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Offlinewilshire
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windows question
    #5313508 - 02/18/06 02:14 PM (17 years, 11 months ago)

i have a friend with a computer with windows xp installed. he has an install disc too. he's got a new hard drive and he wants to run windows from the new hard drive and use the old one for storage.

is there issues with registration\licensing with this, or is it just a straightforward install using the same windows license key?


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OfflineGomp
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Re: windows question [Re: wilshire]
    #5313856 - 02/18/06 04:21 PM (17 years, 11 months ago)

sounds like it would be using the same key.. :wink::thumbup:


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InvisibleOJK
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Re: windows question [Re: wilshire]
    #5315704 - 02/19/06 06:28 AM (17 years, 11 months ago)

There should be no issues with licensing, it should install and activate just fine.

It might be worth formatting the old drive before installing windows on the new one, to make the process simpler (to make it easier to tell which installation is which).


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Offlineouterwave
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Re: windows question [Re: wilshire]
    #5356288 - 03/02/06 03:40 AM (17 years, 10 months ago)

you actaully are allowed something like 10 registration attempts per cd-key... but no, even if you had multiple computers running of the same disk, microsoft servers cannot tell that it is more than one machine.

you could just leave the old drive unformatted just incase, just need to set the filesystem to 'system drive' (/s) and make sure the new one is the master, and the old the slave...

dual boot!


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outerwave


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OfflineSeussA
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Re: windows question [Re: outerwave]
    #5356325 - 03/02/06 04:21 AM (17 years, 10 months ago)

> but no, even if you had multiple computers running of the same disk, microsoft servers cannot tell that it is more than one machine.

I am not sure I understand what you are tyring to say here. Micro$oft uses a system where they measure ten (I think) different metrics (items) inside the computer. If any of these items change, one or two points is added to a score. Something like CPU speed change adds one point, while something like a new MAC address adds two points. If the total number of points exceeds a threshold, then the product fails to start with the assumption that the product has been copied from one install to another.


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Offlineouterwave
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Re: windows question [Re: Seuss]
    #5356338 - 03/02/06 04:33 AM (17 years, 10 months ago)

correct, though i really didn't think it monitored MAc addresses and the like. i've done it without problems at work. i though there was more of the 'privacy invading' issues in the past, like when there was all the hubub about processor serial numbers...

if push came to shove, i don't think they would have a valid argument against someone who loves to swap hardware and overclock, etc. it's my 'test machine'... the one time there was a problem, a year ago, my associate just called, and they read his a universal key over the phone that was free from these restrictions...

at least that is my understanding, but i also used to have seven computers at home and prolly lost track of which had what.


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take care,

outerwave


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OfflineSeussA
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Re: windows question [Re: outerwave]
    #5356403 - 03/02/06 05:40 AM (17 years, 10 months ago)

http://www.microsoft.com/piracy/activation_faq.mspx:

Quote:

How does product activation determine tolerance? In other words, how many components of the PC must change before I am required to reactivate?

Common changes to hardware such as upgrading a video card, adding a second hard disk drive, adding RAM or upgrading a CD-ROM device will not require the system to be reactivated.

Specifically, product activation determines tolerance through a voting mechanism. There are 10 hardware characteristics used in creating the hardware hash. Each characteristic is worth one vote, except the network card which is worth three votes. When thinking of tolerance, it's easiest to think about what has not changed instead of what has changed. When the current hardware hash is compared to the original hardware hash, there must be 7 or more matching points for the two hardware hashes to be considered in tolerance. If the network card is the same, then only 4 additional characteristics must match (because the network card is worth 3, for a total of 7). If the network card is not the same, then a total of 7 characteristics other than the network card must be the same. If the device is a laptop (specifically a dockable device), additional tolerance is allotted and there need be only 4 or more matching points. Therefore, if the device is dockable and the network card is the same, only one other characteristic must be the same for a total vote of 4. If the device is dockable and the network card is not the same, then a total of 4 characteristics other than the network card must be the same.




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Offlineouterwave
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Re: windows question [Re: Seuss]
    #5357029 - 03/02/06 10:56 AM (17 years, 10 months ago)

oh tay... i still stand by the availabilty of keys that bypass the hashing scheme...


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take care,

outerwave


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OfflineSeussA
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Re: windows question [Re: outerwave]
    #5360703 - 03/03/06 05:32 AM (17 years, 10 months ago)

> i still stand by the availabilty of keys that bypass the hashing scheme...

It wouldn't surprise me.

Edit: that they exist, not that you would still stand by the statement. :grin:


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