REQUEST = M9 & M8 bokeh/DOF comparison

[mjh]

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That's not the question the OP asked.

 

"I so i keep hearing from people that the DOF for a given lens (say at f/1.4) is different on a cropped sensor compared to a FF sensor, that actually there is more depth of field on the cropped sensor compared to a FF sensor, even though it is shot with the same lens and at the same aperture."

And that question has since been answered. Yes, depth of field is indeed different, although it’s the other way round: With the same lens/focal length and aperture, a smaller sensor gives you less, not more depth of field.

 

If I understand your setup correctly, your comparison shots answer a different question: If I shoot the same scene with cameras with different sensor sizes and adjust the distance from the subject to compensate so the field of view will be the same, will the depth of field (at the same aperture setting) be different? To which the answer is: Yes, depth of field will be different; specifically, the images taken with a smaller sensor will show more depth of field.

 

So there are two different questions with two different answers.

[rosuna]

Luminous landscape

 

The DoF of a particular camera can be approximated in different ways, but we find particularly useful this expression:

 

DoF = (2 * H * s^2) / (H^2 - s^2) for H > s

 

The DoF depends on the hiperfocal distance (H) and the distance to the subject (s). The hiperfocal distance is the focus distance that maximizes the DoF. The hiperfocal distance and the deph of field have an inverse relationship. When s approaches H, DoF tends to infinite. We can calculate the hiperfocal distance from

 

H = L^2 / (f * CoC)

 

Where L is the focal length, f is the aperture number and CoC is the circle of confusion. The circle of confusion is a conventional value. It depends on the size of the sensor, the size of the print and the particular vision capabilities and subjectivity of the observer.

 

The question is this: assuming the CoC conventions are useful for practical work, how the cropped sensor affects the depth of field? When the format changes the CoC changes too, because the enlargement of the original capture medium needed for a particular print size is altered. If we use a smaller sensor, we must to enlarge the detail captured more than necessary from a bigger sensor. Then, we have CoCf = CoCd*1.33, this is, the circle of confusion of the Leica M8 (CoCd) is 1.33 smaller than the circle of confusion of a film based Leica M (CoCf). Obviously, this affects the hiperfocal distance (Hd = Hf * 1.33) and the depth of field, which has an inverse relationship with the variable H (so DoFd < DoFf). Other variables being the same, the M8 will show a narrower depth of field due to the smaller circle of confusion. However, on the other hand, due to the crop in the sensor, not all variables can be the same. The angle of view for a particular focal length (L) will be different, and if we compensate the distance to the subject or the focal length for obtaining the same angle of view, the DoF will be affected again. An interesting result derived from the previous reasoning is this: when you crop a digital image in the computer you are effectively reducing the DoF in the print!

 

[luigi bertolotti]

Luminous landscape

 

Not dissimilar from Michael straight conclusions : anyway, let me say clearly, Michael "mjh" is one of the best tech divulgators of this Forum.

[BerndReini]

If anyone has any doubt that sensor size influences depth of field, pick up a medium format camera vs. the M8, shoot a close up of a person, framing the person's head and shoulders exactly the same size, shoot the picture at exactly the same aperture, and you will be hit over the head with the results.

 

If technical explanations confuse you, and since they are irrelevant anyway, please go out and do this test. I would say that the depth of field of the M8 at f2 is about the same as the depth of field at about f4 with my 645 with the same focal length if the framing is the same.

 

I expect about a one stop difference between the M8 and the M9.

 

Btw. all the technical explanations are beyond irrelevant for real photography. The fact remains that with a small sensor camera such as the DLux4, the depth of field at say f5.6 is endless, whereas with a 4x5 camera, it is paper thin. That's all that's relevant for photography, the rest is for engineers and forums. This is in part why people tend to emotionally respond so much to large format photography when they see the result. I know that I am oversimplifying here and I am well aware of the calculations etc., but as a working photographer, I could care less about what is supposed to happen on paper because what happens on the real paper is obvious to even an amateur.

