10" induction pans don't work with my GE induction cooktop but 8" and 12" pans work?

Question: I'm happy with my GE portable induction cooktop (single burner). I have two 8" pans (one cast iron, one induction-certified Oxo) and one induction-certified 12" pan (Nuwave) that work great. I've tried three induction-certified 10" pans and none work! Oxo, Sensarte, and Green Pan brands. Is it possible that my induction cooktop is incompatible with 10" pans???
Comment: Have you confirmed the 10" pans are actually induction ready?
Comment: Only ferrous metals will work on induction, have you tested these pans using a magnet?
Comment: What are your 10" pans made of? Brand does not matter, materials and construction very much does.
Comment: I checked the Sensarte and Green Pan skillets. Magnets stick to the bottoms of both. All of the pans are induction certified.
Answer: Induction cookers are tricky. They heat with a donut-shaped antenna under the glass top. You have likely noticed a circular "hot spot" in your pans. This is the size and shape of the underlying antenna. Consumer induction cooktops all have single-donut antennae. Commercial units, which can cost more than 10X as much, usually have two concentric "donuts" for better coverage.
The circular graphic on the stovetop is not necessarily centered over the donut antenna. My Frigidaire induction has one graphic a full inch off-axis. You can tell by putting 1/8" of water in a large cast iron frying pan and putting the burner on "hi". The circle of bubbles will identify the center of the antenna. Center your pot there, not over the graphic.
The burners have sensors to determine if a magnetic pot is correctly located above the antenna. That's why they "know" to turn off if you lift the pot. And why  burners will sometimes refuse to heat small pans, especially cheapies.
Your misbehaving pans may be "induction compatible", not "induction optimal". Laminated pans which are optimized for induction are more expensive to manufacture.
I suspect it is this "safety" sensor which is causing mischief. If a cheapie pan is off-center from the antenna, the cooktop may interpret that as "no pot present".
The magnet test is helpful, but not foolproof. Trust in cast iron. All other pots must prove themselves.
Comment: I'm sure it is the pan sensor failing to detect the pan, as otherwise it would try to heat and you'd see some effect.  But it does seem odd that the sensor can pick up 8" pans and not 12".  Of course the sensor may be off-centre too, so systematically offsetting a 10" pan might be of some use.
Comment: Looking at the bubbles boiling water in the 8" cast iron skillet, It looks like the indiction ring is about 5" in diameter. When I checked the 10" skillets with a magnet I didn't check all over to see if the bottoms were all ferrous or if there was just a ferrous ring. Maybe the non-compatible pans had a ferrous ring more than 6" in diameter? My compatible 12" saute pan has a full ferrous bottom, as do my 8" Oxo skillet and my 8" cast iron skillet.


Add an event to GestureDetector

Question: Actually i have GestureDetector for a WebView, it's works, i can change the page to load by left and right swipe but zoom, pan and active links has stopped to work. This is the listener:
private GestureDetector.SimpleOnGestureListener sogl = new GestureDetector.SimpleOnGestureListener() {

        public boolean onDown(MotionEvent event) {
            Log.d("Gesture","onDown");
            return true;
        }

        public boolean onFling(MotionEvent event1, MotionEvent event2, float velocityX, float velocityY) {
            if (event1.getRawX() > event2.getRawX()) {
                if(currentPage<maxPage)
                    currentPage++;
                webV.loadUrl("file:///sdcard" + Values.APP_FOLDER + "/" + catalogo +"/versione-html/" + currentPage + estensione);
            } else {
                if(currentPage>1)
                    currentPage--;
                webV.loadUrl("file:///sdcard" + Values.APP_FOLDER + "/" + catalogo +"/versione-html/" + currentPage + estensione);
            }
            return true;
        }
    };

The question is: can i add this single event (left and right swipe) to the webview default GestureDetector? I need to zoom and keep active links on the webview and also change page with swipe.
Thank you for your help.
Comment: Try returning false in `onFling`.
Answer: Wild guess, but I would consider sending all the touches you receive inside the GestureDetector down to the WebView via webV.onTouchEvent(event)

Add an event to GestureDetector


About Google Play App Signing

Question: When I login to my Google Play developer Console, I see this:

I understand how Google Play App Signing works, in general it keeps the app signing key on behalf of you, and I need to use another so called "upload" key to sign my app and submit to Google Play Console for app signing.
But, my question is:
I already published my app without using Google Play App Signing half year ago, now, if I opt-in to Google Play App Signing, would it affect my already released app? Or Does it go smoothly with my published app if I switch to Google Play App Signing?
Comment: The Google article answers everything except what it's for. What are some advantages of Google Play App Signing?
Answer:
Would it affect my already released app? Or Does it go smoothly with my published app if I switch to Google Play App Signing?

No, it will not affect your already released app. It will go smoothly with your already published app if you switch to Google Play App Signing.
But once you opt in Important: App Signing opt-in is permanent.
I did the same for few of my apps in the past and all are working fine.
Once you opt-in your original app signing key is uploaded to Google Play and you will be signing your new build with the newly generated upload key.
Further updated build will be signed locally or on dev machine via upload key. After uploading signed locally build the to Google Play, the Google play will remove your key and will sign the apk once again with your original app signing key.
Existing apps
Answer: Google app signing works perfectly with the existing apps and your app doesn't get affected if you enroll now.
I am listing the steps that may help you to enroll without any difficulties.
1) Sign in to your Play Console and select your app.
2) On the left menu, click Release Management > App Signing.
3) Upload your existing key and follow the rest of instructions.
P.S. Once you opt in for Google app signing, you can't opt out.

About Google Play App Signing


rtsp server forwarding

Question: So I am trying to stream a webcam feed from my computer to my android phone.  I am using a simple forwarding server to connect the two so that I don't have to worry about home network IP firewalls ect.  I am using FMJ to capture the video feed and was thinking rtsp protocol would be best because android supports reading it.  My problem is I dont know how to forward the information via the server, is it as simple as forwarding UDP packets?  all programs involved are implemented in java btw.
Answer: Use Netty. It is very easy to create an RTSP server with this library. For hints see the sources here.
Answer: Or you can port my library which is written in C#
http://net7mma.codeplex.com/
CodeProject article @ http://www.codeproject.com/Articles/507218/Managed-Media-Aggregation-using-Rtsp-and-Rtp

rtsp server forwarding


Play button in browser

Question: I want to put songs on a web page and have a little play button, like you can see on Last.fm or Pandora. There can be multiple songs listed on the site, and if you start playing a different song with one already playing, it will pause the first track and begin playing the one you just clicked on. I think they use Flash for this, and I could probably implement it in a few hours, but is there already code I could use for this? Maybe just a flash swf file that you stick hidden on a web page with a basic Javascript API that I can use to stream mp3 files?
Also, what about WMA or AAC files? Is there a universal solution that will play these 3 file types?

http://musicplayer.sourceforge.net/
Answer: There are many flash mp3 players that you can use that do this. Usually, you just have to edit a text file to point at the mp3s you want to have available.
Here is the first one that showed up on a google search for flash mp3 player: http://www.flashmp3player.org/demo.html
Answer: This is fairly simple if you want to embed the WMP you can use all the controls via JavaScript. There is a great MSDN section on it but I cant seem to find it now.Edit: I found this on MSDN it contains the properties that an embeded WMP will accept then all you have to do is call the methods via javascript.
<OBJECT id="VIDEO" width="320" height="240"
    style="position:absolute; left:0;top:0;"
    CLASSID="CLSID:6BF52A52-394A-11d3-B153-00C04F79FAA6"
    type="application/x-oleobject">

    <PARAM NAME="URL" VALUE="your file or url">
    <PARAM NAME="SendPlayStateChangeEvents" VALUE="True">
    <PARAM NAME="AutoStart" VALUE="True">
    <PARAM name="uiMode" value="none">
    <PARAM name="PlayCount" value="9999">
</OBJECT>

Then for the javascript

Somewhere around here I have code to even control the volume.
A while ago I built a custom (looking) player for a client purely in HTML and JavaScript.
Answer: Something I bookmarked long ago, but never got to test so far: http://www.schillmania.com/projects/soundmanager2/
Answer: I second superjoe30's suggestion: I had great success with musicplayer. The only (slight) negative is that it's a little older project and not as well skinnable as some of the alternatives (although you have the full source code, so - given some time - you can make it look exactly as you need it to).

Play button in browser


Why is B flat in this key sigiture when B is never played?

Question: In the below sheet music B is marked as flat however the only notes played are g g g d e c C. Is that just a typo on the part of the person who made this image and it should be on the upper C or am I missing something else?
Answer: Sharps and flats in the key signature are one indication as to the key of a piece. Lacking those notes in the actual music is irrelevant. For example, play all the white notes on a keyboard in order from C to the next C an octave higher. Now do the same from F to F. Hear how it sounds weird on the 4th note (B)? That's because you need to play it in the key of F. Now try playing F to F again but this time play Bb instead of B.
Comment: It also tells accompanists to use chords that do not contain B natural.
Comment: @neilfein I think that's true only for simplistic music. Music that an accompanist would play (songs, for example) modulate. A common modulation is to 5 of 5, which has a raised 4th. In this case, that's to G, which uses B natural.
Comment: Speaking as a working musician who uses notation for traditional folk tunes: Your jab at "simplistic" music aside, modulations away from the basic key *should* be noted in the notation, either in this staff or other staves in the same work. (If they're not, someone did an incomplete job of transcription.)


Is there a philosophical stance concerning justification of decisions?

