High!

After a six-week break, I returned to my self portrait project... as I 
firstly try to do a "cheap" (in terms of rendering time) CSG version, I 
started with a "skull ellipsoid" to which I would like to add face with 
upper jaw and an animatable mandible, as displayed in the three 
schematical drawings below.

The radii of the "skull ellipsoid" (centered at the origin) are as 
follows: <0.115, 0.108, 0.138>

To fit the front parts in, I vertically sliced off the skull shell at 
z=-0.097. Now I have to determine the width of the cut surface at y=0. 
With a non-stretched sphere, that would be easy: 
sqrt(radius^2-0.097^2)*2*radius - but how to transfer this to an 
ellipsoid? To determine the exact radius length at the intersection of x 
axis, cutoff box and ellipsoid, I would have to know its angle with the 
z axis...

See you in Khyberspace!

Yadgar

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4707fde8@news.povray.org...
> High!
>
> After a six-week break, I returned to my self portrait project... as I
> firstly try to do a "cheap" (in terms of rendering time) CSG version, I
> started with a "skull ellipsoid" to which I would like to add face with
> upper jaw and an animatable mandible, as displayed in the three
> schematical drawings below.
>
> The radii of the "skull ellipsoid" (centered at the origin) are as
> follows: <0.115, 0.108, 0.138>
>
> To fit the front parts in, I vertically sliced off the skull shell at
> z=-0.097. Now I have to determine the width of the cut surface at y=0.
> With a non-stretched sphere, that would be easy:
> sqrt(radius^2-0.097^2)*2*radius - but how to transfer this to an
> ellipsoid? To determine the exact radius length at the intersection of x
> axis, cutoff box and ellipsoid, I would have to know its angle with the
> z axis...
>
> See you in Khyberspace!
>
> Yadgar
>

Hi Yadgar.

Doesn't seem to difficult (but maybe I missed something?)

The equation of your ellipsoid is:
(x/0.115)^2 + (y/0.108)^2 + (z/0.138)^2 - 1 = 0
The cut surface z==-0.097 is an ellipse, with equation:
(x/0.115)^2 + (y/0.108)^2 + (0.097 /0.138)^2 - 1 = 0
the width at y=0 is given by:
(x/0.115)^2 + (0.097 /0.138)^2 - 1 = 0
<=> |x| = 0.115 * sqrt ( 1 - (0.097 /0.138)^2)
so width = 2 * 0.115 * sqrt ( 1 - (0.097 /0.138)^2)

Hope this helps...

Xavier

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4707fde8@news.povray.org...
> High!
>
> After a six-week break, I returned to my self portrait project... as I
> firstly try to do a "cheap" (in terms of rendering time) CSG version, I
> started with a "skull ellipsoid" to which I would like to add face with
> upper jaw and an animatable mandible, as displayed in the three
> schematical drawings below.
>
> The radii of the "skull ellipsoid" (centered at the origin) are as
> follows: <0.115, 0.108, 0.138>
>
> To fit the front parts in, I vertically sliced off the skull shell at
> z=-0.097. Now I have to determine the width of the cut surface at y=0.
> With a non-stretched sphere, that would be easy:
> sqrt(radius^2-0.097^2)*2*radius - but how to transfer this to an
> ellipsoid? To determine the exact radius length at the intersection of x
> axis, cutoff box and ellipsoid, I would have to know its angle with the
> z axis...
>
> See you in Khyberspace!
>
> Yadgar
>

Hi Yadgar.

Doesn't seem to difficult (but maybe I missed something?)

The equation of your skull ellipsoid is:
(x/0.115)^2 + (y/0.108)^2 + (z/0.138)^2 - 1 = 0
The cut surface z==-0.097 is an ellipse in the xy plane, with equation:
(x/0.115)^2 + (y/0.108)^2 + (0.097 /0.138)^2 - 1 = 0
the width at y=0 is given by:
(x/0.115)^2 + (0.097 /0.138)^2 - 1 = 0
<=> |x| = 0.115 * sqrt ( 1 - (0.097 /0.138)^2)
so width = 2 * 0.115 * sqrt ( 1 - (0.097 /0.138)^2), no?

Hope this helps...

Xavier

Post a reply to this message

> High!
> 
> After a six-week break, I returned to my self portrait project... as I
> firstly try to do a "cheap" (in terms of rendering time) CSG version, I
> started with a "skull ellipsoid" to which I would like to add face with
> upper jaw and an animatable mandible, as displayed in the three
> schematical drawings below.
> 
> The radii of the "skull ellipsoid" (centered at the origin) are as
> follows: <0.115, 0.108, 0.138>
> 
> To fit the front parts in, I vertically sliced off the skull shell at
> z=-0.097. Now I have to determine the width of the cut surface at y=0.
> With a non-stretched sphere, that would be easy:
> sqrt(radius^2-0.097^2)*2*radius - but how to transfer this to an
> ellipsoid? To determine the exact radius length at the intersection of x
> axis, cutoff box and ellipsoid, I would have to know its angle with the
> z axis...

