Paul Plans His Own Convention
His gathering, also in Minn., takes place on second day of GOP convention
By Jonas Oransky,  Newser Staff
Posted Jun 10, 2008 5:35 PM CDT
Republican presidential hopeful Rep. Ron Paul, R-Texas, greets supporters.   (AP Photo/Thomas Whisenand)
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(Newser) – John McCain won’t be the only GOP candidate throwing himself a party in Minnesota the first week of September—libertarian-leaning Ron Paul will host a mini-convention of his own in Minneapolis. Paul's gathering will take place on the second day of the Republican convention, just a few exits down I-94. He hopes his 11,000-strong meeting will “send a message” to the GOP, a rep told the Pittsburgh Tribune-Review.

The candidate shocked watchers by raising $35 million for his campaign. He won 35 delegates to that other convention but was not invited to speak because, his campaign says, he will not endorse McCain. “There is a growing surge of people out there just craving” a return to “limited government that places personal liberty first," said Paul's spokesman.