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A Minimal Closed-form Solution for the Perspective Three Orthogonal Angles (P3oA) Problem: Application To Visual Odometry

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Abstract

We provide a simple closed-form solution to the Perspective three orthogonal angles (P3oA) problem: given the projection of three orthogonal lines in a calibrated camera, find their 3D directions. Upon this solution, an algorithm for the estimation of the camera relative rotation between two frames is proposed. The key idea is to detect triplets of orthogonal lines in a hypothesize-and-test framework and use all of them to compute the camera rotation in a robust way. This approach is suitable for human-made environments where numerous groups of orthogonal lines exist. We evaluate the numerical stability of the P3oA solution and the estimation of the relative rotation with synthetic and real data, comparing our results to other state-of-the-art approaches.

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Acknowledgments

We would like to thank to Ali Elqursh for providing us the source code of the odometry method of Elqursh and Elgammal [15]. This work has been funded by the Spanish Ministerio de Ciencia e Innovacion under the projects “TAROTH: New developments toward a Robot at Home” (Contract DPI 2011-25483) and “PROMOVE: Advances in mobile robotics for promoting independent life of elders” (Contract DPI 2014-55826-R). Both projects are cofounded by the “European Regional Development Fund ERDF”.

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  1. MAPIR-UMA Group, Departamento de Ingeniería de Sistemas y Automática, Universidad de Malaga, 29071, Malaga, Spain

    Jesus Briales & Javier Gonzalez-Jimenez

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  1. Jesus Briales
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  2. Javier Gonzalez-Jimenez
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Correspondence to Jesus Briales.

Appendices

Appendix 1: Reduction of Pencil Basis Parameter

The bilinear map \( \varvec{B}_{ij} \) is greatly simplified when pencil bases are used. The substitution of \( \varvec{\varOmega }_k \) defined in (7) into (5) gives

since \( \left||\varvec{t}\right|| = 1 \Rightarrow \varvec{t}^\top \varvec{t}= 1 \) and by definition of the cross product \( \varvec{t}^\top \left( \varvec{t}\times \varvec{n}\right) = 0 \) for any vector \( \varvec{n}\).

The remaining parameter

$$\begin{aligned} \alpha _{ij}= - \left( \varvec{t}\times \varvec{n}_i\right) ^\top \left( \varvec{t}\times \varvec{n}_j\right) \end{aligned}$$

can be further simplified to a single scalar product by applying some properties of the cross product. Firstly, the skew matrix representation for the cross product as well as the property \( \left[ \varvec{t}\right] _\times ^\top = - \left[ \varvec{t}\right] _\times \) allows us to write

$$\begin{aligned} \alpha _{ij}= \varvec{n}_i^\top \left[ \varvec{t}\right] _\times \left[ \varvec{t}\right] _\times \varvec{n}_j \end{aligned}$$

and then rewriting the product of skew matrices in the equivalent form

$$\begin{aligned} \left[ \varvec{a}\right] _\times \left[ \varvec{b}\right] _\times = \varvec{b}\varvec{a}^\top -(\varvec{a}\cdot \varvec{b})\mathbf I _3 \end{aligned}$$

the expression finally reduces to

$$\begin{aligned} \alpha _{ij}= \varvec{n}_i^\top \left( \varvec{t}\varvec{t}^\top - (\varvec{t}^\top \varvec{t})\cdot \mathbf I _3 \right) \varvec{n}_j = -\varvec{n}_i^\top \varvec{n}_j \end{aligned}$$

where it is used that \( \varvec{t}\perp \varvec{n}_k \) by the definition in (6).

Appendix 2: Relations between Interpretation Planes for the Degenerate P3oA Problem

Given a trihedron formed by three intersecting orthogonal lines, if the projections of two lines \( \mathbf l _i \) and \( \mathbf l _j \) become the same, the normals of the interpretation planes of the lines fulfill \(\varvec{n}_i \parallel \varvec{n}_j \perp \varvec{n}_k\).

The corresponding proof is developed in a sequence of minor steps:

  • If the lines \(\mathbf l _i\) and \(\mathbf l _j\) are parallel, so are the corresponding normals \(\varvec{n}_i\) and \(\varvec{n}_j\): \( \mathbf l _i \parallel \mathbf l _j \Rightarrow \varvec{n}_i \parallel \varvec{n}_j \).

