Three rotations, and four benches

One chain, four mechanisms names the four machines the slider-crank chain produces when each of its links is bolted to the bench in turn: the engine, the Whitworth quick-return, the oscillating-cylinder engine and the hand pump. Four names, four centuries, one chain.

Four kinds of slider-crank then classified one of them. Carrying the eight regions of four-bar length space into the limit where the output pivot runs off to infinity, it found that four of the eight survive and that the two sums deciding which are $b - a + e$ and $b - a - e$, with $a$ the crank, $b$ the rod and $e$ the offset. It ended with a question: each inversion permutes the three sums and flips some of their signs, so does each inversion have its own four kinds, a different selection from the eight, or the same four relabelled?

The answer is the third, and the reason for it is a fact about the chain rather than about the limit: a slider-crank chain has three relative rotations, not six.

The whole chain in one equation #

Put the crank’s pivot at the origin, let $\theta$ be the crank’s angle and $\varphi$ the rod’s, measured from the crank pin to the slider pin. The slider pin is then at $a e^{i\theta} + b e^{i\varphi}$, and the only thing the slide asks of it is that its horizontal coordinate be the offset:

$$a\cos\theta + b\cos\varphi = e.$$

That is the chain’s entire configuration space. It is a curve on the torus of two angles rather than a function of either, and the slider’s position along its guide, $y = a\sin\theta + b\sin\varphi$, is determined once the two angles are.

Two things about that equation are worth stating before anything is computed. It contains no ground — nothing in it says which link is bolted down — so a classification read off it cannot depend on the choice. And it is symmetric in the two terms: swapping $(a, \theta)$ with $(b, \varphi)$ leaves it unchanged. Whatever is true of the crank with the rod’s length in the condition is true of the rod with the crank’s.

The panels in the figure above are that square of angles, with the curve tracked by arclength on the equation alone — at each step the tangent to the constraint is taken, a step is made along it, and the step is pulled back onto the curve by Newton on the one residual. Nothing in the tracking knows which region the chain is in, which angle is an input, or that there is such a thing as a slider.

Six pairs, and one that cannot turn #

Four links make six pairs, and each pair has a relative rotation. That is where the count of inversions usually stops, and it is where this essay starts, because the six are not six.

The block and the frame are joined by a slide. A prismatic joint permits no relative rotation at all, so that pair’s rotation is identically nought — not small, not restricted to an arc, but zero at every configuration of every slider-crank ever built. Five pairs are left, and each of the five is $\pm\theta$, $\pm\varphi$ or $\pm(\varphi - \theta)$.

So there are three rotations in this chain. The crank against the frame, the rod against the frame, and the rod against the crank — and the third is the difference of the first two, so even three overstates the independence: two numbers fix everything, and one equation ties them, which is why the chain has one freedom.

An inversion does not create a rotation. It bolts down a link and puts the motor at one of that link’s two joints, and the rotation at that joint is one of the three. Four inversions, two joints each, eight motor positions, and all eight are drawn from a pool of three.

The three triples #

A quantity that is an angle on a closed circuit either comes back to where it started or comes back a whole number of turns later, so each of the three has a winding number on each circuit, and the winding is an integer. Followed round every circuit of fifteen hundred chains with random lengths and offsets, the triple of windings is one of exactly three things.

A chain with $b - a$ larger than $|e|$ has two circuits, and on each of them the crank makes one turn, the rod makes none and the rod-against-crank makes one the other way. A chain with $a - b$ larger than $|e|$ has two circuits with the rod turning and the crank not. A chain with $|b - a|$ smaller than $|e|$ has a single circuit on which nothing turns at all.

There is no fourth triple and there cannot be. The crank goes round exactly when the rod can span every horizontal gap the crank opens, which is $b \ge a + |e|$; the rod goes round exactly when the crank can span every gap the rod opens, which is $a \ge b + |e|$; and no chain with a nonzero offset satisfies both. So the three cases are exhaustive, and the argument for that is the equation’s own symmetry rather than the census.

The circuit counts are a second check on the same thing arriving from the four-bar. The two regions where something turns have two circuits each and the two where nothing does have one, which is exactly what the eight-kinds table says of the four regions these came from: a Grashof-like region has two assemblies and a triple-rocker region has one.

