PERSPECTIVE "distortion"

[johnielvis]

here's 1000 words in a pic---the one ruler is at the tip of the nose--the other at the eye socket...shot from like a foot away...cut out of the nose ruler bottom and copied it to the other ruler to compare size LOOK!!!!! the nose ruler is BIGGER than the other ruler!!!!!

that means that there's more magnification on the nose tip than the things behind the nose tip!!!!!

SEE....this was shot from like 1 foot away....with the apple built-in cam

BUT...you see...the short distance causes the magnification to change greatly over the depth of a face....

if you use a longer lens and get farther away, the change in magnification is smaller, therefore you see it in more "proper perspective".....

so there is no "distortion" here--it is pure projection--a pinhole will do this...i think the lens on the apple IS a pinhole actually--this is pure perspective and solely due to distance from the subject--absoluetly NO distortion--this is what happens when you project a 3D object onto a 2d plane---then you look at the 2d plane and your brain thinks there's something screwy going on cause that's now how you "saw it".

so anybody thinking they got 'barrel distortion" or any kind of "distortion" is misguided....it's pure projection/perspective...things closer to the lens are bigger because they are magnified more. PERIOD.

[Paul Fitzgerald]

Correct, that's the 'Z axis' distortion, you can use it to advantage or it can make you crazy.

"if you use a longer lens and get farther away, the change in magnification is smaller, therefore you see it in more "proper perspective"....."

Both reasons why longer lenses have preferred for portraits, the extra distance gives a nicer perspective AND a deeper apparent DOF for the same f/stop=magnification. You do start to run into telephoto compression from 12'(4m) or further.

[ic-racer]

And the effect is independent of the lens. It depends only on subject distance.

[johnielvis]

right--it seems that people see write-ups on perspective...and there this very good one that was on wikipedia with bottles as an example--BUT, this, for some reason, apparently leads people to believe that the effect is only seen for two separate objects.

they seem to miss the fact that the SAME object can have varying magnification over it's surface if it is sufficiently 3-dimensional. One object is NOT all at the same distance from the lens.

This is what I think is going on in the minds of some---they understand it all with the "separate" objects, and the effects, but then the connection is missed when there is one object that looks "distorted"--the implicit assumption is that all parts of the same object are the same distance away, since its all connected on the same object--therefore it should record perfectly like it was all at the same distance---the way you "see it" when you look at it.

Then you get the picture and your brain interprets the projection as 2-d and it looks strange--and you are confused---it don't look like how it was first seen with the eye.

[polyglot]

I assume you're referring to my comments in the barrel distortion thread, my point was that walls are flat, unlike faces. If shot straight-on, they should look square not barreled, whereas the OP there was alleging barrel distortion (I think he imagined it).

My second post even described this effect you've illustrated nicely with the big nose.

The reason I disagreed with you in that thread was because you asserted that a flat wall would appear bulged in the middle because the middle is closer, and that is not the case.

[unixrevolution]

Very interesting blurb. I'm glad I read it...though, the apple iSight's lens is no pinhole. It's got an f/2.8 aperture, with a 3-element glass lens with 2 aspherical elements. The focal length is just unbelieveably short thanks to its 1/4" sensor.

[Paul Fitzgerald]

"The reason I disagreed with you in that thread was because you asserted that a flat wall would appear bulged in the middle because the middle is closer, and that is not the case."

Actually it is the case. Lens companies can adjust the lens to be flat-field but only at it's intended focus distance / magnification, going outside that range the distortion does show.

Draw it out or graph it:
center of lens to center of target = 100%
center of lens to far corner of target = 110%
10% magnification difference
adjusted and/or compounded by:
center of lens to center of film = 100%
center of lens to far corner of film = 110%

the difference from center to corner expands exponentially the closer you get.
'And the effect is independent of the lens. It depends only on subject distance.'

[johnielvis]

wouldn't the extra distance EXTRA magnification on the film side exactly compensate for the extra disttance which reduces magnification towards the subject?...there's some sort of compensation going on there as well...the extra length inside causes the numerator to get bigger (increase magnification), but the extra distance on the outside to the subject causes the denominator to get bigger (decrease magnification)....do these EXACTLY cancel each other out? I'll have to check the calculations--I know I calculated the exponential distance increase in my younger days and even wrote a computer program that printed out what distroted graph paper would look like as you chance the proximity.....I'll see about calculating the %'s but it does seem that they should exactly cancel each other out--so a pinhole, pure projection, wont' distort---but we'll see....busy now to think about it or draw diagrams....

[E. von Hoegh]

Perspective is one thing.

Artifacts due the lens are another issue entirely, better not to confuse the two. Brunelleschi knew all there was to know about perspective 600 or so years ago. :)

[Paul Fitzgerald]

"wouldn't the extra distance EXTRA magnification on the film side exactly compensate for the extra disttance which reduces magnification towards the subject?...there's some sort of compensation going on there as well...the extra length inside causes the numerator to get bigger (increase magnification), but the extra distance on the outside to the subject causes the denominator to get bigger (decrease magnification)....do these EXACTLY cancel each other out?"

Depends on how the lens was optimized at the factory. They can optimize for color correction, flat field and rectilinear at 1 focus distance / magnification. Outside of that all bets are off. :eek:

Tele and aero lens = inf = 1 fl
General purpose = inf - hyperfocal = 1 fl +
Wide angle = hyperfocal = 1 fl ++
Process = 1-1 = 2 fl
Macro = 2-1 = 2 fl+

[Robert Tilden]

So... The bulgey flat wall thing. Wouldn't a pinhole image of the wall settle the question of whether the magnification/perspective cancels out?

[Mark Woods]

Perspective is one thing, lens artifacts are another. BTW, different lenses with the same image size will have the same DOF if the F/Stop is the same. There is no free lunch.

[Paul Fitzgerald]

"Wouldn't a pinhole image of the wall settle the question of whether the magnification/perspective cancels out?"

