Windows 2016: Docker container error

Question: I'm using docker on Windows server 2016, I have created a container using the "microsoft/windowsservercore:latest" image.
On this image i have installed "Print-Server" role but when I try to call "Get-Printer" cmdlet I obtain an error with the spooler service.
These are the commands used to recreate the problem:
docker run -d --name testspoolererror1 microsoft/windowsservercore:latest ping -t localhost
docker exec -it testspoolererror1 powershell
Install-WindowsFeature Print-Server
Set-Service spooler -StartupType Automatic
Start-Service spooler
Get-Service spooler
Get-Printer

This is when I receive the error:

Get-Printer : The spooler service is not reachable.  Ensure the spooler service is running.
  At line:1 char:1
  + Get-Printer
  + ~~~~~~~~~~~
      + CategoryInfo          : NotSpecified: (MSFT_Printer:ROOT/StandardCimv2/MSFT_Printer) [Get-Printer], CimException
      + FullyQualifiedErrorId : HRESULT 0x800706ba,Get-Printer
  In the event viewer i found the error:
  The Print Spooler service terminated unexpectedly.  It has done this 2 time(s).

Can Anyone help me to solve this problem?
Comment: I am having the sample problem. `Get-Service spooler` shows that the spooler service starts but it crashes after running `Get-Printer`. `Get-WinEvent` indicates that the spooler crashed and in addition shows: `The print spooler failed to load a plug-in module WSDMon.dll, error code 0x7E. See the event user data for context`
Answer: Because Windows containers are sharing same kernel with host machine you cannot have spooler running on both same time. So stop and disable spooler from host and you are able use spooler on one container on that server.
Here is fixed set of commands:

Stop-Service spooler
Set-Service spooler -StartupType Disabled

docker run -d --name testspoolererror1 microsoft/windowsservercore:latest ping -t localhost
docker exec -it testspoolererror1 powershell
Install-WindowsFeature Print-Server
Set-Service spooler -StartupType Automatic
Start-Service spooler
Get-Service spooler
Get-Printer
Answer: I'm sorry to hear you're having this issue and I'll be glad to do what I can to help you sort it out :)
For the sake of being thorough, I tried this myself by running the following commands:
docker run -it microsoft/windowsservercore:latest powershell

(Now running powershell from within container)
Install-WindowsFeature Print-Server
Set-Service spooler -StartupType Automatic
Start-Service spooler
Get-Service spooler
Get-Printer

I was able to run these on my system, without an error. So that's a start.
Now, from your error it looks like the spooler service didn't even start. What do you see when you run Get-Service spooler? Will you try running these commands on your system just as I have listed them above then report back with your results?
Also, to clarify, what are you trying to do when you're pinging localhost from the container? Are you trying to ping your container host?
And as a side note, if you're looking for background info on how container networking works on Windows, here's a good place to start: https://learn.microsoft.com/en-us/virtualization/windowscontainers/manage-containers/container-networking
--Kallie B. (Program Manager, Microsoft Networking Team)
Comment: I've made a test with your commands and i have an error: Get-Printer : The spooler service is not reachable.  Ensure the spooler. The spooler service is running before and after the "Get-Printer" call. I've tested on a Windows 2016 DataCenter updated, with docker 1.12 and docker 1.14-dev with the same result.
Comment: I am getting the same problem as Scott. `Get-Service spooler` show that the spooler starts but it crashes after running `Get-Printer`. `Get-WinEvent` indicates that the spooler crashed and in addition shows: `The print spooler failed to load a plug-in module WSDMon.dll, error code 0x7E. See the event user data for context`
Comment: @Kallie-Microsoft Any suggestions?
Answer: The reason that Kallie seems to have been able to get the above steps to work is probably because it is being hosted differently. I tried the above steps via Docker on a Server 2016 box, and hit the same errors. When I tried it via Docker on Windows 10, I was able to launch the spooler successfully and run the above commands, but I couldn't install any drivers successfully which would make it actually useful. Pnputil just throws odd "No Data" errors when attempting to install any .inf's.
My guess is that it works on Windows 10 because it's using hyper-v emulation instead of the native container used when hosting Server 2016 Core on Server 2016. Another thing I noticed was that the drivers are inherited from the base machine when creating a container on Server 2016, but not on Windows 10. I assume that's fairly well-understood behavior by Docker experts, but it does seem like the inherited drivers might be causing the crash. I'm not a Windows expert either, though.
Either way, it seems like something that Microsoft will have to look into and resolve.
Answer:
The network that docker runs on (by default) is not that same network as the host.
Pinging localhost from inside the container is not doing what you think it's doing.
Learn how docker networks as step 1.
Comment: Don't think the issue is network related, as all commands are executed from within the container. Additionally, the `ping` command just causes the container to keep running. The question seems perfectly legit to me.

Windows 2016: Docker container error


Samsung galaxy 3S, battery is good, phone won't charge or power up

Question: I have a samsung galaxy 3S, about a week ago it rebooted itself, since then it will won't charge or turn back on, I have tested the battery to see if it was that, the battery works perfect, I have cleaned the connections to the battery and the battery itself, when i do plug it in the red charge light will come on for about 2 minutes then goes off, but still won't power up, please help, i have lots of picutures i would like to get off the phone
Answer: If your device is not modified, the best option from here may simply be to contact Samsung and ask about using your warranty to get your device fixed. All Samsung mobile phones have a two year warranty from the moment you receive the device, so if your device is still within that period of time you should be able to, for free, have Samsung fix the device and your pictures retrieved.
Use the Samsung website for more information regarding this.
USA: http://www.samsung.com/us/support/
UK: http://www.samsung.com/uk/support/


Pages or posts for reference site

Question: I have a suburban farm website that I made using wordpress. It is really just for fun but I want to build it with best practise in mind in case I ever want to do more with it. To this point I have been writing 'how to' and 'update' articles which are both fine as posts. I have also been writing keeping and growing guides as posts but really, this content is timeless. Before I get too far down the rabbit hole should I make this kind of content as pages? The fear is that eventually I could have hundreds of these pages, is that a problem?
Answer: There are 2 things here:
1.SEO:
There is absolutely no issue in terms of ranking when it comes WP pages vs post. As long as you got rich content either should be fine.
2.Website Structure:
Pages vs Post could come under your website planning and future consideration. For general understanding, pages are generally for website page and posts are for the blog.
Note: Also, it's good to have fewer pages with rich and good length content instead of having 100s of small pages.
Answer:
Before I get too far down the rabbit hole should I make this kind of content as pages?

Doesn't really matter as long as you remain consistent with it.  Personally, I would make this content into a custom post type and that would allow for custom taxonomies to better handle navigation and SEO issues, but this isn't necessary.

The fear is that eventually I could have hundreds of these pages, is that a problem?

Nah. Again, as long as you are consistent about which content container you use, there's no benefit nor penalty to using posts over pages.
Answer: Articles/posts with links to always reputable University Agriculture resources can't hurt. Try Google search [site:.edu filetype:pdf etc terms etc] for key resource docs and work from there to direct traffic/integrate with invites/orders.   In addition, several market farmers/gardeners I've spoken with report earnings near or better from farm holiday/edu-tainment/activity/tourism events than actual field produce sales - which have advantage of dates/times and so always fresh upcoming content. With CMS platforms  custom content forms and filters/list can make events calanders easy.

Pages or posts for reference site


Connecting to the Internet with an iPad or iPod Touch, via public wifi

Question: I was thrilled when the cafe at our local sport center started to provide free wifi. Everything was great for awhile, but a few weeks ago something changed, and now my devices cannot browse or check email etc. Despite being able to detect the wifi network with full bars, my devices tell me they cannot connect to the Internet. The staff do not think they have changed anything. What could be happening, or what could be different? Is there anything I can do about it?
Comment: Have you tried restarting the devices? Have you tried Settings/General/Reset/Reset Network Settings?
Answer: When your iOS device doesn't get an IP address from the router that will route packets to web sites like www.apple.com or some of the other under the hood servers like the iCloud servers and APN (Apple Push Notification) servers it will let you know it's not connected to the internet.
In this case, it's almost always a problem with the wireless router or the internet service in the shop. Even if other people are surfing, the router could be out of new addresses and need a reboot. If no-one in the shop has access, then the causes could be lots of other things and not a temporary outage of DHCP addresses.
Comment: Great comment - in that vein, you also can also request to renew the DHCP information which does the same thing as the above assuming there are not more than one base station broadcasting on that SSID / network name.
Comment: Adding on to this, something quick that might help (but doesn't really justify an answer) go to Settings > Wifi and select the network ( the (>) on the right hand side) and forget the network. Then attempt to re-join - it might solve the problem as it forces the i-device to forget all it knows about the network so far.


Can a Tor relay run on MacOS?

Question: I have a powerful computer and a fast internet connection. I'd like to support the Tor network by operating a relay. I've seen that there are relay packages and instructions for various versions of Linux, BSD and other Unixes, and Windows - but not MacOS. Why not? Or have I missed something?
Answer: There are currently 6 relays running on macOS in the consensus, but I don't think it's an officially supported configuration. There are not many people who run Mac servers, so it probably isn't worth the extra effort to build and maintain a Mac-specific package. I don't think there would be any issues running tor though, just install tor (through homebrew, compile from source, etc), configure your torrc file like on other platforms, and set your Mac to run tor at startup. If you run into any issues, ask a new question here and someone might be able to help.

Can a Tor relay run on MacOS?


Newbie question: how to add block, add transaction and search transaction in blockchain

Question: I am new to this. I wanted to know, how i can add block, add transaction and search for transaction in blockchain?
Thanks.
Answer: I am assuming that you are asking for Ethereum Blockchain.

I wanted to know, how i can add block

Adding a block in Ethereum requires mining. You can find more about mining here.

add transaction

Transaction can be added using eth.sendTransaction more details about it can be found at https://github.com/ethereum/go-ethereum/wiki/Contracts-and-Transactions

search for transaction in blockchain

Transaction details can be found from the transaction hash that we get when we send a transaction. Use eth.getTransaction(txHash), to get the details of the transaction.


DataGridView.HitTestInfo equivalent in Infragistics.Win.UltraWinGrid.UltraGrid?

