Re: Re: Re: Re: Flightpath
Posted by: Madmax (
) on Fri Dec 8 19:33:14 2000
In a way I agree with you. The make better contact you should be inline with the flight of the ball. But what about power, according to a Home Run simulator, if you bat speed is 90MPH, with no wind, and the ball comes off the bat around 11 degrees the ball will only go about 130 feet . So how does McGwire hit high 450 foot homers?
> > Madmax,
> > I am not a physicist, just a high school coach. But if the bat was moving upward at 11 degrees, the ball was moving downward at 11 degrees, and you hit it "dead center" (ball exiting at 11 degrees), you might get the effect that you referred to (ball only traveling 130 feet). But if you hit the ball just under dead center (would depend on how much) the ball would exit at a higher angle and would have backspin, thus creating more lift and distance. I don't know what McGwire's bat speed is, but is it possible that he could achieve those monster home runs in this way? Or, does he have to slightly "cross" the flightpath of the ball (from underneath) to get the lift and distance needed? I'm not sure. That's why I am posing these questions.
> > My argument with others (on another site) had to do with hitting the "top half" of the ball and how that was achieved. We all agreed that line drives gave a hitter the best chance to reach base and unless you were a power guy, they preferred to teach their hitters to hit the top half of the ball (better chance to keep the ball out of the air). Their thinking was that you should take your hands "directly" to the ball and aim for the top half. Seems like to me that this flightpath would produce a "cut" ball (bat's flightpath moving across (from above) the ball's flightpath).
> > Whether you get topspin (hit just above dead center), no spin (dead center), or backspin (just under dead center), my point was; to give yourself the best chance to make solid contact, you would want the flightpath of the bat "inline" with the flightpath of the ball just before and at contact. The farther you stray, the more precision required.
> > Am I way off on my thinking? Are there any studies on this? Is there any video proof? I have read a lot of posts on here from people that use a lot of "physics" terminology. What are your thoughts?
> > there was a book written by bahill & someone called "keep your eye on the ball"...the authors are physicists & much of the science & math was over my head......but there was a particular that really intrigued me....they presented a scientific analysis which showed that a ball struck in the middle would have a certain exit trajectory & a certain amount (or lack of) backspin.....logically enough, a ball struck 1/2 inch above center would be a ground ball, 1 1/2 inch below center would be fouled back, etc....their conclusion was that there is an optimal point where the ball can be struck & have maximum backspin as well as optimal exit trajectory....i think they said this optimal point is 1/2 to 3/4 inches below center................my point is this: (1) is there indeed an optimal point of the ball to contact ( i think there is)......and (2) could it be that certain major leaguers have figured that out AND are more sucessful than others in contacting the ball at that optimal point?........most major leaguers hit the ball hard...some harder than others but nevertheless thay all hit the ball hard............aside from other factors such as rotation, weight shift, brute strength, etc, maybe the singles/doubles hitter aims for the center of the ball......and maybe the homerun hitters aim for slightly below center......afterall, how many times have you seen a powerful shot that was a double...and it was a double instead of a homerun simply because of the trajectory?.....just my two cents....respectfully, grc.........
Curt, grc good posts!
My question is, if the ball goes the furthest if you hit the ball 1/2-3/4 inches below dead centre, using rotation and top hand torque, should we go for dead centre, or slightly under the contact point?
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