【正文】 nd 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(jì) 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(jì) 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(jì) Figure 5 the standard deviation is E = Table 4 the best tolerance for E = Sign Value (mm) ?a1 ?a2 ?a4 Table 5 the best tolerance for E = Sign Value (mm) ?a1 ?a2 ?a4 Figure 6 sigma is E = 液壓支架的優(yōu)化設(shè)計(jì) 5, the summary With an appropriate mathematical model of the system and by means of mathematical programming, the design of hydraulic support was improved, and better performance is achieved. However, as a result of the optimal tolerance, it may be reasonable to consider a new building. This is especially so mechanism AEDB, because very little tolerance for improvement of the cost of production 液壓支架的優(yōu)化設(shè)計(jì) documentation [1] Optimal synthesis of fourbar mechanism . . Faculty of Mechanical Engineering MariborHarl, B. 1998: [2] Stochastic analyses of hydraulic support . Thesis. Faculty of Mechanical Engineering MariborHaug, .。 Dukkipati, . 1989: 液壓支架的優(yōu)化設(shè)計(jì) Optimal design of hydraulic support M. Oblak, B. Harl and B. Butinar Abstract This paper describes a procedure for optimaldetermination of two groups of parameters of a hydraulicsupport employed in the mining industry. The procedureis based on mathematical programming methods. In the?rst step, the optimal values of some parameters of theleading fourbar mechanism are found in order to ensurethe desired motion of the support with minimal transversal displacements. In the second step, maximal tolerances of the optimal values of the leading fourbar mechanismare calculated, so the response of hydraulic support willbe satisfying. Key words fourbar mechanism, optimal design, mathematical programming, approximation method, tolerance The designers39。 Arora, . 1979: [3] Applied optimal design. NewYork: WileyHsieh, C.。 圖 5標(biāo)準(zhǔn)差為 E = 圖 4的飛行軌跡點(diǎn) C 表 4 最佳公差為 E = Sign Value (mm) ?a1 ?a2 ?a4 表 5最佳公差為 E = Sign Value (mm) ?a1 ?a2 ?a4 液壓支架的優(yōu)化設(shè)計(jì) 圖 6 標(biāo)準(zhǔn)差為 E = 總結(jié) 用一個(gè)合適的數(shù)學(xué)模型系統(tǒng)和通過 采用數(shù)學(xué)規(guī)劃、設(shè)計(jì)的液壓支架進(jìn)行改進(jìn) ,和更好的性能實(shí)現(xiàn)了。 E = 7個(gè)迭代。衍生品的設(shè)計(jì)計(jì)算用直接數(shù)值微分方法。這些點(diǎn)對(duì)應(yīng) ?21角度 ,?22,。 表 1的參數(shù)的液壓支架 標(biāo)記 長(zhǎng)度 M 110 N 510 O 640 P