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文獻(xiàn)翻譯 英文原文： NOVEL METHOD OF REALIZING THE OPTIMAL TRANSMISSION OF THE CRANK-AND-ROCKER MECHANISM DESIGN 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 parameters. 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 bars is shown, concisely, conveniently and directly. The method possesses the main characteristic. That it is to achieve the optimal transmission parameters under the transmission angle by directly choosing in the diagram, according to the given requirements. The characteristics of the mechanical transmission can be improved to gain the optimal transmission effect by the method. Especially, the method is simple and convenient in practical use. Keywords： Crank-and-rocker mechanism, Optimal transmission angle, Coefficient of travel speed variation INTRODUCTION 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 figure-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 designed crank-and-rocker mechanism. Thus the optimal transmission can be realized. The concerned designing theory and method, as well as the real cases of its application will be introduced later respectively. 1. ESTABLISHMENT OF DIAGRAM FOR OPTIMAL TRANSMISSION DESIGN It is always one of the most important indexes that designers pursue to improve the efficiency and property 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 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 efficiency and property of the mechanism, which mainly indicates in the evaluation of the minimum transmission angle. The aim realizing the 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. Consequently, the complete constraint field realizing the optimal transmission is established. The following steps are taken in the usual design method. Firstly, the initial values of the length of rocker 3l and the oscillating angle of rocker ? are given. Then the value of the coefficient of travel speed variation K is chosen in the permitted range. Meanwhile, the coordinate of the fixed hinge of crank A possibly realized is calculated corresponding to value K . 1.1 Length of bars of crank and rocker mechanism As shown in Fig.1, left arc GC2 is the permitted field of point A . The coordinates of point A are chosen by small step from point 2C to point G . The coordinates of point A are 02 hyy cA ?? (1) 22 AA yRx ?? (2) where 0h , the step, is increased by small increment within range(0,H ). If the smaller the chosen step is, the higher the computational precision will be. R is the radius of the design circle. d is the distance from 2C to G . 2c o s)2c o s (22c o s 33 ???? ?????? ???? lRld (3) Calculating the length of arc 1AC and 2AC , the length of the bars of the mechanism corresponding to point A is obtained[1,2]. 