Dichoral Angle (Dihedral in polychora)

Higher-dimensional geometry (previously "Polyshapes").

Dichoral Angle (Dihedral in polychora)

Postby darthbeppo » Mon Jan 03, 2011 6:40 am

Maybe this question will be an insult to some of you, so pardon my stupidity. I'm a noob with this stuff :P
I've been trying to find a formula to calculate dichoral angles in regular polychora. My ultimate goal is simply to calculate the hypervolume (bulk?) of a regular polychoron. I've been obsessing over this problem for a month now. I'm only 17 and the highest math I've gone through is AP Statistics and Trigonometry, so go easy on me here. A formula similar to the dihedral angle calculation on wikipedia-- one that uses the schlaefli symbols-- would be ideal (http://en.wikipedia.org/wiki/Platonic_solid#Angles). I've been able to derive my own formula for dihedral angles, but dichoral angles have me stumped. I've tried searching the internet, but I can't find anything. All I know is that the dichoral angle for a pentachoron is arccos(1/4) and the dichoral angle for a tesseract is, by definition, pi/2. Seriously, any help would be greatly appreciated.
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Re: Dichoral Angle (Dihedral in polychora)

Postby wendy » Mon Jan 03, 2011 8:01 am

For any regular polytope (in any dimension), the margin angle [eg dihedron in 3d, dichoron in 4d, diteron in 5d], is the same as the corner angle of a regular polygon whose curcumradius is identical to the diameter of the dual of the polytope.

Once given this, one sees that the diameter2 of a polygon of shortchord2 A is D = 4 /(4-A), or 4-A = 4/D, or A = 4/(D-4)

The shortchord is the chord which forms a triangle with two consecutive edges of the polygon. What you are getting here is

2.cos(theat/2) = a, or theta = 2 acos(a/2), a*a = A

Let's work an example. For {3,3,5} we have the dual is {5,3,3}

the radius of 5,3,3 is f^4 is phi^2 * sqrt(2) = 3.702459.

D = 4R^2 = 8.phi^4 = 54.832 815 719

A = 4/(D-4) gives 0.078689235, a = 0.280516177 -> 163.8744 degrees

It works for all dimensions.
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Re: Dichoral Angle (Dihedral in polychora)

Postby darthbeppo » Mon Jan 03, 2011 12:34 pm

I'm a little confused now, but I'm going to try and draw it out. I'm not familiar with the term 'Margin Angle'. The thing about the circumradius of the regular polygon identical to the diameter of the dual polytope also confused me. The circumradius of any polygon can be identical to the diameter of the dual polytope depending on how big the polygon is. (right?) Also, how do you know the radius of the dual? I actually intend to use the dichoral angle calculation as a step in calculating the inradius of the polychoron. From there, the bulk = 1/4 inradius * surcell volume. I need to be able to calculate that, given the length of one edge, and the type of regular polychoron. No other values are known, and thus need to be calculated. In three dimensional figures where the volume = 1/3 inradius * surface area, I needed to know the dihedral angle to calculate the inradius. Based on that, it is only logical to assume that the dichoral angle is needed to calculate the inradius of a polychoron.
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Re: Dichoral Angle (Dihedral in polychora)

Postby 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}.

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Re: Dichoral Angle (Dihedral in polychora)

Postby darthbeppo » Thu Jan 06, 2011 11:27 pm

I got lost in the undefined variables.
p,q,r appear to be different from P,Q,R
I have no idea what the difference is.
From that, is there a difference between d2 and D2?
All these A's started coming in from somewhere. A, A2, A3, A4, A5
Variable D disappears, and D2 comes in.
Also, the answers you gave for the dichoral angle of {3,3,5} are different between your top and bottom posts. I don't know what's going on there.

FYI: Your site (which I was fortunate enough to stumble upon) is the only site on which I can find the definitions for Margin angle and shortchord
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Re: Dichoral Angle (Dihedral in polychora)

Postby PWrong » Sun Jan 09, 2011 6:20 am

FYI: Your site (which I was fortunate enough to stumble upon) is the only site on which I can find the definitions for Margin angle and shortchord


Wendy invented a lot of these words.
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Re: Dichoral Angle (Dihedral in polychora)

Postby wendy » Sun Jan 09, 2011 8:33 am

The upper-case letters like A,B,C,.. refer to shortchord squares. The lower case letters refer to the numbers like 3,4,5.

A is also a function, which A(x) = X. It is variously first and square. Anytime it ends in 2, it is always the square. This comes from a scheme where the chords of a unit edge polygon are of length 1, A, B, C, ... What we're using here is always A2(), the square: ie A2(5) = 2.61803398875, A1(5)=1.618033.

d2() is the diameter-square. This when multiplied by edge-square, becomes the size of the vertex figure, we read something like (p,q,r) as a series of vertex-figures, being a vf2 of edge P(qr) = P(Q(r) = P(Q(R(line(point))).

The series starts with point (1), and line (2), we then find the d2(r).

we have then p,q,r as the numerics like 3,4,5, and P,Q,R are the matching shortchord-squares 1, 2, 2.618. The functions are evaluated in shortchord squares. This avoids the use of trig functions completely.

