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Welcome to the dA Copyright Club!

Wed May 16, 2007, 9:23 PM
Our mission: To create a community effort to report copyright violations on Deviant Art.

What you can do

You don't need to join [but we'd love it if you did!], just make habits of the following:

1. When browsing dA, set aside a little time (just a little!) to skim for obvious violations.
2. Don't be afraid to report violations when you come across them!
3. Don't harass or try to warn violators, just report!
4. Be familiar with dA policy on copyright violations: [link]
5. Check galleries after reporting to make sure the violator isn't a repeat offender (again, DO NOT HARASS the violator, just check silently).

How to join:

1. Watch the club!
2. Put our icon in your journal or shoutbox, and/or put the following in your signature, with the spaces in the URL and the asterisks removed:
<*a href=*"http ://www .copyright-club .deviantart .com">Respect Copyright!<*/a>
3. We will put you on our list of members (unless you request anonymity).

CLUB RULES

1. Always follow dA policy!
2. When in doubt, follow rule number one!

Any questions? Feel free to note or post on the wall. =) A Club FAQ will be created when it becomes necessary.

  • Mood: Pride

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:iconidiottv:
In soviet Russia Copyright has no respect for you!!!

--
This Is Not Apophysis

I don't apologize for my bad grammar because I have a historical reason to despise grammar Nazis.

Crazy westerners.:no:
:iconthe-final-i:
?

--
:
E^2= (p^2c^2)+([Δm^2]c^4)
which implies that
E= [mc^2]/[<sq. root>1-(v^2/c^2)]
when p=0 (in eq 1, or in 2 rearranged to 1])
E=mc^2
[as rest mass cannot be attained, p can never=0, so E should, not always be calculated as = mc^2].
:iconidiottv:
Don't tell me you haven't heard about the Russian reversal.

--
This Is Not Apophysis

I don't apologize for my bad grammar because I have a historical reason to despise grammar Nazis.

Crazy westerners.:no:
:iconthe-final-i:
ok, I googled it.
:|

--
:
E^2= (p^2c^2)+([Δm^2]c^4)
which implies that
E= [mc^2]/[<sq. root>1-(v^2/c^2)]
when p=0 (in eq 1, or in 2 rearranged to 1])
E=mc^2
[as rest mass cannot be attained, p can never=0, so E should, not always be calculated as = mc^2].
:iconthe-final-i:
I don't think I have, :no:
Do explain.

--
:
E^2= (p^2c^2)+([Δm^2]c^4)
which implies that
E= [mc^2]/[<sq. root>1-(v^2/c^2)]
when p=0 (in eq 1, or in 2 rearranged to 1])
E=mc^2
[as rest mass cannot be attained, p can never=0, so E should, not always be calculated as = mc^2].
:iconflutterings:
In russia a burger eats a tasty you!

--
www.kathrynjeanes.com
[link]
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