已知在坐标系中有一个长方形物体,ABCD分别为4个角,已知ABCD4个点的坐标,现在要在另一坐标系中展示该物体,以A为原点,AB为Y轴,AD为X轴,求原坐标到新坐标的转换公式,用java代码实现

假设原坐标系中A(x1, y1)，B(x2, y2)，C(x3, y3)，D(x4, y4)四个点的坐标已知，新坐标系中A(0, 0)，B(0, h)，D(w, 0)三个点的坐标已知。

首先，我们可以求出AB和AD两个向量：

AB = (0, h) - (0, 0) = (0, h)
AD = (w, 0) - (0, 0) = (w, 0)

然后，我们可以求出四个点相对于A点的坐标：

B' = (x2 - x1, y2 - y1)
C' = (x3 - x1, y3 - y1)
D' = (x4 - x1, y4 - y1)

接着，我们可以将B'和D'分别投影到AB和AD向量上，得到它们在新坐标系中的坐标：

B'' = ((B' · AB) / |AB|^2) * AB = (h * (y2 - y1) / |AB|, h * (y2 - y1) / |AB|)
D'' = ((D' · AD) / |AD|^2) * AD = (w * (x4 - x1) / |AD|, w * (x4 - x1) / |AD|)

最后，我们可以得到C''的坐标，因为C'' = B'' + D'' - A：

C'' = B'' + D'' = (h * (y2 - y1) / |AB| + w * (x4 - x1) / |AD|, h * (y2 - y1) / |AB| + w * (x4 - x1) / |AD|)

Java代码实现如下：

public static double[][] convertCoordinates(double[][] originalCoords, double w, double h) {
    double x1 = originalCoords[0][0];
    double y1 = originalCoords[0][1];
    double x2 = originalCoords[1][0];
    double y2 = originalCoords[1][1];
    double x3 = originalCoords[2][0];
    double y3 = originalCoords[2][1];
    double x4 = originalCoords[3][0];
    double y4 = originalCoords[3][1];

    double[][] newCoords = new double[4][2];

    // Calculate AB and AD vectors
    double abX = 0;
    double abY = h;
    double adX = w;
    double adY = 0;

    // Calculate B', C', and D' coordinates
    double bx = x2 - x1;
    double by = y2 - y1;
    double cx = x3 - x1;
    double cy = y3 - y1;
    double dx = x4 - x1;
    double dy = y4 - y1;

    // Calculate projections of B' and D' onto AB and AD vectors
    double bProj = (bx * abX + by * abY) / (abX * abX + abY * abY);
    double dProj = (dx * adX + dy * adY) / (adX * adX + adY * adY);

    // Calculate B'' and D'' coordinates
    double bx2 = bProj * abX;
    double by2 = bProj * abY;
    double dx2 = dProj * adX;
    double dy2 = dProj * adY;

    // Calculate C'' coordinates
    double cx2 = bx2 + dx2;
    double cy2 = by2 + dy2;

    // Store new coordinates in array
    newCoords[0][0] = 0;
    newCoords[0][1] = 0;
    newCoords[1][0] = bx2;
    newCoords[1][1] = by2;
    newCoords[2][0] = cx2;
    newCoords[2][1] = cy2;
    newCoords[3][0] = dx2;
    newCoords[3][1] = dy2;

    return newCoords;
}