How to actually deploy a live blockchain?

Question: I have been researching a lot about how to build a new blockchains. So far, all guides say very similar things:

Identify use case
Identify consensus method
Design nodes
design blockchain
select a blockchain platform
do it

But I am looking for more detailed information, like what do I actually need if I want to deploy a live blockchain. Let's say I choose EOSIO, or openchain, whatever blockchain platform. But then what do I actually need to do in order to deploy it once it's developed so that people can start using it? How do I make it go live? How do I deploy it into that network? Do I need a public server running a node? Can I run a node just from my computer? Since everything we hear is that it's "descentralized" it wouldn't make too much sense that one would need a server so that people can join, right? What data does the node need to share so that others can join the blockchain?
Also what if i don't choose any of those platforms like openchain, eosio, ethereum, etc. Let's say i want to build a new blockchain in a new network from scratch, like bitcoin or ethereum. What would the requirements be? Does it cost any money?
These are the kind of things I want to know but can't seem to find anywhere.
If there is any blockchain developing expert with some spare time, I would really appreciate it. Thank you
Answer: You seem to be mixing up different things.  Since this is is the EOSIO stack exchange, I will answer for EOSIO:
If you want to deploy an entirely new blockchain using eosio software, then you need to setup server infrastructure to host it.
If you want to deploy a contract to an existing blockchain (such as EOS Mainnet, which is using the eosio software) then you can use the publicly available nodes to deploy your contract. You can also deploy your own node (running the eosio software) and connect it to the EOS Mainnet.


Are there additional benefits to fielding max-rank soldiers?

Question: One of my soldiers is rapidly approaching Colonel, the highest rank. Once he gets there, is there any incentive to keep bringing him for missions, apart from the obvious advantage of having a strong soldier on the field, i.e. will the additional kills advance him in any way?
Answer: As far as I know, once your soldiers reach Colonel rank, they no longer advance (unless you count Psi training as advancement, but as far as I know, that has nothing to do with killing aliens in the field). The only advantage of having elite soldiers is the fact that you have elite soldiers on the field - more powerful, higher HP and more perks.
Comment: There is also an achievement related to having 1 soldier surviving all the missions.
Comment: @IlyaMelamed does that mean that they must _engage_ in all missions or merely that they should survive till the end of the game?
Comment: @Alex, they must **engage** in all missions and survive.
Answer: They will not improve in any way unless they are Psi-gifted and haven't yet advanced down that path (then you will have to field them, to improve them - not via kills, but by using their abilities).
Still, if you play on Classic difficulty, you will rarely have the luxury of leaving your best soldiers at home. You'll need them. Desperatly. Missions keep surprising you with tougher enemies (or even just unexpectly bad RNG-numbers). Good squad members can be the difference between one or two deaths and a wipe.
Usually the downtime from injuries will be (more than) enough to train other rookies. The game sort of forces a rotation on you anyway.


How is distance between sun and earth calculated?

Question: How has the distance between sun and earth been calculated?
Also what is the size of the sun?
Comment: Do you want to know both how the Earth-sun distance is measured and how the speed of light is measured?  Those are completely different things.  As I asked before, separate threads, please.
Comment: Yes, I agree. Just that the single curiosity that how man has acquired knowledge of solar system has forked so many curiosities. And by putting them in separate threads, I didn't want to lose the context. More importantly, in 20 minutes, I can only post one question. So it will take me full day today to address all the queries separately and put them together to make holistic understanding, but yes point taken. I am isolating my questions in different threads as soon as I am permitted to start a new thread, time wise.
Comment: Okay.  I edited out the part about the speed of light because it is off-topic in this question.
Answer: The most precise measures of this distance are from radars in the 1960s. However, the distance has been known, though roughly, since the Ancient Times.
Aristarchus of Samos (310BC - 230BC) used the angle between the Earth-Moon axis and the Earth-Sun when the Moon is in First Quarter (elongation of the Moon, $E$ ) and then, with simple trigonometry, could deduce the distances:
$$ \cos E = \frac {distance (\text{Earth-Moon})} {distance(\text{Earth-Sun})} $$
Since he had already computed the Earth-Moon distance from the duration of lunar eclipses, he could conclude on the Earth-Sun distance. His results were false, because of too loose measure of the angle, but his method was very accurate.
See Wikipedia for more details.
Another method was explored in 1672 by Cassini and Richer: they measured the parallax (i.e. the variation in angle when seen from different places) under which Mars was seen in Cayenne and Paris, at the moment of opposition. From this, they deduced the distance Earth-Mars.
Then, using the Kepler law
$$\frac{a^3}{p^2}= constant$$
(where $a$ is the distance between the planet and the Sun, and $p$ the sideral time)
they could figure out what was the distance to the Sun.
Comment: This is a great answer, but it neglects the Venus transit method, which is basically the same thing as Mars but looking the other way. Trying to observe rare transits of Venus led to some incredible journeys around the world and makes for great stories about how far astronomers will go to get their data.
Answer: Another way of calculating the earth - sun distance is to look at the centrifugal and the gravitational force. This solution assumes that one already knows the mass of the sun, but thats a different problem ;-). One does only need High-School Math and Physics in order to derive a solution.
Thanks to Newton we know
$F_g = -G\frac{Mm}{r^2}$
where $G=6,674\quad10^{-11}$ is the gravitational constant. We also know the centrifugal force to be
$F_z =  \frac{mv^2}{r}$
Putting these two equations together one gets:
$\frac{mv^2}{r} = G \frac{Mm}{r^2}$
$\Rightarrow r =  \frac{GM}{v^2}$
Furthermore we know the duration of a year and therefore we know $v$:
$v = \omega r = 2 \pi f r = \frac{2 \pi r}{T}$
Consequently
$r = \frac{GMT^2}{4 \pi^2 r^2} \Rightarrow r = \sqrt[3]{\frac{GMT^2}{4 \pi^2}} = 149,8 \quad 10^9 \; m$
Which is very close to the real value, which is varying between 147,1 Mio. and 152,1 Mio. km. According to Wikipedia the average distance is 149,6 Mio. km, so our result is actually quite good.
Comment: Do you mean centripetal force?


How does an atom decide what to bond to?

Question: Given for example, a Hydrogen atom (that can only have 1 bond max), that is sitting on a bonder next to two other atoms that are also on bonders and have available bonding slots, how is it determined which atom the Hydrogen ends up bonding to?  I've tried experimenting with it being a priority order of directions, but that doesn't seem to be correct.
Answer: I'm afraid there's no real science behind this. Each bonder has a priority number likely based on the internal data structure (array, linked list, whatever). The game just processes this list in a first through last order. So the bonding order in game is just how the bonders happen to be arranged. I would be nice if they were labeled, but as far as I know there's no way to get their priority in game. So with two bonders linked, it just goes through the list sequentially. For example, with 4 bonders the list of checked links would look something like this:

(1, 2)
(1, 3)
(1, 4)
(2, 3)
(2, 4)
(3, 4)

Which is simply an ordered list of the pairs of priority numbers. With that knowledge, you just need to know which bonder has which priority. Though experimentation I've found that:
For 4 bonders in a fresh level the bonder priority numbers are:
1 - 2
| . |
4 - 3

For a fresh level with 8 bonders the bonder priority numbers are:
1 - 2
| . |
4 - 3
| . |
5 - 6
| . |
8 - 7

Comment: In a game where one does not need to concern themselves with stereochemistry, induction effects, kinetics, or thermodynamics, science doesn't play much of a role.
Comment: An option was added to the options menu that allows you to "Show Bonder Priority" which adds numbers to the bonders and makes this much easier to keep track of now.
Comment: No real science needed, but this looks promising as a definitive answer.  I'll try it out tonight to verify that it matches my experience.
Comment: Great answer, there really should be an in-game way to determine this apart from remembering the original configuration.
Comment: What if the eight bonders are initially in a horizontal layout?
Comment: @ZeroOne Sorry, I haven't seen that configuration, I didn't ever finish the game :/ Let us know what you find out.
Answer: I don't think that there is a strict order of bonder preference; it is based on something else.
I did the following experiment: three bonders in a horizontal row, with an unbonded H at each one. I recorded the results of the six combinations of bonders.
#    order    result    side
a    1 2 3    H-H H      L
b    3 2 1    H-H H      L
c    2 3 1    H H-H      R
d    1 3 2    H-H H      L
e    3 1 2    H H-H      R
f    2 1 3    H H-H      R

I reran this many times, and there was always a 50% split with one group preferring 'L' bonding, one group preferring 'R' bonding, and one group being split. (Each group is identified by the bonder in the middle.)
So there is evidently some structure to the evaluation, but each of these six cases is contradicted by another one (if we assume the bonders evaluate in order); i.e., if (a) is the correct order, then why in (b) does the bonder prefer (2,3) over (1,2)? This is true for every case.
The mouse wheel trick seems to be doing something (it made a click and I assume processed an action), but I could not change the output of (a).
Comment: If your bonders 1, 2 and 3 have priorities 1, 3 and 2, respectively, then Timbo's answer would explain this.
Answer: I used a few hours testing. The result is that there is a hidden priority, but the bonder only checks the bonder right and under it. That means if there are 4 bonders build up the following way:
2 3 1
4

First bonder 1 will check below and right of it and can't find anything to bond.
The second step is that bonder 2 will bond with bonder 3 and 4.
Afterwards bonder 3 bonds with 1 and last bonder 4 doesn't find anything to bond.
If a bonder has two possible connections (right and under it), then it will bond first with the bonder with the higher priority (lower number).
That means the second step bonder 2 will first bond with 3 and then with 4.
Answer: Some of the other answers are wrong, and it bugs me that the rest is split over multiple posts, so based on thorough research here is my definitive list of how bonding works:

Each bonder has an invisible priority number, 1-8. This number is determined by its initial position in the reactor.
Bonding proceeds with an outer loop across all the bonders and an inner loop across all the bonders. So if there are 4 bonders, the order it checks is (1,1), (1,2), (1,3), (1,4), (2,1), (2,2), (2,3), (2,4), (3,1), (3,2), (3,3), (3,4), (4,1), (4,2), (4,3), (4,4).
However, bonding only occurs if the second bonder is directly below or to the right of the first bonder. (Credit to Timbo for figuring this out.) This is why the list above doesn't lead to double-bonding: If the layout is 2-1, then the entry (1,2) in the list above doesn't trigger, since 2 (the second bonder) is to the right of the first bonder.
Bonder order is not affected by mouse-wheeling on the bonders, at least not in the latest version. Yes, it makes a clicking sound, but I could not get the bond order to change on my test molecule of BO2 no matter how I mouse-wheeled. This is supported by the fact that mouse-wheel clicks are not saved in the undo history like everything else.
The bonder order is different for different reactor types:

Standard/Sensor
1 3
2 4

Assembly/Fusion/Nuclear/Research (4)
1 2
3 4

Research (2) (Presumably, can't verify)
1 2

Research (8)
1 2
3 4
5 6
7 8

Super-Bonder/Sandbox
5 1 3 7
6 2 4 8

Yes, there are two different layouts for the "standard" 2x2 square, and yes, the Super-Bonder really is that weird.

