# What exactly is the memory impact on the database when reducing cells size in a Raster GIS?

I understand that the size of the memory of a raster database increases when reducing the cell resolution. I read some about it in the ArcGIS Help.

However, I am wondering about the exact relation on the memory footprint and was hoping that someone with more database-experience can tell me whether my reasoning here makes sense:

Given cells of 10x10 meters, and reducing the resolution to 5x5 meters, like so: The relation for the increase of number of cells would be `n*4`, so I thought that the memory increase in the database file would follow a similar pattern.

Here's my attempt to generalize this:

The amount of cells and the memory size of the database both increase by the square of the number that the sides of the initial raster cells are divided by.

E.g. `10/2` (`2` being the divisor, thus `2^2 = 4` being the factor in this case) results in `*4` the amount of cells - so does it also result in `*4` the amount of memory size of the database?

Given that my assumption is correct, dividing a raster cell by 4, e.g. `10/4` (with `4` being the divisor thus `4^2 = 16` being the factor), would result in `*16` the amount of cells, and `*16` the amount of memory size of the database! That would be a large increase of file size pretty quickly!

Is this generalized formula applicable or does the size of the database increase following a different pattern - and if so, which one?

• The inverse-square law is generally correct. The result could differ depending on how the data are stored. For instance, subdividing a categorical grid that is stored using some form of compression might have no appreciable affect on the actual physical storage used. – whuber Apr 11 '17 at 22:54
• I needed to reproject a 4-bit color-mapped raster, and found that resampling at a 1/5 interval was able to preserve the image annotation legibility after reprojection, and was, on average, smaller than the original, poorly compressed raster, not 50 times larger (25 from pixels times 2 for rotation). There are too many permutations of format and compression algorithm and compressibility to make a generic answer to this question (and a specific case would have no usefulness to others). I would also note that memory is not applicable to rasters; it is disk storage which is significant. – Vince Apr 11 '17 at 23:13