Matrix of reflection in $R^3$

Question: Please, can you explain me how do we get this formula
$$ A = I - 2nn^{T} $$ in $$ R^{3} $$? This should be matrix of reflection, but I don't know how to prove that.
Comment: Welcome to MSE. Your question is phrased as an isolated problem, without any further information or context. This does not match [many users' quality standards](http://goo.gl/mLWc8), so it may attract downvotes, or be put on hold. To prevent that, please [edit] the question. [This](http://goo.gl/PlJyVQ) will help you recognise and resolve the issues. Concretely: please provide context, and include your work and thoughts on the problem. These changes can help in formulating more appropriate answers.
Comment: This form is called "Householder transform" in the domain of Numerical Analysis.
Answer: Let $a,b,n$ be unit vectors orthogonal to each other  - $a,b$ basis for the plane, $n$ orthogonal to the plane.
You can easily check that any vector $v$ is represented in the basis {$a,b,n$} as
$v=a(a^Tv)+b(b^Tv)+n(n^Tv)=(aa^T+bb^T+nn^T)v$
($a^Tv, b^Tv, n^Tv$ are here scalars)
hence $aa^T+bb^T+nn^T=I$.
During the reflection components in the plane are unchanged, component orthogonal to the plane is negated.
Hence we have $w= (aa^T+bb^T-nn^T)v=(I-2nn^T)v$.
Comment: Very nice! Thank you.
Answer: This is maybe more of a trick than of a proof, but I find this approach much easier to memorize.
Let $x_1, x_2, x_3$ be the Cartesian coordinates of the space. It is clear that the reflection around the plane $x_1x_2$ is the map
$$
(x_1, x_2, x_3)\mapsto (x_1, x_2, -x_3), $$
and this is a linear map that can be written in matrix form as
$$
A_{(0,0,1)}\begin{bmatrix} x_1 \\ x_2 \\ x_3\end{bmatrix} = \left( I-\begin{bmatrix} 0 & 0 &0\\ 0 & 0 & 0 \\ 0 & 0 & 2\end{bmatrix}\right)\begin{bmatrix}x_1\\ x_2 \\ x_3\end{bmatrix},$$
that is,
$$
A_{(0,0,1)}=I-2\begin{bmatrix} 0 \\ 0 \\ 1\end{bmatrix}\begin{bmatrix} 0 &0 &1\end{bmatrix}.$$
We write
$$\tag{1}
A_{(0,0,1)}\vec x = \vec y.$$
Now, to obtain the formula for the reflection around the plane having $\vec{n}$ as normal vector, we change variables;
$$
\vec x = R\vec x ', \quad \vec y = R\vec y', $$
where $R$ is a rotation matrix such that
$$
R\vec n = \begin{bmatrix} 0 \\0 \\ 1\end{bmatrix}.$$
Applying this change of variable in (1) yields
$$\tag{2}
A_{\vec n}\vec x ' =\vec y ',$$
where $A_{\vec n}$ is the reflection matrix we want to compute. And now it is a matter of unwrapping $\vec x', \vec y'$ in (2) and computing, using the fact that $R^TR=I$, because $R$ is a rotation.

Matrix of reflection in $R^3$


Can a GFCI plug on the same circuit as a range cause a fault for power getting to the range plug?

Question: I have a standard 220v plug for an electric range. There is also a GFCI plug along the same circuit but it is tripped and I cannot get it to reset. If the GFCI plug is tripped could that cause a reduction in the current to the 220 for the range?
Comment: This sounds like a really poor circuit layout to have a 120v receptacle on the same circuit as a 240v appliance, possibly non-compliant with NEC. The 240v appliance should have a dedicated circuit.
Comment: Have you checked if an upstream circuit breaker is tripped? If the whole circuit is off the GFCI might not reset and of course, the circuit would have no power.
Comment: Can you add your country of origin to your profile, or edit your question to show where you are in the world? We will be able to better diagnose and answer your question because the electrical systems vary quite a bit country to country.
Comment: I updated my info, but for the sake of speed, I am in central Alberta, canada. My house is not a fixed unit, it is a mobile home that is about 2 years old. We bought it brand new however
Answer: You can't do that.   The 240v power to the range must go only to the range.  You can't fork off one leg of the 240 and a neutral to also serve a 120v plug.  Of any kind.  If you need a 120v plug there, you need to either run a new circuit, or add an outlet to an existing circuit.
That, however, does not explain why the GFCI is tripping.   The GFCI would not know it was applied illegally - as long as it sees 120v across its line terminals, it is happy.   The fact that it is tripping suggests to me that either it is defective or you have connected it to 240v.  Or it is working properly and you are plugging in an appliance which has a ground fault, such as a 3-light electricity tester.


Does $\sum_{n=1}^\infty |\sin n|^{cn}$ converge? What about the terms?

Question: Does $\sum_{n=1}^\infty |\sin n|^{cn}$ converge for some $c > 0$? If so, what $c$? And for the $c$ that give divergence, what about the terms of the series? Do they converge? I saw a question where it was $|\sin n|^n$ instead of $|\sin n|^{cn}$, and it was shown that the sequence diverges (hence the series diverges), hence the question. So let's assume $c > 1$.
Answer: There is no $c>0$ such that
$$\sum_{n\geq 1}\left|\sin n\right|^{cn} \tag{0}$$
converges. Since $\pi$ is an irrational number, there exist an infinite number of rational numbers $\frac{p}{q}$ such that:
$$ \left| \frac{\pi}{2}-\frac{p}{q}\right|\leq\frac{1}{q^2},\tag{1}$$
and $q$ is odd, so, since the sine function is a Lipschitz function, there exists an absolute constant $K>0$ such that:
$$ \left|\sin m\right|\geq 1-\frac{K}{m} \tag{2}$$
holds for an infinite number of integer numbers $m\geq 1$. Proof: let we just take $m=p$.
From $\left|p-q\frac{\pi}{2}\right|\leq\frac{1}{q}$ it follows that $\left|\sin(p)-\sin\left(q\frac{\pi}{2}\right)\right|\leq\frac{1}{q}$, so $(2)$ holds for any $K>\frac{2}{\pi}$.
Since
$$ a_m=\left(1-\frac{K}{m}\right)^{cm} \tag{3}$$
is an increasing sequence converging towards $e^{-Kc}$, the general term of $(0)$ is not $o(1)$, so the series is not convergent.
Comment: Thanks, I had a feeling it needed to be something more like $\pi / 2$ with odd $q$, and that assertion must certainly be true. Also the fact that $|\sin (x) - \sin(y)| \leq |x - y|$, so that you can choose $K > 2/\pi$.
Comment: I get the first assertion, but I don't see how the second assertion follows, namely $|\sin m| \geq 1 - K/m$.. Maybe you meant $\pi/2$ instead of $\pi$ in the first assertion? If you don't have time to elaborate more I'll accept if this gets enough upvotes and/or someone can leave a comment or answer that elaborates more on your proof outline.
Comment: @user2566092: proof updated.


in macOS mojave apps freezes and beachball appears for some seconds when doing tasks like loading a conversation in Messages

Question: macOS 10.14.6 on my MacBook Pro early 2015 13" with an i7 and 16GB Ram the active app freezes for some seconds when doing tasks in it, like loading conversation in Messages (happens not on every conversations, more like.. every fifth or sixth I click), clicking into Address Bar in Safari or calling up the context menu and I can't figure out why this happens. It happens at any time, means right after a restart.
It existed before the 10.14.6 update.
I uninstalled a clipboard manager which I use and thought it could have sth. to do with it... still existing.
I always look and try to figure it out via Activity Monitor, but no clues.
I booted into Safe Mode... still existing.
I reinstalled macOS via Recovery Mode... still existing.
I also tested my RAM with a memtest86 boot stick. Nothing.
My guess is it would still be present when I recover from a backup... are there any other clues or things I can try to figure this out without installing everything from scratch? Help would be much appreciated. Thanks in advance.
Comment: To figure out if it's a software issue (which I highly assume) you could try to install a new macOS on to an external drive. Then test if the error still occurs
Answer: This is easy to isolate.
Sign out of Messages and make sure no other app or system is causing the slowdowns. The more you get freezes, the quicker you can know that it's just messages causing the slowdown as opposed to being impacted. YOu might open Activity Monitor and show all processes to see which are using a lot of CPU.
Then you can make a new user account. In that account - only sign in to messages - see if the slowdown is there (assuming you concluded it was the cause above).
If so - then you might check with Apple Support to be sure you don't have an issue server side and possibly consider backing up and restoring your Mac to make sure it's not part of the system that's stuck or causing the slowdown.
Keep in mind, for physical hard disks - the first sign you get of imminent failure can be slow apps, so maybe make a good backup first and then test to isolate what's slow.
Answer: I found the problem by killing one running app/service by another:
I had activated the Accessibilty Keyboard to appear/disappear when in lower left corner for 1 sec with the cursor and also to fade away after 0.25s.
I think that fading caused the system trouble.
There was actually no way I could have realized that (except maybe through the Console, but I don't know how to filter the given information to finde the neadel in the haystak)


Fetch UTXO Bitcoin to address created by rpc importprivkey without rescan like other api

Question: For few reasons, I must use rpc node to handle wallet at this time. In the near future, I will change the flow to use library to create addreses, store privatekey at database and create raw transaction. That means I will use rpc node for broadcasting only, not depend on it too much. But I need time to do, so I will divide this period to few parts, firstly, I will start at creating and store address:
I have both hex private key and wif key, then I use importprivkey to import wif key to rpc node without rescan because I have a large request contacting to rpc. If those addresses are fresh, it's very nice scenario, each new input to them will automatically fetch UTXO by using listunspent command to check.
In other case, the very worst scenario, for maintenance or backup, I import an address which contains old UTXO and wanna fetch these UTXOs to addresses without using rescan command (it takes a long time to complete), like other api just put address and get UTXOs in a blink.
https://testnet.blockexplorer.com/api/addr/myzhMVXT39L7nnco6TtJ2a7sYaEo8ixT4Q/utxo
Is there any ways to fetch old UTXO to address in rpc node without rescan? Because these api are very fast response, I think they're not using rescan. For long term, using other api is not good for my project, I'm using version 0.16 (I know it's very old but the project need time to change it slowly and carefully), thanks for help so much.
Answer:
Because these api are very fast response, I think they're not using rescan.

No, they maintain massive databases that index every transaction and transaction output and their associated addresses. Bitcoin Core does not need this for normal function, and the vast majority of users do not need access to such information. Bitcoin Core does not and cannot build such databases.

