point/</link>
	<description>Scripting is fun like any other hobby</description>
	<lastBuildDate>Wed, 28 Jul 2010 10:17:28 +0000</lastBuildDate>
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		<title>By: Matthew Clifford</title>
		<link>http://keith-hair.net/blog/2009/01/24/matrix-transforms-based-on-an-arbitrary-registration-point/comment-page-1/#comment-601</link>
		<dc:creator>Matthew Clifford</dc:creator>
		<pubDate>Sat, 06 Mar 2010 22:35:24 +0000</pubDate>
		<guid isPermaLink="false">http://keith-hair.net/blog/?p=28#comment-601</guid>
		<description>Hello Keith,

This is an astonishingly good post and fantastic working example...

Would you be willing to post a link to the source code for this, as it would be most appreciated?

Kind regards,

Matthew</description>
		<content:encoded><![CDATA[<p>Hello Keith,</p>
<p>This is an astonishingly good post and fantastic working example&#8230;</p>
<p>Would you be willing to post a link to the source code for this, as it would be most appreciated?</p>
<p>Kind regards,</p>
<p>Matthew</p>
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		<title>By: Keith H</title>
		<link>http://keith-hair.net/blog/2009/01/24/matrix-transforms-based-on-an-arbitrary-registration-point/comment-page-1/#comment-164</link>
		<dc:creator>Keith H</dc:creator>
		<pubDate>Sat, 24 Jan 2009 17:08:13 +0000</pubDate>
		<guid isPermaLink="false">http://keith-hair.net/blog/?p=28#comment-164</guid>
		<description>Thanks Flanture, but I may be describing the kind of skew I want to do incorrectly. I would like to skew a separate value along the horizontal edge of the shape, and also skew a separate value along the vertical edge of the shape. I’ve updated this post with a diagram showing the kind of skew I’d like to do.

To make the skewing run along the edges is what I wish to do. After skewing along the horizontal axis first, the vertical axis is different. Visa versa if I skewed along the vertical axis first.</description>
		<content:encoded><![CDATA[<p>Thanks Flanture, but I may be describing the kind of skew I want to do incorrectly. I would like to skew a separate value along the horizontal edge of the shape, and also skew a separate value along the vertical edge of the shape. I’ve updated this post with a diagram showing the kind of skew I’d like to do.</p>
<p>To make the skewing run along the edges is what I wish to do. After skewing along the horizontal axis first, the vertical axis is different. Visa versa if I skewed along the vertical axis first.</p>
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	<item>
		<title>By: J.</title>
		<link>http://keith-hair.net/blog/2009/01/24/matrix-transforms-based-on-an-arbitrary-registration-point/comment-page-1/#comment-162</link>
		<dc:creator>J.</dc:creator>
		<pubDate>Sat, 24 Jan 2009 16:15:24 +0000</pubDate>
		<guid isPermaLink="false">http://keith-hair.net/blog/?p=28#comment-162</guid>
		<description>simple, direct and clean. Thanks!</description>
		<content:encoded><![CDATA[<p>simple, direct and clean. Thanks!</p>
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	<item>
		<title>By: Flanture</title>
		<link>http://keith-hair.net/blog/2009/01/24/matrix-transforms-based-on-an-arbitrary-registration-point/comment-page-1/#comment-161</link>
		<dc:creator>Flanture</dc:creator>
		<pubDate>Sat, 24 Jan 2009 12:25:36 +0000</pubDate>
		<guid isPermaLink="false">http://keith-hair.net/blog/?p=28#comment-161</guid>
		<description>not sure if this is what you need, but try:
...
skewAmount:Number=0;
...
skewM.c=skewM.b=Math.tan(skewAmount);
...
It performs skew on both axis in the same time</description>
		<content:encoded><![CDATA[<p>not sure if this is what you need, but try:<br />
&#8230;<br />
skewAmount:Number=0;<br />
&#8230;<br />
skewM.c=skewM.b=Math.tan(skewAmount);<br />
&#8230;<br />
It performs skew on both axis in the same time</p>
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