It's all in the title. I'm staring at a tree as I'm walking down the sidewalk. The tree appears to be about 4 inches tall. I walk twenty steps closer, and it looks 5 inches tall. I walk another twenty steps, and now it looks 7 inches. Its appearance is, perceptively and to me, growing on a steepening curve.
Why is this?
It's something I noticed today so I felt like posting here to get an answer. I'm guessing it has to do with our depth perception.
This really made me think, thanks lol. I think it has to do with how what we can see is kind of like in a cone. You can see this by holding up two things up in front of you, moving them to your sides until you can't see them anymore, and moving them away from you until you can see them again. Imagine that cone as a V with your eye being the point. Now Imagine a tree coming closer and closer to that point, the space between the tree and the two sides becomes acceleratingly smaller as the tree dominates your vision. Another way I thought to think about it is that as you get closer to the tree, you get closer to every point of the tree as the detail grows everywhere. To you, the tree is kind of like expanding everywhere as your perception expands into its detail, which would cause it to look like its growth in size accelerates (I'm not sure if exponentially is the word).
I'd never thought of this explicitly before so I did some maf.
The simplest way to think of this is to imagine that your eye is dead level with the surface of a table at the edge (so that the table takes up all of the bottom half of your vision), and there is something like an upright pencil moving towards you.
If you think about it, what you're asking is how much space in your vision the pencil takes up; that is to say, if you draw a line from the bottom of the pencil to your pupil, and likewise from the top of the pencil to the same place on your pupil, what will the angle be between the two lines? This angle essentially is what goes into your brain; when we say an object looks longer, this means the object is taking up a longer stretch of our field of view, and so the angle is larger. A close, small object will appear the same 'size', in the sense of your question, as a far large object, if this angle is the same.
Anyway, all you need for this question is some basic trig. Say the pencil has a constant height h, and the base of the pencil is a distance x from your pupil. Considering also the aforementioned line from the top of the pencil to your pupil, it's easy to see that what we have here is just a right-angled triangle, and all we're looking for is the angle at your eye, which we'll call @. By the definition of tan, tan@ = h/x. Inverting both sides, we have @ = arctan(h/x), which is what we were looking for.
I was interested so I made a lil' graph of what this looks like:
http://img534.imageshack.us/img534/2129/visioneu.png
The horizontal axis is distance of the pencil and the vertical axis is the perceived angle, or 'size'. You can see that far on the right, at greatest distances, the angle is almost zero, but as it gets closer and closer to your eye, and the distance gets closer and closer to zero, the angle increases faster and faster until it finally hits 90 degrees at zero.
It's not really exponential (an exponential graph would grow forever), but superficially it's quite like it... I believe this explains everything you asked about. :)
Thanks Xei, enjoyed learning from that, I'm actually in the middle of taking Trig right now. I knew there was a good explanation, I just thought about it from scratch lol.
I think Xei pretty much covered it, but...
In art, it's all about one or two point perspective. If you were looking at an object head on, and if the back edge could stretch until the end of the horizon, it would grow smaller and smaller until it disappears at exactly one vanishing point on the horizon line, ie. one point perspective.
If you were looking at an object at one of it's corners, the two adjacent sides facing you would stretch to the horizon at exactly two vanishing points.
If real life, that would never happen, as every object has a finite size, yet the points where the object would normally vanish are still there. So, as you walk toward an object, the vanishing point(s) move away from you as you walk toward it and the horizon line. As the vanishing points move away, the object would "grow" larger proportionally to the distance you have moved and to the size of the object. If you move backward from the object, the object would "shrink" smaller. :)
PS. If you Google "point perspective", you could find some good images of that, and wikipedia may be able to tell you more.
Honey if you think that's mind blowing perhaps you should do a little researching into d.m.T.Quote:
Originally Posted by Puffin
Cool, thanks for the answers!
I'm in grade 11 so I'm not extremely knowledgeable about this type of stuff. :P
Mikeac pretty much covered it from the art angle - which is what I was gonna say - but just to take it a little further, here is a sketch of Goku fighting Cell, that I did for my perspective class:
http://i.imgur.com/aqxvrl.jpg
Even though this is drawn on a flat paper, it's made to mimic the perspective of a person (or object) in 3-D space. Even though the arena floor that they are fighting on is a flat plane, in which the grid work on the ground is (maybe not perfectly) symmetrical, the squares beneath their feet get bigger, as they come toward "us." The pillars on the sides of the arena are roughly the same size, however, due to their position in space, the ones that are closer seem larger, and the ones that are further away seem smaller. Even the two fighters in the middle. They are standing on the same plane, but due to the off-center angle at which we are observing them, Cell looks noticeably bigger. It has to do with the 'sides' of every object in our view trailing off toward the horizon - toward the vanishing points, as Mikeac said. The farther away from you that you are looking, the more 'space' you can fit into your field of view. Because of this, the objects in that field of view must become smaller, since more is being perceived in your peripherals.
Xei is right about the angle.
