街道護(hù)欄自動(dòng)清洗機(jī)構(gòu)設(shè)計(jì)【含10張CAD圖紙+PDF圖】

街道護(hù)欄自動(dòng)清洗機(jī)構(gòu)設(shè)計(jì)【含10張CAD圖紙+PDF圖】,含10張CAD圖紙+PDF圖,街道,護(hù)欄,自動(dòng),清洗,機(jī)構(gòu),設(shè)計(jì),10,CAD,圖紙,PDF 文獻(xiàn)翻譯文獻(xiàn)翻譯 英文原文：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 Apossibly 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 02hyycA (1) 22AAyRx (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 s33lRld (3) Calculating the length of arc 1AC and 2AC, the length of the bars of the mechanism corresponding to point A is obtained1,2. 1.2 Minimum transmission angle min Minimum transmission angle min(see Fig.2) is determined by the equations3 322142322m i n2)(c o sllllll (4) 322142322m a x2)(c o sllllll (5) m a xm i n180 (6) where 1lLength of crank(mm) 2lLength of connecting bar(mm) 3lLength of rocker(mm) 4lLength 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 212CCMC, draw NC1, and make angle 9012NCC. 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/ )(211ACACl (8) and the length of the connecting bar 112lACl (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. 1K 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. 0/43ll consulting Fig.3 according to the values of and . If evaluate 503l mm, then we will obtain 76.5393. 0/504l mm. Next, draw sketch(omitted). As result, the length of bars is 161l mm,462l mm,503l mm,76.534l mm. The minimum transmission angle is 3698.462)(arccos322142322minllllll The results obtained by computer are 2227.161l mm, 5093.442l mm, 0000.503l mm, 8986.534l 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：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：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， 這種方法是簡單和實(shí)用的。 關(guān)鍵字：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)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)桿的長度曲柄搖臂機(jī)構(gòu)桿的長度 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 02hyycA (1) 22AAyRx (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 s33lRld (3) 計(jì)算弧1AC和2AC的長度，機(jī)械桿對(duì)應(yīng)于點(diǎn)A的長度是 obtained1,2 。 1.2 極小的傳動(dòng)角度極小的傳動(dòng)角度min 極小的傳動(dòng)角度min (參見 Fig.2) 由 equations3確定 322142322m i n2)(c o sllllll (4) 322142322m a x2)(c o sllllll (5) m a xm i n180 (6) 由于1l曲柄的長度(毫米) 2l連桿的長度（毫米） 3l搖臂的長度（毫米） 4l 機(jī)器的長度（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的長度。 因而我們能獲得最大值min對(duì)應(yīng)于桿的不同長度, 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度3l,最大值min出現(xiàn)的地點(diǎn)，只與搖臂的長度和機(jī)械的長度的比率3l/4l有關(guān), 當(dāng)確定3l時(shí)。 2 設(shè)計(jì)方法設(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度3l， 機(jī)械的長度是從比率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度3l相等 (參見 Fig.4)，令21DCC。 然后做212CCMC， 連接NC1，并且做角度9012NCC。 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, 從曲柄的長度 2/ )(211ACACl (8) 并且連桿的長度 112lACl (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臂的長度認(rèn)識(shí)優(yōu)化傳動(dòng)設(shè)計(jì)下?lián)u臂的長度(或機(jī)械的長度或機(jī)械的長度) 和滑移速度變化系數(shù)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)例子設(shè)計(jì)例子 已知的條件是, 滑移速度變化系數(shù)1818. 1K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. 0/43ll以及和的值。 假設(shè)503lmm, 然后我們將得出76.5393. 0/504lmm。 然后, 做 sketch(omitted) 。 最后, 算出桿的長度分別是161l mm,462l mm,503l mm,76.534l mm. 極小傳動(dòng)角度是 3698.462)(arccos322142322minllllll 結(jié)果由計(jì)算可得2227.161l mm， 5093.442l mm， 0000.503l mm， 8986.534l 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。