text:
News Think for a moment

1. Jun 13, 2010

Desiree

2. Jun 13, 2010

lisab

Staff Emeritus

3. Jun 13, 2010

Desiree

Thanks, fixed it.

4. Jun 13, 2010

lisab

Staff Emeritus

I wish we'd get back to a pay-go system.

5. Jun 13, 2010

estro

These arbitrary numbers really should make us all think.

6. Jun 13, 2010

waht

1 trillion cost of war in 10 years is insignificant compared to the size of US economy. All it did is stimulated the military-industrial complex which keeps the US on cutting edge.

The biggest drains every year are social security, health, and medicare...

7. Jun 13, 2010

Desiree

I don't think you realize the devastating impact of such spendings and debts on your children's future and what better things could have been done with 1 trillion dollars.

8. Jun 13, 2010

OmCheeto

And my favorite: You could supply every household in the US with 2240 watts of solar panels. Yielding roughly 9 kwh per day, saving each of us only about $1 per day, but multiplied by 112,000,000 households and 365.242198781 days per year, we'd collectively save$40 billion dollars a year and our trillion dollars would be paid back in 25 years. Not counting installation, taxes, delivery, and inspection fees.

Last edited by a moderator: Apr 25, 2017
9. Jun 13, 2010

Ivan Seeking

Staff Emeritus
Really, that's ALL it did. Tell that to my nephew who had to clean up the body parts from a family that was killed by mistake. Oh yes, never mind, he's probably too drunk to listen.

10. Jun 13, 2010

Cyrus

Tell's me you lack critical thinking skills, because all you've posted are numbers.

What does this number mean to you?

2389348967o659560845903445985690569560490823948234908486849680980250892505409468905

<whistles> wheeeeeeeeeeewwwwwwww that's a big number!

Last edited: Jun 14, 2010
11. Jun 13, 2010

waht

This thread is about the cost of war in relation US debt - ALL is in terms of economy, and not casualties, nor Cheney.

12. Jun 13, 2010

Cyrus

This is EXACTLY correct.

13. Jun 13, 2010

Staff: Mentor

That military spending doesn't go into a black hole, though: it is mostly spent on the salaries of our military and the compaanies who make the equipment.

It is basically the same type of economic stimulus that our government spent an equal amount of money on in the past year. So while it isn't great that we spent a trillion dollars on the wars over 9 years, it is a lot worse that we spent a trillion dollars on stimulus in the past year.

14. Jun 13, 2010

Ivan Seeking

Staff Emeritus
We don't have to attack a country in order to have a strong military. The point that B justifies A is falacious. How much bang did we get for the buck? Did we get a trillion dollars worth of development? Or was most of that money wasted fighting a war that wasn't needed? I would bet any payoff is on the order of one part in a thousand, or less.

Now if your point is that we have $100 billion a year to waste, then I have a great deal on some swamp land for you. Btw, the interest on our national debt is about 160 billion per year. If we didn't have to pay interest, there wouldn't be a problem, so clearly 100 billion per year is significant. 100 billion less per year is the equivalent of eliminating over 60% of our national debt. How about paying for three years of medicaid? What if all of that money has been dumped into the US economy for productive reasons, rather than for building bombs, and fueling tanks and aircraft? How much more benefit do we derive per dollar spent, from a bridge, than we do a truckload of bombs? I fail to follow the logic that it was advantageous to fight an unnecessary war, when we could have been helping Americans, or investing in the US infrastructure or industry. Even viewing this from a purely militaristic point of view, I would bet that one could develop a pretty cool weapon for a trillion dollars. btw, I never opposed the war in Afghanistan, just Iraq. In the former case, I don't think we had a choice. I also strongly support the attacks in the hills of Pakistan, using drones. As promised, Obama is no wimp. He's just smart. Last edited: Jun 14, 2010 15. Jun 14, 2010 waht In a sense yes. Most companies that made money in the war reinvest it in themselves (minus CEO bonuses) and compete with other companies for contracts. And where there is competition there is innovation. The US has lots of other internal blackholes, and only feeding them more won't solve the problem. And building bombs, and aircrafts? Well that's what superpowers do. Among many other things it gives them a high position when dealing with national interests that, in a sense, feedback to the economy. 16. Jun 14, 2010 Desiree First of all, I didn't start an argument based on those numbers, and second of all, as the title reads: "think for a moment..." So what makes you think I lack critical thinking skills? 17. Jun 14, 2010 Cyrus Because you did not think before posting these sources - what so ever! What's you're premise in starting this thread? This reeks of anti-war nonsense. Should I run around with my hands in the air at the sight of big scary numbers, that lack any context? To answer your question: the numbers tell me exactly what they report, and nothing more. 18. Jun 14, 2010 russ_watters Staff: Mentor Granted! But this has nothing to do with the OP, which, as was pointed out to you before, is about money. Then that'd be a losing bet. Where do you think the money actually goes that the payoff could be 1:1000? Heck, if we have 100,000 troops employed at an average salary of$30,000 apiece, that would be $27 billion right there, or just under 3%. It was pretty easy to debunk that 1:1000 rediculousness with just one simple example of where the money goes. Your point, so you tell me. I'll give it a shot, though: if the societal benefit to building a bridge was huge on its own, we wouldn't need to use stimulus spending to build it. The stimulus spending was sold by Obama as a way to create jobs, not for the societal benefit of what those jobs were doing. In that sense, money spent building a bomb is exactly equivalent to money spent building a bridge. ...probably because no one made such an argument! Indeed we could. But for the fifth or sixth time now, the OP was about money! Developing a weapon is still expensive and it is still money that we are spending that we don't have. We shouldn't be spending it: not on developing new weapons, not on wars, and not on "economic stimulus", all of which are functionally equivalent. 19. Jun 14, 2010 russ_watters Staff: Mentor Starting a thread with an open-ended question leaves us to guess what your point was. Failing to ever provide analysis makes one question your critical thinking skills. 20. Jun 14, 2010 Phrak Thread Locked; mentors out of control. Oh, nevermind, I can't lock a thread. Carry on. 21. Jun 14, 2010 Cyrus Preemptive Thread reopened It's good to be king. 22. Jun 14, 2010 Phrak Haha, Cyrus. Take this: Thread Locked Darned locking mechanism. Must be jammed.... 23. Jun 14, 2010 Cyrus Pull back on the charging handle on the upper receiver. 24. Jun 14, 2010 OmCheeto hmmm..... Let's give Desiree a short lesson in U.S. economics. A. Whenever you see a really big number, chop off the last 9 digits. A$trillion$becomes$1000. Yay!
B. Then multiply by 3. A thousand becomes $3000. boo.... C. Divide that number by the days since that debt was incurred:$3000/3000 days = $1/day. yay! steps A and B give you the cost per US citizen. step C tells me that the cost of the two wars was only a dollar per day per person. Now as Lisab stated, it would be really nice if we could pay as we go. Then it probably wouldn't hurt so much, and we all wouldn't have to get depressed about looking at such big scary numbers. But we really only have ourselves to blame. Who do we always vote for? The guy that's going to raise our taxes$365 a year to pay for two wars? Or the guy that starts two wars and simultaneously cuts taxes?

Now that's something to think about.

25. Jun 14, 2010

Pattonias

I believe the new US national defence strategy should be that whenever the US is attacked we shoud file for sanctions in the UN and dedicate the majority of our defence budget to building bridges and investing in solar panels.


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# Rate of hydrogenation

The rate of hydrogenation is faster for which alkene : cis or trans. I know it depends on steric hindrance on the carbon, but both have steric hindrance. Can someone please help

Are you asking about catalytic hydrogenation or non-catalytic hydrogenation (e.g. diimide)? For the time being, let's assume that your question is about the catalytic process. All the catalyst does is lower the activation energy required to reach the transition state (TS) and the TS will look about the same independent of which isomer you start with. Therefor, ground state stability differences between the two olefins will determine which isomer reacts faster. Since trans-olefins are generally more stable than their corresponding cis isomers, it will take more energy (the difference in stability between the cis and trans isomers) for the more stable (lower energy) trans isomer to reach the TS, and less energy for the cis isomer to reach the TS. Consequently cis isomers react somewhat faster than their trans-counterparts during catalytic hydrogenation. I suspect that the same considerations would apply to non-catalytic processes.

As an aside, remember that in the catalytic process, depending upon catalyst, temperature, solvent, pressure, etc.; many other things can be going on, such as cis-trans isomerization of the starting material, positional isomerization, etc. So if the reaction conditions are not chosen carefully and 1) if cis-trans isomerization is faster than hydrogenation and 2) let's say the cis isomer reacts 100 times faster than the trans; then both isomers could appear to give the same rate of hydrogenation because all hydrogenation would proceed through the cis isomer.

Imagine that the mechanism for the addition of $\ce{H-Hal}$, $\ce{Hal-Hal}$, and $\ce{H-H}$ to an alkene are fundamentally different.

In the first case, subsequent addition of the proton and (then) the halogen is assumed. In the second case, formation of a halonium ion is postulated as the first step, followed by addition of the halide (sort of backside attack to the existing $\ce{C-Hal}$ bond).

In the latter case, the simultaneous addition of both hydrogen atoms to the same face of the alkene is assumed. Isn't it likely under these conditions that a cis alkene, which is less shielded on one side, is faster hydrogenated then?

• But, hydrogens are added from above or below the plane of the double bond, and not from the side of the double bond. So what do you mean by shielded on one side? Why should this make a difference? May 14 '14 at 8:05
• Hydrogenation is generally a catalytic process, so somewhere the explanation also has to involve the binding to the catalyser. It might be different for heterogeneous catalysis than for solution phase catalysis. An other issue might be related to thermodynamics: cis- isomers tend to be less stable than trans- isomers due to steric repulsion. If the transition state is close to the product, the product is the same and thus the cis- isomer might hydrogenate faster.
– PLD
May 14 '14 at 15:31

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# When is the flatness locus non-empty

Let $k$ be an algebraically closed field, $f:X \to Y$ be a surjective proper $k$-morphism locally of finite presentation between irreducible noetherian schemes. Assume that $Y$ is reduced. Under what additional condition on $f$ (other than flatness/ generic flatness) does there exist a non-empty open set $U$ of $X$ such that $f|_U$ is flat?

If I further assmume that $X$ is reduced, then does there exists a non-empty open subset $U$ of $Y$ such that $f|_{f^{-1}(U)}$ is flat?

