The Virtuosi (Posts by Bohn)https://thephysicsvirtuosi.com/enContents © 2019 <a href="mailto:thephysicsvirtuosi@gmail.com">The Virtuosi</a> Thu, 24 Jan 2019 15:05:00 GMTNikola (getnikola.com)http://blogs.law.harvard.edu/tech/rss- Fun with an iPhone Accelerometerhttps://thephysicsvirtuosi.com/posts/old/fun-with-an-iphone-accelerometer/Bohn<p><a href="http://2.bp.blogspot.com/-f7mSv8QGLGs/Tj9_a2qp1SI/AAAAAAAAAEY/Ax5EEw8Cjmg/s1600/accelerometer.jpg"><img alt="image" src="http://2.bp.blogspot.com/-f7mSv8QGLGs/Tj9_a2qp1SI/AAAAAAAAAEY/Ax5EEw8Cjmg/s320/accelerometer.jpg"></a>
The iPhone 3GS has a built-in <a href="http://pdf1.alldatasheet.com/datasheet-pdf/view/236640/STMICROELECTRONICS/LIS302DL.html">accelerometer, the
LIS302DL</a>,
which is primarily used for detecting device orientation. I wanted to
come up with something interesting to do with it, but first I had to see
how it did on some basic tests. It turns out that the tests gave really
interesting results themselves! A drop test gave clean results and a
spring test gave fantastic data; however a pendulum test gave some
problems. You might guess the accelerometer would give a reading of 0 in
all axes when the device is sitting on a desk. However, this
accelerometer measures "proper acceleration," which essentially is a
measure of acceleration relative to free-fall. So the device will read
-1 in the z direction (in units where 1 corresponds to 9.8 m/s^2, the
acceleration due to gravity at the surface of Earth). Armed with this
knowledge, let's take a look at the drop test: To perform this test, I
stood on the couch which was in my office (before it was taken away from
us!), and dropped my phone hopefully into the hands of my officemate. I
suspected that the device would read magnitude 1 before dropping, 0
during the drop, and a large spike for the large deceleration when the
phone was caught.
<a href="http://2.bp.blogspot.com/-g8pt_dT0gR8/Tj9D2GTkc6I/AAAAAAAAADw/1iwHCsPixtY/s1600/DropTest.png"><img alt="image" src="http://2.bp.blogspot.com/-g8pt_dT0gR8/Tj9D2GTkc6I/AAAAAAAAADw/1iwHCsPixtY/s400/DropTest.png"></a>
As you can see, the results were basically as expected. The purple line
shows the magnitude of the acceleration relative to free-fall. Before
the drop, the magnitude bounces around 1, which is due to my inability
to hold something steadily. The drop occurred near time 12.6, but I
wasn't able to move my hand arbitrarily quickly so there's not a sharp
drop to 0 magnitude. The phone fell for around 0.4 seconds corresponding
to $$y = \frac{1}{2} g t^2 = \frac{1}{2} (9.8 \frac{m}{s^2})<em>(0.4
s)^2 = 0.784 m = 2.57 feet $$ As for the spike at 13 seconds, the raw
data shows that the catch occurs in $$ t = 0.02 \pm 0.01 s $$. In order
for the device to come to rest in such a short amount of time, there
needs to be a large deceleration provided by my officemate's hands. Now
the pendulum test consisted of taping my phone to the bottom of a 20
foot pendulum.
<a href="http://2.bp.blogspot.com/-7uO_824O6t4/Tj9Q1crvlUI/AAAAAAAAAD4/uO3YsrXStak/s1600/pendulum.png"><img alt="image" src="http://2.bp.blogspot.com/-7uO_824O6t4/Tj9Q1crvlUI/AAAAAAAAAD4/uO3YsrXStak/s400/pendulum.png"></a>
I didn't think enough about this, but the period of a pendulum, assuming
we have a small amplitude, is given by: $$T = 2 \pi
\sqrt{\frac{L}{g}}$$ which is about 5 seconds. With a relatively small
amplitude, the acceleration in the x direction will be small. Basically
I'm reaching the limit of the resolution of the acceleration device. It
appears that the smallest increment the device can measure is 0.0178 g.
