StackOverflow2013

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plurals

POInverse 3D (triangle) projection
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copied!<p>I have a 3D math problem which I just can't seem to solve.</p> <p>I have data of 3 points. The data is a (2D) coordinate on a plane, floating somewhere in 3D space. I also know the (2D) coordinate of the projection. That results in the following array of data:</p> <pre><code>[[[x1,y1], [px1,py1], [[x2,y2], [px2,py2], [[x3,y3], [px3,py3]] </code></pre> <p>Where the normal (x1 etc.) coordinates stand for the coordinates on the plane and the other (px1 etc.) for the projected coordinates.</p> <p>What I would like to do is project a <em>new</em> 2D coordinate ([x4,y4]).</p> <p>.</p> <p><strong>What I tried so far:</strong></p> <p>Ofcourse you need an eye for projection, so I set that to [xe,ye,-1]. The xe and ye are known. (It is photo referencing, so I just placed the eye in the center of the photograph.)</p> <p>Beneath the eye I placed the projection surface (z=0). That gives the following projection coordinates:</p> <pre><code>[[[x1,y1], [px1,py1,0], [[x2,y2], [px2,py2,0], [[x3,y3], [px3,py3,0]] </code></pre> <p>I can't do the same for the coordinates on the plane, since I don't know anything about that plane.</p> <p>I also figured that I could make a parameterized formula of the lines running from the eye through the projection coordinates. For line1 that would be:</p> <pre><code>line1x = xe+(px1-xe)*t1 line1y = ye+(py1-ye)*t1 line1z = -1+t1 // = -1+(0--1)*t1 </code></pre> <p>I also know the distance between the points in 3D. That's the same as in 2D. That means the distance between point1 and point2 would be sqrt((x1-x2)^2+(y1-y2)^2).</p> <p>I also know the distance between the lines (line1 and line2) at any time. That is sqrt((line1x-line2x)^2+(line1y-line2y)^2+(line1z-line2z)^2).</p> <p>However, I don't really know how to go from here... Or even whether this is the right route to take.</p> <p>.</p> <p>I hope you understand what I want to be able to do, and that you can help me.</p> <p>Thanks in advance!</p>

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