tayho

Thank you Gene for following up on that. I knew they had to both be cropped bodies, but I wasn’t familiar with D2X... bodies. I hope it was an honest mistake made by Rongnongo.

Rongnongno Loc: FL

Echo upon request.

Note my modified version No magnification on the sensor stands.

blackest Loc: Ireland

The Enlargement factor you are using disagrees with

https://en.wikipedia.org/wiki/Circle_of_confusion#Circle_of_confusion_diameter_limit_in_photography

Enlargement from the original image to the final image. If there is no enlargement (e.g., a contact print of an 8×10 original image), the CoC for the original image is the same as that in the final image. But if, for example, the long dimension of a 35 mm original image is enlarged to 25 cm (10 inches), the enlargement is approximately 7×, and the CoC for the original image is 0.2 mm / 7, or 0.029 mm.

http://www.hasselbladdigitalforum.com/index.php?topic=4511.0

The common values for CoC may not be applicable if reproduction or viewing conditions differ significantly from those assumed in determining those values. If the original image will be given greater enlargement, or viewed at a closer distance, then a smaller CoC will be required. All three factors above are accommodated with this formula:

CoC (mm) = viewing distance (cm) / desired final-image resolution (lp/mm) for a 25 cm viewing distance / enlargement / 25

For example, to support a final-image resolution equivalent to 5 lp/mm for a 25 cm viewing distance when the anticipated viewing distance is 50 cm and the anticipated enlargement is 8:

CoC = 50 / 5 / 8 / 25 = 0.05 mm

5lines per mm means 1mm / 5 = 0.2mm width for each line.

The formula above is easiest to work with when planning to produce a large photo.

Since the final-image size is not usually known at the time of taking a photograph, it is common to assume a standard size such as 25 cm width, along with a conventional final-image CoC of 0.2 mm, which is 1/1250 of the image width. Conventions in terms of the diagonal measure are also commonly used. The DoF computed using these conventions will need to be adjusted if the original image is cropped before enlarging to the final image size, or if the size and viewing assumptions are altered.

That works for me :)

Using the “Zeiss formula”, the circle of confusion is sometimes calculated as d/1730 where d is the diagonal measure of the original image (the camera format). For full-frame 35 mm format (24 mm × 36 mm, 43 mm diagonal) this comes out to be 0.025 mm. A more widely used CoC is d/1500, or 0.029 mm for full-frame 35 mm format, which corresponds to resolving 5 lines per millimeter on a print of 30 cm diagonal. Values of 0.030 mm and 0.033 mm are also common for full-frame 35 mm format. For practical purposes, d/1730, a final-image CoC of 0.2 mm, and d/1500 give very similar results.

That last line is giving 3 methods for calculating CoC d/1730, d/1500, final image CoC of 0.2mm (5 lines per mm).
1730/d you can calculate the enlargement values by dividing final image CoC by the sensor size values Zeiss 1730 =8.04 d/1500 = 6.97 and final CoC at 5 Lpm gives 7 for a 35mm negative. choose your poison, the Zeiss formula is the most conservative

Criteria relating CoC to the lens focal length have also been used. Kodak (1972), 5) recommended 2 minutes of arc (the Snellen criterion of 30 cycles/degree for normal vision) for critical viewing, giving CoC ≈ f /1720, where f is the lens focal length. For a 50 mm lens on full-frame 35 mm format, this gave CoC ≈ 0.0291 mm. This criterion evidently assumed that a final image would be viewed at “perspective-correct” distance (i.e., the angle of view would be the same as that of the original image):

Viewing distance = focal length of taking lens × enlargement
However, images seldom are viewed at the “correct” distance; the viewer usually doesn't know the focal length of the taking lens, and the “correct” distance may be uncomfortably short or long. Consequently, criteria based on lens focal length have generally given way to criteria (such as d/1500) related to the camera format.

ok so if we use 7x as enlargement for fx and 10x enlargement for dx
then we are supposed to view a photo taken with a 50mm lens at 35cm for fx and 50cm for dx to get the correct perspective. At 600mm then it would be 4.2 meters and 6 meters (a little under 14 feet and 20 feet). if an uncropped 600mm photo is 144mm wide on your monitor that would be 4x and 6x enlargement 2.4 meters 3.6 meters viewing distance to get the correct perspective :) I may need a bigger room :)

If an image is viewed on a low-resolution display medium such as a computer monitor, the detectability of blur will be limited by the display medium rather than by human vision. For example, the optical blur will be more difficult to detect in an 8″×10″ image displayed on a computer monitor than in an 8″×10″ print of the same original image viewed at the same distance. If the image is to be viewed only on a low-resolution device, a larger CoC may be appropriate; however, if the image may also be viewed in a high-resolution medium such as a print, the criteria discussed above will govern.

