# F-stops Explained



## 480sparky (Feb 7, 2013)

Many of us will readily recognize the following sequence of numbers:

1, 2, 4, 8, 16, 32, 64, 128, 256, 512.

Save for the first digit (1), each number is twice that of the number to it's left.  So you could say that, as you read from left to right, the numbers gets _doubled_.  Conversely, if you were to read the numbers right to left, the numbers get _halved_.

If you've been involved with photography for a significant length of time, you will also recognize some familiar numbers;.. and you usually call them _f-stops_, or _f-numbers_.

But those numbers, when stamped into a lens barrel, have some additional numbers in them.

2, 2.8, 4, 5.6, 8, 11, 16, 22, and maybe 32.







And for some reason, you've been told that this sequence, just like the list above, is either _doubling _or _halving_ depending on which direction you are reading them.

OK, so what gives?

I'll start out with a word of warning;.. if math gives you a headache, you may want to stop here and go do something else.  What I'm going to attempt to do is explain F-stops in a bit more detail.  And that involves math.  No, not rocket-science math or quantum physics.  But math nonetheless.

Starting with this:  8 x 2 = 11.

OK, I cheated.  f8 x 2 = f11.

Make sense now?  I didn't think so.  Nor does f8 / 2 = f5.6, right?

To understand those f-numbers, lets take a look at an ordinary camera lens.  In this case, I'll use my 105mm Nikkor Micro, which is pictured above.  If you want to increase the exposure by one stop;, you've probably been told to use a smaller number.  I'll use going from f8 to f5.6 for this example. F5.6 lets twice as much light in as f8.  But even your kids and grandkids know that 8 / 2 = 4, not 5.6!

The reason behind the odd photographic sequence is because f-numbers are actually the result of a ratio.  Remember the term _quotient_ from your school days?  That's the name given to the results of a simple division problem.  F-numbers are really quotients!

OK, so where do the other two numbers come from (recall _dividend or numerator_ and _divisor or denominator_?  If so, you&#8217;ve probably got a head-ache now!)?  The first number (dividend) is the focal length of your lens.  The second number (divisor) is the diameter of the aperture inside your lens.

So let's say you have a 200mm lens. You turn the aperture ring until the aperture blades create a circle inside the lens that measures 25mm in diameter.  The result is the equation 200 / 25 = 8.  You now have the lens set to f8.  If you were to open the aperture until it measures 50mm in diameter, you have the equation 200 / 50 = 4;.. meaning the lens is set to f4.

In essence, you are letting in four times as much light at f4 as you were with the lens set to f8.  F5.6 would be twice as much light as f8.

So now you're probably wondering why f5.6 lets in twice as much light as f8.  Why isn't it f4 instead?  Well, sorry to say, it's time for some more math; and it's for the same reason 50mm is four times as large as 25mm.  Confused?  Don't feel bad.  It's a concept that takes some people time to wrap their heads around.


Let's take that 25mm opening.  How do you calculate the _area_ of that opening? Time to think back to school;.. remember the formula?  pi x r². I'll use the everyday value of 3.14 for;.  r is for radius, which is half the diameter.  So a 25mm circle will have an area of 3.14 x 12.5² = 3.14 x 156.25 = 490.625 mm².

A 50mm opening would be 3.14 x 25² = 3.14 x 625 = 1962.5 mm².  A 50mm opening, while being _twice the_ _diameter_ as a 25mm opening, has _4 times the area_! (1962.5 / 490.625 = 4)  So a 50mm opening will let in _4 times as much light_!

In order to get twice as much light, you would need a _35.35mm diameter_ opening (3.14 x 17.68  x 17.68 = 3.14 x 981.51 mm²; and 490.625 x 2 is roughly 981.51;..remember I'm using rounded-off numbers for simplicity!).  And where would this size opening fall into the f-stop number?  200 / 35.35 = 5.6.

Does 5.6 sound familiar?

This is why f5.6 lets in twice as much light as f8;.. the *area *is twice the size; not the *number* (quotient!) itself!


OK, so why not just use the areas created by the aperture blades instead of this seemingly long-winded way?  Well, if you told someone you took a shot at ISO 100, 1/640th of a second, and an aperture set to 25 mm²;.. you may be providing _accurate_ information, but you're not providing _complete_ information.

To explain why, I placed two lenses up on a table and set them both to f8.  On the left is a 105mm, on the right is a 28mm.  Remember, they're both set to f8!  Notice how much larger the opening is on the 105mm on the left compared to the 28mm on the right?