I'm using version 0.16

Then there is nothing you can do.
If you were using 0.18, then you could use the scantxoutset command to just scan the UTXO set for UTXOs. This is very quick, and you can specify multiple addresses to do it as single batch scan. scantxoutset will give you all of the UTXOs for the specified addresses but will not add them to any wallets loaded in Bitcoin Core. However, it does not sound like you are using the wallet so that should be fine.
Comment: Thank you, I've understood. But I can not find more detail on this `scantxoutset` command. May you give me a simple format with a valid adress without publickey, example:
`bitcoin-cli scantxoutset start '[{"desc":"MY_ADDRESS", "range":100}]'`
Comment: `bitcoin-cli scantxoutset start '["addr()", "addr()"]'` where `` and  are the addresses you want to scan for. Since it is an array, you can just add more `addr()` style descriptors.


What flavour is quark gluon plasma?

Question: What flavour of quarks make up quark gluon plasma? What would it most likely cool down into, protons, neutrons, strange matter, neutronium or what?
Answer: Quark-gluon plasma is just a phase matter in QCD may enter into.
Just as a solid or a gas may be made out of any kind of atoms or molecules, quark-gluon plasma may be made out of quarks of any flavor. If you make it by heating ordinary matter, it will obviously mostly contain up- and down-quarks. It's just a phase of QCD, and it cools into whatever its constituents can cool into.
Comment: @Jitter: weak interactions don't conserve flavour i.e. they change quarks into quarks of a different type. So in a QGP you can't point to a quark and say "this is an up quark". The quarks will be continuously changing types - actually you'd have to describe them as being in a superposition of all possible types.
Comment: Thankyou. I'm curious about the recent stuff made out of a proton hitting some lead. I assume there was a bit of extra energy involved which may create other flavours? Also aren't up down nuclei full of strange quarks too?
Answer: The quark model has developed to fit a great number of measurements

In particle physics, the quark model is a classification scheme for hadrons in terms of their valence quarks—the quarks and antiquarks which give rise to the quantum numbers of the hadrons. The quark model underlies "flavor SU(3)", or the Eightfold Way, the successful classification scheme organizing the large number of lighter hadrons

.....

Hadrons are not really "elementary", and can be regarded as bound states of their "valence quarks" and antiquarks, which give rise to the quantum numbers of the hadrons. These quantum numbers are labels identifying the hadrons, and are of two kinds. One set comes from the Poincaré symmetry—JPC, where J, P and C stand for the total angular momentum, P-symmetry, and C-symmetry, respectively.
The remaining are flavour quantum numbers such as the isospin, strangeness, charm, and so on. The strong interactions binding the quarks together are insensitive to these quantum numbers, so variation of them leads to systematic mass and coupling relationships among the hadrons in the same flavor multiplet.

Looking at the eightfold way of calassifying into flavor SU(3)  the measured resonances/particles  we note that one of the axis in the plot is proportional to the mass of the resonances.

baryon decuplet -------------------------------- meson octet

The strangeness equal zero state has the lowest masses in both. The quark gluon plasma happens at very high energies, proposed in the hypothesis for the evolution from the singularity of the Big Bang as the energy falls due to the expansion of the universe. As the plasma cools  there will be all flavors (and antiflavors) according to the available energy to generate the masses , further cooling will disfavor the hadrons with strange quantum numbers, which will follow the normal decay paths,  until protons and neutrons can form and the plasma stops existing, at about a microsecond from the Big Bang.
The other way, trying to create a quark gluon plasma is being studied at the LHC at the moment .
So the quark gluon plasma does not have a unique flavor but is a statistical distribution of flavors, of elementary particles and antiparticles , depending on the energy available and the quark model. From conservation of quantum numbers in the creation  of flavors in pairs of particle antiparticle , the overall plasma is flavor neutral.
Comment: It's more than just being studied. It was created last week. https://www.bnl.gov/rhic/news2/news.asp?a=1749&t=pr
Comment: In experiments quarks are produced in particle-antiparticle pairs so no flavor is created as far as we know today and observe in experiment. Even so we do have lots of strangeness (in pairs) - that is a well established signature of QGP formation.
Answer: There are two conserved charges, baryon number and electric charge. This means that the total baryon number and charge of a quark gluon plasma are determined by how it was made. Everything else is determined by thermal equilibrium (if there is enough time), and kinetics (if there is not enough time to fully equilibrate).
In a heavy ion collision there is not enough time for weak interactions to equilibrate, so net flavor (up minus anti-up, down minus anti-down, strange minus anti-stange, charm minus anti-charm, .. ) is effectively conserved as well. In the quark core of a neutron star (if such a thing exists) weak equilibrium is reached.
In a relativistic heavy ion collision, as a first approximation, you can think of all quarks as heaving been pair-produced by gluons. This means that the QGP has no net flavor or baryon number. Of course, the initial nuclei had net baryon number and a slight excess of neutrons over protons (i.e. down over up). Most of this goes down the beam pipe, but some of it ends up in the plasma, so there is a slight excess of quarks over anti-quarks, and a slight excess of down over up.
If you look at the total number (up plus anti-up, down plus anti-down, ..) rather than the net number (up minus anti-up, ..), then in thermal equilibrium there will be a tiny excess of up over down, and a larger excess of up over strange, because of the mass ordering of the quarks.
Answer: Experiments with relativistic nuclear (heavy ion) collisions at CERN-SPS in '90s at BNL-RHIC in first decade of this century and 2nd decade at CERN-LHC have all produced quark-gluon plasma consisting of gluons, up-quark pairs, down-quark pairs and strange-quark pairs. At LHC highest energy also protons can  induce formation of QGP fireball, and in nuclear LHC collisions we see lots of charm pairs, the 4th quark flavor, and even bottom flavor (5th flavor).
At the SPS (fixed target) this new phase of matter shows high additional   up-quark, down-quark contents (not pairs) relating to a fraction of baryon number originating with the colliding  nuclei being stopped in the fireball. Since charges of up and down are not opposite,  aside of baryon number (three quarks=one baryon number) there is also a net fireball electrical charge. These effects are smaller at RHIC (collider) which reaches 15 times CM collision energy of  SPS  and are unobservable at LHC where the energy is up to 25 times higher compared to RHIC.
There is no time in these collisions to convert any flavors of quarks into each other by known weak interaction processes: the QGP fireball explodes too fast, explosively hadronizing i.e. creating matter out of energy of the primordial -soup, comprising many  free streaming new particles  helping us understand QGP properties. A YouTube channel collects the related clips on creation of matter  https://www.youtube.com/channel/UCRYj1UAYtCfXQdLeeiENu8A
Answer: Well it really depends if the things you heated up had mostly nucleons then you can expect a lot of up and down quarks. However because of the Heisenberg's uncertainty principle you can never truly know what particles are inside a proton or neutron. Instead all you know is that there are three valence quarks(the quark composition for proton being up, up, down and quark composition for neutron being down, down ,up quarks). So there is a really low possible you could get a charm quark as long as there is a anticharm quark somewhere. It is okay to have an antidown quark as long as there is an extra down quark on top of what is needed for it to be a proton or neutron. If there is enough energy all these virtual quarks can become real. Then the resulting particles can be anything really. It does not have to end up as protons and neutrons. Instead they can end up as Delta particles. However these particles have a very short lifetime and all decay into protons and neutrons. Even neutrons are unstable on their own and will decay into protons in 15 minutes. There is a also the possibility of pion and antiproton production. Though pions are unstable and antiprotons annihilate with protons into a shower of other particles.
Comment: a) we discussing here QGP and not parton structure of nucleons; b) We are not discussing hadrons which emerge from QGP but properties of QGP; c) the qua ntum nature of particles is  hardly of relevance as the QGP is relatively long lived - etc. All of the above is not to the point.

What flavour is quark gluon plasma?


How could a Yaksha kill the 4 Pandavas in Mahabharat? Was he that mighty? Why was Arjuna/Bhima not able to wage a war with that Yaksha?

Question: In Mahabharat Yakhsha Parva, when the pandavas try to drink water from a pond, a Yakhsha appears. He warns the pandavas not to drink water from that pond before answering his questions. But the 4 pandavas (except Yudhisthira) drank and they died instantly. Was that Yaksha so much mighty that even warriors like Arjuna and Bhima could not at least fight him (if not defeat).
Answer: That was not an Yaksha. It was Yama in the disguise of a yaksha -

'O child, I am even your father, the Lord of justice, possessed of great prowess! Know, bull of the Bharata race, that I came hither
desirous of beholding you! Fame, truth, self-restraint, purity,
candour, modesty, steadiness, charity, austerities and Brahmacarya,
these are my body! And abstention from injury, impartiality, peace,
penances, sanctity, and freedom from malice are the doors (through
which I am accessible).'

https://www.wisdomlib.org/hinduism/book/the-mahabharata-mohan/d/doc7596.html
And even then, it seemed that Pandavas died because water was poisoned. Yama didn't defeat them in any actual battle. Remember that Arjuna earlier had defeated all the gods including Yama at Khandava forest -

But Arjuna and Krishna, fearless and invincible in battle, beholding Sakra and the other celestials prepared for fight, calmly waited, bows
in hands. Skilled in battle, those heroes in wrath assailed the
advancing host of celestials with their own thunderlike arrows. The
celestials repeatedly routed by Krishna and Arjuna, at last left the
field of battle for fear and sought the protection of Indra.

https://www.wisdomlib.org/hinduism/book/the-mahabharata-mohan/d/doc4305.html
Answer: Yaksha/ Yama did not kill the Pandavas.. he merely made Pandavas unconscious with his maya/ power of illusion.


Does IntelliJ have an internal web server to serve static content of a web application?

