Is there any theory that states that the universe is moving?

Question: Is there any theory that states that the universe is moving? By that, I mean the universe is moving relative to a point in a different space containing the universe. By analogy, the universe would then be like a moving train with the planets inside of it its passengers. Is there any theory or idea that expresses this possibility?
Comment: See [brane cosmology](https://en.wikipedia.org/wiki/Brane_cosmology), where our spacetime is embedded into a higher dimensional one.
Comment: The idea of the universe being "into" something is problematic, let alone it moving. If the universe is inside something, wouldn't that something be, in fact, the universe? If not, what is the criteria to fix a border between the universe and its surroundings?
Comment: Can see this. https://en.wikipedia.org/wiki/Gödel_metric#Properties
Comment: @armand: Lots of theories propose *explanatory higher dimensions*, which we think might be impissible to reach or causally disconnected from our future eg 'Our Universe Could Be An Expanding Bubble in an Extra Dimension' https://www.livescience.com/64417-universe-sits-on-extradimensional-bubble.html Given cosmological event horizons, much of what came out of the Big Bang would no longer be in 'our universe' were we to use this type of definition. There are ways to parse it though, is all I'm saying. & universe now conventionally means cosmos we can detect history of.
Comment: Lots of speculations, yes.
Answer: Yes, brane collision in M-theory, where our universe is pictured as the collision 'line' between higher dimensional surfaces.
Compare to angular momentum, which in non-higher dimensional theories is widely considered to only exist relative to the restframe of the whole universe (ie, net zero over all).  Edit: To clarify, in this picture the universe spinning would not create any change, in fact it would be meaningless, and similarly any translational momentum/velocity of the universe as a whole (in an open, 'saddle shaped' cosmological spacetime, this would seem possible).
Our star is slowly orbiting the Milky Way galaxy's centre, so our solarsystem's planetary orbits are spirals not circles, relative to the galaxy as a whole.
Relativity favours no specific frame of reference, it only provides a recipe to convert between different inertial (non-accelerating) rest frames.
You may find this discussion interesting: Can space be distorted without things that occupy the space being distorted?
Comment: Hmmm. If we define universe as everything, ie the multiverse, is that moving? After all your answer needs to redefine "the universe" as only a sub-universe. Not that you're wrong, only that this is subject to interpretation. Not a bad answer though.
Comment: @user4894 for the universe to be moving, it has to be moving *with respect to something*. If the universe is *everything*, then it can't be moving, because there's nothing left for it to be moving with respect to. At best you could argue that it's edges are shrinking or growing.


Is centrifugal force equal in magnitude to the centripetal force in the frame of a body undergoing circular motion?

Question: I was working out the minimum tangential velocity required for a swing to complete a full revolution and assumed the centripetal force is equal to the centrifugal force, so that I could set the weight force equal to the formula for centripetal (i.e. net force 0 at the top so swing won't fall). I'm pretty sure the answer is correct but doubt my reasoning.
Answer: If you set your reference frame to be fixed to the swing (let's call the origin the pivot point of the swing), you are now dealing with a rotating reference frame.
In a rotating reference frame, all objects observe a centrifugal force that pulls them outwards from the origin.  For any stationary object in your rotating reference frame (the seat on the swing), the sum of forces must be equal to 0.  The name we've ascribed to whatever force is equal and opposite to the centrifugal force is "centripetal force".
In your case, this magical "centripetal force" is some combination of the tension in the swing's chain/rope and the force due to gravity acting on whatever is in the seat of the swing.

To answer your question directly:
Your thinking appears correct, but beware that centripetal force is only equal in magnitude to centrifugal force when the object it is acting on is stationary in your rotating frame.  Furthermore, this case is what defines "centripetal force" in the first place.
And don't say "weight force" unless you're explicitly describing the apparent weight of an object (i.e. what the object "feels"; e.g. the weight on the seat of the swing).  It can be confused with the weight of an object (force solely attributed to gravity).
Answer: Your reasoning should be improved. In an inertial system there are no centrifugal force. You may in that system speak about a centripetal force, which is not a real force but an expression for how much net force is needed for the object to perform a circular motion. You probably mean that the expression for the centripetal force should be equal to the physical force at hand, the weight force.
Comment: Centripetal force _is_ a real force. Centrifugal force is an apparent force.
Comment: wgrenad - I try to tell students that the expression for the centripetal force $F_c=mv^2/r$ is not a real force, but rather an expression for how large force perpendicular to the orbit is needed. There is always som real force that create this, e.g. from a string or gravitational force.


