用matlab 画一个椭圆，如何保存椭圆周上的点的坐标，谢谢

Question

如题所述

Answer 1

,y1）和(x2,y2)
求椭圆步骤：
1、求a,b: a=len/2 b=sqr((x1-x2)^2+(y1-y2)^2)
2、求旋转角α ：求出(y1-y2)/(x1-x2)的反正切值即为α
3、求椭圆中心坐标（x0,y0）: x0=(x1+x2)/2 y0=(y1+y2)/2

3、计算椭圆上点的坐标（x,y）：
x=acosθ ， y=bsinθ ( 0<=θ< 2*π)
4、计算图形绕原点旋转α 弧度后的坐标(xx,yy):
xx=x*cos(-α )+y*sin(-α )
yy=-x*sin(-α )+y*cos(-α )
5、计算椭圆中心从原点平移到（x0,y0）后椭圆上点的坐标（xxx,yyy）:
xxx=xx+x0
yyy=yy+y0
6、在坐标（xxx,yyy）处画一各点
7、在( 0<=θ< 2*π)范围内，按一定间隔取值，重复3-7步骤，即得所要求的椭圆。

以下是vb写的简单示例，新建一各工程，把代码粘贴进去替换原来的所有代码，运行即可看效果

Option Explicit
Dim X1, Y1, X0, Y0, X2, Y2 As Double
Dim A, B, PI As Double
Dim F As Boolean

Private Sub Form_Load()
PI = 3.14159265358979
F = False
DrawWidth = 2
Width = 10000
Height = 8000
End Sub

Private Sub Form_MouseDown(Button As Integer, Shift As Integer, X As Single, Y As Single)
X1 = X
Y1 = Y
B = 1000
F = True

End Sub

Public Sub tuoYuan()
Dim Jiao As Double
Dim i, m, n, m1, n1 As Double

Cls
X0 = (X1 + X2) / 2
Y0 = (Y1 + Y2) / 2
A = Sqr((X0 - X2) ^ 2 + (Y0 - Y2) ^ 2)

If X1 <> X2 Then
Jiao = Atn((Y1 - Y2) / (X1 - X2))
Else
Jiao = PI / 2
End If

Form1.PSet (X1, Y1), RGB(255, 0, 0)
Form1.PSet (X0, Y0), RGB(255, 0, 0)
Form1.PSet (X2, Y2), RGB(255, 0, 0)

For i = 0 To PI * 2 Step 0.01
m = A * Cos(i)
n = B * Sin(i)
'Form1.PSet (m + X0, n + Y0), RGB(255, 0, 0)

m1 = m * Cos(-Jiao) + n * Sin(-Jiao)
n1 = -m * Sin(-Jiao) + n * Cos(-Jiao)

Form1.PSet (m1 + X0, n1 + Y0), RGB(0, 255, 0)

Next i

End Sub

Private Sub Form_MouseMove(Button As Integer, Shift As Integer, X As Single, Y As Single)
If F = True Then
X2 = X
Y2 = Y
Call tuoYuan
End If
End Sub

Private Sub Form_MouseUp(Button As Integer, Shift As Integer, X As Single, Y As Single)
F = False
X2 = X
Y2 = Y
Call tuoYuan
End Sub

Answer 2

matlab根据函数画的，你用函数生成坐标就行了啊。

Answer 3

假设椭圆方程为：(x^2)/9+(y^2)/25=1，则

>> clear all;

t=0:0.1:2*pi;

x=3*sin(t);%%x即为椭圆周上的点的横坐标

y=5*cos(t);%%y即为椭圆周上的点的纵坐标

plot(x,y);

grid on