In last month’s Insights puzzle and in my recent Abstractions blog posts, I dissected what we might call “over-projection” in election forecasts. This month, I turn to a different kind of over-projection — the fascinating phenomenon known as “overhang.”

You’ve surely played with overhang as a child. You stack a few identical flat objects such as playing cards, books or Jenga blocks overhanging the edge of a table so that the object at the bottom projects a little over the edge, the one resting on it projects a little more, the next one projects quite a bit more, and so on. The topmost object reaches out across space, a half-bridge to nowhere, defying gravity and seemingly supported by nothing. It looks cool and is fun to make. On a grander scale, cantilevered viewing platforms that overhang yawning chasms such as the Grand Canyon Skywalk and the Glacier Skywalk at Jasper National Park in Alberta, Canada, are truly spectacular and awe-inspiring.

Well, it turns out that the mathematics of overhang is equally entertaining, so let’s build our own overhanging towers using blocks and solve some puzzles about them.

Question 1

The classic overhang problem stipulates that all the blocks must be homogeneous, identical in shape and size and have a length of one unit; there can be only one block at every level of the stack; and none of the blocks can be bound or glued or attached to one another in any way. If you had five such blocks, what is the maximum horizontal distance that the tip of the top block can be made to project past the edge of the table? Can you derive a formula for the maximum overhang possible for

nblocks?

Those who feel intimidated by the physics can relax. The stack of blocks can be modeled with a simple and elegant mathematical series. Here’s how you can think about it: If you start with just one block, it is obvious that the most you can get it to project is up to half its length — an overhang of one-half of a unit. Any more, and it will tip over. Now you can lift this block in place and put another block under it, resting on the table with its end flush with the table’s edge. Now slide both blocks forward as far as they will go without tipping over. The distance can be determined with a simple rule that I’ll state here without proof: The total length of the portions of all the blocks hanging out in space must equal the total length of the portions that are above the table and therefore supported by it. (*Update:* As reader Greg Egan pointed out, this shortcut rule is only accurate for the first three stack levels. This will be enough to give you a sense of the pattern of the mathematical series but not to apply it as stated for the fourth and fifth levels in Question 1. For a more general statement of the shortcut that applies to all levels, see Greg’s comment and my reply.) Now you can add another block below the two and push out all three blocks following the same rule. You can keep doing this indefinitely. That’s all you need to solve this problem.

Next let’s get into some simple variations that will require a deeper understanding of the physics. The key principle is to balance the torque of the supported and unsupported portions. The overhanging portion of the stack has a torque (its weight multiplied by the distance from its center of mass to the edge of the table) that cannot exceed the torque of the supported part of the stack, or the stack will tip over. By applying this physical principle you can derive the rule I stated above and tackle our second question.

Question 2

Imagine that you have the same five blocks as before, and you want to balance a little ornament on the topmost overhanging block, at the point on its center line located a quarter of the block’s length away from the overhanging end. The blocks all weigh one unit and the ornament weighs one-fifth of a unit. What’s the maximum overhang possible now? How does this change the general formula for maximum overhang?

Question 3

Imagine that you are competing against your friend in a multi-round game that involves creating overhanging structures. Initially, the two of you are each given a single block. You each have to place your block over an edge of your respective tables, with any amount of overhang you want. Then, you are each given one to four additional blocks, with the number chosen at random. (You both get the same number of additional blocks.) Each round starts with your initial block as the base, without changing its original position, and with a new random set of one to four blocks to add on top. How far over the edge should you place your initial block so that you have the largest possible average overhang over many rounds of this game?

The maximum-overhang formula yields a mathematical series that diverges. Astoundingly, this means that you can keep adding blocks and the amount of overhang will keep increasing forever, without limit! Obviously, this is not possible with real blocks: At some point the overhanging bridge will come crashing down. What are the physical constraints that will prevent a hypothetical mathematical bridge to infinity?

