Double Link Universal Joint Mechanism: How It Works, Diagram, Parts, Formula and Uses

Two crosses and a phased intermediate link

The animation shows a normalized two-Cardan construction: a blue input yoke, two purple intermediate yokes connected by a shaft, and a gold output yoke. Each pair shares a cross with perpendicular trunnions. The shaft axes remain fixed while the yokes rotate and the crosses articulate.

The entered bend angles are shown in a separate side profile at their actual values. Phasing error rotates the second intermediate yoke around its own shaft relative to the cancelling arrangement. With equal bend angles and zero phase error, output rotation is uniform even though the intermediate shaft accelerates and slows twice per revolution.

The illustration represents ideal kinematics. It does not include a centering device, bearing clearances, lubrication, structural dimensions or collision checking for a particular manufactured joint.

Use the calculator to inspect phasing

Compare the two joint angles, then move the phasing slider to see how their combined output changes. The intermediate and output speed traces share one input-angle axis, so the second joint is evaluated at the intermediate angle it actually receives.

The results describe ideal angular motion. They do not establish an allowable RPM, torque rating, noise level, component life or vehicle-specific suitability.

Compose the two joint motions

Let θ be input angle, β1 and β2 the two bend angles, and φ the intermediate-yoke phasing error. In the animation’s angle convention, ψ=atan2(cosβ1 sinθ, cosθ) is the intermediate angle. Output angle is χ=atan2(sin(ψ+φ), cosβ2 cos(ψ+φ)), with angles unwrapped continuously.

The intermediate speed ratio is cosβ1/(cos²θ+cos²β1 sin²θ). Multiply it by cosβ2/(cos²β2 cos²(ψ+φ)+sin²(ψ+φ)) to obtain output/input speed. Equal β values and φ=0 give exactly one at every input angle.

For each standalone joint, peak deviation above unit speed is secβ−1. Coupled peak deviation is found from the extrema of the composed angular transformation. Reduction compares that coupled deviation with joint 1 alone; a negative percentage means phasing has made the ripple larger. If joint 1 is aligned, that comparison has a zero denominator and is reported as unavailable.

Example: equal 20-degree bends

At 20° in both joints and zero phasing error, each individual joint has a peak speed deviation of about 6.42%, but the coupled output has zero ideal variation. The intermediate shaft still varies in speed.

Set either bend angle to zero while leaving the other at 20° and the remaining joint’s variation appears at the output. Change phasing while both bends remain equal and the speed cancellation is lost.

Angle equality alone is not enough

The intermediate yokes must have the appropriate relative orientation as well as equal operating angles for cancellation. The model assumes fixed, coplanar shaft axes and ideal rigid crosses. A practical double joint may need a centering mechanism to retain the intended geometry as its supports move.

The second joint receives the angle produced by the first joint. Composing those two motions gives the coupled output speed while keeping the two trunnion axes of each cross perpendicular.

Double-link joint questions

Does zero output ripple mean every part rotates uniformly?

No. The intermediate shaft can still vary in speed while the output is uniform.

Why can the reduction be negative?

A poorly phased second joint can increase output fluctuation relative to the first joint alone.

What happens at zero bend?

An aligned joint transmits angular velocity one to one. With both bends zero, the output is uniform at any fixed phasing offset.

Are the depicted parts a manufacturing drawing?

No. They are normalized yokes, shafts and crosses for examining motion, with no load or collision rating.