CoC Management and the Object Field.

[JimKasson]

JimKasson wrote:.

I don't think that applies here. There are only three kernels to convolve with no AA filter, and four with an AA filter. In the case of leaving out lens aberrations, there are one less than those numbers in either case. So, consider a case with no AA filter, and a pillbox of diameter 100 um and an Airy disk with 3 um between the first zeros. My claim is that the convolution of those two is essentially a pillbox, not a Gaussian.

You are right, the convergence to Gaussian is slow, and I wouldn't expect that to work for a small number of variables.

It can go pretty fast if all the kernels are similar. Here's how fast the 1D version of a pillbox converges to a Gaussian with repeated convolutions:

Loading…

oscar6echo.blogspot.com

But, you're right, in general.

As lenses tend to be either over or under corrected, a pure disk and diffraction and AA filter seems reasonable, unless you want to also add a factor for over or under correction of spherical aberration.

What would that kernel look like?

Jim

 

[JimKasson]

I was looking for some formula incorporating the special nature of diffraction CoCs, but that will have to wait. For today, I'm going with the square root of the sum of squares of c and d.

To me, a measure of the size of the combined kernel is of limited interest, since I think that extinction resolution is not a particularly useful measure, although it's relatively easy to calculate. We moved beyond extinction resolution for in-focus images decades ago with SFR, MTF, and SQF. Why do we still stick with it for DOF? Well, I know. It's because the math is easier. But I think that's looking for your keys under the lamppost (I'll be happy to tell you joke if you haven't heard it before).

I know the joke, but I may lack the skill to tell it, except to recall the punchline <spoiler alert>, "because the light is better here." Yeah, we should keep the joke in mind.

If extinction resolution is of limited interest, terrific: Please offer your alternative.

MTF50.

Yes, I recall that you are an MTF50 guy. One thing I don't understand about MTF50: Is the horizontal axis on these curves always the distance from the center of the imaging system?

I think you are thinking of the curves that lens manufacturers publish. I'm thinking of system MTF curves that photographers measure. There are many ways to present them. MTF50 is itself simply a scalar, the spatial frequency at which the modulation transfer function falls to one half. There are lots of measures of spatial frequency. The one that I favor is cycles per picture height.

It's not that MTF50 is such a wonderful measure of sharpness by itself. In fact, I think I remember, but can't access right now, studies that indicated that MTF90 -- the spatial frequency at which the modulation transfer function falls to 0.9, is a better measure. But usually the SNR at MTF50 is high enough that it can be turned into MTF90 in post if need be.

Ed Granger's SQF is a more sophisticated way to turn the spatial frequency response (SFR) into a scalar, or a quasi-scalar, since viewing distance is still a parameter, by weighting parts of the SFR and integrating. I use MTF50 because it's not a bad metric and most people understand it. If I started using SQF, I think I'd lose a lot of my audience.

If there is some more general definition, I would like to see it. Call me a simpleton.

Imatest has a lot of good stuff here:

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www.imatest.com

It's early days for me in this enterprise. I want to understand what's going on first, and then try to figure out ways to simplify it.

Terrific. This seems at odds, though, with your several statements of denigration of the object-field method, even before you fully understand it.

Jerry, if fully understanding the OF method is a requirement for commenting on it, I will have to shut up here and now. From my conversations with you, I believe that I will never fully understand it.

OK, full understanding is rare.

I don't want your consideration of these issues to cease. I'm only hoping that you'll wait for a bit before stating your final conclusion.

I have not yet come close to stating it.

I'm still digesting things you wrote several days ago, hoping for a worthwhile reply.

There are a few areas where your summary of object-field methods is not yet quite accurate in my eyes. I have a two-step approach in mind, and we may yet get both steps today!

Jim

 

[JimKasson]

What distances are the horizontal axis referring to? Are we looking at distances from the center point of the image formed with a full-frame camera?

Maybe I just need a list of assumptions used to get this graph.

I guess when I said this it was too much in short hand:

Take a look at this MTF50 plot for a diffraction-limited 55 mm lens focused at 10m with MTF50 in cy/ph versus actual object distance

The horizontal axis is object distance from the center of the lens, measured along the lens axis. The lens is focused at 10 meters.

Does that help?

Jim

 

[JimKasson]

Take a look at this MTF50 plot for a diffraction-limited 55 mm lens focused at 10m with MTF50 in cy/ph versus actual object distance:



The question is: how should we judge DOF if preserving most of the lans/camera maximum possible resolution is the criterion (for this exercise, forget output size and viewing distance).

