The Neat Alignment of the World's Biggest Antiprism
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- čas přidán 17. 10. 2023
- Check out Jane Street paid internships www.janestreet.com/internships (or pass on to someone you think would benefit from it).
Huge thanks to Laura Taalman for joining me for a day of walking and math. You can see all things Taalman on her website including excellent 3D print files: mathgrrl.com/
Next "An evening of Unnecessary Detail" show is 20 November 2023 in London. More shows in the future in the UK and USA, ticket links always here: fotsn.com/tickets
Thanks to "biludavis" for the 3D model of the WTC: www.thingiverse.com/thing:263478 All other 3D models and prints were designed by Laura.
Huge thanks to my Patreon supporters. They put the twist in my antiprism. / standupmaths
CORRECTIONS
- None yet, let me know if you spot anything!
Filming by Alex Genn-Bash
Math by Laura Taalman
Produced by Nicole Jacobus
Stills photography by Truman Hanks
Editing by Christopher Brooks
Sound mix by Steve Pretty
Music by Howard Carter
Design by Simon Wright and Adam Robinson
Written and performed by Matt Parker
MATT PARKER: Stand-up Mathematician
Website: standupmaths.com/
US book: www.penguinrandomhouse.com/bo...
UK book: mathsgear.co.uk/collections/b... - Zábava
“Proof by not on the internet”
“Proof by thinking about it for a minute”
I’m really liking these in depth proofs that were getting in maths now
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"
Petition to turn "proof by thinking about it for a minute" into a real thing
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.
Architect here. One of the best arguments for making an Anti-Prism shaped building is designing around lateral loads (more specifically wind loads). As you build taller, wind loads becoming a much larger factor in building design. One way to design for the increased wind load without adding more bracing, is to rotate the structure (This is also why a lot of skyscrapers "twist"). With an anti-prism, you break up the large surface of each facade, and achieve a similar effect.
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.
These units are mind-boggling. Matt converting his feet to inches and then back to another kind of feet.
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.
You know, I thought "world's biggest antiprism? how do you know?" and then realized it'd be _really_ obvious if there was a bigger one, so fair enough :D
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!
The anti-prism has more corner offices than the regular prism. It means you can charge more for each square area to tenants.
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
"Proof by thinking about it for a minute" I'm sure as hell going to use this on my next exam
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.
I wonder if the official figure of 200 feet is actually a measure of the internal dimensions of the building, so the extra 5 feet or so that Matt and Laura measure would be the combined thickness of the walls on either end. Or I guess it could just be that 200 feet is a very rounded (or truncated to 1 sig fig) measurement.
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. 😂
Matt, I rarely comment on CZcams but I'd like to thank you for the candid and honest way your videos are made. Other math channels always made me feel a bit stupid and wonder how can they be sooo good at math as to never make any mistakes. You're by far the most sincere math youtuber I've ever watched, for you don't try to hide your mistakes, instead, you show us them and it helps a lot in the learning process!
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.
For the non americans out there, 39,433,333 cubic feet is about 1,116,628 cubic meters.
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
You could make a good guess by considering the n=2 case instead of n=4. A digonal prism has zero volume, while a digonal antiprism is a tetrahedron.
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
Me, during my whole education: "math is something that happen in dark rooms with old (usually beardy) people scribbling on kilometers of blackboards."
The video here: "let's get tons of people from everywhere and make math video in parks, outside buildings, during conventions…"
We should put more fun in math in school.
My first thought was that a true antiprism would have the same volume as a prism of the same height and top/bottom areas. But when Matt and Laura both disagreed with me, I was pretty sure I was wrong.
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 decided to have some fun with the regular anti-prism:
Using some geometry, we can find that the general formula for the cross sectional area of any slice of the square anti-prism is (b^2)*(1+(2sqrt(2)-2)*(h-y)*y/(h^2)), where h is the height of the prism, b is the side length of the base, and y is how high up that cross section is. Note that if y=h/2, this corresponds to the middle cross section (the octagon) and we get that 20.7% figure Matt mentioned in the video. Also note that (1+(2sqrt(2)-2)*(h-y)*y/(h^2)) is always at least 1, which means that any cross sectional slice of the anti-prism will be at least b^2 if not larger. b^2, of course, is the size of any cross section of regular square prism.
