>>> Same chains as today.
>>>
>>>       
>> So, if I place 50 blacklist entries for tun0 and 1 for eth0, then in 
>> order to get a packet through eth0 it has to traverse through 51 entries 
>> in that same chain? "Square pegs in round holes" comes to mind... 
>> Thanks, but no thanks!
>>
>>     
>
> Why? It doing that now.
>   
No, not really. It is only "doing that now" because blacklist entries 
are entered for all interfaces - if/when that changes, I would be able 
to enter blacklist entries for a specific interface.

If you are going to lump up all blacklist entries regardless of which 
interface they are entered for into a single chain (I presume that would 
be the blacklst/blackout chain again as you already pointed out) and you 
reference that chain from all interfaces/zones (as is the case now), 
that means a single packet has to traverse through all entries in the 
blacklst/blackout chain - including entries which have been entered for 
a different interface - before it passes through and that is something I 
am not very keen on doing, quite frankly...Unless I am missing something 
obvious.


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