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A visual demo on the equivalence of small and large sensors
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rjlittlefield
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PostPosted: Fri Aug 31, 2018 5:39 pm    Post subject: Reply with quote

For the benefit of others...

Quote:
You probably did not understand that Dof formula with magnification just ignore the diffraction effect at all.

Justwalking is completely correct that the DOF formula ignores diffraction. What the DOF formula shows is only that you need to scale the effective aperture exactly in proportion to the sensor size, to get the same DOF.

Then it is the diffraction formulas that take over, to add the very important information that scaling effective aperture in this way also automatically causes diffraction effects to match in the captured images.

On the surface, this seems quite a remarkable coincidence. I can easily understand how people miss it when reading most literature. What I do not understand is why Justwalking still misses the concept, after multiple explanations.

Quote:
but F'=8.5 is not the same as F'=47 for
diffraction effect.

Justwalking has once again scrambled something in his concepts. We're talking about f/8.5 on sensor size 4.3 mm, versus f/47 on sensor size 24 mm. In both cases, about 380 Airy disk diameters fit across the frame. So with respect to the final image as seen by the human, these combinations in fact are exactly the same.

Quote:
System with your small sensor become diffraction limited at f/5.6 and FF with f/14.6.

It's annoying that no derivation or source for these numbers is given.

Based on the ratio of the numbers, I think that what Justwalking has done is to compare a 7 megapixel 1/2.5" sensor versus a 36 megapixel FF sensor.

If that's the case, then what he's saying is that it's more challenging to fill 36 megapixels with pixel-sharp detail than it is to fill 7 megapixels with pixel-sharp detail. I would have no disagreement with that statement in isolation, but it's not relevant to the current discussion.

Quote:
It's not idea. It's a math.

It seems you've missed another key concept.

All math is just an idea, until it has been shown to describe reality.

Your math, unfortunately, does not survive that test. So yeah, what you're doing is just an idea.

Quote:
Also i can't trust to your last picture with safety match. They do not looks like your previous in parallel thread, sorry.

Indeed, the images are not identical to what I posted in the other thread.

That's because I looked at comments in the other thread, considered what I could do to further reduce irrelevant distractions, adjusted the test setup accordingly, and shot new pictures. In the current setup, I have taken more care to match the focus point, I have removed a previously unnoticed tilt of one camera to better match the alignment of foreground and background, and I have adjusted the processing of both images to remove some distracting chromatic aberration. None of those things have any effect on the topic of discussion, of course.

By the way, Justwalking -- and now I am talking directly to you -- please take note that accusing another member of dishonesty would be considered very bad form. I'm happy to see that you did not quite do that. I suggest staying safely back of that line in the future.

--Rik
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Justwalking



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PostPosted: Fri Aug 31, 2018 5:41 pm    Post subject: Reply with quote

JohnyM wrote:
Laughing
Do you realise that 16mpx FF pixel is way bigger than 16mpx 1/X" sensor?

Quote:
Yes, small sesor can give in macro same Dof with less diffraction. or more Dof with same diffractioin, or both at same time


Only if you use low resolution sensor that is severly limiting overall resolution, yes. Someting like 1mm diagonal 1 pixel sensor, infinite DOF regardless of magnification!


Even if you get very high resolution sensor (let's say 200 MP) you've got very small information with Feff=100, much less than 1MP.

Let's say i take a lens El-Nikkor 50mm/2.8, set magnification X2 and put f(nom) = 5.6 with crop factor sensor 5.5.
Then Feff will be 5.6(2+1) = 16.8 - still pretty acceptable to see the result and determine the DoF.
To get same DoF on FF f(eff) must be = 16.8x5.5 = 92.4(!) for the same FoV.
Then the DOF on FF will be the same (if both sensor with same MP). Nice! But what happens with picture? Where is this Dof or something else on this picture?

With low magnification it is possible try to say that is no difference in Dof when set optimal F(nom) for FF. But when Magnification will rise the situatioin will be worse.

