# Analytic Geometry and Singularities of Mechanisms

Jean Lerbet

► To cite this version:

Jean Lerbet. Analytic Geometry and Singularities of Mechanisms. Journal of Applied Mathematics and Mechanics / Zeitschrift für Angewandte Mathematik und Mechanik, 1998, 78 (10), pp.687-694. 10.1002/(SICI)1521-4001(199810)78:103.0.CO;2-T. hal-04367002

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Lerbet, J.

Analytic Geometry and Singularities of Mechanisms

In this article, we analyse the geometrical structure of the set M of configurations of a mechanism. Using Lie groups language and without any hypothesis of regularity, one can attach to this set a structure of an analytic variety. We shall then compute the tangent cone at each point of this set without introducing any coordinates and we shall deduce the structure of M whatever the singularity may be. Finally, some examples are studied.

MSC (1991): 70B15, 52C25, 51P05, 70G25, 58A07

Introduction

The most of references concerning the analysis of mechanisms do not use an intrinsic approach except, for example, [7]
or [8] or [3]. Of course, an intrinsic formulation permits to focus on the geometrical aspects which are fundamental in
the analysis of the singularities of mechanisms, and we will adopt such a formulation.

In this article, we show how the concept of tangent cone at a point of an analytic variety, which is naturally
introduced by analytic geometry ([1], [16]) is the correct concept to analyze the singularities of mechanisms. It is worth
noting that, in full generality, the configuration space of a mechanism is a real analytic variety (and not a differential
manifold). The main advantage of the concept of tangent cone at a point of an analytic variety is its great generality:

For analytic varieties, the set of tangent vectors is a cone for regular points as well as for singular points.

The structure of the singularities may be described by this tangent cone.

Of course, it should be possible to describe singular positions without this concept, however, it should be neces
sary to state some additional hypotheses (constant rank of the Jacobian matrix, transversality, and so on [12]). More
precisely, the concept of singular configuration of a mechanism can possess various senses according to the mathemati
cal or mechanical point of view.

If we look for a synthetical analysis of the mechanism, it is necessary that the space spanned by the Jacobian
matrix be a Lie algebra (*type I*). If only a local analysis is necessary, two situations can occur: a regular one, according
to differential geometry when the Jacobian matrix has a maximal rank (*type II*), and a singular one, if the previous
hypothesis is not valid. In this latter case three different situations can still occur.

*Type III:* as for the types I and II, the set of configurations is a submanifold, but more sophisticated concepts of differential geometry are required (for example accurate study of transversality and second order calculations [6], [12]).

*Type IV:* as for types I, II, III the set of configurations is a submanifold, but there is no justification of this fact based on standard results of differential geometry.

*Type V:* there exist more singular situations: the set of configurations is no longer a submanifold and there are configurations where the degree of freedom makes no sense.

For example, the well known four bars mechanism in the singular configuration for which the four joints are on a
same line may not be analysed in a way considering only calculations of ranks of matrices.

We shall start by giving an intrinsic formulation of the problem of singularities of a mechanism which opens the
way to using the general theory of analytic varieties. Therefore we shall state the main results of analytic geometry which
are relevant to our case: notion of tangent cone, and how to calculate it. We prove that, in the case of the configuration
space of a mechanism, the computation of the tangent cone results in an explicit algorithm involving a finite number of
steps depending on the mechanism. In all the known cases, the first two steps are sufficient. In the last part, we shall give
several examples of calculating a tangent cone in order to analyse the corresponding singular configuration.

1. Intrinsic formulation of the problem of singularities

1.1 Some recalls of geometry

To obtain explicit results which may be valid in any kind of situation it is necessary to have a formulation of the
problem which is independent of any choice of coordinate frames. The structure of the Lie group of the group of
Euclidean displacements allows us to obtain such a formulation. More precisely, let $\mathcal{E}$ be the three dimensional affine
space of ordinary Euclidean geometry and $E$ the associated vector space. To each affine map $A: \mathcal{E} \to \mathcal{E}$ usually its
linear part $\mathbf{A}$ is associated such that

$$
\forall m, p \in \mathcal{E} : A(m) = A(p) + \mathbf{A}(\overrightarrow{pm}).
$$

We denote by $\mathbb{D}$ the group of affine mappings $A$ such that $\mathbf{A}$ is an element of the special orthogonal group $\mathrm{SO}(E)$.
$\mathcal{D}$ denotes the six dimensional vector space of skew symmetric vector fields $X: \mathcal{E} \to E$ such that there exists $\mathbf{o}_X$ in $E$ with the following well known property:

$$ \forall a, b \in \mathcal{E}: X(a) = X(b) + \mathbf{o}_X \wedge \vec{ba}. $$

In other words, the linear part $\mathbf{X}$ of $X$ is the following linear operator in $E$:

$$ u \mapsto \mathbf{X}(u) = \mathbf{o}_X \wedge u, $$

and it may be identified with the set of screws.

