Forks on a tree sized by stress

Da Vinci’s rule leaves a free constant in the cost that fixes a fork’s angle, and one fork cannot refute it, because some value of the constant reproduces any angle. Across a tree it can: the constant is one number for the whole tree, a fork’s share of it falls as the square of the fork’s radius, and a tree spanning a factor of ten in radius must fan from 29.4° at its biggest forks to 111.6° at its smallest, while Murray’s rule opens every fork at 74.93°.

That comparison put the fork on area-conserving radii and gave no reason a tree would have them. A cube law with a lever arm gave one: size every branch so that equal loads on the tips bend it to one stress, and a junction conserves $r^p$ with $p = 3/(1 + \log_2(1/\lambda))$, where λ is the length of a branch over its parent’s. A planar crown gets exactly two. The question here is what fork angle the transport cost predicts on a tree sized that way, at every λ.

Flow through a tree sized by stress #

The cost the angle comes from is the cube law’s: pumping per unit length goes as $aQ^2/r^4$ and upkeep as $b\,r^2$, and a fork’s branch point sits where the sum of three weighted lengths is least. The weights are each segment’s cost per unit length.

On a tree sized by stress the radii are not the ones that minimise that cost, so the pumping term does not settle at half the upkeep as it does on Murray’s tree. It has to be carried. Flow is proportional to the tips a branch feeds, and on a tree conserving $r^p$ with equal tips a branch carrying N tips has radius $N^{1/p}$, so the flow in a branch of radius $r$ is $q\,r^p$. The weight is then $b\,r^2 + a\,q^2 r^{2p-4}$, which is $b$ times

$$r^2 + c\,r^{2p-4}, \qquad c = a\,q^2/b$$

and c, like Da Vinci’s constant, is one number for the whole tree: two cost coefficients and the flow a single tip draws.

The constant’s exponent is 2p − 6 #

Scale a fork by s. Every weight becomes $s^2\,(r^2 + c\,s^{2p-6}\,r^{2p-4})$, and a fork’s angle depends only on the ratios of its weights, so the scaled fork has exactly the angle of the unit fork at $c' = c\,s^{2p-6}$.

At p = 2 that exponent is −2, which is Da Vinci’s constant divided by the square of the fork’s size, and the weights are $r^2 + c$ — the same weights, to the last digit, as the rule that predicts everything at one fork. At p = 3 the exponent is nothing, the two terms of the weight go as the same power of the radius, and c cancels out of every ratio: Murray’s angle at every size.

In between, the constant is still a radius axis, but a shorter one. A tree conserving 2.25 moves its constant by a factor of a thousand across a hundredfold range of radius where a planar crown moves it by ten thousand.

Every angle the weights allow #

The trend has two ceilings, and both have closed forms at an even fork. As c′ goes to nothing the weights are the squared radii, the parent’s weight is $2^{2/p}$ times each daughter’s, and the cosine of each half-angle is $2^{2/p - 1}$. As $c'$ grows without bound the weights go as $r^{2p-4}$ and the cosine is $2^{1 - 4/p}$.

For a planar crown those are 0° and 120°, Da Vinci’s whole range. At p = 3 both are Murray’s 74.93°. In between, the range closes on Murray’s angle from both sides as the crown shortens more slowly: 44.40° to 108.64° for a crown filling a volume, 63.26° to 92.79° at a length ratio of 0.9, and 69.57° to 84.33° at 0.95.

Worked at the drawn trees’ length ratio #

The trees drawn beside the cube law shrink each branch to 0.74 of its parent. A stress rule at that ratio conserves $p = 3/(1 + \log_2(1/0.74)) = 3/1.4344 =$ 2.0915.

