Alexander Stohr wrote:
>
> > > (This is based on the idea that a full set of mipmaps packs
> > perfectly to take
> > > up two times the size of the base texture). That's also
> > not true for all
> > > architectures...
> >
> > Ok, that explains a bit. However, in some circumstances we
> > may loose a
> > level. The mipmaps don't double the size, the only increase
> > it by 1/3.
> > Then there are architectures like MGA the can't use all 11 mipmaps.
>
> Okay, me was odd.
> (this always happens when i am thinking with the stomach...)
>
> But now i will try to go into deep and get to final formula of
> (4/3 * Lvl_size)
> for the maximum of expected pixels in the total amount of data
> stored in a squared level stack. Further i will point out to
> the goods and bads of the mentioned bithshift formlua and the
> alternative method of looping. In the end i will point out to
> reverse calculation methods and to specific implementations
> that finally might break or at least complicate the overall
> systematics of pyramid stack calculations.
>
> grafically:
>
> n=2 n=1 n=0
> #### ## #
> #### ##
> ####
> ####
>
> mathematically:
> level width = 2^n
> level size = (2^n)*(2^n) = 2^(2*n)
> size increse ratio per level increment = 4
> = (2^(n+1))*(2^(n+1)) / ((2^n)*(2^n))
> = (2^(n+1-n))*(2^(n+1-n))
> = (2^1)*(2^1)
> = 2*2 = 4
>
> summarized size of all levels = sum(x=0, x=n, 2^(2*x) )
> (sorry, i dont have the math book handy for solving that here)
just FYI the formula is
$$
\sum_{k=0}^n q^k = \frac{q^{n+1}-1}{q-1}
$$
in this case
$$
\sum_{x=0}^n 2^{2x} = \sum_{x=0}^n 4^x = \frac{4^{n+1}-1}{3}
$$
and hence the ratio
$$
r = \frac{4}{3} - \frac{1}{3 4^n}
$$
I hope that this latex code is self-explaining for those who aren't used
to it ;-)
Regards,
Andreas Karrenbauer
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