1. [50%] For the two-stage planetary gear train assigned to you in Table 1, the numbers of tooth of the first and second

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answerhappygod
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1. [50%] For the two-stage planetary gear train assigned to you in Table 1, the numbers of tooth of the first and second

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1 50 For The Two Stage Planetary Gear Train Assigned To You In Table 1 The Numbers Of Tooth Of The First And Second 1
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1. [50%] For the two-stage planetary gear train assigned to you in Table 1, the numbers of tooth of the first and second stages are defined as (N) and (N)2, respectively. Answer the following questions: (a) For each stage, determine the condition (for the numbers of teeth) when it can serve as a differential. Please specify which gear is the input, and which two gears are the outputs to wheels. Planetary gears cannot be input or output. (b) Draw a sketch of the two-stage planetary gear train. (c) For each speed, if the input speed is (1)1, write down and solve the equations (in matrix form) for the motion in terms of x₁, y1, x2, and y2. Then determine the speed ratio from output to input. (d) For each speed, use the tabular method to determine the speed ratio from output to input. (e) Determine one set of tooth numbers to generate speed ratios closest to 0.1 and 0.2 that can be used for the 1st speed and 2nd speed, respectively. All gears are to have at least 12 teeth. (If a positive speed ratio is not possible, please show your reason and then use a negative speed ratio such as -0.1 or -0.2 instead)

1st stage G (5) 00₁ = x+y 0₂2 = x 00₂=x- (4=x- (0)₂ = x + 2nd stage J(7) - N₁ N3 N₁ N₂ N₁N₂ N₂N₁ N Type G Table 1 Connection G1-J1, G2-J2 4 + 5 (G-1) (G-2) (G-3) (G-5) Fixed (1st speed) G5 - w₁ = x+y W₂ = x 003=x- 004=x- 00₂ = x + 006=x+ (0)₂ = x + Fixed (2nd speed) J2 N₁ N₂ 2 Type J N₁ N₂ N₁N₁ N₂N₂ N₁N₁ N₂N₁ N₁N₂N6 NẠNẠN- Input Output GI 37 7 6 (J-1) (J-2) (J-3) (J-4) (J-5) (J-6) (J-7)

(W), Winput = x₁+y₁ = (0₁) 7 (10) = 1(²m) (W₂), (2) (W9), = 0 = el s 2nd, (km) 16+1x =YIM) (will D (NN)2 (N3)₂ (5) 2 (N) 2 -(Ni), (Na) (ws, + XITY, S(WI), Xirg₁=x2+y₁ = (wit) X1 X2 (NI); (N4) (N₂), (NS) 21 92 1 (IM)= ² & trx = 14 +1x X1 = X ₂ = 0 (012) | = (1+)2 = 0 () = 00 01-01 0010 10 (M) (N4) ₂ (NG)₂) (M₂)2 (N5) 2 (N₁) 2 -W₁), (N4) 11/1 + (N5) 1(N5) (0) That relat ME (N1)2(NG) (NG) [N]) = (MS) (MG/L (N₂)i (NS) 322 YI 11 TX th (Ni) (N4) (6) (N3) (MS (₂) ₂ (WI)1 (WI) (Wit) 0 1 (wi) 1 (MU, (4) 7% 0 T 0 -1 D 00 00 J (N₂Js (NS), (M)), (N4)| (W)) 0 (11) X1 ess th %2 th X (ND), (NE): () D (NN 4),(@))) (Mg) (NS)-(MJ (4) EMVICENTI (EN) ¹(EN)*(N7-1(5N) (EM (21) 7(km) 1 (IM). 0 (Mi)(N4) (N3) (MS) 2 (Ng) ₂ NILOATING)2 Z(UN) =(5N) (FN)
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