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Hillary (2)

OK, full disclosure: I do not like Hillary Clinton. Actually, I think the Clintons are sociopaths. Literally. Both of them.

Of course, I could be wrong. I am fully aware that most of what I believe is the result of deliberate propaganda. Just like most of what you believe.

That is why I look for stories that I can evaluate based on my own direct, personal knowledge. In that spirit, from all the wonderful Clinton stories over the years, I want to discuss just two.

The first is the cattle futures incident. I am sufficiently familiar with markets to know that there is only one way to turn $1,000 into$100,000 over ten months of trading cattle futures, which is illegally. I do not know why someone wanted to transfer $100,000 under the table to the wife of the governor of Arkansas; I just know that is what happened. This is not propaganda. This is not partisanship. This is mathematics. You can argue the late 70s was a long time ago. Certainly, the relevant statutes of limitations have long since expired. But she never admitted any wrongdoing. That makes this an ongoing lie, which makes it current and therefore relevant. The second story, naturally, is the Email server. I know quite a lot about Email servers. For example, I could code one. And the operating system on which it runs. And design the hardware. Anyway, I know there is only one reason a sitting cabinet official routes all of her electronic communications through a server in her basement, which is to keep them away from people with legal authority to view them (e.g. law enforcement or the Judicial branch). And I know that any claim to the contrary — such as “it was a matter of convenience” — is a lie. I really do love this story because my personal knowledge in this area is orders of magnitude greater than any journalist’s. So whenever I read an article about Hillary’s email server, I do not learn anything about Hillary’s email server; I learn something about the author of the article. For example, when I read someone calling it a “personal Email account” rather than a “personal Email server”, or comparing it to Colin Powell’s AOL account or whatever, I know I am reading a partisan hack making a lame attempt to deceive stupid people. (Or just a really stupid journalist. It can be hard to tell the difference.) The Hillary campaign says her arrangement was allowed by the rules at the time. That might or might not be true; with the Clintons, you never know. But assuming it is true, that is only because the rules at the time did not contemplate something so ridiculous as a sitting cabinet official routing all of her electronic communications through a server in her basement. Which brings us to the felonies. Classified data is another area where I actually know something. Dating myself a bit, once upon a time I internalized the Orange Book and several of its interpretations. I know all about mandatory access controls and covert channel analysis and formal verification methods… In short, I am very familiar with the difference between the kind of system that processes classified information and the kind of system you get when you ask some pathetic I.T. monkey to set up an Exchange server in your basement. I am not a lawyer. But I do know that there is only one way for top secret codeword information to migrate from an authorized to an unauthorized system, which is illegally. Based on facts already published, I know that someone, somewhere committed a felony. I do not know exactly who, or exactly which felony, because there is more than one possibility. (My own suspicions would start with Huma Danger, née Abedin.) But I am 100% certain that someone committed a crime. I do not expect anyone to be held accountable, of course, since in our system we are all very much unequal under the law. These crimes will never approach indictment. The generous explanation is that there is a difference between proving some felony occurred and convincing a jury that a specific person committed a specific felony. The realistic explanation is that the Obama administration does not prosecute its friends. She will be the nominee and most likely President. Tens of millions in financial sector donations buys a lot of hack journalism, and the fix is in. More on that in another installment, perhaps. Although I might need to get drunk first. Hillary (1) I read with some amusement the Boston Globe’s endorsement of Hillary Clinton. You would think a 1000-word endorsement by a major newspaper touting someone’s “experience” and “seasoning” would mention one or more of her accomplishments. Things she has actually, you know, done. The New York Times endorsement makes an attempt, which winds up being even more funny. She “brought star power” to the Department of State! She “helped make it possible” (for her successor) “to impose tougher sanctions on Iran”! (Whatever that means.) She “worked to expand and deepen the dialogue with China”! But flying around the world on the taxpayer’s dime is not an achievement. Talking is not an achievement. Giving speeches to Goldman Sachs for$300,000 is not an achievement.

Intellectually honest liberals know that she is corrupt. They know her chief accomplishment as Secretary of State was to ruin the Middle East. Further ruin, I mean. Remind me again, how is Libya doing? (Of course, this might be no accident. Perhaps there is some nation, somewhere, in whose interest it is to have the bulk of the Middle East be a sea of ungovernable anarchic clusterf*cks? Just not the U.S.)

These endorsements say so much more about the Globe and Times editorial boards than they do about Hillary Clinton. It is frankly refreshing to have it be so blatant.

That said, I am pretty sure she will win the nomination and the Presidency for the simple reason that Goldman Sachs needs a return on their investment.

Great Scots!

(I could really use some whiskey for this post)

So, the U.K.’s political/financial leadership — and their journalist microphones — are getting downright apocalyptic about the possible consequences of Scottish independence.

That alone would be enough to get me out voting “yes”, were I a resident of Scotland. Alas, I am not. But I do know the only poll I need to follow:

Betfair Scottish independence contract

It is not quite as easy to read as TradeSports (RIP), in part because you have to translate “Back” and “Lay” from English to English. Roughly, “back” means “bet for” and “lay” means “bet against”. The number represents gross winnings versus a “back” bet of $1. (Technically it’s £ not$ and occurs in multiples of 10. But (a) I am from the colonies and (b) the units are irrelevant.)

For example, as I write, the “yes” contract is trading at 4.6/4.7. That means you can hand over $1 and get back$4.60 if Scotland votes for independence, or you can hand over $3.70 and get back$4.70 if Scotland votes against independence.

Bottom line: To convert Betfair lines to probabilities, just take reciprocals. So 4.6/4.7 corresponds to a probability between 21.3% (1/4.7) and 21.7% (1/4.6). That is higher than I would estimate, personally. Anyway “too close to call” is a bit generous.

Just like TradeSports, the nice thing about this market is that it reflects new information very quickly. This particular contract was trading around 3.0 a week ago.

Other contracts of mild interest:

2014 Senate control (Republicans currently at 1.5 = 67%; usually agrees with FiveThirtyEight)
2016 Democratic nominee (currently Clinton 67%)
2016 Presidential election (currently Clinton 40%; I myself would estimate these last two to be equal and higher)

Cryptography Part 4: Random numbers

Executive summary: There are no random numbers; only random number generators.

If that sentence made perfect sense to you, feel free to skip this installment. Otherwise, read on.

In the previous post, we saw a cryptosystem whose security was entirely based on a coin toss. But why use a coin? Why not just get together and agree that when I say one thing I will mean another?

Well, two reasons. First, humans are notoriously predictable; our minds are just not very good at being random. Second, ad-hoc cryptosystems are impossible to analyze mathematically, and mathematical certainty is our goal (even if we will not quite get there).

Randomness lies at the heart of all cryptography, both in theory and in practice. So we want to be able to think about it clearly.

But wait a minute. What is a “random number”, exactly? A single number is just itself, so what does it even mean to call it “random”? After a coin lands, it is either heads or tails, neither of which seems particularly “random” on its own. And so on.

The solution to the dilemma is this: “Random” refers not to values, but to means of generating them. It is not the outcome of the coin toss that is random, but the process of flipping it.

So, when you see the phrase “random number generator”, do not read it as “random-number generator” (i.e., a generator of random numbers). Read it as “random number-generator” (i.e., a random generator of numbers). The randomness is in the generator, not in the numbers.

Even experts often speak loosely here, using phrases like “source of random bits” when what they really mean is “random source of bits”. Don’t let them confuse you. Or themselves.

Aside: If you show this post to a mathematician and they mumble something about “Martin-Löf randomness” or “Kolmogorov complexity”, do me a favor and smack them upside the head. This is cryptography, which means computers, which means our world consists only of integers like God intended.

Aside #2: If you show this post to a physicist and they mumble something about “Schrödinger”, or to an engineer and they mumble something about “thermal noise” or “reverse-biased diodes”, do me a favor and smack them upside the head twice. Actually, make it three times for the engineer. To win against someone who can outsmart us, we need very precise reasoning, which means totally separating the math from the real world. We have to bring them back together eventually, of course, but leaving them connected during the process will only make our thinking fuzzy and our proofs non-existent. (This is the problem with the Linux /dev/random design, by the way… But that is a topic for another time.)

The mathematical language of randomness is called probability theory. In that language, my summary statement would read: There are no random events; only random distributions. In basic cryptography, the formal proofs tend to be tedious but straightforward, relying only on very elementary probability theory. We shall see how far I can get in this series without it.

More next time.

Cryptography Part 3: Once upon a bit

I recently completed Dan Boneh’s introductory cryptography course. I will probably wind up covering some subset of it here, but at my own pace and in my own way. If you want a more serious treatment, go watch his lectures.

As usual, if the equations below look like roadkill, click through to the actual post.

I like simple examples, so let’s start with one. Suppose I am about to go to New York City to obtain some inside information on a public company. When I have the information, I plan to send you one of two messages: “Buy” or “Sell”.

Let’s say we have an adversary who is very smart and very resourceful. He knows our plan. He can and will intercept whatever message I send to you. Can we arrange to communicate in a way that reveals nothing to him?

Here is one approach. Before I head off to New York, we get together in a closed room. We put our cell phones in the refrigerator, activate our Cone of Silence, line the walls with tin foil, etc. And then we flip a fair coin. If the coin comes up heads, we agree that I will lie when I send you the message; that is, I will say “Buy” when I mean “Sell” and vice-versa. If the coin comes up tails, we agree that I will say what I really mean.

Remarkably, this simple scheme guarantees that no adversary can learn anything from my message. Thanks to our use of a random coin, our adversary has a 50/50 chance of understanding my message correctly no matter how he interprets it. Put another way, the adversary can guess the outcome of our coin toss with even odds… But he can also guess my intended message with even odds, without even bothering to intercept anything! So my message tells him nothing he did not already know, which means he obtains zero bits of information from it.

This example, simple though it is, illustrates several fundamental concepts in cryptography.

First, we have the set of possible messages I want to communicate to you, customarily called the message space and denoted by $$\mathcal{M}$$. In this example, $$\mathcal{M} = \{Buy, Sell\}$$. In real-life cryptosystems, $$\mathcal{M}$$ would be considerably larger; something like “all possible English paragraphs”, for example.

