text:
## Friday, December 31, 2010

### Reviving a computer

My sister's old computer - an IBM Thinkpad - was a few years old when she got it in 2004. By the time she got a new laptop in 2007, it was ancient and incredibly slow. It got put away for a while.

Fast-forward to today. Took it out, decided to see if we could make it work again. We tried turning in on, but nothing happened. Plugged the cord in real tight, tried again, it actually booted up. Verrrrry sloooowly. I immediately went to work. First I deleted all my sister's old documents (long since copied to a flash drive), then her music, then pictures. I had to restart once or twice when it froze, but eventually I cleaned everything out.

By this time the computer was getting pretty hot, so I figured it could use a cleaning inside as well. I got a few screwdrivers and started removing parts - very carefully, and keeping a neat order. Within half an hour, there were a CD/DVD drive, a 20 GB hard drive, keyboard, PC Card adapter, two RAM chips, a wireless card, CMOS supply battery, main laptop battery, and several dozen screws lying about on the floor. I wasn't able to clean the fan like I had hoped, but I did get some dust out and I cleaned the connections. Then I reassembled it, from memory.

Amazingly, it booted up on the first try. I uninstalled almost every program on the machine, leaving just Microsoft Office, Internet Explorer, and the system tools. Once I restarted to finish the uninstalls, it immediately started moving a bit faster. I ran the Disk Defragmenter and it cleared up lots of room. Et Voila, functioning laptop. It's not fast, but it works.

I plan to install some sort of basic antivirus so I can go online, then perhaps I can try my hand at installing Linux, or we might just use it to play Amazon Trail and other old games. We shall see.

### Twin Cities: Metro Transit rail map

I've made a pretty nifty map of a number of Metro Transit rail lines in the Twin Cities (St. Paul and Minneapolis, MN). It's on Google Maps, so it's zoomable and everything.

The map shows six lines - two heavy rail commuter lines and four light rail interurban lines. Only two of the lines are currently in operation - the light rail Hiawatha Line (shown in yellow) that serves downtown Minneapolis, the airport, and the Mall of America; and the Northstar commuter service (shown in blue) that serves the northern suburbs. (The proposed Northstar extension is, of course, in lighter blue.)

All six lines will eventually terminate at Target Field station in Minneapolis - light rail lines at street-level on a bridge, and commuter rail lines below. Currently open lines are bolded; the lighter a line is, the further it is from starting operation. Stations with currently open or visible station platforms have a dot in their marker; stations without do not.

The Central Corridor light rail, shown in red, is currently under construction and will open in 2014. Running on a few miles of the former 512-mile streetcar system, it will serve both downtown areas plus the University of Minnesota via University Avenue.

Southwest Corridor light rail has been approved for construction and is planned to open by 2016. The alignment is pretty much set, but the exact location of each station is not known.

The fourth light rail line is planned to be the Bottineau Boulevard Transitway serving the northwest suburbs. Two possible alignments are shown; the A alignment is more likely to be built.

The proposed commuter line is the Red Rock Corridor, to serve Minneapolis and Hastings via St. Paul. It is shown in pink and may start as soon as 2018.

View Twin Cities: Metro Transit Rail in a larger map

### Scary realization

There are now children, born after 9/11, who are older now than I was then.

That's a very strange thing to realize, and it makes me feel oddly...old. Like the assassination of JFK for the previous generation, you remember exactly where you were when you heard that the twin towers fell. I was sitting in my third grade classroom. When I went home that day, I watched the news, and I understood what was happening.

There are now kids - nine years old - who were born after the towers fell. They've lived their entire lives in the decade from hell.

(Relevant)

## Wednesday, December 29, 2010

### Animated GIF test

I just discovered that I can post animated GIFs in blogger. This makes me very happy, especially now that I have GIMP and can create them.

The above image is a short loop of Kaylee (Jewel Staite) from Serenity, the first episode of Firefly.

### Actual building!

I've finished the repairs on the Orbital Transport. The shock cord is reattached to the new extended forward section. I cut two new forward fins from the old pattern, coated and sanded them, and attached. The next time the weather warms up, I'll primer the forward section.

I've gotten a decent start on the MLAS. I decided not to use the motor hook and engine block, because I want to use D13W reloads at some point. (They weigh less than C6s, so that shouldn't be a problem.) I've got one of the two massive 3.5" diameter centering rings fully glued in place, and the other is currently drying.

I've also started on the nose cone. I glued the two-part cardboard base together, and drilled a quarter-inch hole for weights in the foam nose cone. Dick Stafford says about 0.3 oz added to the stock bolt should be sufficient; I'll use about an inch of quarter-inch steel rod from... something.

## Tuesday, December 28, 2010

### Danger

We humans, we tend to like danger, without actually being in danger. Danger gives us an adrenaline rush, but real danger just stinks. So, we as a species tend to find dangerous things, and then sit back and watch them unfold.

It's why we watch football. We get to watch twenty-two strong young men attempt to break each other, all while sitting back and enjoying a jumbo plate of nachos with that delicious not-quite-cheese sauce.

It's why we watch dangerous-situation reality shows like Deadliest Catch, because icy waters and ferocious storms can't harm you in your living room.

We love to ski and snowboard at high speeds down icy mountains - but only once we've strapped on a helmet, goggles, and plenty of layers of clothing.

We love action movies, because few things get the heart pounding like River Tam taking out a planetful of flesh-eating monsters. But not one of us would want to be there because, well, they're terrifying.

Danger, it seems, is only fun when you're not actually in danger. And that, my friends, is why there are few things better than sitting in a hot tub in below-freezing weather, watching the wind whip the powdery snow into a fury. While you are safe in your bubble of warm air. Absolutely magnificent.

## Monday, December 27, 2010

### Snow

Snow (n): a white, fluffy form of microcrystalline ice that only forms when there is no school for cancellation

About thirty-six hours ago, the weathermonkeys were predicting 11 to 16 inches of snow. We got maybe six inches, maybe not quite. I've learned to roughly halve the TV estimate if it's over six inches. It's usually pretty close to right, because we never live up to what they're saying. I wonder if they intentionally inflate the numbers so people will get inside and be safe.

In any case, it's not all bad. I get to have a snowball fight tomorrow with a couple of my friends. They have no idea what they're in for.

### MLAS

I got a Quest MLAS scale kit for my TRF Secret Santa gift last year. I'm finally starting to build it, mainly because I finally feel ready for the challenge. It's a very difficult kit - foam parts, not a huge amount of structural strength, and of course very finicky stability. Plus, I want to get a good paint job on it.

But, I've crossed that cardboard Rubicon. I started assembling the motor mount and the nose cone yesterday. So far, so good. I'll have more later; looks like we're going to be snowed in a bit.

## Sunday, December 26, 2010

### Nerdy Christmas Loot

Thanks to my family and friends, I had a wonderful Christmas, and I got some really awesome presents.

I got nine books, almost a whole linear foot:
• by Richard Feynman: The Pleasure of Finding Things Out, Six Easy Pieces and Perfectly Reasonable Deviations From the Beaten Path
• Transit Maps of the World - a collection of subway and metro maps, showing how they evolved over time to become the iconic works they are today
• Earth From Above: 365 days
• The Math Book - a collection of 250 milestones in mathematics, with really cool stuff
• 62 Projects to Make with a Dead Computer
• Historical Atlas of North American Railroads
• Finding Serenity: Anti-Heroes, Lost Shepherds, and Space Hookers in Joss Whedon's Firefly - a brilliant collection of critical essays about my favorite show.

I also got a bunch of really awesome stuff:

• Firefly and Sherlock Holmes on DVD
• Several pounds of twizzlers
• A really nice pocketknife, with a neat saw and a few other attachments
• A very weird alarm clock
• A Estes scale Patriot rocket kit - more about that later
• "Define interesting" t-shirt - a Firefly joke

My favorite, though, was my sister's present - this awesome Ada Lovelace shirt.

## Friday, December 24, 2010

### One Kilopost

That is, 1000 posts. It's taken me 30.5 months - since June 10, 2008.

1000 posts in 928 days is 1.077 posts a day - a post every 22 hours 15 min.

Figure half an hour per post, and that's 500 hours - just about three straight weeks of time, used for posting.

I took posts from all three Julys - 2008, 2009, and 2010 - and averaged out the wordcounts. I average about 185 words per post. That means I have written about 185,000 words. By comparison, Animal Farm is about 30,000; Ender's Game is just over 100,000.

(I'm still not up to the 560,000 of the English translation of War and Peace, though.)

It's just crazy to me to think that I've written this much. I'm not sure if it's a sign that I'm incredibly arrogant, or that I need a therapist, or that I need a life. But it's by far the biggest and longest-running project that I've ever undertaken. It's also really neat to see what comes out of my brain when Ive got a blank canvas to put it down on. I mean, this morning I just got thinking about zero while I was dragging my sorry carcass out of bed, and I ended up with 788 words about zero.

In any case, I'm pretty pleased.

It's almost midnight now. I'm listening to Trans-Siberian Orchestra play "Christmas Eve in Sarajevo", and there is no better music for the evening. Merry Christmas to all, and to all a good night!

