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Author Topic: visual acuity  (Read 2250 times)

Austin Parker

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visual acuity
« on: October 20, 2007, 10:06:43 am »

Alright, so here my issue.

GENERALLY we can agree that good viewing distances for text and graphics is 6x and 8x the displays image height (respectively). Example: if the displays image height is 10", good viewing for text will be up to 60" away, and graphics 80" away.

Can i in turn use this linear scale to determine the size of an image when moving away from it? IE, a division of 6/image height and 8/image height?

I don't think so...

I was then shown the math to visual acuity and the 5 degree arc test, and then i got lost  Laughing

Plain and simple, whats the math i need to use in order to determine the image size from x distance?
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Austin Parker

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John Birchman, CTS

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Re: visual acuity
« Reply #1 on: October 20, 2007, 12:46:50 pm »

There are several ways to look at this.

Most schools of thought say the ideal viewing distance is 1.5 times the width of the screen.

Another basic rule is the screen height should be approximately equal to 1/6 the distance from the screen to the last row of seats.

As there are many different possible screen sizes, seating configurations, and screen brightness/ambient lighting situations; common sense and experience also factor in.

I do not know that there is "one" formula that fits every single time.

John
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John Birchman, CTS
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Austin Parker

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Re: visual acuity
« Reply #2 on: October 20, 2007, 01:32:39 pm »

Thank John.

I understand the theories behind good view distance, but what im in fact looking for is the math behind determining how big a image will be from 30 feet, then from 60 feet, then 80 feet. As one moved farther from the display, the display appears to decrease in size. How do i calculate this perception?
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Austin Parker

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John Birchman, CTS

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Re: visual acuity
« Reply #3 on: October 20, 2007, 02:53:49 pm »

Inverse Square Law?

Quote:

Suppose you're looking at a sphere of diameter d at a distance r. Its diameter appears to span a certain angle, which we'll call a. If you back away from the sphere, to a distance 2r, the sphere (provided it's small enough for small-angle trig approximations to hold) now appears to span an angle a/2. Since the angular area of the sphere is proportional to a2, there is a simple inverse-square relationship between apparent sphere size and distance.


John

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John Birchman, CTS
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Austin Parker

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Re: visual acuity
« Reply #4 on: October 20, 2007, 03:05:40 pm »

its really that simple?

i thought it had something to do with visual acuity and the arc..??

http://en.wikipedia.org/wiki/Visual_acuity


if its that simple then thanks a bunch!
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Austin Parker

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Brad Weber

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Re: visual acuity
« Reply #5 on: October 20, 2007, 06:42:37 pm »

It does, it has to to with the arc subtended by a character.  The goal is to have characters that are legible and that requires determining how large the character need to be.  Of course, then you have to factor in the image resolution and font size to get to how big the screen has to be, a character that is a fixed pixel height on a 800x600 screen is physically half as high on a 1600x1200 resolution screen and thus if the font size remains constant, the 1600x1200 screen would have to be twice as large to maintain the same character legibility.  So you need to consider font size, resolution to get the character height and then the distance to the viewer to get the angle that represents.  And that is only valid for that font size.  That's why some very simplified and more generic 'rules of thumb' were developed.  Here are some articles

http://www.da-lite.com/education/angles_of_reflections.php?a ction=details&issueid=2

http://www.da-lite.com/education/angles_of_view.php?action=d etails&issueid=34

http://www.da-lite.com/education/angles_of_view.php?action=d etails&issueid=23

http://www.da-lite.com/education/angles_of_view.php?action=d etails&issueid=22

Be careful with any calculations based on screen width as that really has no direct relationship and will give different answers for the same image depending upon whether it is 4:3, 5:4 or 16:9 format, which actually has nothing to do with the screen size required to view text.
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Brad Weber
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Austin Parker

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Re: visual acuity
« Reply #6 on: October 22, 2007, 05:32:05 pm »

ok so inverse square law for perception of decrease, visual acuity for text formatting, x1.5,x6,x8 for legible distance.
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Austin Parker

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Brad Weber

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Re: visual acuity
« Reply #7 on: October 23, 2007, 12:15:10 am »

I'm not sure that I understand what you mean by the perception of decrease, but it seems to all go back to the same thing.  If a character subtends some arc when X feet away and you move to 2X distance, the character height remains the same but because the distance increased, the angle it subtends is smaller.  In a way it is simple geometry.  If a character is 3" high and you are 10' away, it subtends an arc of 1.43 degrees, move to 20' and that becomes 0.716 degrees, at 40' it is 0.358 degrees.  The further you move away at a right angle to and on axis with the screen, the smaller the characters, but in a direct relationship based on the distance to the screen and the height of the character.  If you move off axis the geometry gets a little trickier.  However, the same applies to the overall screen, as you move away it occupies a geometrically smaller angle of sight, you are just replacing the character height with the image height.

The 1.5, 6 and 8 times factors come from assuming typical text sizes as well as when one can start to see individual pixels (with CRT projectors it was discerning scan lines).  Too close, the 1.5 times the height, and you typically start to see individual pixels.  6 times the height is the general guideline for typical graphics, say 20 point font PowerPoint while 8 times the height is for large fonts and motion video.  And 4 times the image height is for critical graphics and text such as 12 point text and CAD.  But these guidelines are actually based on assumed typical text and object sizes.
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Brad Weber
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Austin Parker

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Re: visual acuity
« Reply #8 on: October 23, 2007, 11:15:30 pm »

Brad Weber wrote on Tue, 23 October 2007 00:15

  If a character is 3" high and you are 10' away, it subtends an arc of 1.43 degrees, move to 20' and that becomes 0.716 degrees, at 40' it is 0.358 degrees.  The further you move away at a right angle to and on axis with the screen, the smaller the characters, but in a direct relationship based on the distance to the screen and the height of the character.



aka perceived decrease.
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Austin Parker

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"Are you with me? Do you get it? If you have to ask what "IT" is, you don't get it"
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