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#751
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How to measure ISO
In article , Eric Stevens wrote:
Savageduck: Aaaaagh!!!! Enough already!! Yes. If he doesn't get it this time I wil give. Please just leave the thread, you probably don't want to learn, even though you at times make it seem like you do. Remember, I am the one that have supported everything I've said. With examples, math, web links to other people saying the exact same thing etc etc. You and nospam are the ones that have provided nothing but "nuh-uh" in response. No support, nothing but hot air. So, no, you have nothing more to provide in this thread and you are doing the right thing to leave. -- Sandman |
#752
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How to measure ISO
On 29 Nov 2015 10:23:22 GMT, Sandman wrote:
In article , Eric Stevens wrote: Savageduck: Aaaaagh!!!! Enough already!! Yes. If he doesn't get it this time I wil give. Please just leave the thread, you probably don't want to learn, even though you at times make it seem like you do. Remember, I am the one that have supported everything I've said. With examples, math, web links to other people saying the exact same thing etc etc. You and nospam are the ones that have provided nothing but "nuh-uh" in response. No support, nothing but hot air. So, no, you have nothing more to provide in this thread and you are doing the right thing to leave. I give up. You are a believer, not an understander. -- Regards, Eric Stevens |
#753
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How to measure ISO
On 29 Nov 2015 09:09:55 GMT, Sandman wrote:
This is a hell of a long thread in distinct chunks. About half way along a realised what has really been going on and chopped things off there. I have placed my conclusions at the end for anyone who has the stamina to geth there. In article , Eric Stevens wrote: Eric Stevens: "You have to apply the crop factor (squared) to get the same amount of light into a smaller sensor." The above statement is quite true if you really do want to get the same amount of light into the smaller sensor but I don't know why on earth you would want to do this. Sandman: To create an equal image. With not enough light, the smaller sensor will have to amplify the signal more and this creates more noise. It's just something that most people don't think about. The amount of light entering the camera depends on the area of the entrance pupil. If it's all focussed on the sensor the intensity of the light on the sensor depends on the area of the sensor. If you scale the camera up or down by a factor k the amount of light entering the camera varies as k^2. The area of the sensor also varies as k^2. In other words the area of the sensor varies as the entering light varies. Yes, but as you know, exposure is amount of light per unit area. If the exposure is the same, the amount of light collected by the sensors will be different. The smaller sensor needs more light. The smaller sensor needs more light to do what? http://www.josephjamesphotography.com/equivalence/ "The only factors in the exposure are the scene luminance, f-ratio, shutter speed, and transmissivity of the lens (note that neither sensor size nor ISO are factors in exposure)." Which means that in order to create an equal *image*, you have to adjust the *exposure* to fit the size of the sensor. How on earth can you say that when you have just quoted "note that neither sensor size nor ISO are factors in exposure." What do you mean by "equal image"? Equal in what way? Eric Stevens: The truth of the matter is that if you take a camera and scale it either up or down, the lens f/value stays the same and the level of illumination on the sensor stays the same. If the sensitivity of the sensor has not changed then the invariant level of illumination of the sensor means that exposure time always remains the same. Sandman: But with the same f-stop, the smaller lens gets less light. So you either need a larger f-stop or longer exposure time to match the amount of light. This means you adjust the ISO down to the crop factor square. Who cares about the sensor getting less light? Those that care about noise, since less light means more amplification and more noise. Why should there be more noise? Providing everything has been kept in proportion as the camera has been scaled up or down, the light intensity on the sensor remains the same irrespective of the size of the sensor. Lighting of the sensor remains te same. Sensor remains the same ... The point is that the amount of light falling on each square millimetre (or each square inch) remains exactly the same. As far as the sensor is concerned there has been no change. Not sure what supposed "change" you are talking about here? Changing the size of the sensor means that comparing "exposure" falls apart. The behaviour of a small patch of sensor is not affected by how much more sensor there is around it. Certainly the sensitivity is not affected. As I almost said "the point is that the amount of light falling on each square millimetre remains exactly the same. As far as the sensor is concerned there has been no change." Certainly it needs neither more nor less amplification than it did before". http://www.josephjamesphotography.com/equivalence/#8 "For a given scene, perspective, and framing, the total light depends only on the aperture diameter and shutter speed (as opposed to the f-ratio and shutter speed for exposure). That's fine, if you are talking about a porthole letting light into a room. But we are talking about a lense which takes light from a series of point sources in limited external area and focusses that light onto a series of point images on the sensor. Fully equivalent images on different formats will have the same brightness and be created with the same total amount of light. Not so, in the case of a camera. Thus, the same total amount of light on sensors with different areas will necessarily result in different exposures on different formats, and it is for this reason that exposure is a meaningless measure in cross-format comparisons." Having started with an incorrect premise he has finished with an incorrect conclusion. Eric Stevens: Alternatively, if you have an invariant f/ value, an invariant exposure time and an ivariant level of illumination of the sensor then it follows that the ISO has remained the same. That is, ISO value is not a function of sensor size. Sandman: Of course it isn't. But with the same exposure, sensor amplification is a function of the sensor size, so while the ISO number are similar, the sensor amplification is not. Not so. Go back to my statement above and think about it. You will conclude that amplification has nothing to do with sensor size. http://www.josephjamesphotography.com/equivalence "The exposure (light per area on the sensor) at f/2.8 1/100 ISO 100 is 4x as great as f/5.6 1/100 ISO 400 for a given scene luminance, regardless of the focal length or the sensor size. However, the brightness for the two photos will be the same since the 4x lower exposure is brightened 4x as much by the higher ISO setting. If the sensor that the f/5.6 photo was recorded on has 4x the area as the sensor as the f/2.8 photo (e.g. FF vs mFT), then the same total amount of light will fall on both sensors, which will result in the same noise for equally efficient sensors" Same amount of total light = same amount