3dino

21st June 2002 13:16 UTC

Platonic Solids
try to do an Tetrahedron
but i can only rotate around the Y Axis
and why are the green lines behind of the blue ones?
:igor: :(

3dino

21st June 2002 13:22 UTC

the same prob. with an Octahedron:

UnConeD

21st June 2002 16:45 UTC

You're using replace mode, so the next superscope will overwrite the previous one. You can use additive/maximum to get around this, but to do actual depth-culling is very difficult unless you shade by depth (z) and use maximum mode.

3dino

21st June 2002 18:11 UTC

sometimes i hate avs like this :( :cry: :hang:
:p <( LOL )

Jaheckelsafar

21st June 2002 19:22 UTC

Well, that last one looks pretty good. :up:

UnConeD

21st June 2002 19:35 UTC

Circle segments
I think the 3 partitions would look better as circle segments rather than triangles though...

3dino

21st June 2002 23:06 UTC

i just try it!
here is the work in progess overwiew pac,
btw. same prob. :( (can !only! rotate with x or y Axis)
GEEEKS Help määän! (before i try a supermäään logo!)

Nic01

21st June 2002 23:54 UTC

I like all the Radioaktiv presets except for the first and last one...

I can see the limited rotation...

Your best bet to make it at least more dynamic - Use getosc/getspec on the OnBeat section.
I'll look into the coding some more and see if I can solve the prob...

3dino

22nd June 2002 01:01 UTC

hmm need a onbeat-function to move the 3 triangels around the Y-Axis (10°-20° back and forward (vY^)+(^Yv)).
then I don't need to redesign the 3 triangels! :p
can somone explain getosc/getspec variable settings a bit more plz?!

UnConeD

22nd June 2002 11:30 UTC

You seem to have some trouble getting the 3 side-shapes right. Try this in one superscope:

if (i < 0.5) draw circle-arc (30°) clockwise at radius 90%
if (i > 0.5) draw circle-arc (30°) counter-clockwise at radius 20%

That should get you what you need. On top of that, you can easily rotate this shape by adding a phase-angle to the circle-arcs' begin and endpoints


Here's the scoop on getosc (same applies to getspec):

v = getosc(point,width,channel);

point is a value from 0...1 that defines which oscilloscope point you want

width is a smoothing parameter. a value of 0 means no smoothing (normal behaviour), a value of 0.1 means that the surrounding values that are in the range of 10% will be blended. For example getosc(0.3,0.1,0) will grab all the values between 0.25 and 0.35 and average them. This is useful if you want a shape to be sound responsive, but not get too distorted by high frequencies (rapid changes).

channel sets which sound-channel you need, 0=center, 1=left, 2=right

3dino

22nd June 2002 20:25 UTC

TILT!
sorry, need a break! :o
perhaps I will try to fix this later?!

3dino

24th June 2002 19:39 UTC

its like pot-strike: it becomes warmer :igor: :weird:

3dino

25th June 2002 21:05 UTC

- fixed the Octahedron :)
- make a Cuboctahedron, a Torus + a cell:

peepingdan

30th June 2002 07:34 UTC

All this math is *WHOOSH* above my head :winamp:
I just started learnind Dynamic Movements :)

3dino

1st July 2002 22:35 UTC

yea, do this and tell me how to use them ;)

dirkdeftly

3rd July 2002 20:44 UTC

Well...As for the rotation, you could just apply the scope(s) to a normal 3D rotation (I explained this in my Primer, but no one seems to be downloading it :( it's probably at the bottom of the forum by now :cry: )


BTW, sorry, but I haven't actually looked at the preset, I don't have WA on this comp (nor do I have access to a comp with WA). I'm just catching up with the forums while I'm away...

