The Virtuosi (Posts about baseball)https://thephysicsvirtuosi.com/enContents © 2019 <a href="mailto:thephysicsvirtuosi@gmail.com">The Virtuosi</a> Thu, 24 Jan 2019 15:05:03 GMTNikola (getnikola.com)http://blogs.law.harvard.edu/tech/rss- Coriolis Effect on a Home Runhttps://thephysicsvirtuosi.com/posts/old/coriolis-effect-on-a-home-run/Corky<div><hr>
<p><a href="http://3.bp.blogspot.com/-LxzlQ5iNaaI/Ta458E_AwvI/AAAAAAAAAMc/D0Z4vZS7IzA/s1600/phillies_stadium.jpg"><img alt="image" src="http://3.bp.blogspot.com/-LxzlQ5iNaaI/Ta458E_AwvI/AAAAAAAAAMc/D0Z4vZS7IzA/s320/phillies_stadium.jpg"></a>
Citizen's Bank Park</p>
<hr>
<p>I like baseball. Well, technically, I like <del>laying</del>[3] lying on the
couch for three hours half-awake eating potato chips and mumbling
obscenities at the television. But let's not split hairs here.
Anyway, out of curiosity and in partial atonement for the sins of my
past [1] I would now like to do a quick calculation to see how much
effect the Coriolis force has on a home-run ball.
The Coriolis force is one of the artificial forces we have to put in if
we are going to pretend the Earth is not rotating. For a nice intuitive
explanation of the Coriolis force see <a href="http://www.wired.com/wiredscience/2011/04/coriolis-force-in-a-wipeout-rotating-slide/">this
post</a>
over at Dot Physics.
Let's now consider the following problem. Citizen's Bank Park (home to
the Philadelphia Phillies) is oriented such that the line from home
plate to the foul pole in left field runs essentially South-North.
Imagine now that Ryan Howard hits a hard shot down the third base line
(that is, he hits the ball due North). Assuming it is long enough to be
a home run, how with the Coriolis force effect the ball's trajectory?
This is a well-posed problem and we could solve it as exactly as we
wanted. But please don't <a href="https://docs.google.com/viewer?a=v&pid=explorer&chrome=true&srcid=0Bwd5hrDOxWsrOGZmMWZkYzQtM2M2Ny00NjlmLTgyYmMtNTQwZjI1ODU1NWI4&hl=en_US">make
me</a>.
It's icky and messy and I don't feel like it. So let's do some
dimensional analysis! Hooray for that!
So what are the relevant physical quantities in this problem? Well,
we'll certainly need the angular velocity of the Earth and the speed of
the baseball. We'll also need the acceleration due to gravity. Alright,
so what do we want to get out of this? Well, ideally we'd like to find
the distance the ball is displaced from its current trajectory. So is
there any way we can combine an angular velocity, linear velocity and
acceleration to get a displacement?
Let's see. We can write out the dimensions of each in terms of some
length, L, and some time, T. So:
$$ \left[ \Omega \right] = \frac{1}{T} $$
$$ \left[ v \right] = \frac{L}{T} $$
$$ \left[ g \right] = \frac{L}{T^2} $$
where we have used the notation that [some quantity] = units of that
quantity. Combining these in a general way gives: $$ L = \left[
v^{\alpha} \Omega^{\beta} g^{\gamma} \right] = \left(
\frac{L}{T}\right)^{\alpha}\left(
\frac{1}{T}\right)^{\beta}\left( \frac{L}{T^2}\right)^{\gamma}
= L^{\alpha+\gamma} T^{-(\alpha+\beta+2\gamma)}$$ Since we want
just want a length scale here, we need: $$\alpha+\gamma =
1\~\~\~\mbox{and}\~\~\~\alpha+\beta+2\gamma = 0. $$
We can fiddle around with the above two equations to get two new
equations that are both functions of alpha. This gives: $$\beta =
\alpha - 2\~\~\~\mbox{and}\~\~\~\gamma = 1 - \alpha. $$
Unfortunately, we have two equations and three unknowns, so we have an
infinite number of solutions. I've listed a few of these in the Table
below.</p>
<hr>
<p><a href="http://4.bp.blogspot.com/-_wBdDaNzEP4/Tg_of5pylyI/AAAAAAAAANE/CU2Oe225_UI/s1600/table.png"><img alt="image" src="http://4.bp.blogspot.com/-_wBdDaNzEP4/Tg_of5pylyI/AAAAAAAAANE/CU2Oe225_UI/s320/table.png"></a>
Ways of getting a length</p>
<hr>
<p>At this point, we have taken Math as far as we can. We'll now have to
use some physical intuition to narrow down our infinite number of
solutions to one. Hot dog! One way we can choose from these expressions
is to see which ones have the correct dependencies on each variable. So
let's consider what we would expect to happen to the deflection of our