If I set the 105mm lens' aperture to, let's say 10mm diameter, I would have an f-number of  roughly f10.5. But if I set the 28mm to the same 10mm diameter, it would be set to f2.8!  To get f10.5 on a 28mm lens, the aperture would need a diameter of 2.67 mm.

So even though the areas created by the aperture blades in lenses of different focal lengths are different; _optically they create the same f-number_;. And ultimately the same exposure!  That's why f-numbers are used instead of the areas created by the aperture blades;.. it makes it just that much easier for us to work with.  Otherwise, converting aperture areas from one lens to create an identical exposure in another lens would REALLY give you a headache!

So what the manufacturers do when they put seemingly archaic numbers like 2.8, 5.6 and 11 on the lenses is really just taking the math out of the equation for us!  So f/8 on one lens gives us the same exposure aperture on any other lens!  Whether it's a $50 point-and-shoot, a vintage 8x10 view camera or a 5-digit Hassy dream setup.... f/8 is f/8.


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## oldhippy (Feb 7, 2013)

Wow


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## Josh66 (Feb 7, 2013)

You have succeeded in in making simple things way more complicated than they have to be...

Each stop is a function of the square root of 2...  the square root of 2 is roughly 1.4...  &#8730;2*2=2, &#8730;2*3=4.2 (rounded to 4 on lenses), &#8730;2*4=5.6 (rounded), &#8730;2*5=7.07 (rounded to 8 on lenses), etc...

The reason why is because that is how you figure out the area of a circle - we're not interested in the diameter of the aperture, just the area.


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## 480sparky (Feb 7, 2013)

O|||||||O said:


> ......... we're not interested in the diameter of the aperture,..........




So, without the diameter, how do you figure the f-number?


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## r0adki1l (Feb 7, 2013)

Thank you!!! that was very educational and made a lot of sense.


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## Heitz (Feb 7, 2013)

480sparky said:
			
		

> So, without the diameter, how do you figure the f-number?



I suppose if you found a lens with absolutely no markings you could reverse engineer it using your method.


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## Josh66 (Feb 7, 2013)

480sparky said:


> O|||||||O said:
> 
> 
> > ......... we're not interested in the diameter of the aperture,..........
> ...


You only need to know one - the diameter or the area.  You can find one of you know the other very easily.  "F-Stops" are neither.  It is not a diameter or an area.

When you make simple things complicated, it makes me wonder if you understand the 'simple thing'.  I assume that in this case you do understand, and you are just trying to 'dumb it down' for TPF - but I would argue that TPF would be better off without any 'dumbing down' of information.


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## 480sparky (Feb 7, 2013)

O|||||||O said:


> 480sparky said:
> 
> 
> > O|||||||O said:
> ...




Simple solution.... don't read it.

If anything, your 'square root of 2' simply makes it even MORE complicated.


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## Josh66 (Feb 7, 2013)

480sparky said:


> If anything, your 'square root of 2' simply makes it even MORE complicated.



How so?  That is the foundation of the whole thing.


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## 480sparky (Feb 7, 2013)

O|||||||O said:


> 480sparky said:
> 
> 
> > If anything, your 'square root of 2' simply makes it even MORE complicated.
> ...



I really don't think you're qualified to speak for everyone here on TPF.

If you don't like it, then DON'T READ IT.  Simple as that. I'm not going to argue with you over it.  You are just not worth it.


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## Josh66 (Feb 7, 2013)

480sparky said:


> O|||||||O said:
> 
> 
> > 480sparky said:
> ...


lol

I tried not to read it, but I was compelled to - now I regret it.  I didn't want to comment without being certain that I fully understood what you were trying to say, so I in fact read it more than once...

OK, we'll just ignore ... 'math, and stuff', just so we can make sure everything stays at a 2nd grade reading level.

The first thing I thought when I read it was that you must have been thinking, 'Oh, this will make a great Sticky!'.


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## 480sparky (Feb 7, 2013)




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## Josh66 (Feb 7, 2013)

480sparky said:


> O|||||||O said:
> 
> 
> > 480sparky said:
> ...



If anybody on TPF disputes that the square root of 2 is 1.414213562..., speak up now.

Also, if anybody on TPF disputes that the area of a circle = pi*r^2, speak up now.


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## 480sparky (Feb 7, 2013)

O|||||||O said:


> If anybody on TPF disputes that the square root of 2 is 1.414213562..., speak up now.
> 
> Also, if anybody on TPF disputes that the area of a circle = pi*r^2, speak up now.



How absolutely, positively pathetic. You've managed to take TPF to an incredibly and dismally new low.

I hope you enjoy your self-proclaimed importance.


