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22-Mec-A2 Kinematics and Dynamics of Machines · May 2013

Question 3 of 6: Compound Planetary Gear Reduction

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Notes on this paper

Paper: National Examinations — 07-Mec-A2 Kinematics and Dynamics of Machines, May 2013. Open-book, 3-hour paper; answer FIVE of six questions, all of equal value (20 marks). Part A (mechanisms & machine dynamics, Q1–Q4) and Part B (mechanical vibration, Q5–Q6). All six questions are solved here.

Reference texts: R. L. Norton, Design of Machinery (6th ed., McGraw-Hill) — velocity analysis, cams, gear trains, engine dynamics; C. E. Wilson & J. P. Sadler, Kinematics and Dynamics of Machinery (3rd ed.); S. S. Rao, Mechanical Vibrations (6th ed., Pearson) — Part B; J. E. Shigley, Mechanical Engineering Design — gear geometry.

Note on the figure-based questions. Q1 (press) is a scaled graphical problem (“Scale 1:15”); its link geometry is read from the printed diagram, so the mechanical advantage carries a graphical tolerance of roughly ±5 %. Q3 (planetary box) is solved from the kinematic topology inferred from the sectioned schematic. Both interpretations are stated explicitly in a check callout so the reader can re-scale from the original drawing.

Question 3: Compound Planetary Gear Reduction (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. Two epicyclic stages sharing a common carrier (the “Arm”). Stage A: input sun 1 (28T); compound planet 2 (52T)–3 (18T); ring 4 (94T) held fixed. Stage B: sun 5 (18T) held fixed; planet 6 (35T); ring 7 (88T) is the output. The ring/sun tooth counts confirm coaxial geometry ($z_7=z_5+2z_6\Rightarrow 88=18+2(35)$ &checkmark).

Find. The output speed $\omega_7$ and its direction, given $\omega_1=1800$ rpm ccw.

axis11800 rpm in234 ring (FIXED)Arm (common carrier)55 sun (FIXED)67 ring OUT
Fig. 3 — kinematic skeleton of the two-stage compound planetary reducer. Stage A (sun 1, compound planet 2-3, fixed ring 4) sets the carrier speed; Stage B (fixed sun 5, planet 6, output ring 7) is driven by the same carrier.

Approach. Use the epicyclic train-value equation $\dfrac{\omega_L-\omega_\text{arm}}{\omega_F-\omega_\text{arm}}=e$, where $e$ is the ordinary (arm-held) gear ratio from first to last gear with proper mesh signs. Solve Stage A for the carrier, then Stage B for the output.

  1. Stage A train value (sun 1 → ring 4, arm fixed): external mesh 1–2 contributes $-z_1/z_2$; compound 2≡3; internal mesh 3–4 contributes $+z_3/z_4$: $$e_A=\left(-\frac{z_1}{z_2}\right)\!\left(\frac{z_3}{z_4}\right)=-\frac{28\cdot18}{52\cdot94}=-0.1031.$$
  2. Carrier speed. Ring 4 fixed ($\omega_4=0$): $$\frac{0-\omega_\text{arm}}{\omega_1-\omega_\text{arm}}=e_A\ \Rightarrow\ \omega_\text{arm}=\omega_1\frac{e_A}{e_A-1}=1800\cdot\frac{-0.1031}{-1.1031}=\boxed{168.3\ \text{rpm (ccw)}}.$$
  3. Stage B train value (sun 5 → ring 7, arm fixed): external 5–6 gives $-z_5/z_6$, internal 6–7 gives $+z_6/z_7$; the planet cancels: $$e_B=\left(-\frac{z_5}{z_6}\right)\!\left(\frac{z_6}{z_7}\right)=-\frac{z_5}{z_7}=-\frac{18}{88}=-0.2045.$$
  4. Output speed. Sun 5 fixed ($\omega_5=0$): $$\frac{\omega_7-\omega_\text{arm}}{0-\omega_\text{arm}}=e_B\ \Rightarrow\ \omega_7=\omega_\text{arm}\,(1-e_B)=168.3\,(1.2045)=\boxed{202.7\ \text{rpm (ccw)}}.$$
  5. Overall ratio. $\dfrac{\omega_1}{\omega_7}=\dfrac{1800}{202.7}=8.88{:}1$ reduction; the output turns the same direction (ccw) as the input.
QuantityValue
Stage-A train value $e_A$−0.1031
Carrier (Arm) speed168.3 rpm ccw
Stage-B train value $e_B$−0.2045
Output gear 7 speed202.7 rpm, ccw
Overall reduction8.88 : 1
Check: topology read from the sectioned schematic — input sun 1, compound planet 2-3 with fixed ring 4 (Stage A), and fixed sun 5 with output ring 7 (Stage B), both planet sets on one common carrier. The tooth-count identity 88 = 18 + 2(35) confirms Stage B is a coaxial sun-planet-ring set; if the drawing instead ties sun 5 to the carrier, re-solve Stage B with $\omega_5=\omega_\text{arm}$.