Question: When faced with dilemmas, I always felt I had no solid ground that'd justify my decision. Say I'm considering becoming a lawyer. But I obviously don't know what being lawyer is like. I can ask my friend (let's say Jim), who is a lawyer how it's like to be a lawyer. But all I will know is what's it's like for Jim to be a lawyer, not this hypothetical version of me working as a lawyer.
Let's say that one of my priorities when looking for a job is what the coworkers are like and if my boss is fair. I won't know any of that unless I work there for some time, so the choice of me choosing a job would seem like a guess - perhaps educated guess (I'd try to gather as much information about the firm etc), but a guess nonetheless. So if somebody would ask me "would you like to work there" the only answer seems to be "I don't know, there is too much hidden data to make a decision in one way or the other" which is silly.
Of course, that applies to basically any area of life, private or professional alike. Making decision seems very hard and justification of it seems shaky at best. We seem to have very limited knowledge of future implications of our actions.
I think I actually read some philosopher arguing in similar fashion, which I'd find cool since I'm not the only one who thinks along these lines. But I don't remember the details - perhaps philosophers here could guide me to literature concerning this topic. I'd appreciate any input on this matter.
Comment: There is an entire subfield at the intersection of philosophy and mathematics concerned with justifying decisions, it is called [Decision Theory](https://plato.stanford.edu/entries/decision-theory/). And yes, justifying present decisions in the face of future uncertainty is a big part of its complexity. It is dealt with by using utilities, probability and expected values, quantified when possible, qualitative when not.
Answer: You should try searching this site on a topic you are interested in. I found:

Ethics, axiology, and decision theory
Has anyone claimed that no decision is inherently better or more rational than any other decision?
Overcoming procrastination & the fear of making decision.

were on the first page.
Choosing a career is so complex that chaos theory has been applied to it.
I would argue that our senses of being a willed agent with a character, are exactly heuristics for making decisions with incomplete data: Constraints on free will
The deepest aspect is being able to live with being the kind of person that made a given decision. Otherwise, yes, informed guesswork. Usually with abundant cognitive biases and post hoc reasoning. I argue here wisdom can be understood as our dilemma-solving faculty: Wisdom and John Vervaeke's awakening from the meaning crises?
Comment: As a sidenote, F. Nietzsche had often claimed in his works that people should affirm their lives through their choices. In the final analysis, they should be content with deciding such and such and living such and such..


How to train our brain to see everything in slow motion?

Question: People who had near death experience often found that they were able to see everything in slow motion and capture every details in their surrounding at that particular period. Because the brain process a lot of information at that time due to adrenaline in the blood to aid our chances of survival. So is there a way to train the brain to increase its processing capacity so we can perceive things like mentioned above all the time.
Comment: Often I would prefer to have a "fast forward" feature on my brain. :)
Answer: Here is an article that explains that the time warp during near death experiences is an illusion. http://www.livescience.com/2117-time-slow-emergencies.html In such situations, the mind captures more memories of the event. "And the more memory you have of an event, the longer you believe it took."
Comment: So according to that article we just hold more memories of that incident so we get the illusion of slowing time. (Consider something is going to hit you hard) Does that mean we cannot do anything out of the blue at that instant to dodge or save ourself from getting hurt ,but instead we could do only what it is humanly possible and hold memories of time slowing down?


How did the Alliance find out Liana Hallik was Jyn Erso?

Question: At the beginning of Rogue One, Jyn is imprisoned at the Wobani Labor Camp, but the Empire thinks she is one Liana Hallik. How does the Rebel Alliance find out Jyn hides under this alias? How does the Rebellion find Jyn's location?
Answer: The Rogue One: Rebel Dossier factbook indicates that the Rebellion proper (under Mon Mothma) have a source of information, presumably Tivik, inside Saw Gerrera's rebel cell. This infiltrator has evidently fed them information about the cell's members, identities and aliases.

Our analysis also turned up information about Jyn Erso, Galen Erso's
  daughter and a former fighter for Saw Gerrera (see related file). We
  agreed that Jyn Erso was an essential source: she could identify Galen
  Erso and might know more about his research.

As to how they located her, the Rebellion apparently had operatives tasked with searching public (and semi-public) databases for her known aliases. Her name popped up in an arrest report.

We aren't sure how Jyn Erso wound up in Gerrera's militia, when she
  left, or why she did so. After that we have her Imperial arrest
  records as a street fighter, smuggler, and petty criminal. She drifted
  from planet to planet, acquiring and discarding fake names, until she
  was arrested on Corulag and sentenced to a labor camp on Wobani.
...
We located arrest records for Jyn Erso and discovered that she was
  being held in a labor camp on Wobani under the alias Liana Hallik.

You may want to note that their (in-universe) dossier also contains a considerable number of other aliases that they know her to be using;

RECENT ALIASES: Liana Hallik, Lyra Rallik, Kestrel Dawn, Tanith Ponta, Nari McVee
Comment: Note that since Saw was originally a member of the Rebellion proper, he might have been the source of information about Jyn's other identities or she might be operating under a disposable identity that the Rebellion created.
Comment: I thought Jyn had acquired those fake names _after_ she was abandoned by Saw?
Comment: @Lupin - Presumably it would be moderately simple for an intelligence expert to track someone entering an area using one name but leaving under another. That's even assuming that she's not simply acquiring them from her rebel contacts (which would fool the Imperials but make it trivially simple for the Rebels to find out her aliases).


Salesforce Marketing cloud user creation error

Question: I have an enterprise 2.0 SFMC account and while creating a new user I am facing "Invalid @MemberID. It should be of type Enterprise_2 if business unit user is to be created" error.
I have tried it with parent and child BU both.
Answer: This seems to be a known issue. Please refer this.
Can you logout from SFMC and login again and try to create the user in Parent Business Unit.
Comment: Yes, this is also confirmed to be a known issue, and also listed on [Adam Spriggs](https://salesforce.stackexchange.com/users/5202/adam-spriggs) blog page [SFMC Known Issues](https://sprignaturemoves.com/sfmcki.php)...visit this page and seach for `@MemberID`
Comment: I have tried it also...still not working


Do you have to pay capital gains tax when selling and immediately buying another cryptocurrency in US?

Question: Is it possible to convert eg. BTC to ETH and pay capital gains tax only when you sell and convert crypto into USD?
Comment: I'm voting to close this question because bitcoin.stackexchange is not the appropriate place to ask about taxstion questions (especially when OP doesn't even mention which jurisdiction they reside in).
Comment: Does this answer your question? [How to report the tax if I trade BTC to LTC?](https://bitcoin.stackexchange.com/questions/56597/how-to-report-the-tax-if-i-trade-btc-to-ltc)
Comment: No, it's outdated, 4 years ago.
Answer: https://www.irs.gov/individuals/international-taxpayers/frequently-asked-questions-on-virtual-currency-transactions
(Retrieved 2022-03-16)

Q16.  Will I recognize a gain or loss if I exchange my virtual currency for other property?
A16.  Yes.  If you exchange virtual currency held as a capital asset for other property, including for goods or for another virtual currency, you will recognize a capital gain or loss.  For more information on capital gains and capital losses, see Publication 544, Sales and Other Dispositions of Assets.
Answer: as per my understanding, any sell event is a taxable event. Even if that event is converting a crypto.


Is it ok to say that I burned my hand if it was injured via hot water?

Question: I feel like it should be alright, but thought I'd get your feelings on whether it's appropriate to employ the term "got burned" if no fire was involved, specifically as it pertains to injury to a living thing, or if I am precluded from having been burned by hot water, and have to employ scalded or cooked or some other non flame related term?
Comment: "Scalded" is the more accurate term, but "I burned my hand when I spilled the boiling water." works too.
Comment: Should it be "if it were injured"?
Answer: In my estimation, you can suffer burns from any overly hot substance (boiling water, cooking oil, bacon grease, air, electric stovetop, etc.), not just the direct application of actual flame.
Comment: yeah, I think this is right. I also thought of acid burns after I posted the question, which is entirely correct looking usage, and obviously has no fire involved.
Answer: This definition here seems a good example of why your usage is okay:

to injure or damage by or as if by exposure to fire, heat, or radiation
Answer: burn v "burned my skin with the acid." (TFD)

3. To damage or injure by fire, heat, radiation, electricity, or a caustic agent: burned the toast; burned my skin with the acid.

So, yes, I guess you could say that.

Burns are injuries to tissues caused by heat, friction, electricity, radiation, or chemicals.
The source of this heat may be the sun (causing a sunburn), hot liquids, steam, fire, electricity, friction (causing rug burns and rope burns), and chemicals (causing a caustic burn upon contact).

(src: TFD Med.)


Word for sophisticated art that the masses don't appreciate

Question: Popularity and sophistication don't always overlap. Think of the novels only English teachers love. The music only Mitch Grassi listens to. The paintings you only want to look at after graduating from RISD. The movies they only show at film festivals. Do we have a word or phrase to describe this kind of thing?
Comment: Shakespeare's Hamlet used the metaphor "caviar to the general" (at 2.2.377 or thereabouts) to describe "an excellent play" that "pleased not the million." But then, do you really want an expression that is as culturally highbrow and exclusive as the stuff it is an expression *for?*
Answer: May I offer a phrase, rather than a word? "Caviar to the general" first appeared in Hamlet and has since entered more common currency. (The populace, rather than any military rank, is denoted.)
Or, if it has to be a single word, will the hyphenated "high-brow" do instead?
Comment: In fact, the hyphen may be unnecessary: just "highbrow".
Answer: Indeed, popularity and supposed "sophistication" often do not overlap. The distinction between the two types of art that you are asking about is often described in the following terms:

high culture: "a term now used in a number of different ways in academic discourse, whose most common meaning is the set of cultural products, mainly in the arts, held in the highest esteem by a culture. This is also a term used in the pre WW1 era to describe people of European cultures, as every other culture was considered under developed and of low culture." (from Wikipedia)
pop culture: "the entirety of ideas, perspectives, attitudes, memes,1 images, and other phenomena that are within the mainstream of a given culture, especially Western culture of the early to mid 20th century and the emerging global mainstream of the late 20th and early 21st century. Heavily influenced by mass media, this collection of ideas permeates the everyday lives of the society. Popular culture is often viewed as being trivial and "dumbed down" in order to find consensual acceptance throughout the mainstream. As a result, it comes under heavy criticism from various non-mainstream sources (most notably religious groups and countercultural groups) which deem it superficial, consumerist, sensationalist, or corrupt.

The former describes specifically what you are asking about.
Now, because your question referenced "English teachers" and RISD, for further description and distinction here's the first part of an article from The Guardian entitled "High culture versus pop culture: which is best for engaging students?"

"A high culture is the self-consciousness of a society", Roger Scruton wrote last year in the Guardian. "It contains the works of art, literature, scholarship and philosophy that establish a shared frame of reference among educated people."
As a teacher, the last part of this statement interested me because one of the aims of schools should be nurturing "educated people". Importantly for Scruton, an absence of "high culture is superseded by a culture of fake". This 'culture of fake' consists of many things, including false ideologies, opinions and expertise, but unfortunately for me, Scruton did not really identify what a 'low culture' could be.
In terms of teaching, learning and the curriculum, my dilemma is relatively simple, but not trivial; should I endeavour to reference high culture in my lessons so that students can appreciate cultural life at its finest, or, in order to engage students and make learning enjoyable, should I litter my lessons with references what might be perceived as low culture, which is probably best defined as 'pop culture' in the context of young people.
Pop culture would include just that; culture which is popular, easy to understand and entertaining to the majority of young people. For example, pop music, romantic Hollywood comedies and soap operas. High culture, on the other hand, may include renaissance art, classical music and opera. The latter is arguably more sophisticated, intellectually challenging and intrinsically rewarding.


