## Wednesday, September 26, 2012

### Solving The Raindrop Problem

Have you ever noticed that when you're driving during a rainstorm, the number of raindrops hitting your windshield seems to increase along with your speed? I first noticed it soon after I started driving, and ever since then I've intended to sit down and work out a formula to explain it. (Living in Hilo and having it rain frequently while driving up and down from Mauna Kea has tended to keep “the Raindrop Problem” as I've come to call it fresh in my mind.)

I've toyed with it on occasion, but never definitively solved it, so I finally decided to sit down and work it out rigorously. So, without further ado I shall put my years of mathematical training to the test and attempt to figure out just how the amount of rain hitting your windshield changes as a function of your speed, while simultaneously trying to explain it in terms you can follow. Ideally, I'd like to get a graph out of it.

We start out in the grand tradition of physicist everywhere by considering a very simple, idealized case. Let us assume that for our purposes, the density of raindrops is uniform everywhere that we are considering. On a small enough spatio-temporal scale this is not a bad assumption. Furthermore, assume that the raindrops are falling straight down, with no gusts of wind or other forces acting on them other than gravity and air resistance. Again, a fairly plausible scenario, especially for a lot of the rain we get in Hilo, which often comes without any accompanying wind. A steady or gusty wind such that the raindrops had a set non-zero horizontal velocity or acceleration could also be taken into account, but is more complicated than I'd like to get into right now.

Let's begin by assuming the simplest possible case. Imagine a sheet of glass of width w and height h laying horizontally under a steady, uniform rain with evenly distributed raindrops falling at uniform velocity r (for “raindrop”. I'm saving v for later). Now, we want to know how many raindrops will hit the glass in a time interval $$\Delta t$$ (pronounced "delta-t", if you don't know).

If we know that the raindrops are falling at speed r, then we can multiply by the time interval $$\Delta t$$ to figure out how far they fall during that interval. Thus, any raindrop within a distance  $$d=||r||\cdot\Delta t$$ above the glass will hit it within time interval $$\Delta t$$. (The double vertical bars around the ‘r’ serve to remind us that it is technically a vector quantity and indicate that we want the length [or magnitude] of the vector in this equation.)

Intuitively, this gives us a rectangular box of volume $$V=d\times h\times w$$ over the sheet of glass within which raindrops will be able to hit the glass in time interval $$\Delta t$$. Less intuitively but more rigorously this can be achieved by a double integration of the raindrop fall distance   $$d=||r||\cdot\Delta t$$ over the sheet of glass:
$V=\int_{0}^{h}\int_{0}^{w}||\overrightarrow{r}||\cdot\Delta t\ dx\,dy$
You may refer to the image below to help keep all these symbols and concepts straight:

At this point we've nearly solved the problem of how many raindrops will hit the sheet of glass in time interval $$\Delta t$$, which for our purposes will be 1 second. We just need to know the numerical density N of raindrops per unit volume times the volume where raindrops will be able to hit the glass. Putting everything we have so far into a formula, we have
\begin{align}n&=N\cdot V\\ &=N\cdot h\cdot w\cdot ||r||\cdot\Delta t\end{align}
This is all well and good, but there are two additional factors we must take into account to better approximate a car's windshield. Those factors are the angle of the windshield, and the fact that we are interested in a moving windshield.

We will now consider each effect independently, before adding them together to get a full picture of the situation.

Let's start by introducing a non-zero angle of repose to the glass sheet. Refer to the picture below to see what I mean (I've added a coordinate system for future reference):

Now, the basic problem remains the same: figuring out the volume marked by the blue parallelograms and the glass sheet. This figure is known in geometry as a parallelepiped (PARR-uh-lel-EH-pi-ped), and has the following formula for its volume (from vector calculus)
$V=|\overrightarrow{a}\cdot(\overrightarrow{b}\times \overrightarrow{c})|$
where a, b, and c are the vectors that make up three of the sides that meet at a vertex and the $$\times$$ sign and dot have special meanings because these are vectors. (I'm not being super consistent about notating all my vectors all the time due to the constraints of working in a blog post, but I'll try to keep it clear when the distinction is important.)

