One cam rise, two laws: cycloidal and 3-4-5 polynomial calculations
Two curves can deliver the same lift and look almost identical, yet demand different velocities, accelerations and transitions. This notebook builds a comparison you can repeat on paper and use when interpreting CrucibleCam. These are calculated teaching examples, not equipment measurements.
1. Specify the motion before choosing a curve
A follower rises 30 mm over 90° of cam rotation at a constant 600 rpm. The rise starts and ends next to dwell segments. The remainder of the revolution must contain dwell and return: this is not a complete cycle or a manufacturing profile.
We use lift \(h=30\,\mathrm{mm}\), rise angle \(\beta=\pi/2\,\mathrm{rad}\) and angular speed \(\omega=2\pi n/60=20\pi\,\mathrm{rad/s}\). Rise time is \(T=\beta/\omega=0.025\,\mathrm{s}\). The variable \(u=\theta/\beta\), from zero to one, measures progress through the rise. Angle θ is measured from the start of this segment.
This analysis prescribes follower displacement. It does not include mass, forces, friction, compliance, base radius or roller radius. It compares kinematic laws, but cannot by itself determine which cam can withstand an application.
2. Write the law and differentiate with consistent units
With \(s=h f(u)\), the laws are:
For the cycloidal law, derivatives with respect to u are \(f'=1-\cos(2\pi u)\), \(f''=2\pi\sin(2\pi u)\), and \(f^{(3)}=4\pi^2\cos(2\pi u)\). For the polynomial, they are \(f'=30u^2-60u^3+30u^4\), \(f''=60u-180u^2+120u^3\), and \(f^{(3)}=60-360u+360u^2\).
Follower velocity is not simply the derivative with respect to angle. For constant cam speed, the chain rule gives:
With h in mm and β in radians, the units are mm/s, mm/s² and mm/s³. Entering 90 instead of π/2 in this formula changes the result by a large factor. Variable cam speed introduces additional angular acceleration terms; this calculator does not include them.
3. Compare curves and numerical results
Both curves start at zero displacement, end at full lift and reach half lift at u=0.5. At that point normalized velocity is 2 for the cycloidal law and 1.875 for the polynomial. The small visual difference corresponds to 150 mm/s in this example.
| Quantity | Cycloidal | 3-4-5 polynomial |
|---|---|---|
| Rise time | 0.025 s | 0.025 s |
| Peak velocity | 2,400 mm/s | 2,250 mm/s |
| Peak absolute acceleration | 301,592.895 mm/s² | 277,128.129 mm/s² |
| Jerk at rise endpoints | 75,798,561.800 mm/s³ | 115,200,000 mm/s³ |
The maxima are analytical: \(\max f'\) is 2 and 15/8; \(\max|f''|\) is 2π and \(10\sqrt{3}/3\), respectively. They are not estimates from graph sampling. The calculator uses these same expressions when inputs change.
4. Lower acceleration does not settle every design choice
The polynomial reduces peak velocity by 6.25% and peak acceleration by approximately 8.11% relative to the cycloidal law in this problem. However, its endpoint jerk is larger. During dwell, constant displacement means zero velocity, acceleration and jerk.
Both laws join a dwell without jumps in displacement, velocity or acceleration, but have a jerk jump. Neither eliminates every discontinuity in the SVAJ diagram. Selection depends on the required boundary conditions and system dynamics.
In CrucibleCam, configure an equivalent rise, check lift and angle, and identify whether the scale represents derivatives with respect to angle or time. Inspect joins with dwell and return. Compare signs, units and continuity as well as peak values.
Geometric design must then check pressure angle, curvature, undercutting and roller dimensions where applicable. Even smooth kinematic diagrams can produce an infeasible profile. This notebook does not perform those checks or claim to validate a simulator release.
5. Predict the change before calculating
Exercise 1: keep lift and angle fixed, but increase speed from 600 to 1,200 rpm. What happens to time, velocity, acceleration and jerk?
Show answer and reasoning
Time halves to 0.0125 s. Velocity doubles, acceleration quadruples, and jerk increases eightfold, because they depend on ω, ω² and ω³. Cycloidal peaks become 4,800 mm/s and 1,206,371.579 mm/s². Speed does not change the normalized displacement curve.
Exercise 2: keep 30 mm and 600 rpm, but increase the rise angle from 90° to 180°. Is this equivalent to halving speed for this segment?
Show answer and limits
For this rise's kinematics, yes: ω/β halves. Time doubles, v halves, a becomes one quarter and j one eighth. However, less angle remains for the rest of the cycle. This is not the same change to the complete cycle and does not ensure that dwell and return fit the new specification.
Basis of this example and further reading
The expressions were differentiated and the values calculated for this notebook. For background, read Displacement curves and SVAJ diagrams. For geometry, read Analytical cam design; for joins, continue with Correcting SVAJ discontinuities.
Found an error? Send the inputs and this page address to arturavelar@ufsj.edu.br.