the 'Red-Headed' stepchild of photography

[Paul Fitzgerald]

Hi there,

A recent thread had brought up the 'Red-Headed" stepchild of photography:

the Z - axis

X is horz.; Y is vert.; Z is depth. Read everything you can but noone writes of it, so I just had to do a spreadsheet for it and it shows the problem of "zooming with your feet". The size of the scene you choose is the field and the distance from lens-to-scene is the depth. The optical axis cuts this into 2 right triangles and it's easy to calculate the third side from this. Watch what happens with the PHYSICAL distortion, the change in magnification of the scene as you move closer:


6 Field

Depth Distortion%
1 216.228
2 80.278
3 41.421
4 25.000
5 16.619
6 11.803
7 8.797
8 6.800
9 5.409
10 4.403
11 3.652
12 3.078
13 2.628
14 2.270
15 1.980
16 1.743
17 1.545
18 1.379
19 1.239
20 1.119
25 0.717
30 0.499
35 0.367
40 0.281
50 0.180
100 0.045


For this graph, a 6 ft scene at 6 ft = 11.803%. The measures are which ever you choose; mm., cm., in., ft., meters, the proportions are always the same. This is the distortion from the center of the film to the top, not the corners. Most wedding and portrait shots keep this distortion between 3 - 4%, more will make people look fat thru fish-eye, less than 2% will start to flatten features with tele-photo compression.

Calculating the focal length is easy: depth/field = lens/film size. It works for all formats and sizes. This would never be a problem for landscapes but does show up whenever you shorten the distance like; weddings, portraits, interiors, table-top product shots, macro work.

Have fun with it.

[Scott Fleming]

Does this explain why I look SO bad in the head shot my dentist took with his G-3?

[paulr]

i have no idea what your numbers mean. but they're a sigh of relief .. i always feared that i was the redheaded stepchild of photography.

[Armin Seeholzer]

Now I can sleep very good again ;-))

[Leonard Evens]

Okay. I see what you are calculating. It is the hypotenuse of the right triangle divided by the "depth" less one, given as a percentage. But what in the world does this have to do with anything?

If you are interested in subject magnification as a function of distance from the center of the field, or something like that, then I believe you are going about it wrong. Let's assume no tilts of swings to avoid those complications. Then for a subject in a plane perpendicular to the lens axis, the magnification is constant. It is independent of where in the subject plane you are. If the subject is not in such a plane, it is not completely clear what one would mean by magnification.

Perhaps you are referring to the fact that three dimensional objects at the edges of the field have images which appear to be distorted. Thus a sphere would appear to be an ellipsoid. Its image on film would be an ellipse instead of a circle. These are not true distortions in the geometric sense, and exactly how they appear to your eye when you view a print depends on a variety of factors. One important one would be where you place your eye in relation to the print. But, there are also various features of the visual system which determine how your brain interprets what it sees. I don't see how your calculation helps.

[Paul Fitzgerald]

Good morning,

Scott Fleming,

"Does this explain why I look SO bad in the head shot my dentist took with his G-3?"

yes exactly, an extreme example.

Leonard Evens ,

"Okay. I see what you are calculating. It is the hypotenuse of the right triangle divided by the "depth" less one, given as a percentage. But what in the world does this have to do with anything?"

Yes, precisely. The other side of the spreadsheet shows me which lens to use at x distance for x size film. Everyone has always seen this but I was amazed how fast this builds up under 10 ft.

"If you are interested in subject magnification as a function of distance from the center of the field, or something like that, then I believe you are going about it wrong. Let's assume no tilts of swings to avoid those complications. Then for a subject in a plane perpendicular to the lens axis, the magnification is constant. It is independent of where in the subject plane you are. If the subject is not in such a plane, it is not completely clear what one would mean by magnification. "

I am sure that the lens companies can correct the lenses to be rectilinear and flat field, but only for a flat object, copy work. This distortion % is the difference of distance from the center to the end of the field, then X2, the same distortion % exists inside the camera for each scene.

"Perhaps you are referring to the fact that three dimensional objects at the edges of the field have images which appear to be distorted. Thus a sphere would appear to be an ellipsoid. Its image on film would be an ellipse instead of a circle. These are not true distortions in the geometric sense, and exactly how they appear to your eye when you view a print depends on a variety of factors. One important one would be where you place your eye in relation to the print. But, there are also various features of the visual system which determine how your brain interprets what it sees. I don't see how your calculation helps."

No, I am referring to the perceived depth in the print. We see the 3D world with stereo vision and the only way we perceive depth in a print is the difference in magnification brought by distance. For portraits there is a 'sweet spot' in distance. Move closer in with a shorter lens and you exaggerate the facial features like the dentist office. Move farther back with a longer lens and you flatten the features in reverse manner.

At these distances, inches actually make a difference in the 'look' of the print, make or break the portrait. I think this is the actual reason studio portrait cameras were tail-board style, so the lens-to-subject distance doesn't vary with focusing, holding up the 'brass cannons' was just a happy coincidence. With portraits, unlike landscapes, there is a direct comparison between the person (and how they see themself) and the print. It's either a home-run or you struck out, no wiggle room. The sweet spot is the difference between a portrait and a snap-shot.

