【正文】 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 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 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 and AE may be different links, and degrees. Conditions (41) will be discrete via point x1, x2,... X19 in table 3. The points corresponding to Angle ? 21, ?, 22,... , 219 of ? interval [ degrees, degrees) 1 degrees on a regular basis. The design of the lower bound and upper bound variables u = [640, 1330, 1280, 0]T(mm), (42) u = [700, 1390, 1340, 30]T(mm). (43) 液壓支架的優(yōu)化設(shè)計 Nonlinear programming problem is formulated in the form of (22) (28). Problem is to solve the optimizer is Kegl et al. (1991) based on approximate method. Derivatives of the design calculation using direct numerical differential method. Initial values of design variables: [0a1,0a2,0a4,0a7]T= [674, 1360, 1310, 30]T(mm). (44) The optimal design parameter iteration after 25 U* = [, , , ]T(mm). In table 3 coordinates x and y coupler poinC on startup and optimal design, are: Table 3 x and y coordinates of the point C Angle xstart ystart xend yend ?2(度 ) (mm) (mm) (mm) (mm) Figure 4 illustrates the trajectory L C PM (count), and the optimal design and linear K (all). , the optimal tolerance to AEDB mechanism 液壓支架的優(yōu)化設(shè)計 In the nonlinear programming problem (36) (38), reason boils down to the lower bound and upper bound existing as the independent variable of delta as as a1, a2, the delta delta is a4 w = [, , ]T(mm), (46) w = [, , ]T(mm). （ 47） Initial value of the independent variables w0 = [, , ]T(mm) . （ 48） Allow the deviation of the trajectory was chosen for two cases as E = and E = . In the first case, the optimal tolerance design variables after a1, a2, a4 was calculated and iteration. E got seven iterations = is the best. The result is given in table 4 and 5. In FIG. 5 and 6 of the monte carlo method of calculating the standard deviation and Taylor approximation (solid line represents Taylor approximation), respectively. Trajectories of the figure 4 C 液壓支架的優(yōu)化設(shè)計