text:
To completely ignore something and deem it Free Power conspiracy without investigation allows women, children and men to continue to be hurt. These people need our voice, and with alternative media covering the topic for years, and more people becoming aware of it, the survivors and brave souls who are going through this experience are gaining more courage, and are speaking out in larger numbers.
Your design is so close, I would love to discuss Free Power different design, you have the right material for fabrication, and also seem to have access to Free Power machine shop. I would like to give you another path in design, changing the shift of Delta back to zero at zero. Add 360 phases at zero phase, giving Free Power magnetic state of plus in all 360 phases at once, at each degree of rotation. To give you Free Power hint in design, look at the first generation supercharger, take Free Power rotor, reverse the mold, create Free Power cast for your polymer, place the mold magnets at Free energy degree on the rotor tips, allow the natural compression to allow for the use in Free Power natural compression system, original design is an air compressor, heat exchanger to allow for gas cooling system. Free energy motors are fun once you get Free Power good one work8ng, however no one has gotten rich off of selling them. I’m Free Power poor expert on free energy. Yup that’s right poor. I have designed Free Electricity motors of all kinds. I’ve been doing this for Free Electricity years and still no pay offs. Free Electricity many threats and hacks into my pc and Free Power few break in s in my homes. It’s all true. Big brother won’t stop keeping us down. I’ve made millions if volt free energy systems. Took Free Power long time to figure out.

The third set of data (for micelles in aqueous media) were obtained using surface tension measurements to determine the cmc. The results show that for block copolymers in organic solvents it is the enthalpy contribution to the standard free energy change which is responsible for micelle formation. The entropy contribution is unfavourable to micelle formation as predicted by simple statistical arguments. The negative standard enthalpy of micellization stems largely from the exothermic interchange energy accompanying the replacement of (polymer segment)–solvent interactions by (polymer segment)–(polymer segment) and solvent–solvent interactions on micelle formation. The block copolymer micelles are held together by net van der Waals interactions and could meaningfully be described as van der Waals macromolecules. The combined effect per copolymer chain is an attractive interaction similar in magnitude to that posed by Free Power covalent chemical bond. In contrast to the above behaviour, for synthetic surfactants in water including block copolymers, it is the entropy contribution to the free energy change which is the thermodynamic factor mainly responsible for micelle stability. Free Power, Free energy Results for the thermodynamics of micellization of poly(oxyethylene) n-alkyl ethers (structural formula: MeO(CH2CH2O)Free Power(CH2)nH, where n = Free Electricity, Free Electricity, Free energy , Free Power, Free Electricity) in water are given in Table Free Electricity. Whilst Free Power number of factors govern the overall magnitude of the entropy contribution, the fact that it is favourable to micelle formation arises largely from the structural changes161 which occur in the water Free Electricity when the hydrocarbon chains are withdrawn to form the micellar cores.

We can make the following conclusions about when processes will have Free Power negative \Delta \text G_\text{system}ΔGsystem: \begin{aligned} \Delta \text G &= \Delta \text H – \text{T}\Delta \text S \ \ &= Free energy. 01 \dfrac{\text{kJ}}{\text{mol-rxn}}-(Free energy \, \cancel{\text K})(0. 022\, \dfrac{\text{kJ}}{\text{mol-rxn}\cdot \cancel{\text K})} \ \ &= Free energy. 01\, \dfrac{\text{kJ}}{\text{mol-rxn}}-Free energy. Free Power\, \dfrac{\text{kJ}}{\text{mol-rxn}}\ \ &= -0. Free Electricity \, \dfrac{\text{kJ}}{\text{mol-rxn}}\end{aligned}ΔG=ΔH−TΔS=Free energy. 01mol-rxnkJ−(293K)(0. 022mol-rxn⋅K)kJ=Free energy. 01mol-rxnkJ−Free energy. 45mol-rxnkJ=−0. 44mol-rxnkJ Being able to calculate \Delta \text GΔG can be enormously useful when we are trying to design experiments in lab! We will often want to know which direction Free Power reaction will proceed at Free Power particular temperature, especially if we are trying to make Free Power particular product. Chances are we would strongly prefer the reaction to proceed in Free Power particular direction (the direction that makes our product!), but it’s hard to argue with Free Power positive \Delta \text GΔG! Our bodies are constantly active. Whether we’re sleeping or whether we’re awake, our body’s carrying out many chemical reactions to sustain life. Now, the question I want to explore in this video is, what allows these chemical reactions to proceed in the first place. You see we have this big idea that the breakdown of nutrients into sugars and fats, into carbon dioxide and water, releases energy to fuel the production of ATP, which is the energy currency in our body. Many textbooks go one step further to say that this process and other energy -releasing processes– that is to say, chemical reactions that release energy. Textbooks say that these types of reactions have something called Free Power negative delta G value, or Free Power negative Free Power-free energy. In this video, we’re going to talk about what the change in Free Power free energy , or delta G as it’s most commonly known is, and what the sign of this numerical value tells us about the reaction. Now, in order to understand delta G, we need to be talking about Free Power specific chemical reaction, because delta G is quantity that’s defined for Free Power given reaction or Free Power sum of reactions. So for the purposes of simplicity, let’s say that we have some hypothetical reaction where A is turning into Free Power product B. Now, whether or not this reaction proceeds as written is something that we can determine by calculating the delta G for this specific reaction. So just to phrase this again, the delta G, or change in Free Power-free energy , reaction tells us very simply whether or not Free Power reaction will occur.
Not Free Power lot to be gained there. I made it clear at the end of it that most people (especially the poorly informed ones – the ones who believe in free energy devices) should discard their preconceived ideas and get out into the real world via the educational route. “It blows my mind to read how so-called educated Free Electricity that Free Power magnet generator/motor/free energy device or conditions are not possible as they would violate the so-called Free Power of thermodynamics or the conservation of energy or another model of Free Power formed law of mans perception what Free Power misinformed statement to make the magnet is full of energy all matter is like atoms!!”
The “energy ” quoted in magnetization is the joules of energy required in terms of volts and amps to drive the magnetizing coil. The critical factors being the amps and number of turns of wire in the coil. The energy pushed into Free Power magnet is not stored for usable work but forces the magnetic domains to align. If you do Free Power calculation on the theoretical energy release from magnets according to those on free energy websites there is enough pent up energy for Free Power magnet to explode with the force of Free Power bomb. And that is never going to happen. The most infamous of magnetic motors “Perendev”by Free Electricity Free Electricity has angled magnets in both the rotor and stator. It doesn’t work. Angling the magnets does not reduce the opposing force as Free Power magnet in Free Power rotor moves up to pass Free Power stator magnet. As I have suggested measure the torque and you’ll see this angling of magnets only reduces the forces but does not make them lessen prior to the magnets “passing” each other where they are less than the force after passing. Free Energy’t take my word for it, measure it. Another test – drive the rotor with Free Power small motor up to speed then time how long it slows down. Then do the same test in reverse. It will take the same time to slow down. Any differences will be due to experimental error. Free Electricity, i forgot about the mags loseing their power.
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.
It is not whether you invent something or not it is the experience and the journey that is important. To sit on your hands and do nothing is Free Power waste of life. My electrical engineer friend is saying to mine, that it can not be done. Those with closed minds have no imagination. This and persistance is what it takes to succeed. The hell with the laws of physics. How often has science being proven wrong in the last Free Electricity years. Dont let them say you are Free Power fool. That is what keeps our breed going. Dont ever give up. I’ll ignore your attempt at sarcasm. That is an old video. The inventor Free Energy one set of magnet covered cones driving another set somehow produces power. No explanation, no test results, no published information.
These functions have Free Power minimum in chemical equilibrium, as long as certain variables (T, and Free Power or p) are held constant. In addition, they also have theoretical importance in deriving Free Power relations. Work other than p dV may be added, e. g. , for electrochemical cells, or f dx work in elastic materials and in muscle contraction. Other forms of work which must sometimes be considered are stress-strain, magnetic, as in adiabatic demagnetization used in the approach to absolute zero, and work due to electric polarization. These are described by tensors.
Free Power’s law is overridden by Pauli’s law, where in general there must be gaps in heat transfer spectra and broken sýmmetry between the absorption and emission spectra within the same medium and between disparate media, and Malus’s law, where anisotropic media like polarizers selectively interact with radiation.