The so called Lie bracket defined in $\mathcal{D}$ by $[X, Y](a) = \mathbf{o}_X Y(a) - \mathbf{o}_Y \wedge X(a)$ for all $a$ in $\mathcal{E}$ structures $\mathcal{D}$ as a Lie algebra, and it is then isomorphic to (and identified with) the classical Lie algebra of $\mathcal{D}$ which is geometrically the tangent space of $\mathcal{D}$ at the identity.

The exponential mapping $\exp: \mathcal{D} \to \mathcal{D}$ allows to express a finite displacement by its infinitesimal generator. The last concept that will be used is the adjoint mapping $\mathrm{Ad}: \mathcal{D} \to \mathcal{L}(\mathcal{D})$ which describes how any displacement is acting on the elements of $\mathcal{D}$. For more details about the operations defined on $\mathcal{D}$ and on $\mathcal{D}$, see [3].

## 1.2 Intrinsic formulation of closure equations

It is well known that if $A_k$ denotes a relative displacement of the joint $k$ for $k = 1, \dots, n$ of a mechanism, then the following closure equation must be satisfied:

$$ A_1 \circ \dots \circ A_n = id_{\mathcal{D}}. $$

For simplicity, we suppose that each joint is a rotational or prismatic one and can be described by a one parameter group, that is by one element $X_k$ of $\mathcal{D}: A_k = \exp(q_k X_k)$ with $q_k$ in $\mathbb{R}$. We then have to consider the so called closure function of the mechanism:

$$ f: \mathbb{R}^n \to \mathcal{D}, $$

$$ q = (q_1, \dots, q_n) \mapsto f(q) = \exp(q_1 X_1) \circ \dots \circ \exp(q_n X_n). $$

The closure set or the set of configurations of the mechanism is the following subset $V$ of $\mathbb{R}^n$:

$$ V = \{q \in \mathbb{R}^n \mid f(q) = id_{\mathcal{D}}\} = f^{-1}(id_{\mathcal{D}}). $$

## 1.3 Some remarks about such a formulation

One of the many problems of the theory of mechanisms is to know the geometry of $V$ when the family $(X_1, \dots, X_n)$ describing the mechanism (nature of the joints and reference configuration) is given. Of course, this geometry is hardly linked to the nature of the closure function $f$.

Because of the continuity of $f$, we can deduce that $V$ is a closed topological subspace of $\mathbb{R}^n$. If we consider the differentiability of $f$, we could expect that $V$ should be a submanifold of $\mathbb{R}^n$, and that, if $d$ is its dimension, $d$ should be also the degree of freedom of the mechanism. At each point $q_0$ of $V$, the infinitesimal aspect (set of the parameters of the virtual displacements) should be the tangent vector space $T_{q_0}V$ of dimension $d$. For regular position of types I and II, the computations of $d$ and $T_{q_0}V$ are easy: $d = n - \mathrm{rank}(f'(q_0))$ and $T_{q_0}V = \mathrm{Ker}(f'(q_0))$.

In other situations, where the rank of $f'(q_0)$ is not locally constant (type III), the results can still be explicit but more complicated, and the second derivative of $f$ the appears in the definition of $T_{q_0}V$ [12].

Unfortunately, there are more complicated situations where the concept of degree of freedom itself is no longer clearly defined. The singularity is such that its tangent aspect no longer can be described by a vector space. This means that for such a situation the differential geometry is not applicable, and it is then necessary to use other proper ties of $f$. In fact, analycity will allow us to describe the most general tangent aspect of $V$. Of course, the explicit calculation of the tangent manifold (which in general is not a vector space) can necessitate higher derivatives of $f$.

# 2. Some notions of analytic geometry and applications

## 2.1 Analytic variety Tangent cone

A real analytic variety $V$ is the inverse image of a point by an analytic mapping between two real analytic manifolds (based on charts $V$ appearing as the set of zeros of ordinary real analytic functions). This is our framework since $\mathcal{D}$, $\mathbb{R}^n$ are analytic manifolds and exp, Ad are analytic mappings.

Analogously to a differential manifold and its tangent bundle, it is possible to describe the tangent properties of any analytic variety $V$.

If $q_0$ is an element of $V$, let us consider an analytic curve $\gamma: [0, 1] \to V$ such that $q(0) = q_0$.

