Thanks... I need all the tutorials I can find.  However yours
leaves me a little confused...

In your example you have:

#declare  P = function {x*x + y + z*z - 1}

isosurface {
  function { P(x,y*(1.05-y/5),z) }
  ...

Should P be the formula of a sphere?  In which case shouldn't
the function be x*x+y*y+z*z-1?  If that's not a typo I'm really
confused.  Hmmm... ok after looking closer I see you are substituting
in a y*y term for y when you call the function.  But that leaves me
wondering why you only scaled one of the y's?  I'm now curious how
that differes from:

#declare  P = function {x*x + y*y + z*z - 1}

isosurface {
  function { P(x,(1.05-y/5),z) }
  ...

I'll try it and find out.

Also... a more general question...

Is there a difference between these two as used in an isosurface?

#declare  P = function {x*x + y*y + z*z - 1}
isosurface {
  function { P(x,y,z) }
  threshold 0

and

#declare  P = function {x*x + y*y + z*z}
isosurface {
  function { P(x,y,z) }
  threshold 1

Thanks,
Carl



"Mike Williams" <nos### [at] econymdemoncouk> wrote in message
news:FsgJnBAGpCx$EwP7@econym.demon.co.uk...
> Wasn't it Warp who wrote:
> >  Only linear transformations are possible with transformation matrices
> >(scales, rotates and translates are just shortcuts to equivalent
> >transformation matrices).
> >  What you are trying to achieve is a non-linear transformation which
> >is not possible with matrices.
> >
> >  The way to go is most probably to create the cone as an isosurface
> >and then modify the input variables of the isosurface function
> >appropriately.
>
> There's an example of a non-linearly scaled sphere in my isosurface
> tutorial:
>
> <http://www.econym.demon.co.uk/isotut/substitute.htm#nls>
>
>
>
> -- 
> Mike Williams
> Gentleman of Leisure

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