I understand that this is an All-star game and everyone on the NL roster is very good, but Utley leads the starting roster in batting average, RBIs, and extra base hits. And as stated by John Kruk today on ESPN, "Chase Utley is the toughest to out for the AL pitchers." David Wright (Mets) and Russell Martin (Dodgers) are both hitting ahead of him which is ridiculous. Tony LaRussa must have pick the lineup out of a hat!
2007-07-10 10:11:26 · 18 answers · asked by mattybeck12 1 in Sports ➔ Baseball
Cuz there are 7 better players ahead of him.
2007-07-10 10:14:43 · answer #1 · answered by Dr.Strangelove 2 · 3⤊ 1⤋
Well let me ask you this then.... Why is Palanco hitting 8th for the AL? He is impossible to get out, yet Leyland buries him behind 7 other great players. That's just it, It's an All-Star game. Eveyone in that starting line-up deserves to be there because they are exceptional hitters. You could switch the batting orfer around until you are blue in the face and no matter how you look at it you will still have a great order. I'm in the same thinking as you, Utley is a tremendous player who deserves to get some props, but he still has many years ahead of him. This could be the last time we see Barry in this game. His hometown fans want to see him... maybe batting second he can sneek in 3 AB's. Griffey is almost done too. Utley's time will come, but he's not the name that some of these other players are. A second school of thought is that he is batting 8th to be the second clean-up hitter, someone to bat in front of the pitcher that you know can hit the ball with runners on.
2007-07-10 17:26:05 · answer #2 · answered by lombardiandhispackage 2 · 0⤊ 0⤋
LaRussa is trying to throw the fear of God into the AL with Barry batting second. Also working on the theory that maybe he can get Barry three AB's in front of the home crowd.
2007-07-10 17:29:00 · answer #3 · answered by bobdanailer69 3 · 0⤊ 0⤋
Once the game gets started, it no longer matters what number you are in the order. Utley could wind up being the leadoff hitter in the 2nd or 3rd inning, which could be what LaRussa has in mind.
2007-07-10 18:22:00 · answer #4 · answered by Anonymous · 0⤊ 0⤋
The lineup is alternating left handed and right handed batters. And it's an ALL-STAR game lineup, why do you care what order people bat in? He'll get two at bats and go to the bench, regardless of batting order.
2007-07-10 17:15:01 · answer #5 · answered by dime_western 3 · 2⤊ 0⤋
The reason I think is because Tony wants Chase to be in the bottom of the order to protect Jake Peavy in the 9 spot.
2007-07-10 18:18:14 · answer #6 · answered by Atay31 2 · 0⤊ 0⤋
well yea he is a good batter but he doesn't have that much speed so he cant bat 1st or 2ND because That's for The fast runners and the 3rd -6th batters are power hitters i would have batted him 7th in the all star game
2007-07-10 18:49:42 · answer #7 · answered by mike 1 · 0⤊ 0⤋
fielder has 70 rbi utley 68 their all good ball players and its just the all-star game so who cares
and garret anderson was not voted in
and kruk is just a fat blow hard he don't know squat
2007-07-10 17:17:21 · answer #8 · answered by SLOMO 5 · 1⤊ 0⤋
It's not my man's fault. It is LaRusa's fault. He makes the batting order based on what he feels the need to win. I would just be happy to be on the All-Star team and even more happy to be starting. So don't hate the player, hate the manager.
GO PHILLIES!!!!!!
2007-07-10 17:24:59 · answer #9 · answered by Mr. Pink ♫Lance Uppercut♫ 5 · 1⤊ 0⤋
I agree with what your saying, but I dont understand why Garret Atkins isnt playing at third. The guy is on fire lately. The coach must be an idiot to be playing them like that. Chase Utley is on my fantasy team and is one of my best players. Better than bonds is and griffey was.
2007-07-10 17:15:13 · answer #10 · answered by egethepege 3 · 0⤊ 3⤋
He gets voted to the All-Star team by the fans. Then the coaches decide they want him to bat 8th. He then bats 8th. There you go...that's how he bats 8th.
2007-07-10 17:32:35 · answer #11 · answered by Veritas et Aequitas () 7 · 0⤊ 0⤋