Abstract
Besides enlarging the pitch circle, a cam’s pressure angle can be held down by offsetting the follower’s line of motion from the cam center, which lowers the pressure angle on the rise side. For the four motion curves of Paper 961 (a 10 mm rise over 60 degrees), this paper adds return angles of 60, 90 and 120 degrees and a follower offset, and searches for the offset that makes smallest the pitch circle holding the largest pressure angle to 30 degrees on both rise and return. No new theorem or law is claimed.
First, with zero offset the values of Paper 961 return: with zero offset and a 60 degree return, the needed pitch circle was 21.4575 mm for simple harmonic and 28.3851 mm for the cycloid, within $10^{-3}$ mm of Paper 961 (Section 2).
Second, if the return has the same angle as the rise, offsetting gains nothing: with a 60 degree return, the best offset was $0.0000$ mm for all four curves, and the needed radius stayed at its zero-offset value (Section 2).
Third, the slower the return, the larger the best offset and the smaller the needed radius: for simple harmonic, a 90 degree return gave a best offset of $2.4335$ mm and needed radius $17.4135$ mm, and a 120 degree return gave $3.6194$ mm and $\mathbf{15.6138}$ mm (Section 3).
Fourth, the proportion of reduction is nearly the same for every curve: the needed radius relative to zero offset was $0.811534$ to $0.815958$ for a 90 degree return and $\mathbf{0.725686}$ to $0.732789$ for 120 degrees (Section 3).
Fifth, the separator is whether rise and return run at the same speed or at different speeds: the offset is subtracted in the numerator of the rise pressure angle and added in that of the return. At the same speed, lowering one raised the other by the same amount; with a slower return, its spare margin could be passed to the rise (Section 4).
Keywords: cam, pressure angle, follower offset, return stroke, pitch circle, motion curve.
Scope (what is not claimed)
No new theorem or law is claimed: that offsetting a follower’s line of motion from the cam center lowers the pressure angle on the rise and raises it on the return, and the pressure angle formula for an offset translating follower, are known in cam design. These are set values: the 10 mm rise, 60 degree rise angle and four motion curves are those of Paper 961, and the return angles of 60, 90 and 120 degrees, the allowed pressure angle of 30 degrees and the offset search range of 0 to 15 mm are set here. They correspond to no machine’s cam. Only a translating roller follower is treated: flat-faced followers, swinging-arm followers, dwell segments and reversing the direction of rotation are not included. Profile strength is not treated: contact stress, materials, undercutting (Paper 967) and bearing friction from the side thrust are not treated. Relation to earlier papers: Paper 961, in its scope note, treated only a radial roller follower, leaving out followers offset from the center, flat-faced followers and swinging-arm followers; this paper sits in the seat of followers offset from the center (no earlier occupant). Paper 961 showed that even among cam curves with the same peak velocity, the pitch circle holding the pressure angle changes with where the angle peaks. Here a return and a follower offset are added to the same curves, and how the offset changes the needed radius is counted. Paper 967 showed that the roller size avoiding undercutting is ordered by deceleration. Here only the pressure angle is examined. What is added: reproducing Paper 961 at zero offset, finding best offsets and needed radii for three return angles, showing that the best offset is zero for an equal return angle, showing that the proportion of reduction is nearly independent of the curve, and placing the separator on “same speed, or different speeds”.
1. Notation and the Question
Setting: a rise of $h=10$ mm takes place over $\beta=60$ degrees, and the return follows the same curve backward over an angle $\beta_r$. With the follower’s line of motion offset by $e$ from the cam center, the starting height for pitch circle radius $R_p$ is $y_0=\sqrt{R_p^2-e^2}$, and the pressure angle is
$$\tan\phi=\frac{\pm(h/\beta)\,v(u)-e}{y_0+h\,s(u)}$$($+$ and $\beta$ on the rise, $-$ and $\beta_r$ on the return). With $u$ divided into $4000$ steps, the largest pressure angles on rise and return were found, and $R_p$ holding the larger of them to 30 degrees was found by bisection. The offset $e$ was refined by golden-section search over 0 to 15 mm to make the needed $R_p$ smallest.
Question: how much smaller does offsetting the follower make the pitch circle that holds the pressure angle on both rise and return?
2. Equal Return Angle
Computation 2.1: with zero offset and a 60 degree return, the needed pitch circle was 21.4575 mm for simple harmonic and 28.3851 mm for the cycloid, within $10^{-3}$ mm of Paper 961. Searching the best offset with a 60 degree return gave $0.0000$ mm for all four curves, with needed radii unchanged at $28.0797$ mm (parabola), $21.4575$ mm (simple harmonic), $28.3851$ mm (cycloid) and $26.3648$ mm (3-4-5 polynomial), a ratio of $1.000000$.
