4d sphonges

Higher-dimensional geometry (previously "Polyshapes").

4d sphonges

Postby thigle » Tue May 16, 2006 12:11 pm

i found (surely not first) that when one takes a parallel projection of 4-cube into 3-space that is vertex first and gets a rhombic dodecahedron (that Kepler found when trying to break the secret of the morphogenesis of snowflakes for czech then-king Karol IV) with 4 diagonals of a cube inscribed in it, corresponding to the 4 axies being projected from 4space.this is the same shape as the one that bees use to waxbuild their bee-homes and the angles between the axies are the maraldi angles that all bubbles hold when not alone.

but, this shape packs the 3 space as the 4/cubes pack the 4-space. so projection of spacefilling 4cubes into 3space gives a lattice that has 4 axies through each point, which is the double of diamond carbon-structure lattice.

to get to the point: one can erase or pick out from this spacefilling mass of rhombic dodecahedrons in such a way that there's left one surface that divides the 3 (and supposedly also 4) space into 2 equivolumetric sphonges, interpenetrating through each other.

my question is this: are there any other sphonge-yielding space-fillings in 4d ?

this is to me connected with question of: what amount of anglemeasure is there around a 4vertex ? around 2vertex it is 360, around 3vertex it is 720 according to wendy, but i don't get it (or is each 360 an I2 and both are simply connected along the infinity/horizon so it adds up ?). anyway how much(and what kind of) angular measure is there around a 4-point ?
thigle
Tetronian
 
Posts: 390
Joined: Fri Jul 29, 2005 5:00 pm

Postby wendy » Wed May 17, 2006 8:54 am

You can make a spounge in any dimension, by considering the middle truncate of a cubic. In four dimensions, this yields a partition of the tiling of {3,4,3,3} = 24chora into two distinct regions.

In all dimensions, it is possible to do partitions of any divisor > 1, of n+1, ie a division of a 5-space into three partitions.

One notes in 6d, there is a similar sponge formed of the 2_22, which divides space into 2, bounded by regular figures only.
The dream you dream alone is only a dream
the dream we dream together is reality.

\ ( \(\LaTeX\ \) \ ) [no spaces] at https://greasyfork.org/en/users/188714-wendy-krieger
User avatar
wendy
Pentonian
 
Posts: 2014
Joined: Tue Jan 18, 2005 12:42 pm
Location: Brisbane, Australia

Postby thigle » Wed May 17, 2006 9:04 am

noone :cry:

i try to reformulate through vords the image of the sphonge again:
take a cubic-packing. give each cube its 4 diagonals. now erase all the cubic wireframe. what is left is a lattice in which every corner of every used-to-be-cube, as well as every center of every used-to-be-cube is crossed by exactly 4 axies, in maraldi's angle with each other.

in such a lattice - which is a double diamond lattice, one can see a packing of tesseracts projected into 3space vertex first, parallely. in wire frame.

now one could take away half of the rhobic dodecahedrons in such a way that 2 voids are left, of the same volume, interpenetrating through each other, filling up the whole extent of 3-space.
this can be accomplished by stacking a "plane" of rhobicdodecas in a way that bees do with their honeycombs, looking vertexFirst, hexagons pack to fill up a plane.
one can also see a hexagonal pattern of rhobicdodeca in such a plane, and taking away rhombicdodecas that are in the centers of these hexagons made of rhobicdodeca, one gets a plane of rhombicdodeca full of holes in triangular pattern and hexagonal pattern of full rombicdodecas. next you put a level of rhombicdodecas that were the centers of the previous layer and put them on that layer but with a displacement in such a way that each of these sits in the middle of 3 rhobicdodecas under it that are where the 3 hexagonal formations from the previous layer meet.
then one just continues to alternate these 2 kinds of planes of rhombicdodeca to fill up 3 space.

one gets a sphonge-like infinite polyhedron if one erases all the rhombs that are faces where 2 rhombicdodecas meet.

