Basically:
log2(largetsMipSize)+1
log2(256) + 1 = 8 + 1 = 9
Show differencesHistory of post edits
#4TiagoCosta
Posted 19 January 2013 - 06:59 AM
<p>Basically:</p>
<p> </p>
<p>log<sub>2</sub>(largetsMipSize)+1</p>
<p> </p>
<p>log<sub>2</sub>(256) + 1 = 8 + 1 = 9</p>
<p> </p>
<p>log<sub>2</sub>(largetsMipSize)+1</p>
<p> </p>
<p>log<sub>2</sub>(256) + 1 = 8 + 1 = 9</p>
#1TiagoCosta
Posted 19 January 2013 - 06:55 AM
Basically:<br /><img data-cke-saved-src="http://latex.codecogs.com/eq.latex?log_2(largestMipSize)+1" src="http://latex.codecogs.com/eq.latex?log_2(largestMipSize)+1" /><br /> <br /><img data-cke-saved-src="http://latex.codecogs.com/eq.latex?log_2(256)+1%20=%208+1%20=%209" src="http://latex.codecogs.com/eq.latex?log_2(256)+1%20=%208+1%20=%209" />