Hyperfocal Distance & Circle of Confusion

[Wil_Bloodworth]

As engineers and "thinkers" in general, we are taught to never assume (much). As I was memorizing my hyperfocal distances for various focal lengths at my most-used apertures, I came across the common accepted value of 0.03 for all of the Nikon full-frame sensors.

To recap, hyperfocal distance is (f^2)/(A*c) + f where f is the focal length, A is the aperture used and c is the circle of confusion for the particular camera body you want to use. For example, on the D700 (which an assumed CoC of 0.03) using a 24mm lens stopped down to F8, the hyperfocal distance would be (24*24)/(8*0.03)+24 = 576 / 0.24 = 2400 mm or 7 feet 10.5 inches.

That's all good... except. There are a few different formulas for calculating the circle of confusion. They all involve the length of the diagonal of the sensor or piece of film being used. The Zeiss formula is d/1730 where d is the length of the diagonal. Another accepted forumula is d/1500... and there are a couple of others.

The D700/D3/D3S/D3X has a physical sensor size of 36mm x 23.9mm. That gives a diagonal of roughly 43.21mm. Using that number in the two formulas mentioned gives:

Zeiss CoC: 0.02498
d/1500: 0.02880

Sure, those numbers are "in the ballpark" as far as being close to 0.03. But, when calculating hyperfocal distance, that delta makes a noteworthy difference.

In the first example I used the 0.03 CoC. If we substitue the Zeiss number of 0.02498, then the hyperfocal distance would be 2906.3 mm or 9' 6.4".

Just under 8 feet versus a little over 9.5 feet is quite a difference.

Now, given that the near distance (h/2) effectively halves the delta where it's most important, is a >9-inch delta still important?

Anyone care enough to discuss this?

- Wil

[jeffkohn]

One important thing to remember about DOF is that it's just an approximation of what will look acceptably sharp in a print of a given size. There's still only one distance/plane that is truly in-focus; everthing else is out of focus to varying degrees, just maybe not enough to look that way in the print.

The CoC is the fudge factor that lets you compensate for different print sizes and different standards for sharpness. Larger prints require smaller CoC's. The .03mm CoC is not a very good number to use for full-frame DSLR unless you're only shooting for the web or 8x10 prints. Larger prints will require a smaller CoC.

At the very least, I would recommend using the Zeiss number. And even then you may want to stop down an extra f/stop from what the DOF calculator says to give you some extra DOF for larger prints (assuming this doesn't push you too far into diffraction territory).

On my D300, I used a CoC of .014. For the D3x, I use .018. My thinking is that the D3x pixels are not much larger than the D300 pixels; and while it's true there are more of them, I'm using them to print larger. So the actual enlargement factor for a given pixel is not all that different.

This means that my acceptable DOF is more limited for a given focal length and f/stop than would be traditionally calculated. But I didn't buy the D3x to make 8x10 prints, I bought it for the capability to print big; and .03mm wouldn't give me results that I'm happy with. This can make getting enough DOF a struggle, which is part of why I prefer to use tilt/shift lenses when I can.

Another thing to remember, you don't necessarily want to use the hyperfocal distance unless you really need it. While infinity is supposedly sharp when focused at the hyperfocal distance, it won't be as sharp as if you actually focus at infinity. On the other hand, if you have a prominent foreground subject, making it appear just a little sharper than infinity can sometimes be desirable, because we're used to things getting a little hazy in the distance when looking with our eyes. It really just depends on the composition and what you're trying to achieve.

[Wil_Bloodworth]

Jeff,

Thank you. How did you decide on 0.014 and 0.018? Did you reverse-engineer those from a specific print size the you are targeting?

Regarding your last paragraph: I agree completely, which is why adding a little bit of tilt sharpens everything in the FG & BG much nicer than just stopping down.

- Wil

[jeffkohn]

The .014 was something I picked up a few years ago. I think it was a comment on DPReview by either Thom Hogan or Iliah Borg, but it may have a technical article I came across on the web. It's a pretty pessimistic CoC, but when printing up to 16x24" with 12mp images, you can't use the same CoC you would for an 8x10 and expect optimal results.

The .018 was something I chose just based on the fact that my enlargement factor for 24mp at 20x30" is slightly less than in the 12mp case. It seems to work reasonably well.

[Wil_Bloodworth]

Good to know. Thanks. I will adjust accordingly.

