Cams and Cam-Follower Design: Key Symbols and Relationships

Key Symbols and Relationships

Before diving into the four displacement curves, lock these symbols into your working vocabulary:

The foundational derivatives that connect everything:

$$v = \frac{dy}{dt} = \omega \frac{dy}{d\phi}$$

$$a = \frac{d^2y}{dt^2} = \omega^2 \frac{d^2y}{d\phi^2}$$

Remember: Velocity is the first derivative of displacement. Acceleration is the second derivative. Forces come from acceleration. Design the displacement curve, and you design the forces.

The Four Displacement Curves That Matter Most

These four curves are the foundation of cam design. Each has distinct characteristics, advantages, and limitations. Choosing the wrong one for your application is one of the most common—and most expensive—cam design errors.

. Constant Velocity Motion

Displacement:

$$y = h \cdot \frac{\phi}{\beta}$$

Velocity:

$$v = \frac{h \cdot \omega}{\beta}$$

Acceleration:

$$a = 0 \quad \text{(except at } t = 0 \text{ and } t = T \text{ where } a = \infty\text{)}$$

Displacement      Velocity         Acceleration
  h ┌─────/       h/T ┌─────       ∞ ↑        ↑ ∞
    │    /            │             │        │
    │   /             │             │        │
    │  /              │             │        │
    │ /               │            0├────────┤
    │/                │             │        │
  0 └─────           0└─────       ↓ -∞    -∞ ↓
    0    T            0    T         0       T

Verdict: Theoretically infinite acceleration at both ends of the stroke. Rarely used except in very crude, low-speed devices. However, the principle of uniform velocity is so desirable that modified versions (blended with parabolic curves) are widely used.

. Parabolic Motion (Uniformly Accelerated)

For the first half of the stroke (0 ≤ φ ≤ β/2):

$$y = 2h\left(\frac{\phi}{\beta}\right)^2$$

$$v = \frac{4h\omega\phi}{\beta^2}$$

$$a = 4h\left(\frac{\omega}{\beta}\right)^2$$

For the second half of the stroke (β/2 ≤ φ ≤ β):

$$y = h\left[1 - 2\left(1 - \frac{\phi}{\beta}\right)^2\right]$$

$$v = \frac{4h\omega}{\beta}\left(1 - \frac{\phi}{\beta}\right)$$

$$a = -4h\left(\frac{\omega}{\beta}\right)^2$$

Displacement      Velocity           Acceleration
  h ┌───╮         vmax┌──╲             a ┌──────┐
    │  ╱ │            │╱   ╲              │      │
    │ ╱  │            │     ╲           0 ├──────┤──────
    │╱   │           0└──────╲           │      │
  0 └────┘                               │-a    └──────┘
    0    T            0      T            0      T

The advantage: For a given angle of rotation and rise, parabolic motion produces the smallest possible maximum acceleration.

The trap: Sudden changes in acceleration (called jerk or pulse) occur at the beginning, middle, and end of the stroke. In real machines—which always have some flexibility and backlash—these sudden changes create impact forces that can be two to three times the theoretical values.

Verdict: Suitable for moderate speeds. At high speeds, the jerk problem makes parabolic motion a poor choice despite its low peak acceleration numbers on paper.

. Simple Harmonic Motion

Displacement:

$$y = \frac{h}{2}\left(1 - \cos\frac{180° \cdot \phi}{\beta}\right)$$

Velocity:

$$v = \frac{h}{2} \cdot \frac{\pi\omega}{\beta} \cdot \sin\frac{180° \cdot \phi}{\beta}$$

Acceleration:

$$a = \frac{h}{2} \cdot \left(\frac{\pi\omega}{\beta}\right)^2 \cdot \cos\frac{180° \cdot \phi}{\beta}$$

Displacement         Velocity          Acceleration
  h ┌──╮             vmax             +amax┌╲
    │  │ (S-curve)       ╱╲                │  ╲
    │  │               ╱    ╲            0 ├────╲──
    │  │              ╱      ╲             │      ╲
  0 └──┘            0╱        ╲0     -amax └───────╲
    0   T            0        T              0      T

The advantage: Smoothness in velocity and acceleration during the stroke. The motion follows a sinusoidal pattern, which inherently produces gradual changes.