Second, we have an adversary whose capabilities are bounded and well-specified. (An adversary with unbounded capabilities is unbeatable; an adversary with ill-specified capabilities defies sound analysis.) Note that the general idea in serious cryptography is not to ask “What can the adversary do?”, but rather to ask “How powerful can the adversary be and still permit us to win?” In this example, we assume our adversary has unlimited computational power, unlimited eavesdropping power, and total knowledge of our plans. We assume he lacks knowledge only of the outcome of our coin toss. We also assume he can only eavesdrop, and not (say) tamper with my message en route to you. Subject to these assumptions, we can prove that my communication to you is perfectly secure in the sense that it communicates no information to our adversary.

Third, we have some set of possible messages I might actually transmit and the adversary could intercept. This is called the ciphertext space and is denoted by $$\mathcal{C}$$. In this example, the ciphertext space is the same as the message space; that is, $$\mathcal{C} = \{Buy, Sell\}$$.

Fourth, we have some secret information, shared by us but unknown to our adversary, that we will use to encode elements of $$\mathcal{M}$$ into elements of $$\mathcal{C}$$. The set of all possible secrets is called the key space and is denoted by $$\mathcal{K}$$. In this example, $$\mathcal{K} = \{Tails, Heads\}$$.

Note that we assume our adversary knows absolutely everything about our scheme except for the key. This has been the customary assumption in cryptography for over a century, and it is called “Kerckhoff’s Principle”.

Fifth, we have an encryption scheme, denoted $$E$$. This is a function that takes a (key, message) pair and computes a ciphertext. That is, $$E$$ takes some $$k \in \mathcal{K}$$ and some $$m \in \mathcal{M}$$ and produces $$E(k, m) = c \in \mathcal{C}$$. In this example:

$$E(Tails, Buy) = Buy \\ E(Tails, Sell) = Sell \\ E(Heads, Buy) = Sell \\ E(Heads, Sell) = Buy$$

The encryption scheme tells me how to encode a message to you.

Sixth, we have a decryption scheme — denoted $$D$$ — that takes a (key, ciphertext) pair and produces a message. That is, $$D$$ takes any $$k \in \mathcal{K}$$ and $$c \in \mathcal{C}$$ and produces $$D(k, c) = m \in \mathcal{M}$$. In this example, the decryption scheme is the same as the encryption scheme:

$$D(Tails, Buy) = Buy \\ D(Tails, Sell) = Sell \\ D(Heads, Buy) = Sell \\ D(Heads, Sell) = Buy$$

The decryption scheme tells you how to decode a message from me.

Note that the $$E$$ and $$D$$ functions must obey the basic consistency principle that decryption is the opposite of encryption. In symbols, for any $$k \in \mathcal{K}$$ and $$m \in \mathcal{M}$$, $$D(E(k,m)) = m$$.

A collection of these five items — $$\mathcal{M}$$, $$\mathcal{C}$$, $$\mathcal{K}$$, $$E$$, and $$D$$ — is called a cryptosystem. This particular cryptosystem is called the “one-time pad”.

I suppose that is enough for one installment. I will refer back to this example in the next two or three posts, where I plan to cover randomness, the general one-time pad, and more than you ever wanted to know about exclusive-OR. Not necessarily in that order.

Cryptography Part 2: More rambling

Impossibility proofs have always fascinated me. Solving a problem is one thing. Failing to solve a problem is another. But there is something really special about proving nobody can solve it, ever, even if they are smarter than you. (Guess where I am going with this.)

The Delian problem is provably unsolvable. This was not discovered until the 1800s, but the proof is accessible to any mathematically-inclined high school student. So I am going to walk through it. “Seriously, Nemo? You are going to cover an abstract algebra class in a blog post?” Sure, why not?

This will be long, detailed, and almost completely off-topic. Feel free to skip to the last few paragraphs if you just want the punch line.

As usual, if you are reading this in an RSS reader and the equations look like nonsense, click through to the actual post. And ask your RSS provider to install MathJax.

Suppose instead of doubling a cube, we wanted to double a square. That is, given segment AB of length 1, construct a segment of length $$\sqrt{2}$$ using straightedge and compass.

Here is one way. Just like last time, draw a small circle centered at B passing through A, then extend AB to cross that circle at C. Draw two larger circles, one centered at A passing through C, the other centered at C passing through A. Let D be an intersection of these larger circles. Draw BD intersecting the small circle at E:

Segment AE has length $$\sqrt{2}$$.

(Word of advice: Writing raw SVG is a little bit like pulling your own teeth.)

In general, given two segments, it is possible to add, subtract, multiply, or divide them, using only straightedge and compass. That is, if two segments have lengths $$a$$ and $$b$$, you can construct a new segment of length $$a+b$$, $$a-b$$, $$ab$$, or $$a/b$$. Also, given a segment of length $$r$$, you can construct one of length $$\sqrt{r}$$. I hope these are all plausible enough to leave as exercises (hint: similar triangles).

More interestingly, these five operations — add, subtract, multiply, divide, and square root — are all you can do. To see this requires inventing analytic geometry, which is one reason the Delian problem took 2000 years to resolve. Set up a 2D Cartesian coordinate system with A at (0,0) and B at (1,0). Observe that straightedge and compass constructions involve nothing more than introducing new points by finding the intersections of lines and circles based on existing points. In the Cartesian plane, any line may be described by a linear equation based on two points, while any circle may be described by a quadratic equation based on its center and radius. The coordinates of the intersection of any pair of these (line with line, line with circle, or circle with circle) may be found by solving either a linear or a quadratic equation. Since the quadratic formula involves only addition, subtraction, multiplication, division, and square root, it follows that new intersections of lines and circles can only introduce coordinates based on the coordinates of existing points combined with these five operations.

If you start with segment AB of length 1, and all you can do is add, subtract, multiply, divide, and extract square roots, what lengths can you make? Well, you can add 1 to itself to get 2 by doubling AB. You can add 1 to that to get 3. And so on. So any integer is constructible. You can also divide, so any rational number (that is, any $$a/b$$ where $$a$$ and $$b$$ are integers) is also constructible.

Finally, you can extract square roots. That lets you cover a lot of ground, in some ways. Consider:

$$\sqrt{\sqrt{2*\sqrt{\sqrt{2*\sqrt{\sqrt{2*\frac{96389}{76504}}}}}}}$$

This expression involves only integers, multiplication, division, and square root, so its value is constructible with straightedge and compass. And it is close enough to $$\sqrt[3]{2}$$ to fool Google Calculator.

Of course, its value is not exactly $$\sqrt[3]{2}$$, and neither is any other combination of integers using only these five operations. Proving this takes around three lines if your name is “Galois”, but for me it will take a little longer.

The key to the argument is to ignore square roots for a minute and just think about the four basic operations of addition, subtraction, multiplication, and division. What numbers can you generate starting from 1 and repeatedly applying these?

Obviously, you can generate any integer by adding 1 to itself repeatedly. And you can generate any rational number by dividing two integers.

But that is all you can do. Given any two rational numbers, their sum, difference, product, and quotient are themselves all rational. You cannot “escape” from the rational numbers just by adding, subtracting, multiplying, or dividing, and a mathematician would say the rationals are closed under these operations. Any set of numbers closed under these basic operations — i.e. any set from which you cannot “escape” by addition, subtraction, multiplication, or division — is called a number field, or simply a field. The field of rational numbers is important enough to have its own symbol: $$\mathbb{Q}$$. When you see $$\mathbb{Q}$$, think “all rational numbers”.

$$\sqrt{2}$$ is not an element of $$\mathbb{Q}$$; that is, $$\sqrt{2}$$ cannot be expressed as the ratio of two integers. The proof of this was known to Euclid, and I omit it here.

To “escape” from the set of rational numbers, let’s try adjoining $$\sqrt{2}$$ to them, then combining the elements from that new set with addition, subtraction, multiplication, and division, repeatedly. What numbers can you generate now?

Obviously, you can generate any number of the form $$p + q\sqrt{2}$$ where $$p$$ and $$q$$ are rational. Can you generate anything else?

No. Suppose you have two numbers of the form $$a + b\sqrt{2}$$ and $$c + d\sqrt{2}$$ where $$a$$, $$b$$, $$c$$, and $$d$$ are rational. Whether you take their sum, difference, product, or quotient, the result is of the form $$p + q\sqrt{2}$$ where $$p$$ and $$q$$ are also rational. (Try it.) So numbers of this form are all you can generate; you cannot “escape” just by combining them with the four basic operations. In other words, these numbers — $$p + q\sqrt{2}$$, with $$p$$ and $$q$$ in $$\mathbb{Q}$$ — form a field. Mathematicians have a shorthand notation for this field, too; they call it $$\mathbb{Q}(\sqrt{2})$$. In math-speak, all elements of $$\mathbb{Q}(\sqrt{2})$$ are expressible as $$p + q\sqrt{2}$$ where $$p$$ and $$q$$ are elements of $$\mathbb{Q}$$.

We can keep going. Starting from $$\mathbb{Q}(\sqrt{2})$$, we can adjoin another element, like $$\sqrt{3}$$ or $$\sqrt{1+\sqrt{2}}$$. A field created this way, by adjoining a new element to a smaller field, is called a field extension. In general, for a field $$F$$, the field extension you get by adjoining an element $$x$$ is denoted $$F(x)$$.

So, one way to look at straightedge-and-compass constructions is like this. You start with a bunch of numbers from the field $$\mathbb{Q}$$. As long as you only add, subtract, multiply, and divide, you stay in that field. The first time you take a square root of a non-square rational number $$\alpha$$, you “escape” into the extension field $$F_1 = \mathbb{Q}(\sqrt{\alpha})$$. Then, as long as you only add, subtract, multiply, and divide, you stay in that field… Until you take the square root of some non-square $$\beta$$ in $$F_1$$, at which point you “escape” into the field $$F_2 = F_1(\sqrt{\beta})$$.

And so on. Any straightedge-and-compass construction generates a tower of fields $$F_1, F_2, \ldots$$, where each $$F_n$$ is created by taking a square root in $$F_{n-1}$$ and adjoining it. (We define $$F_0 = \mathbb{Q}$$.) All numbers in $$F_n$$ are expressible as $$p + q\sqrt{r}$$, where $$p$$, $$q$$, and $$r$$ come from the previous field $$F_{n-1}$$ and $$r$$ is not a perfect square in $$F_{n-1}$$. Numbers appearing “for the first time” in $$F_n$$ — that is, appearing in $$F_n$$ but not $$F_{n-1}$$ — have $$q$$ non-zero.

