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FREELessons: 27Length: 4.7 hours

• Overview
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# 4.1 Colors and Typography

When writing the CSS for a webpage, a good place to start is the typography and color scheme. That’s exactly what we’ll cover in this lesson.

## 5.Conclusion1 lesson, 02:03

### 4.1 Colors and Typography

In the previous lesson we finished writing the markup for most of our pages. Now, it's time to begin styling these pages, and a place where I always start doing that is with colors and typography. First, let me show you the SAS structure of foundation. So basically, in the SRC Folder under assets, we find app.scss, which basically loads all the classes, all the mixins I use to buy foundation. Now what I did was right after they imported their settings, I'm importing my own settings. And I'm gonna be using this custom/helper/settings file which can be found right here. I'm gonna be using this to overwrite some of their settings. And also at the very bottom I'm importing custom/custom which is this custom.scss file. Now in here I'm loading my own styles. I'm loading some utilities that currently look something like this just very basic stuff like, defining a class for the icon font, defining a class for vertical centering, one for small text, round corners and now we have to make sense here and so on. Just very basic, basic stuff. And then I am loading my own files typography, navigation, sidebars, headers and so on. And all of these you can find right here. And I prefer using my own files to create my own styling to overwrite and to complete some of the foundation classes. I think that's the best way to do it. So we'll begin in the settings. So if we take a look at the foundation settings, this is what their file looks like. They're very well organized into global, bright points, the grid, so these are all variables, these are all settings. And you can either overwrite these in their own settings file or you can create your own file and overwrite them there, just like I did. So for the colors they're using this foundation palette and they have a mix and cold yet color which queries this map and gets a particular color. Now, what I'm gonna do is simply copy this, and I'm gonna open my own settings file, and I'm gonna say foundation palette, that's fine. And I'm actually gonna keep these values that they have, but I'm gonna add my own, so I'm gonna say custom colors, let's add some comments in here, and the colors will be based on the color palette that I defined in sketch, right? Which is this. So I'm gonna get this as the primary, as the secondary, and this has the black, black 85, black 65, 50, 12, five and so on. And I'm also gonna get these colors right here. So we can simply start like this, ap-Black, and we're gonna set that particular color, which is this, and then we will continue with the rest like that. So we have all the colors that we currently need. Next, I'm gonna define a variable for each of these colors, like this. That way it's much easier to get a color. For example, I can say body color equals AP black or AP black 85. And the values of these colors are taken from this foundation palette map using the get color mixing. Next, I'm gonna define some typography settings. First, the body font family. Setting it to PT sans with the proper fallbacks, the header font family, mully again with the proper fallbacks. And then I'm defining a couple of settings where I'm overwriting a couple of settings from the original foundation like header margin bottom, paragraph line height. And the small font size I believe they had it at about 75%. I'm defining another variable called super small font size. That's gonna be used, I think, for badges. And a border for the code, currently set to none. Then foundation also gives us some method maps for the header styles. And you can very easily use these to define the font sizes for all of your headings. On small screens, these are the font sizes that I want. On medium and large screens or on medium and above, these are the ones that I need. So by doing that, we can already start to see changes in our page. Now let's open up typography.scss and start adding the proper styling for our typography. We'll start for the body, and the body has a color set to $ap-black-65. And I'm just gonna be doing back and forth between the page and the code so you can see the changes live. Next, let's target the links. And the links have a color of$ap-primary, it's that orange color. And also they're bold. So font-weight, set that to bold and also we'll set a position relative because we will be adding a lot of icons to these links. And then on hover, we'll be changing the color to $ap-black-85. And also let's add a transition. So I'm gonna include one of my own mix-in's, which is called ap-transition. So, if we take a look at the utilities, by default, ap-transition, transitions all properties in 200 milliseconds with this particular ease-in. So now, transition, if I can spell this correctly. So now, all of the links have changed. They have a nice transition, they have the proper color both on the default state and on hover, very good start. Next, let's see about the headings. So first of all we'll target H1 through H4, and we'll set the font-weight to black, which is 900, and we'll also set their color to black 85. So$ap-black-85. Okay. Then we'll target the H5 and H6. And we'll set a font weight. These are in bold so it's 700 and we'll also set a color, $ap-black 85, and actually, we can add them right here. And finally, the H6 which is used and these pages right here, that will be a bit lighter so it's gonna be$ap-black-50. It's gonna be uppercase, and also let's do a little bit of letter spacing 1 pixel. Okay, what else? Let's create that ap-muted class. And simply set the color to ap-black-50. And actually, we can also target it here. I can simply say, extend ap-muted. And because of the power of SAS, this particular style will be applied to the H6s. Let's refresh. Okay, perfect. Now, that's pretty much all I'm gonna do at this point. I've created the proper SAS structure. I've defined the proper settings, and I've also written some very basic typography styles. In the next lesson we're gonna start styling this main navigation, and also the off canvas navigation, so I'll see you there.

Back to the top

format: plain


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# 2nd basis function for 2nd order ODE

1. ### X89codered89X

153
i have the first solution y_1(t) = t for (1-t)y'' + ty' - y = 0.

I need to get the 2nd linearly independent using Abels theorem.

the integration is messy but i have it set up (sorry no latex);

y_2 = (t) * integral to t ( 1/s^2 * exp( -integral to t (s(s+1) ds) ) ds.

Could anyone show me how to do this integration step by step?


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# Export CSV from tables

I am reading my csv which looks like:

float,float,float,integer,boolean

table = Take[Import["~/myfile.csv"]];
x = table[[All, 1]];
y = table[[All, 2]];
z = table[[All, 3]];
time = table[[All, 4]];
bool = table[[All, 5]];

Then I do some job with x,y,z and I need to export it back to csv. But I dont know how to get the table to export.

-
Put it in a list like so: list={x, y, z, time, bool}; and then Export["newfile.csv",list,"CSV"] –  Pickett Apr 13 at 20:12
it puts every table into one row. I need it as columns seperated by comma –  SpeedEX505 Apr 13 at 20:19
I what @Pickett suggests is basically correct, but I also that you get an output table more to your liking by using list = Transpose @ {x, y, z, time, bool};. That will put things back into columns. –  m_goldberg Apr 13 at 20:21
BTW, {x, y, z, time, bool} = Transpose[table]; is an easy way to extract column data. –  m_goldberg Apr 13 at 20:24
It works, thank you :) If you write it as an answer I'll mark it as solved. –  SpeedEX505 Apr 13 at 20:25

This is a job for Transpose.
table = Take[Import["~/myfile.csv"]];


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# Basic operations¶

## Supported VSM formats¶

At the moment the following data formats are supported:

• .bin format of word2vec (the file has to be called “vectors.bin”)
• .npy arrays with separate vocab files
• .txt plain-text vectors
• sparse vectors in hp5 format

## Importing vectors¶

VSMlib assumes a one-folder-per-vsm folder structure. All files related to the same vsm - the metadata, vectors, vocab files, etc. - must all be stored in one directory. If the vector files has the correct extension (.npy, .txt, .bin, .hp5), the library will attempt to “guess” the correct module to load it with.

>>> import vsmlib
>>> path_to_vsm = "/path/to/your/model"

The name of the model is the name directory in which the vector files are stored. For models generated with VSMlib, interpretable folder names with parameters are generated automatically.

>>> print(my_vsm.name)
w2v_comb2_w8_n25_i6_d300_skip_300

You can access the VSM metadata (recorded in metadata.json file located in the same directory as the VSM) as a Python dictionary:

>>> print(my_vsm.metadata)
{'size_dimensions': 300, 'dimensions': 300, 'size_window': '8'}

## Getting top similar neighbors of a word¶

>>> my_vsm.get_most_similar_words("apple", cnt=5)
[['apple', 1.0000000999898755],
['fruit', 0.61400752577032369],
['banana', 0.58657183882050712],
['plum', 0.5850951585421692],
['apples', 0.58464719369713347]]

This method takes an optional cnt argument specifying how many top similar neighbors to output (the default is 10). Note that the top similar vector is always the target word itself.

If you need to compute nearest neighbors for many words, this function works faster if the VSM is normalized. If it was generated with vsmlib, the normalization will be recorded in metadata, and can be checked with .normalized. VSMlib will automatically check for normalization and use the faster routine if possible. If not, you can first normalize your VSM with .normalized() method.

If for whatever reasons you need your VSM to not be normalized, you can use .cache_normalized_copy() method to cache normalized copy of embeddings. Please note that latter will consume additional memory.

