LTC4001 Can charge my 1s4p Li-ion battery pack? I was trying to design a battery charger for my application where I have made a pack of 1s4p cells where each cell is having 2600mAH and Charging rating as 4.2V. So can I consider it as 1 cell? Can I use LTC4001 - 2A Synchronous Buck Li-Ion Charger for this pack?
Charging Li-ion batteries in parallel I wanted some opinion on my set up here, just to make sure I got the basics right.  I have a Li-ion battery charging circuit based on the MCP73113. This is designed to be a single-cell battery charger.   The battery itself (3.7V, 650mAh) comes with its own PCB with Schottky diode and current regulators as protection. EDIT: Not a Schottky diode. Current limiter and a Protection IC.  By design, they work together just fine.  I have more batteries from the same manufacturer and wanted to make higher capacity packs by putting two cells in parallel. The two cells come with their own PCB, but I only kept one of them, as I soldered their leads together. It seems to work and they are charged and discharged just like regular batteries (3.7V, now 1300 mAh).  However, not all the packs I made work smoothly and I get a failure rate of 2 in every 10 packs. They all start out normally, but occasionally when the batteries are not charged for some time, or are used up, they tend to not work anymore. So far, in only 1 pack I found the voltage to be different (both now below 2V, aka dead).  I assume my method is not the right approach to ensure 100% success rate, so I was wondering how can this be solved. All the controllers I looked into was designed for multiple cells in series, not parallel. Am I fundamentally missing something here?
Different mechanics of \hyperlink vs. \hyperref I am fiddling around with the hyperref package to build a hypertext structure into my pdf. At first glance, hyperref provides two similar macros to put links between parts of my document: \hyperlink{target}{text} and \hyperref[target]{text}. You might notice their slightly different syntax. What really interests me is their different working mechanics. I will illustrate this by the following example: \documentclass{article}  \usepackage{pdfpages} \usepackage[]{hyperref} \usepackage{mwe}  \begin{document}  \section{Section Heading} \label{sectionheading}  \includepdf[link=true, linkname=pdfpage]{example-image}   Link to section works with \hyperref[sectionheading]{hyperref} and with \hyperlink{sectionheading}{hyperlink} (but with slightly different targets on the page).  Link to pdf works with \hyperlink{pdfpage.1}{hyperlink} but not with \hyperref[pdfpage.1]{hyperref} (and not this way, either: \hyperref[pdfpage]{hyperref}).  \end{document}  To link to the labeled section heading, I may use both commands. While \hyperref directs my pdf viewer directly to the linked section, \hyperlink directs me somewhere slightly different, so I can see more context. To link to the \includepdf of pdfpages however, I may only use \hyperlink. \hyperref does not work at all. I suspect this is because of a subtle difference in their working mechanisms. While checking the aux file, I found a line \newlabel{sectionheading}{{1}{1}{Section Heading}{section.1}{}} which \hyperref seems to use as a target. However, I did not find any similar target for the included pdf. My question: I suspect, I understand the workings of \label and \ref very well. \hyperref seems to work quite similarly with the added clickability. Could you please explain to me how \hyperlink does work differently and how it is implemented? E.g. what mechanics explain the difference when clicked on the link and where and how are the link targets stored in the first place? Running own executable on Android shell I need a specific command line tool and I have made a C program in my Linux shell. I have compiled the program with an ARM cross-compiler. I have then moved the program into the Android file system and tried to run it.  The output is permission denied.  What do I have to do, in order to run my own compiled programs in Android file system? Shell script to send an alert mail if disk usage exceeds 90% Quotient group $G/H$ is abelian iff $[G,G] \subseteq H$ I bought a Minecraft account for my little brother for Christmas, and I'd like to play with him on my server every now and then. If I log in while he's on, however, it kicks him off since we have the same username. Is there any way to change my username so that I can play with him, or anything I can do so that we can both play together using this account on my server?  I'm open to anything that could let us play, whether it be username changes, server property changes, etc. $(this) inside of AJAX success not working LaTeX seems to be having problems handling nested optional arguments. I've included a minimal code example that generates the error:  \documentclass[12pt,oneside]{book} \begin{document}  \newcommand{\termbad}[2][]{#2#1} \newcommand{\termgood}[2][]{#2#1}  \newcommand{\up}[1]{#1} \newcommand{\dd}[1][]{#1}  \termgood{\up{Tart}}{Pop} \termbad[\up{Tart}]{Pop}      \termgood{\dd[Tart]}{Pop} \termbad[{\dd[Tart]}]{Pop} %\termbad[\dd[Tart]]{Pop}   \end{document}   If you run this code, it works. If you remove the % in the second to last line, it fails with:  Runaway argument?  