What a circuit is, here #

A circuit in this picture is a connected piece of the curve, and the number of them is the number of assemblies a builder has to choose between. That is the same object the branches essay identified in the four-bar: an assembly branch is a connected component of the configuration space and not a sign in a formula, and two configurations can be joined by a motion exactly when they lie on the same component.

Drawn on the two angles the distinction becomes something to look at rather than something to compute. A chain whose crank turns has two curves running right across the square, one for each way of assembling the rod; a chain where nothing turns has a single closed loop that does not go round the square in either direction, so its two apparent assemblies are joined and are one circuit after all. The space of configurations makes that argument in general; here it is the difference between two panels of the first figure.

The winding is the extra thing this drawing has that a branch count does not. Two chains can both have two circuits and differ in which angle runs round them, and that difference is the whole of what separates the engine’s region from the hand pump’s.

Four benches, and what each of them can see #

With the three windings in hand the inversion table writes itself. Each inversion’s grounded link has two joints; each joint carries one of the three rotations; the motor at that joint drives continuously exactly when that rotation winds.

Ground the frame and the joints are the crank’s pivot, carrying $\theta$, and the slide, carrying nothing. That is the engine, and it has a rotating input in one region only — the familiar condition that the rod be longer than the crank plus the offset.

Ground the crank and the joints are the crank’s two pins, carrying $\theta$ and $\varphi - \theta$. That is the Whitworth quick-return, and it has a rotating input in two of the four regions, because two different rotations are available at its bench.

Ground the rod and the joints carry $\varphi - \theta$ and $\varphi$: the oscillating-cylinder engine, again with two chances. Ground the block and the joints are the slider pin, carrying $\varphi$, and the slide, carrying nothing: the hand pump, with one chance and it is the one the engine does not have.

Read down a block and the pattern is plain. In the crank-turning region the engine can be driven, the Whitworth can be driven two ways, the oscillating cylinder one way and the hand pump not at all. In the rod-turning region the same four counts appear in the opposite order. In the two regions where nothing turns, nothing can be driven continuously anywhere: all four machines are things that rock.

That symmetry between the first and second blocks is the equation’s symmetry again. Swapping the crank’s length with the rod’s swaps $\theta$ with $\varphi$, which swaps the frame with the block and the crank with the rod — and those are precisely the two pairs of inversions that trade places.

The reading a designer would take from that block is short. A machine that must be driven by a motor turning at a steady rate has four benches available and the region decides which of them will take one. If the rod is the long member, the bench is the frame or the crank; if the crank is, it is the block or the rod; and if neither exceeds the other by more than the offset, no bench will take a motor and the chain has to be driven by something that reciprocates. The offset slider-crank is usually met with the first of those assumed, which is why the other three read as separate inventions.

Grashof’s count, after the limit #

There is a classical count of inversions that this reproduces without being told to.

A four-bar chain satisfying Grashof’s inequality has a fully turning grounded member in three of its four inversions and none in the fourth; a chain failing it has none in any of the four. The inequality is stated about the shortest bar, and in the limit the shortest bar is not what decides anything — the ground and the output have both gone to infinity, and the surviving test is $|b - a|$ against $|e|$.

Three, three, none, none. The count survives the limit intact even though the quantity it is usually stated about has left the mechanism. That is worth more than a tidy coincidence: it says the Grashof count is a statement about how many of a chain’s rotations wind, which is a property of the chain, and the shortest-bar test is one way of computing it for four finite bars rather than what it is about.

Grashof’s rule, swept rather than quoted, is the four-bar’s version of this measurement, and the two can be set side by side: there the input’s turn is measured by driving the crank and counting the angles that assemble, here by following a curve and counting turns. Neither is told the other’s answer.

It is also where the double rocker’s coupler comes back. A four-bar in the double-rocker region has neither grounded member going round and a coupler that does; here the double-rocker region is the one where the rod — the coupler — turns fully while neither the crank nor the block does. The same fact is what gives a double rocker a coupler-turn term in its enclosed area and what gives the slider-crank chain a Whitworth.

