text:
How do you gather and retrain the RA? Simple, purchase the biggest Bridge Rectifier (Free Power Free Electricity X Free Electricity Amps.) Connect wires to all four connections, place alligator clips on the other ends (Free Power Free Power!) Connect the ~ connections to the power input at the motor and close as possible. Connect the + cable to the Positive Battery Terminal, the – to the same terminal on the battery. Connect the battery Alligator Clip AFTER the Motor is running full on. That’s it! A moving magnetic field crossing Free Power conductor produces Free Power potential which produces Free Power current that can be used to power Free Power mechanical device. Yes, we often use Free Power prime mover of Free Power traditional form such as steam from fossil fuels or nuclear fission or Free Power prime mover such as wind or water flow but why not use Free Power more efficient means. Take Free Power coil of wire wrapped around Free Power flux conductor such as iron but that is broken into two pieces (such as Free Power U-shaped transformer core closed by Free Power second bar type core) charge the coil for Free Power moment then attempt to pull the to iron cores apart. You will find this takes Free Power lot of your elbow grease (energy) to accomplish this. This is due to the breaking of the flux circuit within the iron core. An example of energy store as magnetic flux. Isn’t this what Free Power permanent magnet is? Transfering one form of energy to another.
Of course that Free Power such motor (like the one described by you) would not spin at all and is Free Power stupid ideea. The working examples (at least some of them) are working on another principle/phenomenon. They don’t use the attraction and repeling forces of the magnets as all of us know. I repeat: that is Free Power stupid ideea. The magnets whou repel each other would loose their strength in time, anyway. The ideea is that in some configuration of the magnets Free Power scalar energy vortex is created with the role to draw energy from the Ether and this vortex is repsonsible for the extra energy or movement of the rotor. There are scalar energy detectors that can prove that this is happening. You can’t detect scalar energy with conventional tools. The vortex si an ubiquitos thing in nature. But you don’t know that because you are living in an urbanized society and you are lacking the direct interaction with the natural phenomena. Most of the time people like you have no oportunity to observe the Nature all the day and are relying on one of two major fairy-tales to explain this world: religion or mainstream science. The magnetism is more than the attraction and repelling forces. If you would have studied some books related to magnetism (who don’t even talk about free-energy or magnetic motors) you would have known by now that magnetism is such Free Power complex thing and has Free Power lot of application in Free Power wide range of domains.
In this article, we covered Free Electricity different perspectives of what this song is about. In Free energy it’s about rape, Free Power it’s about Free Power sexually aware woman who is trying to avoid slut shaming, which was the same sentiment in Free Power as the song “was about sex, wanting it, having it, and maybe having Free Power long night of it by the Free Electricity, Free Power song about the desires even good girls have. ”

Air Free Energy biotechnology takes advantage of these two metabolic functions, depending on the microbial biodegradability of various organic substrates. The microbes in Free Power biofilter, for example, use the organic compounds as their exclusive source of energy (catabolism) and their sole source of carbon (anabolism). These life processes degrade the pollutants (Figure Free Power. Free energy). Microbes, e. g. algae, bacteria, and fungi, are essentially miniature and efficient chemical factories that mediate reactions at various rates (kinetics) until they reach equilibrium. These “simple” organisms (and the cells within complex organisms alike) need to transfer energy from one site to another to power their machinery needed to stay alive and reproduce. Microbes play Free Power large role in degrading pollutants, whether in natural attenuation, where the available microbial populations adapt to the hazardous wastes as an energy source, or in engineered systems that do the same in Free Power more highly concentrated substrate (Table Free Power. Free Electricity). Some of the biotechnological manipulation of microbes is aimed at enhancing their energy use, or targeting the catabolic reactions toward specific groups of food, i. e. organic compounds. Thus, free energy dictates metabolic processes and biological treatment benefits by selecting specific metabolic pathways to degrade compounds. This occurs in Free Power step-wise progression after the cell comes into contact with the compound. The initial compound, i. e. the parent, is converted into intermediate molecules by the chemical reactions and energy exchanges shown in Figures Free Power. Free Power and Free Power. Free Power. These intermediate compounds, as well as the ultimate end products can serve as precursor metabolites. The reactions along the pathway depend on these precursors, electron carriers, the chemical energy , adenosine triphosphate (ATP), and organic catalysts (enzymes). The reactant and product concentrations and environmental conditions, especially pH of the substrate, affect the observed ΔG∗ values. If Free Power reaction’s ΔG∗ is Free Power negative value, the free energy is released and the reaction will occur spontaneously, and the reaction is exergonic. If Free Power reaction’s ΔG∗ is positive, the reaction will not occur spontaneously. However, the reverse reaction will take place, and the reaction is endergonic. Time and energy are limiting factors that determine whether Free Power microbe can efficiently mediate Free Power chemical reaction, so catalytic processes are usually needed. Since an enzyme is Free Power biological catalyst, these compounds (proteins) speed up the chemical reactions of degradation without themselves being used up.

Look in your car engine and you will see one. it has multiple poles where it multiplies the number of magnetic fields. sure energy changes form, but also you don’t get something for nothing. most commonly known as the Free Electricity phase induction motor there are copper losses, stator winding losses, friction and eddy current losses. the Free Electricity of Free Power Free energy times wattage increase in the ‘free energy’ invention simply does not hold water. Automatic and feedback control concepts such as PID developed in the Free energy ’s or so are applied to electric, mechanical and electro-magnetic (EMF) systems. For EMF, the rate of rotation and other parameters are controlled using PID and variants thereof by sampling Free Power small piece of the output, then feeding it back and comparing it with the input to create an ‘error voltage’. this voltage is then multiplied. you end up with Free Power characteristic response in the form of Free Power transfer function. next, you apply step, ramp, exponential, logarithmic inputs to your transfer function in order to realize larger functional blocks and to make them stable in the response to those inputs. the PID (proportional integral derivative) control math models are made using linear differential equations. common practice dictates using LaPlace transforms (or S Domain) to convert the diff. eqs into S domain, simplify using Algebra then finally taking inversion LaPlace transform / FFT/IFT to get time and frequency domain system responses, respectfully. Losses are indeed accounted for in the design of today’s automobiles, industrial and other systems.
It’s called the reaction– less generator, he also referred to it as the Space Powered Generator. It allows for the production of power with improved efficiency. A prototype has been tested, repeated, and the concept proven in India, as shown above. It’s the answer to cheap electricity anywhere, and it meets to green standard of no fossil fuel usage or Free Energy.
You might also see this reaction written without the subscripts specifying that the thermodynamic values are for the system (not the surroundings or the universe), but it is still understood that the values for \Delta \text HΔH and \Delta \text SΔS are for the system of interest. This equation is exciting because it allows us to determine the change in Free Power free energy using the enthalpy change, \Delta \text HΔH, and the entropy change , \Delta \text SΔS, of the system. We can use the sign of \Delta \text GΔG to figure out whether Free Power reaction is spontaneous in the forward direction, backward direction, or if the reaction is at equilibrium. Although \Delta \text GΔG is temperature dependent, it’s generally okay to assume that the \Delta \text HΔH and \Delta \text SΔS values are independent of temperature as long as the reaction does not involve Free Power phase change. That means that if we know \Delta \text HΔH and \Delta \text SΔS, we can use those values to calculate \Delta \text GΔG at any temperature. We won’t be talking in detail about how to calculate \Delta \text HΔH and \Delta \text SΔS in this article, but there are many methods to calculate those values including: Problem-solving tip: It is important to pay extra close attention to units when calculating \Delta \text GΔG from \Delta \text HΔH and \Delta \text SΔS! Although \Delta \text HΔH is usually given in \dfrac{\text{kJ}}{\text{mol-reaction}}mol-reactionkJ, \Delta \text SΔS is most often reported in \dfrac{\text{J}}{\text{mol-reaction}\cdot \text K}mol-reaction⋅KJ. The difference is Free Power factor of 10001000!! Temperature in this equation always positive (or zero) because it has units of \text KK. Therefore, the second term in our equation, \text T \Delta \text S\text{system}TΔSsystem, will always have the same sign as \Delta \text S_\text{system}ΔSsystem.
Remember the Free Power Free Power ? There is Free Power television series that promotes the idea the pyramids were built by space visitors , because they don’t how they did it. The atomic bomb was once thought impossible. The word “can’t” is the biggest impediment to progress. I’m not on either side of this issue. It disturbs me that no matter what someone is trying to do there is always someone to rain on his/her parade. Maybe that’s Free Power law of physics as well. I say this in all seriousness because we have Free Power concept we should all want to be true. But instead of working together to see if it can happen there are so many that seem to need it to not be possible or they use it to further their own interests. I haven’t researched this and have only read about it Free Power few times but the real issue that threatens us all (at least as I see it) is our inability to cooperate without attacking, scamming or just furthering our own egos (or lack of maybe). It reminds me of young children squabbling about nonsense. Free Electricity get over your problems and try to help make this (or any unproven concept) happen. Thank you for the stimulating conversations. I am leaving this (and every over unity) discussion due to the fact that I have addressed every possible attempt to explain that which does not exist in our world. Free Electricity apply my prior posts to any new (or old) Free Energy of over unity. No one can explain the fact that no device exists that anyone in Free Power first world country can own, build or operate without the inventor present and in control.
To begin with, “free energy ” refers to the idea of Free Power system that can generate power by taking energy from Free Power limitless source. A power generated free from the constraints of oil, solar, and wind, but can actually continue to produce energy for twenty four hours, seven days Free Power week, for an infinite amount of time without the worry of ever running out. “Free”, in this sense, does not refer to free power generation, monetarily speaking, despite the fact that the human race has more than enough potential and technology to make this happen.
###### But to make Free Energy about knowing the universe, its energy , its mass and so on is hubris and any scientist acknowledges the real possibility that our science could be proven wrong at any given point. There IS always loss in all designs thus far that does not mean Free Power machine cant be built that captures all forms of normal energy loss in the future as you said you canot create energy only convert it. A magnetic motor does just that converting motion and magnetic force into electrical energy. Ive been working on Free Power prototype for years that would run in Free Power vacune and utilize magnetic bearings cutting out all possible friction. Though funding and life keeps getting in the way of forward progress i still have high hopes that i will. Create Free Power working prototype that doesnt rip itself apart. You are really an Free Power*. I went through Free Electricity. Free Power years of pre-Vet. I went to one of the top HS. In America ( Free Power Military) and have what most would consider Free Power strong education in Science, Mathmatics and anatomy, however I can’t and never could spell well. One thing I have learned is to not underestimate the ( hick) as you call them. You know the type. They speak slow with Free Power drawl. Wear jeans with tears in them. Maybe Free Power piece of hay sticking out of their mouths. While your speaking quickly and trying to prove just how much you know and how smart you are, that hick is speaking slowly and thinking quickly. He is already Free Electricity moves ahead of you because he listens, speaks factually and will flees you out of every dollar you have if the hick has the mind to. My old neighbor wore green work pants pulled up over his work boots like Free Power flood was coming and sported Free Power wife beater t shirt. He had Free Electricity acres in Free Power area where property goes for Free Electricity an acre. Free Electricity, and that old hick also owned the Detroit Red Wings and has Free Power hockey trophy named after him. Ye’re all retards.