Then $\gamma'(0) = \lim_{t \to 0^+} \frac{\gamma(t) - q_0}{t}$ is a tangent vector of $V$ at $q_0$. A classical result is that at any point $q_0$ of $V$ the set $C_{q_0}V$ of all tangent vectors is a cone, the tangent cone of $V$ at $q_0$. Let us recall that a subset $C$ of $\mathbb{R}^n$ is a cone if and
only if $u \in C \Rightarrow tu \in C$ for all $t \in \mathbb{R}$. Obviously, any vector space is a cone and singular configurations of a mechanism of type V are in correspondence with circumstances where the tangent cone is not degenerated into a vector space. We will see that in many cases the tangent cone is a union of vector spaces corresponding to the fact that V is a union of manifolds.

## 2.2 Applications to the mechanisms

There is no chance to find a general formula explicting $C_{q_0}V$ for all $q_0$ in $V = f^{-1}(id_D)$. But we will see that it is possible to describe the steps of an algorithm for computing this. In the last part of the article, the examples will show the complexity of the computations. But we can already announce that in all of the examples, the first two steps are sufficient.

Let us consider now an analytic curve $\gamma : [0, 1] \to V$ such that $\gamma(0) = q_0 \in V$. It is possible to differentiate the following relation as many times as we want:

$$f(\gamma(t)) = g(t) = id_D \quad \text{for all } t \in [0, 1]. \qquad (1)$$

Before differentiating (1) to construct the tangent cone, let us introduce some notations. At each point $q$ of $\mathbb{R}^n$, let us define:

$$\mathcal{R}_q = \text{Ker } f'(q), \quad \text{where } f'(q) \in \mathcal{L}(\mathbb{R}^n, T_q D),$$

$$F_q = \text{Im}(f'(q) \circ f^{-1}(q)) \subset \mathcal{D} = T_{id_D} D,$$

$$H_q = \text{Lie algebra generated by } F_q.$$

One can prove that $H_q$ is the same for all $q$ in $\mathbb{R}^n$, and we denote by $H$ this subspace of $\mathcal{D}$. Let us put also, for all $q$ in $\mathbb{R}^n$ and for all $k$ in $\{1, \dots, n\}$,

$$\mu_k(q) = \text{Ad}(\exp(q_1 X_1) \circ \dots \circ \exp(q_k X_k)) X_k.$$

At first, we shall express the differentials of the closure function $f$ in a suitable form for our purpose using the Lie group structure of $D$. Secondly, by composition, we will obtain the derivatives along the curve, and finally the equations of the cone.

### 2.2.1 Derivatives of the closure function

For the first derivative, we then have

$$f'(q) : \mathbb{R}^n \to T_{f(q)} D,$$

$$x = (x^1, \dots, x^n) \mapsto \left( \sum_{k=1}^{n} x^k \mu_k(q) \right) \circ f(q) = L(q)(x) \circ f(q).$$

The calculation of further derivatives is equivalent to the calculation of the derivative of $q \mapsto L(q)$, a function valued in $\mathcal{L}(\mathbb{R}^n, \mathcal{D})$. We obtain

$$\mu_k'(q) : \mathbb{R}^n \to \mathcal{D},$$

$$\delta q = (\delta q^1, \dots, \delta q^n) \mapsto \sum_{l<k} \delta q^l [\mu_l(q), \mu_k(q)]$$

$$\text{and}$$
$$L'(q)(\delta q)(x) = H_2(q)(\delta q, x),$$
$$H_2'(q)(\delta q)(x, y) = H_3(q)(\delta q, x, y)$$

and so on, where $H_2, H_3, \dots$ are multilinear functions on $\mathbb{R}^n$, taking their values in $\mathcal{D}$ (note that, for example, $H_2'(q)(\delta q)$ is the value of the differential of $q \mapsto H_2(q)$, a function taking its values in the vector space $\mathcal{L}(\mathbb{R}^n \times \mathbb{R}^n, \mathcal{D})$). These functions have the following expressions:

$$H_2(q)(x, y) = \sum_{l<k} x^l y^k [\mu_l(q), \mu_k(q)], \qquad (2)$$

$$H_3(q)(x, y, z) = H_3^1(q)(x, y, z) + H_3^1(q)(y, x, z) + H_3^2(q)(x, y, z), \qquad (3)$$

and so on, where we have successively

$$H_3^1(q)(x, y, z) = \sum_{i<l<k} x^i y^l z^k [\mu_i(q), [\mu_l(q), \mu_k(q)]],$$

$$H_3^2(q)(x, y, z) = \sum_{l<k} x^l y^l z^k [\mu_l(q), [\mu_l(q), \mu_k(q)]] = H_3^2(q)(y, x, z).$$
Let us mention that the functions $H_k(q)$ possess several properties of symmetry. For example, $H_2(q)$, which is, in general, not symmetric, verifies $H_2(q)(x, y) = H_2(q)(y, x)$ if $x$ or $y$ belong to $\mathcal{R}_q$ and is then a symmetric function from $\mathcal{R}_1 \times \mathcal{R}_q$ to $\mathcal{D}$.

