(************** Content-type: application/mathematica ************** CreatedBy='Mathematica 5.2' Mathematica-Compatible Notebook This notebook can be used with any Mathematica-compatible application, such as Mathematica, MathReader or Publicon. The data for the notebook starts with the line containing stars above. To get the notebook into a Mathematica-compatible application, do one of the following: * Save the data starting with the line of stars above into a file with a name ending in .nb, then open the file inside the application; * Copy the data starting with the line of stars above to the clipboard, then use the Paste menu command inside the application. Data for notebooks contains only printable 7-bit ASCII and can be sent directly in email or through ftp in text mode. Newlines can be CR, LF or CRLF (Unix, Macintosh or MS-DOS style). 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For more information on notebooks and Mathematica-compatible applications, contact Wolfram Research: web: http://www.wolfram.com email: info@wolfram.com phone: +1-217-398-0700 (U.S.) Notebook reader applications are available free of charge from Wolfram Research. *******************************************************************) (*CacheID: 232*) (*NotebookFileLineBreakTest NotebookFileLineBreakTest*) (*NotebookOptionsPosition[ 23736, 761]*) (*NotebookOutlinePosition[ 24444, 785]*) (* CellTagsIndexPosition[ 24400, 781]*) (*WindowFrame->Normal*) Notebook[{ Cell[CellGroupData[{ Cell["Bezier Curves via G.M.T", "Title"], Cell["\<\ Joseph O'Rourke October 23, 2006\ \>", "Subtitle"], Cell["\<\ Illustrates how to compute a Bezier curve from four input points, using a \ matrix formulation. The powers of t for the polynomial are placed in T. The \ coordinates of the four points are placed in the geometric array G. M is a \ \"magic matrix\" that encodes already the solution of the simultaneous \ equations to find the coefficients. Then G.M.T gives the Bezier curves \ directly.\ \>", "Subsubtitle"], Cell["T is a column vector of symbolic powers of t.", "Text", Background->GrayLevel[0.900008]], Cell[CellGroupData[{ Cell[BoxData[{ \(T = {t^3, t^2, t, 1}\), "\[IndentingNewLine]", \(MatrixForm[T]\)}], "Input"], Cell[BoxData[ \({t\^3, t\^2, t, 1}\)], "Output"], Cell[BoxData[ TagBox[ RowBox[{"(", "\[NoBreak]", TagBox[GridBox[{ {\(t\^3\)}, {\(t\^2\)}, {"t"}, {"1"} }, RowSpacings->1, ColumnAlignments->{Left}], Column], "\[NoBreak]", ")"}], Function[ BoxForm`e$, MatrixForm[ BoxForm`e$]]]], "Output"] }, Open ]], Cell["\<\ M is the fixed matrix that encodes the solution of the system of eqs.\ \>", "Text", Background->GrayLevel[0.900008]], Cell[CellGroupData[{ Cell[BoxData[{ \(M = {\[IndentingNewLine]{\(-1\), 3, \(-3\), 1}, \[IndentingNewLine]{3, \(-6\), 3, 0}, \[IndentingNewLine]{\(-3\), 3, 0, 0}, \[IndentingNewLine]{1, 0, 0, 0}\[IndentingNewLine]}\), "\[IndentingNewLine]", \(MatrixForm[M]\)}], "Input"], Cell[BoxData[ \({{\(-1\), 3, \(-3\), 1}, {3, \(-6\), 3, 0}, {\(-3\), 3, 0, 0}, {1, 0, 0, 0}}\)], "Output"], Cell[BoxData[ TagBox[ RowBox[{"(", "\[NoBreak]", GridBox[{ {\(-1\), "3", \(-3\), "1"}, {"3", \(-6\), "3", "0"}, {\(-3\), "3", "0", "0"}, {"1", "0", "0", "0"} }, RowSpacings->1, ColumnSpacings->1, ColumnAlignments->{Left}], "\[NoBreak]", ")"}], Function[ BoxForm`e$, MatrixForm[ BoxForm`e$]]]], "Output"] }, Open ]], Cell["\<\ G is the geometric data, the four points. In this case: p0=(0,0), p1=(0,1),p2=(1,0),p3=(2,1) So curve runs from origin to (2,1). All the x-coords are in the 1st row, all the y-coords in the 2nd row.\ \>", "Text", Background->GrayLevel[0.900008]], Cell[CellGroupData[{ Cell[BoxData[{ \(G\ = \ {\[IndentingNewLine]{0, 0, 1, 2}, \[IndentingNewLine]{0, 1, 0, 1}\[IndentingNewLine]}\), "\[IndentingNewLine]", \(MatrixForm[G]\)}], "Input"], Cell[BoxData[ \({{0, 0, 1, 2}, {0, 1, 0, 1}}\)], "Output"], Cell[BoxData[ TagBox[ RowBox[{"(", "\[NoBreak]", GridBox[{ {"0", "0", "1", "2"}, {"0", "1", "0", "1"} }, RowSpacings->1, ColumnSpacings->1, ColumnAlignments->{Left}], "\[NoBreak]", ")"}], Function[ BoxForm`e$, MatrixForm[ BoxForm`e$]]]], "Output"] }, Open ]], Cell["\<\ If one were doing this by hand, probably easiest to first compute G.M:\ \>", "Text", Background->GrayLevel[0.900008]], Cell[CellGroupData[{ Cell[BoxData[ \(MatrixForm[G . M]\)], "Input"], Cell[BoxData[ TagBox[ RowBox[{"(", "\[NoBreak]", GridBox[{ {\(-1\), "3", "0", "0"}, {"4", \(-6\), "3", "0"} }, RowSpacings->1, ColumnSpacings->1, ColumnAlignments->{Left}], "\[NoBreak]", ")"}], Function[ BoxForm`e$, MatrixForm[ BoxForm`e$]]]], "Output"] }, Open ]], Cell["Then multiply by T:", "Text", Background->GrayLevel[0.900008]], Cell[CellGroupData[{ Cell[BoxData[ \(MatrixForm[\((G . M)\) . T]\)], "Input"], Cell[BoxData[ TagBox[ RowBox[{"(", "\[NoBreak]", TagBox[GridBox[{ {\(3\ t\^2 - t\^3\)}, {\(3\ t - 6\ t\^2 + 4\ t\^3\)} }, RowSpacings->1, ColumnAlignments->{Left}], Column], "\[NoBreak]", ")"}], Function[ BoxForm`e$, MatrixForm[ BoxForm`e$]]]], "Output"] }, Open ]], Cell[TextData[{ "In ", StyleBox["Mathematica", FontSlant->"Italic"], ", just G.M.T suffices (matrix multiplication is associative):" }], "Text", Background->GrayLevel[0.900008]], Cell[CellGroupData[{ Cell[BoxData[ \(MatrixForm[G . M . T]\)], "Input"], Cell[BoxData[ TagBox[ RowBox[{"(", "\[NoBreak]", TagBox[GridBox[{ {\(3\ t\^2 - t\^3\)}, {\(3\ t - 6\ t\^2 + 4\ t\^3\)} }, RowSpacings->1, ColumnAlignments->{Left}], Column], "\[NoBreak]", ")"}], Function[ BoxForm`e$, MatrixForm[ BoxForm`e$]]]], "Output"] }, Open ]], Cell["\<\ Here is a plot of the resulting curve: (AspectRatio->Automatic means draw it without distortion)\ \>", "Text", Background->GrayLevel[0.900008]], Cell[BoxData[ \(\(curve = \ G . M . 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