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0:00
I'm here at Math City. It is the UK's first hands-on maths exploratorium. It's full
0:06
of all sorts of fantastic mathematical hands-on activities. If you're anywhere near Leeds, I
0:12
highly recommend coming along and checking it out. I've been very excited to come and
0:16
play with some of their stuff, particularly this. This is their giant bubble, I don't
0:21
know, contraption. I'm standing in a ring of bubble solution, and if I pull this
0:25
cord, it lifts up this hoop and it brings, look at that, oh, I popped,
0:31
brings the bubble mixture with it. So before we do any maths based on this,
0:34
which we will, I'm gonna do what most kids do when they get in this
0:38
and just see how high I can get the bubble to go. Yes, I'm in
0:43
a bubble. I was in a bubble.
0:49
What I actually wanna look at is the shape that we get if we very
0:53
slowly pull this up so that the bubble mixture reaches an equilibrium. So if I
0:59
hold it there, you can see it's kind of bowing in towards me.
1:05
So it's not forming a cylinder, it's curving in and that's because the bubble film
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wants to minimize its surface area, something-something energy, and instead of just being a
1:16
cylinder, if it comes in a bit, it's going to have a smaller distance around,
1:22
but if it slopes in that takes more bubble mixture. So there's kind of a
1:27
balance where it comes in a bit to minimize the area then curves back out
1:31
again. And we're going to try and mathematically calculate exactly what this shape is.
1:38
I'm going to start by setting up our system once we've drawn the hoop up.
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So we'll have a hoop, let me try and, I'm just going to sketch what
1:45
I can see over there. We've got one hoop at the top like this, not
1:51
a bad hoop, and then we've got one at the bottom, so this one down
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here, this is like the dish that the bubble mixture is all in,
2:00
and then when we pull this one up, it's gonna have some kind of curve
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joining this point to this point, and we already know it's gonna look a bit
2:10
like a curve like this, but let's draw in some kind of exaggerated, oh the
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curve, it might go in, it might come out, it's gonna do whatever it does
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and join on there. Of course this then rotates all the way around. So we're
2:22
going to try and get an equation for the total surface area, minimize it, and
2:27
that should give us the optimal shape, which should be what happened over there. However,
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it's actually slightly easier, if you don't mind, I'll flip it up this way, because
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now what I can do is put in this one here, put in an axis
2:42
here. So if I put in this is my horizontal x-axis right through the center
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and it means up here I can have my vertical y-axis where this I'm
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going to plot some function of x and that's what we're trying to find. Trying
2:59
to find the function of x which gives us this surface such that when it's
3:03
rotated all the way around minimizes the area. So we have to be able to
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calculate that area. Okay, we can do that. So we'll look at just one little
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bit. So I've got an exaggerated bit of the curve. Let's do here, where it
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goes from there to there. And we're going to call that, that's our tiny delta
3:24
S for a small change in length along the curve. But we actually want our
3:29
delta A, which would be the whole band all the way around. So let me
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see if I can draw this in. This is going to be like, and it's
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going to come around here somewhere, I guess. So that's that edge and then it's
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going to come down here and then around and then like up here. So this,
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so that's an adequately drawn band all the way around. So that's going to have
3:53
some area delta A. So to work out what S is, well I'm going to
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draw a zoomed in version here. So here's our delta S and we could have
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gone along some, that's some change in X, very small change along that axis. then
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there's been a change in height of the function and that's our delta, well there's
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delta y but it's whatever the gradient is times the distance and so
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I'm going to write this as delta x times the gradient which is just the
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first derivative of the functions. I'm going to use y dash for that gradient. So
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y, y is actually the radius here, so if I was to draw in the
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radius up to there that radius is getting a little bit crowded but that's whatever
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the value of y is for x at that point, it's all a function of
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x, and then this is the derivative at that point, and so delta s
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equals, well it's Pythagoras right, so that's the square root of delta x squared
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plus delta x squared y dash squared. And actually you may have noticed there's two
5:00
delta x's squared in there so we can simplify that by just bringing that delta
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x out the front. 1 plus first derivative squared. And now the
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area, so our small change in area, is that whole band, so it's all
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the distance, it's like the width times the circumference all the way around which is
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just 2 pi r where r is y. So that equals 2 pi y
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times that, so square root of 1 plus first derivative squared
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times delta x. And already you can see where we're going with this. So we're
5:38
just going to integrate across all of them. So from zero to wherever, I'm not
5:42
even going to bother putting that on there, the total area equals two pi integral
5:48
of the function times square root of one plus its derivative squared
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dx. What we got to do now, solve that, the integral,
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minimize it and we'll get the shape. Sadly, that's super difficult. So here's
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the deal. You need a bunch of identities to work this through. It's a nightmare.