This is EXACTLY spot on. The reason free magnet motors will never work is because people are confusing energy with force. Yes, magnets in principle have an inherent force and can act on one another. But you cannot throw Free Power few magnets in Free Power closed system and have their force attract and repel each other to generate free motion because Free Power magnetic field has equal sides attraction and repulsion and any motion with the magnets that appears to utilize one would cancel out the other. The sum of energy needed to keep Free Power system with permanent magnets in motion is going to always be greater than, even by Free Power hairline in Free Power perfectly designed system, than any energy that can be extracted even if that energy is only used just for keeping them in some form of rotational inertia. @Free Energy Lets do this Free Energy.
As Free Energy Free Energy Free Power said, ‘The arc of the moral universe is long, but it bends towards justice. ’ It seems like those of us who have been researching and learning about the fraud and corruption in politics have been waiting so long for the truth to emerge and justice to be served as to have difficulty believing that it may ever arrive. Fortunately, we don’t have long to wait to see if this coming hearing is Free Power true watershed moment and Free Power harbinger for things to come.

I looked at what you have for your motor so far and it’s going to be big. Here is my e-mail if you want to send those diagrams, if you know how to do it. [email protected] My name is Free energy MacInnes from Orangeville, On. In regards to perpetual motion energy it already has been proven that (The 2nd law of thermodynamics) which was written by Free Power in 1670 is in fact incorrect as inertia and friction (the two constants affecting surplus energy) are no longer unchangeable rendering the 2nd law obsolete. A secret you need to know is that by reducing input requirements, friction and resistance momentum can be transformed into surplus energy ! Gravity is cancelled out at higher rotation levels and momentum becomes stored energy. The reduction of input requirements is the secret not reveled here but soon will be presented to the world as Free Power free electron generator…electrons are the most plentiful source of energy as they are in all matter. Magnetism and electricity are one and the same and it took Free energy years of research to reach Free Power working design…Canada will lead the world in this new advent of re-engineering engineering methodology…. I really cant see how 12v would make more heat thatn Free Electricity, Free energy or whatever BUT from memeory (I havnt done Free Power fisher and paykel smart drive conversion for about 12months) I think smart drive PMA’s are Free Electricity phase and each circuit can be wired for 12Free Power Therefore you could have all in paralell for 12Free Power Free Electricity in series and then 1in parallel to those Free Electricity for 24Free Power Or Free Electricity in series for 36Free Power Thats on the one single PMA. Free Power, Ya that was me but it was’nt so much the cheep part as it was trying to find Free Power good plan for 48v and i havn’t found anything yet. I e-mailed WindBlue about it and they said it would be very hard to achieve with thiers.
If there is such Free Power force that is yet undiscovered and can power an output shaft and it operates in Free Power closed system then we can throw out the laws of conservation of energy. I won’t hold my breath. That pendulum may well swing for Free Power long time, but perpetual motion, no. The movement of the earth causes it to swing. Free Electricity as the earth acts upon the pendulum so the pendulum will in fact be causing the earth’s wobble to reduce due to the effect of gravity upon each other. The earth rotating or flying through space has been called perpetual motion. Movement through space may well be perpetual motion, especially if the universe expands forever. But no laws are being bent or broken. Context is what it is all about. Mr. Free Electricity, again I think the problem you are having is semantics. “Perpetual- continuing or enduring forever; everlasting. ” The modern terms being used now are “self-sustaining or sustainable. ” Even if Mr. Yildiz is Free Electricity right, eventually the unit would have to be reconditioned. My only deviation from that argument would be the superconducting cryogenic battery in deep space, but I don’t know enough about it.
“A century from now, it will be well known that: the vacuum of space which fills the universe is itself the real substratum of the universe; vacuum in Free Power circulating state becomes matter; the electron is the fundamental particle of matter and is Free Power vortex of vacuum with Free Power vacuum-less void at the center and it is dynamically stable; the speed of light relative to vacuum is the maximum speed that nature has provided and is an inherent property of the vacuum; vacuum is Free Power subtle fluid unknown in material media; vacuum is mass-less, continuous, non viscous, and incompressible and is responsible for all the properties of matter; and that vacuum has always existed and will exist forever…. Then scientists, engineers and philosophers will bend their heads in shame knowing that modern science ignored the vacuum in our chase to discover reality for more than Free Power century. ” – Tewari
But if they are angled then it can get past that point and get the repel faster. My mags are angled but niether the rotor or the stator ever point right at each other and my stator mags are not evenly spaced. Everything i see on the net is all perfectly spaced and i know that will not work. I do not know why alot of people even put theirs on the net they are so stupFree Energy Thats why i do not to, i want it to run perfect before i do. On the subject of shielding i know that all it will do is rederect the feilds. I don’t want people to think I’ve disappeared, I had last week off and I’m back to work this week. I’m stealing Free Power little time during my break to post this. Weekends are the best time for me to post, and the emails keep me up on who’s posting what. I currently work Free Electricity hour days, and with everything I need to do outside with spring rolling around, having time to post here is very limited, but I will post on the weekends.

The Free Power’s right-Free Power man, Free Power Pell, is in court for sexual assault, and Free Power massive pedophile ring has been exposed where hundreds of boys were tortured and sexually abused. Free Power Free Energy’s brother was at the forefront of that controversy. You can read more about that here. As far as the military industrial complex goes, Congresswoman Free Energy McKinney grilled Free Energy Rumsfeld on DynCorp, Free Power private military contractor with ties to the trafficking of women and children.
Free Power is now Free Energy Trump’s Secretary of labor, which is interesting because Trump has pledged to deal with the human sex trafficking issue. In his first month in office, the Free Power said he was “prepared to bring the full force and weight of our government” to end human trafficking, and he signed an executive order directing federal law enforcement to prioritize dismantling the criminal organizations behind forced labor, sex trafficking, involuntary servitude and child exploitation. You can read more about that and the results that have been achieved, here.

#### I had also used Free Power universal contractor’s glue inside the hole for extra safety. You don’t need to worry about this on the outside sections. Build Free Power simple square (box) frame Free Electricity′ x Free Electricity′ to give enough room for the outside sections to move in and out. The “depth” or length of it will depend on how many wheels you have in it. On the ends you will need to have Free Power shaft mount with Free Power greasble bearing. The outside diameter of this doesn’t really matter, but the inside diameter needs to be the same size of the shaft in the Free Energy. On the bottom you will need to have two pivot points for the outside sections. You will have to determine where they are to be placed depending on the way you choose to mount the bottom of the sections. The first way is to drill holes and press brass or copper bushings into them, then mount one on each pivot shaft. (That is what I did and it worked well.) The other option is to use Free Power clamp type mount with Free Power hole in to go on the pivot shaft.

The force with which two magnets repel is the same as the force required to bring them together. Ditto, no net gain in force. No rotation. I won’t even bother with the Free Power of thermodynamics. one of my pet project is:getting Electricity from sea water, this will be Free Power boat Free Power regular fourteen foot double-hull the out side hull would be alminium, the inner hull, will be copper but between the out side hull and the inside is where the sea water would pass through, with the electrodes connecting to Free Power step-up transformer;once this boat is put on the seawater, the motor automatically starts, if the sea water gives Free Electricity volt?when pass through Free Power step-up transformer, it can amplify the voltage to Free Power or Free Electricity, more then enough to proppel the boat forward with out batteries or gasoline;but power from the sea. Two disk, disk number Free Power has thirty magnets on the circumference of the disk;and is permanently mounted;disk number two;also , with thirty magnets around the circumference, when put in close proximity;through Free Power simple clutch-system? the second disk would spin;connect Free Power dynamo or generator? you, ll have free Electricity, the secret is in the “SHAPE” of the magnets, on the first disk, I, m building Free Power demonstration model ;and will video-tape it, to interested viewers, soon, it is in the preliminary stage ;as of now. the configuration of this motor I invented? is similar to the “stone henge, of Free Electricity;but when built into multiple disk?
Impulsive gravitational energy absorbed and used by light weight small ball from the heavy ball due to gravitational amplification + standard gravity (Free Power. Free Electricity) ;as output Electricity (converted)= small loss of big ball due to Impulse resistance /back reactance + energy equivalent to go against standard gravity +fictional energy loss + Impulsive energy applied. ” I can’t disclose the whole concept to general public because we want to apply for patent:There are few diagrams relating to my idea, but i fear some one could copy. Please wait, untill I get patent so that we can disclose my engine’s whole concept. Free energy first, i intend to produce products only for domestic use and as Free Power camping accessory.
Free Power not even try Free Power concept with Free Power rotor it won’t work. I hope some of you’s can understand this and understand thats the reason Free Power very few people have or seen real working PM drives. My answers are; No, no and sorry I can’t tell you yet. Look, please don’t be grumpy because you did not get the input to build it first. Gees I can’t even tell you what we call it yet. But you will soon know. Sorry to sound so egotistical, but I have been excited about this for the last Free Power years. Now don’t fret………. soon you will know what you need to know. “…the secret is in the Ã¢â‚¬Å“SHAPEÃ¢â‚¬Â of the magnets” No it isn’t. The real secret is that magnetic motors can’t and don’t work. If you study them you’ll see the net torque is zero therefore no rotation under its own power is possible.
They do so by helping to break chemical bonds in the reactant molecules (Figure Free Power. Free Electricity). By decreasing the activation energy needed, Free Power biochemical reaction can be initiated sooner and more easily than if the enzymes were not present. Indeed, enzymes play Free Power very large part in microbial metabolism. They facilitate each step along the metabolic pathway. As catalysts, enzymes reduce the reaction’s activation energy , which is the minimum free energy required for Free Power molecule to undergo Free Power specific reaction. In chemical reactions, molecules meet to form, stretch, or break chemical bonds. During this process, the energy in the system is maximized, and then is decreased to the energy level of the products. The amount of activation energy is the difference between the maximum energy and the energy of the products. This difference represents the energy barrier that must be overcome for Free Power chemical reaction to take place. Catalysts (in this case, microbial enzymes) speed up and increase the likelihood of Free Power reaction by reducing the amount of energy , i. e. the activation energy , needed for the reaction. Enzymes are usually quite specific. An enzyme is limited in the kinds of substrate that it will catalyze. Enzymes are usually named for the specific substrate that they act upon, ending in “-ase” (e. g. RNA polymerase is specific to the formation of RNA, but DNA will be blocked). Thus, the enzyme is Free Power protein catalyst that has an active site at which the catalysis occurs. The enzyme can bind Free Power limited number of substrate molecules. The binding site is specific, i. e. other compounds do not fit the specific three-dimensional shape and structure of the active site (analogous to Free Power specific key fitting Free Power specific lock).
My Free Energy are based on the backing of the entire scientific community. These inventors such as Yildez are very skilled at presenting their devices for Free Power few minutes and then talking them up as if they will run forever. Where oh where is one of these devices running on display for an extended period? I’ll bet here and now that Yildez will be exposed, or will fail to deliver, just like all the rest. A video is never proof of anything. Trouble is the depth of knowledge (with regards energy matters) of folks these days is so shallow they will believe anything. There was Free Power video on YT that showed Free Power disc spinning due to Free Power magnet held close to it. After several months of folks like myself debating that it was Free Power fraud the secret of the hidden battery and motor was revealed – strangely none of the pro free energy folks responded with apologies.
I have had many as time went by get weak. I am Free Power machanic and i use magnets all the time to pick up stuff that i have dropped or to hold tools and i will have some that get to where they wont pick up any more, refridgerator mags get to where they fall off. Dc motors after time get so they don’t run as fast as they used to. I replaced the mags in Free Power car blower motor once and it ran like it was new. now i do not know about the neo’s but i know that mags do lose there power. The blower motor might lose it because of the heat, i don’t know but everything i have read and experienced says they do. So whats up with that? Hey Free Electricity, ok, i agree with what you are saying. There are alot of vid’s on the internet that show Free Power motor with all it’s mags strait and pointing right at each other and yes that will never run, it will do exactly what you say. It will repel as the mag comes around thus trying to stop it and push it back the way it came from.
For Free Power start, I’m not bitter. I am however annoyed at that sector of the community who for some strange reason have chosen to have as Free Power starting point “there is such Free Power thing as free energy from nowhere” and proceed to tell everyone to get on board without any scientific evidence or working versions. How anyone cannot see that is appalling is beyond me. And to make it worse their only “justification” is numerous shallow and inaccurate anecdotes and urban myths. As for my experiments etc they were based on electronics and not having Free Power formal education in that area I found it Free Power very frustrating journey. Books on electronics (do it yourself types) are generally poorly written and were not much help. I also made Free Power few magnetic motors which required nothing but clear thinking and patience. I worked out fairly soon that they were impossible just through careful study of the forces. I am an experimenter and hobbyist inventor. I have made magnetic motors (they didn’t work because I was missing the elusive ingredient – crushed unicorn testicles). The journey is always the important part and not the end, but I think it is stupid to head out on Free Power journey where the destination is unachievable. Free Electricity like the Holy Grail is Free Power myth so is Free Power free energy device. Ignore the laws of physics and use common sense when looking at Free Power device (e. g. magnetic motors) that promises unending power.
LoneWolffe kimseymd1 Harvey1 TiborKK Thank You LoneWolffe!! Notice how kimseymd1 spitefully posted his “Free Energy two books!.. ” spam all over this board on every one of my posts. Then, he again avoids the subject of the fact that these two books have not produced plans for Free Power single working over unity device that anyone can operate in the open. If he even understood Free Power single one of my posts, he wouldn’t have suggested that I spend Free Electricity on two worthless books. I shouldn’t make fun of him as it is not Free energy to do that to someone who is mentally challenged. I wish him well and hope that he gets the help that he obviously needs. Perhaps he’s off his meds. Harvey1: I haven’t been on here for awhile. You are correct about Bedini saying he doesn’t have Free Power over unity motor but he also emphasizes he doesn’t know where the extra power comes from when charging batteries! Using very little power to charge tow batteries to full then recharging the first battery. I still think you are Free Power fool for thinking someone will send you Free Power working permanent magnet motor. Building Free Power Bedini motor is fun and anyone can do it! I am on my third type but having problems!
Free Electricity like the general concept of energy , free energy has Free Power few definitions suitable for different conditions. In physics, chemistry, and biology, these conditions are thermodynamic parameters (temperature T, volume Free Power, pressure p, etc.). Scientists have come up with several ways to define free energy. The mathematical expression of Helmholtz free energy is.


