# Tag Info

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

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 ...

8

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 ...

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

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

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 ...

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

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 ...

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

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

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, ...

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

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

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

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

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 $0$. ...

1

Martin Gardner's The Ambidextrous Universe discusses this and related issues (mathematical, physical, chemical, anatomical...) in detail, with Gardner's inimitably readable style. Jeffrey Weeks's The Shape of Space investigates the geometry and topology of $3$-manifolds in more mathematical detail, yet in a friendly, engaging way. As Thomas Andrews notes, ...

1

There are also few other options: compute the derivatives using quaternions - this article explains how to differentiate exponential map this way use simpler Jacobian based on angular velocity - note that in minimization, the underlying model should be locally as linear as possible; simpler derivatives lead to more stable method (I think) use numerical ...

1

A recent paper by Guillermo Gallego and Anthony Yezzi suggests a compact formula for deriving the rotation matrix in the exponential map coordinates: http://arxiv.org/pdf/1312.0788v1.pdf Formula (III.7) on page 5 is what you were looking for.

1

question: Not a complete answer, but shouldn't you derive for $p$ and not $\omega$? But you maybe can do the derivative for $\omega$ as a pre-step, and then derive this for $p$ question: The Jacobian is not a $9\times 4$ matrix because the 4 values of a 4-component representation are not independent of each other. Either you take an arbitrary axis with an ...

1

At this point I am quite certain that I computed the first three elements of $J_R$ correctly. So I can continue computing the remaining elements of the matrix. One way to confirm that the analytic gradient is correct, is to look at the numeric gradient and then compare values between the two. For this you might want to check the gradient()-function in ...

1

I believe that rigidity is used in a relatively intuitive way here. However, if I wanted to ignore any pretense of geometric purity and just try to define the concept in a way that works, I would do it like this: it seems like one needs to have a notion of a ''bottom'' edge, which we consider to be fixed. The others should be free to move, and in particular ...

1

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, ...

1

Isometries of $R^n$ are known to be linear functions of the form $T(x)=A(x)+b$ where $A$ is a linear isometry fixing 0 and $b$ is a vector. For an isometry $T$ the displacement $f(x) = T(x) - x$ is a linear vector-valued function of $x$. If $f$ is of constant length then $f$ must be constant, otherwise there is a pair of points $p,q$ with $f(p) \neq f(q)$ ...

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

All isometries (rigid transformations) of a plane can be expressed as a composition of 3 or fewer reflections. A composite of two reflections over intersecting lines is a rotation about the point of intersection of the lines of reflection. A composite of two reflections over parallel lines is a translation perpendicular to the lines of reflection. A ...

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 $3\times ... 1 Choose two vectors$u$and$v$that together with$w=(a,b,c)^\mathrm{T}/x$form an orthonormal basis. The let $$R = \left[\array{u^\mathrm{T}\\v^\mathrm{T}\\w^\mathrm{T}}\right].$$ This gives a "rotation matrix" (orthogonal matrix) that maps$(a,b,c)^\mathrm{T}$to$(0,0,x)^\mathrm{T}$as you wanted. This matrix is not unique and varies by the orthonormal ... 1 I don't understand how you arrived at your rotation matrix, so I'll just go by your description of the transformed system and the image, which I think I understand. If we choose$\phi=0$,$O'$lies on the$x$-axis, so the$z'$-axis should be the negative$x$-axis, the$y'$-axis should be the$z$-axis, and the$x'$-axis should correspondingly ... 1 I think the problem lies in "As soon as we update$\mathtt R^{(m)}$, it no longer corresponds to$\mathbf p = \mathbf 0$". As I understand the answer you're referring to, it does;$\mathbf p$is just the change, not the whole thing. It says$\$ \mathtt J :=\left.\frac{\partial f(\exp(\hat{\mathbf p})\cdot\mathtt R^{(m)})}{\partial \mathbf p} \right|_{\mathbf ...

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