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Sandbox Physics

Spatial mechanisms · Parallel offset

Oldham Coupling

Connect parallel shafts that do not share a center. Inspect the perpendicular tongue-and-groove faces, follow the floating disc around its orbit, and compare constant output speed with changing sliding speed.

Interactive modelOldham Coupling
Output angle00
Output / input speed00
Disc-center speed00
Disc-center acceleration00
Required input torque00
Signed load power00
Slide closure residual00
Quasistatic power residual00

Physics tutorial

Equal rotation, unequal sliding

BackgroundCouplings carry rotation between shafts whose axes do not coincide.

Why it mattersWatch the connecting geometry before reading the speed curve.

Start with the essentials

Focus question
How can offset shafts turn evenly while the disc center moves?
One-sentence intuition
Two perpendicular slides absorb the lateral offset without changing the angular phase.

Core mathematical model

Two perpendicular sliding constraints

u=(cos⁡θ,sin⁡θ),v=(−sin⁡θ,cos⁡θ),C=ecos⁡θ u=(e,0)+esin⁡θ v\mathbf u=(\cos\theta,\sin\theta),\quad\mathbf v=(-\sin\theta,\cos\theta),\quad\mathbf C=e\cos\theta\,\mathbf u=(e,0)+e\sin\theta\,\mathbf v

Coordinates are in the transverse upward–depth plane, with the input center at the origin.

Twice-around center orbit

C=(e2(1+cos⁡2θ),e2sin⁡2θ),∣C˙∣=eω,∣C¨∣=2eω2\mathbf C=\left(\frac e2(1+\cos2\theta),\frac e2\sin2\theta\right),\quad |\dot{\mathbf C}|=e\omega,\quad|\ddot{\mathbf C}|=2e\omega^2

The center orbit has radius half the axis offset; it is not either shaft orbiting.

Constant transmission with changing sliding

ϕ=θ,ϕ˙ω=1,s˙1=−eωsin⁡θ,s˙2=eωcos⁡θ\phi=\theta,\quad\frac{\dot\phi}{\omega}=1,\quad\dot s_1=-e\omega\sin\theta,\quad\dot s_2=e\omega\cos\theta

Travel is signed from the corresponding shaft center along its own groove direction.

Massless quasistatic virtual power

τin=−τodϕdθ,τinω+τoϕ˙=0\tau_{\rm in}=-\tau_o\frac{\mathrm d\phi}{\mathrm d\theta},\qquad\tau_{\rm in}\omega+\tau_o\dot\phi=0

No frictional loss or inertial power is included.

Common difficulties

Equal turns versus equal speed

Typical misconceptionOne output turn per input turn guarantees constant speed.

Better mental modelA universal joint has a varying instantaneous ratio; an ideal parallel Oldham coupling has a constant ratio.

Kinematics versus dynamics

Typical misconceptionA smooth animation establishes acceptable bearing loads and wear.

Better mental modelInput motion is prescribed, and the connecting parts are massless and frictionless. Dynamic inertia, real clearances and contact forces require a different model.

Run the experiment

  1. 01

    Establish the aligned limit

    Select aligned shafts and play one turn.

    What to observe: Both shaft phase markers agree and the speed ratio stays unity.
  2. 02

    Follow the floating center

    Increase the offset and compare the two slide-speed curves.

    What to observe: The center goes around twice per input turn. Sliding speeds are in quadrature while shaft speeds stay equal.
  3. 03

    Inspect the transfer of power

    Reverse the signed load and export the cycle.

    What to observe: Torque and power change sign; the prescribed motion does not.

Theory & apparatus: Carnegie Mellon · Universal Joint · Ruland · Oldham Coupling

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