Question: Does anyone know if the Infragistics UltraGrid control provides functionality similar to that of DataGridView.HitTestInfo?
Answer: Check this out.
They don't convert the coordinates, but they use a special Infragistics grid event (MouseEnterElement) to get the element, which the mouse currently hovers over.
Maybe it helps.
Answer: There's a .MousePosition property which returns System.Drawing.Point and "Gets the position of the mouse cursor in screen coordinates" but I'm using an older version of their UltraWinGrid (2003).
They have a free trial download, so you could see if they've added it to their latest and greatest  :o)
Answer: If you had a MouseEventHandler for the UltraGrid then you can do the following:
UltraGrid grid = (UltraGrid)sender;

UIElement element = grid.DisplayLayout.UIElement.ElementFromPoint(new Point(e.X, e.Y));

You can then cast the element depending on its expected type using element.GetContext():
 UltraGridCell cell = (UltraGridCell)element.GetContext(typeof(UltraGridCell));


Differences from SuperUser

Question: Unfortunately a lot of questions here are about Hardware support as well as shopping. These questions belong on SuperUser. How do we prevent users from asking questions? How can we as a whole improve on this situation?
Answer: I'm an SU mod. And well, I suggest that the best way to deal with this is the same way we would deal with questions that belong on SR or HW - close em. Keep closing em.
If its a great and glorious question that causes the heavens to open up and have a single sunbeam shine down on how awesome it is, vote or flag to migrate. If its a horrible question, don't bother, we'll kick it back.
As a SR mod, and having been through the process of helping bootstrap a new site, closures are an essential part of setting the scope of the site. Sometimes you got to set an example.
Comment: As a NE mod (pro tem and current) where we also get many questions that belong on SU, you can't stop users from asking them. I agree that you just have to keep closing them (and closing them and closing them and closing them and ...). Let me also throw in that you can vote/flag to migrate even a mediocre question where users have already provided good answers that deserve to be preserved.
Answer: We really can't prevent them from asking them. Some (a lot of) people simply don't read the rules. What we can do, though, is close the bad ones and migrate the good ones to Super User (if you think it isn't crap, and won't be closed by SU, flag it with a custom flag and ask us to migrate it. We probably will).
I'm not terribly happy about so many of our questions being closed, but I think that it's just going to be a battle this site fights forever. SR fights the same battle to a lesser extent.

Differences from SuperUser


Is there any app to make/receive calls on iphone from another non-apple device?

Question: I have an iPhone that I use mainly to connect to internet and another device that simply makes calls and not have much features. It becomes frustrating using two devices when answering calls. There is a feature called "Call Divert" which charges for each call diverted, and using a Bluetooth headset is inevitable especially when you are having much less call frequency. So is there any app, without the need to be JAILBROKEN to connect the iPhone with other Bluetooth enabled device and simply act as a sound pathway so that calls and music on the other phone can be heard on this iPhone. More like using an iPhone as Bluetooth headset.
Comment: I'm not sure what carrier options you have available but that's a factor.  I'm in the USA and my carrier allows simultaneous calls and data usage, other US carriers (CDMA) don't allow this.  Verizon apparently recently made this available but only if you have an iPhone 6.
Answer: Yes, on your iPad or iPhone go to settings > facetime > choose the caller id option for your phone number, then an option will appear to allow iphone cellular calls.


Safe distance from a black hole?

Question: Okay so we all know that one planet from the movie "Interstellar" that orbited the black hole Gargantua. How do I NOT get that? What's the minimum safe distance for a planet to be from a black hole ("safe" as in no funky relativistic time dilation and no megatsunamis)?
Comment: "Okay so we all know that one planet...". Actually: no... I have not seen that film. I am most likely not the only one. Kindly edit your question to not assume people know what you are talking about when you refer a specific element to a specific film.
Comment: Interstellar - Another film that was pretending to be scientifically acurate while in reality it features severe pseudoscientific facts further hindered by severe lack of logic. At least it was not so bad as 2012.
Comment: @VictorStafusa The astrophysicists consulting for Interstellar published several papers based on a groundbreaking new method of visualizing a black hole that was developed for use in the film (the most accurate visualization ever). So, while there were some handwavium moments for the magic of film, to take the tone that it was nothing more than pseudo science is ignorant, at best.
Comment: Ah, well, this sounded like a question I might be able to answer but unfortunately I can't understand the question, having not seen the movie that the question is contained in.
Comment: @TylerH I didn't say that it is "nothing more than pseudo science". I said that it "features severe pseudoscientific facts". Those two claims are very different. There is a lot of good science there indeed, but unfortunately, it also got mixed with a large chunk of pseudoscience.
Comment: @MichaelKarnerfors you're right, I haven't seen it either
Comment: https://www.physicsforums.com/threads/minimum-safe-distance-to-black-hole.557292/ this was an amusing read which may be intresting to you, it doesnt give a clear cut answer as such as the size of the black hole does affect the distance and Gargantua isnt at the size they are talking about, but it is close so hey, its an approximation
Comment: How about the nice shower of X and gamma rays that you'd get by staying that close to a black hole with an accretion disk? Isn't that a huge concern too?
Answer: A black hole does not have any magic properties, it does not "radiate" time dilation or any other nonsense like that. Noticable time dilation happens when one observer moves at relativistic speeds with reference to another, or when one observer is under much higher gravitational acceleration than another.
In the movie "Interstellar", the crew of a spaceship landed on a planet that orbited a supermassive black hole. Being on the surface of that planet is said to cause a time dilation of 7 years on Earth for each hour on the planet surface, in other words a Lorentz factor of > 61,000.
The scientific problem with that is that the numbers don't nearly add up.
The planet itself has 1.3g surface gravity. That causes a negligible time dilation compared to Earth.
That only leaves relativistic speed compared to Earth as the source of time dilation. For a Lorentz factor of 61,000 and ignoring the effect of the planet's surface gravity, you would need a relative speed of 99.999999987% of c, the speed of light. That is how fast the planet would have to orbit around its black hole in order to cause the time dilation presented in the story, and is obviously impossible. Even if you had a planet rotating that quickly, in order to land on it, you would have to match its orbit, i.e. you would at least have to accelerate your spaceship to the same speed.
The movie got a lot of things right, and even caused some scientific papers to be written, but that bit about time dilation, while necessary for the story, was the worst kind of nonsense: Scientifically sound, but the numbers were way, way off.
Comment: I thought it was the closeness to the black hole that caused the time dilation, not the gravity of the planet itself...
Comment: This answer completely misrepresents how gravitational time dilation works. Gravitational time dilation is due to gravitational *potential*; it doesn't matter whether you "experience" the gravity as a normal force from whatever surface you're standing on, or if you're in orbit, or even if you're standing between two masses and their pull cancels out.
Comment: I believe the astronauts *are* affected by the black hole's gravity, just as much as the planet is. The only reason the "dominant gravitational force" they experience is the planet's is because their frame of reference (the planet) is accelerated by the same gravitational force as the astronauts on it. So yes, the black hole's gravity should cause time dilation relative to an observer that isn't as close to the black hole as the planet. How much farther away the Endurance was and how much less affected it would be, I can't say though.
Comment: @Selenog My whole point is that the astronauts don't experience the gravity of the black hole and are not affected by it. If they did, time dilation would be the least of their worries. The dominant gravitational force, as depicted in the movie, came from the planet and was 1.3g, which is not nearly enough to cause that much time dilation. And yes, the spacecraft should have experienced pretty much the same time dilation, because it was at most a few hundred kilometers away from the astronauts. Another reason why the science failed on that part.
Comment: @Hackworth The example with the sun doesn't matter. This central body's force on the planet's surface can easily be larger than the planet's own force. The sun's gravitational acceleration on the Earth's surface could be 10g and you still wouldn't feel it or fall off the planet as long as you're orbiting together with the Earth. As long as the gravitational force form the SMBH doesn't *vary* significantly on the length scale of the planet (which would tear it apart), it can be arbitrarily strong without affecting anything but the orbital speed and of course the time dilation.
Comment: As I wrote, it can be an indirect consequence of being near a black hole, if the orbital velocity is relativistic. However, in the movie the dominant force the astronauts experienced were the 1.3g of gravity from the planet. If they had experienced enough gravity to cause that much time dilation, they would have been flattened like a pancake, and if they were moving fast enough, they would be torn apart from tidal forces.
Comment: @Hackworth The astronauts experience the gravitational pull of the black hole  in exactly the same way the planet does, which is why they are even able to stay near each other with only 1.3g of acceleration while orbiting at around (from film) 55% the speed of light.
Comment: @MartinEnder The astronauts and the planet are only affected insofar as they are held in an orbit around the BH. The planet's gravity is still the dominant force, because they are obviously standing on it as opposed to falling into the BH. The Earth is also affected by the Sun's gravity, but that doesn't mean we experience time dilation due to the Sun's gravity. The *net* gravitational force is what matters. If a spacecraft were to sail directly between 2 SMBHs and experience no net force, it would also not experience time dilation, even though it is under the influence of 2 SMBHs.
Comment: It might also help to note that time dilation does not care where the force comes from. For example, a black hole, a neutron star, and an ordinary star of the same mass at the same distance all produce the same gravitational force and, if an observer were to directly experience their gravitational pull, would measure the same amount of time dilation.
Comment: @Selenog Exactly. And in fact using various wikipedia formulas with 100 mil solar masses for gargantua, we get this https://i.imgur.com/xLLHqXV.png which shows that given a possibly unstable circular orbit and an orbital speed of 50-60% c we get the necessary time dilation when combined with the gravitational effects. This answer really shows a misunderstanding of how gravitational time dilation works.
Comment: As an astrophysicist I agree that this answer displays several gross misunderstandings about the concepts of general relativity (GR).  Time dilation is due to the gravitational potential, not "gravitational acceleration" which makes no sense in GR since gravity is not considered a force, but rather space-time curvature.  The planet was deep within a gravitational well, so to an outside observer, clocks on the planet would run more slowly.  I urge viewers to downvote this scientifically incorrect answer as it spreads misinformation.
