The Virtuosi (Posts by Jesse)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- End of the Earth VII: The Big Freezehttps://thephysicsvirtuosi.com/posts/old/end-of-the-earth-vii-the-big-freeze/Jesse<div><hr>
<p><a href="http://1.bp.blogspot.com/-c8vJR4CVwZc/T5R7_62SLuI/AAAAAAAAAHU/POCT5Fhx-CQ/s1600/Space_Scene_Frozen_Earth_WP_BG_by_PimArt.jpg"><img alt="image" src="http://1.bp.blogspot.com/-c8vJR4CVwZc/T5R7_62SLuI/AAAAAAAAAHU/POCT5Fhx-CQ/s320/Space_Scene_Frozen_Earth_WP_BG_by_PimArt.jpg"></a>
http://tinyurl.com/7rdj996</p>
<hr>
<p>It is traditional here at The Virtuosi to
<a href="http://thevirtuosi.blogspot.com/2010/04/end-of-earth-physics-i.html">plot</a>
<a href="http://thevirtuosi.blogspot.com/2010/04/end-of-earth-ii-blaze-of-glory.html">the</a>
<a href="http://thevirtuosi.blogspot.com/2010/04/end-of-earth-physics-iii-asteroids.html">destruction</a>
<a href="http://thevirtuosi.blogspot.com/2011/04/end-of-earth-iv-shocking-destruction.html">of</a>
<a href="http://thevirtuosi.blogspot.com/2011/04/end-of-earth-v-there-goes-sun.html">the</a>
<a href="http://thevirtuosi.blogspot.com/2011/04/end-of-earth-vi-nanobot-destruction.html">earth</a>.
We also are making secret plans for our volcano lair and death ray.
However, since it is earth day, we will only share with you the plans
for the total doom of the earth, not the cybernetically enhanced guard
dogs we're building for our <a href="http://thevirtuosi.blogspot.com/2012/04/earth-day-2012-escape-to-moon.html">moon
base</a>.
The plan I reveal today is elegant in its simplicity. I intend to alter
the orbit of the earth enough to cause the earth to freeze, thus ending
life as we know it. According to the internet at large, the average
surface temperature of the earth is \~15 C. This average surface
temperature is directly related to the power output of the sun. More
precisely, it is directly related to the radiated power from the sun
that the earth absorbs. Assuming that the earth's temperature is not
changing (true enough for our purposes), the then power radiated by the
earth must be equal to the power absorbed from the sun. More precisely
$$ P_{rad,earth}=P_{abs,sun}$$ Now, the radiated power goes as
$$P_{rad}=\epsilon \sigma A_{earth} T^4 $$ where A_earth is the
surface area of the earth, T is the temperature of the earth, and
epsilon and sigma are constants. I'll be conservative and say that I
want to cool the temperature of the earth down to 0 C. The ratio of the
power the earth will emit is
$$\frac{P_{new}}{P_{old}}=\frac{T_{new}^4}{T_{old}^4} \approx
.81$$ Note that the temperature ratio must be done in Kelvin. The power
radiated by the sun (or any star) drops off as the inverse square of the
distance from the sun to the point of interest: $$P_{sun} \sim
\frac{1}{r^2} $$ To reduce the power the earth receives from the sun
to 81% of the current value would require
$$\frac{P_{sun,new}}{P_{sun,old}}=\frac{r_{old}^2}{r_{new}^2}=.81
$$ This tells us that the new earth-sun distance must be larger than the
old (a good sanity check). In fact, it gives $$r_{new}=1.11 r_{old} $$
So I'll need to move the earth by 11% of the current distance from the
earth to the sun. No small task! The earth is in a circular orbit (or
close enough). To change to a circular orbit of larger radius requires
two applications of thrust at opposite points in the orbit It turns out
that the required boost in speed (the ratio of the speeds just before
and after applying thrust) for the first boost of an object changing
orbits is given by
$$\frac{v_{f}}{v_{i}}=\sqrt{\frac{2R_{f}}{R_i+R_f}}=1.026$$ To
move from the transfer orbit to the final circular orbit requires
$$\frac{v_{f}}{v_{i}}=\sqrt{\frac{R_{i}+R_f}{2R_i}}=1.027$$ Note
that despite the fact that we boost the velocity at both points, the
velocity of the final orbit is less than that of the initial. Now, how
could we apply that much thrust? Well, the change in momentum for the
earth from each stage is roughly (ignoring the slight velocity increase
of the transfer orbit) $$\Delta p = .03M_E v_E $$ The mass of the
earth is \~6<em>10^24 kg, the orbital velocity is \~30 km/s, so $$\Delta
p = 5\cdot 10^{27} kg</em>m/s$$ A solid rocket booster (the booster
rocket used for shuttle launches, when those still happened) can apply
about 12 MN of force for 75 s (thank you wikipedia). That's a net
momentum change of \~900 <em>10^9 kg</em>m/s (900 billion!). So we would
only need $$\frac{2*5\cdot 10^{27}}{9\cdot 10^{11}}=12 \cdot
10^{15}$$ That's right, only 12 million billion booster rockets! With
those I can freeze the earth. I assure you that this plan is proceeding
on schedule, and will be ready shortly after we have constructed our
volcano lair.</p></div>https://thephysicsvirtuosi.com/posts/old/end-of-the-earth-vii-the-big-freeze/Sun, 22 Apr 2012 19:34:00 GMT
- Grains of Sandhttps://thephysicsvirtuosi.com/posts/old/grains-of-sand/Jesse<div><p><a href="http://3.bp.blogspot.com/-87-vnzGa9Po/TiRR2qFWprI/AAAAAAAAAF0/KsfRQhoL5Ds/s1600/SandUDunesUSoft.jpg"><img alt="image" src="http://3.bp.blogspot.com/-87-vnzGa9Po/TiRR2qFWprI/AAAAAAAAAF0/KsfRQhoL5Ds/s320/SandUDunesUSoft.jpg"></a></p>
<p>Have you ever sat on a beach and wondered how many grains of sand there
were? I have, but I may be a special case. Today we're going to take
that a step further, and figure out how many grains of sand there are on
the entire earth. (Caveat: I'm only going to consider sand above the
water level, since I don't have any idea what the composition of the
ocean floor is). I'm going to start by figuring out how much beach there
is in the world. If you look at a map of the world, there are four main
coasts that run, essentially, a half circumference of the world. We'll
say the total length of coast the world has is roughly two
circumferences. As an order of magnitude, I would say that the average
beach width is 100 m, and the average depth is 10 m. This gives a total
beach volume of $$ (100 m)(10 m)(4 \pi (6500 km) )= 82 km^3$$ That's
not a whole lot of volume. Let's think about deserts. The Sahara desert
is by far the largest sandy desert in the world. Just as a guess, we'll
assume that the rest of the sandy deserts amount to 20% (arbitrary
number picked staring at a map) as much area as the Sahara. According to
wikipedia the area of the Sahara is 9.4 million km^2. We'll take, to an
order of magnitude, that the sand is 100 m deep. 10 m seems to little,
and 1 km too much. That amounts to \~1 million km^3 of sand. We're
going to assume that a grain of sand is about 1 mm in radius The volume
occupied by a grain of sand is then 1 mm^3. Putting that together with
our previous number for the occupied volume gives $$ \frac{1\cdot
10^6 km^3}{1 mm^3}=\frac{1 \cdot 10^{15}}{1\cdot
10^{-9}}=1\cdot 10^{24}$$ That's a lot of grains of sand. Addendum:
Carl Sagan is quoted as saying</p>
<blockquote>
<p>"The total number of stars in the Universe is larger than all the
grains of sand on all the beaches of the planet Earth"</p>
</blockquote>
<p>If we just use our beach volume, that gives a total number of grains of
sand as \~1*10^20, which is large, but not as large as what we found
above. Is that less than the number of stars in in the universe? Well,
that's a question for another day (or google), but the answer is, to our
best estimate/count, yes.</p></div>carl saganorder of magnitudesandhttps://thephysicsvirtuosi.com/posts/old/grains-of-sand/Mon, 18 Jul 2011 11:33:00 GMT
- Lifetime of Liquid Waterhttps://thephysicsvirtuosi.com/posts/old/lifetime-of-liquid-water/Jesse<p><a href="http://3.bp.blogspot.com/-fyjvPBm_INs/ThpaZFszL5I/AAAAAAAAAFw/6sJBTUj905c/s1600/water_drop.jpg"><img alt="image" src="http://3.bp.blogspot.com/-fyjvPBm_INs/ThpaZFszL5I/AAAAAAAAAFw/6sJBTUj905c/s320/water_drop.jpg"></a>
Apologies for the hiatus recently, it's been a busy time (when isn't
it). I hope to get back to talking about experiments soon, but for now I
wanted to write up a quick problem I thought up a while back. The
question is this: how long does a molecule of H2O on earth remain in the