Can't you just use a sphere, fit the other part to it
and then scale it all along an axis so that the sphere
becomes an ellipsoid ?

-- 
Tor Olav
http://subcube.com

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High!

Xavier M. schrieb:

> The equation of your ellipsoid is:
> (x/0.115)^2 + (y/0.108)^2 + (z/0.138)^2 - 1 = 0
> The cut surface z==-0.097 is an ellipse, with equation:
> (x/0.115)^2 + (y/0.108)^2 + (0.097 /0.138)^2 - 1 = 0
> the width at y=0 is given by:
> (x/0.115)^2 + (0.097 /0.138)^2 - 1 = 0
> <=> |x| = 0.115 * sqrt ( 1 - (0.097 /0.138)^2)
> so width = 2 * 0.115 * sqrt ( 1 - (0.097 /0.138)^2)

Meanwhile, I found out a different solution:
as the length of the secant of a circle or ellipse decreases with 
increasing distance to its center point proportionally with the cosine 
of the arcus-sine of distance/radius, the searched length here is 
0.115*2*cos(asin(0.097/0.138)) = 0.163596726 !

But nevertheless, thank you!

See you in Khyberspace!

Yadgar

Now playing: I Think I'm Going To Kill Myself (Elton John)

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Hi Yadgar,


47091c9b$1@news.povray.org...
> High!
>
> Xavier M. schrieb:
>
>> The equation of your ellipsoid is:
>> (x/0.115)^2 + (y/0.108)^2 + (z/0.138)^2 - 1 = 0
>> The cut surface z==-0.097 is an ellipse, with equation:
>> (x/0.115)^2 + (y/0.108)^2 + (0.097 /0.138)^2 - 1 = 0
>> the width at y=0 is given by:
>> (x/0.115)^2 + (0.097 /0.138)^2 - 1 = 0
>> <=> |x| = 0.115 * sqrt ( 1 - (0.097 /0.138)^2)
>> so width = 2 * 0.115 * sqrt ( 1 - (0.097 /0.138)^2)
>
> Meanwhile, I found out a different solution:
> as the length of the secant of a circle or ellipse decreases with 
> increasing distance to its center point proportionally with the cosine of 
> the arcus-sine of distance/radius, the searched length here is 
> 0.115*2*cos(asin(0.097/0.138)) = 0.163596726 !

a little more complicated formula, but the same result, since 
(cos(asin(x)))^2= 1-x^2

> See you in Khyberspace!

By the way what is Khyberspace? The page on your web site has been "under 
construction" for a long time now...

Xavier

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High!

Xavier M. schrieb:

> By the way what is Khyberspace? The page on your web site has been "under 
> construction" for a long time now...

Under construction? Perhaps you mean the particular link "What is 
Khyberspace?", but the homepage in general is reachable via 
www.khyberspace.de !

Yes, it's surely time to finally add a thorough definition of Khyberspace...

Khyberspace is a general term for "virtual Afghanistan", Afghanistan 
(preferably the pre-war country before 1978, but also alternate history 
concepts or even future extrapolation from the current state) as a 
virtual world, either as a "primitive" text-based version like seen in 
Multi User Dungeons (MUDs), a (more or less) photo-realistic 3D world 
(that's where PoV-Ray would be used) or even a realtime-navigable 3D 
world based on VRML/X3D.

Full-blown Khyberspace would be an immense task, since modeling a whole 
country of about 650,000 square kilometres at a reasonable level of 
detail also in "pedestrian" views would take tens of thousands of 
programming years, let alone populating it with credible Afghan 
characters (ideally avatars... er, afghatars played by human users of 
Khyberspace, not just stupid computer-generated creatures).

Three years ago, I started with Kabul (the capital) and its nearer 
surroundings by manually pixeling a high-resolution heightfield from a 
topographical map (attached here some recent renderings and maps - still 
very crude!)... unless I find people to join me, there would be some 
chance to finish at least the city until 2050...

See you in Khyberspace!

Yadgar

P. S. Earlier this year, I already had these (and some more) images 
included on my homepage, but unexpectedly, my free PHP webspace account 
on kilu had been terminated by the company, so currently, my attempts at 
Khyberspace are not on public display...

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