The good folks on the most-excellent BBC Radio/Open University statistical literacy programme More or Less decided to answer a year-old Reddit argument about how many Lego bricks can be vertically stacked before the bottom one collapses.

They got the OU's Dr Ian Johnston to stress-test a 2X2 Lego in a hydraulic testing machine, increasing the pressure to some 4,000 Newtons, at which point the brick basically melted. Based on this, they calculated the maximum weight a 2X2 brick could bear, and thus the maximum height of a Lego tower:

The average maximum force the bricks can stand is 4,240N. That's equivalent to a mass of 432kg (950lbs). If you divide that by the mass of a single brick, which is 1.152g, then you get the grand total of bricks a single piece of Lego could support: 375,000.

So, 375,000 bricks towering 3.5km (2.17 miles) high is what it would take to break a Lego brick.

"That's taller than the highest mountain in Spain. It's significantly higher than Mount Olympus [tallest mountain in Greece], and it's the typical height at which people ski in the Alps," Ian Johnston says.

"So if the Greek gods wanted to build a new temple on Mount Olympus, and Mount Olympus wasn't available, they could just - but no more - do it with Lego bricks. As long as they don't jump up and down too much."

How tall can a Lego tower get?

More or Less: Opinion polling, Kevin Pietersen, and stacking Lego 30 Nov 2012 [MP3]

### Read more at Boing Boing

Hurray! Artist Bob Staake has a follow-up to his terrific kids' book, Look! A Book! It's called, Look! Another Book!

Quadcopters are easy-to-fly remote control vehicles that have four propellors.

NPR's fantastic Tiny Desk concert by Macklemore and Ryan Lewis has me going back through the entire series as it looks so wonderful.

Mark Frauenfelder at 8:00 am Tue, Dec 4, 2012

Hurray! Artist Bob Staake has a follow-up to his terrific kids' book, Look! A Book! It's called, Look! Another Book!

Mark Frauenfelder at 7:00 am Tue, Dec 4, 2012

Quadcopters are easy-to-fly remote control vehicles that have four propellors.

Jason Weisberger at 10:19 pm Mon, Dec 3, 2012

NPR's fantastic Tiny Desk concert by Macklemore and Ryan Lewis has me going back through the entire series as it looks so wonderful.

Usually when, in response to some information such as this, someone says “These people have too much spare time” it’s said disparagingly.

When I say it it’s because I’m filled with joy, and a little envy, that there are people in this world with the perfect combination of time and knowledge to figure out stuff like this.

Thank you. It was fun and interesting.

Please recalculate taking into account the decreased force of gravity at the top of the tower. TYVM.

I am assuming you’re not serious, but back of the envelope, the gravity at the top will be ~0.1% percent less than at the bottom, so you’ll get perhaps 0.05% more height, which I’d bet is within the strength variation from one block to another.

I was serious, but only because math problems are fun :)

Really, you’d want to set up a function that indicates the weight as a function of height, then integrate (or sum, if you want to keep it discrete) leaving the upper limit of the integral/sum as an unknown, then solve for that equal to 4,240N.

For bonus marks, calculate how much taller you could make the stack if you started at the peak of Mt. Everest, instead of at sea level.

Assumptions & simplifications allowed:

- Treat the mass of the Earth as a point mass at the centre of the earth.

- No need to address tidal forces (assume Earth is not rotating)

- Don’t worry about tower buckling.

- Sea level is 6000km from the centre of the Earth.

- The peak of Mt. Everest is 8,848m above sea level.

As an aside, treating the earth as a point math is the same as assuming it is spherically symmetric- physically, any sphere, ball, or hallow ball is gravitationally identical to a point mass at its center, the integral just works out that way.

And still, if you’re rounding the earth’s radius to one sig fig, then all the effects you’re talking about won’t affect the final answer. :)

“And still, if you’re rounding the earth’s radius to one sig fig, then all the effects you’re talking about won’t affect the final answer.”

Touche. Assume that sea level is 6000.0 km from the centre of the Earth.

What about the force required to press on the top block?

Clearly this was left out of the calculations.

You’re all wrong until you also calculate condensation, atmospheric pressure, etc. ^_^

Well, of course I’m wrong. But I’m working on becoming less wrong :)

This is what I was thinking. The mass of the bricks isn’t the critical number, it’s the weight, isn’t it? On the moon I could make a tower with the same mass and it wouldn’t collapse.

