Did the Buddha ever say excessive meditation is bad?

Question: I believe that I remember once reading that the Buddha abandoned excessive meditation. Is this true?
Comment: The question is very broad in it's current format - do you have any recollection of where in the Tipitaka (Pali canon) it's mentioned? Are you sure you are not thinking of the ascetic practices that the Buddha did before finding the right path?
Comment: Maybe it was at the time before his enlightenment yes
Answer: the buddha says that ''with excessive thinking and pondering I might tire my body, and when the body is tired, the mind becomes strained, and when the mind is strained, it is far from concentration.''
https://suttacentral.net/mn19/en/bodhi
same thing with ''excessive energy'' https://suttacentral.net/mn128/en/sujato
he says meditation must be ''made much of''
https://suttacentral.net/arv12/en/anandajoti
Comment: Ok thanks a lot
Answer: MN8 indicates that blissful meditation is incomplete on its own:

mn8: It's possible that some mendicant, with the giving up of pleasure and pain, and the ending of former happiness and sadness, might enter and remain in the fourth absorption, without pleasure or pain, with pure equanimity and mindfulness. They might think they're practicing self-effacement. But in the training of the noble one these are not called 'self-effacement'; they're called 'blissful meditations in the present life'.

The Buddha goes on to explain that self-effacement is important as well:

mn8: Now, Cunda, you should work on self-effacement in each of the following ways.
mn8: 'Others will be cruel, but here we will not be cruel.'  ...

The exposition is lengthy, but quite clear. Practice goes beyond meditation.
Comment: Ok thanks for that


Creation of polygon in the gap between two existing polygons

Question: If there is a gap between two polygons is it possible to fill it? Much like what happens using the hatch function in AutoCAD. The output would be a new polygon in place of the gap.
Answer: I don't know if QGIS has such tool but you can find that from OpenJUMP. The name of the tool is "Create Polygon from Closed Area.

For using the tool just click inside the closed area. New polygon can be created on any layer and in the image below I put into the red layer.

The tool is also topological and the closed area is searched by analysing all visible layers of the project. Here, find the closed area between green and red polygons and place the new polygon into orange layer.

OpenJUMP can save the result as shapefile or as OpenJUMP JML format. QGIS can read both and you can continue your work with QGIS, which is much more advanced software in many other areas.
Answer: I think this is what you're looking for: QGIS - Creating a new polygon layer from the space between other layers
Otherwise you can use the tracing tool and digitize it yourself.
Answer: In QGIS, I recommend Digitizing Tools plugin.
Once installed, you will see a new DigitizingTools Toolbar.

It has many icons, but what you need is Fill gaps which is 9th icon from the left.
By clicking on a small triangle button, you can select your desired tool from two options:
 Fill gap with a new feature (interactive mode), or
 Fill all gaps between selected polygons with new features.
First one is Interactive mode which allows you to select the gap you want to fill. And the second one will fill all gaps at once.
One good thing about this tool is that the polygons surrounding the gaps do not have to be topologically correct- They can overlap each other. (Just make sure the gap is a closed area).


How do I remove a brick patio embedded in concrete?

Question: How can I remove a brick patio set in concrete and on a concrete slab? Is it possible for a woman to do this herself?
Comment: For what it's worth, I'm a firm believer that a woman can do anything a man can do. Basic strength isn't that different, women may have an edge in stamina, and tools are a great equalizer.
Comment: Are you trying to remove the slab, or just the brick on top?
Comment: Why are you trying to remove it?
Comment: my dad always said: if it's too hard, then you've just got the wrong tool or technique. There was no mention of being a man or a woman.
Answer: If you have plenty of time, and don't want to hire power tools, you can do it a bit at a time over several weeks (months?). It will be hard work but anyone can do it.

That's a brick bolster and a club hammer, which is what I had to hand when I needed to break up some brickwork. You could try some other type of cold chisel or hammer. It depends what you can borrow or what advice you can get from the local store.
The brickwork I was attacking was set with mortar. If your concrete turns out to be too tough you may have to try power tools instead.
It so happens I have an old cheap Chinese-made SDS hammer drill that had been used to break up a large heap of set asphalt (you can buy these very cheaply, they won't be very durable but OK for occasional light use). These things usually have a rotation-stop lever and come with a chisel. I'd probably escalate to that before considering hiring a jackhammer.

A tool-hire shop should be able to advise.
Answer: With a jackhammer. If that woman can use a jackhammer, then she can do it herself.
Comment: Strictly speaking, you don't need a jackhammer, A sledge hammer, a four-pound hammer and a large brick chisel should do fine if you have the time.
Comment: @Edwin - Concrete is incredibly strong in compression and weak in tension.  Breaking with a sledge hammer is really difficult unless you can pry it up off the dirt to put it in tension.
Comment: I'm with @Edwin on this - cancel the gym membership, grab a pair of safety glasses, and think about things that irritate you, imagine them on the patio, and smash them. You can do it yourself and come out in great shape to boot. Having done a bit of it, I'm going to suggest that Comintern's comment appears to come from book knowledge rather than experience. Concrete will break when you ask it to, forcefully. Prying up is rarely practical or necessary.
Comment: One tip would be to start at one corner and break up a brick, then chip away the subsequent bricks in the direction of the empty space. That way rather than having to reduce the bricks to dust you should be able to pop them out whole. I should warn you that you might find a slab underneath the bricks and that will require some different tactics. Let us know if that's the case.
Comment: @Ecnerwal - Comes from experience of breaking up a bunch of sidewalks. Even an 1/8" lift makes a huge difference - I have an old 5' railroad pinch bar I use for this.
Answer: Sounds like it is on a concrete slab. Nevertheless the suggestions of using a small slam hammer first, and see if that is sufficient, is the way to go. The bedding for the brick is mortar so it should break up pretty easily.


Phosphate group acidic or not?

Question: I would like to ask whether or not the "Phosphate Functional Group" itself is acidic. I understand that the compound Phosphoric Acid is very acidic, with several hydrogens. However the Phopshate itself lacks hydrogens to donate. Please clear up any misudnerstandings I am having right now. Thank you so much.
Comment: Hi and welcome to chemistry.stackexchange.com. Feel free to take a [tour] of the site. Visit the [help] to learn more about it.
Answer: No, the phosphate functional group, $\ce{P(O)(OR)_3}$, itself is not acidic. Only protic phosphates are acidic. Phosphates that aren't phosphoric acids (i.e. $\ce{P(O)(OH)_{x}(OR)_{3-x}}$) are actually Lewis bases. The carbonyl oxygen is able to coordinate metals and other Lewis acids.
Comment: Right, but doesn't quite go far enough. In aqueous solution the phosphate ion itself is actually basic, not acidic. Thus $$ \ce{PO4^{3−} + H2O -> HPO4^{2−} + OH^{-}}$$
Comment: OP asked about the phosphate functional group. If asking about the phosphate (tribasic) anion, then what MaxW said is correct.
Comment: I believe you meant $\ce{HPO^{2-}_4}$.
Comment: Curious but pK2 of phosphate is 7.198 according to Wikipedia which would make the $\ce{HPO3^{2-}}$ slightly basic too. but all the second ionization of all(?) $\ce{R-PO(OH)2}$ compounds seems to be acidic. (I'd wonder if there isn't some weird molecule where the second ionization pKa would be more than 7 in aqueous solution.)
Comment: absolutely. thanks for that correction and it should be $\ce{R-O-PO(OH)2}$.

Phosphate group acidic or not?


How would the Earth look without water in the oceans?

Question: Has there ever been a movie, a documentary, or an image that provides a visualization of the oceans without water? I'm writing an article on geology and I thought a visual representation would be of great importance. Any help is appreciated.
Comment: I'm voting to close this question as off-topic due to insufficient research.
Comment: However, an lmgtfy link is a bit rude.
Comment: If there were no water ,you would not be here to wonder.
Answer: National Geographic made a documentary called Drain the Ocean, this is You Tube link to part one. the other three parts are there as well.
Comment: That is a wonderful documentary!
Answer: There is a YouTube clip How the Earth Would Look Like Without Oceans, it is only 1 minute 34 seconds; however, it is very graphical with no commentary, but with information about features of the ocean basins, ridges (and even the location of the Titanic!).
Information presented also includes spreading rates and depths.
Comment: cool! and do me actually know _what_ is there below?


Renewing French ID if living in another EU country

Question: I'm a French national living in another EU country (United Kingdom). My French ID will expire this year. Will I be able to renew it either from abroad, or in person in my old town in France? Or do I need to be a resident in France to renew it?
(I have a passport, as well, so strictly speaking I don't need it, but I find it easier to travel around the EU and within the UK on an ID card)
Answer: You can do it at a consulate, the costs (it's free if you still have the old one, €25 otherwise) and requirements are the same, except that you usually need to make an appointment in advance, whereas it's not always required in small towns. I have done it several times at different consulates with no issues (but it can take a couple of months). If you have a passport that's still valid or an ID card that expired less than 5 years ago, you only need that, two photos and a proof of address (justificatif de domicile).
When applying in France, I am not sure whether living in the town where you apply is actually a requirement but that's what vosdroits.service-public.fr suggests. Still, if you can get some proof of address (utility bill under your name if you still own a house or maybe some family willing to say you live with them…), you could presumably also do it in France without actually residing there but you will need to go back a few weeks later to pick the card up so it's not particularly convenient.
On the other hand, there is nothing special about your former place of residence in this respect. You can vote in local elections there – if you chose to remain registered – but not renew your ID. So to the extent that renewing your ID in France is possible, it should be possible elsewhere too.
Incidentally, registering with a consulate or notifying the authorities of a change of address is definitely not mandatory so even though there is an address on the card and you do have to submit a proof of address to get it, it does not have much evidential value and does not need to be up to date.
Answer: I would add two elements to Gala's great answer:

Go to https://www.service-public.fr/particuliers/vosdroits/F21089 to get the instructions to renew your French ID if living in another country.
France extended the validity of this card and it is now valid for 15 years (instead of 10), and it is retroactive. From https://www.service-public.fr/particuliers/vosdroits/F21089:

Si votre précédente carte a été délivrée entre 2004 et 2013 et que vous étiez majeur lors de sa délivrance, elle reste valide 5 ans après la date d'expiration qui est indiquée.