Though really, distances (as well as sheer sizes) are LINEAR, not exponential. Otherwise, if you stood on parallel railroad tracks, they wouldn't look like two straight lines heading into the horizon (meeting at infinite distance), but two curved lines, due to the exponentialness. But for the reason Xei showed, as an object gets closer (say the train on the tracks :P), the angle from its left to right (or top to bottom, etc.) increases and it appears to grow faster. If I remember right, the size of an object is inversely proportional to the distance (size = (some constant) * (actual size) / (distance from observer)), which if course, is not a straight line, and shoots up as distance goes to wards zero.
I'm not sure what d.m.t. is...but if you're referencing the psychedelic compound N,N-dimethyltryptamine, then you can just call it DMT. Just like LSD isn't l.s.d. ...also probably shouldn't encourage an 11th grader to "research" one of the most mind-altering substances known to man. With that said...I think I might do a bit of research myself this evening. :shadewink:Quote:
Originally Posted by Seroquel
There's something I don't understand about vanishing point art (with respects to using one, two, three, four, etc. vanishing points): which one is correct (in the sense that it matches human perception, or a photograph)? Or are they equivalent?
When you say size here, I suppose you are comparing the object to some standard (such as a metre stick one metre away), as opposed to the absolute measure of angle?Quote:
Originally Posted by Replicon
I have thought a lot about this... Even to the point of wondering if size is possibly related to distance, in a multidimensional sense. I am still coming to a conclusion.
This is just projective geometry. To get projecive 3-space, take euclidian 4-space and pick an orthonormal basis so we can write our coordinates (x, y, z, w). Consider the set of all lines through the origin and the 3-subspace W defined by w= 1. If the line passes through W, then we can write the equation of the line as x = aw, y = bw, z = cw. It will intersect the plane W at the point (x, y, z, w)/w = (x/w, y/w, z/w, 1) = (a, b, c, 1). For a line not through W, w = 0 for all points on the line (otherwise, it would vary continuously and assume the value 1 at some point). There is precisely one line for every direction in the plane w = 0. These are the "vanishing points"/"points at infinity". We think of it as adjoining a "2-subspace at infinity". So there is one "vanishing point"/"point at infinity" for every direction in 3-space. Obviously, this generalizes to n-space where we can get projective n-space by going through the above construction with (n+1)--space resulting in adjoining an "(n-1)-subspace at infinity". Obviously, there's a lot of handwaving there but it should get you started.Quote:
Originally Posted by Xei
Human vision is really just the projection of a 3 dimensional world onto a 2 dimensional surface. Wikipedia has a short article (kinda) on the subject.
Actually I'm fairly certain the function you've got there can be written as an exponential decline for x in the interval [0, ∞), which is the only interval that makes sense in this context.Quote:
Originally Posted by Xei
edit: Never mind the last part. I was wrong.
Why?Quote:
Originally Posted by khh
Because it's late at night and I decided to rely on what my eyes told me. After actually tinkering a bit with it, I find I was wrong.Quote:
Originally Posted by Xei
If I was right then it should be possible to write it like this:
f(x) = k e-(x t)
where k is (lim x -> 0+) arctan(h / x) and t is the solution for x of the equation arctan(h / x) = k(1 - e-1). When I actually tried doing that I got this graph (The blue line is f(x) and the red line is arctan(1 / x)). It clearly shows I was wrong.
Ya, I'm doing a maths degree so I have the advantage of 'just knowing' from experience that those functions are of fundamentally different types, and also the shape of exp would go to 0 much faster. The quickest way to check IMO is to just compare the values at 0, 1, and 2; they should decrease by the same ratio. You raised a weird question for me, though, which is how to know, in general, if two functions are equivalent. I've never thought about that before, and it's kinda weird to me... Ima do some reading.
That would make sense if I started out with the intention "I'm gonna test if these are equal". But I didn't. I started out with the intention "I'm gonna show this ragamuffin I'm right". :cheeky:Quote:
Originally Posted by Xei
(I really should have known better, though. We've argued math before and I have yet to win)
I don't believe there's any way of mathematically proving two functions are the same other than to just show the conversion from one form to the other in small, recognizable steps. Of course, proving them not to be equal can be much simpler. But if you find something interesting on the topic, please post it :)Quote:
Originally Posted by Xei
Hah, oh, I see. I thought the graph was quite an elaborate method! :PQuote:
Originally Posted by khh
Hm well, I just had one of those weird kind of moments were I realised I'd never tried to do this before, and had been using some shaky assumptions.Quote:
I don't believe there's any way of mathematically proving two functions are the same other than to just show the conversion from one form to the other in small, recognizable steps. Of course, proving them not to be equal can be much simpler. But if you find something interesting on the topic, please post it :)
I thought disproving would be easy too, but then I realised this ends up involving, as a random example, showing that arctan(1/2) is not equal to e^-2 or something; and I've been subliminally trained to treat calculators as abhorrent, so it's a lot harder than it seems.