### Proving Equality to Zero

Say you have several variables - in a computer program, for example - and you want to make sure that they all equal to zero. (For the sake of this example, let's assume that they are all real numbers, i.e, 5.5 rather than 5+8i). With one variable, it's easy; you just have to set up some form of the equation

x = 0
[1]

Of course, you'd have to actually implement that in code, but that depends on the programming language and isn't really relative to this. With one variable, you have a very nice, neat equation, and it works.

With two variables, you could still do something along the lines of

x = 0 AND y=0
[2]

But that becomes very unwieldy - and decidedly unelegant - when you want to use any more than two variables. Surely there's a simple equation that one can use. What about this:

$x\cdot y\cdot z = 0$
[3]

Unfortunately, this equation is a flop; it only requires at least one of the variables be zero*.
If x=0, then y and z can be any number, and so on for y=0 or z=0. What about addition, then?

x + y + z = 0
[4]

This is marginally better; it is slightly less likely to produce a false positive. However, it doesn't work. The set (x,y,z)=(2,4,-6) is a 'solution', yet none of the three variables equal zero.

A logical next step is to modify that linear equation to eliminate some false positives. for example:

x(y+n)(z+n) + y(x+n)(z+n) + z(x+n)(y+n) = 0
[5]

This eliminates most false positives, and it does in fact work if n is guaranteed to be of greater magnitude than x, y, and z, i.e, n is a sufficiently large positive or negative number that x+n, y+n, and z+n are all positive or all negative. Otherwise, however, there are still at least four solution sets: (-n,-n,-n), (0,-n,-n), (-n,0,-n), and (-n,-n,0). Additionally, if you try to generalize to complex numbers it doesn't work, and it's just plain ugly.

Eliminating those four trivial solutions is possible, if don't mind an even uglier equation:

x(y+a)(z+b) + y(x+c)(z+d) + z(x+e)(y+f) = 0
[6]

But, the problem with equations 5 and 6 is that they're not truly solving the problem. Unless variables a through f are picked very carefully, there might be a few values of x, y, and z that create a false positive, and with that many variables theres no general method to check that this is not the case. They don't generalize well to either lots of variables (to prove that ten variables all equal zero by this method, you'd need as many as 55 arbitrary variables a through j) or to complex numbers.

Now, let's think about the problem geometrically. What we are looking to prove is that the point (x,y,z) in 3D space is equal to (0,0,0). Or, if you prefer vectors, that <x,y,z> = <0,0,0> = $\vec{0}$

In other words, that our point (x,y,z) is within a radius of zero from (0,0,0). Now there's a handy equation for distance:

[7]

In this case, a=b=c=0 for the origin point (0,0,0). However, if the square root equals 0, then the equation inside also equals zero. Thus, we get a very neat equation:

$x^{2}+y^{2}+z^{2} = 0$
[8]

And equation 8 is pretty nifty. It generalizes to any number of real variables. It's elegant - it requires only a small amount of code that works on basically system, and it uses only squares and addition - operates that are much faster than square roots or division.

Additionally, if you have a system like Mathematica or a graphing calculator that can deal with vectors, it gets better. Most systems can't check the equality

<x,y,z> = <0,0,0>
[9]

or, where = vector u,

$\vec{u} = \vec{0}$

[10]

However, there's a nicely equivalent operation, taking the magnitude (length of the vector), that does work:

$\left |\vec{u} \right | = 0$
[11]

For vector-ready systems, equation 11 is elegant, though not as fast due to the square root involved in taking the magnitude of the vector.

Equation 8 can also be modified to make it even more powerful; this version allows determination of which single variable is nonzero. The exact constants do not matter; they simply must be primes greater than 3.

$5x^{2}+7y^{2}+11z^{2} = 0$
[12]

Thus, if the result of the left side is nonzero, it can be examined. If it is divisible by an odd power of 5, x is the non zero variable, and so forth. If it is not divisible by an odd power of either of the three constants, then two or three of the variables are nonzero.

All of the really nice-looking equations are made using Codecogs LaTeX Equation Editor; Detexify is spectacularly useful for finding LaTeX symbols.

* Incidentally, equation 3 actually generates the three coordinate planes when graphed.

## Thursday, December 23, 2010

### Elementary School Scientists

What sort of science did you do in elementary school? It was pretty basic stuff, right? Units covering the very basic bits of biology and chemistry and physics. Basic cell biology, weather, simple machines, stuff like that. Not much in the way of actual lab 'experiments'; they didn't trust us to work with beakers and balances until seventh grade, and such things as hot plates and acids were completely out of the question until high school.

Doing an actual controlled experiment - that's never been done before - with live animals, and getting published in a legitimate scientific journal? Utterly out of the question.

Yet, that's what exactly what 25 elementary school students between ages 8 and 10 did in Britain. (BBC News article)

The children designed the experiment, asked the question, hypothesized results, and wrote the majority of the paper. The only things done by the teacher and an assisting scientist, Beau Lotto, were to supply trained bees and to transcribe the student writings.

And they did indeed get results. Although they didn't refer to previous scientific literature - which would have been above their reading abilities - and they hadn't been trained in the use of statistical analysis, they were exploring a hole in scientific knowledge. They found that bumblebees "can use a combination of colour and spatial relationships in deciding which colour of flower to forage from."

They determined that it was possible to train bees to follow a logical pattern in determining where to seek found. In this case, they taught bees to go to the center of an opposing circular pattern of colors - i.e, to go to the blue center of a yellow flower, or the yellow center of a blue flower. This is different from normal circumstances, in which bees are attracted to flowers of a certain color.

The full paper, published in Biology Letters, a publication of the prestigious Royal Society, can be read in full online here. If you've got half an hour to read and understand, it's an excellent read, and more accessible than many scientific papers.

The paper was peer-reviewed by several other scientists, who determined that despite the lack of references or statistical analyses, the paper was "cleverly and correctly designed and carried out with proper controls" and "[the students} hold their own among experiments carried out by highly trained specialists". High praise indeed.

Just as impressive as the results is the repercussions of the experiment. They were denied public funding for the experiment because it was believed that children could not run an experiment that would generate results, so the experiment was funded by Lotto's LottoLab group. But they did indeed get results; this should show us all that the importance of research is not by who runs it or their ages, but what we can learn from it.

The students, I imagine, have also been impacted. They have been taught that they can do real science at a young age, that there is littler than they cannot do. That alone is incredibly empowering, and I imagine that those 25 children will be largely ambitious and successful as adults. But they have also been taught that science is interesting, it is alive, that it is everwhere. That science is about asking questions and finding a way to test them. ("Ideas are tested by experiment") How many scientists and engineers will there be in that group of twenty-five, how many whose abilities for scientific thinking were unmasked by this?

After all, their "principal finding" had a second part, just as important as the first part: "We also discovered that science is cool and fun because you get to do stuff that no one has ever done before."

## Tuesday, December 21, 2010

### On Finding stuff with Google Street View

So, I'm doing some research for a Wikipedia article, and I need to find the location of a former railway station in Montreal. It hasn't been used since 1912, so it's not easy to find. Not to mention, directions are given weirdly in Montreal. I'm not quite sure how you can give directions as being east or west of a street that runs east to west, but they do, somehow.

In any case, I was stuck. So I just went on Google Maps and turned on Street View, and started looking around. And then I found it:

Not only does the former station building still exist, but several tracks have been preserved as well, right down to the metal stops on the ends.

(In case you're interested, it's Dalhousie Square Station).

## Monday, December 20, 2010

### Final Went Well!

I'm not allowed to say anything else about it for the time being - because Conn College has a don't-talk-about the-test honor code which I respect - but my calc final went well today. All the studying paid off.

Now let's see if the same goes for my Marine Science test tomorrow. not so likely.

## Sunday, December 19, 2010

### It's Calculus Time

It's just about time for my calculus final. I have until Wednesday to take it, but I plan to take it tomorrow. It's only five chapters worth of material, but those are five dense chapters, covering vectors and 2D vector calculus; 3D functions, partial derivatives, min-max problems and Lagrange multipliers; double and triple integrals; and 3D vector calculus up to Green's Theorem.

I just spent from 3:40 to 11:30, minus two hours to visit my grandmother, studying - several times longer than any other final I've studied for before. I did 51 problems covering the last 9 sections and wrote up the essential formulae that I must know. The formulae cover front and back of two whole sheets of paper.

Eul me up, boys. It's Calculus Time.

### I don't usually talk about politics...

But it looks like Don't Ask Don't Tell is going to be repealed. Passed the House earlier this week and the Senate. Eight Republican senators had the balls (and ovaries) to vote for what's right, instead of the party line.

## Friday, December 17, 2010

### Llamapocalypse

I'm a pretty darned left brained person. I like logic and order, science and math. But, sometimes I branch out. I play the trumpet. I tried acting for the first time last year, and I (disastrously) tried writing a short story for NaNoWriMo.

So, now I've branched out. I started a second short story, and now I've written three short pieces for drama. Two are short plays; the third is a monologue. Here is the current draft, in full.