- Wil

[ggeen]

In the spring 2010 issue of Nikon World, this topic was discussed. Specifically, the author tells us a quick "trick" for finding the hyper focus on /some/ Nikon lenses, specifically those with aperture markings.

Secret Weapons from Nikon

[Jake]

Originally Posted by Wil_Bloodworth

As engineers and "thinkers" in general, we are taught to never assume (much). As I was memorizing my hyperfocal distances for various focal lengths at my most-used apertures, I came across the common accepted value of 0.03 for all of the Nikon full-frame sensors.

To recap, hyperfocal distance is (f^2)/(A*c) + f where f is the focal length, A is the aperture used and c is the circle of confusion for the particular camera body you want to use. For example, on the D700 (which an assumed CoC of 0.03) using a 24mm lens stopped down to F8, the hyperfocal distance would be (24*24)/(8*0.03)+24 = 576 / 0.24 = 2400 mm or 7 feet 10.5 inches.

That's all good... except. There are a few different formulas for calculating the circle of confusion. They all involve the length of the diagonal of the sensor or piece of film being used. The Zeiss formula is d/1730 where d is the length of the diagonal. Another accepted forumula is d/1500... and there are a couple of others.

The D700/D3/D3S/D3X has a physical sensor size of 36mm x 23.9mm. That gives a diagonal of roughly 43.21mm. Using that number in the two formulas mentioned gives:

Zeiss CoC: 0.02498
d/1500: 0.02880

Sure, those numbers are "in the ballpark" as far as being close to 0.03. But, when calculating hyperfocal distance, that delta makes a noteworthy difference.

In the first example I used the 0.03 CoC. If we substitue the Zeiss number of 0.02498, then the hyperfocal distance would be 2906.3 mm or 9' 6.4".

Just under 8 feet versus a little over 9.5 feet is quite a difference.

Now, given that the near distance (h/2) effectively halves the delta where it's most important, is a >9-inch delta still important?

Anyone care enough to discuss this?

- Wil

I couldnt follow anything after the word "As" but thanks for the write up. It sounds right'ish to me!

[Wil_Bloodworth]

LOL Jake! That's awesome.

- Wil

[srwatters]

I've always used a Continuum Transfunctioner to calculate CoC for my work. But then again what do I know?

[PlanetFinder]

Wil wrote: "As I was memorizing my hyperfocal distances for various focal lengths at my most-used apertures, I came across the common accepted value [for CoC] of 0.03 for all of the Nikon full-frame sensors."

Your right, that this is "commonly accepted," for full-frame 35 mm cameras. See Online Depth of Field Calculator , where CoC is given as 0.030 mm, for example, for the Nikon D3, D3x, and D3s.

But there is something wrong here. The D3x has sqrt(2) more rows and columns of pixels than the D3 and D3s. The CoC must depend on the pixel size as long as the pixel size is comparable to the size of the blurring function! We are in that regime when we are seeking the best possible focus (otherwise all those pixels are expensive folly).

Here is my analysis for the Nikon D3x. I'm using the fact that observed onset of blurring due to diffraction provides an experimental estimate of CoC. Based on this, I think CoC = 0.012 mm is more reasonable estimate of the true value.

Step by step argument:

1. The D3x offers a pixel spacing of 36/6048 = 0.006 mm per pixel

2. At f/20, the D3x pixel spacing critically samples a diffraction-limited image for lamda > 600 nm.

3. Critical sampling is defined as two samples per full width at half maximum, FWHM.

3. For N > 20 (N = f-number), I experimentally detect a degradation in the sharpness of an picture. The degradation increases with N. I presume this degradation is due to diffraction.

4. The effect of diffraction is convolution with the Airy function, which has FWHM = lamda*N.

5. The degradation of sharpness due to convolution with all symmetric and centrally peaked blurring functions with the same FWHM will be similar.

6. The circle of confusion (CoC) due to focus error corresponds approximately to the FWHM of the blurring function associated with focus error.

7. Therefore, for lamda << 600 nm (where diffraction can be neglected) CoC = 600 * 20 nm = 0.012 mm

8. For lamda approximately equal to 600 nm, the image is blurred by BOTH diffraction and focus error, which means CoC < 0.012.

I would be very interested to hear your further views on this topic.

[andyz]

As someone who reads banking regulations for a living and quotes cites §226.18(d)(1)(i) for an understated finance charge rule, you guys really need to get a life.

