The load picks the wall

The slide five poses almost have found the best slide for five prescribed poses of a moving body. The slide a fourth pose leaves had found exactly one slide for four; a fifth pose leaves none. No slide reaches all five: each image of the chosen body point is its signed distance off the guide, the residuals obey one linear condition $\sum \lambda_i r_i = F$ whatever slide is chosen, and the best a slide can do is to spread $F$ evenly. At the five poses used there it misses each by 0.0140 of a body about two units long, on alternate sides of the guide: +, −, +, −, +.

That essay read the miss as play. A block with 0.0140 of clearance either side of its guide can put the chosen point anywhere within 0.0140 of the guide’s centre line, so it can reach all five images. The alternating signs said what the block would have to do: bear on one wall at the first pose, the other at the second, the first again at the third. So the block would cross its slot four times on the way. The essay ended by asking whether a block pushed by a crank actually does that, or sits on one wall and misses.

It does not, and the reason is that nothing about the slot decides the wall. The load decides it.

One crank of four drives the poses in order #

To push the block there has to be a crank. Five poses leave Burmester’s points, a crank centre and a body pin whose five images lie on one circle about it. At these poses there are four of them, all real. Five positions and what is left found them. Any one of them, with the compromise slide, makes a slider-crank that carries the body through the five poses. The question is whether it can be driven through them.

Two of the four turn back. Crank 4 goes from −115° to −12° and returns to −51° at the last pose, and crank 2, whose centre is 24 units away, creeps from 121.0° to 121.4° and back to 113°. A crank that must reverse between poses is a rocker driven by something else, and the defect that where a pin becomes a slide screened for at three poses — the right poses in the wrong order — is the same one.

Crank 1 turns one way, from −167° through 157°, but between poses 4 and 5 its rod passes square to the guide. The component of the rod along the guide goes from −1.07 to +0.91 there. At that instant, with the crank held still, the body can still turn about the crank pin while the block slides, so the mechanism has a freedom no crank torque controls, and any load across the slot needs an infinite force to resist. That leaves crank 3, centred at (1.53, −0.73) with a radius of 1.49. It turns steadily from 161° down to 13° through the five poses, and its rod is never square to the guide. It is the only one of the four a motor could drive.

A frictionless slot chooses no wall #

The block has play either side, and whether it bears on one wall or the other is a question about forces. With no friction, the slot can push on the block only across the guide, along its normal $n$, and the block bears on the wall that push comes from. The push is set by equilibrium of the body. The body is held at two places: the crank pin, which can push it any way, and the block, which only across the guide. Take moments about the crank pin and the crank’s push drops out:

$$N\,(d\cdot u) = -M_A$$

where $d$ runs from the crank pin to the block, $u$ is the guide’s direction, and $M_A$ is the moment of the load about the crank pin. So the wall is the sign of the load’s moment about the crank pin, turned over wherever the rod’s projection on the guide changes sign — which crank 3’s never does. The block changes walls only where the load’s line passes through the crank pin.

This is a clearance is a link in its simplest form: the play is a short link, and a short link swings to wherever the force puts it. It is also the reverse of what why a hinge works found. A three-knuckle hinge works because its play is larger than its misfit, whichever way the door is pushed; the knuckles never need to be on particular sides of their pins. Here the play is exactly the misfit, and it works only with the block on particular sides.

The walls two loads choose #

Two loads are natural for a guided body. One is the body’s own weight, hung at the centre of its outline. The other is a working load on the block itself, a force along the guide opposing the block’s motion, as a press or a pump would put there.

The weight’s force changes sign twice: once between the first and second poses and once between the third and fourth, at crank angles of 155° and 83°. It is small, a third of the working load’s at its largest, because the weight’s lever about the crank pin is short, but only its sign matters. The block bears on + − − + + at the five poses.

The working load changes sign once, where the block reverses along the guide between the first and second poses, and the load reverses with it. Its force is large near pose 2, where the rod’s projection on the guide is shortest at 0.50, and the curve is clipped there. That is the transmission angle the slider-crank is judged by, read at the block: a rod that stands nearly across the guide pushes the block along it poorly, and presses it hard against a wall instead. The block bears on + − − − −.

