The Neat Alignment of the World's Biggest Antiprism 

In other walks of life there exists the "proof because I want it to be true".

When I was a little kid I employed what I later realized was "proof by poking you until you agree with me"

And "proof by not on the internet".

a "parker proof", if you will

A petition to completely dismantle science? That could catch on. Oh wait...

I think these would be more like your everyday kinda proof. Nothing you write down in your science paper but something you say when someone just keeps talking nonsense because they acually havent thought about it for a minute

Nah, the UK government would say it would distract drivers too much.

Much nicer than the corkscrew fins found on large chimneys. I hope chimney designers could use these antiprisms.

Architects don’t design big buildings they are unqualified. You need a structural engineer.

@@brendan3603 Different Architect Here: Architects are very much involved with the design of buildings, both big and small, including skyscrapers. However, except for the smallest of buildings, we never design a building alone. There will always be a team of consultants who work together with the architect to make a building, with each consultant focusing on their area of expertise. One of those consultants is always a structural engineer, who is primarily charged with keeping the building standing. Other consultants include electrical engineers, plumbing engineers, HVAC engineers, civil engineers (for site work), landscape architects (for plants and irrigation). Additional consultants can be brought in if a project needs it. The architect works to organize the consultant team and drive the overall design, while also choosing finishes, dealing with life safety items, dealing with accessibility items and numerous other things. The larger and more complicated a building is, the more the architect needs to rely on their consultants and the less likely an architect can just do whatever they want. So an architect likely came up with the design for the anti-prism shape, then worked with the structural engineer to figure out how best to achieve that look.

And seemingly by using an anti-prism you get almost the same internal volume you'd get from just using a rectangle.

​@@brendan3603generally they're designed by architects then the engineers fix the designs.

why obvious? have you kept track of every sky scrapper in the world?

​@NoNameAtAll2 the big ones? Yeah, lots of people keep track of that.

​@@NoNameAtAll2It really depends on what he means by "biggest" In this case, I think he means tallest? And if he means tallest, then you just have to look up a list of buildings that are taller, and verify they are not antiprisms.there probably isn't very many.

@lunasophia9002 it would be on the internet if there was a bigger one. Proof by ‘not on the internet’ ;)

@@NoNameAtAll2 The point was it'd be hard to miss something that big. Also, no, I don't track every sky scraper (or sky scrapper), but the folks on Wikipedia do!

That's a good angle - several good angles, actually! I suppose we could make a building that's all corners by just making a pie crust shape on top and bottom and get super rich! I mean, the majority of those corner offices will be facing each other, but the math doesn't lie!

But the angle would be unlike most corner offices. It just wouldn't be right.

@@vigilantcosmicpenguin8721Just make your tenants sign before they realise that.

@@vigilantcosmicpenguin8721 This assumes that the lower degree corner the more desirable corner office

@@vigilantcosmicpenguin8721 I thought this was acute joke

While writing a paper, I once asked my advisor if I could use "years of learning maths" as a citation for a well-known, often-used formula. He said no.

And I still have no clue what the volume of the building is, 33 million cubic feet? 🤷‍♂️

And thats quite a feat

@@kj_H65f and thats quite a feet

neatly, this is basically one million cubic meter, which is 0.001 cubic kilometer !

@@DantevanGemert It's funny how it round down to a million cubic meter.

Same thought. Their measurements are more trustworthy than a 'front page' published size.

One of the… “fun things” with engineering projects are the as-builts - the drawings that are updated to show what was actually made

@@dobystoneSome say you can hear the architect screaming whenever you view them. 😂

This is crucial. Mistakes and mis-steps are an intrinsic part of the process. You tidy it all up afterwards. Presenting it as a _fait accompli_ does no-one any favours. To take maybe a cliched analogy, it's like presenting a completed jigsaw and pretending you didn't make any guesses at any stage about which piece went where, they all just slotted in first time in the correct place with the correct orientation :-)

That’s the Parker Brand Promise.

I was coming to comment the same thing, but I'll just reply to boost your comment. It is great modeling that even brilliant people make mistakes, and that those mistakes are not the end of the world. You accept the correction, fix your work, and move forward.