### 2.2.2 Derivatives of the closure equation

Now we take the derivatives of (1) by use of the formulas for compounded functions along the curve $\gamma$. For the first derivative we obtain

$$ \frac{d}{dt}(f(\gamma(t))) = f'(\gamma(t)) \circ \gamma'(t) = 0 \quad \text{for all } t \in [0, 1] $$

which is equivalent to

$$ L(\gamma(t))(\gamma'(t)) = 0 \quad \text{for all } t \in [0, 1] \qquad (4) $$

and means that $\gamma'(t) \in \mathcal{R}_{\gamma(t)}$ for all $t \in [0, 1]$.

Taking the derivatives of the previous equation, we then obtain the two following relations:

$$ H_2(\gamma(t))(\gamma'(t), \gamma'(t)) + L(\gamma(t))(\gamma''(t)) = 0 \quad \text{for all } t \in [0, 1], \qquad (5) $$

and using the symmetries of the functions $H_2$ and $H_3$, since $\gamma'(t) \in \mathcal{R}_{\gamma(t)}$ for all $t \in [0, 1]$:

$$ H_3(\gamma(t))(\gamma'(t), \gamma'(t), \gamma''(t)) + 3H_2(\gamma(t))(\gamma'(t), \gamma''(t)) + L(\gamma(t))(\gamma'''(t)) = 0. \qquad (6) $$

The results on higher derivatives are omitted as far as they are not needed for the treatment of the examples. The way to obtain such equations of type (6) is easy to understand (but the calculations may be complicated).

### 2.2.3 Calculation of the tangent cone

We apply now the previous calculations at the point $\gamma(0) = q_0$ which may be singular. To simplify the notation, let us put $L = L(q_0)$, $H_k = H_k(q_0)$ for all $k$. It is then natural to define successively

$$ K_{q_0}^1 V = \mathcal{R}_{q_0} = \text{Ker}(L), $$

$$ K_{q_0}^2 V = \{x \in \mathcal{R}_{q_0} | \exists x_1 \in \mathbb{R}^n | H_2(x, x) + L(x_1) = 0\}, $$

$$ K_{q_0}^3 V = \{x \in \mathcal{R}_{q_0} | \exists x_1, x_2 \in \mathbb{R}^n : \{H_2(x, x) + L(x_1) = 0 \wedge H_3(x, x, x) + 3H_2(x, x_1) + L(x_2) = 0\}, $$

and so on. One defines in this way inductively $K_{q_0}^k V$ for all integers $k \ge 1$. It is obvious that each $K_{q_0}^k V$ is a cone of $\mathbb{R}^n$. Let us see, for example, the case of $K_{q_0}^2 V$.

If $x$ is in $K_{q_0}^2 V$ and if $x_1$ is such that $H_2(x, x) + L(x_1) = 0$ then $tx$ is in $\mathcal{R}_{q_0}$ and $H_2(tx, tx) + L(t^2x_1) = 0$ that is to say $tx$ belongs to $K_{q_0}^2 V$.

We have then the following sequence:

$$ \{0\} \subset C_{q_0} V \subset \dots \subset K_{q_0}^k V \subset \dots \subset K_{q_0}^{k-1} V \subset K_{q_0}^2 V \subset \mathcal{R}_{q_0}, $$

and

$$ C_{q_0} V = \bigcap_k K_{q_0}^k V = \lim_{k \to \infty} K_{q_0}^k V. $$

Of course, the sequence $(K_{q_0}^k V)_{k \ge 1}$ is not strictly decreasing, and it is obvious that there exists, for each mechanism, an integer $k_0$ for which $C_{q_0} V = K_{q_0}^k V$.

Before examining some examples, we shall study now a type of situations which is often encountered in practice and where the tangent cone $C_{q_0} V$ is equal to $K_{q_0}^2 V$. Of course, situations with $C_{q_0} V = K_{q_0}^1 V$ are corresponding with situations for which the set of configurations is locally a submanifold (types I, II, III, IV). But let us recall that, in our knowledge, there does not exist any example for which we have neither $C_{q_0} V = K_{q_0}^1 V$ nor $C_{q_0} V = K_{q_0}^2 V$.

### 2.3 A case where $C_{q_0} V \subset K_{q_0}^2 V$

For simplification we suppose that $(\mu_1(q_0), \dots, \mu_p(q_0))$ is a basis of $F_{q_0}$ and that

$$ \mu_k(q_0) = \sum_{\alpha=1}^{p} C_k^{\alpha}(q_0) \mu_{\alpha}(q_0) $$

if $p+1 \le k \le n$.