Google Translate
:

If your previous card was issued between 2004 and 2013 and you were of age at the time of issue, it remains valid 5 years after the expiry date indicated.

FYI: Which European countries accept an expired national French ID?


Raspberry Pi dnsmasq performance

Question: I want to block ads in my whole network with dnsmasq running on a Raspberry Pi, connected via Ethernet to the router. Is there any reason to take a RPi3? Or is a RPi2 also sufficient for that task?
Answer: An RPi2 is sufficient if your internet connection speed is less than 100mb/s which it is likely to be.  If you have a faster internet connection speed then you would be better off with the latest model of RPi3 (model B+) which has an ethernet speed of about 300mb/s (gigabit port but speed is limited by USB 2 bus).
Having said that I would try to make the change on the router instead.  I have DD-WRT installed on my router although I would recommend OpenWRT instead but either support dnsmasq modifications or even have specific services to block ads.
Comment: Your answer implies that the raspberry pi will be used as some sort of router between the internet and the LAN ... but, `dnsmasq` has nothing to do with routing
Comment: Yes, exactly. I just want to intercept the dns queries using `dnsmasq`. My home internet would be 10gbit synchronous, so the RPi would be a massive performance impact.

Raspberry Pi dnsmasq performance


Calculate the size of things which are present in image

Question: Is there is any software which can calculate the size of things which are present on picture or image? For example, I have an image of furniture I want to calculate its length, width , height size of that furniture.  Is there is any approach or way?
P.S.: Cost of the software is not a concern, however I prefer free software. The operating system which I am using is Windows 7 so it should run on it.
Comment: Welcome to Software Recommendations! We will need much more information to give good recommendations here. Please take a look at [What is required for a question to contain "enough information"?](//softwarerecs.meta.stackexchange.com/q/336/185) Then please [edit] your question and see if you can incorporate some of these improvements. Especially missing: What OS should it run on, and what's your price margin?
Answer: There might be software if you take a more feasible approach. Google Pixel phone apparently uses one camera but uses information from many video frames to determine depth. Depth is certainly required as a minimum for your needs.
There are more options available out there for determining depth if you use more than one camera - stereo vision. Details further away will have less accuracy though.
Even better if you have a radar or lidar.
Ultimately, you want to build a "point cloud" and then a 3d model of a scene. When the user chooses 2D points on the screen, they can be mapped to places in that 3D model and then measured.
You'll probably also need some kind of configuration step which adjusts for a given camera's lens and attributes.
Is there an off the shelf software for this? Not that I'm aware of. But each of the points above should help you break down your problem and find smaller components to use.
Answer: Those available for the PC aren't very useful for your situation. They mainly work to determine angles and areas from maps or angles for smaller objects.
If you can photograph the item with your own phone, there are a number of apps available for both iOS and Android.
EasyMeasure (iOS and Android) bases the measurements on your camera's height off the floor. It works pretty well.
There are several others that base their measurements against the known dimensions of common objects that you can place in the photo.
From reading your question I figure you're a woodworker wanting to build a replica piece. If so, you might swing by one of the larger woodworking forums with a furniture area and ask around there, too. Best of luck on your project.


What is the difference between Discussion board and newsfeed ? notifications can be done fo both?

Question: We are using SharePoint online, and I wanted to know what is the difference between discussion board and newsfeed? and can we activate notifications for both?
Thanks,
Answer: Newsfeeds
At its core the newsfeed allows you to post and reply to users in a format that encourages discussion and participation. The newsfeed allows you to see events, actions and posts from people, sites and documents that you are following.
You can check the colleague's status and post up-to-date. SharePoint newsfeed is where you can see what your co-workers are doing throughout your work day and share ideas and observations as they occur to you. Depending on your newsfeed settings, the feed may also send emails to you to notify the events, such as when someone mentions you in a post, and it may show your activities, such as when you've started following someone. You control email notifications and your activities you want to share with others
Newsfeeds display the most recent microblogs and activities (interactions with specific entities within SharePoint) to users; they are what pulls everything together. Newsfeeds are generically known as activity streams.  SharePoint 2013's activity streams show you all of the recent activities performed on the portal, as well as allowing you to interact with microblogs you see in your activity
Discussion Board
Discussion Board is used for communication among team members. The discussion board shows the most recent discussion first on top, as well as the number of replies for each discussion. User can quickly see which discussions have the most recent activity and which ones are the most popular discussion. Members can also customize their own views of the discussion board.
Key Features of a Discussion Board

Communication between team members
Shown recent discussion of team members
Admin user can set number or limit of discussions and shown on top


Could Gollum have been trying to prevent Sauron from obtaining the One Ring?

Question: I'm not doubting that Gollum was corrupted by the ring and that he wanted it completely for himself.  However, In the movie "The Two Towers", when Frodo and Sam arrive at the black gate, Gollum stops them from entering after the army.  He eventually says something along the lines of "He wants it, always seeking, and the precious wants to go back to him."(I don't remember the exact line).  So clearly, Gollum understands the danger reuniting Sauron and the ring would pose.  So my question is, in Gollum's own narcissistic way, was he actively trying to keep the ring to himself so that it never fell back into the hands of Sauron?  After all, he didn't learn what Frodo was actually trying to do until much later in their journey.
Answer: Yes and no.
This sequence is in the books too, in the Two Towers chapter The Black Gate is Closed, and here Smeagol/Gollum's words are stronger:

'No, no, master!' wailed Gollum; pawing at him, and seeming in great distress. 'No use that way! No use! Don't take the Precious to Him! He'll eat us all, if He gets it, eat all the world.'

It's notable that, while Gollum claims a motive of wanting to prevent danger to the world, this is also exactly the point at which he suggests the alternative route through Cirith Ungol.
Later in the same passage Gollum expands on his motive a little more:

So Smeagol is very afraid. He does not want to lose nice master. And he promised, master made him promise, to save the Precious. But master is going to take it to Him, straight to the Black Hand, if master will go this way. So Smeagol must save them both...

(Note that when Gollum says "Precious" with an upper-case "P" he's referring to the Ring, with a lower-case "p" to himself.)
From this it's clear: Gollum knows the risk of Sauron regaining the Ring and very much does not want that to happen.  As to the reason why: so that he can have it instead, or for a more altruistic motive, we don't really have sufficient information at this point in the narrative (although it's also clear that at a later point it was so he could have it, but this reason could have changed from the earlier time).
Comment: @GemmanAster - you're forgetting that at one stage Gollum *did* come close to genuinely repenting.  Tolkien notes this in Letter 96 - "the tragedy of Gollum who at that moment came within a hair of repentance – but for one rough word from Sam" - and the scene is in *The Stairs of Cirith Ungol*.  This isn't a matter of "opinion" or "interpretation" - it's what Tolkien actually wrote.
Comment: Gollum understands that should Sauron obtain the Ring, Gollum will never see it again. Sauron is far too powerful and far too attached to the Ring to give Gollum any chance of stealing it.
Comment: There is no ambiguity behind his desire for Sauron not to have it.  He is also wily enough and malicious enough to exploit Frodo and Sam in terms they understand and fear.  At the moment he begs them not to 'let him have it' he knows he cannot by force take it - yet - so instead emotionally blackmails them and buys time to work out his plans with the giant spider a few days up ahead.  Every one who dealt with Gollum was to a degree 'taken in' and his blackest actions only turned to good due to the unseen hand of Eru.  This dude **eats babies in their crib**...  You cannot come back from that!
Comment: Nice. I knew capitalisation of certain terms was important, but I never noticed Precious vs precious.
Answer: I don't think so. My impression of Gollum is he is very childlike in his mind. I felt that he just wanted to keep the ring to himself because of a childish possessiveness, rather than some innate knowledge to keep the ring from Sauron.
However you do bring up a good point. Unfortunately the only way we would really know is to ask Tolkien, but that isn't going to happen.
By the way, I'm going from knowledge of the books. I've never seen the films, nor do I desire to.
Comment: You really should see them. For all their faults, they're excellent films.
Comment: I think we share an opinion there @Timothy!!!  Nonetheless I have seen them and bought the various disks in so many different formats...  Jackson has made money from me!!!  However, I have done so because here and there - sometimes only flickeringly - we get a glimpse that IS authentic and makes me think 'Yes, that **is** what it was like.'  Sometimes this makes obscenities like a small company of Sindar fighting at the Hornburg a **little** easier to stomach!!  Also - yes J.R.R. is gone - but remember it is _literature_ we are talking about, not math.  Our own interpretations are the key.
Comment: I agree.  Losing the ring to Sauron would have been as much of a blow to Gollum as it would have been for Sam and Frodo, so in a sense, they were allies.  However that is about where the similarities end.  Gollum would not have hesitated to kill them both and take the ring had he been able, and Sauron would have had nothing to do with it.
Comment: I watched only the first movie at the Cinema, then I read the book. I tried later to watch the trilogy but every time I managed to watch less and less, bored by the numerous (and IMHO unnecessary) changes made by Jackson. Same happened for the Hobbit. :(


