9.2 Cams & Followers, Toothed Gearing & Epicyclic Gear Trains

Cams & Followers, Toothed Gearing & Epicyclic Gear Trains

Quick Reference: For high-speed cams, cycloidal motion avoids dynamic shock by maintaining finite jerk throughout the stroke. The fundamental law of gearing requires the common normal to pass through the pitch point $P$. Minimum pinion teeth to prevent interference with a rack is $Z_{\min} = 2a_r / \sin^2\phi$ ($18$ teeth for standard $20^{\circ}$ full-depth). Epicyclic gear trains are solved systematically via the tabular superposition method.

In mining mechanical engineering, gear transmissions and cam mechanisms drive critical machinery: multi-stage reduction planetary drives in coal shearers, dragline walking mechanisms, heavy conveyor gearboxes, and internal combustion valve trains. Precision kinematic design ensures high power density, reliability, and smooth transmission under extreme operating torques.

---

1. Cam & Follower Kinematics

Cam Terminology & Geometric Parameters

• Base Circle: Smallest circle tangent to the cam profile drawn from the camshaft center. Its radius ($r_b$) dictates the overall size of the cam.
• Prime Circle: Smallest circle drawn from the camshaft center tangent to the pitch curve. For a knife-edge follower, prime circle equals base circle; for a roller follower of radius $r_r$, $r_p = r_b + r_r$.
• Pitch Curve: The locus of the follower trace point as the follower traverses the rotating cam profile.
• Pressure Angle ($\phi$): The angle between the direction of the follower motion and the normal to the pitch curve at the point of contact.

$\tan\phi = \frac{1}{r} \frac{ds}{d\theta}$

Design Rule for Pressure Angle: High pressure angles cause severe lateral side thrust, resulting in follower jamming in its guide bushings. In practice:

• For translating roller followers: Maximum pressure angle $\phi_{\max} \le 30^{\circ}$.
• For translating flat-faced followers: $\phi = 0^{\circ}$ continuously (side thrust is virtually eliminated, but bending moments on the stem increase).
• To decrease pressure angle: Increase base circle radius $r_b$, increase follower lift angle $\theta_o$, or introduce an optimal follower offset $e$.

---

2. Follower Motion Profiles & Dynamic Comparison

Let $h = \text{Total follower stroke (lift)}$, $\theta_o = \text{Total cam angle of ascent/descent}$, and $\omega = d\theta/dt = \text{Cam angular speed}$.

+-----------------------------------------------------------------------------------------+
|                        FOLLOWER MOTION KINEMATIC FORMULAS                               |
+-----------------------------------------------------------------------------------------+
| Motion Profile      | Displacement s(theta)        | Max Velocity v_max   | Max Accel a_max      |
+---------------------+------------------------------+----------------------+---------------------+
| Uniform Velocity    | h * (theta / theta_o)        | (h * w) / theta_o    | Infinite (at ends)  |
| Simple Harmonic     | (h/2)*[1 - cos(pi*th/th_o)]  | (pi*h*w)/(2*th_o)    | (pi^2*h*w^2)/(2*th_o^2)|
| Uniform Accel (UARM)| 2h*(th/th_o)^2 [first half]  | (2*h*w)/th_o         | (4*h*w^2)/th_o^2    |
| Cycloidal           | h*[(th/th_o) - sin(2pi*th/th_o)/(2pi)] | (2*h*w)/th_o | (2pi*h*w^2)/th_o^2  |
+-----------------------------------------------------------------------------------------+

  ACCELERATION CURVES COMPARISON:
  
  1. SHM Accel:             2. Parabolic Accel:       3. Cycloidal Accel:
     +a_max |\              +a |+-------+             +a_max |   /---\
            | \                |        |                    |  /     \
       0 ---+--\-----+--       |--------+-------             +--+-----+--+---
                \    |                  |        |                 \     /   |
                 \   |                  +--------|-a                \---/ -a_max
                  \|-a_max
     (Finite ends -> Inf Jerk)  (Step jumps -> Inf Jerk)   (Zero ends -> Finite Jerk)

