Machine Kinematics
solved, then drawn
Concept

Acceleration — where it appears

The second derivative of a position with respect to the driving parameter, which for a mechanism is what the motion law produces rather than what it was asked for. It is where the cost of a prescribed motion appears: displacement is specified and acceleration is the consequence, and the smoothest law has the largest peak.

Named by 21 essays across 6 fields — each of them below, with the objects they name alongside it.

A cycloidal cam at 60°. A 20-unit rise over 120°, a dwell, a return and a dwell. The dashed curve is the pitch curve — where the roller's centre travels — and the solid one is the surface that has to be cut, which is the pitch curve offset inward by the roller radius along its own normal. The pressure angle at this instant is 25.5°: the angle between the follower's direction of travel and the normal to the surface, and the number that decides whether the follower jams in its guide rather than sliding.

Prescribing motion

A linkage gives the motion its geometry allows. A cam gives the motion it was asked for, which sounds like an improvement and is a trade — the displacement becomes free and the derivatives stop being.

cams · Cam 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.

cams · Cam A 6-slot Geneva wheel. The driving pin enters a radial slot, carries the wheel through 60°, and leaves. The centre distance is not free: it must be crank ÷ sin(180°/6) = 60.00 so that the pin enters along the slot, with the crank and slot perpendicular. That is the mechanism's one design requirement and it is what makes the driven wheel start and stop from rest — measured here at 1.5e-3 against a peak of 1.000. What it does not fix is the acceleration, which peaks at 1.35 and is why film sprocket holes tear.

Stopping thirty times a second

A Geneva wheel turns continuous rotation into steps, and its one design requirement is that the pin enters the slot along the slot so the driven wheel starts and stops from rest. That fixes every dimension from the slot count. What it does not fix is the acceleration, which is why film sprocket holes tear.

cams · Cam A 34-unit roller on a 20-unit base: the cam cannot be made. The outer curve is the pitch curve — where the roller's centre has to go, which is just the base radius plus the follower's displacement, plotted in polar coordinates. The inner curve is what has to be cut: the pitch curve offset inward by the roller's radius along its own normal. Where the pitch curve bends more tightly than the roller is wide, that offset crosses itself, and the metal the profile needs has already been removed. The tightest radius of curvature here is 19.43 at 85°, against a roller of 34 — so the cam is undercut by 14.57 and no amount of care in manufacture recovers it. The pressure angle, meanwhile, peaks at a perfectly comfortable 21.2°: this cam fails a test it never looked like failing.

The cam that cannot be cut

A cam has to be big enough for two reasons, and they are not the same reason. One is that the follower will jam in its guide if the pressure angle is steep. The other is that the roller will gouge the profile if the curvature is tight — and there is a combination where the pressure angle is comfortable and the cam still cannot be manufactured at all.

cams · Cam Slider-crank at 50°. Crank 1, connecting rod 3. The slider's travel is 2.0000 — exactly twice the crank throw, which is the one thing about this mechanism that does not depend on the rod length. Everything else does: the rod length decides how far the piston's motion departs from a sine wave, and that departure is the second harmonic every engine balancer has to deal with.

The slider-crank

Replace one pin of a four-bar with a slide and you get the mechanism in every reciprocating engine ever built. Its stroke is exactly twice the crank throw and does not depend on the connecting rod at all. Everything else about the motion depends on the rod, including the part that is always described as a sine wave and is not.

linkages · Fourbar Where the coupler is pivoting, at 70°. At any instant the coupler is turning about one point — not a pin, and usually not on the mechanism at all. Kennedy's theorem finds it: the crank and coupler share the pin at A, the coupler and rocker share B, so the coupler's centre relative to the frame must lie on both O₂A extended and O₄B extended, and it is where they cross. The dashed lines are that construction. The cross is a completely different route to the same point — the place where the coupler's solved velocity field is zero, computed from the Jacobian and knowing nothing about Kennedy. Across 119 positions the two agree to 2.7e-15. At this instant the centre lies outside the frame — the two construction lines are nearly parallel, the coupler is close to translating, and the pivot has run off rather than gone missing.

Where the coupler is turning

At every instant the coupler of a four-bar is rotating about a single point — not a pin, and usually not on the mechanism at all. Track that point in two different frames and you get two curves which, rolled on each other without slipping, reproduce the coupler's motion exactly. The bars are one way of producing it and not the motion itself.

curves · Coupler A quantity that is not there, refusing to converge. A second central difference of a function with a bounded second derivative settles as the step is halved; across a step in the first derivative it doubles, every time. The rising line is a mutilated gear at the instant its teeth engage, where the driven wheel goes from stopped to full pitch-line speed: its successive estimates grow by a factor of 2.000000, which is the signature and not an accident of the step. The flat line is a six-slot Geneva at the same point in its cycle, whose pin enters along the slot and whose acceleration is a number. This is the only way to report an acceleration that does not exist: not by quoting a large one, but by showing the measurement refuse.

