>> Well, 56k bps = 56 * 1000 bits per second >> 56000 bps/ 8 = 7000 bps >> 7000 bytes / 1024 bytes (1kB)= 6.8 kB/s >Shouldn't it be 56000 bps /10 = 5600 bytes/sec. (allowing for start and >stop bits)?
I was just illustrating the theoretical modem to modem speed -- start/stop bits are an RS232 thing and are so only really relevant when you're talking about PC<->modem transfers (or more strictly, DTE<->DCE transfers). Of course between compression and line quality it's all pretty variable. -i [EMAIL PROTECTED] http://www.wienand.org ********************************************************************** CAUTION: This message may contain confidential information intended only for the use of the addressee named above. If you are not the intended recipient of this message, any use or disclosure of this message is prohibited. If you received this message in error please notify Mail Administrators immediately. You must obtain all necessary intellectual property clearances before doing anything other than displaying this message on your monitor. There is no intellectual property licence. Any views expressed in this message are those of the individual sender and may not necessarily reflect the views of Woolworths Ltd. ********************************************************************** -- SLUG - Sydney Linux User's Group - http://slug.org.au/ More Info: http://lists.slug.org.au/listinfo/slug