A necessary and sufficient condition for a space curve to lie on a ellipsoid

Question: Any (arc-length parametrized) space curve is uniquely determined (up to rigid motion) by its curvature  and its torsion.
For instance we know that a necessary and sufficient condition for a space curve to lie on a sphere is
$R^2+(R')^2T^2=const$, where $R=1/\kappa$, $T=1/\tau$, and $R'$ is the derivative of $R$ relative to $s$.
I want to know if there is a necessary and sufficient condition for  a space curve to  lie on  a ellipsoid (in terms of its curvature and  torsion).
Comment: This is a very natural and interesting question, and definitely constitutes research level mathematics in my opinion. I do not see why it is put on hold at all. Anyone who thinks the answer is trivial should try to solve it. You would be surprised. There are very few curves for which an intrinsic characterization is known, and it would be great if there are some reasonable equations which would characterize curves on an ellipsoid.
Comment: It should be noted that the criterion that you write down is only necessary and sufficient for *fully nondegenerate* space curves, i.e., (connected) curves for which $\kappa$ and $\tau$ are nowhere vanishing.  In fact, no local characterization can be both necessary and sufficient, since there exist (smooth, connected) space curves (even ones with $\kappa$ nowhere vanishing) such that every point of the curve has an open neighborhood within which the curve lies on a sphere, but the whole curve does not lie on any sphere.
Comment: I suggest that you first derive the conditions for a plane curve to be an ellipse.  That will give you a start.  The best way to do this (and to do the higher dimensional case) is to use the moving frame for affine curves to derive the condition for lying on a hyperquadric, and then, use the Euclidean moving frame to compute the affine moving frame.  This will give you the conditions you want.  A good recent source would be J. Clelland's book "From Frenet to Cartan: The Method of Moving Frames".
Comment: @NateEldredge this is not my homework.Since  I've known the result about a curve lie on a sphere, I just want to know if there is a similar result on a ellipsoid.
Comment: I'm voting to reopen per @MohammadGhomi's comment.
Comment: @user122298 The reason characterizing spherical curves is easy is that  one can easily express the center of the osculating sphere in terms of curvature and torsion. Setting the derivative of that to zero, then yields the equations. I do not see an easy way to do it for ellipsoids. It may even be possible that there is no reasonable or pretty answer. At any rate, it is not a trivial problem. Just finding some necessary conditions would be quite interesting.
Comment: Then it would be a good idea if you didn't state it like a homework problem, and explained the context and any known progress or partial results.
Comment: @NateEldredge thank you for your advice
Comment: By the way, you can use [LaTeX-style math formatting](https://math.stackexchange.com/help/notation).
Comment: @NateEldredge sorry,but I'm posting by my smart phone .It seems the app doesn't support type LaTex codes
Comment: @NateEldredge I'll use my computer to recompost this question with LaTeX code  later this day
Comment: Use a diagonal linear (or affine) transformation to translate the conditions for a sphere into that of an ellipsoid.
Comment: @MohammadGhomi thank you for your support,I am trying to solve it since this Friday, but haven't got something important yet. It seems there's a big gap between curves on a ellipsoid and on a sphere. That's not just a question can be easily solved by using affine transformation.
Comment: @PeterMichor since affine transformation doesn't preserve the differential properties of curves, this approach may not be helpful to find the relations between curvature and torsion.
Answer: There is a straightforward way to deduce necessary conditions for a space curve to lie on an ellipsoid, and it's really a matter of calculation to make these conditions explicit in terms of the curvature and torsion.  I'll describe how to do this and the result of the calculation below, but first let me insert a note of caution about the 'necessary and sufficient conditions' that the OP wants.
When a space curve is sufficiently nondegenerate, i.e., its curvature $\kappa$ and torsion $\tau$ and the first derivative of its curvature with respect to arclength $\kappa'$ are nowhere vanishing, the necessary and sufficient condition for the curve to lie on a sphere is that the expression $\bigl((\kappa')^2 + \kappa^2\tau^2\bigr)/(\kappa^4\tau^2)$ be constant.  Indeed, when it is constant under these nondegeneracy hypotheses, this expression is just the square of the radius of the sphere on which the curve lies.  Note that the assumption that $\kappa'$ be nonzero is necessary:  If $\kappa'$ vanishes identically, then the above expression is constant (because the $\tau$-factors cancel), but not every curve with constant $\kappa$ lies on a sphere (for example, consider the circular helices, which have $\kappa$ and $\tau$ constant and nonzero).
Now, any curve that lies on a sphere has positive curvature $\kappa$, but it need not have nonvanishing torsion $\tau$.  Thus, the above criterion does not make sense for all nondegenerate curves, i.e., space curves for which $\kappa$ is positive (which are the curves for which the classical Frenet frame is well-defined).  The following argument shows that one cannot hope to have a necessary and sufficient condition expressed in terms of local conditions for spherical curves that works for all nondegenerate space curves:  Consider two distinct spheres $S_1$ and $S_2$ that intersect along a circle $C$.  It is easy to construct a smooth curve $x(s)$ with positive curvature $\kappa$ that starts out on $S_1$ (but not on $S_2$), runs along $C$ for some interval, and then continues on $S_2$ (after leaving $S_1$).  This curve is locally spherical, but not globally spherical, so no local condition can be necessary and sufficient for all nondegenerate space curves.
Meanwhile, one can get an expression that makes sense and works for all nondegenerate curves by considering the identity
$$
\frac{d\ }{ds}\left(\frac{(\kappa')^2 + \kappa^2\tau^2}{\kappa^4\tau^2}\right)
= \frac{2\kappa'\bigl((\kappa\kappa''-2\kappa'^2-\kappa^2\tau^2)\,\tau - \kappa\kappa'\tau'\bigr)}{\kappa^5\tau^3}.
$$
In fact, any space curve that satisfies the nondegeneracy condition that $\kappa\tau$ be nonvanishing while
$$
P(\kappa,\kappa',\kappa'',\tau,\tau') = (\kappa\kappa''-2\kappa'^2-\kappa^2\tau^2)\,\tau - \kappa\kappa'\tau' = 0
$$
must necessarily lie on a sphere of radius $r$ where $r^2 =\bigl((\kappa')^2 + \kappa^2\tau^2\bigr)/(\kappa^4\tau^2)$ .  This is, in some sense, the correct statement of the classical result.  (Proof:  If the above equation holds and $\kappa\tau$ is nonvanishing, then the corresponding space curve $X:(a,b)\to\mathbb{R}^3$ satisfies the condition that $\bigl((\kappa')^2 + \kappa^2\tau^2\bigr)/(\kappa^4\tau^2)$ be constant.  Moreover, if $T$, $N$, and $B$ are the Frenet frame of $X$, so that $X' = T$, $T' = \kappa N$, $N' = -\kappa T+\tau B$ and $B' = -\tau N$, where the prime denotes differentiation with respect to arclength, then the curve $Y = X + (1/\kappa)N - (\kappa'/(\kappa^2\tau)) B$ satisfies $Y' = 0$.)
The reader may ask, "What about assuming real-analyticity?".  However, real-analyticity is not necessary for a space curve to be spherical, so this does not count as a local criterion that would be be part of a necessary and sufficient condition for any nondegenerate space curve to be spherical.  I think that the reasonable thing to do is to simply restrict to the appropriate class of sufficiently nondegenerate space curves, for which a necessary and sufficient condition to be spherical is available.  (Alternatively, one could restrict to the space of real-analytic nondegenerate space curves, but then one has to make a special exception for the curves with constant $\kappa$, etc.)
The seriousness of this problem becomes evident in the case of what we might call ellipsoidal space curves, i.e., the space curves that lie on some ellipsoid.  Now, even the above nondegeneracy ($\kappa$, $\kappa'$, and $\tau$ be nonvanishing) is not sufficient to avoid the above difficulty, for one can easily write down a pair of ellipsoids that intersect in a space curve that is not planar and does not have constant $\kappa$ and hence contains such nondegenerate arcs.  In particular, one can construct such a nondegenerate space curve that is locally ellipsoidal but is not globally ellipsoidal.  Thus, even the right notion of 'nondegenerate' needs to be specified in order to get anywhere.
Here is what I propose:  Let $Q$ be the 10-dimensional vector space of quadratic functions on $\mathbb{R}^3$, i.e., functions of the form
$q = a_{ij}\,x^ix^j + 2b_i\,x^i + c$ for some constants $a_{ij}=a_{ji}$, $b_i$, and $c$.  For any smooth curve $\gamma:(a,b)\to\mathbb{R}^3$, let $Q^k_\gamma(t)\subset Q$ be the linear subspace consisting of those quadratic functions $q\in Q$ such that $q{\circ}\gamma$ vanishes to order at least $k$ at $t\in(a,b)$.  Then, for obvious reasons, $\dim Q^k_\gamma(t)\ge 10-k$ for $k\ge 0$.
Let us say that $\gamma$ is $Q$-nondegenerate if $\dim Q^9_\gamma(t)=1$ for all $t\in(a,b)$.  It is easy to see that being $Q$-nondegenerate implies that $\gamma$ is fully nondegenerate (but is much stronger) and that being $Q$-nondegenerate can be expressed as the condition of non-simultaneous vanishing of a set of $9$ polynomials in $\kappa$ and $\tau$ and their derivatives with respect to arc-length up to order $6$ in $\kappa$ and $5$ in $\tau$.  Thus, it is an open condition on space curves.  Further, let us say that $\gamma$ is a $Q$-curve if it is $Q$-nondegenerate and $\dim Q^{10}_\gamma(t)=1$ for all $t\in(a,b)$.  This extra condition can be expressed as the vanishing of a certain polynomial $P$ in the $15$ variables $\kappa,\kappa',\ldots,\kappa^{(7)},\tau,\tau',\ldots,\tau^{(6)}$
that is of degree $27$ and has $8882$ monomial terms.
One then has the following result:
Theorem:  A $Q$-nondegenerate space curve $\gamma$ lies on a (necessarily unique) quadric hypersurface if and only if it is a $Q$-curve.  Moreover, if it also satisfies a certain pair of strict inequalities on the curvature and torsion and their derivatives up to order $6$, then the quadric hypersurface on which it lies will be an ellipsoid.
Comment: Is the condition being $Q$-nondegenerate the condition of "non-simultaneous vanishing" of some polynomials, or is it the condition of "simultaneous non-vanishing"? Also, why is the number of derivatives up to order 6? In analogy with the sphere case (which has $Q$ being dimension 4), don't we expect the non-degeneracy condition to be up to $\mathrm{dim} Q - 3 = 7$ derivatives in $\kappa$ and $6$ in $\tau$?
Comment: @WillieWong:  It's 'non-simultaneous vanishing', i.e., I don't want all of them to vanish at once, so, for example, it's OK if all but one of them vanish for the $Q$-nondegeneracy, and the $Q$-nondegeneracy condition only involves the derivatives of $\kappa$ up to order $6$ and $\tau$ up to order $5$.  It is *somewhat* like the sphere case, in that the appropriate nondegeneracy condition there is the nonvanishing of $\kappa\tau$, while the equation $P=0$ that must be satisfied also involves $\kappa''$ and $\tau'$.  (I allow *any* quadric surface for $Q$-curves, not just ellipsoids.)
Comment: Okay, I think I get it now. The idea is that by your definition if $q$ solves $(d/dt)^k q\circ \gamma(t_0) = 0$, so does $\lambda q$. So the one dimensionality is natural. The $Q$-nondegeneracy condition states that based on the data at $t_0$, there is only one quadric hypersurface that can be compatible (depending on the point $t_0$). The condition on $Q^{10}_\gamma$ is an integrability condition saying that the hypersurfaces can be pieced together. // What I did wrong in my previous comment was that I looked at only the 4 dimensional set of polynomials of the form $|x-a|^2 - r^2$, and not...
Comment: ...the five dimensional vector space $a|x|^2 + b\cdot x + c$, which linear structure is used in arguing that the condition on the top level derivative gives an integrability condition. Thanks! (Your parenthetical on allowing any quadric surface is what made it click for me.)
Answer: Robert describes the differential equations which one can write in terms of $\tau$ and $\kappa$, and the inherent limitations in this local approach. But maybe one can find more reasonable or useful conditions in terms of integral equations, and the whole problem could be  more interesting or natural if we consider closed curves. In other words, a global approach could be more enlightening.
For instance, a necessary condition for a closed curve to lie on a sphere is that $\int \tau=0$, e.g. see p.171 of Millman and Parker, which incidentally turns out to characterize spheres. Furthermore, any closed curve lying on a convex surface must have at least $4$ points where $\tau=0$, which is a generalization of the classical four vertex theorem due to Sedykh; see also this paper for another proof, and this paper for a generalization. Another necessary condition for a curve to lie on an ellipsoid is that it have a pair of parallel tangent lines, which turns out to characterize ellipsoids, as described in this paper with Bruce Solomon.
It would be interesting to find more simple or nice necessary conditions for a closed curve to lie on an ellipsoid, and I think it is possible that a collection of these may turn out to be sufficient as well.
Addendum 1: I found a nice paper which seems to be relevant:
Kreyszig, Erwin; Pendl, Alois
Spherical curves and their analogues in affine differential geometry.
Proc. Amer. Math. Soc. 48 (1975), 423–428.
In this paper the authors define a curve to be spherical in the affine sense if all its normal planes pass through a common point. If I am not mistaken these include curves which lie on ellipsoids, but other curves as well. At any rate, they obtain a very nice characterization for these "affine spherical curves":
$$
\left(\frac{1}{\widetilde\tau}\right)''+\frac{\widetilde\kappa}{\widetilde\tau}=0,
$$
where $\widetilde\kappa$, $\widetilde\tau$ are the affine curvature and torsion and diferrentiation is with respect to affine arc length.
Addendum 2: Maybe a necessary condition could be that $\int\widetilde\tau=0$, but this is just a quick guess.
Comment: @MohammadGhomi:  Just as in the Euclidean case, there are *two* affine curvatures of nondegenerate space curves, one of differential order $5$ and one of differential order $6$.  (These are the affine analogs of $\kappa$ and $\tau$.)  The affine arclength itself is of differential order $3$.
Comment: Note that, except for the criterion $\int\tau\,\mathrm{d}s = 0$, the others are actually affinely invariant conditions, which is appropriate, since the condition of lying on an ellipsoid is affinely invariant.  As I pointed out in an earlier comment, it would be more reasonable to seek for conditions in terms of the affine invariants of the curve, and, indeed, the differential equation characterizing $Q$-nondegenerate ellipsoidal curves in terms of affine invariants is *much* simpler than the one in terms of Euclidean invariants $\kappa$ and $\tau$.
Comment: @Wlod The only reference for this that I know is the paper "Scherrer, W. Eine Kennzeichnung der Kugel.  Vierteljschr. Naturforsch. Ges. Zürich 85, (1940)", cited in Millman and Parker.
Comment: I've seen a paper by Karol Borsuk, written in German, in which he proved $\int \tau = 0$ for closed spherical curves. Could you confirm this or was this theorem proven earlier by another geometer?
Comment: @Robert, Yes $\int \tau=0$
 is not affine invariant, since it characterizes only spheres, as opposed to ellipsoids. I think it would be interesting if this has an affine analogue. There are notions such as *affine arc length* and *affine curvature* which could relevant. I do not know if someone has yet developed *affine torsion*.
Comment: @Robert, maybe then the integral of the affine torsion would have to vanish?
Comment: @MohammadGhomi, I am sure that the mentioned Borsuk's paper was written before WWII. Hardly any Polish mathematician would write a paper in German during or soon after WWII for obvious reasons, and later because English took over mathematics (with mostly Germans, French, Soviets, and Japanese making an exception for their own languages).