The gear with its teeth cut away

Leave teeth on part of a gear's circumference and take the rest off, and the output turns for part of the input's revolution and stops for the rest. It is the cheapest intermittent drive there is and it engages at full speed, so its output's velocity has a step and its acceleration is not a large number — it is not a number, and the way to report that is to watch a difference quotient refuse to converge.

timing · Intermittent The transmission angle through one turn. μ is the angle at B between coupler and rocker, computed from each solved position rather than from a formula. It runs from 54.3° to 100.3° for these lengths. The shaded band is the usual design rule — keep μ between 40° and 140° — and this linkage stays inside it throughout. The rule is about geometry alone: nothing here knows about friction, and a mechanism with a comfortable μ can still be a poor machine.

What is still outside

Eight essays here end by saying that clearance, backlash, friction or wear are not modelled. This field took two of those four, because a tolerance is a set of geometries and a clearance is a short link, and both are questions about where a mechanism can be. The other two are not, and this is the page that says exactly where the line falls and why it is where it is.

practice · Practice A flat face on a base circle of 30, where the profile bends tightest. The cycloidal programme — rise 10 over 120°, return over 120° — cut for a flat-faced follower on a base circle of 30, turned to 212.4°, where the profile's radius of curvature is smallest. The contact sits −5.37 from the follower's axis, which is ds/dθ, and the dashed circle is the one that fits the profile there: R₀ + s + s″ = 24.668, and 24.668 fitted through the drawn curve. The face's pressure angle is zero at every angle. What it needs instead is for that radius to stay positive, and it does so on every base circle above 5.332.

A flat face asks for a convex cam

A flat-faced follower is pushed at right angles to its face, so its pressure angle is zero at every angle and the constraint that sizes a roller's cam disappears. What replaces it is convexity: the profile's radius at the contact is R₀ + s + s″, and a base circle below 5.332 leaves a cycloidal cam with a hollow the face cannot reach — held 0.446 high on a base of 2.

cams · Cam A roller in the groove of a quick-rise cam at 600 rpm. The cycloidal programme — rise 20 over 90°, return over 170° — cut as a groove on a prime circle of 40 for a roller of 10, driving a follower of 0.5 kg pressed toward the cam by 20 N, and turned anticlockwise to 59.9° at 600 rpm. The groove must supply m·s″·ω² + F along the follower's travel, −66.6 N at this angle, so the roller bears on the outer wall. The marks on the groove are the four places in a turn where that force changes sign and the roller changes walls: 47.87°, to the outer; 87.13°, to the inner; 151.35°, to the outer; 193.65°, to the inner. The stretches of deceleration begin to cross over at 267.6 and 505.5 rpm.

A roller in a groove changes walls with the speed

A groove holds its roller between two walls and chooses between them by the sign of one force, the follower's mass times its acceleration plus the load pressing it in. Below a threshold speed that force never changes sign and the roller stays on the inner wall; each stretch of deceleration adds two crossovers above its own threshold. On a quick-rise cam at 20 N and half a kilogram the count goes from none to two at 268 rpm and to four at 506. A cycloidal law brings the roller across smoothly; a constant-acceleration law throws it across by a step of 3,200 N at 3,000 rpm.

cams · Cam One pin and two, on the same wheel. A 6-slot Geneva wheel at one instant of its index, driven by a crank carrying one pin and by the same crank carrying two. The wheel, the slots and the centre distance are identical; only the pin count differs. Each pin drives the wheel through one slot pitch while the driver sweeps 120°, so 1 pin gives 1 index a driver turn and leaves 67% of it at rest, 2 pins give 2 indexes a driver turn and leave 33% of it at rest. Nothing else about the mechanism changes, which is why the whole question is arithmetic.

Two pins and no dwell at all

Put a second pin on a Geneva's crank and the wheel indexes twice a turn instead of once, at a quarter of the acceleration for the same output rate. Put a third on a six-slot wheel and it never rests; put a third on an eight-slot wheel and two pins meet in two slots and it jams. Both of those look like separate conditions and are one, and the boundary between them is an equation in integers with exactly three solutions.

timing · Intermittent 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).

timing · Intermittent The roller's flight across a 0.05 mm clearance at every crossover. The cycloidal quick-rise programme carrying 0.5 kg against 20 N at 600 rpm, with 0.05 mm between the roller and the wall it is not bearing on. At each sign change of the groove's force the roller leaves its wall, and each curve is the gap it opens, from nought to the far wall at the top of the plot: leaves at 47.9° and lands 6.6° later at 80 mm/s; leaves at 87.1° and lands 7.1° later at 63 mm/s; leaves at 151.4° and lands 14.6° later at 35 mm/s; leaves at 193.6° and lands 13.4° later at 42 mm/s. Dragging the speed moves the crossings, lengthens or shortens each flight, and just above a stretch's threshold shows a flight that never reaches the far wall.