Question: Right now, the directory of my module is defined as an IIS virtual directory and IIS serves the files.
I was wondering whether IntelliJ has an internal web server, which can serve the files, without the need for any third party. Eclipse does.
Answer: UPDATE: built-in web server is available in the recent IntelliJ IDEA versions (starting from 13). You can find more details in the blog (yes, this feature first appeared in WebStorm).
IntelliJ IDEA has no this feature, you need to install and use any third-party web server that can serve the content from the project folders.
Comment: Thank you for your answer! For future visitors: just watch this [YouTube video](https://youtu.be/EKiP9b8ikTY). You need to add new 'JavaScript Debug' configuration.
Comment: @naXastandswithUkraine, the youtube video link is no more valid
Comment: @rinilnath new video: https://youtu.be/YNNDMpoGV0w (Debugging JS in WebStorm and Chrome)
Answer: A built-in HTTP preview server will be part of Intellij IDEA 13 and is already available in the EAP: http://youtrack.jetbrains.com/issue/WEB-7148

"All existing actions — preview in browser (pop-up over html file or menu action or
  shortcut), open in browser and create/debug html file action now open file on built-in web
  server
  http://localhost:63342/<project name>/<file path relative to source or content root>"

In other words, right-click on an HTML page and select "Debug" or "Open in browser", and IDEA 13+ will serve up that page via port 63342.
Answer: Here's another super simple option, install Python: http://www.python.org/getit/
Then open a shell prompt, navigate to your root web folder (e.g. public) and run python -m SimpleHTTPServer - This starts an HTTP Service on port 8000.
Further reading should you need it: http://www.linuxjournal.com/content/tech-tip-really-simple-http-server-python
I've got mine running on Windows 7 but the above article still applies.
Answer: Another option is is create a PHP project that, starting with v 5.4.0 of PHP includes a built in web server.  This page explains it all ...
http://www.jetbrains.com/idea/webhelp/php-built-in-web-server.html
Answer: IntelliJ IDEA has a built-in web server that can be used to preview and debug your application. Just watch this YouTube video or follow the steps below.
Option 1
You need to add new 'JavaScript Debug' configuration:

Click Add Configuration... or Edit Configurations... in the Navigation bar

Click  button in the toolbar or press Alt + Insert to create a new configuration
Select JavaScript Debug under the Templates node in the tree view of run configurations
Fill in Name, URL, Browser and click [OK] to save the configuration
Use http://localhost:63342/YOUR-PROJECT-NAME/index.html for URL

Now you can run the configuration:

Click run  or debug  button in the Navigation bar (or use Shift + F10 / Shift + F9 hotkeys).

Option 2
Running web page in browser without creating a configuration. Refer to the related IntelliJ IDEA Help article.

In the editor, open the HTML file. This HTML file does not necessarily have to be the one that implements the starting page of the application.
Do one of the following:

Choose View | Open in Browser on the main menu or press Alt+F2. Then select the desired browser from the pop-up menu.
Hover your mouse pointer over the code to show the browser icons bar:  . Click the icon that indicates the desired browser.

Result
Google Chrome browser with a demo web page served by the Intelij IDEA's built-in webserver:
Answer: One simple way is to create a NodeJS / Express project in IntelliJ that is your web server.  You can then use it to serve your static web pages and any other web content. The NodeJS web server is very small and runs fast - noticeably faster than IIS and Apache. Best of all you can just hit the Run button in IntelliJ or WebStorm to start it up.
By default, a NodeJS / Express project includes a public/ directory that you can use to contain your static pages that you can then view from http://localhost:3000/
This explains the steps required to enable NodeJS in IntelliJ and includes links to other Node resources:
http://www.jetbrains.com/idea/webhelp/node-js.html
If you feel the need, you can reconfigure your NodeJS server using server side Javscript code. You can add SSL support or almost any other server side features you care to dream up. Just add NodeJS modules using the npm (Node Package Manager) command line tool included with the install. NPM Registry https://npmjs.org/ indexes all the available modules.
Answer: You can configure IntelliJ to use a lot of different application containers, but each of them must be downloaded and installed separately. I currently have mine configured to serve via jetty, like eclipse, and also tomcat, tc-server, jboss, and node.js. It's pretty easy to set up.


visual studio bug annoying, any fix?

Question: Running visual studio 2008 sp1. When I go to run without debugging a web app, it open 2 IE browser windows? There is nothing in my code that is doing this! I cant seem to find anything on how to fix this. Has anyone run into this? Is there a fix?
Answer: Do you have multiple web projects?
Check the startup options on the Web tab in the Project Settings and see what they're set to.
If that doesn't help, try creating a new project and see if it does the same. If not, try diffing the csproj files (or csproj.user if you've set it to store settings in a separate file).

visual studio bug annoying, any fix?


Netcat as a handler for windows/meterpreter/reverse_tcp payload

Question: How can I use netcat as a handler for windows/meterpreter/reverse_tcp payload? I am getting a connection but am unable to run any command
Answer: You need to use the meterpreter handler rather than netcat. The issue is that meterpreter uses multiple stages. The reverse_tcp payload is only used to connect back to the client and load the additional shellcode needed to spawn the meterpreter. When you are using netcat this doesn't work.
If you want to have a payload that runs with netcat you need to consider using a regular reverse TCP shellcode rather than meterpreter.
Comment: Bro I know that meterpreter is multistaged but I am unable to use metasploit handler.

Actually i am getting the error "Connection reset by peer ssl_accept". On investigation, I came to know that I was running my powershell payload using powershell.exe (32 bit method to run powershell payload) which was causing that error. When i tried that payload using %WinDir%\syswow64\windowspowershell\v1.0\powershell.exe , payload run gracefully.

Now I have deployed my payload on a victim machine (I cant access that machine).
Comment: Actually I had customized my payload and removed 64 bit method which comes with standard .bat file generated by veil-evasion. Can I get connection using netcat or any other technique?


Going back to previous page when you open a new tab in browser?

Question: Why do browsers (Chrome and Firefox at least) not copy over your browsing history when you click on a link to open in a new tab? I often open up a lot of links in new tabs and then am frustrated when I am unable to go back from the newly opened tab.
Comment: I think the only people who could sufficiently answer this question are those who developed said browsers.
Comment: Same problem!  F*rk this. ;D  But it goes back like 2-5 pages, sometimes.
Answer: In Firefox, if you hold down the Control key while dragging the tab to a new position on the tab bar, a small blue vertical arrow will show between the tabs where you are about to drop the tab. This creates a duplicate tab which is an exact duplicate in that the back button works, bringing you back to exactly the same place as the original tab.
You can achieve the same thing in Chrome by right clicking on a tab and choosing duplicate tab.
Comment: Nice! I didn't know that.
Answer: While you can't make the browsers operate like this when you click on a link, you can still make them "clone" the tab's history by middle-clicking on the refresh button (or back button or forward button), at least in Chrome. In these situations, it will create a new tab, completely with cloned history, and then execute the function requested (refresh, back, or forward). It's a bit of a hassle if you want to actually follow a link (Middle-click refresh, then find the link again, then click on the link), but it will achieve the desired result you are looking for.
Comment: That's not what he's talking about.  The browser randomly jumps back several pages for no reason.  Imagine you're just typing an e-mail, and then suddenly BACK BACK BACK BACK!  It destroys your e-mail.
Comment: @Wolfpack'08 What? The OP isn't complaining about that at all. Also, hit the `Forward` button for your issue. I can't speak for Firefox, but Chrome will recover your partially-written email. Also, keep in mind that the `Backspace` key is often bound to the `Back` button in many browsers.
Comment: Come on, you know the Internet.  There are like a billion pages.  And even Chrome won't recover many things.
Answer: because its not the history of that tab(window). Change the settings of your browser so all clicked links will open in the same tab(window).
Comment: My question is about the design decision. When you fork() a process you get a straight copy of the original process, so I don't understand why forking a new tab doesn't give you the history of the original tab. I'm aware that I could just open everything in the same window, but I like the functionality of forking off multiple tabs and don't understand why the new tabs do not copy over the history of the original tab.
Comment: Because creating a new tab doesn't fork() a new process inherently, and that's a bad design decision anyways because you are copying large blocks of memory for every new tab that likely will never, ever be used, further bloating the memory footprint of the browser.


Are the method-of-moments-based normal confidence intervals asymptotically valid and optimal?

Question: A common way of getting confidence intervals for the mean of a normal random variable is to estimate the mean and variance by the method of moments and then compute a confidence interval based on those estimates. Since this is commonly used, I would assume that these are asymptotically valid and (in some sense) optimal. Is this so? If so, is there somewhere where I could read a proof of this?
Comment: This [paper](https://www.jstor.org/stable/1912775) by Lars Hansen establishes large sample properties of generalized method of moments estimators under weak stationarity and ergodicity, which nests the standard IID method of moments case. You can also find basic textbook treatments but I'll suggest trying this on your own (the Taylor expansion method you use to get asymptotic normality for MLE should work here!)
Answer: For a normal random variable, the moment-matching estimator (MME) for the mean is the maximum likelihood estimate (MLE). For the variance, the MME and the MLE differ just by the bias adjustment ( n/(n-1) ), so asymptotically they will coincide, and the MME will have the asymptotic properties of the MLE.
You can observe this in the Wikipedia article about the Normal distribution
The MLE has the properties you mention: consistency, efficiency, asymptotic normality, etc.
But the empirical mean of a sample coming from a normal distribution actually has a normal distribution, this is not true only asymptotically but for any sample size.
You can understand this easily if you recall that the sum of two normal random variables is itself a normal random variable. The mean is a sum of normal random variables, multiplied by a scalar (1/n), so it also is a normal random variables. There should be a wealth of material for a step by step proof, here is one.


How does one 'Extract Com Information' from an OCX without InstallShield?

Question: In one of the projects our team is working on, we are trying to make an automated deployment system for an existing desktop application.  In order to do that we need to understand how InstallShield installs the application to begin with.
We have access to the InstallShield manifest, but there is an OCX file that we cannot figure out how to install manually (without InstallShield). This particular OCX file is set to 'Extract COM Information'.
Here's a screenshot:

The other OCXs in this application are self-registering, so they can be registered with Regsvr32.exe.  But the OCX we are having problems with cannot be registered in that fashion.
How would one manually install an OCX file that is set to 'Extract COM Information' in an InstallShield manifest?
Answer: RegSvr32.exe calls the LoadLibrary API to load your DLL and then invokes the DllRegisterServer entry point inside your DLL.  The code inside that function does the actual COM registration.  If RegSvr32 is failing, that typically means a dependency of your DLL is missing or invalid.
InstallShield does all of this along with some really low level bit hacking to virtualize all of this and then harvest it.  An old article on the subject is:
Spying on Registry Entries
InstallShield doesn't actually use this technique per say ( they have several techniques, most of which is not documented and various filters and transform engines to clean up the data ).    If you are just looking for a way to do it without InstallShield,  then look at Windows Installer XML's  "Heat" command line tool.  This can "harvest"  COM metadata into WxS XML elements.
Also WiX is open source so if you are really curious you could go looking at their code.
Comment: I'm sorry we can't modify the dll, so we have to find another way.
Comment: I don't believe I said anything about modifying any DLL's.
Comment: +1 Thanks for the response, let me clarify.  I'm not sure what "RegSvr32.exe call LoadLibrary to load your DLL and then DllRegisterServer() inside your DLL." means.  Are you saying run DllRegisterServer inside the deployment system?  (we can do things programmatically).  To restate, are saying that inside whatever deploys the app, DllReigsterServer() must be called?
Comment: I cleaned up my wording and added some links, maybe that will clarify it for you.
Answer: As Christopher mentioned, InstallShield extracts COM information from your .ocx by seeing what it registers when invoked similarly to regsvr32.exe will invoke it. Its various forms of redirection (for capturing purposes) have the added benefit of working around several potential permissions problems while the file is registering in your build environment. However if I'm not missing the point of your question, it's "why doesn't regsvr32.exe your.ocx work on the target machine?"
This is a bit of a stab in the dark, as you haven't included enough information. While missing dependencies can cause this, I'm going to guess you only see this failure on Windows Vista/Server 2008 or higher. If this is the case, there's a good chance your application is trying to write to registry keys that are protected by Windows Resource Protection (WRP), or is being tripped up by a per-user typelib registration problem.
When a poorly behaved self-registration routine encounters WRP, it attempts to write to a registry key it lacks permission to modify, then fails the entire registration. I'm uncertain what happens to the keys it wrote before that point, but all ones after it definitely never make it to the machine. You should be able to confirm whether this is the case with a tool like Process Monitor.
What do you do if this is the case? Well, you can stick with an extraction approach like that of InstallShield (which you say you want to leave). You can fix the file to not attempt to write to protected keys (which you say you cannot modify). Or you might be able to use the Application Compatibility Toolkit (ACT) to shim things, but I don't see how you can generally do that downstream. Generally speaking, I would recommend fixing the file, or continuing to use a working approach.
Comment: +1 thanks for the thoughtful comments, what additional information do you think I should include in the question?
Comment: Michael has the unfair advantage of actually working at InstallShield so he knows the code inside and out. :-)  FWIW, what wasn' clear to me about your question was if you wanted to understand at a very low level how the extraction works so you could write your own program or if you were just trying to solve the COM extraction problem for this one time or if you simply didn't want to depend on InstallShield going forward to get the job done.
Comment: If you are just trying to get around a COM extraction problem for this one DLL and you don't mind emailing it to me, I could run it through a number of tricks that I have to try to take a crack at it for you.
Comment: We're just trying to solve the COM extraction so we can rewrite the ocx's install in our own clickonce deployment app, but there is a chance that we may have to do this for more applications.  It's not that we don't want to use installshield, its just that we are trying to wrap a suite of applications that used installshield in Microsoft's ClickOnce deployment system so we have to understand how all the dlls and ocxs are registered and installed so we can duplicate it in .net.  The target audience is considered of low expertise and does not want to relay on an unautomated install.
Comment: Oh, just the usual suspects - like which versions of IS, Windows, etc. you've tried, and what the behavior is with each. For example, if it worked with regsvr32 on Windows XP and failed on Vista, that would lend credence to what I talked about, whereas if it never worked, or worked on Vista but failed on Windows 7, it's probably something else.
Comment: Thank you both for your continued interest, I will post more of the requested data when i return to the office on Monday.
Comment: I think you'll find there are many, many limitations with ClickOnce.  FWIW, there are ways in IS/MSI to do autoupdates and keep the user story simple.
Comment: Thanks for the great responses, we were able to solve the problem.  The ocx was dependent on another microsoft ocx that we missed or were unaware of.


How can I prove that all rational numbers are either terminating decimal or repeating decimal numerals?

Question: I am trying to figure out how to prove that all rational numbers are either terminating decimal or repeating decimal numerals, but I am having a great difficulty in doing so. Any help will be greatly appreciated.
Answer: Recall that in long division, one gets a remainder at each step:
$$
\begin{array}
& & & 0 & . & 2 & 2 & 7 & 2  \\
\hline
22 & ) & 5&.&0&0&0&0&0 \\
& & 4 & &  4 \\
& & & & 6 & 0 & \leftarrow \\
& & & & 4 & 4 \\
& & & & 1 & 6 & 0 \\
& & & & 1 & 5 & 4 \\
& & & & & & 6 & 0 &  \leftarrow & \text{repeating}\\
\end{array}
$$
6 is a remainder.  The next remainder is 16.  Then the next is 6.  This brings us back to where we were at an earlier step: Dividing 60 by 22.  We have to get the same answer we got the previous time.  Hence we have repetition of "27".  The answer is $0.2272727\overline{27}\ldots$, where "27" keeps repeating.
The question then is: Why must we always return to a remainder that we saw earlier?  The answer is that the only possible remainders are $0, 1, 2, 3, \ldots, 21$ (if $22$ is what we're dividing by) and there are only finitely many.  If we get 0, the process terminates.  If we never get 0, we have only 21 possibilities, so we can go at most 21 steps without seeing one that we've seen before.  As soon as we get one that we've seen before, the repetition begins.
A related question worth asking is how you know that every repeating decimal corresponds to a rational number.  E.g., if you're handed $0.2272727\overline{27}\ldots$ with "27" repeating forever, how do you figure out that it's exactly $5/22$?  There's a simple algorithm for that too.
Answer: Hint$\ $ Consider what it means for a real $\rm\ 0 <  \alpha < 1\ $ to have a periodic decimal expansion:
$$\begin{array}{rrl} &\alpha\!\!\! &\rm =\,\ 0\:.\overbrace{a_1a_2\cdots a_n^{\phantom{l}}}^{\textstyle a}\ \overbrace{\overline{c_1c_2\cdots c_k}}^{\textstyle \overline c}\ \ \text{ in radix } 10\\[.2em]
\iff& \beta \!\!&\!:= \rm 10^n\, \alpha\! -\! a = 0\:.\overline{c_1c_2\cdots c_k}\\[.2em]
\iff& 10^k\: \beta \!\!&=\ \rm c + \beta\\[.2em]
\iff& (10^k-1)\ \beta \!\!&=\ \rm c\\[.2em]
\iff&\!\!\!\! \rm (10^k-1)\ 10^n\: \alpha \!\!&\in\ \Bbb Z
\end{array}\qquad$$
Thus to show that a rational $\rm\,\alpha\,$ has such a periodic expansion, it suffices to find $\rm\,k,n\,$ as above, i.e. so that $\rm\,(10^k-1)\,10^n\,$ serves as a denominator for $\rm\,\alpha.\,$ Put $\rm\,\alpha\, = a/b,\,$ and $\rm\, b = 2^i\,5^j\,d,\,$ where $\rm\,2,5\,\nmid d\,.\,$ Choosing $\rm\,n\, >\, i,\,j\,$ ensures that $\rm\,10^n\,\alpha\,$ has no factors of $\rm\,2\,$ or $\rm\,5\,$ in its denominator. Hence it remains to find some $\rm\,k\,$ such that $\rm\,10^k-1\,$ will cancel the remaining factor of $\rm\,d\,$ in the denominator, i.e. such that $\rm\,d\,|\,10^k-1\,,\,$ or $\rm\,10^k\equiv 1\pmod{d}\,.\,$ Since $\rm\,10\,$ is coprime to $\rm\,d,\,$ by the Euler-Fermat theorem we may choose $\rm\,k = \phi(d),\,$ which completes the proof sketch. For the converse, see this answer.
Comment: Although this post is old, I have a question: what if k is not unique, that is what if 10^k = 1 mod d, has more than one solution? This means there are at least two different representations of c, right? This is impossible, hence k is unique right?
Comment: Well, k is not unique since if k_1 is a solution then also 2k_1 is a solution; thus we may choose the least k that satisfy that equation
Comment: so here $\bar{c}$ is a compact form of writing $\overline{c_1 c_2 \dots c_k}$, and represents the purely periodic part? And then $a$ and it's indexed counterpart represent the possibility of a non repeating part that precedes the periodic part? So the second line of your proof eliminates the non-repeating decimal part which has length $n$. Next you form an integer $c$ by shifting left by order $k$. I'm starting to see the underlying theory behind this stuff, but I'm not quite to the point where I can apply it to find $n$ and $k$ here. I'll keep reading. Thank you.
Comment: @rocksNwaves Yes, you've correctly understood it.
Comment: @BillDubuque I am sorry but I don't understand the leap from step 3 to step 4. Is it possible to elaborate it more?
Comment: @BillDubuque I finally understood your proof after reviewing elementary number theory. I should have studied more about it. Sorry for the inconvenience and thanks.
Answer: First, it's clear that you need only look at proper fractions. Now look at the long division algorithm for calculating the decimal expansion of a rational number. At each stage you get a remainder. What happens if you get a remainder of $0$ at some stage? If you don't ever get a remainder of $0$, can you keep getting different remainders forever, or must a remainder repeat at some point? What happens if you do get a repeated remainder?
Answer: Remainders must be less than the divisor.  So, for example, if you divide the numerator by the denominator, n, the only remainders can lie between 0 (in which case the division ends) and n-1.  What happens if and when you run out of remainders (i.e. you've used all of the numbers between 1 and n-1)?  You will be facing division of some remainder, r, by n again.  You've been there and done that so you know what the next remainder (hence, the next dividend) will be.  And so on...


Only 2 photons in the universe?

Question: Actually I'm trying to figure out what is time and how it works, imagine there are only 2 photons in the universe (and assuming they are not traveling in parallel). Since photon do not have any inertial frame, so can I safely say that in this particular scenario time do not exist because otherwise the distance between the photon would change. Since time never exist there shouldn't be space let alone... never mind. My question is can there be a universe whereby there are only 2 photons in existence?
Comment: Related:  ["One-electron universe"](https://en.wikipedia.org/wiki/One-electron_universe), Wikipedia.
Comment: You can say that *clocks* do not exist since there is nothing with which to build a clock in a two-photon universe.
Answer: You are mixing up time and the flow of time. Spacetime is a four dimensional manifold, i.e. we need four coordinates to uniquely identify identify points in spacetime, and this is true whatever is occupying that spacetime. The fact only photons are present doesn't change the dimensionality of the spacetime (though it would make it conformal).
The trajectory of a photon is a 2D line embedded in the 4D spacetime and this line is given by the null geodesic equation. For any particular choice of coordinates it is possible to express the three spatial coordinates as a function of the time coordinate just as we do for moving objects in everyday life.
Comment: @user6760: see [What is time, does it flow, and if so what defines its direction?](https://physics.stackexchange.com/questions/235511/what-is-time-does-it-flow-and-if-so-what-defines-its-direction)
Comment: Can I ask what do time do if it doesn't behave like time in my scenario i.e. flowing?
Comment: I read it and but how do i draw a space-time diagram for my 2 photons? If this get philosophical then I would simply ignore and abandon my question entirely as I'm only interested in science hee hee.
Comment: @user6760: In any spacetime diagram (in Minkowski space at least) the worldline of a photon is a line at 45º to the time axis. So your diagram will have two lines on it at 45º. What you can't do is draw the diagram in the rest frame of one of the photons because photons do not have a rest frame.
Answer: Time , tricky question. The rate that events happen? This would mean that time runs faster in the flame of a candle than it does outside the candle flame. However we evolved to quantify time by the seasons of earth. When to plant crops and when to harvest. For this reason our time is not the rate that events happen rather the rate the earth goes around the sun.
With this in place what if there were only 2 photons in the universe? I must object as I need to be there to make a few measurements. It is now 2 photons plus myself and test equipment. I will measure the speed of both photons. Unfortunately I and my test equipment are in the same space as the 2 photons so I will measure their speed to be c regardless of how fast time is ticking along relative to our crowded universe. Did not mean to weeny out but I see no way to measure the passage of time unless one is outside the system. The cause of the speed of the photons is proportionally effecting my test equipment so I will always say the speed of the 2 photons is c.