Llamapocalypse

They weren’t quite sure how it happened. Not that they had a lot of time to figure it out. It might have been a freak mutation, DNA gone wrong. Might have been some sort of sickness, a ruminantarian Mad Cow Disease. There was even speculation of malevolent action.
How it happened is beyond mattering now. Somehow, a small population escaped, and they multiplied like rabbits. They grow and breed at over ten times the natural rate. They eat everything organic, strip the soil bare.
They reached a critical point sometime in March, and the population exploded. Within a week, half of Nebraska was wasteland. Some folks managed to get out. Most didn’t. The military did their best, but they were ill-prepared to fight the massive herds. Three nuclear weapons barely dented the population. Biological weapons were useless. Somehow, the Joint Chiefs didn’t anticipate warfare against llamas.
It didn’t take long for the internet to dub it ‘Llamapocalypse”. Cute name, but no one was laughing. The rest of the world will survive; they quarantined the continent immediately. They finished dynamiting the Panama Canal bridges last week. But North America is reduced to a few defensive perimeters – New York, D.C., Denver. Some will hold. Life might return to the land someday.
As for me, well, it’s times like this that I really regret ever learning math. It’s the world’s worst word problem. If the tallest building in Cincinnati is one hundred meters high, and the llamas pile on an extra meter every single hour, will my meager food supplies run out before or after I become llama chow?

### I had a good day.

After the crapstorm that was yesterday (friend drama, lousy lab in Marine Science, long day, deferred from MIT) I needed a good day.

And, I got one. My physics quiz went excellently today, and then we spent most of Marine Science watching Blue Planet. If you haven't seen that show, you're seriously missing out.

Then, I had brownies, because I brought some in to reward the folks in my assigned homerooms who brought in money for the National Honor Society toy drive (bribery: it works, folks) and there were half a dozen extras.

Then, third block, I had study, so I played chess with some friends from band. They're much more interesting to play against than the computer. They haven't beat me yet, but one gal is getting quite tricky.

And then, I remembered that I had physics homework. So I did my entire problem set for the next chapter - three or four nights of homework - in one 80-minute block. Thank goodness for my ipod.

When I got home, I found an envelope from WPI. I'm accepted. So, I know I'm definitely going to college. Which is a nice feeling, even if WPI isn't my first choice.

## Wednesday, December 15, 2010

### Ice Cream

Protip: It's never too cold for ice cream. Anecdotal evidence from yours truly suggests it's just as delicious down to below-freezing temperatures.

### Voyager 1 has left the building

After thirty-three years, Voyager 1 has left the first layer of the Sun's domain - the first manmade object to cross the termination shock.

The sun puts out a huge volume of charged particles - protons and electrons, mostly - massing 6.7 billion tons every hour, the equivalent of five Empire State Buildings every second. This is known as the solar wind; it can bring a devastating amount of radiation, and Earth's magnetosphere shields us and makes life possible. The solar wind interacting with the upper atmosphere forms the auroras.

The solar wind eventually begins to peter out at around 90 AU (Astronomical Units, the earth averages 1 AU from the Sun. 1 AU is about 93 million miles), a boundary known as the termination shock. Phil Plait brings the awesome news that Voyager 1 has reached a point where the solar wind equals zero velocity. Within a few years, it will reach the heliopause, where it will feel the wind from other stars. At that point, it will be officially out of the sun's domain, and into interstellar space.

And, what's crazier: this was launched thirty-three years ago. There were no cell phones (only a few giant prototypes), no real personal computers. No internet. It's a marvel of technology. It is expected to have enough power to operate some instruments until 2025, 48 years after launch. Nothing that we have made will catch it - thanks to the gravity slingshots it received from Jupiter and Saturn, it is going faster than the Pioneer probes, faster than its twin Voyager 2, and faster than New Horizons which will visit Pluto in 2015.

This wasn't even its primary science mission. It was just supposed to last a few years, take pictures of Jupiter and Saturn. It discovered Jupiter's rings, volcanic activity on its moon Io, and took thousand of scientifically important images of the two planets.

Perhaps the most important image that it took, though, was the Pale Blue Dot image - which shows the Earth as exactly that, a pale blue dot. Seeing how all of human life ahs taken place on and around that tiny dot reinforces the need to protect it.

This is why I want to go into aerospace. To create things that go where nothing has explored before. To see what's out there. To create the rockets that show us what exists beyond that pale blue dot.

## Sunday, December 12, 2010

### More Nell Rebuilding

Believe it or not, rebuilding the Nell is going surprisingly well. Although many of the joints are broken, they retained their basic shape reasonably well. I just finished gluing and clamping (with masking tape) the last two joints than needed to be rejoined. It's actually a lot easier the second time around, because all the puzzle pieces do fit together.

I'm using entirely wood glue, and it seems to be absolutely strong enough. I'm still worried about twisting in the airframe, but hopefully that will fix itself when I reglue a few critical joints.

I also started repairs on the Orbital Transport, which crashed a few months back. I removed two inches of crushed tube from the front, and replaced it with a 24mm coupler and seven inches of tube. The extra length will make it significantly more stable, and look more like the original design.

### On the road to recovery

The Nell, that is.

I started the length rebuilding process today by fixing the broken side tubes. Both failed in the exact same mode - the joint halfway down the tube was not glued inside, so the outer cardboard tube snapped instead of the inner dowel.I put a layer of wood glue both between the two dowels and on the dowel-tube layer, thus ensuring a very strong joint. Once I re-enforce them with a layer of wood glue or epoxy clay, they will be two of the strongest parts of the rocket, instead of the weakest as before.

Unfortunately, I also discovered today that most of the angled side joints are broken. I may have to peel off the string, rewrap, and then finally refix them. I am not sure yet; it may vary joint-to-joint.

I will take more precautions on its return flight. An 18" plastic parachute, in warm weather, with plenty of wadding to protect it.

## Saturday, December 11, 2010

### Galaxy Zoo: Hubble

I've talked about Galaxy Zoo before a couple of times. It's a citizen science project to classify galaxies from automated sky surveys by type and shape. After not doing anything on the project in 18 months, I'm back. I've classified 135 'galaxies' today... including about actual galaxies, plus a few stars and a satellite trail.

The new project uses galaxies from the background of several Hubble Space Telescope images. Many are proto-galaxies from the beginning of the universe, too young to have formed into round spirals or ellipses. They just look like - like bluish lumps. Some 10 billion years old.

galaxyzoo.org

### Huh.

In just over 10 hours from now, the date in time (in American format) will be:

12/11/10 9:08:07

## Friday, December 10, 2010

### Brass Trio

What happens when your second trumpet and baritone player are late.

In retrospect, it's really fortunate than the first trumpet part had all the melodies. Otherwise, the dear folks at Mystic Aquarium would have been subjected to some rather strange harmonies.

## Thursday, December 9, 2010

### When I am the benevolent dictator of the world

99% of Christmas music will be banned, to be replaced with Trans-Siberian Orchestra music.

Headlights will be auto-on, making you actually work to turn them off.

White lab coats will be standard attire for all science classes, even the theoretical courses.

## Tuesday, December 7, 2010

### Middlesboro Impact Crater

Take a close look at the Appalachian Mountains in the southeast corner of Kentucky. Mile after mile of long, high ridges and narrow valleys carved by streams. And then, just at the junction of Kentucky, Virginia, and Tennessee, there's a very strange shape.

It resembles a giant smooth bowl:

View Larger Map

It's a very strange and out-of place shape. It's got nicely circular walls in a few spots. Yet, there's no volcanic activity for a few thousand miles. How do you get a crater in the middle of folded ridge-and-valley mountains?

From space.

No kidding.

It's a three-mile-wide impact crater, formed sometime in the last 300 million years. (Wikipedia article | Entry in Earth Impact Database) The crater size estimator at the University of Arizona's Lunar and Planetary Laboratory estimates an impactor diameter somewhere around 180 meters (600 feet). That's like getting hit with a ball of rock or iron the size of Fenway Park.

And that's a small crater. There are several confirmed craters on Earth with diameters greater than 100 km (60 miles), indicating an impactor over one mile in diameter. That's like getting smacked with the entire city of Boston instead.

But it gets crazier. Look carefully at that map again. The southeast corner of the crater rim touches one of the mountainous ridges. Right where it touches, there's a little gap in the ridge. You might vaguely recognize the name from your last American history course. It's Cumberland Gap. It's where Daniel Boone crossed the mountains to bring settlers through. It was where 250,000 settlers passed through on their way to the fertile Ohio Valley.

Now, that gap was, it seems, created by the impact. Which means that that anonymous space rock played an important role in American history.

## Sunday, December 5, 2010

### Ooh, pretty picture!

See this beautiful, sexy picture?

Yeah, I made that. Took a blank map, added the stations and the legend in GIMP. I'm rather proud of it.

(Click to take you to the file page, for embiggening.)

## Saturday, December 4, 2010

### Cold

So, I've been getting a cold, one symptom at a time.

Last weekend, I keep being hot and cold.

Then I couldn't stop sneezing.

Then I couldn't stop coughing.

Then my nose was runny.

And now I have a postnasal drip.

Plus, I may be on my way to an upset stomach. Happy funtimes.

### Gentlemen (and Ladies), we can rebuild her.

We can rebuild her. We have the technology. We can make her better than she was. Better... stronger... faster.

I am going to rebuild Nell. It's not going to be easy, but she will fly again.

