Being too curious and not being able to find any more information about the camera, etc. other than hearing it produces "warm... true... honest... life-like" images... I took matters on to my own hands and decided to do some calculations.
I first looked at the (completely assuming that it's to scale) and estimated the relative height of a seated person in pixels... approximately 98. So if we assume an average person is 6ft tall that translates to around 16.3 pixels per foot. The trailer measured out to be almost exactly 570 pixels and this makes total sense because they say it's a "35 ft camera" (570/16.3=35).
Looking at the rotating CGI (0:49 in the video), I noticed the film plane is directly above the rearmost axle. This was the last piece of the puzzle as I could estimate out the focal length of the lens using the thin lens equation as follows:
1/o + 1/i = 1/f
o = 75 pixels = 4.6 ft
i = 300 pixels = 18.4 ft
(Note how i is exactly 4 times o) So, after solving the equation, f or focal length = 3.7 ft = 44 inches = 1120mm
As we know the image plane is "6 feet" (1828.8mm) tall (in portrait orientation), we can calculate the angle of view now:
2 * arctan(1828.8/(2*1120)) = 78.5 degrees
... but that's a classic wide... exactly like a 75mm lens on 4x5 or a 150 on 8x10. And his portraits don't look like they've been shot with a wide.
Also, I consider the "portrait" field of view to as half of "normal" which is approximately 22.5 degrees. In which case, either the film height needs to be 18 inches, instead of 6 ft... or the focal length needs to be around 168 inches instead of 44 but then that would mean the trailer (o+i) needs to be at least 87.5ft!
Since this is an estimation I expect things to be a bit off, but not by this much... especially when the CGI is accurate about the person's height and the trailer length.
What am I missing here?
I'll take a wallet size print please.
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I am here: http://maps.google.com/maps?ll=43.628221,-116.227373
Zak Baker
zakbakerphoto.com
"Sometimes I do get to places just when God's ready to have somebody click the shutter."
Ansel Adams
@genotypewriter. As a novice with just the basic physics knowledge I can follow your calculations. Something is off, but I can't figure out what it is. It's like figuring out a magicians secret.
Same goes for the film he uses. There is not a lot of light going onto the film. So the secitivity would have to be high and the shutter speed long.
In the video, they show a fairly elaborate head and neck armature-brace for the posing chair. I'm sure the exposure times and lighting duration is much longer then what was shown in the video...trade secrets and so on.
Lets see...the exposure compensation for a bellows draw of 20+ feet...would be...?
Stevan and Marc,
Doesn't he use strobes?
Also off the top of my head I can't see how the bellows factor would be related to 20ft, etc. because it's purely related to relative magnification.
So if a 1.5ft subject translate to a 6ft tall image, the magnification is 4x. So it's not too different from a typical LF macro that one might do?
Your first equation is fine. Another way to look at it is to say that the images look like they're around 4x life size on the negatives. That is 1:4 macro. That means the image is going to be 5 x focal length from the lens. That give a focal length of 18.4/5 ft = 44 inches.
The problem is in your second calculation - it assumes that the camera is focused at infinity and that the film plane is a distance f from the lens. In this (macro) case it is not. You need to substitute the actual distance into your equation:
2 * arctan(1828.8/(2*(18.4*12*25.4))) =
2 * arctan(1828.8/(2*5608)) = 18.5 degrees
I wonder what lens they will use - something like a 47.5" Goerz APO Artar would cover nicely. But it might be a bit slow.
Richard
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