Visual Plane
every picture states the point it is correct from
What survives

The conic a circle becomes

A circle photographed is an ellipse, or a parabola, or a hyperbola, and which one is decided by a single incidence: whether the circle reaches the plane through the eye parallel to the picture. Not the lens, not the tilt, not how far away it is. The discriminant of the image agrees with that one test at every point of a sweep, and at the crossing it is zero to 1e-13.
23 min read 17 figures What a projection destroysProjected, not constructedOne machine used twice

Worth reading first: The circle whose centre moves · What a projection destroys.

Everybody knows that a photographed circle is an ellipse. It is one of the few pieces of projective geometry that survives into ordinary speech: a plate seen at an angle is an oval, a coin on a table is an oval, the rim of a cup is an oval.

It is also not true, and the exception is not exotic. Stand in the middle of a traffic roundabout and photograph the kerb, and the picture contains a hyperbola. Not an ellipse so eccentric it looks like one; a genuine hyperbola, with two branches and two asymptotes, one branch below the horizon and its partner above it.

A 10.0 m circle on the ground, seen from outside itThe whole conic is drawn, including the part no camera can photograph. Every point of the circle is at least 3.96 m beyond the plane through the eye, so the image is an ellipse and the camera can see all of it. B² − 4AC = -1.86e-1.horizonthe eye stands 8.78 m from the centrecorrect from 26 cm, at 160 mm wideellipse · nearest point +3.96 m
Fig. 1 A 10.0 m circle painted on the ground, with the camera outside it. The whole conic is drawn — including any part of it no camera could photograph — and here there is none: every point of the circle is at least 3.96 m beyond the plane through the eye, so the picture is an ellipse and the camera sees all of it.

This essay is about what decides which. The answer turns out to be one incidence, it has nothing to do with the lens, and it is exactly the kind of statement projective geometry exists to make: a condition about meeting, checked without measuring anything.

What a projection does to a curve

Start with what does not change. A projection through a point takes lines to lines, so it takes a curve of degree two to a curve of degree two. A circle is a curve of degree two; its image is therefore a conic, always, without qualification.

That much is a statement about degree and it is not the interesting half. Ellipse, parabola and hyperbola are the same object as far as a projection is concerned — one conic, seen three ways — and they are told apart only by a piece of information a projection does not carry: where the line at infinity is.

In a picture, that line has a name. It is the horizon: the image of the ground plane’s points at infinity, and the place every family of horizontal parallels goes to meet.

A family of parallel ground lines at 52°, and where they meetAll five lines pass through one point on the horizon, off the edge of the frame at x = 2965. The point fitted from the drawn lines agrees with the one computed from the direction to 1e-11 px, and the fit's own residual is 1e-12 px.horizon — the image of the line at infinityvanishing point at x = 2965 — off the framecorrect from 26 cm, at 160 mm wide34° across
Fig. 2 The horizon, as the thing the classification is about. Five parallel ground lines at 52° meet at one point on it, and the point fitted from the drawn lines agrees with the one computed from the direction to arithmetic noise. Every direction in the ground plane has its own point on that line, which is what makes the horizon the image of a line rather than a decoration on the sky.

So the three names are three answers to one question: how does the image of the circle meet the horizon?

  • Miss it entirely, and the image is an ellipse.
  • Touch it once, and the image is a parabola.
  • Cut it twice, and the image is a hyperbola — and the two crossings are where the asymptotes’ directions come from.

That is the whole classification, and it is worth noticing what it does not mention. Not the focal length, not the angle of view, not the tilt of the plane, not the distance to the circle. The horizon is where the plane’s infinity went, and the question is whether the circle got there.

Where the circle goes to meet infinity

The horizon is the image of the points at infinity of the plane. Which points of the circle could possibly land on it?

A point images on the horizon when the ray from the eye to it is parallel to the ground — which for a point in the ground plane means the ray never gets there, and that means the point is infinitely far away. On a circle, no point is infinitely far away. So no point of the circle images on the horizon, and every circle is an ellipse.

