(Co-) Homology associated to Waldhausen K-Theory

Question: Waldhausen K-Theory takes as input a Waldhausen category C and produces a spectrum K(C). I would like to know what is known about generalized (co-) homology theories that can be realized by this spectra. I guess that stable homotopy groups are known to be representable this way? Can you help me to get an overview?
Answer: Thomason's paper "Symmetric monoidal categories model all connective spectra" claims to show exactly what the title claims - namely, you can model all generalized homology theories E with En(*) = 0 for n < 0 by taking the spectrum associated to a symmetric monoidal category.  So in principle, these are what you might feel like you should get.
However, Waldhausen categories are more restrictive - they require that the symmetric monoidal structure is actually the underlying categorical coproduct.  I don't know of any results along this line.
It appears that Thomason's proof is something like the category of weakly contractible spaces over X, but - I will be blunt - I have never managed to sort through Thomason's paper.  It seems conceivable that the object he constructs might be equivalent to something coming from a Waldhausen category or its opposite, but this might be optimistic.
Comment: Thanks for the nice reference. I think there might be a good chance to build a Waldhausen category that models Thomason's construction. Does anyone know how to fill the gap?
Answer: The Waldhausen category C which is the category of finite pointed sets, with cofibrations the monomorphisms and weak equivalences the isomorphisms, has K(C) = the sphere spectrum—or so I am told.
Comment: Oblig: This is called the Barrat-Priddy-Quillen theorem.


react-native-reanimated and react-native-gesture-handler X/Y coordinates

Question: I've setup an expo snack here, where you can drag an image and see the X/Y coordinates.
I don't understand why if I drag the image on the top left corner, the coordinates are negative numbers.
You could argue that the position is relative to the starting point. But as you can se in the Cardeditor.js file the cardItems state X and Y are both set to 1.
In my app (not in the example) the user is able to navigate to another screen where the cards is in landscape mode, so I have to reposition the image in the correct spot, for this I need to be sure that the top left corner is always x:0 and y:0.
This issue probably has to do with absolute/relative positioning, but I can't find any source of information on the net.
Answer: try to main container position:'relative' and other components position: 'absolute'
Comment: Can you please look at my code and change it accordingly ? thank you!


Are Battlefield 3 Premium keys region restricted?

Question: I'm considering buying boxed CD key for Battlefield 3 Premium, as it has significantly better price than one in Origin Store. Are these keys region restricted in any way?
Comment: Where would you be buying it from? As far as I knew the only legitimate retail channel was through Origin.
Comment: You know that you don't need five tags by question? In fact, we are trying to [reduces our tags](http://meta.gaming.stackexchange.com/questions/4758/the-great-arqade-clean-up-of-2012).
Comment: @bbodien you can buy a key-in-a-box in many countries, and as I said in my post, they are usually cheaper. In my case 20% cheaper. I had a quick look, but couldnt seen any cheap, grey-market (typically Russian, in my experience) versions of Premium, but I'm sure they are out there.
Comment: @bbodien: http://www.amazon.co.uk/gp/offer-listing/B0089L1X08/ref=dp_olp_new?ie=UTF8&condition=new for example, although I've found even better options for example in Poland for 136zł (32€) for key-in-box, 119zł (28€) for key only -- on Origin I see it for 50€.
Comment: Ah okay, thanks for the info. I assumed any such codes were created to be used in promotions, and weren't intended for retail as standalone SKUs. My advice would be to buy the Amazon UK one as they have a reliable returns procedure in case you run into problems. Some of those other retailers look suspect.
Answer: No, it's not region locked
It's been quite a while, and since I've actually purchased an origin CD-key from third party (not Russian though). It has been marked in store officially as multi-language and region-free (aka region: Worldwide).
Answer: I can tell you from personal experience, I bought a Bf3 CD key from one of the bigger cd key websites that was not a Russian key but a normal English key.
It worked for for about two weeks then my Bf3 got its key disabled and I couldn't start it anymore. I got a refund from the key supplier and purchased a copy from amazon and put in the new key. It accepted the key but it wouldn't run either. When I contacted EA, they said the account was disabled.
I had to argue with the online support for a few hours to get my amazon key working. Finally after being persistent I had to create a new origin account and online support generated a new key for that new account.
So I would be really cautions about using a key from one of these sites. If you do I would advise creating a Origin account Just for the game so that if you key/account get disabled it won't affect any other origin games you have.
Comment: Are you talking about the base game or the Premium pack? The base game is not normally sold as key-in-a-box. It is usually sold via DVD or via Origin. However, Premium IS routinely sold as key-in-a-box. It sounds like your experience is slightly different - though I completely agree with your conclusion. BTW I found Zavvi to be the cheapest in the UK, but as long as you buy from a known reliable seller, you are unlikely to go wrong.
Answer: I'm not sure about Premium specifically, but I know someone who bought an Russian version of the game; beyond having to download a language pack, they had no problem running it. However some games will check your locality when installing, or may have region-specific keys (as you have considered).
I doubt the key is region specific, but you never know. Plus there is always the possibility that they could retrospectively enforce this later, which would be infuriating.
However, I'd completely recommend you don't buy Premium from Origin. I bought a boxed key version for 20% less (£33 vs £40), which is a decent saving for a 2 day wait.
By all means, buy a grey-market version - there is a decent chance that it will work, However, be prepared to suck up the consequences if it doesn't. To quote Clint:

You've gotta ask yourself a question: "Do I feel lucky?" Well, do ya, punk?
Comment: ...since there was no definitive answer forthcoming ...since my recent experience with a grey-market purchase of the base game may be indicative. Besides, my perfectly sensible suggestion is to opt for a more legitimate discount, rather than a purchase of unknown risk.
Comment: Since when was guessing an answer?
Comment: @CJM: I'm not really interested in gray- or black-market, I mean more like legitimate key-in-a-box from legitimate UE retailer. As for experience with the BF3 itself, I did buy BF3LE on Amazon UK, and using it in NL w/o problems. So as far as guessing goes, I'm guessing it should work. But I'd rather be sure.
Comment: @CJM A suggestion is not an answer, it is a suggestion
Comment: @vartec There is absolutely no difference between buying software intended for a foreign market (i.e. grey-market) domestically and buying software in a foreign market and bringing it back 'home'. Some software providers takes steps to prevent this (either method) - in my experience, EA do not do this with the base game, and probably won't with Premium. But it's still a gamble...


Installing a USB device manually on Win 8.1

Question: I have been the not-so-proud owner of a Nokia Lumia 920 for over 2 weeks now. I am also in possession of two PCs, both running Windows 8.1.
I am unable to connect my phone to either PC in any way. Device manager keeps reporting my phone as RM-821|Nokia Lumia 920. I tried most of the solutions available on the web; here's what I've tried so far:
Everything listed on this Microsoft KB2749484 article: searching for the driver on the internet and manually in Windows folder. I have also downloaded the driver itself (.cab file) from the Microsoft support page. I pointed to that folder as the location of the driver and Windows found nothing.
I tried installing the driver manually with pkgmgr.exe. Though since this is Windows 8.1 x64 Professional, received an error message that Pkgmgr is deprecated and cannot be used (Error 0x80070002).
I tried every single USB port on the laptop (Dell XPS 15 L502X) and desktop (Asus P4P800 Motherboard), and I didn't find any USB driver update. Both BIOSs are up to date. I have also tried rebooting with volume down while connected to the PC.
I am confident that this issue is not phone or cable related, as I tried plugging the phone into a friend's PC and it worked instantly. I tried chatting with a Microsoft help employee, who directed me to the Nokia homepage to search there. sigh
I have seen many people encountering this same issue.
Is there a way to install the .cab file manually in any other way?
Comment: Wow, this sucks. I'm a little confused though, what .cab file do you mean?
Answer: I was facing this same problem a long time ago, and I found a solution.
All you have to do is switch off the device before plugging in the USB cable, and then switch it back on after plugging it in. Wait for about 5 minutes until the phone boots up.


Algebraic number field has multiplicative inverse

Question: I am struggling with showing that for algebraic number $\alpha$, the ring generated by $\mathbb{Q}[\alpha]$ is a field. I understand that to do this, I will have to show that any $r+s\alpha, r,s\in \mathbb{Q}$ has an inverse in $\mathbb{Q}[\alpha]$. I'm lost on how to go about doing this, though. Help?
Answer: If $\alpha$ is an algebraic number, then $\mathbb{Q}[\alpha]$ is a finite-dimensional vector space over $\mathbb{Q}$.
The map $x \mapsto \alpha x$ is an injective linear transformation and so is surjective.
This means that $1$ is in the image and so $\alpha$ has an inverse in $\mathbb{Q}[\alpha]$.
(Injectiveness follows from $\mathbb{Q}[\alpha]\subseteq \mathbb{C}$.)
The main advantage of this approach is that it works for finding the inverse of any $\beta \in \mathbb{Q}[\alpha]$ simply by solving a linear system, without any tricks or flashes of insight.
Answer: The ring $\mathbb{Q} [\alpha]$ is the image of $f: \mathbb{Q}[x] \to \mathbb{C}$ where $f: p(x) \mapsto p(\alpha)$ ($f$ is evaluation at $\alpha$.) Since $\alpha$ is algebraic it has a minimal polynomial $q_{\alpha}(x)$. Show the kernel of $f$ is the principal ideal generated by $q_{\alpha}(x)$. Since $q_{\alpha}(x)$ is irreducible its principal ideal is maximal. Thus the image of $f$, the subring $\mathbb{Q}[\alpha]$ generated by $\alpha$, is a field.
Answer:
From the very definition of    $\mathbb{Q}[\alpha]$ (look in your course if you don't understand), this is a ring ($\alpha$ could be transcendental here).

From point $1$, the only thing you need to do is to find an inverse for each element of $\alpha$. Use the fact that $\alpha$ is algebraic to show that $\alpha^{-1}\in \mathbb{Q}[\alpha]$.

Hint $\alpha$ is algebraic, then there is a minimal polynomial (irreducible) $P\neq 0$ so that $P(\alpha)=0$. Write :
$$P(X)=a_kX^k+...+a_0$$
It follows that :
$$\alpha(a_k\alpha^{k-1}+...+a_2\alpha+a_1)=-a_0 $$
After you justified that  $a_0\neq 0$, conclude that the inverse of $\alpha$ is indeed in $\mathbb{Q}[\alpha]$.

Using $2$, show that any element in $\mathbb{Q}[\alpha]$ is invertible in $\mathbb{Q}[\alpha]$.
Comment: This only gives an inverse for $\alpha$. What about the other elements of  $\mathbb{Q}[\alpha]$?
Comment: You only have to justify $a_0 \ne 0$, right? Then $a_k\alpha^{k-1}+...+a_2\alpha+a_1\neq 0$ follows automatically from the equation $\alpha(a_k\alpha^{k-1}+...+a_2\alpha+a_1)=-a_0$.
Comment: I understand 1, it's precisely 2 that I don't understand how to do :(.
Comment: @nargles324, ok I'll give a more precise hint.
Comment: ahh, does the above work because the ring is closed under multiplication and addition? thank you so much, that was silly of me!
Comment: @TonyK, you are right.


Guy with $7$ friends

Question: Some guy has $7$ friends$(A,B,...,G)$. He's making dinner for $3$ of them every day for one week. For how many ways can he invite $3$ of them with condition that no couple won't be more then once on dinner. (When we took $A,B,C$ we cannot take $A$ with $B$ or $C$ and $B$ with $C$). My idea: We have $7$ subset with $3$ elements. So $7!$,but subset $\{A,B,C\}=\{B,A,C\}$ etc. So we have to divide by $3!$, but I don't think that's correct thinking. Should I divide also by couples? ($2!^3$) Any hints?
Answer: Either I am missing something or the book answer is wrong. I take it that when you write "no couple won't be more than once on dinner", you actually mean "no couple will ....." as has been amply clarified in your examples.
Now there can only be $\dbinom72 = 21$ distinct couples, and each day $3$ such couples get eliminated, e.g. if $ABC$ are invited, $AB, BC$ and $AC$ can't be invited again.
Thus there are only $7$ trios to be invited, and if the order in which they are invited is to be considered, there are $7!$ ways of inviting them.
Note
There is no question of dividing by $3!$, because we considered combos of $3$, and for the life of me, I can't understand multiplication by $2$  and division by $2^2$]
Answer: We can chose 3 men from 7 in $7$$C$$3$ many ways. That is $35$. Now, a couple repeats $5$ times. So, the we can chose three men with the criteria in $35/5=7$ many ways. Now, there are $7$ parties in total. So, we can permute the groups across days. That gives a total of $7!$ many ways.
Comment: true blue anil's answer coincides with mine.
Comment: I checked in book and the awnser is $\frac{7!*2}{2!^2*3!}$
Comment: Does the ordering matter? I mean on which day a trio is invited...
Comment: Yes. There is a difference between Monday and Tuesday
Comment: I think his (or her) awnser is correct. I'll ask the authors of book, how did they find this numbers. Thanks ;)
Comment: Hmm... He made a different approach... But, in the end, we both agree to the same answer...

Guy with $7$ friends


Problems with axioms and their potential uses in real life.

Question: Okay, so I am looking for more examples that revolve around this topic.
My topic is: "Is the assumption of basic truth needed in order to create a system of mathematics that is reliable?"
Here, when I reference basic truth I am talking about axiomatic systems. As of now I am going to talk about axioms in euclidean geometry and how they make sense because geometry can measure perfection. What else is there?
Comment: When you say you are looking for examples, what are you looking for examples _of_?  Also, could you clarify the last part: "As of now I am going to talk about axioms in euclidean geometry and how they make sense because geometry can measure perfection. What else is there?"  I gather from this that you are looking for examples from Euclidean geometry, but apart from that I don't understand what you are saying.
Comment: @TrevorWilson I guess I am looking for examples where axioms are in play and very apparent. So in geometry they are apparent and make sense. In many other disciplines like Calculus ZFC is in play at all times.
Comment: I see.  ZFC is the most important example then, in the sense that most mathematical theorems (when phrased appropriately) can be proved from the ZFC axioms.  In many cases they can be proved from a weaker axiom system, however.  Whether or not axioms are "necessary" in order to have a notion of proof is a different question.
Answer: (If I understood the question correctly:) Essentially, yes. You must make some unverifiable assumptions - i.e. assume some 'basic truths'. This is a big topic in the philosophy of mathematics - the most relevant reading on the subject would be on Goedel's Incompleteness Theorems.
You may also find the idea of the Muenchhausen Trilemma relevant, which deals succinctly with what you seem to be finding difficult to accept: it is practically impossible to support every axiom with a proof - to do so would require either circularity or an infinite regress.
Comment: I'm not sure whether Goedel's Incompleteness Theorems are relevant to the question, because even if we had been able to prove the consistency of all mathematical theories from some weak set of axioms (say PA) this still would not mean that axioms were not necessary for proofs.


How should I solve this succession's limit?

Question: I have this succession's limit: $ \lim_{n \to +\infty}{\tfrac{n^{n!}}{(n!)^n}} $ . I've been tried by using root's criterion and I've got $ +\infty $. Can someone show me how to solve this exercise? Just to compare it with mine. Thanks
Comment: Do you know Stirling approximation ?
Answer: Hint.
You have
$$\frac{n^{n!}}{(n!)^n}=e^{n!\log(n)-n\log(n!)}=e^{n((n-1)!\log(n)-\log(n!))}.$$
And recall the fact that
$$\log(n!)=\sum_{k=1}^n \log(k).$$
Answer: The numerator has $n!$ factors all equaling $n$, while the denominator has just $n^2$ of them, most smaller than $n$.
Answer: Probably not the answer expected in the excercise but you can do just rough estimates and it will work in this case. For example from $n!^n\leq (n^n)^n=n^{n^2}$ we have
$$\frac{n^{n!}}{n!^n}\geq \frac{n^{n!}}{n^{n^2}}=n^{n!-n^2}\geq n \to \infty$$
since $n!$ grows much faster than $n^2$ (shouldn't be hard to show $n!-n^2 \geq 1$ for $n \geq 4$).


Writing to the STDIN of a C program from PHP

Question: I literally have almost no PHP experience at all whatsoever. I have a C program that I'd like to put online (it's computationally intense, I'd rather keep it in C), so my question is this: How to I write to the STDIN of a C program using PHP (and read from it's STDOUT)? Like, taking a string from an HTML form, or something similar, and writing that to the STDIN.
Answer: Long story short, check this out: http://php.net/manual/en/function.proc-open.php
Answer: Take a look at the System program execution section of the manual.
Esp. proc_open — Execute a command and open file pointers for input/output
You could also pass the parameter as an argument to the executable and fetch the output which can easily be done via escapeshellarg() and exec()
You might also be interested in http://www.swig.org/
SWIG is a software development tool that connects programs written in C and C++ with a variety of high-level programming languages. SWIG is used with different types of target languages including common scripting languages such as Perl, PHP, Python, Tcl and Ruby.
Answer: Try using fwrite and fget
fwrite(STDOUT, "\n-CMD$: "); //Output
$site = fgets(STDIN)   // Get Input

Answer: If your php program just echoes strings (using echo),
php -f program.php | c_program


How to make a clamp for this motor to move it in two directions?

Question: I'm working on a disc brake balancing machine, as shown below this is a belt pulley drive and for alignment purposes I want motor to be able to move vertically as well as inwards outwards (to tighten the belt). Is there any clamp/adjuster design to achieve this?
Comment: What have you found so far?
Comment: I found to use a slotted hole but this can just translate in one direction.
Comment: Can the pulley on the motor slide up and down? If so, how much? That would solve one direction of adjustment very easily.
Answer: Use a dovetail guide for the vertical movement and place the motor on it using elevating screws for the horizontal movement.

If you place the bottom plate of the contraption above (picture source) on the box and use the threaded holes on the top plate to mount the motor, you can move the motor up or down. Use screws in the holes at the side of the upper plate to adjust friction and lock both in one position.

Next, use something like the screws above (picture source) to fix the motor to the upper plate.
Comment: The dovetail guide Looks to be a very expensive solution...
Answer: For vertical movement you could use a round guide and a tight fit rubber ring inside the holder to enable movement when you want to move it but for friction to keep it in place when you want to keep it still.
For horozontal movement you could use scissor mechanism with screw which could control movement and act as a brake.


What does the 'よく' mean in 'よくわからない'?

Question: I know that よく means 'often' or 'well', but what is the meaning of it when it is used before わからない?
Does it mean 'I often don't understand', or does it mean 'I don't really know'?
Comment: I think it is most likely to mean 'well', but without context it's hard to say for certain. Have you come across a situation where you think it is ambiguous?
Comment: I can also mean the subtly different word 'properly'.
So the meaning of the sentence would be 'Don't know properly'
Answer: The meaning of よく here is "well". よくわかる means to understand something well, or thoroughly. In contrast, よくわからない in speech would indeed be "I don't really know" or "I'm not sure". Here's a helpful link for understanding how to use this in various contexts.
It's worth mentioning that even though it technically means "I don't understand (it) thoroughly", which may give off the impression that the subject does understand the topic on a basic level, many may turn to this expression even when they're totally clueless about a topic e.g. すみません、よくわかりません。
You may also come across よくわかっていない, which is usable in a more academic or report-type context, when something is not well understood in general. e.g. この病の原因はまだよくわかっていない ("the cause of this disease is still largely unknown").
Finally, if you want to strictly say "I don't understand (something) often", you would say something like 分からない(事/ところ/時)がよくある, which can be replaced by よくわからない in most contexts anyway because it makes sense that you would not understand something often when you don't understand something thoroughly.

What does the 'よく' mean in 'よくわからない'?


How do Android control auto-wrapping for TextView?

Question: When text reach the end of the TextView, the next text will wrap to the new line.
How do Android implement/control this(So that we can override)?
I need to override something for my app, which doesn't do auto-wrapping correctly(known issue on Android 3.0+) with ImageSpan. But I don't know where to start.
I asked this question hoping to find a work-around for the issue, but no answer yet.
Answer: You could extend TextView and override the paint() method.


reverse geocoding api with no limitation?

Question: I was wondering for reverse geocoding api, which doesn't apply any per day or monthly limit on the data? Till now all i have got is google api and geonames api.
I need to get postal code from latitude and longitude of location.
Answer: Build your own PostGIS Server that runs on PostgreSQL.

Install PostgreSQL
Install PostGIS
Download a Zipcode file either as points or as geometry

Zipcode information can be downloaded for the US from the US Census Bureau and other sites.  Note that many US Postal Zip Codes don't have a physical location/boundary, which explains why many services could potentially report a different location for some zip codes.


Can't open email attachments sent from Windows

Question: Why does a PDF attachment on an email from a PC end up as a file named winmail.dat when it arrives on my MacBook Pro?
Answer: The sender is using the proprietary email client Microsoft Outlook.  The entire message is sent in a proprietary format to preserve Rich Text Formatting, according to Microsoft.
A Google search for Winmail.dat will reveal several work arounds to deal with these kinds of files.  The simplest, but most annoying, is to ask the sender to resend the email in plain text format.  This will prevent the attachments from being bundled into the winmail.dat file.
From Microsoft:

The data in a winmail.dat file is not usable. To resolve this issue, ask the sender to re-send the message in plain text format.
Answer: Easy: winmail.dat is just a compressed item. There are numerous little apps, that let you "open" them by using it:
I use TNEF's ENOUGH (appstore) but there is also winmail.dat reader on the appstore.
Comment: I have been using TNEF's for years, so that would be my solution of choice.  You can also forward the message to a gmail accoint... gmail is able to handle the outlook format properly...  And if you happen to use the thunderbird mail client, there used to be a plugin (LookOut) that handles the winmail.dat container
Answer: It was likely sent from a Windows PC using some variety of MS Exchange.
There is a fix but it has to take place on the Exchange server. And while it is a simple configuration change most Exchange admins don't see the need unless prodded from above.
You may have to get one of the utilities that decode a winmail.dat files. I thought I saw one the other day in the App Store. A quick search should turn it up...
Answer: I found that there were a few "freemium" programs promising to solve this, but the desired functionality - the ability to open and save the compressed attachments - was only available as an "in-app" purchase of the "premium" version for $10. I had no choice, as I needed the files to finish some time sensitive work from my home using my iMac. Microsoft Exchange has it's good points, but this is not a feature to recommend it. FYI, the app I purchased was called Winmail.dat Viewer - Letter Opener 9, Winmail.dat Premium.


Netty, throttling HTTP requests, is it possible to hold them open for long periods of time from the server side?

Question: I'm working on a server process which receives files via HTTP PUT. The requests come from devices automatically throughout the day, not users. I currently handle everything in memory for performance reasons, but resort to writing the files to disk for later processing when I need to throttle. I'd like to leave writing to disk as a last resort, instead,'d love to have the option to throttle them by holding the connection open, and not receiving any data until I'm ready, for maybe 5-10 minutes. Can I do this from the server side with Netty?
Thanks for any thoughts or ideas
Answer: You can use Channel.setReadable(false) to prevent any new read from the Channel. Once you are ready again you must call Cannel.setReadable(true) again.


My dog ate yeast balls

Question: I was making rolls and the yeast failed and i baked it for 15 minutes at 350F . Will it still hurt my dog? I haven't tried anything yet, he's currently laying down and acting normal.
Answer: Disclaimer: If you're seriously worried or your dog shows any unusual symptoms, call your vet.
That said, he's probably fine.
If your dough didn't rise because the yeast failed and it was exposed to oven heat afterwards, then you can assume that either all or at least the vast majority of the yeast cells were dead by the time your dog munched on your rolls. The rest is maths: even if he consumed a very small amount of miraculously vital yeast, the time these cells would need to multiply to digest the carbohydrates in the bread at a rate that produces an amount of either CO2 or alcohol big enough to cause trouble is counteracted by what the digestive tract of your dog does to both the yeast and the carbohydrates over time.
Note that our bodies (canine and human) usually handle small quantities of yeast just fine. Live wild yeast occurs naturally on plenty of fruit and other plants. And still dogs (and humans) can eat for example an apple and not have any problems.
Comment: Just to add regarding dogs handling yeast: Brewer's yeast is a popular food/barf addition to support coat changes. As such smaller amounts are really a non issue and usually no reason for concern.
Comment: @Mario yep, B vitamins etc. There's conflicting information on bakers yeast and as the ASPCA issued  a warning (although more about dogs and cats getting into the bowl with rising dough, which apparently was a frequent occurrence during Covid lockdown), I am not dismissing the possibility entirely. Plus we have no information on dog size - a toy poodle or a GSD with the same roll are two very different stories.

My dog ate yeast balls


Reporting visibility on a custom child object

Question: I have a head scratcher of a problem. I created a custom object called Event Sales Activity which is child to Campaigns. I added lookups to Lead and Contact to the ESA Object. I've populated the object with some testing data. I can see all of the ESA records in a List View.
When I try to report on the object, I can see records on:
Campaign and ESA
Campaign and Lead
Campaign and Contact
But I can't see records in the following reports:
Leads and ESA
Contacts and ESA
Campaigns with ESA and Lead
Campaigns with ESA and Contact
I've checked the sharing settings. There Campaigns are set to public full access at the moment. Event Sales Activity is Controlled by Parent. Lead is Public Read Write. Contact at the moment is Private. All records are owned by me (it's a sandbox ).
I'm completely flummoxed. Any help would be appreciated.
Answer: If I understand correctly, you have to create a custom report type:
select Campaign as a Primary Object, then ESA Object, and Lead at the end. Do the same for Contact


What can I use to hold an iron rod Torchiere lamp section to the pre-wired lower iron rod threaded bolt that won't tighten securely?

Question: I purchased a pre-wired Torchiere floor lamp that attaches together in sections of hollow iron rods that each have a smooth end and a threaded bolt on the other end. On one of the iron rods the thread on the bolt does not have a sharp edge - like the others - which is preventing the bolt from 'cutting' a thread into the connecting hollow iron rod end when I thread it into the opening. I was already issued a refund by seller who said to just "keep the lamp." My mom loves the lamp so, if possible, I want to try to salvage it. I tried turning it as far as it will turn and then taped a small plastic collar with foam around the rod to hold it in place. But the halo torchiere light makes it top heavy and it is not secure. Is there a collar or something I can use to secure the torchiere section to the lower iron rod since I'm unable to screw the 'threaded bolt" into the rod to tighten securely?
Comment: If there are several sections would it be possible to use a section with sharp threads to cut for the damaged one?
Answer: I solved this by using a pipe repair clamp from the plumbing department. It is meant to seal pinhole leaks in pipe. It looks like a 3-4" long bracket that clamps to the pipe --not in this case the floor lamp tubing-- and has a rubber, inner liner. Get one in the 1" or greater size. It has two parallel bolts that tighten to clamp it all together.
It's not the prettiest solution but it works. I salvaged a perfectly good floor lamp with bad threads this way.
Answer: You could try removing your tape, plastic collar and foam. Then carefully place a forming sleeve over the gap that spans between 270 degrees and 315 degrees of the circumference of the gap around the tubing post. This cover can be made from any one of a number of materials and if it is made long enough beyond the gap width you can use strong tape to hold it in place around the tubes.
If you use a paper or thin paperboard material (a manila file folder material would be ideal) use wide strips of clear plastic packing type tape to line the insides of this piece so that when it is taped in place the very smooth surface of the tape faces into the circumference of the gap area.
Now mix up a good sized batch of epoxy (I recommend to use the grey JB Weld) and stuff it down into the gap through the narrow opening of the wrapped forming material. The packing tape liner that I mentioned above will not bond to JB Weld and leave a very smooth surface on the cured epoxy.
After the JB weld has cured (approx 24 hours) you can remove the wrapped forming sleeve. A second application of epoxy can be applied to fill in the remaining area where the opening was in the forming sleeve. (I leave it to you to decide if you want to try to use more sleeve material to try to form the remaining epoxy after you fill the space or just gob it in and sand it down after it has cured).
If this is done correctly there should be a very secure connection without wobble in the joint.


How to clean dirt embedded in epoxy sealed cement floor

Question: Three days ago I applied a 2 part epoxy sealer over a self leveling cement floor in my laundry room. It was still a bit tacky when I walked and moved some stuff back into it. Now there are some dirt spots that seem to be embedded in the tacky parts of the epoxy. How should I go about cleaning this while also letting the epoxy thoroughly dry? I'm not sure if I should wait or try to clean up the spots now.
Comment: Should have waited - "a bit tacky" is not "cured" so walking on it and moving things into it was premature.
Answer: I have never had success when messing with a partly cured thermoset compound.
You will probably have to wait for it to fully cure then sand and if necessary paint over the marks.
Answer: Dirt embedded in epoxy doesn't clean. You can grind it out, or try to hide it with another coat. If you grind it out, you'll probably have to recoat anyway.


Why doesn't an inductor mind polarities?

Question: I have seen many textbooks that say consider a capacitor which is polarised (DC capacitors) and non-polarised capacitors also exist, but why don't inductors have AC inductors and DC inductors? Sorry for the silly question, but I want to know the answer.
Comment: polarised capacitor exist only because there is a chemical process that forms the insulating layer, in one polarity, and dissolves it in the other polarity. The polarisation is a chemical thing, nothing to do with the desired electrical properties. The capacitor itself is not polarised - it just fails if used for long with the wrong polarity.
Comment: Why *should* it be polarized is the question you should ask. For example, you also wouldn't ask "why do normal cars don't have guns mounted on them, because I saw a tank once, and that tank had a gun"! The principle you should be applying has a name in philosophy: [*Occam's razor*](https://en.wikipedia.org/wiki/Occam%27s_razor) and it means that the general assumptions used to understand something should be minimal. Things needing to be polarized is the opposite of a minimal assumption!
Answer: There are inductors with polarity. They have a permanent magnet to bias the inductor opposite to the expected DC current. It's generally not worth it, but it has been done.
See, for example, DOI: 10.1109/ESTS.2011.5770892 Permanent magnet inductor design.

Abstract:
  Permanent magnet inductors (PMI) are useful in dc biased applications. They utilize the flux produced by a permanent magnet (PM) to partially offset the flux established by current in the inductor winding. The addition of permanent magnets allows for decreased inductor mass for a given inductor current and inductance rating. Herein, the Pareto-optimal front between loss and mass is established for a PMI inductor and is compared to that of a traditional EI core inductor. It is shown that the PMI inductor is significantly less massive than the EI core inductor for a given loss.

Polarized capacitors (which is a significant limitation on their application) are used because a polarized capacitor can be made significantly smaller and lighter than a non-polarized capacitor of the same  value. If non-polarized capacitors were the same size, cost and performance as polarized capacitors of the same capacitance and voltage rating, there would be no reason to use the latter.
Comment: If there is one export from Russia that I like, it's sci-hub for getting articles behind paywalls. Scientists want their work to be seen, never say no when I ask for a copy by email, and broadly speaking do NOT approve of Elsevier's (who over the years has bought itself a solid monopoly) policies much. Now I can just pull it. So I've been asking less, I admit, and reading more.
Comment: @jonk. Thanks. I have a vague recollection such a device was used in a consumer product (Japanese-designed TV set maybe).
Comment: Interesting find on PMI design. Thanks. Also, I think "lighter than a polarized capacitor" might want to be "lighter than a non-polarized capacitor". Typo, I know.
Answer: No this is an interesting question and I learned things myself while answering this question :) There are some instances where inductors may be polar sensitive. For instance, let's look at mutual inductance.

Credit goes to Fundamentals of Electric Circuits (6th edition) by Matthew Sadiku, chapter 13, page 557.
As you can see, polarity does matter when you're talking about transformers and mutual inductance. But this is more of reference polarity and magnetic field polarity. But you place the inductors upside down and it won't change anything. In other words, the inductors themselves don't have polarity.
In the case of regular inductors that are not meant to be used for its mutual inductance, it's just a coil... and no matter what direction that the current travels, it will always follow the proper path that the laws of physics define it to be. You can ask the same question about resistors. Why aren't resistors polar sensitive? It's because current travels the same way regardless how you place the resistor and the current behavior is always the same through the resistor. The same is true about an inductor.
As Iganacio pointed out, the reason why capacitors may have polarity is that of the material inside of a capacitor that defines it as polarization. On one side, there's dielectric material and on the other side, there's electrolytic material... You don't have that with inductors...
Comment: It's worth mentioning the flyback transformer here. It has similar properties to the transformer but includes an airgap where the energy is stored, just like a regular inductor does. You can use a regular transformer/CMC as a flyback but the core does not store much energy so it's suitable for low-power circuits where isolation is desired.
Answer: Maybe you are a bit confused. An inductor has no polarity until a coil of wire around it gives it a DC polarity because it produces a polarized magnetic field by using DC current, or use AC current of a frequency within the cores resonate range and you get an AC magnetic field of the same frequency.
Polarized cores for relays were tried for a few years then pulled from the market, as the pre-existing magnetic field tended to make them change state with extreme vibration on the z-axis.
If a core were pre-magnetized and a strong AC magnetic field was applied it would loose its pre-magnetized field and have no magnetic field unless AC or DC current flowed in the coil around it.
Inductors in general do NOT retain any magnetic field once the current is turned OFF. They are made of different materials than those used to make electromagnets or permanent magnets.
Comment: I think miniature relays still are: `By using the highly efficient polar
magnetic circuit "seesaw balance
mechanism", a nominal operating
power of 140 mW (minimum operating
power of 79 mW) has been achieved `   https://www.panasonic-electric-works.com/pew/de/downloads/ds_61020_en_tq_pd.pdf . I always assumed this meant they have a magnet inside.
Comment: Latching relays are generally magnetised, they are not considered an inductor first though even though they are a solenoid coil, their function may require passing current through in a specific direction and this current may behave the same electrically though the mechanical effect differs depending on direction.
Answer: We can also remember that while capacitors are not theoretically polarised, though electrolytically formed ones fail if incorrectly connected so similarly a general inductor is also theoretically not polarised unless magnetically biased (well described in other answer).
However in practice they may be constructed in ways that require taking note of which terminal is which.  Mostly this relates to the capacitive coupling to cores and cases or the electrostatic shielding from the outside casing or layers.
Ganged capacitors and mutual inductors are also polarised in relation to other sections of the component (as mentioned in other answer).


Drupal site communication with other site

Question: Hello I have a Drupal 7 site.
I want users to login without using the drupal form.
I will silently send username and password credentials from another site and when users click the link to my drupal site they will be already logged in.
So how to communicate with drupal login and send the details of the user from another site or maybe a browser plugin or a PC application etc.
Comment: Do you want to create unique usernames for all those who follow link to your drupal or just use one User say Guest for all ? If its the former case then you will end up with lot of data in user s table.
Comment: Does the other site use Drupal too?
Answer: This is a fairly complicated thing to do, but the "go to" module for handling integration with third-party sites is the services module.


Airport Time Capsule US power cord in Europe?

Question: I have a US power cord for my AirPort Time Capsule (802.11ac 2TB), and I want to use it in Europe (specifically Norway). The power outlets here deliver 220V rather than the 120V a US power outlet delivers. Will a simple plug adapter be enough, or will that fry my AirPort? That is, can the AirPort Time Capsule sold in the US handle 220V as well as 120V?
Answer: I remember reading the tech specs for Timecapsule some time ago and it is specified it has a range of 100V to 240V. So it should be fine.

AirPort Time Capsule 802.11ac - Technical Specifications
Electrical and Environmental Requirements

100-240V AC, 50-60Hz; input current: 1.5 amp

You can check this, too.
Here is a link to the Apple TimeCapsule Technical Specifications - please also check if this is your version of Timecapsule.
Have a good trip!
Comment: Here is a [search query for older Airport TC Specs](https://support.apple.com/kb/index?page=search&q=airport%20time%20capsule%20power&product=&doctype=SPECIFICATIONS&currentPage=1&includeArchived=false&locale=en_US&src=support_searchbox_main&type=suggested)...
Comment: I updated my question. Thanks, this is exactly what I needed!
Answer: From the Apple Support Technical Specifications pages, it appears that Time Capsules (2nd gen, 3rd gen, 4th gen, 802.11ac) all show as capable of accepting voltages between 100V - 240V, AC, 50-60 Hz. So a simple outlet adaptor should work OK for these models.
You did not specify which version you had, so please check the specifications for your particular model.
Edit to add: It may be prudent to set up your Time Capsule with the region set to Norway as there are some slight differences between USA and EU WiFi channels. Using Airport Utility, access the Wireless Options and make sure these suit your local arrangements.
Comment: Updated my question with additional specs.

Can you provide a link? All I could find on Apple's site were discussion posts.


Enable ratings for custom list via JavaScript in a SharePoint Hosted Add-In (App)

Question: let me just start by saying that I've searched for about 8 hours and I've read every accepted answer for CSOM, PowerShell, list settings etc and tried to convert the codes myself into a working JavaScript/jQuery solution with no luck.
I need to enable ratings to my Custom List within the App context when the app deploys. As of now the only way to get it to work is if I manually go to the URL of the list and create a new list item and rate it through the SharePoint UI.
The only way to make this work is to put a star between 1-5 on the newly created list item, liking doesn't trigger the needed feature.
to sum it up, when I deploy my app the list is created and I've added the needed fields to my list. When I create a list item through the App UI (through JavaScript) I've added the code to like/rate the item from the very start to trigger the rating feature but I receive:

"Content reputation feature is not activated."

I've tried to set this property through JSOM:

<code>var context = new SP.ClientContext.get_current();
var list = context.get_web().get_lists().getByTitle("ratingsList");
var properties = list.get_rootFolder().get_properties();

context.load(list);
context.load(properties);
context.executeQueryAsync(function () {

    properties.set_item("Ratings_VotingExperience", "Rating");
    //properties.set_item("Ratings_x005f_VotingExperience", "Likes");
    list.update();

    context.load(list);
    context.load(properties);
    context.executeQueryAsync(function () {

        // Success! the property is there and set to the correct value

        }, DoError.JSOM);
}, DoError.JSOM);
</code>

Well, this "Works" the property is there when I fetch the properties but the feature isn't activated. The reason why I did this is because I started out by loading the properties to see what changes when the list item is rated with a star from the SharePoint UI and figured I could trigger this property through code.
So that doesn't work, the only thing that works is if I deploy the app and then create a list item through the SharePoint UI of the custom list and then rate it with a star between 1-5. After that, I can delete the list item and the code is running perfectly and all is good.
How is it that there is a 90% working solution for this? anyone had any luck/suggestions to this please post a line or two!
Remember that I ONLY need answers for SharePoint-Hosted Apps (Add-Ins) and not ProviderHosted.
Answer: I think 's' is missing form 'Ratings'
properties.set_item("Ratings_VotingExperience", "Ratings");


Banking of roads

Question: When drawing a free body diagram of banking of road, why don't we resolve weight of car into its cos and sin components? It is seen that only normal reaction force and frictional force are resolved
enter link description here
Comment: The component is resolved with respect to axes .... As you see , $mg$ lies on the axis , si no resolution was required...
Comment: @NehalSamee Oh yes right. That point escaped my mind. Thank you.
Comment: @NehalSamee That looks like it should be an answer
Answer: Although the choice of axes is arbitrary in lots of cases, including this one, there is a choice which makes solution of a problem easier.
In this problem you want a car to go around a corner without moving up or down vertically.
So one of the axes is chosen to be vertically upward so that the sum of the resolved forces in that direction is zero and the is no motion in the vertical direct and the other axis is chosen to be horizontally inwards so that the sum of the resolved forces in that direction provide the centripetal acceleration to enable the car to turn the corner.
With that choice of axes you have to resolve the frictional force and the normal reaction into two components but the weight of the car only has one component in the vertically down direction.

Banking of roads


WordPress REST API not working on localhost

Question: I have a WordPress site running WordPress 5.9.3 that is using a React-based theme, which uses the REST API (v1), to retrieve content and menus from the standard WordPress CMS back-end.
However, when trying to install it locally, I can't get the API to work and retrieve the needed data.
I'm using a Docker environment based off this repo: https://github.com/kassambara/wordpress-docker-compose
Since the theme uses React and the REST API to get the site content, trying to view the site gets stuck on the loading screen.
These are the console errors:
<code>GET http://localhost:3000/wp-json/{REDACTED_API_NAME}/v1/site net::ERR_CONNECTION_REFUSED
GET http://localhost:3000/wp-json/{REDACTED_API_NAME}/v1/menus?location=main-menu net::ERR_CONNECTION_REFUSED
GET http://localhost:3000/wp-json/{REDACTED_API_NAME}/v1/menus?location=footer-menu net::ERR_CONNECTION_REFUSED
</code>
And here is a stack trace of the first one, for reference if it helps:
<code>(anonymous) @   VM382:1
fetchSiteInformation    @   bundle.ea81906….js:2499
fetchInitialContent @   bundle.ea81906….js:2535
(anonymous) @   bundle.ea81906….js:5653
(anonymous) @   bundle.ea81906….js:13750
(anonymous) @   bundle.ea81906….js:12811
componentWillMount  @   bundle.ea81906….js:18846
Ng  @   bundle.ea81906….js:12158
fi  @   bundle.ea81906….js:12197
Rj  @   bundle.ea81906….js:12284
Qj  @   bundle.ea81906….js:12267
Kj  @   bundle.ea81906….js:12267
yj  @   bundle.ea81906….js:12260
Ig  @   bundle.ea81906….js:12251
bk  @   bundle.ea81906….js:12302
(anonymous) @   bundle.ea81906….js:12305
Nj  @   bundle.ea81906….js:12261
ik  @   bundle.ea81906….js:12305
exports.render  @   bundle.ea81906….js:12311
(anonymous) @   bundle.ea81906….js:11350
__webpack_require__ @   bundle.ea81906….js:20
(anonymous) @   bundle.ea81906….js:63
(anonymous) @   bundle.ea81906….js:66
</code>
webpack itself compiles with warnings.
The webpack-dev-server is running off port 8081, and phpmyadmin is running off 8080.
I tried to switch the webpack server to use port 3000 instead (to match what the API is calling), but that didn't help.
I can log into WP admin and am having no issues on that side. All content and plugins are present.
This is not an issue on production or UAT environments, nor is it occurring with others on their own local machine.
I'm running Mac Monterey with the Apple M1 chip.
Answer: Just had this same problem and checked other threads and this really worked:
in the AdminPanal => Settings -> Permalinks -> change from 'default' to 'Post name' then Save Permalinks and reload your api url and it should be working now.


How to increase a low recall value?

Question: I am dealing with a HR Attrition Dataset which is highly unbalanced. I used Balancing technique like SMOTE to generate synthetic data and then used Gaussian Naive Bayes to Classify the Attrition. After plotting Confusion Matrix, I see that its giving high precision but low recall. The accuracy and F1 score (after SMOTE) are also very low. Can anyone suggest how to improve recall value?
Answer: Your problem isn't just a low recall value, your problem is your model needs improving.
A high accuracy with a highly unbalanced dataset means practically nothing since simply predicting the most common label will get you a very high accuracy.

With imbalanced classes, it's easy to get a high accuracy without actually making useful predictions. So, accuracy as an evaluation metrics makes sense only if the class labels are uniformly distributed.

(or at least close to uniform)
For example, imagine you have a cancer detecting model. And your dataset is 990,000 images with no sign of a tumor and 10,000 images of tumors. A simple useless model that predicts "NO TUMOR" for every single image will get an incredibly high accuracy of 99%! But a recall of 0%.
Relative to your problem. Imagine if the cancer detecting model was really bad at detecting tumors, but was capable of noticing obvious signs of tumors in images, then it's still a useless model but it will manage to get an accuracy of 99% and a low recall value of 5% for example (because it's only able to detect the obvious signs of tumors). The problem here isn't specifically that we just have a low recall value. The true major problem is that we have a bad model.
For more information on evaluation metrics of unbalanced datasets: http://www.davidsbatista.net/blog/2018/08/19/NLP_Metrics/
For information on how to combat imbalanced datasets: https://machinelearningmastery.com/tactics-to-combat-imbalanced-classes-in-your-machine-learning-dataset/

How to increase a low recall value?


Material to record though so camera isn't visible?

Question: I am building an Arduino (https://www.arduino.cc/en/Main/ArduinoBoardUno) device that will record video based on a motion sensors, as well as other data in the environment. The thing is, I need to record from within a box to the outside, but the camera should not be visible. Any ideas or advice on how and what to use?
Comment: If the camera can see you, you can see the camera (except possibly through a half-silvered mirror). All you can do is disguise it, for example "teddy-bear cams". How big is the lens opening? What are the surroundings? How is the scene lit? More details will help.
Answer: If quality of video is not so important to you, then you can use effect, which is used on public transport windows:

It is when you use lot of dots on the glass. From the outside it looks like semitransparent image. But from inside it will look like that:

With correct focus dots will be less visible on video.
And second option is to use half-silvered mirror.
Answer: Traffic cameras and security cameras shoot through a dark glass, so you could use something like that.
I'm sure you've seen those black domes in stores - it's called a Pan/Tilt/Zoom (PTZ) dome.  You might look in the area of security camera accessories.


How to protect streaming videos from download

Question:
 I want the videos that I play on my site should only be streamed but should not be accessible to download softwares. Can videos be protected from downloading so that no one should be able to record them
Thanks
Comment: Think about it this way: Can you speak to your friend, but prevent him from memorizing what you said?
Comment: possible duplicate of [How to prevent downloading images and video files from my website?](http://stackoverflow.com/questions/1294501/how-to-prevent-downloading-images-and-video-files-from-my-website)
Answer: Can videos be protected from downloading so that no one should be able to record them?
The short answer is: No.
Longer answer: It is impossible to protect anything from downloading unless you don't want anyone to watch it. Remember: if they can see it, they can record it.
Answer: Sure, just don't serve them up. By definition, for someone to view an online video, they have to download it. There's absolutely no way around it, short of forcing them to use a remote desktop type viewer. Once the bytes leave your server, you have absolutely zero control over what happens to them.
Answer: The answer to your question is no.
Why? No matter what sort of protection you put on the actual file, it needs to be transferred to the client computer somehow. And if you put some kind of DRM on the file so that only the person who's authorized to download the file can view it, it still needs to be viewed. This means that an enterprising person can intercept the actual values being displayed somehow and store them elsewhere. It could be as labour-intensive as getting a screenshot of every frame, then stitching that together to form a movie.
If you don't want people to download your videos, don't put them on public servers.


Minecraft limit to book reading

Question: Is there anyway to set it that only a player with a certain level can read a book?
Exp. i have a magic tome and i want it that only players that have 10 level of exp. to open it.
Answer: You can't.
But you can make it so when they reach 10 XP levels, the empty book in their inventory (make it so it was originally empty) is replaced with a written book like this:

Place a command  block with a redstone comparator coming out of it, with the command /testfor @p[lm=10] This will  test if the closest player to the command block has 10 XP levels. If they do, a signal will be sent through the redstone comparator to...
Place another command block in the direction the redstone comparator is facing. This will be the command block that gives the player the book. You can add another command block here too if you want to clear an empty book from the player's inventory. The /give command for the book is: /give KingBspd written_book 1 0 {title:"Book name",author:"Your name",pages:["{text:\"John woke up\nJhon went to work\"}"]}, but you can find more info on that online.

If you need any additional help, leave a comment.

Minecraft limit to book reading


Image of the system to use it on smaller SD card

Question: My raspberry pi runs raspbian. The actual SD card is 32 GB. I'd like to make an image of the system installed (with my configuration and my personal files) in order to install it easily on other SD card. As the card has only 2.5GB capacity, I'd like to obtain an image of less than 4GB, so I can use smaller cards.
I have tried using win32 Disk Imager, but the image is around 30GB, so I can't use it. I also have another computer running Linux, with no SD card reader, but with SSH access. I think I can use dd, but I don't want to obtain a big image.
Here is some information about my system on the raspberry pi:
$fdisk -l
Disk /dev/mmcblk0: 32.3 GB, 32270450688 bytes
4 heads, 16 sectors/track, 984816 cylinders, total 63028224 sectors
Units = sectors of 1 * 512 = 512 bytes
Sector size (logical/physical): 512 bytes / 512 bytes
I/O size (minimum/optimal): 512 bytes / 512 bytes
Disk identifier: 0x000c7b31

        Device Boot      Start         End      Blocks   Id  System
/dev/mmcblk0p1            8192      122879       57344    c  W95 FAT32 (LBA)
/dev/mmcblk0p2          122880    63028223    31452672   83  Linux

$df -h
Sys. fich.     Taille Util. Dispo Uti% Monté sur
rootfs            30G  2,1G   27G   8% /
/dev/root         30G  2,1G   27G   8% /
devtmpfs         212M     0  212M   0% /dev
tmpfs             44M  256K   44M   1% /run
tmpfs            5,0M     0  5,0M   0% /run/lock
tmpfs             88M     0   88M   0% /run/shm
/dev/mmcblk0p1    56M   22M   35M  40% /boot

Answer: You could try to resize your root filesystem (rootfs) so that it is smaller than 4GB, then your image should only copy your new smaller filesystem.
Here is a tutorial on how this can be done.


Low attaching vs high attaching mudguards/fenders on MTB

Question: I am about to buy mudguards for my 29" MTB bike. I will only use them while commuting on wet surface. I considered SKS Shockblade and XBlade set, but noticed they are attached to steering tube and seat post in comparison to others being attached to suspension fork arch and seat stays. The latter attachment allows mudguards to stick close to the tires. Is there any practical difference between mudguards attachment points? For me it looks like closer attachment to tire is better solution? No?
Answer: The practical differences are…
Fork arch based mudguards that attach with zip ties are very light and cheap but are more vulnerable to clearance issues.
Practically this means when riding on sticky flavours of mud that it's possible to get your wheel completely locked up with a layer of mud between the guard and tread that will not clear without removing the wheel.
Fork arch based guards also generally have seal guards that reduce the amount of mud that reaches the legs and wiper seals.

Mudguards that attach to the steerer with a compression plug need to be high quality products as the budget versions have a tendency to flop around and fall out. They don't provide any protection for your fork legs.
Rear guards that attach to the seat post are good at keeping gunk off your back and rear but are not always feasible as they may not work with a dropper. They also can rub the wheel if you need it to work with full suspension. To clear a fully compressed rear wheel the guard needs to sit very high above the wheel which reduces the effectiveness.
From personal experience if you can use a post mounted guard they work really well but again need to be quality versions to prevent breakage and floppyness.
In summary, for your use of commuting you might find that ziptie versions are sufficient and their price is generally low enough (free if you make your own) to make it a suck it and see scenario.
For seatpost/steerer versions you'll get what you pay for.
Comment: E.g., Mughugger rear fenders attach to the rear swing arm and so they move with the rear wheel through its travel. So that's nice.

But the main point of this comment is to remind readers (not the author) that if your fenders are clogging up with mud, turn around and don't further damage the trails :)
Answer: On any bike, the higher the mudguard, the smaller an "angle" is protected.  Thus, the more road water/mud misses the guard and sprays out radially.
Extreme example - I have an "ass saver" on the road bike but it is at the level of the saddle.   So any real water and the underside of my legs still gets wet.
The closer to the tyre, the wider the guard's protection is.   Risk is, too close and debris can wedge in there and slow you down or lock the wheel completely.  For road, 5-10mm space is perfect.  For mud/etc perhaps an inch or so.


Correlation between prices or returns?

Question: If you are interested in determining whether there is a correlation between the Federal Reserve Balance Sheet and PPI, would you calculate the correlation between values (prices) or period-to-period change (returns)?
I've massaged both data sets to be of equal length and same date range and have labeled them WWW (WRESCRT) and PPP (PPIACO). Passing them into R we get the following:
<code>> cor(WWW, PPP)
[1] 0.7879144
</code>
Then applying the Delt() function:
<code>> PPP.d <- Delt(PPP)
</code>
Then applying the na.locf() function:
<code>PPP.D <- na.locf(PPP.d, na.rm=TRUE)
</code>
Then passing it through cor() again:
<code>> cor(WWW.D, PPP.D)
[1] -0.406858
</code>
So, bottom line is that it matters.
NOTE: To view how I created the data view http://snipt.org/wmkpo. Warning: it needs refactoring but good news is that it's only 27 iines.
Comment: If the period-to-period change shows a correlation, you've shown a correlation between the first derivative (with respect to time) of the two quantities. That's a little different than showing correlation between the values themselves (although I'm guessing there's some relation between the two).
Comment: WRT your bottom line: don't forget that you now have quantitative easing, which you should consider as a structural break in the model. Using only data up to 2007 I get a correlation of only -0.1524359
Comment: The blog post: http://www.portfolioprobe.com/2011/01/12/the-number-1-novice-quant-mistake/
shows (and tries to explain) why you generally want to use returns and not prices.
Comment: Clicking the link I get "The resource you are looking for has been removed, had its name changed, or is temporarily unavailable."
Answer: Short answer, you want to use the correlation of returns, since you're typically interested in the returns on your portfolio, rather than the absolute levels.
Also, correlations on price series have very strange properties.  If you think about a time series of prices, you could write it out as [P0,P1,P2,...,PN], or
[P0,P0+R1,P0+R1+R2,...,P0+R1+...+RN], where Ri = Pi-P(i-1).  Written this way you can see that the first return R1, contributes to every entry in the series, whereas the last only contributes to one.  This gives the early values in the correlation of prices more weight than they should have. See the answers in this thread for some more details.
Comment: I would add that one may want to look at either correlation of returns or cointegration of prices rather than correlation on prices
Comment: This answer makes intuitive sense based on the premise that R1 has an inordinate influence.
Answer: It depends, usually you would want to measure correlation between variables that are both stationary, else you would always be able to measure a correlation in the case of variables developing with a trend, even if they are unrelated. In this case I would guess that you should use first differences.
Answer: There is a good explanation here. Summarizing, what we are computing with Pearson correlations are relations between deviations respect to the means, which is no meaningful using prices. So, you should calculate Pearson correlations using returns.

Correlation between prices or returns?


Why do objects float in liquids denser than themselves?

Question: Why do objects float in liquids denser than themselves? I know that a balloon floats on water because it has air in it, but why?
Comment: mass falls down. What configuration has the most mass as far down as possible?
Answer: Objects sitting in a fluid (liquid or gas) experience a pressure on every surface, equal to the pressure of the fluid.
Horizontally these pressures cancel out - the pressure pushing on the left hand side of an object will be the same as on the right, and the object has no net sideways force.
Vertically however is a different story. The pressure of a fluid increases as you move lower (deeper) into it. Going back to our object, its bottom is deeper into the fluid than its top is, so the pressure on the bottom is larger than the pressure on the top. This difference results in a net force upwards.
We call this net upwards force buoyancy, and if you work through the maths it turns out that the upward force is equal to the weight of the amount of fluid that would fit in the same space.
If the object is less dense than the fluid, its own weight is therefore smaller than the buoyancy force upwards, and it floats.
Answer: If an object is completely immersed in a liquid denser than it then the resulting buoyant force will exceed the weight of the object because the weight of the liquid displaced by the object is greater than weight of the object.As a result, the object cannot remain completely submerged and this causes the object to float.
Search up more about buoyant force to learn more about this


Which is more efficient way to generate energy using hydrogen: direct hydrogen combustion or through fuel cell?

Question: Which is more efficient way to generate energy using hydrogen: direct hydrogen combustion or through fuel cell? Please explain so that even a primary school student can understand.
Comment: Note that efficiency is just one of important parameters, when it comes to electrochemical cells. There are at least 6 other important parameters like (available power or stored energy (cycle or lifetime) per ( volume or mass or price(cycle or lifetime) )
Comment: This is mostly an engineering question. To generate heat energy, direct combustion. To generate electrical energy, perhaps a fuel cell.
Answer: If you want energy as heat, the best way to generate energy is combustion. If you want energy as electricity, the best way is a fuel cell (which also produce some heat). If you want mechanical work, using hydrogen like gas or oil in a usual combustion engine, the yield would be a maximum of $30$%. The best yield would be to produce electricity in a fuel cell, then use this electricity to run an electric motor. Here the yield is of the order of $60$%, twice as much as the combustion engine.


Did god create logic?

Question: If God made logic, the process itself requires pre-existing logic. Since logic must be eternal, it's doubtful God created it. Did God create logic, or is it always there?
Comment: God created the universe but logic is humN. It was created by Pamenides, Plato and Aristotle.
Comment: My logic is: no, God does not created logic. Logic is a way to predict behavior of practical experiment. If 1, 2, 3 expressing some truth, then 1+1 = 2 meet some circumstances to be true, but same truth, 1+1 = 3 degenerate in falsity, because do not meet some circumstances. Therefore, even if we have some truth, truth can degenerate in falsity, because does not meet desired [circumstances.](https://medium.com/@gvitalie/zero-is-odd-or-even-number-9472f489bf94) Therefore, logic is relative, even if truth may be absolute, and truth can degenerate in falsity, because logic is relative.
Comment: @I vote to close the question. - Please define first your concept of God. Then 2) pose a question concerning the arguments in favour and against God. If you are convinced that God exists, then 3) pose a question for arguments that God has created logic, and make clear which logic you mean.
Comment: See e.g.  Andrew Pessin, [Divine Simplicity and the Eternal Truths Descartes and the Scholastics (Philosophia, vol. 38, issue 1, 2010)](https://www.academia.edu/104277073/Divine_Simplicity_and_the_Eternal_Truths_Descartes_and_the_Scholastics?uc-sb-sw=57218097) for a discussion of the well-known "Descartes' view that God freely created the eternal truths, such that He could have done otherwise than He did."
Answer: First you will have to define what is God ?Even if we don't define God and  we ask about the origin of logic then rudimentary logic can be understood as an innate ability to satisfy desires. Consider the logic - if you are hungry then you must search for food , or , if you are thirsty then you must search for something to drink.
Cravings are more fundamental than any apparent understanding of logic. You don't need to understand logic , when you are thirsty or hungry. The logic automatically gets executed. Moreover, we are not purely logical beings. We sometimes go mad. We indulge in irrational behaviour. Did God created madness or irrationality? No. He didn't. Did God created cravings to eat and drink ? No. He didn't because it ultimately leads to suffering.
Therefore we can safely say that logic arises , changes and vanishes. But who created the logic does not apply because there is an absence of true mind or at least a mind which is superior than cravings.
Comment: God did not create suffering? That's too facile a solution. In the Christian world [theodicy](https://en.wikipedia.org/wiki/Theodicy) is generally accepted as a fundamental (Christian) question. There are analogous issues/paradoxes across [all religions](https://philosophy.stackexchange.com/a/64065/37256)
Comment: @Rushi Even the God suffers. Why would God create his own and his followers suffering?
Answer:
You say you don't believe in God, yet you believe in grammar
Nietzsche

Logic and grammar are quite similar for the purposes of this question.
And they both point to the paradoxes that arise in the belief in a literal Creator-God.
Also related are the set theory paradoxed of Cantor and Russell which arise due to unrestricted quantification.
Which in turn is related to unrestricted "Creator-ability".
Also related is the quip that used to be directed against the Windows OS: Why does Windows have a shutdown in its startup menu but no startup?
Hint: The frame for the menu — a running OS — does not exist at startup.
All the above are framing issues. What happens without a frame?
Answer: Logic is the (formal) rational representation of the laws of the universe. Logic is moreover created by men, while the laws of the universe are not created by men.

Did god create logic?


Prove regular languages are closed under f(n) = 2^n and f(n) = n^2

Question: Suppose $ R $ is a regular language, let $ f(R) = \{ w \mid \exists x \text{ such that } |x| = f(|w|) \land wx \in R\}$, prove that $ f(R) $ is regular for $ f(n) = 2^n $ and for $ f(n) = n^2$. I've been on this question for a while now, and I'm not really sure how to approach it. I only really know elementary techniques like Myhill-Nerode and constructing DFA/regex, so I probably won't understand solutions using more complex logical models of regular languages.
Comment: A (somewhat technical) characterization of such functions where regularity holds is given in [Closure of regular languages are closed under certain cut operations](https://cs.stackexchange.com/q/71021)
Answer: For $f(n)=2^n$:
Let $M=(Q,\Sigma,\delta,q_0,F)$ be the DFA recognizing $R$. Denote by $\mathcal{G}$ the set of functions from $Q$ to $\mathcal{P}(Q)$ (where $\mathcal{P}(Q)$ means the power set of $Q$), and for any $g,h\in\mathcal{G}$, denote by $gh$ the function defined as $$gh(\cdot)=\bigcup_{q\in h(\cdot)}g(q).$$
Construct a new DFA $M'$ of whose states each is labeled a pair $(q,g)\in Q\times\mathcal{G}$. The transition function $\delta'$ of $M'$ is defined as
$$\delta'((q,g),a)=\left(\delta(q,a),gg\right),$$
for all $q\in Q, g\in\mathcal{G}, a\in\Sigma$. The start state of $M'$ is $(q_0,g_0)$ where $g_0$ is defined as $g_0(\cdot)=\{\delta(\cdot,a)\mid a\in \Sigma\}$. The set of accept states of $M'$ is $\{(q,g)\in Q\times\mathcal{G}\mid g(q)\cap F \neq \emptyset\}$.
Now we claim that

After reading a word $w$, if $M$ reaches state $q$, then $M'$ reaches the state $(q,g)$, where $g(\cdot)$ represents the set of states $M$ can potentially reach when starting at state $\cdot$ then reading a word of length $f(|w|)$.

This cliam can be proven by mathematical induction on $|w|$. With this claim, we can see $M'$ indeed recognizes $f(R)$.

For $f(n)=n^2$, the proof is similar. In this case, each state of $M'$ is labeled a tuple $(q,g_1,g_2)\in Q\times \mathcal{G}^2$, the transition function becomes
$$\delta'((q,g_1,g_2),a)=\left(\delta(q,a),g_1g_0,((g_2g_1)g_1)g_0\right),$$
where $g_0(\cdot)=\{\delta(\cdot,a)\mid a\in \Sigma\}$ as above, the start stae becomes $(q_0,I,I)$ where $I(\cdot)=\{\cdot\}$, and the set of accept states becomes $\{(q,g_1,g_2)\in Q\times\mathcal{G}\mid g_2(q)\cap F \neq \emptyset\}$.
Also, our claim becomes

After reading a word $w$, if $M$ reaches state $q$, then $M'$ reaches the state $(q,g_1,g_2)$, where $g_1(\cdot),g_2(\cdot)$ respectively represent the set of states $M$ can potentially reach when starting at state $\cdot$ then reading a word of length $|w|,|w|^2$.
Comment: Wow that's really clever way to sort of make the DFA "remember" |w|. I was under the impression that an approach like this won't work.
Answer: $f(R)$ is regular since it can be decided by a one-tape TM in linear time.
Do a constant-sized matrix multiplication inside the TM's finite control.
The DFA transition can be viewed as a 0-1 matrix $A$ by forgetting the arc label. First comput $A^n$, then bounce back to compute $A^{n^2}$. Then proceed to run the DFA as usual to know the last state it ends after finishing readinv $x$. Query the computed $A^{n^2}$.
Similarly for $2^n$.
Comment: The language $\{a^nb^n\mid n\in\mathbb{N}\}$ can be decided by a one-tape TM in linear time but it is not a regular language.
Comment: Nope, it is impossible to decide that language by a one-tape TM in linear-time. You are confused with the two-tape one. This is exactly the proof of non-linear lowerbound for that language. In fact, an $\Omega(n^2)$ lowerbound for one-tape can be proved. For 2-tape, you can do it in $O(n\log(n))$.
Comment: And you are repeating my idea in details in your answer. Ok, more TeX typesettings may get accepted easier.
Comment: You are right DFA = linear time TM but the proof seems a bit tricky. In addition $\{a^n b^n\mid n\in\mathbb{N}\}$ can be decided by a one-tape TM in $O(n\log n)$ by moving the counter in the tape.
Answer: For this and similar problems you need to first prove a property of finite state machines: The set of lengths of strings that transition from one state to another is either a finite set of integers, or there is a k >= 1, and two finite sets X and Y such that every possible length is either in X, or is equal to n*k + y, where n >= 0 and y is an element of Y.
Second, if we know the lengths of all inputs going from state S to a fixed accepting state T, can we determine the set of integers k, such that after going from initial state to S, we can reach the accepting state after a total of f(n) steps?
As an example, if we took k steps from initial state to S,and there are 7j + 15 steps needed to reach an accepting state, and f(n)= n^2 then n^2 mod 7 = 0,1,4,2,2,4,1 so k+7j+15 = n^2 If k mod 7 is 7, 0 or 3.
And here it is essential that not only can we solve this for k, but the solution is easy enough that we can modify our stat machine to handle it.


Is it haram to do height increasing meditation?

Question: I am 20 years old and only 4ft 10 inch. I am doing a pituitary gland meditation which increases height. Is it something haram?
Comment: If you have a problem with height due to hormonal imbalance then you should see an endocrinologist. They can prescribe treatment and artificial hormones which will be more effective than meditation.
Answer: Meditation by itself is not haram as long as it is not related to an other religion. However, this may often be the case if it is not explicitly done in the context of Islam belief.
I personally doubt that meditation can make you grow taller and I suspect that those who promise it want to influence you to practice something that may be haram or just make money.
Comment: It doesn't include any mantra and other things. It's like looking into a dot for couple of minutes  with open eyes and then imagine the dot with closed eyes. That's it. I just wanted to make sure if it's okay.
Comment: Then it is probably not a problem with Islam but I think you should follow the hint of UmH and look for something more serious. You won't grow by looking into a dot.


Show that every subspace of $\mathbb{R}^n$ is closed

Question: Show that every subspace of $\mathbb{R^n}$ is closed.
I'm not sure how to do this or even what closed means. I don't even have a starting point. Any hints or solutions are greatly appreciated.
Comment: I assume you mean vector subspace? If it's only a topological subspace then it's extremely false.
Comment: yeah, vector subspace
Answer: Let $S$ be the subspace. Let $v_1,...,v_p$ be a basis for the subspace, and let $v_{p+1},...,v_n$ be such that $v_1,...,v_n$ form a basis for $\mathbb{R}^n$.
Define the linear map $\phi:\mathbb{R}^n \to \mathbb{R}^{n}$ by
$\phi(v_k) = \begin{cases} 0, & k \le p \\
v_k, & k > p\end{cases}$, and note that
$S = \ker \phi$. Since $\phi$ is continuous, it follows that $S$ is closed since $\ker \phi = \phi^{-1} \{0\}$ and $\{0\}$ is closed.
Answer: A closed set contains all of its limit points.
Answer: A set is closed if and only if its complement is open. In $\mathbb{R}^n$ (or any metric space), a set $U$ is open if and only if for any $x\in U$ there exists an $\epsilon>0$ such that all points of distance less than $\epsilon$ from $x$ are contained in $U$.
Answer: If $M$ is a $k$-dimensional subspace of $\mathbb{R}^{n}$, then you can use Gram-Schmidt to find an orthonormal basis $\{e_1,e_2,\cdots,e_k\}$ of $M$, complete this to a full orthonormal basis $\{e_1,e_2,\cdots,e_k,e_{k+1},\cdots,e_{n}\}$ and create an isometric isomorphism
$$
                   U : \mathbb{R}^{n}\rightarrow \mathbb{R}^{n}
$$
that maps $(x_1,x_2,\cdots,x_k,0,0,\cdots,0)$ onto $M$. So the problem reduces to showing that this second description, which is really $\mathbb{R}^{k}$, is closed.
Answer: I think the other answers are a bit high brow. You don't even know what a closed or open subset is.
The question seems to be asking, show that every vector subspace is topologically closed.
In $\mathbb{R}^n$ we may think of topologically closed sets as those that contain their boundary elements. This is a specific case of the definition given earlier talking about containing 'limit points'. An open subset is one that doesn't contain any of it's boundary elements.
An example in $\mathbb{R}$ is the interval $[0,1]$ is closed because it contains $0$ and $1$. These are the boundary elements because any open subset in $\mathbb{R}$ containing 0 must contain points outside of $[0,1]$ and the same is true of $1$.
The set $(0,1)$ containing all points between $0,1$ but not $0,1$ is open because it doesn't contain any of it's boundary points.
In higher dimensions, we can think of an open 'interval' as any 'blob' sitting inside of all of the dimensions of the space. So in $mathbb{R}^2$ this would be any section of the plane enclosed by a line. But not a line or a dot or any combination of lines or dots. An example might be a square on the plane without boundary.
We want to show that a vector subspace is a closed set. So that is, it contains it's boundary elements.
In $\mathbb{R}^n$ a set of boundary elements will itself be a closed set, because any open subset containing elements of this will contain elements of the boundary and elements outside the boundary. Therefore a boundary set is it's own boundary set, and contains itself and so is closed. And we'll show that a vector subspace is it's own boundary set.
Suppose we have $k$ linearly independent vectors $x_1,...,x_k\in \mathbb{R}^n$.
When we talk the span of these vectors we get a vector subspace.
$$S = span(x_1,...,x_k)$$
All vector subspaces are formed with such a span, and this isn't unique.
We take a point in $x \in S$, and by the definition of $span(x_1,...,x_k)$, $x = a_1x_1+...+a_kx_k$, with $a_1,..,a_k \in \mathbb{R}$.
We consider an arbitrary open subset $U_x$ on the point $x$. This contains elements of our subspace obviously.
I'll prove the above statement. We consider a ball centred at $x$ $B_x \subset U_x$ and take the radius of the ball $r$. We take $\epsilon = \frac{r}{2}$ and then consider the point $x+\epsilon x_1 = (a_1+\epsilon)x_1+...+a_kx_k$. This is still in $S$ by definition of a span, and in $U_x$ because it is in $B_x$ because the distance between $x$ and $x+\epsilon x_1$ is less than the radius of $B_x$.
We also want to show that for all $U_x$ there is a point not in our subspace. This is easy. Since $S$ is a strict subspace, we can take $x_{k+1}$ linearly independent from $x_1,...,x_k$. By the same construction we can consider a ball $B_x \subset U_x$ and $x+\epsilon x_{k+1} \not \in S$.
Thus for all $x \in S$ and all $U_x$ such that $x \in U_x$, $x$ is a boundary point of $S$. Therefore $S$ contains it's boundary, and $S$ is closed, for all subspaces of $\mathbb{R}^n$ formed by taking the span of vectors. Which is all vector subspaces.
So all vector subspaces are topologically closed.
Answer: Let $S$ be a linear subspace of $\mathbb{R}^n$. Consider a sequence $\{x_n\}_{n\in\mathbb{N}}$ of points in $S$ which converges to a point $y\in\mathbb{R}^n$. Show that in fact $y$ must lie in $S$. (Here I am using a characterization of "closed" which is equivalent to that described in the other answers).
You could use the continuity of any projection map to the subspace, if that's available to you or easy to prove. You could also show it directly by just doing the analysis. (Hint: the finite-dimensionality should matter)


Can I generate CREATE TABLE from a visio diagram?

Question: I have a database diagram done up in Visio 2003 Professional.
Is there any way to re-engineer the database from the diagram?
Answer: You need Visio for Enterprise Architects, which was previously supplied with some of the higher editions of Visual Studio. It (and only it) had the ability to generated database scripts from a Visio diagram.
Answer: To my knowledge, older versions of Visio could do it in the standard version. Visio 2000 (yes, that old) had very solid forward and reverse engineering functionality. Maybe you can get hold of that version (current versions should still be able to save in that format).


Difference between '-ться', and '-тся'?

Question: 'Тс' already sounds like 'Ть' so why is there a soft sign there? Do I pronounce them both like a soft T?
Comment: Is the question about verb endings?
Comment: The following is incorrect: "Тс' already sounds like 'Ть' " - these two are pronounced as "цц". Смеются произносится как "Смеюцца", смеяться произносится как "смеяцца"
Comment: @Artemix: the original question was: "ть" already sounds like "тс" - because unfortunately some sites teach that soft [t] is pronounced [ts]. So the OP's question began with that misconception. Thus your edit makes no sense and his question (in theory) makes no sense. A less confusing questions (that won't confuse new readers) would be: Why does the verb ending -ться have a soft sign, and how is it pronounced?
Comment: @CocoPop I do not agree with you. I just added dashes to show that question is about endings. In fact it is a common mistake for native speakers as well. Most answers posted prior to my edit are about verb endings. It's a pity that OP does not react nor to comments, nor to answers.
Answer: There is no difference in pronunciation but "ь" is kept merely for the sake of consistency.
You see, most infinitives end in "-ть". And the 3rd person singular/plural end in -т :

to love = любить
he loves = он любит

And now, the magic! To form a reflexive verb form you add 'сь' after a vowel and 'ся' after a consonant. So....

говорить + ся = говориться
говорит + ся = говорится
женят + ся = женятся

The pair "говориться/говорится" indeed sounds exactly the same because it so happens that 3rd person singular for many "-ить" verbs of the И-conjugation differs from the infinitive only by the hardness of the last consonant.
Answer: Actually ь is there to indicate the infinitive. However, there is no distinction in pronunciation between -ться and -тся : they're both pronounced -тца.
смея́ться - [сме•я́т•ца]    infinitive
смеётся - [сме•йо́т•ца]  3rd p. sing.
смеются - [сме•ю́т•ца]   3rd p. pl.
Comment: Thanks. Haven't gotten into verb aspects much but I remember the infinitive part about 'ть'.
Comment: This actually has nothing to do with aspect; it's phonetic, not grammatical - both perfective and imperfective verbs have these endings.
Answer: It's mostly grammatical and as correctly stated above "ться" usually indicates the infinitive. It is relatively simple to catch the rule by using the following cheat:

If the word answers the "что делать" question use "ться".
If the word answers the "что делает / -ют" question you use "тся".

Yes, if there's "ь" in the question word, feel free to use "ться", and the opposite - if there's no "ь" in the question, use тся.
Comment: Yes, this is the way how pupils are taught in Russian schools. But it seems to me that such mnemonics are suitable for native speakers only.
Comment: Ah, I see :) In fact there are many native speakers on Russian.SE (including myself) and I thought you may be one of them. It's good that this "cheat" works for non-native speakers, because there are some language features that are intuitive for natives (like pairs of verbs of different aspect) and should be learnt by heart by others.
Comment: This cheat is more confusing that the original question! This looks like it would only work if you were Russian.
Comment: I am not native speaker and this "cheat" sounds perfectly suitable for me. ;) Requires certain language proficiency, but still.
Comment: Well I am not Russian however I learned it during teenager years and find this explanation clear. I am sorry if it does not work for you.
Answer: Some native russian have troubles with "-тся"/"-ться" too.
So we have special site for that: http://tsya.ru/
Comment: Neither your post nor that site you are linking to answers the question in any way.

Difference between '-ться', and '-тся'?


Formula for non - active power of capacitors connected in a star?

Question: There is a common question in my coursework that we are asked to :
Let's say we have an inductive 3 - phase load that consumes $$S=15 + j10$$MVA
"If , we connect parallel to this load, 3 capacitors that form a star, compute the capacity of each in order to achieve, unit power factor.  "
I know, there is a formula that says $$Q_c = \omega*C*V^2$$ but I don't know the proof( we were just given the formula right away) and I am not able to identify if I am using it correctly.
If V, is the polar voltage ("line voltage" I think it is in English), then does this $$Q_c$$ represent the non - active power that the whole star produces?
So, I can say (??): $$C=\frac{Q_c^{3ph}}{\omega V^2}$$
What, would change if those capacitors were connected such as to form a triangle?
Answer:
I don't know the proof( we were just given the formula right away) and
I am not able to identify if I am using it correctly.

\$Q_c=ω∗C∗V^2\$ is the reactive power (Q) of a capacitor (C) connected across an AC voltage source (V). It's related to real power in a resistor (\$P = \frac{V^2}{R}\$) but, because we know the reactance of a capacitor is \$\frac{1}{ωC}\$, the reactive power formula becomes as you show it.

does Qc represent the non - active power that the whole star produces?

No, it's the single phase reactive power of a capacitor across a voltage source. If you have three capacitors in star formation and you only have the line voltage then, the reactive power for each of those capacitors is found using \$\frac{V}{\sqrt3}\$.
In fact you might as well convert your first formula to a single phase apparent power by dividing it by 3 then solving as a single phase problem.
Answer:
If you have phase-to-phase (Vpp) and phase-to-neutral/ground (Vpn) voltages, you may calculate reactive power for three phases in the following way
$$
Q_{3ph}=3Q_c=3V_{pn}^2\omega C=3\left(\frac{V_{pp}}{\sqrt3}\right)^2\omega C=V_{pp}^2\omega C
$$
By the way, this is the magic of calculating balanced three phase circuits. You draw scheme for just one phase, then pretend you have phase-to-phase voltage instead of phase-to-ground and get three phase powers instead of single phase powers. Very handy. Just remember that currents are square root of 3 times bigger than in reality.
Comment: @brucebanner My formulae are for star connection. So \$Q_{3phy}=380^2\omega C\$ or  \$Q_{3phy}=3*220^2\omega C\$ if you like. For delta connection sum three capacitors at Vpp voltage, thus  \$Q_{3ph\Delta}=3*380^2\omega C\$
Comment: so , if for example $$ V_{pp} = 380V$$ then , $$V_{pn} \approx 220V$$ and $$Q_{3ph}=220^2 * \omega C$$ (for a star connection)  am I right ?
Comment: but if it was a triangle connection: then we would have $$Q_{3ph}=380^2*\omega*C$$?


How can I share save games between multiple PCs?

Question: How can I share a savegame for Dwarf Fortress between multiple computers (laptop and desktop)?  Can one instance of Dwarf Fortress run graphics and the other not?
Answer: Yes.  When you "Save and Quit" the game is saved to the hard drive.  This can be uploaded to other computers.  DwarfFortress.exe uses the folder path specified in its config file to look for saves on your computer.  It's worth noting that you do need identical "world" files as well or the game will crash.
The easiest way I've found to do this is to use a software versioning system (such as SVN or GIT) and version the entire folder.  This will have the added benefit of indexing your saves and allowing you to go back to previous save points.
Your graphics set does not need to be the same.  In fact, on a single computer you do not have to have a consistent graphics set between saves.  You do, however, need to make sure your config file correctly identifies the graphics set you plan to use.
Sharing saves between different people is known as a Bloodline game.  Some famous ones can be found here. Boatmurdered being one of the most famous.
Comment: Shouldn't that be *infamous*?  But yes, I use SVN to keep track of it now.  Smart guy that StrongAxe.
Comment: I used to use SVN, but I've found I like GIT better.  SVN tends to be more painful for merges (ie. I rewrite my config file and want it represented everywhere).
Comment: @C.Ross, @tzenes what hosting provider do you use?
Comment: @antony.trupe I Host it on a Linux server at my house.  This is great for at home, and I don't even try when I'm away.  @tzenes I only version my save folder, so I can run graphics in one place and not the other, etc.
Comment: SVN has the nice property that any subfolder can act as the head of a repository.  So for a while I had the entire thing on an SVN repo so I could bootstrap a new comp, and just updated the save folder.
Answer: I use Dropbox! The easiest way is just to drop the entire Dwarf Fortress folder into the Dropbox folder. This would also make it easy to share with others, since you can make Dropbox folders publicly or small-group accessible. (Of course this doesn't allow you to have different configurations, but you might be able to modify my setup for your needs.)


Returns to scale in perfect competition

Question: In a perfectly competitive market, is it necessary for the production functions to have decreasing returns to scale? Can it have a constant or increasing returns to scale production function like $q(l,k) = l^2k$? Further, is homogeneity of the production function a necessity?
Answer: If the price of the output good is positive, there can be no profit-maximizing profit plan for the production function $q(l,k) = l^2k$, which means that this production function is not compatible with perfect competition.
However, it is in principle possible to have a production function with increasing returns to scale under perfect competition- provided producing positive amounts is never optimal. Here is an example: Let the output price be $p=1$, the input price be $w=1$ and the production function be given by $q(l)=l-\log(1+l)$. This production function is strictly increasing and strictly convex, which means it also has increasing returns to scale. But the slope is always smaller than $1$, so the unique optimal production plan is to produce zero output using zero input.
More importantly, a production function needs to have neither decreasing, constant, or increasing returns to scale if one takes these notions to be global.
Answer: The answers to these questions is "no, yes, no".
Note that if you have increasing RTS like in your example and fixed prices, it is likely that the firm can always increase their profits by doubling input use, thus there won't be a profit maximizing output.
Comment: Yes, there won't be any profit maximizing solution. So in that case, how does the firm pick which quantity and inputs to choose? Can you please explain this in your answer? I learnt that a competitive firm has zero profit in the long run, so this seems contradictory or at least, confusing.
Comment: There is no positive answer to that question, which is why they usually assume CRS or DRS.

Returns to scale in perfect competition


Protective put became ITM before expiration. How to prevent option exercise?

Question: Options beginner here. I have a protective put on a stock and it became ITM 2 days before expiration. I do not want  the option to be exercised, since I'm bullish on the stock, and want to continue holding. I'm afraid that if I leave the put to expire, it will be exercised. I also do not want to sell off the put, because I will then be in obligation to the put buyer to buy stock. What are my choices here?
Comment: Sell to close your position.
Answer: If you sell your Put (closing the position), you are not in any further obligations, you get your sale price in cash and walk away.
I don't understand what makes you think you would be in any relationship with the put buyer - if you don't have or own any Put, you are out of the deal. If your buyer executes the Put, a random put writer is picked to fulfill it - from all the puts still written. You are no longer in that list, as you closed your position.
Comment: I guess the confusion might stem from the fact that the option writer is sometimes also referred to as the option seller. The key here is that it actually doesn't hold the other way around, just selling an option you bought before doesn't make you an option writer.
Comment: Not really a clear explanation. The only obligation that the OP would have would have would be if his put was ITM at expiration because the OCC exercises ITM options then.  Prior to expiration he has the right rather than the obligation to sell the stock and could never be in "in that list".
Comment: Thank you for clearing the air. I assumed an option seller is an option writer!
Answer:
I also do not want to sell off the put, because I will then be in obligation to the put buyer to buy stock.

The above is incorrect and is not how it works.
As the owner of the put, you have the right to sell the stock at the strike price any time before expiration.  It's your decision, not someone else's.
However, the decision would be out of your hands at expiration because the OCC exercises options that expire in-the-money, whether they be long or short.  This is called Exercise By Exception.
If you are long the option, you can designate to the OCC via your broker that you do not want your long option to be auto exercised. This would make sense if it about to expire with the underlying near the strike price (pin risk).
What makes the most sense is if your put is ITM, sell it to close for its gain or its salvage value.  Then, your  option position will cease to exist.
Options 201:  If you would like to continue to own a protective put, if you have the appropriate option level approval, use a spread order to close the existing long put and buy a put for a later expiration.  It's more efficient.


Tab colors in ProdPowerTools using RegEx and full file path

Question: I've just installed PPT on VS2010. There is ability to color tab by project and by reg.ex. (of file name). May be anyone knows is it possible to color also by full file name/path?
for example: I have two files "Index.cshml" one placed in the "User\" subfolder and one in the "Products\".
If I set up color Reg.Ex as ".User." then only those files will be colored that have "User" in their names : "Controolers\UserController.cs" but not an "Views\User\Index.cshml"
Answer: There is Settings to use full document path for regex.