根据两个点旋转图像
本文关键字：两个 旋转 图像 | 更新日期: 2023-09-27 18:34:33
我的表格上有两个点，还有一个图片框，像这样：
*
[^]
[ ]
*
我想将图片框与点对齐，使其看起来像这样：
*
'^'
' '
*
我将如何计算角度以及如何旋转图片框？
目前我正在使用这个：
double xDifference = Math.Abs(point2.X - point1.X);
double yDifference = Math.Abs(point2.Y - point1.Y);
double angle = Math.Atan(yDifference / xDifference) * 180 / Math.PI;
但这不起作用，因为 x 和 y 值是绝对值，因此如果点 2 在点 1 的剩余部分，他们无法计算它。
为了旋转图像，我找到了以下功能：
public Bitmap rotateImage(Image image, PointF offset, float angle) {
// Create a new empty bitmap to hold rotated image
Bitmap rotatedBmp = new Bitmap(image.Width, image.Height);
rotatedBmp.SetResolution(image.HorizontalResolution, image.VerticalResolution);
// Make a graphics object from the empty bitmap
Graphics g = Graphics.FromImage(rotatedBmp);
// Put the rotation point in the center of the image
g.TranslateTransform(offset.X, offset.Y);
// Rotate the image
g.RotateTransform(angle);
// Move the image back
g.TranslateTransform(-offset.X, -offset.Y);
// Draw passed in image onto graphics object
g.DrawImage(image, new PointF(0, 0));
return rotatedBmp;
}
我将如何使用该功能？我不确定要插入哪些值进行偏移。
谢谢
我不喜欢在不必要的情况下使用角度。
在这里，您只想更改正交基础。
从 {X;Y} 移动到 {U;V}，其中 V(范数 1(平行于 AB(或点 1 点 2(。
因为{U;V} 是正交基，Ux = Vy 和 Uy = -Vx。
using System;
using System.Drawing;
using System.Drawing.Drawing2D;
using System.Windows.Forms;
namespace CsiChart
{
public partial class CustomControl1 : Control
{
private const float EPSILON = 1e-6f;
private Image _image;
private ImageLayout _imageLayout = ImageLayout.Center;
private PointF _pointA = new PointF(0, 100);
private PointF _pointB = new PointF(100, 0);
public CustomControl1()
{
InitializeComponent();
}
public Image Image
{
get { return _image; }
set
{
if (Equals(_image, value)) return;
_image = value;
Invalidate();
OnImageChanged(EventArgs.Empty);
}
}
public event EventHandler ImageChanged;
public ImageLayout ImageLayout
{
get { return _imageLayout; }
set
{
if (Equals(_imageLayout, value)) return;
_imageLayout = value;
Invalidate();
OnImageLayoutChanged(EventArgs.Empty);
}
}
public event EventHandler ImageLayoutChanged;
public PointF PointA
{
get { return _pointA; }
set
{
if (Equals(_pointA, value)) return;
_pointA = value;
Invalidate();
OnPointAChanged(EventArgs.Empty);
}
}
public event EventHandler PointAChanged;
public PointF PointB
{
get { return _pointB; }
set
{
if (Equals(_pointB, value)) return;
_pointB = value;
Invalidate();
OnPointBChanged(EventArgs.Empty);
}
}
public event EventHandler PointBChanged;
protected override void OnPaint(PaintEventArgs pe)
{
base.OnPaint(pe);
if (DesignMode) return;
var g = pe.Graphics;
g.Clear(BackColor);
var image = Image;
if (image == null) return;
var clientRectangle = ClientRectangle;
var centerX = clientRectangle.X + clientRectangle.Width / 2;
var centerY = clientRectangle.Y + clientRectangle.Height / 2;
var srcRect = new Rectangle(new Point(0, 0), image.Size);
var pointA = PointA;
var pointB = PointB;
// Compute U, AB vector normalized.
var vx = pointB.X - pointA.X;
var vy = pointB.Y - pointA.Y;
var vLength = (float) Math.Sqrt(vx*vx + vy*vy);
if (vLength < EPSILON)
{
vx = 0;
vy = 1;
}
else
{
vx /= vLength;
vy /= vLength;
}
var oldTransform = g.Transform;
// Change basis to U,V
// We also take into acount the inverted on screen Y.
g.Transform = new Matrix(-vy, vx, -vx, -vy, centerX, centerY);
var imageWidth = image.Width;
var imageHeight = image.Height;
RectangleF destRect;
switch (ImageLayout)
{
case ImageLayout.None:
destRect = new Rectangle(0, 0, imageWidth, imageHeight);
break;
case ImageLayout.Center:
destRect = new Rectangle(-imageWidth/2, -imageHeight/2, imageWidth, imageHeight);
break;
case ImageLayout.Zoom:
// XY aligned bounds size of the image.
var imageXSize = imageWidth*Math.Abs(vy) + imageHeight*Math.Abs(vx);
var imageYSize = imageWidth*Math.Abs(vx) + imageHeight*Math.Abs(vy);
// Get best scale to fit.
var s = Math.Min(clientRectangle.Width/imageXSize, clientRectangle.Height/imageYSize);
destRect = new RectangleF(-imageWidth*s/2, -imageHeight*s/2, imageWidth*s, imageHeight*s);
break;
default:
throw new InvalidOperationException();
}
g.DrawImage(image, destRect, srcRect, GraphicsUnit.Pixel);
g.Transform = oldTransform;
}
protected virtual void OnImageChanged(EventArgs eventArgs)
{
var handler = ImageChanged;
if (handler == null) return;
handler(this, eventArgs);
}
protected virtual void OnImageLayoutChanged(EventArgs eventArgs)
{
var handler = ImageLayoutChanged;
if (handler == null) return;
handler(this, eventArgs);
}
private void OnPointAChanged(EventArgs eventArgs)
{
var handler = PointAChanged;
if (handler == null) return;
handler(this, eventArgs);
}
private void OnPointBChanged(EventArgs eventArgs)
{
var handler = PointBChanged;
if (handler == null) return;
handler(this, eventArgs);
}
}
}
让我们把所有的计算放在一起。
首先，连接两点的线的方向可以计算如下：
double xDifference = point2.X - point1.X;
double yDifference = point2.Y - point1.Y;
double angleRadians = Math.Atan2(yDifference, xDifference);
后垂直方向(90度(必须与上面考虑的方向平行，因此旋转角度为
double rotationAngleRadians = angleDegrees - Math.PI/2;
有了这个角度，我们可以计算边界框的大小：
double newWidth = image.Width * Math.Abs(Math.Cos(rotationAngleRadians)) +
image.Height * Math.Abs(Math.Sin(rotationAngleRadians));
double newHeight = image.Width * Math.Abs(Math.Sin(rotationAngleRadians)) +
image.Height * Math.Abs(Math.Cos(rotationAngleRadians));
现在，我们首先需要以旧图像的中间位于位置 0 的方式进行转换。这使得翻译转换(-image.Width/2, -image.Height/2) .然后，我们应用rotationAngleDegrees旋转(这是rotationAngleRadians * 180 / Math.PI(，因为Graphics旋转期望角度以度为单位。然后，我们将图像移动到新图像的中间，即平移变换 (newWidth/2, newHeight/2) .