of amplification = same amount of noise. I assume you have been using the terms in the same way as James, although you have not previously made that clear. His important definitions a "Exposu The total light per area (photons / mm˛) that falls on the sensor while the shutter is open, which is usually expressed as the product of the illuminance of the sensor and the time the shutter is open (lux · seconds). The only factors in the exposure are the scene luminance, t-stop (where the f-ratio is often a good approximation for the t-stop), and the shutter speed (note that neither sensor size nor ISO are factors in exposure)." Here is a trap: "Brightness: The brightness of an image (what people usually think of as "exposure" -- same units as exposure): Brightness = Exposure x Amplification." Most people would assume that brightness = "The total light per area (photons / mm˛) that falls on the sensor while the shutter is open..." In fact James has been using the term as above, where 'magnification' is a factor of arbitrary value. "Brightness" in this sense is the value in a data array external to the sensor. (it might be part of the sensor but it it is part of the electronics processing the output of the sensor.) "Total Light: The total number of photons that falls on the sensor (lumen·seconds, or, equivalently, photons): Total Light = Exposure x Effective Sensor Area." --- long tail snipped (Savageduck heaves a sigh of relief) --- I've finally realised where the problem lies in all this. A. The terms as defined above are not entirely the same as those used in the wider world. That this discussion was based on certain exact definitions was not made clear from the outset. As James wrote: "Many of the misunderstandings come from people using different definitions for the same words. In particular, "f-ratio" is often confused with "aperture", and "exposure" is confused with "brightness" and "total light". The importance of these distinctions is often overlooked or simply not understood, ... " B. It has been assumed that noise is inversely proportional to photons received: the more photons the less noise. To describe this as a hairy approximation is a gross understatement, but no matter. C. All this pseudo-mathematics is centred about an attempt to set out the rules relating the properties of cameras of different sizes which produce not only the the equivalent optical geometry but also equal image noise. D. If sensors on smaller cameras are required to accept (say) four times as much light as a sensor in the larger camera, they will have to have larger 'buckets' in the individual photosites. In the real world their buckets will not be four times as large so will be overflowed, with resulting highlight burnout. E. Nobody operates a camera so as to obtain constant noise so the conclusions, while interesting, are irrelevant to the real world. I've had more than enough of this. Now that I've finally worked out what Sandman has been talking about I'm happy to leave him to go his own way and achieve his own state. -- Regards, Eric Stevens |
#754
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How to measure ISO
In article , Eric Stevens wrote:
Savageduck: Aaaaagh!!!! Enough already!! Eric Stevens: Yes. If he doesn't get it this time I wil give. Sandman: Please just leave the thread, you probably don't want to learn, even though you at times make it seem like you do. Remember, I am the one that have supported everything I've said. With examples, math, web links to other people saying the exact same thing etc etc. You and nospam are the ones that have provided nothing but "nuh-uh" in response. No support, nothing but hot air. So, no, you have nothing more to provide in this thread and you are doing the right thing to leave. I give up. Finally! -- Sandman |
#755
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How to measure ISO
In article , Eric Stevens wrote:
Eric Stevens: "You have to apply the crop factor (squared) to get the same amount of light into a smaller sensor." The above statement is quite true if you really do want to get the same amount of light into the smaller sensor but I don't know why on earth you would want to do this. Sandman: To create an equal image. With not enough light, the smaller sensor will have to amplify the signal more and this creates more noise. It's just something that most people don't think about. Eric Stevens: The amount of light entering the camera depends on the area of the entrance pupil. If it's all focussed on the sensor the intensity of the light on the sensor depends on the area of the sensor. If you scale the camera up or down by a factor k the amount of light entering the camera varies as k^2. The area of the sensor also varies as k^2. In other words the area of the sensor varies as the entering light varies. Sandman: Yes, but as you know, exposure is amount of light per unit area. If the exposure is the same, the amount of light collected by the sensors will be different. The smaller sensor needs more light. The smaller sensor needs more light to do what? To create an equivalent exposure. Sandman: http://www.josephjamesphotography.com/equivalence "The only factors in the exposure are the scene luminance, f-ratio, shutter speed, and transmissivity of the lens (note that neither sensor size nor ISO are factors in exposure)." Which means that in order to create an equal *image*, you have to adjust the *exposure* to fit the size of the sensor. How on earth can you say that when you have just quoted "note that neither sensor size nor ISO are factors in exposure." What do you mean by "equal image"? Equal in what way? There's a difference between "exposure" and "equal exposure". An exposure is a given amount of light in a given unit area. With a smaller sensor, you create an equal image by making an equal exposure, meaning more light on that smaller area. Eric Stevens: The truth of the matter is that if you take a camera and scale it either up or down, the lens f/value stays the same and the level of illumination on the sensor stays the same. If the sensitivity of the sensor has not changed then the invariant level of illumination of the sensor means that exposure time always remains the same. Sandman: But with the same f-stop, the smaller lens gets less light. So you either need a larger f-stop or longer exposure time to match the amount of light. This means you adjust the ISO down to the crop factor square. Eric Stevens: Who cares about the sensor getting less light? Sandman: Those that care about noise, since less light means more amplification and more noise. Why should there be more noise? Amplification leads no noise. Always. Providing everything has been kept in proportion as the camera has been scaled up or down, the light intensity on the sensor remains the same irrespective of the size of the sensor. Lighting of the sensor remains te same. Sensor remains the same ... If the amount of total light is the same, the exposure is equal, but not the *same* exposure (see above). Using the same amount of light, you adjust the ISO down by the crop factor squared, which means the signal is amplified the same amount (give or take, sensor tech). Eric Stevens: The point is that the amount of light falling on each square millimetre (or each square inch) remains exactly the same. As far as the sensor is concerned there has been no change. Sandman: Not sure what supposed "change" you are talking about here? Changing