.get_most_similar_vectors() enables you to do the same as .get_most_similar_words(), but searching the top neighbors by the vector representation rather than its label.

## Words to vectors and back¶

First, you need to import your model from a directory that holds only that model (.npy, .bin, .hp5 or .txt formats) and any associated files.

getting the vector representation of a word

>>> my_vsm.get_row("apple")
array([-0.17980662,  0.27027196, -0.33250481,  ... -0.22577444], dtype=float32)

You can use the above top-similar function to get the label of the vector most corresponding to your vector in your VSM vocabulary:

>>> vsm.get_most_similar_vectors(vsm.get_row("apple"))

## Filtering the vocabulary of a VSM¶

In certain cases it may be useful to filter the vocabulary of a pre-trained VSM, e.g. to ensure that two models you are comparing have the same vocabulary. VSMlib provides a .filter_by_vocab() method that returns a new model instance, the vocabulary of which contains only the words in the provided Python list of words. The list can be empty.

>>> my_vsm.get_most_similar_words("cat", cnt=5)
[['cat', 1.0],
['monkey', 0.95726192],
['dog', 0.95372206],
['koala', 0.94773519],
['puppy', 0.94360757]]
>>> my_new_vsm = my_vsm.filter_by_vocab(["dog", "hotdog", "zoo", "hammer", "cat"])
>>> my_new_vsm.get_most_similar_words("cat", cnt=5)
[['cat', 1.0],
['dog', 0.95372206],
['hotdog', 0.84262532],
['hammer', 0.80627602],
['zoo', 0.7463485]]


text:
# Get Answers to all your Questions

#### Fill in the blanks to make the statement true:Fraction  $\frac{7}{25}$  is equal to the decimal number ______.

$\frac{7}{25}=\frac{7}{25} \times \frac{4}{4}=\frac{28}{100}=0.28$

Fraction  $\frac{7}{25}$  is equal to the decimal number 0.28.


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### What is the score?

1-2

3-4

stesteste Newbie | Next Rank: 10 Posts
Joined
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Posted:
3 messages

Mon Oct 09, 2017 6:50 am
The following appeared as part of an article in a daily newspaper:

“The computerized on-board warning system that will be installed in commercial airliners will virtually solve the problem of midair plane collisions. One plane's warning system can receive signals from another's transponder-a radio set that signals a plane's course-in order to determine the likelihood of a collision and recommend evasive action.”

Discuss how well reasoned...

THIS is my essay:

The computerized on-board warning system is considered to be the solution of midair plane collisions. This argument is extremely flawed as it stands on a questionable assumption. For instance, the argument assumed that the radio set can receive signals from anywhere, even the highest sky plane can be. Furthermore, it stands on no grounds that recommending evasive action can actually resolve the collision.

To assume that the transponder can receive signals no matter how high one plane is an assumption that cannot be made. There are some occations in which warning system cannot correspond with the airliners because the plane is too high to send and receive signal. If the warning system is designed for receiving signals in a very high place, the argument will be defendable.

Additionally, recommending an evasive action does not necessarily means that the collision will be avoided at once. Sometimes people are too panic to do just simple things to save themselves. Thus, there might be not enough time to get ready to do things advised by the device. This flaw can be fixed if all crew are trained and experienced in facing emergency situations. However, if the collisions are inevitable, even the most sophisticated evasive actions can only stem a devastated effect at best.

Overall, we need to question the ability of computerized on-board warning system on solving midair plane collisions. The assumption on the ability the transponder can get signals is not necessarily true, and so does recommending an evasive action.

----
Notes: I find pretty hard to write within 30 minutes. I wrote this essay in just 30 minutes, and I guess this essay isn't long enough. A good advice to improve my time management will be so helpful. Thanks!

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Why is the hydrolysis of the conjugate base of a weak acid neglected in buffer solutions?

Consider a solution which contains a weak acid and the salt of its conjugate base with a strong base. e.g. $$\ce{CH3COOH + H2O <=>[K_\rm{a}] CH3COO- + H3O+}\tag1$$ $$\ce{CH3COO- + H2O <=>[K_\rm{h}] CH3COOH + OH-}\tag2$$

Now, $\displaystyle{K_\rm{a} = \frac{[\ce{H3O+}] [\ce{CH3COO-}]}{[\ce{CH3COOH}]}}$ and $\displaystyle{K_\mathrm{h} = \frac{K_\rm{w}}{K_\mathrm{a}}}$.

Now, my notes say that for the equilibrium concentration of undissociated acid, the acid coming due to hydrolysis of the salt is neglected and the equilibrium concentration is approximately equal to the initial concentration.

It is also mentioned in a book that

Assume that the extent of protonation of acetate ions and the deprotonation of acetic acid molecules is so small that the concentrations of both species are nearly the same as their initial values.

Why is this so? If $K_\rm{a}$ is low, then $K_\rm{h}$ should be high. I understand that the deprotonation of acid will be low as it is a weak acid so that we can neglect the $\ce{CH3COO-}$ coming through the acid dissociation but how can we neglect the hydrolysis of the $\ce{CH3COO-}$ coming from the salt?

Edit $-$

My calculation for equilibrium concentrations.

$\ce{[CH3COOH]_\mathrm{eq}} =$ initially added acid $-$ deprotonated acid $+$ hydrolysed salt

$\ce{[CH3COO-]_\mathrm{eq}} =$ initially added salt $-$ hydrolysed salt $+$ deprotonated acid

$\ce{[H3O+]}_{\mathrm{eq}} =$ deprotonated acid $-$ hydrolysed salt

Let the concentration of

initially added acid $= c_1$

initially added salt $= c_2$

deprotonated acid at eq. $= x$

hydrolysed salt at eq. $= y$

Thus, $$K_{\mathrm{a}} = \frac{[\ce{H3O+}] [\ce{CH3COO-}]}{[\ce{CH3COOH}]} = \frac{(x - y)(c_2 - y + x)}{(c_1 - x + y)}$$

Now, assuming $x << c_1, \, c_2$,

$$K_{\mathrm{a}} = \frac{(- y)(c_2 - y)}{(c_1 + y)}$$

• You may take it into account and compare the results. – Ivan Neretin Sep 21 '17 at 5:10
• @IvanNeretin Please see my edited question. – Apoorv Potnis Sep 21 '17 at 5:54
• You said before that you neglect $y$, not $x$. – Ivan Neretin Sep 21 '17 at 6:15
• @IvanNeretin I said that I understand why deprotonation is neglected, i.e. why $x$ is neglected. I don't understand why the hydrolysis, i.e. $y$ is neglected. – Apoorv Potnis Sep 21 '17 at 6:21
• Hydrolysis = cleavage of chem. bonds by addition of water (en.wikipedia.org/wiki/Hydrolysis). Where is there a hydrolysis here? How can acetic acid be hydrolyzed? – logical x 2 Sep 21 '17 at 10:40

$\mathrm{p}K_\mathrm{a} = 4.75$, therefore $K_\mathrm{a} = 10^{-4.75}$. Thus,

$$K_\mathrm{a} = \frac{[\ce{AcO^-}][\ce{H+}]}{[\ce{AcOH}]}$$

So, let's assume $[\ce{AcOH}] = \pu{1.0 M}$. Just to get a feel, how far the acid is dissociated:

$$K_\mathrm{a} = \frac{x \cdot x}{\pu{1.0 M}} \to x = \sqrt{10^{-4.75}}~\pu{M} = 0.00422 \approx 0.4\%$$

where we have assumed that the contribution from the dissociation of water is negligible, and therefore $[\ce{H+}] = [\ce{AcO-}] = x$.

Now, what happens if I add $\pu{1.0 M}$ $\ce{AcO-}$? Then, effectively, we have:

$$K_\mathrm{a} = \frac{[\ce{AcO-}][\ce{H+}]}{[\ce{AcOH}]} = \frac{\pu{1.0 M} \cdot [\ce{H+}]}{\pu{1.0 M}}$$

Therefore, we have:

$$[\ce{H+}] = K_\mathrm{a} \frac{\pu{1.0 M}}{\pu{1.0 M}} = \pu{10^{-4.75} M}$$

If we compare the two cases, before addition of acetate, we had a $\mathrm{pH}$ of

$$[\ce{H+}] = \sqrt{10^{-4.75}} \to \mathrm{pH} = 2.375$$

and after addition of acetate, we have

$$[\ce{H+}] = 10^{-4.75} \to \mathrm{pH} = 4.75$$

Meaning that the solution got more basic after the addition of acetate, as excpected.

