A second slide keeps only the rod

A slide exchanges the crank and the rod took Roberts’s three machines to the limit where a four-bar’s rocker grows into a slide. One cognate went to infinity with the rocker. The other stayed, as the slider-crank with its crank and rod exchanged and the whole machine turned and scaled by the tracing point λ. Of the three orientations Roberts’s figure lets turn — the ones a closed form belongs to — two kept turning, one in each member of each machine, and the rocker’s froze.

That essay ended by growing the crank the same way. With its pivot receding square to its own slide, the crank becomes a second slide and the machine becomes the double slider: two blocks on two straight lines through a common point, joined by a rod. With square slides it is the elliptic trammel, and every point of its rod’s plane draws an ellipse — a conic, with no double point left, because the trammel takes both of the slider-crank’s away as the crank grows. The question was whether a second slide keeps anything of the cognates, and which rotation, if any, is left standing.

One rotation is left, the rod’s, since both of the other members have become lines. The cognates are a stranger answer. The exchange that produced the slider-crank’s one cognate stops producing a machine at all. And the ellipse turns out to have not one or two cognates but two continuous families of them, each machine in them carrying the one surviving rotation.

The machine, and what its pins do #

The machine throughout has slides crossing at 70°, a rod of 2 and a tracing point at $\lambda = 0.3 + 0.35i$ in the rod’s own units, 0.3 of the way along the rod and 0.35 of a rod across it. Put the crossing at the origin, the first slide along the real axis, and write the rod as $B - A = L e^{i\varphi}$.

Each pin is then a single sinusoid in the rod’s angle. Solving the triangle of the two pins and the crossing, the pin on the second slide sits $L \sin\varphi / \sin\alpha$ from the crossing and the pin on the first sits $L \sin(\varphi - \alpha)/\sin\alpha$. So every point of the rod’s plane, $P = A + \lambda(B - A)$, is a sum of $e^{i\varphi}$ and $e^{-i\varphi}$ with two complex coefficients:

$$P(\varphi) = p\,e^{i\varphi} + q\,e^{-i\varphi}.$$

That is an ellipse centred at the crossing, with semi-axes $|p| + |q|$ and $\big||p| - |q|\big|$. The machine here draws one with semi-axes 1.587 and 0.542, tilted to neither slide. Exact because two circles roll drew the square-slide case and found that a point on the rod draws an ellipse whose semi-axes add up to the rod; the general statement is that any point of the plane, on any pair of slides, draws some ellipse about the crossing, and the pair $(p, q)$ says which.

The pair says more than the ellipse does, and everything below comes from reading it.

Why the exchange stops producing a machine #

The slider-crank’s exchange built its cognate from two vectors of the original. The cognate’s crank was λ times the original’s rod, $A' = \lambda(B - A)$, which keeps the rod’s length scaled by |λ| and turns with it. The cognate’s rod was λ times the original’s crank vector, $B' - A' = \lambda A$, and that was a rigid member because the crank pin A stayed at a fixed distance from the pivot.

On a double slider A is a block on a line through the crossing. Its position vector keeps its direction and changes its length, passing through nought twice a turn as the block crosses the other slide. So λA keeps a fixed direction and runs from nothing to 0.981 and back. The exchanged crank still has the constant length $|\lambda| L = 0.922$, and $A' + (B' - A')/\lambda$ still reaches the traced point, to $5 \times 10^{-16}$. What the exchange names is the ellipse written as a turning vector plus a sliding one, which is a true statement and not a machine. No member of fixed shape can be a vector whose length passes through nought.

So the first limit exchanged crank and rod because it still had a crank and a rod to exchange. The second limit leaves one rod and two slides. The rod’s rotation has no partner to trade places with, and the construction that depended on trading them returns a sum instead of a mechanism.

Where the cognates are: the rolling circle #

The coefficient q is the one that matters. The term $q\,e^{-i\varphi}$ is a point going round a circle of radius |q| about the crossing, turning the opposite way to the rod. That point is the centre of the circle the trammel essay found as the moving centrode: a circle of radius |q| rolling inside a fixed circle of twice that radius, with the rod’s plane carried on it. This is the Cardan motion, and it is the rod’s whole motion, since every planar motion is one curve rolling on another. The coefficient p is where the tracing point sits in the rod’s plane, measured from the rolling circle’s centre.

A point $w$ of the rod’s plane, measured the same way, traces $w\,e^{i\varphi} + q\,e^{-i\varphi}$. When $|w| = |q|$ the two terms have equal size and turn in opposite senses, so their sum stays on a line through the crossing, running back and forth along it. The points that draw straight lines are exactly the points of the rolling circle. That is the classical fact, and it says something about double sliders that the trammel’s own description hides.