Lastly, here are some practical tips for how to debug a balky design:
Let's say you're trying to make the molecule HP≡CO, in a bent configuration. You make a 2x2 square of bonders and your input looks like this:
P=C

H O

But when you bond it, it comes out like this:
P≡C
  |
H-O

Given how bonding works, we can deduce that the lower-left bonder is running before the upper-left bonder, so swapping them will fix it. The relative priority of the other two bonders doesn't matter.
However, if your failing output was rotated like this instead:
H P
| ⦀
O-C

In this case, the lower-left is before the upper-right, and you have to swap diagonally. The priority of the two bonders on the opposite diagonal doesn't matter. The difference in the two examples comes from the rule that bonds only form to the down and right.
Comment: Thanks for this late but more comprehensive answer, it should be the accepted one. Just a comment -- your steps (2) and (3) can be restated as there is a single loop over all bonders in priority order. For each bonder, it tries to bond with the square to the right and the square below. If there is a bonder in both squares, priority breaks the tie.
Comment: Thanks! Yes, I agree that what you said is equivalent, and it's definitely shorter to say it that way. I laid out the algorithm the way I did because that's the way that's most precise and makes the most sense to me; in particular, I suspect that's how it's actually coded in the game.
Comment: Actually while I agree your way is clearer to understand, I would *hope* that the way it's actually coded in the game is what I said. A double loop over all pairs of reactors would be inefficient, an O(n^2) algorithm, whereas mine is O(n). A trained programmer should know not to do a double loop when a single loop would work.
Comment: This is getting into the weeds, but... It's possible that there isn't a convenient map from position to bonder object. It really depends on how the internals of the game work, which obviously we can only speculate at. What I do know is that an O(N^2) algorithm is no big deal when N has a maximum value of 8.
Comment: Largely agree, but 64 vs 8 is definitely a big deal since this is in the critical section of the code. Could be that that level of optimization was necessary to get the highest speed setting to go as incredibly fast as it does. Anyway let's exit the weeds ;)
Answer: The bonders each have a "hidden priority", where they will try the other bonders in a certain order. If it's not doing what you want, you can swap the bonder's position around.
Sorry, can't find a good source to cite for this.
Comment: @lilserf As far as I can tell, the logic is something like "Determine what the user is trying to do, then come up with a way to do something else". :p
Comment: There appears to be some kind of hidden order, but it's WEIRD.  I laid out C, H, C on 3 bonders horizontally and tried bonding.  For the 6 combinations of those 3 bonders, half bonded C-H C and half bonded C H-C, but any pattern is not clear at all.
Answer: Apparently, if you 'mouse over' the bonder and roll the mouse wheel up/down, the bond order of the bonders changes. It worked for me.


how to display hierarchy in search bar

Question: So we have an internal form creating tool and the names of data here are very repetitive. We decided to bring in a search bar and show the data along with the hierarchy( basically the path of the file). Any clue on how to display it efficiently in a user friendly and pleasing manner?
Comment: Can you show the work you've done so far? It's kind of hard to give any "best practices" if we don't know where to start.
Comment: hello. I've edited the document by adding pictures. the first picture is how it looks when the keywords are entered. there is a small button on the right side of 4th result. On hovering the button the result will expand to a detail explanation of the hierarchy( like the one shown in second picture). Is there any other effective way to do it? thanks in advance.
Comment: I think you are looking for what is called autocomplete.  What environment are you in?
Answer: You might want to look at the answers to this question, and this one.
If you visit https://www.amazon.co.uk/ and type 'truck' in their search bar you get two sections, the first for your query in all the different departments, and the second for wild-card searches. Perhaps this may inform your decision.
A project I'm working on at the moment deals with deep-hierarchy searches like yours. At present I'm looking at providing a quick high-level view of where each node in the results sits:

Clicking on the result will expand a separate hierarchy view, so you can see nodes on the same branch and make your selection from there. I find this better than trying to cram too much into the search results, and am hoping that searching and viewing the expanded hierarchy will be smooth enough to mitigate the lack of definition in the search result description and potential confusion that could result.
That being said, my data may not have as much similarity as yours does, so this may not work in your case. Also I should say that this is quite early-stage and I haven't put it in front of any users yet (that's next week!).
Hope that helps :)
Comment: Thank you for the reply. This did give me a different perspective to approach the problem I am facing :).

how to display hierarchy in search bar


home buying and moving in coronavirus in USA

Question: For UK, I see news that may help people there. Is there any guidelines of USA.
I know a few people in Maryland, Virginia who have closing coming in First week of April, should they close, what about moving to new home?
Comment: should one consider https://money.stackexchange.com/questions/123306/ for credit to offset the moving cost in troubled time ?
Answer: The National Association of Realtors has published guidance for its members at https://www.nar.realtor/coronavirus.  That's probably the nearest you'll get to a set of national guidelines. The rules and regulation for real estate transactions vary from state to state
Answer: Please refer to my answer in the question you referenced for some background on my personal choice. Now on to your questions:
The first is should they close?
My first question would be how much earnest money do they stand to lose? I would assume this is somewhere between $500 and $5,000 depending on whether its based on a flat rate of $500-$1,500 or 1% of the value of the house (i'm assuming $500k).
My second question would be how much do they love the house in question? If we're talking dream home that would be a perfect fit for their family, everything they ever wanted, OMG! its the only one in a 10 mile radius with window seats!, then emotional factors might very well override any logical reasoning we try to carry out here.
My third question would be where do they live now? Do they have the flexibility to either stay in the home they currently own, or extend their lease?
My fourth question would be how secure are they in their source of income as we enter a period when layoffs and unemployment are likely to increase across all industries and skill levels?
Depending on the answers to the above, I would suggest they have a few options:

Ask the seller to delay the closing until they're able to actually move into the home. This is far from standard real-estate contract language, but nothing is standard right now. You could enter a deal in principle with a firm close date/period of the two weeks following the relevant states lifting current shutdown/lockdown/closure of non-essential businesses
Try to renegotiate price to reflect the added risk of the current time and foreseeable future. In the question I linked to you'd see i indicated that this might be a futile exercise both in terms of pricing any asset accurately in this environment, and in terms of getting anywhere with a seller that might think you're trying to take advantage of them in a difficult time (you aren't, just managing your own financial risk)
Pull out of the deal completely and wait until you have a clearer view of the impact of this virus to the real estate market, your health, finances and future. Per the linked question above, this is my personal choice. However, I don't have all the details, and this might not be a fit depending on their unique circumstances.
A softer take on option 2 above would be to go for complete transparency and express your concerns to the seller to see if there's a middle ground that satisfies both parties. I don't think there should be any shame in re-opening discussions in this environment. It is happening in every industry and country you can imagine.

Your second question was should they move to the new home?
I would argue that the question should be "Can they move?". Looking at this Axios article from a few days ago, it appears that COVID-19 peak is expected in MD/VA sometime in early May. Maybe its the skeptic in me, or the fact that every timeline we've all been given so far has been wrong, but I'd bet that its later. Now, given that states have shutdown as we rose to the peak, its reasonable to expect that will remain the same as we descend from it...at least to prevent a quick second wave. This means the peak is just the beginning of when we can start to hope things get back to normal.
So if MD & VA are effectively out of business for all of April, most of May, and potentially part of June one would be unable to move until the latter half of May or early June (at the earliest). This has obvious implications for your first question and should provide one framework within which you can make a decision. If you close and can't move in for a month or two, how much does that cost, and how does that compare to your earnest money + whatever other costs are already sunk into the prospective purchase (e.g. home inspection costs etc)
I know i've probably left you with more questions than answers. The goal here is to guide you to a decision making framework as there isn't one answer for everyone. I hope this helps.


How to make continuous movement of a windshield wiper with Logic Bricks?

Question: With Logic Bricks, how can I make my windshield wiper move back and forth until I turn it off? When I press a key to start the movement, I can move left to right, but not right to left without a different key press.
I'm in Blender 2.79, Blender Game Engine.
Answer: Animate your windshield wipers moving back and forth, just one complete cycle.

Then add an Action actuator to play it. Make sure to set the EndFrame to match the length of your animation. Change the Action Playback Type from Play to Loop Stop, and choose the name of the action (by default is is named "Object name + Action")
The trick here is you want the keyboard sensor triggering a Property actuator, not the action actuator, then the property will trigger the action.
Add a game property, a boolean named "wipers" leave it false by default (blank checkbox).
Connect the keyboard sensor to a Property actuator, set it's mode to Toggle and choose the property you just created (wipers).
Now back on the sensor side, add a Property sensor, connect it to the action actuator. Set like so: Evaluation Type to be Equal, Property to be your game property (wipers), and the Value to be "True".
The full logic setup looks like this.
Comment: I would actually use "loop end" to get them to finish the cycle that they were in rather than stopping in the middle of the windshield. Both are actually valid, because my dad's car is the "loop stop" and it's really annoying, so I'd go with "loop end" just a suggestion
Comment: I'll try this as soon as I get a better graphics card, crashing problems exist right now. Thanks for your help!
Answer: Thanks to Monster for answering my question on multiple pivot points on one object, I was able to use the information to make a mechanically correct operation of windshield wipers. The length of the connecting rod between the motor and the wiper bottom pivot point are important, as is the vertical (Z) position of the motor. If you have problems with the wiper going 360 degrees, move the pivot point on the motor arm closer to the motor shaft (shorten the throw).

It doesn't matter which direction the motor turns, the effect is the same. This isn't a continuous operation, you need to continue a keypress for operation. The time of operation depends on the amount of motor revolutions in degrees.