the size of the sensor means that comparing "exposure" falls apart. The behaviour of a small patch of sensor is not affected by how much more sensor there is around it. Of course not, but on a smaller sensor, that "small patch" is a part of the whole image that is rendered by the camera. On a larger sensor, the same region of the resulting image is made out of a larger "patch" of the sensor, which receives more light. So there is more signal going in to that part of the sensor that makes out that part of the image. Certainly the sensitivity is not affected. No, but to create an equally bright image, the smaller sensor needs to amplify the signal more, since it has received less light to being with. I'm not sure how many times I've said this in various forms. As I almost said "the point is that the amount of light falling on each square millimetre remains exactly the same. As far as the sensor is concerned there has been no change." Certainly it needs neither more nor less amplification than it did before". Take it to the extreme, then. You have two sensors. One is 100x100mm and one is 10x10mm, both are 10MP sensors. Using the same exposure, the smaller sensor will receive one hundredth of the "signal" as the larger sensor. Both generate a 10MP image out of this signal, but the smaller sensor can't possibly make it as bright as the larger sensor without amplification. And that amplification creates noise. Sandman: http://www.josephjamesphotography.com/equivalence/#8 "For a given scene, perspective, and framing, the total light depends only on the aperture diameter and shutter speed (as opposed to the f-ratio and shutter speed for exposure). That's fine, if you are talking about a porthole letting light into a room. But we are talking about a lense which takes light from a series of point sources in limited external area and focusses that light onto a series of point images on the sensor. Yeah? Do you have a point? Sandman: Fully equivalent images on different formats will have the same brightness and be created with the same total amount of light. Not so, in the case of a camera. How so? Be specific. Sandman: Thus, the same total amount of light on sensors with different areas will necessarily result in different exposures on different formats, and it is for this reason that exposure is a meaningless measure in cross-format comparisons." Having started with an incorrect premise he has finished with an incorrect conclusion. Haha. Eric Stevens: Alternatively, if you have an invariant f/ value, an invariant exposure time and an ivariant level of illumination of the sensor then it follows that the ISO has remained the same. That is, ISO value is not a function of sensor size. Sandman: Of course it isn't. But with the same exposure, sensor amplification is a function of the sensor size, so while the ISO number are similar, the sensor amplification is not. Eric Stevens: Not so. Go back to my statement above and think about it. You will conclude that amplification has nothing to do with sensor size. Sandman: http://www.josephjamesphotography.com/equivalence "The exposure (light per area on the sensor) at f/2.8 1/100 ISO 100 is 4x as great as f/5.6 1/100 ISO 400 for a given scene luminance, regardless of the focal length or the sensor size. However, the brightness for the two photos will be the same since the 4x lower exposure is brightened 4x as much by the higher ISO setting. If the sensor that the f/5.6 photo was recorded on has 4x the area as the sensor as the f/2.8 photo (e.g. FF vs mFT), then the same total amount of light will fall on both sensors, which will result in the same noise for equally efficient sensors" Same amount of total light = same amount of amplification = same amount of noise. I assume you have been using the terms in the same way as James, although you have not previously made that clear. I have said so many many many many many times. His important definitions a "Exposu The total light per area (photons / mm˛) that falls on the sensor while the shutter is open, which is usually expressed as the product of the illuminance of the sensor and the time the shutter is open (lux · seconds). The only factors in the exposure are the scene luminance, t-stop (where the f-ratio is often a good approximation for the t-stop), and the shutter speed (note that neither sensor size nor ISO are factors in exposure)." Indeed. Here is a trap: "Brightness: The brightness of an image (what people usually think of as "exposure" -- same units as exposure): Brightness Exposure x Amplification." Exactly like I've said. Most people would assume that brightness = "The total light per area (photons / mm˛) that falls on the sensor while the shutter is open..." In fact James has been using the term as above, where 'magnification' is a factor of arbitrary value. "Brightness" in this sense is the value in a data array external to the sensor. (it might be part of the sensor but it it is part of the electronics processing the output of the sensor.) "Brightness" is in the resulting image, like he said. Brightness = (Exposure + amplification). 100% what I've said this entire time. "Total Light: The total number of photons that falls on the sensor (lumen·seconds, or, equivalently, photons): Total Light = Exposure x Effective Sensor Area." I.e. exactly what I've said. I've finally realised where the problem lies in all this. I won't hold my breath. A. The terms as defined above are not entirely the same as those used in the wider world. That this discussion was based on certain exact "Many of the misunderstandings come from people using different definitions for the same words. In particular, "f-ratio" is often confused with "aperture", and "exposure" is confused with "brightness" and "total light". The importance of these distinctions is often overlooked or simply not understood, ... " Yes, these are some of the terms you and nospam have repeatedly misunderstood, I agree. B. It has been assumed that noise is inversely proportional to photons received: the more photons the less noise. To describe this as a hairy approximation is a gross understatement, but no matter. Noise has nothing to do with amount of photons. It has everything to do with amplification. C. All this pseudo-mathematics is centred about an attempt to set out the rules relating the properties of cameras of different sizes which produce not only the the equivalent optical geometry but also equal image noise. I.e. creating equivalent images using different sensor sizes. D. If sensors on smaller cameras are required to accept (say) four times as much light as a sensor in the larger camera, they will have to have larger 'buckets' in the individual photo sites. This is false unless you're playing with highly overexposed images. A normally exposed image, using the settings I have repeatedly mentioned, will not saturate the photo sites of a MFT camera. In the real world their buckets will not be four times as large so will be overflowed, with resulting highlight burnout. This is a false assumption. We are talking about a MFT sensor that has amplified its signal four times as much as a FF sensor to create a "normal" photo. Lowering the ISO and opening up the aperture will result in the exact same photo, only with the exact same noise as the FF camera. E. Nobody operates a camera so as to obtain constant noise so the conclusions, while interesting, are irrelevant to the real world. It's very relevant to a number of things: 1. The myth that smaller sensors are noisier, they aren't. 2. The myth the "ISO" is a certain "sensitivity" (amplification), it is not. 3. The problem with using "35mm equivalent" terms for the focal length but not for aperture to mislead users. 4. The myth that MFT sensors have larger DOF (a part of #3) I've had more than enough of this. Now that I've finally worked out what Sandman has been talking about I'm happy to leave him to go his own way and achieve his own state. No one is happier than me, believe me. And even if you didn't really understand all of it, you came a long way, and have restored some of your maturity in the process. Well done. -- Sandman |