liquid state, on average? I'm going to treat this purely as an order of
magnitude problem. I'm also going to have to start with one assumption
that is almost certainly inaccurate, but makes things a lot easier. I'm
going to assume perfect mixing of all of the water on earth. Given that
assumption, I really only need to figure out two things. The first is
how much liquid water there is on earth. The second is now much liquid
water leaves the liquid phase each year. Let's start with the total
amount of liquid water on earth. This is relatively easy to estimate. I
happen to know that about 70% of the earth's surface is covered in
water. Most all of that is ocean. To an order of magnitude, the average
depth of the ocean must be 1 km, as it is certainly not 100 m or 10 km
[1]. For a thin spherically shell, the volume of the shell is roughly $$
4 \pi r_e^2 \Delta r $$ where r_e is the radius of the earth. Thus,
the total volume of water on the earth is $$.7<em>4 \pi r_e^2 (1 km)$$
Now, we need to figure out how much H20 leaves the liquid phase every
year. To an order of magnitude, it rains 1 m everywhere on earth each
year, it's not .1 m or 10 m [2]. I'm going to ignore any
freezing/melting in the ice caps, assuming that is small fraction of the
water that leaves the liquid phase each year. Since we have a closed
system, all the water that rains must have left the liquid phase. So, on
average, the total volume of water that leaves the liquid phase is $$4
\pi r_e^2 (1 m) $$ Thus, the fraction of liquid water that changes
phase per year is $$ \frac{4 \pi r_e^2 (1m)}{.7</em>4\pi r_e^2 (1
km)} = .0014 $$ This means that, given my assumption of perfect mixing,
in somewhere around 1/.0014 = 700 yr all of the water on earth will have
cycled through the vapor phase. Since we're only operating to an order
of magnitude, I'll call this 1000 years. This is the answer to our
question if every molecule has been in the vapor phase once in 1000
years, then we expect a molecule to stay in the liquid phase for 1000
years [1] According to wikipedia, this is really about 4 km, so we're
underestimating a bit. [2] According to wikipedia, this is spot on (.99
m on average).</p>funorder of magnitudephysicswaterhttps://thephysicsvirtuosi.com/posts/old/lifetime-of-liquid-water/Sun, 10 Jul 2011 22:07:00 GMT
- Anatomy of an Experiment I - The Questionhttps://thephysicsvirtuosi.com/posts/old/anatomy-of-an-experiment-i-the-question/Jesse<div><hr>
<p><a href="http://4.bp.blogspot.com/-V-68a5ev1W0/TcNG1UXmaBI/AAAAAAAAAEY/0f-Uq2EjMJY/s1600/installation_of_world_largest_silicon_tracking_detector.jpg"><img alt="image" src="http://4.bp.blogspot.com/-V-68a5ev1W0/TcNG1UXmaBI/AAAAAAAAAEY/0f-Uq2EjMJY/s320/installation_of_world_largest_silicon_tracking_detector.jpg"></a>
Warning: picture has little or no relation to this post.</p>
<hr>
<p>I realized the other day that I've seen a lot of people talk about
research results, but it is much more rare that I see someone talk about
how we do research. I think that may be because, to us as scientists,
the process is second nature. We've been doing it for years. Other folks
may be less familiar with the process though. With this in mind, I'm
going to do a short series of posts focused on how we do an experiment.
Not the results, not so much the physics, but the process that we go
through to create, setup, and carry out an experiment. As my example
I'll use a short little experiment that I built from the ground up in
the last few weeks, that I'm currently in the process of (hopefully)
wrapping up. Today I'll talk about the driving force behind almost any
experiment: The Question. It might be argued that there are two types of
experiments. There are those that set out to answer a specific question,
and those that set out to explore what happens under certain conditions
(explore some part of phase space). An example of the first type that
comes immediately to mind is the recently announced results from
<a href="http://science.nasa.gov/science-news/science-at-nasa/2011/04may_epic/">Gravity Probe
B</a>
(GP-B). This was an experiment designed with one goal in mind, to test
the validity of einstein's theory of general relativity, specifically
geodesic precession and frame dragging. They asked the question, built
the apparatus, and then got results. Here's a spoiler from the article:
Einstein was right, to remarkable precision. I'm going to mostly ignore
the second type of experiment. While very important, I argue that those
exploratory experiments are (almost?) always done on experimental
apparatus that was built for another experiment. You don't spend the
time, money, and energy to build an experimental apparatus without
having good evidence that it's worth doing, that is, without expecting
to see something. This brings me to The Question. The name might be
misleading, the motivation for an experiment might not be a question
(though it can usually be phrased as one). One common motivation is to
test theoretical predictions, as was the case with GP-B. Theory without
experimental verification is empty. It may sound nice, but we can't
trust it unless we've tested it against what nature actually does.
Sometimes theory develops because of experimental results, for example
the knowledge of the quantization of light came out of anomalous
experimental effects of the photoelectric effect and blackbody radiation
(among others). Other times, experiment develops to test theory, the
GP-B and the Large Hadron Collider for example. Another common
motivation is a question based on a physical observation, for example:
<a href="http://physicsbuzz.physicscentral.com/2011/04/small-insects-paddle-through-air.html">how does a fly
fly</a>?
That question is, as these things go, very simply stated. For an idea of
how complicated they can get, just take a look at any recent collection
of articles from any physics journal, wherein we find things like the
form and source of 'itinerant magnetism in FeAs' (grabbed from a recent
Phys. Rev. B article). I classify a third type of question, one that is
more process based: "How can we do X?". This third category is where the
question that motivates (at least in the broad sense) the experiment I'm
going to describe comes from. I can phrase it as: "How can we
successfully cryopreserve biological samples?" For those unfamiliar with
biological cryopreservation, this is something I discussed <a href="http://thevirtuosi.blogspot.com/2010/05/cryopreservation.html">almost a
year
ago</a>.
From there, we get into smaller questions, most of those are type two,
based on physical observations. This particular small experiment has
grown out of my work on cryopreservation, and has more to do with the
structure of water on freezing. Over the past year, one of the projects
I've been working on has been to measure the so called critical warming
rate of aqueous solutions. This is the rate at which you have to warm
vitreous aqueous solutions (see my earlier cryopreservation post for
more details) to prevent ice formation on warming. The question that has
grown out of this work is: how does the cooling history of my vitreous
sample affect the critical warming rate? Having arrived at the question,
we'll next discuss the apparatus.</p></div>cryopreservationexperimentphysicsquestionhttps://thephysicsvirtuosi.com/posts/old/anatomy-of-an-experiment-i-the-question/Thu, 05 May 2011 21:06:00 GMT
- End of the Earth IV - Shocking Destructionhttps://thephysicsvirtuosi.com/posts/old/end-of-the-earth-iv-shocking-destruction/Jesse<p><a href="http://4.bp.blogspot.com/-aa4EF60W7m0/TbCwQ8Vc3WI/AAAAAAAAAEU/03HiJiGJ6hc/s1600/exploding-earth11.jpg"><img alt="image" src="http://4.bp.blogspot.com/-aa4EF60W7m0/TbCwQ8Vc3WI/AAAAAAAAAEU/03HiJiGJ6hc/s200/exploding-earth11.jpg"></a>
Earth day is upon us once more. So many other namby-pamby bloggers out
there (don't hurt me!) are writing about how wonderful the earth is and
how great earth day is. We here at The Virtuosi take a more hardline
approach. Today I'm going to tell you how to destroy the earth.
Completely and totally. Unlike
<a href="http://thevirtuosi.blogspot.com/2010/04/end-of-earth-physics-i.html">last</a>
<a href="http://thevirtuosi.blogspot.com/2010/04/end-of-earth-ii-blaze-of-glory.html">year's</a>
<a href="http://thevirtuosi.blogspot.com/2010/04/end-of-earth-physics-iii-asteroids.html">methods</a>,
this one should work. In fact, this method is so simple that I can tell
you what to do right now. Just tweak the charge on the electron so it is
a bit out of balance with the charge on the proton. Just a little bit.
How little a bit, you might ask? A very little bit. Really, this doesn't
sound hard. I mean, sure, you have to do it for all of the electrons in
the earth, but we're talking about a very very small percentage change.
Not convinced? Let me show you just how small a change we're talking. If
there is a charge imbalance in the electron and the proton, this will
give the earth a net charge, throughout it's volume. I've got to make a
few assumptions about the earth here, so hold on. I'm going to assume
that the earth is a uniform density everywhere, and I'm going to assume
that the earth is made entirely of iron.<em> Now, the net charge of any
iron atom will be $$ (q_e-q_p)Z=(q_e-q_p)26$$ where Z is the atomic
number of iron, the number of protons (and electrons) the atom has. The
net charge of the earth, Q, is the number of iron atoms, N, times this
charge, $$Q=(q_e-q_p)ZN$$ I've <a href="http://thevirtuosi.blogspot.com/2010/04/end-of-earth-ii-blaze-of-glory.html">previously
estimated</a>
that N is about 3</em>10^50 atoms. Now, the electric potential energy of a
sphere of radius r with charge q uniformly distributed throughout it's
volume is $$U_e=\frac{3kq^2}{5r}$$ where k is the coulomb constant.
Dissolution of the earth will occur when the electrostatic energy of the
earth equals the gravitational potential energy of the earth. The
gravitational bound energy of the earth is given by
$$U_g=\frac{3GM^2}{5R}$$ Where M is the mass of the earth, G is
Newton's gravitational constant, and R is the radius of the earth.
Setting this equal to the electrostatic energy of the earth,
$$\frac{3GM^2}{5R}=\frac{3kQ^2}{5R}$$ $$Q^2=\frac{G}{k}M^2$$ so
$$(q_e-q_p)ZN=\left(\frac{G}{k}\right)^{1/2}M$$ Now, N is given,
in our approximations, by $$N=\frac{M_{earth}}{m_{iron}}$$ so
$$q_e-q_p=\left(\frac{G}{k}\right)^{1/2}\frac{m_{iron}}{Z_{iron}}$$
Now we can plug in some numbers. G=6.7<em>10^-11 m^3</em>kg^-1<em>s^-2, k=
9</em>10^9 m^3<em>kg</em>s^-2<em>C^-2, m_iron=9</em>10^-26 kg, Z=26. Thus,
$$q_e-q_p=3<em>10^{-37} C$$ To put this in perspective, the charge on
the electron is 1.6</em>10^-19 C, so this is roughly 10^18 times less
than that charge. Put another way, if the charge on the electron was
imbalanced from that of the proton by roughly 1 part in 10^18, the
earth would cease to exist due to electrostatic repulsion.<strong> As I told
you at the beginning, you only have to change the charge by a very small
amount! So get working. There are only about
1000000000000000000000000000000000000000000000000000 electrons you need
to modify! *According to the internet, the density of the earth, on
average, is roughly 5.5 g/cm^3. The density of iron is 7.9 g/cm^3 at
room temperature, and the density of water is 1 g/cm^3 at room
temperature. So, while the earth is not entirely iron (of course), it is
a better approximation to assume the earth is iron than the earth is
water. And those, of course, were really our only two choices. </strong> It
turns out that this is a good argument for the charge balance of the
electron and the proton.</p>chargeelectronend of the earthprotonhttps://thephysicsvirtuosi.com/posts/old/end-of-the-earth-iv-shocking-destruction/Fri, 22 Apr 2011 07:54:00 GMT
- Blown Awayhttps://thephysicsvirtuosi.com/posts/old/blown-away/Jesse<div><p><a href="http://2.bp.blogspot.com/-NFVShpYzscc/TYutYtPSIVI/AAAAAAAAAEM/vRA8KbFV6tc/s1600/wind-turbine-2.jpg"><img alt="image" src="http://2.bp.blogspot.com/-NFVShpYzscc/TYutYtPSIVI/AAAAAAAAAEM/vRA8KbFV6tc/s200/wind-turbine-2.jpg"></a></p>
<p>I was reading a discussion on green energy recently, in particular wind
power, where the following claim was made</p>
<blockquote>
<p><em>enough wind turbines to power the world would cover the surface of
the world.</em></p>
</blockquote>
<p>Now, this was quickly decried by supporters of wind power, but the claim
has stuck with me. The question on my mind today is: How much of the
earth's surface would have to be covered to power the earth with wind
turbines? We can't hope to put an exact number on this, the best we'll
be able to do is an order of magnitude. I also don't know much about
wind turbines, so I'll be making liberal use of
<a href="http://en.wikipedia.org/wiki/Wind_turbine">wikipedia</a> as I go. Let's
start with the size of the wind turbine. According to wikipedia the
largest wind turbine has a rotor sweep diameter of 128 m. To an order of
magnitude, we'll say that our average wind turbine has a diameter of 100
m. Next we need to know how much power this puts out. The maximum power
of this turbine is \~8 MW. However, it certainly wouldn't be producing
that at all times. Current wind farms produce around 20-30% maximum
capacity. However, these turbines are careful placed in areas of high
wind. We're not going to get that lucky with our wind dose when we place
our turbines haphazardly, so we'll assume they produce at 1% maximum
capacity. According to
<a href="http://en.wikipedia.org/wiki/World_energy_resources_and_consumption">wikipedia</a>,
the world energy consumption in 2008 was 474 EJ (exajoules), or an
average power use in 2008 of 15 TW. To an order of magnitude then, the
area we'd have to occupy with wind turbines to power the world would be:
$$\left( \frac{(100\text{ m})^2}{1\text{
turbine}}\right)\left(\frac{1\text{ turbine}}{.01<em>8 \text{
MW}}\right)15 \text{TW} = 2\cdot 10^{12}\text{ m}^2$$ That's
2</em>10^6 km^2, or, in english, 2 million square kilometers. For
comparison, the land area of the united states is roughly 10 millon
square kilometers. So we'd only have to cover 1/5th of the united states
with wind turbines to power the entire world (in 2008, no doubt power
use has risen since then)! While that is a lot of space taken up, it is
nowhere near the entire surface of the world. There are, of course,
<a href="http://xkcd.com/556/">other concerns</a> about wind power. <em>Note: maybe
the wind turbines are less efficient overall. Also, I assumed that the
footprint was just the square area of the turbine diameter. I know this
is the size of the face of the turbine, but to an order of magnitude I
imagine it is correct for the space occupied on the ground.</em></p></div>green powerorder of magnitudewindhttps://thephysicsvirtuosi.com/posts/old/blown-away/Thu, 24 Mar 2011 17:46:00 GMT
- Japan Nuclear Crisishttps://thephysicsvirtuosi.com/posts/old/japan-nuclear-crisis/Jesse<p>Though I know that two posts in one day is recently unprecedented, I've
been meaning to post about the Japan nuclear crisis for a few days. The
various major news outlets are doing a good job, or so it seems, of
keeping us informed of the events going on over there. However, I found
myself rather puzzled over the physics of what was happening. From the
news articles I was unable to figure out what was actually causing the
meltdown, beyond some problem with the cooling. As a postdoc in my lab
asked, "Isn't all they have to do drop the control rods and the reaction
ends?" So I decided to do a little digging. I've found a couple of
places that do a nice job of explain some of the physics of what is
actually happening, <a href="http://blogs.nature.com/news/thegreatbeyond/2011/03/fukushima_crisis_anatomy_of_a.html">nature
news</a>
(not sure if the nature blogs are behind a paywall), and <a href="http://www.scientificamerican.com/article.cfm?id=fukushima-core">scientific
american</a>
(not up on current events, but a nice summary of what can/might go
wrong). I'm sure there are many other places doing a good job of
explaining things, but these are the ones and I found, and hopefully
they help clarify what is actually happening.</p>crisisjapannuclearphysicshttps://thephysicsvirtuosi.com/posts/old/japan-nuclear-crisis/Wed, 16 Mar 2011 10:36:00 GMT
- Falling Icehttps://thephysicsvirtuosi.com/posts/old/falling-ice/Jesse<div><p><a href="http://4.bp.blogspot.com/_SYZpxZOlcb0/TUINK4AQYVI/AAAAAAAAAEA/XWqewl-lS_o/s1600/shattered_windshield.jpg"><img alt="image" src="http://4.bp.blogspot.com/_SYZpxZOlcb0/TUINK4AQYVI/AAAAAAAAAEA/XWqewl-lS_o/s320/shattered_windshield.jpg"></a>
It's been a while since I posted anything, much to my shame. Hopefully
this post marks a change in that streak. Today I'm going to consider a
very practical application of all this physics stuff. One of my
housemates parks his car on the side of the house, with the front of the
car facing the house. Living in Ithaca, NY, the weather has been the
usual cold and snowy, like the rest of the northeast USA this winter.