The narrow limit’s cosine is $2^{2/2.0915 - 1} = 2^{-0.0437} = 0.9701$, a half-angle of 14.03° and a total of 28.07°. The wide limit’s is $2^{1 - 4/2.0915} = 2^{-0.9125} = 0.5316$, a half-angle of 57.89° and a total of 115.82°. So a tree at the drawn ratio can show any angle from about 28° to about 116° depending on the size of the fork, against Da Vinci’s 0° to 120° — a range narrowed by a quarter at the bottom and hardly at all at the top.

The constant that puts its unit fork at Murray’s angle is c = 0.6723, against Da Vinci’s 0.7024. Everything else on that tree is read off one curve and one number.

Worked at a crown that fills a volume #

A crown whose branches fill a volume shortens each generation by $2^{-1/3} = 0.7937$, and a stress rule there conserves exactly p = 9/4. The constant’s exponent is 2p − 6 = −1.5, so a fork a hundred times smaller than another sees its share of the constant grow by 100^1.5, a thousand.

Its narrow limit’s cosine is $2^{2/2.25 - 1} = 2^{-1/9} = 0.9259$, a half-angle of 22.20° and a total of 44.40°. Its wide limit’s is $2^{1 - 4/2.25} = 2^{-7/9} = 0.5832$, a half-angle of 54.32° and a total of 108.64°. The calibrating constant is 0.6281, and the tenfold fan it gives, 53.9° to 97.6°, uses 43.7° of the 64.2° the weights allow at that length ratio, a share of 0.68. A planar crown’s tenfold fan uses 82.1° of 120°, also 0.68; the drawn trees’ 65.3° of 87.8°, 0.74.

Past the volume-filling crown the share falls as well as the range. A crown at 0.87 has 38.7° of room and its tenfold fan uses 19.5° of it, half; at 0.9 the fan uses 12.0° of 29.5°, four tenths. Two things narrow together as a crown shortens more slowly: the weights leave less room for a trend, and the constant’s exponent is shorter, so a tenfold range of sizes walks less far along the room there is.

Across a tenfold tree #

Centre each tree’s constant on the size where an even fork opens at Murray’s angle, and read the forks a factor of $\sqrt{10}$ either side. A planar crown runs from 29.4° to 111.6°, a spread of 82.1°, which is Da Vinci’s tree exactly. At the drawn trees’ 0.74 the fan runs from 41.1° to 106.4°, a spread of 65.3°. A crown filling a volume runs from 53.9° to 97.6°, a spread of 43.7°.

Then the fan closes fast. At a length ratio of 0.84 it is 28.2°, at 0.87 it is 19.5°, at 0.9 12.0°, and at 0.95 just 3.2°. With no shortening at all the angle is Murray’s at every size, because a tree that does not shorten conserves three.

Against one fork’s flatness #

The trend only matters if a fork is held tightly enough to show it, and a fork is held very loosely: one per cent of a network’s cost buys 43.3° of total angle at an even fork. A trend narrower than that could sit inside the looseness of every fork on the tree.

Against that band the answer is sharp. A tenfold tree’s spread equals it at a length ratio of 0.795 for even forks and 0.784 for forks whose daughters are in the ratio one to two. A crown filling a volume shortens by $2^{-1/3} = 0.794$. So only crowns shortening about as fast as a volume-filling crown or faster carry a trend wider than one fork’s flatness across a tenfold range, and a crown shortening by 0.87 fans across less than half of it.

The equality at the volume-filling ratio is a numerical coincidence of this cost and this band rather than a derived result, and it is stated as one.

Forks with unequal daughters #

Lopsided forks carry a slightly smaller trend at every length ratio. With daughters in the ratio one to two, a tenfold tree’s spread is 77.7° on a planar crown, 60.8° at 0.74, 39.8° for a volume-filling crown, 25.3° at 0.84, 17.5° at 0.87, 10.8° at 0.9 and 2.9° at 0.95 — between five and eleven per cent less than an even fork’s at the same ratio, the gap widening as the crown shortens more slowly.