Estimating PCB Trace Delays

Question: In the context of FPGA design, I sometimes need to estimate PCB trace delays between the FPGA and external devices to properly constrain the input/output timing. As those do not need to be accurate to the picosecond, I'm looking for generally accepted rules of the thumb rather than exact formulas. From what I've seen here, 150 ps/in seems to be a reasonable estimate. Which values do you use?
Comment: The speed of light in air is 1 ft / ns. On PCB, it's about half to 2/3rds of that, so lining up with your 150 ps/in. [Grace Hopper](https://en.wikipedia.org/wiki/Grace_Hopper) used to take a 1ft long piece of twisted pair into her lectures and show it to the audience, to show them what 1 ns looked like.
Answer: As the delay is dependent on "many" variables, you should use AppCAD from: https://www.broadcom.com/appcad.
Don't forget "matching" impedances load.
Example:
Comment: Thanks a lot. A former colleague of mine also suggested this tool: https://saturnpcb.com/saturn-pcb-toolkit/
Answer: Well, this dives into the topic of high-speed board design, and that's fine.  (going a bit more basic, also for future readers)
Basically, when we're talking about rapidly changing currents coming out of a IC pin, we cannot (appropriately) model these as the current flowing in the conductor. (A simple example to illustrate that: if you attach a couple cm of trace to a pin that outputs something at 5 GHz, be it a sine oscillation, the 5 GHz fundamental or the 5 GHz harmonic of some digital serial line, and simply let it end, congratulations, you've built a transmitter, and the energy put into the line will just be dissipated as radiated wave. You couldn't explain "energy disappearing from the trace" in a model that things of the signal as current on the trace alone.)
What you do need to do is understand the signal as the changing magnetic field around the conductor, and the changing electrical field between the conductor and ground (or, e.g., its differential complement). Now, Maxwell's Equation link these two: The time-derivative of one field always inherently leads to a rotation of the other. And vice versa!
So, you're changing the current on the line, you cause a changing magnetic field, because wherever there's current there's a magnetic field, and a changing magnetic field leads to a curl in the electric field, which changes, and a change in the electric field leads, just like your original current, to a curl in the magnetic field, and so on.
The fact that this is mutually linked physically means that the change in time of one field leads to a change in space of the other, and vice versa: You causing a current here means you get an electromagnetic field over there, a bit later.
The a bit later is the effect of the factors between the sides of the differential equations described above. In vacuum, that's actually the speed of light!
Now, back to the trace on the PCB: Say, it's a (so-called) microstrip line above a ground plane. Now, because that ground plane is (sufficiently well) a perfect conductor, there cannot be an electrical field within the ground plane (right? if there's an electrical field and something that conducts is in there, a current flows. If you have perfect conductor, an infinitely large current flows instantly, eradicating all the reason there could have been different potentials), and your differential equations above get boundary conditions that say "Hm, OK, so if there's an electrical field in parallel to the surface of the metal, there must be one right inside as well, else we get an infinite difference of energy in a finite (or even zero) thickness, and physics don't like that kind of incontinuities". So, there cannot be a parallel electrical field to the surface of the ground plane. That in turn says you only get that as normal field, pointing towards the conductor in which your current flew.
Now you have vertical electrical field lines, meaning the magnetic field makes "circles" in horizontal planes. Whenever the voltage difference between conductor and ground reduces, the magnitude of the electrical field reduces by the same amount. So does the magnitude of the magnetic field "circles", but: delayed! (same phenomenon as vacuum speed of light). The longer-lasting magnetic curls cause an electric field opposing the original one, again "delayed", and you end up with an alternation of strong electric vertical/weak magnetic circles and strong magnetic circles/weak vertical E-field along the length.
Now, you can influence the speed at which this alternation changes – which is actually the speed at which the signal travels along the trace – by defining thickness and shape of the trace. It turns out that for many useful applications, you will pick it such that in an average glass fiber board, your speed is roughly 2/3 of the speed of light, so 2·10⁸ m/s. (too lazy to convert to ps per inch.)
But: that's really just a thing you'd end up after intentionally designing the trace to be guiding waves in their fundamental mode, with low losses, in the usual board materials. This can, and does, look different for high-performance (RF-rated) ceramic boards, and transmission lines that do not have anything more complicated than the "alternating vertical/horizontal 1:1" setup described above.

Estimating PCB Trace Delays


What traits do the Dwemer have in Elder Scrolls' lore?

Question: I'd like to know what traits the Dwemer have in terms of character creation on the ES games. For example, Wood Elves are good at archery. Is it known what the Dwemer were good at?
Answer: The Dwemer have never been (officially) playable and in the entire series we have only encountered one live Dwemer (and he was suffering from a crippling disease). There are thus no official sources to cite regarding what they would have been good at and what perks they would have imparted from a gameplay perspective.
However, there is a fair bit of lore about them, most of which serves to underscore the fact that they were excellent craftspeople and engineers, both of the physical and of the arcane.
From the UESP wiki:

The Dwemer appear to have been a highly technologically advanced and particularly dogmatic race. Others races have still not cracked the secret behind their metal.

Their creations have survived for hundreds of years after their race vanished, some of them completely intact and functional. Their use of steam and geothermal power has not been widely reproduced in the rest of Tamriel, even at the time of Skyrim, which is (iirc) the most-recent era of Tamriel we've seen so far.
In addition to powering devices through mechanical means, they were also very adept at magical artifice (nearly all of their constructs, the guardians and spider-droids, appear to derive their power from soul gems or similar magical devices).
We can theorize, then, that they would be highly intelligent and proficient with magic, and that from a gameplay perspective one might see boosts to skills that dealt with passive magical abilities and construction skills. That is, they'd probably be good at enchanting and smithing and the like, not necessarily destruction, illusion or conjuration magic.
Comment: Thanks that's great. Is it know whether they would have a crafting advantage in game if they were playable?
Comment: @CJF No, it's not known since they are not playable. Perhaps there is a mod that would make them playable, but that mod would not be canon.
Comment: It is not, as they have never *been* playable. It's probably reasonable to assume as much (it certainly seems to *me* that it would fit with what we know of them from lore) but until the race is made playable by Bethesda the decision is in the hands of their designers.
Comment: Thanks for your replies. I agree with Josh, that it is fair to assume that they would probably have a crafting advantage if they were implemented into the game.


How do I get the menu on the right side of the screen back?

Question: I just started learning blender and was following the donut tutorial on youtube but I closed the menu that shows all the different options and settings on the right by accident. It is there if I start a new project, but on my current donut project, it is still missing. Can someone please tell me how to get this menu back? I included a picture, basically a "before" and I circled the part I am trying to get back. Please help
Answer: You can open a new panel by moving your mouse e.g. to the lower left corner of another panel until a cross appears and then drag it up.
Then change the panel to "Properties"
Comment: ok but if I don't want these floating pink hexagons?
Comment: @moonboots : if you want to have a new panel - you will have that hexagons. You know - Nothing is for free. Everything has its prize.
Comment: And by the way - they are pink and cyan….


How does PK know better than Jaggu?

Question: In PK movie, PK knows about Jaggu's boyfriend that he didn't cheat/left her in Belgium. But PK only read the mind of Jaggu. PK didn't know about Tapasvi's lies. How does he know that Sarfaraz still loves Jaggu in the ending?
Answer: PK did't know about Tapasvi's lies but he had power to know what actually happened by holding person's hand . So he came to know about the lies while holding jaggu's hand which Tapasvi told him on the video call .
Now coming to the Sarfaraz's point , He made that assumption which was made by considering these three situations or facts which actually happened at church on the wedding day .

The boy who came to give the letter was unknown to Jaggu .
There wasn't anybody available there other than a girl who was
holding the basket with the cat .
Nobody's name was written on the letter.

So PK came to conclusion by using this facts that letter might not from him it was a guess which could turned out to negative but considering Sarfaraz's image on Jaggu's mind, PK may have thought that this is possible which actually was .. :)
Answer: He shouldn't have known given the available data.  He guessed and it turned out to be right.  But he acted as if he knew the truth, which is a logical flaw.  He should have acted like he is guessing.   Given that he was guessing, it looked like it was a frivolent wager to prove that the Tapasvi is a shallow Sadhu.  Had the guess gone wrong, both Tapasvi and his remote would have gone other way and the movie would have ended differently.  Sadly, Indian movie goers dont worry about Logic & consistency
Answer: Tapaswi was also very smart. He told Jagat Janani that Sarfaraz was cheating her because he was living in Bruges and working at the Pakistani embassy (as he said). But there is no Pakistani embassy in Bruges, but in Brussels which is 100 km away from that place.

How does PK know better than Jaggu?


How to solve this question based on average speed?

Question: Without stoppage, a train travels a certain distance with an average speed of 60 km/h, and with stoppage, it covers the same distance with an average speed of 40 km/h. On an average, how many minutes per hour does the train stop during the journey?
Comment: Hint. Just imagine that the "certain distance" is 60km, so the trip takes one hour.
Comment: You are assuming that the train travels at either $60$ km/h or $0$ km/h with no acceleration or deceleration
Answer: As distance is same.
D = S * T
Let t = usual time
s = stop time
60t = 40 (t+s).
60t - 40t = 40s
20t = 40s
t = 2s
Here total time is 1 hour or 60 minutes.
So, t + s = 60 minutes
2s + s = 60
3s = 60
s = 20
So train stops for 20 minutes.

One trick that I use to solve this type of question.

Difference in two speeds is 20 kmph. Which is 1/3rd of original speed. So find the 1/3rd of 1 hour or 60 minutes. Which is 20 minutes.


