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Thread: Matrix Sheer Factor

  1. #1
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    Question Matrix Sheer Factor


    Any ideas on the Part 4 the shear question?
    Last edited by Enslaved; 10-26-2012 at 09:47 PM.

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    My method:
    Since the invariant line is Y=X, You find the the Shortest Point and the line Y=X
    Call this Point : P
    For simplicity purposes i used the point (2,0)
    the perpendicualar gradient of Y=X is -X, henceforth the co-incidental point of the line Y=-X+2 and Y=X is (1,1)
    Using Pythagoras' Therom the distance between (1,1) and P is Root 2

    Then you apply the transformation matrix to the point P to get P' = (6,4)
    Via:
    (3 -2)(2) = (6)
    (2 -1)(0) = (4)
    Then you find the distance between P and P'
    giving the distance of Root 32 --> Simplified to 4 Root 2

    Then you divide the distance of P & P' by the distance of P and (1,1)

    Giving: 4 = (4 root 2)/(root 2)

    so the shear factor is 4


    Any comments, not sure if this is right?

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