我正在编写一个简单的光线跟踪器，为了使其简单起见，现在我决定在场景中仅包含球体。我现在处在一个阶段，我只想确认我的光线正确地与场景中的一个球体相交，没有别的。我创建了一个Ray和Sphere类，然后在我的主文件中创建了一个函数，该函数遍历每个像素以查看是否存在交集(相关代码将在下面发布)。问题在于与球体的整个交点的行为都非常奇怪。如果创建一个以中心(0，0，-20)和半径为1的球体，那么我只会得到一个交集，该交集始终位于我的图像的第一个像素处(左上角)。一旦达到15的半径，我突然在左上区域得到三个交叉点。半径18给我六个交点，一旦我达到20+半径，我突然得到一个EACH像素的交点，所以有些事情正在发生，这是不应该做的。
我很怀疑射线球交点代码可能在这里出错，但是仔细检查并在网上查找了更多信息之后，大多数解决方案都描述了我使用的相同方法，因此我认为它不应该是错误的!这里。所以...我不确定自己在做什么错，这可能是我的交集代码，也可能是导致问题的其他原因。我似乎找不到。给出球体和光线的值时，可能是我在想错吗？以下是相关代码
球形类:
Sphere::Sphere(glm::vec3 center, float radius)
: m_center(center), m_radius(radius), m_radiusSquared(radius*radius)
{
}
//Sphere-ray intersection. Equation: (P-C)^2 - R^2 = 0, P = o+t*d
//(P-C)^2 - R^2 => (o+t*d-C)^2-R^2 => o^2+(td)^2+C^2+2td(o-C)-2oC-R^2
//=> at^2+bt+c, a = d*d, b = 2d(o-C), c = (o-C)^2-R^2
//o = ray origin, d = ray direction, C = sphere center, R = sphere radius
bool Sphere::intersection(Ray& ray) const
{
//Squared distance between ray origin and sphere center
float squaredDist = glm::dot(ray.origin()-m_center, ray.origin()-m_center);
//If the distance is less than the squared radius of the sphere...
if(squaredDist <= m_radiusSquared)
{
//Point is in sphere, consider as no intersection existing
//std::cout << "Point inside sphere..." << std::endl;
return false;
}
//Will hold solution to quadratic equation
float t0, t1;
//Calculating the coefficients of the quadratic equation
float a = glm::dot(ray.direction(),ray.direction()); // a = d*d
float b = 2.0f*glm::dot(ray.direction(),ray.origin()-m_center); // b = 2d(o-C)
float c = glm::dot(ray.origin()-m_center, ray.origin()-m_center) - m_radiusSquared; // c = (o-C)^2-R^2
//Calculate discriminant
float disc = (b*b)-(4.0f*a*c);
if(disc < 0) //If discriminant is negative no intersection happens
{
//std::cout << "No intersection with sphere..." << std::endl;
return false;
}
else //If discriminant is positive one or two intersections (two solutions) exists
{
float sqrt_disc = glm::sqrt(disc);
t0 = (-b - sqrt_disc) / (2.0f * a);
t1 = (-b + sqrt_disc) / (2.0f * a);
}
//If the second intersection has a negative value then the intersections
//happen behind the ray origin which is not considered. Otherwise t0 is
//the intersection to be considered
if(t1<0)
{
//std::cout << "No intersection with sphere..." << std::endl;
return false;
}
else
{
//std::cout << "Intersection with sphere..." << std::endl;
return true;
}
}
程序:
#include "Sphere.h"
#include "Ray.h"
void renderScene(const Sphere& s);
const int imageWidth = 400;
const int imageHeight = 400;
int main()
{
//Create sphere with center in (0, 0, -20) and with radius 10
Sphere testSphere(glm::vec3(0.0f, 0.0f, -20.0f), 10.0f);
renderScene(testSphere);
return 0;
}
//Shoots rays through each pixel and check if there's an intersection with
//a given sphere. If an intersection exists then the counter is increased.
void renderScene(const Sphere& s)
{
//Ray r(origin, direction)
Ray r(glm::vec3(0.0f), glm::vec3(0.0f));
//Will hold the total amount of intersections
int counter = 0;
//Loops through each pixel...
for(int y=0; y<imageHeight; y++)
{
for(int x=0; x<imageWidth; x++)
{
//Change ray direction for each pixel being processed
r.setDirection(glm::vec3(((x-imageWidth/2)/(float)imageWidth), ((imageHeight/2-y)/(float)imageHeight), -1.0f));
//If current ray intersects sphere...
if(s.intersection(r))
{
//Increase counter
counter++;
}
}
}
std::cout << counter << std::endl;
}
最佳答案
在t1的情况下，您对二次方程式的第二种解决方案(disc > 0)是错误的，您需要这样的东西:
float sqrt_disc = glm::sqrt(disc);
t0 = (-b - sqrt_disc) / (2 * a);
t1 = (-b + sqrt_disc) / (2 * a);
我认为最好以这种形式写出等式，而不是将2除以0.5乘以，因为代码越类似于数学，则检查起来就越容易。
其他一些小意见:
disc重复使用名称sqrt(disc)似乎令人困惑，因此我在上面使用了新的变量名。 t0 > t1，因为您知道a和sqrt_disc均为正数，因此t1始终大于t0。 t0可能为负，而t1可能为正。您似乎无法处理这种情况。 disc == 0的特殊情况，因为一般情况下计算出的值与特殊情况下相同。 (而且，特殊情况越少，检查代码就越容易。)关于c++ - 射线-球面相交的相交问题，我们在Stack Overflow上找到一个类似的问题：https://stackoverflow.com/questions/12678225/