Machine Kinematics
solved, then drawn
Motion that stops

A curved slot moves the step

A Geneva's pin arrives with an acceleration of tan(π/n) of the crank's speed squared, and no choice of entry direction removes it. A choice of slot shape can: bend the slot's mouth with the crank's own circle and the slot absorbs the pin's acceleration instead of handing it to the wheel. But the pin leaves by the mouth it came in by, and there the same bend adds to the step. Solved at four slot counts and seven curvatures, the entry and exit steps are tan(π/n)(1 ∓ κ̂) and always sum to 2 tan(π/n). A curved slot moves the step from one end of the index to the other, every bend raises the peak, and no slot tried beat the straight one.

Assumes When the index law becomes a choice and Detached, and safe while detached.

15 min read 5 figures Two routes to the same numberExact and approximateThe singularities are the point

When the index law becomes a choice set a Geneva against a cam indexer on the cam’s own terms and found one thing no slot count removes. The pin arrives with an acceleration of exactly tan⁡(π/n)\tan(\pi/n) of the crank’s speed squared. At tangential entry the pin moves along the slot, so its velocity turns the wheel not at all. Its acceleration is the crank’s centripetal acceleration, square to the crank and so square to the slot, and all of it is handed to the wheel. The essay put it as a verdict: choosing the entry direction removed the velocity across the slot, and nothing about the entry direction can remove the crank’s centripetal acceleration.

The direction of the slot is not its only property. A slot is a curve, and a curve has a curvature as well as a direction. A cam’s follower meets a profile whose bend is chosen to shape the motion, and a Geneva’s slot could be bent too. This essay bends it, solves the wheel’s motion over the whole index for slots of many shapes, and asks what the bend can do to the step.

It can move it. It cannot remove it.

A curved slot bends at its mouth with the crank's circle, or against itA six-slot Geneva wheel whose slots are curved: mouth curvature 0.5 of the crank's. The driver's pin runs round the dashed circle and enters each slot along its mouth, which is radial so that the wheel starts from rest. At entry the pin's acceleration is square to the slot, and the part of it the slot's own bend takes up does not turn the wheel. Solved over the index, the wheel starts with an acceleration of 0.289 and stops with a deceleration of 0.866, in units of the crank's speed squared, against 0.577 at both ends for a straight slot; the two always add to 1.155. Its peak is 1.649 against the straight slot's 1.350.law: entry 0.289, exit 0.866dashed: the pin's circle
Fig. 1 A six-slot Geneva wheel with slots bent at their mouths, the driver’s crank and pin, and the pin’s circle dashed. The slider sets the slot’s curvature at its mouth as a share of the crank’s circle, from bending against the circle to bending with it, and last a mouth bent exactly to the circle and straightened inside.

What a bend at the mouth does to the entry

Put the wheel at rest and the pin at the mouth of a slot, moving along it. In the wheel’s frame the pin is constrained to the slot’s curve. So its acceleration relative to the wheel must have a component across the slot equal to its speed squared times the slot’s curvature, whatever the wheel does, because that is what moving along a bent curve at that speed requires.

The pin’s acceleration in the ground is the crank’s centripetal acceleration, aω2a\omega^2 across the slot, which is the speed squared times the crank circle’s curvature 1/a1/a. The difference between that and what the slot’s bend requires is the part the wheel has to supply by turning. So the wheel’s angular acceleration at entry is proportional to the crank’s curvature less the slot’s. A straight slot has no curvature and the wheel gets everything, tan⁡(π/n)\tan(\pi/n). A slot whose mouth bends with the crank’s circle by a share κ^\hat\kappa of its curvature takes up that share, and the wheel starts with

tan⁡(π/n) (1−κ^).\tan(\pi/n)\,(1 - \hat\kappa).

At κ^=1\hat\kappa = 1 the slot’s mouth bends exactly with the pin’s circle, the slot absorbs the whole centripetal acceleration, and the wheel starts with no step at all. The verdict of the earlier essay was right about direction and incomplete about shape.

