MINIMIZATION OF CLOSURE TIME FOR A FAST ACTING SWITCH David B. Clauss Member Tech*Wial Staff Applied Mechanics Division III Sandia National Laboratories Albuquerque, NM 87185 Panos Papalambros Assistant Professor Mechanical Engineering and Applied Mechanics Dept. University of Michigan Ann Arbor, MI 48109 Technical Report UM-MEAM-81-3 *Also appeared as Sandia National Laboratories, Report SAND80-0790J February 1982 Abstract A mechanisms optimization problem involving minimization of closure time is solved in this paper using noniterative global optimization techniques, whereas most mechanisms problems in the literature have focused on path optimization and use iterative. methods. NMonotonicity analysis is used successfully to identify active constaints which reduce the problem to one degree. of.freedom. The solution method may be applicable.to a variety of dynamics problem in which algebraic.relations can be constructed from the governing differential equations. This work was supported by the U. S. Department of Energy Contract DE-AC094-76DP00789

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Introduction Previous optimization work on mechanisms has focused on path optimization problems. The problem presented in this paper represents a different class of mechanisms optimization problems; the objective is to minimize closure time. A simple design for a fast acting electrical switch, [1], is shown in Figure 1. Although this design is perhaps artificially simple, the solution does suggest an optimization method which may be applicable to more complicated designs. The simple slider crank mechanism functions as a fast acting switch in the following way: The helical spring is compressed initially by the piston, which is held in place by a restraining pin. When the pin is released, link 2 swings shut to complete the circuit. The mechanism is assumed to be constructed of rigid parts with uniform cross section. Parts are connected with rotational contacts. Minimization of closure time implies obtaining the maximum possible acceleration over the shortest possible distance. This results in tradeoffs between the initial spring load (which depends on the spring constant and deformed spring length) and the spring shear strength, and also between the length of link 2 necessary to achieve linear motion of part 2 and the maximum width of the mechanism. The optimization problem is originally formulated with thirteen degrees of freedom and twenty-one constraints. Global optimization is achieved using the methods presented in [2], [31, and [4]. The problem is reduced to seven degrees of freedom by direct elmination, which is used due to the large number of equality constraints. Monotonicity analysis [4] and implicit elimination are then used to identify constraint activity, and the problem is successfully reduced to one degree of freedom. The optimum is located by mapping the objective against the remaining degree of freedom on the feasible domain.

Derivation of the Mathematical Model The objective for optimization of the fast acting switch shown in Figure 1 is minimization of the closure time, tf. Initially time is defined only implicity, the motion of the mechanism being time dependent. Thus, the objective function is originally undefined because it must be derived from the differential equations of motion describing the slider-crank mechanism. The slider crank mechanism has only a single degree of freedom and consequently the objective function should be rather straightforward. Typically the important variables are the system mass, damping and stiffness and the initial and final configuration of the mechanism. The optimization problem is originally stated in the form min tf subject to: constraints, where the constraints are grouped in the following categories: (1) differentialI equations of motion, (2) loop equations, (3) design space and geometry, (4) initial - final configuration, (5) helical spring and (6) strength of joints. The differential equations of motion are derived using the constrained D'Alembert method [5], m ape n + 3fk Fj q-k -qi. (1) j=1 k=l In the analysis several forces were treated as negligible. Gravitational forces and mass inertial forces on links 2 and 3 were assumed small compared to the rotational inertial forces, and frictional forces were considered negligible. The model could be further refined by including these forces, although the added complexity would make solution for the objective function very difficult. Details of the derivation of the equations of motion are given in Appendix A. The result is, in matrix form, I2 O 0 -rsinO2 r2cos 1 0 2[ 22 13 0 -rsinO3 r3cos03 3 0 O M4 1 0 J I -k(S-S (2) k S - ud) A.

The loop equations and their derivatives express geometric constraints and result in six equalities: r2sine2 + r3sine3 0, r2cos82 + r3Cs3 + P4 + S g = 0 r20 2sin(02 - 63) - scose3 = 0, r3 3sin(3 82) S- 0cos2 = 0, -r262 r383sin(3 -2 - 2 - ) + scos8 0 -r 22sin( - r2 c- (0-) -e r 2 + 2262c 3) 2 2 2 - 3 3 3 3 - Again, the details are in Appendix A. The design space is limited, in a two-dimensional sense, to a rectangular area with height L and width 2L. The total length of link 2, L, is fixed, but the position of the joint with link 3 is variable. Thus, R1: r2 <L. (4) The total width of the mechanism cannot exceed 2L so R2: g < 2L. (5) The angles 82 and 83, measurer' as shown in Figure 1, are limited by the design space such that 0 < 02< 7/2, 37/2 < 83 < 2r. With respect to this design, the path of the mechanism can be left quite arbitrary, so that the initial and final conditions result in the most important constraints on the problem. In.order for any motion to take place, the spring must be compressed initially, so that R3: s(t = 0) = so < Sud ~ (7) The piston also has zero initial velocity s(t = 0) = s =. (8) The distance between the end of link 2 and the contact must be greater than some distance B to prevent arcing of the electrical signal when the switch is open R4:820 < W. (9)

