Concept

Prediction — where it appears

A statement about a measurement not yet made, recorded before it is. Several here have been made and then tested in a later round, and the ones that failed are kept beside the ones that held.

Named by 33 essays across 5 fields — each of them below, with the objects they name alongside it.

Where the disc's counts change, predicted from a cylinder. The dashed lines are the transition radii the cylinder's ladder gives through h = c²/4πr², with nothing fitted. The dots are what the blind counter returns from the disc: 15 of 16 bands agree, and the ones that do not sit on a transition.

A disc is a cylinder

Vogel's seed head makes the rise fall as one over radius squared, so a disc is not one lattice but a family of them. Feed that into the cylinder's ladder and it predicts where a sunflower's spiral counts change — with nothing fitted, and against a counter that never sees either model.

cylinder · Rise and radius
The step floor derived, against the step floor measured — exact in 12 of 12, with 3.2 over 3.5 turns marked. One step of the dividers subtends half a turn at the inner end when it reaches the square root of the growth factor less one, which puts the floor at the factor to the power of the span, less one, over that. The smallest count at which the factor actually comes back exactly is then found by bisection, and the two agree in 12 of 12 cases: 74 steps for a 3.2 spiral over three and a half turns and 521 for a golden one. The three that appear not to agree are the ones whose floor falls below the fit's own nine-point minimum, where it cannot be observed.

A measurement in steps

Walking a pair of dividers along a shell's spiral is the oldest way to measure it and the best one available once there are enough steps, because it puts the points where the curve is. Under a count that follows exactly from the geometry it inflates the answer instead, and it is the only route measured here that pushes a nautilus towards a golden spiral.

shells · Spiral fit
The three weights at an exponent of 2 and 3: one closes a triangle and one is a straight line. A branch point minimising a weighted sum of three lengths has an interior solution only when the three weights close a triangle, and the weight on a segment here is its own cross-section. At an exponent of 3 the three areas clear that condition by 0.4126, and the triangle they close is what the two fork angles are read off. At an exponent of 2 the parent's area is exactly the daughters' areas summed — that is what area conservation says — so the slack is -4.44e-16, the triangle collapses onto a line, and the fork closes to 0.0000 degrees. Leonardo's rule does not predict a different angle here; it predicts no angle.

A rule that predicts everything

Leonardo's rule says a fork conserves cross-section, and cross-section is exactly the weight the branch point is minimised against. So the three weights land on the boundary of the triangle inequality, the cosine comes out at one to the last bit, and the rule predicts no angle at all — and the free constant its own derivation leaves behind then walks the prediction across every angle a fork could have.

branching · Fork angle
One tree sized for flow and for stress, with each branch 2^(−1/2) the length of its parent. The same symmetric tree, 8 generations deep, each generation's branches 2^(−1/2) the length of the one before and turned 30° at every fork, sized two ways and drawn to one trunk width. On the left each branch's radius cubed is proportional to the tips it feeds — Murray's flow rule — and every junction conserves r³. On the right each branch is sized so that the same load on every tip bends it to the same stress at its base, radius cubed proportional to the sum of its lever arms to its tips; its trunk junction conserves r to the power 1.967, its outermost junctions 1.349, against a deep-tree limit of 2.000. The two trees thin at different rates from the same trunk.

A cube law with a lever arm

Murray's exponent of three comes from moving fluid for the least work, and Da Vinci's two has had no derivation here, only the name of the mechanical answer. Size every branch so that the same wind on every tip bends it to the same stress, and a junction conserves r to the power 3/(1 + log₂(1/λ)), where λ is how much shorter each branch is than its parent. A crown that fills a plane gives exactly two; halving lengths gives one and a half; no shortening gives three. Murray's flow rule gives three at every λ, so the lengths of a tree's branches say which mechanism sized it.

branching · Murray
A deeper rule passes more of a drift, not less. The wander left in a stem's divergences, against how many organs its disturbance stays correlated over, for rules whose neighbourhoods run from 3 organs to 182. The prediction under test said a rule should pass a drift once the drift outlasts its neighbourhood, so the shallow rules should be the leaky ones and each line should turn where its own depth is crossed. Every line rises smoothly and the deepest rule is the highest of them at every correlation length — 82 against 22 at the longest drift. There is no crossover anywhere in the sweep.

