jquery transformable top and left position
<p>i have two images in a page first image is a large image and second image is a small image placed in a div on big image small image can be drag and drop and rotate to any degree on big image after that i want to save these two images as one image in php </p>

<p>here is live code </p>

<p><a href="http://jsfiddle.net/cQyvG/" rel="nofollow">http://jsfiddle.net/cQyvG/</a></p>

<p>but problem is that when i drag small image around i cant get the image position on big image where it is placed </p>

<p>here is my php code
     photoRotation = $_POST["img_angle"];   </p>

<pre><code>$image_2 = imagecreatefrompng($sec);
$image_2 = imagerotate($image_2,$photoRotation, -1,0);
$image_1 = imagecreatefromjpeg($first);

imagecopy($image_1, $image_2,$_POST["img_left"],$_POST["img_top"], 0, 0, $_POST["img_width"],$_POST["img_hight"]);
</code></pre>

<p>this is working but i always get second image on wrong position </p>

<p>jquery.transformable.js
<a href="http://plugins.jquery.com/plugin-tags/jquerytransformablejs" rel="nofollow">http://plugins.jquery.com/plugin-tags/jquerytransformablejs</a></p>
