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On Snapsort I found this explanation:

True resolution

Manufacturers advertise high megapixels to sell their cameras, Snapsort calculates the true resolution of each camera based on the physical limitations of the size of it's sensor.

However, I would like a better explanation.

For example, I have a compact camera that has 16.1 MP, and true resolution is 9.7 MP. Why is that? What are the differences?

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Let me start by saying that the term "true resolution" has no set meaning. It is a term that Snapsort uses to try and simplify the meaningful detail a camera can capture.

Resolution, at its most basic, is the level of granularity of detail that a camera can capture. You could have a 200 mega pixel camera, but if the image was out of focus and you only had a giant brown blob, the meaningful resolution would be pretty much nothing because you can't make out any level of detail.

A large number of factors impact how much detail you can capture, the quality, speed and clarity of the lens, the positions within the frame that you are looking at (the center is generally higher detail than the outside), the size of the sensor (and consequently the diffraction limit), the level of noise on the sensor, even atmospheric conditions can impact the total level of actual meaningful detail that can be captured by the camera.

"True Resolution" is simply Snapsort's attempt to generalize that down in to an easily consumable number, but as it is a gross over-simplification of a complex topic, it is also next to useless. For example, one cheap lens may have a super sharp center but fall apart near the corners. It would end up with a low total resolution because of the average, however another lens that is generally uniform but fairly low quality might end up being marked as higher "total resolution".

The problem is, if you are taking a portrait with the subject in the center for example, that you may not care about edge sharpness since all the detail you want will be in the center. Thus, the "lower resolution" lens would actually be the better choice.

Snapsort is a good source of basic stat comparison's between cameras, but a large amount of their information is overly simplified and thus useless. Don't put a lot of stock in their comparisons as they are not particularly trustworthy or reliable.

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    \$\begingroup\$ Thank you, AJ! So in short what I should take into account: the actual resolution, like 21MP, or the ''true resolution''? \$\endgroup\$
    – Morpho
    Commented Apr 2, 2014 at 20:30
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    \$\begingroup\$ @Morpho - that isn't a one or the other answer. The "true resolution" that Snapsort makes is an oversimplification and is near meaningless, however the concepts that are behind it are extremely important if you are trying to get maximum quality. Extra megapixels won't hurt you, but it might not always be of much benefit either unless you have better optics. \$\endgroup\$
    – AJ Henderson
    Commented Apr 2, 2014 at 20:50
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    \$\begingroup\$ OK, to put it differently: when my compact camera has actual resolution of 16 MP and Snapsort says that its ''true resolution'' is 9.7 MP, how should I address this camera? A camera that has 16 MP or a camera that has a 9.7 MP? I hope you understand my confusion here. \$\endgroup\$
    – Morpho
    Commented Apr 2, 2014 at 20:54
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    \$\begingroup\$ But so is the 9.7. You are still capturing 16 million pixels, but some of them won't be sharply defined. \$\endgroup\$
    – AJ Henderson
    Commented Apr 2, 2014 at 20:57
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    \$\begingroup\$ Because it isn't actually 9.7mp of data. It is still more than 9.7mp, but there is some blurriness and lack of detail, but if you only care about the center of the shot for example, then it might as well be 16mp because you don't care about the detail you are losing. \$\endgroup\$
    – AJ Henderson
    Commented Apr 2, 2014 at 20:58
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"True resolution" is a term that this particular site (snapsort.com) has made up in an attempt to account for the fact that pixel size and density play a factor as well. You can check out their whole page about it here. There is no industry standard term called 'true resolution'.

They're calculating it based on the size of the Airy disk, given a maximum of four pixels per Airy disk at f/5.6 . The concept being that a greater density of pixels beyond four per Airy disk doesn't yield any further detail. So, given the physical size of the sensor and the focal lengths possible on the lenses, they seem to be trying to calculate a separate measure from actual megapixels - more like 'effective' megapixels for measuring the ability of the sensor to capture detail.

