PS printing in 8 or 16 bits

wrote:

When printing a 16-bit depth file from PS (CS or other versions), is it converted to 8-bits first before sending it to a printer (Epson 1280/2200)?

It is.

It is also reduced to 8bit when you view it on the monitor.

Michael

In article , says…

When printing a 16-bit depth file from PS (CS or other versions), is it converted to 8-bits first before sending it to a printer (Epson 1280/2200)?

Ink jet printers can only reproduce about 65 distinct shades of a given color, not even the 255 that an 8 bit image would indicate. So there is no point is sending a 16 bit image to the printer.
16 bit is useful for doing large changes in contrast or color without worrying about posterization.
Take a look at the tips section of my web site for some relevant discussions.

—
Robert D Feinman
Landscapes, Cityscapes and Panoramic Photographs
http://robertdfeinman.com
mail:

When printing a 16-bit depth file from PS (CS or other versions), is it converted to 8-bits first before sending it to a printer (Epson

1280/2200)?

Yes. No consumer printer is capable of reproducing anything even remotely close to 16-bit color under any circumstances; in fact, consumer-grade printers are incapable of reproducing the full tonal range of even an 8-bit image.

—
Art, literature, shareware, polyamory, kink, and more:
http://www.xeromag.com/franklin.html

Tacit wrote:

When printing a 16-bit depth file from PS (CS or other versions), is it converted to 8-bits first before sending it to a printer (Epson

1280/2200)?

Yes. No consumer printer is capable of reproducing anything even remotely close to 16-bit color under any circumstances; in fact, consumer-grade printers are incapable of reproducing the full tonal range of even an 8-bit image.

I print little currently, but plan to do so before long. Sometime ago I was under the impression that many consumer grade printers threw out a lot of info as they really couldn’t handle more than 240 or so dpi. How that translates into throwing out pixel info, I don’t know, but is this still a factor to consider?

—
John Mcwilliams

I print little currently, but plan to do so before long. Sometime ago I was under the impression that many consumer grade printers threw out a lot of info as they really couldn’t handle more than 240 or so dpi. How that translates into throwing out pixel info, I don’t know, but is this still a factor to consider?

Yes.

There are actually two different things going on here; first, consumer inkjet printers have an effective pixel resolution of somewhere between 240 and 280 pixels per inch typically, and images whose resolution is higher do not add to the quality of the image; and second, consumer inkjet printers can only reproduce about 60-100 shades of any particular hue, which means that they can’t produce the full tonal range of an 8-bit image, much less a 16-bit image.

These two factors aren’t directly related; image resolution and bit depth are two separate aspects of a digital image.

—
Art, literature, shareware, polyamory, kink, and more:
http://www.xeromag.com/franklin.html

In message ,
Robert Feinman wrote:

Ink jet printers can only reproduce about 65 distinct shades of a given color, not even the 255 that an 8 bit image would indicate.

Perhaps, but many printer dots usually are used to make one image pixel, with dithering, so far more than 65 colors are perceived in the print.

So there is
no point is sending a 16 bit image to the printer.

There’s no point in converting it to 8 bit, just to print it, either. —

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John P Sheehy

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wrote:

In message ,
Robert Feinman wrote:

Ink jet printers can only reproduce about 65 distinct shades of a given color, not even the 255 that an 8 bit image would indicate.

Perhaps, but many printer dots usually are used to make one image pixel, with dithering, so far more than 65 colors are perceived in the print.

Even with the multitude of dots used for shading, there are no more than 65 shades of <inkcolor> possible with current inkjet printers. So 65^<number of inks> is the number of colors the printer could create if overprinting would not result in something near black too.

More shades require more space – resulting in lower ppi.

So there is
no point is sending a 16 bit image to the printer.

There’s no point in converting it to 8 bit, just to print it, either.

You do not need to convert manually. Conversion to printer compatible ranges is done automatically, even if you still work in 16bit mode (maybe PS needs a little more memory and time when printing 16bit images).

Michael

Yes – because there are no APIs for printing more than 8 bits/channel.

Chris

In article wrote:

When printing a 16-bit depth file from PS (CS or other versions), is it converted to 8-bits first before sending it to a printer (Epson 1280/2200)?

Robert Feinman …

Ink jet printers can only reproduce about 65 distinct shades of a given color, not even the 255 that an 8 bit image would indicate. So there is no point is sending a 16 bit image to the printer.

Where does this number, 65, come from?

I can print 256 patches from paper white to full cyan ink on my Epson 2200, measure them with the eye-one spectro and get distinct XYZ, L*a*b* or spectral reflectance values for each patch.

Is there some deltaE limit on what is considered a distinct shade?

–Tom

In article ,
says…

Robert Feinman …

Ink jet printers can only reproduce about 65 distinct shades of a given color, not even the 255 that an 8 bit image would indicate. So there is no point is sending a 16 bit image to the printer.

Where does this number, 65, come from?

I can print 256 patches from paper white to full cyan ink on my Epson 2200, measure them with the eye-one spectro and get distinct XYZ, L*a*b* or spectral reflectance values for each patch.

