A piano hinge is not forty door hinges

Why a hinge works turned a verdict into a number of microns. Bennett’s linkage is called unbuildable because its condition on four twists and four lengths is exact and no bored hole has ever met an exact condition; the misfit a detune creates was measured at 0.507 times the detune, shared among four joints is 0.127 each, and an ordinary running fit is larger than that — which is a clearance behaving as a short link rather than as an error. A door hinge was the example the argument used — three knuckles, three axes imposed where one would do — and it worked for the same reason.

Three was never varied. A piano hinge has forty knuckles, thirty-nine of which are redundant by any count, and the reading that follows from the arithmetic alone is alarming: if three axes cost a certain misfit, forty should cost far more, and a long hinge should be a far harder thing to make than a short one.

It is not, and the reason is not about hinges. A rigid leaf carries its pins on one straight line. A line in space has four parameters — two of position and two of direction — and a hinge assembles exactly when some line passes within the clearance of every bore. That is a Chebyshev fit of a line to $n$ points, and everything the count does to a hinge is what adding a point does to a line fit.

The model has one idea in it #

The hinge is $n$ bores at axial positions $z_i$, nominally coaxial, each bored where the machine put it — somewhere within the bore tolerance $\delta$ of where the drawing said. The errors are taken uniform over the disc of radius $\delta$, which is what a position tolerance on a drawing means.

The leaf’s pins are on one axis $L(z) = \mathbf{a} + \mathbf{b}z$. It goes together when every bore admits its pin, which with a radial clearance $c$ is $|L(z_i) - p_i| \le c$ for every $i$. So the play the hinge needs is

$$\text{misfit}(n) \;=\; \min_{L}\; \max_i\; |L(z_i) - p_i| ,$$

the smallest largest miss over all lines. Nothing about hinges is in that statement, and in particular no mechanism: this is the same kind of object as the band a tolerance puts on an output and is computed the same way, by asking what the geometry admits rather than what the nominal does. What is in it is that the leaf is rigid and the bores are not obliged to be anywhere in particular.

The objective is a maximum of norms of affine functions of four parameters, so it is convex and has no local minimum to be trapped in — and it is not differentiable exactly where the answer is, because that is where two or more bores tie for worst. A gradient method walks into the kink and stops moving. A shrinking coordinate search does not care: it proposes a step, keeps it when the largest miss falls, and halves the step when a whole round is refused.

The nominal axis is kept in the race throughout, and that is not tidiness. It is a bound: a line through the middle of the error disc misses every bore by at most the largest error, so the misfit can never exceed $\delta$. Without it in the race, a search that stalls at eighty bores reports a mean misfit of 1.01 $\delta$ — a number larger than the errors it is fitting, which is impossible and looks like a result.

Two knuckles are free #

The first row of the answer is the one that explains the rest.

A hinge of two knuckles needs no play at all. A line goes through two points exactly, so whatever the two bores did, the leaf’s axis can be put through both of them, and there is nothing to absorb. Counted the usual way a two-knuckle hinge is overconstrained — two revolutes removing ten of a body’s six freedoms — and it is not, because the redundancy is a coincidence the geometry always supplies rather than one it has to be given.

That is a sharper version of the point the hinge argument already made. The count says overconstrained; the geometry says whether anything actually has to give. A roller is not a slider and a count that cannot tell them apart cannot tell these apart either.

Three knuckles need 0.409 of the bore tolerance on average, which is where the hinge’s own number comes from: the middle bore’s departure from the chord through the other two, halved, because the best line splits the difference. Four need 0.591, five 0.683, six 0.741.

The requirement saturates #

Past six the increments collapse, and the reason is the bound.

Twelve knuckles need 0.878 of the tolerance. Twenty need 0.929. Forty need 0.966. Eighty need 0.984. Each doubling adds less than the one before, and the sequence is climbing towards one and cannot pass it — because the nominal axis is always available and always within $\delta$ of every bore.