3dino

9th October 2002 19:14 UTC

my last work: (sorry, it is not perfect!)
http://mitglied.lycos.de/llldino/Bilder/NUKE.jpg
<<!nuke.ace>>

UnConeD

9th October 2002 19:55 UTC

Your radioactive three-foil sign looks perfect :)

Xion(810)

9th October 2002 22:09 UTC

Originally posted by 3dino
my last work: (sorry, it is not perfect!)
http://mitglied.lycos.de/llldino/Bilder/NUKE.jpg
<<!nuke.ace>>

jheriko

9th October 2002 23:38 UTC

WTF? How the hell did I miss this one?

Anyway, I better check out these presets.

Xion, .ace is a zip alternative.

3dino

10th October 2002 17:47 UTC

>Xion got to => http://www.winace.com/ ;) (i think, winzip also support ace-files)

>jheriko, do you make your icosahedron pic with an ssc?

>UnConeD: "Your radioactive three-foil sign looks perfect :)"

Thanks, Yes but i like to cange:
1. the rotation, (x/y/z & reaction on beat)
2. add yellow or green Fog,
3. filled shapes (dark-green or black).
please, give me some tips! :)

>http://cage.rug.ac.be/~hs/polyhedra/6to12anim.gif
who will try to do this?

>If you have VRML installed, check this: Tetraeder Spirale

jheriko

10th October 2002 19:05 UTC

Yup, its a ssc.

The reason I was a bit surprised at seeing this thread is because I released a minipack containing presets centred around the Platonic solids which contains a preset for each of them, 4 of them use filled faces made from scopes and one of them uses wireframe. If you are interested I have also made a rhombic dodecahedron which features in my Pack VI. My next pack will probably feature a couple of other solids, I did make a cuboctahedron but that ones already been done so I'm working on a icosidodecahedron (triacontrahedron with pentagonal and hexagonal faces - like a soccer ball) by using the golden ratio do determine its vertices and a rhombic triacontrahedron by using powers of the golden ratio.

Here are links to my the packs I mentioned above:

Purely Platonic
Jheriko - Pack VI

If you really want that animated gif as an AVS preset, I could do probably do it in wireframe, doing it solid would require that the vertices were ordered in a specific way (clockwise or anti-clockwise) but it is possible, I may give it a pop if I'm feeling mathematical later on...

Yeah your radioactive symbol was very good, if you want to fill it in I can give you some recommendations. To fill the circle the easiest way is probably to make your ssc go from the centre to one of the circle points, then back again, then to the next point on the circle... by using an alternating variable per point. I'd recommend that if you are going to do it, NOT to do it that way since although it is easy to code it will look like poo without a very high number of points, it would be better to make lines that zigzag across the circle by spliting it into two equal curves (vertical semi-circle then mirror it with x=-x would proably be easiest) and use an alternating variable to make it draw lines from the first point on one curve to the second point on the other back to the third on the first... and so on. You could fill the other shapes by mathematically defining them in a similar way since they are all symetrical, and since they are made of lines and circle arcs it shouldn't be too difficult to figure out.

:)

3dino

11th October 2002 00:28 UTC

jheriko :up:
very nice pacs!
hope, i will get more time for avs (shool work sucks! :mad: )

some nice links:
http://www.polyedergarten.de/klintro.htm
http://www.georgehart.com/virtual-polyhedra/vp.html
http://www.georgehart.com/
http://astronomy.swin.edu.au/~pbourke/

jheriko

11th October 2002 00:50 UTC

I don't suppose you happen to have any links which list polyhedral vertex co-ordinates, I'm gettig really sick of working them out, some of the larger ones take ages, especially when they have very little symmetry since you have to work out practically every vertex individually. :(

UnConeD

11th October 2002 01:34 UTC

Maybe you could model them in CAD/3DSMAX and export the coordinates as a textfile?

jheriko

11th October 2002 02:10 UTC

Sounds like a good idea, but would those co-ords be nice numbers? Probably some hashed up messy skank using square roots every which way...

