View Full Version : DoF Charts and f/stops

Mark Sawyer

25-Nov-2018, 23:17

I never use Depth of Field charts, tables, calculators, etc., but a question had me lying in bed awake and wondering last night...

Say you're doing close up work at 1:1, and you set your f/stop to f/16. With the bellows/focal length at twice the normal extension, your effective f/stop is f/32. Which do you use for the chart/table/calculator, f/16 or f/32?

DOF is determined by subject magnification at the film / sensor plane, at any given f stop.

The f-number of an optical system (such as a camera lens) is the ratio of the system's focal length to the diameter of the entrance pupil. For exposure purposes we call the adjusted f stop at close distances an effective f stop, but for optical purposes it is still bound by the original definition. Take a look at the barrel of an older 35mm lens and see how the focus numbers relate to the DOF scale that used to be engraved on each lens.

So in your case use f 16.

Pere Casals

26-Nov-2018, 06:11

but for optical purposes it is still bound by the original definition. Take a look at the barrel of an older 35mm lens and see how the focus numbers relate to the DOF scale that used to be engraved on each lens.

So is the rule using engraved f/ and magnification...

Bob Salomon

26-Nov-2018, 06:18

The Rodenstock DOF/Scheimpflug pocket calculator tells you f stop, focus point, tilt amount, etc for every format from 35mm to 810 and fits flatly in your shirt pocket assuring you of a good night’s sleep!

Pere Casals

26-Nov-2018, 06:48

Being DOF scales of great help, IMHO it should be noted that all DOF scales are approximate, because formulas have simplifications and limitations.

Proficient cinematographers use Through Focus MTF charts for the specific lenses.

184771

Bob Salomon

26-Nov-2018, 06:50

Being DOF scales of great help, IMHO it should be noted that all DOF scales are approximate, because formulas have simplifications and limitations.

Proficient cinematographers use Through Focus MTF charts for the specific lenses.

184770

The Rodenstock one lets you set for different CoC sizes so that is no problem.

Are you two channeling Abbot and Costello or Laurel and Hardy?

John Layton

26-Nov-2018, 08:11

...or Penn and Teller?

Pere Casals

26-Nov-2018, 08:36

Bob, I'm not talking about the CoC one may require, but about how CoC actually behaves for each particular lens compared with theoric/generic calculations. Still the pocket guide is of great help, no doubt.

Are you two channeling Abbot and Costello or Laurel and Hardy?

...or Penn and Teller?

Hey guys, TF MTF charts are serious stuff, explaining why one has to learn how his lens behaves beyond what stated in generic DOF calculators.

You may find several proficient members here explaining why DOF calculations have remarkable flaws in the theoric CoC/DOF calculations for particular glasses... I you need the links just tell it.

Or just understand what this curve plots:

184772

This is specially critical for macro, a particular lens may behave better or worse in the front or in the rear of the focus plane. I was thinking that this was well known.

... and let me add that bokeh's nature also may vary front vs rear.

Dan Fromm

26-Nov-2018, 09:19

Papi, I think you're confusing CoC, an arbitrary number chosen by the photographer based on how much the negative is to be enlarged, and the blur circle, which is given by the lens.

Pere Casals

26-Nov-2018, 09:49

Papi, I think you're confusing CoC, an arbitrary number chosen by the photographer based on how much the negative is to be enlarged, and the blur circle, which is given by the lens.

Dan, to me "In optics, a Circle of Confusion is an optical spot caused by a cone of light rays from a lens not coming to a perfect focus when imaging a point source".

"It is also known as disk of confusion, circle of indistinctness, blur circle, or blur spot."

My understanding is that what a photograher picks is a maximum allowed CoC to say that something is in the DOF range... but each spot in the scene has its own CoC on film....

(https://en.wikipedia.org/wiki/Circle_of_confusion)

Is there another definition ?

_____

(of course it has to be pointed that confusion is not always in a circle, other shapes are there for coma, astigmatism...)

ic-racer

26-Nov-2018, 10:47

I never use Depth of Field charts, tables, calculators, etc., but a question had me lying in bed awake and wondering last night...

Say you're doing close up work at 1:1, and you set your f/stop to f/16. With the bellows/focal length at twice the normal extension, your effective f/stop is f/32. Which do you use for the chart/table/calculator, f/16 or f/32?

If the lens markings show f16 when at 1:1 your effective aperture is f/32. Use f/32 for calculations of exposure, CoC and Airy disk size.

