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Phil Hughes Should be Traded -- Now

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If this be a lost year, let us make the most of it.

Nick Laham

I'm not going to write a huge exegesis on this. This can be boiled down to three points:

1. Hughes is in his third year of arbitration eligibility. He's about to get expensive for his level of production, and he'll be a free agent at the end of the season. Given the team's current austerity kick and Hughes' inconsistent record, are you really going to pony up for the kind of multi-year contract that is going to keep this guy from exploring his options? I don't think so.

2. At the moment, he holds the crown for the most extreme fly-ball guy in the bigs and is a home run machine. That's not a new thing, and it won't change so long as Yankee Stadium plays the way it does. Parenthetically, Hughes was much better at home than on the road this year, but that was a fluke -- his home-run rate was still crazy high, but his BABIP at home was .251. It was .321 on the road. He's going to give some of that back, and what were solo shots in the Bronx will become three-run shots. (He may also get a few of those home runs back as his home-runs-per-fly-ball rate peaked last year, but the two regressions may wash each other out.) Hughes would be a whole other pitcher in one of the league's bigger parks.

3.Given the inconsistency that results from Hughes' up-up-and-away approach, he might not ever be more than a league-average pitcher for the Yankees. Yet, if the perception of his upside possibilities still exists around the league -- that Hughes is a quintessential "change of scenery" guy -- the Yankees might be able to get something or something(s) good for him. The Yankees are not exactly deep in starting pitching just now, but if 2013 is going to be a transition year, why not make the most of it?

What I don't know is what a reasonable return on Hughes would be. I mentioned this story to a colleague and he immediately said, "Michael Morse?" who is clearly the soup of the day now that Adam LaRoche has re-signed with the Nationals. That seems to me to be an insufficient return given Morse's limitations, but I could be wrong.