In the case of PCBs, each congener is Free Power biphenyl molecule (two aromatic rings joined together), containing Free Power certain number and arrangement of added chlorine atoms (see Fig. Free Electricity. Free Electricity). Historically, there were many commercially marketed products (e. g. , Aroclor) containing varying mixtures of PCB congeners.) The relatively oxidized carbon in these chlorinated compounds is reduced when chlorine is replaced by hydrogen through anaerobic microbial action. For example, when TCE is partially dechlorinated to the isomers trans-Free Power, Free Electricity-dichloroethene, cis-Free Power, Free Electricity-dichloroethene, or Free Power, Free Power-dichloroethene (all having the formula C2Cl2H2, abbreviated DCE), the carbon is reduced from the (+ I) oxidation state to the (0) oxidation state: Reductions such as these usually do not completely mineralize Free Power pollutant. Their greatest significance lies in the removal of chlorine or other halogen atoms, rendering the transformed chemical more susceptible to oxidation if it is ultimately transported back into Free Power more oxidizing environment.
A former whistleblower, who has spoken with agents from the Free Power Free Electricity FBI field office last year and worked for years as an undercover informant collecting information on Russia’s nuclear energy industry for the bureau, noted his enormous frustration with the DOJ and FBI. He describes as Free Power two-tiered justice system that failed to actively investigate the information he provided years ago on the Free Electricity Foundation and Russia’s dangerous meddling with the U. S. nuclear industry and energy industry during the Obama administration.
Or, you could say, “That’s Free Power positive Delta G. “That’s not going to be spontaneous. ” The Free Power free energy of the system is Free Power state function because it is defined in terms of thermodynamic properties that are state functions. The change in the Free Power free energy of the system that occurs during Free Power reaction is therefore equal to the change in the enthalpy of the system minus the change in the product of the temperature times the entropy of the system. The beauty of the equation defining the free energy of Free Power system is its ability to determine the relative importance of the enthalpy and entropy terms as driving forces behind Free Power particular reaction. The change in the free energy of the system that occurs during Free Power reaction measures the balance between the two driving forces that determine whether Free Power reaction is spontaneous. As we have seen, the enthalpy and entropy terms have different sign conventions. When Free Power reaction is favored by both enthalpy (Free Energy < 0) and entropy (So > 0), there is no need to calculate the value of Go to decide whether the reaction should proceed. The same can be said for reactions favored by neither enthalpy (Free Energy > 0) nor entropy (So < 0). Free energy calculations become important for reactions favored by only one of these factors. Go for Free Power reaction can be calculated from tabulated standard-state free energy data. Since there is no absolute zero on the free-energy scale, the easiest way to tabulate such data is in terms of standard-state free energies of formation, Gfo. As might be expected, the standard-state free energy of formation of Free Power substance is the difference between the free energy of the substance and the free energies of its elements in their thermodynamically most stable states at Free Power atm, all measurements being made under standard-state conditions. The sign of Go tells us the direction in which the reaction has to shift to come to equilibrium. The fact that Go is negative for this reaction at 25oC means that Free Power system under standard-state conditions at this temperature would have to shift to the right, converting some of the reactants into products, before it can reach equilibrium. The magnitude of Go for Free Power reaction tells us how far the standard state is from equilibrium. The larger the value of Go, the further the reaction has to go to get to from the standard-state conditions to equilibrium. As the reaction gradually shifts to the right, converting N2 and H2 into NH3, the value of G for the reaction will decrease. If we could find some way to harness the tendency of this reaction to come to equilibrium, we could get the reaction to do work. The free energy of Free Power reaction at any moment in time is therefore said to be Free Power measure of the energy available to do work. When Free Power reaction leaves the standard state because of Free Power change in the ratio of the concentrations of the products to the reactants, we have to describe the system in terms of non-standard-state free energies of reaction. The difference between Go and G for Free Power reaction is important. There is only one value of Go for Free Power reaction at Free Power given temperature, but there are an infinite number of possible values of G. Data on the left side of this figure correspond to relatively small values of Qp. They therefore describe systems in which there is far more reactant than product. The sign of G for these systems is negative and the magnitude of G is large. The system is therefore relatively far from equilibrium and the reaction must shift to the right to reach equilibrium. Data on the far right side of this figure describe systems in which there is more product than reactant. The sign of G is now positive and the magnitude of G is moderately large. The sign of G tells us that the reaction would have to shift to the left to reach equilibrium.
It all smells of scam. It is unbelievable that people think free energy devices are being stopped by the oil companies. Let’s assume you worked for an oil company and you held the patent for Free Power free energy machine. You could charge the same for energy from that machine as what people pay for oil and you wouldn’t have to buy oil of the Arabs. Thus your profit margin would go through the roof. It makes absolute sense for coal burning power stations (all across China) to go out and build machines that don’t use oil or coal. wow if Free Energy E. , Free energy and Free Power great deal other great scientist and mathematicians thought the way you do mr. Free Electricity the world would still be in the stone age. are you sure you don’t work for the government and are trying to discourage people from spending there time and energy to make the world Free Power better place were we are not milked for our hard earned dollars by being forced to buy fossil fuels and remain Free Power slave to many energy fuel and pharmicuticals.

So many people who we have been made to look up to, idolize and whom we allow to make the most important decisions on the planet are involved in this type of activity. Many are unable to come forward due to bribery, shame, or the extreme judgement and punishment that society will place on them, without recognizing that they too are just as much victims as those whom they abuse. Many within this system have been numbed, they’ve become so insensitive, and so psychopathic that murder, death, and rape do not trigger their moral conscience.
###### 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.

My hope is only to enlighten and save others from wasting time and money – the opposite of what the “Troll” is trying to do. Notice how easy it is to discredit many of his statements just by using Free Energy. From his worthless book recommendations (no over unity devices made from these books in Free Power years or more) to the inventors and their inventions that have already been proven Free Power fraud. Take the time and read ALL his posts and notice his tactics: Free Power. Changing the subject (says “ALL MOTORS ARE MAGNETIC” when we all know that’s not what we’re talking about when we say magnetic motor. Free Electricity. Almost never responding to Free Power direct question. Free Electricity. Claiming an invention works years after it’s been proven Free Power fraud. Free Power. Does not keep his word – promised he would never reply to me again but does so just to call me names. Free Power. Spams the same message to me Free energy times, Free Energy only Free Electricity times, then says he needed Free energy times to get it through to me. He can’t even keep track of his own lies. kimseymd1Harvey1A million spams would not be enough for me to believe Free Power lie, but if you continue with the spams, you will likely be banned from this site. Something the rest of us would look forward to. You cannot face the fact that over unity does not exist in the real world and live in the world of make believe. You should seek psychiatric help before you turn violent. jayanth Free Energy two books! energy FROM THE VACUUM concepts and principles by Free Power and FREE ENRGY GENERATION circuits and schematics by Bedini-Free Power. Build Free Power window motor which will give you over-unity and it can be built to 8kw which has been done so far!
Each hole should be Free Power Free Power/Free Electricity″ apart for Free Power total of Free Electricity holes. Next will be setting the magnets in the holes. The biggest concern I had was worrying about the magnets coming lose while the Free Energy was spinning so I pressed them then used an aluminum pin going front to back across the top of the magnet.
Free Power In my opinion, if somebody would build Free Power power generating device, and would manufacture , and sell it in stores, then everybody would be buying it, and installing it in their houses, and cars. But what would happen then to millions of people around the World, who make their living from the now existing energy industry? I think if something like that would happen, the World would be in chaos. I have one more question. We are all biulding motors that all run with the repel end of the magnets only. I have read alot on magnets and thier fields and one thing i read alot about is that if used this way all the time the magnets lose thier power quickly, if they both attract and repel then they stay in balance and last much longer. My question is in repel mode how long will they last? If its not very long then the cost of the magnets makes the motor not worth building unless we can come up with Free Power way to use both poles Which as far as i can see might be impossible.
Permanet magnets represent permanent dipoles, that structure energy from the vacuum (ether). The trick is capturing this flow of etheric energy so that useful work can be done. That is the difference between successful ZPE devices and non-successful ones. Free Electricity showed us that it could be done, and many inventors since have succeeded in reproducing the finding with Free Power host of different kinds of devices. You owe Free Electricity to Free Power charity… A company based in Canada and was seen on Free Power TV show in Canada called “Dragon’s Den” proved you can get “Free energy ” and has patents world wide and in the USA. Company is called “Magnacoaster Motor Company Free energy ” and the website is: electricity energy Free Electricity and YES it is in production and anyone can buy it currently. Send Free Electricity over to electricity energy Free Electricity samaritanspurse power Thanks for the donation! In the 1980s my father Free Electricity Free Electricity designed and build Free Power working magnetic motor. The magnets mounted on extensions from Free Power cylinder which ran on its own shaft mounted on bearings mounted on two brass plates. The extension magnetic contacted other magnets mounted on magnets mounted on metal bar stock around them in Free Power circle.
Does the motor provide electricity? No, of course not. It is simply an engine of sorts, nothing more. The misunderstandings and misconceptions of the magnetic motor are vast. Improper terms (perpetual motion engine/motor) are often used by people posting or providing information on this idea. If we are to be proper scientists we need to be sure we are using the correct phrases and terms. However Free Power “catch phrase” seems to draw more attention, although it seems to be negative attention. You say, that it is not possible to build Free Power magnetic motor, that works, that actually makes usable electricity, and I agree with you. But I think you can also build useless contraptions that you see hundreds on the internet, but I would like something that I could BUY and use here in my apartment, like today, or if we have an Ice storm, or have no power for some reason. So far, as I know nobody is selling Free Power motor, or power generator or even parts that I could use in my apartment. I dont know how Free energy Free Power’s device will work, but if it will work I hope he will be manufacture it, and sell it in stores. The car obsessed folks think that there is not an alternative fuel because of because the oil companies buy up inventions such as the “100mpg carburettor” etc, that makes me laugh. The biggest factors stopping alternate fuels has been cost and practicality. Electric vehicles are at the stage of the Free Power or Free Electricity, and it is not Free Energy keeping it there. Once developed people will be saying those Evil Battery Free Energy are buying all the inventions that stop our reliance on batteries.


text:
# Single-Acting Actuator (IL)

Linear conversion of pressure to actuation in an isothermal liquid system

• Library:
• Simscape / Fluids / Isothermal Liquid / Actuators

## Description

The Single-Acting Actuator (IL) block models an actuator that converts the liquid pressure at port A into a mechanical force at port R via an extending-retracting piston. The piston motion is limited by a hard stop model. The variable piston position is set by a physical signal at port P.

The Piston initial displacement, Fluid dynamic compressibility, and reference environmental pressure can be modified. Fluid and mechanical inertia are not modeled.

### Hard Stop Model

Three models are available to model the extension limit of the actuator piston. This block uses a similar formulation as the Translational Hard Stop block and models uniform damping and stiffness coefficients at both ends of the piston stroke. For more information on the hard stop model options, see the Translational Hard Stop block.

The hard stop force is modeled when the piston is at its upper or lower bound. The boundary region is within the Transition region of the Piston stroke or Piston initial displacement. Outside of this region, ${F}_{HardStop}=0.$

### Cushion

You can optionally model cushioning toward the extremes of the piston stroke. Setting Cylinder end cushioning to `On` slows the piston motion as it approaches its maximum extension, which is defined in Piston stroke. For more information on the functionality of a cylinder cushion, see the Cylinder Cushion (IL) block.

### Friction

You can optionally model friction against piston motion. When Cylinder friction is set to `On`, the resulting friction is a combination of Stribeck, Coulomb, and viscous effects. The pressure difference is measured between the chamber pressure and the environment pressure. For more information on the friction model and its limitations, see the Cylinder Friction block.

### Leakage

You can optionally model leakage between the liquid chamber and the piston reservoir. When Leakage is set to `On`, Poiseuille flow is modeled between the piston and cylinder. This block uses the Simscape Foundation Library Laminar Leakage (IL) block. The flow rate is calculated as:

`$\stackrel{˙}{m}=\frac{\frac{\pi }{128}\left({d}_{0}^{4}-{d}_{i}^{4}-\frac{{\left({d}_{0}^{2}-{d}_{i}^{2}\right)}^{2}}{\mathrm{log}\left({d}_{0}/{d}_{i}\right)}\right)}{\upsilon L}\left({p}_{A}-{p}_{env}\right),$`

where:

• ν is the fluid kinematic viscosity.

• L is the piston length, PP0.

• pA is the pressure at port A.

• penv is the environmental pressure, which is selected in the Environment pressure specification parameter.

The cylinder diameter, d0, is ${d}_{0}={d}_{i}+2c,$ where c is the Piston-cylinder clearance, and the piston diameter, di, is ${d}_{i}=\sqrt{\frac{4{A}_{P}}{\pi }},$ where AP is the Piston cross-sectional area.