With X3 and F/5.6(nom) system with small sensor will have Feff =22.4
On FF Feff=123.2 (!!!)


Last edited by Justwalking on Fri Aug 31, 2018 6:10 pm; edited 1 time in total
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rjlittlefield
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PostPosted: Fri Aug 31, 2018 6:10 pm    Post subject: Reply with quote

Justwalking wrote:
Let's say i take a lens El-Nikkor 50mm/2.8, set magnification X2 and put f(nom) = 5.6 with crop factor sensor 5.5.
Then Feff will be 5.6(2+1) = 16.8 - still pretty acceptable to see the result and determine the DoF.
To get same DoF on FF f(eff) must be = 16.8x5.5 = 92.4(!) for the same FoV.
Then the DOF on FF will be the same (if both sensor with same MP). Nice! But what happens with picture? Where is this Dof or something else on this picture?

You have a remarkable ability to ignore the results of your own calculations.

Effective f/16.8 across a 4.3 mm sensor gives 190 Airy disk diameters per frame width, corresponding to 930 usable pixels (using Nikon's rule).

Effective f/92.4 across a 24 mm sensor again gives 190 Airy disk diameters per frame width, again corresponding to 930 usable pixels (using Nikon's rule).

The image is the same on both size sensors, effectively about 1 MP.

--Rik
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ray_parkhurst



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PostPosted: Fri Aug 31, 2018 6:14 pm    Post subject: Reply with quote

rjlittlefield wrote:
ray_parkhurst wrote:
It looks to me like the equivalent image concept is getting in the way of understanding and agreement.

I disagree.


I still think this is the problem, especially given the last response from JW.

It seems that the number of MP in the sensor is constantly referred to, yet this is not relevant if both images are viewed at the same output size. What actually matters is the number of Airy disk diameters across the image. The sensor could be 10MP or 100MP (or stitched 1GP) and it would not make a bit of difference to the result when downsized and viewed on a monitor.
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Justwalking



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PostPosted: Fri Aug 31, 2018 6:17 pm    Post subject: Reply with quote

rjlittlefield wrote:
Justwalking wrote:
Let's say i take a lens El-Nikkor 50mm/2.8, set magnification X2 and put f(nom) = 5.6 with crop factor sensor 5.5.
Then Feff will be 5.6(2+1) = 16.8 - still pretty acceptable to see the result and determine the DoF.
To get same DoF on FF f(eff) must be = 16.8x5.5 = 92.4(!) for the same FoV.
Then the DOF on FF will be the same (if both sensor with same MP). Nice! But what happens with picture? Where is this Dof or something else on this picture?

You have a remarkable ability to ignore the results of your own calculations.

Effective f/16.8 across a 4.3 mm sensor gives 190 Airy disk diameters per frame width, corresponding to 930 usable pixels (using Nikon's rule).

Effective f/92.4 across a 24 mm sensor again gives 190 Airy disk diameters per frame width, again corresponding to 930 usable pixels (using Nikon's rule).

The image is the same on both size sensors, effectively about 1 MP.

--Rik


It is wrong, Rik
The resolution will be far far worse with F92 on FF than at f16 on crop

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JohnyM



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PostPosted: Fri Aug 31, 2018 6:31 pm    Post subject: Reply with quote

Rik already answered to your reply to my post, i'll skip that.
That table is correct, but you're interpreting it incorrectly. It is correct for nominal aperture, and in your special comparision we're adjusting EffF and Magnification to match your criteria, so table should be adjusted along that.
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rjlittlefield
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PostPosted: Fri Aug 31, 2018 6:44 pm    Post subject: Reply with quote

The table also does not include a column for 1/2.5" sensor, and the MP values that it quotes are for a different criterion than used elsewhere in this thread.

Note the part about "2 pixels per Airy disk diameter", versus 4.88 as used by Nikon's rule (2 pixels per cycle at cutoff).

These differences make it very simple for Justwalking to continue misinterpreting numbers and formulas so as to be consistent with his imagination.

