I may be having a math problem here. Can someone help me out?
Have you seen anything giving the conventional wisdom of what size an area the eye can see best at once is? The below quote from Bates says that the textbooks say it's 1/2" at 20 feet. But the only info I can find online says it's 1-2 degrees of the visual field, which would mean about 4-8" at 20 feet, with this math:
40 ft * pi * 1/360 * 12 in/ft = 4 in.
If the visual field is a 360 degree circle, 20 feet away from the edge is a 40 foot diameter.
Circumference = diameter * pi.
1 degree of the visual field is 1/360 circumference.
Bates says it's really less than 1/4" at 20 feet, and the smaller the better, and he's right about that, but I can't find anything online that suggests it's smaller than 1 degree of the visual field. 1 degree, or 4" at 20ft, is a huge area, but I can't find a problem with my math. If my math is right, I'm in disbelief that the authors of the material I've found could make such a huge error that's so easily disproved subjectively.
Quote:The text-books say that at twenty feet an area having a diameter of half an inch can be seen with maximum vision, but anyone who tries at this distance to see every part of even the smallest letters of the Snellen test card - the diameter of which may be less than a quarter of an inch - equally well at one time will immediately become myopic. The fact is that the nearer the point of maximum vision approaches a mathematical point, which has no area, the better the sight. PSWG, ch 11
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"Half of our funny, heathen lives, we are bent double to gather things we have tossed away." - George Meredith
Without getting into the formula, notice that he said that the best spot has virtually no area. A Bates teacher from the 1950s said that it was less than 1/16 of an inch on the eye (even more odd).
I'm not the best at trying to figure out flaws in formulas, but there's something to remember from the section you quoted.
Yes, I know. What I'm trying to determine is why Bates got a different answer from his sources of conventional wisdom than I'm getting. He got 1/2", and I'm getting 4-8". I can't find any source that actually gives it in terms of an area in inches wide at a certain distance, corresponding to either size of the fovea or else a size smaller than the fovea in recognition of the cones getting smaller and denser deeper in the fovea. It's only given in terms of degrees (1-2). It seems like the topic of "how large of an area can be seen best and equally" has been ignored by vision science, unless I'm looking in the wrong places and need to get some actual textbooks.
Site Administrator
"Half of our funny, heathen lives, we are bent double to gather things we have tossed away." - George Meredith
Looking in old optometry, ophthalmology books, if find something will post. Have you checked Thomas Q's. book?
I 'think' Bates said the 20/20 letter, 3/4 " at 20 feet is the basic central field size created by the macula which contains the fovea in the center, stated by most eye doctors but, it is really smaller... clearest central vision as comes to a small point. It can't be too small or the fovea would not exist to produce anything? Maybe my memory is incorrect but I think it's right, just too busy to search in Bates Mags today.
Are you looking for that smallest central point the fovea produces in the center of the fovea or the whole fovea or the fovea and mac combined?
Maybe there's a way use math to determine what size central field each of these produces at 20 feet? I cannot do math so can't begin to figure this.
David, sorry I posted this on wrong thread and had re-post here. Busy moving...
Got some pics for you, might help, will send.
They say the foveola (center of the fovea) is .35mm diameter, which apparently is 1 degree of the visual field. I don't know the math to verify that, but if it's 1 degree then my math is right as far as I can tell.
Site Administrator
"Half of our funny, heathen lives, we are bent double to gather things we have tossed away." - George Meredith
Okay I don't have enough knowledge on fovea and stuff
But asumming the lense to fovea distance is 1.5cm and the image with a retinal distance of 3*10^-6m is the clearest.. (its minimum required to reslove image)
I get an area of perfect vision of 0.2 Inches.
I get a different value for the angle aswell...
Do you understand the assumptions I made?
What I wrote is probably useless, but what's perfect distance between to retinal cells for clearest vision? Minimum is 3*10^-6 m logic tells me the smaller the more perfect the image right? Just like a laptop is clearer than when you project it to the screen?
Also angle of clearest vision I got was around 0.11 degrees
Okay a retinal cell has a radius if 1.5*10^-6 m, the lenses im our eye is convex so it converges the image which becomes inverted and diminished in size. The minimum diameter of the image has to be 3*10^-6 to be resolved. Which is the distance between the mid point of two retinal cells.
Assuming that this value gives the clearest image.... Which is a guess in all fairness.
Okay david, I sent you an email, I took a picture of it with my phone. I don't know how to attach images here. I don't ever use my computer anymore either... I'm sending this message thru my phone to.
Not sure if I'm right might be fluke I'm using high school maths and also its an estimate could be +- 0.1 inches for all I know...
Thanks, I replied to the email. It gets a little complex. One factor I don't think we've considered is how much the refraction of the lens and cornea affect the angles.
Site Administrator
"Half of our funny, heathen lives, we are bent double to gather things we have tossed away." - George Meredith
As a frame of reference for pilots, the FAA pilot safety guide describes this one-degree conical area of foveal vision as equivalent to the size of a quarter at 4.5 feet away:
"This foveal field of vision represents a small conical area of only about 1 degree. To fully appreciate how small a one-degree field is, and to demonstrate foveal field, take a quarter from your pocket and tape it to a flat piece of glass, such as a window. Now back off 4 ½ feet from the mounted quarter and close one eye. The area of your field of view covered by the quarter is a one-degree field, similar to your foveal vision."
A quarter is almost an inch in diameter (about 15/16ths). That's a lot larger area than 1/4 inch at 20 feet -
I don't actually know what I calculated I think it is actually the smallest possible thing that you can register at 20ft. I guess I then decided to make a circle out of it.
arocarty Wrote:As a frame of reference for pilots, the FAA pilot safety guide describes this one-degree conical area of foveal vision as equivalent to the size of a quarter at 4.5 feet away:
"This foveal field of vision represents a small conical area of only about 1 degree. To fully appreciate how small a one-degree field is, and to demonstrate foveal field, take a quarter from your pocket and tape it to a flat piece of glass, such as a window. Now back off 4 ½ feet from the mounted quarter and close one eye. The area of your field of view covered by the quarter is a one-degree field, similar to your foveal vision."
A quarter is almost an inch in diameter (about 15/16ths). That's a lot larger area than 1/4 inch at 20 feet -
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I think it is actually the smallest possible thing that you can register at 20ft. I guess I then decided to make a circle out of it.