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## Josh66 (Feb 7, 2013)

480sparky said:


> I hope you enjoy your self-proclaimed importance.



I do.  Thank you.  Though, I did nothing to promote this - that was all you.


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## Bitter Jeweler (Feb 7, 2013)




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## bigtwinky (Feb 7, 2013)

The exposure on that eagle seems a bit low Bitter


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## NickStevens (Feb 7, 2013)

I think my head just exploded, far too much thinking for this time of the night/morning :madmad:

Night people


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## Josh66 (Feb 7, 2013)

480sparky said:


> You've managed to take TPF to an incredibly and dismally new low.



So many things I want to say about that, but I won't because it would probably mean a ban (nothing about you, Sparky - believe it or not, we might actually get along 'in the real world')...  I agree that TPF is at a low point, but I think you give me WAY too much credit to say that I have any part of it.


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## Bitter Jeweler (Feb 7, 2013)

bigtwinky said:


> The exposure on that eagle seems a bit low Bitter


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## bigtwinky (Feb 7, 2013)

BWAHAHAHAHAHA :lmao:


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## laynea24 (Feb 7, 2013)

I thoroughly enjoyed reading this and learning! Thanks.


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## dmunsie (Feb 8, 2013)

"1, 2, 4, 8, 11, 16, 32, 64, 128, 256, 512.

Save for the first digit (1), each number is twice that of the number to its left."

I was right there with ya, till I got to the 11.


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## 480sparky (Feb 8, 2013)

dmunsie said:


> "1, 2, 4, 8, 11, 16, 32, 64, 128, 256, 512.
> 
> Save for the first digit (1), each number is twice that of the number to it&#8217;s left."
> 
> I was right there with ya, till I got to the 11.



Corrected!


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## ChrisCalvin (Feb 8, 2013)

Thanks, useful info.


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## Tiberius47 (Feb 8, 2013)

I gotta agree that the OP makes it needlessly complicated.

It's like this...

The F stop refers to the size of the hole that the light goes through.  Being a hole, it has two dimensions - height and width.

If you take any shape and enlarge it so it becomes 1.4 times as wide and 1.4 times as high, then it's area will roughly double.  So, for example, you can have a square that is 1 foot wide by 1 foot high, and it has an area of 1 square foot.  If you want to have a square with an area of two square feet, you can't just double the width and height - then you get four square feet.  Instead of multiplying the dimensions by 2, you multiply them by 1.4.  A square that is 1.4 feet by 1.4 feet is about two square feet in area.

And if you have a hole that is letting light through, and you increase its size by 1.4 like this, then it will let about twice as much light through.  In other words, it lets in an extra stop of light.

And that's why we use that progression of numbers in f-stops.

Easy.


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## NickStevens (Feb 8, 2013)

Tiberius47 said:


> I gotta agree that the OP makes it needlessly complicated.
> 
> It's like this...
> 
> ...



Yes, I think lol

Thing is does it actually make any difference to any of us why the numbers are the way they are? 

As long as we remember the numbers then I can't see it matters to much. 

UNLESS Im missing something glaringly obvious?


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## O'Rork (Feb 8, 2013)

It appears to me we have all been cooped up by winter long enough. Hey I'm fighting a squirrel! And it looks like I'm winning.

6 of 1 is half a dozen of another. 

You can turn left or you can take 3 rights.


If you adjust by  *full* stops you're going to half or double appreciated light.  < Now wasn't that easy?


The above are true. What's below is an opinion.

You can pick your friends and you can pick your nose, but you can't pick your friends nose.


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## Blitz55 (Feb 8, 2013)

Good info in this thread. Thanks for the posts.


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## pixmedic (Feb 8, 2013)

that was an awesome read sparky!
honestly though....I originally clicked on it because i THOUGHT it said F-Bombs explained....
still a good read. thanks!


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## thunderkyss (Feb 8, 2013)

480sparky said:


> If I set the 105mm lens' aperture to, lets say 10mm diameter, I would have an f-number of  roughly f10.5. But if I set the 28mm to the same 10mm diameter, it would be set to f2.8!  To get f10.5 on a 28mm lens, the aperture would need a diameter of 2.67 mm.
> 
> So even though the areas created by the aperture blades in lenses of different focal lengths are different _optically they create the same f-number_. And ultimately the same exposure!  Thats why f-numbers are used instead of the areas created by the aperture blades.. it makes it just that much easier for us to work with.  Otherwise, converting aperture areas from one lens to create an identical exposure in another lens would REALLY give you a headache!
> 
> So what the manufacturers do when they put seemingly archaic numbers like 2.8, 5.6 and 11 on the lenses is really just taking the math out of the equation for us!  So f8 on one lens gives us the same exposure aperture on any other lens!  Whether it's a $50 point-and-shoot, a vintage 8x10 view camera or a 5-digit Hassy dream setup.... f8 is f8.