Yet, early last week, we had some unusually warm weather, in the 30s
(fahrenheit). A few days later, my housemate went out to his car, and
discovered that falling chunks of ice had broken his windshield! Now, to
be clear here, I'm not talking about icicles, I'm talking about large,
block-like, chunks. My best guess is that during the warm days, snow on
the roof turned into chunks of ice, and slid off the roof. The question
I'm going to try to answer today is: How far from the house could these
chunks possibly land? Put another way, what I want to know is, how far
from the house would we have to park our cars to not risk broken
windshields from falling ice? <strong>The First Attempt</strong> We'll start with the
simplest assumptions we can think of. First, we'll assume that there is
no friction on the ice block as it slides down the roof. We'll also
assume there's no air resistance slowing down the ice in the air. The
maximum range will be given by a block of ice sliding from the top of
the roof. Taking the height of the peak of the roof as h, relative to
the edge of the roof, we can write down the magnitude of the velocity of
the ice chunk when it reaches the edge of the roof. We start by setting
the change in gravitational potential energy equal to the change in
kinetic energy. Recalling the form for both of these, $$PE=mgh$$
$$KE=\tfrac{1}{2}mv^2$$ we can set these equal and solve for v, $$mgh
= \tfrac{1}{2}mv^2$$ so $$|v|=\sqrt{2gh}$$ This should be a familiar
expression to anyone who went through introductory mechanics. Now, given
that the roof is at an angle theta, we can write down the x (horizontal)
and y (vertical) components of velocity, $$v_x=|v|\cos\theta$$
$$v_y=-|v|\sin \theta$$ where I've introduced a minus sign in the y
component of velocity to indicate that the ice chunk is falling. Now
that we have the velocity, we have to call upon some more kinematics. To
figure out how far the ice flies, we have to know how long it is in the
air. So we start by considering the vertical motion. The distance
traveled by an object with an initial velocity, v_0, and a constant
acceleration, a, is given by $$\Delta y=\tfrac{1}{2}at^2+v_0t$$ In
our case, the distance traveled is the height of the first two floors of
my house. The acceleration is that of gravity, g, and the initial
velocity is the y component of velocity we found above. We'd like to
find the time it takes to travel this distance. We have to be a little
careful with our minus signs, by our convention the acceleration is in
the negative direction, and the change in position is negative. Working
all of that out, and plugging in our known values, we get
$$\tfrac{1}{2}gt^2+|v|\sin \theta t - l =0$$ where l is the height
of the house. We can solve this for t, finding $$t=\frac{-|v|\sin
\theta + \sqrt{(|v|\sin \theta)^2+2gl}}{g}$$ The horizontal
distance traveled is simply the horizontal velocity times the time,
$$x=\frac{|v|\cos\theta}{g}(-|v|\sin \theta + \sqrt{(|v|\sin
\theta)^2+2gl})$$ a result that you may recognize as the 'projectile
range formula' (particularly if I brought the minus on the v sine theta
term into the sine, indicating that I'm firing at a negative angle, that
is, downwards). Having found that result, lets plug in our velocity, and
then some numbers. First,
$$x=\frac{\sqrt{2gh}\cos\theta}{g}(\sqrt{2gh}\sin \theta +
\sqrt{(2gh\sin^2 \theta+2gl})$$ Now, for some estimation. I'd say
that the height of the roof peak is 10 ft, the height of the first two
floors of the house is 20 ft, and the angle of the roof is 30 degrees.
Having made those estimates, now I just have to plug in all the numbers,
yielding $$x=5.2 m=17 ft$$ That's a very long range! Now, I didn't see
any chunks of ice that were more than about 7 ft from the house. So we
have to question what went wrong with the above derivation. Well, maybe
nothing went wrong. I did calculate the <em>maximum</em> range. It's quite
possible none of these ice chunks were from the very top of the roof.
Still, I'm inclined to think we may have overestimated. I'd say that our
initial velocity was too high. The ice, as it comes down the roof, will
have to push a bunch of snow out of the way. Even though it may not have
much friction with the roof, all that snow will slow it down, and reduce
the velocity with which it comes off. I'm just going to guess that about
half of the potential energy it had is lost to the snow and roof, as a
rough estimation. That would give a velocity $$|v| = \sqrt{gh}$$ and a
maximum distance of $$x= 4m = 13ft$$ which is closer to what I observed.
<strong>The Second Attempt</strong> I'm still not completely satisfied with the
previous work, the answer doesn't match my observation. As a wise man
(Einstein) once said, "make things as simple as possible, but no
simpler." I may be guilty of making the problem too simple here. So I'm
going to add back in air resistance. In general, we physicists like to
avoid this because it usually means we can't get nice, analytic
expressions as answers (like the one I have above). Instead, we usually
just have to calculate the result numerically. This isn't the end of the
world, and often times it is actually a bit easier, but it's not as
pretty looking. Still, to satisfy myself, and you, gentle reader, I will
step into that realm. We start by writing down the force on our block of
ice once it is falling. We've got gravity, and air resistance. Thus
$$\vec{F}=-mg\hat{y}-bv^2\hat{v}$$ I've input a drag force that goes
as v^2, and is in the opposite direction of v. The 'v direction' is a
cop out, because I didn't want to do the explicit direction, so lets fix
that. We'll have x and y components, and we note that the magnitude of v
times the direction of v is the velocity vector. So,
$$\vec{F}=-mg\hat{y}-bv\hat{v}_x-bv\hat{v}_y$$ Breaking this up
into components we get $$a_x=-\frac{bv}{m}v_x$$
$$a_y=-g-\frac{bv}{m}v_y$$ This is as far as we can take this work
analytically. I'll say a little more about the coefficient b. This
depends on the exact size and shape of the object, as well as the medium
it is moving through. I'm going to use $$b=.4\rho A$$ because that's
what we used for hay bales in my classical mechanics class years ago.
Here, rho is the density of air, and A is the surface area of the
object. I would estimate that the large face of the ice chunk is roughly
one square foot, or .1 m^2. I'd estimate the mass of the ice was around
2 kg. Now, for some magic. I've put all of this into mathematica, and
asked it to solve this numerically. First we have the plot for the full
initial velocity, $$v=\sqrt{2gh}$$</p>
<hr>
<p><a href="http://1.bp.blogspot.com/_SYZpxZOlcb0/TUILYLcLr3I/AAAAAAAAAD4/q_j8djBIPP0/s1600/Falling+Ice.jpg"><img alt="image" src="http://1.bp.blogspot.com/_SYZpxZOlcb0/TUILYLcLr3I/AAAAAAAAAD4/q_j8djBIPP0/s320/Falling+Ice.jpg"></a>
The solid line is with air resistance, the dashed line without air resistance. The plot shows vertical vs. horizontal distance, and the units are meters. (click to enlarge)</p>
<hr>
<p>Next we have the plot for the half initial velocity, $$v=\sqrt{gh}$$</p>
<hr>
<p><a href="http://3.bp.blogspot.com/_SYZpxZOlcb0/TUIMIxdlzLI/AAAAAAAAAD8/G3P9KWUB3mw/s1600/Falling+Ice2.jpg"><img alt="image" src="http://3.bp.blogspot.com/_SYZpxZOlcb0/TUIMIxdlzLI/AAAAAAAAAD8/G3P9KWUB3mw/s320/Falling+Ice2.jpg"></a>
The solid line is with air resistance, the dashed line without air resistance. The plot shows vertical vs. horizontal distance, and the units are meters. (click to enlarge)</p>
<hr>
<p>As you can see from the plots, in neither case does it make a large
difference, about .2 m. <strong>The Third Round</strong> The final thought that
occurs to me is that perhaps I got the angle of the roof wrong. That
would be quite easy. Humans are notoriously bad at estimating angles.
