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If the iPhone and iPad are both retina display, where retina display means resolution sharp enough to be indistinguishable to the human eye, is there a way to tell if other tablets are retina display quality? Is there a formula?

For example, to be retina quality at 3.5" you need to have 326 PPI correct? Given the width and height and / or screen size in inches and PPI how can you determine it?

For example is Nexus 7 retina quality, is the Samsung Galaxy SIII retina quality, is the Playbook retina quality etc.

More context: I'm using a iPad 3 for work and it's resolution is compelling. I like Android as well and want to know how to figure out if other tablets (at whatever size they are) can match the quality. I want to know the minimum DPI for any given size for personal use and development purposes.

~~ UPDATE ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
After reading the post given in the comments I realized the PPI is itself a constant value useful by it's self. Distance does come into the equation but I'm not sure how yet.

Quoting a commenter at that URL,

...I also use a formula to calculate the distance at which the eye can no longer distinguish pixels. Perhaps I'm biased, but I think this formula is more practical. With the formula I use, pixel density (ppi) is the only information that is necessary.

I'm not great at math but I think the forumula would be something like:

distance in inches*80=minimum dpi quality

4*80=324dpi

So at 10" away the dpi would have to be: 

10*80=800dpi
1.21 gigawatts
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    http://www.tuaw.com/2012/03/01/retina-display-macs-ipads-and-hidpi-doing-the-math/ – Robert Harvey Sep 26 '12 at 02:56
  • Great read. I realized some part of the way through that the PPI already are the answer I need (in this case). I then read the commenter coming to a similar conclusion (updated in main question). – 1.21 gigawatts Sep 26 '12 at 03:47

2 Answers2

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I'm going to work through the math to demystify it a bit.

Human visual acuity is measured in terms of the minimum resolvable distance between features. The distance is measured in arcminutes (1/60th of a degree), as a distance-independent measurement of acuity.

We're going to define "retina" as meaning "at least 1 pixel per arcminute". Some sources claim that the human eye is better than this (one source says 3 pixels per arcminute is necessary); some worse. The calculation used in this TUAW page assumes that "retina" is 1 pixel per arcminute.

Then, all we have to do is figure out how many pixels you need to get this resolution, equivalent to 60 pixels per degree of visual field. At a distance of x inches, one degree is x tan(1°), or about 0.01746x. The PPI requirement is thus 60 pixels in 0.01746x inches, or 3437/x PPI.

So, at a distance of 12 inches, we require 3437/12 = 286 PPI, and at a distance of 24 inches, we'd need 143 PPI. Note that the minimum PPI required decreases as distance increases, which is what we'd expect (e.g. faraway billboards have great resolution even if their PPI is very poor).

Hope that helps. Note that if you assume a higher visual acuity (e.g. y pixels per arcminute), then you have to multiply the resulting PPI minimums by y.

Alex78191
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nneonneo
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When introducing the iPhone 4, Steve Jobs said the number of pixels needed for a Retina display is about 300 PPI for a device held 10 to 12 inches from the eye

So the formula is DISTANCE = 11.5*300 / PPI, where

PPI - PPI of a screen

DISTANCE - distance from screen where screen becomes Retina.

Alex78191
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