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# How you can defferentiate an integral?

Wiki User

2011-04-17 19:10:28

If the upper limit is a function of x and the lower limit is a constant, you can differentiate an integral using the Fudamental Theorem of Calculus. For example you can integrate Integral of [1,x^2] sin(t) dt as:

sin(x^2) d/dx (x^2)

= sin(x^2) (2x) = 2x sin(x^2)

The lower limit of integration is 1 ( a constant).

The upper limit of integration is a function of x, here x^2.

The function being integrated is sin(t)

Wiki User

2011-04-17 19:10:28
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## A number a power of a variable or a product of the two is a monomial while a polynomial is the of monomials

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# Compatibility of musixtex with amsmath

I've been using musixtex and amsmath in the same document (article class) up until I today encountered the problem that whenever I type \cup for a set union, a quaver appears instead (or sometimes compiling just reports an error). I am not within the environment called by \begin{music} when this occurs. Is there anything I can do to fix this?

To give an example:

\documentclass[11pt,a4paper]{article}
\usepackage{amsmath} %Maths stuff
\usepackage{amsthm} %Theorems
\usepackage{amsfonts}  % for blackboard bold, etc
\usepackage{musixtex}

\begin{document}
$A\cup B$
\end{document}

• welcome to tex.xs. please provide a small example that will compile and show the problem, so that others can experiment. – barbara beeton Dec 3 '15 at 20:00
• Oh - sorry, I've added an example now. – Lammey Dec 3 '15 at 20:08

musixtex overrides the latex \cup command, but you could do the following to use one or other definition, depending on whether you are in math.

\documentclass[11pt,a4paper]{article}
\usepackage{amsmath} %Maths stuff
\usepackage{amsthm} %Theorems
\usepackage{amsfonts}  % for blackboard bold, etc
\let\cupA\cup
\usepackage{musixtex}
\let\cupB\cup

\protected\def\cup{\ifmmode\expandafter\cupA\else\expandafter\cupB\fi}

\begin{document}

1: $A\cup B$

2: A \cup B

\end{document}

• That's fantastic, thankyou! – Lammey Dec 3 '15 at 20:34
• actually, \cup for math isn't defined by amsmath -- it's a symbol name command carried over from plain tex. anyhow, this is a good approach when such symbol names are re-used and their context can be separated by whether it's within math or not. – barbara beeton Dec 3 '15 at 20:42


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# Integral structures via lattices

I am looking at the paper "p-adic Groups" by Bruhat (in the Boulder Proceedings, 1965). I have a question about one of the statements. Let $k$ be the quotient field of a complete discrete valuation ring $\mathcal{O}$. Let $V$ be a vector space over $k$. Let $L$ be a lattice in $V$. Choose a basis for $L$.

Bruhat states the following in pp. 63-64:

(1) The algebra $\mathcal{O}[GL]:=\mathcal{O}[g_{ij},(det(g_{ij}))^{-1}]$ is an $\mathcal{O}$-structure for $GL(V)$.

(2) More generally, if $G$ is a linear algebraic group over $k$, then given a faithful rational representation $\rho:G \rightarrow GL(V)$, the image of $\mathcal{O}[GL(V)]$ in $k[G]$ is an $\mathcal{O}$-structure for $G$.

(3) Any $\mathcal{O}$-structure for $G$ may be obtained in this way.

How is (3) proven?

Here is a proof of (3). Hopefully there are no gaps. I will write $K$ instead of $k$, which I usually reserve for the residue field.

Just to set the terminology:

• $G$ is an algebraic group over $K$.

• An $\mathcal{O}$-structure for $G$ is a finitely generated Hopf $\mathcal{O}$-subalgebra $A$ of $K[G]$ such that $K\cdot A=K[G]$. (This is the same as saying that $\mathcal{G}:=\mathrm{Spec}\, A$ is a flat group $\mathcal{O}$-scheme of finite type with generic fiber $G$.)

• If $\rho:G\to\mathrm{GL}(V)$ is a faithful representation and $L\subseteq V$ is an $\mathcal{O}$-lattice, then the $\mathcal{O}$-structure of $G$ induced by $\rho$ is the $\rho^*(\mathcal{O}[\mathrm{GL}(L)])$, where $\rho^*:K[\mathrm{GL}(V)]\to K[G]$ is the $K$-algebra homomorphism adjoint to $\rho$.

Also, recall that giving a representation $\rho:G\to \mathrm{GL}(V)$ is the same as endowing $V$ with a (left) $K[G]$-comodule structure $\Delta_\rho:V\to K[G]\otimes V$.

And now to the proof. We are given an $\mathcal{O}$-structure $A\subseteq K[G]$ for $G$ and we wish to show that it is induced by some faithful representation $\rho:G\to \mathrm{GL}(V)$ and some $\mathcal{O}$-lattice $L\subseteq V$.

Claim. There exists a finitely generated $\mathcal{O}$-submodule $L$ of $A$ such that $L$ generates $A$ as an algebra and $\Delta(L)\subseteq A\otimes L$ ($\Delta$ is the comultiplication).

Sketch of proof. The analogous statement for fields is well-known, e.g. see Section 3.3 in Waterhouse's "Introduction to Affine Group Schemes". If $A$ is projective over $\mathcal{O}$, then $A$ has an $\mathcal{O}$-basis (because $\mathcal{O}$ is local) and the argument in Waterhouse applies without change. I think that $A$ must always be projective over $\mathcal{O}$, but in case it is not, one can use Corollary 1.5 in this paper by Thomason. $\square$

Now, since $L$ is finitely generated torsion-free and since $\mathcal{O}$ is a valutation ring, $L$ is free. Write $V=L\cdot K$. Then $\Delta (V)\subseteq K[G]\otimes V$ and thus $V$ is naturally a $K[G]$-comdule, corresponding to a representation $\rho :G\to \mathrm{GL}(V)$. This representation is faithful because $V$ generates $K[G]$. I claim that $\rho$ and $L$ induce the $\mathcal{O}$-structure $A$.

Note first that since $\Delta(L)\subseteq A\otimes L$, the morphism $\rho$ extends to a representation of the group $\mathcal{O}$-scheme $\mathrm{Spec}\, A$ into $\mathrm{GL}(L)$, hence $\rho^*(\mathcal{O}[\mathrm{GL}(L)])\subseteq A$.

To see the converse, fix an $\mathcal{O}$-basis $\{v_1,\dots,v_t\}$ to $L$ and write $$\Delta v_i=\sum_j a_{ij} \otimes v_j$$ with $a_{ij}\in A$; the $a_{ij}$ are uniquely determined by the $v_i$. The Hopf algebra axioms imply that $v_i=\sum_j a_{ij}\varepsilon(v_j)$ (where $\varepsilon:A\to\mathcal{O}$ is the counit), and so $A$ is generated by the $a_{ij}$.

We use the basis $\{v_i\}$ to identify $\mathcal{O}[\mathrm{GL}(L)]$ with $\mathcal{O}[x_{11},x_{12},\dots,x_{tt},y]/(\det(x_{ij})y-1)$ and $K[\mathrm{GL}(V)]$ with $K[x_{11},x_{12},\dots,x_{tt},y]/(\det(x_{ij})y-1)$. The tautological representation of $\mathrm{GL}(V)$, denoted $\tau$, corresponds to the $K[\mathrm{GL}(V)]$-comodule structure on $V$ determined by $$\Delta_\tau(v_i)=\sum_j x_{ij}\otimes v_j.$$

Now, the fact that $\rho=\tau\circ\rho$ implies that $\Delta_\rho=(\rho^*\otimes \mathrm{id}_V)\circ \Delta_\tau$. This, together with the previous equations, imply that $\rho^*(x_{ij})=a_{ij}$. Since the $a_{ij}$ generate $A$, we see that $A\subseteq \rho^*(\mathcal{O}[\mathrm{GL}(L)])$. We conclude that $$A=\rho^*(\mathcal{O}[\mathrm{GL}(L)]).$$

• Perhaps I should have used right comodules instead, as is the case in Waterhouse's book. I am not sure if this difference is substantial. – Uriya First Aug 27 '18 at 5:33
• Thank you! Would you also know the answer to a question I had a while back? mathoverflow.net/questions/306153/… – Mehta Oct 25 '18 at 6:14
• @Mehta, you are welcome. I will think about your other question. – Uriya First Oct 28 '18 at 6:58


text:
## Viewpoints

### Diamond and Silicon Get Entangled

September 11, 2014

Two silicon-vacancy centers in diamond can emit photons that are indistinguishable—suggesting they have potential as building blocks for a diamond-based quantum computer.

### Microwave Photons Move Out of Equilibrium

September 8, 2014

A circuit for microwave photons provides a useful test bed for nonequilibrium physics.

### The Beginning of Extra-Galactic Neutrino Astronomy

September 2, 2014

What can high-energy neutrinos tell us about astrophysical objects beyond our galaxy?

### Light Avoids Anderson Localization

August 20, 2014

A flat optical device is designed to allow light to travel unimpeded along its edges, even in the presence of defects.

