Rigid Body Displacements


 Isabel Hamilton
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1 Chapter 2 Rigid Body Displacements 2.1 The Isometry Group A rigid body displacement can be described geometrically as an Isometry: A bijective linear mapping of E 3 onto itself which leaves the distance between every pair of distinct points, and the angle between every pair of distinct lines, and planes, invariant. A mapping is a functional relation from the set of elements of the domain to the set of elements in the image. Consider the mapping from set A to set B (a are the elements of {A}, while b are the elements of {B}). The mapping, which can be represented algebraically as a linear transformation, can be one of the following three types: Figure 2.1: Types of Mappings. 13
2 14 CHAPTER 2. RIGID BODY DISPLACEMENTS Although there is a motion associated with an isometry, the isometry does not represent a motion: it is the correspondence between an initial and a final position of a set of points. A motion is a continuous series of infinitesimal displacements. Because an isometry maps collinear points into collinear points, it transforms lines into lines, and hence is a collineation. The invariance of distance also ensures that triangle vertices are transformed into congruent triangle vertices. Thus, isometries preserve angle and are, therefore, also conformal transformations. The word set so far has been used to mean a collection for geometric objects, such as points, or lines. It may, however, be used more broadly to mean a collection of any sort. The set of isometries includes the following transformations: rotation; translation; screw; reflection (in a plane); central inversion (reflection in a point). It is easy to show that the set of isometries, together with a binary operator that combines any two of them, called product, defined on the set, constitutes a group, G. The elements of G, {x, y, z,...} and the product operator must satisfy all the properties listed in Table (2.1) in order to constitute a group. Note that commutativity is, in general, not a group property. Table 2.1: Group properties i Closure (x y)ɛg (x, y) ɛ G ii Associativity (x y) z = x (y z) (x, y, z) ɛ G iii Identity I ɛ G I x + x I = x, x ɛ G iv Inverse x 1 ɛ G x x 1 = x 1 x = I, x ɛ G The Isometry group of the Euclidean plane E 2 is a subgroup of the isometry group of E 3. Every isometry is the product of, at most, four reflections. In E 2, four is replaced by three. Since a reflection reverses sense, an isometry is direct or opposite according to whether it is the product of an even or odd number of reflections. The set of direct isometries form a subgroup. This is because any product of direct isometries is another direct isometry. Whereas, the same does not hold for opposite isometries: the product of two opposite isometries is a direct isometry, violating the closure property. Hence, the opposite isometries are not a subgroup. The subgroup of direct isometries is also known as the group of Euclidean Displacements, G 6. The subscript 6 refers to the fact that 6 generalized coordinates are required to specify a displacement. The subgroup is also called SE3 (Special 3D Euclidean Group). In turn, the isometries are a subgroup of the Euclidean Similarity Transformations (the principal group, G 7 ). Seven parameters determine a similarity transformation, the additional one being a magnification factor to uniformly scale distances. The G 7 transformations are conformal collineations, but the distance between two points is not, in general, invariant.
3 2.2. ISOMETRY IN THE EUCLIDEAN PLANE Isometry in the Euclidean Plane An isometry in E 2 is a bijective mapping of the plane onto itself which preserves distance, and can be represented algebraically by a congruent linear transformation. There are four isometries in the plane Reflection R in line r Let r be any line in E 2. A reflection in r leaves all points on r invariant, every other point of E 2 goes into the symmetrical point on a line right bisected by r. The reflection R in line r reverses the order of the triangle ABC. That is, the clockwise circulation of vertices ABC is transformed to a counterclockwise (CCW) circulation A B C. The sense of the angles between edges, AC and AB for instance, are reversed. Hence, reflections are opposite isometries. Figure 2.2: A reflection R in line r. It is easy to show that a reflection in an arbitrary line r has the equations: x = x cos(2τ) + y sin(2τ) + 2p cos(θ), y = x sin(2τ) y cos(2τ) + 2p sin(θ), where τ is the angle r makes with the xaxis, CCW being positive, φ is the length of the perpendicular from the origin of the referenced coordinate system, θ is the angle of the vector represented by this directed perpendicular connecting line
4 16 CHAPTER 2. RIGID BODY DISPLACEMENTS segment, CCW being positive. Note: line r is called a mirror line. (Convention: the positive sense for angles is CCW.) Rotation S θ, Center S Through Angle θ Let S be any point in E 2, and θ be any measure of angle (CCW being positive). A rotation in S through θ maps S onto itself, and any other point P onto P such that distance SP = SP and P SP = θ. The rotation about point S through angle θ preserves distance, angle, and sense. That is, vertices of triangle ABC are mapped to A B C. The order of the vertices is preserved as well as the sense of the angles between the edges. Hence, rotations are direct or positive isometries. Figure 2.3: Rotation S θ. An arbitrary rotation has the following equations: x = x cos θ y sin θ + S x, y = x sin θ + y cos θ + S y, where S x, S y are the (x,y) coordinates of center S.