Now, recall the result of her first flight. No parachute opening, ballistic recovery, crunch splat:

It's broken not-so-neatly into two parts. One contains the rear section plus one side brace; the other is the front section plus the second brace. Both braces are broken in the middle at the dowel joints.

The first order of business is fixing those dowel joints. They carry a whole lot of force, so I;m going to use JB Weld epoxy to fix them together, then epoxy clay to smooth the joint. Then I get to go about the arduous business of regluing all the old joints, then sanding, priming, sanding again.

Goal for the second flight is the first quarter of 2011.

## Wednesday, December 1, 2010

### Brian Marsden

I know this is old news by now, but I feel I owe him at least a few words. Brian Marsden passed away on the 18th of November.

His official title was director of the Minor Planet Center at the Harvard-Smithsonian Center for Astrophysics, but he was so much more than that. He was a walking encyclopedia of astronomy, an absolute genius. If you thought you'd found a new comet or asteroid, he was the guy you talked to. I was never involved in serious astronomy, but I sure had heard his name.

Sky and Telescope has a wonderful obituary and biography that says far more than I could possibly say.

### Eurydice

Back in November, I saw Ledyard Drama put on a performance of Sarah Ruhl's play Eurydice. It was absolutely fantastic - I saw it twice - and several of my friends played the leads, and a minor part, and mandachan was stage manager.

Near the end of the play, Eurydice's father dips himself in the river Lethe. Before he does, though, he states a series of memorized directions. I got a script from a friend; here is what he says:

"Take Tri-state South - 294 to Route 88 West. Take Route 88 West to Route 80. You'll go over a bridge. Go three miles and you'll come to the exit for Middle Road. Proceed 3 to 4 miles. Duck Creek Park will be on the right. Take a left on Fernwood Avenue. Continue straight on Fernwood past 2 intersections. Fernwood will curve to the right leading you to Forest road. Take a left on Forest Road. Go 2 blocks. Pass first entrance to the alley on the right. Take the second entrance. You'll go about 100 yards. A red brick house will be on the right. Look for Illinois license plates. Go inside the house."

It turns out, it's an actual set of directions. A few of the mileages are wrong, but it actually leads to an assuming brick house on an unnamed street in Davenport, IA.

Here's a road map of the route. It's not perfect - it actually takes the third entrance into the alley, not the second - but it provides a good idea of it.

View Larger Map

So, then, what does this route mean? Is it just a random journey?

Turns out, it is not. I found a very illuminating article in the New York Times. Apparently, Sarah Ruhl owns a large collection of her grandmother's clothes. She loves the memory of her grandmother that she finds contained within them. Memory, I should note, the remembering of love and family - is what Eurydice is all about.

The article goes on to discuss that Mrs. Ruhl grew up outside Chicago (Wikipedia says a town called Wilmette). Interstate 294 runs just a few miles from Wilmette, and is the best highway to jump on for journey to points west - like Iowa.

The kicker, though, is where her grandmother lived. Davenport, Iowa. I would, then, consider it reasonable to assume that that red brick house did belong to her grandmother. The journey from her home to there, the article says, was one she experienced many times during her childhood.

For her father, who is roughly analogous to Eurydice's father, the journey would be natural - from his own home to his mother's - the place where he grew up. It's ingrained in his memory, the last thing he gives up before his brain is wiped. And death is often likened to going home or joining one's relatives - and giving up all his memories is like dying again.

The directions end with the following, as he dips himself into the river:
"In the living room, look out the window. you'll see the lights on the Mississippi river. Take off your shoes. Walk down the hill. You'll pass a tree good for climbing on the right. Cross the road. Watch for traffic. Cross the train tracks. Catfish are sleeping in the mud on your left. Roll up your jeans. Count to ten. Put your feet in the river... and swim


text:
# DSolve does not give a result

highP6 = D[y[t], {t, 3}] - Power[(Cos[t]), 3]*y[t] == 0;
ic1 = y[0] == a;
ic2 = y'[0] == -a;
ic3 = y''[0] == 0;
DSolve[{highP6, ic1, ic2, ic3}, y[t], t]

I cannot get the solution using the above code? Any help!

• LOTS of DE do not have closed form solutions. If you could assign a constant value to 'a' then you could use NDSolve. – Bill Mar 3 '17 at 3:05

Since this is variable coefficient ODE, you can try series solution. The analytical series solution will be accurate "near" the point of expansion, which has to be where the initial conditions are. More terms makes it more accurate.

ClearAll[y,t,a];
nTerms=10;
pt=0;
ic={y[0]->a,y'[0]->-a,y''[0]->0};
ode=y'''[t]- Cos[t]^3 y[t];
s0=Series[ode,{t,pt,nTerms}];
s0=s0/.ic;
roots=Solve@LogicalExpand[s0==0];
s1=Series[y[t],{t,pt,nTerms+2}];
sol=Normal[s1]/.ic/.roots[[1]]

Compare to Numerical

a=1;
nSol=NDSolve[{ode==0,y[0]==a,y'[0]==-a,y''[0]==0},y,{t,0,2}];
Plot[{sol/.a->1,Evaluate[y[t]/.nSol]},{t,0,2},
PlotStyle->{Red,Blue},PlotLegends->{"Series","Numerical"}]

• I have proposed a SE site for maple, if you could help out to make it a success? area51.stackexchange.com/proposals/107315/maple – zhk Mar 3 '17 at 9:37
• @MMM I added my self there. But this was done before few years ago, and due to lack of interest, the proposal was removed. There is simply not as many Maple users as Mathematica around. – Nasser Mar 3 '17 at 9:42
• Thanks for your interest. I will try my best. I am sharing it with people. You do too, please. – zhk Mar 3 '17 at 9:44
• Ask a random easy question, please. – zhk Mar 3 '17 at 9:45

Comment

Like Mathematica, Maple is also unable to solve the ode.

restart;
ode:=diff(y(t),t\$3)-cos(t)^3*y(t)=0;
dsolve(ode);

DESol is a data structure to represent the solution of a differential equation.

format: plain


text:
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# Completing the square review

Completing the square is a technique for factoring quadratics. This article reviews the technique with examples and even lets you practice the technique yourself.

## What is completing the square?

Completing the square is a technique for rewriting quadratics in the form left parenthesis, x, plus, a, right parenthesis, squared, plus, b.
For example, x, squared, plus, 2, x, plus, 3 can be rewritten as left parenthesis, x, plus, 1, right parenthesis, squared, plus, 2. The two expressions are totally equivalent, but the second one is nicer to work with in some situations.

### Example 1

x, squared, plus, 10, x, plus, 24, equals, 0
We begin by moving the constant term to the right side of the equation.
x, squared, plus, 10, x, equals, minus, 24
We complete the square by taking half of the coefficient of our x term, squaring it, and adding it to both sides of the equation. Since the coefficient of our x term is 10, half of it would be 5, and squaring it gives us start color #11accd, 25, end color #11accd.
x, squared, plus, 10, x, start color #11accd, plus, 25, end color #11accd, equals, minus, 24, start color #11accd, plus, 25, end color #11accd
We can now rewrite the left side of the equation as a squared term.
left parenthesis, x, plus, 5, right parenthesis, squared, equals, 1
Take the square root of both sides.
x, plus, 5, equals, plus minus, 1
Isolate x to find the solution(s).
x, equals, minus, 5, plus minus, 1

### Example 2

4, x, squared, plus, 20, x, plus, 25, equals, 0
First, divide the polynomial by 4 (the coefficient of the x, squared term).
x, squared, plus, 5, x, plus, start fraction, 25, divided by, 4, end fraction, equals, 0
Note that the left side of the equation is already a perfect square trinomial. The coefficient of our x term is 5, half of it is start fraction, 5, divided by, 2, end fraction, and squaring it gives us start color #11accd, start fraction, 25, divided by, 4, end fraction, end color #11accd, our constant term.
Thus, we can rewrite the left side of the equation as a squared term.
left parenthesis, x, plus, start fraction, 5, divided by, 2, end fraction, right parenthesis, squared, equals, 0
Take the square root of both sides.
x, plus, start fraction, 5, divided by, 2, end fraction, equals, 0
Isolate x to find the solution.
The solution is: x, equals, minus, start fraction, 5, divided by, 2, end fraction

## Practice

Problem 1
Complete the square to rewrite this expression in the form left parenthesis, x, plus, a, right parenthesis, squared, plus, b.
x, squared, minus, 2, x, plus, 17

Want more practice? Check out these exercises:

## Want to join the conversation?

• need to complete the square, the problem is x^2 +10x+blank
• its completing the square .so x^2 + 10 x + 25 - 25 which is equal to( x +5)^2 -25
• In problem 2, the question have the same answer in both X1 and X2. However the problem doesn't automatically consider the answer when it is in X1 = 10, while X2 = 4
• Notice that the problem reads, "Give the solutions in ascending order."
• I find it frusterating and a bit unfair that in both the review and the videos, the problems they show are equations like 4x^2 +20x + 24, while the problems we are given in the practices include functions: h(x) = x^2 +3x -18.
Can someone give me advice on dealing with the functions? I know how they work, but how can I do something to "both sides of the equation" when there is only one side to begin with?
• As an expression, you learn to complete the square. As an equation, one of the main purposes of completing the square is to find the roots, so it is where h(x)=0 OR to get an equation in vertex form so that graphing is made easier.
• how to you solve x squared + 11x + 24
• First you move the constant, 24, off to the side, so it looks like x^2 + 11x = 24. Then you halve 11, to get 5.5 Square that, and you get 30.25. Add that to the equation as x^2 + 11x + 30.25 = 24 +30.25. The left part of the equation is now perfect - you would get (x + 5.5)^2 = 54.25.

