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Possible with an isosurface function?
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Am 12.06.2010 18:40, schrieb Shay:
> Possible with an isosurface function?
I bet the mathematicians have some f(x) = 0 formula to define an
elongated torus, so yes, it should be possible.
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Shay wrote:
> Possible with an isosurface function?
Elongated in what way?
Regards,
John
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John VanSickle wrote:
> Shay wrote:
>> Possible with an isosurface function?
>
> Elongated in what way?
>
> Regards,
> John
Like a sphere_sweep through a perfect ellipse.
-Shay
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Le 13/06/2010 21:02, Shay a écrit :
> John VanSickle wrote:
>> Shay wrote:
>>> Possible with an isosurface function?
>>
>> Elongated in what way?
>>
>> Regards,
>> John
>
> Like a sphere_sweep through a perfect ellipse.
A torus is a 4th order polynomial, so it should be possible to get an
elongated torus (as revolution of a circle along an ellipse) with a 4th
order polynomial, but it might not be as simple for the equation.
Hint: it's a quartic (and povray has a syntax for them)
explicit torus quartic is:
(x²+y²+z²+R²-r²)² - 4R²(x²+y²) = 0
--
A good Manager will take you
through the forest, no mater what.
A Leader will take time to climb on a
Tree and say 'This is the wrong forest'.
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Le_Forgeron <lef### [at] free fr> wrote:
> A torus is a 4th order polynomial, so it should be possible to get an
> elongated torus (as revolution of a circle along an ellipse) with a 4th
> order polynomial, but it might not be as simple for the equation.
Note that it's not enough to simply use an ellipse as the major radius
because that probably won't keep the minor radius constant throughout
(in the same way as simply scaling the torus unevenly doesn't).
--
- Warp
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"Shay" <n@n.n> schreef in bericht news:4c13b889$1@news.povray.org...
> Possible with an isosurface function?
Shay,
Years ago I needed a formula for the same shape you now request and, beleive
me, it took me several month to figure it out.
You start with:
x = (A + a.cos(alpha)).cos(phy)
y = b.sin(alpha)
z = (B + a.cos(alpha)).sin(phy)
A and B are the different main radii in x and z direction,
and a and b are the different radii in de pipe of the torus.
So the major radius R is now changed in A and B
and the minor radius r is now changed in a and b.
When you rotate alpha and phy over 360 degrees, you get the torus you want.
If you want to translate this to the f(x,y,z) = .... you need much patience
and time and
at the end (couple of days at least) you get your variables you can put in
the
poly shape.
In this range t = a/b so this is the ratio of the minor radii.
If you want a circular pipe, then t = 1.
Constant:
powers:
+t^4
x^4y^4
+2t^2(B^2-a^2)
x^4y^2
+(B^2-a^2) ^2
x^4
+2t^6
x^2y^6
+2t^4
x^2y^4z^2
-2t^4{(A^2+B^2)-3(B^2-a^2)} x^2y^4
-2t^2{(A^2-a^2)+(B^2-a^2)-4(AB-a^2)} x^2y^2z^2
+2t^2{(B^2-a^2)^2-2(B^2-a^2)(A^2+a^2)} x^2y^2
+2{(A^2-a^2)(B^2-a^2)+2(A-B)^2a^2 x^2z^2
-2(B^2-a^2)^2(A^2+a^2)
x^2
+t^8
y^8
+2t^6
y^6z^2
+2t^6{(A^2-a^2)+(B^2-a^2)} y^6
+t^4
y^4z^4
-2t^4{(A^2+B^2)-3(A^2-a^2)} y^4z^2
+t^4{(A^2-a^2)^2+(B^2-a^2)^2+4(A^2-a^2)(B^2-a^2)} y^4
+2t^2(A^2-a^2)
y^2z^4
+2t^2{(A^2-a^2)^2+2(A^2-a^2)(B^2+a^2)} y^2z^2
+2t^2{(A^2-a^2)^2(B^2-a^2)+(A^2-a^2)(B^2-a^2)^2} y^2
+(A^2-a^2) ^2
z^4
-2(A^2-a^2)^2(B^2+a^2)
z^2
(B^2-a^2)^2(A^2-a^2)^2
constant
This equation works fine if the poly shape power 8 is possible, but for some
reason today this is limited to power 7. There was a short period of time
that power 15 was alowed, but alas, not in this times. For me it would be
fine if this limitation is raised again.