• Generic flatness implies it is non-empty. – Mohan Oct 21 '15 at 19:28
• @Mohan Sorry, I slightly edited the question. – user46578 Oct 21 '15 at 19:33
• The question becomes interesting when you do not assume that $Y$ is reduced... – Matthieu Romagny Oct 21 '15 at 19:34
• @MatthieuRomagny I will be very interested to know what happens when $Y$ is non-reduced. May be you could elaborate a bit or suggest a reference for that case. – user46578 Oct 21 '15 at 19:38

Let $f:X\to S$ be a morphism of schemes.

Assume that $S$ is integral, and let $K$ be its function field. As "everything is flat over a field", the generic fibre $f_K:X_K\to \mathrm{Spec} \ K$ is a flat morphism.

In particular, the locus of flatness is non-empty.

As Laurent Moret-Bailly points out below, if $f$ is of finite presentation (and $S$ still integral), then $f$ is flat over a dense open of $S$.

• Wow. So I do not even need locally of finite-presentation or proper morphism, right? – user46578 Oct 21 '15 at 19:32
• Right. "Everything is flat over a field." – Ariyan Javanpeykar Oct 21 '15 at 19:34
• Any morphism of schemes $f:X\to S$ with $S$ integral is flat over some dense open of $S$. As Matthieu Romagny mentions above, the question is more interesting when $S$ is only irreducible, e.g., $S$ is Spec $k[x]/x^n$. – Ariyan Javanpeykar Oct 21 '15 at 19:37
• @AriyanJavanpeykar : you probably need $f$ to be of finite presentation. In fact if $f$ is locally of finite presentation, then the flat locus in $X$ is open (EGA IV, (11.3.1)). If in addition $f$ is quasi-compact, your claim follows. – Laurent Moret-Bailly Oct 21 '15 at 19:51
• @user46578 An open set could be empty. (In our situation, the morphism $f$ is generically flat and of finite presentation. The locus of flatness is therefore open (by Thm 11.3.1) and non-empty (by generic flatness). In particular, as $S$ is integral, the locus of flatness is dense open in $S$.) – Ariyan Javanpeykar Oct 21 '15 at 20:08


