Two stations share a cut when they hand over together

A cam machine is cut at its handovers found the rule for how many times a region test must cut a cam machine’s drive. A region test sweeps each part over an arc of the drive and flags any ground two parts both cover within that arc. Over a whole turn it flags parts that pass through the same place at different times, so the drive is cut into arcs, and each cut must fall between two parts’ visits to their shared ground. On a work station that four cam-driven tools take in turn, the fewest cuts are the handovers: the number of times the station passes from one tool to the next after the turn’s seam. Each cut can move only within the time the station stands empty.

That station had one piece of shared ground. The essay ended with a cell that has two: a work point with a press, another with tools of its own, and a transfer arm that carries the part from one to the other. Each piece of ground has its own handovers. The question was whether the cell’s count is the total of the two stations’ handovers, so that a region test on a cell is a sum over its stations, or whether a handover at one station and a handover at the other can share a cut.

They can, and whether they do is a matter of timing that the designer controls.

The cell #

The cell has two stations 3 apart. Station A is fed from the left and pressed from above. Station B is pressed from above and emptied by an ejector to the right. An arm pivoted below and between them swings to A, picks up the part, swings it to B, puts it down, and swings back to park midway. Every tool is a bar 0.3 wide whose front, or tip, passes 0.15 beyond its station’s centre, as on the single station. So the only ground any two tools share is a square 0.3 across at each station. Station A’s square is shared by the feed, the press and the arm, and station B’s by the arm, the press and the ejector. No pair of tools shares ground at both stations, which the check confirms for every pair.

Every slide is driven by its own cam through a cycloidal rise of 30°, a dwell and a cycloidal return of 30°, the law the law that costs least found to have finite jerk everywhere. A tool’s front is inside its station’s square from 19.5° into its rise to 10.5° into its return, the crossings the single-station essay computed. The arm swings in 30°, with its tip reaching a station about 20° into the swing that brings it and leaving about 8° into the swing that takes it away. The dwells are longer than on the single station, so that each station is busy most of the turn and stands empty between visits for only about 20°. That makes the timing tight enough for it to matter which way the two stations are phased.

Every tool is rasterised at each of the turn’s 360 degrees on one grid, 230 cells across the cell. Each pair of tools that shares a cell gives an interval from one tool’s last visit to it before the other arrives to the other’s first. The fewest cuts that stab every interval are found greedily, and the intervals the greedy rule cut at are disjoint, which proves that no partition does with fewer. That is the single-station method unchanged. All that is new is a second piece of shared ground, and an arm that belongs to both.

The grid’s resolution matters more here than on the single station, because the windows are narrower. A window of 20° is twenty samples of the turn, and a cut is placed half a sample inside the interval it stabs. A sweep that missed nothing found that a sampled sweep can miss a crossing between samples, and the same caution applies to a window. So the windows were also read off the schedule directly, as the stretches when no tool covers any of a station’s shared cells, and every cut the greedy rule placed was checked to fall within two samples of one of them. The room a machine sweeps measured the ground each part covers; here only the ground two parts both cover counts, and on this cell it is the two squares.

Out of step: one cut for every handover #

In the first schedule station A runs feed, press, arm. The feed is on its square from 248° to 344°, the press from 4° to 100°, the arm from about 120° to 228°. Station B runs arm, press, ejector, with the arm there from about 242° to 330°. The turn’s seam is put at 235°, while the arm is swinging from A to B. Neither station is occupied then, so the seam falls in an empty time of both, and the handover the arm makes at each station, at A from arm to feed and at B from ejector to arm, costs no cut. That leaves two handovers at each station: at A, feed to press and press to arm; at B, arm to press and press to ejector.

Each station alone needs two cuts, its two handovers, proved least by its own disjoint intervals. The whole cell needs four, the sum. Station B’s empty times, 330° to 339° and 41° to 60°, fall inside station A’s visits, the feed’s and the press’s. Station A’s empty times, 344° to 3° and 101° to 120°, fall inside station B’s press and ejector visits. At no moment are both stations empty except in the arm’s swing, which the seam has already taken. So every cut separates the handover of one station only, and four are needed.

In step: one cut for each pair of handovers #

The second schedule keeps every tool and the arm as they were and moves only station B’s press and ejector. Its press now arrives at 350° instead of 340°, and its handover to the ejector falls at 110° instead of 50°. Station B’s empty times become 330° to 349° and 101° to 120°. Each overlaps one of station A’s, 344° to 3° and 101° to 120°. A cut at 348.5° separates station A’s feed from its press and station B’s arm from its press at once. A cut at 119.5° separates A’s press from the arm and B’s press from its ejector.

The whole cell needs two cuts, each station’s own count, and both cuts serve both stations. The cell’s region test costs exactly what its busier station costs alone. Nothing about the machine’s work changed: each tool makes the same stroke, the arm the same swings, and each station the same sequence. Only where in the turn station B does its work changed.

The general rule, and its range #

Read across the two schedules, the rule from the single station extends directly. Every handover at every station has a window: the stretch when its station stands empty between the two tools. A cut serves a handover exactly when it falls in that window. So the fewest cuts for the whole cell are the fewest moments that fall in at least one window of each handover, the same interval-stabbing problem, now over the windows of every station together. Four arcs and the false alarm is gone first solved it on one pair of parts, and nothing about the greedy rule or its proof cares which station an interval belongs to.

That puts the cell’s count in a definite range. It is never less than the busier station’s own count, since one station’s windows never overlap each other: the station hands over to one tool at a time. And it is never more than the sum, since each station’s own cuts serve it. Within that range the phases decide. Two handovers at different stations can share a cut if and only if their windows overlap, and windows of widths $w_A$ and $w_B$ overlap when their centres are less than $(w_A + w_B)/2$ apart.

The figure tests that statement by moving one handover. Station B’s handover from its press to its ejector was slid through the turn in steps of 10°, with its other handover already on top of station A’s. While B’s empty time overlaps A’s press-to-arm empty time, from 100° to 120°, the cell needs two cuts. Everywhere else it needs three. Each window is about 20° wide, so the overlap condition allows centres up to 20° apart, a band 40° wide out of the 360°. The count jumps exactly where the windows stop overlapping, and nowhere else, which is what the interval picture predicts.

The arm’s swing is a shared cut for free #

One kind of sharing needs no design. While the transfer arm swings from A to B it is in neither station. The handover at A that ends its visit there, from arm to feed, has an empty time that starts as the arm’s tip leaves A’s square, about 8° into the swing. The handover at B that begins its visit, from ejector to arm, has an empty time that ends as the tip reaches B’s square, about 20° into the swing. Unless the feed reaches A or the ejector is still at B through the whole swing, the two empty times overlap during it, and one cut there serves both stations.

In the schedules above, the seam took that cut, which is why the arm’s handovers did not appear. Put the seam somewhere else, at 300°, inside the feed’s visit to A and the arm’s visit to B, and each station needs three cuts. The stations out of step then need five, not six. Four cuts each serve one station and one, in the arm’s swing, serves both. In step they need three, every cut serving both. The arm makes the cell cheaper to check than its two stations taken apart, simply because it carries the part from one to the other.

The same holds for any tool that moves directly between two stations. It leaves one and arrives at the other in a single motion, and during that motion both are short of the same tool. A cell laid out so that one part travels from station to station, as most production cells are, has its stations already partly in step. The handovers the travelling part causes come in pairs that share a cut.

The ledger #

The four rows put the numbers side by side. With the seam in the swing, the stations need two cuts each. The cell needs four out of step and two in step, and the two in step both serve both stations. With the seam inside a visit, the stations need three each. The cell needs five out of step, one cut fewer than the sum because of the arm, and three in step, all three serving both. Every count in the table is proved least by the disjoint intervals its greedy cuts were placed at, and every set of cuts clears every pair of tools on the grid.

What a designer can take from it #

The single-station essay found that a region test’s cuts are not a bookkeeping device, because each cut’s window is the station’s timing margin. With two stations the cuts say something more about the cell’s layout. A cell whose stations hand over together has fewer distinct moments at which anything changes hands. A designer laying out its cams to bring that about is, in effect, giving the whole cell a common rhythm. Its handovers then fall into a few shared instants, and the rest of the turn is all work.

There are ordinary reasons to want that besides a cheaper region test. Every handover is a moment when a tool and its successor are both moving near the same ground, and the margin is the empty time. A cell whose handovers coincide can put all its timing checks in a few places, and the margins those checks inspect are the overlaps of the stations’ empty times. A plane is a colour found that the conflicts between parts decide how many layers a machine must be built in. For a cam cell the shared moments decide how many arcs its motion must be checked in, and the designer can make them fewer.

The rule also has a limit worth stating. Sharing cuts never takes the cell below its busier station. A station that changes hands five times a turn needs five cuts whatever the rest of the cell does, so a cell is only as cheap to check as its busiest piece of shared ground. Four cuts serve the whole cell found Peaucellier’s linkage at that lower limit by symmetry. Its parts move as one, so all their crossings coincide and the cell needs only its worst pair’s count. A cam cell can reach the same limit only by design.

Why the stations do not interfere otherwise #

One thing the count does not show is that the two stations could, in principle, interact through a tool that reaches both. In this cell no tool’s body covers the other station’s square: the presses and slides come from outside, and the arm’s bar points along its swing. So every pair’s shared ground is at one station, and the problem separates into two sets of windows on one turn. A tool whose body swept across a second station on its way to the first would add shared ground at both. The regions overlap and the parts never meet found such crossings to be the ordinary case in a linkage, and they would make a pair’s intervals belong to two stations at once. The windows would then be those of the pair rather than of a station, and the cell’s count would be a stabbing problem over pairs again, as three arcs, cut in the right places set it out.

What the cell does not settle #

More stations. Two stations give a range from the larger count to the sum. With three or more the range widens, and whether a cell can be laid out to reach its busiest station’s count depends on whether every other station’s windows can be placed on that station’s. That is a question of fitting intervals of fixed width round a circle, and it is not answered here.

Speed. The windows are widths in cam angle. A machine run faster keeps them in angle and shortens them in time. A cut shared by two stations has the narrower of their two windows as its margin, so phasing for a cheaper check can trade away timing margin. That trade is not measured.

The arm’s own clearances. The arm is a bar swung on a cycloidal law between dwells, and only its tip is ever at a station. A real transfer arm with a gripper and a part in it is a larger body, and the part it carries sweeps ground of its own. Whether the part shares ground with a press on its way out of a station is not modelled.

Still open: the cell that can be checked in its busiest station’s count #

The cell here was put in step by hand, by moving one station’s tools until its windows lay on the other’s. A cell with more stations, or with stations that hand over different numbers of times, would need its phases chosen by something more systematic.

The distinct argument would be that choice made as a problem. Given each station’s sequence and the widths of its empty times, find phases for the stations that put every handover’s window on a window of the busiest station, or show that none exists. The measurement would take three stations with two, three and four handovers, search their relative phases, and report the fewest cuts the cell can be brought to and how much timing margin, the overlap of the shared windows, the best phasing leaves. If the busiest station’s count is reachable with useful margins, a region test on a cam cell is as cheap as on its busiest station whenever the designer wants it to be. If the margins collapse, a cheap check and a robust schedule are in tension, and the cuts measure the price.