Course:Harris, Fall 08: Diary Week 11

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.