# Is my thinking about equivalence right?

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Re: Is my thinking about equivalence right?
11

rogerstpierre wrote:

bobn2 wrote:

rogerstpierre wrote:

bobn2 wrote:

Olympus has to produce double the resolution of the Sony just to match it in a final image, due to being magnified twice as much, then there's not much of a comparison. You have to compare the Olympus lines with the lined for half the lp/mm on the Sony. On FF 10 lp/mm corresponds to 215 lp/ph, whilst on FF that final resolution is given by 20 lp/mm, so if we compare the two at the same final resolution, we find that the Sony gives an MTF of 0.93 in the centre and 0.85 at the edge, whilst the Olympus gives 0.77 in the centre and 0.69 at the edge.

You can't talk "magnification" with a digital sensor like you do with a film substrate. On a digital sensor the number of pixels, or data points you may want to call it, defines the size of the image, not the size of the sensor itself. Hence a sensor 1/2 the size with twice the number of pixels will produce an image twice larger when viewed at the same "magnification". This is where I don't understand when magnification is used as a variable to compare DoF between different sensor size. It has no relevance in a digital world, rather the total number of pixels that makes up an image does.

Call it 'enlargement', or what you want. Whether the mechanism is chemical or electronic. the image that the lens projects has to be enlarged to the size that you want ti view it. If your image frame is 17.3x13mm it needs to be enlarged twice as much to get any given size output as it does if it is 36x24mm. Your point has no impact on what I was saying, that the image the lens projects must be enlarged twice as much on mFT as on FF. It also has no impact on DoF, where the important factor is the saze of the blur in the viewed image. To take into account the doubled enlargement DoF calculations for mFT use a CoC that is half the size as they do for FF, resulting in the same size CoC in the viewed image.

You obviously don't understand how a digital image can be viewed.

With respect, I don't think that the problem here is with my understanding.

The size of the photosite that makes up the data point has no relevance when it come to the representation of a single RGB value. 1pixel is 1 pixel regardless of how big the photosite that captured the data is.

Erm, yes...and?

A 20mpix image viewed by any means will always be the same size regardless of how large the sensor that captured the image is.

ah.. You're assuming that images never get resampled on the way to being viewed? That's incorrect. They almost invariably get resampled.

The only difference from the sensor's pov will be in the signal to noise ratio.

We aren't talking about the sensor's point of view. The only points of view that matter with respect to what I was talking about is how large is the camera's image frame and how large is the size at which you view the photo. As I said, whatever is the mechanism used, and image that is tiny has to be enlarged to one that is big enough for you to view. The size of the pixels (or grains) is irrelevant to this. The only thing that they affect is the reproduction quality that you will get after enlargement.

The larger the photosite, the stronger the signal.

First, that's simply not true. The 'strength' of the signal depends on the amount of light energy collected by the photosite, which is only loosely coupled to its size. Second, even were it true, it's completely irrelevant to this discussion. The 'strength' of the 'signal' has no bearing at all on how much the image that the lens projects onto the sensor needs to be enlarged before viewing.

The quality of the lens optical components however, will dictate the resolution of the projected image, hence M43 lenses have to be optically superior to equal the same resolution with an image circle that is 1/2 the size indeed but that has nothing to do with the visualization of the image captured once it has been captured.

It has everything to do with that. It is precisely caused by the fact that to view the mFT image the same size it needs to be enlarged twice as much, where 'enlargement' means the ratio of the linear sizes of the camera's image frame and the image display frame.

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Re: Is my thinking about equivalence right?
10

rogerstpierre wrote:

the number of pixels, or data points you may want to call it, defines the size of the image

As any scanner operator worth his salt will tell you, no. The size of the image depends on the number of pixels and output resolution (which, effectively, is the space between the adjacent pixels at the output, or the ratio between the spacing in input to the spacing in output, and that is the magnification).

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Re: Is my thinking about equivalence right?
8

Serguei Palto wrote:

Mark Ransom wrote:

RobBobW wrote:

- the argument about total light is pointless as what is important is light density. Yes FF will bring in 4 times the light, but FF also has 4 times the surface area of sensor to illuminate, so it is a wash. Faster lenses bring in more light per unit area, period.

This is simply false. The reason apertures are measured by F-stops is because this equalizes the light density per unit area between lenses of different characteristics. A lens at F/1.2 will produce the same light density, no matter the focal length of the lens or the size of the sensor behind it. This means that a sensor with 4x the area really will collect 4x the light, when measured over the whole image, as long as the F-stops/T-stops of the lenses are the same.