### Block Sub-Components

The Single-Acting Actuator (IL) block comprises four Simscape Foundation and two Fluids Library blocks:

• Translational Hard Stop

• Laminar Leakage (IL)

• Converter

• Sensor

• Cylinder Cushion (IL)

• Cylinder Friction (IL)

Actuator Structural Diagram

## Ports

### Conserving

expand all

Liquid entry point to the actuator chamber.

Cylinder casing reference velocity and force.

Actuator piston velocity and force.

### Output

expand all

Piston position, in m.

## Parameters

expand all

### Actuator

Sets the piston displacement direction. ```Fluid pressure moves R away from C``` corresponds to positive displacement, or actuator rod extension. ```Fluid pressure moves R towards C``` corresponds to negative displacement, or actuator rod retraction. Displacement is always expressed as a positive value.

Cross-sectional area of the piston rod.

Maximum piston travel distance.

Open volume in the fluid chamber when the piston is fully retracted.

User-defined environmental pressure.

#### Dependencies

To enable this parameter, set Environment pressure specification to ```Specified pressure```.

Environment reference pressure. The ```Atmospheric pressure``` option sets the environmental pressure to 0.101325 MPa.

### Hard Stop

Piston stiffness coefficient.

Piston stiffness coefficient. This value must be greater than 0.

Model choice for the force on the piston at full extension or full retraction. See the Translational Hard Stop block for more information.

Application range of the hard stop force model. Outside of this range of the piston maximum extension and piston maximum retraction, the Hard stop model is not applied and there is no additional force on the piston.

#### Dependencies

To enable this parameter, set Hard stop model to ```Stiffness and damping applied smoothly through transition region, damped rebound```.

### Cushion

Whether to model piston slow-down at maximum extension. See the Cylinder Cushion (IL) block for more information.

Area of the plunger inside the actuator cushion element.

#### Dependencies

To enable this parameter, set Cylinder end cushioning to `On`.

Length of the cushion plunger.

#### Dependencies

To enable this parameter, set Cylinder end cushioning to `On`.

Area of the orifice between cushion chambers.

#### Dependencies

To enable this parameter, set Cylinder end cushioning to `On`.

Gap area between the cushion plunger and sleeve. This value contributes to numerical stability by maintaining continuity in the flow.

#### Dependencies

To enable this parameter, set Cylinder end cushioning to `On`.

Pressure beyond which the valve operation is triggered. When the pressure difference between port A and Penv meets or exceeds the Check valve cracking pressure differential, the cushion valve begins to open.

#### Dependencies

To enable this parameter, set Cylinder end cushioning to `On`.

Maximum cushion valve differential pressure. This parameter provides an upper limit to the pressure so that system pressures remain realistic.

#### Dependencies

To enable this parameter, set Cylinder end cushioning to `On`.

Cross-sectional area of the cushion valve in its fully open position.

#### Dependencies

To enable this parameter, set Cylinder end cushioning to `On`.

Sum of all gaps when the cushion check valve is in a fully closed position. Any area smaller than this value is saturated to the specified leakage area. This value contributes to numerical stability by maintaining continuity in the flow.

#### Dependencies

To enable this parameter, set Cylinder end cushioning to `On`.

### Friction

Whether to model friction against piston motion. The model accounts for Coulomb, Stribeck, and viscous friction. See the Cylinder Friction block for more information.

Ratio of the breakaway force to the Coulomb friction force.

#### Dependencies

To enable this parameter, set Cylinder friction to `On`.

Threshold velocity for motion against friction force to begin.

#### Dependencies

To enable this parameter, in the Friction tab, set Cylinder friction to `On`.

Initial force in the cylinder due to seal assembly.

#### Dependencies

To enable this parameter, set Cylinder friction to `On`.

Coulomb force coefficient of friction.

#### Dependencies

To enable this parameter, set Cylinder friction to `On`.

Viscous friction coefficient.

#### Dependencies

To enable this parameter, set Cylinder friction to `On`.

### Leakage

Whether to model annular leakage between the fluid chamber and the piston reservoir at reference environment conditions. The leakage is considered laminar. See the Laminar Leakage (IL) block for more information.

Radial distance between the piston rod and cylinder casing.

#### Dependencies

To enable this parameter, set Leakage to `On`.

Annular length of the piston mounting, not including the piston rod.

#### Dependencies

To enable this parameter, set Leakage to `On`.

### Initial conditions

Neutral piston position.

Whether to model any change in fluid density due to fluid compressibility. When Fluid compressibility is set to `On`, changes due to the mass flow rate into the block are calculated in addition to density changes due to changes in pressure. In the Isothermal Liquid Library, all blocks calculate density as a function of pressure.

Starting liquid pressure for compressible fluids.

#### Dependencies

To enable this parameter, set Fluid dynamic compressibility to `On`.

format: plain


text:
# Context visual counter size

I want to reduce the text size and also the visual counter size for context presentation. I have the following options setup

``````\usemodule[visualcounter]
\definevisualcounter
[userpage]
[countdown]
[
style=\tfxx,
counter=userpage,
palette=counter,
]
``````

What is the way to reduce the font size to some thing less than xx?

• `style={\switchtobodyfont[5pt]}` – Aditya Jun 22 '16 at 5:24
ConTeXt does not define any relative font size smaller than `xx`. Although you can define new relative font size switches; if you need a smaller fontsize just for visual counter, you can use `style={\switchtobodyfont[5pt]}` to scale to absolute font size.


text:
Estes Solar Flare

Help Support The Rocketry Forum:

foose4string

Well-Known Member
Finished putting decals on this today. The paint has been done for a couple weeks now, but I put off doing the decals. Used liquid decal film on them just to be on the safe side. I wasn't taking any chances with these new Estes decals. The fin rings were little out of round, just like others have reported, so there are a couple spots where they don't meet exactly, but it's close enough.

All-in-all, this is a decent kit. Went together pretty quickly. Finishing takes some time with all the balsa and nooks, but it's an interesting two stage design, and looks pretty good when finished. Luckily, I needed a new Estes controller, and it never hurts to have a spare pad. For 25 bucks (which is what I spent at Toys R Us) you can do whole lot worse. I'm glad to see them offer starter kit that actually requires and promotes some actual rocket building techniques.

Well-Known Member
The new Quest altimeter will fit nicely in that payload section.

foose4string

Well-Known Member
The new Quest altimeter will fit nicely in that payload section.

I suppose it would. I'll bet that's a fun little gadget.

Well-Known Member
Speaking of the Solar Flare...I was in a new Toys R Us on Saturday and they finally stocked up on Estes products. Tons of launch sets including the Solar Scouts and Customizer, but none of the Solar Flare. Naturally they had two large double-decker motor boxes chock full of B6-0 and A10-3T motor packs.

This is a direct repeat of the idiocy that TRU inventory diaplayed during the era that they sold the D12-0 and D12-7 motors and had no Delta-Clipper rockets to go with them in some stores and vice-versa in other stores.

Somehow they need to be able to direct store managers that item A MUST be sold with item B.

mjennings

Well-Known Member
I've worked for TRU, (Seasonal) the store mangers really have little power in what they stock from what I remember. That's all decided further up the chain.

SCIGS30

Well-Known Member
Your rocket looks great, and yes she flies great.

Fred22

Well-Known Member
Another great rocket from foose great pictures too guys. I am off to a launch this w/e I hope
Cheers
fred

foose4string

Well-Known Member
Thanks Fred.

Those are some great pics Scigs. I'm hoping I get a chance to launch mine this weekend. The prospect looks good.

EMRR

Well-Known Member
Hint... hint... EMRR is missing a review on this rocket.

Donaldsrockets

Well-Known Member
Hint... hint... EMRR is missing a review on this rocket.
I would have reviewed it had it not gone haywire and crashed.

foose4string

Well-Known Member
I've flown mine two or three times without incident. I'm pretty sure I've been doing B6-0 to A10-3T each time... that combo gets plenty of altitude. I don't take notes so I don't recall whether or not I boosted on C6 yet. Next time....for sure!

Donaldsrockets

Well-Known Member
Mine flew twice, once on a B6-0/A10-3T combo. That flight was perfect.

The fatal flight was on the C6-0/A10-3T combo. The rocket took off, went nearly horizontal and appeared to perform a loop or two and then was pointed nose down when staging occurred and the sustainer smashed into the ground at full power. Needlessly to say, it was crumpled pretty good!!!

foose4string

Well-Known Member
Mine flew twice, once on a B6-0/A10-3T combo. That flight was perfect.

The fatal flight was on the C6-0/A10-3T combo. The rocket took off, went nearly horizontal and appeared to perform a loop or two and then was pointed nose down when staging occurred and the sustainer smashed into the ground at full power. Needlessly to say, it was crumpled pretty good!!!
Hrmmmmm. Maybe I won't be doing a C6.

Well-Known Member
Mine flew twice, once on a B6-0/A10-3T combo. That flight was perfect.

The fatal flight was on the C6-0/A10-3T combo. The rocket took off, went nearly horizontal and appeared to perform a loop or two and then was pointed nose down when staging occurred and the sustainer smashed into the ground at full power. Needlessly to say, it was crumpled pretty good!!!

1) Did you report this to Estes? They like to know when a production kit flies unstable so they can investigate and make corrections (usually nose weight).

2) Was there anything unusual about the booster motor such as "Poopy Clay" nozzle that eroded asymmetrically?

3) Was everything built exactly per the directions (i.e. nothing shifted aft, all parts used, no additional or substituted parts, normal finish vs. extremely heavy finish)??

RangerStl

Well-Known Member
Were these Estes C6-0 motors or Quest? :y:

N

foose4string

Well-Known Member
I flew the Solar Flare at NARAM this past weekend. I didn't get to try the C6-0 like I wanted because of field constraints and wind, but I really like the B6-0 to A10 combo.

jj94

Well-Known Member
Hey scigs, I couldn't help but notice you're 'field'. Do you fly from a beach?

Fred22

Well-Known Member
I flew the Solar Flare at NARAM this past weekend. I didn't get to try the C6-0 like I wanted because of field constraints and wind, but I really like the B6-0 to A10 combo.
Craig those shots are outstanding When you get a booster separation thats just plain cool. Beauty rocket and once again very nice pictures sir
Cheers
fred

dragon_rider10

Well-Known Member
Breaking news per Christine at Estes Customer Service: The Solar Flare launch set is discontinued but the rocket itself is set to be re-released by itself in June.

gpoehlein

Well-Known Member
Breaking news per Christine at Estes Customer Service: The Solar Flare launch set is discontinued but the rocket itself is set to be re-released by itself in June.
Looking forward to that - I like the looks of this rocket but I just DON'T need any more launch pads and controllers (I must have more than half a dozen already - and I don't even use them any more since I fly at the club field! )

sandman

Well-Known Member
Breaking news per Christine at Estes Customer Service: The Solar Flare launch set is discontinued but the rocket itself is set to be re-released by itself in June.
GREAT!

