Four cuts serve the whole cell

A region test checks two moving parts for interference by sweeping each over the machine’s whole drive and asking whether the regions they cover share any ground. It is cheap and never misses a collision, but it raises false alarms. The regions overlap and the parts never meet found two parts that pass over the same ground at different moments reported as colliding. Four arcs and the false alarm is gone cut the drive into equal arcs and compared only same-arc regions. Three arcs, cut in the right places then placed the cuts. Each cell of shared ground asks for one cut between the two parts’ visits, and the fewest cuts that answer every cell can be read off a single sweep and proved least.

All of that was done one pair at a time. A clearance check covers a whole machine, every pair of its parts on one drive, and the last essay ended by asking what that costs. If each pair’s crossings happen at their own moments, a machine needs a partition per pair, and the cuts add up. If a machine’s crossings cluster at a few shared moments, one partition serves every pair.

On the three straight-line linkages measured, one partition serves, and it needs no more cuts than the machine’s worst pair.

Which pairs a machine test is about #

A machine’s pairs of parts fall into three kinds, and only one of them is a region test’s business. A pair joined by a pin touches at the pin all the time. A pair that actually overlaps somewhere in the drive — as two bars that have to cross found some pairs must — has to be built in different planes, which a plane is a colour turned into a colouring of a conflict graph. What is left are the pairs that never touch at all, and those are the ones a region test can get wrong: if it cannot separate their visits to shared ground, it calls them a conflict.

Every part of each machine was rasterised at each of 360 configurations of the drive, on one grid covering the whole machine, so that any two parts’ cells line up. Chebyshev’s straight-line linkage has two pairs that never touch, and one of them — its crossing arms — shares ground. Watt’s has one, its two arms. Peaucellier’s cell has eight parts, twelve pairs that never touch, and six of them share ground: each long arm with the two rhombus bars across from it, and the rhombus bars with each other diagonally. None of the eighteen pairs has a cell both parts cover at the same configuration, so none of them is a collision the grid has mislabelled.

The pairs that never touch and share no ground either are free. Chebyshev’s coupler and its frame never cover the same cell anywhere in the drive, and neither do six of Peaucellier’s twelve: every rhombus bar against the frame, and the two rhombus bars meeting at P against the crank, which is pinned at the rhombus’s other corner. A whole-drive region test already clears them, and they need no cut. The machine’s partition is decided entirely by the pairs whose regions overlap while the parts never meet, which is the false alarm this series began with, multiplied by the number of pairs a machine has.

Sixteen cuts, or four #

For each pair, the cells both parts cover give intervals of the drive, one per shared cell. Each runs from one part’s visit to the other’s, and each must hold a cut. Alone, the pairs of Peaucellier’s cell need two or four cuts each: each long arm against the rhombus bar across from it needs four, as the last essay found with its own grid, and the other four pairs need two. Cut one at a time, that is sixteen cuts.

The machine’s question is the fewest cuts that fall inside every interval of every pair at once, and the greedy rule answers it the same way: take all the intervals together, in order of where they end, and cut at the first one not yet stabbed. It places four cuts, at −1.132, −0.290, 0.658 and 1.360 radians of a drive that runs from −1.57 to 1.57. The intervals it cut at do not overlap each other, so no partition of the drive clears the cell with fewer. Handed back to the grid test pair by pair, the five arcs flag nothing on any of the six.

The other two machines say the same thing more simply. Chebyshev’s linkage needs three arcs, from its one pair that shares ground, and Watt’s four, from its one. Neither has a second pair to add anything. On all three, the whole machine needs exactly as many arcs as its worst pair.

Watt’s arms show what a regular crossing looks like. They share 903 cells of ground over a drive of 1.43 radians, and the two arms visit those cells nearly the same stretch of drive apart: the intervals run from 0.287 to 0.366 radians, with a median of 0.307 and nineteen in twenty under 0.339. That is the shape of two parts sweeping past each other at a steady rate. Equal arcs must each be shorter than the shortest interval, so they need five, which is 1.43 over 0.287 rounded up. Placed cuts need four arcs: three cuts at −0.418, −0.127 and 0.179, the first three arcs each about 0.3 radians long and the last 0.54, where the arms no longer share anything. The saving over equal arcs is one arc, and it comes from the end of the drive where no crossing happens.

