But how much will the Nationals benefit having Werth bat second? Dave Sheinin, on a whim, tackled that question last night. He looked up how many plate appearances each spot in the top half of the Nationals’ lineup received in 2010. This is what he found:

1: 743

2: 722

3: 713

4: 697

5: 682

6: 668

First, an aside: this explains why speed should be considered a bonus, and not even close to a determining factor, when choosing a leadoff hitter. If you put a hitter with a low on-base percentage first, you are only giving him more chances to make outs.

Now, the real point of Sheinin’s research. Over a full season, Werth would receive 25 or 30 more plate appearances hitting second as opposed to say, fourth. That works out to about one per week.

So here’s a question: how many outs will the Nationals save by Ankiel hitting sixth instead of second?

Over the past two seasons combined, Ankiel punched up a .298 on-base percentage, which is 1) pretty much unplayable and 2) four points higher than Ivan Rodriguez’s last season. The Nationals are certainly counting on Ankiel to improve that mark, but let’s wok with .298.

Now, Ankiel will not play everyday, since Jerry Hairston will play center against lefties, roughly one-third of the games. So let’s reduce the No. 2 plate appearances to 477 and No. 6 plate appearances to 440. Now divide those by .702 (which is 1-.298).

So by our methodology, Ankiel would make 25 fewer outs by batting sixth rather than second over a full season.

I am assuming that math is terribly misguided. Someone smarter than me can probably make this point in a more sophisticated way. But this is why batting Werth second seems like a good idea: It gives a hitter who is good at getting on base more chances to do that, and it gives a below-average on-base hitter less chances to make outs.


Remember Jack McGeary, the draft pick who chose a $1.8 million signing bonus AND college? Now he’s done with school, recovering from Tommy John surgery and ready for the next chapter.