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

Discussion in 'Leica and Rangefinders' started by travis|1, Apr 10, 2003.

1. ?

2. 33.3%

3. I would have answered 2x1/3 = 66% ... am I wrong?
Alain

4. 1 stop = 100% = x2 = double etc
therfore 1/3rd of this = 33.3% =x.33

5. 26 % (2 at the power of 1/3, minus 1)

6. Jean-Christophe Barnoud is correct.

7. looks like it too. btw, whats the value of 2^1/3?

8. 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

9. <<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?

10. Photons live by the inverse square law. Famous equation 1/r squared.
r is the radius of the distance from the light source. Works for X-Rays, infra-red, visible light, gamma rays, but doesn't work for Manta Rays.

As mentioned, Jean-Christophe has it correct.

11. phew!

anyone else wanna challenge 26%? 12. 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.
What's with the higher math is that they're 100% right and you're 100% wrong.

13. 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 ,

14. Pardon -- my last word should be 1/2 stop.

15. 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.

16. Jack, Travis isn't asking what percentage larger the aperture is getting, he's asking how much percentage increase of light exposure the film is getting.

17. Yea Jack. 26% increase in exposure (light) based on the formula 2^1/3-2^0. Not aperture size.

2^x-2^0: where x=stop difference either way.

18. How could it be 26%? It's 33.3%, that's why they call it ONE THIRD stop. If it was 26% you'd only end up with 78% increase in exposure over a stop.

19. 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.

20. 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,

21. 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.

22. Jack, where it the world do you get that bizarre interpretation of the question? He said nothing about aperture.

23. Jack, I get what you are saying.

Could you point to us where that 1:1 correlation between F STOP and amount of light comes from?

24. And, by the way, none of this is "higher math," unless you consider the sort of math used to compute compound interest at a trade school math-for-dummies class to be "higher."

25. 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.

26. "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.... "

Travis, the problem is your math is wrong... If you increse your exposure by 2 stops using your formula you get the following: x(2)^2 - x(2)^0 = x4 - x1 = 3; or three times the light -- we all know you get 4 times the light with a 2 stop change.

Cheers,

27. Times like these make me glad I'm just a picture-taker.

28. Mark, ouch!

In any case 1/3 stop could be enough to make that baby really glow.... The high math guys could be doing their calculations while we dummies just pushing the shutter button and clicking f/stops make masterpieces, or so my simple mind imagines..... 29. Jack, you get 3x MORE, not 3x. That is, the difference is 3x, but the total amount of light is 4x. 2 stops we are talking here.

31. Jack, my last post, "x" means an amount, not multiply(times)! boy my brains are cramped.

32. You are discussing about a maths question but you are not considering the fact that the exposure value depends on the lens you are using. A 90mm is not a 28 mm and the weight (!) of light you are introducing into the camera is completly different.
Secondly: it depends even from the distance you are measuring the light. 0,30 m is not 30 m or infinitive.....
To shoot a landscape is not as shooting in macro, because the distance make the difference in the quantity of the light.
By
Joe

33. Not to mention flare.. 34. If you spend too much time wondering about 1/3 stops, you will miss half of your picture-taking opportunities, so the light entering the camera will decrease by half.

35. "Jack, I get what you are saying.

Could you point to us where that 1:1 correlation between F STOP and amount of light comes from?"

To clarify, I should have stated there is a 1:1 correlation relatively between single stops.

An f-stop is already funtionalized by the formula F/a where F = Focal length of lens and a = Absolute Aperture DIAMETER. Hence a Noctilux has an aperture that measures 50mm in diameter when it is set to f1.0 (f = 50mm/50mm = 1). Now when I stop it down to f2, the diameter of the aperture is 25mm, so f = 50mm/25mm = 2. So LIGHT changes with respect to f-stop change at the rate of 1/((2)^x) where x = the number of stops changed. So to change any SINGLE aperture by 33%, I must multiply the DIFFERENCE in their f-number by 1/((2)^.33) or 0.796. Thus to increase f11 by 33% I get f10.3 -- 11-8 = 3, 3x0.796 = 2.27, 8+2.27 = f10.3

Try it this way: When I increase (open up) from an original stop to the next in line, I am always doubling the light, regardless of what aperture I start from. Re-read that last sentence. Doubling the light is the same as increasing it by 100%. Please re-read that last sentence. Hence, increasing by 1/3 stop is increasing LIGHT FLOW by 33%, regardless of where I start because I am basing it RELATIVELY to the original stop I chose, NOT ABSOLUTELY to it.

36. x2^(opportunity loss)+ x2^(leica glass)= good picture.

37. "Jack, you get 3x MORE, not 3x. That is, the difference is 3x, but the total amount of light is 4x. 2 stops we are talking here." ...and... "Jack, my last post, "x" means an amount, not multiply(times)! " ...and... "boy my brains are cramped."

You said it, I didn't Travis, last one -- How can you get 3 times more anything with 2 stops??? Is the "3" from your formula 3 stops, or 3 times the light? Either way it's simply wrong, because with 2 stops you can only have have 4 times the light, or 1/2 the light, or 2 stops, or a factor of 2 on the aperture diameter! ,

38. \$hit! Should have read 1/4 the light...

39. Jack thx.

You see,

Say at f1, I get T amount of light. At f1.4(1 stop more), I get 2T.

At f2(2 stops more), I get 2(2T)=4T amount of light.

4T is 4 times of T (i.e at f2, I have 3T amount more than at f1).

It has always been a factor of 2.

Maybe we should close the quiz? ;00

40. Jack, you have 2 bucks, I have 8 bucks. I have 4 times the money you have, but I have 3 times(6 bucks) more.

41. "Jack, you have 2 bucks, I have 8 bucks. I have 4 times the money you have, but I have 3 times(6 bucks) more."

Okay, really the last time -- maybe First off you have 4 times more, period. You have 6 bucks over what I have, but it is still 4 times what I have. You are trying to say you have 3 times what I have ADDED to what I have, and while that is accurate it is nothing more than a complicated way to say 4 times what I have... Now relate this to stops in photography: f2 to f4 is two stops. You are now trying to claim f2 has 3 times more light than f4. But we all know that isn't the case...

