A third slide moves only through the crossing

A double slider is a rod whose two ends run in two straight slides that cross. Exact because two circles roll described its motion as a circle rolling inside another of twice its radius, with the rod’s plane carried on the smaller. A second slide keeps only the rod used that picture to find every double slider drawing the same ellipse, and it rested on one fact: every point of the rolling circle runs back and forth on a straight line through the slides’ crossing. Any two such points joined by a rod are a double slider.

That essay ended on the fact’s other consequence. If every point of the rolling circle runs on a line through the crossing, a third point can carry a third block on a third slide along its line, and the rod still turns. The count forbids it. A rod and three blocks on a frame are five bodies. Three pins join the blocks to the rod and three slides join them to the frame, six joints of one freedom each, and $3 \cdot 4 - 2 \cdot 6 = 0$.

So the machine is one of the mechanisms the mechanism Grübler says cannot move began with, a count of nought that moves anyway. The open question was what makes it move and what stops it: how an error in the third slide, in its offset from the crossing or in its direction, locks the rod, and how much clearance in the third block buys back.

Why the count is wrong #

The machine here has slides crossing at 70° and a rod of 2, the same double slider as before. Its rolling circle has radius 1.064, and it passes through both pins and through the crossing at every instant. The third pin was put on the rod at a point of that circle, and its slide along the line that point draws.

Turned through a whole revolution, the rod keeps its third pin on the third slide to $9 \times 10^{-16}$ at every one of 3,600 angles, for the pin at six different places on the circle. The motion is real and it is the double slider’s own, with the third block riding along.

The count is wrong because it treats the six joints as independent, and they are not. Fix the frame and the rod has three coordinates: two for where it is and one for its angle. Each slide imposes one condition, that its pin stay on its line. Three conditions on three coordinates would fix the rod if they were independent, but they are not. The Jacobian of the three conditions — how each changes as the rod’s coordinates change — was computed at four instants for each of the six pin places. Its determinant is nought to within $3.3 \times 10^{-9}$ at all twenty-four configurations, while the largest two-by-two minor never falls below 0.94. The conditions have rank two, one short of three. The third is implied by the first two, since any rod whose first two pins are on their slides is a double slider, and a double slider carries every point of its rolling circle on a line through the crossing.

Adding more slides does not change that. A rod on k blocks, all on the rolling circle and all on concurrent slides, has a count of $3 - k$: one, nought, minus one and minus two for two to five blocks. The rank of the k conditions is two in every case, and the rod has one freedom in every case. Counting and measuring mobility set Grübler’s count beside the rank of the constraint Jacobian as two routes to one number, and showed that when they disagree, the rank is right. Here they disagree by $k - 2$, one for each slide past the second.

Where the third pin is along its slide #

The third pin’s motion along its slide has a simple form, and it decides everything that follows. A point of the rolling circle is the sum of two equal circular motions turning in opposite senses, so it moves on its line as $s = 2r\cos(\varphi - \varphi_3)$ from the crossing, r being the rolling circle’s radius and $\varphi_3$ set by where on the circle the pin sits. It swings from $2r$ on one side of the crossing to $2r$ on the other, 2.128 each way for this machine, and passes through the crossing twice a turn.

That is the whole of what a misplaced third slide can act on. The rod’s motion is fixed by the first two slides. A misplaced third slide can only allow or forbid it, according to how far the third pin, carried by the double slider’s motion, strays from the misplaced line.

Two kinds of error, two kinds of answer #

Give the third block a clearance: its pin can sit up to c from the slide’s centre line and still fit. Here c is 0.01, a hundredth of the rod’s half-length. Then misplace the slide in the two simplest ways.

Moved parallel. A slide moved a distance δ off the pin’s line, keeping its direction, is δ from the pin at every instant, because the pin runs along the true line and the true line is parallel to the false one. So the answer is all or nothing. At an offset of 0.0099 the rod turns freely all the way round; at 0.0101 it does not assemble. The clearance buys back exactly its own width of offset, and not a hair more. Nothing about the pin’s travel enters, because the pin moves along the false slide’s direction the whole time and never towards it. In the machine here that means the third slide may sit anywhere within 0.01 of its true line, parallel to it, for the whole of the rod’s turn.

