Free to turn and unable to

Everything in this field so far is a statement about a linearisation. A contact contributes a linear inequality in the twist, the cone is an intersection of half-spaces, and the margin is a distance in the space those rows live in. It is worth finding out what that costs before trusting any of it.

The case that shows it is not a degenerate drawing and not a mechanism at a singularity. It is an ordinary part in an ordinary pocket.

Rank two, and what that means here #

Take an ellipse of semi-axes $a$ and $b$ and put four flat walls against it, touching at the ends of its own axes. Four contacts, each with its normal pointing at the ellipse’s centre.

A normal through the centre has $(p \times n)_z = 0$: the moment entry of its row is exactly nought, not nearly so. So all four rows lie in the plane $\omega = 0$ of the twist space, the rank is two rather than three, and the cone $\{t : At \ge 0\}$ contains the whole spin axis — a line through the origin rather than any number of rays.

Read as a region of centres, that is a uniformly shaded plane: every point, in both senses, permitted.

The first-order answer is therefore that the part spins freely, both ways, without limit. The answer is wrong, and it is wrong for a reason no amount of care with the rank decision could have caught: the decision is not close.

The support function does the work #

Rotate the ellipse by $\theta$ about its centre and ask how far it now reaches toward the top wall. That is its support function in the direction $(0, 1)$, evaluated on the rotated shape:

$$h(\theta) \;=\; \sqrt{a^2 \sin^2\theta + b^2 \cos^2\theta}\;.$$

At $\theta = 0$ that is $b$, which is where the wall is. And $b$ is the minimum of $h$, since $a > b$. So any rotation whatever — either sense — increases the reach and drives the ellipse into both the top and the bottom wall at once. Expanding,

$$h(\theta) - b \;=\; \frac{a^2 - b^2}{2b}\,\theta^2 + O(\theta^4),$$

quadratic, with a strictly positive coefficient, and with no linear term at all. The missing linear term is the first-order analysis being right: to first order nothing happens, which is exactly what a rank of two said.

Measured rather than expanded: at a hundredth of a radian the deepest penetration is $6.4 \times 10^{-5}$; at a sixteenth, $1.6 \times 10^{-2}$; and a straight-line fit through five decades of angle gives a slope of 1.9944 against a predicted 2.

The two side walls do the opposite. The reach toward $(1,0)$ after a rotation is $\sqrt{a^2\cos^2\theta + b^2\sin^2\theta}$, which decreases from $a$, so the ellipse pulls away from them. Only two of the four contacts do the blocking, and which two is decided by which axis is shorter.

The circle, which is the whole argument #

The same computation on a circle in the same pocket returns exactly nought at every angle, because a circle’s support function is constant.

So here are two arrangements with:

• the same number of contacts, four;
• the same normals, all pointing at the centre;
• the same rank, two;
• the same escape cone, the whole spin axis;
• the same margin, nought;

and opposite behaviour. One turns for ever and the other cannot turn by a thousandth of a radian. The entire difference is a second derivative, and every instrument in the four rungs before this one returns identical output for both.

That is a stronger statement than the corresponding one in the network field, and worth comparing. There, a flex the rank permits may be blocked, and the obstruction is a number computed from the dependencies among the constraints — a framework with no dependencies has no obstruction, which covers every mechanism in eighteen of this site’s fields. Here the obstruction has nothing to do with dependencies: the rows of the ellipse’s arrangement are independent in pairs and their deficiency is a rank deficiency, and the thing that blocks the motion is the curvature of the part, which is not in the constraint matrix at all.

Where the curvature went #

That is worth saying twice, because it is the sharpest thing in the rung.

A contact’s row is built from a point and a normal. Nothing about the surface’s shape enters it — not its curvature, not its extent, not whether it is a flat or an arc. Two parts with the same contact points and the same normals produce identical matrices and identical cones, however differently they are shaped between those points.

So the first-order analysis is not merely a linearisation of the geometry; it is an analysis that has thrown the geometry away except at the contact points. Most of the time that is harmless, because the linear answer is already trivial and there is nothing left for curvature to do. It stops being harmless exactly when the linear answer is not trivial and the contacts are curved, which is what a rank deficiency on a smooth part looks like.