• Hopefully that provides a good starting point for understanding. – logical x 2 Sep 21 '17 at 12:35
• @deuxexmachina After you added 1.0 M acetate ions, how did you take equilibrium concentration to be 1.0 M? How is initial concentration equal to equilibrium concentration? – Apoorv Potnis Sep 24 '17 at 4:51
• @ApoorvPotnis Yeah, I figured that this point wasn't really clear yet. So first of all, you can neglect that the acid CH3COOH will dissociate to CH3COO-, as by the above argument, it only dissociates to a very little extent. However, your concern is the other way around, and you argue that CH3COO-Na+ could be protonated to form CH3COOH + NaOH. But: look at the pKa of NaOH and CH3OO-Na: NaOH is a much stronger base! Therefore, you can assume that $\mathrm{[NaCH3COO] = [NaCH3COO]_0}$. – logical x 2 Sep 24 '17 at 7:38
• @ApoorvPotnis You can also find a similar description here: study.com/academy/lesson/… for a similar explanation – logical x 2 Sep 24 '17 at 7:40
• Can you add your comment about $\ce{NaOH}$ being a stronger base into the answer so that I could accept it? – Apoorv Potnis Nov 6 '17 at 12:50


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# Predicting the 2018 World Cup Winner Using Machine Learning

With the start of the 2018 FIFA World Cup approaching, soccer fans around the world is trying to figure out who will win the tournament? If you’re a techie and a soccer fan, you’ll probably want to figure out more quantitiative methods of figuring out the answer to that question.

## Goal

The goal of this project is established as follows:

• Use Machine Learning to predict who will win the 2018 FIFA World Cup.
• Predict the outcome of individual matches for the entire competition.
• Run simulation of the next matches i.e. quarter final, semi finals and finals.

These goals present a unique real-world Machine Learning prediction problem and involve solving various Machine Learning task: data integration, feature modeling, and outcome prediction.

## Data

I used two data sets from Kaggle. You can find them here. We will use results of historical matches since the beginning of the 1930s FIFA Championship for all participating teams.

We have opted not to use FIFA national team rankings, due to the limitations of FIFA rankings being created in the 90’s. Considering this limitation, we will stick to the historical match records.

First, I’m going to conduct some exploratory analysis on the two datasets. Next, I will use a feature engine to select the most relevant feature for my prediction. Afterwards, I will attempt to manipulate the data. Finally, I will choose a Machine Learning model and deploy it on the dataset.

#### First Thing’s First

We need to import the necessary libaries and load the datasets into a Dataframe. We will be using the following python libraries.

• Pandas: Provides in-memory 2d table objectd referred to as ‘Dataframes.’
• Numpy: Allows use to conduct fast mathematical computation on arrays and matrices.
• Matplotlib: A Python SD plotting library.
• Seaborn: A data visualization library based on matplotlib.
• Scikit-Learn: A machine learning library.

Ensure the information is imported by calling world_cup.head() and results.head() for both datasets. The Dataframes should look something like the following:

## Exploratory Analysis

Exploratory analysis and feature engineering is the most time consuming part of any Data Science project.

After analyzing both datasets, the resulting dataset has information on previous matches. The new (resulting) dataset will be useful for analyzing and predicting future mathces.

We begin our exporatory anaylsis by added goal differentials and match outcomes to the results Dataframe.

The outcome of the new Dataframe has the new results with the goal differential visible.

Now I’ll work on a subset of the data; one that includes games played by a participating team. I wanted to analyze the US Mens National Team, but unforunately, they were eliminated by Trinidad & Tobago in the qualifying rounds.

The Danish Mens National Team (NKVD) was also eliminated by The Swedish Mens National Team. Because of this, I have chosen to analyze the German Mens National Team.

Analyzing the subset of the data will help us determine which features are more relevant for our analysis.

As you can see, the dataset has matches dating back as early as 1908! However, we’re only interested in looking at matches from the World Cup era, which started in 1930. We then create a column for year and pick all the games played 1930.

We can now visualize the most common match outcome for Germany throughout the years.

FIFA has recorded 863 matches from the German Mens National Team since 1930, of which Germany has won over 500 matches with ~175 losses. Slightly more matches has ended in a draw. Getting the win-rate for every country is a useful metric. We could use it to predict the most likely outcome of each match in the tournament.

#### Analyzing Teams Participating in the World Cup

We can start off by creating a list of teams participating in the 2018 World Cup. The following teams participating are as follows:

• Group A: Uruguay, Russia, Saudi Arabia, Egypt
• Group B: Spain, Portugal, Iran, Morocco
• Group C: France, Denmark, Peru, Australia
• Group D: Croatia, Argentina, Nigeria, Iceland
• Group E: Brazil, Switzerland, Serbia, Costa Rica
• Group F: Sweden, Mexico, South Korea, Germany
• Group G: Belgium, England, Tunisia, Panama
• Group H: Columbia, Japan, Senegal, Poland

The following code will show only a dataframe with team wordl cup results starting from 1930, while dropping the duplicates (dropping games before 1930). We will also drop the date, home_score, away_score, tournament, city, country, goal_difference, and match_year columns from the df_teams dataframe and assign it to a variable named df_teams_1930. This will help us create a prediction label to simplify and process our model.

For our prediction labels, we need to create a system to help our algorithm to determine a “winning” outcome versus a “losing” outcome. During group play, FIFA rewards 2 points to the winner of the match; 0 points to the loser of the match; and 1 point to both teams if the match results in a draw. We will adopt the same system to our labels dataset.

From there, the home_team and away_team columns will be converted from categorical variables to continuous inputs, by setting dummy variables using Pandas get_dummies() function. It replaces categorical columns with quantitative representations, which enable them to be utilized in the Scikit-learn model.

We we separate the labels and features, the train_test_split function will split the data into 70 percent training and 30 percetn testing.

#### Logistic Regression

The Scikit-learn package allows us to use machine learning packages such as Linear Regression. The only problem is that we can’t use linear regression on a categorical dependent variable. Instead, in such situations, we should try using algorithms such as Logistic Regression.

Logistic regression is a Machine Learning classification algorithm that is used to predict the probability of a categorical dependent variable. In logistic regression, the dependent variable is a binary variable that contains data coded as 1 for success or 0 for failure. In other words, the logistic regression model predicts P(Y=1) as a function of X.

Logistic regression is one of the most popular ways to fit models for categorical data, especially for binary response data in Data Modeling. It is the most important (and probably the most used) member of a class of models called generalized linear models. Unlike linear regression, logistic regression can directly predict probabilities (values that are restricted to the (0,1) interval).

Logistic Regression is used when the dependent variable (target) is categorical. For example:

• To predict whether an email is spam (1) or not spam (0)
• Whether a company is a bankrupt (1) or not bankrupt (0)

Logistic regression is generally used where the dependent variable is Binary or Dichotomous. That means the dependent variable can take only two possible values, such as “Yes” or “No,” “Default” or “No Default,” “Living” or “Dead,” etc. Independent factors or variables can be categorical or numerical variables.

In our case, logistic regression attempts to predict an outcome (a win or loss) given a set of data points (stats) that likely influence that outcome. The way this works in practice is you feed the algorithm one game at a time, with both the aforementioned “set of data” and the actual outcome of the match. The model then learns how each piece of data you feed it influences the result of the game positively, negatively, and to what extent.

## Machine Learning: Match Prediction

Passing our features and labels into our algorithm, we recieved a training set accuracy of 0.571 and a testing set accuracy of 0.564, which isn’t the greatest, but we will continue to use these features. At this point, we will create a dataframe that we will deploy out model.

We will start by loading the FIFA ranking as of April 2018 data set and a dataset containing the fixtures of the group stages of the tournament obtained from here. The team which is positioned higher on the FIFA Ranking will be considered the “favorite” of the match, and therefore, will be positioned under the “home_teams” column, since obviously the only home team in this world cup is Russia. We will then add teams to the new prediction dataset based on the ranking position of each team. The next step will be to create dummy variables and then deploy the machine learning model.

We can start with deploying the model to the group matches. The follow shows the match results for the Group of Death, which contains Germany, Mexico, Sweden and South Korea.