#756
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How to measure ISO
On 29 Nov 2015 22:42:49 GMT, Sandman wrote:
In article , Eric Stevens wrote: Eric Stevens: "You have to apply the crop factor (squared) to get the same amount of light into a smaller sensor." The above statement is quite true if you really do want to get the same amount of light into the smaller sensor but I don't know why on earth you would want to do this. Sandman: To create an equal image. With not enough light, the smaller sensor will have to amplify the signal more and this creates more noise. It's just something that most people don't think about. Eric Stevens: The amount of light entering the camera depends on the area of the entrance pupil. If it's all focussed on the sensor the intensity of the light on the sensor depends on the area of the sensor. If you scale the camera up or down by a factor k the amount of light entering the camera varies as k^2. The area of the sensor also varies as k^2. In other words the area of the sensor varies as the entering light varies. Sandman: Yes, but as you know, exposure is amount of light per unit area. If the exposure is the same, the amount of light collected by the sensors will be different. The smaller sensor needs more light. The smaller sensor needs more light to do what? To create an equivalent exposure. Sandman: http://www.josephjamesphotography.com/equivalence "The only factors in the exposure are the scene luminance, f-ratio, shutter speed, and transmissivity of the lens (note that neither sensor size nor ISO are factors in exposure)." Which means that in order to create an equal *image*, you have to adjust the *exposure* to fit the size of the sensor. How on earth can you say that when you have just quoted "note that neither sensor size nor ISO are factors in exposure." What do you mean by "equal image"? Equal in what way? There's a difference between "exposure" and "equal exposure". An exposure is a given amount of light in a given unit area. With a smaller sensor, you create an equal image by making an equal exposure, meaning more light on that smaller area. I'm breaking my vows but I sense progress. You may be saying that the amount of information carried to the camera depends upon the number of photons which reach it. From this it follows that the small sensor has to handle the same amount of light as the larger sensor. Have I got it right? Eric Stevens: The truth of the matter is that if you take a camera and scale it either up or down, the lens f/value stays the same and the level of illumination on the sensor stays the same. If the sensitivity of the sensor has not changed then the invariant level of illumination of the sensor means that exposure time always remains the same. Sandman: But with the same f-stop, the smaller lens gets less light. So you either need a larger f-stop or longer exposure time to match the amount of light. This means you adjust the ISO down to the crop factor square. Eric Stevens: Who cares about the sensor getting less light? Sandman: Those that care about noise, since less light means more amplification and more noise. Why should there be more noise? Amplification leads no noise. Always. Noise is always generated in a camera. The camera deals with this, in part, by setting a noise floor. It sets a minimum level below which it will ignore whatever is emerging from the sensor and it's amplification. Of course this can never be entirely successful. Providing everything has been kept in proportion as the camera has been scaled up or down, the light intensity on the sensor remains the same irrespective of the size of the sensor. Lighting of the sensor remains te same. Sensor remains the same ... If the amount of total light is the same, the exposure is equal, but not the *same* exposure (see above). It's this sort of statement which makes it hard to follow exactly what you are saying. James defines exposure as (see below) "The total light per area (photons / mm˛) that falls on the sensor while the shutter is open". In the scaling up/down situation, if the total light is the same, the total light per area (photons / mm˛) cannot be the same. There is something wrong with your terminology. Using the same amount of light, you adjust the ISO down by the crop factor squared, which means the signal is amplified the same amount (give or take, sensor tech). I understand you better now that I know that by 'brightness' you don't mean the density of the light which falls on the sensor, but instead the amplified output of the sensor. At least James does. Do you mean that also? Eric Stevens: The point is that the amount of light falling on each square millimetre (or each square inch) remains exactly the same. As far as the sensor is concerned there has been no change. Sandman: Not sure what supposed "change" you are talking about here? Changing the size of the sensor means that comparing "exposure" falls apart. The behaviour of a small patch of sensor is not affected by how much more sensor there is around it. Of course not, but on a smaller sensor, that "small patch" is a part of the whole image that is rendered by the camera. On a larger sensor, the same region of the resulting image is made out of a larger "patch" of the sensor, which receives more light. So there is more signal going in to that part of the sensor that makes out that part of the image. There comes a point where the message has got through, no matter how many extra photons you have poured through. Certainly the sensitivity is not affected. No, but to create an equally bright image, the smaller sensor needs to amplify the signal more, since it has received less light to being with. I'm not sure how many times I've said this in various forms. And the statement seemed meaningless, mad or daft before I realised what you meant by brightness. As I almost said "the point is that the amount of light falling on each square millimetre remains exactly the same. As far as the sensor is concerned there has been no change." Certainly it needs neither more nor less amplification than it did before". Take it to the extreme, then. You have two sensors. One is 100x100mm and one is 10x10mm, both are 10MP sensors. Using the same exposure, the smaller sensor will receive one hundredth of the "signal" as the larger sensor. Both generate a 10MP image out of this signal, but the smaller sensor can't possibly make it as bright as the larger sensor without amplification. And that amplification creates noise. It is likely that even at their native ISO both sensors will require amplificaton. It is also a big jump to assume that noise is proportional to amplification, but never mind. Now that the terminology is being defined I am beginning to understand you. Sandman: http://www.josephjamesphotography.com/equivalence/#8 "For a given scene, perspective, and framing, the total light depends only on the aperture diameter and shutter speed (as opposed to the f-ratio and shutter speed for exposure). That's fine, if you are talking about a porthole letting light into a room. But we are talking about a lense which takes light from a series of point sources in limited external area and focusses that light onto a series of point images on the sensor. Yeah? Do you have a point? Providing everything has been scaled correctly, the intensity of that point of light on the sensor remains the same. The image is constructed by the array of those points of light on the sensor. Despite the fact that there are fewer photons finding the way to the smaller sensor the intensity of the light (what I thought you meant by brightness) remains the same across all sizes of cameras. If that is the case, and I believe it is, all sizes of sensors will require the same amount of amplification to reach the same brightness as defined by James. Sandman: Fully equivalent images on different formats will have the same brightness and be created with the same total amount of light. Not so, in the case of a camera. How so? Be specific. I hope I just did. However I expect that my explanation will not satisfy James' definition of equivalence. Sandman: Thus, the same total amount of light on sensors with different areas will necessarily result in different exposures on different formats, and it is for this reason that exposure is a meaningless measure in cross-format comparisons." Having started with an incorrect premise he has finished with an incorrect conclusion. Haha. I think the chain of logic will have to be reworked with some care given to definitions if this is going to be resolved. Eric Stevens: Alternatively, if you have an invariant f/ value, an invariant exposure time and an ivariant level of illumination of the sensor then it follows that the ISO has remained the same. That is, ISO value is not a function of sensor size. Sandman: Of course it isn't. But with the same exposure, sensor amplification is a function of the sensor size, so while the ISO number are similar, the sensor amplification is not. Eric Stevens: Not so. Go back to my statement above and think about it. You will conclude that amplification has nothing to do with sensor size. Sandman: http://www.josephjamesphotography.com/equivalence "The exposure (light per area on the sensor) at f/2.8 1/100 ISO 100 is 4x as great as f/5.6 1/100 ISO 400 for a given scene luminance, regardless of the focal length or the sensor size. However, the brightness for the two photos will be the same since the 4x lower exposure is brightened 4x as much by the higher ISO setting. If the sensor that the f/5.6 photo was recorded on has 4x the area as the sensor as the f/2.8 photo (e.g. FF vs mFT), then the same total amount of light will fall on both sensors, which will result in the same noise for equally efficient sensors" Same amount of total light = same amount of amplification = same amount of noise. I assume you have been using the terms in the same way as James, although you have not previously made that clear. I have said so many many many many many times. Yes you have said that about total light many many times but you have never explained it. At several points in this thread I variously enquired after or taken you to task over the terms you have been using and you have responded by accusing me of semantic trolling and similar. It wasn't semantic trolling: it was me trying to find out *exactly* what you were trying to tell me. His important definitions a "Exposu The total light per area (photons / mm˛) that falls on the sensor while the shutter is open, which is usually expressed as the product of the illuminance of the sensor and the time the shutter is open (lux · seconds). The only factors in the exposure are the scene luminance, t-stop (where the f-ratio is often a good approximation for the t-stop), and the shutter speed (note that neither sensor size nor ISO are factors in exposure)." Indeed. Here is a trap: "Brightness: The brightness of an image (what people usually think of as "exposure" -- same units as exposure): Brightness Exposure x Amplification." Exactly like I've said. But you have never explained that by 'brightness' you meant the output of the amplifier. Most people would assume that brightness = "The total light per area (photons / mm˛) that falls on the sensor while the shutter is open..." In fact James has been using the term as above, where 'magnification' is a factor of arbitrary value. "Brightness" in this sense is the value in a data array external to the sensor. (it might be part of the sensor but it it is part of the electronics processing the output of the sensor.) "Brightness" is in the resulting image, like he said. Brightness = (Exposure + amplification). 100% what I've said this entire time. The only place in the camera where there is an image is on the sensor. After the sensor is scanned the image is converted into a stream of data which takes whatever form the camera designer has thought of. "Total Light: The total number of photons that falls on the sensor (lumen·seconds, or, equivalently, photons): Total Light = Exposure x Effective Sensor Area." I.e. exactly what I've said. I think you will find you have talked about 'total light' only once, im message I've finally realised where the problem lies in all this. I won't hold my breath. A. The terms as defined above are not entirely the same as those used in the wider world. That this discussion was based on certain exact You have missed the end of the sentence which originally read "That this discussion was based on certain exact definitions was not made clear from the outset. As James wrote:" "Many of the misunderstandings come from people using different definitions for the same words. In particular, "f-ratio" is often confused with "aperture", and "exposure" is confused with "brightness" and "total light". The importance of these distinctions is often overlooked or simply not understood, ... " Yes, these are some of the terms you and nospam have repeatedly misunderstood, I agree. And you consistently failed to explain *exactly* what you meant to many more people than me and nospam. B. It has been assumed that noise is inversely proportional to photons received: the more photons the less noise. To describe this as a hairy approximation is a gross understatement, but no matter. Noise has nothing to do with amount of photons. It has everything to do with amplification. Which depends on the number of photons received. C. All this pseudo-mathematics is centred about an attempt to set out the rules relating the properties of cameras of different sizes which produce not only the the equivalent optical geometry but also equal image noise. I.e. creating equivalent images using different sensor sizes. And lens sizes also. D. If sensors on smaller cameras are required to accept (say) four times as much light as a sensor in the larger camera, they will have to have larger 'buckets' in the individual photo sites. This is false unless you're playing with highly overexposed images. A normally exposed image, using the settings I have repeatedly mentioned, will not saturate the photo sites of a MFT camera. I haven't got a MFT camera but one of these days I will try comparing the Nikon D750 with my wife's Canon G12 (7.6mm x 5.7mm). In the real world their buckets will not be four times as large so will be overflowed, with resulting highlight burnout. This is a false assumption. We are talking about a MFT sensor that has amplified its signal four times as much as a FF sensor to create a "normal" photo. Lowering the ISO and opening up the aperture will result in the exact same photo, only with the exact same noise as the FF camera. Here is something you can't prove. It is probably an assumption. E. Nobody operates a camera so as to obtain constant noise so the conclusions, while interesting, are irrelevant to the real world. It's very relevant to a number of things: 1. The myth that smaller sensors are noisier, they aren't. 2. The myth the "ISO" is a certain "sensitivity" (amplification), it is not. The concept that ISO is a certain amplification is a myth (I hope). 