On Wed, 18 Aug 2004, Daniele Guazzoni wrote:

> - with 802.11b you have a RF bitrate of 11 Mbps wich means a pure IP
> throughput of about 4 Mbps.

The phrase "pure IP" isn't entirely clear since it doesn't say whether the
header is included.

Personally, I like specs of this form to be for the lowest layer where the
overhead is predictable.  For example, with wired Ethernet, each IP packet
(including header) has to add 18 bytes of visible overhead (14 header, 4
CRC) and 20 bytes of invisible overhead (8 preamble, 12 byte times minimum
interframe gap).  That's precisely the structure that runs at the link's
bitrate (except that the interframe gap gets larger at Gb speeds), and one
can use that to compute the theoretical maximum rate for any kind of
packet.

A more complicated example would be DSL as used here (RFC1483 bridging).  
To each IP packet (with header) you add 14 bytes of Ethernet header (no
CRC), and 18 bytes of bridging overhead (32 total).  Divide by 48
(rounding up) to get the number of ATM cells, and multiply by 53 to get
the actual number of raw ATM bytes.  This is what gets sent at the
provisioned line rate (what the SpeedTouch calls "used Atm rate").

So, for 802.11 I'd at least go down to the pseudo-Ethernet link layer, and
perhaps to some lower layer if the encapsulation is well-defined.

> - my personal legal record (within the +20dBm limit) is 50km with 802.11g
> using 1.2m dishes and a 25dB 2.4GHz RX-preamp.

I know someone who's successfully *eavesdropped* on 802.11 from about 25mi
(40km) away, which obviously didn't involve any dish at the sending end.
As Murphy would have it, 802.11 range limitations are much more severe for
the desired parties than for the spies and pirates. :-)

When the crackers can send as well as receive, things can get much
nastier:

	http://www.evilscheme.org/defcon/

					Fred Wright