Further constraints are placed on 020 (although the above will likely be the least U!pper bound) and 83o by applying the design space criteria (6) at time t = 0 R5: 20 <'u/2, (10) 30 - 2 R7: 30 < 2r. (12) Note that satisfactory closure of the switch implies the final angular position of link 2, 02f, is 82f = w/2. (13) Using equation (13), application of the first two loop equations (3) at the initial and final time leads to R8 r2sin2 + r sin0 =0 (14) r2s 2o 3 3o R9: r2cos0 2o + r3cos30 P + p g = 0, (15) R10: r2 + r3sin3f = 0(16) R11: rcos3f + P4 + Sf - g = 0 (17) Finally note that Sf < smax (18) For an undamped system the spring will oscillate such that max - Sud Sud o (19) Rearranging (19) and substitution into (18) yields the approximate relation R12: sf < 2s s (20) f ud o The only variables with respect to the helical spring are the spring constant, k, the number of coils, N, and the undeformed spring length sud; all other quantities which affect the spring are treated as parameters. The spring constant and the number of coils are related by d 4G R13: k = (21) 8D N

The allowable deformation of the spring is limited by the spring wire shear strength [6] such that KsDk(sud - so) R14:..... < 1, (22) 0.17Ad3m and K Dk(sf - s ) R15: <-m 1, (23) 0.17Ad 3-m Geometrical considerations lead to R16: so > Nd (24) which is the final constraint derived from consideration of the helical spring. Typically in the design of simple mechanisms such as considered here the joints-between links will be the weakest part. The reaction forces can be calculated using the method of virtual work and equilibrium ideas; they must not exceed the joint strength. The result is 2 F (25':M4 - d r cose 4 u- 3 rated' M4s + k(s + r cos8 (M4 + k(s - Ud <Frate (25 The details of the determination of reaction force is shown in Appendix B. At this point the objective function must be derived, and the constraints should be put in a form such that the problem is well posed. This requires elimination of all the differential quantities in constraints such as equation (25), and further solving the differential equations of motion (2) for the objective function tf. Consider the third of equations (2) Mq4 + k(s - sud) + i =, (26) and note that Xi can be solved for in terms of e2 and 83 by writing a virtual work expression (see Appendix B) 123 c2s 3 i383cose2 2 2s3 3... i =-r2sin (2-3 ) r3sin(83 -e2) (27) Then.using the last four of equations (3) to eliminate e82 and 03 in (27), and substituting the result into (26) m's + cs + k(s - Sud) = (28)

where 2 2 I2cos M3 I3cos 02 m= M4 +2 2+ 2.2 r2 (62 -803) r3sin (83-02) (29) 3 3 [I2 3 3 2 cos(3 2) r2 r3 sin (03 92) (30) I cos 233 I2cs2 3cs 03 cose2cose3 + 2 4., r 2 r3 r2r3 sin (02 3) Since 02 and 03 are functions of time (theoretically they can be solved in terms of s), and since the s term is squared, the differential equation (28) is highly nonlinear. In order to simplify the solution of the problem, the equation is approximated by m s + cs + k(s-sud) = 0 (31) where m* and c are treated as constants with respect to time (see Appendix C): m' = M4 + 083M2.021M3 (32) r2.0083M2L.0042M3.033M2L.067M3 23r 3r (33) r r3 r2 r3 2 Equation (31) together with the initial conditions (7) and (8) has a straightforward solution _o Sud -~ t t _ 2 s - Sud e n sin w t + tan (34) where wnc mn ~.0083M2.0042M1.033M2L.067M3 R17- =n 2 - 2k 3 + + 2r3 r2 1(35) L2 + + r21 (35)

and 1/2.083M 2L -1/2 R18: n = = k -M4 + +.021M (36) 2m.. Note that we are concerned with the motion during, at most, one-half cycle and thus the decay should be relatively small, so that e nt, 0 < t < tf.(37) using the above result and solving for sf in (34) s S sin _ 2utf + tan f - ud = n-f tan (38) Finally, the objective function is obtained by solving the above for tf: ~ ~ ~..........1 sf Sud min tf - 1 2 i 1 2j tan 1 (39) In the analysis a sinusoidal response has been assumed which implies R19: i < 1. (40) Only equation (25) contains differentials; these can be eliminated by differentiating (38) twice to obtain s and solving for 03 in terms of s. The approximate result is Sud So 2 R20: (M k) Frated (41) Mathematical Model Substituting symbols xi, i = 1,..., 13 for the corresponding physical variables, the problem can now be stated in normalized form as: (X x (1 - X 21/2 min tF = sin-1 3 4(1 - 1) -tan1 (1 - x12) 1/2 xl (1 x ~2)1/2 ~ 1 ~~J (1 X.