Let us put then:

• $\mathcal{S}_{q_0}$ the complementary space of $\mathcal{R}_{q_0}$ in $\mathbb{R}^n$ defined by

$$ \mathcal{S}_{q_0} = \{x = (x^1, \dots, x^n) \in \mathbb{R}^n | x = (x^1, \dots, x^p, 0, \dots, 0)\}. $$
• $L_{\mathcal{S}_{q_0}}^1$ the isomorphism from $F_{q_0}$ onto $\mathcal{S}_{q_0}$ defined by

$$U = \sum_{\alpha=1}^{p} u^{\alpha} \mu_{\alpha} \mapsto L_{\mathcal{S}_{q_0}}^{1}(U) = (u^1, \dots, u^p, 0, \dots, 0).$$

Equation $L(v) = U$ with $v$ in $\mathbb{R}^n$ and $U$ in $F_{q_0}$ is obviously solved like this:

$$v = v_{\mathcal{S}_{q_0}} + L_{\mathcal{S}_{q_0}}^{1}(U),$$

where $v_{\mathcal{S}_{q_0}}$ is any vector of $\mathcal{S}_{q_0}$. The following result holds:

**Proposition:** For all $m \ge 3$, there exists a homogeneous mapping $T_m$ from $\mathbb{R}^n$ to $\mathcal{D}$ such that

$$x \in K_{q_0}^m \Leftrightarrow \exists v \in \mathcal{S}_{q_0} : T_m(x) + \alpha_m H_2(x, v) \in F_{q_0}.$$

In particular if $G_{q_0}(x) = \text{space}(\{H_2(x, u) | u \in \mathcal{S}_{q_0}\})$ is a complementary subspace of $F_{q_0}$ in $H$ for all $x$ no null in $K_{q_0}^2$ then $C_{q_0}V = K_{q_0}^2$.

• Let us prove a weaker result demonstrating only the case $m = 3$. The general case is similar. Denoting $\gamma'(0)$ and $\gamma''(0)$ by $x$ resp. $y$, equation (6) delivers

$$H_3(x, x, x) + 3H_2(x, y) \in F_{q_0}$$

with $H_2(x, x) + L(y) = 0$. According to the previous remarks, we get: $y = y_{\mathcal{S}_{q_0}} + L_{\mathcal{S}_{q_0}}^1(H_2(x, x))$ and $\alpha_3 = 3$, and finally

$$T_3(x) = H_3(x, x, x) + 3H_2(x, L_{\mathcal{S}_{q_0}}^1(H_2(x, x))).$$

### 3. Examples

For each example, we study the configuration $q_0 = 0$. We begin by giving the family $(X_1, \dots, X_n)$ of 6 vectors. The coordinates of each vector are given with respect to a basis of $\mathcal{D}$ which is deduced from a fixed coordinates frame $(0; \vec{i}, \vec{j}, \vec{k})$ of $\mathcal{E}$ in the following way:

$$i(m) = \vec{i}, \quad j(m) = \vec{j}, \quad k(m) = \vec{k}, \quad \xi(m) = \vec{i} \wedge \overrightarrow{0m}, \quad \eta(m) = \vec{j} \wedge \overrightarrow{0m}, \quad \zeta(m) = \vec{k} \wedge \overrightarrow{0m}$$

for all $m$ in $\mathcal{E}$.

The basis of $\mathcal{D}$ is then $(i, j, k, \xi, \eta, \zeta)$, and we then shall give successively for each example:

* The rank $p$ of $(X_1, \dots, X_n)$ and the set $F_0$.

* The maximal rank $r_m$ of $f'$.

* The Lie algebra $H$ generated by $F_0$ and its dimension $d$.

* A basis $(X_\alpha)_\alpha$ of $F_0$ and the family $(C_i^\alpha)_{\substack{i=1,\dots,p \\ i \neq 1,\dots,n}}$.

* The kernel $\mathcal{R}_0$ of $f'(0)$.

* The bilinear mapping $H_2(x, y)$ for $x, y \in \mathbb{R}^n$.

* The cone $K_0^2V$.

* The space $G_0(x)$ for all $x$ in $K_0^2V$.

* The cone $C_0V$.

* A local analysis of $V$ in a neighbourhood of 0.

We will examine successively the Bricard mechanism, the X mechanism, the four bar planar mechanism, and the five bar planar mechanism.