Using some calculus, you can then get the following formula for the volume of the anti prism: (b^2)*h*(sqrt(2) - (2sqrt(2)-2)/3). Since (b^2)*h is the volume of a regular square prism and (sqrt(2) - (2sqrt(2)-2)/3) is equal to about 1.138, a square anti-prism is 13.8% more voluminous than the corresponding square prism!
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
"proof by thinking about it for a minute"
exactly how I do proofs
This is how I proved One World Trade Center is rather large
New York... the city of architexture
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
I was waiting for them to check the math by putting the model in water and seeing the displacement.
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.
I love seeing math nerds work together. It doesn't have to such rigorous tedious work to just figure something out for fun with a friend. Great video.
I've not been so pleased with myself in ages, as when I figured out the area of an octagon from first principles in my head.
I used a calculator to find out if it was bigger or smaller than a square of twice the side length, since it wasn't a nice neat formula. I was so pleased with myself when I checked with the formula online and got the exact same result.
Maths is great like that! :D
My mental approach was to imagine that the number of sides of the prism (N) increased then the counterpoint of the the anti-prism would have 2N sides and thus be a closer approximation of a circle. As a circle has the highest surface area to circumference ratio the mid point of the anti-prism would therefore have a higher surface area to circumference ratio than the floors at either end.
I would also suspect that as N increases for the starting polygons then the boost in volume/area gained for making an anti-prism (over a regular boring prism) decreases.
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.
You could solve this problem by constructing two watertight models, one being a cube and one being the antiprism. Fill them up with water and then measure the volume of the water.
“No views 46 seconds ago”
I feel so cutting-edge.
Great video! I love that you showed the process you went through to calculate the volume!
as compared to (5/6)A*h, I computed that if the building was a frustum with bottom area A and top area A/2, then the volume would be (1/2+sqrt(2)/6)*A*h ~= 0.7357 A*h
88.3% the volume of the anti-frustum
More collabs with Laura Taalman please.
++
It actually seems like the volume of the proper antiprism would be harder to work out than the "antifrustum", I think you'd have to work out the area of those octagonal cross-sections and then integrate it along the length of the antiprism.
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
Proofoid by inspection: Halfway up must be an regular octagon, with side length = 1/2 the base, which then requires the perimeters to be identical. The octagon is closer to a circle, which is the shape with the most area for a given perimeter.
Now to see if Matt notices the same thing.
nice idea!
Interesting fact that the more aerodynamic a city's buildings, the cooker the city will be. Blocking wind helps contribute to the urban heat islands.
"the cooker the city will be" - I think that is the opposite of what you meant to say.
Love how excited you are! Wishing you luck on your presentations!
The anti-frustum volume was surprisingly easier to calculate for the special case the One World Trade center is, since you can join the negative space in the corners of the building into a pyramid with a side of (a/2)*sqrt2 (half the length of the base's diagonal) and the height of h, and then you can simply calculate the volume as V=a^2*h-(((a/2)*sqrt2)^2*h)/3. Sorry if you said it in the video, I paused it at 8:22 when I thought of this interpretation. Thanks for the amazing vid and an interesting problem to solve!
Edit: Yay, we found the same solution! :D
or
Why this video could be 30 seconds long.
My instincts were telling me that going from the prism to the antiprism made the shape closer to a cylinder with diameter square root 2, which is larger than the normal rectangular prism, thus my guess was that the antiprism was larger.
I was waiting for someone to cover this for a long time, so great video!
Completely tangential anecdote - as a hobby I used to do 3d modelling and the like, and sometimes built things for custom games in WarCraft III. For a game like that, you want the models to be as low-polygon as possible, which presents an issue for, of course, round objects, such as cylinders.
A very clever solution that I noticed some of the base game models used for small cylindrical details, such as the hilts of weapons, or polearms, flagpoles, fences, etc, was that instead of the normal intuition of making, for example, a hexagon and extruding it to the other side, which results in an ok-looking roundness compared to the terrible long triangular prism or square, they used a triangular or square anti-prism, exactly for some of the reasons mentioned in the video - the middle cross-section of the antiprism has double the visual edges, which gives it a round look, but uses half as many triangles around the sides to do it! (remember that while most artists use quads for modelling conceptually, everything is ultimately triangles in polygonal 3d art, so a square is actually 2 triangles).
