Friday 26 May 2017 11:37, UK
The New England Patriots have restructured tight end Rob Gronkowski's deal for 2017, giving him the chance to become the NFL's highest-paid player at the position.
The restructure offers Gronkowski the chance to boost his salary for this season from £4m ($5.25m) to as much as £8.3m ($10.75m), depending on how he performs this season.
Gronkowski can earn £8.3m ($10.75m) if he is named an All Pro, plays 90 per cent of the Patriots offensive snaps, has 80 catches or 1,200 receiving yards.
Gronkowski can earn £6.7m ($8.75m) if he plays 80 per cent of the Patriots offensive snaps, has 70 catches, 1,000 receiving yards or 12 touchdowns.
Gronkowski can earn £5.2m ($6.75m) if he plays 70 per cent of the Patriots offensive snaps, has 60 catches, 800 receiving yards or 10 touchdowns.
The 28-year-old is entering the fourth season of a six-year, £42m ($54m) contract extension which keeps him tied to the Patriots through the 2019 season. His contract restructure applies only to the 2017 season.
According to Spotrac, Gronkowski's average salary of $9m per season was the fourth highest in the NFL among tight ends prior to the restructure, trailing Jimmy Graham $10m, Travis Kelce $9.37m and Jordan Reed $9.35m.
When healthy, Gronkowski is widely viewed as the NFL's premier tight end but he has not managed to play a full season since 2011, missing 32 of the Patriots' 92 games over the past five seasons (including playoffs).
His 2016 season ended prematurely when he suffered a lower-back injury against the New York Jets on November 27, with New England going on to win the Super Bowl without him.
The terms of the contract restructure were initially reported by ESPN.
