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"Le_Forgeron" <lef### [at] free fr> wrote in message
news:4cc67e2d$1@news.povray.org...
> Le 26/10/2010 05:09, Steve Martell a écrit :
>> So...
>>
>> All I want to do is start out at <1,0,0> and spiral up to the top of a
>> hemisphere. I want to be able to vary the vertical step up (via angle,
>> of
>> course) and the horizontal step. I can make plain old vertical columnar
>> spirals
>> all day long, but I'm just not getting the translation for the
>> hemisphere. The
>> eqs I have found online aren't quite doing what I want - I get fancy loxo
>> things, but not just a nice spiral.
>
> loxodrom is the spiral on a sphere when driven by only the spiral
> constant angle.
> So if your only constant parameter is the angle, you would get a
> loxodrom (which also mean it never reach the pole exactly)
>
>> If I leave the Phi out of it all, just using the X and Z, I get my nice
>> columnar
>> spirals. Nice if I want to make car springs, that is! I am going to use
>> this
>> to make indents on a golfball, but it is really just a concept struggle
>> at this
>> point. I just want a nice, adjustable, hemispherical spiral! Argh!
>
What you want is this:
***********************************
light_source {
0*x // light's position (translated below)
color rgb <1,1,1> // light's color
translate <-20, 40, -20>
}
camera {
location <0.0, 2.0, -5.0>
look_at <0.0, 0.0, 0.0>
right x*image_width/image_height
}
#declare Theta = 0;
#declare Phi = 0;
#while (Theta < 2*pi*10 )
sphere { <0,0,0>, .05
texture { pigment {color rgb 1 }}
translate <cos(Theta)*cos(Phi),abs(sin(Phi)),sin(Theta)*cos(Phi)> }
#declare Theta = Theta + pi/10;
#declare Phi = Phi + pi/400;
#end
***********************************
You exchanged cos and sin for the Phi.
Further sin(Phi) must not exceed 90 degrees or 0.5*pi.
The only problem with this aproach is that the distance higher up the
hemisphere becomes smaller and smaller.
The solution of Le_Forgeron hasn't that problem.
Succes,
Jaap Frank
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