The problem: Instantaneous changes in acceleration at the beginning and end of the stroke—where the acceleration jumps from zero to its maximum value (and back to zero). These jumps cause vibration, noise, and wear, particularly if the inertia loads are significant.

Verdict: Good for moderate to moderately high speeds. The acceleration discontinuities at the ends limit its use in very high-speed applications.

. Cycloidal Motion

Displacement:

$$y = h\left(\frac{\phi}{\beta} - \frac{1}{2\pi}\sin\frac{360° \cdot \phi}{\beta}\right)$$

Velocity:

$$v = \frac{h\omega}{\beta}\left(1 - \cos\frac{360° \cdot \phi}{\beta}\right)$$

Acceleration:

$$a = \frac{2\pi h\omega^2}{\beta^2}\sin\frac{360° \cdot \phi}{\beta}$$

Displacement         Velocity         Acceleration
  h ┌──╮                              +amax   ╱╲
    │  │ (S-curve)     ╱──╲                  ╱    ╲
    │  │             ╱      ╲          0 ──╱──────╲──
    │  │           ╱          ╲             ╲      ╱
  0 └──┘          0            0    -amax    ╲──╱
    0   T          0          T              0    T

This is the curve that saved the practitioner Engström's production line.

The cycloidal motion curve has no abrupt changes in acceleration at any point. The acceleration starts at zero, rises smoothly to a maximum, returns to zero at the midpoint, reaches a maximum in the opposite direction, and returns smoothly to zero at the end.

The maximum acceleration is somewhat higher than simple harmonic motion for the same rise and time. But this is where the critical insight lies:

Because cycloidal motion has no sudden changes in acceleration (no jerk), the actual dynamic forces in a real machine are only slightly higher than the theoretical values. Multiply by a safety factor of just 1.05.

Compare this to parabolic motion, where the sudden jerk means you must multiply calculated acceleration forces by a factor of 2 or more to account for dynamic pulses.

Verdict: The preferred choice for high-speed machinery. Results in low noise, low vibration, and low wear. The slightly higher peak acceleration is more than offset by the dramatically lower dynamic forces in practice.

Displacement Curve Comparison Table

Displacement Diagram Synthesis: The Art of Blending Curves

Pure constant velocity has the advantage of uniform speed. Pure parabolic has the advantage of zero starting velocity. What if you could combine them?

This is exactly what displacement diagram synthesis achieves. By matching a parabolic curve to the beginning and end of a constant-velocity (straight-line) displacement, you eliminate the infinite accelerations while preserving the uniform velocity through the middle of the stroke.

How the Matching Works

Consider a parabola with its vertex at point O. The tangent to the curve at any point P intersects the baseline at the midpoint of the horizontal distance to P. This geometric property means the tangent represents the velocity—and if you match this tangent to the constant-velocity line, the transition from rest to full speed happens smoothly.

Worked Example: Modified Constant Velocity Cam

Problem: Design a cam where the follower:

• Rises 0.25 units with constant acceleration (parabolic)
• Rises 1.25 units with constant velocity over 50° of cam rotation
• Rises 0.50 units with constant deceleration (parabolic)

Total rise: h = 0.25 + 1.25 + 0.50 = 2.00 units

Step 1—Find the blending angles:

Using the matching relationship where the tangent at the blend point bisects the horizontal distance:

$$\frac{\phi_1}{\phi_2} = \frac{y_1}{y_2} \quad \Rightarrow \quad \frac{\frac{1}{2}\phi_1}{50°} = \frac{0.25}{1.25} \quad \Rightarrow \quad \phi_1 = 20°$$

$$\frac{\phi_3}{\phi_2} = \frac{y_3}{y_2} \quad \Rightarrow \quad \frac{\frac{1}{2}\phi_3}{50°} = \frac{0.50}{1.25} \quad \Rightarrow \quad \phi_3 = 40°$$

Total rise angle: β = 20° + 50° + 40° = 110°

Step 2—Calculate displacement values:

For the acceleration phase (0 ≤ φ ≤ 20°), using parabolic formula with substituted values (2y₁ for h, 2φ₁ for β):

$$y = \frac{2(0.500)}{(40°)^2} \cdot \phi^2 = 0.000625 \cdot \phi^2$$

For the constant velocity phase (20° ≤ φ ≤ 70°), the 1.250 units of rise are divided uniformly across 50° of rotation.

