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A’Shawn Robinson agrees to 1-year deal worth up to $8M with NY Giants

Rams lose another veteran along the defensive line to free agency

Tampa Bay Buccaneers v Los Angeles Rams Photo by Katelyn Mulcahy/Getty Images

The Los Angeles Rams have lost another veteran defender to free agency. Defensive tackle A’Shawn Robinson has agreed to terms with the New York Giants on a one-year deal worth up to $8M, per Ian Rapoport of NFL Network:

Robinson joined the Rams from the Detroit Lions during the 2020 free agency period. He initially chose to opt out of the season due to COVID-19 concerns, but Los Angeles was able to rework his contract and bring him into the fold mid-season that year. He played a pivotal role in the middle of the Rams’ defensive line during their 2021 Super Bowl campaign.

While the NFL Network reports Robinson’s deal with the Giants as worth “up to $8M”, we may need more details before we can discern what this means for the compensatory pick formula. The reported figure could be inflated by incentive-driven compensation, and these are measured against the salary cap and compensatory picks based on the likelihood that the incentives could be met.

While unofficial, the Rams are slated to receive the following compensatory picks based on their free agent losses so far in 2023. Teams can only receive a maximum of four picks, so Robinson would likely replace one of the lower picks and possibly give the Rams a selection in an earlier round:

  • Matt Gay (Colts); Sixth Round
  • Baker Mayfield (Buccaneers); Sixth Round
  • Nick Scott (Bengals); Seventh Round
  • Greg Gaines (Buccaneers); Seventh Round