Finding loops and double edges ASAP in configuration model random graph

Question: A common approach (at least in theory) to generating a random $n$ vertex graph  uniformly subject to having a given (feasible) degree sequence $(d_i)_{i = 1}^n$ is to use the configuration model, i.e. just attach $d_i$ half-edges to vertex $i$ and then pair up all the half-edges uniformly at random.
Loops or double edges can occur- if they do, repeat the procedure until you obtain a simple graph.
(The probability of obtaining a simple graph is usually small in practice so the expected number of restarts is large, and this algorithm isn't used much in practice for $n$ large.)
Now, the pairings in the configuration model can performed sequentially:

Select a half-edge arbitrarily.  (SELECT)
Sample a remaining half-edge uniformly at random.  (SAMPLE)
Pair up the SELECT'd half-edge and SAMPLE'd half-edge (to form or 'reveal' an edge).
If there are half-edges remaining then return to 1.

Of course, in practice, you'd also restart the whole procedure when the first loop or double edge is discovered.
You want to find any loop or double edge as quickly as possible.
What is the best (adapted) strategy for SELECT'ing half-edges in step 1?
Easy guess: SELECT any half-edge from a vertex with maximal (conditional) probability of being incident to a double edge or loop.  I.e. letting $H_i$ be the number of half-edges remaining at vertex $i$, then we choose the vertex that maximises $B_i = (H_i - 1) + \sum_{j\sim i} H_j$, where the sum is over vertices $j$ that already have an edge revealed to vertex $i$.  (B_i is the number of remaining half-edges which, if SAMPLE'd, would lead to a loop/double edge).  So you'd begin by SELECT'ing a half-edge from a vertex of maximal degree.
Is this strategy known to minimise the expected number of pairings needed to find the  first loop/double edge (assuming there is one)?
N.b. I'm specifically asking about the best way to select the half-edge in step 1 of the algorithm given above, (just for curiosity, not because I think it will yield a practical improvement over other approaches to simulating random graphs)
Small update:

I implemented this algorithm (c++), and it does offer a marked improvement over naively pairing up half-edges from vertices chosen in a lexicographic way.  The speed up is not as much as I'd hoped though (less than a factor of 1/2 for the degree sequences I looked at).
Computing the "badness" scores $B_i$ above isn't too computationally intensive.  In fact the scores can be updated on the fly and this seems pretty quick.
For anyone generally interested in efficiently sampling graphs Igor's link below or Blitzstein-Diaconis survey some methods (probably better).
Answer: Check out this work of Kim and Vu and references therein. WARNING: their algorithm is of very questionable use in practice, since the distribution is only asymptotically uniform, and since in the limit the Sun turns nova, this is not useful without bounds, which are sadly lacking.
Comment: Hi Igor, thanks, I'm aware these MCMC / rewiring type algorithms may be better than that described in the question.  I'm interested in the problem as stated just for curiosity, and don't really plan to use it in practice.


How can I duplicate a model and it's armature, so the rig still works?

Question: I need to make a duplicate of a character with its armature. When I try to duplicate the character, the duplicate remains associated with the original armature. When I make a duplicate of the armature, and try to associate the duplicate character with it, it fails. How can I accomplish this? Or is there a way to make duplicates of the character and the armature together, making the duplicate character associated with the duplicate armature?
Comment: this is done ^^
Comment: Thank you. So how do you combine the armature ? =)
They told me to move the model and the frame in a null block
And the create a group =/
Comment: I didn't actually know how to do this, so I had to experiment. The answer I posted is based on my understanding of your question, but since you also mention a frame, I'm not 100% certain. If it doesn't solve your issue, you should ask someone who knows both French and English to translate your original question.
Answer: Make sure you are in Object mode, then select both the character and the armature. First select one, then hold Shift while selecting the other, then duplicate with ShiftD
Comment: @TheMinecraftMan757 Yes, you have to be in object mode. In Edit mode, you only affect the object that you're editing, and in pose mode only bones can be selected. Thanks for pointing it out. I've edited my answer.
Comment: Do you have to be in object mode? That should probably be in your answer. (Unless my assumption is wrong.)
Answer: Update in Blender 2.8.
In Object mode, duplicate (Shift-D) the mesh and rig. In Object Data tab, in Armature field, with the new armature selected, click on the small icon on the right with the tootip "Display number of the users of this data (click to make a single user copy)". This lets you have a unique armature, that you can rename and manipulate without affecting the main one.
Comment: Can you please provide a screenshot of this? I'm looking at the `Object Data Properties` and not seeing anything related to Armature
Answer: I'm a noob, so this is more a workaround than anything. No matter how many times I tried to duplicate an object with it's rig (and maintain parent/child relationships) I failed. Finally I created a new file and used file -> append to bring in two copies of the target object. Had to redo the parenting (that doesn't seem to survive append) but I did wind up with two objects each parented to its own rig. This probably wouldn't be practical for an extremely complex file, but it worked for a simple one..


How can I speed up my old computer?