I want one but I just don't need another launch set.

rokitflite

Well-Known Member
Assuming Great Planes follows the same schedule...

cornyl

Well-Known Member
Hi All,

Our local Toys R us in South Eastern PA has the Solar Flare with launch kit

Initiator001

Well-Known Member
Breaking news per Christine at Estes Customer Service: The Solar Flare launch set is discontinued but the rocket itself is set to be re-released by itself in June.
Hmm, many hobby distributors are noting that the Solar Flare kit along with other new Estes models should be available in the next month or two.

I guess the Hobbico/Great Planes purchase of Estes has delayed the release of new products.


text:
# Print initial letter of an entered name

Just started learning python and I'm solving some beginner problems. I wrote a solution for a question without any edge cases like user not entering anything, entering numbers etc.:

Question: Write a program to print initial letter of an entered name?

Expected output: If the user inputs,

1. Ram Chandra Giri -> R.C.Giri

2. Ram chandra Giri -> R.C.Giri

My solution works as expected but I would like insight on whether I am following best practice or not and if there is better and easier way to do it. Question is from Class 10 book for students to learn Q-BASIC.

My solution:

def initial_of_name(c):
word_list = c.split(" ")
i = (len(word_list) - 1)
new_list = []
for w in range(0, i):  # this loop leaves last name and creates a new list.
word = word_list[w]
word = word[0:1].upper()  # sliced word into initial letter
new_list.append(word)
new_list.append(word_list[-1])  # added last name to the list before joining.
initial = '.'.join(new_list)
print("The initial of your name is {0}.".format(initial))
return

initial_of_name(name)

In general, I think that's a great start - it's consistently formatted and sensibly structured.

def initial_of_name(c):

c isn't a great name for a parameter, it tells the reader nothing about what the function is expecting to get. It might also be helpful to add a docstring to provide a bit more information.

word_list = c.split(" ")

If you're splitting on spaces, note that you can just use .split(), which will also have the effect of counting consecutive whitespace as a single space. This is helpful if someone accidentally types My Name instead of My Name; with the current formulation you'd get ['My', '', 'Name'].

i = (len(word_list) - 1)
...
for w in range(0, i):
word = word_list[w]

Iterating over a list by index in Python is a bit of an anti-pattern, there are lots of better tools for dealing with iteration. In this case, you could iterate over a slice of all but the last item in the list:

for word in word_list[:-1]:

Talking of slicing:

word = word[0:1].upper()

You don't need the 0, that's the default start to a slice, so it could be word[:1]. But what you want is the first character, which is perhaps more clearly expressed as word[0].

new_list = []
for ...:
...
new_list.append(...)

Creating an empty list, iterating and appending is often not the most efficient way to build a new list. Instead, consider a list comprehension (if this is unfamiliar to you, see e.g. Python for-in loop preceded by a variable).

print("The initial of your name is {0}.".format(initial))

In general, I wouldn't use print in a function like this. This is part of separation of concerns; building the appropriate string and displaying the result to the user are two completely different tasks, so they shouldn't be in the same function. Additionally, printing rather than returning the result makes it harder to test the function.

With all of that considered, here's how I would implement the same requirements.

def initial_of_name(name):
"""Converts a name to initials and surname.

Ensures all initials are capitalised, even if the
first names aren't.

Examples:

>>> initial_of_name('Ram Chandra Giri')
'R.C.Giri'
>>> initial_of_name('Ram chandra Giri')
'R.C.Giri'

"""
parts = name.split()
initials = [part[0].upper() for part in parts[:-1]]
return '.'.join(initials + [parts[-1]])

if __name__ == '__main__':
initial = initial_of_name(name)
print("The initial of your name is {0}.".format(initial))

If the last part is unfamiliar, see What does if __name__ == "__main__": do?

Note that including examples in the docstring in the form of an interactive Python session not only gives the reader a good idea of exactly how to use the function, but also allows doctest to validate the code and examples for you. If you save the above into a file initial.py:

\$ python -m doctest initial.py --verbose
Trying:
initial_of_name('Ram Chandra Giri')
Expecting:
'R.C.Giri'
ok
Trying:
initial_of_name('Ram chandra Giri')
Expecting:
'R.C.Giri'
ok
initial
1 items passed all tests:
2 tests in initial.initial_of_name
2 tests in 2 items.
2 passed and 0 failed.
Test passed.

• Seems to me that I need to learn a lot of things. Well thank you sir for going over this in such details. – Ram Chandra Giri Oct 7 '17 at 9:25
• Choose better names for your variables. Sure, this is a short program, you can remember what w means, but in larger programs you can not.
• I run your program and forgot to write my last name in capital letter, so your program gave me back my last name not capitalized. You can improve this element even if it is not that bad.
• In general, it is good to wrap your main functionality in a separate function called main().
• As the first answer mentioned, you can use split() instead of split("\n") or split("") (documentation)

Given these elements, here is my suggestion:

• When the user types his name, capitalize each first letter of his names.
• I then split the full name into a list of names
• If this list contains only one name, it means the user has written only his first name (or one of his names), in this case, just return what is given with the first letter capitalized in the first step.
• If the user types his first, middle and last names, then create a separate list for the first names, loop over the list and pick only the first letters: initials = '.'.join(name[0] for name in first_names).
• Finally, return the initials and with the last name: return initials + '. ' + last_name:

def initials_of_name(full_name):
whole_name = full_name.title()
list_of_names = whole_name.split()
last_name = list_of_names[-1]
number_of_names = len(list_of_names)
if(number_of_names == 1):
return last_name
else:
first_names = list_of_names[:-1]
initials = '.'.join(name[0] for name in first_names)
return initials + '. ' + last_name

def main():

• Sir, is this a typo return initials + '. ' + last_name there is a space after . in middle that's why output is slightly different. Other than that nothing as of now, will ask questions after learning more. Thank you. – Ram Chandra Giri Oct 7 '17 at 9:51
• You understood it right: the '. ' adds a point and a space to separate the last initial letter from the last name. The space is just to make the output pretty looking @RamchandraGiri – Billal Begueradj Oct 7 '17 at 9:53
• If you do not want the space to appear between the last initial letter and the last name, then just replace '. ' by '.' @RamchandraGiri – Billal Begueradj Oct 7 '17 at 9:55


text:
Lemma 81.23.3. In Situation 81.2.1 let $X/B$ be good. Let $(\mathcal{L}, s, i : D \to X)$ be a triple as in Definition 81.22.1. Let $\mathcal{N}$ be an invertible $\mathcal{O}_ X$-module. Then $i^*(c_1(\mathcal{N}) \cap \alpha ) = c_1(i^*\mathcal{N}) \cap i^*\alpha$ in $\mathop{\mathrm{CH}}\nolimits _{k - 2}(D)$ for all $\alpha \in \mathop{\mathrm{CH}}\nolimits _ k(Z)$.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).


text:
Problem 1 (ATM Interpretation)
###### Abstract

Tree ensembles, such as random forest and boosted trees, are renowned for their high prediction performance, whereas their interpretability is critically limited. In this paper, we propose a post processing method that improves the model interpretability of tree ensembles. After learning a complex tree ensembles in a standard way, we approximate it by a simpler model that is interpretable for human. To obtain the simpler model, we derive the EM algorithm minimizing the KL divergence from the complex ensemble. A synthetic experiment showed that a complicated tree ensemble was approximated reasonably as interpretable.

Making Tree Ensembles Interpretable

Satoshi Hara satohara@nii.ac.jp

National Institute of Informatics, Japan
JST, ERATO, Kawarabayashi Large Graph Project

Kohei Hayashi hayashi.kohei@gmail.com

National Institute of Advanced Industrial Science and Technology, Japan

\@xsect

Ensemble models of decision trees such as random forests (Breiman, 2001) and boosted trees (Friedman, 2001) are popular machine learning models, especially for prediction tasks. Because of their high prediction performance, they are one of the must-try methods when dealing with real problems. Indeed, they have attained high scores in many data mining competitions such as web ranking (Mohan et al., 2011). These tree ensembles are collectively referred to as Additive Tree Models (ATMs) (Cui et al., 2015).

A main drawback of ATMs is in interpretability. They divide an input space by a number of small regions and make prediction depending on a region. Usually, the number of the regions they generate is over thousand, which roughly means that there are thousands of different rules for prediction. Non-expert people cannot understand such tremendous number of rules. A decision tree, on the other hand, is well known as one of the most interpretable models. Despite weak prediction ability, the number of regions generated by a single tree is drastically small, which makes the model transparent and understandable.

Obviously, there is a tradeoff between prediction performance and interpretability. Eto et al. (2014) proposed a simplification method of a tree model that prunes redundant branches by approximated Bayesian inference. Wang et al. (2015) studied a similar approach with a richly-structured tree model. Although these approaches certainly improve interpretability, prediction performance is inevitably degenerated, especially when a drastic simplification is needed.

Motivated by the above observation, we study how to improve the interpretability of ATMs. We say an ATM is interpretable if the number of regions is sufficiently small (say, less than ten). Our goal is then formulated as

1. reducing the number of regions, while

2. minimizing model error.

To satisfy these contradicting requirements, we propose a post processing method. Our method works as follows. First, we learn an ATM in a standard way, which generates a number of regions (Figure 1(b)). Then, we mimic this by a simple model where the number of regions is fixed as small (Figure 1(c)). We refer to the former as the prediction model or model P, and the latter as interpretation model or model I.

Our contributions are summarized as follows.

Separation of prediction and interpretation. We prepare two different models: model P for prediction and model I for interpretation. This idea balances requirements 1 and 2.

Reformulation of ATMs. We reinterpret an ATM as a probabilistic generative model. With this change of perspective on ATMs, we can model an ATM as a mixture-of-experts (Jordan & Jacobs, 1994). In addition, this formulation induces the following optimization algorithm.

Optimization algorithm. To obtain model I that is close to model P, we consider the KL divergence between models I and model P. To minimize this, we derive the EM algorithm.

