Logo by Maya - Contribute your own Logo!

END OF AN ERA, FRACTALFORUMS.COM IS CONTINUED ON FRACTALFORUMS.ORG

it was a great time but no longer maintainable by c.Kleinhuis contact him for any data retrieval,
thanks and see you perhaps in 10 years again

this forum will stay online for reference
News: Visit the official fractalforums.com Youtube Channel
 
*
Welcome, Guest. Please login or register. February 21, 2019, 05:59:34 PM


Login with username, password and session length


The All New FractalForums is now in Public Beta Testing! Visit FractalForums.org and check it out!


Pages: 1 ... 5 6 [7] 8 9   Go Down
  Print  
Share this topic on DiggShare this topic on FacebookShare this topic on GoogleShare this topic on RedditShare this topic on StumbleUponShare this topic on Twitter
Author Topic: Polyhedrons, many many polyhedrons...  (Read 22619 times)
0 Members and 1 Guest are viewing this topic.
DarkBeam
Global Moderator
Fractal Senior
******
Posts: 2512


Fragments of the fractal -like the tip of it


« Reply #90 on: March 25, 2012, 10:53:33 PM »

Keep doing good don't stop cheesy
Logged

No sweat, guardian of wisdom!
JosLeys
Strange Attractor
***
Posts: 258


WWW
« Reply #91 on: March 25, 2012, 11:27:55 PM »

Knighty, would this not also be the case with your magic formula?
A point is folded until it is in the fundamental domain, and a distance estimate follows from an angle measured within that domain. Could not a feature point in a neighbouring domain be closer after stereographic projection?
Logged
knighty
Fractal Iambus
***
Posts: 819


« Reply #92 on: March 26, 2012, 10:19:56 PM »

No, on the sphere, the distance from a test point to a feature inside the same cell is always less than the distance to the features that are in the neighbouring cells (ie the kaleidoscopic images of the feature). The stereographic projection of the corresponding circle is still a circle but it's center is not on the projection of the test point. it is on the line from the origin to the projection of the test point and is farther from the origin than the projection of the test point. That projected circle obviousely won't contain any neighbour's projected features and because the circle corresponding to the DE is inside we won't get overstepping smiley. The picture below illustrates approximatively what happens. (sorry I used Windows paint)


* Sdodeca02.JPG (63.62 KB, 800x800 - viewed 194 times.)
Logged
JosLeys
Strange Attractor
***
Posts: 258


WWW
« Reply #93 on: March 27, 2012, 11:32:51 AM »

Ok I see why I would overstep if I do what I had in mind.
Thanks for your patience!
Logged
knighty
Fractal Iambus
***
Posts: 819


« Reply #94 on: March 28, 2012, 08:30:47 PM »

Thank you too. Thanks to your questions, I am having a better understanding of the subject.  smiley

This subject is very rich, so I think it won't stop here. Maybe just take some time to complete.

The idea of keeping track of the applied transformations will be useful to handle the subgroups (the crystallographic groups for example) of the symmetry groups used so far. I think one still have to consider neighbouring cells in order to get consistant DE. That way the snub variations could be possible.

It is possible to render those tessellations with direct raytracing within the projective model. The rays travel in the 4D space instead. It is much like what is described here  grin. The advantage of using the projective model is linearity (and correct "deformations"). Maybe it will be better and faster to use it for the crystallographic groups (and their spherical and hyperbolic couterparts).

But before, I want to do the euclidean case for the honeycombs. find a better "magic formula" for the hyperbolic case to handle points at infinity. And of course, a DE for the stellations (the data kindly provided by punto will be very useful).

