Question about apertures and stops and f/0.9 lenses!

foodpoison

foodpoison

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So I was always under the impression that a double or a half was a stop.

For example, a shutter speed of 1/100, a one stop change would be 1/50 or 1/200.

However is it the same for apertures?

People say, for example a difference between 70-200 f/4 and f/2.8 they say "Is the one stop change really important to you?"

I thought a one stop change would be f/5.6?

Lens question here
 
One stop below f4 is 2.8, it goes 2.8,4,5.6

the apertures work on a log scale so the numbers dont double, but the size of the opening does

It would have been much easier to just go 1,2,4,8,16 instead of using the log scale but then togs tended (and some still do) to be pretty anal about using the right terminology so we got the system we use.

Incidentally there were a few aperture scales and this is just the one that survived, much like ASA/ISO is the one that survived out of multiple speed rating systems
 
Ohhh.... okay jay that makes sense. Easy way to remember. However I'm assume all the 1/3 / 1/2 stops have been added in the digital age? i.e. 1.8, 2.2, 2.5, 4.5, 5, 6.3, 7.1, 9, 10, etc etc?
Never seen them on the 645 or the canonette :)

Also.

I was always under the impression that f/1 meant that the aperture blades opened into a full circle and so that basically related to the percentage of opening. i.e. f/1 is 100%.

I've seen lenses which are f/0.9...
HELP!

It's a rip in the space time continuum!
 
The steps between an aperture and the next 'stop' is 'the square root of 2', which is approx 1.414. The reason is that the amount of light passing through the aperture is proportional to the area of the aperture opening, and the area of the aperture doubles when its diameter increases to the square root of 2 times its original size.

So :-

-to go from f4 to one stop smaller, you multiply 4 x 1.414 = 5.656 (or approx 5.6);
to go from f5.6 to one stop smaller = 5.656 * 1.414 = 7.9184 (or approx 8)

Therefore from f4 to f8 is 2 stops, and 2 stops gives 4 x the amount of light passing through.

Or summat like that NEways
 
foodpoison said:
Ohhh.... okay jay that makes sense. Easy way to remember. However I'm assume all the 1/3 / 1/2 stops have been added in the digital age? i.e. 1.8, 2.2, 2.5, 4.5, 5, 6.3, 7.1, 9, 10, etc etc?
Never seen them on the 645 or the canonette :)

Also.

I was always under the impression that f/1 meant that the aperture blades opened into a full circle and so that basically related to the percentage of opening. i.e. f/1 is 100%.

I've seen lenses which are f/0.9...
HELP!

It's a rip in the space time continuum!
Click to expand...


No it's nothing to do with the shape of the blades, it's the area of the (near) circle.
It's just a mathematical equation. Bit of algebra if you like.

f= focal length
/= divided by

100mm prime at f/2 would have an aperture of 50mm.
I think the 50m refers to the area of the opening, not the diameter.

The astronauts that didn't landed on the moon had f/0.7 lenses.
 
cyclone said:
I think the 50m refers to the area of the opening, not the diameter.
Click to expand...

No. It is the diameter. If it was the area, the aperture scale would be linear not logarithmic.

It is logarithmic because if you double the diameter, the area increases by a factor of four - as does the amount of light passed.

f stop = focal length / aperture diameter.


Steve.
 
Steve Smith said:
No. It is the diameter. If it was the area, the aperture scale would be linear not logarithmic.

It is logarithmic because if you double the diameter, the area increases by a factor of four - as does the amount of light passed.

f stop = focal length / aperture diameter.


Steve.
Click to expand...

That's why I used the word "think".
Thanks for the clarification though :thumbs:
 
Yeah lol, plus the casing!

Divide by root 2, (1.414 ish as stated before) for even wider apertures and fatter lenses.
 
I'm sure you've seen this before.
Must be around 200mm diameter or more, compare that to the Bigma that's sat in your shop window. Same max focal length.
 
cyclone said:
I'm sure you've seen this before.
Must be around 200mm diameter or more, compare that to the Bigma that's sat in your shop window. Same max focal length.
Click to expand...

