Two-View Triangulation: A Novel Approach Using Sampson’s Distance

Abstract

With the increase in the need for video-based navigation, the estimation of 3D coordinates of a point in space, using images, is one of the most challenging tasks in the field of computer vision. In this work, we propose a novel approach to formulate the triangulation problem using Sampson’s distance, and have shown that the approach theoretically converges toward an existing state-of-the-art algorithm. The theoretical formulation required for achieving optimal solution is presented along with its comparison with the existing algorithm. Based on the presented solution, it has been shown that the proposed approach converges closely to Kanatani–Sugaya–Niitsuma algorithm. The purpose of this research is to open a new frontier to view the problem in a novel way and further work on this approach may lead to some new findings to the triangulation problem.

The work was carried out during the first author’s affiliation with CSIR-Central Scientific Instruments Organisation, Chandigarh, India.

Appendix 1: Derivation of the Lagrangian

To start with, we consider the Lagrangian in (10)

$$ f = f_s + \lambda (\mathbf{x '}-\mathbf{S ^\top \varDelta \mathbf x '})^\top \mathbf F (\mathbf{x }-\mathbf{S ^\top \varDelta \mathbf x }) $$

where, $f_s$ is from (9).

Computing the partial derivatives of f with respect to ${\varDelta \mathbf x }$ and ${\varDelta \mathbf x '}$ yields,

$$\begin{aligned} \begin{aligned} \dfrac{\partial f}{\partial {\varDelta \mathbf x }}=\quad -\lambda \mathbf s _2{{\hat{\mathbf{x }}'}}\ +\ \dfrac{2a(\mathbf s _2{{\hat{\mathbf{x }}'}} + \mathbf s _2\mathbf S ^\top \varDelta \mathbf x ' )b}{b^2}\ \\ -\ \dfrac{2[{(\mathbf F \mathbf{x })_1}\ \mathbf s _2(:,1) + {(\mathbf F \mathbf{x })_2}\ \mathbf s _2(:,2)]a^2}{b^2}, \end{aligned} \end{aligned}$$

(18)

$$\begin{aligned} \begin{aligned} \dfrac{\partial f}{\partial {\varDelta \mathbf x '}} =\qquad -\lambda \mathbf s _1{{\hat{\mathbf{x }}}}\ +\ \dfrac{2a(\mathbf s _1{{\hat{\mathbf{x }}}} + \mathbf s _1\mathbf S ^\top \varDelta \mathbf x )b}{b^2}\ \\ -\ \dfrac{2[{(\mathbf F ^\top \mathbf{x '})_1}\ \mathbf s _1(:,1) + {(\mathbf F ^\top \mathbf{x '})_2}\ \mathbf s _1(:,2)]a^2}{b^2} \end{aligned} \end{aligned}$$

(19)

respectively. Here, we have considered

$$\begin{aligned} (\mathbf F \ ({{\hat{\mathbf{x }}}}+\mathbf{S ^\top \varDelta \mathbf x }))_1=(1,0,0)\ \mathbf F \ ({{\hat{\mathbf{x }}}}+\mathbf{S ^\top \varDelta \mathbf x }) \end{aligned}$$

(20)

and,

$$\begin{aligned} (\mathbf F \ ({{\hat{\mathbf{x }}}}+\mathbf{S ^\top \varDelta \mathbf x }))_2=(0,1,0)\ \mathbf F \ ({{\hat{\mathbf{x }}}}+\mathbf{S ^\top \varDelta \mathbf x }) \end{aligned}$$

(21)

Therefore, their derivatives will become

$$ \mathbf S {} \mathbf F (1,0,0)^\top = \mathbf s _1 (:,1),$$

$$\mathbf S {} \mathbf F (0,1,0)^\top =\mathbf s _1(:,2)$$

respectively. Similarly for $\mathbf F ^\top \ ({{\hat{\mathbf{x }}'}}+\mathbf{S ^\top \varDelta \mathbf x '})_1$ and $\mathbf F ^\top \ ({{\hat{\mathbf{x }}'}}+\mathbf{S ^\top \varDelta \mathbf x '})_2$. This will lead to (18) and (19).

Now, equating $\dfrac{\partial f}{\partial {\varDelta \mathbf x }}$ and $\dfrac{\partial f}{\partial {\varDelta \mathbf x '}}$ to zero, we get (11) and (12). From here, the solution given in (16) and (17) can be obtained, by keeping the value of $\lambda$, as in (13).