Warp wrote:
> 
> Christopher James Huff <chr### [at] maccom> wrote:
> > You are correct (assuming no uneven scaling or shearing), but I think a
> > box would be a better choice. A sphere doesn't fit a tetrahedron that
> > closely, a box isn't much if at all better but is faster to test.
> 
>   Have you actually made the math, which shape "wastes" more space when
> bounding (optimally) a regular tetrahedron ("regular" meaning all sides
> have the same length), the sphere or the box?
>   I wouldn't be so sure that the optimal box is smaller than the optimal
> sphere ("smaller" meaning that its volume is smaller), although I can't
> be sure of the contrary either.

In my example one of the tetrahedrons is regular, and exactly fits
inside a 2X2X2 box.  Both exactly fit in a sphere with radius sqrt(3).  

object{Tet_prism(<1,1,1>,<-1,1,-1>,<1,-1,-1>,<-1,-1,1>) pigment{Green}
finish{Dull}}
box{-1,1 pigment{color rgbft<1,0,0,0,.6>}finish{Dull}}
sphere{0,pow(3,(1/2)) pigment{color rgbft<0,0,1,0,.8>}finish{Dull}}
sphere {<1,1,1>,.1 pigment{Yellow}}
sphere {<-1,1,-1>,.1 pigment{Yellow}}
sphere {<1,-1,-1>,.1 pigment{Yellow}}
sphere {<-1,-1,1>,.1 pigment{Yellow}}

Don't know about using a box to bound an irregular tetrahedron.  Sounds
relatively tricky.  

>   Also I think that a ray-sphere intersection is faster than a ray-box
> intersection, so the sphere is in that sense a better bounding object.
> 
> --
> #macro N(D)#if(D>99)cylinder{M()#local D=div(D,104);M().5,2pigment{rgb M()}}
> N(D)#end#end#macro M()<mod(D,13)-6mod(div(D,13)8)-3,10>#end blob{
> N(11117333955)N(4254934330)N(3900569407)N(7382340)N(3358)N(970)}//  - Warp -


-- 
Dan Johnson 
http://www.livejournal.com/userinfo.bml?user=teknotus
http://www.geocities.com/zapob

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