On Thu, Jan 30, 2014 at 10:58 AM, Alexandros Drymonitis <adr...@gmail.com>wrote:
> > On Thu, Jan 30, 2014 at 6:36 PM, Charles Z Henry <czhe...@gmail.com>wrote: > >> If you want to use a contribution from both of your axes, you can just >> sum them together. (x+y)*sqrt(2)/2 is just a projection along the line >> x-y=0 >> > Let me correct myself: the line is x+y=0 > Can't really try it right now, but just to be sure, the last equation is > to be interpreted like this: (x+y)*(sqrt(2)/2) or like this: > ((x+y)*sqrt(2))/2? > > It's the same either way. That's only needed if you want to be exact in the projection value.
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