An Explanation of the Rendering Equation

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  • Опубліковано 19 гру 2024

КОМЕНТАРІ • 36

  • @SaHaRaSquad
    @SaHaRaSquad 6 років тому +12

    Finally an explanation that's understandable for non-mathematicians, thank you! I've pieced together the basic concept in the past by looking at countless different sources, but this gets to the point like none other.

  • @arsnakehert
    @arsnakehert 3 роки тому +1

    This is beautiful, the best explanation I've seen of this so far

    • @arsnakehert
      @arsnakehert 3 роки тому +1

      Come to think of it, it only makes sense to use graphics to explain graphics

  • @mortenbrodersen8664
    @mortenbrodersen8664 6 років тому +4

    GREAT explanation. Very much appreciated.

  • @Knnorek
    @Knnorek 3 роки тому

    If we do not want to spread a confusion, there seems to be a small problem in this equation.
    In recursive part where we calculate Rendering equation again, we do it for the new point (y), given by our hemisphere in direction -omega_i, not again for point x from direction omega_i.
    This way, in theory, we would end up calculating radiance for the same point from every possible angle over and over again, completely omitting the nature of reflectance in global illumination.
    So, instead of L_i(x, omega_i) it should be written L_i(y, -omega_i), it's just a notation method that is less confusing, but the explanation still applies correctly.

  • @MrJoeYellow
    @MrJoeYellow 6 років тому +1

    I looking forward for the next video like this one

  • @tigerspidey123
    @tigerspidey123 4 роки тому +1

    wouldn't the hemisphere ray been shoot out at 1:33 be highlighted inside intergral as Li(x, wi)?

  • @bimDe2024
    @bimDe2024 3 роки тому

    @Eric Arneback, please illustrate the Monte Carlo approximation. One question, how the ray determining whether it is hitting a light source or not? For example, in 1:31 the shadow ray is hitting the light (emitting object) and determining it is a light source, how?

  • @FlyingTurtleLP
    @FlyingTurtleLP 4 роки тому

    This is top notch material!

  • @victoravilaparcet904
    @victoravilaparcet904 6 років тому +6

    Very good, simple and concise explanation, great job :^)

  • @FrancisChenSF
    @FrancisChenSF 6 років тому

    Subscribing. Such a beautiful integral representing all that radiance.

  • @ranam
    @ranam 5 років тому +1

    Can you please explain Monte Carlo equation satisfying the rendering equation sir please I çan not find any material on internet

  • @andreiblackfish1185
    @andreiblackfish1185 10 місяців тому

    Hi! Thank you for the great explanation! However, i still dont understand how in order to get the Li you need to calculate its own Rendering equation, and so on.. so basically its an infinite loop?

  • @lawrencedoliveiro9104
    @lawrencedoliveiro9104 6 років тому +1

    So, no taking account of transparency, translucency or SSS?

  • @zhenhuili4965
    @zhenhuili4965 5 років тому

    Thank you for your explanation, it really helps!

  • @amitsheokand011
    @amitsheokand011 6 років тому +2

    Thank you, i finally understand it.

  • @ribarmilos
    @ribarmilos 5 років тому +5

    Hello, can I pleaaaaaase use this in my masters presentation?
    This video really explains it very well.

  • @houdinik
    @houdinik 5 років тому

    From the hemisphere vector we just need to select a single random vector to do the recursion or a few of those?

    • @AnsonSavage
      @AnsonSavage 3 роки тому

      If I understand correctly, because computers are limited, the integral in practice is actually just a summation. So then there will be one vector from the hemisphere for each ray calculated in the summation.

  • @T2000-f3j
    @T2000-f3j 6 років тому

    yes im looking for this for a longgggg time! thanks dude

  • @kangChihLun
    @kangChihLun 6 років тому

    Thank you so much for such clear explanation!

  • @lucien5112
    @lucien5112 3 роки тому

    What is a gray expression?

  • @ancarofl1
    @ancarofl1 5 років тому

    Nice & clean, thanks!

  • @AnsonSavage
    @AnsonSavage 3 роки тому

    Really well done! One question: does the BRDF simply need omega_i to get the color of whatever omega_i points to so that it can properly reflect it? It seems like the overall brightness comes from the BRDF being multiplied by L_i(x, omega_i).

    • @sychuan3729
      @sychuan3729 3 роки тому

      BRDF (and similiar functions BSDF etc) describe what amount of light remains when it reaches point x. If x is mirror all light will go in one direction, if x is soot all light would be absorbed and 0 of it will go to the camera, etc.

  • @degiatronglang6103
    @degiatronglang6103 5 років тому

    what happens when you hit the light, you return the radiance or keep going

    • @benstech726
      @benstech726 4 роки тому +2

      Very late reply but from graphics lectures essentially you can keep going forever. Obviously a never ending amount of samples would be not something that may be implemented so a limited sample size is taken. Which samples are taken and how and when to choose when to stop are all part of optimizing efficiency and final image quality given the limitations.

    • @mehmetakyuz5290
      @mehmetakyuz5290 4 роки тому

      @@benstech726 Was wondering the same question, thank you for the clear answer.

  • @김준호-o6m2b
    @김준호-o6m2b 4 роки тому

    It really helps! Thank you !

  • @seyedmortezakamali2597
    @seyedmortezakamali2597 5 років тому +2

    Nice job! please add voice

  • @jaimeplt
    @jaimeplt 2 роки тому

    you're the best

  • @_zproxy
    @_zproxy 6 років тому

    Nice. Should have mathml and glsl side by side.

  • @kingflight6721
    @kingflight6721 4 роки тому +1

    It explains so well. Thank you so much. I also found a c++ example in github, which implements this idea. github.com/ssloy/tinyraytracer.git