View Poll Results: What is less 1/3 stop compared to the original exposure?

Voters
12. You may not vote on this poll
  • 67%

    3 25.00%
  • 74%

    0 0%
  • 79%

    8 66.67%
  • 84%

    0 0%
  • Others

    1 8.33%
Results 1 to 15 of 15

Thread: Brain Teaser

  1. #1

    Lightbulb Brain Teaser

    If you reduce your exposure by 1/3 stop, how much is the final exposure compared to the original?

    [list=1][*]67%[*]74%[*]79%[*]Others (Please state)[/list=1]

  2. #2

    Default

    84%?
    Last edited by Zerstorer; 9th April 2003 at 05:38 PM.

  3. #3

    Default

    Engineer's recommended solution

    1/3 stop equivalent 33.33333....%

    Answer = 100% - 33.33333....% = 66.66666....% (QED)

    PROVEN WRONG by Zerstorer
    Last edited by mervlam; 9th April 2003 at 05:55 PM.

  4. #4

    Default

    Originally posted by mervlam
    1/3 stop = 33.33333....%

    Answer = 100% - 33.33333....% = 66.66666....% (QED)
    Don't think its so straight forward.

    1stop difference is 2xthe exposure.

  5. #5

    Default

    read carefully...

    "less 1/3 stop compared to the original exposure?"

  6. #6

    Default

    Originally posted by mervlam
    read carefully...

    "less 1/3 stop compared to the original exposure?"
    Yep.

    Less 1 stop would be 50% of the exposure.

    So less 1/3 stop should be 1-(0.33x0.5)=0.835->83.5%.

    At least to my understand of how it goes. Certainly not a straight forward 100-33%.

  7. #7

    Default

    Originally posted by Zerstorer
    Yep.

    Less 1 stop would be 50% of the exposure.

    So less 1/3 stop should be 1-(0.33x0.5)=0.835->83.5%.

    At least to my understand of how it goes. Certainly not a straight forward 100-33%.

    sounds logical....

  8. #8

    Default

    Originally posted by mervlam
    sounds logical....
    Should be the case. Coz setting -0.3EV on any camera only results in a very minor effect.

  9. #9

    Default

    how abt 1/3 stop over?

  10. #10

    Default

    then thinking again

    1 stop over means double exposure = 200% of original
    1 stop under means half exposure = 50% of original

    thus,

    1/3 stop over = 133% of original??
    1/3 stop under = 83.33% of original??

    a bit confusing.

  11. #11

    Default

    Should be 133% I presume.

  12. #12

    Default

    Originally posted by Zerstorer
    Should be 133% I presume.
    it's logarithmic in nature.... aha!

    it's a brain teaser for the rest to read through and understand our posts even.
    Last edited by mervlam; 9th April 2003 at 06:19 PM.

  13. #13
    Member
    Join Date
    Dec 2002
    Location
    Singapore
    Posts
    64

    Default

    +1/3 EV = 2^(1/3) = 1.26 = 126% of original exposure

    - 1/3 EV = 1/[2^(1/3)] = 2^(-1/3) = 0.79 = 79% of original exposure

    Similarly,

    +1/2 EV = 2^(1/2) = 1.41 = 141% of original exposure

    -1/2 EV = 2^(-1/2) = 0.71 = 71% of original exposure

    +2/3 EV = 2^(2/3) = 1.59 = 159% of original exposure

    and of course

    +1 EV = 2^1 = 2 = 200% of original exposure

    -1 EV = 2^(-1) = .5 = 50% of original exposure
    Last edited by ckhaos; 9th April 2003 at 06:55 PM.

  14. #14

    Thumbs up

    Originally posted by ckhaos
    +1/3 EV = 2^(1/3) = 1.26 = 126% of original exposure

    - 1/3 EV = 1/[2^(1/3)] = 2^(-1/3) = 0.79 = 79% of original exposure

    Similarly,

    +1/2 EV = 2^(1/2) = 1.41 = 141% of original exposure

    -1/2 EV = 2^(-1/2) = 0.71 = 71% of original exposure

    +2/3 EV = 2^(2/3) = 1.59 = 159% of original exposure

    and of course

    +1 EV = 2^1 = 2 = 200% of original exposure

    -1 EV = 2^(-1) = .5 = 50% of original exposure
    Yep..I think you nailed it correctly.

  15. #15

    Default

    Just use what has already been calculated for you by film speeds. The calculations have already been done, so just substitute the values.

    e.g. using ISO values for film.
    iso125 film is a third of a stop slower than 160, so 125/160 approx. equals 78% with respect to the original value. similarly, 160/200 is 80%, 50/64=78%, 1250/1600=78%.
    You can also use aperture values, or just use the logarithmic function to calculate and prove it to yourself.

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