If we're very particular, we might say that MTF50 = 1200 cy/ph is our criterion, and there's almost a meter of DOF behind the object (and a little less than that in front of it, although that's not shown on the graph), and you should stop down to f/4 or f/5.6 to get all that DOF. If we're not as choosy, we'd say that 1000 cy/[h is good enough, and now there's almost 3.5 m of DOF behind the object, and we can get it at f/8 or f/11.

Note that in the 1200 cy/ph case, f/8 through f/16 have zero DOF. In the 1000 cy/ph case, f/16 has zero DOF.

I'm open to other ways to interpret this data.

Anyone?

I find it difficult to make a choice without first determining a criterion for the CoC based on viewing distance and size.

I think doing that is pretty well worked out. You've done a great job on your blog.

For instance, how do you know that 1000 cy/ph is good enough without knowing what those are? Good enough for what exactly? DOF depends on viewing distance and size.

There are many photographers -- and I am one of them a lot of the time -- who don't have a final use (or at least not a complete list of possible final uses) in mind at the instant of exposure. Many of them us want everything in a distance range to be pretty nearly as sharp as it can be given the camera and the lens. Then the issue of how far down to stop turns into an issue of how much loss in resolution is tolerable.

I for one am much more comfortable making those tradeoffs in MTF50 cy/ph than in CoC. I think of MTF50 all the time I'm choosing a lens or an aperture without regard to DOF. Curves like the above, if generated or looked up quickly rather than in hours of 12-core CPU time, would help me in ways that CoC don't. For example, balancing diffraction and DOF is a piece of cake, because we see the effects of each in the same units.

On the other hand I have a pretty good idea how to choose the CoC (hence related near/far/DOF): acuity of viewer at set distance projected onto the sensing plane via the displayed photograph size. From there it's just a hop, skip and a jump to MTF50 and back ;-)

Let me work through a few of these experiments and I'll get back to you on that. SQF?

Jim

--
http://blog.kasson.com

 

[alanr0]

JimKasson wrote:.

I don't think that applies here. There are only three kernels to convolve with no AA filter, and four with an AA filter. In the case of leaving out lens aberrations, there are one less than those numbers in either case. So, consider a case with no AA filter, and a pillbox of diameter 100 um and an Airy disk with 3 um between the first zeros. My claim is that the convolution of those two is essentially a pillbox, not a Gaussian.

You are right, the convergence to Gaussian is slow, and I wouldn't expect that to work for a small number of variables.

Agreed. Regarding Jerry's CoC formula request, root sum of squares addition can work reasonably well to estimate the radius, provided you figure out what measure of radius is appropriate for distributions with completely different shapes.

Once the central limit theorem kicks in, and everything is Gaussian, RMS radius works nicely. For a slightly blurred pill-box, I would probably go for width at half maximum intensity, but I haven't checked how well that works with an Airy disk.

It can go pretty fast if all the kernels are similar. Here's how fast the 1D version of a pillbox converges to a Gaussian with repeated convolutions:

http://oscar6echo.blogspot.com/2012/10/convolve-n-square-pulses-to-gaussian.html

But, you're right, in general.

As lenses tend to be either over or under corrected, a pure disk and diffraction and AA filter seems reasonable, unless you want to also add a factor for over or under correction of spherical aberration.

What would that kernel look like?

The kernel varies with defocus plane.

For positive spherical aberration, the near (lens) side of focus has a blurred version of your pill box with a dip at the centre ("doughnut bokeh"). On the far side, the intensity peaks on-axis but spreads out farther than at an equivalent defocus distance on the short side.

One way to calculate this is to model the optical field incident on a circular aperture in a 2-D array, and Fourier (or Hankel) transform to estimate the point-spread function. Image plane amplitude squared gives the intensity. Constant amplitude and phase across the lens aperture gives you the diffraction-limited Airy disk. Phase proportional to square of radius within the aperture simulates defocus. An additional phase error proportional to fourth power of radius represents spherical aberration.

This is essentially a Fresnel diffraction calculation, simplified by omitting most of the inclination and distance factors, and corresponds to coherent monochromatic illumination. You would repeat and sum over multiple wavelengths for a realistic picture of the incoherent intensity distribution.

Arguably over the top for your blur kernel, but potentially more accurate than assuming the Airy disk diffracted intensity profile is unchanged with defocus, and a reasonable route to analysing spherical aberration, coma and other aberrations.

