John D. Norton
Department of History and Philosophy of Science
University of Pittsburgh
https://sites.pitt.edu/~jdnorton/jdnorton.html
February 21, 2025
In the block stacking problem, a collection of blocks are to be stacked at a table edge. If we stack the blocks so that they protrude past the edge of the table, how far can they go horizontally? The curious and then surprizing result is that we can extend the stack horizontally arbitrarily far. We just need to have a high enough stack of many blocks. Until we think more about it and perhaps do some sums, that just seems wrong. Surely, we expect that there is a limit to how far the stacks can go before the whole stack topples off the edge.
There is a long standing literature on the problem that extends back to the mid-19th century. This literature has given the physics in much detail for both the simple case to be dealt with here and for more complicated cases.
We can do the calculations that establish the stability of the various stacks and thereby affirm that the stacks are stable and have the properties claimed. However it can still remain puzzling that the stack just does not fall since it extends so far past the edge of the table. My goal here is to develop an intuitive sense of comfort with the behaviors of these stacks. If I succeed, you will not just understand that the physics allows the stacks to be stable, but you will feel that it is proper and just.
There will be a small bonus. Once we are comfortable with stacks of blocks that extend arbitrarily far beyond the table edge, we might think that we can just "take the limit to infinity" and have an infinite stack that extends infinitely far beyond the table edge. Limits to infinity, we shall see, are curious procedures. In this case they produce the oddest result: either no stack of blocks at all; or one that does not extend past the table edge at all.
How can that be? All will be revealed below!
We will consider blocks of width two units, since that will simplify the formulae. They are homogeneous, which entails that their centers of mass, indicated by the black spots, are at their geometric centers.
The first puzzle is that merely stacking four blocks as shown below enables us to locate the fourth, colored block such that it is displaced so far horizontally that no part of it is above the table.
This is possible without any artifice. There is no glue holding the blocks in place and no nails or screws or locking pins somehow hidden from view. Each block just rests on the block beneath and the lowest block just rests on the table.
Why doesn't this last block fall? Why doesn't the whole stack just topple over the edge of the table?
The figure above shows the idealized case in which the blocks have been moved over to the very last configuration in which stability is possible. I tried to come as close to this configuration as I could, using books from my book shelf. Here's the result. I came very close to this idealized case. The slight compressibility of the books led to some tilting of the books.
The next puzzle is that this displacement of the fourth block is just the beginning. If we start with a larger stack of blocks, then there is no limit to how far the stack can extend beyond the table edge.
Here is the case of ten block. It is extended as far as this type of construction allows:
This construction can continue indefinitely. The more blocks we are willing to have in the stack, the more the blocks will be displaced past the edge of the table. They will still be properly supported.
The puzzling part is that, no matter how far the stack projects beyond the table edge, its support on the table is limited to just one block width on the table (indicated by the colored shading).
We can stack up n blocks in this configuration, where n is any number, no matter how large: 10, 25, 50, 100, ... The topmost blocks will be arbitrarily displaced beyond the table edge. Here is how the stack looks with 100 blocks (drawn correctly to scale).
This figure shows only 100 blocks in the stack. What if we consider stacks with 500, 1,000, 1,000,000, ... blocks and so on; and take the limit as n goes to infinity? We would now have infinitely many blocks, all of them located past the end of the table and extending an infinite distance? Would the narrow part of the table of one block width be able to support this infinite extension?
If you are impatient, here are the answers that will be developed below:
Puzzle 1 and 2: Simple principles of balance, extending back to Archimedes, affirm the stability all the stacks as long as they have finitely many blocks.
Puzzle 3: The infinite case fails. Taking the limit to infinity does not produce the expected infinite, leaning stack. Depending on how the limit is taken, it produces either nothing at all, or a stack of blocks that does not lean.
The principles that determine whether a stack of blocks is stable are elementary in physics and were known already to Archimedes. It is, however, helpful to review them in order to see that nothing fancy is being used to establish the stability of the stacks. Archimedes could have justified the stability.