I'll plot the results (with air resistance) for 15, 30, and 45 degree
angles and the lower velocity.</p>
<hr>
<p><a href="http://3.bp.blogspot.com/_SYZpxZOlcb0/TUI5Q4qB5DI/AAAAAAAAAEE/9ydcM8EXzxA/s1600/Falling+Ice3.jpg"><img alt="image" src="http://3.bp.blogspot.com/_SYZpxZOlcb0/TUI5Q4qB5DI/AAAAAAAAAEE/9ydcM8EXzxA/s320/Falling+Ice3.jpg"></a>
The plot shows vertical vs. horizontal distance, and the units are meters. The red line is 15 degrees, the blue line is 30 degrees, and the black line is 45 degrees. (click to enlarge)</p>
<hr>
<p>In summary, the answer is unclear. What I really need to do is measure
the angle of my roof better, because there's a significant angle
dependence. It's also quite possible that we didn't see a maximum
distance hit (thankfully!). In addition, air resistance doesn't seem to
matter much in this particular problem, probably because the distance
the thing falls is short enough that terminal velocity is not reached.
Hopefully this gave you a bit of a taste of a more practical physics
problem, and how to approach air resistance (if you want to see the
mathematica code, let me know). The lesson here seems to be either don't
park too close to roofs, or have insurance for your windshield!</p></div>air resistanceicephysicshttps://thephysicsvirtuosi.com/posts/old/falling-ice/Thu, 27 Jan 2011 22:41:00 GMT
- Caught In The Rain IIhttps://thephysicsvirtuosi.com/posts/old/caught-in-the-rain-ii/Jesse<div><p><a href="http://1.bp.blogspot.com/_SYZpxZOlcb0/TLIIqVYyp-I/AAAAAAAAADo/KOuEncc3zFo/s1600/Rain.jpg"><img alt="image" src="http://1.bp.blogspot.com/_SYZpxZOlcb0/TLIIqVYyp-I/AAAAAAAAADo/KOuEncc3zFo/s200/Rain.jpg"></a>
I was rather proud of my last post about being <a href="http://thevirtuosi.blogspot.com/2010/09/caught-in-rain.html">caught in the
rain</a>. In
that post, I concluded that you were better off running in the rain, but
that the net effect wasn't incredibly great. However, when I told people
about it, the question I inevitably got asked was: What if the rain
isn't vertical? That's what I'd like to look at today, and it turns out
to be a much more challenging question. I'm still going to assume that
the rain is falling at a constant rate. Furthermore, I'm going to assume
that the angle of the rain doesn't change. With those two assumptions
stated, let me remind you of the definitions we used last time. $$\rho
- \text{the density of water in the air in liters per cubic meter}$$
$$A_t - \text{top area of a person}$$ $$\Delta t - \text{time
elapsed}$$ $$d - \text{distance we have to travel in the rain}$$ $$v_r
- \text{raindrop velocity}$$ $$A_f - \text{front area of a person}$$
$$W_{tot} - \text{total amount of water in liters we get hit with}$$
As a reminder, our result from last time was: $$W_{tot}= \rho d (A_t
\frac{v_r}{v} + A_f)$$ Now, let's look at the new analysis. As
before, let us consider the stationary state first. Our velocity now has
two components, horizontal and vertical. Analogous to the purely
vertical situation, we can write down the stationary state, but now we
have rain hitting both our top and front (or back). I'm going to define
the angle, theta, as the angle the rain makes with the vertical (check
out figure 1 below). this gives $$W = \rho d A_t v_r \cos(\theta)
\Delta t+\rho A_f v_r \sin(\theta) \Delta t$$ Let's check our
limits. As theta goes to zero (vertical rain), we only get rain on top
of us, and as theta goes to 90 (horizontal rain), we only get rain on
the front of us. Makes sense! Alright, so let's add in the effect of
motion now. This is going to be more challenging than in the vertical
rain situation. We're going to examine two separate cases</p>
<hr>
<p><a href="http://1.bp.blogspot.com/_SYZpxZOlcb0/TLIARlf0FBI/AAAAAAAAADY/uyx083CikOU/s1600/CITR+II+-+against.jpg"><img alt="image" src="http://1.bp.blogspot.com/_SYZpxZOlcb0/TLIARlf0FBI/AAAAAAAAADY/uyx083CikOU/s320/CITR+II+-+against.jpg"></a>
Fig. 1 - The rain, and our angle.</p>
<hr>
<p><strong>Case 1: Running Against The Rain</strong> This is the easier of the two
cases. After thinking about it for a while, I believe that it is the
same as when the rain is vertical. Let me explain why. If you are moving
with some velocity v, in a time t you will cover a distance x=v<em>t. Now,
suppose we paused the rain, so it is no longer moving, then moved you a
distance x, turned the rain back on, and had you wait for a time t. And
repeated this over and over until you got to where you were going. This
would result in an </em>average<em> velocity equal to v, even though it is not
a smooth motion. However, my claim is that in the limit that t and x go
to zero, this is a productive way of considering our situation. We note
that v=x/t, and in the limit that both x and t go to zero, that is the
</em>definition* of instantaneous velocity. The recap is, that my 'pausing
the rain' scheme of thinking about things is fine, as long as we
consider moving ourselves only very small distances over very short
times. Using this construction, we have an additional amount of rain
absorbed by moving the distance delta x of:
$$ \Delta W = \rho A_f \Delta x $$
$$ \Delta W = \rho A_f v \Delta t $$
This gives a net expression of
$$\Delta W = \rho A_t v_r \cos(\theta) \Delta t+\rho A_f v_r
\sin(\theta) \Delta t+\rho A_f v \Delta t $$
$$\Delta W = \rho A_f v \Delta t \left( \left(
\frac{A_t}{A_f}\right)
\left(\frac{v_r}{v}\right)\cos(\theta)+\left(\frac{v_r}{v}\right)\sin(\theta)
+ 1 \right)$$
As before, turning the deltas into differentials and integrating yields
$$W = \rho A_f v t \left( \left( \frac{A_t}{A_f} \right)
\left(\frac{v_r}{v}\right)\cos(\theta)+\left(\frac{v_r}{v}\right)\sin(\theta)
+ 1\right)$$
$$W=\rho A_f d \left( \left( \frac{A_t}{A_f}\right)
\left(\frac{v_r}{v}\right)\cos(\theta)+
\left(\frac{v_r}{v}\right)\sin(\theta) + 1 \right)$$
Note that when theta is zero, our vertical rain result gives the same
thing as we found in the last post (the first term lives, the second
term goes to zero, the third term lives). I'm going to use the
reasonable numbers I came up with in the last post. However, since we
have wind, we'll have to modify our rain velocity. More specifically,
we'll assume the rain has the same vertical component of velocity in all
cases. Then the wind speed, v_w, will be what controls the angle. More
exactly, the magnitude of the raindrop velocity will be
$$v_r=\sqrt{(6 m/s)^2+v_w^2}$$
While the angle will be
$$\theta=\tan^{-1}(v_w/ (6 m/s))$$
Next we note that
$$v_r\cos\theta = 6 m/s$$
which is just the vertical component of our rain. Similarly, the other
term is just the horizontal component of our rain. So we can write our
as a function of our velocity and the wind speed (the angle and wind
speed is interchangeable):
$$W = \rho A_f d\left( \left( \frac{A_t}{A_f}\right)
\left(\frac{6 m/s}{v}\right) +\left(\frac{v_w}{v}\right) +
1\right)$$
Using the reasonable numbers I came up with in my last post yields (with
a distance of 100m)
$$W = .2 liters \left( \left(\frac{.72
m/s}{v}\right)&+\left(\frac{v_w}{v}\right) + 1\right)$$
Once again, we have a least wet asymptote, which is the same as before.