Accounting for the difference in criterion, that underlined 4 in the table at 35mm f/22 would become 4*2.44*2.44 = 24 MP, and then extrapolating from f/22 to f/92.4 would give 1.35 MP. A corresponding column for 1/2.5" would give the same MP for f/16.8.

--Rik
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rjlittlefield
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PostPosted: Fri Aug 31, 2018 7:07 pm    Post subject: Reply with quote

ray_parkhurst wrote:
rjlittlefield wrote:
ray_parkhurst wrote:
It looks to me like the equivalent image concept is getting in the way of understanding and agreement.

I disagree.

I still think this is the problem, especially given the [earlier] response from JW.

It could be a factor, but I note that in posts after the one you've referred to, JW takes the step himself of computing the MP content of images.

The problem is that he then manages to either screw up the computation or ignore its results, and continues to express the opinion that the FF image will look more blurred, despite having the same number of Airy disks across the frame.

I still think the problem is that JW just cannot accept that somehow he got attached to the wrong idea, and now he's deadset on justifying the erroneous visions offered up by his experience-free imagination.

--Rik
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ray_parkhurst



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PostPosted: Fri Aug 31, 2018 7:51 pm    Post subject: Reply with quote

rjlittlefield wrote:
ray_parkhurst wrote:
rjlittlefield wrote:
ray_parkhurst wrote:
It looks to me like the equivalent image concept is getting in the way of understanding and agreement.

I disagree.

I still think this is the problem, especially given the [earlier] response from JW.

It could be a factor, but I note that in posts after the one you've referred to, JW takes the step himself of computing the MP content of images.


True, he said this:

"'Even if you get very high resolution sensor (let's say 200 MP) you've got very small information with Feff=100, much less than 1MP. "

This shows he does understand.
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rjlittlefield
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PostPosted: Sat Sep 01, 2018 2:39 pm    Post subject: Reply with quote

My equipment was still mostly set up, so I thought it would be worth the effort to show people what f/94 on fullframe actually looks like.

Setup was the same as before, except that I reduced the size of the aperture in the infinity space from 4 mm to 2 mm. That reduces the subject side NA from 0.04 to 0.02, and increases the sensor side f-number from f/47 to f/94.

For comparison purposes, I'll show the f/94 first, then the older f/47. As usual, make your browser window wider or zoom out to see the images side by side.

Whole frame, slightly cropped to 3:2 aspect ratio:



100% crops from 3072x2304 renditions:


The crops clearly show that in and near the plane of focus, the f/94 image is a lot less sharp than the f/47 image. That's exactly as theory predicts, because the f/94 Airy disk is twice as big as f/47.

At the same time, it is apparent at both scales that farther away from the plane of focus, the f/94 image is a lot more sharp than the f/47 image. Again that's exactly as theory predicts, because the geometric blur circles grow only half as fast at f/94 as they do at f/47.

And finally, it is apparent from the whole frame images that the f/94 image is certainly not a hopeless jumble of blur.

Again this is completely consistent with theory. At lambda=0.55 micron, the f/94 Airy disk is 126.1 microns in diameter. That means the largest pixel size that can possibly catch all the detail in the optical image (2 pixels per cycle at cutoff) is 25.85 microns, or 928 x 1238 pixels in an FF 4:3 crop as shown here (928x1393 in uncropped full frame).

So, the rendition shown here at 768x1024 is not quite enough pixels to show detail clear down to cutoff. It is nominally a pretty good match to 4 pixels per Airy disk diameter, which is the minimum needed to resolve subject features spaced at the Rayleigh separation of 1/2 Airy disk diameter between peaks.

By the way, if there's anybody reading this who still believes that 2 pixels per Airy disk diameter is enough to capture all the detail in an optical image, then I strongly recommend that they run the experiment of trying to shoehorn the whole frame f/94 image shown above into half as many pixels on each axis. The loss of detail is striking.

I hope this is helpful.

