A bar between two midpoints

The previous rung left a machine with sixteen assemblies, four of which draw a curve that is not the one asked for. The problem is discrete: a parallelogram closes two ways, and the two closures are separate configurations rather than two ends of a continuum.

A discrete problem admits a discrete repair. Find a quantity that is constant in one closure and not in the other, and make it a bar.

The quantity #

Take a parallelogram $P$, $X$, $Y$, $R$, with $Y - R = X - P$. Let $M_1$ be the midpoint of $XY$ and $M_2$ the midpoint of $PR$. Then

$$M_1 - M_2 = \left(X + \tfrac{R - P}{2}\right) - \tfrac{P + R}{2} = X - P,$$

so the two midpoints are exactly one side apart, and they stay that way as the parallelogram moves, because $|X - P|$ is a bar.

In the crossed assembly the identity fails, because $Y$ is no longer $X + R - P$.

So: attach a point rigidly at the midpoint of each of two opposite sides, and put a bar between them. Three parts, and the crossed closure is no longer a configuration the mechanism has.

What it costs #

Two rigid attachments and one bar. Each attachment is one new joint — two unknowns — and two equations. The bar is one equation and no unknowns.

Four unknowns, five equations. One equation more than the parts it brings, which means the brace adds a redundant equation: a condition the machine’s other constraints already imply, at every configuration on the intended branch.

That is the definition of overconstraint, and this site has a field’s worth of it. A ring of hinges closes at every size because its constraints are not independent. Sarrus’s linkage moves although Kutzbach declares it immobile. A door hinge with three knuckles is overconstrained by design and works because it is made to a condition.

What is different here is the direction. Everywhere else on this site, overconstraint is something a mechanism turns out to have, and the interesting question is why the count is wrong about it. Here it is put in deliberately, one equation at a time, for a stated purpose: not to change what the machine can do, but to change what it can be.

The count and the rank, on a large machine #

Take the quintic’s machine. Unbraced: four hundred and eighty unknowns, four hundred and seventy-nine equations, one degree of freedom by the count and one by the rank of the constraint Jacobian. The count is exactly right, which is unusual on this site and is a consequence of every gadget being determinate.

Braced: a hundred and fifty-four braces, one per parallelogram. One thousand and ninety-six unknowns, one thousand two hundred and forty-nine equations, and a Grübler count of minus a hundred and fifty-three.

By the count that machine is a structure a hundred and fifty-three times over. It is not. The Jacobian’s rank is one thousand and ninety-five, so the mobility measured is one, exactly what it was before, and the hundred and fifty-four surplus equations are precisely the hundred and fifty-four braces.

The site’s oldest instrument comparison, Grübler against a rank, at a scale nobody would attempt by hand and with an answer nobody would be tempted to doubt: the machine visibly turns.

What the brace removes #

On the hyperbola’s machine: sixteen seeds unbraced, eight of which close, four on the curve and four off it. Sixteen seeds braced, four of which close, all four on the curve, and all four at the same tracing point.

Both halves of that are the claim, and the second half is the one an over-eager repair would fail. A brace that removed the spurious branches by removing every branch would satisfy the first requirement perfectly and leave a structure. The gate for this field asserts both: the unbraced machine must have assemblies off its own curve, and the braced one must have none and still assemble.

Why the rank does not move, spelled out #

The two numbers in that table — a count of minus a hundred and fifty-three and a measured mobility of one — are far enough apart to deserve an argument rather than an assertion.

Take one braced parallelogram in isolation. Its four bars, plus the two attachments and the brace bar, are seven equations. Its four moving joints plus the two midpoints are twelve unknowns. So the piece has five freedoms: two for where the whole thing is, one for its orientation, one for its shape as a parallelogram — and one left over, which is the input direction it is carrying.

Now count the same piece without the brace: five equations, eight unknowns, three freedoms, plus the two the midpoints would have added, which is five again. The brace changed the count and not the freedom, because it constrained a quantity that was already determined.