subject to: 1 -1 x7L < 1 7:L R10: -x -1xinxll = 1 10 - -1 -1 x5x4 1 R: x [x6cosxll + P4 + x3] = 1 L; x13 W -1< 1 R12: 3[2 ] < 1 x31 W - x5]- < <1 2 3 4 1 x:< R13: 8D3X d4G1 1 1 -1 m-3 2 X12 < 1 R14: 5.88A d KsD x(X4 - x) 1 7: x 1 R15: 5.88A d m-3KsDx8(x3 - x4) < 1 2ir 12- 8 3 4 -1 -1 -1 B: -x7(sinx) (sinx13) R16 ( x95 d < 1 9: x10 [x7cOsx13 + X6 C~SX12 + P4 + x5] 1 ~~-1 -3 2 -1 -1 -2 2 -1 R17: 2xlx2 x8 [.0083M2x7 L +.0042M3x6 +.033M2X6 7 L +.067M3x7 ]=1 R18: xx8 1 [M4 +.083M2X7 L +.021M3]1/2 1. R19: x1 < 1 2 -1/2 2 -1 R20: 1.2(x4 - x5) (1 - x1 ) (M4x2 8 Frated < 1 - ~~~~~(M4x22 - x8) Frated -

Direct Elimination Many of the constraints in the mathematical model are equalities and can be used to eliminate appropriate variables. In many cases the elimination leads to the derivation of additional inequality constraints from the equalities in order to preserve the original feasible domain. Constraint R10 can be used to eliminate xll, the final angle of part 3 -1 -1 Xll = sin (-x7x6 ) which also implies RlOa: x7x6 < 1. (43) Similarly R8 and R9 are used to solve for X12 and X13, the initial angle of parts 3 and 2, respecively. Solving for these v-riables exactly is fairly difficult, however, adequate bounds can be imposed. Consider R8: -X7(sinxl13)x6 (sinx12)- 1. (44) Since x7x6 < 1, sinx13 < -sinxl2so that R8a: x13(2r - x12)- 1 (45) Similarly, R9 implies R9a: x 10[x7 + x6 + p4 + x5]-i < 1. (46) In order to bound x12 and x13- R8 and R9 must be solved simultaneously. Use R8 to solve for cosx13, cosxl3 = [1 - (x6x7 1) sin x12), (47) and substitute into R9 x10 [X((x6x7 1) 2sin 2x2]1/2 + x(1 - sin2x12) 1/2 +p4 + x5} 1 (48) x10 7 6 7 12 6 or, rearranging, -12 2 21/2 +.2 l/2 x7(1-(x6x7 ) 6(sin x12 - x10-P4-X5 (49) Since x7 < x6, it follows that 2 2 2 1/2 2 2 2 1/2 (x6 - x, sin xi) > (X - x sin x ), (50) 6 Ae-l. 7 6 12 (50) and 2x7[1 - (xsx1 ) 2sin2x]] 1/2< xo - p4 - x5. (51)

Finally, the following results are obtained: -1 -l x13 > cos [(X10 - P4 - x5) (2x7) ]' 13 (52) 12 < sin1 - {(x7x612 -[ (x10 - P4 - x:5) (2x6) -1] 1/2... 7x 6 10 P4 5 6(53) x12 > sin {1 - [(x P4 - x5)(2x6) -1 21/2 1210 (54) which can be substituted into R6, R8 and R9, respectively, to effectively eliminate xl2 and x13. Variables xg and x10 can be eliminated easily using the equalities R13 and Rll. From R13 -3 -l 4 =.125D x8d4G (55) and from Rll, using equation (42), = (X62 72 1/2 (56) It is advantageous to eliminate x2 from the problem using equality constraint R18: X2 = x1/2[M4 +.083M2x7 -2L2 +.021M3]1/2 (57) Although this makes the objective function more complicated R18 is a difficult equality to direct, and so using R18 to eliminate x2 is the best approach. The problem can now be reformulated as a seven degreee of freedom problem in a form much more suitable for monotonicity analysis. Equality constraints have been used to eliminate six degrees of freedom. In the reformulation R5 is omitted since it is apparent that R4 is the least upper bound. Constraint R15 is redundant (R15 is active if R12 and R14 are both active) and it is also dropped. Using equations (54) - (57) to eliminate x2, and x9to x13, the problem can now be stated as: min tF = fsinl[ ( ( ) 4 tan1 (lx2) 1/2 M+.083M2 x72L2 +.021M3)/2 X 1-x1,2 1/2