Since the same triangle that gives the dichoral angle is also the edge-angle of the dual, we calculate the D2 of the dual, the dihedral angle becomes then the acos of sqrt(1/D2).

for 3,3,5, we find the d2 of 5,3,3 as

1
2
1.382 2*2 - 1*1.618
0.764 2*1.382 - 1*2
0.146 2*0.762-1.382

The required d2 is 2*0.764/0.146.
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Re: Dichoral Angle (Dihedral in polychora)

Postby quickfur » Tue Jan 25, 2011 3:48 am

The dichoral angle in the 120-cell is surprisingly simple. Consider my favorite projection:

Image

This shows half of the 120-cell, the half that's facing the 4D viewpoint. Start with the blue dodecahedron in the middle. Select one of the green dodecahedron next to it. Select a second green dodecahedron directly opposite this one. Now you have a segment of a great circle running around the girth of the 120-cell. Along this great circle, you have two red dodecahedra next to either of the green dodecahedra (they appear flattened 'cos they are almost 90° w.r.t. the 4D viewpoint). On the other side of the 120-cell are another 5 dodecahedra in the same arrangement, along the same great circle, connected to the first 5 dodecahedra at the red cells. So there are exactly 10 dodecahedra along this great circle. Now, since cells are flat in 4D, this means that the great circle is exactly a decagon. Therefore, the dichoral angle is exactly 360/10 = 36°, or pi/5 for you radian fans. EDIT CORRECTION: This is wrong, it should be 360 - 360/10 because the dihedral angle is internal, so the real dihedral angle is 4pi/5. Sorry about that!!

Another way to see this is looking at the parallel projection of the cells that lie on the "equator" (i.e., at 90° from the 4D viewpoint):

Image

Notice that the cells form rings that wrap around the limb of the projection. There are 10 dodecahedra in each ring. Therefore their dichoral angle is precisely 360/10 = 36°. EDIT CORRECTION: This is wrong, it should be 360 - 360/10 because the dihedral angle is internal, so the real dihedral angle is 4pi/5. Sorry about that!!
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Re: Dichoral Angle (Dihedral in polychora)

Postby darthbeppo » Tue Feb 14, 2012 9:33 pm

Okay. I figured out the answer. (no offense, but you guys weren't making sense to me. haha)
θ = 2arcsin[(cos(π/r) / sin(π/q)) / (sin(2arcsin(cos(π/q) / sin(π/p)))]
Where θ is the dichoral angle, and p, q, and r are the components of the Schläfli symbol.
I'm gonna try to simplify that equation for clarity.
cos(π/r) / sin(π/q) = A
cos(π/q) / sin(π/p) = B
θ = 2arcsin[A / sin(2arcsin(B))]
I derived this equation from info on this site:
http://davidf.faricy.net/polyhedra/Platonic_Calc.html
Since this is 4D,
ω = (π/2) - (Dihedral angle of cell)
β = ½ (Dihedral angle of cell)
It all makes sense to me now.
... And I didn't make up any words to do it. None of that shortchord, supplementchord stuff. The greatest source of confusion was all the undefined variables and terms. Not everybody knows your site, and more importantly, your definitions are not widely used enough to warrant imposing them on people. I come from a programming background, so I understand the importance of being explicit in instructions. If nothing else, use pictures. If you don't feel like drawing a picture, find one that somebody else has drawn. The bottom line is, if you can't explain your language to anybody, you're going to be the only one who understands it.
Here's an example from PolyGloss v0.05
"Shortchord
The base of the isosceles triangle formed by two consecutive edges of a triangle, usually taken to be the circumdiameter of the vertex figure. As a number, written as L1, and its square is L2."
What?(!)."Two consecutive edges of a triangle" You mean two adjacent sides? "The base of the isosceles triangle" What isosceles triangle? What the heck are you talking about?
Supplementchord? That's not even in there.
"Wendy invented a lot of these words."
Well, ain't that convenient.
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Re: Dichoral Angle (Dihedral in polychora)

Postby darthbeppo » Tue Feb 14, 2012 10:39 pm

Ditopeal Angle: The angle between two intersecting "topes." (ie: dihedral angle in polyhedra or dichoral angle in polychora)
a = ditopeal angle of the (n−1)polytopes making up the n-polytope. (ie: dihedral angle of the cells that make up the polychoron)
b = ditopeal angle of the vertex figure
β = ½(b)
ω = (π/2) − (a)
y = sin(β)
z = cos(ω)
θ = 2arcsin(y / z)
-:- Works for all dimensions
-:- Unknown terms defined
-:- Undefined terms are universally used (you can find the definitions on Wikipedia)
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Re: Dichoral Angle (Dihedral in polychora)

Postby darthbeppo » Thu Feb 16, 2012 4:22 am

Look, I'm sorry I got so frustrated. Well, for one thing, I found the updated polygloss.
"shortchord *
The base of the isoceles triangle formed by two consecutive sides of a polygon. The number is much used in circle-drawing geometry, and also somewhat in other geometries, the values designated as L1 for the value, and L2 for the square.
Certian shortchords are widely used, and have convenient single letter forms. For polygons with an infinite number of sides, the style is to write {wL2}, where L2 is the shortchord-square,"
That makes sense to me. You've put a lot of work into that, and I shouldn't have spat in your face to the degree I did. I wasn't thinking clearly enough to comprehend just how undeveloped this area of math is. Somebody has to invent words sooner or later. I guess I just got overwhelmed. I'm a noob at this stuff. I don't know all your sites, and all the little things that have evolved here. Please pardon my ignorance.
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Re: Dichoral Angle (Dihedral in polychora)

Postby darthbeppo » Thu Mar 01, 2012 8:59 pm

I MADE AN ERROR!
The correct formula is:
θ = 2arcsin[(cos(π/r) / sin(π/p)) / (sin(2arcsin(cos(π/q) / sin(π/p)))]
Broken down for clarity:
cos(π/r) / sin(π/p) = A
cos(π/q) / sin(π/p) = B
θ = 2arcsin[A / sin(2arcsin(B))]
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Re: Dichoral Angle (Dihedral in polychora)

Postby Klitzing » Wed May 29, 2013 3:04 pm

While googling found a nice site providing a similar formula (for the cos of that angle), which also includes its derivation...
cf. http://gregegan.customer.netspace.net.au/SCIENCE/Dihedral/Dihedral.html

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