[Plochmann]

I'm writing this in response to a trollish comment on another site's forum. The post was on the announced Voigtlander 17 0.95. Though many Mft users are happy with the prospect of owning one, someone made the comment that the f-stop was only so fast because the puny sensor is so small that it needs less light. <img src='http://forum.photozone.de/public/style_emoticons/<#EMO_DIR#>/huh.gif' class='bbc_emoticon' alt='' /> I think an f-stop is just an f-stop and sensor sizes are not part of the equation. But maybe I'm wrong. Can 4/3 lenses be faster because the imaging circle is smaller and therefore more focused? But then why aren't point and shoot cameras ridiculously fast? And then we have really fast ff lenses. But I can't think of any super fast medium format lenses. And I also notice that rangefinder lenses tend to be faster, does the shorter flange depth help with this f-stop equation? Any experts?

[PuxaVida]

[quote name='Plochmann' timestamp='1333691598' post='17341']

I'm writing this in response to a trollish comment on another site's forum. The post was on the announced Voigtlander 17 0.95. Though many Mft users are happy with the prospect of owning one, someone made the comment that the f-stop was only so fast because the puny sensor is so small that it needs less light. <img src='http://forum.photozone.de/public/style_emoticons/<#EMO_DIR#>/huh.gif' class='bbc_emoticon' alt='' /> I think an f-stop is just an f-stop and sensor sizes are not part of the equation. But maybe I'm wrong. Can 4/3 lenses be faster because the imaging circle is smaller and therefore more focused? But then why aren't point and shoot cameras ridiculously fast? And then we have really fast ff lenses. But I can't think of any super fast medium format lenses. And I also notice that rangefinder lenses tend to be faster, does the shorter flange depth help with this f-stop equation? Any experts?

[/quote]



I'm not an expert... But you gave me the chance to make the experts here confirm the knowledge I have <img src='http://forum.photozone.de/public/style_emoticons/<#EMO_DIR#>/smile.gif' class='bbc_emoticon' alt='' />...



To my knowledge, a fast lens is called "fast" because it allows faster shutter speeds due to the maximum aperture compared to the slower ones. And if one wants to make fast lenses one must allow more light to reach the sensor plane. And if the sensor area is smaller, then there's the advantage that with relatively smaller glasses (compared to FF for example), one can produce fast lenses just because the image circle required is smaller. But of course, all these arguments are about the fastness in terms of shutter speeds, and have nothing to do with the DoF you can have on the final image.



F-stop values are calculated as FL / diameter of the entrance pupil. There's no direct relationship with the sensor sizes.



Further reading:



http://toothwalker.org/optics/cop.html (down below the page)

http://toothwalker.org/optics/dofderivation.html#fig1



I find Mr. van Walree a very knowledgable guy... And there's also a windows executable tool he shared in his site for such optical calculations.



Serkan

[Brightcolours]

[quote name='Plochmann' timestamp='1333691598' post='17341']

I'm writing this in response to a trollish comment on another site's forum. The post was on the announced Voigtlander 17 0.95. Though many Mft users are happy with the prospect of owning one, someone made the comment that the f-stop was only so fast because the puny sensor is so small that it needs less light. <img src='http://forum.photozone.de/public/style_emoticons/<#EMO_DIR#>/huh.gif' class='bbc_emoticon' alt='' /> I think an f-stop is just an f-stop and sensor sizes are not part of the equation. But maybe I'm wrong. Can 4/3 lenses be faster because the imaging circle is smaller and therefore more focused? But then why aren't point and shoot cameras ridiculously fast? And then we have really fast ff lenses. But I can't think of any super fast medium format lenses. And I also notice that rangefinder lenses tend to be faster, does the shorter flange depth help with this f-stop equation? Any experts?

[/quote]

I do not understand how you can write a response to a to us unknown comment. Why not post a link to the "comment", so we can see you interpret it right?



4/3rds lenses can NOT be smaller because they are "more focussed", that is just nonsense. They can just be smaller because of the small imager circle. What is so hard to understand about that, though?



Now lets look at the lens in question. It is a 17mm f0.95 lens, according to the manufacturer. If one were to make a 17mm f0.95 lens for for instance 135 format, the image circle would have to be MUCH bigger. In case of such a wide angle lens, it is not a question of just making it 2x as big, as on 4/3rds it is not very wide angle at all. So... the only reason that 17mm f0.95 is possible for the money and size is indeed because of the small sensor of 4/3rds, and no other reason.



