The art of drawing circles

Higher-dimensional geometry (previously "Polyshapes").

The art of drawing circles

Postby wendy » Sat Aug 24, 2013 10:20 am

This is an approach to the non-euclidean geometries by using euclidean polytopes. The name itself comes from that a circle is the intersection of a euclidean and a non-euclidean hedrix (2-space), and that we transfer lengths around by way of drawing circles. Polytopes are a kind of circle-array, each face presents vertices on both the flat surface they fall, and the sphere that contains the vertices.

In the euclidean geometry, the ruler of zero curvature is the same as the one of minimal curvature (or straight ruler). In non-euclidean geometries, these are different curves. While most geometries use the straight ruler, which leads to a thicket of sines and cosines, the circle-drawing uses the zero-curvature ruler, which does not lead to this thicket, and is more general. In particular, a straight line is a particular instance of a curved equidistant.

For example, the polytope o3x5o occurs as both the mid-section of x3o3o5o, and as a small facd on x3o3x5o. If one were to deal with the various chords of this polyhedron in terms of the 'flat ruler', then the ratios of edges change from size to size. When one uses the zero-curvature ruler, the chords maintain their same ratios, which is why people deal with {5,3,3} as a series of polytopes, rather than as spherical tilings.

The constant of curvature corresponds to 1/r^2, which may be positive or negative. Minimal curvature is a straight line, is represented as a curve the same radius as the containing space. For example, a great circle has the same radius as the sphere it falls on, and is thus a 'straight line' in S2. The maximum curvature has 1/r^2 = inf, ie r=0. This is a point: all spaces contain this. A space may not contain a curve less curved than itself, ie 'hyper-straight'.

With hyperbolic geometry, we suppose that 1/r^2 is a negative, like -36, which makes a notional radius of i/6. A larger curvature means "more in the direction of +\infty, which means that a curvature of -9 is an equidistant in a curvature of -36, but the radius is now 2i/6 = i/3. That is, the equidistant is twice the size. Here is an example using a row of cells of x5o4o, which can be no clearer.

Code: Select all
      A     B     C          A row of cells of x5o4o
      +     +     +          Note that ABC matches abc
     / \   / \   / \         on an equidistant, but the
   +-   -+-   -+-   -+       edges are twice the angle
   |     |     |     |       of abc.  Also, it has
   |     |     |     |       right-angles.
   +-----+-----+-----+
   a     b     c



Here we see three pentagons of x5o4o, the straight line at the base abc is run equidistant by a right-angle construction of ABC, the edge AB is exactly that of two edges of the base (ie ac), although it projects by perbpendiculars frm abc onto a.5 to b.5. The equidistant ABC has a 'larger curvature' which means that the curvature is less -infty than the base, but this means -1/R^2 is larger than -1/r^2.

Just as equidistants of the sphere are reduced by cos L, the equidistants of the hyperbolic are increased by cosh L.

What we shall show in later episodes is thae right-angle rule is c²=a²+b² - a²b²/2r², where a, b, c are the chords of the edges of the rightangle triangle, the angle is opposite c, and r is the radius of the sphere containing the right-angle. When r is infinite, then c²=a²+b² gives the euclidean rule. When a²=2r² (that is, the quadrant of a sphere), then we see that c²=a², is independent of b. This is the double-right angle. But we allow r² to be negative too, and this formula continues to work in hyperbolic geometry.

The sort of ruler that corresponds to 'zero-curvature' in hyperbolic space, might be written with the fibonacci series.
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Re: The art of drawing circles

Postby Klitzing » Sun Aug 25, 2013 9:23 am

Hmm, so far did not find a consistent reasoning to use K = 1/r² as curvature throughout for ANY dimension, cf. my previous mail in the other thread.

But your rewriting of the Pythagorean equation (aka Cosinus thoerem in spherical spaces), i.e. c² = a² + b² - 2ab [(a/2r)(b/2r)] = a² + b² - a²b²/2r², perhaps might serve for some reasoning to use K as provided above in general with respect to polytopal contexts...