In the short period that power 8 was alowed, it traces fine and not too
slow. The slowing down happened with textures. Then it slowed down
considerable. But I think that with our modern fast computers this is no
problem anymore.
You can still use this with the parametric Object, but then the shape is
made out of triangles and is not the pure mathematic form.
Maybe this mail causes that the max power is raised to 8 so this shape can
be used in his pure mathematical form again.
Jaap Frank
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Thank you!
The formatting of your message is a bit skewed on my end, but I will
attempt to retrace your steps myself and check my work against yours.
-Shay
Jaap Frank wrote:
> "Shay" <n@n.n> schreef in bericht news:4c13b889$1@news.povray.org...
>> Possible with an isosurface function?
>
> Shay,
>
> Years ago I needed a formula for the same shape you now request and, beleive
> me, it took me several month to figure it out.
>
> You start with:
>
> x = (A + a.cos(alpha)).cos(phy)
> y = b.sin(alpha)
> z = (B + a.cos(alpha)).sin(phy)
>
> A and B are the different main radii in x and z direction,
> and a and b are the different radii in de pipe of the torus.
> So the major radius R is now changed in A and B
> and the minor radius r is now changed in a and b.
> When you rotate alpha and phy over 360 degrees, you get the torus you want.
> If you want to translate this to the f(x,y,z) = .... you need much patience
> and time and
> at the end (couple of days at least) you get your variables you can put in
> the
> poly shape.
> In this range t = a/b so this is the ratio of the minor radii.
> If you want a circular pipe, then t = 1.
> Constant:
> powers:
> +t^4
> x^4y^4
> +2t^2(B^2-a^2)
> x^4y^2
> +(B^2-a^2) ^2
> x^4
> +2t^6
> x^2y^6
> +2t^4
> x^2y^4z^2
> -2t^4{(A^2+B^2)-3(B^2-a^2)} x^2y^4
> -2t^2{(A^2-a^2)+(B^2-a^2)-4(AB-a^2)} x^2y^2z^2
> +2t^2{(B^2-a^2)^2-2(B^2-a^2)(A^2+a^2)} x^2y^2
> +2{(A^2-a^2)(B^2-a^2)+2(A-B)^2a^2 x^2z^2
> -2(B^2-a^2)^2(A^2+a^2)
> x^2
> +t^8
> y^8
> +2t^6
> y^6z^2
> +2t^6{(A^2-a^2)+(B^2-a^2)} y^6
> +t^4
> y^4z^4
> -2t^4{(A^2+B^2)-3(A^2-a^2)} y^4z^2
> +t^4{(A^2-a^2)^2+(B^2-a^2)^2+4(A^2-a^2)(B^2-a^2)} y^4
> +2t^2(A^2-a^2)
> y^2z^4
> +2t^2{(A^2-a^2)^2+2(A^2-a^2)(B^2+a^2)} y^2z^2
> +2t^2{(A^2-a^2)^2(B^2-a^2)+(A^2-a^2)(B^2-a^2)^2} y^2
> +(A^2-a^2) ^2
> z^4
> -2(A^2-a^2)^2(B^2+a^2)
> z^2
> (B^2-a^2)^2(A^2-a^2)^2
> constant
>
> This equation works fine if the poly shape power 8 is possible, but for some
> reason today this is limited to power 7. There was a short period of time
> that power 15 was alowed, but alas, not in this times. For me it would be
> fine if this limitation is raised again.