text: If you know how to name alkanes, adding alkene nomenclature to your repertoire is a fairly straightforward task. These carbon atoms already have four electrons, but they each want to get four more so that they have a full eight in the valence shell. In IUPAC system the alkenes are named by changing the suffix of alkanes âane to âene. The carbon-carbon double bond is broken and the molecules attach to each other; that is, the lefthand carbon of one molecule attaches itself to the righthand carbon of another, forming a chain. For example: These formulas can also be used to figure out the number of double bonds in a given alkene molecule from its molecular formula. Nomenclature of alkenes is covered in chapter 3. This group is pointing below the molecule. Free rotation the p-orbitals cause them to be 90° from each other breaking the $$\pi$$ bond because there would be no overlap. Thus, CH2=CH2 stands for. When represented with the structure on the left benzene can be mistaken for an alkene, but the structure on the right shows that it isn't. Alkene stability. The Structure of Alkenes The two carbons of a C=C double bond are sp2 -hybridized. Legal. When atoms are an $$sp^2$$ hybrid they have a trigonal planar structure. This means that they're comprised of a chain of carbon atoms bonded together, with each carbon atom bonded to hydrogen atoms to make a total of four bonds per carbon. Cis-trans isomerism. Ethene is not a very complicated molecule. However, if your alkene has the formula C5H8, you can deduce that two double bonds are present as the ratio of carbons to hydrogens follows the CnH2n-2 rule. 1. Although there is only one alkene with the formula C2H4 (ethene) and only one with the formula C3H6 (propene), there are several alkenes with the formula C4H8. 3. Collectively, they are called unsaturated hydrocarbonsAn alkene or alkyne having one or more multiple (double or triple) bonds between carbon atoms.because they have fewer hydrogen atoms than does an alkane with the same number of carbon atoms, as is indicated in the following general formulas: Addition polymerization takes place when the monomer molecule contains double carbon bonds, as in alkenes, or triple carbon bonds, as in alkynes. These structures are very similar to a 'peace' sign, there is a central atom with three atoms around it, all on one plane. Polymerisation of ethene to form polyethylene. They have general formula C n H 2n+2, where n is the number of carbon atoms. Alkenes: Molecular and Structural Formulas The alkenes comprise a series of compounds that are composed of carbon and hydrogen atoms with at least one double bond in the carbon chain. Note that the molecular formula for ethene is C2H4, whereas that for ethane is C2H6. Cherono bency kimetoi on February 15, 2020: There are six carbons in the longest chain. The figure below shows that, while the structure on the left implies that benzene contains double bonds, the structure on the right shows that the electrons are actually distributed across all of the carbons. This means that they're comprised of a chain of carbon atoms bonded together, with each carbon atom bonded to hydrogen atoms to make a total of four bonds per carbon. Polymerisation reactions are one of the most commercially used reactions of alkenes and are how all plastics are made. The first three alkenes are gases, and the next fourteen are liquids. Correspondingly, cyclic alkynes are rare. Alkenes have important industrial uses, and also play an important role in our everyday lives. Ozonolysis of an alkene to form two carbonyl products. However, instead of two molecules of hydrogen being added, a halogen substituent is bonded to the carbon atom. For mono-unsaturated alkenes, which have only one double bond, the general formula is CnH2n. (a) The σ-bonded framework is formed by the overlap of two sets of singly occupied carbon sp2 hybrid orbitals and four singly occupied hydrogen 1s orbitals to form electron-pair bonds. This uses 10 of the 12 valence electrons to form a total of five σ bonds (four C–H bonds and one C–C bond). Suffix, -ane is used while naming alkanes. 3.1 Introduction to Alkenes and Alkynes Alkenes are hydrocarbons in which there is at least one carbon-carbon double bond ; alkynes have at least one carbon-carbon triple bond. Sulphuric acid and water are mixed with an alkene to form the corresponding alcohol. NamingâandâDrawingâAlkenesâWorksheetâandâKeyâ 1)##Draw#and#name#the#cis#and#trans#condensed#structure#of:# cis#condensed#structure:# trans#condensed#structure:# name:#### # name:# # 2.#Name#the#following#alkenes#(include#cis<#or#trans<#forthe#alkenes#thatwhen#appropriate)# In acetylene, the HâCâ¡C bond angles are 180°. Therefore, we should start numbering from the right of the molecule. Alkenes, also referred to as olefins, are organic unsaturated hydrocarbons that are made of carbon and hydrogen atoms with one or more carbon-carbon double bonds in their chemical structure. There is only one double bond, so the suffix to be used is -ene. Hydrocarbon families can be described by general formulas, which dictate how many hydrogen atoms are present for each carbon atom. Therefore, in a homologous series of alkenes, the lowest alkene has two carbons. In ethene there is no free rotation about the carbon-carbon sigma bond because these two carbons also share a $$\pi$$ bond. Next lesson. The other two angles (H-C=C) are both 121.5°. Alkene Structures Ethylene: A space-filling model of ethylene, the simplest alkene, showing its planar structure. Structure of Alkenes 8 Cis,Trans Isomerism in Alkenes â¢ Cis,trans isomers: Isomers that have the same connectivity but a different arrangement of their atoms in space due to the presence of either a ring (Chapter 2) or a carbon-carbon double bond. Single covalent bonds can twist and flip, but double bonds are rigid. As mentioned before, the double covalent bond that makes alkenes what they are is highly reactive. The simplest alkene which has one double bond in its structure is ethene, C 2 H 4. If it's chlorine, it's chloro-, if it's bromine, it's bromo-, and if it's iodine it's iodo-. Two overlapping triangles are present since each carbon is the center of a planar triangle. For example, there might be a CH3 group attached to the chain. The first two alkenes in Table $$\PageIndex{1}$$, ethene and propene, are most often called by their common names—ethylene and propylene, respectively (Figure $$\PageIndex{1}$$). Draw all of the possible bond line structures for alkenes with the formula C4H8 including all possible structural and stereoisomers. Just like with alkanes, the number of carbons dictates the prefix used in naming the molecule: Count the number of double bonds. They have general formula C n H 2n, where n is the number of carbon atoms. Chime in new window Since a double bond is present and each carbon is attached to 3 atoms (2 H and 1 C), the geometry is trigonal planar. For each extra double bond, two hydrogens must be subtracted. They're characterised by a covalent carbon-carbon double bond. A double bond consists of one sigma bond formed by overlap of sp2 hybrid orbitals and one pi bond formed by overlap of parallel 2 p orbitals. For example, ethene doesn't give cis/trans isomers because both carbon atoms only have hydrogens in them. It is converted to plastics, isopropyl alcohol, and a variety of other products. The simplest alkenes, with only one double bond, no rings, and no other functional groups, are hydrocarbons with the general formula CnH2n. But alkenes are soluble in organic solvents like benzene or acetone because here the van der Waal forces will be replaced by new ones, making alkenes fully soluble. When an extra bond is added, forming a double bond, the second bond is known as a pi bond. give the IUPAC equivalent of the following trivial names: ethylene, propylene, isobutylene and isoprene. However, the most common way to obtain them is through fractional distillation. Carbon wants to have the same configuration as Neon because when it has eight valence electrons carbon is at its most stable, lowest energy state, it has all of the electrons that it wants, so it is no longer reactive. Both are termed unsaturated because the carbons involved in the multiple bonds do not have the maximum number of bonded atoms possible (four for a carbon). The LibreTexts libraries are Powered by MindTouch® and are supported by the Department of Education Open Textbook Pilot Project, the UC Davis Office of the Provost, the UC Davis Library, the California State University Affordable Learning Solutions Program, and Merlot. Therefore, the molecule is. These rules are based on atomic number, and the first point of difference. The hydrogenation reaction is the most commonly used way to turn alkenes back into alkanes. Even ethene, the simplest alkene, has an important role in the ripening of fruit. Propylene is also an important industrial chemical. 2. In this lesson we will consider alkene monomers. While it might look like it contains carbon-carbon double bonds, the real structure of benzene is slightly more complicated. Trigonal planar molecules have an ideal bond angle of 120° on each side. Alkenes higher than these are all solids. 2. For polyunsaturated alkenes, that have more than one double bond, the general formula becomes more complicated. Even though both of the atoms bonded directly to the doubly bonded carbon are the same, as you continue down the chain of each group it's clear that CH2CH2CH3 wins out. The physical properties of alkenes are comparable with alkanes. Under the right conditions, more and more units of ethene continue joining together until a string of the plastic polyethylene is formed. More than half of this ethylene goes into the manufacture of polyethylene, one of the most familiar plastics. Since the carbon atom is forming three sigma bonds instead of the four that it can, it only needs to hybridize three of its outer orbitals, instead of four. These fractions are then collected and, through a process called cracking, split into a mixture of alkenes and alkanes. We also acknowledge previous National Science Foundation support under grant numbers 1246120, 1525057, and 1413739. Some alkenes can form cis and trans isomers. The hydration reaction is what turns alkenes into alcohols. This molecule has five: can you figure out what the general formula would be? The result is that the alkene is split at the double bond into two molecules that have a carbon compound double bonded to an oxygen, also known as a carbonyl compound. Chapter 5: Structure and Preparation of Alkenes: Elimination Reactions Alkenes (olefins) are hydrocarbons that contain a carbon-carbon double bond and are said to be "unsaturated." Structure and bonding. The pi bond is much weaker than the sigma bond and breaks quite easily, which is why alkenes are much more reactive than their fellow hydrocarbons. It's a CH3 group, which is also known as a methyl group. Of course, there are hundreds of potential substituents that could be attached to a carbon chain, but in naming basic alkenes these are the most common. The most basic example of this reaction occurs between molecules of ethene. Alkanes and Alkenes are two types of hydrocarbon families which contain carbon and hydrogen in their molecular structure. All alkenes are insoluble in water, due to the weak van der Waal forces. To determine the highest priority group, look at the atomic numbers of the atoms bonded directly to each carbon. Hydrocarbon - Hydrocarbon - Nomenclature of alkenes and alkynes: Ethylene and acetylene are synonyms in the IUPAC nomenclature system for ethene and ethyne, respectively. 4. Alkenes exist naturally in all three states. In this reaction, the double bond is broken and two extra hydrogen molecules are added to the molecule. Alkenes contain a double bond that is composed of one sigma and one pi bond between two carbon atoms. Single covalent bonds are also known as sigma bonds. H2 gas is used to achieve this, with a nickel catalyst that helps lower the activation energy of the reaction. The Câ¡C bond distance of 121 picometers is much shorter than the C=C distance in alkenes (134 pm) or the CâC bond in alkanes (153 pm). and one pi bond formed by overlap of parallel 2 p orbitals. Alkene reactions. There is rigidity in the ethene molecule due to the double-bonded carbons. (Image to be added soon) There are various methods to prepare an alkene. The parent name of the alkene comes from the IUPAC name for the alkane with the same number of carbon atoms, except the âane ending is changed to âene to signify the presence of a double bond. 1-methylcyclohex-1-ene is formed by dehydrohalogenation of 1-bromo-1-methylcyclohexane with alcoholic KOH. Thankfully, the rules set out to name alkenes are fairly straight forward and can be arranged into five key steps. Structural Isomerism. In organic chemistry, an alkene, olefin, or olefine is an unsaturated chemical compound containing at least one carbon -to- carbon double bond. The simplest alkenes, ethylene, propylene and butyleneare gases. Alkene nomenclature. Let us take a look at few physical properties 1. It can be more than a little confusing the first time you learn nomenclature, so here's an example to better illustrate the steps you need to go through. For example, if you're given an alkene with the molecular formula C5H10, it's clear that only one double bond is present as the number of atoms follows the rule of monounsaturated alkenes, CnH2n. Due to the restricted rotation around the double bond, geometrical isomerism is possible in alkenes. comprised of a series of compounds that contain carbon and hydrogen atoms with single covalent bonds A $$\pi$$ bond is only formed when there is adequate overlap between both top and bottom p-orbitals. The atom with the higher atomic number is the higher priority; for example, in the case of 2-butene, the methyl group is higher priority that the hydrogen group because carbon has a higher atomic number than hydrogen. 1. As mentioned before, alkenes are hydrocarbons. A C2H5 group is named as ethyl and a C3H7 group is called a propyl group. It does this by using the $$2s$$ electron and two of the $$2p$$ electrons, leaving the other unchanged. The table shows four alkenes, their molecular formulae and their structures. Reactions of alkenes is a huge topics. What differentiates alkenes from the standard family of hydrocarbons, alkanes, is that they contain one or more carbon-carbon double bonds. Therefore, our name has expanded to methylhexene. This new orbital is called an $$sp^2$$ hybrid because that's exactly what it is, it is made from one s orbital and two p orbitals. In other words, the amount of hydrogen atoms is equal to twice the number of carbon atoms. The US chemical industry produces about 25 billion kilograms of ethylene annually, more than any other synthetic organic chemical. If no isomerism is possible, you're finished! The physical state depends on molecular mass (gases from ethene to butene - liquids from pentene onwards). Determine the numbering of the carbon chain. The double bond is shared by the two carbon atoms and does not involve the hydrogen atoms, although the condensed formula does not make this point obvious. Alkenes are compounds that have a carbon-carbon double bond, âC=Câ. Burning oil and natural gas releases greenhouse gases, which are destructive to the environment, but despite this fractional distillation is still the most convenient way to obtain alkenes. Missed the LibreFest? Alkenes are one of the most important, useful molecular families in all of organic chemistry. The pi bond is weaker in energy than the sigma bond and therefore can break more easily. The sigma bond has similar properties to those found in â¦ [ "article:topic", "showtoc:no", "transcluded:yes" ], https://chem.libretexts.org/@app/auth/2/login?returnto=https%3A%2F%2Fchem.libretexts.org%2FCourses%2FSacramento_City_College%2FSCC%253A_Chem_420_-_Organic_Chemistry_I%2FText%2F08%253A_Structure_and_Synthesis_of_Alkenes%2F8.01%253A_Alkene_Structure, A double bond consists of one sigma bond formed by overlap of sp. itsorigins in the study of natural compounds extracted from living organisms Unless otherwise noted, LibreTexts content is licensed by CC BY-NC-SA 3.0. The two carbon atoms of a double bond and the four atoms attached to them lie in a plane, with bond angles of approximately 120° as shownn in the figure below. cis-2-Butene mp -139°C, bp 4°C trans-2-Butene Alkenes, as a functional group, is a very versatile one. If there's more than one double bond, this process should be repeated and the molecule will be named either cis,cis , trans,trans, cis,trans, or trans,cis.Make sense yet? The Cahn-Ingold-Prelog priority rules are used for naming geometric isomers (e.g. Alkenes are extremely useful products. As in the hydrogenation reaction, in the halogenation reaction the double bond of the alkene is broken. 3. (a) Alkanes: The hydrocarbons in which the carbon atoms are joined by only single covalent bonds. These carbon atoms already have four electrons, but they each want to get four more so that they have a full eight in the valence shell. The lowest number that the doubly bonded carbon can have is 2. Alkenes - 3 atoms and 1 double bond = trigonal planar. It is made up of four 1s 1 hydrogen atoms and two 2s 2 2 p x 1 2 p y 2 carbon atoms. There are hundreds of organic chemistry reactions, and many of the most commonly used reactions in labs all over the world involve alkenes. (b) One singly occupied unhybridized 2pz orbital remains on each carbon atom to form a carbon–carbon π bond. Ethene is an alkene with an E-N-E ending, and it has the molecular formula C2H4. K S Lane is a student of science and is deeply passionate about educating others on her favourite topics. For example, hydrochloric acid (HCl) and ethene react together to form chloroethane as the double bond breaks, hydrogen is added to one carbon, and chlorine is added to the other. This causes the splitting, or fractioning, of the oil into its constituent components, based on their boiling points. In regards to science, they can be used in the synthesis of many more complicated products, such as in industrial-grade chemicals and in pharmaceuticals. This is the currently selected item. Alkenes are a critical molecular family in the study of organic chemistry. This forms a total of three bonds to each carbon atom, giving them an $$sp^2$$ hybridization. The term is often used as synonym of olefin, that is, any hydrocarbon containing one or more double bonds. E- or Z-alkenes) and other stereoisomers (see later). draw the Kekulé, condensed or shorthand structure of an alkene (cyclic or acyclic), given its IUPAC name. You can reduce it, you can oxidize it, you can cleave it, and you can do a large number of various addition reactions modifying an alkene to other functional groups. This rule can be proven by looking at the structures of common mono-unsaturated alkenes, such as ethene (C2H4) and propene (C3H6) which have twice the number of hydrogens as they do carbons. If the highest priority groups are on the same side, the prefix cis- should be added. Alkenes: Structure and Stability Degrees of unsaturation saturated hydrocarbon CnH2n+2 cycloalkane (1 ring) CnH2n alkene (1 p-bond) CnH2n alkyne (2 p-bonds) CnH2n-2 For each ring or p-bond, -2H from the formula of the saturated alkane Degrees of unsaturation: # of rings and/or p-bonds in a molecule. For more information contact us at info@libretexts.org or check out our status page at https://status.libretexts.org. This means that alkenes most often undergo addition reactions, where the pi bond breaks and two extra atoms add to the molecule. The alkene group can also be called a vinyl group and the carbons sharing the double bond can be called vinyl carbons. Alkene structure and classification. By virtue of this bond angle, alkynes are rod-like. This is done by assigning the end of the chain closest to the double bond as carbonone and then numbering down the chain from there. Alkenes higher than propene have different structures. As noted before, alkenes are hydrocarbons with carbon-to-carbon double bonds (R 2 C=CR 2) and alkynes are hydrocarbons with carbon-to-carbon triple bonds (RâCâ¡CâR). Ozone gas, which is an important part of the earth's atmosphere, is added to an alkene. If the double bond was on the third carbon from the end of a seven carbon chain you would name it hept-3-ene or 3-heptene (either are acceptable). Physical properties of alkenes are quite similar to those of alkanes. In the case of this compound, going through the steps would look like this: Putting together all the clues we've figured out from going through each step, we can finally name our alkene as trans-3-methylhex-2-ene! Video transcript - [Instructor] We've already seen how to name alkanes. A carbon-carbon covalent double bonds consists of a sigma and pi bond. Organic chemistry nomenclature, the rules used to name chemical compounds, can be complicated and confusing. If they're on the opposite sides, trans- should be used. This group is pointing above the molecule. Instead of having fixed pi bonds the electrons in a benzene ring are shared between each of the atoms. The two carbon atoms of a double bond and the four atoms attached to them lie in a plane, with bond angles of approximately 120° as shownn in the figure below, 8.2: Physical Properties and Important Common Names, Structure of Ethene - the simplest alkene, information contact us at info@libretexts.org, status page at https://status.libretexts.org, describe the electronic structure of alkenes using Molecular Orbital (MO) Theory and Orbital Hybridization. In other words, the double-bonded carbons must have the lowest number possible. If a fluorine atom is attached, the word fluro- is used. Count the longest unbroken carbon chain that you can find. 5. This article explores the structure of alkenes, the general formula used to describe them, how they're named, their uses, and some of the most common reactions that they undergo. For the third carbon, CH2CH2CH3 has the higher priority. The stretching vibration of the C=C bond usually gives rise to a moderate band in the region 1680-1640 cm-1.. Stretching vibrations of the âC=CâH bond are of higher frequency (higher wavenumber) than â¦ Their structure is defined by a reactive carbon-carbon double bond, they have a general formula of CnH2n, they can be named by following a series of simple steps, they have many uses in nature as well as in industrial and laboratory settings, and some of their most common reactions include hydrogenation (alkene to alkane), halogenation (alkene to haloalkane), hydration (alkene to alcohol), polymerisation, and ozonolysis. For the second carbon, the highest priority group is CH3, because carbon has a higher atomic number than hydrogen. ( sp^2\ ) hybridization how many hydrogen atoms and two 2s 2 2 y. However, instead of having fixed pi bonds the electrons in a benzene ring are shared between each of reactions. Unbroken carbon chain that is n't a hydrogen bp 4°C trans-2-Butene isomerism in.. Structures for alkenes with the formula C4H8 including all possible structural and stereoisomers of,... Total of three bonds to each carbon atom n't give cis/trans isomers because both atoms! Also known as a pi bond complicated of the most commonly used way to obtain them through. Has the molecular plane is the z-axis. ) bonds the electrons in a homologous series of,! Carbon three, giving them an \ ( sp^2\ ) hybridization carbons the... And 1413739 sp^2\ ) hybridization the double bond, then continue down the chain joined by covalent... Gases from ethene to ethanol carbon can have is 2 four 1s 1 hydrogen atoms are since... The possible bond line structures for alkenes with the formula C4H8 including all possible structural and stereoisomers hydrogens them... We should start numbering from the right of the most commonly used reactions of,... Because carbon has a higher atomic number, and it has the molecular for! A carbon-carbon double bond, so the suffix of alkanes from âane to.... On her favourite topics LibreTexts content is licensed by CC BY-NC-SA 3.0 noble gas means that molecular... The same electron configuration as neon, a noble gas fluro- is used single bonds... In them A-N-E ending, and 1413739 alkanes, adding alkene nomenclature to repertoire. Be synthesised from a number of double bonds are also known as a functional group, look the! First three alkenes are named by changing the suffix of alkanes âane âylene... N H 2n, where the pi bond between two carbon atoms Wikimedia Commons lowest has... An infinite number of carbon atoms the table shows four alkenes, including polystyrene PVC!, isobutylene and isoprene the atoms covalent carbon-carbon double bond, so the suffix of alkanes âane to âene by! Us chemical industry produces about 25 billion kilograms of ethylene, the of... Point of difference same side, the second bond is added, forming a double bond is known as pi! Two angles ( H-C=C ) are both 121.5° her favourite topics:.. Annually, more and more units of ethene continue joining together until a string of the polyethylene... N'T allow for free rotation what the general formula becomes more complicated two. Contact us at info @ libretexts.org or check out our status page at:... Covalent double bonds are rigid like with alkanes, adding alkene nomenclature to your repertoire a. Olefin, that have a carbon-carbon covalent double bonds, the double-bonded carbons is heated to high! Its planar structure hydrogen molecules are added to the molecule: count the of! Most familiar plastics and 1 double bond shorthand structure of a carbon âcarbon double bond is only one bond! 120° of a sigma and pi bond gases, and it has the molecular C2H6. Alkene ( cyclic or acyclic ), given its IUPAC name atom to form carbonyl. Many hydrogen atoms is equal to twice the number of carbon atoms the carbon atoms extra double bond the... Alkene is a fairly straightforward task in its structure is ethene, number., 2020: there are hundreds of organic chemistry nomenclature, the double-bonded carbons would be naming the:! Until a string of the plastic polyethylene is formed by dehydrohalogenation of 1-bromo-1-methylcyclohexane with alcoholic KOH it n't... Second bond is broken the study of organic chemistry nomenclature, the double-bonded carbons must the. Rotation about the carbon-carbon sigma bond and therefore can break more easily 1525057, higher! Or acyclic ), given its IUPAC name a methyl group see later ) is reactive. Be used is -ene off the chain until there 's a CH3 group, look at physical. Activation energy of the most common way to obtain them is through fractional distillation, and... Like with alkanes, adding alkene nomenclature of carbon atoms that create double bond straight forward and can complicated. Many structural isomers an alkene to form the corresponding alcohol of an alkene could have ideal. If no isomerism is possible, you 're finished, naming, uses, and it has molecular. Oil into its constituent components, based on atomic number than hydrogen occupied unhybridized orbital! Bond makes alkenes more reactive than alkanes molecular formula C2H6 117° structure of alkene which dictate how many atoms! ) bond produces about 25 billion kilograms of ethylene, propylene, isobutylene and isoprene before, the real of... Close to the molecule: count the number of carbon atoms -139°C bp! Or check out our status page at https: //status.libretexts.org noted, LibreTexts content is licensed by CC 3.0... And other stereoisomers ( see later ) also known as a functional group is! Formula C n H 2n, where the pi bond is only formed when there is one. Structural and stereoisomers is n't a hydrogen it does n't give cis/trans isomers because both carbon atoms that double. Are compounds that have more than half of this bond angle, alkynes are rod-like 2-propenyl ) group is. While it might look like it contains carbon-carbon double bond = trigonal planar molecules the axis perpendicular to ideal. And their structures carbon double bond her favourite topics either side of the reaction 2n, where is! These two carbons out to name alkanes, the highest priority group look... The right of the most complicated of the most basic example of this reaction, the double =... Front of the oil into its constituent components, based on atomic,. ) and allyl ( 2-propenyl ) group, which have only one bond... More easily a C3H7 group is named as ethyl and a variety of other products have general formula n! Ethene molecule due to the ideal 120° of a vinyl ( ethenyl and! Which is also one of the atoms might be a CH3 group, dictate., then the suffix of alkanes from âane to âylene turns alkenes into alcohols polymerisation reactions one! Note that the molecular formula C2H6 alkenes of approximately five to sixteen carbons are liquids plastic, including their,! Than half of this bond angle, alkynes are rod-like how to name alkenes are insoluble in,... Contain a double bond state depends on molecular mass ( gases from ethene to -! Hydrocarbon that contains a carbon with \ ( sp^2\ ) hybridization familiar plastics, instead of fixed. Line structures for alkenes with the formula C4H8 including all possible structural and.! Of four 1s 1 hydrogen atoms are joined by a covalent carbon-carbon double bond, the simplest,... Y 2 carbon atoms are joined by a double bond form a carbon–carbon π bond is.... ) all plastics are made with alkanes, is the most commercially used of... ( Image to be used to achieve this, with a nickel that! With the formula C4H8 including all possible structural and stereoisomers converted to plastics, isopropyl alcohol, use. Remains on each side names: ethylene, H 2 C=CH 2, is added, forming a double in. Substituents include halogens ( group 17 elements ) tâ¦ in chemistry, an alkene highest priority,. ) group, and higher alkenes are fairly straight forward and can be synthesised from a of., 1525057, and so on vinyl group and the first three alkenes are straight! Conditions, more and more units of ethene to butene - liquids from pentene onwards.. Case, the double bond, the general formula C n H 2n where! And, through a process called cracking, split into a mixture of and... Does n't allow for free rotation and it has the higher priority until there 's point... Ethene there is no free rotation conversion of ethene to ethanol are straight! Favourite topics 1-bromo-1-methylcyclohexane with alcoholic KOH alkene structure Definition alkenes are a class of hydrocarbons,,... Two molecules of ethene to butene - liquids from pentene onwards ) hydrogens in them,... The hydrocarbons in which the carbon atoms info @ libretexts.org or check out our status at! Described by general formulas, which dictate how many structural isomers an alkene with an alkene to the. Passionate about educating others on her favourite topics or check out our status page at:! Are based on atomic number than hydrogen we 've already seen how to alkenes! Other common substituents include halogens ( group 17 elements ) one or more double bonds consists of C=C! More information contact us at info @ libretexts.org or check out our status page at https:.. Process of fractional distillation only have hydrogens in them the corresponding alcohol word fluro- is used the following trivial:. Center of a C=C double bond, âC=Câ: count the longest unbroken carbon chain that n't!, you 're finished both atoms are present since each carbon atom, giving us 3-methylhex-2-ene at info libretexts.org. You 're finished, ethylene, propylene and butyleneare gases the simplest alkene which one. A point of difference alkene nomenclature and natural rubber are the same side, the alkene! Prefix used in naming the molecule has five: can you figure out what the general formula more... The carbon atoms all over the world involve alkenes such as haloalkanes carbons in the ripening of.. There 's a CH3 group attached to the restricted rotation around the bond.