  • If the normals \(\varvec{n}_i\) and \(\varvec{n}_j\) are parallel, the camera center must lie in the IJ plane defined by the lines \(L_i\) and \(L_j\).

  • If the camera center lies in the IJ plane, the normal \(\varvec{n}_k\) is orthogonal to \(\varvec{n}_i=\varvec{n}_j\).

1.1 (a) Parallelism of \(n_i\) and \(n_j\)

Let \( \mathbf l _i \) and \( \mathbf l _j \) be the homogeneous vectors corresponding to the projection of the lines into the camera image. Since the projective entities \( \mathbf l _i, \mathbf l _j \in \mathbb {P}^2 \) are defined up to scale, we use the similarity operator \( \mathbf l _i \sim \mathbf l _j \) to represent the equality of both variables in the projective space \( \mathbb {P}^2 \). Two vectors are then equivalent if they are parallel, or stated otherwise,

$$\begin{aligned} \mathbf l _i \times \mathbf l _j = \varvec{0}. \end{aligned}$$

The homogeneous lines in the image are related to the normal of the corresponding interpretation plane through the intrinsic calibration matrix \(\varvec{K}\) [27]

$$\begin{aligned} \varvec{n}\sim \varvec{K}^\top \mathbf l , \end{aligned}$$

so the equivalency relation is transmitted to the normals:

$$\begin{aligned} \varvec{n}_i \times \varvec{n}_j&= (\varvec{K}^\top \mathbf l _i) \times (\varvec{K}^\top \mathbf l _j)\\&= \det (\varvec{K}) \varvec{K}^{-\top } (\mathbf l _i \times \mathbf l _j) = \varvec{0}\Rightarrow \varvec{n}_i \sim \varvec{n}_j \end{aligned}$$

1.2 (b) Position of the Camera to Fulfill \(n_i \parallel n_j\)

Let us assume, without loss of generality, that the three intersecting lines fit the axes of the canonical coordinate system. As a result, the direction of the lines in this particular case is given by the canonical vectors \( \{\varvec{e}_k\}_{k=1}^3 \). Denote the position and orientation of the camera as seen from this coordinate system as \( \varvec{t}\) and \( \varvec{R}\), respectively. The normal to the interpretation plane of each line \( L_k \), as seen from the camera, is then equal (up to scale) to

$$\begin{aligned} \varvec{n}_k \sim \varvec{R}^\top (\varvec{e}_k \times \varvec{t}). \end{aligned}$$
(42)

From the equivalency of the lines i and j , it follows then that

$$\begin{aligned} \varvec{n}_i \sim \varvec{n}_j&\Rightarrow (\varvec{R}^\top (\varvec{e}_i \times \varvec{t})) \times (\varvec{R}^\top (\varvec{e}_j \times \varvec{t}))\\&= \varvec{R}^\top \left( (\varvec{e}_i \times \varvec{t}) \times (\varvec{e}_j \times \varvec{t})\right) \\&= \varvec{0}\Rightarrow (\varvec{e}_i \times \varvec{t}) \times (\varvec{e}_j \times \varvec{t}) = \varvec{0}\end{aligned}$$

and this expression is symbolically equivalent to

$$\begin{aligned} (\varvec{e}_i \times \varvec{t}) \times (\varvec{e}_j \times \varvec{t})&= \left[ \varvec{e}_i \times \varvec{t}\right] _\times \left[ \varvec{e}_j\right] _\times \varvec{t}\\&= ( \varvec{t}\varvec{e}_i^\top - \varvec{e}_i \varvec{t}^\top ) \left[ \varvec{e}_j\right] _\times \varvec{t}\\&= (\varvec{e}_k^\top \varvec{t}) \; \varvec{t}\end{aligned}$$

So, it is concluded that the equivalency \(\varvec{n}_i \sim \varvec{n}_j\) is only possible if \(\varvec{t}= \varvec{0}\) or \(\varvec{e}_k^\top \varvec{t}= 0\). The first solution forces the camera to be in the point of intersection of the three lines, but this makes no sense. The second solution implies that, for \( L_i \) and \( L_j \) to project into the same image line \( \mathbf l _i \sim \mathbf l _j \), the camera center must lie in the plane defined by lines \( L_i \) and \( L_j \).

