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Author Topic: Lame Klein bottle  (Read 839 times)
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cKleinhuis
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« Reply #15 on: January 27, 2012, 08:38:35 PM »

from the kleinbottle guys, it looks like it could keep something "inside" i really need to talk to those guys ....
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knighty
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« Reply #16 on: January 27, 2012, 09:21:50 PM »

Wow! Very nice klein bottle. Not lame anymore. Please change the title.  grin

Well, here you have an implicit formulation of the symmetric projection of the Klein bottle
http://www.wolframalpha.com/input/?i=klein+bottle

If you explicitly want the bottle-like one, I'm pretty sure, that requires yet another Polynomial of degree 12. Most likely one that, due to lack of symmetry, is even more complicated than the one given there...
Here is a degree 6 formula :

((x^2+y^2+z^2)+2*y-1)*(((x^2+y^2+z^2)-2*y-1)^2-8*z^2)+16*x*z*((x^2+y^2+z^2)-2*y-1);

You can play with the numerical parameters. It's remarquable that doing so won't change the topology of the surface. It's in the structure of the polynomial not its parameters.

(I don't remember were I found the formula. That was some time ago.)


* algebraic.zip (2.12 KB - downloaded 19 times.)
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DarkBeam
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« Reply #17 on: January 27, 2012, 11:31:40 PM »

from the kleinbottle guys, it looks like it could keep something "inside" i really need to talk to those guys ....


Also the mobius scarf and metallic platonic solids for sale smiley
explore the site
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Alef
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« Reply #18 on: January 29, 2012, 02:12:00 PM »

I would prefare some darker glass for A Beer Cup A Beer Cup A Beer Cup Are you using fragmentarium?
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DarkBeam
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« Reply #19 on: January 30, 2012, 10:47:50 AM »

Wow! Very nice klein bottle. Not lame anymore. Please change the title.  grin

Well, here you have an implicit formulation of the symmetric projection of the Klein bottle
http://www.wolframalpha.com/input/?i=klein+bottle

If you explicitly want the bottle-like one, I'm pretty sure, that requires yet another Polynomial of degree 12. Most likely one that, due to lack of symmetry, is even more complicated than the one given there...
Here is a degree 6 formula :

((x^2+y^2+z^2)+2*y-1)*(((x^2+y^2+z^2)-2*y-1)^2-8*z^2)+16*x*z*((x^2+y^2+z^2)-2*y-1);

You can play with the numerical parameters. It's remarquable that doing so won't change the topology of the surface. It's in the structure of the polynomial not its parameters.

(I don't remember were I found the formula. That was some time ago.)



My god Knighty!!!!! I casually opened the zip and another surprise! You are... Ran out of words! very surprised very surprised very surprised
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DarkBeam
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« Reply #20 on: January 30, 2012, 12:10:19 PM »

I would like to discuss also my method for finding a DE, without derivatives (that's the method that Knighty followed in its script)

Too bad, it is a bit too complicated for some surfaces, because it's needed to know the signs of every "polynomial component" of the surface.

Let's examine a very very simple example.

Let's define;
x2 = x*x;
y2 = y*y + z*z;
And examine the "lathed" Scarabaeus curve expression;

(x2 + y2) (x2 + y2 + R1 x)^2 - R2^2 (x2 - y2)^2 = 0

The problem for DE is that the field needs to be "approximately linear", on x,y,z. For example a sphere equation is;

x*x + y*y + z*z - R*R = 0

To make it "linear", we can take square root. It does not exist for negative values? Oh it's simple, bring R*R to the right.

x*x + y*y + z*z = R*R
Then;
sqrt(x*x + y*y + z*z) - abs(R) = 0
This gives a very good sphere DE.

Okay, applying the same method to the scarab curve is not so plain; (x2 + y2) (x2 + y2 + R1 x)^2 is always positive (a sum of squares) but (x2 - y2)^2 has a variable sign. If we expand we have (x2 - y2)^2 = (x^4 + y^4) - 2*(x*x*y*y)
Again we can separate positive and negative, because x*x*y*y is always >0 ;

ssqrt((x2 + y2) (x2 + y2 + R1 x)^2 + 2*R2*R2* (x*x*y*y))) - sqrt(R2)*ssqrt(x^4+y^4) where ssqrt(a) is sqrt(sqrt(a)) to make the vector field of power 1 because the highest power is 4. smiley

That seems to works like a charm! cheesy

The problem is that many surfaces have lots mixed terms like x*y, or even power terms like x, x^3 ... , so separate those correctly is a trouble.

In particular Enneper implicit equation...





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knighty
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« Reply #21 on: January 30, 2012, 03:23:34 PM »

That's interesting! unfortunately beside some 'simple' cases (like the sphere) the straighten operation is not (can't ?) be perfect. But you could also use derivatives of the straightened formula to get even better DE.

I had the same idea but in the case of escape time fractal distance estimation that i discussed some time ago with syntopia in his blog.
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DarkBeam
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« Reply #22 on: January 30, 2012, 05:22:48 PM »

That's interesting! unfortunately beside some 'simple' cases (like the sphere) the straighten operation is not (can't ?) be perfect. But you could also use derivatives of the straightened formula to get even better DE.

I had the same idea but in the case of escape time fractal distance estimation that i discussed some time ago with syntopia in his blog.

For me your formula is wonderful but calculates three+1 times the function and it's hard to code (and slower than taking some roots). So wondering about a workaround. Later. embarrass
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DarkBeam
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« Reply #23 on: January 30, 2012, 05:26:52 PM »

Uh, in any case; Ennepers is 4-way symm. and in fact, look closer at the expressions.

z appears at even & odd powers.
x&y appear at even powers only.


This explains the symmetry and also reduces the problem a lot, because we must only pay attention to z sign. Let me work on this. wink
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knighty
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« Reply #24 on: January 30, 2012, 09:17:00 PM »

You are right. It's should be faster without using derivatives. smiley
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DarkBeam
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« Reply #25 on: February 09, 2012, 08:08:50 PM »

Another merchandise idea

http://www.etsy.com/listing/66353455/worlds-smallest-klein-bottle

 smooth....
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