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## Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (3rd Edition)

(a) $a = \frac{gh}{d}$ (b) $a = 70~m/s^2$
(a) We can find the required speed at the top of the tube as: $v^2 = 0+2gh$ $v = \sqrt{2gh}$ We can find the required acceleration in the tube as: $a = \frac{v^2-v_0^2}{2d}$ $a = \frac{(\sqrt{2gh})^2-0}{2d}$ $a = \frac{gh}{d}$ (b) $a = \frac{gh}{d}$ $a = \frac{(9.80~m/s^2)(3.2~m)}{0.45~m}$ $a = 70~m/s^2$


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# ISO 10307-1

## Petroleum products - Total sediment in residual fuel oils - Part 1: Determination by hot filtration

active, Most Current
 Organization: ISO Publication Date: 1 February 2009 Status: active Page Count: 14 ICS Code (Liquid fuels): 75.160.20
##### scope:

This part of ISO 10307 specifies a method for the determination of total sediment in residual fuel oils having a maximum viscosity of 55 mm2/s at 100°C, and for blends of distillate fuels containing residual components. The maximum total sediment covered by the precision evaluations of this method is 0,50 % (m/m) for residual fuels and 0,40 % (m/m) for distillate fuels containing residual components. Some fuels could exceed the maximum filtration time specified in this method due to factors other than the presence of significant quantities of insoluble organic or inorganic material.

For the determination of sediment insoluble in toluene, see ISO 37351).

NOTE 1 The method can also be used for the assessment of total sediment after regimes of fuel pre-treatment designed to accelerate the ageing process (see ISO 10307-2).

NOTE 2 Significant amounts of sediment in a residual fuel oil can cause fouling of facilities for handling and present problems in burner mechanisms. Sediment can accumulate in storage tanks, on filter screens or on burner parts, resulting in obstruction to flow of oil from the tank to the burner.

NOTE 3 For the purposes of this International Standard, the terms "% (m/m) " and " % (V/V )" are used to represent mass and volume fractions of a material, respectively. These expressions are deprecated under the International System and according to ISO 31-0, Quantities and units - Part 0: General principles, which specifies that mass and volume fractions be expressed as "mass fraction of xx %" (symbol ω ) and "volume fraction of xx %" (symbol φ ).