For the deceleration phase (70° ≤ φ ≤ 110°), using the inverted parabolic formula:

$$y = 2.000 - 0.0003125 \cdot (110° - \phi)^2$$

Complete Displacement Table: Modified Constant Velocity Cam

Important note: The matching procedure is identical when using cycloidal motion instead of parabolic, because both have the same maximum velocity for equal rise and lift angle.

Cam Profile Determination: From Diagram to Physical Shape

The displacement diagram tells you what the follower does. Now you must translate that into the physical shape of the cam. This is where a powerful construction technique called inversion comes into play.

The Inversion Concept

Constructing a cam profile requires drawing the cam in many rotational positions with the follower in each related location. This is cumbersome.

The inversion trick: Instead of rotating the cam and keeping the follower fixed, you keep the cam fixed and rotate the follower around it. This dramatically simplifies the graphical construction.

Critical rule: To preserve the correct sequence of events, the artificial rotation of the follower must be the reverse of the cam's prescribed rotation. If the cam rotates counterclockwise, the follower positions are laid out clockwise.

Construction for Radial Translating Roller Follower

Step-by-step process:

1. Draw the base circle with radius R_min centered on the cam shaft
2. Draw the outer circle with radius R_max (= R_min + h)
3. Divide the full 360° into increments matching your displacement diagram
4. For each angular position, measure the corresponding displacement y from the displacement diagram
5. Mark each point on the pitch curve by measuring y radially outward from R_min at the corresponding (reversed) angle
6. Connect all points to form the smooth pitch curve (the path of the roller center)
7. Draw a series of circles with radius r_f (roller radius) centered on the pitch curve points
8. The inner envelope tangent to these circles is the actual cam working surface

Construction for Offset Translating Roller Follower

The construction is similar to the radial case, with one key difference: the angular position lines are not drawn radially from the cam shaft center. Instead, they are drawn tangent to a circle whose radius equals the offset distance e.

This offset shifts the follower's line of action away from the cam center, which changes the pressure angles and can significantly improve cam performance.

Construction for Swinging Roller Follower

For swinging followers, the displacement h represents movement along a circular arc (not a straight line). The construction requires:

1. Establish the pivot point M and the follower arm length L_f
2. Draw R_min and R_max circles from the cam shaft center through the lowest and highest positions of the roller center
3. Rotate the pivot point M to successive angular positions (reversed from cam rotation)
4. At each rotated pivot position, swing an arc of radius L_f between the R_min and R_max circles
5. Mark the displacement y along each arc, measured from the R_min circle
6. Connect these points to form the pitch curve

Note: If the angular displacement φ₀ of the swinging arm is known, the linear displacement is:

$$h = \frac{\pi \cdot \phi_0 \cdot L_f}{180°}$$

Pressure Angle and Radius of Curvature: The Two Parameters That Make or Break Your Cam

If displacement curves are the brain of cam design, pressure angle and radius of curvature are the cardiovascular system. Get them wrong, and your cam mechanism will suffer chronic stress, excessive wear, and eventual failure.

Pressure Angle: Definition and Significance

The pressure angle at any point on a cam profile is the angle between:

• The direction the follower wants to go (the tangent to the follower's path of motion)
• The direction the cam pushes it (the line perpendicular to the tangent of the cam profile at the contact point)

         Direction follower
         wants to go
              ↑
              |
              | α ← Pressure Angle
              |╱
              ●─────→ Direction cam pushes
         (Contact Point)

Why Pressure Angle Matters

Increasing the pressure angle increases side thrust. The normal force from the cam splits into two components: a useful component that moves the follower, and a wasteful side component that creates friction in the follower guides.

