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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
Post a reply to this message
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Download 'schaedelkonstruktion_grob_seitlich.png' (43 KB)
Download 'schaedelkonstruktion_grob_frontal.png' (48 KB)
Download 'schaedelkonstruktion_grob_aufsicht.png' (5 KB)
Preview of image 'schaedelkonstruktion_grob_seitlich.png'
Preview of image 'schaedelkonstruktion_grob_frontal.png'
Preview of image 'schaedelkonstruktion_grob_aufsicht.png'
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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
Post a reply to this message
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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
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> 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)
Post a reply to this message
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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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Attachments:
Download '2007-02-13_kabulmodul-00001smooth.jpg' (29 KB)
Download '2007-02-13_kabulmodul-00002smooth.jpg' (25 KB)
Download '2007-02-13_kabulmodul-00003smooth.jpg' (74 KB)
Download '2007-02-13_kabulmodul-00004smooth.jpg' (49 KB)
Download '2007-02-13_kabulmodul-00005smooth.jpg' (35 KB)
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Preview of image '2007-02-13_kabulmodul-00005smooth.jpg'
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From: Jörg 'Yadgar' Bleimann
Subject: The Khyberspace maps... (Re: Geometrical Problem)
Date: 7 Oct 2007 19:05:20
Message: <47096630@news.povray.org>
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...I had to split the message due to the 1 MB limit!
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Attachments:
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Download 'kblmap00001.png' (139 KB)
Download 'kblmap00002.png' (139 KB)
Download 'kblmap00004.png' (137 KB)
Download 'kblmap00005.png' (135 KB)
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Preview of image 'kblmap00001.png'
Preview of image 'kblmap00002.png'
Preview of image 'kblmap00004.png'
Preview of image 'kblmap00005.png'
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