What is the optimal distance to sit from a TV? I was wondering how to calculate the optimal distance to sit away from a TV. I don't quite know the full set of parameters it will depend on, I would suspect it to include the following:


*

*Size of the screen

*pixel size/resolution of the screen

*I've probably forgotten something...


There will probably also be a biological factor, as in a property of the eye that matters. To take any biology out of the problem and make it purely physics let's assume any parameter necessary (probably the focal length of the eye? not sure though) without determining its value (although it would be great if someone knew it). I am more interested in which parameters determine this problem and the associated formula rather than actual values for my TV though.
For simplicity we shall also assume the TV screen to be square (if it is easier you may take the freedom to change this to circular or any shape you like).
 A: So, the question is sort of broad but I'm bored enough. I describe my reasoning bellow. 
When we watch TV we are constantly looking at a screen, which I assumed to be circular with radius $L$. This is represented in this beautiful figure:

Now, to make sense of what you're talking about I needed to compute some personal, yet generalizable, estimates. My criterion was really simple: the optimal angle 
$$\theta_{opt} = tan^{-1} \left( \frac{L}{d_{opt}} \right)$$
is defined the following way: from very near start walking away from the screen, stopping at each step to look at two different points, each located on the circle and diametrically opposed (that is, these two points must be $2L$ away from each other). At the minute you feel comfortable to look at both points without moving your head, than this is the optimal distance. This means you can watch TV by just moving your eyes and keeping your head still, which is my way of defining optimal. I did some testing for myself and noticed that 
$$\theta_{opt} = (16 \pm 1)º  \, .$$
So you can apply this to any TV size you want. If your hypothetical round TV from the future has a $50 \, \text{cm}$ radius, then the optimal distance is $d_{opt} \approx 174 \pm 11 \, \text{cm}$. 
P.S.: I'm assuming perfect pixel separation and variable luminosity, which is pretty much the scenario with actual technology.
P.S.S.: I checked and, strangely, this result pretty much mimics the official comfort distances provided by most of the sellers. That's nice!
A: Whatever your personal definition of "optimal", I will go ahead and assume that among other things you're aiming for is seeing as much image detail as possible - and get as immersive, or "large", an image as possible - without starting to perceive the pixels distinctly (or more precisely to perceive the mesh of thin dark lines separating the pixels, which has been dubbed the "screen door effect").
The simplest way to calculate this is to consider what viewing angle is going to be occupied by a central pixel on the screen and compare that to the viewing angle taken up by the smallest dot you can perceive distinctly given your eyes' power of resolution (which depends on genetics, health etc.).
The visual angle subtending the smallest detail that can be perceived distinctly by someone with 20/20 vision - which may or may not be your case but is highly probable if you're wearing recently (re-)prescribed correction eyeglasses - is 1 arcminute or 1/60 of a degree (this value is the basis for setting the smallest font size at the bottom of the Snellen chart you see in the eye doctor's office). So when your TV is at the desired distance from you a central pixel should occupy no more than, but still close to, 1 arcminute of your viewing angle if you want to maximize perceived image quality while avoiding the screen door effect.
The reason I keep insisting on a central pixel is that we can simplify the calculation by thinking of the right triangle formed by the pixel's width and the lines from the left and right edges of the pixel to the center of your eye. In this triangle, the angle we want to set at 1 arcminute is the small one between your viewing lines toward the pixel edges and since we know it's a right triangle we can easily calculate the ratio of the pixel's size to the viewing distance as equal to tan(1 arcminute) = 0,000290888. (The triangle would look something like the longest one at the bottom of this image, but much longer, having your eye at the red corner, and with "a" being the pixel size.)
So since you asked "what distance" it means you want to consider a TV with a known physical pixel size and work out the viewing distance from that. Then you need the inverse of the value we got above, i.e. 1 / 0,000290888 = about 3438. This is how much longer your viewing distance needs to be than your pixel width or height (assuming square pixels).
So for a TV with a pixel size of 0.29 mm you would need to sit at least 0.29 * 3438 = 997 mm or about 1 meter away in order to avoid the screen door effect, but not too much farther away if you don't want to waste any image definition or "quality". (This is based on a random 4K TV's "pixel pitch" spec. For a 1080p TV of the same diagonal I'd expect this minimum viewing distance to go to about double, so 2 meters.)
Of course, if you don't wear corrective eyeglasses and happen to have better-than-20/20 vision this calculation will not apply directly to you and you will need to sit farther back than the distance determined in this way.
