[ENet-discuss] Shooter Game

[Daniel Aquino]

So is my calculation correct that for up/down together from
perspective of one client you would use,  2 * (   (N-1)*20*30  )  ?

On Wed, Apr 15, 2009 at 4:53 AM, Lee Salzman <lsalzman1 at cox.net> wrote:
> The ENet header overhead is not bad, per se, but keep in mind if you are not
> bulking stuff into packets, the overhead for my case would have been
> N*(N-1)*30 packet headers, times whatever the size of the headers is. That
> adds up quickly, especially if the headers are >= 20 bytes in size all
> total, so that it would almost double (or worse) the bandwidth usage the
> server was sending out!
>
> Now, with respect to the second question, ISPs place much harsher caps on
> upstream bandwidth than they do downstream, so upstream and downstream bytes
> are not created equal. :)
>
> Lee
>
> Daniel Aquino wrote:
>>
>> On Tue, Apr 14, 2009 at 10:52 PM, Lee Salzman <lsalzman1 at cox.net> wrote:
>>
>>
>>>
>>> Regardless, the biggest consumers of bandwidth and perhaps the most
>>> latency
>>> sensitive data is position packets. In Cube 1/2, each position packet
>>> runs
>>> about 20 bytes for each player (since it carries position, velocity, any
>>> physics/animation state, etc), and it is sent unreliably at a fixed rate
>>> of
>>> 30 times a second. So you end up with a situation where the bandwidth
>>> usage
>>> is  (for N=number of players): N*(N-1)*20*30 bytes  per second, since
>>> each
>>> player must send its  position to every other player.
>>>
>>
>>
>> Do you know how much the overhead there is for udp and enet headers?
>>
>>
>>
>>>
>>> If you went with a peer-peer scheme, the individual load
>>> per peer would then only be (N-1)*20*30, but with greater overhead in
>>> packets/headers, since they can't be bulked together into a single packet
>>> -
>>> each update goes to a different client. But for each peer, this is all
>>> upstream bandwidth.
>>>
>>
>>
>> What did you mean by this?
>>
>> wouldn't the full upstream/downstream bandwidth be   2 * (   (N-1)*20*30
>>  )  ??
>>
>>
>
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