Mark Sawyer

26-Nov-2018, 11:13

If the lens markings show f16 when at 1:1 your effective aperture is f/32. Use f/32 for calculations of exposure, CoC and Airy disk size.

That's what common sense tells me: figure what your f/stop actually is, and that's what your f/stop actually is, so use it. But cowanw posted:

So in your case use f 16.

...and not being familiar with how DoF calculators calculate, I'm still not sure, maybe there's a built-in compensation I don't know about. Some have a magnification factor built in, but that's more for the inherent lack of DoF close up, not an f/stop compensation.

Just a curiosity, really. Like the Scheimpflug Principle, there's always some party of reality that mucks up the theory, and you're better off trusting a good loupe, a fine ground glass, and the patience to check things thoroughly. But I'd still like to know...

Pere Casals

26-Nov-2018, 11:14

If the lens markings show f16 when at 1:1 your effective aperture is f/32. Use f/32 for calculations of exposure, CoC and Airy disk size.

racer, are you sure ?

Rodenstock DOF/Scheimpflug pocket calculator does not ask for the effective aperture... it looks...

For LF lenses effective aperture is lower (at 1:1) but perhaps we also have a more enlarged CoC because of bellows draw, so one factor may just compensate the other in the formula.

of

The "pocket calculator" does not ask is the lens is "unit focus" or "internal focus"...

cowanw says we have to use the engraved aperture (I thing that's right) and you say that we have to use the effective one, have we available literature on that ?

Added:

But I'd still like to know...

me also

Mark Sawyer

26-Nov-2018, 11:21

cowanw says we have to use the engraved aperture (I thing that's right) and you say that we have to use the effective one, have we available literature on that ?

That's what I've hunted for, and found nothing. I think conanw is wrong, and the f/stop you're at is the f/stop you're at. Why would the circumstances of being focused at infinity apply to a completely different set of circumstances like being focused at 1:1? That's a whole different set of numbers. But again, I'm not sure...

Bob Salomon

26-Nov-2018, 11:23

racer, are you sure ?

Rodenstock DOF/Scheimpflug pocket calculator does not ask for the effective aperture... it looks...

For LF lenses effective aperture is lower (at 1:1) but perhaps we also have a more enlarged CoC because bellows draw, so one factor may just compensate the other in the formula.

The "pocket calculator" does not ask is the lens is "unit focus" or "internal focus"...

cowanw says we have to use the engraved aperture (I thing that's right) and you say that we have to use the effective one, have we available literature on that ?

Added:

me also

The calculator does not ask for your aperture. It tells you what aperture is required for the near and far points required and for the magnification that results in, it then tells you where to place the rear standard for those conditions and, if required, the angle of the camera to the subject and lens tilt.

Pere Casals

26-Nov-2018, 11:51

It tells you what aperture is required

Bob, OK, but what aperture tells? the effective one stated by racer ? or the engraved one stated by cowanw ?

at 1:1 we have a difference of 1 stop...

That's what I've hunted for, and found nothing. I think conanw is wrong, and the f/stop you're at is the f/stop you're at. Why would the circumstances of being focused at infinity apply to a completely different set of circumstances like being focused at 1:1? That's a whole different set of numbers. But again, I'm not sure...

Also not sure, but instead I find more plausible cowanw's way. Rodenstock does not ask if the lens is "unit focus" or "internal focus", for any format, this suggests me that there is an invariant relationship for the engraved aperture. Consider that the effective aperture variation also involves an apparent focal variation, all linked with bellows draw.

If we use the effective aperture perhaps we have to also use effective (aparent) focal in a certain formula.... and perhaps this ends in the same than always using always the engraved aperture... just guessing possibilities.

Please call me Bill.

As we all know DOF diminishes as the subject gets closer than infinity. If we were to substitute a smaller F stop as our number, then the image should be as sharp (risking diffraction) and DOF as great or greater (smaller Fstop= more DOF). A lens has an effective Fstop as regards exposure at closeup because of the inverse square law of light. This has nothing to do with the optical performance of a lens.

Mark Sawyer

26-Nov-2018, 14:10

cowanw says we have to use the engraved aperture (I thing that's right) and you say that we have to use the effective one, have we available literature on that ?

The calculator does not ask for your aperture. It tells you what aperture is required for the near and far points required and for the magnification that results in, it then tells you where to place the rear standard for those conditions and, if required, the angle of the camera to the subject and lens tilt.

