DoF - different formats?

RaymondC · July 4, 2014

Hi all - I had a look online with DoF calculators but they ask for a distance to your subject.

With 35mm people use F8 or F11. I tend to use the latter more. Is F8 ok for landscapes? When we move to 6x6 or 4x5 what is generally the equivalent f stop?

craigd · July 4, 2014

Is there a problem with providing a distance? The calcuator needs a distance because DoF varies with focus distance as well as focal length, aperture, and circle of confusion (CoC). If you're mostly shooting landscapes, you can probably just use the hyperfocal distance.

If you want to make "the same picture" (that is, similar framing, DoF, and exposure, printed at the same size) with two different formats of camera (say 35mm and 6x7), shooting from the same location, you will need to use a longer focal length with the larger format to match the framing, and because of the longer focal length, you will need a smaller aperture to match the DoF. At the same time, though, one typically uses a larger CoC with a larger format because the image won't need to be enlarged as much to reach the same print size (that is, to make, say, an 8x10" print, a 36x24mm negative needs to be blown up more than a 70x56mm negative), which will tend to partly counteract the need for a smaller aperture. You can fiddle with the different parameters to try to get the final images from both cameras to match as closely as possible, but it's really kind of an academic exercise because nobody would bother doing this in the real world other than as a technical experiment.

As a general rule of thumb, when shooting landscapes or other distant subjects, I would tend to use f/8 with a 35mm camera, f/11 or f/16 with medium format, and f/16 or f/22 with large format.

William Michael · July 4, 2014

The (approximate) “Equivalence” in F/ Stops correct to ⅓Stop, in respect of Depth of Field between Formats / Negative Size is:

135 Format to 6 x 4.5 Format: increase 1 Stop

135 Format to 6 x 6 Format: increase 1⅔ Stops

For clarity “increase” means to stop down to a SMALLER aperture. e.g. if you use F/8 on a EOS 5D and you want to maintain “Equivalence” then you need to use F/11 on a 6 x 4.5 and F/14 on 6 x 6.

WW

Andrew Garrard · July 4, 2014

Depth of field is determined by the absolute size of the aperture and the distance to the subject - there's a "cone of confusion" heading out into the scene from the aperture to any point on the focal plane. The relative aperture (I may be abusing terminology here, but I hope unambiguously...) determined by the f-stop is the focal length divided by the absolute aperture - that's why f-stops are described as fractions such as "f/8": "f" is the focal length (so a 200mm f/2 lens has an absolute aperture size of 100mm - 10cm - which is why the front element is so big).

If you're keeping the same field of view as you enlarge the format, the focal length increases proportionately to the change in film (or sensor) size. Therefore, at the same f-stop, the absolute aperture increases proportionately. To balance things out, reduce the relative aperture by the amount you're enlarging the film size: 10x8 is twice the (linear) size of 5x4, so you need half the relative aperture - remembering that the aperture is a fraction, so half of f/8 is f/16. 135 film and 5x4 aren't the same aspect ratio, but if we match the angle of view across the diagonal, the size difference is sqrt(52 + 42) * 25.4 / sqrt(362 + 242) = roughly 3.75, so f/8 would become f/(8 x 3.75) = f/30 - assuming you're using a (3.75 x 50 =) 187.5mm lens to match the angle of view of the 50mm lens you were using on 35mm film.

Having said all that, bear in mind that diffraction kicks in and reduces resolution. Depending on what I'm shooting, for maximum quality, I quite often try to stick at f/8 or even f/5.6 for this reason. To get the foreground and background in focus, there's always focus stacking. On a large format camera, hopefully you have lens movements to put the focal plane in the right place to solve this. Diffraction also has less of an effect as the format gets larger (or the same amount constitutes less of the image area), hence the large format "f/64" club actually got useful images - though that's possibly going a bit far for general use.

