The storage of a mipmap is around 4/3 of the original size. Take a 256 * 256 map for example, the lower level mipmaps are from 128 * 128 to 1 * 1. The ratio of the area between levels is 1/4. So the amount of the mipmap is (1 + 1/4 + 1/16 + ... + 1/2^n) = 4/3 (here 2^n is the original size).

upasanachatterjee

Hey just to clarify this point, the sum that you're looking for is actually 4^(-n) not 2^(-n), because each level is split into 4 parts. That way you'll get the desired summation.

The storage of a mipmap is around 4/3 of the original size. Take a 256 * 256 map for example, the lower level mipmaps are from 128 * 128 to 1 * 1. The ratio of the area between levels is 1/4. So the amount of the mipmap is (1 + 1/4 + 1/16 + ... + 1/2^n) = 4/3 (here 2^n is the original size).

Hey just to clarify this point, the sum that you're looking for is actually 4^(-n) not 2^(-n), because each level is split into 4 parts. That way you'll get the desired summation.