subject to: R1: x7L 1 R2: 0. 5 ( X6-X7 )1/2 + p4 + X] L I R3: x - R3 x5x4 1 1 R4: 2x7CosW[(x6-x7)1/ + x- x3 ] 11 R6: 1.5r/sin1 - {(x7x61) (x10-p4-x5)(2x6) / 1 R, ~~oo~ x~-~~~/~+x~- 46 R8a: cos [(x0-p4-x5) (2x7) ] [2 r-7 1 -1 sinl1 - 1i- [(x10-p4-X5) (2x6) 27 1 /2]-i i R9a: x0 (x7+x6+p4+x5) 1 < 1 R10a: x7x6 < 1 R12: x3(2x4-x5) 1 1 R14: 5.88A d KsDx8 (x4-x5) < 1 R16: -1 -1 5 R16:.125D3 x x5 d G < 1 -32 0042M 1 -1 -2 -1 [.0083M2x7 L + 0042 3x6 +. 033M2x6 x7 L +. 067M3x7 R17: 2 1/2 083M -22 1/2 2 [M4+.083M2x7 L +.021M3] R19: x1 < 1 2 l/2 x[M. -2 2 -1 -1 R20: 1.2(x.-x-)(l-x1 [M(M4+.083M2x7 L +.021M3) - ]Frated 1.

Monotonicity Analysis The monotonicities of the variables in the constraints Can be determined by inspection. However, the objective function depends on the sine and tangent, which are double valued functions. Thus a branch and bound technique must be used to determine the monotonicities. With respect to the tan-1 term no branching need be done since the feasible domain is limited to the first quadrant and tan- is a strictly increasing function in this range. However, for the sin- term the feasible domain includes the second and third quadrants, and the monotonicities of the inverse sine are opposite each other in these quadrants. Due to constraint R3 (x5 -x4) is always negative, so that the sign of (x3 -x4) will determine the monotonicity. Thus, two cases must be considered: Case I with x3 > x4 and case II with x3 < x4. Consider case I first: An additional constraint is implied by the branching, R21a: x3 x4 < 1(59) For this case the inverse sine is increasing. To check the monotonicity of x4 in the objective it is sufficient to show the monotonicity of x4 with respect to sin (x (X "4) sinX4) (1 x12~1/2 5 -1 -l (60) g = sin (x3 - x4) (1 X l2)1/2 (x5 x4) = il (60) Note that 1/ (61) ax4 av ax4 (1 - V) 1/2 X5 1 X4) (x5 4 Since x3 - x5 > 0, it follows that g > ~ (62) and the objective is increasing with respect to x4, everywhere.

The monotonicity of x1 in the objective is more difficult to show. However, note that if 2X 12 2 1/2 -1 (x3-x4) (1-X tan 1 (63) w = sin tan (63) (X-x4) X is increasing or stationary with respect to x1 then the objective is increasing with respect to x1. The details are in Appendix D; the result is aw 1 A > 0 x 1 (lx 21/ 21/2 - (64) 1 (-1 12-(l-xl) Thus, x5 is the only variable in the objective whose monotonicity depends on the sign of (x3 -x4). Recalling equation (60) note that the monotonicity of x5 in the objective depends on the sign of __ _as av _ 1 1-(x3-x4)( aX5 aV ax 5 (1_v2)1/2 (X -x4) 2 J (65) ag For case I x3 - x4 > 0 and thus x5 < 0 and x5 is decreasing The monotonicities of x3, x7 and x8 in the objective can be determined by inspection. Before the monotonicities in R17 can be determined, it is necessary to direct the equality. Since x1, is increasing inthe objective it must be decreasing in at least one constraint. Note that xl is not decreasing in any of the inequality constraints, and thus the equality can be directed as follows [0083M2x7 2L2 +.0042M3X6 27 L +.067M X R:1/2 -2 2 2xX /2(M+.083M2X7 L +.021M3) Note further that R7, R8a, and R9a are assumed inactive since they are dominated by RlOa. For case I the problem can now be expressed in the form: min tf (xj 4 x, X7 8 - )