1.3 (c) Orthogonality of Normals when the Camera Lies in the IJ Plane

Now, we will prove that if the camera center lies in the IJ plane, the normal of the interpretation plane for the remaining line \( L_k \) is orthogonal to both \( \varvec{n}_i \sim \varvec{n}_j \). Say \( \varvec{n}_k \) and \( \varvec{n}_i \) are orthogonal, which is equivalent to state \(\varvec{n}_k^\top \varvec{n}_i = 0\). Using the relation in (42)

$$\begin{aligned} \varvec{n}_k^\top \varvec{n}_i&= (\varvec{e}_k \times \varvec{t}) \varvec{R}\varvec{R}^\top (\varvec{e}_i \times \varvec{t})\\&= \varvec{t}^\top \left[ \varvec{e}_k\right] _\times ^\top \left[ \varvec{e}_i\right] _\times \varvec{t}\\&= - \varvec{t}^\top ( \varvec{e}_i \varvec{e}_k^\top - (\varvec{e}_k^\top \varvec{e}_i) \mathbf I _3 ) \varvec{t}\\&= - (\varvec{t}^\top \varvec{e}_i) (\varvec{e}_k^\top \varvec{t})\\&= - (\varvec{e}_i^\top \varvec{t}) (\varvec{e}_k^\top \varvec{t}) = 0 \end{aligned}$$

and the condition above is then fulfilled only if the camera center lies in the JK plane, the IJ plane, or both. Similarly, if \( \varvec{n}_k \) and \( \varvec{n}_j \) are orthogonal (necessary from \( \varvec{n}_i \sim \varvec{n}_j \)), the camera center lies in the IJ plane, the IK plane, or both. As a result, we see that if the camera lies in the IJ plane as assumed, the orthogonality constraints are fulfilled.

In conclusion, we see that if the projections of two lines, \( \mathbf l _i \) and \( \mathbf l _j \), are coincident, then the camera center lies in the plane formed by \( L_i \) and \( L_j \). This, at the same time, provokes that \( \varvec{n}_k \perp \varvec{n}_i \), or equivalently, \( \varvec{n}_k \perp \varvec{n}_j \).

Appendix 3: Reparameterization of Radical Equation as a Quadratic Form

The equation

$$\begin{aligned} \varvec{n}_k^\top \varvec{t}\cdot \bar{\alpha }+ \alpha _{ij} \cdot \varvec{n}_k^\top \left[ \varvec{t}\right] _\times \varvec{n}_* = 0 \end{aligned}$$

defined in (18) is non-linear wrt \( \varvec{n}_* \) due to the square root operation in

$$\begin{aligned} \bar{\alpha }&= \pm \,\sqrt{ \alpha _{*i}\alpha _{ij}\alpha _{j*}}\\&= \pm \,\sqrt{ - \left( \varvec{n}_*^\top \varvec{n}_i\right) \left( \varvec{n}_i^\top \varvec{n}_j\right) \left( \varvec{n}_j^\top \varvec{n}_*\right) } \end{aligned}$$

which makes the equation radical. As usual for these equations, we separate both terms in the sum and square them to get an almost-equivalent quadratic equation

$$\begin{aligned} \left( \varvec{n}_k^\top \varvec{t}\right) ^2 \cdot \left( \pm \sqrt{\alpha _{*i} \alpha _{ij} \alpha _{j*}} \right) ^2&= \alpha _{ij}^2 \cdot (\varvec{n}_k^\top \left[ \varvec{t}\right] _\times \varvec{n}_*)^2\\ \frac{\left( \varvec{n}_k^\top \varvec{t}\right) ^2}{\alpha _{ij}} \cdot \alpha _{*i} \alpha _{j*}&= (\varvec{n}_k^\top \left[ \varvec{t}\right] _\times \varvec{n}_*)^2 \end{aligned}$$

Then, we substitute \( \alpha _{*i} \) and \( \alpha _{j*} \) and arrange the matrix operations

$$\begin{aligned} \frac{(\varvec{n}_k^\top \varvec{t})^2}{\alpha _{ij}} \cdot \varvec{n}_*^\top \left( \varvec{n}_i \varvec{n}_j^\top \right) \varvec{n}_*&= (\varvec{n}_k^\top \left[ \varvec{t}\right] _\times \varvec{n}_*)^\top \varvec{n}_k^\top \left[ \varvec{t}\right] _\times \varvec{n}_*\\&= \varvec{n}_*^\top \left[ \varvec{t}\right] _\times ^\top \left( \varvec{n}_k \varvec{n}_k^\top \right) \left[ \varvec{t}\right] _\times \varvec{n}_* \end{aligned}$$

so that the condition above can be encoded by a quadratic form

$$\begin{aligned} \varvec{n}_*^\top \left( \frac{(\varvec{n}_k^\top \varvec{t})^2}{\alpha _{ij}} \cdot \varvec{n}_i \varvec{n}_j^\top - \left[ \varvec{t}\right] _\times ^\top \varvec{n}_k \varvec{n}_k^\top \left[ \varvec{t}\right] _\times \right) \varvec{n}_* = 0 \end{aligned}$$