That argument is wrong, and finding out where it is wrong is the content of this essay.

It is wrong because the image of a point is not defined by drawing a line from the eye to the point and seeing where it crosses the picture. It is defined by the ray direction, and a direction has an image whether or not the point in it is in front of the camera. The plane through the eye parallel to the picture — call it the vanishing plane, because its image is the vanishing line — is where the two accounts come apart. A point on it has a ray parallel to the picture, so it images infinitely far away in the picture; a point on the far side of it images in the ordinary way; a point on the near side images too, on the opposite side of the picture, which is a perfectly well-defined thing to do and is what a camera refuses to photograph.

So the question “does the image of the circle meet the horizon” has an answer in the world: does the circle reach the vanishing plane?

Three circles on one ground, and the three conics they drawThe same camera and the same ground. The only thing that differs between the rows is how far the nearest point of the circle is from the plane through the eye — 4.95 m, 0.37 m, -2.98 m — and that alone decides whether the picture is an ellipse, a parabola or a hyperbola.circlenearest point, past the eye planeB² − 4ACthe picture isradius 4.00 m, wholly beyond the eye+4.951 m-1.61e-1ellipseradius 8.62 m, just touching it+0.372 m-1.58e+0ellipseradius 12.00 m, crossing it-2.978 m1.65e-1hyperbolaone camera, 34° across, eye 8.78 m from the centreellipse · ellipse · hyperbola
Fig. 3 Three circles on one ground, seen by one camera. The only thing that differs between the rows is how far the nearest point of each circle is from the plane through the eye, and that alone decides whether the picture is an ellipse, a parabola or a hyperbola. The parabola’s row is the knife edge: its nearest point is a third of a metre past the plane and its discriminant is three orders of magnitude smaller than either neighbour’s.

Computing the image without photographing it

There is an obstacle to measuring any of this the obvious way, and it is worth naming because it shaped how the figures here are built.

The obvious way is to take a few hundred points on the circle, project each one, and fit a conic through the results. That is exactly what the essay on the circle whose centre moves does, and for a circle in front of the camera it is right and it is enough.

It cannot reach the case this essay is about. A camera refuses a point at or behind the vanishing plane — correctly, since there is no ray from the eye through such a point that crosses the picture in front of the eye — so a circle that crosses that plane comes back as a broken arc with the interesting part missing. A conic fitted to what is left is a conic fitted to an arc.

The repair is to stop sampling. A plane in the world carries a 3×33\times3 homography to the picture: three columns, being the two directions in the plane and the offset to its origin, all written in the camera’s own basis and multiplied by the calibration. A conic in that plane is a 3×33\times3 symmetric matrix. The image of the conic is then

C′=H−⊤C H−1,C' = H^{-\top} C\, H^{-1},

one matrix product, valid for every point of the circle including the ones no camera can see. Whether the result is an ellipse, a parabola or a hyperbola is the sign of B2−4ACB^2 - 4AC read off its coefficients.

Two checks keep that route honest, and both are made rather than assumed. The plane’s vanishing line, computed as the bottom row of H−1H^{-1}, has to be the camera’s own horizon — which is derived a completely different way, from the ground normal written in camera coordinates — and it is, to about 10−1410^{-14} of a pixel. And where the sampled route can be taken, the two must agree: sample a circle wholly in front of the camera, fit the conic, normalise both sets of coefficients, and they match to 1.1×10−161.1\times10^{-16}.

A circle on the ground, and the two points that get called its centreThe image of the centre and the centre of the image ellipse are 27.7px apart — 6.2% of the ellipse's own width. The third mark is constructed from the drawn ellipse and the horizon alone, with no access to the circle: it lands 6e-14 px from the image of the centre and 27.70 px from the ellipse's own.centre of the ellipseimage of the centrepole of the horizon — 6e-14 px awaycorrect from 22 cm, at 160 mm widepole 6e-14 px from the truth
Fig. 4 The sampled route, which this essay leaves behind and does not contradict. A circle in front of the camera, projected point by point, with the offset between the image of its centre and the centre of its image measured — and the pole of the horizon drawn, which is where the image of the centre actually is.