The same two sums, in different slots #

One thread from the question is still loose. Re-grounding a four-bar reads the same loop from the next link round, which takes $(g, a, b, c)$ to $(a, b, c, g)$ and so takes the three sums to $(-T_3, -T_2, T_1)$. Applied four times that is the identity, as it must be. So each inversion’s own three sums are different expressions in the four lengths, and the question was whether they cut length space up differently.

They do not. The limit sends $g$ and $c$ to infinity together, so $T_1$ runs away with them and $T_2$ and $T_3$ stay put. Under the permutation the infinite sum alternates between the first slot and the third, and the two finite ones come back as $\pm T_2$ and $\pm T_3$ in the other two — the same two quantities in every row, differing by a sign and by which slot they are in.

The middle slot is never the infinite one, and that is not an accident of where the cycle starts. $T_2 = g + b - a - c$ holds the ground and the output with opposite signs, and the limit grows those two together, so $T_2$ is finite whichever link is bolted down. One of the three sums is immune to the limit by construction.

So the four inversions are classified by one pair of expressions, and the answer to the question is that each inversion’s kinds are the same four relabelled — with the relabelling being a sign and a position rather than a different selection from the eight.

Where the sums and the windings meet #

The two halves of this can be put together in one sentence, and it is the sentence the essay exists for. The windings are a property of the chain because the closure equation has no ground in it. The sums are a property of the chain because the limit leaves the same two of them whichever link is grounded. Those are two independent routes to one statement — that a slider-crank chain has one classification and not four — and they share nothing past the four lengths: one follows curves on a torus and never writes a sum, the other rewrites three expressions and never solves anything.

What an inversion decides is not what the chain does but what a motor at its bench can see. That distinction is invisible while only the engine is being classified, because the engine has a rotation at one of its two ground joints and nothing at the other, so its bench sees exactly one of the three and the classification looks like a property of the mechanism. The Whitworth is the case that separates them: two rotations at its bench, so it can be driven in two regions where the engine cannot be driven in one.

What this leaves out #

The slide is square to the ground line. That is what the limit $c = g - e$ produces, and an inclined slide is a different limit along a different line in length space. Whether the two surviving sums are still $b - a \pm e$ with $e$ measured square to the slide is the question the essay below left open and this one does not answer.

The two joints of a bench are not equally useful. The table counts a joint as drivable when its rotation winds, and says nothing about the torque needed to do it or about the transmission angle at which the force arrives. A Whitworth driven at the pin rather than at the pivot is a different machine to build even though both entries read the same here.

Nothing here is a time ratio. The table says which inversions can be driven round, not how unevenly. A Whitworth is used because its output’s two strokes take unequal times, and the size of that inequality is a continuous quantity inside a region which no winding number can see. The quick-return essay computes one; this computes none.

The census does not sample the walls. Chains within four per cent of $b = a \pm e$, or of the edge of assembly, are redrawn. On a wall the two circuits of a turning region meet, the winding of a circuit through a change point is not defined by the tracking used here, and what the inversion table should say there is not examined.

Forces are absent, and they are where three of the four names come from. An oscillating-cylinder engine is chosen because the cylinder’s rocking works the valves, a hand pump because the operator’s arm is the frame. Nothing in a winding number says which of a chain’s four benches is the useful one.

Still open: the six-bar’s own pool of rotations #

The count that did the work here was six pairs, one of them nought, five rotations, three quantities. A six-bar has six links and fifteen pairs, and its inversions are six rather than four. The same question can be asked of it: how many distinct relative rotations does a given six-bar chain have, how many of its fifteen pairs are prismatic and therefore nought, and is the pool again much smaller than the pair count suggests?

Its distinct argument would be that count made for the two six-bar chains — Watt’s and Stephenson’s — with the windings followed on their own closure equations as they are here, and the inversion table built from the pool rather than from six separate classifications. Two things would come out of it that this essay cannot produce. Whether the Grashof count generalises, which for a six-bar would be a statement about how many of its six inversions have a turning input; and whether the pool of distinct rotations is small enough that a six-bar’s inversions are also readings of one classification, or whether the extra loop is exactly what makes them genuinely different mechanisms.