Two shared twists lie on loops, not at points

Two legs of three revolute joints each, holding one platform, share a twist when all six joint axes cross one line: the line of a wrench that both legs can carry. A shared load is a shared motion found that, and that the count behind it is exact. Three constraints and three make six, so a shared load and a shared motion come in equal numbers. Pushed along that twist, the platform shakes rather than moves. A shake has an order measured how: the gap between the two legs’ nearest poses grows as the square of the push. Turning a joint of the second leg about the shared line raises that to the cube, and a second joint to the fourth power. Three tuned joints stop at the fifth order turned all three and followed a whole curve of settings in the three tuning angles on which the second- and third-order coefficients both vanish. On it the fifth order is reached at two points and no further.

That curve did not close. In both directions it ran towards settings where every coefficient grew without bound. There the fifth singular value of the two legs’ six joint screws fell to a few thousandths, as though the legs were about to share a second twist. The essay ended by asking for those ends exactly. It expected isolated settings, one condition more than the curve, and it asked what the pair does there: a second shake, of some order, or a finite motion that the diverging coefficients had been approaching.

The expectation was off by one condition, and the rest of the answer follows from correcting it.

Six screws that already live in five dimensions #

A generic set of six screws spans all six dimensions of screw space. Two legs that share one twist span five, and the counting argument for a second shared twist was the obvious one. Dropping from five to four looks like one more condition on the three tuning angles, so it should happen at isolated points. That count treats the six screws as free vectors in six dimensions, and they are not.

Every joint axis of both legs crosses the shared line $\ell$. A revolute screw whose axis meets a line is reciprocal to a pure force along that line: the force does no work on a turn about an axis it intersects. So all six joint screws are reciprocal to one fixed wrench and lie in the five-dimensional subspace of screws reciprocal to it, $\ell^\perp$. Inside that subspace, generic means spanning five, which is all of it. The shared twist is not a coincidence of the tuning; it is built into the construction, and it is why the lines a leg turns about and the lines it is pushed along could give every three-joint leg built through the line a share in the load.

Losing one more rank is now a statement about six vectors in five dimensions: the 5 × 6 matrix of their coordinates has rank four. A matrix of that shape drops from rank five to rank four when every one of its six 5 × 5 minors vanishes. Those six conditions are not independent; the set of such matrices has codimension $(6-4)(5-4) = 2$. Two conditions on three angles leave a curve, not points.

The tuning does not move a screw out of $\ell^\perp$. Turning a joint about $\ell$ keeps its axis crossing $\ell$, so the five-dimensional home of the six screws is fixed while the tuning angles move three of them around inside it. That is why the codimension count holds over every setting of the angles rather than only near one.

Three ellipses and a line through the origin #

The curve has a construction, and it is plain enough to draw.

The first leg’s three screws span a three-dimensional subspace $A$ of $\ell^\perp$, the three-system that what a leg of three joints leaves free drew, fixed whatever the tuning angles are. Divide it out. What is left of $\ell^\perp$ is a plane, $Q = \ell^\perp / A$, and each joint screw of the second leg lands in it as a point $p_i$. The six screws together span $3 + \operatorname{rank}[\,p_1\ p_2\ p_3\,]$ dimensions. Generically the three points span the plane, and the six screws span five. They span four exactly when the three points lie on one line through the origin of $Q$.

Turning joint $i$ about $\ell$ by $\psi_i$ is a rigid rotation, and a rigid rotation acts on screws through its adjoint, whose entries are built from 1, $\cos\psi_i$ and $\sin\psi_i$. So the joint’s point in the plane is

$$p_i(\psi_i) = c_i + d_i\cos\psi_i + e_i\sin\psi_i,$$

an ellipse, which three settings determine. Read off at 0, 90° and 180° and then checked at thirty-seven other settings, each of the three ellipses fits its joint’s computed point to $3.5\times10^{-16}$.