Only when all the planets align :D

The difference from image center to image far corner <=> film center to far corner will only cancel themselves out at 1 ( image size-image distance-focal length-format).

As you get closer you move the lens farther from the film plane. The difference at the image plane grows but the difference at the film plane shrinks, they no longer cancel each other. That is why they optimizes lenses for different distances.

A 14" Ektar should be 14" from the film, a 14" Artar should be 28" from the film.
Just pick the correct lens for the distance, they already did the math. :eek:

[Bill_1856]

You've got too much spare time on your hands.

[johnielvis]

YESSSSS...THAT was the key with the real lens

for a pinhole--zero distortion, because the pinhole never moves--it is always at the same image distance, f, "the focal length", so no matter what the object distance, s, or image distance, f, it receives a constant magnificaiton, M=s/f....BUT
....
IF you have to FOCUS your lens, then when you focus closer on something, the lens moves out---this changes the magnification from what the pinhole would be...the pinhole would still be at the focus distance, f away from the image plane, but a focused lens...focused on something a distance, s, away would now be a distance s' given by lens equation:
1/s' + 1/s = 1/f....so you are now with an image distance s' = 1/(1/f - 1/s)....and no longer the fixed distance, f....

RIGHT...

BUt yes--for a straight pinhole that is NOT FOCUSED....just a simple projection point with a location that never MOVES, then the image is constant magnification with absoluetly no distortion...yes...that brings it back....I remember starting off a LONG time ago when I was calculating view angles for certain format sizes based on focal length and distance of the object focused on---the viewing angle changes when the lens effectively changes it's focal length--when you move it out to focus....RIGHT...I forgot about that bit....gonna go through the transformation again..Ihave an old fortran program that maps the grids to a certain sized format...shows that "fiseye" effect....now to try to translate that to something I have now ......no fortran anymore on the apple...that was YEARS ago....

[Leigh]

If shot straight-on, they should look square not barreled, whereas the OP there was alleging barrel distortion (I think he imagined it).
That is true iff* the imaging system is absolutely square with the wall, with the lens axis absolutely perpendicular to both the film and the wall.


The reason I disagreed with you in that thread was because you asserted that a flat wall would appear bulged in the middle because the middle is closer, and that is not the case.
Whether or not the wall appears distorted depends entirely on the distance of the film/lens from the wall. If it's short, such that the distance from the front lens node to the center of the subject is significantly shorter than from the node to the extreme corners, then some distortion could be introduced.

- Leigh

*n.b. iff means "if and only if". It's a term from formal logic.

[polyglot]

Whether or not the wall appears distorted depends entirely on the distance of the film/lens from the wall. If it's short, such that the distance from the front lens node to the center of the subject is significantly shorter than from the node to the extreme corners, then some distortion could be introduced.


No, not unless the lens itself introduces distortion due to it being used outside of its optimised range.

[Leigh]

No, not unless the lens itself introduces distortion due to it being used outside of its optimised range.
Perhaps you should review some basic laws of optics and geometry.

- Leigh

[EdWorkman]

Poly
You need to actually go out and measure from a camera position.
There is no way on THIS planet that the wall at the edge of the frame isn't farther away from the wall at the center. By going backbackback the difference can be made small, maybe even smaller than the increments on your tape, if you have a really long tape.
And since it's farther, by geometry it must be projected on the film plane as a shorter line.
Now instead of some snappy comebacks, come back with a plan sketch of your dimensions actually measured- you might even plot them to scale.
You can even add dimensions from the top to the lens and from the bottom to the lens .
Then draw the projections thru the lens onto the film plane so you'll SEE what happens
Start reeeeal simple by measuring on a horizontal plane at the midheight of your vertical lines. Say the center of the wall is 4 units away, then go at a right angle 3 units perpendicular. You'll find the flat wall from that point back to the camera is 5 units . Nice convenient triangle and you can use yard as the unit , or a meter, or use 4 foot strips cut from a sheet of plywood .
By similar triangles a vertical on the wall at the edge of the frame is close to 4/5 as tall on the film plane as a line of identical height at the center of the wall.
If that's too simple, measure the lengths of the rays from the top and bottom points of the lines to the center of the lens. [ hint , more than 4 and 5 units, oh but wait, we're measuring to the same points on those vertical lines and....]
You can , if you measure very carefully and with really small incements, see how far off the 4/5 is, or use lots of significant figures and calculate it.
Now the lines from the tops of [all] the verts get longer, as do the lines from the bottoms of the lines. Goodbye simple 3-4-5 triangles and hello trig functions, but the relationship is ever the same: center-taller than sides measured on the film plane
When do the distances get to be the same? NEVER.
And no fair using a rubber tape measure.
I suggest a 25 foot Stanley, but you pick.
Let us know how it works out with actual dimensions.
If you prefer not to make measurements, come back with a dead horse and we'll beat it.

[Leigh]

A 14" Ektar should be 14" from the film, a 14" Artar should be 28" from the film.
Huh???

The rear node of a 14" lens is 14 inches from the center of the film when focused at infinity.

That's true for any/every 14" lens. It's the definition of focal length.

That's true for Aardvarkars and zzenzenars and every -ar in between.

- Leigh

[E. von Hoegh]

I think his point was that the lenses are optimised for different reproduction ratios.

[Paul Fitzgerald]

"Originally Posted by Paul Fitzgerald
A 14" Ektar should be 14" from the film, a 14" Artar should be 28" from the film.

Huh???
The rear node of a 14" lens is 14 inches from the center of the film when focused at infinity.

That's true for any/every 14" lens. It's the definition of focal length.
(No, the definition for focal length is the difference from inf. to 1-1 reproduction)

That's true for Aardvarkars and zzenzenars and every -ar in between.
- Leigh"
**********************
"I think his point was that the lenses are optimised for different reproduction ratios."

