The stack that leans past its own table

Slide or topple found the condition that decides whether a body resting on a surface stays where it is: the line of action of its weight, and of whatever else pushes on it, must pass through the region where it touches the surface. A wardrobe pushed high topples because the push swings that line outside its base; a brick pushed low slides because it never does. The spool that rolls towards the hand pulling it used the same reasoning to decide which way a pulled spool turns. Both were about a single body and a single surface.

The oldest puzzle in statics stacks the bodies. Take a set of identical rectangular blocks — books, bricks, playing cards, dominoes — and stack them one per level at the edge of a table, each pushed out over the one beneath. How far out can the top block be made to reach? The answer surprises almost everyone who has not seen it, in two opposite ways: the stack can reach further than seems possible, and getting it there takes far more blocks than seems reasonable.

Building from the top down #

The condition for the stack to stand is the one from slide or topple, applied at every interface at once. Consider the boundary between any block and the one beneath it. Everything above that boundary acts on the block beneath only through their contact, so the stack above stands if and only if its combined centre of mass lies over the block beneath — not over the table, which the upper blocks never touch. The same must be true at every interface, down to the table edge itself, where the centre of mass of the whole stack must lie over the table.

That makes the problem easy to solve from the top down. The top block can project half its length beyond the second, putting its centre exactly over the second’s edge. The top two blocks together then have their centre of mass three-quarters of a length in from the top block’s outer end, and the third block can be placed with its edge exactly there. Each step places the next block’s edge under the centre of mass of everything already placed.

The offsets that come out are a half, a quarter, a sixth, an eighth: the $k$-th block from the top overhangs the one beneath it by $1/2k$. The reason is a balance of moments. When a new block is slid under a stack of $k$ blocks whose centre of mass sits at its edge, the combined centre of mass of the $k+1$ blocks lies between the old centre, weighted $k$, and the new block’s centre, half a length further in, weighted one. It is therefore $\tfrac12/(k+1)$ in from the edge, and that is how far the next block down can be set back. The sum of the offsets over $n$ blocks is

$$D_n = \frac12\left(1 + \frac12 + \frac13 + \cdots + \frac1n\right) = \tfrac12 H_n,$$

half the $n$-th harmonic number. For four blocks it is $25/24 = 1.042$: the top block’s inner end has passed the table’s edge, and nothing beneath it touches the table at all. Eight blocks reach 1.359 lengths.

Every interface in this stack is exactly balanced. Each partial stack’s centre of mass sits on an edge, which is the condition for being on the point of toppling. Push the top block a fraction of a millimetre further and it falls; push any block further out and everything above it goes. The harmonic stack is the limit of what one block per level can do, and it has no margin anywhere.

The offset each block can take #

The shrinking offsets have a direct physical reading.

The top block carries nothing and can project half its length. The block beneath it carries one block’s weight on its upper face, concentrated at its outer edge, and can project only a quarter, because now two blocks’ worth of weight must be balanced about its own support. The $k$-th block carries $k-1$ blocks above it plus its own weight, and the lever it has to balance them on shrinks in proportion. Each new block added at the bottom must hold the entire stack above it out over its own edge, with a moment arm that grows only as fast as one new block can supply, and so it contributes less than the block above it did.

This is why the stack is built from the top. Built from the bottom, the first block could be pushed out half its length — but then a second block on top of it could not be pushed out at all, since the first block’s own centre of mass is already on the table’s edge, and any weight placed beyond it tips both. The optimal arrangement is fixed by the last block placed, not the first, which is the general shape of problems in statics where every part supports everything above it.

A sum that never stops growing #

The harmonic series diverges. Nicole Oresme proved it around 1350 by grouping its terms: $\tfrac13 + \tfrac14 > \tfrac12$, $\tfrac15 + \cdots + \tfrac18 > \tfrac12$, and so on for ever, so the sum passes any number given enough terms. In the stack, divergence means that there is no limit to the overhang. A single stack of identical blocks can reach as far beyond the table as anyone chooses.

The growth is logarithmic: $H_n$ is close to $\ln n + 0.5772$, so the overhang is close to $\tfrac12\ln n + 0.29$. Doubling the overhang requires squaring the number of blocks, roughly. One block length takes four blocks; two take 31; three take 227; four take 1,674; five take 12,367. Each additional block length costs about $e^2 \approx 7.4$ times as many blocks as the previous one. A stack reaching ten block lengths would need about $2.7\times10^8$ blocks, and for blocks the thickness of a brick it would stand seventeen thousand kilometres tall.

The contrast between “unlimited” and “logarithmic” is the whole point of the puzzle. A mathematician says the stack can reach any distance; a builder says it reaches nowhere. Both are reading the same sum. It is the same contrast as in the collisions that count the digits of pi, where an innocent mechanical arrangement turns out to compute a famous number, except that here the number grows so slowly that no physical stack ever gets to show it off.

A careful stack is a smaller perfect one #

A real stack cannot be built with zero margin. Blocks are not perfectly rectangular, tables are not perfectly level, and the stack will be disturbed. The natural precaution is to keep the centre of mass above each interface a little way inside the edge beneath, by some safety margin $\varepsilon$. It is tempting to think this caps the stack: the offsets shrink as $1/2k$, the margin does not shrink, and eventually the margin should swallow the offset.

It does not. The reason is that a margin left at one interface is inherited by every block beneath it. Setting the second block back by $\varepsilon$ moves the top two blocks’ centre of mass, which moves where the third block can go, which moves the fourth, and so on down. Worked through, the margin at each interface is spread down the stack exactly as the offsets themselves are, and $n$ blocks with a margin $\varepsilon$ at every interface reach

$$D_n(\varepsilon) = H_n\left(\tfrac12 - \varepsilon\right),$$

the perfect stack scaled down by the factor $1 - 2\varepsilon$. A margin of a twentieth of a block costs a tenth of the overhang at every height and changes nothing about its divergence. The careful stack is not a different kind of object from the perfect one; it is a smaller copy.

There is a practical consequence. A careful stack of books keeps the characteristic harmonic shape — long steps at the top, tiny ones at the bottom — because the margin its builder leaves at each interface shrinks the whole stack in proportion rather than flattening its profile.

The rule a whole stack shares with one wardrobe #

The condition used at every interface is the only one there is for a rigid body at rest on a support: the resultant of everything above must pass through the region of contact. For a single wardrobe it decides tipping. For a ladder leaning on a wall it decides, together with friction, whether the foot slides out — the ladder that lets go of the wall found the moment its top leaves the wall from exactly this kind of balance of moments about a contact. For the stack it is applied once per interface, and the stack’s shape is what results when every one of those conditions is met with equality.

The centre of mass enters because it is where the weight of a set of bodies acts as a whole. The point that keeps moving found that the centre of mass of a system moves as if all its mass were there and all the external forces acted on it, and that internal forces, however complicated, cancel in pairs. The stack exploits the same cancellation. The blocks above an interface push on each other in complicated ways, but none of that matters to the interface: only their total weight and where it acts, their centre of mass, reaches the block beneath.

A tower is the extreme case, a stack of a single block, and it shows the condition from the other side. The bell tower of Pisa leans about four degrees from the vertical, its top displaced nearly four metres over a base about fifteen and a half metres across. Its centre of mass, about twenty metres up, lies only about one and a half metres off-centre, well inside its base, so by the condition above it is in no danger of toppling as a rigid body. What threatened it was the soil: the tower’s weight presses harder on the side towards which it leans, the soft clay there compresses more, and the lean grows, a feedback between tilt and ground that a rigid support would not have. The engineers who stabilised it around the year 2000 removed soil from under the higher side and brought its top back by about forty-five centimetres. For the stack of blocks on a rigid table no such feedback exists, and the balance at each edge is the whole story.

More than one block per level #

The harmonic stack is the best that can be done with one block on each level. Allowing several blocks side by side, or blocks used as counterweights on the back of the stack, changes the answer completely, and the full problem was solved only in this century.

The first improvement is easy to see. With three blocks, the harmonic stack reaches $11/12$ of a block length. Place one block on the table edge, and two blocks side by side on top of it, one leaning outward and one inward as a counterweight, and the outer one reaches a full block length. The counterweight pays for itself, because it allows the block beneath to carry its load nearer its centre. Larger stacks of this kind become structures — walls, with blocks arranged like brickwork and counterweights piled at the back — and the forces between blocks in them are no longer fixed by statics alone. The table statics cannot settle found that a top on four legs has infinitely many possible sets of reactions; a brick wall with more contacts than equations has the same freedom, and the question becomes whether any set of contact forces, all pushing rather than pulling, can balance every block.