baseball by the Coriolis force if we changed each variable. What happens
if we were to "turn up" the gravity and make g larger? If we make g much
larger, then a baseball hit at a given velocity will not be in the air
as long. If the ball isn't in the air as long, then it won't have as
much time to be deflected. So we would expect the deflection to decrease
if we were to increase g. This suggests that g should be in the
denominator of our final expression. What happens if we turn up the
velocity of the baseball? If we hit the ball harder, then it will be in
the air longer and thus we would expect it to have more time to be
deflected. Since increasing the velocity would increase the deflection,
we would expect v to be in the numerator. What happens if we turn up the
rotation of the Earth? Well, if the Earth is spinning faster, it's able
to rotate more while the ball is in the air. This would result in a
greater deflection in the baseball's path. Thus, we would expect this
term to be in the numerator. So, using the above criteria, we have
eliminated everything on that table with alpha less than 3 based on
physical intuition. Unfortunately, we still have an infinite number of
solutions to choose from (i.e. all those with alpha greater than or
equal to 3). But, we DO have a candidate for the "simplest" solution
available, namely the case where alpha = 3. Since we have exhausted are
means of winnowing down our solutions, let's just go with the alpha = 3
case. Our dimensional analysis expression for the deflection of a
baseball is then $$ \Delta x \sim \frac{v^3 \Omega}{g^2} $$
Plugging in typical values of $$ v =
50\~\mbox{m/s}\~\~\~(110\~\mbox{mi/hr}) $$ $$ \Omega = 7 \times
10^{-5}\~\mbox{rad/s} $$ $$ g = 9.8\~\mbox{m/s}^2 $$ we get $$
\Delta x \approx 0.1\~\mbox{m} = 10\~\mbox{cm}. $$ That's all fine
and good, but which way does the ball get deflected? Is it fair or foul?
Well, remembering that the Coriolis force is given by: $$ {\bf F} =
-2m{\bf \Omega} \times {\bf v} $$ and utilizing Ye Olde Right Hand
Rule, we see that a ball hit due north will be deflected to the East. In
the case of Citizen's Bank Park, that is fair territory. But how good is
our estimate? Well, I did the full calculation (which you can find
<a href="https://docs.google.com/viewer?a=v&pid=explorer&chrome=true&srcid=0Bwd5hrDOxWsrOGZmMWZkYzQtM2M2Ny00NjlmLTgyYmMtNTQwZjI1ODU1NWI4&hl=en_US">here</a>)
and found that the deflection due to the Coriolis force is given by $$
\Delta x =-\frac{4}{3}\frac{\Omega v^3_0}{g^2} \cos \phi
\sin^3 \alpha \left[1 -3 \tan \phi \cot \alpha \right] $$
where phi is the latitude and alpha is the launch angle of the ball. We
see that this is essentially what we found by dimensional analysis up to
that factor of 4/3 and some geometrical terms. Not bad! Plugging in the
same numbers we used before, along with the appropriate latitude and a
45 degree launch angle we find that the ball is deflected by: $$ \Delta
x = 5\~\mbox{cm}. $$ For comparison, we note that the diameter of a
baseball is 7.5 cm. So in the grand scheme of things, this effect is
essentially negligible. [2] That wraps up the calculation, but I'm
certain that many of you are still a little wary of this voodoo
calculating style. And you should be! Although dimensional analysis will
give you a result with the proper units and will <em>often</em> give you
approximately the right scale, it is not perfect. But, it can be
formalized and made rigorous. The rigorous demonstration for dimensional
analysis is due to Buckingham and his famous pi-theorem. The original
paper can be found behind a pay-wall
<a href="http://prola.aps.org/abstract/PR/v4/i4/p345_1">here</a> and a really nice
set of notes can be found
<a href="http://www.math.ntnu.no/~hanche/notes/buckingham/buckingham-a4.pdf">here</a>.
It's a pretty neat idea and I highly recommend you check it out!
Unnecessary Footnotes:
[1] Once in college I argued with a meteorologist named Dr. Thunder over
the direction of the Coriolis force on a golf ball for the better half
of the front nine at Penn State's golf course. I was wrong. Moral of the
story: don't play golf with meteorologists.