Travis, trust me on this one: the difference between single f-stops is a direct linear relationship when you compare increasing that stop by a given PERCENTAGE to the PERCENTAGE of additional light it actually transmits. The only things not changing linearly are the actual f-number and the actual diameter of the aperture.

Cheers,

42. Without changing the lens aperture, the difference between a 1 second exposure and a 2 second exposure is one full stop, right? Anybody disagree with that? I hope not.

The difference bewteen a 1 second exposure and a 2 second exposure is also...1 second. Anybody disagree with *that*? I hope not.

So, in this example if a one-stop increase = 1 second, a 1/3 stop increase must = 1/3 of a second. 1.33 seconds. 133% of 1 second. A 33% increase over 1 second. Any way you fry it it's still chicken.

43. Jack..k 44. "Sorry guys, but if you want to double the light -- a 100% increase -- you add 1 stop ..."---Jack.

Say I have T light at f2.8. At f2, I would have 2T. I could say I have "2 times the light at one stop more" OR I have " T more light at one stop more".

"You are now trying to claim f2 has 3 times more light than f4. But we all know that isn't the case... "---Jack.

This is the case. At f4 , say I have T light, at f2 I would have 4T light. You agreed to that Jack, it's how you put it in words. I could say at F2, I indeed have "3 times MORE light than at f4".(4T-T = 3T)

But sadly, I couldn't do anything with Jay's shutter counter.;(

cheers.

45. I'm surprized nobody has brought up T stops yet... :*)

46. Of course all of this ignores the intentions of the lens maker, the crudity of the linkages between the components, and where the diaphragm end up when they creak open and close ....

47. As my computation seems to have stirred some controversy, let me state that I have never seen the difference on a picture between +26% and +33% and in most case that is all that counts. Now, if -due to an improbable set of circumstances- you have to deal with a number of simultaneous corrections (macro+filters+reciprocity+whatever else) and do not have a reliable TTL meter (we are not speaking M Leicas here, more likely large format !!!), it may add up to a significant error.

48. "3 times MORE light than at f4".(4T-T = 3T)"

Jack gets what you are saying Travis, but your wording is screwing it up. Capitalising the word "more" doesn't help either because the word that's making your sentance wrong is the word "times". "Times" implicitly means MULTIPLICATION - hence "Times tables" - so 3 TIMES more light will ALWAYS mean 3T.

Now if you say that T is one unit of light and by opening up 2stops you have 4T, you can state that you have 3 ADDITIONAL units of light (1+3=4)

I THINK. moiz

49. By the time you guys are done arguing the picture opportunity is long gone. If I had to figure an exposure that close I'd start using film with more latitude.

50. If you open up by 1/3 stop, you increase the exposure by "a little bit". I'm not sure why everybody else's answer was so long. I'd hate to see you guys try to add a "heaping tablespoon" while following a recipe. 51. Moiz, that's exactly what Im saying. 3T units more light, which is 3 TIMES more than T. Where is the screw up? Not that it matters with the picture anyway.. )

52. \$hit, that should be just "3 TIMES of T..", no "More than".

No more.

53. "The light shineth in darkness, and the darkness comprehendeth it not."
Fiat lux.</i

54. </i><p>

55. As I was a liberal arts major, could somebody please explain this stuff to me? I just shoot my M7 on Auto.

56. Steven, the quiz is closed. The answer to the question is :

(26%+33%)/2=29.9% increase, plus minus 2% with/without flare..

57. \$hit, that should be 29.5%.

58. This is a fascinating anthropological thread. The fact is that 1/3 stop is by definition 2^1/3 - 1, because 1/3 stop is defined by the non-linear decrease in shutter diameter. Just look at a Copal shutter and convince yourself of this. However it is a little easier to think of 1/3 stop intervals as linear in exposure and it's quite adequate to make one's mental calculations simpler.
What is fascinating to me is the number of people who take an almost ethical stand about what is purely a scientific fact. It shows, I think, the common but false belief, that the science is opposed to the humanities. Science is as humane as Art, in that they both appeals to fundamental human mental needs.
My apologies, but I couldn't help but comment.
Blatant Plug

59. Further apologies for my frightful proofreading. I am capable of matching noun and verb number and generally writing coherent English 60. I is lost. Lessee now: we agrees that 1/2 stop + 1/2 stop = 1 stop. And according to Jack, 1/2 stop maketh 50% change in exposure. So from f1.4 to F1.0 is a 1 stop change, a 100% percent change in exposure.

Now if I moves in 1/2 stop increment from f1.4 i.e. according to Jack, I maketh a 50% incremental change, I gez to f1.2. And then I makes another 1/2 stop change from f1.2 and according to Jack's mathematics, that is a 50% incremental change from f1.2, I gez to....wait a minute, I gez to f1.1. Okay, if I makes another 1/2stop change from f1.1, I get to f1.05 and if I... Wait a cotton-pickin minute again. 1/2 + 1/2 + 1/2 = 1 and a little less? Okay, golly, I supposes that I ken always round it off to f1.0. But three halves making a whole is a whole nuther realm of higher mathematics to me.

61. Anatole Freedom wrote: "Now if I moves in 1/2 stop increment from f1.4 i.e. according to Jack, I maketh a 50% incremental change, I gez to f1.2. And then I makes another 1/2 stop change from f1.2 and according to Jack's mathematics, that is a 50% incremental change from f1.2, I gez to....wait a minute, I gez to f1.1. Okay, if I makes another 1/2stop change from f1.1, I get to f1.05 and if I... Wait a cotton-pickin minute again. 1/2 + 1/2 + 1/2 = 1 and a little less? Okay, golly, I supposes that I ken always round it off to f1.0. But three halves making a whole is a whole nuther realm of higher mathematics to me.

JHEYSUS FREAKING CHRISTOBAL Anatole! So if your car gets 20 MPG (or 8 KPL or .125 L/K) and you buy a new one that gets 50% BETTER mileage you now get 10 MPG...