Question: I run Ubuntu on an old Dell Dimension 2400 (with only 2d accel.) and it runs really slow when opening up certain programs, like the Ubuntu Software Center.  So I was just wondering if there is any software that can speed it up.  Any response at all is appreciated.
Answer: You can use some less powerful DE's such as
Xubuntu/XFCE
Xubuntu is lightweight and contains a basic set of applications, Looks similar to gnome-panel .To install it
sudo apt-get install xubuntu-desktop
Lubuntu/LXDE
Lubuntu is very lightweight and is very basic. Resembles gnome-panel . To install it
sudo apt-get install lubuntu-desktop
Answer: I would use the Xubuntu distro as it has been designed to have a small footprint as far as performance goes.

How can I speed up my old computer?


What is the shorter way of saying - "I am sharing my understanding, in a hope to get corrected"?

Question: I am looking for a concise way of saying or writing, "I am sharing my understanding about blah, in a hope to get corrected". The phrase "IMHO" expresses part of the idea, but does not necessarily imply that the speaker is seeking correction.
Comment: It does, on the surface,  imply that the speaker wants to be corrected. Speakers can say a lot about of things without really meaning them. Speakers can also make indirect requests. *I hope you correct me* is an indirect request. It can be said ingeniously by someone who  really doesn't want correcting, but then so can a lot of things. One can also say the imperative *Correct me if I'm wrong,  but...* with zero interest in people actually correcting them.
Comment: What's more important than actual wording is perceived speaker intention. If you think a  person is sincerely inviting correction, then by all means do so, gently if necessary.
Comment: If I'm not sure I'm correct I add the question "What do you think?" after my statement.
Comment: What @AlMaki said.  "Here's my take.  What do you think?"
Answer: A common approach is to open with "Is this right?" and/or to close with "Is that right?"
Answer: You could say "... and I'm speaking under correction." To indicate that you may be wrong when you state something, another approach could be to say x,y, and z but to add "but don't quote me on this" at the end.
Comment: What is the opposite of preface?, I could not think of the word when you add something to the end of a sentence.
Comment: Your suggestion doesn't seem idiomatic to me. Is it? Maybe if you included supporting evidence or a reference, if it's not too much trouble.
Comment: @NeilMeyer I've seen the term "postface" used but only a few times.


How to get correct ferroelectric polarization from a supercell calculation?

Question: Although the ferroelectric polarization is usually calculated with primitive cell using the Berry phase method, supercells are sometimes needed for particular applications, e.g., defects, doping, etc. As polarization is defined as the dipole moment divided by the volume, using larger supercell would not change the polarization in principle. However, due to the multi-value nature of the polarization according to the modern theory of polarization, the polarization of crystals is only defined modulo the polarization quanta
$$\mathbf{P} = \mathbf{P}_0 + \sum_{i=a,b,c} n_i \frac{e\mathbf{R}_i}{\Omega}$$
where $n_i$ is an integer, $\mathbf{R}_i$ is the lattice vector, and $\Omega$ the volume. As can be seen, supercell would change the polarization quanta ${e\mathbf{R}_i}/{\Omega}$, thus change the calculated polarization $\mathbf{P}_0$. So I'm wondering if there is any recipe or remedy on   the polarization calculations with supercells.
Answer: I'm not sure if I understand your question completely, especially this part:

As can be seen, supercell would change the polarization quanta $e\textbf{R}i/\Omega$, thus change the calculated polarization $\textbf{P}_0$.

I'm guessing $\textbf{P}_0$ here stands for the polarization modulo $e\textbf{R}i/\Omega$. If that's the case, then I don't think using a supercell instead of a primitive cell will change the value of $\textbf{P}_0$.
The macroscopic polarization in the framework of the "modern theory of polarization" is calculated by taking the difference of values calculated with centrosymmetric structure and ferroelectric structure, assuming they are on the same "branch".
Using a supercell can indeed change the polarization quantum, but these changes take place in both centrosymmetric and ferroelectric structures and because the polarization of these two structures needs to be on the same "branch", they have the same changes. In the end, the effect of using a supercell is canceled out.
I hope this is helpful =)
Comment: I think you are right. If the polarization is calculated along the branch, the changes should keep same.
Comment: +1 Not a bad first answer. Welcome to our community! Thanks for your contribution and we hope to see much more of you!!! Hopefully the user that asked the question can clarify what that sentence means.


Does total reflection occur when the wave is incident at a grazing incidence (angle of incidence equal to 90 degrees)?

Question: I've posted the picture below from Introduction to Electrodynamics by Griffiths, 3rd edition. Is the phenomenon described talking about total internal reflection? Is it necessary for the incident angle to be equal to 90 degrees for total internal reflection to occur?
Answer: No and no: this seems to be considering "external" reflection, where the incoming and reflected ray travel in the medium with the lower index of refraction (in the air in the driving example); total internal reflection occurs rather when the ray travels in the medium with the higher index, and grazing incidence is not required, only an incidence angle greater than what would be required to refract the transmitted ray to 90 degrees.


Is this an example of the third person?