Derivation on torque required to raise load, isn't there a force disapearing?

Question: I drew a little picture:

Try to use you imagination to view these pictures as profile of bolt threads. In derivation on torque required to raise load, the angle of friction (ro) is calculated like in the left picture. Then it says that due to the angle of the 'tooth', friction-angle is adjusted, like in the right picture.
The problem is, a force F2 should exist, but they just ignore it. Why?
I opend Karl-Heinz Deckers famous Maschinenelemente book, where it shows how to get the formula required to raise load, and also at one point it just ignores the F2 force (at least it acknowledges its existence in one of the earlier steps of the derivation, but then just ignores it).
Can someone explain this?
Answer: From the diagram of the tooth profile in the upper right corner I see FN, FN' and F2. FN coincides with the lifting force so it is relevant to the solution. FN' results in friction which will affect the required torque. But F2 does not affect either. F2 might be viewed as a component of FN', but it doesn't uniquely affect the solution in a way not already accounted for by FN'.
If F2 resulted from, for example, the bolt being too large for the hole, then it, F2, would be a unique force not derived from FN or FN', and would need to be factored in. But that is not the case here.


Does prayer timetable stay same every year?

Question: Through out the year prayer times change, but is it the same every year? For example on the 04/07/15 morning prayer is xx:xx, does it stay the same next year at same day and month (04/07/16)? Given that the location is the same.
Comment: Yes, if you have the timetables for the entire year, then they shouldn't change much for the next year, unless the earth shifted.
Comment: Depends wherr on the earth you're located and how many seasons occur.
Comment: Given that the location is exactly the same, let's say somewhere in London, or New York, the same exact location, would the prayer time table remain the same every year?
Answer: The earth takes just under 365.25 days to orbit the sun; the calendar year was designed to approximate this, but typically only has 365 days (because whole numbers). As such, every solar year, the calendar year will get slightly more and more out of sync: This is why leap years were invented.
Since each calendar day actually occurs about six hours earlier in the solar cycle every year, the prayer times which are tied to direct solar position will also be slightly out-of-sync from year to year; these will mostly (but still not perfectly) sync up every four years, after a leap day has re-synced the calendar.
(It'll sync up even better every four hundred years because of how the current Gregorian leap cycle is calculated.)
The actual difference would be negligible in most cases; unlikely to exceed a minute either way. But no, you can't expect them to match up perfectly between years.
Answer: There is a very small change in the prayer times, even when following the solar calendar. Look at the following prayer timetables for different years.
July 2010
July 2020
Answer: They don't change, unless you are using a lunar calendar instead.Perhaps that is why are you asking this question.
For instance Ramadhan this year started on June 17th, next year it will start 11 days earlier and therefore the timetable for the first day of Ramadhan will change while if Ramadhan was following a solar system then it would have been the same every solar year.


Get multiple list-items by URL with one Rest API request

Question: I have an array with multiple SP list-item URLs.
I want to get all data from every list-item .
I solved the problem with a loop that does a Rest API Request for every single URL.
I don't like this solution, because it is really slow to do a request for every item.
Is it possible to get this done with a single Rest API request to get a better performance?
Comment: why you are not just sending that whole array of list Url to server side instead of for loop for every item in an array and get items from the lists in an array from the backside.in this way, you can get all items in one shot!
Answer: What @Ekta said:
Just get all the items, then in javascript you can parse out the array by id. Let's say you had an array of all the listitem ids named itemidarray such as
var itemidarray = [1, 5, 10, 4, 29, 40];
$.ajax({
    url: url + "/_api/web/lists/getbytitle('" + listname + "')/items",
    method: "GET",
    headers: { "Accept": "application/json; odata=verbose" },
    success: function (data) {
        if(itemidarray.indexOf(data.d.results.Id) > -1){
            //do something with the item
        }
    },
    error: function (data) {
        alert(data);
    }
});

This is pseudocode, so I haven't tested the syntax of it.
Comment: this does only work if all items are within the same List,  right ?
Comment: correct. If you wanted multiple items from multiple lists, you may have to look into CSOM batching.
Answer: There are 2 different approaches to read items from List using REST API

If you are reading items from single list with different conditions matches your results then use filter operation as shown in below example
$filter=(Id eq 1) or startsWith(Title,'abc') or (Department eq 'XYZ')
If you want to read items from multiple lists or in any scenario to read items from single lists then you can also use batch operation. You can refer the sample below link
https://msdn.microsoft.com/en-us/library/office/dn903506.aspx
Answer: Combining filters: $filter=((ZIP eq 12345) or (ZIP eq 22222))
you can make huge strings of combined filters I experienced


Why is centrifugal force considered fictitious, when it's the one that feels real to us when we are moving in a circle?

Question: Why is centrifugal force considered fictitious, when it's the one that feels real to us when we are moving in a circle? I understand the explanation regarding the reference frames: if our body is the reference frame, and it is rotating, a fictitious centrifugal force needs to be made up to cancel the centripetal force and explain why we appear to be stationary in relation to ourselves. However, if we are in a round up ride, we feel pushed againsts the wall, rather than towards the center of the ride, which would be the centripetal force. So if centrifugal force is the fictitious one, wouldn't that be like saying that the force we feel pushing us againsts the wall is only in our minds?
Comment: Obligatory https://xkcd.com/123/
Comment: You may not have given enough thought to what "we feel pushed" means. If you think through some more day-to-day instances you'll find that there are two different ways to interpret that phrase. One that is consistent with the way we talk about gravity and one consistent with the way we talk about *everything else*. There is a reason for this. How does it feel if a friend pushes you unexpectedly from behind; how does your body move and how do your loose limbs move. What happens if a sliding sidewalk stops unexpectedly (noting that it is only in contact with your feet).
Answer: Suppose you're in a fast car and you stomp on the accelerator. You feel pressed into the back of the seat. In which direction are you accelerating? Forward, obviously, but you feel a force pushing against your back. Now you turn a corner. Your seatbelt, and maybe the door next to you, press against your side. In which direction are you accelerating? In this case, it's not so obvious, but it's inward, not outward.  There is no centrifugal force here.
In both cases, the side on which you feel the force is opposite the direction in which you are accelerating.
Comment: So which force makes us feel pushed againsts the wall in a round up ride? From your answer I understand it is not the centrifugal force. Is that a third law partner to the centripetal force? I'm a little confused about how to name the force acting on our body outward from the the center of rotation, which feels like a real force.
Comment: @Auggie -- There is no outward force acting on you. The forces acting on you are gravitation, which pulls you downward, friction against the wall, which opposes gravity, and the normal force by the wall, which pushes you toward the center of the ride.
Comment: So, just to clarify, when you say there is no outward force acting on us, you mean in an inertial reference frame, right? what about in a rotating reference frame? And then, going back to my original question, is the force that we feel in the outward direction only an illusion? I'm a beginner and an amature, so I appreciate your patience :)
Comment: @Auggie - What you feel in that ride is very similar to what you feel in an accelerating car when the back seat of the car pushes on you. I hope you agree that there's no force pushing you into the seat. It's the same for the ride. That push by the wall of the ride (or by the back of the car seat) against you propagates through your body. Your body is a mix of bones, cartilage, and soft tissues that respond differently to that push. You feel that push internally as well as externally.
Answer: The reason we feel that we are pushed outwards is due to inertia. Inertia is the resistance to movement. It is measured by mass. When we have more movement, it makes it harder to get us moving.
In a car that is on a curve, for example, our inertia makes us want to keep on going forward. Going forward in this case would make us feel that we are being pushed outwards. Therefore, there is no centrifugal force in this case.
Sometimes, the centrifugal force is referred to the reaction force of the centripetal force. But usually, it refers to a ficitious force used to simplify thr mathematics.
Comment: Why the downvote? How can I improve my answer?
Comment: The idea of centrifugal force is not used to simplify the mathematics.  Just the opposite.  Centrifugal force exists only in the rotating frame of reference, and it _complicates_ things.  Much simpler to analyze the situation from a non-rotating frame of reference, where it becomes obvious that a body can not revolve around the center unless a _real_ force continually pushes it _toward_ the center.
Answer: We say fictitious because the actual source of the centrifugal acceleration is somewhat indirect and the experience one has results from the unbalanced forces acting on the reference frame, not a force.  Note, it is an acceleration not a force.
For instance, imagine yourself on a swing.  The swing seat is constrained to move in a circular arc by two opposing forces, gravity and tension.  Those are the only two forces acting on the seat, yet while swinging, the seat is not in an inertial reference frame.  So if you sit in the seat and constrain yourself to the seat (i.e., you don't fall off), you will be accelerated just like the seat because the tension and gravitational forces do not balance.  The only place in the arc where you do not accelerate is same place where the seat would rest in equilibrium (i.e., just hanging there if left alone).  Note that there is no centripetal or centrifugal force here, just tension and gravity.  These two terms only apply to discussions of the acceleration of the reference frame or object.  They are not forces and should not be called forces.
The point is that centrifugal acceleration only exists in non-inertial reference frames, namely frames of reference that are accelerating.  It is not a force, only a term in the acceleration vector resulting from the frame of reference in which you calculate the acceleration.  There is a useful animation found on the Coriolis effect Wikipedia page.


Dehumidify a room for drying laundry

Question: I currently don't have a dryer, so I have to hang my clothes on drying racks. I also have to do this inside, since I don't have much of a yard outside. How can I make a room less humid, especially on rainy or generally humid days,  so the clothes dry faster?
Comment: Welcome to Lifehacks S.E! If you need any assistance with the site, please visit [The Help Center](http://lifehacks.stackexchange.com/help).
Answer: Removing humidity from the air is an energy intensive process.  Further, while your clothes will dry faster in a lower humidity environment, they will still dry in a humid environment.
Increasing the airflow, even on humid days, will significantly speed up drying.  The clothes are still more damp than the air, so moving more air past them will remove water from them more quickly.
Fans are much less expensive to operate than dehumidifiers and heaters.  Put the clothes in an area where you can move air into and back out of the room.  If you have two windows in the drying room, that would be best - have one fan pulling air into the room via one window, and the other fan blowing air out of the room via the other window.
If you only have one window, pull air from the door into the room, and then blow it out the window.
If your room doesn't have two openings to increase the airflow, set up an oscillating fan near the one opening that alternately blows air out of the room and pulls air into the room.
Alternately, if you have forced air heating/cooling, you can turn the fan of that system on and it will circulate the air through the room into the rest of the house, where you can ventilate the more humid air outside.
Answer: An electric dehumidifier would reduce humidity, but would not necessarily be more efficient than a clothes dryer.
Calcium chloride is sold as a dehumidifier, but it quickly "fills up" with water, forming a semi-liquid mess, so it would be of no help for your application.
If you have a furnace-room where there is a hot water heater or furnace that runs most of the time, humidity is usually lower there because of the heat and the airflow.
Comment: Even condenser driers tend to increase the humidity in the room, so an electric dehumidifier is better than that (though much slower).  If you're heating the house anyway the waste heat from the motor/fan and latent heat recovered from condensing the water contribute to heating the house.  If not you're better off opening the window.