A moment or turning moment is the basic quantity on which stability calculations depend. Each mass exerts such a moment about any nominated point. Its magnitude is given by the product of the distance to the mass and the component of the weight force exerted by the mass perpediculuar to the distance. In the case of horizontal beam balances, this moment simplifies to just the distance to the mass multiplied by the mass. For example, here is a mass of 2 units on a weightless beam that is a single unit distance from a fulcrum:
It exerts a moment of 1x2 = 2 about the fulcrum. If it is not counteracted in some way, it will lead the beam to tip.
For a system of masses to be in balance, this first mass' moment must be balanced by a second. Such a balance is provided by an equal mass of size 2 located at one unit distant from the fulcrum on the other side.
Since the position is in the opposite direction, we compute its moment using a negative number, -1. The moment is -1x2=-2.
The total moment about the fulcrum is:
-1x2 + 1x2 = -2 + 2 = 0
and the beam is balanced.
A balanced beam need not have the balancing weights placed symmetrically. A heavier weight closer to the fulcrum can be balanced by a lighter weight farther from the fulcrum. Here a unit mass at -3/4 distance from the fulcrum balances a heavier mass of 3 units that is one third the distance closer to the fulcrum, that is, as a distance of 1/4.
if we recall that the beam itself is massless, then there are two moments: -(3/4)x1 from unit mass and +(1/4)x3 from the mass of size 3. The total moment about the fulcrum is:
-(3/4)x1 + (1/4)x3 = 0
We can also balance a beam with more than just two masses. Here, for example, we distribute four, unit masses over the beam. We designate the ___location of the fulcrum with the x coordinate as x=0 and orient the positive x values to the right. The four masses are located at
x = -9/12, x = -5/12, x = 1/12, x = 13/12
The total moment about the fulcrum at x=0 is
(-9/12)x1 + (-5/12)x1
for masses to the left of the fulcrum; and
(1/12)x1 + (13/12)x1
for masses to the right of the fulcrum. The total moment sums to zero, so that the beam is in balance:
-9/12 + -5/12 + 1/12 + 13/12 = 0
The moment of a set of masses on the beam about any point is the same as the moment that would be produced by all the masses if they were concentrated at a point. That point is the center of mass.
To see this, imagine that we have masses m1, m2, m3, ... located at positions
x1, x2, x3, ...
on the beam. The moment they produce is
x1m1 + x2m2 + ...
We can rewrite this moment as
(x1m1 + x2m2 + ...)
------------------------ x (m1 + m2 + ...)
(m1 + m2 + ...)
That is, the moment is the same as produced by the sum of the masses (m1 + m2 + ...) if all the masses were located at what is called the "center of mass," defined as
(x1m1 + x2m2 + ...)
center of mass = -------------------------
(m1 + m2 + ...)
We now have a simple criterion for when a system of masses on a beam is in balance. If the center of mass is located at the fulcrum, then the masses have no tendency to tilt the beam in either direction. All the moments cancel and the beam is in balance.
Here is how it works with a triangular shaped mass. We imagine the triangular mass to be divided into very many, thin, vertical slivers. These slivers correspond to the masses m1 + m2 + ...
If we now compute the center of mass by the formula above, we find the center of mass is at a position 2/3rd removed from the apex of the triangle on its axis of symmetry.
If we locate the fulcrum at this 2/3 point, the triangle will be in perfect balance.
Those of you who have seen examples of centers of mass will know that the center of mass is usually found within the mass itself. That need not be the case. In the example of the four masses above, the center of mass is at the marked position x=0, which is outside the masses.
To resolve Puzzle 1 and see how the stack of four blocks above conforms with the principles of balance just sketched, we can step through the sub-stacks that form the full stack and confirm that each is stable. We start at the outermost block, block 1, and work backwards towards the table edge.
In each case we find that the block or sub-stack of blocks considered are stable, since their centers of mass are supported by blocks underneath or by the table. In each case, however, the support is provided right at the block's or table's edge. That means that the slightest nudge past the table edge would destabilize them. They are all at the last stable position.
The first or topmost block is block number 1. Its farthest edge is located at:
x = 1 + 1/2 + 1/3 + 1/4
Since the block is two units wide, its center of mass is located one unit of distance closer to the table edge at:
x = 1 + 1/2 + 1/3 + 1/4 - 1 = 1/2 + 1/3 + 1/4
This locates the center of mass immediately above the outermost edge of the second block in the stack, block 2, whose edge is at position x = 1/2 + 1/3 + 1/4. We can conclude that this first block is placed stably as long as the second block underneath it is stable.
The second block can be stable if the mass of the first block does not topple it, in relation to the support it provided from the third block below. To show this stability, we compute the center of mass of the sub-stack consisting of the blocks 1 and 2 and find that it is located above the edge of the supporting third block.
That is, the center of mass of block 1 is at position x = 1 + 1/2 + 1/3 + 1/4 - 1; and the center of mass of block 2 is at position x = 1/2 + 1/3 + 1/4 - 1. The center of mass of the two together is at:
x = 1/2 (1 + 1/2 + 1/3 + 1/4 - 1
+1/2 + 1/3 + 1/4 - 1)
= 1/2 (1 + 2/2 + 2/3 + 2/4 - 2)
= 1/2 (2/3 + 2/4) = 1/3 + 1/4
That is the center of mass of blocks 1 and 2 is located over the outer edge of block 3 at x = 1/3 + 1/4. These two blocks are supported.
The analysis that establishes the stability of the top three blocks 1, 2, and 3 proceeds analogously to the one that establishes the stability of the top two blocks 1 and 2. It shows that the center of mass of the three blocks taken together is located above the end of the fourth block.
The center of mass of these three blocks is given by
x = 1/3 ( 1 + 1/2 + 1/3 + 1/4 - 1
+1/2 + 1/3 + 1/4 - 1)
1/3 + 1/4 - 1)
= 1/3 ( 1 + 2/2 + 3/3 + 3/4 - 3)
= 1/3 (3/4) = 1/4
That is, the center of mass of the three blocks taken together is located above the end of the fourth block at x = 1/4.
The stability of the four blocks of the stack taken together is again demonstrated by the same means. That is, we compute the center of mass of the four blocks 1, 2, 3 and 4 taken together and find that it is located exactly above the table edge.
The center of mass of these four blocks is given by
x = 1/4 (1 + 1/2 + 1/3 + 1/4 - 1
+1/2 + 1/3 + 1/4 - 1)
1/3 + 1/4 - 1)
+ 1/4 - 1)
= 1/4 (1 + 2/2 + 3/3 + 4/4 - 4)
= 1/4 (4-4) = 0
It follows that the full stack of four blocks is stable. Its center of mass lies exactly at the table edge. It is the last point on the table that can provide support. The slightest movement of the stack away from the edge would leave the center of mass unsupported and the stack would topple.
It might be helpful here to see how leverage plays a role in holding up the stack. The center of mass of the top three blocks 1, 2 and 3 are held up by block 4. That block of unit mass must have enough leverage to support the greater mass of blocks 1, 2 and 3. The moment of these three blocks about the table edge is 3x(1/4). To prevent them toppling, the fourth block must supply a matching, stabilizing moment. It does, as the figure indicates. Its moment about the table edge is 1x(-3/4)
If we compute total moment of all four blocks about the table edge, it is:
-(3/4)x1 + (1/4)x3 = 0
The stack is stable.
Finally, we can also compute directly all the moments associated with each block individually about the table edge.
The summation is:
1x(-9/12) + 1x(-5/12) + 1x(1/12) + 1x(13/12) = 0.
The resolution of Puzzle 1 is, in summary, that the configuration of stacked blocks shown is fully compatible with the principles of balance. This holds both of the stack as a whole, whose center of mass is located above the table edge; and of the individual sub-stacks, each of whose centers of mass are located above the edge of the supporting block below.
The remaining question is why the stability of this particular stack was surprizing in the first place. Perhaps the simple answer is that this sort of configuration is unfamiliar. It is so because it is precarious. The slightest bump is all that is needed to topple it. Stable stacks of ordinary life commonly have all the blocks located above the bottom-most supporting block. Are we thereby inclined to the false generalization that all stable stacks must be so?
The second puzzle is solved in a similar manner. Take a stack of as many blocks as you like. Call the number n. Then a similar analysis, using the principles of balance, will show that there is a configuration with the topmost block moved farther from the edge. We can have a stack with the topmost block moved as far away from the edge as we like. All we need to do is select a high enough stack, that is, a large enough n.
To see this, it is convenient to find a formula for the distance from the edge of the farthest ends of the blocks in an n block stack. The formula is already suggested by the figure for a four block stack:
The rightmost edges of the blocks are displaced to ___location x as:
Block 1: x = 1 + 1/2 + 1/3 + 1/4
Block 2: x = 1/2 + 1/3 + 1/4
Block 3: x = 1/3 + 1/4
Block 4: x = 1/4
It is easy to guess now what the general result is for n blocks.
Here are these displacements for the case of a ten block stack:
That this is the right general formula follows from simple summations, similar to those given above for the four block stack.
Since each block is two units wide, its center of mass is one unit removed towards the table edge from the outermost edge of the block. That is, they are at positions:
Block 1: x = 1 + 1/2 + 1/3 + ... + 1/(n-1) + 1/n - 1
Block 2: x = 1/2 + 1/3 + ... + 1/(n-1) + 1/n - 1
etc.
Since there are n blocks of unit mass, the center of mass of the whole stack is given by
(1/n) (1 + 1/2 + 1/3 + ... + 1/(n-1) + 1/n - 1
+ 1/2 + 1/3 + ... + 1/(n-1) + 1/n - 1
+ 1/3 + ... + 1/(n-1) + 1/n - 1
...
+ 1/(n-1) + 1/n - 1
+ 1/n - 1 )
If we collect all like fractions, this sum reduces to:
(1/n) (1 + 2/2 + 3/3 + ... + (n-1)/(n-1) + n/n - n)
= (1/n)(n - n) = 0
It follows that the center of mass of the entire stack is located over the table edge at x=0.
This last result is necessary for the stability of the entire stack. If the center of mass of the entire stack were located past the table edge, the entire stack would topple over. We need more. We must also establish that each of the sub-stacks are stable. Take the substack of blocks 1, 2 and 3. Its center of mass must also be supported so that this substack does not topple. This stability is demonstrated in the Appendix.
With a stack of sufficiently large n, we can displace the topmost block as far from the table edge as we like. This follows since the outermost edge of the top block is displaced by:
1 + 1/2 + 1/3 + ... + 1/(n-1) + 1/n
This expression is the summation in the first n terms of the harmonic series, which diverges to infinity as we complete the series. The standard way to show this divergence is to group terms in the series so that their sums are each greater than or equal to 1/2:
| 1 | ≥ | 1 |
| 1/2 | ≥ | 1/2 |
| 1/3 + 1/4 | > | 1/4 + 1/4 = 1/2 |
| 1/5 + 1/6 + 1/7 + 1/8 | > | 1/8 + 1/8 +1/8 + 1/8 = 1/2 |
| 1/9 + 1/10 + 1/11 + 1/12 + 1/13 + 1/14 + 1/15 + 1/16 | > | 1/16 + 1/16 +1/16 + 1/16 1/16 + 1/16 +1/16 + 1/16 = 1/2 |
| ... | > | ... |
It now follows that, as we proceed with the summation of terms in the harmonic series, we keep adding terms that sum to greater than 1/2. The fully summation is greater than infinitely many terms of 1/2:
1/2 + 1/2 + 1/2 + 1/2 + ... = ∞
The topmost block of a stack of n blocks can be located as far past the edge of the stack as we like. We just need to choose a large enough n.
The partial sums of the harmonic series are cumbersome to work with. A simple formula, using the natural logarithm ln(.), gives a surprizingly good approximation for the partial sums:
ln(n) + γ ≅ 1 + 1/2 + 1/3 + ... + 1/n
where the Euler constant ≅ 0.57721...
The approximation is very good. For n≥9, the error is less than 2%. That makes the approximation quite adequate for the analysis to follow.