[2] For a counterargument, see Fisk et al. (1975) [3] Text has been
corrected to illustrate our enlightenment by a former English major as
to the difference between 'lay' and 'lie' through the following story:
'Once in a college psych class, a young student said "It's too hot.
Let's lay down." A mature student, a journalist, asked, "Who's Down?" '</p></div>baseballcoriolisscott bakulahttps://thephysicsvirtuosi.com/posts/old/coriolis-effect-on-a-home-run/Sun, 03 Jul 2011 11:52:00 GMT
- My Pepsi* Challengehttps://thephysicsvirtuosi.com/posts/old/my-pepsi-challenge/Corky<div><p><a href="http://1.bp.blogspot.com/_fa6AZDCsHnY/TAz2-SbRtAI/AAAAAAAAAFY/eaUWEFQHTDs/s1600/bottles.png"><img alt="image" src="http://1.bp.blogspot.com/_fa6AZDCsHnY/TAz2-SbRtAI/AAAAAAAAAFY/eaUWEFQHTDs/s200/bottles.png"></a>The
basement of the Physics building has a Pepsi machine. Over the course of
two semesters Alemi and I have deposited roughly the equivalent of the
GDP of, say, Monaco to this very same Pepsi machine (see left, with most
of Landau and Lifshitz to scale). It just so happens that Pepsi is now
having a contest, called "Caps for Caps," in which it is possible to win
a baseball hat. There are several nice things about this contest.
Firstly, I drink a lot of soda. Secondly, I like baseball hats. So far
so good. Lastly (and most important for this post), is that it is fairly
straightforward to calculate the statistics of winning (or at least
simulate them).
So how does the game work? Well, on each soda cap the name of a Major
League Baseball team is printed. All thirty teams are (supposedly)
printed with the same frequency, so the odds of getting any particular
team are 1/30. You can win a hat by collecting three caps with the same
team printed on them. So if I had five caps, the following would be a
win:
Phillies Cubs Tigers Phillies Phillies
whereas the following would not win me anything:
Yankees Rays Blue Jays Orioles Royals
and I would also lose if I had:
Mets Nationals Braves Braves Mets.
In addition, one in eight caps gets you 15% off on some $50 dollar or
more purchase to MLB.com or something like that. For simplicity, I
ignored these 15% off guys, but all they will do is push back the number
of caps you need by one for every eight purchased. It should not be too
difficult to factor these ones in, but I was lazy and I already made all
these nice graphs, so...
The first thing I tried to do was just simulate the contest. I wrote a
little Python script that randomly generates a team for each cap and
counts my wins over a given number of caps purchased. Running this about
100,000 times for all number of caps between 1 and 61 (with 61
guaranteed to win) and averaging over the number of wins, I could
determine both the expected number of wins per cap value and the
probability of winning at least once. The results are shown below.
<a href="http://4.bp.blogspot.com/_fa6AZDCsHnY/TAwPaC7VV-I/AAAAAAAAAEA/ei6DPQ09bmI/s1600/CapsforCaps(sim1).png"><img alt="image" src="http://4.bp.blogspot.com/_fa6AZDCsHnY/TAwPaC7VV-I/AAAAAAAAAEA/ei6DPQ09bmI/s400/CapsforCaps(sim1">.png)</a>
<a href="http://1.bp.blogspot.com/_fa6AZDCsHnY/TAxwinoE0FI/AAAAAAAAAFA/VTOnToOXcmM/s1600/WinsvsSodas.png"><img alt="image" src="http://1.bp.blogspot.com/_fa6AZDCsHnY/TAxwinoE0FI/AAAAAAAAAFA/VTOnToOXcmM/s400/WinsvsSodas.png"></a>
But we can also exactly solve this game. This turned out to take longer
for me (I'm bad at probability) than just simulating the darn thing. I
had initially included my derivation in the post but it was long,
muddled, and none too illuminating, so I took it out. But I super-duper
promise I did it and can post if you really really want to know.
Otherwise, I have just plotted the predicted results below (as a red
curve) along with the simulated data (blue dots). Turns out they agree
pretty well!
