[Images derived from Times_Square_Ball_2010.jpg by Susan Serra, CKD from Long Island, USA - 222 (2), CC BY-SA 2.0, https://commons.wikimedia.org/w/index.php?curid=23313305]
It is described as having "a diameter of 12 feet and weighing 11,575 pounds" and is "covered with 2,688 Waterford Crystals", also called "triangular crystal panels". It has a total of "32,256 Philips Luxeon Rebel energy-saving LEDs", and from this I figured that there are 12 LEDs per triangular crystal panel.
There are 672 "LED units", each having 48 LED lights, 12 each of "red, blue, green and white colors". 672 times 48 is 32256, confirming the total LEDs. We can also conclude that each "LED unit" has 4 triangular crystal panels, because there are 12 LEDs per crystal panel.
In the photo of the Times Square Ball, we can see triangles framed by the black aluminum struts; we will call these 'frame' triangles. Each frame triangle is divided into four 'color' triangles which apparently have independently selectable colors. And if you look carefully, each color triangle is divided into four 'tiny' triangles. So which of these triangle sizes is a "crystal panel"?
Geodesic spheres are nearly always based on the geometry of an icosahedron, generally by dividing its triangular faces into smaller triangles. There are 20 triangles in a icosahedron; 12 vertices, and 30 edges, and 5 edges meet at each vertex. A typical geodesic sphere is full of triangles, and in most places, a cluster of six triangles around a vertex form a hexagon. But in rare places, a cluster of five triangles around a vertex will form a pentagon.
Finding these pentagons is the trick to discovering the icosahedron from which the geodesic sphere was formed, because the centers of these pentagons are the vertices of the icosahedron. You can impress your friends by 'counting' the number of triangles in the complete 'sphere' that you see (even though it is not actually complete or not all visible) in two or three seconds, as follows: Find two pentagons and along the line that connects their centers, count the number of edges of the small triangles. Square this number and multiply by 20. That's it. If you counted 10 edges, the answer is 10x10x20=2000. In a few seconds, you have 'counted' 2000 triangles!
In the photo at right, we have located the centers of three pentagons, and thus one triangle of the icosahedron. Three struts make one side of this triangle of the icosahedron, making 3x3=9 frame triangles in one triangle of the icosahedron, or 9x20=180 frame triangles in the complete geodesic sphere, not enough for "672 LED units". But there are four color triangles in each frame triangle, so there are 4x180=720 color triangles in the complete geodesic sphere, 48 more than the "672 LED units". But there must be two holes in the sphere to allow it to slide down the supporting pole, so we can guess that each hole must account for 24 color triangles, or 24 "LED units", or 24/4=6 frame triangles. A hexagon of 6 frame triangles is marked on the next photo below, which looks plausibly the right size for a hole. A study of the symmetries of the icosahedron confirms that such a hexagon is matched by one on the opposite side of the sphere.
So we conclude that each color triangle corresponds to an LED unit, which has 48 LED lights. And each tiny triangle must correspond to a Waterford crystal panel, which has 12 LED lights, 3 LEDs of each color, which can be neatly arranged in a triangle. I have seen a number of products using LEDs which embed LEDs in a plastic sheet to distribute the light over the area; so I guess that LEDs are similarly embedded in the Waterford Crystals.
We have been ignoring one small but interesting detail. When a flat triangle of an icosahedron is divided into smaller triangles (144 Waterford Crystals in this case) by dividing the edge length by an integer value (12 in this case), the big triangle remains flat. What is actually done is that the smaller triangles are made a tiny bit larger; making the big triangle bulge out just enough so that all the triangle corners (vertices) lie perfectly on a spherical surface; which is how geodesic spheres look almost like real spheres. The calculation of this adjustment involves trigonometry; need I say more?
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