Bat Speed -- Baseball Hitting Forum

Posted by: Mike Myers (mike.myers@bellnexxia.com) on Mon Jul 7 10:02:59 2003

I am coaching a coach pitch team, and one of the other coaches and I have different thoughts on the speed of the ball being pitched. We have several kids that are use to the ball being thrown fast and they seem to hit it further, but the other coach says that the speed of the pitch has nothing to do with it.
> > > >
> > > > Give me some opinion of what you guys think. I read some where that there was a formula that proved it, something like speed x volicity = distance.
> > > >
> > > > thanks in advance for any help
> > > > Jim
> > >
> > > I believe the speed of the pitch has some to do with how far or hard a ball is pitched. Bat speed is most important to how far a ball travels. Pitch does a little. Think aboutit, could you hit as far in BP as hitting a ball off a tee?
> > +++++++++++++++++++++++++++++++
> > The answer is simple physics. With the same bat speed, a ball pitched faster will travel farther because you are adding or multiplying the speed of the ball to the speed of the bat.
> >
> > coach joed
>
> What do you mean? i thought that the faster the ball hits the bat the more compression of both the ball and the bat creating more of a trampolin effect... this is just conjecture on my part. take care.

Net momentum (mass x velocity) of the bat and ball is the same before and after contact:

Therefore,
mballi x vballi + mbati x vbati = mballf x vballf + mbatf x vbatf

However, mass of bat (mbat) and ball (mball) are the same before and after contact. Also, assume all bat energy is transferred to ball at contact (assume vbatf = 0). Although the bat continues to move through the strike zone, this is mainly because body momentum pulls it through:

Then, mball x vballi + mbat x vbati = mball x vballf + 0
vballf = (mball x vballi + mbat x vbati) / mball

Typically, mbat is about 6 x mball.

Therefore, vballf = (mball x vball + 6 x mball x vbat) / mball
= vball + 6 x vbat

So, the final velocity of the ball (after contact) is about 6 times more related to batspeed, then pitch speed.

Regards,
Mike.

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> I am coaching a coach pitch team, and one of the other coaches and I have different thoughts on the speed of the ball being pitched. We have several kids that are use to the ball being thrown fast and they seem to hit it further, but the other coach says that the speed of the pitch has nothing to do with it. 
> > > > > 
> > > > > Give me some opinion of what you guys think. I read some where that there was a formula that proved it, something like speed x volicity = distance.
> > > > > 
> > > > > thanks in advance for any help
> > > > > Jim
> > > > 
> > > > I believe the speed of the pitch has some to do with how far or hard a ball is pitched.  Bat speed is most important to how far a ball travels.  Pitch does a little. Think aboutit, could you hit as far in BP as hitting a ball off a tee?
> > > +++++++++++++++++++++++++++++++
> > > The answer is simple physics. With the same bat speed, a ball pitched faster will travel farther because you are adding or multiplying the speed of the ball to the speed of the bat. 
> > > 
> > > coach joed
> > 
> > What do you mean?  i thought that the faster the ball hits the bat the more compression of both the ball and the bat creating more of a trampolin effect... this is just conjecture on my part. take care.
> 
> Net momentum (mass x velocity) of the bat and ball is the same before and after contact:
> 
> Therefore, 
> mballi x vballi + mbati x vbati = mballf x vballf + mbatf x vbatf
> 
> However, mass of bat (mbat) and ball (mball) are the same before and after contact.  Also, assume all bat energy is transferred to ball at contact (assume vbatf = 0).  Although the bat continues to move through the strike zone, this is mainly because body momentum pulls it through:
> 
> Then, mball x vballi + mbat x vbati = mball x vballf + 0
>       vballf = (mball x vballi + mbat x vbati) / mball
> 
>       Typically, mbat is about 6 x mball.
> 
> Therefore, vballf = (mball x vball + 6 x mball x vbat) / mball
>                   = vball + 6 x vbat
> 
> So, the final velocity of the ball (after contact) is about 6 times more related to batspeed, then pitch speed.
> 
> Regards,
> Mike.

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