## [PyX-checkins] pyx/design beziers.tex,1.1,1.2

 [PyX-checkins] pyx/design beziers.tex,1.1,1.2 From: Michael Schindler - 2005-08-30 19:12:48 Update of /cvsroot/pyx/pyx/design In directory sc8-pr-cvs1.sourceforge.net:/tmp/cvs-serv30368/design Modified Files: beziers.tex Log Message: adapt the parameterization in controldists_from_endpoints_pt to the derivation in design/beziers.tex Index: beziers.tex =================================================================== RCS file: /cvsroot/pyx/pyx/design/beziers.tex,v retrieving revision 1.1 retrieving revision 1.2 diff -C2 -d -r1.1 -r1.2 *** beziers.tex 30 Aug 2005 18:31:49 -0000 1.1 --- beziers.tex 30 Aug 2005 19:12:40 -0000 1.2 *************** *** 45,58 **** \begin{align} \left(\begin{array}{cc} \dot x(0) \\ \dot y(0) \end{array}\right) ! = \alpha \left(\begin{array}{cc} t_x(0) \\ t_y(0) ! \end{array}\right) \qquad \left(\begin{array}{cc} \dot x(1) \\ \dot y(1) \end{array}\right) ! = \beta \left(\begin{array}{cc} t_x(1) \\ t_y(1) --- 45,66 ---- \begin{align} \left(\begin{array}{cc} + x_1-x_0 \\ y_1-y_0 + \end{array}\right) + &= \frac{1}{3} + \left(\begin{array}{cc} \dot x(0) \\ \dot y(0) \end{array}\right) ! =: \alpha \left(\begin{array}{cc} t_x(0) \\ t_y(0) ! \end{array}\right) \\ ! \left(\begin{array}{cc} ! x_3-x_2 \\ y_3-y_2 ! \end{array}\right) ! &= \frac{1}{3} \left(\begin{array}{cc} \dot x(1) \\ \dot y(1) \end{array}\right) ! =: \beta \left(\begin{array}{cc} t_x(1) \\ t_y(1) *************** *** 62,66 **** Here, the externally prescribed tangent vectors are expected to be parallel to the tangent vectors of the parameterization and normalized to 1. This implies ! $\alpha>0$ and $\beta>0$. The problem to now to find proper parameters $\alpha>0$ and $\beta>0$ for a given set of --- 70,75 ---- Here, the externally prescribed tangent vectors are expected to be parallel to the tangent vectors of the parameterization and normalized to 1. This implies ! $\alpha>0$ and $\beta>0$ are the distances between the endpoints and their ! corresponding control points. The problem to now to find proper parameters $\alpha>0$ and $\beta>0$ for a given set of *************** *** 71,76 **** % $$! x_1 = x_0 + \frac{\alpha}{3} t_x(0) \quad\text{and}\quad ! x_2 = x_3 - \frac{\beta}{3} t_x(1).$$ % --- 80,85 ---- % $$! x_1 = x_0 + \alpha t_x(0) \quad\text{and}\quad ! x_2 = x_3 - \beta t_x(1).