Question: Suppose Alice says to Bob, "I cook dinner with Bob." Is Alice talking about Bob in the third person?
Comment: Your example is a bad one. Why would she say this to Bob? _I cook dinner with Paul_ is what she could say to Bob. Better is to not mention who she said it to.
Comment: To clarify, Alice is referring to the same Bob; Alice means "I cook dinner with you".
Comment: _You_ is second person. That's why she wouldn't say it to Bob; while he's the addressee, _you_ is used. When she's talking to somebody else, like on the phone, even if he's still in the room, she can say that and it's third person, because it's **referential** -- it refers to something or someone. First and second persons are **deictic** -- their reference depends on the current context only and can change with the context.
Answer: Grammatically, yes Bob is third person. Hence even if Alice is speaking to Bob, then he will interpret her sentence as referring to some other person named Bob.
Comment: Unless you're addressing somebody by name (_Say, Bob, are you coming over tonight?_), all English nouns are third person. All pronouns except _I, you,_ and _we_ are also third person.

Is this an example of the third person?


Connecting points - plugin problem QGIS

Question: I have a question related to connecting points in QGIS. I am doing a project relating to transport. I have a layer of points (bus stops) and I have to connect them according to a timetable (as runs a bus line). Lines must proceed on the road (I have layer of roads combined with OpenStreetMap). I found the plugin Online Routing Mapper, but in this plugin are only two points (start and stop) and making line where each has 15 or more points is time-consuming. Can someone please me to prompt another method of how to do it? The plugin of which I mentioned would meet my needs if I could add more points.
Answer: There is an experimental QGIS plugin (you need to mark "Show also experimental plugins" in Plugins|Settings) named PointsToPaths that converts points to lines with verticies grouped by a text or integer field and ordered by an integer or date string field. You can try out it.
Answer: You can also try Points2One plugin. It converts points to lines.
Answer: Thanks for your advice. However, the lines must run along existing roads (marking the buses lines) and this is my main problem. Using these plug-ins I combine points with stright lines.

Connecting points - plugin problem QGIS


A Cramer's rule for $m \times n$ systems with unique solution?

Question: I have a system of $n+1$ equations with $n$ unknowns for which I know the existence of a unique solution. I am interested in writing out an explicit formula for this solution (I am not concerned with speed of computation). For $n \times n$ systems, one could use Cramer's rule (http://mathworld.wolfram.com/CramersRule.html). Is there an analogous result for $m \times n$ systems with a unique solution?
Comment: Are these all linear equations? If so, one of them is redundant and you can get rid of it
Comment: crammers rule is useful when case has as many equations as number of unknowns , where you have a formula.Roughly speaking it is for square matrices .

In other cases use  rank method [gaussian elimination](http://en.wikipedia.org/wiki/Gaussian_elimination)  to solve .

The above is for $m \times n $ system , if you can reduce equations back to $n \times n $ then you can thereafter use crammers rule
Answer: Yes. If you know the existence of a unique solution then what you can do is to calculate the determinant of submatrices until you find a nonzero one. In particular, select $n$ of the rows of your $m \times n$ matrix and compute the determinant. If it is nonzero then those rows are linearly independent and those $n$-equations suffice to determine the system. At that point, you can use Cramer's Rule on that sub-system.
All of this said, you could just use $\text{rref[ A|b]}$ and read off the answer just the same.


Do the D&D source books assume any particular ethical system?

