如何找到一条线与一个轴对齐的框相交的位置，其中该线的至少一个点始终在该框内

Question

我有一个数据类型，表示在我制作的游戏中，每个轴可以施加到机器上的最大力。此数据类型分别存储六个轴——X、Y和Z，分别为正轴和负轴。这用于定义设备在3D空间中给定方向上可以施加的最大力。由于所有可以对这台机器施加力的零件都是轴对齐的(扩展到只能对一个或多个世界空间法线施加力(，因此这可以完美地表示其极限。

利用这些信息，我构建了一种边界框。此边界框旨在将移动机器所需的力限制为一个较小的值，该值表示机器实际输出的力。

我目前的方法是根据这些数据创建这个虚拟边界框，并查看我所需的力(可视化为线段，其中第一点是3D空间的原点，第二点是所需力向量本身(与这个虚拟矩形棱镜上的至少一个面相交的位置。我还没有完全使用这种方法，因为我在理解如何在我这边解决这个问题方面存在一些问题，这也是我提出这个问题的目的之一。边界框始终包含三维空间的原点，但不能保证框本身的原点等于三维空间的起点。

我认为这个盒子的类比是迄今为止最好的方法，因为我的有限力最好用一条线与这个虚拟盒子相交的地方来表示。有没有更好的方法可以按照我需要的的方式约束框内的点，而不必简单地按轴约束它？以下是问题的常数：

• 表示力的线段的原点将始终等于三维空间的原点。因此，表示力的线段的组合方向*大小将始终等于力本身

三维空间的原点将永远不在此虚拟框之外

我认为我的想法是正确的，但感觉太复杂了，我认为有很多东西我可以删掉，让计算更容易，但问题是，我不太确定从哪里开始，让它尽可能干净高效，更不用说真正解决这个问题的最佳方法了。

这是我试过的代码。

public Vector3f ConstrainedWithinScaled(Vector3f value) {
// Vector3f is a simple class that contains x, y, z, basic arithmetic operators (*/+-), Magnitude/Normalized properties, and Dot/Cross methods.
// The code that this method exists in is a class called DualVector3f which is the class described above. It contains properties PosX, PosY, PosZ, NegX, NegY, and NegZ -- All of these properties have positive values as they describe magnitude on that specific face.
if (IsInBounds(value)) {
// The input value is already within the constraints of the virtual box.
return value;
}
// Get all intersections
// The first returned intersection that resides on this box's surface is correct.
// Only in cases where the point reside on a corner or edge will result in multiple of these conditions being true, however in these cases the returned point will be identical in all 2 or 3 cases.
Vector3f center = Center;
Vector3f size = Size;
// Cache these so that I don't calculate them every single time.
// These are calculated from the minimum/maximum coordinates to create the center of the virtual box and the size of the virtual box respectively.
// public Vector3f Size => Negative + Positive; (Negative and Positive both only have positive components, as outlined up top)
// public Vector3f Center => Positive - (Size / 2); (Positive is always the maximum)
// As should be evident, Positive is composed of PosX, PosY, and PosZ
// Likewise, Negative is composed of NegX, NegY, and NegZ
Vector3f topIntersection = IntersectPoint(value, new Vector3f(0, 1, 0), center + new Vector3f(0, size.y, 0));
if (topIntersection.y == PosY) return topIntersection;
Vector3f bottomIntersection = IntersectPoint(value, new Vector3f(0, 1, 0), center - new Vector3f(0, size.y, 0));
if (bottomIntersection.y == -NegY) return bottomIntersection;
Vector3f leftIntersection = IntersectPoint(value, new Vector3f(-1, 0, 0), center - new Vector3f(size.x, 0, 0));
if (leftIntersection.x == -NegX) return leftIntersection;
Vector3f rightIntersection = IntersectPoint(value, new Vector3f(1, 0, 0), center + new Vector3f(size.x, 0, 0));
if (rightIntersection.x == PosX) return rightIntersection;
Vector3f frontIntersection = IntersectPoint(value, new Vector3f(0, 0, 1), center + new Vector3f(0, 0, size.z));
if (frontIntersection.z == PosZ) return frontIntersection;
Vector3f backIntersection = IntersectPoint(value, new Vector3f(0, 0, -1), center - new Vector3f(0, 0, size.z));
if (backIntersection.z == -NegZ) return backIntersection;
return new Vector3f(); // Fallback. This should theoretically never occur under any condition, so this simply satisfies the need to return.
}
// This was derived from https://rosettacode.org/wiki/Find_the_intersection_of_a_line_with_a_plane#C.23 and omits the "rayOrigin" parameter as this is always a zero vector.
private static Vector3f IntersectPoint(Vector3f dirWithMag, Vector3f planeNormal, Vector3f planeCenter) {
Vector3f diff = -planeCenter;
float prod1 = diff.Dot(planeNormal);
float prod2 = dirWithMag.Dot(planeNormal);
float prod3 = prod1 / prod2;
return dirWithMag * -prod3;
}

谢谢。

Answer 1

我发现解决这个问题的最好方法是取向量，找出哪个分量离边界最远。在我找到这个组件后，我将所有组件除以表示其越界程度的百分比，这会将所有组件缩小到我的框内。这满足了解决方案，因为这将点放置在我的长方体的边界上。

这是实现它的代码：

public Vector3f ConstrainedWithinScaled(Vector3f value) {
if (IsInBounds(value)) {
// The input value is already within the constraints of the virtual box.
return value;
}
float factorX = 1;
float factorY = 1;
float factorZ = 1;
if (value.x > PosX) {
factorX = value.x / PosX;
} else if (value.x < -NegX) {
factorX = value.x / -NegX;
}
if (value.y > PosY) {
factorY = value.y / PosY;
} else if (value.y < -NegY) {
factorY = value.y / -NegY;
}
if (value.z > PosZ) {
factorZ = value.z / PosZ;
} else if (value.z < -NegZ) {
factorZ = value.z / -NegZ;
}
float largestFactor = Mathf.Max(factorX, factorY, factorZ);
// Catch case: Box has zero size.
if (largestFactor == 0) return Vector3f.Zero;
return value / largestFactor;
}