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10

Wells’s explanation seems perfectly correct to me. Think of $\mathbb R^3$ as embedded into $\mathbb R^4$ by $(x,y,z)\mapsto(x,y,z,0)$. Then apply, in four-space, the rigid rotation $$ R_\theta\colon\quad \pmatrix{1&0&0&0\\0&\cos\theta&0&-\sin\theta\\0&0&1&0\\0&\sin\theta&0&\cos\theta}\,. $$ Here $\theta=0$ ...


9

I don't think you need the general Jacobian $\frac{\partial}{\partial \mathbf p}\exp(\hat{\mathbf p})$, but only the much simpler Jacobian $\left.\frac{\partial}{\partial \mathbf p}\exp(\hat{\mathbf p})\right|_{\mathbf p=\mathbf 0}$ with $\mathbf p$ being at the identity. Background The group of 3d rotations (SO3) is a matrix lie group. Thus, in general we ...


9

No, not by far. The matrix needs to be orthogonal, which means that $A^tA=I$ where $A^t$ is the transposed matrix -- and then it also has to have determinant 1. (You can think of orthogonality by considering how the matrix acts on the standard basis vectors -- since they were orthogonal to each other and had length 1 before the rotation, this must also be ...


7

My guess is that this is a very classic error in computer graphics implementation; you have to make sure that you don't accidentally use the new value of X when computing the new value of y! For instance, this code: x = x*cos(theta) - y*sin(theta); y = x*sin(theta) + y*cos(theta); will actually perform the following operation: x' =x cos θ - y ...


4

This reminds me of this answer of mine, which explores chirality in higher (and lower) dimensions. Firstly, let's use the definition of dimension1 of an object as the last value of $n$ such that the object or a rotated copy of it can be said to cover an open subset of $\mathbb R^n$. Now, the concept of "chirality" is where a reflection cannot be composed ...


4

I think the key here is that a flat object living purely in the $xy$-plane is not affected by reflection in $z$. Thus, an apparent reflection across the $x$-axis can be achieved by reflecting across both the $x$ and $z$ axes — this is a net rotation and hence can be achieved by continuous transformation without breaking orientability. The same principle ...


4

The problem is called Wahba's problem (though it likely has other names too) and seeks to minimise the following: $J(\mathbf{R}) = \frac{1}{2} \sum_{k=1}^{N} a_k|| \mathbf{w}_k - \mathbf{R} \mathbf{v}_k ||^2$ where $a_k$ are weights, $\mathbf{w}_k$ are the vectors in one frame, $\mathbf{v}_k$ are the vectors in the other frame and $\mathbf{R}$ is the ...


4

Rigidity of closed surfaces of positive curvature in $R^3$ is a theorem by S. Cohn-Vossen from 1927. See for instance "Isometric Embedding of Riemannian Manifolds in Euclidean Spaces" by Q.Han, theorem 8.1.2. As for $C^2$-smooth closed surfaces of negative curvature in $R^3$, they do not exist; hence, one can say that they all are rigid.


3

Not it is not. A rotation matrix is a square matrix with orthonormal columns and determinant 1. The set of all $n\times $n such matrices is commonly referred to as the special orthogonal group, and is denoted $SO(n)$.


3

Minimizing reprojection error using a sparse Levenberghe Marquardt algorithm is discussed in detail in an appendix to Hartley and Zisserman. Have you tried looking there?


3

Call the two transformations $a$ and $b$. Then conjugating, $b^{-1}ab$ is a rotation of 1 radian about a new point $(1-1/\sqrt{2},-1/\sqrt{2})$ distinct from the first two. Call this conjugate $c$. Then $bc^{-1}$ is a translation along the line perpendicular to their two fixed points, like waddling along with a board glued on your feet.


3

Consider the matrix $M_a$ that gives you the first frame representation of vectors $P_1, P_2, P_3$: $P_{a1}=M_a \cdot P_1, P_{a2}=M_a \cdot P_2, P_{a3}=M_a \cdot P_3$. Similarly, you've got some matrix $M_b$ that gives you the second frame representation: $P_{b1}=M_b \cdot P_1, P_{b2}=M_b \cdot P_2, P_{b3}=M_b \cdot P_3$. Since any $P_i$ is a column-vector, ...


3

I posted this answer to a similar question on sci.math. I will transcribe the question and the summary of the solution below. For this problem, we don't need to compute $r$, just set it to $1$. Least-Squares Conformal Multilinear Regression Given $\{ P_j : 1 \le j \le m \}$ and $\{ Q_j : 1 \le j \le m \}$, two sets of points, we want to find a conformal ...


3

Let $G$ be the group of rigid motions over $\mathbb{R}^3$. Topologize $G$ in any ways you like. As long as $G$ remains to be a topological transformation group over $\mathbb{R}^3$, the evaluation map at origin. i.e. the map defined by $$G \ni g \quad\mapsto\quad g(\vec{0}) \in \mathbb{R}^3$$ is a continuous surjection. If $G$ is compact, so does its image ...


2

With no coordinate system, or at least, as long as possible without one. An affine transformation of $\mathbb{R}^n$ consists of a linear transformation followed by a translation $x \mapsto A x+ b$, which we denote by $(A,b)$. The transformation $(A,b)$ is a rotation if and only if $A$ is an element of the special orthogonal group $SO(n)$ and $A$ is not ...


2

Background First, let us try to understand the problem better. What does a set of approximations $x_1,...,x_n$ of an unknown correct entity $\hat{x}$ mean? Well, this can be understood as a set of samples from a Gaussian distribution: $$x_1,...,x_n \quad\tilde{}\quad {\cal N}(\hat{x},\Sigma).$$ Here, $\Sigma$ is the sample covariance, thus the assumed ...


