【正文】 Harl, B. 1998: [7] The optimal synthesis ofhydraulic support. ZAMM 3, 1027–1028Rao, .。 A nonlinear programming problem mainly four link mechanism AEDB therefore can be defined as Min a7, (22) constrained (a3 + a4) ? (a1 + a2) ≤ 0 , (23) （ a2 + a3) ? (a1 + a4) ≤ 0 , （ 24） a1 ≤ a1 ≤ a1 ,a2 ≤a2 ≤ a2 a4≤ a4≤ a4, （ 25） [g0(y) ? f0(y)]2? a7≤ 0 ,(y ∈ y, y ) , （ 26) And the vector of response variables (x ? a5cos Θ)2+ (y ? a5sin Θ)2? a22= 0, （ 27） [x ? a6 cos(Θ + γ) ? a1]2+[y ? a6 sin(Θ + γ)]2? a24= 0 .（ 28） This scheme is to minimize, indispensable is the transverse displacement between points C and the trajectory of k. The result is the optimal values of the parameters a1, a2, a4. 3, stochastic model of hydraulic support Mathematical model (22) (28) can be used to calculate the values of the parameters a1, a2, a4, the difference between l and K is the smallest. However, the real trajectory L C point deviates from the value of the calculation, can be used for different effects. Suitable mathematical model of the deviation can be calculated in the parameter tolerance of a1, a2, a4. 液壓支架的優(yōu)化設(shè)計(jì) Response equation (27) (28) allows us to calculate the vector of response variables v depends on the vector of design variables in the u. This means that the v = ? h (u). ? function h (U) is the basic mathematical model (22) (28), because they are the relationship between the reaction of the vector of design variables U and v of the mechanical system. Same function h enables us to calculate the maximum allowable tolerance as value delta as as a1, a2, the delta delta a4 and the parameters a1, a2, a4. In stochastic model vector u = [U1... Un) falling fish at the top of the design variable as a random vector u = [U1... Un] T, which means that the vector v = [v1... vm) landed fish at the top of the response variable is a random vector v = [v1... vm) T, V = ?h(U) . (29) It should be design variables U1,... , the Un is independent from the Angle of probability, they showed a normal distribution, the Uk ~ N? K, sigma (k) (k = 1, 2,..., N). Main parameters k and sigma k (k = 1, 2,..., n) can be binding technology concepts as nominal measures, such as the UK = UK and tolerance, and as such as delta UK = 3 sigma k, so: ?k ? ?uk ≤ Uk ≤ ?k +?uk , k = 1, 2, . . . , n , (30) So will appear with the selected probability. Probability distribution function of random vector v, so look for depends on the probability distribution function of random vectors U and other puting is almost impossible. Therefore, random vector v characteristics will be described with the help of digital, Taylor can estimate function approximation ? h in point u = [U1... Un] or in Oblak (1982) and Harl (1998) paper, with the help of using monte carlo method. 4, mathematical model Optimal tolerance mathematical model calculation of hydraulic support is formulated as an independent variable of nonlinear programming w = [?a1?a2?a4]T, (31) And the vector of response variables 液壓支架的優(yōu)化設(shè)計(jì) f(w)=1/?a1+1/?a2+1/?a4 (32) And conditions σY ? E ≤ 0 （ 33） ?a1 ≤ ?a1 ≤ ?a1 , ?a2 ≤ ?a2 ≤ ?a2 ,?a4≤ ?a4≤ ?a4 （ 34） In (33) E is the maximum allowable standard deviation sigma Y coordinates of the point C and x A = {1, 2, 4} . （ 35） lculation of the nonlinear programming problem, therefore, the optimal tolerance can be defined as constrained σY?E≤0 (37) ?a1≤?a1≤?a1,?a2≤?a2≤?a2 ?a4≤?a4≤?a4 (38) Carrying capacity is 1600 kN hydraulic support. Two AEDB and FEDG fourbar mechanism must meet the following requirements: 1. They must allow a minimum of lateral displacement of the point C 2. They must provide adequate lateral stability The parameters of the hydraulic support (figure 2) is given in table 1. Incentive mechanism FEDG is specified in the vector [b1, b2, b3, b4]T= [400, (1325 + d), 1251, 1310]T(mm) , (39) And mechanism AEDB [a1, a2, a3, a4]T= [674, 1360, 382, 1310]T(mm) . (40) In (39), the parameter d is a walking support maximum 925 mm. Parameters of shaft AEDB mechanism is given in table 2. Table 1 the parameters of the hydraulic support Mark length M 110 液壓支架的優(yōu)化設(shè)計(jì) N 510 O 640 P 430 Q 200 S 1415 T 380 Mechanism AEDB axis table 2 parameters Mark length a5 a6 Α 179034 Β Γ 0..14 、 The optimal mechanism AEDB link The mathematical model of the data of the four connecting rod mechanism AEDB (22) can be written form (28). A straight line is x = 65 (mm) (FIG. 3) the required trajectory point c. This is why conditions (26) (x ? 65) ? a7 ≤ 0 . (41) The angles between the AB a