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... with a different twist this time. Original macro/source by Rich
Allen <ric### [at] ricosweb com> I've only added phi to the equation:
#declare phi = (sqrt(5)-1)/2; // The golden ratio
It was originally traced at 1024x768, it took 4 days 8 hours on a 1Ghz
PIII w/256 Mb RAM. The final photons file was 630 Mb large...
The full size image [about 541k] can be found at
http://home.no.net/azoth/spheres_big.png
Enjoy,
--
------------------------------------------------------------------------
+47 90 62 71 78 DoD#2101, DoDRT#017, NIC#015, PJ#006, OGM#007
azo### [at] dodno, Ducati M600, Clementine Ubesudlet: Aldri eid en J&%#PS.
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Attachments:
Download 'us-ascii' (4 KB)
Download 'glasspheres.png' (186 KB)
Preview of image 'glasspheres.png'
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> ... with a different twist this time. Original macro/source by Rich
> Allen <ric### [at] ricoswebcom> I've only added phi to the equation:
>
> #declare phi = (sqrt(5)-1)/2; // The golden ratio
Hey, neat! I haven't played with this code in a while, can you post your
source so I can see how you've used phi in the equation?
--
Rich Allen
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wrote:
>... with a different twist this time. Original macro/source by Rich
>Allen <ric### [at] ricoswebcom> I've only added phi to the equation:
>
>#declare phi = (sqrt(5)-1)/2; // The golden ratio
Hmm...
I'm certainly not a math expert, but the formula you give above
disagrees with the "Golden Ratio" web pages I've visited. The formula
I see on the web is: Golden Ratio = ( 1 + sqr(5) ) / 2
Your formula subtracts, where this one adds. It seems that someone has
a typo on their hands. I hope it isn't a typo in your original POV-Ray
scene code.
Other than that potential flaw, it's a nice image. I even downloaded
the larger version.
Later,
Glen
7no### [at] ezwvcom (Remove the numeral "7")
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> I'm certainly not a math expert, but the formula you give above
> disagrees with the "Golden Ratio" web pages I've visited. The formula
> I see on the web is: Golden Ratio = ( 1 + sqr(5) ) / 2
Sigh.
The idea of the golden ratio is that it's the number that satisfies the
equation
1/x = x - 1
This is, of course, a quadratic equation, meaning that it has two solutions:
(1+sqrt(5))/2 and (1-sqrt(5))/2. Plug either of these into the above
equation and it is satisfied.
Notice, however, that one of these is the negative reciprocal of the other
one. The useful one is the one that's positive ( (1+sqrt(5)/2 ).
Now, if you have a ratio, taking the reciprocal of that ratio is still
useful; you just have to use it slightly differently in whatever
calculations you do (meaning, divide by it where you would have multiplied
by it and vice versa).
Borge merely took the reciprocal of (1+sqrt(5))/2, which is (sqrt(5)-1)/2
(do the math).
- Slime
[ http://www.slimeland.com/ ]
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On Sat, 25 May 2002 01:45:47 -0400, "Slime" <slm### [at] slimelandcom> wrote:
>Sigh.
>The idea of the golden ratio is that it's the number that satisfies the
>equation
>
>1/x = x - 1
>
>This is, of course, a quadratic equation, meaning that it has two solutions:
>(1+sqrt(5))/2 and (1-sqrt(5))/2. Plug either of these into the above
>equation and it is satisfied.
>
>Notice, however, that one of these is the negative reciprocal of the other
>one. The useful one is the one that's positive ( (1+sqrt(5)/2 ).
>
>Now, if you have a ratio, taking the reciprocal of that ratio is still
>useful; you just have to use it slightly differently in whatever
>calculations you do (meaning, divide by it where you would have multiplied
>by it and vice versa).
>
>Borge merely took the reciprocal of (1+sqrt(5))/2, which is (sqrt(5)-1)/2
>(do the math).
Yeah, but Benge had declared the following, and I quote:
"#declare phi = (sqrt(5)-1)/2; // The golden ratio "
This doesn't give Phi, but it's reciprocal, just as you have said
yourself. All the definitions of Phi that I have seen, declare it to
be equal to about 1.618033989... You get that with this formula:
(1+sqrt(5))/2
The formula (1-sqrt(5))/2 doesn't give the value of Phi. My point was
valid, in that there is a typo here. Did it really screw up his POV
scene? I don't really think so. I assume he treated his declaration of
"Phi" as the reciprocal it really is, and not as actual "Phi", which
it isn't. (I hope that made sense.)
Later,
Glen
7no### [at] ezwvcom (Remove the numeral "7")
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Glen Berry wrote:
> Yeah, but Benge had declared the following, and I quote:
>
> "#declare phi = (sqrt(5)-1)/2; // The golden ratio "
>
> This doesn't give Phi, but it's reciprocal, just as you have said
> yourself. All the definitions of Phi that I have seen, declare it to
> be equal to about 1.618033989... You get that with this formula:
>
> (1+sqrt(5))/2
>
> The formula (1-sqrt(5))/2 doesn't give the value of Phi. My point was
> valid, in that there is a typo here.