To close, I encourage you to create the largest overhanging stack of Jenga blocks that you can. Theoretically, it is possible to get an overhang of two unit lengths using 31 blocks. In practice you will probably need more. There are also more efficient solutions that we didn’t consider here, ones that do not follow the constraint of one block per level. Let us know what your personal best is, with and without the one block per level constraint. You can even include links to pictures of your gravity-defying constructions (you can upload pictures to internet image storage sites like Imgur and Flickr). We will include images of the most magnificent overhanging towers in our solution column.

Have fun stacking and puzzling. Just don’t go too far over the edge.

*Editor’s note: The reader who submits the most interesting, creative or insightful solution (as judged by the columnist) in the comments section will receive a *Quanta Magazine* T-shirt. ( Update: The solution is now available here.) And if you’d like to suggest a favorite puzzle for a future Insights column, submit it as a comment below, clearly marked “NEW PUZZLE SUGGESTION” (it will not appear online, so solutions to the puzzle above should be submitted separately).*

*Note that we may hold comments for the first day or two to allow for independent contributions by readers.*

The whole structure is really just a series of nested first-class levers. For the topmost block, the fulcrum is the edge of the block below it, the effort is the weight supported by the block (or table) below, and the load is the weight of the overhanging section. The next lever consists of both the top two blocks with the fulcrum being the edge below them, the third lever consists of the top three blocks with the fulcrum being the edge below them, and so on. If any one of these levers has an effort less than the load the structure will fail.

So the top block can have up to 1/2 its mass overhang the edge below it:

|||| <— block – each line represents an equal amount of mass

| <— edge below block

The second-from-top block's end is the edge for the top block, so we can expand the above diagram like so:

||||

||||

To determine how far out over the edge below these two blocks will be, just find the point with an equal amount of mass on both sides.

That gives you the maximum overhang formula for one blokc per layer. However, a perfectly balanced overhang would be destabilized by the tiniest force in reality – tilting the block at all will shift the distribution of mass over the fulcrum. Only in a perfectly isolated environment could you actually build a "perfect" overhang. The "local" overhang formula is determined by how much mass you need to counteract outside forces.

Modeling the blocks as levers lets us apply the same model to overhangs not limited to one block per layer. You can even use this method to work with asymmetrical blocks!

So to answer each question:

1: The formula for the "perfect" maximum overhang for n blocks is 1-(1/(2^n)). For 5 = n, the maximum is 31/32.

2: The new formula is 1-(1/((2^n)-0.2(n-1))). Since the ornament is on the center of the top block, it only affects the overhang for n>1.

3: Exactly 1/4 block-lengths over the edge.

The main physical limitation is the strength of your starting surface. All the potential energy can be channeled there by a perfect structure if you have access to enough blocks, but eventually the weight will be too much for the table itself.

The spacing on my diagrams didn't show up correctly in the comment format. Here's how they're supposed to look using . instead of spaces:

||||

…|

and

…||||

||||

Apparently the font for the comments section is different than the one used in the submission box. never mind about the diagrams…

Pradeep,

This is personal between you and me and I expect no response. I am no more than a disgruntled, former reader.

Early on, I enjoyed your puzzles and did, on one early occasion, offer a solution. In the final analysis, glory was given to others whose solutions were, indeed, polished. Measure upon measure, my analysis had been a reasonable explanation which satisfied the problem outlay. And I felt shunted aside because my explanation lacked savoir-faire. I was personally affronted. I have not submitted solutions ever since because the ‘winners’ are clearly those with advanced degrees and/or experience. Lay respondents ignored. Examine your list of ‘winners’. Do many of the names seem familiar? Ans: Most. Then, to whom have future problems been directed? Either you welcome a wider audience, like me, or do you warm up to a more exclusive audience?

My argument is that others, like me, deserve to be recognized. I am an academician, teaching at high school level. If, on occasion, my/our analyses seem plausible, our offerings deserve to be recognized.