Germany and Mexico
Winner: Germany
Probability of Draw: 0.267
Probability of Mexico winning: 0.147

Sweden and Korea Republic
Winner: Sweden
Probability of Draw: 0.308
Probability of Korea Republic winning: 0.170

Mexico and Korea Republic
Winner: Mexico
Probability of Draw: 0.295
Probability of Korea Republic winning: 0.218

Germany and Sweden
Winner: Germany
Probability of Draw: 0.243
Probability of Sweden winning: 0.189

Mexico and Sweden
Winner: Mexico
Probability of Draw: 0.253
Probability of Sweden winning: 0.317

Germany and Korea Republic
Winner: Germany
Probability of Draw: 0.277
Probability of Korea Republic winning: 0.119

Based on this information, it looks like the former champions will be making it out of the Group of Death. Unless the World Cup Champion’s curse has anything to say about it, prehaps Mexico and Sweden will both make it out of group stages.

##### Knockout Stage (Group of 16)

Based on the model, the following matchups are generated.

Our model has generated the following winners from the knockout stage of the 2018 World Cup:

• Portugal: 44.1% Probability of Winning
• France: 47.6% Probability of Winning
• Brazil: 70.2% Probability of Winning
• England: 51.3% Probability of Winning
• Spain: 52% Probability of Winning
• Germany: 66.8% Probability of Winning
• Belgium: 50.3% Probability of Winning
##### Quarter-Finals

Our model predicted a quarter final matches of the following:

• Portugal vs. France
• Spain vs. Argentina
• Brazil vs. England
• Germany vs. Belgium

Based on the matchups, our algorithm has predicted the winners of their quarter final matches:

• France: 42.9% Probabillity of Winning
• Argentina: 52% Probabillity of Winning
• Brazil: 52.5% Probabillity of Winning
• Germany: 58% Probabillity of Winning

This time, the algorithm has less than a 60% certainity for most of our match-ups. This is what we can expect as we reach a higher level of play from our competitors.

##### Semi-Finals

The model predicted the following semi-final matches:

• Brazil vs. France
• Germany vs. Argentina

Germany and Argentina are once again matched up with one another, except this time it is in the semi-finals instead of the grand finals of the 2014 world cup. Based on the matchups, our algorithm has predicted the winners of their semi-final matches:

• France: 69.3% Probabillity of Winning
• Germany: 52% Probabillity of Winning

Our algorithm once again has Germany emerging has the victory of the Germany vs. Argentina match. Our model also estimates that Brazil will end up losing to France. Unforunately, the 5 time World Cup champs will have to miss out on another Final.

##### Finals

Finally, we have the grand finals with France vs. Germany. What outcome does our model give us?

• Germany: 52.6% Probabillity of Winning

According to the model, Germany is likely to win the World Cup, although, it isn’t possible for repeat champions to emerge. The last time this occur was during the 1962 in Chile.

## Areas of Further Research/Improvement

Some areas for improvement:

• For improvements of the datasets, we could use FIFA (the game, not the organization), to assess the quality of each team player
• A confusion maxtrix would be great to analyze
• Using more models together to improve accuracy.

## Conclusion

There is so much that can be improved upon here. For now, lets see if we get any lucky.

The full code can be found here.

##### Andre Sealy
###### Data Scientist / Researcher / Writer

My research interests involves Quantitative Finance, Mathematics, Data Science and Machine Learning.


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2019
Том 71
№ 7

# On the theory of $\mathcal{PT}$-symmetric operatorss

Abstract

This article develops a general theory of $\mathcal{PT}$-symmetric operators. Special attention is given to $\mathcal{PT}$-symmetric quasi-self-adjoint extensions of symmetric operator with deficiency indices 〈 2, 2 〉. For these extensions, the possibility of their interpretation as self-adjoint operators in Krein spaces is investigated, and a description of nonreal eigenvalues is given. These abstract results are applied to the Schrodinger operator with Coulomb potential on the real axis.

English version (Springer): Ukrainian Mathematical Journal 64 (2012), no. 1, pp 35-55.

Citation Example: Kuzhel' S. A., Patsyuk O. M. On the theory of $\mathcal{PT}$-symmetric operatorss // Ukr. Mat. Zh. - 2012. - 64, № 1. - pp. 32-49.

Full text


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SkyCivドキュメント

SkyCivソフトウェアのガイド - チュートリアル, ハウツーガイドと技術記事

1. ホーム
2. TechNotes
3. 構造3D
4. SkyCivを使用したモーメントフレームの設計 (AISC 360-10)

# SkyCivを使用したモーメントフレームの設計 (AISC 360-10)

## How to model, analyse and design a Steel Moment Frame using SkyCiv

This tutorial is a simple guide of the moment frame design process using the integrated SkyCiv modules: 構造3D, メンバーデザイン, and Connection Design.

Moment frames are combinations of columns and beams connected with partially or fully restrained connections. These connections experience moment forces and cannot be considered simply supported conditions. They are used in similarity to shear walls and braced frames; they resist lateral forces in conjuncture with vertical forces. 固定された性質のため, フレーム メンバー’ 剛性は、フレームの横方向の剛性の主な原因です. ブレース付きフレームは通常、必要な材料に基づいて安価なオプションです, but moment frames gives architects and designers more space between columns and therefore give a better aesthetic. 次の記事は次の内容で構成されています:

## SkyCiv構造3D (S3D): Modeling and Analyzing our Moment Frame

### モデリング

For modeling, we will be using SkyCiv構造3D 分析ソフトウェア (get started for free at SkyCiv’s homepage). But before we get into modeling, make sure to confirm what units you are working in. This can be done by going to 設定 on the top right of the toolbar. この例では, we will be operating with Imperial units. We will assume our single bay moment frame has a centroid beam height of 18 feet and a 30 foot span. また, we will assume our moment frame is made up entirely of W-shapes from the AISC database and are fabricated from ASTM A992 Gr. 50 鋼. 最後に, the trial column and beam size of W12x40 そして W18x65 使用されます, それぞれ,

To start the workflow, lets input the nodes of our moment frame. 右から左に、選択ボックス内に完全に含まれている要素のみが選択されます ノード on the input tab (left hand side of the screen). The nodes of our frame will be (バツ,そして): (0,0), (30,0), (0,18), そして (30,18).

Now lets generate the two columns and beam that make up our frame. これをする, 2つのオプションがあります: 1) に行く 会員 in the input tab and type in the end nodes for each of our members or 2) draw the members in using the ペンツール located on the right vertical toolbar in the model space. 瞬間フレームで作業しているため, the end releases of each member at each connection location needs to be fully fixed, denoted as “FFFFFF” and highlighted as “フレーム” when creating members within the 会員 タブ. これまでの瞬間フレームを見てみましょう:

Click submit to import the material. After importing the wanted material, lets assign our member sizes. 右から左に、選択ボックス内に完全に含まれている要素のみが選択されます セクション on the input tab and click on Builder. 今, are operating within the SkyCivセクションビルダー モジュール, directly integrated with our current project in S3D. While in Section Builder, lets import the two W-shapes by going to the データベース portion of the section builder and selecting each shape. Before clicking 繰り返し間の距離, make sure that the material on the right hand side is set to the desired material that we want associated with both members: ASTM A992 Steel.

After bringing the two sections into the model, assign them to each member using the セクションID field on the input tab while each member is selected. For a more in-depth look at SkyCivセクションビルダー and its capabilities, check out its Software Documentation.

The last thing we need to do is apply a support condition. Unlike gravity columns or columns part of a braced frame, our column supports need to be fully fixed and therefore will experience moment at the base. サポートを追加するには, NS + 両方のノードをクリックします, 右クリックしてから左クリック サポートを追加. 入力タブに移動します – すでにノードが選択されています – 固定支持条件を選択. これは、に行くことによっても行うことができます サポート in the input tab and applying the supports to each node.

Our stick model should now show two different members for the columns and beam in addition to our supports:

For our moment frame, we also need to make sure that the orientation of our columns is correct and favorable for lateral action. Because the moment frames will be experiencing lateral forces, the strong axis of the column is almost always perpendicular to the load direction. この場合, our lateral force will be in the X-direction, so the column flanges should be perpendicular to the X direction. A good way to verify the orientation of your members is by entering the 3D rendering. これをする, go to the right toolbar in the model space and click on the shape that looks like a 3D cube, will bring you into the 3D render of your model. Take a look at our moment frame in 3D:

You can see that our columns are orientated the correct direction. With a more complicated 3D structure you can rotate the orientation of any member by selecting the member, clicking on 高度な, and then inputting the intended rotation in degrees into the 回転角 分野.

Before you load your structure, double check that you have the correct supports, セクション, and materials for each section.

### 負荷と負荷の組み合わせ

$$側面:Wind\:Load=8\:キップ:@\:top\:elevation$$

$$Dead\:Load=40\:psf) \(40\:psf*8\:ft=320\:lb/ft=0.320\:k/ft$$

$$住む:Load=100\:psf) \(100\:psf*8\:ft=800\:lb/ft=0.800\:k/ft$$

モーメントフレームが対称であるため, 1つの風向だけを見ていきます. フレームがさまざまな方向にさまざまな大きさの風荷重を見る場合, その風荷重ケースも追加する必要があります.

Notice that you need to assign a load case to the self-weight (SW1) 同様に. To generate load combinations go to コンボをロードする in the input tab and click on デザインコード. We will be using LRFD design in this case. Display the load combinations and you should see this:

If you forgot to assign load cases before the import, you can do that here as well. Our load combination generator will pull any ASCE 7-10 load combinations that include at least one of any of the load cases your project. 私たちの場合には, we do not have seismic loads, so we will uncheck ULS: 5 and ULS: 6. We also do not have any Roof Live or Snow loads, and we know that the 1.2D + 1.6L load combination will govern for vertical forces, so we will uncheck ULS 3. Once we have the desired load combinations checked, クリック インポート to import them into our project.

Now that we have generated our nodes, 会員, サポート, 負荷, 荷重ケース, and load combinations, we can commence our analysis. For any information about any of these steps or about using S3D, 私たちをチェックしてください ソフトウェアのドキュメント.

### 分析, 結果, and Post-processing

It is good practice to go to 編集する – 修理モデル to repair any inconsistencies or merge nodes in your model before beginning the analysis of your structure. また, make sure your project is stable by checking end releases and support conditions throughout. 最後に, に行く 設定 – ソルバー to look at and edit any pertinent analysis settings, such as convergence tolerance during P-delta analysis or the number of evaluation points per member.

Because we did not include any dynamic loading into this example, there are three solution methods that we can use:

1. 線形静的
2. 線形静的 + 座屈
3. 非線形静的 + Pデルタ

At a minimum it is suggested to run the Linear Static + Buckling analysis to see if any members in your structure are subject to buckling. We will use Non-Linear Static + P-delta analysis due to the second-order effects caused by the lateral wind load. Find the analysis options on the drop-down after hovering on 解決する, located on the upper toolbar. After generating the results, we found that there is no concern for buckling of our frame, thanks to the buckling analysis prompt:

On the left side of the results page you can see the different types of forces and results that can be shown. Any individual load combination can be evaluated, as well as any individual load case or group. For design purposes, we will be using the 封筒絶対最大 because it will show us the worst forces that each member experiences. 注意, the envelope is not showing the worst load combination, but the worst combination of all of the load combinations at once. This loading condition is not realistic, but itenvelopesall of the possible forces that each member できる 見る. See the figures below for the pertinent force and deflection diagrams for our moment frame:

Reaction Envelope Absolute Max

Shear Envelope Absolute Max

Moment Envelope Absolute Max

Axial Envelope Absolute Max

Deflection Envelope Absolute Max

While there are no direct code requirements for story drift not due to seismic loading, it is recommended to keep story drift of structures between 1/500 そして 1/200 the story height (エリングウッド). For this frame, the story height is 18 フィート, so our drift should be between (18 ft/500) そして (18 ft/5\200), or between 0.43 そして 1.0 インチ. Our envelope max drift is shown as 0.55 インチ, これは実際には H/500 の下限に近い, which satisfies our engineering judgement.

ボタン, 結果概要, メンバーのストレス限界をすばやく確認するために、大多数のユーザーが使用しています, 特に鋼の場合. このウィンドウには、応力容量に最も近い部材のスナップショットが表示されます. Users can customize stress and deflection limits for their project using the toolbar on right side of the modeling space. Here is a glimpse of our frame using the 結果概要: Deflection/Span そして Member Stress are the customizable fields. 最後に, don’t forget that Structural 3D gives its users the ability to print reports with almost unlimited functionality and customization.

One of the biggest advantages of using the SkyCiv suite is that although all of the modules can used as standalone, most of them are integrated directly and are better suited to be used directly from Structural 3D. こちらです, all of the analytical results we have just calculated can be pulled directly into adjunct modules for design of members and connections.

## SkyCivメンバーデザイン: Checking and Selecting a Design our Members

The majority of the work done throughout SkyCiv is the modeling, 読み込み中, and analysis phases of a 3D model. SkyCiv Member Design follows a similar style in that it majors in user-friendliness and simplicity. It’s ease of use and integration with S3D gives users the ability to seamlessly switch between the model and the design module. We recommend that users spend some time going through each design module to see this themselves. 今, lets take a look at the design of the members in our frame; we will be using AISC 360-10 LRFD for our design code.

When opening the Member Design module, your projects job details will be brought in where you can change and confirm them. The next input tab lets you customize the phi-factors for steel. We will keep the defaults as laid out by AISC. の 会員 input tab is a good place to stop and look at your list of design-able members. Because of SkyCiv’s integrated ecosystem, it will import all of the steel members in your 3D model and calculate unbraced lengths, 有効長さ係数, and slenderness ratios for them. また、モデルに存在する材料とさまざまなセクションもリストします. フォース gives you the option to design for any single load combination, but is defaulted to the envelope forces of your structure. You can view the absolute worst case forces for your structure for any load combination as well.

This window is the main results page that will show you the unity, or capacity ratios, of each design check. Don’t forget to hover over the tooltips (“私” within a circle) if you are not exactly sure which each variable stands for. You can see that the Combined Forces limit state is governing for all three members.

Design Summary Report is similar to the 結果概要 そして 報告する for the analysis phase, it gives a succinct look at the forces that the members are experiencing and whether or not capacity has been reached.

One of the most exclusive functions to SkyCiv design modules is the individual member design reports that are available by clicking one of the report icons under the Report column for every member; this can be printed to a PDF or HTML. These reports give much more detailed hand calculations of the checks that are being done for the member. 例えば, lets take a look at a few snippets of our beam (メンバー 2):

The start of the axial strength check

The start of the flexural strength check

Our governing limit states check:

If you look in the modeling window, you will see that the frame shows the design ratios of each member as well as color coded. Because all three of the membersPassed”, here is what our current frame looks like:

メンバーデザインモジュールの詳細については、モジュールがサポートするコードバージョンを参照してください, see our ソフトウェアのドキュメント.

### 次は何ですか?

SkyCiv is always updating and adding features to our platform, with software updates almost every 2 数週間! 他のほとんどの構造工学ソフトウェアは、年に1回だけ更新されます。. この記事の時点で現在開発中の主要な設計更新には、次のものがあります。:

• オプティマイザ: integrated tool with Structural 3D that will optimize member designs for users automatically, 設定またはカスタマイズ可能な基準に基づく.
• More design codes and modules
• Need some other features? Request them today at [email protected]

## SkyCiv接続設計: Designing a Moment Connection

Now lets look at designing the two beam-column connections for our frame. They will both be designed as moment connections due to the presence of moment at the connection location. Make sure that the 封筒絶対最大 is shown in the analysis portion of the workflow because the connection module will bring the forces that are shown at the time you enter the module. メンバーデザインに似ている, to enter the Connection module, クリックしてください 設計 button and Change Module in the drop-down menu. Add a connection by pressing Add Connection Assembly. Toggle to モーメント接続 and find the Welded Flanges moment connection type. 接続選択ウィンドウは次のようになります。:

With the connection type chosen, 接続の親メンバーと子メンバーを識別する必要があります. 設計を簡素化するために、両方の梁の端に同じ接続を使用します。. If we look back on our analysis, 右側の接続では、モーメントの大きさが大きくなっていることがわかります, so we will design for this connection. この場合, 右側の列が親メンバーになります (メンバー 1), ビームは子メンバーになります (メンバー 2). Once this has been edited in the correct input area, the model space will identify the connection and should look like this:

This gives the user a visual representation and confirmation of the specific connection location(s) they are designing for their structure. To enter the design, click on the pencil image on the input tab. 注意, the connection can and will be rendered in 3D for you to look at while the inputs are being changed. Each time there is a change to the connection, クリックしてください レンダリングする button to refresh the rendered image. フォース can be customized, but thanks to the connection module’s integration with S3D, それらは自動的に入力されます. 前述のとおり, the forces listed will be taken from whatever load combination/envelope was being viewed in the analysis portion of S3D.

Click on 実行設計. Results for each portion of the connection should be shown on the left. Similar to the Member Design module, limit states are color coded to tell the user if their connection design is under or over capacity. Fortunately for us, the connection that we manufactured has passed each design check and will work for our frame!

Following the theme of transparency and accuracy, SkyCiv Connection has the power to print full hand calculations for each steel connection in accordance with AISC 360-10 Chapter J. The limit states that are shown in the hand calculations are summarized in the results after each design is ran. For continuity, take a look at a couple of snippets from this connections report:

Welded Flanges Tensile Yielding (AISC Sec. J4)

デッドロードからの反応 (AISC Sec. J4):

### 次は何ですか?

SkyCiv is always updating and adding features to our platform, with software updates almost every 2 数週間! 他のほとんどの構造工学ソフトウェアは、年に1回だけ更新されます。. この記事の時点で現在開発中の主要な設計更新には、次のものがあります。:

• Baseplate and Anchor Design: analyze and design baseplates and anchors at the base of columns in S3D
• More design codes (ユーロコードは近日公開予定)
• UI Improvements to simplify and speed up user workflow
• 基礎デザイン: design isolated and combined footings and piles for columns, directly integrated with your 3D model in Structural 3D

## 最終的な考え

We were able to go through the design of a simple 2D moment frame from conception, 接続設計から詳細まで. Throughout this process we never had to leave the program or load multiple modules separately. 各モジュールは、利用可能な他の構造工学ソフトウェアとは異なり、互いにシームレスにつながります. その間ずっと, あなたは任意のブラウザを介して操作しています, 任意のデバイスで, すべてのファイルがクラウドに保存されている. SkyCivスイートで可能な限り機能を拡張!

## 参考文献

1. 構造用鋼製建物の仕様. American Institute of Steel Construction, 2010.
2. エリングウッド, ブルース. “Serviceability Guidelines for Steel Structures.” www.aisc.org, 1989, www.aisc.org/globalassets/aisc/awards/tr-higgins/past-winners/serviceability-guidelines-for-steel-structures.pdf.
この記事は役に立ちましたか?


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# Thermal Conductivity Paper Database

## Recommended Papers for: Metal foams

Total Papers Found: 2

#### EMI shielding and thermal conductivity of a high porosity reticular titanium foam

As titanium foams are generally not as widely studied as aluminum foams, the electromagnetic interference (EMI) shielding performance and thermal conductivity of a high porosity reticular titanium foam were determined to further study the performance of titanium foams. From the experimental results, the samples had ...

Author(s):

#### Effective thermal conductivity of sintered metal foams: experiments and a model proposal

It is important that porous metal foams used in gas turbines do not exceed their upper temperature threshold. In this paper, the Transient Plane Source (TPS) technique was used to measure the thermal conductivity of metal foams up to 1000 °C. Results showed that the thermal ...

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### does the amplitude affect the period of a pendulum

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I said that the scale invariance of periods implies linear differential equations' periodic solutions have amplitude-independent periods (the result to be explained). As the amplitude of the pendulum increases, the period increases. The pendulum equation is difficult because of its nonlinear factor $$\sin$$. The black pendulum is the small angle approximation, and the lighter gray pendulum (initially hidden behind) is the exact solution. Compared to the period at zero amplitude, a $$10^{o}$$ amplitude produces a fractional increase of roughly $$_{0}^{2}/12 0.0025$$ or 0.25%. Electromagnets Lesson for Kids: Definition, Facts & Uses. Restoring forces create oscillation in a system. Blamed in front of coworkers for "skipping hierarchy". This is an intuitive way to explain why a circular pendulum approximately has this behaviour with small angles. Discover examples of hysteria in ''The Crucible'', see quotes from ''The Crucible'', and explore ''The Crucible'' themes. The two effects cancel each other out (at least if the amplitude of vibration isn't too large). Revelation 21:5 - Behold, I am making all things new?. Then the pendulum equation becomes, \[\frac{d^{2}}{dt^{2}} + \frac{g}{l}f(_{0}) = 0. It seemed to be an appropriate way to add more intuition to this). For the pendulum equation, the corresponding period is, \[T = 2\pi\sqrt{\frac{l}{g}} \text{ (for small amplitudes). Electromagnets attract pieces of metals through the power of electricity. Use that idea along with Newtons laws of motion to explain the $$2\pi$$. As is often the case, a changing process is difficult to analyze for example, see the awful integrals in Problem 3.31. For a small initial angle, it takes a rather large number of oscillations before the difference between the small angle approximation (dark blue) and the exact solution (light blue) begin to noticeable diverge. All other trademarks and copyrights are the property of their respective owners. Discover how to find sine using the periodic function and unit circles. What Led to the Storming of the Bastille? The frequency of the spring-mass system is $$w = \sqrt{k/m}$$, and its period is $$T = 2\pi/ = 2\pi \sqrt{m/k}$$. You can say it's because average speed (in the RMS sense) scales in the same way as displacement does: double the amplitude, and the RMS speed will also double. Practice calculating rotational kinematics using examples. Daniel A. Russell, Graduate Program in Acoustics, The Pennsylvania State University. The $$f() f(0)$$ lumping approximation, which predicts $$T = 2\pi\sqrt{l/g}$$, underestimates the period. a-S%f*YpzL{[p+]8RwiDU80&%S+"j>ldNIfAtY'q{o-}Y8+1-f! Explore the characteristic of harmonic oscillations, and follow examples to calculate the amplitude and period of a harmonic oscillation. Learn about hysteria in ''The Crucible''. How much larger than the period at zero amplitude is the period at $$10^{}$$ amplitude? AP Environmental Science: Homeschool Curriculum, Ohio State Test - Physical Science: Practice & Study Guide, ICAS Science - Paper G: Test Prep & Practice, SAT Subject Test Physics: Practice and Study Guide, SAT Subject Test Biology: Practice and Study Guide, NY Regents Exam - Chemistry: Test Prep & Practice, NY Regents Exam - Earth Science: Test Prep & Practice, DSST Health & Human Development: Study Guide & Test Prep, UExcel Science of Nutrition: Study Guide & Test Prep, Prentice Hall Biology: Online Textbook Help, High School Physical Science: Homework Help Resource, High School Physical Science: Tutoring Solution, All Teacher Certification Test Prep Courses, Pendulums in Physics: Definition & Equations, Working Scholars Bringing Tuition-Free College to the Community. @PeterASchneider I disagree. How to help player quickly make a decision when they have no way of knowing which option is best. Learn what is meant by the term simple harmonic motion. The best answers are voted up and rise to the top, Start here for a quick overview of the site, Detailed answers to any questions you might have, Discuss the workings and policies of this site, Learn more about Stack Overflow the company, There is no intuition, since in fact the period. When the angular displacement amplitude of the pendulum is large enough that the small angle approximation no longer holds, then the equation of motion must remain in its nonlinear form The mechanical equivalent of heat developed by physicist James Prescott Joule states that an amount of motion generates the same amount of heat. However, at $$\pi/2$$ the exact $$h$$ is the following awful expression (Problem 3.31): \[h(\pi/2) = \frac{\sqrt{2}}{\pi} \int_{0}^{pi/2} \frac{d}{\sqrt{\cos }}\label{3.39}$. Because $$f(_{0}$$) is a constant, this equation is linear! \label{3.52} \]. Just read Enn's answer in more detail, and my answer is, more or less, a more intuitive presentation of his. At zero amplitude ($$_{0} = 0$$), does $$h(_{0})$$ have zero slope (curve A) or positive slope (curve B)? The integral is likely to have no closed form and to require numerical evaluation (Problem 3.32). The two groups $$T/\sqrt{l/g}$$ and $$_{0}$$ are independent and fully describe the problem (Problem 3.30). The Taylor series for $$\sin$$ begins $$^{3}/6$$, so, $f(_{0} = \frac{\sin_{0}}{_{0}} 1 - \frac{_{0}^{2}}{6}. Sine Function, Formula & Examples | What is Sine? Why do the displayed ticks from a Plot of a function not match the ones extracted through ChartingFindTicks in this case? The amplitude of a pendulum is the maximum height it reaches. Use the preceding result for $$h(_{0})$$ to check your conclusion in Problem 3.33 about the slope of $$h(_0)$$ at $$_{0} = 0$$. As the amplitude approaches , the dimensionless period h diverges to infinity; at zero amplitude, $$h = 1$$. But what about the derivative of $$h$$? copyright 2003-2022 Study.com. In constructing useful groups for analyzing the period, why should $$T$$ appear in only one group? Here's an intuitive and non-rigorous answer. >> For small angular displacements, the pendulum differential equation is effectively linear and as such, the amplitude of the oscillation must be independent of the period. \[T 2\pi\sqrt{\frac{l}{g}} (1 + \frac{_{0}^{2}}{12}). Suppose a pendulum with length L (meters) has A physical pendulum consists of a meter stick Two pendulums have the same dimensions (length A torsion pendulum is made from a disk of mass Kinematics, the study of the displacement, velocity, and acceleration of objects in motion, involves simple harmonic motion whereby an equal and opposite force is applied to an object's motion. The dimensionless factor of $$2\pi$$ can be derived using an in-sight from Huygens [15, p. 79]: to analyze the motion of a pendulum moving in a horizontal circle (a conical pendulum). Check that period $$T$$, length $$l$$, gravitational strength $$g$$, and amplitude $$_{0}$$ produce two independent dimensionless groups. If the bob is connected to the pivot point by a string, however, a vertical release would mean that the bob falls straight down instead of oscillating. Learn about Newton's Force Laws and practice using force and acceleration formulas to solve the provided sample problems. This lesson explores differential calculus. In this lesson, learn what is elastic collision and find elastic collision examples for better understanding. A $$10^o$$ amplitude is roughly 0.17 rad, a moderate angle, so the approximate prediction for $$h$$ can itself accurately be approximated using a Taylor series. The larger the swing, the more vertical is the angle at which the pendulum bob starts falling at, hence the faster it accelerates at the start, and the increase in speed exactly balances the longer swing length. Why does the period not depend on the amplitude? Use conservation of energy to show that the period is, \[T(_{0}) = s\sqrt{2}\sqrt{\frac{l}{g}} \int_{0}^{_{0}} \frac{d}{\sqrt{cos - cos_{0}}} \label{3.40}$, Confirm that the equivalent dimensionless statement is, For horizontal release, $$_{0} = \pi/2$$, and, $h(\pi/2) = \frac{\sqrt{2}}{\pi} \int_{0}^{\pi/2} \frac{d}{\sqrt{cos}}. \label{3.51}$. @CarlWitthoft I have elaborated a bit on that in my comment here; but "for a small perturbation" the math does state that they will be the exact same, as long as that assumption holds. Pendulums in Physics | Calculation, Potential Energy & Kinetic Energy. Learn the equations of this principle demonstrated in the case of a pendulum. Therefore, the average f is greater than $$f(_{0}$$). Equivalently, it assumed that the mass always remained at the endpoints of the motion where $$|| = _{0}$$. And why should $$_{0}$$ not appear in the same group as $$T$$? The preceding results might change if the amplitude $$_{0}$$ is no longer small. One could continue but the key feature is already there. I think one of the important things to note about this is that near the end of the curve, the path looks. In the animation below right, the initial angle is large. If the amplitude doubled so would the distance covered in a given time. Do weekend days count as part of a vacation? As $$_{0}$$ increases, does the period increase, remain constant, or decrease? The dark blue pendulum is the small angle approximation, and the light blue pendulum (initially hidden behind) is the exact solution. Get access to this video and our entire Q&A library. This equation is, again, the ideal-spring equation. However, the pendulum spends more time toward the extremes (where $$f() = f(_{0})$$) than it spends near the equilibrium position (where $$f() = f(0)$$). \label{3.53} \]. xXKNlU-jM|TRl<78IFd-FC?(8D!eIo%-]oWJVo8{-pGS?vSvCOkizQX*W+X]_ WMwARxi:o\ Therefore, \[h(_{0}) 1 + \frac{_{0}^{2}}{12}. Use dimensional analysis to show that the equation cannot contain the mass of the bob (except as a common factor that divides out). In this approximation, period does not depend on amplitude, so h = 1 for all amplitudes. Why does a simple pendulum or a spring-mass system show simple harmonic motion only for small amplitudes?