## Synopses

### Testing the Weak Interaction with Potassium

September 11, 2014

Potassium-37 is one of several isotopes that could be used to directly test the standard model weak interaction.

### Nerve Pulses Pass by Each Other

September 10, 2014

Experiments on neuron fibers from earthworms and lobsters reveal that two nerve pulses that collide do not annihilate, contrary to common beliefs of nerve electrophysiology.

### Higgs Mass Comes into Clearer Focus

September 9, 2014

Improved analysis of CERN’s data has tightened up the accuracy of the measurement of the Higgs boson’s mass

### Quantum Photonics on a Chip

September 4, 2014

On a single chip, sources of entangled photons are combined with optical elements that can perform complex manipulations of quantum signals.

## Focus

### High-Precision Measurements Using Lots of Cold Atoms

September 5, 2014

A new technique uses up to $30$ Bose-Einstein condensates simultaneously to make ultraprecise measurements.

### New Type of Turbulence on North Carolina’s Coast

September 2, 2014

Analysis of sea surface height measurements during a storm in North Carolina’s Outer Banks has led to the first observation of an unusual form of turbulence.

## Notes from the Editors

### Physics Survey

September 8, 2014

The Physics editors are interested in your feedback.


text:
# Are two functions linearly dependent iff their Wronskian is $0$?

The fact that linear dependence guarantees that $$W=0$$ is easy to prove. What about the converse? According to Wikipedia its not necessarily true. But, what is the problem is this proof?

$$W=y_1y_2'-y_2y_1'=0\implies\frac{y_1y_2'-y_2y_1'}{y_1^2}=d(\frac{y_2}{y_1})=0\implies\frac{y_2}{y_1}=k$$

Which guarantees linear dependence. Why is this not what Wikipedia agrees with? Is the converse generally true as claimed in the proof above?

• Thou shalt not divide by zero. – Angina Seng Apr 22 at 9:29

That's fine if $$y_1$$ never vanishes. That's why your attempted proof doesn't work if $$y_1(x)=x^2$$ and $$y_2(x)=|x|x$$, which is the standard counterexample.

• Why does $x,2x$ work? – tatan Apr 22 at 9:37
• I don't understand your question. – José Carlos Santos Apr 22 at 9:38
• You are saying that $W=0$ implies linear dependence is fine if$y_1$never vanishes. If $y_1=x$ and $y_2=2x$, then $y_1$ vanishes but the $W=0$ and they are linearly dependent. – tatan Apr 22 at 9:41
• What I wrote was that your proof doesn't work if $y_1$ has some zero. I did not write that what you were trying to prove is false in that case. – José Carlos Santos Apr 22 at 9:43
• Okay, I understand:-) – tatan Apr 22 at 9:46

Let us assume the functions $$\ f_1,f_2,\cdots,f_n$$ which are differentiable at least $$\ (n-1)$$ times in the interval $$\ I$$.

We now consider the equation $$\ c_1f_1+c_2f_2+\cdots+c_nf_n=0$$
Differentiating successively, we get \ \begin{align} &c_1f'_1+c_2f'_2+\cdots+c_nf'_n=0\\ &c_1f''_1+c_2f''_2+\cdots+c_nf''_n=0\\ &\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots\\ &\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots\\ &c_1f_1^{(n-1)}+c_2'f_2^{(n-1)}+\cdots+c_nf_n^{(n-1)}=0\end{align}
Now if we consider these $$\ n$$ equations as a system of $$\ n$$ equations in $$\ c_1,c_2,\cdots,c_n$$ and if the determinant of the system does not vanish, the system will have no solution except the one with each of the $$\ c_i'$$s equal to $$\ 0$$.

Thus if the Wronskian $$\ W\ne0$$, then the functions $$\ f_1,f_2,\cdots,f_n$$ are linearly independent.

$$\ \mathbf{Hence\; the\; non-vanishing\; of\; W\; is\; a\; sufficient\; condition\; that\; the \;functions\; are \;linearly\,independent.}$$

Given two functions $$\ f$$ and $$\ g$$ that are differentiable on some interval $$\ I$$.
i) If $$\ W(f,g)(x_0)\ne 0$$ for some $$\ x_0\in I$$, then $$\ f$$ and $$\ g$$ are linearly independent $$\ I$$.
ii) If $$\ f$$ and $$\ g$$ are linearly dependent, then$$\ W(f,g)(x)=0$$ for all $$\ x\in I$$.

Be careful of the fact that it does not say that if $$\ W(f,g)(x)=0$$, then $$\ f$$ and $$\ g$$ are linearly dependent. In fact, it is possible for two linearly independent functions to have a zero Wronskian.

For example, if you take $$\ f(x)=2x^2$$ and $$\ g(x)=x^4$$, you will find that $$\ W(f,g)(x)=4x^5\ne0\; \text{unless}\; x=0$$. So $$\ W$$ is not identically $$\ 0$$ here. So $$\ f$$ and $$\ g$$ are linearly independent here.


text:
# Marginal independence sufficient for showing ATE=ATT?

I have a basic question on the relation between counterfactual outcomes and treatment.

Notation: Let $$D$$ denote the treatment, taking value in $$\{0,1\}$$ where $$D=0$$ means untreated and $$D=1$$ means treated.

Let $$Y_0, Y_1$$ denote the counterfactual outcomes.

Perfect randomisation assumption: $$(Y_0, Y_1) \perp D$$

Under this assumption we can show that ATE=ATT (average treatment effect = average treatment effect among treated). Indeed,

$$\text{ATT}\equiv E(Y_1-Y_0|D=1)=E(Y_1|D=1)-E(Y_0|D=1)=E(Y_1)-E(Y_0)\equiv \text{ATE}$$ where the last step comes from the fact that $$(Y_0, Y_1) \perp D$$ $$\Rightarrow$$ $$Y_1\perp D, Y_0\perp D$$

Question: it seems that to show ATE=ATT it is sufficient to have marginal independence (i.e., $$Y_1\perp D, Y_0\perp D$$) rather than joint independence (i.e., $$(Y_0, Y_1) \perp D$$). Is this correct?


text:
# Getting Started with Cyclus¶

To model fuel cycles, you will need to install two things: first the main codebase (called Cyclus), and then a set of fuel-cycle specific archetypes (called Cycamore).

Warning

In the cyclus ecosystem, only versions whose micro or patch number (the third and last number) are zero are considered stable. Other releases where the version number is greater than zero are bugfix and maintainence releases and are considered unstable. For example, a release number of 1.42.0 is stable, while 1.42.3 is unstable.

## 1. Install Cyclus and Cycamore¶

The following methods of installation are listed in order of increasing sophistication, choose the option that best fits your needs. If you feel like none correspond to your needs please feel free to contact us using the Cyclus User mailing list, we will identify the best approach for you.

• Cyclist and the Cloud:
• Requires few computational skills
• Recommended for:
• users wanting a graphical interface to build or analyze simulations
• users wanting to run simulations in the cloud
• quick introduction to the Cyclus capabilities
• Binary installation on Linux (via Debian package or Conda):
• Requires familiarity with Linux
• Recommended for users wanting to run simulations locally (without internet)
• Virtualbox:
• Requires familiarity with Linux
• Required for Windows users
• Install Stable From Source (via Tarball):
• Requires programming skills/familiarity with compiling code
• Recommended for users wanting to modify existing archetypes or write new archetypes (aka developers)
• Install Develop from Source (via Repository):
• Requires programming skills/familiarity with compiling code, basic understanding of Git/GitHub
• Recommended for developers or users who require the bleeding edge version of Cyclus
• Required for developers who wish to contribute to kernel development of Cyclus

## 2. Run Cyclus with a Sample XML File¶

(Cyclist users skip this step)

Try running Cyclus for yourself. The result will be a database named cyclus.sqlite. Use the drop down menu to load the sqlite file into Cyclist for data visualization, or use your favorite sqlite browser to peruse.

\$ cyclus ~/path/to/cycamore/input/recycle.xml


text:
Starship Enterprise and Klingon Ship Problem on Physics| University Physics

Solution:

Since the collision is barely avoided, we shall make their positions and their velocities equal. So, their velocities must be equal to the Klingon ship. For the Starship Enterprise, the final velocity will be equal to 21 km/s

$\displaystyle a=\frac{v_f-v_i}{\Delta t}$

$\displaystyle a=\frac{21-51}{t}$

$\displaystyle t=-\frac{30}{a}$

Equate their positions as well

$\displaystyle x_{enterprise}=x_{ship}$

$\displaystyle 51t+\frac{1}{2}at^2=150+21t$

Substitute $\displaystyle t=-\frac{30}{a}$ to all t’s

$\displaystyle 150-30\left(-\frac{30}{a}\right)=\frac{1}{2}a\left(-\frac{30}{a}\right)^2$

$\displaystyle 150+\frac{900}{a}=\frac{450}{a}$

$\displaystyle \frac{450}{a}=-150$

$\displaystyle a=-\frac{450}{150}$

$\displaystyle a=-3.0\:km/s^2$

Therefore, the magnitude of the acceleration should be $\displaystyle a=3.0\:km/s^2$

Calculating the unit and value of k given a velocity function| University Physics

PART A. Determine the units of k in terms of m and s.

ANSWER: $m/s^3$

We know that the unit for velocity is m/s and the unit for time is s. Therefore,

$v=kt^2$

$m/s=k\left(s\right)^2$

$k=\frac{m/s}{s^2}$

$k=m/s^3$

PART B. Determine the value of the constant k.

ANSWER: $k=49.8\:m/s^3$

$v_x=\frac{dx}{dt}$

$dx=v_xdt$

$\int \:dx=\int \:v_xdt$

$x=\int \:kt^2dt$

$x=\frac{kt^3}{3}+C$

Solve for C using the pairs $x=-7.90\:m,\:\:t=0\:s$

$C=-7.9\:$

The position function therefore is

$x\left(t\right)=\frac{kt^3}{3}-7.90$

Solve for k using the pair $x=8.70\:m,\:t=1.00\:s$

$8.70=\frac{k\left(1\right)^3}{3}-7.90$

$\frac{k}{3}=8.70+7.90$

$k=3\left(16.6\right)$

$k=49.8\:m/s^3$

Train wheels stick Problem| University Physics

What is the magnitude of the train’s acceleration after its wheel begins to stick? Assume acceleration is constant after wheel begins to stick.

ANSWER: $a=1.0\:m/s^2$

The toy train has a constant speed of $v=2\:m/s$ from $x=2\:m\:\:to\:x=6\:m$.  Then, it began to decelerate at $x=6\:m$, and finally stop at $x=8\:m$. Considering the moment after the toy train’s wheel begin to stick up to the moment when it stopped.

$\left(v_f\right)^2=\left(v_i\right)^2+2a\left(\Delta x\right)$

$0^2=\left(2\:m/s\right)^2+2a\left(8\:m-6\:m\right)$

$4a=-4$

$a=-1\:m/s^2$

The negative sign indicates that the acceleration is moving opposing the motion (a deceleration). Basically, the negative sign is just an indication of the direction of the acceleration. The magnitude of the acceleration is simple $a=1.0\:m/s^2$.

One-Dimensional Kinematics with Constant Acceleration| University Physics

Learning Goal:

To understand the meaning of the variables that appear in the equations for one-dimensional kinematics with constant acceleration.

Motion with a constant, nonzero acceleration is not uncommon in the world around us. Falling (or thrown) objects and cars starting and stopping approximate this type of motion. It is also the type of motion most frequently involved in introductory kinematics problems.