The drift goes the other way

A rule that corrects what its neighbourhood shares should let through any disturbance slower than its own reach, and should suppress anything faster — a crossover, tracking the depth. Swept over a neighbourhood that changes by a factor of sixty, there is no crossover anywhere, and the deep rule passes nearly four times as much as the shallow one. The prediction is not weakly supported; it is backwards.

mechanism · Noise colour
The fork angle Da Vinci's rule predicts, against the size of the fork, in one tree. A tree has one value of the constant Da Vinci's rule leaves free, so a fork's share of it falls as the square of the fork's size and the angle the rule predicts changes with size. For even forks it is 119.1° at a relative size of 0.1, 101.7° at a relative size of 0.5, 74.9° at a relative size of 1, 44.6° at a relative size of 2, 9.6° at a relative size of 10, where Murray's rule gives 74.93° at every size; for daughter ratio 0.5 it is 118.9° at a relative size of 0.1, 101.3° at a relative size of 0.5, 77.6° at a relative size of 1, 48.9° at a relative size of 2, 10.9° at a relative size of 10, where Murray's rule gives 77.58° at every size. From a total of 100° to 20° at an even fork is a factor of 8.93 in radius.

One constant for every fork

Da Vinci's rule leaves a free constant in the cost that sets a fork's angle, and running it over its range walks the predicted angle from nothing to 120 degrees, through Murray's 74.93. So no single fork can refute the rule. But the constant is one number for a whole tree, and a fork's share of it falls as the square of the fork's size — the constant is a radius axis. A tree spanning a factor of ten in radius must show forks from 29.4 degrees at its biggest to 111.6 at its smallest, a spread wider than one fork's flatness can hide, while Murray's angle is the same at every size.

branching · Fork angle
Which lattices survive a fifth of a degree of noise. The share of runs that still have a lattice. The prediction was that a cut-off at this range would be as fragile as the truncation it replaces; it is not. Stating the neighbourhood as a function of distance did not merely make the old result honest — 100% of runs survive against 33%, at a scatter an inverse-cube rule cannot be told from.

The fragility belonged to the window

A pattern that exists only because the rule cannot see far was expected to be held together by that cut, and to fall over when nudged. It does — while the cut is a loop bound. Written down as a falloff at the same range, the same rule keeps every run under the same nudge, at a scatter an inverse-cube rule cannot be told from.

emergence · Noise amplitude
The fork angle the transport cost predicts on a tree sized by stress, against the size of the fork. A tree sized so that equal loads on its tips bend every branch to one stress conserves rᵖ with p set by how much shorter each branch is than its parent, λ. On such a tree the cost that fixes a fork's angle carries one constant, entering a fork of size s as s^(2p − 6), so the predicted angle changes with size unless p is three. With the constant set so that a fork of relative size one opens at Murray's 74.9°: At λ = 0.707 (p = 2.000) an even fork opens at 119.1°, 111.6°, 74.9°, 29.4°, 9.6° at relative sizes of 0.1, 0.32, 1, 3.2 and 10; at λ = 0.74 (p = 2.091) an even fork opens at 114.5°, 106.4°, 74.9°, 41.1°, 30.1° at relative sizes of 0.1, 0.32, 1, 3.2 and 10; at λ = 0.794 (p = 2.250) an even fork opens at 106.3°, 97.6°, 74.9°, 53.9°, 46.4° at relative sizes of 0.1, 0.32, 1, 3.2 and 10; at λ = 0.87 (p = 2.498) an even fork opens at 92.7°, 85.5°, 74.9°, 66.0°, 61.5° at relative sizes of 0.1, 0.32, 1, 3.2 and 10; at λ = 0.95 (p = 2.793) an even fork opens at 78.4°, 76.6°, 74.9°, 73.4°, 72.2° at relative sizes of 0.1, 0.32, 1, 3.2 and 10. Murray's rule opens every fork at 74.9°.