Industrial Selection Criterion:

1. Uniform Velocity: Unacceptable in dynamic machinery due to infinite accelerations (infinite impact force $F = m \cdot a$) at stroke boundaries.
2. Simple Harmonic Motion (SHM): Excellent for low-to-moderate speed machinery. However, at $\theta = 0$ and $\theta = \theta_o$, acceleration has finite non-zero values that jump instantly to zero during dwell, causing infinite jerk ($j = da/dt \to \infty$) and severe acoustic chatter.
3. Uniform Acceleration and Retardation (UARM / Parabolic): Yields the minimum possible peak acceleration for a given lift and angle, but exhibits step changes in acceleration at start, midpoint, and end.
4. Cycloidal Motion: Acceleration starts from zero, varies smoothly as a sine curve, and returns to zero at the end of the stroke. Jerk is finite everywhere. This makes cycloidal profiles mandatory for high-speed automated systems, IC engine valve trains, and heavy mining vibratory feeders.

---

3. Toothed Gearing & The Law of Gearing

The Fundamental Law of Gearing

Law of Gearing: For two meshing gear teeth to transmit a constant angular velocity ratio, the common normal to the tooth profiles at the point of contact must, in all positions of contact, pass through a fixed point on the line of centers known as the pitch point ($P$).

$\frac{\omega_1}{\omega_2} = \frac{O_2 P}{O_1 P} = \frac{R_2}{R_1} = \frac{Z_2}{Z_1} = \text{Constant}$

          O_1 (Pinion Center)
           o
            \ 
             \  R_1
              \ 
               o P (Pitch Point) ======= Common Normal (Line of Action)
              /                          passes through P at angle phi
             / 
            /  R_2
           o
          O_2 (Gear Center)

Involute vs. Cycloidal Profiles

---

4. Interference & Undercutting in Involute Gears

Mechanism of Interference

Involute profile geometry exists only outside the base circle. If the addendum of the mating gear is excessively large, the tip of the gear tooth makes contact with the pinion tooth flank below the pinion's base circle on the radial non-involute portion. During cutting or operation, this causes the gear tip to gouge out material from the pinion root flank—a destructive phenomenon termed undercutting that severely weakens tooth root strength.

                   Interference Boundary Condition:
                   
    Base Circle Pinion o-----------------------------o Base Circle Gear
                       \             |             /
                        \          Line of        /
                         \         Action        /
                          \          |          /
                   Point of Tangency (K)      Point of Tangency (L)
                          
    Interference occurs if tip circle of gear crosses tangency point K!

Contact Parameters:

1. Path of Contact ($KL$): $KL = \text{Path of Approach } (KP) + \text{Path of Recess } (PL)$ $KL = \sqrt{R_A^2 - R^2 \cos^2\phi} + \sqrt{r_A^2 - r^2 \cos^2\phi} - (R + r)\sin\phi$ Where $R_A, r_A$ are addendum circle radii, $R, r$ are pitch circle radii, and $\phi$ is pressure angle.
2. Arc of Contact: $\text{Arc of Contact} = \frac{\text{Path of Contact}}{\cos\phi}$
3. Contact Ratio ($\text{CR}$): $\text{CR} = \frac{\text{Arc of Contact}}{p_c} = \frac{KL}{\pi m \cos\phi} \ge 1.2\text{ to }1.4$

Minimum Number of Teeth to Avoid Interference:

1. Pinion Meshing with an Identical Rack: $Z_{\min} = \frac{2 a_r}{\sin^2\phi}$ Where $a_r$ is the rack addendum fraction ($a_r = 1.0$ for standard full-depth teeth).