Jay, we need to establish an ESL forum here on PN! <HUGE> Travis: When was the last time you went out on a date? (with a girl?)...(and one that had reached the age of consent?) I am beginning to worry about you! Cheers ))))

62. Annatoal! O mie goodnez, I'm ineth the sem bote az you I feareth. I los' thu thred ov thiz thread along tyme a go, an' now on tob of evereethin elz I sea Eyem en anthru-polug-ikal spesimen, oppozed ta sighenz. I suppoz I em, tho I harber no ill feelins fer id reely.

63. Ah, it's all a tricky subject, isn't it?

There are several issues that add up and end up being confusing:

One of them is that stops are additive, but the exposure is multiplicative. Stops were "invented" because it's easier to add than to multiply.

Another one is that people are used to dealing with percentages as if they could be "added", even though this is only an approximation, and the approximation is only valid for small percentages.

Before we talk about partial stops, let's talk about full stops. We all agree that "adding" one f-stop will multiply the exposure by two, right?

It's reasonably logical, then, that adding two stops will multiply the exposure by four, isn't it? After all, if I add one stop then add one stop, I've multiplied by two then multiplied by two. Similarly, adding three stops multiplies the exposure by eight, and so on.

We can also subtract stops instead of adding them. Subtracting one stop divides the exposure by two. It makes sense: if I add then subtract one stop, I've multiplied my exposure by two then divided it by two - I've come back to the same exposure. As we did above, we can subtract two stops and divide exposure by four, we can subtract three stops and divide exposure by eight, and so on.

In mathematics, the relationship between stops and exposure is called a logarithm. Since adding a stop multiplies the exposure by two, mathematicians say that the "base" of the logarithm is two. Other people like to use other logarithms. In physics, almost everything uses ten as a base. In mathematics, almost everything uses a special number called "e", whose value is approximately 2.7 (the exact value doesn't matter for us photographers). We also use base ten in photography, when talking about density of film.

So, what happens if I try to talk about half a stop? Well, if I add half a stop, I'm going to multiply my exposure by a certain number. If I then add an extra half a stop, I'm going to multiply my exposure by the same certain number. Overall, I've added one stop, so I have multiplied my exposure by 2. What is this certain number? In mathematics, it is called the square root of 2, it is the number which, when multiplied by itself becomes 2. Its value is approximately 1.41. Does it work? Yes, it does. If I multiply by 1.41 then by 1.41 again, I've multiplied by almost exactly 2. Try it!

It also works if I remove half a stop. Removing half a stop will divide my exposure by 1.41 - so that when I remove half a stop then half a stop again my exposure gets divided by 2.

We can do the same with thirds of stops. If I add a third of a stop, I multiply my exposure by a certain number. Add another third of a stop, I multiply again by this same number. Add yet another third of a stop, and I multiply by this number again. Overall, I've added one stop, so I have multiplied my exposure by two. The special number we're looking for here is called the cubic root of 2, it is approximately 1.26. Yes, it works. Take your favorite number, multiply by 1.26, then multiply the result by 1.26, then multiply the result by 1.26. You'll get twice your original number.

It also works when removing a third of a stop: doing so will divide my exposure by 1.26.

Still here? OK, let's play with percentages now, that's where it becomes confusing...

Percentages are confusing beasts. Here's why: we very often talk about percentages as if we were adding them, but in fact "adding" a percentage is fundamentally a multiplication...

If I add 10% to a number, it means that I take my number, multiply by 10, divide by 100, and add my original number. In the end, what I've done is multiplying by 1.1.
Similarly, if I add 20% to a number, it means that I take my number, multiply by 20, divide by 100, and add my original number. In the end, I've multiplied by 1.2.
What if I add 10% then add 10% again, what do I get. I take my original number, multiply by 10, divide by 100, and add my original number. This is my intermediate number, and I've added 10% once. Let's add another 10%. I take my intermediate number, multiply by 10, divide by 100, and add my intermediate number. That's my final number. If my original number was 1000, my intermediate number is 1100, and my final number is 1210. It makes sense, we've multiplied by 1.1 then by 1.1 again, and 1.1 times 1.1 is 1.21. So, adding 10% then 10% again is the same as adding 21%.

Where does the extra 1% come from? 1% is 10% of 10%. My intermediate number above was my original number plus 10%. So when I added 10% of all that, I added 10% of my original number plus 10% of the 10% that I had added before...

Why do we usually add percentages and get the results almost right? Well, when the percentages are very small, the "percentage of the percentage" is going to be very small. If I have 0.5% of interest on my checking account every year, the value in my account gets multiplied by 1.005 every year. After two years, it gets multiplied by 1.005 and by 1.005 again, i.e. by 1.010025. For \$10000, the difference is 25 cents, not even enough to buy a postage stamp, which is why we usually just ignore it entirely. This works fine because 0.5% of 0.5% is a tiny tiny value.

For photography, though, it doesn't work well. We've seen that adding half a stop would multiply the exposure by 1.41 - i.e. it "adds" 41%. If I add half a stop twice, I'm going to add 41% then add 41% again. The first time, I add 41% of my original exposure. The second time, I add 41% of the original exposure plus 41% of the 41% I've just added. 41% of 41% is about 17%, and 41% plus 41% plus 17% is almost 100% (it's not exact, because the exact value of the square root of 2 is not 1.41, it's approximately 1.4142135... don't ask me do give you all the digits, there's an infinity of them, and they don't repeat - you don't want me to give you a mathematical proof).

Not entirely lost?

A few more percentages?

I find all those percentages confusing and annoying, which is why I avoid them. They get even worse when you start to subtract percentages. Let's see what happens if I try to subtract half a stop... We've seen above that subtracting half a stop was the same as dividing the exposure by 1.41, which is the same as multiplying is by 0.71 (0.71 is the reciprocal of 1.41, also called it multiplicative inverse). So, subtracting half a stop is the same as multiplying the exposure by 0.71, i.e. the same as subtracting 29%. Does this sound right? Yes, it does. If I add 41%, I end up with my original number plus 41% of my original number. If I then subtract 29% of all that, I end up with my original number plus 41% of my original number minus 29% of my original number minus 29% of 41% of my original number. 41% of 29% is 12%, so overall I've added 41%, subtracted 29% and subtracted 12%, and it all cancels out almost like magic... You can try to subtract 29% first then to add 41%, you'll see, it works exactly the same, and the same 12% come in just to fill the gap.