The shaky assumptions I realised I had were that certain functions are simply fundamentally different from others, and so equivalences between them, or functions that involve them, can be dismissed; this only works if you know all of the important relationships between functions (which I'm sure I do by now), and are hence able to see when there is incompatibility. For instance, I 'know' that for real x, there are no identities linking trig, exponential, or algebraic functions. That's why I could discount your relationship.
However, for something like
f(x) = cos(x)
g(x) = 1/2*( exp(ix) + exp(-ix) )
for complex x, it's not clear at all that the functions are identical (they are), and if you hadn't learn Euler's formula you'd probably think they were totally different, and discount this relationship on the same basis as me. It's never actually been proven to me that I can't turn a trig into an exponential or whatever, it's just that it'd be so fundamental that I would definitely know it.
Anyway, I think the proper, general basis for this kind of thing is Taylor series, which at a higher level are actually usually said to define exp, trig, etc. (rather than being derived). If you determine the Taylor series of two given functions and compare coefficients, that'll determine it for you, as Taylor series are unique.
Well, if it gives you pleasure, your little mistake opened up a whole can of worms for me and I've spent quite a long time Googling stuff now. The above is a basic answer, but then there's the problem of the fact that
- The problem of checking coefficients may well be no easier than checking values of the function in the first place, as you may get two different looking formulae for the coefficients. If your functions are the same, no algorithm will ever be able to tell you, because it'll be checking forever.
- You may also have to check infinite intervals of the function because some Taylor series only work for finite distances from a point, and I've just discovered that there are even functions where the Taylor series only converges for the point about which it is taken, along with a host of other analysis concepts I've never seen before...
Right, bed now.
I always thought it was because more light was getting to you faster from closer objects and so you would take in more light from closer objects and therefore more detail leaving less room for far objects in your eyes before you had to blink. But that's probably incorrect.
Ah, those moments are fun.Quote:
Originally Posted by Xei
Actually that there is one of the main reasons I thought as I did. The first thing I tried when I wanted to show I could be right was to start messing about with (exp(i h / x) - exp(-i h / x) )/( exp(i h / x) + exp(-i h / x) ) with the assumptions that h and x are always positive, real numbers. Needless to say, I didn't get anywhere I wanted with it.Quote:
Originally Posted by Xei
Ah, that makes sense.Quote:
Originally Posted by Xei
It does. Cool :D And thanks for taking the time to write that reply.Quote:
Originally Posted by Xei
edit:
Yeah, that's incorrect for several reasons.Quote:
Originally Posted by ooflendoodle
1. If less light was coming to your eyes, the object would not seem smaller it would seem dimmer. If this was correct, turning the dimmer on a light bulb would make it seem smaller.
2. There is a steady stream of light being reflected of objects, so (assuming as much light is being sent out in our direction) the same amount of light will hit your eyes from a nearby object as from a distant object. But the light from the distant object will have been emitted earlier, so the more distant an object you look at the longer back in time you see. When you look up at night, the image you see of the moon is not the moon as it is now, but the moon as it was roughly 12 seconds ago. When you look at the sun, that's the sun 8 minutes ago. And with the stars it's been years (four for the closest). But this doesn't affect how bright they seem.
Showing that two functions are different is normally pretty easy. Take two functions f(x), g(x). Then if they're the same, f' = g' and the integral of f-g over any arbitrary interval where they're both defined would be zero. Hence a failure for either of these two conditions suffices to prove that they're distinct. The converse is not true in either case though without certain assumptions on f and g.
Here we can assume equality for some t, k, and h and take derivatives to get (Hi Slash. How's the LaTeX installation coming?)
tke-xt = h/(x2 + h2)
Then multiply across by [(x2 + h2)/tk]ext to get
x2 + h2 = (h/tk)ext
and differentiate twice, throwing away constants to conclude that we must have t such that
ext = 0
for all t. That's ridiculous enough for me to conclude that they're not the same.
I think the number of vanishing points determines the apparent FOV (field of view), with more vanishing points simulating wider FOVs.Quote:
Originally Posted by Xei
Termespheres- Art that captures the up, down and all around visual world from one revolving point in space
I had a whole bunch of shit typed up, but then googled perspective and this came up, which better explained what I was gonna say.
I'm not sure if PhilStoned answered your question, coz I don't understand what he wrote.Quote:
Originally Posted by Xei
But basically, they are all correct.
You just use them for different angles.
One point is for drawing objects from pretty much any direction/angle that has a side of the object facing you.
Two point is when you want to have a horizon line and you have to have one of the corners of the object facing you. For example, looking at the corner of a building.
Three point is when you want to have the object appear to be "distorted" from every angle.
For example, looking at a building from the horizon line, showing the side's distortion and also the top's distortion. (By distortion I just mean getting smaller).
Basically, you add a point for every plane which needs to appear as if it's going in to the distance.
Four point would be if you wanted to be looking at the middle of the building, from a corner, so it's getting smaller on all planes.