The reason this matters is the nature of noise. The majority of noise in today's cameras is from photon shot noise, which is a property of the light itself and not of the lens or sensor or any other camera electronics. The only way to reduce shot noise is to collect more light. Whether you do this with a larger sensor, a larger aperture, or a slower shutter speed is immaterial.

RobBobW is right!

What you are saying on F-numbers is correct, but there is no contradiction with what RobBobW is saying: "Faster lenses bring in more light per unit area, period."

I only want to add that in addition to the light intensity the pixels size is another important factor.

For example, when we check for the noise we usually go to 1:1 view to resolve individual pixels. At this moment there is nothing to do with the total light captured by the sensor. It is just evident that what you see on your display at 1:1 view is tipically a small fraction of a whole sensor. If you use the lenses of the same focal length and sensors with the same pixel density then independently on the sensor size you will get the same S/N ratio at the same F-number (of course, same sensor technology and lens quality are assumed).

We can tell nothing on S/N ratio if we look just on a value captured by a single pixel. It is because a single pixel can't form an image. So we need many pixels. But we also are not able to judge on S/N if we look at the whole image at conditions when the individual pixels are not resolved. For example, it is well known that visible S/N depends on a distance from a viewer to a printed photo. The larger distance - the higher S/N. It is because we loose the resolution with increasing the distance. The human vision itself works as a low pass filter. The larger distance results in filtration the higher frequency components. So with increasing the distance we are narrawing the spectral pass band.

In other words the S/N must be defined at conditions when there are no additional filters which can influence the spectral band width of an image. Moreover, in case of different sensors we must compare the S/N at the same spectral band width defined explicitely by a sensor properties. The spectral band width is the difference between the highest frequency the sensor can capture and the lowest frequency. In ideal case the highest frequency is the half of the Nyquist frequency that is defined by the pixel size, while the lowest frequency is defined by the sensor size. In case of a sensor size is significantly higher than the pixel size, the lowest frequency can be considered as zero. Thus the sensor frequency band is just defined by the pixel size which is the main factor responsible for true (whole band) S/N .

In spectral terms, the total light captured by a sensor is just a magnitude of a single spectral component at zero frequency in an image, while an image consists of a huge number of the frequency components. Actually, the number of frequency components in the image is equal to number of pixels in a sensor. For that reason the so-called EQ-"theory" based on a total light or just a magnitude of a single frequency component at zero frequency is a missleading concept.

In terms of spectral approach the FF-sensor performs better than m43 only because of the larger pixel size (the same number of the pixels is assumed).

If someone wants to learn something on S/N then EQ must be forgotten as a "theory". This concept has been created just as a practical tool for quick recalculating FOV and DOF in case someone uses cameras of different formats, and it can't be used for something else.

All what I am saying is implemented in my iWE raw processing software, which is free .

Best regards, SP

OMG, throwback Tuesday! This was so thoroughly debunked more than a decade ago that it is a bit embarrassing to see it still propagated here.

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Re: Is my thinking about equivalence right?

bobn2 wrote:

RobBobW wrote:

What I meant by “it is a wash” is that more light is required to illuminate a full frame sensor to the desired light intensity. The only relevance of total light is in obtaining the desired intensity.

The 'desired intensity' is set by your requirements for image quality (noise), and for any given requirement will be different full-frame and mFT.

Actually, I was thinking about 'desired intensity' referring to the exposure to give a good Zone system type of overall tonal range.  No ETTR or stuff like that.  My film days are showing!

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Re: Is my thinking about equivalence right?
3

RobBobW wrote:

bobn2 wrote:

RobBobW wrote:

What I meant by “it is a wash” is that more light is required to illuminate a full frame sensor to the desired light intensity. The only relevance of total light is in obtaining the desired intensity.

The 'desired intensity' is set by your requirements for image quality (noise), and for any given requirement will be different full-frame and mFT.

Actually, I was thinking about 'desired intensity' referring to the exposure to give a good Zone system type of overall tonal range. No ETTR or stuff like that. My film days are showing!

Film days were much like digital days. The 'overall tonal range' was determined mostly by processing, and, if you were making prints, but the printing. The zone system was about getting the most information into the negative, not about setting the tonal range. Remember, Adams would quite frequently reprint his negatives with completely different tonal ranges.

You have come to think that setting exposure is entirely about setting the output lightness, whilst on digital the ISO control is effectively the lightness control. If your exposure is not 'correct' according to lightness, you can just make it 'correct' by changing the ISO.