The same mouth at the other end

The trouble is that a slot is one channel. The pin enters at the mouth, runs in, turns the wheel through its 2π/n2\pi/n and comes back out through the same mouth. At the exit, the mouth has turned with the wheel to the mirror-image position, and the pin is moving outward along it. Seen from the slot, the crank’s circle now bends the other way. A mouth that bent with the circle at entry bends against it at exit, the slot’s curvature now adds to the crank’s instead of cancelling it, and the wheel stops with a deceleration of

tan⁡(π/n) (1+κ^).\tan(\pi/n)\,(1 + \hat\kappa).

The two steps together are 2tan⁡(π/n)2\tan(\pi/n) whatever the bend. The mouth’s curvature decides only how that total is shared between the start of the index and the end of it.

The mouth's bend takes the step from the entry and gives it to the exit. The wheel's acceleration step at entry (circles) and its deceleration step at exit (squares), each as a share of tan(π/n), against the slot's curvature at its mouth as a share of the crank's circle, for Genevas of 4, 6, 8, 12 slots with slots of constant curvature, solved over the whole index. Every point lies on 1 − κ̂ (entry) or 1 + κ̂ (exit) to 3e-6 of tan(π/n): the four slot counts collapse onto the two lines, and the two steps always add to 2 tan(π/n). A bend with the crank's circle lowers the entry step and raises the exit step by the same amount; a bend against it does the reverse.
Fig. 2 The wheel’s acceleration step at entry and deceleration step at exit, each as a share of tan(π/n), against the slot’s curvature at its mouth as a share of the crank’s circle, for 4, 6, 8 and 12 slots with slots of constant curvature, each solved over the whole index, beside the two lines 1 − κ̂ and 1 + κ̂.

That argument is about two instants, and it was checked against the whole motion. For slots of constant curvature at 4, 6, 8 and 12 slots and seven curvatures from −0.75 to 0.75, the wheel’s angle was solved at twelve hundred crank angles across the index from the one condition that the pin lies on the slot. The acceleration was then taken by second differences, and the steps read off at the two ends. Every step lies on its line to 3×10−63 \times 10^{-6} of tan⁡(π/n)\tan(\pi/n). Scaled by tan⁡(π/n)\tan(\pi/n), the four slot counts fall onto the same two lines. The entry step falls as the mouth bends with the crank’s circle, the exit step rises by exactly as much, and a bend against the circle does the reverse.

The index itself does not change. The mouth sits on the wheel’s rim, the pin crosses the rim at the same two places whatever the slot’s shape, and so the wheel turns through 2π/n2\pi/n for every slot solved. What the shape decides is how the wheel gets there.

Where the step goes

A step moved is still a step. Its effect depends on which end it lands at.

Where the step goes when the slot is bent. A six-slot wheel's angular acceleration through the index, in units of the crank's speed squared, for three slots: straight slot, starting at 0.577 and ending at −0.577 with a peak of 1.350; mouth bent half the crank's, starting at 0.289 and ending at −0.866 with a peak of 1.649; mouth bent to the crank's, then straight, starting at nought and ending at −1.155 with a peak of 1.611. The straight slot jumps by tan 30° at both ends. The bent slots take the jump off the entry and put it on the exit, where it arrives doubled for the slot bent all the way to the crank's circle, and both bent slots peak higher than the straight one.
Fig. 3 The six-slot wheel’s angular acceleration through the index, in units of the crank’s speed squared, for the straight slot, a slot bent at its mouth by half the crank’s curvature, and a slot bent at its mouth exactly with the crank’s circle and straightening within half a crank radius.

The straight slot jumps by 0.577 at entry, tan⁡30°\tan 30°, and by the same at exit. The half-bent slot starts at 0.289 and finishes by dropping from 0.866. The last slot is bent exactly to the crank’s circle at its mouth, its curvature fading to nothing half a crank radius in. It starts from nought and finishes by dropping from 1.155, twice the straight slot’s exit step.

Which end is worse is a question about the machine. The entry step is felt as the pin hits the slot’s wall and the load is set going from rest. A pin that arrives with a sudden acceleration demand sets the driven mass ringing, and in a film projector or an indexing table the start is where the shock is noticed. The exit step is a sudden release. The pin stops decelerating the wheel all at once, the wheel is about to be held by its lock, and whatever the lock’s play allows the wheel then does. A V holds the wheel to first order found the lock that keeps the wheel’s rock proportional to its clearance. A wheel handed to that lock with twice the usual deceleration step arrives with the same speed, nought, and a larger change in load on the lock’s faces.

