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Topic: Apparent size with respect to position function (Read 1978 times) |
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jarls
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Apparent size with respect to position function
« on: May 18th, 2009, 3:20pm » |
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I'd like to know what is thought of the solution to a problem I thought of I've got. Not sure if its correct.
The problem is as follows.
An observer look is a direction with their eyes fixed. A line which if superimposed over their face would pass through both of their eyes moves away from the observer along their line of sight with a position function x(t). What function, with respect to position, would give the apparent length of the line?
My solution:
Imagine an infinitely long train track. Each rung is the same distance from its preceding or succeeding rung. Dividing the apparent distance between any two adjacent rungs by the distance between the preceding pair of rungs with always yield the same number the modulus of which will always be less than one. this number will be called 'r'.
Arbitrarily choose a rung to designate as the first (n=0) and the distance between it and the next to be 'd(0)'. The distance between the two rails here at the intersection between it and the first rung will be called 's(0)'. This will be the initial length of a line which lies on the track as it first begins to move away from an observer.
The track is now to be divided along all midpoints between the two rails. This yields two right triangles the tops vertices of which exist at a singularity, where the apparent distance between the rails is zero, the track is infinitely far away and the apparent height of the track is the height of the observers eyes. The longer side of one of these right triangles is the infinite sum of a geometric series (d(0))x(1/(1-r) --> d(0)/(1-r) and the shorter side of which is half the apparent length of a line traveling along the track to an observer. If a length from the first rung to some m th rung is to be hacked off, the ratio of the new shorter distance from this m th rung and the singularity at the horizon would be the infinite sum subtracted by the partial sum from n=0 to n=m d(0)/(1-r) - d(0)(1-r^(n+1)/(1-r) which is equal to d(0)(r^(n+1))/(1-r). The new shorter triangle is similar to original triangle with maximized side length. Therefore the ratio of the new longer side length and the half-size at the n th rung is the same as the ratio of the original longer side of the right triangle and the initial half-size.
d(0)/(1-r))/(s(0)/2)=((d(0)r^(n+1))/(1-r))/(s(n)/2)
(1-r)s, d(0)s and twos cancel
1/(s(0))=r^(n+1))/(s(n))
cross multiply
s(n)=s(0)x (d(0)r^(n+1))
divide by d(0)
s(n)=(s(0) x d(0)r^(n+1))/d(0)
d(0)s cancel
s(n)=s(0)r^(n+1)
What is n in therms of distance from the observer to an nth rung? It is the hypotenuse of s triangle whose one side is the height of the observers eyes and whose other side is the actual distance from the first rung to the nth rung. Therefore
distance^2=h^2+(cn)^2
where h is the height of the observers eyes and c is the actual distance between any two adjacent rungs.
Therefore n^2=(distance^2 - h^2)/c^2
so n=square root((distance^2 - h^2)/c^2)
This should be plugged into the previously derived equation for the apparent size with respect to n ,s(n)=s(0)r^(n+1), as n.
Which yields
s(x)=s(0)r^(((x^2 - h^2)/c^2)+1) therefore s(x(t))=s(0)r^(((x(t))^2- h^2)/c^2)+1)
Does this mean that for an arbitrary track with arbitrary distance between two adjacent rungs, arbitrary height of the observers eyes, arbitrary distance between the two rails, if those parameters are plugged into the above equation the solution for a certain x will be the same as the solution for this same x of the same equation with entirely different parameters but consistent? If any test track is made and analyzed, would the resultant function be a generalized one which give an objects apparent size with respect to distance for any configuration?
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| « Last Edit: May 18th, 2009, 8:07pm by jarls » |
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Obob
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Re: Apparent size with respect to position functio
« Reply #1 on: May 18th, 2009, 8:10pm » |
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I think you need to explain more about what the original problem is. Is the line finite in length? What do you mean by "apparent length"? Doesn't our stereoscopic vision just mean that apparent length is the same thing as the actual length?
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jarls
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Re: Apparent size with respect to position functio
« Reply #2 on: May 19th, 2009, 12:00am » |
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I should have said line segment. As an object moves farther from an observer it appears to become smaller. Analogously as a line segment moves farther from an observer it should appear to become shorter.
By apparent size I meant, if you were to set a up a piece of glass an arbitrary fixed distance away, the area an object seen through the glass would occupy if traced. I misstated the problem. The problem I thought of was the relationship between the apparent size of an object and distance from the observer. I treated an object as a collection of line segment and with the train track analysis, I think, determined the relationship between its apparent length and distance from the observer.