Mike Paterson and Uri Zwick showed in 2006 that with such stacks the overhang can grow as the cube root of the number of blocks, rather than as its logarithm, and in 2009, with Yuval Peres, Mikkel Thorup and Peter Winkler, they proved that nothing does better: no stack of $n$ blocks, however arranged, overhangs by more than about $6n^{1/3}$ block lengths. A cube root still grows slowly, but compared with a logarithm it is fast, and for large stacks the difference between the two is a factor of several. Twenty thousand blocks in a single harmonic stack lean out about five lengths; arranged as a wall with counterweights, the same blocks do substantially better, and the advantage widens without limit as the number grows.

The reason more blocks per level help is visible in the force figure. In the harmonic stack each block’s job is to hold the whole weight above it out over its edge, with a lever of only half a block. Give the stack blocks it can place behind the overhang, pushing down at the back, and the weight that has to be balanced at the front no longer grows with every level. The structure stops being a column leaning outward and becomes something closer to a cantilever.

A block held down at its back #

The cantilever idea is ancient, and it is how builders without arches spanned spaces.

A block resting on an edge with nothing on it can project half its length. Press its back end down with a weight equal to its own and it can project three-quarters; with ten times its weight, more than ninety-five per cent. In a corbelled vault each course of stone steps inward over the one beneath, and the wall or earth piled on its back end acts as the counterweight, so each course can step a substantial fraction of its length without any mortar in tension. The Treasury of Atreus at Mycenae, built around 1250 BC, closes a dome thirteen metres high and fourteen across with nothing but corbelled courses held down by the earth mound around it, and it was the largest dome in the world for more than a thousand years. The overhang of a corbelled vault grows in proportion to its height, not as the logarithm of its stones, because each course is held from behind rather than balanced from below.

The same principle stops a cantilevered balcony falling off a building, and it explains a cheap trick with playing cards or dominoes: a stack built with a few counterweight pieces at the back reaches visibly further than the harmonic stack of the same number of pieces. What it cannot do is reach far with few pieces. A counterweight has to be heavy, and in a stack of identical blocks it is made of blocks, which is why the best possible growth, even with counterweights, is only a cube root.

Rigid blocks, sharp edges and no wind #

Every figure treats the blocks as identical, perfectly rigid, perfectly rectangular and of uniform density, resting on each other along their full faces with sharp, unrounded edges, under gravity alone. Real blocks compress at their edges, which moves the effective edge inward and is part of why the measured overhang of a stack of real books or bricks falls a little short of $\tfrac12H_n$ even when it is built carefully. Friction plays no part, because a stack balanced on its edges is not pushed sideways; but it would matter if the table tilted or the stack were jarred, and then slide or topple would decide which failure came first. The stack is two-dimensional — every block the full width of the one beneath — and the analysis of stacks with several blocks per level is described rather than drawn. The corbel figure assumes the counterweight presses exactly at the block’s back end. The domain is stacks from one block to tens of thousands, with blocks rigid at the scale of the overhang and no load beyond their own weight.

Still open: what the best stacks look like for every number of blocks #

The asymptotic answer is known: the best possible overhang grows as $n^{1/3}$, up to a constant known only to lie below about 6. The exact optimum for a given number of blocks is not. Paterson and Zwick found the best stacks for small numbers by computer search, and their shapes are surprising — irregular arrangements of blocks, some of them tilting the forces through several layers, that look nothing like the neat brick-wall constructions that achieve the right growth for large numbers. Whether the optimal stacks settle into a regular pattern as the number of blocks grows, and what the constant in front of $n^{1/3}$ really is, has not been settled. The problem is a linear programme for any fixed arrangement, but the number of arrangements grows so fast that the search runs out long before the pattern, if there is one, appears.

The single stack is settled completely. With one block per level, each partial stack’s centre of mass must sit over the block beneath, which makes the $k$-th block’s offset $1/2k$ and the total overhang $\tfrac12H_n$ — unbounded, but reaching one block length at 4 blocks, two at 31, three at 227 and five at 12,367; a margin $\varepsilon$ at every interface scales this to $H_n(\tfrac12 - \varepsilon)$ without capping it, and only counterweights, holding blocks down from behind, can beat the logarithm, up to a cube root. A stack can lean as far as anyone likes. It just has to be unreasonably tall to do it.