Christoph Hormann wrote:
> 
> Achill wrote:
> 
>>[...]
>>
>>>
>>>So the vector sum of an arbitrary surface and a sphere at <0,0,0> with
>>>radius 1 is the same as with any other sphere with the radius 1?  Sounds
>>>strange.  Furthermore this means that this sum is not commutative.
>>>
>>
>>The sum of a set with any sphere of the same radius is the same, up to a
>>translation. Moreover, the operation is commutative.
> 
> 
> So the sum of a 
> 
> sphere { <1,0,0>, 0.1 }
> 
> and a
> 
> sphere { <0,0,2>, 0.1 }
> 
> would be a 
> 
> sphere { <0,0,0>, 0.2 }
> 
> ???
> 
> Now that's a really strange idea for combining shapes...


It would give a

sphere { <1,0,2>, 0.2 }

Generally, if you have a convex (see 
http://mathworld.wolfram.com/Convex.html) solid A and skalar values x,y, 
then x*A + y*A = (x+y)*A.

Note also that a translation of a set A by a vector t may be regarded as 
a special sum {t}+A.

Achill

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