On 1/7/2013 6:05 PM, Henry Baker wrote:
I was thinking today that I need to make more precise my definition of "approximation" to a 3D surface.
In the case of a sphere, I think that 1) the limit of the series should have the same _volume_ as the sphere; and 2) the limit of the series should have the same _surface area_ as the sphere.
Wouldn't you want to require that for any distance d, the construction produce an approximating solid with quadrangular faces such that every point of the surface is within d of a point on a quadrangle and conversely every point on a quadrangle is within d of the surface? Brent Meeker
The construction I presented approaches the same volume, but not the same surface area. In fact, due to all the "crinkles", I think that the surface area is significantly larger than that of the sphere.
Marc LeBrun's construction also approaches the same _volume_ as the sphere, but his surface area is the same as the enclosing cube (I think), which is also significantly larger than the surface area of the sphere.
At 05:36 PM 1/6/2013, Henry Baker wrote:
The vertices aren't the problem: they can all lie on the sphere. The problem is the faces that share one of the early edges become almost vertical (parallel with a radius vector). The surface becomes exceedingly crinkled.
As I concluded in another post: we need to find a way to break the early edges, while preserving the properties that the quadrilaterals are planar and the faces all have 4 edges. While the corners are made up exclusively of only 3 faces coming together, this is not a requirement; perhaps in breaking an edge, we can construct corners with 4 (or more) faces.
At 04:55 PM 1/6/2013, meekerdb wrote:
On 1/6/2013 2:33 PM, Henry Baker wrote:
Note that my "pushing out" algorithm approximates the sphere_from the inside_, in the sense that the 8 vertices of the cube lie on the surface of the sphere, and each new "pushed out" face will also have all of its vertices on the surface of the sphere. Is that true? What about the trapezoidal faces that are pushed out, not just the central square one? I don't think their vertices can all lie on a sphere.
Brent Meeker
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