Where a pin becomes a slide

Three positions and a circumcentre is a construction with no failure in it. Take any point of a body that has to occupy three prescribed poses, place it in all three, and the three places it lands are three points; three points have one circle through them, and the circle’s centre is where a fixed pivot has to go. Every point of the body works. The difficulty that essay names is abundance — a two-parameter family of exact answers per dyad and nothing in the problem to choose between them.

There is one exception, stated there and set aside. When the three images happen to fall in a line, no circle passes through them, and the construction has nothing to return. The right reading is not that it has failed but that the answer has moved to infinity: a point that has to visit three places in a line wants to be carried along that line, by a block in a guide, and the dyad it asks for is a slide rather than a crank.

That leaves a question the construction itself never asks. Which points are those? They might be scattered, or confined to some awkward curve, or a set of measure nothing that a designer will never land on by accident. The answer is none of these, and it is tidy enough to be suspicious of. They lie on a circle, and the circle is one a compass could draw before any linkage exists.

A line of algebra that says it must be a circle #

The reason is short enough to give whole, and it is worth giving because it is what makes the figure a check rather than a discovery.

Write a body point as $x$ in the body’s own frame. In pose $i$ it lands at $X_i = R_i x + d_i$, where $R_i$ is the pose’s rotation and $d_i$ its translation. Three images are in a line when the triangle they make has no area:

$$(X_2 - X_1) \times (X_3 - X_1) = 0.$$

Each difference is affine in $x$ — a matrix times $x$ plus a constant — so the whole left side is a quadratic in the two coordinates of $x$. What decides the shape of a quadratic curve is its second-degree part, and here that part is $\big((R_2 - R_1)x\big) \times \big((R_3 - R_1)x\big)$.

The one fact needed is that a difference of two plane rotations is itself a rotation times a scale. A rotation matrix has the form $\begin{pmatrix} a & -b \\ b & a \end{pmatrix}$, a difference of two of them has the same form, and every matrix of that form turns a vector and stretches it without distorting it. So both factors are $x$ turned and stretched, by different amounts, and the cross product of two such copies of one vector is its length squared times a constant — the product of the two stretches and the sine of the angle between the two turns.

A quadratic whose second-degree part is a multiple of $x^2 + y^2$ is a circle. There is no case analysis in that and no special pose: whatever three poses are prescribed, provided no two share an orientation, the slider points form a circle.

Which circle, without solving anything #

A circle is fixed by three of its points, and three are available for nothing.

Every pair of poses has a pole — the point about which a single rotation carries the body from one pose to the other, the same object that where the coupler is turning follows continuously as an instant centre. The body point that sits on the pole $P_{12}$ when the body is in pose one is still there in pose two, because the pole is the point the displacement does not move. So two of that point’s three images coincide, and a triangle with two coincident corners has no area. That body point is a slider point, trivially.

The same holds for $P_{13}$, and for $P_{23}$ — with one change of bookkeeping. The body point that sits on $P_{23}$ is the one that is there in pose two, and in the body’s own frame it is somewhere other than where $P_{23}$ is drawn. Expressed in the frame of pose one it is the classical image pole $P_{23}'$.

So the circle of slider points is the circle through $P_{12}$, $P_{13}$ and $P_{23}'$. The first figure does not draw that circle. It measures, at each point of a grid of 110 by 110 over the moving body, the signed height of the image triangle over its longest side — a length in the drawing’s own units — and contours the measurement at zero. Points taken off that contour and refined onto it by bisection lie on the circle through the image poles to $2 \times 10^{-15}$. The circle through the three poles themselves, which passes through two of the same points and has the same radius, misses them by up to 1.98.

That second number matters more than the first. A check that tested only whether the traced points lay on a circle through $P_{12}$ and $P_{13}$, or on a circle of the right size, would have passed on the wrong one.

A point on the circle, and the guide it gives #

Pick a point on the circle and the construction that used to produce a pivot produces a line instead: the three images lie on it, and a block running along it carries the point through all three poses. The guide’s direction is not a design choice. It is the line the images give, just as a crank’s pivot was the circumcentre they gave.

Dragging the point along the circle shows two things together. The guide turns as the point moves — every slider point has its own line — and the three images stay on it at every stop. And the images crowd together as the point approaches a pole: at $P_{12}$ two of them meet, the line is decided by the third alone, and at the pole itself the direction of the guide is no longer determined by anything. The slider circle is a smooth curve, but the guide it hands out is not defined at the three points that fixed it.