Where else do we use the square root of 2 in photography?

The aperture f-stops have those nice "round" values 2, 4, 8, 16, 32, which seem natural when I multiply 2 by 2, then by 2 again, then by 2 again. But there are also those bizarre values in the middle, 2.8, 5.6, 11, 22... Where do those come from? Remember, the value of the aperture f-stop is the diameter of the diaphragm divided by the focal length. When you increase by one stop, you want to multiply your exposure by 2, by means of letting twice as much light shine through the diaphragm. You do that by doubling the area of the diaphragm, and doubling the area of the diaphragm is achieved by multiplying the diameter of the diaphragm by 1.41 (remember, the area of a circle is pi times the radius times the radius - if you multiply the radius by 1.41, you multiply the area by 1.41 times 1.41, i.e. by 2). 5.6 is 4 multiplied by 1.41.

Where else do we use the cubic root of 2 in photography?

Yes, we do use it, even though it's hiding and masquerading so that we can't recognize it. It's hiding in the sensitivity of some specialized films, e.g. films that are rated at ISO 125 or ISO 160 (125 is approximately 100 multiplied by 1.26, and 160 is approximately 100 multiplied by 1.26 multiplied by 1.26).

PS: please excuse my poor English, I am not a native English speaker. On top of that, I'm very tired so that I probably left lots of typos even though I tried to proofread my post. Finally, let me apologize for writing such a long post, and for probably still being too technical for some people, I spent 8 years of my life in college studying math, physics and computer science and this has left some irreversible damage on my brain...)

64. Homage to thu creater of Krazy Kat, uv corz, Mizter George Herriman.

65. mine mine..what a bomb.

Jack, I think the cofusion arose(sp?) because of my grammar. Im not sure when to use "times" and "more than". I realised one cannot use "Xtimes more than.." or else one end up with a double addition.

Can we at least agree on this?:

at f1 say you get T units of light. At f1.4(one stop more), you would get 2t units of light and at f2(two stops more), you get 4T units of light?

If that is agreed on, then the formula 2^x: where x=stop change would denote the amount of exposure(amount of light) change.

Im not sure why the linear change in f stops=linear change in exposure as you stated. You could be right and Im not arguing Jack the last time I went on a date was with Tommy, my M6. He's good.

66. Travis wrote: "Can we at least agree on this?: at f1 say you get T units of light. At f1.4(one stop more), you would get 2t units of light and at f2(two stops more), you get 4T units of light? "

Uhhhh.... Sorry Travis, but alas, no we cannot agree on that You see, if you have T units of light at f1, you will only have 1/2T units of light at f1.4, not 2T units. Re your date you took Tommy on: Yeah, but did you get any pics of the girl?

Cheerio good buddy 67. \$*X&\$#**X&\$#**X&\$#**X&\$#*, I got it all reversed! Sorry Jack! i meant f2 = T, f1.4=2T, f1=4T.

sorry!

68. \$hit again, I just realised some of my previous posts were reversed as well!!

69. travis wrote: "\$hit again, I just realised some of my previous posts were reversed as well!!"

Exactly what I've been trying to tell you all along! Not to mention your math is screwed up in most of them too 70. Jack, now the "reversal" were made out of hasty typing. I do of course know f1 receives more light than at f1.4, 2T vs T. That's not what you have been trying to tell me. Those were my own mistake and I made it clear just.

at F1 you get 2T, at f1.4 you get T, a difference of T units of light, hence T2^1-T2^0=T.
That is the formula. so a 1/3 stop is T2^1/3-T2^0=1.26-1=0.26=26% difference.

I may not be good with my English as you but Im not screwed either.

71. Travis:

Your problem is you insist that 1/3 of the way betwen f(whatever) and the next higher stop is in the proprtion of 2^(1/3) when in fact it is not... It is simply 1/3 of the way between them LINEARLY speaking, as Jay pointed out in his shutter speed example. You can insist that the in-between f-stops increase geometrically, but in simple fact they do not; rather they increase linearly (as I've been saying all along), while the aperture's diameter is changing geometrically.

So yes, the cube root of 2 is 1.26~, and the square root of 2 is 1.41~, but neither of the have ANY bearing on the value of light transmitted. Again, as I've already stated, they DO have bearing on the absolute aperture diameter, but NOT on the amount of light -- By stating you incresed your exposure 33%, you added 33% more to where you started from period, and that required you to increase the diameter of your aperture by something less than 33% of the way between the two, but nonetheless you added 1/3 stop more light.

You can insist on looking at it your way, but it will never change the reality of the physics that is happening inside your camera.

Cheers,

72. Jack, let me strongly disagree with you, as your statement contradicts the reciprocity law that we all hold dear.

Let's imagine that I have my camera set to 1/8s plus half a stop (I mean, *more* exposure, i.e. half-way between 1/4s and 1/8s) and f/16.

According to you, the time I have set is linearly half-way between 125ms and 250ms, i.e. 187.5ms, or 1/5.33s.

Now, I realize that I need a bit more light, half an EV. I have two primary choices, which (according to the reciprocity law) are equivalent: I can open my aperture by half a stop, or I can lengthen my exposure time by half a stop.

According to you, between "whole" stops the aperture increases linearly, i.e. if I open my aperture by half a stop from a whole stop, I'm going to increse the amount of light hitting my film by 50%. On the other hand, if I take a longer exposure at my same aperture, I'm gonna expose 250ms instead of 187.5ms, i.e. an increase of only 33%. So much for reciprocity.