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I'll link below to the Math World page that does it. And right now we
6:18
would just skip to the punchline. So it turns out that function y equals,
6:24
it's cosh, right? So it's a times cosh of cosh.
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your x coordinate minus b divided by a
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again. And that's it. So, cosh is just the hyperbolic
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co-sign. It's a bit of a shame, it's like we've got sine, and then you've
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got the complement sign, and then you've got the hyperbolic version, and that's kosh. Don't
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be fooled by its two degrees of separation. It is a fantastic function in its
6:54
own right. You may recognize it from when you hang a chain. So a catenary,
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which is like a suspended chain under the force of gravity, forms a
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kosh function. So the shape we should get on the bubble mixture
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is a catenary, but then on its side, However, we still need to work out
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what a and b are, and these are just some constants because you've got different
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shaped cosh plots. You've got this one, you've got this one, you've got all sorts
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of weird things, and so there'll be some way that this will match up for
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the specific, you know, situation we've got here. It's not like a parabola where there's
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only one, two parabola. No, there are lots of continuary plots, so we
7:43
need to work out what a and b are, and sadly, It can't be done.
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There is no like proper way of doing it. The only way you can do
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it is because we happen to know the point here where it
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starts on this ring, we know what its coordinates are exactly, and we know what
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the coordinates are here, that gives us two points which will give us two
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equations so we can solve for these two unknowns, but you can't solve it algebraically,
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you've just got to do it numerically. So I wrote a little bit of terrible
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Python code and it starts with arbitrary values for A and B and then
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gradually iterates them until it has less and less error between where it thinks
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these two points should be and where they should actually be. And so that way
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it's able to like home in the optimal solution and give us values for
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A and B, but they will of course be slightly different depending on the actual
8:45
size of our setup because if we want to get the equation which gives us
8:50
this specific bubble film shape we need to know
8:56
exactly how big our hoop is and then we put those values into my code
9:02
and we should get the equation. 1.2
9:13
0.16, pull this up. 1.12, we'll do that, 1.12. Mild problem, the hoop at one
9:17
end is a slightly different size to the hoop at the other end, because it
9:20
has to fit inside it, and there's a bit of give. So I've just averaged
9:25
them. I've taken an average. This height here is 1.0688.
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meters. So that's the value I'm going to use in all my calculations and we
9:38
need to now decide how far apart they are. Of course in our situation they
9:41
can move but I'm just going to assume that they're let's say half a meter
9:47
apart, right? Just for argument's sake we're going to call that 0.5 and I can
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put those into my code. So I've got some terrible, terrible code, which will do
9:59
these calculations. I'm going to say it's got a diameter of 1.068.
10:05
Actually, that's a good point. That's all the way across. So that is
10:11
divided by two for that specific height. Thankfully, my code uses the full diameter. And
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then I'm going to say we've got 0.5 length. Okay, we run
10:23
that and oh, okay, here we go. So I've now got my values for a
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and b. So in this specific case a equals 0.46519 0.46519
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That's exciting and b equals 0.25 b equals 0.22
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Ta-da! It's a bit underwhelming, I'll give you that. So I've also got this to
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spit out some values. So I can tell you that it should have a surface
11:03
area of 1.61 meters squared when we pull it up by half a meter. More
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importantly, I predict that the diameter
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inside this, when it curls in, that mid diameter should be 93 centimeters.
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So I think that's what we'll do. We will try pulling it up half a
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meter and I should be able to measure a gap of 93 centimeters. So this
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kind of, I'm drawing all over this now, I do apologize. Once you curl that
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in, that mid throat all the way across there should be 93
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centimeters. And we'll see how close that gets. And if we go up to, you
11:49
know what, while we're here, Let's do 60 centimeters as well, just so we've got
11:55
that. And then that should be 85. So if we get excited, we can try
12:00
for two. Okay, first of all, I'm gonna pull this up
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very slowly until it's 50 centimeters off the top of the
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rim, which I think is gonna be about, actually, if I hold this out at
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the same time, it's gonna be about
12:23
is roughly a little bit more there.
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Okay, now without moving this, huh, I'm
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gonna stand up. Here we go. Yes.