5 2.2. ISOMETRY IN THE EUCLIDEAN PLANE Translation τ Let τ be a directed linesegment. A translation τ maps any point P onto P in the direction of τ a distance equal to the length of τ. The translation has no finite invariant points. All points of E 2 have different images under τ. Moreover, translations are clearly direct isometries. They preserve distance, angle, sense, and orientation. Figure 2.4: Translation τ. An arbitrary translation in E 2 has the following linear equations: x = x + τ x, y = y + τ y, where τ x and τ y are the directed length of τ projected on the x and y axis, respectively GlideReflection G A glidereflection G is an opposite isometry that has no invariant point, but has a unique fixed line, called the axis of G. The glidereflection can be represented as the unique product of a translation and parallel reflection. See Figure 2.5. In subsequent sections we will drop the product operator symbol (*) to simplify equations and save ink. Thus: G = τ R = τr. G is represented by a half arrow the length of τ along R
6 18 CHAPTER 2. RIGID BODY DISPLACEMENTS Figure 2.5: Glidereflection G = τr. 2.3 Every Isometry in E 2 is the Product of At Most Three Reflections Every isometry can be represented, in many ways, as the products of different isometries. For the glidereflection this is selfevident by definition. The reflection may be considered as the fundamental isometry as all isometries may be represented by the product of, at most, three reflections (in E 3 replace three with four). Before proceeding, recall the group properties of isometries: Closure Let us simply accept that the product of any two isometries results in another isometry. Associativity Consider three isometries A, B, C. (AB)C = A(BC). proof: (AB)C = (A)(B)(C), = (A)BC, = A(BC). Identity Let the isometry I be the one resulting in the following: x = x, y = y. The identity leaves all points in E 2 invariant. The identity can be constructed as the product of two successive reflections in the same line. First application of R maps P to P. The second application maps P to P, which is P itself. See Figure 2.6. Inverse Clearly, if an isometry can be done, it can always be undone. The inverse is indicated with a (1) right superscript: S 1. This leads directly to the result: SS 1 = S 1 S = I.
7 2.3. EVERY ISOMETRY IN E 2 IS THE PRODUCT OF AT MOST THREE REFLECTIONS19 Figure 2.6: RR = I, the identity. The only commutative isometries are the product of an isometry with its inverse. The product is otherwise, in general, not commutative. Consider the situation in Figure 2.7. Evidently, a translation results when R 1 and R 2 are Figure 2.7: Product of Two Parallel reflections. parallel, but concatenation in different order produces different translations. Displacements produced by planar mechanisms must be direct, not opposite, isometries. Opposite isometries reverse sense, implying that the associated mo
8 20 CHAPTER 2. RIGID BODY DISPLACEMENTS tion required the mechanism to leave the plane. Since this violates the planar condition, we need only consider the direct isometries for planar mechanisms. We now consider the following product theorems for the products of two reflections: Theorem: The product of two reflections in parallel lines is a translation in a direction perpendicular to the lines, and whose magnitude is twice the distance between the mirror lines. Proof: The product R 1 R 2 operates on all points in the plane, including the mirror lines. The magnitude must be the same for all points. The points on r 1 stay on r 1 under R 1, but are reflected in r 2 on lines right bisected by r 2. The magnitude is clearly twice the distance between r 1 and r 2. Consider R 1 R 2 in Figure 2.8. Figure 2.8: Product of parallel reflections. Note: The mirror lines are not unique!!! The product of two reflections in any two lines parallel to r 1 and r 2 separated by distance d will yield τ. Theorem: The product of two reflections in two finitely intersecting lines is a rotation about the point of intersection through twice the angle between the mirror lines. See Figure 2.9. Proof: Clearly, SP = SP = SP. The angle between SP and SP is: 2(α + β) = 2φ = θ Note: The same rotation results for every pair of mirror lines through the same point with angle φ between them. 2.4 Products of Three Reflections While not important for planar mechanisms, the following theorem are important for spacial mechanism in particular, and for spatial kinematics in general.