Any two points of the rolling circle are a double slider drawing the same ellipse. Join them with a rod and put each on a block in the line it traces. The rod’s plane then moves exactly as before, since two of its points are held to the motion, and the tracing point p goes with it. The new machine’s slides lie along $(\arg w + \arg q)/2$ for each pin. Its tracing point, in its own rod’s units, is $\lambda' = (p - w_1)/(w_2 - w_1)$.

The machine’s own pins, the crossing and the rolling circle are therefore one circle seen twice. The crossing is always on the rolling circle, since the rolling circle’s radius equals its centre’s distance from the crossing. The two pins are on it by construction. And the angle the rod subtends at the crossing is the angle between the slides. By the inscribed-angle theorem, then, a rod on a rolling circle of radius r with slides at α has length $2r\sin\alpha$. The original’s circle has radius $2/(2 \sin 70°) = 1.064$.

That fixes what is free, and it is more than a four-bar’s cognates ever leave: three machines there, a continuum here. Pick any slide angle, and the rod’s length follows. Pick where on the circle the pair sits, and the pair of slides turns as a whole about the crossing. Two numbers, and every choice of them is a double slider on the same ellipse. The four drawn over the hero figure have rods of 1.73 and 2.08 on slides at 54° and 77° from the original’s circle, and every other pair of that circle would have done as well.

Two circles for every ellipse #

Nothing in the argument preferred p to q. Write the same ellipse with the rod’s angle reversed, $\varphi' = -\varphi$:

$$P = q\,e^{i\varphi'} + p\,e^{-i\varphi'},$$

and the roles swap. Now $p\,e^{-i\varphi'}$ is the centre of a rolling circle of radius |p|, the tracing point sits at q in its plane, and any two points of that circle are a double slider on the same ellipse. Their rods turn with $\varphi'$, the other way. So every ellipse has two rolling circles. Since the semi-axes are $a = |p| + |q|$ and $b = \big||p| - |q|\big|$, the two radii are $(a + b)/2$ and $(a - b)/2$. On this machine they are 1.064, the original’s own, and 0.522.

The two circles have a familiar pair of members. Put the pins at the two ends of a diameter and the slides are square, since the inscribed angle on a diameter is a right angle. Choose the diameter through the tracing point and the tracing point lies on the rod’s line. On the larger circle the rod is a diameter of $2 \cdot 1.064 = 2.128$, which is $a + b$, with the tracing point between the blocks. On the smaller it is 1.045, which is $a - b$, with the tracing point beyond one block. Both pairs of square slides lie along the ellipse’s own axes. These are the two paper-strip constructions of an ellipse that drafting manuals give, the long strip with the pencil between the marks and the short strip with the pencil beyond them. They are one member of each family, each the member on square slides with the pencil on the strip. The drawing office’s two methods are the two sheets of one construction.

The count agrees with the construction. A double slider placed in the plane has five numbers: the two slides’ directions, the rod’s length, and the two coordinates of λ. An ellipse about a given centre has three. So if the ellipse depends on the five in a generic way, each ellipse is drawn by a two-parameter family, which is what each rolling circle supplies. The ellipse’s three shape numbers, differentiated against the five parameters at 200 random machines, have rank three at every one. Nothing is lost to a hidden degeneracy, so two parameters are left over for each sheet.

The census #

The construction was derived, and derivations of this kind can be wrong in a sign or a factor of two, so each cognate was built from its pins and then forgotten. It was traced as a machine in its own right, by solving its own slides and rod at each angle, and checked against the original’s ellipse rather than against the construction that made it.

The two hundred machines have slides crossing anywhere from 15° to 165°, rods from 0.5 to 3, and tracing points anywhere in the rod’s plane up to 1.2 rods either side. Machines too near a circle or a segment were left out: a circle has only one rolling circle worth the name, and a segment is on the boundary between sheets. For each machine, four random pairs of pins were taken on each circle, kept at least 20° apart. The cognates’ slides cross at anything from 10° to 170°.

Every one of the 1,600 lies on the original’s ellipse to $3 \times 10^{-15}$ of its semi-major axis, and encloses its area to $7 \times 10^{-14}$. The coefficients p and q, computed in closed form, agree with a discrete Fourier transform of the traced points to $3 \times 10^{-15}$. None of these numbers is a tolerance; they are the size of rounding in double precision. The claim is exact.

Only the rod turns, and every cognate carries it #

The slider-crank’s two machines split its two surviving rotations between their members: the original’s crank turned with the cognate’s rod and the original’s rod with the cognate’s crank. A double slider has one member that can turn. Every cognate is also a double slider, so every cognate has exactly one turning member, its rod, and the question is only how the rods’ turning is related.