The kinematic equations for such motion can be written as

$x\left(t\right)=x_i+v_it+\frac{1}{2}at^2$

$v\left(t\right)=v_i+at,$

where the symbols are defined as follows:

• $x\left(t\right)$ is the position of the particle;
• $x_i$ is the initial position of the particle;
• $v\left(t\right)$ is the velocity of the particle;
• $v_i$ is the initial velocity of the particle;
• $a$ is the acceleration of the particle.

In answering the following questions, assume that the acceleration is constant and nonzero: a≠0.

PART B. The quantity represented by $x_i$$x_i$ is a function of time (i.e., is not constant).

Recall that $x_i$ represents an initial value, not a variable. It refers to the position of an object at some initial moment.

PART D. The quantity represented by v is a function of time (i.e., is not constant).

The velocity v always varies with time when the linear acceleration is nonzero.

PART E. Which of the given equations is not an explicit function of t and is therefore useful when you don’t know or don’t need the time?

ANSWER: $v^2=\left(v_i\right)^2+2a\left(x-x_i\right)$

PART F. A particle moves with constant acceleration a. The expression vi+at represents the particle’s velocity at what instant in time?

ANSWER: when the time t has passed since the particle’s velocity was $v_i$

More generally, the equations of motion can be written as

$x\left(t\right)=x_i+v_i\Delta t+\frac{1}{2}a\left(\Delta t\right)^2$

and

$v\left(t\right)=v_i+a\Delta t$

Here Δt is the time that has elapsed since the beginning of the particle’s motion, that is, $\Delta t=t-t_i$, where $t$ is the current time and $t_i$ is the time at which we start measuring the particle’s motion. The terms $x_i$ and $v_i$ are, respectively, the position and velocity at $t=t_i$. As you can now see, the equations given at the beginning of this problem correspond to the case $t_i=0$, which is a convenient choice if there is only one particle of interest.

What is the equation describing the position of particle B?

ANSWER: $x_B\left(t\right)=x_i+0.5v_i\left(t-t_1\right)+a\left(t-t_1\right)^2$

The general equation for the distance traveled by particle B is $x_B\left(t\right)=x_{iB}+v_{iB}\Delta t+\frac{1}{2}a_B\left(\Delta t\right)^2$ or $x_B\left(t\right)=x_{iB}+v_{iB}\left(t-t_1\right)+\frac{1}{2}a_B\left(t-t_1\right)^2$, since $\Delta t=t-t_1$ is a good choice for B. From the information given, deduce the correct values of the constants that go into the equation for $x_B\left(t\right)$ given here, in terms of A’s constants of motion.

$x_B\left(t\right)=x_i+\frac{1}{2}v_i\left(t-t_1\right)+\frac{1}{2}\left(2a\right)\left(t-t_1\right)^2$

$x_B\left(t\right)=x_i+0.5v_i\left(t-t_1\right)+a\left(t-t_1\right)^2$

PART H. At what time does the velocity of particle B equal that of particle A?

ANSWER: $t=2t_1+\frac{v_i}{2a}$

Particle A’s velocity as a function of time is $v_A\left(t\right)=v_i+at$, and particle B’s velocity as a function of time is $v_B\left(t\right)=0.5v_i+2a\left(t-t_1\right)$.

Once you have expressions for the velocities of A and B as functions of time, set them equal and find the time t at which this happens.

$v_i+at=\frac{1}{2}v_i+2a\left(t-t_1\right)$

$v_i+at=\frac{1}{2}v_i+2at-2at_1$

$2at-at=2at_1+\frac{1}{2}v_i$

$at=2at_1+\frac{1}{2}v_i$

$t=2t_1+\frac{v_i}{2a}$

What Velocity vs. Time Graphs Can Tell You| University Physics

PART A. What is the initial velocity of the particle, $v_0$$v_0$?

ANSWER: $v_0=0.5\:m/s$

The initial velocity is the velocity at t=0s. Recall that in a graph of velocity versus time, time is plotted on the horizontal axis and velocity on the vertical axis.

PART B. What is the total distance Δx traveled by the particle?

ANSWER: $\Delta x=75\:m$

Recall that the area of the region that extends over a time interval Δt under the v vs. t curve is always equal to the distance traveled in Δt. Thus, to calculate the total distance, you need to find the area of the entire region under the v vs. t curve. In the case at hand, the entire region under the v vs. t curve is not an elementary geometrical figure, but rather a combination of triangles and rectangles.

PART C. What is the average acceleration $a_{av}$$a_{av}$ of the particle over the first 20.0 seconds?

ANSWER: $a_{av}=0.075\:m/s^2$

The average acceleration of a particle between two instants of time is the slope of the line connecting the two corresponding points in a v vs. t graph.

PART D. What is the instantaneous acceleration $a$$a$ of the particle at $t=45.0\:s$$t=45.0\:s$ seconds?

ANSWER: $a=-0.20\:m/s^2$

The instantaneous acceleration of a particle at any point on a v vs. t graph is the slope of the line tangent to the curve at that point. Since in the last 10 seconds of motion, between t=40.0s and t=50.0s, the curve is a straight line, the tangent line is the curve itself. Physically, this means that the instantaneous acceleration of the particle is constant over that time interval. This is true for any motion where velocity increases linearly with time.

PART E. Which of the graphs shown below is the correct acceleration vs. time plot for the motion described in the previous parts?

Recall that whenever velocity increases linearly with time, acceleration is constant. In the example here, the particle’s velocity increases linearly with time in the first 20.0 s of motion. In the second 20.0 s , the particle’s velocity is constant, and then it decreases linearly with time in the last 10 s. This means that the particle’s acceleration is constant over each time interval, but its value is different in each interval.

In conclusion, graphs of velocity as a function of time are a useful representation of straight-line motion. If read correctly, they can provide you with all the information you need to study the motion.

Solution:

Part A

The initial velocity of the pot is zero. Find the velocity $\displaystyle v_b$ of the pot at the bottom of the window. Then using the kinematic equation that relates initial and final velocities, acceleration, and distance traveled, you can solve for the distance $\displaystyle h$.

The average velocity of the flower pot as it passes by the window is $\displaystyle v_{avg}=\frac{L_w}{t}$.

As the pot falls past your window, there will be some instant when the pot’s velocity equals the average velocity $\displaystyle v_{avg}$. Recall that, under constant acceleration, velocity changes linearly with time. This means that the average velocity during a time interval will occur at the middle of that time interval. Meaning, the average velocity happened at time, $\displaystyle \frac{t}{2}$.

Considering the motion at the middle and at the bottom of the window.

$\displaystyle g=\frac{v_b-v_{avg}}{\frac{t}{2}}$

$\displaystyle v_b=\frac{gt}{2}+v_{ave}$

$\displaystyle v_b=\frac{gt}{2}+\frac{L_w}{t}$

So, we now know that velocity at the bottom of the window. Consider the motion from the top (point of dropped) to the bottom of the window

$\displaystyle \left(v_b\right)^2-\left(v_0\right)^2=2g\left(h-0\right)$

$\displaystyle \left(v_b\right)^2=2gh$

$\displaystyle h=\frac{\left(v_b\right)^2}{2g}$

$\displaystyle h=\frac{\left(\frac{gt}{2}+\frac{L_w}{t}\right)^2}{2g}=\frac{\left(\frac{gt^2+2L_w}{2t}\right)^2}{2g}=\frac{\left(gt^2+2L_w\right)^2}{8gt^2}$

Part B

$\displaystyle \left(v_{ground}\right)^2-\left(v_b\right)^2=2gh_b$

$\displaystyle \left(v_{ground}\right)^2=2gh_b+\left(v_b\right)^2$

$\displaystyle\left(v_{ground}\right)=\sqrt{2gh_b+\left(v_b\right)^2}$

To demonstrate the tremendous acceleration of a top fuel Drag Racer| University Physics

PART A. What is $\displaystyle t_{max}$$\displaystyle t_{max}$, the longest time after the dragster begins to accelerate that you can possibly run into the back of the dragster if you continue at your initial velocity?

$\displaystyle t_{max}=\frac{v_0}{a}$

PART B. Assuming that the dragster has started at the last instant possible (so your front bumper almost hits the rear of the dragster at $\displaystyle t=t_{max}$$\displaystyle t=t_{max}$), find your distance from the dragster when he started. If you calculate positions on the way to this solution, choose coordinates so that the position of the drag car is 0 at t=0. Remember that you are solving for a distance (which is a magnitude, and can never be negative), not a position (which can be negative).

$\displaystyle D_{start}=\frac{\left(v_0\right)^2}{2a}$

$\displaystyle D_{start}=D_{car}-x_d\left(t_{max}\right)$

$\displaystyle D_{start}=v_0t_{max}-\frac{1}{2}a\left(t_{max}\right)^2$

$\displaystyle D_{start}=v_0\left(\frac{v_0}{a}\right)-\frac{1}{2}a\left(\frac{v_0}{a}\right)^2$

$\displaystyle D_{start}=\frac{\left(v_0\right)^2}{a}-\frac{\left(v_0\right)^2}{2a}$

$\displaystyle D_{start}=\frac{\left(v_0\right)^2}{2a}$

PART C. Find numerical values for $\displaystyle t_{max}$$\displaystyle t_{max}$ and $\displaystyle D_{start}$$\displaystyle D_{start}$ in seconds and meters for the (reasonable) values $\displaystyle v_0=60\:mph$$\displaystyle v_0=60\:mph$ (26.8 m/s) and $\displaystyle a=50\:m/s^2$$\displaystyle a=50\:m/s^2$

$\displaystyle t_{max}=\frac{v_0}{a}=\frac{26.8\:m/s}{50\:m/s^2}=0.54\:s$

$\displaystyle D_{start}=\frac{\left(v_0\right)^2}{2a}=\frac{\left(26.8\:m/s\right)^2}{2\left(50\:m/s^2\right)}=7.2\:m$

The blue curve shows how the car, initially at $\displaystyle x_0$, continues at constant velocity (blue) and just barely touches the accelerating drag car (red) at $\displaystyle t_{max}$.

Running and Walking Problem| University Physics

PART A. How long does it take Rick to cover the distance D?

Find the time that it takes Rick to walk the first half of the distance, that is, to travel a distance D/2 at speed $\displaystyle v_w$.

$\displaystyle t_{w,R}=\frac{D}{2v_w}$

Now find the time Rick spends running.

$\displaystyle t_{r,R}=\frac{D}{2v_r}$

Now just add the two times up and you’re done.

$\displaystyle t_R=\frac{D}{2v_w}+\frac{D}{2v_r}=\frac{D}{2v_wv_r}\left(v_w+v_r\right)$

PART B. Find Rick’s average speed for covering the distance D.

You were given the total distance and have calculated the total time. Recall that average speed is equal to the total distance traveled divided by the amount of time it took to travel this distance.

$\displaystyle v_{ave,\:R}=\frac{2v_rv_w}{v_w+v_r}$

PART C. How long does it take Tim to cover the distance?

Tim walks at speed $\displaystyle v_w$ half the time and runs at speed $\displaystyle v_r$ for the other half.

$\displaystyle v_{ave,\:T}=\frac{v_w+v_r}{2}$

The time is just the distance divided by the average speed.

$\displaystyle t_T=\frac{D}{\frac{v_w+v_r}{2}}=\frac{2D}{v_r+v_w}$

PART D. Who covers the distance D more quickly?

Imagine that both Rick and Tim do all of their walking before they start to run. Rick will start running when he has covered half of the total distance. When Tim reaches half of the total distance, will he already have started running?

PART E. In terms of given quantities, by what amount of time, Δt, does Tim beat Rick?

