Fit every extra slide at one pose

A fourth block can only take freedom away put two extra blocks on a double slider’s rod, one at a point of its rolling circle and one at the opposite point. Each block runs on the straight line its pin draws through the crossing of the double slider’s own two slides, so the rod turns although the count gives it minus one freedom. Build the extra slides slightly wrong and the count takes over. With every slide turned by the same small angle about the crossing, each block lets the rod through only in two windows a turn, where its pin passes the crossing. The opposite pins pass the crossing a quarter of a turn apart, so their windows never meet, and from a turn of 0.0066 rad the pair locks the rod.

A third slide moves only through the crossing had found that a turn about some other point moves a block’s windows. The fourth-block essay ended by asking whether a pair of such turns could put two blocks’ windows on top of each other. If so, a fourth block would cost nothing beyond the third, and the count’s minus one would turn out to be a rule about where a fitter measures from.

Half of that is right. The windows can always be made to share a pose, and the rule for doing it is one a fitter already uses. They can never share more than one window a turn, so a fourth block always costs something.

A turn about a point frees a band round it #

With the double slider’s own two blocks exact, the rod’s pose at every angle is fixed by them, and an extra pin stays exactly on its true line, at a place $s$ along it that runs between $\pm 2r$ as the rod turns, $r$ being the rolling circle’s radius. A slide turned by $\varepsilon$ about the point at place $d$ of that line misses the pin by

$$(s - d)\sin\varepsilon,$$

exactly, since the pin is on the true line and the turned line crosses it at $d$. So the block is free while $|s - d| \le c/\sin\varepsilon$ for a clearance $c$. That is a band of the slide centred on the pivot, and the rod turns through the block’s windows while the pin is inside the band.

A window is therefore centred where the pin passes the pivot. Its width is the band’s width divided by how fast the pin moves along the slide there. Turned about the crossing, a slide’s pivot is where its pin is fastest, and the pin passes it twice a turn: two narrow windows. The third-slide essay found that moving the pivot towards the end of the travel widens them, until near the end the two merge into one wide window round the instant the pin turns back. For one extra block alone, the best pivot sits a little short of the travel’s end, where the band just reaches it.

The law was checked against the pins themselves. At 72 angles of the rod, a slide turned 0.01 rad about a point 0.7 along the third’s line misses the third pin by $(s - 0.7)\sin 0.01$ to $10^{-12}$.

What is best for each block is worst for the pair #

Give each of the two extra slides its own best pivot, and the pair behaves well while the errors are small. Each band is then so wide that it covers most of its pin’s travel, and the two merged windows overlap. At a turn of 0.01 rad the rod is free over 23.1% of a turn, where the pair about the crossing is locked. But each block’s merged window is centred on the instant its own pin reaches the end of its travel. The opposite pins reach their ends a quarter of a turn apart, exactly as they pass the crossing a quarter of a turn apart, so as the errors grow and the windows narrow they separate again. At 0.02 rad the rod is free over 7.2%, at 0.05 nowhere.

That is the fourth-block essay’s quadrature returning in another form. The two pins are the ends of a diameter of the rolling circle, so every event in one pin’s travel happens to the other a quarter of a turn later or earlier. Any rule that places each slide’s pivot by reference to its own pin’s travel alone puts the two windows a quarter of a turn apart. Turned about the crossing, about the ends, or about any other point chosen from the pin’s own travel, the same quadrature applies. Such rules differ only in how wide the windows are, and so in the error at which the separation becomes a lock.

Set every slide through its pin at one pose #

The way out is to choose the pivots by reference to the rod rather than to the pins: pick one pose of the rod and pivot every extra slide at the place its pin has at that pose. Then at that pose every pin is exactly at its pivot, every misfit is nought, and every block’s window contains the pose, whatever the errors are. The rod can never be locked.

That rule is not a mathematical device. It is what a fitter does when setting a slide to its pin with the rod clamped in one position. A slide adjusted until it passes exactly through its pin at that position can still be wrong in angle, but only by a turn about the point it was set at. So fitting every extra slide with the rod clamped in the same position makes every error a turn about the pins’ places at that pose. Fitting each slide with the rod wherever it happened to be, or setting each slide from a datum on the frame such as the crossing, gives pivots that belong to no common pose.

The figure shows what the common pose buys. About the crossing, the third’s windows and the fourth’s sit a quarter of a turn apart and the rod is locked. About the places at one pose, both blocks have a window at that pose and the rod turns through it, over 17.4% of a turn.

One shared window, never two #

It cannot be more than one window a turn, and the reason is again the diameter. Measure the rod’s angle $\theta$ from the instant the third pin reaches the end of its travel, so that the third is at $2r\cos\theta$ and the opposite fourth at $2r\sin\theta$. Pivoted at their places at the pose $\theta^*$, the third’s windows are where $\cos\theta = \cos\theta^*$, at $\pm\theta^*$, and the fourth’s where $\sin\theta = \sin\theta^*$, at $\theta^*$ and $180° - \theta^*$. The two sets share $\theta^*$ and nothing else at any pose. A third block alone about the crossing keeps two windows; the posed pair keeps one.

Within that window the narrower block decides, and a pin moving fast along its slide at the shared pose has the narrower window. The third moves at a rate proportional to $|\sin\theta^*|$ and the fourth to $|\cos\theta^*|$. The slower of the two fastest is at $\theta^* = 45°$, where both pins are equally far from the crossing and moving equally fast. There, once the windows are thin, the pair’s share of a turn is

$$\frac{c}{2\pi r \sin\varepsilon \cdot \max(|\sin\theta^*|, |\cos\theta^*|)},$$

which at $45°$ is $1/\sqrt 2$ of what the third alone keeps about the crossing.

Swept through 3,600 angles at each of 37 poses and three errors, no pose locks the rod. The choice of pose matters less than the rule. At 0.01 rad the share runs from 15.6% at the poses where one pin is at the crossing to 18.0% near 30° and 60°, where the wide windows’ curvature still counts. From 0.02 rad the best is 45°, 9.4% at 0.02 and 4.0% at 0.05. The share at 45° over the third alone’s rises towards the thin-window limit as the error grows: 0.56 at 0.01, 0.62 at 0.02, 0.67 at 0.05 and 0.69 at 0.1, against $1/\sqrt 2 = 0.71$. Turning the fourth slide the other way, by $-\varepsilon$, changes nothing, because only the size of a misfit decides a window.

The rules against the best pivots there are #

To see how good the rule is, the best pair of pivots was searched for directly. Both pivots were run over the whole of their pins’ travel, 161 places each, and the share computed from the closed bands at 1,440 angles. The best pair found was then swept on the pins themselves, and the swept share agreed with the bands’ to within a twentieth of a per cent of a turn.

At small errors the best pivots are the ones each slide would choose alone. At 0.005 rad they free 68.2% of a turn and each-alone pivots 68.4%, the search’s grid being a little coarse. At 0.01 rad the figures are 22.9% and 23.1%. In that range the bands are wide enough to cover most of each pin’s travel, so their windows overlap whatever their quadrature. At 0.02 rad the best is 10.7%; each-alone pivots fall to 7.2% and the common pose gives 9.3%. From 0.05 rad the best pivots put both pins at almost the same distance from the crossing, 1.49 against the 45° pose’s 1.505, and that pose is within a tenth of the best: 4.0% against 4.2%, and 2.1% against 2.1% at 0.1. The pair about the crossing is locked from 0.0066 rad on; the posed pair overtakes it at 0.0055.

So there are two good rules, and they fail differently. Each slide set where it alone is freest is better while the errors are small and locks once they are not. Every slide set at one pose is somewhat worse while the errors are small and never locks. A fitter who does not know the size of the errors, which is the usual position, should take the second.

Reading the error from the window #

A posed drive also tells the fitter something the drive about the crossing could not. With every slide set at one pose, the rod always turns some way either side of it before it binds, and how far depends on the errors. Clamped at the pose and then released, the rod turns until the faster-moving block reaches the edge of its band. The window’s half-width is about $c/(2r\sin\varepsilon\max(|\sin\theta^*|, |\cos\theta^*|))$, so the angle the rod reaches before binding gives the error of whichever block binds first. Which block that is says which slide is responsible, since the binding block is the one whose pin has left its band. The fitter can correct that slide, leave the rest, and the window widens.