For this pair the first and third ellipses enclose the origin and the second does not. The second passes 0.020 from the origin at its nearest, against sizes of order one; seen from the origin, its directions cover 168° and no more. That settles the shape of the whole rank-four set. For every setting of the middle joint, its point fixes a line through the origin. An ellipse that encloses the origin meets every line through it exactly twice, once on each side. So the first joint has one setting on each side of the line, and so does the third. Choosing a side for each gives four ways to complete the line, and as $\psi_2$ runs round its circle each choice traces a closed loop.

Four loops, each a graph over the whole circle of the middle joint. Every one of the 2,880 settings sampled on them has the quotient gap, the smaller singular value of $[\,p_1\ p_2\ p_3\,]$, below $5\times10^{-16}$. The six screws’ fourth singular value stays at 0.37 or more on every loop. So the screws lose exactly one rank there, never two. The two loops that put the first joint on the second joint’s side keep $\psi_1$ between −165° and −16°, and the other two between 0° and 171°.

This is a statement about this pair, not every pair. An ellipse that missed the origin for the first or third joint would meet a line only over a cone of directions, and the loops would break into arcs joined where that line is tangent to the ellipse. The construction covers that case too: draw three ellipses and turn a line about the origin.

The tuning curve runs into two of them #

The earlier curve has two ends, and each is now near something definite. One heads for loop 4 and the other for loop 3. Over the last forty traced points before each end, the distance to the nearest loop falls steadily, by about three-quarters of the step length at every step. The last points sit 0.013 and 0.015 radians short.

The fourth-order coefficient grows as the inverse square of that distance: fitted where $|\sigma_4| > 1$, the slopes are −1.98 and −1.91. The earlier essay fitted the same growth against the fifth singular value and found −1.94 and −1.92. The two fits agree because, near a loop, the distance and the fifth singular value are proportional.

Whether the curve actually reaches the loop cannot be decided by tracing it further. Each correction onto the curve solves for $\sigma_3$ through equations whose conditioning worsens as the square of the distance. A trace forced onward closes to a quotient gap of 0.0015 and then stops converging.

So the landing is found from the other side. Near a loop, the curve’s one shared twist and one shared wrench tend to members of the loop’s plane of twists and pencil of wrenches. Which members they tend to depends on the direction from which the loop is approached. $\sigma_2$ tends to a finite value, and $\sigma_3$ grows as one over the distance $d$. So a setting at distance $d$, in direction $\theta$ across the loop and at position $\psi_2$ along it, has two finite limits, $\sigma_2$ and $d\,\sigma_3$. The curve can end on the loop only where both are nought: two equations in two unknowns, $\psi_2$ and $\theta$. Solved at $d = 10^{-3}$ and again at $d = 10^{-4}$, starting from the nearest loop point to each traced end, the two solutions agree to $7.7\times10^{-4}$ radians. The curve lands on loop 4 at $(\psi_1, \psi_2, \psi_3) = (83.9^\circ, -24.3^\circ, 23.3^\circ)$ and on loop 3 at $(164.0^\circ, 110.8^\circ, 45.3^\circ)$. Both lie within half a degree of the point each trace reached before its corrections failed.

What the curve brings with it explains why every coefficient diverged on the way in. At each landing, the curve’s limiting twist lies in the loop’s plane and its limiting wrench in the loop’s pencil. Pushed along that twist, the second-order obstruction is not small: 0.186 at one landing and 0.255 at the other. Its component along the curve’s own wrench is nought, to the $10^{-5}$ the limit is good for. All of it lies along the other shared wrench, the one that exists only on the loop.

Off the loop, the series that defines $\sigma_3$, $\sigma_4$ and $\sigma_5$ has to absorb that obstruction somewhere. At each higher order it can cancel only what lies in the span of the six screws. The direction about to become a second shared wrench is almost out of that span, so cancelling along it divides by the vanishing singular value, once more at each order. The tuning curve is a path along which a second-order shake is hidden from the one wrench the analysis can see. The divergence is that shake coming back.

