文字随弧排列，高分求算法

Question

在椭圆内写入汉字！汉字随椭圆弧度排列！汉字旋转的角度要与该点于水平夹角相等！
当椭圆的纵轴或短轴变化的时候文字的位置跟着变化！
希望能得到解答！谢谢！

Answer 1

//看以下我写的代码能否满足楼主需求（可是原创呀！^_^)
unit Unit1;interfaceuses
  Windows, Messages, SysUtils, Variants, Classes, Graphics, Controls, Forms,
  Dialogs, StdCtrls;type
  TForm1 = class(TForm)
    Button1: TButton;
    procedure Button1Click(Sender: TObject);
  private
    { Private declarations }
  public
    { Public declarations }
  end;var
  Form1: TForm1;implementation{$R *.dfm}
procedure MyTextOut(Ca:TCanvas;X,Y,Angle:Integer;Text:String);
var
  lf : TLogFont;
begin
  with Ca do
  begin
    GetObject(Font.Handle, sizeof(lf), @lf);
    lf.lfEscapement := Angle*10;
    Font.Handle := CreateFontIndirect(lf);
    TextOut(X,Y,Text);
    lf.lfEscapement := 0;
    lf.lfOrientation := 0;
    Font.Handle := CreateFontIndirect(lf);
  end;
end;
procedure CircleText(Ca:TCanvas;X0,Y0,RX,RY,BeginAngle,EndAngle:Integer;Text:WideString);
const
  aq=180/3.1415926;
  qa=3.1415926/180;
  procedure _CircleAlpha(Ca:TCanvas;X0,Y0,RX,RY,Angle:Integer;Alpha:WideChar);
  var
    x,y:Integer;
    tw,th:Double;
  begin
    tw:=Ca.TextWidth(Alpha) / 2;
    th:=Ca.TextWidth(Alpha) / 2;
    x:=Round(Cos((Angle-90)*qa)*tw)+X0;
    y:=Round(Sin((Angle-90)*qa)*tw)+Y0;
    x:=Round(Cos(Angle*qa)*RX)+x;
    y:=Round(Sin(Angle*qa)*RY)+y;
    MyTextOut(Ca,x,y,270-Angle,Alpha);
  end;
var
  Angle,Step:Double;
  n:Integer;
begin
  Step:=(EndAngle-BeginAngle)/Length(Text);
  Angle:=BeginAngle;
  for n:=1 to Length(Text) do
  begin
    _CircleAlpha(Ca,X0,Y0,RX,RY,Round(Angle),Text[n]);
    Angle:=Angle+Step;
  end;
end;procedure TForm1.Button1Click(Sender: TObject);
begin
  Canvas.Font.Name:='宋体';
  Canvas.Font.Color:=clBlue;
  Canvas.Font.Size:=12;
  CircleText(Canvas,200,200,150,100,180,360,'I am Wizardqi(男巫)!');
end;end.

Answer 2

参考：http://community.csdn.net/Expert/topic/4417/4417147.xml?temp=.731518