Ok we don't have much to worry about on screen. A retina screen has 264 pixels per inch /25.4 = 10.4 pixels per mm. you might be able to see 5lpm :)

Depth of field formulas derived from geometrical optics imply that any arbitrary DoF can be achieved by using a sufficiently small CoC. Because of diffraction, however, this isn't quite true. Using a smaller CoC requires increasing the lens f-number to achieve the same DOF, and if the lens is stopped down sufficiently far, the reduction in defocus blur is offset by the increased blur from diffraction. See the Depth of field article for a more detailed discussion.

What does that mean?

5Lpm = 127 lines per inch so maybe thats why to print 300 dpi is preferred but 150 dpi can do in a pinch...

when light passes through an aperture it spreads the smaller the aperture the wider the spread (the frequency of the light is also a factor here).
So the size of the Airy Disk depends on aperture if its smaller than a pixel its not going to have any effect as it gets bigger though it will spill over and detail turns to mush.

It's not as simple as just overlapping into an adjacent pixel most of a pixels color will be from the right light waves but the overall value will be influenced by neighboring pixels. Steve Perrys rather excellent video on diffraction demonstrated that you can sharpen (and this really means increase the contrast between pixels) and get the detail back. However and this is also shown in the video it only works to a point once the airy disk is too big you can't recover.

https://www.cambridgeincolour.com/tutorials/diffraction-photography.htm

As a result of the sensor's anti-aliasing filter (and the Rayleigh criterion above), an airy disk can have a diameter of about 2-3 pixels before diffraction limits resolution (assuming an otherwise perfect lens). However, diffraction will likely have a visual impact prior to reaching this diameter.

So this basically is telling us there is a minimum aperture size thats useful and that is related to pixel size on your camera. It's still better to have more than pixels than less to capture detail but diffraction limiting comes in at a larger aperture. the calculator on the cambridge in color link lets you set enlargement size and viewing distance as well as everything else.
It showed for example that if you keep the viewing distance at the same as a 10 by 8 and enlarged to 20 by 16 that you would be diffraction limited at f8 on a dx camera with 24 Mpix and not at f5.6.

Well you should be viewing from further away! But if you are cropping so you are only using a 1/4 of your camera's pixels you get a 10 by 8 and you would be viewing at the normal distance (that's not the correct distance for perspective according to kodak).

So if you are going to crop it's pretty much the same as using a smaller sensor and while f8 would be good without cropping f5.6 would be better if you do. You will have better Image Quality.

If there is anything to be gained from this is that bigger sensors perform better than smaller ones due to less enlargement for a given size.
A high pixel count helps increase detail but can limit your minimum aperture.
Cropping is no substitute for a longer focal length lens
Faster lenses beat slower lenses pretty much all things being equal, although better lens design will reduce aberration.

Of course this all stands for nothing if you can't take a decent photo in the first place. :)

Garyminor Loc: Raleigh, NC

That should be 5 lp/mm (line pairs per millimeter).
This gives 0.1mm width for each line, and 0.1mm width for each corresponding space.

blackest Loc: Ireland

Thank you for your correction, I missed that. Maybe thats why 300dpi is preferred minimum for printing 127 lp per inch is 254 ... retina displays are 264 dpi too.

Garyminor Loc: Raleigh, NC

Ron,
You are correct! The DoF doesn’t depend on the size of the sensor.

I see that there are 13 pages of claims and counterclaims. Since this is a technical discussion, there should be only one correct set of facts. I’ve been reading this thread, and trying to understand the logic from both claims. Both are valid, each in their own way.

First, let’s be sure that we understand, that in photography, CoC describes the largest blur circle that is indistinguishable from a point, not the size of the of the blur circle.

If you are going to compare DoF between two photographs, then the lens needs to have the same focal length, aperture, and plane of focus.