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| « Last Edit: May 19th, 2009, 12:00am by jarls » |
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towr
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Re: Apparent size with respect to position functio
« Reply #3 on: May 19th, 2009, 12:33am » |
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The apparent size is inversely proportional to the distance, because you see along a cone (or triangle in 2D)
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Something twice as far away occupies half as much of the field of vision.
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jarls
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Re: Apparent size with respect to position functio
« Reply #4 on: May 19th, 2009, 1:24am » |
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I see what you're saying. Eyes can be considered to be huge virtual funnels. If you cut the funnel at a certain distance everything entering the mouth of the funnel would have to be squeezed into the retina. Therefore Thus if the distance is great The mouth of the funnel is large and when squeezed into the eye everything will be small.
But would this not depend on the angle of the funnel tip (correlating to field of vision) not determine this relationship though? If that were the case, if I had a narrower field of vision than you would an object a certain distance away appear larger to me than to you?
I see your point. but I don't see how my solution is wrong.
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towr
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Re: Apparent size with respect to position functio
« Reply #5 on: May 19th, 2009, 1:51am » |
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on May 19th, 2009, 1:24am, jarls wrote:| But would this not depend on the angle of the funnel tip (correlating to field of vision) not determine this relationship though? |
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No, because whatever the angle, it remains a triangle/cone; and therefor something twice as far away seems half the size.
Quote:| If that were the case, if I had a narrower field of vision than you would an object a certain distance away appear larger to me than to you? |
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It would occupy a greater part of your field of vision, yes.
Quote:| I see your point. but I don't see how my solution is wrong. |
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I haven't looked closely yet at what you're doing. But for one, I haven't factored in the height of the observer, since I measure distances from the eye; you seem to measure distance along the ground.
Also, I'm not sure "Dividing the apparent distance between any two adjacent rungs by the distance between the preceding pair of rungs with always yield the same number" is true. Using a straight floor, and height h, the apparant distance (measured as angle) between rung n-1 and n is atan(n/h)-atan((n-1)/h), and [atan(n/h)-atan((n-1)/h)]/[atan((n-1/h))-atan((n-2)/h)] isn't constant. Although it's quite possible I'm mistaken about what you're doing because I haven't looked it it very closely.
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| « Last Edit: May 19th, 2009, 2:21am by towr » |
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Grimbal
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Re: Apparent size with respect to position functio
« Reply #6 on: May 19th, 2009, 2:28am » |
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I agree with towr.
The apparent size doesn't decrease exponentially as your assumption would imply. It decreases as 1/d where d is the distance from the eye.
If you are standing on rung 0 and you place a vertical stick on rung 1, then the 2nd rung will visually cross the stick at 1/2 your height, the 3rd crosses at 2/3 your heigth, etc, the nth rug will visually cross the stick at (1-1/n) of your height The apparent distance between rung n and n+1, as measured on the stick, is 1/n - 1/(n+1) = 1/(n(n+1)) of your size. The ratio between the apparent distance between the spaces before and after rung n is 1/(n(n+1)) / (1/((n-1)(n))) = (n-1)/(n+1) and is not constant.
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| « Last Edit: May 19th, 2009, 2:29am by Grimbal » |
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Obob
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Re: Apparent size with respect to position functio
« Reply #7 on: May 19th, 2009, 7:41am » |
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I still don't think I'm getting what apparent size is. Are we just saying that the apparent size is the proportion of the field of vision which is occupied by the object, so that it is a dimensionless quantity?
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towr
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Re: Apparent size with respect to position functio
« Reply #8 on: May 19th, 2009, 8:01am » |
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on May 19th, 2009, 7:41am, Obob wrote:| I still don't think I'm getting what apparent size is. Are we just saying that the apparent size is the proportion of the field of vision which is occupied by the object, so that it is a dimensionless quantity? |
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Jarls said: "By apparent size I meant, if you were to set a up a piece of glass an arbitrary fixed distance away, the area an object seen through the glass would occupy if traced."
So, apparant size is the size of the projection on a plane at a fixed distance. Of course the absolute measure will depend on where we place the projection plane. But for the relative (dimensionless) measure this wouldn't matter; there's a fixed scaling difference between any two choices of planes. So it's equivalent to the size of the projection on the retina, and also to the proportion of the field of vision it takes.