Question: AceCalhoon's answer to this question got me thinking, is there any reason to believe that the D&D source books assume a specific system of ethics, such as deontology, virtue ethics, or consequentialism? Are these terms or ideas ever mentioned in a Player's Handbook or Dungeon Master's Guide? Do the D&D source books ever attempt to ground their alignment systems in a particular ethical system, or is it purely a mechanical distinction?
Answer: No, they don't. The origin of the ethical system encoded in the alignment system is teleological: different moral approaches to reality exist because cosmic forces created them, and your alignment is literally which cosmic forces you are aligned (aka "allied") with and working to further in the Cosmic War (whether you know it or not).
The Cosmic War is not a setting assumption that survived beyond 1st edition AD&D. Every edition after just held on to the idea alignments (which no longer indicate alignment with anything), with various tweaks to the specific list, because of tradition. No attempt to ground them in a real or reasonable ethical system has ever been attempted; only post hoc justifications and analogies have been attempted, and then only for the pragmatic goal of explaining them to a reader who has very likely never read Three Hearts and Three Lions or anything by Michael Moorcock regarding Cosmic Law and Chaos.
Comment: No detail on a "Cosmic War", but this is from the 4E Player's Handbook: If you choose an alignment, you're indicating your character's dedication to a set of moral principles... In a cosmic sense, it's the team you believe in and fight for...Alignments are tied to universal forces bigger than deities...with a lawful good alignment, you're on the same team as Bahamut, regardless of whether you worship that deity. Bahamut is not..the captain...just a...important player...Most people...haven't signed up to play on any team-they're unaligned. Picking...an alignment represents a distinct choice.
Comment: @DCShannon Yeah, 4e throws back to the original concept of what alignment means, but ironically without the underlying teleological justification for there even being "teams". It's a neat detail though, for sure.
Comment: I agree that its a teleological system, but not because the game setting had the need, but because the game itself had the need.
Answer: No.  The alignment system is largely a hold-over from early D&D that was evolved into it's own new beast, separate from it's original inspirations.  Oddly, the D&D alignment system is well-known even outside of RPG circles as a simple method of describing ethical caractitures.
In OD&D it was theoretically tied to the forces of Law/Good, Evil/Chaos, and Balance, which warred constantly in what was called the 'Cosmic War', though many people, myself included, have never played the game that way or had a campaign world with that kind of set up.
In game (in 1st through 3rd edition D&D, and I will also 5e if I ever play it again and use the actual alignment system), I always tell my players (or ask my GM) what ethical system we will be using for the campaign.  I have played with multiple different implementations of all the ethical starting points you mention, as well as other, and they all work quite well.  Utilitarianism tends to make for a bleak, awful existence as man's alignment and ultimate fate is a matter of luck (and skill, but that just makes it worse).  Deontology works well, and can generally just fade into the background if you don't want to focus on alignment too much.  Virtue ethics is awesome and totally captures the medieval genre and feel.  Psychological Egoism is hilarious.
The point I'm trying to make here is that once upon a time it was initially kinda grounded in a particular fantastical ethical system, but then it stopped being that way.  Furthermore, the way it evolved was a good thing and allows you to tie it easily to any ethical system you'd like in your particular game world.  Some particularly strange and poorly thought out systems-- like moral relativism-- don't work very well, but the systems you listed and many others are fully compatible with all editions of D&D (not including 4e).
4e decided Law=Good and Chaos=Bad, and thus is a little different.  This sort-of re-grounds it in the original inspirations for the system, but because people are no longer familiar with the particular tropes necessary to engage with the game in that alignment system it usually just ends up either ignored or Hobbesian.  Certainly the Law=Ultimate Good thing supports a Hobbesian approach.
Comment: luck *and skill*.  Don't go devaluing the self-replicating intelligent machines.  Also it's far less bloody bleak than virtue ethics I tell you what
Comment: @JackLesnie XD fair enough.  Virtue ethics is more fun though.  And the skill component just makes things even bleaker, but that's the part that make Utilitarianism fun so...

(Really it's only *super* bleak in high-op games as a non-caster, because it just reinforces that nothing you do really matters)
Comment: @Miniman Alignment goes Lawful<--Good<--Unaligned-->Evil-->Chaotic

Oh hey!  Nevermind, I was wrong, my GM must just have been a former 4e player or something.  Sorry, haven't played 5e much, nor studied the rules ^^;
Answer: No, that's up to the players involved in the campaign.  The alignment system in D&D is not meant to be a psychological profiling tool.  Its simply meant to give some ability for players and DMs to generate scenarios where morality of an action can come into question.  There may be no question that Lancelot the Paladin is a Lawful good Scout in good standing.  What happens to him when he is forced to kill a hostage in order to avoid epic destruction and loss of life?  The "Answer" is irrelevant because the answer I give in my campaign, is different than in yours.  Certainly Michael Moorcock's books are a source of inspiration for the original concept (and if memory serves was mentioned in the DMG).  I believe that alignment was added to the game not to inject a "cosmic war" concept but to provide a framework for players to have an action boundary to allow for dramatic actions to occur.  It should be pretty obvious (even if the debates on an action define it during the game ) where each of the alignments stand to the players. I've likely spent no more than 2mins explaining the idea to new players.    One thing that has been removed from the game (but I firmly believe in) is the XP penalty for out of alignment actions.  This is not a punishment penalty in my view but a way to reflect that out of alignment actions would have real consequences.  Thus when Lancelot puts a sword thru Princess Penny's tearful blue eye to stop Dastardly Dan from destroying the kingdom, it's an actual sacrifice that made Lancelot spend quality time thinking about his commitment to lawful good (and therefore drop a little XP).


Own compiled linux kernel does not recognize partitions

Question: I have compiled a linux kernel 5.4.81 as monolytic kernel.
I added drivers to access the disks and I added drivers to access msdos partitions and ext4 filesystem.
I installed it on /dev/sdb6 by copying it to /dev/sdb6/boot and configured grub with:
menuentry "Teddix" {
    echo 'Loading Teddix'
    insmod gzio
    insmod part_msdos
    insmod ext2
    set root=(hd1,msdos6)
    search --no-floppy --fs-uuid --set=root 5312030c-4457-4ad0-88cb-bf3488886e90
    linux (hd1,msdos6)/boot/linux-5.4.81 root=UUID=5312030c-4457-4ad0-88cb-bf3488886e90 ro
}