If you're just comparing the formats rather than trying to work out a specific image, this relationship applies. If you try to work out the absolute depth of field, that depends on the subject distance, the amount you're enlarging the result by, how closely you're looking at the image... which I always find to be a faff. Which is why I prefer knowing the maths to using a DoF calculator. (While I believe the common ones are accurate, there are also a few out there which get the maths wrong, which is another reason to do it yourself!)

Hope that helps.

Alan Marcus · July 4, 2014

Your question gyrates around “hyperfocal distance”:

Setting your camera to a specific distance called the hyperfocal point delivers the maximum span of depth of field that just kisses off infinity ∞. Fixed focus “Brownie” cameras are set to this value and landscape photographers routinely set focus to this value. This is because the span of acceptable focus starts at ½ the hyperfocal distance and stretches to infinity ∞. Let me add that it is a special case of depth of field calculations. You will find it listed on line with all the depth of field calculators. You can make your own calculations; it’s easy!

First we must agree on the criteria that constitutes optical sharp. We are talking about the size of the circles of confusion. These are super tiny disks of light projected by the lens onto film or digital sensor. They must remain super tiny so they are perceived as points of light and not disks by the observer. That means they must be tiny enough to withstand enlargement. Generally charts and tables are based on making a final print or display that measures 8x10 inches. Most will be based on a circle size that is 1/1000 of the focal length. For scientific and serious work Leica uses 1/1750 and Kodak 1/1500. It is customary to use 1/1000. The basis is: the circles remain just below the appearance of a disk at a viewing distance of 20 inches (500mm). This dictates the final circle size shall be no greater than 1/50 inch or 0.5mm.

Now for the math:

f = focal length

a = aperture (f/#)

d = diameter of aperture

c = circle of confusion diameter

Say you mount an 80mm lens set to f/8

The aperture diameter is 80 ÷ 8 = 10mm thus d=10

c = 80 ÷ 1000 = 0.08mm

Formula = fd/c

80 x 10 ÷ 0.08 = 10,000 (this is the hyperfocal distance in millimeters

10,000 ÷ 2.5 = 394 inches

394 ÷ 12 = 33 feet.

Set your camera to 33 feet and everything will be in focus from ½ the hyperfocal distance to infinity ∞.

Thus the range of acceptable focus is 16 ½ feet to infinity ∞.

Once more for a 100mm lens set to f/11

The aperture diameter is 100 ÷ 11 = 9mm thus d=9

c = 100 ÷ 1000 = 0.10mm

Formula = fd/c

100x9÷0.10=9,000mm

9,000 ÷ 25.4 = 354 inches

354 ÷ 12 = 29 ½ feet

Set to this distance, everything will be in acceptable focus 15 feet to infinity ∞.

Andrew Garrard · July 4, 2014

<blockquote>First we must agree on the criteria that constitutes optical sharp.</blockquote>

As Alan showed, the maths are not too difficult to apply. (And I trust him - I've not checked the reasoning!) Hyperfocal settings are useful. However, I do think the agreement about what's optically sharp is not a "one size fits all" conclusion. Take a wide-angle shot, print it huge so you get an immersive effect when viewed at a reasonable distance, and suddenly things are more demanding. Know you're targetting a thumbnail for the web, and you can be more lax. If you're aiming at the 8x10 print that Alan mentions and a normal viewing distance, I'm sure the numbers work - just be wary that they don't continue to apply afterwards. Or if you decide to make an 8x10 from a crop of the image, for example. It's good to know in an absolute sense what would constitute "sharp", but I'd tend to remember it by "f/8 will do", for example. (And besides, it may be worth some diffraction if you really need something nearer to the camera to appear in focus.) Then knowing the relationship between film formats allows you to apply that knowledge. YMMV.

Another calculation you can consider with a digital sensor is the size of the circle of confusion compared with the size of the sensor sites. If your image is as sharp as the sensor can resolve, there's no argument about the region that's "in focus". However, with modern high-density sensors, there's a good chance of running out of lens resolution, or the eye's ability to determine sharpness, first.

