将三角形转换为另一个三角形 [英] Transform a triangle to another triangle

问题描述

您好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三角形。我所拥有的是2个三角形的坐标。你能帮助我吗？

Hi i am trying to create the affine transform that will allow me to transform a triangle into another one. What i have are the coordinates for the 2 triangles. Can you help me?

在亚当罗森菲尔德的回答之后我想出了这个代码，以防有人厌倦自己解决这个问题：

Following the answer by Adam Rosenfield i came up with this code in case anyone is bored to solve the equation himself :

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);
}

推荐答案

我要假设你在这里谈论2D。仿射变换矩阵中有9个值：

I'm going to assume you're talking about 2D here. An affine transformation matrix has 9 values in it:

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

有3个输入顶点 x1 ， x2 和 x3 ，转换后应变为 y1 ， y2 ， y3 。但是，由于我们在齐次坐标系中工作，因此将 A 应用于 x1 并不一定会给出 y1 - 它给出了 y1 的倍数。所以，我们还有未知的乘数 k1 ， k2 ，以及 k3 ，方程式：

There are 3 input vertices x1, x2, and x3, which when transformed should become y1, y2, y3. However, since we're working in homogeneous coordinates, applying A to x1 doesn't necessarily give y1 -- it gives a multiple of y1. So, we also have the unknown multipliers k1, k2, and k3, with the equations:

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

每个都是向量，所以我们在12个未知数中确实有9个方程，所以解决方案将受到不足。如果我们要求 a7 = 0 ， a8 = 0 ，并且 a9 = 1 ，那么解决方案将是唯一的（这个选择很自然，因为它意味着输入点是否为（ x ， y ，1），那么输出点将始终具有齐次坐标1，因此得到的变换只是一个2x2变换加上一个平移。）

Each of those is a vector, so we really have 9 equations in 12 unknowns, so the solution is going to be underconstrained. If we require that a7=0, a8=0, and a9=1, then the solution will be unique (this choice is natural, since it means if the input point is (x, y, 1), then the output point will always have homogeneous coordinate 1, so the resulting transform is just a 2x2 transform plus a translation).

因此，这将方程式减少为：

Hence, this reduces the equations to:

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得出：

So, k1 = k2 = k3 = 1. Plugging these in and converting to matrix form yields:

| 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，它才会有一个独特的解决方案。

Solving this 6x6 system of equations yields you your affine transformation matrix A. It will have a unique solution if and only if the 3 points of your source triangle are not collinear.

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