2

The approach of Least square optimization on matrix Lie groups is also explained in a technical report on Minimization on the Lie Group SO(3) and Related Manifolds.


2

I am trying to solve the confusion, not using a rigorous mathematical argument, but rather by illustrating the similarity of standard Gauss Newton over the Euclidean Space and Gauss Newton wrt. to a matrix Lie Group. Let us first look at the derivative of a scalar function $f$: $$\frac{\partial f(x)}{\partial x} := \lim_{y \to x} \frac{f(y)-f(x)}{y-x}$$ ...


2

I'm assuming you are asking that given $\vec{x} = (x_k)_{k=1}^3, \vec{y} = (y_k)_{k=1}^3 \in \mathbb{R}^3$, how do you find $A = (a_{i,j})_{i,j=1}^3 \in \mathbb{R}^{3 \times 3}$ so that $A \vec{x} = \vec{y}$. In case $\{x_i\}_{i=1}^3$ are all non-zero, a simple diagonal matrix would do: define $a_{i,i} = y_i/x_i$ and let all off-diagonal elements be $0$. ...


2

The reason you are finding these ideas hard to reconcile is because the diagram is referring to a physical notion of rigidity, while the wiki page you're reading is centered around a geometrical notion of rigidity. (I don't find the article you linked to be particularly clearly written either.) Geometry In geometry, we don't talk about rigid shapes really, ...


2

Yeah, I needed to rewrite everything and start on a new piece of paper before I realized the EDITed reverse transformation was relevant. The proof is as follows: $$\begin{align*} \ell_{ki}\ell_{kj}&=(\hat{\mathbf{e}}_k'\cdot\hat{\mathbf{e}}_i)\ell_{kj}\\ &=(\ell_{kj}\hat{\mathbf{e}}_k')\cdot\hat{\mathbf{e}}_i\\ ...


2

Let's say your tetrahedron with the four known vertices (Lets call it $T_A$) is given by four points $\{\mathbf a_1, \mathbf a_2, \mathbf a_3, \mathbf a_4\}$ and the other (Lets call it $T_B$) has a face corresponding to the first three points defined by $\{\mathbf b_1, \mathbf b_2, \mathbf b_3 \}$. You want to find $\mathbf b_4$ which corresponds to the ...


2

As remarked by Daniel Rust the area of the hexagon in question is the sum of the projected areas of the three "kinds" of cube facets. When a piece of a plane $\Sigma$ is orthogonally projected onto another plane $\Pi$ then the area is multiplied by $|\cos\phi|$, where $\phi$ is the angle between the planes, or equivalently: the angle between the ...


2

Let $G$ denote the group of isometries of $ R^n$ and let $H$ denote the stabilizer of your point configuration $p$. Then $M(p)$ is naturally diffeomorphic to $G/H$. Note that both $G$ is a Lie group and $H$ is its closed Lie subgroup. Hence, $G/H$ is a smooth manifold.


1

Let $\vec{v}_p=(x_p,y_p,z_p)$ and $\vec{v}_r=(x_r,y_r,z_r)$. Denote $\vec{v}_{p'}=(x_{p'},y_{p'},z_{p'})$ the coordinates of P after rotation. Then we have: $$(\vec{v}_{p'}-\vec{v}_r)=R_x(\phi)R_y(\theta)R_z(\psi)(\vec{v}_{p}-\vec{v}_r)$$ So $$\vec{v}_{p'}=\vec{v}_r+R_x(\phi)R_y(\theta)R_z(\psi)(\vec{v}_{p}-\vec{v}_r)$$


1

In the projective plane $\mathbb{RP}^2$ there is no notion of rotation about the $x$- or $y$-axis (nor of any other line in the plane). Though homographies in $\mathbb{RP}^2$ are represented by $3\times 3$-matrices, they are not to be confused with the linear transformations $\mathbb{R}^3\rightarrow\mathbb{R}^3$. The latter are also represented by $3\times ...


1

Let $\mathbf{n}$ be a unit vector in $\mathbb{R}^3$ be normal to the plane $P$. The projection $p$ of a vector $v\in\mathbb{R}^3$ onto the plane $P$ is given by $p(v)=v-(v\cdot\mathbf{n})\mathbf{n}$. The projection of the unit cube onto $P$ along the vector $\mathbf{n}$ is a hexagon whose interior has preimage which intersects the boundary of the unit cube ...


1

(This is not an answer) This is a very good, and "open", problem. The objective function should be invariant with respect to reparametrizing the boundary curve $t\mapsto \alpha(t)$; but your proposed $e$ does not have this property. Even insisting on $t$ being arc length is not enough, because arc length is not affine invariant. Here is a pointer to ...


1

Assuming all three are rigid transformations: If such a $C$ exists, then $T$ and $P$ are said to be orthogonally similar. In fact, given that $T$ and $P$ are rigid transformations, we know that the following are equivalent: $T$ and $P$ are orthogonally similar $T$ and $P$ are similar $T$ and $P$ are rotations by the same angle If $T$ and $P$ are ...


1

There is a simpler proof. Since you are transforming from one set of orthonormal Cartesian coordinates to another, your change of basis matrix $[l]$ is orthogonal (its transpose is also its inverse). Thus, $[l][l]^T = [l]^T[l]=[I]$. Clearly then, $l_{kj}l_{ki}$ = $l^T_{ik}l_{kj} = \delta_{ij}$, since this is just the matrix product $[l]^T [l]$ in ...



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