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`