Case in point, the current problem: The key principle is to balance the torque of the supported and unsupported portions. The overhanging portion of the stack has a torque (its weight multiplied by the distance from its center of mass to the edge of the table) that cannot exceed the torque of the supported part of the stack, or the stack will tip over. By applying this physical principle you can derive the rule I stated above and tackle our second question.

November 18th. The puzzle: How to Hang Far Out Over the Edge. Your commentary is deluxe, as usual, however, all of you reasoning seems to overlook one significant element.

Let’s say you have a 100 m square surface of some thickness. Upon this, put another piece of equal size, overlapping by 50 m. If you add to the upper surface . . . almost at any point . . . additional mass, will it always ‘tip over’? And I argue, No. Certainly not instantaneously.

The reason being that, assuming all surfaces are clean and et cetera, that, in order for the upper piece to ‘roll over’, you must have enough ‘tipping force’ above that calculated . . . so as to overcome the instant requirement to fill a volume created by a separation of these two surfaces. This may be only instantaneous, however it is real and ordinarily ignored. It depends upon several factors all of which relate to the environment surrounding the two plates.

Rotation will occur . . . but not instantaneously. You may expect this to be accompanied by a characteristic sound! So, it seems to me that something beyond your discussion needs to be taken into consideration.

Have I missed something here? Am I being overly stupid?

Pradeep wrote: "The distance can be determined with a simple rule that I’ll state here without proof: The total length of the portions of all the blocks hanging out in space must equal the total length of the portions that are above the table and therefore supported by it. "

I don't believe that's correct, and though I think I can see what was intended, readers who take this passage literally might be confused.

To balance the torque, the sum of the signed distances from the edge of the table to the centre of each block must come to zero. But since the signed distance to the centre is just the average of the signed distances to the two edges of the block, this can also be stated by saying that the sum of the signed distances from the edge of the table to both edges of each block must come to zero.

This is only the same as saying that the total length of the portions of the blocks that lie on either side of the table's edge must be equal if none of the blocks lie fully on one side of the edge. As soon as one block lies entirely above empty space, the formulation in the quoted passage will not yield balanced torques.

A good reference for the "other" solutions to creating an overhang is the paper by

Mike Paterson, Yuval Peres, Mikkel Thorup, Peter Winkler, and Uri Zwick in the American Mathematical Monthly:

http://www.maa.org/sites/default/files/pdf/upload_library/22/Robbins/Patterson2.pdf

Question 1: 5 blocks give a maximum distance of 137/120 units.

General formula for n blocks: h(n) = Sum(1/(2 i), {i, 1, n})

Question 2: 5 blocks + ornament give h = 101579/96096

General formula: h(n) = Sum((i/2 + 1/20 – Sum((j + 1/5) * h(j), {j, 1, i – 1}))/(i + 1/5), {i, 1, n})

Question 3: According to solution 1, the selected distance should be in the range {1/10 .. 1/4}. If the selected distance is shorter than possible then no remedy exists. If it is longer than possible then the best remedy is to shorten the distance of the second block, counting bottom up. Since every number of blocks from 2 to 5 has the same probability, the best chance is correlated to the maximum sum over the four cases. Due to the case distinction, the graph of the total distance over the selected distance has two kinks. On the left of the second kink is the gradient positive, on the right it is negative. So the kink is the global maximum. Since the kink is contributed by the third block (counting bottom up), it is located at the horizontal length 1/6. So, 1/6 is the best choice.

Oops, I forgot the physical constraints! Surprinsigly, the real geometry of the gravitational field permits even larger distances since

a) gravitation near the table is stronger than high above it, and

b) due to the radial field, the torque of high blocks is smaller because the distance from the edge of the less than in a parallel field.

If external disturbances and accuracy of placement would provide no problem, the main reason for not reaching infinity is the elasticity of the blocks, the table and the ground supporting the table.

