Visual Acuity, DPI, and Resolution
Recently at work I was asked to investigate various display technologies (like visualization walls and large format printing) and I thought I would take a slightly different approach to the problem.
Warning: There is a lot of math in this article but you can skip the equations along the way and still follow along.
Whenever we talk about displaying things we invariably talk about the size of the display and the resolution.
In my multimedia class I define resolution as the number of pixels wide something is by the number of pixels tall something is – it is the absolute number of pixels that we have.
The trick when displaying or printing is what to do with these pixels.
Pixels vs Dots
When asked by a student: How big is a pixel? I always answer: A pixel is. A pixel is a unit of measurement. A pixel simply defines a single piece of data. A “dot” on the other hand is a pixel in physical form. A “dot” can be measured. On the screen one pixel of information can be represented by a dot – this dot has a physical dimension. When we print we use the same terminology but it gets a little murkier. When we talk about how we distribute a resolution across a given space we use the term DPI or dots per inch. This is a measurement of density. Are the dots spread out or are they closer together. The reason this is murky is that printers use even smaller dots (of primary print colors) to create a single dot. So a single pixel printed as a single dot is really made up of even smaller dots. This bothers a lot of people but it shouldn’t because each dot on the screen is also made up of smaller dots (of primary light colors). In the end – these smaller dots whether on the screen or on the page are supposed to be so small that our visual acuity cannot discern them individually and they blend into a whole – a single dot for each pixel of information.
People tend to use the term DPI for print and PPI (pixels per inch) for screens – I feel DPI is the more appropriate term as we are talking about measurable dots – and I will use the term DPI for the rest of this article.
A common misconception is that 300DPI is ideal for print and 72DPI for the screen. This is a gross oversimplification and too often wrong. While 300DPI does tend to work well for print – display technologies have changed drastically and it is a rare occasion that the 72DPI concept is useful.
So what I want to do here is flip the conversation. Instead of talking about printers and screens – I want to talk about what a human being can see. The perceptual side of things.
Visual Acuity
A circle is made of of 360 degrees. A single degree is made of of 60 arc minutes. So 1 arc minute is 1/21,600 of a full circle.
When we say a person has 20/20 vision we mean they have the acuity to discern a detail of 1 arc minute.
Some people have worse vision such as 20/40 and they can only discern at 2 arc minutes – some have better vision such as 20/10 and they can discern 0.5 arc minutes.
There is a physical limit to human visual acuity as there are only so many cones (color sensors) in the eye and while 20/8 is possible it is rare – the kind of thing you see in fighter pilots.
Size, Distance, and Density
So when we look at picture, poster, or monitor we are really talking about 3 different elements.
The size of the thing – and based on the size of the thing we choose a distance to be from it when we view the thing, and then from there we can determine the density of dots that our visual acuity can actually resolve.
For my first example – let me start with my TV. I have a 55″ 1080p HD LCD TV.
TVs are measured diagonally and in reality the TV is approximately 50″x30″.
So based on the size – how far away do I want to sit? Well that depends on how much of my Field of View (FOV) I want to fill.
While the human Field of View is very wide – a comfortable (practical) Field of View is 40 degrees. Pick too wide a FOV and yours eyes dart around – too narrow and the image just isn’t large enough.
With these 2 variables – Width of target and Field of View – I can calculate an ideal viewing distance.
Distance = (Width /2) / Tan(Viewing Angle/2)
Distance = (50/2) / Tan (40/2) = 68.7″ or 5.7′
So now I know that the ideal distance from my couch to my TV is 68.7″ or about 5 1/2′.
My TV is 1080p which means it has a resolution of 1920×1080 pixels.
1920 pixels spread over 50″ results in 38.4 DPI – this seems really low at first – but how does this line up with my visual acuity.
To answer this – I have to ask: If I am sitting 68.7″ away from something – what is the smallest sized dot that my visual acuity can resolve?
For this I need just 2 variables – the Distance from the target – and my own Visual Acuity which I will give as 20/20 (I wish).
From my Visual Acuity I can calculate my Visual Resolution – the smallest number of visual degrees I can detect.
Visual Resolution = (1 / Visual Acuity) * (1 / 60)
Visual Resolution = (1 / (20 / 20)) * (1 / 60) = 0.0166667 degrees
(1 / Visual Acuity) tells me how many arc minutes I can detect so (1 / (20/20)) = 1, (1 / (20/10)) = 0.5, and (1 / (20/8)) = 0.4 (which I said before is as good as it gets).
Multiplying the Number of Arc Minutes * (1 / 60) converts the number from arc minutes into degrees.
Smallest Dot Size = 2 * Distance * Tan (Visual Resolution/2)
Smallest Dot Size = 2 * 68.7 * Tan (0.0166667/2) = 0.019984 inches
Dots Per Inch = 1 / Smallest Dot Size
Dots Per Inch = 1 / 0.019984 = 50 DPI
So with my 20/20 vision I cannot resolve more than 50DPI on something that is about 5 1/2′ away. So my TV’s 38.4 DPI isn’t too bad after all.
And let us not forget the reality:
1) I am not looking for individual dots on the screen
2) I have worse that 20/20 vision
3) I sit more than 5 1/2′ from my TV
Lessons
So basically here are the two lessons:
1) The DPI that we can actually perceive is a factor of our Visual Acuity and the Distance we are from an object. As with my previous example if I am 5 1/2′ away from something I cannot resolve more than 50DPI.
2) The distance we are away from something is a choice determined by our ideal FOV.
Other Examples
iPhone 4S
I have an iPhone 4s with a ‘retina’ display. The resolution of the screen is 640 x 960 pixels and the screen size is about 2″x3″. This means the screen is 320DPI.
The question – how far away from my face do I need to hold the phone for me to ‘not see the dots’ or at what distance is my visual acuity 320DPI?
Pixel Size = 1 / DPI
Pixel Size = 1 / 320 = 0.003125″
Distance = Pixel Size / (2 * Tan (Visual Resolution/2))
Distance = 0.003125 / (2 * Tan (0.0166667/2)) = 10.7″
So as long as I hold the phone 11″ away from my face I shouldn’t see individual dots. Seems about right.
Poster Printing
I want a 24″x36″ poster for my office wall about 10′ from my desk. What DPI do I need to make it look good?
In this case we basically want to make sure we have at least the same amount of DPI as our visual acuity.
Using our equation we find:
Smallest Dot Size = 2 * Distance * Tan (Visual Resolution/2)
Smallest Dot Size = 2 * 120 * Tan (0.0166667/2) = 0.0349 inches
Dots Per Inch = 1 / Smallest Dot Size
Dots Per Inch = 1 / 0.0349 = 28 DPI
28 DPI – seems awfully low right?
Well it is – it is because there is another factor when it comes to photography and that is we don’t only view things hanging on a wall from a distance – people walk up to posters and photographs and look at them closer.
If this was a billboard – it would work great because you can’t get near a billboard – but with something hanging on the wall have to assume that people will get closer.
How close? Let’s say instead of 10 feet I choose 1 foot.
Smallest Dot Size = 2 * Distance * Tan (Visual Resolution/2)
Smallest Dot Size = 2 * 12 * Tan (0.0166667/2) = 0.00349 inches
Dots Per Inch = 1 / Smallest Dot Size
Dots Per Inch = 1 / 0.00349 = 286 DPI
And this is where the 300DPI rule comes from. Because at about a foot away – everything looks good at 300DPI.
Hope this helps.