Dehumidify a room for drying laundry


Difference between $\mathbb C$ and $\mathbb R^2$

Question: What are the basic differences between $\mathbb C$ and $\mathbb R^2$?
The points in these two sets are written as ordered pairs, I mean the structure looks similar to me. So what is the reason to denote these two sets differently?
Comment: That MUST be a duplicate.
Comment: As sets, they are identical. As groups, vector spaces, also. The point is that we denote and use $\mathbb{C}$ rather than $\mathbb{R}^2$ when we need or refer to its field structure.
Comment: If you are familiar with the language of abstract algebra, $\mathbb{C}\cong \mathbb{R} \times\mathbb{R}.$
Answer: The answer to this question depends on what you mean by $\mathbf{R}^2$. You can write $\mathbf{R}\times\mathbf{R}$, but the "$\times$" can have several different meanings depending on which category you are working in.
As sets:
If you view $\mathbf{R}\times\mathbf{R}$ and $\mathbf{C}$ as sets, this means you are ignoring any possible structure on these things besides their elements (so forget about multiplication, addition, anything.) From this perspective, $\mathbf{R}\times\mathbf{R}$ and $\mathbf{C}$ are the same object (in technical terms they are isomorphic in the category of sets) because there is a bijection
$$(a,b)\ \leftrightarrow\ a+bi.$$
A key thing to note about this is that "$\times$" refers to a direct product of sets, i.e. the Cartesian product. This will change as we add more structure.
As real vector spaces:
You can give each of $\mathbf{R}\times\mathbf{R}$ and $\mathbf{C}$ the structure of a real vector space, meaning you can add vectors and multiply by real numbers. Then from the theory of linear algebra, we know that $\mathbf{R}\times\mathbf{R}$ has a basis of $\{(1,0), (0,1)\}$ and $\mathbf{C}$ has a (real) basis of $\{1, i\}$. Since these real vector spaces both have dimension 2, they are isomorphic (in the linear algebra sense, i.e. in the category of $\mathbf{R}$-modules). So from this perspective they are again the same object.
Note here that $\mathbf{R}\times\mathbf{R}$ can be interpreted as $\mathbf{R}\oplus \mathbf{R}$ which may be more familiar to linear algebra students. The point is that now we are requiring more from the operation (it has to preserve addition of vectors now).
As rings:
Here is where the difference comes in. We can think of $\mathbf{R}$ and $\mathbf{C}$ as rings, meaning we can add and multiply elements together according to some axioms. Then if you write $\mathbf{R}\times\mathbf{R}$, you mean a direct product in the category of rings, so now multiplication in $\mathbf{R}\times\mathbf{R}$ has to satisfy
$$(a,b)\cdot (c,d)=(ac,bd).$$
But in particular this means things like
$$(1,0)\cdot (0,1)=(0,0),$$
which means it is possible for two nonzero things to have a product zero. In contrast, if $z_1z_2=0$ in $\mathbf{C}$, then either $z_1=0$ or $z_2=0$. In this way, $\mathbf{R}\times\mathbf{R}$ and $\mathbf{C}$ have fundamentally different behavior as rings. Because of this, there is no isomorphism of rings between the two objects.
As fields:
A field is a commutative ring with more structure (we can invert multiplication for nonzero things). It turns out that $\mathbf{C}$ can be given the structure of a field because $z^{-1}$ exists for any nonzero $z\in\mathbf{C}$, but $\mathbf{R}\times\mathbf{R}$ cannot be a field because equations like $(1,0)\cdot (0,1)=(0,0)$ mess everything up (try to cancel something from the left side).

tl;dr -- You have to specify what you mean by "$\times$". $\mathbf{C}$ and $\mathbf{R}\times\mathbf{R}$ are exactly the same until you start saying you want to do things like multiply elements together.
Comment: @analysis89 Right, and because all fields are integral domains, this means it's impossible for something not an integral domain to be a field. $\mathbf{R}^n$ is just shorthand for $\mathbf{R}\times\cdots\times\mathbf{R}$.
Comment: That means the main difference is that , $\mathbb C$ is an integral domain but  $\mathbb R \times \mathbb R$ is not an integral domain.
Comment: is there any difference between  $\mathbb R^n$ and $\mathbb R \times \mathbb R \times....\times\mathbb R$ (n times)
Comment: @Germain that is its definition https://en.wikipedia.org/wiki/Cartesian_product#n-ary_Cartesian_product
Answer: Well $\mathbb{R}^2$ is (by default) interpreted as a vectorspace. So normally you only expect addition and scalar multiplication in it. Which give different algebraic structures.
And I want to add something. The points of $\mathbb{R}^2$ are written normally as tupels, namely $x\in \mathbb{R}^2$ then $ x=(x_1,x_2)$ with $x_1,x_2\in \mathbb{R}$.
The elements of $\mathbb{C}$ have (at least) 3 common ways to denote.
\begin{align*}
z&=a+i\cdot b \quad a,b\in \mathbb{R}\\
z&=(a,b)\\
z&=|z|\cdot \exp(i\varphi)
\end{align*}
The first and the second look pretty much the same, but in $\mathbb{C}$ the multiplication is a function
$$ \mathbb{C} \times \mathbb{C} \to \mathbb{C}$$
while the scalar multiplication in $\mathbb{R}^2$ is a function
$$ \mathbb{R} \times \mathbb{R}^2 \to \mathbb{R}^2$$
You can copy the structure of $\mathbb{C}$ and put it on $\mathbb{R}^2$, to make it a field for example.
But for example the differentiability is different in $\mathbb{R}^2$ and in $\mathbb{C}$ namely a function which is differentiable on $\mathbb{R}^2$ need to fulfill the Cauchy Riemann differential equations to be complex differentiable.
Comment: To say it in an easy english, $\mathbb{R}^2$ and $\mathbb{C}$ are like twins, even though they are not the same they look pretty much the same
Answer: $\Bbb C$ is a field in its own right with an operation of multiplication.  $\Bbb R^2$ does not have that.
Answer: To put it succinctly, $\mathbb{R}^2$ is a vector space while $\mathbb{C}$ is a field. One of the first difference to be noticed is that you can divide two complex numbers but you cannot divide two points in $\mathbb{R}^2$.
Comment: Yep every field is a VS over itself but every VS certainly isn't a field. So that's a (big) difference between $\mathbb{R}^2$ and $\mathbb{C}$. $\mathbb{C}$ is both but $\mathbb{R}^2$ is *only* a VS.
Comment: Some would argue that every field is a vector space over itself.
Answer: If we desire that every integer has an inverse element, we have to invent rational numbers and many things become much simpler.
If we desire every polynomial equation to have a root, we have to extend the real number field $\mathbb{R}$ to a larger field $\mathbb{C}$ of 'complex numbers', and many statements become more homogeneous. To construct a complex number, we associate with each real number $a$ second real number $b$.
A complex number is then an ordered pair of real numbers (a,b).
We write that new number as $a+bi$.
Complex numbers contain solutions to all polynomial equations and hence are an algebraically closed field unlike the real numbers. The complex numbers are not an ordered field. $\mathbb{C}$ is geometrically represented by identifying it with $\mathbb{R^2}$. (This is sometimes called the
Argand diagram).


Why many professional Xiangqi players cannot pronounce 車 correctly?

Question: When referring to a piece in Chinese chess (Xiangqi), 車 should be pronounced as jū or ㄐㄩ, instead of chē (ㄔㄜ). It is understandable that not every native speaker knows this pronunciation. However, I was surprised to hear professional Xiangqi players call the piece the wrong way. After all, if Xiangqi is their profession, how come that they do not even know how to pronounce the piece correctly?
Comment: Colloquially, some pronounce 车 as chē though they know the correct pronunciation is jū. This phenomenon occurs on other characters too. E.g. 钴 is gǔ according to dictionaries, but it's often pronounced as gū, even on TV.
Comment: See https://chinese.stackexchange.com/questions/22333/pinyin-of-%e8%bb%8a-in-idioms maybe those pro players are the younger generation that only learned only one pinyin for 车 /chē/
Comment: Why should they? Is it a requirement? Is there any time penalty if they don't? Is even the very knowledge of Chinese language required? Anyway, dictionaries and Xiangqi boards have something in common: contrasting colors. Some people evidently chose the board.
Comment: This may be one of the products from the Simplification - eliminate 破音字.
Comment: Many professional Xianqi player can't pronounce "車" correctly?
I don't think so, it's a false proposition, in China, even a junior high school student knowns to correctly pronounce "車" in Xianqi.
Answer: Because in 廣韻 (a Chinese rime dictionary from 1000 years ago), 車 has two different pronunciations, 1: kio and 2: chja. As time goes by, 1 becomes ju and 2 becomes che. In ancient Chinese, 1(ju) refers to combat vehicles, 2(che) indicates the means of transport for carrying heavy objects. Although these two different meanings are now included in 2, but the ancient pronunciation is still retained in the Xiangqi pieces 車


What is the badge for a person who viewed a Tumbleweed's question?

Question: Seriously in lighter vein: I have acquired the Tumbleweed badge for asking a question on astyle, as it had "attracted" low number (actually zero) of views, comments or answers. Almost after three months, I got my reputation topped by five because somebody had viewed it. Should there be a badge too for viewing a Tumbleweed question?
Comment: I'd be rich....
Comment: You actually got +5 rep because someone up voted your question, not because someone viewed it.
Answer: No, I don't think there should be a badge for viewing a tumbleweed question.
The tumbleweed badge is mostly a consolation prize.  It could be a really great question, a diamond in the rough, or it could be a really poor question that slipped through the cracks of community moderation.  It could also have been posted on a low-traffic tag.
What would be the criteria for awarding?  Could it be awarded more than once?  Would users abuse this badge by digging up old questions without votes?
Additionally, if a question has low activity like that for a year, it is automatically deleted.
Answer:
What is the badge for a person who viewed a Tumbleweed's question?