A slide is a different pair from a pin, and the block in the guide has a length is a reminder that it is a different object to build: a guide has to be as long as the stroke plus the block. Here the three images span 2.27 of the line on a body two units wide, which is a respectable guide and not an absurd one.

Near the circle, a pin still works — on a crank as long as one over the miss #

The construction never refuses a point that is merely close to the circle. It returns a circumcentre and a radius and says nothing about either. What it returns is worth measuring.

Move a body point off the circle along the circle’s normal, by a distance $\delta$, and ask the construction for its crank. The image triangle’s area is a smooth function of position with a simple zero on the circle, so it grows in proportion to $\delta$, while the triangle’s sides barely change. A circumradius is the product of the three sides over four times the area, so it runs as $1/\delta$.

Measured over eight distances from 0.3 down to 0.0001, the fitted slope is −0.994 and the product $R\,\delta$ settles to 0.8552, varying by three parts in ten thousand over the last four. At a thousandth of a unit off the circle the construction’s answer is exact, satisfies all three poses, and puts the fixed pivot 855 units from a body two units across.

This is the practical content of the result, and it is sharper than “avoid degenerate points”. A designer choosing coupler points to keep the ground pivots somewhere buildable is not steering away from a few isolated bad choices. There is a whole curve across the body along which the pivots run off to infinity, and a band either side of it in which they are merely very far away, with the width of the band set by how far away is too far: a pivot no more than ten body-widths off requires a point at least $0.855/20 \approx 0.04$ from the circle. The circle is a thing to draw on the body before choosing anything, because it is where the choice stops being a crank.

A slider-crank through three prescribed poses #

With one point on the circle and one anywhere else, the two dyads make a linkage — not a four-bar this time but a slider-crank, a crank dyad from the ordinary construction and a slide from the circle, joined by the rod between the two body points.

The claim that this linkage reaches its poses is not taken from the construction. Each pose is set up separately as a mechanism with three constraints — the crank pin at its angle, the rod at its length, the block on a fixed line — seeded at the pose and solved by Newton’s method, for the same reason the four-bar version is checked one pose at a time: three prescribed poses are three questions, not a sweep. All three are reached to $2 \times 10^{-15}$.

The same machine with its guide moved sideways by a hundredth reaches none of them and misses by 0.018. The miss is the more informative of the two numbers: it says the verification is capable of failing, and it says by roughly how much a guide mounted a hundredth out will spoil every pose, which is the tolerance a builder would actually need to hold.

What the verification does not say is whether the slider-crank can travel from one pose to the next without its rod passing through a dead centre or its crank changing branch. That is the defect exactly right and unbuildable is about, and a slider-crank has the same two assemblies a four-bar has. Nothing about choosing the slide from a circle protects against it.

The same question with the slot on the body #

A slide has two halves, and the question so far has put the moving half on the body: a point of the body runs along a line of the frame. The other way round is just as buildable. Fix a pin to the frame and cut a slot in the body, and ask which pins a slot can pass over in all three poses.

A fixed pin $X$ lies on a line of the body in every pose when its three positions as seen from the body are in a line. Seen from the body, the frame is what moves, through the three inverse poses. That is the slider question with the two planes exchanged — the same move that turns one chain into four mechanisms and turns function generation into motion generation — so the answer is again a circle through three poles. In the fixed frame the poles need no translating, and the circle is simply the pole triangle’s own circumcircle.

Built this way, the slot in each pose passes through the pin to $10^{-15}$. A pin taken from the other circle — the circle of slider points, carried into the fixed frame — misses at least one of the three slots by 0.55 or more. So the two circles answer different questions and cannot be swapped, which matters because they are easy to confuse: each is the circumcircle of a triangle of poles, and on the prescribed poses here they are the same size.

Two circles of one radius, and why #

The equal radii are not a coincidence of these poses. The image-pole triangle and the pole triangle share two corners, $P_{12}$ and $P_{13}$, and the third corners are mirror images of each other in the side that joins those two. That is a classical property of the image pole, and it is measured rather than quoted: reflecting $P_{23}$ in the line $P_{12}P_{13}$ lands on $P_{23}'$ to $10^{-15}$, while reflecting in either of the other sides misses by 6.7.

Two congruent triangles have congruent circumcircles, so the radii agree — to $7 \times 10^{-15}$, at 8.7667 — and the two circles are mirror images of each other in the same line.