And for n=infinity they're equal! So it's a good guess that it would never be less.

​@@MrCheeze cylinder

@@Foxxey Yes, both the prism and the antiprism would be the exact same cylinder and thus have the same volume.

Could you say that as n goes to infinity, the volumes of the prism and antiprism get closer to one another? Where n=2 has the highest difference and at n=infinity they're equal.

​@@eutiyes

which is about 1.116 cubic hectometres

Can I just point out that both of those numbers are so large that they are meaningless to me.

While I prefer metric, I think it's perfectly okay to use feet and inches since it's an American building

Thank you!

Which is 0.001116628 cubic kilometers

Actually, my very first thought upon first seeing Matt was "I really need to shave my head."

my first thought was - if i twist something, it contracts either in length or diameter. if you keep the length and diameter the same, the volume must increase.

@@supremecommander2398 I had a similar thought - the volume of a wire frame prism shrinks when you twist it since it get shorter, and even stretched its thinner in the middle because the faces aren't planes. But when you add a diagonal wire to each face then it rotates into an antiprism with thicker cross sections meaning more volume. neat.

@@supremecommander2398 And if you keep twisting it eventually becomes a cylinder of the same length.

I went a bit further and calculated the algebraic formula for the volume of an anti-frustum and it comes out to (2 + sqrt(2))(a^2 + b^2)/6 where a and be are the sides lengths of the two bases. In the case of the building in question it comes out to about 0.85355 as opposed to 5/6=0.8333, so this is another Parker calculation.

Yep. I got the factor of ((sqrt(2)+2)/3) I liked the intermediate formula of A=w^2+2sqrt(2)ww'+w'^2 for the cross sectional area. Where w=one side of octagon and w' is the other. A pretty parabola.

Just to add: given you factored out _b²/h_ at the end, I note that your formula is more intuitive if you just factor it out at the beginning, i.e., scale your area formula to b=1, h=1: _A(y) = 1 + 2(√2-1)y(1-y)_ _V = ∫₀¹ A(y) dy_ which simplifies a bit further than you did, to _V = (√2 + 2)/3_

@@viliml2763 I think you are missing the h factor in your formula. So, if we wanted to know what should the top side b be for the anti-frustum volume to be equal to the volume of a square prism with side a and same height, it would come out as b = sqrt( ((4 -sqrt(2))a^2) / sqrt(2 + sqrt(2)) ) or approximately 0.87a

Yeah, if that's a scale model then Archimedes seems like a great way to get/check an answer - although the physical geometry of re-arranging the quarters is a cool realisation too!

Way too physics for this maths class haha

3d prints are mostly hollow, so it would just float and youd have a hard time getting an accurate measurement

@@sachathehuman4234 It would also fill with water so it wouldn't displace all that much...

You could also check the volume of each using the slicer software used to print it. At least PrusaSlicer tells you the volume, and I assume others do too.

exactly how I do proofs

This is how I proved One World Trade Center is rather large

Normal people: New York City Americans: Nooo yourkh siddee"

And at 1:00 he was "joint" by his friend.

That non metric system sounds feudal.

​@@mildlydispleased3221L

Maths is great like that! :D

I had same exact intuition, my guess is that’s what Matt thought too but didn’t know how to say it.

I also suspected this, and the limiting case is obvious since it doesn't matter how you twist a cyclinder so you get 0 volune gained.

I thought of that too, but the perimeter isn't constant, so there's no reason the perimeter-to-area ratio would necessarily be relevant. The central octagon should actually have the smallest perimeter of any cross-section.

@@evanhoffman7995 As you correctly identified, the perimeter of any cross section being constant is a big part in the proof. But that perimeter IS indeed constant: All the faces that make up the perimeter are triangles. If you go X% up the structure, then the length contributed to the perimeter from the triangles that start at the bottom is X% of the perimeter of the lowest slice. And the perimeter contributed from triangles that start at the top is 1-X% of the perimeter of the highest slice. If you assume that the top and bottom face are identical, then the perimeter stays constant. Otherwise the perimeter varies linearly from bottom to top.

