Describing a cubic spline segment with cubic polynomials We showed in the previous lecture that we can use cubic spline segments, laid end-to-end, to create smooth curves of arbitrary complexity. Those spline curves can be used in various ways, including the creation of geometric shapes and motion paths for animation. There are various ways that a user might specify a cubic spline segment. One way is as an Hermite spline, in which both the position and the tangent direction at the beginning and end of each cubic spline segment is specified. Another way is as a Bezier spline, in which two end positions and two intermediate guide points for cubic spline segment is specified. We show below that these methods are merely different ways of specifying a value for each coordinate as a cubic polynomial, which varies as parameter t goes from 0 to 1: x(t) = ax t3 + bx t2 + cx t + dx y(t) = ay t3 + by t2 + cy t + dy z(t) = az t3 + bz t2 + cz t + dz     // Include this only if the curve is in 3D   From polynomial to dot product to matrix multiplication We can think of a cubic polynomial as a dot product between a row vector containing different powers of t and a column vector containing coefficients for each of those powers of t: at3 + bt2 + ct + d   ←   t3 t2 t  1   ●   a b c d Describing a cubic polynomial in terms of those four coefficients a,b,c,d makes it very easy for the computer to evaluate a cubic spline, given any value of t. But those coefficientsare very difficult for most users to work with. Fortunately, once we have the polynomial in dot product form, we can use matrix transformations to allow the user to describe the cubic curve in ways that are easier for them to think about. The general idea is that the user can pick four values A,B,C,D that make sense to her. Then we use some 4x4 matrix to convert from those four values to the raw polynomial coefficients a,b,c,d: a b c d   ←     ●   A B C D Below we show how to use matrix transformations to go from an Hermite or Bezier description of a cubic spline to a description consisting of cubic polynomials. In each case, we need to work out the correct basis matrix for that particular way of describing a cubic spline.   The Hermite basis matrix Consider the positions P1 and P2 at the beginning and end of a cubic segment, respectively, and the derivatives T1 and T2 at the beginning and end of a cubic segment, respectively. We can think of P1,P2,T1 and T2 as each varying the weight of a cubic function which affects only that one value, not the other three. For example, the top function to the right affects the value of P1, but not P2,T1 or T2. We can describe the sum of these four basis functions, weight respectively by the values P1,P2,T1 and T2, as a multiplication of the column vector [P1,P2,T1,T2] by a 4x4 matrix: a b c d ←  2 -2  1  1 -3  3 -2 -1  0  0  1  0  1  0  0  0 ● P1 P2 T1 T2 The above 4x4 matrix is the Hermite basis matrix.   Hermite curves in multiple coordinates Remember that everything we have just described is done independently for each coordinate. So, for example, if we have a spline in 3D, we start with two 3D positions P1x,y,z and P1x,y,z, as well as two 3D tangents T1x,y,z and T2x,y,z. We need to multiply by the Hermite basis matrix to get cubic polynomomial coefficients [a,b,c,d] in each of the three dimensions: ax bx cx dx ←  2 -2  1  1 -3  3 -2 -1  0  0  1  0  1  0  0  0 ● P1x P2x T1x T2x ay by cy dy ←  2 -2  1  1 -3  3 -2 -1  0  0  1  0  1  0  0  0 ● P1y P2y T1y T2y az bz cz dz ←  2 -2  1  1 -3  3 -2 -1  0  0  1  0  1  0  0  0 ● P1z P2z T1z T2z The three above multiplications give us the coefficients needed for three cubic polynomials, one in each dimension: x(t) = ax t3 + bx t2 + cx t + dx y(t) = ay t3 + by t2 + cy t + dy z(t) = az t3 + bz t2 + cz t + dz       The Bezier basis matrix The principle of Bezier interpolation Suppose we start instead with the user view of end positions A and D, with intermediate "steering" points B and C. This produces the cubic curve which results from three levels of linear interpolation, as shown in the figure to the right. The cubic curve (shown in purple) is the result of varying t between 0 and 1 while linearly interpolating A→B and B→C and C→D (shown in blue) to produce three points, followed by two more linear interpolations between those points to produce two points (shown in green), followed by a final linear interpolation (shown in red).     Defining the Bezier basis matrix Mathematically, the three stages of linear interpolation that we described above looks like this: (1-t) * ( (1-t) * ( (1-t) * A + t * B ) + t * ( (1-t) * B + t * C ) ) + t * ( (1-t) * ( (1-t) * B + t * C ) + t * ( (1-t) * C + t * D ) ) If we multiply out the above expression to separate A,B,C and D, we get: (1-t)3A + 3(1-t)2tB + 3(1-t)t2C + t3D If we then multiply out the expressions of t within each term, we get: (-t3 + 3t2 - 3t + 1) A + (3t3t - 6t2 + 3t) B + (-33 + 32) C + t3 D We can describe the above polynomial as a matrix/vector multiplication, where the four rows represent the weights a,b,c and d that we want to apply to t3,t2,t and 1: a b c d ← -1  3 -3  1  3 -6  3  0 -3  3  0  0  1  0  0  0 ● A B C D The above 4x4 matrix is the Bezier basis matrix.   Bezier curves in multiple coordinates As was true for the Hermite spline, everything we have just described for defining Bezier splines is done independently for each coordinate. So, for example, if we have a spline in 3D, we start with four 3D positions Ax,y,z, Bx,y,z, Cx,y,z, and Dx,y,z. We need to multiply by the Bezier basis matrix to get cubic polynomomial coefficients [a,b,c,d] in each of the three dimensions: ax bx cx dx ← -1  3 -3  1  3 -6  3  0 -3  3  0  0  1  0  0  0 ● Ax Bx Cx Dx ay by cy dy ← -1  3 -3  1  3 -6  3  0 -3  3  0  0  1  0  0  0 ● Ay By Cy Dy az bz cz dz ← -1  3 -3  1  3 -6  3  0 -3  3  0  0  1  0  0  0 ● Az Bz Cz Dz The three above multiplications give us the coefficients needed for three cubic polynomials, one in each dimension: x(t) = ax t3 + bx t2 + cx t + dx y(t) = ay t3 + by t2 + cy t + dy z(t) = az t3 + bz t2 + cz t + dz       Homework To give some students a chance to catch up, I have decided not to give any homework assignment this week. If you have fallen behind, please use this time to finish and post any assignments that you have yet to complete. If you are already up to date with your work, feel free to add features to your last assignment, or to create another fun and interesting and challenging scene for extra credit. I have posted your midterm grades, which are based on your cumulative work on the hw1 through hw5.