As the pressure angle grows:

• Side thrust increases → more friction force in guide bushings
• Follower rod bending increases → risk of jamming
• Required cam driving torque increases → larger drive motors needed
• Overall mechanism efficiency drops

Maximum Recommended Pressure Angles

These values are conservative. In many applications they can be exceeded, but beyond these limits, trouble can develop and detailed analysis becomes mandatory.

The Pressure Angle Dilemma

Here is the engineering tension you must resolve:

Reducing the pressure angle requires increasing the cam size. But larger cams bring their own problems:

The design challenge: Find the smallest cam that keeps the pressure angle within acceptable limits. This is a constrained optimization problem, and the graphical and analytical methods below solve it.

Graphical Method for Determining Cam Size (Translating Follower)

This method allows you to find the minimum cam size for specified maximum pressure angles:

Step 1: From the displacement diagram, measure the total length L of the abscissa (0 to 360°) and calculate:

$$k = \frac{L}{2\pi}$$

Step 2: Locate the two points P₁ and P₂ on the displacement diagram having the maximum angles of slope (τ₁ and τ₂).

Step 3: Calculate the critical distances:

$$k \tan \tau_1 \quad \text{and} \quad k \tan \tau_2$$

Step 4: Construct a vertical line of length h (the total stroke). Lay out the y₁ and y₂ positions and the k tan τ values as horizontal offsets.

Step 5: Draw rays at the specified maximum pressure angles (α₁ and α₂) from the offset points. The intersection area of these rays defines all valid cam shaft center locations.

Key outcomes:

• Any point inside the intersection area gives a cam with pressure angles not exceeding the specified values
• The point on the boundary gives the smallest possible cam for the given requirements
• Points directly above the follower line give radial followers (zero offset)
• Points to the side give offset followers, with the offset distance e determined by the horizontal displacement

Analytical Formulas for Pressure Angles

For standard cam profiles with radial translating roller followers, you can calculate minimum cam size directly.

Uniform Velocity Motion

$$\alpha = \arctan\frac{180° h}{\pi \beta R_\alpha}$$

$$R_{min} = \frac{180° h}{\pi \beta \tan \alpha_{max}} \quad \text{(at } \phi = 0°\text{)}$$

Parabolic Motion

$$\alpha_{max} = \arctan\frac{360° h}{\pi \beta R_\alpha} \quad \text{(occurs at } \phi = \beta/2\text{)}$$

$$R_{min} = \frac{360° h}{\pi \beta \tan \alpha_{max}} - \frac{h}{2}$$

Simple Harmonic Motion

$$\alpha = \arctan\frac{90° h}{\beta R_\alpha} \cdot \sin\frac{180° \phi}{\beta}$$

The rise angle φ_p where maximum pressure angle occurs:

$$\phi_p = \frac{\beta}{180°} \cdot \text{arccot}\left(\frac{\beta}{180°} \cdot \tan \alpha_{max}\right)$$

Cycloidal Motion

$$\alpha = \arctan\frac{\frac{180° h}{\pi \beta}\left(1 - \cos\frac{360° \phi}{\beta}\right)}{R_\alpha}$$

The rise angle φ_p where maximum pressure angle occurs:

$$\phi_p = \frac{\beta}{180°} \cdot \text{arccot}\left(\frac{\beta}{180°} \cdot \tan \alpha_{max}\right)$$

Radius of Curvature: Preventing Undercutting and Surface Failure

The minimum radius of curvature of the cam profile determines two critical things:

1. Whether undercutting occurs (making the cam impossible to manufacture or function correctly)
2. Whether surface stresses are acceptable (preventing premature wear and fatigue failure)

The Three Cases of Curvature

Case 1: Normal (ρ_min > r_f) The radius of curvature of the pitch curve is greater than the roller radius. The cam surface has a well-defined convex profile with:

$$R_c = \rho_{min} - r_f$$

No problems. This is the design target.

Case 2: Sharp Corner (ρ_min = r_f) The cam will have a sharp corner (R_c = 0) at that point. Surface stresses become theoretically infinite. This is unacceptable in virtually all applications.

Case 3: Undercutting (ρ_min < r_f) This case is physically impossible to manufacture correctly. The roller follower would deviate from its intended path, and the actual motion would differ from the designed displacement diagram.

Case 1: Normal          Case 2: Sharp Corner     Case 3: Undercutting
   ╱─╲                      ╱╲                     ╱ ╲
  ╱   ╲   ← Smooth         ╱  ╲  ← Point          ╱   ╲ ← Interference
 ╱ ○   ╲  profile          ╱ ○  ╲                  ╱  ○  ╲
           ρ > rf              ρ = rf                ρ < rf

General Radius of Curvature Formula

$$\rho = \frac{\left(r^2 + \left(\frac{dr}{d\phi}\right)^2\right)^{3/2}}{r^2 + 2\left(\frac{dr}{d\phi}\right)^2 - r\frac{d^2r}{d\phi^2}}$$

Where r is the radial distance from the cam center to the pitch curve at angle φ.

Radius of Curvature Formulas in the supplied reference

Parabolic Motion (deceleration portion, β/2 ≤ φ ≤ β):

$$r = R_{min} + h - 2h\left(1 - \frac{\phi}{\beta}\right)^2$$

$$\frac{dr}{d\phi} = \frac{720° h}{\pi\beta}\left(1 - \frac{\phi}{\beta}\right)$$

$$\frac{d^2r}{d\phi^2} = \frac{-4(180°)^2 h}{\pi^2 \beta^2}$$

Simple Harmonic Motion:

$$r = R_{min} + \frac{h}{2}\left(1 - \cos\frac{180° \phi}{\beta}\right)$$

$$\frac{dr}{d\phi} = \frac{180° h}{2\beta}\sin\frac{180° \phi}{\beta}$$

$$\frac{d^2r}{d\phi^2} = \frac{(180°)^2 h}{2\beta^2}\cos\frac{180° \phi}{\beta}$$

Cycloidal Motion:

$$r = R_{min} + h\left(\frac{\phi}{\beta} - \frac{1}{2\pi}\sin\frac{360° \phi}{\beta}\right)$$

$$\frac{dr}{d\phi} = \frac{180° h}{\pi\beta}\left(1 - \cos\frac{360° \phi}{\beta}\right)$$

$$\frac{d^2r}{d\phi^2} = \frac{2(180°)^2 h}{\pi\beta^2}\sin\frac{360° \phi}{\beta}$$

For cycloidal motion, ρ_min occurs near φ = 0.75β and has a dedicated formula:

$$\rho_{min} = \frac{(R_{min} + 0.91h)^2 + \left(\frac{180° h}{\pi\beta}\right)^2]^{3/2}}{(R_{min} + 0.91h)^2 + 2\left(\frac{180° h}{\pi\beta}\right)^2 + (R_{min} + 0.91h)\frac{2(180°)^2 h}{\pi\beta^2}}$$

Minimum Radius of Curvature Comparison

Given: h = 1 unit, R_min = 2.9 units, β = 60°

Takeaway: Cycloidal motion produces the smallest radius of curvature for the same parameters—meaning you may need a slightly larger base circle to keep surface stresses acceptable. This is the trade-off for its superior dynamic characteristics.

Cam Forces, Contact Stresses, and Materials: The Engineering That Prevents Catastrophic Failure

After determining the cam geometry, the next step is calculating the forces acting on the system. This is where the practitioner Engström's analysis began when she redesigned the failed cam on Unit 7.

Acceleration Forces

The force acting on a translating body given an acceleration a is:

$$R = \frac{W \cdot a}{g}$$

Where:

• g = gravitational constant (386 in/sec² or 9,807 mm/sec²)
• W = effective weight of the system
• a = acceleration of W

The effective weight combines all moving masses:

$$W = W_f + \frac{1}{3}W_s + W_e$$

Where:

• W_f = weight of follower
• W_s = weight of return spring (only 1/3 contributes to effective mass)
• W_e = weight of external mechanism

Spring Forces

The return spring must be strong enough to hold the follower against the cam surface at all times. The critical point is where the maximum negative acceleration occurs—this is where the follower most wants to separate from the cam.

Spring force required:

$$F_s = R - W_f - F_e - F_f$$

Where:

• F_e = external force resisting motion
• F_f = friction force from guide bushings

Dynamic safety factors:

Spring constant:

$$K_s = \frac{F_s - \text{preload}}{y_a}$$

Where y_a is the cam rise from R_min to the height at which maximum negative acceleration occurs.

Spring force at any height y:

$$F_y = y \cdot K_s + \text{preload}$$

Pressure Angle and Friction Forces

The pressure angle creates a sideways force component that generates friction in the follower guide bushings. The complete normal force equation accounting for friction is:

$$F_n = \frac{P}{\cos\alpha - \mu \sin\alpha \cdot \frac{2l_1 + l_2}{l_2} - \mu \cdot d}$$

Where:

• P = sum of all forces (acceleration + spring + follower weight + external)
• α = pressure angle
• μ = coefficient of friction in bushings
• l₁, l₂ = guide bushing dimensions
• d = follower stem diameter

$$P = \frac{W \cdot a}{386} + y \cdot K_s + \text{preload} + W_f + F_e$$

Critical observation: If the coefficient of friction is zero, F_n is simply P/cos α. But if the follower is too flexible, causing sideways bending and jamming, the effective friction coefficient can rise to 0.5 or more—potentially doubling the normal force on the cam surface.

Cam Torque

The instantaneous torque required to drive the cam:

$$T_o = (R_{min} + y) \cdot F_n \cdot \sin\alpha$$

The maximum resisting torque determines the cam drive motor requirements.

Complete Force Analysis: Worked Example

Let's walk through the same analysis the practitioner performed, step by step.

Given:

• Cycloidal motion, rise h = 1 unit, lift angle β = 100°
• Cam speed N = 900 RPM
• Follower weight W_f = 2 units force
• Spring and external weights negligible
• Follower stem diameter d = 0.75 units
• Guide dimensions: l₁ = 1.5, l₂ = 4 units
• Coefficient of friction μ = 0.05
• External force F_e = 10 units force
• Maximum pressure angle not to exceed 30°

(a) Minimum Pitch Curve Radius

Using the cycloidal formula, the rise angle where α_max occurs:

$$\phi_p = \frac{100°}{180°} \cdot \text{arccot}\left(\frac{100°}{360°} \cdot \tan 30°\right) \approx 45°$$

The radius at that point:

$$R_{\alpha_{max}} = \frac{\frac{1}{2\pi}\left(1 - \cos\frac{360° \times 45°}{100°}\right)^2}{2 \cdot \sin\frac{360° \times 45°}{100°}} = 1.96 \text{ units}$$

The minimum base circle radius:

$$R_{min} = 1.96 - 1 \times \left(\frac{45°}{100°} - \frac{1}{2\pi}\sin\frac{360° \times 45°}{100°}\right) = 1.56 \text{ units}$$

(b) Spring Constant

Maximum negative acceleration occurs at φ = ¾β = 75°. Using cycloidal acceleration:

$$a = \frac{2\pi \times 1 \times (6 \times 900)^2}{(100°)^2} \sin\frac{360° \times 75°}{100°} = -18{,}300 \text{ length/sec}^2$$

Acceleration force:

$$R = \frac{(2 + 0 + 0) \times (-18{,}300)}{386} = -95 \text{ units force (upward)}$$

Required spring force (with 1.05 safety factor for cycloidal motion):

$$F_s = 1.05 \times 95 - 2 - 10 - 0 = 88 \text{ units force}$$

The rise at φ = 75°:

$$y_a = 1 \times \left(\frac{75°}{100°} - \frac{1}{2\pi}\sin\frac{360° \times 75°}{100°}\right) = 0.909 \text{ units}$$

Spring constant:

$$K_s = \frac{88 - 36}{0.909} = 57 \text{ force/length}$$

(Where 36 is the specified preload.)

