Thoughts to ponder

[pnodrog]

I don't think I belong to them.

George

Really? ....

[george013]

+1. I also thought the OP meant "perspective distortion" when he used "compression". When people talk about "background compression", which is defined by the angle of view, they mean that in photo #3 you see less of the wall behind the plant than in #1.

No. Read post 12 again. Or tell me where I'm wrong.

There are two definitions of DoF - traditional based on a 10x8 printout (for which pixel count and pixel density are irrelevant) and pixel-based (for which everything is relevant). So my answer to the second question is

A1) full frame camera will have greater DoF (traditional definition)
A2) the camera with lower pixel density, most likely full frame, will have greater DoF (pixel-based definition)

I don't know the definition of pixel based dof.
Optical there's no difference between a FF or crop camera, meaning the same brand, when used on a same focus distance. Focus distance is equal, image distance is equal etc. Only the framing will be different. The difference is in the used CoC, for Nikon FF it's 0.03 and for the DX 0.02. Based on that traditional printout.
Richard meant a comparison with the same distance, a different framing.

George

[george013]

Perhaps a clear definition of;



To me this implies that the 'subject'' (what you see recorded on the sensor) would fill lets say 75% of the FF result and also 75% of the Crop result.

That means another focus distance. And then things are changing.

George

[Stagecoach]

That means another focus distance. And then things are changing.

George

And that is how I read it.

[george013]

And that is how I read it.

In post 4 he's mentioning the same distance and a different framing. Maybe it has changed afterwards.

George

[Stagecoach]

In post 4 he's mentioning the same distance and a different framing. Maybe it has changed afterwards.

George

Richard's statement in Post 4 is correct in that for a given subject when shot with a lens giving a magnification of 1:1 will have the same linear size within the image whether shot with a full or crop frame.

But to me the 'subject to image ratio' will be different.

[dem]

No. Read post 12 again. Or tell me where I'm wrong.

George, you are correct within your definition of "compression" - the size of the background features relative to the size of the subject changes as one steps further away from the subject while using progressively longer focal lengths to keep the framing constant:

https://digital-photography-school.c...e-your-photos/

However, the longer the focal length, the LARGER these background features become - we can call it "background expansion" if you want. Why do then people talk about "background compression" associated with telephoto lenses as in the link above? I guess this is because what people are mainly interested in is to narrow the field of view to keep the background free of clutter - minimise/compress background. I prefer your definition by the way.

I don't know the definition of pixel based dof.

The idea is that the CoC is related to the resolution of the sensor rather than that of human eye. Recent Fuji 24 MPx APS-C cameras have the option of displaying either "paper-based DoF" (CoC=0.02 mm) or "sensor-based DOF" (CoC = 0.008 mm).

This is quite a neat app that can calculate either DOF and also takes diffraction into account:

http://www.dl-c.com/DoF

[george013]

George, you are correct within your definition of "compression" - the size of the background features relative to the size of the subject changes as one steps further away from the subject while using progressively longer focal lengths to keep the framing constant:

https://digital-photography-school.c...e-your-photos/

However, the longer the focal length, the LARGER these background features become - we can call it "background expansion" if you want. Why do then people talk about "background compression" associated with telephoto lenses as in the link above? I guess this is because what people are mainly interested in is to narrow the field of view to keep the background free of clutter - minimise/compress background. I prefer your definition by the way.


The idea is that the CoC is related to the resolution of the sensor rather than that of human eye. Recent Fuji 24 MPx APS-C cameras have the option of displaying either "paper-based DoF" (CoC=0.02 mm) or "sensor-based DOF" (CoC = 0.008 mm).

This is quite a neat app that can calculate either DOF and also takes diffraction into account:

http://www.dl-c.com/DoF

It's here where I heard first of background compression.

Your link shows exactly what I mean and what's going on. However, it's totally independent of the focal length. Solely on the distance. It's just that the distances are mostly bigger when you use a long focal length. You can get the same result with a shorter lens and a long distance. You only have to enlarge the picture more to get the same frame.

I've been googling "sensor based dof" and "pixel based dof". The only result is that Fuji has something as pixel based. Reading this article I still don't know what's meant. http://fujifilm-x.com/de/x-stories/d...field-in-deep/. To many nonlogical conclusions, for me anyway.

George

[DanK]

However, the longer the focal length, the LARGER these background features become - we can call it "background expansion" if you want. Why do then people talk about "background compression" associated with telephoto lenses as in the link above?

Dem, they are talking about a plane perpendicular to the sensor. You are talking about a plane parallel to the sensor. Making background objects larger ("background expansion" in your terminology) makes the background appear closer ("compression"0.

However, it's totally independent of the focal length. Solely on the distance.

