The slide a fourth pose leaves

Where a pin becomes a slide found which points of a moving body want a slide rather than a crank when three poses are prescribed. Three-position synthesis gives every body point a fixed pivot, the centre of the circle through its three images, except when the three images are in a line. Then the pivot is at infinity, and the dyad that point wants is a block running in a straight guide. The points with that property lie on one circle, the circle through the three image poles, which one line of algebra predicts and a contour of a measured length draws.

A fourth pose asks each point for four images in a line. That is two conditions on the two coordinates of a point, so the answer is no longer a curve but a finite set of points. What the fourth position costs found the crank’s version of this: a fourth pose shrinks the plane of usable pivots to Burmester’s cubic curves. The slide’s version was left as a count to be made, with a prediction attached. A slide is a crank whose pivot is at infinity, a real cubic meets the line at infinity in one real point or three, so there should be one slider point or three depending on the poses.

There is exactly one, always. The prediction’s arithmetic was right, and one fact about the cubics took two of the three points away.

Two circles and a shared point #

Four poses can be split into triples four ways, and each triple has its slider circle. A body point with four images in a line has any three of them in a line, so it lies on all four circles at once. The count is where they all meet.

Take the triples (1, 2, 3) and (1, 2, 4), which share poses 1 and 2. Both of their circles pass through one particular body point: the one that sits on the pole $P_{12}$ in pose one. That point is also at $P_{12}$ in pose two — that is what the pole of two poses is — so its first two images coincide, and two coinciding points and any third are trivially in a line. So the two circles share that point, and two circles through a common point meet in exactly one other.

Call the other meeting Q. Its first three images are in a line, because it is on the first circle. Its first two and fourth images are in a line, because it is on the second. Images 1 and 2 are two different points, since Q is not on the pole, so both lines are the line through them, and all four images are on it. Q is a slider point, and it is the only one: any slider point is on both circles, and the only other point on both is the pole point, whose images 3 and 4 are not in general on the line of its images 1 and 2.

In the figure, the two solid circles are the triples through poses 1 and 2, and they meet at the small dot on the pole point and at Q. The two dashed circles are the other two triples, and they pass through Q and not through the pole point. On the right, Q’s four images lie on one line to 3 × 10⁻¹⁶ of its length. A block on the body at Q, running in a straight guide along that line, carries the body through all four poses.

The construction is compasses and nothing else. Each slider circle is the circle through three image poles, each pole is the intersection of two perpendicular bisectors, and Q is where two circles cross. Five positions needed two cubics intersected, which is algebra rather than construction. Four positions, for a slide, need none.

Why not one or three #

The prediction reasoned from Burmester’s other curve. With four poses, a crank’s fixed pivot has to lie on the centre-point cubic. A slide is a crank whose pivot has gone to infinity, so the slides should be where the centre-point cubic meets the line at infinity. A real cubic meets any line in three points counted properly, and complex ones come in pairs, so one or three of them are real.

That is correct for a general cubic. It is wrong for this one, because this one is not general.

The cubics were not taken from a formula. Each is the determinant that says four points lie on one circle — four rows of |X|², x, y and 1 — with the four points being the images of a body point for the circle-point cubic, or a fixed point’s four positions seen from the body for the centre-point cubic. That determinant is a polynomial of degree three in the point, since the |q|² every image shares cancels between the rows. Its ten coefficients come from interpolating it at forty scattered points, exact to 2 × 10⁻¹³.

The cubic part of each — the terms in $x^3$, $x^2y$, $xy^2$ and $y^3$, which decide where the curve meets the line at infinity — divides by $x^2 + y^2$ with a remainder of 10⁻¹⁵. So each cubic passes through the two circular points, the complex points at infinity that every circle passes through. Nine times through each circular point found the same property in the curves a pin-jointed machine draws, and the reason here is related: every row of the determinant begins with a squared distance, and $x^2 + y^2$ is the circular points’ own equation. A cubic through the circular points meets the line at infinity in those two and one more. Two of the three meetings are complex whatever the poses, so exactly one is real.

That one is the slide. The centre-point cubic’s real direction to infinity is 121.858° at these four poses, and Q’s guide runs at 31.858°. The direction to infinity is square to the guide to 10⁻¹¹. That is where a crank’s pivot goes when its radius grows without bound and its arc becomes the straight guide’s line.

The two cubics, and where the slide sits on them #

The picture also answers the second part of the open question: whether the slider points are where the circle-point cubic’s branches run off the drawing. They are not. The slider point is an ordinary finite point on the circle-point cubic. Four images in a line are four images on a circle of infinite radius, and the determinant does not distinguish a line from a circle. What runs off the drawing is the centre-point cubic, whose branches head away parallel to one another, square to the slide.

A designer tracing the centre-point cubic by hand would see its branches leave the page in one direction and could read the slide’s direction off them without knowing why. The finite point on the other cubic that goes with that pivot at infinity is Q. The rest of the circle-point cubic is where a crank’s moving pin can go, and there is a curve’s worth of those. There is one slide.

The planes exchanged #

Everything above has a counterpart with the two planes exchanged. A slide can also be built the other way round: a pin fixed to the ground, running in a slot cut in the moving body. That needs a fixed point whose four positions, seen from the body’s own frame, lie on a line.

Where a pin becomes a slide found the three-pose version: in the fixed frame, the slot pins lie on the circle through the three poles themselves. With four poses, the fixed-frame pole circles of (1, 2, 3) and (1, 2, 4) share the pole $P_{12}$ itself, and their second meeting is the slot pin. Its four positions in the body’s frame are in a line to 3 × 10⁻¹⁵. So every four-pose problem has one slide in each form: a block on the body in a fixed guide, and a fixed pin in a slot in the body. Both are found by compasses. At these poses the slot runs well outside the body’s own outline, so a real body would have to be extended to carry it. That is a practical cost the geometry does not show until it is drawn.