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jarls
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Re: Apparent size with respect to position functio
« Reply #9 on: May 19th, 2009, 1:56pm » |
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on May 19th, 2009, 2:28am, Grimbal wrote:I agree with towr.
The apparent size doesn't decrease exponentially as your assumption would imply. It decreases as 1/d where d is the distance from the eye.
If you are standing on rung 0 and you place a vertical stick on rung 1, then the 2nd rung will visually cross the stick at 1/2 your height, the 3rd crosses at 2/3 your heigth, etc, the nth rug will visually cross the stick at (1-1/n) of your height The apparent distance between rung n and n+1, as measured on the stick, is 1/n - 1/(n+1) = 1/(n(n+1)) of your size. The ratio between the apparent distance between the spaces before and after rung n is 1/(n(n+1)) / (1/((n-1)(n))) = (n-1)/(n+1) and is not constant.
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Imagine a glass-world. This is a world that would be created if all things seen through the glass were traced.
The properties of glass-world are different than the properties of the real world. Moving outward along a flat surface in the real world is moving up in glass-world. Maintaining constant velocity in glass-world does not correlate to maintaing constant velocity in glass-world. Moving out at constant velocity in the real world would correlate to moving up with a velocity that asymptotically approached zero.
In glass-world an infinite train track is a triangle with the same properties as a triangle is 2-dimensional euclidean space.
If a train track were to be traced, rungs and all, in order to determine the apparent distance between any two adjacent rungs could be determined by simply consulting the glass-world of this configuration.
The apparent dimensional properties of objects are identical to the real dimensional properties of objects is glass world. The triangle in glass-world with its exponential geometric properties would apply also to the apparent dimensions of an object.
To glass-world by height does not exist. This is because my height is a poperty of the real (not apparent) world.
My height is used only to determine the actual distance between my eyes and a rung that I'm looking at. In the actual world my actual height and the actual distance between the zeroth rung and the nth rung are the two legs of a new right triangle (looks like a fin pointing upward) and the actual distance between my eyes and the rung is question is the hypotenuse of this triangle. so one leg is my height, the other is the actual distance between two adjacent rungs multiplied by 'n' and the sum of the squares of these values is the square of the distance between my eyes and the rung in question.
In my equations 'h' is my height, 'c' is the actual distance between two adjacent rungs, 'n' is the number of rungs and 'x' is the distance between my eyes and the nth rung.
I am not saying that you are wrong. I'm just having a lot of trouble seeing why I'm wrong.
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| « Last Edit: May 19th, 2009, 2:13pm by jarls » |
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towr
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Re: Apparent size with respect to position functio  tracks.jpg
« Reply #10 on: May 19th, 2009, 2:22pm » |
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While a train track will form a triangle when you make a projection, the rungs won't be in geometric progression.
Just look at the attached picture; the ratio of distances between rungs aren't constant.
61/110 ~= 0.55, 14/18 ~= 0.78
And while there's bound to be some measuring error, it's not that big.
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rmsgrey
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Re: Apparent size with respect to position functio
« Reply #11 on: May 20th, 2009, 12:16pm » |
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on May 19th, 2009, 1:56pm, jarls wrote:| The properties of glass-world are different than the properties of the real world. Moving outward along a flat surface in the real world is moving up in glass-world. |
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Not always. In general, lines perpendicular to the glass in the real world will will appear to converge on the perpendicular that passes through the centre of projection (which itself projects to a point) as they move away. If you're in a long corridor, then lines on the ceiling will move downwards in glass world to meet the rising floor and the walls which are moving left and right...
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jarls
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Re: Apparent size with respect to position functio
« Reply #12 on: May 20th, 2009, 6:00pm » |
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on May 20th, 2009, 12:16pm, rmsgrey wrote:
Not always. In general, lines perpendicular to the glass in the real world will will appear to converge on the perpendicular that passes through the centre of projection (which itself projects to a point) as they move away. If you're in a long corridor, then lines on the ceiling will move downwards in glass world to meet the rising floor and the walls which are moving left and right... |
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You're right. I should have said "...could be moving up in the real world". All real world motion perpendicular to and away from the glass correlates to glass-world motion which converges to the eyes of the tracer.
More formally: All real world perpendicular (to the glass) motion will be in accordance with a radial gradient whose center is the projection of the eyes of the tracer. The glass-world gradient will be inward if the real-world motion is away from the glass and outward if the real-world motion is towards the glass.
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| « Last Edit: May 20th, 2009, 6:01pm by jarls » |
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