Now what does 17mm f0.95 mean? It means the lens gives (on 4/3rds) a similar field of view as a 34mm lens on 135 format, also known as "full frame" 35mm. It also means that the aperture is 17 / 0.95 = 17.9 mm. A 34mm lens with the same aperture will have a f-value (focal length / aperture ration) of: f1.9



Not very impressive, is it? Simply put: a 17mm f0.95 lens is equivalent to a 35mm f2 or f1.8 lens on full frame.



Simply put: it is easy to make a short focal length lens for 4/3rds with big focal length / aperture ratio, because on 4/3rds 17mm is not very wide angle at all. It crops so much that the image circle is very small, and the glass elements design and manufacturing is relatively cheap. It is all about the view angle captured by the sensor/film.



An f-value is an f-value. Is an f-value. Sure. However, an angle of view is not an angle of view. F-values ONLY say something about the ratio between focal length and aperture. They mean nothing more to us, the photographer. What does matter to us:

The angle of view

The size of aperture (which determines the depth of focus)

And maybe, just maybe, the amount of light reaching the sensor.



In all 3 points above, the 17mm f0.95 is equivalent to a 35mm f2 lens on FF. Or a 24mm f1.4 lens on APS-C. 35mm f2 lenses (135 format) and smaller and lighter than the voigtlander 17mm f0.95, though.



For 135 format (FF) there are also 35mm f1.4 lenses. An equivalent for 4/3rds or mft would have to be a 17mm f0.75 lens.



You also state that range finder lenses tend to be faster? I have no idea where you got that from. Which range finder lenses are so fast that there are no SLR equivalents for? The fastest I know is the very expensive Leica 50mm f1 and f0.95. For SLRs there are the Canon 50mm f1's..... I have not heard about a range finder 85mm f1.2, nor a 200mm f1.8...



Another point: there ARE compact digital cameras with very fast f-ratios. Only possible because of their super tiny sensors.

[popo]

Thanks to Brightcolours for saving me a lot of typing on this one <img src='http://forum.photozone.de/public/style_emoticons/<#EMO_DIR#>/smile.gif' class='bbc_emoticon' alt='' /> Even down to making the same typing error I do sometimes (ratio - ration).



Speaking of small sensor fast lenses, I'm still kicking myself for missing out on a Sony C-mount TV lens. I can't remember the spec but it was a 4x or 5x zoom at constant f/2 or faster. By the time I queried about the image circle coverage it was sold. It wouldn't have covered m4/3 anyway, but it would still have been fun.

[Rainer]

[quote name='Plochmann' timestamp='1333691598' post='17341']

I think an f-stop is just an f-stop and sensor sizes are not part of the equation.

[/quote]



You are correct.



A given lens has a certain maximum aperture even when absolutely no sensor or

camera is attached to it.



The maximum aperture as specified on a lens simply describes the ratio between

focal length and diameter of the entrance pupil. With that ... a lens with 17mm focal length

and an an entrance pupil of 18mm has a maximum aperture of f/0.95. Since both,

the focal length as well as the entrance pupil are attributes of the lens (and do

not change with the camera the lens is mounted to), the f/0.95 always applies to such a lens.



Nothing in this definition tells us, that the imagecricle produced by this lens will cover

Mft, crop-1.6 or even FF (135-film). The image-circle produced is another attribute of the

lens, and here, you would find that while it is possible to produce a 17mm f/0.95 for Mft, it

is extremly hard to construct one with an imagecircle covering FF.



Just my thoughts ... Rainer

[Plochmann]

[quote name='Brightcolours' timestamp='1333703652' post='17344']

I do not understand how you can write a response to a to us unknown comment. Why not post a link to the "comment", so we can see you interpret it right?



4/3rds lenses can NOT be smaller because they are "more focussed", that is just nonsense. They can just be smaller because of the small imager circle. What is so hard to understand about that, though?



Now lets look at the lens in question. It is a 17mm f0.95 lens, according to the manufacturer. If one were to make a 17mm f0.95 lens for for instance 135 format, the image circle would have to be MUCH bigger. In case of such a wide angle lens, it is not a question of just making it 2x as big, as on 4/3rds it is not very wide angle at all. So... the only reason that 17mm f0.95 is possible for the money and size is indeed because of the small sensor of 4/3rds, and no other reason.