“But wait a minute,” you may be thinking to yourself at this point. “Wouldn't it be easier in this case, in order to find the volume, to simply multiply the height h by a factor of $$\cos(\theta)$$ to account for the diminished surface area as seen from above (where $$\theta$$ runs between $$0^\circ$$ for a flat sheet and $$90^\circ$$ for a vertical one), and then multiply by d and w?”

Indeed it would, astute reader. In this case, such a formula would be simpler. In fact, the formula for the volume would be simply $$V=w\cdot\cos(\theta)\cdot h\cdot||r||\cdot\Delta t$$.

However, the second effect we will be considering is the velocity of a moving car and attached windshield, and since I foresee vector addition on the horizon I think it would be prudent to begin incorporating vectors into the picture now.

That brings us to considering the velocity of the car (and by extension windshield) intself. Let's assume that the car is moving with a constant horizontal velocity in the positive x-direction at velocity v, as per the picture below.

The nice thing about using vectors to find the volume of the parallelepiped is that it's very easy to find the length of one of the blue lines (what we were calling d before) in the above picture: it's simply the sum of the vectors r and v. Let's call it g (for no particular reason), and we can define it as
$\overrightarrow{g}=\overrightarrow{v}+\overrightarrow{r}$
We can break vectors into their component parts along each axis, and in this case we have $$g_x=||\overrightarrow{v}||$$ (since the x-component of g is coming from the velocity of the car), $$g_z=||\overrightarrow{r}||$$ (since the z-component is coming from the velocity of the rain), and $$g_y=0$$ (since we are assuming the rain is falling straight down and the car is traveling only in the x-direction).

Using the volume formula for a parallelepiped from before, the volume can be found by
$V=|\overrightarrow{g}\cdot(\overrightarrow{h}\times\overrightarrow{w})|$
as long as we can find a vector for g, h, and w. Once we have, the volume is given by the absolute value of the determinant of a $$3\times3$$ matrix like so:
\begin{align}V&=\begin{Vmatrix}g_x &g_y &g_z\\ h_x &h_y &h_z\\ w_x &w_y &w_z\end{Vmatrix}\\ &=|g_x\begin{vmatrix}h_y &h_z\\w_y &w_z\end{vmatrix}-g_y\begin{vmatrix}h_x &h_z\\w_x &w_z\end{vmatrix}+g_z\begin{vmatrix}h_x &h_y\\w_x &w_y\end{vmatrix}|\\ &=|g_x(h_yw_z-h_zw_y)-g_y(h_xw_z-h_zw_x)+g_z(h_xw_y-h_yw_x)|\end{align}
At this point it's an easy matter of plugging in the appropriate numbers to get the volume, after which we can use $$n=N\cdot V$$ to get the number of raindrops. And to make it even better, a little reflection shows that we can simplify this symbolic equation even further.

We already know that $$g_y=0$$ from above, and we can remove several other quantities upon analysis. Remember, the vector w runs only in the y-direction, so $$w_x=w_z=0$$. Similarly, g and h run only the x- and z-directions, so $$g_y=h_y=0$$. Upon dropping all the zero terms, the equation simplifies nicely into
$V=|g_zh_xw_y-g_xh_zw_y|$
We can now replace the vector components with the original quantities they stand for to remind ourselves what everything is.
\begin{align}g_z&=||\overrightarrow{r}||\\ h_x&=-\cos(\theta)||\overrightarrow{h}||\\ w_y&=||\overrightarrow{w}||\\ g_x&=||\overrightarrow{v}||\\ h_z&=\sin(\theta)||\overrightarrow{h}||\end{align}
($$h_x$$ is equal to negative $$\cos(\theta)$$ because I'm taking all vectors to start at the origin, and the windshield extends slightly backwards into the negative x-axis.)