Bah, in the end the difference between all those definitions is far less than a tenth of a stop, regardless of what definition you choose to pick, and now that much bigger than the approximations that are already done here and there throughout the photographic world. They're probably about as big as the tolerance that most equipment will use, and very small compared to e.g. the effect of using an uncoated lens against a multi-coated lens when using an external meter.

I'd be outside taking pictures instead of writing all that if it wasn't raining unusually hard on the SF bay area right now.

Cheers!

73. Jack,

You've made an interesting point in that those who argue for 2^1/3 - 1 are not addressing what actually happens when you change an aperture on a physical lens. I just examined one of my Copal shutters and noted that the aperture scale is linear in the normal stop progression (5.6, 8, 11, 16, ...) and therefore exponential in exposure and therefore 2^1/3 - 1 is the only internally consistent answer for that shutter. Now looking at my Leica lenses I note that there too the aperture scale is linear in the stop progression and therefore exponential in exposure and therefore 2^1/2 - 1 is the correct answer for the half stops that a Leica lens has clicks for.

As another way to look at it, I hope that we can all agree that the EV scale is exponential, by definition. I'm sure that there is an ISO document that defines it as such. Is it not true that one treats 1 EV unit as a stop unit in one's photography? If you answer yes then, Q.E.D. 2^1/3 - 1. If you don't agree that 1 EV unit is 1 stop then I'd be delighted to hear an explanation.

Finally (for this post at least) an experiment could answer this question. If there is someone out there with a meter that accurately measures flux in foot-candles, with a probe that can be inserted in front of a film holder in a view camera, then a simple set of measurements at all the marked stops on the shutter and a plot should answer the question empirically. I do not have the required meter, or I would do it myself.

74. I think we all agree that this notion of "stops" measures out exposure along an exponential scale. Each "major point" along this scale is a "full stop". As you make 1 step along the scale, you double or halve the associated luminance, and assuming reciprocity, you double or halve the required shutter speed.
Now, given such a scale with "major points" arranged exponentially, you now have to decide where to place the "intermediate" points between the major points, i.e. the half stops or the third stops. Say what you will about physics or diaphragms, in reality this location of "sub-stops" is a matter of convention. It would be possible to define the convention exponentially, with respect to luminance and shutter speeds, or linearly. But given that the major points (the full stops) are clearly laid out exponentially, it wouldn't make much sense to lay out the intermediate points (the half and third stops) linearly, that wouldn't be internally consistent.
As I understand it, the conventional system of stops, half-stops, third-stops, etc., is laid out consistently exponentially such that each step (or "sub-step") along the "f number" scale requires multiplication/division by a single constant, and the corresponding step along the shutter speed scale requires multiplication/division by a different constant.
So unless I flubbed my math, to walk along the progression of stops, half stops, and third stops, you'd do something like this (assuming, for instance, EV 10):
Full stops:
Multiply/divide "f number" by sqrt(2) = 1.414
Multiply/divide shutter speed by 2
{f/1, 1/1000}, {f/1.414, 1/500}, {f/2, 1/250}

Half stops:
Multiply/divide "f number" by (sqrt(2) ^ 1/2) = 1.189
Multiply/divide shutter speed by (2 ^ 1/2) = 1.414
{f/1, 1/1000}, {f/1.189, 1/707}, {f/1.414, 1/500}

Third stops:
Multiply/divide "f number" by (sqrt(2) ^ 1/3) = 1.122
Multiply/divide shutter speed by (2 ^ 1/3) = 1.260
{f/1, 1/1000}, {f/1.122, 1/794}, {f/1.260, 1/630}, {f/1.414, 1/500}

75. Harry, you have got it! And offered perhaps a better visual explanation than my verbal one The F-number does in fact change by sqrt 2^1/3 for 1/3 stops because the f-number is determined by dividing the focal length of the lens by the diameter of the aperture, as I have been stating all along.

However, Travis (and all the others who disagreed with me) is (are) still adding 1/3 or 33% more light, not 1.26 (or 1.22) more light, when they increase their exposure by 1/3 stop. MY point all along has been that you are adding 1/3 stop more period, and that is in fact 33 1/3% more light, but the diameter of the aperture is the only thing changing geometrically at a lesser rate -- in this case by a factor of 1.22 for every 1/3 stop.

Cheers,

76. Jack I get what you are saying. Just for discussion sake and let's not get agitated over this, let me rephrase again:

We AGREED that exposure increases in this geometric progression, for sunsequent stops: T, 2T,4T,8T,16T...

If you are saying that stops changes should be linear with exposure changes, then the above is wrong! Because then you should get the below:

..for subsequent stop change, t,2t,3t,4t,5t,6t...etc, an additional progression which is not how light exposure works through a lens.

So if like you said Jack, f stop changes should be linear with exposure changes, then when you increase 2 stops, you should by your definition get 3T and not 4T!

The exponential curve for light exposure vs f stop is there for you to plot. It can never be a straight line.

77. and you don't need to bring the diameter changes of the aperture into play here because that's not in question.

78. The exponential curve for light exposure vs f stop is there for you to plot. It can never be a straight line.
It can, if you do a funky mathematical thing to one of the axes... ;-)

79. "The exponential curve for light exposure vs f stop is there for you to plot. It can never be a straight line.
It can, if you do a funky mathematical thing to one of the axes... ;-)
"

Oh yeah... I really have to get funky to do this: Plot the relative percentages of light transmission for each f-stop increase against the corresponding shutter speed required to maintain equivalent exposure... In other words, assume your meter reads 1/1000th at f2.0 and you are using your 50mm Summicron. You have 100% transmission at f2, 50% at f2.8, 25% at 4, etc. My graph would look like this: 100%@1/1000th; 50%@1/500th; 25%@1/250th... Really tough math and a straight-line graph too Now if you persist in plotting shutter speed against f-stop, then you will get a curve -- But that is simply the wrong way to look at it because light TRANSMISSION is what we are talking about, and transmission is dependant on the AREA of the circle the aperture inscribes (and shutter open time) -- NOT the absolute f-number!