Turned about the crossing. A slide turned by a small angle ε about the crossing is $s \sin\varepsilon$ from the pin, since the pin moves along the true line and the false one diverges from it through the crossing. That distance is nought where the pin passes the crossing and largest at the ends of the pin’s travel, $2r\sin\varepsilon$. So the rod moves freely only while $|2r\cos(\varphi - \varphi_3)|\sin\varepsilon \le c$, and is held everywhere else. At a turn of 0.01 rad the pin strays up to 0.021 from the false slide, twice the clearance, and the rod is free only in the two stretches of the turn where the pin is near the crossing — 31% of the turn in all.

The difference between the two answers is the difference between a stack of errors and a mechanism’s geometry. Fragility has a direction found a Sarrus linkage stopped dead by one tilt of an axis and untroubled by a tilt two hundred times larger in another direction. The third slide shows the same asymmetry in the plane. An error that keeps the false slide parallel to the pin’s path costs nothing until it exceeds the clearance. An error that crosses the path costs something at once, and how much depends on how far the pin travels.

The free share of a turn #

For a turn error, the rod is free in two windows of rod angle, each $2\arcsin\!\big(c/(2r\sin\varepsilon)\big)$ wide and centred on an instant the third pin passes through the crossing. Together they are a share

$$\frac{4\arcsin\!\big(c/(2r\sin\varepsilon)\big)}{2\pi}$$

of a turn, or the whole turn when $2r\sin\varepsilon \le c$.

The second route computed the same thing without the closed form. The rod’s pose was solved from the first two slides alone at 7,200 rod angles. The third pin was placed on the rod at its fixed position in the rod’s plane, and its distance from the turned slide measured directly. The free share is the fraction of angles at which that distance is within the clearance. For turns of 0.002, 0.005, 0.01, 0.03, 0.1 and 0.3 rad the closed form gives 1, 0.778, 0.311, 0.100, 0.030 and 0.010. The measurements at all six places agree with it to within two samples of the 7,200.

The whole turn survives only while $\sin\varepsilon \le c/2r$, a turn of 0.0047 rad for this machine: the third slide must be aimed at the crossing to about a quarter of a degree. Past that, the free share falls about as $c/(\pi r \varepsilon)$ — halving with every doubling of the error. It depends on the clearance and the error only through their ratio, and on the machine only through the rolling circle’s radius. A larger double slider is less forgiving of the same turn, since its third pin travels further from the crossing.

Where on the circle does not matter #

The six pin places give windows of exactly the same width, 18° each at a turn of 0.03 rad, and differ only in when they fall. Each pair is half a turn apart and centred where that place’s pin passes the crossing, and the places are spread round the turn in step with their positions on the circle. Every point of the rolling circle passes through the crossing twice a turn; that is what running on a diameter of the fixed circle means. So a designer choosing where to put the third block chooses when in the turn the rod will be free, and cannot choose how much.

That has a consequence as soon as there are two redundant blocks. A rod on four blocks, with a third and a fourth slide each turned by the same small error, is free only where both windows are open at once, and the two pins’ windows fall at different moments. With the extra pins at 2.0 and 3.4 rad on the circle and each slide turned 0.03 rad, each slide alone leaves the rod free for 10% of a turn and together they leave it nothing: the windows never overlap, and the rod is locked at every angle. Pins close together share more of their windows — 5.3% free with the pins at 2.0 and 2.3 rad, 8.4% at 2.0 and 2.1 — and smaller errors widen both: at 0.005 rad the far-apart pair still leaves 56% of the turn. Every redundant slide past the third narrows the free share further unless its pin travels in step with the others’.

Pivot the error where the pin is slowest #

A real slide is rarely turned about the crossing exactly. It is mounted at some datum and turned about that, which is a turn about a point off the crossing. That is a turn and an offset together, and it moves where along the pin’s line the false slide passes closest to the true one.

