外文翻譯--認(rèn)識曲柄搖臂機(jī)構(gòu)設(shè)計的最優(yōu)傳動方法.doc (2)

1、<p><b>　　文獻(xiàn)翻譯</b></p><p><b>　　英文原文：</b></p><p>　　NOVEL METHOD OF REALIZING THE OPTIMAL TRANSMISSION</p><p>　　OF THE CRANK-AND-ROCKER MECHANISM DESIGN<

2、;/p><p>　　Abstract: A novel method of realizing the optimal transmission of the crank-and-rocker mechanism is presented. The optimal combination design is made by finding the related optimal transmission parame

3、ters. The diagram of the optimal transmission is drawn. In the diagram, the relation among minimum transmission angle, the coefficient of travel speed variation, the oscillating angle of the rocker and the length of the

4、bars is shown, concisely, conveniently and directly. The method possesses the ma</p><p>　　Keywords：Crank-and-rocker mechanism, Optimal transmission angle, Coefficient of travel speed variation</p><

5、;p>　　INTRODUCTION</p><p>　　By conventional method of the crank-and-rocker design, it is very difficult to realize the optimal combination between the various parameters for optimal transmission. The figur

6、e-table design method introduced in this paper can help achieve this goal. With given conditions, we can, by only consulting the designing figures and tables, get the relations between every parameter and another of the

7、designed crank-and-rocker mechanism. Thus the optimal transmission can be realized.</p><p>　　The concerned designing theory and method, as well as the real cases of its application will be introduced later r

8、espectively.</p><p>　　ESTABLISHMENT OF DIAGRAM FOR OPTIMAL TRANSMISSION DESIGN </p><p>　　It is always one of the most important indexes that designers pursue to improve the efficiency and proper

9、ty of the transmission. The crank-and-rocker mechanism is widely used in the mechanical transmission. How to improve work ability and reduce unnecessary power losses is directly related to the coefficient of travel speed

10、 variation, the oscillating angle of the rocker and the ratio of the crank and rocker. The reasonable combination of these parameters takes an important effect on the efficien</p><p>　　The aim realizing the

11、optimal transmission of the mechanism is how to find the maximum of the minimum transmission angle. The design parameters are reasonably combined by the method of lessening constraints gradually and optimizing separately

12、. Consequently, the complete constraint field realizing the optimal transmission is established.</p><p>　　The following steps are taken in the usual design method. Firstly, the initial values of the length o

13、f rocker and the oscillating angle of rocker are given. Then the value of the coefficient of travel speed variation is chosen in the permitted range. Meanwhile, the coordinate of the fixed hinge of crank possibly real

14、ized is calculated corresponding to value .</p><p>　　Length of bars of crank and rocker mechanism</p><p>　　As shown in Fig.1, left arc is the permitted field of point . The coordinates of point

15、 are chosen by small step from point to point .</p><p>　　The coordinates of point are</p><p><b>　　(1)</b></p><p><b>　　(2)</b></p><p>　　where

16、, the step, is increased by small increment within range(0,). If the smaller the chosen step is, the higher the computational precision will be. is the radius of the design circle. is the distance from to .</p>

17、<p><b>　　(3)</b></p><p>　　Calculating the length of arc and , the length of the bars of the mechanism corresponding to point is obtained[1,2].</p><p>　　1.2 Minimum transmissi

18、on angle </p><p>　　Minimum transmission angle (see Fig.2) is determined by the equations[3]</p><p><b>　　(4)</b></p><p><b>　　(5)</b></p><p><b

19、>　　(6)</b></p><p>　　where ——Length of crank(mm)</p><p>　　——Length of connecting bar(mm)</p><p>　　——Length of rocker(mm)</p><p>　　——Length of machine frame(mm)

20、</p><p>　　Firstly, we choose minimum comparing with . And then we record all values of greater than or equal to and choose the maximum of them.</p><p>　　Secondly, we find the maximum of corr

21、esponding to any oscillating angle which is chosen by small step in the permitted range (maximum of is different oscillating angle and the coefficient of travel speed variation ).</p><p>　　Finally, we cha