1.2 Minimum transmission angle min? Minimum transmission angle min? (see Fig.2) is determined by the equations[3] 32 2142322 m i n 2 )(c o s ll llll ????? (4) 32 2142322 m a x 2 )(c o s ll llll ????? (5) maxmin 180 ?? ???? (6) where 1l —— Length of crank(mm) 2l —— Length of connecting bar(mm) 3l —— Length of rocker(mm) 4l —— Length of machine frame(mm) Firstly, we choose minimum comparing min? with min?? . And then we record all values of min? greater than or equal to ?40 and choose the maximum of them. Secondly, we find the maximum of min? corresponding to any oscillating angle ? which is chosen by small step in the permitted range (maximum of min? is different oscillating angle ? and the coefficient of travel speed variation K ). Finally, we change the length of rocker 3l by small step similarly. Thus we may obtain the maximum of min? corresponding to the different length of bars, different oscillating angle ? and the coefficient of travel speed variation K . Fig.3 is accomplished from Table for the purpose of diagram design. It is worth pointing out that whatever the length of rocker 3l is evaluated, the location that the maximum of min? arises is only related to the ratio of the length of rocker and the length of machine frame 3l / 4l , while independent of 3l . 2. DESIGN METHOD 2.1 Realizing the optimal transmission design given the coefficient of travel speed variation and the maximum oscillating angle of the rocker The design procedure is as follows. (1) According to given K and ? , taken account to the formula the extreme included angle ? is found. The corresponding ratio of the length of bars 3l / 4l is obtained consulting Fig.3. ????? 18011KK? (7) (2) Choose the length of rocker 3l according to the work requirement, the length of the machine frame is obtained from the ratio 3l / 4l . (3) Choose the centre of fixed hinge D as the vertex arbitrarily, and plot an isosceles triangle, the side of which is equal to the length of rocker 3l (see Fig.4), and ??? 21DCC . Then plot 212 CCMC ? , draw NC1 , and make angle ????? 9012 NCC . Thus the point of intersection of MC2 and NC1 is gained. Finally, draw the circumcircle of triangle 21CPC? . (4) Plot an arc with point D as the centre of the circle, 4l as the radius. The arc intersections arc GC2 at point A . Point A is just the centre of the fixed hinge of the crank. Therefore, from the length of the crank 2/)( 211 ACACl ?? (8) and the length of the connecting bar 112 lACl ?? (9) we will obtain the crank and rocker mechanism consisted of 1l , 2l , 3l , and 4l .Thus the optimal transmission property is realized under given conditions. 