Randomoneh is right. Viewing distance isn't chosen like that. You don't move half as close to the screen when watching a DVD (instead of a BD). Rather than asking yourself "at what viewing distance does the size and resolution of my TV approach 'perfection'?", you ask yourself "at what viewing distance does the size of my TV offer the highest immersion, in spite of other limitations?". You watch from the distance that 'feels' best to you, or provides optimal 'immersion', as Randomoneh put it.
Yes, to calculate the number of dots needed on a piece of paper (I will use the words "pixels" and "screen"), you follow the same procedure as I did for 16:9 screens. Only this time you know the height and width of the screen beforehand, so you won't have to bother with the Pythagorean theorem and aspect ratios.
First, you calculate the desired pixel pitch, of the pixel which is the very closest to your eyes, based on the viewing distance. Then, you go on to calculate the target PPI (pixels per inch) or target resolution from that. The only problem I can see is that, on the diagonal pixel pitch will be wider (due to the Pythagorean theorem). Again, it's an approximation.
I only used Dr. Clark's sources, not his calculations, so I haven't really looked at them. But, I can make my own:
To calculate the desired pixel pitch p when viewed from distance y:
Example: The desired pixel pitch when viewed from a distance of 20 inches:
( 20 in ) * tan( 0.3 arcminutes ) = 44.3313631 microns
From this, the desired PPI can be calculated:
Which is the same as:
Example: The desired PPI when viewed from a distance of 20 inches:
((( 20 in ) * tan( 0.3 arcminutes ))^(-1))*(1 in) = 572.958
That is very close to the standard 600 DPI resolution of printers at home.
To calculate the desired resolution r for a specific length (dimension) w of screen, replace "1 in" with w:
Example: The 13.3 inches from a distance of 20 inches in Clark's example:
( 13.3 in ) / ( ( 20 in ) * tan( 0.3 arcminutes ) ) = 7620.34
Now, you asked, what if we only know the field of view the paper or screen occupies?
Well, the FoV really only lets you calculate the size of the sceen, so you can replace the screen size with that calculation in the above formula.
If the screen occupies an angle of q when centered, perpendicular to your line of vision, and viewed from distance y, its length is:
Example: The length of a screen occupying 35.3 degrees when viewed from 20 inches:
(2 * tan( 35.3 degrees / 2 ) * 20 in) in inches = 12.7271772 inches
Now we can just merge the two formulae into one:
And simplify:
Example: Pixels needed to approach theoretical "perfection" on a screen occupying 35.3 degrees:
2 * tan( (35.3 degrees) / 2 ) / tan( 0.3 arcminutes ) = 7292.14
Hope that helps.
Edited by MisterMuppet - 8/7/12 at 11:08am
Quote:
I think your chart looks very professional, much more advanced than mine! Originally Posted by Randomoneh 
Warning: Spoiler! (Click to show)
Warning: Spoiler! (Click to show)
No, everything is fine. Somehow I missed "vertical" resolution.
Now, there is an error with how Mr. Clark at ClarkVision calculates needed resolution and "resolution of the eye". He simply multiplies number of pixels per degree (200 for 0.3 arcmin-per-pixel) with number of degrees that print / display is occupying. Try doing that and you'll see how wrong it is. Displays and printed materials are straight and one degree of FOV might occupy 1 inch in the center of the display and 2 inches at the edge. I've made that same mistake before.
So, his example is this: "Consider a 20 x 13.3-inch print viewed at 20 inches. The Print subtends an angle of 53 x 35.3 degrees, thus requiring 53*60/.3 = 10600 x 35*60/.3 = 7000 pixels, for a total of ~74 megapixels to show detail at the limits of human visual acuity."
At 20 inches, apparent size of every pixel should be 0.3 arcminutes, that is 0.0017453292531 inches. That makes 572.957795 pixels per inch or 11459 x 7620 for whole print. Center of his image (10060 x 7000 doesn't match 200 pixels per degree / 0.3 arcminutes per pixel. His error would even higher if imaginary print occupies higher angle of viewer's field of view.
Here's number of pixels needed for some viewing angles.
Degrees of field of view / needed number of pixels:
1 = 200
5 = 1000.6097
10 = 2005.04157
15 = 3017.1764
20 = 4041.01414
25 = 5080.73843
30 = 6140.78725
35 = 7225.93252
40 = 8341.37157
45 = 9492.8346
50 = 10686.713
55 = 11930.2151
60 = 13231.5576
65 = 14600.2042
70 = 16047.1673
75 = 17585.3928
80 = 19230.2588
85 = 21000.2305
90 = 22917.73
95 = 25010.3136
100 = 27312.2871
105 = 29866.9673
110 = 32729.9105
115 = 35973.6303
120 = 39694.6728
125 = 44024.5498
130 = 49147.2306
135 = 55328.2946
140 = 62965.9458
145 = 72685.7534
150 = 85530.1329
155 = 103375.2
160 = 129972.906
165 = 174077.442
170 = 261950.853
Cotangent of 0.3 arcminutes can be used for easy calculation of needed viewing distance (in inches) or needed ppi value of display / print.
Needed viewing distance = 11459.155895344 / PPI
Needed PPI value = 11459.155895344 / distance in inches
For 7680 x 4320 displays, viewing distance = 1.29923485 x diagonal measurement. Or 2.65258238 x image height.
As for your graphs, I like them. I've made something similar before, based on 0.3 arcminute per pixel value, too.