That is the whole of it, and the reason it scales: every brace is locally redundant, and locally redundant equations remain redundant when the pieces are assembled, because assembly only adds conditions. So the rank deficiency of the braced machine is exactly the number of braces, which is what the measurement finds — a hundred and fifty-four braces and a hundred and fifty-four redundant equations, at every size tried.

The site’s gate asserts that ratio rather than the totals, for a good reason: the totals depend on the curve, and the ratio is the claim. One brace, one redundant equation, no change in mobility.

Why a midpoint rather than a diagonal #

The obvious alternative brace is a bar across a diagonal, and it does not work. A parallelogram’s diagonals change length as it moves — that is what makes it a mechanism rather than a rigid frame — so a bar across one would freeze it.

The midpoints are the right choice because the quantity they define is invariant under the parallelogram’s own motion and different in the crossed assembly. Finding such a quantity is the whole art of bracing, and there is not always one: a four-bar with general lengths has two assemblies and no invariant separates them, which is why bracing is available for parallelograms and not for four-bars in general.

The parallelogram is special because its two assemblies are related by a reflection that the midpoint construction is not symmetric under. That is a small piece of luck and the whole repair rests on it.

The solver has to be told, in effect #

A redundant system is not the ordinary case for a Newton solve, and it is worth saying what happens when one is handed to this site’s solver, because the answer is a piece of machinery the fleet already carries.

With more equations than unknowns the normal equations are singular. A plain Newton step is then undefined and a lightly damped least-squares step comes out as numerical noise of enormous magnitude. The solver in lib/mechanism.js handles it by escalating its damping — trying a nearly undamped step, and if that does not reduce the residual, damping harder and trying again — which is Levenberg–Marquardt, and it was added to this site for the parallelogram with a third parallel bar, a mechanism with seven equations, six unknowns and rank five.

So the machinery for solving braced machines was already here, put in three phases ago for a three-bar example, and a machine with a hundred and fifty-four redundant equations uses it unchanged. That is worth recording because it is the ordinary way a shared piece earns its keep: the case it was written for was small and the case that needed it later was not.

What bracing does not fix #

Two things, and they are the honest limits.

It does not fix the reflectors or the means. Only the parallelograms are braced. A reflector’s rhombus has its own second placement — the far vertex at the pivot instead of out along the mirror — and no bar between midpoints separates those, because the degenerate placement is not a reflection of the working one but a collapse of it. The census therefore reports a lower bound on the assemblies, braced or not.

It does not fix a singularity. A braced machine still passes through configurations where two placements of some joint merge, and passing through one is a continuous event that no discrete constraint prevents. That is the next rung, and it is the failure that has no repair.

The difference between the two is worth stating plainly because they are easy to conflate. A branch is where the machine was put; a singularity is where the machine goes. Bracing decides the first and cannot touch the second.

What it would take to brace everything #

The census counts sixteen assemblies and says the real number is larger, because only the parallelograms were flipped. It is fair to ask what a complete repair would look like.

The reflector’s second placement is the degenerate one: its rhombus collapsed with the far vertex at the pivot. That configuration satisfies both of the vertex’s bars — the two circles that place it meet at the pivot and at the working point — and it is not a reflection of the working configuration but a collapse of it, so no midpoint identity separates the two. A brace against it would have to be an inequality — the vertex must be at least some distance from the pivot — and a bar cannot say that. Bars state equalities; the machinery for a constraint that only pushes belongs to the holding field and is not part of a linkage.

The mean’s degenerate placement is the same shape: its far vertex at the pivot, which happens when its two inputs are opposite.

So a complete repair is not available with bars alone, and what is available is the parallelogram brace, which handles the assemblies that are reflections. The reflector’s degenerate placements are not reachable from a working configuration without passing through a singularity, which is the honest reason they are not enumerated: they are a continuous failure wearing a discrete failure’s clothes, and they belong to the next rung.

The cost, put beside the machine #

A brace is three parts on a machine whose smallest interesting version has twenty. On the quintic there are a hundred and fifty-four of them, so the braced machine has one thousand and ninety-six unknowns against four hundred and eighty — more than twice the size.