:eek: Well, I hadn't seen that before. Holy **** sauce - what a beast !!!

14kgs ...ouch :( !?

Interesting thread, BTW ;).
 
Great thread. Very interesting all that and just what this site is about. Fantastic.
 
Steve Smith said:
No. It is the diameter. If it was the area, the aperture scale would be linear not logarithmic.

It is logarithmic because if you double the diameter, the area increases by a factor of four - as does the amount of light passed.

f stop = focal length / aperture diameter.


Steve.
Click to expand...

Not logarithmic, a power series :D
 
Now someone is bound to ask the questions. If a zoom is increased from 50mm to 100mm using the same aperture, the area of the aperture increases by a factor of 4. why doesn`t the sensor /film get twice as much light as it should. If the focal length is constant and the aperture is increased from f8 to f16 the area decreases by a factor of 4. 30.67 to 7.67
 
classcams said:
Now someone is bound to ask the questions. If a zoom is increased from 50mm to 100mm using the same aperture, the area of the aperture increases by a factor of 4. why doesn`t the sensor /film get twice as much light as it should. If the focal length is constant and the aperture is increased from f8 to f16 the area decreases by a factor of 4. 30.67 to 7.67
Click to expand...
Let's see if I can paraphrase that question before I answer it.

I think what you're trying to say is this. A 50mm lens at say f/4 has an aperture of 12.5mm. A 100mm lens at f/4 has an aperture of 25mm, which means the area of the aperture is 4 times as big. So why is the exposure the same? You'd have thought that with an aperture 4 times as big, there would be 4 times a smuch light falling on the sensor, you you'd get 2 stops difference in exposure.

Is that right?

If so, here's the answer. The 100mm lens has a smaller field of view than the 50mm lens. In fact it's 1/4 the size. So the 50mm lens is gathering light from 4 times as much subject matter as the 100mm lens, and that exactly compensates for the difference in aperture sizes.
 
I've got a "table" that relates aperture to shutter speed for a given amount of light, if that's of any interest to anyone.

Basically it works by demonstrating that changing aperture one stop in one direction requires you to change your shutter speed one stop in the other if the light passing to the sensor/film is to remain constant.

I found it very useful for getting my head around manual shooting. Just don't mention changing ISO :lol:







I said don't mention ISO ! :bat:


Steve
 
Suvv said:
Not logarithmic, a power series :D
Click to expand...

Yes but there is a logarithmic relationship between f number and stops.

As you imply f number is a geometric progression of root 2

As logs reduce exponentiation to multiplication and map multiplication to addition then using Log base 2 you can derive a linear progression of stops.

The number of stops that any given f number is away from f1 is the number of times that you must raise 2 to the power of 0.5 to arrive at the given f number.

f1 to f1 -> 2^(0*0.5)
f1 to f1.4 -> 2^(1*0.5)
f1 to f2 -> 2^(2*0.5)
f1 to f2.8 -> 2^(3*0.5)

From this we can see that the increase in stops is logarithmic with respect to the increase in f number.

Disclaimer: It has been a number of decades since I last did this sort of math.
 
Steve Smith said:
No. It is the diameter. If it was the area, the aperture scale would be linear not logarithmic.
Click to expand...

Suvv said:
Not logarithmic, a power series :D
Click to expand...

DavidMarq said:
Yes but there is a logarithmic relationship between f number and stops.

As you imply f number is a geometric progression of root 2

As logs reduce exponentiation to multiplication and map multiplication to addition then using Log base 2 you can derive a linear progression of stops.

The number of stops that any given f number is away from f1 is the number of times that you must raise 2 to the power of 0.5 to arrive at the given f number.

f1 to f1 -> 2^(0*0.5)
f1 to f1.4 -> 2^(1*0.5)
f1 to f2 -> 2^(2*0.5)
f1 to f2.8 -> 2^(3*0.5)

From this we can see that the increase in stops is logarithmic with respect to the increase in f number.

Disclaimer: It has been a number of decades since I last did this sort of math.
Click to expand...


I was going to say that!!!


Steve.
 
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