Posted by: Teacherman () on Thu Apr 22 06:13:15 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

Did you ever stop to think that the swing arc is not a circle. What does that do to your calculations?