Reading it the other way: the field’s whole machinery works because most parts are polyhedral. A polygon has flat faces, a flat face has zero curvature, and a flat contact’s second-order term vanishes — so for a polygonal part the first-order answer is the answer.

Measured rather than argued #

The expansion above is exact and the site’s habit is to have a second route, so the numbers here come from a walk rather than from a series.

The part is rotated about the twist’s own centre by a finite angle — a rigid motion, not a step of one — and the depth is how far past each obstacle plane the part now reaches, computed from its support function in the direction that plane is being pushed. A genuine motion gives a depth at the arithmetic’s floor at every angle and no exponent worth fitting; a blocked one gives a depth growing as $\theta^2$, and the exponent is the evidence rather than the assumption.

That is the same instrument the network field walks a flex with, and it is here for the same reason: a cone is a candidate list and a walk is what settles it.

There is one difference and it is in the field’s favour. A framework’s walk needs a Newton projection at every step, because the constraints are equations and a step off the manifold has to be corrected. Here the motion is a rigid rotation and there is nothing to correct — the part is rigid, the walk is exact, and the only question is whether the moved part is inside the obstacle. So the second-order test in this field costs a support function evaluation and is exact at any angle, which is why the figures can put a slider on it.

Second-order form closure, and what it is not #

An arrangement that is not form-closed at first order and cannot move at all is called second-order form closure, and the name is worth handling carefully because it invites two wrong readings.

It is not a weaker kind of hold. The ellipse in its box cannot rotate by any amount, at all, ever; there is no motion, not a small one. What is weaker is the test, not the restraint.

And it is not friction. Nothing above needs a coefficient, a normal force or a material. The blocking is done by the shapes: the pocket is the ellipse’s bounding box and the ellipse does not fit into it turned.

What is genuinely weaker is the robustness. A first-order hold survives a perturbation: move the contacts a little and the margin changes a little and stays positive. A second-order hold does not, in one direction — give the pocket a clearance and the ellipse can turn, by an angle that goes as the square root of the clearance, since the penetration is quadratic. A clearance of a micron on a part 10 mm across buys about 0.03 radians of rotation, which is nearly two degrees. That is the same square-root relationship the network field measured on a blocked flex, fitted there at 0.512 against a half, and it arrives here for the same reason: a quadratic obstruction crossed at a given tolerance is crossed at the square root of it.

So a second-order hold is a hold that a tolerance destroys much faster than a first-order one, and a designer who has one should know which kind they have.

The pocket nobody would draw, and the one everybody does #

The ellipse in its bounding box reads as a contrived example, and it is worth saying why the situation is not.

Every part with a smooth boundary held in a pocket cut to fit is this arrangement. A bearing race in a housing, a lens in a cell, a cam follower roller between its cheeks, a shaft in a bush — each is a curved part against a curved or flat obstacle, touching where the two are tangent, with every normal running along the line of centres. The moment entries are nought or nearly so, the rank is short, and whether the part turns is decided by the relative curvature and not by anything the contact rows contain.

For a circular part in a circular pocket that spin is real and is the point of the arrangement: a bearing is exactly a second-order failure to hold, deliberately, and it is the same observation the counting field makes about a pin from the other side. For anything not circular it is not, and the difference is invisible to the first-order test.

Which suggests a way to read the whole rung. The first-order test asks whether the contacts surround the part in direction; the second-order test asks whether they surround it in shape. A polygon has no shape between its contacts and the first question is the whole question. A smooth part has shape everywhere, and the arrangements where that matters are exactly the ones where the normals line up.

Where a reader has met this before #

Two places on this site, and they are opposite cases, which is why neither of them prepared anybody for this one.

A four-bar at a toggle has a momentarily degenerate Jacobian and a perfectly good motion straight through it. The rank is short and the flex is real. A parallel platform at a direct singularity gains a freedom with every actuator locked, and that freedom is real too — the platform genuinely moves.

In both, a rank deficiency was a warning that the mechanism was about to do something. Here it is a rank deficiency that means nothing at all, on a part that is going nowhere, and the sign of it is indistinguishable.

Two rank deficiencies, and how to tell them apart #

Since the first-order test cannot separate the two cases, it is worth saying what can, cheaply, before anybody reaches for a walk.