Comment: There is also the gravity of the black hole itself to explain the time-dilation between Earth and the Scientists and it is the most likely cause for the time-dilation. However what I never quite got is how the space-craft Endurance did not experiance the same levels of time-dilation given they must have been close to the planet (likely orbiting it).
Comment: The movie got a lot of things right, but I just can't look past the crop duster looking at a screen and declaring he'll do a slingshot by a convienietly placed neutron star to get up to speed to go into orbit around the near speed of light planet...
Answer: Orbiting a black hole and you don't want extreme tidal forces? This depends on several things, including:

The mass of the black hole; the Interstellar one looks like it's in the largest category, a supermassive black hole, because you can actually see the the black hole is much larger than a planet in front of it. This puts in anywhere from $10^{5}$ to $10^{10}$ solar masses. (If it was about the same size as the planet, it would still weigh 1000 ($10^3$) times as heavy as our sun!)
The mass of the planet in question is pretty unknown, but I'm assuming earth-like mass. ($~5.7 * 10^{24} kg$)

Now, let's talk about tidal forces. Wikipedia goes through some derivations of the tidal force, but eventually ends up with
$$\vec{a}_g \approx \hat{r}*2 \Delta r* G*\frac{M_{bh}}{R_{bh}^3} $$
with:

$a$ being the acceleration of the tidal force
$G$ being the universal gravitational constant
$\Delta r$ being the distance from the planet's center to the the surface
$M_{bh}$ is the mass of the black hole
$R_{bh}$ is the orbital radius of the planet to the black hole
$\hat{r}$ just lets us know the force is in the direction of the black hole.

but wait, this gets easier! We only want the orbital radius, and Earth-like tidal forces on an Earth-sized planet. That means we want to solve the following for $R_{bh}$
$$\frac{M_{bh}}{R_{bh}^3} = \frac{M_{sun}}{R_{earth}^3}$$
Just looking at this, I can tell you that $R_{bh}^3$ needs to scale with the size of your black hole, producing an extra $10^5$ times $R_{earth}^3$. $\sqrt[3]{10^5} \approx 46$, so the planet is about 46 AU out from the middle of the black hole.
Comment: Thanks mate! That's really helpful. How would those numbers change for a planet slightly smaller than Earth orbiting a stellar mass (315 sol) black hole? Or rather, a planet slightly smaller than Earth orbiting a star that orbits a stellar mass black hole? :)
Comment: @Z.Schroeder Yep, the numbers would change; the equations hold for all two-body systems. You could assume a Star+Small black hole acts like a really massive thing at the center of mass for the star+black hole system, reducing that three body problem to a two body. This only works if the sun/black hole are really far away.
Comment: @Z.Schroeder, if I've done the math correctly, a 315 sol black hole would have Earth-like tidal forces at an orbital radius of 6.8 AU.
Answer: Orbiting a black hole is no more or less dangerous than orbiting a small sun: You get too close, bad things happen. If your orbit decays without correction, you are going to be obliterated.
As long as you properly insert yourself into an orbit at a safe distance at the appropriate velocity, there is no particular danger from a black hole.
And leaving is no special challenge either. By being in orbit you are already at escape velocity, so you don't need any particularly special 'oomph' to get back home. [EDIT: Sorry, I was wrong here. Your speed is about 0.7 of what you would need for escape, but it's in the ball-park!]
Now a planet close to a black hole might be a more complex story.
But don't believe the hype of 80's SciFi - You won't inevitably have to hold the rails of your starship as your head gets stretched. Unless of course, you get too close...
There was an episode of Battlestar Galactica where the Cylons were orbiting a black hole and my brief reaction was "How ridiculous!", but then I thought about it, and concluded "Why not!?!".
I will trust @PipperChip on the math!
Comment: "By being in orbit you are already at escape velocity" - no you're not. If you were at escape velocity, you would be on an escape trajectory, not orbit.
Comment: @user2357112 - I am no orbital mechanics expert, but I can read Wikipedia: "this object's speed at any point in the orbit will be equal to the escape velocity at that point". There is a difference between "at escape velocity" and "above escape velocity". In stable orbit, you are precisely "at". In mathematical terms - 0.001 joule less energy and you are decaying, 0.001 joule more energy and you are escaping (I believe...) That is what is special about a stable orbit - You are sitting on the knife-edge of escape velocity. Neither decaying, nor escaping.
Comment: No, you've completely misread that section of the article. [That section](https://en.wikipedia.org/wiki/Escape_velocity#Orbit) describes a particular type of escape trajectory sometimes referred to as an escape orbit, which occurs in an ideal 2-body system when there is exactly enough energy in the system for escape. With less than escape velocity, you are not decaying; orbital mechanics do not work that way. Less than escape velocity produces an elliptical or circular orbit (or a crash if your path goes through the central body).
Comment: (I don't know whether using the term "escape orbit" to refer to such parabolic trajectories is actually common; I could only find [one source](https://books.google.com/books?id=O_7TCgAAQBAJ&pg=PA83&lpg=PA81&dq=%22escape+orbit%22#v=onepage&q=%22escape%20orbit%22&f=false) using such terminology, which the Wikipedia article seems to have plagiarized.)
Comment: OK, I concede. But is it true that in a stable circular orbit, your "speed" is equivalent to the "speed of escape velocity at that same altitude" - Just in the wrong direction? (90 degrees off) That's what I read from that Wikipedia section. My main point still stands though - There is nothing magical about orbiting a mass after it tips over to being a black hole.
Comment: No, in a stable circular orbit, your speed is roughly 1/sqrt(2) times the speed necessary for escape. [Escape velocity is really a speed; it matters very little which direction you're pointing, as long as you're not pointing right at the thing you want to escape.](http://physics.stackexchange.com/questions/193497/is-escape-velocity-really-a-velocity-rather-than-a-speed)
Answer: Yes, orbiting a two solar-mass black home should be no different to orbiting a two solar-mass star. Stay at a sensible planetary distance and you will be OK ...
... unless or until something falls into the back hole. The process of its being chewed up  and eaten will convert a good fraction of its mass into energy, much of it gamma rays. Most of that energy will be tightly beamed more or less perpendicular to the plane of the accretion disk into which that object will be transformed. So an orbit at right angles to the disk would be especially unhealthy.
If the black hole has "hoovered" its stellar neighbourhood many aeons ago and there's no active accretion disk when you arrive, you are probably safe.
Edit ... should have said, something large falls into it. The sun converts 5 million tons of mass into energy per second(!), most of which comes out as heat and light rather than gamma rays. But at a planetary distance I don't think a few tons of rock being converted into gamma rays will be lethal.

Safe distance from a black hole?


Timing of Birchas Illonos

Question: From when until when are you allowed to make the bracha,and what is optimum?
Comment: Here's a list of trees (probably not exhaustive) in local communities where one can recite the Berachah.  According to the creator of the list, anytime in Nisan seems to be acceptable (see the Hebrew comments on the right).  http://www.misaskim.org/bracha.php?page=1
Answer: The bracha can be made from the time the tree starts to bloom, regardless of the month (Mishna Berura quoting "Acharonim").  Lechatechita it should be said before the fruit grows.
The Mishna Berurah sides that, bediavad, if you did not see the blossoming and the fruit started to grow, you can rely on those who say you can make a bracha until the fruit is fully complete (minimally edible?).  It sounds like if you saw the blossom and passed up the opportunity, you should not rely on this opinion.  The Aruch haShulchan does not side with that qualification.
Answer: A lot more sources posted at Dose of Halacha

According to some poskim (Halachos Ketanos 2:28; Kaf Hachaim OC 226:1) the beracha should only be recited in the month of Nissan, though most poskim (Ritva, Rosh Hashana 11a; Mishna Berura 226:1; Aruch Hashulchan OC 226:1; Tzitz Eliezer 12:20:3) write that one should say it whenever when one first sees the blossom. R' Ephraim Greenblatt (Rivevos Ephraim 5:173) writes that he witnessed R' Moshe Feinstein say the beracha in Iyar.
Thus, R' Ovadia Yosef (Yechave Daas 1:1; Yabia Omer OC 8:8:4) writes that one should ideally wait until Nissan to recite the beracha, though one can still do so afterwards. R' Yitzchok Yaakov Weiss (Minchas Yitzchak 10:16) writes that where trees blossom at different times, such as in Australia, one recites the beracha then (See Har Tzvi OC 1:118; Teshuvos Vehanhagos 1:190).
While the Kaf Hachaim (OC 226:9) writes that one can only recite the beracha upon first seeing the blossom, most poskim (Machatzis Hashekel 226; Shaar Hatziun 226:3; Halichos Shlomo 1:2:5) write that one can recite it later if necessary.

Timing of Birchas Illonos


Who is Morris Ayin?

Question: I frequently hear that certain things should not be done because of Morris Ayin. People of all backgrounds seem to hold by him, yet I have been unable to find any sforim by Rav Morris. Who is he?

This question is Purim Torah and is not intended to be taken completely seriously. See the Purim Torah policy.
Comment: The name is not Morris Ayin but Maurice Ayin. Maybe that might help in your quest to find his seforim.
Answer: According to Rav Yehuda Amar rav (Shabbat 64b)  he is "Big Brother" because he is present even bechadre Chadarim.
Answer: Note that many things are forbidden because of Morris Ayin. Thus, we see that he is regarded as a min (kind) of heretic who is attempting to get (other) people to think that good people are doing the wrong thing. This means that we do not want to read what he says about us and avoid him.
Answer: No, the halachic concern is that although you might see nobody eyeing you, remember the possibility of Morris eyeing.
Answer: You've been hearing it incorrectly. The name is Moreh Zayin - the teacher of armaments. This is a metaphor for all those teachers who tell you not to do things because someone else may misunderstand you. These teachers stand with "virtual weapons" and give you a guilty look, so that you don't do what's forbidden.
Moreh Zayin come in different forms and they aren't necessarily school teachers. A parent who gives you a guilty look is also a Moreh zayin.

Who is Morris Ayin?


USA tourist visa application for a student

Question: I am still a student but wanted to apply for a tourist visa, because my best friend wants me to be at her birthday. My childhood best friend's mom would be taking care of all the expenses; since I don't have any proof of income tax since I'm still a student, they will be the ones to cover my expenses. Am I allowed? They are now US citizens.
