How do I upgrade Tom Nook's store in Animal Crossing for the Gamecube?

Question: I made my house bigger in the Gamecube version of Animal Crossing, but Tom Nook's store is still Nook's Cranny. What should I do?
Comment: Welcome to Arqade! If one of the answers answered your question, you can click the checkmark below its score to mark it as accepted and earn 2 reputation.
Answer: In Animal Crossing for Nintendo GameCube, Animal Crossing: Wild World for Nintendo DS, and Animal Crossing: City Folk for Wii, you upgrade Tom Nook's store by buying and selling things. Every Bell that a player spends or earns in the store increases a counter. (The counter is shared among all human villagers; NPC villagers do not affect it.) Once this counter passes a certain amount, the store will close the next day for an upgrade. On the GameCube and DS versions (not the Wii version), the final upgrade (Nookington's) requires someone from another town to buy something from your Nookway store after reaching the threshold, so you may need to buy another GameCube memory card.
The quickest legit ways to increase this counter are:

Orchard Shake one-third of the fruit trees in your town each day (face a tree and press A) and sell the fruit. If an acre has fewer than 13 trees, bury fruit to make more fruit trees. Fruit species other than the one your town came with are more valuable; owning another memory card helps.
Fishing Equip a fishing rod, throw the lure in front of a fish with A, and when it's pulled all the way under, press A to reel it in. More valuable fish come out on rainy days.
Selling fossils Talk to Blathers to get Museum in Farway added to your contacts. Then grab paper from the dump, dig up unidentified fossils by using the shovel on asterisk marks, attach them to letters to Museum, and mail up to three per day. Once you get them back, you can donate the fossils to Blathers and sell anything he already has to Tom Nook.

Paying back your loan does not by itself increase this counter. But if you're paying back your loan, you probably sold things to Tom Nook to get the money, and that increased the counter.
Answer: To upgrade your Nook's Cranny to a Nook 'n' Go, you need to have bought or sold at least 25,000 Bells worth of stock and have played the game for at least 8 days. Additional information about it (time it is open and what it has) can be found here.


If the tensor product of two $kG$-modules is projeсtive, does either of them have to be projective?

Question: Let $G$ be a finite group and $k$ a field, let us assume that char($k$) divides the group order. Let $kG$-mod denote the category of fintely generated $kG$-modules.
This category has as a tensor product $\otimes_{k}$ with diagonal $G$-action.
Given now $M,N\in kG$-mod such that $M\otimes_{k}N$ is projective, can we then conclude that either $M$ or $N$ had to be projective?
If not, can we ask for certain conditions on the field $k$ or the group $G$ such that the statement holds?
Answer: The answer to the first question is no.  Let $G$ be the group $C_2 \times C_2$ and let $k$ be a field of characteristic 2.  Then $kG \cong k[x,y]/(x^2, y^2)$.  Let $M = k[x]/(x^2)$ with $y$ acting trivially, and let $N = k[y]/(y^2)$ with $x$ acting trivially.  Then neither $M$ nor $N$ is projective (for modules over a $p$-group in characteristic $p$, projective is the same as free), but $M \otimes_k N \cong kG$.
Edit: I think that if the Sylow $p$-subgroup of $G$ is cyclic, then you can use Avrunin and Scott's tensor product theorem to prove that whenever $M \otimes_k N$ is projective, then so is either $M$ or $N$, but that seems like a big tool to use for this.
Comment: Ah, OK. Thank you. I'd already be happy with a (concrete) example of a group $G$ and a field $k$ such that char($k$) divides the order of $G$ and such that $M\otimes_{k}N$ projective, implies $M$ or $N$ projective.
Comment: This is a guess, but perhaps if the Sylow $p$-subgroup of $G$ is cyclic of order $p$, or maybe even just cyclic?
Comment: @John: `$SL_2(\mathbb{F}_p)$` is an interesting example to try out in this spirit.   Unlike most groups of Lie type, all of its (finitely many) indecomposable modules are known due to the special nature of its Sylow `$p$`-subgroup, as well as all projective modules.  This probably gives the kind of concrete example Heskie is asking for.
Answer: There are interesting but isolated examples for finite groups of Lie type to consider.   Here the simple Lie type and the characteristic of the defining field combine to give factorizations of some projective modules for finite Chevalley groups as tensor products of two non-projective simple modules.   This combines work of Chevalley on special (or exceptional) isogenies between simple algebraic groups with work of Steinberg on related simple modules and their twisted tensor products.
It's all a bit complicated to explain, but relates to the simple groups discovered by Suzuki and by Ree.   For instance, in type $G_2$ with $p=3$ the usual Chevalley group over a field of order $q=p^r$ has a smallest possible projective module of dimension $q^6$ (the Steinberg module) which factors when $r$ is odd as the tensor product of two non-projective simple modules whose highest weights are supported on long (resp. short) simple roots; take $r>1$ here to get an associated simple Ree group.
Besides the original literature, there is a short account in my LMS Lecture Note volume Modular representations of finite groups of Lie type (see 20.3).
Obviously what is going on in such examples requires specific primes for the specific groups involved as well as specific simple or projective modules.  But these examples indicate some of the complicated behavior found in modular representation theory and caution against oversimplified expectations based on older characteristic 0 theory.
Comment: Jim, might you have any specific references for those tensor product factorization results in the twisted groups? It sounds like it may be related to some things I've been working on recently.
Comment: My 2005 Cambridge book mentioned above gives the details along with the basic references.  The most relevant sections are 5.4 and 20.3 along with the short Chapter 18 summarizing results for type `$G_2$`.   Following the 1956-58 Chevalley seminar which introduced special isogenies, Steinberg's 1963 paper and 1967-68 Yale lectures deal explicitly with representations but are very concisely written.
Comment: P.S. The special tensor product factorizations of simple modules including the Steinberg module occur at the algebraic group level first, but then these simple modules remain simple for certain of the finite groups of Lie type (with the Steinberg module then being projective as well).   The Suzuki and Ree groups correlate with the "special" here, but only indirectly.   As I said, it's all a bit complicated.   I guess `$p$`-groups are a more direct laboratory to work in.
Answer: If $G$ is a $p$-group, this is almost never true (and I believe but may be wrong that for general $G$, it's completely governed by the $p$-Sylow).  In that case, we can understand modules in terms of the support variety $V={\rm Proj} H^{2*}(G,k)$, the projective variety associated to the (even-dimensional) cohomology ring of $G$.  To a finite $kG$-module $M$ we can associate its support in $V$, namely the support of Ext(M,M) as a graded $H^{2*}(G,k)$-module.  This support is a closed subset of $V$, and conversely every closed subset is the support of some finite $kG$-module.  Finally, the support of $M\otimes N$ is the intersection of the support of $M$ and the support of $N$, and a module is projective iff its support is empty.  Thus $M\otimes N$ is projective iff $M$ and $N$ have disjoint support.
Thus unless $V$ is just a single point, it is possible to have non-projective modules whose tensor product is projective.  For $V$ to be a point, the cohomology of $G$ must be a polynomial ring in one variable, up to nilpotent elements.  By Quillen's theorem, this is the case iff all elementary abelian subgroups of $G$ are conjugate and rank 1.
In particular, it is true for cyclic groups, but otherwise it is almost always false.  There's a simple argument to see directly that it holds for cyclic groups of order $p$: in that case, $kG$ can be identified with $k[x]/x^p$, and every indecomposable module is of the form $M_i=k[x]/x^i$ for some $i\leq p$.  Such a module $M_i$ is projective iff $i=p$.  If $M_i$ and $M_j$ are not projective, then $M_i\otimes M_j$ has dimension $ij$, which is not divisible by $p$.  Thus $M_i \otimes M_j$ cannot be a sum of copies of $M_p$ and is hence not projective.
Comment: Right, this is what I was referring to in my answer. Is there a way to approach this without using support varieties?
Comment: Great. I'm not quite sure why in the example, the indecomposable modules are exactly of the form $M_{i}$. Could elaborate on that or give a reference? Thank you
Comment: Any $k[x]/x^p$-module is in particular a $k[x]$-module, so it follows from the classification of finitely generated modules over a PID.
Comment: LOL...Oops. Sorry, I should have thought about this a bit longer ;).
Thanks.


Ubuntu on Laptop with switchable Graphics Card (AMD)

Question: After I got problems with Windows again, i switched to Ubuntu on my old Laptop (Acer Aspire 7750G). Im very happy with Ubuntu and how user friendly and light it is. I mainly use it to browse the web or watch videos on YouTube or Netflix and casually play on it. Here's the thing. I liked to play CS:GO with my friends, but I cant get my AMD graphics card to work (Game uses onboard IntelHD 3000). My question is, how do I install/utilize my graphics card(AMD Radeon HD 7670M) because all the tutorials/links on google are for older Versions and now on the AMD site there are so many packages, where I don't now which one to chose. Please help me :/
English isn't my native language, so sorry for any grammatical mistake.
Comment: Tutorials for older versions of Ubuntu usually work fine. To get the official AMD drivers you can open the `Additional Drivers` program and install them from there. To make sure you don't use the Intel card, you might have to disable it in the BIOS.
Comment: Check this article for driver installation [How do I install additional drivers](http://askubuntu.com/questions/47506/how-do-i-install-additional-drivers), and this article for switchable graphics [Disable AMD graphic card to save battery](http://askubuntu.com/questions/525108/disable-amd-graphic-card-to-save-battery/525127#525127)
Answer: Ubuntu will not automatically choose the VGA for you, you should do it manually.
first, install AMD Catalyst by running this command:
sudo apt-get install fglrx

After that, reboot your computer. Then you can access the aticonfig command. First, it needs to be initialized:
sudo aticonfig --initial

reboot your PC again, then Run AMD Catalyst Control Centre and you will have the option of switchable graphics like Windows.
EDIT: it's always the best practice to install additional drivers using GUI because you can see if the driver is compatible with your system before actually installing it. Using the command line will install the driver anyway, it does not look if your hardware supports the driver or not.
Comment: @AliRezaMosajjal think it's better to encourage people to install proprietary drivers form the GUI (the "Additional Drivers" tool), so that you could know if the card is supported and detected correctly. By using the command line command, people can just install that package anyway.
Comment: Thanks for the fast response!
After the installation and the reboot, i tried the command 'sudo aticonfig --initian' but it says " no supported adapters found" ....
Comment: if the answer helped you, you can mark it as correct so others can benefit from it as well @HuyTaoDuc


Formula for comparing Date Fields

Question: I have a report which has a filter contact: created date not equal to website inquiry: created date, however it is not working. I created a formula on the website object Contact__r.CreatedDate  <>  CreatedDate however it is not working. Could someone please assist me in figuring this out because For some reason, SFDC does not support comparisons between two fields.
Comment: CreatedDate stores more than just the Date. It goes all the way down to milliseconds. Is your filter returning all records?
Answer: My guess is that the CreatedDate fields are never matching because they store the time part as well. If you just want to compare the date parts, try this in your formula:
DATEVALUE(Contact__r.CreatedDate) <> DATEVALUE(CreatedDate)
Comment: Hey Daniel, yeah my filter was returning all records. I will try this out and let you know how it works. Thanks very much

Formula for comparing Date Fields


can we store values in constant at the time of program deployment?

Question: I'm working on a Solana program and would like to understand if it's possible to set constant values during program deployment, similar to constructor parameters in other smart contract platforms. Additionally, I'm curious to know if there is a way to update or change these constant values after the program is deployed.
Answer: You can use Rust feature flags for this use-case. These can be used as environment variables, which allow you to set deployment environments, config, and more.
For example, you'll set them in your Cargo.toml file like the following:
[features]
feature-one = []
feature-two = []

and use them in your program like this:
#[cfg(feature = "testing")]
pub const SOME_VARIABLE: usize = 100;

#[cfg(not(feature = "testing"))]
pub const ANOTHER_VARIABLE: usize = 50;

More comprehensive resources on this:

https://doc.rust-lang.org/cargo/reference/features.html
https://www.soldev.app/course/env-variables (examples were taken from here)


release coco2dx application in vs2015 to win7 msvcr110.dll is missing dll

Question: for debug version, missing MSVCP140D.DLL
for relase verson, missing MSVCR110.DLL
could any body tell me how to release coco app in vs 2015?
I've set all build settings to MTd for debug, and MT for release, and use of MFC: use static MFC.
and tried c++ redistributable for vs 2015/2012/2010/2008, no any luck.
Answer:
release coco2dx application in vs2015 to win7 msvcr110.dll is missing
  dll  So if you have such a problem with this error you should remove it quickly and to do it properly you simply must install such tool like fix dll one for example from this url: http://fix4dll.com/msvcr100_dll
Answer: To fix this problem it is enough to download msvcr110.dll , we sure that file is not corrupted. Sometimes this type of issue is arise because it's main reason is that file is crashed or damaged.After that you need to reboot the computer.


How to get a good start on being a CEO?

Question: Alright, so I am just starting to want to be a CEO in Grand Theft Auto Online and I was wondering, how do I get a good start to this job? I will have a total of 1,450,000 (or maybe slightly above, just a couple hundreds above if so), and will this be enough to get a good start on being a CEO? Will I NEED more money, or will that be enough in the beginning? What will I need to do to get started on being a CEO and increasing my money amounts, warehouses, etc.
Comment: I consider this a game strategy question and think that it should be left open.
Answer: Alright. To start off, you are going to want quite a large amount of money. The cheapest CEO building cost around $1M, so seeing as you say you have about $1.4M, you should be able to get it and still have a large amount of money left over. If you want to get a gun rack and safe, however, this will increase the price of the property you are buying. I suggest not buying those unless you are able to buy some more money or get it with missions, because you will need the extra money to buy warehouses and crates for items to sell later, etc. Hope this will help, and good luck on being a CEO on GTA Online, and  I hope you do well and get some good associates to work with (yes you will probably want those).
Comment: Wow, lot's of help there! Makes sense though. Any more opinions would be great on how to get a good start as a CEO, and any help is greatly appreciated. Thanks again Frankenyuyu!

How to get a good start on being a CEO?


Correct Screw length I should be buying

Question: Can you help me out. I am planning to install this (https://www.homedepot.com/p/Rubbermaid-70-in-White-Twin-Track-Upright-for-Wood-or-Wire-Shelving-FG4B8900WHT/100047070) in my cabinet to add organizer inside my closet. I am not sure though what should be the lenght of the wood screw that I will be using if I will install it in the stud. Note that depth of that material is around 1 inches. Thank you
Answer: Just add up all the parts:

Bracket - your questions says "about 1 inches" but I usually find these types of brackets (but haven't used exactly this one) are a bit thinner than 1 inch
Cabinet back - varies, might not be any at all
Drywall or other wall material - typically 1/2"
Between 1 inch and 2 inches into the stud. Studs are nominally 3-1/2" thick, but if there is any possibility of wires going through the studs then you don't want to go more than 1-1/2" in.

So that makes a total of 2-1/2" to 3" typical.
As noted elsewhere, don't use drywall screws. They are handy, but they are not designed to hold a heavy load, which is important for shelving brackets.
Comment: IMHO, wood screws or deck screws would both be fine.
Comment: I think deck screws would be fine.
Comment: I was planning to use wood screw is that good one? Can I use deck crew? I am going to home depot tomorrow to buy screw
Comment: Just a question do you think there would be negative effect of using deck screw instead of wood screw? Cant find answer on google if there is a negative effect of using deck screw. I got could of deck screw that fits the length not sure if it will bite me after couple of years
Comment: I'd go with deck screws or chipboard screws, traditional wood scrws are too much hassle and not as strong.

Correct Screw length I should be buying


Is there any point to breaking the surveillance cameras?

Question: I believe in the original Portal there was an achievement for knocking down all the cameras. It doesn't look like there's a related achievement in Portal 2. Does knocking down the cameras have any effect (aside from nostalgia?)
Answer: There doesn't appear to be an achievement for removing cameras in Portal 2. I wondered the same thing. It is, however, fun to watch them burn with the lasers. I don't think there was a fun way to destroy cameras and other objects in the original Portal.
Comment: I once gathered all boxes, milk cartons and tin cans from a level to drop them down the incinerator. Then I fell down myself and died. Guess what, the items weren't destroyed.
Does that mean her pieces weren't destroyed either? You simply cannot trust anything in that game.
Comment: Other than the material emancipation grid and or incinerator?
Comment: Yeah but it's just walking through the grid and then disintegration. With incineration, it just disappears. Now with the lasers there's a bit of burning before an explosion for those of us who enjoy the dark pleasure of destroying turrets and cameras. Still in the middle of the game somewhere so I can't comment on any other fun destroyable objects that may lie in the levels ahead. :P
Answer: It has no effect, and as far as I can tell GLaDOS never reacts to you breaking the cameras; there is a related achievement in Chapter 8 where you're challenged to break monitors, however.
Comment: I guess Glados is beyond caring about cameras at this point :)


Removing overshots (dangles) using FME or FME/MRF Transformers

Question: I have a network of many connected lines where there is overshoot topological errors. Is there any way to remove them using FME Desktop?
Answer: For FME2018 I would strongly recommend the Intersector transformer. It has a new tolerance parameter and it can quite easily clean up a lot of issues like this.
But it also partially depends on what you are going to do with the data afterwards. Most techniques will split the line at the intersection, leaving the overshoot as a small piece of linework.
If you are building polygons, then it's easy to ignore the overshoots (they appear out of the unused port). Otherwise you could apply a LengthCalculator transformer and simply remove lines less than x length (maybe 0.5 metres or however big the overshoots are).
Comment: Thank you for your answer.
The problem is that when I used the intersector, not only overshoots are extracted, but also all other intersections. This adds lots of new features to the network which were supposed to stay only as one feature.
Is there any way to deal with this ?
I also don't understand well the Intersector tolerance, can you explain it to me ?
Answer: You can use the Matcher Transformer to remove overlapping topologic features.
Otherwise use the ChangeDetector Transformer to detect overlapping features but to have more power in which to keep.
Answer: The Snapper transformer is useful in this scenario

"Brings lines, segments, end points or vertex points of features
  together if they are within a certain distance of each other and
  (optionally) if they have one or more attributes in common.
The difference between the AnchoredSnapper and the Snapper is that
  anchor features are considered to be accurate and consequently do not
  move."

(Image from SafeSoftware)

http://docs.safe.com/fme/2018.0/html/FME_Desktop_Documentation/FME_Transformers/Transformers/snapper.htm
But consider using a combination of transformers to to improve your chances of removing under and over-shoots.
Note there is a module called MRF2DCleaner http://docs.safe.com/fme/2018.0/html/FME_Desktop_Documentation/FME_Transformers/Transformers/mrf2dcleaner.htm?Highlight=mrfc%20cleaner but this is at additional cost and only on the 32 bit version of FME. Contact the vendor for a cost.


Does a PNP transistor pull up itself?

Question: Blog schematics for PNP transistors sometimes connect +5V to the emitter and a voltage divider between the base and an open drain output, and sometimes only connect a base resistor to the low side switch. In the simpler configuration will the current flowing from the emitter to the base pull up the base and turn off the transistor? How long will this take?
Comment: Bloggers might not be representatives for good engineering practices.
Comment: Sample schematic, please.
Comment: PNPs don't have drain terminals. Do you mean open collector?
Comment: The low side switch is a fet.
Comment: Schematic given
Answer: That's actually pretty easy. You have a schematic editor you could use here. I'll use it now. I've paired them, side by side, to match up with your pictures in your question:

simulate this circuit – Schematic created using CircuitLab
As you probably already know, \$R_1\$ is used to sink the needed base recombination current for \$Q_1\$, when it's activated as a switch by \$SW_1\$. Your question, as I read it, asks about \$R_2\$ in the right-hand case.
\$R_2\$ is used here as a "pull-up" of \$Q_1\$'s base. It's added security of a sort. For example, assume for a moment that there was an accumulated of dirt on the switch contacts which allowed a very slight current to flow across its contacts. Or that someone might be able to "brush by" and accidentally touch something that was connected to \$R_1\$ or directly to the base of \$Q_1\$ through dirt accumulation or other unintended means. Or also just realize that there may be very small capacitances which can hold a slight charge that might, for a moment, provide enough current at \$Q_1\$'s base to cause it to turn on momentarily.
To cheaply provide some added insurance against all of those cases, a "weak" pull-up resistor can be added to the circuit. This provides a way of pulling unintended charges away from the base of \$Q_1\$. Only when \$SW_1\$ is engaged fully and with definite intention will (then) enough current be supplied to overwhelm \$R_2\$'s weak action and turn the BJT on.
But both work when used in ideal situations. But adding the pull-up is cheap and it helps reduce the likelihood of unintended switching events when the situations are not ideal. (Which is most of the time.)

In normal situations (manual switch), there is no meaningful difference in speed between the two schematics. But even when a microcontroller is used to activate \$Q_1\$ as a switching device, there will usually be no meaningful difference in speed. When using a BJT as a highly saturated switch, like this, there is charge storage in the BJT that takes a certain amount of time to remove and a weak pull-up doesn't change it a lot. So for most uses, you can consider them of equivalent speed.
However, \$R_2\$ does improve the speed a tiny bit in practice. The stronger you make it (up to a point), the more improvement in speed you get. (This improvement will be in the turn-off time and NOT the turn-on time, which remains fast either way.) It still won't be that much of an improvement, though. In the end, you might normally see a few microseconds of delay in a turn-off and the improvement might be 10% or 20% of that time. (The actual percent of improvement will depend on details about the BJT device, itself.)
Here's a simulation to show you what's happening on the turn-off side:

There are both cases shown above. The green trace shows you the base current of the right-side PNP BJT and a dark blue trace shows you the base current of the left-side PNP BJT. The dark blue trace is almost invisible because it is everywhere exactly the same as the green trace, except for a tiny bit that you may be able to see near a "porch" that appears when both BJTs are being turned off. This porch takes less time for the green trace (faster circuit) than for the dark blue trace (slower, left-side circuit.)
The red trace is for the faster (right-side) schematic's output. As you can see, it still takes a while for the BJT to come out of saturation. The light blue trace is for the slower (left-side) schematic's output. And as you can see here, it also still takes a while for the BJT to come out of saturation. But as you can see it takes even longer because \$R_2\$ is not present to help. Either way, it's not a LOT of difference.

In most cases, this is just added insurance and has nothing to do with improving speed. In some rare circumstances using a BJT as a switch like this (and as a hobbyist life it has never been a consideration), you might need that extra few hundred nanoseconds of speed in the turn-off time. (Perhaps to slightly improve the accuracy of the output duty cycle?) But there are better methods then.
Comment: I happened to have a low side MOSFET on the board, but I'm sure the 5V tolerant I/O pin would have worked too.
Comment: I built the circuit to do high side switching on LEDs with a micro and it works great.
Comment: @joeforker Glad to hear it!! (It's nice when things that are supposed to work, actually do work when you build them.) In the case where the high side voltage is higher than the MCU's I/O pin voltage, you'll need another BJT. But that's a different story.

Does a PNP transistor pull up itself?


Card Attachments disappeared after Trello Update this morning!

Question: I updated my Trello today per the Notification, afterwards all my Card Attachments were gone.  Icon on the Card still shows the number of attachments for each card, however when I open a card, it states no attachments.  Anyone know of a bug?
Comment: Update on this.  I went back to start re-adding the attachments and when I upload 1 on a card, all missing attachments for that card reappear.  But I must then delete the one I just added as it is now a duplicate.
Answer: I'm guessing you saw this bug, which has since been fixed


Great Smoky Mountains National Park

Question: Why is it that Spring in the Great Smoky Mountains National Park is the peak season for forest fires both man made and lightning caused when it rains so much in the spring and it is so wet there in the Spring so how can forest fires possibly form and even be the peak season for fires there?
Comment: Possibly because all of the park may not be wet at the same time.
Comment: Do they have a dry winter? If so, it could be like California: the fires are at the beginning of a wetter season due to lots of dry/dead vegetation + stronger winds & lightening.
Comment: No they don't have a dry winter. It rains there a lot in the spring and summer a little more so in the spring than the summer but winters aren't dry even though it rains less there than in the spring and summer. Fall is the driest season there with the least amount of rain. But even then it still rains a good amount in the Fall there
Answer: With fires, the forest cover and climate generally control what are the worse seasons for fires, add to this the human factor, accidental and prescribed burning.  For purposely ignited fires spring and fall are the major seasons, spring because the ground is often still damp but the fine fuels on the surface are dry enough to burn, making the fire easier to control, fall because the fuel is generally dry but the weather is getting colder and wetter with winter approaching making the fire easier to extinguish and reducing the time required to monitor the fire.
In the case of the Smoky mountains, it appears that the weather is more damp, humid, in the winter and summer, additionally, summer has more moderating green moist vegetation that is not present in the spring or fall.  The following excerpt from the Great Smokey Mountains Fire management plan directly mentions the spring and fall seasons (See page 16).  Great Smoke Mountains Fire Management Plan

The annual fire weather cycle is more a function of seasonality and temperature than
  precipitation. Warm temperatures and exposed fuels during the dormant seasons of spring and
  fall condition the fuels for easy ignition and spread. In contrast, during the winter, cold
  temperatures prolong the wetting effect of the precipitation, and in the summer, heavy sheltering
  of fuels by tree foliage (coupled with high humidity) makes for poor ignition conditions under
  normal circumstances. Exceptions to the norm have occurred. Atypical dry spells in any
  season can result in more fires and/or unusually higher fire intensities. In 1987, and again in
  2007, extreme summer-time droughts made the normally fire-resistant hardwood forest
  susceptible to fire. Numerous lightning- and human-caused fires in and near the Park burned
  thousands of acres during summer months in those years.

Great Smoky Mountains National Park


Auto login via email / newsletter

Question: I'd like to send customers an email with a link to their (private) profile/account page.
This link should contain login credits so they can click to go to their profile page and log in at the same time.
How can I add login info to a url?
Answer: In the admin, if you go to the page of the user (members -> view all -> [select desired user]) You can copy the url of that page and send that to them. This allows them to log in and go directly to their account page.
As far as automatically logging them in, that is dangerous and would be a big security concern. You should never automatically log someone in via and email link. But if you are willing to try you need to enable html in your emails (not all clients support it) and you would have to make hidden inputs to tie directly to your cp. (please do not do that though)
Edit: I do not believe that you can tie a form directly to the login of EE. There are session keys for the form that are for authentication to check that you are in a browser and on that specific page.
Comment: Thanks, i wouldn't have them log into the CP, but to a personalized, private page.
Answer: Expressionengine logs you in by checking your username and password in the database and then creating a session (a cookie and a corresponding database entry).
It would be pretty simple to add another field in the members table and create a GUID for each member that you use to identify and authenticate the member.  You can create either an EE plugin or a CI or raw PHP script that provides a landing page that accepts the GUID as a url parameter and performs the pseudo login.  You can then send the url with the GUID via a mailmerge, mailchimp or whatever to your users.
I'm sure you know, but you should do your own risk analysis to be sure that this method is secure enough.  Make sure whatever group these members are in doesn't have control panel access etc.
Answer: You can use the add-on LogMeIn to do this exact thing. All it needs is the username to feed into the template. You can achieve that in a variety of ways, some being more secure than others.
The most direct (not so secure) method would be to do the following:
1) Create a template, for example profile/logmein, and put the LogMeIn tag in it like this: {exp:logmein:now username="{segment_3}"}
2) Then have the email you send include a link to the logmein template with their username in the last segment: site.com/profile/logmein/username
Once the user hits that page it'll redirect them to wherever you have that set up.
It's really too simple.
Comment: It almost sounds too good to be true... Thanks!

Auto login via email / newsletter


Symmetric functions on three parameters being perfect squares

Question: Is it possible for $x+y+z, xy+yz+zx$, and $xyz$ to be perfect squares at the same time for positive integer values of $x,y,z$?
Comment: Yes.  $(x,y,z) = (80,225,320)$ works, giving $25^2$, $340^2$, $2400^2$.  There are infinitely many other (non-proportional) examples.
Comment: Thanks Noam. How do you find them?
Comment: You're welcome.  The initial solution is found by exhaustive search (scaling from $(16,45,64)$, though this turns out not to be minimal).  For further solutions and details, see the answer I just posted [which also explains why I'm editing to include the "k3-surfaces" tag].
Answer: Yes.  By straightforward search the smallest example is
$\lbrace x,y,z \rbrace = \lbrace 45,64,180 \rbrace$, with
$$
(t+45) (t+64) (t+180) = t^3 + 17^2 t^2 + 150^2 t + 720^2.
$$
Given any solution $(x,y,z)$ we may produce infinitely many others
(other than the trivial scaling $(c^2 x, c^2 y, c^2 z)$) by using
the theory of elliptic curves to find rational $z'$ such that
$x+y+z'$, $xy+yz'+z'x$, and $xyz'$ are all squares, at which point
$(d^2 x, d^2 y, d^2 z')$ works for any integer $d>0$ such that
$d^2 z' \in {\bf Z}$.  For example, in $\lbrace 45,64,180 \rbrace$
we may replace $64$ by $(460163992/28591599)^2$, and then multiply
through by $28591599^2$ to obtain the new solution
$$
\lbrace
 28591599^2 \cdot 45, \phantom{+} 460163992^2, \phantom{+} 28591599^2 \cdot 180 \rbrace .
$$
A complete parametrization is not possible, because it would be
tantamount to a rational parametrization of the surface
$$
S: xy + yz + zx = r^2, \phantom{and} (x+y+z)xyz = s^2
$$
in projective $(1:1:1:1:2)$ space, and that surface is K3.
If I did this right, $S$ is a "singular" K3 surface, i.e. has Picard number
$20$ which is maximal for a K3 surface in characteristic zero, and the
Néron-Severi group ${\rm NS}(S)$ has rank $20$ and discriminant $-48$,
and consists of (classes of) divisors defined over ${\bf Q}(i)$.
It is actually quite common for
natural Diophantine equations to give rise to K3 surfaces of
maximal or nearly-maximal Picard number, but that's a story for
another time.
Comment: At an appropriate time, I'd love to hear the story of why natural Diophantine equations often give rise to K3 surfaces of maximal or nearly-maximal Picard number.
Comment: It's mostly a surprisingly long collection of examples; I don't claim to have a structural "why".  Heuristically what seems to happen is that — assuming the Diophantine equation is of the right dimension and complexity to yield a K3 surface in the first place — there's enough divisors coming from trivial solutions, and often enough symmetry, that there's barely enough room for the Néron-Severi lattice to accommodate them all under the constraint of rank at most $20$.
Comment: (And often there are also singular points whose resolution contributes to NS.)


What are some options for a 90 degree turn in a long hallway for laminate flooring?

Question: We are continuing down the hallway from the living room. There's a 90 degree turn to go past the four bedrooms. How do we make that turn or do we go cross wise in that part of the hallway? I was thinking a staggered corner by running all the way then turn the corner and go down that section. Then repeat the second row staggering at the corner.
Answer: It's really a matter of personal taste and the other aesthetic aspects of your home. There are basically 3 options, as I see it:

Keep the flooring oriented the same direction throughout. Obviously cross-hall boards may look odd depending on the style of the flooring. In a very short hallway run it may not be a concern.
Use a T molding to transition over a change in direction, orienting the seam perpendicular to the hallway. This is probably best done inline with the view from the living room, where it'll be less noticeable.
Use a T molding at a 45 degree angle between the wall corners to essentially miter the entire hallway (like a picture frame). You'll see this more from the living room, but it gives more symmetry from the perspective of the hallway itself.

I don't think a herringbone turn is possible with most laminates due to the disparate tongue and groove shapes. That might look nice, though.
You'll need to decide which approach results in the least disturbance to the ambiance created by your new floor.


Is "Attero Denarios" correct for "I destroy coins"?

Question: I am writing a series and looking for a title particularly in Latin. Got the idea from Sabaton's 'Attero Dominatus' and have read that thread.
'Attero Denarios', seems be correct for 'I destroy coins' or 'I weaken coins'. But open for correctness or something better.
For coins is short for cryptocurrency coins, so is of the cyber realm.
Comment: A denarius is a specific coin; you'd want *nummus* (acc. pl. *nummos*). *Attero* is also a weird verb to use.
Comment: Yeah... nummus though correct just doesn't sound fitting, especially to a person not knowing Latin. For example,  seems like related to being numb.
Answer: As denarius is a specific coin and and atterere only means destroying in the sense of rubbing, attero denarios would mean primarily something like "I rub dimes against each other" and can be stretched to "I ruin dimes by rubbing them".
In order to give an impression of the tone, I translated denarius to "dime" just because they are both specific coins; the two coins are unrelated.
My answer to the question is no: attero denarios is not correct for "I destroy coins".
The words have meanings in the right general direction, but they don't add up to a good translation.
Be careful with what you consider synonyms.
The most straightforward way to say "I destroy coins" is nummos deleo.
The word nummus is a general word for a coin, and delere is good general word for destroying.
(It is not the only verb for destroying; see this answer for a list or consult any good online Latin dictionary.)
In a comment you objected to nummus as it sounds a bit like the English "numb".
Such coincidences happen when translating between languages, and I don't see any reason to be alarmed unless it suggests something particularly obscene or counterproductive.
If you want to worry about such things, you need to know your target audience's languages (not only English!) and have several experts help you out.
I see nothing wrong with using nummus.
Comment: *Deleo* is off course fine; if you are looking for other choices for "destroy," Latin does not let you down; I recommend [this old answer](https://latin.stackexchange.com/questions/13649/how-do-you-say-this-phrase-in-latin-venisti-vedisti-you-destroyed/13653#13653) of mine.
Comment: @SebastianKoppehel Good point! I added that and a dictionary link for more breadth in vocabulary.
Comment: Thank you for that explanation, it has given me something to chew on for my long drive. I think I can see how it may look to have more than one meaning.  As is for, destroying coins, but also as, numerous destruction, kind of playing off it as a pun in a way. I do like when a title may appear to have more than one meaning.   I do agree rubbing coins doesn't sound right when translated, just seems to sound decent.   I may go with Nummos Destruo.... or similar,  Nummo Destruo, Nummos Destruos... or just as it was stated. Got to think on it. Thanks!
Comment: @MrPatrick I'm glad to be able to help! *Destruo* is certainly a possible verb. Regarding forms of the noun: Use either *nummos* (plural) or *nummum* (singular) but not other forms like *nummo*. If you use other cases than accusative, you get things like "I destroy with coins" or "I destroy for coins" or "I am a coin and I destroy". Also, I recommend taking a look at [our quick tour](https://latin.stackexchange.com/tour) at some point to see more functionalities.
Comment: Thanks again. If I ever write a satire, I would definitely go with "I rub dimes." Or, "I am a coin and I destroy" (Translated)


Sellotape caused discoloration on the wooden surface

Question: As seen in the picture it seems that the sellotape removed some of the color on the furniture surface. There doesn't seem to be any adhesive left on the surface. What are my options to repair this?
Wood
Comment: Looks like they used a tinted varnish rather than staining the wood. You could try layering on very thin coats of another tinted varish, feathering the edges and progressively working toward an approximate mstch..
Comment: would sth like restor a finish help with this?
Answer: hard to tell from the photo, but I think you have old varnish, just aged and oxidized.  take some acetone and wipe the area at the edge of the ripped area.  if you wipe it and get some yellowy residue, you probably have old varnish or lacquer.  if not, wipe with methanol or methyl hydrate.  if this produces yellowy residue, you have old shellac.  if it doesn't wipe with either process you may have a urethane that's tinted and it would need sanding and refinishing.
either way, if it wipes off with either chemical, just wet feather the edge and then let dry.  you should see a lighter area when it dries, but with a smooth transition.  then you can topcoat.  don't try to match colours as the area will naturally darken in the stripped wood section when you coat it, and then further as it ages.
Answer: I have had great luck with this product. I usually do not advocate specific brands but I know of no other product like this. It just wipes on, it's like an oil polish which blends away imperfections and discoloration. You should apply it to the whole area, but you will not notice that color change afterwards. They make it for dark wood or med./light wood, your's looks like med./light.


Why does Yukio Okumura sacrifice Rin?

Question: Title says it all: why does Yukio sacrifice Rin? It was in the last couple episodes, when Yukio became the Paladin.
Answer: Because Okumura was told by his grandpa that if they destroy "the other world", he and his brother can turn back to human. And to open the gate, they need a lot of blood of the devil, so he sacrifice Rin. This is revealed in episode 22.
Answer: Yukio didn't know that Rin was going to be sacrificed. As you can see in the second last episode of Season 1, Yukio was surprised that Rin's blood was going and he had said yes because he wanted to protect Rin by turning him back into a human, but when he saw that Rin was going to die, he stepped in.

Why does Yukio Okumura sacrifice Rin?


Subfloors and stairs

Question: It seems to me that many people put subfloors in their basements (and other floors), but what do they do when it comes to the existing stairs?  I'm code says that you can't have more than 3/8" difference in the steps, and most subfloors are more than 3/8" thick (especially with flooring on top of it).  What do most people do to fix their stairs?
Comment: Honestly I have done the exact same thing more than a few times and have never had an inspector say anything about the "short" landing.  Why don't you call the inspector and ask him what he wants done - if anything?
Answer: Technically the floor below the last step is considered a landing, and stair risers may be measured from tread to tread, not landing to tread. The 3/8", depending on your exact local codes, may not apply to the tread to landing at the top and bottom of the stairs. It may have a different standard, or it may have the same standard.  Of course if the height difference is large it's going to cause tripping and stumbling, but for minor changes a little more than the 3/8" code it may not make much of a difference or present a safety hazard.
However this can be fixed by putting a thin board on the bottom tread, then half that thickness on the next tread, then half that thickness on the next tread until the requirement is met. If your subfloor is under 3/4 inch higher than it used to be, then it should only take one 3/8 piece of wood on the bottom step to make the first two risers only 3/8 shorter than the rest of the risers, for instance.
You can, of course, replace the stairs or stringers entirely to meet the code requirements and account for the new floor height at the bottom landing.  If the height difference is large enough that you'd have to modify more than half the stairs you'll probably have less work replacing the stringers than adding height to each tread.
Comment: Adding to the stairs in that manner seems like it'd be a real pain.  For one, you have to cut enough wood to cover all of the steps, plus what do you do with the existing nosing?  Also, won't all of the stairs be of slightly different heights?  That'd be weird.  I was thinking that I could cut new stringers.  Seems like a pain, but at least I could re-use the existing stairs and they'd all be the same height.
Comment: @User58220 Your method is a linear change in riser height, which may be better simply from the standpoint of maintaining an even riser height top to bottom with no variation.  My method will cause the bottom few treads to be a different height, but still within code, and require fewer modifications overall.  If you're going to go to the trouble of adding height to every tread, it'd probably be easier simply to recut the stringers anyway.  Still, 1.5" height adjustment for such a short stairway would suggest replacement of the stairs.
Comment: It seems like if you didn't adjust the stairs than that last step-to-landing would be really short.  Let's say that I put in 1" of XPS insulation + 2x4 (1.5") sleepers + 3/4" plywood or OSB.  That's 3 1/4" just for the sub-floor!  If the existing risers are 7 3/4", that's cutting the stairs 40%.
Comment: @Adam Davis:  I think your corrections are a little off.  Say you have 6 risers, and put 1.5 inches on the bottom "landing".  Then you need 0.25 inches on the first tread from the top, 0.50 inches on the next one down, and 0.75 on the next.  Each new riser will be equal, and 0.25 inches less than before...  In a real case, the numbers will never be that neat...
Comment: @2LeftThumbs Cutting new stringers is an obvious fix which I didn't mention, so I'll add that in.
Comment: What do you mean by "replacement of the stairs"?  And with the additive method, how do you account for the rounded nosing of each tread?

Subfloors and stairs


Bash script to remove the oldest file from from a folder

Question: I want to make a bash script to delete the older file form a folder. Every time when I run the script will be deleted only one file, the older one. Can you help me with this?
Thanks
Comment: It is worth reading [this](https://www.ostechnix.com/find-delete-oldest-file-x-files-directory) article. It explains the command `ls -1t | tail -n +11 | xargs rm -f`
Answer: As Kos pointed out, It might not be possible to know the oldest file (as per creation date).
If modification time are good for you, and if file name have no new line:
rm -- "$(ls -rt | head -n 1)"

(if you don't have control over the file names beware of parsing ls output)
Add the -A option to ls if hidden files are also to be considered.
Since GNU coreutils 9.0, GNU ls now has a --zero option to list file names NUL-delimited, so with those recent versions and bash or zsh as the shell, a version that would work with any file name would be:
IFS= read -rd '' file < <(ls --zero -rt) &&
  rm -- "$file"

Those newer versions also have a --time=birth which can sort by birth time on some systems and filesystems (including recent GNU/Linux ones on most native filesystems).
Comment: what if I have space or newline in file name?! nope it will crash /-:
Comment: How to delete last 10 oldest files? as the above comment gives error for more than 1 file
Answer: It looks like you're fine with deleting the oldest modified file instead of the oldest created file;
Wherever the GNU implementations of the stat, sort, sed and xargs utilities are available, I consider this to be the safest method, as it won't break on filenames containing newlines:
stat --printf='%Y %n\0' ./* | sort -zn | sed -z 's/[^ ]\{1,\} //;q' | xargs -r0 rm

stat --printf='%.Y %n\0' ./*: prints a NUL-separated list of the last modification's time (with as much precision as available) followed by the file's path for each non-hidden file in the current working directory, prefixed with ./ to avoid problems with filenames starting with - (-- wouldn't work for GNU stat where a file called - would still be interpreted as meaning stdin);
sort -zn: sorts the list numerically using NUL as the line separator;
sed -z 's/[^ ]\{1,\} //;q': removes the first occurrence of a string containing one or more characters other than space followed by a space from the first NUL-terminated line and quits after processing and printing the first line;
xargs -r0 rm: passes the NUL-terminated line to rm as an argument, only if there's such a line;

% touch file1
% touch file2
% stat -c '%.Y %n' ./*
1447318965.123211222 ./file1
1447318965.784234234 ./file2
% stat --printf='%.Y %n\0' ./* | sort -zn | sed -z 's/[^ ]\{1,\} //;q' | xargs -r0 rm
% ls
file2

Answer: To determine the oldest file according to "Access Time" or "Modify Time"?
If it is the second, then just use the following command:
rm  ls -l --sort=time  | sed -n 2p  | awk '{print $NF}'

Comment: As you noted already, `ls`'s output can be sorted either by atime or mtime, as creation time is not supported by the kernel (and a method to remove the last created file is what the question is asking for by the way); but mostly this can break 1) On a file containing a newline in `ls`'s output. *Never parse `ls`*. Use `find`. 2) On a file containing a space or a tabulation, since `awk '{print $NF}'` will only print the last space/tab-separated word.
Comment: The command is incorrect (tries to execute "rm" first).
Answer: In zsh, working regardless of what character or non-character file names may contain:
rm -- *(Om[1])

Where Om Orders in reverse by last modification time (from oldest to newest like for ls -rt), and [1] limits the expansion to the first match.
Or:
rm -- *(-Om[1])

To consider the modification time after symlink resolution (like with the -L option of ls).
rm -- *(Om[1]^/)

to exclude the files of type directory (which rm can't remove without -r anyway), with again a - prefix if you want that type and the age to be considered after symlink resolution. See also the D qualifier to also consider hidden files, or . for regular files only.
In bash or any other shell, just use zsh -c 'those codes above'.


Why is the wave equation applicable to the EM wave?

Question: I was first exposed to the wave equation in which it was derived from a model of 1-D string. But afterwards the derived wave equation somehow was universal and thus applicable to EM waves.
But because wave equations was found out of physical relationship of materials involving mass and tension, I don't see how it is naturally applicable to EM waves which don't have qualities of materials.
So, if wave equation is universal, it seems that it must be speaking of geometric nature. So is there a more general derivation of wave equation that justifies its generality?
Answer:
The wave equation is an important second-order linear partial differential equation for the description of waves—as they occur in classical physics—such as mechanical waves (e.g. water waves, sound waves and seismic waves) or light waves. It arises in fields like acoustics, electromagnetics, and fluid dynamics.

The study of electricity and magnetism took some century. During that time "laws" were found that were dependent  on the experimental observations. Then came Maxwell's equations, which used those laws as axioms to develop his theory, and that theory came up with the wave equation for light.  So the simple answer is : "the wave equation for light depends on experimental observations" on electricity and magnetism, and is continually validated with new observations and experiments.
Now your:

But because wave equations was found out of physical relationship of materials involving mass and tension, I don't see how it is naturally applicable to EM waves which don't have qualities of materials.

Was a question for the first researchers too, who wanted a medium for those waves to propagate, the luminiferous aether. That brought the light differential equations in line with your thinking. The Michelson Morley experiment showed that light waves propagated in vacuum.

So, if wave equation is universal, it seems that it must be speaking of geometric nature.So is there a more general derivation of wave equation that justifies its generality?

It is a second derivative differential equation. If we accept that geometry was the start of mathematics all differential equations describe a geometry. How and in what spaces depends on the use that is made for the various models.
Answer: They all come from Maxwell equations.  Solutions of Maxwell equations provide oscillations of both electric fields and magnetic fields (not propagation).   The questioner stated in the first sentence that one dimensional string problem gives wave differential equation, for which solution need not give a propagating wave but only an oscillation.  Since Maxwell equations also provide wave equations, em waves also satisfy wave equations, if waves are produced.  So, answer is ready.  However, I ask the questioner to assume the followings (for which a basic theory may be found in a book entitled "Planets and electromagnetic waves"), for more understandings.  The followings are based on mathematical interpretations for solutions of Maxwell equations.
(1) If current exceeds some limit (depending on source of production), then magnetic field waves are produced.  These waves are called in general radio waves.
(2) If voltage exceeds some limit (depending on source of production), then electric field waves are produced.  These waves are called in general light waves.
(3) If current exceeds the corresponding limit and the voltage also exceeds the corresponding limit then both magnetic field waves and electric field waves exist separately; not in combined form (that is, not as everyone gives a figure for em waves with two component waves).
A peculiar situation for (3) is the Hertz's experiment.  It may also happen that both types of waves may not be produced, and only oscillations may happen when the limits are not reached.
Answer: The answer to your title question is that it follows from Maxwell's equations.
The answer to the question why the wave equation is universal is that it follows from special relativity. The wave equation is the wave form of the special relativistic energy momentum relation $E^2=m^2+p^2$ (or $E^2=m^2c^4+p^2c^2$ if you don't like setting $c = 1$). Inspired by this and the massless wave equation following from Maxwell's equations, De Broglie proposed to make the replacement $$E \rightarrow \frac{\hbar}{i} \frac{\partial}{\partial_t}$$ and $$\vec p \rightarrow \frac{\hbar}{i} \vec \nabla ~.$$
This gave birth to wave mechanics and the wave equation, to which any non-interaction free particle wave function is a solution.


Is there much snow around the Adirondack railroad in the end of February?

Question: I'm planning to take a railroad trip from New York City to Albany via Amtrak (Adirondack route) end of February, for scenery purposes, but I am a bit worried I will only see lots of snow, so is it snowy on that route this time of the year?
https://www.amtrak.com/routes/adirondack-train.html
Comment: In my view a rail trip is worth it any time of the year and snow covered landscape may well be great for you as well.  That is the opinion based answer.
Comment: Seeing snow for 3-5 hours would be awesome for many people. Me included. I may pay more to see more snow - just saying :-)
Comment: yeah I like rail trips too but with 3 to 5 hours of snow? not sure it would be worth the price, I might just cut it short and turn back :)
Comment: That is why I edited the header of the question for you, so you get more relevant information.
Answer: In February in upstate New York, you will very likely see snow on the ground. The 2022-2023 winter has been a little warmer than normal, but expect some snow.
If you are concerned that all you will see out the windows of the train is a solid wall of snow, that is very unlikely. That would mean that there would have to be about 7 feet (2.1 meters) of snow on the ground. Even in the snowiest part of the winter that is an extraordinary amount of snow for the Hudson Valley.
In late February I would expect it to look a bit like this:

You may have between 2 – 3 inches (5 – 7 cm) on the ground.
There are a few options in heading north from New York City along the Hudson. You can take Metro North (commuter rail) as far north as Poughkeepsie. You will see the southern part of the Hudson Valley and there are some things to see and do in the area, such as Hyde Park (FDR Presidential Library) or the Culinary Institute of America. Since this is commuter rail the trains run often.
Albany is further north, Amtrak runs regular service here. On average it will be about 1 train per hour each way so you can be a little flexible with the travel. The Adirondack goes through Albany, as well as the Ethan Allen Express. Empire service also serves Albany.
Once you go north of Albany on the train you do need to be aware of the schedule. The Adirondack runs once a day each way. If you go north of Albany on the train, you will be spending the night since the next train will be on the following day.
It really depends on what you want to see, how long you want to take and how much snow is too much.
Comment: Thank you, I decided to just ride the train straight to Washington DC, less snow, less time, and sort of cheaper :)


How does The Expanse solve the big problems of initially settling on Mars?

Question: The big problems of settling on Mars are high radiation (worse than anywhere on Earth) and short period extreme radiation, moderate and extreme cold, need all basics of life and energy supply and I read its wiki and it says nothing about that. Do they say anything about solving these fundamental problems?
Comment: "*about three generations after the first colonists dug into the rock and soil*" - So they dug down to avoid the radiation. This would also help deal with the heating issue; https://expanse.fandom.com/wiki/Mars
Comment: Currently Martians all live under domes, and many necessary resources are shipped in from elsewhere, like from the belt. The first colonists probably lived in some sort of semi-temporary shelters while building the permanent domes, but that's just a guess.
Comment: How the current domes (assuming first domes were the same)  solve the problems- seem pretty useless against radiation bursts and provide no energy solution?
Answer: Radiation: the domes were built to keep out radiation, and cities were built underground while terraforming was ongoing.
Energy is already wholly satisfied by fusion reactors.
From Cibolas Burn:
Prologue quote confirming no magnetosphere was set up on Mars:

"It's going to be generations beefore anyone walks on the surface here! We don't have a fucking magnetosphere!" - Bobbie Draper

Chapter Two quote from Elvi's thoughts, emphasis mine:

When the first colonies had begun on Mars, the perimeter domes had been a question of survival. Something to hold in air and keep out radiation.

So you can see that the domes were in fact built to keep out radiation, and the ones they brought along for the expedition to Ilus/New Terra were very similar to the ones originally used on Mars.
Chapter Twenty-Five, emphasis mine:

"The newer areas, the ones built after we had some idea of what folks need to feel happy, were built different. No more narrow stone corridors." - Alex

Epilogue, Avasarala's thoughts:

the entire city--the entire planetary network--was built like a fucking tomb. The absence of a magnetosphere had made Mars' first priority protection from the radiation.

So in the initial build on Mars, they tunneled into the ground, and continued doing so to hide from radiation -- the rest of that quote describes how they figured out how to make it feel less horrible. It should be noted that they did the same on Ganymede by burrowing most of the place underground, though with the exception they weren't attempting to terraform.
The rest of this is speculation but:
Food is shown to be pretty simple to make with rice and yeasts and mushrooms and other such stuff being the staple of outer planets diets, it's likely, though I can't prove it, that they used that and imports from Earth to start -- at the time of colonization they were still a UN colony so they would have a vested interest in keeping them alive, and since they were a portion of the population from Earth, the capacity to do it.
Water is tricky, reclamation systems are shown to be pretty solid by the first book, but who knows what of that was advancements after the initial colonization. There are also some ice deposits on Mars that could help.
Comment: Bobbie, being a simple marine, must be unaware that the existence of a magnetosphere is irrelevant on human timescales. It's the mass of the atmosphere that's going to block most of the radiation. Atmospheric loss due to solar wind erosion takes place on geologic timescales, so if you can thicken it in the first place, topping it off isn't an issue.
Comment: @Harabeck I thought magnetospheres were important for blocking radiation? I admit I'm fairly ignorant to the real world problems of colonizing mars.
Comment: It does block some, but as I said, the greater part of the effect comes from the mass of the atmosphere itself. The current very thin atmosphere on Mars is already thick enough such that a hypothetical human on the surface would get a radiation dose about equal to being on the ISS: https://www.jpl.nasa.gov/images/estimated-radiation-dosage-on-mars/
Comment: @Harabeck good reading, thanks, I think it might be the Author's ignorance rather than Bobbie, they talk about it at least once with Avasarala


Static Methods and Multi-Threading

Question: Can there be any problems in having a static class with only static methods (no properties, fields or anything else) and having several threads accessing those methods at the same time?
Answer: As long as there is no shared state or external resources, then: no risk whatsoever. You've mentioned "no fields" etc, which is good - so as long as they aren't indirectly talking shared state (perhaps via some common argument that isn't thread-safe in this scenario, or via some other utility methods that it calls, which share state in messy ways), you should be fine.
Answer: Yes, there can. It depends how your methods are written. Now, if there are only static methods, this would theoreitcally mean fully reentrant code without problems.
Answer: Potentially, yes.
Imagine:
<code>class Logger
{
    WriteToLogs(string msg)
    {
        // Write data to file1
        ...

        // Write data to file2
        ...
    }
}
</code>
Now imagine you have multiple threads trying to call WriteToLog() simultaneously. What happens when thread2 gets scheduled before thread1 is done writing to the log files? You could end up with all manner of data corruption in this case.
Static classes and methods do NOT inherently provide any form of thread synchronization, locking, etc. That's for YOU to design and implement.
Answer: Yes, for instance singleton implementations can be static but they have to implement mechanisms such as double checked locking to prevent multithreading issues.

Static Methods and Multi-Threading


What is a good strategy to make sure my Hiragana proportions are correct?

Question: I am teaching myself Japanese. To begin learning how to read and write in Japanese, I have started my studies with Hiragana. I know that I need to have specific proportions to make sure what I am writing is accurate and legible. However, the only paper I have at hand is lined, and I am not sure how I should know if my proportions are correct. Should I start getting graph paper to practice?
Answer: Graph paper might indeed be helpful, but just by being mindful of the proportions and flow you have already made a step that few ever do.
Ultimately, for legibility, the most important thing is that the characters are distinct, not so much that they adhere to form, but what I would recommend is getting any one of a great many books that focus on writing legible kana.
Simply getting graph paper is unlikely to help much if your form is sloppy to begin with.


My phone auto corrects things wrong

Question: How do I fix what words my phone autocorrects. For example, when I type the word "my" it corrects it to a capital M and when I type "lost" it corrects it to Los. I find this annoying and I have to go back and correct it every time.
Comment: How are you typing? WordFlow?
Comment: What's your phone's language?
Answer: I have the same problem, it's so annoying. I don't want to turn off autocorrect, I just want it to stop autocorrecting correct words to stupid words, eg long to Lon, if to I.
Just typing normally, not WordFlow.
Language is English (New Zealand)
Keyboard is English (UK)
I've just reset suggestions under keyboard\advanced, and retyped the same message (in messenger app) and it didn't correct long to Lon. Maybe that's fixed it, but I have done that before.
Answer: go to settings > keyboard > select language (keyboard) > turn off "correct misspelled words" DONE!
Comment: I think the OP wants to only keep certain words from being corrected, not all of them.

My phone auto corrects things wrong


Determination of displacement value in work formula

Question: A net force will always create displacement which approaches infinity if left undisturbed. So how to fix a value of displacement in work formula? Isn't a mention time interval necessary?  Or do we measure displacement only while force is acting on the body i.e the moment we stop applying force we also stop measuring displacement.
Comment: Related question by OP: http://physics.stackexchange.com/q/106570/2451
Answer: Usually problems are framed with  "before" and "after" states.  A book was here, a force applied, and later the book was there.  So there might be an implicit time interval, but that interval is not used in calculating the work.  Also, frequently, the object starts at rest and ends at rest, also providing a time interval, which again is not used in calculating work.  And the question can be phrased where the position of the object changes and continues to change.  In that case the question must specify an initial position and ending position between which work is calculated.  If you know how the position varies with time, $\vec{r}(t)$ then those positions can be found from a starting time and ending time.
But in any event, it is always the displacement that counts.  Time would be involved only as a means to identify the starting position and the ending position.
Comment: Summing up, you're almost always going to calculate work done between points A and B. If you want to know how much work is being done every second, that's called "power".


How to decide earth trace width in protection circuit?

Question: I am using the earth trace for protection circuit in telecom system. The protection circuit includes MOV, sidactor & fuse.  So please suggest How to calculate the Earth trace width.
Answer: The Track current capacity should be more than the MOV rating. So whenever the MOV acts, the fuse will open up safely without burning the tracks. You can use the resistance formula rL/A to calculate the current capability/dissipation of the copper track/conductor. You can improve the PCB track capacity by removing solder mask at that particular area.
Comment: Is it same in sisactor case also? what happen in sidactor case. or high voltage surge/spike case.
Comment: Yes. sidactor is also just like a back to back zener. The fuse should blown before the sidactor melts.


The biggest difference between medical statistics and biological statistics

Question: I am turning my work from medical statistics into biology statistics (fishery). I am wondering if there is a big differences in methodology or focus between these two fields? In medical field, we do a lot clinical trials or patient cohort studies, and we try to find associations by applying different types of significance tests and regressions. In the case of biology, there is very limited control of the sampling method. I have a feeling that the statistician's main focus would be to find the best way to sampling the data. Since I am really fresh in this field, can someone give me some ideas about the main focus, common methodologies in biology statistics? Is it really different from medical statistics? Do you have any nice books to recommend?
Thanks
Comment: One thing that they do in fisheries is to estimate the population size of a species using capture - recapture sampling methodology.  This is certainly different from what we do in medicine and as you point out sampling methodology is the key to this.
Answer: A couple things come to mind. As a disclaimer, my exposure to the statistics of Biology come from undergrad, and the occasional times my research overlaps with Ecologists.

First, "biological statistics" is a staggeringly broad definition. What might be routine for the microbiology people may be impossible for the population ecologists, for example.
As mentioned by Michael Chernick, there may be some sampling methods that fisheries research relies on that don't generally show up in clinical or population-based health research. Capture-Recapture comes to mind as one. The sheer difficulty in observing animals, who by their nature tend to be poor research subjects, and sample size issues. To take a somewhat extreme example, if one is doing Black Rhino research, your maximum N will be 4,240. But even for more abundant species, sampling large numbers can be extremely difficult.
Multi-species analysis is generally of greater interest to Biologists than medical researchers. While one could make an argument that some infectious disease research is multi-species analysis, attempting to estimate species diversity, predator-prey relationships, ecological networks etc. are fairly field specific.
While spatial statistics are becoming a thing in population health research, they're already a major aspect of biological research.

I suspect one of the hardest transitions will just be figuring out what counts as "standard" in the field in terms of analysis.


Hall-effect switch or photointerrupter

Question: I'm dealing with a small automation project, a rack and pinion system controlled by an Arduino card with two limit switches to ensure that the rack moves within the boundaries.
I have been trying both photointerrupters and hall-effect switches and both of them seems working without problems.
Which sensor should I use if I need to prioritize reliability and correct functioning in time?
I heard that optical sensors are sensible to environmental light, dust and aging. Which are the minus points of an hall-effect switch?
In this case I mean a unipolar hall-effect switch, will all the logic already build in, and that it can be connected to a microcontroller with a simple pull-up resistor.
Answer: I'm not hall sensor export but I can tell you I worked on a lot of consumer printers and scanners, and every time we did some kind of limit switch it was a hall effect.   So that's got to tell you something ;)
With a little online research I found this paper that talks about the hall effect and it's disadvantages.   They don't seem like they would negatively affect your project as you described it though.
Basically the limitations they mentioned are distance that the sensor works at (not really a problem for your limit switch), sensitivity change over temperature, and the presence of an offset voltage.   Again if you're just detecting when something hits an edge I'd guess it wouldn't be a problem.
Here's the article:
http://tec.upc.es/ie/transp_2010-11/HallEffect.pdf
Answer: Since you mention mechanical systems like rack and pinion, it sounds like there is dirt, dust, and grease in the environment.  In this case I'd definitely go for Hall sensors first.  Grease, dust, and dirt don't effect them.  Metal shavings can, but no worse than they would effect a optical sensor either.
Hall sensors can be fooled by external magnetic fields, but fortunately those fall off fast with distance.  Usually the magnet a Hall sensor is trying to detect is much much closer, and therefore produces a much stronger field than something like a motor that is multiple inches away.
Answer: Both are reasonably good solutions.
 You have identified the main factors that affect optical sensors. Worse case you may also manage interruption from spider webs and insects generally.
Hall sensors can be affected by extraneous external magnetic fields - but this is unlikely to be a problem in most situations. Occasionally you can get  unexpected fields of some magnitude at a distance from the source. I long ago saw a situation where a CRT display  would distort slightly whenever an elevator's power cable loop being trailed by an elevator on the other side of the wall swept past the rear of the screen. Separation was probably one metre or more.
Rats seem to like to eat PVC (specifically) cable sheath - regardless of the sensor or power system it feeds.
Either type of sensor will have a degree of hysteresis. This will depend on the specific design but the Hall sensor will probably have more hysteresis.  Hysteresis is useful in reducing the effect of operation/ de-operation "bounce" but may need to be allowed for. Worst case an excessively astute control system may shut a motor down if it does not detect sensor operation when motor power is applied.
An early US manned space launch*  was aborted when an umbilical cable pulled clear as the rocket started to lift and one pin pulled out before the other, creating an anomalous circuit configuration. The computer decided to abort the launch , turned off the motors and the rocket settled back onto the pad without disaster. The astronaut who had control of the escape tower firing kept his cool and did not trigger the eject system. Had he done so the mission would have been much delayed. As it was they launched successfully a few days later. You should be able to better than NASA when it comes to sensor bounce :-).
*- Gemini VI - The man in the high castle who made the brave and technically improper but  mission saving decision not to eject was Walter Schirra

Hall-effect switch or photointerrupter


Running Selenium on a server such as AWS Lightsail for ETL pipeline (MageAI)

Question: Quick introduction:
MageAI is an ETL pipeline tool that is similar to Airflow. I'm using MageAI to run cronjobs daily to crawl websites using Selenium. As of now, MageAI offers installation via docker-compose, kubernetes, and via pip install. When I install it via docker-compose or kubernetes, Selenium driver will not start. But when I use pip install, selenium works.
Scenario of how it is working via pip install:
I'm using AWS Lightsail to run my ubuntu instance. Within this server, I did pip3 install mage-ai selenium. This works and the selenium driver was able to start. The problem of this approach is that, since it is running on my server as a python module, it can be unstable as the server may go down.
Ideal scenario:
If I were to implement this via docker-compose or even kubernetes, it will become even more stable and scalable. But all the approaches I've done so far will result in the same error which is selenium chrome driver failed to start.
What I've done so far:

I tried extending from the MageAI docker image and installed all the necessary dependencies to run Selenium and Mage, it gave me the same error.
I tried extending from a ubuntu docker image, installed MageAI and Selenium via pip3 install, and it still didn't work.

Anyone would know how to successfully run selenium via docker-compose or even kubernetes?
Answer: version: '3'
services:
  magic:
    image: mageai/mageai:latest
    <mage-ai stuff here>

  selenium:
    image: selenium/standalone-chrome:latest
    environment:
      - SE_NODE_OVERRIDE_MAX_SESSIONS=true
      - SE_NODE_MAX_SESSIONS=10
      - SE_NODE_GRID_URL=https://0.0.0.0:4444
    ports:
      - 4444:4444

Use the selenium docker image and have it as a service.
Call within Python like so:
selenium_server_url = "http://<ip>:4444/wd/hub"

driver = webdriver.Remote(command_executor=selenium_server_url, options=chrome_options)

Make sure to add this after crawling to free up instances:
driver.quit()


How do I log in with a different account?

Question: My sister downloaded Draw Something on my phone. She thought she could log in as herself, play, then I could log in and play. Since she logged in first it will not let me log in. I have uninstalled app and reinstalled but it always pulls up her account.  This has gone on for months. She has it on her phone now and all her drawing drawings come up on my phone too.  I want to play too! How can I fix this?
Comment: Have you tried logging out by clicking on the cog in the top-right, then Account and then Log Out?
Answer:
On the top right of your screen there is an orance little options - wheel
Click on that and then tap the Account option.
Then scroll down and you will find a logout button.
Then you will be redirected to the main screen again and you will have to option to log in using either your email or a facebook account etc.
Comment: We have done that. Somehow her sisters account got attached to her Facebook account and we can not separate them. This happened with her old phone, now she has a new phone and we downloaded game and still same thing!
Comment: This game should really fix it's log in system...

Did you try logging out of your facebook application first?
Then go in drawsomething and log out and log in again WITHOUT using facebook. Just normal log in.


Windows Deployment Services cannot add localhost

Question: I have two servers running Windows Deployment Services (WDS) Servers running. This past week they both 'lost' themselves as WDS servers. What I mean by that is under the Servers group in the WDS role there are no machines listed. When I try to add localhost I get "Access is denied." After enabling tracing, I got a log with the error below.
[3812] 09:40:49: Started the asyncCreateServer in the background thread.
[3992] 09:40:49: -> CDataServer::CreateServer
[3992] 09:40:49: -> CDataCache::GetWdsServer
[3992] 09:40:49: Failed to get the WDS Server
[3992] 09:40:49: <- CDataCache::GetWdsServer=80070005
[3992] 09:40:49: Failed to create IWdsServer object
[3992] 09:40:49: Done Getting the Status of the Server.
[3992] 09:40:49: <- CDataServer::CreateServer=80070005
[3812] 09:40:49: There was some error during the Creating the Server 0x80070005

I'm at a loss for what caused this and how to bring the servers back up. What causes this error?
Edit: It turns out there was a DNS issue with the DCs. There weren't syncing correctly and causing authentication errors.
Answer: Check your services.  Specifically the Windows Deployment Services Server service.  If it's not running try and restart it.  I have seen issues where if WDS is running on a domain controller it will have problems starting on reboot.  Generally if I start it manually after the machine is stable it will fix the problem.  I'm not sure what the default start type is, but mine is "Automatic Delayed Start".


Two qubits system in polar co-ordinates

Question: I know that I can write a single qubit state in terms of polar co-ordinates $(r,\theta,\phi)$ on a Bloch sphere.
\begin{equation}
\rho =
\begin{pmatrix}
\frac{1+r \cos\theta}{2} &\frac{r \exp(-i\phi)\sin\theta}{2} \\
\frac{r \exp(i\phi)\sin\theta}{2} &\frac{1-r \cos\theta}{2} \nonumber
\end{pmatrix}
\end{equation}
Are this kind of polar decomposition exist for two qubits system?
Answer: A valid density operator is any Hermitian, trace 1, matrix (with complex entries) and all eigenvalues between 0 and 1. Any two qubit system may be represented therefore by a Hermitian, trace 1 4x4 matrix.
Your qubit representation could be rewritten, more suggestively as:
\begin{align}
   \rho &= \frac{1}{2}\left(\operatorname{I} + a_1 \sigma_x + a_2 \sigma_y + a_3\sigma_z \right)\\
&= \frac{1}{2}\left(\operatorname{I} + \vec{a} \cdot\vec{\sigma}\right)
\end{align}
Where $\operatorname{I}$ is the 2x2 identity matrix and the $\sigma_i$ are the Pauli matrices which are a basis for the space of Hermitian, trace 0 2x2 complex matrices.
Similarly to qubits we can decompose any n-dimensional quantum system using the identity and any basis for the space of Hermitian, traceless nxn complex matrices, a good choice of basis is the Gell-Mann matrices, and a construction in dimension n is given here.
Since they are Hilbert-Schmidt orthogonal you can find the coefficients $a_i$ for your state $\rho$ by taking the inner product:
$$a_i = \operatorname{tr}\sigma_i \rho$$
If you like you can then rewrite the vector $\vec{a}$ in any coordinate system you like (including polars).
There is a problem however because although every valid density operator may be rewritten in this way (with $\vec{a}$ being a n-dimensional sub-normalised vector) it is not the case that every operator of that form will be a valid density operator, the problem is that some of your eigenvalues will become negative and others will become greater than 1 to balance this out. This doesn't happen for qubits (any operator of the form $\rho$ I wrote above will be a valid density operator) and it makes working in higher dimensions fairly annoying sometimes.
Answer: Yes.  We can do this for any number of qubits  using N dimensional spherical coordinates.  For two qubits  we can write a general density matrix as a linear combination of direct products of Pauli matrices and identity , $$  \rho = \sum_{ij=0}^3 a_{ij}~ \sigma_{i} \otimes \sigma_{j}$$.
Here $\sigma_{0} = I$, and the rest are the usual Pauli matrices. For a valid density matrix $a_{00} = 1/4$. It can be shown that all valid density matrices lie inside a higher dimensional sphere centered at the coordinate corresponding to $a_{00}$. And these points can be represented using higher dimensional spherical coordinates. But unlike the Bloch sphere for a single qubit, all points do not represent valid density matrices as there can be matrices with negative eigenvalues. In all higher dimensions it is a convex body inside this sphere. A detailed explanation can be found here and here.
Answer: A one qubit state may be written in general as
\begin{equation}
|\psi_1\rangle=\alpha|0\rangle+\beta|1\rangle
\end{equation}
where $\alpha,\beta \in \mathbb C$ and there is the further restriction that $\langle \psi_1|\psi_1\rangle=|\alpha|^2+|\beta|^2=1$. However, only the relative phase between $|0\rangle$ and $|1\rangle$ is physically meaningful (quantum states are rays in Hilbert space), so we may factor out a phase $e^{i\theta}$.
Let's choose that phase so that $a=e^{-i\vartheta}\alpha$ is real while $b=e^{-i\vartheta} \beta$ can remain complex. Our state is then
\begin{equation}
|\psi_1\rangle=e^{i\vartheta}(a|0\rangle+b|1\rangle).
\end{equation}
The previous condition is now $\langle \psi_1|\psi_1\rangle=|a|^2+|b|^2=1$. We have three real parameters and one constraint and so we may say that our state $|\psi_1\rangle$ lives on the unit 2-sphere.
Analagously, a two qubit state may be written in general as
\begin{equation}
|\psi_2\rangle=a|00\rangle+b|01\rangle+c|10\rangle+d|11\rangle
\end{equation}
where we can choose $a\in \mathbb R$ and $b,c,d \in \mathbb C$ with the further restriction that $|a|^2 + |b|^2 + |c|^2 + |d|^2 =1$. In this case there are seven free parameters and one constraint so a two-qubit system would live on the unit 6-sphere and have a corresponding higher dimensional polar decomposition.
Comment: This answer is for pure states only, for general mixed states the problem is more complicated.

Two qubits system in polar co-ordinates


How to cite the author of an image if you're using numbers for citation?

Question: Like instead of saying: "image taken from [1]"
Is there a better way?
I'm using Zotero linked to Overleaf. Just in case this information is somehow relevant.
Comment: Why do you think this is not sufficient?
Comment: What's correct depends on the style guide for the publisher.
Comment: I just wanted to make sure this was the correct way to cite an image. Or if there was a more "formal" way.
Answer: Usually, "image taken from [1]" is sufficient.
However, you can be more specific within your citation style. For example, in IEEE style, you can add a particular detail, like a lemma, page numbers, equations, a specific figure, etc.:

[3, Th. 1]; [3, Lemma 2]; [3, pp. 5–10]; [3, eq. (2)]; [3, Fig. 1]; [3, Appendix I]; [3, Sec. 4.5]; [3, Ch. 2, pp. 5–10];
[3, Algorithm 5].

This can be useful when the reference is very large or contains several images that could be confused.
Anyway, you should consult the citation style and reference guide for your particular publisher.
NB: for images, ensure that you actually can reproduce an image in your publication (if that is your intention), as often the copyright might prohibit you from doing it without explicitly asking for permissions (IEEE case).
Answer: It will very much depend on the who/what you're writing for.  If this is towards your dissertation, you may have a lot of stylistic freedom concerning such things, or you may not.  Look up your school's dissertation policy.
If this is for a journal article, read whatever "Instructions to Authors" the journal offers.  If it links you to a style guide, read it.  If you still can't find a good answer, "image taken from [1]" is probably good enough.  If the journal feels that this isn't good enough, then it's the copy editor's job to correct it, either through an author's query in the gallies or just fixing it on their own to match their own style guide.
Comment: I have complete freedom. It's my Master's dissertation and no specifications regarding this were given.
Comment: Note that for copyright reasons, it's sometimes easier to "recreate" the figure instead of blatantly copying it, even with citation.  You would then say "adapted from [1]", but you may not need to get permission from the copyright holder to do so.
Answer: If you use natbib you can use \citeauthor or \citet to get alternative presentations that include (or are only) author names.


What to do when you have massive amount of data but you don't have enough computation power for training a machine learning model?

Question: For example, I have a massive amount of data, but I have limited computational resources and time to train on the full data. Other cases may include, I have huge amounts of 360-degree images, where I need to train on full-size images (without cropping down), but I have limited computation power (GPU, RAM, etc.), what can I do in those cases?
Answer: It's hard to answer this question without knowing what your goal is, but if your data is extensive, high quality, especially if it is labelled, and no similar dataset is publicly available, then publishing it freely with some kind of challenge could be very helpful if that's an option. Many organisations have the opposite problem: available computing resources but lack of data. If your data are new and interesting, I could imagine researchers wanting to use it. If people get good results, they may publish them, good for AI generally, and presumably also useful to you. If you do this of course you would need to publicise it a bit so people become aware of it.


Why isn't shutter priority mode automatically adjusting my aperture on my Nikon Coolpix P520?

Question: I've tried different ISO settings as well as auto ISO, but whatever the ISO, when I'm in shutter priority mode, the camera isn't automatically adjusting the aperture as I change shutter speed. I'm just ending up with very dark pictures. Is this a problem with the camera or am I missing something? Help!
Comment: Did you read the manual about it?
Comment: What is recorded in the EXIF info of the pictures? What ISO, what aperture, and what shutter speed? What were the lighting conditions where you took the photos?
Comment: Can you post an example image? Maybe you have exposure compensation dialed way down by mistake? Does the camera have any minimums or maximums that it's not allowed to go past (say, 1600ISO) - and you're simply in too dark a scene for the current settings?
Answer: It sounds like you might be selecting a shutter value that is too short (too "fast") for the amount of light in the scene you are photographing, even when the lens is opened to its maximum aperture and the camera is set to its highest ISO value.
Answer: Page 70 of the manual states:

When the subject is too dark or too bright, it may not be possible to obtain the appropriate exposure. In such cases, the shutter speed indicator or aperture value indicator flashes when the shutter-release button is pressed halfway (except when the M mode is used). Change the shutter speed setting or aperture value [or ISO value].

Emphasis mine, words in brackets are condensed information from the rest of the original paragraph.

Technical explanation:
For better explanations, please see "What is the relationship between ISO, aperture, and shutter speed?"
There are three values that have influence on the exposure: ISO, shutter speed and aperture. You chose to set ISO and shutter speed, which leaves the aperture to the camera's CPU.
The aperture is limited by the lens's design - your camera has a variable maximum aperture of f/3-5.9 (and a maximum ISO of 6400).
Think, for example, of the following situation:
The scene is very cloudy and therefore, you would need something like ISO 400 @ 1/100s @ f/4. You, however, have decided to use ISO 100 @ 1/200s - ISO alone would require 2 full stops, and the shorter shutter speed is 1 additional stop, so your aperture would need to be opened by 3 full stops, thus requiring f/1.4 instead of f/4.
Your camera, however, can only deliver f/3, resulting in a picture about 2 full stops too dark.
God - I hope that these calculations are correct. This always gives me a headache...


will 32 bit ubuntu 16.04 be wiped out if i install 64 bit ubuntu 16.04

Question: I am using 32 bit ubuntu 16.04.  however a program I wish to use is only for 64 bit machines.  How may I install the 64bit ubuntu 16.04 without wiping all my current files.  I am new to this so if it is possible, could you use a step by step answer to help me overcome this problem.
I am using a desktop computer and have been told it will handle 64 bit programs.
Answer: You would have to create a different partition for and alternate Operating System.  You can do this manually or allow the installer to create the new partition by splitting the space you're already using.
To do the latter, run the Install process the same way you did when you originally installed the 32bit version.  Then choose the option Install Alongside Other OS
The install process will give you a graphic screen where you can use a slider to determine how big for you want allocate for Linux.
Your current 32 bit installation would be over written if you choose to install on the same partition as it.  I'll overwrite the system files, but leave your /home folder.  This latter choice would leave you with one OS, the 64-bit version along with the files that you have in your /home/username folder.
The installation details from the official Canonical Site are here.
Creating the Live boot media
Steps for creating a USB or DVD installer are below.  Some people find one of the methods easier than the other.  But they both work fine.
Making a USB installer

1: Insert the USB stick with at least 2GB of space.
2: Type Startup Disk Creator in the Ubuntu Search Button and start the Creator   app.
3: (a) In the CD-Drive/Image field of the program select the ISO image.
   (b) In the Disk to use field select the USB device.
4: Click the Make Startup Disk button and follow the prompts.

Details are here.
Making a DVD installer
While there are many CD/DVD creators I use Brasero.
You can install it from the repository with:
$ apt install brasero

Once installed, perform these steps:

1: Insert a blank DVD into the drive.
2: Type brasero in the Ubuntu Search Button and launch the app.
3: Click Burn Image
4: Select the ISO image for the Select a disk image to write field.
5: Select your Blank DVD disc for the Select a disc to write to field.
6: Click Burn and follow the prompts.
Comment: That was fantastic, thank you for the help. especially the step by step installation guide,
Comment: Thank you again, I didn´t know how to mark a question answered.     P.S.  I down loaded the 64 bit version as an  .ISO file and after a day of Googling I still have no idea how to make it an installable.  (.EXE)
Comment: I'll update my answer with easy steps.  Look at my answer in about 15 minutes to see the steps.
Comment: Thank you yet again L.D. James.  I would be totally lost without this step by step information from you.
Comment: All set again L.D. James,  works well and thank you.


TYPO3 scheduler task with Extbase

Question: When creating a scheduler task with Extbase CommandController, how can I tell the scheduler extension that a task failed?
When creating tasks with scheduler (without Extbase), this is done with "return false", but this has no effect in tasks with Extbase CommandController.
Answer: You can throw an Exception. That will mark the scheduler task as failed.
Comment: One might add the following comment to Daniels answer:

Basicly if a CommandController::command finishes its marked completed, if it fails during execution it'll be marked incomplete, and if an exception is thrown the exception message will be added to the task:info. Not really any reason to add anything else, and no special config for it.

TYPO3 scheduler task with Extbase


How do I manually reorder program tabs on the bottom panel taskbar of Centos 7?

Question: I would like to be able to reorder the program tabs found at the bottom panel, the taskbar of my Centos desktop. Similar to how you can reoderder tabs when using Firefox, I would like to have this functionality for the programs I open.
Comment: This seems to be more like a GNOME Desktop or KDE Desktop related question than a CentOS specific question? GNOME/KDE runs on Debian, too.
Answer: I'm using KDE on Centos7, so I can explain it only on KDE.
(Right click on task bar) -> Task Manager Settings -> Sorting: Manually -> Apply

And then, you can drag and drop a task on task bar to reorder as you wish.
Comment: Is there any other way to get to the "Task Manager Settings" as I can't seem to right click on the task bar. I looked into the "Settings" application as well as the "KDE System Settings" application and cannot seem to find it there.
Comment: Nvm. I just check my system using the command 'echo $DESKTOP_SESSION' and it outputs gnome-classic so I think this is why I cannot right-click on the taskbar but why would I have an application called KDE System Settings if I'm on gnome?
Comment: My desktop is KDE and yours are Gnome classic. They may have different taskbars. Currently I don't have gnome classic desktop and don't know it support manual sorting. If I find how to set it manually, I will add a comment here.
Comment: I set Gnome-classic and tried to find how to reorder tasks, but I couldn't find it. I think Gnome-classic may not support it. If you really want it, I suggest that you should change your desktop manager such as KDE.


Want to make 3D wall art how to project/bake texture from image plane to the walls of a room?

Question:
I want to project a texture from plane to camera view and bake the texture to the room walls so the texture will be distorted properly for me to print.
as people suggest to use project UV from view right now its work in 3D but texture(in image editor are still the same as we distorted UV).To be able to print fit to the wall texture need to distorted to fit rectangle plane How I would do that?
Comment: Have you tried UV unwrapping your destination wall, using the "Project From View" option?
Answer: As Rob says, you should use the Project From View mode when you unwrap.

Import your image in the the UV Editor. Move your camera in front of your box until you get the view you want. In Edit mode make sure your box is subdivided enough, otherwise the projection will bug. Press U > Project From View.

Move or scale your UV as you want in the UV Editor. Create your material, load the image in the Image Texture node.

If you stay in the camera view you'll see your image as it is, i.e. it will appear flat. If you move your view you'll see how the image has been projected on the surfaces.
Comment: Thanks guys this is work in 3D as we distorted UV but I need distorted texture right now the image texture is still the same and cannot be print fit to the wall I need its distorted to fit every rectangle plane how I would do that.