22、nge the length of rocker by small step similarly. Thus we may obtain the maximum of corresponding to the different length of bars, different oscillating angle and the coefficient of travel speed variation .</p>

23、<p>　　Fig.3 is accomplished from Table for the purpose of diagram design.</p><p>　　It is worth pointing out that whatever the length of rocker is evaluated, the location that the maximum of arises is

24、 only related to the ratio of the length of rocker and the length of machine frame /, while independent of .</p><p>　　DESIGN METHOD</p><p>　　Realizing the optimal transmission design given the c

25、oefficient of travel speed variation and the maximum oscillating angle of the rocker</p><p>　　The design procedure is as follows.</p><p>　　(1) According to given and , taken account to the form

26、ula the extreme included angle is found. The corresponding ratio of the length of bars / is obtained consulting Fig.3.</p><p><b>　　(7)</b></p><p>　　(2) Choose the length of rocker

27、according to the work requirement, the length of the machine frame is obtained from the ratio /.</p><p>　　(3) Choose the centre of fixed hinge as the vertex arbitrarily, and plot an isosceles triangle, the

28、side of which is equal to the length of rocker (see Fig.4), and . Then plot , draw , and make angle . Thus the point of intersection of and is gained. Finally, draw the circumcircle of triangle .</p><p>　　

29、(4) Plot an arc with point as the centre of the circle, as the radius. The arc intersections arc at point . Point is just the centre of the fixed hinge of the crank.</p><p>　　Therefore, from the length o

30、f the crank</p><p><b>　　(8)</b></p><p>　　and the length of the connecting bar</p><p><b>　　(9)</b></p><p>　　we will obtain the crank and rocker m

31、echanism consisted of , , , and .Thus the optimal transmission property is realized under given conditions.</p><p>　　2.2 Realizing the optimal transmission design given the length of the rocker (or the lengt

32、h of the machine frame) and the coefficient of travel speed variation</p><p>　　We take the following steps.</p><p>　　(1) The appropriate ratio of the bars / can be chosen according to given . Fu

33、rthermore, we find the length of machine frame (the length of rocker ).</p><p>　　(2) The corresponding oscillating angle of the rocker can be obtained consulting Fig.3. And we calculate the extreme included

34、angle .</p><p>　　Then repeat (3) and (4) in section 2.1</p><p>　　DESIGN EXAMPLE</p><p>　　The known conditions are that the coefficient of travel speed variation and maximum oscilla

35、ting angle . The crankandrocker mechanism realizing the optimal transmission is designed by the diagram solution method presented above.</p><p>　　First, with Eq.(7), we can calculate the extreme included ang

36、le . Then, we find consulting Fig.3 according to the values of and .</p><p>　　If evaluate mm, then we will obtain mm.</p><p>　　Next, draw sketch(omitted).</p><p>　　As result, th

37、e length of bars is mm, mm, mm, mm.</p><p>　　The minimum transmission angle is</p><p>　　The results obtained by computer are mm, mm, mm, mm.</p><p>　　Provided that the figure d

38、esign is carried under the condition of the Auto CAD circumstances, very precise design results can be achieved.</p><p>　　CONCLUSIONS </p><p>　　A novel approach of diagram solution can realize

39、the optimal transmission of the crank-and-rocker mechanism. The method is simple and convenient in the practical use. In conventional design of mechanism, taking 0.1 mm as the value of effective the precision of the comp

40、onent sizes will be enough.</p><p><b>　　譯文：</b></p><p>　　認(rèn)識曲柄搖臂機(jī)構(gòu)設(shè)計的最優(yōu)傳動方法</p><p>　　摘要：一種曲柄搖臂機(jī)構(gòu)設(shè)計的最優(yōu)傳動的方法被提出。這種優(yōu)化組合設(shè)計被用來找出最優(yōu)的傳遞參數(shù)。得出最優(yōu)傳遞圖。在圖中，在極小的傳動角度之間, 滑移速度變化系數(shù)，搖臂的擺動角