9 2.4. PRODUCTS OF THREE REFLECTIONS 21 Figure 2.9: Product of two intersecting reflections. There are four distinct ways to arrange three lines in the plane. Or course in all cases, the product of three reflections must be another opposite isometry Case 1: All three mirror lines intersect in one point Theorem: The product of three such reflections R 1 R 2 R 3 is a single reflection R. See Figure Proof: Select r such that r 2 r 3 = r 1 r. This condition means R 2 R 3 = R 1 R. Then R 1 (R 2 R 3 ) = R 1 (R 1 R) = IR = R Case 2: All three mirror lines are parallel Theorem: The product of three reflections R 1, R 2, R 3 whose mirror lines are all parallel is a single reflection, R, parallel to the given three. Proof: Select r to be parallel to r 3 such that the directed distance from r to r 3 is identical to the directed distance from R 1 to r 2, giving: R 1 R 2 = RR 3, (R 1 R 2 )R 3 = (RR 3 )R 3, = RI, = R.
10 22 CHAPTER 2. RIGID BODY DISPLACEMENTS Figure 2.10: Three reflections intersecting in one point. (2.1) Figure 2.11: Three reflections with parallel mirror lines Case 3: Three reflections whose mirror lines intersect in three distinct points Before progressing to the theorem we must discuss two special products: 1. The product of two reflections intersecting in perpendicular mirror lines is a rotation through 180 degrees, called a halfturn, H = R 1 R 2.
11 2.4. PRODUCTS OF THREE REFLECTIONS The product of a reflection and a half turn whose center is not on the mirror line is a glide reflection. The proof of 1 follows from definition. The proof of 2 uses the fact that a glide is the product os a translation and a reflection in a line parallel to the direction of the translation, and a translation is the product of two parallel reflections. See Figure Figure 2.12: GlideReflection. A rotation of 180 degrees about center H can be represented in infinitely many ways as the product of two reflections whose mirror lines intersect at H an are 90 degrees apart. Now consider halfturn H and reflection R whose mirror line does not pass through H. Decompose H into the product of R 1 R 2 where r 2 is parallel to r and r 1 is perpendicular to r. Now HR = R 1 R 2 R = R 1 τ = G. See Figures 2.13 and Theorem The product of three reflections R 1 R 2 R 3 whose three mirror lines r 1, r 2, r 3 intersect in three distinct finite points is a glidereflection. Proof Select r 4 and r 5 such that r 4 r 1 and r 2 r 3 = r 4 r 5 and r 4 and r 5 are incident on the intersection of r 2 and r 3, S. See Figure R 1 R 2 R 3 = R 1 R 4 R 5 since R 1 R 4, = HR 5 since R 1 R 4, = R 7 τ (2d), = G. The three reflections are equivalent to the glidereflection, which is also equivalent to the product of HR 5, which can be decomposed into R 7 R 6 R 5
12 24 CHAPTER 2. RIGID BODY DISPLACEMENTS Figure 2.13: Rotations can be represented as the product of any two mirror lines intersecting on the rotation center separated by half the rotation angle. The halfturn is no different. such that (see Figure 2.16): G = HR 5, = R 7 R 6 R 5, = R 7 τ (2d), = R 1 R 2 R 3, = G. Figure 2.14: Product of halfturns and nonincident reflections are glide reflections.
13 2.4. PRODUCTS OF THREE REFLECTIONS 25 Three reflections in three distinct intersections are a glide Figure 2.15: reflection. Figure 2.16: The product of HR 5.
14 26 CHAPTER 2. RIGID BODY DISPLACEMENTS Case 4: Three reflections with two parallel mirrors Theorem: The product of three reflections where only two mirror lines are parallel is a single glidereflection. Proof: The product (see Figure 2.17) R 1 R 2 = S 2θ = R 4 R 5 R 1 R 2 R 3 = R 4 R 5 R 3 = R 4 H = G. Figure 2.17: Three reflections with only two parallel mirror lines is a glidereflection. Which is, of course, equivalent to (See Figure 2.18): G = R 4 R 6 R 7 = τ 2d R 7. Note that we could have simply asked the question: what is the product of R 1 R 2 R 3? Knowing the above proof, we could select r 4 and r 5 such that the angle between r 4 and r 5 is the same as the angle between r 1 and r 2, and with
15 2.5. THE POLE OF A PLANAR DISPLACEMENT 27 Figure 2.18: Equivalent to three reflections in two parallel mirrors. r 5 perpendicular to r 1. Again we have R 1 R 2 R 3 = R 4 R 5 R 3 = R 4 H = G. Since G is equivalent to R 4 R 6 R 7, select r 6 parallel to r 4 and r 7 perpendicular to r 6. Then R 4 R 6 is a translation from r 4 towards r Summary Any isometry in E 2 is, or can always be expressed as the product of, at most, three reflections. For a summary, see Figure The Pole of a Planar Displacement The group of planar displacements in E 2 are the subgroup of the direct isometries in E 2. An important theorem, which we will use later on with kinematic mapping, is that any general planar displacement can always be represented by a single rotation. Even a pure translation may, if we bound E 2 with the line at infinity, L (thereby creating the projective plane P 2 ), be considered a rotation through an infinitesimal angle about the point of intersection between L and the normal to the direction of translation. The coordinates of this rotation center are invariant under the rotation. This unique point is the pole of the displacement.
