Am I write in thinking a 3D display with no textures will require...
a) 800 x 600 16 bit = 800x600x2 bytes = 960,000 bytes
b) 800 x 600 32 bit = 800x600x4 bytes = 1,920,000 bytes
c) 1024 x 768 16 bit = 1024x768x2 bytes = 1,572,864 bytes
d) 1024 x 768 32 bit = 1024x768x4 bytes = 3,145,728 bytes
As such, a requires a 1 MB graphics card, b and c need 2 meg cards, and d needs a 4 MB card? If I add a texture with mipmapping, that amount of VRAM will be consumed too in the processing of rendering, so a 2 MB graphics card at 800x600 32 bit will have all of 100kb for textures?
Also has anyone an idea of when 4 MBs became the minimum, and PCs after that time period should have 4 MBs?
a) 800 x 600 16 bit = 800x600x2 bytes = 960,000 bytes
b) 800 x 600 32 bit = 800x600x4 bytes = 1,920,000 bytes
c) 1024 x 768 16 bit = 1024x768x2 bytes = 1,572,864 bytes
d) 1024 x 768 32 bit = 1024x768x4 bytes = 3,145,728 bytes
As such, a requires a 1 MB graphics card, b and c need 2 meg cards, and d needs a 4 MB card? If I add a texture with mipmapping, that amount of VRAM will be consumed too in the processing of rendering, so a 2 MB graphics card at 800x600 32 bit will have all of 100kb for textures?
Also has anyone an idea of when 4 MBs became the minimum, and PCs after that time period should have 4 MBs?