Alan - because I've not checked your maths... can I confirm how to interpret the focal length term in your formula as the film format changes? I suspect - for the same final image size - this focal length needs to be scaled by the sensor size (assuming that the relative aperture has been selected to give the same absolute aperture and we're using a longer focal length on the larger format). I'm too tired on a Friday to confirm that, though - I suspect you can get there faster.

john_wheeler6 · July 4, 2014

Hi Ray

While I love the equations here are some short answers

- DOF does depend on subject distance

- The table below gives you a "generally" equivalent DOF for the same angle of view (same scene with shot from same location). Both the focal length and the Aperture scale approximately by the diagonal of the film size.

For the three film formats in question I show the ~film diagonal, the ~focal length lens for same angle of view (50mm used as base for 35mm film), and the ~Aperture for the same DOF (with f/8 use as base for 35mm film).

Hope that gives you a quick feel without all the equations:

JDMvW · July 4, 2014

Good old Wikipedia is always a good place to start. http://en.wikipedia.org/wiki/Depth_of_focus

As a social scientist, I am amazed at how good Wikipedia usually is. My personal, rather glum, assessment of the human collective intelligence may have to be revised.

BTW, not the least of the uses of manual-focus lenses is that they usually, at least in the day when, had DoF markings on the distance scale. Not always exact, but eminently usable.

Alan Marcus · July 4, 2014

To Andrew

The size of the circle of confusion is the key to this discussion. It must be perceived as a point without dimension or the image appears unsharp. This stuff is all based on the fact that a coin viewed from 1/3000 of its diameter is perceived as a point without dimension to a person with 20/20 vision. For photographic purposes the industry adopts 1/1000 of the viewing distance which is equivalent 1/100 of an inch in diameter viewed from 10 inches or 2/100 of an inch viewed from 20 inches (standard reading distance). This is the criterion: 2/100 of an inch disk size is 1/50 of an inch = 0.02 of an inch x 25.4 = 0.5mm. This lower standard of 3.4 minutes of an ark is adapted because of the contrast of photographic image and uncorrected aberrations and the presence of flare. Elongate the viewing distance and the larger disk diameter is acceptable.

Because of all the different format sizes it has become customary to use 1/1000 of the focal length for the size of the disk at the focal plane. This assumes the image will be enlarged for viewing. As an example, a 35mm full frame image is enlarged 10X. A 50mm is mounted. Using the 1/1000 rule of thumb, the permissible size of the disk at the focal plane is 50 ÷ 1000 = 0.05mm. Now if we enlarge 10X making a print 9 ½ x 14 inches, the disk on the final image is 0.05 x 10 = 0.5mm. Thus the disk size meets the criterion.

The 1/1000 of the focal length is crude and assumed to take into account the fact that longer lenses will be mounted if the format size is large, and shorter lenses will be mounted if the format size is smaller. One guideline fits all. Again precision work requires 1/1500 or 1/1750, however most tables charts and online calculators use 1/1000.

AlanKlein · July 4, 2014

I shoot 6x7 120 medium format film. I don't start with the f/stop first but rather determine what DOF I want in focus. Looking at the tables or DOF range marks on my lens, I read the f stop required to handle that DOF but stop down one additional stop for good measure to assure the DOF will be satisfactory. I don't worry about refraction, because if the DOF focus isn't right, refraction won't matter. Also I've been told that refraction is not as big an issue as with 35mm or FF.

William Michael · July 4, 2014

JDM, the wiki link you supplied is to “Depth of Focus”

Here is the wiki link to “Depth of Field”: <a href="http://en.wikipedia.org/wiki/Depth_of_field">http://en.wikipedia.org/wiki/Depth_of_field</a>

***

As a social scientist, I am amazed at how good Wikipedia usually is

</blockquote>

I think it is a good starting point for many and a good general reference point for most. I think that un-referenced / un-footnoted claims need to be investigated and not just taken as gospel. I think that accepting unsubstantiated claims can re-write history and be a launching pad for untruths to become truths - and whilst such might not be serious when we are discussing the minutia of photographic optics: the principle does have a serious social impact of which we should all be aware.

