So we're going to start by drawing the 8CP structure. I like to draw two squished hexagons and attach them with lines. I'll show you what I mean. So this is my squished hexagon. And this is my other squished. And these lines just connect. It's not always perfect. All right, there you go. Now I'm going to draw in the atoms. I know that Professor Ramsey likes to draw in a quarter or a third, like drawing the slice of it. I'm just going to draw in the full atom. It's kind of complicated on the board. So we know that every atom goes in each corner. OK. And one atom fits in the center of each hexagon. Three atoms are in between these two planes. But I'm going to draw it so that one is facing towards me and two is in the back. So two in the back, one in the front. I did this for a very important reason. And that's because it forms a tetrahedral site right here. I'm going to draw this tetrahedral site bigger because that's what's going to help us find the 8C ratio. So a simple tetrahedral site has is composed of many triangles. So the height here, it goes down the center right in the middle of this bottom plane. That is what I call the height. And this little buddy here, that's our x. And every single side on this tetrahedral site is 8. It's an equilateral triangle. What we want to find is this 8C ratio. But what we know is that we have an h and an x, both unknown, except we want to relate these things. So how I would relate this, because this is a right angle triangle, I would say that 8 squared plus x squared is equal to a squared. And that would just be the interior triangle of this tetrahedral site. Now what we don't have is h and we don't have x. And I'm going to show you how you find that. H is actually really simple. If you look at the 8CP structure, this little line here would be the h. And if you can see that, it's actually the half, halfway point. So h is equal to c over 2. Good. We got that down. Now we want to find the x, because it's missing from our equation. I'm going to draw this bottom plane of this tetrahedral site bigger over here. So we know that a is on either every single side. And the x is actually right here. To find, oh yeah. And also, this is a 60 degree angle, which means that this is a 30 degree angle. So to find what this is, I'm going to extend the line halfway through the a towards the center, making this a right triangle. And if you know your sine cosine rules, you can see that cos 30 is actually, so katoa a over 2, which is this little section, all over x, adjacent over hypotenuse. And if you know what cos 30 is, which you all should, it's actually root 3 over 2. If you just solve for x, so this is root 3 over 2, it's equal to a over 2x. That's an a. So solve for x and you get a over root 3. So now we have x, lovely. So this initial equation is what we're going to continue over here. k squared c over 2 squared plus x squared, so a over 3 squared is a squared. Now we only have a's in season our equation. We can solve for the c to a ratio, kind of easy. You got to carry it through. It's kind of mathy. So c over 4 a squared over 3 is a squared. Take this to the other side. c squared over 4 is equal to, let's say, 3 over 3, 3 minus 1, 2 over 3 a squared. All right. So I'm going to solve for c over here. And then I will say that c is equal to, so 2 root 2.