A designer who cares about the start more than the finish can therefore bend the slots with the crank’s circle and trade the entry’s shock for a sharper release. A designer with a delicate lock and a robust start can bend them the other way. What neither can have is both ends smooth. The mechanism that waits opened this field with machines that are still for part of each turn and moving for the rest. For the Geneva the moving part begins and ends with a jump that can be divided between its two ends but not avoided.

What the bend costs in between

The steps are set at the mouth. The whole motion depends on the slot’s whole shape, and the next question is whether a bent slot at least lowers the wheel’s peak acceleration, which is where the earlier comparison with cams found the Geneva doing well.

Every bend raises the peak, and the same either way. The peak angular acceleration of the wheel over the index, as a share of the straight slot's, for slots of constant curvature, against the mouth's bend as a share of the crank's circle, at 4, 6, 8, 12 slots. 4 slots: 1.071 at 0.25, 1.288 at 0.5, 1.903 at 0.75; 6 slots: 1.068 at 0.25, 1.222 at 0.5, 1.826 at 0.75; 8 slots: 1.109 at 0.25, 1.281 at 0.5, 1.755 at 0.75; 12 slots: 1.162 at 0.25, 1.368 at 0.5, 1.644 at 0.75. A bend and its opposite peak equally, since one is the other's index run backwards. None is below one.
Fig. 4 The wheel’s peak angular acceleration over the index, as a share of the straight slot’s, against the mouth’s curvature, for slots of constant curvature at 4, 6, 8 and 12 slots.

It does not. On six slots a bend of a quarter of the crank’s curvature raises the peak by 6.8%, a half by 22% and three quarters by 83%. The other slot counts follow the same pattern, from 7% to 16% at a quarter and from 64% to 90% at three quarters. A bend and its opposite peak equally, since a slot bent one way run from exit to entry is the slot bent the other way run from entry to exit.

The reason is visible in the motion. A straight slot’s index is symmetric: the wheel accelerates through the first half as it decelerates through the second, and the largest acceleration is shared evenly between them. A bent slot is asymmetric, since it moves part of the step from one end to the other, and the rest of the index has to make up for the difference. In the half-bent slot the wheel starts gently and must then turn the same 60° in the same crank angle, so it has to accelerate harder later. The peak rises to 1.649 against the straight slot’s 1.350.

The slot bent only at its mouth fares better than a slot bent all the way. It peaks at 1.611 while taking the entry step to nought. A designer who wants the smooth start should put the bend only where the step is decided and straighten the slot as soon as possible afterwards.

What the bend costs the wheel

A bent slot also has to be longer. The pin runs deepest at the middle of the index, and a straight slot on a six-slot wheel with its centres a unit apart reaches 0.366 in from the rim, which is the rim’s radius less the distance of the crank’s circle from the wheel’s centre. A bent slot carries the pin along a curve rather than a line and makes the index lopsided, the wheel turning less early and more late or the reverse, and in either case the pin gets further in before it turns back. Solved, the depth grows to 0.369 at a quarter of the crank’s curvature, 0.380 at a half and 0.405 at three quarters, the same for a bend either way, and on four slots from 0.414 to 0.451. The slot bent only at its mouth reaches 0.380, as deep as the half-bent one, for the whole of the entry step’s removal.

That is material the wheel does not have. The disc decides the pin count found the locking disc’s cut-away, not the slots, deciding how many pins a driver can carry. A deeper slot reaches further towards the wheel’s hub, and it leaves the lock between slots on a thinner neck of material. Neither the lock nor the wheel’s strength is measured against the depth here, but both move in the wrong direction as the slot bends.

A slot straight at the mouth

That leaves one freedom unexamined. A slot straight and radial at its mouth keeps both steps at tan⁡(π/n)\tan(\pi/n), and its shape inside is then free, since the steps do not depend on it. If any shape could lower the peak, it would be one of these.