I didn't say it was the golden number, but the golden ratio - and the
ratio stays the same if you use '(1+sqrt(5))/2' or '(1-sqrt(5))/2'
That I merely named the variable phi was out of convenience than
anything else.
That said, it really makes a difference of the result if you use
'(1+sqrt(5))/2' rather than '(1+sqrt(5))/2', which may explain why I got
the funny results when trying to use the macro for I really wanted to
use it for. :-)
--
------------------------------------------------------------------------
+47 90 62 71 78 DoD#2101, DoDRT#017, NIC#015, PJ#006, OGM#007
azo### [at] dodno, Ducati M600, Clementine Ubesudlet: Aldri eid en J&%#PS.
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Rich wrote:
>
>
> > ... with a different twist this time. Original macro/source by Rich
> > Allen <ric### [at] ricoswebcom> I've only added phi to the equation:
> >
> > #declare phi = (sqrt(5)-1)/2; // The golden ratio
>
> Hey, neat! I haven't played with this code in a while, can you post your
> source so I can see how you've used phi in the equation?
I only subsituted your #declare rad with #declare phi, and by doing that
the spiral automagically became what you see in the image, but with only
one call to the macro, wher you had three and two of the three were
rotated.
// Original bit by Rich Allen <ric### [at] ricoswebcom>
// I've only added phi to the equation
#declare phi = (sqrt(5)-1)/2; // The golden ratio
#declare dots = 168; // number of spheres in spiral
#declare rstart = 5; // starting radius of spiral
#declare rstop = 0; // ending radius of spiral
#declare loops = 1; // number of times spiral circles the
center
#declare cirstart = 0.3; // starting radius of the spheres
(outer sphere)
#declare cirstop = .05; // ending radius of the spheres
(inner sphere)
#declare tdegs = 360*loops;
#declare dstep = 360/dots;
#declare rstep = (rstart-rstop)/(tdegs/dstep);
#declare cstep = (cirstart-cirstop)/(tdegs/dstep);
#declare r1 = rstart;
#declare c1 = cirstart;
#declare y1 = c1;
#declare deg=1;
#declare spiral = union {
#while (deg < tdegs)
#declare x1 = r1*cos(deg*phi);
#declare z1 = r1*sin(deg*phi);
sphere { <x1,y1,z1>, c1
#if(!FastTexture)
material { M_Glass }
photons { // photon block for an object
target 1.0
refraction on
reflection on
}
#end
#if(FastTexture)
texture { fastTex }
#end
}
#declare c1 = c1 - cstep;
#declare r1 = r1 - rstep;
#declare y1 = c1;
#declare deg=deg+dstep;
#end
}
object {
#object {spiral}
translate <0,0,2.5>
}
--
------------------------------------------------------------------------
+47 90 62 71 78 DoD#2101, DoDRT#017, NIC#015, PJ#006, OGM#007
azo### [at] dodno, Ducati M600, Clementine Ubesudlet: Aldri eid en J&%#PS.
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> Yeah, but Benge had declared the following, and I quote:
>
> "#declare phi = (sqrt(5)-1)/2; // The golden ratio "
>
> This doesn't give Phi, but it's reciprocal, just as you have said
> yourself. All the definitions of Phi that I have seen, declare it to
> be equal to about 1.618033989
Ah. OK. Gotcha.
- Slime
[ http://www.slimeland.com/ ]
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Makes me wish I had paid closer attention in High School. <smile> Thanks
for posting your source!
--
Rich Allen
(Remove SPAM from my address to reply by e-mail)
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Rich wrote:
>
> Makes me wish I had paid closer attention in High School. <smile> Thanks
> for posting your source!
no problem, it really was yours... :-) btw, you should try to render it
with these values:
#declare phi = (sqrt(5)+1)/2; // The golden ratio
#declare dots = 49; // number of spheres in spiral
#declare rstart = 3; // starting radius of spiral
#declare rstop = 0; // ending radius of spiral
#declare loops = 1; // number of times spiral circles the
center
#declare cirstart = 0.9; // starting radius of the spheres
(outer sphere)
#declare cirstop = .5; // ending radius of the spheres (inner
sphere)
Then change the number of dots to, say 41, and marvel at the difference.
Try also 86 and then 89. Weirdness. I am sure there are other numbers
that give the same weirdness.
For the really fun stuff start at 2, 5, 7, 11, 13 dots... see what
happens. The really weird stuff happens at 17, 19, 21 and then, at 23
beauty of math reveals itself. Why the huge difference between 22 and
23?! I cannot explain it, anyone care to take a stab at it?
--
------------------------------------------------------------------------
+47 90 62 71 78 DoD#2101, DoDRT#017, NIC#015, PJ#006, OGM#007
azo### [at] dodno, Ducati M600, Clementine Ubesudlet: Aldri eid en J&%#PS.
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