So, let's go back to the question of f-number and its effect on the image. If you're using f/4 at 200 ISO on mFT, you can get the same results with the same shutter speed by using f/8 and 800 ISO on FF. That's as much the 'same result' as you would get using two different mFT cameras at f/4 and 200ISO.

That's why your 'period' was a bit premature. It ignores the fact that what is the 'correct' exposure tonality wise is determined by the ISO setting, and with different format cameras you'll use different ISO settings, and different exposures, to get the same results.

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This month’s very own equivalence thread
7

Ok, it can be very informative and we can let it run, but how many times have we discussed this very point?

Some fully understand, others come from a more limited understanding of what it specifically means to them.  Others simply don’t  care and it it is the image that is caught that counts and not the technicalities of physics.

Myself I like to compare the equivalent field of view (only) and refer it to the common denominator of what a FF fov “looks like” in my mind. Luckily the 4/3 sensor allows this with a simple 2x multiplier that even my mental mathematics can cope with.  I found it so much more of an issue when using a Pentax Q with its tiny odd-sized sensor.

As far as I am concerned a “fast” M4/3 lens is a “fast lens” and I care nothing about what it might be equivalent to in FF sensor terms.  If it makes images that I like I need not know  what the equivalence might be in order that I can be assured that my personal appreciation is “correct”.

In any case I am sure that we appreciate these regular updates as they always seem to be popular.

But I would  not be so relaxed if the thread became personal or in any other respect ragged around the edges.  Nor need we have a continuator thread to follow up the seemingly endless argument.

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Re: This month’s very own equivalence thread

I completely understand and in retrospect should have

1. kept the original post much smaller in scope and just talked about the result of my calculations regarding how vingetting affects total light falling on the sensor (which it is my understanding is what equiv. is all about).
2. posted this in the science and tech forum not here as it's not really about photography in anyway, just a critique of a math formula I often see used.
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Re: This month’s very own equivalence thread
2

Muster Mark wrote:

I completely understand and in retrospect should have

1. kept the original post much smaller in scope and just talked about the result of my calculations regarding how vingetting affects total light falling on the sensor (which it is my understanding is what equiv. is all about).
2. posted this in the science and tech forum not here as it's not really about photography in anyway, just a critique of a math formula I often see used.

Not an issue Ian. I have no problem with starting this discussion, nor with those that choose to respond to it.  I was just summarising the feelings on the equivalence issue and noting that sometimes it can get quite heated.

Equivalence threads at one stage threatened to choke the forum but there is nothing wrong with a refresher course every now and then.  It is obvious that many need to agree to disagree on the subject.

I don’t profess to understand equivalence fully myself nor is it my just to adjudicate.  I only relate to equivalence in a limited way which I explained  - that is all that I need and as long as I like the images that I capture then that is sufficient.

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Re: Is my thinking about equivalence right?

Iliah Borg wrote:

rogerstpierre wrote:

the number of pixels, or data points you may want to call it, defines the size of the image

As any scanner operator worth his salt will tell you, no. The size of the image depends on the number of pixels and output resolution (which, effectively, is the space between the adjacent pixels at the output, or the ratio between the spacing in input to the spacing in output, and that is the magnification).

Absolutely, and a 20Mpix M43 image will print/display the same size as a 20Mpix FF one for any given print/display resolution, so if you want to talk enlargement of the image in a digital world you can not refer to frame (sensor) size but rather to the resolution of the image.

From an enlargement perspective the actual physical size of the sensor is irrelevant and to say that the image captured by a sensor 1/2 the physical size that of another needs be enlarged twice as much to achieve the same viewing size is simply wrong. If that were true, a 20Mpix image captured by an M43 sensor printed at 150dpi would be the same size as a 20Mpix one captured by a FF sensor printed at 300dpi.

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Re: Is my thinking about equivalence right?
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Based on that argument the image captured by my 20mp phone needs enlarging the same as my m43 20mp image if I am viewing the results at the same size, which is not correct.

The size of the image captured will determine how much enlargement is required to view the image at a given output size.

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Re: Is my thinking about equivalence right?
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Olymore wrote:

Based on that argument the image captured by my 20mp phone needs enlarging the same as my m43 20mp image if I am viewing the results at the same size, which is not correct.

The size of the image captured will determine how much enlargement is required to view the image at a given output size.

What does it mean in practice?

Hypothetically, if I shoot a scene with m43 at f/2.8, 1/100, and FF at f/5.6, 1/100, both 20MP, what kind of theoretical difference should I expect, assuming the necessary enlargement is the only factor?

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Re: Is my thinking about equivalence right?
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Olymore wrote:

Based on that argument the image captured by my 20mp phone needs enlarging the same as my m43 20mp image if I am viewing the results at the same size, which is not correct.

