There has been plenty of talk about which player is the best prospect with the 2018 NFL Draft just over a month away. The names usually thrown into the mix are USC quarterback Sam Darnold and Penn State running back Saquon Barkley.
However, Jacksonville Jaguars cornerback Jalen Ramsey has given that distinction to Florida State defensive back Derwin James, according to NFL Network's Tom Pelissero
“(He's the) top player in this draft this year. Should go No. 1 overall, but you know how things go in the draft,” Ramsey said of James. “You never know where he could go … top five, top 10, top 15.”
“If I had the power, it would be done, but the Jags don't have the No. 1 pick,” Ramsey said. “It needs to get done before he gets to the Jags.”
It should come as no surprise to hear Ramsey's pick for the best player in this draft. The two were teammates at Florida State and formed quite the formidable secondary back in 2015 where James had eventually worked his way up to a starting cornerback spot opposite of Ramsey.
Aside from being a fellow alumnus, there is good reason for Ramsey's confidence in James. He registered 84 tackles, 2 interceptions, 11 passes defended last season en route to his second consecutive First-team All-ACC honors.
While he is listed as a safety, James possesses a versatile skill set in terms of coverage. This means he could ultimately wound up switching back to cornerback for his next team. That said, the league would be wise to heed Ramsey's advice going into the draft.
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