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

[travis1]

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..;))

[travis1]

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

 

 

No more.

[Guest]

"The light shineth

in darkness, and the darkness comprehendeth it not."<p><i>Fiat lux.</i

[Guest]

</i><p>

[steve_brantley2]

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

[travis1]

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

 

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

[travis1]

$hit, that should be 29.5%.

[britt_park]

<p>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. </p>

 

<p>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.</p>

 

<p>My apologies, but I couldn't help but comment.</p>

 

<p><a href="http://www.sciencething.org/photos/photos.html">Blatant Plug</a></p>

[britt_park]

Further apologies for my frightful proofreading. I am capable of matching noun and verb number and generally writing coherent English ;)

[_________1]

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.

 

Jack, Jay, please help me. I is all confused and shit.

[jackflesher]

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.

 

 

Jack, Jay, please help me. I is all confused and shit."

 

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 :)))))

[ray .]

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.

[jbq]

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?

 

Let's try our thirds of stops. We've seen that adding one third of a stop multiplies the exposure by 1.26. With the vocabulary we've used, it adds 26% (hey, that's the value that some people have used as an answer!). Let's try it. I'll start with 1000. Add 26%, I get 1260. Add 26% another time, I get 1587.6. Add 26% yet another time, and I get 2000.376 (yes, not exactly 2000, because the cubic root of 2 is not 1.26, it's about 1.259921). You can break down in two intermediate numbers and take percentages of percentages of percentages, trust me, it work (you'll add approximately 26% plus 26% plus 26% plus 7% plus 7% plus 7% plus 2% of your original number, which adds up to about 100%. 7% is approximately 26% of 26%, and 2% is approximately 26% of 26% of 26%).

 

 

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...)

[ray .]

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

[travis1]

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.

[jackflesher]

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 :)

[travis1]

$shit, I got it all reversed! Sorry Jack! ;)

 

i meant f2 = T, f1.4=2T, f1=4T.

 

sorry!

[travis1]

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

[jackflesher]

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 ;)

[travis1]

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.

[jackflesher]

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,

[jbq]

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!

[britt_park]

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.

[hm]

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.

<p>

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.

<p>

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.

<p>

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):

<p>

<pre>

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}

</pre>

[jackflesher]

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,