That is not a small overhead and it should be quoted with the other counts rather than tucked into a footnote. The cost table reports unbraced machines, because the unbraced count is the count of the construction; the braced count is the count of a machine that would actually be right when built, and it is a little over double.

Doubling a four-hundred-bar machine to remove assemblies nobody wants is the sort of overhead that would be intolerable if any of this were being built, and it is the correct decision anyway, because the alternative is a machine that draws a plausible wrong curve.

What a builder would have to hold to #

The brace is exact, and the machine it repairs is made of parts. The two facts have to meet somewhere, and where they meet is worth stating because it is the practical residue of this rung.

The brace bar’s length is $|X - P|$, one of the parallelogram’s own sides. If it is made a little long, the parallelogram is no longer exactly a parallelogram at any configuration, and the direction being carried is off by an amount that grows with the error. If it is made a little short, the same. There is no configuration at which a mis-made brace is harmless, because the brace is holding an identity rather than a limit.

What that buys, against the alternative, is worth being clear about. An unbraced machine made perfectly is right until it is assembled wrongly or driven through a flat parallelogram, and then it is wrong by a number of order one. A braced machine made to a tolerance is wrong by a number of order the tolerance, always, and cannot be assembled wrongly at all.

Trading a rare catastrophic error for a constant small one is the ordinary bargain of the practice field, and it is the right way round here: a machine that draws the wrong curve is useless, and a machine that draws the right curve to a thousandth is a machine.

Where overconstraint changes character #

There is a general point here that belongs to the constraint field as much as to this one.

Overconstraint on this site has always been a fact about a mechanism: a discovery, usually surprising, that a count is wrong about something that works. The spatial field separates two kinds of it and names them. The practice field asks what it costs to make something whose motion depends on an exact condition.

Bracing is overconstraint as a tool. The redundant equations are added because they are redundant — a constraint that removed a freedom would be no use — and what they buy is the removal of a configuration rather than of a motion.

That is a use of the concept the site has not had before, and it reframes the earlier fields’ examples slightly. A door hinge with three knuckles is overconstrained, and the usual account is that the redundancy costs manufacturing precision and buys stiffness. It also buys something this field’s language names better: it removes assemblies. A hinge with one knuckle can be put together with the leaves crossed; a hinge with three cannot.

A second use for the same bar #

There is a side effect of bracing that has nothing to do with assemblies and is worth mentioning because a builder would find it first.

An unbraced parallelogram passing near its flat configuration is badly conditioned: its two bars become nearly parallel, the Jacobian’s smallest singular value approaches zero, and the position of its output joint becomes very sensitive to small errors in its bar lengths. A braced one is not, because the brace constrains the output from a third direction.

This field does not exploit that — the tolerance rung measures unbraced machines, so its numbers are the pessimistic ones — and it is worth flagging as an unexplored consequence rather than letting the omission pass silently. Whether bracing improves a compiled machine’s sensitivity as well as its branch structure is a question nothing here asks, and the two effects are independent enough that the answer is not obvious either way.

What the machine looks like now #

Braced, the compiled machine has one assembly at its starting configuration, one degree of freedom, and a tracing point that satisfies its polynomial to $10^{-13}$ over its working arc.

Everything in the construction is now settled except how far it turns, and that is the last thing this half of the field has to report. The answer is not encouraging — the quintic’s machine works over a tenth of a radian — and the reason has nothing to do with anything that can be braced.

The brace deserves one more description, because what it is doing is unusual for a bar. An ordinary bar removes a freedom: it takes a configuration space and cuts it down. This one removes nothing — the rank does not move, as the essay measures — and what it does instead is refuse an assembly. The correct parallelogram satisfies the midpoint condition exactly and the crossed one does not, so a bar of that length can be fitted to the first and cannot be fitted to the second. It is a discriminator rather than a constraint. That is a genuinely different use for a member, and it is the mechanical form of something this site has met as an abstraction: a mechanism’s assemblies are the components of its configuration space, and a bar that admits one component and refuses another is a physical way of choosing a component at build time. The change-point linkage’s whole difficulty is that it can flip between components while running; a brace that cannot be fitted in the wrong one settles the question in the assembly shop, permanently, with a part rather than with a control system.