3. The problem with using "35mm equivalent" terms for the focal length but not for aperture to mislead users. Look up the definition of f/ number (and think about it). 4. The myth that MFT sensors have larger DOF (a part of #3) I've had more than enough of this. Now that I've finally worked out what Sandman has been talking about I'm happy to leave him to go his own way and achieve his own state. No one is happier than me, believe me. And even if you didn't really understand all of it, you came a long way, and have restored some of your maturity in the process. Well done. Thank you for those kind words. Now, please go and peddle your nostrums somewhere else. I'm not buying. -- Regards, Eric Stevens |
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How to measure ISO
In article , Eric Stevens wrote:
Sandman: There's a difference between "exposure" and "equal exposure". An exposure is a given amount of light in a given unit area. With a smaller sensor, you create an equal image by making an equal exposure, meaning more light on that smaller area. I'm breaking my vows but I sense progress. You may be saying that the amount of information carried to the camera depends upon the number of photons which reach it. From this it follows that the small sensor has to handle the same amount of light as the larger sensor. Have I got it right? Pretty much. In order to create an equal exposure, the smaller sensor needs the same amount of total light (equal exposure), not the same amount of light per unit area (same exposure). Sandman: Those that care about noise, since less light means more amplification and more noise. Eric Stevens: Why should there be more noise? Sandman: Amplification leads no noise. Always. Noise is always generated in a camera. As a product of the signal amplification, i.e. the ISO setting. Eric Stevens: Providing everything has been kept in proportion as the camera has been scaled up or down, the light intensity on the sensor remains the same irrespective of the size of the sensor. Lighting of the sensor remains te same. Sensor remains the same ... Sandman: If the amount of total light is the same, the exposure is equal, but not the *same* exposure (see above). It's this sort of statement which makes it hard to follow exactly what you are saying. James defines exposure as (see below) "The total light per area (photons / mm˛) that falls on the sensor while the shutter is open". In the scaling up/down situation, if the total light is the same, the total light per area (photons / mm˛) cannot be the same. Exactly. "Same exposure" = "Same amount of light pr unit area" "Equal exposure" = "Same amount of light on whole sensor" Making an equal exposure means you're not using the same light per unit area. This is what I have been saying the entire time. You need to give the smaller sensor the same amount of *total light*, not the same exposure. This is the same exposu FF: 1/250, f5.6, ISO 800 MFT: 1/250, f5.6, ISO 800 That will create two images that are equally bright. But they are ONLY equally bright because the MFT sensor has amplified its signal MORE than the FF sensor has, since it has received less light. This is an equal exposu FF: 1/250, f5.6, ISO 800 MFT: 1/250, f2.8, ISO 200 Now we have given the MFT sensor the same amount of *total* light, not light per unit area. This means that during the exposure, it has received as many photons totally as the FF sensor (using an equivalent aperture). To balance this, we have to adjust the ISO by the crop factor squared, to amplify the signal LESS. This means that at ISO 200, the MFT sensor is amplifying the signal equally to the FF sensor at ISO 800. This tells us a couple of things: 1. "ISO" says nothing about sensor amplification (called "sensitivity" to most people) when comparing between sensor formats 2. Smaller sensors aren't noisier, unlike what most people think. They just need the same amount of light to create an equal image Sandman: Using the same amount of light, you adjust the ISO down by the crop factor squared, which means the signal is amplified the same amount (give or take, sensor tech). I understand you better now that I know that by 'brightness' you don't mean the density of the light which falls on the sensor, but instead the amplified output of the sensor. At least James does. Do you mean that also? I have only ever talked about brightness in relation to the generated image. I.e. I have mentioned several times that the ISO setting is not so much a way to know how sensitive (amplified) the sensor is, but rather what brightness level to expect. So yes, "Brightness" has nothing to do with exposure (other than being a result of it). It is how bright the image created by the camera is. Brightness is exposure + amplification. Eric Stevens: The point is that the amount of light falling on each square millimetre (or each square inch) remains exactly the same. As far as the sensor is concerned there has been no change. Sandman: Not sure what supposed "change" you are talking about here? Changing the size of the sensor means that comparing "exposure" falls apart. Eric Stevens: The behaviour of a small patch of sensor is not affected by how much more sensor there is around it. Sandman: Of course not, but on a smaller sensor, that "small patch" is a part of the whole image that is rendered by the camera. On a larger sensor, the same region of the resulting image is made out of a larger "patch" of the sensor, which receives more light. So there is more signal going in to that part of the sensor that makes out that part of the image. There comes a point where the message has got through, no matter how many extra photons you have poured through. Yes, but the fewer the photons, the "weaker" the message, and without more amplification, it will not be as bright. Eric Stevens: Certainly the sensitivity is not affected. Sandman: No, but to create an equally bright image, the smaller sensor needs to amplify the signal more, since it has received less light to being with. I'm not sure how many times I've said this in various forms. And the statement seemed meaningless, mad or daft before I realised what you meant by brightness. Well, good we got that sorted out, then. Eric Stevens: As I almost said "the point is that the amount of light falling on each square millimetre remains exactly the same. As far as the sensor is concerned there has been no change." Certainly it needs neither more nor less amplification than it did before". Sandman: Take it to the extreme, then. You have two sensors. One is 100x100mm and one is 10x10mm, both