gpt4 book ai didi

c++ - 分离轴定理错误触发

转载 作者:行者123 更新时间:2023-11-27 23:42:12 32 4
gpt4 key购买 nike

我正在用 C++ 编写一些代码,用于使用分离轴定理测试碰撞,并且在某些方向上它错误地触发了碰撞的发生

我正在关注 this教程,但是教程只有 2D,我正在尝试以 3D 实现它,尽管我认为它应该仍然是一样的。

我现在拥有的算法不会错过任何碰撞,但对于两个盒子的某些方向,它认为它们正在碰撞,而实际上它们并没有。可以在这里看到一个例子,根据下面的代码,这两个盒子显然发生了碰撞。

False collision

代码是用C++写的

BoxCollider.h

class BoxCollider :
public Collider
{
public:
BoxCollider(Vector3 position, Vector3 rotation, Vector3 size);


~BoxCollider();

void Update();

public:
Vector3 rotation;
Vector3 size;
Matrix transformMatrix;

std::vector<Vector3> points;

Vector3 normals[3];
};

方 block 对撞机.cpp

BoxCollider::BoxCollider(Vector3 position, Vector3 rotation, Vector3 size) : rotation(rotation), size(size)
{
this->position = position;
points.resize(8);

}

BoxCollider::~BoxCollider()
{
}

void BoxCollider::Update()
{
Transform* eTransform = m_entity->GetComponent<Transform>();
transformMatrix.RotateYawPitchRoll(rotation + eTransform->rotation);
Vector3 ePos = eTransform->position;

points[0] = transformMatrix * (Vector3( 0.5, -0.5, -0.5) * size) + position + ePos;
points[1] = transformMatrix * (Vector3( 0.5, 0.5, -0.5) * size) + position + ePos;
points[2] = transformMatrix * (Vector3( 0.5, -0.5, 0.5) * size) + position + ePos;
points[3] = transformMatrix * (Vector3( 0.5, 0.5, 0.5) * size) + position + ePos;
points[4] = transformMatrix * (Vector3(-0.5, -0.5, -0.5) * size) + position + ePos;
points[5] = transformMatrix * (Vector3(-0.5, 0.5, -0.5) * size) + position + ePos;
points[6] = transformMatrix * (Vector3(-0.5, -0.5, 0.5) * size) + position + ePos;
points[7] = transformMatrix * (Vector3(-0.5, 0.5, 0.5) * size) + position + ePos;

normals[0] = transformMatrix * Vector3(1, 0, 0);
normals[1] = transformMatrix * Vector3(0, 1, 0);
normals[2] = transformMatrix * Vector3(0, 0, 1);
}

算法:

void EntityManager::CheckCollision(BoxCollider * col0, BoxCollider * col1)
{
for (int i = 0; i < 3; i++) //First cube
{
Vector3 axis = col0->normals[i];
axis = Vector3(axis.z, -axis.x, axis.y);


Projection proj1 = GetProjection(col0->points, axis);
Projection proj2 = GetProjection(col1->points, axis);

float overlap = GetOverlap(proj1, proj2);

if (overlap > 0.0) //The projections do not overlap
return;
}

for (int i = 0; i < 3; i++) //First cube
{
Vector3 axis = col1->normals[i];
axis = Vector3(axis.z, -axis.x, axis.y);

Projection proj1 = GetProjection(col0->points, axis);
Projection proj2 = GetProjection(col1->points, axis);

float overlap = GetOverlap(proj1, proj2);

if (overlap > 0.0) //The projections do not overlap
return;
}
}

float GetOverlap(Projection proj1, Projection proj2)
{
float a = proj2.left - proj1.right;
float b = proj1.left - proj2.right;

return a > b ? a : b;
}

Projection GetProjection(std::vector<Vector3> points, Vector3 axis)
{
float tmp = 0;
float left = D3D10_FLOAT32_MAX, right = -D3D10_FLOAT32_MAX;

for (int i = 0; i < points.size(); i++)
{
tmp = DotProduct(points[i], axis.Normalize());
if (tmp < left)
{
left = tmp;
}

if (tmp > right)
{
right = tmp;
}
}

return Projection(left, right, axis);
}

最佳答案

tutorial is only in 2D and i'm trying to implement it in 3D although i think it should still be the same

不幸的是,事实并非如此。 3D 案例有点复杂。要检查两个复杂形状是否在 3D 中发生碰撞,您需要检查每个面的法线(您这样做)和垂直于每个对象边缘的方向(您错过了这些)。

因此,对于边方向为 A0、A1、A2 和 B0、B1、B2 的框(A 和 B),我们有:

  • A 的 3 个法线
  • B的3条法线
  • 9 个方向:A0 x B0、A0 x B1、A0 x B2、A1 x B0、A1 x B1、A1 x B2、A2 x B0、A2 x B1、A2 x B2

所以你只需要添加缺少的 9 个检查。

进一步注意:您不需要调整法线。我的意思是不需要这一行:

    axis = Vector3(axis.z, -axis.x, axis.y);

在这种情况下,它不会造成任何伤害。但对于更复杂的形状,它实际上可能会使测试不正确。

关于c++ - 分离轴定理错误触发,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/53795758/

32 4 0
Copyright 2021 - 2024 cfsdn All Rights Reserved 蜀ICP备2022000587号
广告合作:1813099741@qq.com 6ren.com