A Geometry/Perspective Question

[nycplayer]

I am 100% sure that it is unsolvable given just 3 points, because the 4th point can be almost anywhere and create a valid perspective, even given a fixed aspect ratio.
This is because there are two parameters controlling the location of the 4th point - camera location AND viewing angle.

Hitchcock used a camera technique that illustrates this in his movie 'Vertigo'. He moves the camera back, while zooming in on his subject. The subject stays the same size in the frame, but the perspective changes wildly around him to create a feeling of disorientation.

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

If you could fix either one of these params, then with aspect ratio and 3 points, you might be able to solve for the other parameter, and thus the 4th point.

[ICantRaed]

That is 100% true if you wanted to find the point in 3D space - but not if you wanted to find the point in that photo (2D space). As I read the post, he is interested only in the 2D space and not the 3D space.

If the point were higher up, then the line connecting it on the bottom would be at a higher angle. If the point were further to the right or left, then the line connecting it would be at a different angle.

The only thing which would break that for 2D space is if there are more than 4 sides to it (we can already see there are at least 4) and that would then allow the angles at the corners to add up to more than 360 degrees - meaning it is no longer a trapezoid and instead an irregular shape.

So while with the information given to us in that photo (due to compression and the nature of diagonal lines on a grid of pixels) it is hard to give an exact location of the point - you can get close.

[ICantRaed]

I have to go out to dinner with my wife soon, but I will try to write up a full proof and accompanying Photoshop image with diagrams - but basically you connect the top right and the bottom left corners via a diagonal.

That breaks the image up into two triangles - the bottom one with its lower right angle and point hidden.

We know that a triangle has to have its three angles add up to 180 degrees. So from there and with the coordinate data we can get the lengths of all three sides of the top triangle and with that the angles as well.

From there, we can then get the two visible angles of the other triangle and also the length of the hypotenuse (the line we drew through the center) and from there calculate out the hidden angle and with all of the angles and the length of the hypotenuse, we can get the length of all sides.
Once you have that, you can then determine the hidden point.

But I still would say that it is easier/shorthand just to do my original suggestion of extending out the lines.

The only way that would fail is if in the 3D space which was observed to create the 2D image, the corner of the screen was folded up. That would then cause the issue I was stating before where this is no longer a 4 sided object, but instead something with more sides and therefore we can't find the point.

[MarkD]

edit: my bad. I was thinking of the problem wrong.

[ethan]

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please elaborate on the meaning of these pics???

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Given ONLY the coordinates of the three visible corners of the green screen, how can we calculate the coordinates of the lower right corner?

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It's the same envelope in both. The top left, bottom left and top right corners have the same pixel coordinates. The bottom right's different. Thus, the problem as stated isn't solvable.

Sorry, thought it'd be clearer.

[ICantRaed]

They are not the same pixels, not to mention that the line on the top edge of the top image is relatively flat, and the same top edge in the bottom image is clearly not flat.

Looking at the edges, even if you couldn't see the lower right corner, it is obvious that they are different pixel coordinates.

Are you even talking about 2D pixels?

[ethan]

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They are not the same pixels, not to mention that the line on the top edge of the top image is relatively flat, and the same top edge in the bottom image is clearly not flat.

Looking at the edges, even if you couldn't see the lower right corner, it is obvious that they are different pixel coordinates.

Are you even talking about 2D pixels?

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I held up an envelope and took a picture. No, the pixel coordinates aren't exactly the same. But they're as close as I was going to get, and (IMO) close enough to illustrate the concept. If it'll make you happier, rotate the bottom picture 2 degrees CCW. It'll make the left/top edges line up better, and the bottom right corner will still be way way off.

I'm sorta surprised there's still confusion.

[Siegmund]

ethan's pictures did a nice job of illustrating how to prove there is not a unique solution. They just didn't spell it out explicitly enough for some of you, I guess.

Let me spell it out with some actual numbers for y'all.

Picture a rigid rectangular "flag" free to rotate around the axis of a flagpole.

The lower left corner of the flag is the origin. The upper left corner of the flag is (0,0,1). The upper and lower right corners of the flag are (x,y,1) and (x,y,0) where x^2+y^2=aspectratio^2. For the sake of this example let's take a 2:1 aspect ratio, and say the outside edge of the flag lies somewhere on x^2+y^2=4.

Now place a camera at (3,0,0). Since (3,0,0), (x,y,0), and (0,3y/(3-x),0) are collinear, the camera will "see" a plane image of the flag, in the y-z plane with fixed corners at (0,0,0), (0,0,1), and projections of the movable corners at (0,3y/(3-x),0) and (0,3y/(3-x),3/(3-x)).

Given a picture of a flag showing the bottom right corner at (0,u,0), there are TWO (x,y) pairs such that x^2+y^2=4 and 3y/(3-x)=u, except in the rare instance of a "fully extended" flag.

BUT, when (3,0,0), (x1,y1,0), (0,u,0), and (x2,y2,0) are collinear, (3,0,0), (x1,y1,1), and (x2,y2,1) are not.

So, NO, the projections of three corners and the aspect ratio, and the position of the camera are not enough to always uniquely determing where the fourth corner should be projected. (Knowing all those things AND whether the flag is tilted toward or away the observer - in the original photo we can tell it must be 'farther away on the right' - IS enough. But in the original question, we weren't given the position of the camera either.)

[gaming_mouse]

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It's the same envelope in both. The top left, bottom left and top right corners have the same pixel coordinates. The bottom right's different. Thus, the problem as stated isn't solvable.

Sorry, thought it'd be clearer.

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Okay, now I'm really feeling retarded because I still don't see it. In the upper picture the top edge has positive slope. In the lower picture it has negative slope. So how can the points align with one another?

What am I missing?

[ethan]

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It's the same envelope in both. The top left, bottom left and top right corners have the same pixel coordinates. The bottom right's different. Thus, the problem as stated isn't solvable.

Sorry, thought it'd be clearer.

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Okay, now I'm really feeling retarded because I still don't see it. In the upper picture the top edge has positive slope. In the lower picture it has negative slope. So how can the points align with one another?

What am I missing?

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I meant for the top edge's slope to be more negative in both pictures - its rightmost end is about 1 pixel below its left, but the slope looks positive because of an optical illusion.

Would replacing the top one with this picture help?

Here's the bottom pic from my first post for easier comparison:


So, if we're given the bottom-left, top-left and top-right corners we don't know whether the bottom of the envelope will have a positive or negative (or zero) slope. Does that help?