Using 7A power cord on 10A laptop

Question: Does using a 7A power cord on a laptop that needs 10A damage components or mess something up? I know I need a new one but I have already used the old power cord. This laptop is a Compaq Armada 1500 computer (20+ years old).
Comment: The manual said it needed 10A for some reason. Everything makes sense now though. Thanks for your help.
Comment: Are you referring to the mains lead or the lead from the mains adapter to the computer? If the former, it will be ok as the adapter will not be drawing even 5 A from the mains; if the latter I would be concerned about the lead overheating and melting or catching fire.
Comment: Well the adapter is built inside the computer. It is just a cord with no adapter that goes from the wall tongue computer. The adapter inside the computer needs at least 10A and the cord only supports 7A.
Comment: If the laptop is drawing even 5 A from the mains then it will be generating 600 W of heat (assuming 120 V mains) - think of a small electric fan heater - which is unlikely. Are you asking because it has a connector rated at 10 A?
Answer: Assuming the computer really does draw more then 7 amps (which be very unlikely - even at only 100 volts that would be 770 watts  which would drain the battery in seconds to minutes and the laptop would need to disipate an impractical amount if heat ) this could theoretically be dangerous as the cord would heat up and possibly short out - so its a fire risk.
If this is a British plug with a 5 amp or 7 amp fuse in the base, it is ok to use. If not, the risk is very small [but not non existant. It would be a question if " why did the manufacturer specify a 10 amp cable?" My guess us that they were cheap and standard  and not to do with the laptop requirements.

Using 7A power cord on 10A laptop


how can I connect a scanner with Centronics 50 output to USB?

Question: How can I connect a Legacy scanner with a Centronics 50 output to a laptop with USB ?
Comment: If it's really a 50 pin Centronics connector, then it's SCSI. You've actually got two problems here: Even if you get a USB to SCSI adapter, you still need to find drivers for your laptop that will talk to that scanner. Any scanner that is old enough to be SCSI may well not have drivers available for a modern OS. Please let us know the OS on the laptop as well as the scanner's model.
Comment: Do you know what protocol it speaks? It sounds like maybe SCSI.
Comment: With information about mark and model someone might be able to give a correct answer. With the current information it will only be more or less qualified guesses.
Answer: It depends what protocol the scanner uses. If it is SCSI you need a USB to SCSI converter. If it is parallel port you need a USB to parallel port converter, although be aware that sometimes they are not compatible with scanners.


Metrizable Topological Space In Many Ways

Question: Prove that if a topological space $(X, T)$ is metrizable then it is metrizable in infinitely many ways.
$$$$As the given topological space is metrizable so there exists a metric $d$ on the set $X$ such that the it can generate a class of open subsets which is the topology $T$. Now as we know that whenever $d$ is a metric on $X$, then the function satisfying $$d'(x, y)=\frac{d(x, y)}{1+d(x, y)}$$, is also a metric on $X$. Now suppose $A$ be an open subset of $X$ under the metric $d$. Now chose a $x \in A$, then there exists some $\epsilon$ such that for all $y$ satisfying $$d(x, y) < \epsilon$$ lie in $A$. Now for the metric $d'$ we see that $\frac{\epsilon}{1+\epsilon}$ works and for all $y$ satisfying $$d'(x, y) < \frac{\epsilon}{1+\epsilon}$$ satisfy the above equation and hence lie in the set $A$ and hence $A$ is also open under the metric $d'$. So the class of open sets generated by the metric $d$ can also be generated by the metric $d'$ and hence $d'$ can also induce the topology $T$. Similarly we can find infinitely many metric like $d''$ satisfying $$d''(x, y)=\frac{d'(x, y)}{1+d'(x, y)}$$. And hence the topological space $(X, T)$ is metrizable in many ways.
$$$$Is the proof Correct??
Answer: The statement is false: a set with a single element admits one and only one metric.
If we assume that $X$ has more than one element, then, although your proof works, I think that it is simpler to say that, if $d$ is a metric on $X$, then, for each $k>0$, $kd$ is another metric on $X$ which induces the same topology.
Comment: @user728159: No, it can't: the only metric on $\{x\}$ is the function $$d:\{x\}\times\{x\}\to\Bbb R:\langle x,x\rangle\mapsto 0\,.$$
Comment: A singleton set can have many metrics
Answer: You proved that open sets in $d$ are open in $d'$. But you also have to prove the converse. For this you just have to replace $\frac {\epsilon} {1+\epsilon}$ by $\frac {\epsilon} {1-\epsilon}$ in your argument (taking $\epsilon <1)$. Except for this your construction of the metrics is  is fine.
Answer: The idea is fine, but you really should include an actual demonstration that if $d'(x,y)<\frac{\epsilon}{1+\epsilon}$, then $d(x,y)<\epsilon$ and hence $y\in A$. If
$$d'(x,y)=\frac{d(x,y)}{1+d(x,y)}<\frac{\epsilon}{1+\epsilon}\,,$$
then
$$(1+\epsilon)d(x,y)<\epsilon\big(1+d(x,y)\big)\,,$$
so
$$d(x,y)+\epsilon d(x,y)<\epsilon+\epsilon d(x,y)\,,$$
and hence $d(x,y)<\epsilon$.
More important, you also need to show that $d$-open sets are $d'$-open. If we let $\epsilon'=\frac{\epsilon}{1+\epsilon}$, we can solve for $\epsilon$ to find that $\epsilon=\frac{\epsilon'}{1-\epsilon'}$, a fact that should suggest how to do this.
There is, however, an easier way to get infinitely many different equivalent metrics. (It does require that $X$ have at least two points, but so does any approach.) Let $x$ and $y$ be two distinct points of $X$, and let $r=d(x,y)$. For each $s\in(0,r)$ define a metric $d_s$ on $X$ by setting $$d_s(u,v)=\min\{d(u,v),s\}$$ for all $u,v\in X$. It's easy to verify that $d_s$ and $d$ generate the same topology, since they have the same open balls of all radii less than $s$, and they are clearly distinct, because $d_s(x,y)=s$ for each $s\in(0,r)$.


Drawing lines from points in QGIS

Question: I have a point layer and need to draw a line passing through or in the vicinity of the points (up to 0.5m distance between the point and the line).
The nodes of the line will receive the attributes of the points closest to them.
Comment: Please, do not forget about ["What should I do when someone answers my question?"](https://stackoverflow.com/help/someone-answers)
Answer: The simplest way is to use the points to create the line. To do this in QGIS use the Points2One plugin. Install this plugin (available via Plugins > Manage and install plugins) and follow the dialog to create a line from your points.

If you have multiple lines, then make sure your points data has a suitable ID field to identify the lines they belong to.
Alternatively you can use v.build.polylines in GRASS (available through QGIS) or the SAGA GIS tool Convert points to line(s) Available through the Processing Toolbox (Ctrl+Shift+T).
Comment: The Points2One plugin is not available for the current release of QGIS (3.4.4).
Comment: points2one fails for hard to tell reasons in 3.22.3, and needs to be installed from .zip.
So the inbuilt alternatives (processing toolbox) are to be preferred, but it is surprisingly hard to define a line outside a feature set for (in my case) debugging purposes.
Answer: To create polygons from points, you can also use

"Points to path", then
"Lines to polygons" tools without need to install any plugin
Comment: Go to processing and click Toolbox  or "Ctrl + Alt +T"  to search tools , then type "Points to path" and you will see the tool.
Comment: Where do you find the "points to path" tool?
Comment: Not available in `2.8.6-Wien`
Answer: In QGIS 3 instead of the plugin mentioned above, you can find the "Points to path" tool in your Processing Toolbox (Ctrl+Shift+T). Works exactly the same.
Answer: You can use the "Nearest Points To Path" algorithm from ProcessX Plug-In. You will find it in your Processing Toolbox (Ctrl+Shift+T) in ProcessX -> Vector - Creation -> Nearest Points To Path.
It works similar to the native Points To Path algorithm, but instead of an attribute it takes the distance of the points as order for creating the path.

Disclaimer: I am the author of this Plug-In.
Answer: The Points2One plugin last version is from 2015 and as has been commented, isn't working in the current QGIS 3.x versions. I'm posting an alternative solution that I have also answered here.
One way to do this would be using a "Virtual Layer".

Create "X" and "Y" fields in your point layer containing the x and y coordinates
Create a constant field with the same value
Save a copy of the point layer
Add a virtual layer joining by the constant value of both layers (original and copy) using the following expression:

SELECT *
FROM points
JOIN points_copy
    ON points.constant = points_copy.constant

This will create a Virtual Layer that will contain every combination of X,Y origin and destination coordinates. In my example I used 29 points thus resulting in a virtual layer of 841 points (29²).

Just for symbolization purposes the point layer can be symbolized into lines using a geometry generator expression:
make_line(make_point("cx", "cy"), make_point("cx:1", "cy:1"))

To make a permanent layer I recommend using the tool XY to line from the plugin Shape Tools.
In this particular case the OP asked to limit the distance to 0.5 m which can be accomplished filtering the lines by length.
Answer: I used the Points2Path feature, but the lines never appeared.  I am using an unnamed CRS due to the coordinate system of the maps that I received.  I right clicked on the layer that was created by Points2Path, and selected "Source".  The "Assigned Coordinate Reference System (CRS)" for the points was set to  "unnamed".  When I changed it to "Project CRS: unnamed", the line appeared.

Drawing lines from points in QGIS


Why is the temporal element of the FRW metric tensor constant?

Question: When written in conventional coordinates, the temporal element of the FRW metric tensor is a constant over all space-time, and the Friedman equations that result from the tensor describe a universe that corresponds to the observed expansion of space.  Assuming that Energy/Momentum can be viewed as a fluid whose pressure/density is conserved in this expanding space, I would expect this pressure/density to correspondingly decrease with time. I am under the impression that in a volume of space having higher Energy/Momentum, a clock will run slower than in a volume having lower density. If this is correct, the temporal element of the FRW metric tensor cannot be a constant over the age of the universe. Can someone explain where my reasoning is incorrect?
Answer: You can have some $g_{00}(t)$, but you can always redefine your time coordinate so that you recover the (cosmic time) FLRW form of metric (where $g_{00}= $ constant).
Say you have $g_{00} = f(t)$ and you want to redefine coordinates into FLRW form. This just means solving
$$
\int \sqrt{f(t)} \ dt = \int dT
$$
so that the metric expressed in the new time coordinate $T$ has the FLRW form.
Also, why would energy density ($\rho$) be conserved during the expansion of the universe? There is no timelike Killing vector; from Noether's theorem, the lack of time-translation symmetry implies that energy is not conserved during the universe's expansion. Perhaps what you're missing are the Christoffel symbols. These have non-trivial dependence on the scale factor $a(t)$, and you get the usual relation that $\rho \sim a^{-3}$ for matter and $\rho \sim a^{-4}$ for radiation.
Comment: Your link explains how to move from physical to comoving coordinates, but this is not what the OP is asking. His point is that the formula 1.1.16 is not logically justified in your link (which is true). In fact, it is not justified for both temporal and spatial parts. Consider a hypothetical example of this uniform universe: https://physics.stackexchange.com/questions/424982/what-is-this-metrics-scale-factor - What is the scale factor there?
Comment: @safesphere It's just a coordinate transformation. These cosmology notes by Daniel Baumann are nice. See page 7, for instance - http://theory.uchicago.edu/~liantaow/my-teaching/dark-matter-472/lectures.pdf