Bending the inside of a straight-mouthed slot never lowers the peak. 300 six-slot Geneva slots, radial and straight at the mouth so both steps stay at tan 30°, with the curvature inside a random cubic in the depth, seeded; 246 index cleanly and turn the wheel 60°. Their peak angular accelerations, as shares of the straight slot's 1.3496, ordered: the least is 1.0155, the median 1.283, and none is below one. On this evidence the straight slot is the least-peak slot with a straight mouth, not a convenient special case.
Fig. 5 Three hundred six-slot slots, straight and radial at the mouth, with their curvature inside a random cubic in the depth, each solved over the index; the peak of each that indexes cleanly, as a share of the straight slot’s, in order.

Three hundred such slots were drawn, with the curvature inside a random cubic in the depth, seeded so the draw can be repeated. Of them 246 index cleanly, the pin staying in the slot and the wheel turning its 60°; the others bend so hard that the pin would leave the slot’s end or the constraint folds. Every one of the 246 peaks higher than the straight slot. The least is 1.0155 times it and the median 1.28, and none falls below one. On slots of constant curvature and slots bent only inside, the straight slot has the lowest peak of every shape tried.

A random search is not a proof, and this is stated as what was measured: the straight slot beat 246 out of 246 on its own terms. It does agree with the symmetry argument. Any bend makes the index asymmetric, and an asymmetric index spends its largest acceleration where a symmetric one would have shared it.

What the Geneva’s law is, then

The earlier comparison called the Geneva’s motion law forced by its slot count, where a cam’s is chosen. Bending the slot adds exactly one choice to the forced law, where the step goes. The total step is fixed at 2tan⁡(π/n)2\tan(\pi/n). The peak is least with the straight slot, and the index is fixed at 2π/n2\pi/n in the crank’s π−2π/n\pi - 2\pi/n. Everything else is paid for.

That is a sharper version of the earlier conclusion rather than a reversal of it. The gear with its teeth cut away has a step it cannot even report as a number; the Geneva’s is a number, and now a number that can be divided. Putting several pins on the crank shortens the time between steps and changes nothing about each one, since every engagement is the one-pin engagement with the same mouth. A cam indexer can start and finish with no step at all, because its follower’s law is chosen point by point. The law that costs least priced exactly that freedom for cams. A Geneva’s pin is on a circle and its slot is one channel, and between them they fix the step’s total. Changing that needs one of those two things changed. The pin could follow a path that is not a circle, or the slot could be two channels, an entry and an exit that are not the same curve.

Why the slot cannot also be the lock

The same one-channel geometry settles a question the lock essays left open, whether the pin in its slot could hold the wheel between indexes, doing without the locking disc and its cut-away altogether. A V holds the wheel to first order ended by proposing it: the slot’s walls have normals that miss the wheel’s centre by the pin’s own distance from it, so a pin at rest in a slot would hold to first order.

With the pin on a crank it cannot happen, for the reason that decided the steps. While the pin is in a slot, the one condition that it lies on the slot fixes the wheel’s angle for every crank angle. The wheel is still only when the pin is moving along the slot, towards or away from the wheel’s centre, and a pin on a circle moves that way only at the instants of entry and exit. So a crank pin in a slot is always turning the wheel. A pin that held the wheel from inside a slot would need a driver whose pin runs straight along the slot for a stretch of the turn, which is a different mechanism and a different essay.

What the slot shape does not settle

Two channels. Everything here uses one slot for entry and exit. A forked slot, with the pin entering by one branch and leaving by another, would not be bound by the sum, since its entry and exit mouths could bend differently. It would also change where the pin crosses the rim, and so the index and the wheel’s material between slots.

Wear and contact. A curved slot puts the pin’s contact on a curved wall, and the contact force and sliding speed along it change with the bend. Neither is computed, and a slot bent to lower the entry step may wear faster at its mouth.

The pin’s size. The pin is a point on the slot’s centreline. A real pin has a radius and a real slot two walls, and the slot’s two walls are offsets of its centreline by that radius. A bend tighter than the pin’s radius would undercut one wall. None of the bends here comes close, but the offset is not drawn.

Dynamics. The steps and peaks are kinematic, in units of the crank’s speed squared. What they do to a real driven load, with the driver slowing as it is loaded, is a dynamics question.