> In the short period that power 8 was alowed, it traces fine and not too
> slow. The slowing down happened with textures. Then it slowed down
> considerable. But I think that with our modern fast computers this is no
> problem anymore.
>
> You can still use this with the parametric Object, but then the shape is
> made out of triangles and is not the pure mathematic form.
>
> Maybe this mail causes that the max power is raised to 8 so this shape can
> be used in his pure mathematical form again.
>
> Jaap Frank
>
>
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Hello Shay,
"Shay" <n@n.n> schreef in bericht news:4c22b7da@news.povray.org...
> Thank you!
Your welcom.
>
> The formatting of your message is a bit skewed on my end,
I noticed when I read it back from the news side.
If you want I can do it in a word document, so there is no misunderstanding.
> but I will attempt to retrace your steps myself and check my work against
> yours.
>
> -Shay
If you are interested in the whole mathematical derivation,
then I can scan it for you, because I don't have it electronically.
It's hand written in a blank book with some other mathematical things
I did need. I didn't exaggerate that it takes some days. It's a colossal
derivation because you have to square a rather big equation twice
and then sort everything on A, B, a and b.
If you like to do it yourself, first try the normal torus, so you know
how to do that. The oval_oval_torus starts the same way, but half way
I subtracted two equations, otherwise you end up with a power 16 result.
I will reorganise the variables and powers for you so you can read it
better:
Constants for their powers:
+t^4 x^4y^4
+2t^2(B^2-a^2) x^4y^2
+(B^2-a^2) ^2 x^4
+2t^6 x^2y^6
+2t^4 x^2y^4z^2
-2t^4{(A^2+B^2)+
-3(B^2-a^2)} x^2y^4
-2t^2{(A^2-a^2)+
+(B^2-a^2)+
-4(AB-a^2)} x^2y^2z^2
+2t^2{(B^2-a^2)^2+
-2(B^2-a^2)(A^2+a^2)} x^2y^2
+2{(A^2-a^2)(B^2-a^2)+
+2(A-B)^2a^2 x^2z^2
-2(B^2-a^2)^2(A^2+a^2) x^2
+t^8 y^8
+2t^6 y^6z^2
+2t^6{(A^2-a^2)+(B^2-a^2)} y^6
+t^4 y^4z^4
-2t^4{(A^2+B^2)-3(A^2-a^2)} y^4z^2
+t^4{(A^2-a^2)^2+(B^2-a^2)^2+
+4(A^2-a^2)(B^2-a^2)} y^4
+2t^2(A^2-a^2) y^2z^4
+2t^2{(A^2-a^2)^2+
+2(A^2-a^2)(B^2+a^2)} y^2z^2
+2t^2{(A^2-a^2)^2(B^2-a^2)+
+(A^2-a^2)(B^2-a^2)^2} y^2
+(A^2-a^2) ^2 z^4
-2(A^2-a^2)^2(B^2+a^2) z^2
(B^2-a^2)^2(A^2-a^2)^2
The last one is the constant or for power x^0y^0z^0 if you wish.
>> This equation works fine if the poly shape power 8 is possible, but for
>> some reason today this is limited to power 7. There was a short period of
>> time that power 15 was alowed, but alas, not in this times. For me it
>> would be fine if this limitation is raised again.
>> In the short period that power 8 was alowed, it traces fine and not too
>> slow. The slowing down happened with textures. Then it slowed down
>> considerable. But I think that with our modern fast computers this is no
>> problem anymore.
>>
>> You can still use this with the parametric Object, but then the shape is
>> made out of triangles and is not the pure mathematic form.
>>
>> Maybe this mail causes that the max power is raised to 8 so this shape
>> can be used in his pure mathematical form again.
>>
>> Jaap Frank
Because of this a request to Christof Lipka or Thorsten Froehlich;
Is it possible to raise the poly shape to power eight?
Power seven is just one power to low. If you mesh around with
these equations, then it always ends up with power four, eight or even
sixteen.