Observation 2.2 if the return has the same angle as the rise, offsetting gains nothing. Returning along the same curve over the same angle makes the return pressure angle a mirror of the rise: an offset that lowers one raises the other by the same amount, so the best offset was zero.
3. Slower Returns
Computation 3.1: best offsets and needed pitch circles for 90 and 120 degree returns.
| curve, return | zero offset (mm) | best offset (mm) | needed radius (mm) | ratio to zero offset |
|---|---|---|---|---|
| parabola, 90 | $28.0797$ | $3.1831$ | $22.7898$ | $0.811611$ |
| simple harmonic, 90 | $21.4575$ | $2.4335$ | $17.4135$ | $0.811534$ |
| cycloid, 90 | $28.3851$ | $3.1395$ | $23.1610$ | $0.815958$ |
| 3-4-5 polynomial, 90 | $26.3648$ | $2.9341$ | $21.4840$ | $0.814875$ |
| parabola, 120 | $28.0797$ | $4.7746$ | $20.3771$ | $\mathbf{0.725686}$ |
| simple harmonic, 120 | $21.4575$ | $3.6194$ | $\mathbf{15.6138}$ | $0.727660$ |
| cycloid, 120 | $28.3851$ | $4.6881$ | $20.8003$ | $0.732789$ |
| 3-4-5 polynomial, 120 | $26.3648$ | $4.3772$ | $19.2865$ | $0.731525$ |
For all four curves, offsetting made the needed radius more than 10 percent smaller with a 120 degree return.
Observation 3.2 the slower the return, the more offsetting gains. Going from a 90 to a 120 degree return, the best offset grew and the radius ratio fell for all four curves: the ratio was $0.811534$ to $0.815958$ at 90 degrees and $0.725686$ to $0.732789$ at 120 degrees, nearly the same for every curve.
4. The Separator
Observation 4.1 the separator is whether rise and return run at the same speed or at different speeds. The offset $e$ is subtracted in the numerator of the rise pressure angle and added in that of the return: at the same speed, spare margin on one side is shortfall on the other, and the best offset was zero. With a slower return, the return pressure angle has spare margin, and offsetting passes it to the rise, turning 21.4575 mm into $15.6138$ mm for simple harmonic.
Observation 4.2 what is not said. Flat-faced or swinging-arm followers, dwell segments, direction of rotation, contact stress, undercutting and which offset to use are not treated. What is shown is that for the four curves of Paper 961, the gain from offsetting appears only with a slower return, in a proportion nearly independent of the curve.
5. Roll Call and Closing
- pressure angle of an offset follower (Section 1): subtracted on the rise, added on the return.
- zero offset reproduces Paper 961 (Computation 2.1): 21.4575 and 28.3851 mm.
- best offset 0 for a 60 degree return (Computation 2.1): offsetting gains nothing.
- $\mathbf{15.6138}$ mm for simple harmonic with a 120 degree return (Computation 3.1): a slow return gains.
- ratios $\mathbf{0.725686}$ to 0.732789 (Computation 3.1): nearly the same for every curve.
- same speed, or different speeds (Observation 4.1): the separator; the scope note of Paper 961.
Closing: Offsetting a cam follower pays only when the return is slower than the rise. For the four curves of Paper 961, the best offset was zero with a return equal to the 60 degree rise, but with a 120 degree return it became $3.6194$ to $4.7746$ mm and the needed radius fell to $0.725686$ to $0.732789$ of zero offset: for simple harmonic, 21.4575 mm became $15.6138$ mm. The separator is whether rise and return run at the same speed or at different speeds. Placed among the earlier papers: to the pressure angle Paper 961 counted with a follower through the center, the offset and return Paper 961 did not include were added. To be honest: neither contact strength nor undercutting was examined. Where the gain from offsetting appears and how large it is, and no further.
References
- Norton, R. L. (2009). Cam Design and Manufacturing Handbook. Industrial Press.
- 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
- Yamagishi, Y. (2026). The Roller Size That Avoids Cam Undercutting Is Ordered by Deceleration, Not by Pressure Angle [Paper 967]. YOI: https://www.yamagishiyuuki.com/yoi/967
Reference
Yamagishi, Y. (2026). Offsetting a Cam Follower Pays Only When the Return Is Slower Than the Rise [Paper 972]. YOI: https://www.yamagishiyuuki.com/yoi/972
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 972. This paper is a preprint and has not been peer reviewed.