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Old Apr 2, 2007, 4:09 PM   #1
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2.8 and above I have know what the full stops were for years on film cameras, but had forgotten below.... but after looking at some old lenses on EBay, just did a quick goolgle and found

"Since f/stops can be confusing, here's the sequence of full f/stops. Remember, each of these lets in twice as much/half as much light as the adjacent numbers:

f/1.01.42.02.84.05.6811162232 "

For all intents and purposes you are never going to see a f/1.0 lens... even 1.2 are rare and expensive except in the "normal" 35-55mm prime range .

One reason I stuck this up is digital cameras can make it confusing as they usually step in half or third stop steps of both shutter and aperture. Shutter is pretty easy to calculate as its just simple doubling/halving, the the f/stops can get rather confusing though.

PS Though just noticed a pattern I don't think had ever really sunk in before.... every two stops doubles.... so I should have been able to interpolate that f/2 and 1.4 from f/4 and 2.8.... seems so much more OBVIOUS with those low ratio extra few stops sequence below 2.8...... oh well learning is a life long process.
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Old Apr 2, 2007, 6:28 PM   #2
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actually i think canon has a f1 lens


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Old Apr 2, 2007, 10:55 PM   #3
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OOPS a double post delay hiccup.
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Old Apr 2, 2007, 11:15 PM   #4
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Quote:
actually i think canon has a f1 lens
Yes.... but how many need them such as:
http://www.stevesforums.com/forums/view_topic.php?id=121144&forum_id=94

But you got the money or writeoff... why not?...you have the need. Certainly most really don't/can't.

And how many ZILLION for 1/2 a stop???

(COMPARED TO EVEN A MINOR PRO'S BUDGET?... without real (now, job cost out... soon after) need?)

As I said "for all intents.... never" not absolutely or coundn't.
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Old Apr 3, 2007, 12:02 PM   #5
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FWIW

On a 100mm lens, an aperture setting of f/4 means that the actual opening is 25mm (100/4), on a 28mm lens set to f/4 the opening will be 7mm wide (28/4). The use of these relative apertures enables you to change lenses, yet, keeping the f/stop constant, giving the same exposure for both lenses since both let in the same amount of light at f/4.

On faster lenses, you are paying for more than just the opening, you are paying for special glass formulas, and larger glass, which is the main reason for the extra cost. Professional wildlife photographers will pay as much as $10,000 (Nikon 600mm f/4 AF-S) for a high end telephoto lens that will mean getting shots they would not be able to get with lesser glass.

It is simply a business expense, nothing else.

Tom
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Old Apr 3, 2007, 6:13 PM   #6
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Maybe mention for people new to photography that any lens with an f number of 2.8 or lower, is referred to as a "fast" lens. (referring to the faster shutter speed you can use when you have more light available)


Also, check this out: http://www.visual-memory.co.uk/sk/ac/len/page1.htm

it's an f/0.7 (!!!) lens Stanley Kubrick used to film by candlelight.
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Old Apr 3, 2007, 6:36 PM   #7
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I thought I'd read that the pattern is a full stop is 1.4x the previous one... its pretty close.


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Old Apr 4, 2007, 1:32 AM   #8
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Still not an easy factor to manipuate in your head, like just halving/doubling shutter is.

The the 1.4 thing is close but not exact... but every other full stop does double.

But knowing what the full stops are... makes it easier to quickly go from the midsteps say one stop to the same mid step, without using EC or its not available.
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Old Apr 4, 2007, 5:23 PM   #9
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actually, in theory the multiplying factor should be the square root of 2, which is 1.4142135623730950488016887242097...

why? simple.

The amount of light is proportional to the area of the opening of the lens, right?

So an opening that's wice as wide, will let in twice the amount of light.

if we call the f-stop a (so we get f/a for the diameter of the opening), the area is equal to

(f/a)²
xPi

If we want to let half the light in, we need to divide that area by 2., so we get

1/2x(f/a)²xPi

if we bring the 2 inside the brackets we get:

(f/(sqrt(2)xa))²
xPi

if we state that sqrt(2)xa = a' , we see that the area of the new opening (with half the amount of light coming trough and as an f-stop a') is:


(f/a')²xPi

and that a' = sqrt(2)xa


so to let half the amount of light in, we need to multiply the f-stop with sqrt(2)


so if we do this for every stop, when we do it twice to go 2 stops down, we need to multiply twice with sqrt(2), which is of course 2




I guess it isnt exact on the aperture ring, because I think they'd rather write "4" isntead of "3,9597979746446661366447284277872"
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Old Apr 4, 2007, 6:00 PM   #10
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Whenever I want to do some calculating (estimating, really) having to do with f numbers, I like to use this site: http://www.robert-barrett.com/photo/...alculator.html
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