by **wendy** » Tue Jan 04, 2011 8:19 am

"Margin angle" is the angle around a margin (the thing between 'facets'). The notation here is that the margin is the second-order boundary (dividing facets), and that the margin-angle is the fraction of space occupied by the polytope at the margin: ie dihedral angle in 3d, dichoral angle in 4d etc.

The thing about polygons is like this:

1. Start with your polytope, and draw a dot at the centre, and the dot at the centres of the faces (n-1 things). Connecting face-centres up, is the way of getting the dual of the polytope (ie for 5,3,3, connecting face-centres gives 3,3,5). An face-centre face-centre polytope-centre will give a triangle, that forms one slice of a pie, whose vertex-angle is the sought-after angle.

2. One can then determine either the length of arc or something else of this triangle, but it involves finding the radius of the dual. You can do this by using a Schläfli sequence, which depends entirely on the string of digits in the symbol.

3. The easiest way to calculate the Diameter2 of a given polytope, is to use shortchord2. This is a number corresponding to sch²(pi/n), where sch = supplement chord (ie chord (180-angle), and pi = 180 deg, and n is the number of sides. This is then, eg 4 cos²(pi/p) or 2+2.cos(pi/p) does the trick.

eg the shortchord² of 5/2, 3, 4, 5, 6 become 0.381, 1, 2, 2.618, 3.

4. The diameter2 D of a polygon is derived from its shortchord A, by D = 4/(4-A). The regular polyhedron have vertexes that connect to a vertex by way of a circle that has a diameter that can be read as a shortchord of a polygon. That is, for p,q, we get D2(q) * A2(p) = A2(p,q), from which D2(p,q) comes,

A2(p,q) = A2(p) * D2(q), or 4P/(4-Q)

Since 4/D = 4-A, or D = 4/(4-A)

we get D(p,q) = 4 / 4 - (4P/4-Q)

= 4(4-Q) / 16-4P-4Q or (4-Q)/(4-P-Q)

eg cube, p=4, q=3, P=2, Q=1, we get (4-1)/(4-2-1) = 3

For (P,Q,R), one feeds these values again, to get

A2(p,q,r) = P(4-R)/(4-Q-R), and D2(p,q,r) = ( 16-4Q-4R )/ (16-4P-4Q-4R+PR).

putting eg p=5, q=3, r=3 gives P=2.618, Q=R=1, (16-4 - 4)/16-4*2.618-4-4+2.618) or 54.832815

Of course, one can do this by way of iteration, eg

d2() = 2(1) / (2)

d2(R) = 2(2)/(4-R) : 4-R = 2(2) - R(1)

d2(Q,R) = 2(4-R) / (8-2Q-2R) : (8-2Q-2R) = 2(4-R) - Q(2)

d2(P,Q,R) = 2(8-2Q-2R) / (16-4P-4Q-4R+PR) ; 16-4P-4Q-4R+PR = 2(8-2Q-2R) - P(4-R)

etc

The trick here is to write the schlafli symbol in reverse, eg

1 (always)

2 (always)

3 3 = 2*2 - A(3)*1 A3 = 1 Also A4 = 2

3 4 = 2*3 - A(3)*2 A3 = 1 A(5/2) = 0.381 = 2-1.61803398875

5 0.1458 = 2*4 - A(5)*3 A5 = 2.61803398875

D2 of {5,3,3} is then 2*4/0.1458 = 54.831&c.

Given this value, one then draws a triangle, the sides-square are 1 (the edge), D2 (the hypot), and (D2-1), or in normal form

1/sqrt(D2) gives the cosine of the half-angle. ie 2.acos(1/sqrt(54.831)) = 164.477, which is the dichoral angle of {3,3,5}.

W