Message-Id: <200105301537.IAA22782@dns.ccit.arizona.edu> Date: Wed, 30 May 2001 11:31:29 -0400 From: Mary Winter <mailto:mary.winter@MAIL.STATE.KY.US> Subject: Re: information contained in surrogates To: mailto:IMAGELIB@LISTSERV.ARIZONA.EDU
<pre>
Our photographer, Nathan Prichard, has been looking over my shoulder on this
discussion. I include his thoughts on this below:
Mary,
Well, I see a Hollinger box of weasels has been opened.
I derived my very approximate figures based on the normal
levels of resolution for various film formats, if the
photographer is doing his part. These are the resolutions
on the film, which are almost always less than what the
lenses are actually capable of producing, if that nasty old
physical stuff with emulsion were not involved. Resolution
is in line pairs per millimeter (often listed as lines per
millimeter or lpm). The lines are separated by a space equal
to the width of the line, so the pair actually made up of a
black line and a white one (| (sp) | (sp) |). In digital
terms, that means two pixels (one for the black and one for
white/blank space). So a resolution in lpm's is doubled for
pixels (or dots) per millimeter and then multiplied by 25.4
(millimeters per inch) for dpi (dots per inch).
lpm x 2 x 25.4 = dpi
Film size Nominal lpm Approx. DPI
35mm 80 4,000
120 60 3,000
4x5 40 + 2,000
[8x10 uses some of the same or similar lenses as 4x5.
Resolution would be about the same, at least for many
modern lenses. Some lenses can produce greater resolution
on some formats, as with macro/copy applications.]
Uncompressed file size can be calculated by multiplying the
format dimensions (in inches) by the dpi squared for B & W.
For color, multiply that result by three (assuming 8 bit
color depth in both cases).
Nathan
I think this is pertinent to the original question about stitching files.
Please note, however, that the issue of resolution was not the primary
factor in our decision to incorporate analog surrogates into our
preservation/access workflow. But then, we have Nathan . . .
Thanks, Mary
Mary E. Winter
Special Collections Manager
Kentucky Historical Society
100 W. Broadway
Frankfort, KY 40601
tel.: (502) 564-1792 ext. 4428; fax. (502) 564-4701
-----Original Message-----
From: Guenter Waibel [mailto:mailto:guenter@UCLINK4.BERKELEY.EDU]
Sent: Tuesday, May 29, 2001 6:05 PM
To: mailto:IMAGELIB@LISTSERV.ARIZONA.EDU
Subject: information contained in surrogates
Hi list,
this is a question tangential to Mary Winter's post. Mary wrote:
At 12:25 PM -0400 5/29/01, Mary Winter wrote:
>Basically, you can capture about 3 times the surface area on 4x5 (about a
>240 meg. file), which could be scanned at 2000dpi (fairly normal for a 4x5
>film scanner).
If I understand her correctly (please chime in if I'm misreading you,
Mary!), she says that the data contained in a 4x5 transparency is
equivalent to 240 MB (I take it at 24 bit RGB color).
Now this raises an interesting question for me: how much data do
analog surrogates contain? Or in other words, at what capture
resolution do we exceed the source resolution of the surrogate? Does
anybody have information, or leads to information, that spells out
the amount of data contained in different types of surrogates in
digital terms (resolution of capture, pixel dimension, filesize)? I'd
be especially interested in data pertaining to 4x5 transparencies, 35
mm slides, 8x10 prints and also filmstock (16 mm and 35 mm).
All comments and insights are very much appreciated.
Thanks,
Guenter
-- ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ Guenter Waibel Berkeley Art Museum & Pacific Film Archive Digital Media Developer http://www.bampfa.berkeley.edu/ Digital Imaging SIG Chair, MCN http://www.mcn.edu/visig_subscribe.taf mailto:guenter@uclink4.berkeley.edu Phone 510-643-8655 Fax 510-642-4889 ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~</pre>
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