This happens to match the specifications from the spec sheet I linked at
the top of the page, where they specify a minimum of 0.0162 g, and a
typical sensitivity of 0.018 g! Now we come to the most exciting test,
the spring test! Setup: I taped my phone to the end of a spring and let
it go. Ok. Here is the actual acceleration data:
<a href="http://3.bp.blogspot.com/-3qbU9p2I3sE/Tj9a7-eOk-I/AAAAAAAAAEA/0ByUQ039dio/s1600/springdata.png"><img alt="image" src="http://3.bp.blogspot.com/-3qbU9p2I3sE/Tj9a7-eOk-I/AAAAAAAAAEA/0ByUQ039dio/s400/springdata.png"></a>
The first thing I see is that the oscillation frequency looks constant,
as it should be for a simple harmonic oscillator. There is also a decay
which looks exponential! Let's see how well the data fits if we have a
frictional term proportional to the velocity of the phone. This gives is
a differential equation which looks like this: $$ m \ddot{x} +
F\dot{x} + k x = 0 $$ Now we can plug in an ansatz (educated guess) to
solve this equation: $$ x(t) = A</em>e^{i b t} $$ $$-b^2 mx(t) + i b
Fx(t) + kx(t) = 0$$ $$-m b^2+iFb+k = 0$$ We can solve this equation for
b with the quadratic equation: $$ b = \frac{\sqrt{4km - F^2}}{2m} +
i\frac{F}{2m} \equiv \omega + i \gamma $$ where I defined two new
constants here. So we see that our ansatz does solve the differential
equation. Now we want acceleration, which is the second time derivative
of position with respect to time. $$a(t) \equiv \ddot{x} = -b^2 A
e^{ibt} $$ Now are only interested in the real part of this solution,
which gives us (adding in a couple of constants to make the solution
more general): $$a(t) = -(\omega^2 - \gamma^2) A e^{-\gamma t}
cos(\omega t + \phi) + C $$ Let's redefine the coefficient of this
acceleration to make things a little cleaner! $$a(t) = B e^{-\gamma t}
cos(\omega t + \phi) + C $$ Ok, with that math out of the way (for
now), we can try to fit this data. I actually used Excel to fit this
data using a not-so-well-known tool called Solver. This allows you to
maximize or minimize one cell while Excel varies other cells. In this
case, I defined a cell which is the <a href="http://en.wikipedia.org/wiki/Residual_sum_of_squares">Residual Sum of
Squares</a> of my fit
versus the actual data, and I tell Excel to vary the 5 constants which
make the fit! The values jump around for a little while then it gives up
when it thinks it converged to a solution. Using this you can fit
arbitrary functions, neato! With this, I come up with the following
plot:
<a href="http://1.bp.blogspot.com/-23lXSPVOwuY/Tj9a8Curm3I/AAAAAAAAAEI/F5Fb9dxdvDs/s1600/springlineardamp.png"><img alt="image" src="http://1.bp.blogspot.com/-23lXSPVOwuY/Tj9a8Curm3I/AAAAAAAAAEI/F5Fb9dxdvDs/s400/springlineardamp.png"></a>
$$B = 0.633740943$$ $$\gamma = 0.012097581 $$ $$\omega = 8.599670376
$$ $$\phi = 0.693075811 $$ $$C =-1.004454967 $$ with an R^2 value of
0.968! At this point it should be noted that if I discretize my smooth
fit to have the same resolution (0.0178 g) as the accelerometer, then
see what the error is comparing the smooth fit to its own
discretization, I get an R^2 of 0.967! This means that there is a
decent amount of built-in error to these fits due to discretization on
the order of the error we're seeing for our actual fits. Immediately we
can recognize that C should be -1, since this is just a factor relating
"free-fall" acceleration to actual iPhone acceleration. If we wanted, we
could solve for the ratio of the spring constant to the mass, but I'll
leave that as <a href="http://www.amazon.com/Classical-Electrodynamics-Third-David-Jackson/dp/047130932X">an exercise for the
reader.</a>
If you look closely, you can see that the frequency appears to match
very well. The two lines don't go out of phase. One problem with the fit
is the decay. The beginning and the end of the data are too high