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  • \$\begingroup\$ Thank you, rfusca! However, I am looking for the simplest answer. What is the main difference between resolution like 21 MP and true resolution? What I need to pay attention at when purchasing a camera? \$\endgroup\$
    – Morpho
    Commented Apr 2, 2014 at 18:13
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    \$\begingroup\$ @Morpho - None of snapsorts numbers should be the only factor in a buying decision. And generally these days, on the list of things to look for in a new camera, super high megapixels shouldn't be near the top for most people. We could debate the methodology behind snapsort's 'true resolution' and say "they're not taking this or that into account" but overall, they're probably in the realm of a good measure for this. So, if you're looking at two cameras and everything else is equal, but its a different 'true resolution', get the higher 'true resolution'. \$\endgroup\$
    – rfusca
    Commented Apr 2, 2014 at 18:18
  • \$\begingroup\$ The listed resolution of '21 megapixels' references to the physical pixel count on your sensor. The 'true resolution' that Snapsort offers, trying to account for the fact that all 21 megapixels can't be capturing detail because they're too closely packed together. \$\endgroup\$
    – rfusca
    Commented Apr 2, 2014 at 18:22
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This is a contentious issue, and I'm afraid that you won't find a single, universally-accepted definition anywhere. The website in question is using a relatively simple calculation that doesn't really cover all of the variables involved (and rfusca's answer addresses that).

The "most correct" answer (if this was one of those confusing multiple-choice questions with several answers that are partly right) looks at the modulation transfer function (MTF); that is, what size details at what contrast level can the sensor record and translate into pixels. That is, the answer is come by experimentally by taking test pictures (or projecting images directly onto the sensor) and determining what size a pattern needs to be before it's rendered at an acceptable level of contrast and detail.

With a typical Bayer-pattern sensor (or with similar colour-array sensors), this number can never be the same as the number of sensor elements. Since each sensor element records only one colour, its neighbours need to be consulted for colour information before any one pixel's value can be determined. At best, you can expect a "true" resolution that is approximately 1/sqrt(2) of the number of sensor elements/pixels. (The obvious exception here is a multishot studio camera like the Hasselblad 50MS back, which has a Bayer filter but takes four sequential images, each shifted by one pixel, so that every pixel in the image has its own complete colour information recorded along with the luminance info.)

There is also the antialiasing (optical low-pass) filter to consider, when there is one in place. Its job is to deliberately blur the image by a controlled amount in controlled directions to prevent image artifacts (like moiré patterns) from appearing in the image when the size and pattern of the details approach the size and pattern of the sensor elements. That is to say that the amount of detail you are able to record is deliberately limited to less than the bare sensor can theoretically record (the Nyquist limit) by some amount in order to prevent false details from appearing in the output image. This overlaps somewhat with the resolution you're losing due to the colour array filter, so the effect is less than cumulative. (That is, you can't just multiply the colour filter array loss by the optical low-pass filter loss and come up with a number.)

At best, a Bayer-pattern sensor will only have about a 70% data-to-pixel ratio. Monochrome sensors, whether manufactured that way or as the result of an aftermarket modification, as well as Foveon-type sensors, when not "choked" by an optical low-pass filter, approach 100%. (At exactly 100%, you can never be sure whether you're seeing the real data or an aliasing artifact. That's a fundamental problem with discrete data; all you can do about it is hope - or ensure - that the data you are recording is "bigger" than the buckets you are recording it in. And that's why very high-resolution sensors can get away without optical low-pass filters in most instances - you are rarely recording anything that has a repeating pattern small enough to cause a problem on the one hand, and the lens itself will lend a certain amount of low-pass filtering to anything that isn't very sharply in focus.)

There are other things that will influence the amount of real detail you can record as well, such as the inherent noise of the sensor and reading circuitry. Since the "effective megapixels" depends on you actually being able to see details in the image, anything that can't be easily distinguished from noise doesn't really count. With a very noisy sensor, it may take the cumulative data of several neighbouring pixels before you can objectively determine what constitutes image information. That's not a horrible thing, necessarily; Nokia is using a tiny 41MP sensor in some of its devices to produce 5MP images, which allows it to have an effective "digital zoom" while embracing all of the data losses.

