Posted by: Jack Mankin (MrBatspeed@aol.com) on Fri Dec 31 11:42:02 1999

> Hello, Everyone
>
> I promised some more info on the bat-ball impact issue, so I figured it's about time I posted it (been a little pre-occupied.)
>
> The basic formula for ball-exit velocity is as follows:
>
> Velocity (MPH)= M1/(M1+M2)*((M2/M1-COR*(-V2)+(1+COR*V1)
>
> Where:
>
> M1= Bat Mass (oz)
> M2= Ball Mass (oz)
> V1= Bat Velocity (MPH)
> V2=Ball Velocity (MPH)
> COR= ball coefficient of restitution (0.55@ 60 MPH.)
>
> The COR is not constant, and in fact will be reduced considerably from its nominal value at high impact speeds and as a result of impact with a round object (bat.)
>
> Taking for example Mark McGwire's 538- foot shot off Randy Johnson a couple of years ago, Mac's batspeed has been clocked at 98, his bat is typically 33 oz., and Randy's fastball was about 97 (crosses plate at around 90.) The effective mass of the bat is somewhat less than 33 oz, but the exact number depends on the model. For this calculation, I'm using 29 ounces.
>
> This yields an exit velocity of 140.75 MPH. This assumes NO interaction from Mac's massive body, but merely a result of conservation of momentum. At a launch angle of 35 degrees the ball would travel about 530 feet at sea level, given typical values of ball rotation, humidity, and no wind (I think it was in the Kingdome.) Sounds about right.
>
> Comments? (I have more if needed- the effects of adding in body mass to eliminate the bat "recoil")
>
> Question: How far could Mac have hit the ball with an 18 ounce little-league bat? What if his body mass played a role?
>

Hi Steve

Thank you for your hard work on this. Wish you nothing but the best in 2000.

Jack Mankin