Posted by: grc () on Thu Apr 22 06:49:59 2004

Hi Jeff M. and GRC,
>
> All my calculations take into account the angular displacement which the bat must traverse during the swing. The formula of finding the diameter of a circle in the following:
>
> d = 2(pi)r
>
> where d represents 360 degrees of angular displacement, assuming that the center remains fixed, and r represents the initial bat length.
>
> If the bat-head rotates in a semicircle to a centerfield contact position, we must calculate a fractional part of the entire angular traversion (i.e., 360 degrees), leaving us with the following:
>
> d = 2(pi)r(x/360)
>
> where x represents a fractional part of the entire diameter (i.e., x equals 180 degrees).
>
> Now, suppose we make x a constant, and use y to represent the additional degrees that are needed to rotate past x to induce "dead" pull-hitting:
>
> d = 2(pi)r[(x+y)/(360)].
>
> As for my calculation on percentages, I will crystalize them so I do not appear too cryptic in explanation.
>
> Note: these calculations also assume a fixed center, and uses only two bat lengths--one, a 34-inch bat, and, the other, a 42-inch bat.
>
> On inside pitches, assuming that collision occurs at the same x+y degree location for all pitch locations, we can calculate the constant, or the percent increase of using the longer lever over the shorter one:
>
> 42 in. - 34 in. = 8 in. ; (8 in./34 in.)(100)=23.53%
>
> This means that, on all pitches of angular displacement, the 42-inch bat will sweep in a 23.53% wider arc than the 34-inch bat.
>
> Since the entire plate is 17 inches, hitting pitches over the plate requires increasing the radial arm to cover half the plate size, or by 8 1/2 inches (i.e., 17/2 inches):
>
> 34 in. + 8 1/2 = 42 1/2 in. ; 42 in. + 8 1/2 in. = 50 1/2 in.
> 50 1/2 in. - 42 1/2 in. = 8 in.
> (8 in./42 in.)(100) = 18.82%
>
> This means that, on pitches over the middle of the plate, on all points of angular displacement, the 42-inch bat will sweep in an 18.82% wider arc than the 34-inch bat.
>
> Finally, outside pitches require a lever arm of the entire 17 inches, or the length of the plate:
>
> 34 in. + 17 in. = 51 in. ; 42 in. +17 in. = 59 in.
> 59 in. - 51 in. = 8 in.
> (8 in./51 in.)(100) = 15.69%
>
> This means that, on pitches on the outside corner, on all points of angular displacement, the 42 inch bat will sweep in a 15.69% wider arc than the 34 inch bat.
>
> BHL

bhl, i don't have the math background so i will defer to your calculations and take them at face value...my questions are these: in referring to one of my previous posts sing that you stand 3 feet from the plate and hit an outside "pitch" off the tee.....

1) do you agree or disagree that you (A) require a wider arc and (B) this results in a flater arc?

2) how much less, if any force do you seem to be applying?

3) using the notion of the ball being "farther in" vs "out in front" as a reference point, do you see the point where the wrists start wanting to roll changing relative to this reference point as compared to moving another 1 foot closer to the plate and making the same evaluation?

i think i understand your overall point: longer bat enables a wider arc...my questions focus on the notion that while a wider arc enables the bat to more easily "reach" the ball, i submit to you that there may be negatives to having a wider arc........eg., hands farther from body = less force applied because the hands are doing more of the work and the body less of the work....eg.,a flater arc makes it more difficult to contact the ball on the "upswing" portion of the swing....eg., ball has to be contacted "farther in", because the farther the hands are from the ball, the sooner the wrists will want to roll....this necessitates hitting the ball when it is farther in, and at that moment of the swing, the swing is still in it's "downswing" mode.....