lines per mm vs PPI and DPI
Bill Wehrmacher
I have just joined this NG and thought with all the conversations about
National Geographic someone could help me with these concepts.
I understand what Pixels Per Inch means. I thought I understood what
lpmm implied, but now I am not sure. I assumed for a long time that 25
lpmm would imply something like 25 PPI. Another reference I read
convinced me that in order to see a line, you need a space next to it,
therefore 25lpmm would translate to about 50PPI. Any truth here?
Finally, I wonder about DPI. It was explained to me at one time that
Dots Per Inch is a printing term and that it takes about 4 dots to make
one pixel. As a result, to print 300 PPI, one would need a printer that
provides at least 1200 DPI. Any truth here?
In some of these threads it seems as if the terms DPI and PPI are used
interchangeably. What is the straight story here.
I do appreciate all your efforts. I am getting an education that has
long escaped me.
Thanks,
Bill
Godfrey DiGiorgi
the "To," "Cc," and "Newsgroups" headers for details. ]]
A pixel is a picture element, each one comprised of Hue, Saturation and
Value. "Pixels Per Inch" is a natural way to measure digital image
spatial resolution.
Lines per Millimeter, or more accurately, Line Pairs per Millimeter
(lp@mm) is a measure of resolution, most useful in analog imaging.
Basically, it says some number of black/white lines fitted into a
millimeter measure can be discerned, usually by eye or with some kind
of optical sensor... In a digital world, tricks of discrete
mathematical entities make it less useful as a measure of resolution
although it's ultimately pretty simple if you don't consider the hard
cases of "partial pairs per millimeter" .. If you can discern a white
and a black space in the space of a millimeter, that's 2 lines per
millimeter or 1 lp@mm. (The reason LP@MM is more accurate is that you
cannot discern lines as distinct entities unless they alternate in
color, white/black, it is also easier to consider the alternating lines
to be all the same width.)
Dots per inch is an measure of spatial resolution used in reference to
marking engines (printers) which lay down ink in discrete spots. Hue,
Saturation and Value are comprised of multiple dots of inks laid down
together.
Digital cameras make images sized in pixel dimensions, and the physical
sizes of those pixels are related to the specific CCD or CMOS sensor
you have. Taking that pixel array and applying it to an output device,
you size the pixels to a ppi standard.
DPI is often confused with PPI because the people who wrote the spec
sheets and manuals were more familiar with printer terminology than
with digital imaging terminology. Photoshop and other image processing
software perpetuates this confusion by presenting image size tools that
express PPI values as DPI.
Example for illustration (and I ask for pardon for simple math errors
in advance ... ;-):
My camera makes an image of 1920x2560 pixels (4.7 Mpixels) and I want
to make a highest quality print on an Epson 1270 printer, which is
capable of 1440x1440 dpi marking resolution. My experience shows that
the Epson 1270 printer and its driver achieves the best quality image
it can make when it is rendering pixels which are at a 1:4 ratio of
pixel resolution to dots per inch marking capability, so my output
print file should be set at 360 PPI and the printer set to 1440 ppi.
That will achieve a print image size of 5.3x7.1 inches dimension.
If the image was a macro photo of a grid at a 1:1 image:subject
magnification, the finest grid that I could image accurately would be
constrained by the physical size of the CCD sensor in my camera and the
actual pixel dimensions. I don't know those dimensions off the top of
my head, but let's conjecture that the sensor is 12x16mm in size and
the effective pixel area represents the whole area of the sensor. That
makes each pixel .00625mm wide. Each pixel in the sensor can only take
on one value at a time, so the minimum width of a black-white line pair
has to be double that or .0125mm. So the sensor is limited to being
able image a grid with a line spacing of 80 LP@MM at most. Imaged as a
5.3x7.1 inch print at maximum resolution, that 80LP@MM grid is
magnified approximately 15 times, resulting in a calculated 5.33 LP@MM
grid on paper.
This is simplistic as it assumes a b&w grid and a sensor in which every
pixel is used, not your typical case, as well as a lot of other
simplifications but I hope it illustrates the relationship of pixels,
PPI, DPI and LP@MM.