The time a crossover takes

A roller in a groove changes walls where the groove's force changes sign, and it gets there by flying across the clearance. How hard it lands depends on the clearance through an exponent the motion law decides — two thirds where the force passes through nought, one half where it steps — a flight that ends in a dwell lands at √(2cF/m) whatever the speed, and just above each threshold the force changes sign and the roller never arrives at all.

cams · Cam The acceleration pole walks round the inflection circle. The four-bar of this field — ground 4, crank 1, coupler 3.5, rocker 3 — at a crank angle of 240°, with its inflection circle, the locus of coupler points whose paths are momentarily straight, of diameter 9.23. The cross is the velocity pole and the ring is the far end of the diameter from it, the inflection pole. The square is the acceleration pole when the crank is driven with an angular acceleration of 0.323 times its angular speed squared. It is on the circle, at 30.0° from the diameter, which is arctan of the coupler's angular acceleration over its angular speed squared. The circle does not move when the drive changes; only the square does. Dragging changes the crank's acceleration.

The other pole is on the circle

The point of a moving plane with no acceleration depends on how fast the input is driven, which is why it was set apart from the rest of this field as the one object that is not geometry. It is not free, though. It always lies on the inflection circle — the geometric locus of points going straight — at an angle from the circle's diameter whose tangent is the plane's angular acceleration over its angular speed squared. A drive chooses one angle, and the geometry has already chosen everything else.

curvature · Curvature The same cycloidal rise behind a bent joint. A cycloidal rise over 120° of cam angle, with the camshaft driven at constant speed through a Hooke's joint bent at 30°. Across: the fraction of the rise; up: the follower's acceleration in time, divided by its peak on a steady shaft. The joint turns the camshaft unevenly with a period of half a turn, so where the rise falls in that period — the clocking, the cam angle at which the rise begins, measured from where the shaft turns fastest — decides what the follower feels. At the worst clocking, 100°, the peak is 1.352 times the steady one; at the best, 30°, it is 0.828 — lower than on a steady shaft. Every curve still starts and ends at nought. Dragging changes the bend.

A cam behind a bent joint

A cam's motion law is written in cam angle and delivered in time, and the two are the same only on a shaft that turns steadily. Drive the camshaft through a Hooke's joint and the follower's acceleration gains a second term — the law's velocity times the shaft's angular acceleration — that a steady shaft does not have. Where the rise falls in the joint's half-turn then decides everything: at a 30° bend the same cycloidal rise peaks 35% above its steady value or 17% below it. And past a bend of 34.5° a well-clocked cycloidal cam gives its follower less peak acceleration than the constant-acceleration law, the one designed to minimise it.

cams · Cam The third-order pole goes where the drive puts it. The four-bar of this field at a crank angle of 60°, with its inflection circle. The cross is the velocity pole and the ring Ball's point, where the circle meets the cubic of stationary curvature. The square is the third-order pole — the coupler point whose jerk is nought — when the crank is driven with an angular acceleration of 1 and a jerk of 6, each per the right power of its speed. It is at (0.492, 10.663). The faint curve is where it goes as the jerk alone is varied: a circle through the velocity pole. Every point off the inflection circle is the third-order pole of exactly one drive; on the circle only the velocity pole and Ball's point can be. Dragging changes the drive.

The third pole goes anywhere but the circle

The velocity pole is fixed by the geometry and the acceleration pole by the geometry up to one number, its place on the inflection circle. The next one, the point of a moving plane whose jerk is nought, depends on two numbers of the drive — and two numbers are enough to put it anywhere. Every point of the plane off the inflection circle is the third-order pole of exactly one drive, because a body point's jerk is affine in the crank's acceleration and jerk. On the circle only two points qualify: the velocity pole, and Ball's point, where the path is straight to third order.

curvature · Curvature A cycloidal rise on a chain-driven camshaft. A cycloidal rise over 120° of cam angle, the camshaft driven by a roller chain moving at steady speed over an 11-tooth sprocket. The chain leaves the sprocket at a radius that varies as each pin seats, so the shaft speeds up and slows once per tooth, and its angular acceleration jumps from one sign to the other at every seating. The follower's acceleration (solid) therefore steps at every tooth — 4 or so inside this rise — by up to 0.414 of the steady peak, and the cycloidal law's continuous acceleration, the reason it is chosen, is lost. At the worst clocking against the teeth the peak is 1.165 times the steady shaft's (dashed). Dragging changes the tooth count.

