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Take a piece of rigid cardboard. Draw a perfect circle on it. Hold it up, and take a picture, with the cardboard held perpendicular to the direction we're looking. You get a photo that looks like this:

circle viewed from straight on

Notice: it looks like a perfect circle in the photograph.

Now tilt the cardboard to the right, or tilt it back, so we're no longer viewing it straight on:

viewed, after tilting to the right   viewed, after tilting back

Notice that in the photograph the black ink has the shape of an ellipse now, instead of a circle.

What if we tilt it to the right and then tilt it back?

viewed, after tilting to the left and also back

Visually, it still looks like an ellipse to me. Is it?

Conjecture. The shape of the black in in the photograph will always be a perfect ellipse, no matter what orientation the cardboard is held in.

Is this conjecture true? Can we prove it?

I think I can prove it is true if the cardboard is tilted to the left/right or front/back. However, I can't see how to prove it for a combination of those two operations.

If you don't to think about how cameras work, you can think of the problem like this: We stand facing a wall (which is perpendicular to the direction we're looking). We hold the cardboard in front of us in some orientation. Then, we project each speck of black ink onto the wall behind us, by tracing a line from our eye to the speck of ink and continuing until it hits the wall; then we draw a dot there on the wall. Consider the locus of points on the wall obtained in this way. What shape does this locus have? Is it always an ellipse?

Or, if you prefer: hold up a coin in a dark room. Shine a flashlight towards the coin. What is the shape of the shadow on the wall? Is it always an ellipse, no matter what orientation we hold the coin?

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    $\begingroup$ What do you know about projective transformations? It’s quite easy to show that the image of a conic under a planar homography (which is what you’re describing) is another conic. The type depends on the spatial relationships between the source plane, image plane, and viewpoint. $\endgroup$ – amd Feb 6 at 20:54
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    $\begingroup$ depending on viewing angle it's possible to see just a line segment or nothing at all $\endgroup$ – Vasya Feb 6 at 20:56
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    $\begingroup$ @amd, I currently know nothing about projective transformations, but that sounds like a perfect type of answer -- feel free to write an answer based on those concepts and I'll go read up on projective transformations to understand it. Thank you! $\endgroup$ – D.W. Feb 6 at 21:36
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    $\begingroup$ @timtfj, interesting. I'm afraid I can't see the difference in curvature with my eyes. Perhaps you could write up an answer to explain why? That sounds interesting. (For the flashlight, let's assume the flashlight is a point source of light, emitting light in all directions -- e.g., a tiny light bulb, not a laser.) $\endgroup$ – D.W. Feb 6 at 21:38
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    $\begingroup$ See my answer to nearly the same question here: math.stackexchange.com/questions/3102248/… $\endgroup$ – Aretino Feb 6 at 21:56
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This is a fairly standard result in computer vision and projective geometry: the image of any conic under a perspective transformation is another conic. If the image of a circle is a closed curve, then it must be an ellipse. Aretino gives a lovely classically-inspired proof of your particular case in this answer to a very closely-related question. I’ll offer a construction that uses machinery you’re more likely to see in the computer vision literature. It’s related to the argument I present in my answer to that same question that the intersection of any quadric surface with a plane is some sort of conic.

Without going into details of the construction, which you can find in any standard reference, the perspective projection in your question can be represented by a full-rank $3\times 4$ matrix $\mathtt P$: If $\mathbf X$ is the homogeneous coordinate vector of a point in the scene, then the homogeneous coordinates of the corresponding point in the image are $\mathtt P\mathbf X$. We impose a coordinate system on the plane of the conic via a $4\times 3$ matrix $\mathtt M$ that maps the homogeneous coordinates $\mathbf x$ of a point on the plane to the world point $\mathtt M\mathbf x$. This source plane and the image plane are then related by the homography (planar projective transformation) $\mathtt H=\mathtt{PM}$. If the camera doesn’t lie on the source plane, then $\mathtt H$ is invertible. (Otherwise, we’re viewing the paper edge-on, and it all collapses to a line segment.)

If we have a conic on the source plane given by the equation $\mathbf x^T\mathtt C\mathbf x=0$, its image is then the conic defined by the matrix $\mathtt H^{-T}\mathtt C\mathtt H^{-1}$, which can be verified by direct substitution. What kind of conic is it, though? Well, nondegenerate conics can be distinguished by the number of their real intersections with the line at infinity. The line at infinity in the image is the image under $\mathtt P$ of the camera’s principal plane: the plane normal to the camera’s axis on which the camera lies. So, if the original circle doesn’t intersect the principal plane, its image is an ellipse; if it intersects at one point (is tangent to the plane), it’s a parabola; and if the principal plane cuts the circle, its image is hyperbolic. A similar thing occurs with your flashlight beam: as you tilt the flashlight more and more, you eventually get a parabola and after that one lobe of a hyperbola.

It’s also possible to work directly in world coordinates to show that the image is an ellipse. Parameterize the circle as $\mathbf c+\mathbf u\cos t+\mathbf v\sin t$, with $\|\mathbf u\|=\|\mathbf v\|$ the radius of the circle and $\mathbf u\perp\mathbf v$. Its projection onto the image plane will end up being of the form $\mathbf c'+\mathbf u'\cos t+\mathbf v'\sin t$. Then, show that under the conditions of your question this parametric curve is an ellipse.

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The answer is yes: the intersection of an oblique circular cone with any plane intersecting all generatrices is always an ellipse. This was proven by Apollonius of Perga in his treatise on conic sections, circa 200 b.C.

I wrote here a (slightly modernised) version of his proof, in the simple case when a base diameter, perpendicular to the intersection between the plane and the base of the cone, forms with the vertex a plane perpendicular to the base.

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    $\begingroup$ Can you clarify why that is the correct geometrical setup? The elongation towards the viewer can be quite pronounced. I don't think what's actually seen is a conic section, but I can't quite see why. $\endgroup$ – timtfj Feb 6 at 22:17
  • $\begingroup$ I don't care about real cameras here, I'm sticking to the setup suggested in the two last periods of the question. In real cameras lens distortion can play a role, I suppose. $\endgroup$ – Aretino Feb 6 at 22:21
  • $\begingroup$ @timtfj You’ve projecting onto the visual sphere. The OP is projecting onto a flat image plane. $\endgroup$ – amd Feb 6 at 22:29
  • $\begingroup$ For an ellipse, the distances between the major axis and the centre of the front and back edges would be equal. But at the eye/lens/pinhole, those distances subtend different angles. There's something wrong with the geometry. $\endgroup$ – timtfj Feb 6 at 22:30
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    $\begingroup$ @timtfj +1 for pursuing this to the point of measuring the images. Perception and belief are difficult things to overcome. $\endgroup$ – amd Feb 7 at 0:25

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