*

The first line of the second paragraph at the wiki link on Depth of Focus is -

“While the phrase depth of focus was historically used, and is sometimes still used, to mean depth of field (DOF),”

That statement (my bolding for emphasis) does not have any reference in the footnotes. I have researched quite a few (older) texts and not yet ever noted that ‘Depth of Focus’ was historically used to mean ‘Depth of Field’.

I have noted that the phrases are certainly sometimes interchanged nowadays, though, which causes confusion, especially when technical detail requires an exact separation of meaning, or when the correct answer depends on an exam credit.

WW

Alan Marcus · July 4, 2014

Alan Klein,

Diffraction is a big problem regardless of format, it an optical phenomenon that plagues all lens and we can’t improve as per the well-studied works of Nobel laureate Lord Ryleigh, England Astronomer and Physicist published as the Rayleigh Criterion and it still stands today.

Resolving power decreases with aperture: Resolving power = 1392 ÷ f/#

f/1 = 1392 lines/mm

f/2 = 696 lines/mm

f/2.8 = 497 lines/mm

f/4 = 348 lines/mm

f/5.6 = 249 lines/mm

f/8 = 174 lines/mm

f/11 = 127 lines/mm

f/22 = 63 lines/mm

f/32 = 44 lines/mm

Note: The resolving power at apertures larger that f/8 is higher than that of pictorially useful film emulsions.

q.g._de_bakker · July 4, 2014

Those figures produced by the Rayleigh criterion give us the maximum achievable resolution at each f-stop, assuming everything else that could limit resolution is not present. So the fact that these figures for apertures of f/8 and wider are beyond the capability of film and digital sensors should not be understood as implying that lenses do actually project images having such a high resolution. Only that - if they were perfect, and everything else too - they could (and then do no better than that). The thing with DoF and sharpness is not that points must be seen as points, without dimension per se. But that two adjacent points should be seen as two separate points, even though they are two blobs instead, that do overlap/merge to a quite considerable degree. There are a number of things that do have an influence on the shape and energy distribution inside such a blob (and thus on how close two of them can come before merging into one) besides aperture size and diffraction. So it's not a simple matter. But to keep it simple, i think the best way (and fully correct as well) to 'circumscribe' that thing called depth of field is to say that it depends on two things: f-stop and (inversely) magnification. Given the same angle of view in two pictures, made using the same f-stop, one produced using a 35 mm camera, the other using a 6x9 camera, the one on 6x9 will be larger, i.e. have greater magnification. So the 6x9 image will show less DoF. Until, that is, you enlarge the 35 mm film image to 6x9. Then the difference is gone. It's a simple matter (ignoring all other factors): the closer you look, i.e the higher the magnification, the better you will see that things are not sharp, but only appeared sharp because you weren't looking close enough. Because DoF is complex, involves more than just magnification and f-stop (our eyes, for instance, play a large role) mathematics may help to give an idea of whereabouts the limits could be, but no more than that. Maths and formulae do however create an impresson that there is something exact about DoF. There isn't. Too many largely unknown variables. So use the calculators. But be aware that none of them has the final word. DoF changes in ways not taken account of in those formulae. Hyperfocal distance is not the right approach to DoF. It describes one special instance (one that involves that 'rather precise' notion of infinity) and doesn't transfer well to any other instances. Using hyperfocal distance formulae is also responsible for the misconception that DoF also depends on focal length. It does, in as far as focal length and distance together make up magnification. But not as an extra, separate factor. Maths that is based on focal length, f-stop, aperture size and permissable size of the circle of confusion is wrong. Whatever it calculates, it is not DoF. We sometimes need DoF in our pictures. But since it is variable, depends on things beyond our control (such as the distance someone will be viewing the print at) a good (best) approach is to make sure that the best focus is on the part of the scene that needs to be sharp (always, no matter what) and then stop down until the rest also looks (as near as possible) the way we want. Using hyperfocal focussing will let the formulae decide where in the scene best sharpness will fall. Come too close to such a photo, or enlarge it more than the formulae assumed you would, and you will not only see that DoF is far less than was promised by the maths, but also that something that is not the main subject of the photo is sharp, while the main subject is not. So do focus on your subject, and do not let mathematics decide where focus will be. Confusing DoF is (even without that possible confusion between 'field' and 'focus'). One thing that is very clear about it though is that the name for that blob, circle of confusion, is very apt. ;-)