Correct :D , each lens is optimised for 1 magnification / focus distance, just pick the right lens for the job.

The 3-4-5 geometric 20% difference is ever present and ever changing, the lens companies can design for perfection at 1 magnification. Once you move to focus outside their design range, all bets are off.

[Tim Meisburger]

I'm sorry Paul, I'm a bit simple. This means the Ektar will photograph a wall as a flat field 14" from the lens, while the Artar yields a flat field at 28"? If my understanding is correct, that sounds like it would be very useful for something like product photography, and perhaps tight portraits.

[E. von Hoegh]

I'm sorry Paul, I'm a bit simple. This means the Ektar will photograph a wall as a flat field 14" from the lens, while the Artar yields a flat field at 28"? If my understanding is correct, that sounds like it would be very useful for something like product photography, and perhaps tight portraits.

The Artar is a process lens, optimised for close-ratio reproduction. Great for portraits, but your subjects may not like the results, it will show every last little detail.
Artars were very popular with product photographers. They are quite useable at infinity, and the ones mounted in shutter by Goerz were optimised for longer working distances.

[Tim Meisburger]

Thanks. That makes sense. A process lens would be a lens for an enlarger, a copy camera, or macro work. A portrait lens would be optimized to yield a flat field at portrait distance, and a landscape lens would be optimized for infinity. Is that about right?

Tim (learning about lenses) Meisburger

[E. von Hoegh]

Only a process lens would be a truly flat field lens. A general purpose lens, say a Symmar, would be less so, and a portrait lens least of all.

[Paul Fitzgerald]

"I'm sorry Paul, I'm a bit simple. This means the Ektar will photograph a wall as a flat field 14" from the lens, while the Artar yields a flat field at 28"? "

NO, from the film, not to the target.

The Z-axis distortion, from the optical center to far corner, is ever present and ever changing, the difference expands exponentially as you get closer, so a lens can only be perfected for 1 distance from the film plane. Most lens companies produced different lines of lenses optimized for different magnification ranges, just check their web site for specs and pick the lens closest to your needs.

Is this splitting hairs? Yes, it is literally 'splitting hairs', just set-up exactly perfect then use any tilt or swing and see how fast it goes sideways and how much fun it is to re-correct.

For Z-axis distortion see the link below, any movement of the lens or GG changes the distortion and is INSTANTLY visable of the GG. Here the camera is at ground level, GG square to the building and uses front rise, changing the magnification from top to bottom and he gets a 'mushroom head' building, all from Z-axis distortion. He should have used some camera tilt and some rise to equalize the building.

mushroom head (http://www.shorpy.com/node/11374?size=_original)

[Tim Meisburger]

Who? Me?

[Doremus Scudder]

We should remember that a lens's rectilinearity is not directly related to having a flat field of focus. The ability to render a square subject exactly square on the film is what is meant by rectilinear. Flat-field refers to a lens being able to focus on specific subject plane exactly and project this plane in-focus on the film plane (as opposed to curved field of focus). The two are not mutually exclusive, but are not a function of each other. One could have rectilinearity, but a curved field of focus, and vice-versa.

Best,

Doremus Scudder

[polyglot]

You would benefit by paying closer attention in your ninth grade English classes.

Hear hear.

I'm glad to hear that everyone seems to have an enlarger that is producing fisheye images (what's that? no?), what with the lens being closer to the middle of the paper and all.

Go learn some optics and geometry instead of spouting bullshit on the internet, people.

[Tim Meisburger]

Wow! This is a really touchy subject!

Paul, how can a 14" lens be optimized to yield a flat field at 14" from the film plane? Wouldn't that be where the lens is when focused at infinity? Or are you saying that the designation of the lens as 14" refers to flat field distance rather than focal length?

[E. von Hoegh]

Wow! This is a really touchy subject!

Paul, how can a 14" lens be optimized to yield a flat field at 14" from the film plane? Wouldn't that be where the lens is when focused at infinity? Or are you saying that the designation of the lens as 14" refers to flat field distance rather than focal length?

14" is a designation of the focal length when focussed at infinity. Period. The reproduction ratio the lens is designed for determines it's optimum working distance.

[Tim Meisburger]

Okay. That's what I would have said. But then what does the quote below mean? I thought he was trying to say that some lenses are designed to reproduce a flat object parallel to the film plane as flat at a particular distance from the lens, and at all other distances the object will be reproduced with some distortion. The reproduction ratio you mention could also be given as a distance from the lens. If that is not the case, then the sentence below makes no sense, as a 14" lens could never photograph something 14" from the film.




A 14" Ektar should be 14" from the film, a 14" Artar should be 28" from the film.
Just pick the correct lens for the distance, they already did the math. :eek:

[E. von Hoegh]

He's talking about lens to film distance.

[Paul Fitzgerald]

As people are still reading this, let's continue:

If you wish to see a perfect example of "Perspective Distortion" owing to varying distance only, go to www.shorpy.com and scroll down to 'Toledo Panorama: 1909'.

View it on the main page to see the 'smile', then click on 'view full size' to see there are no lens induced distortions visible, if there were they could not have stitched the 5 plates together seamlessly.

To my eyes the FOV looks like a 300mm on 8x10, set for hyper-focal distance, letting DOF carry it front and back. I'd love to know which lens, it's beautifully corrected for a general purpose taking lens.
_________________________________________________

Plugging the formulas into a spread sheet is simple, let the computer do the math. The Z-axis is a simple right triangle converted to difference %. You'll notice that the focus displacement does not proceed in a linear fashion and the in-camera Z-axis % does not proceed in a linear fashion by focus distance or across the format.

The changing FOV by focus displacement does not proceed in a linear fashion and the object side Z-axis % does not proceed in a linear fashion either. Moving up in format or shorter with focal length compounds these problems. The Z-axis % also effects exposure difference across the film compounded by the inverse square rule, that is why they make center filters.

