The slide five poses almost have

The slide a fourth pose leaves found that four prescribed poses of a moving body always leave exactly one slider point: a point of the body whose four images lie on one line, so that a block at that point, running in a straight guide along that line, carries the body through all four poses. It is found with compasses, as the second meeting of two circles. The count behind it was one of conditions. A point of the plane has two coordinates. A guide line has two numbers of its own. Each pose after the first two asks one image to land on the line the first two define, so four poses ask two conditions of the point and the point meets them at one place.

A fifth pose asks a third, and a point of the plane has only two coordinates to give. So five poses in general have no slide at all — the same arithmetic of conditions against freedoms that three problems called synthesis used to tell its three kinds of problem apart. Five positions and what is left found that the same five poses still leave a crank a finite set to choose from — Burmester’s points, of which the five poses in the figures below have four, all real — because a crank’s circle has one number more to spend than a slide’s line — the count what the fourth position costs began. The slide has run out one pose early.

That essay closed on the question a designer would then ask. If five poses admit no slide, how close does the nearest thing to one come, and should a designer who needs five poses and a slide give up the slide, or give up exactness in one pose? This essay measures that miss, and it turns out that the miss does not have to be searched for. Five compass constructions already price it.

Five slides that each give up one pose #

Five poses contain five four-pose problems, one for each pose left out, and each has its one slider point. Call the point that leaves out pose $k$ by the name $Q_k$. Its four images other than the $k$-th lie on a line to the last digit of the arithmetic, since it is an exact four-pose slide. Its $k$-th image lies off that line by some distance $m_k$, and $m_k$ is what a designer pays for giving up pose $k$: a block in that guide carries the body exactly through four of the poses and misses the fifth sideways by $m_k$.

The poses in the figures are the site’s own five, the ones Burmester’s four points were found for. The body is about two units long, and the five one-pose misses are 0.5304, 0.0836, 0.0383, 0.0422 and 0.1288. Giving up the first pose is expensive; giving up the third is cheapest. The five slider points are scattered across a region about two and a half units tall in the body’s frame, and nothing about their positions says the misses should be related to one another.

They are, and the reason is the same count of conditions.

One obstruction, five places to put it #

A slide is four numbers: two for the body point $q$, and two for the guide — the angle $t$ of its direction and its offset $c$ from the origin. For any choice of the four, each pose gives a residual $r_i$, the signed distance of the $i$-th image of $q$ from the guide. A slide that fits the problem exactly makes all five residuals nought. That is five equations in four unknowns.

Near a slide, the five residuals depend on the four numbers linearly, through a matrix with five rows and four columns. The residuals any choice can produce therefore fill a four-dimensional subspace of the five-dimensional space of residuals, and a four-dimensional subspace of a five-dimensional space is the set where one linear combination is fixed. So there are five weights $\lambda_i$, fixed by the poses, for which

$$\sum_{i=1}^{5} \lambda_i\, r_i = F$$

whatever slide is chosen. $F$ is the problem’s obstruction: it is nought when the five poses have a slide and not otherwise, and no choice of body point or guide can change it. Every slide the designer might build is a way of distributing $F$ among the five poses.

A four-pose slide makes four of the residuals nought, so it puts the whole of $F$ on the one pose it left out. Its miss is then

$$m_k = \frac{|F|}{|\lambda_k|}.$$

The five misses are the one obstruction divided by five different weights. A pose with a small weight is cheap to fit and expensive to give up; a pose with a large weight is the reverse.

Where each slide puts its miss #

The picture of each slide’s five residuals shows what the equation says. Each of the five upper rows has one bar and four blanks: the whole miss sits on the pose that was given up. The obvious alternative is to share the miss, and the figure’s bottom row is the slide that does so in the way a designer with a tolerance would want. Its largest residual is as small as any slide’s can be. That slide — the minimax slide, called the compromise here — has five residuals of exactly the same size, 0.0140, alternating in sign from pose to pose.

That the five come out equal is not a coincidence of these poses. With four numbers to choose and five residuals to control — how many points may be prescribed is the same count made for a four-bar — a slide whose worst residual cannot be reduced any further has all five at that worst value. If one were smaller, the four numbers would have a spare direction in which to shrink the other four. So the compromise is found by solving five equations — each residual equal to $\pm e$ — for five unknowns, the four numbers and $e$, and keeping the smallest $e$ over the sign patterns. Started from each of the five four-pose slides and tried with every sign pattern, the least $e$ found is the compromise, and its residuals are equal to $3 \times 10^{-14}$.

The misses add like resistances in parallel #

The linear picture predicts the compromise without solving for it. If all five residuals have size $e$ and signs matching the weights, the obstruction equation becomes $e \sum_i |\lambda_i| = |F|$, so

$$e = \frac{|F|}{\sum_i |\lambda_i|} = \frac{1}{\sum_k 1/m_k}.$$

The compromise’s miss is one over the sum of the reciprocals of the five one-pose misses. That is the rule for resistances in parallel, and the analogy is exact in the sense that matters. Each one-pose miss is the whole obstruction forced through one pose. The compromise lets it through all five at once, and the easy poses — those with large reciprocals — carry the most of it.

