# The math behind visual acuity

The number of megapixels required to print something, or view a television is ultimately determined by the human eye’s visual acuity, and the distance the object is viewed from. For someone with average vision (i.e. 20/20), their acuity would be defined as one arcminute, or 1/60th of a degree. For comparison, a full moon in the sky appears about 31 arcminutes (1/2 a degree) across (Figure 1).

Now generally, some descriptions skip from talking about arcminutes to describing how the distance between an observer and an object can calculated given the resolution of the object. For example, the distance (`d`, in inches) at which the eye reaches its resolution limit is often calculated using:

`d = 3438 / h`

Where `h`, is the resolution, and can be ppi for screens, and dpi for prints. So if `h`=300, then d=11.46 inches. Now to calculate the optimal viewing distance involves a magic number – 3438. Where does this number come from? Few descriptions actually give any insights, but we can can start with some basic trigonometry. Consider the diagram in Figure 2, where `h` is the pixel pitch, `d` is the viewing distance, and `θ` is the angle of viewing.

Now we can use the basic equation for calculating an angle, Theta (`θ`), given the length of the opposite and adjacent sides:

`tan(θ) = opposite/adjacent`

In order to apply this formula to the diagram in Figure 2, only `θ/2` and `h/2` are used.

`tan(θ/2) = (h/2)/d`

So now, we can solve for `h`.

```d tan(θ/2) = h/2
2d⋅tan(θ/2) = h```

Now if we use visual acuity as 1 arcminute, this is equivalent to 0.000290888 radians. Therefore:

```h = 2d⋅tan(0.000290888/2)
= 2d⋅0.000145444```

So for `d`=24”, `h`= 0.00698 inches, or converted to mm (by multiplying by 25.4), `h`=0.177mm. To convert this into PPI/DPI, we simply take the inverse, so 1/0.00698 = 143 ppi/dpi. How do we turn this equation into one with the value 3438 in it? Well, given that the resolution can be calculated by taking the inverse, we can modify the previous equation:

```h = 1/(2d⋅0.000145444)
= 1/d * 1/2 * 1/0.000145444
= 1/d * 1/2 * 6875.49847
= 1/d * 3437.749
= 3438/d```

So for a poster viewed at `d`=36″, the value of `h`=95dpi (which is the minimum). The viewing distance can be calculated by rearranging the equation above to:

`d = 3438 / h`

As an example, consider the Apple Watch Series 8, whose screen has a resolution of 326ppi. Performing the calculation gives d=3438/326 = 10.55”. So the watch should be held 10.55” from one’s face. For a poster printed at 300dpi, d=11.46”, and for a poster printed at 180dpi, d=19.1”. This is independent of the size of the poster, just printing resolution, and represents the minimum resolution at a particular distance – only if you move closer do you need a higher resolution. This is why billboards can be printed at a low resolution, even 1dpi, because when viewed from a distance it doesn’t really matter how low the resolution is.

Note that there are many different variables at play when it comes to acuity. These calculations provide the simplest case scenario. For eyes outside the normal range, visual acuity is different, which will change the calculations (i.e. radians expressed in θ). The differing values for the arcminutes are: 0.75 (20/15), 1.5 (20/30), 2.0 (20/40), etc. There are also factors such as lighting, how eye prescriptions modify acuity, etc. to take into account. Finally, it should be added that these acuity calculations only take into account what is directly in front of our eyes, i.e. the narrow, sharp, vision provided by the foveola in the eye – all other parts of a scene, will have slightly less acuity moving out from this central point.

p.s. The same system can be used to calculate ideal monitor and TV sizes. For a 24″ viewing distance, the pixel pitch is `h`= 0.177mm. For a 4K (3840×2160) monitor, this would mean 3840*0.177=680mm, and 2160*0.177=382mm which after calculating the diagonal results in a 30.7″ monitor.

p.p.s. If using cm, the formula becomes: `d = 8595 / h`

# Resolution of the human eye (ii) – visual acuity

In the previous post, from a pure pixel viewpoint, we got a mixed bag of numbers to represent the human eye in terms of megapixels. One of the caveats was that not all pixels are created equal. A sensor in a camera has a certain resolution, and each pixel has the same visual acuity – whether a pixel becomes sharp or blurry is dependent on characteristics such as the lens, and depth-of-field. The human eye does not have a uniform acuity.

But resolution is about more than just how many pixels – it is about determining fine details. As noted in the last post, the information from the rods in coupled together, whereas the information from each cone has a direct link to the ganglion cells. Cones are therefore extremely important in vision, because without them we would view everything as we do in our peripheral vision, oh and without colour (people who can’t see colour have a condition called achromatopsia).

Cones are however, not uniformly distributed throughout the retina – they are packed more tightly in the centre of the eyes visual field, in a place known as the fovea. So how does this effect the resolution of the eye? The fovea (which means pit), or fovea centralis, is located in the centre of a region known as the macula lutea, a small oval region located exactly in the centre of the posterior portion of the retina. The macula lutea is 4.5-5.5mm in diameter, and the fovea lies directly in the centre. The arrangement of these components of the retina is shown below.

The fovea has a diameter of 1.5mm (although it varies slightly based on the study), and a field of view of approximately 5°. Therefore the fovea has an area of approximately 1.77mm². The fovea has roughly 158,000 cones per mm ² (see note). The density in the remainder of the retina is 9,000 cones per mm². So, the resolution of the human eye is much greater in the centre, than on the periphery. This high density of cones is achieved by decreasing the diameter of the cone outer segments such that foveal cones resemble rods in their appearance. The increased density of cones in the fovea is accompanied by a decrease in the density of rods. Within the fovea is a region called the foveola which has a diameter of about 0.3mm, and a field of view of 1° – this region contains only cones. The figure below (from 1935) shows the density of rods and cones in the retina.

The fovea has the highest visual acuity in the eye. Why? One reason may be the concentration of colour-sensitive cones. Most photoreceptors in the retina are located behind retinal blood vessels and cells which absorb light before it reaches the photoreceptor cells. The fovea lacks the supporting cells and blood vessels, and only contains photoreceptors. This means that visual acuity is sharpest there, and drops significantly moving away from this central region.

For example pick a paragraph of text, and stare at a word in the middle of it. The visual stimulus in the middle of the field of view falls in the fovea and is in the sharpest focus. Without moving your eye, notice that the words on the periphery of the paragraph are not in complete focus. The images in the peripheral vision have a “blurred” appearance, and the words cannot be clearly identified (although we can’t see this properly, obviously). The eyes receive data from a field of view of 190-200°, but acuity of most of that range is quite poor. If you view the word from approximately 50cm away, then the field of view is about ±2.2cm from the word – beyond that things get fuzzier. Note that each eye obviously has its own fovea, but when you focus on a point both fovea overlap, but the resolution doesn’t increase.

The restriction of highest acuity vision to the fovea is the main reason we  spend so much time moving our eyes (and heads) around. From a processing perspective, the fovea represents 1% of the retina, but the brains visual cortex devotes 50% of its computation to input from the fovea. So in the fovea, resolution is equivalent of a TIFF, whereas elsewhere it’s a JPEG. So, if the sharpest and most brilliantly coloured human vision comes from the fovea, what is its resolution?

Again this is a somewhat loaded question, but let’s attempt it anyway. If the fovea has a field of view of 5°, and assuming a circular region, we can create a circular region with a radius 2.5 degrees = 19.635 degrees2, and 60×60 = 3600 arcmin2/degree2. Assume a “pixel” acuity  of 0.3×0.3=0.09 arcmin2. This gives us 19.635*3600 / 0.09 = 785,400 pixels. Even if we round up we get a resolution of about 1MP for the fovea. And honestly, the actual point of highest acuity may be even smaller than that  – if we considered the foveola, we’re looking at a mere 125,000 pixels.

NOTE
Note: There are many studies relating to the size of the fovea, and the density of photoreceptors, given that each human is a distinct being, there is no one exact number.

Jonas, J.B., Schneider, U., Naumann, G.O.H., “Count and density of human retinal photoreceptors”, Graefe’s Archive for Clinical and Experimental Ophthalmology, 230(6), pp.505-510 (1992).