subject to: R10a: (X6, x7 ) < 1 R1: (x7+) < 1 R12: (x3, x4 ) < 1 R2: (x3, x7 ) < 1 R_: +' 6' ~ R14: (x4,5, xR) < 1 R3: (X4, x5 ) < 1 < R4: x) 1 R16: (5, x8 ) < 1 + R20: (x,, ) < 1 R + < 1 R8a: (x5, x6, x7, x10 ) 4 1 R9a: (X5, x6 x7, x10) < 1 R2la: (x3, x4~) < 1 Inactive constraints will not be written in remaining formulations. Constraint Activity Rules of monotonicity analysis state that for a monotonic variable in the objective there must be at least one active constraint with opposite monotonicity in that variable (for minimization) for a well posed problem. If a variable does not appear in the objective then at least two constraints with opposite monotonicities in that variable are active, or none of the constraints containing that variable are active. Thus, constraint R2 must be active since R17 is decreasing wrt X6 and only R2 is increasing wrt x6. Since tf is increasing wrt X3, R4 and/or R21a are active. R12 must be active because of x4, and R14 or R20 must be active because of x8 in the objective. The problem appears completely bounded by these constraints, and thus the activity of the remaining inequalities cannot be determined directly. Implicit elimination is the next step in attempting to reduce the problem further.

Implicit Elimination An appropriate branching must be considered here so that all possible solutions are covered. It -has already been shown that either R4 or R21a is active; now assume R4 is active. R4 is then used to eliminate x3 + + 2 2 1/2 3 = 3(x5, x6, x7 2x7cosW-(x6 x7 ) (67) In order to find the monotonicities in the objective and in R2, x3 must be solved for explicitly and substituted into these constraints. Substituting (67) into the objective 2 2 1/2 (1 2 2x cosW-( -x )22 = sin 7-1 2x7csW-(X6-x7 ) +5-4 Xl ) -f sin x5-x4 - tan 2 f(xlx7,x8) X5 Only the monotonicity of x5 in the objective cannot be determined implicity; x5 depends only on the sin-l term, which can be rearranged as i-i i- 2x7cosW-(x -x ) 7 ~ g sin V= sin [ =7 +. (69) Then ag ag av - L ) [-2X7 (x5-x6 2 ]7(70) ax.5=v =[-v / (x5_x 4).2....' (70) and since 2x coswx2 2 1/2 2x7cosW- (x62 x72) />, (71) ag < (72) ax5 and x5 is still decreasing in the objective. Substitution of X3 from (67) into R2 yields 0.5[2x7cosW + P4 + X5]L- < 1 (73) which is increasing for x5 and x7 as expected, but independent of x6. The remaining constraints yield to implicit elimination. The resulting problem is Min tf l+ + Min tf(xl X4 x5 x7 x8 ),

subject to: R1 (x7 ) < 1 R16 (x5, x8 ) < 1 + X +~< R 7 X R2 (X5 x6, x7 ) < Rl7 (x, x6, x7 x <R10a 6 7 < R2 (xx6, x7 ) < 1 (74)x5 R12 (x4, x5+, x, x) < R21a (, x, x x) < 1 R14 (x4+ x5, x8+) < 1 A significant result arises from the elimination: R21a must be active due to x6. Thus R21a is always active, since either R4 or R21a is active and when R4 is assumed active R21a must also be active. The activity of R21a has a major impact on the problem (case I); without assuming R4 active, use R21a to eliminate x3 3 = x4 = (x4 (75) and substitution into the objective yields 1 (1-x1 (M4+.083M2x7 L +.0213)1/2 an 2 2) 12 1/2 (76) X1 ~ l-x 8 The problem (case I) is now formulated as: m- in tf (X1 x7 X8) subject to: Rl: (x7) < 1 R14: (x4, x8 ) < 1 R2: (x4,x6,X7) < 1 R16: (x5, x8 ) < 1 R4: (x4, X 5 x 6 x7 ) < 1 R17: ( < 177) 1' x6' x7' x8 ) < 1 (77) R10a- (x6, X7 ) < 1 - + + RlR20 (xl, x4, x5, x, x87) < 1 R12: (x4, x5 ) < 1