JDMvW · July 4, 2014

DoF

Close enough for government work, I say.

It seems to me that most of the discussion has been confus(ed/ing) about what exactly is asked, anyhow.

William Michael · July 4, 2014

"Close enough for government work, I say."

</blockquote>

haha. That's funny. I'll use that, please.

*

I understood the question was general in nature, kind of "chuck it out there I want to know what are your opinions and views"

Maybe I misinterpreted.

WW

Andrew Garrard · July 8, 2014

Oops, sorry, meant to respond to this sooner. I'm not sure how appropriate the below formulae are to the forum, but they're probably useful to have discussed somewhere, and "beginner at photography" doesn't necessarily mean "bad at maths". So I hope this is of interest to someone.

<blockquote>The 1/1000 of the focal length is crude and assumed to take into account the fact that longer lenses will be mounted if the format size is large, and shorter lenses will be mounted if the format size is smaller. One guideline fits all.</blockquote>

Assuming you're creating a print of the same size (and yes, there are different relationships if you make a different print size or move the camera, but at that point I tend to think "equivalence" is the least of your problems), I'm confused by this. I'm going to have to re-derive in order to work out where I'm confused...

Let's start with the <a href="http://en.wikipedia.org/wiki/Thin_lens_formula#Imaging_properties">thin lens formula</a>:

1/(lens-to-image distance) + 1/(lens-to-subject distance) = 1/(lens focal length)

where the former two distances refer to the optical centre of the lens. I'll use that Wikipedia article's terminology, S1 for the lens-to-object and S2 for the lens-to-image. You've called the (absolute) aperture diameter d, which I'll use as well.

On to our circle of confusion. If we want this to correspond to 0.5mm in a 10x8 print, we're looking at... well, actually a 10x8 isn't a direct map to most common sensors. Let's work with film formats instead. Let's say we have a larger film area of 10"x8", and a smaller film area of 5"x4". (We could work with digital sensor areas, but it doesn't do much but confuse the numbers.) The 10x8 capture is printed at 1:1, so the circle of confusion at the film plane is 0.5mm. The 5x4 capture is printed at 2:1, so the circle of confusion at the film plane is 0.25mm (so that it's 0.5mm when we scale up by two for the final print). I'll call this circle of confusion diameter c, as Alan did.

For hyperfocal distance, we want infinity to focus such that the circle of confusion is this large. We're focussing closer to the camera than infinity, so the image of infinity falls in front of the sensor (closer to the lens than the sensor) - lenses rack out as they focus closer. The image of infinity falls at the focal length (we'll call the lens's focal length f), by definition. By similar triangles, we can say:

(image-of-hyperfocal-distance - image-of-infinity)/c = image-of-infinity/d

or

(S2 - f)/c = f/d

...where S2 is the distance from the lens to the image of the hyperfocal distance. Rearranging, we can say:

dS2 - df = cf

and...

dS2 = cf + df

(wow, it's hard to rearrange formulae when editing raw html... please excuse the baby steps, but I want to check I've got this right), so...

