A compound epicyclic train can produce a slow output by subtracting two nearly equal relative rotations. This calculator defines two ring-gear planes and two rigidly joined planet halves. It computes signed output motion and makes both mesh centre distances agree.
Epicyclic Train Slow-Motion Interactive Calculator
Compare fixed and output ring tooth counts in a defined compound train. Inspect both gear planes, signed output direction and the module adjustment that makes their centres agree.
Equation Used
- Carrier input, first ring fixed, second ring output.
- Two rigidly joined25-tooth planet halves.
- Second module adjusted to match carrier radius.
- Ideal gearing; illustrative stub profiles and no losses.
Two axial planes and one compound planet. Module adjustment is mathematical, not a stock-gear selection.
Two gear planes share one planet shaft
The first internal ring is fixed. An orange carrier moves a blue planet around it. A second planet half is rigidly attached to the same shaft in a neighbouring axial plane and drives the output ring.
Both planet halves have 25 teeth in this defined example. When the ring counts differ, their required modules also differ: the second mesh uses the module ratio shown below the diagram. This keeps both planet centres at the same carrier radius.
The two end views use a common scale. Their orange arms occupy the same angle, and both planet halves have the same absolute spin. The brown marker on the output ring reveals its small accumulated movement.
Compare small tooth-count differences
Enter the fixed-ring and output-ring tooth counts. Nearby counts produce a small output advance per carrier revolution. A larger output ring count gives output in the carrier’s direction; a smaller count reverses it. Equal counts hold the output stationary.
This is a defined compound-gear example, not a claim that one unchanged standard-module planet can bridge two rings of different tooth counts. Selecting standard modules, tooth modifications and manufacturing details is a separate design task.
The two original tooth-count inputs and four results remain, but direction is retained and the reduction formula is corrected.
Derive the signed output ratio
For fixed ring F, output ring O and equal 25-tooth planet halves, the first internal mesh gives ωp−ωc=−Fωc/25. The second gives O(ωo−ωc)=25(ωp−ωc). Combining them yields ωo/ωc=1−F/O=(O−F)/O.
The output advances 360(O−F)/O degrees per carrier turn. Reduction magnitude is |O/(O−F)| when the difference is nonzero. At equal counts, output is stationary and no finite reduction is reported.
The common-centre condition is m1(F−25)/2=m2(O−25)/2, so m2/m1=(F−25)/(O−25). The diagram uses that same relation. Its illustrative 25° stub outlines have 0.75-module addendum and simplified roots.
100 fixed teeth and 101 output teeth
The output/carrier ratio is 1/101, giving a 101:1 reduction and 3.564° output advance per carrier turn. The prior calculator used 100 in the denominator and lost the sign; both issues are corrected.
The second/first module ratio is 75/76, approximately 0.98684. Thus the two 25-tooth planet halves have slightly different diameters while sharing a rigid shaft and the same angular speed.
Changing the output ring to 99 teeth gives −1/99 output/carrier motion: a 99:1 magnitude with reversed output. Choosing 100 teeth for both rings leaves the output stationary.
A ratio is not a finished gearbox
The module adjustment gives compatible pitch-circle centre distances. It does not select standard stock gears or certify tooth interference, contact stress, backlash, efficiency or load capacity.
One compound planet is shown. The illustration therefore does not imply a multi-planet assembly phase, balance condition or equal load sharing. The two views expose different axial planes rather than placing all four gears in one plane.
The earlier claims about routine enormous single-stage ratios, universal tolerances and guaranteed named-machine applications have been removed. The current model states its geometry and mathematical limits directly.
Slow-motion epicyclic questions
Why is the default reduction 101:1 instead of 100:1?
The output ring has 101 teeth, and that count is the denominator in the derived output/carrier relation.
Why are the two modules different?
With equal planet tooth counts and different ring counts, different modules are needed to share the same carrier radius in this construction.
What happens at equal ring counts?
The output remains stationary while the carrier and planet shaft continue moving.
Why does the output barely move?
At the default setting, each full carrier turn advances it only about 3.564°. The angle readout and ring marker show that small movement without exaggerating it.
Reference
- KHK Internal Gears — Technical Information — internal-mesh direction, centre-distance relations and examples of planetary and mechanical-paradox gearing. The two-ring, different-module construction here is explicitly defined by its own equations.
Building or designing a mechanism like this?
Explore the precision-engineered motion control hardware used by mechanical engineers, makers, and product designers.