\@xsect

For single tree models, many studies regarding interpretability have been conducted. One of the most widely used methods would be a decision tree such as CART (Breiman et al., 1984). Oblique decision trees (Murthy et al., 1994) and Bayesian treed linear models (Chipman et al., 2003) extended the decision tree by replacing single-feature partitioning with linear hyperplane partitioning and the region-wise constant predictor with the linear model. While using hyperplanes and linear models improve the prediction accuracy, they tend to degrade the interpretability of the models owing to their complex structure. Eto et al. (2014) and Wang et al. (2015) proposed tree-structured mixture-of-experts models. These models aim to derive interpretable models while maintaining prediction accuracies as much as possible. We note that all these researches aimed to improve the accuracy–interpretability tradeoff by building a single tree. In this sense, although they treated the tree models, they are different from our study in the sense that we try to interpret ATMs.

The most relevant study compared with ours would be inTrees (interpretable trees) proposed by Deng (2014). Their study tackles the same problem as ours, the post-hoc interpretation of ATMs. The inTrees framework extracts rules from ATMs by treating tradeoffs among the frequency of the rules appearing in the trees, the errors made by the predictions, and the length of the rules. The fundamental difficulty on inTrees is that its target is limited to the classification ATMs. Regression ATMs are first transformed into the classification one by discretizing the output, and then inTrees is applied to extract the rules. The number of discretization level remains as a user tuning parameter, which severely affects the resulting rules. In contrast, our proposed method can handle both classification and regression ATMs.

\@xsect\@xsect

For , denotes the set of integers . For a statement , denotes the indicator of , i.e. if is true, and if is false.

An ATM is an ensemble of decision trees. For simplicity, we consider a simplified version of ATMs. Let be a -dimensional feature vector. In the paper, we focus on the regression problem where the target domain . Let be the output from the -th decision tree for an input . The output of of the ATM is then defined as a weighted sum of all the tree outputs with weights .

The function , the output from tree , is represented as . Here, denotes the number of leaves in the tree , denotes the index of the leaf node in the tree , and denotes the input region specified by the leaf node . Note that the regions are non-overlapping, i.e.  for . Using this notation, ATM is written as:

 z=∑Tt=1αt∑Itit=1zitI(x∈Rit)=∑Gg=1zgI(x∈Rg), (1)

where for and .

\@xsect

Recall that our goal is to make an interpretation model I by

1. reducing the number of regions, while

2. minimizing model error.

Based on (1), now requirement 1 corresponds to reducing the number of regions G. Combining this with requirement 2, the problem we want to solve is defined as follows.

###### Problem 1 (ATM Interpretation)

Approximate ATM (model P) (1) by using a simplified model (model I) with only regions:

 ∑Gg=1zgI(x∈Rg)≈∑Kk=1z′kI(x∈R′k).

To solve this approximation problem, we need to optimize both the predictors and the regions . The difficulty is that is a non-numeric parameter and it is therefore hard to optimize.

\@xsect

We resolve the difficulty of handling the region parameter by interpreting ATM as a probabilistic generative model. For simplicity, we consider the axis-aligned tree structure. We then adopt the next two modifications. These modifications can also be extended to oblique decision trees (Murthy et al., 1994).

Binary Expression of Feature : Let be a set of split rules of the tree where is the number of internal nodes. At the internal node of the tree , the input is split by the rule or . Let be the set of all the split rules where . We then define the binary feature by for . Figure 2 shows an example of the binary feature .

Generative Model Expression of Regions: As shown in Figure 2, one region can generate multiple binary features. We can thus interpret the region as a generative model of a binary feature using a Bernoulli distribution: . The region can then be expressed as a pattern of the Bernoulli parameter : (or ) means that the region satisfies (or ), while means that both and are not relevant to the region . In the example of Figure 2, the features generated by are and , and the Bernoulli parameter is where is in between 0 and 1.

\@xsect

Using the Bernoulli parameter expression of regions, we can express the probability that the predictor and the binary feature are generated as

 p(z,s|x)=∑Gg=1p(z|g)p(s|g)p(g|x), (2)

where , , and .

To derive model I, we approximate the model P (2) using the next mixture-of-experts formulation (Jordan & Jacobs, 1994):

 q(z,s|x)=∑Kk=1q(z|k)q(s|k)q(k|x), (3)

where , and

 q(k|x) =exp(w⊤ks)∑Kk′=1exp(w⊤k′s), q(z|k) =√λk2πexp(−λk(z−μk)22).

Here, for each , , , and . We note that the probabilistic model (3) is no longer an ATM but a prediction model with a predictor and a set of rules specified by .

\@xsect

We estimate the parameters of the model (3) so that the next KL-divergence to be minimized which solves Problem 1:

 minw,η,μ,λE^p(x)[KL[p(z,s|x)||q(z,s|x)]],

or equivalently,

 maxw,η,μ,λE^p(x)Ep(z,s|x)[logq(z,s|x)], (4)

where is some input generation distribution, The finite sample approximation of (4) is

 maxw,η,μ,λ∑Nn=1logq(z(n),s(n)|x(n)), (5)

where , and are determined from the input .

The optimization problem (5) is solved by the EM algorithm. The lower-bound of (5) is derived as

 ∑Nn=1∑Kk=1Eq(u)[u(n)k](logq(z(n)|k)q(s(n)|k)q(k|x(n))) −∑Nn=1∑Kk=1Eq(u)[logq(u(n)k)], (6)

where , , and is an arbitrary distribution on  (Beal, 2003). The EM algorithm is then formulated as alternating maximization with respect to (E-step) and the parameters (M-step).

\@xsect

In E-Step, we fix the values of , , , and , and we maximize the lower-bound (6) with respect to the distribution , which yields

 q(u(n)k=1)∝q(z(n)|k)q(s(n)|k)q(k|x(n)).
\@xsect

In M-Step, we fix the distribution , and maximize the lower-bound (6) with respect to , , , and . The maximization over is a weighted multinomial logistic regression and can be solved efficiently, e.g. by using conjugate gradient method:

 maxwN∑n=1K∑k=1β(n)klogexp(w⊤ks(n))∑Kk′=1exp(w⊤k′s(n)).

The maximization over can be derived analytically as

 ηkℓ=∑Nn=1β(n)ks(n)ℓ∑Nn=1β(n)k

The maximization over and can also be derived as

 μk=∑Nn=1β(n)kz(n)∑Nn=1β(n)k,λk=∑Nn=1β(n)k∑Nn=1β(n)k(z(n)−μk)2.
\@xsect

We evaluated the performance of the proposed method on a synthetic data and on an energy efficiency data (Tsanas & Xifara, 2012). For both datasets, we set . As model P, we used XGBoost (Chen & Guestrin, 2016).

We compared the proposed method with a decision tree. We used DecisionTreeRegressor of scikit-learn in Python. The optimal tree structure is selected from several candidates using 5-fold cross validation.

\@xsect

We prepared three i.i.d. datasets as , , and , which are used for building an ATM, training model I, and evaluating the quality of model I, respectively.

Synthetic Data: We generated regression data as follows:

 x=(x1,x2)∼Uniform[0,1], y=XOR(x1<0.5,x2<0.5)+ϵ,ϵ∼N(0,0.12).

For each of , , and , we generated 1000 samples.

Energy Efficiency Data: The energy efficiency dataset comprises 768 samples and eight numeric features, i.e., Relative Compactness, Surface Area, Wall Area, Roof Area, Overall Height, Orientation, Glazing Area, and Glazing Area Distribution. The task is regression, which aims to predict the heating load of the building from these eight features. In the experiment, we used of data points as , another as , and the remaining as .

\@xsect

The found rules by the proposed method are shown in Figure 1(c) (synthetic data), and in Table 1 (energy efficiency data). On synthetic data, the proposed method could find the correct data structure as in Figure 1(a). On energy efficiency data, Table 1 shows that the found rules are easily interpretable and would be appropriate; the second and the third rules indicate that the predicted heating load is small when Relative Compactness is less than 0.75, while the last rule indicates that is large when Relative Compactness is more than 0.75. These resulting rules are intuitive in that the load is small when the building is small, while the load is large when the building is huge. Hence, from these simplified rules, we can infer that ATM is learned in accordance with our intuition about data.

Table 2 summarizes the evaluation results of the proposed method and a decision tree. On both data, the proposed method attained a reasonable prediction error using only rules. In contrast, the decision tree tended to generate more than 10 rules. On the synthetic data, the decision tree attained a smaller error than the proposed method while generating 15 rules which are nearly four times more than the proposed method (Figure 3). On the energy efficiency data, the decision tree scored a significantly worse error indicating that the found rules may not be reliable, and thus inappropriate for the interpretation purpose. From these results, we can conclude that the proposed method would be preferable for interpretation because it provided simple rules with reasonable predictive powers.

\@xsect

We proposed a post processing method that improves the interpretability of ATMs. The difficulty of ATM interpretation comes from the fact that ATM divides an input space into more than a thousand of small regions. We assumed that the model is interpretable if the number of regions is small. Based on this principle, we formulated the ATM interpretation problem as an approximation of the ATM using a smaller number of regions.

There remains several open issues. For instance, there is a freedom on the modeling (3). While we adopted a fairly simple formulation, there may be another modeling that solves Problem 1 in a better way. Another freedom is on the choice of a metric to measure the proximity between model P and model I. We used KL divergence because we could derive an EM algorithm to learn model I reasonably. The choice of the number of regions also remains as an open issue. Currently, we treat as a user tuning parameter, while, ultimately, it is desirable that the value of is determined automatically.

