Abstract
When the angle between the direction in which a cam pushes its follower and the direction in which the follower moves (the pressure angle) is large, more of the push goes sideways on the follower, and in design the cam is made larger to hold the angle down. For the four motion curves of Paper 929, with a radially moving roller follower rising $10$ mm over $60$ degrees of cam angle, this paper finds the pitch circle radius needed to hold the largest pressure angle to $30$ degrees (a set value). No new theorem or law is claimed.
First, the needed pitch circle differs by $6.9276$ mm between curves: the needed radius was $21.4575$ mm for simple harmonic, $26.3648$ mm for the 3-4-5 polynomial, $28.0797$ mm for the parabola and $\mathbf{28.3851}$ mm for the cycloid (Section 2).
Second, the order is the order of peak velocity: the peak velocity factor was $1.570796$ for simple harmonic, $1.875000$ for the 3-4-5 polynomial and $2.000000$ for both parabola and cycloid, and the needed radius also rose in the order simple harmonic, 3-4-5 polynomial, parabola, cycloid (Section 2).
Third, two curves with the same peak velocity need different radii: parabola and cycloid both have a peak velocity factor of $2.000000$, but their needed radii differed by $\mathbf{0.3054}$ mm. The parabola’s angle peaks at half the rise ($u=0.5$), matching the closed form $28.0797$ mm. The cycloid’s angle peaked at $u=0.4694$, with lift $0.4391$ and velocity factor $1.981634$ (Section 3).
Fourth, the angle formula was checked by another construction: the largest pressure angle at a radius of $30$ mm was $28.6202$, $23.4132$, $28.8218$ and $27.3039$ degrees from the angle formula, and $28.6201$, $23.4132$, $28.8218$ and $27.3039$ degrees from the slope of the pitch curve drawn in Cartesian coordinates, agreeing within $10^{-3}$ degrees (Section 2).
Fifth, the separator is whether the follower’s velocity is read alone or divided by the cam’s radius at that moment: read alone, parabola and cycloid share the same $2.000000$. Divided by the radius at that moment, the cycloid’s angle peaks earlier, while the radius is still small, and the needed radius became $0.3054$ mm larger (Section 4).
Keywords: cam, pressure angle, pitch circle, follower, motion curve, cycloid.
Scope (what is not claimed)
No new theorem or law is claimed: the definition of a cam’s pressure angle, enlarging the pitch circle (the base circle plus the roller radius) to hold the pressure angle down, rules of thumb for the allowed pressure angle, and the peak velocity of each motion curve are all known in cam design. These are set values: the $10$ mm rise, $60$ degrees of cam angle and $30$ degrees of allowed pressure angle are set here. They correspond to no particular machine. Only a radial roller follower is treated: the follower’s line of motion passes through the cam center; offset followers, flat-faced followers and swinging-arm followers are not included. Profile strength is not treated: contact stress, materials, undercutting where the roller radius exceeds the profile curvature, and the return segment are not treated. Relation to earlier papers: Paper 929, in its scope note, did not design cams, leaving out cam profile, pressure angle, contact stress and materials; this paper sits in the seat of pressure angle (no earlier occupant). Paper 929 set out the peak velocities of the four motion curves as $2$, $1.570796$, $2$ and $1.875$, and showed that the cycloid, despite its large peak acceleration, leaves less residual vibration because its acceleration does not jump at the ends. Here, for the same curves, the pressure angle that sets the size of the cam is counted. Paper 557 showed that a gear’s pressure angle works in opposite directions for strengthening the tooth and for smoothing the drive: the pressure angle here is not that of a gear tooth but the angle at which a cam pushes its follower. What is added: finding the needed pitch circle radius for four curves, setting out where the angle peaks and the lift there, checking the parabola against a closed form, checking the angle formula against Cartesian slopes, and placing the separator on “velocity alone, or divided by the radius”.
1. Notation and the Question
Setting: a rise of $h=10$ mm takes place over a cam angle of $\beta=60$ degrees ($\pi/3$ radians). The lift at dimensionless angle $u\in[0,1]$ is $s(u)$, as in Paper 929, and the velocity factor is $v(u)=ds/du$. The follower is a radially moving roller; the pitch curve traced by the roller center has radius $r(u)=R_p+hs(u)$, where $R_p$ is the pitch circle radius. The pressure angle $\phi$ is
$$\tan\phi(u)=\frac{1}{r}\frac{dr}{d\theta}=\frac{(h/\beta)\,v(u)}{R_p+h\,s(u)}$$Its maximum was taken over $u$ divided into $20000$ steps, and $R_p$ giving a maximum of $30$ degrees was found by bisection. As another construction, the pitch curve was drawn in Cartesian coordinates and the largest pressure angle was also found from the angle between the curve’s tangent and the circumferential direction.
Question: how much does the pitch circle that holds the largest pressure angle to $30$ degrees differ between curves, and what sets it?
2. Needed Pitch Circles for Four Curves
Computation 2.1: needed pitch circle radius and where the angle peaks.
| curve | peak velocity factor | needed radius (mm) | $u$ at peak angle | lift there | velocity factor there |
|---|---|---|---|---|---|
| simple harmonic | $1.570796$ | $21.4575$ | $0.4395$ | $0.4055$ | $1.542509$ |
| 3-4-5 polynomial | $1.875000$ | $26.3648$ | $0.4625$ | $0.4299$ | $1.853910$ |
| parabola | $2.000000$ | $28.0797$ | $0.5000$ | $0.5000$ | $2.000000$ |
| cycloid | $2.000000$ | $\mathbf{28.3851}$ | $0.4694$ | $0.4391$ | $1.981634$ |
Computation 2.2: largest pressure angle (degrees) at a pitch circle radius of $30$ mm.
| curve | angle formula | Cartesian slope |
|---|---|---|
| parabola | $28.6202$ | $28.6201$ |
| simple harmonic | $23.4132$ | $23.4132$ |
| cycloid | $28.8218$ | $28.8218$ |
| 3-4-5 polynomial | $27.3039$ | $27.3039$ |
All four agreed within $10^{-3}$ degrees. The peak velocity factors agreed with Paper 929’s $2$, $1.570796$, $2$ and $1.875$ within $10^{-6}$.
