Confusion between intersect, intersects, intersection

Confusion between intersect, intersects, intersection, and intersections.




Page 59 introduces a method named "intersect" that returns a collection of double t values.



<list> t intersect(sphere, ray)



On page 63 you mention an "intersection" method. I'm not really sure if this is a different method than "intersect", or a modified and renamed version of "intersect". This is what I imagined that would look like:


double intersection(double, struct sphere)


On page 64 there is a new method called "intersections" that does not seem to have a description at all.



My tests do not seem to pass because the algorithm in the lower part of page 62 is probably is not implemented correctly.



This is currently what I have for this method:




bool intersect(ray ray, sphere sphere, List_Head* intersection_list) {

 tuple sphereToRay = tupleSub(ray.point, createPoint(0.0f, 0.0f, 0.0f));

 double a = dot(ray.vector, ray.vector);

 double b = 2 * dot(ray.vector, sphereToRay);

 double c = dot(sphereToRay, sphereToRay) - 1;

 double discriminant = pow(b, 2) - 4 * a * c;

 if (discriminant < 0) {

   return false;

 }

 intersection *intersect1 = generateIntersectWithSentinalValues();

 intersection *intersect2 = generateIntersectWithSentinalValues();



 intersect1->object_id = sphere.id;

 intersect2->object_id = sphere.id;



 intersect1->t = -b - sqrt(discriminant) / (2 * a);

 intersect2->t = -b + sqrt(discriminant) / (2 * a);

 if (intersect1->t < intersect2->t) {

   addIntersectionToList(intersection_list, intersect1);

   addIntersectionToList(intersection_list, intersect2);

 }

 else {

   addIntersectionToList(intersection_list, intersect2);

   addIntersectionToList(intersection_list, intersect1);

 }

 return true;

}

I have removed the parenthesis on the intersect1->t lines to make the equation get the same answer as Matlab (Octave).



One thing that would be tremendously helpful would be to have a couple of known inputs for this method, along with the answers for a, b, c, discriminant and the intersect->t values.



This way it would not be an all or nothing situation and I could narrow down where things start to break.



Thanks for your help.