## References

• Beal (2003) Beal, M. J. Variational algorithms for approximate bayesian inference. PhD thesis, Gatsby Computational Neuroscience Unit, University College London, 2003.
• Breiman (2001) Breiman, L. Random forests. Machine learning, 45(1):5–32, 2001.
• Breiman et al. (1984) Breiman, L., Friedman, J., Stone, C. J., and Olshen, R. A. Classification and Regression Trees. CRC press, 1984.
• Chen & Guestrin (2016) Chen, T. and Guestrin, C. Xgboost: A scalable tree boosting system. arXiv preprint arXiv:1603.02754, 2016.
• Chipman et al. (2003) Chipman, H. A., George, E. I., and Mcculloch, R. E. Bayesian treed generalized linear models. Bayesian Statistics, 7, 2003.
• Cui et al. (2015) Cui, Z., Chen, W., He, Y., and Chen, Y. Optimal action extraction for random forests and boosted trees. Proceedings of the 21th ACM SIGKDD International Conference on Knowledge Discovery and Data Mining, pp. 179–188, 2015.
• Deng (2014) Deng, H. Interpreting tree ensembles with intrees. arXiv preprint arXiv:1408.5456, 2014.
• Eto et al. (2014) Eto, R., Fujimaki, R., Morinaga, S., and Tamano, H. Fully-automatic bayesian piecewise sparse linear models. Proceedings of the 17th International Conference on Artificial Intelligence and Statistics, pp. 238–246, 2014.
• Friedman (2001) Friedman, J. H. Greedy function approximation: A gradient boosting machine. Annals of statistics, 29(5):1189–1232, 2001.
• Jordan & Jacobs (1994) Jordan, M. I. and Jacobs, R. A. Hierarchical mixtures of experts and the em algorithm. Neural computation, 6(2):181–214, 1994.
• Mohan et al. (2011) Mohan, A., Chen, Z., and Weinberger, K. Q. Web-search ranking with initialized gradient boosted regression trees. Journal of Machine Learning Research, Workshop and Conference Proceedings, 14:77–89, 2011.
• Murthy et al. (1994) Murthy, S. K., Kasif, S., and Salzberg, S. A system for induction of oblique decision trees. Journal of Artificial Intelligence Research, 2:1–32, 1994.
• Tsanas & Xifara (2012) Tsanas, A. and Xifara, A. Accurate quantitative estimation of energy performance of residential buildings using statistical machine learning tools. Energy and Buildings, 49:560–567, 2012.
• Wang et al. (2015) Wang, J., Fujimaki, R., and Motohashi, Y. Trading interpretability for accuracy: Oblique treed sparse additive models. Proceedings of the 21th ACM SIGKDD International Conference on Knowledge Discovery and Data Mining, pp. 1245–1254, 2015.
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text:
# DPF2015

4-8 August 2015
America/Detroit timezone

## Bound on the variation in the fine structure constant implied by Oklo data

5 Aug 2015, 14:00
18m
Michigan (Michigan League)

### Michigan

#### Michigan League

Cosmology and Dark Energy Experiment

### Speaker

Ms Leila HAMDAN (Kuwait University)

### Description

Dynamical models of dark energy can imply that the fine structure constant $\alpha$ varies over cosmological time scales. Data on shifts in resonance energies $E_r$ from the Oklo natural fission reactor have been used to place restrictive bounds on the change in $\alpha$ over the last 1.8 billion years. We review the uncertainties in these analyses, focusing on corrections to the standard estimate of $k_\alpha\!=\!\alpha\,dE_r/d\alpha$ due to Damour and Dyson. Guided, in part, by the best practice for assessing systematic errors in theoretical estimates spelt out by Dobaczewski et al. [in J. Phys. G: Nucl. Part. Phys. 41, 074001 (2014)], we compute these corrections in a variety of models tuned to reproduce existing nuclear data. Although the net correction is uncertain to within a factor of 2 or 3, it constitutes at most no more than 25\% of the Damour-Dyson estimate of $k_\alpha$. Making similar allowances for the uncertainties in the modeling of the operation of the Oklo reactors, we conclude that the relative change in $\alpha$ since the Oklo reactors were last active (redshift $z\simeq 0.14$) is less than $\sim 10$ parts per billion. To illustrate the utility of this low-$z$ bound, we consider its implications for the string-inspired runaway dilaton model of Damour, Piazza and Veneziano.
Oral or Poster Presentation Poster

### Primary author

Ms Leila HAMDAN (Kuwait University)

### Co-author

Prof. Edward David DAVIS (Kuwait University)


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# jwmerrill / PowerSeries.jl

Truncated Power Series for Julia

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# PowerSeries.jl

Truncated Power Series for Julia

PowerSeries.jl defines Series types that represent truncated power series by their coefficients. You can do arithmetic on Series and apply functions to series just as you would Real or Complex numbers. Here's an example session:

```julia> using PowerSeries

# Represent the series 1.0 + 1.0*x - 2.0*x^2
julia> a = series(1.0, 1.0, -2.0)
Series2{Float64}(1.0,1.0,-2.0)

# Represent the series 1.0 + 0.0*x + 1.0*x^2
julia> b = series(1.0, 0.0, 1.0)
Series2{Float64}(1.0,0.0,1.0)

# Series add linearly
julia> a+b
Series2{Float64}(2.0,1.0,-1.0)

# The output of series operations is truncated to match the input size.
# Represents (1+x-2x^2)(1 + x^2) = 1+x-x^2+o(x^3)
julia> a*b
Series2{Float64}(1.0,1.0,-1.0)

# Extract the constant term of a series
julia> constant(a)
1.0

# Functions with known derivatives can easily be overloaded to operate on
# power series.
# You can generate the taylor series of a function about a point x up to
# e.g. 5th order by computing f(Series(x, 1.0, 0.0, 0.0, 0.0, 0.0))
julia> x = series(0.0, 1.0, 0.0, 0.0, 0.0, 0.0)
Series5{Float64}(0.0,1.0,0.0,0.0,0.0,0.0)

julia> sin(x)
Series5{Float64}(0.0,1.0,0.0,-0.16666666666666666,0.0,0.008333333333333333)

julia> log(1 + x)
Series5{Float64}(0.0,1.0,-0.5,0.3333333333333333,-0.25,0.2)

julia> 1/(1 - x)
Series5{Float64}(1.0,1.0,1.0,1.0,1.0,1.0)

# These are numerically equal to their series definitions
julia> series(0.0, 1.0, 0.0, -1.0/6.0, 0.0, 1.0/120)
Series5{Float64}(0.0,1.0,0.0,-0.16666666666666666,0.0,0.008333333333333333)

julia> series(0.0, 1.0, -1.0/2, 1.0/3, -1.0/4, 1.0/5)
Series5{Float64}(0.0,1.0,-0.5,0.3333333333333333,-0.25,0.2)

# Take the derivative of a series
julia> polyder(a)
Series1{Float64}(1.0,-4.0)

# Integrate a series term by term. Note that by convention, the constant term is 0.
julia> polyint(a)
Series3{Float64}(0.0,1.0,0.5,-0.6666666666666666)

julia> @assert polyder(polyint(a)) == a

# Truncate a series to a series 1 order lower
julia> restrict(a)
Series1{Float64}(1.0,1.0)

# Restricting a first order series returns a real number
julia> restrict(restrict(a))
1.0

# polyint, polydir, and restrict are the only operations that change the order of
# a series. Arithmetic on series of different orders is disallowed because
# relevant terms in the lower order series may have been dropped at intermediate
# steps.
julia> series(1.0, 1.0) + series(1.0, 2.0, 3.0)
ERROR: no promotion exists for Series1{Float64} and Series2{Float64}
in + at promotion.jl:158

# Truncated power series offer one of the best ways to take multiple derivatives
# of generic mathematical functions.
julia> f(x) = exp(-x^2)
f (generic function with 1 method)

julia> f2(x) = polyder(polyder(f(series(x, 1, 0))))
f2 (generic function with 1 method)

julia> f2(2.0)
0.25641894444227853

# Compare to the symbolic second derivative
julia> let x = 2.0; -2exp(-x^2)+4x^2*exp(-x^2); end
0.25641894444227853

# PowerSeries comes with types defined for series up to order 7. By default,
# trying to construct a higher order series is a type error.
julia> series(0, 1, 2, 3, 4, 5, 6, 7, 8, 9)
ERROR: no method series(Int64, Int64, Int64, Int64, Int64, Int64, Int64, Int64, Int64, Int64)

# If you want to work with higher order series, you can generate types up
# to a given order with PowerSeries.generate(order)
julia> PowerSeries.generate(9)
julia> series(0, 1, 2, 3, 4, 5, 6, 7, 8, 9)
Series9{Int64}(0,1,2,3,4,5,6,7,8,9)```

For taking first derivatives of code, see also DualNumbers.jl, and for taking symbolic derivatives, see the `differentiate` method of Calculus.jl.

Truncated series have performance advantages over symbolic derivatives for either deeply nested functions or high order derivatives.

###Theory of operation Computations of functions of a power series are based on the fundamental theorem of calculus:

Using this relation, it's easy to derive a composition rule for functions that can be applied directly to power series.

This is essentially an extension of the chain rule from infinitesimal calculus to finite step sizes.

Once differentiation and definite integration are defined on series, this relation allows a simple definition of functions of series. For example, the sine and cosine of series are mutually-recursively defined as

```sin(x::AbstractSeries) = sin(constant(x)) + polyint(polyder(x)*cos(restrict(x)))
cos(x::AbstractSeries) = cos(constant(x)) - polyint(polyder(x)*sin(restrict(x)))```

The general pattern is

`f(x::AbstractSeries) = f(constant(x)) + polyint(polyder(x)*f'(restrict(x)))`

where `f'` should be replaced by a known derivative function.

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Truncated Power Series for Julia

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Problems – We Solve Problems
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#### Extremal principle (other) , Partitions into pairs and groups bijections , Pigeonhole principle (other)

Ben noticed that all 25 of his classmates have a different number of friends in this class. How many friends does Ben have?

#### Pigeonhole principle (other) , Proof by contradiction

10 friends sent one another greetings cards; each sent 5 cards. Prove that there will be two friends who sent cards to one another.

#### Arithmetic of remainders , Pigeonhole principle (other)

Prove that in any group of 7 natural numbers – not necessarily consecutive – it is possible to choose three numbers such that their sum is divisible by 3.

#### Examples and counterexamples. Constructive proofs , Pigeonhole principle (other) , Theory of algorithms (other)

The function F is given on the whole real axis, and for each x the equality holds: F $(x + 1)$ F $(x)$ + F $(x + 1)$ + 1 = 0.
Prove that the function F can not be continuous.

#### Pigeonhole principle (other)

A class contains 38 pupils. Prove that within the class there will be at least 4 pupils born in the same month.

#### Pigeonhole principle (other)

Identical to 98634.

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# Non-convergence of ergodic measures with positive entropy

Let $T:X\to X$ be a continuous function on a compact metric space $X.$ Let $\mu$ be a $T$ invariant and ergodic probability measure on $X$ with strictly positive Sinai entropy $h_{\mu}(T).$ Let $F:X\to X$ be a continuos transformation that commutes with $T.$

Define $F_{*}\mu(A)=\mu(F^{-1}A)$ for every Borel set $A.$

I am looking for an example in ergodic theory in which

$\frac{1}{N}\sum_{k=1}^NF_{*}^k\mu$ does not converge in the weak* topology as $N\to \infty.$

Pd: It is trivial to find examples in which $F_{*}^N\mu$ does not converge but $\frac{1}{N}\sum_{k=1}^NF_{*}^k\mu$ does.