Just food for thought

[Leonard Evens]

I still don't think I understand what you are getting at.

It is true that one factor affecting how the print appears is the differences in magnification in different parts of the scene. It seems to me that this is purely a function of two numbers, the ratios of the distances to the nearest and the furthest points of interest to the focal length. Distances should be measured along the lens axis to the planes perpendicular to that axis containing the relevant points. Diagonal distances are not relevant. This is all assuming no tilts or swings. Issues like field curvature should not materially affect the calculations although they can affect what is in focus.

Another important factor when viewing a print is the position of your eye. Ideally, to see what the camera sees, it should be place at the proper center of perspective. If the lens was at a certain distance from the film when the picture was taken, the eye should be set further back by the factor of the enlargement of the film to produce the print. But we usually view prints from a distance determined by the dimensions of the print, usually at the diagonal of the print. That means that the eye will be placed properly for a normal lens, too far away for a short focal length lens, and too close for a long focal length lens. In addition, there are some features of the visual system which make certain adjustments to the strict optical image when we look at a scene. One is size constancy which makes important objects appear larger than they are optically. That is the reason why longer lenses are often recommended for portraiture. If the subject is placed at a proper position for pleasing perspective (in depth), the image will be too small if a short or normal length lens is used. If a long lens is used, the image size is greater, thus at least partially compensating for the size constancy of the visual system.

[Paul Fitzgerald]

Hi there,

Leonard Evens;

"I still don't think I understand what you are getting at."

The ever elusive 'WOW' factor. A portrait is a landscape of a face. Foreground, midfield and background of only 4 inches depth from chin to ears. You can get everything technically perfect and still get a school photo (yeah, SO), no WOW factor.

The diagonal % is an easy way to shown what I am driving at, the distortion a spatial perception, 3-dimensionality, at this distance builds-up so fast that inches in placement of the lens to sitter make a obvious difference in the print. The Z-axis distortion, the apparent exaggerating of facial features, by being either too close or too far makes or breaks the photo. It also tends to make or break the photographer, it's not the lens it's the distance.

the hunt continues

[Scott Fleming]

Hmmmmmmmmmmmm. This is interesting. A couple years ago my mother in law was given an 80th birthday party and it was a fairly large affair at a country club. I don't do much people photography but just for fun I bought a flash bracket and diffuzer for my 10D and speedlight and invited myself to be the official photographer. I read up a bit on flash photogrphy dropped my 50mm f/1.4 which would be like a 80mm or so lens on the crop factor 10D and proceded to snap away all night. I was amazed and surprised how much success I had. I got a lot of shots that looked pretty good.

Reading this post I'm guessing right now that the lens I used on the crop factor camera I had .... filling the frame with four folks properly bunched in for a shot .... was the exact right lens to subject distance. The people all looked really great.

[Armin Seeholzer]

Sorry Paul I don't get it!
Maybe its because english is not my motherlanguage and math is also not my thing.
Give please some examples in 4x5 and 8x10 what would be the right lens and distance for a head shoulder portrait!
Thanks in advance, Armin

[Paul Fitzgerald]

Good morning,

JohnArs, here are the numbers for 8x10 for an 18" head shot:


1.5 Field ft. Lens mm:
10in 8in
Distance ft. Distortion %
0.5 80.28 84.7 67.7
0.75 41.42 127.0 101.6
1 25.00 169.3 135.5
1.25 16.62 211.7 169.3
1.5 11.80 254.0 203.2
1.75 8.80 296.3 237.1
2 6.80 338.7 270.9
2.25 5.41 381.0 304.8
2.5 4.40 423.3 338.7
2.75 3.65 465.7 372.5
3 3.08 508.0 406.4
3.25 2.63 550.3 440.3
3.5 2.27 592.7 474.1
3.75 1.98 635.0 508.0
4 1.74 677.3 541.9
4.25 1.55 719.7 575.7
4.5 1.38 762.0 609.6
4.75 1.24 804.3 643.5
5 1.12 846.7 677.3
5.25 1.02 889.0 711.2
5.5 0.93 931.3 745.1
5.75 0.85 973.7 778.9
6 0.78 1,016.0 812.8


The z-axis distortion makes the picture appear to bow towards the camera in the center. For a head shot you want some of this distortion for depth BUT not too little or too much.The lens distances are to the optical center, not the marked focal length. Like I said, inches make a difference, see how fast the distortion changes.

Have fun with it

[Paul Fitzgerald]

Hi there,

Just to be obsessive, compulsive, anal retentive and relentless, I just set this up with a 305mm B&L tessar on 8x10. Focused to get both ends of an 18 inch SS. ruler, the waterhouse slot is 33 in.(2.75ft.) and 440mm from GG.. Yes, the spreadsheet numbers do work out in the real world but I guess this waterhouse slot is not at the optical center of the lens.

Have fun with it.