Machine Kinematics
solved, then drawn
Concept

Trammel — where it appears

A rod with its two ends constrained to two intersecting lines, whose points trace ellipses and whose centrodes are two circles. It is the field's degenerate case: its angular rate is exactly constant, so the fifth-order condition holds identically and the search for special points returns everything.

Named by 6 essays across 2 fields — each of them below, with the objects they name alongside it.

The coupler's motion at 66°. Every point drawn as a stub is a point of the coupler's own plane, and the stub is that point's velocity — solved, not sketched. They all point different ways and they are all consistent with one statement: at this instant the whole plane is turning about a single point, the pole, marked with a cross. It is off this frame at 1.5 coupler lengths from the crank pin, which happens whenever the coupler is close to translating. The arrow through it is the pole's own velocity, which is a quantity about the motion rather than about any point of it, and half of everything in this field follows from its direction and its size. positioned by solving, not by drawing.

The mechanism drops out

Every other field here is about a machine. This one is about the motion a machine makes — a plane sliding over a plane — and near any instant that motion is a handful of numbers with no linkage in them. Two mechanisms that agree on those numbers make the same motion, and one of them can always be thrown away.

curvature · Curvature Two circles, and an exact straight line. The moving centrode of this motion is a circle of radius 1.5 and the fixed one is a circle of radius 3.0, measured to 8.9e-16. The small circle rolls inside the large one, and a point on its rim traces a diameter of the large one — exactly, with no error term. Watt's linkage is straight to nine per cent of its span and Chebyshev's to twelve; this is straight to 0.0e+0, and the difference is not one of degree. It is the difference between a curve that approximates a line and two centrodes whose rolling produces one. positioned by solving, not by drawing.

Exact because two circles roll

Watt's straight line is straight to nine parts in a hundred and Chebyshev's to twelve. Here is one that is straight to nothing at all — no error term, no working range, no approximation anywhere — and the reason is that its two centrodes are circles, one rolling inside the other at exactly half its radius.

curvature · Curvature One ellipse, drawn at once by four double sliders on four different pairs of slides. The ellipse traced by a double slider with slides at 70°, a rod of 2 and a tracing point off the rod at λ = 0.3 + 0.35i: semi-axes 1.587 and 0.542. Drawn over it at one instant, four more double sliders holding the same point, each on its own pair of slides through the ellipse's centre, with rods of 1.73, 2.08, 0.93, 1.03 and slides at 54°, 77°, 63°, 80°. The two in the first colour are from the original's own rolling circle and turn with it; the two in the second are from the other and turn against it. Any two points of either circle would do as well.

A second slide keeps only the rod

A slide turned a four-bar's rocker into a line and left one cognate, the slider-crank with its crank and rod exchanged. Turn the crank into a slide as well and the exchange stops producing a machine, since λ times a sliding pin's position is a vector that shrinks to nothing twice a turn. The cognates do not vanish; they multiply. Every double slider's ellipse is drawn by any two points of either of two rolling circles — a two-parameter family on each — and the one rotation left, the rod's, is carried by every one of them, with the original on one circle and against it on the other.

curves · Coupler A rod on three blocks, on three slides through one point, and it moves. The double slider of slides at 70° and a rod of 2, with a third block on the rod at a point of its rolling circle (radius 1.064, drawn at this instant) and a third slide along the line that point draws, through the same crossing. Five bodies and six joints of one freedom each: the count gives 3·4 − 2·6 = 0. The rod turns anyway, all the way round, every pin on its slide, because each point of the rolling circle runs on a line through the crossing and the third slide is that line.

A third slide moves only through the crossing

Every point of a double slider's rolling circle runs on a line through the slides' crossing, so a rod can ride a third block on a third slide there, and a fourth and a fifth. The count gives nought, then minus one, then minus two; the rod turns freely every time, because each slide past the second is implied by the first two. Build the third slide slightly wrong and the count wins, in a way that depends on the error's kind: moved parallel, the rod turns freely inside the block's clearance and does not assemble outside it; turned by ε about the crossing, it moves only in two windows a turn, of 2·arcsin(c/(2r sin ε)) each, where the third pin passes through the crossing.

curves · Coupler The pin opposite the third runs on the perpendicular slide, a quarter turn behind. The double slider of slides at 70° and a rod of 2, with a third block at a point of its rolling circle (radius 1.064) and a fourth at the opposite point, each on the slide its pin draws through the crossing. The two extra pins are the ends of a diameter of the rolling circle, so their slides are at right angles and their positions along them are 2r cos(φ − φ₃) and −2r sin(φ − φ₃): here 1.864 and 1.027. Four blocks and the count gives minus one; the rod turns, and the fourth pin passes the crossing a quarter turn after the third.

A fourth block can only take freedom away

A rod on a double slider can carry a third block on a third slide through the crossing and still turn, and a slide turned slightly wrong leaves it free only in two windows a turn. The hope was that a fourth block, at the point of the rolling circle opposite the third, would travel in opposition and cancel the error. It cannot, for two reasons. The opposite pin is the other end of a diameter: it runs on the perpendicular slide, a quarter of a turn out of phase, so its windows fall exactly between the third's. And whatever its place, a block only adds a condition the rod must also meet, so the rod's freedom is an intersection and can only shrink. With the double slider's own blocks exact, the third and the opposite fourth turned 0.0066 rad each lock the rod, where the third alone still moves at any error.

curves · Coupler A slide turned about a point is free in a band round that point. The double slider of slides at 70° and a rod of 2, with a third block and a fourth at the opposite point of the rolling circle (radius 1.064), each on the line its pin draws through the crossing. Each extra slide is set to pass through its pin at one common pose of the rod, 45° from either pin's crossing, so a turn error of 0.01 rad pivots it about that pin's place then: -1.505 along the third's line and -1.505 along the fourth's. With a clearance of 0.01 the block is free while its pin is within c/sin ε = 1.000 of the pivot (the shaded bands). At the angle shown the third pin is at -1.505 and the fourth at -1.505: the rod is free.

Fit every extra slide at one pose

A double slider's extra slide, turned slightly wrong, leaves its block free in a band of the slide centred on the point it was turned about, and the rod free in windows where the pin passes through that band. Turned about the crossing, a third and an opposite fourth block have their windows a quarter of a turn apart and lock the rod from 0.0066 rad. Set each slide through its own pin at one common pose of the rod and every error becomes a turn about the place its pin has then, so every window contains that pose and the rod can never lock, with two blocks or six. Each slide pivoted where it alone is freest does better while the errors are small and locks from between 0.02 and 0.05 rad, where the common pose 45° from either crossing comes within a tenth of the best pivots there are.

curves · Coupler

Named alongside it

The objects these essays reach for when they reach for this one.

CentrodeClearanceMobilityRedundant constraintGrübler's criterionInstantaneous centreClosed formCognate linkageConstraint jacobianCuspDesign ruleExact arithmetic

All concepts