btw, is there a name already for this infinite polyhedron ?
thigle
Tetronian
 
Posts: 390
Joined: Fri Jul 29, 2005 5:00 pm

Postby thigle » Wed May 17, 2006 4:12 pm

:shock: i see that while i was typing the previous, wendy already replied.
the danger of slowMotion :D

wendy wrote:
considering the middle truncate of a cubic

in 6d, there is a similar sponge formed of the 2_22, which divides space into 2, bounded by regular figures only.

by cubic you mean cubic lattice ? or something like quadrics but one less degree ?
what is 2_22 notation for ?
thigle
Tetronian
 
Posts: 390
Joined: Fri Jul 29, 2005 5:00 pm

Re: 4d sphonges

Postby pat » Thu May 18, 2006 6:20 am

thigle wrote:this is to me connected with question of: what amount of anglemeasure is there around a 4vertex ? around 2vertex it is 360, around 3vertex it is 720 according to wendy, but i don't get it (or is each 360 an I2 and both are simply connected along the infinity/horizon so it adds up ?). anyway how much(and what kind of) angular measure is there around a 4-point ?


For the 2vertex, consider the unit circle. The circumference of the circle is 2pi. An angle of k-degrees (with vertex at the center of the circle) would encompass pi k / 180 of the surface area of the circle. Arguing in the inverse.... if an angle encompasses some part of the surface of the circle with size x, then we say that the angle is 180 x / pi degrees. For example, if the angle encompasses the whole 2 pi of circumference, then the angle is 360 degrees.

For the analogous situation for a 3vertex, we consider a unit sphere. If the angle encompasses some part of the surface of the sphere with size x, then we say that the angle is 180 x / pi degrees. For example, if the angle encompasses the whole 4 pi (the surface area of a sphere is 4 pi r<sup>2</sup>) of surface area, then it is 720 degrees.

It's easier though to work in radians. You just measure the encompassed surface area on the unit-{circle, sphere, glome, ...} and call that the angle in radians.

Technically, however, the angles aren't at all comparable. The circle-based angles are measured in radians. The sphere-based angles are measured in steradians. You can't directly compare them since the circle's surface area is proportional to the radius while the sphere's surface areas is proportional to the square of the radius. In other words, the circle-based angles enclose a two-dimensional submanifold of R<sup>n</sup> whilst the sphere-based angles enclose a three-dimensional submanifold of R<sup>n</sup>. For the glome, you'll need a whole new type of angle taking into account that the surface area is proportional to the cube of the radius.

Also, for extra grins, the surface area of a glome is 2 pi<sup>2</sup>, so there are 360*pi degrees around a 4vertex. :twisted:
pat
Tetronian
 
Posts: 563
Joined: Tue Dec 02, 2003 5:30 pm
Location: Minneapolis, MN

Postby wendy » Thu May 18, 2006 10:29 am

For all dimensions, my preferred angle system is to use all-space, divided into powers of 120. In practice, this is treated is a twelfty-wise angle.

The notion that degrees are related to the radian is something to do with modern mathematics, but is implemented by placing pi = 180 degrees. For the glome, one then has a surface of 64800 sq degrees.

This works in 3d as well, ie

circle = 2pi deg^0 = 360 deg^1

sphere = 4pi deg^0 = 720 deg^1 = 129600/pi deg^2

glome = 2pi2 deg^0 = 360 pi deg^1 = 64800 deg^2 = 11664000/pi deg^3

One notes that astronomers would use deg^(n-1), since the sphere is regarded as a large thing, and degrees are a kind of mile. Constellations are typically given in sq degrees.

When i first enumerated the solid angles around the vertices of the regular solids, i did not know how many radians make the glome, so i used fractions of all-space.

