How does an online shooter decide whether a shot hit when the two players have different pings? Three views of the same firefight, Alice at 250 ms, the server, and Bob at 50 ms, compare three kinds of lag compensation. Based on Lee and Chang (MMSys ’18).
Key results
Traditional lag compensation rewinds the world by the shooter’s ping, so a low-ping victim can be hit after reaching cover.
Ping-limited compensation stops that for shooters over the limit, but they then have to lead a moving target by about their ping.
Advanced lag compensation lets low-ping victims appeal a hit. In the authors’ 12-player study, shots behind cover fell from 85 to 5, with no significant change in hit accuracy.
Can a network deliver messages using only coordinates in hyperbolic space? This re-runs Kleinberg’s 2007 experiment and also covers the Poincaré disk and ball, coloring curved planes, and frequency reuse.
Key results
Greedy routing on Kleinberg’s hyperbolic coordinates always delivered. On physical coordinates it failed about 7–10% of the time in typical runs.
Hyperbolic routes stay short: 80th-percentile stretch near 1.25, against 4.0 for routing along the tree.
A hyperbolic circle of radius 6 is about 33 times longer than a flat one. That extra room is why trees fit.
Context
This started with OpenAI’s preprint of September 23, 2026, which shows the flat plane can’t be colored with 5 colors, so the answer to the Hadwiger–Nelson problem is 6 or 7. Flat-plane coloring is the picture behind cellular channel planning, so the next question was how the problem changes when the plane is curved.
Following that question went through the Shannon–Hartley theorem, Wi-Fi BSS coloring, routing tables and AI-RAN, and arrived at hyperbolic geometry and Kleinberg’s paper, the best-known result that routes a network with hyperbolic coordinates. The lab was built to see greedy routing, the Poincaré disk and ball, Möbius maps, Kleinberg’s numbers and frequency reuse on a curved plane working, not just described.
Kleinberg’s paper
Robert Kleinberg, “Geographic Routing Using Hyperbolic Space,” IEEE INFOCOM 2007. Greedy routing sends each packet to whichever neighbor is closest to the destination, using only coordinates and no routing tables. With ordinary map coordinates it can get stuck at a node that has no closer neighbor. Kleinberg proved that if every node gets a coordinate in the hyperbolic plane, computed from a spanning tree of the network using Möbius maps, some neighbor is always strictly closer, so greedy routing delivers on any connected network.
His test used 50 random nodes in a circle of radius √5, linked when at most 1 apart, about 9.8 links per node. Euclidean greedy routing failed on 9.3% of routes (23.3% with a barrier in the way). Hyperbolic greedy routing failed on 0%, which is the theorem. Its stretch was 1.2 at the 80th percentile, against 4.1 for routing only along the tree, and its busiest nodes carried 9.1% more load than with Euclidean routing, against 233% more for the tree. Both stay low because a dense network offers short cuts off the tree.
What the Poincaré disk and a Möbius map look like
Poincaré disk. The whole hyperbolic plane drawn inside a circle. The rim is infinitely far away, and straight lines are arcs that meet the rim at right angles. Every triangle in this tiling is the same size: the shaded one is no bigger than the slivers near the rim, because lengths are stretched by ds = 2|dz| / (1 − |z|²).Möbius map. A rigid motion of the disk, z ↦ (αz + β) / (β̄z + ᾱ) with |α|² − |β|² = 1. As you scroll, this one slides the tiling around the top of the rim. The shaded triangle changes shape on the page but keeps its true size, and no distance changes. Composing two maps is multiplying their 2×2 matrices; in Kleinberg’s algorithm a parent hands each child one matrix, and the child computes its own coordinate from it.
Hyperbolic geometry and greedy routing
How does greedy routing work?
Each node knows its own coordinate and its neighbors’ coordinates. It hands the packet to whichever neighbor is closest to the destination, and stops if none is closer. In physical coordinates that can get stuck at a node with no closer neighbor. Kleinberg embeds a spanning tree into the hyperbolic plane so that some neighbor is always strictly closer, so delivery never fails.
What coordinates does hyperbolic geometry use?
The same point can be written in several systems: the Poincaré disk (inside a unit circle), the Klein disk, the upper half-plane, the hyperboloid (one sheet of t² − x² − y² = 1 in 3D), or polar form with true distance from the center. Routing code usually uses the disk or the hyperboloid.
How do the saddle and the Poincaré disk relate?
They are two pictures of the same geometry. A saddle shows negative curvature locally: circles have circumference 2π·sinh r, and triangles add up to less than 180°, the shortfall equal to their area. Hilbert proved the whole plane can’t sit smoothly in 3D space, so the Poincaré disk draws all of it inside a unit circle whose rim is infinitely far away.
Why do trees fit so well?
With curvature −1 a circle of radius r has circumference 2π·sinh r, which grows exponentially; at r = 6 that’s 33 times the flat value. A tree’s levels also grow exponentially, so there is room for every branch.
What is the Poincaré ball?
The same model in more dimensions. Izgorodin embeds WordNet’s 82,115 noun synsets in a 10-dimensional ball and averages points through flat tangent spaces. In his tests the hyperbolic distances mattered most for choosing nearest neighbors, worth about 5.7 points of mid-level clustering accuracy over a Euclidean embedding.
Why were Kleinberg’s percentages so low?
For three different reasons. Hyperbolic failure was 0% because it is a theorem. Stretch and congestion stayed low because a dense network, about 9.8 links per node, offers short cuts off the tree that greedy routing takes.
Does hyperbolic space help pack data or reuse frequencies?
Packing works; frequency reuse doesn’t carry over to real radios. Packing: a hub with n spokes needs every leaf closer to the hub than to any other leaf, so the flat plane fits at most 5 spokes, while the hyperbolic plane fits any n once leaves sit at radius about 2 ln(n/π). Frequency reuse: interference depends on real distances in flat space. Even on paper, a hyperbolic tiling with 6 neighbors per cell needs the same 7 channel groups as hexagons, and puts 30 co-channel cells at the reuse distance instead of 6.
Routing tables and hash tables
Are routing tables hash tables?
Partly. IP routers match each destination against stored prefixes and take the longest match, using tries or special memory called TCAM. Exact-match hash tables are used elsewhere, for example in Ethernet switches’ MAC address tables.
Do routing tables still exist in the hyperbolic approach?
Big tables aren’t needed for forwarding: each node keeps only its neighbors’ coordinates. Two lookups remain. A sender must learn the destination’s coordinate, much as DNS turns names into addresses. And Kleinberg’s paper builds a distributed hash table on top: a key is hashed to a point in the hyperbolic plane and stored at the node nearest it, which his Theorem 3 shows is the same node from any starting point. Hash tables move up a layer; they don’t disappear.
Does an IP address say nothing about where a destination is?
IP addresses are hierarchical: one prefix covers a whole network. That grouping keeps the global routing table near a million prefixes instead of billions of devices.
Does every router update its tables when a phone moves?
Not in mobile networks. A phone’s data travels in a tunnel from the base station to the core; when it hands over to another cell, only the tunnel’s end moves. Base stations don’t hold global tables.
Are hyperbolic coordinates permanent?
No. They come from the spanning tree, so a device that attaches somewhere new gets a new coordinate that senders must learn, and each node joining or leaving costs O(n) operations on average to repair the embedding.
Does a node only need its neighbors’ coordinates?
Yes, and that’s the main saving. With Kleinberg’s embedding some neighbor is always strictly closer to any destination, so a packet never gets stuck.
Is the distance calculation Pythagoras?
No. The distance in the Poincaré disk is d(z, w) = arcosh(1 + 2|z − w|² / ((1 − |z|²)(1 − |w|²))). It’s cheap to compute, but it isn’t Pythagoras.
Does machine learning choose the coordinates?
Kleinberg’s coordinates come from a simple deterministic algorithm, a spanning tree plus Möbius maps, and the delivery guarantee depends on that exact construction. Hyperbolic graph neural networks are research models that learn embeddings for prediction tasks such as delay; they carry no delivery guarantee.
Do satellite networks look like trees?
A satellite laser mesh is usually modeled as a grid wrapped around the planet: each satellite links to neighbors in its own orbit and the adjacent ones. That isn’t tree-like. Because the motion is predictable, routes can be computed ahead of time.
Hash keys and elliptic-curve cryptography
Is a hash-table key “hyperbolic” the way elliptic-curve cryptography is?
No. Hash tables use ordinary fast hash functions, which turn a key into a bucket number and are designed to scatter nearby keys far apart. Elliptic-curve cryptography is public-key cryptography, used for signatures and key exchange: a public key is a secret number times a base point on a curve y² = x³ + ax + b over a finite field.
Does “elliptic” mean elliptic geometry?
No. “Elliptic” in elliptic curves comes from elliptic integrals, the formulas for the arc length of an ellipse. It has nothing to do with elliptic (spherical) geometry, and so nothing to do with its opposite, hyperbolic geometry. Over the complex numbers an elliptic curve is a torus, which is flat. ECC works with points over a finite field, where curvature has no meaning.
Do hashes and hyperbolic coordinates work alike?
They do opposite jobs. A hash scatters on purpose. A hyperbolic coordinate keeps structure on purpose, so that distance reflects the network and a router can measure progress toward a destination.
Does changing one character move an ECC key somewhere else?
That’s the avalanche property of hash functions. For ECC the closer statement is that a public key reveals nothing usable about the secret number behind it.
Do hashing and hyperbolic routing ever meet?
Once: Kleinberg’s distributed hash table hashes a key to a point in the hyperbolic plane, then routes greedily to it. A hash used directly as an address can’t tell a router which neighbor is closer.
Wi-Fi BSS coloring and hyperbolic routing
What is BSS coloring?
Wi-Fi 6 (802.11ax) adds a 6-bit “color”, 1 to 63, to each frame header. Neighboring networks still share the same channel; the color only tells a device whose frame it is hearing.
Does a device ignore a neighbor’s signal and transmit anyway?
Only under conditions. For a frame with its own color it holds off as usual. For a different color it may raise its detection threshold, anywhere from −82 up to −62 dBm, and transmit if the neighbor’s signal is weaker than that, while lowering its own transmit power in step.
Does BSS coloring cut interference?
It works the other way round: it accepts a bit more interference in exchange for more chances to transmit. Throughput in a crowded building goes up; interference doesn’t go down.
Do the Internet and Starlink route with hyperbolic geometry?
The Internet routes with tables: BGP between networks and OSPF inside them. Hyperbolic routing is research: Kleinberg proved in 2007 that it can’t get stuck, and Boguñá, Papadopoulos and Krioukov proposed mapping the Internet this way in 2010. SpaceX hasn’t published how Starlink routes.
Are hyperbolic coordinates physical positions?
No. They’re computed from the network’s spanning tree and have nothing to do with where the equipment is.
Can spreading nodes apart in hyperbolic space reduce interference?
No. Interference depends on real distances between real antennas. Virtual coordinates can spread nodes apart on paper, but the radios stay where they are.
Does hyperbolic routing avoid traffic jams?
That’s about congestion, not noise. In Kleinberg’s test, hyperbolic greedy routing put 9.1% more load on its busiest nodes than Euclidean greedy routing, and far less than routing only along the tree, which added 233%.
RAN and AI-RAN
The radio access network (RAN) is the radios, base stations and links back to the core; user data goes phone → radio link → base station → transport → core → Internet. The AI-RAN Alliance, launched in February 2024 with over 130 members by early 2026, works in three areas: AI-for-RAN uses machine learning to run the radio better, AI-and-RAN shares GPU servers between radio processing and AI workloads, and AI-on-RAN runs AI applications at the network edge.
Is a cellular network a tree?
Mostly, as a logical picture. Small cells usually connect over fiber or wireless backhaul to an aggregation point rather than through a macro tower, but the network is a hierarchy, and that hierarchy is where hyperbolic ideas could fit.
Do base stations keep lists of every node?
No. A base station tunnels each phone’s traffic to the core network. The RAN does no per-destination routing.
Does AI-RAN embed the network in hyperbolic space?
That isn’t part of AI-RAN as the Alliance defines it; at most it’s a research idea. The claim traced back to mnemoverse.com, the company behind Izgorodin’s paper, which is about knowledge graphs, not RAN. Plausible research uses are placing AI workloads across edge GPUs, routing over multi-hop links such as wireless backhaul meshes and satellite links, and reasoning over network telemetry at the right level of detail.
Does greedy routing give near-zero latency?
It cuts the state each node stores, not the time a packet takes. Routes run a little longer than the shortest: in Kleinberg’s test, 80% were within 1.2× the shortest path.
Can hyperbolic coordinates reuse spectrum exponentially more?
Virtual coordinates don’t move antennas, so they don’t change interference. As a thought experiment, a hyperbolic tiling with 6 neighbors per cell and a reuse distance of 3 tiles still needs 7 channel groups, the same as hexagons, and each cell has 30 co-channel cells at the reuse distance instead of 6.
Do networks still plan frequencies like the coloring problem?
Early cellular networks used fixed channel plans like that. LTE and 5G mostly reuse every frequency in every cell and manage interference with scheduling, power control and beamforming.
The Shannon–Hartley theorem
C = B · log₂(1 + S/N)
C is the most bits per second a channel can carry with errors kept as rare as you like. B is the bandwidth in hertz. S is the received signal power and N the noise power in that bandwidth, both as plain ratios, not decibels. Bandwidth multiplies capacity directly; signal strength only works through the logarithm, so at high SNR doubling the power adds about one bit per second per hertz. Example: 20 MHz at an SNR of 20 dB (a ratio of 100) gives 20 MHz × log₂(101) ≈ 133 Mbit/s.
What does “error-free below C” mean?
If the data rate is below C, errors can be made as rare as you want by using longer and longer codes. It doesn’t mean one particular code is perfect. Above C, reliable communication is impossible.
Does N include interference?
In the theorem, N is noise only, spread evenly across the band, N = N₀·B. Engineers fold interference in by using SINR = S / (N + I) in place of S/N, an approximation that works well when interference comes from many sources at once.
Does infinite bandwidth give infinite capacity?
No. As B grows, N = N₀·B grows with it and the signal-to-noise ratio falls. Capacity levels off at S / (N₀ ln 2), about 1.44 × S/N₀.
Does it set how far data can be compressed?
Compression limits come from two other Shannon results, the source-coding theorem (entropy) and rate–distortion theory. Shannon–Hartley is about sending data through noise, though reading bits back off a disk or tape is itself a noisy channel.
Is the theorem 2D or 3D?
Neither. The formula has no physical space in it. Distance and layout only enter through S and N: received power falls with distance, and interference depends on where other transmitters are. Calling it 1D is loose too: it describes one link, with time and frequency as its variables.
Where is the geometry, then?
A signal of bandwidth B lasting T seconds is pinned down by about 2BT samples, so it is one point in a space of about 2BT dimensions; one megahertz for one millisecond is already about 2,000. Shannon proved the theorem by packing small noise spheres inside a big sphere in that space. Capacity is how many fit.
Takeaways
Hyperbolic geometry’s one big property is exponential room. Almost every real use traces back to it.
Hyperbolic coordinates are virtual. They change how you address and route, never where antennas are or how radio waves interfere.
Greedy routing cuts the state each node keeps. It doesn’t cut latency, and you still need a lookup to learn where a destination is.
Plane coloring and frequency reuse share the 7-cell hexagon picture, but they are different math problems, and modern networks reuse every frequency everywhere.
Coloring can lower interference, never noise, and it always costs bandwidth.
Similar names hide different ideas: elliptic curves aren’t elliptic geometry, tanh activations aren’t Poincaré projections, and antenna hyperbolas aren’t light cones.
Does the interference pattern of two antennas look the same as a spacetime diagram?
Key results
Every peak and null curve is a hyperbola with the two antennas as its foci.
Each curve has its own asymptote angle, sin θ = k/D, so there is no single shared X like the light cone in a spacetime diagram.
Context
This checks whether the interference pattern of two antennas looks the same as the hyperbolas of a spacetime diagram and the Penrose diamond built from them.
Are both governed by the same geometry?
They look alike but are built differently. An antenna hyperbola is the set of points with the same difference of distances to two towers. A spacetime hyperbola is the set of events with the same t² − x², a difference of squares.
Do antenna hyperbolas share one X like a spacetime diagram?
No. Spacetime hyperbolas all share the 45° light cone. Each antenna hyperbola has its own asymptote angle, sin θ = k/D, where k is the path difference and D the antenna spacing, so they fan out instead of sharing one X.
Antenna hyperbolas and spacetime hyperbolas both keep a constant distance difference between two foci. Does that difference mean the same thing in both?
Key results
For antennas it is a real path difference between two towers, the time difference of arrival × c. Changing it changes the curve’s shape.
In spacetime the foci are only points on the drawing. The quantity that matters is the interval s = √(t² − x²), and changing s rescales the curve without changing its shape.