Cheers,

80. If f stop changes is linear with exposure changes, why do you get T,2T,4T,8T,16T etc. where T=units of light, as the F stop increases in subsequent full stops?

f1,f1.4,f2,f2.8,f4<=> 16T, 8T, 4T,2T,T. Anyone disagree?

How can one agree on this and then say the relationship is not geometrical? And that the f Stop number is in linear relationship with amount of light units?

If you plot the above : T vs f stop, you get an exponential curve. Agree? y=2^x

SO if you choose any 1/3 increase in x starting from any f stop number, the y increase is 2^1/3=26%. Isn't this what the question was asking? What has aperture size got to do with the question in this case?

When you click on your aperture ring, every click follows this exponential curve. Every lens does this.

Really, what Jean said.

81. Hello,

My comment about the inverse square law was to show that the change in light hitting the film plotted against fstop/exposure/shutter speed is exponential. Hence the r squared.

Certainly you may convert exponential curves to a straight line by using a log scale. Jack is "proving" his point by describing his logic using a log scale. Problem is, it only represents a straight line. Hence the thought connecting the dots between 100% and 50% and 25% is a straight line. Yes it is--when graphed on log paper. The physics however, as pointed out by many, is really curved lines (not straight) mathematically symbolized by squares or square root's of the power of 2.

No matter how Jack plots it, Jean Christophe is still correct.

82. In Electrical Engineering; the cube root of 2 is used alot in motor design; and wire sizing.<BR><BR>Each wire guage increases in cross sectional areas by 1.26 ; Three guages vary by a factor of 2 in area. 36 guage wire is exactly 0.005 inch diameter; 0000 wire is exactly 0.4600 inches in diameter. ALL OTHER GUAGES ARE DETERNINED BY THESE TWO DEFINITIONS. <br><br>WIRE diameters vary almost as the 1/3 root of two for the difference in two guages; ie 30 guage is about 0.010; and 28 guage is about 0.0126 <BR><BR>In Photography; a 1/3 stop is 1.26; 2/3rds varyis 1.59 ; 3/3 stop is 2.00 <br><BR>The 26% number is correct for a 1/3 stop; and the 33% a grave error; at least if this was a math test!.

83. "In Electrical Engineering; the cube root of 2 is used alot in motor design; and wire sizing. Each wire guage increases in cross sectional areas by 1.26 ; Three guages vary by a factor of 2 in area. 36 guage wire is exactly 0.005 inch diameter; 0000 wire is exactly 0.4600 inches in diameter. ALL OTHER GUAGES ARE DETERNINED BY THESE TWO DEFINITIONS. WIRE diameters vary almost as the 1/3 root of two for the difference in two guages; ie 30 guage is about 0.010; and 28 guage is about 0.0126 In Photography; a 1/3 stop is 1.26; 2/3rds varyis 1.59 ; 3/3 stop is 2.00 The 26% number is correct for a 1/3 stop; and the 33% a grave error; at least if this was a math test!."

Except we ARE NOT talking about DIAMETER when we are discussing TRANSMISSION capability of a given f-stop -- we are talking AREA of the circle that f-stop covers!!! Read that last sentence again! Which is PRECISELY why wire is "guaged", because as you increase the cross-sectional AREA you increase the "flow". BUT, you only need to increase the DIAMETER of the wire by root2 to DOUBLE the area, and hence DOUBLE the transmission capability. And it's the same with lens apertures. F-stops only represent the aperture's DIAMETER, while the light TRANSMISSION of any given f-stop is a function of the aperture's AREA --

AREA, NOT DIAMETER!!! RELATIVE LIGHT TRANSMISSION QUANTITIES, NOT JUST F-NUMBERs!

Cheers,

84. Exactly, it is a function of area which is why 1/3 stop more is 26% more (1.26x) in area of light transmission and 1/2 stop more is 41% more (1.41x) in area of light transmission. Do you still not see why, Jack? You really need to sit down and think about it more instead of making physics conform to your own wrong assumptions. First off, your assumption that 1/2 stop more corresponds to 50% increase in area or 1/3 stop more corresponding to 33% more in area is wrong. Disabuse yourself of that wrong assumption and it will be clear to you.

As a function of area, exposures multiply, not add. F-stops add up because they are the exponents of base 2.

Your mistake lies in adding. If we follow your line of reasoning i.e. one 1/3 stop = 1.33 (33% more), then 1/3 stop + 1/3 stop + 1/3 stop = 1 stop = 1.33 x 1.33 x 1.33 = 2.35x! And 1/2 stop + 1/2 stop = 1 stop = 1.50% x 1.50% = 2.25x!! You are going to insist that 1/4 stop is 25% more so let us do the maths your way: 1/4 stop + 1/4 stop + 1/4 stop + 1/4 stop = 1 stop = 1.25 x 1.25 x 1.25 x 1.25 = 2.44.

Sp, according to your mathematics, 1 stop can be 2x, 2.25x, 2.35x or 2.44x. It simply cannot be.

Now, as what we have said in this thread all along:
1/3 stop + 1/3 stop + 1/3 stop = 1 stop = 1.26 x 1.26 x 1.26 = 2.
And 1/2 stop + 1/2 stop = 1 stop = 1.41 x 1.41 = 2.
And 1/4 stop + 1/4 stop + 1/4 stop + 1/4 stop = 1 stop = 1.189 x 1.189 x .189 x 1.189 = 2!

That is exactly what 1/3 stop more means: 1.26x more in area than the base and so on for the rest, no matter if it is 1/2 stop or 1/24th of a stop. The area of light transmission which you have maintained increases or decreases in this fashion.

I wonder how you are going to spin this and still insist that we are all mistaken.

Yes, you can get 33% and 50% more in area of light transmisson or exposure in other words but they are not 1/3 stop more and 1/2 stop more respectively.

Work it out for yourself, Jack. Start with the formula for area of a disc and then do the mathematics.

85. Jack ;
1/3 stop is 1 26% increase; do this again and it is a 59% total increase; do this again and you are at 1 full stop; or a 100% increase.