miniature parabolic antenna

Question: I have to build a 2.4ghz wifi antenna with one SMA feedpoint. The rules are that it must fit inside a 15cm cube shaped box. For testing, the antenna will be pointed at a laptop that is moved to various distances however remains in the same direction. Therefore, I want to build the most directional antenna possible. Upon a quick google search, I started looking into making a parabolic antenna. However, since most parabolic antennas I have seen are at least 1 or 2 feet in diameter, I was wondering if making such a small parabolic antenna would still be effective and better than other types of antennas. If not, what other types of antennas should I look into? Yagi was one option, but I thought parabolic might be better.
Answer: From this equation, for a passive antenna of 15 x 20 cm, you would expect a gain of under 14 dBi.
$$ A_e = 0.15 * 0.2 = {G\lambda^2\over{4\pi}}$$
I don't recommend trying to make a reflector antenna this small. There are two main problems:
First, you have to illuminate them properly. For highest gain, you want uniform illumination, this means the feed provides a constant power across the surface of the dish, and nothing outside that. This is difficult in practice, so there's usually a smooth taper from centre to edge, where the power density at the edge is about 6 to 10 dB less than the centre. The rest spills over the edge and is lost. This leads to practical (large) dishes having an effective area only about 60% of their physical area. This will reduce the gain of your antenna by ~ 3 dB.
Second, the feed itself is not small. At 2.4 GHz it might be a patch or a yagi - both are in the order of 5 x 5 cm at least. This is hard to fit in the box, and the feed will block some of the signal.
Finally, given the size constraints, the "dish antenna" will be almost indistinguishable from a "corner reflector" antenna.

The great thing about building wifi antennas is that you have quite good quality test equipment available, in the form of your laptop or phone.
Find a wifi signal strength app (ubiquiti wifiman is excellent). Set up your antenna on an access point, old router etc. In an open space, walk slowly away from the antenna and measure the signal strength with your phone. (rotate your phone around to get the strongest signal, and hold it that way). This will let you compare the performance of different versions of your antenna, without any special test equipment.

Start by considering a corner reflector antenna - this will fit nicely into the box. I've linked to the original 1940 patent but there are many websites about it.

You could make the reflector from aluminium foil taped to cardboard (as it doesn't have to connect to anything). The dipole could be a simple wire dipole; the absolute simplest would be to salvage a dipole from an old wifi router, as that already comes with a connector and cable...

The best gain will probably be from a 2 x 2 patch array. You'll need a PCB to split the power and feed the four patches, this would need to be microwave board to keep the losses down. The patches themselves should be air spaced, at least 5 mm above the PCB, fed with a pin or an L.
If you don't have access to microwave PCB manufacturing, but you can simulate the structure, you will probably do almost as well with a single patch, and a groundplane that is somewhat cupped. The ground can be 15 x 19 cm with a ~ 3 cm sides. The patch needs to be $\lambda/2$ on the active side, but can be $3\lambda/4$ on the other side, so say 61 x 90 mm in air. Mount it about 6 mm from the ground. Feed it in the centre of one edge, directly with the inner wire of the SMA connector.
Comment: Nice alternatives. I'd suggest to the OP that yagi types tend to be narrow-band, requiring precise construction. Corner reflector is less fussy. Arrays are nice if you can split power evenly and maintain phase coherence.
Comment: yes! Patches with wide air spacing are also super-forgiving. A yagi would definitely require a network analyser to tune, especially to fully cover 2.4-2.5.
Comment: For a corner reflector, would it be worth it to make  it a cone shape? (revolve the corner reflector around its axis) I have access to a 3d printer, so shape isn't a problem.
Comment: Also, does the half wavelength antenna begin where it says antenna, or is that where it ends.
Comment: A corner reflector is two flat plates behind a dipole antenna. I don't think there's any benefit to curving it. Look at the 3D sketches in the patent. The dipole is vertical in the figure in my answer, i.e. into the page.
Answer: Try a Kraus helical antenna.

The Kraus end-fire helical has wide bandwidth, circular polarization, and is very non-critical in its dimensions.
At 2.4 GHz your wavelength $\lambda = 12.5\text{ cm}$. Then the dimensions of your helical antenna would be
$$
\text{circumference} = \lambda = 12.5 \text{ cm}\\
\text{diameter } D = \lambda/\pi = 4 \text{ cm}\\
\text{turn spacing } S = \lambda/4 = 3.1\text{ cm}
$$
Groundplane should be at least $\lambda/2 = 6.25 \text{ cm}$ across (square or circular).
Inside your $A = 15\text{ cm}$ box you could fit a helix with almost five turns. This would give you a gain of perhaps 10 dBi.
This is a circularly polarized antenna. For Wi-Fi you want right-hand circular polarization (RHCP). Make sure you wind the helix in the correct direction! As your right thumb points along the axis and toward the target, then your right fingers curl in the direction of the twist.
If you can measure the matching, you can lower the input impedance by tweaking the first half-turn of the helix closer to the groundplane.
In contrast to a Yagi-Uda antenna, the Kraus helical is quite forgiving in its dimensions. Here's an example improvised Wi-Fi antenna which actually works fairly well:

Another fun hack: build a LHCP version, and bounce the signal off a wall. The reflection will swap the polarization from LH to RH.
Comment: If the other end of the link is linearly polarized, then yes either RH or LH would work. But there do exist circularly polarized Wi-Fi antennas, and invariably they are RHCP. So build yours RHCP to stay compatible.
Comment: @MarcusMüller I think the CP WiFi antennas are mostly intended for highly directional point-to-point links. CP antenna on both ends gives you some nice gain.

If the base station has typical ~omnidirectional linear polarization, then CP antenna on the remote station will be advantageous because you won't get as much polarization mismatch.
Comment: Good idea, and easy to build. Why specifically RHCP though? Wifi antennas are generally linear so either sense will work well.
Comment: @SteveGolson-W1SEG there's circularly polarized Wifi – wow, didn't know that. Also feels a bit counter-intuitive, wave-wise, since Wifi's signal design (OFDM) assumes a multipath channel, and thus reflections, and reflections can (and will) have the opposite polarization!
Comment: Ah. Makes sense!

miniature parabolic antenna


related to fail safe on linux suse machines

Question: Mechanism for swithing virtual ip if one server goes down to other server sharing same virtual ip as secondary ip
Answer: You might want to check out CARP or Heartbeat. I have experience of CARP and it works really well, Heartbeat is extremely popular as well.
Answer: Heartbeat has been working very well for me. In addition to IP addresses it can take care of other resources (processes, disk mounts etc) as well. And as 3molo said, CARP is a good one, too.


Is there any cloud service has end to end encryption plus binary diff sync?

Question: I like Dropbox's binary diff sync as we don't need to upload the entire file when there's a minor change, especially useful for large files. But Dropbox doesn't support end to end encryption which is privacy concern. So, I wonder if there is a cloud service that has both ?
Answer: To get end-to-end encryption it is inevitable to install a piece of software on your client. Such tools exist and often work with more than one cloud. I would recommend to have a look at Boxcryptor, as far as I know it also supports differential synchronisation.


Exchange 2010: Receiving "You can't send a message on behalf of this user..." error when trying to configure delegate access

Question: I am trying to give someone (John Doe) delegate access to another account (Jane Doe) in our test environment. However, I receive the following error from Exchange:
*Subject:* Undeliverable: You have been designated as a delegate for Jane Dow
*To:* Jane Dow

Delivery has failed to these recipients or groups:

John Doe
You can't send a message on behalf of this user unless you have permission to do so. Please make sure you're sending on behalf of the correct sender, or request the necessary permission. If the problem continues, please contact your helpdesk.

Diagnostic information for administrators:

Generating server:

/O=UNIONCO/OU=EXCHANGE ADMINISTRATIVE GROUP (FYD132341234)/CN=RECIPIENTS/CN=jdoe
#MSEXCH:MSExchangeIS:/DC=local/DC=unionco:MAILBOX-1[578:0x000004DC:0x0000001D] #EX#

Can someone help me troubleshoot this?
Answer: In your question, you need to clearly define what you want to achieve.
There is a conflict between 'delegate rights' and 'Send As'.
Delegating rights can be anything from send-as, send-on-behalf, control calenders etc.
You need to ensure only either the client-side changes are made or the server-side changes are made, otherwise will cause you more trouble then ever.
I also recall a long time ago (might not be applicable now) that if you made changes to the client-side (aka Outlook), you need both users to be using the same version.
Personally any changes like these, i would do it on the Exchange server. The error message provides implies you are attempting to 'Send on Behalf'; therefore you need to:
1) Go to Recipient Management and choose Mailbox
2) Choose the mailbox and choose Properties in Action Pane
3) Go to the Mail Flow Settings Tab and choose Delivery Options
4) Click the Add button
5) Select the user who wish to grant access to and click Add and then Ok
Answer: In Exchange 2010's Exchange Management Console, under Recipient Configuration you can right click on a mailbox to "Manage Send As Permission". This will directly give a user permissions to send on behalf of another user. I believe any other non-CLI method is deprecated in Exchange 2010.
Comment: But why would I need to have Send-As permission for a user that I'm trying to give delegate rights? Could, perhaps, some other setting not be set correctly?


VBA function that sets a range depending on adjacent cells

Question: This question is a little tricky for me to explain but I hope it will somehow make sense to someone.
At the moment in terms of setting ranges on my cells, I have always given it some form of a limit (e.g Range("A1:A10")) before adding my functions, but I now want to make a macro that is more flexible whereby the limit is based on the last cell row that is recorded on a column in  the spreadsheet. The spreadsheet I use in question is linked to a database in which i refresh on a weekly basis and the range of rows I get can vary from one day being 250 rows, the next being 300 and sometimes back to 275 for example.
There must be an easier way around this where I dont have to change the Range limit on vba macro every time I refresh the spreadsheet. I can imagine it being such a simple solution though I cant figure out how to go about this without me reading too much into it and further confusing myself. Am I looking at using a form of a loop here in this case and if so how do I go about it?
Answer: To get the last row you can do something like this
<code>Dim last_row as Integer, input_sheet as Worksheet, col_number as Integer
col_number = 1
Set input_sheet.Sheets(1) ' change it accordingly
last_row = input_sheet.Cells(input_sheet.Rows.Count, col_number).End(xlUp).Row
</code>
What this does basically is select the last row of column A of the worksheet and replicates the action of "Ctrl" + "Up direction". So you end up at the last row. You could modify this accordingly according to your needs.