Context
How do the variables and the distance difference change between the two kinds of hyperbola?
On both, every point has the same difference of distances to the two foci. What changes is what that difference means.
Antennas
The variables are positions on the ground, the foci are the real towers, and the difference k is a real path difference. Divided by the speed of light it is an arrival-time difference, which is how hyperbolic positioning locates a receiver. Changing k keeps the towers fixed and reshapes the curve.
Spacetime
The variables are an event’s time t and position x. The quantity that matters is the interval s = √(t² − x²), the time a clock shows after travelling from the origin to that event at constant speed. The difference of distances to the foci is 2s, with foci at t = ±s√2, but those foci are only points on paper. Changing s rescales the curve without changing its shape.
What happens to a spacetime diagram when the whole infinite plane is squeezed into a finite picture?
Key results
Light rays stay at 45° the whole way, and the light-cone X ends at the midpoints of the diamond’s edges.
Lines of constant t pinch to the side corners i⁰. Lines of constant x pinch to the top and bottom corners i⁺ and i⁻.
The hyperbolas bunch up along the edges ℐ⁺ and ℐ⁻.
Context
Do the X and the hyperbolas curve inward to become the Penrose diagram?
The hyperbolas do curve inward. The X doesn’t: the light-cone lines stay straight at 45° and only get shorter. Penrose squeezes the diagram by applying arctan to the light-ray coordinates t − x and t + x, which keeps light rays at 45°.
Do the hyperbolas end at the corners?
No. The corners collect the grid lines: lines of constant position run to the top and bottom corners (i⁺, i⁻), and lines of constant time run to the side corners (i⁰). The hyperbolas end on the diagonal edges ℐ⁺ and ℐ⁻.
How does a spacetime-style hyperbola turn into the Poincaré disk?
Key results
Projecting t² − x² = 1 from the point (0, −1) puts the point at hyperbolic distance r at tanh(r/2).
So the whole infinite curve fits between −1 and 1. In two dimensions the same projection turns the hyperboloid t² − x² − y² = 1 into the Poincaré disk.
Context
Do a Penrose diagram and the Poincaré disk do the same thing?
Both squeeze infinity into a finite picture, but of different spaces. Minkowski spacetime is flat, with a minus sign in its distance. The exact bridge to hyperbolic geometry is one sheet of t² − x² − y² = 1 inside it: that sheet is the hyperbolic plane, and projecting it from (0, 0, −1) gives the Poincaré disk, with a point at distance r landing at tanh(r/2).
Is the tanh activation function in neural networks a Poincaré projection?
It shares the formula, not the geometry. tanh does squeeze the whole real line into (−1, 1), and tanh(r/2) is how the Poincaré disk places a point at distance r, but the activation function isn’t projecting anything.
Each level of a branching tree sits closer to the rim, yet every edge has the same hyperbolic length.
The rim is infinitely far away, so the crowding near it is only in the picture.
Kleinberg (2007) uses this to give every node in a network a routing address.
Context
Why is there room for a whole tree?
With curvature −1 a circle of radius r has circumference 2π·sinh r, which grows exponentially; at r = 6 that’s 33 times the flat value. A tree’s levels also grow exponentially, so every branch fits.
Where do concepts sit in a hyperbolic embedding?
General concepts tend to sit near the center and specific ones near the rim. The papers don’t describe layers of constant Shannon information or word frequency, and standard LLM embeddings are Euclidean, not hyperbolic.
One step, a 60° turn and another step leave you √3·d ≈ 1.732d from the start.
The 60° is the exterior angle you turn through. The two steps enclose 120°.
Context
Where is the two-step walk with the 60° turn?
The centers of the hexagons form a grid of equilateral triangles with step d = √3·s, where s is the hexagon’s side. One step, a 60° turn, then a second step lands √3·d from the start.
Is the 60° inside the triangle or outside it?
Outside. The 60° is the exterior angle, how far you turn off your heading. The two legs enclose an interior angle of 120°, making an isosceles triangle with sides d and d.
Why is the straight line √3·d?
Dropping a perpendicular from the turning point splits the triangle into two 30-60-90 halves, each with base √3/2·d, so the straight line is √3·d.
Why does the hexagon coloring of the plane need 7 colors instead of 3 or 4?
Key results
A hex pattern whose nearest same-colored tile is √K·d away uses K colors (K = 1, 3, 4, 7, 9, 12…).
