Curt,
        Ok... so you dont own the network, thats the issue, you own 3 out
of 128 ip's. In this case definetly use your isp's gateway (whatever they
suggest you use (usually x.x.x.1), your broadcast will probably be
x.x.x.127, network address would be x.x.x.0, but you dont need to worry
about any of that, and that doenst solve your problem. so... i guess i
forgot your problem? do you want to setup a network radio device using a
machine on one of your static ip's? 

Jamie

On Thu, 25 May 2000 [EMAIL PROTECTED] wrote:

> 
> Well, in my particular case, I have three static IPs in a
> class C block, but my subnet is 255.255.255.128 so it's
> not within the entire class C block.  However, my IPs aren't
> sequential and are as much as 50 apart (sequentially speaking)
> - which leads me to assume that someone else has IPs assigned 
> to them that are in between two of mine.  Does that complicate 
> anything?  Can I still arbitrarily choose one of my three as 
> a gateway for the other two without having to tell my ISP anything?
> 
> Curt
> 
> On Thu, May 25, 2000 at 03:00:43PM -0700, Seth Cohn wrote:
> > At 11:38 AM 05/25/2000 -0700, you wrote:
> > > > Ah.  The three IPs are in a block.  I sit corrected.
> > >I dont believe so... There is no such thing as a block of 3 IP's AFAIK,
> > >you can make a subnet of 8 ip's, inwhich you use 3 of them for your net
> > >(network, gateway, and broadcast, addresses,
> > 
> > Not quite:
> > 
> > on 8, you lose just 2.  that's 6.
> > 
> > check out
> > 
> > http://www.agt.net/public/sparkman/netcalc.htm
> > 
> > that's /29 netting
> > 
> > >         So... probably he has 3 ip's in a block of 256, with a netmask of
> > >255.255.255.0 (which means he has 3 ip's in someone elses network, which
> > >is really quite different!)
> > 
> > 
> > maybe not.  He might have 3 in a a block of 6, or might have 3 in a
> > block of 14, etc...
> > 
> > 2 ips can be blocked as /30 btw, with 4 ips used total.
> > 
> > I guess the real question to ask is, your existing netmask is WHAT?
> > that would answer the question.
> > 
> > Seth
> > 
> 

Reply via email to