Correct. In zero gravity you can make your tower as large as you want. (Until it starts to collapse under it’s

owngravity, which takes a whole different order of magnitude of Lego blocks.)Obviously, you can do better than 3.5km if you use actual building techniques and create a load-bearing base, instead of just a big column. (New dream job… chief architect for Lego Olympus.)

My math thinks that in the absence of an external gravitational field, even an infinite stack of lego bricks does not collapse under it’s own gravity. This is because the force drops off as 1/r^2.

When summing the total force on any particular piece, you have a bunch of constants and then the infinite sum of 1/r^2 – the second Riemann Zeta Function, which converges to pi^2/6, rather than diverging to infinity.

Unless you jigger with the mass of the bricks, their thickness, or the universal gravitational constant, lego bricks are not dense enough to collapse under their own gravity.

Rats! I was going to build a black hole. Out of black bricks, obviously.

I almost proved this. I got to about 350,000 bricks high and then my brother knocked it down. What a dick.

That’s worse than the time I asked Mr. Owl how many licks it took to get to the center of a Tootsie Roll Tootsie Pop.

They’ll love this over at Lego HQ – I hope they’ve been informed!

Umm to clarify the brick did nothing like melt.

It simply failed slowly due to buckling of the walls and then a tear.

I think the presented wanted to convey that it didn’t shatter but that it failed progressively.

This is very different to melting which involves a change of state.

Sorry I have had my pedant engineers hat on too long I fear.

Also I wonder if the weight distribution would have an effect. I assume the all weight from the press was on the nubs whereas in a tower a significant portion would flow down the sides of the bricks.

That shouldn’t matter, as you can see from the photo that the brick failed at the edges, so the nubs didn’t factor into its failure. They’re lot stronger than the sides, anyways.

Also, we don’t have a description of how the test was performed. They may have used a short stack of bricks.

Looking at the original BBC article it seems they had a flat sheet brick on the top and bottom which would probably stiffen the top and bottom.

Given that the failure is in the walls where I would expect a brick loaded by another brick to fail I think this is probably a pretty good bit of work.

What, you expect me to read the article??!?

No that is why I clarified for you. :D

The top and bottom, including the nubs, just hold the bricks together. It’s the walls on the side that carry the weight. There are no other vertical regions connecting the top of the brick to the bottom of the brick, so all vertical loads must go through the walls. So that’s where failure would happen. The question is: Did they apply the force mostly along the perimeter (as if another 2×2 brick is on top, which is accurate) or down the nubs in the middle (which would bend the top and be less accurate)? I think the answer is the latter, but even then, the failure was on the side, so the top is stronger in bending than the sides are in compression.

See my previous reply. There is another vertical load-bearing region – the tube down the middle. However, it’s very thin compared to the walls and I don’t think contributes much.

We used buffer plates top and bottom to simulate real life loading as closely as possible.

jpgsawyer: Would this be an example of creep, then?

No. Creep is when you hold a stress in a material and it gradually reduces over time. Usually the effect isn’t very large for room temperature materials. (Paper and some plastics being the only one I know off.)

What we see here isn’t time dependant just load dependant.

The walls will gradually bulge out until they become unstable and buckle and they probably stretch permanently before they tear. I say probably as I haven’t seen the test. Depending on the plastic they may just fail.

Oh dear. This is much more complex that I can explain which is why I don’t write textbooks.

Wikipedia has articles on creep https://en.wikipedia.org/wiki/Creep_%28deformation%29 and buckling https://en.wikipedia.org/wiki/Buckling.

I hope this explains things.

We didn’t see much creep in this case. We ran tests at different strain rates and the loads were very similar.

Indeed. As someone who does fracture mechanics research and who tests structures to understand failure modes, I’m not even going to begin criticizing this description of how the brick failed. (But yes, it was “due to buckling of the walls and then a tear”, which I would think is both obvious and very different from melting. Does an empty soda can “melt” when you flatten it?)

The failed bricks did indeed look as if they had melted, although of course it wasn’t a melting process. Remember that we did this for a radio programme, so the reporter needed a good verbal description of the results.

My comment wasn’t a complaint that the presenter said that the brick looked like it melted I think that is a very good description.

It was more that Cory then described the process as “brick basically melted” which was not what the presenter had said or what happened.

Nice work by the way.

Thanks. Academic life has its entertaining moments.

There might be a few other problems with a 375,000 brick tall, one brick wide tower.

Also, no brickmason in their right mind would build a strictly vertical tower. All tall brick buildings have wider walls at the base for load spreading, to permit a taller structure than a pure vertical stack would allow.

So the appropriate question, really, is, what is the tallest possible Lego ziggurat?

I’ll have a go!