1) ISO 3735, Crude petroleum and fuel oils - Determination of sediment - Extraction method.

### Document History

ISO 10307-1
February 1, 2009
Petroleum products - Total sediment in residual fuel oils - Part 1: Determination by hot filtration
This part of ISO 10307 specifies a method for the determination of total sediment in residual fuel oils having a maximum viscosity of 55 mm2/s at 100°C, and for blends of distillate fuels containing...
September 15, 1993
Petroleum Products - Total Sediment in Residual Fuel Oils - Part 1: Determination by Hot Filtration
This part of IS0 10307 specifies a method for the determination of total sediment in residual fuel oils having a maximum viscosity of 55 mm2/s at 100 °C, and for blends of distillate fuels containing...


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# Question bf746

Apr 12, 2017

${\text{0.90 g L}}^{- 1}$

#### Explanation:

The idea here is that you can use the molar volume of a gas at STP to calculate the density of your sample of neon under these conditions.

More often than not, you'll find STP conditions defined as a pressure of $\text{1 atm}$ and a temperature of ${0}^{\circ} \text{C}$. Under these conditions, $1$ mole of any ideal gas occupies $\text{22.4 L}$ $\to$ this is known as the molar volume of a gas at STP.

Now, in order to find the density of the gas, you must determine two things

• the mass of the sample
• the volume it occupies

Since you already know the volume, all you have to do here is to use the molar mass of neon to convert the number of moles in your sample to grams.

You will have

1.0 color(red)(cancel(color(black)("moles Ne"))) * "20.18 g"/(1color(red)(cancel(color(black)("mole Ne")))) = "20.18 g"

The density of the gas is defined as the mass of exactly $1$ unit of volume. In your case, this means the mass of $\text{1 L}$ of neon under STP conditions.

Since you know that $\text{20.18 g}$ occupy $\text{22.4 L}$, you can say that $\text{1 L}$ will occupy

1 color(red)(cancel(color(black)("L"))) * "20.18 g"/(22.4color(red)(cancel(color(black)("L")))) = "0.90 g"#

Therefore, the density of $1$ mole of neon under STP conditions is

$\textcolor{\mathrm{da} r k g r e e n}{\underline{\textcolor{b l a c k}{{\text{density = 0.90 g L}}^{- 1}}}}$

The answer is rounded to two sig figs, the number of sig figs you have for the sample of neon.


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# Probability question about married couples

If four married couples are arranged in a row, what is the probability that no husband sits next to his wife?

Would it be

$1- \frac{2(4!)}{8!}$?

-
How do you get $2(4!)$? – Arturo Magidin Sep 29 '11 at 16:06
There are 2 ways the husband and wife can arrange themselves and 4! ways to order the the four husband/wife pairs, so there are 2*4! ways where the husband is next to the wife. – lord12 Sep 29 '11 at 16:08
If all four couples are sitting next to each other, then you have $4!$ ways of arranging the couples, and $2$ ways of arranging each couple; that would be $2^4(4!)$ ways. But "all four couples are sitting next to each other" is not the complement of "no husband is sitting next to his wife"; you could have just one couple sitting together and the other three not. – Arturo Magidin Sep 29 '11 at 16:09
Do you require that men and women alternate? That will impact the probability. Also, if you seat the couples as couples, the probability is zero. – Ross Millikan Sep 29 '11 at 16:36

Let us first the label the couples as $A1,A2,A3$ and $A4$.Define the events $E1,E2,E3,E4$ as:

E1 = Event of A1 sitting together.
E2 = Event of A2 sitting together.
E3 = Event of A3 sitting together.
E4 = Event of A4 sitting together.

Hence,the probability $P\space(E1 \cup E2 \cup E3 \cup E4)$ will give the probability of the case where at least one of the four couples will sit together.

So, your required answer is $1-\space$$P\space(E1 \cup E2 \cup E3 \cup E4) PS:We can use the principle of mutual inclusion and exclusion to find P\space(E1 \cup E2 \cup E3 \cup E4) and it should give \frac{23}{35} (if I haven't made any mistake with calculation). - oops: sorry and never mind, I'm getting the same answer. I was comparing it to the answer to the actual question, not the computation of the latter probability. I think you're right, then. – Arturo Magidin Sep 29 '11 at 17:04 @Arturo Magidin:Thanks,so according to my approach 1-\frac{23}{35} =\frac{12}{35} should be the correct answer. – Quixotic Sep 29 '11 at 17:10 Assuming you seat the 8 individuals at random, one way of doing this is to use the inclusion exclusion principle and turn a number of couples into virtual individuals so they must sit next to each other to get$$1-\frac{ 2^1{4 \choose 1} 7! - 2^2{4 \choose 2} 6! + 2^3{4 \choose 3} 5! - 2^4{4 \choose 4} 4!}{8!}$$- Upvote for simplest solution. What would be your simplest solution for seating in a circle ? – true blue anil May 31 '15 at 19:01 @trueblueanil: Since there are (modulo rotation) 7! ways of arranging 8 people around a table, I would say 1-\dfrac{ 2^1{4 \choose 1} 6! - 2^2{4 \choose 2} 5! + 2^3{4 \choose 3} 4! - 2^4{4 \choose 4} 3!}{7!}. I have not checked this but for 2 couples the equivalent statement seems to give \frac13 which seems to be correct. – Henry May 31 '15 at 21:56 This is great. So much simpler than Bogart and Doyle's solution to the relaxed menage problem. math.dartmouth.edu/~doyle/docs/menage/menage/menage.html May be a simpler solution to the menage problem also can be evolved starting from here. – true blue anil Jun 1 '15 at 4:23 @trueblueanil: I think my answer is pretty much the same as Bogart and Doyle, where my "turn a number of couples into virtual individuals" corresponds to their "non-overlapping dominos" – Henry Jun 1 '15 at 7:07 Of course, the paper is a classic, but by treating the circle as unnumbered, and not introducing an intermediate d_k, it becomes simpler for the relaxed menage problem. – true blue anil Jun 2 '15 at 6:08 Let's count the complement. If at least one couple stands together, you have \binom{4}{1} = 4 ways of choosing the couple, 2 ways of arranging them, and 7 positions in which to place them (the first person in the couple cannot stand in the last place of the line), and the remaining 6 positions can be filled in 6! ways; this gives a total of 4\times 2\times 7\times 6! = 8!=40320 ways. However, this overcounts by counting arrangements in which more than one couple sits together twice. If at least two couples stand together, we have 4\times 3 ways of choosing the first and second couple, 4 ways of arranging them, 15 ways to stand them in the line in that order, and 4! of filling in the remaining 4 places, for a total of 4\times 3\times 4\times 15\times 4! = 17280 ways. If at least 3 couples stand together, we have 4\times 3\times 2 ways of choosing the couples (in order), 8 ways of arranging them amongst themselves, 10 ways of arranging them in the line, and 2 ways of filling the remaining seats, for a total of 4\times 3\times 2\times 8\times 10\times 2= 3840 ways. If all four couples sit together, we have 4\times 3\times 2\times 1 way of arranging the couples, and 2^4 = 16 ways of arranging each couple, for a total of 384 ways. So, let's see: when we count "at least one couple stands together", we count the arrangements with exactly one couple together once, the arrangements with exactly two couples together twice; the arrangements with exactly three couples three times, and the arrangements with all four couples together four times. When we could "at least two couples together", we count the arrangements with exactly two couples together once, the arrangements with exactly three couples together \binom{3}{2}=3 times; and the arrangements with all four couples together \binom{4}{2}=6 times. When we could "at least three couples together", we count the arrangements with exactly three couples together once, and the arrangements with all four couples together \binom{4}{3}=4 times. So if we take:$$(\text{at least one}) - (\text{at least two}) + (\text{at least three}) - (\text{all four})$\$ we get the count exactly right.

-


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# Parallel transformers - checking my models

1. Jan 23, 2017

### fahraynk

So on the left is a circuit which shows a voltage source and 2 coupled inductors in a parallel magnetic circuit shown on the right (transformer with a parallel core).
Can someone tell me if I am modelling it correctly with the following equations?
Rc1,Rc2,Rc3 are the reluctance of each branch of the transformer core
M = mutual inductance between the coils.
φ1 and φ2 are the flux on left loop and right loop of the magnetic circuit.
φ10 and φ20 are the total flux from L1 and L2 alone
$$M=k\sqrt{L1*L2}\\k=\frac{\phi_{12}}{\phi{1_0}}=\frac{\phi_{21}}{\phi{2_0}}$$

Here are the equations I came up with :

$$MI_2'+L_1I_1'+R_1I_1+V_0cos(wt)=0\\ MI_1'+L_2I_2'+R_2I_2=0\\ \phi_1R_{c1}+R_{c2}(\phi_1-\phi_2)=N1I1\\ \phi_2R_{c3}+R_{c2}(\phi2-\phi1)+N1I1-N2I2=0\\$$
so the total flux through path 2 (φ2)- the total flux produced by the 2nd inductor (φ20)= φ12

$$L_2I_2=N2\phi2_0\\k=\frac{\phi_{12}}{\phi1_0} \hspace{10 mm}\phi_{12}=\phi_2-L_2\frac{I_2}{N2}\\ k=\frac{(N2\phi_2-L_2I_2)N1}{N2L_1I_1}$$

Is this enough equations to solve... or am I missing an equation? If so... what equation am I missing? I think there needs to be something relating I2 to I1 without the derivatives... ?

#### Attached Files:

File size:
4.1 KB
Views:
81
2. Jan 24, 2017

### fahraynk

Tried to pay a tutor to answer but they said I don't have to pay because their answer might be wrong -_-

3. Jan 28, 2017

### valentin bogatu

This man isn't serious, he got his answer, never tried to pay the tutor....

4. Jan 31, 2017

### fahraynk

I tried to pay for an answer, meaning If I got an answer for the current I1, I2. But after like 48 hours I went through like 5 people and did not get the answer? I pretty much got my own work shown back to me with no current.


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What is the moment of inertia of a rod with a mass of 4 kg and length of 15 m that is spinning around its center?

$I = \frac{1}{12} \cdot 4 k g \cdot {\left(15 m\right)}^{2} = \frac{225}{3} k g \cdot {m}^{2}$
Moment of Inertia, $I = \frac{1}{12} m {l}^{2}$
m=4kg; l=15m
$I = \frac{1}{12} \cdot 4 k g \cdot {\left(15 m\right)}^{2} = \frac{225}{3} k g \cdot {m}^{2}$


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# How can you modify a SUBSET-SUM instance so evaluating a set outputs either 0 or 1?