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    \$\begingroup\$ I' just looking for a more basic answer. Is it possible to have a simpler answer? \$\endgroup\$
    – Morpho
    Commented Apr 2, 2014 at 17:53
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    \$\begingroup\$ @Morpho - No, it isn't. \$\endgroup\$
    – user2719
    Commented Apr 2, 2014 at 17:53
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    \$\begingroup\$ How is it possible to explain this to students? \$\endgroup\$
    – Morpho
    Commented Apr 2, 2014 at 17:56
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    \$\begingroup\$ @rfusca - No, the question asked "what is true resolution", not "what does Snapsort mean by 'true resolution'". And since your answer to that question is apparently not meeting the "simplest answer" criterion, simply answering the Snapsort question is apparently not doing the job either. In any case, I quit. \$\endgroup\$
    – user2719
    Commented Apr 2, 2014 at 18:17
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    \$\begingroup\$ @rfusca - you can't really get in to the snapsort numbers in detail without understanding what they are talking about though. I agree that it is possible to skim the details and explain the gist of it, but it isn't a complete answer to someone else who may want a technical answer. Ideally, having both an in-depth answer like Stan's as well as a spectrum of different levels of detail in other answers works out for the best in terms of anyone who comes to this question being able to walk away with a good answer. \$\endgroup\$
    – AJ Henderson
    Commented Apr 2, 2014 at 20:25
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Let's start with resolution, which is not the same as the sensor pixel density which is usually called resolution.

To get the actual resolution of a camera and lens combination, you would photograph a resolution chart, and use the resulting image to determine how many lines can be reproduced. You would measure the resolution for primary horizontal, vertical and diagonal lines.

Ref: Camera testing resolution charts explained

It sounds like that website does some calculation based on the sensor size and sensor type, to determine what the maximum usable pixel density could possibly be with a theoritical ideal lens. The "true resolution" would be lower than the sensor pixel density if the sensor has more pixels than it can actually make use of.

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  • \$\begingroup\$ Could you please explain it to me in more basic lines? \$\endgroup\$
    – Morpho
    Commented Apr 2, 2014 at 17:49
  • \$\begingroup\$ Why the downvote? If you don't explain what it is that you think is wrong, it can't improve the answer. \$\endgroup\$
    – Guffa
    Commented Oct 23, 2014 at 11:16
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They are simplifying a much more complex world. your true resolution will be a product of lens and sensor. but the lens resolution (lp/mm) will depend on its settings as well as its quality. so to do those conversions they need to make a lot of assumptions. To spread light on the complexity of resolution:

First of all, MP is not a measure of resolution. It starts in the real world a line pairs (black and white adjacent lines), and the get projected and squeezed onto the sensor plane. a dot in the real world wont end up on a dot, but a circle, which is wavelength dependent, too. your sensor samples these in a certain digital resolution. The result of that depends on the sensor size and pixel count in each direction. the resolution is typically close to (not always equal) eachother on the X and Y axis. The lens should also be close to eachother in each direction. that means a 4000x3000 sensor 36*24mm will have the same resolution in X and Y, but not diagonally! Lets say a lens with 120lp/mm projects those lines so they are perfectly aligned on that 4000x3000 sensor. Then you get a perfect picture - but only if it is a monochrome camera! If the lines are not aligned, you get moire. So the producer adds a optical blur filter. then the neat image gets screwed up. now you need to increase the resolution of the lens, or move closer for better magnification and lose half the image or project on a larger sensor to spread those "Airy disks" apart. add bayer pattern interpolation to the mix and you need to double or quadruple your resolution ( on each axis, not the MP which will be 4x - 16x as high).

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A lot of the other answers made this unnecessarily complex and talked about things irrelevant to the OP, so let me try to be clearer:

Many people think that a higher megapixel camera produces sharper photos. However, for a given sensor size, there's a limit on how many megapixels of actual information can be captured. Exceeding that limit doesn't help. If anything, it hurts, because you have bigger file sizes for no discernible benefit. Companies do this because they can advertise a higher megapixel number to fool people who don't know any better.

Snapsort tries to capture this number as "true resolution". In your case, you have a camera advertised as 16 megapixels, but its true resolution is only 9.7 megapixels. This means that your camera can't capture any more detail than it could if it were equipped with a 9.7 megapixel sensor.

If you bought this 16 megapixel camera over an otherwise identical 12 megapixel camera, say, thinking you're going to get more detailed photos, you were fooled :) Both the 16 megapixel and the 12 cameras are in effect 9.7 megapixel cameras.

Note that this is all theoretical -- Snapsort doesn't actually go and measure the performance of the camera by taking photos using it. Instead, it does some mathematical calculations based on the sensor size to determine its "true resolution".

The "true resolution" is also an upper limit. You may not actually get 9.7 megapixels worth of detail from this camera, but you're certainly not going to get more.