Only 2 photons in the universe?


After upgrading magento 2.3 to 2.4.6, the media does not work

Question: When performing setup:upgrade, it returns Media files stored outside the "Allowed Media Gallery" folders will not be available for the media gallery. Please refer to the developer's guide for more details.  and when I want to upload an image for a product, the error appears "Something went wrong when saving the file(s).
`
please help me, I don't understand why the media didn't work
Answer: i found solutions for the upload image error, but still i need to know how add Media files to"Allowed Media Gallery" if any solution ?
Thanks
bellow the solution upload file error
in the  path  veondor/magento/framework/Image/Adapter in Gd2.php find
function validateURLScheme and replace
if ($url && isset($url['scheme']) && !in_array($url['scheme'], $allowed_schemes)) {
to
if ($url && isset($url['scheme']) && strlen($url['scheme']) > 1 && !in_array($url['scheme'], $allowed_schemes)) {


What is the Christian understanding of the vision of Obadiah?

Question: How do Christians understand the vision of Obadiah? Is this an end time prophecy? If so who is Edom today? If they believe this was already fulfilled in the past and Edom no longer exists then why do we not see that the kingdom has already been established as the vision concludes?
Comment: Welcome to Christianity.SE! and thank you for your contribution. When you get a chance, please take the [tour] to understand how the site works and [how it is different than others](https://christianity.meta.stackexchange.com/a/1809/50259). I would also recommend reading the [Help Center's](https://christianity.stackexchange.com/help) sections on asking and answering questions.
Comment: This article is helpful: https://www.gotquestions.org/Edom-Obadiah.html
Comment: @PaulChernoch given the content of the webpage you linked to I'm guessing you ascribe to the view that Edom has already been destroyed. But you didn't answer the second half of the question. How do you explain that the kingdom has not come? The destruction of Edom is intrinsically linked with the coming of the kingdom in Obadiah.
Answer: Historical Overview: The Edomites were descendants of Esau, the twin brother of Jacob and grandson of Abraham.  In the fifth century B.C., a people called the Nabateans defeated the Edomites and forced them from the city of Petra.  Some first-century leaders, such as Herod the Great, still traced their ancestry to Edom, but all mention of Edomites fades after the Jewish Wars of that era. At the end of the 4th century, Jerome referenced the land of Idumea (Edom), but the people of the region had long since disappeared.
This source provides extensive background information on Edom as well as references to the Bible and the Tanakh:

The Edomites first established a kingdom ("Edom") in the southern area of modern-day Jordan and later migrated into the southern parts of the Kingdom of Judah ("Idumea", modern-day Mount Hebron) when Judah was first weakened and then destroyed by the Babylonians in the 6th century BC...  Most of its former territory is now divided between present-day southern Jordan and Israel.

You ask if Christians think Obadiah is an end time prophecy.  Not to my knowledge, but then I've only been a Christian for 40 years.  Although I am aware that one group of Christians think that Obadiah verse 21 points to the coming of the Christ and the deliverance of Zion in the last days, they are very much in the minority.  As far as I am aware this is not a view upheld by mainstream Christian denomination.
There is one reference to the coming "day of the Lord" in verse 15:

The day of the LORD is near for all nations. As you have done, it will be done to you; your deeds will return upon your own head.

This warning applies to ALL nations, and not just to Edom.  There was a literal fulfilment with regard to Edom but the end-time day of the Lord and the establishment of His kingdom is some time future.
Answer: On 14th December 2002 I took shorthand notes of a lecture on the book of Obadiah, given by John L. Mackay. He was a graduate of Glasgow, Oxford and London universities, and became an internationally respected Old Testament scholar, providing practical exposition and application of doctrine in his lectures. I transcribed my notes, which are now before me. This answer is extracted from the relevant bits. First, a bit of historic background, with the prophecy having two possible dates as to when it was written, either 480 B.C. or around 586 B.C.:

"The focus is on the future life of Judah in the south that prevailed
for many centuries. Edom was constantly hostile to Israel, right from
coming out of Egypt. David had brought them into the kingdom but they
asserted their independence. At the time of the Babylonian invasion of
Judah, before the siege of Jerusalem started, they devastated the land
in the south of Judah. After that, the Edomites moved into that area.
They were almost like Babylon's allies. Idumea became the name of that
land."

The prophecy only has 21 verses, yet contains four significant themes:

1.Divine condemnation of pride: Edom trusts in her sages, her wise men.
This pride will lead to her destruction. Pride distorts human
perceptions of the present and the future.
2.The Law and retribution: Vs. 15, "As you have done, it will be done to
you; your deeds will return upon your own head." God's judgment is
never arbitrary. It matches in severity and kind to the offence
perpetrated. "The fruit of your doings" is what vs. 15 means by "your
deeds". We reap what we sow - Gal. 6:7. Obadiah repeats the word 'day'
nine times (vss. 11-14). The Day of the Lord in vs. 15 is compared
with those "days" of the nations, reinforcing the message of
retribution.
3.The Day of the Lord: God will decisively intervene in the nation'
affairs. Each time he does, that is "a day of the Lord" and there will
be a final, ultimate Day of the Lord.
4.The restoration of the Lord's people: He is never prepared to let
their sin have the last say on the day of their destiny. Is this an
O.T. picture of the future?
Just as the Day of the Lord spoke of a particular time in history past
when God intervened in judgement, yet the Day of the Lord will be a
final coming of the Great Day, so here is a prophecy that at one level
looks to the physical land of Palestine - is that an aspect of the
meek inheriting the earth? Covenant theology favours those prophecies
having a measure of validity in olden times to finding their
fulfilment in the Church of Christ."

He leaves the question unanswered. Yet elsewhere in the notes he makes the point that Obadiah is giving this oracle of God against the nations - plural. The recipients of the oracle were "God's people, to encourage them in the face of opposition and hostility. Doom is to come on the nations [plural] because of their treatment of God's people. It was also a bold assertion of the Lord's universal dominion which was not an obvious thing in those days where each nation had its own array of deities responsible for particular areas. This reminded God's people of his covenant love for them and his commitment to them."
Not contained in my notes, but thrown in here at no extra charge, is that national groups can change their names, merge, move to different geographic locations, or even be subsumed in myriad other nations so that they are no longer identifiable as a distinct nation. The name "Edom" with respect to any coming fulfilment on the future Day of the Lord might well stand for all and any peoples (plural) who show ancient Edom's characteristics in the time of Obadiah. To get hung up on identifying just one nation by its name is to miss the final point of the future Day of the Lord. Why? Because as the ultimate end-time prophecy in the book of Revelation shows, names come to represent identifying characteristics. And, by then, things have gone global, with no group on Earth free from the wrath of the Lamb when he returns to judge the whole world. Nobody will escape divine retribution, apart from those whom he identifies as belonging to him - his people, called in the New Testament, spiritual Israel, which is made up of both Jews and Gentiles.
This answers the second query raised. It's not a literal earthly kingdom with nations called Israel, or Edom, or whatever, that the final fulfilment points to. The Kingdom of God is heavenly, with Christ as its King. Presently, he is enthroned in heaven, but when he comes to usher in the Day of Resurrection and Judgment, the entire world, with all its nations, will so tremble that they will cry out for the mountains to fall over them, to hide them from the wrath of the Lamb. But those who already belong to him by faith are spiritual Israel, and they will be delivered. Only after that will the old Earth be consumed with fire, then "a new heavens and a new earth, in which dwells righteousness" will replace the old Earth. This future righteous judgment of God is explained in 1st Thessalonians 4:14-5:3; 2nd Thessalonians 1:4-10 & 2 Peter 3:7-13. No nations are named. All the nations will be judged, and - don't forget - "Judgment must begin at the house of God" 1 Peter 4:17.
Comment: Up-voted +1. Excellent point that prophecies may prophesy of _characteristics_ and may label those characteristics with the _current and contemporary_ nation displaying them.
Comment: So you ascribe to replacement theology? The creator has gone back on his word to the patriarchs and made the citizens of Rome his chosen people rather than the descendants of Yaakov?
Comment: @Yaakov Tzir  No, I don't. Kindly stop trying to pigeon-hole people like that. I am not here to defend 'replacement theology' let alone promote it. God will never go back on his promises, either in the Old Testament or the New. "Citizens of Rome" is your phrase, not mine, and I do not identify with that. I am a citizen of the Kingdom of God, with Christ as its King, and it embraces both Jews and Gentiles.
Comment: What would be the law code of this kingdom? Would it be an anarchic society? Or would Rome or whatever you see ans its successors get to dictate the new law code? I am presuming you believe that the Torah has been nullified.
Comment: @YaakovTzir As you are new to this site, please note that comments are not for starting debates, arguments, or trying to prove the OPs point of view to be correct. That sort of thing can get Qs closed down. Therefore, I will not respond to your further Qs to me personally, because I don't want to see your main Q closed down. I hope you soon get into the swing of how this site works. Click the links that explain that.


Centroid of a spherical triangle.