I've plotted this function for various values of theta, and, more
intuitively, for various wind speeds (measured in mph, as we're used to
here in the US), and the plots are shown below (click to enlarge).
Unsurprisingly, you get the most wet when the rain is near horizontal,
but interestingly enough you can get the most percentage change from a
walk to a run when the rain is near horizontal. All angles are in
degrees.</p>
<hr>
<p><a href="http://3.bp.blogspot.com/_SYZpxZOlcb0/TLCFzqCBV2I/AAAAAAAAADI/iUpSnEgsrkM/s1600/Caught+in+Rain+II+-+running+against+theta.jpg"><img alt="image" src="http://3.bp.blogspot.com/_SYZpxZOlcb0/TLCFzqCBV2I/AAAAAAAAADI/iUpSnEgsrkM/s400/Caught+in+Rain+II+-+running+against+theta.jpg"></a>
Fig. 2 - How wet you get vs. how fast you run for various wind angles.</p>
<hr>
<hr>
<p><a href="http://2.bp.blogspot.com/_SYZpxZOlcb0/TK-CNTq4DTI/AAAAAAAAADE/wx39XVNB-f0/s1600/Caught+in+Rain+II+-+running+against+vw.jpg"><img alt="image" src="http://2.bp.blogspot.com/_SYZpxZOlcb0/TK-CNTq4DTI/AAAAAAAAADE/wx39XVNB-f0/s400/Caught+in+Rain+II+-+running+against+vw.jpg"></a>
Fig. 3 - How wet you get vs. how fast you run for various wind speeds in mph.</p>
<hr>
<p><strong>Case 2: Running With The Rain</strong>
This is the potentially harder case. We've got two obvious limiting
cases. If you run with the exact velocity of the rain and the rain is
horizontal, you shouldn't get wet. If the rain is vertical, it should
reduce to the result from my first post. We'll start with the stationary
case. This should be identical to case 1, if you're stationary it
doesn't matter if the rain is blowing on your front or back. That means
that for v=0, we should have $$\Delta W = \rho A_t v_r
\cos(\theta) \Delta t + \rho A_f v_r \sin(\theta) \Delta t$$
Now, let's use the same method as before, pausing the rain, advancing in
x, then letting time run. First we'll deal with our front side. Consider
figure 4.</p>
<hr>
<p><a href="http://3.bp.blogspot.com/_SYZpxZOlcb0/TLIDC2OwATI/AAAAAAAAADc/TfIMFDmOl7Y/s1600/CITR+II+-+with1.jpg"><img alt="image" src="http://3.bp.blogspot.com/_SYZpxZOlcb0/TLIDC2OwATI/AAAAAAAAADc/TfIMFDmOl7Y/s320/CITR+II+-+with1.jpg"></a>
Fig. 4 - Geometry for small delta x.</p>
<hr>
<p>Note that in front of us there is a rainless area, which we'll be
advancing into. Consider a delta x less than the length of the base of
that triangle. If we advance that delta x, we'll carve out a triangle of
rain as indicated, which, by some simple geometry, contains an amount of
rain $$\rho w \frac{(\Delta x)^2}{2 \tan(\theta)} = \rho w
\frac{v^2 (\Delta t)^2}{2 \tan(\theta)}$$ where w is the width of
our front. Now, consider if delta x is longer than the base of the
rainless triangle, as shown in figure 5.</p>
<hr>
<p><a href="http://1.bp.blogspot.com/_SYZpxZOlcb0/TLID3V-ITNI/AAAAAAAAADg/V2oek4ZMsC4/s1600/CITR+II+-+with2.jpg"><img alt="image" src="http://1.bp.blogspot.com/_SYZpxZOlcb0/TLID3V-ITNI/AAAAAAAAADg/V2oek4ZMsC4/s320/CITR+II+-+with2.jpg"></a>
Fig. 5 - Geometry for large delta x.</p>
<hr>
<p>We'll carve out an amount of rain equal to the indicated triangle plus
the rectangle. From the diagram we see this gives an amount of water
$$A_f \rho (\Delta x - h \tan(\theta)) + A_f \rho h
\tan(\theta)/2 = A_f \rho (\Delta x - \frac{h
\tan(\theta)}{2})$$ We could write two separate equations for these
two cases, but that's rather inefficient notation. I'm going to use the
<a href="http://en.wikipedia.org/wiki/Heaviside_step_function">Heaviside step
function</a>, H(x).
This is a function that is zero whenever the argument is negative, and 1
whenever the argument is positive. That means that for our front side,
$$\Delta W_f=\rho w \frac{v^2 (\Delta t)^2}{2 \tan(\theta)} H(
h\tan(\theta) - \Delta x) $$ $$+A_f \rho \left(\Delta x -
\frac{h \tan(\theta)}{2}\right)H(\Delta x - h \tan(\theta))$$
Note that I've written my step function in terms of the relative length
of delta x and the base of the rainless triangle. We get the first term
when delta x is less than the base length, and the second term when
delta x is more than the base length. Now, let us consider the rain
hitting our back. There are two cases here as well. First consider the
case where we're running with a velocity less than that of the rain. See
figure 6..</p>
<hr>
<p><a href="http://4.bp.blogspot.com/_SYZpxZOlcb0/TLIFgzj3yNI/AAAAAAAAADk/C5ImwwcoYME/s1600/CITR+II+-+with3.jpg"><img alt="image" src="http://4.bp.blogspot.com/_SYZpxZOlcb0/TLIFgzj3yNI/AAAAAAAAADk/C5ImwwcoYME/s320/CITR+II+-+with3.jpg"></a>
Fig. 6 - The back.</p>
<hr>
<p>We get two terms. There's the triangle of rain that moves down and hits
our back, shown above. Hopefully it is apparent that this is the same as
the triangle of rain we carved out with our front, and so will
contribute a volume of water $$\rho w \frac{v^2 (\Delta t)^2}{2
\tan(\theta)}$$ There's also the rain that manages to catch up with
us, $$A_f \rho (v_r \sin(\theta) \Delta t - \Delta x) =A_f \rho
\Delta t (v_r \sin(\theta) - v)$$ In the case where we outrun the
rain, we don't want this term, and our triangle gains a maximal length
of the horizontal and vertical components of the rain velocity times
delta t. We can write this backside term using a step function as
$$\Delta W_b =A_f \rho \Delta t \left(v_r \sin(\theta) - v +w
\frac{v^2 (\Delta t)^2}{2 \tan(\theta)}\right)H( v_r
\sin(\theta) - v)$$ $$+\rho w v_r^2 \Delta t^2
\frac{\sin(\theta)\cos(\theta)}{2} H(v-v_r\sin(\theta))$$ We can
combine these terms, with our usual top term, to get $$\Delta W =A_f
\rho \Delta t \left[ \left(v_r \sin(\theta) - v +\frac{w}{A_f}
\frac{v^2 (\Delta t)}{2 \tan(\theta)}\right)H(v_r \sin(\theta)
- v)$$ $$+ \frac{w}{A_f} v_r^2 \Delta t
\frac{\sin(\theta)\cos(\theta)}{2} H(v-v_r\sin(\theta) $$ $$+
\frac{w}{A_f} \frac{v^2 (\Delta t)}{2 \tan(\theta)} H(
h\tan(\theta) - \Delta x)$$ $$+\left(\frac{\Delta x}{\Delta t} -
\frac{h \tan(\theta)}{2 \Delta t}\right)H(\Delta x - h
\tan(\theta))+\frac{A_t}{A_f} v_r \cos(\theta) ] $$ I'm sure
this four line equation looks intimidating (I'm also sure that it is the
longest equation we've written here on the virtuosi!). But it'll
simplify when we take our limit as delta t goes to zero. Let's do this a
little more carefully than usual. $$\lim_{\Delta t \to
0}\frac{\Delta W}{\Delta t} =\lim_{\Delta t \to 0}A_f \rho
\left[ \left(v_r \sin(\theta) - v +\frac{w}{A_f} \frac{v^2
(\Delta t)}{2 \tan(\theta)}\right)$$ $$*H(v_r \sin(\theta) - v)+
\frac{w}{A_f} v_r^2 \Delta t
\frac{\sin(\theta)\cos(\theta)}{2} H(v-v_r\sin(\theta) $$ $$+
\frac{w}{A_f} \frac{v^2 (\Delta t)}{2 \tan(\theta)} H(
h\tan(\theta) - v \Delta t)$$ $$+\left(v - \frac{h
\tan(\theta)}{2 \Delta t}\right)H(v \Delta t - h
\tan(\theta))+\frac{A_t}{A_f} v_r \cos(\theta) ] $$ We'll take
this term by term. On the left side of our equality, we recognize the
definition of a differential of W with respect to t. Any term on the
right without a delta t we can ignore. The first term with a delta t is
$$\frac{w}{A_f} \frac{v^2 (\Delta t)}{2 \tan(\theta)}H(v_r
\sin(\theta) - v)$$ In all cases except when theta = 0, this term goes
to zero. Now, when theta = 0, tan(theta) = 0, so our limit gives zero
over zero, which is a number (note, I'm not being extremely careful. If
you'd like, tangent goes as the argument to leading order, so we have
two things going to zero linearly, hence getting a number back out).
