Goal

Given $N \geq 2$ motion pairs $(A_i, B_i)$ of rigid transforms — $A_i \in SE(3)$ the gripper-to-gripper displacement between two robot stations and $B_i \in SE(3)$ the camera-to-camera displacement observed at the same pair of stations — compute the constant camera-to-gripper transform $X \in SE(3)$ that satisfies $A_i X = X B_i$ for every $i$ . Rotation and translation are solved jointly in a single SVD, without the two-stage decoupling used by Tsai-Lenz.

Algorithm

Write a rigid transform $H = \begin{bmatrix} R & t \\ 0 & 1 \end{bmatrix} \in SE(3)$ with rotation $R \in SO(3)$ and translation $t \in \mathbb{R}^3$ . For a vector $v \in \mathbb{R}^3$ let $[v]_\times$ denote the skew-symmetric matrix such that $[v]_\times w = v \times w$ . Let $\otimes$ denote the Hamilton product of quaternions.

Every rigid motion admits a screw decomposition: there exists an axis line $\ell$ in space, a rotation angle $\theta$ about that axis, and a translation distance $d$ along it (the pitch) such that $H$ is the composition of a rotation by $\theta$ around $\ell$ and a translation by $d$ along $\ell$ . The line $\ell$ is specified by a unit direction $n \in S^2$ and its moment $m = p \times n$ for any point $p$ on the line.

Definition

Unit dual quaternion \hat q

A dual quaternion is a pair $\hat q = q + \varepsilon q'$ of ordinary quaternions, with $\varepsilon^2 = 0$ . It is a unit dual quaternion when $|q|^2 = 1$ and $q \cdot q' = 0$ . Rigid motions are in one-to-one correspondence with unit dual quaternions up to sign.

q = \cos\tfrac{\theta}{2} + \sin\tfrac{\theta}{2}\,n, \qquad q' = \tfrac{1}{2}\,t \otimes q,

where $t = (0, t_x, t_y, t_z)$ is the translation as a pure quaternion.

Definition

Screw form

A compact equivalent representation in which the dual angle $\hat\theta$ and dual axis $\hat\ell$ are Study's dual-number extensions of the real angle and axis.

\hat q = \cos\tfrac{\hat\theta}{2} + \sin\tfrac{\hat\theta}{2}\,\hat\ell, \qquad \hat\theta = \theta + \varepsilon d, \qquad \hat\ell = n + \varepsilon m.

Write $\hat a$ , $\hat b$ , $\hat x$ for the unit dual quaternions of $A$ , $B$ , $X$ . The hand-eye equation $A X = X B$ lifts to

\hat a\,\hat x \;=\; \hat x\,\hat b.

Definition

Screw congruence

Corresponding hand and eye motions share the same screw angle and pitch; $X$ rotates the hand screw axis into the eye screw axis and offsets its moment.

\theta_a = \theta_b, \qquad d_a = d_b.

The scalar parts of $\hat a$ and $\hat b$ coincide because they encode $\cos(\hat\theta/2)$ . Split each quaternion into scalar and vector imaginary parts: $q = (q_0, \mathbf q)$ , and write $\hat a = (a_0, \mathbf a) + \varepsilon(a_0', \mathbf a')$ , similarly for $\hat b$ . Subtracting $\hat x \hat b$ from $\hat a \hat x$ and keeping only the imaginary parts yields six scalar equations per motion pair, linear in the eight unknowns $(q_0, \mathbf q, q_0', \mathbf q')$ .

Definition

Hand-eye constraint matrix

For each motion pair $(A_i, B_i)$ , the 6 × 8 matrix below annihilates the unknown dual quaternion of $X$ .

S_i \;=\; \begin{bmatrix} \mathbf a_i - \mathbf b_i & [\mathbf a_i + \mathbf b_i]_\times & \mathbf 0 & 0_{3 \times 3} \\ \mathbf a'_i - \mathbf b'_i & [\mathbf a'_i + \mathbf b'_i]_\times & \mathbf a_i - \mathbf b_i & [\mathbf a_i + \mathbf b_i]_\times \end{bmatrix}.

S_i \begin{pmatrix} q_0 \\ \mathbf q \\ q'_0 \\ \mathbf q' \end{pmatrix} = \mathbf 0_6.