--Rik
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rjlittlefield
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PostPosted: Sat Sep 01, 2018 10:47 pm    Post subject: Reply with quote

Justwalking wrote:
It is wrong, Rik
The resolution will be far far worse with F92 [92.4] on FF than at f16 [16.8] on [5.5] crop

In a word, "nonsense".

The table you reference does not have a column for the 5.5 crop sensor. So you're comparing some number for FF, with nothing at all for 5.5 crop.

Do the calculation yourself, for effective MP at those f-numbers and sensor sizes. Show your work.

--Rik
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Justwalking



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PostPosted: Sun Sep 02, 2018 1:03 pm    Post subject: Reply with quote

rjlittlefield wrote:


Accounting for the difference in criterion, that underlined 4 in the table at 35mm f/22 would become 4*2.44*2.44 = 24 MP, and then extrapolating from f/22 to f/92.4 would give 1.35 MP. A corresponding column for 1/2.5" would give the same MP for f/16.8.

--Rik

Unfortunatelly we can't just change the pixels pitch on the sensor.

Indeed it's more complex task.
Because of the colour filter array on top of the sensor each pixel responds to only part of the light spectrum. Usually half of the pixels on the sensor respond to green light, thus we need to consider this with our calculations. The easy way is just to do an ad multiplication of the radius of the Airy disk by factor of square root of two (X1,414) – this is because that is the closest distance between two “green” pixels (in pixels). For red and blue light you’d use a multiplier of two instead.

Let's count more convenience 50% MTF criterion for color sensor Then it usually be 4 pixels per diameter.

Letr's take crop 5.5 sensor about 1/2.3"
m=1, F='4
2.44*0.55um*4=5,368um 4364/5,368 = 812 airy disk diameter across short side of the sensor crop 5.5 and 1219 across long side
m=5.5 F'=22.
2.44*0.55um*22=29.524um 24000/29,524 = 812 airy disk diameter across short side of the FF sensor.

So 1219*812 In airy diameters or we need 4876*3248 pixels (~15.84MP).

It is about optimal aperture to take maximum resolution of the crop sensor at green light considering the 50% MTF criterion.
The resolving ability of FF will be exacnly the same about 16MP but at F'22 not F'4.
Indeed, according to math for same DoF that F' is inversely proportional to crop factor 4*5.5=22.
Meanwhile you need to set F nom more at FF to achieve the same DoF.

So if i set F(nom) = 5.6 for crop sensor and take magnification of m=0.2
It will be F'eff=6.72.
On the FF to take same Dof you need m=1.1 and to set F(nom) = 17.6! (3.14more than crop) to take Feff=36.96.

Quote:

In a word, "nonsense".

The table you reference.The table you reference does not have a column for the 5.5 crop sensor.

It is my inconvenience that i used "resolution" word. I want to say mean "same looking".

The Dof at crop 5.5 at m=0.5 at Feff=15 will be same as on FF but FF must be Fnom= 22(!) and F'=82.5(!)
Do you think that they will be looks the same?
Depth of field calculators don’t quantify the subjective difference between high contrast at high resolution, and low contrast at high resolution.

Stopping down beyond the optimal aperture continues to increase depth of field, but quickly reduces optical performance (first contrast, then sharpness) everywhere in the image.
The aperture you need to shot to get same Dof as crop is far away from optimal sweet spot of any lens.


Last edited by Justwalking on Sun Sep 02, 2018 4:30 pm; edited 1 time in total
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Justwalking



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PostPosted: Sun Sep 02, 2018 1:49 pm    Post subject: Reply with quote

rjlittlefield wrote:
My equipment was still mostly set up, so I thought it would be worth the effort to show people what f/94 on fullframe actually looks like.

I hope this is helpful.

--Rik


Also people can take close view of the F/100 vs F/16 on APS-C at Magnification m=3 here
http://extreme-macro.co.uk/microscope-objectives/
Why Use Microscope Objectives For Macro?

rjlittlefield wrote:

At the same time, it is apparent at both scales that farther away from the plane of focus, the f/94 image is a lot more sharp than the f/47 image. Again that's exactly as theory predicts, because the geometric blur circles grow only half as fast at f/94 as they do at f/47.