#### 3.1 The Bricard-mechanism

This well known linkage is constituted by six rotary joints which for a moment successively point to three orthogonal directions in the space. These directions are associated with the orthogonal basis $(\vec{i}, \vec{j}, \vec{k})$ of $E$, and the point 0 is chosen in the axis of the first joint. We then obtain:

$$X_1 = \begin{pmatrix} 0 \\ 0 \\ 0 \\ 1 \\ 0 \\ 0 \end{pmatrix}, \quad X_2 = \begin{pmatrix} 1 \\ 0 \\ 0 \\ 0 \\ 1 \\ 0 \end{pmatrix}, \quad X_3 = \begin{pmatrix} 0 \\ -1 \\ 0 \\ 0 \\ 0 \\ 1 \end{pmatrix}, \quad X_4 = \begin{pmatrix} 0 \\ -1 \\ -1 \\ 1 \\ 0 \\ 0 \end{pmatrix}, \quad X_5 = \begin{pmatrix} 0 \\ 0 \\ 1 \\ 0 \\ 1 \\ 0 \end{pmatrix}, \quad X_6 = \begin{pmatrix} 1 \\ 0 \\ 0 \\ 0 \\ 0 \\ 1 \end{pmatrix}.$$
* $p = 5$ and $F_0 = \langle X_1, \dots, X_5 \rangle$.

* $r_m = 6$.

* $H = \mathcal{D}$ and $d = 6$.

* $(X_a)_a = (X_1, X_2, \dots, X_5)$, $X_6 = X_1 + X_2 + X_3 - X_4 - X_5$, and $(C_i^a)_{\substack{a \\ i}}^{1, \dots, 5} = (1, 1, 1, -1, -1)$.

* $\mathcal{R}_0 = \{(-t, -t, -t, t, t, t) \mid t \in \mathbb{R}\}$ and $\dim \mathcal{R}_0 = 1$.

* $H_2(x, y) = (x^1y^4 - x^2y^4 + x^2y^5 - x^3y^4 + x^4y^5 - x^4y^6) i + (-x^1y^5 - x^2y^3 - x^2y^6 - x^3y^5 - x^4y^5) j + (-x^1y^3 - x^1y^4 + x^2y^5 - x^2y^6 + x^3y^4 - x^3y^6 - x^5y^6) k + (+x^2y^3 + x^2y^6 - x^3y^3 + x^5y^6) \xi + (-x^1y^3 - x^1y^6 + x^3y^4 - x^4y^6) \eta + (x^1y^2 + x^1y^5 + x^2y^4 + x^4y^5) \zeta$

* Because $H_2(x, x)$ belongs to $F_0$ if $x$ is in $\mathcal{R}_0$ then $K_0^2V = K_0^1V = \mathcal{R}_0$. Using transversality of $L$ ([6], [12]), it can be proved that $C_0V = \mathcal{R}_0 = T_0V$ and that $V$ is a one dimensional submanifold.

## 3.2 The X-mechanism

It is constituted by four rotary joints in the space. In the singular configuration, which is studied here, the mechanism is planar and the axes are as in figure 1.

Let be $\vec{i}$ and $\vec{j}$ like on figure 1, and $\vec{k} = \vec{i} \wedge \vec{j}$. We obtain:

$$X_1 = \begin{pmatrix} 0 \\ 0 \\ 0 \\ 1 \\ 0 \\ 0 \end{pmatrix}, \quad X_2 = \begin{pmatrix} 0 \\ 0 \\ 0 \\ 0 \\ 1 \\ 0 \end{pmatrix}, \quad X_3 = \begin{pmatrix} 0 \\ 0 \\ 0 \\ 1 \\ 0 \\ 0 \end{pmatrix}, \quad X_4 = \begin{pmatrix} 0 \\ 0 \\ 0 \\ 0 \\ 1 \\ 0 \end{pmatrix}.$$

* $p = 2$ and $F_0 = \langle X_1, X_2 \rangle = \langle \xi, \eta \rangle$.

* $r_m = 3$.

* $H = \langle \xi, \eta, \zeta \rangle$ and $d = 3$.

* $(X_a)_a = (X_1, X_2)$, $X_3 = X_1$, $X_4 = X_2$, and $(C_i^a)_{\substack{a \\ i}}^{1, 2} = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}$.

* $\mathcal{R}_0 = \{(-s, -t, s, t) \mid (s, t) \in \mathbb{R}^2\}$ and $\dim \mathcal{R}_0 = 2$.

* $H_2(x, y) = (x^1y^2 - x^2y^3 + x^1y^4 + x^3y^4) \zeta$.

* $K_0^2V = \{x = (x^1, x^2, -x^1, -x^2) \mid x^1 = 0 \text{ or } x^2 = 0\}$ that is to say $K_0^2V = K_0^2'V \cup K_0^2''V$, where $K_0^2'V = \{x = (-u, 0, u, 0) \mid u \in \mathbb{R}\}$ and $K_0^2''V = \{x = (0, -v, 0, v) \mid v \in \mathbb{R}\}$ which are two vector spaces of one dimension.