$$ % *************** *** 80,84 **** \kappa(0) (\dot x^2(0) + \dot y^2(0))^{3/2} &= \dot x(0) \ddot y(0) - \ddot x(0) \dot y(0) \\ ! = \kappa(0) |\alpha|^3 &= \begin{aligned}[t] +\dot x(0) &\bigl[- 4\dot y(0) - 2\dot y(1) + 6(y_3-y_0)\bigr] \\ --- 89,93 ---- \kappa(0) (\dot x^2(0) + \dot y^2(0))^{3/2} &= \dot x(0) \ddot y(0) - \ddot x(0) \dot y(0) \\ ! = 27 \kappa(0) |\alpha|^3 &= \begin{aligned}[t] +\dot x(0) &\bigl[- 4\dot y(0) - 2\dot y(1) + 6(y_3-y_0)\bigr] \\ *************** *** 90,104 **** \end{aligned}\\ &= \begin{aligned}[t] ! &-2 \alpha\beta \bigl[t_x(0)t_y(1) - t_y(0)t_x(1)\bigr] \\ ! &+6 \alpha \bigl[t_x(0) (y_3-y_0) - t_y(0) (x_3-x_0)\bigr] \end{aligned} \end{align} % ! And in short, % $$! 0 = \kappa(0) \alpha^2 \sign(\alpha) ! + 2\beta \bigl[t_x(0)t_y(1) - t_y(0)t_x(1)\bigr] ! - 6 \bigl[t_x(0) (y_3-y_0) - t_y(0) (x_3-x_0)\bigr]$$ % --- 99,113 ---- \end{aligned}\\ &= \begin{aligned}[t] ! &-18 \alpha\beta \bigl[t_x(0)t_y(1) - t_y(0)t_x(1)\bigr] \\ ! &+18 \alpha \bigl[t_x(0) (y_3-y_0) - t_y(0) (x_3-x_0)\bigr] \end{aligned} \end{align} % ! And short, % $$! 0 = \frac{3}{2}\kappa(0) \alpha^2 \sign(\alpha) ! + \beta \bigl[t_x(0)t_y(1) - t_y(0)t_x(1)\bigr] ! - \bigl[t_x(0) (y_3-y_0) - t_y(0) (x_3-x_0)\bigr]$$ % *************** *** 108,112 **** \kappa(1) (\dot x^2(1) + \dot y^2(1))^{3/2} &= \dot x(1) \ddot y(1) - \ddot x(1) \dot y(1) \\ ! = \kappa(1) |\beta|^3 &= \begin{aligned}[t] \dot x(1) &\bigl[ 4\dot y(1) + 2\dot y(0) - 6(y_3-y_1)\bigr] \\ --- 117,121 ---- \kappa(1) (\dot x^2(1) + \dot y^2(1))^{3/2} &= \dot x(1) \ddot y(1) - \ddot x(1) \dot y(1) \\ ! = 27 \kappa(1) |\beta|^3 &= \begin{aligned}[t] \dot x(1) &\bigl[ 4\dot y(1) + 2\dot y(0) - 6(y_3-y_1)\bigr] \\ *************** *** 118,129 **** \end{aligned}\\ &= \begin{aligned}[t] ! &-2 \alpha\beta \bigl[t_x(0)t_y(1) - t_y(0)t_x(1)\bigr] \\ ! &-6 \beta \bigl[t_x(1) (y_3-y_1) - t_y(1) (x_3-x_1)\bigr] \end{aligned}\\ \end{align} $$! 0 = \kappa(1) \beta^2 \sign(\beta) ! + 2\alpha \bigl[t_x(0)t_y(1) - t_y(0)t_x(1)\bigr] ! + 6 \bigl[t_x(1) (y_3-y_1) - t_y(1) (x_3-x_1)\bigr]$$ % --- 127,138 ---- \end{aligned}\\ &= \begin{aligned}[t] ! &-18 \alpha\beta \bigl[t_x(0)t_y(1) - t_y(0)t_x(1)\bigr] \\ ! &-18 \beta \bigl[t_x(1) (y_3-y_1) - t_y(1) (x_3-x_1)\bigr] \end{aligned}\\ \end{align} $$! 0 = \frac{3}{2}\kappa(1) \beta^2 \sign(\beta) ! + \alpha \bigl[t_x(0)t_y(1) - t_y(0)t_x(1)\bigr] ! + \bigl[t_x(1) (y_3-y_1) - t_y(1) (x_3-x_1)\bigr]$$ % *************** *** 132,142 **** \begin{gather} \begin{aligned} ! 0 &= \frac{1}{2} \kappa(0) \alpha^2 \sign(\alpha) + \beta T - D \\ ! 