are 10MP sensors. Using the same exposure, the smaller sensor will receive one hundredth of the "signal" as the larger sensor. Both generate a 10MP image out of this signal, but the smaller sensor can't possibly make it as bright as the larger sensor without amplification. And that amplification creates noise. It is likely that even at their native ISO both sensors will require amplification. No, that's not "likely" at all. That would only be "likely" if you knew exactly how many photons that were transmitted, which wasn't a factor in the input. It is also a big jump to assume that noise is proportional to amplification, You have different kind of noise, of course. Usually grouped together by "photon noise" (different kind of noise from how the light hits the sensor, travels through the lens etc) and "read noise" (noise created by the sensor and hardware). But even though photon noise affect the end result, the read noise is usually a lot higher and affects the resulting image much more. Sandman: Sandman: http://www.josephjamesphotography.com/equivalence/#8 "For a given scene, perspective, and framing, the total light depends only on the aperture diameter and shutter speed (as opposed to the f-ratio and shutter speed for exposure). Eric Stevens: That's fine, if you are talking about a porthole letting light into a room. But we are talking about a lense which takes light from a series of point sources in limited external area and focusses that light onto a series of point images on the sensor. Sandman: Yeah? Do you have a point? Providing everything has been scaled correctly, the intensity of that point of light on the sensor remains the same. But that's the thing, using the same exposure *doesn't* scale everything correctly. Using a 25mm/f1.4 lens on MFT is usually called "equivalent" to a 50mm on FF, but it's actually equivalent to a 50mm/f2.8 if scaled "correctly". So: FF: 50mm, f2.8, 1/250, ISO 800 MFT: 25mm, f2.8, 1/250, ISO 800 Will result in the *same* exposure, i.e the same amount of light per unit area. But: FF: 50mm, f2.8, 1/250, ISO 800 MFT: 25mm, f1.4, 1/250, ISO 200 Will result in an *equal* exposure, meaning you get the same amount of total light through the lens on the sensor, you get the same depth of field, you get the same signal amplification and the same field of view (i.e. collecting the same scene). This is what I have been saying for the last two weeks. The image is constructed by the array of those points of light on the sensor. Despite the fact that there are fewer photons finding the way to the smaller sensor the intensity of the light (what I thought you meant by brightness) remains the same across all sizes of cameras. *only* if you adjust the exposure as outlined above. If that is the case, and I believe it is, all sizes of sensors will require the same amount of amplification to reach the same brightness as defined by James. Absolutely, which is what I've said this entire time. Eric Stevens: I assume you have been using the terms in the same way as James, although you have not previously made that clear. Sandman: I have said so many many many many many times. Yes you have said that about total light many many times but you have never explained it. At several points in this thread I variously enquired after or taken you to task over the terms you have been using and you have responded by accusing me of semantic trolling and similar. It wasn't semantic trolling: it was me trying to find out *exactly* what you were trying to tell me. I "accused" you of semantic trolling when you cut out the context and claimed that you can't change the sensitivity of the sensor. Which, while technically true, was beside the point. A lot of people call the ISO setting on the camera as a way to change the sensor sensitivity. If you hadn't cut out the context, and just made a remark but signaled that you understood what I meant even so, then you wouldn't have been trolling. Eric Stevens: His important definitions a "Exposu The total light per area (photons / mm˛) that falls on the sensor while the shutter is open, which is usually expressed as the product of the illuminance of the sensor and the time the shutter is open (lux · seconds). The only factors in the exposure are the scene luminance, t-stop (where the f-ratio is often a good approximation for the t-stop), and the shutter speed (note that neither sensor size nor ISO are factors in exposure)." Sandman: Indeed. Eric Stevens: Here is a trap: "Brightness: The brightness of an image (what people usually think of as "exposure" -- same units as exposure): Brightness Exposure x Amplification." Sandman: Exactly like I've said. But you have never explained that by 'brightness' you meant the output of the amplifier. I was not aware that it was unclear. This is the first time I used it: Sandman How to measure ISO 11/11/2015 "ISO was created decades ago measures amount of light gathered per square inch on film. On a smaller sensor, ISO breaks down because there are less such "inches". Using the same ISO setting may result in similar brightness, but much higher signal to noise ratio. To get the same picture with two different size sensors, you need to adjust the total amount of light, which ISO can not do, it can only handle the amount of light given to it. Smaller sensors needs a lower ISO to gather the same amount of light as a larger sensor." Note the date, that was 19 days ago. This is the very first time I wrote it in a reply to you: "A smaller sensor needs a higher sensitivity to produce the same level of brightness as a larger sensor, meaning that ISO 400 on MFT is more sensitive (more amplified) than ISO 400 on FF." / Sandman- 11/14/2015 Note the "to *produce* the same level of brightness". To my knowledge, I have only ever used it in relation to the result, i.e. the generated photo. Eric Stevens: "Total Light: The total number of photons that falls on the sensor (lumen·seconds, or, equivalently, photons): Total Light Exposure x Effective Sensor Area." Sandman: I.e. exactly what I've said. I think you will find you have talked about 'total light' only once, im message Of course not, I've talked about this many many times: Sandman 11/11/2015 "With less total amount of light, the signal to noise ratio differs between sensor sizes, meaning that ISO 200 on MFT has the same s/n ratio as ISO 800 on FF." Sandman 11/14/2015 "Now, adjust the ISO by the crop factor squared, AND give each sensor the exact same *total* amount of light:" Sandman 11/14/2015 "So suddenly, higher pixel density isn't noisier at all. because if you give the sensors the same amount of *total* light, the noise level is more or less the same. So with the same total amount of light, the end result is roughly the same." Sandman 11/15/2015 "Each square cm is hit by 10 photons in a given exposure. The smaller sensor is hit by fewer *total* photos, regardless of how many photo sites it has. It could be a 10MP MFT or a 24MP MFT, it matters not. The total amount of light that hits the sensor is what is different." And so on, and so on. A quick search found more then thirty occupancies of me talking about the total amount of light. I didn't want to list them all here. Eric Stevens: "Many of the misunderstandings come from people using different definitions for the same words. In particular, "f-ratio" is often confused with "aperture", and "exposure" is confused with "brightness" and "total light". The importance of these