Greetings,
Jaap Frank
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I would very much like to see that scan. I would teach me a lot, I'm
sure. My email address is myn### [at] hotmailcom.
Thank you.
Jaap Frank wrote:
> Hello Shay,
>
> "Shay" <n@n.n> schreef in bericht news:4c22b7da@news.povray.org...
>> Thank you!
> Your welcom.
>
>> The formatting of your message is a bit skewed on my end,
>
> I noticed when I read it back from the news side.
> If you want I can do it in a word document, so there is no misunderstanding.
>
>> but I will attempt to retrace your steps myself and check my work against
>> yours.
>>
>> -Shay
>
> If you are interested in the whole mathematical derivation,
> then I can scan it for you, because I don't have it electronically.
> It's hand written in a blank book with some other mathematical things
> I did need. I didn't exaggerate that it takes some days. It's a colossal
> derivation because you have to square a rather big equation twice
> and then sort everything on A, B, a and b.
> If you like to do it yourself, first try the normal torus, so you know
> how to do that. The oval_oval_torus starts the same way, but half way
> I subtracted two equations, otherwise you end up with a power 16 result.
>
>
> I will reorganise the variables and powers for you so you can read it
> better:
>
> Constants for their powers:
> +t^4 x^4y^4
> +2t^2(B^2-a^2) x^4y^2
> +(B^2-a^2) ^2 x^4
> +2t^6 x^2y^6
> +2t^4 x^2y^4z^2
> -2t^4{(A^2+B^2)+
> -3(B^2-a^2)} x^2y^4
> -2t^2{(A^2-a^2)+
> +(B^2-a^2)+
> -4(AB-a^2)} x^2y^2z^2
> +2t^2{(B^2-a^2)^2+
> -2(B^2-a^2)(A^2+a^2)} x^2y^2
> +2{(A^2-a^2)(B^2-a^2)+
> +2(A-B)^2a^2 x^2z^2
> -2(B^2-a^2)^2(A^2+a^2) x^2
> +t^8 y^8
> +2t^6 y^6z^2
> +2t^6{(A^2-a^2)+(B^2-a^2)} y^6
> +t^4 y^4z^4
> -2t^4{(A^2+B^2)-3(A^2-a^2)} y^4z^2
> +t^4{(A^2-a^2)^2+(B^2-a^2)^2+
> +4(A^2-a^2)(B^2-a^2)} y^4
> +2t^2(A^2-a^2) y^2z^4
> +2t^2{(A^2-a^2)^2+
> +2(A^2-a^2)(B^2+a^2)} y^2z^2
> +2t^2{(A^2-a^2)^2(B^2-a^2)+
> +(A^2-a^2)(B^2-a^2)^2} y^2
> +(A^2-a^2) ^2 z^4
> -2(A^2-a^2)^2(B^2+a^2) z^2
> (B^2-a^2)^2(A^2-a^2)^2
> The last one is the constant or for power x^0y^0z^0 if you wish.
>
>>> This equation works fine if the poly shape power 8 is possible, but for
>>> some reason today this is limited to power 7. There was a short period of
>>> time that power 15 was alowed, but alas, not in this times. For me it
>>> would be fine if this limitation is raised again.
>>> In the short period that power 8 was alowed, it traces fine and not too
>>> slow. The slowing down happened with textures. Then it slowed down
>>> considerable. But I think that with our modern fast computers this is no
>>> problem anymore.
>>>
>>> You can still use this with the parametric Object, but then the shape is
>>> made out of triangles and is not the pure mathematic form.
>>>
>>> Maybe this mail causes that the max power is raised to 8 so this shape
>>> can be used in his pure mathematical form again.
>>>
>>> Jaap Frank
>
> Because of this a request to Christof Lipka or Thorsten Froehlich;
> Is it possible to raise the poly shape to power eight?
> Power seven is just one power to low. If you mesh around with
> these equations, then it always ends up with power four, eight or even
> sixteen.
>
> Greetings,
>
> Jaap Frank
>
>
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