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Re: The art of drawing circles

Postby wendy » Sun Aug 25, 2013 9:53 am

We now turn to an isoceles triangle, of edges E:E:A, and try to derive the radius. This is doubled, to give give a right-angle coordinate system.

Code: Select all
            o                        Capitals are squares of lower case
           /| --    d  = 2r          1   4E = A+B
        2e/ a    ---                 2   D = D + B - 2rd + A
         /  |        ---                  2d  = 4E
        o-b-+--------------o--            bd  = 2E
         \  |     d-b                     BD  = 4EE
          \ |                          (4E-A)R = EE
           \|                             E/R  = 4 - A/E
            o                             4 = E/R + A/E



With polyhedra, it is normally usual to give R/E, that is, the radius, given an edge. With bollohedra, one gives usually E/R. In either case, this equation defines the curvature (1/R) of the containing space.

For polygons, the formula takes the form D = 4/(4-A), where the edge is 1, and A is the 'short-chord square of the polygon'. D is the square of the diameter.

The general polyhedron {p,q} have shortchords P,Q, and the value of A/E is then the diameter of a polygon pQo. This leads directly to A = 4P/(4-Q), and hence the diameter of {p,q} gives D = (4-Q)/(4-P-Q).

For a polychoron, one sees this gives, by the same process, D(p,q,r) = 2(8-2P-2Q)/(16-4P-4Q-4R+PR). In general, the process is iterative, based on

D (p,q,r,...) = 2 F(q,r,s...) / (p,q,r,s...).

Where F(p,q,r,s...) = 2 F(q,r,s...) - P. F(r,s,...) The function is iterative, and derives exactly to the same value for a given symmetry. In short, it's a kind of Schläfli symbol function. That is, eg F(p,q,r,...) = F(...,q,r,p).

With a bit of calculations, one can show it does not just apply at the ends, but for all single-node polytopes, with F(p, 2, r) = F(p)F(r). So for

F(3,5) = 4-2f. F(5) = 3-f, F(3) = 3. F() = 2

We get then, the dimaeter2 of x3o5o = 2 (3-f) / (4-2f) = f+2. For x3o5o, we have 2. (2) (2) / (4-2f) = 4 (f+1) and for o3o5x, we have 2 (3) / (4-2f) = 3(f+1).

The closest we come to exactly using 1/r² for curvature is the measure "b" above. If you think of the circle as being "buckled" or bent away from the straight line, the maximum it comes to is b = e²/r, and because we're using squares, B=E²/R. So the curvature of the curve becomes a measure of the buckling it can support in E². But we don't directly measure the buckling, but the measure of the difference between the diameter-square less 4*radius-square of the vertex-figure. This gives the buckling (which still has an angle component in it), and R = E²/B.
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Re: The art of drawing circles

Postby wendy » Mon Aug 26, 2013 9:59 am

The notation for infinite polygons, is to use the shortchord square, after the letter 'W'. The polygon that girths a figure has a shortchord equal to the diameter of the vertex figure, which then becomes the regular polygon, multiplied by the shortchord of the edge-becomes. So the {5,4} has a vertex-figure of f4o, the diameter2 of which is D = 2.618034 * 2 = 5.236067. So W5.236067 is both the straight-polygon formed by the edges of {5,4}, and the right polygon formed by pairs of edges. Since the girthing-polygon is as good a measure as curvature (it gives A/E = E/R+4), it is a useful thing to find.

Note that the W notation can be used for polygons less than W4. For example, Coxeter's "regular polytopes" gives a polygon which makes {3,K,3} into a euclidean tiling (it's given in the tables at the back, for example). We find K = W2.25.

The 'D' notation can be used to space out the vertices, in the same way that {5/2} is P5D2. We shall find formulae for writing things like W5.236067D2 etc.
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