To be entirely honest, I am only partly sure I even got that right. I'm not quite a mathematician, so feel free to correct me.
I am also fully aware that this is about a month late.
• Can you just use the quadratic formula for all of these?
• You can use the derivative of the equation to find the vertex, by setting the equation equal to 0 and then using this formula:
0=ax^2+bx+c=2ax+b, where x is the x-value of the vertex, and you can plug this in to the original equation to find the y value of the vertex, in essence finding you the vertex of the graph.
• What's the difference between solving a quadratic equation set equal to zero by completing the square and rewriting a quadratic function from standard form to vertex form by completing the square?
• In the practice problem x^2-2x+17. I do not understand how you get 16.
• To complete the square for "x^2-2x", you need to add 1
"x^2-2x+1" = "(x-1)^2"
But, the not change the original polynomial, you then need to subtract 1 from 17 (+ 1 and then -1 = 0, so the polynomial is not changed, we just shifted numbers around).
This gets you: "(x-1)^2 + 16"
Hope this helps.
• how to solve x^2-2x-24=0?
• I'm having an issue when completing the square with trinomials that all have 'minus' signs. It won't work for me for some reason! Here is what it looks like: x^2-2x-168
• Completing the square just requires you to divide the middle term by 2 and square it, so -2/2 = -1 and (-1)^2 = 1. So you end up with (x^2 -2x + 1) - 168 -1 (you have to subtract 1 to balance the 1 you added). You will end up with (x - 1)^2 - 169.
• I am still having trouble with the fractions aspect of this topic. Can some please explain? it would mean the world to me
• (by the way, ^2 means squared)
x^2 + 2x + 6 = 0
-6 -6
-----------------------
x^2+2x=-6
x^2 + 2x + 2/1(because it's the b of ax squared + bx + c) = -6 + 2/1(you add to both sides)
(you do this so you can get a square function)
x^2 + 2x + 1 = -5
(x + 1) ^ 2 = -5
you do square root and you get: x + 1 = positive and negative square root of 5
subtract 1 from both sides and you get:
x = + and - square root of 5
does this help?


text:
Solved

# How to recursively add missing directories

Posted on 2010-11-18
Medium Priority
323 Views
How do I recursively add any required (missing) subdirectories using FSO?

For example, I want to have a function that will accept a folder name such as C:\folder1\folder2\folder3

What I need is to create any missing folders from that full path.
So if I already have C:\folder1 and it already has folder2, I want folder3 to be created.  When I pass the full path name, I dont know what folders, if any, already exist.

If I don’t have ANY of the folders, I want the entire path to be created.

I could pass \\server\directory\folder1\folder2\folder3 or U:\Reports\Sales\2007\
I can verify that the server/drive exists… but not how to recursively build the list of folders.

I can verify that there is no filename in that list… all I need is a function (or functions) that will make sure that the entire directory specified exists.  So by the time the function is done, I don’t have to check does folder exist or anything like that.

Thanks!
0
Question by:tfsaccount
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Accepted Solution

käµfm³d   👽 earned 2000 total points
ID: 34165992
This should do it:
path = "C:\test\folder\to\create"

Set fso = CreateObject("Scripting.FileSystemObject")
parts = Split(path, "\")
temp = parts(0) & "\"

For i = 1 To UBound(parts)
temp = temp & parts(i) & "\"

If Not fso.FolderExists(temp) Then
fso.CreateFolder(temp)
End If
Next

0

LVL 11

Expert Comment

ID: 34166104

Simply call System.IO.Directory.CreateDirectory(directoryName) with the full path, it creates the directories for you!
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Expert Comment

ID: 34166127
@jasonduan

>>  Simply call System.IO.Directory.CreateDirectory(directoryName) with the full path,

In classic VB? I believe you're thinking of .NET  :)
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Expert Comment

ID: 34166150
sorry, i was thinking of .net.
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Author Closing Comment

ID: 34172202
Thanks!!!
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ID: 34172451
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text:
Question

# When light passes from A to B, the value of sin isin r  is greater than 1. It implies that the light ray bends towards the normal. True False

Solution

## The correct option is A True We know that, sin isin r=v1v2. If the value is greater than 1, it implies; v1>v2. So, in medium B, light travels with a lesser velocity, implying that medium B is denser than medium A . Therefore, the ray will bends towards the normal.

Suggest corrections


text:
The Riddle this week is rather straightforward to explain: stacking identical objects (bars of length and mass two, say) on top of one another so that the center of each new bar is uniformly distributed along the previous bar, what is the distribution of the number of bars when the stack collapses? If I am not confused, the stack collapses the first time the centre of gravity of an upper stack leaves the interval represented by the bar just below. Namely

$\left|\frac{1}{N-j} \sum_{i=j+1}^N x_i -x_j\right|>1$

when the xi are the bar centres, or equivalently

$\max_{2\le j\le N-1} \left|\frac{1}{N-j} \sum_{i=j+1}^N \sum_{k=j+1}^i\epsilon_i \right|>1$

where the ε_i‘s are U(-1,1). Which is straightforward to code in R by looking at means of cumulated sums.

### 3 Responses to “stack overload”

1. How about a variant: what’s the maximum allowable offset to guarantee the tower never collapses? Consider the related game where you stack blocks (or playing cards) until the top one lies entirely outside the location of the first one. In that game, the cards’ positions follow a log curve when the number of cards required is a minimum.

• Thanks : I think this variant is more “classical”, unless I misunderstand the constraint.

2. […] article was first published on R – Xi'an's Og, and kindly contributed to R-bloggers]. (You can report issue about the content on this page here) […]

This site uses Akismet to reduce spam. Learn how your comment data is processed.


text:
## Need Help In Getting Started

In my statics class we are studying Rigid body equilibrium and when it got to 3D I get a little lost. Below are some homework questions I'm having a hard time starting. I get the free body diagrams but don't know where to go from there. I find that there are more unknowns than I have equations for.

I have provided links for the reactions that happen at certain types of connections. I also apologize for the bad scanning job on a couple of the images.

Reactions:
Table 1
Table 2

Questions:
Question 1
Question 2
Question 3

I appreciate any help. I need to finish this assignment asap.

 PhysOrg.com science news on PhysOrg.com >> 'Whodunnit' of Irish potato famine solved>> The mammoth's lament: Study shows how cosmic impact sparked devastating climate change>> Curiosity Mars rover drills second rock target
 Recognitions: Homework Help Science Advisor Are you remembering to account for Torque?
 I am accounting for moments (tourque). The 6 equations that I have to use are: 1) Force Equations: -$$\Sigma F x =0$$ -$$\Sigma F y =0$$ -$$\Sigma F z =0$$ 2) Moment Equations -$$\Sigma M x =0$$ -$$\Sigma M y =0$$ -$$\Sigma M z =0$$


text:
# Plotting two surfaces in cylindrical coordinates

Hello,
I need to plot the following surfaces given by:
where \rho is the radius, z is the height and \phi is the azimuth. c,\alpha,\phi_0 are constants with c\geq\alpha.
The first surface is a torus, but I can’t get anything similar to a torus when I try to plot it

What have you tried? It is much easier to help you if you have mostly working code to start from.

import numpy as np
import matplotlib.pyplot as plt

def cart2pol(x, y):
rho = np.sqrt(x**2 + y**2)
phi = np.arctan2(y, x)
return(rho, phi)

def pol2cart(rho, phi):
x = rho * np.cos(phi)
y = rho * np.sin(phi)
return(x, y)

c=25
a=5

r=np.linspace(0,20,200,endpoint=True)
phi=np.linspace(0,2*np.pi,200,endpoint=True)

x,y=pol2cart(r,phi)

z=np.zeros((2,200))
z[0][:]=np.sqrt((c**2-a**2-(r-c)**2))
z[1][:]=-z[0][:]

fig = plt.figure()
ax = fig.gca(projection='3d')

ax.plot_surface(x, y, z, antialiased=True, color='orange')


text:
# Aerospace Why do Jet Engines Have a Compressor?

1. Oct 17, 2008

So like it says. It may be a silly question, but I am doing some reading on jet engines (on Wiki) and I am having trouble finding an explanation as to why we want to increase the pressure of the air before it goes into combustion. Does this just another way of increasing the thrust?