* Let be $x = (-u, -v, u, v)$ in $K_0^2V$, and $y = (-s, -t, s, t)$ in $\mathcal{R}_0$. Thus we get

$$H_2(x, y) = (ut - vs) \zeta.$$

If $x$ is in $K_0^2'V$, then by choosing $y = (-1, 0, 1, 0)$, we obtain $H_2(x, y) = u\zeta$ and $G_0(x) = \mathbb{R}\zeta$. If $x$ is in $K_0^2''V$, then by choosing $y = (0, -1, 0, 1)$, we obtain $H_2(x, y) = -v\zeta$ and $G_0(x) = \mathbb{R}\zeta$. In both cases, $G_0(x) = \mathbb{R}\zeta$ is a complete mentary subspace of $F_0$ in $H$.

* $C_0V = K_0^2V = K_0^2'V \cup K_0^2''V$ which corresponds to the global equality $V = V' \cup V''$ with $T_0V' = K_0^2'V$ and $T_0V'' = K_0^2''V$ (see figure 2).

<figure><img src="image_2.png" /><figcaption>Fig. 1</figcaption></figure>

<caption>Fig. 2</caption>

## 3.3 The four-bar planar mechanism

Like the previous one, it is constituted by four rotary joints. But they are here perpendicular to a common plane $(\vec{i}, \vec{j})$. We consider a singular configuration where the four axes are on a same line $(0, \vec{i})$, and 0 is on the axis of $X_1$. Then, supposing the distance between two successive axes to be equal to unity, we obtain:

$$X_1 = \begin{pmatrix} 0 \\ 0 \\ 0 \\ 0 \\ 0 \\ 1 \end{pmatrix}, \quad X_2 = \begin{pmatrix} 0 \\ -1 \\ 0 \\ 0 \\ 0 \\ 1 \end{pmatrix}, \quad X_3 = \begin{pmatrix} 0 \\ -2 \\ 0 \\ 0 \\ 0 \\ 1 \end{pmatrix}, \quad X_4 = \begin{pmatrix} 0 \\ -1 \\ 0 \\ 0 \\ 0 \\ 1 \end{pmatrix}.$$
* $p=2$ and $F_0 = \langle X_1, X_2 \rangle = \langle j, \zeta \rangle$.

* $r_m = 3$.

* $H = \langle i, j, \zeta \rangle$ and $d=3$.

* $(X_a)_a = (X_1, X_2)$, $X_3 = -X_1 + 2X_2$, $X_4 = X_2$, and $(C_i^a)_{\substack{a \\ 1,2}} = \begin{pmatrix} -1 & 0 \\ 2 & 1 \end{pmatrix}$.

* $\mathcal{R}_0 = \{(u, -2u - v, u, v) \mid u, v \in \mathbb{R}\}$ and $\dim \mathcal{R}_0 = 2$.

* $H_2(x, y) = (x^1y^2 + 2x^1y^3 + x^1y^4 + x^2y^3 - x^3y^4) i$.

* $K_0^2V = \{x = (x^3, -2x^3 - x^4, x^3, x^4) \mid x^3 = 0 \text{ or } x^3 = -x^4\} = K_0^2V \cup K_0^{2''}V$ with

$$K_0^{2'}V = \{x = (0, t, 0, t) \mid t \in \mathbb{R}\}, \quad K_0^{2''}V = \{x = (u, -u, u, -u) \mid u \in \mathbb{R}\},$$

which are two one dimensional vector spaces.

* Let be $x = (-u, -2u - v, u, v)$ in $K_0^2V$ and $y = (-s, -2s - t, s, t)$ in $\mathcal{R}_0$. We have then

$$H_2(x, y) = (-u(2s + t) - vs) i.$$

If $x$ is in $K_0^2V$, then by choosing $y = (-1, -2, 1, 0)$, we obtain $H_2(x, y) = -vi$ and $G_0(x) = \mathbb{R}i$. If $x$ is in $K_0^{2''}V$, then by choosing $y = (-1, -3, 1, 1)$, we obtain $H_2(x, y) = -2ui$ and $G_0(x) = \mathbb{R}i$. In both cases, $G_0(x) = \mathbb{R}i$ is a complete mentary subspace of $F_0$ in $H$.

* $C_0V = K_0^2V \cup K_0^{2''}V$ which corresponds to the global equality $V = V' \cup V''$ with $T_0V' = K_0^2V$ and $T_0V'' = K_0^{2''}V$ (see figure 2).

