P3D Printer resolution (was Re: P3D Re: Lenticulars)

[Brian Reynolds <reynolds@xxxxxxxxx>]

Tony Alderson wrote:
> I must confess I don't have a good understanding of the issues of
> dpi and pitch and sizing for printing and all that. I pretty much
> followed a receipe by Steve Aubrey and Bob Mannle, or sent the
> images off to them for interlacing and prep for printing. Perhaps
> someone else can explain.
> 

David Fokos has written a very good paper on how to digitally enlarge
negatives for use with contact printing processes.  Part of the
process involves getting a half-tone negative made by a service
bureau.  These negatives are normally used for printing on a press.
David's description of the various issues involved in digital output
is excellent.  The paper can be found on Bostick & Sullivan's home
page <URL:http://www.bostick-sullivan.com/> under the section for
technical papers.

Basically there are two things to remember.  First, in order be able
to properly reproduce an analog signal (e.g., a photographic image)
with a digital signal you must sample the original analog signal at
twice the highest frequency available in the original analog signal.
This is known as the Nyquist sampling theorem (from communications
theory).  For example, if your original image contains a series of
vertical 1/4 inch wide lines you must sample the image at least every
1/8 inch in order to properly reproduce the lines.  If you sample less
frequently than every 1/8 inch you will most likely miss some of the
edges of the lines.  This will create an artifact (e.g., one of the
lines may look wider than the others) known as aliasing which is most
often seen as the jagged stair-step edges of lines that are not
perfectly vertical or horizontal on a computer display.

The second thing to know is that printing is a binary operation.
There is either ink at a particular location on the page or there
isn't.  For the types of printers being discussed there are no shades
of gray (or whatever the base ink color is).  Printers produce shades
of gray by taking a small area and dividing it up into a grid.  The
number of locations on this grid that get ink depends on the shade of
gray that that area is supposed to have.  The number of shades of gray
depends on how finely the area is divided into a grid.  A 2x2 grid can
produce 4 shades of grey, a 4x4 grid can produce 16 shades of gray,
and a NxN grid can produce N^2 (N squared).  To get 256 shades of grey
you need a 16x16 grid.  The small area is a pixel, and a location on
the grid is a dot.

Note that pixels are not dots!  The commercial side of the computer
industry is very inconsistent about the definition of a pixel (or a
dot), and hasn't bothered to check the definitions as used in the
printing industry.  Published specifications have generally been put
out by the marketing side and tend to use which ever is the most
favorable definition for the device involved.

>From these two pieces of information we can now figure out what the
limits for computer output are.  Lets assume I'm going to output to a
600dpi laser printer.  (Color is similar, and would just be a
complication in this example.)  The original is a 35mm full frame
negative (36mmx24mm).  The output will be 8x10 inch on a 8-1/2x11 inch
sheet of paper (missing the 1/4 inch dead zone at the edges).  In
order to get nice gradations of tone I want 256 shades of gray.  Given
600dpi across the 8x10 inch printable area of the page that works out
to 4800x6000 dots (8*600x10*600), or 300x375 pixels (4800/16x6000/16)
or 37.5 pixels per inch (ppi).

That's not a lot of pixels!  This is where having the image on a
computer is a plus.  Many images will not use all the available gray
values.  By studying the histogram (a graph of gray value vs. number
of pixels with that gray value) for the image you might be able to
determine that you don't need 256 shades of gray.  Normally you would
stretch the image to use all available gray values, but here you want
to use as few gray values as you can.  If you only needed 121 shades
(a 11x11 grid) the output would have 436x545 pixels (54.5ppi).  If you
only needed 64 shades of gray (a 8x8 grid) the output would have
600x750 pixels (75ppi).

Finally you need to figure out what resolution to scan the original
at.  First figure out the magnification factor.  For this example it
will be about 8.5 (8 inches / (24mm / 25.4mm/inch)).  Now multiple
this by the ppi of the output.  For 256 shades of gray (37.5ppi) this
works out to 317.5.  You have to scan the original negative at at
least 318 dpi.  After reading this file into photoshop (or some
similar program) you would adjust the output pitch to match that of
the printer (37.5ppi in this example) and let the program resample the
image.

By the way, this is were using a larger negative is really an
advantage.  A 6x6cm negative would only need to be scanned at 170dpi
and a 4x5inch negative would only need to be scanned at 80dpi to
produce the same level of detail at the 35mm negative for the same
size output.

If anything I've said doesn't make sense please read David Fokos'
paper before responding.  I know he explains all of this much better
than I do.

-- 
Brian Reynolds                  | "Dee Dee!  Don't touch that button!"
reynolds@xxxxxxxxx              | "Oooh!"
http://www.panix.com/~reynolds  |    -- Dexter and Dee Dee
NAR# 54438                      |       "Dexter's Laboratory"


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