I've read a little on the CoC and DoF calculations, but I've never really wondered THAT much about it. Thats why we have iPhone apps.

[Wil_Bloodworth]

Originally Posted by PlanetFinder

But there is something wrong here. The D3x has sqrt(2) more rows and columns of pixels than the D3 and D3s. The CoC must depend on the pixel size as long as the pixel size is comparable to the size of the blurring function! We are in that regime when we are seeking the best possible focus (otherwise all those pixels are expensive folly).

Absolutely!

Originally Posted by PlanetFinder

Here is my analysis for the Nikon D3x. I'm using the fact that observed onset of blurring due to diffraction provides an experimental estimate of CoC. Based on this, I think CoC = 0.012 mm is more reasonable estimate of the true value.

Step by step argument:

1. The D3x offers a pixel spacing of 36/6048 = 0.006 mm per pixel

Technically, it's 35.9 / 6048 = 0.00594... but that's just being nit-picky. And this also assumes that the pixels take up 100% of the sensor space... as we know they do not. There are spaces between each pixel well. Each of the D3X pixel wells is (supposedly) 5.49 microns in diameter. I believe the space between the pixel wells is 0.25 microns. That's not much but it does play a part when you multiply that by the total number of rows (minus 1). 6047 * 0.25 = 1511.75 microns worth of space.

Of course, since 1 millimeter = 1000 microns, we're really only losing 1.51175 millimeters. But since we're having a technical discussion, it may be important. Now, the sensor size, sans spacing, is really 35.9mm minus 1.51175mm or 34.38825mm.

So, now we're down to 34.38825 / 6048 = 0.005686... Is that important? Maybe; maybe not.

Originally Posted by PlanetFinder

2. At f/20, the D3x pixel spacing critically samples a diffraction-limited image for lamda > 600 nm.

3. Critical sampling is defined as two samples per full width at half maximum, FWHM.

Yes.

Originally Posted by PlanetFinder

3. For N > 20 (N = f-number), I experimentally detect a degradation in the sharpness of an picture. The degradation increases with N. I presume this degradation is due to diffraction.

Hmmm. From the test chart samples I've seen, for the D3X, on the DigLloyd site, diffraction seems to set in right as you get to f/11 and dramatically negatively affects image quality thereafter.

Originally Posted by PlanetFinder

4. The effect of diffraction is convolution with the Airy function, which has FWHM = lamda*N.

5. The degradation of sharpness due to convolution with all symmetric and centrally peaked blurring functions with the same FWHM will be similar.

6. The circle of confusion (CoC) due to focus error corresponds approximately to the FWHM of the blurring function associated with focus error.

7. Therefore, for lamda << 600 nm (where diffraction can be neglected) CoC = 600 * 20 nm = 0.012 mm

8. For lamda approximately equal to 600 nm, the image is blurred by BOTH diffraction and focus error, which means CoC < 0.012.

I would be very interested to hear your further views on this topic.

This is fairly close to what Jeff Kohn suggested with 0.018mm... and definitely A LOT closer than 0.03mm.

While it would be easy to "overdo it", I think anywhere between your number and the 0.018mm number is much better than the "wild ass guess" that is 0.03. Thanks for your contribution Robert!

- Wil

[donlfaulkner]

Originally Posted by thejakestir

I couldnt follow anything after the word "As" but thanks for the write up. It sounds right'ish to me!

I'm with Jake.

[jeffkohn]

Robert,

I don't question the theory behind your calculations, but I think the CoC you arrived at may be a little pessimistic for real-world photographs.

Also keep in mind, when calculating DOF based on pixel pitch, you still need to factor in print size. It's true the D3x will show focus blur sooner than a D3s at 100% pixel view, but that will only show up in print if you're blowing up the D3x image to a larger print size than the D3s image. If they're both being printed at 12x18", for instance, the D3x doesn't need to use a smaller CoC than the D3s.

[jeffkohn]

Originally Posted by andyz

As someone who reads banking regulations for a living and quotes cites §226.18(d)(1)(i) for an understated finance charge rule, you guys really need to get a life.

I've read a little on the CoC and DoF calculations, but I've never really wondered THAT much about it. Thats why we have iPhone apps.

Somebody's got to worry about it, or there wouldn't be any iPhone apps. It may not be of much concern if shooting for the web or small prints; but the larger the print the more you have to worry about this stuff (assuming you care about print quality).