Mth-227-lab-computer animation PDF

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The lab about how computer animation is related to linear algebra with the answers. ...

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MTH-227 Section

Graded Lab on Computer Animations – Answer Sheet Question / Problem #1)  a b c   xold   axold  byold  c      The matrix product is  d e f   yold    dxold  eyold  f   0 0 1   1    1

It relates to the system of equations

xnew  axold  byold  c y new  dx old  ey old  f

because it is simply a matrix equation

equivalent (Ax  b ) for that linear system of equations a b c Where A   d e f  , x   0 0 1 

 xold   axold  byold  c   y  , and b  dx    old  eyold  f   old   1    1

I believe that the function of the "1" in the third column in the component vector x is the keep the figure flat on the plane or lock it to the plane (i.e. if it was not there then figure could be translated in three dimensions, that is not only up, down, right, and left, but forwards and backwards). In other words, it keeps the third component of the resulting vector always the same (i.e. at one). Question / Problem #2) The figure has been reflected of the x2-axis.  1 0 0 The transformation matrix that does this is:  0 1 0  0 0 1

Question / Problem #3)

2 cos( 4 )  sin( 4 ) 0  2    The matrix of transformation the produces the first image is  sin(4 ) cos(4 ) 0   22  0 0 1  0  



2 2 2 2

0

0  0  1

cos( )  sin( ) 0 This is found by substituting for  in  sin( ) cos( ) 0 4  0 0 1



The second image is produced from the first by rotating all the points in the image by another 4 radians, the third is produced from the second by rotating all the points in the image by another from the third by rotating all the points in the image by yet another

 4

 4

radians, and the fourth is produced

radians.

cos( 2 )  sin( 2 ) 0  0  1 0     The matrix of transformation the produces the second image is sin( 2) cos( 2 ) 0    1 0 0  0 0 1  0 0 1  2 2 cos( 34 )  sin( 34 ) 0  2  2    2 3 3 The matrix of transformation the produces the third image is  sin( 4 ) cos( 4 ) 0    2  22  0 0 1   0 0   cos(  )  sin( ) 0   1 0 The matrix of transformation the produces the fourth image is  sin( ) cos( ) 0   0  1  0 0 1   0 0

0  0  1  0 0 1 

Question / Problem #4)

1 0 0  1) The translation matrix that will make the figure to correspond to the sequence in the first image is  0 1  1 0 0 1  The translation matrix moves all the points in the figure down one unit. 1 0 0  2) The translation matrix that will make the figure to correspond to the sequence in the second image is  0 1  2 0 0 1  The translation matrix moves all the points in the figure down oneaddition al unit.

2 0 0  3) The translation matrix that will make the figure to correspond to the sequence in the third image is  0 2  2 0 0 1  The translation matrix dilates all the points in the figure by a factor twounits, then moves the figure down two additional units. Bonus) The single translation matrix that would perform the above three steps is found by multiplying the three matrices in order of 1, 2, & 3.  The single translation matrix that would perform the above three steps is 1 0 0   1 0 0   2 0 0   2 0 0  0 1  1  0 1  2   0 2  4   0 2  7       0 0 1   0 0 1   0 0 1   0 0 1 

Question / Problem #5) Part 1) It is obviously a shear transformation. The points are not scaled up or down vertically they are only scaled or moved horizontally. 1 1 0  The transformation matrix that would do this is:  0 1 0 0 0 1 

Part 2)

left Image) It is obviously a vertical shear transformation. The points are not scaled up or down horizontally they are only scaled vertically.  1 0 0 The standard transformation matrix for a vertical shear is (in this case):  k 1 0  0 0 1 1 It looks like k  for the above picture on the left 2 1 0   The standard transformation matrix that would acheive the left image is:  12 1  0 0

Where 0  k  1.

0  0  1

Right Image) It is obviously another vertical shear transformation. The points are not scaled up or down horizontally they are only scaled vertically. 1 0 0  The standard transformation matrix for a vertical shear is (in this case): k 1 0 0 0 1  1 It looks like k  for the above picture on the left 4 1 0   The standard transformation matrix that would acheive the right image is:  14 1  0 0

Where 0  k  1.

0  0  1

Question / Problem #6)

Scene 1 - Commentary : Watch out Wooody here comes a car you better move out of the way! How To Illustrate : 1 0 2  Use the transformation matrix 0 1 0  to move Wooody two units to the right. 0 0 1 

Scene 2 - Commentary : Watch out Wooody here comes another car from the other way you better move! How To Illustrate : 1 0 2    Use the trans. matrix 0 1 0  to move Wooody 2 units to the left of the origin. 0 0 1 

Scene 3 - Commentary : Watch out Wooody here comes a car from the other direction you better move out! How To Illustrate : 1 0 2  Use the transformation matrix 0 1  1 to move Wooody two units to the right. 0 0 1 

Scene 4 - Commentary : Woody was not fast enough, poor Woody, he got ran over. How To Illustrate : 1 2  2  Use the transformation matrix 0 .25  1 to skew Woody into a flattened elongation 0 0 1  postion....