Cheers,

 

[JimKasson]

JimKasson wrote:.

I don't think that applies here. There are only three kernels to convolve with no AA filter, and four with an AA filter. In the case of leaving out lens aberrations, there are one less than those numbers in either case. So, consider a case with no AA filter, and a pillbox of diameter 100 um and an Airy disk with 3 um between the first zeros. My claim is that the convolution of those two is essentially a pillbox, not a Gaussian.

You are right, the convergence to Gaussian is slow, and I wouldn't expect that to work for a small number of variables.

Agreed. Regarding Jerry's CoC formula request, root sum of squares addition can work reasonably well to estimate the radius, provided you figure out what measure of radius is appropriate for distributions with completely different shapes.

Once the central limit theorem kicks in, and everything is Gaussian, RMS radius works nicely. For a slightly blurred pill-box, I would probably go for width at half maximum intensity, but I haven't checked how well that works with an Airy disk.

It can go pretty fast if all the kernels are similar. Here's how fast the 1D version of a pillbox converges to a Gaussian with repeated convolutions:

http://oscar6echo.blogspot.com/2012/10/convolve-n-square-pulses-to-gaussian.html

But, you're right, in general.

As lenses tend to be either over or under corrected, a pure disk and diffraction and AA filter seems reasonable, unless you want to also add a factor for over or under correction of spherical aberration.

What would that kernel look like?

The kernel varies with defocus plane.

For positive spherical aberration, the near (lens) side of focus has a blurred version of your pill box with a dip at the centre ("doughnut bokeh"). On the far side, the intensity peaks on-axis but spreads out farther than at an equivalent defocus distance on the short side.

One way to calculate this is to model the optical field incident on a circular aperture in a 2-D array, and Fourier (or Hankel) transform to estimate the point-spread function. Image plane amplitude squared gives the intensity. Constant amplitude and phase across the lens aperture gives you the diffraction-limited Airy disk. Phase proportional to square of radius within the aperture simulates defocus. An additional phase error proportional to fourth power of radius represents spherical aberration.

This is essentially a Fresnel diffraction calculation, simplified by omitting most of the inclination and distance factors, and corresponds to coherent monochromatic illumination. You would repeat and sum over multiple wavelengths for a realistic picture of the incoherent intensity distribution.

Arguably over the top for your blur kernel, but potentially more accurate than assuming the Airy disk diffracted intensity profile is unchanged with defocus, and a reasonable route to analysing spherical aberration, coma and other aberrations.

Cheers,

 

[JimKasson]

J

Yes, and note the quote, "(with σ equal to the square root of the sum of squares of the original two sigmas)."

This (It seems to me) supports the Image Clarity version with the square root of the sum of squares. Their formula is the one that also satisfies the Goldilocks principle: It is the neither the floor of max(c,d) nor the ceiling of c+d. It is between them.

Yes, an approximation.

I was looking for some formula incorporating the special nature of diffraction CoCs, but that will have to wait. For today, I'm going with the square root of the sum of squares of c and d.

Here is knowledgeable support for your approach:

http://www.dpreview.com/forums/post/57847956

Although, to apply it, you'll need to come up with a way to decide what the radius of the Airy solid is. The usual numbers are to the first zero, but Alan's suggestion of the half-peak intensity would probably be better if you're looking for the half-peak intensity of the result. Of course, the half-peak intensity of a pillbox is the same as its radius (or it doesn't exist, if you're being pedantic).

On another front, I Iooked over the sim code (it's a big, complex program and I forget some of the details), and found that I trim the combined kernel, discarding small-valued peripheral elements to speed the computations. With a little tweaking, I can capture its size, once we've decided on how to measure the size.

Jim

--
http://blog.kasson.com

 

[bclaff]

However that's not a normal situation for the average photographer. A more typical situation is displayed size, viewing distance and viewer dependent. The smallest discernible detail on the output photograph is related to the smallest discernible detail on the retina of the observer. We can estimate the smallest discernible detail on the retina of a typical photographer and project it onto the sensing plane to determine the CoC to use in DOF calculations. Here is how it works with the following assumptions:

• maximum visual acuity of va cycles/degree on the retina for the typical viewer
• printed photograph of height ph
• viewed from distance d
• sensor height sh

Then CoC diameter on sensing plane = 2 * d * tan(1/(2*va)) * sh/ph

in the same units as d, as better explained here. With minimum CoC equal to pitch.