So a diameter of 2.67mm on a 28 millimeter lens lets in just as much light as a 10mm diameter on a 105mm lens? 

How is that possible?


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## IByte (Feb 8, 2013)

Sparks I love ya man lol, but I'm going to come back to this after a few pints.


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## cwcaesar (Feb 8, 2013)

The very definition of f-stop is focal length divided by diameter, or F/D. 

Very well explained OP! :hail:


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## Helen B (Feb 8, 2013)

thunderkyss said:


> 480sparky said:
> 
> 
> > If I set the 105mm lens' aperture to, lets say 10mm diameter, I would have an f-number of  roughly f10.5. But if I set the 28mm to the same 10mm diameter, it would be set to f2.8!  To get f10.5 on a 28mm lens, the aperture would need a diameter of 2.67 mm.
> ...



Obviously it lets in less light. The image of an object, however, is smaller and that exactly compensates for the lesser amount of light - less light spread over a smaller area, to give the same light/area (in simple terms).


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## thunderkyss (Feb 8, 2013)

O|||||||O said:


> You have succeeded in in making simple things way more complicated than they have to be...
> 
> Each stop is a function of the square root of 2...  the square root of 2 is roughly 1.4...  &#8730;2*2=2, &#8730;2*3=4.2 (rounded to 4 on lenses), &#8730;2*4=5.6 (rounded), &#8730;2*5=7.07 (rounded to 8 on lenses), etc...
> 
> The reason why is because that is how you figure out the area of a circle - we're not interested in the diameter of the aperture, just the area.



If the square root of 2 is roughly 1.4, how is &#8730;2*2=2; wouldn't it be 2.8? 

Why do we round 4.2 to 4, but don't round 2.8 to 3? 

How is f4 on 50mm lens equivalent to f4 on a 300mm lens? Or why? 

Your explanation makes it _easy_ to figure out the number, but doesn't explain how you get that number or what that number represents. It's nice to know that the ratio of the area of the aperture opening to the focal length of the lens produces a quotient that is common between lenses & that the f-stop is that quotient.


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## pixmedic (Feb 8, 2013)

thunderkyss said:


> O|||||||O said:
> 
> 
> > You have succeeded in in making simple things way more complicated than they have to be...
> ...



its magic. just turn the dial and let the magic work.


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## thunderkyss (Feb 8, 2013)

pixmedic said:


> thunderkyss said:
> 
> 
> > O|||||||O said:
> ...



No offense, I know you're shooting for levity, but this is the very definition of "dumbing down"

The OP obviously wanted to go beyond the basic, "A full stop up is twice as much light & a full stop down is half as much." & while it is true, that is all that a beginner "needs" to know, there are some people who would like to know why a full stop up lets in twice as much light. The simple answer of doubling the area doesn't fully explain as it is the ratio of the focal length to the aperture diameter. 

& I appreciate it.


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## thunderkyss (Feb 8, 2013)

Helen B said:


> thunderkyss said:
> 
> 
> > So a diameter of 2.67mm on a 28 millimeter lens lets in just as much light as a 10mm diameter on a 105mm lens?
> ...



Thank you. I think the OP would greatly benefit if that info was included.


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## JayJeep (Feb 8, 2013)

O|||||||O said:


> You have succeeded in in making simple things way more complicated than they have to be...
> 
> Each stop is a function of the square root of 2...  the square root of 2 is roughly 1.4...  &#8730;2*2=2, &#8730;2*3=4.2 (rounded to 4 on lenses), &#8730;2*4=5.6 (rounded), &#8730;2*5=7.07 (rounded to 8 on lenses), etc...
> 
> The reason why is because that is how you figure out the area of a circle - we're not interested in the diameter of the aperture, just the area.



&#8730;2*2=2 is 2.8 and why would 7.07 get rounded to 8.  and &#8730;2*6=8.48 do we round that to 8 or 9 or 11.

You have it wrong I think.  It should be the &#8730;2^2=2, &#8730;2^3=2.8 and &#8730;2^6=8.  You take it to the next power not multiply it by the next integer.


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## Helen B (Feb 8, 2013)

thunderkyss said:


> O|||||||O said:
> 
> 
> > You have succeeded in in making simple things way more complicated than they have to be...
> ...