compared to the fit, which is a problem. This implies that there is some
other kind of friction at work. Some larger objects or faster moving
objects tend to experience a frictional force proportional to the square
of the velocity. I don't think my iPhone is large or fast (compared to a
plane for example), but I'll try it anyway. The differential equation
is: $$ m \ddot{x} + F\dot{x}^2 + k x = 0 $$ yikes. This is a tough
one because of the velocity squared term. <a href="http://www.jstor.org/pss/3620747">One trick I found
here</a> attempts a general solution for
a similar equation. They make an approximation in order to solve it, but
the approximation is pretty good in our case. Take a look at the paper
if you're interested. The basic idea is to note that the friction term
is the only one that affects the energy. So, assuming that the energy
losses are small in a cycle, we can look at a small change in energy
with respect to a small change in time due to this force term. This
gives us an equation which can let us solve for the amplitude as a
function of time approximately! Really interesting idea. So I plugged
the following equation into the Excel Solver: $$a(t) = \frac{A
cos(\omega t + \phi)}{\gamma t + 1} + B$$ Here's the fit:
<a href="http://2.bp.blogspot.com/-fCXr2CkPGE4/Tj9a8pH2KFI/AAAAAAAAAEQ/XaulleF3xK4/s1600/springsqdamp.png"><img alt="image" src="http://2.bp.blogspot.com/-fCXr2CkPGE4/Tj9a8pH2KFI/AAAAAAAAAEQ/XaulleF3xK4/s400/springsqdamp.png"></a>
Which uses these values: $$A = 0.772773705 $$ $$\gamma = 0.029745368 $$
$$\omega = 8.600177692 $$ $$\phi = 0.688610161 $$ $$B = -1.004530009
$$ with an R^2 value of 0.964! This fit seems to have the opposite
effect. The middle of the data is too high compared to the fit, while
the beginning and end of the data seems too low. This makes me think
that the actual friction terms involved in this problem are possibly a
sum of a linear term and a squared term. I don't know how to make
progress on that differential equation, so I wasn't able to fit
anything. If you try the same trick I mentioned earlier, you run into a
problem where you can't separate some variables which you need to
separate in the derivation unfortunately. So there you have it, I wanted
to find something neat to do, and I got really cool data from just
testing the accelerometer. Stay tuned for an interesting challenge
involving some physical data from my accelerometer!</p>accelerometerBohnfree falliphonependulumSHOhttps://thephysicsvirtuosi.com/posts/old/fun-with-an-iphone-accelerometer/Wed, 03 Aug 2011 03:24:00 GMT
- Physics in Sports: The Fosbury Flophttps://thephysicsvirtuosi.com/posts/old/physics-in-sports-the-fosbury-flop/Bohn<p><a href="http://1.bp.blogspot.com/-slmXXaMCcMI/TjXnJ3qe-kI/AAAAAAAAADI/PdIuocXmC5w/s1600/Fosbury.jpg"><img alt="image" src="http://1.bp.blogspot.com/-slmXXaMCcMI/TjXnJ3qe-kI/AAAAAAAAADI/PdIuocXmC5w/s320/Fosbury.jpg"></a>Physics
has greatly influenced the progress of most sports. There have been
continual improvements in equipment for safety or performance as well as
improvements in technique. I'd like to talk about some physics in sports
over a series of posts. Here I'll talk about a technique improvement in
High Jumping, the Fosbury Flop. The Fosbury Flop came into the High
Jumping scene in the 1968 Olympics, where <a href="http://en.wikipedia.org/wiki/Dick_Fosbury">Dick
Fosbury</a> used the technique
to win the gold medal. The biggest difference between the Flop and
previous methods is that the jumper goes over the bar upside down
(facing the sky). This allows the jumper to bend their back so that
their arms and legs drape below the bar which lowers the center of mass
(See the picture above). Here is a video of the <a href="http://www.youtube.com/watch?v=_bgVgFwoQVE">Fosbury
Flop</a> executed very well.
