What's Going On zshearer commented on slide_039 of Dynamics and Time Integration ()

In response to the first question, it's when the total energy of the system decreases in a way not modeled by the system. Using the pendulum example, the pendulum will decrease in speed over time with backward euler, and eventually come to rest at 0, which is not the expected behavior of a frictionless system with no air resistance.

For the second question: you're correct. Misaka-10032 commented on slide_039 of Dynamics and Time Integration ()

Also, just to make sure I understand the difference between Forward Euler and Symplectic Euler.

$$\ddot{u} = \Delta u$$

• Forward Euler

$$u_{k+1} = u_k + \tau v_{k} \\ v_{k+1} = v_k + \tau \Delta u_k$$

• Symplectic Euler

$$v_{k+1} = v_k + \tau \Delta u_k \\ u_{k+1} = u_k + \tau v_{k+1}$$

It seems the only difference is the order to update $u$ and $v$? Misaka-10032 commented on slide_039 of Dynamics and Time Integration ()

What is numerical damping? Haboric commented on slide_044 of Variance Reduction ()

I'm assuming $A(S)$ is the normalized area of $S$? Haboric commented on slide_016 of Variance Reduction ()

Got a bit confused. I understand this is inconsistent, but why is it unbiased? Can anyone explain? taoy1 commented on slide_002 of The Rendering Equation ()

I guess that's a typo. taoy1 commented on slide_005 of The Rendering Equation ()

Why there isn't a $cos$ term for the outgoing radiance similar to the $cos \theta$ term for the incoming radiance? Haboric commented on slide_002 of The Rendering Equation ()

Shouldn't $\Phi$ be radiant flux? Haboric commented on slide_015 of Radiometry ()

I think the visible light wavelength should be $390 - 700$ nm, which converts to $390 - 700 \times 10^{-9}$ m. pchatrat commented on slide_026 of The Rendering Equation ()

Irradiance is number of hits/ (given_time * given_area). So irradiance is total hits from all direction. Why in the above equation is irradiance said to be a function of direction? taoy1 commented on slide_056 of Radiometry () keenan commented on slide_059 of Color ()

In general, you can't cover the entire gamut of human color vision with three sources. So, you try to pick the triangle that covers the largest gamut (or satisfies other criteria). Some good discussion here. genericname commented on slide_059 of Color ()

This subset seems concentrated outside of green, but don't humans perceive green the best? keenan commented on slide_021 of Color () keenan commented on slide_025 of Color ()

Yes, you're both absolutely right---this is the "bug" I mentioned in class. Everything (including the picture on the bottom) makes sense if you think of g as the percent reflected, rather than the percent absorbed. (Of course, these two quantities are complementary, as pointed out by @shhhh.) keenan commented on slide_039 of Monte Carlo Integration ()

@Haboric: Correct; this algorithm will generate too many samples in the direction of the corners, and too few in the direction of the axes. keenan commented on slide_022 of Monte Carlo Integration ()

Looks like a typo; probably this slide was created by duplicating the previous slide, and someone (not me!) forgot to delete the "=1". shhhh commented on slide_025 of Color ()

Shouldn't g(v) be 1 - absorbtion function in order to calculate how much light is reflected? stutiRastogi commented on slide_021 of Color ()

Why is a "choppy" spectrum bad? Is there any reason we always want to match the daylight spectrum as closely as possible? Haboric commented on slide_025 of Color ()

I am not sure I understand "what remains is the product $f(v)g(v)$" means. Is it intensity absorbed by the object as a function of frequency? Haboric commented on slide_039 of Monte Carlo Integration ()

Given a direction, the number of points within the bounding box lying in this direction is different. Haboric commented on slide_022 of Monte Carlo Integration ()

Why does $CDF(b)$ equal 1? haojun commented on slide_047 of Monte Carlo Integration ()

Is p(w) a pdf here? What is the range of w ? stutiRastogi commented on slide_026 of Perspective Transforms and Texture Mapping ()

Thanks! It's because it makes clipping easy, i.e., because it makes it easier to determine if a triangle is outside the view frustum, and also much easier to actually compute the geometry of clipped (i.e., partial) triangles. keenan commented on slide_025 of Geometry Processing ()

Correct. :-) keenan commented on slide_055 of Rasterization Pipeline ()

No, it's probably because blending is not the most common mode for drawing. So, (i) the user likely doesn't expect it to be turned on initially, and (ii) drawing operations with blending operations turned on still costs more (the blending operations are computed, even though the result is the same). Why do we transform view frustum to unit cube? Is it because it will make rasterization easy? Misaka-10032 commented on slide_025 of Geometry Processing ()

The split edges shouldn't be counted as new edges :-) Misaka-10032 commented on slide_055 of Rasterization Pipeline ()

Is it because the rasterization pipeline does not support the penetration of transparent triangles well that OpenGL turns off GL_BLEND by default? Yes, if the point is inside the triangle then it's a convex combination. In general, it's an affine combination. keenan commented on slide_051 of Sampling ()

Bingo. keenan commented on slide_023 of 3D Transformations ()

Great! stutiRastogi commented on slide_026 of Perspective Transforms and Texture Mapping ()

I might be a little pedantic here, but technically it is a convex combination right? Or is there a case where $\alpha, \beta$ and $\gamma$ can be negative? I thought since they are ratios of areas they have to be non-negative. jellybean commented on slide_051 of Sampling ()

I would guess that the lines relate to the rectangular-ness of the images. Since pixels are square, possibly there is a stronger signal exactly in the x and y dimensions, causing these lines. stutiRastogi commented on slide_023 of 3D Transformations ()

Worked it out, and got it finally! Thank you :) jellybean commented on slide_027 of Sampling ()

We should limit sampling rate to decrease the amount of storage and compute power needed to represent the graphic. keenan commented on slide_045 of Introduction ()

@nmrrs: Modern GPUs are responsible for implementing some algorithm whose output passes the diamond test. That does not mean this algorithm has to proceed by walking through the image pixel-by-pixel checking the diamond geometry explicitly.

The reality of the situation is that there is still a little flexibility in what vendors are forced to implement. Roughly speaking, anything that's not spelled out in black-and-white in the OpenGL spec (somewhat like our own Scotty3D User Guide) is up to the developer to decide (much like your last assignment!). keenan commented on slide_039 of Introduction to Geometry ()

Distance functions that are based on closed-form expressions (like $x^2 + y^2 + z^2 - r^2$) are very uncommon, because it is hard to describe interesting geometry this way. The example above is perhaps an exception, but it takes a huge amount of work to design scenes this way, and the rendering performance is still rather poor compared to an optimized mesh-based renderer.

However, the advantage is that there are potentially fewer "geometric aliasing" artifacts. For instance, if you zoom in on a mesh of a sphere, eventually you will see a blocky polygonal boundary. Not true with the implicit description above. Also, it is easier to draw fractal surfaces with infinite detail if you're willing to ray trace implicit surfaces. (But again, controlling this fractal detail is very difficult, and performance is typically not great.) keenan commented on slide_037 of Introduction to Geometry ()

Yes. Thank you.