On Jan 7, 1:17 pm, Markw65 <[email protected]> wrote:
>
> But the CofG as computed by the above is the same for every one of
> those triangles - so I believe the above *is* exact (feel free to
> point out the flaw in my reasoning!).
>
Hmmm... I do see a flaw in the reasoning.
I dont exactly see why that statement is true. Its obviously true if
you *dont* do the normalization step after computing the midpoints.
Not so obvious (to me at any rate!) when you *do* do it.
Thats not to say its false either, but I can see the proof needs some
more work :-)
Mark
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