Tart{Pop} \par \par \end {document}  !File ended while scanning use of \dd.   After further tinkering, I have discovered the problem: LaTeX is reading the ] for the inner optional argument as the end of the outer optional argument and then passing a malformed argument in. So it is passing in \dd[Tart as the optional argument to \termbad instead of \dd[Tart].  This still feels like a bug to me. Nesting braces in arguments works, brackets should as well. The following are workarounds:   Enclose the inner optional argument in braces (as in the line before %). Redefine your inner function to require its argument instead of using optional arguments. Redefine your outer function to require its argument instead of using optional arguments.   Any one of the three works around the issue.  (Reposted here; question was originally posted to stackexchange and then closed instead of being migrated.) I would like to redirect a URL with uppercase characters http://localhost/A/B to a lowercase version http://localhost/a/b using the .htaccess file. Using regex I can capture A and B but is it possible to convert them to lowercase? Is stopwatch benchmarking acceptable? Does anyone ever use stopwatch benchmarking, or should a performance tool always be used?  Are there any good free tools available for Java?  What tools do you use?  To clarify my concerns, stopwatch benchmarking is subject to error due to operating system scheduling.  On a given run of your program the OS might schedule another process (or several) in the middle of the function you're timing.  In Java, things are even a little bit worse if you're trying to time a threaded application, as the JVM scheduler throws even a little bit more randomness into the mix.  How do you address operating system scheduling when benchmarking? How do you overlay an image on a shader in Cycles? I checked yesterday and today the new homepage in the Android app. Very surprised I saw that I would almost reach a silver tag badge, that confused be because I always take a look on possible badges I can get. However I compared the score with the website and I got a total different results. See this two screenshots: How do I transcode a Javascript string to ISO-8859-1? Can't format / delete locked partition from GParted? The sum of powers of two and two's complement – is there a deeper meaning behind this? Probably everyone has once come across the following "theorem" with corresponding "proof": $$\sum_{n=0}^\infty 2^n = -1$$ Proof: $\sum_{n=0}^\infty q^n = 1/(1-q)$. Insert $q=2$ to get the result.  Of course the "proof" neglects the condition on $q$ for this formula, and the sum really diverges. However I now noticed an interesting fact:  If you use two's complement to represent negative numbers on computers, $-1$ is represented by all bits set. Also, sign extending to a larger number of bits (that is, getting the same number in two's complement representation on more bits) works by copying the left-most bit (also known as sign bit) into the additional bits on the left.  Now imagine that formally you sign-extend the number $-1$ to infinitely many bits. What you get is an infinite-to-the-left string of $1$s. Which, using the normal base-2 formula $n = \sum_k b_k 2^k$ (where $b_k$ is the bit k positions from the right, i.e. $b_0$ is the rightmost bit), that infinite string of $1$s translates into exactly the sum above! So in some sense we have an independent re-derivation of that equation.  Now my question is: Is there something deeper behind this? Somehow I cannot imagine it is just coincidential. Unable to change payment information to NONE in iOS Appstore Intuitive explanation for formula of maximum length of a pipe moving around a corner? For one of my homework problems, we had to try and find the maximum possible length $L$ of a pipe (indicated in red) such that it can be moved around a corner with corridor lengths $A$ and $B$ (assuming everything is 2d, not 3d):    My professor walked us through how to derive a formula for the maximum possible length of the pipe, ultimately arriving at the equation $L = (A^{2/3} + B^{2/3})^{3/2}$.  The issue I have is understanding intuitively why this formula works, and exactly what it's doing. I understand the steps taken to get to this point, but there's an odd symmetry to the end result -- for example, is the fact that $\frac{2}{3}$ and its inverse are the only constants used just a coincidence, or indicative of some deeper relationship?  I also don't quite understand how the formula relates, geometrically, to the diagram. If I hadn't traced the steps myself, I would have never guessed that the formula was in any way related to the original problem.  