I hope that helps, and I truly hope I have not lost a decimal point or
made some other stupid error... ;-)
Godfrey
In article <3E15142F...@integraonline.com>, Bill Wehrmacher
JohnP
John Houghton
news:3E15142F...@integraonline.com...
> In some of these threads it seems as if the terms DPI and PPI are used
> interchangeably. What is the straight story here.
>
of us here believe that the distinction you observe between dpi and ppi is a
useful one and we like to encourage it.
John
Ian Burley
Congratulations!
Ian
--
Digital Photography Now
UK-based digital photo Web magazine
http://www.dp-now.com
"Godfrey DiGiorgi" <rama...@bayarea.net> wrote in message
news:020120032143518766%rama...@bayarea.net...
Bit Bucket
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"Bill Wehrmacher" <w_wehr...@integraonline.com> wrote in message
news:3E15142F...@integraonline.com...
Jim V
I have a Minolta D7i 5MP that puts out a file with 72DPI, Now I have
enlargements made at 36x44 inch that are absolutely beautiful. How can
that be?
Check my pbase site. I have also had a couple of panoramics made 1 13"
x 6ft.
Jim Venner
See what I am Saying?
http://www.pbase.com/jimven/galleries
John Houghton
news:020120032143518766%rama...@bayarea.net...
> sheets and manuals were more familiar with printer terminology than
> with digital imaging terminology. Photoshop and other image processing
> software perpetuates this confusion by presenting image size tools that
> express PPI values as DPI.
>
dialog. Moreover, the help page for image size and resolution makes clear
the difference between ppi and dpi.
John
Povl H. Pedersen
> I'm sorry, I don't understand any of this!
> I have a Minolta D7i 5MP that puts out a file with 72DPI, Now I have
> enlargements made at 36x44 inch that are absolutely beautiful. How can
> that be?
Because you are not trying to pick out details from 2 inches away.
a 4 mpixel picture scales quite well if you consider that the
person looking at it will usually be at a larger distance if the
print is huge.
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Bit Bucket
news:v1bsoqs...@news.supernews.com...
> Jim V wrote:
>
> > I'm sorry, I don't understand any of this!
> > I have a Minolta D7i 5MP that puts out a file with 72DPI, Now I have
> > enlargements made at 36x44 inch that are absolutely beautiful. How can
> > that be?
>
2560
> pixels across 44 inches when you print, then you are printing at 58 pixels
per
> inch. It's simple math.. 2560 pixels divided by 44 inches = 58 PPI.
>
> If this 44 inch wide print is absolutely beautiful then you must be
looking at
> it from *quite* a distance away. If 58 PPI can produce a great picture,
then
> my little 2 Megapixel ELPH should be able to produce great 27" x 20"
photos
> :-)
>
Around here we usually forget something very basic: Every time you double
the viewing distance, you double the *effective* resolution. Not directed
at you, Jim or Jim, but I guess there is often little interest in
effectiveness when this issue comes up. ;-)
Godfrey DiGiorgi
how the dialogs work. I just checked PS 7 and you're right; I wonder
when they corrected it. Some other applications still erroneously use
dpi where ppi would be appropriate.
Godfrey
In article <U4kR9.3913$YQ6....@newsfep1-gui.server.ntli.net>, John
Dave Martindale
>LOL ... You know, I hadn't looked lately, I've just gotten too used to
>how the dialogs work. I just checked PS 7 and you're right; I wonder
>when they corrected it.
It must have been a long time ago. I fired up Photoshop 4, and it used
pixels/inch in the resize dialog.
Dave
Dave Martindale
>I'm sorry, I don't understand any of this!
>I have a Minolta D7i 5MP that puts out a file with 72DPI, Now I have
>enlargements made at 36x44 inch that are absolutely beautiful. How can
>that be?
First, your camera does not put out a "72 DPI" file, it just produces a
file with a certain number of pixels. The "72 DPI" value is just
Photoshop's default when you read in a file that lacks a DPI figure in
the file header. You can change the "72" to any other value you want.
This changes the printed image size without changing the number of
pixels (as long as the "resample image" box is unchecked).