A plane of twists and two quadratic forms #

On a loop, the three-and-three count of the shared-load essay gives two shared twists and two shared wrenches. The wrenches are the reciprocal complement of everything the six screws span, the half of the matrix that what a mechanism cannot do showed is computed in the same act as the freedoms. The legs share a plane $\Pi$ of twists, any $s = \cos\theta\, S_1 + \sin\theta\, S_2$. The platform can be pushed in any direction in that plane at first order, so the question the earlier essay asked about the curve’s end becomes a question about every direction at once.

The second-order coefficient is the same object a shake has an order found: half the difference of the two legs’ bracket terms, $\tfrac12(B_A - B_B)$, where a leg’s term is $\sum_{i<j} c_i c_j [L_i, L_j]$ over its joints’ rates $c_i$ for the push. With one shared wrench, the obstruction was the reciprocal product of that twist with the wrench, one number. With two, it is two numbers, one per shared wrench: $q(s) = (q_1(s), q_2(s))$. Each is quadratic in the push direction, a binary quadratic form in $\cos\theta$ and $\sin\theta$. The platform passes second order along $s$ only if both vanish there.

Two real binary quadratics have no common real root in general; a shared root is one condition on them. So at an ordinary setting of a loop no direction of the plane escapes. At $(\psi_1, \psi_2, \psi_3) = (1.2^\circ, -80.0^\circ, 41.6^\circ)$ on loop 4, the obstruction’s size runs between 0.171 in the weakest direction and 0.306 in the strongest. Every direction is a shake as the square of the push, and the plane of first-order freedom is a plane of shake.

The common-root condition is the vanishing of the resultant of $q_1$ and $q_2$, a polynomial in their six coefficients. Changing the basis of the plane or of the pencil only multiplies it by a positive factor. So its sign at each setting is a property of the pair, and where the sign changes, round a loop, two forms share a real root.

Round the four loops it changes sign 6, 2, 4 and 0 times. Each count is even, as it must be for a sign carried round a closed loop. At each of the twelve roots, refined by bisection along $\psi_2$, the weakest direction’s obstruction is below $2\times10^{-15}$ of the strongest. Loop 4, where one end of the curve lands, has no root at all. Its weakest direction never falls below 0.022 of its strongest, so every setting on it shakes as the square in every direction. That also places the landing on loop 4. The curve’s limiting twist arrives with an obstruction of 0.186, in a direction that was never going to escape.

So a second shared twist makes the pair more mobile only at first order. Three angles have now been spent on two conditions for the loop and one for the resultant. They buy twelve settings with one direction that clears the square, and no setting where any direction clears the cube. That is the same bargain the fifth-order tuning struck, one angle for one order. The loops start the count lower because their plane of directions is two conditions deep already.

A push that cannot slide #

The bracket says where the obstruction vanishes. A push test says whether the platform actually behaves that way. The push test of the earlier essays needs changing first, and the reason is worth stating.

The old test held one joint of the first leg at a value proportional to the push, solved the other five joints to bring the two legs’ poses as close as possible, and read the gap left. With one shared twist, holding one joint pins the push. With a plane of them it does not. The remaining joints can slide the platform along the second shared twist, to wherever the obstruction is weakest. Every direction then reports some mixture of that weakest value, and at a root it reports the cube for directions whose own obstruction is 0.15. A test that cannot hold a push direction cannot measure a direction.

So the push is held in the whole plane. The six joint angles are set to the push times the joint rates of one direction, plus a correction kept square to the rates of both shared twists. The platform cannot drift along $\Pi$ at first order, and Gauss–Newton finds the correction in the four directions left.

At the ordinary setting of loop 4, even the weakest direction leaves a gap falling with slope 2.000. At the root on loop 1, at $(-106.2^\circ, -155.4^\circ, -158.7^\circ)$, a direction 69° from the null one falls with slope 2.000. The null direction falls with slope 3.001: a cube, not a motion. The gap there at a push of 0.0025 is $2.7\times10^{-10}$, so the cube’s coefficient is about 0.017. Nothing else about the pair is free to clear it.

The two routes also agree on size, not only on order. Over 96 cases, six settings on each loop and four directions at each, the push test’s gap divided by the square of the push matches $|q(s)|$ from the brackets. The largest relative difference is 0.0038, of the size the third-order term has at that push.