These are each and all 'perspective distortions' relying on distance only. They are directly and continuously related to each other to the point you trip the shutter. They do not include any of the optical abberations the lens companies deal with.

The men and women at the lens companies are very good at their jobs BUT unless and until they install floating aspheric elements into large format prime lenses they no longer manufacture I guess we're stuck with 'Pick the right lens for the job'.

(I hope will display correctly)

focal length (mm) ____ 300
c. of c. (mm) __ 00.174418605 ___ (f/1720)

distance (feet)
hyper __ 100 ___ 50 ___ 30 ____ 15 ____ 10 ____ 8 ____ 6 ____ 3 ___ (1-1)

Movement from INF (mm)

0.8 ____ 3.0 ___ 6.0 __ 10.2 __ 21.1 __ 32.8 __ 42.2 __ 59.1 __ 147.1 __ 300.0

Z axis distortion %

13.3 __ 13.1 __ 12.8 __ 12.5 __ 11.7 __ 11.0 __ 10.4 ___ 9.5 ___ 6.2 ___ 3.5 ___ 8x10 _ (1/2 dia. = 160mm)

5.9 ____ 5.8 ___ 5.7 ___ 5.6 ___ 5.2 ___ 4.9 ___ 4.6 ___ 4.2 ___ 2.7 ___ 1.5 ___ 5X7 __ (1/2 dia. = 105mm)

3.2 ____ 3.2 ___ 3.1 ___ 3.0 ___ 2.8 ___ 2.6 ___ 2.5 ___ 2.3 ___ 1.5 ___ 0.8 ___ 4X5 __ (1/2 dia. = 77mm)

*************************************************

focal length (mm) ___ 210
c. of c. (mm __ 00.122093023 ___ (f/1720)

distance (feet)
hyper __ 100 ___ 50 ____ 30 ____ 15 ____ 10 ____ 8 ____ 6 ____ 3 ___ (1-1)

Movement from inf (mm)
0.6 ____ 1.5 ___ 2.9 ___ 4.9 __ 10.1 __ 15.6 __ 19.8 __ 27.3 __ 62.8 __ 210.0

Z axis distortion %

25.6 __ 25.4 __ 25.1 __ 24.7 __ 23.6 __ 22.6 __ 21.8 __ 20.6 __ 15.9 ___ 7.0 ___ 8x10 _ (1/2 dia. = 160mm)

11.7 __ 11.6 __ 11.5 __ 11.3 __ 10.8 __ 10.3 ___ 9.9 ___ 9.4 ___ 7.2 ___ 3.1 ___ 5X7 __ (1/2 dia. = 105mm)

6.5 ____ 6.4 ___ 6.3 ___ 6.2 ___ 5.9 ___ 5.7 ___ 5.5 ___ 5.1 ___ 3.9 ___ 1.7 ___ 4X5 __ (1/2 dia. = 77mm)

[Mark Woods]

Leigh I agree with most of what you say. Retrofocus lenses are different. Sorry to be a pedant. I like your posts, btw.

[rdenney]

As people are still reading this, let's continue:

If you wish to see a perfect example of "Perspective Distortion" owing to varying distance only, go to www.shorpy.com and scroll down to 'Toledo Panorama: 1909'.

View it on the main page to see the 'smile', then click on 'view full size' to see there are no lens induced distortions visible, if there were they could not have stitched the 5 plates together seamlessly.

Each of the five plates was pointed in a different direction, and when stitched, a cylindrical projection was selected in the stitching software. PanoTools, Panavue, and most other panoramic stitchers provide that feature. Had a rectilinear view been selected, you would not see the "smile" distortion. This distortion is exactly what you get with a swing-lens panoramic camera, which is not rectilinear at all in the horizontal direction.

Such software usually has features to correct minor geometric distortions in lenses, too. In Panavue, for example, I could dial in a lens correction separately from the geometric conversion of the component images when assembling the panorama.

A fisheye produces a spherical projection, a swing-lens camera produces cylindrical projection, and a rectilinear lens (which includes nearly all regular camera lenses) produces rectilinear projection. If a lens is rectilinear, then no matter how short the lens is, the lines will be straight.

What makes a very wide image look very wide is the usually close camera position to foreground subject material, which causes that material to loom large with respect to the background. A box which is showing its front and side will look distorted because the near part looms so large with respect to the far part that the perspective convergence is exaggerated. But if you back up to the same position you'd use with a longer lens, you could extract a longer-lens image from the middle of the short-lens image, and it would look the same.

If you want 180-degree coverage from a stitched panorama, you must choose a cylindrical or spherical projection. Choosing a rectilinear projection becomes computationally impossible--there is no rectilinear view of an object at 90 degrees to the line of sight. That limits how short a rectilinear lens can be. The current maximum field of view for a rectilinear lens is 121 degrees, offered by the 12mm Voightlander lens on the Bessa 35mm camera (and by the 12mm end of the Sigma 12-24 zoom lens for other 35mm cameras). That lens is about the same across the width of the frame as a 42mm rectilinear lens would be on 4x5 (or an 84mm lens on 8x10). The widest angle of view provided in a large-format lens is the 110-degree Super Angulon XL--which barely covers 4x5 at 47mm. The 110-degree 90 does not cover 8x10, thought it will cover the 10" width through a narrow stripe in the middle of the frame.

Rick "a wide-angle freak who has used all three projection methods" Denney

[Paul Fitzgerald]

"Each of the five plates was pointed in a different direction, and when stitched, a cylindrical projection was selected in the stitching software. PanoTools, Panavue, and most other panoramic stitchers provide that feature. Had a rectilinear view been selected, you would not see the "smile" distortion. This distortion is exactly what you get with a swing-lens panoramic camera, which is not rectilinear at all in the horizontal direction.