As for "absolutely beautiful" enlargements at any particular DPI value,
that depends on how closely you look at the englargements. There is a
rule of thumb that you should have a pixel density of at least 200 DPI
when printing, but that applies to prints that you hold about as close
as is reasonable to examine them - about 10 inches from your eye. Thus,
for tiny prints up through 5x7 and maybe 8x10, you view the prints at
the same distance and you want about 200 DPI.
But a 16x20 print is often viewed from further than 10 inches. If, in
fact, you view a 16x20 from twice the distance of an 8x10, it needs only
half the DPI to have *the same* resolution at the eye. And if you are
viewing your 36x44 inch print from 30 inches or more, you only need
about 67 DPI for an equal impression of sharpness.
Looked at another way, if your viewing distance is proportional to print
width, the same number of pixels will look equally sharp at any size,
no matter what the DPI value.
On the other hand, if you're looking at your large print from 10 inches
away (people might do this if it's a panorama with a very wide field of
view), then you still need 200 DPI or so, and 72 DPI would look soft.
Dave
Alan Sherwood
<w_wehr...@integraonline.com> wrote:
>I understand what Pixels Per Inch means. I thought I understood what
>lpmm implied, but now I am not sure. I assumed for a long time that 25
>lpmm would imply something like 25 PPI. Another reference I read
>convinced me that in order to see a line, you need a space next to it,
>therefore 25lpmm would translate to about 50PPI. Any truth here?
>
A line pair is a spacial cycle so you could think of lp/mm like
electrical cycles per second.
Nyquist sampling theorem applies to both: two digital samples per
cycle are required. In actual practice, about 2.8 to 3.0 samples per
cycle are required for good quality. A pixel is a digital picture
sample.
The 35 mm film negative that you have processed at Walmart will
probably result in about 25 lp/mm resolution at best.
25 lp/mm X 2.8 samples/lp = 70 samples/mm
or 70 pixels/mm which is 1,778 samples/inch.
If the active area on the 35 mm negative is about 28 mm X 35 mm, how
many overall pixels are required to preserve the assumed 25 lp/mm?
(28 mm X 25 lp/mm X 2.8 pixels/lp) X (35 mm X 25 lp/mm X 2.8
pixels/lp) = 4,802,000 pixels.
For prints, anything more than about 10 lp/mm is waisted on the
unaided human eye. That is a little over 700 pixels per inch in both
dimensions. For an 8 X 10 print, this is about 4 megapixels overall.
Dave Martindale
>A line pair is a spacial cycle so you could think of lp/mm like
>electrical cycles per second.
>Nyquist sampling theorem applies to both: two digital samples per
>cycle are required. In actual practice, about 2.8 to 3.0 samples per
>cycle are required for good quality. A pixel is a digital picture
>sample.
So far so good...
>The 35 mm film negative that you have processed at Walmart will
>probably result in about 25 lp/mm resolution at best.
There's not much Wal-mart can do to reduce negative resolution. That is
determined pretty much entirely by the film you chose, the quality of
the lens, whether you used a tripod, the shutter speed, whether the
autofocus worked, and so on. However, most any film and half-decent
lens ought to be capable of 50 lp/mm at least. 25 lp/mm is a very
pessimistic estimate of the film resolution. (On the other hand, the
print could be awful - that depends on Wal-mart's enlarger).
>25 lp/mm X 2.8 samples/lp = 70 samples/mm
>or 70 pixels/mm which is 1,778 samples/inch.
Double these values for 50 lp/mm.
>If the active area on the 35 mm negative is about 28 mm X 35 mm, how
>many overall pixels are required to preserve the assumed 25 lp/mm?
>(28 mm X 25 lp/mm X 2.8 pixels/lp) X (35 mm X 25 lp/mm X 2.8
>pixels/lp) = 4,802,000 pixels.
The negative is actually 24x36 mm nominal. That changes the answer to
4.2 megapixels - not a very big change.
However, assuming 50 lp/mm instead of 25 means you need 16.9
megapixels.