Why a second shared twist is still not a mechanism #

It moves to first order and not at all separated two things a count cannot. One is a mechanism whose freedom holds at every order. The other is a structure that is merely flexible at the first. The loops add a third case: a structure flexible at first order in a whole plane of directions, and at the second in none of them. The rank of the joint screws counts first-order freedom and nothing more. Two shared twists double that count and leave the motion where it was.

A finite motion needs every order of closure to hold along some path. On a loop the second order already fails in every direction except at twelve settings, and there the third fails. The structures that do move, two ways to be overconstrained showed, hold all orders at once through a symmetry or through a freedom that is a group. Two wrists are the standing example: each leg’s three axes meet at one point, so each leg’s freedom is the rotations about it, and the shared rotations close under composition. Tuning a few angles on two arbitrary legs reaches neither. What the loops show is that a coincidence in the joint screws, even one that holds along a whole curve of settings, says nothing about the orders beyond the first.

The loops also put the earlier divergence in proportion. It looked as though the fifth-order tuning curve might be running into a finite motion. It runs into a plane of second-order shake that its own single wrench could not see.

What the loops give a designer #

A rank count overstates mobility by the size of the null space, and here by a whole dimension. The formula is repaired by the thing it replaced found that the rank of the joint screws repairs Kutzbach’s count; it repairs it only to first order. On a loop the legs share two twists, and a Jacobian rank test calls the platform two-degrees-of-freedom at first order. In joints with play it behaves as a second-order shake in every direction of that plane. It is not a mechanism with a second freedom. The plane push test is how to tell the two apart. It holds the push in the plane, which a test that pins one joint cannot.

Near a loop, high-order tunings are fragile in a way the coefficients announce. The fourth-order coefficient grows as the inverse square of the distance to the loop, so a tuning set close to one carries large high-order terms. With play, as a shake travels as a root of the play found, the travel is a root of the play over the coefficient. A large coefficient cancels much of what a high order buys.

The rank-four set is drawn in a plane, not searched for. Dividing the first leg out of the screws crossing the shared line leaves one ellipse per tuned joint and a line turning about the origin. Whether a family of tunings meets a second shared twist is then decided by whether each ellipse encloses the origin, without any search of the three angles.

What the loops leave unmeasured #

Other pairs. The four loops belong to these two legs. A different first leg, or a second leg built through the line at other points, moves the ellipses. An ellipse that missed the origin for the first or third joint would break the loops into arcs. The construction covers every case, but only this one was computed.

The approach. The landing was solved in the limit, from two distances that agree to a few ten-thousandths of a radian. The curve between its last traced point and the loop was not followed, because the equations that define it degrade as the square of the distance.

The order at a root beyond the cube. At the twelve roots the null direction’s gap is a cube; at the one measured closely its coefficient is about 0.017. Its size was read off the push test, not from a series expansion with two push directions, which would also give the cubic form in the other directions.

Still open: how far a plane of shake travels #

At a setting on a loop, the platform can be pushed in any direction of a plane. Each direction meets its own second-order coefficient $|q(\theta)|$, which runs between 0.17 and 0.31 at the setting measured. With play $c$ in every joint, of the kind why a hinge works measured in a single pin, a push in direction $\theta$ uses up the play at $\varepsilon \approx (cG/|q(\theta)|)^{1/2}$. So the set of poses the play allows is not a segment along one twist, as it was for a single shared twist, but a region in the plane, bounded by that radius in every direction.

The distinct argument would be that region measured. Its shape is set by the pencil, an oval whose radius is $|q|^{-1/2}$. At a root of the resultant, along the null direction, the radius stops being a square root and becomes a cube root. The region grows an arm along one line, longer than the rest by the ratio of a cube root to a square root of the play. At a millionth of play the cube root is ten times the square root. The measurement would give every joint play, push in many directions of the plane until the play runs out, and check whether the boundary follows $|q(\theta)|^{-1/2}$ and where the arm appears. It would also check whether a pair set at a root behaves, in tight joints, like a mechanism with one freedom over a short range and none in the other direction.