Now what does 17mm f0.95 mean? It means the lens gives (on 4/3rds) a similar field of view as a 34mm lens on 135 format, also known as "full frame" 35mm. It also means that the aperture is 17 / 0.95 = 17.9 mm. A 34mm lens with the same aperture will have a f-value (focal length / aperture ration) of: f1.9



Not very impressive, is it? Simply put: a 17mm f0.95 lens is equivalent to a 35mm f2 or f1.8 lens on full frame.



Simply put: it is easy to make a short focal length lens for 4/3rds with big focal length / aperture ratio, because on 4/3rds 17mm is not very wide angle at all. It crops so much that the image circle is very small, and the glass elements design and manufacturing is relatively cheap. It is all about the view angle captured by the sensor/film.



An f-value is an f-value. Is an f-value. Sure. However, an angle of view is not an angle of view. F-values ONLY say something about the ratio between focal length and aperture. They mean nothing more to us, the photographer. What does matter to us:

The angle of view

The size of aperture (which determines the depth of focus)

And maybe, just maybe, the amount of light reaching the sensor.



In all 3 points above, the 17mm f0.95 is equivalent to a 35mm f2 lens on FF. Or a 24mm f1.4 lens on APS-C. 35mm f2 lenses (135 format) and smaller and lighter than the voigtlander 17mm f0.95, though.



For 135 format (FF) there are also 35mm f1.4 lenses. An equivalent for 4/3rds or mft would have to be a 17mm f0.75 lens.



You also state that range finder lenses tend to be faster? I have no idea where you got that from. Which range finder lenses are so fast that there are no SLR equivalents for? The fastest I know is the very expensive Leica 50mm f1 and f0.95. For SLRs there are the Canon 50mm f1's..... I have not heard about a range finder 85mm f1.2, nor a 200mm f1.8...



Another point: there ARE compact digital cameras with very fast f-ratios. Only possible because of their super tiny sensors.

[/quote]



You do not understand for you do not ask. But why ask something if the answer does not help you. The Buddha once said... <img src='http://forum.photozone.de/public/style_emoticons/<#EMO_DIR#>/tongue.gif' class='bbc_emoticon' alt='' /> But there is nothing to interpret, the troll only said, the f-stop is so fast because the 4/3 sensor is so puny!! It was many a month ago.

And also you've read to swiftly. I never said anything about 4/3 lenses being smaller because they are more focused. You were right, that is nonsense <img src='http://forum.photozone.de/public/style_emoticons/<#EMO_DIR#>/biggrin.gif' class='bbc_emoticon' alt='' />

The rest is fine, but I don't think technically answers my question. Why is it that a smaller imaging circle allows for a lower f-stop number? The equation is the same mathematically. Focal length divided by diameter of iris opening. This common and overly simplified equation which I've read doesn't even actually relate to modern optics, is what is confusing. I've read that you have to use a trigonometric equation when dealing with complex elements. I've also read that a lens element's Focal length is something fixed by the shape of a the lens element and determined in factory, and the iris diameter would depend on the dimensions of a lens, neither of these say anything about the image circle.

What I'm trying to say is, if I take that 17mm 0.95 lens and somehow put it on a 135 camera and then go and take a light meter and read a gray card in a dark place, and it says expose to 0.95, so I go and open it all the way up and take a picture, would not it even on 135 film expose the gray card to be 40% when properly printed? I think so. It comes down to that maybe section you mention, the exposure.

In fact, whether or not you were trying to say I was wrong, which I can not figure from the semantics of what you wrote, I think it is a matter of focus. Focusing in the most simple terms is the converging of light rays to a point. If we were to take that lens, with a 0.95 aperture, but have to spread it out over a full frame the depth of field would be a lot less and the aperture value seems to also be affected somehow(for whatever mathematical reason or equation that we do not possess). How are size of the aperture opening and image circle related, is that the missing link.

Ahhh... I'll have to reread and edit this if it makes no sense. Wife and kid are having a hoot'nanny next to me. <img src='http://forum.photozone.de/public/style_emoticons/<#EMO_DIR#>/blink.gif' class='bbc_emoticon' alt=':blink:' />

[popo]

[quote name='Plochmann' timestamp='1333774268' post='17352']

What I'm trying to say is, if I take that 17mm 0.95 lens and somehow put it on a 135 camera and then go and take a light meter and read a gray card in a dark place, and it says expose to 0.95, so I go and open it all the way up and take a picture, would not it even on 135 film expose the gray card to be 40% when properly printed? I think so. It comes down to that maybe section you mention, the exposure.[/quote]

Exposure wise, a f/0.95 lens is near enough a f/0.95 lens. But in the above scenario, you'd only get a little circle of exposed image on the sensor. Plus that lens probably wouldn't be able to give a focused image at all due to extension, but that's another story...



