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)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:

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