计算转换球体的AABB

本文介绍了计算转换球体的AABB的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！ 问题描述 29岁程序员，3月因学历无情被辞！ 我有一个由对象空间中心点和半径表示的球体。这个球体被转换成世界空间，其中有一个转换矩阵，其中可能包括尺度，旋转和平移。我需要为世界空间中的球体构建一个轴对齐的边界框，但我不知道如何去做。 这是我目前的方法，对于一些情况： public void computeBoundingBox（）{ // center是球体的中间 // averagePosition是AABB 的中间// getObjToWorldTransform（）是从obj到世界空间的矩阵 getObjToWorldTransform（）。rightMultiply（center，averagePosition）; Point3 onSphere =新的Point3（center）; onSphere.scaleAdd（radius，new Vector3（1,1,1））; getObjToWorldTransform（）。rightMultiply（onSphere）; //但你怎么知道变换的半径是统一的？ double transformedRadius = onGeneration.distance（averagePosition）; // maxBound是AABB 的上限maxBound.set（averagePosition）; maxBound.scaleAdd（transformedRadius，new Vector3（1,1,1））; // minBound是AABB minBound.set（averagePosition）的下限; minBound.scaleAdd（transformedRadius，new Vector3（-1，-1，-1））; } 然而，我怀疑这总会奏效。它不应该失败的非均匀缩放？解决方案一般来说，转换后的球体将是某种椭球体。为它确定一个确切的边界框并不难;如果你不想完成所有的数学运算： 注意 M 是您的转换矩阵（比例尺，旋转，翻译等） 读取 S 下面的定义 按照最后描述计算 R 计算 x ， y 和 z 基于 R 如上所述。 一般的圆锥曲线（包括球体及其变换）可以被表示为对称的4×4矩阵：当时，同构点 p 位于圆锥曲线 S p ^ t S p 。通过矩阵M变换你的空间如下变换S矩阵（下面的惯例是点是列向量）： A以原点为单位的半径球体表示为： S = [1 0 0 0] [0 1 0 0] [0 0 1 0] [0 0 0 -1] 点p在圆锥曲面上时： 0 = p ^ t S p = p ^ t M ^ t M ^ t ^ -1 SM ^ M =（M p）^ t（M ^ -1 ^ t SM ^ -1）（M p） 变换点（M p）应保持发生率 - >由矩阵M转换的圆锥曲线S是：（M ^ -1 ^ t SM ^ -1） 用于飞机而不是点的圆锥曲线的对偶表示为S的倒数;对于平面q（表示为行向量）： 平面q在以下情况下与锥体相切： 0 = （q M ^ -1）（MS ^ - 1）（1）其中， 1 M ^ t）（q M ^ -1）^ t 变换平面（q M ^ -1）应保持发生率 - >由矩阵M转换的双圆锥是：（MS ^ -1 M ^ t） 重新寻找与变换圆锥相切的轴对齐平面： $ $ $ $ $ $ $ $ $ R = [r11 r12 r13 r14]（注意R是对称的：R = R ^ t） [r12 r22 r23 r24] [r13 r23 r33 r34] [r14 r24 r34 r44 ] 轴对齐的平面是： xy平面：[0 0 1 -z] xz平面：[0 1 0 -y] yz平面： [1 0 0 -x] 要找到与R相切的xy平面的平面： [0 0 1 -z] R [0] = r33 - 2 r34 z + r44 z ^ 2 = 0 [0 ] [1] [-z] so，z =（2 r34 +/- sqrt（4 r34 ^ 2 - 4 r44 r33））/（2 r44 ） =（r34 +/- sqrt（r34 ^ 2 - r44 r33））/ r44 和yz对齐的平面： x =（r14 +/- sqrt（r14 ^ 2 - r44 r11））/ r44 这为转换后的球体提供了精确的边界框。 I have a sphere represented in object space by a center point and a radius. The sphere is transformed into world space with a transformation matrix that may include scales, rotations, and translations. I need to build a axis aligned bounding box for the sphere in world space, but I'm not sure how to do it.Here is my current approach, that works for some cases:public void computeBoundingBox() { // center is the middle of the sphere // averagePosition is the middle of the AABB // getObjToWorldTransform() is a matrix from obj to world space getObjToWorldTransform().rightMultiply(center, averagePosition); Point3 onSphere = new Point3(center); onSphere.scaleAdd(radius, new Vector3(1, 1, 1)); getObjToWorldTransform().rightMultiply(onSphere); // but how do you know that the transformed radius is uniform? double transformedRadius = onSphere.distance(averagePosition); // maxBound is the upper limit of the AABB maxBound.set(averagePosition); maxBound.scaleAdd(transformedRadius, new Vector3(1, 1, 1)); // minBound is the lower limit of the AABB minBound.set(averagePosition); minBound.scaleAdd(transformedRadius, new Vector3(-1,-1,-1));}However, I am skeptical that this would always work. Shouldn't it fail for non-uniform scaling? 解决方案 In general, a transformed sphere will be an ellipsoid of some sort. It's not too hard to get an exact bounding box for it; if you don't want go through all the math:note that M is your transformation matrix (scales, rotations, translations, etc.)read the definition of S belowcompute R as described towards the endcompute the x, y, and z bounds based on R as described last.A general conic (which includes spheres and their transforms) can be represented as a symmetric 4x4 matrix: a homogeneous point p is inside the conic S when p^t S p < 0. Transforming your space by the matrix M transforms the S matrix as follows (the convention below is that points are column vectors):A unit-radius sphere about the origin is represented by:S = [ 1 0 0 0 ] [ 0 1 0 0 ] [ 0 0 1 0 ] [ 0 0 0 -1 ]point p is on the conic surface when:0 = p^t S p = p^t M^t M^t^-1 S M^-1 M p = (M p)^t (M^-1^t S M^-1) (M p)transformed point (M p) should preserve incidence-> conic S transformed by matrix M is: (M^-1^t S M^-1)The dual of the conic, which applies to planes instead of points, is represented by the inverse of S; for plane q (represented as a row vector):plane q is tangent to the conic when:0 = q S^-1 q^t = q M^-1 M S^-1 M^t M^t^-1 q^t = (q M^-1) (M S^-1 M^t) (q M^-1)^ttransformed plane (q M^-1) should preserve incidence-> dual conic transformed by matrix M is: (M S^-1 M^t)So, you're looking for axis-aligned planes that are tangent to the transformed conic:let (M S^-1 M^t) = R = [ r11 r12 r13 r14 ] (note that R is symmetric: R=R^t) [ r12 r22 r23 r24 ] [ r13 r23 r33 r34 ] [ r14 r24 r34 r44 ]axis-aligned planes are: xy planes: [ 0 0 1 -z ] xz planes: [ 0 1 0 -y ] yz planes: [ 1 0 0 -x ]To find xy-aligned planes tangent to R: [0 0 1 -z] R [ 0 ] = r33 - 2 r34 z + r44 z^2 = 0 [ 0 ] [ 1 ] [-z ] so, z = ( 2 r34 +/- sqrt(4 r34^2 - 4 r44 r33) ) / ( 2 r44 ) = (r34 +/- sqrt(r34^2 - r44 r33) ) / r44Similarly, for xz-aligned planes: y = (r24 +/- sqrt(r24^2 - r44 r22) ) / r44and yz-aligned planes: x = (r14 +/- sqrt(r14^2 - r44 r11) ) / r44This gives you an exact bounding box for the transformed sphere. 这篇关于计算转换球体的AABB的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持！ 上岸，阿里云！