The Lombardi Award is an honor handed out annually to either a lineman on either side of the ball or a linebacker who lines up no deeper than five yards off the line of scrimmage. This year, the finalists will have a decided end lean to them.
Tuesday, The Rotary Club of Houston announced the four finalists for this year’s award, and all four are defensive ends: Ohio State’s Joey Bosa, Texas A&M’s Myles Garrett, Clemson’s Shaq Lawson and Penn State’s Carl Nassib. Garrett is the only sophomore in the group, while Bosa and Lawson are juniors. Nassib is a senior who is a former walk-on for the Nittany Lions.
Nassib leads the country in sacks with 15.5, while Garrett is tied for fourth with 10.5. Nassib also is tops in tackles for loss with 19.5.
All four are actually in the top 10 in tackles for loss. In addition to Nassib, Lawson is tied for second (17.5), Garrett tied for fourth (17) and Bosa sixth (15).
Bosa is a finalist for the second consecutive year, and, as the OSU sports information department points out, he’s the fifth Buckeyes to be a two-time Lombardi finalist: linebacker Chris Spielman (1986-87), offensive tackle Orlando Pace (1995-96), linebacker Andy Katzenmoyer (1997-98) and linebacker James Laurinaitis (2007-08). The last OSU winner was A.J. Hawk in 2005.
Clemson has a finalist for the second straight year, with Lawson joining former teammate and fellow defensive end Vic Beasley. Nassib is the first finalist from Penn State since Paul Posluszny in 2005 and 2006.
Garrett is looking to become A&M’s second Lombardi winner, joining Dat Nguyen in 1998. Three other Aggies have been named finalists: Ed Simonini in 1975, Robert Jackson in 1976 and Sam Adams in 1993.
Arizona linebacker Scooby Wright won last year’s award.