++

Or, since you know the exact center is the widest point, you could probably get away with doing something similar to the negative pyramid trick.

true - you simply cut it in easy to calculate geometric volumes and calculate their volumes... just like a 6 grader learns in math lessons

"the cooker the city will be" - I think that is the opposite of what you meant to say.

Would think so, each set of four sides of the octagonal intersection changes continuously (and in opposite directions) from zero at one end to the nonzero side length at the other, so somewhere in between there must be an intersection where they're all the same length...

Yes, it smoothly goes from an octagon with diagonal length of 0 to an octagon with orthogonal length 0. In fact, since the missing shape is a pyramid, the change is linear (since the side length of a pyramid changes linearly with height), so: d=1/2w*x o=w*(1-x) Where d is the diagonal length, o is the orthogonal length, w is the width of the base and x is the percentage up the tower, set d=o and 1/2x =1-x, or x=2/3, so at 2/3rd up the tower the floor area is a perfect octagon.

Four sides go outwards, but other four go inwards. If you see from the bottom, *other* four sides go outwards. Eight walls in total, completely symmetric. So it is not so obvious.

Above and below, spooky

​@@ch347But they only go inward the point of the original prism. If you were to construct this out of two sheets of construction paper, all 8 triangles would be angled out word relative to the base square they were attached to

This was my thinking as well, in super naïve terms the convex hull of the anti-prism is larger and I just had a gut feeling the "truncated" corners wouldn't reduce the volume by more than was gained

@@KrishnaKumar_Profile_Denver Yes, but also all 8 walls go inwards, if we follow each triangular side from tip to base. But this fact does not mean, that volume of anti-prism is obviously *smaller* then regular prism.

No. In the antipirism, the line connecting a vertex on the base to a top vertex is not perpendicular to the base, it goes outward from it a bit. If you rotate the top more, past 90 degrees, that line would now be crossing over the base to connect to the vertex, and becomes a concave polyhedron. And You couldn't possibly end up with a cylinder, because there are two square faces that stay squares. How many square faces does a cylinder have?

@@LeoStaley That's why I said to continue subdividing faces to keep it convex. While it's true that the ends will always be square, the ends are also two-dimensional objects with no volume. The volume can approach that of a cylinder without any faces being circular.

its not twisting its two rotating faces connected by straight lines if it was twisting smoothly it would have the same cross section throughout just rotated and have a constant volume

But if you always keep all the edges while twisting the top square, you get an hourglass kinda shape everytime it is 180° rotated. And thats obviously gonna have less volume. And the volume of any prism or antprism like shape with non-circle top and bottom faces can only converge to the volume of a cylinder if you kinda inflate it and it bulges outwards so that a part of the side of the shape is becoming the top/bottom and is filling the circle there

​@@LeoStaley the angle of the line approaches 90° as the number of sides of the polygon approaches Infinity, it's very easy to visualise it yourself. Obviously the lines are never gonna cross the base side to side because we're talking about anti-prisms and not any ordinary solid, that was well defined from the beginning and it's not something one should be questioning so far into the problem.

nice idea!

or Why this video could be 30 seconds long.

But in France, you also have _le mètre_ to make the calculation so much easier .....

But it’s an irregular antiprism (top and bottom are different sizes). Is not a frustum at all. The faces are complete isosceles triangles not trapezoids.

The Twin Towers (the original WTC buildings) were simple cuboids. Square footprints (now big memorial fountain pools), and rising ~1360 feet above the sidewalk. I imagine it was mostly a pretty interesting place to work, except for a couple of days here and there. And that last one, of course.

antiprism always bigger no matter how big and how many size they both have. This can be proved easily by calculus.

I mean the shape that is like an antiprism, but where one of the bases is smaller than the other, like one world trade.

