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copied!<p>Approximate the whole thing to be a sphere and apply the following formula multiplied divided by pixel-to-model-unit ratio. You can calculate the pixel-to-model-unit ratio by rendering a unit sphere at known z coordinate. Good thing about a sphere is that its width is always going to be diameter regardless of where you look from. See <a href="http://en.wikipedia.org/wiki/3D_projection#Perspective_projection" rel="nofollow noreferrer">perspective projection</a> for details.</p> <p><img src="https://upload.wikimedia.org/math/b/0/a/b0a6e114c9ef0dc4e2e4181534f54cb1.png" alt="alt text"></p> <p><strong>Edit</strong>: The good thing about sphere is that it's independent from directions. Regardless of box or sphere you need to specify the direction of camera vector and the <em>up</em> vector. Once you have that, the only unknown is the depth z from the eye's point of view. From there you can calculate the position of the eye by:</p> <pre><code> eye := (center.x, center.y, center.z) - depth * (camera.x, camera.y, camera.z) </code></pre> <p>From the formula, <code>Bx := Ax * Bz / Az</code> You can derive <code>Bz := Bx * Az / Ax</code>. </p> <p>Let me define a model coordinate 1.0 to be 1.0m. Suppose sphere of radius 1.0m from distance 100.0m looks 20 pixels wide.</p> <pre><code>Bz := 1.0 * 100.0 / 10 Bz := 10.0 </code></pre> <p>Now let's say your bounding box is 200.0m wide, which makes Ax to be 100.0, and your screen width is 1024pixel, which makes Bx 512pixel.</p> <pre><code>Az := (Ax / Bx) * Bz Az := (100.0/512.0) * 10.0 Az := 1.953m </code></pre> <p>In the above the unit conversion betwen m and pixel is being absorbed by Bz.</p> <p>Suppose the up vector is (0, 0, 1), camera vector is (1, 0, 0), and the center is (0, 0, 0).</p> <pre><code>eye := center - depth * camera eye := (0, 0, 0) - 1.953 * (1, 0, 0) eye := (-1.953, 0, 0) </code></pre>

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