So far, we've talked only about the sensor, but the camera's lens can and does reduce the quality of photos. Going back to our previous example, your camera with a "true resolution" of 9.7 megapixels might be equipped with a lens that lets the sensor capture only 5 megapixels worth of information. This is similar to looking through a pair of blurry binoculars -- even if you have excellent eyesight, you're not going to be able to see the detail you could otherwise. By the same analogy, maybe the sensor can capture 9.7 megapixels of information, but not when looking through this blurry lens.

A company named Dxomark tries to capture this in a metric called "perceptual megapixels". As an example, the Sony E-mount 35mm F1.8 lens has a perceptual megapixels count of 11. This means that whether you mount this lens on a camera advertised as 11 megapixels, or 24 megapixels, or 200 megapixels, you're not going to get any clearer photos out of it.

So, if you're comparing cameras or lenses, you have three measurements you could use to tell how sharp they are:

  1. Advertised megapixels.
  2. Snapsort's "true resolution"
  3. Dxomark's "perceptual megapixels".

Of these, perceptual megapixels are the most accurate measurement, since it accounts for both the camera and the lens, and is derived from real-world measurements. The second-best metric is Snapsort's "true resolution". Ignore the megapixels advertised by the manufacturer, because that's just a number on paper, which might never be achieved in practice by your camera.

More important than all is that sharpness / resolution is only one aspect of choosing a camera. Don't blindly go with the sharpest camera. There are many factors, such as low-light performance, autofocus accuracy and speed, battery life, whether the camera supports interchangeable lenses, sensor size, camera type (SLR, mirrorless camera, superzoom, phone, compact), whether it fits in your pocket, and so on. Don't go just by the resolution of the camera when you make a decision.

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For my camera, it lists the true megapixel number as the same as the actual megapixel number, so I don't think we should make too much of this number. The site also lists sensor size and pixel size. You can calculate the pixel size also by dividing the sensor area by the megapixel number. If you tak the square root of this number, you get the pixel size as the distance between two neighboring pixels. You can then calculate the maximum resolution you can theoretically obtain by dividing this by the focal length of the lens. E.g. the distance between the pixels on the sensor of my camera is 4.2 micrometers, dividing this by a lens focal length of 50 mm gives a resolution of 8.4*10^(-5) radians = 4.8*10^(-3) degrees = 17 arcseconds.

In practice, the resolution will be worse than 17 arcseconds, due to noise, lens imperfections, low pass filter blurring the image slightly to remove artifacts. This is discussed in detail in the other answers given. But with a lot of effort you could come close to the theoretical limit if you can take multiple shots and combine them in post processing (e.g. in strophotography or landcape photography).

Diffraction will become inportant above F numbers of about 7. A point source of wavelength lambda will have an angular spread theta (measured from the direction of the source) of theta = 1.22 lambda/d where d is the lens diameter. If 2*theta equals the angular size of the pixel of 8.4*10^(-5) radians then the part of the diffraction pattern till its first minimum approximately completely covers one pixel. If the diffraction pattern is made wider such that theta equals the angular size of the pixel then the neigboring pixel will still not receive much light, because it is at the minimum intensity of the diffraction pattern and near that point the intensity variation is not large. For green light lambda = 500 nanometers, so for my camera this happens at d = 7.3 mm, which corresponds to an F-number of 50/7.3 = 6.9. Note that this doesn't depend on the focal length because theta also depends on the focal length, the focal length then drops out of the equation for the critical F-number above which diffraction starts to limit the resolution. In general, the criticial F-number is given by:

Critical F-number = r/(1.22*lambda)

where r is the distance between neighboring pixels on the sensor.

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Simply you can process a 4mp image using photoshop and upscale it to 8mp image. Then there will be duplicated pixels , but theoretically it's an 8mp image now. I believe snapsort's "true resolution" means an image's actual pixel resolution without having such duplicated pixels.

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  • \$\begingroup\$ Although it's less common now (as very cheap yet reasonably high-res sensors are readily available), it used to be that low-end phones and cameras would list a megapixel value with a *, and that * said "interpolated" — meaning the resulting images where upscaled, just like you said. So, yes, that's a thing. However, that's not what Snapsort means in this case. (See the other answers.) \$\endgroup\$
    – mattdm
    Commented Nov 7, 2015 at 22:07

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