A shake has an order

A shared load is a shared motion put two legs of three revolute joints under one platform and let them share a line of force. A force along that line does no work on any joint of either leg, so both legs resist it, and the count that gives the platform its freedoms gives it one more in the same act. Three constraints from each leg make six, the shared one is counted twice, and the platform keeps one twist both legs permit. Pushed along that twist it does not go anywhere. The two legs part, and the gap between their nearest poses grows as the square of the push. The platform can move to first order and not at all, which is a shake.

A third leg that keeps the shake built a third leg whose freedom contains the shared twist, and found the same square. It ended on the question of what would make the square vanish. Two wrists, legs whose three joints meet at a point, share the line through their two centres, and the platform turns about that line for as far as it is pushed. Two general legs through a shared line do not. Somewhere between them is a condition, and the question was what it is, and whether meeting it makes the twist a motion.

The answer has two parts, and the second is not the one that was expected. The condition is one number, a bracket, readable off the joints without pushing anything. Meeting it does not make a motion. It raises the shake by one order.

The gap as a series in the push #

A leg’s pose is the product of its three joints’ exponentials. The platform follows the shared twist $S$ if each leg’s joint angles can be chosen so that both legs put it at the same pose. To first order that is how $S$ was found: $S = \sum c_i L_i$ for each leg’s joint screws $L_i$, so moving each joint by $c_i t$ moves the platform by $tS$.

That first-order term is all a count of freedoms or a rank of screws can see, and it is the same for a shake and for a motion. To second order it is a question about the joints’ brackets. Move each joint by $c_i t + d_i t^2$ and the product of exponentials has a $t^2$ term

$$\sum_i d_i L_i + \tfrac{1}{2}\sum_{i<j} c_i c_j [L_i, L_j],$$

by the Baker–Campbell–Hausdorff formula, where $[L_i, L_j]$ is the Lie bracket of two twists. It moves to first order and not at all met the same structure in a planar framework, where the second-order term of the bars’ lengths was the obstruction. Here the $d_i$ are free: each leg can add any combination of its own joints at second order. So the two legs agree at second order exactly when half the difference of their bracket terms, $\tfrac{1}{2}(B_A - B_B)$ with $B = \sum_{i<j} c_i c_j [L_i, L_j]$, is something the two legs’ joints can make up between them.

The two legs’ freedoms together span five dimensions of the six, since they share only $S$. So the difference must lie in a five-dimensional space, and whether it does is one number: its reciprocal product with the one wrench both legs resist, the shared line of force. Call it $\sigma_2$. If it is not nought, no choice of second-order joint motion closes the gap, and the gap grows as $\sigma_2$ times the square of the push.

What a bracket of two joints is #

The bracket of two twists has a picture. Every motion is a screw, and turning a little about one joint, then a little about another, then back about the first and back about the second does not return a body to where it started. It leaves a small displacement of second order, and that displacement is the bracket. For two revolute joints whose lines cross, it is a turn about a line through the crossing point, square to both. For two parallel joints it is a slide square to both. For two skew joints it is a screw along their common perpendicular, with a pitch set by their distance apart.

A leg’s bracket term $B$ adds these up with the weights $c_i c_j$ the shared twist gives the joints. So $B$ is the second-order drift a leg produces when it is made to follow $S$ with its joints moving in proportion, the part of its motion the first-order picture throws away. The lines a leg turns about and the lines it is pushed along found a leg’s freedom and its constraint sharing a centre and three axes. The drift is a property of the joints that that picture does not contain, and two legs with identical freedoms can drift differently.

Each leg can absorb any drift that lies in its own freedom, by adjusting its joints at second order. What neither can absorb is drift along the shared wrench, because that is the one direction both legs resist. So the shake’s coefficient is the component, along the shared line of force, of the difference between what the two legs drift.