Comment: If you are not from a country that's in the Visa Waiver Programme, before applying for a tourist visa make sure you read and understand all the eligibility and document requirements. Your residence and ties to your home country will be an important factor in the decision https://travel.state.gov/content/travel/en/us-visas/tourism-visit/visitor.html
Answer: Yes, anyone can apply for a visa. First, check whether you need one at all: if you're from a country that's in the Visa Waiver Program, you don't need a visa for short visits. If you do need a visa, you can find the application procedure from the website of the US embassy in your country. Make sure you fill in all the forms accurately and truthfully and follow all the instructions carefully and accurately.
Answer: I have a Tunisian citizenship.
In 2013 I had worker visa and I travelled to canada to work.
I had a permanent job (CAD $65K salary per year)
I had an apartment.
I filled all the required forms truthfully and accurately.
I applied in the US embassy in Ottawa (Canada) : Yet I got refused the visa : It depends on the agent that you will give your documents too. Depends on his mood too, the dude didn't even bother to check my work contract, pay-stubs or any other proof : DONT ARGUE, ANSWER THE QUESTIONS DONT ELABORATE
And depends on the country of birth too.
Got refused the visa for : not enough economic and family ties with Canada.
I applied again 1 month later ... got my Visa.
And I'm talking here about my personal experience.
Ps : Once you get refused the visa, and even if you get if afterwards , every time you try to enter the USA you will be redirected to the secondary check.
Having the VISA in your passport doesn't guarantee the entry to the country : if you have the visa ... make sure always to have proof of residency in your origin country and self sufficient financially and your phone fully charged.


How to judge accuracy and precision of measurement devices?

Question: I am unsure about the accuracy and/or precision of measuring instruments.
For example, if we have a graduated cylinder, a pipette and a beaker, how and why can we assert them that they can be used with accuracy, precision, or both?
Answer: Beakers, Erlenmeyer flasks, dropping funnels and the like are not volumetric instruments. They are not precisely calibrated, and their scales serve only as approximate guides. Therefore, the beaker that is mentioned in the question is right out.
Graduated cylinders, volumetric flasks, bulb pipettes, graduated pipettes, and burettes are volumetric instruments. They are calibrated; however, like any other measuring instrument, they only have a limited accuracy and precision.
The accuracy and precision do not only depend on the type of the instrument (e.g. graduated cylinder or pipette). They also depend on the dimensions, graduating divisions, and accuracy class of the instrument, which are usually internationally standardized.
In principle, volumetric instruments with a small diameter have a better precision because the same difference of the liquid level (i.e. the meniscus) corresponds to a smaller difference in volume. Therefore, typical long and thin pipettes are usually better than short and thick graduated cylinders. For example:

A standardized graduated pipette (Class A or AS) with a nominal capacity of 10 ml and an overall length of about 360 mm has a maximum permissible error of ±0.05 ml.
A standardized graduated measuring cylinder (Class A) with a nominal capacity of 10 ml and a maximum overall height of 140 mm (tall form) has a maximum permissible error of ±0.1 ml.
If you use an even larger graduated measuring cylinder (Class A) with a nominal capacity of 100 ml and fill it just to the 10 ml mark, the maximum permissible error is ±0.5 ml.

These errors represent the maximum permissible error at any point on the scale, and also the maximum permissible difference between the errors at any two points.
Since the same sources of error are, naturally, inherent both in calibration and use, a volumetric instrument should have a good precision in order to be able to be calibrated with a good accuracy.
However, note that graduated pipettes are usually calibrated 'to deliver' (TD, Ex), whereas graduated measuring cylinders are calibrated 'to contain' (TC, In). Hence, graduatted pipettes and graduated measuring cylinders may not be interchangeable at will.
Answer: There are too many measurement devices to give an answer for all types. However, assuming the scope of your question is limited to measuring fluid volume the way this can be done in a professional laboratory is simple.
You will need:
(1) high purity water
(2) a high precision balance which has been recently calibrated
(3) a type of container called a weigh boat

Tare the weigh boat. This means place the weigh boat on the balance
and (if the balance is electronic) press the tare button.  This will
cause the balance to read zero.  (With the old pivot beam style
balance another identical weigh boat could have been placed on the
other end of the balance to level it.  These pivot beam balances are
rarely used in modern lab practice as most balances are electronic.)
Using the measuring tool in question, measure a certain volume of
high purity water.  Pour the measured quantity of high purity water
into the weigh boat on the electronic balance.  The balance should
read the mass of the water.  Assuming the temperature of the lab is
close to 20°C divide the mass of the water by 0.9982 g/ml.  This is
an accurate way to tell how many ml were actually measured using the
tool.
Repeat the measurement three times.  Three true measurements can be used to determine an average and standard deviation.  The difference between what was measured and the average of the three true measurements is related to the accuracy of the measurement.  The standard deviation is related to the precision of the measurement.

If a pipet is supposed to dispense 10 mL and it actually dispenses 15, 16, and 15 mL then it has high precision (because the numbers are close.)  Its accuracy is way off because the average of 15.3 mL is half of an order of magnitude different than 10.  It could be used, though it would need to be either calibrated so the measurement is correct or the end numbers adjusted for the inaccuracy.
If a pipet is supposed to dispense 10 mL and it actually dispenses 5, 9, and 16 mL then despite being accurate, its precision is off.  Because the precision is off, this pipet is too unreliable to be used in scientific practice.
Realistically, if a pipet were supposed to dispense 10  mL and it dispenses 10.0010, 10.0000, and 9.9998 mL then it would be suitable for lab work.  Because both the accuracy and precision is high, this measurement is said to have high trueness.  Trueness -- which is both accuracy and precision -- is desirable in a good measurement.
It is always good practice to verify the calibration of any measurement before using the measurement to generate publishable data.  Recently published data in high-quality publications is a good source of information to judge if a certain set of data has acceptable trueness.
Answer: The term precision means two different things. The number 1.23456789 has more "precision" than the number 1.23. That is, there is a mathematical (and computer science) meaning that is different than the one used by physical scientists.
In physical science, precision means the property of reproducibility of a value or result. If I am precise, I get the same answer every time (or nearly so). If I am accurate, I get the correct answer. Here is one example: I have a machine which counts people going through the door. Whenever someone walks through, it adds 1.1 to its total. This machine is precise but not very accurate. I modify it and now it adds between 0.95 and 1.05 to the count each time someone walks thru. It is definitely not more precise, but may or may not be more accurate.


Is it possible to detect DPI (Deep Packet Inspection)?

Question: If one suspects one's packets are subject to DPI, how can it be detected?
Comment: I am wondering if this can be detected by roundtrip timing. Even a passive device that inspects before forwarding will add to the latency.
Answer: It depends on the kind of DPI you are talking about. If it is purely passive DPI (i.e. only looking at the packets) then you will not be able to detect it. If this is instead DPI which can modify the data then you might be able to detect it depending on the kind of access and knowledge you have.
Some typical examples on how active DPI devices change traffic, i.e. what you can look for:

If you transfer specific data (like known malware) the transfer will fail because the IDS detected and blocked the possible attack. One could use for example the innocent EICAR test virus.
DPI using SSL interception can be detected because the certificate is issued by a different CA.
Some DPI devices strip STARTTLS from SMTP traffic so that the traffic will not be encrypted and can be analyzed. Such change to the traffic can be observed on the client side.
Some devices modify the "Accept-Encoding" header of a HTTP request to strip unsupported compression schemes. This can be observed on the server side or sometimes on the client side too because the content is not transferred compressed.


Does a two-way fixed-length hash function exist?

Question: Just wondering if a two-way fixed-length hash function exists. I'm not talking about encryption where your output is the same size if not larger than your input. I guess the best way I can put it is "like SHA or MD5, but also reversible." (I know SHA and MD5 aren't reversible by design, but you get the point).
Comment: Does it need to be reversible for everyone, or reversible in secret by few/one person? i.e. Are you looking for [trapdoor one way functions](https://en.wikipedia.org/wiki/Trapdoor_function)?
Comment: Well, it depends on the given secret. I'm working on a project where everything **needs** to be small, so say I want to turn a 512-character input into a 32-character hash. If I also needed to give 256 characters to reverse it, it still fits along the lines of my project's purpose, but not to the level I'd like.
Comment: I am confused. What security objectives would such a function have and what of these functionalities is not fullfilled by taking a block cipher and fixing the key (or some other permutation)?
Answer: What I understood from your question is that you want to map your data uniquely such that the output has less bits relative to the input. You can use data compression techniques to compress your data. Then, if you need the security, use cryptography. Although hash functions are able to compress your data, but they can't be reversed by everyone. This goal is opposite of their nature and their duties.
Answer: No. This is answered by the
pigeonhole principal.
It is fundamentally hard to do.
The problem is, if you reduce the input into smaller numbers, it could be reversed back either into the message you want, or any other weird and meaningless number.
2274866217183956372711949572
Could reduce into 3726 theoretically, but from there it could be turned back into the above number, or a totally different one.
It's like burning up a log.
You're gonna get ash.
Theoretically, if you grab all the molecules and painstakingly reverse them, you would get the wood back.
But not without knowledge of what the wood looked like.
You could take all those molecules and reassemble them into something totally different, and that would also be possible.
So, in the end, this question's answer is yes, but only if you know a thing about the original message.


How does calcium carbide artificially ripen fruits?

Question: Lately, use of chemicals to artificially ripen fruits is in full swing. Firstly, how do they do it? Next, Are there any ways to distinguish between those that are naturally ripened and those that are artificially ripened?
Comment: If it's cheap/not pricey and available off-season, then it's likely to be artificially ripened (<----common sense) ;P
Comment: Note that there is significant list of fruits, including apples and most berries that do not ripen at all after being picked
Comment: @paracetamol Then does famous soft drinks(like maaza )use artificially ripened fruits?
Comment: Chitra, I don't think they would tell me that  ;)
Comment: But can anybody make a guess.
Answer: Technically, there is no difference between naturally ripened fruit and artificially ripened fruit... they're both ripened fruit. I suppose you were looking for some sort of simple chemical test to distinguish between them; but like I said, there's no point in it since in both cases you end up with ripe fruit.
