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Two cam motion laws can produce exactly the same lift through exactly the same rise angle at exactly the same camshaft speed—and still behave very differently dynamically.
Consider a follower rise bounded by dwells. Compare simple harmonic motion (SHM) with cycloidal motion using the same lift h, rise angle β, and constant camshaft speed ω.
Which is smoother?
The short answer: cycloidal motion gives better
continuity at the dwell boundaries, but it does not produce lower
kinematic peaks. For the same lift, rise angle, and cam speed,
cycloidal motion has about 27.3% higher peak velocity and 27.3%
higher peak acceleration than SHM.
Normalize the Cam Rise First
Let the normalized cam coordinate during the rise be
u = θ / β, 0 ≤ u ≤ 1
The comparison below assumes an ideal translating follower, constant camshaft angular velocity, and a dwell immediately before and after the rise.
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Simple Harmonic Motion
For simple harmonic motion, normalized displacement is
s / h = [1 − cos(Ï€u)] / 2
Differentiating with respect to time gives
v / (hω/β) = (π/2) sin(πu)
a / (hω²/β²) = (Ï€²/2) cos(Ï€u)
and, within the rise,
j / (hω³/β³) = −(Ï€³/2) sin(Ï€u)
The peak normalized velocity is therefore
vmax / (hω/β) = π/2 = 1.5708
and the peak acceleration magnitude is
|a|max / (hω²/β²) = Ï€²/2 = 4.9348
At first glance, SHM appears quite smooth: displacement is continuous and velocity starts and ends at zero.
The problem appears one derivative higher.
What Happens When SHM Leaves a Dwell?
During the dwell, follower acceleration is zero. But immediately after the SHM rise begins,
a(0+) = (Ï€²/2)hω²/β²
So acceleration jumps directly from zero to 4.9348 in normalized units. At the end of the rise, an equivalent acceleration discontinuity occurs when the motion rejoins the dwell.
This is the important SHM boundary condition:
displacement is continuous;
velocity is continuous;
acceleration is discontinuous.
displacement is continuous;
velocity is continuous;
acceleration is discontinuous.
The finite jerk function inside the SHM rise has an amplitude
Ï€³/2 = 15.503
but that number does not describe the complete jerk behavior of a dwell-bounded SHM motion.
In the ideal mathematical motion law, differentiating the acceleration step produces an impulsive contribution to jerk at each dwell boundary. The physical machine will not reproduce an ideal mathematical impulse: structural compliance, damping, clearances, contact stiffness, and other real dynamics alter the response. But the acceleration discontinuity remains an important indicator of dynamic excitation.
Cycloidal Motion
For cycloidal rise,
s / h = u − sin(2Ï€u)/(2Ï€)
with normalized velocity, acceleration, and jerk
v / (hω/β) = 1 − cos(2Ï€u)
a / (hω²/β²) = 2Ï€ sin(2Ï€u)
j / (hω³/β³) = 4Ï€² cos(2Ï€u)
At both ends of the rise,
v = 0 and a = 0
That means velocity and acceleration both join the zero-motion dwell continuously.
Cycloidal improves the dwell transition:
displacement is continuous;
velocity is continuous;
acceleration is continuous.
displacement is continuous;
velocity is continuous;
acceleration is continuous.
But cycloidal motion is not infinitely smooth.
Its normalized jerk immediately inside the rise is
4Ï€² = 39.478
while jerk in the dwell is zero. Jerk therefore changes by a finite step at the boundary.
Cycloidal motion has not eliminated the boundary discontinuity. Compared with dwell-bounded SHM, it has moved that discontinuity one derivative higher—from acceleration to jerk.
See the Difference in the SVAJ Curves
Figure 1: Normalized displacement, velocity, acceleration, and jerk for simple harmonic and cycloidal rises. The shaded regions are dwells. Both laws use the same lift, rise angle, and camshaft speed.
The displacement curves look similar. The dynamic difference becomes much more obvious as the derivatives are examined.
SHM leaves the dwell with a nonzero acceleration. Cycloidal leaves with zero acceleration, but its jerk changes immediately from the dwell value of zero to a finite value.
Better Boundary Continuity Has a Price
Cycloidal motion achieves its improved boundary behavior by reshaping the motion inside the available cam angle.
Its peak normalized velocity is
2.000
compared with
Ï€/2 = 1.5708
for SHM.
