On Wed, Dec 26, 2012 at 10:27 AM, Robert Hansen <firstname.lastname@example.org> wrote:
> Let's look at it this way. Consider a pizza and you are going to slice this pizza into 1, 2, 3, 4, ... slices. For each n, where n = 1, 2, 3, ... the interior angle of each slice will be 360/n, correct? And if for each n we sum up the interior angles we will get 360, correct?. >
In the algorithm I'm talking about, we triangulate the sphere such that each vertex has either 6 or 5 spokes coming out from each vertex, making a wheel-like pattern.
The perimeter is either a hexagon or a pentagon. If we bashed out the spokes, we would have like a soccer ball but with a vast number of hexagons, yet still only 12 pentagons.
This is de-generalizing just a little, in that "all hexagons and pentagons" (were the spokes removed) only occurs in multiples of 3, counting the number of intervals (edges) from one pentagon center to the next (this is called the "frequency" of the ball / sphere).
That's OK (to de-generalize a little) if it helps with the visualizations.
All very apropos of New Year's by the way, as the Times Square ball is one of these.
The higher the frequency, the more triangles, but it's still all six slice and five slice pizzas, just more of them.
> As n -> infinity, the interior angles approach 0. The limit of 360/n as n approaches infinity is 0. Yet, the sum of the interior angles is always 360, or stated another way, n * 360/n = 360, a constant. >
The apex of each hexagon makes the pizza slightly convex / concave i.e. the center point is a not in plane with the points around the rim (the rim doesn't have to be all in a plane either). Every vertex is a local apex i.e. is surround by a little less than 360 degrees.
The number of degrees around each vertex is never quite 360, because we're on a ball. But the triangles are so small, relatively, that it's as flat as a bathroom floor (triangular tiles) except mathematically (and with careful measurement) we see that it isn't. Picture a ball the size of the Earth with triangular bathroom floor tiles smoothly covering it -- they look almost perfectly equilateral except around pentagons (which may be hard to find are there are only 12 of them).
Every vertex contributes some small positive finite amount to the grand total of 720. 720 degrees apportioned to all these vertexes, really would create perfect flatness around each of them. The skin would no any longer have convexity and this would not be a polyhedron.
> But you are missing a step when you stated your contradiction. > > You are asking why does limit(n) * limit(360/n) as n->infinity not equal the limit(n * 360/n) as n->infinity. >
360 is divided by either 6 or 5, but in such a way that | 360 - d | where d is the total degrees around v (vertex) > 0.
I can make | 360 - d | arbitrarily close to 0 as I increase the frequency of the triangular mesh (more pizzas, but not more slices around each vertex -- that's 6 or 5).
Give me an epsilon, like .000000000000001 degrees, and I can give you a frequency that is less than epsilon, i.e. | 360 - d | < e.
> This is because the limit(n) as n->infinity does not exist, and thus, you can't multiply the limits. >
As the number of vertexes increases, | 360 - d | decreases as close to 0 as we like. But it can never be 0, because all the vertexes together contribute a grand total of 720 degrees (unchanging constant).
This is what's known as Descartes Deficit, i.e. if you add up the number of degrees round each corner of a cube (90 + 90 + 90), the local deficit (the difference from 360) is 90 and 90 times the number of vertexes is 720. This will be true of your octahedron, icosahedron, dodecahedron etc. etc., and true of your super high frequency geodesic sphere.
> You did show that limit(360/n) exists and is zero. > You did show that limit(n * 360/n) exists and is 360. > > But your final statement actually involves limit(n), which you never showed existed, and indeed, doesn't exist. > > Bob Hansen >
n = number of vertexes on the sphere (note: if p = n - 2, then the ratio of p:faces:edges = 1:2:3 if omni-triangulated) 720 = constant angular deficit (all local deficits added) -- per Descartes d = number of degrees around each vertex (< 360 in all cases) but arbitrarily close |360 - d| = local angular deficit (a given vertex)
Sigma [local angular deficit] == 720 all v
As the limit n increases without limit, local angular deficit -> 0, but sum total angular deficit remains fixed at 720.
How could the sum of all the |360 - d| amounts be 720 (a constant) and yet each have 0 as a limit?
I guess you could say 720 * n * (1/n) --> 720 for any n (which isn't quite the same thing), but it's interesting nonetheless how you can argue a limit of 0 at each vertex (perfect flatness at the limit), and yet know there's a fraction of 720 that's always keeping us from true 0 (720/n approaches 0 too, as n -> infinity, I guess one would say).
Not a show stopper, but in the process we learn something about subdividing a sphere, and maybe about calculus.