   • For $\phi = 14.5^{\circ}$: $Z_{\min} = \frac{2 \times 1}{\sin^2(14.5^{\circ})} = 31.9 \approx 32\text{ teeth}$.
   • For $\phi = 20.0^{\circ}$: $Z_{\min} = \frac{2 \times 1}{\sin^2(20^{\circ})} = \frac{2}{0.11698} = 17.1 \approx 18\text{ teeth}$.
   • For $\phi = 25.0^{\circ}$: $Z_{\min} = \frac{2 \times 1}{\sin^2(25^{\circ})} = 11.2 \approx 12\text{ teeth}$.

2. Pinion Meshing with a Gear of Gear Ratio $G = Z_G / Z_P = T / t$: $Z_{\min} = \frac{2 a_p}{\sqrt{1 + \frac{1}{G}\left(\frac{1}{G} + 2\right)\sin^2\phi} - 1}$

Remedial Measures to Eliminate Interference:

1. Increase the number of teeth on the pinion ($Z_P \ge Z_{\min}$).
2. Increase the pressure angle (e.g., transition from $14.5^{\circ}$ to $20^{\circ}$ or $25^{\circ}$).
3. Stub the tooth addendum ($a_p = 0.8m$ instead of $1.0m$).
4. Apply profile shift (addendum modification) by displacing the cutter rack outwards.

5. Gear Trains & Epicyclic Gear Systems

Classification of Gear Trains

1. Simple Gear Train: Each shaft carries only one gear. Intermediate idler gears change rotational direction but do not influence the overall velocity ratio.
2. Compound Gear Train: At least one intermediate shaft carries two or more rigidly keyed gears. Enables massive velocity reductions in compact spaces.
3. Reverted Gear Train: The input driving shaft and the output driven shaft are coaxial (share the same centerline). Essential for automotive manual transmissions, machine tool speed boxes, and industrial clocks.
   • Center distance requirement for identical module $m$: $r_1 + r_2 = r_3 + r_4 \implies Z_1 + Z_2 = Z_3 + Z_4$
4. Epicyclic (Planetary) Gear Train: One or more gear axes rotate relative to a fixed frame around the central axis of another gear.

Tabular Method for Epicyclic Gear Trains

Consider a planetary system consisting of a Sun gear $S$ ($Z_S$), Planet gear $P$ ($Z_P$), Internal Ring gear $R$ ($Z_R$), and Carrier Arm $A$.

Geometric Relationship: $d_R = d_S + 2 d_P \implies Z_R = Z_S + 2 Z_P \quad (\text{for identical module})$

Torque Relationships in Epicyclic Gear Trains

In an ideal lossless epicyclic gear train in steady state:

1. Static Torque Equilibrium: $\sum T = 0 \implies T_S + T_A + T_R = 0$
2. Power Balance (Conservation of Energy): $\sum T_i \omega_i = 0 \implies T_S \omega_S + T_A \omega_A + T_R \omega_R = 0$

---

6. Worked Numerical Examples

Example 1: Planetary Reduction Drive in a Mining Hauler

Problem: An epicyclic gear reduction box has a sun gear $S$ with 24 teeth and an internal ring gear $R$ with 96 teeth. The planet gear $P$ meshes internally with $R$ and externally with $S$. The sun gear is driven by an electric motor at $+1200\text{ rpm}$ clockwise, and the ring gear $R$ is held stationary ($N_R = 0$). Calculate:

1. The number of teeth on the planet gear $P$.
2. The rotational speed and direction of the carrier arm $A$.
3. The torque on the carrier arm $A$ if the input torque on the sun gear is $T_S = 250\text{ N}\cdot\text{m}$ (assuming $100%$ transmission efficiency).

