## 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%

1. ## Brain Teaser

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

2. 84%?

3. Engineer's recommended solution

1/3 stop equivalent 33.33333....%

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

PROVEN WRONG by Zerstorer

4. 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.

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

6. Originally posted by mervlam

"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. 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. 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. how abt 1/3 stop over?

10. 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. Should be 133% I presume.

12. 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.

13. +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

14. 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. 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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