New answers tagged

5

An easy way to parallelize a path tracer is to have different threads (or machines!) work on different parts of the image. For instance, you could break the image into a 20x20 grid and have each thread grab a tile, render it, and try to grab another tile. A way I like to divy out this work is to use an atomic integer. Each thread atomically increments the ...


0

I have to answer my question. Consider that all patches have the same value. Then choosing a patch uniformly and selecting a random point uniformly on that patch is as the same as sampling the hemisphere uniformly. If $p_s=Pr(Q_k \in Q)=1/n$, which n is the number of patches, I can conclude that $p_\omega=p_s*(n/2\pi)=1/2\pi$ . And if patches don't have ...


0

Disclosure: I work for IMG/PowerVR Indeed there are rendering systems that can do this: Imagination demonstrated hybrid rendering on their "Wizard" Ray Tracing/Rasterisation system. Scroll down to "Making ray tracing happen" and there is a screen grab of a hybrid demo. As others have mentioned, there are pros and cons: Some advantages include that ...


3

Rasterization is sometimes used for primary rays. However, it limits greatly what you can do - depth of field, motion blur, participating media, refraction, only basic camera models, no adaptive or custom sampling, etc.


0

Because when you test the visibility of a fragment during rasterization you already know (or can easily calculate) all the details that raytracing will give you. So doing the ray-triangle collision calculation is redundant. But it is the additional lighting beyond the first hit where raytracing is helpful. However each ray is only a small part of what ...


1

There are renderes that do this. Most notably Pixars Renderman prefers to work this way. Hybrid renderers can work in other ways too... It used to be that hybrid renderers like this were quite common, they still are somewhat in software only implementations. They are slowly being replaced though by a new breed of trace only renderes though. Mostly the ...


2

The issue is, that you considered that the rays coming from the camera to be the light carrying rays. Instead, the rays bounce off the surface, and return to the light source (which you have decided to position at the camera). This means that your light direction is not $\pmb{d}$ but rather $-\pmb{d}$, then $\cos\theta = - \pmb{d} \cdot \pmb{n}$, provided ...


0

When you take the dot product of two unit vectors, you are essentially calculating the cosine of the angle between them. This value can become negative when the two vectors are less than 90 degrees apart (which they are when you have outward normals). Negate the normal before dotting it. That way you can have outward facing normals and also correct shading.


0

Let the horizontal FOV be given as theta (in radians), and the vertical FOV be given as phi (in radians). The we know that: $$\tan\frac{\theta}{2}=\frac{w}{f}, \tan\frac{\phi}{2}=\frac{h}{f}$$ Let us pick $f = 1$, then $w = \tan\frac{\theta}{2}, h = \tan\frac{\phi}{2}$. Now your ray generation direction generation formula looks like: $$\pmb{d} = x w\pmb{u}...


Top 50 recent answers are included