Question: I'm trying to find the centroid of a spherical triangle over a unit sphere centered at the origin. I have already seen that I can obtain the centroid of the planar triangle (which are formed by the three position vectors of the triangle vertices) and then project the vector to the sphere surface. But what I'm looking is actually to find the centroid by using spherical trigonometry. Any help is appreciated!
Comment: "..project the vector to the sphere surface "  : how do you define the centroid of a spherical triangle ?
Comment: In agreement with @GCab, I'd like to know what the definition of centroid is to be. I'm moderately handy with spherical trigonometry, but would need the definition.
Comment: Thank you for your replies. What I meant by centroid is the intersection of the three medians or analogous to them in spherical trigonometry. This because I'm trying to find the centroid/center of mass of a spherical triangle over a sphere. Any help is appreciated
Comment: One has to be careful here; the medians of a spherical triangle do not divide its area in half. See https://en.wikipedia.org/wiki/Lexell%27s_theorem#Steiner's_theorem_on_area_bisectors.
Answer: This is a really good question. Firstly google spherical trigonometry & look at the wiki page. Make sure you are familiar with the cosine rule, sine rule & supplemental cosine law. It maybe an idea to look at the first ref (Todhunter) In chapter 7 he answers two similar problems to yours.  He does the circumcenter and incenter ... these are good examples for you to sharpen your teeth on.Figures 1 & 2
right in the first figure we have drawn the medians (concurring.) This is  slightly cheating , it remians to be seen that they really do concur.
In the next figure we have drawn half of the triangle with one median drawn. Two sides & the angle between them are known (the side b, the angle C & the side a/2). We have named the other side x & the angle at vertex A A_1 & the final angle A_+. You need to calculate the sin & cos of all three of these quantities
\
\begin{eqnarray*}
\cos x&=&\frac{...}{...}  &\sin  x=&\frac{...}{...}\\
\cos A_1&=&\frac{...}{...} & \sin  A_1=&\frac{...}{...}\\
\cos A_+&=&\frac{\sin c\cos B -\sin b\cos C}{\sqrt{\sin^2 b +2\sin b\sin c \cos A +\sin^2c}} & \sin  A_+=&\frac{...}{...}\\
\end{eqnarray*}
I have left some blanks for you to fill in.
Figure 3
In figure 3 we have drawn the triangle formed by two medians & half a side ("one sixth of the original triangle"). We know two angles and one side of this triangle, the angle ${A_+}$, the side $a/2$ & the angle ${C_1}$. (${C_1}$ is the same formula we have for ${A_1}$ with a,b & c & A,B & C cyclically permuted.)
We need to calculate the cos and sin of the other 3 unknowns
\begin{eqnarray*}
\cos x_1&=&\frac{...}{...}  &\sin  x_1=&\frac{...}{...}\\
\cos z_2&=&\frac{...}{...} & \sin  z_2=&\frac{...}{...}\\
\cos B_*&=&\frac{\sin^2b+\sin b(\sin a\cos C + \sin c\cos A) -\sin a\sin c\cos B}{\sqrt{(\sin^2a +2\sin a\sin b \cos C +\sin^2 a)(\sin^2b +2\sin b\sin c \cos A +\sin^2c)}} & \sin  B_*=&\frac{...}{...}\\
\end{eqnarray*}
Again we have left lots of blanks for you to fill in. (The calculation of $\sin  B_*$ is sublime !)
Now note that $\cos  B_*$ is invariant in the interchange of a & c and A & C  (together.) The lengths $x_1$ or $z_2$ could have been calculated using any other pair of medians & exactly the same formulea would have been derived; this shows that the three medians concur & therefore the centroid of a spherical triangle exists.
It remains only to give the trilinear coordinates of the point.
[Figure 4][3]
See figure 4, we need the perpendicular distance  $t_a$ to each side.
\begin{eqnarray*}
\sin t_a&=&\frac{\sin b\sin C}{\sqrt{(3+2(\cos a+\cos b+\cos c))}} \\
\end{eqnarray*}
$t_b$ & $t_c$ can be obtained similarly. We can rescale these by any symmetric function in $a,b$ & c ... & thus the sin of the trilinear coordinates for the centroid of a spherical triangle is
\begin{eqnarray*}
\sin b\sin C : \sin c\sin A :\sin a\sin B. \\
\end{eqnarray*}
I am slightly saddened that my original comment (I made a couple of days ago has been removed. I do agree that it was not an answer but I had no other means to post my comment.) I am very pleased that I have caused at least 8 other people to revisit this problem :-)

Centroid of a spherical triangle.


Cross Database with Application Role

Question: I have a requirement to access (read) a table/view in database B from database A. The application that needs to execute the query however uses an Application Role. From what I've read and tried, doing Module Signing is not an option.
What would be the next best approach?
Is that setting TRUSTWORTHY on for database A and use EXECUTE AS?
Answer: Just found the answer:

Create SQL login with minimal privileges
Execute function from source database under that user (WITH EXECUTE AS)
Module sign object
GRANT AUTHENTICATE to user created from certificate (this is the one I missed)

Cross Database with Application Role


Why do I feel as if Allah doesn't listen to me

Question: I have made countless duas and but none get accepted. At first it was of grades then to get accepted into a uni. After I didn't get accepted I researched on duas and figured it may have not been best for me. Then I made dua to pass in my uni and that didn't happen I got offered another chance and didn't pass that as well after so much dua and I didn't just make dua I also tried and studied. I pray and ask for constant forgiveness but now I feel helpless I don't know what I'm doing wrong or why nothing works out for me even if I have so much hope I cry to Allah but I don't feel as if I am being heard why?
Answer: This is a very real and heartfelt question. I hope I'm able to offer you some feedback that answers your question sufficiently AND soothes your aching heart. Here goes....
Firstly, let me assure you that Allah definitely listens and hears your duas.
"When My servants ask you concerning Me, I am indeed close.I LISTEN TO THE DUA OF EVERY SUPPLICANT WHEN HE CALLS ON ME...." Surat-al-Baqara, ayat 186 (2:186)
So, yes, He's listening to your duas. The question is, "why won't Allah give you what you want?" That's really what's vexing you,so that is what I will address.
Everyone knows of the Sunnah (or well-known practices) of the Messenger of Allah (SAWS), but a lot of times we forget that Allah also has a Sunnah, or way of doing things, that He does not deviate from.
"The Sunnah of Allah among those who lived aforetime. No change will you find in the Sunnah of Allah." Surat-al-Ahzab, ayat 62 (33:62)
"Then no change will you find in Allah's Sunnah. No turning off will you find in Allah's Sunnah." Surat-al-Fatir, ayat 43 (35:43)
Now, with regard to our duas, Allah's Sunnah is that 1) we ask Him for what we want; and then 2) we strive RELENTLESSLY in pursuit of what we want.
"Man can have nothing but what he strives for, and the fruit of his striving will soon come in sight." Surat-an-Najm,ayah 39-40 (53:39-40)
Dear brother (or sister) it is Allah Who feeds the birds, BUT HE DOES NOT PUT THE FOOD IN THEIR NESTS. No, they have to get up everyday and go looking for something to eat. Likewise, we are obliged to GO GET IT --WHATEVER IT IS WE DESIRE.
Now, yes, you said you did try. You studied. You worked hard. All of that. Still you were unsuccessful in your endeavors. Okay, listen to me now....
Thomas Edison is the man who invented the lightbulb. It is said that he FAILED in trying to make his invention work something like 1,000 times, but eventually he got it to work. When asked why he refused to give up after over 900 failed attempts to make his invention work, he said, "because I knew I only had to get it right once."
That is the kind of tenacity and determination you have to have when pursuing your goals and aspirations. The Arabic word for it is 'AZM = resolve, determination, firmness of purpose.
To put it simply, you have to try. If you fail, then try again. Then try again. Then step back, re-think and revise your strategy, AND THEN TRY AGAIN. Keep trying. Do not give up. Keep trying, striving, struggling towards your goals until you get what you want, OR UNTIL IT IS REVEALED THAT IT'S NOT MEANT FOR YOU TO HAVE WHAT YOU WANT.
The Qada of Allah is always a factor, even though I don't personally concern myself with it when I'm in pursuit of my own goals. Let me explain why.
Qada is what has been written for you in the Decree of Allah.Your money, your possessions, your good times, and your rough times, all of that has already been measured out for you. I accept that as true. But it is also true that NONE OF US KNOWS WHAT HAS BEEN WRITTEN IN THE DECREE OF ALLAH FOR US. We just don't. And I don't take failures (not even repeated failures) as some kind mystic message from Allah that He doesn't want me to have the object of my desire. I see failures as feedback from my efforts that tells me that I just used the wrong method to do whatever I'm trying to do.
That's how I see it, and it works for me. There have been many times when I've tried to get something done and I failed. Then I tried and tried and tried, and finally I succeeded. Alhamdulillah!
Allah is Al-Ghani (Rich) and Al-Kareem (Generous). He is the Ultimate Cause behind all effects in this Dunya. When we try something and it works for us, it is Allah Who made it work. He granted our efforts EFFICACY. THIS IS THE TRUTH.
This world operates according to laws and principles. Once you understand the laws that govern a thing you can make that thing happen anytime you want.
For example, the three elements of fire is fuel, heat, and and oxygen. If you want fire, bring these three things together. If you want to destroy fire then take one of them away. This is law.
Likewise, success is also governed by laws. Learn them, and then use them to your advantage.
I advise you to find someone who has done what you desire to do, and then ask them exactly what they did to attain the goal. They can give you a lot of good insights that you may not even be aware of. It's possible that you're overlooking something or leaving our a necessary ingredient for success in your endeavors. At any rate, modeling successful people is an excellent way to pursue your own success. The Messenger of Allah is a model for us to follow to be successful on the Day of Judgment. You need a role model for getting into the University. Try it. It can't hurt.
In closing, I leave you with this thought:
Allah answers our duas with His Guidance, His Grace, and His Blessings.
His Guidance (al-hudaa) is His telling us what we should or shouldn't do, or sending someone to us with the knowledge
we may be in need of.
His Grace (an-ni'mat) is sending us the people and things we need in our lives, and His removing of obstacles that impede our progress.
His Blessings (al-barakat) is the granting of EFFICACY to our efforts so that they bear the fruit that we desire.
Guidance, Grace, and Blessings are all different expressions of His Compassion (RAH'MAT) upon us. Never EVER give up hope for the RAH'MAT of your Lord, for Allah is Ar-Rah'maan (Most Gracious) and Ar-Raheem (Most Merciful).
Praise be to Allah, the Lord of all the world's!
As'salaamu alaykum.
Answer: Bismillahir Rahmanir Rahim. Alhamdulillahi rabbil-'alamin. Was-Salatu was-Salam 'ala Sayyidina wa Azimina, Wa Habibi Qulubina wa Shafii Nufusina, Abul Qasim Muhammad. Wa 'ala Ahli Bayti tayyibina tahireen.