The compromise needs + − + − +: four changes, one between every pair of poses. Neither load gives more than two, and neither gives the right ones. The figure below shows the result pose by pose. The hollow ring is the wall the compromise needs, the filled dot is the wall a working load presses, and at poses 3 and 5 they are on opposite sides of the slot.

Where the weight’s line crosses the crank pin #

The weight’s walls can be read off the drawing without solving anything. The weight acts straight down through the centre of the body’s outline, so its moment about the crank pin is the weight times the horizontal distance from the pin to that centre, and the wall follows its sign. At the five poses that distance is −0.043, +0.156, +0.012, −0.185 and −0.354. The centre starts a little to the left of the pin, moves to the right as the body rises through the second pose, and passes back to the left between the third and fourth, where the crank has turned to 83°.

The third pose is the delicate one. There the body’s centre is 0.012 to the right of the crank pin, less than the width of the pin itself on any real machine. The weight’s wall at pose 3 is decided by six thousandths of the body’s length, and pose 3 is the pose that carries 37% of the obstruction. A body with its mass a little further left, or a crank pin placed a little differently among the few the synthesis offers, would press the other wall there. That would change the weight’s pattern from + − − + + to + − + + +, which still gets pose 4 wrong, and would move $\sum\lambda_i s_i$ from −0.396 to +0.335.

That sensitivity is general, not a property of this body. Wherever the load’s line passes close to the crank pin at a prescribed pose, the wall at that pose is a coin toss decided by manufacturing. The compromise then has two numbers that matter where it had one. Its play must be the right size, and the load’s moment must have a comfortable margin of the right sign at every pose where the obstruction’s weight is large.

The play a pattern of walls needs #

What a wrong wall costs follows from the same condition that priced the compromise. If the block bears on wall $s_i = \pm 1$ at pose $i$, a play $e$ puts the chosen point at $s_i e$ from the guide’s centre line, so a slide reaches all five poses exactly when its residuals are $r_i = s_i e$. Put that into $\sum \lambda_i r_i = F$ and the play is fixed:

$$e = \frac{F}{\sum_i \lambda_i s_i}$$

The play must come out positive, because a play is a clearance. The compromise is the pattern that makes it least, $s_i$ taking the signs of $\lambda_i$, since then the denominator is $\sum|\lambda_i|$, as large as it can be.

The weights at these poses are 0.026, −0.168, 0.366, −0.332 and 0.109, normalised so that their magnitudes sum to one. The working load gets poses 3 and 5 wrong, the largest weight and a moderate one, and $\sum\lambda_i s_i$ falls from 1 to 0.051. The play it needs is 0.274, nineteen times the compromise’s. The weight gets poses 3 and 4 wrong, the two largest weights, and the sum goes negative: −0.396. No play reaches the five poses with those walls, because the play the condition asks for would be a negative clearance.

Every pattern has its price #

There are sixteen patterns of walls up to swapping the sides, and the law prices each. Each can be checked by solving for the exact slide whose residuals are $s_i e$, starting from the compromise and from every four-pose slide.

The law and the solved slides agree to within 3.3% on all fifteen, closely where the play is small and loosest at the two largest, where the linear condition is being asked to hold a quarter of a unit from the compromise. The ordering follows the weights. Getting pose 1’s wall wrong, a pose that carries 2.6% of the obstruction, costs almost nothing: 0.0148 against 0.0140. Getting pose 3’s wrong costs most, since pose 3 carries 37% of it. The walls a working load presses are the most expensive pattern of the fifteen.

The pattern + − + + + that the weight would give with its centre slightly further left is among them, at 0.0418, three times the compromise: better than the walls the weight actually gives, which have no price at all, and much worse than the walls the compromise wants.