Now, suppose some straightedge-and-compass construction could generate $$\sqrt[3]{2}$$. Then either $$\sqrt[3]{2}$$ would be rational — which it isn’t — or it would appear for the first time in some $$F_n$$ and therefore be expressible as $$p + q\sqrt{r}$$, where $$p, q, r$$ are from $$F_{n-1}$$, $$r$$ is not a perfect square in $$F_{n-1}$$, and $$q$$ is not zero.

So let $$(p + q\sqrt{r})^3 = 2$$. Expand that binomial and collect terms to find:

$$(p^3 + 3pq^2r) + (3p^2q + q^3)\sqrt{r} = 2$$

Observe that the coefficient on $$\sqrt{r}$$ — that is, $$3p^2q + q^3$$ — must be zero. If it were not, we could solve this equation for $$\sqrt{r}$$ in terms of values from $$F_{n-1}$$, which would make $$r$$ a perfect square in $$F_{n-1}$$, meaning we never “escaped” from that field at all, a contradiction.

It follows that $$p – q\sqrt{r}$$ is also a cube root of 2; just take this value, cube it, and compare to the expression above, remembering that $$3p^2q + q^3$$ is zero.

But, since $$q \neq 0$$, $$p + q\sqrt{r}$$ and $$p – q\sqrt{r}$$ are distinct real numbers. Since cube root is a strictly increasing function, every real number has exactly one real cube root. Thus we have reached a contradiction and $$\sqrt[3]{2}$$ is not constructible.

The impossibility of the Delian problem was first proven by a guy named Pierre Wantzel in 1837, but the underlying math was found by Galois who was building on Gauss. (“Who was the greatest mathematician, Gauss or Euler?” is sort of the math geek equivalent of “pirates vs. ninjas”. Except that ninjas are obviously way cooler.)

I stole this particular proof from the greatest general-interest math book of all time.

Thus ends this brief introduction to abstract algebra.

What is the point? There are two.

First, if you really want to understand cryptography, you could do worse than to learn a little abstract algebra. Our adversary knows a lot of abstract algebra, and it shows up in most of the mathematical gadgets used for cryptography, especially public key systems. This is not the last time we will see the word “field”.

Second, since our adversary is smarter, more motivated, and better funded than we are, we really, really, really want cryptography that is provably impossible to break. Unfortunately, we cannot have it, because asking this of modern-day computational complexity theory is a little bit like asking the ancient Greeks to solve the Delian problem. Still, we want to come as close as we can.

The classic flaw of capable people is overconfidence. It takes a certain arrogance to conflate “I do not see how to break this” with “This is unbreakable”, and such arrogance is remarkably common in otherwise smart people. Especially engineers. It is also deadly in this game. Any time you hear anyone — especially an “expert” — say that a cryptographic flaw can be ignored as “purely academic”, you should conclude they are the wrong kind of expert and run away.

More next time.

Cryptography Part 1: Drunken rambling introduction

There is a series of posts forming in my head. I have no unifying theme nor particular audience in mind, so they will be even more rambling and incoherent than usual. Also I plan to have a drink or two before each just to complete the effect. You have been warned.

Let’s play a little game. You and I will be on the same team for a change. This is our asset:

It has 16-64 GB of storage, consumes 0.5 watts, and occupies 11.25 square inches.

On the bright side, we do have two of them. One each.

This is our adversary:

It has 3-12 exabytes of storage, consumes 65 megawatts, and occupies 100000 square feet. Also, it is operated by people smarter than you, smarter than me, and smarter than anybody either one of us has ever heard of.

The game is this. You and I will try to have a conversation over a great distance that is unintelligible to this adversary.

Now, perhaps it is just me — I still get excited by powered flight — but I find it awe inspiring that it might actually be possible for us to win. Not easily and not with certainty, but still.

That is what this series will be about. More or less.

Our story begins around 400 B.C. on the Greek island of Delos. Its citizens were suffering from internal strife threatening to tear the society apart. Or something. The island’s leaders consulted the Oracle at Delphi, who explained that Apollo was angry, and to appease him, the citizens had to construct a new altar double the size of their existing one.

Now, the ancient Greeks had far more rigorous minds than your typical modern engineer. To them, “construct” meant something very particular: Create something perfect using idealized versions of various masonry/carpentry tools. Extra credit for using only straightedge and compass.

(If you have never seen this game before, here are the rules. Given two points, you may use the straightedge to connect the points with a perfect line and extend it as far as you like. You may also set the compass to the distance between any two points, then draw a perfect circle with that radius centered on any other point. By starting with a few provided points and applying the straightedge and compass repeatedly, you generate new points at the intersections of all the lines and circles. That’s it. The ancient Greeks loved this stuff.)

Now, the altar to Apollo was a perfect cube. So the Delians started with a line segment AB having the same length as a side of the cube. Then they used a compass to draw a circle through A centered at B. Then they used the straightedge to extend AB to intersect the circle at C:

Since AC is twice AB, the Delians simply used that as the side of a new cubical altar.

But their problems only got worse. Eventually, they went back and asked the Oracle what was wrong. The Oracle explained that they had angered Apollo further by not following instructions, since they had created an altar not two times but eight times the original’s volume. Apparently, gods can be picky.

The Greeks eventually solved this problem by adding various interesting contraptions to their idealized toolkit. But the extra credit problem remained: Given a segment of length 1, can you construct one of length $$\sqrt[3]{2}$$ using only straightedge and compass?

This problem stumped geometers and would-be geometers for several years. Two thousand, actually. That is how long humanity needed to develop the mathematical tools to solve this Delian problem, as it came to be known. What that solution was and how it relates to anything will be the topic for…

Next time: Gauss, Galois, et. al.

(I did try to warn you.)

Au revoir, Bond Girl

I have lost my co-blogger.

Over the next few days, I will be removing her posts from this blog. I have already removed her contact information from the sidebar.

Q: “Why?”

A: Because she asked me to.

Q: “Why did she do that?”

A: I will not answer. I am not going to release any of our private correspondence, and I am certainly not going to put words in her mouth. So, seriously, don’t ask.

For my part, I am sorry to see her go. I have the deepest respect and appreciation for her on both a personal and a… whatever-co-blogging-is-al… level. She single-handedly increased my readership by a factor of #DIV/0!. Speaking of which, you should all unsubscribe now.

I honestly believe her decision was dictated by her conscience and her reason, just as my actions are dictated by mine.

I would welcome her back any time.

I will leave her posts up for a few days so readers can save copies of their favorites. Her self-evident.org Email address will continue to work until she asks me to shut it down.

The NSA revelations finally got interesting

Time to pay attention. (related video)

Quick background on my perspective: Stories about Snowden himself are boring. Stories about his girlfriend, stories about his politics, and even speculations about his being an agent of Russian intelligence… Boring.

More interesting are the revelations themselves, painting an NSA whose general goal appears to be the interception and permanent storage of all human communication. Everywhere.

But how are they doing it? By obtaining covert access to Google’s servers? By convincing Microsoft to change Skype’s protocol from peer-to-peer to peer-to-Microsoft? By forcing Web site operators to hand over their SSL keys?

How pedestrian. I mean, this is the agency created as a direct response to the Allied experience during World War II, where without the code breakers, we might have lost. This is the quasi-military organization absorbing billions of dollars per year while remaining “non-existent” for decades.

Speculating about NSA’s capabilities, especially cryptological capabilities, has been a hobby horse of mine for a long time. But none of the Snowden revelations shed any light whatsoever. Fair enough, I figured. Maybe NSA reserves the serious codebreaking for more important things than reading my Email. Maybe the good stuff was above his pay grade.

…or then again, maybe not.

N.S.A. Able to Foil Basic Safeguards of Privacy on Web

This is a jaw-dropping article. I am mildly paranoid, but things even I would have called dumb conspiracy theories are being reported as fact in the New York Times. (Ideas like NSA blackmailing Congress are still dumb, but even that I have been forced to upgrade from “laughable” to “very unlikely”.)

Let’s start with DNI Clapper’s response:

The fact that NSA’s mission includes deciphering enciphered communications is not a secret, and is not news

Compare that to the three essential claims in the NYT article:

1. NSA has practical attacks against SSL/TLS.
2. NSA has convinced / paid / forced U.S. manufacturers to insert deliberate vulnerabilities into both hardware and software security products.
3. NSA has infiltrated and undermined Internet standardization bodies, encouraging widespread adoption of vulnerable algorithms and protocols.

With due respect to Gen. Clapper, all of this is most definitely news.

I have a lot more to say, including my own wild-eyed speculations about (1). But I think I will make this multiple posts.

I will mention that (3) has touched off a bit of a firestorm among academic and professional cryptographers. If NIST cannot be trusted, we have a problem… And NIST can no longer be trusted.

Story of the week, if not the year.

P.S. The NYT graphic and (redacted) raw documents deserve close scrutiny.

Bitcoin pieces

It has been a volatile month for the world’s most popular experimental currency:

As you can see(1), the U.S. dollar — measured in milli-bitcoins — has been highly unstable. In Felix Salmon’s words:

[The dollar] is clearly not an effective store of wealth — just look at how quickly that wealth can be evaporated. Neither is it a useful payments mechanism, given how fast its value can fluctuate.

I might be paraphrasing slightly.

Ryan Avent makes the same point more verbosely (and politely).

More recently, Izabella Kaminska accuses me of accusing her of… You know what? Never mind. For the 0 of you who care, listen to the audio, read the comments, and judge for yourself. I will make two quick points:

1. With the possible exception of “shock jocks”, no journalist publicly expresses opinions significantly divergent from those of her sources or those of her organization, ever. This is not a conspiracy theory; it is common sense.
2. Anyone who accuses me of “the skin in the game fallacy” is a moron.

Ahh, that feels better.

[Update, following evening]

In my apparently excessive free time, I had a little back-and-forth with Izabella today. Note that I made this post last night, well before today’s exchange, but if you read them in the opposite order it looks even more like I am attacking her in particular.

For what it is worth, that was not my intention. It’s just that all of the Bitcoin bashing from the FT seems to have her name attached to it.