With 3 or 4 colors a tile cannot be narrower than 1 and also far enough from the next tile of the same color.
With 7 colors any hexagon side between 1/√7 ≈ 0.378 and 0.5 works. Isbell’s side of 0.4 leaves a gap of about 1.058.
Context
OpenAI’s preprint of September 23, 2026 shows the flat plane can’t be colored with 5 colors, so the answer is 6 or 7. It is checked in the Lean proof assistant but not yet peer reviewed. Flat hexagon layouts are used for wireless channel planning, and flat is the right model there, because radio travels through ordinary space and Earth’s curvature is negligible across a cell. That raised the question of how the coloring changes when the plane is curved, and of where the √3, √4 and √7 distances come from.
The 6–7 coloring problem
What is the problem?
The Hadwiger–Nelson problem asks for the fewest colors that can paint every point of the flat plane so that no two points exactly 1 apart share a color. From 1950 the answer was known to be between 4 and 7; Aubrey de Grey raised the lower bound to 5 in 2018, and OpenAI’s preprint rules out 5. Seven always works: Isbell’s hexagons of side s need 2s < 1 and a gap of √7·s > 1, which holds for any s between 0.378 and 0.5.
Does curving the plane change it?
Yes. On a sphere the answer depends on how the forbidden distance compares with the radius. On the hyperbolic plane, with forbidden distance d, small d behaves like the flat plane, but the known bounds differ: at least 4 colors for every d, at least 6 for large d when color regions are measurable (DeCorte and Golubev), and best-known colorings whose color count grows with d (Parlier and Petit). Curving the plane changes the problem without making it easier.
Do more hexagons fit at a corner on the hyperbolic plane, and does that change the color count?
The first part is true: four or more regular hexagons can meet at a corner, where the flat plane allows exactly three. Whether that changes the color count isn’t known; see the bounds above.
Coloring and wireless interference
Is the coloring problem applied to wireless interference?
A close cousin is. Assigning channels to cells forbids every distance below a reuse distance, not one exact distance, so it is a different and better-understood problem: coloring a graph of cells. The two share one picture, the 7-color hexagon pattern.
Is the unit distance the interference radius of a transmitter?
No. The math problem forbids exactly one distance; two points 0.5 apart may share a color. Interference forbids everything closer than the reuse distance.
Does a 7-frequency plan remove interference?
It lowers it. The 7-cell pattern spaces out cells that share a channel; the textbook estimate of the signal-to-interference ratio is (√(3K))γ/6, about 18 dB for K = 7 and path-loss exponent γ = 4. It never removes interference.
Does curving the map change interference?
Drawing a map differently doesn’t move towers, and radio waves travel through ordinary flat space. Earth’s curvature only limits how far a tower can see over the horizon. Also, interference is I, not N.
Does a good coloring drop noise to zero and maximize capacity?
Noise N is set by temperature and bandwidth; no channel plan touches it. Coloring lowers interference I, and it costs bandwidth: with K channel groups each cell gets B/K. Capacity per cell is roughly (B/K)·log₂(1 + SINR), and for realistic propagation small K wins. That’s why LTE and 5G reuse every frequency in every cell.
The √3, √4 and √7 walks
Where does √7 come from?
Walk 2 steps straight ahead, turn 60° and take 1 step. The law of cosines gives c² = (i² + ij + j²)·d² with i = 2 and j = 1, so c² = 7d². In general a hex coloring whose nearest same-colored tile sits at that distance uses K = i² + ij + j² colors, so K can be 1, 3, 4, 7, 9, 12, 13.
Is there a 6-color hexagon pattern?
Not of this kind: 6 isn’t of the form i² + ij + j². The distances √3·d and √4·d belong to the 3-color and 4-color patterns.
What makes 7 colors work?
The gap. The nearest same-colored tiles are s apart with 3 colors, √3·s with 4 and √7·s with 7. A valid coloring needs tiles narrower than 1 (2s < 1) and a gap above 1. With 3 or 4 colors those contradict each other; with 7 they hold for any s between 1/√7 ≈ 0.378 and 0.5. It isn’t about rotating tiles out of alignment.
Did Isbell use a side of 2/5?
That couldn’t be confirmed. A side of 0.4 sits inside the working range, so it works, but so does any side between 0.378 and 0.5.