S2 = f(c/d + 1)

We really want to know S1, the distance from the lens to the hyperfocal plane in the scene, not the distance from the lens to the film plane. We can get this from S2 with the thin lens formula:

1/S1 = 1/f - 1/S2

Deep breath... substituting the above value of S2, we can say:

S1 = 1/((1/f) - 1/(f(c/d + 1)))

...and that rearranges to:

S1 = 1/((1/f)(1 - 1/(c/d + 1)))

...and...

S1 = f/(1 - 1/(c/d + 1))

Woohoo. So, we have a hyperfocal distance that's proportional to the focal length, as Alan started out by saying. Conceptually, this makes sense - enlargement is proportional to focal length, and since the circle of confusion needs to be the image of the aperture projected to infinity, a longer focal length enlarges the background more and we need a more acute cone of confusion (and more distant focal plane) for the same results. However, it's also dependent on c and d. What happens when we change sensor format?

Remember that the circle of confusion at the sensor is larger for a larger, 10x8", film than for a smaller 5x4" film by a factor of 2; to produce the same final image, we need to magnify the captured image by 2x less to get the same result. Changing c modifies S1.

However, remember that the larger sensor also captures a wider angle of view, so it's conventional to use a lens that's 2x shorter for "equivalence". We can't just change f and assume everything stays the same, but one consequence of f changing is that, at the same relative aperture (a.k.a. f-stop), d changes proportionately to f. As the c term shrinks, if we've also reduced the focal length and kept the f-stop the same, d changes in proportion, meaning that - for a given f-stop - the hyperfocal distance is proportional to the focal length, irrespective of film or sensor format. Particularly, this means that the "hyperfocal" approach of positioning the far DoF marker (if the lens has them) for the current aperture on the infinity point is a valid approach irrespective of whether you're using a crop sensor DSLR or a full-frame one (though whether the depth of field itself was valid for your enlargement is another matter).

But we're not getting the same depth of field from the two cameras. To achieve that, we still need to scale f by 2 between the small and large films to compensate for the change in the field of view, but we need the (1 - 1/(c/d + 1)) term to compensate. Fortunately, we can simplify things:

S1 = f/((1 - 1/(c/d + 1))(c/d + 1)/(c/d + 1))

S1 = f/((c/d)/(c/d + 1))

S1 = f(c/d + 1)/(c/d)

S1 = f(1 + 1/(c/d))

S1 = f(d/c + 1)

Much better. But we can do more: unless we're doing something very wrong, the physical aperture of the lens is going to be a lot bigger than the circle of confusion. Therefore d/c should be a lot bigger than 1, and we can ignore the "1" in the above equation, to a good approximation. (And we started with the thin lens formula, which is already an approximation.)

S1 = fd/c (roughly)

If we've scaled the film size by ½ (let's say we're going from 10x8 to 5x4), c is smaller by 2 - but f is also smaller by 2, because we used a shorter lens to keep the same angle of view. For S1 to be unchanged, that means we can't change d when we switch formats. Reassuringly, that's the rule for most depth of field equivalency - you need to keep the physical aperture the same, because it should be possible to think of depth of field in terms of a cone of confusion projected in front of the camera, and what happens behind it shouldn't matter so long as the angle of view is unchanged. So with a shorter lens, you need to increase the relative aperture by the same factor as the focal length shrank in order to retain the same d - in this case, going from f/8 to f/4, for example. Although I'm curious that there's an extra "+1" in there.

In summary: I believe Alan - everything (approximately) cancels out so long as you're keeping the same physical aperture size. I'm slightly surprised that it doesn't exactly cancel out, but there could always be a bug in my maths.

What I still don't understand is what c has to do with focal length, as such. I get that it scales with captured image size, and that the captured image size for the same field of view scales with focal length, so equivalency sorts itself out. But for one camera, the size of the circle of confusion depends only on the perceived viewing angle of the print.