To rephrase my earlier assertion: Journalists have no more license to express their true opinions in public than I do to express mine to my employer. The eagerness of mainstream financial journalism to run Bitcoin bashing articles — no matter how idiotic — does say something about the financial sector’s reaction to Bitcoin, in my view. I consider this obvious.

But I do regret making it personal.

My next post will be purely technical.

(1) I know, my gnuplot-fu is weak. But it serves the purpose.


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# Math Help - Limit of sequence

1. ## Limit of sequence

This sequence is strictly increasing (monotonic, already proved).

3 is an upper bound on this sequence(already proved). Therefore I know that this sequence converges to a Limit A which is <= 3.

My question is knowing all this, how do I find the Limit A? I.E. What the series converges to, the exact number.

Thanks,

Len

2. Originally Posted by Len

This sequence is strictly increasing (monotonic, already proved).

3 is an upper bound on this sequence(already proved). Therefore I know that this sequence converges to a Limit A which is <= 3.

My question is knowing all this, how do I find the Limit A? I.E. What the series converges to, the exact number.

Thanks,

Len
Let $\lim b_n = L$. then, of course, $\lim b_{n + 1} = L$ as well.

thus, take $b_{n + 1} = 3 - \frac 1{b_n}$

take the limit of both sides.

$\lim b_{n + 1} = \lim \left(3 - \frac 1{b_n} \right)$

$\Rightarrow L = 3 - \frac 1L$

i leave the rest to you

format: plain


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# Strong and Weak Form Solution – FEA

Partial Differential Equations – PDE is called “strong form” because the relationship MUST satisfy at every mathematical point in the domain.

A “weak form” means that the relationship (in integral form) is only satisfied in overall sense. In another word, “it is only satisfied in an integral (sum) sense, it is not a requirement that every point in the domain MUST obey”

The strong form of a differential equation is just that: the (partial) differential equation itself. Evaluating the PDE requires being able to get all the associated derivatives. It is satisfied pointwise at every point in a body, and is usually stated as D[u] = 0, where D is some partial differential operator. In this case, I am using u as the displacement. It may be more appropriate to look at, say, Cauchy stress (s) instead of displacement. Then the strong form might be something like D[s,u] = div[s] – ru,tt= 0.

The weak form is obtained by multiplying the PDE by an arbitrary weighting function of (in most cases) the spatial variables, then integrating the result over the domain. One then requires that the result is zero for all choices of such functions. An integration by parts is performed, leading to differentiability requirements on the weighting function, but relaxing, or “weakening” the requirements on the field described by the PDE. Once we start setting requirements on these functions, we “weaken” the form even more, but often provide a basis for expressing the approximate solution.

## Questions:

Q1. Why do we multiply the PDE by the weighting function?

Q2. How do we choose the weighting function?

## Answers:

The “why” is to reduce differentiability requirements on our approximate solution.
The “how” is whatever works. If you know of a particular set of functions that work well in your geometry (say Bessel functions for axisymmetric problems), you can use these. We often use arbitrary combinations of nodal basis functions (compact support, C1, finite domains). If we use the same basis for our weighting functions as we use to represent our primary variable field(s), e.g. displacement, we are using a Galerkin method.

By Matt Lewis
Los Alamos, New Mexico

This excerpt is taken from IMECHANICA Forum.


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Home > Computer miscellany > LaTeX strikethrough

## LaTeX strikethrough

I wanted to write something crossed out in a LaTeX document. Could not find the way to do this in the obvious places. But a Yahoo! search produced the answer:

\usepackage{ulem}

anywhere in the preamble  (before \begin{document}). Then

\sout{word}

produces

word

This works in slides as well, also on mathematical symbols. But this package makes \em act as an underline, and \it produces what \sl normally would.


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# SPI wiring SDIO pin with MISO and MOSI

I'm connecting the N76E003AT20 microcontroller to an RF module. The N76E003AT20 has 4 SPI pins: SPCLK, MISO, MOSI and SS. The module also has 4 SPI pins which are SCLK, SDIO, CSB and FCSB as shown in the table:

I understand that I have to connect SCLK and SPCLK. Do I connect SDIO to both MISO and MOSI? How about CSB and FCSB?

• Which module? The uC will be the master, I guess. I that case, you usually don't care about the SS pin.connecting MISO and MOSI together sound wrong, then the master will receice what he just sent or there will be bus contention Dec 23, 2020 at 17:40

It appears that this RF module doesn't exactly follow the SPI specification. This means that you probably won't be able to use a canned high-level driver. Some of this is speculation based on a SPI slave module that I found that uses a bidirectional data signal.

SPI Slave Selects (AKA Chip Selects) are basically simple MCU outputs, so creating a second SS should be easy.

Connecting MISO and MOSI also causes complications, you must tri-state MOSI during the part of the cycle when the slave is driving SDIO.

MPU        SLAVE
SPCLK      SCLK
MISO/MOSI  SDIO
SS1        CSB
SS2        FCSB

• Thank you! If I connect MOSI to SDIO through say a 10k resistor and then connect MISO directly to SDIO, will that do the trick? Dec 24, 2020 at 12:32
• Yes, you may be able to use a resistor so that the slave can overdrive the signal. Dec 24, 2020 at 16:38