<a href="http://4.bp.blogspot.com/_fa6AZDCsHnY/TAxxKz3m-MI/AAAAAAAAAFI/S6YpmCcO1Fs/s1600/CapsforCapsSimandPred.png"><img alt="image" src="http://4.bp.blogspot.com/_fa6AZDCsHnY/TAxxKz3m-MI/AAAAAAAAAFI/S6YpmCcO1Fs/s400/CapsforCapsSimandPred.png"></a>
Just eyeballing the graph, we see that after 18 or 19 sodas the chances
of winning are about a half. Beyond about 25 or so it appears to be
almost 90% that you'll win at least once. In reality, these percentages
would occur about 2 or 3 caps later to compensate for the 15% off
thingies. So now that we have some numbers and can trust our model a
bit, let's see how worth it this contest is for us.
First, we can ask: Is this a good way to get a hat cheaper than retail
value (about $15)? To quantify "worth it" I have chosen to find the
value of winnings (price of hat times expected number of wins) minus the
cost of caps (how much I spend on soda). I am fairly embarrassed to say
that the cost of each soda is $1.75. See plot below.
<a href="http://1.bp.blogspot.com/_fa6AZDCsHnY/TAxzBhvw-4I/AAAAAAAAAFQ/RiKEfFBDqig/s1600/ValueDifferential.png"><img alt="image" src="http://1.bp.blogspot.com/_fa6AZDCsHnY/TAxzBhvw-4I/AAAAAAAAAFQ/RiKEfFBDqig/s400/ValueDifferential.png"></a>
From this plot, we see that it doesn't become "worth it" (that is, value
of winnings is greater than cost of sodas) until about 40 sodas
purchased. That's a lot! In fact, we see that just when I start feeling
pretty confident I'll win something (around 20-25 sodas), I'm right in a
big valley of "totally not worth it." So if I just want a baseball hat,
I'm better off forking over the $15 dollars.
Although, one does see from this plot that once I get above about 40 or
so sodas, it becomes much more cost effective to just keep buying sodas
and winning hats. However, Pepsi tries to stifle this a bit in the
<a href="http://www.pepsiusa.com/capsforcaps/">rules</a>, stating that "Limit one
(1) Official MLB® baseball cap per name, address or household." Unless I
either make a lot of friends real soon or develop a creative definition
of my address, it looks like I'm out of luck.
But what if I want a hat but I don't want to actually buy soda like a
chump? This contest, like many others, needs to have a "No Purchase
Necessary" clause for some legal reason or another (so they aren't
lotteries or gambling or something). I had assumed they (the nameless
overlords at Pepsi) would limit the number of caps possible from just
mailing in, but it doesn't seem that way. From the Official rules,
Chapter Two, verses nine to twenty-one:
"Limit one (1) free game piece per request, per stamped outer envelope."
That sounds to me like you could get as many as you want, as long as you
use different envelopes. So we can redo our cost analysis with the cost
of getting one cap as the cost of a stamp. Putting the value of a cap
now at the cost of a stamp (44 cents), we get the following:
<a href="http://2.bp.blogspot.com/_fa6AZDCsHnY/TAz3jfb-KcI/AAAAAAAAAFg/erU1nZU7ZCg/s1600/mailincaps.png"><img alt="image" src="http://2.bp.blogspot.com/_fa6AZDCsHnY/TAz3jfb-KcI/AAAAAAAAAFg/erU1nZU7ZCg/s400/mailincaps.png"></a>
Zooming in:
<a href="http://1.bp.blogspot.com/_fa6AZDCsHnY/TAz3tj6kg8I/AAAAAAAAAFo/MDHgjjjS2J0/s1600/mailinZoom.png"><img alt="image" src="http://1.bp.blogspot.com/_fa6AZDCsHnY/TAz3tj6kg8I/AAAAAAAAAFo/MDHgjjjS2J0/s400/mailinZoom.png"></a>
Hey, that seems worth it! And it should, since from above we saw that
the probability of winning after about 30 caps was in the high 90% 's.
The cost of getting 30 caps this way is the cost of 30 stamps, which is
less than the $15 that the hat is (supposedly) worth. So if I really
wanted a hat from this contest and didn't feel like drinking all my
money away, I'd just send away for the mail-in pieces.
I may try this method, since it seems to be allowed under the rules.
Although, even a strict constructionist reading of the contest rules
pretty much allows Pepsi to do whatever the heck it wants. Either way,
I'll be sure to update to see how well my model holds up!</p>
<hr>
<p>*NOTE: In no way is The Virtuosi blog affiliated in any way with Pepsi.