[luigi bertolotti]

....Btw. all the technical explanations are beyond irrelevant for real photography. The fact remains that with a small sensor camera such as the DLux4, the depth of field at say f5.6 is endless, whereas with a 4x5 camera, it is paper thin. That's all that's relevant for photography, the rest is for engineers and forums. This is in part why people tend to emotionally respond so much to large format photography when they see the result. I know that I am oversimplifying here and I am well aware of the calculations etc., but as a working photographer, I could care less about what is supposed to happen on paper because what happens on the real paper is obvious to even an amateur.[/quote]

 

Personally, I like technical explantions... but your conclusion is spot on target : in user terms one has simply to take into account that with a larger sensor (or film) he simply and generally will have less DOF, period.

[ampguy]

.015 might be right, but the Osther Gunterlach (sp?) book states that Leica's lenses (this was late '80s film era) was designed for 0.023 in his Leica M Advanced book.

 

I note that dofmaster online calc uses different coc values for m8 and m9?? which is giving some folks grief.

 

Best thing is to test a 35 with M8, and a 50 on M9 at 1, 2, 5, 10, 20, and 100 meters and determine for yourself.

 

And subject distance, don't forget that! Having said that, if you go for the least simplified formula (which isn't in the Wiki article), there is a parameter for receptor size introduced. However it is better ignored.

More relevant is that the standarized COC value is far too large for present-day digital photography. I would advocate 0.015 for full format.

[lct]

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.015 might be right, but the Osther Gunterlach (sp?) book states that Leica's lenses (this was late '80s film era) was designed for 0.023 in his Leica M Advanced book...

Too many lenses would be impossible or difficult to focus with those CoC values, reason why i consider the classical 0.030mm as the best approximation for FF cameras personally.

For instance with a FF CoC value of 0.015mm, 90mm lenses could not be focused at f/2.8 on the M9 nor at f/4 on the M8.

[lars_bergquist]

What do you mean? Would for instance my 90mm Elmarit-M be more difficult to focus by the rangefinder, if the engraving on the lens barrel was different?

 

Of course not. But engraving showing realistic figures for depth of field would show that with a lens of that focal length, you would have to stop down beyond f:8 or even f:11 to obtain any appreciable depth to your field. *And that would be true*. Scale-focusing a 90mm lens, by the d.o.f. scale as well as by the distance scale, is self-delusion. In practice, no lens longer than 35mm has any depth of field that you can reliably operate with.

 

And it had been so ever since Paul Wolff and the boys started to print Leica negatives larger than 6x9cm way back in the 1930's. For any larger enlargement, 1920's d.o.f. scales are comically out of date, and except for wide angle lenses, depth of field even is an outdated concept.

 

The old man from the Age of Max Berek.

[lct]

What do you mean?....

I mean that with that small value, 90mm lenses would be out of the accuracy range of the rangefinder at f/2.8 for the M9 and f/4 for the M8.

There is a requested base length for a rangefinder to focus a lens accurately and that legth is based on the circle of confusion as well.

If memory serves, it is calculated through the formula b' = (e * f^2) / (k * z) where b' is the requested base length, e the visual acuity (0.0003 at approx. 1 arcmin), f the focal length, k the aperture and z the circle of confusion.

By example, at the usual CoC value of 0.030mm, the requested base length of the M9 is 28.93mm for a 90mm lens at f/2.8, based on that formula. Then the M9 is accurate enough given that its effective base length (47.09 mm) is longer than 28.93mm.

Now if you take 0.015mm as CoC value, the requested base length becomes 57.86mm, based on that same formula, and the effective base length (again 47.09mm) is not long enough to focus the lens any more.

[artur5]

May I add my worthless 5 cents to this Byzantine thread ?

You're all right and you're all wrong. DOF doesn't exists, It's merely a philosophical concept, an illusion, an opinion. It can be whatever you like.

Just get a M8 and a M9. Mount the same lens on both cameras and shoot from the same distance with the same aperture on lens.

Now make a 18x27 cmtr. print of the M8 file and a 24x36 cmtr. print of the M9 file.

The smaller M8 print will look exactly the same as a 18x27 cmt. central crop of the M9 print. Same perspective, sharpness, bokeh, DOF. So we can conclude that a given lens will have equal DOF on the M8 and the M9 ?

Now, if you take another route and enlarge both prints to the same final size, things would change of course and so they will if you shoot from different distances each time to obtain the same FOV on both pictures or if you use a 28mm. lens on the M9 and a 21mm. on the M8...