Fruits normally emit ethene (ethylene) which is essential to the ripening process. Which is why fruits picked before they're ripe are sometimes wrapped in a bag that seals in whatever little ethene they emit, thereby accelerating the ripening process.
As you would know, calcium carbide does not ripen fruit per se. Rather, it is the reaction of calcium carbide with water/moisture to release ethyne (acetylene) that's important here. Though I do not know of their mode(s) of action, apparently ethyne is almost a effective as ethylene in ripening fruits (I still need to find a reliable reference for this; though it is supposed to be a fairly well known fact).
If you let a fruit ripen naturally, or if you use ethyne instead... there's really no chemical difference at the end (apart from the usual arsenic contaminants that's associated with commercial-grade calcium-carbide).
The actual mechanism by which ethene (ethylene) and ethyne (acetylene) initiate, and thereafter, accelerate fruit-ripening is not really understood. Pectinases play a major role in the ripening process (they break down a structural component of plant cell walls called "pectin", which is why ripe fruit tend to be soft), however the scientific community has not yet arrived at a consensus as to how ethene and ethyne are linked to the activation/production of these enzymes.

However, there are a couple of subtle (non-chemical) difference between naturally and artificially ripened fruit:
Fruit that has ripened prior to plucking tend to be "fuller" than artificially ripened ones. This is because the fruit still gets an influx of nutrients while it's still attached to the tree...in other words; in the time it takes to ripen, the fruit grows as well. Artificially ripened fruit however, tend to be smaller, as they pretty much stop growing as they ripen (off the tree). ["Small", "big", "fuller" are crudely defined terms...you need some "experience" with fruits if you want to put this tip to good use].
Additionally, if you find the fruit being sold off-season (i.e- At a time of the year when it doesn't usually grow/ripen),  for a normal or somewhat cheaper price...then in all likelihood you have an artificially ripened fruit in your hand.

I reiterate:
It doesn't matter how it was ripened. It's ripe, that's all there is to it (from a consumer's PoV). Don't forget to wash and peel it before taking a bite :-)
Comment: @David But there's no significant change in the *fruit* itself. The arsenic contaminants stick to the surface...hence the "wash and peel" tip at the end of the post ;)
Comment: @Chit Commercial grade calcium-carbide usually contains arsenic impurities : https://www.researchgate.net/publication/268226372_Quantification_of_arsenic_and_phosphorus_in_calcium_carbide_treated_mangoes
Comment: @ChitraLekha Calcium carbide is made by heating coke and lime in an electric arc furnace. Presumably the raw materials contain traces of arsenic -- nothing is ever pure. (I'd just drink the coke and lime but...)
Comment: "there's really no chemical difference at the end (apart from the usual arsenic contamina[tion]...)" That seems like... quite a significant difference.
Comment: @DavidRicherby how does arsenic contamination occur here? since we aren't using the element anywhere?
Answer: Ethylene (C2H4) triggers enzymes including kinase (neutralizes acid), amylase (converts starch to sugar), hydrolase (degrades chlorophyll) and others to start ripening fruit. Calcium carbide, when in contact with moisture (such as on the surface of the fruit) produces acetylene (C2H2) which is quite similar to ethylene in starting ripening. For more information, visit Wikipedia.
Answer: One of the most simple tests include putting the mangoes in a bucket of water . The naturally ripened mangoes will sink while those that are artificially ripened would float. (because the density of artificially ripened would be lower than those that are naturally ripened.
And the color of the mangoes also indicate their ripening process . In some places there would be patches of green.
And I have also experienced burning taste while eating
And after we cut these mangoes certain places would be very ripe and other areas would still be hard(somewhat).
For more information click link https://www.youtube.com/watch?v=cwk-rtHPdes&t=42s
Comment: Is this specifically for mangoes? (Your question didn't specify "mango".)
Comment: yes this is for mangoes. And the color test may apply to bananas (where there would be PATCHES of green)


Reducing the bandwidth of non-symmetric matrix

Question: Is there an efficient algorithm to reduce the bandwidth of a directed graph's adjacency matrix? Something like the reverse Cuthill-McKee, but for non-symmetric matrices.
Answer: The following article discusses various approaches to reducing the bandwidth of unsymmetric matrices.
J.K. Reid, J. A. Scott: Reducing the total bandwidth of a sparse unsymmetric matrix, SIAM Journal on Matrix Analysis and Applications 28(3):805–821.
The technical report version of the article is available here:
J. K. Reid and J. A. Scott, Reducing the total bandwidth of a sparse unsymmetric matrix, Technical Report RAL-TR-2005-001, STFC Rutherford Appleton Laboratory. http://www.numerical.rl.ac.uk/reports/rsRAL2005001.pdf
Comment: Whoops, thanks for the edit! And for the answer, that seems to be exactly what I'm looking for :)
Comment: You're welcome! I'm glad that you found the answer useful.


Why do ethanethiol molecules have stronger London dispersion forces than those of ethanol?

Question: I know that, molecules with higher molecular weight tend to form London forces. Is there any other reason which makes Ethanethiol to have strong London dispersion forces as compared to Ethanol?
Answer: Oooh, interesting question. Let me have a go at this and we'll see if there's someone else who can put it in better terms.
So, the way I tried to reason this out is to remind myself that every single bond, in each of these two molecules, is the same except the ones where C is bonded to O or S and the ones where H is bonded to O or S. We then pay attention to these bonds, particularly.
Now, at this point, we also have to remind ourselves that O is more electronegative than S and therefore, pulls electrons towards itself much more than S does and both of these atoms are a tad bit more electronegative than C or H. Therefore, the region of higher electron density in ethanol is O and the region of higher electron density in ethanethiol is S. We also have to remember that O has a more negative partial charge than S does. We also need to remind ourselves that S is larger than O which, by the way, is also a contributing factor as to why O is more electronegative than S.
I then uses these facts and try to make sure my reasoning is sound; If we look at the way london dispersion forces work, we realize that it is easier for electrons to fluctuate between molecules when they are within the bonds than when they are within the atoms constituting the molecules. In other words, when the electron density is concentrated in the bond, as is the case when there is 0 electronegativity difference, then london dispersion forces are stronger. (Electron fluctuation, by the way, is the basis for dispersion forces)
Using the reasoning above, it becomes natural to think that the dispersion forces in ethanthiol are stronger, since the electron density is more concentrated in the bonds than it is when you consider ethanol. Therefore, it is easier for the electrons to fluctuate between two ethanethiol molecules than between two ethanol molecules, which is why ethanethiol molecules have stronger dispersion forces between themselves as opposed to the ethanol molecules.
Note: However, ethanol molecules have a much more powerful intermolecular force available to them, which is hydrogen bonding. So, clearly, ethanol molecules have a higher boiling point or whatnot than ethanethiol. Just to make sure that this is clear enough.


Ford Ranger clutch pedal won't depress

Question: I just had the clutch replaced 2 months ago on my 1995 Ford Ranger.  I went to the store yesterday and when I went to leave, the clutch pedal wouldn't depress.
Answer: Check the clutch line at the quick disconnect (source).  If it has come loose that would explain not being able to depress the pedal.

Ford Ranger clutch pedal won't depress


Clicking NXT Brick

Question: Help! I have recently installed leJOS NXJ on to my NXT brick, and soon after my batteries died. I inserted new ones, and now I cant start my brick up. When I press the startup button(orange) it makes a clicking sound and when I let go it stops. I have tried reflashing the brick with both leJOS NXJ and the NXT software and both programs say something along the lines of "unable to locate brick." Any suggetions?
Comment: Possible suggestion: do a hard reset (using the reset button in the Technic pin hole on the bottom of the unit), *then* attempt to re-upload firmware from the NXT software.
Answer: The clicking indicates the brick is waiting for firmware download.
Double check device manager and it should also be listed there as waiting firmware.
Use either the LEGO software or another utility (eg RobotC, LeJOS) to download the firmware of your choice.
Big if ->  If the device does not show up as Nxt Waiting firmware in device manager, there may be some conflict with USB drivers often installed with Arduino.  It is a mess at the moment and there are no simple solutions. Keep Uninstalling (Arduino first), rebooting, re-installing (phantom drivers) until the NXT shows up as waiting firmware.
Comment: I blogged about the driver conflict. There are some screenshots there that might help; http://www.spiked3.com/?p=2911

Clicking NXT Brick


G - forces imparted by launchers

Question: How can one know the G forces imparted from the technical data presented on the websites of launch vehicles? I am interested in finding a way to theorise the max G-forces imparted by researching mission profiles and engine data, or anything else that may be public.
I have tried going through some manuals but that did not help me.
Comment: Look up TWR, then subtract 1 for G-force added by the launcher. Or don't subtract, for the G-force experienced by the astronaut (acceleration + Earth gravity). If you don't have TWR, you have thrust and mass, it's the simple (thrust to weight) ratio. Obviously both change over time, so you can't just happily insert launchpad data; you'll need tables/graphs over time.
Comment: @SF. TWR (thrust to weight ratio) is unitless so subtracting 9.81 m/s^2 doesn't work. Subtracting 1 g from *thrust* doesn't work either. You'd need to know the ratio of maximum thrust to mass, but mass is constantly changing. And for part of the launch there is quite a significant *drag* force that you'd have to subtract from thrust first.
Comment: @uhoh Correct, but we should note that thrust to *weight* is nonsense, and thrust-to-mass is not unitless. But as you note, both mass and thrust vary over time; it's not a fixed ratio. Even g varies a bit with altitude if you want to be precise.

You can get away with a fixed TWR for terrestrial aircraft if you ignore cargo and fuel, but commercial pilots have to include those in their takeoff and landing calculations as well.
Answer: This is easy to calculate using our old friend $F = ma$ and knowing the thrust and mass at any given point in the launch:

Take the combined thrust of all the engines $F$ in Newtons, remembering that some of them may be throttled down.
Take the mass of the entire vehicle $m$ in kg.
Use $a = \frac{F}{m}$ to find the acceleration in ms$^{-2}$.
Divide by $g \approx 9.81ms^{-2}$ to get the scalar gees.

