Resolution for print

[MarioHorvat]

Hello everyone, I'm new here and I hope to get some help from experienced photographers.

I plan to view my print from distance of 20 inches and I want apparent size of pixel / dot to be 0.3 arcminutes (0.005 degrees). I want to know how big single pixel / dot have to be to meet that criteria?

Here I've found angular diameter calculator and if I'm using it right it says dot should be 0.0017453292531 inches. Am I right? Thank you very much!

[Chris Ober]

moving to workflow forum

[Tom Graham]

Hope you've checked our Roger Clark's web site where he discusses print sharpness. He discusses the eye, arc minutes, distances, DPI. It might not answer your specific question, but I'd start there.
http://www.clarkvision.com/articles/...ppi/index.html
Tom

[Roger Clark]

Mario,

Here is a simple computation: 0.3 arc minute= 0.3/60 = 0.005 degrees. 20 inches * Tangent (0.005) = 0.00175 inches for one pixel at 20 inches.
Thus 1/0.00175 inch/pixel = 573 pixels/inch.

Roger

[Mike Milicia]

If I understand your question correctly and you are talking about inkjet printing, I don't think you can really do what you are trying to achieve.

Each pixel in your image gets converted to a pattern of microscopic droplets of ink by the printer device driver.
You have very limited control of the density of these ink droplets.
For example, Epson printers give you the option of 1440 or 2880 dots per inch or dpi which refers to the individual ink droplets and has nothing to do with the pixels in your image. See here for more details :

http://people.csail.mit.edu/ericchan...1440_2880_diff

Inkjet printers also have a "native resolution", specified in pixels per inch which is a measure of how many image pixels are REQUIRED (i.e. no more and no less) in order to generate one inch of physical output on the paper. The "native resolution" is a fixed attribute of the printer and you have very limited control over it. On Epson printers, it is 360 PPI unless you check the "Finest Detail" box, in which case it is 720 PPI, but you would never check this box for printing images. The "Finest Detail" option is intended for text or vector graphics. See here for more details :

http://people.csail.mit.edu/ericchan...tml#native_res

If you do not check "Finest Detail" and hand an Epson printer anything other than 360 pixels of image data for every physical inch of desired output, the printer driver will automatically upsize or downsize the image file so that it has exactly 360 PPI before proceeding to convert each image pixel to a pattern of ink droplets.

So, in effect, you have no control over the density of pixels on the page and very limited control over the density of ink droplets.
If you hand the driver an image with a resolution attribute of 573 PPI, the first thing the driver will do is downsize it to 360 PPI for the same physical size print.
If you hand the driver 150 PPI, the first things it will do is upsize it to 360 PPI.
Most people recommend that you size the image to the native resolution of the printer yourself in Photoshop or other specialized image processing software before handing it to the printer.

[arash_hazeghi]

Mike is right, the DPI resolution gets internally mapped to a sophisticated microscopic droplet pattern which is usually beyond 1440dpi (up to 5760 dpi for pro Epson models) and the size of the droplets is not uniform. It is adjusted dynamically based on many factors. So you can't map pixels to dots on paper. you won't see the dots either, the texture you might see in a modern high quality inkjet print is from the paper not the ink.

[MarioHorvat]

Mario,

Here is a simple computation: 0.3 arc minute= 0.3/60 = 0.005 degrees. 20 inches * Tangent (0.005) = 0.00175 inches for one pixel at 20 inches.
Thus 1/0.00175 inch/pixel = 573 pixels/inch.

Roger

573 pixels per inch. That would make 11460 x 7621 for print size 20 x 13.3 inches. This site says 10060 x 7000 but that would make apparent size of closest pixel larger than 0.3 arcminutes. It would be 0.34 arcminures per pixel and 503 pixels / inch.

[MarioHorvat]

Mario,

Here is a simple computation: 0.3 arc minute= 0.3/60 = 0.005 degrees. 20 inches * Tangent (0.005) = 0.00175 inches for one pixel at 20 inches.
Thus 1/0.00175 inch/pixel = 573 pixels/inch.

Roger

573 pixels per degree. That would be 11460 x 7621 for 20 x 13.3 in print. However, this website says it's 10600 x 7000 for 20 x 13.3 print. 10060 pixels over 20 inches would make 530 pixels / inch, making apparent size of pixel 0.324 arcminutes at 20 inches viewing distance.

[MarioHorvat]

Mario,

Here is a simple computation: 0.3 arc minute= 0.3/60 = 0.005 degrees. 20 inches * Tangent (0.005) = 0.00175 inches for one pixel at 20 inches.
Thus 1/0.00175 inch/pixel = 573 pixels/inch.

Roger

573 pixels per inch. That would make 11460 x 7621 for 20 x 13.3 print. However, this website says it's 10600 x 7000 for 20 x 13.3 print. In that case, apparent size of the closest pixel / dot would be 0.324 instead of 0.3 arcminutes. Edges of the image would match "0.3 arcminutes or less" but not the center.

Am I right?

[MarioHorvat]

Mario,

Here is a simple computation: 0.3 arc minute= 0.3/60 = 0.005 degrees. 20 inches * Tangent (0.005) = 0.00175 inches for one pixel at 20 inches.
Thus 1/0.00175 inch/pixel = 573 pixels/inch.

Roger

573 pixels per inch. That would make 11460 x 7621 for 20 x 13.3 print. However, this website says it's 10600 x 7000 for 20 x 13.3 print. In that case, apparent size of the closest pixel / dot would be 0.324 instead of 0.3 arcminutes. Edges of the image would match "0.3 arcminutes or less" but not the center.

Am I right?