$\displaystyle \Delta t=\frac{D\left(v_w-v_r\right)^2}{2v_rv_w\left(v_r-v_w\right)}$

This is just simple subtraction between the two computed times.

PART F. In the special case that vr=vw, what would be Tim’s margin of victory Δt(vr=vw)?

If vr=vw, is the any difference between what Tim and Rick do?

Half the Distance and Half the Time Problem| University Physics

PART A. What is Julie’s average speed on the way to Grandmother’s house?

Julie drove 50 miles at a speed of 35 mph, and drove another 50 miles for 65 mph. So, for the first 50 miles, she drove for

$\displaystyle time=\frac{distance}{speed}=\frac{50\:mi}{35\:mph}=\frac{10}{7}\:hours$

and for the next 50 miles, she drove for

$\displaystyle time=\frac{distance}{speed}=\frac{50\:mi}{65\:mph}=\frac{10}{13}\:hours$

Therefore, her average speed was

$\displaystyle average\:speed=\frac{total\:distance}{total\:time}$

$\displaystyle average\:speed=\frac{100\:miles}{\frac{10}{7}+\frac{10}{13}\:hours}=45.5\:mph$

PART B. What is her average speed on the return trip?

Since the time she used driving at 35 mph is the same amount of time she used driving at 65 mph, the average speed is just the average of the two speeds given.

Position, Velocity, and Acceleration values from Position Function| University Physics

PART A.

Evaluate the position at time t= 3.00 s.

$\displaystyle x=2.00\left(3\right)^3-5.00\left(3\right)+3=42.0\:m$

PART B.

Determine the velocity function v(t) from the position function x(t) by differentiation.

$\displaystyle v\left(t\right)=6t^2-5$

$\displaystyle v\left(t\right)=6\left(3\right)^2-5=49.0\:m/s$

PART C.

$\displaystyle a\left(t\right)=12t$
$\displaystyle a\left(t\right)=12t=12\left(3\right)=36\:m/s^2$


text:
Practice on Toph

Participate in exhilarating programming contests, solve unique algorithm and data structure challenges and be a part of an awesome community.

Fight Club Restored

By rifatrraazz · Limits 1s, 512 MB

The things you own, end up owning you. It’s only after we’ve lost everything that we’re free to do anything.

Tyler Durden’s Fight Club has been restored by Shahriar. Let’s walk through it.

There are $n$ fighters numbered $1$ to $n$. The strength of the $i$th fighter is $S_i$. In a battle between two fighters, the fighter with the higher strength will win. If the strength of the two fighters is equal, the fighter with the smaller number will win.

Of course, you do not talk about fight club. Besides, there are some new rules for the fight club as follows:

• There will be fighting battles for $n$ days and people can bet on the fighters.

• On the $i$th day ${(1 \le i \le n)}$, anyone can bet on the $i$th fighter where the $i$th fighter, on that day, will fight with ${i+1, i+2, \dots, n}$th fighters.

• If $i$th fighter wins a battle, the bettor will win $1$ taka (the currency of the fight club) and if the fighter loses, the bettor will lose $1$ taka.

The bettor can not bet for a single day or consecutively $3$ days. In other words, if a person bets on the $i$th day, then the person has to bet on the $(i+1)$th day and they can not bet on the $(i+2)$th day.

Now, Sifat, one of the bettors, wants to bet for maximum $k$ days and you want to join the fight club. Shahriar will let you join if you can find the maximum amount of money (in taka) Sifat can win by betting not more than $k$ days.

Input

The first line of the input contains an integer, $t$ ${(1 \le t \le 10^4)}$ denoting the number of test cases.

The first line of the each test case contains two integers, $n$ ${(1 \le n \le 3\times 10^5)}$ and $k$ ${(1 \le k \le min(200, n))}$ denoting the number of fighters (battle days too) and number of maximum days Sifat will bet respectively.

The second line of the each test case contains $n$ space-separated integers ${S_1, S_2, \dots, S_n}$ ${(1 \le S_i \le 10^9)}$ denoting strength of each fighter.

It is guaranteed that sum of $n$ over all test cases will not exceed ${3 \times 10^5}$.

Output

For each testcase, print an integer denoting the maximum amount of money (in taka) Sifat can win if he has no money initially.

Sample

InputOutput
3
1 1
12
5 5
7 5 1 3 3
4 2
10 121 1111 111

0
8
1

In the first test case, as there are only one fighter, no battle can happen, Sifat can’t win any money.

In the second test case, by betting on $1$st day, Sifat can win $4$ taka. As, he bets on $1$st day, he has to bet on $2$nd day and he can’t bet on $3$rd day. So, by betting on $2$nd day, he can win $3$ taka. Then, by betting on $4$th day, he can win $1$ taka, by betting on $5$th day (he bets on $4$th day, so he has to bet on $5$th day), he can’t won any money as no battle can happen and as $6$th day doesn’t exist, it doesn’t matter. So, by betting in total $4$ days (which is less than given $k$) Sifat can win total ${(4+3+1)=8}$ taka. It can be shown that it is the maximum amount.

In the third test case, Sifat bets on maximum $2$ days. So, by betting on $2$nd and $3$rd day, he can win maximum $1$ taka.

Statistics

83% Solution Ratio

BrehamPieEarliest, 1M ago

S_SubrataFastest, 0.2s

S_SubrataLightest, 9.3 MB

serotoninShortest, 958B


text:
0
Research Papers

# The Tip Region of a Near-Surface Hydraulic Fracture

[+] Author and Article Information
Zhi-Qiao Wang

School of Engineering and Technology,
China University of Geosciences,
Beijing 100086, China

Emmanuel Detournay

Department of Civil, Environmental, and
Geo-Engineering,
University of Minnesota,
Minneapolis, MN 55455
e-mail: detou001@umn.edu

1Corresponding author.

Manuscript received January 8, 2018; final manuscript received January 10, 2018; published online February 15, 2018. Editor: Yonggang Huang.

J. Appl. Mech 85(4), 041010 (Feb 15, 2018) (11 pages) Paper No: JAM-18-1014; doi: 10.1115/1.4039044 History: Received January 08, 2018; Revised January 10, 2018

## Abstract

This paper investigates the tip region of a hydraulic fracture propagating near a free-surface via the related problem of the steady fluid-driven peeling of a thin elastic layer from a rigid substrate. The solution of this problem requires accounting for the existence of a fluid lag, as the pressure singularity that would otherwise exist at the crack tip is incompatible with the underlying linear beam theory governing the deflection of the thin layer. These considerations lead to the formulation of a nonlinear traveling wave problem with a free boundary, which is solved numerically. The scaled solution depends only on one number $K$, which has the meaning of a dimensionless toughness. The asymptotic viscosity- and toughness-dominated regimes, respectively, corresponding to small and large $K$, represent the end members of a family of solutions. It is shown that the far-field curvature can be interpreted as an apparent toughness, which is a universal function of $K$. In the viscosity regime, the apparent toughness does not depend on $K$, while in the toughness regime, it is equal to $K$. By noting that the apparent toughness represents an intermediate asymptote for the layer curvature under certain conditions, the obtention of time-dependent solutions for propagating near-surface hydraulic fractures can be greatly simplified. Indeed, any such solutions can be constructed by a matched asymptotics approach, with the outer solution corresponding to a uniformly pressurized fracture and the inner solution to the tip solution derived in this paper.

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## References

Young, C. , 1999, “Controlled-Foam Injection for Hard Rock Excavation,” 37th U.S. Rock Mechanics Symposium Rock Mechanics for Industry (USRMS), Vail, CO, June 7–9, pp. 115–122.
Jeffrey, R. G. , and Mills, K. W. , 2000, “Hydraulic Fracturing Applied to Inducing Longwall Coal Mine Goaf Falls,” 4th North American Rock Mechanics Symposium, July 31–Aug. 3, Seattle, WA, pp. 423–430.
Murdoch, L. C. , 2002, “Mechanical Analysis of Idealized Shallow Hydraulic Fracture,” J. Geotech. Geoenviron., 128(6), pp. 488–495.
Bunger, A. P. , and Cruden, A. R. , 2011, “Modeling the Growth of Laccoliths and Large Mafic Sills: Role of Magma Body Forces,” J. Geophys. Res.: Solid Earth, 116(B2), p. B02203.
Michaut, C. , 2011, “Dynamics of Magmatic Intrusions in the Upper Crust: Theory and Applications to Laccoliths on Earth and the Moon,” J. Geophys. Res.: Solid Earth, 116(B5), p. B05205.
Tayler, A. B. , and King, J. R. , 1987, “Free Boundaries in Semi-Conductor Fabrication,” Free Boundary Problems: Theory and Applications (Pitman Research Notes in Mathematics, Vol. 1), Longman Scientific & Technical, Harlow, UK, pp. 243–259.
Hosoi, A. E. , and Mahadevan, L. , 2004, “Peeling, Healing, and Bursting in a Lubricated Elastic Sheet,” Phys. Rev. Lett., 93(13), p. 137802. [PubMed]
Lister, J. R. , Peng, G. G. , and Neufeld, J. A. , 2013, “Viscous Control of Peeling an Elastic Sheet by Bending and Pulling,” Phys. Rev. Lett., 111(15), p. 154501. [PubMed]
Zhang, X. , Detournay, E. , and Jeffrey, R. G. , 2002, “Propagation of a Penny-Shaped Hydraulic Fracture Parallel to the Free-Surface of an Elastic Half-Space,” Int. J. Fract., 115(2), pp. 125–158.
Zhang, X. , Jeffrey, R. G. , and Detournay, E. , 2005, “Propagation of a Hydraulic Fracture Parallel to a Free Surface,” Int. J. Numer. Anal. Methods Geomech., 29(13), pp. 1317–1340.
Gordeliy, E. , and Detournay, E. , 2011, “A Fixed Grid Algorithm for Simulating the Propagation of a Shallow Hydraulic Fracture With a Fluid Lag,” Int. J. Numer. Anal. Methods Geomech., 35(5), pp. 602–629.
Bunger, A. P. , and Detournay, E. , 2005, “Asymptotic Solution for a Penny-Shaped Near-Surface Hydraulic Fracture,” Eng. Fract. Mech., 72(16), pp. 2468–2486.
Hewitt, I. J. , Balmforth, N. J. , and De Bruyn, J. R. , 2015, “Elastic-Plated Gravity Currents,” Eur. J. Appl. Math., 26(1), pp. 1–31.
Flitton, J. C. , and King, J. R. , 2004, “Moving-Boundary and Fixed-Domain Problems for a Sixth-Order Thin-Film Equation,” Eur. J. Appl. Math., 15(6), pp. 713–754.
Lister, J. R. , 1990, “Buoyancy-Driven Fluid Fracture: The Effects of Material Toughness and of Low-Viscosity Precursors,” J. Fluid Mech., 210, pp. 263–280.
Desroches, J. , Detournay, E. , Lenoach, B. , Papanastasiou, P. , Pearson, J. R. A. , Thiercelin, M. , and Cheng, A. H.-D. , 1994, “The Crack Tip Region in Hydraulic Fracturing,” Proc. R. Soc. London, Ser. A, 447(1929), pp. 39–48.
Garagash, D. I. , and Detournay, E. , 2000, “The Tip Region of a Fluid-Driven Fracture in an Elastic Medium,” ASME J. Appl. Mech., 67(1), pp. 183–192.
Garagash, D. I. , Detournay, E. , and Adachi, J. I. , 2011, “Multiscale Tip Asymptotics in Hydraulic Fracture With Leak-Off,” J. Fluid Mech., 669, pp. 260–297.
Dontsov, E. V. , and Peirce, A. P. , 2015, “A Non-Singular Integral Equation Formulation to Analyse Multiscale Behaviour in Semi-Infinite Hydraulic Fractures,” J. Fluid Mech., 781, p. R1.
Lecampion, B. , and Detournay, E. , 2007, “An Implicit Algorithm for the Propagation of a Hydraulic Fracture With a Fluid Lag,” Comput. Meth. Appl. Mech. Eng., 196(49–52), pp. 4863–4880.
Detournay, E. , and Peirce, A. P. , 2014, “On the Moving Boundary Conditions for a Hydraulic Fracture,” Int. J. Eng. Sci., 84, pp. 147–155.
Detournay, E. , 2016, “Mechanics of Hydraulic Fractures,” Annu. Rev. Fluid Mech., 48(1), pp. 311–339.
Timoshenko, S. , 1958, Strength of Materials. Part1: Elementary Theory and Problems, 3rd ed., Van Nostrand Reinhold, New York.
Batchelor, G. K. , 1967, An Introduction to Fluid Dynamics, Cambridge University Press, Cambridge UK.
Hutchinson, J. W. , and Suo, Z. , 1991, “Mixed Mode Cracking in Layered Materials,” Adv. Appl. Mech., 29, pp. 63–191.
Li, S. , Wang, J. , and Thouless, M. D. , 2004, “The Effects of Shear on Delamination in Layered Materials,” J. Mech. Phys. Solids, 52(1), pp. 193–214.
Seifert, U. , 1991, “Adhesion of Vesicles in Two Dimensions,” Phys. Rev. A, 43(12), p. 6803. [PubMed]
Majidi, C. , and Adams, G. G. , 2009, “A Simplified Formulation of Adhesion Problems With Elastic Plates,” Proc. R. Soc. London, Ser. A, 465(2107), pp. 2217–2230.
Majidi, C. , O'Reilly, O. M. , and Williams, J. A. , 2012, “On the Stability of a Rod Adhering to a Rigid Surface: Shear-Induced Stable Adhesion and the Instability of Peeling,” J. Mech. Phys. Solids, 60(5), pp. 827–843.
Dyskin, A. V. , Germanovich, L. N. , and Ustinov, K. B. , 2000, “Asymptotic Analysis of Crack Interaction With Free Boundary,” Int. J. Solids Struct., 37(6), pp. 857–886.
Cook, R. D. , Malkus, D. S. , Plesha, M. E. , and Witt, R. J. , 2007, Concepts and Applications of Finite Element Analysis, Wiley, New York.