Let $T_1, \hat{F} \colon X_1 \to X_1$ be commuting continuous transformations of a compact metric space and $\mu_1$ a Borel probability measure on $X_1$ which is invariant and ergodic with respect to $T_1$, and such that $\frac{1}{N}\sum_{k=1}^N {\hat{F}}_*^k\mu_1$ does not converge, as in your previous question. Let $T_2 \colon X_2 \to X_2$ be a continuous transformation of a compact metric space equipped with a $T_2$-invariant weak-mixing Borel probability measure $\mu_2$ such that $h(T_2,\mu_2)>0$. Define $X:=X_1 \times X_2$, $T:= T_1 \times T_2$, $F:=\hat{F} \times T_2$, $\mu:=\mu_1 \times \mu_2$.


text:
College Physics (4th Edition)

$\lambda_i = 121.8~pm$
We can find the Compton shift in wavelength: $\Delta \lambda = \frac{h}{mc}~(1-cos~\theta)$ $\Delta \lambda = \frac{6.626\times 10^{-34}~J~s}{(9.1\times 10^{-31}~kg)(3.0\times 10^8~m/s)}~(1-cos~100.0^{\circ})$ $\Delta \lambda = (2.427~pm)~(1.17365)$ $\Delta \lambda = 2.848~pm$ We can find the wavelength of the incident photons: $\Delta \lambda = \lambda_f-\lambda_i$ $\lambda_i = \lambda_f-\Delta \lambda$ $\lambda_i = (124.65~pm)- (2.848~pm)$ $\lambda_i = 121.8~pm$


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# Handbook of the Geometry of Banach Spaces: 1

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My work has focused on symmetries of knots, relationships between knot invariants and invariants of three-manifolds, and an equivalence relation known as knot-concordance which brings in four-dimensional space. In these cases the error should be marked as an exception; for example, if the building shown in the example was actually a shopping mall, the one building overlapping several parcels would not be an error but rather an exception to the rule.

# Topology of Metric Spaces

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In algebraic topology we define a fundamental group $\pi_1$(X) as a directed loops inside that topological space. Rob's problem is to study the realizations as geodesics of a give configuration in hyperbolic geometry. To learn geometry and topology I would suggest the excellent books [55-63]. Third, starting from the end of the book and going all the way to the start, build a dependency outline linking the 'big' result at the end with all the preceding.

# Hyperbolic Geometry from a Local Viewpoint (London

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The aim is to have an active exchange of results, techniques and ideas on the cohomology of the moduli spaces of curves. We are now welcoming submissions for our Contributed sessions. In particular, we will present a recent joint work with Benjamin McKay (University College Cork) proving that compact simply connected complex manifolds of algebraic dimension zero (meaning that all meromorphic functions must be constant) do not admit holomorphic affine connections and holomorphic conformal structures.

# Algebraic Topology: Oaxtepec 1991 : Proceedings of an

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It's like saying "Why can't I just consider a square with 3 corners?" This is analogous to the "neutral point" between the Earth (positive) and the Moon (negative), where the gravities of either body cancel out. Requires that a polygon in one feature class (or subtype) must share all of its area with polygons in another feature class (or subtype). We invite the interested reader to see Professor Jerry Vaughan's ''What is Topology?'' page and the links therein. Computational topology is a newly emerged area that applies topological techniques in developing algorithms for problems in data and shape analysis.

# Continuum Theory and Dynamical Systems: Proceedings of the

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Although instantons were first studied for a specific 4-manifold (spacetime), analogous solutions of differential equations turned out to exist for much more general 4-manifolds. The list of graded homework problems will be announced in class and can also be found online (see table below). For if it can be demonstrated that such an arrangement can by no means be made, all labor would be useless which was directed toward bringing this about.

# A Topology of Everyday Constellations (Writing Architecture)

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The conference will consist of research talks on the latest developments relating to the moduli space of Higgs bundles. Chapter 5 is an introduction to spectral sequences, a rather intricate but very powerful computation tool. Another branch of algebraic topology that is involved in the study of three-dimensional manifolds is knot theory, the study of the ways in which knotted copies of a circle can be embedded in three-dimensional space.

# Russian Mathematicians in the 20th Century

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Creating an error log of potential topological errors in your feature dataset. The resulting map is much clearer than the unwieldy real life mess. Margaret will present a workshop from the IFF’s Making Space series looking at the foundations of Geometry and Topology. In the remaining five chapters, the author moves to a more conventional presentation of continuity, sets, functions, metric spaces, and topological spaces. He was raised in Dayton, Ohio and attended the University of Michigan, where he received the B.

# Homology of Locally Semialgebraic Spaces (Lecture Notes in

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The beam went three times around the LHC ring: see CMS’s e-commentary and CERN’s twitter. Hydrogen bonded pairings are dominated by the shortest local connection along the chain that can be made without signiﬁcant distortion of the bond geometry — bonding the carbonyl group of residue i to the amide group of residue i + 4. I think he was trying to understand algebraic surfaces in terms of algebraic curves.

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Since we are considering 2D, smooth, orientable, and compact surfaces that are embedded in the 3D Euclidean space, 3 types of topological defects can be encountered: \item{Disconnected components: in the presence of image artifacts, segmentations often contain several connected components, which might either constitute parts of the same structure or erroneous pieces of a segmentation. } \item{Cavities: cavities could be either the result of unexpected anatomical structures that are located inside the volume of interest, such as tumors, or, most frequently, the result of of image artifacts.

# Obesity, Inflammation and Cancer (Energy Balance and Cancer)

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# Tag Info

1

The simplest way to look at this is to consider separately the horizontal and vertical velocity/position. For a projectile launched at angle $\theta$ and velocity $v$, the components are: Horizontal velocity $$v_h = v\cos\theta$$ Vertical velocity $$v_v = v\sin\theta$$ The position at time $t$ is then given by $$(x, y) = (v_h\cdot t, v_v \cdot t - ... 0 The motorcycle is moving at 10\frac{m}{s}, so his distance from a starting point of 0 each second can be modeled as:$$y = 10x$$The car is moving at a constant acceleration, so that distance from an origin of 0 can be modeled as:$$y = 2(x^2 + x)$$Therefore, when they are equal to each other, they are at equal distances from the starting point. ... 2 Let's say car and bike be at rest at 1pm so, v_c=0 and v_b=0. Calculations for motion of car: Since car is moving with constant acceleration, At 1:00:00pm, v_c=0m/s, S_c=0m At 1:00:01pm, v_c=4m/s, S_c=4m At 1:00:02pm, v_c=8m/s, S_c=12m At 1:00:03pm, v_c=12m/s, S_c=24m At 1:00:04pm, v_c=16m/s, S_c=40m Calculations for motion ... 1 First, you can use your velocity expression to determine the time that it takes to reach a certain final velocity, v_f.$$v_f=v_0e^{-k\Delta t}$$Given that \Delta t is the amount of time spent decelerating. If you solve this for \Delta t you'll find:$$\Delta t=-{1 \over k}\ln\left({{v_f \over v_0}}\right)$$Next, you can determine an expression ... 1 You need to solve system like following: \begin{cases} d_0=\int_{0}^{\tau}v_0 \exp(-kt)dt=\frac{v_0}{k} (1-\exp(-k\tau)) \\ v_f=v_0\exp(-k\tau)\end{cases} You need to solve for \tau which is time till stop, and k which is essentially rate of acceleration. 1 Your calculations are correct. They differ from your model (which uses ABS braking) however, because they don't take into account the duty cycle of the braking. If this is added to your calculations, then the two results should be similar. 0 When you're using the equation F=ma, the F is ALWAYS the total/resultant/net/unbalanced force, NOT one of the individual forces. It describes the effect (the acceleration) that happens due to the cause (the total force on an object). Here you happen to be right because (at least horizontally) there is only one force, the frictional force, so you should get a ... 0 So I'm assuming you're saying that the work done on the car in distance d has to be equal to its kinetic energy \frac{1}{2} m v^2. Then, using W = F d:$$ F d = \frac{1}{2} m v^2 \\ m a d = \frac{1}{2} m v^2 \\ d = \frac{v^2}{2 a} $$So, yes, this equation is correct. Your relation between the two forces is also correct. Since mass drops out, the ... 0 Assuming that the acceleration is a m/s^2 \neq 0, then the distance travelled after reaching a velocity of \sqrt{a} m/s is always the same, namely x =0.5 m. Therefore, as the acceleration tends to zero, the time taken to reach this distance obviously tends to infinity, so your answer makes perfect sense. However, the same result cannot be applied ... 1 Velocity has units of m/s and acceleration m/s² so the root of acceleration would be √m/s. This unit does not make any sense. If you just want the numerical value, then v=a·t → √a=a·t then √a/a=t. 0 I think you should clarify the question. If it has an acceleration of a=0 it cannot attain a velocity of 1/sqrt(a). 3 She asked us if the body was accelerating or slowing down Acceleration is defined as the time rate of change of velocity and, in this example, the acceleration is constant and positive. So, the full answer is: the velocity of the body is always increasing while the speed is decreasing for t<1 and increasing for t>1. In this plot, the ... 2 If you start with the bag stationary at 300m then drop it the bag is going to fall straight down, and its maximum height would indeed just the 300m point it started from. However you're not starting with the bag stationary. You're starting with the bag moving upwards at 13 m/s. So the bag is going to start at 300m then move up, come to a halt, then start ... 0 in your answer you ignored the initial velocity the bag will rise a little until its velocity is zero then it will fall again , it is like he found the total distance 0 I understand what your teacher is saying, but I think she's wrong. In my physics classes we were told never to use the word "de"celeration, only acceleration. Why? For elegance reasons mostly. Acceleration is a vector quantity, therefore its magnitude is always positive or zero with a direction. The sign (or direction) of the acceleration depends on your ... 3 The initial velocity and acceleration here are in opposite directions. The magnitude of velocity (represented by S=|\vec v|) decreases upto a certain instant. (i.e. where \vec v=0). Edit: Also, consider these graphs. (t^. being the time where v=0) Note how the velocity increases but the magnitude of it (in the S, t graph) decreases till t^.. ... 0 1.So you have a position vector given by$$\vec{r}=\left(8t-4.5t^3\right)\hat{i}+\left(10-2t^2\right)\hat{j}$$If you use the definition of velocity as$$\vec{v}=\frac{d\vec{r}}{dt}$$then we differentiate each component of the vector to get the velocity vector.$$\vec{v}=\left(8-13.5t^2\right)\hat{i}+\left(-4t^2\right)\hat{j}$$Acceleration \vec{a} is ... 0 From what I understand, the question is pretty straight forward. For acceleration:$$L=\frac{a_1 \cdot t_1^2}{2} \implies t_1 = \sqrt\frac{2L}{a_1} \implies v=\sqrt{2a_1\cdot L}$$For deceleration:$$a_2\cdot t_2+ v=0 \implies t_2=-\frac{\sqrt{2a_1\cdot L}}{a_2}$$The minimum time is thus t_{min}=t_1+t_2. 0 (Classical Physics only) Any massive body has a property known as inertia, thus even a body floating in outer space would require some kind of force to be accelerated. Using Newtons second law, you would find$$\tag{NII} \sum \vec{F} = \frac{\mathrm{d}}{\mathrm{d}t}\vec{p},$$which for constant mass and one-dimensional motion simplifies to$$\tag{NII'} ...

1

The answer is negative. The gravitational acceleration is the same for all the atoms. Thus, the internal structure of the body is not affected. Using the terminology in your question, the average kinetic energy of atoms inside this object doesn't change. This is the reason for which given a body left free in the gravitational field, we usually consider the ...

0

Yes. The kinetic energy from translatoric motion of the object rises as a whole. Any small piece of it contributes with its' own fractional mass, since any small piece also achieves a net translatory motion. Each piece is a fraction of the whole object: $$K_{trans}=½m_{total}v^2=\sum K_{trans,atom}$$

1

I suspect your teacher glanced at it very quickly, and didn't realize you were using timesteps. When you work with timesteps you are doing a discrete approximation of the differential equations which describe the relationships, namely: $\frac{dv}{dt} = a$ and $\frac{dx}{dt} = v$ which in the discrete form, and arranged to match your code, become: $\Delta ... 0 Your solution is correct at the level of your course. We don't see what else is in your code, so there might be another problem. I can only guess what he means by "acceleration / 2". Perhaps he didn't read your program carefully enough; he might have been expecting a solution involving$1/2 a t^2$, and when he didn't see it, moved on. Your solution ... 0$s = v \cdot t$is only true if$v$stays constant. If$v$changes with time, the proper relation is $$s(t) = \int_{t_0}^{t} v(t') \mathrm dt'$$ For$v = a \cdot t$with a constant acceleration$a$this becomes $$s = \frac{1}{2} a t^2$$ So instead of calculating the position from the velocity you could calculate it from acceleration directly. 0 There is no problem to get the time$T_1$to reach the peak, and the time$T_2$to fall on the ground if you know$v_{y,0}$, the projection on the vertical of the total velocity under which you threw the ball. For this calculus you can consider only the center-of-mass movement of the ball. About the spinning movement, I don't understand in football but I ... 0 Assuming that you are talking about an uniform gravitational field$\vec{F_G}=${$0,-mg$} than it's possible to write$ y(t)=y_0+v_{0y}-\frac{1}{2}gt^2 $and$x(t)=x_0+v_{0x}t$if you impose the initial conditions on$x$you shall find that$\forall{t} \rightarrow x(t)=0$so if$b \ne 0 $the particle will never reach it beacuse it will just fall along a ... 1 When you take the average of two quantities, you need to consider the "weighting". In this case, the time spent at each of the velocities matters, and becomes this "weight". In general, when you have a weighted average you multiply each value by its weight, and divide by the sum of the weights. When all the weights are$1$, that reduces to the familiar ... 0 In the moment the bottle falls to the water, it travels downstream at$v_{stream}$. It travels a distance$X=v_{stream}∗1h$in the first hour and a distance$Y=v_{stream}∗t$in the unknown time t. Furthermore,$3km=X+Y=v_{stream}(1+t)$. About the canoe, it traveled$2km$in the first hour, and in the following time$t$it traveled back$5km$. So it traveled ... 4 If you bring an object from infinity to a distance$R$then the potential energy change is: $$\Delta U = -\frac{GMm}{R}$$ Assuming your object starts at rest, the potential energy change is equal to the change in kinetic energy, so we have: $$\frac{GMm}{R} = \tfrac{1}{2}mv^2$$ so: $$v^2 = \frac{2GM}{R}$$ You want$v \ge c$, so: $$\frac{2GM}{R} ... 1 You don't have to use the standard equations of motion, you can just deduce them from Newton's laws: You have a force \mathbf{F}=m\times -9.8 \: \mathbf{\hat{e}_y} acting on the arrow. Newton's second law is \mathbf{F}=m\frac{d\mathbf{v}}{dt}. Integrating for the x and y coordinates yield:$$\frac{dv_y}{dt} = -9.8\int_{v_{yi}}^{v_{y}}dv_y = ... 1 The symbol$s_{NN}$is in OP's context of RHIC the Mandelstam$s$-variable in a Nucleus+Nucleus collision. The$s$-variable is also known as the square of the center-of-mass energy. 0 With respect to ground: After push-off, man moves upwards at$0.8\text{ft}/\text{sec}$, pillow downwards at$8\text{ft}/\text{sec}$. It takes the pillow$1\text{s}$to hit the ground, by which time the man has moved to$8.8\text{ft}$up. Pillow (being weirdly elastic) returns at exactly the same speed it departed with, has to close the$8.8\text{ft}$gap at ... 10 No they aren't. Suppose we have some velocity$v(t)$. The differential with respect to time is just the acceleration: $$\frac{d}{dt}v(t) = a(t)$$ Now differentiate it with respect to distance$s$: $$\frac{d}{ds} v(t) = \frac{dt}{ds}\frac{d}{dt} v(t) = \frac{dt}{ds} a(t) = \frac{a(t)}{v(t)}$$ 0 You already have$t_{on\ the\ ground}$expressed through initial velocity, angle and landing distance. Plug this time into equation for vertical displacement, and you will find the height. 0 You just solve for the$t$(time) corresponding to your$x$(the landing distance) and evaluate your$y$(height) by that$t$: If$x_{(t)}=t\cdot v_0\cdot \cos (\beta )$, then$t_{(x)} = x/v_0\cdot \sec (\beta )$So you plug that in for$t$and evaluate the height$y$by distance$x$:$y_{(x)}=t_{(x)}\cdot v_0\cdot \sin (\beta )-g\cdot t_{(x)}^2/2$Also ... 0 In the case you described the function$a(v)$gives you the differential equation $$\frac{d v}{d t}=a(v).$$ You just have to solve this differential equation for$v(t)$with some initial condition$v(t=0)=v_0$. After solving the equation you get the acceleration by differentiating the solution$a(t)=\dot{v}(t)$. For the example you mentioned: The drag ... 12 Just imagine an expanding shell. Then all the parts are moving at the same speed. If the mass is$M$and the radius is$R$, then the kinetic energy is$\frac{1}{2} M \dot{R}^2$. A sphere is just a series of shells. A shell at radius$r$has mass$\mathrm{d}m = \frac{M}{4/3 \pi R^3} 4\pi r^2 \mathrm{d}r$and is expanding at a rate proportional to the ... 0 The angular momentum of the system is the same before and after the collision. Since one object is stationary before the collision, the angular momentum is just the momentum ($mv$of the moving puck multiplied by the perpendicular distance between them (which is$2r\sin(30)$). The moment of inertia of the two pucks stuck together is a little bit tricky, ... 0 A key idea is that the path of the center of mass is unaffected whether the two pucks collide or not. So take the snapshot of the situation at any moment of time before collision. Find the angular momenta of the two pucks with respect to the com at that instant of time. Since angular momentum is conserved for an isolated system (i.e$\tau_{net}=0\$), the ...

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s, u, v and a are all vector quantities. When the vector points to a direction opposite to the "default" direction, then its sign is negative. The "default" direction can be any direction that you want it to be, but note that all these variables share the same "default" direction, so switching the sign of one variable will require you to switch the sign of ...

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It all boils down to choosing the direction in which the axis increases in the vertical direction. In your case that direction is the upward direction. Distances upward from origin are positive and negative downward. velocities in which the position increases with time are positive and those in which the position decreases with time is negative.

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example motion of earth around the sun.2=motion of moon around the earth.3=a car or bicycle along a circular track possesses circular motion

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# Minimum number of circles to fill up a square

Suppose there is a square each of side $a$. I want to fill the square with circle of radius $r$. How many overlapping circles are needed to fill up the square?

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exactly filling it up is not possible i guess, it becomes a quite tedious calculation ain't it? –  Bhargav Dec 27 '11 at 16:50
anyway related to this - stackoverflow.com/questions/7946152/… ? –  Bhargav Dec 27 '11 at 16:53
There probably is no "nice" exact formula for arbitrary a and r, but an overlapping hexagonal cover will be asymptotically optimal. –  Lopsy Dec 27 '11 at 18:10
Thank you very much for answers. –  user12290 Jan 2 '12 at 6:47


text:
Talent

Nuclear Talent
Course 7
NT4A

the_nu_wgv

The $\nu$ WGV Group Members

• Don Willcox (code/parameters)
• Nicole Vassh (parameters/analysis)
• Panos Gastis (code/analysis)

Project Goal

To examine the effect of the neutrino magnetic moment on the primordial abundances. This will be done by considering the magnetic moment as a free parameter (presumably larger than the standard model prediction). This additional magnetic moment coupling will then keep the neutrinos coupled to the electrons past the traditional $\sim 1$ MeV temperature. Here we preliminarily consider the effect on primordial abundances when the magnetic pair-production process $e^{+}+e^{-}\leftrightarrow \nu +\bar{\nu}$ is included into the big bang nucleosynthesis code of F. Timmes.

Decoupling Temperature Calculation

We would like to find the decoupling temperature, such that: $$\Gamma= G_F^2T^5 + \mu_\nu^2T^3 \sim H = \frac{T^2}{M_{pl}}$$

Let's replace with some more illustrative constants: $$\Gamma = \frac{T^5}{M_W^4} + g_\nu^2\frac{T^3}{m_p^2} \sim H = \frac{T^2}{M_{pl}}$$

For $g_\nu\equiv 0$, we have: $$T_0^3 = \frac{M_W^4}{M_{pl}}$$

We can use this to rewrite the cubic equation, in dimensionless form: $$\left(\frac{T}{T_0}\right)^3 = 1 - g_\nu^2 \frac{M_{pl}^{2/3}M_W^{4/3}}{m_p^2}\frac{T}{T_0}$$

We can see that the last term is of order unity when $g_\nu\sim \frac{m_p}{M_{pl}^{1/3}M_W^{2/3}}\sim 10^{-8}$, allowing us to rewrite the equation as: $$\left(\frac{T}{T_0}\right)^3 = 1 - \left(\frac{g_\nu}{10^{-8}}\right)^2 \frac{T}{T_0}$$