Tutorial, Solutions
More on Parametric Curves, Parametric Surfaces

1. 1. P(0) = 1/6 [ 0 0 0 1 ] M G
           = 1/6 [ 1 4 1 0 ] G
           = 1/6 ( 5*P1 + P2 )
      

      The curve does not pass through point P1.

      P'(0) = 1/6 [ 3*t^2 2*t 1 0 ] M G
            = 1/6 [ 0 0 1 0 ] M G
            = 1/6 [ -3 0 3 0 ] G
            = 1/6 ( -3*P1 + 3*P3 )
      

      The tangent of the curve at the beginning of the curve is parallel to the line connecting points P1 and P3. From the above derivation, this is true for any B-spline curve.
      
      
   2. Choosing P1=P2=P3 gives:

      P(t) = 1/6 [ (6-t^3)*P1 + t^3*P4 ]
           = P1 + 1/6 t^3 [ P4 - P1 ]
      

      This is a straight line segment starting at point P1 and proceeding one sixth of the way towards point P4.
      
      

2.   P'(t) = [ 3t^2 2t 1 0 ] M G
     P''(t) = [ 6t 2 0 0 ] M G
     Pa''(1) = [ 6 2 0 0 ] M G
             = 1/6 [ 0 6 -12 6 ] G
             = P2 - 2*P3 + P4
     Pb''(0) = [ 0 2 0 0 ] M G
             = 1/6 [ 6 -12 6 0 ] G
             = P2 - 2*P3 + P4
   

   We should also show C1 continuity; this is a procedure similar to the one shown above, only for the first derivatives.
   
   
3. If the surface patch satisfies the convex hull property, as Bezier and B-spline patches do, we can do the following
   • generate outcodes for all 16 control points with respect to view volume
   • do trivial accept test by ORing all 16 outcodes together, seeing if result is 0 as required for the trivial accept
   • do trival reject test by ANDing all 16 outcodes together, seeing if results is non-zero as required for trivial reject
   • generate quadrilaterals and clip the individual quadrilaterals to the view volume in the regular way, i.e., the following steps
   • generate outcodes for polygon, check for trivial accept or reject
   • clip polygon against the six view-volume planes