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|05-26-2009, 10:07 AM||#1|
Got confused again.. with the viewing distance
For example, I have a 40" LCD TV which is capable of Full 1080p. So the recommended viewing distance for that size is 5 feet away from the TV right.
Here are my questions:
1) What if I sit closer than the recommended viewing distance, what benefits would be lost/would I not get?
2) What if I sit beyond the recommended viewing distance, what benefits would be lost/would I not get?
3) Why is that if you're screen is larger, for example, 52", you'll need to sit around 6.5 feet? Why not like the 40" which is closer? I thought the bigger size of your screen, the nearer you can watch but I was wrong. It was otherwise.
4) So the the purpose of bigger screens like 50" is to let you sit farther from the TV so that your eyes would not hurt and you can still get the full benefits of a Full HD 1080p? Unlike the smaller ones, 40" which can hurt your eyes because of sitting close to the TV just to experience its full benefits OR you're going to sit beyond the recommended viewing distance so you wouldn't hurt your eyes but you would not experience the true essence of Full HD 1080p because, yes you can still see more vibrant colors compared to lower resolutions but you won't anymore notice the difference of the detail between a 1080p and 720p or 1080p and 480p because you're already sitting at the recommended distance for the 720p/480p resolutions. It's just like you're going to watch a 720p/480p movie. Am I right?
Blu-rays: 5 movies and 10 PS3 games.
(I still have few BD but I'm getting addicted to it. :) )
|05-26-2009, 02:06 PM||#3|
good job...my input exactly.
|05-27-2009, 02:32 AM||#4|
I am sure that you have read about "recommended viewing distance" but for the most part it is all BS. Pretty much (within reason) sit as close to as big a screen with as much resolution as you can.
let's pretend those letters where not on a chart but on the screen (in the movie) let's pretend all the rows are the same size, why do they look different? because they are on smaller TVs. so The F in row two is on(for example) a 70" TV, on row 5 42", on row 6 37"... if you sit farther away the size/detail in row 2 eventually becomes like the row 5 when you are closer.
1) not everyone has the same visual acuity, so how can one distance (a line on a graph) be right for everyone.
2) look at the eye chart, look at the last line the letters are hard to read from close by, but as you get farther more and more of the letters are hard to differentiate, but the o is the easiest to see from relatively far away.
here is an other example, look at the pic close and then far (other side of room) before reading the rest.
look at it from close by and your brain focuses on the detail and you see Einstein, but your brain is also telling you "there is something wrong with it", go far away, now you see the shades much more then the detail and it looks like Marilyn, but you are also registering the details and think again "there is something wrong with it". Consciously your brain is not paying attention to the detail from far away in this pic (namely because it contradicts what it thinks it should see), but subconsciously it is registering that detail and telling you there is something wrong with it.
|05-29-2009, 06:46 PM||#5|
The Viewing Distance Post.
Increases in Resolution gives us more viewing options; not all of them will allow your eyes completely resolve what your display is capable of.
Viewing Distances based on Visual Acuity
The distance we need to be from the set is based on spatial resolution (how closely lines can be resolved), screen size and visual acuity. Nominal (Snellen, 1862) acuity is 20/20 (feet) or 6/6 (meters) where the optotype glyphs (sloan letters, tumble E, numbers or geometric symbols) would subtend at 5' of arc on the retina, and the strokes, 1' of arc. 20/40 (0.5) is 85% the effiency (not 50%) of 20/20 (1.0) which means someone with this vision is taking in less information, as defined by the Inverse Square Law for point sources. At any rate, 20/20 determines the angular size, 1 arcminute or 1/60 of a degree - the smallest detail one can see having this level of acuity. This works out to be 0.0698inch/1.77mm at 20 feet and can be used to determine the pixel size we can see at some distance n. Contrast also plays a role: the lower the contrast, the less detail we can perceive.
Details size of up to 1.5 arcminutes are cosidered sharp to the viewer and so this is what determines the distance from and/or the size of the set. For 1080p, this will be 1.1 times image width or 32.86 deg; for 720p, it is 1.75 times image width; and for 480p, 3 times image width. So if the resolution of program material changes, you are going to have to keep moving you chair if you want to sit at the optimum distance when viewing 480/720/1080p sources. Also, since our eyes have different visual acuity, the Viewing Distance Chart are only a guide, not an absolute. You are going to have to pick what is good for you and your eyes.
Viewing Distance and the Pythagorean theorem: aē + bē = cē
A few steps are necessary as we are only given a value for the diagonal or hypotenuse and a ratio of 16:9 that is shared by all widescreen TV regardless of size. We also don't know any angles in the right triangle other then 45 deg. Let a = adj, b = opp.
We start by letting a = 16 and b = 9 and creating the ratio b/a = 9/16;
We then use arctan x = θ; arctan(9/16) = 34.22 degrees
40 inch TV: width 35", height 20".
42 inch TV: width 37.5", height 21":
width: 42cos(34.22) = 34.729; height 42sin(34.22) = 23.61
TV Size and Resolution Info
Inverse Square Law
Where some physical quantity is inversely proportional to the square of the distance from the source.
1 arcminute of resolution, The typical resolving power of the human eye,
Screen Format, Aspect ratio
Throw Ratio, A ratio between projection distance and width of image. distance(D)/screen width(W).
SMPTE 30, 1.6263 times the screen size in a 16:9 TV.
THX 40, Divide the diagonal screen measurement by .84
THX Range, Minimum 40 deg, Mximum 28 deg.
NTSC, 10 deg.
sin θ = opp/hyp = b/c; csc θ = 1/sin θ = 1/(b/c) = c/b
cos θ = adj/hyp = a/c; sec θ = 1/cos θ = 1/(a/c) = c/a
tan θ = opp/adj (tan θ = sin θ/cos θ); cot θ = cos θ/sin θ
Pythagorean trigonometric identity, sinē θ + cosē θ = 1
(sin θ)ē = sinē θ; (cos)ē = cosē θ;
sin θ = b/c; sinē θ = bē/cē
cos θ = a/c; cosē θ = aē/cē
bē/cē + aē/cē = (bē + aē)/cē
sinē θ + cosē θ = 1ē or sinē θ + cosē θ = 1 for angles between 0 and π/2.
sin' x = cos x, csc' x = -csc x cot x
cos' x = -sin x, sec' x tan x
tan' x = secēx, cot'x = -cscēx
∫ cos x dx = sinx + C
∫ sin x dx = - cosx + C
∫ tan x dx = ln |sec x| + C
Back to Getting Started HD POST
Last edited by U4K61; 06-13-2012 at 09:56 PM.
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