Eccentric Spur-gear with Link Control Mechanism Explained: How It Works, Parts, Diagram & Uses

Why the gears remain meshed

A is the fixed input shaft. Circular gear B is rigidly attached to its driver, with its pitch-circle center offset from A. When A turns, B rotates at the input speed while its center follows a small circle around A.

The first pivoting arm joins the centers of B and idler C. The second joins C to output gear D, whose shaft is fixed to the frame. Each arm has a length equal to the sum of the pitch radii of its two meshing gears. These constraints locate C at a circle intersection and maintain the correct tooth-center distances.

The output shaft does not move. Its speed changes because the arms change angle as B moves eccentrically. The two mesh equations include those arm rotations; a fixed-center gear-ratio calculation by itself would miss that variation.

Use the calculator to explore the arrangement

Gear module scales every length. The chosen gears have 40, 24 and 40 teeth, giving pitch radii 20m, 12m and 20m. Both arms are 32m long, and the fixed shafts are 53m apart.

The offset slider sets e/R, where R is the pitch radius of B. It ranges from zero to 0.35. These bounds retain the selected continuous upper assembly branch and allow the input to make a complete revolution without the two arms reaching a straight-line toggle.

At zero offset, the gear centers stop moving. The train behaves as ordinary fixed-center gearing: C reverses direction and D runs at the input speed. Increasing offset creates speed variation. Increasing input rpm changes the time scale and all angular speeds without changing the geometry.

Link closure and the two gear equations

Use coordinates in the plane with A=(0,0), D=(53m,0), and B=(e cosθ,e sinθ). The equal arm length is L=32m. Let d=|D−B| and h=√(L²−d²/4). The upper intersection is C=(B+D)/2+h·perp(D−B)/d.

Let α be the angle of B→C and β the angle of C→D. For external gears on a rotating center line, the no-slip pitch relation is z₁ω₁+z₂ω₂=(z₁+z₂)ωarm.

Consequently, 40ωB+24ωC=64α̇ and 24ωC+40ωD=64β̇. Eliminating C gives ωD=ωB+1.6(β̇−α̇). The calculator differentiates the rigid-link constraints to obtain the arm velocities.

For the selected equal arms, ωD/ωB=1+1.6 d′/[L√(1−d²/(4L²))], with d′=(53m)e sinθ/d.

The displayed minimum and maximum speeds are numerical estimates from 1,440 equally spaced positions in a full input revolution. Both arms return to their starting angles after a turn, so the equal first and last tooth counts give exactly one output revolution per input revolution.

Example dimensions

With module 2 mm, B and D each have a 40 mm pitch radius, C has a 24 mm radius, both arms are 64 mm long and the fixed shafts are 106 mm apart. An offset fraction of 0.20 places B’s pitch center 8 mm from input A.

At the reset position, B is closest to D. The rate of change of their separation is zero, so D has the same instantaneous speed as the input. During the next half-turn it speeds up, then slows below the input speed during the return half.

Choose zero offset to check the ordinary-gear limit. At a 6 rpm input, the idler runs at −10 rpm and D runs at +6 rpm throughout the cycle.

What this model represents

The geometry is an explicit reconstruction of the historical arrangement, not a claim that these tooth counts were used in the original. All gears are circular and the two links keep their centers at constant mesh distances. The eccentric motion belongs to B’s center relative to input shaft A.

There is no independent dwell control. The output motion follows the selected geometry and remains forward throughout the allowed range. This train is not an indexer with a locked rest period.

The tooth outlines illustrate standard involute gearing with 20° pressure angle. The calculation covers ideal geometry and speed. It does not establish load capacity, backlash, shaft strength, bearing life, balance or manufacturing tolerances.

Questions about the linked eccentric gear

Which shaft is eccentric?

The first circular gear’s center B is offset from its fixed driving shaft A. The last gear’s shaft D stays fixed.

Why are there two arms?

One maintains the B–C gear-center distance and the other maintains C–D. Their common pivot carries the idler.

Does a circular gear have to run out of mesh when mounted eccentrically?

It would if all the other centers were fixed at unsuitable distances. Here the linked idler center moves to preserve both meshes.

Can I set a dwell angle independently?

No. Speed variation is a consequence of the geometry. The arbitrary dwell slider in the earlier page has been replaced by the actual offset input.

Why does module not change the speed ratios?

It uniformly scales the gears, offset, links and shaft spacing. Their dimensionless proportions remain the same.