Why do I feel as if Allah doesn't listen to me
I have made countless duas and but none get accepted.
I cry to Allah but I don't feel as if I am being heard why?
Feeling as if Allah does not hear is a sign of Kufr (Disbelief). And one who believes that Allah does not hear is a Kafeer (Disbeliever).
Allah has given clear signs in Quran that He hears His servants. Statements like "Allah hears" or "Allah is All-Hearing" or "He is Hearing" and alike, all together, mentioned in the scripture at least 8 times (according to my own counting on search page at legacy.quran.com). The "All-Hearing", Arabic ٱلْسَّمِيعُ, is one of the 99 attributes of Allah. Thus, one who believes that Allah does not hear is rejecting this attribute of Allah and the verses of Qur'an.
This problem, feeling as if Allah doesn't hear, has been given answer in Quranic verse 10:100

And it is not for a soul [i.e., anyone] to believe except by
permission of Allāh, and He will place defilement (Arabic الرجس -
uncleanness) upon those who will not use reason [original Arabic is
"La Ya'Qiloon" - do not use A'ql (intellect)].
Quran 10:100

Imam Jafar As-Sadiq explains the defilement

(It has been narrated) from Abu Abdullah (asws) having said: 'The
uncleanness (الرجس– ) it is the doubt. By Allah (azwj), we (asws)
do not doubt regarding our (asws) Lord (azwj), ever'.
TafseerHub-e-Ali, page 58

So we have the story of a person who made Dua to Allah, asking certain things to happen, but it didn't happen and the person, as a result, is having doubts - feeling as if Allah doesn't hear him. The verse 10:100 says that doubts are put by Allah if servant does not use A'ql (intellect).
What does it mean? It means the protagonist of our story should use his A'ql (intellect), reason, understand why what happened happened, in order to avoid doubts which eventually will lead to Kufr. It means if one makes a conclusion without pondering over the event he will make a mistake.
We need to consider a number of  questions in order to understand and ponder the event then make best, correct conclusons. For instance, "What is Dua?", "How one makes Dua?", "What are the prerequisites for making dua?", "What should we ask in Dua?", "When does Allah accept Dua?", "How Allah responds to Dua?", "What to do after receiving response to Dua?", "What to do if Dua is not accepted?", "What to do if Dua is not responded?", and so on and so forth.
One may think what's the importance of having answers to these questions.
Let's consider the first question "What is Dua?", for example. Most people have wrong understanding of this Islamic concept and what's in their mind is that making Dua is similar to making an online order to a delivery service to get pizza, thus, if one makes Dua to Allah then that's it (Bob's your uncle), what you asked magically gets done. And what people do is that, based on this false incorrect understanding, they put too much hope to Duas they make, expecting surprising phenomenal outcomes, and eventually get frustrated when they end up with not getting what they expected.
Or let's consider the question "What should we ask in Dua?". This is one of the fateful questions that, unfortunately, most people do not know the answer to and possess relevant knowledge. Quran 2 says

200.And among the people is he who says, "Our Lord, give us in this world," and he will have in the Hereafter no share.
Quran 2:200

So a person who asks Allah to give in this world is making the biggest mistake, price for which will be payed in hereafter and this is what most of the people do we see in everyday life. So to having answer to the question "What should we ask in Dua?" depends our fate in eternal life in Hereafter. And the next verses, 201-202, explains what is correct to ask in duas we make.

201.But among them is he who says, "Our Lord, give us in this world [that which is] good and in the Hereafter [that which is] good and
protect us from the punishment of the Fire." 202.Those will have a
share of what they have earned, and Allāh is swift in account.
2:201-202

So based on these verses it is possible that Allah didn't answer the Dua out of His mercy as He declares Himself mercy and didn't want the servant to be deprived from share in Hereafter.
Each the rest of the questions listed above, regarding the concept of Dua , should be considered and thoroughly studied in similar way and I will leave it as a homework to the OP.

Why do I feel as if Allah doesn’t listen to me


Problem with RVM use

Question: I am having trouble with using rvm use in zsh. Whenever I do rvm use 1.9.2 or other ruby version, it gives confirmation as Using /home/kxhitiz/.rvm/gems/ruby-1.9.2-p290
But when I do rvm list ruby 1.9.2 is not selected. This works fine in bash. That means I can go to bash and select ruby version I need, and I can come back to zsh to use it.
So that means if I have to change ruby from list of rubies I have installed, I have to go to bash to select it and once selected in bash, I can see it selected in zsh as well.
Any solution would be highly appreciated.
Comment: Do you call $HOME/.rvm/scripts/rvm somewhere in your .zshrc? Also, see this: http://beginrescueend.com/integration/zsh/
Answer: Ok, I did fixed it by adding following line at the end of my ~/.zshrc file
[[ -s "$HOME/.rvm/scripts/rvm" ]] && source "$HOME/.rvm/scripts/rvm" # Load RVM into a shell session *as a function*
Cheers!

Problem with RVM use


How to have an infinitely big canvas?

Question: I am building a web app that uses a canvas to draw on. I would like to have an infinitly big drawing space (you can scroll as long as you want in any direction) AND save the data / images to a database.
Something has been done similar to that here: http://wordsquared.com/, where the more people that play it, the larger it gets.
I know this would require somehow tiling images, saving them to a db, then only loding the ones within the viewport. How would this and the infinite scrolling be accomplished / where should I start?
Comment: You can't have it be infinite, not properly, and you don't want to just load the next chunk in a direction, as you'll end up hitting the limits of your coordinate system. What you want to do is constantly recenter the canvas and keep track of that, loading new chunks when the center reaches the edge of the loaded area.
Comment: What I mean is the more you scroll, the larger it gets. Of course I would need to set a limit EVENTUALLY, but I would like something that get scale to as large as I set it.
Comment: Possible duplicate of [Make a canvas infinite](http://stackoverflow.com/questions/38589304/make-a-canvas-infinite)
Answer: This is a bit unconventional a method, but just struck me as potentially (in theory, with unbound storage space) infinite.
You will need to store a currently loaded chunk of the board, given as a unique ID of some sort. The chunks and their data need to be in a table with the following columns:

Chunk ID
Chunk Data
ID of chunk to the north
ID of south
ID of east
ID of west

You need to make the canvas draggable, perhaps by using jQuery or similar (this question has some info on that).
Now, create an event listener for the canvas to be dragged, and track the distance it moves. When released, if the canvas has not changed to a different chunk, do nothing.
If the canvas has left the current chunk, use the stored ID to find the next chunk to load. If the ID is 0, assume the chunk does not yet exist, and create it. Otherwise, load the chunk, replacing the existing chunk. Set the canvas back to centered.
With long enough IDs and enough storage space, this will give you an infinite canvas, as no coordinate system is used. It also allows for wrapping the edges, or creating wormholes.
How to implement it, I'm not quite sure, but you only need to track how far the canvas has been moved. Google Maps does something similar, so I would look at how they handle it (I will do so shortly and see if I can add some implementation details to this answer).
This may not be the most practical or simplest method, but it was fun to come up with.
Edit: I believe this is along the lines of the basic functionality:
http://candrews.net/blog/2010/10/introducing-sprymap/
It is a light-weight draggable javascript map. You simply need to track how far, then.
Comment: Hmm.. that certainly isnt conventional, but its probably the most practical and simplest method i've heard yet. Im thinking it would probablt be best to have each tile be an entire large html canvas like (1000px x 1000px) so thats it is easy to save and load all the data
Answer: I have made a library to deal with this called TiledCanvas
It provides interfaces to zoom and move. And draw in an infinite space using all the canvas apis.
It lets you load chunks dynamically based on coordinates, you just give it a function that gets as paramters a x, y and callback and you simply give it back something that can be drawn on a canvas.
https://github.com/Squarific/TiledCanvas

How to have an infinitely big canvas?


Is the launch of a rocket slow because the change in velocity (delta_v) should be as small as possible because of energetic reasons?

Question: Does it make sense to keep the launch of a rocket slow because the change in velocity (a.k.a. delta_v) should be as small as possible because of energetic reasons?
Comment: The delta-v should be what is required for the mission.
Comment: Until recently, heavy launches (30-ton +) did not have many options in terms of initial launch speed. But Elon Musk did mention the idea of a rocket on rails where engines would ignite when the rocket has already achieved 20+ km/h delta-v. And propellant savings were expected to be significant.
Answer: No. Nor the opposite.
Smallest possible launch useful speed is barely >0 m/s.  The rocket will mostly hover, spending all of its fuel just fighting gravity.  Inefficient.
The other extreme is almost an explosion. Tons of engines (weight!), huge acceleration and within seconds huge speed. And that means huge aerodynamic drag.
So we balance.
Launch as fast as reasonably possible, which means take in account

Weigh of the additional engines, or bigger engines to accelerate faster
Structural supports needed for high acceleration.
The aerodynamic drag

Which most of the time results in:

Launch fast. Engines at max.
Reach high speed (at still low altitude, so still high drag)
Throttle engines down to keep drag in check
Once out of most of the atmosphere throttle back up.

And of course, all of this varies per mission with different rockets, different payload, different target orbit, etc. etc.
Comment: Another reason why real rockets launch slowly: If the initial acceleration is high, you can load more fuel until it isn't. That extra fuel is burned inefficiently, but still better than no extra fuel.
Answer: It makes no sense to deliberately keep launches slow. The slowest possible "launch" is a hovering rocket. It is 100% efficient at converting propellant into sound and fury, but 0% efficient in converting it to lift.
The faster the launch (up to the tolerance of airframe, payload and crew), the more efficient. In the case of airframe, this is "Max Q". For payload, electronics can withstand the acceleration in artillery shells. For crewed launches, maximum G-forces at launch are usually well below what is tolerated at re-entry.
Answer: Fundamentally, this comes down to fuel being cheap and rockets being a lot more expensive than fuel tanks.
If your rocket takes off fast adding a bit more fuel tank to it isn't going to cost very much, the additional fuel is even cheaper.  You lowered the cost per kilogram of payload.  There is an issue of diminishing returns but the optimum point is with the acceleration pretty low at first stage ignition.
There is also the issue that acceleration goes up as the fuel burns off.  The higher the maximum acceleration the stronger everything above it must be, thus you have an incentive to keep the acceleration down.  Few rockets have enough engines that you could keep the acceleration in check by shutting off some of them early.
Answer: Something that has been implied but not spelled out in the other answers. If you launched going 0 m/s upwards, you would literally be wasting all of your fuel fighting gravity, and go nowhere at all. Therefore, higher accelerations are better. Of course, there is a reason that rockets down accelerate at 1000G:

Your fuel supply will be burnt very quickly
You will lose a lot to drag
You will destroy your payload
You will destroy any crew
The rocket will blow up from the aero-forces
You will need a lot of mass for the engines

See Hennes answer for what we actually do and why. I won't repeat that here.
Answer: The other answers have quite clearly explained that, for a launch vehicle, launching as slowly as possible is inefficient.
But you mention 'for energetic reasons' in your question title:  if you are, in fact, looking for the most energy-efficient rocket (rather than the most practical), the answer is slightly different.
The answers to this question explain this far better than I have been able to.  My attempt to explain it is included below.