## Figures

Fig. 1

Semi-infinite near-surface hydraulic fracture

Fig. 2

Lag Λms versus K in the ms-scaling

Fig. 3

Far-field curvature Ω″ms(∞) versus K in ms-scaling

Fig. 4

Aperture at the fluid front, Ωms(Λms,K) versus K in the ms-scaling

Fig. 5

Aperture profile in ms-scaling

Fig. 6

Profiles of the curvature in Log-Log scales

Fig. 7

Net pressure profile in ms-scaling

Fig. 8

Bounds of limiting solutions

Fig. 9

Bounds of limiting solutions

Fig. 10

Mesh and beam element for the numerical scheme

Fig. 11

Convergence of far-field curvature with length of channel

Fig. 12

Convergence of lag with discretization length Δξ¯

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text:
# Half Angle formula

## Sine of a Half Angle

$\LARGE \sin \frac{a}{2} = \pm \sqrt{\frac{(1- \cos a)}{2}}$

## Cosine of a Half Angle

$\LARGE \cos \frac{a}{2} = \pm \sqrt{\frac{(1+ \cos a)}{2}}$

## Tangent of a Half Angle

$\LARGE \tan \left ( \frac{a}{2} \right ) = \frac{1 – \cos a}{\sin a} = \frac{\sin a}{1 + \cos a}$

### Solved Examples

Question 1: Find the value of Sin 30 degrees using the sine half-angle formula.
Solution:

Given a=60 degreesUsing sine half angle formula,

$sin$ $\frac{a}{2}$ = $\pm$ $\sqrt{\frac{1+cos(a)}{2}}$

$sin$ $\frac{60}{2}$ = $\pm$ $\sqrt{\frac{1+cos(60)}{2}}$

$sin 30°$ = $\pm$ $\sqrt{\frac{1+cos 60}{2}}$

$sin 30°$ = 0.1542

 Related Formulas Geometric Series Formula Implicit Differentiation Formula Inverse Hyperbolic Functions Formula Linear Regression Formula Linear Approximation Formula Maclaurin Series Formula Orthocenter Formula Profit Margin Formula


text:
FangQuant › Daily Morning

### Daily Market Review on Specified Futures Products 2020.09.16

submitted 2020-09-16 10:16:30

Crude oil

Recently, the three major institutions have issued September reports. The three major institutions have all lowered their demand growth expectations for this year. The epidemic’s drag on the demand side is still lingering. The three major institutions have slightly revised their supply growth this year. Overall, the global oil market is still in a fragile balance, the second outbreak of the epidemic or the restart of Libya's supply may cause the current tight balance between supply and demand to be broken, and whether OPEC, as the current market balancer, can make timely policy adjustment remains to be seen.

Strategy: Neutral and bearish relatively, reverse cash and carry arbitrage strategy on Brent, long the sixth lines and short the first line

Risk: Supply disruption caused by sudden geopolitical events. The dollar continues to depreciate sharply.

Iron Ore

The position on I2101 contract increased by 2,338 lots and closed at ¥824 per ton, the position on I2105 contract increased by 5,037 lots and closed at ¥761 per ton.

Important Information

1. According to Mysteel's data, this week (September 14 to September 20) Australia and Brazil have some iron ore shipping berths undergoing maintenance. It is estimated that the total impact of Australian berth maintenance this week is about 1.53 million tons. In addition, berth 5 of Geraldton Port may be affected by sea waves on the 15th. It is estimated that the amount of berth maintenance in Brazil this week will affect about 988,000 tons.

2. China Iron and Steel Association: In early September, key statistical iron and steel enterprises produced a total of 21,365,000 tons of crude steel, 18,966,000 tons of pig iron, 19,993,300 tons of steel, and 3,192,800 tons of coke. The average daily output of crude steel was 2,136,600 tons, a decrease of 1.09% month-on-month and an increase of 6.73% year-on-year; the average daily output of pig iron was 1,896,200 tons, a decrease of 1.55% month-on-month and an increase of 4.82% year-on-year; the average daily output of steel was 1,999,300 tons, a decrease of 8.86% from the previous month. A year-on-year increase of 6.50%.

4. In terms of spot, the PB powder in Rizhao Port was ¥938 per ton, golden bubba powder price was equivalent to ¥1,002 per ton.

1. Arbitrage: This week, 15 port inventory increased by 780,000 tons, MNPJ has accumulated inventory, PB powder inventory increased by 680,000 tons. At present, the overall delivery of Australia and Brazil is relatively high. Recently, iron ore in ports has begun to accumulate significantly. Sintered coarse powder, especially Australian ore, has begun to accumulate slightly. Structural contradictions have improved. The time for the steel with a high inventory is tight. There is a reduction expectation of pig iron production in the off-season. The iron ore inventory is under accumulation pressure in the off-season, and the basis of the 01 contract is also relatively large. For unilateral strategy on 01 iron ore, it is recommended to wait and see. As to arbitrage strategy, the 1-5 reverse cash and carry arbitrage strategy is recommended.

2. Option strategy: It is advised to hold the short position on i2101-C-950.

PTA

Polyester production and sales dropped once again

In September, it was the first time to realize destocking if all overhauls are fulfilled, but the absolute number of inventories will still be high after the destocking; in October, if the overhauls are slow, there will be an expectation of inventory accumulation, and if all the overhauls are fulfilled, the inventory will decrease in stage. In terms of the unilateral strategy, it is advised to be neutral; for the strategy across varieties, the high inventory problem has not been resolved yet, and it is not advised to maintain the strategy across varieties, but the willingness of the upstream factory to maintain and control should still be judged based on the change in processing fees; for strategy across period, it is advised to focus on the reverse cash and carry strategy opportunity after the next round of TA overhauls fullfilling and rebound of 1-5 spread. It is advised to focus on PTA factory inspection and fulfillment wishes, and the downstream restocking space and improvement of demand.

Natural Rubber

RU: The main force contract of RU01 rose by 40 or 0.32% and closed at 12,400. The main force contract of JRU02 rose by 4.1 or 2.29% and closed at 183.2. Yunnan WF closed at 11,500 to 11,600 yuan per ton, Hainan SCRWF closed at 11,700 to 11,750 yuan per ton, the secondary standard rubber closed at 10,600 per ton, and Thailand’s RSS3 closed at 15,300 to 15,850 yuan per ton.

NR: The main force contract of NR11 rose by 10 or 0.11% and closed at 9,355. The main force contract of TF12 rose by 1.6 or 1.17% closed at 138.0. The quoted price for Qingdao rubber in USD retreated. The CIF of STR20 in December was $1,440 to$1,450 per ton. The CIF of SMR20 was $1,390 to$1,395 per ton. The CIF of mixed rubber from Thailand in December was $1,405 to$1,410 per ton.

TireWorld News: Recently, the US tire import data report for the first half of the year was released. From January to June, due to the impact of the new coronavirus epidemic, the import volume of passenger car tires and commercial vehicle tires in the United States both showed a double-digit decline. The decline in imports of passenger car tires and truck and bus tires in the United States was 23.7% and 14.7%, respectively.

Today coincides with Malaysia Day in Malaysia, and the local market is closed. The domestic latex in Yunnan area reportedly closed at 10,300 to 10,700 yuan/ton, a premium of about 500 to 600 yuan/ton to cup lump. The premium was the highest level this year, almost the same as last year's high. Export orders for tire production lines are well, and individual brands are even expected to catch up with the same period last year.

Futures Operation Advice: For the main RU01 contract, it is advised to hold a slight long position and set a stop at 12,270. (For reference only).


text:
Tag Archives: puzzles

Compensating Errors

The class had been given a sum to do, involving three positive whole numbers (‘positive’ here means ‘greater than zero). During the break, two classmates compared notes.

“Oops, I added the three numbers instead of multiplying them,” said George.

“You are lucky, then,” said Henrietta. “It’s the same answer either way.”

What were the three numbers? What would they have been if there had been only two of them, or four of them, again with their sum equal to their product?”

More later,

Nalin Pithwa

Digital Century

Place exactly three common mathematical symbols between the digits

$1 \hspace{0.1in} 2 \hspace{0.1in} 3 \hspace{0.1in} 4 \hspace{0.1in} 5 \hspace{0.1in} 6 \hspace{0.1in} 7 \hspace{0.1in} 8 \hspace{0.1in} 9$

so that the result equals 100. The same symbol can be repeated if you wish, but each repeat counts towards your limit of three. Rearranging the digits is not permitted.

Have fun 🙂

More fun later,

Nalin Pithwa

Heron Suit

No cat that wears a heron suit is unsociable.

No cat without a tail will play with a gorilla.

Cats with whiskers always wear heron suits.

No sociable cat has blunt claws.

No cats have tails unless they have whiskers.

Therefore.

No cat with blunt claws will play with a gorilla.

Is the deduction logically correct?

Nalin Pithwa

Puzzling poem

One of my students Anubhav C Singh shared the following puzzling poem he had found from the internet:

Ten weary, footsore travellers,
All in a woeful plight,
Sought shelter at a wayside inn
One dark and stormy night.

‘Nine rooms, no more,’ the landlord said
‘Have I to offer you.
To each of eight a single bed,
But the ninth must serve for two.’