That is exactly what that means. 20Mpix is 20Mpix regardless of the size of the photosites, and a single pixel will always represent one pixel.

The size of the image captured will determine how much enlargement is required to view the image at a given output size.

The resolution of a digital image determines its representation size at a given ouput device's resolution. A digital image has no physical visual size until it is rendered either in print or display, it is just a bunch of data describing the attributes (color and brightness) of every pixels. An output device may use one or more "dots" to represent a single pixel, and the number of dots per physical unit of distance (mm, inch, whatever) the device is capable of rendering will determine how large the resulting physical rendering of the image data will be. It has nothing to do with how large the sensor was and that is my point. A tiny pixel or a large pixel is still only one pixel and can only have one color and brightness value. If a sensor as 20Mpix, then 20Mpix will be rendered and whether a pixel is the size of a few microns or the size of a baseball matters not in the determination of how big the rendered image will be, it is still 20Mpix.

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Re: Is my thinking about equivalence right?
9

rogerstpierre wrote:

Iliah Borg wrote:

rogerstpierre wrote:

the number of pixels, or data points you may want to call it, defines the size of the image

As any scanner operator worth his salt will tell you, no. The size of the image depends on the number of pixels and output resolution (which, effectively, is the space between the adjacent pixels at the output, or the ratio between the spacing in input to the spacing in output, and that is the magnification).

Absolutely, and a 20Mpix M43 image will print/display the same size as a 20Mpix FF one for any given print/display resolution, so if you want to talk enlargement of the image in a digital world you can not refer to frame (sensor) size but rather to the resolution of the image.

You don't see that you contradict yourself.

From an enlargement perspective the actual physical size of the sensor is irrelevant

From the "enlargement perspective" it's not, because if you think about it, you are enlarging what is captured, and what is captured depends on the capture system as a whole, which is more than just a sensor.

BTW. Do you think that the taking lens resolution and aberrations don't matter? How do you maintain CoC on the enlarged image? Imagine that pixels are infinitely small, and thus the "resolution" of the sensor is infinite. What's the limiting factor for the resolution of the captured image?

and to say that the image captured by a sensor 1/2 the physical size that of another needs be enlarged twice as much to achieve the same viewing size is simply wrong. If that were true, a 20Mpix image captured by an M43 sensor printed at 150dpi would be the same size as a 20Mpix one captured by a FF sensor printed at 300dpi.

OMG.

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Re: Is my thinking about equivalence right?
1

Never said that IQ would be similar. I am only saying that there is no correlation between the physical size of the sensor and the size of the rendered image. That is all.

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Re: This month’s very own equivalence thread
2

Tom Caldwell wrote:

Ok, it can be very informative and we can let it run, but how many times have we discussed this very point?

Some fully understand, others come from a more limited understanding of what it specifically means to them. Others simply don’t care and it it is the image that is caught that counts and not the technicalities of physics.

Myself I like to compare the equivalent field of view (only) and refer it to the common denominator of what a FF fov “looks like” in my mind. Luckily the 4/3 sensor allows this with a simple 2x multiplier that even my mental mathematics can cope with. I found it so much more of an issue when using a Pentax Q with its tiny odd-sized sensor.

As far as I am concerned a “fast” M4/3 lens is a “fast lens” and I care nothing about what it might be equivalent to in FF sensor terms. If it makes images that I like I need not know what the equivalence might be in order that I can be assured that my personal appreciation is “correct”.

Best and most sensible comment yet regarding this equivalency crap on this thread. Back in the days of 110, 35mm,120, 135, etc. no one complains about which format one is using. Like you, I treat my two format (FF and m4/30) in similar fashion. I care about the outcome than what the sensor size equivalent is.

In any case I am sure that we appreciate these regular updates as they always seem to be popular.

But I would not be so relaxed if the thread became personal or in any other respect ragged around the edges. Nor need we have a continuator thread to follow up the seemingly endless argument.

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Re: Is my thinking about equivalence right?
6

rogerstpierre wrote:

Never said that IQ would be similar. I am only saying that there is no correlation between the physical size of the sensor and the size of the rendered image. That is all.

Well, before you were saying that there is no such thing as 'magnification' or 'enlargement' in a digital image. Isn't it very clear that if a lens projects an image which is 17.3 by 13mm and you view it at 173 by 130mm, there is a relationship between those two sizes, like one is ten times the other. Now you decide to view it at 346 x 260mm. The relationship is now that one is 20 times the other. So, let's agree that relationship could be called 'enbigification', however the enbigification happens.