The first consequence that can be read from this logarithmic approximation is that extending the stack horizontally is quite hard. If, for example, we increase the height of the stack from 100 blocks to 1000 blocks, then we increase the horizontal extent of the stack by a meager:
(ln (1000) - γ) - (ln (100) - γ) = ln(1000/100) = ln(10) = 2.3
That is, if we multiply the height of the stack by a factor of 10, we only add to the horizontal extension by 2.3. Successive muliplications of the height by 10: x10x10x...; only yield additive effects for the width: +2.3+2.3+...
 |
This last relationship enables us to develop a better picture of how the large stacks of blocks look. As we consider taller stacks, they become thinner. Very tall stacks become very slender. Taller stacks are wider than shorter stacks, but the ratio of their width to their height becomes smaller. The formula, using the logarithmic approximation, is: width/height = (2.57721 + ln(n)) / n This ratio approaches zero as n grows arbitrarily large. Outlines for stacks with n=100 and n=500 are drawn as well-scaled as I can manage on the left. The block height is taken to be 0.1 units of distance and the block widths are two units. |
That the stacks become more slender moderates our initial surprize at how wide the stacks can become. There is a catch, however, in this moderation. The two figures above assume that the block thicknesses, vertically, are the same. That means that the stack with 500 blocks is five times taller than the stack with only 100 blocks.
The catch is that the thicknesses of the blocks play no role in the determination of the stability of the stacks. It is natural to keep the block thicknesses the same when we move our consideration from a stack of 10 blocks to a stack of 100. We could, instead, diminish the block thicknesses in order to keep the overall height of the stacks the same.
The figure below shows stacks of 10 and 100 blocks. The thickness of the blocks in the stack of 100 has been reduced to one tenth the thickness of the blocks in the stack of 10 blocks, so that both stacks have the same height.
The comparison restores the sense that this stacking of the blocks really does allow the top of the stack to extend surprizing far from the table edge. The device of reducing the thickness of the blocks can be repeated for stacks of 1,000 blocks and of 10,000 blocks. In each case, we reduce the thickness of the blocks by successive, multiplicative factors of ten. The vertical height of the stacks remains the same, but the horizontal extent of the stack increases by an additive distance of 2.3.
In these idealized calculations of the balance of the stacks, we have not considered the mechanical stresses in the blocks. The reduction of the block thicknesses would result in corresponding increases in the shear and bending stresses within the blocks. A thick block can withstand such stresses better than a very thin one. If we model the stacks more realistically, the properties of the materials used would lead to larger stacks eventually fracturing under these stresses; and the fractures would come sooner with very thin blocks.
In the case of a four-block stack, the topmost block 1 was the only block fully past the edge of the table. If we have a large stack of blocks, what portion of the blocks are similarly fully past the edge of the table?
The logarithmic approximation, as we shall now see, makes it easy to answer: roughly 14%.
Since each block is two units wide, the innermost edge of the mth block is located at position:
x = 1/m + 1/(m+1) ... + 1/n - 2
= 1 + 1/2 + ... + 1/(m-1) + 1/m + 1/(m+1) ... + 1/n
- (1 + 1/2 + ... + 1/(m-1))
- 2
Applying the logarithmic approximation to these two series, we find that:
x ≅ ln(n) + γ - (ln(m-1) + γ) - 2 ≅ ln(n/(m-1)) - 2
The condition that a position on the block is past the table edge is that x > 0. That is,
ln(n/(m-1)) > 2.
Taking exponentials of both sides of this inequality, we find:
m/n < 1/e2 + 1/n ≅ 0.1353 + 1/n
For a stack of 100 block for which 1/n = 0.01, we can approximate the condition as m < 0.145. This conforms with an exact calculation of the series that shows that the topmost 14 blocks are fully past the edge.
The logarithmic approximation tells us that this fraction is quite stable. The 1/n term is already small at 0.01 in the case of a stack of 100 blocks. For larger stacks, it will be even smaller. We can conclude that for large stacks of blocks, roughly 14% of the blocks will be fully past the table edge, no matter how large the stack.
The small proportion of blocks fully past the table edge for stacks of all sizes gives some indication of how the tall stacks can be stable. They may extend in width well beyond the table edge. However the bulk of the mass of the blocks stays close to the table edge. This larger mass counter-balances the smaller mass of blocks farther removed from the table edge.
We can arrive at a quantitative assessment of this effect by dividing the blocks into two parts. There are those whose centers of mass lie beyond the table edge. They act to topple the tower. Call them the "toppling blocks." There are those whose centers of mass remain above the table. They counter-balance the tipping and keep the stack stable. Call them the "stabilizing blocks."