2.2 Realizing the optimal transmission design given the length of the rocker (or the length of the machine frame) and the coefficient of travel speed variation We take the following steps. (1) The appropriate ratio of the bars 3l / 4l can be chosen according to given K . Furthermore, we find the length of machine frame 4l (the length of rocker 3l ). (2) The corresponding oscillating angle of the rocker can be obtained consulting Fig.3. And we calculate the extreme included angle ? . Then repeat (3) and (4) in section 2.1 3. DESIGN EXAMPLE The known conditions are that the coefficient of travel speed variation 1818.1?K and maximum oscillating angle ??40? . The crankandrocker mechanism realizing the optimal transmission is designed by the diagram solution method presented above. First, with Eq.(7), we can calculate the extreme included angle ??15? . Then, we find 93.0/ 43 ?ll consulting Fig.3 according to the values of ? and ? . If evaluate 503?l mm, then we will obtain 76.5393.0/504 ??l mm. Next, draw sketch(omitted). As result, the length of bars is 161?l mm, 462?l mm, 503?l mm, 76.534 ?l mm. The minimum transmission angle is ?????? 3698.46 2 )(a r c c o s 32 2142322 m in ll llll? The results obtained by computer are 2227.161 ?l mm, 5093.442 ?l mm, 0000.503 ?l mm, 8986.534 ?l mm. Provided that the figure design is carried under the condition of the Auto CAD circumstances, very precise design results can be achieved. 4. CONCLUSIONS A novel approach of diagram solution can realize 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 component sizes will be enough. 譯文： 認(rèn)識(shí)曲柄搖臂機(jī)構(gòu)設(shè)計(jì)的最優(yōu)傳動(dòng)方法 摘要： 一種曲柄搖臂機(jī)構(gòu)設(shè)計(jì)的最優(yōu)傳動(dòng)的方法被提出。這種優(yōu)化組合設(shè)計(jì)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(dòng)角度之間 , 滑 移速度變化系數(shù)，搖臂的擺動(dòng)角度和桿的長(zhǎng)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(dòng)角度之下的最優(yōu)傳動(dòng)參數(shù)直接地表達(dá) 在圖上。通過這種方法，機(jī)械傳動(dòng)的特性能用以獲取最優(yōu)傳動(dòng)效果。特別是， 這 種方法是簡(jiǎn)單和實(shí)用的。 關(guān)鍵字： 曲柄搖臂機(jī)構(gòu) 最優(yōu)傳動(dòng)角度 滑移速度變化系數(shù) 0 介紹 由曲柄搖臂機(jī)構(gòu)設(shè)計(jì)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(dòng)。通過本文介紹的圖面設(shè)計(jì)方法可以幫助達(dá)到這個(gè)目的。在指定的情 況下，通過觀查設(shè)計(jì)圖面 , 我們就能得到每個(gè)參量和另外一個(gè)曲柄搖臂機(jī)構(gòu)設(shè)計(jì) 之間的聯(lián)系。由因認(rèn)識(shí)最優(yōu)傳動(dòng)。 具體的設(shè)計(jì)的理論和方法， 以及它們各自的應(yīng)用事例將在以下介紹。 1 優(yōu)化傳動(dòng)設(shè)計(jì)的建立 優(yōu)化傳動(dòng)的設(shè)計(jì)一直是設(shè)計(jì)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(gè)。曲柄搖臂機(jī)構(gòu)被廣泛應(yīng)用在機(jī)械傳動(dòng)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(dòng)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(duì)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。 認(rèn)識(shí)機(jī)械優(yōu)化傳動(dòng)目的是找到極小的傳輸角度的最大值。 設(shè)計(jì)參數(shù)是適度地 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(dòng)建立 了。 以下步驟被采用在通常的設(shè)計(jì)方法。 首先，測(cè)量出搖臂的長(zhǎng)度 3l 和搖臂的 擺動(dòng)角度 ? 的初始值。 然后滑移速度變化系數(shù) K 的值被定在允許的范圍內(nèi)。 同 時(shí)，曲柄固定的鉸接座標(biāo) A 可能被認(rèn)為是任意值 K 。 