Such software usually has features to correct minor geometric distortions in lenses, too. In Panavue, for example, I could dial in a lens correction separately from the geometric conversion of the component images when assembling the panorama."

Nice to know, have never used it.

Only problem is that this was 1909, with glass plates being contact printed, intended for manual stitching on a production line. These would normally have been displayed on a flat wall. Nowhere along the line could they have corrected for distortions.

It does look like the photographer had it 'dialed-in' to get it right in-camera. Perfectly level, normal focal length lens, hyper-focal distance (100 ft. f/16), 1/15 sec. exposure (look at the water). It also looks like he was pivoting at or very near the rear lens node.

"What makes a very wide image look very wide is the usually close camera position to foreground subject material, which causes that material to loom large with respect to the background. A box which is showing its front and side will look distorted because the near part looms so large with respect to the far part that the perspective convergence is exaggerated. But if you back up to the same position you'd use with a longer lens, you could extract a longer-lens image from the middle of the short-lens image, and it would look the same."

Exactly, you just described 'Z-axis distortion' to a tee.:D

As you get closer to the object, the Z-axis distortion expands exponentially on the object side and continues to reduce on the film side. Once you are too close for the designed range of the lens the flat wall appears to bow out, the center being measurably closer than the edges and corners of the frame.

[Leigh]

I'm glad to hear that everyone seems to have an enlarger that is producing fisheye images...
Go learn some optics and geometry instead of spouting bullshit on the internet, people.
Yes, you should, since you obviously don't know the difference between a fisheye lens and a regular lens.

- Leigh

[Paul Fitzgerald]

www.shorpy.com just put up another 8x10 panorama 'Atlantic City Panorama: 1910'.
Marvelous and yes, the ocean does have a slight curve to it naturally.
I still want to know which lens they were using.

[rdenney]

Only problem is that this was 1909, with glass plates being contact printed, intended for manual stitching on a production line. These would normally have been displayed on a flat wall. Nowhere along the line could they have corrected for distortions.

It was not my impression that the assembly of the images was done to achieve this projection back in 1909. They bragged about the pixels dimensions, which suggested to me that the plates were made in 1909 but that this particular panorama was assembled recently, using software. Hence the sentence: "Humongous 40,000-pixel-wide panorama made from five 8x10 glass negatives". If using flat glass plates, they would have made five separate images and hung them side by side. If you look at the writing in the lower right corner, you'll note that what little of it that can be seen is curved as well. Had they made this print back in 1909, that writing would be straight.

Here are several early panoramas made in that way (public-domain images from Wikipedia):

http://upload.wikimedia.org/wikipedia/commons/thumb/a/a1/Panoramic_San_Francisco_from_Rincon_Hill_c.1851.jpg/798px-Panoramic_San_Francisco_from_Rincon_Hill_c.1851.jpg

http://upload.wikimedia.org/wikipedia/commons/thumb/8/8f/Panoramic_from_Lookout_Mountain_Tenn.%2C_1864.jpg/800px-Panoramic_from_Lookout_Mountain_Tenn.%2C_1864.jpg

This one, on the other hand, was made with a swing-lens panoramic camera using curved film:

http://upload.wikimedia.org/wikipedia/commons/thumb/d/de/Downtown_Philadelphia_Pano_1913.jpg/800px-Downtown_Philadelphia_Pano_1913.jpg

The first is an 1851 Daguerreotype, the second was made in 1864, and the last was made in 1913, well after the invention of flexible film. Swing lens cameras, of course, arrange the film in a curve--that is the only way to get the cylindrical projection--and that type of photo is not possible with flat glass plates without a cylindrical optic. (I suppose it is mechanically possible to slide a flat plate in the opposite direction of the rotation, so that the portion of the film being exposed is continuously behind the swinging lens, but I've never heard of such a thing.) The earliest swing-lens cameras that I've read about, which date back into the middle 1800s, used special curved-glass plates.

Rick "wondering if a cylindrical lens was ever used to remove the lens swing requirement, as with Cinemascope" Denney

[johnielvis]

as for "fisheye" enlargers---there is lens induced distortion with any real lens--including enlarger lenses...

BUT....

actually---for PERFECT no LENS distortion on enlargements---take the camera lens and put THAT in the enlarger---just enlarge to the same scale as the photo--whatever lens induced distortion there was in the film image is then REVERSED in the enlargement---ZERO distortion with a for REAL lens---hey hey!

[Leigh]

... for PERFECT no LENS distortion on enlargements---take the camera lens and put THAT in the enlarger ...
That's quite an innovative idea...

At least it was back when the Graflarger was introduced for the Graflex cameras about 60 years ago.

- Leigh

[E. von Hoegh]

That's quite an innovative idea...

At least it was back when the Graflarger was introduced for the Graflex cameras about 60 years ago.

- Leigh

It's also incorrect. "Enlarge to the same scale as the photo".

[johnielvis]

nope--it's correct. 100% correct.

[Leigh]

It's also incorrect.
What is incorrect?

- Leigh

[E. von Hoegh]

How do you "enlarge to the same scale as the photo"?

[johnielvis]

OH---I forgot...not everyone is as clever as I am...
believe me,...I can and have done it---don't trouble youself witht he complex details...just accept that it is done....you'lll figure it out some day

[E. von Hoegh]

OH---I forgot...not everyone is as clever as I am...
believe me,...I can and have done it---don't trouble youself witht he complex details...just accept that it is done....you'lll figure it out some day

I just had to quote this, to preserve it.

[Louis Pacilla]

OH---I forgot...not everyone is as clever as I am...
believe me,...I can and have done it---don't trouble youself witht he complex details...just accept that it is done....you'lll figure it out some day

W.A.D.

[Paul Fitzgerald]

Rick "a wide-angle freak who has used all three projection methods" Denney

"The current maximum field of view for a rectilinear lens is 121 degrees, offered by the 12mm Voightlander lens on the Bessa 35mm camera (and by the 12mm end of the Sigma 12-24 zoom lens for other 35mm cameras)."