(c) Normal Force on Cam at φ = 45°

At φ = 45°, rise y = 0.40 units, acceleration a = 5,660 length/sec². Using the full friction-inclusive formula:

$$F_n = \frac{\frac{2 \times 5660}{386} + 0.4 \times 57 + 36 + 2 + 10}{\cos 30° - 0.05 \sin 30° \cdot \frac{2(1.5) + 4}{4} - 0.05 \times 0.75} = 110 \text{ units force}$$

Sensitivity to friction:

• If μ = 0 → F_n = 104 (only 5% less)
• If μ = 0.5 (jammed follower) → F_n = 200 (nearly double!)

This demonstrates why follower rigidity and guide bushing condition are critical.

(d) Effect of Manufacturing Error

Suppose a 0.001 unit "bump" occurs over 1° of cam rotation (from a chatter mark or poor surface blending). The change in acceleration:

$$\Delta a = \frac{2 \times 0.001 \times (6 \times 900)^2}{(1°)^2} = 58{,}320 \text{ length/sec}^2$$

This is more than 10 times the acceleration on a perfect cam and would generate sufficient force to damage the cam surface.

The lesson: On high-speed cams, manufacturing accuracy is critically important. A tiny surface imperfection can multiply dynamic forces by an order of magnitude.

(e) Cam Torque at φ = 45°

$$T_o = (1.56 + 0.4) \times 110 \times \sin 30° = 108 \text{ torque units}$$

(f) Radius of Curvature at φ = 45°

Using the cycloidal curvature formulas:

• r = 1.96
• dr/dφ = 1.12
• d²r/dφ² = 0.64

$$\rho = \frac{(1.96^2 + 1.12^2)^{3/2}}{1.96^2 + 2(1.12)^2 - 1.96 \times 0.64} = 2.26 \text{ units}$$

Calculation of Contact Stresses

When a roller follower presses against a cam, the contact produces compressive stress at the surface. This stress determines whether the cam will survive millions of cycles or fail from fatigue.

The Contact Stress Formula

For a steel roller against a steel cam:

$$S_c = 2290\sqrt{\frac{F_n}{b}\left(\frac{1}{r_f} \pm \frac{1}{R_c}\right)}$$

For a steel roller against a cast iron cam, replace 2290 with 1850.

Where:

• S_c = maximum calculated compressive stress
• F_n = normal load
• b = width of cam
• r_f = radius of roller follower
• R_c = radius of curvature of cam surface

Sign convention:

• Plus (+): roller on convex portion of cam (most critical)
• Minus (−): roller on concave portion
• Flat surface: R_c = ∞, so 1/R_c = 0

Worked Contact Stress Example

Given: r_f = 0.25 units, R_c = 2.26 − 0.25 = 2.01 units (convex), b = 0.3 units, F_n = 110 units force.

$$S_c = 2290\sqrt{\frac{110}{0.3}\left(\frac{1}{0.25} + \frac{1}{2.01}\right)} = 93{,}000 \text{ stress units}$$

This calculated stress must be compared against the allowable stress for the selected cam material.

Cam Materials: Selecting for Survival

The failure of a cam or roller is almost always due to fatigue. The critical factor is the surface endurance limit—the maximum contact stress the material can withstand for 100,000,000 or more cycles.

Cam Materials Reference Table

(Maximum allowable compressive stress values for use with hardened steel rollers, based on 100,000,000 cycles of pure rolling)

Important notes:

• Where stress repetitions significantly exceed 100,000,000 cycles, use more conservative stress figures
• Where appreciable misalignment exists, reduce allowable stress
• Where there is sliding (not pure rolling), reduce allowable stress
• The phosphate-coated SAE 4150 at only 20,000 is not a typo—the coating dramatically reduces the endurance limit compared to uncoated