Patriots' Rob Gronkowski could be NFL's highest-paid tight end if he hits new contract incentives

The New England Patriots, again, sweetened the deal for all-pro tight end Rob Gronkowski.

Patriots TE Rob Gronkowski is entering his ninth NFL season.

The maximum value of his contract for 2018 now sits at $13.05 million after New England added $1 million in per-game bonuses and $3.3 million in potential incentives. If Gronkowski meets or surpasses three of the following four targets, he will earn $1.1 million for each: 80% of playing time, 70 receptions, 1,085 yards, and/or nine touchdowns. Those terms were first reported by ESPN and confirmed by USA TODAY Sports.

Gronkowski’s agent, Drew Rosenhaus, confirmed the revised contract on Instagram, posting a photo tagged at New Jersey’s Newark Liberty International Airport. The Patriots were in the area ahead of their Week 4 preseason game against the New York Giants.

More:Jaguars' Jalen Ramsey on Patriots' Rob Gronkowski: 'I don't think Gronk's good'

More:Eight NFL coaches who could quickly land on the hot seat in 2018

The numbers for those incentives are not coincidental. Last season, Gronkowski caught 69 passes for 1,084 yards and eight touchdowns.

In May 2017, after an injury-shortened 2016 season, Gronkowski agreed to a similarly structured revision, boosted with multiple tiers of incentives.

Gronkowski, 29 and entering his ninth season in the NFL, remains one of the toughest matchups in the league. If he reaches the newly agreed incentives, he will be the highest-paid tight end in the NFL.


Follow Lorenzo Reyes on Twitter @LorenzoGReyes