Here you go, a 162 degree 3.7mm Pacific Optical in Nikon mount. Have no idea if it is rectilinear but I do remember 8mm/180 degree fisheye lenses so I'm guessing this is something trick. :D

3.7mm-F1.5-165 degree Pacific-Optical Nikon-Lens (http://www.ebay.com/itm/3-7mm-F15-165-Degree-Pacific-Optical-L-A-CA-Nikon-Lens-/300614038874?_trksid=p3286.m7&_trkparms=algo%3DLVI%26itu%3DUCI%26otn%3D3%26po%3DLVI%26ps%3D63%26clkid%3D3706148031844355838)

(usual disclaimers - not involved)

[Vaughn]

W.A.D.

I had to look that one up! :rolleyes:

[E. von Hoegh]

"Working As Designed".

Right?

[Vaughn]

"Working As Designed".

Right?

Yes, of course! :D

[polyglot]

I just had to quote this, to preserve it.

Indeed.

johnielvis, you're being wilfully obtuse. His point was that the use of the objective lens as an enlarger lens will only correct for distortions if the magnification is the same, since lens aberrations are magnification-dependent. So there's little point in taking a scene focused at 50 feet and then enlarging onto an 8x10" with the same lens because the magnifications are totally different and therefore you get different distortions from the lens. Nothing cancels out. It's better to use an objective lens well-corrected for the magnifications you actually use it at and then a separate enlarger lens that's well-corrected for use at enlarging magnifications.


Yes, you should, since you obviously don't know the difference between a fisheye lens and a regular lens.

- Leigh

Well, I've posted the projection formulae for rectilinear lenses. I'll repeat them here for your edification:

Ximage = F * Xsubject / Zsubject
Yimage = F * Ysubject / Zsubject

where X and Y are parallel to the film plane and Z is perpendicular. If you draw out a little diagram with a pinhole, film and subject then draw the rays and apply the Rule of Similar Triangles, you will arrive at the same formula. Go ahead, I'll wait...

Likewise, if you follow through with some algebra, you will find that straight lines in the scene remain straight on the image, no matter where you place them.

That's what's known as a "falsifiable claim", i.e. an implicit invitation to prove me wrong. So you have the following options:

- empirically disprove it, i.e. produce a photo from a properly corrected rectilinear lens that contains curved lines from a straight subject[1];
- provide an alternate theory of projection[2];
- find a citation that disproves it, i.e. appeal to authority; or
- keep on with the hand-waving with no actual evidence to back it up.

[1] note that this would really only prove that you found a lens with some barrel distortion, of which there are plenty (my 20/2.8 is spectacularly bendy). The rectilinear projection is a mathematical abstraction to which real lenses may aspire; enlarger lenses get really really close, some other lenses not so much. Hence the original post to the previous thread asking if he was observing barrel distortion with that particular lens, which is a reasonable question.

[2] is quite possible, there are many many projections to choose from and they all have some form of undesirable behaviour or artefacts in extremis. Only rectilinear is rectilinear though.

This whole thread seems to be posted on the mistaken notion that rectilinear projections contain barrel distortion, which is demonstrably false. If you go read up the definition of rectilinear projection or, god forbid actually take a photo with a rectilinear lens, you would see this.

As for the thing with stitches, I invite you again to go read up on stitching engines (many of them have excellent webpages and wikis) and the mathematics behind them. A stitcher performs at least the following transformations:
- per-lens corrections for barrel, pincushion, etc,
- adjustments to account for light falloff,
- transformation from rectilinear to azimuth/elevation,
- transformation to final chosen projection, and
- blending.

Note the three transformations; they imply that the final image is not rectilinear unless you choose for it to be. You can't have a rectilinear image covering 180 degrees as it violates the mathematics of the projection and would have to have greater than infinite width... so most people stitching a wide landscape will use a cylindrical projection, which is what shorpy did. rdenney's post explains it extremely well and shows why the "smile distortion" is a product of the cylindrical projection.

You can't just take multiple rectilinear images taken at different angles and paste them on top of each other as they won't line up. That will only work if you shoot cylindrical projection and rotate about the cylindrical axis, as per cameras like the Cirkut which do so mechanically.

[Leigh]

Ximage = F * Xsubject / Zsubject
Yimage = F * Ysubject / Zsubject
where X and Y are parallel to the film plane and Z is perpendicular.
In post #16 I said:
"Whether or not the wall appears distorted depends entirely on the distance of the film/lens from the wall. If it's short, such that the distance from the front lens node to the center of the subject is significantly shorter than from the node to the extreme corners, then some distortion could be introduced."

to which you replied:
"No, not unless the lens itself introduces distortion due to it being used outside of its optimised range."

So, using your formula: Yimage = F * Ysubject / Zsubject

If the subject contains two elements of equal size (cynder blocks, for example), one at a distance F from the front lens node and the other at 2F:
Yimage1 = F * Ysubject / Zsubject and
Yimage2 = F * Ysubject /( 2 * Zsubject)

So Yimage2 = 1/2 Yimage1, which is exactly what I said previously.

Thank you for proving me correct.

Nowhere in this thread did I ever mention rectilinearity. That's an argument among you, Paul, Rick, and Doremus (perhaps others). I am not found in that group.

My only comment that might be related is as quoted above: "... some distortion could be introduced.". Whether it is or not depends on the lens design.

- Leigh

[polyglot]

Z is not the distance from the lens to the subject, it is the perpendicular distance from the subject to the lens' plane. If the wall is parallel to the film (and lens) plane, the whole wall has the same Z value and therefore the same magnification, therefore the bricks in the middle do not look larger. Camera lenses (other than fisheyes) aim to be rectilinear, whether you like it or not.