<a href="http://1.bp.blogspot.com/-UYbUVO1G8JM/TjYpGgCGOxI/AAAAAAAAADQ/wKjdCsuwLB0/s1600/flopdiagram.jpg"><img alt="image" src="http://1.bp.blogspot.com/-UYbUVO1G8JM/TjYpGgCGOxI/AAAAAAAAADQ/wKjdCsuwLB0/s320/flopdiagram.jpg"></a>Let’s
assume Dick Fosbury is shaped like a semi-circle as he moves over the
bar. The bar is indicated as a red circle, as this is a side view. From
this diagram, we can guess his center of mass is probably near the
marked 'x', since most of his mass is below the bar. It is important to
recall the definition of center of mass, which is the average location
of all of the mass in an object. $$ \vec{R} = \frac{1}{M} \int
\vec{r} dm $$ Note that this is a vector equation, and the integral
should be over all of the mass elements. This integral gets easier
because I'm going to assume that Dick Fosbury is a constant density
semi-circle. This means that $$ M = C<em>h $$ where C is a constant equal
to the ratio of the mass to the height, and $$dm = C * dh $$. This is a
vector equation, so in principle we need to solve the x integral and the
y integral; however, due to the symmetry about the y-axis, the x
integral is zero. Finally we'll convert to polar coordinates, leaving us
with: $$ y = \frac{1}{C \pi R} \int_0^\pi R\sin{\theta} C R
d\theta = \frac{1}{C \pi R} R (-\cos{\theta}) C R \bigg|_0^\pi
= \frac{2R}{\pi} $$ Ok, so this is the y-coordinate of the center of
mass of our jumper relative to the bottom of the semi-circle. Now we
need to calculate relative to the top of the bar, which is roughly the
location of the top of the circle. We just need to subtract from R: $$ R
- \frac{2}{\pi} R = R * (1 - \frac{2}{\pi}) = \frac{h}{\pi} * (1
- \frac{2}{\pi}) $$ Now Dick Fosbury was 1.95m tall, which gives us a
distance of 22.6 cm BELOW the bar! Of course he's not a semi-circle, but
this isn't a terrible approximation, as you can see from the video
linked above. Further, wikipedia mentions that some proficient jumpers
can get their center of mass 20 cm below the bar, which matches pretty
well with our guess. A nifty technique in physics is looking at the
point-particle system, which allows us to see the underlying motion of a
system. If you’re not familiar with this method, you collect any given
number of objects and replace them with a single point at the center of
mass of the object. We can use energy conservation now for our
point-mass instead of the entire body of the jumper.<a href="https://thephysicsvirtuosi.com/posts/old/physics-in-sports-the-fosbury-flop/#note1">^note^</a> In
this case, we can simply deal with the center of mass motion of the
jumper. All of my kinetic energy will be converted to gravitational
potential energy. Again this is an approximation because some energy is
spent on forward motion, as well as the slight twisting motion which
I'll ignore. $$E = \frac{1}{2} mv^2 = mgh$$ Now let’s look at some
data. Here is a plot of each world record in the high jump.
<a href="http://3.bp.blogspot.com/-DVfHxJG5b-U/TjY-0WQ1YkI/AAAAAAAAADg/GDFfNr0KiBo/s1600/worldrecords.png"><img alt="image" src="http://3.bp.blogspot.com/-DVfHxJG5b-U/TjY-0WQ1YkI/AAAAAAAAADg/GDFfNr0KiBo/s400/worldrecords.png"></a>The
blue data show jumps before the Flop, and the red data show records
after the Flop. <strong>Note: In 1978, the straddle technique broke the
world record, being the only non-flop technique to do so since 1968.