The sweep, and the knife edge

With the algebra in place the claim can be put to a test that could fail it.

Take one camera and one ground plane. Grow a circle centred where the camera is looking, from a radius small enough to sit entirely in front of the eye to one large enough to swallow it, and at each radius compute two things independently: the depth of the nearest point of the circle relative to the vanishing plane, and the discriminant of the image conic.

The discriminant of the image, as the circle grows past the eyeNegative is an ellipse and positive is a hyperbola, and the crossing is at 8.995 m, found by bisection rather than by landing on it: there the discriminant is 1.0e-13 and the picture is a parabola. The camera stands 8.78 m from the centre, so "the circle reaches the plane through the eye" and "the camera is inside the circle" differ by 2.4% here — the gap is the camera's 7.7° of downward tilt, and it closes to nothing for a level camera.-4-20251015radius of the circle on the ground (metres)B² − 4AC of the image, scaled by the coefficientsthe eye planeellipsehyperbolaeye 8.78 m from the centre, 34° acrossparabola at 9.00 m, to 1e-13
Fig. 5 The discriminant across the sweep. Negative is an ellipse, positive is a hyperbola, and the crossing is where the nearest point of the circle reaches the plane through the eye. The crossing radius is found by bisection rather than by landing on it, because a parabola is one value out of a continuum and a sweep that happened to include it would be reporting its own step size.

At every sample the two agree. That is the first half, and on its own it is not worth much: a sweep that stayed on one side of the plane would have every sample agree and would show nothing. So the sweep is required to visit both types, which it does.

The knife edge is the part worth the arithmetic. The parabola is a single value of the radius out of a continuum, so it is found by bisecting on the depth of the nearest point — ninety halvings, driven entirely by the geometric condition — and only then is the discriminant computed there. It comes out at 10−1310^{-13} of the coefficients’ own size. The geometric test located the parabola without ever looking at the algebra, and the algebra agreed.

There is a small honest discrepancy in that figure and it is worth the sentence. The crossing is at a radius of 8.995 m and the camera stands 8.78 m from the centre of the circle, a difference of 2.4%. “The circle reaches the plane through the eye” and “the camera is inside the circle” are the same condition only for a level camera; this one is pointed down by 7.7°, which tilts the vanishing plane, and the gap is exactly that. For a level camera it closes to nothing.

Standing inside it

The hyperbola is worth drawing, because the picture is not what most readers expect from the words.

A 24.0 m circle on the ground, seen from inside itThe whole conic is drawn, including the part no camera can photograph. The nearest point of the circle is 2.98 m behind the plane through the eye, so the image is a hyperbola: one branch below the horizon, its partner above, and the two asymptotes are the images of the two points where the circle crosses that plane. B² − 4AC = 1.65e-1.horizonthe eye stands 8.78 m from the centrecorrect from 26 cm, at 160 mm widehyperbola · nearest point -2.98 m
Fig. 6 A 24.0 m circle with the camera standing inside it. The nearest point of the circle is 2.98 m behind the plane through the eye, so the image is a hyperbola: the branch below the horizon is the part of the kerb in front of the camera, and the branch above it is the part behind — drawn where its rays’ directions say it goes, which is a place no photograph shows.

The branch below the horizon is what the camera actually records: the near arc of the kerb, curving away. The branch above is the far arc, the part behind the photographer, imaged at the place its ray directions land — which is above the horizon and, in a real photograph, simply absent. The asymptotes are the images of the two points where the circle crosses the vanishing plane, which are exactly the two points whose rays are parallel to the picture.

So the everyday statement is nearly right and its exception is the ordinary case of standing in the middle of something round. The ellipse is the view from outside; the hyperbola is the view from inside; and the parabola is the moment of stepping over the kerb.