For a wrist that component is nought, and not by accident. A wrist’s joints all pass through its centre, so every bracket among them is a turn about a line through the centre, and its drift lies inside its own freedom, the rotations about that point. Two wrists each absorb their own drift completely and there is nothing left to disagree about.

A bracket that prices the shake without pushing #

That is a prediction that can be checked against the push test the earlier essays used: hold one joint where the push puts it, solve the other five joints of the two legs to bring the platform’s two poses as close as they will come, and measure what is left.

Over 24 second legs, each built through the shared line with a different seed, the bracket and the push test’s coefficient agree to 0.33%, over coefficients from 0.0065 to 0.27. At a push of 0.01 they agree less closely, and the difference shrinks in proportion to the push as it falls, which is the next term of the series. The bracket is the limit, and it needs no solving at all: three joint screws per leg, their brackets, and the shared wrench.

That makes it a design quantity. What a leg of three joints leaves free drew the freedom of a leg as a three-system of screws; the bracket is what the leg does at second order, which the three-system alone cannot say. Two legs with the same shared twist and the same shared wrench can have very different brackets. The census here runs over a factor of forty.

Turning one joint to cancel it #

A number that varies from leg to leg can be tuned to nought. The second leg was built with each of its three joints crossing the shared line, and any joint can be turned about that line and still cross it, so the pair keeps its shared line, its shared twist and its shake. Only the bracket changes.

As the first joint turns through half a turn, $\sigma_2$ falls from 0.026, crosses nought at 36.49°, and crosses back at 109.61° on its way to a peak of 0.21. The figure below shows the pair at those turns. At either root the pair’s second-order obstruction is gone, to $10^{-17}$, and by the reading the earlier essay proposed, the pair should now move.

It does not. The push test at the first root gives a gap that falls as the cube of the push, slope 3.01. The shake is still there, one order down.

Why one joint’s turn cancels exactly one order #

Turning a joint about the shared line is a one-parameter family, and a one-parameter family generically passes through the nought of one number and no more. That is the whole reason the first root cancels the second order and leaves the third. The third-order coefficient is a different function of the same joints, and there is no reason for it to vanish where the second does. At the first root it is 0.015, comparable to the 0.026 the second-order coefficient had before it was tuned.

The count is also why a single tuned joint cannot be expected to do better by being chosen cleverly among the two roots. At the second root, 109.61°, the gap is a cube as well. The two roots are two places on a surface in the space of second legs, and both leave the third order untouched. So a designer with one free parameter can buy exactly one order, and every further order costs a further parameter. Here that parameter is the turn of the second joint, and the fourth order came from it.

A second joint, a third condition #

The same move works again. Turning the second joint about the shared line as well, and re-solving the first joint’s turn to keep $\sigma_2$ at nought each time, traces a curve of second legs on which the second-order coefficient is nought. Along it the gap is a cube, and the cube’s coefficient varies.

The third-order coefficient is read from the push test rather than from a formula. The third-order term of the product of exponentials has nested brackets and cross terms that are no longer one number, so it is measured, at two pushes and extrapolated to none. It starts at 0.015, crosses nought at a turn of 23.62° with the first joint at 50.77°, reaches −0.005, and crosses back near 80°. At the first crossing both the second- and third-order coefficients are nought.

The slopes in that figure are the orders. The pair built through the shared line has slope 1.99, a second-order shake. With one joint turned to cancel the bracket the slope is 3.01, and with two turned it is 4.01. At a push of 0.1 the three gaps are $4.1 \times 10^{-4}$, $3.1 \times 10^{-5}$ and $3.0 \times 10^{-6}$. Each tuned joint buys a factor of about ten at that push and a factor of the push at every smaller one. Two wrists have no slope at all. Their gap stays at the arithmetic’s floor, $10^{-17}$, at every push, which is what a motion looks like.