I can boot this kernel but then I get the following error:
ata5.00: supports DRM functions and may not be fully accessible
ata1.00: supports DRM functions and may not be fully accessible
ata5.00: configured for UDMA/133
ata1.00: configured for UDMA/133
scsi 0:0:0:0: Direct-Access     ATA       Samsung SSD 860   3B6Q PQ: 0 ANSI: 5
scsi 4:0:0:0: Direct-Access     ATA       Samsung SSD 860   3B6Q PQ: 0 ANSI: 5
VFS: Cannot open root device "UUID=5312030c-4457-4ad0-88cb-bf3488886e90" or unknown-block(0,0): error -6
Please append a correct "root=" boot option; here are the available partitions:
Kernel panic - not syncing: VFS: Unable to mount root fs on unknown-block(0,0)
Kernel Offset: disabled
---[ end Kernel panic - not syncing: VFS: Unable to mount root fs on unknown-block(0,0) ]---

I also tried with /dev/sdb6 as root partition, but there is no difference.
Did I miss some drivers or modules?
Edit: What I do not understand too is: Why does this kernel not list all possible root partitions? I would expect a list there before the kernel panic.
Answer: You will need an initramfs if you want to use UUID=, LABEL=, or /dev/disk/by-* device names.
The Linux kernel natively supports only three types of root device names: PARTUUID= (using the GPT partition GUID, not the filesystem UUID), PARTLABEL= (again using the GPT partition label, not the filesystem label) and the traditional /dev/sdX. Anything else has to be handled by initramfs.
(For msdos disks, as long as they have a non-empty "NT disk signature", you can still use a pseudo-PARTUUID which you can find out from lsblk -o +partuuid or blkid. You can use fdisk expert-mode command i to set the disk identifier to some random number if it's currently blank.)
Comment: Thanks for your answer. So UUID can't work for me. So I also tried with PARTUUID= but this gave me the same effect: Kernel panic.
Answer: One has to activate the following drivers:

SCSI-hardware-driver (hardware dependent, correct here)
SCSI-disk-driver (CONFIG_BLK_DEV_SD=y, missing)
Partitiontable-driver (CONFIG_MSDOS_PARTITION=y)
Filesystem-driver (CONFIG_EXT4_FS=y)

Recompile and reinstall linux kernel.


Whether or not to use a multilevel (hierarchical) model

Question: I have a dataset, and I am not sure whether I have to use a multilevel(hierarchical) model or not. Suppose I have 10 dishes with a cell culture in each dish, and of these 10 cell cultures, 5 are treated in a certain way (e.g. pharmaceutically) and the other 5 are control. Now I take each cell culture, and measure for each culture lets say the diameter of 20 cells. I want to examine the effect of the treatment on the cell diameter. My first approach would be to conduct a simple t-test with the 100 cells from 5 treated cultures one the one and 100 cells from 5 control cultures on the other hand. However, I now came across multilevel models and I am not sure, if a multilevel model is better, as cell diameters are more similar within each culture. One the other hand, the independent variable (treatment) does not differ within, but only between cell cultures, so I am not sure, whether I have to use a multilevel model (with cells on level 1 and cultures on level 2) at all. I would be thankful for your advice.
Comment: Thank you all very much for your kind and helpful answers. I have to admit, that I changed the description of my experiment a bit to clearly elaborate on my problem - so my cultures are alright, but thanks for concern. Again, thanks a lot.
Answer: For the situation you describe you don't need to consider a multilevel model. Each dish only sees a single drug treatment so there is no way to distinguish whether cells are small because of the dish; all you have to compare is drug/control. If you care about differences in cell size among dishes as an issue in its own right, then you could include dishes as a random factor in your model. I'm a bit worried about why your cell diameters seem to differ among dishes. Examine the reproducibility of your culture conditions; it's very important to get them under control.
The classical case for a multilevel model would be if there is a comparison of treatment and control within each of multiple plots of land; then plots would be the higher level, with treatment/control below. If you had some fancier experimental design (say, all dishes got both experimental and control treatments but in different orders) then you might consider multilevel. But not in this case.
Comment: Another way to put this is nested ANOVA.


Power factor leading and lagging effect on power

Question: Why doesn't power factor leading or lagging bring any effect to apparent power? While for reactive power, lagging will result in a positive answer and leading will result in a negative answer?
Answer:
Why does power factor leading or lagging did not bring any effect to
apparent power ?

Apparent power is \$V_{RMS}\times I_{RMS}\$ hence, PF does not affect it.

While for reactive power, lagging will result in a positive answer and
leading will result in a negative answer?

Image above from wiki power factor.
Answer: To do this, you have to know what the apparent power is. The apparent power in no way takes into account the phase position of the current in relation to the voltage. It is read and calculated from 2 different measuring devices. The specification of the apparent power alone has as little informative value as, for example, the specification of the screen diagonal for televisions.
The reactive power is the power that is returned to the network as soon as it has been withdrawn from it. For this you need the simultaneous consideration of current and voltage. You have to multiply the voltage value and the current value with each other at every moment. The negative power is the reactive power.
The same applies to the real power, only that the sign of the real power is positive. This means that the product of the voltage and the current is positive at every moment.


