Two planes meeting in a line

Sarrus’s linkage is six links and six pin joints. Kutzbach’s count gives it zero degrees of freedom. It moves, and what it does is carry a platform up and down along an exactly straight line.

The spatial field established that by solving it: sixty configurations, each a converged root of the loop-closure equations, and the platform’s path measured against a straight line to a departure of $9.8 \times 10^{-16}$ of its span — fourteen orders of magnitude better than Watt’s approximation, which is 9% out over its stroke.

This rung derives the same fact in two lines and no solve.

The argument #

Each arm of Sarrus’s linkage is three revolute joints with parallel axes.

Three parallel revolutes are three one-dimensional groups all lying inside one planar group — the one whose normal is that arm’s axis direction — so the arm’s product lies inside it too. Whatever the platform does, one arm holds it inside planar motion with normal $\mathbf u$ and the other holds it inside planar motion with normal $\mathbf v$.

So the platform’s displacements are in $G(\mathbf u) \cap G(\mathbf v)$, and an intersection of groups is a group.

The intersection of two planar groups with non-parallel normals is the one-dimensional group of translations along $\mathbf u \times \mathbf v$. Computed as an intersection of algebras it comes back one-dimensional, of type $T$, with its direction agreeing with the cross product to the last bit.

The platform translates. It does not turn, it does not go sideways, and its straightness is not an approximation to anything.

What is not in the argument #

The striking thing about that derivation is what it does not mention.

No length appears. Not the arm links, not the platform, not where the pivots are. Multiply every dimension of one arm by seven and the argument is unchanged.

No symmetry appears. The two arms are usually drawn identical and they need not be. One may be short and one long, one may have its pivots spread and the other bunched, the platform may be attached anywhere. The only requirement is that the two axis directions are not parallel.

No configuration appears. The argument is about the whole motion at once rather than about a position, so there is nothing to sweep and no residual to report.

That is a strong claim and the site’s habit is to test a strong claim rather than admire it. The test is the mechanism itself, solved: with the arms at different proportions, the platform’s path is still straight to the solver’s own floor. What the group argument predicts about any Sarrus linkage the solve confirms about this one.

What the two routes each buy #

It would be easy to read the group argument as a replacement for the measurement, and it is not. The site runs two routes wherever two exist, and here they answer different questions.

The group argument says the straight line is exact and says why. It is a statement about the ideal geometry, it has no tolerance in it, and it explains rather than reports: a translation group’s orbits are lines, and the platform is in a translation group.

The measurement says this mechanism delivers it. The linkage as built has to assemble; its Newton solve has to converge; the joint angles it needs have to be reachable; nothing in the group argument says a Sarrus linkage exists at all, and a group argument for a mechanism that jams is worthless.

The two are also sensitive to different mistakes. A wrong group argument would give a wrong motion type and the solve would catch it. A wrong solve — a sign error, a branch guard misfiring, the double-negated right-hand side the foundation found — would give a wrong path and the group argument would catch it. Neither is a check on itself.

The arms need not be arms #

A version of the argument that makes its indifference to detail obvious: replace one of the two three-revolute arms with something else entirely.

Any chain whose product lies inside the planar group $G(\mathbf u)$ will do. Three parallel revolutes is the usual one. Two parallel revolutes and a slide perpendicular to them is another. A single planar pair — two flat faces resting on each other, one lower pair rather than three joints — is a third, and it confines the platform to exactly the same group in one part instead of three.

So a “Sarrus linkage” made of one planar pair and one three-revolute arm has the same straight line, and so does one made of two planar pairs. Nobody builds those, for reasons that are about friction and manufacture rather than kinematics, and the fact that the kinematics is unchanged is what shows the argument to be about the group and not about the linkage.

That freedom is the design rule of the previous rung at work: what is required of a leg is which group it confines the platform to, and how the leg achieves it is a separate decision belonging to a different set of trade-offs.

The flat configuration #

The intersection is one-dimensional at every angle between the arms except one, and the exception is the whole of what can go wrong.

At zero — the two arms’ axis directions parallel, which is the linkage built flat — the two planar groups are the same group, and the intersection is the whole of it: three dimensions instead of one.

The mechanism does not lose its straight line gradually as the arms are brought together. It has a straight line at eighty-nine degrees, at forty-five, at five, and at $10^{-6}$ degrees, and at exactly zero it has a planar mechanism that folds sideways.