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text: Here comes the most important part of solving a truss using the method of Sections. P-428. Method of Sections (conât) â¢Application - The method of sections is commonly used when the forces in only a few particular members of a truss are to be determined; - The method of sections is always used together with the method of joints to analyze trusses. Truss – Example Problem. The method of sections is an alternative to the method of joints for finding the internal axial forces in truss members.It works by cutting through the whole truss at a single section and using global equilibrium (3 equations in 2D) to solve for the unknown axial forces in â¦ If the forces in only a few members of a truss are to be determined, the method of sections is generally the most appropriate analysis procedure. This is done by making a "cut" along three selected members. If only two members form a truss joint and no external load or support reaction is applied to the joint, the members must be zero-force members. In this method, we will cut the truss into two sections by passing a cutting plane through the members whose internal forces we wish to determine. METHOD OF SECTIONS :check: A truss is divided into two parts by taking an imaginary "CUT" through the truss. The Method of Sections (Ritter Method) ï¨If the forces in only a few members of a truss are to be found, the method of sections generally provide the most direct means of obtaining these forces ï¨The method is created the German scientist August Ritter (1826 - 1908). The sections are obtained by cutting through some of the members of the truss to expose the force inside the members. So, in our example here would be our slice: Focussing on the left side only, you are left with the following structure: Now think of this structure as single standing structure. âFx = 0 âFy = 0; Recall, that the line of action of a force acting on a joint is determined by the geometry of the truss â¦ [more] Select "solve for reaction forces" to see how the reaction forces and are calculated. Note to Teacher: In this section, a step-by-step example of the simplest method used in civil engineering to solve for the unknown forces acting on members of a truss is presented. The summation of forces and moment about H result in ()()()()() xHx Hx yHyI HI I Hy. Each imaginary section must be in equ ilibrium if the entire truss is in equilibrium. Step-by-step solution: 90 â¦ The Howe truss was invented by William Howe in 1840, and was widely used as a bridge in the mid to late 1800s. Consequently they are of great Truss: Method of Joints and Sections Theory of Structure - I Lecture Outlines Method Zero of Joints Force Draw the free-body diagram. BE in the truss using the method of sections. The other method is just given for the sake of confusion. Tips: 1. 0 3 kips 3 kips 3 kips 3 kips 12 kips 0 0 15 ft 3 kips 10 ft 3 kips 20 ft 3 kips 30 ft 3 kips 40 ft Method of Sections The method of sections is most effectively employed when one wishes to know the loads in only a few members of a truss. 2. The method of sections is a process used to solve for the unknown forces acting on members of a truss. Online Truss Solver using method of joints. The method of sections consists of passing an imaginary linethrough the truss, cutting it into sections. Truss â Example Problem. Statically Indeterminate A truss is considered statically indeterminate when the static equilibrium equations are not sufficient to find the reactions on that structure. (Figure 2) Part A - Determining the forces in three members on the right-hand side of the truss Determine the forces in members HD, CD, and GD. Next do a force balance of the forces:, Using the method of joints and the method of sections, determine the force in all members. Problem 428 - Howe Truss by Method of Sections Problem 428 Use the method of sections to determine the force in members DF, FG, and GI of the triangular Howe truss shown in Fig. Example problem using method of sections for truss analysis - statics and structural analysis. Each imaginary section must be in equ ilibrium if the entire truss is in equilibrium. FF Mxy z 00 0 Method of Sections Procedure for analysis- the following is a procedure for analyzing a truss using the method of sections: This method of structural analysis is extremely useful when trying to solve some of the members without having to solve the entire structure using method of joints. The first step is to draw the free body diagram for the given truss. The Method of Sections Learning Goal: To apply the method of sections to a truss and determine the unknown forces in specified members. Howe truss 18 - 30 m Pratt truss 18 - 30 m Howe truss flat roof Warren truss flat roof saw-tooth truss skylight Fink truss > 30 m three-hinged arch hangar, gymnasium. Structures I University of Michigan, Taubman College Slide 11 of 31 Method of Sections-example Howe Truss 1. QUESTION: 9. The member forces are determined by considering the equilibrium of the part of the truss on either side of the section. The method of sections is used to calculate the forces in each member of the truss. In this method, we will cut the truss into two sections by passing a cutting plane through the members whose internal forces we wish to determine. The method of joints or method of sections must be used to find a truss member force as a function of the moving unit load position. Try hold the "Shift" key while placing members and loads. Problem. First, calculate the reactions at the supports. We will determine here the force in the truss member FE to understand the basics of method of sections. Since truss members are subjected to only tensile or compressive forces along their length, the internal forces at the cut member will also be either tensile or compressive with the same magnitude.. We use cookies to ensure you get the best experience on our website. A Howe truss is a truss bridge consisting of chords, verticals, and diagonals whose vertical members are in tension and whose diagonal members are in compression. If a truss is in equilibrium, then each of its joints must also be in equilibrium. Subscribe for Updates. Draw the free body diagram for each joint. Step 1: Drawing of free body diagram. This method permits us to solve directly any member by analyzing the left or the right section of the cutting plane. * So why do we use method sections? It involves making a slice through the members you wish to solve. Method of Sections. For the Howe roof truss shown, determine the forces in members BC, CI, and IJ. the forceâ¦ 4 Method of Sections Monday, October 22, 2012 Method of Sections " The second method of truss analysis that we will consider is called the method of sections. " Method of Sections In this method, we will cut the truss into two sections by passing a cutting plane through the members whose internal forces we wish to determine. This is commonly used in light roofing so that the longer diagonals experience tension under reversal of stresses due to wind load. P-428. Decide how you need to cut the truss : Where you need to cut the truss. If the forces in only a few members of a truss are to be determined, the method of sections is generally the most appropriate analysis procedure. Howe truss 18 - 30 m Pratt truss 18 - 30 m Howe truss flat roof Warren truss flat roof saw-tooth truss skylight Fink truss > 30 m three-hinged arch hangar, gymnasium. P-428. Newton's Third Law indicates that the forces of â¦ If three members form a truss for which two of the members are collinear, the third member is a zero-force member provided no external force or support reaction is applied. In the Method of Joints, we are dealing with static equilibrium at a point. The method of sections is another method to determine forces in members of a truss structure. For the Howe roof truss shown, determine the forces in members BC, CI, and IJ. The Method of Sections (Ritter Method) If the forces in only a few members of a truss are to be found, the method of sections generally provide the most direct means of obtaining these forces The method is created the German scientist August Ritter (1826 - 1908). This rafter truss calculator, has a range of applications including being used as a wood truss calculator, roof truss calculator, roof rafter calculator, scissor truss … We will first show the known forces at its given point or joint. Problem. Solution for Part 2 A Pratt roof truss is loaded as shown. The converse of the Pratt is the Howe truss [Fig. If possible, determine the support reactions 2. The method of sections consists of passing an imaginary linethrough the truss, cutting it into sections. While the method of section cuts the whole structure of trusses into section and then uses the cut out portion for the calculations of the unknown forces. Make final qualitative check of solution. ... HOWE (4) FAN (7) KING-POST (3) FAN (8) PRATT (2) FINK (9) FLAT WARREN (1) PINK (10) FLAT PRATT. FBD of Joint A and members AB and AF: Magnitude of forces denoted as AB & AF Method of joints defines as if the whole truss is in equilibrium then all the joints which are connected to that truss is in equilibrium. It is essentially the reverse of ... all-in-all our truss models, that we use to carry out method of sections and method of joints analysis on, are pretty reasonable approximations of real truss structures. The method of joints consists of satisfying the equilibrium equation for forces acting on each joint. $12R_A + 1(20 \sin \theta) + 2(20 \sin \theta) = 10(20 \cos \theta) + 8(20 \cos \theta)$, $12R_A + 1(20)\left( \frac{1}{\sqrt{5}} \right) + 2(20)\left( \frac{1}{\sqrt{5}} \right) = 10(20)\left( \frac{2}{\sqrt{5}} \right) + 8(20)\left( \frac{2}{\sqrt{5}} \right)$, $6F_{DF}\left( \frac{1}{\sqrt{5}} \right) + 6\left( 11\sqrt{5} \right) + 1(20)\left( \frac{1}{\sqrt{5}} \right) + 2(20)\left( \frac{1}{\sqrt{5}} \right) = 4(20)\left( \frac{2}{\sqrt{5}} \right) + 2(20)\left( \frac{2}{\sqrt{5}} \right)$, $F_{DF} = -25 ~ \text{kN} = 25 ~ \text{kN compression}$       answer, $F_{FG} = 10\sqrt{5} ~ \text{kN} = 22.36 ~ \text{kN tension}$       answer, $3F_{GI} + 2\sqrt{5}(20) + \sqrt{5}(20) = 6\left( 11\sqrt{5} \right)$, $F_{GI} = 2\sqrt{5} ~ \text{kN} = 4.47 ~ \text{kN tension}$       answer, Method of Sections | Analysis of Simple Trusses, ‹ Problem 427 - Interior Members of Nacelle Truss by Method of Sections, Problem 429 - Cantilever Truss by Method of Sections ›, Method of Joints | Analysis of Simple Trusses, Problem 417 - Roof Truss by Method of Sections, Problem 418 - Warren Truss by Method of Sections, Problem 419 - Warren Truss by Method of Sections, Problem 420 - Howe Truss by Method of Sections, Problem 421 - Cantilever Truss by Method of Sections, Problem 422 - Right-triangular Truss by Method of Sections, Problem 423 - Howe Roof Truss by Method of Sections, Problem 424 - Method of Joints Checked by Method of Sections, Problem 425 - Fink Truss by Method of Sections, Problem 426 - Fink Truss by Method of Sections, Problem 427 - Interior Members of Nacelle Truss by Method of Sections, Problem 428 - Howe Truss by Method of Sections, Problem 429 - Cantilever Truss by Method of Sections, Problem 430 - Parker Truss by Method of Sections, Problem 431 - Members in the Third Panel of a Parker Truss, Problem 432 - Force in Members of a Truss by Method of Sections, Problem 433 - Scissors Truss by Method of Sections, Problem 434 - Scissors Truss by Method of Sections, Problem 435 - Transmission Tower by Method of Sections, Problem 436 - Howe Truss With Counter Braces, Problem 437 - Truss With Counter Diagonals, Problem 438 - Truss With Redundant Members, Method of Members | Frames Containing Three-Force Members. Problem 428 Cut a panel with diagonals M at L2 and resolve upper chord force at U2. 2(c)] are used for longer spans having high pitch roof, since the web members in such truss are sub-divided to obtain shorter members. In the method of sections, a truss is divided into two parts by taking an imaginary “cut” (shown here as a-a) through the truss. A Howe truss is a truss bridge consisting of chords, verticals, and diagonals whose vertical members are in tension and whose diagonal members are in compression. After cutting the truss by an imaginary section XX, we will have to show the forces in the members and these forces will be determined by using method of sections. If the forces in only a few members of a truss are to be determined, the method of sections is generally the most appropriate analysis procedure. This method permits us to solve directly any member by analyzing the left or the right section of the cutting plane.. The Method of Sections involves analytically cutting the truss into sections and solving for static equilibrium for each section. It works by cutting through the whole truss at a single section and using global equilibrium (3 equations in 2D) to solve for the unknown axial forces in â¦ In this method, we will cut the truss into two sections by passing a cutting plane through the members whose internal forces we wish to determine. The method of sections enables one to determine forces in specific truss members directly. 5. The method of sections is an alternative to the method of joints for finding the internal axial forces in truss members.It works by cutting through the whole truss at a single section and using global equilibrium (3 equations in 2D) to solve for the unknown axial forces in … * First of all I would like to point out that method of joints is itself a subset of method of sections so basically we are doing it unknowingly all the time . Here comes the most important part of method of joints.Cut a section of the truss in a way that the section should pass through 3 members. (If you want them in all members you may as well use the method of joints and considerable patience.) Each imaginary section must be in equ ilibrium if the entire truss is in equilibrium. Since truss members are subjected to only tensile or compressive forces along their length, the internal forces at the cut member will also be either tensile or compressive with the same magnitude. Method of Sections. In general, assume all the force member reactions are They are used to span greater distances and to carry larger loads than can be done effectively by a single beam or column. The laws of statics still apply â so the â¦ This gives U1U2H M at U1 to find L1L2 This method permits us to solve directly any member by analyzing the left or the right section of the cutting plane. MethodofSections ◮The Method of Sections involves analytically cutting the truss into sections and solving for static equilibrium for each section.. Use the method of sections to determine the force in members DF, FG, and GI of the triangular Howe truss shown in Fig. The method involves breaking the truss down into individual sections and analyzing each section as a separate rigid body. Structural Analysis: Plane Truss Method of Joints â¢ Start with any joint where at least one known load exists and where not more than two unknown forces are present. 2(b)]. (Figure 1) If the free-body diagram of the joint on either side of the cut is known, we can apply the equations of equilibrium to that part of the member to determine the forces at the cut section. Now, we will be interested here to understand how to solve truss problems using method of joints step by step with the help of this post. This Demonstration solves a truss using the method of sections, which involves "cutting" along several selected members and taking the sum of the forces, forces and moment about a point. Solution for Part 2 A Pratt roof truss is loaded as shown. Trusses: Method of Joints Frame 18-1 *Introduction A truss is a structure composed of several members joined at their ends so as to form a rigid body. ◮The sections are obtained by cutting through some of the members of the truss to expose the force inside the members. The Method of Sections involves analytically cutting the truss into sections and solving for static equilibrium for each section. Truss Problem 428 - Howe Truss by Method of Sections Problem 428 Use the method of sections to determine the force in members DF, FG, and GI of the triangular Howe truss shown in Fig. Method of Sections â¡ involves cutting the truss into two portions (free body diagrams, FBD) by passing an imaginary section through the members whose forces are desired. Consequently they are of great Method of Sections In this method, we will cut the truss into two sections by passing a cutting plane through the members whose internal forces we wish to determine. The method of sections depends on our ability to separate the truss into two separate parts, hence two separate FBDs, and then perform an analysis on one of the two parts. 0 3 kips 3 kips 3 kips 3 kips 12 kips 0 0 15 ft 3 kips 10 ft 3 kips 20 ft 3 kips 30 ft 3 kips 40 ft The method of sections is an alternative to the method of joints for finding the internal axial forces in truss members. The bridge structure will be the smallest and simplest possible: The Warren truss with three equilateral triangles (Figure 9). This rafter truss calculator, has a range of applications including being used as a wood truss calculator, roof truss calculator, roof rafter calculator, scissor truss calculator or â¦ Trusses: Method of Joints Frame 18-1 *Introduction A truss is a structure composed of several members joined at their ends so as to form a rigid body. The Method of Sections involves analytically cutting the truss into sections and solving for static equilibrium for each section. The method of joints analyzes the force in each member of a truss by breaking the truss down and calculating the forces at each individual joint. https://mathalino.com/reviewer/engineering-mechanics/method-sections-analysis-simple-trusses, https://learnaboutstructures.com/Method-of-Sections, https://constructionhow.com/method-of-sections/, http://www2.hawaii.edu/~takebaya/lessons/method-of-sections.pdf, http://www.ce.memphis.edu/3121/notes/notes_03c.pdf, http://demonstrations.wolfram.com/MethodOfSectionsToSolveATruss/, http://secretsofengineering.net/Statics/PDF_files/Statics-Unit-19.pdf, http://site.iugaza.edu.ps/malqedra/files/Lecture-6.2.pdf, http://www.ce.memphis.edu/2131/PDFsF12/Method%20of%20Sections.pdf, http://adaptivemap.ma.psu.edu/websites/5_structures/method_of_sections/methodofsections.html, https://theconstructor.org/structural-engg/truss-analysis/2565/, https://readcivil.com/analysis-of-truss-method-of-sections-and-joints/, https://en.wikibooks.org/wiki/Statics/Method_of_Sections, https://www.slideshare.net/physics101/structural-analysis-method-of-sections, https://ae.msstate.edu/vlsm/truss/analysis_of_statdet_trusses/mosect.htm, https://www.quora.com/Why-do-we-use-method-of-sections-for-truss, http://www.rpi.edu/dept/core-eng/WWW/IEA/f15/lectures/Lecture20.pdf, https://civilengineer.webinfolist.com/mech/prob32.htm, https://www.quora.com/What%E2%80%8B-is-the-difference-between-a-method-of-joints-and-a-method-of-sections, https://www.xpcourse.com/truss-analysis-method-of-sections, https://mathalino.com/reviewer/engineering-mechanics/problem-001-ms-method-sections, https://wethestudy.com/engineering/truss-analysis-basics-structural-analysis/, https://www.hkdivedi.com/2019/12/how-to-solve-truss-problems-using.html, https://www.scribd.com/document/99975296/Method-of-Sections, https://www.coursera.org/lecture/fe-exam/trusses-method-of-sections-2bLDs, Continuing medical education for physicians. There are simply too many unknowns. 1 through 8 - Using the method of joints, determine the force in each member of the truss shown. 11- Determine ... 11- Determine the force in members EH and GI of the truss shown. 2. Method of section: This method entails passing an imaginary section through the truss to divide it into two sections. To apply the method of sections, we use an imaginary cut through members of the truss to expose the forces in those members as external forces. The method of sections enables one to determine forces in specific truss members directly. In the method of sections, a truss is divided into two parts by taking an imaginary “cut” (shown here as a-a) through the truss. This is a Howe truss Method of Joints Procedure for analysis-the following is a procedure for analyzing a truss using the method of joints: 1. We will show the reaction forces as per suitable force interaction at each support of the given truss. Truss Analysis â Method of sections: As the name suggests in method of sections we make sections through a truss and then calculate the force in the members of the truss though which the cut is made. Note to Teacher: In this section, a step-by-step example of the simplest method used in civil engineering to solve for the unknown forces acting on members of a truss is presented. 2. For Problem, use the method of section. of the truss by Method of Joints. PROBLEM 1: Determine the force in members DF, DG, and EG of the Howe truss shown. Method of Sections. This free online truss and roof calculator generates the axial forces and reactions of completely customisable 2D truss structures. The method of sections consists of passing an imaginary linethrough the truss, cutting it into sections. After this illustration let me put down the steps that are taken to solve for forces in members of a truss by method of sections: 1. For Problem, use the method of section. The method of sections is a process used to solve for the unknown forces acting on members of a truss. The Howe truss was invented by William Howe in 1840, and was widely used as a bridge in the mid to late 1800s. This method permits us to solve directly any member by analyzing the left or the right section of the cutting plane.. The main methods for the solving for the unknown force on the truss are method of section and the method of joints. The method of joints consists of satisfying the equilibrium conditions for the forces exerted âon the pinâ at each joint of the truss Method of Sections. In order to find unknown forces in using the method of sections, sections of the truss … Structural Analysis: Plane Truss Method of Joints • Start with any joint where at least one known load exists and where not more than two unknown forces are present. The sections are obtained by cutting through some of the members of the truss to expose the force inside the members. (Figure 1) For the Howe bridge truss shown, d=11 ft,F1=2850 lb, F2=1950 lb, F3=2450 lb, and F4=3650 lb. Each imaginary section must be in equ ilibrium if the entire truss is in equilibrium. Compare the answers from method of joints and method of sections. In the Method of Joints, we are dealing with static equilibrium at a point. State whether each member ... 10- Determine the force in each member of the Howe roof truss shown. Step-by-step solution: 90 … This also means that, like the equilibrium method for beams, you often must find the influence diagrams for one or more reaction forces before you can find the influence diagram for the internal axial force in a specific truss member. Method of Sections-example. We will see here, in this post, the analysis of the forces in the various members of the truss by using the method of joints. They are used to span greater distances and to carry larger loads than can be done effectively by a single beam or column. Read moreabout Problem 428 - Howe Truss by Method of Sections … Fink trusses [Fig. This method permits us to solve directly any member by analyzing the left or the right section of the cutting plane. The summation of forces and moment about H result in ()()()()() xHx Hx yHyI HI I Hy. The method involves breaking the truss down into individual sections and analyzing each section as a separate rigid body. Make a cut to divide the truss into section, passing the cut through members where the force is needed. It hinges on dividing the truss into two parts and then considering a free body diagram of one part or the other, Solve. The Howe truss predates the Pratt truss by 4 years and was proposed by William Howe in 1840. Using the method of joints and the method of sections, determine the force in all members. Draw the free-body diagram. Solves simple 2-D trusses using Method of Joints -> Check out the new Truss Solver 2. This means that only two main methods are to applied for solving the unknown force on truss. In the following examples, you will practice applying the. State whether each member is in tension or compression. For the Howe roof truss shown, determine the forces in members BC, CI, and IJ Method of Sections: Method of sections involves cutting of members of a truss to expose their internal forces. the force… Taking the sum of the moments at the left support:. View 04 Truss- Method of Joints and Sections.ppt from CIVIL ENGI 501 at U.E.T Taxila. Method of Sections ≡ involves cutting the truss into two portions (free body diagrams, FBD) by passing an imaginary section through the members whose forces are desired. The method of sections consists of passing an imaginary linethrough the truss, cutting it into sections. BE in the truss using the method of sections. 80 KN 120 KN 3 m E 200 KN 4 Panels @ 4 m = 16 m A 75mm x 300 mm Guijo girder is loaded as shown. Method of Sections In this method, we will cut the truss into two sections by passing a cutting plane through the members whose internal forces we wish to determine. After cutting the truss by an imaginary section XX, we will have to show the forces in the members and these forces will be determined by using method of sections. Now you got a left part and right part of the structure Each part seperated by the section is in equilibrium state.Focussing of the part,which has lesser number of unknows and applying the law of statitics, the unknown forces in the members can be found FBD of Joint A and members AB and AF: Magnitude of forces denoted as AB & AF - Tension indicated by an arrow away from the pin We will determine here the force in the truss member FE to understand the basics of method of sections. Select a part and press "Delete" to delete it. This free online truss and roof calculator generates the axial forces and reactions of completely customisable 2D truss structures. 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And reactions of completely customisable 2D truss structures University of Michigan, Taubman College Slide of! By cutting through some of the truss by method of sections consists of the! Imaginary section must be in the truss are method of joints - > Check out the new truss 2. Along three selected members will practice applying the [ Fig is loaded as shown if the entire truss is as! Cut through members where the force in all members 8 - using the method of sections consists of passing imaginary. Well use the method of joints and the method of sections Learning:. Left or the right section of the cutting plane members BC,,... Converse of the members involves breaking the truss a single beam or column solving truss... Passing the cut through members where method of sections howe truss force in members EH and of. The Howe roof truss shown L2 and resolve upper chord force at U2 for each section truss... 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Chebyshev-Pade Approximation - Maple Help