Can a capillary opening lift a huge load?

Question: The problem I pose is about an application of the Pascal law and capillary action. It is based on a plot which I describe hereafter.
Imagine a one side sealed cylinder with a piston on the other side. The sealed side is up, so the piston can move downwards. The cylinder is filled with water. Now the piston must move down and fall out due to hydraulic pressure p=rogh.
But suppose now there is a tiny opening of diameter of the order of few nanometers on the top. It delivers a curved surface which leads to a Laplace pressure under a curved surface p1=2 s/R where s is sigma – the surface tension (it can be around 10 atm if the opening is 0.3 micrometer). This pressure must be delivered everywhere in the water according to Pascal low and hence on the surface of the piston. This pressure is negative against the hydraulic pressure and hence the full pressure will be P=p-p1. While p1 is much greater than p the piston would not fall down and even would be able to support a huge load.
Now while it seems reasonable I haven't seen it described in textbooks. My questions are:
1.  Is this physically true and ---if yes
2.  Has anyone seen this described in a textbook and ----
3.  Has anyone seen the experiment described to be performed ---
4.  If we have N openings with the same radius will this give p1=2sN/R
Comment: This is a solid block of text, not easy to read.
Answer: I have no idea about actual calculations you are getting into, but all the trees grow via capillary action. That means, it lifts tons of weight, dozens of meters above the ground, but over a period of time.
If you sit on a tree, the tree is not going to stop growing. It will lift you but very slowly.
Comment: My concern are the actual calculations and not observations from nature.
Answer: In the first case when the cylinder is closed, a sub-atmospheric pressure must develop inside the cylinder so that pressure difference across the piston may be able to support the weight of water+piston. When you open a hole into it, you provide a vent to atmosphere, and because of this sub-atmospheric pressure cannot develop inside the piston (if it did, air would rush in through the vent), and so the piston would continue to fall forever (along with water on top of it). So, contrary to what you said, if you wanted to support large weight by this method you must take care to seal off any vent to atmosphere.
Comment: I am not interested in the first case at all. Nevertheless you neglect the weight of the water. Water + piston will eventually make more pressure than athmospheric pressure from below,  even than vacuum is build above.
Comment: Unable to understand your comment.
Comment: What I mean is that pressure from the opening of 0.3 micrometer is 10 atm if it is transmitted according to Pascal law in every part of the liquid. Plus the 1 atm from below of the piston its 11 atm up. 1 atm from the air through the opening its 11-1 = 10 atm up. When you have no opening you have just 1 atm from below and zero from above = 1 atm up. So the meaning of all this is that the opening can hold a tenfold load compared to cylinder without opening. And more I think that 2 openings can hold 20 times more.
Comment: Maybe I should mention that when I mean water I mean that it is free from dissolved air. Otherwise there should be cracking under the pressure. This as I think is the idea of the so called cohesion tension theory (about the ascent of water in the plants) devised 100 ago by Dixon and Joly but it was not based strictly on any physical law (just qualitative suggestions and experimental).


Why do VPN connections break when I sleep my computer?

Question: On the limited (Cisco) VPN systems that I've used, the VPN connectivity is broken when my computer is put to sleep and must be re-established on wakeup. Is this a necessary property of VPN?
Comment: ** Disable Power Saving mode**
all I had to do was go into my ethernet adapter setting a disable the power/energy saving mode. it has been renamed once before so.
navigate to the adapter and then into "Configure" where you should find a advanced tab with a ton of different switches.
** Disable Power Saving mode** This will literally stop the hardware adapter from switching off during nap time.
it has nothing to do with the VPN. VPN cannot go through a powered down adapter.
Answer: Because VPN is a not a stateless communication system like HTTP, it needs to have continuous connection. And since the OS is sleeping, it'll no longer maintain the connection.
Comment: Is there a specific reason it can't maintain the session though a sleep cycle? Is it a question of sleep duration?
Comment: Note that you can probably handle a quick sleep; it can be handled like any other network glitch.
Comment: I wonder if this is going to change somehow. Where I work, the power settings on all of the PCs are set to "never sleep" so users can VPN to their machines which isn't very "green."
Answer: I think Bob Somers hit the answer, but there's another possibility. You could be sleeping your computer through the rekey interval, which breaks the connection.
VPNs negotiate a session key for every connection. This key gets "stale" after a while, and may be vulnerable to attack if enough data has been passed with it. To maintain the security of the VPN channel, there is a rekey interval where both ends automatically renegotiate a fresh session key. This is usually transparent to both sides of the tunnel, assuming both sides are connected. When you put your computer to sleep, it's possible the rekey interval is up during that time, and the VPN host closes the connection because it can't renegotiate the session key.
Answer: I'm not exactly sure but my guess is because the network card (NIC) has been disconnected from the operating system stack. The fact that a VPN connection remains open is not so much hardware related but more by the software layers (think back to your 7 layer architecture).
Comment: It is hardware...
VPN will stop when the adapter setting of "Power Saving Mode" is enabled.
this literally is the config of the hardware adapter.
and this kills VPNs when sleep enables
Answer: Bingo. Most VPN software is implemented using the TCP/IP stack, which is not stateless. They have timeouts associated with the connections and when you put your computer to sleep there's no way to keep that connection alive.