41、度和桿的長度被直觀地顯示。 這是這種方法擁有的主要特征。根據(jù)指定的要求，它將傳動角度之下的最優(yōu)傳動參數(shù)直接地表達(dá)在圖上。通過這種方法，機(jī)械傳動的特性能用以獲取最優(yōu)傳動效果。特別是， 這種方法是簡單和實(shí)用的。</p><p>　　關(guān)鍵字：曲柄搖臂機(jī)構(gòu) 最優(yōu)傳動角度 滑移速度變化系數(shù)</p><p><b>　　0 介紹</b></p><

42、p>　　由曲柄搖臂機(jī)構(gòu)設(shè)計的常規(guī)方法, 在各種各樣的參量之間很難找出優(yōu)化組合的最優(yōu)傳動。通過本文介紹的圖面設(shè)計方法可以幫助達(dá)到這個目的。在指定的情況下，通過觀查設(shè)計圖面, 我們就能得到每個參量和另外一個曲柄搖臂機(jī)構(gòu)設(shè)計之間的聯(lián)系。由因認(rèn)識最優(yōu)傳動。</p><p>　　具體的設(shè)計的理論和方法， 以及它們各自的應(yīng)用事例將在以下介紹。</p><p>　　1 優(yōu)化傳動設(shè)計的建立 <

43、;/p><p>　　優(yōu)化傳動的設(shè)計一直是設(shè)計師改進(jìn)傳輸效率和追求產(chǎn)量的最重要的索引的當(dāng)中一個。曲柄搖臂機(jī)構(gòu)被廣泛應(yīng)用在機(jī)械傳動中。如何改進(jìn)工作效率和減少多余的功率損失直接地與滑移速度變化系數(shù)，搖臂的擺動角度和曲柄搖臂的比率有關(guān)系。這些參數(shù)的合理組合采用對機(jī)械效率和產(chǎn)量有重要作用, 這些主要體現(xiàn)在極小的傳輸角度上。</p><p>　　認(rèn)識機(jī)械優(yōu)化傳動目的是找到極小的傳輸角度的最大值。設(shè)計參數(shù)是

44、適度地減少限制而且分開的合理優(yōu)化方法的結(jié)合。因此，完全限制領(lǐng)域的優(yōu)化傳動建立了。</p><p>　　以下步驟被采用在通常的設(shè)計方法。 首先，測量出搖臂的長度和搖臂的擺動角度的初始值。 然后滑移速度變化系數(shù)的值被定在允許的范圍內(nèi)。 同時，曲柄固定的鉸接座標(biāo)可能被認(rèn)為是任意值。</p><p>　　1.1 曲柄搖臂機(jī)構(gòu)桿的長度</p><p>　　由圖Fig.1，左

45、弧是點(diǎn)被允許的領(lǐng)域。點(diǎn)的座標(biāo)的選擇從點(diǎn)到點(diǎn)。</p><p><b>　　點(diǎn)的座標(biāo)是</b></p><p><b>　　(1)</b></p><p><b>　　(2)</b></p><p>　　當(dāng)，高度，在range(0 ，) 被逐漸增加。如果選的越小,計算精度將越高。

46、 是設(shè)計圓的半徑。是從到的距離。</p><p><b>　　(3)</b></p><p>　　計算弧和的長度，機(jī)械桿對應(yīng)于點(diǎn)的長度是obtained[1,2 ] 。</p><p>　　1.2 極小的傳動角度 </p><p>　　極小的傳動角度 (參見Fig.2) 由equations[3]確定</p>

47、;<p><b>　　(4)</b></p><p><b>　　(5)</b></p><p><b>　　(6)</b></p><p>　　由于——曲柄的長度(毫米)</p><p>　　——連桿的長度（毫米）</p><p>　　—