Forks on a tree sized by stress

Da Vinci's rule predicts that a tree's forks open wider as they get smaller, because the constant in its angle cost is one number for a tree and enters each fork scaled by its size. A tree sized for equal bending stress conserves an exponent set by how much shorter each branch is than its parent, and on such a tree the same constant enters each fork as its radius to the power 2p − 6. The trend survives at every length ratio short of one and shrinks with it: 105 degrees a decade of radius for a crown filling a plane, 83 at a length ratio of 0.74, 53 for a crown filling a volume, 13 at 0.9. Only a crown shortening about as fast as a volume-filling one fans wider across a tenfold range than one fork's flatness can hide.

branching · Fork angle
One planar crown sized for a load on its tips and for the weight of its own wood. The same symmetric crown, 9 generations deep, each branch 2^(−1/2) the length of its parent and turned 30° at every fork, sized so that every branch is bent to one stress, drawn to one trunk width. On the left the load is on the tips and the trunk junction conserves r to the power 1.980; on the right the load is the weight of the wood, found by iterating the radii until they stop moving, and the trunk junction conserves r to the power 0.969. A crown sized for its own weight thins much faster from the trunk, because a branch's weight grows with the square of its radius.

A crown that carries its own wood

Sizing every branch so that equal loads on the tips bend it to one stress gives a crown filling a plane Da Vinci's exponent of two. Move the load onto the wood and the sizing becomes a fixed point, because a branch's load now depends on the radii being solved for. Under the wind on its wood a planar crown still conserves two, but only as a limit its trunk is two tenths short of at fifteen generations. Under its own weight it conserves one — radius rather than area, the stress-similarity law that radius goes as length squared — and a crown carrying leaves and wood reads the leaves' two near its twigs and the wood's one at its trunk, with the handover set by how much of the trunk's load the wood carries.

branching · Murray
The number of growing points after each season, for buds that wait no season, one, two, three or four. From one mature apex, each season every mature apex makes a new bud, and a bud branches only after it has waited its delay. With no delay the counts run 1, 2, 4, 8, 16, 32, 64, 128, 256, 512 and settle into growing by 2.0000 a season; with one season the counts run 1, 2, 3, 5, 8, 13, 21, 34, 55, 89 and settle into growing by 1.6180 a season; with two seasons the counts run 1, 2, 3, 4, 6, 9, 13, 19, 28, 41 and settle into growing by 1.4656 a season; with three seasons the counts run 1, 2, 3, 4, 5, 7, 10, 14, 19, 26 and settle into growing by 1.3803 a season; with four seasons the counts run 1, 2, 3, 4, 5, 6, 8, 11, 15, 20 and settle into growing by 1.3247 a season. On a logarithmic axis each settles into a straight line whose slope is its growth rate, the positive root of x^(d+1) = x^d + 1.

A count set by a delay

An L-system describes a plant and forbids nothing, because none of its parameters is anything a plant has. One branching grammar is the exception: a mature apex makes a new bud every season, and a bud waits d seasons before it branches. Its counts grow at the root of x^(d+1) = x^d + 1, a delay of one season gives Fibonacci's numbers and nothing else does, and the fourth count already separates a one-season wait from every longer one. So a Fibonacci count in a branching plant is a measurement of how long its buds wait. It is also a fragile one: if one bud in ten waits two seasons instead, eleven counts in a row come out Fibonacci's three times in a thousand.

branching · Lsystem
Which rule passes more drift — exponent 1.5 against exponent 6. The same disturbance is given to two placement rules, one with a falloff exponent of 1.5 and one of 6, and the bar counts how many of six seeds let more of it through under the deeper rule. Above the line means the deeper rule passed more. Reading left to right the disturbance is given a longer memory, from white noise to a correlation length of 99 organs. The two change hands: the shallower rule passes more up to a correlation length of 1.4 organs and the deeper one from 2.8 organs onward. The comparison is made seed by seed rather than between two averages, because the spread between seeds at one setting is larger than the difference being measured.