What the type is, and is not, a fact about

Three consequences follow that are worth stating separately, because each is a thing readers reliably attribute to the wrong parameter.

It is not the lens. Change the focal length and the picture scales about the principal point; a scaling cannot turn an ellipse into a hyperbola, because it cannot move a curve across the line at infinity. A wide lens makes an oval more eccentric and a long lens makes it rounder, and neither changes the name.

A 10.0 m circle on the ground, seen from outside itThe whole conic is drawn, including the part no camera can photograph. Every point of the circle is at least 3.96 m beyond the plane through the eye, so the image is an ellipse and the camera can see all of it. B² − 4AC = -1.86e-1.horizonthe eye stands 8.78 m from the centrecorrect from 11 cm, at 160 mm wideellipse · nearest point +3.96 m
Fig. 7 The same circle from the same eye at a 74° field of view instead of 34°. Everything about the drawing changes and the classification does not: the nearest point of the circle is where it was, so the picture is still an ellipse and the discriminant still has the sign it had.

It is not the tilt of the circle’s plane. A circle standing upright on a wall, a circle lying on the floor and a circle leaning at 40° all obey the same rule. What matters is whether the circle reaches the vanishing plane, and a circle can do that from any orientation.

And it is not distance in the ordinary sense. A small circle a long way off and a huge circle nearby are both ellipses if both are wholly beyond the eye plane. The relevant distance is signed and is measured to one particular plane, not to the eye.

Seven identical spheres across a 84° frameThe outer sphere images 27% wider than the central one. That is what a correct rectilinear projection does, and it vanishes if the picture is viewed from 9 cm.54 px69 px84° across27% wider at the edge
Fig. 8 The neighbouring case, from a field where it does depend on the lens. A sphere images as an ellipse whose eccentricity grows toward the edge of a wide frame — a shape effect, continuous in the field of view, and a completely different phenomenon from the one this essay is about even though both are described as “circles becoming ovals”.

Where the conic’s centre went

Once the image is a conic in its own right, a second question becomes askable: what happened to the circle’s centre?

The image of the centre is not the centre of the image. That is the older result and it has its own essay; what is worth adding here is where the image of the centre actually is, in the language of the conic.

It is the pole of the horizon. The circle’s centre is the pole of the line at infinity with respect to the circle, and a projection carries poles to poles and polars to polars, so the image of the centre is the pole of the image of the line at infinity — which is the horizon. That relation holds for all three types, and it is the reason the classification and the centre question are the same subject.

The centre offset against distance, for two circle sizesThe offset is largest for a near, large circle and never reaches zero until the circle's plane is parallel to the picture.0501003456distance from the eye to the circle (m)offset between the two centres (% of the ellipse's width)r = 0.60 mr = 1.20 mr = 2.00 mr = 3.00 mr = 4.00 mmeasured from fitted ellipses9.9% at 2.5 m
Fig. 9 How far apart the two centres are, as a function of the circle’s size, at one camera. The offset is not a small correction that vanishes with care: it grows with the circle, because the near half of the circle is magnified more than the far half.
A circle on the ground, and the two points that get called its centreThe image of the centre and the centre of the image ellipse are 9.6px apart — 3.0% of the ellipse's own width. The third mark is constructed from the drawn ellipse and the horizon alone, with no access to the circle: it lands 6e-14 px from the image of the centre and 9.55 px from the ellipse's own.centre of the ellipseimage of the centrepole of the horizon — 6e-14 px awaycorrect from 22 cm, at 160 mm widepole 6e-14 px from the truth
Fig. 10 The same construction with the circle further away. The gap shrinks and does not close: the pole of the horizon is still the image of the centre, and it is still not the centre of the drawn curve.

The two points that are always there

There is a second intersection with the horizon that has nothing to do with the classification and is worth naming here, because it is where the next rung of this ladder starts.

Every circle in a plane passes through two fixed points of that plane’s line at infinity — the circular points, which are complex. They are on the line at infinity, so their images are on the horizon; they are on every circle, so their images are on the image of every circle. Intersect an imaged circle with the horizon and the pair comes back, with no fitting and no choice.