Also, the print needs to be the same size, be viewed from the same distance, and the objects in the photo must be the same size. How does that happen?

a) The size of the blur circles on the sensor doesn’t depend on anything but the lens settings and the distance from the plane of focus. There is a different size blur circle for each point, depending on the distance from the plane of focus. That is, when you take a photo of an object, both the size of the object and the size of the blur circles on the sensor will be the same, regardless of the sensor size.

b) Therefore, a FX lens used at 10 meters with the same f11 aperture (example) has the same size blur circles on the sensor, regardless if the sensor is on a FX or DX body.

c) When the image is transformed from the sensor to a viewable image, whether it is a print, projection on a screen, or displayed on a monitor, there is a magnification factor. In the case of using a 4x5 negative to produce an 8x10 print, it is 2:1. This will magnify the size of the object and the blur circles by the same amount.

d) If you produce a print (of the same size) from each of two different size sensors, such that the object is the same size on both prints, then the magnification factor and the size of the blur circles will be the same size on both prints. When viewed from the same distance, the CoC and the DoF will be the same.

Congratulations Ron, you are correct. Does that mean that all of the DoF calculator apps are wrong? Not at all!

The DoF calculators expect that you will make a print that includes the entire image from the sensor. If you produce a full image print (of the same size) from each of two different size sensors, then the print from the larger sensor will have a wider field of view. The objects will be smaller, and the blur circles will be smaller. When viewed from the same distance, the CoC and the DoF will be the different. The DoF calculator doesn’t know that you want to compare two photos that need to have the same object size.

I hope that this helps everyone involved.

Gary Minor

carl hervol Loc: jacksonville florida

Dose it really make a difference it not going to change your pictures.

blackest Loc: Ireland

Depends doesn't it, the simplest take on this is avoid cropping as much as possible if you want the best IQ, If you are printing large you might want to choose a slightly wider aperture. (see cambridge in colors article on diffraction). The bigger the sensor the better the IQ will appear to be (essentially because you just enlarge less for any given size). If you look at large and medium format cameras the price is still in the thousands for anything better than a pinhole camera.

JohnFrim Loc: Somewhere in the Great White North.

Gary, you are completely missing the point. The whole argument centers around the crop factor, which is pertinent because most people will be printing or viewing images at the same size, such as a full screen monitor. You have said as much in your last paragraph. It is NOT about object size, but rather about print size.

I fully agree that the blur spot on the sensor is the same regardless of sensor size, and if you compare a FF image to a crop sensor image when objects are the same size there is no difference in DoF. But images from both cameras are normally printed or viewed at the same dimensions, and that is where the magnification comes into play in altering the DoF.

Garyminor Loc: Raleigh, NC

John, You are right on all counts. If the whole argument centers around the crop factor, then I've missed the point. If it is not about object size, but rather about print size, then I've missed the point.

My point is exactly what you've said in your last paragraph. You've summed up 13 pages of disscussion (at least on the topic of equal DoF) in one neat paragraph. Well done!

Gary

Rongnongno Loc: FL

Then any sensor that has a greater pixel density can be considered as a magnifier.

Why are the manufacturers not using this as a sales argument?

They are advertising as cameras being able to capture more details out of the same lens, create larger prints. You are confusing input with output.

That folks use this as a magnifying option is way beside the point.

If you were dealing with the same sensor density on both sides FX/DX you would not even consider this as anything but a washout.

The only decisive interest of a FX lens on a DX body is the use of the lens center.

As stated in this particular case everything is the same.

As usual potatoes/tomatoes.

I am just tired of this crap.

TriX Loc: Raleigh, NC

The main reason to use an FX lens on a DX body (at least in the Canon world) is because the highest performance lenses are generally FX (or EF) with a very few exceptions.

rehess Loc: South Bend, Indiana, USA

Or because you are thinking of moving to FF

Rongnongno Loc: FL

Oh and also love this statement:

that is where the magnification comes into play in altering the DoF

DoF alters the field of sharpness. At the same time the aperture adjustment alters the size of the airy disk which affects the capture sharpness at the sensor level.

DoF does not magnify anything as far as the image goes.

Once again you confuse input with output.

Take a flea. It has a set size. Now take the same flea, squash it between to glass plates. Using a projector display squashed flea onto a wall. Now you have a freaking monster. Yet the flea you squashed is still the same size. That is what you call 'magnification'. I call that fertilizer reasoning.

TriX Loc: Raleigh, NC

That too!