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  1. PO
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    <p>Let's assume that you can use the accelerometers to correctly determine which way is 'up'. We can call that vector <strong>N</strong>. It seems to me that you want the 'up' direction of your cube to align with <strong>N</strong>. As already mentioned, that leaves the cube free to spin, but you can still find a rotation matrix that might accomplish what you're going for, but you need to account for two separate cases or else you'll have a singularity in the solution. I'll assume that 'Z' is 'up' on the cube.</p> <p>If you treat the three accelerometer values as a vector and normalize it (to get <strong>N</strong>), you've got the new 'Z' axis portion of your rotation matrix so that a vector pointed in the z direction will now align with the 'up' vector.</p> <pre><code>| a d N.x | |0| |N.x| | b e N.y | * |0| = |N.y| | c f N.z | |1| |N.z| </code></pre> <p>So we need to decide what to do with a-f. One common thing to do is this: if <strong>N</strong> is pointing mostly along the original 'Z' axis, then make the new 'Y' axis portion of the matrix be <strong>M</strong> = <strong>N</strong> cross <strong>X</strong>:</p> <pre><code>d = 0 e = N.z f = -N.y </code></pre> <p>Normalize <strong>M</strong> and then find the 'X' axis portion of the matrix: <strong>L</strong> = <strong>M</strong> cross <strong>N</strong>. Normalize <strong>L</strong>.</p> <p>If <strong>N</strong> is not pointing mostly along the 'Z' axis (N.z &lt; .707), then you find the new 'Y' axis portion as <strong>M</strong> = <strong>N</strong> cross <strong>Z</strong>. Normalize <strong>M</strong> and find <strong>L</strong> = <strong>M</strong> cross <strong>N</strong> and, finally, normalize <strong>L</strong>.</p> <hr> <p>Edit:</p> <p>So we have our three accelerometer values: A.x, A.y, A.z. First step is to normalize them: </p> <pre><code>a = sqrt(A.x*A.x + A.y*Ay + A.z*A.z); and then N.x = A.x/a; N.y = A.y/a; N.z = A.z/a; </code></pre> <p>We assume that if <code>N == [0, 0, 1]</code> then the correct rotation matrix is the identity matrix. If <em>N</em> doesn't point directly along the <em>z-axis</em>, then we want to form a matrix that will rotate the z-axis of the cube so that it lines up with <em>N</em>.</p>
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    1. POConverting G Force data to rotation?
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    1. USJCooper
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    1. COHi JCooper, I've been looking at your answer over and over again over the last few days. But I am just getting my head around vectors and I'm finding it really hard to understand! Would you be able to post some example code, I might be able to understand that better :)
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      1. USWillDonohoe
    2. COI started trying to clarify, but I'm not sure exactly what you don't understand. Do you know how to compute a cross product?
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