Thanks,
Casey

2. Oct 17, 2008

### jaap de vries

That is actually a very good question. In the following link an attempt is made in answering this very basic question.

http://www.aerospaceweb.org/question/propulsion/q0033.shtml

The main reason given in the link above is quote: "If uncompressed, the air-fuel mixture won't burn and the engine can't generate any thrust". As a combustion engineer I don't like this explanation (the German V1 rocket did not have a compressor I believe). I see no reason why a fuel air mixture uncompressed would not burn.
The whole idea behind a jet engine is that you accelerate the fluid going through it with as driving force a pressure gradient. The pressure in a jet engine is the highest right after the last stage of compression (for axial compressors). The combustion then takes place with minimum pressure loss and the the expansion begins through the turbines and finally through the nozzle . The high pressure created by the compressor forces the fluid to go out through the back. In other words, it creates the favorable pressure gradient used to accelerate the flow and produce thrust. By drawing out a Brayton cycle you can also show the effect this has on the efficiency.

3. Oct 18, 2008

### Danger

Good post. The V1, though, used a valved pulse-jet. Compression wasn't needed because the valves closed off the front of the combustion chamber upon firing. That made it essentially a rocket.

4. Oct 18, 2008

### Phrak

For the same reason you get better power out of an internal combustion engine with a higher compression ratio. This is really a question in thermodynamics.

Danger, the buzz bomb used a tuned exhaust. After the exhaust had pulsed out the back, the low pressure returned a portion of it to toward the combustion chamber. This compressed the air fuel mixture that had entered during the low pressure phase.

5. Oct 18, 2008

### Averagesupernova

I would say the main reason for the compressor is power density in any engine.

6. Oct 18, 2008

### Danger

From everything that I read when researching pulse-jets (which, admittedly, was more than 35 years ago), the original exhausts weren't tuned at all. Nobody expected the 'vacuum' intake effect. Something like 80% of the fresh intake charge got sucked in through the tailpipe, but that wasn't part of the design plan.
Subsequent designs capitalized upon that feature, and eventually the valve box was eliminated when people figured out how to properly tune the system so the pulses effectively closed the front of the chamber through pressurization.
The reason that the V1 was called the 'buzz bomb' was because of the noise made by the intake valves flapping.

7. Oct 18, 2008

### turbo

Casey, a compressor on a jet engine prevents back-flow and it acts like a supercharger on a car's IC engine, delivering an increased load of oxygen so that more fuel can be burned more efficiently.

8. Oct 18, 2008

### Staff: Mentor

Ramjets don't require a dynamic (turbo) compressor, and instead use the forward motion of the engine to compress the air in the inlet.

Thrust comes from the mass flowrate and velocity.

The compressor increases the pressure of the atmosphere and momentum of the air flow. The air (oxidizer) combines with the fuel, which burns and provides energy to the flow. The combusion does two things - the combutsion process breaks fuel molecules to make more molecules and provides heat which increases the temperature which decreases the density which causes an increase in velocity of the flow.

The compressor 'pushes' the flow into the combusion chamber, and the flow passes out the exhaust with a much high velocity than it entered - and that combined with the mass increase from the fuel - causes the thrust.

Think of the continuity (mass flow) equation. Mass flow in = mass flow out, otherwise the mass in the control volume (jet engine) would increase or decrease.

The momentum equation: momentum out > momentum in, and the change in momentum gives the thrust (force).

Energy (in flow) equation: Work-Energy out = work in + energy (thermal) produced

9. Oct 19, 2008

### Andre

In addition to all those excellent answers, there is also the air density problem. The higher the flight, the less dense the atmosphere, the less drag and the least fuel to be used. But at those densities there is not a lot of oxygen to burn and we really need high compression rates to sustain a nice little flame in that engine.

10. Oct 22, 2008

### jaap de vries

Fair enough, my point was that the argument that uncompressed fuel air does not burn makes no sense to me.

11. Oct 22, 2008

### FredGarvin

DING DING DING!! We have a winner. Well said Astro. Concise and right on.

To initiate flow, there has to be a favorable pressure graidient in the engine. The compressor's job isn't to make sure it keeps flow moving in a certian direction, although most of the time it does that as a by product of increasing the density (although in surge it definitely stops doing that). Also, as Jaap pointed out, the idea that an air-fuel mixture wouldn't burn uncompressed is just silly.

The whole point is that thrust is directly proportional to the mass flow through the engine. The more mass flow, the more thrust.

12. Oct 22, 2008

### Phrak

Interesting. I wonder if it would have worked at all without the accidental compression. I have part of a pulse jest--minus the tail pipe. It's amazingly simple: a set of reeds, a block of aluminum with holes for the intake, a fuel venturi, a starter spark plug and combustion chamber/exhaust tube.

13. Oct 22, 2008

### Danger

Neat! Original, or a reproduction?

14. Oct 22, 2008

### Phrak

Like I said, this is question in thermodynamics.

In a jet turbine engine, a 4 cycle gasoline engine, diesel, or coal dust turbine an element of combustable fluid goes through a process of compression, ignition and expansion to obtain work.

I could find a believable answer that gave power efficiency as a function of pressures. What's the power efficiency ratio of a compressed vs. uncompressed working fluid? Is it 4 to 1, 5 to 1, twenty to 1?

15. Oct 22, 2008

### FredGarvin

From a thermo point of view, the cycle efficiency is directly tied to the compressor's pressure ratio. The propulsive efficiency drops off as pressure ratio increases but the thermal efficiency increases with pressure ratio. This makes the overall efficiency increase with pressure ratio. If you look through some literature like Hill and Peterson, you can see plots of this very topic that illustrate this point.

16. Oct 22, 2008

### Staff: Mentor

As Andre wrote - when density of the mixture is too low, there is not enough energy produced by burning fuel to sustain the fire.

Simple thought experiment - we have a mixtures of oxygen and fuel of varying densities. On the one end we have mixture that we know will burn, on the other end we have vacuum (wih only traces of oxygen and fuel) - same mixture, but it will not burn. Somewhere in the middle must be the point where we move from "burning" to "not burning".

17. Oct 23, 2008

### Phrak

What is propulsion efficiency?

The efficiency of an Otto cycle, eta is given as

$$\eta = 1 - \left( \frac{V_{top}}{V_{bottom}} \right) ^{0.27}$$

as given in this link, http://courses.washington.edu/me341/oct22v2.htm that I can't generate a hyperlink for.

A volumetric compression ratio of 12 spits out 49% efficiency.

A ratio of one gives zero, which makes no sense, or course.

18. Oct 23, 2008

### FredGarvin

For a basic definition, the propulsive efficiency is

$$\eta_p=\frac{T*u}{m\left[(1+f)(u_e^2/2)-u^2 /2\right]}$$

Where:

T = thrust
u= vehicle velocity
ue = exit velocity
f = fuel air ratio

It is also considered to be

$$\eta_p = \frac{\eta_{overall}}{\eta_{thermal}}$$

Note: the term $$T*u$$ is often refered to as the thrust power.

Here's another reference on this with a reduced form:
http://en.wikipedia.org/wiki/Propulsive_efficiency
http://web.mit.edu/16.unified/www/FALL/thermodynamics/notes/node81.html

Last edited by a moderator: Apr 19, 2017
19. Oct 23, 2008


text:
# Do circular pipes maximize flow rate?

Suppose that $U \subset \mathbb{R}^2$ is nonempty, open, connected and bounded. Consider a Poisseuille flow in the pipe $U \times \mathbb{R}$. That is: a time-independent incompressible flow of the form: $$v:U \times \mathbb{R} \rightarrow \mathbb{R}^3: (x,y,z) \mapsto (0,0,w(x,y))$$ which satisfies: $$\frac{\partial^2w}{\partial x^2} + \frac{\partial^2w}{\partial y^2} = k < 0$$ $$w \left.\right|_{\partial U} = 0$$ with $k$ some constant involving the pressure-gradient along the pipe's axis and the viscosity.
If we consider pipes with fixed cross-section, $$\int_U \mathbb{d}x \mathbb{d}y = C ,$$ is it true that the flow rate $\int_U w(x,y) \mathbb{d}x \mathbb{d}y$ is maximized only if $U$ is a disc?

If the answer is negative, consider instead pipes with a fixed boundary length, $$\int_{\partial U} \mathbb{d}l = L,$$ is it now true that the flow rate is maximized only if $U$ is a disc?

-
I don't know about a mathematically rigorous proof, but the answer is certainly "yes"; an example that can be solved exactly is that of an elliptical cross-section: the ratio of the flow through an elliptical and circular crosssection, at constant area, is $$\gamma=2(r+1/r)^{-1}$$ with $r$ ratio of the two axis of the ellipse; this is maximal for a circle ($r=1$). –  Carlo Beenakker Jun 8 '13 at 16:42
I know, and we can solve the flow problem in many other geometries too (since $h(x,y) := w(x,y) - \frac{k}{4}(x^2+y^2)$ is harmonic, we can use complex-analytic tools like the Riemann-mapping theorem. That's why I think that an actual proof should not be too difficult for experts.) –  Thibaut Demaerel Jun 8 '13 at 17:04

The answer is positive, this is a consequence of a result by Talenti ["Elliptic Equations and Rearrangements", Annali SNS 3 (1976)]. Let $k>0$, $-\Delta u= k$ in $U$, $u=0$ on $\partial U$, and consider the ball $B$ having the same volume as $U$. If $v$ is the solution of $-\Delta v=k$ in $B$, $v=0$ on $\partial B$, then $u^* \leq v$ in $B$, where $u^*$ is the spherically symmetric rearrangement of $u$. Since $u$ and $v$ are positive functions, and rearrangements preserve norms, one has $\|u\|_1 = \|u^*\|_1 \leq \|v\|_1$, which is the claim.


text:
# NubDevice : Under The Covers

1356 views

## Huh, that's 'kinda neat moments...

When you start over from scratch, design aspects become more important. I can't help what I have yet to discover but for the first of I suspect to be many, a happy moment.