Symfony on Godaddy windows hosting

Question: I uploaded all symfony framework files to my godaddy windows shared hosting that supports php5 but I cannot access when I try to type the folder path? It shows a 401 authentication error but when I uploaded a normal php file it was working perfectly. Could it be an issue with .htaccess file as I found that there is not htaccess in windows hosting. Someone please help!
Answer: I'm not sure if you can use Symfony (v2 or v1, BTW?) without modification VirtHost settings for development.
Htaccess is useful, but nor virthost spec nor htaccess contain nothing for basic http-auth.
I have a working (WAMP) Symfony 1.x host and can show (and understand) it's config.
In common, on-site instruction for starting Symfony is rather full and detailed, without Alias /sf  ... it will not work (based on own experience).

Symfony on Godaddy windows hosting


What is the difference between data leakage and endogeneity?

Question: I have the impression the former is used in ML whereas the latter is used in econometrics. They both carry the idea that information from the target is "leaking" in explanatory variables.
Is there any difference between those two notions?
Answer: Endogeneity refers to explanatory variables correlated with error term because of a missing variable or measurement error. Data Leakage is introduction of spurious explainability in the model because of bringing in new data than ground truth like synthetic data (say we used SMOTE). In the former case, we will see that the model is biased too much for some feature coefficients whose explainability seeped into the error term. In the latter case we will see that the model will have high variance and lacks generalization which makes it useless for production/test data. Typical example to introduce data leakage is by SMOTING (yeah using the word as a verb) before you do validation split.
Answer: No, data leakage is an example of simultaneity, which is a form of endogeneity.


What would be the maximum stable mass of an artificially created sphere of very cold, very pure hydrogen-1?

Question: The Chandrasekhar limit assumes the white dwarf is made of carbon, but hydrogen has a much lower ratio of mass to electrons, so the limit should be different in that case. And electron degeneracy pressure might not even be the limiting factor, if neutronization or cosmic ray initiated fusion chain reactions are possible.
Answer: The "standard" Chandrasekhar mass is
$$M_{\rm Ch} = 1.44\left(\frac{\mu_e}{2}\right)^{-2} M_{\odot}\  ,$$
where $\mu_e$ is the number of mass units per electron in the gas.
For pure hydrogen $\mu_e= 1$, so  $M_{\rm Ch}=5.76M_{\odot}.$
At this mass, a Newtonian star supported by ideal electron degeneracy pressure would shrink to zero size and infinite density.
Would such an object ever exist? No, because it would become unstable due to General Relativity or Inverse Beta Decay or because nuclear reactions fuse the hydrogen into helium (with $\mu_e = 2$). All of the above will commence at finite densities and hence lower masses.
In the real universe (as opposed to a hypothetical one), you cannot assemble a cold hydrogen white dwarf without it being much hotter in its earlier life. We know that if such an object exceeds about $0.08 M_{\odot}$ (aka the brown dwarf limit), then eventually, its contracting core will commence thermonuclear hydrogen fusion.
So my real answer would be $0.08M_{\odot}$.
But you are going to ask, what if I could somehow extract all the potential energy and not allow it to heat the interior? Well, you would still get pycnonuclear reactions caused by zero point energy and quantum tunneling in a solid proton lattice at around $10^{9}$ kg/m$^3$, even at cold temperatures. This will occur well before inverse beta decay, which would require electron Fermi (kinetic) energies of 0.78 MeV and hence densities of $1.2\times 10^{10}$ kg/m$^3$.
Back of the envelope: if we use non-relativistic degeneracy pressure, the mass-radius relation of an ideal, cold white dwarf is
$$ R \simeq 0.013\left(\frac{\mu_e}{2}\right)^{-5/3} \left(\frac{M}{M_{\odot}}\right)^{-1/3} R_{\odot}$$
The average density for $\mu_e=1$ is therefore
$$\rho \simeq 2\times 10^7 \left(\frac{M}{M_{\odot}}\right)^2$$
and the central density will be about 6 times higher than this average density (for a $n=1.5$ polytrope).
Thus a central density of $10^9$ kg/m$^3$ will not be reached before the electrons become (mildly) relativistic. Unfortunately, that means running a numerical model to find the density at a given mass. However, roughly scaling the well-known equivalent model for a carbon white dwarf I would estimate that this density is reached at a hypothetical mass of $\sim 0.25 M_{\rm Ch}\simeq 1.4M_{\odot}$.
Comment: " ... you cannot assemble a cold hydrogen white dwarf without it being much hotter in its earlier life. " Simple, ultra slow accretion of material (e.g. one Jupiter mass every quintillion years.)
Comment: @blademan9999  I have addressed this hypothetical possibility in the second part of my answer. However, I note that I forgot that the central density is about 6 times the average density. This lowers the pycnonuclear reaction threshold mass a bit. I'll edit.
Comment: I was pointing out that it is possible to assemble a cold hydrogen white dwarf.


Certificate not found error trying to set up PEAP-MS-Chap v2 authentication

Question: I am trying to set up PEAP-MS-Chap v2 authentication for my VPN on a Windows 2003 SP2 server using a self-signed certificate. I have meticulously followed the PEAP certificate requirements when issuing the certificate as outlined here (i.e. subject name not empty, public key algorithm of RSA, minimum key length 2048 and an extended key usage of server authentication). More specifically, using the MakeCert utility of the Windows SDK, this is the command that I used to create the server certificate:
Makecert.exe -r -pe -n CN="WV21066" -eku 188.8.131.52.184.108.40.206.1 -sky exchange -sp "Microsoft RSA SChannel Cryptographic Provider" -sy 12 -len 2048 server.cer

I then installed the certificate in the server's certificate store (localmachine) under "Trusted Root Certificate Authorities". However when I now try to select the certificate for PEAP in the Routing and Remote Access Snap-in, I keep getting an error message saying there is no suitable certificate installed ("A certificate could not be found that can be used with this Extensible Authentication protocol", see here).
Examining the certificate in Windows Explorer, I can see that it lists Server Authentication (22.214.171.124.184.108.40.206.1) under the "Enhanced Key Usage" property and that its public key (RSA) has a length of 2048 and that the subject property corresponds to CN = WV21066.
Is there anything I'm missing?
Answer: Sorry for answering my own question, but I hope it helps somebody else. The problem was that I only added the certificate to the "Trusted Root Certificate Authorities" store whereas it must also be added to the "Personal" store as well.


iPhone 3GS stuck in apple logo

Question: My friend owns an old iPhone 3GS(I know it's vintage) that entered a boot loop a couple of days after it was jailbroken. So how I tried tackling the problem was by performing a restore through iTunes(since it is the only way) but unfortunately that didn't work. I tried 5 different ipsw files and I even tried clicking on restore to allow iTunes to fetch the newest ipsw for me but still no luck. I tried using a different computer, still no luck. I even tried using an older version of iTunes since the iPhone 3GS's release but guess what, still no luck. Any ideas on how to tackle this problem and do you reckon the phone is soft bricked or hard bricked?
Answer: Firm-Bricked I'm afraid. It happens a lot with those.

iPhone 3GS stuck in apple logo


Pi power from a PC?

Question: Given the type of USB port that the Raspberry Pi 2's DC input has, is it possible to just plug the Pi into a powered device with USB ports (such as a PC) and power it from that device?. Thought it'd be a neat idea since my PC is an all-in-one (an HP Pavilion).
Answer: The Pi doesn't care where you get the power from.
All will be okay as long as you can supply sufficient power to the microUSB socket for the Pi's needs and the needs of any peripherals you connect to the Pi.
Comment: That was one fast down vote that I do not understand. I'll vote up.
Comment: I'm not the downvoter... but it is probably worth noting that the USB3 standard only specifies that the port can output 900 mA. Many USB ports will output more (especially if they have a lightning bolt symbol), but it's not guaranteed.
Comment: @Jacobm001 Good point. I usually want to give my Pis at least 2A. A Model 2 or 3 are speced to pull 4W, which should keep them at 800 mA, so the Pi on its own should be fine. Not very much room for any peripherals, though. Model Zero should be no problem at all at 160 mA.
Comment: I've seen a 500 mA limit mentioned a lot for USB1/2.  The Pi Zero/B+/Pi2 (not sure about Pi3) will work quite happily powered from my laptop USB.  I generally only have a wireless dongle attached though.
Answer: Yes, it is possible and should work fine (with one caveat).  From a current perspective, the Pi2 draws 230ma at idle and under some load scenarios it stays below the 500ma rating of USB2 ports which are most common on PC's unless you have a newer PC that supports USB3, rated at 900ma (reference: http://raspi.tv/wp-content/uploads/2015/02/Pi2-power-table-1024x338.png ).  As long as the current draw of the Pi2 plus all your connected peripherals remains below the rating of the USB power supply, you are OK from a power perspective.
Now here's the caveat, which takes a moment to explain and may or may not be a concern for you.  In the normal case, where you power the Pi from a conventional USB power adapter (wall wart), it's not earth grounded to your AC main.  The power adapter has a 2-wire plug and does not connect you to the AC ground (3rd pin).  In this case, the Pi is said to have a floating ground.
Your PC, along with its USB bus, is earth grounded (with the exception of a laptop running on battery power) because it uses a 3-wire plug to connect to AC.  Once you connect an earth grounded USB line to your Pi (could be a USB power supply or a USB peripheral like a printer or monitor), you have a common ground with earth.
When your Pi is earth grounded, you have the risk of a ground loop short-circuit when there is accidental contact with AC-powered devices (e.g., an oscilloscope for testing) on the Pi.  The advantage of the Pi with a floating ground is that it is isolated from AC earth ground and will not pose this ground loop short circuit risk.  There is a whole field of study about power isolation for safety as well as to eliminate noise that can occur due to a ground loop.
Admittedly, this ground loop risk only exists for some narrow use cases but you should be aware of it.

Pi power from a PC?


How does the Firewind Blade's bonus stack?

Question: The Firewind Blade's property states,
"Whenever you hit one or more targets with a fire power using this weapon, one enemy adjacent to you takes fire damage equal to 1+the weapon's enchantment bonus."
How often does this damage activate? If I use a close blast power that hits three targets, would the damage instance occur once for each target, or once total for the power?
Answer: It would only trigger once. This is because the trigger is written "one or more" and 3 is one or more. If it had instead been "Whenever you hit a target" it would indeed trigger 3 times.
Comment: There are absolutely people on the charOp boards who disagree with diego here, but even there, they're in the minority. The clear intent is 1/power. In any discussion of the Firewind Blade, it's important to note that it's a separate damage instance, which potentially pings vulnerabilities twice.