distinctions is often overlooked or simply not understood, ... " Sandman: Yes, these are some of the terms you and nospam have repeatedly misunderstood, I agree. And you consistently failed to explain *exactly* what you meant to many more people than me and nospam. Maybe, and that's on me, for sure. But there's a fair level of stubbornness in play here as well, especially when it comes to nospam. I.e. there is nothing really "new" brought to the table in these last posts than haven't already been said by me many many times. You have ignored the support, snipped the support or just flat out denied the support without offering up a reason (substantiation) for why the support isn't valid. Eric Stevens: B. It has been assumed that noise is inversely proportional to photons received: the more photons the less noise. To describe this as a hairy approximation is a gross understatement, but no matter. Sandman: Noise has nothing to do with amount of photons. It has everything to do with amplification. Which depends on the number of photons received. Only if the number of photons require amplification. Eric Stevens: C. All this pseudo-mathematics is centred about an attempt to set out the rules relating the properties of cameras of different sizes which produce not only the the equivalent optical geometry but also equal image noise. Sandman: I.e. creating equivalent images using different sensor sizes. And lens sizes also. Indeed. I talked about that above as well. Eric Stevens: In the real world their buckets will not be four times as large so will be overflowed, with resulting highlight burnout. Sandman: This is a false assumption. We are talking about a MFT sensor that has amplified its signal four times as much as a FF sensor to create a "normal" photo. Lowering the ISO and opening up the aperture will result in the exact same photo, only with the exact same noise as the FF camera. Here is something you can't prove. It is probably an assumption. I have proven it, remember: http://jonaseklundh.se/files/iso_adjusted.png Eric Stevens: E. Nobody operates a camera so as to obtain constant noise so the conclusions, while interesting, are irrelevant to the real world. Sandman: It's very relevant to a number of things: 1. The myth that smaller sensors are noisier, they aren't. 2. The myth the "ISO" is a certain "sensitivity" (amplification), it is not. The concept that ISO is a certain amplification is a myth (I hope). Yes, that's what I just said Sandman: 3. The problem with using "35mm equivalent" terms for the focal length but not for aperture to mislead users. Look up the definition of f/ number (and think about it). No need, the f-number is a physical function of the lens, but the f-ratio is what is important when making equivalent comparison between cameras. See above. -- Sandman |
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How to measure ISO
In article , Whisky-dave
wrote: Sandman: To create an equal image. With not enough light, the smaller sensor will have to amplify the signal more and this creates more noise. It's just something that most people don't think about. Those that started taking pictures using film know this No, because if you shot with a 35 mm camera, all "sensors" were the same size. It wasn't an issue back then. It's only an issue now. ISO 400 was ISO 400 regardless of camera used. Eric Stevens: The truth of the matter is that if you take a camera and scale it either up or down, the lens f/value stays the same and the level of illumination on the sensor stays the same. If the sensitivity of the sensor has not changed then the invariant level of illumination of the sensor means that exposure time always remains the same. Sandman: But with the same f-stop, the smaller lens gets less light. So you either need a larger f-stop or longer exposure time to match the amount of light. This means you adjust the ISO down to the crop factor square. Crop factor is irrelivent. Not sure why you're getting so confused by this. Whoosh. Eric Stevens: Alternatively, if you have an invariant f/ value, an invariant exposure time and an ivariant level of illumination of the sensor then it follows that the ISO has remained the same. That is, ISO value is not a function of sensor size. Sandman: Of course it isn't. But with the same exposure, sensor amplification is a function of the sensor size, so while the ISO number are similar, the sensor amplification is not. So, to repeat: FF: 1/250, f5.6, ISO 800 MFT: 1/250, f5.6, ISO 800 so no differnce then. Great way to snip out the other example that illustrates the difference, troll. What you snipped: FF: 1/250, f5.6, ISO 800 MFT: 1/250, f5.6, ISO 800 The above is the exact same *exposure*, but with different signal amplification, the smaller sensor needs to amplify the signal more than the FF sensor to present an equally bright image. FF: 1/250, f5.6, ISO 800 MFT: 1/250, f2.8, ISO 200 The above is the same *amount of light* on the sensor, which creates as identical image as possible using different sensor technologies. Also, you adjust the ISO by the crop factor squared to match the signal amplification of the larger sensor, so you will get very equivalent noise. -- Sandman |
#759
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How to measure ISO
In article , Whisky-dave
wrote: To create an equal image. With not enough light, the smaller sensor will have to amplify the signal more and this creates more noise. It's just something that most people don't think about. Whisky-dave: Those that started taking pictures using film know this Sandman: No, because if you shot with a 35 mm camera, all "sensors" were the same size. They didn;t have sensors, although some had inbuilt meters. Whoosh Sandman: It wasn't an issue back then. It's only an issue now. It's not an issue now, if anything it was more of an issue in the past. That's why ISO came about. Whoosh Sandman: ISO 400 was ISO 400 regardless of camera used. Still the same today. ASA400 is ISO 400 which is 27 DIN. Whoosh -- Sandman |
#760
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How to measure ISO
In article , Whisky-dave
wrote: Great way to snip out the other example that illustrates the difference, troll. What you snipped: FF: 1/250, f5.6, ISO 800 MFT: 1/250, f5.6, ISO 800 The above is the exact same *exposure*, but with different signal amplification, How do you know what signal amplification is being applied, you don't. Simple math and observed result. Sandman: the smaller sensor needs to amplify the signal more than the FF sensor to present an equally bright image. No it does NOT. Incorrect. Sandman: FF: 1/250, f5.6, ISO 800 MFT: 1/250, f2.8, ISO 200 wrong. Incorrect. Sandman: The above is the same *amount of light* on the sensor, which creates as identical image as possible using different sensor technologies. Also, you adjust the ISO by the crop factor squared to match the signal amplification of the larger sensor, so you will get very equivalent noise. You're talking bull****. Incorrect. Take a sensor full frame cut it in two. They don;t halve in sensitivity because you've cut it in two ! Since the photosites remain the same size, but will generate a photo with half the fidelity (resolution) instead of more noise. MFT sensors aren't one fourth the resolution of a FF sensor. Simple physics, simple math. Sensor size has NOTHING to do with the ISO rating. Whoosh. -- Sandman |
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