They are mostly rounded to two significant figures. The 0.5 - 1 - 2 - 4 - 8  series do not need to be rounded because they are exact numbers, 0.5 being the lowest f-number theoretically possible. The others usually get rounded to a nominal f-number with no more than two significant figures, such as 1.4, 2.8. 5.6, 11 etc.

The sequence is (&#8730;2)^x, where x is a positive or negative integer equal or greater than -2, or zero.

F/4 on a 50 mm lens is equivalent to f/4 on a 300 mm lens in terms of image brightness. As I mentioned earlier, the 300 mm lets more light in from an object, but it forms a larger image, so the brightness is the same. How much detail do you want on this? How good are you at maths?


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## ClickAddict (Feb 8, 2013)

Both Sparky and O|||||||O have the same math. Just one is longer form.
Sparky's math Breaking it down: 

focal lens = m 
diameter = d 
aperture = f 
He states m/d = f........................... (His 200mm lense to 25 diameter = f8 example) 
which means m=df ...........................(multiply both sides by d) 
and d=m/f .....................................(divide both sides by f) 

opening area = A 
A = pr^2 .............................(This is a known calculation for area of circle) 
p=3.14 ..................................(rounded) 
r = radius 

r= 1/2 d .....................................(By definition radius = 1/2 diameter) 

r= 1/2X(m/f)................................ (replacing d with m/f as per above) 
r= m/2f........................................ (result of 1/2 X m/f) 

A1 ...................area of lens at a given aperture 
A2 .................area of same lense at a different aperture 
r1............... radius of lense to get Area = A1 
r2.................... radius of lense to get Area = A2 

if A1 = 1/2 of A2 then ............................(looking for next stop which is half the area) 

A1 = A2/2 
pr1^2=(pr2^2)/2................................... (A1 = pr1^2 as per area formula, same for A2)) 
p(m1/2f1)^2 = p(m2/2f2)^2/2 ............................(replacing r with m/2f) 
(m1/2f1)^2 = (m2/2f2)^2/2 ...........................divide both sides by p 
2(m1/2f1)^2 = (m2/2f2)^2 ........................multiply both sides by2 
2 = ((m2/2f2)^2)/(m1/2f1)^2...................... divedi both sides by 
2 = (m2^2/(2f2)^2)/(m1^2)/(2f1)^2 ......................(square the fractions) 
2= (m2^2)(2f1)^2)/((2f2)^2)(m1^2) ......................(a/b)/(c/d) = a/b X d/c = ad/bc 
2= (m1^2)(2f1)^2)/((2f2)^2)(m1^2) .....................since were keeping m the same and only changing f, m1=m2) 
2=(2f1)^2)/((2f2)^2) ...................................(xy/zx = y/x) 
2= 4f1^2 / 4f2^2 ........................................(ab)^2 = (a^2)(b^2) 
2=f1^2 / f2^2................................ (4A/4B = A/B) 
sqrt(2) = f1/f2 ..................................sqrt = square root) 
1.4 = f1/f2 

so to get the half amount of light as per sparkys math AND O||||||O you need to have that ration 

Essentially many of the variables eliminate themselves to get to the 1.4 value


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## bhop (Feb 8, 2013)




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## JayJeep (Feb 8, 2013)

ClickAddict said:


> Both Sparky and O|||||||O have the same math. Just one is longer form.
> Sparky's math Breaking it down:
> 
> focal lens = m
> ...




Except  O||||||O is wrong.  He says to multiply 1.4*(integer) to get 2, 2.8, 4, 5.6, 8, 11.  You don't multiply you take it to the power of the integers.

sqrt(2)^0=1, sqrt(2)^1=1.41, sqrt(2)^2=2, sqrt(2)^3=2.82, sprt(2)^4=4

If I used his math of rounding 7.07 to 8 and 8.48 to 11 in my line of work I would most likely be fired and endangering lives.

The problem I see is that he insisted that he was right and the OP was wrong to post a complicated explanation for something I know has been asked.

"where do they get f/5.6 as double the light of f/8"  and "where do those numbers come from"


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## jwbryson1 (Feb 8, 2013)

O|||||||O said:


> 480sparky said:
> 
> 
> > O|||||||O said:
> ...



I won't dispute those but I will dispute the following:



O|||||||O said:


> Each stop is a function of the square root of 2...  the square root of 2  is roughly 1.4...  _*&#8730;2*2=2*_, &#8730;2*3=4.2 (rounded to 4 on lenses), &#8730;2*4=5.6  (rounded), &#8730;2*5=7.07 (rounded to 8 on lenses), etc...


