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quiz_two_04

# quiz_two_04 - Massachusetts Institute of Technology...

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Massachusetts Institute of Technology Department of Computer Science and Electrical Engineering 6.801/6.866 Machine Vision QUIZ II Handed out: 2004 Dec. 2nd Due on: 2004 Dec. 9th Problem 1: Suppose we allow for a scale factor in the problem of absolute orientation—in addition to translation and rotation—so that the transformation from the “left’’ to the “right’’ coordinate systems becomes r r = sR( r l ) + r 0 Then the absolute orientation problem becomes one of minimizing n r r,i sR( r l,i ) r 0 2 i = 1 (a) By differentiating w.r.t. to r 0 , dividing by n , and setting the result equal to zero, find a formula for the best fit translation involving the centroids ¯ r l and ¯ r r of the two sets of measurements (b) By shifting the origin of each set of measurements to its centroid, show that the translation r 0 drops out and the error to be minimized simplifies to n r r,i sR( r l,i ) 2 i = 1 where r l,i = r l,i − ¯ r l and r r,i = r r,i − ¯ r r . (c) Show that this can be written in the form S r 2sD rl + s 2 S l where S r and S l are sums of squares of the lengths r r,i and r l,i respectively, while D rl is the sum of dot-products of r r,i and R( r l,i ) . (c) Conclude that the best fit scale is given by s = D rl /S l (d) Now suppose we instead find the best fit transformation r l = s R ( r r ) + r 0 from “right’’ to “left’’ coordinate system. We might expect that this trans- formation is the inverse of the earlier one, that is, s = 1/s , R = R 1 , and r = (1/s)R 1 ( r 0 ) . Unfortunately this is not so. In particular show that 0 s = D lr /S r where D lr is the sum of dot-products of r l,i and R ( r r,i ) . In general s = 1/s .

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2 6.801/6.866 Quiz #2 (e) Show that this asymmetry can be removed by instead minimizing 2
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