The Reversal badge.
Yeah, I know.  You have to do more than just look at the question.  Yeah, I know.  It requires answering a question with score of -5.  I'm proud of my reversal badge.  There's only been 257 handed out in SO's entire history.
The point, of course, is that I actually had to do something to earn that badge.
Comment: The Reversal badge requires the question to have a score of -5 of lower, while the Tumblewood requires the question to have a score of 0. While it's not impossible, I doubt that this happened at all in the SO history.
Comment: @Glorfindel: Fixed.  Don't bother with the SEDE query.
Comment: I can't check ^^ in SEDE because it doesn't store the post ID of a badge :(


Allow users to view all processes

Question: I want to allow accounts with only user permissions on a Windows Server 2003 machine to be able to view all processes.  I simply want them to be able to open task manager and be able to check the box that says "Show processes from all users".  How can I enable that?
Answer: First option is to use process explorer instead of task manager.

Allow users to view all processes


Publishing copyrighted material on GitHub for phishing

Question: I want to publish some demo code on github that deals with a new type of phishing attack. However, I've used a Google-branded sign-in page for the demo. Will this be a problem with copyright or any legal issues?
Comment: why use a specific company branding? publish a generic template and the person using it can re-brand as they wish
Comment: I wanted to show the 'effeciveness' of the code in a related blog post
Comment: Branding is not code. Branding shows the effectiveness of social engineering. That's a completely different matter.
Answer: Obviously, Google owns their copyrights. I can't say they will for sure sue you for it, but it's their property and they can make that decision if they choose to.


Is sheafification always an inclusion?

Question: Let $\mathscr{F}$ be a presheaf and $\mathscr{F}^+$ its sheafification, with the universal morphism $\theta:\mathscr{F}\rightarrow\mathscr{F}^+$. Question is: is $\theta$ always an inclusion? I'm pretty sure it isn't, but in many cases it seems that it is. For example, if $\phi:\mathscr{F}\rightarrow\mathscr{G}$ is a morphism of sheaves, then $\operatorname{ker}\phi$, $\operatorname{im}\phi$, $\operatorname{cok}\phi$ all seem to have this property. If it isn't always the case, then is there any characterization of such presheaves? A "subpresheaf" of a sheaf certainly has this property (e.g.,  $\operatorname{im}\phi$), but what about $\operatorname{cok}\phi$?
Comment: Dual question [here](http://math.stackexchange.com/questions/122523/why-isnt-the-associated-sheaf-transformation-onto).
Comment: Just to make sure, you are asking if the kernel of $\theta$ is always trivial?
Comment: @M Turgeon: Yes.
Comment: Another example : consider the _presheaf_ $F : U \mapsto O_X(1)(U) \otimes_{O_X(U)} O_X(1)(U)$ where $(X, O_X) = \Bbb P^1_k$. Then $F^+ \cong O_X(2)$ but $F \to F^+$ is not injective, since $F(X) \cong k^4$ but $F^+(X) \cong k^3$. The key notion is _separated_ presheaf ; see [here](https://mathoverflow.net/questions/105232).
Comment: It might be relevant to note that I've seen presheaves with this property be called "separated".  It's equivalent to half of the definition of sheaf: that if $\{ U_i : i \in I \}$ is a cover of $U$, $f,g \in \mathscr{F}(U)$, and $f \mid_{U_i} = g \mid_{U_i}$ for each $i \in I$, then $f = g$.
Answer: No, $\mathcal F \to \mathcal F^+$ is not injective in general.
Consider the sheaf $\mathcal C$ of continuous functions on $\mathbb R$, its subpresheaf $\mathcal C_b\subset \mathcal C$ of bounded continuous functions and the quotient presheaf $\mathcal F$, characterized by $\mathcal F(U)= \mathcal C (U)/\mathcal C_b(U)$.
For every open subset $U\subset \mathbb R$, we have $\mathcal F(U)\neq 0$ but the associated sheaf is $\mathcal F^+=0$, so that
the morphism $\mathcal F \to \mathcal F^+=0$ is definitely not injective.
Injectivity of  $\mathcal F \to \mathcal F^+$ is equivalent to requesting that whenever you have compatible gluing data $s_i\in \mathcal F(U_i)$ on an open  covering $(U_i)$ of an open $U$ of your space, they can glue to  at most one $s\in \mathcal F(U)$ : one half of the conditions for a presheaf to be a sheaf must be satisfied (the other half is to require that $s$ always  exist)
Answer: The answer to the question is NO since the global information about a presheaf can not be determined by its local information. In fact, this is the essential point of the  definition of a sheaf. In above, Georges Elencwajg has given a nice counter-example which is quite natural. We can also contruct a counter-example directly as follows.
Take a topological space $X$ which is not empty, define a presheaf $\mathcal{F}$ of abelian groups in the following way: for any proper open subset $U$ define $\mathcal{F}(U)$ to be zero while define $\mathcal{F}(X)$ to be any nontrival abelian group, say $\mathbb{Z}$. Then the sheafication $\mathcal{F}^+$ is zero. Consider the global sections we will find it is not an inclusion.
Comment: Ah, yes:  yours  is a great, simple and amusing example. Bravo: +1
Comment: @Georges Elencwajg: Is it enough to show that $\mathcal{F}^{+}(X)=0 $ ? so that $\mathcal{F}(X) \to \mathcal{F}^{+}(X)$ is not injective group homomorphism.
Answer: Cokernels do not have this property. The example to look at is the exponential map from the sheaf of holomorphic functions on $\mathbf C \setminus \{0\}$ to the sheaf of non vanishing holomorphic functions on that same domain. This is not surjective on global sections, since for example the identity map does not have a logarithm, but it is surjective as a morphism of sheaves, so the sheaf cokernel is trivial.
What distinguishes the kernel and image sheaves, I think, is that they satisfy the identity axiom [I think this is Hartshorne's (5)] because they sit inside of sheaves.
Comment: Ah, sorry, I was thinking of $\mathscr{F}/\mathscr{G}$, where $\mathscr{F}\subset\mathscr{G}$ is a subsheaf.

Is sheafification always an inclusion?


Addition of hops to double IPA

Question: When brewing a double IPA do I have to put the hops in steeping bag or do I just toss toss them in. This question applies to both the boiling phase and when I have to dry hop later!
Answer: This will depend on what kind of hops you're using and personal preference.  Let's start with the boil:
In general, hops in a bag are going to give you less flavor/bitterness than the same quantity floating freely in the kettle.  You can weight the bag, to mitigate this somewhat. (with, say, a stainless steel ball).  I don't have numbers for this; I can only tell you "less."
When using pellet hops, I will typically just throw them in and let them dissolve.  They'll eventually end up on the bottom of the fermentor.  Some folks like to avoid as much of this material as possible, so they use bags.
If you are using whole leaf hops, I highly recommend using a bag in the boil.  When I first used whole hops, I just tossed them into the kettle. This led to a variety of problems, including a plugged sight glass and plugged dip tube (which I cleared with a sanitized latex glove).  They also soak up a surprising amount of wort.  If I were to use them again, I would bag them and use a sanitized latex glove to squeeze some of the wort out of them.
As for dry hops:
I like to bag all dry hops, to aid in removal.  Some people like to see whole hops floating at the top of the carboy.  Again, you'll get less aroma when using a bag, so you may want to weight the bag or compensate by using extra hops.
I didn't address plug hops, both because I have no personal experience, and they tend to be less popular than other forms.  I would expect them to behave similar to whole hops, but perhaps others can correct this assumption.
Comment: So if I do a secondary fermentation and transfer it to a carboy(better bottle) I should us a steeping?, or is it better to keep it in the primary fermenter and add the dry hops there and not use secondary fermentation?
Comment: Personally, I dry-hop in the keg.  Before I bought a kegging system, I dry-hopped in primary, but this is because I never use a secondary.  I find that my beer does great with a long primary (3-4 weeks), and it's less trouble than transferring.
Answer: For both boil and fermenter additions, using a steeping bag:

Reduces chances of clogging during transfers (+)
Gives lower hop utilization (-)
Provides for easier removal of the hops when they've done their job (+)
Keeps hops out of the next phase (the fermenter / your glass) (+)
May allow for more wort/beer to be transferred (+)
More trouble/expense at the point of the hop addition (-)

If you want your transfers to be smooth, you need a plan for managing hop solids.  Many brewers have plans that manage hop residues without putting hops in a bag.  But unless you have such a plan, item 1 alone, should be enough to get you to use a bag.  Number 2 can be easily compensated for.  As for 3 & 4, you probably don't want bittering hops in your fermenter or chunks in your glass.  If you've pulled the hops out (especially pellets), you might be inclined to transfer a little more volume (5), and not leave so much wort/beer behind.  But you do need to buy nylon or many muslin bags (6).

Addition of hops to double IPA


Problem of the Google play store

Question: As soon as I get into Google Play store and search for an app, it automatically comes to the home screen of the Android device and a notification appears asking whether you want to close the app or open it once again.Will this problem will be solved?
Answer: This could just be that you are running out of RAM. There is more information on this here, if you need it.

Problem of the Google play store


What is this flicker I get while tracking a camera?

Question: After having my camera tracked, I decided to put in a cube for testing. But then I've noticed this strange artifact where every object kinda flickers and changes it's rotation.
Comment: note: some retracking and resolving seems to have fixed it, but i still wanna know what the problem was
Answer: You can clearly see the tracker shifting on the video posted on the question, and it happens at the moment where the reconstruction fails.
Successful reconstruction can only be achieved with accurate tracking
You should aim to have a solve error of less than 0.3 pixels.
There are plenty of tools to help you get more accurate tracking.
Read:
Solve error high with good track?
Motion Tracking. What do the graphs represent and how are they useful?
Does the motion tracker follow a point if it changes colour throughout the video?

For a comprehensive post on motion tracking tips read
How can I get better results when doing camera motion tracking?
Comment: will a better solve also help me with my world alignment issues? i've been having this problem where the trackers and the camera are slightly tilted on Y and X axis and they're not on Z = 0, so I have to rotate my model for it to look good
Comment: different questions should be asked in a new post, not as comments... but before you post you can also use the big "search on blender" box at the top.


