POV-Ray : Newsgroups : povray.general : A little trig help : Re: A little trig help Server Time
4 Aug 2024 14:27:35 EDT (-0400)
  Re: A little trig help  
From: Alan Walkington
Date: 26 Apr 2003 04:06:17
Message: <3eaa3df9@news.povray.org>
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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