When the image is an ellipse those two intersections are complex conjugates, which is the ordinary case and is the whole metric upgrade in one line. When the image is a hyperbola they are real — which is exactly the case this essay has been drawing, and it means something has been lost.

Two points, and everything metric followsThe imaged circular points are where the horizon meets the image of any circle in the plane, and they are a conjugate pair — the first coordinate here is 169.5 − 446.0i. A rectification built from them and nothing else returns the world's angles to 2.0e-13° and its length ratios to 2.9e-15, and no length at all.horizonthe horizon does not cut the circle — the pair is complexrectified from the two points aloneangles: 2.0e-13°ratios: 2.9e-15length: —circle of radius 2.40 ma dash is a quantity two points cannot buy
Fig. 11 The two points a picture hides, in the case where they are complex: the imaged circle misses the horizon, the intersection has no real solution, and the pair that comes back is the image of the circular points. That pair is what a rectification is computed from.
Projective, affine, metric — what each stage buysThe photograph fixes the plane only up to a projectivity: the midpoint of a receding side lands 0.3970 of the way along. Supplying the plane's vanishing line buys the midpoint back exactly and nothing else. Supplying the image of one circle buys the last three numbers, at which point the right angle is 90.000° and two equal sides measure 1.000000. The cross-ratio is 1.333333 in all three, because it was never lost.projectiveaffinemetricmidpointtwo equal sidesa right anglecross-ratioprojective———1.333333333affine0.500000——1.333333333metric0.5000001.00000090.000°1.333333333— means the stage does not determine it at allcross-ratio 1.333333 throughout
Fig. 12 And what they buy. The chain from a projective picture to a metric one, with each rung’s price marked — the circular points are the last step, and the reason a photographed circle is worth so much more than a photographed square.

Reading the type off a photograph

The classification is not only true, it is checkable by a reader with a picture and a straightedge, which is the test any statement here is asked to pass.

Find the horizon. Two families of parallel ground lines give two vanishing points, and the line through them is the horizon. Any photograph with a floor and a couple of rectangular objects supplies it.

Extend the drawn curve toward that line. If the curve closes without reaching it, the picture holds an ellipse. If it runs off toward the horizon and appears to approach a straight line as it goes, the picture holds a hyperbola and that line is an asymptote.

And check by walking. A hyperbola on the picture means the eye was inside the circle when the picture was taken, allowing for tilt. That is a fact about where the photographer stood, recovered from the shape of a curve — which is the same sort of trade this site makes everywhere: something about the picture in exchange for something about the room.

A box drawn from a known camera, and the camera recovered from the drawingThree vanishing points found from the twelve drawn edges alone give back the focal length to 4e-15 relative.recovered principal pointused to drawrecoveredgapfocal length812.77812.774e-15principal x345.0345.02e-12angle46.0°46.0°—correct from 19 cm, at 160 mm wide46° across
Fig. 13 The trade in its most familiar form: three vanishing points found in a picture’s own edges give back the focal length and the principal point of the camera that made it, with nothing about the scene supplied. The classification above is the same kind of statement made about a curve instead of a bundle of edges.
The vanishing point runs to infinity and the measurement does not careAs the camera comes level the vertical vanishing point leaves the canvas, the page and eventually the plausible — 7.2 × 10⁹ px at a tilt of one part in eight million. The recovered height stays exact to 2e-16 relative the whole way. At exactly level the method has nothing to work with and refuses.0510-4-20how far the camera looks down, over eight metres (metres, log scale)where the vertical vanishing point falls (log₁₀ pixels)the vanishing pointthe error in the recovered heightthe error curve is offset by 17 decades to be visiblea flat line at machine precision
Fig. 14 And the practical caution that goes with it. As a direction approaches parallel to the picture plane its vanishing point runs away, so a horizon found from two nearly-parallel families is found from an intersection nobody can locate. The same fragility sits under an asymptote read by eye.