A note that none of the calculators or charts I've seen has "aperture" listed on them, and the raw aperture is almost never engraved on lenses, (though I often wish it were). They all deal with f/stops, which is focal length divided by aperture. Aperture is only half the f/stop ratio.

Just nit-picking the technical points in a technical discussion...

Emmanuel BIGLER

26-Nov-2018, 14:50

Hello from France!

I'm coming very late to this discussion, in the OP's question there is the notion of engraved f-number vs. effective f-number for computing depth of field (DoF) with conventional, well-known formulae.

If N is the engraved f-number, the effective f-number is simply Neff = N(1+M) where M is the magnification ratio, M = (image size)/(object size)

Hence for a given magnification factor M, you can compute depth field as you wish, in terms of N or Neff.

It happens that when the magnification factor M is larger than about 0.1, i.e. when the object to lens distance is shorter than about 10 times the focal lenght, general DoF formulae can be simplified under the following form, where the total DoF is p2 - p1 i.e. the difference between the far (p2) and near (p1) limits of sharpness:

p2 - p1 = ((M+1)/M2) 2 N c = (1/M2) 2 Neff c

In macro work, the model based on geometrical optics suggests that DoF is distributed equally in front of the subject and behind the subject, the half- DoF beeing simply equal to (p2 - p1)/2.

This, of course, is plain wrong for subjects at large distances, larger that about 10 times the focal length.

So as soon as you know the magnification factor M, in the case of M>0.1, you can compute your DoF limits (according to the well-known model of geometrical optics, in its simplified form) without the help of any computer, without need of any application running on your smartphone: this can be done by hand with a pencil on the back of an envelope ;)

The inputs are: the magnification factor M, the f-number, either engraved N or effective N(1+M), and the circle of confusion, a parameter which, unfortunately, is not intrinsic to any given lens, but depends on the film size and conditions under which you examine a final print of a given size.

The circle of confusion for a pre-WWII 6x9 cm - 2"x3"image was about 100 microns in old DoF charts for amateur 2"x3" rollfilm cameras (actually, c was taken as f/1000, hence about 100 microns for a standard focal lens of 100 mm - 4" for the 6x9 cm - 2"x3" film format), at the time, amateurs only looked at contact prints of their 2"x3" images from a distance of about 12"...

Pere Casals

26-Nov-2018, 15:18

Thanks for the clarification, really a very good explanation.

So can we guess that usual dof scales and calculators are using the engraved aperture as input data?

...as N is asked, or delivered...

Mark Sawyer

26-Nov-2018, 15:27

I just spent a while reading the DoF articles on the LF Home Page, and they confirmed that I'm not as fluent in math as I once was. The article at http://www.largeformatphotography.info/dofknob/#how to use it contained the following passage, (the underlining is mine):

"Good things to know

The min. f-numbers calibrated are correct at magnification M = 0, infinity focus. However, at any magnification, the min. f-number required for a given delta will be less than that at infinity focus. Therefore you're always guaranteed to get min. DOF for a given delta at any magnification.

At magnification M = 1 (life-size, 1:1), the min. f-numbers N on scale are half. At any magnification in between 0 and 1, the min. f-number required will fall in between the f-number scales and their corresponding halves. For an arbitrary N, use N(M) = 1/(1+M)*N(M=0) to figure out exact min. f-number required at M. For example, at M = 0.5, N(M=0.5) = 2/3*N(M=0). So multiply the f-numbers on scale by 2/3."

From this, I'm presuming that one uses the effective, not marked, f/stop. But from Emmanuel's post above, and from looking at varying results given by on-line calculators, I'm guessing some calculators/charts may figure this in, and some don't.

BTW, here's a deeper look at some of the math...

http://www.largeformatphotography.info/articles/DoFinDepth.pdf

Emmanuel BIGLER

26-Nov-2018, 16:09

So can we guess that usual dof scales and calculators are using the engraved aperture as input data?

Hi, Pere! In short, yes, exactly, at least this is how I feel things.

Well, it is always uncomfortable when DoF calculators do not precisely define their inputs, or worse, do not tell which formulae are actually in use, and the limits of validity or relevance (for example, the good old model for DoF is based on geometrical optics assumes that diffraction is negligible).

However, it is reasonable to think that authors of those calculators aim at a general public who never heard about anything else than the engraved f-number, which by nowadays standards, denotes a very deep knowledge in photographic optics: look at modern digital cameras, most have no F-number scale engraved at all!