16 28 CHAPTER 2. RIGID BODY DISPLACEMENTS Figure 2.19: Summary of Isometries. Before pursuing this topic, we must carefully discuss two different interpretations of a displacement. One is a vector transformation where displacements transform point position vectors. The other is a coordinate transformation (which we will use almost exclusively). Here, the position vector is unmoving, but the bases change relative to it Vector Transformation v is the image of v under a CCW rotation through angle θ, centered at the origin. See Figure Given (x, y) and θ, what are (x, y )? x = v cos φ, y = v sin φ x = v cos(φ + θ), y = v sin(φ + θ). We use the fact that v = v, and the trigonometric identities: gives: cos(φ + θ) = cos φ cos θ sin φ sin θ, sin(φ + θ) = cos φ sin θ + sin φ cos θ, x = v [cos φ cos θ sin φ sin θ], y = v [cos φ sin θ + sin φ cos θ].
17 2.5. THE POLE OF A PLANAR DISPLACEMENT 29 Figure 2.20: Vector transformation. Now, using: gives: x = v cos φ v = x/ cos φ, y = v sin φ v = y/ sin φ, x = v cos φ cos θ v sin φ sin θ = x cos θ y sin θ, y = v cos φ cos θ + v sin φ cos θ = x sin θ + y cos θ. Assembling in matrix form yields the general vector transformation for rotation: [ x y ] = [ cos θ sin θ sin θ cos θ ] [ x y ]. (2.2) Coordinate Transformation Given (x, y) and θ, what are (x, y ), the coordinates of the same point in the transformed coordinate system? See Figure By inspection, we have: P x = P x cos θ + P y sin θ, P y = P y cos θ P x sin θ. Assembling in matrix form: [ ] [ x cos θ sin θ = sin θ cos θ y ] [ x y ], (2.3)
18 30 CHAPTER 2. RIGID BODY DISPLACEMENTS Figure 2.21: Coordinate transformation. It is clear the rotation matrices in equations (2.2) and (2.3) are each other s inverse. Hence, the interpretations are also inverse. In kinematics we usually know the coordinates of points in the transformed frame and need to know their coordinates in the original frame. For the rotation matrix in equation (2.3), this is simply the inverse: [ ] [ ] [ ] x cos θ sin θ x = y sin θ cos θ y, (2.4) General Planar Displacements It is convenient to consider a general displacement of a rigid body in E 2 as the displacement of a reference coordinate system E that moves with the rigid body relative to a fixed reference frame Σ. Moreover, let the coordinates of points in E be described by the lowercase pairs (x, y) and those in Σ by the uppercase pairs (X, Y ). A general displacement of E relative to Σ is described by three numbers (a, b, φ), where (a, b) are the coordinates of O E (the origin of E) expressed in Σ, and φ is the angle the xaxis makes with respect to the Xaxis, with CCW rotations considered as positive. The position of a point in E described in Σ can be given by: [ ] [ ] [ ] [ ] X cos φ sin φ x a Y = sin φ cos φ y +. (2.5) b Equation (2.5) is not a linear transformation because the translation of the sum of two vectors x and y by the amount d is x + y + d, and not the sum of the translation of each vector separately, which is ( x + d) + ( y + d) = x + y + 2 d. This violates the definition of a linear transformation.