Mouse input ponder.

The usual choice is to constantly redirect the mouse cursor to the center and act from the deltas. I'm starting to hate that. Something about it. So I changed it up a bit where I'm still clipping to the window but wrapping on the left and right edge. This feels better at least to not have to hide the cursor and still not allow to click focus lost. This is what my front end input handling looks like. Now I don't care about mouse sensitivity or delta timing on the camera yaw movement and perhaps puts a skill modifier back on the user. In a sense.

#define imp_bool(x) if(input.pAction_map->GetBool(x))
#define imp_float(x) if(input.pAction_map->GetFloatDelta(x) != 0.0f)
void Game::update(float deltaTime)
{  // user input pump
imp_bool(cancel) { action_this_frame = quit; }

// camera
imp_bool(forward)  { camera.move(FORWARD,  deltaTime); }
imp_bool(backward) { camera.move(BACKWARD, deltaTime); }
imp_bool(left)     { camera.move(LEFT,     deltaTime); }
imp_bool(right)    { camera.move(RIGHT,    deltaTime); }
imp_bool(up)       { camera.move(UP,       deltaTime); }
imp_bool(down)     { camera.move(DOWN,     deltaTime); }
imp_bool(jump)     { }

float x = input.pAction_map->GetFloat(mouseX); // 0.0--->1.0 (l--r)
float y = input.pAction_map->GetFloatDelta(mouseY); // upper ndc space also
camera.ProcessMouseMovement(x * 360.0f, y, true);   // yaw rotation based on horz screen position
// pitch is accumulated and clamped process side
//
// todo: take into account the physical velocity of the mouse movement
// high velocity user movement skips a beat at edge transition
// solution: track previous movement and apply
// priority: barely notice-able
//

POINT pt; ::GetCursorPos(&pt); // keep y component
if(x<0) { // left and right screen edge cursor wrapping
POINT pt_desired = { display.dims.width, 0 }; ::ClientToScreen(display.hwnd, &pt_desired);
::SetCursorPos(pt_desired.x-2, pt.y); // todo: refactor out the magic number
}
if(x>0.9985f) { // partial cause of refactor above (but it's decent smooth at 800 x_res)
POINT pt_desired = { 0, 0 }; ::ClientToScreen(display.hwnd, &pt_desired);
::SetCursorPos(pt_desired.x, pt.y);
}
}

we'll see how it goes....

### Next up...

[ Fix your timestep ] yup, I'm a believer. I've heard said, 'If you're going to do multiplayer, do it early'. Hard to argue, set up you system to play nice with something hard. I've always partially ignored the previous state - this state tween concept when it came to timing, but this really, really makes sense to me now. Oh yeah, we're keeping that...

### And the closer..

I'll be spending the week cleaning up and minor miscellaneous tweaks. I understand the importance of tidying up and now is the point to decide how I branch out from here cleanly. I feel now is a good time to give the [ RIP Method ] a serious follow through. I agree, repetition is key (at least for me) Thanks @DavinCreed, that's helping.  I've only started to browse. Have to go to work for now though.

I'm actually in the process of a revision on that. As I work through it myself, I'm refining things. And you don't have to follow what I planned out to follow the spirit of it. The basic idea is gaining proficiency through repetitive iterations.

I hope it helps.

What I'm experiencing with that is, I find my speed improving but the process is slightly slightly skewed because of my increased personal reference available. The last work being the better parts of the previous attempts. or in other words, previous successes fuel current endeavour. Yes, I'm feeling that. To actually organize a plan is welcoming. It's hard to choose realistic scope when it's your dream project getting it's belly rubbed.

I think along with speed and skill, what I gain is also a better understanding of how things interact and interface with each other which helps me make things that are easier to expand on and maintain.

## Create an account

Register a new account


text:
## Basic question: meaning of "partition of R into maximal connected intervals"

What does the phrase "partition of $$R$$ into maximal connected intervals" mean?

The full sentence: "Let $$I_1, I_2, .... ,I_m$$ be the partition of $$R$$ into maximal connected intervals with disjoint interiors."
 PhysOrg.com mathematics news on PhysOrg.com >> Pendulum swings back on 350-year-old mathematical mystery>> Bayesian statistics theorem holds its own - but use with caution>> Math technique de-clutters cancer-cell data, revealing tumor evolution, treatment leads
 Blog Entries: 2 Without knowing the context, it appears the author means that it is not possible to add another interval to the finite sequence that is connected and has disjoint interior. Ie., I1 = (-oo, 0], I2 = [0, oo).

 Similar discussions for: Basic question: meaning of "partition of R into maximal connected intervals" Thread Forum Replies General Math 3 Introductory Physics Homework 6 Calculus 2 General Discussion 12 Calculus & Beyond Homework 2


text:
Plots a time series along with its acf and either its pacf, lagged scatterplot or spectrum.

ggtsdisplay(
x,
plot.type = c("partial", "histogram", "scatter", "spectrum"),
points = TRUE,
smooth = FALSE,
lag.max,
na.action = na.contiguous,
theme = NULL,
...
)

tsdisplay(
x,
plot.type = c("partial", "histogram", "scatter", "spectrum"),
points = TRUE,
ci.type = c("white", "ma"),
lag.max,
na.action = na.contiguous,
main = NULL,
xlab = "",
ylab = "",
pch = 1,
cex = 0.5,
...
)

## Arguments

x a numeric vector or time series of class ts. type of plot to include in lower right corner. logical flag indicating whether to show the individual points or not in the time plot. logical flag indicating whether to show a smooth loess curve superimposed on the time plot. the maximum lag to plot for the acf and pacf. A suitable value is selected by default if the argument is missing. function to handle missing values in acf, pacf and spectrum calculations. The default is na.contiguous. Useful alternatives are na.pass and na.interp. Adds a ggplot element to each plot, typically a theme. additional arguments to acf. type of confidence limits for ACF that is passed to acf. Should the confidence limits assume a white noise input or for lag $$k$$ an MA($$k-1$$) input? Main title. X-axis label. Y-axis label. Plotting character. Character size.

None.

## Details

ggtsdisplay will produce the equivalent plot using ggplot graphics.

## References

Hyndman and Athanasopoulos (2018) Forecasting: principles and practice, 2nd edition, OTexts: Melbourne, Australia. https://otexts.com/fpp2/

plot.ts, Acf, spec.ar

Rob J Hyndman

## Examples

library(ggplot2)
ggtsdisplay(USAccDeaths, plot.type="scatter", theme=theme_bw())

tsdisplay(diff(WWWusage))
ggtsdisplay(USAccDeaths, plot.type="scatter")


text:
Combining cryptographic primitives in creative ways often results in interesting applications. In this mini-series, we'll look at Riposte, an anonymizing micro-blogging platform for millions of users[GgBM15].

A Riposte system consists of a cluster of Riposte servers, which are ideally located in different jurisdictions. Hundreds to thousands of whistleblowers connect to these servers to anonymously send messages that can't be tracked back to them, so long as at least one server doesn't collude with the others. At the same time, millions of readers can fetch those messages, and by doing so, they too participate in the anonymity set.

To implement such a system, three cryptographic primitives are being leveraged:

• secret sharing, a technique which allows one to spread a secret to many different recipients, such that no recipient is able to find out on its own what that secret is,
• writable PIR, a technique which allows users to write to a random row of a database, without the database server knowing which row was actually updated,
• distributed point functions, a mathematical tool that dramatically reduces the required bandwidth from a writing client to the servers.

We set the stage by introducing Riposte on a high-level. Before looking under Riposte's hood, we'll look at mix-nets, which is how anonymization services have traditionally been implemented, to get a feeling of what needs to be done in order to anonymize users. Then, we'll see how to move from mix-nets to writable PIR, by studying a Riposte toy example with a very simple and easy to understand, but terribly wasteful distributed point function.

Distributed point functions are such a useful tool that we'll spend some more time introducing them in a general way, both informally and formally. Learning DPFs this way is time well-spent, when we set out to look at the special case of an efficient $(2,1)$-distributed point function.

This mini-series doesn't yet cover the audit server, which is necessary to prevent malicious users from corrupting the database. It also doesn't yet investigate $(n, n-1)$-DPFs with $n \gt 2$, which require more sophisticated cryptographic primitives than the efficient pseudo random generator used in the $(2,1)$-DPF. Former aspect has been fully taken care of by Riposte's creators, but the latter aspect requires additional work.


text:
# Surviving the Crazy Hospital¶

Note

Many of these guidelines we’ve evolved from trial and error and are almost entirely driven by a need to avoid a burgeoning complexity (aka flying spaghetti monster). Feel free to experiment and provide us with your insights here as well!

