Spatial mechanisms · Uneven rotation

Universal Joint

Why does a universal joint set at an angle make the output speed up and slow down twice each turn? Because the cross in the middle keeps its two pins at right angles while the two forks turn in planes tilted by the bend, so equal turns of the input cannot pass through as equal turns of the output. The output speed rises and falls every half turn; at a 30 degree bend it swings between about 0.87 and 1.15 times the input speed. Both shafts still finish each full turn together.

Two shafts meet at an angle, joined by a gold cross whose arms are at right angles. Increase the bend, compare the input and output phase markers, and measure how the output speed rises and falls within each full turn.

Output angle $0$

Output / input speed $0$

Output phase error $0$

Output acceleration $0$

Required input torque $0$

Signed load power $0$

Cross orthogonality residual $0$

Quasistatic power residual $0$

Turn & compare

Follow what changes between the shafts.

Input phase

\theta

40.00\,{}^\circ

Prescribed input speed

\omega

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

Shaft intersection angle

\beta

35.00\,{}^\circ

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 tilts upward in the plane of the world horizontal and vertical axes. Each fork turns about its own shaft, and the two arms of the cross stay perpendicular. In the end view, each phase marker is drawn in the plane perpendicular to its own shaft; the two are not projected onto one common plane.

A single joint set at an angle does not transmit constant speed. The output makes one turn for each input turn, but it speeds up and slows down twice along the way. Lining the shafts up removes the ripple. The phase relation in the plot follows from the right-angled trunnions and from where zero phase is defined.

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.

This is a single rigid Hooke–Cardan joint with ideal intersecting axes, driven at a set input speed. The motion is exact, and the load passes through massless parts in static balance at each instant. Inertia, friction, bearing forces, flexibility and interference ratings are not modeled.

Start with the essentials

Focus question: Why does a bent universal joint change output speed twice each turn?
One-sentence intuition: The cross keeps its two arms perpendicular, which ties the two forks together, so equal steps of input angle do not come out as equal steps of output angle.

Core mathematical model

Constraint from the right-angle cross

\tan\phi=\cos\beta\tan\theta,\qquad\phi=\operatorname{unwrap}\!\left[\operatorname{atan2}(\cos\beta\sin\theta,\cos\theta)\right]

The tangent relation alone cannot tell which quadrant the output is in. The two-argument arctangent, with the angle unwrapped, follows the output through a full turn.

Instantaneous speed ratio

R=\frac{\dot\phi}{\omega}=\frac{\cos\beta}{1-\sin^2\beta\sin^2\theta},\quad R_{\min}=\cos\beta,\quad R_{\max}=\frac1{\cos\beta}

The slowest and fastest points are a quarter input turn apart, and the pattern repeats every half turn.

Output acceleration

\ddot\phi=\omega^2\frac{\cos\beta\sin^2\beta\sin2\theta}{(1-\sin^2\beta\sin^2\theta)^2}

The input turns at constant speed, yet the output can still have angular acceleration.

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
Increase the bend
Compare the slowest and fastest output presets, then inspect the cross from different views.
What to observe: The output speed rises and falls twice per turn, and the cross arms stay perpendicular to within numerical precision.
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