At the site’s five poses the reciprocals are 1.89, 11.96, 26.08, 23.71 and 7.77. They sum to 71.40, and one over that is 0.01401. The compromise, solved, misses every pose by 0.01402. The two routes share nothing but the poses: one is five compass constructions and a sum, the other a Newton solve of five non-linear equations. They agree to a tenth of a per cent, and the tenth of a per cent is the part of the problem the linear picture leaves out, since at these poses the miss is a few thousandths of the images’ spread rather than nought.

The weights can be measured directly as well. They are the one combination of the five rows of the residuals’ derivative that vanishes, and computed from the problem’s own equations at the compromise they are 0.0256, 0.1676, 0.3656, 0.3324 and 0.1089, as fractions of their total. The shares of the reciprocals are 0.0264, 0.1675, 0.3653, 0.3321 and 0.1088. They agree to $9 \times 10^{-4}$. Pose 1, the expensive one to give up, is the one the obstruction barely touches, and the compromise is free to leave it where the other four put it.

A designer who wants least squares instead — the smallest sum of squared residuals rather than the smallest worst — gets the same kind of answer. With the residuals proportional to the weights, the linear picture gives a root-sum-square miss of $1/\sqrt{\sum_k 1/m_k^2}$. At these poses that is 0.02627, and least squares, solved by descent on the body point with the best line fitted each time, gives 0.02627. The linkage that is only nearly right found the function-generation version of this choice, three linkages each best by a different measure of error. Here the worst-case and the least-squares compromise are both priced by the same five numbers, before either is solved.

Giving up a pose costs two to four times more #

The comparison a designer came for is now arithmetic. Giving up the cheapest pose costs the smallest of the $m_k$; the compromise costs $e$; and the ratio of the two is

$$\frac{\min_k m_k}{e} = \frac{\sum_i |\lambda_i|}{\max_i |\lambda_i|}.$$

At the site’s poses it is 2.73. The cheapest pose to give up, the third, costs 0.0383, and the compromise costs 0.0140.

That ratio has bounds, and they come from a fact about the weights that has nothing to do with these poses. One of the four numbers of a slide is the guide’s offset, and moving the guide sideways by a small amount moves every one of the five residuals by that same amount. A change the designer can always make cannot change the obstruction, so the weights must sum to nought:

$$\sum_i \lambda_i = 0.$$

Half their total magnitude is therefore positive and half negative. No single weight can exceed half the total, so the ratio is at least two. And five weights cannot split evenly into two signs, so one sign has at most two of them sharing half the total, and one of those is at least a quarter of it: the ratio is at most four.

Across 560 five-pose problems, 360 built near a slide and disturbed and 200 drawn at random, the ratio runs from 2.011 to 3.997. It sits against both bounds without crossing either. So the answer to the designer’s question is general and does not depend on the problem. Giving up a pose is never less than twice as bad as sharing the miss, and never more than four times.

The parallel law as a limit, and as a floor #

The parallel law is exact only for a problem at a slide, where there is nothing to share. It is worth knowing how quickly it stops being exact.

The census builds 360 problems round an exact slide — a body point, a guide, five images placed along it at random, random body angles — and then moves each image off the guide and along it by up to a set amount, from 0.001 to 0.3 units. For the problems disturbed by 0.003 or less, the solved compromise and the parallel law agree to about $10^{-4}$. The error of the law grows as the square of the miss, which is the order a linearisation leaves out. At 0.3 units it reaches 23%, and on the 200 random problems, whose median miss is 6% of their images’ spread, it reaches 50%.

In every one of the 560, the ratio is at least one. The five compass constructions never promise a compromise better than the solved one, so the sum of reciprocals is a floor on what can be had, as well as the value it tends to. That is measured, not proved: the ratio’s smallest value over the census falls short of one by $10^{-10}$, which is the tolerance the compromise is solved to, and none falls further. A designer who computes the five one-pose misses, which takes ten circle constructions, therefore knows two things before solving anything. There is no slide better than $1/\sum 1/m_k$, and near a slide there is one that good.

Five poses of one motion brought together #

The slide a fourth pose leaves checked its slider point against something computed another way: when four poses are four nearby positions of one moving body, the slider point tends to Ball’s point, the one point of a four-bar’s coupler whose path is straight to third order. Five nearby poses ask for a point straight to fourth order, and there is none — so the question is how fast the miss falls as the poses close up.

It falls as the fourth power of the spacing. The miss is $1.4 \times 10^{-4}$ at a spacing of 0.2 radians of crank and $8.9 \times 10^{-10}$ at 0.01, and between neighbouring spacings the measured order is 3.954, 3.989, 3.998 and 4.000.

