Okay, now it's official: ~250dpi is eye resolution or “retina” resolution, (with a non-capital ‘r’, mind!), if viewed from a 30cm distance. Can we calculate this in the opposite direction? For sure. Here is the main formula that this post is about:

Let's say the manual for a LED panel specifies a pixel pitch of 3.36mm. What's the optimal viewing distance so that we can claim eye resolution?

$$ \begin{align} d &= r / \tan{θ} \\ d &= 3.36mm / \tan{0.02°} \\ &= 9625.69mm \end{align} $$

Woah, that's almost 10 meters distance. You're gonna need a bigger boat (or a smaller pixel pitch), or we'll have to sacrifice some quality.

## Pixel Size(in mm) × 1.5 = acceptable viewing distance in meters

Sometimes, a resolution of 72dpi for 50cm is still acceptable. After all, this is quite close to a common resolution used with CRT (=cathode ray tube, you millennial 😉!) displays for decades. What kind of angular resolution does this relate to?

72dpi means a pixel size of \( 25.4mm/72 = 0.35mm \), which in turn, viewed at a distance of 500mm, leads us to an angular resolution:

$$ \begin{align} θ &= \arctan{ \frac{0.35mm} { 500mm } } \\ & = 0.040425349° \end{align} $$
Let's see what optimal viewing distance we would get for this new angular resolution, and a pixel pitch of 3.36mm:

$$ \begin{align} d &= r / \tan{θ} \\ d &= 3.36mm / 0.000739845 \\ &= 4762.204724409mm \end{align} $$

About 4.8 meters? Yes that could work.

## Simplify: No need for Pythagoras, let's stick with Thales