19 2.5. THE POLE OF A PLANAR DISPLACEMENT 31 If T : V W is a function from the vector space V to the vector space W, then T is a linear transformation iff: 1. T ( u + v) = T ( u) + T ( v), u and v V ; 2. T (k u) = T k( u) u V and all scalars. Clearly, equation (2.5) violates condition 1. Moreover it is not a linear transformation because equation (2.5) cannot be represented by an n n matrix. This situation can be remedied by the use of homogenous coordinates. They will be discussed in detail later, but we can start using them now. They replace cartesian coordinate pairs (x, y) with triples of ratios (x : y : z) such that: x = x z, y = y z, X = X Z, Y = Y Z. Substituting these into equation (2.5) gives: X Z = x z cos φ y sin φ + a, z Y Z = x z sin φ + y cos φ + b. z (2.6) The third coordinate, Z and z may be thought of simply as scaling factors. As long as Z 0 and z 0 we can set Z = z, multiply equations (2.6) through by Z and write equation (2.5) as a linear transformation, which is computationally extremely convenient: X Y Z = cos φ sin φ a sin φ cos φ b x y z, (2.7) Equation (2.7) represents a displacement of reference frame E with respect to Σ. See Figure Displacements of E relative to Σ are completely described by the three numbers a, b,φ where: (x : y : z) are the homogenous coordinates of points in E; (X : Y : Z) are the coordinates of the same point in Σ; (a, b) are the cartesian coordinates of the origin of E, O E measured in Σ, i.e. the position vector of O E in Σ; φ is the rotation angle measured CCW from the Xaxis towards the xaxis, (CCW being considered positive).
20 32 CHAPTER 2. RIGID BODY DISPLACEMENTS Figure 2.22: A general displacement of E with respect to Σ in terms of a, b, φ. The pole of a displacement is the coordinates of the invariant point, which is the eigenvector corresponding to the real eigenvalue of the matrix in equation (2.7). Eigenvalues represent matrix invariants and are roots of the characteristic equation of the following: λ x = A x, (A λi) x = 0. (2.8) With λ being the scalar leading to nontrivial solutions of equation (2.8). Nontrivial solutions exist iff det(a λi) = 0. For equation (2.7) the characteristic polynomial is third order and factors to: The three eigenvalues are: (1 λ)(λ 2 2λ cos φ + 1) = 0 λ 1 = 1, λ 2 = e iφ, λ 3 = e iφ. The eigenvector associated with λ = 1 is the pole of the displacement. We can compute it by setting Z = z = 1 and expanding equation (2.7), revealing two equations in two unknowns; the pole coordinates. From: x p y p 1 = cos φ sin φ a sin φ cos φ b x p y p 1.
21 2.6. OBTAINING THE POLES GEOMETRICALLY 33 We get: x p x p cos φ + y p sin φ a = 0, Solving using Cramer s rule, we obtain: y p x p sin φ y p cos φ b = 0. (2.9) x p = 1 ( ) a sin φ b(1 + cos φ), 2 sin φ y p = 1 ( ) a(1 + cos φ) + b sin φ). (2.10) 2 sin φ A much more symmetric, and as it turns out useful, representation of the pole coordinates are obtained if we use the halfangle substitutions in equations (2.10): giving: sin φ = 2 sin(φ/2) cos(φ/2), cos φ = cos 2 (φ/2) sin 2 (φ/2), x p = y p = a sin(φ/2) b cos(φ/2), 2 sin(φ/2) a cos(φ/2) + b sin(φ/2). (2.11) 2 sin(φ/2) The value of the homogenizing coordinate, which we choose to be Z = z, is arbitrary and we are free to set it to be 2 sin(φ/2), which gives us the homogenous coordinates of the pole (for now we ignore rotations of φ = 180 degrees. X p = x p = a sin(φ/2) b cos(φ/2), Y p = y p = a cos(φ/2) + b sin(φ/2), Z p = z p = 2 sin(φ/2). (2.12) The displacement which ends up translating O E to a, b and rotating the xaxis by φ can be represented by a single rotation about point P by φ degrees. See Figure The pole coordinates of a displacement are used in the kinematic mapping we shall be working with. But there is much ground to cover between here and there. 2.6 Obtaining the Poles Geometrically Obtaining the pole amounts to determining the products of the subgroup of direct isometries. This requires looking at the products of two translations, two rotations, and the product of a translation and rotation.
22 34 CHAPTER 2. RIGID BODY DISPLACEMENTS Figure 2.23: Pole of a displacement Products of Translations Because a translation involves no change in orientation, the product of any number of sequential translations is a translation. Translations are represented by free vectors, i.e. no specific line of action, only direction and magnitude. Hence, the can always be arranged tiptotail, and the product of n translations is the vector joining the tail of τ to the tip of τ n. See Figure The product Figure 2.24: The product of two translations is always a translation. Moreover, translations are a commutative subgroup of isometries.
