A fourth block can only take freedom away

A third slide moves only through the crossing put a third block on a double slider’s rod. Every point of the rod’s rolling circle runs on a straight line through the crossing of the two slides, so a block at such a point can ride a third slide along that line, and the rod still turns all the way round, though the count says five bodies and six one-freedom joints should give nought freedoms. The three slide conditions are dependent: any two imply the third. Build the third slide slightly wrong and the count takes over. Turned by an angle $\varepsilon$ about the crossing, the slide leaves the rod free only in two windows per turn, where the third pin passes through the crossing and the error does not matter. Each window is $2\arcsin(c/(2r\sin\varepsilon))$ wide for a clearance $c$ and a rolling circle of radius $r$.

That essay ended with a hope. A fourth block at the point of the rolling circle opposite the third would travel, it seemed, in exact opposition to it. A fourth slide turned by the same error might then have a misfit equal and opposite to the third’s at every instant. Could the fourth slide be placed to cancel the third’s error, making a redundant pair self-aligning, or would every added slide only narrow the windows further?

Every added slide only narrows them. The premise of opposition is wrong too, and in the worst direction.

The opposite pin runs at right angles #

A point $w$ of the rolling circle draws the line through the crossing at an angle $(\arg w + \arg q)/2$, where $q$ fixes the circle’s centre, and its place along that line is $2r\cos(\varphi - \varphi_w)$ as the rod turns through $\varphi$. The opposite point, $-w$, has its argument increased by $\pi$. So its line is turned by half of that, a right angle, and its place along it is $2r\cos(\varphi - \varphi_w + \pi/2) = -2r\sin(\varphi - \varphi_w)$.

The two extra pins are the two ends of a diameter of the rolling circle, and a diameter of a circle rolling inside one twice its size has its ends on two perpendicular lines. That is Archimedes’ trammel, the simplest double slider of all. The third and fourth pins do not move in opposition. They move in quadrature: when one is at the crossing, the other is at the far end of its travel. Measured on the site’s double slider, slides at 70° and a rod of 2, the two lines are at right angles to $10^{-15}$ and the fourth pin’s place matches $-2r\sin(\varphi - \varphi_3)$ to $10^{-15}$ through the whole turn.

Each block’s windows are centred where its own pin passes the crossing, and those instants are a quarter of a turn apart. With both extra slides turned 0.005 rad and a clearance of 0.01, each block alone lets the rod through 77.8% of the turn. The windows overlap over 55.6%, so the pair leaves the rod free there and nowhere else. At a turn of 0.007 rad each block alone still leaves 46.9%, but the windows have shrunk to less than half a quarter-turn each and no longer meet at all.

The windows can only intersect #

The deeper reason the hope fails holds for every place a fourth block could go.

With the double slider’s own two blocks exact, they fix the rod’s position at every angle of its turn. Each further block then asks one thing, that its pin be within its clearance of its slide, and the rod can turn through an angle only if every block’s condition holds there. The rod’s freedom is the intersection of each block’s windows. A fourth block’s windows can coincide with the third’s, at best, and can never add angles the third forbade. Nothing a fourth slide does can cancel a third slide’s error, because the conditions are not added together; they must all be met.

For the third and the opposite fourth there is a closed form. Write $x = c/(2r\sin\varepsilon)$ for each extra slide. The third is free while $|\cos(\varphi - \varphi_3)| \le x_3$ and the fourth while $|\sin(\varphi - \varphi_3)| \le x_4$. Both hold somewhere only if $x_3^2 + x_4^2 \ge 1$, since a cosine and a sine cannot both be small at once, and then the rod is free over

$$\frac{4\left(\arcsin x_3 + \arcsin x_4 - \pi/2\right)}{2\pi}$$

of a turn. A single extra block moves at any error, in two thin windows; the pair locks the rod outright once $x_3^2 + x_4^2 < 1$.