### 3.4 The five-bar planar mechanism

The last mechanism which is studied here is constituted by five rotary joints perpendicular to a common plane $(\vec{i}, \vec{j})$. In the singular configuration, the five axes of the joints are on a same line $(0, \vec{j})$, 0 being on the axis of $X_1$. We obtain:

$$X_1 = \begin{pmatrix} 0 \\ 0 \\ 0 \\ 0 \\ 0 \\ 1 \end{pmatrix}, \quad X_2 = \begin{pmatrix} 0 \\ -1 \\ 0 \\ 0 \\ 0 \\ 1 \end{pmatrix}, \quad X_3 = \begin{pmatrix} 0 \\ -2 \\ 0 \\ 0 \\ 0 \\ 1 \end{pmatrix}, \quad X_4 = \begin{pmatrix} 0 \\ -3 \\ 0 \\ 0 \\ 0 \\ 1 \end{pmatrix}, \quad X_5 = X_2.$$

* $p=2$ and $F_0 = \langle X_1, X_2 \rangle = \langle j, \zeta \rangle$.

* $r_m = 3$.

* $H = \langle i, j, \zeta \rangle$ and $d=3$.

* $(X_a)_a = (X_1, X_2)$, $X_3 = -X_1 + 2X_2$, $X_4 = -2X_1 + 3X_2$, $X_5 = X_2$, and $(C_i^a)_{\substack{a \\ 1,2}} = \begin{pmatrix} -1 & -2 & 0 \\ 2 & 3 & 1 \end{pmatrix}$.

* $\mathcal{R}_0 = \{(u + 2v, -2u - 3v - w, u, v, w) \mid u, v, w \in \mathbb{R}\}$ and $\dim \mathcal{R}_0 = 3$.

* $H_2(x, y) = (x^1y^2 + 2x^1y^3 + 3x^1y^4 + x^1y^5 + x^2y^3 + 2x^2y^4 + x^3y^4 - x^3y^5 - 2x^4y^5) i$.

* $K_0^2V = \{(x^3 + 2x^4 - 2x^3 - 3x^4 - x^5, x^3, x^4, x^5) \mid (x^3)^2 + 3(x^4)^2 + 3x^3x^4 + x^3x^2 + 2x^4x^5 = 0\}$.

* Let be $x = (u + 2v, -2u - 3v - w, u, v, w)$ in $K_0^2V$ and $y = (r + 2s, -2r - 3s - t, r, s, t)$ in $\mathcal{R}_0$. We then have

$$H_2(x, y) = ((-2u - 3v - w) r - (3u + 6v + 2w) s - (u + 2v) t) i.$$

If $x$ is such that $-2u - 3v - w = 0$, then by choosing $y = (1, -2, 1, 0, 0)$, we obtain $H_2(x, y) = (-2u - 3v - w) i$ and $G_0(x) = \mathbb{R}i$.

If $x$ is such that $-2u - 3v - w = 0$ and $u + 2v = 0$, then by choosing $y = (1, -3, 1, 0, 1)$, we obtain $H_2(x, y) = (-u - 2v) i$ and $G_0(x) = \mathbb{R}i$.

If $x$ is such that $-2u - 3v - w = 0$ and $u + 2v = 0$, then choosing $y = (3, -6, 1, 1, 1)$, we obtain $H_2(x, y) = (-2v) i$ and $G_0(x) = \mathbb{R}i$.

In all cases, $G_0(x) = \mathbb{R}i$ is a complementary subspace of $F_0$ in $H$.

* $C_0V = K_0^2V$ which is not the union of two vector spaces but an irreducible cone ...

**Remark:** If we study the three bar planar mechanism in a singular position in a similar way, we obtain $C_0V = K_0^2V = \{0\}$ which corresponds to the case $V = \{0\}$ that is to say to an isolated configuration. It is well known that such a mechanism cannot move.

## Conclusion

To analyse the configurations of a mechanism which are singular in the most pathological way as possible (type V), the structure of analytic variety of the set of configurations has been used. Although the cone of tangent vectors is no more in general case a vector space, its computation remains always possible by induction.

But, for all examples that we have discussed, the first two steps are sufficient. We do not know if it is the general situation and then why, or if there exists an example with $C_{q_0}V$ strictly included in $K_{q_0}^2V$.
If such an example does not exist, it will be necessary to prove that the second order is always sufficient. Surely the result is linked to the structure of $F_{q_0}$ of $H$, and more generally to the particular structure of $\mathcal{D}$ (see [3] for example).

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Address: Prof. Dr. JEAN LERBET, CERMICS, Ecole Nationale des Ponts et Chaussées, 6 et 8 avenue Blaise Pascal, Cité Descartes Champs sur Marne, F 77455 Marne la Vallée CEDEX 2, France