I can't understand this argument. It's obviously not independent of focal length, which is why everyone observes the phenomenon when long lenses are used. It boils down to a few straightforward things:

--AOV (notice, I did not say focal length) determines how far away background objects are, relative to a foreground object of a given size.
--If you frame to have a near object the same size, longer focal lengths will create an impression that backgrounds are closer because of their narrower AOV. This of course requires standing different distances from the near subject.
--if you stand in the same location, shoot with a wide lens and a long lens and then crop the image from the wide lens so that the near subject is the same size as in the image from the long lens, then the cropped image will be identical to that from the long lens, and there will be no difference in compression. Why? Because cropping the image from the wide angle lens creates an image with the same AOV as the uncropped image from the longer lens.

That is why both distance and focal length matter.

[george013]

Dem, they are talking about a plane perpendicular to the sensor. You are talking about a plane parallel to the sensor. Making background objects larger ("background expansion" in your terminology) makes the background appear closer ("compression"0.



I can't understand this argument. It's obviously not independent of focal length, which is why everyone observes the phenomenon when long lenses are used. It boils down to a few straightforward things:

--AOV (notice, I did not say focal length) determines how far away background objects are, relative to a foreground object of a given size.
--If you frame to have a near object the same size, longer focal lengths will create an impression that backgrounds are closer because of their narrower AOV. This of course requires standing different distances from the near subject.
--if you stand in the same location, shoot with a wide lens and a long lens and then crop the image from the wide lens so that the near subject is the same size as in the image from the long lens, then the cropped image will be identical to that from the long lens, and there will be no difference in compression. Why? Because cropping the image from the wide angle lens creates an image with the same AOV as the uncropped image from the longer lens.

That is why both distance and focal length matter.

Read post 12 again. It's with rather simple calculations. The difference of magnification is dependent on the difference in distances. The further away, the less difference. The focal length is just enlarging the image. A thing you can do in pp.
I'll try to make a drawing later that will make it easier to see.

George

[Black Pearl]

In macro photography, which 1:1 ratio image has a greater DOF: a full frame camera image or a crop camera image? I contend that the images from either camera are exactly the same if the subject to image ratio is the same...

I should be able to test this. My X-T1 with its 1.5x crop sensor has 16mp and a friend's Df with its FX sensor also has 16mp. We both have a range of macro lenses so next time we're together I'll run some comparisons.

Could be a week or two so might need to resurrect this thread...

[george013]

A simple drawing. I made it orthogonal so it's easier to explain.
Camera at A. An angle of view of a degrees. The lines h represent the width of the aov at that distance, so at distance d3 the sensor covers the real world image of h3. I don't know how to express exactly.
The tangens is the ratio between h and d: tg(a)=h1/d1=h2/d2=h3/d3=h4/d4.
The s lines are equal subjects with a realworld size of s. The magnification M = s/h, since s doesn't change, M does change. M1=s/h1, M2=s/h2,etc.

What happens when the aov changes? The distances don't change, the s sizes don't change, but the h sizes do change. If the aov changes so that h1 is changed with a factor x, then all the other h sizes are changed with a factor of x. Since M=s/h, all the M are changed with a factor x. A smaller aov is essential cropping out a part of the former picture and enlarge that to the sensor size. Figuratively.

How does that feeling of "compression" develops.
I did draw the subjects on even distances to make the calculations easier.
M1=s/h1, M2=s/h2. Since h2=2.h1 M2=0.5M1
But what for the last 2?
M3=s/h3 and M4=s/h4. We can express h3 in3h1 and h4 in 4h1. M3=s/3h1 and M4=s/4h1. M4=0.75 of M3. A smaller difference in M but with the same distance.
There is no aov involved.

Always hoping I didn't make a mistake.

George

ps.
Changed some typo that nobody saw.

[pnodrog]

Another thought:

In macro photography, which 1:1 ratio image has a greater DOF: a full frame camera image or a crop camera image? I contend that the images from either camera are exactly the same if the subject to image ratio is the same...

The DOF of the larger format will be greater unless the larger format is cropped to match the smaller format or the overall enlargement scale is identical. e.g if the subject was a 12mm coin and prints made from any format display it at say 180mm (x15).

"subject to image ratio is the same" ? = 1:15?

[JohnRostron]

An alternative interpretation of ' compression' is the rate of change of magnification with distance. Using the standard symbols:

U = subject distance from lens centre
V = image distance from lens centre
F = focal length of lens

And using the standard optical formulae:

1/U + 1/V = 1/F
Magnification M = V/U

Gives us

M = F/(U - F)

Applying a bit of calculus gives the rate of change of M with U (call it P for perspective) gives us:

P = -F/(U - F)^2 or approximately -F/U^2.

This definition of compression (or perspective) says that it is directly proportional to the focal length F, and inversely proportional to the subject distance squared.

John

[george013]

An alternative interpretation of ' compression' is the rate of change of magnification with distance. Using the standard symbols:

U = subject distance from lens centre
V = image distance from lens centre
F = focal length of lens

And using the standard optical formulae:

1/U + 1/V = 1/F
Magnification M = V/U

Gives us

M = F/(U - F)

Applying a bit of calculus gives the rate of change of M with U (call it P for perspective) gives us:

P = -F/(U - F)^2 or approximately -F/U^2.