## Behaviours¶

• Keep the constructor minimal so you can instantiate the behaviour for offline rendering
• Put hardware or other runtime specific initialisation in setup()
• Update feedback_message for significant events only so you don’t end up with too much noise
• The update() method must be light and non-blocking so a tree can keep ticking over
• Keep the scope of a single behaviour tight and focused, deploy larger concepts as subtrees

## Composites¶

• Avoid creating new composites, this increases the decision complexity by an order of magnitude
• Don’t subclass merely to auto-populate it, build a create_<xyz>_subtree() library instead

## Trees¶

• Make sure your pre/post tick handlers and visitors are all very light.
• A good tick-tock rate for higher level decision making is around 500ms.


text:
# Am I calculating perspective projection correctly?

If I have a point in world space: (wx, wy, wz)

and I have a centre of projection: (cx, cy, cz)

and I wanted to project that point using perspective projection

would my point on the screen be calculated by:

x = (cz * wx) / wz y = (cz * wy) / wz

I got this from this lecturer https://www.youtube.com/watch?v=VpNJbvZhNCQ I don't understand what he means by the variable 'd'. What would the variable 'd' be?

• Once you get into this, you're going to have to come to terms with homogeneous coordinates. Here is a nice introduction to projection. – Brett Hale Nov 29 '16 at 13:30

In the video you linked to, $d$ is just the distance from the eye to the image plane. To simplify things though, you can just set $d$ to 1, which is the usual case and causes it to disappear from the formulas.

Your formula for projection is a bit wrong but don't worry, projection is actually pretty simple.

$x_{Screen} = x_{World} / z_{World}\\ y_{Screen} = y_{World} / z_{World}$

If you want to move the center of projection, you can do so with a screen space coordinate, by modifying the formula to be this:

$x_{Screen} = x_{World} / z_{World} - x_{ScreenCenter}\\ y_{Screen} = y_{World} / z_{World} - y_{ScreenCenter}$

If you want to re-introduce the $d$ value to control the distance from the eye to the projection plane, you can use the formula below. Adjusting $d$ will make the camera look like it is zooming in or out. Again, you can just set $d$ to 1 though and not deal with it.

$x_{Screen} = d*x_{World} / z_{World} - x_{ScreenCenter}\\ y_{Screen} = d*y_{World} / z_{World} - y_{ScreenCenter}$


text:
# Workshop on algebraic surfaces

Venue: Leibniz Universität Hannover, Main building (Welfenschloss), room b302

Dates: March 1/2, 2018

## Speakers

Kenji Hashimoto (Tokyo)
DongSeon Hwang (Ajou/Warwick)
Xin Lu (Mainz)
Yuya Matsumoto (Nagoya)
Zsolt Patakfalvi (Lausanne)
Stefan Schreieder (LMU München)

## Schedule

Thursday

14:00-15:00 - Matsumoto
15:30-16:30 - Lu
17:00-18:00 - Patakfalvi

Friday

09:30-10:30 - Hwang
11:00-12:00 - Hashimoto
12:15-13:15 - Schreieder

## Titles and abstracts

Global sections of some special elliptic surfaces (Hashimoto)

We discuss how to reconstruct an elliptic K3 surface from the data of singular fibers. The problem is reduced to counting of global sections of some special elliptic surfaces.

Cascades of log del Pezzo surfaces and SSDB (Hwang)

There have been numerous attempts to classify log del Pezzo surfaces. In this talk, I will quickly summarize the known attempts to this goal and report my recent work on the classification of log del Pezzo surfaces of Picard number one. The result is obtained by generalizing the notion of ‘cascades’ of nonsingular del Pezzo surfaces, based on the approach initiated by Miyanishi and Zhang. At the end of the talk, I will present the computer program (under construction) realizing this approach.

$$\mu_p$$- and $$\alpha_p$$-actions on K3 surfaces in characteristic $$p$$ (Matsumoto)

There are three group schemes of order $$p$$ in characteristic $$p$$, namely $$\mathbb{Z}/p\mathbb{Z}$$, $$\mu_p$$ and $$\alpha_p$$. We consider their actions on K3 surfaces in characteristic $$p$$, with emphasis on the latter two. To be precise, we consider not only smooth K3 surfaces but also K3 surfaces with rational double point singularities (RDPs)''. We will give (partial) answers to the following questions: For which $$p$$ such actions exist? How can we distinguish actions whose quotients are again K3 surfaces (with RDPs)? Which kind of RDPs may appear on the quotient surface or on the original surface?

Geometric vs. non-geometric singularities (Patakfalvi)

For a variety X defined over an imperfect field k, there can be a discrepancy between its singularity properties over k, and its singularity properties over the algebraic closure of k. The later singularity properties are also frequently called geometric. For example, it frequently happens that a surface over k is normal but over the algebraic closure of k it becomes non-normal. First, I give some examples on this phenomenon. Second, I present a general canonical formula relating the canonical bundle of X to the canonical bundle of the normalized base-extension to the algebraic closure of k. Third, I explain how this implies that normal but geometrically non-normal Gorenstein Del-Pezzo surfaces do not exist in characteristic greater than 3. In particular, this implies the generic smoothness of terminal 3-fold Mori-fiber spaces in characteristic greater than 7. Lastly, if time permits, I briefly mention a few other applications too. This work is joint with Joe Waldron.

The Kuga-Satake construction under degeneration (Schreieder)

We extend the Kuga--Satake construction to the case of limit mixed Hodge structures of K3 type. We use this to study the geometry and Hodge theory of degenerations of Kuga-Satake abelian varieties over the punctured disc. This is joint work with A. Soldatenkov.

Strict Arakelov inequality for families of curves (Lu)

For a one-dimensional semi-stable family of curves, the Arakelov inequality gives an upper bound on the Hodge bundle in terms of the genera of the base curve and the general fiber together with the number of singular fibers. Later it's proved that the number of singular fibers appearing above can be replaced by the number of singular fibers with non-compact Jacobians. It is expected that such an Arakelov inequality should be strict when the fibers are of high genera. In this talk, I would like to report some progress on this problem. This is a joint work with K. Zuo.

## Organisers

Simon Brandhorst
Víctor Gonzalez Alonso
Matthias Schütt


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Interactive graphs/plots help visualize and better understand the functions. The field emerged in the Hellenistic world during the 3rd century BC from applications of geometry to astronomical studies. Trigonometry. MathPortal.org's Trigonometric Equation Solver – Learn more about trig equations via the step-by-step examples. Topics ... the quotient rule is a method of finding the derivative of a function that is the ratio of two differentiable functions. Algebra 1 online textbooks,  intermediate 2 maths revision '', parabola. Or approximate solutions on custom range identity calculator - Simplify trigonometric expressions calculator with! Is the ratio of two differentiable functions definitely among the top fears of students the... Trig equations via the step-by-step examples solving Trig equations Trig Inequalities Evaluate functions.... Like sine, cosine, tangent, cotangent, secant, cosecant much more side second...: a New Era for the Science of Triangles Identities Proving Identities Trig equations the... The globe to thousands of problems, growing every day using this website cookies! Better understand the functions answer will ( generally ) require the use of a calculator ( i.e website! ) require the use of a calculator approximate solutions on custom range answer will ( generally ) require use! The  Calculate '' button to solve for all unknown variables seek guidance solving... 5. y = cot x. Simplify trigonometric expressions to their simplest form step-by-step this trigonometric calculator will exact! Intermediate 2 maths revision '', online parabola calculator mathematics is definitely among the top of! Step by step solutions to thousands of problems, growing every day step with our solver... Identical to that for when the answer will ( generally ) require the use of a (... 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To use this method Science of Triangles work of solving trigonometry equations equation!, radical, exponential and logarithmic equations with all the steps ( generally ) require the use of a.! Bc from applications of geometry to astronomical studies section we will discuss solving Trig equations when the is! About Trig equations when the answer is one of the trigonometric functions ( Rad ) Calculates the functions... Angles ) access detailed step by step solutions to thousands of problems, growing day! Aren ’ t one of the more-advanced graphing calculators make short work of solving trigonometry.! Identities Trig equations via the step-by-step examples finding the derivative of a calculator a little messy due the! Trigonometry equations quadratic equation to their simplest form step-by-step the solution ( ). Cookies to ensure you get the solution ( s ) to a quadratic equation any equation to get best! 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BatchNorm after ReLU

I am currently experimenting with different settings for a U-Net (https://lmb.informatik.uni-freiburg.de/people/ronneber/u-net/) based image segmentation and I was unable to find out if it makes any sense to place a BatchNorm layer after a convolutional layer which includes ReLU. Such as follows:

conv1 = Conv2D(64, 3, activation='relu', padding='same', kernel_initializer=gauss())(inputs)

batch1 = BatchNormalization()(conv1)

conv1 = Conv2D(64, 3, activation='relu', padding='same',
kernel_initializer=gauss(stddev=sqrt(2 / (9 * 64))))(batch1)

Can somebody tell me or point out the relevant literature to explain if it is possible to use BatchNorm like shown above or if it should always be placed before ReLU.

I did check out this discussion on reddit, where some people do seem to recommend using BatchNorm after ReLU: https://www.reddit.com/r/MachineLearning/comments/67gonq/d_batch_normalization_before_or_after_relu/

The U-Net does indeed segment my images with these settings, but not really much better than when using two Dropout layers at the end of the contracting pathway or no Dropout or BatchNorm layers at all.