Thanks Janne!</strong> The Flop was revealed in 1968, so I’ll assume that all
jumps before this year used a method where the center of mass of the
jumper was roughly even with the bar, while all jumps after this year
used the flop (see the previous note). Clearly something happened just
before the Flop came out, and this is something called <a href="http://en.wikipedia.org/wiki/Straddle_technique">the Straddle
technique</a>. I want to
know the percent difference in the initial energies required, so I will
calculate $$ 100\% * \frac{E_0-E_f}{E_0} = 100\% </em>
\frac{mgh_0-mgh_f}{mgh_0} = 100\% * \frac{h_0-h_f}{h_0} $$
where $$E_0$$ is the initial energy without the force, err the flop,
and $$ E_f $$ is the initial energy using the flop. Since we are using
the point-particle system, the gravitational potential energy only cares
about the center of mass of the flopper, and we need to know the height
of the center of mass for a 2.45m flop, which is the current world
record. This corresponds to a flop center of mass height of 2.25m, which
gives us an 8.2% decrease in energy using the flop (versus a method
where the center of mass is even with the bar)! The current world record
is roughly 20 cm higher than it was when the flop came out. This could
be due to athletes getting stronger, but this physics tells us that some
of the height increase could have been from the technique change. To sum
up, the high jump competition, along with many other sports, is being
exploited by physics! [note] Here we're relying on the center of mass
being equal to something called the center of gravity of the jumper. The
center of mass is as defined above. The center of gravity is the average
location of the gravitational force on the body. This happens to be the
same as the center of mass if you assume we are in a uniform
gravitational field, which is essentially true on the surface of the
Earth.</p>BohnFosbury flophigh jumpphysics in sportssportshttps://thephysicsvirtuosi.com/posts/old/physics-in-sports-the-fosbury-flop/Mon, 01 Aug 2011 01:30:00 GMT
- Fun Fact: Lebron James Plays Basketballhttps://thephysicsvirtuosi.com/posts/old/fun-fact-lebron-james-plays-basketball/Bohn<div><p><a href="http://turbo.inquisitr.com/wp-content/2010/07/lebron-james.jpg"><img alt="image" src="http://turbo.inquisitr.com/wp-content/2010/07/lebron-james.jpg"></a>
Between building airplanes and playfully destroying everyone else in my
apartment at Super Smash Brothers, my roommate Nathan brought up an
interesting recent fact about LeBron James. He told me that LeBron
scored 11 consecutive field goals (not in football... you know who you
are) in one game. Apparently this was a pretty special event, but how
rare is it for a player of LeBron's caliber? TO THE SCIENCE-MOBILE! The
Problem! <a href="http://www.youtube.com/watch?v=kHltCzuwlOs&feature=related">ESPN 8, The
Ocho</a> tells
me that LeBron's career field goal percentage is 47.5%. Considering the
number of shots he takes, this is a pretty good number. To compare, the
highest field goal percentage for a single season was Wilt Chamberlin
with 72.7%, but eye witness testimony says he was around 10 feet tall
and would wait in the offensive paint all game. Let's see how improbable
this 11 in a row streak is. The generic question we are going to need to
answer is as follows: If a basketball player takes N shots in one game,
with a shooting probability of q, what is the probability that the
player will make AT LEAST k shots in a row? We'll call this probability
P(N) This turns out to be a tricky problem, but let's take a shot (awful
pun... I sincerely apologize). We can take care of simple cases: If N <
k, then P(N) = 0. This tells us you can't have a streak of k if you
don't take k shots! If N = k, P(N) = q^k. This is the probability of
getting k in a row if you take k shots, not too surprising yet. When N
> k, things get more interesting. Finding the Recurrence Relation Our
goal is to write a relationship that has this form: P(N) = P(N-1) +
blank What's blank? "You don't worry about blank... let me worry about
blank!" We'll need to look at the <a href="http://en.wikipedia.org/wiki/Inclusion_exclusion_principle">inclusion-exclusion
principle</a>.
This principle basically says that when we want to take all DISTINCT
items in two sets, we need to take all of the elements in one set, and
add all elements in the second set which are DISTINCT from the first.
For example, if A = {0, 1, 2, 3, 4} and B = {3, 4, 5, 6, 7, 8}, then the
union of A and B is {0, 1, 2, 3, 4, 5, 6, 7, 8}. Note that I did not
include 3 and 4 twice. Let's take a look at the expertly designed (5
minutes before class) google docs drawing below:
<a href="https://docs.google.com/drawings/pub?id=1Ef34hZJ9mtF-GDUSpmJA4Ke2mS3BAHLsDwAk1GX19Dc&w=1122&h=485"><img alt="image" src="https://docs.google.com/drawings/pub?id=1Ef34hZJ9mtF-GDUSpmJA4Ke2mS3BAHLsDwAk1GX19Dc&w=1122&h=485"></a>
The entire line represents N shots being taken. Each shot gets its own
little column (not all columns shown). Using the inclusion-exclusion
principle with the following sets will give us the answer. Choose A to
be the first N-1 shots, and B to be all N shots. The principle tells us
first to take everything from A, which is the probability P(N-1) shown
in red. B will be the entire line, but the principle tells us to only
add DISTINCT chances from B. Since the only difference in B is one more
shot than A, the only distinct chance for a streak of k shots will be in
the last k shots, shown in yellow as P(k). This is only distinct if the
(k+1)th to last shot shown in green is missed! Otherwise a streak of k
would have been included in A already. There is one more place for a
streak to be already included in A. If there was a streak in the blue
section, we must not include the B streak so we don't double count.