Correction to my solution: The second last sentence should read: Since the kink is contributed by the three-block-case, …

The formula for the i-block-case is:

if(l <= 1 / (2 * i))

then h = hmax + l – 1 / (2 * i)

else h = hmax + (l – 1 / (2 * i)) * (1 – i / (i – 1))

This paper describes a natural generalization of the overhang problem in which there may be multiple blocks at the same level. What I find remarkable is that this question can be posed to a young child playing with blocks and yet is challenging enough to challenge mathematicians.

https://math.dartmouth.edu/~pw/papers/maxover.pdf

#1. Maximum projection of 5th block's tip is 137/120 = 1.141666666

in general, formula is 0.5 * sum{i=1 to n} (1/(n))

it's half the sum of the harmonic series to that point

#2. Maximum overhang of 5th block's tip is now 1.057058

in general, formula is 1/24 + 0.5 * sum{i=1 to n} (1/(n+0.2))

The denominator is the total mass of the system to that point (i.e. the 3rd term is 3.2 which is the mass of 3 blocks plus the token) I feel like there's a solid explanation for the 1/24, but I don't have it

#3.

overhang of first block | average overhang of last block

———————————————————————–

0.25 | 0.927083

0.2 | 0.933333

0.175 | 0.9338543

0.1666667 | 0.934028

0.15 | 0.928125

0.125 | 0.919271

0.1 | 0.902083

It's entirely possible that 0.16666 is NOT the maximum, it's simply very close to it, but I don't want to test closer and closer values to verify that 0.1666666 is indeed the maximum. I calculated the average max overhang by summing up the best overhangs, given the initial overhang of the bottom block, for n = 1,2,3,or 4 more blocks, and dividing by 4.

A real bridge couldn't reach infinity for a bunch of reasons.

1. Wind. or any outside force that would easily disturb the bridge.

2. Jenga blocks, bricks, etc, are not perfectly uniform – they're simply very close.

3. We don't have infinite precision. Eventually, the overhang of the next block would be so tiny that we can't extend it that minute amount over the edge.

Correction@Greg Egan

Thank you for pointing this out. Yes, Greg correctly states "As soon as one block lies entirely above empty space, the formulation in the quoted passage will not yield balanced torques." So my attempted shortcut works only for the first three levels, which might be enough to give an indication of what the pattern may be like, but no more.

Greg's formulation "The sum of the signed distances from the edge of the table to both edges of each block must come to zero" is a much better statement of the shortcut. That is, for the supported portion, you add the supported lengths, minus the distances to the table ends of the blocks completely off the table. For the unsupported portion, you add the distances of all the outer ends from the edge of the table.

Thanks, Greg.

I'll make sure that an appropriate update is made in the original text with a link to Greg's comment.

@Michael

Our solutions are identical, just two comments:

Question 2: When writing the terms for the torque, the ornament contributes (1/5) * (h – 1/4) with mass 1/4 and distance from edge h – 1/4. This yields (among h/5) the constant 1/20 which ends up in the numerator of the series (with reversed sign after solving for h(i)). Since the denominator for i = 1 is 6/5 this yields 1/20 * 5/6 = 5/120 = 1/24. In my "raw" form of the series one sees that for each h(i) with i > 1 this amount is added but also subtracted (in the summation over j). Therefore, instead of the series 5/12 + 5/22 + 5/32 + … we get 11/24 + 5/22 + 5/32 … with 11/24 = 5/12 + 1/24.

Question 3: As outlined in my first comment, for each of the four cases one has to distinguish the cases "distance smaller than optimal" and "distance larger than optimal". The first case cannot be amended, in the second case decreasing the distance of the second block has the best effect because the largest amount of mass is shifted. In the following graph, the contributions of the for cases are labelled "2B" = 2 blocks to "5B" = 5 blocks, and the bold line stands for the sum:

http://www.em-slipstream.info/QuantaMag1.jpg

One sees that the kink of the 3B case coincides with the maximum of the sum, and therefore the maximum is exactly at 0.166666… = 1/6, as seen from the solution of question 1.