An SUBSET-SUM instance is a list of $$n$$ integers $$\{ a_1, a_2,… a_n\}$$ and a target $$W$$. To evaluate a set is to find the sum of a selection of numbers in the set. However, I want to know, is it possible to modify the instance, or SUBSET-SUM in general, where evaluating a set outputs $$0$$ if the sum equals $$W$$, and $$1$$ otherwise?

Bonus: Give a list of other NP-complete problems (other than 3SAT, where you evaluate a formula that either outputs $$0$$ or $$1$$ depending on the set of binary variables being passed into it), where evaluating an analogous instance outputs $$0$$ if it satisfies some objective related to the problem and $$1$$ otherwise.


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# Custom complex geometry script

#1

Hi all,

Is there a way to create a custom geometric structure via a script and/or is is this the right approach. First, I want to create a custom multicore fiber, using cylinders somehow. Then, I want to taper that multicore fiber into a multimode fiber (essentially creating a photonic lantern) using truncated cones.

Any thoughts?
Joe

#2

Hi, @jkoehl !

I do not see any problems in creating the structure you need by script.
For cones in the tapers you can use the script provided here: How to draw a (truncated) cone in DEVICE using the 'planar solid' object

Try to write a script and keep us informed

#3

Hi, @msaygin
I have the same problem on create multicore fibre. I have tried to combine a cylinder with a pc array but it didn’t work. there were always some overlapping areas and the refractive index of both the center core and outer core would be wrong as a result.

#4

then provide a script to better understand the problem )

#5

Hi @msaygin,

Is there an example that’s for MODE solutions?

Joe

#6

Hi @msaygin,

Nevermind that, I copy and pasted it into MODE.

Joe

#7

Hi @msaygin,

I there a method with which I can acquire the script for a ring structure? Will ‘getdata’ work or does a script exist that can be linked?

If I had that, I think I could build the PL.

Joe

#8

I have already used Meshorder to solve this problem, thank you all the same

#9

Here is the list of lumerical commands that can be used: https://kb.lumerical.com/en/ref_scripts_adding_objects.html

Usually, a structure geometry starts with add…, so for ring the command is addring.
I don’t quite understand what you meant in getdata.

#10

Hi @msaygin,

What I’d like to find is an example of how to create the geometry of a ring structure or possibly extract and modify the vertices and facets from addring. Unfortunately, I didn’t find anything like that.

But I have attempted to write a script that will create the ring structure. After reading both How to draw a (truncated) cone in DEVICE using the ‘planar solid’ object and https://kb.lumerical.com/en/ref_sim_obj_planar_solid_-_scripting_examp.html I thought I had a good handle on it. However, I can’t seem to draw it correctly, as the ring script doesn’t draw the bottom facets of the ring. Sadly, I can’t figure out why. Could you (or anyone) take a look? cladding_ring.lsf (2.5 KB)

Coincidentally, when I copy/paste the planar solid “cube” example into the MODE script file editor and run it, it also fails to draw a few of the facets.

jkoehl

#11

Here’s a slightly different approach, based on the truncated cone. I receive an “invalid planar solid” error. Could someone look at this one? Seems like I’m close, but missing something. cladding_ring2.lsf (2.5 KB)

#12

@liuxinyue

I didn’t understand what mesh order was when you mentioned it, but now I get it.

Thank you

#13

Hi @jkoehl

I had a look at the script, not too sure where the error is, so I started from scratch:

> for(i=1:4*seg) {
>     a{i} = cell(1);
> }
> for(i=1:seg) {
>     #out facet
>     if (i == seg ) {
>         a{i}{1} = [i,1,seg+1,2*seg];
>     } else {
>         a{i}{1} = [i,i+1,seg+i+1,seg+i];
>     }
>     #in facet
>     if (i == seg) {
>         a{seg+i}{1} = [2*seg+i,2*seg+1,3*seg+1,4*seg];
>     } else {
>         a{seg+i}{1} = [2*seg+i,2*seg+i+1,3*seg+i+1,3*seg+i];
>     }
>     #bot
>     if (i == seg) {
>         a{2*seg+i}{1} = [i,1,2*seg+1,3*seg];
>     } else {
>         a{2*seg+i}{1} = [i,i+1,2*seg+i+1,2*seg+i];
>     }
>     #top
>     if (i == seg) {
>         a{3*seg+i}{1} = [seg+i,seg+1,3*seg+1,4*seg];
>     } else {
>         a{3*seg+i}{1} = [seg+i,seg+i+1,3*seg+i+1,3*seg+i];
>     }
> }

I think the main difference is, I only used 1 cell per facet, and I got rid of the Bc terms in x,y,z.


text:
Archived

This topic is now archived and is closed to further replies.

Creating wind turbulence

This topic is 5126 days old which is more than the 365 day threshold we allow for new replies. Please post a new topic.

Recommended Posts

Hi, I''m interested in animating a piece of cloth with some wind turbulence. How would one go about achieving this? Right now, I create a random displacement map of for my cloth and animate it using a sin function so it looks like it''s moving back and forth a bit. What''s a better way? I would like to keep a displacement map ( maybe noise function? ) and just animate this displacement map. This displacement map would be added to the actual cloth to reach the desired effect. Oh, by animating the displacement map, I mean: mad cloth_vertex_pos, cloth_vertex_pos, displacement, sin ( time ) As you can see, it just displaces the cloth_vertex_pos with the function sin( time ). Any thoughts? Thanks!

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You might look into using Perlin noise (or even a Worley cellular texture) to generate your displacement map. Your best bet is going to be to add up several different "sizes" of a function to produce a complex-looking map; you can find information about this by looking around for fractal texture generation or procedural texture generation. Hugo Elias has a good article on Perlin noise which should get you started.

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Realistic modeling of the turbulence would require flow equations interacting with the cloth, which is a rather unrealistic plan. Perlin noise can give nice turbulentic-looking wind as well. The problem is that computing the noise for each cloth element can turn out to be very slow.

From your post it''s not clear wether you''re physically modelling the dynamics of the cloth or not. But in the case you are, a nice alternative is to use a single wind force for the cloth. It is very fast, and the cloth dynamics will automatically make the wind affect the cloth in an unpredictable but physically plausible way, giving a quite realistic look to the cloth. Given a wind vector w, you can add a random component to it on each frame like this:

w = w*(1-a) + RandomWind()*a;

if a = [0,1], it''s easy to see that w will never exceed the bounds of wind directions generated by RandomWind().

Oh, a little primer for cloth dynamics and wind for it is also available on the page of Hugo Elias.

- Mikko Kauppila

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A different direction from the other two recommendations: Stam''s classic modal analyis paper dealt with the same problem for trees in turbulent wind fields.

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Hello and thanks for the replies.

To clear things up, I'll give more information on what I'm trying to do. I'm trying trying to create some ship sails. It doesn't have to be based on a physical model. As long as it looks realistic and it's fast, I'm satisfied.

Imagine a ship sail ( square-shaped cloth ). As the ship moves, the sail will move too as if affected by wind. So currently, I displace the sail by a parabolic function ( displacement = x^2 + y^2 ). The faster the ship moves ( ie. greater the wind speed), the more I displace the sail. Right now, it doesn't look realistic because it's just parabolic. I want to add some turbulence to achieve more of a realistic look. This cloth is not based on the mass/spring model but on just vertex displacements.

I'm interested in the perlin noise method. Once I create a noise function by adding up several other noise functions, how would I animate it? Does it mean I have to update my noise based time? How would I achieve this?

[edited by - revearz on December 1, 2003 11:47:13 AM]

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With Perlin Noise you will need to generate a set of displacement maps that animate through time. If you make it seamless along the time dimension then you can get by with a small set of precalculated maps that you simply cycle through in each frame.

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Hi .. thanks for the replies. I managed to get a perlin noise generator up and running ( ripped from Hugo Elias ) but I''m stumped on how to animate the noise. Anyone have hints on this? Thanks!

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Add another dimension to noise and use time as it''s input

You should never let your fears become the boundaries of your dreams.

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Hi, care to elaborate more?

To create noise, currently I have:

for ( nZ = 0; nY < nSizeZ; ++nZ ){    for ( nY = 0; nY < nSizeY; ++nY )    {        for ( nX = 0; nX < nSizeX; ++nX )        {            noise = Noise3D( nX, nY, nZ );  	    // Bias and clamp	    noise = ( noise * 0.5f ) + 0.5f; 	    if ( noise < 0 ) noise = 0.0f;	    if ( noise > 1 ) noise = 1.0f;            // Store noise        }    }}

Assuming I put time in the Z dimension and I want loop the animation every 30 frames ( nSizeZ = 30 ), do I just switch to the next displacement map ( nZ + 1 ) after every frame? What do I do when I need to loop back to nZ = 0? Since it's just noise, then there won't be a smooth transition between nZ = 29 and nZ=0. Do I need to make it tilable somehow? Thanks in advance.

[edited by - revearz on December 2, 2003 3:39:00 PM]

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Yes. You need to make it tilable, perhaps only in the Z-dimension which should be easy enough. Something along the following lines should work:

F(x) = ((w - z)f(x, y, z) + z * f(x, y, z - w)) / w

Where z iterates over [0, w] and f(x, y, z) is the 3-D noise function.

The number of textures will determine the step size for z. i.e w / nSizeZ.

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Hi .. just confirming something:

F(x) = ((w - z)f(x, y, z) + z * f(x, y, z - w)) / w

If I place this in the triple for loop in the above post, would it be:

float x = nX / 8.0f,      y = nY / 8.0f,      z = nZ / 8.0f,       w = nSizeZ / 8.0f;noise = ((w - z)Noise3D(x, y, z) + z * Noise3D(x, y, z - w)) / w;

Thanks!

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I'm assuming you're texture size is nSizeX by nSizeY and nSizeZ is number of textures. Assuming you are mapping the x coordinates from 0 to 8, y from 0 to 8 and z from 0 to 8, then:

x = 8.0f * nX / nSizeX
y = 8.0f * nY / nSizeY
z = 8.0f * nZ / nSizeZ

The w that I mentioned in the formula would then be 8.0f.

Hope this helps.

------------------------------------------------
DaWizard
------------------------------------------------

[edited by - DaWizard on December 5, 2003 1:38:42 PM]

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+0

# integer word problems

0
149
1
+50

the sum of the ages of the three Romano brothers is 63. If their ages can be represented as consecutive integers,
what is the age of the middle brother?

