Course:Harris, Fall 08: Diary Week 11

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Mon:

  • Escher Art Project:
    • Art project due Friday, this week.
    • Written report due Monday, next week
      • I pointed out some additional details for the written report, specified on the schedule for this week.
      • Written report should normally be a few pages long.
  • There are a couple general types of geometry:
    • geometry of a surface
      • Geodesic--the counterpart for a line--means a curve which is the shortest path between any two of its nearby points.
      • Measurements of lengths and angles are fairly obvious.
      • Examples:
        • plane (Euclidean geometry)
        • sphere (sometimes called elliptical geometry)
        • any surface you can imagine in 3-space
    • abstractly defined geometry
      • Measurement must be explained for the particular model.
        • Once lengths are defined, a geodesics is, again, a curve which is the shortest path between any two of its nearby points.
        • Sometimes geodesics are (for convenience) defined separately from distance measurements, but the result should be the same.
      • Example:
        • Poincare disk (hyperbolic geometry)
          • geodesics: circular arcs intersecting the boundary of the disk at right angles (or diameters of the disk)
          • length measurement: hyperbolic units get Euclidean-smaller as one approaches the boundary
          • angle measurements: same as Euclidean measurement
  • Groups took about 25 minutes to do the first Hyperbolic Geometry exploration.
  • Left over from spherical geometry: regular spherical tessellations
    • This means tessellating a sphere with
      • regular polyogons
      • all congruent to one another
      • with all vertices alike.
    • Can one do this on a sphere? Do regular tessellations exist?


Wed:

  • We spent the whole period on regular tessellations of the sphere.
    • Recall regular tessellations in general:
      • The tessellation is by regular polygons:
        • All sides have the same length.
        • All angles are congruent.
      • All the polygons are congruent to one another.
      • All vertices look the same.
        • (Maybe that follows automatically? Not sure.)
    • On the plane, there are 3 regular tessellations:
      • by triangles (6 at a vertex)
      • by squares (4 at a vertex)
      • by hexagons (3 at a vertex)
    • On the sphere:
      • Totally unrelated to plane tessellations, we can tessellate a sphere by biangles (2-gons).
        • We can have any number <math>k</math> of biangles (all the same size), all fitted around the same pair of antipodal points as vertices:
          • Then, since <math>k</math> fit around a vertex, each biangle has an angle of <math>A = 360/k</math> degrees.
          • When <math>k = {2}</math>, the biangles are also 1-gons (same as 180-degree 2-gon).
          • When <math>k = 1</math>, it's sort of a degenerate situation, with the two sides coinciding and producing a 360-degree 2-gon.
      • Tessellation by triangles:
        • Again, let <math>k</math> = # of triangles at a vertex.
        • Let <math>A</math> = the angle in each triangle.
          • Since <math>k</math> angles fit around a vertex, the again have <math>A = 360/k</math> degrees, or <math>A = \frac{2\pi}{k}</math> in radians.
        • Let <math>{\Sigma}</math> = angle-sum of each triangle.
          • Then <math>\Sigma = 3A = \frac{6\pi}{k}</math> radians.
        • Let <math>{\delta}</math> = defect.
          • Then <math>{\delta} = {\Sigma} - \pi = \frac{6\pi}{k} - \pi = (\frac{6}{k} - 1)\pi</math> radians.
        • Let <math>n</math> = total # of triangles in the tessellation. How can we find <math>n</math>?
          • We know <math>\delta = \frac{Area(triangle)}{R^2}</math> (so long as we measure angles in radians)
            • or <math>Area(triangle)</math> = <math>{\delta}R^2</math>
          • We also known <math>Area(sphere) = 4\pi{R^2}</math>.
          • We also know <math>n</math> x <math>Area(triangle)</math> = <math>Area(sphere)</math> (because there are <math>n</math> triangles covering the sphere).
          • Putting those together: <math>n</math> = <math>Area(sphere)/Area(triangle)</math> = <math>\frac{4{\pi}R^2}{\delta R^2}</math> = <math>\frac{4\pi}{\delta}</math>
        • Then putting together formulas for <math>{\delta}</math> and <math>n</math>, we get
          • <math>n = \frac{4}{\frac{6}{k} - 1}</math>
        • For 2 triangles at a vertex (i.e., <math>k = {2}</math>), this gives a total of <math>n = \frac{4}{3-1} = {2}</math> triangles.
          • This is really just a tessellation by 2 biangles.
        • For 3 triangles at a vertex (i.e., <math>k = {3}</math>), this gives a total of <math>n = \frac{4}{2-1} = 4</math> triangles.
        • For 4 triangles at a vertex (i.e., <math>k = 4</math>), this gives a total of <math>n = \frac{4}{\frac{3}{2}-1} = 8</math> triangles.
        • For 5 triangles at a vertex (i.e., <math>k = {5}</math>), this gives a total of <math>n = \frac{4}{\frac{6}{5}-1} = 20</math> triangles.
        • For 6 triangles at a vertex (i.e., <math>k = 6</math>), this gives a total of <math>n = \frac{4}{1-1}</math> = infinity triangles.
          • This is really just a reflection of what happens in the plane, where 6 triangles around a vertex works very well.
        • For <math>k = 7</math> or larger, we get a negative number for <math>n</math>, so those are all impossible.
        • Thus, the only possible combinations for triangles tessellating the sphere are
          • 4 triangles, 3 at a vertex
          • 8 triangles, 4 at a vertex
          • 20 triangles, 5 at a vertex
      • What about tessellating by squares?
        • Defect changes in that <math>\delta = \Sigma - 2\pi</math> (since Euclidean angle-sum for quadrilateral is 360 degrees).
        • But we again have <math>\delta = \frac{Area(square)}{R^2}</math> (since both defect and area come from two triangles making up the quadrilateral).
        • What else changes from triangles?
        • What remains the same as with triangles?


Fri:

  • Art Projects were turned in.
    • Reports are due Monday. Things to look for including:
      • the base polygonal tessellation (by parallelograms, for instance) and its symmetry group
      • the means by which the base polygonal tessellation was transformed
        • Was a changed line on one side simply translated to the next side?
        • Was a changed line glide-reflected to the next side?
        • Was a changed line rotated about an endpoint?
        • Was a changed half of a line rotated about its midpoint?
      • How did the artistic vision influence the choice of base polygonal tessellation, method of changing it, symmetries in finished product?
      • How did symmetries in polygonal pattern influence the development of the art?
  • We spent most of the time on looking at spherical regular tessellations by polygons with 4 or more sides. With <math>k</math> = number of polygons at each vertex and <math>n</math> = total number of polygons on the sphere:
    • We did squares on the board.
      • We found <math>n = \frac{4}{\frac{8}{k} - {2}}</math>.
        • only one possible:
          • 6 squares, 3 at a vertex
        • 4 at a vertex is the plane.
        • 5 or more at vertex is impossible.
    • Groups worked on pentagons.
      • We found <math>n = \frac{4}{{\frac{10}{k}} - 3}</math>.
        • Possible:
          • 12 pentagons, 3 at a vertex
        • 4 or more at a vertex is impossible.
    • We looked at hexagons on the board.
      • We found <math>n = \frac{4}{\frac{12}{k} - 4}</math>.
        • No possibles:
          • 3 at a vertex is the plane.
          • 4 or more at a vertex is impossible.
    • Polygons of 7 or more sides will have nothing possible.
  • No exercises for this weekend. We'll go back to hyperbolic geometry next week.