The obvious way to build it is to have one of these bricks on top, with 4 (2 by 2) for the next layer, 9 on the next layer etc.

So let’s make the ziggurat n bricks tall, and look for how high we can make it before the bottom layer can no longer support the rest of it (the top n-1 layers).

The top n-1 layers contain a total of (n-1) * n * (2n-1) / 6 bricks.

Assuming a large n, this is approximately n^3 / 3.

The bottom layer contains n^2 bricks. Each of those therefore has to support the weight of n/3 other bricks above it. So the maximum possible value of n has n/3 = 375000, or n = 1.125 million.

So the ziggurat can be three times taller than the tower. Which is rather disappointing seeing as it requires a third of a quintillion bricks (or 1/3 of an exabrick) for its construction.

As the gravitational forces approximately all act in the same direct, we need only consider the bricks directly above the brick at the bottom, much like hydrostatic forces, as there are no sideways forces in the idealised pyramid. Thus, for the brick with the most above it (i.e. at the centre of the ziggurat), it will still have n-1 bricks above it, thus the shape of the construction doesn’t change the maximum height if the structure is filled in all the way beneath.

Well that only really works if lego are considered a fluid. They aren’t and have shear stiffness and so the shape makes a difference.

But the failure mode of the brick will change when there are bricks on each side preventing the sides from buckling outwards. If each side of each brick can only buckle inwards, then each brick gets a lot stronger. In other words, a whole bunch of 375000-brick-tall towers all stacked against each other would be able to take quite a bit more stuff on top, even though a single 375000-brick-tall tower is on the verge of collapse. (And that’s not even pointing out the obvious simplification that global buckling is being ignored as the most likely failure mode. In reality, a 375000-brick-tall tower would collapse for the same reason that a human hair could not be balanced on its end, for reasons other than compressive forces destroying the material at the base).

“increasing the pressure to some 4,000 Newtons”

Newtons is a unit of force, not pressure. Perhaps “increasing the pressure until a force of some 4000 Newtons was exerted”, given that it was a hydraulic testing machine. (The main article doesn’t make the basic error of confusing force and pressure.)

And yes, as jpgsawyer points out, the brick failed, but certainly didn’t “melt”.

Actually, it would take 375,001 bricks to break a Lego brick.

please recalculate to account for brick separation in middle of tower due to shear force from different angular velocities between base and top.

How many bricks to build a space elevator? ;]

don’t you have to also take the weight of the air contain in-between each block to precisely approximate the total block it would take before the first one collapse ?

No.

Air presses equally in all directions, as it is a gas. So it doesn’t weigh things down. If you pull a vacuum on the bottom of the brick, then it would affect it.

ah i see thx

If you start building down from a geostationary orbit you could theoretically build a Lego tower that is a minimum of 35,786 kilometres (22,236 mi) high — using the gravitational tug from this tethered position should offset the crushing mass (some superglue might be necessary to keep it all together).

I never really like it when people compare the height of something to the elevation of a mountain. Usually the base of the mountain is not at sea level and therefore not truly representing a prospective size.

Add to all the other complaints the fact that the “research” doesn’t answer the basic question of tower height, just compressive strength. Such a tower would surely fail from column buckling before the bottom failed in compression. Yet there was no attempt to calculate that failure mode nor even determine the modulus of elasticity of lego… I suspect the creator doesn’t know crap about how actual structures fail, just an overconfidence in what little they do know. If THIS is what passes for either technically interesting or educational then God help us. How about they go ask some civil engineers who… I don’t know… design towers to help them. Have them talk about why towers really fail and how they design acutal towers so they aren’t just giant piles of steel blocks.

Before you get all snooty about the answer to a fun and obviously silly question, you should perhaps re-read the question they were trying to answer.

You’ve assumed the question is something that it isn’t; i.e. the question was not how tall you can build the tower before it fails, but rather

“how many Lego bricks can be vertically stackedbefore the bottom one collapses.“Thank you for your comments. Having researched and taught in materials and structural engineering for twenty five years, I have picked up a few basic ideas about the processes involved.

My brief in this case was to answer a simple question about the crush strength of a Lego brick. If you listen to the BBC podcast, or read the original BBC article, you will see that I clearly made the point that structural failure through buckling would occur first, and this was backed up by a professional Lego builder who was also interviewed.

Incidentally, buckling of a Lego column is significantly more complicated than simple elastic / Euler buckling, because of the possibility of bricks pulling apart under tension. It’s more like buckling of concrete columns, which crack under relatively low tensile stress but can take very high compressive ones.

