I searched for an isosurface formula that describes a "3D version" of a
1/x hyperbola, but not just rotating them or the like (hyperboloids are
nice but too common)

After some mistakes I found it and here is the result. Does this surface
has a special name?

I hope, you enjoy it.

Lars R.

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Download 'iso-grid.jpg' (123 KB) Download 'iso-grid.pov.txt' (1 KB)

Preview of image 'iso-grid.jpg'

On 24.04.2014 12:05, Lars R. wrote:

> After some mistakes I found it and here is the result. Does this surface
> has a special name?

Don't know that but it is a nice solution :)

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On Thu, 24 Apr 2014 12:05:10 +0200, Lars R. <rou### [at] gmxnet> wrote:

> I searched for an isosurface formula that describes a "3D version" of a
> 1/x hyperbola, but not just rotating them or the like (hyperboloids are
> nice but too common)
>
> After some mistakes I found it and here is the result. Does this surface
> has a special name?
>
> I hope, you enjoy it.
>
> Lars R.

I think it is called the Rouker Hyperboloid Matrix or something ;^>

-- 
-Nekar Xenos-

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"Lars R." <rou### [at] gmxnet> wrote:

> After some mistakes I found it and here is the result. Does this surface
> has a special name?
>
I don't know the name, but it will have hyperboloid in it, as suggested.
>
> I hope, you enjoy it.
>
I did enjoy it. So much so that I took the liberty of tweaking your function to
give first derivative continuity at the join.

For y = 0, the hyperbola looks like x^2z^2 = 0.25 or z = sqrt(0.25)/x.

Tweaking this as z = sqrt(0.25)(1/x + epsilon*x) and cutting off at x = X gives
epsilon = 1/X^2, so the new function becomes, in 3D:

function { x*x*y*y + x*x*z*z + y*y*z*z - 0.25*pow(((x*x+y*y+z*z)/(3*3)+1),2) }

I'm not too familiar with user-defined isosurfaces but will need to become so to
do some of the things I am planning, so thanks for the tutorial.

JimT

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Attachments:
Download 'iso-grid_v2.png' (275 KB)

Preview of image 'iso-grid_v2.png'

"JimT" <nomail@nomail> wrote:
> I did enjoy it. So much so that I took the liberty of tweaking your function to
> give first derivative continuity at the join.
>
> [snip]
>
> I'm not too familiar with user-defined isosurfaces but will need to become so to
> do some of the things I am planning, so thanks for the tutorial.

You seem to be well on your way.  In my experience, the trickiest part was
figuring out the max_gradient, but once you understand conceptually what's going
on, that becomes fairly easy to deal with.

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> > I'm not too familiar with user-defined isosurfaces but will need to become so to
> > do some of the things I am planning, so thanks for the tutorial.
>
> You seem to be well on your way.  In my experience, the trickiest part was
> figuring out the max_gradient, but once you understand conceptually what's going
> on, that becomes fairly easy to deal with.

I have used POV-Ray for a while, but this is the first time I have posted an
image of any kind.

LarsR has provided the checkered plane and a reflective surface. My tweak is
spherically symmetric. Does that count?

JimT.

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On 05/29/2014 07:14 AM, JimT wrote:
>
> I have used POV-Ray for a while, but this is the first time I have posted an
> image of any kind.
>
> LarsR has provided the checkered plane and a reflective surface. My tweak is
> spherically symmetric. Does that count?
>
Absolutely. :-)

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On 29/05/2014 12:38, William F Pokorny wrote:
> On 05/29/2014 07:14 AM, JimT wrote:
>>
>> I have used POV-Ray for a while, but this is the first time I have
>> posted an
>> image of any kind.
>>
>> LarsR has provided the checkered plane and a reflective surface. My
>> tweak is
>> spherically symmetric. Does that count?
>>
> Absolutely. :-)

Gone are the days when a first post was a RSOACP

Nice one! Jim

-- 

Regards
     Stephen

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