The sweep and the closed form agree at every error, to the sweep’s resolution. Below 0.003 rad both extra blocks allow the whole turn. At 0.005 the third alone allows 77.8% and the pair 55.6%. From 0.0066 rad each, the threshold the closed form gives for equal errors, the pair allows nothing, while the third alone still allows 46.9% at 0.007, 31.1% at 0.01 and 6.0% at 0.05. Turning the fourth slide the other way, by $-\varepsilon$, changes nothing, since only the size of each pin’s misfit decides its windows. That is the last of the hope of cancellation: an error and its opposite cost the same.

Where a fourth block costs least #

If a fourth block is wanted anyway, as a spare or to share a load, its place round the rolling circle decides how much it costs. A pin $\beta$ round the circle from the third passes the crossing $\beta/2$ later in the rod’s turn, so its windows are offset from the third’s by half the separation. Beside the third, 7° round, the pair keeps 29.1% of a turn against the third’s 31.1% alone. The share then falls in proportion to the separation. From 113° to 247° round, the windows no longer overlap and the rod is locked, the opposite point, 180°, at the middle of that range.

The figure follows from the windows’ geometry alone. Each extra block has two windows a turn, half a turn apart and each $2a$ wide, with $a = \arcsin x$. A second block’s windows are shifted by half its pins’ separation round the circle, and only that shift modulo half a turn matters. The two blocks’ windows then overlap by $2a$ less the shift, or not at all, so the pair keeps a share $(2a - \Delta)/\pi$ of the turn, $\Delta$ being the shift folded into a quarter turn. At 0.01 rad, $x = 0.47$ and $2a = 0.98$ rad, so the pair locks once the pins are more than 1.96 rad, 112°, apart round the circle. The scan reads its first locked place at 113°.

Quadrature is therefore not one bad case among several but the worst case. Half a turn of separation round the circle is a quarter of a turn of shift in the rod’s angle, the largest shift there can be between two sets of windows half a turn apart. The point the hope picked as the natural partner for the third block is the one place round the circle where a fourth does most damage.

So the best place for a fourth block is next to the third, where it duplicates the third’s windows and adds almost no new condition. That is the place where it is least a fourth slide at all. A designer who wants the count’s minus one to cost nothing must put the extra pins close together on the rolling circle. Spreading them round the circle, as one would to share load across the rod, is exactly what locks it.

With clearance in every block #

The double slider’s own blocks are not exact either, and with clearance in every block the rod can shift to share an error out. At a fixed rod angle each pin sits at a fixed offset from the rod’s reference point, so each pin’s misfit from its slide is a straight-line function of the rod’s position. Whether some position puts every pin within its clearance is then a linear program in two unknowns, solved exactly with no small-error approximation.

For three blocks a second route gives the same answer from one line of algebra. Three slide normals in a plane are dependent, so one combination of the three misfits does not change however the rod is moved. By Farkas’s lemma the three blocks can all be satisfied exactly when that combination is within the combined clearance its weights allow. The third block’s error is therefore shared out over all three clearances, in proportion to each block’s weight in the combination. The third block carries 0.37 of it on average, so the rod tolerates nearly three times the error it could with the first two blocks exact. At a turn of 0.01 rad the rod becomes free for the whole turn instead of 31%; at 0.02, 43.6% instead of 15.1%; at 0.05, 16.1% instead of 6.0%. The linear program and the combination agree at every angle sampled.

A fourth block still only narrows. With every block loose, a fourth opposite the third, turned the same way or the other, leaves the rod free over the whole turn up to 0.01 rad, where the third alone also did, and then locks it: nothing at 0.02 and 0.05, against the third’s 43.6% and 16.1% alone. A fourth beside it, 17° along, costs a little: 38.3% at 0.02 and 11.1% at 0.05. Clearance everywhere widens every window and leaves their intersection the only thing that counts.

What this means for assembling one #

A double slider with extra blocks is built in some order, and the windows say what the order should be. The double slider’s own two slides are aligned first and fix the rod’s motion. A third slide is then aligned by turning the rod and adjusting the slide until its block runs freely the whole way round. That succeeds once the slide is within about $c/(2r)$ of its true line, 0.27° for a clearance of 0.01 on a rolling circle of 1.064. Before then the block binds everywhere except in its two windows, and the windows’ width tells the fitter how far the slide is out, though not which way, since an error and its opposite give the same windows.

