Three tuned joints stop at the fifth order

A shared load is a shared motion found that two legs of three revolute joints each, holding one platform, share a twist exactly when all six joint axes cross one line, the line of a wrench both legs can carry. Pushed along that twist the platform moves at first order and no further. A shake has an order found how much further: the gap between the two legs’ nearest poses grows as the square of the push, and turning one joint of the second leg about the shared line to the right angle raises it to the cube, and turning a second joint raises it to the fourth power. A shake travels as a root of the play turned each order into a distance: in joints with play, a shake of order $k$ moves until its gap uses up what the play can take, a $k$-th root of the play, and at a millionth of play the second-, third- and fourth-order pairs travel 0.013, 0.066 and 0.139.

That essay ended on the fifth order. A third joint turned about the shared line gives a third free angle, and the fourth-order coefficient might be driven to nought with it. It asked whether that works, what the fifth-order gap buys, and whether two legs of three joints can be tuned to every order — which would make them a mechanism, as two wrists are — or whether the conditions run out somewhere.

They run out at the fifth.

Every order from the legs’ own series #

The earlier essays read each coefficient off a push test: push the two legs a little along their shared twist, find the gap between their nearest poses by a Newton solve, and divide by a power of the push. That works for the square and the cube. At the fourth order the gap is a hundred-millionth at a push of a hundredth, and getting a coefficient from it means extrapolating several pushes, each with its own solve; the earlier essay’s two-joint tuning was located that way, to a precision that left the third-order coefficient at $7\times10^{-6}$ rather than nought. A third angle needs a whole curve of such points, and the push test cannot supply one.

So the coefficients are read another way. A leg’s pose is a product of three exponentials of its joint screws. Write each joint angle as a power series in the push $t$, and every exponential, every product, and the logarithm of the first leg’s pose inverted against the second’s are power series too, with 4 × 4 matrices for coefficients. At order one, both legs’ joint rates are fixed by the shared twist. At each higher order $m$, the coefficient of $t^m$ in that logarithm is a twist; the two legs’ order-$m$ joint coefficients can cancel any part of it that lies in the span of their six joint screws, which is five-dimensional because the legs share one twist, and what they cannot cancel is its component along the one wrench reciprocal to both legs. That number is $\sigma_m$. A pair shakes at order $m$ when $\sigma_2$ to $\sigma_{m-1}$ vanish and $\sigma_m$ does not.

One detail decides whether this works. At every order the two legs can also re-time the push, replacing $t$ by $t + \lambda t^m$, which changes nothing geometric. Left free, a least-squares solve picks that λ out of rounding error and the noise leaks into every later order; fixed at nought, the series route agrees with the bracket formula of the earlier essay to $1.2\times10^{-14}$ at $\sigma_2$ over all twenty-four legs of its census.

The series route also says how good the earlier tuning was. The push test located the two-joint tuning at $\psi_1$ = 50.770°, $\psi_2$ = 23.622°; with the series the exact point is $\psi_1$ = 50.761°, $\psi_2$ = 23.607°, a hundredth of a degree away. That difference left $\sigma_3 = -6.8\times10^{-6}$ behind, and a leftover third-order term overtakes the fourth-order one at pushes below $|\sigma_3/\sigma_4| = 6\times10^{-4}$: for the very smallest pushes the earlier essay’s fourth-order pair, as built, was a third-order pair again. At the pushes its test used it made no difference, and the fourth-order slope it measured was real.

The series route reproduces the earlier tuning’s numbers where they can be compared. At the two-joint tuning it gives $\sigma_3 = -6.8\times10^{-6}$, the push test’s leftover, and $\sigma_4 = -0.0109$, against the earlier push-test estimate of about 0.012. Computed to order six, a whole set of coefficients costs four milliseconds, and to order three under one.

Where two coefficients vanish, a curve #

Three angles and two conditions leave a curve. Starting from the earlier two-joint tuning, where the third joint is untouched, the set where $\sigma_2 = \sigma_3 = 0$ was followed by continuation: step along the curve’s tangent, correct back onto it by Newton’s method with the step held square to the tangent, and refuse any correction that lands further than twice the step from its guess, so that the trace cannot jump to another piece of the curve.