However, to get approximately the same image as from a m4/3 camera with 17mm f/0.95 lens on it, you would still use a 34mm f/1.9 lens on a full frame body (or nearest). Use this from the same place, giving you the same field of view, the same depth of field. You keep the shutter time the same through using a higher ISO setting, as the increased noise from a higher ISO setting is cancelled out from having lower overall noise per normalised output area from a bigger sensor in the first place.

How are size of the aperture opening and image circle related, is that the missing link.

Short answer: they're not. I'm not aware of any easy way to tell the image circle size from easily available lens specifications, other than of course what the manufacturer says it will do!





Back to the original question, why *can* lenses be smaller for a given specification if you also have a smaller image circle? Look at it this way, you have light going into the front of the lens, this gets refracted around, and you get light coming out of the back of the lens. There are lots of things that are not ideal when passing through lens elements, so you use multiple elements (sometimes of different optical material) to better correct for flaws. The better the correction, the more elements you would tend to use. A bigger image circle is harder to maintain image quality over, so would tend to requiring bigger or more elements.



Note the above only considers the difference due to image circle keeping the lens focal length and focal ratio the same. If you are trying to compare m4/3 to full frame then of course you also have different focal lengths for a given field of view, and lens mount distances factoring into the design considerations.

[Brightcolours]

[quote name='Plochmann' timestamp='1333774268' post='17352']

You do not understand for you do not ask. But why ask something if the answer does not help you. The Buddha once said... <img src='http://forum.photozone.de/public/style_emoticons/<#EMO_DIR#>/tongue.gif' class='bbc_emoticon' alt='' /> But there is nothing to interpret, the troll only said, the f-stop is so fast because the 4/3 sensor is so puny!! It was many a month ago.

And also you've read to swiftly. I never said anything about 4/3 lenses being smaller because they are more focused. You were right, that is nonsense <img src='http://forum.photozone.de/public/style_emoticons/<#EMO_DIR#>/biggrin.gif' class='bbc_emoticon' alt='' />

[/quote]

"Can 4/3 lenses be faster because the imaging circle is smaller and therefore more focused?"

[quote name='Plochmann' timestamp='1333774268' post='17352']

The rest is fine, but I don't think technically answers my question. Why is it that a smaller imaging circle allows for a lower f-stop number?

[/quote]

It does not "allow for" a lower f-value. The small sensor crops so much, that the view angle is a LOT smaller. a 17mm lens on MFT has about 54 degrees a view angle. On 135 format, the view angle of a 17mm lens is about 93%. That is a LOT wider. To design a lens with such a wide view angle and then such a big aperture will be so difficult and expensive, that that is the reason why you will never see a 17mm f0.95 FF lens. Likewise, you will never see a 8mm f0.5 lens on MFT (which would be the equivalent of thet 17mm f0.95 lens on FF).

[quote name='Plochmann' timestamp='1333774268' post='17352']

The equation is the same mathematically. Focal length divided by diameter of iris opening. This common and overly simplified equation which I've read doesn't even actually relate to modern optics, is what is confusing. I've read that you have to use a trigonometric equation when dealing with complex elements. I've also read that a lens element's Focal length is something fixed by the shape of a the lens element and determined in factory, and the iris diameter would depend on the dimensions of a lens, neither of these say anything about the image circle.

[/quote]

Aperture is not the size of the opening in the aperture mechanism, but how big the hole looks through the optics. That is meant with that it is complex to calculate. Not the size on the aperture mechanism. So, for the 17mm f0.95 the apparent aperture is 17 / 0.95 = ~17.9mm. Which is NOT the hole size in the aperture mechanism, but how big the hole appears for the light through the aperture.

[quote name='Plochmann' timestamp='1333774268' post='17352']

What I'm trying to say is, if I take that 17mm 0.95 lens and somehow put it on a 135 camera and then go and take a light meter and read a gray card in a dark place, and it says expose to 0.95, so I go and open it all the way up and take a picture, would not it even on 135 film expose the gray card to be 40% when properly printed? I think so. It comes down to that maybe section you mention, the exposure.