Then again I could just divide them all by varying powers of phi to make them nicer. Thats mainly so tht the code is easier to type in and nicer to look at otherwise I'd end up with something like:

x1=if(equal(q,3),1.078912323,x1);
y1=if(equal(q,3),2.414276302,y1);
...

which is hassle if you have to type out numbers like that 90 times. :p

UnConeD

11th October 2002 09:19 UTC

3 decimals should be enough...

jheriko

11th October 2002 10:43 UTC

I guess as a mathematician I have some kind of bizarre desire to see numbers being expressed as accurately as possible... I did a few web searches to find co-ordinates and I managed to find a few sites with co-ordinates for the platonic solids and some other shapes defined in terms of numbers like sqrt(2) and phi so I'm going to be lazy and use some of them. Thanks for the tip though.

If any one else is interested:

http://www.rwgrayprojects.com/Lynn/C...s/coord01.html

This one saved me the trouble of figuring out the rest of the rhombic triacontahedron. It also has a few more co-ords.

UnConeD

11th October 2002 11:23 UTC

As an engineering student I see math as the tool I need to solve certain problems. If you don't need anymore accuracy, why bother? :)

At 320 width, one pixel is 0.00625 AVS units wide. So for a 2D scope you don't need any more than 3 decimals in your constants. Given the fact that in 3D scopes, you divide by Z (> 1) most of the time, usually you can settle for 2 decimals.

Warrior of the Light

12th October 2002 15:08 UTC

Originally posted by jheriko
I guess as a mathematician I have some kind of bizarre desire to see numbers being expressed as accurately as possible...

dirkdeftly

13th October 2002 08:49 UTC

31,769 signifigant digits. DAMN. :D

jheriko

13th October 2002 10:55 UTC

Pi and e to 100000 decimal places, courtesy of Mathematica. It took about 3 seconds to calculate it. I would have done them to 1000000 but that would have probably taken about half an hour.

:p

jheriko

13th October 2002 11:02 UTC

:igor:

I could have sworn that I attached it...

UnConeD

13th October 2002 13:13 UTC

The only reason that huge amount of significant digits is useful is for statistical analysis for randomness and such. For most practical purposes, 3.1415 is enough precision :) (whenever I say this to a mathematician, they usually attempt to come up with some excuse why a larger precision would really be better!).

ujay

13th October 2002 14:32 UTC

Don't usually find myself over here, but my eye was caught by the title.

Has anyone tried any concave polyhedra yet.

May I suggest the Great Dodecahedron - it is formed from just 12 intersecting pentagons and looks easy ( but what do I know ).

It also breaks the Euler formula that links vertices, edges and faces.

Sorry to intrude, please carry on :)

UJ

UnConeD

13th October 2002 15:24 UTC

The only problem is that you need a good way of depth-culling each face against the others. This isn't easy in AVS. You could simulate a z-buffer as color value (using maximum blend), but then you'd need to tesselate each face into short line segments to get any sort of accuracy.
Or you could split it up in non-intersecting triangles, guaranteed to be fun :P.

jheriko

14th October 2002 01:26 UTC

A relatively simple way to not draw hidden faces is to make sure that the co-ordinates are entered in a certain order, clockwise or anti-clockwise, then to see if from the perspective of the viewport the co-ordinates are being place clockwise or anticlockwise, if they are in the wrong order then you don't draw the face. The only problem with that method is that it only works 100% for convex objects.

As for pi I can remember being told something like 'if the universe were a perfect sphere and its radius was the distance from us to the furthest known object, then you could calculate it volume to within a cubic meter of accuracy by using pi to 10dp'. I can't remeber the exact number, it could have been 15 or 20dp but it was definately a lot less than 30000.

EDIT: The other thing is that if you are doing a calculation where you need to multiply pi several times then the error will increase more and more.

UnConeD

14th October 2002 08:56 UTC

Jher: I guess I should've used a different term. If you look at that great dodecahedron you can see every face intersects practically every other face. Traditional culling isn't going to work here, you need z-buffer-ish techniques to handle that. And getting a smoothly shaded pentagon isn't easy either.

By the way, I haven't actually encountered powers of pi in any formula. I know about pi^2/6 as the limit of some expression, but that's about it.

jheriko

14th October 2002 09:12 UTC

Yeah, but you could split it into triangles like you said. I'm sure that the normal back face culling method would work fine then, although i remember reading somewhere that it fails for certain convex objects.