And after... back to fractals. But for now, back to work cry hurt
Logged
matsoljare
Fractal Lover
**
Posts: 215



WWW
« Reply #95 on: March 28, 2012, 11:48:37 PM »

That's a really beautiful picture, any more like that?
Logged
puntopunto
Alien
***
Posts: 28

keep it simple


« Reply #96 on: March 29, 2012, 09:27:12 AM »

Hallo Knighty:

Although I don't understand  how you render the polyhedra, you cannot be far away from rendering the Archimedian duals. There are a few possible methods. Using the vertex figure or using the property that duals are derived from interchanging the vertices and faces of a polyhedron. You will find sufficient info on the wikipedia page. Anyway here are the vertex coordinates.



http://www.stitchstitch.info/m3d/Archimedian%20duals.docx

One question: Which basic information of the polyhedra do you use. The reflections in the symmetry group, together with what?

Next: some compounds. They can't be far away for you either
Logged
DarkBeam
Global Moderator
Fractal Senior
******
Posts: 2512


Fragments of the fractal -like the tip of it


« Reply #97 on: March 29, 2012, 11:41:53 AM »

It's already possible to render zillions of truncations with his scripts. And with some nice colors and stuff, too! grin
Logged

No sweat, guardian of wisdom!
puntopunto
Alien
***
Posts: 28

keep it simple


« Reply #98 on: March 29, 2012, 12:58:30 PM »

Dualisation is a general concept. If you succeed to implement this general concept, then you will have also the duals for the non convex polyhedra yet to come. And you cannot generate the duals by truncation.
« Last Edit: March 29, 2012, 04:54:02 PM by puntopunto » Logged
knighty
Fractal Iambus
***
Posts: 819


« Reply #99 on: March 29, 2012, 10:54:02 PM »

Yes, there are many stellations and duals that are possible with that script (http://www.fractalforums.com/index.php?topic=10166.msg41859#msg41859)  but not all of them, just a tiny part  wink. IIRC, the polyhedras presented in the previous picture are already possible with that script except the last two in the second row (they don't have full mirror symmetry).

One question: Which basic information of the polyhedra do you use. The reflections in the symmetry group, together with what?
just the position of a vertex inside the fundamental domain. The normal to the 1 to 3 faces that are adjacent to that vertex are simply the intersection lines of the three symmetry planes.

Next: some compounds. They can't be far away for you either
Ah yes, I forgot about them. Thank you. Not complcated indeed: Fold about one set of symmetry planes then draw one rotated polyhedra. Another possibility is to simply make rotated copies.

That's a really beautiful picture, any more like that?
Thank you  smiley, It was made with an evaldraw script that you can find here:www.fractalforums.com/new-theories-and-research/the-power-of-fold/msg36045/.
« Last Edit: March 30, 2012, 12:08:48 AM by knighty » Logged
JosLeys
Strange Attractor
***
Posts: 258


WWW
« Reply #100 on: April 01, 2012, 12:14:09 AM »

Knighty, thinking about about rays traveling in 4D...

In the simple case of stereographic projection from the 2-sphere to a 2D plane : imagine we are shooting rays within this 2D plane. A point on such a ray, elevated to the sphere, would be on a circle through the north pole (the projection center) and the inverse projection of two points on the ray.
The point would then progress along this circle until it encounters a vertex or a segment, or until it encounters the inverse projection of the target point of the ray. The progress along the circle is then a rotation around the normal on the plane of the circle, and the rotation angle can be found by the folding operation.

Now, how can this be translated to the case of a ray traveling in 3D space and inverse stereographic projection to the 3-sphere?
I have a hard time visualising this in my head... angry

As the ray in 3D is a straight line, then in 4D, on the 3-sphere it would also be a circle through the projection center. (a line is a circle with infinite radius, stereographic projection projects circles as circles...).
If we take two points A and B on the ray in 3D, and we call P the projection pole and A' and B' the inverse projection on S^3 of A and B, then this circle would be the intersection of a plane defined by the vectors PA' and PB' and the sphere.
That's about as far as I got. How do we find the normal on this plane...