0 &= \frac{1}{2} \kappa(1) \beta^2 \sign(\beta) + \alpha T + E \end{aligned}\\[\medskipamount] \begin{aligned} T &:= t_x(0)t_y(1) - t_y(0)t_x(1) \\ ! D &:= 3 \bigl[t_x(0) (y_3-y_0) - t_y(0) (x_3-x_0)\bigr]\\ ! E &:= 3 \bigl[t_x(1) (y_3-y_0) - t_y(1) (x_3-x_0)\bigr] \end{aligned} \end{gather} --- 141,151 ---- \begin{gather} \begin{aligned} ! 0 &= \frac{3}{2} \kappa(0) \alpha^2 \sign(\alpha) + \beta T - D \\ ! 0 &= \frac{3}{2} \kappa(1) \beta^2 \sign(\beta) + \alpha T + E \end{aligned}\\[\medskipamount] \begin{aligned} T &:= t_x(0)t_y(1) - t_y(0)t_x(1) \\ ! D &:= \bigl[t_x(0) (y_3-y_0) - t_y(0) (x_3-x_0)\bigr]\\ ! E &:= \bigl[t_x(1) (y_3-y_0) - t_y(1) (x_3-x_0)\bigr] \end{aligned} \end{gather} *************** *** 145,156 **** % \begin{align} ! 0 &= \frac{1}{8} \kappa(0)^2\kappa(1) \alpha^4 ! - \frac{1}{2}D\kappa(0)\kappa(1)\alpha^2 + T^3 \alpha ! + E T^2 + \frac{1}{2}\kappa(1) D^2 \\ ! 0 &= \frac{1}{8} \kappa(0)\kappa(1)^2\beta^4 ! + \frac{1}{2}E\kappa(0)\kappa(1)\beta^2 + T^3 \beta ! - D T^2 + \frac{1}{2}\kappa(0) E^2 \\ \end{align} % --- 154,165 ---- % \begin{align} ! 0 &= \frac{27}{8} \kappa(0)^2\kappa(1) \alpha^4 ! - \frac{9}{2}D\kappa(0)\kappa(1)\alpha^2 + T^3 \alpha ! + E T^2 + \frac{3}{2}\kappa(1) D^2 \\ ! 0 &= \frac{27}{8} \kappa(0)\kappa(1)^2\beta^4 ! + \frac{9}{2}E\kappa(0)\kappa(1)\beta^2 + T^3 \beta ! - D T^2 + \frac{3}{2}\kappa(0) E^2 \\ \end{align} %

 [PyX-checkins] pyx/design beziers.tex,1.1,1.2 From: Michael Schindler - 2005-08-30 19:12:48 Update of /cvsroot/pyx/pyx/design In directory sc8-pr-cvs1.sourceforge.net:/tmp/cvs-serv30368/design Modified Files: beziers.tex Log Message: adapt the parameterization in controldists_from_endpoints_pt to the derivation in design/beziers.tex Index: beziers.tex =================================================================== RCS file: /cvsroot/pyx/pyx/design/beziers.tex,v retrieving revision 1.1 retrieving revision 1.2 diff -C2 -d -r1.1 -r1.2 *** beziers.tex 30 Aug 2005 18:31:49 -0000 1.1 --- beziers.tex 30 Aug 2005 19:12:40 -0000 1.2 *************** *** 45,58 **** \begin{align} \left(\begin{array}{cc} \dot x(0) \\ \dot y(0) \end{array}\right) ! = \alpha \left(\begin{array}{cc} t_x(0) \\ t_y(0) ! \end{array}\right) \qquad \left(\begin{array}{cc} \dot x(1) \\ \dot y(1) \end{array}\right) ! = \beta \left(\begin{array}{cc} t_x(1) \\ t_y(1) --- 45,66 ---- \begin{align} \left(\begin{array}{cc} + x_1-x_0 \\ y_1-y_0 + \end{array}\right) + &= \frac{1}{3} + \left(\begin{array}{cc} \dot x(0) \\ \dot y(0) \end{array}\right) ! =: \alpha \left(\begin{array}{cc} t_x(0) \\ t_y(0) ! \end{array}\right) \\ ! \left(\begin{array}{cc} ! x_3-x_2 \\ y_3-y_2 ! \end{array}\right) ! &= \frac{1}{3} \left(\begin{array}{cc} \dot x(1) \\ \dot y(1) \end{array}\right) ! =: \beta \left(\begin{array}{cc} t_x(1) \\ t_y(1) *************** *** 62,66 **** Here, the externally prescribed tangent vectors are expected to be parallel to the tangent vectors of the parameterization and normalized to 1. This implies ! $\alpha>0$ and $\beta>0$. The problem to now to find proper parameters $\alpha>0$ and $\beta>0$ for a given set of --- 70,75 ---- Here, the externally prescribed tangent vectors are expected to be parallel to the tangent vectors of the parameterization and normalized to 1. This implies ! $\alpha>0$ and $\beta>0$ are the distances between the endpoints and their ! corresponding control points. The problem to now to find proper parameters $\alpha>0$ and $\beta>0$ for a given set of *************** *** 71,76 **** % $$! x_1 = x_0 + \frac{\alpha}{3} t_x(0) \quad\text{and}\quad ! x_2 = x_3 - \frac{\beta}{3} t_x(1).$$ % --- 80,85 ---- % $$! x_1 = x_0 + \alpha t_x(0) \quad\text{and}\quad ! x_2 = x_3 - \beta t_x(1).$$ % *************** *** 80,84 **** \kappa(0) (\dot x^2(0) + \dot y^2(0))^{3/2} &= \dot x(0) \ddot y(0) - \ddot x(0) \dot y(0) \\ ! = \kappa(0) |\alpha|^3 &= \begin{aligned}[t] +\dot x(0) &\bigl[- 4\dot y(0) - 2\dot y(1) + 6(y_3-y_0)\bigr] \\ --- 89,93 ---- \kappa(0) (\dot x^2(0) + \dot y^2(0))^{3/2} &= \dot x(0) \ddot y(0) - \ddot x(0) \dot y(0) \\ ! = 27 \kappa(0) |\alpha|^3 &= \begin{aligned}[t] +\dot x(0) &\bigl[- 4\dot y(0) - 2\dot y(1) + 6(y_3-y_0)\bigr] \\ *************** *** 90,104 **** \end{aligned}\\ &= \begin{aligned}[t] ! &-2 \alpha\beta \bigl[t_x(0)t_y(1) - t_y(0)t_x(1)\bigr] \\ ! &+6 \alpha \bigl[t_x(0) (y_3-y_0) - t_y(0) (x_3-x_0)\bigr] \end{aligned} \end{align} % ! And in short, % $$! 0 = \kappa(0) \alpha^2 \sign(\alpha) ! + 2\beta \bigl[t_x(0)t_y(1) - t_y(0)t_x(1)\bigr] ! - 6 \bigl[t_x(0) (y_3-y_0) - t_y(0) (x_3-x_0)\bigr]$$ % --- 99,113 ---- \end{aligned}\\ &= \begin{aligned}[t] ! &-18 \alpha\beta \bigl[t_x(0)t_y(1) - t_y(0)t_x(1)\bigr] \\ ! &+18 \alpha \bigl[t_x(0) (y_3-y_0) - t_y(0) (x_3-x_0)\bigr] \end{aligned} \end{align} % ! And short, % $$! 