The pin is then free while its position along its line, s, is within $c/\sin\varepsilon$ of the point $s_0$ about which the slide was turned: a band of fixed width, now centred on $s_0$ rather than on the crossing. The width in s is the same wherever the band sits. What changes is how much of the turn the pin spends in it, because the pin’s position is $2r\cos\psi$ and it moves fastest through the crossing and slowest at the ends of its travel.

So the band buys most when it sits at the end of the travel. For a turn of 0.03 rad the rod is free for 10.0% of a turn when the error is pivoted at the crossing, 14.6% when it is pivoted 1.5 along the line, and 25.9% at best. The best is when the band’s far edge just reaches the end of the pin’s travel, at $s_0 = 2r - c/\sin\varepsilon$, 1.795 here. There the two windows about the pin’s turning point have merged into one, and the free share is

$$\frac{1}{\pi}\arccos\!\left(1 - \frac{c}{r\sin\varepsilon}\right).$$

For a turn of 0.1 rad that is 13.9% against 3.0% at the crossing, four and a half times as much. The best share falls as $\sqrt{c/\varepsilon}$ rather than as $c/\varepsilon$, so the larger the error, the more it matters where it is pivoted. Pivoting it exactly at the end of the travel does slightly worse again, 18.1% at 0.03 rad, since then half the band hangs past a point the pin never reaches.

The rule for a designer who cannot aim the third slide perfectly is to make its datum, the point it is set from, lie a little short of the end of its pin’s travel rather than at the crossing where the slides meet. The measurements at 7,200 rod angles agree with the closed form at every pivot position tried, to two samples.

What a clearance buys, and what it does not #

Why a hinge works found a three-knuckle door hinge surviving its own overconstraint because the misfit its bore errors create, 0.507 times the error, is smaller than the play already in its pins. A piano hinge is not forty door hinges followed the same comparison along a much longer row. Their test was one comparison, misfit against play.

The third slide splits that test in two. For an error whose misfit is the same at every instant — the parallel offset — it is exactly the hinge’s comparison: play at least as large as the misfit, and the mechanism works for its whole motion. For an error whose misfit changes with the motion — any turn — the comparison has to be made instant by instant, and play buys a share of the motion rather than all or none of it. Which kind an error is depends on the mechanism: the third slide’s misfit grows with its pin’s travel, and that travel is the whole diameter of the fixed circle every half turn. An overconstrained mechanism whose redundant joint travels far from the place its error vanishes will behave like the turned slide, and the only ways to widen its windows are shorter travel, more clearance, or an error pivoted where the travel is slowest.

What the planar model leaves out #

A rigid rod. A real rod and real blocks bend. Past the window a real rod would not lock dead: it would be forced through, bending as the third pin is dragged off the false slide, at a cost the rigid model does not see.

Clearance in the first two blocks. Only the third block was given play. Play in all three would let the rod shift a little in its own plane and trade misfit among the three pins, widening the windows. That is the case a clearance is a link modelled for pins.

Errors in more than one slide. Only the third slide was misplaced. A turn error in the first or second slide changes the double slider’s own motion, and the rolling circle with it.

Still open: a fourth block that fixes the third #

The windows are set by the third pin’s travel along its slide. A fourth block at the point of the rolling circle opposite the third pin travels in exact opposition to it, so its misfit from a slide turned by the same error would be equal and opposite at every instant. A rod on four blocks has a count of minus one, and the question is whether the fourth slide can be placed to cancel the third’s error rather than add to it.

The distinct argument would be that cancellation. The measurement would turn both the third and a fourth slide, the fourth’s pin diametrically opposite, by errors of the same or opposite sign. It would ask whether the free share recovers a whole turn when the turns match, and how fast it falls as they part. If matched turns free the rod, a redundant slide is a way to make a pair of slides self-aligning. If they do not, every added slide can only narrow the windows further, and the count’s minus one is a fair warning after all.