Is there an easy proof that every closed curve is contained in a ball?

Question: I actually wanted to generalize this to higher dimensions (and prove something akin to the statement "every $\bf{n}$ dimensional closed manifold is contained in an $\bf{n}$ dimensional ball), but I'm very new at this yet so I don't really have a clue. It seems obvious enough (I mean, take any closed planar curve and there exists a circle in which it is contained... the same goes for closed surfaces, I can always find a sphere big enough), but I see no clear path on proving it.
Answer: I assume closed means compact without boundary and your manifold is inside euclidean space. Since compact sets of a metric space are bounded, you have your ball for free.
Comment: I think this answer could be improved: Is it by definition? By a theorem? Etc.
Answer: How is your definition of closed curve? If it just a continuous image of $S^1$ then it is compact, so bounded in $\mathbb{R}^n$
Comment: Hello to the anonymous downvoter who failed to tell me what his problem with my answer is. That way (s)he missed the oppurtunity to tell me how to improve my answers.
Answer: Maybe too late, but this is a quite simple explanation:
Consider a closed curve as a function:
$f:[0,1]\to\mathbb{R}^n$.
Next, this function must be continuous on the entire domain $[0,1]$ (since it's a curve), and the $f(0) = f(1)$ equality must hold (since it's closed, having no endpoints).
Continuousness in $\mathbb{R}^n$ means that for each $\varepsilon>0$ exists such $\delta>0$ that if $|x_2-x_1|<\varepsilon$, then $D(f(x_2),f(x_1)) < \delta$ in some metric (distance function $D$) defined in $\mathbb{R}^n$.
Of course, this must hold for $\varepsilon=1$, so we get that for any $x_1, x_2 \in [0,1]$ must be $D(f(x_2),f(x_1)) < D_{max}$, where $D_{max}$ is $\delta$ from above, defined for $\varepsilon=1$ (since $|x_2-x_1|<1$ holds for any $x_1, x_2 \in [0,1]$, except 0 and 1 themselves, but we know that $f(0)=f(1)$, so $D(f(1), f(0))=0$).
So, recalling definition of a ball
$$B_d(c)=\{x \in \mathbb{R}^n | D(c,x)<d\}$$
(a ball with center $c$ and radius $d$), we can fit our curve in a ball (for example, taking $c=f(0)$ and $d=D_{max}$)


If I transfer my character to Xbox One, can I continue playing on the 360?

Question: If I transfer my GTA 5 profile from an Xbox One to an Xbox 360, can I still play on the 360 and still get money and RP?
Answer: You can still play on your 360, and rank up and get money, yes, but what you do on your 360 after the point of transfer, will not count for your Xbox One character.
For example, if you transfer your 360 level 50 character to Xbox One, and then play your Xbox one character for a while, ranking it up to 60, your Xbox 360 character will still be level 50. This goes both ways.


Do distinguishable fermions obey the Pauli exclusion principle?

Question: We know that fermions are identical particles and obey Pauli exclusion principle. But what is meant by distinguishable fermions? Does that mean, like proton and electron both are fermions but they are distinguishable because of charge? And if we put together both distinguishable fermions, will they obey Pauli exclusion principle ?
Comment: You may want to fix the first sentence, as it seems to assert that fermions are indistinguishable by definition, which makes the rest of the question difficult to parse.
Answer: The simple way to think about it is to imagine that all fermions are excitations of a single field. These excitations can differ in their position, spin, charge, mass, and so on, and the Pauli exclusion principle applies to all of them. Mathematically, this is just the fact that all fermionic creation operators anticommute; the joint wavefunction of all fermions is antisymmetric.
So fermions that are far apart aren't affected, because they differ in position space. Fermions that have different spins aren't affected, because they differ in spin. And protons and neutrons don't affect each other because they differ in mass and charge. The Pauli exclusion principle always applies and there are no exceptions. Any two fermions must be different in some way to coexist.
Because spin and position are easily changed, and mass and charge aren't, sometimes people break the rule into two cases. They say that Pauli exclusion only looks at spin and position, and doesn't apply to things with different masses and charges (because they're automatically different). This works for simple situations, but it's dangerous, because if you take it too literally you'll get the wrong answer when constructing baryon wavefunctions. There you really do have to antisymmetrize over all degrees of freedom, including the type of quark. You cannot just apply it to each quark flavor individually.
Answer: This is the table of elementary particles, and if you read it carefully you will see that there are a number of quantum numbers, not only charge and mass, the make them individual and distinguishable.
Distinguishable fermions do not fall into the Pauli exclusion principle.
The proton is composite , and also in addition to charge has baryon number 1, the positron ( antiparticle of the electron ) has baryon number zero and lepton number 1, also their mass is very different. So they are distinguishable and do not obey the Pauli exclusion.
Comment: They obey Pauli exclusion, it's just that Pauli exclusion doesn't prohibit the proton and electron from co-existing.
Comment: @Acccumulation maybe my engish is at fault. I though exclusion comes from exclude ,and that means that the state is excluded if it follows the principle.
Comment: When you have a rule "If A, then B", then in cases where A is false, the rule is being followed. For instance, if there's a rule that says that vehicles with more than two axles can't drive more than 55 mph, and you're driving a vehicle with two axles at 60 mph, you're not violating the rule.
Answer: The Pauli Exclusion Principle is that no two indistinguishable fermions may occupy the same quantum state. It does not apply to pairs of distinguishable fermions (e.g. a neutron and a proton). If it did, then nuclear physics would be very different. See Are protons and neutrons affected by the Pauli Exclusion Principle?
Distinguishable fermions may be distinguished from each other by their mass, charge, spin, isopin etc.
Comment: there's some semantic distinction between identical and indistinguishable.  In the case where fermions have different masses etc, they are not *identical* and thus cannot possibly be indistinguishable.  Two fermions could be identical yet distinguishable if they were so far apart that their wavefunction would practically not overlap.