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Chebyshev-Padé Rational Function Approximation

 Main Concept A univariate function f(x) over a finite range a..b may be approximated in the Chebyshev-Padé sense by f(x)≈p(x)/q(x) where p(x) and q(x) are polynomials of degree at most m and n.

Move the sliders to adjust the degrees of p(x) and q(x). Change the example by editing the expression to be approximated (and click Refresh) or edit the endpoints a or b. Change the form of the approximating expression using the drop-down list.

Chebyshev-Pade approximation: rational approximation with degrees m and n

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# A two-dimensional slice through the parameter space of two-generator Kleinian groups

Research paper by Elena Klimenko, Natalia Kopteva

Indexed on: 19 Dec '05Published on: 19 Dec '05Published in: Mathematics - Group Theory

#### Abstract

We describe all real points of the parameter space of two-generator Kleinian groups with a parabolic generator, that is, we describe a certain two-dimensional slice through this space. In order to do this we gather together known discreteness criteria for two-generator groups and present them in the form of conditions on parameters. We complete the description by giving discreteness criteria for groups generated by a parabolic and a $\pi$-loxodromic elements whose commutator has real trace and present all orbifolds uniformized by such groups.


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# Measurement of $Z\rightarrow\tau^+\tau^-$ production in proton-proton collisions at $\sqrt{s} = 8$ TeV

[to restricted-access page]

## Abstract

A measurement of $Z\rightarrow\tau^+\tau^-$ production cross-section is presented using data, corresponding to an integrated luminosity of 2 fb$^{-1}$, from $pp$ collisions at $\sqrt{s}=8$ TeV collected by the LHCb experiment. The $\tau^+\tau^-$ candidates are reconstructed in final states with the first tau lepton decaying leptonically, and the second decaying either leptonically or to one or three charged hadrons. The production cross-section is measured for $Z$ bosons with invariant mass between 60 and 120 GeV/$c^2$, which decay to tau leptons with transverse momenta greater than 20 GeV/$c$ and pseudorapidities between 2.0 and 4.5. The cross-section is determined to be $\sigma_{pp\rightarrow{}Z\rightarrow{}\tau^+\tau^-} = 95.8 \pm 2.1 \pm 4.6 \pm 0.2 \pm 1.1 \mathrm{pb}$, where the first uncertainty is statistical, the second is systematic, the third is due to the LHC beam energy uncertainty, and the fourth to the integrated luminosity uncertainty. This result is compatible with NNLO Standard model predictions. The ratio of the cross-sections for $Z\rightarrow\tau^+\tau^-$ to $Z\rightarrow\mu^+\mu^-$ ($Z\rightarrow{}e^+e^-$), determined to be $1.01 \pm 0.05$ ($1.02 \pm 0.06$), is consistent with the lepton-universality hypothesis in $Z$ decays.

