改变PrintEventArgs的屏幕原点.c#中的图形
本文关键字:图形 屏幕 PrintEventArgs 改变 原点 | 更新日期: 2023-09-27 18:10:31
我一直在c#中使用PaintEventArgs绘制3D对象。图形和以下代码
public void findPoints()
{
xp1 = (p1x * 1000) / (p1z + 1000);
xp2 = (p2x * 1000) / (p2z + 1000);
xp3 = (p3x * 1000) / (p3z + 1000);
xp4 = (p4x * 1000) / (p4z + 1000);
yp1 = (p1y * 1000) / (p1z + 1000);
yp2 = (p2y * 1000) / (p2z + 1000);
yp3 = (p3y * 1000) / (p3z + 1000);
yp4 = (p4y * 1000) / (p4z + 1000);
xp5 = (p5x * 1000) / (p5z + 1000);
xp6 = (p6x * 1000) / (p6z + 1000);
xp7 = (p7x * 1000) / (p7z + 1000);
xp8 = (p8x * 1000) / (p8z + 1000);
yp5 = (p5y * 1000) / (p5z + 1000);
yp6 = (p6y * 1000) / (p6z + 1000);
yp7 = (p7y * 1000) / (p7z + 1000);
yp8 = (p8y * 1000) / (p8z + 1000);
}
private void drawCube(PaintEventArgs e)
{
/*
* Bottom left = 1
* Bottom right = 2
* Top right = 3
* Top left = 4
* Back Top left = 5
* Back Top right = 6
* Back bottom right = 7
* Back bottom left = 8
*/
findPoints();
Graphics g = e.Graphics;
g.DrawLine(pen, xp1, yp1, xp2, yp2);
g.DrawLine(pen, xp2, yp2, xp3, yp3);
g.DrawLine(pen, xp3, yp3, xp4, yp4);
g.DrawLine(pen, xp4, yp4, xp1, yp1);
g.DrawLine(pen, xp1, yp1, xp8, yp8);
g.DrawLine(pen, xp8, yp8, xp7, yp7);
g.DrawLine(pen, xp7, yp7, xp2, yp2);
g.DrawLine(pen, xp2, yp2, xp7, yp7);
g.DrawLine(pen, xp7, yp7, xp6, yp6);
g.DrawLine(pen, xp6, yp6, xp3, yp3);
g.DrawLine(pen, xp3, yp3, xp6, yp6);
g.DrawLine(pen, xp6, yp6, xp5, yp5);
g.DrawLine(pen, xp5, yp5, xp4, yp4);
g.DrawLine(pen, xp4, yp4, xp5, yp5);
g.DrawLine(pen, xp5, yp5, xp8, yp8);
}
问题来了,当我试图操纵使用矩阵绘制的立方体。旋转和嵌套for循环。
例如,我可以用下面的代码创建一个圆。
private void rotateCube(PaintEventArgs e, int value)
{
Matrix myMatrix = new Matrix();
myMatrix.Rotate(value, MatrixOrder.Append);
e.Graphics.Transform = myMatrix;
}
private void Form1_Paint(object sender, PaintEventArgs e)
{
for (int x = 0; x < 360; ++x)
{
rotateCube(e, x);
Thread.Sleep(3);
drawCube(e);
}
}
当我使用该代码画一个圆圈时,我只能得到圆圈的右下方部分,因为屏幕的原点(和圆圈的中心)在坐标系统上是0,0或显示器的左上方。我的问题是……有没有一种方法可以将屏幕或圆圈的原点移动到屏幕的中间?
将旋转矩阵乘以平移矩阵。
请记住,将相机移动(100px,100px)相当于将模型移动(-100px,-100px)。