That moves the crossing with the one-per-cent band from 0.795 to 0.784, so a crown whose forks are mostly lopsided needs to shorten a little faster than a volume-filling one before its trend clears one fork’s flatness. The fork-constant measurement found that forks of two shapes sit on one trend displaced along it in size, and a test that recorded each fork’s daughter ratio beside its size and angle would carry that displacement rather than being blurred by it. The same holds here, with a displacement computable from the shape and the length ratio together.

A decade of radius #

The steepness of the trend puts a number on how much of a tree a test needs. At the size where a tree’s constant gives Murray’s angle, the predicted total angle falls by 105.1° for each factor of ten in radius on a planar crown, 83.2° at 0.74, 52.8° for a volume-filling crown, 32.1° at 0.84, 21.4° at 0.87, 12.8° at 0.9 and 3.2° at 0.95.

Da Vinci’s tree was steep enough that two forks a decade apart, scattered by twenty degrees, resolved its slope. A crown at 0.87 moves a fifth as far per decade, and at the same scatter it would need about twenty-four times as many forks for the same significance, since the forks needed go as the square of the scatter over the slope, and $(105.1/21.4)^2$ is 24.1.

Reading a length ratio off a trend #

The slope is a steady function of the length ratio, falling at every step from 105.1° a decade to nothing, so it runs both ways. A crown whose length ratio has been measured predicts a slope; a crown whose fork angles have been measured against their sizes reports one; and the two can disagree. The slope is steepest where the forks open near Murray’s angle and flattens towards both ends of the curve, so the comparison belongs there. A crown measured at 0.74 whose forks around 75° change by 20° a decade of size, or by 100°, is a crown the stress rule does not describe, whatever its exponent turns out to be. That is a condition under which this account is withdrawn for that tree, and it needs no value of the constant.

Fork size need not be read with calipers either. On a tree conserving $r^p$ a fork’s radius is its tip count to the power $1/p$, so a decade of radius is p decades of tips, and a count of tips has no measurement error in it. On a planar crown a decade of radius is a hundredfold range of tips; at 0.74 it is a range of 123; for a volume-filling crown, of 178. The trend is a statement about tip counts as much as about radii, and tip counts are the easier half to take.

Why the planar crown is Da Vinci’s tree #

The coincidence at p = 2 is not a coincidence. A planar crown sized by stress conserves cross-section, so it is an area-conserving tree, and on an area-conserving tree the flow per unit of cross-section is the same in every branch. The pumping term $aQ^2/r^4$ is then the same in every tube, which is the additive constant the fork-constant measurement found.

So the stress rule does not merely reproduce Da Vinci’s exponent at a planar crown; it reproduces Da Vinci’s angle cost too, constant and all, and every result about Da Vinci’s fork angles applies to a planar stress-sized crown unchanged. What the stress rule adds is the rest of the length-ratio axis, where neither Da Vinci’s rule nor Murray’s has anything to say.

What hiding the trend costs #

An angle spread and a cost band are different quantities, and the direct question is what a tree would pay to hide its trend by holding one angle at every size. At the biggest forks of a tenfold planar crown, holding Murray’s angle costs 1.69 per cent above that fork’s own optimum, and at its smallest forks 4.39 per cent — Da Vinci’s numbers.

At 0.74 the costs are 1.34 and 3.06 per cent; for a volume-filling crown 0.66 and 1.43; at 0.87, 0.19 and 0.27; at 0.9, 0.06 and 0.10. Every tree past the volume-filling ratio could show Murray’s angle at every size and pay less than a third of a per cent for it — an amount no fork measurement could see.

A hundredfold tree #

A larger range of sizes buys more trend, but slowly on a slowly shortening crown. Across a factor of a hundred in radius a planar crown spreads its forks over 109.5°, the drawn trees’ ratio over 84.4°, a volume-filling crown over 59.9°. A crown at 0.87 spreads over only 31.2° — still inside a single fork’s flatness across a hundredfold range.

