This arrangement converts continuous cam rotation into straight-line reciprocation of a bar. Two rollers attached to the same rigid yoke contact opposite sides of a three-lobed cam. The calculator uses a harmonic three-lobe profile, producing three bar cycles per shaft revolution.
Continuous Rotary Cam to Alternating Bar Interactive Calculator
Explore a three-lobed cam driving a guided bar through two opposed rollers. Adjust stroke, shaft speed, moving mass and maximum pressure angle. The calculator derives the pitch radius and calculates three bar cycles per shaft revolution.
Equation Used
- Three identical harmonic lobes and constant cam angular speed. One cam revolution gives three full bar cycles.
- Ideal rigid yoke, no clearance, no friction and no external or gravity load along the bar.
- The selected maximum pressure angle determines the pitch radius; it is not a constant angle throughout the motion.
- Guide reaction is the net transverse contact-force component. Individual bearing reactions and overturning moment depend on support geometry and are not calculated.
- Roller radius is 0.3a and the cam is the normal-offset envelope of its centre curve. No material strength, contact stress, manufacturing clearance or spring preload is sized.
- At zero pressure angle no finite cam can provide the entered stroke; the displayed preview is explicitly marked.
The illustration follows the opposed-roller arrangement in Brown’s plate, using the harmonic three-lobe profile defined below. A zero pressure angle with nonzero stroke cannot give a finite cam; the animation then shows a labelled 30° preview.
Three lobes, two rollers and one guided bar
Brown movement 276 shows a three-lobed cam between rollers attached to a translating bar. The rollers and guides follow that arrangement: the rollers belong to one yoke, their spacing stays fixed, and the guide blocks hold the bar on a straight path.
For a reproducible calculation, the roller-centre pitch curve is defined here as r=R+a cos(3φ). Its two opposite radii add to 2R. As it rotates, the right roller centre is at R+a cos(3θ) and the left is at −R+a cos(3θ). Their common translation is therefore x=a cos(3θ).
A roller has finite radius. Drawing the pitch curve as the actual cam would make the rollers overlap it. Instead, the cam boundary is offset inward along the pitch-curve normal by 0.3a. The drawing uses this same geometry to locate each contact.
What to compare
Use this model to compare how a harmonic three-lobe drive responds to stroke, shaft speed, moving mass and a pressure-angle limit. A smaller allowable angle increases the required pitch radius, which can make the mechanism much larger relative to its stroke.
The display fits the entire mechanism into the window. Always read its calculated dimensions: visual size on the screen is not a fixed scale. At a small pressure-angle limit, the lobes and bar movement are correctly small relative to the large cam.
Motion and force calculation
Write a=S/2 and ω=2πN/60. The displacement is x=a cos(3θ), so velocity is −3aω sin(3θ) and acceleration is −9aω² cos(3θ). Peak speed is 3aω; peak inertia force is 9maω². The cycle rate is 3N cycles/min.
For the radial pitch curve r=R+a cos(3φ), the pressure-angle magnitude obeys tan α=|r′/r|. Its maximum is 3a/sqrt(R²−a²), giving R/a=sqrt(1+9/tan² αmax). This requires a positive angle for a finite nonzero stroke.
Guide reaction is calculated at the same pose as inertia force. The model maximizes 9maω²×|cos u|×3|sin u|/(k−|cos u|), where u=3θ and k=R/a. The roller on the accelerating side carries the ideal inertia load. If z=|cos u|, its stationary point satisfies z³−2kz²+k=0 with z between 1/sqrt(2) and 1. Bisection locates that point.
This differs from multiplying peak inertia by the maximum tangent of pressure angle: those maxima do not generally occur at the same pose. The result is a net guide-side reaction, not a complete bearing-load calculation.
Example at the initial settings
A 50 mm stroke gives a=25 mm. At 200 shaft rpm, ω is approximately 20.944 rad/s and the bar completes 600 cycles/min. Peak bar speed is 1,570.8 mm/s.
For a moving mass of 1 kg, peak inertia force is 98.696 N. A maximum pressure angle of 30° requires R/a=sqrt(28), so R is approximately 132.29 mm and the roller-centre spacing is 264.58 mm.
Doubling shaft speed doubles cycle rate and peak speed but quadruples inertia and guide reaction. Doubling moving mass doubles both force results without changing motion. Changing the pressure-angle limit changes cam radius and guide reaction; it does not change the prescribed stroke law.
Limits and interpretation
The historical plate does not provide a manufactured profile or load specification. The selected harmonic pitch curve and roller size define this model. The supports are schematic, and the profile is not a manufacturing drawing of the historical mechanism.
The results exclude process loads, friction, clearance, deformation and component inertia beyond the entered translating mass. They do not establish a permissible running speed, service life, hardness, spring setting or contact-stress limit.
At the zero-angle endpoint, the theoretical radius tends to infinity. Motion quantities remain available for the specified harmonic law, but the guide-reaction result is marked unavailable and the mechanism is a clearly labelled preview rather than a finite design.
Questions about the cam
Why are there three cycles per revolution?
The selected harmonic profile has three identical lobes. Each lobe produces one complete outward-and-return bar cycle.
Are the two rollers connected?
Yes. They are carried by one rigid translating yoke and stay 2R apart.
Why does a smaller pressure angle need a larger cam?
The prescribed stroke must occur over a shallower contact slope. The stated pitch-curve equation relates that slope to R/a.
Why is zero pressure angle unavailable?
A finite cam cannot provide a nonzero stroke at zero pressure angle throughout this motion. The required radius tends to infinity.
Does moving mass change the motion?
No. Shaft speed and stroke prescribe motion. Mass changes the required inertia force and associated guide reaction.
Does the side-load result size the guide bearings?
No. It gives the net transverse reaction from ideal contact. Bearing spacing, applied moments, friction and other loads require further analysis.
Construction reference
Henry T. Brown, Five Hundred and Seven Mechanical Movements, movement 276, printed pages 68–69 (PDF pages 15–16 of the linked part). The plate establishes the three-lobe, opposed-roller arrangement. The harmonic pitch curve and equations define the ideal model used by this calculator.
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