This tread could go and on, just for fun, and we'll never agree.

[Shootist]

Ok I have something to add.

If you take a shot with a M8 using a 50 f/1.4, @ a f/stop of 1.4, at a distance of 20 feet and then take that same shot using a M9 using the same lens and f/stop at the same distance the DOF, bokeh, will be exactly the same. The difference is the M9 image will cover (capture) a wider area then the M8 image.

 

The deal is with smaller sensor cameras they use very wide lenses to capture the same area as a 35mm camera would capture. That is why you see references to 7mm to 40mm zoom lens (equivalent to a 28mm to 160mm lens on 35mm). That means that on this small senosr camera the widest part of the zoom lens is 7mm equaling a 28mm lens on a 35mm camera. Since the lens is 4 times as wide on this small sensor camera it has much greater DOF. And so does the long end, 40mm equal to a 160mm on a 35mm camera, but since the lens is really a 40mm lens you will get greater DOF then a 35mm camera with a 160mm lens.

 

A 50mm lens is still a 50mm lens no matter what sensor size camera you are using it on.

No need for complex equations.

[lars_bergquist]

May I add my worthless 5 cents to this Byzantine thread ?

You're all right and you're all wrong. DOF doesn't exists, It's merely a philosophical concept, an illusion, an opinion. It can be whatever you like.

Just get a M8 and a M9. Mount the same lens on both cameras and shoot from the same distance with the same aperture on lens.

Now make a 18x27 cmtr. print of the M8 file and a 24x36 cmtr. print of the M9 file.

The smaller M8 print will look exactly the same as a 18x27 cmt. central crop of the M9 print. Same perspective, sharpness, bokeh, DOF. So we can conclude that a given lens will have equal DOF on the M8 and the M9 ?

Now, if you take another route and enlarge both prints to the same final size, things would change of course and so they will if you shoot from different distances each time to obtain the same FOV on both pictures or if you use a 28mm. lens on the M9 and a 21mm. on the M8...

This tread could go and on, just for fun, and we'll never agree.

You are completely right. As I have pointed out before, we should distinguish between d.o.f. on the sensor/film, and d.o.f. in the final print, which is dependent on the enlargement. But shoot a subject to the same final reproduction in the print, with the M8 and the M9 at the same aperture, even with lenses of different focal length, and print d.o.f. will be the same.

 

To repeat a recent and hopefully immortal coinage in this thread, there is indeed much 'confusion of circles' here.

 

The old man from the Age of the 3.5cm Elmar

[Shootist]

You are completely right. As I have pointed out before, we should distinguish between d.o.f. on the sensor/film, and d.o.f. in the final print, which is dependent on the enlargement. But shoot a subject to the same final reproduction in the print, with the M8 and the M9 at the same aperture, even with lenses of different focal length, and print d.o.f. will be the same.

 

To repeat a recent and hopefully immortal coinage in this thread, there is indeed much 'confusion of circles' here.

 

The old man from the Age of the 3.5cm Elmar

 

I don't think, in fact I know, that is not correct. If you shoot a subject with a 35mm lens on the M8 and a 50mm lens on the M9 using the same f/stop on each lens and capturing the same area, IE shooting from the same distance, the M8 file/image will have more DOF.

It has to because you are using a wider lens on the M8. Wider lens = greater DOF. Doesn't really matter that the M8 is not using the whole image circle the 35mm lens can deliver.

[lars_bergquist]

I don't think, in fact I know, that is not correct. If you shoot a subject with a 35mm lens on the M8 and a 50mm lens on the M9 using the same f/stop on each lens and capturing the same area, IE shooting from the same distance, the M8 file/image will have more DOF.

It has to because you are using a wider lens on the M8. Wider lens = greater DOF. Doesn't really matter that the M8 is not using the whole image circle the 35mm lens can deliver.

Exactly: ".. capturing the same area". That was NOT what I was saying. I said "to the same final reproduction [ratio] in the print". Admittedly, the word in brackets was accidentally omitted -- mea culpa -- but the meaning should have been crystal clear nevertheless.

 

'Confusion of circles' indeed.

 

The old man from the Age of the 3.5cm Elmar