Why the cuts coincide #

The lanes in the first figure show where the sixteen go. Each pair’s own cuts, placed for that pair alone, wander: the greedy rule puts each just inside the end of the first interval it has not yet stabbed, and each pair’s intervals end in different places. The machine’s four cuts are placed for all the intervals at once, and they need not coincide with any pair’s own. What matters is only that every interval of every pair holds one of them.

Two things make that possible, and both can be read off the measurements.

A symmetry that runs the drive backwards. Peaucellier’s cell is symmetric about the line through its fixed pivots, but its crank is not on that line, so the mirror image of the cell at one drive angle is the cell at the opposite angle. Long arm A against rhombus bar QB at a drive angle θ is the mirror image of long arm B against rhombus bar AQ at −θ. The two pairs’ intervals are the same lengths — the shortest 0.439 radians in both, the median 1.632 — and run in opposite orders. The first pair leans on the machine’s first two cuts: 710 and 772 of its intervals hold no other cut. The second leans on the last two: 805 and 275. The pairs of opposite rhombus sides, AQ against PB and QB against AP, are a mirror pair in the same way, and so are the two long arms against the far rhombus bars. So the six pairs are three, each met once forwards and once backwards.

Long intervals. A pair whose parts pass over the same ground far apart in the drive gives long intervals, and a long interval is easy to stab. The four-cut pairs’ intervals have a median of 1.632 radians, half the drive. The two-cut pairs’ are shorter, a median of about 1.0 and never under 0.746, but they sit at one end of the drive or the other. Long arm A against PB, and AQ against PB, need only the machine’s first two cuts: none of their intervals depends on the third or fourth. Their mirror images need only the third, with a few intervals held by the second or the fourth alone. The machine’s cuts at the ends of the drive were placed for the four-cut pairs, and they fall inside the two-cut pairs’ intervals because those pairs cross in the same part of the drive, as the rhombus folds and the arms sweep past it.

So the machine’s crossings do cluster, as the question hoped. The cell takes turns over shared ground at four moments, and every pair’s turn falls on one side of each. The four disjoint intervals that prove it are −1.575 to −1.128, −1.119 to −0.285, −0.276 to 0.663 and 0.671 to 1.365 radians. Each is one pair’s interval between two visits to a single cell, and no two of them share a moment, so each needs a cut of its own.

On the cell itself the four cuts divide the shared ground into four patches of 4.82, 6.36, 5.33 and 0.62 square units on the machine’s grid. The last is the small patch at the rhombus’s far corner that the last essay found costing one pair a fifth arc. Here it costs the machine nothing it was not already paying for that pair.

What a false alarm costs in planes #

So far the cost of a false alarm has been checking effort: a region test’s verdict read as though it were a measurement, which six things a body is not put first among the things said wrongly about bodies. It has a cost in the machine too, because planes are assigned from conflicts. A plane is a colour assigned each part of a machine a plane so that no two conflicting parts share one and no pin has to pass through a part between the two it joins. A designer who reads the conflicts from a region test gets every false alarm as an extra conflict, and extra conflicts can need extra planes.

They do, on all three machines. Read from a whole-drive region test, Chebyshev’s linkage needs three planes where two will do, Watt’s four where three will do, and Peaucellier’s cell six where four will do. Equal arcs recover the true count only at four arcs for Chebyshev’s linkage, five for Watt’s and eight for Peaucellier’s cell, which sits at six planes until five arcs and at five until eight. The machine’s own cuts give the true count at three, four and five arcs, the least any partition can.