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Re: This month’s very own equivalence thread
2

Tom Caldwell wrote:

Ok, it can be very informative and we can let it run, but how many times have we discussed this very point?

Some fully understand, others come from a more limited understanding of what it specifically means to them. Others simply don’t care and it it is the image that is caught that counts and not the technicalities of physics.

Myself I like to compare the equivalent field of view (only) and refer it to the common denominator of what a FF fov “looks like” in my mind. Luckily the 4/3 sensor allows this with a simple 2x multiplier that even my mental mathematics can cope with. I found it so much more of an issue when using a Pentax Q with its tiny odd-sized sensor.

Some could argue that the DoF also plays a role in what an image looks like. Luckily, to get the same DoF you can use the same 2x multiplier to get the equivalent f-number which will give the same DoF (and diffraction and noise)

and yes, most people will probably find it easier to multiply or divide by 2 rather than 5.6.

As far as I am concerned a “fast” M4/3 lens is a “fast lens” and I care nothing about what it might be equivalent to in FF sensor terms. If it makes images that I like I need not know what the equivalence might be in order that I can be assured that my personal appreciation is “correct”.

supposedly a fast lens offers a fast shutter speed. Perhaps consider what sets the limit of which shutter speed you wish to use.

In any case I am sure that we appreciate these regular updates as they always seem to be popular.

But I would not be so relaxed if the thread became personal or in any other respect ragged around the edges. Nor need we have a continuator thread to follow up the seemingly endless argument.

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Tom Caldwell

there really is no argument as equivalence is a set of very basic  facts. Of course that will not stop some from calling these facts “crap” for reasons which I think will continue to be mysterious (and keep these threads going)

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Re: This month’s very own equivalence thread

Tom Caldwell wrote:

Myself I like to compare the equivalent field of view (only) and refer it to the common denominator of what a FF fov “looks like” in my mind. Luckily the 4/3 sensor allows this with a simple 2x multiplier that even my mental mathematics can cope with.

for those, how never had FF - "150mm" is a lens for birds, "25mm" - standard portrait lens, etc

But you do it for the same reason, as I do it for aperture - I shoot portraits and landscapes keeping 2x in my mind. Same reason for the 2x noise, as it is acceptable up to ...

As far as I am concerned a “fast” M4/3 lens is a “fast lens” and I care nothing about what it might be equivalent to in FF sensor terms. If it makes images that I like I need not know what the equivalence might be in order that I can be assured that my personal appreciation is “correct”.

is this forum for owners only? every new camera review of DPR has comparison tool to other cameras in term of noise/DR.

As a buyer, when I'm choosing camera+lens for portraits, do I need to be aware of equivalence? Do I need to be warned that “fast” \$1200 M4/3 lens is "f/2.4 fast" in term of DOF or SS/ISO?

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Re: This month’s very own equivalence thread
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Tom Caldwell wrote:

I found it so much more of an issue when using a Pentax Q with its tiny odd-sized sensor.

As a Pentax Q owner, I must take issue with that. First, the Q's sensor is not 'odd-sized'. There are many more cameras in the world with 1/2.3" sensors than there are with 4/3" sensors. And us Q owners are just fed up with 4/3 owners telling us that our sensors are 'tiny' and no good for photography. You forgot to mention all the advantages of the Q system:

• Much more depth of field than the clunky old m4/3 system.
• Much smaller lenses with no meaningful drop in image quality.
• Smaller cameras.
• Ability to adapt all kinds of legacy lenses, including many that won't go on mFT.
• Not based on a legacy film size.

If it wasn't for all the naysayers, Q would be the dominant camera system in the world.

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Re: This month’s very own equivalence thread
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bobn2 wrote:

Tom Caldwell wrote:

I found it so much more of an issue when using a Pentax Q with its tiny odd-sized sensor.

As a Pentax Q owner, I must take issue with that. First, the Q's sensor is not 'odd-sized'. There are many more cameras in the world with 1/2.3" sensors than there are with 4/3" sensors. And us Q owners are just fed up with 4/3 owners telling us that our sensors are 'tiny' and no good for photography. You forgot to mention all the advantages of the Q system:

• Much more depth of field than the clunky old m4/3 system.
• Much smaller lenses with no meaningful drop in image quality.
• Smaller cameras.
• Ability to adapt all kinds of legacy lenses, including many that won't go on mFT.
• Not based on a legacy film size.

If it wasn't for all the naysayers, Q would be the dominant camera system in the world.

us 1 inch sensor uses just smile smugly to ourselves knowing what the others don't.....

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