[/quote]

A light meter does not say expose to 0.95. You have to tell a light meter which ISO sensitivity of film you are using. A light meter will tell totally different exposure settings for ISO 25 than for ISO 800. That is a key issue if you want to understand exposure.

Take 12mp mft and 12mp FF sensors as example. They both will have a similar number of photo sensitive diodes which collect light. If we put each of these sensors under the same lens, say a 50mm f2 lens with big enough image circle to cover the biggest sensor, for the same duration of time... The photo sensitive diodes of the FF sensor will collect 4 times as much light! Why? Because they are covering 4x the surface of the mft ones. This makes the FF sensor much more sensitive than the MFT sensor. If one would rate the mft sensor ISO 100, the FF sensor would be rated ISO 400. Similar to film really, a bigger grain is more sensitive to a finer grain. Not only the pixels themselves of the FF sensor collect more light per exposure in this example, even the whole image gets formed with 4x more light than the MFT one.

And that is the tricky bit no exposure meter will tell you. Sensors have no fixed sensitivity, though. In fact, their sensitivity is never published. The signal received by the sensor can be amplified to whatever the maker wishes. This happens in two stages. One stage to "equalize" sensors, so they all will produce more or less similar exposure times to make things simple for the user, and one the user has control over, the so called ISO setting (which is not about sensitivity anymore with sensors, as you now should realize).

[quote name='Plochmann' timestamp='1333774268' post='17352']

In fact, whether or not you were trying to say I was wrong, which I can not figure from the semantics of what you wrote, I think it is a matter of focus. Focusing in the most simple terms is the converging of light rays to a point. If we were to take that lens, with a 0.95 aperture, but have to spread it out over a full frame the depth of field would be a lot less and the aperture value seems to also be affected somehow(for whatever mathematical reason or equation that we do not possess). How are size of the aperture opening and image circle related, is that the missing link.

[/quote]

You mix things up and get a totally wrong idea of what happens. Lets take a 50mm f2 with big enough image circle to cover the biggest sensor, as example.

The lens just does what the lens does. It captures light from as wide an angle as its image circle allows. How wide that is is only limited by its image circle. When we lay the FF sensor under it, NOTHING what so ever changes in how light converges (how can it, the sensor does not pull light towards it like a magnet). The only thing that happens is that the FF sensor ONLY captures the part of the projected image that gets projected on its surface. Whatever falls outside its surface gets "cropped".

When we now lay an MFT sensor in its place, the MFT sensor ONLY captures the part of the projected image that gets projected on its surface. The rest of the image gets "cropped".

Nothing got spread out more on the bigger sensor, the aperture did not get affected (still 25mm for both sensors). The focal length of the lens did not change either (remained a steady 50mm). Focus did not change either. The image circle of the lens did not change either. No matter how small a sensor we lay under that 50mm f2 lens, with image circle big enough to cover the biggest sensor, nothing changes in focus, in focal length, in aperture, in image circle of the lens.

The only thing that changes is how much of the projected image gets captured by the sensor, how much of it gets cropped.

What DOES change is the angle of view, then. For the MFT sensor, the angle of view with this 50mm lens will be about 20 degrees. The rest of the lens' angle of view remains UNUSED. For the FF sensor, the angle of view used is about 40 degrees. It collects light from a MUCH wider angle.



So, we are back from where we started. The reason you see a lens with a focal length of 17mm with an f-ratio of 0.95 on MFT lays in the very small image circle. A bigger image circle would mean MUCH larger elements having to refract light from a MUCH wider angle, making it a super complex and super heavy and expensive lens.

What you have to remember then is that focal length on its own does not mean anything. What actually matters is the angle of view captured, NOT the focal length number.

The 17mm f0.95 has to be compared to a lens on FF which provides the same angle of view for FF, and with a similar aperture. This then means it should be compared to a 35mm f1.8-f2 lens. As you might be well aware of, there is no shortage of 35mm f2 lenses for 135 format FF cameras.

[Rainer]

[quote name='Plochmann' timestamp='1333774268' post='17352']

... but I don't think technically answers my question.

Why is it that a smaller imaging circle allows for a lower f-stop number?

[/quote]



1) Lenses with a focal length relatively near to the register distance can be build

without great afford even with larger focal lengths. Typical SLR cameras have

register distances around 40-45mm ... that is why you find a cheap 50/1.4, and that

is also why a 24/1.4 or a 85/1.4 already costs more.