I ain't going to post my whole previous post here again, that would be pointless. What you seem to be thinking of is

magnification = F / distance
or stated alternatively
Ximage = F * Theta_subject

which is a Linear Scaled Fisheye (http://en.wikipedia.org/wiki/Fisheye_lens). Go to that link, scroll to the bottom and read the formulae.

Edit: seriously, draw the diagram out.

[Leigh]

Z is not the distance from the lens to the subject, it is the perpendicular distance from the subject to the lens' plane. If the wall is parallel to the film (and lens) plane, the whole wall has the same Z value and therefore the same magnification, therefore the bricks in the middle do not look larger.
Wow. You obviously live on a different planet than the rest of us.

That statement is so preposterous it doesn't deserve a comment.

- Leigh

[polyglot]

Wow. You obviously live on a different planet than the rest of us.

That statement is so preposterous it doesn't deserve a comment.

- Leigh

Back to the hand-waving and ad-hominem again.

Range = sqrt(X^2+Y^2+Z^2)

Which is clearly not Z. Z is the separation between the wall's plane and the lens' plane. Range is the distance from the lens to a particular brick (each of which have different X and Y values but all the same Z; it's the identical Z values that make them co-planar!). You're asserting that magnification is a function of range, which is true only for fisheyes; I'm asserting that it's a function of Z, which is true for rectilinear projection, i.e. your average objective lens.

Or did you have something substantive to add? "Preposterous" and spluttering doesn't count.

Perhaps a photograph showing this effect you have such unshakeable faith in despite all evidence to the contrary?

No?

[polyglot]

I'm as guilty of this as anyone I know (http://xkcd.com/386/)

[Leigh]

You're asserting that magnification is a function of range, which is true only for fisheyes;
So you would have us believe that a car photographed with a non-fisheye would be the same size at a mile away as one at ten feet away.

- Leigh

[Paul Fitzgerald]

(1)"Z is not the distance from the lens to the subject, it is the perpendicular distance from the subject to the lens' plane."
(2)" If the wall is parallel to the film (and lens) plane, the whole wall has the same Z value and therefore the same magnification, therefore the bricks in the middle do not look larger."
(3)"Camera lenses (other than fisheyes) aim to be rectilinear, whether you like it or not."

This is actually too funny. :D

(1) there is the crux of your problem, it's NOT a plane it's a POINT along the lens axis called the front node (or rear node back to the film)

(2) Welcome back to grade school math class.

Hypotenuse = sqrt(base2 + height2)
Z-axis distortion = hypotenuse / height (converted to %)

On the film side the base is set at 1/2 diagonal of film holder, the height is set by focus displacement for the lens.
On the object side height is set by focus distance to the front node, base is set by FOV for the lens/format/focus displacement, again 1/2 diagonal (FOV changes with lens displacement)

Now you can directly compare the Z-axis distortion at the film plane to the Z-axis distortion at the object plane and see how the difference of the 2 changes by focus distance. Unless it has floating / aspheric elements, a prime lens can only be perfected for 1 magnification range.

(3) Persective Distortion is by distance only. Optical abberations are an entirely different kettle of fish.

[rdenney]

Here you go, a 162 degree 3.7mm Pacific Optical in Nikon mount. Have no idea if it is rectilinear but I do remember 8mm/180 degree fisheye lenses so I'm guessing this is something trick. :D

It's a fisheye.

Rick "fisheye lenses make 180-degree images about three times their focal length in diameter" Denney

[Leigh]

polyglot,

Perhaps some basic geometry is in order, since you're fond of equations.

Given an object 1 meter tall at a distance of 1 meter from the front lens node, and a second object 2 meters tall moved off to the side such that it is 2 meters from the front node:

When viewed, object #1 subtends an angle = arctan(1/1) = 45°.
Using the same formula, object #2 subtends an angle = arctan(2/2) = 45°.

Note that object #2 appears exactly the same size as object #1, even though object #2 is twice the height.

This is what happens with a brick in the middle of a wall v. a brick at the end of the wall.

It's simply an example of 'vanishing point', which is well-known and accepted in any type of image rendering.

- Leigh

[polyglot]

So you would have us believe that a car photographed with a non-fisheye would be the same size at a mile away as one at ten feet away.

- Leigh

Not at all. Assuming you're pointing the lens directly at the cars, range = Z (because X = Y = 0, i.e. the subject is on the lens' axis). A car at ten feet and a car at a mile clearly do not lie in a plane parallel to the lens and film, therefore they have very different magnifications.

I'm saying that the bricks in a wall parallel to the lens (or even two cars parked parallel to the lens plane, though there are issues there with looking at the front of one car and the side of the other because they are not 2D obejcts) will be of the same size on the image. You can see that for yourself (http://images.google.com/search?tbm=isch&q=brick+wall&biw=1920&bih=1010) if you ignore some wonky walls.

If you put a car in front of the wall or a column that just out of the wall (as per one of the example photos in the first thread), that will have greater magnification... but those things are not part of the planar wall.

Paul: I'm talking about the projection. As I've said several times, actual physical lenses don't quite achieve a perfect rectilinear projection and in fact can be corrected well only for some magnifications; at other times they will have barrel, pincushion or even wave-like distortions. Hence my point that using an objective lens as an enlarging lens is a futile approach to correcting distortions because the magnification is very different for each use.

So consider a rectilinear lens, being used at a magnification for which it was designed, the projection will be very close to rectilinear, no? And in that particular case, will the bricks in the middle of the wall be larger? Do your prints bulge in the middle because your enlarger lens is closer to the centre of the paper than its corners? Of course not.


(1) there is the crux of your problem, it's NOT a plane it's a POINT along the lens axis called the front node (or rear node back to the film)

While the front node of the lens is indeed a point (actually a disc - if it were a point then it'd be a pinhole camera and there would be no issues with range-dependent corrections), the concept of lens plane (http://www.google.com.au/search?q=%22lens+plane%22) is not something I just made up on the spot. The Scheimpflug Rule is expressed in terms of it for example.