1.1 曲柄搖臂機(jī)構(gòu)桿的長(zhǎng)度 由圖 Fig.1，左弧 GC2 是點(diǎn) A 被允許的領(lǐng)域。點(diǎn) A 的座標(biāo)的選擇從點(diǎn) 2C 到點(diǎn) G 。 點(diǎn) A 的座標(biāo)是 02 hyy cA ?? (1) 22 AA yRx ?? (2) 當(dāng) 0h ，高度，在 range(0 ， H ) 被逐漸增加。如果選的越小 ,計(jì)算精度將越高。 R 是設(shè)計(jì)圓的半徑。 d 是從 2C 到 G 的距離。 2c o s)2c o s (22c o s 33 ???? ?????? ???? lRld (3) 計(jì)算弧 1AC 和 2AC 的長(zhǎng)度，機(jī)械桿對(duì)應(yīng)于點(diǎn) A 的長(zhǎng)度是 obtained[1,2 ] 。 1.2 極小的傳動(dòng)角度 min? 極小的傳動(dòng)角度 min? (參見 Fig.2) 由 equations[3]確定 32 2142322 m i n 2 )(c o s ll llll ????? (4) 32 2142322 m a x 2 )(c o s ll llll ????? (5) maxmin 180 ?? ???? (6) 由于 1l —— 曲柄的長(zhǎng)度 (毫米 ) 2l —— 連桿的長(zhǎng)度（毫米） 3l —— 搖臂的長(zhǎng)度（毫米） 4l —— 機(jī)器的長(zhǎng)度（毫米） 首先 , 我們比較極小值 min? 和 min?? 。 并且我們記錄所有 min? 的值大于或等于 ?40 ，然后選擇他們之間的最大值。 第二 , 我們發(fā)現(xiàn)最大值 min? 對(duì)應(yīng)于一個(gè)逐漸變小的范圍的任一個(gè)擺動(dòng)的角度 ? (最大值 min? 是不同于擺動(dòng)的角度和滑移速度變化系數(shù) K ) 。 最后 , 我們相似地慢慢縮小搖臂 3l 的長(zhǎng)度。 因而我們能獲得最大值 min? 對(duì)應(yīng) 于桿的不同長(zhǎng)度 , 另外擺動(dòng)的角度 ? 和滑移速度變化系數(shù) K 。 Fig.3成功的表達(dá)設(shè)計(jì)的目的。 它確定了無論是搖臂的長(zhǎng)度 3l ,最大值 min? 出現(xiàn)的地點(diǎn)，只與搖臂的長(zhǎng)度和機(jī) 械的長(zhǎng)度的比率 3l / 4l 有關(guān) , 當(dāng)確定 3l 時(shí)。 2 設(shè)計(jì)方法 2.1 認(rèn)識(shí)最優(yōu)傳動(dòng)設(shè)計(jì)下滑移速度變化系數(shù)和搖臂的最大擺動(dòng)的角度 設(shè)計(jì)步驟如下。 (1) 根據(jù)所給的 K 和 ? ， 通常采取對(duì)發(fā)現(xiàn)極限角度 ? 的解釋。 桿的長(zhǎng)度的 對(duì)應(yīng)的比率 3l / 4l 是從圖 Fig.3獲得的 。 ????? 18011KK? (7) (2) 根據(jù)工作要求選擇搖臂的長(zhǎng)度 3l ， 機(jī)械的長(zhǎng)度是從比率 3l / 4l 獲得的。 (3) 任意地選擇固定的鉸接的中心 D 作為端點(diǎn)，并且做一個(gè)等腰三角形 ,令一 條邊與搖臂的長(zhǎng)度 3l 相等 (參見 Fig.4)，令 ??? 21DCC 。 然后做 212 CCMC ? ， 連接 NC1 ，并且做角度 ????? 9012 NCC 。 因而增加了交點(diǎn) MC2 和 NC1 。 最 后， 畫三角形 21CPC? 。 (4)以點(diǎn) D 作為圓的中心， 4l 為半徑畫圓弧。 弧 GC2 交點(diǎn)在 A 點(diǎn)。 點(diǎn) A 是曲 柄的固定鉸接的中 心。 所以 , 從曲柄的長(zhǎng)度 2/)( 211 ACACl ?? (8) 并且連桿的長(zhǎng)度 112 lACl ?? (9) 我們將獲得曲柄搖臂機(jī)構(gòu)包括 1l ， 2l ， 3l 和 4l 。因而優(yōu)化傳動(dòng)加工會(huì)在指定的情 況下進(jìn)行。 2.2 認(rèn)識(shí)優(yōu)化傳動(dòng)設(shè)計(jì)下?lián)u臂的長(zhǎng)度 (或機(jī)械的長(zhǎng)度 ) 和滑移速度變化系數(shù) 我們采取以下步驟。 (1)根據(jù)選擇的 K 確定桿的適當(dāng)比率 3l /4l 。 此外，我們得出機(jī)械 4l (搖臂的 長(zhǎng)度 3l ) 。 (2) 搖臂對(duì)應(yīng)的擺動(dòng)的角度可以從圖 Fig.3 獲得。 并且我們計(jì)算出極限角 度。 然后根據(jù) 2.1重覆 (3) 和 (4) 3 設(shè)計(jì)例子 已知的條件是 , 滑移速度變化系數(shù) 1818.1?K 和最大擺動(dòng)角度 ??40? 。 提 出曲柄搖臂機(jī)械優(yōu)化傳動(dòng)圖方法設(shè)計(jì)方案。 首先， 通過公式 (7)，我們能計(jì)算出極限角度 ??15? 。 然后，我們通過表 格 Fig.3 查出 93.0/ 43 ?ll 以及 ? 和 ? 的值。 假設(shè) 503?l mm, 然后我們將得出 76.5393.0/504 ??l mm。 然后 , 做 sketch(omitted) 。 最后 , 算出桿的長(zhǎng)度分別是 161?l mm, 462?l mm, 503?l mm, 76.534 ?l mm. 極小傳動(dòng)角度是 ?????? 3698.462 )(a r c c o s 32 2142322m in ll llll? 結(jié)果由計(jì)算可得 2227.161 ?l mm， 5093.442 ?l mm， 0000.503 ?l mm， 8986.534 ?l mm。 在運(yùn)用 Auto CAD 制圖設(shè)計(jì)的情況 , 可達(dá)到非常精確設(shè)計(jì)結(jié)果。 4結(jié)論 認(rèn)識(shí)圖解法解答曲柄搖臂機(jī)構(gòu)的最優(yōu)傳動(dòng)。這種方法是簡(jiǎn)單和實(shí)用的。通常 在機(jī)械設(shè)計(jì)中 , 將 0.1 毫米作為最小有效精度是足夠的。