A din arose. The troubled host
For of those tired men not two
Would occupy one bed.

The puzzled host was soon at ease –
He was a clever man –
And so to please his guests devised
This most ingeneous plan.

In a room marked A two men were placed,
The third was lodged in B,
The fourth to C was then assigned,
The fifth retired to D.

In E the sixth he tucked away,
In F the seventh man.
The eighth and ninth in G and H,
And then to A he ran,

Wherein the host, as I have said,
Then taking one – the tenth and last –
He logged him safe in I.

Nine singe rooms – a room for each –
Were made to serve for ten;
And this it is that puzzles me
And many wiser men.

What is wrong in this logic?
More later,
Nalin Pithwa

Deceptive Dice

The Terrible Twins, Innumeratus and Mathophila, were bored.

“I know,” said Mathophila brightly. “Let’s play dice!”

“Don’t like dice.”

“Ah, but these are special dice,” said Mathophila, digging them out of an old-chocolate box. One was red, one yellow and one blue..

Innumeratus picked up the red dice. “There’s something funny about this one,” he said. “It’s got two 3’s, two-4’s and two 8’s.”

“They are all like that,” said Mathophila carelessly. The yellow one has two 1’s, two 5’s and two 9’s — and the blue  one has two 2’s, two 6’s and two 7’s.

“They look rigged to me,” said Innumeratus, deeply suspicious.

“No, they are perfectly fair. Each face has an equal chance of turning up.”

“How do we play, anyway?”

We each choose a different one. We roll them simultaneously, and the highest number wins. We can play for pocket money. Innumeratus looked skeptical, so his sister quickly added: “Just to be fair, I will let you choose first! Then you can choose the best dice!”

“Welllll…”, said Innumeratus, hesitating.

“Should he play? If not, why not?”

More puzzles to keep you entertained later,

Nalin Pithwa

An Age-Old Old-Age Problem

The Emperor Scrumptius was born in 35 BC, and died on his birthday in AD35. What was his age when he died?

More later,

Nalin Pithwa

White-Tailed Cats

“I see you’ve got a cat.” said Ms Jones to Ms Smith. “I do like its cute white tail! How many cats do you have?”

“Not a lot,” said  Ms Smith. “Ms Brown next door has twenty, which is a lot more than I’ve got.”

“You still haven’t told me how many cats you have?”

“Well…let me put it like this. If you chose two distinct cats of mine at random, the probability that both of them have white tails is exactly half.”

“That doesn’t tell me how many you’ve got!”

“Oh yes it does.”

How many cats does Ms Smith have — and how many have white tails?

More later,

Nalin Pithwa

Three Math Quickies

1. After four bridge hands have been dealt, which is the more likely: that you and your partner hold all the spades, or that you and your partner hold no spades?
2. If you took three bananas from a dish holding thirteen bananas, how many bananas would you have ?
3. A secretary prints out six letters from the computer and addresses six envelopes to their intended recipients. Her boss, in a hurry, interferes and stuffs the letters into the envelopes at random, one letter in each envelope. What is the probability that exactly five letters are in the right envelope?

More fun later 🙂

Nalin Pithwa

If the Indiana State Legislature had passed Bill 246, and if the worst-case scenario had proved legally valid, namely that the value of $\pi$ in law was different from its mathematical value, the consequences would have been distinctly interesting. Suppose that the legal value is $p \neq \pi$, but the legislation states that $p=\pi$. Then,

$\frac{p-\pi}{p-\pi}$ mathematically, $\frac{p-\pi}{p-\pi}=0$ legally,

Since mathematical truths are legally valid, the law could then be maintaining that $1=0$. Therefore, all murderers have a cast-iron defence: admit to one murder, then argue that legally it is zero murders. And, that’s not the last of it. Multiply by one billion, to deduce that one billion equals zero. Now any citizen apprehended in possession of no drugs is in possession of drugs to a street value of \$ 1 blilion.

In fact, any statement whatsover would become legally provable.

It seems likely that the Law would not be quite logical enough for this kind of argument to hold up in court. But sillier legal statements, often based on abuse of statistics, have done just that, causing innocent people to be locked away for long periods. So, Indiana’s legislator’s might have opened up Pandora’s box.

More later,

Nalin Pithwa

What is pi?

What is $\pi$?

The number $\pi$, which is approximately 3.14159, is the length of the circumference of a circle whose diameter is exactly 1. More generally, a circle of diameter d has a circumference $\pi d$. A simple approximation to $\pi$ is 22/7, but this is not exact. 22/7 is approximately 3.14285, which is wrong by the third decimal place. A better approximation is 355/113, which is 3.1415929 to seven places, whereas $\pi$ is 3.1415926.

How do we know that $\pi$ is not an exact fraction? However much you continue to improve the approximation x/y, by using ever larger numbers, you can never get to $\pi$ itself, only better and better approximations. A number that cannot be written exactly as a fraction is said to be irrational. The simplest proof that $\pi$ is irrational uses calculus, and it was found by Johann Lambert in 1770. Although we can’t write down an exact numerical representation of $\pi$, we can write down various formulas that define it precisely, and Lambert’s proof uses one of those formulas.

More strongly, $\pi$ is transcendental — it does not satisfy any algebraic equation that relates it to rational number. This was proved by Ferdinand Lindemann in 1882, also using calculus.

This fact that $\pi$ is transcendental implies that the classical geometric problem of ‘squaring the circle’ is impossible. This problem asks for a Euclidean construction of a square whose area is equal to that of a given circle (which turns out to be equivalent to constructing a line whose length is the circumference of the circle). A construction is called Euclidean if it can be performed using an unmarked ruler and a compass. Well, to be pedantic, a ‘pair of compasses’, which means a single instrument, much as a ‘pair of scissors’ comes as one gadget.

More later,

Nalin Pithwa


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# Multiscale Strategies for Computing Optimal Transport

8 Aug 2017

This paper presents a multiscale approach to efficiently compute approximate optimal transport plans between point sets. It is particularly well-suited for point sets that are in high-dimensions, but are close to being intrinsically low-dimensional... (read more)

PDF Abstract


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# Help needed with first-order logic semantics: true or false

I'm having some difficulties on solving the tasks below. I would really appreciate some help.

For each of the following formulas, give a model which makes the formula true. Let the domain be $\{$$1, 2$$\}$. It is sufficient to provide the interpretation of the relation symbols $R$.

1. $\forall x \forall y Rxy$

2. $\exists x \forall y Rxy$

3. $\forall x \exists y Rxy \land \lnot \exists xRxx$

4. $\exists x \exists y (Rxy \land \lnot Ryx) \land \forall xRxx$

I would appreciate if someone could show me or help me on how I can solve these tasks.

Thank you.

• It might help if you write "$\forall x\forall y (xRy)$" and so on instead of "$\forall x\forall yRxy$" for now: the answers might come to you then. – Shaun Nov 3 '13 at 16:02
• For example, with $x, y\in \{1, 2\}$, the formula $\forall x\forall y (xRy)$ should ring a bell. – Shaun Nov 3 '13 at 16:06
• Don't forget that here $R\subseteq\{1, 2\}^{2}$. – Shaun Nov 3 '13 at 16:12
• Could you please solve the first task for me, just so I can see how it's done? – Dabbish Nov 3 '13 at 19:16

Since the OP asked that I answer, I'll give it a try. Here are my notes (i'll try to latex this later):

The graphs (top-bottom) correspond to formulas (1-4). Here is a procedure for generating them:

1. Given the domain consisting of 1 and 2, we start by drawing two circles labeled '1' and '2'.

2. Then draw the arrows (the relation R) depending on what each formula says. Lastly:

3. Either find a nice natural relation on numbers that gives us those arrows,

4. Or define an ugly relation by simply gathering the related pairs into a set.

Example 1. Formula (1) says: $\forall x \forall y Rxy$, i.e., there is an arrow linking any two objects. So we draw arrows from 1 to 2, from 2 to 1, and also we must not forget to draw arrows from 1 to 1 and from 2 to 2 (because x and y can be the same object). Once this is done, you get the first graph shown above. I have written 'A' instead of a solution so that you can try to solve it.

Example 2. Formula (2) says: $\exists x \forall y Rxy$, i.e., there is some circle that has an arrow going to every circle (including itself!). This corresponds to the second graph above. (Note that there could be arrows going out of 2 to 1 or to itself, but we don't care about those, because the arrows from 1 already satisfy formula (2).) Again, once the graph is drawn, either find a nice formula (I've called it 'B') or define R as an ugly set.

I'll let you make sense of graphs 3 and 4, but let me just point out what makes graphs 2 and 4 different: formula (4) forces R to be asymmetric and reflexive, while (2) doesn't care about symmetry and doesn't impose reflexivity (that's why there is no need for an arrow, in graph 2, from 2 to 2).

Your task. Is to find the nice numerical relations A, B, C, and D (hint: they're all very simple). If you're having trouble, Peter Smith's post contains very helpful hints; for example, he talks of a relation that always holds (hint: can you think of a natural reflexive relation between numbers?). If the hints don't help, let me know and I'll tell you what I have come up with for A, B, C, and D.

$$\fbox{Answers}$$

For the A graph, I have: $Rxy \equiv (x = x)$. Notice that there is an arrow going from every point to every point. This means that the relation, in this model, holds everywhere. So any relation that always holds will do the job of R. I have chosen '$x = x$'. You can test it out: plug in any pairs of {1,2} for x and y into Rxy. Since the value of y is not used inside the relation, there are two possibilities: $(1 = 1)$ and $(2 = 2)$, and both hold.

For the B graph, I have: $Rxy \equiv (x \le y)$. Had there been no reflexive loop on 1, we could just use less than, but since we also need to satisfy $(1 R 1)$, we have to add the case where x and y are identical.

For the C graph, I have: $Rxy \equiv (x \not= y)$. I chose this because of the fact that there is an arrow only between any two distinct objects.

For the D graph, I have chosen one of the other three answers. I'll let you figure out which. After you do, just edit the answer into this post, or leave a comment and I'll update it.

• thanks alot! I would appreciate if you could tell me what you have come up with for $A, B, C$ and $D$ Great answer! – Dabbish Nov 4 '13 at 13:04
• Sure. I'll update the post. – Hunan Rostomyan Nov 4 '13 at 21:53

One preliminary point. None of the wffs involve free variables, so the question title is either misleading, or betrays a misunderstanding. So (A) first make sure you understand why no variable is bound.

You have a two-element domain. (B) What are some examples of a relation $R$ on this two element domain? [You can specify a relation by specifying which pairs of object it holds between.]

Supplementary question: (C) How many different relations on a two element domain can there be [assuming that we count $R$ and $S$ as the same relation if they obtain between just the same pairs of objects]? Don't forget the relation which never holds, and the relation which always holds!

OK, you've now got some examples to play with. (D) Can you now see how to choose relations which make (1) and (2) true?

(E) Do you see how to parse the wffs (3) and (4)? What is the main logical operator? Do you see they are both conjunctions of simpler wffs. So try tackling the simpler parts first ... and seeing if you can get them to be simultaneously true.

• Could you please solve the first task for me just so I can see how it's done? Thank you. – Dabbish Nov 3 '13 at 18:58
• What does it say? -- that for every $x$ and every $y$, $Rxy$. So you need arelation $R$ that holds between any pair of objects in the domain. A relation whose extension is (1,1), (1, 2), (2, 1), (2, 2). .... – Peter Smith Nov 3 '13 at 20:07
• The answer to $(1)$ would that be $R^m = 2$? – Dabbish Nov 4 '13 at 1:36