Still open: a pin that runs straight along its slot

The step’s total and the impossibility of a slot lock both came from one fact, that the pin is on a circle. A pin on a crank of variable length, carried in a radial guide on the driver and steered by a cam, can follow any path.

The distinct argument would be a driver whose pin path has a straight stretch along the slot at the end of each index. The pin would run inwards along the slot and out again with the wheel held still by the slot’s walls, then leave. The measurement would find the pin paths that do this while keeping tangential entry. It would ask how much of the dwell the slot can hold the wheel for, against the rest, where a disc must, and whether a path that is not a circle also frees the step, since the argument above used the circle’s curvature and a path can have any. It would also say what the cam that steers the pin costs, compared with a cam indexer that dispenses with the Geneva altogether.

Where this goes next

A Geneva's acceleration against three cam laws, at the same index angle. The output's acceleration through one index, divided by the step over the square of the input angle it takes — the unit in which a cam law's acceleration coefficient is quoted — so every law is one fixed curve whatever the step. The dark curve is a Geneva of 6 slots: its step is 360°/6 and its input turns 120° while indexing, and a cam indexer is given exactly those two numbers. The Geneva's coefficient is 5.653 against cycloidal 6.283, modified sine 5.528 and simple harmonic 4.935, so at 6 slots modified sine and simple harmonic have the lower peak. The Geneva's curve starts and ends away from nought, at tan(π/6) = 0.577 in absolute terms: its acceleration steps the instant the pin enters. Dragging the slot count moves only the Geneva. When the index law becomes a choice A Geneva's motion law is forced by its slot count and a cam indexer's is chosen, so the fair comparison gives the cam the Geneva's own index angle and step. On peak acceleration the cam wins only below a slot count that depends on the law — 5.19 for cycloidal, 6.23 for modified sine, 8.06 for simple harmonic — and above it the Geneva does. What no slot count removes is the step: the pin arrives with an acceleration of exactly tan(π/n). The disc, the wheel, and what has to be cut out of it. A 6-slot Geneva at 0° of driver, with the driver's locking disc of radius 18.8 drawn about its axis and the wheel drawn as the material it actually has — inside its rim, outside the 6 V notches cut into it, each face tilted 14° off the chord, and clear of the 6 slots. The disc and the wheel share the region near the line of centres, so the disc has to be cut away wherever the wheel is ever there while the pin is driving. Swept over the whole index that cut-away spans 159.8°, against an index sweep of 120° — it is wider, because the rim is still swinging through the disc's circle after the pin has left the slot. A V holds the wheel to first order A Geneva wheel locked on concave arcs holds to first order and carries one pin; locked on flat chords it carries two and holds only to second order. Cut the lock as a shallow V instead — two faces tilted a few degrees off the chord — and it keeps both. Its disc is the crank times sin(β + γ), small enough for a second pin up to 34° at four slots and 5.6° at twelve. It rocks as acos(cos γ − c/a) − γ, which is linear in the clearance below 2a sin²γ, and at a running clearance of 0.1 on a crank of 30 it holds five to eight times tighter than the flat. The trade was between two shapes, not a law. The same rise, three ways. A 20-unit rise over 120°, then a dwell. Above, the displacement — all three do the same job and are hard to tell apart. Below, the acceleration, differentiated from those same curves. Constant acceleration peaks lowest at 5.556e-3 per degree² and arrives at the dwell with a step; cycloidal peaks highest at 8.727e-3 and arrives at zero. The smoothest law has the largest peak, which is the trade the whole of cam design turns on and which the displacement plot gives no hint of. The law that costs least is not the smoothest Constant acceleration gives the lowest peak acceleration of any motion law and an impulsive jerk. Cycloidal motion has finite jerk everywhere and a peak acceleration 57% higher. The trade is real, it is measurable, and the displacement curves that everyone plots give no hint of it.

What this makes readable

Essays that name this one as a prerequisite.

About the same objects

Not linked from either essay — found by the objects both name.

What links here

Essays that link to this one from their own argument.

The objects this essay names

Each one links to every other essay that touches it.

Accelerationthe Geneva mechanismIndexingJerkMotion lawPin in slotTangential entry