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Oh, I can probably hold that between my knees. Come on.
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Here we go. All I have to do now, thank goodness for long arms, down
12:54
here, just doing some serious maths here, and measure this. It's moving a
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little, but it's fairly stable. I think it's about 93
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centimeters. That's insane. Maths. I mean, I shouldn't be so surprised it worked,
13:13
but if people are familiar with my back catalog, I think I'm allowed to be
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surprised that works so well. Okay, let's go two for two. Let's see if we
13:21
can get a 60 centimeter height and go for an 85 centimeter diameter. Here we
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go. Again, we'll bring it, actually, I don't think I'm gonna fit squatting down this
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time. Where's 60? Slowly.
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Going past 50. Okay. 55. 60 centimeters.
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Okay, I'm gonna lock that. Come on. Got it.
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Oof. Oh. It's moving around a lot. Oh no, okay, it's settling down.
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I'm trying to not pull on the cable by accident. The width is
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roughly... Oh, just over 80. I think I bumped it too soon. I'm currently getting
14:12
a measurement of 83. with the wobbling, I'm gonna call
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that 85. 100% success. Thank you very much. Those two values
14:24
work to treat. Question now is, what about all these other values, huh? All of
14:28
these? You may have spotted,
14:34
so far, I've only calculated the values of A and B, hence the shape of
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the curve for one, meh, kind of two specific cases of the height of
14:44
the ring, but in theory I could do it for all of them and put
14:47
together a nice animation of what it looks like as the rings move apart. So
14:52
of course I will do that. Future Matt here and I've plotted the hyperbolic
14:58
cosine function for those values of a and b when the hoops are half a
15:03
metre apart and actually I've plotted more than just this one. I can now animate
15:08
as the distance increases. And I'll stop it here. Actually, a little bit too far.
15:11
I'll go back. I've done every single millimeter. So there it is, exactly 0.6 meters
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apart. And you can see at the top there, that 56.1%, that's just comparing the
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separation between the hoops to the diameter of the hoop. And so in this case,
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the diameter of the hoop is just over a meter. And actually, because I've saved
15:28
these all out, we can now slide through any separation we want. Before we ever...
15:33
too close to look at that. I just want to remind you of how I
15:36
actually worked these out. So here is half a meter, but I've reset it to
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my first guess. A and B both equals 1. And the error value of 1.6,
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I'm just adding the total distance accumulative of both ends from where
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they should be. And then what I do is I just iterate each value of
15:56
a and b a little bit at a time and here we go, then it
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comes flying in. You know, I'm gonna back that up a bit. So what I'm
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actually doing here is you can see when they're both equal to 1, it's so
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bad. Like, they're 1.6 meters combined, like, each end, I guess, on average is 0.8
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meters away. You can't even see it on the plot. But then my code, first
16:14
of all, realizes a and b both have to be smaller. and then smaller, and
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then smaller, and smaller, and eventually, oh, here we go, it's coming in. So now
16:21
one end's close, but the other end's pretty bad. So it realizes it can keep
16:26
A still and tweak B to bring the other one down. Still a lot of
16:29
error, and it just keeps doing, and it does it in smaller and smaller steps
16:33
until the ends get closer and closer to where they should be. And then I've
16:36
had like a cutoff. I say, keep going until the error is smaller than 10
16:41
to the negative seven. And so eventually it settles in on these values and then
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it locks in that plot. So now I can show you, actually if we go
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right back to the beginning, I can start it at zero and then just set
16:53
it running. So this is where we're gradually drawing the hoops apart. The blue line's
16:57
a little bit jumpy just because of the way that I've plotted it. It's trying
17:01
to stay as close as it can, um, to that point but that's not the
17:04
maths that's just the way I've done this visualization you can see here as the
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hoops are getting further and further apart it's starting to hang down I mean sagging
17:13
like a chain but of course this is the soap film on the side until
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suddenly oh what happened there okay this really threw me when I was trying to
17:21
work that originally because if I back that up again to here it's great and
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you can see here it really looks like the soap film moving quite organically but
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then around about here watch that error The error is tiny, 10 to the negative
17:35
8 kind of level. And then suddenly here, I mean the error is so small
17:39
here, registers as zero. And then, boom, it's off. Look at those ends, they're flying
17:44
away. And that really confused me. And this is not because my technique of finding
17:50
the value of A and B has broken. That's just the best A and B
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possible. And the error is huge. So over here, great. And then suddenly, where does
18:00
it break? It's like... here somewhere. So it's tiny, tiny, tiny, massive.