Stacking $M$ motion pairs gives a $6M \times 8$ system $T = [S_1^\top \ldots S_M^\top]^\top$ with $T\,\hat x = 0$ . Each pair contributes a rank-6 block in generic position; two pairs with non-parallel screw axes are enough to cut the null space of $T$ down to two dimensions. The two vectors $v_7, v_8$ spanning that null space are the right singular vectors of $T$ associated with the two zero singular values.

The physical solution is a linear combination $\hat x = \lambda_1 v_7 + \lambda_2 v_8$ subject to the two dual-quaternion unit constraints:

|q|^2 = 1, \qquad q \cdot q' = 0.

Split each $v_k$ into its $q$ -half $u_k \in \mathbb{R}^4$ (rows 1–4) and its $q'$ -half $w_k \in \mathbb{R}^4$ (rows 5–8). The second constraint is quadratic in $s = \lambda_1 / \lambda_2$ :

(u_1 \cdot w_1)\,s^2 + (u_1 \cdot w_2 + u_2 \cdot w_1)\,s + (u_2 \cdot w_2) = 0.

Of its two real roots, pick the one that maximises $s^2\,|u_1|^2 + 2s\,u_1 \cdot u_2 + |u_2|^2$ ; the reciprocal square root of that value fixes $\lambda_2$ , and $\lambda_1 = s\,\lambda_2$ follows.

Algorithm

Daniilidis dual-quaternion hand-eye

Input: Motion pairs

\{(A_i, B_i)\}_{i=1}^{M}

with

M \geq 2

and non-parallel screw axes across pairs.

Output: Camera-to-gripper transform

X = \begin{bmatrix} R_X & t_X \\ 0 & 1 \end{bmatrix}

Convert each $A_i$ , $B_i$ to its unit dual quaternion $\hat a_i = (a_{0,i}, \mathbf a_i) + \varepsilon(a'_{0,i}, \mathbf a'_i)$ and $\hat b_i$ .
Assemble $S_i$ from the imaginary parts $(\mathbf a_i, \mathbf a'_i, \mathbf b_i, \mathbf b'_i)$ and stack into $T \in \mathbb{R}^{6M \times 8}$ .
Compute the SVD $T = U \Sigma V^\top$ ; take $v_7, v_8$ as the last two columns of $V$ .
Split each $v_k$ into $u_k, w_k \in \mathbb{R}^4$ . Solve the quadratic $(u_1 \cdot w_1) s^2 + (u_1 \cdot w_2 + u_2 \cdot w_1) s + (u_2 \cdot w_2) = 0$ for $s$ .
Choose the root maximising $s^2 |u_1|^2 + 2 s\,u_1 \cdot u_2 + |u_2|^2$ ; set $\lambda_2 = (s^2 |u_1|^2 + 2 s\,u_1 \cdot u_2 + |u_2|^2)^{-1/2}$ , $\lambda_1 = s\,\lambda_2$ .
Assemble $q = \lambda_1 u_1 + \lambda_2 u_2$ and $q' = \lambda_1 w_1 + \lambda_2 w_2$ ; recover $R_X = R(q)$ and $t_X = 2\,(q' \otimes q^*)_{\text{vec}}$ .

Implementation

The per-pair block $S_i$ of Eq. 31 in Rust. A stacked $T \in \mathbb{R}^{6M \times 8}$ is filled by calling fill_block for every motion pair.

use nalgebra::{DMatrix, Matrix3, Vector3};

fn skew(v: &Vector3<f64>) -> Matrix3<f64> {
    Matrix3::new(0.0, -v.z,  v.y,
                 v.z,  0.0, -v.x,
                -v.y,  v.x,  0.0)
}

fn fill_block(t_mat: &mut DMatrix<f64>, k: usize,
              a: Vector3<f64>, ap: Vector3<f64>,
              b: Vector3<f64>, bp: Vector3<f64>)
{
    let (d, dp) = (a - b, ap - bp);
    let (sx, spx) = (skew(&(a + b)), skew(&(ap + bp)));
    t_mat.fixed_view_mut::<3, 1>(6 * k,     0).copy_from(&d);
    t_mat.fixed_view_mut::<3, 3>(6 * k,     1).copy_from(&sx);
    t_mat.fixed_view_mut::<3, 1>(6 * k + 3, 0).copy_from(&dp);
    t_mat.fixed_view_mut::<3, 3>(6 * k + 3, 1).copy_from(&spx);
    t_mat.fixed_view_mut::<3, 1>(6 * k + 3, 4).copy_from(&d);
    t_mat.fixed_view_mut::<3, 3>(6 * k + 3, 5).copy_from(&sx);
}