Four hundred problems #

The argument that there is always exactly one is short, and it uses one fact about the pole point that was worth checking rather than assuming: that its images 3 and 4 are not accidentally in line with $P_{12}$.

The census draws four poses with positions within two units of the origin and angles anywhere. In all four hundred, the second meeting is a slider point, with its four images in a line to 2 × 10⁻¹². It lies on the other two triples’ circles to 8 × 10⁻¹². The slot pin is collinear to 10⁻¹². Both cubics are circular to 2 × 10⁻¹⁴, and the centre-point cubic’s real direction is square to the slide to 3 × 10⁻¹¹ rad. The first meeting, on the pole, is never a slider point: its images miss a line by at least 1.7 × 10⁻³ of their spread. Nothing in the census comes near a third.

Can there be none? Two circles through one real point meet in a second real point unless they are tangent there, and then the second meeting has merged into the first. Tangency at the pole point is a single condition on the poses, so it happens only on a curve in the space of four-pose problems, never in a random one. At it, the slider point runs into the pole point, and a block would have to sit on the pole — a body point whose first two images coincide, so that those two poses no longer tell the guide which way to run. A four-pose problem without a usable slide is therefore a special problem. The census, drawing at random, never found one.

Four poses brought together #

There is a second check on Q that uses nothing in this essay. Where a pin becomes a slide found that when three prescribed poses are three nearby positions of one moving body, the slider circle becomes the circle of points going straight: the inflection circle, whose points are momentarily moving in a straight line. The four-pose version should go one step further. A body point with four nearby positions in a line is a point whose path is straight to third order, not just at an instant. The straightest point there is found exactly one such point on a four-bar’s coupler, Ball’s point, where the inflection circle meets the cubic of stationary curvature.

It does. With the four poses spaced by h along the crank, the slider point sits 3.8 × 10⁻² from Ball’s point at h = 0.1, and 9.3 × 10⁻⁵ at h = 0.005. The measured order between neighbouring spacings is 1.99, 2.00, 2.00 and 2.02: the gap closes as $h^2$, the same order at which the three-pose slider circle closes on the inflection circle. The compass construction of Q and the curvature theory that finds Ball’s point were computed separately, and in the limit they name the same point. Four positions brought together found the crank’s version of this, Burmester’s curve becoming the cubic of stationary curvature. This is the slide’s.

It also says what Q is for a designer who has a motion rather than four poses: the slide that best carries a moving body through four closely spaced positions is a block at Ball’s point, and Ball’s point is where Watt and Chebyshev did not put their tracing points.

What this changes for a designer #

A four-pose crank-and-slider mechanism — a slider-crank that carries a body through four prescribed poses — needs a crank dyad and a slide dyad. The crank’s moving pin must be on the circle-point cubic, and there is a curve of choices. The slide’s block must be at Q, and there is no choice at all. So the designer’s freedom in a four-pose slider-crank is one parameter, the position of the crank along the cubic. It is not two, as it would be for a four-bar built from two cranks on the cubic.

It also reverses the usual order of work. With four poses, a designer who wants a slide in the mechanism should construct Q first. Its guide direction then fixes the slide, and the question is only whether some point of the circle-point cubic makes a crank that goes with it. The slide cannot be traded against anything. The crank can. The slider-crank is usually introduced with its crank first and its slide as a consequence; for four prescribed poses, the slide is the given and the crank is the choice.

The slide runs out one pose early #

Set the two dyads side by side and the pattern is a count of conditions. A crank asks a body point to keep its images on a circle, and a circle in the plane has three numbers to choose: two for its centre and one for its radius. A slide asks for a line, which has two. Each prescribed pose beyond the first adds one condition on the images. So with three poses a crank has every point of the plane to choose from and a slide has a circle. With four, a crank has a cubic curve and a slide one point. With five, a crank has Burmester’s four points, and the slide has run out: two coordinates cannot meet three conditions.

That is the same arithmetic three problems called synthesis used to tell path, motion and function generation apart, applied to the two kinds of dyad instead of the three kinds of problem. It also accounts for the construction. The slide is one pose ahead of the crank in how little it leaves, so at four poses it has reached the finite set the crank only reaches at five. And the finite set is small enough — one point — for circles to find, where the crank’s four points need cubics.

What this does not settle #

Whether the mechanism works between the poses. A slider-crank built from Q and a point of the cubic reaches the four poses, but whether it reaches them in order and on one circuit — the branch and order defects that three-position synthesis had to screen — is not checked for any crank here.

The special problems. The problems where the two circles are tangent at the pole point, or where Q lands at infinity, form a set of measure nought. What their slide looks like — a slide of no length, or a guide at infinity — is described, not measured.

Five poses. Five poses would ask for five collinear images, three conditions on two coordinates, and a random five-pose problem should have no slider point at all. Whether that is true, and how close the nearest thing to a slide comes, is not measured here.

Still open: the five-pose slide that almost exists #

With five poses a slide needs five collinear images, one condition more than a point of the plane can meet, so a general five-pose problem has no slide. But each of its five four-pose subproblems has exactly one slider point, found as above.

The distinct argument there would be how far apart those five points are, and what that distance means. If the five slider points of the five subproblems cluster, a five-pose problem has a near-slide: a block that misses one pose by a measurable amount. The spread of the five points, set against how far the fifth pose is from the four-pose problem’s own guide, would say whether a designer who needs five poses and a slide should give up the slide or give up exactness in one pose. Burmester’s five points for cranks are finite; for a slide the answer should be none, with a miss that can be computed.