The corner that does not move

Read as degrees of drift getting through rather than as a ratio, and compared seed by seed, the deep and shallow rules change hands. The share that goes to the deeper rule climbs from twenty-three per cent under white noise to ninety-four at a correlation length of a hundred organs — and the crossing sits at two or three organs whether the two rules differ by a factor of four or sixty-one.

mechanism · Noise colour
The rate a branching count grows at, against the chance a growing point dies. Every point dies with probability q each season and the survivors rewrite as before, so the expected counts obey x^(d+1) = (1 − q)·x^d + (1 − q)^(d+1). Substituting x = (1 − q)y returns the deathless equation exactly, which makes every line here straight: the rate is the deathless root multiplied by the survival. No delay runs from 2.0000 to one at q = 0.5000; one season runs from 1.6180 to one at q = 0.3820; two seasons runs from 1.4656 to one at q = 0.3177; three seasons runs from 1.3803 to one at q = 0.2755; four seasons runs from 1.3247 to one at q = 0.2451. Below the marked line a lineage shrinks.

A count that loses its growing points

The branching grammar behind the Fibonacci claim has no deaths in it, and a stem that loses shoots is the common case. Giving every growing point a chance q of dying each season leaves the counts a linear recurrence and does exactly one thing to it: the growth rate becomes the deathless root multiplied by 1 − q, at every delay and every death chance, to the last bit a double holds. So each waiting time has a death chance above which its lineage shrinks — a half with no wait, 0.3820 at one season, 0.2451 at four — and a longer wait tolerates less. What does not survive is the count itself: a plant losing one growing point in ten a season shows eight Fibonacci counts in a row one time in ten thousand, against one time in eight for a bud that occasionally waits an extra season.

branching · Lsystem
What a removal costs the next organ. One mark per wrecked cut in the census: how far the first organ placed after the removal ended up from where the control put it. The rows split by which organ was taken. Removing a direct chain-neighbour of the growing tip — an organ at a multiple of one of the two counted numbers — moves the next organ by between 8.9 and 30.7 degrees. Removing anything else inside the front moves it by between 62.8 and 167.6. Nothing lands between the two groups and the ratio across the gap is 2.05, so the line is a gap rather than a threshold. Taking away a neighbour is the cheap removal, which is the opposite of what the words suggest.

Removing a neighbour costs least

Take away an organ that is a direct chain-neighbour of the growing tip and the next organ moves by under thirty-one degrees. Take away anything else inside the front and it moves by at least sixty-three. Thirty cuts, two groups, a factor of two between them and nothing in the gap.

mechanism · Ablation
Sizing for equal bending and sizing for equal stress cross at one length ratio. Equal stress holds r³ against the sum of a load's arms and gives 3/(1 + ℓ); equal deflection holds r⁴ against the sum of the arms squared and gives 4/(1 + 2ℓ), with ℓ = log₂(1/λ). Setting them equal gives 3(1 + 2ℓ) = 4(1 + ℓ), whose only root is ℓ = 1/2 — the crown that fills a plane, λ = 0.707107 — and there both are exactly two. Below that ratio the stiffness rule reads the lower exponent of the two and above it the higher, so the two criteria size the same crown at one length ratio in the whole family and it is the one Da Vinci's rule names.