23 2.6. OBTAINING THE POLES GEOMETRICALLY 35 τ is also a free vector: it is a directed linesegment that can be placed anywhere in the plane. These line segments all point towards the same point on L (so, we are really talking about P 2, the projective plane... more on that later). Similarly, the normals to τ all intersect a different, but unique, point on L. See Figure The projective plane P 2 is bounded by a line that is infinitely Figure 2.25: Model of the line at infinity, L. distant. This line is the intersection of P 2 and the plane at infinity, π, and hence must be a line. However, it is a line with some interesting properties. Two parallel lines intersect in only one point on L, regardless of which direction that L is approached. So, for illustration, is is usually shown as a circle, though it s equation is linear. With the above in mind, the product of two translations can be represented by a rotation about a unique center S on L through an infinitesimally small angle. Thus, the pole of a translation is a point on L Products of Rotations Clearly, the product of any number of rotations about a single center is a rotation about that same center through the sum of the angles of the individual rotations. For two rotations with distinct centers, there are three cases to consider. Using there, the product of any number of rotations about arbitrary centers can be determined. This product will be a rotation (plus/minus). Case 1: Same rotation angle, two distinct centers. This product will be written as A θ B θ. Recall, we can represent a rotation A θ in many ways by two
24 36 CHAPTER 2. RIGID BODY DISPLACEMENTS intersecting mirror lines. Actually, they are a pencil of line pairs on the rotation center separated by angle θ/2. See Figure Figure 2.26: Pencil of lines; range of points. Select r to contain both centers A and B. Then, A θ = R 1 R and B θ = RR 2. The product of the two rotations A and B is another rotation, C φ : A θ B θ = R 1 RRR 2 = R 1 R 2 = C φ. (Note: The two identical rotations, R, combine as the identity, I) The rotation angle is easy to obtain from a basic triangle theorem: the sum of the interior angles equals 180 degrees ( or π radians). See Figure Hence, we can write: Figure 2.27: Two rotations A θ B θ, same angle, distinct centers. θ 2 + θ ( φ ) = φ = 2θ. The rotation centre of C is the invariant point of the product of A θ and B θ.
25 2.6. OBTAINING THE POLES GEOMETRICALLY 37 Case 2: Same angle, but opposite sense, two centers. We proceed as in Case 1. See Figure Select r to contain both centers A and B. Figure 2.28: Different angles and centers. The distance d is simply A θ B θ = R 1 RRR 2 = R 1 R 2 = τ ( d = AB sin 180 θ ) 2 But as we discussed earlier, this translation in E 2 can be considered a rotation in P 2. Case 3 Different angles, distinct centers. See Figure Figure 2.29: Three different rotation centres and angles. A θ B φ = R 1 RRR 2 = R 1 R 2 = C ψ
26 38 CHAPTER 2. RIGID BODY DISPLACEMENTS The angle ψ is obtained as: θ 2 + φ 2 + (180 ψ 2 ) = 180 ψ = θ + φ This rotation center C can be obtained (for both Case 1 and 3) using the Law of Sines, as shown in Figure Figure 2.30: Law of sines. sin α a = sin β b = sin γ c c = sin γ sin β b Or, as we have for the Case 3, triangle CB = sin(θ/2) sin(180 ψ/2) AB which gives the coordinates of C in reference frame Σ. We see immediately that the products of rotations are not a commutative subgroup Products of Translations and Rotations Here we can have τs and Sτ, which are in general different displacements. Hence, these products are generally not commutative. TranslationRotation τs θ : We decompose the product as in Figure 2.31: τs θ = R 1 RRR 2 = R 1 R 2 = P θ. This product is a rotation through the same angle θ but about a translated center. The coordinates of P relative to S are simply the basis elements of the position vector of distance SP given by: SP = d sin(θ/2)
27 2.6. OBTAINING THE POLES GEOMETRICALLY 39 Figure 2.31: Product of translation and rotation. Figure 2.32: Product of a rotation and a translation. Rotation center P is the pole of the product of displacements. RotationTranslation S θ τ: We take the same displacement but in opposite order, as shown in Figure 2.32 and write: S θ τ = R 1 RRR 2 = R 1 R 2 = Q θ, which is again a rotation, but about a different center. Hence these products are not commutative, but still a subgroup of the group of planar isometries.