Fugitive Gbrowse while being ssh on a server

Question: I work remotely via ssh on my dev machine which runs a tmux. I want to pipe the output of the Gbrowse back to my laptop. Is there any reasonable way to do that.
Answer: You can use Gbrowse! to get the url copied into the clipboard and then either paste it in the local browser or fire another :!ssh command from the remote machine to the local machine, like (from vim in the ssh session)
:Gbrowse! | exe "!ssh $LOCAL 'nohup google-chrome " . @+ . "' &"

change google-chrome to other browsers or xdg-open to use the default if you are in ubuntu.


For pose estimation, what is the typical range of the rotation values that are estimated? Is it from -180 to 180 degrees?

Question: I am currently working with the Rutgers APC 2015 data set for pose estimation, and see that the pose estimates, specifically the rotation estimates, have seemingly been rescaled to range between -1 and 1. Does anyone know if these values were arrived at by dividing by 180?
Answer: Check whether they are in radians. Most algorithms use radians for pose estimation.
Answer: The range between -1 or +1 is a normalized range like you said. The question is the following: Were those values divided by 180 (degrees) or 2*pi (radians)? You should check any existant documentation regarding Rutgers APC 2015. Usually, the convention is to have the angles bounded between -pi/pi or -180/180, because expressions like atan2 or atan2d returns angles in that range. The range 0-360 might lead to tricky bugs to catch and error during the 360,0 crossover.


What happens if we add denormalized number and normalized number?

Question: Denormalized numbers can represent numbers smaller than 2^(-1022) whereas normalized number cannot. So I'm curious what happens if we add denormalized number and normalized number. Actually, I have tried some tests and it seems that the computer converts denormalized number to 0. Is my thought correct?
Answer: The result depends on the difference between the normal and denormal numbers you are adding.
 $1 + [denormal]$ is rounded to $1$.
But if the range difference between your numbers is inferior to $2^{52}$ for doubles, or $2^{23}$ for singles, you will get a real addition. There is nothing special about denormals here : $1 + 2^{-200}$ is rounded to $1$ as well.
To implement a floating point adder (as hardware or software), you shift right the smaller number to equalize the exponents before doing an integer addition to the aligned mantissas. If the range difference is too large, the smaller number is entirely shifted out.
Comment: Can you explain me why the range difference determines everything?
Comment: The result must fit in a fixed number of bits, 52 for doubles and 23 for singles. You can try doing the addition manually (with singles, it is easier).


Preferred current flow in relay contacts?

Question: Yet another relay question. In this document the moving and fixed part of a relay contact are named anode and cathode. This suggests that there's a preferred current flow through the contact in DC applications. Is this taken in consideration if both contact parts have different contact materials? For instance, if contacts are tungsten and silver-nickel for anode and cathode resp. (just an example, not necessarily realistic), are you supposed to have the anode more positive than the cathode?
Comment: +1 for a good question. I never gave it much attention, and I think few people do, but I can imagine that for a DC application there could be a preferred polarisation if the contact has different materials for anode vs cathode. (But then I switch mostly AC.)
Answer: With high current contactors (big relays) the stationary contact is usually the most positive terminal. The load gets connected to the armature (movable) contact. Very high current contactors, or large DC circuit breakers ~4KA-20KA will have a "Blow out coil" or a permanent magnet of sorts which will direct the arc discharge upwards usually through an "Arc chute" to dissipate the arc into the atmosphere. If the polarity of such a contactor were reversed, the arc would deflect down and into the closing coil likely burning it up.
EDIT:
I stated that the stationary contact is usually the most positive, but this is not always the case. One needs to follow the manufacture's guidelines. I have seen 2 pole contactors with both a positive and negative arc chute, showing that the stationary contact can be either polarity. However the polarity is still defined, and must not be reversed. AC contactors also use arc chutes, and often (not always) the line side is the stationary contact.
Comment: +1 Interesting. I always thought it was heat which caused the upward motion of the arc.
Comment: Do you have any information I could access to read more about this?
Comment: @Kortuk - Unfortunately no, I can not quote a source directly. My statement is based on my many years as a DC Substation engineer. My comments are common knowledge within this field. Much of what I learned came from documents from GE, Westinghouse, and a few others who were (and still are) big players in the industry. However these documents were written in the 40's and 50's (some even older!) way before the internet. I am since retired and no longer have access to those documents, otherwise I would offer to scan them. I did see a lot of hits on Google however, searching for "DC contactors".
Comment: I just meant if you had anymore to read for someone like myself whom is forever curious it is nice. Your answer made logical sense and I have already upvoted you. Thank you for sharing your experience. Glad to hear we have a few power guys. I do not go higher then 13kV at work. The 300V switches have polarity also though.


Linear Discriminant Analysis, which parameters can be tunned in cross validation set up?

Question: I am implementing Linear Discriminant Analysis in R, which parameters can be tunned in cross validation set up? In regularized mode called penalizedLDA there are parameters which are optimised but I want to know which parameters are turned in case of simple LDA method?
Answer: LDA has a closed-form solution and therefore has no hyperparameters. The solution can be obtained using the empirical sample class covariance matrix. Shrinkage is used when there are not enough samples. In that case the empirical covariance matrix is often not a very good estimator.
Comment: Yes, compare it on the same test set. And yes, regularization is recommended if you have more features than samples. Whether LDA is appropriate for your dataset also depends on how well your data fits the assumptions of LDA.
Comment: I made a SVM classifier where I have a nested cross-validation setup for hyper-parameter running. so If I want to compare the accuracy, It is a recommended practice to build the same framework for all the classifiers when predictive performance has to be compared. So, If I use LDA then I can compare it with SVM performance with nested C.V for parameter running?
Comment: I have a dataset with 55 observations and 180 features which are non-collinear so can I use PenalisedLDA?


How can 16 Psyche be the core of a protoplanet?

Question: The recent approval of a NASA mission to study the asteroid 16 Psyche brought to my attention that Psyche (about 200 km in diameter) is thought to be the exposed iron core left over from a protoplanet (theorized as being about 500 km in diameter). My question is: what is the mechanism by which a body of that size can accrete, differentiate as thoroughly as Psyche apparently did, and then completely lose its mantle? I have a basic understanding of planetary differentiation and I had the impression that the migration of iron and nickel to the core took a long time, and that by that time major impacts capable of shattering large bodies must have been rare.
Comment: How long are you talking about? The protosolar nebular dissipated on a scale of 3-10 Myrs, the Earth-Moon system was formed via giant impact after 60Myrs, Earth's core formation was finished at 100Myrs, while the dynamic evolution of the solar system, including giant impacts possibly went on until ~900Myrs after solar ignition. There's plenty of time and material.
Comment: As I said, I had the impression that differentiation took longer. But my question was more oriented toward how a large differentiated body could be shattered and lose its mantle.
Answer: If this hypothesis is correct, the body from which 16 Psyche formed was a 500km planetesimal (about half the diameter of Ceres) It would have been destroyed in a collision with another body. The other body would have had to have been large enough (300km) to complete fragment the asteroid. The problem with this hypothesis is that no other pieces of this planetesimal have been found (they should still have similar orbits as Psyche).
Another hypothesis is that Psyche suffered from a series of smaller impact, that fractured and disrupted the parent body, but did not destroy it. This would give Psyche a composition similar to the stony-iron mesosiderite meteorites.
The Psyche mission is intended to answer these questions.


Learning F# with Prime Factorization

Question: In my quest to learn F#, I implemented Prime Factorization.
First, the code:
let primefactors x =

    let check = seq { let limit = uint64((ceil(sqrt(float(x)))));
                      for x in Seq.concat [seq { yield 2UL }; { 3UL .. 2UL .. limit }] do
                        yield x }

    let getFirstOrDefault def ns =
        if ns |> Seq.isEmpty then
            def
        else
            ns |> Seq.head

    let nextfactor x =
        match x with
        |1UL -> x
        |_ -> check
              |> Seq.skipWhile(fun idx -> x % idx <> 0UL)
              |> getFirstOrDefault x

    let rec getfactors x factors =
        match nextfactor x  with
        |1UL -> factors
        |factor -> (factors |> List.append [factor]) |> getfactors (x / factor)

    [] |> getfactors x

let smallFactors = primefactors 3672215407307975928UL
let bigPrimeFactor = primefactors 18446744073709551556UL
let biggestuint64Prime = primefactors 18446744073709551557UL

Im satisfied with the performance, but have some gripes with this code that I'd like feedback on:
I'm checking for 1UL in two places. In nextfactor it's there to skip the last run through the numbers; in getfactors it's the break condition. I don't feel like this is as elegant as it could be.
Another thing I don't like is the getFirstOrDefault function. At first I simply had a Seq.head after the Seq.skipWhile but that throws if the number itself is prime and so the sequence is empty. I'd like to do pattern matching with that as I would with lists but in that case I'd have to materialize the whole sequence which I'd like to avoid. Is there something like |[] -> ... for sequences?
I would also like to hear suggestions on how to construct the check sequence in a more elegant way.
I am aware that the algorithm itself is a pretty naive implementation of the problem, but I'm ok with that as it's (for me) sufficiently fast in the range of uint64. If you do have suggestions on how to improve performance while keeping with this simple algorithm, I'd love to hear them, though!
I'm also interested in hearing where I'm missing functional concepts that might be applicable here.
Answer: I think you show a good understanding of F# as a language and functional programming in general as far as I can see. A couple of things though:
You use x as variable name on multiple levels of function definitions. It makes it hard to read.
IMO match .. with is a very readable construct for more than two matches. But for only two I prefer if-statements.
The check function/sequence is a kind of double sequence.
I would do it this way:
let primefactors target =
    let check ta = Seq.concat [seq { yield 2UL }; { 3UL .. 2UL .. uint64(ceil(sqrt(float(ta))))}]

and then call it like this:
let nextfactor x =
    match x with
    |1UL -> x
    |_ -> check x
          |> Seq.skipWhile(fun idx -> x % idx <> 0UL)
          |> getFirstOrDefault x

It will make the function much faster.
The getfactors - function is not tail-recursive, but can easily be changed to be:
let rec getfactors x factors =
    match nextfactor x  with
    |1UL -> factors
    // Here the factors list is handled before the recursive call.
    |factor -> getfactors (x / factor) (factor :: factors)

Learning F# with Prime Factorization


radio mics, what is in the future

Question: what is the future of radio mics, is it worth spending thousands of euros (or dollars) now for devices that soon may become antiquate? are we expecting some total change or improvement in wireless technology? I am refering at a pro range.
Comment: This is an opinion-based question and probably not within the range of what this forum was intended for.
Answer: "Change"?  Certainly. Governments are re-purposing vast swaths of bandwidth from traditional users (which permitted or tolerated small-beans users like wireless mics) to new users like "land-mobile" (i.e. cell phones and other consumer wireless) which do NOT tolerate "white-space" unofficial users like wireless mics.  Nobody knows exactly where this is going or where it will end up.
"Improvement"? That is a different matter. And rather subjective depending on how you are defining "improvement".  Certainly we have much better and cheaper technology with spread-spectrum, frequency-hopping, auto-pairing digital wireless technology (courtesy of WiFi, BlueTooth, et.al.)
While this new digital technology offers all the aforementioned benefits, it also brings with it the disadvantages of latency (which is not an issue with digital communication like WiFi, BlueTooth, or even cell phones).  And the disadvantage of shrinking available RF spectrum (which is the same problem whether analog or digital.)

radio mics, what is in the future


Are resonance and resonant frequency the same?