Anyway, small anecdote on a somewhat practical use of antiprisms I learned about some time around middle school, haha. It's a bit funny because while I like the design of the WTC1 building, it always to me looks like the hilt of a WarCraft III sword stuck in the ground :P
A subsection of the Shanghai tower might qualify as anti prism and would certainly be taller than the anti prism subsection of the owtc
Great video as always, thanks to all involved! I'm not particularly maths inclined, but I enjoy learning a little, even if much of what I retain is just "wow, how cool is maths?!" :)
For the anti-prism, you could also look from directly above and directly below and see that all of the sides expand outwards from the perimeter of either base. You can see this in the graphic at @8:04. Since all of the sides move outwards from their initial border to reach the corresponding corner at the opposite end, it would have to be larger.
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.
I watched the building go up, and have walked the underground tunnel adjacent to it, and I've sort of built a model using Magnetiles, but I'd never heard what the shape was called, and I did wonder about it.
So thanks, Matt, from a grateful New Yorker, who is now miffed he didn't know you were in town to attend your lecture.
Yey I guessed it right. Because every triangle tilts outwards it must cover more space.
ive always loved the simple complexity of the shape of that building, and im glad i now know the name of it
I like the idea of continuing to twist it, but I think Laura stopped too soon. If you consider the limit as you twist it to infinity, continuing to subdivide the vertical faces as you go to keep it convex, it seems natural to me that you'll eventually end up filling the space of a cylinder.
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.
I've found that in so many things, just breaking down whatever it is into smaller, more manageable chunks (think: simpler shapes here), makes figuring out the "big picture" much easier.
Thanks Matt, thanks Laura, helps a lot!
Should've waited till November 9th to post a very British video about the American One World Trade Center 😂😂😂
In the tapered case, can you use the intermediate value theorem to show there is always exactly one horizontal slice which is a perfect octagon?
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.
I loved how you searched for a solution, I loved how you dealed with your small mistakes (keeping them in the final cut). We have to improve those 2 points here in France.
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.
Super clever idea by Laura to split the prism into four quadrants like that(!)
I feel like you could have emailed the architects for the total volume.
A way to quickly intuitively grasp why the antiprism is larger in volume than the prism if you have 3d modelling software handy (or if you are very good at picturing solids in space) is to intersect the two - the parts of the antiprism that "stick out" are visually obviously larger than those of the prism...
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. 😅
So, you have an anti-frustum with a larger square at one end and a smaller square at the other. To find the volume, you take two perfect anti-prisms, one with the larger square at both ends and one with the smaller square at both ends. You find the volume of each and then average together the two volumes.
One in Hong Kong, one in London, one in New York.... *Setting reminder to investigate the basement of Jane Street office buildings in search of any wierd fiery portals, whips, giant creatures or flying red capes.*
An interesting fact you missed about anti prisms is that even though the volume is larger (~21% in this case) the surface area is the same:
cuboid: B * H * 4 + B^2 * 2
Antiprism: area of a triangle * # triangles + ends = 1/2 * B * H * 8 + B^2 * 2 = B * H * 4 + B^2 * 2
The height of the anti-prism's triangles is slightly larger than H, because the face of the triangle is not perfectly perpendicular to the ground. So the surface area would be a bit greater than the prism.
Curses! You are correct
Can you imagine working in a simple cuboid?
That must be so depressing...
😂
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.
The octagon in the middle is just the square base with its corners cut off... Since the outer walls are plumb, four of the octagon sides are directly above segments of the base square.
Great vid as always.
Conductivity enters the chat to speak to the Fresnel reflection and transmission coefficients to ruin the fun.
The perimeter is independent of the height.
Therefore the octagonal cross section is bigger than the square section, because it has the same perimeter and more sides.
My first solution was, you already have a 3D print, submerge it and a normal prism with the same sized bases. water displaced will be the volume. But you are Stand-up Maths and I expect you'll create a math proof either way.
My other question is how this effect structural integrity as a building. (With the Frustum version, since a true anti-prism wouldn't be made because of the overhang.)
Thanks for the fine and fun analysis from first principles.
For those who would like to have it handy, I've long enjoyed having the useful and versatile formula for the volume of a prismatoid:
V = (h/6)(T + 4M + B), where T is the area of the top, M the midsection and B the bottom.
It took me a while, but I managed to follow the proof in an old geometry book.
For the case in the video, some simple geometry shows that T = B/2 and M = 7B/8, so the expression for V reduces nicely to the value (5/6)Bh.
