04-07-2012, 04:51 AM
[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:' />
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:' />