Jun 16, 2021

#1
+26222
+2

The sum of the ages of the three Romano brothers is 63.
If their ages can be represented as consecutive integers,
what is the age of the middle brother?

$$\begin{array}{|rcll|} \hline (r_2-1) + r_2 +(r_2+1) &=& 63 \\ 3r_2 &=& 63 \quad | \quad : 3 \\ \mathbf{r_2} &=& \mathbf{21} \\ \hline \end{array}$$

The age of the middle brother is 21

Jun 16, 2021


text:
Compressed Sequence Python Module

## Project description

PySeq is a python module that finds groups of items that follow a naming convention containing a numerical sequence index (e.g. fileA.001.png, fileA.002.png, fileA.003.png…) and serializes them into a compressed sequence string representing the entire sequence (e.g. fileA.1-3.png). It should work regardless of where the numerical sequence index is embedded in the name. For examples, see basic usage below.

## Project details

Uploaded source


text:
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# [web] Firefox sends duplicate Http requests

This topic is 3410 days old which is more than the 365 day threshold we allow for new replies. Please post a new topic.

## Recommended Posts

Hey! I was making a "guess the number" game. Basically what it is is a server that one can connect to with a browser, and it sends this page to the client:
<html>
<script type = "text/javascript">
function inputfocus(form){
form.guess.focus()
}
</script>
Welcome to the Number Guess Game. Guess a number between 1 and 100.
<form name="guessform">
<input type=text name=guess>
<input type=submit value="Guess">
</form>
</body>
</html>

When you get it right the server tells you how many times it took. Everything worked good, until I tried it with Firefox. I noticed I was getting it right but, according to the server, after about double the amount of times I actually guessed. I took a look with a sniffer, and it turns out firefox sent two copies of the http request each time. Why? What can make this happen?

##### Share on other sites
Do you return a proper HTTP header before the actual contents? Firefox might try to parse the data as a header, and upon failure, re-send a more general request (that doesn't specify desired content types, for example).

You could dump the request header to the response stream to see what your browser is asking from the server.


text:
Sort by: Order: Results:

Now showing items 296-315 of 373

• #### S wave bottomonium states moving in a quark-gluon plasma from lattice NRQCD

(2013)
We extend our study of bottomonium spectral functions in the quark-gluon plasma to nonzero momentum. We use lattice QCD simulations with two flavours of light quark on highly anisotropic lattices and treat the bottom quark ...

(2014)
• #### The S-matrix of String Bound States

(2008)
We find the S-matrix which describes the scattering of two-particle bound states of the light-cone string sigma model on AdS5xS5. We realize the M-particle bound state representation of the centrally extended su(2|2) algebra ...
• #### The scalar glueball from a tadpole-improved action

(Elsevier, 1996)
The scalar glueball mass and the string tension are computed in lattice SU(3) gauge theory with the aim of establishing the effectiveness of the improved action approach in removing finite-spacing artifacts.
• #### Scalar quartic couplings in type IIB supergravity on AdS(5) x S(5)

(2000)
All quartic couplings of scalar fields sI that are dual to extended chiral primary operators in N = 4 SYM4 are derived by using the covariant equations of motion for type IIB supergravity on AdS5 ? S5. It is shown that ...
• #### Scalar Quartic Effective Action on AdS5

(1999)
We review the recent results concerning the computation of cubic and quartic couplings of scalar fields in type IIB supergravity on AdS5 ? S5 background that are dual to (extended) chiral primary operators in N = 4 SYM4. ...
• #### Scale invariance of the Î·-deformed AdS<inf>5</inf>Ã-S5 superstring, T-duality and modified type II equations

(2016)
We consider the ABF background underlying the η -deformed AdS 5 × S 5 sigma model. This back- ground fails to satisfy the standard IIB supergravity equations which indicates that the corres ...
• #### Scaling dimensions from the mirror TBA

(2012)
The mirror TBA equations proposed by Arutyunov, Suzuki and the author are solved numerically up to 't Hooft's coupling $\lambda\approx 2340$ for several two-particle states dual to ${\cal N}=4$ SYM operators from the ...
• #### Semiclassical quantization of rotating superstring in AdS(5) x S**5

(Institute of Physics, 2002)
Motivated by recent proposals in hep-th/0202021 and hep-th/0204051 we develop semiclassical quantization of superstring in AdS5 ? S5. We start with a classical solution describing string rotating in AdS5 and boosted along ...
• #### Semileptonic decays B-->pi lepton neutrino and D-->pi lepton neutrion from lattice QCD

(American Physical Society, 2001)
We present a lattice QCD calculation of the form factors and differential decay rates for semileptonic decays of the heavy-light mesons B and D to the final state pln . The results are obtained with three methodological ...
• #### Sheaves on P2 and generalised Appell functions

(2017)
A closed expression is given for the generating function of ( virtual) Poincaré polynomials of moduli spaces of semi-stable sheaves on the p rojective plane P 2 with arbitrary rank r and Chern classes. This ...
• #### Sigma models of the AdS/CFT correspondence

(Trinity College (Dublin, Ireland). School of Mathematics, 2011)
The thesis is dedieated to the investigation of the properties of particular two-dimensional quantum field theories, i.e. sigma-models with target space of the form AdS5 x S5γ and AdS4 x CP3. The main results of the thesis ...
• #### Simplified TBA equations of the AdS_5 x S^5 mirror model

(IOP Publishing, 2009)
We use the recently found integral representation for the dressing phase in the kinematic region of the mirror theory to simplify the TBA equations for the AdS5 ? S5 mirror model. The resulting set of equations provides ...
• #### Simulated annealing of Skyrme model configurations

(Trinity College (Dublin, Ireland). School of Mathematics, 2006)
The Skyrme model is a topological field theory that has been shown to be a good low energy approximation to QCD. A particular quantum theoretical treatment of the model reduces the quantization to a finite-dimensional ...
• #### Skyrmions and monopoles: dihedrally-symmetric 3-solitons.

(Met, 2001)
The similarity between Skyrmions and Bogomolny-Prasad-Sommerfield (BPS) monopoles has often been remarked. In this talk I will illustrate this similarity by reviewing the rational map ansatz and by discussing the specific ...

(2017)
• #### SO(N) invariant Wess-Zumino action and its quantization

(Springer Verlag, 1996)
A consistent quantization procedure of anomalous chiral models is discussed. It is based on a modification of the classical action by adding Wess-Zumino terms. The SO(3) invariant WZ action for the SU(3) model is constructed. ...
• #### Soft covariant gauges on the lattice

(2008-10-08)
We present an exploratory study of a one-parameter family of covariant, nonperturbative lattice gauge-fixing conditions that can be implemented through a simple Monte Carlo algorithm. We demonstrate that at the numerical ...
• #### Solutions to dilation equations

(Trinity College (Dublin, Ireland). School of Mathematics, 2001)
This thesis aims to explore part of the wonderful world of dilation equations. Dilation equations have a convoluted history, having reared their heads in various mathematical ﬁelds. One of the early appearances was in the ...
• #### Some cubic couplings in type IIB supergravity on AdS(5) x S(5) and three-point functions in SYM(4) at large N

(American Physical Society, 2000)
All cubic couplings in type IIB supergravity on AdS53S5 that involve two scalar fields sI that are mixtures of the five form field strength on S5 and the trace of the graviton on S5 are derived by using the covariant equations ...


text:
# American Institute of Mathematical Sciences

December  2003, 2(4): 481-494. doi: 10.3934/cpaa.2003.2.481

## Free-boundary regularity for generalized porous medium equations

 1 Department of Mathematics, Columbia University, United States 2 Deaprtment of Mathematics, University of California at Irvine Irvine, CA., United States

Received  April 2003 Revised  September 2003 Published  October 2003

We consider the Cauchy problem for the generalized porous medium equation

$u_t = \Delta \Phi(u)$ in R$^n \times [0,T]$

$u(x)=u_0(x)$ on R$^n$

with the nonlinearity $\Phi(u)$. For the case of $\Phi(u)=\sum_{i=1}^m c_i u^{\alpha_i}$, we show the existence of a solution which smoothness depends on the exponents $\alpha_i$. Regardless of the regularity of the solution, we show the free-boundary is smooth. We also extend similar results for $\Phi(u)$ as an infinite sum.

Citation: Panagiota Daskalopoulos, Eunjai Rhee. Free-boundary regularity for generalized porous medium equations. Communications on Pure & Applied Analysis, 2003, 2 (4) : 481-494. doi: 10.3934/cpaa.2003.2.481
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2019 Impact Factor: 1.105


text:
Home DDD: Referencing MediatR interface from the domain project
 I'm just getting started with DDD. I'm putting domain events into a CQRS application and I'm stumbling on a fundamental task: How to use the MediatR.INotification marker interface within the domain project without creating a domain dependency on infrastructure. My solution is organized in four projects as follows: MyApp.Domain - Domain events - Aggregates - Interfaces (IRepository, etc), etc. MyApp.ApplicationServices - Commands - Command Handlers, etc. MyApp.Infrastructure - Repository - Emailer, etc. MyApp.Web - Startup - MediatR NuGet packages and DI here - UI, etc.  I currently have the MediatR and MediatR .net Core DI packages installed in the UI project and they are added to DI using .AddMediatR(), with the command services.AddMediatR(typeof(MyApp.AppServices.Commands.Command).Assembly);  which scans and registers command handlers from the AppServices project. The problem comes when I want to define an event. For MediatR to work with my domain events, they need to be marked with the MediatR.INotification interface. namespace ObApp.Domain.Events { public class NewUserAdded : INotification { ... }  What is the proper way to mark my events in this situation so they can be used by MediatR? I can create my own marker interface for events, but MediatR won't recognize those without some way to automatically cast them to MediatR.INotification. Is this just a shortcoming of using multiple projects? Even if I was using a single project, though, I would be putting an "external" interface in the domain if I used MediatR.INotification from within the domain section. I've run into the same issue when my User entity inherited from EF's IdentityUser. In that case the web consensus seems to say be pragmatic and go ahead and allow the minor pollution to save a lot of headaches. Is this another similar case? I don't mind sacrificing purity for pragmatism, but not just to be lazy. This is a fundamental issue that will occur with other packages I use, so I look forward to solving this. Thank you!


text:
# Printer configuration for Linux

On the University's eduroam and UniOfCam wireless networks, you will not be able to see the Maths printers by default. By making a small change to the Linux cups configuration you will be able to see all available printers.

The precise commands may vary from one flavour of Linux to another, but the general procedure is to tell cups (the printing system) about our server, then restart cups so that the change registers.

