将三角形转换为另一个三角形

Question

嗨,我正在尝试创建仿射变换,这将允许我将三角形转换为另一个.我所拥有的是2个三角形的坐标.你能帮助我吗？

按照亚当罗森菲尔德的回答,我想出了这个代码,万一有人无聊自己解决方程:

public static AffineTransform createTransform(ThreePointSystem source,
            ThreePointSystem dest) {        
    double x11 = source.point1.getX();
    double x12 = source.point1.getY();
    double x21 = source.point2.getX();
    double x22 = source.point2.getY();
    double x31 = source.point3.getX();
    double x32 = source.point3.getY();
    double y11 = dest.point1.getX();
    double y12 = dest.point1.getY();
    double y21 = dest.point2.getX();
    double y22 = dest.point2.getY();
    double y31 = dest.point3.getX();
    double y32 = dest.point3.getY();

    double a1 = ((y11-y21)*(x12-x32)-(y11-y31)*(x12-x22))/
                ((x11-x21)*(x12-x32)-(x11-x31)*(x12-x22));
    double a2 = ((y11-y21)*(x11-x31)-(y11-y31)*(x11-x21))/
                ((x12-x22)*(x11-x31)-(x12-x32)*(x11-x21));
    double a3 = y11-a1*x11-a2*x12;
    double a4 = ((y12-y22)*(x12-x32)-(y12-y32)*(x12-x22))/
                ((x11-x21)*(x12-x32)-(x11-x31)*(x12-x22));
    double a5 = ((y12-y22)*(x11-x31)-(y12-y32)*(x11-x21))/
                ((x12-x22)*(x11-x31)-(x12-x32)*(x11-x21));
    double a6 = y12-a4*x11-a5*x12;
    return new AffineTransform(a1, a4, a2, a5, a3, a6);
}

Answer 1

我假设你在这里谈论2D.仿射变换矩阵中有9个值:

    | a1 a2 a3 |
A = | a4 a5 a6 |
    | a7 a8 a9 |

有3个顶点输入x1,x2和x3,当其转化应该成为y1,y2,y3.然而,由于我们在齐次坐标的工作,应用A到x1不一定给y1-它给人的倍数y1.所以,我们也有未知的乘法器k1,k2以及k3,用公式:

A*x1 = k1*y1
A*x2 = k2*y2
A*x3 = k3*y3

每个都是一个向量,所以我们确实有12个未知数的9个方程,因此解决方案将受到不足.如果我们要求a7=0,a8=0以及a9=1,然后将溶液将是唯一的(这种选择是自然的,因为这意味着如果输入点为(x,y,1),则输出点将总是有齐次坐标1,因此所得到的变换是只是2x2变换加上翻译).

因此,这将方程式减少为:

a1*x11 + a2*x12 + a3 = k1*y11
a4*x11 + a5*x12 + a6 = k1*y12
                   1 = k1
a1*x21 + a2*x22 + a3 = k2*y21
a4*x21 + a5*x22 + a6 = k2*y22
                   1 = k2
a1*x31 + a2*x32 + a3 = k3*y31
a4*x31 + a5*x32 + a6 = k3*y32
                   1 = k3

所以,k1= k2= k3= 1.将这些插入并转换为矩阵形式会产生:

| x11 x12   1   0   0   0 |   | a1 |   | y11 |
| x21 x22   1   0   0   0 |   | a2 |   | y21 |
| x31 x32   1   0   0   0 | * | a3 | = | y31 |
|   0   0   0 x11 x12   1 |   | a4 |   | y12 |
|   0   0   0 x21 x22   1 |   | a5 |   | y22 |
|   0   0   0 x31 x32   1 |   | a6 |   | y32 |

求解这个6x6方程组可以得到你的仿射变换矩阵A.当且仅当源三角形的3个点不共线时,它将具有唯一的解决方案.