Phew... Let's put this all together by multiplying the probabilities of
each of those events: P(N) = P(N-1) + (probability of yellow
streak)<em>(probability we miss green)</em>(probability of no streak in blue)
$$P(N) = P(N-1) + q^k \times (1-q) \times (1 - P(N-k-1))$$ This gives
us a recurrence relation for the probabilities! This is a general
statement about the probability of at least one streak of length k out
of N chances, given each has a probability q. Since I'm just going to
plug this into Python anyway to handle the data, this equation is good
enough. The expectation value of an event is the probability multiplied
by the number of chances. For example, the expectation value of getting
heads with 2 tosses is just (1/2)*2 = 1. The plan is to compile a list
of his field goal attempts in every game LeBron has played in the NBA,
and sum the expectation values for each N. $$ \mbox{Expectation} =
\sum_i P(i) \times \mbox{(number of games with i shots)} $$ Using
LeBron's actual field goal attempt data for each game (up to February 4,
2011), we find that LeBron is expected number of games with at least a
streak of 11 in a row is 1.128. This is a higher expectation value than
the number of heads in 2 coin flips! So this is MORE expected than the
number of heads we would see with 2 coin flips. This isn't very exciting
given the number of shots he has taken and his shooting percentage. Data
Tables</p>
<p>Consecutive Shots</p>
<p>Expected out of 667</p>
<p>Percent of Games</p>
<p>1</p>
<p>666.9</p>
<p>99.99</p>
<p>2</p>
<p>643.6</p>
<p>96.49</p>
<p>3</p>
<p>489.0</p>
<p>73.32</p>
<p>4</p>
<p>281.8</p>
<p>42.25</p>
<p>5</p>
<p>140.0</p>
<p>21.00</p>
<p>6</p>
<p>65.23</p>
<p>9.780</p>
<p>7</p>
<p>29.55</p>
<p>4.431</p>
<p>8</p>
<p>13.20</p>
<p>1.980</p>
<p>9</p>
<p>5.854</p>
<p>0.8778</p>
<p>10</p>
<p>2.578</p>
<p>0.3866</p>
<p>11</p>
<p>1.128</p>
<p>0.1691</p>
<p>12</p>
<p>0.4898</p>
<p>0.07344</p>
<p>13</p>
<p>0.2110</p>
<p>0.03164</p>
<p>14</p>
<p>0.09011</p>
<p>0.01351</p>
<p>15</p>
<p>0.03807</p>
<p>0.005709</p>
<p>16</p>
<p>0.01592</p>
<p>0.002387</p>
<p>17</p>
<p>0.006560</p>
<p>0.0009839</p>
<p>18</p>
<p>0.002660</p>
<p>0.0003995</p>
<p>19</p>
<p>0.001067</p>
<p>0.0001600</p>
<p>20</p>
<p>0.0004180</p>
<p>6.267E-05</p>
<p>21</p>
<p>0.0001599</p>
<p>2.397E-05</p>
<p>22</p>
<p>6.03E-05</p>
<p>9.033E-06</p>
<p>23</p>
<p>2.22E-05</p>
<p>3.328E-06</p>
<p>24</p>
<p>8.06E-06</p>
<p>1.208E-06</p>
<p>25</p>
<p>2.84E-06</p>
<p>4.259E-07</p>
<p>26</p>
<p>9.98E-07</p>
<p>1.495E-07</p>
<p>27</p>
<p>3.51E-07</p>
<p>5.255E-08</p>
<p>28</p>
<p>1.20E-07</p>
<p>1.797E-08</p>
<p>29</p>
<p>3.92E-08</p>
<p>5.870E-09</p>
<p>30</p>
<p>1.15E-08</p>
<p>1.717E-09</p>
<p>31</p>
<p>3.45E-09</p>
<p>5.171E-10</p>
<p>32</p>
<p>1.06E-09</p>
<p>1.589E-10</p>
<p>33</p>
<p>3.37E-10</p>
<p>5.050E-11</p>
<p>34</p>
<p>7.23E-11</p>
<p>1.083E-11</p>
<p>35</p>
<p>1.70E-11</p>
<p>2.554E-12</p>
<p>36</p>
<p>2.30E-12</p>
<p>3.442E-13</p>
<p>The streak in question is highlighted in red, so it appears we expect it
to happen 0.169% of his games. The Realization Of course I did all of
this before looking up the <a href="http://espn.go.com/nba/truehoop/miamiheat/notebook/_/page/heatreaction-110203/miami-heat-orlando-magic">actual
article</a>.
