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Ok .. I want to distribute 'things' equaly spaced around a streched disk.
I can do it on the plain disk easily enough:
#local R = 1.75;
#local V =.4;
#local Count = 0;
#local Max = 4;
#while (Count < Max)
#local Theta = Count * (360/Max);
#local X = -R * cos(radians(Theta));
#local Z = -R * sin(radians(Theta));
#local Count = Count + 1;
sphere{<0,0,0>.15 translate <X,V,Z>}
#end
but I really want an oval? .. elipse? .. created by stretching the disk:
cylinder {<0,0,0><0,.5,0> 2 scale <1.5,1,1>}
I can place the 'decorations' equally spaced by degrees on a plain disk, but
what I want is to place them equally spaced by distance on a stretched disk.
Any suggestions appreciated
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Alan Walkington wrote:
>Ok .. I want to distribute 'things' equaly spaced around a streched disk.
>
>I can do it on the plain disk easily enough:
> #local R = 1.75;
> #local V =.4;
> #local Count = 0;
> #local Max = 4;
>
> #while (Count < Max)
> #local Theta = Count * (360/Max);
> #local X = -R * cos(radians(Theta));
> #local Z = -R * sin(radians(Theta));
> #local Count = Count + 1;
>
> sphere{<0,0,0>.15 translate <X,V,Z>}
> #end
>
>but I really want an oval? .. elipse? .. created by stretching the disk:
>
>cylinder {<0,0,0><0,.5,0> 2 scale <1.5,1,1>}
>
>I can place the 'decorations' equally spaced by degrees on a plain disk, but
>what I want is to place them equally spaced by distance on a stretched disk.
>
>Any suggestions appreciated
>
I know this code looks bizarre, but it works pretty good. Try this inside
your loop:
#local Theta = radians(Count * (360/Max));
#local Theta = Theta + 0.1*sin(2*Theta);
#local X = -1.5*R * cos(Theta);
#local Z = -R * sin(Theta);
#local Count = Count + 1;
sphere{<0,0,0>.15 translate <X,V,Z> pigment {color rgb 1}}
Now, if you want an explanation, I can't offer a very good one, but I'll
try.
Basically, you have to compensate for the fact that near Theta=0 and
Theta=pi, the points are too close together.
Near Theta=0, adding 0.1*sin(2*Theta) is pretty close to adding 0.2*Theta.
This means that near Theta=0, Theta is really 1.2*Theta, so the points will
be spread out.
Near Theta=pi/2, adding 0.1*sin(2*Theta) is really close to subtracting
0.2*(Theta-pi/2). So, near Theta=pi/2, Theta is really 0.8*Theta + some
constant.
The effect is, Theta increases at a rate of between 0.8 and 1.2, instead of
at a constant rate of 1. 1.2/0.8 is 1.5, which is how skewed your ellipse
is. So the points should be equally spaced near Theta =0,pi/2, pi, and
3/2*pi. In between, there will be some variance, but it shouldn't be too
bad.
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Thanks
Alan
"Retsam" <nomail@nomail> wrote in message
news:web.3eaa18733bdd876d34dff4bb0@news.povray.org...
> Alan Walkington wrote:
> >Ok .. I want to distribute 'things' equaly spaced around a streched disk.
> >
> >I can do it on the plain disk easily enough:
> > #local R = 1.75;
> > #local V =.4;
> > #local Count = 0;
> > #local Max = 4;
> >
> > #while (Count < Max)
> > #local Theta = Count * (360/Max);
> > #local X = -R * cos(radians(Theta));
> > #local Z = -R * sin(radians(Theta));
> > #local Count = Count + 1;
> >
> > sphere{<0,0,0>.15 translate <X,V,Z>}
> > #end
> >
> >but I really want an oval? .. elipse? .. created by stretching the disk:
> >
> >cylinder {<0,0,0><0,.5,0> 2 scale <1.5,1,1>}
> >
> >I can place the 'decorations' equally spaced by degrees on a plain disk,
but
> >what I want is to place them equally spaced by distance on a stretched
disk.
> >
> >Any suggestions appreciated
> >
>
>
> I know this code looks bizarre, but it works pretty good. Try this inside
> your loop:
>
> #local Theta = radians(Count * (360/Max));
> #local Theta = Theta + 0.1*sin(2*Theta);
> #local X = -1.5*R * cos(Theta);
> #local Z = -R * sin(Theta);
> #local Count = Count + 1;
>
> sphere{<0,0,0>.15 translate <X,V,Z> pigment {color rgb 1}}
>
> Now, if you want an explanation, I can't offer a very good one, but I'll
> try.
>
> Basically, you have to compensate for the fact that near Theta=0 and
> Theta=pi, the points are too close together.
>
> Near Theta=0, adding 0.1*sin(2*Theta) is pretty close to adding 0.2*Theta.
> This means that near Theta=0, Theta is really 1.2*Theta, so the points
will
> be spread out.
>
> Near Theta=pi/2, adding 0.1*sin(2*Theta) is really close to subtracting
> 0.2*(Theta-pi/2). So, near Theta=pi/2, Theta is really 0.8*Theta + some
> constant.
>
> The effect is, Theta increases at a rate of between 0.8 and 1.2, instead
of
> at a constant rate of 1. 1.2/0.8 is 1.5, which is how skewed your ellipse
> is. So the points should be equally spaced near Theta =0,pi/2, pi, and
> 3/2*pi. In between, there will be some variance, but it shouldn't be too
> bad.
>
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