The ratio is
2 / (Ï€/2) = 4/Ï€ = 1.2732
so cycloidal peak velocity is approximately 27.3% higher.
The same ratio appears in peak acceleration:
(2Ï€) / (Ï€²/2) = 4/Ï€ = 1.2732
Therefore cycloidal peak acceleration is also approximately 27.3% higher.
| Characteristic | Simple Harmonic | Cycloidal |
|---|---|---|
| Peak normalized velocity | 1.5708 | 2.000 |
| Peak normalized acceleration | 4.9348 | 6.2832 |
| Velocity at dwell boundary | 0 | 0 |
| Acceleration at dwell boundary | Discontinuous | 0 and continuous |
| Jerk at dwell boundary | Impulsive contribution | Finite step |
| Peak velocity vs SHM | Reference | +27.3% |
| Peak acceleration vs SHM | Reference | +27.3% |
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Why the Jerk Numbers Can Be Misleading
A simple comparison of the finite jerk amplitudes would give
SHM interior jerk amplitude = 15.503
Cycloidal peak finite jerk magnitude = 39.478
Looking only at those numbers could suggest that SHM has the lower jerk. That conclusion would be incomplete.
The SHM value of 15.503 describes only the ordinary jerk function inside the rise. It does not include the impulsive contributions at the dwell boundaries created by its acceleration discontinuities.
Cycloidal jerk, by contrast, remains finite. It has a finite jump between the dwell and rise rather than an acceleration step producing an impulse.
Do not compare 15.5 with 39.5 as though they were two ordinary
peak-jerk values. They describe fundamentally different
boundary behavior.
What Does “Smoother” Actually Mean?
The word smooth can hide two different engineering questions.
1. How Smoothly Does the Motion Join the Dwell?
In this sense, cycloidal motion is smoother.
SHM reaches the dwell with continuous displacement and velocity, but acceleration jumps. Cycloidal also makes acceleration continuous, leaving jerk as the first discontinuous derivative.
2. How Large Are the Kinematic Peaks?
In this sense, SHM has an advantage for the conditions considered here. Its peak velocity and peak acceleration are both lower.
So the statement “cycloidal is smoother” is incomplete unless smoother is defined.
Better boundary continuity does not mean lower peaks.
Acceleration Continuity and Inertia Force
For a fixed translating or equivalent moving mass,
Finertia = m a
The acceleration discontinuity of SHM therefore produces an idealized step in follower inertia force as the rise leaves or enters a dwell.
Cycloidal acceleration instead starts at zero and increases continuously, so its inertia-force component also joins the dwell continuously.
Later in the stroke, however, cycloidal acceleration reaches a larger magnitude. With the same moving mass, its peak inertia-force magnitude is therefore approximately 27.3% higher than for SHM.
That statement applies to the inertia-force component. It does not mean that cam normal force, bearing load, drive torque, Hertz contact stress, or total machine load automatically increases by 27.3%. Those depend on the complete mechanism, including pressure angle, springs, gravity, external process forces, effective inertia, contact geometry, and structural dynamics.
So Which Motion Law Should You Choose?
There is no universal winner.
If a mechanism is sensitive to abrupt changes in inertia force or to excitation at a dwell transition, acceleration continuity can be very important. Cycloidal motion improves that boundary behavior.
If available drive torque, follower inertia, or peak speed is the dominant constraint, the lower velocity and acceleration peaks of SHM may be important.
As machine speed increases, higher derivatives and structural dynamics become progressively more significant, and neither displacement nor peak acceleration alone is sufficient for judging the motion.
Pressure angle and contact force also remain separate design checks. A motion law with desirable SVAJ characteristics does not automatically produce acceptable cam contact geometry.
The design lesson:
Choosing a cam motion law is not about finding the curve with the smallest-looking SVAJ plot. It is about deciding which dynamic behavior the mechanism can tolerate.
Choosing a cam motion law is not about finding the curve with the smallest-looking SVAJ plot. It is about deciding which dynamic behavior the mechanism can tolerate.
When additional endpoint derivatives need to be controlled, more advanced polynomial or modified motion laws can provide other compromises. The important point is that every motion law redistributes kinematic demand; none makes the dynamics disappear.
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About the Author: This article is written by a senior engineering leader with over 25 years of experience in high-mix low-volume (HMLV) industrial automation, process optimization, and custom machine design.
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