Solution:

1. Determine Planet Teeth ($Z_P$): $Z_R = Z_S + 2 Z_P \implies 96 = 24 + 2 Z_P \implies 2 Z_P = 72 \implies Z_P = 36\text{ teeth}$

2. Kinematic Equations from Tabular Method:

   • Sun speed: $N_S = y + x = +1200\text{ rpm}$
   • Ring speed: $N_R = y - x \left(\frac{Z_S}{Z_R}\right) = 0$

   Substitute $Z_S / Z_R = 24 / 96 = 1/4$ into the ring equation: $y - \frac{1}{4}x = 0 \implies x = 4y$

   Substitute $x = 4y$ into the sun equation: $y + 4y = 1200 \implies 5y = 1200 \implies y = +240\text{ rpm}$ $x = 4(240) = +960\text{ rpm}$

   Therefore, carrier arm speed $N_A = y = +240\text{ rpm}$ (Clockwise). Gear reduction ratio $= 1200 / 240 = 5:1$.

3. Torque Calculation:

   • Power in $=$ Power out $\implies T_S \omega_S + T_A \omega_A + T_R \omega_R = 0$ Since $N_R = 0$, $T_R \omega_R = 0$: $T_S N_S + T_A N_A = 0 \implies (250) (1200) + T_A (240) = 0$ $T_A = -\frac{250 \times 1200}{240} = -1250\text{ N}\cdot\text{m}$ (The negative sign indicates the resisting torque on the arm opposes its motion, so output drive torque $= +1250\text{ N}\cdot\text{m}$ in clockwise direction).

   • Holding Torque on Fixed Ring ($T_R$): $T_S + T_A + T_R = 0 \implies 250 - 1250 + T_R = 0 \implies T_R = +1000\text{ N}\cdot\text{m}$

---

Example 2: Involute Gear Interference Sizing

Problem: A $20^{\circ}$ full-depth pinion of module $m = 8\text{ mm}$ with 20 teeth drives a gear of 60 teeth. Determine whether interference will occur, and calculate the maximum permissible addendum on the gear to prevent interference.

Solution:

1. Gear Parameters:

   • Pressure angle $\phi = 20^{\circ}$, Pinion pitch radius $r = \frac{m Z_P}{2} = \frac{8 \times 20}{2} = 80\text{ mm}$
   • Gear pitch radius $R = \frac{m Z_G}{2} = \frac{8 \times 60}{2} = 240\text{ mm}$
   • Gear ratio $G = Z_G / Z_P = 60 / 20 = 3$

2. Minimum Pinion Teeth for $G = 3$ without Interference: $Z_{\min} = \frac{2 a_p}{\sqrt{1 + \frac{1}{G}\left(\frac{1}{G} + 2\right)\sin^2\phi} - 1}$ For standard addendum $a_p = 1.0$: $Z_{\min} = \frac{2}{\sqrt{1 + \frac{1}{3}\left(\frac{7}{3}\right)\sin^2(20^{\circ})} - 1} = \frac{2}{\sqrt{1 + (0.7778)(0.11698)} - 1} = \frac{2}{\sqrt{1.09098} - 1} = \frac{2}{1.0445 - 1} = \frac{2}{0.0445} \approx 44.9$ Because the actual pinion has only 20 teeth ($20 < 44.9$), standard addendum ($a_g = 1.0m = 8\text{ mm}$) will produce severe interference.

3. Maximum Addendum Radius ($R_{A,\max}$) to Prevent Interference: The tip of the gear tooth must not cross the tangent point $K$ on the pinion base circle: $R_{A,\max} = \sqrt{R^2 \cos^2\phi + (R + r)^2 \sin^2\phi}$ $R_{A,\max} = \sqrt{(240)^2 \cos^2(20^{\circ}) + (240 + 80)^2 \sin^2(20^{\circ})}$ $R_{A,\max} = \sqrt{57600(0.88302) + (320)^2(0.11698)} = \sqrt{50862 + 102400(0.11698)} = \sqrt{50862 + 11979} = \sqrt{62841} \approx 250.68\text{ mm}$

   Maximum permissible gear addendum: $h_{a,\max} = R_{A,\max} - R = 250.68 - 240 = 10.68\text{ mm}$