Trailing slash dups

Question: My short research on trailing slashes duplicate On page issues, covers only the dir & dir/ cases.
But what about the /file.html & /file.html/.
Are they acceptable both or one of them should be redirected?
Answer: You should always have 1 version of the page to prevent duplicate content being flagged.
People structure their URL's differently, some people never have trailing slashes, some always have trailing slashes and some only have trailing slashes on directories and non on files. I believe the last option is to do with a slight improvement to server speed (will need to look this up further)
Make sure you always use a canonical or a 301 redirect and be consistent.
Comment: "trailing slashes on directories ... a slight improvement to server speed" -  normal Apache behaviour is for mod_dir to "fix" the URL and append a trailing slash on a physical directory if it is omitted. Apache does this with a 301 redirect. So, yes, linking to a directory without a trailing slash would indeed see a performance penalty, in the same way that linking to a URL that redirects would.
Answer: Regardless of what you are linking to (physical directory, file or something entirely virtual), the URL with and without a trailing slash are two different URLs, so ideally they should be canonicalised.
These are two different URLs and could potentially return different content:

example.com/<whatever>
example.com/<whatever>/

As GrapeSoda suggests, whether you use trailing slashes or not on your URLs is largely down to personal preference. However (on Apache) when referencing a physical directory then a trailing slash is required in order to correctly serve the directory index.

But what about the /file.html & /file.html/

If file.html is a physical file on the filesystem then /file.html/ (with a trailing slash) will, by default, trigger a 404 because the default handler that serves text/html files does not permit path-info (the part after the filename, ie. just / in this case).
Comment: For a directory, a trailing slash is technically correct regardless of the web server. This is largely because `in the beginning, there was the directory and the file` (biblical) meaning that any reference made to the web server had to match the path from the web root on the file system exactly. Today that is not exactly always the case for everything, however, the rules still apply when you are referencing anything to be served from the file system or assumed to be served by the file system whether that is actually the case or not. File system rules are presumed historically. Cheers!!
Answer: Two Pages
Clearly, a slash or no slash define two different resources, except at the root level because the HTTP request can't access the root without a slash.
The HTTP request for the root of a website looks like this:
GET / HTTP/1.1
Host: www.example.com

It's just not technically possible to distinguish a version without a slash even if the URL syntax allows you to write it...
About a website hierarchy Google says:

When referring to the homepage, a trailing slash after the hostname is optional since it leads to the same content (https://example.com/ is the same as https://example.com). For the path and filename, a trailing slash would be seen as a different URL (signaling either a file or a directory), for example, https://example.com/fish is not the same as https://example.com/fish/.

Notice that here the author says "signaling either a file or a directory". Another hint of the thought about slashes at Google.
Filename, most commonly no slash when there is an extension
Since file.html is referencing a file, it generally should not have a slash at the end (i.e. on a hard drive, you don't put slashes at the end of filenames.) However, on the web you see it all and both appear here and there. I still think no slash is more common when you have an extension such as the .html.
I've seen CMS that place files attached to a page in a virtual sub-folder under that very page. So for example, if you had a recipe.pdf file, it could be:
http://www.example.com/path/file.html/recipe.pdf
  or
http://www.example.com/path/file.html/recipe.pdf/

Personally, I think it is weird to have a sub-directory under a .html because you can't replicate that in a file system (in case you wanted later to create a static version of your website... you'd be in trouble with such virtual folders.)
A Google Example
Google had some examples about this problem (I'll add the link if I find that page...) Their examples removed the extension as in:
http://www.example.com/path/file/recipe.pdf

(Note: Google says that they use the extension as a hint of the page's contents, so file.html tells them that the content is most certainly HTML.)
So no slash at the end of what represents files, but when no extension is present, we get a folder and it makes sense to have a slash. So a complete list of the files could look like this:
http://www.example.com/path/file.html
http://www.example.com/path/file/
http://www.example.com/path/file/recipe.pdf

Note: In the past, Google said that they would test parent folders automatically, even if the parent page is not defined in the sitemap.xml or in a link. So GoogleBot would check out all of the following:
# Found document:
http://www.example.com/path/file/recipe.pdf

# Also check parents:
http://www.example.com/path/file
http://www.example.com/path
http://www.example.com/

However, I don't recall ever seeing such hits in my logs. I never really look to prove that statement either. Yet, this means a functional parent is a good idea. Actually, as a user, I do that once in while, because when I see I'm in a certain sub-folder and think that I should be able to find things of interest one directory up, I manually try to go there by changing the URL. Website that break when you do that are definitely annoying.
Browser's Saved Page
When you save a webpage from Firefox or other browser, they create a sub-directory for all the files attached to the page (CSS, JS, images, etc.) and name it with the extension but change the period with an underscore. So something like this:
http://www.example.com/path/file_html/recipe.pdf

Redirection & Canonical
As for redirecting, it is not mandatory if your sitemap.xml and canonical are correct. In other words, if the search engines goes to:
http://www.example.com/path/file.html/

but finds the following link meta tag:
<link rel="canonical" href="http://www.example.com/path/file.html"/>

then no 301 is required. Google & Co. will use the version without the slash.
The opposite is true too, of course. With the following, the slash version is saved and still no 301 is required:
http://www.example.com/path/file.html
<link rel="canonical" href="http://www.example.com/path/file.html/"/>

If you don't have a proper canonical (or as I've seen on some CMS, if the code generating the canonical copies the URI as is--so you get two different pages in this case,) a redirect is probably the easiest way to fix the problem.
The sitemap.xml file must reference the page using the canonical URL of the page. So if the canonical uses the slash, the sitemap.xml URL will include the slash, and vice versa.

Trailing slash dups


How can I make my apache configuration file visible in a web page?

Question: How can I make my apache configuration file visible in a web page ?
Is there a short way to do it ?
Is that safe? I mean.. it should be visible only to me ? Or it is reasonably ok, if nobody knows which page it is ?
Answer: Whether it is safe or not depends on your specific Apache configuration.  But really you shouldn't expose configuration details of any system to the public at large, unless you have a very specific need to do so (i.e. you're running a website about Apache security and using the website itself for an example).
Nonetheless, if you really want to do it, just make a symbolic link in your web directory to your configuration file.  Something like:
ln -s /etc/apache2/apache2.conf /var/www/apache2.conf

Then users can enter http://www.yourdomain.example/apache2.conf and see it.  Of course your configuration file has to be world-readable.
Note that there is usually more to the Apache configuration than just that file, you'd have to expose the entire /etc/apache2 conf (ln -s /etc/apache2 /var/www/my-apache-configuration-directory) if you want the full configuration visible.
Now, if what you meant is you want the configuration file to show up as part of a web page, basically you'll need a CGI script or other executable that builds the web page and includes the contents of that file as part of that process.  Or you can create a external script that builds that web page and then just dumps an HTML file somewhere.  You can have this script run via cron every so often (once a day?) to check for updates to your file, or if you're using Linux you can do some elaborate stuff with inotify and have it call the rebuild script whenever you modify the file.  Or you can go the easy route and use server side includes.
But really, it's a bad idea.  Don't do it.
Answer: No it's not safe, the www (or whatever user apache is running as) user should not be able to read files outside the web tree.
The best way would be to copy the file to the a sub dir of the web-root (perhaps doing it periodical with cron) and protect it using .htaccess
Comment: what about logs ? Can I use link and then remove permissions ? So, I can read them only when I need ?
Answer: My question would be why would you need to expose this externally?  Apache pretty much should be a 'set and forget' server.  I don't want to presume too much, but my first guess is, if you want to do this there's something wrong in your Apache model.  Why would you need to see it (yet not be able to change it) at a distance, or often enough to make this worth the work needed to get it out.  Config files can include other config files and you'd need to make all of these exposed to make sure you can see everything involved in configuration.
And yes, this can be dangerous.  It's a HUGE information leak to attackers on what your config is.  If you have an insecure config, you'd expose it.  If you want to do this, please lock it down.  Don't assume that by not linking to this file people won't find it.  If you do anything like this, create a directory for it and lock it down with passwords, IP restrictions, or both.
I have two suggestions for you.  One is to expose the subset of config that's available through the module mod_info.  There's a config already in the default httpd.conf that maps the URL /server-info to mod_info module, but it's commented out by default.  Be sure to understand the Allow/Deny configs for it and lock it down to where just you can see the data. It doesn't expose as much as the true configs would, but it should be locked down.
Also, look at Webmin, its a way of configuring a UNIX machine, including Apache.  You'd be able to configure remotely.  Obviously this takes a lot more to get to work.


Do I need to contact the author before using data available in supplementary material or open data linked to article?

Question: More and more journals require authors to make their data available, as an increasing trend to open science and reproducibility. If I want to use some data openly accessible that is linked to a publication, do I just need to cite the data source in my manuscript or will I also need to contact the authors to let them know I am using their data or to ask for permission?
Comment: It could be useful if you could provide some more details on your field. Is it computing, maths, biology, etc?
Answer: The future of scholarly research will be based on open data to ensure accountability, transparency and independent replication (e.g. Warren et al, New Engl J Med 2016). Thus, we need to be ready for similar scenarios.

Do I just need to cite the data source in my manuscript or will I also
  need to contact the authors to let them know I am using their data or
  to ask for permission?

Unless there is an explicit policy you need to follow (quite uncommon indeed), you just need to provide the precise citation to what you accessed. However, contacting the authors could be a smart choice: they could provide you additional data or collaborate with you more actively.
Answer: You can, at your choice. It depends on whether it is required by the authors (that is unlikely given that the data is already public), and your intentions. Do you need their inputs beyond the public data? In that case, feel free to contact them.
Otherwise, just cite them (the data sets as well as their primary research paper that describes the data sets).
ONCE your paper is accepted for publication, you can send a copy of your paper indicating that you have used their data sets in your work. They will be delighted to hear that their data set has contributed for yet another research work. This will be a win-win, as they will read your paper, and you can become collaborators.