That is the same cliff the chain rung found in a serial arm, and it is worth noting that it points the opposite way. There, a coincidence had to hold exactly for the group to exist and any perturbation destroyed it. Here the group exists on an open set of geometries and one exact value destroys it. A design condition that holds on an open set is one a machine shop can meet, which is why Sarrus’s linkage is a practical mechanism and a mechanism relying on exact parallelism is a nuisance.

Reading the degeneracy as a design margin #

The cliff at zero is not just a curiosity, because a built mechanism has its arms at a nominal angle with a tolerance on it, and the question a designer actually asks is how close to zero is too close.

The group answer is any angle but zero, which is true and useless on its own. The useful version is the same one the four-joint rung arrives at from the other side: the dimension is a threshold applied to a continuous quantity, and the continuous quantity is what to specify. Here it is the angle between the normals, and what degrades as it shrinks is not the type of the motion but the conditioning of the mechanism — the platform’s one permitted direction is the cross product of two nearly-parallel vectors, so the constraint that holds it there gets weaker in proportion.

That is a statement the group argument cannot make and the solve can: sweep a near-flat Sarrus and the Newton iteration takes more steps, the Jacobian’s smallest singular value falls, and the mechanism becomes the kind of thing this site calls a mechanism that moves to first order and not at all. Two routes again, each answering the half the other cannot.

Why the count is nought #

The count’s failure on this mechanism is worth restating in the new vocabulary, because the two explanations fit together.

Kutzbach’s arithmetic says $6(n-1) - 5j$: six links, six revolutes, $6 \times 5 - 5 \times 6 = 0$. The rank of the constraint Jacobian says five of six, which is one short of the six independent conditions the count assumed, so one of the constraints is redundant and the mechanism has the one freedom the count lost.

Both of those are statements about dimensions. What the group view adds is the reason the constraint is redundant: each arm imposes the same three conditions. An arm of three parallel revolutes forbids exactly the three displacements outside its planar group, and the two arms’ forbidden sets overlap in the two conditions that any planar constraint imposes about the shared direction. Two arms, three conditions each, five independent — which is the rank, arrived at from the geometry rather than from a matrix.

What the intersection is computed from #

The computation behind the two-line argument is worth showing, because it is short enough that a reader can check it and because it is where the “no lengths” claim becomes concrete.

A planar group with normal $\mathbf u$ has a three-dimensional algebra: a rotation about $\mathbf u$, and two translations perpendicular to it. Written as six-vectors that is three rows. Do the same for $\mathbf v$ and there are three more.

The intersection of the two subspaces is the null space of the two orthogonal complements stacked — six numbers in, a rank decision, and a basis out. For $\mathbf u$ and $\mathbf v$ perpendicular the answer is one-dimensional, its angular part is exactly zero, and its linear part is $\mathbf u \times \mathbf v$ to the last bit. The classifier reads the three integers off it and returns $T$: a translation.

Six-vectors, a null space, and a classification. No mechanism, no configuration, no Newton step, no tolerance beyond the rank decision — and the rank decision has the whole of double precision on either side of it, because a plane’s normal is a plane’s normal.

The comparison with the site’s usual route is stark. The straightness measurement is sixty converged solves, each a Newton iteration against an analytic Jacobian with Levenberg escalation, and it produces one number: $9.8 \times 10^{-16}$. The group computation is one rank decision and it produces a type. Neither could have produced the other’s answer.

Where else this argument works #

The same shape of derivation covers most of the site’s overconstrained mechanisms, and stating them as a family is worth the paragraph.

A planar four-bar built as a spatial loop. Four revolutes with parallel axes: every joint’s group is inside one planar group, so the whole mechanism is, and Kutzbach’s $-2$ is a count applied to a mechanism whose motion is three-dimensional in a six-dimensional space. Closure reports three, type $G$.

A spherical four-bar, and the universal joint. Four revolutes whose axes meet at a point: every group is inside one spherical group. Closure reports three, type $S$, and the universal joint being a spherical four-bar in disguise is exactly the statement that the two have the same group.

Sarrus. Two planar groups intersected. Closure reports one, type $T$.

In every case the count is wrong for the same reason and the group is the reason. An overconstrained mechanism of this kind is one whose joints all lie in a proper subgroup, and the count is an arithmetic that assumes they do not.