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# The Southern HII Region Discovery Survey (SHRDS): Pilot Survey [GA]

The Southern HII Region Discovery Survey (SHRDS) is a survey of the third and fourth quadrants of the Galactic plane that will detect radio recombination line and continuum emission at cm-wavelengths from several hundred HII region candidates using the Australia Telescope Compact Array. The targets for this survey come from the WISE Catalog of Galactic HII Regions, and were identified based on mid-infrared and radio continuum emission. In this pilot project, two different configurations of the Compact Array Broad Band receiver and spectrometer system were used for short test observations. The pilot surveys detected radio recombination line emission from 36 of 53 HII region candidates, as well as seven known \hii regions that were included for calibration. These 36 recombination line detections confirm that the candidates are true HII regions, and allow us to estimate their distances.

C. Brown, C. Jordan, J. Dickey, et. al.
Thu, 25 May 17
1/44

Comments: 19 pages, 5 figures, Astronomical Journal in press

# Variable stars in the northern Galactic plane from KISOGP [SSA]

We have conducted a large-scale survey of the northern plane using Kiso Wide Field Camera attached to Schmidt telescope at Kiso observatory. The KISOGP (KWFC Intensive Survey of the Galactic Plane) project have made 40-70 epoch observations in I band of about 320 sq. degrees for 5 years starting in 2012. The limiting magnitude is around 16.5 in I. In the data analysis so far, we detected a couple of thousands of variable stars including approximately 100 Cepheids and more than 700 Miras. Roughly 90 percent of them were not previously reported as variable stars, indicating that there are still many relatively bright variables to be found in the Galactic plane.

N. Matsunaga
Thu, 25 May 17
2/44

Comments: Conference on wide field variability surveys: a 21st-century perspective, 2 pages in press

# Modelling polarized light from dust shells surrounding asymptotic giant branch stars [SSA]

Winds of asymptotic giant branch (AGB) stars are commonly assumed to be driven by radiative acceleration of dust grains. For M-type AGB stars, the nature of the wind-driving dust species has been a matter of intense debate. A proposed source of the radiation pressure triggering the outflows is photon scattering on Fe-free silicate grains. This wind-driving mechanism requires grain radii of about 0.1 – 1 micron in order to make the dust particles efficient at scattering radiation around the stellar flux maximum. Grain size is therefore an important parameter for understanding the physics behind the winds of M-type AGB stars. We seek to investigate the diagnostic potential of scattered polarized light for determining dust grain sizes. We have developed a new tool for computing synthetic images of scattered light in dust and gas shells around AGB stars, which can be applied to detailed models of dynamical atmospheres and dust-driven winds. We present maps of polarized light using dynamical models computed with the DARWIN code. The synthetic images clearly show that the intensity of the polarized light, the position of the inner edge of the dust shell, and the size of the dust grains near the inner edge are all changing with the luminosity phase. Non-spherical structures in the dust shells can also have an impact on the polarized light. We simulate this effect by combining different pulsation phases into a single 3D structure before computing synthetic images. An asymmetry of the circumstellar envelope can create a net polarization, which can be used as diagnostics for the grain size. The ratio between the size of the scattering particles and the observed wavelength determines at what wavelengths net polarization switches direction. If observed, this can be used to constrain average particle sizes.

E. Aronson, S. Bladh and S. Hofner
Thu, 25 May 17
3/44

# Discovery of a new branch of the Taurid meteoroid stream as a real source of potentially hazardous bodies [EPA]

Taurid meteor shower produces prolonged but usually low activity every October and November. In some years, however, the activity is significantly enhanced. Previous studies based on long-term activity statistics concluded that the enhancement is caused by a swarm of meteoroids locked in 7:2 resonance with Jupiter. Here we present precise data on 144 Taurid fireballs observed by new digital cameras of the European Fireball Network in the enhanced activity year 2015. Orbits of 113 fireballs show common characteristics and form together a well defined orbital structure, which we call new branch. We found that this branch is characterized by longitudes of perihelia lying between 155.9-160o and latitudes of perihelia between 4.2-5.7o. Semimajor axes are between 2.23-2.28 AU and indeed overlap with the 7:2 resonance. Eccentricities are in wide range 0.80-0.90. The orbits form a concentric ring in the inner solar system. The masses of the observed meteoroids were in a wide range from 0.1 g to more than 1000 kg. We found that all meteoroids larger than 300 g were very fragile, while those smaller than 30 g were much more compact. Based on orbital characteristics, we argue that asteroids 2015 TX24 and 2005 UR, both of diameters 200-300 meters, are direct members of the new branch. It is therefore very likely that the new branch contains also numerous still not discovered objects of decameter or even larger size. Since asteroids of sizes of tens to hundreds meters pose a treat to the ground even if they are intrinsically weak, impact hazard increases significantly when the Earth encounters the Taurid new branch every few years. Further studies leading to better description of this real source of potentially hazardous objects, which can be large enough to cause significant regional or even continental damage on the Earth, are therefore extremely important.

