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help with aperture math



 
 
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  #1  
Old February 9th 06, 09:59 PM posted to rec.photo.digital.slr-systems
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Default help with aperture math

I'm trying to understand apertures. I know that if F2 @ 1/500s then F2.8 @
1/250s and F4 @ 1/125s and F5.6 @ 1/60s..

Where do the seemingly odd numbers 2.8 and 5.6 come from? This must have
something to do with PI and area of a circle?

--
Mark

Photos, Ideas & Opinions
http://www.marklauter.com/gallery


  #2  
Old February 9th 06, 10:12 PM posted to rec.photo.digital.slr-systems
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On Thu, 09 Feb 2006 21:59:13 GMT, "Beach Bum"
wrote:

I'm trying to understand apertures. I know that if F2 @ 1/500s then F2.8 @
1/250s and F4 @ 1/125s and F5.6 @ 1/60s..

Where do the seemingly odd numbers 2.8 and 5.6 come from? This must have
something to do with PI and area of a circle?



Has more to do with the square root of two, 1.414, or thereabouts.

since the area of the opening is pi d squared, each time d is double,
four times the light gets in. so, for double the light we need to
multiply the size of the opening by the square root of two.
  #3  
Old February 9th 06, 10:13 PM posted to rec.photo.digital.slr-systems
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Beach Bum skrev:

I'm trying to understand apertures. I know that if F2 @ 1/500s then F2.8@
1/250s and F4 @ 1/125s and F5.6 @ 1/60s..

Where do the seemingly odd numbers 2.8 and 5.6 come from? This must have
something to do with PI and area of a circle?


No. They are powers of the square root of 2. The diameter of a circle
varies with the square root of its area. So a circle with twice as
large area has a diameter 1.414 (rounded off to three decimal values)
times as long.

Jan Böhme

  #4  
Old February 9th 06, 10:58 PM posted to rec.photo.digital.slr-systems
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Default help with aperture math

"Charles" wrote in message

Has more to do with the square root of two, 1.414, or thereabouts.

since the area of the opening is pi d squared, each time d is double,
four times the light gets in. so, for double the light we need to
multiply the size of the opening by the square root of two.


Ah HAH! Thanks.



--
Mark

Photos, Ideas & Opinions
http://www.marklauter.com/gallery


  #5  
Old February 9th 06, 11:40 PM posted to rec.photo.digital.slr-systems
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Default help with aperture math

"Beach Bum" writes:

I'm trying to understand apertures. I know that if F2 @ 1/500s then F2.8 @
1/250s and F4 @ 1/125s and F5.6 @ 1/60s..

Where do the seemingly odd numbers 2.8 and 5.6 come from? This must have
something to do with PI and area of a circle?


Close -- they have to do with the square root of two, from the formula
for the area of a circle.
--
David Dyer-Bennet, , http://www.dd-b.net/dd-b/
RKBA: http://noguns-nomoney.com/ http://www.dd-b.net/carry/
Pics: http://dd-b.lighthunters.net/ http://www.dd-b.net/dd-b/SnapshotAlbum/
Dragaera/Steven Brust: http://dragaera.info/
  #6  
Old February 9th 06, 11:44 PM posted to rec.photo.digital.slr-systems
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In message ,
"Beach Bum" wrote:

I'm trying to understand apertures. I know that if F2 @ 1/500s then F2.8 @
1/250s and F4 @ 1/125s and F5.6 @ 1/60s..

Where do the seemingly odd numbers 2.8 and 5.6 come from? This must have
something to do with PI and area of a circle?


Those numbers are the focal length divided by the effective aperture.
They have traditionally been based on powers of the square root of two,
but that is totally arbitrary. 1.414 is the square root of 2; 2 is
1.414 squared; 2.828 is 1.414 to the third power, etc.

For any given focal length, the amount of light on the focal plane is
inversely proportional to that number squared (all other things being
equal). This is awkward, for instance, if you're trying to trade f-stop
against ISO, but is very practical in other regards, like setting
aperture based on the guide number of a flash (GN/distance =
needed_f-stop).
--


John P Sheehy

  #7  
Old February 9th 06, 11:47 PM posted to rec.photo.digital.slr-systems
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Default help with aperture math


I'm trying to understand apertures. I know that if F2 @ 1/500s then F2.8
@
1/250s and F4 @ 1/125s and F5.6 @ 1/60s..

Where do the seemingly odd numbers 2.8 and 5.6 come from? This must have
something to do with PI and area of a circle?

Mark, yes, the numbers for aperture settings are functions of the area of a
circle as others have pointed out. But perhaps an easier way to think of
the "f stops" or aperture settings is that each number is actually a
fraction indicating the ratio of the "circle" of your lens opening versus
the maximum area of your lens mount (not the maximum opening of your lense
aperture). Thus a setting of f/4.0 means the aperture on your lens is
opened 1/4th of the area of the lens mount on your camera, f/8.0 is 1/8th of
the maximum opening, f/11 is 1/11th, etc....the "f" numbers in between the
whole stops are still fractions of the maximum area of the mount.