That is a complete account of one kind of overconstraint. It is not a complete account of overconstraint, and the mechanism in the fourth panel is why: Bennett’s linkage has the same count, the same rank and the same one redundant constraint, and there is no subgroup for its joints to lie in. What happens when this argument is run on it is the next rung, and it is where the field earns its place.

A translation group is a prismatic pair #

The last thing to note about the answer is what kind of thing it is.

The platform’s group is a one-dimensional translation group — which is exactly the group a prismatic pair gives, the one whose surface is a prism and whose orbit is a straight line. So Sarrus’s linkage is, kinematically, a slide: six links and six pins that between them deliver precisely what one prism sliding in another delivers.

That is not a deflation. It is the reason the mechanism exists. A prismatic pair needs two long accurately-made surfaces in sliding contact, and everything that makes a slideway expensive — straightness over its length, wear, swarf, lubrication, stiction — is a consequence of that contact. Sarrus’s linkage gives the same group out of six rotating joints, each of which is a bearing rather than a way.

Replacing a lower pair by a chain that produces the same group is a move this field can now name, and it is one of the most common in machine design: a wrist replaces a ball joint, a Sarrus linkage or a Roberts mechanism replaces a slide, a four-bar’s parallelogram replaces a way. In every case the group is what has to match, and the classification says when two things can be swapped.

What the angle between the arms is worth #

The group argument answers any angle but zero and the essay calls that true and useless on its own. It is worth putting a number to it, because the number is available from the same two lines and it turns the cliff into a design margin.

The intersection’s direction is $\mathbf{n}_1 \times \mathbf{n}_2$, whose magnitude is $\sin\theta$ for $\theta$ the angle between the two arms’ normals. So the direction is perfectly well defined for any $\theta \ne 0$, and the conditioning of the computation that finds it degrades exactly as $\sin\theta$: the two three-dimensional algebras approach coincidence, the smallest principal angle between them goes to zero linearly, and the null space that picks out the translation becomes ill-determined at the same rate.

That gives a penalty with a formula. Working at $\theta$ rather than at a right angle costs a factor of $1/\sin\theta$ in how precisely the translation direction is determined — 1.15 at 60°, 2.0 at 30°, 5.8 at 10°, and 19 at three degrees. A mechanism built with its arms nearly coplanar has a translation direction that is a group-theoretic certainty and a numerical guess.

The physical consequence follows the same factor. Every source of error in the arms — a bore out of parallel, a pin with clearance, a link a fraction long — perturbs the two normals, and the induced error in the platform’s direction of travel goes as $1/\sin\theta$. So a near-flat Sarrus does not lose its straight line, it loses its aim, and it loses it in proportion to how flat it is.

Which explains why Sarrus’s linkage is always drawn with its two arms perpendicular, and gives the reason as a maximum rather than as a convention: $\sin\theta$ is greatest at ninety degrees, so the perpendicular arrangement is the best-conditioned member of the family, and every departure from it is paid for at a known rate.

It also completes the division of labour between the two routes one last time. The group argument says the straight line exists for every $\theta \ne 0$ — a statement with a discontinuity in it. The conditioning says how much the straight line is worth at each $\theta$ — a continuous quantity that the group argument has no way to produce, and that is the one a designer choosing an angle actually needs.

The straight line, once more #

One last observation, because it is the sort the site collects.

Every exact straight-line mechanism this site has drawn produces its line by a different route. Peaucellier’s cell inverts a circle through a point on it, which is a statement about an inversive transformation and needs eight links. Hart’s contraparallelogram does the same inversion with fewer. Sarrus intersects two planar groups, needs six links, and leaves the plane to do it — the only one of the three that is not planar at all.

The three have nothing in common as constructions. What they share is that each is exact rather than approximate, and each is exact because of a structural fact rather than a tuned dimension — which is the distinction the site keeps returning to, and which the group vocabulary states most cleanly of the three: the platform’s displacements are a translation group, and a translation group’s orbits are lines.

Which is also why a mechanism that looks like Sarrus’s and has one axis a little out of parallel is not a straight-line mechanism at all. The arm no longer confines the platform to a planar group, there is no group for the other arm to intersect with, and the platform’s path is a curve with no name — straight to a few parts in a thousand rather than to a few parts in $10^{16}$. Fourteen orders of magnitude are riding on an alignment, which is the practical form of everything this field has to say.