For instance for a photograph displayed with height 300mm, viewed from 400mm, with va = 50 c/d (campbell-Robson), captured by a FF sensor 24mm high the CoC would be 11.2um. Different assumptions, different CoC. Incidentally I believe this is the logic by which Bill Claff arrives at his PDR.

Yes. In fact take your equation and divide through by 'sh'
Then (CoC diameter on the sensing plane)/sh = 2 * tan(1/(2*va)) * (d / ph)
So for a given d/ph the left hand side is a constant.
In my case the constant is 1/800 and (CoC diameter on the sensing plane) = sh / 800

Regards,

 

[ThrillaMozilla]

Nice. It works out pretty close to the classical calculation except close to the focal plane where diffraction kicks in around f/8 and above.

 

[Jerry Fusselman]

What distances are the horizontal axis referring to? Are we looking at distances from the center point of the image formed with a full-frame camera?

Maybe I just need a list of assumptions used to get this graph.

I guess when I said this it was too much in short hand:

Take a look at this MTF50 plot for a diffraction-limited 55 mm lens focused at 10m with MTF50 in cy/ph versus actual object distance

The horizontal axis is object distance from the center of the lens, measured along the lens axis. The lens is focused at 10 meters.

Does that help?

Jim

 

[ThrillaMozilla]

Also, you said that you can convert this information into depth of field. How would you do that?

Well, 24 mm height divided by 0.030 mm circle of confusion = 800 cycles/image height, assuming I didn't make a mistake of a factor of 2. Oooh.

 

[bclaff]

Also, you said that you can convert this information into depth of field. How would you do that?

Well, 24 mm height divided by 0.030 mm circle of confusion = 800 cycles/image height, assuming I didn't make a mistake of a factor of 2. Oooh.

Yup.

 

[Jerry Fusselman]

JimKasson wrote:.

I don't think that applies here. There are only three kernels to convolve with no AA filter, and four with an AA filter. In the case of leaving out lens aberrations, there are one less than those numbers in either case. So, consider a case with no AA filter, and a pillbox of diameter 100 um and an Airy disk with 3 um between the first zeros. My claim is that the convolution of those two is essentially a pillbox, not a Gaussian.

You are right, the convergence to Gaussian is slow, and I wouldn't expect that to work for a small number of variables.

Agreed. Regarding Jerry's CoC formula request, root sum of squares addition can work reasonably well to estimate the radius, provided you figure out what measure of radius is appropriate for distributions with completely different shapes.

Once the central limit theorem kicks in, and everything is Gaussian, RMS radius works nicely. For a slightly blurred pill-box, I would probably go for width at half maximum intensity, but I haven't checked how well that works with an Airy disk.

Terrific! For a formula involving c and d, is Merkinger's (pp. 31, 49--51) d = 5 um at f/8 reasonable? That is, I would use d = (5 um)*N/8, where N is the f number.

 

[JimKasson]

Also, you said that you can convert this information into depth of field. How would you do that? Maybe an example.

I took a crack at that here:

Loading…

www.dpreview.com

Jim

 

[JimKasson]

What distances are the horizontal axis referring to? Are we looking at distances from the center point of the image formed with a full-frame camera?

Maybe I just need a list of assumptions used to get this graph.

I guess when I said this it was too much in short hand:

Take a look at this MTF50 plot for a diffraction-limited 55 mm lens focused at 10m with MTF50 in cy/ph versus actual object distance

The horizontal axis is object distance from the center of the lens, measured along the lens axis. The lens is focused at 10 meters.

Does that help?

m

Yes, thanks.

What made the difference? Was it that part about measuring the distance alone the lens axis?

I'll keep sharing my ignorance:

Still, a list of assumptions would be appreciated.

Diffraction-limited lens

Diffraction calculated at 450, 550, and 650 nm

Bayer CFA, FF 24 MP

No aberrations

AHD demosaicing

Zero photon noise

Zero read noise

Zero pattern noise

Zero PRNU

Jim

--
http://blog.kasson.com

 

[Jerry Fusselman]

Also, you said that you can convert this information into depth of field. How would you do that? Maybe an example.

I took a crack at that here:

http://www.dpreview.com/forums/post/57847189

Jim

--
http://blog.kasson.com

Yes, thanks, I see. What kind of results might you get if you focused at infinity? (I don't know hard these kinds of changes are to make in your model.)


Also, what about the other side of 10 meters?

I don't want to delay your schedule, but these are the two cases I'm now curious about, if you get around to it.