Thus, R2 must be active because of x6 in R17. Since R2 is active, either R4 or R12 must be active due to x4. Either R14 or R20 ic active due to x8. Following the same approach as before, assume R12 is active and eliminate x4. This yields a physically absurd result, i.e. x4 =x5 = (x (78) The problem becomes: min tf(xl x7 x8 ) subject to: R1: (X7+) < 1 R14: (x8 ) = O R2: (X5, x6, x7) < 1 R16: (x5 x8) < 1 R4': (X6, + < R16: (x, x7 ) < 1 6 7 Rl~a: (x6, x7) < 1 R20: (x0, x7 0, x8 ) = 0 If R12 is active, x8 cannot be bounded from below as required by the objective. Thus, the assumption that R12 is active must be false and thus R4 must be active. There is one feasible branch for case I left to consider: either R14 or R20 must be active. Again, we assume the activity of one of the constraints, in this case R20, so that 1 (x4, x5 7, x8 ) (80) and the problem becomes: min tf(x 4, x5, 7, X8 ), subject to: + + R2: (x 6 X7 ) < 1 R16: (X5, x8 ) < 1 R2: (x4+ + - x x + x x5'' + < 1 R17: x +x x x 7 R4: X4, 5) ( 4, 5 6 7 ) < RlOa: (x6, x7 ) < 1 The result is similar to the previous result; R14 must be active because of x8, so that R14 is always active.

Without assuming R20 active, use R14 to eliminate x4: x4 = 5.88A dm 3KsDX8 + (82) s 8 5 0P4(X5 1X8) The problem is stated as: min tf (X+ X7 x8 ) subject to: R: 7) <1 R16: (x5, x8 ) < 1 R2: (x5 X6 7 8 x) < 1, x8) < R4: (x5 X8 < R20 (x 5' 6 7 1 (x0 x7 +) < 1 RlOa: (x6, x7 +) < 1 Since R2 is active, x5 must also be bounded from below. Thus R16 must be active. Case II, that is, the range for x3 < x4; can be reduced to exactly the same set of active constraints. For case II, the problem is initially stated as: m+ + + + - _ tf(1 x3 X 4 5 7 subject to: R1: (x ) < 1 R14: (x4+ x5, x8) R2: (x 6 < 1'R16: (x5 x 8 ) < 1 R4: (x3, x56 x7 ) 1 R16: (x5, x68 ) x 1 R4: (X + X + < R20: < RlOa (x6, x7) < 1 R20: (x1, x4, x5, x7, x8+) < 1 R12: (3, X x5 < 1 R2b (3, x R12: (x3'4' 5 - 4R21b- (x x4 R12 cannot be active since case II implies x3 < x4, and thus R21b is active because of x4 in the objective. Using R21b to eliminate x3: x3 = x4 = ~3(x4+) (84) and the problem becomes: min tf (x1, x7, x8 )

subject to: R1: (x7 1 R12 (x4, x5 ) < 1 R1: (x7+) < 1 - + R1: + R2: (x4, x, x ) < 1 R14: (x4 x5,x8), 6 7 x7 (x, x+ x6, x7 < 1 R16: (x5 x8 ) < 1 R4: (x+, x6, x 7(8) < 1 1 x6, x7' x8 ) <! RO1a: (x6, x7 ) < 1R20: (x x 1, x4 x5 7, ) < 1 This is exactly the form in which case I was reduced after it was determined that R21a was active, and so the two cases must reduce to the same optimum. Monotonicity analysis has successfully identified five active constraints: R2, R4, R14, R16, and R21, with the result that the problem has now one degroe of freedom, which can be easily mapped to locate the optimum design point. Generating an Optimum The problem has been reduced to a single case with one degree of freedom. Thus, the optimal solution can be obtained by plotting the objective against the independent variable. Note, however, that xg (the number of spring coils) must be an integer value, and in addition the number of coils on a spring generally should be greater than 3 (practical constraint) so that xg > 3. Treating x5 as the independent variable; and substituting values for the parameters gives R16: x =.125D 3d5G x1 103.5 X 8 5 5 R14: x= [588A- dm-3K Dx + x 4.739 x s 8 5.739 x5 R21a: x = x= 4739x (86) 33952 2. 4L(x3-x5) + 2p4(x5-x3) + x5 - x3 R4: x7 = (2L - p4 - x5) (2cosW) R2: x6 = [(2L-p4-X3)2 + x72 ]1/2 The feasible range can be reduced by noting that x7 < L so that from R4 X5 > 1.732(.202-L) =.00346m (87) and also that x3 < 2L so that from R21a x5 < 4.-3 =.0844m. (88) 4., A 739I

