# HDTV and the Resolving Power of the Eye

A principal advantage of HD over SD video is the ability to view larger images from closer distances.

The design goal for HDTV was to facilitate viewing at a distance of three picture heights from the screen as opposed to the five- to seven-picture-height design goal for NTSC. What is the relationship between resolving power, screen resolution and viewing distance from the screen? **READ THE FIRST LINE**

Resolving power is the ability of the eye, or an optical instrument, to differentiate between two points or lines. It is typically expressed as an angle of separation radiating outward from the eye or instrument. When resolving power is expressed as an angle, simple geometry permits us to calculate the relationships between viewing distance and line or point separations.

The theoretical resolving power of the human eye, based solely on the size of its aperture, may be calculated using the formula:*(click thumbnail)*

where *a* is the minimum separation angle in arc seconds (1 arc second = 1/3,600 degree), l is the wavelength of light in nanometers (1 nanometer =10-9meter), and *D* is the diameter of the aperture (the pupil of the eye) in centimeters.

If the diameter of the pupil is 5 mm and a light wavelength of 500 nanometers (green light--the color to which the eye is most sensitive) is used, solving for ? gives us:*(click thumbnail)*

This is the theoretical resolving power of the eye based on its light-admitting aperture. It does not take into account such optical aberrations as misconvergence or astigmatism, nor does it take into account how well light sensors in the eye function.

To determine at what distance this theoretical eye aperture can resolve two parallel lines spaced 1 mm apart, consider an isosceles triangle in which the short side--L--is a distance of 1 mm between the lines. The two long sides, d, the distance from each line to the pupil's center, are unknown; the angle, a, is 20 arc seconds, or 20/3,600 = 0.00556 degree. Because L is negligible relative to d, we may assume that d represents the distance from the eye to the plane of the parallel lines. Solving for d: *(click thumbnail)**10,313 mm, or about 34 feet*

It is doubtful that anyone's real eye would be capable of resolving two lines 1 mm apart from a distance 34 feet. The actual resolving power of the human eye with 20/20 vision is typically considered to be about one arc minute or 60 arc seconds, which is about one-third of the theoretical resolution we just calculated based solely on the diameter of the pupil. In fact, it is a little more complicated.**ACUITY IN PRACTICE**

A single human eye sees roughly a 140-degree field horizontally and a 90-degree field vertically. The eye's resolution capability varies over the area of the retina and therefore with viewing angle. It is highest in the foveal, or straight-ahead, viewing region of the retina, which lies within a 2- to 4-degree radius of the optical axis of the eye, where an angular separation of about one arc minute may be detected under ideal conditions.

The eye can rotate 45 degrees vertical, so the foveal region may be directed to view any part of a 90- to 95-degree section of the horizontal monocular field of vision. Solving the previous equation using an angle of 60 arc seconds or 0.01667 degree, we find that the distance at which the foveal region of the 20/20 eye may resolve lines 1 mm apart is 3,438 mm or about 11.3 feet.

You can check the resolving power of your own eyes by drawing two lines spaced 1 mm apart on a piece of white paper, using black ink to maximize contrast. Hang the paper on the wall, back up until you reach that point at which you can still just discern two lines, but stepping further back results in your seeing a single line. Measure the distance from your eyes to the paper. Check with the lines oriented both horizontally and vertically to verify that resolution is virtually the same in both dimensions.**HUMAN VISION AND HDTV**

What does all this mean with respect to HDTV and SDTV pictures? Consider display sizes, with a 20/20 viewer located about 11.3 feet (3,438 mm) from the screen. The design goal for HDTV was to facilitate viewing at a distance of three picture heights, so if three picture heights is 3,438 mm, then one picture height is 1,146 mm, or about 45 inches (3.76 feet). Since the HDTV aspect ratio is 16:9, the width of the image is 2,037 mm, or about 80 inches (6.7 feet).

The horizontal viewing field for this screen may be calculated by computing the length of the hypotenuse of right triangles formed by drawing a line from the viewing spot to the center of the screen and to the left and right edges of the screen. Where c is the distance from the viewing point to the left or right edge of the screen, a is the distance from the center of the screen to either edge and b is the distance from the viewing point to the plane of the screen; c2 = a2 + b2, therefore:*(click thumbnail)*

The derived dimensions are used to solve the previous equation for a: *(click thumbnail)*

We calculate that the horizontal viewing field is about 32.6 degrees. If we plug screen height in place of screen width, we calculate that the vertical viewing field is about 18.8 degrees. Obviously, the viewer's eyes must move around to see the entire picture at maximum resolving power. **720p & 1080i**

Now, let's calculate the distance between scanning line centers and pixel centers (the line and pixel pitches) for the two ATSC HD scanning formats for these screens.

To remove the factor of CRT dot pitch, we'll assume that we are projecting these pictures.

For 1,080 lines, dividing 1,146/ 1,080 gives us a distance of 1.06 mm between line centers; for 1,920 horizontal samples, 2,037/1,080 gives us a distance of 1.06 mm between horizontal pixel centers.

In this case, a 20/20 viewer who can resolve lines 1 mm apart will just be able to see both the line structure and the horizontal pixel structure from a distance of three picture heights.

For 1,280 x 720, the distance between line centers and pixel centers is about 1.59 mm, in which case the line and pixel structure would be somewhat more visible.**4:3 SDTV**

Let us now look at what size of standard-definition 4:3 screen would be optimal for the same viewing distance.

In this case, the recommended viewing distance is five to seven picture heights, with six picture heights being midrange. This would call for a screen height of 573 mm, or about 22.6 inches (1.9 feet). Since the aspect ratio is 4:3, the screen width would be 764 mm, or about 30 inches (2.5 feet).

Not surprisingly, this is a much smaller screen than the HDTV screen. For 525/60 SDTV, the distance between line centers is 573/483, or about 1.18 mm; the distance between horizontal pixel centers is 764/720, or about 1.06 mm.

Unlike the HD formats, 525/60 doesn't have square pixels, so the distance between horizontal pixels is not precisely equal to the distance between lines, but the two are reasonably close.

As we see, this smaller screen provides essentially the same viewing condition in terms of resolution at six picture heights as the larger HD screen did at three picture heights, in that the line-and-pixel structures are just visible to the 20/20 viewer. In this case, the horizontal viewing field is about 12.7 degrees, and the vertical viewing field is about 9.5 degrees.

This exercise does not take dynamic or temporal resolution into account. Considering only static spatial resolution, it is intended to determine the screen sizes that may effectively be used with HD and SD scanning formats, consistent with the resolving power of the eye.

It serves to illustrate rather well the advantage of the increased resolution afforded by HDTV.