The sixteenth pattern is the block on one wall at every pose, and it has no price at all. The weights sum to nought — the guide’s offset enters every residual equally — so $\sum\lambda_i s_i$ is nought and no play reaches the poses. That is obvious once it is said. A block that never leaves one wall is a block in a narrower guide moved sideways, and a guide moved sideways is still a slide, which five poses do not have.

What play does to the miss #

A designer who cannot choose the walls can still choose the play, and ask what it buys. It is the same trade the linkage that is only nearly right made between exactness at chosen points and error elsewhere, with the play as a free parameter the synthesis did not have. With walls $s$ fixed and a play $e$ that is not the pattern’s own, the best slide cannot reach the poses exactly. The same condition says what it can do: the residuals less $s_i e$ still obey one condition, and spreading it evenly gives a worst miss of $|F - e\sum\lambda_i s_i| / \sum|\lambda_i|$.

The solved slides lie on the law at every play tried, to the fifth decimal. With the right walls the miss falls linearly to nought at a play of 0.0140 and rises again past it. Too much play is as bad as too little, because the block’s point then overshoots the poses. With the working load’s walls the miss falls from 0.0140 to 0.0133 at the same play, a twentieth, and it would take 0.274 to reach nought. With the weight’s walls the miss grows: 0.0196 at a play of 0.0140 and 0.0259 at 0.030. The play that the compromise was supposed to use makes this slide 40% worse than the same slide built with no play at all.

So the play reading was right about where the block may be and silent about where it goes. A rigid compromise slide misses every pose by 0.0140 whatever loads it. The same slide with play reaches every pose only under a load that turns at exactly the right places, and under the loads a body usually carries it misses by more.

A block that rocks, and when it does #

The block does rock in its slot. It crosses from wall to wall wherever the load’s moment about the crank pin passes through nought: twice in this travel under the weight, once under a working load. With no friction and no mass the crossing is instantaneous, a knock as the play is taken up the other way. What the measurements say is that the crossings belong to the load, not to the prescribed poses. The compromise asks for four, at places the obstruction’s weights decide, and a load gives what its own geometry gives.

The practical reading has two halves. A designer who needs five poses from a slide, and whose load is known, can price the play for that load’s walls with one division, $F / \sum \lambda_i s_i$, before building anything. If the sign comes out wrong, the right answer is no play: a rigid slide at the compromise. And a designer who can choose the load — a spring on the body, a weight hung in a chosen place — has one more design variable than the synthesis used: where the moment about the crank pin should change sign.

What the walls do not settle #

Friction. A slot with friction holds its block on either wall until the force across the slot has changed sign by enough to overcome the friction. So the crossings move later along the travel, and the block’s wall at a pose depends on the direction of travel. Nothing here includes friction, and a slot run both ways would have two patterns, one for each direction.

Mass. The block and the body have no inertia here, so a crossing takes no time and knocks with no energy. At speed the block crosses in a finite time, and its flight across the play is part of the motion between poses.

One crank. Only one of the four cranks could be driven, so the loads’ walls were measured on it alone. A different set of five poses would have its own usable cranks, and the walls a weight presses depend on where the crank pin is.

The linear condition. The price $F / \sum\lambda_i s_i$ comes from linearising about the compromise and is within 3.3% of the exact slide on every pattern here. At plays of a quarter of the body’s scale it is a first approximation, and the exact slides are the measurement.

Still open: a load chosen for the walls #

The walls follow the sign of the load’s moment about the crank pin, and a load can be designed. A spring from the frame to the body, or a counterweight placed on it, adds a moment of its own, and the combined moment changes sign where the design puts it.

The distinct argument there would be whether some simple load — one spring of no free length between the frame and a body point, whose moment about the crank pin is a smooth function of the crank angle — can be placed so that the combined moment changes sign exactly between poses 1 and 2, 2 and 3, 3 and 4, and 4 and 5. That would make the compromise’s play reach all five poses, and it is four conditions on the spring’s few numbers. The measurement would search the spring’s anchors and stiffness for the pattern + − + − + under a given working load. It would report whether one spring suffices or two are needed, and how large the spring must be against the load before its moment dominates where it has to.