If I get it correctly, using mip map normaly require a 33% increase in storage space, as the sum of the areas is 1/4 + 1/16 + 1/64 + ... which converges to 1/3.
rsha256
That math checks out to me, as we sum over 4−n.
armaangoel78
And I imagine that if textures are sitting on disk and not in use, we can probably compress them pretty efficiently, which could maybe get us to less than 33%. Though I guess realistically in this case might as well delete them entirely and recalculate when you need them.
If I get it correctly, using mip map normaly require a 33% increase in storage space, as the sum of the areas is 1/4 + 1/16 + 1/64 + ... which converges to 1/3.
That math checks out to me, as we sum over 4−n.
And I imagine that if textures are sitting on disk and not in use, we can probably compress them pretty efficiently, which could maybe get us to less than 33%. Though I guess realistically in this case might as well delete them entirely and recalculate when you need them.