The cognates on the original’s circle share its rolling plane, since their pins are points of the same moving plane. Their rods are chords of the same rolling circle, fixed in that plane, and they turn exactly with the original’s, keeping the angle between the chords. The cognates on the other circle live in the other plane, which turns with $\varphi' = -\varphi$, so their rods turn one full turn the other way for every trip round the ellipse. The census confirms it by the sign of each cognate’s enclosed area as its own rod turns forward: all 800 on the original’s circle go round the ellipse the same way as the original, and all 800 on the other circle go round it the other way.

So the one rotation is carried by every machine, in both senses. The earlier essay found that a slide freezes a rotation rather than handing it to another member. Here two slides freeze two, and the one left turns in every cognate, forwards in one family and backwards in the other. For a designer replacing one double slider with another that fits the space better, the choice of circle is a choice of which way the rod must turn to draw the ellipse in a given direction.

What the area keeps #

The area formula lost a term at each limit. For the slider-crank it read, in the earlier essay’s notation, the crank’s term, the rod’s term and a mixed integral, with the rocker’s term gone because a slide encloses nothing. Now the crank pin also runs on a line, so its term goes as well, and what is left is the rod’s term and the mixed integral:

$$S = (|\lambda|^2 - u)\,\pi L^2 \;-\; v \oint B \cdot dA.$$

For slides crossing at α the mixed integral has a closed form. The pins are the two sinusoids already written down, and the integral of their product over a turn is $\pi L^2 \cot\alpha$, so

$$S = \pi L^2\,\big(|\lambda|^2 - u - v\cot\alpha\big) = \pi\big(|p|^2 - |q|^2\big).$$

The census checks both the two-term formula and the closed form against the traced area, to $3 \times 10^{-14}$ and $5 \times 10^{-15}$. It also finds that neither pin encloses anything, to $2 \times 10^{-17}$ of πL².

On the rod itself v is nought and the area is the rod’s term alone, $-u(1 - u)\,\pi L^2$ between the pins. In size that is the trammel essay’s ellipse area, π times the product of the two pieces into which the tracing point divides the rod, with the sign saying which way round it goes. Off the rod the two terms move differently on the two sides. The rod’s term gains $v^2$ whichever side the tracing point is on, while the mixed integral changes sign with v, so a line 0.6 to one side of the rod and a line 0.6 to the other give different curves, unless the slides are square and $\cot\alpha$ is nought.

The zeros are where the construction said they would be. $|\lambda|^2 - u - v\cot\alpha = 0$ is a circle in the rod’s plane through $\lambda = 0$ and $\lambda = 1$, the two pins, and through whichever point of the plane is passing over the slides’ crossing: the rolling circle, whose points draw segments and enclose nothing. The line 0.6 to one side crosses it twice. The line 0.6 to the other misses it altogether, and every point on it draws a proper ellipse. The area’s closed form and the cognates’ construction are the same circle seen twice.

What the second slide leaves unmeasured #

Motion, not force. Every cognate draws the same ellipse, but the forces in them differ. A block is hardest to drive where its stroke reverses, and each cognate’s blocks reverse at different moments of the trip round the ellipse, since each pin’s line is a different diameter. How the angle between rod and slide varies across the family, and which cognate transmits best for a given drive, is not measured.

The circle and the segment. A circle is drawn when p is nought, and then the second rolling circle has radius nought: its pins would coincide, and the family collapses. A segment is drawn when |p| equals |q|, and then the two families meet. The census kept away from both, and neither boundary was traced.

Slides that do not cross. Two parallel slides hold the rod at a fixed angle, and the machine translates. It is not the limit of anything here and is left out.

Still open: a third slide that the count forbids #

Every point of the rolling circle runs on a line through the crossing, so a third point of it could carry a third block on a third slide through the same crossing. The rod would then ride three blocks on three concurrent lines. The count says that machine cannot move — five bodies and six joints of one freedom each give $3 \cdot 4 - 2 \cdot 6 = 0$ — and it moves anyway, as every point of the circle keeps to its line.

The distinct argument there would be what makes it move and what stops it. The third slide must pass exactly through the crossing and lie exactly along the direction its pin’s line takes. The measurement would be how a small error in either, an offset of the third slide from the crossing or a turn of its direction, locks the rod. It would also ask how much clearance in the third block buys back how much travel, and whether that trade depends on where on the circle the third pin sits. It is the double slider’s version of the frameworks whose count says rigid and whose geometry says otherwise.