@@VanjaPejovic *tl;dr:* Ratio of top square to bottom square dimensions: 87.4%. Ratio of areas is 87.4%² ≅ 76.4%. Let's work through it: First, make the maths easy: set the base square to 1×1 and the height to 1. Volume is the integral of cross-sectional area over the range of height. Given an area function A(h): _V = ∫A(z) dz_ integrated from _z=0 to h_ Define _x_ as the top square side length. We're looking for _x_ such that _V = 1._ (Start with a sanity check from what we know already: if _x = 1,_ this is an actual antiprism, with _V > 1._ If _x = √2/2 ≅ 0.707,_ that's the One World Trade Center with _V < 1._ So we know _0.707 < x < 1_ and we'll confirm this at the end.) So the plan is: 1. Find a function for cross-sectional area: the area cut by a horizontal plane at height _z._ 2. To get volume, integrate that with respect to _z_ from height 0 to 1. 3. Set the volume equal to 1, and solve for _x._ *1.* The cross sections are octagons with 45° angles, with alternating side lengths. As _z_ goes from bottom to top, the 4 sides parallel to the _bottom_ square shrink from 1 to 0 with the formula _1 - z,_ while the 4 sides parallel to the _top_ square grow from 0 to _x_ with the formula _xz._ If you draw an octagon with 45° angles and alternating side lengths _a_ and _b,_ you can derive its area _A_ = _a² + b² + (2√2)ab,_ either by adding up rectangles and triangles, or by subtracting the diagonal corner triangles from an enclosing square. Now substitute _a = xz_ and _b = 1 - z,_ then simplify the algebra and so on: _A(z)_ = _x²z² + (1 - z)² + (2√2)(xz)(1 - z)_ _A(z)_ = _x²z² + 1 - 2z + z² + (2√2)xz - (2√2)xz²_ _A(z)_ = _(x² - (2√2)x + 1)z² + 2((√2)x - 1)z + 1_ *2.* Integrate with respect to _z._ _V_ = _∫A(z) dz_ = _(⅓z³)(x² - (2√2)x + 1) + (½z²)2((√2)x - 1) + z_ with _z_ from 0 to 1 Now for the easy bit: take that with _z=1_ minus that with _z=0._ _V_ = _⅓(x² - (2√2)x + 1) + ((√2)x - 1) + 1_ minus … zero. The _z=0_ half is just zero. _V_ = _⅓(x² + (3√2 - 2√2)x + 1 - 3 + 3)_ _V_ = _⅓(x² + (√2)x + 1)_ *3.* Final step: set the volume to 1, simplify the algebra, and solve for _x._ _V_ = _⅓(x² + (√2)x + 1)_ = _1_ _x² + (√2)x + 1_ = _3_ _x² + (√2)x - 2_ = _0_ Quadratic formula time! _x_ = _(-√2 ± √(2 + 8)) / 2_ _x_ = _(-√2 ± √2√5)/2_ _x_ = _(√2/2)(-1 ± √5)_ So there are 2 roots for _x,_ (√2/2)(√5 - 1) ≅ 0.874 and (√2/2)(-√5 - 1) ≅ -2.288. But side length _x < 0_ does not make sense with our model, so we choose the first solution. *And, sanity check: 0.874 is between 0.707 and 1.* *Exercise for the reader:* Figure out what the geometry of a negative side length _x ≅ -2.288_ would actually look like. Apparently it has a volume of 1, but what is it? Perhaps it is completely nonsensical and does not have those octagonal cross-sections at all.

Have you checked this? Just curious and without the software. "My proof of thinking about it for a minute", is that I'd expect that there are also parts of the prism that would stick out too. 😅

Or a square _minus_ four isosceles right triangles, if your octagon is based on a known outer dimension, rather than a known side length.

No, I didn't

That's crazy, I would have thought one of them would have larger volume...

How did you get to that conclusion?

@jansalomon5749 It's on the internet, it must be true. (Disclaimer: this can indeed be true, but only if the prism and antiprism have bases of differing sizes)

@@Maelwys Well, all Matt said was: if it's not on the internet, then it's not true. It does not follow that the inverse is true.

Excellent observation! The area function turns out to be _A(z) = -2(√2 - 1)z² + 2(√2 - 1)z + 1_ at height _z = [0, 1]_ So if you graph {x=height from 0 to 1, y=area}, you will see _a parabola_ going through points (0,1), (0.5, 1.207), (1,1).