I'll let the powers of pi thing slip though, and just pretend that you didn't say that (any trig function working in degrees/graduations). I'll admit that having pi to any excessively large accuracy serves no practical purpose yet.

Zevensoft

14th October 2002 10:17 UTC

Number of parsecs (Smallest possible unit, WITHOUT involving quantum physics) between 2 edges of universe = Digits required.

jheriko

14th October 2002 10:37 UTC

Thats a huge number of digits... several millions of billions.

Zevensoft

14th October 2002 11:47 UTC

I'd say in the order of a googleplex or more.

UnConeD

14th October 2002 12:12 UTC

I think it's googolplex though. The search engine Google is derived from that, but it's misspelled on purpose.

I'll let the powers of pi thing slip though, and just pretend that you didn't say that (any trig function working in degrees/graduations).

jheriko

18th October 2002 09:54 UTC

cos x= 1 + (x^2)/2 + (x^4)/4! + (x^6)/6!...

You plug anything involving pi into that and you get powers of pi. You can't convert the value of cos x afterwards, not without using another infinite series (as far as I know).

dirkdeftly

18th October 2002 20:53 UTC

I believe you're mistaken, Jheriko. The series is closer to:

inf
E (x^(4n))/((4n)!)-(x^(4n+2))/((4n+2)!) = cos x
n=0

(E being the closest thing to epsilon I could think of and inf being infinity)

jheriko

18th October 2002 21:48 UTC

Oh crap.. i missed some more minuses. There is a better way to express that series.

BTW: E should be a capital sigma to represent summation, epsilon is used to represent arbitary small values that tend to zero. You can use the symbol font for those greek letters.

Anyway, what you want is:

n->inf
S(-1^(n) * x^(2n)/(2n)!) = cos(x)
n=0

and replacing 2n with 2n+1 for sin(x).

You can also express them as a product of a series or in terms of hypergeometric functions, either way you still end up with a power series of some kind.

UnConeD

18th October 2002 22:08 UTC

The taylor series expansion of cos x can hardly be seen as a formula :). Besides, it still has nothing to do with the accuracy of pi itself. You'll never fill in pi in that series, because cos and sin of pi are well known values :p.

jheriko

18th October 2002 22:11 UTC

Lets say that you want to find cos of any number in degrees then
cos(x) becomes cos(x*pi/180). or you might want to do cos(pi/7) or something.

nixa

19th October 2002 20:20 UTC

Jheriko I think that the value of cos(pi*n) whith n being an element of Q can be calculated using formulas for sin(x+y),cos(x+y),sin(x/2),cos(x/3) and such.

UnConeD

20th October 2002 03:20 UTC

Ugh...consider the pi issue closed :P.

Warrior of the Light

20th October 2002 14:35 UTC

sorry to have started it... :(

dirkdeftly

22nd October 2002 05:02 UTC

Aww crap...I'm taking Greek, I should know that shit :P

And one of the series for pi is something like:

inf
E 1/(2n)-1/(2n+2)
n=0

I believe that converges on pi/4

NOW it's closed ;)

*mumbles, 'alpha, vita, ghamma, thelta, epsilon, zita, ita, thita, yota...'*

Zevensoft

22nd October 2002 09:15 UTC

Bah, f(x) = x/((1-x^2)^.5) is the inverse function of a sigmoid. It's also the derivative of a semi circle. Circles use pi. I'm sure there's some way of calculating pi by using this relationship.

jheriko

22nd October 2002 10:11 UTC

Completely off topic, if you want to find the derivative of a whole circle you can use a parametric equation like x=cos t, y=sin t and find the derivative by using:

dy/dx=(dy/dt)/(dx/dt)

This will give an expression for dy/dx in terms of t (for a circle its -cot t), which you can replace with x by doing some substitution. You can find higher derivatives by using the chain rule and solving some simple differential equations.

3dino

25th October 2002 22:20 UTC

:hang: hmm... only an exercise: