行星齒輪減速器設(shè)計(jì)畢業(yè)論文-預(yù)覽頁
【正文】 ，外形結(jié)構(gòu)如圖所示：齒輪聯(lián)軸器的齒形為漸開線，幾何計(jì)算見下表（非變位）：(mm)項(xiàng)目代號(hào)計(jì)算公式結(jié)果分度圓直徑dd=mzd=272基圓直徑dd=dcosαd=齒頂圓直徑dd=d＋2hmd=d—2hmd=288d=256齒根圓直徑dd=d—2hmd=d＋2hmd=262d=282齒頂高h(yuǎn)h=hmh=8齒根高h(yuǎn)h=hm=(h＋c)mh=10齒寬bb=120,b=65嚙合角αα=20176?！嗨詽M足使用要求。模塊二設(shè)計(jì)的是行星齒輪傳動(dòng)在高速級(jí)與低速級(jí)時(shí)強(qiáng)度的校核，強(qiáng)度校核也是從接觸疲勞強(qiáng)度與彎曲疲勞強(qiáng)度兩方面進(jìn)行校核。s Mechanical Engineer39。 they may be considered as one in describing the action of the gearing. The sun gear, the arm, or the ring gear may be input or output links. The trick is that any motion of the gear train can carried out by first holding the arm fixed and rotating the gears relative to one another, and then locking the train and rotating it about the fixed axis. The net motion is the sum or difference of multiples of the two separate motions that satisfies the conditions of the problem. To carry out this program, construct a table in which the angular velocities of the gears and arm are listed for each, for each of the two cases. The locked train gives 1 for arm, gear 1, gear 2 and gear 3. Arm fixed gives 0,1,N1/N2, N1/N3. Suppose we want the velocity ration between the arm and gear 1, when gear 3 is fixed. Multiply the first row by a constant so that when it is added to the second row, the velocity of gear 3 will be zero. This constant is N1/N3. Now ,doing one displacement and then the other corresponds to adding the two rows. We find N1/N3, 1+N1/N3, N1/N3N1/N2. The first number is the arm velocity ,the second the velocity of gear 1, so the velocity ratio between them is N1/(N1+N2), after multiplying through by N3. This is the velocity ratio we need for the Tamiya gearbox, where the ring gear does not rotate, the sun gear is the input, and the arm is the output. The procedure is general, however, and will work for any epicyclic train. One of the Tamiya planetary gear assemblies has N1=N2=16,N3=48, while the other has N1=12,N2=18, N3=48. Because the planetary gears must fit between the sun and ring gears ,the condition N3=N1+2N2must be satisfied. It is indeed satisfied for the numbers of teeth given. The velocity ratio of the first set will be 16/(48+16)=1/4. The velocity ratio of the second set will be 12/(48+12)=1/5. Both ratios are as advertised. Note that the sun gear and arm will rotate in the same direction. The best general method for solving epicyclic gear trains is the tabular method, since it does not contain hidden assumptions like formulas, nor require the work of the vector method. The first step is to isolate the epicyclic train, separating the gear trains for inputs and outputs form it. Find the input speeds or turns, using the input gear trains. There are, in general, two inputs, one of which may be zero in simple problems. Now prepare two rows of the table of turns or angular velocities. The first row corresponds to rotating around the epicyclic axis once, and consists of all’s. Write down the second row assuming that the arm velocity is zero, using the known gear ratios. The row that you want is a linear bination of these two rows, with unknown multipliers x and y . Summing the entries for the input gears gives two simultaneous linear equations for x and y in terms of the known input velocities. Now the sum of the two rows multiplied by their respective multipliers gives the speeds of all the gears of interest. Finally , find th output speed with the aid of the output gear train. Be careful to get the directions of rotation correct, with respect to a direction taken as positive. 行星齒輪介紹Tamiya的行星齒輪箱是由一個(gè)小型直流電機(jī)運(yùn)行在大約10500 r/min ,。齒輪箱是一個(gè)設(shè)計(jì)非常精心的塑料套件,可以在大約一個(gè)小時(shí)用很少的工具組裝。如果節(jié)圓的半徑是a和b,齒輪軸之間的距離是a + b。在這里我們不會(huì)進(jìn)一步談?wù)擙X輪輪齒，以及上述有提到的傳動(dòng)裝置的基本原理。如果你知道兩個(gè)齒輪的節(jié)圓直徑，那么你就能夠得出兩齒輪軸之間的距離。如果以一齒輪的旋轉(zhuǎn)方向?yàn)檎?，此時(shí)另外一個(gè)的方向則為負(fù)。齒輪的傳動(dòng)比并不依賴于該軸的精確的間距，而是輪齒或者節(jié)圓諸如此類之間的安裝。一個(gè)移動(dòng)的手臂或有關(guān)該軸以及齒輪自己可以旋轉(zhuǎn)的齒輪軸。這是三個(gè)行星齒輪輪系用于機(jī)械領(lǐng)域的原因；他們可能被認(rèn)為是在描述該傳動(dòng)裝置的操作之一。若要進(jìn)行此程序，構(gòu)造的齒輪和轉(zhuǎn)臂的角速度列出兩例的每個(gè)表。此常量為N1/N3。這就是我們需要的變速器的速度比，在變速器里面，環(huán)齒輪不會(huì)旋轉(zhuǎn)，太陽輪是輸入端，揮臂速度則是輸出值。事實(shí)上，這個(gè)條件得滿足給定齒輪的數(shù)目。請(qǐng)注意，太陽輪和揮臂將向同一個(gè)方向旋轉(zhuǎn)。現(xiàn)在準(zhǔn)備兩行關(guān)于轉(zhuǎn)速或角速度的圖表。把輸入的齒輪值相加，根據(jù)已知的輸入速度，同時(shí)產(chǎn)生兩個(gè)關(guān)于x和y的兩種線性方程組。
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