In both directions the curve runs until it meets the same kind of end. As it approaches its ends every coefficient grows without bound, $\sigma_4$ reaching −101 at one end and −151 at the other. At those points the six joint screws of the two legs nearly lose a rank: their fifth singular value falls to 0.0046 and 0.0058, against 0.24 in the middle of the curve. The legs are close to sharing a second twist, and the order-by-order solve divides by that small singular value. How fast the coefficients grow says which power of it they carry. Fitted on log axes over the points near each end, $|\sigma_4|$ grows as the singular value to the power −1.94 at one end and −1.92 at the other, about its inverse square, and $|\sigma_5|$ faster, as the power −2.85 and −2.41. Each order further from the first draws once more on the direction the legs are about to share, which is what a near-degeneracy in a series expansion looks like, and why the ends are where the method, rather than the shake, stops being meaningful. Between the ends the curve is 7.25 radians long in the three angles, and the hero figure shows it in two projections. One way from the two-joint tuning it bends round and ends at $\psi_1$ = 83°, $\psi_2$ = −24°; the other way it runs out through both roots to $\psi_1$ = 164°, with $\psi_3$ never further than 45° from where it started.

Where three vanish, two points #

At the two-joint tuning $\sigma_4$ is −0.0109. Moving along the curve it climbs slowly through −0.003 for two radians of arc, crosses nought, rises to about +0.03, and crosses nought again before running off to −151 at the far end. The two crossings, refined by bisection along the curve, are at

$$(\psi_1, \psi_2, \psi_3) = (106.31^\circ, 140.35^\circ, 24.66^\circ)\quad\text{and}\quad(136.39^\circ, 57.24^\circ, 18.34^\circ),$$

and at both $\sigma_2$, $\sigma_3$ and $\sigma_4$ are below $10^{-14}$. A third root found in an earlier trace of the curve, near (163°, 116°, 61°), lay beyond the near-singular end, reached by a correction that had jumped across it; the trace that refuses jumps does not reach it, and it is not counted.

That two points exist is what the count predicts: three conditions on three angles, so isolated solutions. What the count cannot say is how many, and the answer here, on the curve through the two-joint tuning, is two. They are not near each other or alike: the first turns the third joint by 24.7° and the second by 18.3°, and the first and second joints differ between them by thirty and eighty-three degrees.

The fifth power, checked by the push #

The series says $\sigma_4 = 0$ at both points. The push test, which knows nothing about series, has to agree.

It does. The untuned pair’s gap falls with slope 1.999, one tuned joint’s with 3.009, two tuned joints’ with 4.036, and the two three-joint tunings’ with 4.971 and 4.894. The last two are fifth powers, short of five by the sixth-order term at the larger pushes, which is larger for the second tuning because its coefficients are larger. The gaps at the smallest push are $4\times10^{-11}$ and $10^{-9}$, near where a push test stops being able to see anything, which is the reason the series route was needed to find the points in the first place.

And no further #

The table reads down a diagonal. Untuned, every coefficient from $\sigma_2$ on is non-zero. Each tuned joint clears one more, and at every stage the next is there: $\sigma_3 = 0.0154$ with one joint tuned, $\sigma_4 = -0.0109$ with two, and $\sigma_5 = -0.0042$ and $0.109$ at the two three-joint tunings, with $\sigma_6$ at 0.0029 and −0.165. No coefficient vanishes by accident. With three joints tuned the three angles are spent, and clearing $\sigma_5$ would need a fourth.

This is the same bargain two ways to be overconstrained described for mechanisms that move when their counts say they cannot: a finite motion needs every order of the closure to hold, and a finite set of exact dimensions buys only finitely many. Bennett’s linkage escapes it because its one condition, on the ratio of its lengths to the sines of its twists, holds the loop’s closure at every order at once, and twelve bars and a symmetry found a half-turn symmetry doing the same for a framework that Maxwell’s count calls rigid. A shaking pair tuned joint by joint has neither a condition of that kind nor a symmetry, and each angle buys exactly one order.

So the answer to the earlier essay’s last question is no. Two legs of three revolute joints, sharing a line of force, with the second leg’s joints turned about that line, can be made to shake at the fifth order and no higher. To go further, something else about the legs has to be free: where a joint crosses the shared line, or the first leg’s joints, or the shared line itself. Each would add parameters and so further orders, but each is a redesign of the pair, not a tuning of it. A pair that shakes at every order is a pair that moves, and it moves to first order and not at all is the general reason that needs a structure, like two wrists whose freedoms are groups, rather than a finite number of angles set exactly.