1/3 reduction in stop requires a 26% more exposure; a 2/3 stop down requires a 59% more exposure; a full stop closure requires a 100% more exposure. These concepts have been around for about 150 years. Also; the legal definition of wire gauges is by diameter; in larger sizes circular mils is used.

The basic flaw many people have in this thread; is they dont know what a cube root is. This is quite sad; we learned this in 6th grade in Indiana; my teacher probably made 1k per year; and wasnt even a college graduate. The school was poor; and built in 1909. They used a paddle; and ran a tight ship. People today are chicken to give teachers any power; the lawyers have ruined education; by pure fear and lawsuits.

The 3 geometric increments from going from 1 to 2 by 3 parts are roughly 1.26 ; 1.59; 2.00. This was known probably several thousand years ago; but seems to confuse some here. The progression is the cube root of 2; which is the number; multiplied by itself three times equals 2.00. This number is about 1.259921050; squaring this number is 1.587401051; cubing it is about 2.00000.

A 1/3 stop increase is a 26% increase; a 2/3 stop increase is a 59% increase; a full stop is a 100% increase; this is old stuff.

A doubleing in area in wire cross section is 3 full wire guages; a doubling in area of a lens's opening is only 1 stop . 30 guage is about 0.010 diameter; 27 guage is about 0.0142 diameter; this 3 guage jump has double the cross section; since area varies as the square of the diameter.

1/3 stop Differences in exposure; are like 1 guage differences in wire guage.

1/3 stop difference is a 26% difference in exposure; a 1 guage difference in wire guage; is about a 13% difference in diameter; or a 26% difference in area; the same as the 1/3 stop in exposure.

Being a registered Electrical Engineer; these analogies are second nature; the cube root of 3 has been a part of electical Engineering for well over 100 years; the jumps of 1; 1.26, 1.59., 2.0 are etched in most electrical Engineers who have designed solenoids; and dc motors.

86. Okay Travis and Anatole, here is the math with a Lieca example to keep this ON topic The area of a circle is pi x r^2. A Noctiluxs aperture is 50mm in diameter at f1 (50mm/50mm = f1). Its radius is 50mm/2 or 25mm, so its area is 25^2 x pi or roughly 1963 sq. mm. At f1.4 its diameter is 50/1.414 or 35.36mm. Its radius is now 35.36/2 or 17.68mm. The new area of the circle is 17.68mm^2 x 3.141 or roughly 982sq mm Which is essentially EXACTLY 1/2 the original 1963sq mm area. One down...

Now, 1/3 of a stop over f1.4 has to add in 1/3 more area. 982sq mm x 0.333 = 327 sq mm. 327sq mm + 982sq mm = 1309sq mm. The radius^2 of that circle equals 1309sq mm/3.141 or 416.746sq mm. The sqrt of that number is (416.746sq mm)^1/2 = 20.41mm. SO the new diameter of the aperture for f1.4 =1/3 stop is 20.41mm x 2 or 40.82mm. (BTW, this numerical value of this new aperture would be 50/40.82 or f1.23.) Two down.

The difference between f1.4+1/3 at 40.82mm and f 1.4 at 35.36mm is 1  (35.36mm/40.82) = .134 or 13.4%. NOTE THAT THIS CHANGE IS ESSENTIALLY EQUAL TO HARRYS EARLIER CALCULATION FOR 1/3 STOP CHANGES AT (sqrt(2) ^ 1/3) -- I obviously suffered from rounding error by to only using 2 decimal points in my above arithmetic. Three down.

So, I did the math, and it proves what I have been saying all along. I stand by my claim that you increase the AREA of the aperture by 33% when you increase an exposure by 1/3 stop!

Cheers,

87. 13.4%...how does this correlate with your 33% that you claimed?

88. Like Kelly said, 13% change in diameter<=> 26% change in exposure.<=>1/3 stop.

89. 1309/982=1.33=33% <=> areas of aperture at (f1.4+1/3 ) area of aperture at(f1.4).

But you are ASSUMING THAT for a 1/3 stop increase, the area increases by 1/3!! from the start.

"Now, 1/3 of a stop over f1.4 has to add in 1/3 more area..."--- Jack.

How can you assume that to be true when you are supposed to proved it in the first place?!

cheers.

90. Kelly, you posted as I was writing. Yes I understand that wire sizes go up by the cube root. And I do understand that if you want to multiply by some number three times to get to the next one you have to use cube roots.

And, a 1/3 increase in the aperture's area only requires a 13% increase in the aperture's diameter. And yes, the f-NUMBER changes by 13% too when you increase the exposure by 33%, but this is because it is a number derived using the aperture's diameter. In either case the area has increased by 33%. See the math above.

You do not increase the area by 26% and gain 33% more transmission, and you do not increase the area by 13% and gain 33% more transmission -- aperture areas to transmission percentages are a direct linear relationship; double the area, double the light; quadruple the aperture area, quadruple the light; increase the aperture area by 33%, increase light flow by 33%.

As I said from the start, f-stops are based on the aperture's diameter, while light tranmission is based on the aperture's area.

You guys keep arguing percents the DIAMETER changes or the percent the f-number changes when the percent the AREA is changing is what is relevant to the original question: "Subject: quiz: if you increase by 1/3 stop, how many % of total exposure are you increasing?" Answer: A 1/3 stop increase is a 33% increase in the exposure value, and also a 33% increase in aperture area. And yes, the 33% exposure increase only requires a 13% increase in the aperture's diameter, and the f-number only changes by 13%, but these numbers have no direct relation to the increase in the exposure VALUE -- they only represent the new diameter and the new f-number.

91. Travis wrote: "Like Kelly said, 13% change in diameter<=> 26% change in exposure.<=>1/3 stop."

I'll correct you slightly and say a 13% change in diameter increases area by 26%. BUT this is only true when you are going UP in number...

To INCREASE light flow through an F-stop we have to use a larger aperture, which is a SMALLER f-number... (This is not an assumption, this is standard photographic knowledge ) And guess what happens next? We DIVIDE the LARGER number by 1.13 to get the new SMALLER f-number. And now if you do the math, this 13% smaller f-number will generate an aperture diameter that is 13% smaller and an aperture area that is 33% LARGER than your original -- NOT 26% larger!

