We’ve already seen a startling example of a surefire Hall of Fame player leaving the only team he’s ever played for in free agency. Depending on how Dwyane Wade‘s meeting with the Denver Nuggets goes today, we could see another before the week is over.
Adrian Wojnarowski of Yahoo! Sports is reporting this morning that Wade, who recently cancelled a meeting with the Milwaukee Bucks, will in fact meet with the Denver Nuggets in New York City today, and the Nuggets will have significant ammo. Denver is prepared to offer Wade a 2-year deal worth $52 million, which is a couple million north of the $50 million Wade has adamantly been seeking, primarily from the team he’s played for his entire career, the Miami Heat.
After rigorous negotiations, Pat Riley and the Heat offered Wade a 2-year deal worth $40 million, but it’s a deal Wade and his team are not tempted by. Wade has never been the highest paid player on his team, and considering the pay cut he took two summers ago to make re-signing LeBron James a possibility–James, as we know, returned home to Cleveland instead–the 3-time champ is likely making up for lost time. Or rather, lost money.
Though Denver is considered a small market, there’s potential there for Wade and the Nuggets front office to give the Marquette University alum one last shot at building a championship contender. Wade would get his money (and then some), and would immediately make Denver an attractive destination for next year’s marquee free agents, like Russell Westbrook and Blake Griffin. (It wouldn’t be the first time Blake Griffin to Denver rumors have swirled.) In addition, Denver’s young core is strong. Last year’s lottery pick, Emmanuel Mudiay, had a promising rookie year, averaging 13 points and a shade under 6 assists per game. There are also high hopes for this year’s lottery pick, Jamal Murray.
No timetable has been revealed yet for today’s meeting, but we’ll be sure to keep you posted on developments.