Why the wrists move and the tuned legs do not #

The count explains the difference. A shake of order $k$ means the two legs agree to order $k - 1$ in the push and no further. Each order past the first asks one more thing of the legs: at second order the bracket difference must lie in the five-dimensional span, and at third order a nested version of it must, again one number, since the span is the same. So a finite motion, which agrees at every order, asks infinitely many conditions of a leg geometry with finitely many numbers.

Legs in general cannot meet that. Legs with structure can meet all of the conditions at once. A wrist’s three joints meet at a point, so its freedom is the group of rotations about that point, and every bracket of its joints is another rotation about the same point, inside the freedom. Its bracket term lies in its own freedom at every order, and the product of exponentials can follow any rotation about its centre exactly. Two wrists share the rotation about the line through their centres, and both follow it for as far as it goes. There is no condition to meet, because none can fail.

So the question the earlier essay asked has a sharper answer than it hoped for. The families of legs on which the second-order coefficient is nought are one condition thick, a surface in the space of leg geometries. They are not the families in which each leg’s freedom contains the whole one-parameter motion. Those are far thinner, and the ones found here are groups: wrists, and by the same argument three parallel joints, whose freedom is the planar group. A tuned leg has removed one obstruction out of a sequence with no end.

What it means for a machine #

A shake of higher order is still a shake, and for a machine it is a better one. The gap at a small push is what the legs must absorb in their joints’ clearances, and it falls from the square to the cube to the fourth power. At a push of a hundredth of the platform’s scale, the three pairs here leave gaps of about $3 \times 10^{-6}$, $2 \times 10^{-8}$ and $10^{-10}$ in the same units. A designer who needs two legs to share a line of force, as twelve bars and a symmetry needed its bars to, can make the resulting shake as small as the clearances by tuning one or two joints, and can price the first tuning without building anything.

The two roots are not equally good. Both cancel the second order, but the cube they leave has a coefficient of 0.016 at the first and 0.139 at the second, nine times larger. The next order is where the choice between roots is made. So a designer who has tuned one joint to cancel the bracket should still compare the roots by the third-order gap, which the push test gives in a few solves.

What a designer cannot do by tuning is get a motion. Two ways to be overconstrained separated overconstraint that is exact everywhere from overconstraint that holds at one ratio only. A shake of any finite order belongs to neither: the mechanism is not a mechanism at any ratio, only a structure that is stiff to a high power.

What the orders do not settle #

Two legs. Every pair here is two legs sharing one line. With the third leg of the earlier essay the second-order obstruction has more than one component, and whether one tuned joint per component still raises the order by one each is not measured. A fourth leg, the case the earlier essay first proposed, adds components again: every leg added to a shared twist adds conditions at every order, which makes a motion rarer, never commoner.

The third coefficient by measurement. The second-order coefficient has a formula and the third was measured by extrapolating two pushes. A closed form for it, from the nested brackets, would make the fourth-order pair a construction rather than a search.

One family of tunings. Every tuning here turns a joint about the shared line, which keeps the shared line and the shared twist. Other tunings exist, and whether some other single change removes two orders at once was not searched for.

Pushes, not loads. The gap is geometric. How a loaded platform with clearance in its joints actually moves along a shake of order three rather than two, and how much a small load moves it, is not computed.

Still open: how far a tuned pair will go #

A shake of order $k$ lets the platform move a push $\epsilon$ at the cost of a gap of order $\epsilon^k$, and joints with clearance $c$ absorb a gap of about $c$. So a pair whose shake is of order $k$, built with clearance $c$, should move about $c^{1/k}$ before the clearance is used up: the square root, the cube root, the fourth root.

The distinct argument there would be that travel measured rather than estimated. The two legs of each order would be given clearance in every joint, as a hinge has, and pushed along the shared twist until the clearance runs out. The measurement would ask whether the travel follows $c^{1/k}$. If it does, a higher-order shake with realistic play behaves as a mechanism over a range a designer can compute. A third tuned joint might then buy a travel that is useful rather than merely larger.