Using the logarithmic approximation, we recover two results. (Proofs are given in the Appendix.)
First, the toppling blocks have numbers m circumscribed by:
m/n < 1/e + 1/n = 0.3679 + 1/n
That is, for all higher stacks, roughly 37% of the blocks have their centers of mass past the table edge and these blocks act to topple the stack. For the case of a 100 block stack, this relation tells us that toppling blocks satisfy m<37.8. This logarithmic approximation matches well with a precise summation of the relevant series, which shows that toppling blocks are numbered from 1 to 37.
Second, the center of mass of the toppling blocks is approximated up to terms in 1/n by:
1 - 1/m = 1 - 1/(n/e + 1) ≅ 1 - e/n
That is, this center of mass is always less that one unit of distance past the table edge. As n becomes large, the center of mass will approach arbitrarily close to a position one unit past the table edge. For a stack of 100 blocks, shown in the figure below, this approximate formula yields 0.9728. A direct computation yields 0.9858. The agreement is close enough given the approximations used.
The center of mass of the stabilizing blocks is given in an approximation that neglects terms in 1/n, for all higher stacks, as:
x ≅ -1/(e-1) = -0.5819
That is, the center of mass is displaced away from the edge by about 0.58 of a unit distance.
The approximation with a term in 1/n included is
x ≅ -1/(e-1) [1 + (e/(e-1))(1/n)]
This approximation may not have correctly collected all the terms in 1/n since the added term diminishes slightly the agreement with the direct computation of the n=100 block stack.
For the case of a 100 block stack, shown in the figure below, a direct computation of the center of mass of the stabilizing blocks locates it at x = 0.5789, which is in agreement with the general formulae, up to the approximations imposed. The stabilizing action of the center of mass of 63 blocks at position x= -0.58 counter-balances the toppling action of the center of mass of 37 blocks at the position x = 0.98.
These results give a clear indication of how the stacks of blocks can remain stable, even when they become very high and the topmost blocks are arbitrarily far removed from the table edge. The key fact is that the massing of the blocks that tend to topple stack are always close to the table edge. Their center of mass never passes beyond one unit distance from the edge, no matter how many blocks are stacked. As a result, it is always possible for stabilizing blocks on the table to exercise enough leverage to keep the whole stack stable. There are always more stabilizing blocks to apply this stabilizing leverage than there are blocks that act to topple the stack. For stacks of many blocks, the ratio of stabilizing blocks to toppling blocks is roughly constant at 67/37.
So far, the narrative has been therapeutic. It may seem that the stacks of blocks described cannot be stable. However a closer examination of the blocks, in the light of principles of balance, shows that they are stable after all. In this last section, the direction of the therapy reverses. It will be argued that something that may now seem plausible is inadmissible.
We have seen that stacks of leaning blocks of arbitrary size can be stable. That is so, no matter how many blocks we stack up. 100, 1,000, 1,000,000 and so on. It seems natural that we can, loosely speaking, "take the limit to infinity" and recover the stability of a leaning stack of infinitely many blocks. We shall now see that this is not so. The process of taking this limit either leads to no stack at all, or just to one that does not lean at all.
As a preliminary, we may consider what is required for the stability an infinite stack of blocks that lean as do the finite stacks. The moment of the stabilizing blocks would have to balance exactly the moment of the toppling blocks. However each of these moments would be infinite.
The condition for stability is that these two infinities would exactly cancel. That is:
∞ - ∞ = 0
Just this sort of subtraction of infinity from infinity is ill-defined. It has no assured value. A more precise analysis should show that different limiting procedures that lead to this same subtraction would assign different values to the subtraction, affirming that it is not well defined. We need not explore this problem any further since, prior to considerations of balance, the limiting procedures already produce problematic results.
As a warm up for what is to follow, consider what happens when we take the limit of circles of infinite size in a Euclidean space. What can go wrong in such a simple case? Everything! Depending on how we take the limit we either get nothing at all; or something we did not expect.
Start by considering circles of finite radius. Take the limit as their radii become infinite. if we do it such that the centers of the circes remain at some fixed point in space, then the limit is nothing at all. We would like to say that we have "an infinite circle." That expression is, literally, nonsense. That is, it is a term that has no referent. The points on the circumference of the infinite circle are, we are to suppose, infinitely far away from the central point. But there are no points in Euclidean space infinitely far away. Every point in Euclidean space is some finite distance from the center. There is no infinite circle in a Euclidean space.
We can use a different sequence to take the limit to an infinite circle. This time, we keep the left edge of the circle at a fixed point and move the center of the circle to successively greater distances. In the limit, we might say that this center is now "infinitely far away." That, again, is literally nonsense. There is no place in Euclidean space that is infinitely far away. However something does remain. The portion of the circles passing through the point we kept fixed has now converged to a limit, which is just a straight line. The limit procedure has produced something, but it is not the infinite circle anticipated. It is just a straight line.
We shall see now that some similar happens with the stacks of blocks as we try to take the infinite limit. Depending on how we take the limit, we either end up with nothing at all or a stack that does not lean at all.
The direct way to take the limit of infinitely many blocks is to extend the stacks as we have been numbering them. We start at the top and number down. This immediately produces a failure. The topmost block is now at an infinitely high elevation above the table. However there is no such elevation. All heights above some fixed point in a Euclidean space are finite, even if they can be arbitrarily large. It follows that, in this limit, the topmost block, numbered 1, is nowhere.
The same is true of block number 2. It is one unit in distance lower than block 1. One unit lower still leaves it at an infinite height. We might write "∞ - 1 =∞."
The same is true of all the remaining blocks numbered 3, 4, 5, ..., n, ...: "∞ - n =∞." That is, the limit of the positions of all the blocks in nowhere.
Purely figuratively speaking, the stack has flown off to Plato's ideal heaven, somewhere outside of our ordinary space.
We might try to avoid this last problem by a simple expedient. In the limit state, block 1 is sent to an infinitely high elevation because all the blocks have the same thickness. What if the block thicknesses diminish on some schedule in which an infinity of blocks occupies only finite total height. A simple scheme would diminish the thicknesses of the blocks geometrically, such as in the ratios 1 : 1/2 : 1/4 : .... We will keep the mass of the blocks individually at unity.
The problem of blocks sent to infinite heights would be resolved. However, an equivalent problem would not. The topmost block 1 would be displaced infinitely far past the table edge. As before, it would be sent to a place that does not exist in the Euclidean space. The same is true of all the other blocks. Block 2 is located a single unit distance closer to the table. Block 3 is 1/2 of distance closer again. And so on. All these distances are still infinite.
That is, the positions of the farthest ends of the blocks are:
Block 1: x = 1 + 1/2 + 1/3 + 1/4 + 1/5 + ... = ∞
Block 2: x = 1/2 + 1/3 + 1/4 + 1/5 + ... = "∞-1" = ∞
Block 3: x = 1/3 + 1/4 + 1/5 + ... = "∞ - 1 - 1/2" = ∞
...
These infinite displacements can be represented figuratively as:
We might diagnose that the problem with these limits arises because we are numbering the blocks, starting with 1, from the top of the stack. Might we recover something more regular, if we number the blocks from the bottom?
This renumbering does help. We do not end up banishing the blocks to an out-of-space infinity of heights. However we also do not get a stack that leans. Each stack with a finite number of blocks n extends horizontally and farther so, the greater the value of n. The effect of taking the infinite limit is to eradicate all extensions beyond the table edge. As the stacks become larger, the blocks in the base of stack are shifted towards the table. In the limit of infinitely many blocks, this shift ends up applying to all the blocks. They are all relocated to positions fully above the table.
If we recall that we are numbering from the bottom of the stack, this is easy to see from the expressions for the outermost edge of each block:
Block 1 at x = 1/n → 1/∞ = 0
Block 2 at x = 1/n + 1/(n-1) → 1/∞ + 1/(∞-1) = 0 + 0 = 0
Block 3 at x = 1/n + 1/(n-1) + 1/(n-2)
→ 1/∞ + 1/(∞-1) + 1/(∞-2) = 0 + 0 + 0 = 0
...
The figure shows the block configuration for n = 4, n = 10 and the final configuration after the limit of n → ∞ is taken.
What have we found?
First, there is the sheer fun of figuring out something initially perplexing.
Second, a moral can be derived from how we found it perplexing. We tend to extrapolate from the familiar and expect that things will continue to behave in familiar ways. We are used to stable stacks of blocks having their blocks always aligned with the base. So we mistakenly expect that all stable stacks will be like this.
We are used to making things bigger and bigger and, in the process, they retain their qualitiative features. Ever bigger circles are always round and enclose a central point. Taking limits to infinity is treacherous. They can behave in ways quite unlike our expectations. We do not expect circles to become straight lines when they grow large. We do not expect leaning stacks no longer to lean in the infinite limit or to disappear entirely in this limit. But they do. And learning that this is so is, in the end, more of the fun.
To complete the demonstration of the stability of stacks with n blocks described above, we have to show that any sub-stack is stable. That is:
Block 1 must not be so far removed that it falls off the stack.
Blocks 1 and 2 together must not be so far removed
that they fall off the stack.
And so on for the remaining sub-stacks.
The stability follows from suitable summations.
We can check for the stability of all these sub-stacks by computing the center of mass of a sub-stack of size m < n. Since there are m unit masses in the stack, we find the center of mass is:
(1/m) (1 + 1/2 + 1/3 + ... + 1/(m-1) + 1/m + ... + 1/(n-1) + 1/n - 1
+ 1/2 + 1/3 + ... + 1/(m-1) + 1/m + ... + 1/(n-1) + 1/n - 1
...
1/m + ... + 1/(n-1) + 1/n - 1 )
If we collect like terms we recover:
(1/m) (1 + 2/2 + ... + (m-1)/(m-1) + m/m - m
+ m/(m+1) + m/(m+2) + ... + m/(n-1) + m/n )
This expression simplifies to:
(1/m) (m - m + m(1/(m+1) + 1/(m+2) + ... + 1/(n-1) + 1/n ))
= 1/(m+1) + 1/(m+2) + ... + 1/(n-1) + 1/n
This last sum is the ___location of the outermost edge of block m+1, which is the block supporting the stack of m blocks. That is, each substack has its center of mass over the outermost edge of the block supporting it. All of the sub-stacks are supported stably.
The center of mass of block m is one unit of distance closer to the table edge than the farthest edge of the block. That is, its center of mass is at:
x = 1/m + 1/(m+1) + ... +1/n - 1
The condition that this center of mass is past the table edge is that this quantity is greater than zero. The application of the logarithmic approximation to this inequality proceeds almost identically with the same calculation that determined which blocks m are fully past the table edge. The difference is that the factor of -2 that locates the nearest edge to the table of the block is replaced by -1, which locates the center of mass.
We find:
ln(n/(m-1)) > 1.
Taking exponentials of both sides of this inequality, we find:
m/n < 1/e + 1/n ≅ 0.3679 + 1/n
We found earlier that the the center of mass of the first m blocks taken together is:
1/(m+1) + 1/(m+2) + ... + 1/(n-1) + 1/n
Applying the logarithmic approximation, this center of mass is closely approximated for larger m by
ln(n) + γ - ln(m) - γ = ln(n/m).
The value of m to be substituted into this formula is deterimed by the range of values of m for blocks whose centers of mass are past the table. That range is delimited by the inequality above: ln(n/(m-1)) > 1. The largest value of m in this range is well approximated by equality in this relation:
ln(n/(m-1)) = 1
Combining these last two equations, we recover the center of mass of the totality of blocks whose individual centers of mass are past the table edge:
ln(n/m)) = ln(n/(m-1)) + ln((m-1)/m) ≅ 1 - 1/m
The last equality above uses the approximation that ln(1-1/m) ≅ -1/m, for large m. To substitute for m in this last equality, we take the exponentiated form of ln(n/(m-1)) = 1, which is:
m/n = 1/e + 1/n = 0.368 + 1/n
The center of mass is finally given in good approximation (to first order in e/n) as:
1 - 1/m = 1 - 1/(n/e + 1) ≅ 1 - e/n
Copyright: John D. Norton, February, 2025.