find vanishing line (horizon) in image without parallel lines

Question

I have the image of a vase from which I would like to extract the horizon of the plane on which the vase lays.

How could I achieve this supposing the vase is a right cone and I can extract the two ellipses corresponding to two circular cross sections of the cone?

Usually to find the horizon in a scene you have to find two pairs of parallel lines and compute their intersection, finding two vanishing points and computing the line through them. But in this scene we have no parallel lines :(

Answer 1

The following figure shows a cone, along with two ellipses that correspond to cross sections of the cone. Some initial remarks:

• the cone need not be right circular, and the sections need not be circular sections.
• the diagram can be interpreted in either two dimensions or three.
• in 2D, we see two ellipses $c,d$ and two common tangents that meet at point $X$.
• in 3D, we see the cone with apex $X$ and two planar sections corresponding to planes $P_c,P_d$.
• we'll show how to construct the line $EF$, which corresponds to the intersection of planes $P_c,P_d$.
• in 2D the line $EF$ is known as a common chord. Common chords are to conics as the radical axis is to two circles.
• in the special case of a right circular cone and circular sections, $EF$ is the horizon (aka vanishing line) sought in the OP.

The construction works as follows.

1. Select four points $A,B,C,D$ on ellipse $d$. Then the four points $A',B',C',D'$ are the respective intersections of ellipse $c$ with the lines $AX,BX,CX,DX$. (The line $AX$ intersects $c$ in two points, a "near" point and a "far" point. We choose the point $A'$ depending on whether $A$ is "near" or "far". Same for $B',C',D'$)
2. Let $E$ be the intersection of lines $AB,A'B'$ and $F$ be the intersection of lines $CD,C'D'$.
3. Then line $EF$ is the intersection of planes $P_c,P_d$.

If the cone and its sections are circular, then $P_c,P_d$ are parallel to each other and the base of the cone. Then $AB$ and $A'B'$ are parallel (in 3D, not 2D) and $E$ is the vanishing point in the direction of $AB$. Similarly $F$ is the vanishing point in direction $CD$. So $EF$ represents the horizon of $P_c, P_d,$ and the base of the cone.

Remark 1: We have constructed $EF$ assuming that we are looking at both $c,d$ from above, and that e.g. $A$ and $A'$ are on the far side of the cone. But we can also imagine that we are looking at $c$ from below, in which case the $A'$ in the figure is on the near side of the cone. In this case we should have chosen the other intersection of $AX$ with $c$. In fact we would get four different points $A',B',C',D'$ and end up with a different horizon line that is between $c$ and $d$. (left as an exercise for the interested reader). This would be the second common chord for $c,d$. (Fun fact: two circles also have two radical axes - the one described in geometry textbooks, and also the line at infinity).

Remark 2: Those familiar with Desargues' theorem will recognize an alternative construction. We can get away with three points $A,B,C$ on $c$ and their corresponding points $A',B',C'$ on $d$. These define two triangles $ABC, A'B'C'$ that are in perspective in 2D (and 3D). By Desargues' Theorem, corresponding sides intersect on a common line which is the sought horizon.