Power transmission calculator
Cycloidal Disc Generator
Generate a cycloidal disc profile from pin count, pin circle, pin diameter, and eccentricity, and export the disc as STEP, DXF, or SVG.
Design a laser-cut cycloidal drive
Set the ring pins and eccentricity, size the output pins, then choose the plates and material.
Ring pins and disc
Output pins
Plates, holes, and material
Drive dimensions
- Reduction ratio Output turns opposite the input
- 11 : 1
- Disc lobes
- 11
- Disc lobe diameter
- 74.9 mm
- Disc valley diameter
- 68.9 mm
- Output hole diameter Output pin + 2 × eccentricity + clearance
- 9.1 mm
- Curtate ratio Eccentricity × pins ÷ pin circle radius
- 0.45
- Pin spacing Center to center
- 20.706 mm
Drive preview
The disc profile is the envelope of the ring pins as the disc rolls around them on the eccentric. Run it to watch the output pins turn once for every 11 turns of the input. The dashed circle is the pin circle.
Download the manufacturing files
100% in-browserSTEP solids are built and checked in your browser. DXF and SVG drawings use lines and true arcs at full scale. Nothing is uploaded until you order.
73.8 mm × 74.3 mm
⌀ 116.0 mm · qty 2
⌀ 64.0 mm
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Quick answer
How this cycloidal disc generator calculator works
A cycloidal disc profile is x = R cos t − Rr cos(t + ψ) − E cos(Nt), y = −R sin t + Rr sin(t + ψ) + E sin(Nt), with N ring pins of radius Rr on radius R and eccentricity E; it has N − 1 lobes. Enter those values to generate the disc with its bearing bore and output holes, plus the pin ring and output plates.
How to use the calculator
- 1Enter ring pin count N, pin circle diameter 2R, and pin diameter 2Rr.
- 2Enter eccentricity E, keeping E·N/R below 1.
- 3Set clearance and the output pin pattern; output holes are sized to pin diameter + 2E.
- 4Download the disc (and a second disc rotated 180°) as STEP, DXF, or SVG, or send it to a Fabworks quote.
Key formulas
Cycloidal disc generator formulas
- Disc profile, x
- x = R·cos t − Rr·cos(t + ψ) − E·cos(N·t)
- Disc profile, y
- y = −R·sin t + Rr·sin(t + ψ) + E·sin(N·t)
- Contact angle term
- ψ = atan2(sin((1 − N)·t), R ÷ (E·N) − cos((1 − N)·t))
- Shape condition
- E·N ÷ R < 1
- Output hole diameter
- D_hole = d_output pin + 2E
Step by step
How to draw a cycloidal disc in CAD
- 1Choose N, R, Rr, and E, and confirm E·N/R < 1.
- 2Evaluate ψ, x, and y for t from 0 to 2π in small steps (a few thousand points).
- 3Join the points with a closed curve; it should show N − 1 lobes.
- 4Offset the curve inward by the running clearance.
- 5Add the center bore for the eccentric bearing and output holes of d_pin + 2E.
- 6Copy the disc and rotate the copy 180° about the input shaft for the second disc.
Worked example
11-lobe disc for 12 pins
- Inputs
- N = 12, R = 40 mm, Rr = 4 mm, E = 1.5 mm.
- Result
- 11 lobes, disc profile from 69 mm to 75 mm diameter, ratio 11:1.
The profile radius ranges from R − E − Rr = 34.5 mm to R + E − Rr = 37.5 mm. E·N/R = 0.45, well inside the limit of 1.
Common questions
What to know before using the result
- How do you draw a cycloidal disc?
- Plot the parametric equations for x and y with ψ over t from 0 to 2π, join the points into a closed curve, and offset for clearance. The generator does this and exports clean arcs within 0.005 mm of the curve.
- How many lobes does a cycloidal disc have?
- One fewer than the number of ring pins, N − 1. Twelve pins need an 11-lobe disc and give an 11:1 reduction.
- What limits the eccentricity?
- E·N/R must stay below 1 or the lobes self-intersect. For R = 40 mm and 12 pins, E must be under 3.33 mm; the default 1.5 mm gives 0.45.
- How big is the cycloidal disc?
- Its profile radius runs from R − E − Rr to R + E − Rr. For R = 40, E = 1.5, and Rr = 4 mm, that is 34.5 to 37.5 mm, or 69 to 75 mm in diameter.
- Can I export the disc as DXF?
- Yes. The disc, pin ring plate, and output plate export as STEP, DXF in millimeters, or SVG, and Order sends the STEP files to a Fabworks quote.
Reference
Cycloidal reduction by ring pin count
One-lobe-difference disc on an 80 mm pin circle (R = 40 mm). E·N/R must stay below 1, so at E = 1.5 mm a 30-pin ring needs a smaller eccentricity.
| Ring pins N | Disc lobes | Reduction | Pin spacing (mm) | E·N/R at E = 1.5 mm | Max E (mm) |
|---|---|---|---|---|---|
| 8 | 7 | 7:1 | 30.61 | 0.300 | 5.00 |
| 10 | 9 | 9:1 | 24.72 | 0.375 | 4.00 |
| 12 | 11 | 11:1 | 20.71 | 0.450 | 3.33 |
| 16 | 15 | 15:1 | 15.61 | 0.600 | 2.50 |
| 20 | 19 | 19:1 | 12.51 | 0.750 | 2.00 |
| 24 | 23 | 23:1 | 10.44 | 0.900 | 1.67 |
| 30 | 29 | 29:1 | 8.36 | 1.125 | 1.33 |
Formula
i = (N − 1) : 1 · r = R ± E − R_r
A disc with N − 1 lobes inside N ring pins reduces speed by (N − 1):1 and reverses the output. The disc profile is the pin-center path offset by the pin radius, x = R cos t − R_r cos(t + ψ) − E cos(Nt) and y = −R sin t + R_r sin(t + ψ) + E sin(Nt).
Engineering references
Assumptions and limits
- The disc has one fewer lobe than there are ring pins.
- E·N/R must be below 1 and ring pins must not overlap.
- Output holes are sized to output pin diameter plus 2E.
- Profile clearance is a uniform inward offset of the theoretical profile.
- Two discs 180° apart are recommended for balance; bearings and pins are not included.
Keep designing
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