Theory predict that the diameter in microns of Airy disk is about exactly 1.342 times the f/number (using the green wavelength).
For the f/94 it must be twice large than at f/47. DoF have rised twice also, but even focused area will be blurred.

Quote:

By the way, if there's anybody reading this who still believes that 2 pixels per Airy disk diameter is enough to capture all the detail in an optical image, then I strongly recommend that they run the experiment of trying to shoehorn the whole frame f/94 image shown above into half as many pixels on each axis. The loss of detail is striking.


For the B&W camera and Rayleigh criterion (MTF 9%) it is 2, for color sensor is usually 4, but we can choose any criterion what we want.
Here is interesting thread
https://www.cloudynights.com/topic/564374-digital-image-resolution/
Rik, what do you think about?
Quote:
So, a good Rule of Thumb for Digital Image Resolution, want f/# / pixel size = or > 3.5.
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rjlittlefield
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PostPosted: Sun Sep 02, 2018 9:23 pm    Post subject: Reply with quote

Justwalking, I'm not sure I understand how you're thinking about this.

You wrote:
Quote:
Letr's take crop 5.5 sensor about 1/2.3"
m=1, F='4
2.44*0.55um*4=5,368um 4364/5,368 = 812 airy disk diameter across short side of the sensor crop 5.5 ...
m=5.5 F'=22.
2.44*0.55um*22=29.524um 24000/29,524 = 812 airy disk diameter across short side of the FF sensor.
...
Depth of field calculators don’t quantify the subjective difference between high contrast at high resolution, and low contrast at high resolution.

Up to now, I have been trying to get you to realize that at same DOF and same FOV, the number of Airy disk diameters across the frame will be the same for all format sizes.

Now, it sounds like you agree that the number of Airy disk diameters across the frame will be the same, but you're thinking that the small sensor will somehow see higher contrast than the larger sensor.

Is that what you're thinking?

If so, then what is your mathematical justification for that idea?

--Rik
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Justwalking



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PostPosted: Mon Sep 03, 2018 10:43 am    Post subject: Reply with quote

rjlittlefield wrote:
Justwalking, I'm not sure I understand how you're thinking about this.
Up to now, I have been trying to get you to realize that at same DOF and same FOV, the number of Airy disk diameters across the frame will be the same for all format sizes.

Now, it sounds like you agree that the number of Airy disk diameters across the frame will be the same, but you're thinking that the small sensor will somehow see higher contrast than the larger sensor.

Is that what you're thinking?
If so, then what is your mathematical justification for that idea?
--Rik


This arithmetic method to count airy disks diameter across sides to find oiptimum pixels is useless for practice where we can't do anything with pixel pitch.
It simplified count that can't compute many other factors.


The highest spatial frequency a sensor can resolve is its Nyquist frequency, equal to 0.5/(pixel spacing). When a lens is stopped down so its Rayleigh limit is below the Nyquist frequency, the camera is limited by the lens rather than the sensor
For optimum quality (when extreme depth of field is not required), the aperture should be set at least one stop larger than the aperture where the Rayleigh limit equals the Nyquist frequency:
NR=N = 3.2 * pixel spacing (um).

http://www.normankoren.com/digital_cameras.html#Diffraction

For a pixel spacing of 1.4 microns, NR=N = 3.2 * 1.4 = f/4.48, so the aperture should be set at f/2.8-f/3.2 or larger.

With small sensor i can leave this "small" DoF as is and it will be not the extreme Dof for my sensor, so i can choose optimum quality for the Dof, but for the FF it is not so to achieve the same Dof.

Very small formats - for compact digital cameras with 11 mm diagonal or smaller sensors (1/4 the size of 35mm) are severely diffraction-limited at f/8, where DOF is equivalent to f/32 or more in 35mm.
But tiny digital cameras still produce very sharp images at f/4 and f/5.6 because their tiny pixels with no anti-aliasing filters — have far better lp/mm resolution than 35mm.
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