Cancel thread with read() operation on serial port

Question: in my Cocoa project, I communicate with a device connected to a serial port. Now, I am waiting for the serial device to send a particular message of some bytes. For the read operation (and the reaction for once the desired message has been received), I created a new thread. On user request, I want to be able to cancel the thread.
As Apple suggests in the docs, I added a flag to the thread dictionary, periodically check if the flag has been set and if so, call [NSThread exit]. This works fine.
Now, the thread may be stuck waiting for the serial device to finally send the 12 byte message. The read call looks like this:
numBytes = read(fileDescriptor, buffer, 12);

Once the thread starts reading from the device, but no data comes in, I can set the flag to tell the thread to finish, but the thread is not going to read the flag unless it finally received at least 12 bytes of data and continues processing.
Is there a way to kill a thread that currently performs a read operation on a serial device?
Edit for clarification:
I do not insist in creating a separate thread for the I/O operations with the serial device. If there is a way to encapsulate the operations such that I am able to "kill" them if the user presses a cancel button, I am perfectly happy.
I am developing a Cocoa application for desktop Mac OS X, so no restrictions regarding mobile devices and their capabilities apply.
A workaround would be to make the read function return immediately if there are no bytes to read. How can I do this?
Answer: Use select or poll with a timeout to detect when the descriptor is ready for reading.
Set the timeout to (say) half a second and call it in a loop while checking to see if your thread should exit.
Asynchronous thread cancellation is almost always a bad idea.  Try to stick with event-driven interfaces (and, if necessary, timeouts).
Comment: This is a very bad approach. You should never write applications that constantly wake up and poll a variable. This destroys battery life, and it sounds like OP is writing this for a mobile device, making that a big issue... You could achieve the same result eliminating the timeout/wakeups by adding a pipe to your `select`/`poll` call, and sending the abort request over the pipe.
Comment: I am developing for desktop Mac OS X, so no mobile device restrictions.
Answer: This is exactly what the pthread_cancel interface was designed for. You'll want to wrap the block with read in pthread_cleanup_push and pthread_cleanup_pop in order that you can safely clean up if the thread is cancelled, and also disable cancellation (with pthread_setcancelstate) in other code that runs in this thread that you don't want to be cancellable. This can be a pain if proper cleanup would involve multiple call frames; it essentially forces you to use pthread_cleanup_push at every call level and structure your thread code like C++ or Java with try/catch style exception handling.
An alternative approach would be to install a signal handler for an otherwise-unused signal (like SIGUSR1 or one of the realtime signals) without the SA_RESTART flag, so that it interrupts syscalls with EINTR. The signal handler itself can be a complete no-op; the only purpose of it is to interrupt things. Then you can use pthread_kill to interrupt the read (or any other syscall) in a particular thread. This has the advantage that you don't have to switch your code to using C++/Java-type idioms. You can handle the EINTR error by checking a flag (indicating whether the thread was requested to abort) and resume the read if the flag is not set, or return an error code that causes the caller to clean up and eventually pthread_exit.
If you do use interrupting signal handlers, make sure all your syscalls that can return EINTR are wrapped in loops that retry (or check the abort flag and optionally retry) on EINTR. Otherwise things can break badly.
Comment: This is a very, very bad approach.  Using a signal and `EINTR` creates an unavoidable race condition, because the signal might be delivered just before the thread enters read().  As a result, the read will not be interrupted and will block forever, but only when the race condition triggers...  Making the bug extremely hard to reproduce and fix.  Polling is approximately infinitely preferable.  (That said, I agree the best answer is to add a pipe to the select/poll.)
Comment: In such an interactive use as OP wanted, the "block forever" would really be "block until the user presses the cancel button again". I agree there's a race condition here, which may be nontrivial to fix. For the general case, I can think of a complex framework with a nontrivial signal handler and longjmps out of signal handlers that could meet the requirements without making any async-signal-safety violations or other UB, but it's probably not what OP wants...


Can the decimal point values of fractions be predicted?

Question: Is there anyway to know what the decimal point value of a given fraction or a string of fractions will be?
For example if you know 21/4 = 5.25 and 21/5 = 4.2 can you predict that 21/6 = 3.5? With some sort of equation or a function? Or do you just have to actually divide it and see the result? It's not the whole value I'm interested in just the decimal points but anything would be helpful
Comment: Decimal point values? You mean the nonintegral part?
Comment: With the identity $\frac{a}{b} - \frac{a}{b+1}$ = $\frac{a}{b^2+b}$, you can find the difference between fractions with the same numerator and denominators differing by 1, and then subtract it from the larger fraction
Comment: Yes, the nonintegral portion but something that predicts the whole value is also great
Answer: Consider p/q, q > 1, 0 < p < q.
If the denominator, q, is relatively prime to both 2 and 5, then for some natural n and m, mq = 10^n-1. The decimal form of the fraction will then be mp repeated every n digits.
If the denominator, q, only has factors of 2 and 5, then it terminates.
Else, the fraction can be split into 2 fractions(p/q=a/b+c/d), where b is relatively prime to both 2 and 5, and d has only factors of 2 and 5.
Answer: The fractional part $.25$ is the same for all of the following:
$$
1/4, \quad 5/4, \quad 9/4, \quad 13/4, \quad 17/4, \quad 21/4, \quad 25/4 \ldots
$$
In this sense, it can indeed be predicted:
$\quad$The fractional part of $x$ is the same as the fractional part of $x-1$.
$\quad$The fractional part of $x$ is the same as the fractional part of $x-[x]$.
(Here $[x]$ denotes the integer part of $x$, i.e. the greatest integer not exceeding $x$.)
In some cases we can predict that the fractional part is precisely zero using divisibility rules; for example, the fractional part of 90108/9 is zero because 90108 is divisible by 9.
Also, we can predict with 100% certainty that for any ordinary fraction, the decimal will either be repeating or terminate.
In addition, we can always predict the result when the denominator of your fraction is 10, 100, 1000, etc. For example:
$$
12345/10 = 1234.5, \qquad 12345/100 = 123.45, \qquad 12345/1000 = 12.345 \ldots
$$
Otherwise, you just have to actually divide it and see the result.
Comment: If the denominator of the fraction (reduced to simplest terms) has only powers of 2 and 5 then the decimal form will be terminate.  Otherwise It will be a repeating decimal.  Further, the number of digits in the repeating part will be smaller than the denominator.


Cable modem and wireless routers set up

Question: I have cable internet, 60meg. I have just bought a Asus rt-ac56r wireless router. I also have a Linksys wrt320N. I tried to get the Linksys to be a repeater by downloading and installing dd-wrt but could never seem to get it to work right. I have purchased a long ethernet cable and have run it from my Asus router to the Linsys and set up the Linsys to use in my bedroom. I can only seem to get 30meg on my iPad and PC that are using wireless on either router. What is the best setup for this equipment? Should i get a Ethernet splitter and connect both straight to the modem or is the way i have it ok? Want to get as much speed as possible.
Comment: Is your internet speed 60Mb/s measured or is that what is advertised by the provider?
Comment: Advertised and measured 60Mb/s PLUS!!!!
Answer: I'd suggest linking the Asus Router to the Cable Modem with that long Ethernet cable you have and place it in your room, leaving the Linksys out altogether.
Then, the best speed you'll be able to get to any of your devices would be through Ethernet, plugged into the Asus.
Speed and throughput over WIFI would be dependant on too many factors, but in any case, it won't beat the Ethernet cable.
Edit to reflect comment from OP:
You can't connect both Routers to the Modem with an "Ethernet Splitter", it doesn't work like that. You can use a switch, which depends on the lay out of your premises, may make it easier. Or you can simply connect the Linksys to the Asus, configure it to be an AP, place it near to any devices that can use cable, to take advantage of the Ethernet ports. Perhaps use a different SSID to that of the Aus's, and manually connect to the nearest AP as you move around with your mobile devices.
This should achieve your goal for best possible speed, out of what you already have.
Comment: I know it wont beat the ethernet cable but I need the extra wireless for all devices so i would like to use both wireless routers.
Comment: But by adding a switch to the modem and connecting the Linksys router to it, you will be using more than one IP address (especially if it's configured as an AP), which will be in violation of the OP's TOS. Most home subscriptions are only eligible for one IP address (either public or private) from the ISP's IP address pool.
Answer: If you're set on having two WAP's in your house then there are a few things to consider.

Network congestion:

Position your access points away from obstructions and other electronic devices.
Ensure there are no other wifi signals on your channel. Otherwise, you could be experiencing a high amount of collisions. When a collision occurs the devices must retransmit the packet which is very costly on network throughput. Use a wifi analysis tool like the free version of Acrylic to find other wifi signals in your area.
Try to use the 5GHz band if your devices support it. In my area anyhow there are far fewer 5GHz devices than there are 2.4GHz.
If you're not competing with other signals increase the channel width to 40MHz. This will increase the bandwidth.

Power

Adjust your antenna's power to suit the environment. DD-WRT specifically has a lot of adjustment capabilities for your antennas. You may find that increasing the TX power will give you better results. Be careful not to push it too far though, you'll need to consider heat dissipation. Also, often reducing the power can increase your speeds, have a play around with it.

Configuration

You will want to setup one of your WAP's in Bridge mode. This basically joins one WAP to the other alleviating the need for your devices to jump between the different wifi networks. It can also reduce the cost of control type transmissions like authentication.
You can prefer to have two separate networks if you like but keep in mind all the extra overhead of having two default gateways both performing routing functions that could be done by just one of them.

If I was you I would do some testing to see exactly what's happening. Wireshark is a great tool to analyse your network. You'll be able to see if there are excessive retransmissions and it may give you a better idea of what the problem is. GLHF

Cable modem and wireless routers set up


How would you explain the general trend in melting point between Group 1 and Group 2 chlorides?