More seriously, general formulae valid at any distance are not much complicated, even when insisting on using exotic formulae --that actually nobody uses-- with lenses exhibitig non-unit pupillar magnification ratio.

Not very complicated, I mean: taking into account that anybody who carries a smartphone with him 24/7 has access to a computer more powerful that the ones in use for the Manhattan Project ;)

General formulae do not tell us immediately that in macro, DoF no longer depends directly on the focal length, which is of course somewhat counter-intuitive but very interesting in practice, immediately applicable to LF photography in the macro range.

The interest of deriving approximate formulae valid in the macro range is that the maths eventually explain why the focal length magically dissappears from the simplified version of the formulae, and yields a very simple expression, that you can compute as you wish, no software needed.

Pere Casals

26-Nov-2018, 16:44

Well, it is always uncomfortable when DoF calculators do not precisely define their inputs, or worse, do not tell which formulae are actually in use, and the limits of validity or relevance (for example, the good old model for DoF is based on geometrical optics assumes that diffraction is negligible).

A calculator also doesn't ask if the lens is unit focus or internal focus, I'd ask... may this be a concern?

Dan Fromm

26-Nov-2018, 16:50

A calculator also doesn't ask if the lens is unit focus or internal focus, I'd ask... may this be a concern?

Focal length is focal length.

Some lenses that are internal focusing -- my infernal 200/4 MicroNikkor AIS, for example, and not a lens for LF -- focus closer by reducing focal length. With them knowing the focal length when the lens is focused closer than infinity is a little difficult. Focal length engraved on, e.g., the trim ring, can be quite different from actual focal length.

Are there any internal focusing lenses for LF?

Bob Salomon

26-Nov-2018, 17:32

A calculator also doesn't ask if the lens is unit focus or internal focus, I'd ask... may this be a concern?

Maybe you should read the instructions for the Rodenstock/Linhof/Sinar calculator.

It doesn’t care what the focal length is. It does ask what the difference is between the near focus point on the lens or the rail and the far point, also the angle of the camera to the subject and the magnification. It then tells you the required marked f stop and where to focus the lens.

Mark Sawyer

26-Nov-2018, 18:23

From: http://www.largeformatphotography.info/articles/DoFinDepth.pdf

"Lenses are assumed unit focusing, and except for the section Depth of Field for an Asymmetrical Lens, are assumed symmetrical. Large-format lenses are unit focusing, and except for telephotos, are nearly symmetrical. Because the effects of asymmetry are minor unless the asymmetry is substantial and the magnification approaching unity or greater, this simplified treatment is justified in most cases for large-format lenses.

Many, if not most, small-format lenses of other than normal focal length are asymmetrical, and many are not unit focusing. However, for other than closeup lenses, the effect of asymmetry is minimal, and close focus usually is approximately 10× focal length; consequently, the change in focal length from the non-unit focusing is minimal, and the simplifying assumptions give reasonable results. However, lens asymmetry must be considered when determining the DoF of an internal-focusing macro lens, as discussed in the section Effect of Lens Asymmetry."

and

"Most treatments of depth of field assume a symmetrical lens for which the entrance and exit pupils are the same size, and for which the pupils coincide with the object and image principal planes. Although this assumption is reasonable for most large-format lenses of short and medium focal length, it may not be appropriate for telephoto designs."

Jac@stafford.net

26-Nov-2018, 18:48

Of course, CoC presumes a lens can actually produce a point of focus. CoC is a variable human-determined abstraction.

Dan Fromm

26-Nov-2018, 19:31

Of course, CoC presumes a lens can actually produce a point of focus. CoC is a variable human-determined abstraction.

Jac, I made this point earlier in this discussion and Pere Casals corrected me. Correctly, as it turns out.

It seems that circle of confusion has two meanings. The one that you and I like and use, especially when calculating depth of field, is the largest blur circle that will allow the desired enlargement. The other, that some in this discussion have used, is the blur circle the lens produces.

Pere Casals

26-Nov-2018, 19:33

From: http://www.largeformatphotography.info/articles/DoFinDepth.pdf

Although this assumption is reasonable for most large-format lenses of short and medium focal length, it may not be appropriate for telephoto designs."

