我使用此过程在WPF中旋转4个边框的矩形作为角:
'find the center
Dim center As New Point(((topRight.Margin.Left - topLeft.Margin.Left) / 2) + topLeft.Margin.Left,
((topLeft.Margin.Top - bottomLeft.Margin.Top) / 2) + bottomLeft.Margin.Top)
'shift the points to center and calculate the rotation
Dim tl As Point = getRotatedPoint(New Point(topLeft.Margin.Left - center.X,
topLeft.Margin.Top - center.Y), 1)
Dim tr As Point = getRotatedPoint(New Point(topRight.Margin.Left - center.X,
topRight.Margin.Top - center.Y), 1)
Dim bl As Point = getRotatedPoint(New Point(bottomLeft.Margin.Left - center.X,
bottomLeft.Margin.Top - center.Y), 1)
Dim br As Point = getRotatedPoint(New Point(bottomRight.Margin.Left - center.X,
bottomRight.Margin.Top - center.Y), 1)
'shift the points back from center and move
topLeft.Margin = New Thickness(tl.X + center.X, tl.Y + center.Y, 0, 0)
topRight.Margin = New Thickness(tr.X + center.X, tr.Y + center.Y, 0, 0)
bottomLeft.Margin = New Thickness(bl.X + center.X, bl.Y + center.Y, 0, 0)
bottomRight.Margin = New Thickness(br.X + center.X, +br.Y + center.Y, 0, 0)
getRotatedPoint 函数是:
'rotating the borders
Private Function getRotatedPoint(ByVal pnt As Point, ByVal degrees As Double)
Dim rAngle As Double = degrees * (Math.PI / 180)
Dim x As Double = pnt.X * Math.Cos(rAngle) - pnt.Y * Math.Sin(rAngle)
Dim y As Double = pnt.X * Math.Sin(rAngle) + pnt.Y * Math.Cos(rAngle)
Return New Point(x, y)
End Function
但是我得到非常混乱的结果,我不知道,任何想法都会受到欢迎:)
edit1:我将getRotatedPoint函数更改为双精度,并添加了弧度到度数转换。
编辑2:修正弧度转换函数。
edit3:更正了中心坐标,但仍会发生一些偏移。
Edit4:这是测试人员的示例项目: http://dl.dropbox.com/u/33417300/testapp.zip
如前所述,目前还不清楚您所说的"混乱的结果"是什么意思,因此我想到了几种解释:
如果旋转不正确,可能是因为您传递的角度以度为单位,而 SIN 和 COS 函数需要弧度。 这似乎是可能的,因为您将角度作为整数而不是浮点表示形式传递。 如果要将它们作为度数传递,请在将它们传递给三角函数之前转换为弧度。
其次,如果混乱的结果意味着代码很复杂,您可能希望将仿射变换表示为矩阵。 对于 2-D 情况,您可以通过添加 1 作为第三个元素将 2-D 向量扩展到 3-D。 您的变换是 3x3 矩阵,您可以将其相乘以创建旋转、缩放、倾斜、平移等序列。 生成的变换可以应用为向量 - 矩阵乘法。
更新:
此函数存在一些问题:
Private Function getRotatedPoint(ByVal pnt As Point, ByVal angle As Integer)
Dim x As Integer = pnt.X * Math.Cos(angle) - pnt.Y * Math.Sin(angle)
Dim y As Integer = pnt.X * Math.Sin(angle) + pnt.Y * Math.Cos(angle)
Return New Point(x, y)
End Function
除了度/弧度混淆的问题外,整数和浮点之间的类型转换留给编译器和 .Net 框架,这可能与您的意图不一致。 我会重写如下:
Private Function getRotatedPoint(ByVal pnt As Point, ByVal radians As Double)
Dim x As Double = CDbl(pnt.X) * Math.Cos(radians) - CDbl(pnt.Y) * Math.Sin(radians)
Dim y As Double = CDbl(pnt.X) * Math.Sin(radians) + CDbl(pnt.Y) * Math.Cos(radians)
Return New Point(CInt(x), CInt(y))
End Function
不过,我仍然建议使用矩阵库来执行此操作。 它将让您专注于开发程序的核心功能,而不是调试和排除已编写和重写多次的内容。
更新 2:
这也是有问题的:
Dim center As New Point((topLeft.Margin.Left + topRight.Margin.Left) / 2,
(topLeft.Margin.Left + bottomLeft.Margin.Left) / 2)
这不应该是:
Dim center As New Point((topLeft.Margin.Left + topRight.Margin.Right) / 2,
(topLeft.Margin.Top + bottomLeft.Margin.Bottom) / 2)
如果我错了,那么你需要详细解释你的左上角、右上角、左下角和右下角的值代表什么,足以让任何人帮助你......参见Kendall Frey对SSCCE的评论。