We have then:

{3,3,3} = pentachoron 0:01.20.V8 = 120.9 f
{3,3,4} = 16-choron 0:05.00.00 = 600 f
{4,3,3} = tesseract 0:07.60.00 = 900 f
{3,4,3} = 24choron 0:15.00.00 = 1800 f
{3,3,5} = fifhundchoron (600ch) 0:33.15.60 = 3975.5 f
{5,3,3} = twelftychoron = 0:38.24.00 = 4584 f

All-space gives 1:00.00.00 = 14400 f

The cubical radian gives 0:06.09.61.60 = 729.512


The name 'degree excess' refers to the spherical excess theorem, which says that the area of a polygon on 2-fabric is proportional to the excess angle that it has. This is true regardless of the fabric as a sphere or H2.


W
The dream you dream alone is only a dream
the dream we dream together is reality.

\ ( \(\LaTeX\ \) \ ) [no spaces] at https://greasyfork.org/en/users/188714-wendy-krieger
User avatar
wendy
Pentonian
 
Posts: 2014
Joined: Tue Jan 18, 2005 12:42 pm
Location: Brisbane, Australia

Re: 4d sphonges

Postby Nick » Thu May 18, 2006 10:58 am

It's 720 degrees for a sphere? I thought it was 129600. A circle has 360 degrees, and you have to rotate the circle 360 degrees three dimensionally to get a sphere; so 360 * 360 = 129600. Can you show me a derivation for 720, Wendy (or Pat)?
I am the Nick formerly known as irockyou.
postcount++;
"All evidence of truth comes only from the senses" - Friedrich Nietzsche

Image
Nick
Tetronian
 
Posts: 841
Joined: Sun Feb 19, 2006 8:47 pm
Location: New Jersey, USA

Postby thigle » Thu May 18, 2006 2:53 pm

i share the question with irockyou. i'll call 4d radians tetradians (fused from from 'tetra-radians'. or should thses be called quadRadians ? sounds better to me.)

pat writes:
For example, if the angle encompasses the whole 4 pi (the surface area of a sphere is 4 pi r^2) of surface area, then it is 720 degrees.
...
...360*pi degrees around a 4vertex.

wendy writes:
sphere = 4pi deg^0 = 720 deg^1 = 129600/pi deg^2

wendy, 129600/pi deg^2 = 720 deg^2. how can that equal 720 deg^1 ? how do you go from deg^0 to deg^2 ? how do you change from deg^n to deg^(n+k) ?
pat, you meant deg^2 and deg^3 respectivelly, perhaps ?

you 2 together wrote:
Technically, the angles aren't at all directly comparable since:

circle's surface area is proportional to the radius, (2p r)
sphere's surface area is proportional to the square of the radius, (4 pi r^2)
glome's hypersurface is proportional to the cube of the radius. (2 pi^2 r^3)

circle-based angles are measured in radians, enclose a 2-dimensional submanifold of Rn.
sphere-based angles are measured in steradians, enclose a 3-dimensional submanifold of Rn.
sphere-based angles are measured in tetradians, enclose a 4-dimensional submanifold of Rn.


and pat wrote:
...work in radians. ...measure the encompassed surface area on the unit-{circle, sphere, glome, ...} and call that the angle in radians.
...if angle encompasses some part of the sphere-surface of the sphere with size x, then angle is 180 x / pi degrees.

to sum up what i still don't grasp operativelly: a general transform between degrees^n and radians^n.

k deg^1 = pi k/180 radians
k deg^2 = ? steradians
k deg^3 = ? quadradians
k deg^n = ? hyperadians
thigle
Tetronian
 
Posts: 390
Joined: Fri Jul 29, 2005 5:00 pm

Re: 4d sphonges

Postby pat » Thu May 18, 2006 6:42 pm

irockyou wrote:It's 720 degrees for a sphere? I thought it was 129600. A circle has 360 degrees, and you have to rotate the circle 360 degrees three dimensionally to get a sphere; so 360 * 360 = 129600. Can you show me a derivation for 720, Wendy (or Pat)?


First, I should mention that you only have to rotate a circle 180 degrees to sweep out a sphere. So, if you're going to multiply the degrees, there would only be half that number.