Cheers,

92. My last sentence should read: "And now if you do the math, this 13% smaller f-number will generate an aperture diameter that is 13% LARGER and an aperture area that is 33% LARGER than your original -- NOT 26% larger!"

93. ...and I bolded the wrong LARGER -- sighhhhh. I'm tired and I'm going to bed. Hopefully I now have you all straightened out on the math area= pie (Diameter/2)^2

Therefore,
area proportional to Diameter^2
F stop proportional to Diameter
Therefore,
area proportional to (F stop)^2

Light transmission proportional to area
therefore, Light transmission proportional to F stop^2

Increase/decrease in F stop will not equal Linear decrease/increase in area/light transmission.

AM I right so far?

So by your LINEAR definition Jack, 1/3 stop increase=33%, 1/2 stop=50%, 1 stop=100%. Isn't this Linear?

How can you say Light transmission is only based on area when in fact area is based on diameter^2 and diameter is proportional to f stop ?

cheers.

95. Area is directly proportional to the change in radius. A=PI.R^2. Everything else is a constant. So, how does area change by 33% with a 13% change in radius or diameter? Area changes by 26% with a 13% change in radius or diameter.

"Now, 1/3 of a stop over f1.4 has to add in 1/3 more area." This is flawed. You have made an assertion without proving it. No, it does not follow that if 1/2 the area of a disc equals 1 stop change then 1/3 stop equals 33% change. You have proved nothing. So, nothing down. And this is what we've been trying to prove: that 1/3 stop change is NOT 1/3 more area but 26% more area.

Start with variables like X and Y and prove them for all cases. If you do not, then that is sloppy mathematics, sloppy physics and sloppy logic.

Remember that mathematics has to be consistent and hold for all cases.
You cannot pull figures out from the air and say that that is the truth.

I have proved it for myself and as a mathematician, I damn well should as this is basic mathematics.

96. Area=pi(R^2)
If R increases by 13%, new r=13/100R +R=113/100R.

New area therefore = pi (113/100R)^2=pi(1.2769R^2)

Change in area= 1.2769-1= 27.7%

Am I right?

97. Travis:

Yes, from a math point ov view, you're right. 1/3 stop increases the diameter of the aperture by (approximately) 12.2% (which is the sixth root of 2). And mathematicians prefer to approximate to the closest value, and to be explicit whenever they approximate.

In physics, though, (and hence in optics) it's usual to factor in an implicit approximation and to say that the implicit precision interval of any expressed result is a variation of plus or minus one on the last digit. i.e. 13% is a valid approximation of the sixth root of 2 for a physicist, so are 12%, 12.2% and 12.3%, but not 13.0%.

98. Ok, where did this 13% thing first come about in this thread? I believe Kelly was the first to mention this 13% increase in radius/diameter<=> 26% increase in area/exposure for a 1/3 stop

My question is, isn't this working backwards? From 26% increase in area, you derive 13% increase in diameter/radius?

Kelly, how did you come out with the 13% thingy? 99. Ah, Jean-Baptiste; your sly dig at mathematicians is duly noted and not forgotten Just using the numbers provided by the respondents so as not to confuse the issue further, significant decimal places be damned.

It is clear that you can drag a horse to the edge of the water but you cannot make it drink. Oh well, more power to Jack and his logic then. Let's hope that he is better at photography than mathematics.

Before you try to prove an erroneous point, Jack, please consult a mathematics teacher or textbook. Your argument and assumption are obviously unsound, trust me on this one. Do not make yourself look anymore foolish by arguing.

100. Anatole,

I don't suppose you'd be willing to share the proof you have. Not that i doubt you, i'm just interested. If you think it will bore the living crap out of everyone here then you can email it to me.

Thanks,

Moiz

101. Moiz: these guys all agree that f-stops are logarithmic: http://www.wikipedia.org/wiki/F-stop http://members.rogers.com/davesphoto/id12.htm http://www.eyeconvideo.com/glossary_of_terms.htm http://www.cliffshade.com/dpfwiw/exposure.htm

Logarithms (by definition) are such that log(a*b)=log(a)+log(b), where the star & denotes a multiplication. They are also such that any two different number have different logarithms (this is enough of a definition to go and prove all the other properties of logarithms).

From here on, we know that log(2)=1 (an increase of one stop multiplies the exposure by 2). Let R be the multiplying factor that matches one third of a stop (mathematicians please don't chuckle). log(R)=1/3, meaning that log(R)+log(R)+log(R)=1, i.e. log(R*R*R)=1 (by definition of logarithms), i.e. R*R*R=2. R is the cubic root of 2.

Yes, I've cut a few corners, and I've taken the definition of logarithms that was most convenient for this "proof". Explaining with more details would take just too long and would bo so totally off-topic that it's not worth my time.

102. Im actually very afraid everytime a new message appears on this thread. Thx for the memories guys. ))

Im planning the next quiz.

103. Maybe y'all should look at it from a different angle, to avoid the weirdness of square roots and aperture radius vs. area, etc.

Take a nice exact M7 electronic SHUTTER. Assume it lets through 1000 units of light in 1/500 second and 500 units of light in 1/1000 second (in the middle of the frame - no quibbles, PLEEZE!)

So, what shutter speed would give 1/3 more exposure than 1/1000 sec.? Or 2/3 more? Is 1/750 sec. actually giving the exact equivalent of 1/2 stop more exposure? Or should it be 1/707 second?

If we can get values for those shutter speeds (which should be easier), then we can come back to apertures-f/stops.

MY math sez the 1/3rd-stop shutter speeds should be 1/794 and 1/630, which follow the 1.26x rule. And that the correct 1/2-stop speed is 1/707 (and you always thought those readouts of 1/700 were just roundoff error from 1/750!).

1000:707 :: 707:500

- and -

1000:794 :: 794:630 :: 630:500

(I hope the spacing held up in those equivalences.)

104. Jean-Baptiste, thanks.