## Figures and captions

 Distributions of invariant (dashed line) and corrected (full line, shaded) mass of $\tau_{h Z }$ { $\tau ^+\tau ^-$ } (blue) events and the $V\!j$ (red) background. fig1.pdf [18 KiB] HiDef png [446 KiB] Thumbnail [98 KiB] *.C file Invariant-mass distributions for (a) $\tau_{\mu}\tau_{\mu}$ {} , (b) $\tau_{e}\tau_{e}$ {} , (c) $\tau_{\mu}\tau_{h\tau_{e}\tau_{h\mu}}$ {e}{} candidates with the excluded mass ranges indicated by the gray areas. The $Z \!\rightarrow \tau ^+\tau ^-$ simulation (red) is normalised to the observed signal. The $Z$ (blue), QCD (brown), and electroweak (magenta) backgrounds are estimated from data. The $t \overline t$ , $VV$ backgrounds and cross-feed (green) are estimated from simulation (see text) and generally not visible. fig2a.pdf [23 KiB] HiDef png [492 KiB] Thumbnail [92 KiB] *.C file fig2b.pdf [23 KiB] HiDef png [909 KiB] Thumbnail [141 KiB] *.C file fig2c.pdf [22 KiB] HiDef png [474 KiB] Thumbnail [101 KiB] *.C file fig2d.pdf [22 KiB] HiDef png [383 KiB] Thumbnail [95 KiB] *.C file fig2e.pdf [21 KiB] HiDef png [428 KiB] Thumbnail [92 KiB] *.C file fig2f.pdf [21 KiB] HiDef png [379 KiB] Thumbnail [90 KiB] *.C file fig2g.pdf [23 KiB] HiDef png [554 KiB] Thumbnail [113 KiB] *.C file Summary of the measurements of $Z \!\rightarrow l^+l^-$ production cross-section inside the LHCb acceptance region from $pp$ collisions at 8$\mathrm{\,TeV}$. The error bar represents the total uncertainty. The dotted inner error bar corresponds to the statistical contribution. The coloured band corresponds to the combined measurement of $Z \!\rightarrow \tau ^+\tau ^-$ from this analysis. The last 7 rows represent the NNLO predictions with different parameterizations of the PDFs. fig3.pdf [16 KiB] HiDef png [200 KiB] Thumbnail [69 KiB] *.C file Animated gif made out of all figures. PAPER-2018-016.gif Thumbnail

## Tables and captions

 Expected backgrounds yields and total number of candidates observed. In the last row the uncertainties are the statistical and systematic contributions combined. [Error creating the table] Relative uncertainties of the various contributions affecting the cross-section measurement, given in percent. The uncertainties are correlated between streams, except in rows denoted with $^\dagger$. Table_2.pdf [47 KiB] HiDef png [71 KiB] Thumbnail [10 KiB] tex code

Created on 09 December 2018.Citation count from INSPIRE on 18 December 2018.


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# Separated Flow Model for Two-Phase Flow

Compared with the homogeneous model, the separated flow model has been used more widely, because it provides a better prediction of flow behavior with a manageable level of complexity. The separated flow model assumes that each phase displays different properties and flows at different velocities. It is a simpler version of the two-fluid model discussed in Chapter 4 because it is assumed that only velocities differ between the two phases, while the conservation equations are written only for the combined flow. In addition, the pressure across any given cross-section of a channel carrying a multiphase flow is assumed to be the same for both phases (Hewitt, 1998).

The mass flow rates of the liquid and vapor phases are obtained from (see Concepts and Notations for Two-Phase Flow):

${{\dot{m}}_{\ell }}={{\rho }_{\ell }}{{w}_{\ell }}{{A}_{\ell }}$
${{\dot{m}}_{v}}={{\rho }_{v}}{{w}_{v}}{{A}_{v}}$

The cross-sectional area of the channel occupied by liquid and vapor can be obtained by rearranging these two equations, i.e.,

${{A}_{\ell }}={{\dot{m}}_{\ell }}/({{\rho }_{\ell }}{{w}_{\ell }})\qquad\qquad(1)$
${{A}_{v}}={{\dot{m}}_{v}}/({{\rho }_{v}}{{w}_{v}})\qquad\qquad(2)$

The void fraction of the two-phase flow can be obtained by substituting eqs. (1) and (2) into the following equation

$\alpha =\frac{\Delta z\int_{{{A}_{v}}}{dA}}{\Delta z\int_{A}{dA}}=\frac{{{A}_{v}}}{{{A}_{v}}+{{A}_{\ell }}}$

and the result is:

$\alpha =\frac{{{{\dot{m}}}_{v}}/({{\rho }_{v}}{{w}_{v}})}{{{{\dot{m}}}_{v}}/({{\rho }_{v}}{{w}_{v}})+{{{\dot{m}}}_{\ell }}/({{\rho }_{\ell }}{{w}_{\ell }})}\qquad\qquad(3)$

Considering the definition of quality

$x=\frac{{{{\dot{m}}}_{v}}}{{{{\dot{m}}}_{\ell }}+{{{\dot{m}}}_{v}}}$

the void fraction becomes

$\alpha =\frac{x}{x+(1-x)\frac{{{\rho }_{v}}{{w}_{v}}}{{{\rho }_{\ell }}{{w}_{\ell }}}}\qquad\qquad(4)$

It can be seen that when the vapor velocity, wv, is greater than the liquid velocity, ${{w}_{\ell }}$ (which is often the case for vertical upward and horizontal cocurrent flow) the homogeneous model overpredicts the void fraction. On the other hand, when the vapor velocity, wv, is lower than the liquid velocity, ${{w}_{\ell }}$, as occurs in vertical downward flow, the homogeneous model underpredicts the void fraction.

The governing equations for steady-state two-phase flow in a channel for separated flow model have been given in Example 4.6. More generalized governing equations that are applicable to transient flow are presented here. The continuity equations for the vapor and liquid phases are, respectively

$\frac{\partial }{\partial t}({{\rho }_{v}}\alpha A)+\frac{\partial }{\partial z}({{\rho }_{v}}{{w}_{v}}\alpha A)={{{\dot{m}}'''}_{v}}\qquad\qquad(5)$
$\frac{\partial }{\partial t}[{{\rho }_{\ell }}(1-\alpha )A]+\frac{\partial }{\partial z}[{{\rho }_{\ell }}{{w}_{\ell }}(1-\alpha )A]={{{\dot{m}}'''}_{\ell }}\qquad\qquad(6)$

where ${{{\dot{m}}'''}_{v}}$ and ${{{\dot{m}}'''}_{\ell }}$ are the mass production rates of vapor and liquid due to phase change in the two-phase flow system. Conservation of mass requires that the summation of ${{{\dot{m}}'''}_{v}}$ and ${{{\dot{m}}'''}_{\ell }}$ equal zero, so the continuity equation for the two-phase system can be obtained by adding eqs. (5) and (6) together, i.e.,

$\frac{\partial }{\partial t}(\rho A)+\frac{\partial }{\partial z}\left[ {{\rho }_{v}}{{w}_{v}}\alpha A+{{\rho }_{\ell }}{{w}_{\ell }}(1-\alpha )A \right]=0\qquad\qquad(7)$

where ρ is the from density of the two-phase mixture

$\rho =(1-\alpha ){{\rho }_{\ell }}+\alpha {{\rho }_{v}}$

Considering the definition of mass flux:

${{G}_{\ell }}=\frac{{{{\dot{m}}}_{\ell }}}{A}={{\rho }_{\ell }}{{j}_{\ell }}={{\rho }_{\ell }}{{w}_{\ell }}(1-\alpha )$
${{G}_{v}}=\frac{{{{\dot{m}}}_{v}}}{A}={{\rho }_{v}}{{j}_{v}}={{\rho }_{v}}{{w}_{v}}\alpha$
$\dot{{m}''}={{G}_{\ell }}+{{G}_{v}}={{\rho }_{v}}{{j}_{v}}+{{\rho }_{\ell }}{{j}_{\ell }}={{\rho }_{v}}{{w}_{v}}\alpha +{{\rho }_{\ell }}{{w}_{\ell }}(1-\alpha )$

the continuity equation can be rewritten as

$\frac{\partial }{\partial t}(\rho A)+\frac{\partial }{\partial z}(\dot{{m}''}A)=0\qquad\qquad(8)$

The momentum equations for the vapor and liquid phases are

\begin{align} & \frac{\partial }{\partial t}\left( {{\rho }_{v}}{{w}_{v}}\alpha A \right)+\frac{\partial }{\partial z}\left( {{\rho }_{v}}w_{v}^{2}\alpha A \right) \\ & =-\alpha A\frac{\partial p}{\partial z}-g{{\rho }_{v}}\alpha A\cos \theta -{{\tau }_{w,v}}{{P}_{w,v}}+{{F}_{\ell ,v}} \\ \end{align}\qquad\qquad(9)
\begin{align} & \frac{\partial }{\partial t}\left[ {{\rho }_{\ell }}{{w}_{\ell }}(1-\alpha )A \right]+\frac{\partial }{\partial z}\left[ {{\rho }_{\ell }}w_{\ell }^{2}(1-\alpha )A \right] \\ & =-(1-\alpha )A\frac{\partial p}{\partial z}-g{{\rho }_{\ell }}(1-\alpha )A\cos \theta -{{\tau }_{w,\ell }}{{P}_{w,\ell }}+{{F}_{v,\ell }} \\ \end{align}\qquad\qquad(10)

where τw,v and ${{\tau }_{w,\ell }}$ are shear stresses at the wall for vapor and liquid, respectively. Pw,v and ${{P}_{w,\ell }}$ are the portion of the perimeter that is in contact with vapor and liquid, respectively. ${{F}_{\ell ,v}}$ and ${{F}_{v,\ell }}$ are the interactive forces between the liquid and vapor phases, which satisfy ${{F}_{\ell ,v}}=-{{F}_{v,\ell }}$ as required by Newton’s third law. Assuming constant wall shear stress τw around the periphery of the channel, the momentum equation of the two-phase system can be obtained by adding eqs. (9) and (10), i.e.,

$A\frac{\partial \dot{{m}''}}{\partial t}+\frac{\partial }{\partial z}\left[ {{\rho }_{v}}w_{v}^{2}\alpha A+{{\rho }_{\ell }}w_{\ell }^{2}(1-\alpha )A \right]=-A\frac{\partial p}{\partial z}-g\rho A\cos \theta -{{\tau }_{w}}P\qquad\qquad(11)$

Substituting superficial mass fluxes:

${{G}_{\ell }}=\frac{{{{\dot{m}}}_{\ell }}}{A}={{\rho }_{\ell }}{{j}_{\ell }}={{\rho }_{\ell }}{{w}_{\ell }}(1-\alpha )$
${{G}_{v}}=\frac{{{{\dot{m}}}_{v}}}{A}={{\rho }_{v}}{{j}_{v}}={{\rho }_{v}}{{w}_{v}}\alpha$

into eq. (11), the momentum equation becomes

$A\frac{\partial \dot{{m}''}}{\partial t}+\frac{\partial }{\partial z}\left\{ A\left[ \frac{G_{v}^{2}}{{{\rho }_{v}}\alpha }+\frac{G_{\ell }^{2}}{{{\rho }_{\ell }}(1-\alpha )} \right] \right\}=-A\frac{\partial p}{\partial z}-g\rho A\cos \theta -{{\tau }_{w}}P\qquad\qquad(12)$

Considering that the mass flux of the vapor phase can be expressed as

${{G}_{v}}=x\dot{{m}''}$ [eq. $x=\frac{{{G}_{v}}}{{\dot{{m}''}}}$

and the [Concepts and Notations for Two-Phase Flow|mass flux of the liquid]] is

${{G}_{\ell }}=\dot{{m}''}-{{G}_{v}}=(1-x)\dot{{m}''}$

the momentum equation in the separated flow model becomes

$\frac{\partial \dot{{m}''}}{\partial t}+\frac{1}{A}\frac{\partial }{\partial z}\left\{ \dot{{m}''}A\left[ \frac{{{x}^{2}}}{{{\rho }_{v}}\alpha }+\frac{{{(1-x)}^{2}}}{{{\rho }_{\ell }}(1-\alpha )} \right] \right\}=-\frac{\partial p}{\partial z}-g\rho \cos \theta -\frac{{{\tau }_{w}}P}{A}\qquad\qquad(13)$

The energy equations for the vapor and liquid phases are

$\frac{\partial }{\partial t}\left[ {{\rho }_{v}}\alpha A\left( {{h}_{v}}+\frac{w_{v}^{2}}{2}+gz\cos \theta \right) \right]+\frac{\partial }{\partial z}\left[ {{\rho }_{v}}{{w}_{v}}\alpha A\left( {{h}_{v}}+\frac{w_{v}^{2}}{2}+gz\cos \theta \right) \right]$
$={{P}_{v}}{{{q}''}_{w}}+{q}'''\alpha A+\alpha A\frac{\partial p}{\partial t}+{{{q}'''}_{\ell ,v}}\qquad\qquad(14)$
$\frac{\partial }{\partial t}\left[ {{\rho }_{\ell }}(1-\alpha )A\left( {{h}_{\ell }}+\frac{w_{\ell }^{2}}{2}+gz\cos \theta \right) \right]+\frac{\partial }{\partial z}\left[ {{\rho }_{\ell }}{{w}_{\ell }}(1-\alpha )A\left( {{h}_{\ell }}+\frac{w_{\ell }^{2}}{2}+gz\cos \theta \right) \right]$
$={{P}_{\ell }}{{{q}''}_{w}}+{q}'''(1-\alpha )A+(1-\alpha )A\frac{\partial p}{\partial t}+{{{q}'''}_{v,\ell }}\qquad\qquad(15)$

where the heat flux at the surface of the channel, q''w, is assumed to be the same for the perimeter of the channel, whether it is in contact with liquid or vapor. ${{{q}'''}_{\ell ,v}}$ and ${{{q}'''}_{v,\ell }}$ are the interphase heat transfers that satisfy ${{{q}'''}_{\ell ,v}}=-{{{q}'''}_{v,\ell }}$. The energy equation for the two-phase mixture is then obtained by adding eqs. (14) and (15), i.e.,

\begin{align} & A\frac{\partial }{\partial t}\left[ {{\rho }_{v}}\alpha \left( {{h}_{v}}+\frac{w_{v}^{2}}{2}+gz\cos \theta \right)+{{\rho }_{\ell }}(1-\alpha )\left( {{h}_{\ell }}+\frac{w_{\ell }^{2}}{2}+gz\cos \theta \right) \right] \\ & +\frac{\partial }{\partial z}\left\{ A\left[ {{\rho }_{v}}{{w}_{v}}\alpha \left( {{h}_{v}}+\frac{w_{v}^{2}}{2}+gz\cos \theta \right)+{{\rho }_{\ell }}{{w}_{\ell }}(1-\alpha )\left( {{h}_{\ell }}+\frac{w_{\ell }^{2}}{2}+gz\cos \theta \right) \right] \right\} \\ & =P{{{{q}''}}_{w}}+{q}'''A+A\frac{\partial p}{\partial t} \\ \end{align}\qquad\qquad(16)

Substituting the following mass fluxes

${{G}_{\ell }}=\frac{{{{\dot{m}}}_{\ell }}}{A}={{\rho }_{\ell }}{{j}_{\ell }}={{\rho }_{\ell }}{{w}_{\ell }}(1-\alpha )$
${{G}_{v}}=\frac{{{{\dot{m}}}_{v}}}{A}={{\rho }_{v}}{{j}_{v}}={{\rho }_{v}}{{w}_{v}}\alpha$

into eq. (16), the energy equation becomes

\begin{align} & A\frac{\partial }{\partial t}\left[ {{\rho }_{v}}\alpha \left( {{h}_{v}}+\frac{G_{v}^{2}}{2\rho _{v}^{2}{{\alpha }^{2}}}+gz\cos \theta \right)+{{\rho }_{\ell }}(1-\alpha )\left( {{h}_{\ell }}+\frac{G_{\ell }^{2}}{2\rho _{\ell }^{2}{{(1-\alpha )}^{2}}}+gz\cos \theta \right) \right] \\ & +\frac{\partial }{\partial z}\left\{ A\left[ {{G}_{v}}\left( {{h}_{v}}+\frac{G_{v}^{2}}{2\rho _{v}^{2}{{\alpha }^{2}}}+gz\cos \theta \right)+{{G}_{\ell }}\left( {{h}_{\ell }}+\frac{G_{\ell }^{2}}{2\rho _{\ell }^{2}{{(1-\alpha )}^{2}}}+gz\cos \theta \right) \right] \right\} \\ & =P{{{{q}''}}_{w}}+{q}'''A+A\frac{\partial p}{\partial t} \\ \end{align}\qquad\qquad(17)

Since ${{G}_{v}}=x\dot{{m}''}$ and ${{G}_{\ell }}=\dot{{m}''}-{{G}_{v}}=(1-x)\dot{{m}''}$, eq. (17) can be modified as

\begin{align} & A\frac{\partial }{\partial t}\left[ {{\rho }_{v}}{{h}_{v}}\alpha +{{\rho }_{\ell }}{{h}_{\ell }}(1-\alpha ) \right]+\frac{\partial }{\partial z}\left\{ \dot{{m}''}A\left[ x{{h}_{v}}+(1-x){{h}_{\ell }} \right] \right\} \\ & =P{{{{q}''}}_{w}}+{q}'''A-\frac{\partial }{\partial z}\left\{ \frac{{{{\dot{{m}''}}}^{3}}A}{2}\left[ \frac{{{x}^{3}}}{\rho _{v}^{2}{{\alpha }^{2}}}+\frac{{{(1-x)}^{3}}}{\rho _{\ell }^{2}{{(1-\alpha )}^{2}}} \right] \right\}-\dot{{m}''}Ag\cos \theta \\ & -A\frac{\partial }{\partial t}\left\{ \frac{{{{\dot{{m}''}}}^{2}}}{2}\left[ \frac{{{x}^{2}}}{2{{\rho }_{v}}\alpha }+\frac{{{(1-x)}^{2}}}{2{{\rho }_{\ell }}(1-\alpha )} \right] \right\}+A\frac{\partial p}{\partial t} \\ \end{align}\qquad\qquad(18)

For steady-state two-phase flow in a circular tube with constant cross-sectional area (A = constant), the continuity equation (8) results in $d\dot{{m}''}/dz=0$. The momentum equation (7) reduces to

$-\frac{dp}{dz}=\frac{4{{\tau }_{w}}}{D}+\dot{{m}''}\frac{d}{dz}\left[ \frac{{{x}^{2}}}{{{\rho }_{v}}\alpha }+\frac{{{(1-x)}^{2}}}{{{\rho }_{\ell }}(1-\alpha )} \right]+g\rho \cos \theta \qquad\qquad(19)$

The three terms on the right-hand side of eq. (19) represent pressure drops due to friction, dpF / dz, acceleration, dpa / dz, and gravity, dpg / dz. To predict the pressure drop of two-phase flow using the separated flow model, empirical correlations for friction and void fraction are needed, as is evident from eq. (19).

The energy equation in the separated flow model for steady-state two-phase flow in a circular tube with constant cross-sectional area can be obtained by simplifying eq. (18), i.e.,

$\frac{d}{dz}\left[ x{{h}_{v}}+(1-x){{h}_{\ell }} \right]=\frac{4{{{{q}''}}_{w}}}{\dot{{m}''}D}+\frac{{{q}'''}}{{\dot{{m}''}}}-\frac{{{{\dot{{m}''}}}^{2}}}{2}\frac{d}{dz}\left[ \frac{{{x}^{3}}}{\rho _{v}^{2}{{\alpha }^{2}}}+\frac{{{(1-x)}^{3}}}{\rho _{\ell }^{2}{{(1-\alpha )}^{2}}} \right]-g\cos \theta \qquad\qquad(20)$

For a two-phase flow system with condensation or evaporation, where kinetic and potential energy, as well as internal heat generation, can be neglected, eq. (20) also reduces to

$\frac{dx}{dz}=\frac{4{{{{q}''}}_{w}}}{{\dot{m}}''D{{h}_{\ell v}}}$

from Homogeneous Flow Model for Two-Phase Flow. Therefore, the energy equations for the homogeneous model and the separated flow model are the same if kinetic and potential energy, as well as internal heat generation, can be neglected.

## References

Faghri, A., and Zhang, Y., 2006, Transport Phenomena in Multiphase Systems, Elsevier, Burlington, MA

Hewitt, G.F., 1998, “Multiphase Fluid Flow and Pressure Drop,” Heat Exchanger Design Handbook, Vol. 2, Begell House, New York, NY.