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## ratssass (Feb 8, 2013)

i like turtles


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## ClickAddict (Feb 8, 2013)

JayJeep said:


> ClickAddict said:
> 
> 
> > Both Sparky and O|||||||O have the same math. Just one is longer form.
> ...



You are correct.  I thought he had said it was 1.4 X previous number.  (approximately)


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## Helen B (Feb 8, 2013)

ClickAddict said:


> Both Sparky and O|||||||O have the same math. Just one is longer form.
> Sparky's math Breaking it down:
> 
> focal lens = m
> ...



You can leave the unneccessary stuff out at the beginning if you want cleaner workings.

N=f/D

(which means f-number = focal length divided by entrance pupil diameter)

therefore 

N&#8733;1/D  (1)

A = (pi D^2)/4

therefore

A&#8733;D^2 and thus D&#8733;A^0.5

Substituting in (1) 

N&#8733;1/A^0.5

which leads to 

N2/N1 = (A1/A2)^0.5


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## thunderkyss (Feb 8, 2013)

This is turning into a very informative thread.


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## bunny99123 (Feb 8, 2013)

OMG.  I am going to have nightmares from all those physic courses and teaching it for so many years. You guys have triggered my PTSD caused from Math. I am too old for this intelligent thinking, so I am just going to turn my zoom and let the light flow in or shut some out. Seriously, you guys have done an excellent job on explaining it simply to college level


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## IByte (Feb 8, 2013)

Helen B said:


> They are mostly rounded to two significant figures. The 0.5 - 1 - 2 - 4 - 8  series do not need to be rounded because they are exact numbers, 0.5 being the lowest f-number theoretically possible. The others usually get rounded to a nominal f-number with no more than two significant figures, such as 1.4, 2.8. 5.6, 11 etc.
> 
> The sequence is (&radic;2)^x, where x is a positive or negative integer equal or greater than -2, or zero.
> 
> F/4 on a 50 mm lens is equivalent to f/4 on a 300 mm lens in terms of image brightness. As I mentioned earlier, the 300 mm lets more light in from an object, but it forms a larger image, so the brightness is the same. How much detail do you want on this? How good are you at maths?



It's the weekend and I have nothing else but crunching the numbers


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## IByte (Feb 8, 2013)

Lol


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## JacaRanda (Feb 8, 2013)

My wife would surely bash me in the face with her camera if I even hinted at trying to teach/show/explain any of this to her. It's dangerous enough telling her the magic trackpad is easier to deal with than that stupid mouse she has to pick up and put down every couple of seconds


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## kundalini (Feb 8, 2013)

While it's interesting to see how passionate some of you are with the minute (pronounced my-noot) intricacies for the math involved, it seems a bit over the top.  With time behind the camera, the results of your settings becomes intuitive and the mathematical formulas become impractical in the field.  The substance is retained, but the details are of no consequence, other than in a classroom environment.


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## Helen B (Feb 8, 2013)

kundalini said:


> While it's interesting to see how passionate some of you are with the minute (pronounced my-noot) intricacies for the math involved, it seems a bit over the top.  With time behind the camera, the results of your settings becomes intuitive and the mathematical formulas become impractical in the field.  The substance is retained, but the details are of no consequence, other than in a classroom environment.



I have no idea who in particular you are dissing here, but I do wonder why you make such assumptions about other people's practice? Do you really think we worry about this stuff when shooting? It is possible to have decades behind the camera and also to understand the trivial physics being discussed here. If someone is curious, should we just ignore them?


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## shefjr (Feb 8, 2013)

I just memorized the f-stops. Is it bad that I don't know the precise why it is what it is? I mean before this post, because I have an understanding now.


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## Helen B (Feb 8, 2013)

shefjr said:


> I just memorized the f-stops. Is it bad that I don't know the precise why it is what it is? I mean before this post, because I have an understanding now.



I'm pretty sure that anyone could be an excellent photographer without knowing this sort of stuff, but it doesn't seem to hurt too much to understand it either. For some of us it is much easier to understand than to try to remember by rote.


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## 480sparky (Feb 8, 2013)

I may have a terrible memory, but I'm fairly certain I never stated that it's _required_ to know this in order to operate a camera.  There's not going to be a pop quiz tomorrow.  No test at the end of the month.  No Finals in order to 'graduate'.


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## shefjr (Feb 8, 2013)

480sparky said:


> I may have a terrible memory, but I'm fairly certain I never stated that it's _required_ to know this in order to operate a camera.  There's not going to be a pop quiz tomorrow.  No test at the end of the month.  No Finals in order to 'graduate'.


