Thursday, June 09, 2005

On Viewing Distance

"Resolution starts (and ends) with the eye," writes Geoffrey Morrison in "GearWorks: Viewing Distance vs. Resolution," covering an important topic in the January 2005 issue of Home Theater magazine (pp. 38-39). "If you're one of the lucky few with 20/20 vision ... your eye can discern one-sixtieth of a degree of arc ... ." So "objects roughly 0.067 inches wide" are the smallest we can distinguish at 20 feet.

At a more typical screen-to-retina viewing distance of 10 feet — half of 20 — we need objects to be at least 0.067 ÷ 2 = 0.033 inches wide in order to be able to distinguish them. At 10 feet, an object 0.033 inches wide subtends one-sixtieth of a degree: one minute of arc.

The "objects" Morrison has in mind are screen pixels. If we are sitting 10 feet from a TV, the smallest pixels we can discriminate from adjacent pixels of sufficiently different light intensity are 0.033 inches wide. If we move up to 5 feet away from the same screen, then we can "see" individual image pixels that are only 0.033 ÷ 2 = 0.0165 inches wide.

My Samsung 61" DLP rear projector has a 720p-based pixel grid that is 1,280 pixels wide by 720 pixels tall. To find the width of a single pixel, I can simply multiply the screen's diagonal measurement, 61 inches, by 0.87. (For calculator geeks, 0.87 is cos[tan-1(9/16)], 9/16 being the inverse of the screen's 16:9 aspect ratio.) Then I divide that result by 1,280, the number of pixels of horizontal resolution the screen provides.

It all comes to [(61 * 0.87) / 1280], giving my Samsung a pixel width of approximately 0.042 inches.

How far back from the Samsung can I sit before those 0.042-inch pixels begin to fuzz together? The formula for maximum viewing distance (MVD) as a function of pixel width (PW) is:

MVD = PW / tan(1/60°)

Calculators ready? Mine reports that tan(1/60°) is 0.0002909, so:

MVD = PW / 0.0002909

When PW is 0.042 inches, the formula gives an MVD of roughly 141 inches, or about 12 feet, give or take.

What would be my maximum viewing distance if my 61" TV were not 720p-native but 1080i? Well, 1080i puts 1,920 pixels across the screen instead of 1,280. 1,920 divided by 1,280 yields 1.5:1, or 3:2. So the 1080i pixel width would be the inverse of that, 2/3, times the width of the 720p-style pixels.

Which means I'd have to sit at 2/3 of the 720p viewing distance to avoid any loss of what I like to call "retinal resolution": 8 feet.

The lesson here is that for any screen size and horizontal resolution there is a maximal viewing distance beyond which "retinal resolution" suffers. This maximal distance can be expressed in screen heights, rather than inches or feet, thus factoring out the specific diagonal screen measure in inches.

For my 61" 720p-native TV, the screen height is 61 x 0.49, which is about 30 inches. (0.49 is sin[tan-1(9/16)].) My maximal viewing distance, brought down from above, is 141 in. Dividing that figure by the screen's height in inches, I get a maximal viewing distance of 4.7 screen heights.

If the TV were 1080i, my maximum viewing distance would be 2/3 that, or 3.1 screen heights.

So if you get much further away from a 1080i display than 3 screen heights — a "3x seating distance" — you start to lose "retinal resolution." But 720p is more forgiving. It lets you back up to a 4.5x or 5x viewing distance.

There is a handy-dandy calculator for some of these values on this web page. (And on this page there is yet more discussion of the viewing distance topic.) The calculator tells me on its very bottom line that, at 12 feet, my 61" 16:9 display is too far away, under the assumption that I want to see a fully resolved 1080i image. It implies I ought to move up to an 8-foot viewing distance.

It doesn't tell me that 12 feet is just right for 720p, but I can tell this by multiplying 8 feet by 3/2.

The calculator also tells me my "viewing angle," which is the angle subtended by the full width of the screen when I sit at my customary 12-foot viewing distance, is 20.9 degrees. In his article, Morrison says the best full-screen viewing angle is either 30 or 33 degrees, depending on which of two sentences you read. The handy-dandy web calculator says that in order for me to get an ideal 30-degree viewing angle, I'd have to sit 8.3 feet away from my diaplay ... which is basically the same viewing distance required for 1080i resolution.

Moral: when we move up to "just the right" (3x) seating distance from a 1080i display, we get ideal "retinal resolution" and, furthermore, we get the best viewing angle.

Or if, with a 720p display, we back up to 4.5-to-5 screen heights away — tolerable with 720p — we lose no "retinal resolution." However, we compromise the viewing angle that brings (in Morrison's phrase) "optimum viewer enjoyment."

But Morrison also points out tht the material we're watching need not have "anything close to that level of detail" — 1,920 distinct pixels across the screen. Moreover, "video noise, artifacts, and poor-quality low-resolution sources ... are all too noticeable at [such] close distances." So he recommends we adopt a blanket 5x optimum seating distance, even for 1080i displays.

Truth to be told, though, I personally find even 5x tough to achieve, given the exigencies of space, cost, and furniture arrangement. I actually sit more like 14-15 feet, not 12, from my Samsung. That's more like 6x than 5x.

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