--
Jerry Fusselman

 

[Jerry Fusselman]

What distances are the horizontal axis referring to? Are we looking at distances from the center point of the image formed with a full-frame camera?

Maybe I just need a list of assumptions used to get this graph.

I guess when I said this it was too much in short hand:

Take a look at this MTF50 plot for a diffraction-limited 55 mm lens focused at 10m with MTF50 in cy/ph versus actual object distance

The horizontal axis is object distance from the center of the lens, measured along the lens axis. The lens is focused at 10 meters.

Does that help?

m

Yes, thanks.

What made the difference? Was it that part about measuring the distance alone the lens axis?

I could be the only one who did not instantly understand "with MTF50 in cy/ph versus actual object distance". I did not immediately understand that your "versus" meant "with the following on the horizontal axis." Not a big deal, but a lot of lingo strung together can be hard to read by the less initiated. It can seem dense, and the reader might be dense.

I'll keep sharing my ignorance:

Still, a list of assumptions would be appreciated.

Diffraction-limited lens

Diffraction calculated at 450, 550, and 650 nm

Bayer CFA, FF 24 MP

Ah, this is the assumption I most wanted to see.

No aberrations

AHD demosaicing

Zero photon noise

Zero read noise

Zero pattern noise

Zero PRNU

Jim

 

[JimKasson]

What distances are the horizontal axis referring to? Are we looking at distances from the center point of the image formed with a full-frame camera?

Maybe I just need a list of assumptions used to get this graph.

I guess when I said this it was too much in short hand:

Take a look at this MTF50 plot for a diffraction-limited 55 mm lens focused at 10m with MTF50 in cy/ph versus actual object distance

The horizontal axis is object distance from the center of the lens, measured along the lens axis. The lens is focused at 10 meters.

Does that help?

m

Yes, thanks.

What made the difference? Was it that part about measuring the distance alone the lens axis?

I could be the only one who did not instantly understand "with MTF50 in cy/ph versus actual object distance". I did not immediately understand that your "versus" meant "with the following on the horizontal axis." Not a big deal, but a lot of lingo strung together can be hard to read by the less initiated. It can seem dense, and the reader might be dense.

I'll keep sharing my ignorance:

Still, a list of assumptions would be appreciated.

Diffraction-limited lens

Diffraction calculated at 450, 550, and 650 nm

Bayer CFA, FF 24 MP

Ah, this is the assumption I most wanted to see.

No aberrations

AHD demosaicing

Zero photon noise

Zero read noise

Zero pattern noise

Zero PRNU

Jim

 

[Jack Hogan]

JimKasson wrote: It's not that MTF50 is such a wonderful measure of sharpness by itself. In fact, I think I remember, but can't access right now, studies that indicated that MTF90 -- the spatial frequency at which the modulation transfer function falls to 0.9, is a better measure. But usually the SNR at MTF50 is high enough that it can be turned into MTF90 in post if need be.

I don't know if one could call it a 'study', but here is a stab at it.

 

[Jack Hogan]

There are many photographers -- and I am one of them a lot of the time -- who don't have a final use (or at least not a complete list of possible final uses) in mind at the instant of exposure. Many of them us want everything in a distance range to be pretty nearly as sharp as it can be given the camera and the lens. Then the issue of how far down to stop turns into an issue of how much loss in resolution is tolerable.

Now I understand what you are driving at (duh!)

I first started thinking about this when you produced the first loCA plots a few months ago. If you take the peak of the luminance curve to be in acceptable focus, with minimal trig you can easily calculate depth of focus from the plot, hence depth of field, based on an acceptable loss of 'sharpness' - say 10% lower MTF50. The problem I had then, and that I have had in answering your question now, is that we do not know what peak focus represents to the viewer. It could mean tack sharp to unwatchably blurry based on photograph size and viewing distance. So what does 10% less than that mean?

Now that I better understand your question, here is one way to start looking at it:

1. calculate the minimum acceptable MTFxx criterion for sharpness given typical print size and viewing conditions (say 1000 lp/ph below).
2. read off the DOF as a function of f-number based on the MTFxx plot of the lens obtained at the desired subject distance

3. Come up with a model so that one does not need to measure and produce such laborious MTFxx plots for every lens in their possession at every distance - and hopefully develop a few rules of thumb

As long as one is not too finicky I think (3) can be done relatively easily only for well corrected lenses near the center of the FOV. After that we need alanr0 or Brandon. Here is a start. If you think it might be helpful I have a more complex version of it that includes terms for defocus, AA and generic aberrations as shown here .

Jack