Question

我在弄清楚2个矩形棱镜是否接触或重叠时遇到了一些麻烦。我只有两个矩形棱镜的高X，Y，Z和低X，Y，Z。以下是我到目前为止的情况：

    public boolean overlaps(AreaSelection other) {
    boolean Xs = (lowX <= other.getHighestX()) && (other.getLowestX() <= highX);
    boolean Ys = (lowY <= other.getHighestY()) && (other.getLowestY() <= highY); 
    boolean Zs = (lowZ <= other.getHighestZ()) && (other.getLowestZ() <= highZ);
    return (Xs && Ys && Zs);        
}

有谁知道这是否正确？如果没有，解决方案是什么？谢谢！

Answer 1

当且仅当坐标轴上的三对投影间隔全部重叠时，两个边缘平行于坐标轴的矩形棱镜重叠。

因此有一个“实用程序”方法检查间隔的重叠是有意义的，我们调用三次以检查棱镜的重叠。我们假设将使用正确排序的间隔'端点来调用此方法：

public boolean overlap_1d(double aLow, double aHigh, double bLow, double bHigh)
{
    if (aLow <= bLow) return (bLow <= aHigh);
    /* else aLow > bLow */
    return (aLow <= bHigh);
}

原始代码将成为：

public boolean overlaps(AreaSelection other)
{
    boolean Xs = overlap_1D(lowX,highX,other.getLowestX(),other.getHighestX());
    boolean Ys = overlap_1D(lowY,highY,other.getLowestY(),other.getHighestY());
    boolean Zs = overlap_1D(lowZ,highZ,other.getLowestZ(),other.getHighestZ());
    return (Xs && Ys && Zs);        
}

请注意，在此方法中，重叠可能包含单个交点（无论是一维还是三维）。

Answer 2

如果棱镜没有旋转，你的方法是正确的，否则没有办法计算它们是否与这些值重叠。

确定2个矩形棱镜是否接触/交叉

2 个答案: