*****ing an e-mail route

[Unregistered]

Hello, I have been receiving some abusive e-mails from this route. Is it possible to ***** the origin from the following header:

===============
X-Apparently-To: via 202.4*.2**.*72;

Authentication-Results: mta***.mail.in.********** from=yahoo.co.in;
domainkeys=neutral (no sig)

X-Originating-IP: [202.4*.2**.*57]

Return-Path: <im4uprism@yahoo.co.in>

Received: from 202.4*.2**.*57 (HELO web840*.mail.in.**********)
(202.4*.2**.*57) by mta***.mail.in.********** with SMTP;

Received:(qmail 82448 invoked by uid 6000*);
-0000

Message-ID: <20050*20*5*5*2.82446.qmail@web840*.mail.in.**********>
Received: by web840*.mail.in.********** via

===============

PLEASE HELP.

Much Appreciated.

[Unregistered]

there's a couple of ways you can do it ... perform a Trace Route or you can ping the IP. Follow these steps and you'll know where it came from:

start -> run -> type "cmd" or "command" (if that doesn't work then look for the MS Dos link and click it. You'll need to enter the following in the MS Dos screen;

c:\tracert (senders IP)

press enter

once you've done this IP's will show up ... if you have your's and the senders then you'll know where the e-mails are coming from.

However, there is a much easier way for you to find out who this person is ... go to "Tool Box" on the Net Tools Menu Bar and enter the IP into the "Whois" function and click go. You'll get the name,phone number, and where they are from.

Hope this helps.
TBevand

[Unregistered]

the search turns up this info:

===========

web840*.mail.in.********** (202.4*.2**.*57)

202.4*.**2.0 - 202.4*.22*.255
Internet content provider

Kirk Williams
[email]kirkw@yahoo-inc.com[/email]
70* First Avenue, Sunnyvale, CA *408* USA
+*-408-*4*-64**
+*-408-*4*-**0*

============

which is basically leading nowhere.

However, a search with ARIN Whois turns up this:

=============
OrgName: Asia Pacific Network Information Centre
OrgID: APNIC
Address: PO Box 2***
City: Milton
StateProv: QLD
PostalCode: 4064
Country: AU

ReferralServer: whois://whois.apnic.net

NetRange: 202.0.0.0 - 20*.255.255.255
CIDR: 202.0.0.0/7
NetName: APNIC-CIDR-BLK
NetHandle: NET-202-0-0-0-*
Parent:
NetType: Allocated to APNIC
NameServer: NS*.APNIC.NET
NameServer: NS*.APNIC.NET
NameServer: NS4.APNIC.NET
NameServer: TINNIE.ARIN.NET
NameServer: NS.RIPE.NET
NameServer: DNS*.TELSTRA.NET
Comment: This IP address range is not registered in the ARIN database.
Comment: For details, refer to the APNIC Whois Database via
Comment: WHOIS.APNIC.NET or [url]http://www.apnic.net/apnic-bin/whois2.pl[/url]
Comment: ** IMPORTANT NOTE: APNIC is the Regional Internet Registry
Comment: for the Asia Pacific region. APNIC does not operate networks
Comment: using this IP address range and is not able to investigate
Comment: spam or abuse reports relating to these addresses. For more
Comment: help, refer to [url]http://www.apnic.net/info/faq/abuse[/url]
Comment:
RegDate: ***4-04-05
Updated: 2004-0*-*0

OrgTechHandle: AWC*2-ARIN
OrgTechName: APNIC Whois Contact
OrgTechPhone: +6* 7 *858 **00
OrgTechEmail: [email]search-apnic-not-arin@apnic.net[/email]

================

Which still does not lead to the ORIGIN.

Any advise will be appreciated.

Thanks and Regards.

[Unregistered]

go to the link "getting someones IP address" and post a reply to SynTaX****** and give him your e-mail he'll send you a program that will:

I have written a fun little windows batch script... All you have to know is the ISP of your friend, the State they live in, and either one shared folder on their computer, their computers name, or their printers name.

Not only can you find their computer but you can also drop them a little note on their desktop and freak them out~

If anyone would like the program leave your email here -

hope this helps you better than my last advice.

Regards,
TBevand

[ABHIS]

You're waisting your time & effort. This will lead ABSOLUTELY NOWHERE.
Are you really expecting to proceed further and find names & address?
At most you'll be able to resolve the IPv4 into an Host_address, and then what?
The question that you are posting has been repeated hundred of times on this forum and thousand of times on other different forums.
Unfortunately, all those repeated questions have ONE answer, read it above