Calculate the base level
Multiply width by height to obtain pixel count, then multiply by bytes per pixel. Divide bytes by 1,048,576 for MiB. In this guide, 1K means 1024 pixels on each side. The square RGBA8 sequence is 4 MiB at 1K, 16 MiB at 2K, 64 MiB at 4K and 256 MiB at 8K.
Doubling both dimensions quadruples the pixel count. Moving from 4K to 2K therefore reduces this base allocation by 75 percent. That is a mathematical size relationship, not a promised frame-rate improvement.
Account for mipmaps and formats
A full mip chain stores successively smaller versions. The series 1 + 1/4 + 1/16 and so on approaches 4/3, so the overhead is approximately one third for a large square texture. A partial chain has different overhead.
RGBA8 is only one representation. Floating-point textures, compressed GPU formats, driver allocation and application copies change actual usage. Keep the format assumption next to every estimate and measure the target application when memory is critical.
Choose detail from the final camera
A texture does not need the same resolution for a thumbnail and a close-up of an engraved label. Inspect the most demanding intended view and the portion of UV space occupied by that detail. A large image can still look soft when the visible label uses only a tiny region.
Compare two versions with identical camera, lighting and compression settings. Inspect diagonals, small lettering and specular movement. Keep the smaller version only if it preserves the detail needed for that delivery. Avoid using a single arbitrary resolution for every map.
Keep download and runtime budgets separate
PNG and JPEG compress the transferred image. The runtime decodes it, so a low-megabyte download can expand considerably. Conversely, GPU texture compression can change the runtime footprint independently of the source image dimensions.
Record downloaded bytes, decoded-format estimate and observed device behaviour as separate columns. This prevents a smaller file from being mistaken for proof that a mobile rendering failure has been solved.