Vito and Luca question

Question: Can anyone tell me why Vito would send Luca to Tattaglia's night club under the ruse of being a traitor in The Godfather? The relationship between Vito and Luca must have been well known in the mob community and Vito would have known that. Are you going to tell me that Vito didn't consider the fact that Tattaglia wouldn't buy that his story of being unhappy with the family? If Vito is as sharp as they would have us believe, why then would he send his most trusted soldier to his death? I believe this is a flaw in the plot.
Answer: There was no one in the Corleone crime family more trustworthy and loyal to the Godfather than Luca.  Luca almost never appeared in public with the Godfather, or really any of the rest of the family members. This was because Luca wanted to ensure Vito's reputation was not ruined from association with him (Luca.) Because of this, it would be believable that the other Bosses and their Capos would not be completely knowledgeable of the extent of Luca's loyalty, or why it was so.
Who else could Vito send to entreat with Sollozzo? He might have been able to trust his own family, but really at the time this all happened, he was still trying to keep Michael away from the family business, Sonny was too hot headed (and spoke out of turn) and Fredo was ... well weak.
No, the only person Vito could completely and implicitly trust was Luca Brasi.
Remember also, that the things we know, as the voyeurs into the mob as watchers of the movies/readers of the novels, are not known by the characters themselves. Sollozzo and Tattaglia don't necessarily know (for sure, with absoluteness) the extent of Luca's loyalty. Luca went to one of the Tattaglia nightclubs and complained about lack of money. This would have been a sure signal about perhaps turning Luca.
Compare this to the (real life) Iceman, Richard Kuklinski. While he never had the loyalty to his boss that Luca was supposed to have to the Godfather, he was a ruthless mob enforcer. While the movie (Iceman) is based on the true story of Kuklinski, I am sure much of it is (unsubstantiated) drama. However, we do see in that story that when the boss, DeMeo, is forced to curtail his family's illegal activities, he benches Kuklinski, which causes K a great deal of financial distress, directly leading to his (K's) eventual downfall and arrest. So, you could see that, from the outside, Luca, like Kuklinski, might have a weakness (debt).
At the time Vito sent Luca to Sollozzo, he (Vito) was 'under fire' from Sollozzo and some of the rest of the bosses for his refusal to embrace drugs. Vito Corleone's power was in his ties into corrupt politicians and the police. Sollozzo (and his Capo Tattaglia) needed Vito's politico connections in order to more successfully promote their drug trade. Remember that the whole mess contained in Godfather (1) started with the meeting between Vito and Sonny and the other bosses (wasn't it Sollozzo?) where they needed his help. Sonny spoke out of turn during it. The whole feud would have caused a drop in the Corleone family businesses and perhaps as a result, Luca might be driven to find outside income.
(All of the above is based on the novels, the movies and the wikis).
Answer: I have always felt that Vito was trying to get rid of Luca.  If you recall, Vito acts in disgust when told that Luca was there to congratualte him on Connie's wedding day.  As Luca stumbles through a carefully rehearsed congratulation, it becomes apparent that he's a little "slow".  Like Fredo, but Vito would never do anything to Fredo.  I always felt like it was a double move; he got rid of Luca and planted a seed in Tattaglia's mind that Tattaglia can't quickly recruit because he might end up recruiting another Corleone member accidentally.
Comment: It wasn't that Luca was "slow". Luca was, as Michael put it, "a very scary man". He was very violent and ruthless and (in the book) single handedly ended two gang wars in favor of the Corleones. Yet his unsavory reputation necessitated that he and Vito are never seen in public. Him having to rehearse his lines (and flubbing them) was a stroke of genius by Coppola, where he turned a nervous actor messing up his lines into Luca being nervous since he never expected to be invited to the wedding. Vito was nervous because he wasn't sure what the savage Luca would ask for.
Comment: In of the sequel books it explains that Luca is an intelligent man but took a overdose of a drug after killing his mistress and child this causes his (to the general population viewing him) look of a simple and scary man . before the accident he was very scary and not simple acting, so even more intimidating!
Answer: Though in the movie it seemed that Luca went straight to Tattaglia's, but its not the fact. As per the book he stayed on Tattaglia's hotel and the word was on the air that the relationship between Corleone family and Luca isn't so well. Then the Tattaglia's called Luca to meet them.
And Godfather sent Luca because he was the strongest and fearing. Anybody would consider twice to kill him and Luca isn't dumb either. He sent the strongest to deal with the enemy. Its not  bad decision to make.
Answer: I think Vito didn't want to send his immediate family. Combative Sony, precious Micheak and incompetent Fredo to the invaluable Tom Hagan. I think he wanted to test how serious Solozzo was and sent someone he knew would march into certain death without question. Luca was more loyal than dumb and Vito knew if anyone can walk away from the meeting going sour, it would be Luca. He rolled the dice but before he could deal with the seriousness of it, there was a hit on him Iit's chess not checkers.(damn Fredo, damn Pauly)

Vito and Luca question


Melted butter or not for pancake batter?

Question: I was making pancakes that called for melted butter in the recipe and when I added cold milk the butter re-solidified. How did that mess up the pancakes? Was the texture more dense and less fluffy? So I presume this batter could not be made ahead of time and refrigerated?
Comment: If you combine your dry ingredients with near room temperature milk, and add melted butter only afterwards, it works out well. Otherwise use oil in the butter and then butter the pan a little more heavily to get that rich flavour.
Answer: The butter needs to be melted so it will emulsify with the eggs and milk. This makes a better pancake texture but your pancakes will still be recognizable and tasty if the butter did not fully emulsify.
As Max suggested in a comment, you can make your milk warmer. You can also make your butter warmer- short of burning it- to give yourself more time. You can replace some of the melted butter with oil.
Once the batter is assembled at room temperature there is less risk of the butter separating. Therefore, mixing the wet and dry ingredients before the butter can solidify is useful. In this case the flour would slightly warm the milk.
However you do it, making pancake batter ahead of time is generally a bad idea because the chemical leavening in the batter will activate and be expended. You might get a second lift from double acting baking powder but it still won't be as fluffy.


Would Amy and Rory be Cybermen?

Question: If Missy made all dead people Cybermen, does that mean people like Amy, Rory, and even Harriet Jones would be Cybermen?
Comment: I guess, but we wouldn't actually see them because they're buried in New York. As for Harriet Jones, maybe...didn't the Daleks vaporize her?
Comment: In Amy and Rory's case, there's really no reason to think that they wouldn't be converted. However, we do not know the final fate of Harriet's body, so we cannot say with her.
Comment: Maybe [Cyberwoman](http://en.wikipedia.org/wiki/Cyberwoman) got her wish?
Comment: That depends on exactly when Missy began her plan, but Amy and Rory would certainly be candidates, albeit unaware ones.
Answer: Assuming that they were dead by this point in time, then yes they would be Cybermen.  We know that they die prior to 2012, and if we assume that Clara/12 are in 2014 Earth (I was unable to find anything that explicitly stated this), then we know that their bodies are in New York and their "souls" are in the Nethersphere.  As such, they would have come back just like everyone else.
However, Harriet Jones is a bit more unclear (still assuming in 2014).  She was killed in The Stolen Earth (about 1:30):

However, we do not see what happens to her body.  Even if she was not vaporized, it was possible that she was later cremated.  Ultimately, there is no way to know.
Answer: I don't think the clouds had yet rained enough "cyber-pollen" to convert all dead bodies to cybermen. On board the UNIT plane, they did say that every place with lots of dead bodies had been rained on, here's the line from the transcript:

NEWS REPORT: Localised rain in the cemeteries has resulted in what can
  only be described as disturbances to the soil. Extraordinary
  eyewitness accounts are claiming that silver creatures are climbing
  from the graves.
KATE: These scenes are being repeated everywhere. Every cemetery,
  every mortuary, every funeral home, every hospital, the dead are
  returning to life as Cybermen.

But later Kate showed a computer graphic of what their sensors had detected underneath a graveyard, showing a large number of coffins but only a small fraction of them glowing with cybermen bodies inside:

And while the image was being displayed, Kate said:

KATE: We've done heat scans of some of the cemeteries and, in each
  case only a handful of Cybermen have so far emerged. But every
  individual burial site is active.
AHMED: Active?
DOCTOR: Hatching.
KATE: More are coming. Potentially millions.
AHMED: So the rain caused all that in just a few hours?
DOCTOR: It wasn't rain, Man Scout. It was pollen. Cyber-pollen. Every
  tiny particle of a Cyberman contains the plans to make another
  Cyberman. All it has to do is to make a contact with compatible living
  organic matter and bang! Full conversion. But if they have learned how
  to convert the dead.

Perhaps the existing clouds, which had been apparently been created by only ninety-one cybermen flying into the air and "pollinating", didn't contain enough cyber-pollen to convert more than a small fraction of the dead, but the newly-created cybermen would be able to add to the pollen in the clouds. Or perhaps the the pollen was self-replicating within the clouds themselves. Either way, the characters spoke of a second rain from the clouds as a future threat which needed to be averted. For example, after cyber-Danny deleted his emotions so that he'd be able to gain access to the cyberman hive-mind, there were these lines:

DOCTOR: Clara, no! Step away! He's activating! Clara, step away now!
  Don't. Danny. Danny, if you can hear me, if you're still there, what
  are the clouds going to do?
CYBER-DANNY: The rain will fall again. All humanity will die.
DOCTOR: And rise again as Cybermen.

And later Missy suggests that the Doctor needs to take control of the existing cyberman because that will allow him to stop the clouds from a second rain which would result in the death of humanity:

MISSY: You will, because you don't have a choice. There's only way
  you can stop these clouds from opening up and killing all your little
  pets down here. Conquer the universe, Mister President. Show a bad
  girl how it's done.

And then cyber-Danny indicates his plan is to stop the clouds from raining their cyber-pollen by burning the clouds:

CYBER-DANNY: The rain will not fall. (He lets go of Clara and puts on
  the control bracelet as he walks towards Missy.)
MISSY: Oh? Why won't it?
CYBER-DANNY: The clouds will burn.
MISSY: And who'll burn them?
CYBER-DANNY: I will burn them.