48、—搖臂的長度（毫米）</p><p>　　——機(jī)器的長度（毫米）</p><p>　　首先, 我們比較極小值和。 并且我們記錄所有的值大于或等于，然后選擇他們之間的最大值。</p><p>　　第二, 我們發(fā)現(xiàn)最大值對應(yīng)于一個逐漸變小的范圍的任一個擺動的角度 (最大值是不同于擺動的角度和滑移速度變化系數(shù)) 。</p><p>　　最后, 我們

49、相似地慢慢縮小搖臂的長度。 因而我們能獲得最大值對應(yīng)于桿的不同長度, 另外擺動的角度和滑移速度變化系數(shù)。</p><p>　　Fig.3成功的表達(dá)設(shè)計的目的。</p><p>　　它確定了無論是搖臂的長度,最大值出現(xiàn)的地點(diǎn)，只與搖臂的長度和機(jī)械的長度的比率/有關(guān), 當(dāng)確定時。</p><p>　　2 設(shè)計方法</p><p>　　2.

50、1 認(rèn)識最優(yōu)傳動設(shè)計下滑移速度變化系數(shù)和搖臂的最大擺動的角度</p><p><b>　　設(shè)計步驟如下。</b></p><p>　　(1) 根據(jù)所給的和， 通常采取對發(fā)現(xiàn)極限角度的解釋。 桿的長度的對應(yīng)的比率/是從圖Fig.3獲得的 。</p><p><b>　　(7)</b></p><p&g

51、t;　　(2) 根據(jù)工作要求選擇搖臂的長度， 機(jī)械的長度是從比率/獲得的。</p><p>　　(3) 任意地選擇固定的鉸接的中心作為端點(diǎn)，并且做一個等腰三角形,令一條邊與搖臂的長度相等 (參見Fig.4)，令。 然后做， 連接，并且做角度。 因而增加了交點(diǎn)和。 最后， 畫三角形。</p><p>　　(4)以點(diǎn)作為圓的中心，為半徑畫圓弧。 弧交點(diǎn)在點(diǎn)。 點(diǎn)是曲柄的固定鉸接的中心。<

52、/p><p>　　所以, 從曲柄的長度</p><p><b>　　(8)</b></p><p><b>　　并且連桿的長度</b></p><p><b>　　(9)</b></p><p>　　我們將獲得曲柄搖臂機(jī)構(gòu)包括，，和。因而優(yōu)化傳動加工會在指定

53、的情況下進(jìn)行。</p><p>　　2.2 認(rèn)識優(yōu)化傳動設(shè)計下?lián)u臂的長度(或機(jī)械的長度) 和滑移速度變化系數(shù)</p><p><b>　　我們采取以下步驟。</b></p><p>　　(1)根據(jù)選擇的確定桿的適當(dāng)比率/。 此外，我們得出機(jī)械 (搖臂的長度) 。</p><p>　　(2) 搖臂對應(yīng)的擺動的角度可以從圖

54、Fig.3 獲得。 并且我們計算出極限角度。</p><p>　　然后根據(jù)2.1重覆(3) 和(4)</p><p><b>　　3 設(shè)計例子</b></p><p>　　已知的條件是, 滑移速度變化系數(shù)和最大擺動角度。 提出曲柄搖臂機(jī)械優(yōu)化傳動圖方法設(shè)計方案。</p><p>　　首先， 通過公式(7)，我們能計算出

55、極限角度。 然后，我們通過表格Fig.3 查出以及和的值。</p><p>　　假設(shè)mm, 然后我們將得出mm。</p><p>　　然后, 做sketch(omitted) 。</p><p>　　最后, 算出桿的長度分別是 mm, mm, mm, mm.</p><p><b>　　極小傳動角度是</b></p

56、><p>　　結(jié)果由計算可得 mm， mm， mm， mm。</p><p>　　在運(yùn)用Auto CAD 制圖設(shè)計的情況, 可達(dá)到非常精確設(shè)計結(jié)果。</p><p><b>　　4結(jié)論</b></p><p>　　認(rèn)識圖解法解答曲柄搖臂機(jī)構(gòu)的最優(yōu)傳動。這種方法是簡單和實(shí)用的。通常在機(jī)械設(shè)計中, 將0.1 毫米作為最小有