We may occasionally purchase Pepsi products (like sweet tasting Wild
Cherry Pepsi!), but we don't do it because we think it makes us look
"cool" or "hip" or "rad" (we KNOW it does). In fact, drinking too much
soda can have certain adverse health effects (like making you stronger,
faster, and in general more attractive). So if you want to have a Pepsi
product (like sweet tasting Wild Cherry Pepsi!) every now and then
(literally, EVERY INSTANT), go ahead. But drinking too many Pepsi
products (like sweet tasting Wild Cherry Pepsi!) could make you sick
(with awesome-itis).</p></div>baseballhow do i get free stuff?pepsiscott bakulahttps://thephysicsvirtuosi.com/posts/old/my-pepsi-challenge/Tue, 08 Jun 2010 01:30:00 GMT
- Physics of Baseball: Battinghttps://thephysicsvirtuosi.com/posts/old/physics-of-baseball-batting/Alemi<div><p><a href="http://4.bp.blogspot.com/_YOjDhtygcuA/S_RkXXs4DpI/AAAAAAAAAKo/jPSgwpl4qHA/s1600/baseball.jpg"><img alt="image" src="http://4.bp.blogspot.com/_YOjDhtygcuA/S_RkXXs4DpI/AAAAAAAAAKo/jPSgwpl4qHA/s320/baseball.jpg"></a></p>
<p>Summer is upon us, and so that means that we here at the Virtuosi have
started talking about baseball. In fact, Corky and I did some simple
calculations that illuminate just how impressive batting in baseball can
be. We were interested in just how hard it is to hit a pitch with the
bat. So we thought we'd model hitting the ball with a rather simple
approximation of a robot swinging a cylindrical bat, horizontally with
some rotational speed and at a random height. The question then becomes,
if the robot chooses a random height and a random time to swing, what
are the chances that it gets a hit?</p>
<h4>Spatial Resolution</h4>
<p>So the first thing to consider is how much of the strike zone the bat
takes up. In order to be a strike, the ball needs to be over home plate,
which is 17 inches wide, and between the knees and logo on the batters
jersey. Estimating this height as 0.7 m or 28 inches or so, we have the
area of the strike zone $$ A_S = (17") \times (0.7 m) = 0.3 \text{
m}^2 $$ when you swing, how much of this area does the bat take up?
Well, treat it as a cylinder, with a diameter of 10 cm, and assume it
runs the length of the strike zone, when the area that the bat takes up
is $$ A_B = (10\text{ cm}) \times (17" ) = 0.043 \text{ m}^2 $$ So
that the fractional area that the bat takes up during our idealized
swing is $$ \frac{A_B}{A_S} \approx 14\% $$ So already, if our
robot is guessing where inside the strike zone to place the bat, and
doing so randomly, assuming the pitch is a strike to begin with, it will
be able to bunt successfully about 14% of the time.</p>
<h4>Time Resolution</h4>
<p>But getting a hit on a swing is different than getting a bunt. Not only
do you have to have your bat at the right height, but you need to time
the swing correctly. Lets first look at how much time we are dealing
with here. Most major league pitchers pitch the ball at about 90 mph or
so. The pitchers mound is 60.5 feet away from home base. This means that
the pitch is in the air for $$ t = \frac{ 60.5 \text{ ft} }{ 90
\text{ mph} } \approx \frac{1}{2} \text{ second} $$ i.e. from the
time the pitcher releases the ball to the time it crosses home plate is
only about half a second. Compare this with human reaction times. My
drivers ed course told me that human reaction times are typically a
third of a second or so. So, baseball happens quick! Alright, but we
were interested in how well you have to time your swing. Successfully
hitting the ball means that you've made contact with the ball such that
it lands somewhere in the field. I.e. you've got a 90 degree play in
when you make contact. How does this translate to time? We would need to
know how fast you swing.</p>
<h5>Estimating the speed of a swing</h5>
<p>I don't know how fast you can swing a baseball bat, but I can estimate
it. I know that if you land your swing just right, you have a pretty
good shot at a home run. Fields are typically 300 feet long. So, I can
ask, if I launch a projectile at a 45 degree angle, how fast does it
need to be going in order to make it 300 feet. Well, we can solve this
projectile problem if we remember some of our introductory physics. We
decouple the horizontal and vertical motions of the ball, the ball
travels horizontally 300 feet, so we know $$ v_x t = 300 \text{ ft} $$
where t is the time the ball is in the air, similarly we know that it is
gravity that makes the ball fall, and so as far the vertical motion is
concerned, in half the total flight time, we need the vertical velocity
to go from its initial value to zero, i.e. $$ g \frac{t}{2} = v_y $$
where g is the acceleration due to gravity. Furthermore, I'm assuming
that I am launching this projectile at a 45 degree angle, for which I
know from trig that $$ v_x = v_y = \frac{v}{\sqrt 2} $$ So I can
stick these equations into one another and solve for the velocity needed
to get the ball going 300 feet. $$ \frac{v^2}{ g} = 300 \text{ ft} =
\frac{ v^2}{ 9.8 \text{ m/s}^2 } $$ $$ v \approx 30 \text{ m/s}
\sim 67 \text{ mph}$$ So it looks like the ball needs to leave the bat
going about 70 mph in order to clear the park. ( This was of course
neglecting air resistance, which ought to be important for baseballs ).