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So around there, about points, just over 0.7 meters, suddenly my mass falls
18:12
apart. That bit where the curve separates off from the fixed points, which by
18:18
definition should be on the curve, really confused me. And when I was debugging my
18:22
code, I'd accidentally picked values, which were one of those points. And then suddenly it
18:27
dawned on me. I'm like, I'm just getting crazy errors here but if I
18:33
change the separation I get much much smaller errors. So I
18:39
quickly plotted in Excel the magnitude of the error. So this is a log plot
18:45
and so as you go up you're going up in orders of magnitude and for
18:48
these ratios and so that's like the diameter of the hoop as a ratio of
18:53
how far apart the hoops are, down here, fine, we're like at 10 to the
18:58
negative 8, 10 to the negative 7, tiny, tiny errors. And that's just because I
19:03
only worked out A and B to a certain level of precision. And then suddenly
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here, bam, we jump up multiple orders of magnitude. So we were suddenly
19:13
way off and it happened so abruptly and I accidentally started working on this side
19:19
of this discontinuity and over here is where it actually works and that point is
19:24
a very significant value. There are actually two key points as you lift up our
19:30
two hoops. The first of them occurs at roughly 52.77%
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of the way up. It's called the Goldschmidt ratio and that is
19:42
where, here we go, this ceases to become the globally
19:48
minimum surface area. So for this set up, let's be honest, it's about a metre
19:53
across on the ring, so it's gonna be about 53 centimetres. It's about there. So
19:59
here, this is where this arrangement of the bubble film ceases to be
20:05
the global minimum, because this is where it exactly equals just the surface area
20:11
the two disks. So technically you could fulfill the same boundary conditions by
20:17
just having a flat disk of film at the top and a flat disk at
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the bottom. And at this point here if I go any further there this
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surface area is now bigger. But this is still a local minimum. So it's
20:32
stable. It'll stay here even though technically if it collapse into two discs that will
20:37
be better. The next point occurs at 66.274%
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of the distance up compared to the distance across here and that's where it
20:49
ceases to be stable and it will collapse into the other solution. So what I'm
20:54
going to do is I'm going to put my measuring tape here and I'm very
20:58
very slowly I'm gonna inch my way up and I'm gonna see at what height
21:03
this ceases to be stable, which is about 70 centimeters. Oh, so we're now at
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66 centimeters. Here it goes. No,
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it's still stable. I'm trying to not breathe on it. I'm worried it's wobbling a
21:21
lot, so we may accidentally perturb ourself into the unstable region.
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So far so good, a little bit higher. Okay, it should be unstable about
21:33
now. Here it goes.
21:39
You can see it's contracting in. And it's gonna flip to the
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other solution. Here we go. And now we're two discs! And it happened
21:51
right on, let me get my exact distance, 71 centimeters.
21:58
That is so close to the point where my error exploded when I was
22:04
doing the numerical solutions. That's amazing. So what I love about this is I worked
22:10
out theoretically how to model this as an equation and then just by playing with
22:16
it I didn't know what was happening but my numerical solution broke. So me doing
22:21
the maths fell apart and it fell apart right at a ratio of 66% which
22:28
in reality is exactly when the bubble falls apart. So because the maths
22:35
stopped working is exactly the same place where reality stopped working. So good.
22:41
Thanks for watching the video. If you're ever anywhere near Leeds in the UK, you
22:46
must come and see Maths City. It's a huge amount of fun. It's in the
22:49
Trinity shopping arcade thing up on the first floor, just past the food court.
22:55
They're gonna be open for at least a year. So if you're watching this and
22:58
it's before October 2022, they're definitely still open and maybe after that. Google it.
23:06
you'll work it out. And I'm actually in Leeds, partly for this and partly tomorrow
23:10
night, a friend of mine, Seb Lee Delisle, is doing a laser installation. And so
23:14
I'm making a video about the maths of that. It's gonna be super great. That
23:18
video is either already out or will be out soon. I don't know. My video
23:24
release schedule is comprehended by no one. So keep an eye out for it. or
23:29
go watch it. And finally, thank you to Nico, who's filmed both this video and
23:34
the one I'm doing with Seb tomorrow night, which may already exist. And they've got
23:37
a channel called Live to Ride, all about mountain biking. Now, I tend to bike
23:41
somewhere between road and gravel, but I will accept that mountains are also an acceptable
23:45
place for bikes. Link in the description.
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