Question: Resonance is when applied oscillation is in phase with natural oscillation. Then, what is resonant frequency?
Answer: Resonance is the name for the phenomenon.
The resonant frequency is the frequency at which this phenomenon occurs. The frequency of the oscillation when you observe resonance.
Frequency is just one of the characteristics of a resonance. Another one is Q factor. Others might be the mode structure of the resonant wave.
Comment: It is possibly worth noting that people will often say a system is "at resonance" to mean "at its resonant frequency"
Answer: Resonance is the process of driving a system to oscillate at a resonant frequency. The property of these frequencies is that they can store vibrational energy or vibrational quanta depending on the description you choose. This means that driving the system at these frequencies can create much larger amplitude oscillations than other frequencies, which disperse energy much more quickly.
A system may support many resonant frequencies, and driving the system at any of those frequencies would be cause resonance. That is, 1 system can have many resonances, each of which is associated with a resonant frequency.
Comment: Why does a system stores more vibrational energy at some particular frequency?
Comment: This is a good question, I think you'd be better off asking it on the main site.


Should I ask this as 3 separate questions?

Question: I asked this yesterday, but I started wondering since long questions are not very popular, if I should ask it as 3 different questions or leave it as it is now.
Comment: Split up the questions. I, and I suspect many others, tend to browse math stackexchange with a short attention span--that is, unless I get hooked by an interesting problem.
Answer: A few points.

Separate, related questions can certainly be asked in the same question, if you so desire.  So there's not necessarily anything wrong with the way you did it.
It's true that long questions are not always accepted as well as concise questions.  Therefore, no one will blame you if you decide to split up the three questions.  Alternatively, you can set up a bounty to give your question more attention, and if you do it will very likely get a good response.
If you do ask related questions separately, make sure you link to the older questions in the later questions.

I also agree with Asaf's comment below: unless the questions are strongly related (i.e. in order to understand the later questions you need to have read the earlier questions) ask them separately.
Comment: The first one should be *quite* related, not just related. One should also not throw in too many questions into the same thread. For example, a list of seven sets with the sole question of "what is their cardinality" (even if all sets have the same cardinality) is certainly not a good fit for this site. Asking two questions which are natural extensions of one another, and are strongly of the same block of wood, that *might* be acceptable, but depends on the situation.
Comment: A person with a long list of questions can hopefully solve the remaining questions themselves after seeing some of them being solved.
Comment: Thanks for the answer, I'll keep that in mind. By now I've already _solve_ the question, except for one part that has been bothering me a lot. I've been thinking in posting an answer, and hope someone will correct/extend it.


Quarterly regression, year-fixed effects

Question: I am trying to inspect the heterogeneous effects of a USA-based shock on a panel of countries using quarterly panel data. My variables of interest are the USA-based shock (which has only temporal variation and is mildly serially correlated) and the interaction of this variable with another variable that has cross-sectional and temporal variation too. My question would be whether it makes sense to include year-fixed effects in this regression?
I mainly worry about something like controlling for a downstream variable (subtracting part of the effect of the US shock with the year-fixed effects).
Answer: I guess there's no easy answer. It depends on what you think the data generating process is.
Let $y$ be year, $q$ be the quarter and $c$ be the country.
Then $S_{y,q}$ is the USA shock at year $y$ and quarter $q$, $X_{c,y,q}$ is the observable covariates of country $c$, at year $y$ and quarter $q$ and let $Y_{c,y,q}$ be the variable of interest.
From reading your question, your basic regression looks like this:
$$
Y_{c,y,q} = \alpha_0 + \alpha_1 S_{y,q} + \alpha_2 X_{c,y,q} + \alpha_3 (X_{c,y,q} \times S_{y,q}) + \varepsilon_{c,y,q}.
$$
The underlying assumption is here that:
$$
\mathbb{E}(\varepsilon_{c,y,q}|S_{y,q}) = \mathbb{E}(\varepsilon_{c,y,q}|X_{c,y,q}) = 0
$$

A first thing you might be worried about is that $Y_{c,y,q}$ might be (systematically) different depending on the country in addition to what you observe in $X_{c,y,q}$. In this case, you should probably add country fixed effects.
If you think that $Y_{c,y,q}$ might differ according to the year $y$ in addition to what is explained by $S_{y,q}$ and $X_{c,y,q}$ then you should add year fixed effects.
If you think that $Y_{c,y,q}$ might change with $q$ (e.g. seasonality) in addition to $S_{y,q}$ and $X_{c,y,q}$ then you can add quarter fixed effects.
In principle you coud add country$\times$year fixed effects or country$\times$quarter fixed effects if you are worried about a combination of these problems. If you add year$\times$quarter fixed effects you loose identification for $S_{y,q}$.

Which of these options is best will depend on what exactly you are looking at (i.e. what is $Y_{c,y,q}$) and how you think this changes by factors not included in your regression.
If you decide to add year fe, then the identification of the coefficient on the shock $S_{y,q}$ (and the interaction) only relies on intra-year variation in $S_{y,q}$ and $Y_{c,y,q}$. So if you think there is a lot of yearly variation in $Y_{c,t,q}$ explained by $S_{y,q}$ then you will not be able to capture those when using year fixed effects.
There is also the (subtle) issue of how you define the year. For example, you could define a year from "Januari year $x$" to "December year $x$". An alternative could be, for example, to define the year from "April year $x$" to "March year $x +1$". Depending on this choice, your results might vary.
Comment: Thank you very much for your really helpful answer tdm! If you don't mind, I'd ask one  follow-up: you didn't touch on this, but my main worry is that adding year fixed effects may introduce inconsistency into my estimates for the coefficient of S and the S x X interaction. I don't have a formal argument but adding fixed effects here feels analogous to adding year fixed effects in a simple 1 country quarterly time series, which (again) feels suspicious. Would part of the common 'global' yearly effect that the year fixed effect subtracts be part of the effect of S as it affects quarters 1,2,..?
Comment: Suppose I assume that S and X are completely exogenous and my motivation for adding year fixed effects is to soak up some variation from Y and increase the precision of estimates.
Thank you very much for you help!
Comment: If the true model doesn't has fixed effects. Then adding fixed effects does not make the estimates inconsistent, although the estimates will be less efficient in the sense of having larger asymptotic variance (this is the whole argument of the Hausman test for FE versus RE).
Comment: To add a bit more explanation about my worry: suppose there is a big US shock that makes it a good year for every country in the sample, not only for the 1st quarter but for all subsequent quarters. That means that yearly average Y is high for all countries, which means that I subtract a large number with year fixed effects. This relationship feels strange, but it may just be due to some bad intuition about FE
Comment: Hmm okay. Thank you very much then. That's really helpful


Self policing incentive in Goods and Services Tax

Question: Amongst several advantages of GST (Goods and Services Tax) I read one as that it prevents tax evasion through self policing incentive. But self policing incentive is already provided in VAT (Value Added Tax). So what's new in GST that improves the current self policing incentive system?
Comment: Would you please link to where you came across said concept so that we may share the joy of exploration? Also, perhaps editthe definition of GST into the body of the question, so that a would be answerer does not have to read this discussion.
Comment: Why do you think there is a self policing incentive for VAT and what is GST? Please go into much more detail. (Otherwise these seem like questions copied from a textbook.)
Comment: @denesp I think my question is more clear now.I was not asking about textbook definitions of GST and VAT
Comment: Unfortunately I still do not know what GST is or what the source of your claim is.
Comment: @denesp By GST I meant a uniform Goods and Services Tax throughout a country.While exploring its pros and cons I came across this concept
Comment: @denesp I read it in a newspaper editorial describing benefits of GST from India's perspective.I tried to explore a lot but couldn't find any information on what's the difference in self policing between VAT and GST because both involve input tax credit.So as far as tax evasion is concerned I found no major change.
Comment: @user6479280 As Denesp said, your question is vague and cannot be answered as written since we don't know what you're referring to. If you're asking people to put effort into answering your question, the onus is also on you to likewise put effort in helping yourself. "I read it somewhere" is not helpful.
Answer: Self-policing incentive, in theory, works similarly for both Value Added Tax (VAT) and Goods and Services Tax (GST). The improvement, however, could come from the reduction in the number of instances where suppliers can circumvent the self-policing system, to evade taxes.
VAT is collected by every supplier in the supply chain and paid by every buyer in that supply chain. At every incidence of tax there is an opportunity for tax fraud. One example of VAT fraud is: the supplier provides an invalid VAT number on the invoice and collects the appropriate amount of VAT from the buyer and instead of transferring the collected tax to the government, the supplier pockets it. Of course, such fraud can be detected by the customs department and when audited the supplier may claim that there was a printing error in the VAT number on the invoice. But if suppliers are not audited (perhaps because the cost outweighs the benefit) then they can plausibly pocket all the tax they collected from the buyer. It is also plausible that such a fraud can be carried out by many suppliers, if not every supplier, in the supply chain. Hence, even though there is self-policing mechanism fraud can still be committed by one or more suppliers in the supply chain under the VAT regime. In contrast, under the GST regime there is only one supplier in the supply chain who can theoretically commit that tax fraud. Since there are less instances under GST to evade taxes it is plausible that the amount evaded will be less and that it would be easier to audit.
Comment: Not VAT per se, but sales tax. Under the GST regime only the final seller collects the tax on sold goods and services. VAT has a cascading tax regime that taxes value added at every level in the supply chain.
Comment: So in GST only the final supplier charges and collects VAT?
Answer: If your allusion is to tax evasion…  Unlike the value added tax, a sales tax is imposed only at the retail level(at the time of sale). If goods are sold at retail more than once (used items, cars etc) the sales tax can be charged on the same item indefinitely.
These 2 'systems' of indirect taxes  are quite similar (as without taxation states and some entities would not make profit HAHA), the main difference is VAT (like known in European countries) is collected at the same level rate for certain product or service in one country, but with GST is collected depending on the region, ZIP code etc so it can happen that in NY (even Brooklyner can pay for 1%more than people from Brox...but usually it varies from rougly 3 to 7%) citizens pay more sales tax than in LA, as vary by state, there is no unified national sales tax, but at least more sophisticated dispute system and more customer protection you can find in US.
For VAT there are carried out audits to prevent frauds. Nobody would risk to not claim the correct accounting procedures and credit notes, invoices and transactions to authorities. In GST only the final seller collect the tax on goods and services (only one entitity involved), whereas in VAT regime there come into play all the participants in supply chain.
 The tax is levied on a value which includes tax paid by the previous buyer (tax on tax levied on a product)is more accurate for VAT or so called cascade effect. And one of the major issues with GST is the removal of this tax on tax effect.


What can we do to encourage people to participate here in meta?

Question: The title says it all. Do you not participate because you have nothing to say? Do you simply not know about meta? (In which case we'll see no answers to this meta question.) Are you simply not interested in participating?
Answer: Compared to stackoverflow which currently has 1.5% as many question on meta as the main site, this site has got 10.1% meta questions.
So, there's proportionally a very high amount of meta'ing going on. What we need is less meta and more real questions.
Comment: To be fair, this is a new community that's still trying to figure out things that SO figured out years ago.  Fairer to compare us to one of the newer stackexchange sites.
Answer: I think maybe we shouldn't particularly try to...  I think meta may already have a healthy level of activity...
95% (99%?) of users of bicycles.SE are going to be coming in because they got a search engine hit.  Hopefully in a way that answers the question they were googling for...
The 1%-5% of users that actually create an account, ask questions, answer questions, etc are the community we're interested in maintaining...  But I expect that most of that community is either interested in sharing their expertise and/or in getting answers to their questions, not specifically in maintaining a community...
And if that community is happy on the main site, maybe they don't need to bother with meta that often...  There's already mechanisms (downvoting, flagging, flagging for moderator attention) they can use for problems with individual questions or answers...  We are getting (excluding questions from moderators) one or two questions on meta every week, and only a couple dozen main site questions every week...
I think us moderators could use the "featured" tag on meta for things we think might be of general interest to users, and if we feel the need to ask about how to handle something (or explain why we did something) and comments aren't enough room, a post on meta and a link to the meta post could be good...  Basically, make it easy to find meta, but don't worry if not too many people go there...
I'm hoping that if there's questions about or problems on the main site, most users know they can use meta...  And if it's really bad (a sudden onslaught of spam posts, for instance), they can also post to meta.stackoverflow.com in case the bicycles moderators are all asleep.
Comment: "because they got a search engine hit" -- Yes that's why/how I arrived; it wasn't because "I'm active on SO and I want to know about bikes: so, I'll look see if there's a SE bike-related community" ... it was via Google, responding to a query (though I recognised SE when I saw it as a Google result to my query). And after I asked a question and people gave me useful info on bicycles, then (I already knew about meta.SO) I began to look for other ways to try to encourage, interact with, and repay the bicycles community (including by visiting meta.bicycles).
Comment: You might find it interesting to look at how many users are also users on other SE sites. And, of the people who sign up: where do they come from? From Google, or from another SE page?
Answer: The way to encourage more meta participation is to treat every question like gold. When someone does come in here with a question, provide the best, most helpful answer you can muster.
Ask interesting questions. Meta is more than talking about the problems with the site. Meta is for talking about promotional activities, and that means finding cool resources and projects you guys can be working on. We have been sending top users from various sites to take at conferences and we're always looking for excellent opportunities to let people know about this site.
When this becomes a dynamic and interesting place to be, the people will gravitate naturally towards participation.
Answer: I check meta regularly, vote up questions & answers I agree with, down the ones I don't.  Most of the time my opinion lines up with an existing one, no need to duplicate it.
I think you'll find most users are interested in bicycle Q&A, not discussing tags or what's on-topic or other minutiae.
Answer: Didn't realize there was a meta until I accidentally hit a linked post on the sidebar.


Bridged DSL connection works for a bit, then fails bizarrely

Question: I've got a BellSouth FastAccess DSL subscription that has run on an ISP-provided 2wire HomePortal 1700HW combination modem/router for a few years now with no significant issues. I recently acquired a brand new Western Digital My Net N600 router in order to boost range and provide access for network storage (the 2wire has no USB ports).
I've tried setting up the new network exactly like so:

factory reset both devices to wipe out any lingering config issues
plug in the 2wire to power and a filtered phone jack and boot
log into its Wifi using the serial on the bottom
under the management and diagnostic console (192.168.1.254/mdc), set the 2wire to bridge mode, to disable its routing capabilities
run an Ethernet cable from one of the 2wire's LAN ports to the N600's WAN
plug the N600 into power and boot
log into its Wifi using default information ("admin"/"password")
attempt to connect to the internet, which fails
when prompted, enter ISP's PPPoE login information
connection success!
set up network security, SSIDs, network storage, etc.

And it all works fine. But within 3-6 hours, the connection drops, with a variety of strange side effects:

WinXP laptop used to set up network can browse the web, and can ping the router, but can't load the router settings page, not even through a direct wired connection to the router LAN ports.
Win7 laptop appear connected in the status bar, and can ping the router, but can't browse the web or load the router settings. If disconnected from the network, it cannot reconnect. Attempting a wired LAN connection returns a junk 169.xxx.xxx.xxx IP address.
a visiting Chromebook appears connected, but cannot load any websites... except for the webmail interface at mail.aol.com, which receives new messages just fine. Aol.com won't load, nor Gmail, nor Live mail, etc. This is purely over Wifi, no cell connection.
iDevices cannot connect to the network.

Power cycling the N600 fixes everything... until the next failure a few hours later.
I've tried a few methods to solve:

when setting up the 2wire, connect via PPPoE and establish a working connection before switching back to the default Direct Connect
set VPI/VCI to 8/35 and disable PVC Search, as recommended by a few online setup guides
when setting up the N600, clone the MAC address of the XP laptop used to set up the 2wire
changing the N600's IP to 192.168.1.10 and the allowed range from .11-.250, to avoid any possible IP conflicts
used Google's DNS (184.108.40.206, 18.104.22.168) as well as the automatic ISP option

Everything leads to the same result: a few hours of working network, followed by this mysterious semi-dropped connection. I'm baffled at why one machine would be able to browse but not connect to the network, or why another would be able to load realtime webmail from one site only.
Does this sound like a problem with the N600, the 2wire, the ISP, the network configuration? I'd hate to spring for a new modem or an Airport Extreme if it's not actually going to fix the problem. I'd love to hear some extremely basic step-by-step methods for getting a network like this established properly, in case I'm missing a simple step somewhere. Here's documentation on the devices for reference:
2wire HomePortal: http://support.2wire.com/view/768/1000_series_Gateway_Install_Guide.pdf
Western Digital N600: http://www.wdc.com/wdproducts/library/UM/ENG/4779-705072.pdf
Caveat: Resetting the router fixes the issue for a few hours, so I won't be able to test the effects of anything that requires restarting it in real time. This has made it frustrating to test myself, but hopefully there will be some good ideas here that won't require it.
Comment: Not that anyone's reading this, but apparently the Spotify desktop player still works on computers with a dropped connection, even while Chrome, Firefox, IE, and even iTunes pull up no data. Oh, and the one XP laptop that used to work works no longer, for no apparent reason. Tearing my hair out over here.
Comment: Just to clarify, did you set the PPPoE settings in the _router_ or on individual computers?
Answer: sounds to me as if you bought a router from a company that specializes in hard drives, i would return it saying it's defective, get yourself a nice Linksys/Cisco E4200 and put dd-wrt. that would be my choice for a wifi relay/booster.


Getting YouTube back on Apple TV

Question: As of the latest Apple TV software update, YouTube is no longer showing up on the home screen, apparently replaced by Vimeo (which I like, but which has a very limited video selection). Is there any way to get YouTube back on the Apple TV home screen?
Answer: You might have turned off YouTube under Settings -> General -> Privacy.  Check there, and then toggle it back to SHOW.  If this was the problem, please "Accept" the answer.
Answer: I don't think Apple has removed YouTube.  Try logging out of your iTunes account, then back in.
Answer: Which generation Apple TV is this on? YouTube has been removed from all but the 3rd gen hardware as Google retired the API in use as of April 20 2015   (Announcement)
Comment: This question is almost three years old and no longer relevant. I no longer have the AppleTV in question.
Answer: You can't get this YouTube channel back until Google re opens its streaming feed API for older devices.  They probably won't until they can rewrite it to incorporate in-stream ads.
Comment: Take a look at the already-posted answers and you'll notice there's an accepted solution to the problem already.

Getting YouTube back on Apple TV


QGIS 3.0.2 NNJoin doesn't complete. Crashes when close status bar

Question: I have run a large spatial point to polygon join of 2 million points to 657 polygons. I used NNJoin so that the closest polygon will be assigned to a point if the point does not lie within one.
The process ran for around 6 hours. The status bar showed 100%, but no 'finished' message or result output file was added to QGIS or can be found in my directory.
If I close the status bar on QGIS, I get a Python error "RuntimeError: wrapped C/C++ object of type QProgressBar has been deleted.
Any ideas what the problem is?
Answer: You need to log a bug report in the tracker for the plugin:
https://github.com/havatv/qgisnnjoinplugin/issues?q=is%3Aopen+is%3Aissue
Comment: The issue can be found on https://github.com/havatv/qgisnnjoinplugin/issues/17. It was fixed 15 May (it was a problem with the progress bar).
Comment: Thanks, I have done that now. Seems to work fine for smaller files.