I like your attention to details.
Now, there is an error with how Mr. Clark at ClarkVision calculates needed resolution and "resolution of the eye". He simply multiplies number of pixels per degree (200 for 0.3 arcmin-per-pixel) with number of degrees that print / display is occupying. Try doing that and you'll see how wrong it is. Displays and printed materials are straight and one degree of FOV might occupy 1 inch in the center of the display and 2 inches at the edge. I've made that same mistake before.
So, his example is this: "Consider a 20 x 13.3-inch print viewed at 20 inches. The Print subtends an angle of 53 x 35.3 degrees, thus requiring 53*60/.3 = 10600 x 35*60/.3 = 7000 pixels, for a total of ~74 megapixels to show detail at the limits of human visual acuity."
At 20 inches, apparent size of every pixel should be 0.3 arcminutes, that is 0.0017453292531 inches. That makes 572.957795 pixels per inch or 11459 x 7620 for whole print. Center of his image (10060 x 7000 doesn't match 200 pixels per degree / 0.3 arcminutes per pixel. His error would even higher if imaginary print occupies higher angle of viewer's field of view.
Here's number of pixels needed for some viewing angles.
Degrees of field of view / needed number of pixels:
1 = 200
5 = 1000.6097
10 = 2005.04157
15 = 3017.1764
20 = 4041.01414
25 = 5080.73843
30 = 6140.78725
35 = 7225.93252
40 = 8341.37157
45 = 9492.8346
50 = 10686.713
55 = 11930.2151
60 = 13231.5576
65 = 14600.2042
70 = 16047.1673
75 = 17585.3928
80 = 19230.2588
85 = 21000.2305
90 = 22917.73
95 = 25010.3136
100 = 27312.2871
105 = 29866.9673
110 = 32729.9105
115 = 35973.6303
120 = 39694.6728
125 = 44024.5498
130 = 49147.2306
135 = 55328.2946
140 = 62965.9458
145 = 72685.7534
150 = 85530.1329
155 = 103375.2
160 = 129972.906
165 = 174077.442
170 = 261950.853
Cotangent of 0.3 arcminutes can be used for easy calculation of needed viewing distance (in inches) or needed ppi value of display / print.
Needed viewing distance = 11459.155895344 / PPI
Needed PPI value = 11459.155895344 / distance in inches
For 7680 x 4320 displays, viewing distance = 1.29923485 x diagonal measurement. Or 2.65258238 x image height.
As for your graphs, I like them. I've made something similar before, based on 0.3 arcminute per pixel value, too.
I like your attention to details.
Yes, to calculate the number of dots needed on a piece of paper (I will use the words "pixels" and "screen"), you follow the same procedure as I did for 16:9 screens. Only this time you know the height and width of the screen beforehand, so you won't have to bother with the Pythagorean theorem and aspect ratios.
First, you calculate the desired pixel pitch, of the pixel which is the very closest to your eyes, based on the viewing distance. Then, you go on to calculate the target PPI (pixels per inch) or target resolution from that. The only problem I can see is that, on the diagonal pixel pitch will be wider (due to the Pythagorean theorem). Again, it's an approximation.
I only used Dr. Clark's sources, not his calculations, so I haven't really looked at them. But, I can make my own:
To calculate the desired pixel pitch p when viewed from distance y:
Code:
p = y * tan( 0.3 arcminutes )
Example: The desired pixel pitch when viewed from a distance of 20 inches:
( 20 in ) * tan( 0.3 arcminutes ) = 44.3313631 microns
From this, the desired PPI can be calculated:
Code:
ppi = ( 1 in ) / ( y * tan( 0.3 arcminutes ) )
Which is the same as:
Code:
((( y ) * tan( 0.3 arcminutes ))^(-1))*(1 in)
Example: The desired PPI when viewed from a distance of 20 inches:
((( 20 in ) * tan( 0.3 arcminutes ))^(-1))*(1 in) = 572.958
That is very close to the standard 600 DPI resolution of printers at home.
To calculate the desired resolution r for a specific length (dimension) w of screen, replace "1 in" with w:
Code:
r = w / ( y * tan( 0.3 arcminutes ) )
Example: The 13.3 inches from a distance of 20 inches in Clark's example:
( 13.3 in ) / ( ( 20 in ) * tan( 0.3 arcminutes ) ) = 7620.34
Now, you asked, what if we only know the field of view the paper or screen occupies?
Well, the FoV really only lets you calculate the size of the sceen, so you can replace the screen size with that calculation in the above formula.
If the screen occupies an angle of q when centered, perpendicular to your line of vision, and viewed from distance y, its length is:
Code:
2 * tan( q / 2 ) * y
Example: The length of a screen occupying 35.3 degrees when viewed from 20 inches:
(2 * tan( 35.3 degrees / 2 ) * 20 in) in inches = 12.7271772 inches
Now we can just merge the two formulae into one:
Code:
( 2 * tan( q / 2 ) * y ) / ( y * tan( 0.3 arcminutes ) )
And simplify:
Code:
2 * tan( q / 2 ) / tan( 0.3 arcminutes )
Example: Pixels needed to approach theoretical "perfection" on a screen occupying 35.3 degrees:
2 * tan( (35.3 degrees) / 2 ) / tan( 0.3 arcminutes ) = 7292.14
Hope that helps.
Edited by MisterMuppet - 8/7/12 at 11:08am