Z-axis distortion = hypotenuse / height (converted to %)

Do you have a definition for this distortion? Perhaps in terms of magnification error? Anything at all that says that there will be barrel distortion inherent in the projection? I'm gonna get back to that equation at the end.

Check this out for the diagrams (http://www.jordicenzano.name/projects/fvv-free-view-video/camera-model), specifically Figures 7 and 8. To define some terms from my previous post in terms of these diagrams I just googled up:
Ysubject = Oy
Zsubject = Oz
Yimage = poy
Magnification = poy / Oy
poy = Fz * Oy / Oz
Magnification = Fz / Oz

Hell, I'll link 'em. Figure 7:
http://www.jordicenzano.name/_/rsrc/1302709761551/projects/fvv-free-view-video/camera-model/Cam_model_fig7.jpg
Figure 8a:
http://www.jordicenzano.name/_/rsrc/1302711198473/projects/fvv-free-view-video/camera-model/Cam_model_fig8.jpg
Figure 8b:
http://www.jordicenzano.name/_/rsrc/1302711387559/projects/fvv-free-view-video/camera-model/Cam_model_fig8b.jpg

Note that Fz is the effective focal length, not the nominal focal length. For a typical lens, Fz > Fnominal when you focus closer than infinity due to bellows draw.

Now pythagoras. Range (hypotenuse as per Paul) is the distance along that red line in Figure 7 from the lens' optical centre to the top of the tree. It's the distance from the front nodal point to each brick; yes it's different for all the bricks.

Range = sqrt (Oz^2+Oy^2+Ox^2).

Note that magnification is defined in terms of Oz (via the similar triangles thing), and not Range. If you replace that tree in the diagram by a brick wall, the value Oz does not vary across the wall while range does, i.e. the middle of the wall is closer. Yet magnification is constant across the wall.

To be even more rigorous in the definition of magnification
M = d(poy) / d(Oy)
In other words, if you step 1mm along the subject at some point, that will cause an M mm step along the image. This definition allows for magnification to vary across the frame (which is what you're asserting). So we perform the derivative:
poy = Fz * Oy / Oz
and get d(poy)/d(Oy) = Fz/Oz, i.e. magnification does not vary across the frame with respect to a parallel-planar (constant Oz) subject. That means identical magnification for all points on the wall, even though the bricks in the middle are closer.


(2) Welcome back to grade school math class.

Indeed. Please state any disagreement clearly and attempt to prove it. With mathematics and geometry, as per grade school.

Paul, I suspect that when you speak of perspective distortion and this "Z distortion" term, you're referring to the inverse of the effect that I'm talking about here, i.e. that items on the edge of the frame are unexpectedly larger due to rectilinear projection.

In other words, if you take the "obvious" (but incorrect) view that M = F / Range then you will have things at the edge of the frame being rendered larger (stretched) than you "expect" because their Z value is smaller than their range.

In other words if I rearrange your "Z distortion" (let's call it D) formula, what I think you're saying is this:
D = increase in apparent magnification = Range / Oz

At the centre, Oz = Range, so D=1 (no issue)
At the frame edge, D = sqrt(Ox^2+Oy^2+Oz^2) / Oz
which is greater than one. In other words, subjects at the frame edge are stretched by D compared to the case where you point the lens directly at those subjects. In other words, when you turn the camera, the size of stuff changes. This is because the size is dependent on Oz (varies with camera pointing-angle) not Range (depends only on relative position of camera and subject).

Further rearranging,
Oz = Range * cos (theta)
D = 1/cos(theta)
where theta is the angle between lens axis the vector from front node to subject, i.e. it's how far off-axis the subject it.

[polyglot]

polyglot,

Perhaps some basic geometry is in order, since you're fond of equations.

Given an object 1 meter tall at a distance of 1 meter from the front lens node, and a second object 2 meters tall moved off to the side such that it is 2 meters from the front node:

When viewed, object #1 subtends an angle = arctan(1/1) = 45°.
Using the same formula, object #2 subtends an angle = arctan(2/2) = 45°.

Note that object #2 appears exactly the same size as object #1, even though object #2 is twice the height.

This is what happens with a brick in the middle of a wall v. a brick at the end of the wall.

It's simply an example of 'vanishing point', which is well-known and accepted in any type of image rendering.

- Leigh

That is correct for the case where you have one object in front of another (in a direct line from lens, through A to B), then you have the perspective relationship you've described. If A and B are the same size and B is at twice the depth, B will have half the magnification. No disagreement there.

However, you're missing the subtlety that Z and Range are two different values; Range depends on how far the camera is from A and B, Z depends also on the direction that the camera is facing. See the diagrams I've just posted; we were concurrently typing and I had lunch in the middle.

[johnielvis]

allow me to answer for Leigh---using he exact words from a different thread:

"Nice equations. Too bad they're not accurate.

They're derived from thin lens equations. The significant error is that thin lens equations are based on the first and second principle planes being coincident. That's seldom true with a compound lens."

[polyglot]

So state what they should be then, and derive your barrel distortion from it. While I realise that some real lenses have barrel distortion, the point of you starting this thread was to state that the perspective projection itself causes barrel distortion (because the bricks in the middle of the wall are closer than the ones on the outside), which is demonstrably false.

There are other geometrical distortions, but barrel is the one thing it doesn't do.

Edit: a little googling reveals this discussion (http://toothwalker.org/optics/cop.html) from Van Walree that seems to imply that even in the case of a compound asymmetric lens focused closer than infinity, the form of the projection does not change, i.e. it is still rectilinear. You get an increase in effective focal length of course, and the centre of perspective is not the front node but the entrance pupil. Nothing that implies barrel distortions though.

[johnielvis]

HA HA!!!!!!

THATS EXACTLY what I told him too!!!!!!

har har har....