Question: This question is based off of the attached chart. I notice how there's a general increase in melting point down group 2 chlorides and a general decrease (except for Lithium) down the group 1 chlorides. So, is the melting points of group 2 chlorides greater generally, or vice versa? Also, how would you explain these trends using electronegativity and intermolecular forces?
Comment: Does this answer your question? [How do we predict the order of melting points of group 1 and group 2 halides?](https://chemistry.stackexchange.com/questions/27879/how-do-we-predict-the-order-of-melting-points-of-group-1-and-group-2-halides) just check and let us know. If your question is different, please edit it.
Comment: Not exactly, I want to know specifically why group 2 halide trend is increasing where group 1 decreases, and if you could say generally which group has higher melting points
Answer: First off, we need to give magnesium chloride its due. Wikipedia reports a melting point of 714°C for the anhydrous salt (the lower figures are for hydrates losing water molecules); apparently the 415°C figure was wrongly recopied. Putting the magnedium chloride point at the correct position makes the trends much more consistent with a sharp change in both groups from  Period 1 to Period 2.
If the chlorides were 100% ionic, we should expect melting points to drop off with heavier metals as electrostatically driven lattice energies drop off with larger atoms. But only alkali metal chlorides from sodium onwards follow this model. For the rest, we should expect less than 100% ionic character to favor easier formation of small molecules/clusters in the liquid phase, leading to a lower melting point.
Fajans' Rules are relevant here. They favor more covalent character if the cation in an ionic model is small or multiply charged. "Small" implies that lithium chloride might have significant covalent character where the chlorides with larger alkali metal atoms would not, while "multiply charged" implies the Group 2 chlorides would have more covalent character (and thus would melt more easily with molecules/clusters) overall. Beryllium, with both small size and multiple charge in the ionic model, gives an especially low-melting chloride.
The covalency in the Group 2 case never appears to reach zero but does decrease with larger ions, so heavier Group 2 chlorides tend to melt at higher temperatures; whereas with the singly charged alkali metal ions only the lightest/smallest-atom one deviates from the trend expected from an ionic model.
With lithium chloride melting at a temperature closer to (corrected) magnesium chloride than to any other on the list aside from the much heavier caesium compound, we see yet another example of lithium resembling alkaline earth metals (apart from valency/oxidation state), especially magnesium, more than other alkali metals.
The reader should contrast the Group 2 behavior shown here with the corresponding results for oxides. Oxides, with a more electronegative nonmetal and a more compact anion, are more highly ionic than chlorides, and the decreasing trend we expect to see from the ionic model sets in for that case from magnesium onwards. See here.


Question of declaration in Texas Hold em

Question: If a player whom is considering a call says, "I want to call..." ambiguously as he is deliberating, is this considered a binding declaration in Hold em?
Answer: This kind of thing is irritating at best and angle shooting at worst.
I suspect whether this is considered a binding action will depend on lots of context, such as the frequency he's been doing this and the intonation of his voice and the level of confusion of the other players in the hand after one of these incidents. Not all "I want to call" statements are created equal. For example, if he's tanking after a big bet into him, and he slowly says "I want to call" as part of some lengthy deliberations, I think you might reasonably assume that this doesn't qualify as stating an action. But if he's just casually saying "I wanna call" when any action's on him and he's thrown in chips after each statement, it might reasonably be interpreted as a call.
The real problem exists if this player is using this ambiguity to gain some sort of advantage, as in trying to discern the actions of players behind him while still trying to give himself a way out of a binding action. A good dealer and floorperson would help put a stop to any such shenanigans. As another player at the table here, you should be careful that you understand the real action in front of you before you act.
So, I think my answer here is a big "it depends." But the fact that you're even asking this question implies that this player was using this as a strategic tactic, which in my mind put it in the angle shooting category. When in doubt, clarify the action with the dealer, and if the player keeps it up, you might discuss it with the floorman.
Answer: If I was called over to the table to make a decision on this it could go either way. If the players were head up, they guy says I want to call ambiguously, and this causes the other player to turn up his cards, I would rule it a call. If their were other players in the hand and the next player to act after hearing the comment says I call to, I would tend to rule it a call.
If it was just ambiguous but did not cause anybody to act in some way, no harm was done I would rule it not a call most of the time. I might rule it a call depending on the factors of the player, like Chris said, how often, how ambiguous, the inflection in the voice etc.
As a player I don't mind a player pulling this kind of angle to much. I am seasoned player, and they are going to have more downside then up. Most of these moves are weak, and the weaker the player is the better the moves work against them. Not to say they do not get irritating, and old.


in Budhism, what would happen with a soul in a eventually case of succesfull cryogenic reanimation?

Question: there is some text that speaks something related about, what would happen with a soul in a eventually case of succesfull cryogenic resucitation? i know this is a very new theme, but, if reincarnation is possible, what happens when that last body reach a reanimation?
Answer: This is not possible. Once you die the stream of consciousness arises in another body. This is another person. You cannot bring back this stream of consciousness to the same / previous body.`
Once the consciousness has left the body regardless of it being frozen or not the body is dead. The body cannot be made alive again. In believe in the process of freezing the body would at some point die as heat is needed to sustain life.
Also in Buddhism there is concept of no soul. So what question like what happens to the soul is irrelevant. I believe once procedure is done you are simply dead and just a lump of matter devoid of any life faculty.
There are stories of dead getting reanimated, but this is due to other entities repossessing the body and not that the person has become alive again. This appears in the Vinaya, exact location I forget, but there is a rule not to take clothes from a fresh dead body for robes until some lapse of time, if I remember right. See note on Vetala-panka-vimsati BRAHMA-GALA SUTTA and p252, THE IMPERIAL GAZETTEER OF INDIA: THE INDIAN EMPIRE VOL II HISTORICAL.
In case of these possessions the body does not live again as the entity which walks in does have a own subtle body but is controlling the dead body. For this to happen is very rare as you have to be a very powerful entity to do something like this. There are some less powerful entities which can do an illusion of being actually moving. That is why when you are meditating on a corpse it is advised to concentrate also on position of the corpse relative to objects around the corpse before starting so you can reassess that this is not the case such illusion comes to your mind.
Also following question might be of interest: Where in the Pali Canon are there Discussions on Reanimated Corpse?
Comment: Suminda, the asker's question is not very clear but I feel like we should assume that if a body is cryogenically frozen successfully, then it is not death.
Comment: See if this is now more incline from what can be inferred from the question?
Answer: If someone was somehow reanimated then I suppose it would be like there was never a death. The resurrected person would be born in the human realm yet again then again and again in every  moment because a being is always changing in every moment, moment by moment. The last moment person is dead and the new present moment person is reborn. The whole frozen death thing would have been just another moment.


How do DVD players know how fast to spin?

Question: Not sure if this is the right place to ask this, but here it goes: Different optical discs have different rotational speeds, based on the size of the pits and the bitrate. Newer disks have smaller pits, so the disks can rotate slower at the same bitrate, and even a single disk can have different speeds on different sections, usually going faster near the center. Now, how does that work? A regular disk drive can usually support multiple types of disks, it doesn't really matter wether you are dealing with a DVD, a CD, or whatever. Do these disks maybe have markers that dictate a set speed/acceleration at the very beginning? Does it simply calculate the size of the pits and adapt to that speed? or is it something completely different? Thanks in advance.
Comment: You could also ask how it knows it's capacity, whether it is R or RW and so on... While I don't know the answer, I would assume that there is some record containing the media information that is being read by the drive at the minimum safe speed, then it is adjusting accordingly.
Comment: so as soon as the laser can focus and read data, it should know wether its a CD, DVD,.. then it will search for an initial mark at an known position I guess... at the higher levels then a header of the format is present. And estimating the rotation speed while reading is easy. if you have a format demanding a constant bitrate, a pll controling the motor speed can measure the drift of the bitrate and compensate for it. If its a variable bitrate format, you simply control speed according to the max-speed of the disc and an error-rate threshold. Should not be to hard to implement...
Comment: Disks don't have one specific rotation speed, rather the device decides how fast it wants to read the disk.  If it wants to read faster (for example to buffer video for decode) it'll spin faster.  If it wants to read slower (for example, an audio only disk) it can choose to spin slower.  In the old days you used to pay more more for CD/DVD drives with higher maximum read rates with the fastest devices being quite loud because of how fast they would spin.
Answer: How is the disc playback rate determined?
CDs and DVDs use a recording technique called Constant Linear Velocity, or CLV. The pit-and-land data on the disc have a channel coding rate for the raw signal. The read back system locks onto that signal, recovers data and clock from it, and uses that in a control loop to maintain servo lock, including focus, tracking and the disc spin rate.
So ultimately it's the controller that sets the pace, and the disc follows along in servo lock. Audio CDs and movie DVDs are played back at their '1x' rate; while a CD- or DVD-ROM data drive is free to read them much faster (and spin them faster too, sounding much like a noisy fan in the process.)
How does the linear track carry a clock?
CD and DVD track pit-and-land coding is designed to have enough transitions in it to recover clock. This is assured by a number of methods, including data scrambling, and by balanced run-length controlled 1/0 encoding into larger data symbols. On CD, 14 channel bits are used per byte; on DVD it's 16 bits per byte.
The disc media pit-and-land dimensions and track pitch are carefully controlled to ensure compatibility with any player.
How does a drive identify the media?
DVD optics and electronics are a bit more complex than CD. Mainly, this is because DVD drives are required to be backwards compatible with CD, which uses different laser wavelength and numerical aperture (NA) than DVD:

CD: laser 780nm (near-infrared), NA = 0.45
DVD: laser 650nm (red), NA = 0.63

And for the sake of completeness:

Blu-Ray: Laser 150nm (blue), NA = 0.85

DVD optics thus have to support two optical paths: one for CD and one for DVD. This makes DVD pickups more complex, although economies of scale and mature technology ensure that the cost difference is tolerable.
Blu-Ray pickups are even more complex, with three optical paths.
And just because, a BD-drive pickup teardown: http://repairfaq.cis.upenn.edu/Misc/Blu-ray/site1/optics.html
When the disc is mounted in a DVD drive, the controller first starts at the middle to identify the basic media type: it tries each optical pickup path and attempts to achieve a valid read channel signal to choose the appropriate optics for disc.
With the optics locked on, the controller then gathers more information from the Table of Contents, or TOC, to determine the data format, and reports that back to the host.
Answer: You simply spin the disk at whatever speed is required to maintain your data stream. A CD output for example produces 1.4 Mb/s. You stream data from the disk into a buffer at a somewhat variable rate and from the buffer at a constant rate to the output. If your buffer is running low you speed up the disk, if its getting full you slow it down.
Comment: There's probably a phase-locked-loop in there keeping the data rate constant.
Comment: No need. The output is simply driven by the audio master clock that clocks both the DAC and the SPDIF output. The input buffer takes up some amount of slack, so the disk speed can vary and you can re-read sections that had bit errors during the first attempt.