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

"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,

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

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..... ;)

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.

I need an Advil.

Jack, my last post, "x" means an amount, not multiply(times)! ;)

 

boy my brains are cramped.

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

Not to mention flare..;)

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.

"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.

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

"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!

 

;),

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

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

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.

"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,

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.

Jack..k ;)

"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.

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

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 ....

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.

"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

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.

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. :)