Blend Modes in Detail: How Colour Burn Works

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  • เผยแพร่เมื่อ 9 ม.ค. 2025

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  • @JMJWill
    @JMJWill 4 ปีที่แล้ว +7

    Please continue as your tutorials are appreciated. Nice to know “Why”, not just another tutorial telling you to do this and this to get this result.

  • @dro7684
    @dro7684 4 ปีที่แล้ว +1

    Thank you. Very instructive

  • @stevehallam0850
    @stevehallam0850 4 ปีที่แล้ว +2

    Good to know the theory as well as the practice!

  • @alekseyabramov9259
    @alekseyabramov9259 ปีที่แล้ว +1

    Very clear explanation, sir. Thank you!

  • @PierGiorgioZuccaro
    @PierGiorgioZuccaro 4 ปีที่แล้ว +2

    Very very good

  • @christianbauermeister6266
    @christianbauermeister6266 4 ปีที่แล้ว +1

    Wonderful tutorial - again 😀👍

  • @pedropuckerstein4670
    @pedropuckerstein4670 4 ปีที่แล้ว +1

    As always a wonderful video I love the mathematical approach to everything !!

  • @stefanotoneatto
    @stefanotoneatto 4 ปีที่แล้ว +1

    You could be my new era Bob Ross, such a quiet voice

  • @Vincent112june
    @Vincent112june 4 ปีที่แล้ว +1

    I usually run through just to see what comes out. But I don't think I have ever used it. It always seems more extreme and I bring in the black in levels when I want some of that. I find the areas I have a hardship with is the 180, with things being too bright.

  • @TV-tq5ed
    @TV-tq5ed 4 ปีที่แล้ว +1

    Math is fun!

  • @mjfolkerts
    @mjfolkerts 3 ปีที่แล้ว +1

    Technically, a/b where a>0, b=0 is undefined. Then, a/b where a=0 and b=0 is indeterminate. Neither is infinity.

    • @inaffinityforaffinityphoto3220
      @inaffinityforaffinityphoto3220  3 ปีที่แล้ว +1

      Good point. Considering this through the process of limits, when a and b both approach 0, then indeed the limit is indeterminate.
      If a is significantly greater than b and b approaches 0, then a/b becomes larger as b, so a/b in this case is in infinity. Even when a is miniscule, as zero is an effectively 'infinite' value, a/b with b=0 surely still evaluates to infinity?
      In a computer, dividing by zero will throw up a 'divide by zero' exception, as it has no way of representing infinity or indeterminateness, which is why in Procedural Texture programs (which are effectively a series of constant declarations), I tend to add 0.001 to denominators. This also handles the 0/0 dilemma.

  • @roohawkins2400
    @roohawkins2400 4 หลายเดือนก่อน

    "computers have no idea what to do with infinity"
    Who does?