A fourth block opposite the third cannot be fitted the same way unless the third is already within the threshold. Fitted while the third is still a little wrong, the fourth binds the rod at the angles where the third was free, and the rod moves nowhere at all. The fitter then has no window to read, and no way to tell which slide is at fault. Fitting extra blocks one at a time, each to full freedom before the next, is the only order that leaves a signal to adjust by.

A second slide keeps only the rod found that a double slider’s second slide fixes everything about the rod’s path except the rod itself. The extra blocks inherit that: once the first two slides are set, every further block’s slide has exactly one right line, the one its pin draws, and the windows measure the distance from it.

What the count’s minus one was saying #

The count gave a rod on three concurrent blocks nought freedoms and on four minus one. Counting and measuring mobility and a constraint that has been said already found that such counts are wrong about exact mechanisms, because redundant constraints are implied by the rest. The double slider’s extra blocks are the cleanest example: with perfect slides they cost nothing at all.

What the count is right about is tolerance. Each redundant block turns into a real constraint the moment its slide is wrong. Every one added is a further condition the rod must meet at every angle, and the precision the slides must be made to rises with each. One extra block needs its slide within $c/(2r)$ of the true line for a full turn, about 0.27° here, and can be anywhere between that and a few degrees for a usable part of the turn. Two in quadrature need both within $\sqrt{2}\,c/(2r)$ or nothing moves. The minus one was not wrong about the rod; it was a warning about the workshop.

The seventh contact found a redundant contact that constrains nothing until something deforms, and a clearance is a link treated a joint’s play as a small extra freedom. Both pictures hold here. With clearance, each extra block is a small extra freedom that its slide’s error uses up, and when it is used up the block becomes a constraint. The rod turns while every block still has some of its clearance left.

What the windows do not settle #

Load. A rod moving inside its windows is free kinematically, but the blocks whose clearance is used up press on their slides. Which block carries how much, as the rod passes through a window’s edge, is a question about stiffness that this essay does not ask.

Other errors. Every error here is a turn about the crossing. The third slide’s essay found an offset parallel to the slide all or nothing, and an error that combines a turn and an offset moves the windows’ centres. A combination of errors could in principle put two blocks’ windows on top of each other where their pure turns would not. That case is not measured.

More blocks. A fifth and a sixth block, spread round the rolling circle, add their windows to the intersection in the same way, each shifted by half its pin’s separation from the others. Three extra blocks evenly spread put windows a third of a turn apart in the circle and a sixth of a turn apart in the rod’s angle, and at a turn of 0.01 rad, whose windows are 56° wide in the rod’s angle, any two of them already lock the rod, since their windows sit 60° apart. The census of three or more erroneous extra blocks is not run.

Which two blocks are the double slider. The rigid model holds the first two blocks exact and lets the others carry the error, which is a choice. In the model with clearance everywhere no block is special, and the third block’s weight of 0.37 is its share against the other two; the rigid model’s numbers are the limit in which the first two blocks are made much more precisely than the rest.

The rolling circle’s own error. The extra pins are placed exactly on the rolling circle. A pin a little off it draws not a line but a thin ellipse, and its slide then fits it nowhere exactly, a different kind of misfit with windows of its own.

Still open: errors that put the windows together #

The windows of two extra blocks fall where their pins pass the crossing, and a turn about the crossing cannot move that. A turn about a different point, as the third slide’s essay found, moves a block’s free band along its slide and so moves its windows in the turn.

The distinct argument would be the pair of errors, a turn of each extra slide about its own chosen point, that puts the two blocks’ windows on top of each other. A fourth block then costs nothing beyond the third, and the pair of misplaced slides behaves like one. The measurement would take the opposite fourth and find the pivot for its slide that centres its windows on the third’s, for each turn error. It would ask how much of the third’s free share the pair then keeps, and whether the pivot that does it is a practical place to align a slide from. If it is, an overconstrained double slider can be assembled to run by aligning its extra slides about the right points, and the count’s minus one is a rule about where to measure from.