Drawn at full size the picture also explains why the first figure showed only a gentle arc. The circles are four body-widths across. What a designer sees on a body two units wide is a short piece of each, nearly straight, crossing the body close to where the poses overlap. That is typical rather than special: poles of modest rotations lie near the body, but not all three of them do, and one far-off pole is enough to make the circle large.

Three positions brought together #

The strongest test of a claim like this is to push it somewhere its author did not have in mind, and a limit does that.

Take the three poses not from a designer’s specification but from a running four-bar: the coupler at crank angles $60° - h$, $60°$ and $60° + h$, for a linkage with ground 4, crank 1, coupler 3.5 and rocker 3. As $h$ shrinks, three positions of one body point collapse onto one position and two derivatives. A point whose three images are in a line becomes, in the limit, a point whose path has no curvature at that instant — a point momentarily travelling straight.

Those points are already known to form a circle. The circle of points going straight is the inflection circle, and the curvature field computes it from the motion’s first and second derivatives with no positions in the calculation at all. If the slider circle is what it is claimed to be, it has to converge on that circle.

It does, and at a definite rate. The gap between the two centres falls from 0.059 at $h = 0.2$ to $3.5 \times 10^{-5}$ at $h = 0.005$, against a circle of radius 18.08, and the measured order between successive spreads is 2.04, 2.01, 2.00, 2.00 and 2.01. It is the square of the spread rather than the first power because the three angles are symmetric about the instant they are compared at, so the odd terms of the error cancel; taking the three at $60°$, $60° + h$ and $60° + 2h$ instead gives a gap that closes only in proportion to $h$.

Two quite different routes therefore meet. One knows about finite positions, circumcentres and poles and nothing about derivatives; the other knows about derivatives and nothing about finite positions. They agree to the order that a Taylor expansion says they should, which is a stronger statement than agreement at one spread. It is the same shape of result as four positions brought together, where Burmester’s circle-point cubic closes on the cubic of stationary curvature — one order of contact lower, and one figure of the construction simpler.

What the circle does and does not decide #

It decides where the kind of dyad changes. Off the circle a body point wants a crank; on it, a slide; near it, a crank long enough to be a slide in everything but name. That makes the circle the first thing worth drawing in a three-position problem where a slider is acceptable, because it shows the designer where the one-parameter family of slides is before any choice about cranks is made.

It does not decide whether a slide is wanted. Nothing in three poses prefers a slide to a crank, and nothing here does either. A slide costs a guide and a guide has a length; a crank costs a bearing and a pivot location. Which is cheaper is a question about the machine and the frame it is mounted in.

It leaves the count unchanged. Three poses give each crank dyad a two-parameter family and each slide a one-parameter family. That is the right difference: a slide has one fewer number to spend than a crank, because a line through a known point needs only a direction where a circle needed a centre. How many points may be prescribed derives such counts in general, and the slide is the case where the arithmetic and the geometry visibly agree — a curve’s worth of answers where a crank had a plane’s worth.

It does not survive a fourth pose as a circle. Four images in a line is two conditions on two coordinates, so for four poses the slider points are generically finitely many, not a curve. That is where Burmester’s cubic meets the line at infinity, and what the fourth position costs is where the cubic is drawn.

And it says nothing about order or branch. A slider-crank built from a point on the circle reaches each prescribed pose, and whether it reaches them one after another without being taken apart is a separate question that the circle is no help with.

Still open: the slide that a fourth pose leaves behind #

With four prescribed poses a crank dyad needs its body point on Burmester’s circle-point cubic. A slide needs its point’s four images in a line, which is two conditions on the two coordinates of the point, and so it is not a curve any more but a finite set.

Its distinct argument would be those points counted and found. The circle-point cubic has to pass through them — a slide is a crank whose pivot is at infinity, and every point whose four images are collinear is a point whose four images lie on a circle of infinite radius — and a real cubic meets the line at infinity in one or three real points, so the count is either one or three depending on the poses. Two things would come out of it. Whether a four-pose problem ever admits no slide at all, which would follow if the only real meeting at infinity is a point with no finite body point attached to it; and whether the slider points are always the places where the circle-point cubic’s branches run off the drawing, which is what a designer tracing the cubic by hand would see without knowing it. The instrument is already here: the circle of slider points for each of the four triples of poses, since a point with four collinear images lies on all four circles at once.