28 40 CHAPTER 2. RIGID BODY DISPLACEMENTS 2.7 Computing DOF Employing Group Concepts The relative motion associated with each of the lower pairs (shown again on the following page) constitute a subgroup of the group of Euclidean displacements G 6 under the binary product operator (i.e. the composition of two displacements). In other words, the motions of all Rpairs are a subgroup of G 6, the motions of all Epairs are a subgroup of G 6, etc... The dimension of these subgroups is defined to be the DOF of the relative motion permitted by the lower pair. The dimension is indicated by dim(g S ), where G S G 6. These subgroups, together with their corresponding dimension are listed in Table (2.2). Table 2.2: Lower Pair SubGroups and Their Dimension Lower Pair G S dim(g S ) E E 3 S S 3 C C 2 H H 1 P P 1 R R 1 Let the product of two subgroups indicated by G = G 1 G 2 be the composition of the displacements they represent. Let dim(g ) = d. In the following equation d represents the maximum possible motion group dimension. In E 3, 3D Euclidean space, d = 6. In E 2, the absolute maximum is d = 3. Consider the RPR linkage shown in Figure 2.33 (implying a serial chain of four links connected to ground by a sequence of revoluteprismaticrevolute joints). Reference frame E has, at most, 3 DOF relative to the groundfixed reference frame Σ, hence d = 3. In E 3, d = 6. The i th kinematic pair imposes µ i constraints on the two links it couples. For example, in E 3 R,P, and Hpairs impose five constraints, a Cpair imposes four constraints, while S and Epairs impose three The ChebyshevGublerKutzbach (CGK) Formula Clearly, l unconstrained rigid links have d(l 1) relative DOF, given that one of the links is designated as a nonmoving reference link. Any joint connecting two neighboring rigid bodies removes at least one relative DOF. If the joint removes no DOF then the bodies are not connected. If the joint removes 3 DOF in the plane, or 6 DOF in E 3 the two bodies are a rigid structure. Summarizing this
29 2.7. COMPUTING DOF EMPLOYING GROUP CONCEPTS 41 Figure 2.33: Reference frame E has 3 DOF relative to Σ d = 3. discussion, the DOF of a kinematic chain, relative to one fixed link in the chain, can be expressed as: d(l 1) j µ i m = DOF, (2.13) i=1 where d = dim(g ), l is the number of links including the fixed link, µ i is the number of constraints imposed by the i th joint, j is the number of joints, and m represents the number of idle DOF of the chain. The idle DOF of a chain are the number of independent single DOF motions that do not affect the transmission of motion from the input to the output links of the chain. Consider the RSSR linkage shown in Figure The coupler is free to spin about its own longitudinal axis between S 1 and S 2. The input link drives the linkage by rotating about axis R1. The output link rotates about axis R2. Clearly, the free spinning of the coupler does not contribute to the DOF of the chain. Hence, in this case, m = 1. This is an example in E 3 so d = 6. There are four links, including the rigid frame to which R 1 and R 2 are fixed, thus l = 4. Each Rpair imposes µ = 5, while each Spair imposes µ = 3. Thus: 6(4 1) (2(5) + 2(3)) 1 = DOF, = 1 DOF. The idle DOF indeed exists, but does not contribute to the DOF of the RSSR chain. Equation (2.13) is known as the ChebyshevGublerKutzbach (CGK) formula. These three kinematicians independently developed it.
30 42 CHAPTER 2. RIGID BODY DISPLACEMENTS Figure 2.34: An RSSR spatial linkage has 1 DOF Examples 1. Dump Truck Mechanism: (simple closed chain). See Figure d = 3, l = 4, µ i = 2, j = 4, m = 0. Figure 2.35: Dumptruck linkage is a planar RPRR mechanism. j d(l 1) µ i m = 3(4 1) 4(2) 0 = 9 8 = 1 DOF. i= Bar Peaucellier Inversor: (complex closed chain). See Figure d = 3, l = 8, µ i = 2, j = 10, m = 0. j d(l 1) µ i m = 3(8 1) 10(2) 0 = = 1 DOF. i=1 3. 3R Serial 3D Robot Arm: See Figure d = 6, l = 4, µ i = 5, j = 3, m = 0.
31 2.7. COMPUTING DOF EMPLOYING GROUP CONCEPTS 43 Figure 2.36: 8bar Peaucellier inversor, a complex closed chain with 1 DOF. Figure 2.37: 3R serial spatial robot. j d(l 1) µ i m = 6(4 1) 3(5) 0 = = 3 DOF. i=1 4. Landing Gear Downlock: See Figure d = 3, l = 5, µ i = 2, j = 5, m = 1. j d(l 1) µ i m = 3(5 1) 5(2) 1 = = 1 DOF. i=1 5. Planar 3Legged Platform: See Figure d = 3, l = 8, µ i = 2, j = 9, m = 0.