From all infinite matrix representations available for the quadratic form given above, we choose the symmetric one

$$\begin{aligned} \varvec{Q}= \frac{1}{2} \frac{(\varvec{n}_k^\top \varvec{t})^2}{\alpha _{ij}} \left( \varvec{n}_i \varvec{n}_j^\top + \varvec{n}_j \varvec{n}_i^\top \right) + \left[ \varvec{t}\right] _\times \left( \varvec{n}_k \varvec{n}_k^\top \right) \left[ \varvec{t}\right] _\times \end{aligned}$$

which permits us to readily diagonalize the quadratic form by eigenvalue decomposition in a unique way.

Then, the original problem in (18) is equivalent to solving

$$\begin{aligned} \varvec{n}_*^\top \varvec{Q}\,\varvec{n}_* = 0 \end{aligned}$$

for the defined \( \varvec{Q}\).

Appendix 4: Necker Duality and Parallax Effect

The two distinct solutions of the P3oA problem for the case of meeting lines are mirrored wrt the plane whose normal is the back-projection direction \( \varvec{t}\) through the intersection point. This relation can be easily expressed through a reflection or Householder matrix [27]. Let \( \varvec{V}\) be the real solution and \( \varvec{V}^* \) the mirrored one. The following relation stands:

$$\begin{aligned} \varvec{V}= \varvec{H}_{\varvec{t}} \, \varvec{V}^* \end{aligned}$$
(43)

with \( \varvec{H}_{\varvec{t}} = \mathbf I _3 -2\,\varvec{t}\,\varvec{t}^\top /\varvec{t}^\top \varvec{t}\). This relation only stands when the three lines meet in a single point, otherwise there exists no reflection fulfilling the relation (43), even though this duality still exists (see green configuration in Fig. 4).

Without loss of generality, we will use the case of meeting lines to prove that both dual solutions provide the same rotation in the case of zero baseline (pure rotation). In such cases, taking the pair of false configurations will produce the relative rotation

$$\begin{aligned} \varvec{R}^*&= \varvec{V}_1^* \, (\varvec{V}_2^*)^\top \nonumber \\&= \varvec{H}_{p_1} \varvec{V}_1 \, \varvec{V}_2^\top \varvec{H}_{p_2}^\top \nonumber \\&= \varvec{H}_{p_1} \, \varvec{R}\, \varvec{H}_{p_2}^\top \end{aligned}$$
(44)

where \( p_i \) stands for the 3D coordinates of intersection point in i -th image wrt the camera frame. A rigid transformation exists between \( p_1 \) and \( p_2 \)

$$\begin{aligned} p_1 = \varvec{R}\, p_2 + \varvec{t}_{\text {rel}}\end{aligned}$$

so that, in the case of zero baseline we have

$$\begin{aligned} \varvec{H}_{p_1} = \mathbf I -2 \frac{\varvec{R}p_2 p_2^\top \varvec{R}^\top }{p_2^\top p_2} = \varvec{R}\varvec{H}_{p_2} \varvec{R}^\top \end{aligned}$$

and substituting in expression (44) reveals that

$$\begin{aligned} \varvec{R}^* = \varvec{R}\varvec{H}_{p_2} \varvec{R}^\top \varvec{R}\, \varvec{H}_{p_2}^\top = \varvec{R}\end{aligned}$$

As a conclusion, in the case of pure rotation, the true and false solutions are the same. So, the greater the parallax effect due to non zero baseline \( \left||\varvec{t}_{\text {rel}}\right|| \), the bigger the difference between both possible solutions.

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Briales, J., Gonzalez-Jimenez, J. A Minimal Closed-form Solution for the Perspective Three Orthogonal Angles (P3oA) Problem: Application To Visual Odometry. J Math Imaging Vis 55, 266–283 (2016). https://doi.org/10.1007/s10851-015-0620-x

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