Is there some deltaE limit on what is considered a distinct shade?
–Tom

It comes from info found on Epson’s web site. The reason for the limited number of values has to do with the size of the square used to create the various densities. For an 8×8 square you can get 65 distinct values. If you print at 1440 dpi then an 8×8 square reduces the effective resolution to 180 dpi, for example. Each of these 8×8 squares represents one pixel in the original image in the simplest case.
Modern printers get more shades by using more colors of inks and some also use variable size dots.
This is the standard tradeoff between smoothness of gradiation and resolution that all halftoning printing systems face.
It’s possible that the 2200 by using a light cyan get generate values "in between" the standard ones and thus get 128 values. It may also use a bigger than 8×8 grid thus allowing more values. A careful study of their technical specs should reveal the details.

Whatever the value is (even 256) it is still much less than what can be obtained with 16 bits and thus the advice to shift to 8 bits for output still is valid.
I have some discussions of this and other 16 bit matters on my web site. Just follow the tips link on the home page.

—
Robert D Feinman
Landscapes, Cityscapes and Panoramic Photographs
http://robertdfeinman.com
mail:

Robert Feinman …

Where does this number, 65, come from?

It comes from info found on Epson’s web site. The reason for the limited number of values has to do with the size of the square used to create the various densities. For an 8×8 square you can get 65 distinct values.

Robert, thank you for explaining that. If you have a link to the info on the Epson site, I would appreciate it.

Given that at 1440dpi, an 8×8 square of ink droplets represents one pixel in a 180ppi image. That 8×8 cell gives 64 possible ink droplet locations. In a four-color printer, if you count up how many combinations of ink droplets can populate that 8×8 area, you get 864497. This number goes up approximately exponentially as you add inks to the printer:

4 color 864497
5 color 12103009
6 color 143218993
7 color 1473109697

As soon as you get to the 6 color printer, there are more possible combinations of ink droplets possible in that 8×8 square than an 8-bit RGB pixel can represent.

Of course, these numbers assume the perfect printer that makes the perfect ink droplets. In reality there are many other factors that affect how these ink droplets combine spectrally to create a visible color. Mechanical dot gain, optical dot gain, Kubelka-Munk theory.

I suspect that 65 is a number that is over-simplified and outdated. As you point out, variable dot size in the newer printers is one factor that will affect this. Also, I believe the newer printers can place ink droplets over previously placed ones which changes things significantly.

So, it seems there might be something to gain from newer printers by driving them with 16-bit data. In practice, I don’t know how much of that could be realized. I do know that ImagePrint can drive the 2200 with 16-bit data. I’ll have to run some tests to see if I can notice a difference.

Tom

—
"They’re my friggin’ pixels!" http://rawformat.com

Thomas Fors wrote:

Robert Feinman wrote in message
news:…

Where does this number, 65, come from?

It comes from info found on Epson’s web site. The reason for the limited number of values has to do with the size of the square used to create the various densities. For an 8×8 square you can get 65 distinct values.

Robert, thank you for explaining that. If you have a link to the info on the Epson site, I would appreciate it.

Given that at 1440dpi, an 8×8 square of ink droplets represents one pixel in a 180ppi image. That 8×8 cell gives 64 possible ink droplet locations. In a four-color printer, if you count up how many combinations of ink droplets can populate that 8×8 area, you get 864497. This number goes up approximately exponentially as you add inks to the printer:

4 color 864497
5 color 12103009
6 color 143218993
7 color 1473109697

The number of dot combinations is irrelevant for the number of shades. If you use a 8×8 square you can cover from zero to 64 of the squares so the intensity of ink in this 8×8 area varies in 65 steps.

If you take in account, that the size of an ink drop is larger than the positioning precision (=dpi value) you understand that if you fill more than 50% of the possible positions in the 8×8 square you already cover the full area.

As soon as you get to the 6 color printer, there are more possible combinations of ink droplets possible in that 8×8 square than an 8-bit RGB pixel can represent.

There are 64 ways to place one dot in an 8×8 area but all of them are the same shade.
The same is true for the 64*63 ways to place two dots, and so on.

I suspect that 65 is a number that is over-simplified and outdated.

65 is the number of possible shades in an 8×8 square and will never change. The arrangement of the 32 dots for a 50% fill does not affect the fact that it is a 50% fill.

As you point out, variable dot size in the newer printers is one factor that will affect this. Also, I believe the newer printers can place ink droplets over previously placed ones which changes things significantly.

Variable dot size is irrelevant for resolution as the smallest ink drop is still bigger than the (virtual) square it should color.

Overprinting has been part of the calculation but its effect is not too big as mixing colors leads to black

So, it seems there might be something to gain from newer printers by

driving them with 16-bit data. In practice, I don’t know how much of that could be realized. I do know that ImagePrint can drive the 2200 with 16-bit data. I’ll have to run some tests to see if I can notice a difference.

To resolve for 16 bits you would need a 256x256dot square or get a ppi value that is 1/256th of the mechanical dpi.

Michael

"Xalinai" …

To resolve for 16 bits you would need a 256x256dot square or get a ppi value that is 1/256th of the mechanical dpi.

Michael

Michael,

Thank you for taking the time to explain this. I was thinking only in terms of combinations of the various colored ink droplets and ignoring the fact that the 8×8 cell is used to create an effective halftone dot. I believe I understand now.

Tom

—
6.3 million pixels held captive. http://rawformat.com