The weights say why. At the smallest spacing, times sixteen, they are 0.990, 3.981, 6.000, 4.019 and 1.010, with alternating signs: the binomial coefficients 1, 4, 6, 4, 1 of a fourth difference. The obstruction of five equally spaced poses is the fourth difference of the path’s sideways deviation from a line, a quantity a surveyor would recognise, and it is of order $h^4$ because the fourth derivative is the first one that no body point can make vanish. The compromise spreads that difference in the proportions its weights set. Its miss is the fourth difference over sixteen, and the five residuals alternate in sign, + − + − +.

The alternation is the pattern Chebyshev spacing produces in function generation, where the best approximation’s error touches its bound with alternating signs. Here it arises from a slide, with no polynomial in the problem. The compromise point itself closes on Ball’s point: at the middle pose it sits 0.084 units from it at a spacing of 0.1 and $8.4 \times 10^{-4}$ at 0.01, falling as the square of the spacing. So the best near-slide of a motion is a block at Ball’s point, which is what the four-pose limit already said, and the same kind of limit four positions brought together found for the crank’s cubic. The fifth pose adds one fact: how much that block misses.

Why the easiest poses carry the most #

The resistances picture says something a designer would not guess from the one-pose misses alone.

Pose 1 at the site’s poses is by far the most expensive to give up, at 0.530, fourteen times the cheapest. The natural inference is that pose 1 is the difficult one and should be protected. The weights say the opposite. Pose 1 carries 2.6% of the obstruction, so it is barely involved in the conflict at all: the other four poses nearly agree with each other about a slide, and pose 1 is expensive to give up only because the four-pose slide that ignores it is forced to put the whole obstruction on it. In the compromise it is missed by the same 0.0140 as every other pose.

The poses that are cheap to give up — 3 and 4, at 0.0383 and 0.0422 — are the ones the obstruction runs through. Between them they carry 70% of it. They are the poses the problem’s conflict is about, and giving up either one is attractive precisely because either one, removed, takes most of the conflict with it. That is the parallel picture read the right way round: a low resistance carries the most current.

What the miss means for a block in a guide #

A miss is a distance across the guide, and a guide that is a slot with a block in it has a natural place to put such a distance: clearance. A block with a clearance of $e$ on either side — a slot $2e$ wider than the block — can place its reference point anywhere within $e$ of the guide’s centre line, so a block with that much play reaches all five images of the compromise point.

So the compromise’s miss is also the least play across the guide at which a slide reaches five poses: 0.0140 either side at the site’s poses, 0.7% of the body’s length. A four-pose slide needs the same play only to reach the one pose it gave up, and at 0.0383 it needs more than twice as much to do it. The alternating signs describe the physical block. At the compromise it bears on one side of its slot at poses 1, 3 and 5 and on the other at poses 2 and 4, so a slide reaching five poses through clearance changes sides four times on the way. A clearance is a link found that a joint’s play behaves as a short link of its own. Here that link’s length is set by the poses, before anything is built.

What five poses do not settle #

The mechanism between the poses. The compromise is a block and a guide for the five poses, with a crank still to be chosen from Burmester’s four points. Whether that crank and the compromise slide assemble into a slider-crank that passes through the five poses in order and on one circuit — the branch and order defects three-position synthesis had to screen — is not checked here.

Clearance as a free parameter. The play reading assumes the block sits wherever its crank puts it within the slot. A real block is pushed against one wall or the other by the load, and which wall depends on the load’s direction, not only on the poses.

Six poses and more. A sixth pose gives six residuals and two obstructions, and the one-number picture becomes a two-dimensional one. Whether the parallel law generalises — a compromise priced by the six five-pose compromises — is not measured.

The planes exchanged. A fixed pin in a slot cut in the body, the other form of slide, obeys the same count with the roles of the two frames swapped, and its five four-pose slot pins should price its compromise the same way. That is not computed here.

Still open: the slide that reaches five poses by rocking in its slot #

The alternating signs say that a block reaching five poses through clearance changes sides of its slot between each pair of poses. That makes the compromise slide a mechanism with an intermittent contact: across each interval the block leaves one wall, crosses the play and meets the other, and while it crosses it is not guided at all.

The distinct argument would be that crossing, followed through a real slider-crank. Take a crank from Burmester’s points for the five poses, pair it with the compromise slide and a slot of width $2e$, and drive it as the slider-crank is driven. Two things would come out. Whether the block, pushed by the crank rather than placed, actually meets its alternating walls at the five poses or instead sits on one wall throughout and misses two of them — which would mean the play reading describes where the block may be, not where it goes. And how much the body wanders from the prescribed motion between poses while the block is in free flight, which is the cost of a compromise that no residual at the five poses can see.