Now, it's true that to get an undistorted view, one should view the print from a distance that's proportional to the focal length - if you look at a shot taken with a 14mm lens from a "normal" viewing distance, you get "distorted corners", but if you print large and press your nose against the print it looks fine - except that you need a very high image quality for the middle of the frame not to look bad if you do that. The effect is minor enough at longer lengths that images don't tend to look weird, so we don't hear much about "telephoto distortion" and nobody talks about focal length when discussing the distance at which people view prints. Still, is this why c is supposed to be proportional to focal length?

That was fun. I may just have written one of the longest ever posts to get the answer "yes", if Alan is still reading.

<blockquote>Resolving power decreases with aperture: Resolving power = 1392 ÷ f/#</blockquote>

Rayleigh's criterion applies when viewing an image - it indicates that the first "well" of the diffraction pattern from one point coincides with the peak from an adjacent point, and therefore the boundary between the two is indistiguishable. (The 1392 figure is wavelength-specific, just completeness.) However, it's possible to apply a (de)convolution based on the expected diffraction pattern - at the cost, as ever, of enhancing noise, since this depends on the accuracy of the pixel data - to invert the diffraction. Several cameras have started doing this. As such, there's an argument for using even apertures that show diffraction, if needed for depth of field. Of course, you're getting less light in, so there may be consequences in terms of noise (from a longer exposure or higher ISO), camera shake (from a longer exposure), and visible stuff on the sensor (a major reason I like larger apertures).

<blockquote>Note: The resolving power at apertures larger that f/8 is higher than that of pictorially useful film emulsions.</blockquote>

I believe that. However, the difference between a D800 and a D800e - which is more than the sensor resolution - has disappeared well above f/8. There's an interesting (I think) article <a href="http://www.luminous-landscape.com/tutorials/resolution.shtml">here</a> on the effects of diffraction and the limits of resolution for each sensor size. This is why I'm uncomfortable that my 14-24 needs to be stopped down to f/7-ish before I can control its field curvature!

john_wheeler6 · July 10, 2014

The effect is minor enough at longer lengths that images don't tend to look weird, so we don't hear much about "telephoto distortion" and nobody talks about focal length when discussing the distance at which people view prints. Still, is this why c is supposed to be proportional to focal length?

</blockquote>

Via email Alan Marcus shared several references to that rule of thumb. Here is a quote from his email than lended some good info and insight:

From Photographic Lenses by C. B. Neblette and Allen E. Murray first edition 1965 Page 24 "The diameter of the circle of confusion may also be expressed as a fraction of the focal length of the lens. This has the advantage of taking the degree of enlargement into consideration, since in this case the image will appear sharp whatever the degree of magnification, provided the print is viewed from the correct distance for proper perspective, i.e. the focal length of the camera lens X the degree of enlargement.

</blockquote>

So with the rule of thumb he presented, there is the built in assumption that the viewing distance is not constant yet should be scaled to the focal length lens used.

Most DOF tables/calculators used today as well as the DOF markings on lenses (mostly from times past) that I have seen assume as a baseline viewing assumption of 10 inches where the image on the film/sensor was enlarged to an 8x10 print (and so do your calculations).

Neither DOF approach is right or wrong, they are just different approaches that will yield different answers because they are based on different assumptions of viewing distance (fixed vs varying with focal length) from the print.

Andrew Garrard · July 11, 2014

Thanks, John. (And thanks Alan - I'd been meaning to reply to your email! I guess John and I got the same one.)

So, yes, defining the circle of confusion in terms of its apparent size from an ideal viewing distance does make sense - I've done similar calculations when doing large prints of wide-angle objects. In retrospect, I'm not sure why I was struggling with that, or why I felt the need to prove the maths. Thanks to everyone for pitying me. :-)

Just to clarify: I think what was throwing me was that the focal length and imaging area need to be factored into the viewing distance, and that the latter cancels out the larger relative CoF you'd need to get the same image from a smaller sensor.

Put another way: this image is (from a) small sensor; this image is (viewed from) far away.