So a piano hinge needs about the bore tolerance. A door hinge needs about four tenths of it. In absolute terms both are small: bores positioned to a hundredth of a millimetre — which is an ordinary expectation of a machine that bores a row of holes in one setup — put the piano hinge’s requirement at ten microns of radial clearance and the door hinge’s at four, against running fits that are measured in tens. A hinge of forty knuckles is two and a half times as demanding as a hinge of three, not thirteen times, and there is no length at which it becomes more than twice as demanding again.

That is the answer the count could not give, and it is worth saying why the count was misleading rather than merely wrong. The redundancy grows linearly with the knuckles — thirty-nine redundant constraints at forty knuckles against two at three — and nothing about a redundant constraint says it must be violated. What matters is how far the geometry departs from the coincidence the constraint assumes, and a line fitted to many points departs from the extreme ones by no more than the extremes themselves depart from the middle. Redundancy counts constraints; misfit measures geometry; the two do not scale together.

Why the count is the wrong instrument here #

It is worth being precise about the disagreement between the two readings, because the count is not being dismissed — it answers a different question correctly.

A mobility count asks how many constraints a mechanism imposes against how many freedoms it has, and for a stack of $n$ coaxial revolutes it returns $6 - 5n$: negative from two knuckles onward and more negative at every one. That number is right. It says that the equations the mechanism writes down have more of them than unknowns, which is true, and it is the number Kutzbach’s criterion is for.

What it does not say is whether the extra equations are consistent. A redundant constraint costs nothing when the geometry it assumes happens to hold, and the whole subject of this field is that geometry never exactly holds and that the interesting quantity is by how much. Two bores are always consistent, because a line goes through two points; forty bores are inconsistent by an amount that depends on how far from a line they fell, which is bounded by how far from the nominal axis each of them is allowed to be.

So the count grows linearly and the misfit saturates, and neither is wrong. This is the same split fragility has a direction found in a Sarrus linkage — where an error of the same size cost three orders of magnitude more in one direction than in another, and no count could distinguish them. There the direction was named by a reciprocal screw system. Here it is named by something simpler: the error that costs nothing is the one a straight line can follow.

One number, and it is a ratio #

The measurement is worth the four decades it was checked over, because it establishes that there is only one number to carry.

Divide the play a six-knuckle hinge needs by the bore tolerance it was made to, and the answer is 0.7356 at $\delta = 10^{-4}$, at $10^{-3}$, at $10^{-2}$, at $10^{-1}$ and at $1$ — the same to nine figures. The whole problem is a line fitted to points, and scaling the points scales the fit; there is no length in the answer, only a proportion.

That is what makes the knuckle count the only axis worth charting, and what makes the result usable. A hinge’s requirement is a pure number times whatever its bores were bored to. A designer who tightens the bores by a factor of ten tightens the play needed by exactly ten, at every count.

Three bores out of forty #

The last measurement is the one that says why a long hinge is not a harder thing to make, as opposed to merely not a more demanding one.

A minimax fit is held by a small set of points and the rest are slack — the same structure an allocation over feature positions has, where most of a budget is spent on features that turn out not to be binding. It takes two to fix a line’s direction and one or two more to balance it, and that is what the measurement finds: averaged over twenty-four stacks, the number of bores at the worst distance is 2.08 at four knuckles, 2.08 at twelve, 2.25 at forty — and never more than three, at any count.

So a hinge of forty bores has its play set by three of them. The other thirty-seven could have been bored anywhere inside the answer without changing it. That is why a long hinge is not harder to machine: thirty-seven of its features have a tolerance that is not binding, and a machine that is repeatable enough to bore three of them well will bore them all well.

It also says what to do when one does not assemble. The bore to look for is not the worst-looking one and not the one at the end; it is one of the two or three holding the fit, and which those are is a property of that particular stack rather than of the design.

The table a drawing needs #

Put together, the specification is four numbers wide and it saturates in every column.