It’s pretty easy to figure out the formula for the area of an octagon by dividing it into a square, four rectangles, and four isosceles right triangles. It’s also a nice and fairly easy calculus exercise to compute the area of the anti-prism and anti-frustum.
Or a square _minus_ four isosceles right triangles, if your octagon is based on a known outer dimension, rather than a known side length.
Instinctively, it makes sense the antiprism would have more volume, because if you break the normal prism into a stack of infinitely thin layers and give that stack a smooth twist so the top is offset by 45 degrees (or the appropriate angle for a different polygon), the vertical sides of the stack become concave. Antiprisms are convex.
Fantastic guest and episode!
I like the figuring out process, despite the fact that the cad program would instantly give the answer.
My initial thought was similar to matt thinking of the octagon.
That's the offices where Sam Bankman-Fried worked at before moving on to FTX! (Jane Street NY)
The fastest ethos to calculate the volume is the create it in autocad (or solids if you’re into that) and calculate the volume and other useful geometric properties
My approach for the guestimation of the volume was "if the top surface has to fit inside of the top surface, it's area would need to be exactly half of the bottom. The center piece is an octagon which shares four sides with the bottom square. The area of this octagon is more than half the area larger than the smaller square. Thus, the vplume must increase".
Also, the "20% bigger" number you calculated is only correcr for antiprisms of squares. For antiprisms with n=infinity, the volume is the same as that of the prism (a cylinder).
that was a great show!
I had a slightly different intuition for it being bigger with the octagon middle - the closer a thing is to a circle, the smaller it's surface area to volume ratio is, and then the more volume it has relative to surface area (i'm a biologist, so this is like, the one math thing i know). Since antiprisims will always have a more circular middle than top or bottom (since the middle profile is always a 2n polygon from the top and bottom surfaces) it should have a greater volume with roughly the same surface area
Great video
We're doing an egg drop competition at work and the maths for it is interesting answering questions like: what will the impact velocity be when dropped from X height? What is a safe velocity for the egg to impact the surface? What is the maximum safe pressure to apply to the egg shell? With Y crumple zone dimensions, what material properties are required to dissipate the energy and ensure a safe egg?
The results of the calculations have steered us towards an unexpected solution. I realise this is like applied maths/physics, but it is 95% maths but knowing what maths to do.
If you were to take a stick model of a prism and twist it, you would notice that the parallel faces at each end would get closer together. It's actually a different problem entirely compared to creating an antiprism of the same height.
Proof by Δx (that the normal sized anti-prism is larger than the prism):
Consider a slice just above the base. It's less an octagon than it is a square slightly larger than the base, with a tiny bit of the corners snipped off. So, compare the area of the bit added to one side of the square (which would be close to s*Δx, where s is the side length) to the bit snipped off one corner (which would be close to Δx*Δx/2).
I initially agreed with Laura's guess and thought it was obvious. When it was shown to be wrong I realised that my mistake was probably that because the top is basically the bottom twisted 45 degrees, my instinct was to think of how if you actually twist the regular prism, the connecting vertical edges will cut into it and reduce the volume. I didn't think about there being two connecting edges for each vertex, not one, and how the second edges would add volume back in.
As a european person, when I hear or read “feet cubed” my mind is only capable of picturing some weird Minecraft-style foot
Never thought I'd say this, but that is actually a pretty neat shape
When you twist the prism the sides of the rectangles connecting the n-gons are no longer going straight down and thus longer than in the prism. Not a proof but that's how I intuited the volume increasing.
One interesting fact I found: Starting from the base, the first 185 feet (85 m) of One World Trade Center are actually a perfect cuboid. Only after that does the actual anti-prism (and tapering) start
It makes a lot of sense when you think about the limits. As you start going up, you barely take anything out of the corners, but you're adding a whole lot of side. Therefore thee area of a slice must be bigger.
Now I want to see a building that's a reverse antiprism. Where the antiprisn is cut in half and then the ends joined together to make the middle.
Using feet, inches and feet^3? Matt you've been in America too long, it's getting to you
I knew that the surface area was affected by twisting, from noticing that when you rotate a bread loaf bag, the bag's height reduces. So for the top to stay stationary, it would need more bag height and surface area to have the same untwisted height. So it also extends to the volume of the shape up to a point (like Laura said) where the straight edges formed still "aim" outward from base shape.
Drawing the shapes in SketchUp and having it calculate the volumes told me that the volume of the anti-prism is larger by 13.807%.
