Appendix A Solving Constraint Equations A.1 The 4R Linkage Constraint The analysis of planar and spherical 4R linkages, the spatial RSSR, and Bennett’s linkage all yield a constraint equation between the output angle y and the input angle q that takes the form A(q)cosy + B(q)siny = C(q): (A.1) There are two ways to solve this equation the trigonometric solution and the tan- half-angle technique. p The trigonometric solution begins by dividing both sides of (A.1) by A2 + B2. This allows us to introduce the angle d such that A B cosd = p and sind = p : (A.2) A2 + B2 A2 + B2 Notice that d = arctan(B=A). The left side of (A.1) takes the form cosd cosy + sind siny = cos(y − d): (A.3) Thus, the right side must be the cosine of an angle k, C cosk = p : (A.4) A2 + B2 Because k and its negative have the same cosine, we have that d + k and d − k are both solutions to (A.1). This combines with the definition of d to yield B C y = arctan ± arccos p : (A.5) A A2 + B2 J.M. McCarthy and G.S. Soh, Geometric Design of Linkages, Interdisciplinary Applied 411 Mathematics 11, DOI 10.1007/978-1-4419-7892-9, © Springer Science+Business Media, LLC 2011 412 A Solving Constraint Equations Notice that the angle k exists only if −1 ≤ cosk ≤ 1. Therefore, C2 ≤ A2 + B2, or equivalently, A2 + B2 −C2 ≥ 0 must be satisfied for a solution to exist. In this formulation d is the angle to a diagonal of the quadrilateral formed by the linkages. The angle k is measured on either side of this diagonal to define the output angle y. The tan-half-angle technique uses a transformation of variables to convert siny and cosy into algebraic functions. Introduce the parameter y = tan(y=2), which allows us to define 1 − y2 2y cosy = and siny = : (A.6) 1 + y2 1 + y2 Substitute this into (A.1) to obtain (A +C)y2 − 2By − (A −C) = 0: (A.7) This equation is solved using the quadratic formula to obtain p y B ± A2 + B2 −C2 tan = : (A.8) 2 A +C In order to have a real solution we must have A2 + B2 −C2 ≥ 0. This is the same condition obtained above for the trigonometric solution. The constraint equations for these linkages are each special cases of a general equation that we can write as (a1 cosq+a2 sinq − a3)cosy + (b1 cosq + b2 sinq − b3)siny −(c1 cosq + c2 sinq − c3) = 0; (A.9) where ai, bi, and ci are constants. Introduce the tan-half-angle parameters x = tan(q=2) and y = tan(y=2) so we have 2 2 ((a1+c1 + a3 + c3)x − 2(a2 + c2)x − (a1 + c1 − a3 − c3))y 2 −2((b1 + b3)x − 2b2x − (b1 − b3))y 2 −((a1 − c1 + a3 − c3)x − 2(a2 − c2)x − (a1 − c1 − a3 + c3)) = 0: (A.10) This is a biquadratic equation in the unknowns x and y. A.2 The Platform Constraint Equations The 3RR planar and 3RR spherical platforms can be viewed formed from two 4R linkages OAB1C1 and OAB2C2 driven by the same crank OA. Let q be the angle of the input crank OA and let f be the angle of the coupler at A. The constraint equations for the two 4R linkages yield A.2 The Platform Constraint Equations 413 A1(q)cosf + B1(q)sinf = C1(q); A2(q)cosf + B2(q)sinf = C2(q): (A.11) We now present two ways to solve these equations. The first eliminates sinf and cosf linearly, and the second uses the resultant to solve simultaneous biquadratic equations. For the first solution let x = cosf and y = sinf and solve the resulting linear equations using Cramer’s rule to obtain C B −C B A C − A C x = 1 2 2 1 and y = 2 1 1 2 : (A.12) A1B2 − A2B1 A1B2 − A2B1 In order for these equations to define a solution f they must satisfy the identity x2 + y2 = 1. This yields an equation in cosq and sinq, given by 2 2 2 (C1B2 −C2B1) + (A2C1 − A1C2) − (A1B2 − A2B1) = 0: (A.13) Introduce the tan-half-angle parameter x = tan(q=2) so this equation becomes the polynomial P(x) = 0. For each root x j of P(x), compute q j and determine the coef- ficients Ai j, Bi j, and Ci j of the platform equations (A.11). Solve either one equations determine f j. An alternative approach transforms (A.11) into a pair of biquadratic equations by introducing the tan-half-angle parameters x = tan(q=2) and y = tan(y=2). This yields the equations 2 D1y + E1y + F1 = 0; 2 D2y + E2y + F2 = 0; (A.14) where 2 Di = d1ix + d2ix + d3i; 2 Ei = e1ix + e2ix + e3i; 2 Fi = f1ix + f2ix + f3i: (A.15) To solve these equations, we introduce a second pair of equations obtained by mul- tiplying both by y and assemble the four equations into the matrix equation 2 38 39 8 9 0 D1 E1 F1 y 0 > 2> > > 6 0 D2 E2 F27<y = <0= 6 7 = : (A.16) 4D1 E1 F1 0 5 y 0 > > > > D2 E2 F2 0 : 1 ; :0; This equation can be solved for the vector (y3;y2;y;1)T only if the coefficient ma- trix [M] has determinant zero. Expand this determinant to obtain an eighth-degree polynomial P(x) in the parameter x. The roots of this polynomial define x j for which (A.16) can be solved to determine y j. Appendix B Graphical Constructions The following constructions use a straightedge to draw lines and a compass to con- struct circles and measure distances. They are useful in the graphical synthesis of planar RR chains. B.1 Perpendicular Bisector Given two points P1 and P2, we construct the perpendicular bisector L of the seg- ment P1P2, Figure B.1, as follows: 1 2 1. Construct circles C1 and C2 centered on P and P with radii equal to or greater than one-half the length of P1P2. 2. C1 and C2 intersect in two points. Join these points to form the perpendicular bisector L. L C 2 P2 P1 C1 Fig. B.1 Construction of the perpendicular bisector L of the segment P1P2. 415 416 B Graphical Constructions B.2 Circle Through Three Points Given the triangle 4P1P2P3, the center C of the circle that circumscribes this trian- gle, Figure B.2, is given by the construction: 1 2 2 3 1. Construct the perpendicular bisectors L1 and L2 to the segments P P and P P . 2. The intersection of the lines L1 and L2 defines the center C of the circle through the three points. L2 P3 P2 L1 P1 C Fig. B.2 Construction of the circle through three points P1, P2, and P3. B.3 Duplication of an Angle Consider two lines L1 and L2 that intersect in an angle a at the point P. The line M2 that makes the same angle with another line M1 about a point Q, Figure B.3, is constructed as follows: 1. Draw a circle CP such that it intersects L1 and L2 in points S and T. 2. Construct a circle CQ with the same radius about Q and denote by U one of the intersections with M1. 3. Measure the distance ST and construct the circle with this radius about U. Its intersection with CQ is a point V. Join Q to V to define the line M2. The circles intersect in two points, so choose the one that provides the same orientation for M2 as L2 relative to L1. B.4 Reflection of a Point Through a Line 417 L1 CP L2 S α CQ T U P Q α M1 V M2 Fig. B.3 Duplication of the angle a about the point P from the line M1 to determine the line M2. L C C R S R S * Q Q Fig. B.4 Reflection of the point Q through the line L. B.4 Reflection of a Point Through a Line In order to find the reflection of the point Q through the line L, Figure B.4, we use the construction: 1. Select two points R and S on L and construct the circles CR and CS with radii RQ and SQ, respectively. 2. The circles CR and CS intersect in two points. One is Q and the other is its reflec- tion Q∗. Appendix C Spherical Trigonometry Consider the spherical triangle 4S1S2S3, where the axes are labeled in a counter- clockwise sense around the triangle, see Figure C.1. Associated with the side SiS j we can define the normal vector Ni j = Si ×S j=jSi ×S jj. The angular dimension ai j of this side is defined by the equations cosai j = Si · S j: (C.1) Thus, we can compute the three angles a12, a23, and a31, which we consider to have a positive magnitude between 0 and p, thus ai j = a ji. The sense of the angles ai j will be determined as needed relative to the normal vector Ni j. N12 3 S3 α31 α x 23 z 2 S1 F1 1 S2 α12 Fig. C.1 The frame F1 has its x-axis along N12 and its z-axis along S1. At each vertex Si we denote the exterior dihedral angle by fi, which is defined by the formula Nki × Ni j · Si Sk × Si · S j tanfi = = : (C.2) Nki · Ni j (Sk × Si) · (Si × S j) 419 420 C Spherical Trigonometry The indices (i; j;k) in this equation are any one of the three cyclic permutations (1;2;3), (2;3;1), or (3;1;2).