A cam on a chain steps at every tooth

A roller chain leaves its sprocket at a radius that shrinks and recovers as each pin seats, so a chain-driven camshaft speeds up through every tooth and its angular acceleration jumps back at every seating. That puts a step into the follower's acceleration once per tooth, whatever the law — a cycloidal rise, chosen for its continuous acceleration, gets four steps inside a 120° rise on an eleven-tooth sprocket, each 0.41 of its peak. The harm falls with the tooth count, as one over it, not up as its square, and the simple harmonic law, whose peak sits where its velocity is nought, is hurt least of all.

cams · Cam A curved slot bends at its mouth with the crank's circle, or against it. A 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.

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.

timing · Intermittent The crank's acceleration fixes its jerk, so the two numbers a point needs cannot both be met. For the coupler point (0.45, 0.5) of the standard four-bar, over the crank's stroke from 30° to 150°: the second schedule number λ = θ⃛/θ̇³ that, with the first κ = θ̈/θ̇², makes the point's jerk nought at each angle; and the λ that a crank actually following that κ has, κ′ + 2κ², since both numbers come from one function, the crank's speed against its angle. Their difference, the mismatch, reaches 1.77 and is nought only at 39.8°. Wherever it is not nought no crank holds the point's jerk at nought.

No drive holds the third pole

At one crank angle a drive's acceleration and jerk can put the coupler's third-order pole on almost any point, making that point's jerk nought. Over a stretch of the stroke they cannot. A crank's schedule is one function, its speed against its angle, and its jerk number is fixed by its acceleration number: λ = κ′ + 2κ². So the two numbers a point needs at every angle must satisfy that relation, and on the standard four-bar they miss it by as much as the numbers themselves, agreeing only at isolated angles. The best schedule a crank can follow takes a fifth to a third off a coupler point's jerk, and no more.

curvature · Curvature Where on the coupler a scheduled crank is smoothest. The standard four-bar's coupler plane, A at the origin and B at (1, 0), shaded darker where a point's jerk per unit of its own travel over the stroke from 30° to 150° is smaller, with the crank's speed scheduled to make that point's jerk least: from 1.34 to 5.12, measured in time over a stroke of fixed duration. The smoothest scheduled point is (−0.02, −0.80), at 1.263 against 4.555 with a steady crank; the smoothest with a steady crank is (−1.00, 1.25), at 3.397. Ball's point at the stroke's middle lies at (1.55, 3.27), off the map, where it scores 1.702 scheduled.

The smoothest coupler point is chosen in time

A crank's speed can be scheduled to cut a coupler point's jerk, though never to nought. Which point should a designer free to choose it pick? Measured as the schedule essays measured jerk, per crank speed cubed, the answer is a point whose schedule swings the crank's speed four to one, and it is a cheat: that measure divides by the speed it changes. Measured in time over a stroke of fixed duration, per unit of the point's own travel, the smoothest point of the standard four-bar is (−0.02, −0.80) in the coupler's frame. Its schedule cuts its jerk by 72% with a speed swing under two to one, and it is 63% smoother than the best a steady crank finds anywhere. Ball's point, the instant's straightest, lies in a second, shallower valley four coupler lengths away.

curvature · Curvature A crank that must come round: slow through the stroke, quick through the return. The crank's speed over a whole turn, as a share of its mean, under periodic schedules that leave it at the speed it started with and give the working stroke, 30° to 150° (shaded), a third of the cycle: smoothest stroke point, return nearly free (−0.02, −0.80), from 0.66 to 1.87; smoothest stroke point, return weighted a thousandth (−0.02, −0.80), from 0.83 to 1.46; the turn's own point, return weighted a thousandth (1.44, 1.89), from 0.81 to 1.35. Highlighted: the turn's own point, return weighted a thousandth.

A crank that must come round picks another point

A crank scheduled over one stroke can cut the smoothest coupler point's jerk by 72%. A crank running continuously must come back to the speed it started at, and its working stroke must keep its share of the cycle. Held to both, the same point keeps two-thirds of its gain only by putting five and a half times a steady crank's jerk into the return, and no schedule beats a steady crank on both strokes. Choose the point for the whole turn instead and a gentle schedule, its speed swinging only 1.66 to one, cuts the working stroke's jerk by 74% and the return's by 18% together. That point is three coupler lengths from the first.

curvature · Curvature

Named alongside it

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

Motion lawDwellJerkFollowerthe Geneva mechanismcycloidal motionIndexingMoving planeAcceleration poleCamConstraintInflection circle

All concepts