A crown sized for how far it bends

Stress is one criterion for sizing a branch and stiffness is another. Holding every branch to the same deflection as a share of its own length sizes r to the fourth against the sum of each load's arm squared, where equal stress sized r cubed against the arm, and the junctions of a deep crown then conserve 4/(1 + 2·log2(1/λ)). A single cantilever under its own weight comes out at radius as length to the three halves — McMahon's elastic similarity, fitted here rather than assumed — against the square that equal stress asks for. And the two criteria agree at exactly one length ratio out of the whole family: λ = 2 to the minus a half, the crown that fills a plane, where both give exactly two.

branching · Murray
A period of 8, and the two classes that are not with the rest. The same wrecked stem, folded on the lag it kept: one row per residue class, each drawn at the mean displacement of its own organs against the level the rest of them share. The bar through each row is the spread inside that class, and the widest of them is 0.87° — so within a class the displacement is a constant. six classes sit at the common level. The two that do not sit at 134.3° and -134.2°, equal and opposite to within 0.0 per cent, and they are neighbouring residues. The stem's own divergence is 137.44°, so an exception is one organ's step.

One level and two exceptions

Inside a wrecked stem's period most residue classes sit at one level and a couple do not. On seventeen of the thirty cuts the exceptions are exactly two, equal and opposite to within five per cent — and on all seventeen they are neighbouring residues, which was not looked for.

mechanism · Damage shape
The exceptional pair, measured in divergences. One row per wrecked cut whose profile has exactly one pair of exceptional classes, drawn at the size of that pair divided by the stem's own settled divergence. Every row sits between 0.882 and 1.076, so the two chains that came apart moved by one organ's step rather than by two or by half of one. The residual is not scatter: rows are grouped by the lag the stem kept, and every lag sits wholly above the line or wholly below it. Why a surviving 5 or 7 overshoots and a surviving 4 or 8 falls short is not answered here.

A step of one organ

The balanced pair inside a wrecked stem's period measures 88.0° to 147.2° against divergences of 99.1° to 138.0° — one organ's step, to within twelve per cent on every row. The residual is not scatter: every stem keeping a 5 or a 7 overshoots and every stem keeping a 4 or an 8 falls short.

mechanism · Damage shape
How constant the displacement is inside one residue class. One row per wrecked cut in the census, drawn at the widest spread found inside any one residue class when the profile is folded on the lag that stem kept. 25 of 30 rows sit between 0.12 and 6.09 degrees, which on a quantity whose between-class differences run past a hundred and fifty degrees is a constant. The five that do not sit from 10.3° up. There is nothing in between, so the line drawn at 10° could have been drawn anywhere in a wide interval.

The plateau was a prediction

The search for a reference organ found that the largest displacement above a hole is a plateau rather than a peak, and reported it as a failure. A profile constant on each of k residue classes has exactly k levels, so its maximum is attained by a whole class — a ninth to a quarter of every window, forever.

mechanism · Damage shape
Fork angles on crowns sized by the larger of flow and stress, handing over at six different generations. Thirteen-generation crowns at a length ratio of 0.707, every branch sized by the larger of flow and stress, twigs at the radius the transport cost prefers, with the two radii equal at generations 3 to 8. Handing over at 3, the forks from the trunk open at 38, 51, 67, 75°; handing over at 4, the forks from the trunk open at 27, 38, 51, 67, 75°; handing over at 5, the forks from the trunk open at 19, 27, 37, 51, 67, 75°; handing over at 6, the forks from the trunk open at 13, 18, 26, 36, 50, 66, 75°; handing over at 7, the forks from the trunk open at 8, 12, 17, 24, 35, 49, 66, 75°; handing over at 8, the forks from the trunk open at 4, 6, 10, 15, 22, 32, 46, 65, 75°. Every fork from the handover outward opens at Murray's 74.9°. No fork on any of the six opens wider than that, and none turns back: the angle rises through the stress-sized generations and stops.

