Gronkowski's restructured contract could make him NFL's highest-paid TE

The era of Gronk rolls on. New England Patriots tight end Rob Gronkowski has agreed to a contract restructuring that gives him the opportunity to become the NFL’s highest-paid tight end, according to ESPN. Gronkowski could more than double his current salary of $5.25 million, to $10.75 million, if he hits all his incentives.

The contract breaks down into three tiers, as follows:

• First, a $10.75 million payout if Gronkowski records 90 percent playing time, 80 catches, 1,200 receiving yards, or is named to the All-Pro team.

• Second, $8.75 million if Gronkowski totals 80 percent play time, 70 catches, 1,000 receiving yards, or 12 touchdowns.

• Third, $6.75 million if Gronkowski hits the marks of 70 percent playing time, 60 receptions, 800 receiving yards, or 10 touchdowns.

Gronkowski has averaged 73.6 receptions, 1,108 yards, and 12.4 touchdowns per 16 games over the course of his career. He’s been named to four All-Pro teams.

If Gronkowski has a weakness, it’s staying on the field. He’s played in all 16 games only twice in his career, none since the 2011 season. He played in 15 games in 2014 and ’15, but missed significant time with injury last season and in two previous ones. Including the playoffs, Gronkowski has missed 30 of the team’s 92 games over the past five seasons.

Gronkowski’s current total cash value prior to the restructuring placed him 18th among tight ends. Jimmy Graham of Seattle ranks first at $10 million, with Arizona’s Jermaine Gresham and Green Bay’s Martellus Bennett, Gronkowski’s old teammate, right behind.

Rob Gronkowski of the New England Patriots. (Getty Images)

Jay Busbee is a writer for Yahoo Sports and the author of EARNHARDT NATION, on sale now at Amazon or wherever books are sold. Contact him at or find him on Twitter or on Facebook.