Spatial mechanisms · Parallel offset

Oldham Coupling

Can two offset shafts turn at exactly the same speed while the middle disc slides? Yes. The middle disc has a groove on each face, the two at right angles, and each shaft's tongue fits one of them. The disc turns through exactly the same angle as both shafts, so the output speed equals the input speed at every instant. Sliding takes up the offset: the disc center runs twice per turn around a small circle that passes through both shaft centers.

Join two parallel shafts whose centers do not line up. Look at the tongue-and-groove faces that cross at right angles, follow the floating middle disc around its orbit, and compare the steady output speed with the changing sliding speed.

Output angle $0$

Output / input speed $0$

Disc-center speed $0$

Disc-center acceleration $0$

Required input torque $0$

Signed load power $0$

Slide closure residual $0$

Quasistatic power residual $0$

Turn & compare

Follow what changes between the shafts.

Input phase

\theta

35.00\,{}^\circ

Prescribed input speed

\omega

60.00\,{}^\circ/\mathrm s

Parallel axis offset

e

0.22\,\mathrm m

Signed external output torque

\tau_o

-1.00\,\mathrm{N\,m}

Blue is the input, coral the output and gold the connecting part. The world axes point right, up and toward you. The input shaft points right, and positive input rotation follows the right-hand rule. To set the phase, drag the gold marker in the separate end view or use the phase slider with the keyboard.

The output axis is parallel to the input and shifted upward. Both shafts and the middle disc turn through the same angle, and the tongues and grooves on its two faces are perpendicular. The disc centre sits exactly where the two sliding directions cross. It goes round a small circle twice for every input turn.

Separating the mating faces gives an exploded view for inspection. The hubs move apart only so you can see the faces, which are then out of engagement. The model, plots and readouts still describe the assembled ideal coupling. Neither the offset nor the clearances on screen indicate a safe working limit.

The input speed is prescribed. In this rigid, massless, frictionless model, the required input torque balances a constant external torque on the output, which can have either sign. The torque comes from a virtual-power calculation and is not a motor-sizing result. The load can supply or absorb power, and the sliding contacts lose no energy.

Playback turns the input once and stops; replaying resets the phase. Lengths are in metres and the sliders use degrees. The CSV holds 721 kinematic samples, one every half degree, with angles in radians and rates in SI units. Part thicknesses, bearings and supports are schematic, and there is no solver for collisions, forces, fatigue or wear.

The shafts are ideally parallel, and the two perpendicular slides have no friction. The offset and part sizes shown are schematic, not manufacturer ratings. Angular misalignment, wear, contact force, inertia and backlash are not modeled.

Start with the essentials

Focus question: How can offset shafts turn evenly while the disc center moves?
One-sentence intuition: Two slides at right angles take up the sideways offset, so the output turns through exactly the same angle as the input.

Core mathematical model

Two sliding constraints at right angles

\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 lie in the cross-section plane spanned by the up and depth directions, with the input shaft center at the origin.

The disc center circles twice per turn

\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 disc center moves on a circle whose radius is half the shaft offset. Neither shaft leaves its own axis.

Constant speed ratio, changing sliding speed

\phi=\theta,\quad\frac{\dot\phi}{\omega}=1,\quad\dot s_1=-e\omega\sin\theta,\quad\dot s_2=e\omega\cos\theta

Each sliding distance is measured, with its sign, from its own shaft center along its own groove.

Power balance for massless parts

\tau_{\rm in}=-\tau_o\frac{\mathrm d\phi}{\mathrm d\theta},\qquad\tau_{\rm in}\omega+\tau_o\dot\phi=0

Friction losses and inertial power are left out.

Common difficulties

Same number of turns is not same speed

Typical misconceptionIf the output makes one turn for every input turn, its speed must be constant.

Better mental modelA universal joint's speed ratio changes from moment to moment. An ideal Oldham coupling between parallel shafts keeps the ratio constant.

Kinematics versus dynamics

Typical misconceptionIf the animation runs smoothly, the bearing loads and wear must be acceptable.

Better mental modelHere the input motion is given, and the connecting parts have no mass and no friction. Inertia, real clearances and contact forces need a different model.

Run the experiment

01
Start with aligned shafts
Select aligned shafts and play one turn.
What to observe: The two phase markers stay together, and the speed ratio stays at one.
02
Follow the floating center
Increase the offset and compare the two slide-speed curves.
What to observe: The disc center goes around twice for each input turn. The two sliding speeds are a quarter cycle out of step, and the shaft speeds stay equal.
03
Check how power passes through
Reverse the sign of the output torque and export one cycle.
What to observe: The torque and power change sign, but the motion stays the same because it is prescribed.

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