A trend that stops at Murray's angle

A crown sized by whichever of flow and bending stress asks for the thicker branch is sized by stress at its trunk end and by flow at its twigs, and the twigs fix the constant that was free in the fork-angle prediction: a twig at the radius the transport cost prefers spends exactly half its upkeep on pumping. On such a crown the fork angle does not change sign at the handover. It rises through every stress-sized generation, meets Murray's 74.93° at the handover and stays there, and no fork anywhere opens wider. The trend turns back only when the twigs are thinner than the cost wants — past a pumping share of (λ^(−2/3) − 1)/(1 − λ^(4/3)), 0.70 at the planar crown and closing on one half as branches stop shortening.

branching · Fork angle
Removing one wall of the slot, then the other, then both. The growing tip sits in a slot between its two chain-neighbours — the organ 5 places back and the organ 8 places back. Each bar is how far the next organ placed moves when those are removed, against a control sharing the same history. Either alone is a cheap removal: 26.3° and 4.9°, both inside the 45° that separates taking a neighbour from taking anything else. Together they move it 164.1°, against 31.2° for the two effects added, so the interaction is +132.9°. The slot is not two independent walls.

Both walls of the slot

The growing tip sits between its two chain-neighbours. Removing either alone is a cheap removal on all six lattices — 2.3° to 41.7°. Removing both together throws the next organ past the expensive line on three of them, and the interaction runs from −25.8° to +132.9°.

mechanism · Both walls
Which hops survive one wall, the other, and both. One row per lattice. The last three columns are the lags whose hop the cut stem still holds, unchanged from a control that shares its history — the measurement that identifies what a wrecked stem has become. Removing a single wall always leaves something standing, which is what every single-organ cut in this collection does. Removing both leaves nothing at all on two of six lattices, including the coarse rung that no single removal can wreck. A stem that keeps no rigid hop is not a wrecked lattice with a slip in it; it is a stem that is no longer a lattice.

The rung that two organs wreck

On the coarse 3/5 stem both walls of the slot heal when removed alone and wreck when removed together — and the wreck keeps no rigid hop at all. Two of the six pairs in the design end at a destination single removals almost never reach.

mechanism · Both walls
What a two-organ cut does at each rise of the 2/3 rung. At every rise the coarse rung is a lattice on, all 36 arrangements of two organs removed, with the ones that never repair counted and split by where they end up. 22 of 324 cuts across the rung reverse the stem's handedness onto the mirror of the divergence they were cut from. 40 fall instead into a cycle whose mean is half a turn, which the lattice they came from has no number for. 4 rises give only the first, 4 give only the second, and at a rise of 0.075 both happen in the same table, which is what says the fate belongs to the cut and not to the rise.

The shallower front turns over

If reversing a stem means rearranging its whole front, then a stem with a shallow front should reverse more often. Measured across three rungs and four hundred and seventy-three cuts: 6.8 per cent at a front of three, 4.7 at five, and none at all at eight — where the nearest approach is two tenths of a degree away and stays there.

cylinder · Coarse rung
A period of 8, with the hole's own chain at the top. Each mark is one residue class of the displacement profile, placed round a ring at its own residue, with the chain the removed organ sat on at the top. The radius is how far that class sits from the level the rest of them share. six of the eight classes sit together at the middle ring; two do not, and on this row they are one pair, equal and opposite to within a twentieth. The forward one is chain 7 and the backward one is chain 0, one residue above it, which is the order every row of the census puts them in.

One way round, seventeen times

The two chains that change places in a wrecked stem are adjacent, which is symmetric and says nothing about direction. Label them by lag from the hole and the one displaced forwards is always the lower of the two — on every row of the census, without an exception.

mechanism · Damage shape
Two lines across the 5/8 rung, crossing once. The divergence the rule settles on, against the divergence at which the two contact steps would be exactly the same length. The second is arithmetic on the lattice and no stem is grown for it. Across this rung the balanced line moves 2.281 degrees and the rule's own line moves 1.262, so the shallower line crosses the steeper one, and it does so exactly once at a rise of 0.0154 — 16 per cent of the way down from the coarse end. That crossing is the handover: above it one family has the shorter step and below it the other does. So a rung has one handover, its position is fixed by the arithmetic rather than by any experiment, and a sweep of the rise carries a stem across it at a place nobody chose.