The ends of the curve say the same thing from the other side. Where the coefficients blow up, the two legs are close to sharing a second twist; at an exact second shared twist the pair would have a first-order freedom of two, and what a leg of three joints leaves free is a three-system that two legs share in a line only generically. The tuning curve runs from one near-degenerate pair to another, and the fifth-order pairs sit on it well away from both.

A higher order is not always more travel #

The travel follows the earlier essay’s account of what play does: a joint with radial play $c$ can shift its outer link by up to $c$ square to its axis, as why a hinge works found for a single pin, and with play $c$ in every joint the gap is absorbed until $|\sigma_k| \varepsilon^k = cG$, where $G = 4.517$ is what the six joints’ radial play can do along the shared wrench, the same for every tuning because turning a joint about the shared line does not change its lean against it. So the travel is $\varepsilon = (cG/|\sigma_k|)^{1/k}$. In a millionth of play the untuned pair travels 0.013, one tuned joint 0.067 and two 0.143, the earlier essay’s numbers to within the push test’s estimate of the fourth-order coefficient. The first fifth-order tuning, with $|\sigma_5| = 0.0042$, travels 0.255, nearly twice the fourth-order pair. The second, with $|\sigma_5| = 0.109$, travels 0.133, less than the fourth-order pair.

The order sets how the travel scales with the play, the slope of each line; the coefficient sets where the line sits. At a small enough play every fifth-order pair travels further than every fourth-order one, since a fifth root of a small number is larger than a fourth root of it. But the second tuning’s coefficient is ten times the fourth-order pair’s, and the two lines cross where $(cG/0.109)^{1/5} = (cG/0.0109)^{1/4}$, at a play of about $2.4\times10^{-7}$. Below that the fifth-order pair travels further; at the earlier essay’s millionth it does not.

What a designer can take from it #

A tuned pair is a precise object with a short list of tunings. Each order is one equation in the angles, and with three angles the fifth order is reached at isolated settings, here two on the curve through the earlier two-joint tuning. Missing those settings by a little costs the order: the gap goes back to the fourth power with a coefficient in proportion to the miss.

Set the angles to the play, not to the last digit. A tuning angle missed by a small amount leaves the cleared coefficient a little away from nought, and the pair falls back to the lower order only at pushes smaller than the ratio of the leftover coefficient to the next one. The earlier two-joint tuning, a hundredth of a degree off, fell back below a push of $6\times10^{-4}$, while in a millionth of play it travels 0.14. What matters is that the fallback push stays well below the travel the play allows, so the tolerance on a tuning angle tightens as the play shrinks: a pair meant to use a higher order in very tight joints has to have its angles set correspondingly closer, since its travel there is small and the fallback push must sit under it.

Choose among the tunings by the coefficient, not only the order. The two fifth-order pairs differ by a factor of twenty-six in $\sigma_5$, and in a millionth of play one travels nearly twice as far as the other.

Read high orders off the series, not the push. The bracket gives the second order in closed form; the series gives every order to rounding error, in milliseconds, where a push test at the fifth order is measuring gaps of $10^{-11}$.

What the tuning does not settle #

Other branches. The curve was followed through the earlier two-joint tuning. A third root seen beyond its near-singular end suggests that $\sigma_2 = \sigma_3 = 0$ has other pieces in the three angles, with their own roots, and none was traced.

Finite play. The travel uses the leading term only. Near a fifth-order tuning the sixth-order coefficient is as large as the fifth at the second root, so at larger plays the travel would differ from the root law, as it did for the lower orders in the earlier essay.

The pole. The curve’s ends are configurations where the legs nearly share a second twist, where every coefficient here diverges. What the pair does there, whether a nearby exact second shared twist exists and what kind of motion it gives, was not followed.

Still open: what the curve’s ends are #

Both ends of the tuning curve sit where the two legs’ six joint screws nearly lose a rank, their fifth singular value falling to a few thousandths. At an exact loss of rank the legs would share two twists, a first-order freedom of two, and the shake analysis would have to be redone with two directions to push in.

The distinct argument would be that end found exactly: the configurations of the three tuning angles where the two legs’ joint screws span only four dimensions, which is one more condition than the curve and so isolated points if they exist. The measurement would locate them by driving the fifth singular value to nought, check whether the curve actually reaches them or only passes close, and ask what the pair does there: whether its second shared twist is also a shake, of what order, or a finite motion the tuning curve has been approaching all along.