The false conflicts are not scattered. In the true four-plane build each long arm shares a plane with the rhombus bar across from it: long arm A with QB in the first plane, long arm B with AQ in the second. Those are exactly the two pairs the region test flags most, the four-cut pairs, and once they are called conflicts the arms and the far bars must separate. The assignment then spreads the rhombus over four planes of its own, with the crank sharing a plane with rhombus bar AP and the frame with AQ, and needs six.

A plane is a layer of material and a clearance gap in the finished machine. Two extra planes in Peaucellier’s cell make it half as thick again as it needs to be, and every pin through it longer and more bent under load. The region test was chosen because it is cheap; built from its whole-drive form, the machine is not.

What one sweep gives a designer #

Put together, the three essays on arcs reduce a machine’s region test to a procedure with one expensive step. Rasterise every part at every sampled configuration once, and keep each configuration’s cells rather than merging them into a region. For every pair of parts that never touch, read off the intervals between visits to cells both cover. Take all the intervals of all the pairs together and cut greedily. The cuts are the fewest any partition can use, the intervals they were placed at prove it, and with them the region test reports exactly the conflicts that exist. The planes follow from those conflicts.

The sweeping is the same single pass a whole-drive region test makes. What the procedure adds is memory, a list of cells per part per configuration, and a sort of the intervals. On Peaucellier’s cell that is eight parts at 360 configurations, and the six pairs that share ground give 20,162 intervals. Against that, the whole-drive test’s answer would have built the cell in two planes too many.

What the procedure does not do is replace a check of the true conflicts. The pairs that actually overlap somewhere, and the joined pairs, were set aside before any of this began, using the parts’ own gaps at every configuration. A region test with the right cuts agrees with that check on every pair that never touches. It does not make the check unnecessary for the pairs that do.

Checked two ways #

The machine’s cuts were placed from the intervals, and they were checked by a route that does not use intervals: each pair’s two parts were rasterised over each of the five arcs on their own, and the cells both covered in the same arc were counted. For every pair on every machine that count is nought. The planes were checked the same way round. The count from the placed partition is the count from the true conflicts, which were read from the parts’ own gaps at every configuration rather than from any region. Both routes share the grid and the sampling of the drive, whose limits a sweep that missed nothing and the room a machine sweeps measured, and nothing else.

What the three machines leave open #

Three machines, all straight-line linkages. All three are symmetric or nearly so, and their crossings are organised by that symmetry. A machine with many parts moving on unrelated schedules — a loom, a packaging machine with several cams — might need a partition much closer to the sum of its pairs’. That is not measured.

Other grids and samplings. Every count here was taken at one resolution, 200 cells across each machine and 360 configurations of its drive. The cut counts agree with the pair-by-pair essay’s, taken on grids fitted to each pair, and the interval lengths are long against the sampling step of about 0.009 radians on Peaucellier’s cell. Neither agreement is a proof that a finer grid would find no new sliver needing a cut of its own, and a sliver like the 0.62 square units at the rhombus’s corner is where one would appear first.

One drive. Each machine here has one input. A machine with two independent inputs has a configuration surface rather than a drive, and its crossings are curves on it, not moments.

The grid decides what counts as shared. The machine-wide grid is coarser for each pair than a grid fitted to that pair alone, so small patches of shared ground can appear or vanish with the grid. The counts of cuts agree with the pair-by-pair essay where they overlap, which says the grid is fine enough here and not in general.

Still open: a machine whose parts keep their own time #

The three machines here share their crossings because their parts are driven by one symmetric linkage. A machine whose moving parts are driven by separate cams on one shaft keeps several independent schedules on the same drive, and each cam’s follower crosses the others’ ground whenever its own profile says.

The distinct argument would be such a machine: three or four followers of different lifts and timings on one camshaft, each sweeping ground the others reach. The measurement would be each pair’s fewest cuts, the machine’s, and the planes a region test asks for. If the machine’s count approaches the sum of its pairs’, the one-partition result belongs to symmetric linkages and a cam machine needs a partition per pair. If it stays near the worst pair’s, a common drive shaft is enough to make parts take turns.