Thus, for the final volume equation we have
$V=|\big(||\overrightarrow{r}||\cdot -\cos(\theta)||\overrightarrow{h}||\cdot ||\overrightarrow{w}||\big)-\big(||\overrightarrow{v}||\cdot \sin(\theta)||\overrightarrow{h}||\cdot ||\overrightarrow{w}||\big)|$
At this point we are nearly ready to begin making graphs. Since we are assuming that h, w, $$\theta$$, and r are all constant, we are left with an equation in v, which lends itself well to plotting.

We just need to put numbers to all our variables. Let's assume that our hypothetical windshield has a width of $$w=2$$ meters and a height of $$h=0.5$$ meters, giving it a surface area of $$0.5\times2=1$$ m$$^2$$. A little searching on the Internet finds that a “typical” raindrop has a terminal velocity of $$r=9$$ meters per second. One source I found suggested a value of about $$780$$ cubic millimeters of water per cubic meter of atmosphere for the density of rain. Raindrops may have diameters anywhere between 0.5 and 5 millimeters (any larger and it breaks up on the way down, any smaller and it's technically not rain, but drizzle) so the volume of an “average” raindrop of diameter 3 millimeters is
$V=\frac{2}{3}\cdot\tau\cdot(1.5\,\text{mm})^3=14.14\,\text{mm}^3$
Thus, on average, a cubic meter of atmosphere contains $$780\,\text{mm}^3\div14.14\,\text{mm}^3\approx55$$ raindrops. So $$N=55$$ for our example here.

Plugging all of this into a Python script I wrote, I was able to use the matplotlib graphing package to generate the graph below with multiple plots for different windshield angles.

The results are, perhaps, not too surprising, but still interesting (and make a rather pretty graph). The number of raindrops appears to increase linearly with speed, with steeper windshield angles (like those found in trucks or large vans) having a higher rate of increase than lower angles (like those found more in cars). The limiting cases of $$0^\circ$$ and $$90^\circ$$ are illustrative; a flat surface (like on the roof of vehicles) would have no change whatsoever with changing speed, while a vertical surface would start out with no raindrops hitting it (as expected), but would eventually have the most raindrops hitting it if you could go fast enough (45 meters per second is about 100 mph, so it would have to be pretty fast, but you can see that even by about 75 mph it has surpassed nearly all other angle inclinations but the $$75^\circ$$ one.) I'd estimate that most cars have windshield angles around $$45^\circ\pm15^\circ$$, which is best represented by the cyan line on the graph.

So, now you know (or at least have a good idea) why you need to run the windshield wipers at faster speeds when going faster. And it only took three semesters of calculus to do! (Granted, I could have done it algebraically for this simple case, but why pass up the chance to do some exciting vector calculus?) It might be interesting for a follow-up post to consider time of travel, and how many raindrops you would actually encounter at different speeds for a given trip length. Anyway, a hui hou!

## Tuesday, September 25, 2012

### Music of the Sphere-Studiers

In my last post I mentioned William Herschel, well-known in astronomy circles as the discoverer of the planet Uranus. What is probably not as well known, however (I didn't know it until last week), is that Herschel, before he became interested in astronomy, was actually a professional musician and composer, writing a total of 24 symphonies and numerous concertos and other works.

Herschel lived from 1738 to 1822, pretty much the entirety of the Classical period (so-called to distinguish it from the preceding Baroque and the following Romantic periods). Although born in what would become Germany, he moved to England at the age of nineteen where he spent over a dozen years performing and composing music before becoming interested in astronomy. Recordings of his work are very rare; at the moment there are only one or two CDs of his music available online, but I was able to pick one up on Amazon that contained 6 of his symphonies.

Listening to them, I can say that he was quite an accomplished composer. It's always impossible to say with these things, but I think that if he hadn't become interested in astronomy he might have become famous as a composer. His music is definitely of the Classical type, a bit less complex than the Baroque, though it still sounds somewhat Baroque in places (which makes sense since he was writing most of it near the transition from Baroque to Classical). Strangely, it often reminds me in places of Handel, another German-born musician who later immigrated to England. (Given that Handel is probably my favorite composer, I'm just fine with the occasional resemblance.) All in all, Herschel's music (or at least symphonies 2, 8, 12-14, and 17) is quite good, and I would highly recommend it to any lovers of good music out there.