The column to specify from is the second. An average is the wrong statistic for an assembly requirement, because half the hinges are worse than it; the value nineteen stacks in twenty stay under is 0.775 of the tolerance at three knuckles and 0.993 at forty. So the honest rule is simple enough to remember: give a hinge a radial clearance equal to its bore tolerance and it will assemble whatever its length.

That is a rule with a mechanism under it rather than a margin. It is not conservative by a factor nobody has computed; it is the bound, and the bound is reached by a hinge of infinite length.

What a knuckle is actually for #

If a hinge of two knuckles needs no play and a hinge of forty needs only the bore tolerance, the obvious question is what the other thirty-eight are doing, and the answer is not kinematic at all.

A hinge carries a load. A door on two knuckles has its weight taken at two places, and the leaf between them is a beam in bending with a span of the whole door; a door on forty has it taken at forty, with a span of a few millimetres. What the knuckles buy is stiffness and bearing area, which are quantities nothing here computes and which have nothing to do with the mobility count that condemns them.

That is the honest shape of the whole result. The redundancy is not a cost being paid for a benefit somewhere else — it is not a cost at all beyond a clearance bounded by the bore tolerance, and the benefit is real. A designer choosing forty knuckles over two is not trading buildability against strength. They are buying strength, and the buildability they appear to be spending turns out to be almost free.

It is worth contrasting with the case where the redundancy genuinely does cost. Bennett’s linkage has four joints, not forty, and its misfit is 0.507 of a dimensional detune rather than a bounded fraction of a positional one — because the coincidence it needs is a relation between lengths and twists, and nothing in the mechanism’s geometry supplies that relation for free. A hinge’s coincidence is that points lie on a line, and a line has four parameters to spend in getting close to them. The difference between an overconstraint that is free and one that is not is whether the assembly has parameters left with which to chase it.

What this does not settle #

The bores are round and the pins are round. A real knuckle is a rolled loop with a seam, and its bore is closer to a slot than a circle — the same distinction a clearance inside a tolerance box draws between a round allowance and a rectangular one. What that does is make the clearance direction-dependent, which turns the fit from a Chebyshev problem in the Euclidean norm into one in some other norm — a different computation with the same shape.

Tilt is not modelled separately. Each bore is treated as a point the axis must pass near, which is right when the bore is short compared with the hinge. A long knuckle also constrains the axis’s direction, and a bore drilled at an angle is an error this model cannot express.

The errors are independent and uniform. Bores made on one setup are not independent — a machine with a worn way puts a systematic tilt into every one of them — and a systematic error is exactly the kind a line fit absorbs for nothing. The numbers here are therefore the pessimistic case, and a real hinge bored in one pass should do better than the table says.

No force anywhere. A long hinge that does not quite assemble is assembled anyway, by springing the leaf. Whether that is acceptable is a stress question, and what this computes is the geometry that decides whether springing is needed at all.

The clearance is the same in every knuckle. A hinge with one loose knuckle and the rest tight is a different problem and an interesting one, since the loose one could be put where the fit is held.

Averages over four hundred stacks. The ratios are sampled, so the third figure moves a little between runs; the two statements that carry the argument — nought at two knuckles, and bounded by one — are exact and not sampled.

Still open: the knuckle that is deliberately loose #

The measurement above gives every knuckle the same clearance, and then finds that two or three of them decide the answer. Those two facts sit oddly together: most of a hinge’s play is being provided where it is not needed.

Its distinct argument would be the clearance as something to allocate rather than to specify. If one knuckle in a stack were bored a little large, the fit would be held by the remaining ones and the requirement on all of them would fall — so the question is how much a single loose knuckle buys, and whether it matters which one is loose. Two things would come out of it. The best position for a relieved knuckle, which the touching-set measurement suggests is wherever the fit is currently held and is therefore not a fixed place; and whether a hinge with one loose knuckle per few becomes insensitive to its length entirely, which would turn the bound from a limit into a design.