However, looking at the step function, when theta goes to zero, we
likewise require v to be zero to get a value. However, our term goes as
v^2, so we conclude that in our limit, this term goes to zero. Next we
have $$\frac{w}{A_f} v_r^2 \Delta t
\frac{\sin(\theta)\cos(\theta)}{2} H(v-v_r\sin(\theta)$$ This
obviously goes to zero, no mitigating circumstances like a division by
zero. The next term is $$\frac{w}{A_f} \frac{v^2 (\Delta t)}{2
\tan(\theta)} H( h\tan(\theta) - v \Delta t)$$ This term presents
the same theta = 0 issues as the first term. The resolution is slightly
more subtle and less mathematical than before. Remember that this term
physically represents the rain that hits us when we move forward through
the section that our body hasn't shielded from the rain (see the drawing
above). I argue from a physical standpoint that when the rain is
vertical, this term would double count the rain we absorb with the next
term (which doesn't go to zero). I'm going to send this term to zero on
physical principles, even though the mathematics are not explicit about
what should happen. Next we have $$vH(v \Delta t - h \tan(\theta))$$
The argument of the step function makes it clear that to have any chance
at a non-zero value we need theta = 0. The mathematics isn't completely
clear here, as the value of a step function at zero is usually a matter
of convention (typically .5). Let's think physically about what this
term represents. This is the rain we absorb beyond the shielded region
(see above figure). This is the term I said the previous term would
double count with when the rain is vertical, so we're required to keep
it. However, only when theta = 0. I'm going to use another special
function to write that mathematically, the <a href="http://en.wikipedia.org/wiki/Kronecker_delta">Kronecker
delta</a>, which is 1 when
the subscript is zero, and zero otherwise. This is a bit of an odd use
of the Kronecker delta, because it's typically only used for integers,
but for those purists out there, there is an integral definition which
has the same properties for any (non-integer) value. Thus $$vH(v \Delta
t - h \tan(\theta))=v\delta_{\theta}$$ The last term we have to
concern ourselves with is $$- \frac{h \tan(\theta)}{2 \Delta t}H(v
\Delta t - h \tan(\theta))$$ Again, there is some mathematical
confusion when theta = 0, so we think physically again. This term
represents the rain in the unblocked triangle (see above). Obviously,
there is no rain in the triangle when theta is zero, because there is no
triangle! We set this term to zero as well. This gives us a much simler
expression than before,
$$\frac{dW}{dt} =A_f \rho \left[ (v_r \sin(\theta) - v)H(v_r
\sin(\theta) - v)+v\delta_{\theta}+\frac{A_t}{A_f} v_r
\cos(\theta) \right]$$
We can pull out a v and integrate with respect to t, giving
$$W=A_f \rho v t \left[ (\frac{v_r \sin(\theta)}{v} - 1)H(v_r
\sin(\theta) - v)+\delta_{\theta}+\frac{A_t}{A_f} \frac{v_r
\cos(\theta)}{v} \right]$$
As before, we can write this in terms of the wind velocity and the
vertical rain velocity,
$$W=A_f \rho d \left[ (\frac{v_w}{v} - 1)H(v_w -
v)+\delta_{v_w}+\frac{A_t}{A_f} \frac{v_{r,vert}}{v} \right]$$
This is a nice, simple expression that we can easily plot. There is one
thing that bothers me, I feel like there should be another step function
term that kicks in when your velocity exceeds the horizontal rain
velocity, and you start getting more rain on your front. But I'm going
to trust my analysis, and assert that such a term would be at least
second order in our work. If someone does find it, let me know! Using
the reasonable numbers from my last post gives $$W=.2 liters \left[
(\frac{v_w}{v} - 1)H(v_w - v)+\delta_{v_w}+\frac{.72 m/s}{v}
\right]$$ Because this post is long enough already, I've gone ahead and
plotted this only vs. wind velocity. I've also plotted the former least
wet asymptote. Most interesting (and you'll probably have to click on
the graph to enlarge to see this) is that there no longer is a least wet
asymptote! In theory if you run fast enough you can stay as dry as you
want.</p>
<hr>
<p><a href="http://1.bp.blogspot.com/_SYZpxZOlcb0/TLH8aqysIkI/AAAAAAAAADQ/onOsQT_VMvA/s1600/Caught+in+Rain+II+-+running+with.jpg"><img alt="image" src="http://1.bp.blogspot.com/_SYZpxZOlcb0/TLH8aqysIkI/AAAAAAAAADQ/onOsQT_VMvA/s400/Caught+in+Rain+II+-+running+with.jpg"></a>
Fig. 6 - How wet you get vs. how fast you run for various wind speeds in mph.</p>
<hr>
<p><strong>Comparison</strong>
I will conclude with a comparison of the two results, to each other and
to the vertical case. First, lets take the appropriate limits.
$$W_{with}=A_f \rho d \left[ (\frac{v_w}{v} - 1)H(v_w -
v)+\delta_{v_w}+\frac{A_t}{A_f} \frac{v_{r,vert}}{v} \right]$$
$$W_{against} = \rho A_f d\left( \left( \frac{A_t}{A_f}\right)
\left(\frac{v_{r,vert}}{v}\right) +\left(\frac{v_w}{v}\right) +
1\right)$$
$$W_{stationary} = \rho t A_f \left(\frac{A_t}{A_f}
v_{r,vert}+v_w\right)$$
$$W_{vert}= \rho d A_f \left(\frac{A_t}{A_f} \frac{v_r}{v} + 1
\right)$$
In the stationary limit, we have to break up the d in our equations into
v t, and that gives
$$\lim_{v \to 0}W_{with}= \lim_{v \to 0} W_{against}=\rho t
A_f \left(\frac{A_t}{A_f} v_{r_vert}+v_w\right)$$
While in the vertical rain limit
$$\lim_{v_w \to 0}W_{with}= \lim_{v_w \to 0} W_{against}
=\rho d A_f \left(\frac{A_t}{A_f} \frac{v_r}{v} + 1 \right)$$
So our limits work. Finally, it's a little hard to tell the difference
between the forward and backwards case, so I've plotted the two lines
together for a few values of v_w. You'll notice that for zero wind
speed they have the same result (which is good, since our limit was the
same), but for the other wind speeds they are remarkably divergent, more
so as you run faster! (again, click to enlarge)</p>
<hr>
<p><a href="http://3.bp.blogspot.com/_SYZpxZOlcb0/TLH87WlO8aI/AAAAAAAAADU/pOq_rEz2wkI/s1600/Caught+in+Rain+II+-+compare.jpg"><img alt="image" src="http://3.bp.blogspot.com/_SYZpxZOlcb0/TLH87WlO8aI/AAAAAAAAADU/pOq_rEz2wkI/s400/Caught+in+Rain+II+-+compare.jpg"></a>
Fig. 7 - Solid lines are running with the rain, dashed lines are running against the rain.</p>
<hr>
<p><strong>Conclusions</strong>
Hopefully this has been an interesting exercise for you. I know it
certainly took me longer to work and write than I initially thought.