The pair about the crossing gives no such reading once it has locked. Nothing moves, so there is no angle to measure and no binding block to identify. Why a hinge works found that a joint’s clearance can be read as play in the motion it allows. A posed double slider keeps a play of that kind available whatever its errors, and reads them out through it. That is the practical case for the rule beyond its guarantee.

Any number of blocks #

The guarantee does not depend on how many blocks there are. Spread two, three, four or six extra blocks evenly round the rolling circle, turn every slide by 0.01 rad, and the rod is locked if every slide was turned about the crossing. Set every slide at the pose where the fastest of the pins passes its place slowest, and the rod turns over 17.4%, 15.7%, 15.3% and 15.1% of a turn. The share falls only slowly as blocks are added. Every window contains the pose and the narrowest decides, and with many pins spread round the circle the fastest one at any pose moves at close to the most any pin can.

The seventh contact found redundant contacts that constrain nothing until something deforms. A redundant double slider with every extra slide set at one pose is the same kind of object. Its redundancy costs width of motion and never motion itself.

With clearance in every block #

The double slider’s own blocks have clearance too, and with clearance everywhere the rod can shift to share an error out among all of them. At a fixed rod angle every pin’s misfit is a straight-line function of the rod’s position, so whether some position fits every block is a linear program in two unknowns, solved exactly at 720 angles.

Clearance in every block moves the crossover but not the conclusion. At 0.01 rad the loose rod with both slides turned about the crossing is free for the whole turn, because the double slider’s own clearances absorb the misfit, while set at one pose it is free over 39.7%. At 0.02 rad the crossing pair locks and the posed pair keeps 18.9%; at 0.05, 7.4% against nothing. With every block loose, the common pose costs freedom up to an error near 0.01 and saves the rod beyond it, and it never frees more than the third alone, since a fourth block can still only narrow.

What the count’s minus one was about #

Counting and measuring mobility and a constraint that has been said already showed that such counts are wrong about exact mechanisms. The fourth-block essay read the minus one as a warning about the workshop: every redundant block turns into a constraint the moment its slide is wrong. This essay finds that the warning has a precise remedy. Whether the redundant constraints lock the mechanism or merely narrow its motion depends on where the errors are pivoted. Where they are pivoted depends on where each slide was set from.

A second slide keeps only the rod found that once a double slider’s two slides are set, every further slide has exactly one right line. This essay adds that it also has a right place to be measured from: the point its pin occupies at the pose every other extra slide was measured at. A clearance is a link treated a joint’s play as a small extra freedom. The common pose is what guarantees that every block’s small freedom is still available at the same instant.

What the pose does not settle #

Offsets as well as turns. A slide set through its pin at one pose has no offset error at that pose, by construction, so its whole error is a turn about the pin’s place there. A slide set any other way can carry an offset and a turn together. The third-slide essay found an offset alone to be all or nothing, and the combination is not measured.

Which pose a real fitter can clamp. The rule says any pose keeps the rod turning and 45° from either pin’s crossing keeps most once the errors are large. Whether the rod can be clamped there, with both extra pins accessible, depends on the machine around it.

Load. A rod moving inside its window is free kinematically. How much each block presses on its slide near a window’s edge is a question of stiffness, not answered here.

The pins off the circle. An extra pin not exactly on the rolling circle draws a thin ellipse rather than a line, and no turn of its slide fits it everywhere. Its misfit is not of the form $(s - d)\sin\varepsilon$, and its windows are not centred on a pivot.

Still open: the pose that keeps the most for loads that come back #

The common pose keeps one window a turn, centred on the pose the slides were set at. A machine that does not turn its rod right round, a double slider used as an ellipse-tracing or straight-line mechanism over part of a turn, needs freedom only over the part it uses.

The distinct argument would be a redundant double slider asked to sweep a stated arc of the turn rather than all of it, with its extra slides set at one pose inside that arc. The measurement would find, for each arc and error, the pose that keeps the whole arc free if any does, and the largest error at which some pose still does. A pose at the arc’s middle keeps the window centred, but the pins’ speeds there may not be the slowest the arc offers. The question is whether the fitter’s rule becomes “set the slides at the middle of the stroke”, or whether the stroke’s own geometry, how fast each pin crosses its slide within it, picks a different pose. If it does, the right place to measure from belongs to the machine’s duty as well as to its geometry.