Olympus Monsby one measure, it has a height of nearly 22 km (14 mi). Not 2917m.

And Felix jump was from 39 Km. Much higer that the brick tower.

Scale was set to altitude at which Felix’s chute was opened.

And they’re referring to Mount Olympus in Greece, not the big-ass volcano on Mars. They are, in fact, two different mountains, not easily confused with each other except for the name.

There is also the lesser-known Mount Olympus in Washington state… The only one of those that I have actually seen. (Although it’s possible that some photons from the one on Mars have hit my eye in the few times when I looked at Mars in the sky). A quick visit to Wikipedia surprises me in revealing that the “real” Mount Olympus is only a little taller than the one over here.

A tall skinny column such as this tower fails due to buckling (eg. like folding in half along the length), not pure compression – so it would fail long before that height. Need to use the slenderness ratio to calculate the maximum unbraced height allowed, and design a braced tower. I think the above comment about the pyramid would solve it, though the base would be enormous.

Yes, but as stated above, when the tower would fail was never the question being asked.

true. tower was only implied, or even imposed on the question by my reading of it (as well as many others). also, its rather well covered anyway on some comments further up the chain here. I would like to see a giant Lego ziggurat tho. that might be limited by lego brick supplies though

Not only that, but each brick becomes stronger (can take more compression before it buckles locally) when it’s surrounded by other bricks (since the brick in the photo failed by having its walls buckle outwards, where other bricks could have been placed). By how much? I don’t know. Time to squash a Lego brick constrained inside a rectangular tube with thick steel walls…

Overall the walls buckled outwards about 75% of the time and inwards 25% of the time. I would only expect significant stiffening when two adjacent walls tried to buckle outwards into each other, which would occur about 9/16 of the time.

I guess the calculation assumes earth has lost its atmosphere. The air above the tower would weight a hell of a lot less at 2 miles up than at sea level. Still it would make the theoretical tower shorter than the calculation shows.

So basically we’ve just stress-tested a lego. We haven’t come close to answering the question of what’s the tallest lego tower that COULD be built, since there are multiple other dimensions to a problem like that.

What’s interesting about answering one dimension of a problem when all the other dimensions are arbitrarily assumed?

“What’s interesting about answering one dimension of a problem when all the other dimensions are arbitrarily assumed?”

Holding all but one variable constant is a way of simplifying a complex problem so that bounds can be placed on the solution space before staring on the much more complex task of simulating the full system. When some of the bounds are known you have a way of detecting when your more complex model is being used outside its valid range.

I think some people here are making a big thing out of a small question which was: “how many Lego bricks, stacked one on top of the other, it would take to destroy the bottom brick?” It’s a simple question and was given a simple answer. The original question did not ask how tall a column could be built, although that may well have been in the mind of the poster.

I suspect that in the real world if you phrased the question so as to require a single brick wide column that the answer might well be that it would not be possible to destroy the bottom brick because the column would fail long before you reached a high enough force. The question is so vague that one has to make a lot of assumptions including which of the many Lego bricks to test, that there are no transverse forces, no wind load, and many others.

For a moment there I thought

Mount Olympuswas Olympus Mons on Mars.A more interesting question (at least to me and probably the olympian gods keen on dancing) would be “how high a lego pyramid can be built?”

I suppose that it might be quite a bit taller, considering how the weight spreads and all… I’m sure any architect with a geeky bent (or a mathematician) can easily provide us with an answer to this burning question.

(all the necessary data is in the article; 2×2 bricks stacked in the obvious pyramid pattern..)

Oh, sorry I see the question’s been answered already (scanned the comments for “pyramid” and so missed “ziggurat”…)

Three times as much? 10,777 m… Now we’re talking!

Actually, the question has not been answered. When surrounded by other bricks, the bottom bricks become stronger, because they’re forced to buckle inwards rather than outwards, and they can take more force that way (presumably from the fact that the test brick failed by buckling outwards). So a Lego pyramid could be EVEN TALLER than the 10777m.

Well it can be much higher if you start with a wide radius column and end it with a one block column, like they build the real buildings!

Yeah, well who’s gonna put the top ones on? Not Felix Baumgartner, he’s over a thousand meters too low.

Find a place on earth with absolutely zero wind at every height along the way, so that it doesn’t get blown over. Good luck with that.

Since they are material testing Legos, they should go ahead and test the shear and tensile strength too – of individual Legos, and then for connected Legos. I imaging they would only do slightly better than dry stacked masonry in shear though, and have no practical strength in tension. Then we could get to work on testing some Lego building assemblies, and incorporate them into the building code.