I thought I did derive the 720?
For the 2vertex, consider the unit circle. The circumference of the circle is 2pi. An angle of k-degrees (with vertex at the center of the circle) would encompass pi k / 180 of the surface area of the circle. Arguing in the inverse.... if an angle encompasses some part of the surface of the circle with size x, then we say that the angle is 180 x / pi degrees. For example, if the angle encompasses the whole 2 pi of circumference, then the angle is 360 degrees.

For the analogous situation for a 3vertex, we consider a unit sphere. If the angle encompasses some part of the surface of the sphere with size x, then we say that the angle is 180 x / pi degrees. For example, if the angle encompasses the whole 4 pi (the surface area of a sphere is 4 pi r2) of surface area, then it is 720 degrees.


This all brings up a different question.... what do you want to do with these degrees? Do you want them to measure the proportion of space enclosed? or do you want them to measure a directed angle? In 2-D, those concepts are interchangeable. In 3-D (and higher), those are distinct concepts.

If you want to measure the proportion of space enclosed, then you should use proportions of all-space or "amount of surface area on the unit sphere encompassed by the angle". If you're going to use proportions of all-space, then I'd recommend starting with the "amount of surface area on the unit sphere encompassed by the angle" and dividing by the surface area of the unit sphere.

If you want to measure directed angles, then an ordered pair (a,b) of angles is fine.... with the caveat that (a,b) = (-a,b+180).... or just restrict b to a 180-degree range. So, there are 360 * 180 = 64800 of them. But, you can't really specify an ordering of them in any way. It doesn't make sense to ask whether (a,b) is bigger than (c,d).... because (a,b) (and, of course (c,d)) specifies both magnitude and direction. You could ask whether |(a,b)| is bigger than |(c,d)|. But, you won't be able to totally order the angles in this way.... since you'll have lots of cases where |(a,b)| = |(c,d)| while (a,b) != (c,d).
Last edited by pat on Fri May 19, 2006 5:59 pm, edited 1 time in total.
pat
Tetronian
 
Posts: 563
Joined: Tue Dec 02, 2003 5:30 pm
Location: Minneapolis, MN

Postby thigle » Fri May 19, 2006 9:33 am

I thought I did derive the 720?

yep. i didnt get the second part of it. now that you mentioned that one needs just half pi of rotation for a circle to generate a sphere, its clear.
This all brings up a different question.... what do you want to do with these degrees?

i wanted to have knowledge of how these higher dimensional angles behave because of trying to bring some order into my understanding of what happens around a vertex in nspace. like with the sphonge that started this thread: i wanted to know how much angle there is around its latticepoints. so just 'measuring the proportion of the space enclosed'

thanx for clearing this all up. :wink:
thigle
Tetronian
 
Posts: 390
Joined: Fri Jul 29, 2005 5:00 pm

Postby wendy » Sat May 20, 2006 8:39 am

The measure in terms of degree^x, really represents degree^x.radians^y in terms of x+y+1 dimensions. But since in mathematics, radian=1, then we have deg^x rad^y = deg^x.

One measures a circle in degrees or radians.

One measures a sphere in steradians, degrees excess or square degrees, these can each be represented as rad^2, rad.deg and deg^2.

One measures a glome in solid radians, degrees, square degrees, or cubic degrees, really r^3, r^2.d , r^1.d^2, and d^3 respectively.

Since all of these respectively represents a multiplication by k=d/r, one can then use as many k's as one requires, for each k, to increase the power on degree by one.



So we have 2pi**2 = 360pi deg = 64800 deg^2 = 11664000/pi deg^3.

The metric weights and measurements act of 1963 states that angles shall be measured in powers of 120, where a unit represents full space. Such an act, even though local, has force in all juridictions.
The dream you dream alone is only a dream
the dream we dream together is reality.

\ ( \(\LaTeX\ \) \ ) [no spaces] at https://greasyfork.org/en/users/188714-wendy-krieger
User avatar
wendy
Pentonian
 
Posts: 2014
Joined: Tue Jan 18, 2005 12:42 pm
Location: Brisbane, Australia


Return to Other Geometry

Who is online

Users browsing this forum: No registered users and 7 guests

cron