I'll quote the blurb here:</p>
<blockquote>
<p>LeBron James set a personal record by making his first 11 field goals
to start the game. His previous career-high was 10 straight field
goals after tip-off, recorded against Chicago in 2008. After hitting
his first 11 field goal attempts on Thursday night, James shot
6-for-14 thereafter.</p>
</blockquote>
<p>Well... this calculation just got a bit easier. He has played 667 games,
and the probability of getting 11 straight off the bat is q^k =
0.475^11 = 0.0002777. Multiply this by 667 games to get the expected
value of 0.185. Sure this is 6 times smaller than our previous
calculation; however it's still not statistically that impressive. How
would LeBron's expected number change if he shot the same percentage
(72.7%) as Wilt for his record breaking season? The expected number of
games with 11 in a row during the game would be 72.38 games!! So this is
incredibly dependent on the shooting percentage. We have a factor of
q^k everywhere! Certainly it's dependent on the number of shots taken
in a game too. The probability P(N) is a monotonically increasing
function! Moral Given LeBron's shooting percentage and high number of
shots per game, we expect that he would have at least 1 of these streak
of 11 games so far in his career. This is certainly not to diminish this
feat though. You still need to take 20 some shots a game in the NBA with
nearly 50% shooting accuracy! We also have a nice formula to apply to
more sports streaks! More to come...</p></div>basketballBohnlebron jamesstatisticshttps://thephysicsvirtuosi.com/posts/old/fun-fact-lebron-james-plays-basketball/Mon, 07 Feb 2011 12:37:00 GMT
- Breaking Intuitionhttps://thephysicsvirtuosi.com/posts/old/breaking-intuition/Bohn<p>When I walked into my first day of physics class in high school, I
carried with me a set of ideas which I learned from simply observing and
interacting with the world. In fact everyone builds up what they believe
to be intuitive concepts, whether it be in science, math, or any other
field. Without any scientific training whatsoever, we begin to build
intuition. If you let go of a ball in the air, what will happen? If you
try to run on the ice of a frozen lake, will it be easier than running
on the sidewalk? If you stand in the sun and on the ground you see a
strange dark misshapen copy of yourself imitating your every move... who
is following you? Unfortunately we run into an issue when our intuition
disagrees with experimental results or someone else’s intuition. At that
point, it is essential to break down and analyze our intuition to find
where any problems in our logic may exist. This process of continually
breaking down and analyzing intuition is key to progressing in science.
<a href="http://4.bp.blogspot.com/_CPJjnXOJ-mQ/TJZ5DVIGsPI/AAAAAAAAABU/-22lS-Cq6_U/s1600/ThreeDice.jpg"><img alt="image" src="http://4.bp.blogspot.com/_CPJjnXOJ-mQ/TJZ5DVIGsPI/AAAAAAAAABU/-22lS-Cq6_U/s320/ThreeDice.jpg"></a>Let's
take a look at a simple dice game. The rules of the game dictate that
you pick a die first, then I pick a die, then we roll together 100 times
(we’re really bored, apparently). The winner is the person who rolls a
higher number more times in 100 rolls. The catch is that the numbers are
not the standard 1-6 on each die, but a magic set of numbers which may
repeat any number from 1-6 as many times as desired, for example {1, 2,
3, 4, 5, 5}. "Sounds easy," you say, as you pick up the yellow die. I
choose blue. We roll, and I win 74 out of 100 times. "Obviously the blue
die is better, give me that one," you say. I proceed to pick up the
green die, lo and behold, I win 63 out of 100 times. "Okay okay, I've
got the hang of it now. Clearly the green die is better than all of the
rest." I choose the yellow die and win 65 out of 100 times. In a fit of
rage you proclaim "witchcraft" and storm off for your witch-hunt gear.