For example, let's take rough values of the Saturn V just after liftoff:
$$gees \approx \frac{5 \times 7000000N}{3000000kg \times 9.81ms^{-2}}g \approx 1.2g$$
For a later point in the flight, you just need to know how much fuel has been spent and whether the thrust has changed. If no engines have been discarded or throttled, the acceleration will increase as fuel is burnt. Also note that atmospheric drag will influence the acceleration felt by the vehicle.
Alternatively, you can often find graphs in the vehicle documentation which give acceleration profiles of the launches. For example, this one from the Apollo 11 launch vehicle evaluation:
Comment: Thank you for helping with the valuable input. There are a few manuals which show the mission profile but no acceleration chart, which makes it hard to figure out a max G at any point. The mass profile is also not so obvious.
Comment: `+1` for science and math!
Answer: The best way to know is to check the payload planning guide for a particular launcher. Many modern launchers throttle down over the course of a flight to limit g-loading, so a simple rated-thrust divided by mass-at-burnout calculation will not give the correct answer.
For example, the 2010 version of the Atlas V Launch Services Guide says (of the 400-series configurations):

Near the end of the booster phase, the RD-180 engine is continuously throttled so that axial acceleration levels are not exceeded. These g-levels may be a function of payload weight and do not exceed 5.0 g steady state.
Comment: That seems improbable, as forces the payload is subjected to is one of the most fundamental and important things a payload operator needs to know. It may be described as "acceleration limits" rather than g-force, but it should be in there; do you have an example of a payload guide which doesn't include that information?
Comment: Thank you for your Answer, but there are payload guides which dont talk about G - forces experienced. I was looking to see if there was any way to figure it our using engine data and mission profile
Comment: I found a document for the GSLV, which isn't a payload guide, but I'm looking for the G-forces for the GSLV Mk-II . I have no way to find the G-forces imparted by it. CHeck here https://www.isro.gov.in/sites/default/files/pdf/gslv-brochures/gslv-d5.pdf
Comment: GSLV II has modest peak accelerations, because it retains the mass of the solid-fuel core stage (which burns out first) through the entire burn of the liquid fueled boosters. I believe the Vikas engines do *not* throttle down, so @Jack's method, taking the stage thrust and dividing by the remaining mass of the rocket in a given situation, is valid. With a 2.5 ton payload, I get a peak of about 5.3g at core burnout, 2.7g at booster burnout, 3.7g at second-stage burnout. Computing core burnout is trickiest; you have to pro-rate the booster fuel remaining based on the stage burn times.
Comment: @RajathPai I recommend you choose a specific example where applying answers here don't seem to work, and ask that as a *new question*. Something like "I've applied answers to my previous [question](https://space.stackexchange.com/q/30091/12102) to Rocket X and still can't figure out what the maximum acceleration loading on the payload could be."

G - forces imparted by launchers


Why was it so important to have a victor in the Hunger Games?

Question: In the first Hunger Games book/movie, why was President Snow so insistent on letting both Katniss and Peta live in order for there to be a victor? This seems like a major plot hole to me, as the entire entry to the series depicts how little the Capitol values a district person's life, e.g.: the poor living conditions, the scarce resources available to them, the annual tribute of a teenager for showcase in a public death match. I don't really see it plausible that allowing them both to die would have caused a revolution against the Capitol, and I would think killing both remaining participants would be a bigger display of power than allowing one to live. District citizens are used to death, they would probably see it as just another injustice but carry on with their lives.
Comment: Disagree with one of the points in the question. Snow did not want two victors. In fact he saw to it that Crane was killed for his failure to deliver one victor as normal...
Comment: I read that and I don't think so. It addresses what happens if there are no victors, but not why they need a victor. The top answer states they need their victor because the Victory Tour promotes hopelessness, but I really don't see it.
Comment: @The Giant Of Lannister. I didn't say he wanted two victors. He was forced to have two victors because Katniss and PETA threatened to leave them with no victors otherwise. He made it perfectly clear he needed one victor over none
Comment: I still disagree. Your sentence is worded awkwardly. Makes it sound like Snow personally intervened to save Peeta and Katniss. If so, that's interesting, but is it cannon? I don't think it is in the books or film.
Comment: For instance Snow was quoted as saying Crane should have "blown both Peeta and Katniss to pieces" (sic, probably not those exact words) the moment they went to eat the berries. He wanted them both dead. You however are positing that he didn't.
Comment: To keep their hope alive...
Comment: Because there was a always a single victor in Battle Royale, a little girl even, and Hunger Games is a complete and total rip off of Battle Royale. Simple enough.
Answer: For the Districts
The Hunger Games were started in response to the Districts rebelling. The intention was to have a permanent reminder of what happened and, more importantly, what happens when the Districts lose. If that's the point of the Hunger Games, why do they need a Victor? Isn't it enough just to have them all kill each other?
I don't think so. The Victor is a constant reminder that no matter what you manage to do or to achieve, the Capitol controls you. The Capitol is effectively saying "We can take your children and we can make them kill each other. Then even if they win we still own them. We can drag them around Panem for the Victory Tour. We can parade them in front of the crowds in all the subsequent Hunger Games. We can, whenever we like, force them to relive the single worst experience of their life. Even the strongest and best amongst you are powerless to stand up to the Capitol."
For the Capitol
For the citizens of the Capitol the Hunger Games are an incredibly big deal. It's the main event of the year and they go absolutely nuts for it. They have their favourite Tributes. They spend money to help them win and they bet on the outcome. Tributes who win are viewed as major celebrities within the Capitol.
The Capitol's citizens invest both financially and emotionally in the outcome of the Games, and while they might not be prepared to riot if there was no Victor, they'd definitely be disappointed. It would be like the Super Bowl ending in a tie - a massive anti-climax.
For Panem
Panem was fragile. President Snow admitted as much himself to Katniss. Her relatively simple act of youthful defiance was enough to light a spark of rebellion that resulted in the government being overthrown. The entire basis of their power was in keeping the Districts downtrodden and the Capitol happy, and in proving that they could deliver on the promises they make. The Hunger Games were a relatively simple set of promises: there will be a Games every year, the children of the Districts will fight and die, and there will be a Victor.
With Katniss, they were ultimately in a lose-lose situation when it came to the Districts. If they killed her and Peeta they'd turn them into martyrs. If they backtrack then they look weak, and show the Districts that they aren't completely and utterly powerless. With the Capitol, they only lose by killing Katniss and Peeta because everybody there loved them.
Comment: @wizloc Yep, the relationship between Capitol citizens and Hunger Games contestants is like us picking someone on a reality show to root for. People are disappointed when their pick loses, and could be *really* unhappy if nobody won. Then the Victors become celebrities, and become beloved by the Capitol people. I don't know if the movies had it, but in the books, the 75th Hunger Games angered a lot of the Capitol people, because suddenly it wasn't some no-name kids dying, it was *their* beloved celebrities, people who were now more Capitol than District.
Comment: I like the mention of how people in the Capitol pick their favorites, and would be disappointed if they died. I hadn't thought of that
Answer: President Snow actually addresses this question during his conversation with Seneca (scene from the Hunger Games movie).

Snow asks Seneca "Why do you think we have a winner?" Because another alternative is that the Capitol could simply randomly round up 24 District kids between ages 12-18 and execute them together. The answer, according to Snow, is "Hope. It is the only thing stronger than fear. A lot of hope is dangerous. A little hope is effective. A spark is fine, as long as it's contained," before Snow sternly warns Seneca to "contain it!"
Answer: There's a point no other answers have made, based on the psychology of rebellions in general.
It's actually not true that oppressors try to make life as bad as possible for the oppressed. If everything is as bad as it could possibly be, then the people will rebel, even if they think they have no chance, because they have nothing to lose. Oppressors need to provide some sort of hope under their rule, so that potential rebels realize a rebellion will make things worse. That's one of the many points of the games - a hope that if your district wins the games, you will get good food for a year, not just starvation rations. That also motivates people to keep competing. If there is no victor there is no reward, and people might refuse to compete.
The suicide of both remaining competitors will also be an act of rebellion against the Capitol. In a sense, by suiciding, Katniss and Peeta would have found a way to not give the Capitol what it wanted, prevented the Capitol from controlling at least one part of their lives - an example that the Capitol didn't want people to see.
Answer: They need a winner to maintain interest in the Hunger Games.  "Bread and Circuses" is an important theme in the first book, and if the games don't have a winner, they won't fulfill their purpose of placating the proles.  There are some other sci-fi dystopias, like Young Rissa, in which a state lottery gives hope to a downtrodden people.  In the real world, events like the North Korean mass games serve the same purpose.
Aside from the districts, Snow also has the population of the capitol to please.  Capitol residents placed bets on different tributes, and sent gifts to their favorites in the arena.  The games must be very expensive, too: aside from the year-long victory tour and television coverage, the arena is enormous and is substantially redesigned every year.  The upper class demands a winner.
Answer: Snow explains it in the first movie
It's why we have a lottery, or stories of starting from the bottom and getting rich. While simultaneously any company that gets too big gets bought out
Give them hope, but maintain control.
The quote is below
President snow: seneca, why do you think we have a winner?
Seneca crane: what do you mean?
President snow: i mean, why do we have a winner? I mean if we just wanted to intimidate the districts why not round up twenty four at random, and execute them all at once? It would be a lot faster.
President snow: hope.
Seneca crane: hope?
President snow: hope. It is the only thing stronger than fear. A little hope is effective. A lot of hope is dangerous. Spark is fine, as long as it's contained.
Comment: While true, this doesn't add much as an earlier answer already quoted that part...


Parity account addresses all having prefix 0x00

Question: After creating multiple new accounts with the Parity wallet (v1.6.6), I notice they all have the address prefix 0x00 (the 0x part being the obvious hexadecimal prefix), for example 0x00A85d2d62E1851ad5702e31619B10c7E4308522. What is the reason behind this address scheme?
Comment: How many times did this happen? When you say "they all have" how many do you mean? Just curious. The address you provided is 20 bytes long, so it looks legit. (The '0x' is not part of the address as you suggest.)
Answer: This is to follow the proposed design of IBAN-compatible Ethereum addresses. (See also What is an ICAP address?)
Any random 160-bit Ethereum private key can be converted to any basic ICAP address. However, not all ICAP addresses are IBANs.
Only direct ICAP addresses follow the IBAN definition which includes the identifier, a checksum, and the account:
XE73 38O0 73KY GTWW ZN0F 2WZ0 R8PX 5ZPP ZS

This, however, limits the length of the underlying private key to 154-bit and thus, the range of compatible Ethereum private keys is limited from 0x0000000000000000000000000000000000000000 to 0x03ffffffffffffffffffffffffffffffffffffff.
Parity ensures future IBAN compatibility of your addresses by only using addresses from this range (actually only up to 0x00ff...f for the sake of simplicity). And that's why all addresses start with 0x00... in your Parity wallet.
Disclosure: I work for Parity.

Edit: By the way, the same is true for the cpp-ethereum client (eth).


Slow flash when shooting a D90 and SB 700

Question: I'm experiencing slow flash recycle times. What can I do to troubleshoot or speed up the flash? I'm using a Nikon D90 with an SB700.
Comment: Ray, your un-edited question said "Why, Nikon doesn't know!" I take this to mean that you contacted them and they weren't helpful. Can you explain what you said to them and what they told you? Also, we're assuming you mean flash recycle time. Is that correct?
Answer: There are a few primary influences of cycle speed:
Batteries
You'll experience the most inconsistent recharge rates with alkaline batteries. You'll want to make sure you are using high quality batteries which can drastically improve your recharge rate and consistency throughout their charge. There is a good discussion on battery selection here.
Flash power
Too often I find people using a higher power flash setting than necessary. With the D90, you should be stepping up to 200-400 ISO to give the flash a break.
Heat
The SB700 knows to slow down if it starts to get too hot. If you're using the flash in warm temperatures or rapidly flashing at high power, the flash will begin to limit itself. This can become permanent if the flash has been brought to extreme temperatures.
Bad batteries
There have been multiple reports of NiZn batteries permanently damaging flashes. I hope you didn't use those...
Short
If there is a short in the flash somewhere, then the camera may be getting very hot in a single flash, triggering the governor that slows it down.
Comment: Actually, fully charged alkaline battery recharge blazing fast.  They just have a much quicker falloff in recharge time than other chemistries.
Comment: True, out the gate they are quick


Too much sitting as bad as smoking?

Question: I have heard that when you sit very long times every day that it is as bad for your health as smoking cigarettes. Especially, it is bad for your metabolism and especially the lungs get into a bad condition on an equal level. Is that true?
Comment: I think the worst part is sitting on your ass constricts the blood flow. Also, it may actually be significantly worse than smoking.
Answer: If we compare the average increase in all-cause mortality due to sitting (approximately 50% according to meta-study published on NHS) and the average all-cause mortality due to smoking (aproximately 300% increase according to CDC data), siting does not appear to be nearly as bad.
Yet, because of how broadly you worded the question it is possible that smoking is not as bad as sitting. According to the same publication by CDC that I linked above:

Quitting smoking before the age of 40 reduces the risk of dying from smoking-related disease by about 90%.

Therefore, for smokers who quit before the age of 40 on average the all-cause mortality due to smoking would be just 30%, lower than 50% for caused by sitting.
Regarding your questions of metabolism and harm to lungs. Bad posture while siting restricts maximum lung capacity therefore limiting oxygen intake. Additionally, siting decreases insulin action in blood, decreasing metabolism.
Answer: Research has linked sitting for long periods of time with a number of health concerns, including obesity and metabolic syndrome — a cluster of conditions that includes increased blood pressure, high blood sugar, excess body fat around the waist and abnormal cholesterol levels. Too much sitting also seems to increase the risk of death from cardiovascular disease and cancer.
http://www.cheatsheet.com/life/side-effects-of-sitting-all-day.html/?a=viewall

Too much sitting as bad as smoking?


why is a companies implied rate of return the correct discount rate for projected cash flows

Question: Let's say we wish to value a company using the DCF method. To determine which discount rate to use for the cash flows we choose to use the CAPM methodology. The estimated beta of the company can tell us how its price today relates to the expected value of its price in the next period. I am having trouble seeing why this implied rate of return is also applicable for discounting the projected cash flows. Why should a company´s cash flows have the same beta as the companie´s value?
Comment: It is only true for projects whose risk is similar to the overall existing business(es) of the company. If a company is getting involved in a completely new business the discount rate to use may indeed be different from this. (But if a car company builds a new car factory, the Beta of the car company is probably the correct risk measure for the new project).
Answer:
To determine which discount rate to use for the cash flows we choose
to use the CAPM methodology.

The determination of the discount rate to use when discounting cash flows depends on what you're valuing. If it's the entire company, and not just it's equity, the discount rate is the WACC (i.e., weighted average cost of capital), that accounts for both debt and equity financing. It is nothing more than a weighted average of the required return on debt and the required return on equity.
CAPM is properly used to estimate the required rate on equity. In other words, what return does an investor who invests in the company's equity demands. It is generally used as the discount rate when you're valuing the company's equity, and therefore, the cash flows are the Free Cash Flows to Equity (in contrast to the Free Cash Flows to the Firm).
Comment: this really doesn't answer my question
Answer: Discount Rate Matches the Cashflow Type
For DCF valuation, there is a simple rule: the type of cashflow dictates the type of discount rate to use. Dividends are discounted at the CAPM rate (although one can also use other equity factor models). Free cash flow to the firm is discounted using the WACC; and, free cash flow to equity is discounted using the CAPM (or other equity factor model) rate after correcting for firm leverage.
I'm going to assume (maybe incorrectly) that the cashflows in your question are dividends. I said we could use any equity factor model, but let's stick to the CAPM.
How CAPM Discounting Affects the Value of Dividends
We use the CAPM for discounting future dividends because it gives us the equilibrium expected return for investing in that firm. If the firm pays us dividends, that money will not grow at the same rate as the firm (unless we reinvest dividends). In some cases, that is good and we will redeploy that money to earn more elsewhere; in some cases, that is bad and we will not want to redeploy that money since it will likely earn less elewhere.
It helps to think of the perpetuity form for a dividend discount model with dividend growth (the Gordon model). That form gives us the equity value by $E=\frac{D}{k-g}$ where $k$ is the CAPM-implied discount rate and $g$ is the dividend growth rate.
If we rewrite the growth rate and dividends to account for $ROE$ and some fraction $b$ of dividends being reinvested in the firm, we get that equity is worth $E=\frac{(1-b)D}{k-b\cdot ROE}$. If $ROE<k$, the optimal policy is to not reinvest dividends. If $ROE>k$, we can grow the firm value without bound as $b\uparrow\frac{k}{ROE}$. (This is unrealistic, but it does give us an upper bound on the optimal reinvestment fraction $b$.)
Therefore, firms with a low ROE (slow growing firms with a low beta) should not keep excess cash. Those firms should pay that cash out to investors who can do better redeploying the funds to other firms.
Firms with a high ROE (above $k$, fast growing firms with a high beta) should reinvest some of their excess cash. Now, this model breaks down because it may even make sense for a firm to pay out no dividends while the firm is growing rapidly (when $ROE>k$). Ostensibly, the firm will eventually grow less quickly and then begin paying out dividends. We saw this happen with Microsoft which paid no dividend from its IPO in 1986 until 2003, when the firm began paying dividends.
Why Use the CAPM for Discounting?
As for why to use the CAPM to discount projected cashflows: the CAPM is a forward-looking expectation of the return on the firm's stock. When we are thinking about the opportunity cost of getting dividends instead of having that money invested in the firm\s stock, a forward-looking estimate of the firm's expected stock return is a good estimate of that opportunity cost.
This is why we could also use other equity factor models: those also yield model-driven estimates of the firm's expected stock return.
Beta of Firm's Cashflows?
Finally, you asked:

Why should a company´s cash flows have the same beta as the companie´s value?

Firm cashflows do not have a beta. I suppose you could compute a cashflow beta, but I have never seen that done and I'm not sure it is very informative.
We refer to beta to describe a firm's stock returns as a multiplier of "the market's" excess returns. (We could fill a whole topic with "What is 'the market,' but that is another question.)
Comment: "The firm should not keep money and invest it in their company. Investors could do better redeploying those funds to other firms". This thinking seems off to me. In the CAPM world each company's returns are competitive at their specific beta.
Comment: It is easier to see this if you look at the Gordon DDM and express the growth rate as a function of ROE and the fraction of dividends reinvested $b$. Depending on ROE, there are different optimal reinvestment policies. Returns are competitive; that does not mean dividend policy is irrelevant.
Comment: Updated the above to use the Gordon+ROE dividend reinvestment thinking.
Comment: Thanks for the edit, your answer makes much more sense now to me. I guess this is the capital markets way of incentivizing company`s to pursue projects which will help diversify the market as a whole.


Why does spiral or annular thickening occur in water conducting plant vessels?

Question: From what I behold, spiral and annular thickening of xylem and trachied cell walls leaves a lot of not thickened regions of cell wall. Lignin is the material which prevents water from escaping these water conduits. If lignin is absent from certain regions, does not that mean that water flows out of the xylem vessels and trachieds through their cell walls? If so, is not this disadvantageous to a plant? Why can't a plant simply thicken the cell walls with uniform lignin deposition, only sparing pits to allow lateral flow into the required tissue? I would appreciate your kind response.
Comment: Lignin adds mechanical strength. Leakage from one xylem lumen to another can be seen to be advantageous as it lets water tension to equalize radially - else a plant would tend toward one-leaf-to-one-root (I exaggerate since there are also pits to distribute water). Trachieds are dead cells. Additions to their walls come from nearby parenchyma. Wouldn't it instead be curiously remarkable if the lignin depostion was uniform?.
Comment: Exactly, if pits have the function of lateral flow then why is not lignin deposition uniform? That is what I am curious about.
Comment: I think pits tend to be located at the ends of the trachaeids. So lateral flow is minor compared to axially. Water can flow laterally thru the unlignified cellulose walls by osmosis. Anyway, you pose an interesting question I cannot answer, though is does boarder on asking why there is air, IMHO. :)
Answer: Non uniform lignification of xylem vessels is beneficial to a plant in that it enables growth. It is at the unlignified parts of plants that growth occurs since fully lignified tissues are dead and so can not grow. Spiral and annular thickening allows for vessels to elongate and thus allows tissues to expand without breaking the vessel walls.
Water cannot just go on flowing from plant cells because they are not lignified, plant cells have cellulose cell walls and cell membranes which prevent water loss. In addition, water in the xylem is not under pressure so I do not think water would be able to flow out of the xylem vessels except in cases where water is forced to move to adjacent phloem tissues by osmosis during phloem translocation.
Comment: Welcome to Biology.SE! Your answer sounds reasonable, however, it is likely to receive a more favorable response if you include supporting references (primary literature is best, but a textbook available online would seem sufficient for this question).  You also should be able to easily remove the pictures by editing your post ...——— Please take the [tour] and then consult the help pages for additional advice on [Answer] effectively on this site and then edit your answer accordingly. Thanks!


Why User Instances (also known as RANU) available with SQL Server Express Edition, are NOT RECOMMENDED for hosting environments?

Question: Quoted from http://blogs.msdn.com:

User Instances (also known as RANU) available with SQL Server Express Edition, are NOT RECOMMENDED for hosting environments.

I have a server running Server 2008 R2 Web Edition with IIS 7.5 and Sql Server 2008 R2 Express installed.
I have only 3 friends as my hosting clients. Each has a restricted Local User Account on the server.
What is the drawback of using user instance in production environment?
Edit
Rather than considering the limitation in memory usage and database size that I know, let us consider the user instance only, why isn't it recommended?
Comment: @Recycle: It's just a close vote.  It will age away after a couple of days.  Nothing to worry about.
Comment: @Robert: did you have a reason to do a close vote?
Comment: I didn't close vote.  Someone else did.  It's OK, it takes 5 votes to close a question.  If there are no more close votes, this one will disappear after a couple of days.
Comment: @Robert: I am so sorry. I misunderstood your comment.
Comment: @Recycle, Robert Harvey is a moderator, so if he had voted to close this question, it would be closed.  It only takes 5 votes when it's us lesser mortals. :)     And, no I didn't vote to close, I followed a link from your meta question.
Comment: Also, to the anonymous close-voter, this is on topic for StackOverflow.
Answer: I think that the answer to your question comes with the definition of the RANU
What is RANU
and also read the http://msdn.microsoft.com/en-us/library/bb264564(SQL.90).aspx
The RANU starts a seperate instance Process of SQL Express for every user, so in a share enviroment with let say 30 sites that needs access to SQL, then SQL run 30 instances at the same time, that is takes as you understand a lot of recource/memory etc. Also as say in the article the first time SQL Express run for every user is make copies of all system database to create this user instance.
In a single user machine to run one process is normal, but if SQL Express starts to run many instances then you take a lot of resource, memory, and this is not easy to administrate, update etc.
In the article on the link there are more detail informations about this.
Comment: "with let say 30 sites that needs access to SQL, then SQL run 30 instances at the same time" ==> 30 SQL Server Express instances?
Comment: @Recycle yes, 30 sql server express instances, with 30 different user login, and 30 different master databases. Did your computer is xeon 4 core with 16Gb Ram Running 64Bit ? even if it is, its not recommended.
Comment: In my understanding, the SQL Server Instance is what we install on the computer using the installer provided by Microsoft. If using `User Instance = True` will install a new SQL Server Instance, I will see the SQL Server Instance when SQL Server Management Studio prompts us to login. Or the term `user instance` does not equal to `Sql Server instance`?
Comment: @Recycle Let me put this way. if the same think happens for iis, then you have to run 30 times the iis, each for every web site. Actually IIS runs one time and handle all site. non express SQL Server runs one time and handle all database.


Anonymity with Tor in wifi connection (student house)

Question: I just moved in a student flat where several students are using the same WiFi connection. A few days ago the internet was very slow and a guy (son of the owner) came in to ask if I was downloading. He could see I was using 'something that looked like a steaming or downloading site'. It really annoyed me, because I believe I just don't want others checking up on me. I've downloaded Tor, but never really used it. Now I wonder how anonymous I am in the local WiFi spot at home if I use it. I think they can check up on IPs or something, he could see I have a Macbook and I'm the only one with a Macbook in the building. I want to know that if I use a secured connection via Tor other within the network can't see what sites I'm visiting, whatever sites that my be. I don't care if the internet provider knows what I'm doing, but rather not my neighbors since its just none of their business.
Answer: If you use Tor or a VPN they cannot detect easily (that is without compromising your computer) which sites you visit, but they can still see how much bandwidth you use. And while they will not see the sites you visit they can just suspect the visiting of special sites from the fact that you try to hide it.
Comment: Besides the bandwith, they can also tell at what times and for how long you're using the wifi, if that matters to you. There's no way to keep that information from them.


Leaning timber retaining wall

Question: I have a timber retaining wall about 4 feet tall builting with 6x6 which is leaning  probably due to the downspout draining inside. I'm wondering if using an earth auger to relieve some of the pressure and then using a sledge hammer to knock it back straight would work?
Comment: A picture would help us understand the problem. However, if you remove all the dirt behind the wall, then yes, I think you could knock it back plumb...if the vertical supports are not rotten.
Comment: Could you add pictures? Did you build that wall? More information might be helpful to get a better picture
Answer: That is not unusual in our neighborhood. A sledge will do nothing but break wood. Repairs that have worked are to dig out soil behind the tipped area and replace the vertical posts with longer ( deeper) ones. The larger the concrete "footing" for the vertical posts , the more resistance to tipping. Additional support can be added by driving metal bars into the ground a few feet behind the wall and attach these with steel cables to the vertical posts. The bars and cables will be under the ground surface so not seen , Yes, they will corrode in a few years but by then the soil will have settled down and be stable ( hopefully). I did this with 3/8" steel cable and it was still holding when I moved 5 years later. The digging sounds worse than it is , or contract it.You only need to remove a narrow trench like 2 ft wide and not necessarily all the way to the bottom. An Auger would be a help. I have also seen a new stronger wall built where boundaries permit ; then fill in the narrow gap between the original wall and new wall .
Comment: Recently I used a different solution; The wall was tipping but mostly rotten. I had room so i got about 12 yards of fill dumped in front of the wall and made a slope to the former top of the wall. The location required me to move over half the fill with a shovel ; working an hour at a time for several days, it was no problem.

Leaning timber retaining wall


Practical use of RNA structure

Question: I have spent some time studying RNA structures. While there is a range of interesting phenomena and functions that certainly deserve understanding from the scientific point of view, I have never encountered a practical use of this knowledge (in terms of disease diagnostics, medications, etc.) Have you? If a practical use of RNA structures is still in the future, what could that be?
Comment: Hi @Vadim, the Bioinformatics Stack Exchange system doesn't work well for open-ended, or opinion-based questions. If you have a specific problem, please state that instead. Otherwise, I'd recommend asking this question on the [Bioinformatics subreddit](https://reddit.com/r/bioinformatics).
Answer: Please look up flavivirus 'double loops' as you described them previously (post for "Coronavirus RNA') and associated RNA secondary structure anomalies for dengue virus and associated vaccine (Butantan) and the yellow fever virus and its vaccine (17D). If you are aware of "double loops", we must be aware of the association of RNA secondardy structure and attenuated strains, i.e. live vaccines. The point is that modification of the RNA secondary structure is a key strategy in flavivirus attenuation.
Comment: When I have a minute I will edit the post to describe the attenuation of flaviviruses via the 3' UTR (untranslated region), this was traditionally about  RNA 2ndary structure modelling of pseudoknots.
Comment: This is a good point. Is this a widely used technique or is it limited to flaviviruses?
Comment: I am not aware of this outside flaviviruses, but my interests are restricted to arboviruses, whilst yours are broader.
Comment: I am more on the mathematical/computational side, and I worked mainly on HIV.

Practical use of RNA structure


Light speed after the Big Bang

Question: We know that immediately after the Big Bang space was opaque to light. As space expanded, light could travel. However I would think that as space expanded at some de sites the light would be much slower than it is today so that the speed of light was much lesser at the beginning?
If so when we peer at the edge of the visible universe won't this "slower" light interfere with the "faster" light and give larger readings for the size of the universe?
Comment: Possible duplicates: http://physics.stackexchange.com/q/21721/2451 , http://physics.stackexchange.com/q/144262/2451 , and links therein.
Answer: I think there is no more "slow" light from the begining od the universe remaining because photon has accelerated after expansion to reach their current speed. Maybe I'm wrong but to me it seems to be the right way to think.
Answer: As far as I know the Speed of light has Always been constant. But some scientists suggest that the physical constants might Change slightly over time. The Situation in Inflation seems to be a bit different.

Light speed after the Big Bang


Why is it vital that atoms in the same sample are all given the same charge? Mass spectrometry

Question: When being tested in the spectrometer why is it vital that the atoms in the same sample are given the same electrical charge e.g all are c+1 or all are +2 etc
Comment: As far as I know, the degree of ionization of particles in mass spectrometry depends on the energy supplied. In other words, it's tunable. However, you would want to use your electron beam (let's say it's electric ionization) as an analytical tool, not a sledgehammer. It makes the MS result analysis more feasible if the ionization energy is just enough to remove as few electrons as possible. Simplifies the spectrum, so to speak. Is that what you're asking about?
Answer: This is a misconception that during ionization, each ion must get a +1 or +2 charge. It is desirable to have +1 so m/z corresponds to the molecular mass of the ion or its fragment. The state of ionization heavily depends on what you are using to ionize your target analyte. For example if you use electron impact, you will see fragments mostly with +1 charge. However in other techniques called matrix assisted laser desorption (MALDI), the state charge can be as high as +12 and so on. They are detected as such. In electrospray ionization, +2 is also common.
There is no way a mass spectrometer "knows" a priori the state of charge of an ion. All charge ratios are detected. However it can distinguish the sign. It depends on the user whether negative or positively charged ions reach the detector. Some ions are better detected in positive mode and some in negative mode.
Answer: In mass spectrometry, the position of each peak you observe depends on the value m/z (i.e. mass over charge). If you have only z=1, then the position will depend only on the mass (m/1 = m).(1)
In a mass spectrometer, you will generate a mixture of charges at first but ions with a charge greater than 1 are filtered out.
(1) Obviously, it also works with a -1 charge.