Then he took control of all the existing cybermen (using the bracelet) and commanded them to fly up into the clouds and explode, which burned them off. If the clouds had already delivered enough cyber-pollen to convert all dead bodies to cybermen, I don't see what the point of this would have been, he could have just ordered all the cybermen to deactivate or self-destruct without bothering with the clouds, which would have already done their dirty work.
So, I think all this indicates that most of the dead had not yet been converted to cybermen, and they needed to destroy the clouds to prevent this from happening. Thus, the chances are that most characters that would have been dead at the time of the episode, like Amy and Rory, were not cybermen (their minds probably would have been in the nethersphere, since the Doctor said "she's been upgrading dying minds to a hard drive for a long time" and indicated that with Missy's time travel capability she may have been doing this throughout humanity's entire history; but at the end of the episode Danny contacted Clara from the nethersphere and said "This place is dying now", indicating that it was on the verge of shutting down).
Edited to add: I suppose it's also possible the second rain was going to kill humanity in some way distinct from just creating more cybermen to cyber-convert them--maybe the second rain was going to fill the atmophere with poison, for example. But given the computer graphic showing that most of the bodies in graves had not yet turned to cybermen, and also given my assumption that Danny's orders succeeded in destroying the cyberman threat, meaning there aren't going to be more cybermen rising that didn't respond to Danny's self-destruct order because they weren't yet fully formed (it's probably safe to say the next season won't open with lots more cybermen emerging from graves), I think one of the following has to be true:
A) Danny's destruction of the clouds would prevent any further cybermen from rising, meaning the second rain must have had something to do with allowing more cybermen to rise, even if it could have had additional deadly effects on living humans.
B) Between the scene on the UNIT plane and the Danny's destruction of the clouds, all the remaining bodies with enough organic material to be converted to cyberman had been fully converted, and had "hatched", so that when Danny gave the order they all flew up and were destroyed. If this were the case, then other buried people like Amy and Rory probably would have risen as cybermen.
I think A) is a lot more likely the intent though, just because of the timing. It seems like the scenes are edited in such a way as to suggest there's not much time between the Doctor's conversation with Kate about what's happening in graveyards and Missy breaking free and causing the Doctor and the TARDIS to fall out of the plane, and after he catches up with the TARDIS in mid-air he immediately materializes in the graveyard and has the dialogue with Danny, and then Missy, and then Danny blows up all the cybermen. Not a lot of time for all the non-glowing coffins in the UNIT graphic to grow cyber-exoskeletons and claw their way up out of the ground.
Answer: https://m.youtube.com/watch?v=mLkfV8M0WFo
Towards the end it talks about the amy and Rory cybermen situation
Comment: Can you expand this answer so it's not just a link to an external resource?
Answer: It depends on whether Amy and Rory's bodies are still intact during the events of Dark Water/Death in Heaven.

Would Amy and Rory be Cybermen?


Volume Limiter for android?

Question: Is there a way to limit volume with a parental passcode on android?  I know there are apps for limiting volume, but is there any way to secure those limits so a child cannot change them?
Answer: If you are on a tablet running recent version of Android OS (4.3 and above), you can create a restricted user profile for your child.  There you can simply disable access to the app which you used to limit the volume.
For older versions of Android OS on tablets and on phones (restricted profiles are disabled for phones even on latest Android OS versions,) you can try various "app protector" type of apps, that provide password-based mechanisms for restricting access to various installed apps.
Comment: I assume the volume limiter app will do that.  Neither restricted profiles nor app protectors do anything for volume controls, as far as I know.
Comment: Can it prevent the volume from changing when Vol Down / Vol Up is pressed (stock behavior)?
Comment: Ah, that's right... I didn't think of that. :)
Answer: If you have Tasker, you could create a task that detects when the volume is changed. If it goes higher than X, it will run a task to reduce the volume back to X.
Tasker also lets you password protect groups of events, and you can enable Tasker to be an Administrator on the device.
Your child would be able to breifly set the volume higher, but it would always reduce back down. They couldn't disable the tasker event without the passcode, and they couldn't uninstall or disable Tasker since it is a device admin.
Of course, they could always get around it somehow, such as clearing the data on the app or figuring out how to remove the Administrator flag so that Tasker could be removed. It is a cat and mouse game. But this solution should suffice for people who aren't as technically minded.
Comment: And if you have Android 4.3 or higher, then Chahk's answer is preferable. Things built-in to the OS are always going to be better than stapled-on solutions.
Answer: Kind of late, but I found one I really liked: Max Volume Control. It allows you to lock the controls and starts up when you restart the device. So far my kid hasn't been able to circumvent it.
Comment: Are you affiliated to the app? There is only review for that app and that appears to be from AI T.
Comment: Also in at least two other blogs/Forums an Al T described that app as useful. All posts has been made around new year's Eve 2016.

Volume Limiter for android?


How to install the latest bfgminer binaries directly from terminal without 'aptitude' on a raspberry pi 4 8gb running raspbian?

Question: months ago I found an entry on StackExchange or somewhere else on how to install the latest (!) bfgminer binaries directly from terminal. I can not find the stackexchange post, no matter how deep I search here / on Google.
I do not remember if it was a download directly from luke-jr's website via terminal or from somewhere else. I only remember that it was done via the terminal and it created an executeable shortcut in the /bin/ and/or /usr/bin/ folder.
Visiting the website from luke-jr triggers my anti virus software telling me the website is infected or hosts infected files. So I can't visit it directly sadly.
Months ago I tried it out on my old RPi and it worked just fine. But now that I had to switch to a newer RPi I can not find or remember the correct terminal command for installing bfgminer binaries directly.
More details about why I need the binaries:
Compiling prints out LOTS of errors for me.
Using 'sudo aptitude install bfgminer' needs "aptitude" installed, so I tried to install it and it returned lots of errors, most of the package files where not downloadable (my Internet connection is fine by the way).
Can you help me out with a terminal command that let's me successfully install bfgminer on raspbian?
On my old Pi I managed to install it without aptitude and without compiling it first.
Thanks so far!
Answer: Update: sudo apt-get install bfgminer failed several times, as the same for sudo apt-get install cgminer.
After trying numerous times it finally worked!
No more need to download it from this website, got it.
Thread can be closed. Thank you anyway for your time! :)


What is the symbolism of rats in George Orwell's "1984"?

Question: Why are rats used as Winston's greatest fear? What do they represent?
Comment: Sometimes a rat is just a rat.
Comment: http://www.quickmeme.com/img/d2/d2e9a5b46fa5f97ca3dc7289543796cd27f38faee5cc26cfb0310cbd2a6606b5.jpg
Comment: I'm not an expert on this, but the English seem to dread rats more than the rest of the English-speaking world. Maybe it's a European thing, involving the memory of the Black Death, and the horrors of the trenches of the two world wars. They're also associated with squalid poverty, starvation, and being trapped, which dominate Winston's life.
Comment: Rats have been used in the USA as an execution device in the last century. See interviews with Richard Kuklinski.
Answer: Rat torture is a common form of punishment and that's alluded to in the book:

"It was a common punishment in Imperial China", said O'Brien as didactically as ever.

I think part of the symbolism of rats is simply that they make for a realistic torture scene. Rat phobia isn't uncommon, so readers can potentially identify with it.
In addition to that, I think one of the biggest issues Winston wrestles with is the sameness of his daily life, the routine that he cannot break out of and nobody around him wants to break out of. He's stuck in this "rat race", if you will, and there's nothing more horrifying for him than becoming just another person like everybody else.
When he's threatened by rats, he gives up the last part of him that clings to independent thought: he betrays Julia, effectively abandoning the person he loves for his personal gain. Just like everyone else would.
There's also an interesting paragraph on this on NovelGuide.com, which seems to fit with this theory:

Winston learns the meaning of Room 101 when O'Brien tortures him with rats.  Room 101 represents a person's worst fear and Winston's worst fear is rats.  So on one level rats represent fear.  On another level, however, rats represent depravity.  Throughout history, humans have associated rats with squalor and pestilence.  Rats carry disease and thrive on human garbage.  Rats rank among the world's most "beastlike" (as opposed to "humanlike") creatures.  Winston's universe is filled with humans who act like and are treated like beasts.  Outer Party members and Proles all eventually become drones-meaningless, inhuman cogs in the Party's machine.  In essence, Winston and his fellow citizens become rats, trapped in Big Brother's cage.  If people allow forces such as those represented by Big Brother to rule, then they will become no better than mindless, multiplying rats.
Comment: @Tim "Rat race" is a common term, not something I made up to fit the theory: ["In an analogy to the modern city, many rats in a single maze expend a lot of effort running around, but ultimately achieve nothing (meaningful) either collectively or individually."](http://en.wikipedia.org/wiki/Rat_race) I think the paragraph I quoted from NovelGuide says more or less the same thing, just without calling it a rat race.
Comment: "Rat race" was in use at the time Orwell published the book, fwiw.  Note also this apropos quote from [Wikipedia](http://en.wikipedia.org/wiki/Rat_race): "This term presumably alludes to the rat's desperate struggle for survival. [Colloquial; first half of 1900s]"  Regardless of whether Orwell had it mind specifically, the symbolism sure does ring true for me.
Comment: +1 for the "rat race" analogy. I thought both snakes or spiders could be used as a common human phobia(perhaps even more common than rats?), but your answer explains it well
Comment: Do you have any kind of support for the "rat race" theory? It could just as well mean that totalitarian society is a "rat trap" or that Orwell would "rather" be a movie star.
Answer: The Rat: A Very Short Essay on Nineteen Eighty-Four and its Central Metaphor
Nineteen Eighty Four is a great book with great themes: truth, lies, history, memory, love, sex, and betrayal.  It is the last that concerns us here in the figure of the rat, the only animal to appear in the novel (if we discount the droning of the ring doves).  The rat is the fellow-traveller of death, destitution and figuratively, betrayal.  Winston's fear of rats is in essence a psychopathological fear of what he takes to be the greatest evil – self-betrayal, betrayal of others ("Do it to Julia!"), ratting or being a rat in general (especially a greedy one).
In his youth he steals a 2oz slab of chocolate from his mother and younger sister. It is a nightmare of guilty betrayal that haunts Winston's night-times.