It's reasonable for most plasmats, these are a lot of cases and most may be quasi-symmetric... but beyond all teles... a verito is not symetrical, nor raptars, then we have heliars and tessars(Nikkor M), Ektars ? and all modern wide "view" angles derived from biogons sporting tilting pupils... there are many LF glasses that are not near to symmetry, including new MF Digarons etc for digital backs that are mounted in optical benches replacing Pro film LF.

Sadly Through Focus MTF charts are scarce for LF products, IMHO that feature is a key factor to realize the presonality of a glass.

Pere Casals

26-Nov-2018, 19:39

Jac, I made this point earlier in this discussion and Pere Casals corrected me. Correctly, as it turns out.

Dan, I'm honored by correcting you in nomenclature this time, but I've to say that you are correcting me x10 more times in concepts, so learning a lot.

Maybe you should read the instructions for the Rodenstock/Linhof/Sinar calculator.

I'll read it, but I think you explained it to me perfectly, I find the included tilt feature pretty powerful...

Emmanuel BIGLER

27-Nov-2018, 00:57

from Pere Casals:

A calculator also doesn't ask if the lens is unit focus or internal focus, I'd ask... may this be a concern?

Well, providing that you have access to the true F-number N, and that you can estimate your magnification ratio M, focal length does not play a direct role in traditional DoF for magnification M greater than about 0.1

The real question, for a lens with internal focus, is to have access to the true f-number, as discussed below.

And Dan says:

Focal length is focal length.

Yes, of course for a LF lens with fixed elements.

But "modern" macro lenses with internal focusing in use un 35 mm, medium format and DSLRs are actually a kind of a zoom. The "commercial" focal length, to the best of my knowledge, corresponds to the use in infinity-focus setting.

When the lens is used down to 1:1, the focal length may shorten significantly, but in this position, nobody has access to the focal point which is recessed somewhere inside the lens mount, and plain users cannot easily estimate the actual focal length of a macro lens with floating elements outside the infinity-focus position.

And since DoF in macro does not depend on the focal length, users of those lenses, in principle, should not have to care for the actual focal length.

But the real question is: when my lens with floating elements is set for 1:1 ratio, what is my actual f-number N or or effective f-number N(1+M) (i.e. effective = 2N at 1:1 for M=1) ???

I do use such a macro lens from time to time [100 OFF TOPIC !!], but I realize that I did not pay attention to the meaning of f-numbers that show up in the view finder on the liquid crystal display!

Regarding the use of general formulae valid for asymmetric lenses with non-unit pupillar magnification ratio, I tried to play with them and compare with traditional formulae valid in principle only for quasi-symmetrical lenses.

I never found any practical differences at large distances, between general formulae and simplified formulae valid for symmetrical lenses.

One of the reasons is that the hyperfocal distance, i.e. the setting distance for which DoF extends to infinity, is the same for assymetric lenses, i.e. at large distances both formulae yield almost the same result for DoF limits.

And since most applications in macro work rely on quasi-symmetrical lenses, at least with non-floating element lenses, is is likely that nobody has ever experienced a huge discrepancy between actual, real-world DoF and simulated DoF based on simplified formulae valid for quasi-symmetical lenses.

Either you use any lens type including a telephoto at large distances, and in this situation the hyperfocal distance is the same regardless of the lens formula, or you work in macro with a symmetrical lens.

Mark Sawyer

27-Nov-2018, 00:59

I think I withdraw my original question. This is making my head hurt... :(

Pere Casals

27-Nov-2018, 02:58

and that you can estimate your magnification ratio M, focal length does not play a direct role in traditional DoF for magnification M greater than about 0.1

In a calculator, when focal does not play a direct role it still serves to calculate magification M, operating focal and subject distance... for those calculators asking focal and distance.

The real question, for a lens with internal focus, is to have access to the true f-number, as discussed below.

In general IF lenses (prime) are designed to avoid focus breathing, so image circle size can be considered constant and true aperture should be mostly invariant (vs focus range) and equal to the engraved one, I guess...

My understanding is that what is engraved is the true optical aperture and not the light transmision.

________________

I think I withdraw my original question. This is making my head hurt... :(

But returning to the original question, if calculation can be done with either N or Neff it would we weird asking Neff to users, under the KISS design principle :), ( noted by the U.S. Navy in 1960, coined by Kelly Johnson at Skunk Works) https://en.wikipedia.org/wiki/KISS_principle

The other challenge is that with N input (or output) the calculator has to deal with IF and UF lenses at the same time, but IMHO Emmanuel illustrated that this is possible with posted formulas, if accepting that IF glass holds aperture across focus setting.

