Question

用VB画的凸轮，程序如下，怎么让他以（Form1.Width / 2，Form1.Height / 2）为中心逆时针转动？请教大神

Private Sub Command1_Click()
pi = 3.14159265
Dim i As Integer
Dim x(360) As Single
Dim y(360) As Single
Show
Line (Form1.Width / 2 + 1000, Form1.Height / 2)-(Form1.Width / 2 + 1000, Form1.Height / 2)
For i = 1 To 206
x(i) = Form1.Width / 2 + (1000 + 7.8 * i) * Cos(i * pi / 180)
y(i) = Form1.Height / 2 + (1000 + 7.8 * i) * Sin(i * pi / 180)
Line -(x(i), y(i))
Next i
For i = 207 To 309
x(i) = Form1.Width / 2 + (2600 - 15.6 * (i - 206)) * Cos(i * pi / 180)
y(i) = Form1.Height / 2 + (2600 - 15.6 * (i - 206)) * Sin(i * pi / 180)
Line -(x(i), y(i))
Next i
For i = 310 To 360
x(i) = Form1.Width / 2 + 1000 * Cos(i * pi / 180)
y(i) = Form1.Height / 2 + 1000 * Sin(i * pi / 180)
Line -(x(i), y(i))
Next i
End Sub

Answer 1

窗体放一个command1,一个tiemr1
Const pi = 3.14159265
Dim agl As Double
Private Sub draw()
Dim i As Integer
Show
Line (Form1.Width / 2 + 1000 * Cos(agl), Form1.Height / 2 + 1000 * Sin(agl))-(Form1.Width / 2 + 1000 * Cos(agl), Form1.Height / 2 + 1000 * Sin(agl))
For i = 1 To 206
x = Form1.Width / 2 + (1000 + 7.8 * i) * Cos(i * pi / 180 + agl)
y = Form1.Height / 2 + (1000 + 7.8 * i) * Sin(i * pi / 180 + agl)
Line -(x, y)
Next i
For i = 207 To 309
x = Form1.Width / 2 + (2600 - 15.6 * (i - 206)) * Cos(i * pi / 180 + agl)
y = Form1.Height / 2 + (2600 - 15.6 * (i - 206)) * Sin(i * pi / 180 + agl)
Line -(x, y)
Next i
For i = 310 To 360
x = Form1.Width / 2 + 1000 * Cos(i * pi / 180 + agl)
y = Form1.Height / 2 + 1000 * Sin(i * pi / 180 + agl)
Line -(x, y)
Next i
End Sub

Private Sub Command1_Click()
If Timer1.Enabled = False Then
Timer1.Enabled = True
Else
Timer1.Enabled = False
End If
End Sub

Private Sub Form_Load()
Timer1.Interval = 30
End Sub

Private Sub Timer1_Timer()
agl = agl + pi / 50
If agl >= 2 * pi Then agl = 0
Form1.Cls
draw
End Sub

追问

您好，怎么让他逆时针转呢？

Answer 2

？请教大神