What is settled and what is not

The type of the image is decided by one incidence and nothing else. That is a complete answer to a question the everyday statement gets nearly right, and it costs one signed distance to check.

What it does not settle is the circle. Knowing that a photographed curve is the image of some circle, and even knowing exactly which conic it is, leaves the circle itself undetermined — its size, its distance and even the tilt of its plane. Five marks on a photograph fix the conic completely, with nothing left over; the next essays are about how much they leave free after that, and the answer is more than it looks.

Five marks fix the conic, and the sixth is a predictionFive marks on a photographed circle determine one conic — five points and five coefficients, with no fitting left over. The sixth mark was withheld from the fit and the conic passes 1.9e-13 px from it. Nothing about the camera, the circle's size or the plane it lies in was used.horizon25withheldfive marks fitted, one withheldcorrect from 26 cm, at 160 mm widethe withheld mark is 2e-13 px off the fitted conic
Fig. 15 Where that goes. Five marks on a photographed circle determine one conic exactly, and the sixth mark — withheld from the fit — lands on it. What the five do not determine is which circle in the world drew them.
A family of parallel ground lines at 14°, and where they meetAll five lines pass through one point on the horizon, off the edge of the frame at x = 813. The point fitted from the drawn lines agrees with the one computed from the direction to 7e-13 px, and the fit's own residual is 4e-13 px.horizon — the image of the line at infinityvanishing point at x = 813 — off the framecorrect from 20 cm, at 160 mm wide44° across
Fig. 16 And the fragility that goes with reading a horizon off a picture. A direction nearly parallel to the picture plane sends its vanishing point far off the canvas, so the line through two such points is found from an intersection nobody can locate accurately.
Four constructions, three of them the same mapA shadow, a floor anamorph, a mirror and a rectification, each decomposed into its fixed points and lines. Three are central collineations with a line of fixed points; the fourth is not, and that is the difference between changing a picture and changing where it is seen from.constructionfixed structurea shadow, ground to floorhomology · ratio 0.6719a floor anamorphhomology · ratio -1.4815a mirror in a vertical planehomology · ratio -1.0000a rectificationgeneral · three fixed points3 of 4 are centrala line of fixed points is what they share
Fig. 17 And the wider census this belongs to. The maps between a plane and a picture of it, sorted by the structure they leave fixed — the frame the classification above is a small corner of.

Where this goes next

centre of the ellipseimage of the centrecorrect from 22 cm, at 160 mm wideoffset 21.1 px The circle whose centre moves The image of a circle is an ellipse, and the image of the circle's centre is not the centre of that ellipse. The gap is 3 or 4% of the ellipse's width in an ordinary view — large enough to see, small enough to be dismissed as a slip by anyone not expecting it, and the reason drawn wheels look wrong. centre of the ellipseimage of the centrepole of the horizon — 1e-13 px awaycorrect from 22 cm, at 160 mm widepole 1e-13 px from the truth The centre, got back out of the picture The image of a circle's centre is not the centre of the image ellipse — this site measured the gap four phases ago and left the obvious question unanswered. It has an exact answer, it needs a straightedge and the horizon, and it is one sentence: the image of the centre is the pole of the vanishing line with respect to the image conic. horizon — the image of the line at infinityvanishing point at x = 1464 — off the framecorrect from 26 cm, at 160 mm wide34° across Where parallel lines meet They meet at a point that is not in the plane, and the horizon is the line made of all such points. Treating that as bookkeeping misses what it buys — a vanishing point becomes an ordinary point of the picture, findable from the drawn lines, checkable, and enterable into any calculation the others are.

Reads more easily once this is understood

Essays that name this one as worth reading first.

Shares its objects with

Essays that name at least two of the same things, and that neither author linked.

Named objects

A flat tag is an object no other essay names yet.

centre of projectionCircleConicDiscriminantEllipseHorizonHyperbolaIncidenceParabolaPicture planepoint at infinityProjective mapVanishing line