A slide exchanges the crank and the rod

A closed form belongs to a rotation read the three cognate four-bars of one coupler curve through the rotations of the figure that Roberts’s construction draws. The figure has exactly three orientations that can turn independently. Each machine gives one of its three moving members to each. The three closed-form terms of the area formula belong to those rotations rather than to columns of the formula, so the three machines carry the same three values, each on the member that belongs to its rotation.

The essay ended on the slider-crank. It is the limit of a four-bar whose rocker has grown without bound, and in that limit two things change. One of Roberts’s machines goes to infinity. And the rocker’s pin, which carried one of the three closed forms, runs on a straight line that encloses nothing. The open question was whether the rotations survive. If the rocker’s rotation disappears with the rocker, something has to happen to its share of the area.

What happens is simpler than either possibility the question named. One cognate leaves. The one that stays is the original machine with its crank and rod exchanged. The third rotation is not carried by anything: it is frozen, and the only four-bars that ever turned it are among those that do not survive the limit.

One cognate, drawing the same curve #

The machine throughout is a slider-crank with a crank of 1, a rod of 3, its slide 0.5 from the crank’s pivot, and a tracing point $\lambda = 0.4 + 0.5i$ in the rod’s own units: 0.4 of the way along the rod and 0.5 of a rod across it.

The cognate on the right has a crank of 1.921, a rod of 0.640 and a slide 0.320 from the crank’s pivot, turned by 51.3° from the original’s. It is not a scaled copy of the original, since its crank is three times its rod where the original’s rod is three times its crank. It draws the same two ovals. And the colours say how: the original’s crank is rotation 1, which the cognate gives to its rod; the original’s rod is rotation 2, which the cognate gives to its crank; the slide is rotation 3 in both, and it does not turn.

The numbers under the names are the closed-form area terms of the oval being drawn. The original’s crank goes all the way round, so its crank term is $(1 - u)\pi a^2 = 1.885$. The cognate carries the same 1.885 on its rod. It is the same value on a different member, exactly as in the four-bar’s triple, but with two machines instead of three and one rotation that can never switch on.

The exchange #

The cognate can be written down in one line, and the line explains everything the figure shows. Put the crank’s pivot at the origin and write points as complex numbers: A is the crank pin, B the slider pin, and the tracing point is $P = A + \lambda(B - A)$. Then the cognate’s pins are

$$A' = \lambda(B - A), \qquad B' = \lambda B,$$

with its tracing point at $1/\lambda$ along its rod. That is Roberts’s parallelogram: O, A, P and A′ are its four corners, since $A' - O = \lambda(B - A) = P - A$.

The cognate’s crank vector is $A' = \lambda(B - A)$, the original’s rod vector multiplied by λ. Its rod vector is $B' - A' = \lambda A$, the original’s crank vector multiplied by λ. So the cognate is the original machine with the two vectors exchanged, and then the whole turned by arg λ and scaled by |λ|. Multiplication by λ is a rotation and a scaling about the crank’s pivot. It carries the slide, the line on which B runs, onto a line on which B′ runs, turned by arg λ and |λ| times as far from the pivot. That is where the cognate’s lengths come from: crank |λ|b, rod |λ|a, offset |λ|e.

Its tracing point checks in one step: $A' + (1/\lambda)(B' - A') = \lambda(B - A) + A = P$. The two machines hold the same point whenever the cognate’s pins are the exchange of the original’s.

There is a condition the exchange has to satisfy that is not automatic. The cognate is a machine in its own right: its crank of |λ|b turns about the pivot and its slider pin runs on its own slide. For it to draw the curve, every configuration it can reach, with its own lengths, has to be the exchange of a configuration of the original. The census below solves the cognate from its lengths alone and checks exactly that.

Where the other cognate went #

The slider-crank is reached from a four-bar by growing the rocker. With a rocker of length c pivoted c below the slide, the four-bar’s rocker pin moves on a circle that touches the slide, and Roberts’s construction gives that four-bar two cognates. The double point a slide cannot keep followed the same family and found one double point leaving linearly in c. Here the whole of one machine does.

The first cognate keeps the crank’s pivot. Its crank is |λ|b and its coupler |λ|a whatever c is, and its rocker is |λ|c, pivoted at the third pivot λ times the rocker’s pivot. As c grows, that rocker’s circle flattens onto a line, and the solid line measures how far the first cognate’s rocker pin is from the exchanged machine’s slide at worst over a turn of the crank. It falls with slope −1.001, from 2.9 at a rocker of 3 to 1.4 × 10⁻³ at a rocker of 10,000. The first cognate becomes the exchanged slider-crank.

The second cognate has both of its ground pivots on the side that recedes: the rocker’s own pivot and the third pivot. Its coupler is $|1 - \lambda|c$ long, 0.781 times the rocker, and its crank pins stay exactly a rocker’s length from the tracing point it shares with the others. That is not a fit. One of the construction’s parallelograms has the rocker as a side and the line from that crank pin to the tracing point as the side opposite, so the dashed line is c itself. The whole machine except its tracing point goes to infinity with the rocker.