## Wednesday, September 19, 2012

### Globular Cluster Series (Part 25): NGC 6293

Wow, number twenty-five already. It seems like just yesterday that I decided to make a photographic catalog of the Milky Way's globular clusters, but it's already been a year. Today's picture is the moderately-sized cluster NGC 6293 in the constellation Ophiuchus, the Serpent Bearer. NGC 6293 is pretty similar to the cluster I showed off last time, NGC 6541. NGC 6293 is a bit further away at 31,000 light-years (compared to 22,800), and also a bit physically smaller at 71 light-years in diameter (compared to 100), which combine to give it much smaller size on the sky, only 7.9 arc-minutes compared to 15.0.

 Globular cluster NGC 6293 in Ophiuchus.

Both NGC 6293 and 6541 are at about the same distance from the galactic center: 6,200 light-years for NGC 6293, and 6,800 for NGC 6541. Interestingly, despite the fact that NGC 6293 is both fainter and smaller than 6541, it was discovered first, in 1785 by William Herschel (discoverer of the planet Uranus), while NGC 6541 wasn't discovered until 1826. This may have something to do with the fact that NGC 6541 is much more southerly than 6293, which would make it appear much fainter and harder to see for the northern European astronomers who discovered both of them.

## Sunday, September 16, 2012

### Globular Cluster Series (Part 24): NGC 6541

Boy, it seems like I've been showcasing globular clusters for a long time now (and in a sense I have, stretching back to June of last year), but I'm still only up to twenty-four so far. Even accounting for the fact that from my location I can see "only" about 80% of the sky, and that there are some globular clusters that are realistically too small and faint for me to capture, there are probably still at least 50–75 globular clusters I can feasibly hope to capture. So, I'm barely a third done at this point, at most.

Anyway, the globular cluster I want to showcase today is called NGC 6541. It's the only globular cluster in the small southern constellation Corona Australis (the Southern Crown), and it turns out to be a lovely little gem of a cluster about 22,800 light-years away from us. Despite this great distance it appears quite large on the sky at 15.0 arc-minutes in diameter, about half the size of the full Moon. This puts its physical size at just about 100 light-years, making it in the upper 50% of globular clusters.

 NGC 6541 in Corona Australis.

Although far from the Sun, NGC 6541 is pretty close to the galactic center, only about 6,800 light-years away. For such a large and pretty cluster I wasn't able to find too much information about it other than that it was discovered in 1826, which strikes me as fascinating. I mean, we've known about the existence of this cluster for less than 200 years, less time than the United States has been a country.

In many ways, astronomy, despite being the oldest science, is still completely fresh and new. We've only been able to build telescopes for the past 400 years, and it wasn't until the last century that we've been able to create telescopes capable of exploring more than the minuscule sliver of the electromagnetic spectrum that is able to penetrate our atmosphere. We still know so little about things in our universe, or even our own galaxy, because it's only been within the last 50 years or so that we've developed the ability to even detect them. So much to learn and discover...enough to last many lifetimes.

And that's why I love astronomy.

## Tuesday, September 11, 2012

### The Fireworks Galaxy

Today I have something besides another globular cluster picture for you: a picture of a galaxy. This galaxy is officially named NGC 6946, though it unofficially goes by the name of the Fireworks Galaxy. It is located on the border between Cepheus and Cygnus, which puts it very close to the plane of the Milky Way. This fact means that it is obscured rather heavily by the dust in our galaxy, making it appear much fainter than it otherwise would.