In this case, while you may still lift-off quickly, the velocity of your rocket engine exhaust should not (initially) be high.  Decreasing this at takeoff is actually beneficial for efficiency (given a constant power) - so, in some abstract, tangential sense, your intuition about this is correct.
Consider:  your rocket exhaust is moving very fast, therefore it has a lot of energy.  That energy is wasted - we don't care about the exhaust once it has exited the rocket.  If you tune your exhaust velocity so that the motion of the exhaust after leaving your rocket is negligible -  in other words, the exhaust velocity is essentially equal to the rocket's velocity - more of the power put into the exhaust goes to the rocket, because the exhaust has minimum energy.  For an explanation using maths, see this answer.
A more mathematical approach to this argument can be found in this NASASpaceflight forum thread, regarding exactly what effect a VASIMR-like engine has on the performance of a rocket.
Do note, however, that modern variable-isp engines are not particularly useful for launching a rocket from Earth.  This is because launch vehicles are very mass-constrained, wanting to get as much energy out of each Kg as possible - they are not especially worried about being 100% efficient with that energy.
Comment: So at liftoff your exit velocity should be zero? I think there is a flaw in that reasoning.
Comment: There are related concepts where this is correct--for example, a turboprop or turbofan are more efficient than a turbojet because they have lower exit velocities, but only because they can also move more mass.  Rocket exhaust is already mass-limited, and so we go to great pains with nozzle design to keep exhaust velocity and specific impulse high.
Comment: This is incorrect.  Momentum is conserved.  If the propellant leaves the rocket at zero velocity, it has zero momentum, and thus no momentum is imparted to the rocket.
Comment: The problem with this reasoning (apart from the rocket starting at zero velocity) is that there is a finite supply of reaction mass available to the rocket, and for chemical rockets this supply is also directly linked to the available energy for propulsion. It is not useful to eject it at a lower velocity to "save energy" that can't be stored or used in any other way. Energy efficiency just isn't a very useful concept for rockets, you use the energy you have available.
Comment: A zero-velocity exhaust has high *energy* efficiency (which is why airplane engines try to lower the exhaust velocity -- they have access to unlimited mass, but finite energy).  Rockets, on the other hand, are concerned with *mass* efficiency, since they need to carry all their mass with them.  In a mass-constrained situation, you want the exhaust to go as fast as possible to get as much momentum change out of it as you can.
Comment: So at liftoff your exit velocity should be  ̶z̶e̶r̶o̶ very small? I think there is a flaw in that reasoning.
Comment: @Mark  Thanks for that example!  I knew this reasoning worked for energy efficiency, which is why I decided to answer it here - since the original question asks "...because of energetic reasons."  I forgot to consider that, when used as launch vehicles, rockets are much more constrained by mass than energy.  The original thread is for electric engines, which _are_ (often) energy-constrained.


iTunes online store architecture

Question: These are days when many major websites are sharing details about their internal architecture.
I am curious if anyone has some idea about the architecture of the iTunes online store.
Is there any published material,video or folklore about the architecture of the iTunes online store?
In particular I would love to know what Apple uses to store so much of the audio and video content.
Any pointers?
Answer: AFIK it's one of the few sites that are based on WebObjects (Apple's own framework like PHP).

iTunes online store architecture


zk SNARKs based anonymous voting

Question: I want to implement zk-SNARKs based anonymous voting. Are there examples or packages available for substrate? Where can I get started.
Comment: There are plenty of zk-SNARK libraries written in rust. Not sure if they are usable under no_std environment. As long as it works, it should be pretty straightforward to use that inside a pallet. In the early days there's a project called ZeroChain developed by the LayerX team, but its development was halt for some time. It's still a good code reference though.
Answer: I know that https://manta.network uses ZK for their solution, so you could see how they have gone about things: https://github.com/Manta-Network/manta-signer. They did a sub0 talk.
I am a beginner in this space, but plonky2 might be fast enough in wasm if it switches to using portable simd (and that allows wasm simd to happen).
For ZK specific stuff, https://crypto.stackexchange.com/ is also a good place to ask questions.

zk SNARKs based anonymous voting


How to tell if I'm using a Server Dell PowerEdge 2900 I or II?

Question: How to find out the exact version of this old server?
It doesn't seem to be any mark or number printed on it (I was looking for a roman numeral).
Thanks
Answer: The service tag is listed in the server's BIOS setup and is likely echoed to the screen during boot.
Additionally, you can get this info by installing OMSA on the system.
Answer: If the Dell Diagnostics are installed, there may be a way to run the interactively from the start menu or just try running DellSystemInfo.exe from an Admin Command Prompt:
<code>C:\Program Files (x86)\Dell\AdvDiags\DSET\bin   DellSystemInfo.exe
</code>
It should output a zip file with instructions on how to extract and open the report.
Answer: This page indicates that there are three generations of PowerEdge 2900, which can be identified by the motherboard model:

Gen I: J7551 or TM757
Gen II: YM158
Gen III: NX642

The motherboard model can be found by looking up your service tag at http://www.dell.com/support, and going to System configuration -> Components -> Other parts.  One of those model numbers should feature there.


Blender Accessibility Features

Question: I'm teaching Blender to my high school students this year and it is a main piece of our worldbuilding course. One of my students is visually impaired (and hearing impaired) and I'm wondering if there are any accessibility features or add-ons for the program. Right now, it seems like there's nothing available, which surprises me. I searched Google and all I've found are a couple of posts where other people are inquiring about this. Can anyone point me in a useful direction for this sort of topic. My student is willing to work hard and she has enough vision to navigate on a computer, but she also emphasized that her vision "fluctuates" and Blender's complexity and small text makes it very tough. Any assistance or insight would be greatly appreciated. Thanks.
best,
Kas
P.S.: I am new to this site, so please forgive me if I've missed something.
Comment: You can increase the UI scale in the preference to help with small text, or personalize the theme to increase for example the contrast between elements. There is also an upcoming project call Blender Apps (https://code.blender.org/2022/11/blender-apps/) to make blender more customizable (for example by removing whole part of UI). Can you expand a bit about what kind of features you are looking for?
Answer: I looked around for some resources for disability features in Blender and came up against some similar problems you did. Looking through this forum thread came up with some interesting routes to make your students Blender experience a little bit easier. Notably this Print-Friendly Theme that overhauls Blenders UI and should hopefully make it a bit easier to use.
There's also a lot of settings you can change inside of Blender that can make it a lot easier to use and more focused on hotkeys and typing which might be easier than navigating with the mouse. For one, a big setting I always like to change is making the Spacebar set to open a search menu that lets you have super easy access to a lot of features that are usually pretty hard to find or remember. To activate it, go to the top left of Blender, click Edit, and from the drop down select Preferences at the bottom. From the Blender Preferences menu, you'll want to go to the section named Keymap, and from the preferences, change Spacebar to Search. Now whenever you need something, you can press Spacebar, type a part of the function you need, and then let autocomplete show you a bunch of options. This feature does replace having Spacebar as the play button for the animation timeline though, just keep in mind that Shift+Spacebar will work for that instead.

Another very handy setting to change would be under Interface, it's the setting called Resolution Scale. By increasing it you can make the icons a bit easier to read, and it'll increase the size of the text as well. There's even an option to change the font to something your student prefers.

You can change the colors of pretty much every single UI element in Blender with the Themes section of the preferences menu, and its where you can also install the previously mentioned Print-Friendly Theme. Hopefully changing the color scheme will make it easier.
I'm afraid I wasn't able to find any way to use a screen reader to make navigating the UI easier through audio, I really hope they add options for that in the future, as it's a real shame there isn't a framework for it already.
Comment: Thank you so much.... this was so helpful! I will be seeing her again tomorrow so we can try these things out. The Resolution Scale really looks like it should make at least some difference.
Answer: As an addendum to @NascentSpace's answer, note that you can set the resolution preposterously high (e.g., 3.0 or all the way up to 4.0) and -- while you're at it -- set the line width to "Thick", so grid lines and active object outlines are easier to see:

In the default workspace (the "Layout" workspace), resolutions this high typically make all of the areas except the 3D Viewport nearly unusable, BUT if you hover the mouse cursor over an area (e.g., the Properties editor) and hit CtrlSpace, it'll toggle that area in and out of fullscreen mode, like so:

Your student may find it useful to use the default "Layout" workspace at very high resolution as a kind of spatial main menu, jumping into areas like the Outliner or Properties with CtrlSpace, taking advantage of the full screen display as she moves between them.  There's also the alternative fullscreen toggle CtrlAltSpace which fullscreens an area without any associated menus (technically, it fullscreens the "main region" of an "editor" within an "area").  This can be useful for distraction-free zooming of the 3D Viewport panel in particular.
Alternatively, if she reconfigures her workspace into a single big area, there are hot keys for switching the area between all those different editors:  ShiftF5 for 3D Viewport, ShiftF9 for Outliner, ShiftF7 for Properties, etc.
At high resolutions, any component that gets cut off is scrollable, and that includes menus, whether horizontal or vertical.  You can scroll them by either middle-click dragging, or by using the mouse wheel when the cursor is within the menu.  Again, the mouse wheel works whether the menu is vertical (like the tools on the left of the 3D viewport) or horizontal (like a truncated menu bar at the top).  There are more obscure controls available as well.  In the Properties editor, for example, Ctrl-middle-click dragging will zoom the properties display itself in and out.
Also for the 3D Viewport, she should learn the T and N hotkeys to quickly toggle the toolbar and sidebar on and off, as they take up a lot of room at high resolutions.  She can also disable the "Navigate" gizmo (either by toggling all gizmos off with Ctrl` or by unchecking "Navigate" in the "Show Gizmos" pulldown).  This gizmo takes up a lot of screen real estate, and all its important functionality is easier to access with numpad hotkeys.

If she finds the grid floor distracting in perspective view, that can be disabled in the "Show Overlays" pulldown:

With the 3D Viewport fullscreened and most of the distractions removed (everything above plus disabling "Tool Settings" in the "View" menu), you can get a pretty clean 3D Viewport display:

If she prefers dark text on a white background, she can switch from the "Blender Dark" to the "Blender Light" preset theme in "Preferences -> Themes".  Those are the only two built-in themes, unfortunately.  Finding alternatives can be a little difficult, as themes tend to be pretty version-dependent, and they aren't always kept up to date.

Finally, you might want to take some time to review the User Interface section of the Blender manual, especially the "Window System" section.  No human being can keep track of all the UI controls Blender offers, so I guarantee you'll find some tricks in there that your student will find useful.
Even more finally, note that everything in the "Preferences" menu gets saved automatically.  Changes to the UI layout won't, but you can work with your student to configure a good starting layout with the venerable default cube and then use "File -> Defaults -> Save Startup File" to save that whole layout as her default "new file" layout.  (Saving the desired default as a normal ".blend" file is helpful, too, so you don't accidentally overwrite your desired startup file, and so you can try out a few different alternative defaults.)
Comment: Thank you so much! A lot of great insights here... looking forward to trying them out with my student. Much appreciated!

Blender Accessibility Features