Great that tells us how fast the ball needs to be going when it leaves
the bat, but how fast was the bat going in order to get the ball going
that fast? Well, lets work worst case and assume that the baseball - bat
interaction is inelastic. I.e. I reckon that if I throw a baseball at
about 100 mph towards a wooden wall, it doesn't bounce a whole lot. In
that case, the bat needs to take the ball from coming at it at 90 mph to
leaving at 70 mph or so, i.e. the place where the ball hits the bat
needs to be going at about 160 mph. That seems fast, but when you think
about it, if a pitcher can pitch a ball at 90 mph, that means their hand
is moving at 90 mph during the last bits of the pitch, so you expect
that a batter can move their hands about that fast, and we have the
added advantage of the bat being a lever.</p>
<h5>Coming back to timing</h5>
<p>So, we have an estimate for how fast the bat is going. Knowing this and
estimating the length between the sweet spot and the pivot point of the
bat to be about 0.75 m or so, we can obtain the angular frequency of the
bat. $$ v = \omega r $$ $$ \omega = \frac{ 160 \text{ mph} }{ 0.75
\text{ m} } \approx 100 \text{ Hz} $$ So, if we need to have a 90
degree resolution in our swing timing to hit the ball in the park, this
means that if our swing near the end is happening at 100 \text{ Hz}, we
need to get the timing down to within $$ t = \frac{ 90 \text{
degree}}{ 100 \text{ Hz} } \sim 15 \text{ ms} $$ So we need to get
the timing of our swing to within about 15 milliseconds to land the hit.
So if our robot randomly swung at some point during the duration of the
pitch, it would only hit with probability $$ \frac{\text{time to land
hit} }{ \text{time of pitch}} = \frac{ 15 \text{ ms}}{500 \text{
ms}} \sim 3\% $$ or only 3% of the time. If we take both the spatial
placement, and timing of the swing as independent, the probability that
our robot gets a hit would be something about $$ p = 0.03 \times 0.14 =
0.004 = 0.4 \% $$ or our robot would only get a hit 1 time out of 250
tries. Suddenly hitting looks pretty impressive.</p>
<h4>Experiment</h4>
<p>Saying that the robot swings at some random time during the duration of
the pitch is pretty bad. So I decided to do a little experiment to see
how good people are at judging times on half second scales. I had some
friends of mine start a stop watch and while looking try to stop it as
close as they could at the half second mark. Collecting their
deviations, I obtained a standard deviation of about 41 milliseconds,
which suggests a window of about 100 milliseconds over which people can
reliably judge half second intervals. Now, I have to admit, this wasn't
done in any very rigorous sort of way, I had them do this while walking
to dinner, but it ought to give a rough estimate of the relevant time
scale for landing a hit. So instead of comparing our 15 millisecond 'get
a hit' window to the full half second pitch duration, lets compare it
instead to the 100 'humans trying to judge when to hit' window. This
gives us a temporal resolution of about $$ p = \frac{ 15}{100} = 15
\%. $$ So that now we obtain an overall hit probability of $$ p = 0.15
* 0.14 = 0.021 = 2 \% $$ So that it seems like a poor baseball player,
more or less randomly swinging should have a batting average of about
2\%. Compare this with typical baseball batting averages of 250 or so,
denoting 0.25 or 25% probability of a hit. I think this is a much better
estimate of how much better over random baseball players can do with
training. So it looks like practice can improve your ability to do a
task by about an order of magnitude or so. Either way, baseball is
pretty darn impressive when you think about it.</p></div>baseballfunorder of magnitudehttps://thephysicsvirtuosi.com/posts/old/physics-of-baseball-batting/Wed, 19 May 2010 18:21:00 GMT