Observation 2.3 the order of needed pitch circles is the order of peak velocity. Both by peak velocity factor and by needed radius, the order from smallest was simple harmonic, 3-4-5 polynomial, parabola, cycloid: between the smallest, simple harmonic at $21.4575$ mm, and the largest, cycloid at $28.3851$ mm, a difference of $6.9276$ mm, came the 3-4-5 polynomial and the parabola.
3. Two Curves with the Same Peak Velocity
Computation 3.1: for $u\le0.5$ the parabola has $v=4u$ and $s=2u^2$; its angle peaks at $u=0.5$, and the needed radius is
$$R_p=\frac{2(h/\beta)}{\tan30^\circ}-\frac h2=28.0797\ \text{mm}$$agreeing with the searched $28.0797$ mm within $10^{-3}$ mm. The cycloid’s needed radius was $0.3054$ mm larger than the parabola’s, and its angle peaked at $u=0.4694$, before $0.5$.
Observation 3.2 with the same peak velocity, a different position of the peak angle means a different needed radius. The parabola’s angle peaks at $u=0.5$, where its velocity also peaks, with lift $0.5000$: the cycloid’s angle peaked at $u=0.4694$, where the lift was $0.4391$, not yet half, and the velocity factor was $1.981634$. For simple harmonic and the 3-4-5 polynomial too, the angle peaked at $u=0.4395$ and $0.4625$, before the velocity peak at $u=0.5$.
4. The Separator
Observation 4.1 the separator is whether the follower’s velocity is read alone or divided by the cam’s radius at that moment. Read alone, parabola and cycloid share the same peak of $2.000000$ and cannot be told apart: the pressure angle is the velocity factor divided by $R_p+hs$, so at the same velocity the angle is larger where the lift is smaller. The cycloid’s angle peaked at a lift of $0.4391$, and its needed radius of $28.3851$ mm exceeded the parabola’s $28.0797$ mm.
Observation 4.2 what is not said. Contact stress, materials, undercutting, the return segment, offset or flat-faced followers, and what the allowed pressure angle should be are not treated. What is shown is how far, for a set rise and cam angle, the pitch circle holding the largest pressure angle to $30$ degrees differs between curves, and that even two curves with the same peak velocity differ by where the angle peaks.
5. Roll Call and Closing
- the pressure angle formula (Section 1): velocity factor divided by the radius.
- needed radius from $21.4575$ to $\mathbf{28.3851}$ mm (Computation 2.1): same order as peak velocity.
- two constructions within $10^{-3}$ degrees (Computation 2.2): check of the angle formula.
- parabola and cycloid differ by $\mathbf{0.3054}$ mm (Computation 3.1): same peak velocity $2$.
- cycloid’s angle peaks at $u=0.4694$, lift $0.4391$ (Observation 3.2): before half.
- velocity alone, or divided by the radius (Observation 4.1): the separator; the scope note of Paper 929.
Closing: Cam curves with the same peak velocity need different pitch circles, set by where the pressure angle peaks. For a $10$ mm rise over $60$ degrees of cam angle, the pitch circle radius holding the largest pressure angle to $30$ degrees was $21.4575$ mm for simple harmonic, $26.3648$ mm for the 3-4-5 polynomial, $28.0797$ mm for the parabola and $28.3851$ mm for the cycloid: parabola and cycloid both have a peak velocity factor of $2.000000$, but the cycloid’s angle peaked at a lift of $0.4391$ and its needed radius was $0.3054$ mm larger. The separator is whether the follower’s velocity is read alone or divided by the cam’s radius at that moment. Placed among the earlier papers: the four curves that Paper 929 compared by vibration were compared by pressure angle and cam size, which Paper 929 did not treat. To be honest: the allowed pressure angle and the rise are set values, and contact strength was not examined. For four curves, how much the cam size holding the angle differs, and no further.
References
- Norton, R. L. (2009). Cam Design and Manufacturing Handbook. Industrial Press.
- Yamagishi, Y. (2026). The Angle That Strengthens a Tooth Points Against the Angle That Smooths the Drive [Paper 557]. YOI: https://www.yamagishiyuuki.com/yoi/557; DOI: 10.5281/zenodo.23014224
- Yamagishi, Y. (2026). A Cycloidal Cam Has the Largest Peak Acceleration, Yet Leaves Less Vibration at Speed Because Its Acceleration Does Not Jump at the Ends [Paper 929]. YOI: https://www.yamagishiyuuki.com/yoi/929
Reference
Yamagishi, Y. (2026). Cam Curves with the Same Peak Velocity Need Different Pitch Circles, Set by Where the Pressure Angle Peaks [Paper 961]. YOI: https://www.yamagishiyuuki.com/yoi/961
ORCID: 0009-0006-8587-3450
PDF (English) · PDF (Japanese) · Date proof of the English PDF (OpenTimestamps .ots) · SHA256SUMS
The record page for this paper (in Japanese, with the PDFs in each language and the date proof): YOI 961. This paper is a preprint and has not been peer reviewed.