While you can't see it in the post, there were a lot of scribblings and
thinking going on before I came to these conclusions. Most of it went
something like: "No, that can't be right, it doesn't have the right
(zero velocity/zero angle) limit!". I think this concludes all of the
running in the rain that I want to do, but if you have more followup
questions, post them below, and I'll do my best to answer. Also, I admit
that my analysis may be a bit rough, so if you have other approaches,
let me know. Finally, note that everything I've found favors running in
the rain, so get yourself some exercise and stay dry!</p></div>fun water rain humanhttps://thephysicsvirtuosi.com/posts/old/caught-in-the-rain-ii/Sun, 10 Oct 2010 14:41:00 GMT
- Caught In The Rainhttps://thephysicsvirtuosi.com/posts/old/caught-in-the-rain/Jesse<div><p><a href="http://1.bp.blogspot.com/_SYZpxZOlcb0/TH8RX3wsh6I/AAAAAAAAAC0/fIrDNg5flzY/s1600/boy+rain.gif"><img alt="image" src="http://1.bp.blogspot.com/_SYZpxZOlcb0/TH8RX3wsh6I/AAAAAAAAAC0/fIrDNg5flzY/s200/boy+rain.gif"></a>
There's an age old question that mankind has pondered. I'm sure that
noble heads such as Aristotle, Newton, and Einstein have pondered it. I
myself have raised it a few times. The question is: do you get more wet
running or walking through the rain? Now, I know that this question was
<a href="http://mythbustersresults.com/episode38">mythbusted</a> a while back. So
this is one of those situations where I know the result I want to get to
with my calculation: according to mythbusters running is better. Still,
I think formulating the question mathematically will be fun, plus if I
fail to agree with experiment everyone can mock me mercilessly. I'll
begin by stating a few assumptions. I'm going to assuming that the rain
is falling straight down, at a constant rate. I'm also going to assume
that if we are standing still, only our head and shoulders get wet, not
our front or back. With those in place, lets start by formulating the
expression for how wet we would get if we stood still. Well, take
$$\Delta W_{top} - \text{the change in water (in liters) on a person}
$$ $$\rho - \text{the density of water in the air in liters per cubic
meter}$$ $$A_t - \text{top area of a person}$$ $$\Delta t -
\text{time elapsed}$$ Intuition suggests that the rate at which
raindrops hit our top, times the area of our top, times the time we
stand in the rain, will give us the change in water. In an equation, $$
\Delta W_{top} = \rho A_t v_r \Delta t $$ Note that whatever
expression we generate for how wet we get when moving will have to
reduce to this form in the limit that we're not moving. This will be a
good check for us. Next, we need to define a few additional measures:
$$d - \text{distance we have to travel in the rain}$$
$$v_r - \text{raindrop velocity}$$
$$A_f - \text{front area of a person}$$
$$W_{tot} - \text{total amount of water in liters we get hit with} $$
Well, no mater how fast we run the rain will keep hitting us on the top
of our heads, so we're going to have our standing still term, plus
another term for how much hits us when running. How do we consider that?
Well, when we run, we're cutting into the swath of rainy air in front of
ourselves. We'll get hit on our frontside by all the additional
raindrops in that stretch we carve out. Mathematically, if we travel
some distance delta x in a time delta t, we'll get hit with an
additional amount of water
$$ \Delta W = \rho A_f \Delta x $$
$$ \Delta W = \rho A_f v \Delta t $$
We combine our two terms to get
$$\Delta W_{tot} = \rho \Delta t (A_t v_r + A_f v)$$
Note that if we stop walking (v goes to zero) we'll return to our
stationary expression. Next I'll take the delta t over to the other
side, turn that into a derivative, and integrate to get the total water
hitting us, not just the change for some delta t. Of course, since
everything else in the equation is constant, this is the equivalent of
dropping the deltas,
$$W_{tot} = \rho (A_t v_r + A_f v) t $$
$$W_{tot} = \rho (A_t v_r + A_f v)\frac{d}{v}$$
$$W_{tot}= \rho d (A_t \frac{v_r}{v} + A_f)$$
where I substituted t = d/v, and did some simplification. Now, lets look
at the qualitative features of this result. First, we have two terms, a
constant and a term that depends inversely on the velocity of motion.
This means that the faster we go, the less wet we get (I'll plot this in
a bit), but also that there's a threshold wetness you cannot avoid. This
threshold represents the amount of rain in a human sized channel between
where you start and where you end. Also note that as velocity goes to
zero, i.e. we stop moving, how wet we get goes to infinity. That is, if
we're going to stand in the rain forever we're going to keep getting
wet. What is the term in 1/v? It's the amount of rain that hits you on
the top of your head! So what we've derived is that for a fixed distance
how wet you get on the front is fixed, and by moving faster you can make
less rain hit you on the top of your head.
Now, lets figure out what some reasonable numbers are, and plot this
function. A few months back when discussion <a href="http://thevirtuosi.blogspot.com/2010/05/human-radiation.html">human
radiation</a>
I estimated my area as a cylinder with a height of 1.8 m and a radius of
.14 m. This gives a top area of A_t = .06 m^2 and a front cross
section area (note, this is not the cylinder area, but my cross section
that will be exposed to the rain!) of .5 m^2. As for raindrop velocity,
well in my <a href="http://thevirtuosi.blogspot.com/2010/04/falling-water-hot-or-cold.html">first
post</a>
on this blog I calculated the terminal velocity of what I described as a
medium sized raindrop as 6 m/s, and since water drops reach that while
going over niagara falls, we can assume that our raindrops are falling
at terminal velocity.
Finally, I need to estimate the water density. In a medium-to-heavy rain
I would say it takes about 45 s to get a sidewalk square totally wet.
Let's assume a raindrop wets an area of sidewalk equal to twice the
cross section of the raindrop. I used a raindrop of 1.5mm radius, so
that's 7<em>10^-6 m^2 cross section. Now, a sidewalk square is about 2/3
m x 2/3m (about 2 ft x 2ft), so we need \~32000 drops! The volume of a
1.5mm drop is 1.4^-8 m^3, so we have a volume of 4.5</em>10^-4 m^3 =
.45 liters. Now, take our stationary expression from above. This allows
us to solve for \rho. Set delta W equal to .45 liters and substitute
the rest of the numbers we've generated.
$$ \frac{\Delta W_{top}}{A_t v_r \Delta t}= \rho $$
$$ \rho = \frac{.45 liters}{(.44 m^2)( 6 m/s )(45 s)} = .004
liters/m^3 $$
We've found our water density, .004 liters/m^3. Having done this, we
can plug numbers into our final equation above and find
$$W_{tot}= d (\frac{(.001 liters/s)}{v} + .002 liters/m)$$
This scales linearly with distance, so lets pick something reasonable,
say 100m, and if you want the result for another other distance just
scale the results appropriately. Thus
$$W_{tot}= (\frac{(.1 liters\text{*}m/s)}{v} + .2 liters)$$
Finally we can plot this.</p>
<hr>
<p><a href="http://1.bp.blogspot.com/_SYZpxZOlcb0/TH8Mj6XEkVI/AAAAAAAAACs/xiGu976NrOs/s1600/rain+graph.jpg"><img alt="image" src="http://1.bp.blogspot.com/_SYZpxZOlcb0/TH8Mj6XEkVI/AAAAAAAAACs/xiGu976NrOs/s400/rain+graph.jpg"></a>
Plot of how wet you get vs. how fast you run. The blue line is the actual curve and the red line is the theoretical least wet asymptote.</p>
<hr>
<p>I've chosen .5 m/s (\~1mph, a meander) and 11 m/s (slightly faster than
the world record for the 100m dash) as my starting and ending points on
the velocity. The blue line is the curve I calculated, and the red line
represents the theoretical minimum, the 'wetness threshold' if you will.
So you see that if you are Usain Bolt, you can reduce how wet you get by
almost a factor of two by going from a meander to a sprint!
Now, there's more I could say about this (what if the rain isn't coming
straight down? What is my best speed if I have an umbrella?), but I
think that's enough for tonight. I've come out with a theoretically
satisfying result that concurs with experiment. Anytime that happens
that's a good day for the theorists.</p></div>fun water rain humanhttps://thephysicsvirtuosi.com/posts/old/caught-in-the-rain/Wed, 01 Sep 2010 22:53:00 GMT