There is no deception here with the exception of logic, the younger
sister of witchcraft. It is actually an interesting challenge to try to
come up with a set of numbers which will yield the following result: The
probability of the value on the blue die being higher than the value on
the yellow die is greater than 1/2. The probability of the value on the
green die being higher than the value on the blue die is greater than
1/2. The probability of the value on the yellow die being higher than
the value on the green die is greater than 1/2. There are definitely a
multitude of possible solutions, so I encourage you to attempt to find
one using only numbers 1-6 before scrolling down. Got a solution? Let's
take a look at the following dice: Yellow : {1, 4, 4, 4, 4, 4} Blue :
{2, 2, 2, 5, 5, 5} Green : {3, 3, 3, 3, 3, 6} I should note that my
solution is set up to have no ties, which makes the analysis a bit more
straightforward. It is certainly possible to come up with interesting
solutions which allow ties.
<a href="http://1.bp.blogspot.com/_CPJjnXOJ-mQ/TJZ436UNErI/AAAAAAAAABM/m-BpJNofQFk/s1600/Screen+shot+2010-09-19+at+4.10.27+PM.png"><img alt="image" src="http://1.bp.blogspot.com/_CPJjnXOJ-mQ/TJZ436UNErI/AAAAAAAAABM/m-BpJNofQFk/s320/Screen+shot+2010-09-19+at+4.10.27+PM.png"></a>The
chart on the right shows how each die compares to the others. The color
of each square indicates the winner when the number of the same row and
column are compared. We can see that blue beats yellow 21 out of 36
times, green beats blue 21 out of 36 times, and yellow beats green 25
out of 36 times. So this combination of dice will show the
non-transitive effect we were looking for. So I explain this “sorcery”
to you, before you try to burn me at the stake for being a witch, and
you calm down. Now I tell you that I’d like to try a new game. I select
two dice of the same color, then you get to select two dice of the same
color, then we roll both pairs 100 times. The winner this time is the
person who rolls a higher total, the sum of their two dice, more times
in 100 rolls. I select two yellow dice. After learning of my trick, you
decide to pick two blues and proceed to lose 60 out of 100 times. You
declare, “‘tis but a statistical error, let’s have another go!” I select
two blues and you, two greens. I win again! Just to rub it in, I choose
green and you choose yellow, and I win once again. Softly weeping, you
listen as I explain that the probabilities have now switched! The chart
on the right shows the different sums that are possible for a given set
of colored dice. When you
look<a href="http://3.bp.blogspot.com/_CPJjnXOJ-mQ/TJZ4wlSQh4I/AAAAAAAAAA0/p69QSdtqFdc/s1600/Screen+shot+2010-09-19+at+4.10.38+PM.png"><img alt="image" src="http://3.bp.blogspot.com/_CPJjnXOJ-mQ/TJZ4wlSQh4I/AAAAAAAAAA0/p69QSdtqFdc/s320/Screen+shot+2010-09-19+at+4.10.38+PM.png"></a>
at the possible sum 4 for the blue dice, you see that 4 can meet up with
2 a total of nine times, with blue winning each, 4 can meet up with 5 a
total of 90 times with yellow winning each, and can meet up with 8 a
total of 225 times with yellow winning each. So the value in each cell
is the number of times each match-up can occur, with the color of the
cell showing who will win each match-up. There are 6^4 = 1296
possibilities, so winning half corresponds to 648. This dice trick is an
example of non-transitive logic, which can certainly be a non-intuitive
topic (Stay tuned for some non-transitive logic involving coins!). In
this case, you must break your intuition that there must be one “best”
die. In science, it’s a great idea to try to look for other examples of
the behavior you are observing to help reinforce what you’ve learned. It
turns out that one of the most basic schoolyard games involves
non-transitive logic! In the game of rock, paper, scissors, we find that
rock crushes scissors, scissors cuts paper, and paper covers rock. This
is analogous to the behavior of our special dice, and I believe makes
the logic much easier to understand. Compare against your intuition,
break down and analyze, build up and reinforce.</p>Bohndiceintuitionlogicnontransitivewitchcrafthttps://thephysicsvirtuosi.com/posts/old/breaking-intuition/Sun, 19 Sep 2010 16:52:00 GMT