Finally, recall the original form of R16: -1X5 d~l, (89) x9x5 1d < 1.(89) Thus, substituting (87) and (88) into (89) 1.73 < x9 < 42.2; (90) however, xg must be greater than 3 due to the practical constraint, so the feasible domain for xg consists of all integer values in the range 3 < x9 < 42 (91) This list can be exhaustively searched; the calculations were coded due to the large number of feasible designs. The results are tabulated in Table 1. The minimum closure time is tf = 18.9 msec at the design point: =.0147 = 14mm, X 7 oils, X 1 x0147 coils, X2 = 84.1 rad/scc, x = 289mm, x 400mm, x3 =66mm, 7 = 56mm, = 348.9 (92) 4 66mm X 7,400 N/m, x = 35440 4 =66mm, 8 12 x = 300 13 The optimum design for the fast acting switch is shown, in scale, for the open and closed positions in Figures 2 and 3, respectively. Conclusions Monotonicity analysis identifies enough active constraints so that the global optimum is easily determined. The problem is successfully reduced to one degree of freedom with a feasible domain that consists of a discrete number of points. The global optimum was found by exhaustively checking the feasible domain - a simple task for one degree of freedom. The active constraints lead to an interesting set of design rules. Not surprisingly, link 2 initially is positioned at as large an angle as possible without the risk of arcing (R4). The active constraints R14, R18 and R21 give information about the spring design: The spring should be designed to its ultimate strength and it should be compressed so that the wires

are flush. The final spring length is equal to the undeformed length so that the spring force always acts in a direction which tends to shut the switch. The interesting aspect of the design is that part 3 should be a very long part such that it is nearly horizontal and its motion is nearly linear; this is done at the expense of building a longer spring with the capability for greater deformation and thus greater spring force (R2). Similarly the joint between parts 2 and 3 is located very near the pivot point of link 2, again so that the length of link 3 is maximized. By designing part 3 so that its motion is nearly linear and in the horizontal plane, the spring force is not "wasted" in driving part 3 vertically. Typically mechanisms optimization problems involve path optimization and must be solved numerically. In this paper a mechanisms problem involving minimization of time has been successfully solved in closed-form using monotonicity analysis. Other classes of problems in dynamics could he solved using the approach in this paper, specifically problems in which the governing differential equations can be used to derive algebraic equations for the objective function and constraints. Possible examples of such problems include those in which the frequencies and/or mode shapes are of principal concern, and those for which an approximate solution to the differential equations can be constructed.

Nomenclature Parameters: L = Length of link 2 = 200 mm p4= One-half piston length = 50 mm M2= Mass of part 2 = 0.5 kg M3= Mass of part 3 = 0.75 kg M4= Mass of part 4 = 0.5 kg D = Spring diameter = 20 mm d = Spring were diameter = 2 mm A = Spring wire strength coefficient = 1880 MPa m = Spring wire strength exponent = 0.186 G = Spring wire shear modulus = 207.0 GPa W = Maximum initial angle on part 2 = 300(to prevent arcing) K = Wahl correction factor = 1 + 0.5 d = 1.05 D Frated = Maximum joint load = 1500N Variables: Mathematical Physical model notation Definition X1 5 Damping coefficient x 2 oWn Natural frequency X3 Sf Final spring length X4 Sud Undeformed spring length X5 so Initial spring length X6 1 r3 Length of part 3 x7 r2 Length to joint on part 2 ~X8 ~k Spring constant

Variables (cont.) Mathematical Physical model notation Definition x9 N Number of spring coils X10 g Width of mechanism X11 08f Final angle of part 3 x12 030 Initial angle of part 3 1x3 02o Initial angle of part 2 2 = 02(t) Angular orientation of part 2 02 Angular velocity of part 2 02 Angular velocity of part 2 03 = 63(t) Angular orientation of part 3 03 Angular velocity of part 3 03 Angular acceleration of part 3 s = s(t) Length of spring at time t Velocity of piston s Acceleration of piston

APPENDIX A Equationsof Motion The constrained D'Alembert Method [51 is mj j=l k=l (93) where F' are the m applied forces, j. are vectors from ground jo the point of application oi Fj, qi are the generalized coordinates 02, 03 and s, Xk are th~ chord forces necessary to maintain superposition and ~k are the k independent closed loops. The only forces considered are the spring force and the D'Alembert forces on each part. Three equations are derived as follows: qi. = 0 26 + X cosr X sin 0(94) 1 2 2 lj 2cos02 - ir2 2 =9~ q. =83: I 383 + X jr3 cosO - X r sin83 0 (95) 1. 3 3 3 1j 3 3 1i 3 3 =95 qi= s: M4s + k(s - ud) + Xli (96) Equations (94 - 96) are easily assembled into the matrix equations (2). Loop Equations The loop equations are dervied from a vector sum around the closed loop in the mechanism. This vector sum can be differentrated twice to yield additional equalities. Geometrical considerations lead to r2 + r3 + P4 + s + g = (97) Taking the dot product of (97) with i and j yields the first two of equations (3). Differentiation of (97) gives u2(k x r2) + 03(k x r3) + ss = 0 (98) Dotting (98) with r2 and r3 results in the third and fourth of equations (3), respectively. Differentiating (97) a second time 2(k x 2) 02r2 + 03(k x r) - 3r3 + ss = 0 (99) The last two of equations (3) are derived by taking the dot product of (99) with r2 and r3, respectively.