My instinct about the anti-prism vs cuboid was that it was the same size, but I knew I wouldn't trust my mental calculations. So I paused, and opened geogebra, learned to use it, and found the area grew quite a lot in the central octagon.
Fun video! My opinion as an experimentalist: You should have dipped your model into the water and measured the displacement afterwards... The perfect marriage of theory and experiment.
Using coordinate geometry, I found that for rectangular antiprisms with two congruent parallel bases of length l, width w, height h (perpendicular to the base; the area doesn't change if the bases are offset), and base displacement angle θ (where θ = 0 is the standard orientation where the bases are at a 45° angle. θ = 45° gives a rectangular prism, while θ = -45° has the triangular faces flatten into four trapezoids), the area of a cross section at height z is:
(2lw(1 - cos(θ - 45°)) + (l^2 + w^2)sin(θ - 45°)) * ((z/h)^2 - z/h) + lw
Integrating over z from 0 to h, the volume is then:
lwh/3 * (cos(θ - 45°) + 2) - (l^2 + w^2)h/6 * sin(θ - 45°)
For a square antiprism, where l = w, these simplify to:
Area = 2lw(1 - sqrt(2)cos(θ)) * ((z/h)^2 - z/h) + lw
Volume = (sqrt(2)cos(θ) + 2)/3 * lwh
You could also argue by preservation of perimeter. Two of the triangles in the antiprism = 1 rectangle in the prism so the perimeter of the cross-section must remain the same. Given fixed perimeter, the max area is a circle, and an octagon is much more circular than a square.
You can prove that the antiprism has greater volume without using the formula for the area of an octagon. The similar triangles argument can be used not only to find the side lengths in the middle cross-section, but it further shows that the cross-sectional perimeters are constant! And of course such an octagon will have greater area than a square with the same perimeter.
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).
So to definitely answer the question raised at 3:00, one needs to find the area of the pyramidal square frustum with height H, base side length L and top side length L/sqrt(2), which is H/3* (L^2+L^2/sqrt(2)+L^2/2) or about 0.7357 times the base area times the height, i.e. less than the volume of the frustum antiprism found to be 5/6 times the base area times the height. Almost 12% less in fact.
I was guessing more, based on this:
The shortest distance between two points is a straight line. In a cuboid building, the corners are connected with straight lines. In order to connect it as an antiprism, the lines have to be angled away from vertical, which makes them longer. That made me think it's probably larger. But I don't know if that description is true or coincidental
13:23 I was always taught that unit-squared is not the same as square-unit and unit-cubed is not the same as cubic-unit. I know it's probably semantics for some, but it seems that there could be confusion, for example (4 meters) squared means a square with sides of 4 meters, whereas 4 square meters, means a square with sides of 2 meters and an area of 4 square meters.
Can't believe Matt didn't measure the sides of the anti prism in cubits
And the volume in cubic cubits
He chose to measure it in feet, the old-fashioned kind.
* It is not an square-antiprism frustum.
A frustum of a regular antiprism has trapezoid faces not triangular faces, and the top has eight sides of alternating lengths (or a regular octagon if bisected).
That hilarious, the two of you nerding out on the balcony
If it's more volume (more floor space) and obviously looks cooler, then why aren't more buildings anti-prisms?
When Matt was off by 3 orders of magnitude, all i could think was "not only is it an antiprism, it's also the anti-tardis."
He could have left the wrong answer in and almost nobody would even have been any the wiser, if it was all in American units.
Feet? Inches? I never imagined seeing a full Matt Parker video using Imperial units! :D
My very non-mathematical intuition was that if you rotate the top square by 45 degrees, then its corners jut out beyond the area of the bottom square. That extra space (4 triangles' worth) should be included in the volume somehow, so the volume should be larger than a regular prism! It's nice to be right about something mathematical for once. Now it makes sense why the building tapers at the top, too.
39million cubic feet... great... I literally have no reference for a cubic foot. Can we have that again in global units please? :D
I think it’s easy to prove if you imagine the volume of the same building with parallelogram sides. Which would be equivalent to the cuboid volume. So by cutting the the parallelograms across the diagonal, you create a bulge whose edge bulges out more than the planar surface of the parallelogram.
well well well, so nice of you to use my name for the explanation 🤭
How big does the top square need to be, relative to the bottom one, so that the antiprism has the same volume as the prism?
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.