The null direction of a rank-two contact arrangement in the plane is always a pure rotation about some point — that is what a deficient moment column means. So the test is: rotate the part about that point and ask whether its support function in each blocked contact’s normal direction is at a minimum there. If it is, any rotation increases the reach and the part is blocked. If it is at a maximum the part pulls away and the contact was never doing anything. If it is stationary to all orders, the part is a circle about that point and the spin is real.

That is a statement about one scalar function of one variable and it costs nothing. It also explains the ledger’s two smooth rows in one sentence each: the ellipse’s support function toward its walls is at a minimum, so it is blocked; the disc’s is constant, so it turns.

The reason the field does not build that test into the routine is that it is planar and specific, and the walk is neither. A walk works in space, works for any twist, and returns a number with an exponent attached rather than a verdict — which is what the site’s habit asks for.

The coefficient, in closed form #

The penetration is measured as quadratic with a fitted exponent of 1.9944, and the coefficient has a closed form worth writing down, because it says exactly what the second-order hold depends on and when it disappears.

The ellipse’s support function in the direction of the top wall, after a rotation by $\theta$, is $\sqrt{a^2\sin^2\theta + b^2\cos^2\theta}$. At $\theta = 0$ that is $b$, which is where the wall is. Expanding,

$$h(\theta) \approx b + \frac{a^2 - b^2}{2b}\,\theta^2 ,$$

so the penetration is $\tfrac{1}{2}\kappa\theta^2$ with $\kappa = (a^2-b^2)/b$ — no linear term, a strictly positive coefficient, and everything in it a function of the two semi-axes alone.

Check it against the measurement. With $a = 1.4$ and $b = 0.9$ the coefficient $(a^2-b^2)/2b$ is $1.15/1.8 = 0.6389$, and at a hundredth of a radian that gives a penetration of $6.39\times10^{-5}$ — against the $6.4\times10^{-5}$ the sweep reports. The closed form and the rotated-and-measured penetration are the same number, which is the second route this claim was owed.

The expression settles the circle in one line as well. At $a = b$ the numerator is zero, the coefficient vanishes identically, and the penetration is nought at every angle rather than small — which is the circle’s free spin, arriving as a special case rather than as a separate computation.

And it says how strong the hold is, which the exponent alone cannot. The coefficient is $(a^2-b^2)/2b$, so it grows with how far the part is from circular and collapses as it approaches a circle. A part with semi-axes differing by one per cent has a coefficient two orders of magnitude below this ellipse’s, so the angle at which it penetrates by a given amount is ten times larger. A second-order hold on a nearly round part is nearly no hold at all, and the arithmetic says precisely how nearly.

That gives the robustness objection its number. A clearance $c$ is defeated when the penetration reaches it, at $\theta = \sqrt{2c/\kappa}$ — so the permitted rotation goes as the square root of the clearance and the inverse square root of the coefficient. Ten times the clearance buys about three times the rotation; a part ten times rounder buys about three times the rotation as well. Both are gentle dependences, which is exactly why a second-order hold degrades so much faster than a first-order one, where the permitted motion is linear in the clearance and vanishes with it.

What is not settled #

Second order is not the end. An arrangement that clears the second-order test can still be blocked at third, and the general question of whether a first-order escape extends to a finite motion is not decided by any finite number of derivatives. The walk is the instrument that settles it and a walk that goes a long way is evidence rather than proof — the same caution a search that has stopped finding new answers earns on this site.

And nothing here is a general test. What is computed is the penetration of one particular part along one particular twist, from that part’s own support function. That works because every obstacle in this field is a plane and every part is convex. A concave part against a concave obstacle has a relative curvature that can go either way, the blocking condition involves both surfaces, and this field does not build one — which is a statement about scope rather than about difficulty.

And the sign of the second-order term is the whole answer. For the ellipse in its bounding box it is positive at both blocking contacts and there is nothing to trade against, which is why the block is unconditional. An arrangement whose second-order terms have mixed signs permits some finite motion and refuses others, and the boundary between them is a curve rather than a cone. That case exists, it is not in the ledger, and the honest reason is that nothing this field is about produces one.