Two lines that cross once

The divergence at which a lattice's two contact steps are exactly equal is a curve across each rung, computable from the geometry with nothing grown. The rule's own settled divergence is a second, shallower curve, and where they cross is where the step ordering changes hands.

cylinder · Rung interior
Four accounts of one angle, scored on the same 17 rows. Each bar is how far an account of the exchanged pair's size sits from the measurement, as a share of the control's divergence, averaged over the census. A step of the wrecked stem's own divergence is the obvious candidate and the worst of the four. The surviving family's own step is not the unit at all: that angle is a fifth of the exchange. A step of the control's divergence is close, and correcting it by a fifth of the surviving hop takes the worst row from 11.8 per cent to 4.0.

Four accounts of one angle

The exchanged pair in a wrecked stem is about one divergence step, and about is doing twelve per cent of work. Four candidate units were written down and scored on the same seventeen rows: the cut stem's own step, the surviving family's step, the control's step, and the control's corrected.

mechanism · Damage shape

A fifth of the hop

The exchanged pair misses one divergence step by up to twelve per cent, and the miss is not scatter: every row keeping a lag of 5 or 7 overshoots and every row keeping a 4 or an 8 falls short. Subtract a fifth of the surviving hop's own angle and the worst row is four per cent.

mechanism · Damage shape

The rung decides the sign

Twenty-four lattices where both walls of the slot are really there. Thirteen give a strongly positive interaction, at 85° to 135°; eleven give a negative or null one, at −25° to −0.5°. Nothing lies between. Every rung's lattices fall on the same side as each other.

mechanism · Both walls

Where the survivors meet

At a matched pair the two stems keep exactly the counted numbers their two pairs have in common — the 5 where 3/5 meets 5/8, the 8 where 5/8 meets 8/13, the 7 where 4/7 meets 7/11, and nothing at all where 3/5 meets 8/13. Four rows, including the empty one.

cylinder · Same angle

How wide a band should be

A band ends where the divergence has slid a twentieth of a degree, so its width should follow from how fast the divergence slides. Predicted from the rung's slope that is wrong by factors of 0.20 to 5.92; predicted from a stationary point it is 0.41 to 1.08, and the outlier is the rung that has no stationary point.

cylinder · Rung interior

A destination or a refusal

Sixty-three wrecked runs were regrown to twelve hundred organs and put to the settling table's own criterion, unchanged in every tolerance. The prediction written down before the sweep said twenty-nine would settle; seventeen do, and the prediction is wrong on its own side of the table as well as in its total.

mechanism · Settling

A steeper rule walls nowhere else

The account of the wall at the fine end was that the basin narrows because the neighbourhood deepens, which predicts a steeper falloff walling somewhere else. Grown at four exponents, the four columns settle 30, 31, 29 and 27 of 72 — a spread of 0.056 against an error of 0.058.

emergence · Settling

Two refinements that do not multiply

The design that located the wall did two things at once — doubled the starting angles and halved the rise spacing — and the arithmetic behind it assumed each would buy about a factor of two. The finer rises did ninety-nine per cent of the narrowing and the doubled angles added under one, because a bracket's ends are rises and no error bar can move them.

mechanism · Falloff exponent

The fourth band, cut whole

Three bands cut at every rise left one account of which bands change their answer standing, and the account was the one nobody had a reason to prefer. The band that would have killed it has now been cut, and it did not kill it.

cylinder · Rung interior

Six bands, one table

Every rung of this ladder that carries a handover now has a band grown on it and cut at every rise it holds, and four accounts of which bands change their answer are scored on all six at once. The survivor is right on every band that can test it, and the same table read one cell differently kills it.

cylinder · Rung interior

Named alongside it

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

Claim testingHonest limitsNegative resultMeasurementAblationControlFalsifiabilityDivergence angleLattice offsetParastichy pairRigid hopDa Vinci's rule

All concepts