32 44 CHAPTER 2. RIGID BODY DISPLACEMENTS Figure 2.38: Landing gear downlock. Figure 2.39: Planar 3legged manipulator (complex closed chain). d(l 1) j µ i m = 3(8 1) 9(2) 0 = = 3 DOF. i=1 6. StewartGough Flight Simulator Platform: See Figures 2.40 and d = 6, l = 14, µ i = 5, j = 18, m = 6. j d(l 1) µ i m = 6(14 1) 6(5) 12(3) 6 = = 6 DOF. i=1
33 2.7. COMPUTING DOF EMPLOYING GROUP CONCEPTS 45 Figure 2.40: StewartGough platform (flight simulator). Figure 2.41: A StewartGough platform. 7. Bench Vice: The bench vice presents some interesting diversion. At first glance, it is planar, so we immediately set d = 3 without hesitation: d(l 1) = d = 3, l = 3, µ i = 2, j = 3, m = 0. j µ i m = 3(3 1) 3(2) 0 = 6 6 = 0 DOF!!! i=1 Bench vice have 1 DOF. What gives? This seeming anomaly is an artifact of representation. The R and H axes are parallel while the translation direction of the Ppair is also parallel. See Figure 2.42 and The P and R joints are kinematically equivalent to a single Cpair. Moreover, the axis of the Hpair is parallel to the axis of the Cpair. Therefore, the dimension of the motion subgroup represented by the common bench vice
34 46 CHAPTER 2. RIGID BODY DISPLACEMENTS Figure 2.42: Bench vice schematic. Figure 2.43: Bench vice. is d = 2, not d = 3. Thus: d(l 1) d = 2, l = 3, µ i = 1, j = 3, m = 0. j µ i m = 2(3 1) 3(1) 0 = 4 3 = 1 DOF. i=1 So one must be careful blindly applying the CGKformula. Still, note that we had to treat the Cpair as a Ppair and an Rpair separately. Be warned though that the subgroup dimension argument is controversial. The geometric model of the CGKformula is incomplete. Certain geometric mechanism properties are not modeled by it. As a result, there
35 2.8. A NOTE ON THE AXES OF RPAIRS AND PPAIRS 47 are many mechanisms with 1 DOF that are not identified by the CGKformula. The Bennett 4R Linkage is one of the best known examples, as shown in Figure Opposite links are twisted by the same angle Figure 2.44: A Bennett mechanism. and have the same length. Twist angles α 1 and α 2 are proportional to lengths a 1 and a 2. Similarly, α 3 and α 4 are proportional to a 3 and a 4. The proportionality is governed by: Using the CGKformula we get: d(l 1) a 1 sin α 1 = ± a 2 sin α 2. d = 6, l = 4, µ i = 5, j = 4, m = 0. j µ i m = 6(4 1) 4(5) 0 = = 2 DOF!!! i=1 The CGKformula predicts a hyperstaticstructure, when the Bennett mechanism actually has 1 DOF. Refining the CGKformula to model such geometric conditions is still an open problem. 2.8 A Note on the Axes of RPairs and Ppairs The axis of an Rpair is a line through the invariant center point of rotation perpendicular to the plane of motion. But it does not make sense, from a
36 48 CHAPTER 2. RIGID BODY DISPLACEMENTS mechanical engineering point of view, to speak about the axis of a Ppair in the same way, as no real points in E 2 are invariant under a translation. Ppairs Figure 2.45: Ppair axis. permit translations parallel to one direction. One such translation, indicated by τ, is shown in Figure Mathematically, the axis of a Ppair could be described as the line at infinity, N, of all planes normal to the direction of τ. This is illustrated in Figure 2.45 where Σ is the plane containing the Ppair, τ is a particular translation effected by the Ppair, N 1 and N 2 are normal to Σ, and Ω is the plane at infinity. The two planes Σ and Ω intersect in L. Lines in the direction of τ intersect L in point P 1. Lines normal to τ in Σ, indicated by η, intersect L in P 2. The line N is the intersection of all planes normal to Σ and parallel to η. Moreover, all normals to Σ, N 1, and N 2 being two of them, intersect N in the point P 3. The join of P 2 and P 3 is N, which is the axis of the particular prismatic joint. In other words, the axis of a Ppair is the absolute polar line to the point at infinity of the direction of translation. More on polar lines when we discuss projective geometry. Regardless, Ppairs would be impossible to manufacture if they had no longitudinal axis of symmetry to establish the direction of translation, i.e. no longitudinal centerline. One must not confuse this centerline with the joint axis, which is, for mechanical reasons, inaccessible.
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