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Chapter 11 Gear Trains §11-1 Gear Trains and Their Classification §11-2 Train Ratio of a Gear Train With Fixed Axes §11-3 Train Ratio of Elementary Epicyclic Gear Train §11-4 Train Ratio of a Combined Gear Train §11-5 Applications of Gear Trains §11-1 Gear Trains and Their Classification 3 Gear Train——A transmission system by 2 3′ more than one pair of gears. 1 Classification of Gear Trains 4 一、 Gear train with fixed axes 4′ (Ordinary gear train) 5 Definition—The positions of all gear axes in a gear train are fixed O1 H 二、 Epicyclic gear train 2 O1 2 O O Definition—At least one gear axis in a gear train rotates the other axis. 3 H 1 3 1 Differential gear train: F=2 三、Combined gear train Definition—is a combination of several elementary epicyclic gear trains or a combination of at least one elementary epicyclic gear train with at least one gear train with fixed axes. O1 H 2 O1 3 2 O O 1 H 1 3 planetary gear train: F=1 §11-2 Train Ratio of a Gear Train With Fixed Axes The train ratio—the ratio of the angular velocities of input and output members in the gear train. 3 Train ratio includes two factors, 1) magnitude and 2)relative rotating direction of the two members. 2 3′ 1 一、Calculations of train ratio Given:the number of all gears, Find train ratio i15 = 1 / 5 。 z2 z3 z4 z5 1 i15 i12 i2 3 i3' 4 i4 5 5 z1 z2 z3' z4 Train ratio of gear train with fixed axes Product of tooth numbers of all the driven gears = Product of tooth numbers of all the driving gears 4 4′ 5 1 z 2 2 z1 z 2 3 3 z2 i12 i23 3 z 4 4 z3 z 4 5 5 z4' i3' 4 i4' 5 二、Relative rotating directions of gears The rotating direction of the input gear 1 is given as shown in Fig. Where the direction of the arrow indicates that of the peripheral velocity of the gear on the visible side. 1) The axis of the input gear is parallel to the axis of the output gear The directions of the input gear and the output gear are the same,train ratio is“+”; the 3 2 3′ directions are opposite, train ratio is“-”. 2) The axis of the input gear is not parallel to the axis of the output gear It can only be determined by drawing arrows. z2 z3 z4 z5 1 i15 5 z1 z2 z3' z4 1 4 4′ 5 idle wheel——The function of gear 2 is to change the rotating direction of the output gear , not the magnitude of the train ratio. §11-3 Train Ratio of Elementary Epicyclic Gear Train Shown in a is a typical elementary epicyclic gear train. It consists of two sun gears , one planet gear , one planet carrier H, and the frame. Suppose that the planet carrier H rotates in an angular velocity wH. The train ratio of this gear train cannot be calculated simply as a gear train with fixed axes because of the rotation of the planet carrier. -ωH Imagine that we add an angular velocity (- wH) to the whole O gear train. It will keep the relative motion between any two links unchanged. Now the angular velocity of the planet carrier H is wH - wH =0. This makes the planet carrier H fixed, as shown in Fig. O1 H 2 O1 3 2 O 1 ω3 ω2 ωH H 1 ω1 3 To add a common angular velocity “-ωH” to the whole epicyclic gear train, the changes of angular velocity of each component are shown in the table below: Components The former angular Angular velocities in the converted velocities gear trains Gear 1 ω1 1H 1 H Gear 2 ω2 2H 2 H Gear 3 ω3 3H 3 H H ωH HH H H 0 Since the converted gear train is a gear train with fixed axes, its train ratio can be calculated as for an ordinary gear train. H 1 H H 1 i13 H 3 3 H z2 z3 z3 z1 z2 z1 O1 H 2 O1 3 2 H O O 1 3 1 Where the“-” sign demonstrates that gears 1 and 3 in the converted gear train rotate oppositely. H m H H m imn H n n H product of tooth numbers of all the driven gears in the converted gear train product of tooth numbers of all the driving gears in the converted gear train When using the equation, please notice the following: 1)The equation above are only fit for the case that axes of the sun gear m, n and the planet carrier H are parallel. 2)When determining the train ratio of an epicyclic gear train,we enter the values along with the signs of angular velocities of ωm,ωn andωH for the equation above. If we designate some direction of rotation as the positive one, then the opposite direction as negative. 3)Judging the “+” “-” signs, we consider the planet carrier H as stationary, then the method same to the ordinary gear trains can be used. As regards planetary gear trains, that is ωn=0, the Eq. can be rewritten as: H mn i m H imH 1 0 H imH 1 i H mn Example11-1 In the gear train in Fig., z1 = z2 = 30, z3=90, n1 = 1 r/min counterclockwise, n3 = 1 r/min clockwise. (the value of angular velocity is defined as positive when counterclockwise). Find nH. Solution: z2 z3 z3 n1 nH i n3 nH z1 z2 z1 z3 1 nH 90 3 1 nH z1 30 H 13 then nH= -1/2 O1 H 2 O1 O O i1H = n1 / nH = -2 3 1 Example 11-2 In the planetary gear train shown in figure, z1 =100, z2 = 101, z2’=100, z3= 99. Find the train ratio iH1. Solution: 2 2’ For the planetary gear train shown in the figure, z 2 z3 i1H 1 i 1 z2 z2' 1 101 99 1 100 100 10000 H 13 Therefore, iH1= 1 / i1H = 10000 H 1 3 §11-4 Train Ratio of a Combined Gear Train z1= 20 z3= 30 Calculating the train ratio of a combined gear train: z4= 80 1.To divide the combined gear train into several elementary epicyclic gear trains and ordinary gear trains; z’2= 20 2. to write their train ratio equations independently. Solution:planet gears z2= 40 planet carriers sun gears Example: Shown in the gear train, the numbers of all gears are given.Calculate the train ratio i1H . Solution ordinary gear trains: Elementary epicyclic gear trains: gears 1、2 z i12 1 2 2 z1 40 2 20 i1H i12i2' H 2 5 10 gears 3、H、 2’、and 4 i2' H 1 i H 2'4 1 z4 1 z 2' 80 5 20 §11-5 Applications of Gear Trains 1. Branching transmission 2. To get a large train ratio 3. To change the speed of rotation 4. To change the direction of rotation 5. To combine the motion 6. To resolve the motion 7. To get a large power transmission Fig. shows differential mechanism of car, where: Z1= Z3 ,nH= n4 n1 nH z3 =-1 i n3 nH z1 H 13 nH 1 n1 n3 2 The motor drives gear 5 through the transmission shaft. When a car move forward straight, n1 =n3 P r v3 v1 ω When a car turns left, the car rotates around the point P: r-radius V1=(r-L) ω 2L-distance between two wheels V3=(r+L) ω n1 /n3 = V1 / V3 = (r-L) / (r+L) rL n1 nH r 2L n3 rL nH r move straight turn corner 5 2 4 1 3 2 H Differential mechanism