Though that +1 term still surprises me a bit. I would have expected everything to be exactly equivalent.

q.g._de_bakker · July 12, 2014

Viewing distances are usually fixed. People don't change viewing distance according to what focal length lens the image they are viewing was produced with. But anyway: lots of math on display above. I'll repeat my caution: there is nothing precise, nothing calculable about DoF. All you can arrive at are approximations, build on (too) many assumptions that may or may not (mostly not) be correct.

AlanKlein · July 12, 2014

That's why I stop down one extra stop after figuring the DOF I need. It covers the bases of viewing distances, enlargements, etc.

Alan Marcus · July 12, 2014

Studies show that people tend to view image at a distance about equal to the diagonal measure of the display be it print or monitor.

William Michael · July 12, 2014

"Studies show that people tend to view image at a distance about equal to the diagonal measure of the display"

</blockquote>

I didn’t know that and obviously I haven't seen the data, but pausing to think about it - that seems logical, because that distance would be the 'closest comfortable' - based upon the AoV of their eyes.

WW

q.g._de_bakker · July 13, 2014

People tend to view images at whatever distance they are presented. When reading a magazine, for instance, people don't stretch their arms or bring the magazine closer depending on how big the different pictures on the magazine pages are. People walking along walls that have billboards do not adjust the distance they walk along that wall according to how big those billboards are. Same in galleries: people tend to come closer than they should when small pictures are presented, the small format inviting them to seek a close, 'intimate' contact with the photo. They do back up for larger prints, but can't resist the temptation to come in (too) close as well. And the space to back up to the 'correct' distance may not be available. Show someone your picture wallet, and they wil look at the pics at a normal 'reading' distance, determined by what is comfortable (sitting down at a table, or keeping the pile of pics on their knees, or [...]), no matter whether they are 6x8s or 8x10s, and not by some relation between angles of view, print size and focal length. Etcetera. In short: the entire DoF theory is full of assumptions, explicit and (mostly) implicit ones, that describe some idealised world in which it could make sense to treat DoF as some fixed entity that we can grasp (even mathematically) instead of the illusive and illusory thing it always is.

Andrew Garrard · July 14, 2014

<blockquote>People tend to view images at whatever distance they are presented.</blockquote>

Agreed - which has something to do with complaints about "distortion". I've been known to print wide-angle images big for my own use so that I can view them without distortion more comfortably, but I agree that people can't be constrained to stand where the photographer wants. Fortunately this only really matters much for wider angles. It's up to the presentation to encourage viewing from a suitable distance, and up to the photographer to use any ensuing distortion creatively.

<blockquote>In short: the entire DoF theory is full of assumptions, explicit and (mostly) implicit ones, that describe some idealised world in which it could make sense to treat DoF as some fixed entity that we can grasp (even mathematically) instead of the illusive and illusory thing it always is.</blockquote>

In an absolute sense, depth of field does depend on a lot of things that are more in control of the viewer than the photographer - hence I dislike any absolute depth of field measure. However, it is absolutely possible to compare the effect of depth of field between multiple formats (the last part of Ray's question), because all of the assumptions cancel out. In other words, you can say that if X was "good enough", Y will also be "good enough"; what you can't say, in an absolute sense, was whether X was "good enough". I find this easier to establish from the mathematics than by putting numbers into a DoF calculator, but then I'm a geek. (You may be able to tell.)

I am, however, frustrated when people say "you can't compare medium format with 35mm", or similar statements (I'm not saying that Q.G. is making that statement, but others have). You absolutely can make the comparison, with a good enough approximation not to be an issue and in a way that allows useful interpretation of what's going on.

john_wheeler6 · July 17, 2014

However, it is absolutely possible to compare the effect of depth of field between multiple formats (the last part of Ray's question), because all of the assumptions cancel

</blockquote>

+1

Thanks for that response. I was almost starting to believe that DOF was as illusive or illusory as Santa Claus, the Easter Bunny, or Big Foot and everything was becoming just one big blur ;)

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DoF - different formats?

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