So on a slowly shortening crown the trend is not a matter of measuring more of the tree. It is too shallow at every range a tree has, and the only honest reading is that such a tree cannot tell a stress-sized crown from a flow-sized one by its angles.

The drawn tree, fork by fork #

Everything so far used forks on the interior exponent. A real stress-sized tree does not conserve that exponent near its tips: the outer generations fall away, and a planar crown’s outermost junctions conserve as little as 1.35. So each fork’s weights are better read from its own radii and its own tip count than from a formula.

On an eleven-generation planar crown, with the constant set so that the fork six generations from the tips opens at Murray’s angle, the forks from the trunk outwards open at 13.0°, 19.2°, 27.9°, 39.9° and 55.8°, then 74.9°, then 95.3°, 113.9°, and 120° at the last two. The interior formula at the same sizes predicts 16.2°, 22.8°, 31.8°, 43.6° and 58.4° inward, and 90.8°, 103.6°, 112.1° and 117.0° outward.

Why the outer forks overshoot #

The inner forks sit between 2.6° and 3.9° inside the prediction, and the outer forks run past it to the ceiling, and both have the same cause. Near the tips a stress-sized branch is thinner than the interior exponent says for the flow it carries, so its pumping term is larger than the formula assumes and its weight is nearer the equal-weight case that gives 120°.

The same happens on every crown measured. At 0.74 the forks run 31.5° at the trunk to 120° at the last two generations; for a volume-filling crown 46.9° to 120°; at 0.87, 60.0° to 115.4° and then 120°. And the control shows that it is the sizing and nothing about the procedure: the same tree sized for flow opens every fork at 74.934622°, to every digit printed, whatever the constant.

Drawn #

Drawn with branch lengths shrinking by their length ratio and each fork opened at its own predicted angle, a planar crown closes at its trunk and flares at its twigs, and a crown at 0.87 keeps its forks inside a narrower range until its outermost generations throw them wide.

The drawing is a picture of the prediction, not of a tree, and the trees drawn at no angle are the reminder of what happens when a picture is mistaken for the other thing. Nothing about the two crowns below was chosen to look right; every angle in them is a weight triangle.

Which trees could test it #

The test needs three measurements on one crown: the length ratio, fork angles across a range of fork sizes, and the size of each fork. The predicted trend is steep enough to see only on crowns shortening by about 0.8 a generation or faster, and on such a crown two decades of fork size would put a spread of 60° to 110° beside a flatness of 43°.

On crowns shortening more slowly a flat set of angles is not evidence for Murray’s rule over stress sizing. Both predict it. What a mechanism would have to show is a derivation whose parameters are measured, and here the parameter is λ: without it, a trend in fork angle cannot be read against any rule at all.

What it assumes #

That the flow each tip draws is the same across the tree, so that c is one number. That radii set by stress and angles set by the transport cost belong together, which is the combination the fork-constant measurement accepted for area-conserving radii and is carried here unchanged rather than defended. And that the tree is symmetric and self-similar, with forks of equal daughters or of one fixed ratio.

A real crown varies its length ratio from trunk to twig, carries loads on its wood as well as its tips, and has forks whose daughters differ in both size and length. Nothing here measures one.

What would withdraw it #

A fork scaled by s whose angle is not the unit fork’s at $c\,s^{2p-6}$. A planar crown whose angles differ from Da Vinci’s, or a crown conserving three whose angle depends on the constant. A tenfold spread that fails to fall as the length ratio rises, or crosses the one-per-cent band far from 0.795. A tree sized for flow whose forks do not all open at one angle. Each is checked every time the measurement runs.

Still open: a crown sized two ways at once #

A trunk carries sap and bends in wind at the same time, so a real crown is sized by something the two requirements agree on, and the exponent it conserves can change from trunk to twig as one requirement hands over to the other. The next test is a tree sized by the larger of the two — flow where it binds, stress where it binds — and whether the fork-angle trend changes sign at the generation where the sizing hands over.