0 = \frac{3}{2}\kappa(0) \alpha^2 \sign(\alpha) ! + \beta \bigl[t_x(0)t_y(1) - t_y(0)t_x(1)\bigr] ! - \bigl[t_x(0) (y_3-y_0) - t_y(0) (x_3-x_0)\bigr]$$ % *************** *** 108,112 **** \kappa(1) (\dot x^2(1) + \dot y^2(1))^{3/2} &= \dot x(1) \ddot y(1) - \ddot x(1) \dot y(1) \\ ! = \kappa(1) |\beta|^3 &= \begin{aligned}[t] \dot x(1) &\bigl[ 4\dot y(1) + 2\dot y(0) - 6(y_3-y_1)\bigr] \\ --- 117,121 ---- \kappa(1) (\dot x^2(1) + \dot y^2(1))^{3/2} &= \dot x(1) \ddot y(1) - \ddot x(1) \dot y(1) \\ ! = 27 \kappa(1) |\beta|^3 &= \begin{aligned}[t] \dot x(1) &\bigl[ 4\dot y(1) + 2\dot y(0) - 6(y_3-y_1)\bigr] \\ *************** *** 118,129 **** \end{aligned}\\ &= \begin{aligned}[t] ! &-2 \alpha\beta \bigl[t_x(0)t_y(1) - t_y(0)t_x(1)\bigr] \\ ! &-6 \beta \bigl[t_x(1) (y_3-y_1) - t_y(1) (x_3-x_1)\bigr] \end{aligned}\\ \end{align} $$! 0 = \kappa(1) \beta^2 \sign(\beta) ! + 2\alpha \bigl[t_x(0)t_y(1) - t_y(0)t_x(1)\bigr] ! + 6 \bigl[t_x(1) (y_3-y_1) - t_y(1) (x_3-x_1)\bigr]$$ % --- 127,138 ---- \end{aligned}\\ &= \begin{aligned}[t] ! &-18 \alpha\beta \bigl[t_x(0)t_y(1) - t_y(0)t_x(1)\bigr] \\ ! &-18 \beta \bigl[t_x(1) (y_3-y_1) - t_y(1) (x_3-x_1)\bigr] \end{aligned}\\ \end{align} $$! 0 = \frac{3}{2}\kappa(1) \beta^2 \sign(\beta) ! + \alpha \bigl[t_x(0)t_y(1) - t_y(0)t_x(1)\bigr] ! + \bigl[t_x(1) (y_3-y_1) - t_y(1) (x_3-x_1)\bigr]$$ % *************** *** 132,142 **** \begin{gather} \begin{aligned} ! 0 &= \frac{1}{2} \kappa(0) \alpha^2 \sign(\alpha) + \beta T - D \\ ! 0 &= \frac{1}{2} \kappa(1) \beta^2 \sign(\beta) + \alpha T + E \end{aligned}\\[\medskipamount] \begin{aligned} T &:= t_x(0)t_y(1) - t_y(0)t_x(1) \\ ! D &:= 3 \bigl[t_x(0) (y_3-y_0) - t_y(0) (x_3-x_0)\bigr]\\ ! E &:= 3 \bigl[t_x(1) (y_3-y_0) - t_y(1) (x_3-x_0)\bigr] \end{aligned} \end{gather} --- 141,151 ---- \begin{gather} \begin{aligned} ! 0 &= \frac{3}{2} \kappa(0) \alpha^2 \sign(\alpha) + \beta T - D \\ ! 0 &= \frac{3}{2} \kappa(1) \beta^2 \sign(\beta) + \alpha T + E \end{aligned}\\[\medskipamount] \begin{aligned} T &:= t_x(0)t_y(1) - t_y(0)t_x(1) \\ ! D &:= \bigl[t_x(0) (y_3-y_0) - t_y(0) (x_3-x_0)\bigr]\\ ! E &:= \bigl[t_x(1) (y_3-y_0) - t_y(1) (x_3-x_0)\bigr] \end{aligned} \end{gather} *************** *** 145,156 **** % \begin{align} ! 0 &= \frac{1}{8} \kappa(0)^2\kappa(1) \alpha^4 ! - \frac{1}{2}D\kappa(0)\kappa(1)\alpha^2 + T^3 \alpha ! + E T^2 + \frac{1}{2}\kappa(1) D^2 \\ ! 0 &= \frac{1}{8} \kappa(0)\kappa(1)^2\beta^4 ! + \frac{1}{2}E\kappa(0)\kappa(1)\beta^2 + T^3 \beta ! - D T^2 + \frac{1}{2}\kappa(0) E^2 \\ \end{align} % --- 154,165 ---- % \begin{align} ! 0 &= \frac{27}{8} \kappa(0)^2\kappa(1) \alpha^4 ! - \frac{9}{2}D\kappa(0)\kappa(1)\alpha^2 + T^3 \alpha ! + E T^2 + \frac{3}{2}\kappa(1) D^2 \\ ! 0 &= \frac{27}{8} \kappa(0)\kappa(1)^2\beta^4 ! + \frac{9}{2}E\kappa(0)\kappa(1)\beta^2 + T^3 \beta ! - D T^2 + \frac{3}{2}\kappa(0) E^2 \\ \end{align} %