In the four-bar’s vocabulary the machine that leaves is the one that carried rotation 1 on its rocker. For a crank-rocker’s triple, a closed form belongs to a rotation named the three members of rotation 1 as the original’s crank, the double rocker’s coupler and the rocker-crank’s output. With the slide, the rocker-crank is the one that goes. The double rocker is the one that stays, and it becomes a slider-crank whose rod turns.

A census over every kind #

Four kinds of slider-crank sorted slider-cranks by which member can go all the way round. The crank turns when $b \ge a + |e|$. The rod turns when $a \ge b + |e|$, and the crank then swings in two separate arcs. In the two bands between, neither turns. So the census draws slider-cranks from all of them.

Crank and rod are drawn from 0.2 to 3, the offset up to 1.5 either side, and the tracing point anywhere in a band off the rod. Machines within 3% of a wall between kinds are left out, since the census is about kinds and a machine on a wall is two at once. For each machine the cognate is built from its own lengths and slide — crank |λ|b, rod |λ|a, offset |λ|e, slide turned by arg λ — and solved and traced on its own, with no reference to the original’s configurations. Each circuit is paired with the cognate circuit that draws the same oval, and the columns of the table are four separate checks.

Carried back. Every configuration the cognate reaches is undone by the inverse exchange, $R = A'/\lambda$ and $B = B'/\lambda$, and the result is checked against the original machine: its crank pin on the crank’s circle, its slider pin on the slide, its rod the rod’s length, and its tracing point the cognate’s. The worst residual over all 376 circuits is 9 × 10⁻¹⁵. So the cognate reaches nothing the original does not, which is the condition the exchange could not guarantee by itself.

By rotation. Each circuit is read the way round that makes its traced area positive, so that the two machines go round the shared oval in the same sense. Then the original’s crank term against the cognate’s rod term, and the original’s rod term against the cognate’s crank term, agree with their signs to 2 × 10⁻¹⁵ of the largest. Every circuit’s whole turns match by rotation too: when the original’s crank goes round, the cognate’s rod does, 376 circuits out of 376.

By column. Compared crank with crank, the two machines disagree by the whole value, a relative difference of 1.00, wherever anything turns. Only in the bands, where no member goes round and every closed form is nought, do they agree, and they agree only because both are zero.

Mixed. The fourth term of the area, $-v\oint B \cdot dA$, has no closed form, and it agrees between the two machines to 4 × 10⁻¹¹. That too is a consequence of the exchange: the cognate’s term is $-v' \oint B' \cdot dA'$, where $v' = -v/|\lambda|^2$, $B' = \lambda B$ and $dA' = \lambda\,dR$, and it comes back to the original’s once $\oint B \cdot dB = 0$ is used, as it is for any closed loop.

The cognate column is the table’s other result. Every machine whose crank turns has a cognate whose rod turns, 74 of 74. Every machine whose rod turns has a cognate whose crank turns, 62 of 62. Every one of the 104 in which neither turns has a cognate in which neither turns.

Two rotations turning together, and one that cannot #

The census compares closed forms, which are only ever a whole turn or nothing. The rotations themselves can be watched through a circuit.

The original’s crank climbs through one full turn, and the cognate’s rod, dotted, lies on top of it for the whole circuit. The original’s rod rocks between about 150° and 190° and comes back, and the cognate’s crank lies on top of that. The slide is a flat line. The two machines’ members of each rotation differ by less than the resolution of matching a sampled configuration, a few hundredths of a degree.

The reason is the same parallelogram as in the four-bar’s triple. The cognate’s crank, from O to A′, is a side of the parallelogram opposite the original’s rod from A to P, so the two stay parallel. The cognate’s rod, from A′ to B′, is λ times the original’s crank, so it keeps the fixed angle arg λ to it. Two orientations can turn, and each machine gives one member to each.

The third orientation of Roberts’s figure was the rocker’s, with the first cognate’s rocker and the second cognate’s coupler turning with it. In the limit the rocker’s direction is fixed square to the slide, the first cognate’s rocker has become its slide, and the second cognate has gone. There is nothing left to turn. The rotation is not handed to some other member. It stops.

What happened to the rocker’s share #

The question left open was what carries the rocker’s share of the area if its rotation disappears. The area formula for a slider-crank answers it directly. Of the four-bar’s four terms, the rocker’s is $u\,S_B$, with $S_B$ the area the rocker pin encloses. A slider pin runs back and forth on a line and encloses exactly nothing, so the term is nought in every slider-crank and in its cognate. A slider-crank’s tracing point encloses

$$(1 - u)\,S_A \;-\; (u - |\lambda|^2)\,\pi b^2 N \;-\; v \oint B \cdot dA,$$

three terms instead of four, with $S_A$ the crank pin’s enclosed area and N the rod’s whole turns. The census computes all of them from the traced pins and finds their sum equal to the traced area of the tracing point to 7 × 10⁻¹².