So IMHO I'd say that there is no reason for a DOF calculator to torturate a user with a Neff values... is it there?

Dan Fromm

27-Nov-2018, 06:32

I think I withdraw my original question. This is making my head hurt... :(

Mark, your question was answered early in this discussion. Depth of field at 1:1? There is none, don't bother calculating it.

This is, of course, a frivolous answer but it gets at the big problem with conventional closeup photography. There's no way to get enough depth of field. This is why focus stacking, which is practical only with digital and static subjects, is so widely used.

Mark Sawyer

27-Nov-2018, 12:11

I frivolously disagree with your frivolous answer, Dan. I do mostly close-up work these days, enough to know there's usually something in focus, but yes, it may be a bit of a scavenger hunt finding it...

In playing with plugging the same 1:1 values into various internet DoF calculators, I received differing answers, confirming 1.) different calculators use different formulae (and I saw a few variations in my readings), and 2.) yup, at 1:1 with longer focal lengths and wide apertures, DoF is scary shallow no matter how you calculate it. Fortunately, my calculator is a loupe, an eyeball, and a ground glass, so for me, the original question just curiosity.

But I still haven't been able to pick out any authoritative documentation on whether one should use the engraved or actual f/stop, and I'm still curious...

Mark Sawyer

27-Nov-2018, 12:22

In a calculator, when focal does not play a direct role it still serves to calculate magification M, operating focal and subject distance... for those calculators asking focal and distance.

But does inputting the magnification or subject distance in the calculations compensate for the Neff value being substantially different than the N value? Or is it simply allowing for the inherently shallower DoF at close focusing distances?

So IMHO I'd say that there is no reason for a DOF calculator to torture a user with a Neff values... is it there?

The reason would be that the Neff is the actual f/stop you're working at. And to me it makes sense that as such, this is the number you should use. But I'm also stupid and rather ignorant, and like to make things complicated...

Pere Casals

28-Nov-2018, 02:16

But does inputting the magnification or subject distance in the calculations compensate for the Neff value being substantially different than the N value? Or is it simply allowing for the inherently shallower DoF at close focusing distances?

Mark, my guess is that it is possible to calculate DOF of a "unit focus" lens (LF lenses...) by operating N, or calculating N, and never using Neff for it. At the end we can find a law that links N to Neff from magnification, so we can make a sustitution of any Neff variable by an N variable (https://en.wikipedia.org/wiki/Change_of_variables).

If we compare what happens with an Internal Focus lens, the Unit Focus at 1:1 has one half of the aperture but it also has twice Nodal to film distance, so enlarging by x2 the confusion circles, and one thing may excatly compensate the other... just a possibility I guess.

So I bet that the Rodenstock pocket DOF calculator uses the the engraved aperture to predict well DOF also for LF, if not they would be saying it in the instructions, would Rodenstock missled people by one stop in a calculator for a Sironar S? I don't think that about Rodenstock...

Other calculators... who knows ??? but it can be checked !

The reason would be that the Neff is the actual f/stop you're working at. And to me it makes sense that as such, this is the number you should use.

Let me suggest a test, attach a DSLR in the back of the view camera without the lens, take a USAF 1951 glass slide or a target sporting Group 6, we need that at 1:1.

Then displace the target (back and forward) from perfect focus at 1:1, perhaps a micrometric rail for macro would be useful.

Then plot the through focus lp/mm graphs, measuring from the DSLR images in the back, this is the same graph we plot to get the optimal holder height in a V750 scanner, you will get a graph similar (but using lp/mm rather than um) to this:

184831

This would be first hand trustful information, and you would find if that matches the DOF calculated by introducing N or Neff in a particular calculator... IMHO this is the way, testing !

But I'm also stupid and rather ignorant, and like to make things complicated...

:) Mark, the Kelly Johnson's KISS calls "stupid" to designers that complicate things for end users, not calling "stupid" to end users, but defending them from "stupid" engineers.

"The principle is best exemplified by the story of Johnson handing a team of design engineers a handful of tools, with the challenge that the jet aircraft they were designing must be repairable by an average mechanic in the field under combat conditions with only these tools. Hence, the "stupid" refers to the relationship between the way things break and the sophistication available to repair them."

Since then KISS is a principle applied to many manufactured goods. Exception are cars, that are designed to invoice customers with spare ECU boards :)

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