APPEND IX B From equilibrium considerations it is easily shown that the magnitude of the reaction force is the same for every joint. To determine the reaction force between parts 3 and 4, hold part 4 fixed and consider a virtual displacement of part 3, 6B1. To compute the horizontal component of the reaction force, the virtual work done is summed to zero -2026 2 I35363 + H6 = 0 (100) where 602 and 603 are related to 6B by -6Ucose3 6e2 r2sin (e2- 3) (101) -6Bcos82 63 r3sin(83-02) (102) Substituting (101) and (102) into (100) and, cancelling 6$ and solving for H the result is I2cos e 3 303cose 2 H - 3(103) r2sin(2-3) r3sin( 3-2) (103) A similar analysis to determine the normal component of the reaction force yields -I 02 sinD3 33 sin2 N = r2sin(e 3) r3 sin(3-2) (104) Note further that H and N are just equal to Xi and Xj, respectively. The magnitude of the reaction force is simply FR = (H + N 2 (105) R~~~~~~~~~~~15

APPENDIX C The differential equation (31) can be simplified by treating the coefficients m' and c as constants with respect to time. These coefficients are in fact complicated functions of time as given by equations (29) and (30). By considering the expected variations in 82 and 03, m' and c can be parameterized by making an average approximation over the time range of interest. The angles 02 and 03 can be expected to vary from 30~ to 90~ and 350~ to 3150 respectively. Also note that 1 12 = 12 M2L2''1 2 (106) I3 ='7 M3r3 Substituting equations (106) into equations (29) and (30) and replacing the 8 dependent terms with a qualitative approximation for the average value, the result for m' and c is M2L2 (107) mi M4 + 3 ~ (1) + 1 M3(.25) r 2 2 1 M2L 1 3 05) 1M2 1 3 (8) 1-2 3 (.1) +r - 12 12 2 4) 12 r' ~-2 3r2r3 It is important to note that the accuracy of the average values used to eliminate the e dependence does not greatly affect the solution since it-does not affect the monotonicity of the variables, and consequently the active constraints are not changed.

APPENDIX D The monotonicity of x1 in the objective depends on -1 (X3-X) (x2) 1/2 2 1/2 (w sin-x31 -) - -1 (-x1 (x5-x4) X sin-1 - =1 k(109) sin v- tan y_ k Partial differentiation yields aw k av a y (110) ax av ay ax, 1.=1 y x2- - /2 - A < 12 12 1- - -x (x5-x4) ( -xxl )113 /2 Therefore > 0 if everywh 1/2 ere 1 Constraint R14 can be manipulated to give V-3- 4/ so that X 1 Therefore -> 0 and the objective is increasing wrt x1 everywhere~

Acknowledgement This work was initiated as a class project. Support for further development was provided to the first author by Sandia National Laboratories, and to the second author by NSF Grant No. CME80-06687. This support is gratefully acknowledged. References [1] Suh, C. and Radcliffe, C., Kinematics and Machine Design, Wiley and Sons, N.Y. (1978), p 430. [2] Papalambros, P., ME566 Global Optimization in Design, Lectures, University of Michigan, Winter 1980. [3] Wilde, D. J., Globally Optimal Design, Wiley Interscience, N. Y. (1978) [4] Papalambros, P. and Wilde, D. J., "Global Non-Iterative Design Optimization Using Monotonicity Analysis," Trans. ASME, J. of Mech. Design, Vol. 101, No. 4, (1979) pp 645-649. [5] Chace, M. A., ME542 CAD-Mechanisms Lectures, University of Michigan, Winter 1980. [6] Shigley, J. E., Mechanical Engineering Design, 3rd edition, McGraw Hill, N.Y. (1977), pp 295-308.

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FAST ACTING SWITCH - NOTATION

j3 ~~~~~~~~~~~~~~~~~-w r6 2 r3. /!/., zi'/ J//// / /J'"-/?'/~J /// p4 s a Figure 1 Claus s

FIGURE 2, OPTIMUM DESIGN - OPEN POSITION

Clauss Figure 2 Clauss

FIGURE 3. OPTIMUM DESIGN - CLOSED POSITION

Figure 3 Clauss