NGC 6946 is located about $$22.5\pm7.8$$ million light-years away, which actually makes it one of the closest galaxies outside the Local Group. As you can see from the picture below, it's fairly large, with a diameter of about 11.5 by 9.8 arcminutes. This gives it a physical size of about 75,000 by 63,000 light-years, about two-thirds to three-quarters the size of the Milky Way galaxy. You can easily make out several spiral arms wrapping around the core, and what appears to be a faint ring around the nucleus.

 NGC 6946 (the Fireworks Galaxy) on the border between Cygnus and Cepheus.
NGC 6946 currently holds the record for the largest number of supernovae observed in a galaxy (and earns it nickname in the process): nine so far, in 1917, 1939, 1948, 1968, 1969, 1980, 2002, 2004, and 2008. Typically in a galaxy this size you'd expect to see one every fifty years or so on average, which gives you an idea just how extraordinary this galaxy is.

## Tuesday, September 4, 2012

### Globular Cluster Photo Series (Part 23): NGC 6356

Well. This is a first. Today's globular cluster is the first one I've showcased here on my blog that doesn't have an article about it on Wikipedia. I don't use Wikipedia for all my information but it does make a nice central repository of info, so I had to do a little more digging around tonight. This sort of thing will probably start happening more and more as I exhaust the Messier objects and the brighter or more famous NGC ones. (I already talked a bit about the New General Catalogue in my post about NGC 3201, the first NGC globular cluster I showcased back in March.)

Anyway, NGC 6356. It's located in the constellation Ophiucus at the rather large distance of 49,200 light-years from us. At this distance, I calculate its diameter to be about 115 light-years (since I couldn't find a number anywhere) given its size of about 8.0 arc-minutes on the sky (full Moon is about 30). This would actually make it among the larger clusters of the Milky Way, but its great distance makes it appear small (though it's still among the top 50% of clusters by area on the sky).

 NGC 6356 in Ophiucus.

NGC 6356 is also located pretty far from the galactic core outside the galactic plane, at about 24,400 light-years (that's about a quarter of the way across the entire Milky Way galaxy). Most globular clusters are closer to the core (and disk) of the galaxy.

Another interesting thing about NGC 6356 is that it is located about 80 arc-minutes from another globular cluster, Messier 9 (which I wrote about on August 5th). That distance is small enough that I could actually catch both clusters in one picture with a well-placed shot, though I didn't know about it until tonight. Once I found out, however, I was able to match up star patterns seen in both pictures and put them together to make the picture below. Messier 9 is the cluster on the left, while NGC 6356 is the one on the right.

 Messier 9 (left) and NGC 6356 (right) in Ophiucus.

While looking at this picture, keep in mind that M9 is about 90 light-years across while NGC 6356 is about 115; it's only NGC 6356's greater distance that makes it look about the same size. Also, since this is two pictures reduced and composited separately, the brightness scale between them is not uniform.

So, all in all NGC 6356 turns out to be a rather interesting cluster for an object not interesting enough to have a Wikipedia page. A hui hou!

## Monday, September 3, 2012

### Globular Cluster Photo Series (Part 22): M68

Today I have a picture of the globular cluster Messier 68, located in the constellation Hydra.

Since globular clusters orbit the center of the galaxy like everything else in it, the vast majority of them are found in the hemisphere containing the galactic core. Messier 68 is an oddity, a large globular cluster found in the hemisphere opposite the galactic center. Physically, it is about 106 light-years in diameter, and appears about 11.0 arc-minutes across on the sky (about a third of the width of the full Moon). It is located at the not-too-shabby distance of 33,300 light-years from us, and given that it is further from the core than we are, it is perhaps not too surprising that it is approaching us at 112 kilometers/second. M68 contains a fairly average number of 42 variable stars (discovered so far).

 Messier 68 in Hydra.
Hydra is pretty far south for a Messier object (declination –26° 44′ 38.6″), and as a result observers in the mid- to far-northern hemisphere tend to see it through a lot of atmosphere low on the horizon, which often led past observers to estimate its brightness as fainter than it really was. Thankfully, being in Hawai‘i means that I can see it pretty well. And other than that, there really isn't too much to say about it. A hui hou!