On 4/26/02 5:08 AM, "Simon Lamb" <[EMAIL PROTECTED]> wrote:
> My mind has gone to mush and I cannot work out the following: > > I scan a 6X6 frame at 4000ppi and select 48-bit output in Silverfast. I > assume the file size will be: > > (((2.25x4000)x(2.25x4000))x3)/1024 = 237Mb > > However, I am getting a file size of 406Mb. Where is my calculation wrong? Simon. Nothing wrong with your math except it is for 8 bits per channel. 8 bits = 1 byte so the total pixels in and 8 bit/channel file (8 bits/pixel) x channels is the size in bytes. However, 48 bit output is 16 bits per channel or 2 bytes (2 bytes/pixel for each channel). For 16 bit multiply by 2 (237 x 2 = 468 MB in your example). 406 is probably the file saved to disk with compression - the disk storage size will not match the file size in RAM except by coincidence. By the way thatıs MB not Mb, megabytes vs megabits. Wayne -- Wayne Simpkins The Univ of Memphis Dept of Art [EMAIL PROTECTED] ---------------------------------------------------------------------------------------- Unsubscribe by mail to [EMAIL PROTECTED], with 'unsubscribe filmscanners' or 'unsubscribe filmscanners_digest' (as appropriate) in the message title or body