Null-space resolution after $V = \mathrm{svd}(T).V$ . Let $v_7, v_8$ be the right singular vectors at the two smallest singular values; split each into an upper $q$ -half and lower $q'$ -half of length 4.

use nalgebra::{SVector, Vector4};

fn resolve(v7: &SVector<f64, 8>, v8: &SVector<f64, 8>)
    -> (Vector4<f64>, Vector4<f64>)
{
    let uq:  Vector4<f64> = v7.fixed_rows::<4>(0).into_owned();
    let uqp: Vector4<f64> = v7.fixed_rows::<4>(4).into_owned();
    let wq:  Vector4<f64> = v8.fixed_rows::<4>(0).into_owned();
    let wqp: Vector4<f64> = v8.fixed_rows::<4>(4).into_owned();

    let (a, b, c) = (uq.dot(&uqp),
                     uq.dot(&wqp) + wq.dot(&uqp),
                     wq.dot(&wqp));
    let disc = (b * b - 4.0 * a * c).max(0.0).sqrt();
    let norm = |s: f64| s * s * uq.dot(&uq)
                        + 2.0 * s * uq.dot(&wq)
                        + wq.dot(&wq);
    let s = [(-b + disc) / (2.0 * a), (-b - disc) / (2.0 * a)]
        .into_iter()
        .max_by(|x, y| norm(*x).partial_cmp(&norm(*y)).unwrap())
        .unwrap();
    let lam2 = norm(s).sqrt().recip();
    let lam1 = s * lam2;
    (lam1 * uq + lam2 * wq, lam1 * uqp + lam2 * wqp)
}

The rigid transform is then $R_X = R(q)$ and $t_X = 2\,(q' \otimes q^*)_{\text{vec}}$ by standard quaternion-to-matrix and quaternion multiplication.

Remarks

Joint rotation-and-translation solution avoids the error propagation of two-stage decoupled solvers: any bias in the rotation estimate of Tsai-Lenz feeds directly into the translation residual through the $R_X T_{c_{ij}}$ term, while here rotation and translation are co-estimated from the same null space.
Minimum two motion pairs; uniqueness requires the screw axes of the two (or more) pairs to be non-parallel. Degenerates when all motions share a common screw axis (planar or pure-rotation trajectories) — $T$ loses rank and the null space grows beyond two dimensions.
Computational cost is dominated by one SVD of a $6M \times 8$ matrix, $O(M)$ in the number of motion pairs.
The root-selection step is a standard dual-quaternion trick: of the two real roots of the quadratic in $s$ , only one yields a non-degenerate $|q|^2$ under the $|q|^2 = 1$ normalisation. A purely imaginary or negative value flags either a sign flip on one of the input quaternions or ill-conditioned motions.
Unit dual quaternions are a double cover of $SE(3)$ : $\hat x$ and $-\hat x$ represent the same rigid transform. The solver returns whichever sign the SVD produces; downstream code should canonicalise by requiring $q_0 \geq 0$ .
Compared with Tsai-Lenz: see When to choose Tsai-Lenz over Daniilidis on the Tsai-Lenz page, which hosts the comparison per the older-paper-hosts rule.

References

K. Daniilidis. Hand-Eye Calibration Using Dual Quaternions. International Journal of Robotics Research 18(3)
–298, 1999. doi
R. Y. Tsai and R. K. Lenz. A new technique for fully autonomous and efficient 3D robotics hand/eye calibration. IEEE Transactions on Robotics and Automation 5(3)
–358, 1989. pdf
Y. C. Shiu and S. Ahmad. Calibration of wrist-mounted robotic sensors by solving homogeneous transform equations of the form AX=XB. IEEE Transactions on Robotics and Automation 5(1)
–29, 1989. doi

Daniilidis Dual-Quaternion Hand-Eye Calibration

Goal

Algorithm

Implementation

Remarks

References

Alternative formulation of