If possible, can somebody give an intuitive explanation as to why this formula works, and how to interpret it geometrically?    Here's how he found the formula, if it's useful:  The formula is found by finding the maximum possible length of the pipe by expressing the length in terms of the angle $\theta$ formed between the pipe and the wall, and by taking the derivative to find when $\frac{dL}{d\theta} = 0$, which is the minimum of $\frac{dL}{d\theta}$ and is therefore when $L$ is the smallest:  $$ L = \min_{0 \leq \theta \leq \frac{\pi}{2}} \frac{A}{\cos{\theta}} + \frac{B}{\sin{\theta}} \\ 0 = \frac{dL}{d\theta} = \frac{A\sin{\theta}}{\cos^2{\theta}} - \frac{B\cos{\theta}}{\sin^2{\theta}} \\ 0 = \frac{A\sin^3{\theta} - B\cos^3{\theta}}{\sin^2{\theta}\cos^2{\theta}} \\ 0 = A\sin^3{\theta} - B\cos^3{\theta} \\ \frac{B}{A} = \tan^3{\theta} \\ \theta = \arctan{\left( \frac{B}{A} \right)^{\frac{1}{3}}} \\ $$  At this point, we can substitute $\theta$ back into the original equation for $L$ by interpreting $A^{1/3}$ and $B^{1/3}$ as sides of a triangle with angle $\theta$ and hypotenuse $\sqrt{A^{2/3} + B^{2/3} }$:  $$ \cos{\theta} = \frac{A^{1/3}}{ \sqrt{A^{2/3} + B^{2/3} }} \\ \sin{\theta} = \frac{B^{1/3}}{ \sqrt{A^{2/3} + B^{2/3} }} \\ \therefore L = A^{2/3} \sqrt{A^{2/3} + B^{2/3} } + B^{2/3} \sqrt{A^{2/3} + B^{2/3} } \\ L = (A^{2/3} + B^{2/3}) \sqrt{A^{2/3} + B^{2/3} } \\ L = (A^{2/3} + B^{2/3})^{3/2} \\ $$  The equation for the formula for the maximum length of the pipe is therefore $L = (A^{2/3} + B^{2/3})^{3/2}$. How to prove $\det(I+uv^\intercal)=1+v^\intercal u$ Let be $u,v\in\mathbb{R}^n$, then $\det(I+uv^\intercal)=1+v^\intercal u $ where $I$ denotes the identity matrix of order $n$. How to prove this?   what I did:   let be $A=\{n\in\mathbb{N}: \forall u,v \in \mathbb{R^n}, \det(I+uv^\intercal)\neq1+v^\intercal u  \} $, and suppose $A\neq \varnothing $, then for the well-orderer-principle there exists $n_0\in A$ such $n_0\leq n,\;\forall n\in A$. since $1\notin A,\; n_0\neq 1 \rightarrow n_0-1\in\mathbb{N}\setminus A  $ so: $$\forall u,v\in\mathbb{R^{n_0-1}}: \det(I+uv^\intercal)=1+v^\intercal u $$ Then, let be $u,v\in\mathbb{R}^{n_0}$, so $$\det(I+uv^\intercal)= \displaystyle\sum_{j=1}^{n_0}(-)^{1+j}a_{1j}\det(A_{1j})$$ where $a_{ij}=\begin{cases} u_iv_j+1 &amp; i=j \\ u_iv_j &amp; i\neq j \end{cases}$, and $A_{1j}$ is the  submatrix of $I+uv^\intercal$ that results deleting the i-file and j-column. Then I try to open the term for $j=1$ and try to use the relation for the new matrix of $n_0-1$ order with the determinant because it holds the same form but I have other terms which difficult me the work. Do you know other method?.  Ps. , in page 2 I've found something similar but I don't understand what it means PS. I'm taking a course of numerical analisys I have a variable, x, and I want to know whether it is pointing to a function or not.  I had hoped I could do something like:  &gt;&gt;&gt; isinstance(x, function)   But that gives me:  Traceback (most recent call last):   File "&lt;stdin&gt;", line 1, in ? NameError: name 'function' is not defined   The reason I picked that is because  &gt;&gt;&gt; type(x) &lt;type 'function'&gt; I have been running Ubuntu 12.04 for the past few weeks. I recently tried to download and install some files through the Update Manager. It located a few files, downloaded them, but then failed to install them. It gave the error:  installArchives() failed: (Reading database ...  (Reading database ... 80% dpkg: unrecoverable fatal error, aborting:  reading files list for package 'libasn1-8-heimdal': Input/output error   I tried using apt-get upgrade but a similar error was produced:  Reading package lists... Done Building dependency tree        Reading state information... Done The following packages will be upgraded:   icedtea-6-jre-cacao icedtea-6-jre-jamvm openjdk-6-jre openjdk-6-jre-headless   openjdk-6-jre-lib 5 upgraded, 0 newly installed, 0 to remove and 0 not upgraded. Need to get 0 B/34.6 MB of archives. After this operation, 1,024 B of additional disk space will be used. Do you want to continue [Y/n]? y (Reading database ... 80% dpkg: unrecoverable fatal error, aborting:  reading files list for package 'libasn1-8-heimdal': Input/output error E: Sub-process /usr/bin/dpkg returned an error code (2)   Any help would be greatly appreciated. Thanks in advance.  Edit: Here's some extra information that may be useful. My original problem was that the Ubuntu Software Centre would not open. I found that the Update Manager and Synaptic Package Manager also failed to open. They all gave an error along the lines of:   "installArchives() failed: dpkg: error: reading package info file '/var/lib/dpkg/available': Input/output errorError in function:"   After much googling I seemed to fix that problem as the package managers would open again. When I then tried to download and install any updates from the Update Manager I got the error at the beginning of this post. Defining starred versions of commands (* macro)