This definition of compression (or perspective) says that it is directly proportional to the focal length F, and inversely proportional to the subject distance squared.

John

Your "applying a bit of calculus" is a mystery for me.

Did you read my post and arguments? If so can you tell me why you've got a different result as I?

I can give you another one, one I posted not so long ago. And also without reaction.

M= magnification, di = image distance, ds=subject distance

M=di/ds A second subject on a distance x from the first will have a M=di/(ds+x). As you see there is no linearity in it. Nearby ds will be relative small to x, further away x will be relative small to ds.

George

[JohnRostron]

I did read your arguments, along with the rest. I will need to take these away to contemplate. I was taking an alternative definition of 'compression' which may not coincide with yours, but which was amenable to my mathematical approach.

I took your estimated distances for the plant and the wall of 8m and 10m, translating them into millimetres and plugging them into my formula: M = F/(U-F)

For the 17mm lens, M at 8m is 0.00213; M at 10M is 0.00170. The difference is 0.00043
For the 70mm lens, M at 8m is 0.00883; M at 10M is 0.00705. The difference is 0.00175
For the 200mm lens, M at 8m is 0.02564; M at 10M is 0.02041. The difference is 0.00523

Thus the differences in perceived magnification are proportional to the focal length.

This is all that I was trying to demonstrate.

John

[dem]

Dem, they are talking about a plane perpendicular to the sensor. You are talking about a plane parallel to the sensor.

Quite right. I am happy with the fact that a telephoto lens allows one to bring the background closer to the subject (only because you have to step further back, nothing to do with a narrow AoV).

--AOV (notice, I did not say focal length) determines how far away background objects are, relative to a foreground object of a given size.

No, not relative to a foreground object. Varying AOV changes magnification of both foreground and background objects by the same amount. This is one of the points George is trying to make and this is also what follows from your third bullet point.

--If you frame to have a near object the same size, longer focal lengths will create an impression that backgrounds are closer because of their narrower AOV. This of course requires standing different distances from the near subject.

It is the standing back that is the deciding factor here, not the narrow AoV.

--if you stand in the same location, shoot with a wide lens and a long lens and then crop the image from the wide lens so that the near subject is the same size as in the image from the long lens, then the cropped image will be identical to that from the long lens, and there will be no difference in compression. Why? Because cropping the image from the wide angle lens creates an image with the same AOV as the uncropped image from the longer lens.

Agreed. That's why focal length does not affect compression, camera-subject and camera-background distances do.

Now I have said that, let's look at the formulas... They say that a longer lens is going to move the lens position further away from the sensor and therefore change both camera-subject (the distances in post #34 are measured not from the sensor but from the lens, rear focal plane for a thick lens) and camera-background distances. The effect on compression will be negligible in most cases apart from macro.

I took your estimated distances for the plant and the wall of 8m and 10m, translating them into millimetres and plugging them into my formula: M = F/(U-F)

For the 17mm lens, M at 8m is 0.00213; M at 10M is 0.00170. The difference is 0.00043
For the 70mm lens, M at 8m is 0.00883; M at 10M is 0.00705. The difference is 0.00175
For the 200mm lens, M at 8m is 0.02564; M at 10M is 0.02041. The difference is 0.00523

John, looking at differences in magnification is not informative - these differences mainly reflects the fact that you get greater magnification when using longer focal lengths. To assess compression, one needs to look at the ratios M_background / M_subject:

For the 17mm lens, M at 8m is 0.00213; M at 10M is 0.00170. The ratio is 0.798
For the 70mm lens, M at 8m is 0.00883; M at 10M is 0.00705. The ratio is 0.798
For the 200mm lens, M at 8m is 0.02564; M at 10M is 0.02041. The ratio is 0.796

There is a difference of about 0.2% in compression between 17mm and 200mm. I think we can safely ignore this difference for this exercise.

[pnodrog]

When I was a kid we used to look down a tube because it seemed to act as a telescope. The answer has almost nothing to do with optics but a great deal to do with perception.

[Stagecoach]

Exactly, and we will all have different perceptions with a plethora of descriptions to explain what is a very simple fact

Drg No 1 : Basics


and I believe the FL/AOV of the lens used at any distance will affect the ratio in size between near and far subjects recorded on the sensor additional to that ratio variation which is due to sensor to subject distance.

Drg No 2 : The affect of FL/AOV change from the same camera position

[george013]

Exactly, and we will all have different perceptions with a plethora of descriptions to explain what is a very simple fact

Basics


and I believe the FL/AOV of the lens used at any distance will affect the ratio in size between near and far subjects recorded on the sensor additional to that ratio variation which is due to sensor to subject distance.

You see the light but have some remarks about the switch?
In Richards example #3 it would mean that the camera has moved forward 193mm. Correcting that by moving the camera backwards would have enlarged this difference in M. With an amount you probably won't see.
Do you correct your dof-tables with that information.?

Dem,
I'm glad I found somebody who understands what I'm trying to explain.

George