"One day a chocolate-ration was issued. There had been no such issue for weeks or months past. He remembered quite clearly that precious little morsel of chocolate. It was a two-ounce slab (they still talked about ounces in those days) between the three of them. It was obvious that it ought to be divided into three equal parts. Suddenly, as though he were listening to somebody else, Winston heard himself demanding in a loud booming voice that he should be given the whole piece. His mother told him not to be greedy. There was a long, nagging argument that went round and round, with shouts, whines, tears, remonstrances, bargainings. His tiny sister, clinging to her mother with both hands, exactly like a baby monkey, sat looking over her shoulder at him with large, mournful eyes. In the end his mother broke off three-quarters of the chocolate and gave it to Winston, giving the other quarter to his sister. The little girl took hold of it and looked at it dully, perhaps not knowing what it was. Winston stood watching her for a moment. Then with a sudden swift spring he had snatched the piece of chocolate out of his sister's hand and was fleeing for the door.
'Winston, Winston!' his mother called after him. 'Come back! Give your sister back her chocolate!'
He stopped, but did not come back. His mother's anxious eyes were fixed on his face. Even now he was thinking about the thing, he did not know what it was that was on the point of happening. His sister, conscious of having been robbed of something, had set up a feeble wail. His mother drew her arm round the child and pressed its face against her breast. Something in the gesture told him that his sister was dying. He turned and fled down the stairs, with the chocolate growing sticky in his hand.
He never saw his mother again. After he had devoured the chocolate he felt somewhat ashamed of himself and hung about in the streets for several hours, until hunger drove him home. When he came back his mother had disappeared."

Later a rat intrudes into Winston's and Julia's love-nest (prefiguring the intrusion of the Thought Police).

"'It's twenty-three at the hostel. But you have to get in earlier than that, because — Hi! Get out, you filthy brute!'
She suddenly twisted herself over in the bed, seized a shoe from the floor, and sent it hurtling into the corner with a boyish jerk of her arm, exactly as he had seen her fling the dictionary at Goldstein, that morning during the Two Minutes Hate.
'What was it?' he said in surprise.
'A rat. I saw him stick his beastly nose out of the wainscoting. There's a hole down there. I gave him a good fright, anyway.'
'Rats!' murmured Winston. 'In this room!'
'They're all over the place,' said Julia indifferently as she lay down again. 'We've even got them in the kitchen at the hostel. Some parts of London are swarming with them. Did you know they attack children? Yes, they do. In some of these streets a woman daren't leave a baby alone for two minutes. It's the great huge brown ones that do it. And the nasty thing is that the brutes always—'
'Don't go on!' said Winston, with his eyes tightly shut.
'Dearest! You've gone quite pale. What's the matter? Do they make you feel sick?'
'Of all horrors in the world — a rat!'
She pressed herself against him and wound her limbs round him, as though to reassure him with the warmth of her body. He did not reopen his eyes immediately. For several moments he had had the feeling of being back in a nightmare which had recurred from time to time throughout his life. It was always very much the same. He was standing in front of a wall of darkness, and on the other side of it there was something unendurable, something too dreadful to be faced. In the dream his deepest feeling was always one of self-deception, because he did in fact know what was behind the wall of darkness. With a deadly effort, like wrenching a piece out of his own brain, he could even have dragged the thing into the open. He always woke up without discovering what it was: but somehow it was connected with what Julia had been saying when he cut her short.
'I'm sorry,' he said, 'it's nothing. I don't like rats, that's all.'"

The rat stands for betrayal – not only the original act (it is clear that Smith has formed a connection between the rats, the women and their babies and his betrayal of his mother and sister), but also both the 'self-deception' and the dragging the' thing into the open.' The 'rat' is an instance of self-betrayal and threatened self-exposure (or betrayal).
O'Brien has made the connection too. His exposition on rats is eerily familiar.

"'The rat,' said O'Brien, still addressing his invisible audience, 'although a rodent, is carnivorous. You are aware of that. You will have heard of the things that happen in the poor quarters of this town. In some streets a woman dare not leave her baby alone in the house, even for five minutes. The rats are certain to attack it. Within quite a small time they will strip it to the bones. They also attack sick or dying people. They show astonishing intelligence in knowing when a human being is helpless.'"

And then there is the final terrible betrayal of Julia, the only person he could usefully interpose between himself and the rats.

"'Do it to Julia! Do it to Julia! Not me! Julia! I don't care what you do to her. Tear her face off, strip her to the bones. Not me! Julia! Not me!'"

Betrayal or 'ratting' provides the very core to the book, with each of the major themes an offshoot of this powerful, central idea. From the revolution itself down to the sinister 'Spies', all is betrayal: the revolution betrayed by the Party; Parsons by his own children; History is betrayed by the Ministry of Truth; natural, more or less autonomous human thoughts and feelings are betrayed by torture and brainwashing. Nothing is left untouched by the Rat.
Comment: Is it not also something as simple as in 1948 (when the book was written) with most of Europe was still recovering from the Second World War, rats would have been a common sight on bomb sites etc (and so an easy fear to relate to)?
Comment: I think '1984' is difficult to understand because it is hard to to realise that  what O'Brian says about Power is sadly true - nothing to do with extra 'wine' or turning down the volume - just power.-
Comment: You are quite right - rats everywhere - real and symbolic. Newton's Apple is a real thing and a metaphor - except in 1984 this one has  a worm within.
Comment: https://en.wikipedia.org/wiki/Informant
Answer: Brewer's Dictionary of Phrase and Fable tells us a lot about rats, notably:

Pliny tells us (VIII, lvii) that the Romans drew prresages from rats and to see a white rat foreboded good fortune. Clothing or equipment gnawed by rats presaged ill fortune. It was an old superstition among ailsors that rats deserted a ship before she set out on a voyage that was to end in her loss. Similarly rats were said to leave a falling house.

& he quotes The Tempest, I, ii, where "the very rats / Instinctively have quit it [a ship]".
Rats are also common laboratory animals--pawns in an experiment not unlike the social "experiment" of 1984.
Answer: Rats represent depravity more so than snakes and spiders. The proles live in conditions more likened to those in which rats are associated with. Winston believed that the proles were the only hope left, quote "If there is hope, it lies with the proles". The rats in room 101 turn on him, forced upon him by the all-encompassing Party, and Winston's resolve crumbles as he realizes that The Party has set him up with false hope simply so that The Party could destroy and change his mind.
Answer: At the beginning of the novel, there is some word about Winston's childhood remembers. There is a point where he remembers some dead bodies, with rats around them.
I think probably it was the point where the "normal" world ended for the children Winston. From that point he lives in a constant fear and fight for survival. It is the "entry point" for him into the world of the 1984.
This is why he has such a fear from the rats.
I can get the citation from the novel for ask.
Answer: But Orwell did well to lay emphasis on what the rats were capable of doing even among the proles:

P.181: Did you know they attack children? Yes, they do. In some of these streets a woman daren't leave a baby alone for two minutes. It's the great huge brown ones that do it. And the nasty thing is that the brutes always——' (P.181)

it was his fear of rats:

P.358: In your case,' said O'Brien, 'the worst thing in the world happens to be rats...
P.362: He was falling backwards, into enormous depths, away from the rats. He was still strapped in the chair, but he had fallen through the floor, through the walls of the building, through the earth, through the oceans, through the atmosphere, into outer space, into the gulfs between the stars—always away, away, away from the rats. He was light years distant,

Room 101 was every interrogated party member's worst fear.
Answer: 1984 was written in 1947, just after the war, so whatever rats may symbolize, they also represent a contemporary fear for Orwell and the people he was writing for.
When London was being bombed, rats were a familiar thing - rats infesting the Underground shelters, and rats attacking the bodies of the victims of bombing. This book documents that

In one notable instance, thousands of them [rats] swarmed out of a
soap factory set ablaze during the bombing.

It's natural that Orwell was transfer this image of the horror that Londoners were then feeling about the possibility that they could be trapped in rubble and eaten alive by rats to Winston Smith.
Answer: In 1984 book, the rats represent Winston's deepest fears because he is more afraid of them than of anything else. On a deeper level, however, the rats also symbolize the extent of the Party's control over the people of Oceania. You can find them on Bansky drawings, they have suits and can represent control through government or corporations.
Comment: Do you have any evidence to back up your first claim that they are what he is most afraid of? That seems to be the bit that directly addresses the question but it isn't more than a sentence so would be good if you could [edit] this to expand on it a bit more.
Answer: Social contract theory is based on the idea that in exchange for giving up some liberties, the state, or collective group provides some benefits, such as protection from violence, to the individual. One may rebel against society's paradigm, but frequently they will find something that makes them reintegrate and sacrifice their liberties for the benefits provided by the state.
It is not the fact that they are rats that is important. It is the fact that everyone has something that gets to them. For almost anyone there is something they fear enough that they will abandon their nonconfoming values, some liberties,  and submit to the state, ruling class, societal paradigm, etc.
In the end one loves big brother. They need/appreciate/love the benefits that come from engaging in the social contract, or submitting to state in exchange for the benefits provided.
There are lots of people who don't like the idea selling someone else the best 8 hours of their day so that the employer can make more money off their time. That is, until rent is due the next week ;).
Answer: I show great appreciation for the reference... but for me it ties into Chinese astrology, the year 1984 (or rather the year 4682 in chinese calendar) is the year of the rat... and if one is familiar with chinese astrology, you'd know the rat is the FIRST animal amongst the zodiac.
Interestingly enough, in the chinese calendar i was born 4th of April 4682 (1984).
Comment: Whilst I don't particularly doubt that 1984 was the year of the rat, do you have any kind of references to support the idea that that was why Orwell chose to use rats as Winston's greatest fear? Also, I'm not really sure this answers the question of what, if anything, they *symbolise* in the novel
Answer: Rats refers to his  experience in the Spanish civil war. Check  homage to catalonia.
Comment: Could you please expand on this?