>For prints, anything more than about 10 lp/mm is waisted on the
>unaided human eye. That is a little over 700 pixels per inch in both
>dimensions. For an 8 X 10 print, this is about 4 megapixels overall.
You dropped a power of 10. 8 x 10 x 700^2 is 39.2 megapixels, not 4.
However, the 10 lp/mm figure is higher than what is recommended by
the sources I've seen. Recommendations for sharp prints range from 4-8
lp/mm. At 4 lp/mm minimum, you need about 285 PPI, which means 6.5
megapixels for an 8x10.
Dave
Stewart Pinkerton
>I'm sorry, I don't understand any of this!
>I have a Minolta D7i 5MP that puts out a file with 72DPI, Now I have
>enlargements made at 36x44 inch that are absolutely beautiful. How can
>that be?
>Check my pbase site. I have also had a couple of panoramics made 1 13"
>x 6ft.
A lot of this is down to viewing distance. Pros insist on 300dpi
because their work must withstand very close inspection and still
appear flawless.
300dpi from your D7i will give you an image around 9x7 inches, which
will appear 'flawless' with close inspection, let's say a reasonable
10 inch viewing distance. If you expand the image to your 44x36 inch
size, that's only about 60dpi, so you will need to use a viewing
distance of 50 inches to achieve the same perceived resolution in
terms of the 'circle of confusion' angle subtended at the eye. Now,
you might consider that it's perfectly reasonable to judge the quality
of a 44x36 print from a distance of 4 feet, rather than sticking your
nose on it, and I'm not going to argue with that at all.
--
Stewart Pinkerton | Music is Art - Audio is Engineering
Jim V
saying, everyone is saying it's not possible. If you are closer than 2
ft you will see a little noise, but who looks at a 36x44 from 10 in. ?
Your eyes will cross.
All I am saying is you can get great enlargements with digital
cameras and the newer high resolution printers, even though the math
says otherwise.
Jim Venner
See what I am Saying?
http://www.pbase.com/jimven/galleries
da...@cs.ubc.ca (Dave Martindale) wrote in message news:<av6hqm$bto$1...@mughi.cs.ubc.ca>...
Bit Bucket
the number of responses that follow attempting to say it does not. ;-)
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"Jim V" <jimv...@yahoo.com> wrote in message
news:623fb113.03010...@posting.google.com...
Dave Martindale
>Did you guys look at my pbase site? You are not hearing what I am
>saying, everyone is saying it's not possible. If you are closer than 2
>ft you will see a little noise, but who looks at a 36x44 from 10 in. ?
>Your eyes will cross.
You don't look at the whole 36x44 from 10 inches away - you look at
parts of it, in exactly the same way that when you drive over the top of
a hill and see a new chunk of landscape you look at it with a sequence
of glances at different parts of it, or the way you gradually look at
the contents of a room that you've never been in before.
I *have* seen large prints (about 4x6 feet in size) made from
large-format negatives. You can stand back 6 feet and take in the whole
scene, but you can also walk up so you're 15 inches away and examine all
the detailed things in the image - and I did do that. A large format
negative has enough resolution to do that. It doesn't matter whether
the print was made chemically, or whether the negative was scanned on a
drum scanner and then printed on a large-format digital printer.
A 6 megapixel image printed 4x6 feet large will look great when viewed
from 6 feet, but not at 15 inches. It's the viewing distance that
matters.
> All I am saying is you can get great enlargements with digital
>cameras and the newer high resolution printers, even though the math
>says otherwise.
But that math doesn't "say otherwise". The math says that what you see
depends on pixels per degree of angle, not pixels per inch. If you're
viewing something from 10 inches (a pretty standard assumption for 8x10
and smaller size prints), then you need 200-400 PPI at the print for a
sharp-appearing image. But a 44 inch print *viewed from 44 inches*
needs only 45-90 PPI for the same apparent sharpness.
The math says you need the same number of pixels for the same visual
angle. The print can be any size, as long as you maintain the same
ratio of viewing distance to image width. On the other hand, if you are
using the larger print to cover a larger visual angle (e.g. the 6 foot
print viewed from 15 inches), you need more pixels.
Dave