Logged
JosLeys
Strange Attractor
***
Posts: 258


WWW
« Reply #101 on: April 10, 2012, 07:18:23 PM »

I implemented Knighty's code on 4D polychora in Ultrafractal.
See an example below.
The code in UF is based on Knighty's in Fragmentarium, but with an important difference, which I think is an improvement: instead of the "magic formula", here is what I do :
- A point on the ray is 'promoted' to 4D : the inverse stereographic projection to the 3-sphere.
- The promoted point is folded by the planes that make up the fundamental tetrahedron a sufficient number of times.
- Determine the distance to a vertex and a segment, and take the smaller of the two, expressed as an angle, that we will call theta.
.....so far, nothing new.
- What we know now is that our point on our ray can move forward by a distance d, so that the angle between the 'promoted ' new point and the old point, both on the 3-sphere, is not larger than theta. If A is the old point, A' the promoted old point and B the new point, whereby B=A+v.d, v the direction vector of the ray , and B' the promoted new point, then the dot product A'.B' equals cos(theta). This produces a quadratic equation in d. So we can solve for d, and in principle step forward on the ray with the exact distance to a vertex or a segment.
As it is a quadratic equation, there are either 2,1 or no solutions. We simply always take the absolute of the solution as we do not want to step backward. If there is no solution (a negative determinant in the solution of the quadratic equation) then we can progress rapidly on the ray: it will not encounter a vertex or a segment.

So this replaces the 'magic formula' by another one (which is not simple either when written out as a formula) but that has the advantage to give the 'exact' distance. In pratice I still use a fudge factor, but 0.95*DE gives no artifacts...

Here is an example in UF: a 4D soccer ball.
 It is obviously not as fast as Fragmentarium, as UF works on the CPU, but an on-screen render in UF takes less than 30 seconds on my machine.

 


* Kaleido_4D_013.jpg (213.09 KB, 1024x1024 - viewed 236 times.)
Logged
knighty
Fractal Iambus
***
Posts: 819


« Reply #102 on: April 10, 2012, 09:18:35 PM »

Hi,

If I understand correctly this should be computationally equivalent to post #85. I don't think that the distance you get is exact in the projection space but it is on the hypersphere anyway smiley. It would be nice to compare the speeds of both methods.
Logged
JosLeys
Strange Attractor
***
Posts: 258


WWW
« Reply #103 on: April 10, 2012, 10:00:55 PM »

The ray we're following translates into a circle on the sphere through the projection pole.
The distance measured as an angle on the sphere translates into another circle on the sphere, centered on the actual point of the ray, promoted to the sphere, with a 'radius' equal to the calculated distance.
If the border of the object (vertex or segment) happens to be on the intersection of these two circles, meaning that the ray is heading straight towards it, we should have the exact distance, don't you think?
Logged
knighty
Fractal Iambus
***
Posts: 819


« Reply #104 on: April 11, 2012, 05:51:52 PM »

IMHO, no.

Let's say we have a point P on the sphere and a circle c (on the sphere) centered on that point.
Let P' abd c' be the stereographic projections of P and c on the plane respectively.
c' is also a circle but it's center doesn't coincide with P'.

If we take three points on the sphere A,B and C and their projections A', B' and C', there are (many) cases where the (spherical) distanc between A and B is greater than the distance between A and C and where the (euclidean) distance between A' and B' is smaller than the distance between A' and C'.

Because any point on the plane (augmented with the point at infinity) can be identified to a point on the sphere (there is a bijection) we can pretend that they are the same point. The only thing that changes is the metric. by changing the metric we also change the definition of the distance.

In the picture below, all the points on the red circle are at the same spherical distance to A (that is the distance "defined" by a spherical metric). Also, all the points on the red circle are at the same euclidean distance to A.

So you can say that you have the exact distance but with respect to spherical metric. smiley


* dist-sph-stereo.png (57.63 KB, 1019x699 - viewed 223 times.)
* dist_sph_stereographique.zip (7.04 KB - downloaded 96 times.)
Logged
Pages: 1 ... 5 6 [7] 8 9   Go Down
  Print  
 
Jump to:  


Powered by MySQL Powered by PHP Powered by SMF 1.1.21 | SMF © 2015, Simple Machines

Valid XHTML 1.0! Valid CSS! Dilber MC Theme by HarzeM
Page created in 0.131 seconds with 27 queries. (Pretty URLs adds 0.009s, 2q)