So there is no share to carry, and there never was on the machines that survive. The four-bar’s rocker term is switched on only when the rocker goes all the way round. The two four-bar kinds in which it does are the rocker-crank, which turns rotation 3 alone, and the double crank, which turns all three. Eight kinds of four-bar and the slider-crank’s own classification put both of them among the four kinds that vanish when the rocker grows: their first sign is negative, and it turns positive on the way to the limit. A rocker cannot go round a circle whose radius has gone to infinity. Every surviving kind already had nothing on that rotation.

The kinds, as a reflection #

The exchange gives the cognate’s kind before anything is traced, and it gives it as a picture.

Measure every slider-crank by $a/(a + b)$ across and $e/(a + b)$ up. Every machine that assembles lies in a square, since the rod can reach the slide only if $|e| \le a + b$. The crank turns in the left triangle, $|e| \le b - a$, and the rod turns in the right one, $|e| \le a - b$. The bands above and below are the two kinds in which neither turns.

The cognate has crank |λ|b, rod |λ|a and offset |λ|e. The common factor |λ| cancels from both coordinates, so the cognate sits at $b/(a + b)$ across and $e/(a + b)$ up: the mirror image in the vertical line through the middle of the square. The two triangles are exchanged, and each band is carried onto itself. The tracing point decides the cognate’s size and the angle of its slide. It has no say in its kind.

That settles a practical question in one look. A motor turns a crank. A slider-crank whose crank turns can be driven from it, and its cognate, in the right triangle, has a crank that swings in two separate arcs and can never be driven through a turn. A cognate is usually wanted in order to move the machine’s pivots or change its size while keeping the curve, as three machines, one curve did for four-bars. For a slider-crank the only cognate there is gives that up. It can still be driven from its slider, as an engine is, through the two dead points where its rod lines up with its crank.

The four-bar’s triple was more generous. The kind is decided before the lengths are found that a crank-rocker’s triple has two machines a motor can drive through a turn, the original and the rocker-crank. The rocker-crank is precisely the cognate that leaves with the rocker, so a crank-driven slider-crank is left with one motor-drivable machine for its curve: itself.

Two routes, and what each one proves #

The two routes here do different work, and it is worth separating them.

The exchange is algebra, and it proves the terms match. Substituting the cognate’s lengths and tracing point into the potential closed forms, per whole turn and without π, the crank term $(1 - u')\,|\lambda|^2 b^2$ becomes $(|\lambda|^2 - u)\,b^2$, which is the original’s rod term exactly. The rod term $-(u' - |\lambda'|^2)\,|\lambda|^2 a^2$ becomes $(1 - u)\,a^2$, the original’s crank term exactly. Checked on two thousand random machines without tracing anything, the two agree to 9 × 10⁻¹⁵. That is the same substitution the four-bar’s essay made for its first cognate, with the rocker’s column struck out.

The census is kinematics, and it proves the right members turn. Nothing in it assumes the exchange: the cognate is built from lengths and a slide, solved at its own crank angles, and traced round its own circuits. What the census adds is the part the algebra cannot see: which members go round, whether the cognate reaches configurations the original does not, and whether its circuits correspond one to one. On all three it agrees with the exchange, on every circuit drawn.

What this does not settle #

Which way round each machine goes. The census reads every circuit anticlockwise before comparing, and with that done the signed terms agree. Whether a cognate driven forwards goes round the shared oval in the same sense as the original driven forwards, which is what a designer replacing one with the other would need, is a separate question about the two machines’ input directions and is not measured here.

The machines on the walls. Machines within 3% of a wall between kinds were left out. On a wall a member reaches exactly square to the slide at the end of its swing, and the census’s circuits would have a corner. The exchange holds there as everywhere, but nothing was traced.

The second cognate’s limit as a machine. Its pins recede exactly as fast as the rocker, so it has no finite limit. Whether some rescaled version of it survives, with its pivots at infinity treated as directions, is not examined.

Still open: whether a second slide keeps anything #

Grow the slider-crank’s crank the same way, with its pivot receding at right angles to its own slide, and the machine becomes a double slider: two pins on two lines, the elliptic trammel, whose rod points draw ellipses and whose meetings with the line at infinity are already all away from the circular points. The exchange says what happens to its cognate before anything is computed. The crank has become a slide, so the cognate’s rod, which is λ times the crank, becomes a member of fixed direction too.

The distinct argument there would be the trammel’s own cognate: whether it exists, whether it is the trammel with its two slides exchanged, and which rotation, if any, survives a second limit. A double slider’s rod can turn all the way round and nothing else can, so the question is whether rotation 2 is the last one standing. The area formula would lose the crank’s closed form as it lost the rocker’s, and the ellipse’s area would be the rod’s term and the mixed integral alone. The measurement would be the same two routes: the exchange taken to its limit, and a census of trammels solved on their own.