P. Spurny, J. Borovicka, H. Mucke, et. al.
Thu, 25 May 17
4/44

Comments: 24 pages, 22 figures, 5 tables. Accepted in Astronomy and Astrophysics

# Far-Ultraviolet Observation of the Globular Cluster NGC 6397 [SSA]

We present an observational far-UV (FUV) and near-UV (NUV) study of the core region of the globular cluster NGC 6397. The observations were obtained with the Space Telescope Imaging Spectrograph (STIS, FUV), and the Wide Field Camera 3 (WFC3, NUV) on board the Hubble Space Telescope. Here, we focus on the UV bright stellar populations such as blue stragglers (BSs), white dwarfs (WDs) and cataclysmic variables (CVs). We present the first FUV-NUV color-magnitude diagram (CMD) for this cluster. To support our classification of the stellar populations, we compare our FUV-NUV CMD with optical data from the ACS Survey of Galactic Globular Clusters. The FUV-NUV CMD indicates 16 sources located in the WD area, and ten BSs within the 25″x 25″ of the STIS FUV data. Eighteen Chandra X-ray sources are located within the FUV field of view. Thirteen of those have a NUV counterpart, of which nine sources also have a FUV counterpart. Out of those, five sources are previously suggested CVs, and indeed all five are located in the WD/CV region in our FUV-NUV CMD. Another CV only has a FUV but no NUV counterpart. We also detect a NUV (but no FUV) counterpart to the MSP located in the core of this cluster. The NUV lightcurves of the CVs and MSP show flickering behaviour typical of CVs. We found that the BSs and CVs are the most centrally concentrated population. This might be an effect of mass segregation or indicate the preferred birth place of BSs and CVs via dynamical interactions in the dense core region of GCs. HB stars are the least centrally concentrated population and absent in the innermost area of the core.

A. Dieball, A. Rasekh, C. Knigge, et. al.
Thu, 25 May 17
5/44

# Rapidly star-forming galaxies adjacent to quasars at redshifts exceeding 6 [GA]

The existence of massive ($10^{11}$ solar masses) elliptical galaxies by redshift z~4 (when the Universe was 1.5 billion years old) necessitates the presence of galaxies with star-formation rates exceeding 100 solar masses per year at z>6 (corresponding to an age of the Universe of less than 1 billion years). Surveys have discovered hundreds of galaxies at these early cosmic epochs, but their star-formation rates are more than an order of magnitude lower. The only known galaxies with very high star-formation rates at z>6 are, with only one exception, the host galaxies of quasars, but these galaxies also host accreting supermassive (more than $10^9$ solar masses) black holes, which probably affect the properties of the galaxies. Here we report observations of an emission line of singly ionized carbon ([CII] at a wavelength of 158 micrometres) in four galaxies at z>6 that are companions of quasars, with velocity offsets of less than 600 kilometers per second and linear offsets of less than 600 kiloparsecs. The discovery of these four galaxies was serendipitous; they are close to their companion quasars and appear bright in the far-infrared. On the basis of the [CII] measurements, we estimate star-formation rates in the companions of more than 100 solar masses per year. These sources are similar to the host galaxies of the quasars in [CII] brightness, linewidth and implied dynamical masses, but do not show evidence for accreting supermassive black holes. Similar systems have previously been found at lower redshift. We find such close companions in four out of twenty-five z>6 quasars surveyed, a fraction that needs to be accounted for in simulations. If they are representative of the bright end of the [CII] luminosity function, then they can account for the population of massive elliptical galaxies at z~4 in terms of cosmic space density.

R. Decarli, F. Walter, B. Venemans, et. al.
Thu, 25 May 17
6/44

Observations of the solar butterfly diagram from sunspot records suggest persistent fluctuation in parity, away from the overall, approximately dipolar structure. We use a simple mean-field dynamo model with a solar-like rotation law, and perturb the $\alpha$-effect. We find that the parity of the magnetic field with respect to the rotational equator can demonstrate what we describe as resonant behaviour, while the magnetic energy behaves in a more or less expected way. We discuss possible applications of the phenomena in the context of various deviations of the solar magnetic field from dipolar symmetry, as reported from analysis of archival sunspot data. We deduce that our model produces fluctuations in field parity, and hence in the butterfly diagram, that are consistent with observed fluctaions in solar behaviour.


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## Bitcoin Tumbles while Gold Steady as China Data Shows Growth

The bitcoin price on Monday slipped below the $10,000 level as benchmark China data eased concerns about gloomy monetary outlook. The spot bitcoin rate was down 3.2 percent at$9,857.27 as of 02:15 UTC. The losses came as a part of a more significant downside action that began soon after the US President Donald Trump criticized bitcoin on Thursday. The rate since had dropped by as much as 25.34 percent on San Francisco-based Coinbase exchange.

Bitcoin Price Down by 4.02% in the Last 24 Hours | Image Credits: TradingView.com

In comparison, Gold – bitcoin’s rival asset – fared better. The precious metal’s spot rate on Monday dropped by 0.2 percent to $1,413.20 per ounce. The asset remained firm towards its interim bullish as it held above a critical pivot of$1,400 per ounce. It also received competition from a slightly stronger US dollar index that crawled higher against a basket of rival fiat currencies on Monday.

## Healthy China Data

Bitcoin’s exponential and gold’s meager fall followed the release of China’s economic data. It showed a 6.3 percent growth in the industrial output against a 5.2 percent forecast. Moreover, Chinese investment into fixed-assets rose by 5.8 percent, surpassing analysts’ estimate of 5.5 percent growth.

The data sent Chinese shares and Yuan higher on Monday. Following the midday break, the Shanghai Composite Index was 2,952.85 points, up 0.76 percent since the open. The CSI300 index also surged by 0.87 percent.

Meanwhile, the Chinese Yuan surged against the US dollar to trade at 6.8721 by 07:00 UTC.

The market believed both gold and bitcoin would surge against a string of supportive economic events, including an ongoing trade struggle between the US and China. A better-than-expected report from China allowed investors to do an intraday profit-taking but, according to Lukman Otunuga of FXT, at least gold is expecting to bounce back. The research analyst said:

“Gold is waiting for a fresh directional catalyst. Today is a slow start for gold, but that doesn’t mean it’s going to be a slow week.”

## Bitcoin Still Strong

Bitcoin, on the other hand, is faring better than gold in terms of year-to-date performance. The world’s leading cryptocurrency is currently up by more than 160 percent since January 1, which means its recent downside action could be a broader bearish correction adjusting an overbought bias.

That is at least what Scott Melker of TexasWest Capital believes. The cryptocurrency trader said on Monday:

“BTC is above 10k and everyone here is acting like it’s worth 10 dollars and a half-eaten happy meal. A few weeks ago 10k was a dream. We are over 3x from the bottom. Have some perspective.”

The Wolf Of All Streets✔@scottmelker

· 17h

3) They feel they should have somehow known to "sell the top" at 14K and are beating themselves up for no doing so.
4) They don't understand that a retrace is not bearish.

The Wolf Of All Streets✔@scottmelker

$BTC Perspective Edition This is the$BTC weekly chart. We have been in a parabolic uptrend. Price went up a few hundred percent. That is represented in the blue box. The red box is the part of the chart that is sending everyone into a panic. Does this look so bad to you?

140

6:39 PM - Jul 15, 2019

As of 11:42 UTC, the bitcoin price was trading 5.02 percent higher at $10,350 on Coinbase exchange. Tags: • 29 | expand thanks expand Mining support expand Thank you for sharing. expand nice expand support expand keep on expand great expand support expand mining expand Come and support expand Great sharing. expand Support Landlord expand Thanks your sharing expand support~ expand Mining~ expand thanks for sharing. expand Support expand Mining expand Nice sharing expand Good afternoon. expand Support you expand Thank you foryour share expand good expand thanks expand come and support expand happy tuesday expand support expand mine expand You should login to reply You will reward to {{ username }} Available Balance: {{balance}} ≈$ {{usdtAmount.toFixed(2)}} (The reward commission rate is 10.0%)


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# How many electrons would have to be removed from a coin to leave it with a charge of +1.6 xx 10^-7 C?

We assume that the coin has a net charge of zero. Recall that electrons have a negative charge, and specifically, they have a charge of $- 1.602 \times {10}^{- 19} \text{C}$ (coulombs).
+1.6xx10^(-7) cancel"C" xx "1 electron"/(-(-1.6xx10^(-19) cancel"C"))
$= \textcolor{b l u e}{{10}^{12} \text{electrons}}$