I meant no offense. I just honestly wondered if there was something that I was missing that would eventually lead me to understand further the importance of the f-stops. I have found at times I think I know it all until someone points something out that corrects me.


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## snowbear (Feb 8, 2013)

thunderkyss said:


> Why do we round 4.2 to 4, but don't round 2.8 to 3?


 Because we don't like the number three.


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## NickStevens (Feb 8, 2013)

Still don't think it matters.... 

Two stops down on shutter two stops up on appeture..... Simples


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## TCampbell (Feb 8, 2013)

When I explain f-stops (aperture), I sometimes point out that while it does seem confusing that the dial has these funny and somewhat arbitrary progression of values (1.4, 2, 2.8, 4, 5.6, 8, 11, 16, 22 ... ) there is a reason for them that makes sense once you know it.  It might seem easier to just use values like 1, 2, 3, 4... but those would actually be arbitrary values that don't have any true meaning.

Sparky is right in that the numbers indicate the focal ratio... the ratio of the effective aperture divided into the effective focal length (and we say "effective" and not "physical" because modern lens optics allows for effective focal lengths that don't match physical lengths.)

But Josh is right in that reason we don't just pick *any* arbitrary ratios is because those particular ratios all happen to allow twice (or half) of the light through because the area of the opening is changing by that amount.  Josh did make the error in that he multiplied the square root of 2, but it's actually being raised by "powers".

&#8730;2 is approximately 1.4.  
&#8730;2[SUP]2[/SUP] is simply "2" again.

In photography, the values are all rounded for simplicity.

The progression is:
&#8730;2[SUP]0[/SUP] = 1 
&#8730;2[SUP]1[/SUP] = 1.4 
&#8730;2[SUP]2[/SUP] = 2 
&#8730;2[SUP]3[/SUP] = 2.8  
&#8730;2[SUP]4[/SUP] = 4 
&#8730;2[SUP]5[/SUP] = 5.6 
&#8730;2[SUP]6[/SUP] = 8
&#8730;2[SUP]7[/SUP] = 11 (really it'd be 11.2 but it's rounded because they never use more than 2 numerals in an f-stop)
&#8730;2[SUP]8[/SUP] = 16
&#8730;2[SUP]9[/SUP] = 22

You can keep going... large view cameras often have f-stops all the way to f/64.   Most DSLR lenses go up to at least f/22.  Some go up to f/32.  I usually don't see f/44 or f/64 on a camera unless it's a large format camera.


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## NickStevens (Feb 8, 2013)

I must be missing something huge here can somebody explain why it matters them being numbered like that. 

I mean is that their actualy a bigger difference between say f11 to f16 than their is between 7.1 and 8......
Does that matter...... I mean I think to myself that I'm gonna stop down by two stops two clicks of the dial is two stops...is that all that's important? 

Is it doing to improve my photography any by knowing why the numbers are like that?


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## shefjr (Feb 8, 2013)

Well 2 clicks isn't two stops. Most cameras also allow for half stops and thirds of a stop.
For instance, 2.8-4 is a full stop
2.8-3.2 is 1/3 a stop 2.8-3.5 is 2/3 and 2.8-3.4 is 1/2 a stop. At least for Nikon
So that is actually four clicks for one full stop.


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## NickStevens (Feb 8, 2013)

Ummm ok then we'll I still don't get it.... But I don't think it's ever affected my photos lol


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## 480sparky (Feb 8, 2013)

NickStevens said:


> ..........Is it doing to improve my photography any by knowing why the numbers are like that?



Did anyone ever claim it would?


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## Helen B (Feb 8, 2013)

TCampbell said:


> Sparky is right in that the numbers indicate the focal ratio... the ratio of the effective aperture divided into the effective focal length (and we say "effective" and not "physical" because modern lens optics allows for effective focal lengths that don't match physical lengths.).



In case this causes confusion, 'effective focal length' is just another way of saying 'focal length'. They are the same thing in photography. The 'effective aperture' Tim refers to is actually the entrance pupil diameter, not the effective aperture, which is something quite different. The effective aperture is the aperture (calculated from the focal length and the entrance pupil diameter) with an adjustment for image magnification.


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## NickStevens (Feb 8, 2013)

480sparky said:


> NickStevens said:
> 
> 
> > ..........Is it doing to improve my photography any by knowing why the numbers are like that?
> ...



Nope they didn't...... If theirs any benefit I'll try to learn it, if not then I'd rather read a thread where I will learn things


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## LungFish (Feb 8, 2013)

NickStevens said:


> I must be missing something huge here can somebody explain why it matters them being numbered like that.
> 
> I mean is that their actualy a bigger difference between say f11 to f16 than their is between 7.1 and 8......



Yes, that's why on most cameras (all?) it would take a bigger adjustment to go from 11 to 16. I actually think it is important to know the difference between big changes and small changes, saves you a lot of time in settings changes.

When I first read Sparkys post I actually got a bit lost because he starts by asking a question about why the common f-stops aren't multiples of 2, then heads in a different direction about what an f-stop actually means. I was expecting something a bit shorter like "area is proportional to the square of the diameter". I read through the rest of the thread nodding at the responses which said it was too complicated. I then went back to read the first post again, and it actually does make perfect sense and isn't that complicated, you've just got to read it without anticipating what the next sentence will be.


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## 480sparky (Feb 8, 2013)

NickStevens said:


> Nope they didn't...... If theirs any benefit I'll try to learn it, if not then I'd rather read a thread where I will learn things



Ever wonder why pianists learn and practice scales?


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## JayJeep (Feb 8, 2013)

NickStevens said:


> 480sparky said:
> 
> 
> > NickStevens said:
> ...



Well I liked reading through it.  It basically comes down whether or not you want to know where the f stop numbers come from.  If you didn't know and don't care to know, don't read it.  I've always wondered where the numbers came from and now I know.  Does it change anything about how I take photos? No not really, but now I know what the numbers mean.  Just because it doesn't interest you doesn't mean you should jump all over the OP about posting something you don't care about.


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## IByte (Feb 8, 2013)

480sparky said:


> I may have a terrible memory, but I'm fairly certain I never stated that it's required to know this in order to operate a camera.  There's not going to be a pop quiz tomorrow.  No test at the end of the month.  No Finals in order to 'graduate'.


But I studied sooo hard >.<


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## 480sparky (Feb 8, 2013)

IByte said:


> 480sparky said:
> 
> 
> > I may have a terrible memory, but I'm fairly certain I never stated that it's required to know this in order to operate a camera.  There's not going to be a pop quiz tomorrow.  No test at the end of the month.  No Finals in order to 'graduate'.
> ...



OK.. pop quiz just for you.

1. What is the capital of Cleveland?
2. Is there ever a time mattresses are NOT on sale?
3. Where does the white go when the snow melts?
4. When do fish sleep?
5. How come we never hear father-in-law jokes?
6. Why is Greenland called Greenland when it's mostly covered with ice, and Iceland is called Iceland when it's mostly green?
7. If a tree falls in the forest, and it&#8217;s not posted on MySpace, YouTube, Facebook or Twitter, did it really happen?
8. Is it true that the only difference between a yard sale and a trash pickup is how close to the street the stuff is placed?
9. Why is it that when you transport something by truck, it's called a shipment, and when you transport it by ship, it's called a cargo?
10. If "practice makes perfect" and "nobody is perfect", why bother practicing?


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## thunderkyss (Feb 8, 2013)

I think understanding the math behind f-stops helps understand why an image taken with a 200mm lens would look different on a 100mm lens even if they were framed the same. Knowing the math also helps *me *understand depth of field. It also gives me something else to consider when shopping for lenses instead of the, "the more expensive one is better" or, "Nikon lenses are just better" line of thinking.


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## 480sparky (Feb 8, 2013)

thunderkyss said:


> .......It also gives me something else to consider when shopping for lenses instead of the, "the more expensive one is better" or, "Nikon lenses are just better" line of thinking.



Maybe I'm missing something here...... What does the math behind f-stop have to do with lens quality?


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## thunderkyss (Feb 9, 2013)

480sparky said:


> thunderkyss said:
> 
> 
> > .......It also gives me something else to consider when shopping for lenses instead of the, "the more expensive one is better" or, "Nikon lenses are just better" line of thinking.
> ...



I didn't say quality. 

But understanding f-stop I know why an f1.8 55mm lens would cost more than an f3.5 55mm lens. Sure, knowing the f1.8 lets in more light than the f3.5 should be enough. But I also know how difficult it would be, & why, to build a 100mm f1.8 or f3.5


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## MiFleur (Feb 9, 2013)

Thanks Sparky for your explanation, math is not my strength and I understood everything you explained, I specially love your example with the 2 lenses.


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## Rick50 (Feb 9, 2013)

This was a good read. Sure made my 1st cup of coffee this morning enjoyable. Thanks to all.


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## Tiberius47 (Feb 19, 2013)

O'Rork said:


> You can pick your friends and you can pick your nose, but you can't pick your friends nose.



It depends on your friends, really...


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