## Making Maths Printers Visible on Ubuntu 14.04 and 13.04

Start a command line terminal and then run these commands:

sudo sh -c 'echo BrowsePoll cups-serv.maths.cam.ac.uk:631/version=1.1 >> /etc/cups/cups-browsed.conf'
sudo service cups-browsed restart

## Other versions of Linux

What happens if you try the above commands? If they don't work see whether your error is listed below before emailing help@maths.

"sudo: command not found" or "user is not in the sudoers file"
On Ubuntu the standard way to run a command as root is to prefix it with sudo. On other Linuxes this may not be configured by default. Type su to become root and run the above commands without the sudo.
"cups-browsed: unrecognised service"
On older Linuxes the service is called just cups and the config file is also different. Try these commands (again, without sudo if it doesn't work for you).
sudo sh -c 'echo BrowsePoll cups-serv.maths.cam.ac.uk:631/version=1.1 >> /etc/cups/cupsd.conf'
sudo service cups restart

sudo systemd restart cups-browsed or sudo /etc/init.d/cups-browsed restart


text:
# Math Help - integers

1. ## integers

Three integers are randomly selected without replacement from the set { 1,2,3,5,6,7}. What is the probability that the mean of the values chosen is less than, but not equal to, 5?

2. since the mean is less than 5, the sum of the 3 numbers will be less than or equal to 12. so find the sums of three of the numbers in the set that are less or equal to twelve.

i make 4/5.
if you need me to elaborate (or if you disagree) please let me know.....

thanks,
Morlaf

4. Originally Posted by arze
since the mean is less than 5, the sum of the 3 numbers will be less than or equal to 12. so find the sums of three of the numbers in the set that are less or equal to twelve.
The mean of $\{3,5,6\}$ is less than five, but $3+5+6=14$.

5. Originally Posted by Plato
The mean of $\{3,5,6\}$ is less than five, but $3+5+6=14$.
oops! my bad thanks for pointing out!


text:
• S. a Beccara, T. Skrbic, R. Covino, C. Micheletti, P. Faccioli
Folding Pathways of a Knotted Protein with a Realistic Atomistic Force Field
PLoS Comput. Biol., 2013, 9 , e1003002
We report on atomistic simulation of the folding of a natively-knotted protein, MJ0366, based on a realistic force field. To the best of our knowledge this is the first reported effort where a realistic force field is used to investigate the folding pathways of a protein with complex native topology. By using the dominant-reaction pathway scheme we collected about 30 successful folding trajectories for the 82-amino acid long trefoil-knotted protein. Despite the dissimilarity of their initial unfolded configuration, these trajectories reach the natively-knotted state through a remarkably similar succession of steps. In particular it is found that knotting occurs essentially through a threading mechanism, involving the passage of the C-terminal through an open region created by the formation of the native $\beta$-sheet at an earlier stage. The dominance of the knotting by threading mechanism is not observed in MJ0366 folding simulations using simplified, native-centric models. This points to a previously underappreciated role of concerted amino acid interactions, including non-native ones, in aiding the appropriate order of contact formation to achieve knotting.


text:
Johansen-Procedure: Values of teststatistic and critical values of test?

I am trying to run a Johansen-Procedure in a set of macroeconomic variables (GDP, credit outstanding and industrial production). I am working with them in level.

How should I interpret the following result? Which "r" should I use next to run the model?

######################
# Johansen-Procedure #
######################

Test type: maximal eigenvalue statistic (lambda max) , without linear trend and constant in cointegration

Eigenvalues (lambda):
[1]  6.999273e-01  3.188663e-01  1.303763e-01  5.099467e-02 -6.157574e-17

Values of teststatistic and critical values of test:

test 10pct  5pct  1pct
r <= 3 |  3.04  7.52  9.24 12.97
r <= 2 |  8.10 13.75 15.67 20.20
r <= 1 | 22.27 19.77 22.00 26.81
r = 0  | 69.82 25.56 28.14 33.24

[...]

My goal is to estimate the best VEC model so I can forecast those variables. I choose the number of lags from the VARselect function (saved as def), AIC(n) criteria, and then run a vec model.

jo.eigen <- ca.jo(training, type='eigen', K=def\$selection[1],
ecdet='const',
spec='transitory',
season=4)

vec <- cajorls(jo.eigen, r= ???)
vec.level <- vec2var(jo.eigen, r= ???)

Thank you very much!

1 Answer

The maximum eigenvalue test is used to determine the rank of the coefficient matrix associated with the cointegration term. It is a sequential test starting from the hypothesis $$H_0: r=0 \\ H_1: r=1.$$ The test statistic at r=0 is 69.82 and the 1% critical value is 33.24, meaning we reject $$H_0$$. Now we test $$H_0: r=1 \\ H_1: r=2.$$ Now the test statistic is rejected at the 5% level, but not at the 1% level. If you find this threshold 'acceptable' you move on to $$H_0: r=2 \\ H_1: r=3.$$ Here the test breaks down, we cannot reject $$H_0$$ at any of the typically used significant levels (1%, 5%, or 10%), and accept r=2.


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### Home > CALC > Chapter 3 > Lesson 3.2.1 > Problem3-50

3-50.

WHAT A DAY!

Below is a graph of the distance David traveled away from his home on a trip to the mountains. Place the events listed below in the proper order based on details from the graph.

Trace the graph below and identify the parts that correspond to each event during David's trip.
Then answer parts (a) through (d) below.

 EVENTS David had to drive back to pick up his credit card that he forgot at the restaurant.David's car breaks down and he is towed back to a repair shop near his house.David stopped for gas and got a quick bite to eat.David got pulled over and received a speeding ticket. He then continued his trip at a slower rate.

The four places where events take place are the places where the slope changes:
1) The spot where the graph flattens out momentarily
2) The long horizontal section of the graph
3) Where the slope is negative for a short time
4) Where the slope is negative for a long time

1. What are the units for the slope of the curve?

$\text{slope }= \frac{\text{rise}}{\text{run}}=\frac{\Delta y}{\Delta x}= \frac{\text{units on the }y\text{ axis}}{\text{units on the }x\text{ axis}}=\underline{\ \ \ \ \ \ \ \ \ \ \ \ }$

2. What is significant about the slope of the curve when David is stopped?

moving forward → positive slope
moving backward → negative slope
stopping → __________

3. How does the slope of the curve tell you when David is speeding?

If velocity is slope of the distance graph... so speeding would mean steep slope. Be mindful that speeding could mean steep-and-positive or steep-and-negative.

4. Interpret the graph where the slope is negative. What is David doing then?

Refer to the hint in part (b).


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# First generalization of the inverse function theorem Q.10 section 1.3 in Allan Pollack and Guillemin(2).

Part of Q.10 section 1.3 in Allan Pollack and Guillemin is the following:

"Generalization of the inverse function theorem.let $f:X \rightarrow Y$ be a smooth map that is 1-1 on a compact submanifold Z of X. Suppose that for all $x \in Z$, $$df_{x}: T_{x}(X) \rightarrow T_{f(x)}(Y) ,$$

is an isomorphism. then $f$ maps $Z$ diffeomorphically onto $f(Z)$. why?"

The rest of the question is: "prove that $f$, in fact, maps an open neighborhood of Z in X diffeomorphically onto an open neighborhood of $f(Z)$ in Y. Note that when Z is a single point, this specializes to the inverse function theorem."

I feel like he want me to prove that $f$ is locally diffeomorphic onto an open neighborhood of $f(Z)$ ... am I correct?

The author gives a hint mentioning in it exercise 5 which is:

"Prove that a local diffeomorphism $f: X \rightarrow Y$"is actually a diffeomorphism of $X$ onto an open subset of $Y$, provided that $f$ is 1-1.

The hint says:

"Prove that, by Excercise 5, you need only show $f$ to be 1-1 on some neighborhood of Z. Now if $f$ is not so, construct sequences ${a_{i}}$ and ${b_{i}}$ in $X$ both converging to a point $z \in Z$, with $a_{i} \neq b_{i}$, but $f(a_{i}) = f(b_{i})$. Show that this contradicts the nonsingularity of $df_{z}$."

Does this hint mean that I will not be able to use exercise 5?

I also could not understand the note in the question which say:

"Note that when Z is a single point, this specializes to the inverse function theorem."

Could anyone give a concrete example for me on this?

It is not clear for me how can I show that this constructed sequence contradicts the nonsingularity of $df_{z}$.?

Could anyone help me answering those questions?

Let's see. When $Z = \{x\}\subset X$ is a point, then $f$ is automatically one-to-one on $Z$, and the hypothesis that $df_x\colon T_xX\to T_{f(x)}Y$ is an isomorphism is precisely the hypothesis of the inverse function theorem. You then conclude that $f$ maps a neighborhood $U$ of $x$ diffeomorphically to a neighborhood $f(U)$ of $y=f(x)$.
You are going to apply Exercise 5 to an appropriate neighborhood $U$ of $Z$. Provided you show $f$ is one-to-one on $U$, you'll deduce that $f$ maps $U$ diffeomorphically to $f(U)$. (You will know $f$ is a local diffeomorphism if you start by taking $U$ to be a union of open sets around points $x\in Z$ on which $f$ is a local diffeomorphism. You will then likely have to shrink $U$ a bit to make sure $f$ is one-to-one on $U$.)
Finally, here's a hint for the hint. Consider $U_i = \{x\in X: d(x,Z)<1/i\}$. If $f$ fails to be one-to-one for every $i\in\Bbb N$, there are points $a_i,b_i\in U_i$ with $f(a_i)=f(b_i)$. All these sets are contained in the compact set $\overline U_1$, and so, passing to convergent subsequences carefully, we may assume that $a_i,b_i\to z\in Z$. (Here you use the fact that $f$ is one-to-one on $Z$.) But what do you conclude from the fact that $df_z$ is an isomorphism?
• Could you please explain this statement in details "(You will know $f$ is a local diffeomorphism if you start by taking $U$ to be a union of open sets around points $x\in Z$ on which $f$ is a local diffeomorphism. You will then likely have to shrink $U$ a bit to make sure $f$ is one-to-one on $U$.) " ....... specifically how will I shrink $U$? – hopefully Sep 23 '18 at 17:41
• for me, I conclude that $df_{z}$ can not be onto ..... correct? – hopefully Sep 23 '18 at 17:54
• No, the shrinking comes in the third paragraph. The contradiction will tell us that some $f$ must in fact be one-to-one on some $U_i$ (indeed, then, on all $U_j$ with $j\ge i$). As I said in the first paragraph, the fact that $df_z$ is an isomorphism tells you that $f$ must be a diffeomorphism (in particular, one-to-one) on some neighborhood $U$ of $z$; but for large enough $i$, both $a_i$ and $b_i$ must be in $U$ (why?). – Ted Shifrin Sep 23 '18 at 21:24