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quiz: if you increase by 1/3 stop, how many % of total exposure are you increasing?


travis1

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Travis:

 

The cube root of 2 is approx (to 3 zeros) 1.26, so the cube root of 2, minus 1, is 26%, as noted above. If you multiply 1.26 by itself 3 times, you will get 2, plus or minus 3 parts in 10,000, which to my eye, but maybe not all Leicaphiles, is negligible.

 

This is similar to the progression of f stops, as they change by multiplying each stop by the square root of 2 (1.41) to get the next full stop. Thus 1 x 1.41= 1.41 and 1.41 x 1.4 = 2, and 2 x 1.4 = 2.8 etc etc up to 16 x 1.4 = 22 and 22 x 1.4 = 32.

 

The progression in Bach's tempered scale of notes is each frequency times the 12th root (there are twelve notes between octaves)of 2 (1.0548), gives you the next note, resulting in an octave (as we standardize it today) having twice the frequency of the note an octave below it: viz bottom A on the piano has a freqency of 55, then up to 110, then 220 then 440, for todays standard A=440).

 

Geometric progressions proceed by multiplying the first member of a series by some other number, and arithmetic progressions, such as calendar years, are one number added to each successive one.

 

Enough pedantry. Time to develop 3 rolls of TRI X from Malaysia this week.

 

Cheers

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<<if you increase by 1/3 stop, how many %>> What's with all the higher math? A 1 stop increase is a 100% increase. 1/2 stop increase is 1/2 of that, or 50%. 1/3 stop increase is approx 33%. 2/3 stop increase is 66% and so it goes. Ever notice how exposure compensation on LCD's (cameras and flashes) in 1/3-stops reads 0.3-0.7-1.0-1.3-1.7 etc?
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<i>

What's with all the higher math? A 1 stop increase is a 100% increase. 1/2 stop increase is 1/2 of that, or 50%. 1/3 stop increase is approx 33%. 2/3 stop increase is 66% and so it goes.

</i>

<p>

What's with the higher math is that they're 100% right and you're 100% wrong.

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The problem is in the way the question is worded -- it can be interpreted two ways, so in fact BOTH camps are correct :)

 

The confusion is comparing light transmission with aperature size percentages. The inverse square law as it relates to fall-off over distance is inconsequential here. The light loss due to aperture reduction is relevant however, and also follows the inverse square law. But the percentage of light loss is LINEAR with respect to the stop. Hence an increase in the size (diameter) of the aperture by 26% (2^1/3 - 1) increases the volume of light by 33%; AND also increasing the aperture by 1/2 stop is increasing the amount of light by 50% but we only need to increase the aperture size by 41% to achieve the extra stop.

 

Nuff said ;),

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Gee, when I went to school 1 stop was equal to changing the exposure time one notch. So f/5.6 to f/8 is the same as going from 1/250th second to 1/125 or cutting the exposure in half. f/2 to f/2.8 would be the same as 1/1000 to 1/500, etc. Increasing by 1/3 stop would be a 33.3% difference. Either I'm simple minded, or these higher math guys are pulling your leg.
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Let me try..;) (deduced from another forumer R.L).

 

Say x is the current amount of light.

 

1 stop more= 2x

2 stops more= 4x

3 stops more= 8x

 

SO the general formula for "amount of light reaching film" is :

 

x2^y: where x=current amount of light, y= stop difference.

 

so say at f1, I had x(2)^0 amount of light= x amount of light.

so at f 1.4, I would have x(2)^1 amount of light=2x amount of light, because f1.4 is 1 stop more than f1.

 

So in general, for any y amount of stop change, I get a difference of:

 

x(2)^y minus x(2)^0 amount of light reaching film....

 

PS: you take 0 as the base when there's no stop changes.

 

 

SO, 1/3 stop increase= x(2)^1/3- x(2)^0= 1.26-1=0.26=26% change in amount of light reaching film= exposure change in %.

 

 

cheers.

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So if we take Travis' question literally, all of the "higher math" math guys are simply wrong...

 

Sorry guys, but if you want to double the light -- a 100% increase -- you add 1 stop or increase the AREA of the aperture by 100% or increase the DIAMETER (not the f-stop) of the aperture by a factor of 1.41 (square root of 2) or 41% -- simple. If you want to increase the amount of light hitting the film by 33%, you increse the APERTURE by 1/3 STOP -- but this is not the same as increasing the diameter of the aperture by 33%. There IS a 1:1 correlation between f-stop and the the amount of light flowing through that aperture. HOWEVER the absolute DIAMETER of that aperture does NOT change linearly with amount of light it allows to pass; here it changes as the square of the diameter since the aperture is gaining AREA at the square of the rate of its radius increase.

 

So, increasing your APERTURE by 1/3 STOP is NOT the same as increasing its diameter by 33%, but it DOES increase your EXPOSURE by 33%.

 

Cheers,

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Yes, Ray, you appear to be simple-minded.

 

If you increase exposure by third stops successively you multiply, not add. 1.26 * 1.26 * 1.26 = 2.0. Increasing exposure 33% three times would be 1.33 * 1.33 * 1.33 = 2.25, which is not right.

 

This is the same reason the aperture midway between f/2 and f/4 is f/2.8, not f/3.

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Ray and others, adding 26% to 100+26% i.e. (100+26%)+26% is not the same as 100+52%.

 

Percentages don't add like that, because percentages are relative. For an extreme example begin with lets say 7, add 100%, then subtract 100%. That is, add 100% to 7, giving 14, then subtract 100% from 14 to get 0. I.e. (7+100%)-100%=0.

 

Don't feel bad about it, most politicians, journalists and lawyers can't do percentages either.

 

If still in doubt, look at the shutter time scale on your camera. For every stop, 100% is added or 50% subtracted, depending on which way you go, yet the series of numbers is not linear.

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