2) The limitations for fast lenses in the telerange are mainly given by the increasing

diameter of the first optical group. A 500mm f/4 already needs a diameter of 130mm.

The optical construction of a 500mm f/2 is not much harder, but results in a lens that

is 4 times heavier. This is obviously more demanding for the mechanical construction

as well as on the stability of the single glass elements.



3) The story is a bit different on the wide side, where the keyfactor is the increasing

angle of view. And here comes the main difference ... a 17mm lens for MFT catches a

much smaller angle of view than the 17mm lens for FF does. The larger angle of view

is one penalty ... the other is the larger register distance of SLRs compared to MFT

which requires a different construction of the lens. With that, you still could build a

17mm f/0.95 for FF, but it would be a very mediocre lens. Nobody would want a lens, that

is not sharp at all (not even in the image center) wide open.



just my 2cts ... Rainer

[Guest]

You said some things here that confuse me (i'm easily confused so no offense intended). A 0.95 lens is in fact a 0.95 lens whether it be on 4/3 camera or a 35mm camera. The thing is that if you put a lens designed for a 4/3 camera on a 35mm camera; the middle will read the same (I think); but the edges will read significantly less (I would say pitch black but that is not a given). The total amount of light if we include total area the light covered woudl be less with a lens designed for 4/3 but I believe the center would equally well light.

-

I.e, it is not the smaller circle that allows for more light; though it is certainly easier to design for a bright small circle than a large bright circle <img src='http://forum.photozone.de/public/style_emoticons/<#EMO_DIR#>/smile.gif' class='bbc_emoticon' alt='' />

-

The trick is when folks start throwing in worlds like dof, perspective, effective sensitivity and similar that the issue becomes more muddy. While a .95 lens on 4/3 allows for a faster shutter speed than a 1.4 lens on 35mm; does the larger sensor perform better with less light ? Does the larger lens allow for shallower dof (if that is your objective); Does ...

-

Anyways I believe that this is why these conversations are a rat hole. The argument usually drifts from one comparison to another rather than the direct light/shutter speed ratio and because of this drift it becomes somewhat frustrating to some people.



[quote name='Plochmann' timestamp='1333774268' post='17352']

You do not understand for you do not ask. But why ask something if the answer does not help you. The Buddha once said... <img src='http://forum.photozone.de/public/style_emoticons/<#EMO_DIR#>/tongue.gif' class='bbc_emoticon' alt='' /> But there is nothing to interpret, the troll only said, the f-stop is so fast because the 4/3 sensor is so puny!! It was many a month ago.

And also you've read to swiftly. I never said anything about 4/3 lenses being smaller because they are more focused. You were right, that is nonsense <img src='http://forum.photozone.de/public/style_emoticons/<#EMO_DIR#>/biggrin.gif' class='bbc_emoticon' alt='' />

The rest is fine, but I don't think technically answers my question. Why is it that a smaller imaging circle allows for a lower f-stop number? The equation is the same mathematically. Focal length divided by diameter of iris opening. This common and overly simplified equation which I've read doesn't even actually relate to modern optics, is what is confusing. I've read that you have to use a trigonometric equation when dealing with complex elements. I've also read that a lens element's Focal length is something fixed by the shape of a the lens element and determined in factory, and the iris diameter would depend on the dimensions of a lens, neither of these say anything about the image circle.

What I'm trying to say is, if I take that 17mm 0.95 lens and somehow put it on a 135 camera and then go and take a light meter and read a gray card in a dark place, and it says expose to 0.95, so I go and open it all the way up and take a picture, would not it even on 135 film expose the gray card to be 40% when properly printed? I think so. It comes down to that maybe section you mention, the exposure.

In fact, whether or not you were trying to say I was wrong, which I can not figure from the semantics of what you wrote, I think it is a matter of focus. Focusing in the most simple terms is the converging of light rays to a point. If we were to take that lens, with a 0.95 aperture, but have to spread it out over a full frame the depth of field would be a lot less and the aperture value seems to also be affected somehow(for whatever mathematical reason or equation that we do not possess). How are size of the aperture opening and image circle related, is that the missing link.

Ahhh... I'll have to reread and edit this if it makes no sense. Wife and kid are having a hoot'nanny next to me. <img src='http://forum.photozone.de/public/style_emoticons/<#EMO_DIR#>/blink.gif' class='bbc_emoticon' alt=':blink:' />

[/quote]