Am 28.12.2013 14:27, schrieb Louis:
> "Louis" <nomail@nomail> wrote:
>> I also realized that z.z in my example would be a constant 1 instead of the
>> z-input into the function.
>
> I'm confused, will z.z be a constant one in this context, or will it be the the
> z- input just like in an ISO-surface-function?

Outside functions, "z" is a predefined vector constant evaluating to 
<0,0,1>; thus, "z.z" is the z-component of that vector, i.e. 1.

Inside functions without an explicit parameter list, "z" is the third of 
three parameters, which in an isosurface object will be the z-coordinate 
of whatever point is currently being examined. Note that with function 
parameters always being scalars, "z.z" will generate a parse error when 
used inside such a function.

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