It's also why as the aperture numbers get "bigger", the opening to let the
light in is actually getting smaller -- because they're fractions, not whole
numbers. The "1/x.0" notation has just been dropped from common usage over
the years.

Almost counter intuitive, isn't it -- the bigger the number the smaller the
opening?

Some think of this relationship for aperture area versus mount area as a
ratio -- 1:4 instead of 1/4th, 1:2.8 instead of 1/2.8th, etc. That's why
you'll see a notation on a lens like "1:4" or "1:2.8" -- that information is
describing the maximum aperture of that particular lens -- f/4.0, f/2.8,
etc. Bottom line, whatever floats your boat to make sense of it.

And you've probably understand this but because the opening is getting
smaller, you have to increase the amount of time the shutter is open to
allow the necessary amount of light to strike the recording medium (film,
sensor, whatever) for a correct exposure (and "correct exposure" is another
subjective phrase and an whole different discussion).

Keep in mind that the inverse correlation between the aperture setting and
the shutter speed is a constant -- thus, for a correct exposure, each time
you open up or close down the aperture, a corresponding, equal, adjustment
needs to be made with the shutter speed. As the aperture gets smaller, the
shutter speed has to slow down a corresponding equal amount. Conversely, as
the shutter speed is made faster or slower, there has to be a corresponding
and equal change in the aperture setting.

The old analogy is a bucket and a funnel -- the correct exposure is the size
of the bucket. How you fill that bucket (with light) is dependent on the
size of the funnel and how fast you pour. The funnel is the aperture and
how fast is the shutter speed. The bigger the funnel, the slower you need
to pour.......as I said, it's an old analogy.

Some photography historian (or some one with a better memory) can probably
recall that there use to be a lens that had a maximum of f/1.0 -- I think
Canon made one some years back???

Hope this doesn't confuse things even more.

Mike


  #8  
Old February 10th 06, 12:38 AM posted to rec.photo.digital.slr-systems
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Default help with aperture math

"Mike Rooney" wrote in message news:ewQGf.24707

Mark, yes, the numbers for aperture settings are functions of the area of

a
circle as others have pointed out.


snip

Thanks Mike.

I know how apertures work - just needed some math to go with it and couldn't
find it in any of my books with quick glance.

The old analogy is a bucket and a funnel


Hadn't heard this before - I like it. I'll use it when explaining exposure
to beginners.

--
Mark

Photos, Ideas & Opinions
http://www.marklauter.com/gallery


  #9  
Old February 10th 06, 01:17 AM posted to rec.photo.digital.slr-systems
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Default help with aperture math

Mike Rooney wrote:

...each number is actually a
fraction indicating the ratio of the "circle" of your lens opening versus
the maximum area of your lens mount (not the maximum opening of your lense
aperture). Thus a setting of f/4.0 means the aperture on your lens is
opened 1/4th of the area of the lens mount on your camera, f/8.0 is 1/8th of
the maximum opening, f/11 is 1/11th, etc....the "f" numbers in between the
whole stops are still fractions of the maximum area of the mount.
...
Some think of this relationship for aperture area versus mount area as a
ratio -- 1:4 instead of 1/4th, 1:2.8 instead of 1/2.8th, etc. That's why
you'll see a notation on a lens like "1:4" or "1:2.8" -- that information is
describing the maximum aperture of that particular lens -- f/4.0, f/2.8,
etc. Bottom line, whatever floats your boat to make sense of it.



Ah, thanks for this added explanation. Ratio to the area of the lens
mount. Hmm, now I'm confused about the relationship to the focal length
though?

So Mark, what f/stop are you getting with the pinhole setup, f/280 or
something? That 1ds web page said a body cap is about 50mm focal length.
Let's just say the pinhole is 1mm sq. for easy math.
  #10  
Old February 10th 06, 01:47 AM posted to rec.photo.digital.slr-systems
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Default help with aperture math

Mike Rooney wrote:

But perhaps an easier way to think of the "f stops" or aperture settings
is that each number is actually a fraction indicating the ratio of the
"circle" of your lens opening versus the maximum area of your lens mount
(not the maximum opening of your lense aperture).


No... it is a ratio of the focal length to the aperture size.

That is, on a 50mm lens, f/4 means an effective aperture opening of 12.5mm.

This is "effective" and not actual in the same way that the focal length
is effective rather than the actual, physical length of the lens -- the
"50mm" is the length the lens *would* be if it were a simple system (a
pinhole), and the aperture size is the size the aperture *would* be if it
were at the front of the lens rather than the back.

--
Jeremy |
 




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