Remove and add markers with zoom in and out

Question: In one of my project (IOS) I need to add functionality that when user zoom out the google maps it adds marks within that coordinates. Right now I am zooming the map on user current location and the nearest location. So when user zoom out it will add more locations that are in that region. I dont know how to achieve that please help me in this.
Answer: As it looks like, GoogleMaps SDK leaves the responsibility for storing and managing markers entirely to you.
So what you need to do is keep a separate storage (like NSMutableArray or NSMutableSet or whatever best suits you) for all your markers and using projection.visibleRegion property on GMSMapView to decide which to remove (or add).
Check out GMSProjection reference on GoogleMaps SDK docs
In order to catch up with zooming in and out, you have to declare your GMSMapViewDelegate and implement this method:

- (void)mapView:(GMSMapView *)mapView didChangeCameraPosition:(GMSCameraPosition *)position

That's about it.


Who is the legal owner of code in a bitbucket repo?

Question: Who is the owner of code uploaded from a personal account to a private Bitbucket repository and there is no license defined in the code, or there is only the name of the author in the header of the code? (based on that there is no any signed contract between the repository owner and the code owner)
Comment: Who actually wrote the code that was uploaded?
Comment: If you have no license, and don't have the copyright, you can't use the code anyway.
Comment: The code created by the commiter (the person uploaded the code)
Answer: The actual creator(s) of the code own the copyrights, unless a written contract transferring the copyright was signed, or unless the work was done by an employee (or employees) within the scope of his or her or their employment, or Unless the code was a "work-made-for-hire" and is so specified in a written contract signed by the creator(s) that confirms with the jurisdiction's law.
This is always true, no matter who uploads code, or what repositories it may come from or be sent to. This is basic copyright law and is true in all Bern Convention countries, except that the provisions on employer ownership may vary somewhat. They are correct for US law.
It is best practice to always include a copyright notice or statement in code being released or placed in a repository, to make it clear just who owns the rights, even when this is not required. Some open source licenses require such a statement.
It does not really make sense to upload code to a repository without a license, except perhaps as an educational example that may not be reused. Only the rights-holder, or someone with permission from the rights-holder,(including an appropriate license) should upload such code in the first place.


What is the sum of all 4 digit numbers with distinct digits?

Question: I tried solving this problem and got into a little trouble.I had done questions similar to this one before where the digits(that I could use) were specified so I thought maybe I could solve it like them. so I started writing all 4 digit numbers with distinct digits using the digits:  $${0,1,2,3}$$
and I found the number of occurrence of each digit in units place, then the 10's place ,100's place and finally the 1000's place and I got this as the sum of all numbers : $$4\times(0+1+2+3)\times(1+10+100) + 6 \times (1+2+3) \times 1000$$
and with some more thought into it I found out that I could Write the sum of the numbers in the form $$(A\times(0+1+...+9) \times 111)+(B\times (1+2+...+9)\times 1000)$$
now my question arises I'm able to find the value of B which is just $9 \times9\times8\times7$ devided by $9$ which is $9\times8\times7$ but I really can't find the Value of A because it seems to have no relations with  the value of $9 \times9\times8\times7$ so I could use some help here; or maybe what I've done is not the best way of doing it. I will appreciate your answers.
Comment: Should i be more elaborate in my answer?
Comment: @AkshajBansal that would help a lot especially if you consider all digits in there!
Comment: Do u wnt an answer when we consider $0,1,2,3,4,5,6,7,8,9$ as digits for the 4 digit numbers?
Comment: @AkshajBansal yes just as the title suggests, I used 4 digits so that I could write down all of the digits myself and use it to find the formula for the general case.
Comment: Repetition not allowed ,right?
Comment: @AkshajBansal Right.
Comment: Hi emad sorry for being late i just edited my answer because earlier some miatake were commited the solution refers your method and gives the correct answer thank you.
Answer: Think like this
For the sum of all digits at thousands place the expression would be
$1000(1+2+3+4+5+6+7+8+9)×{9\choose3}×3!.$
Explanation-
You surely dont have $0$ therefore now only $9$ possibilities left for thousands place suppose you take $1$ at thousands place now you need to see how many times does it repeat i.e $9\choose3$ since you need to select $3$ digits out of $9$(0,2,3...9) possibilites also you need to see in how many ways those $3$ would be arranged that is $3!$
Now-
For hundreds, tens, units place there would be two separate cases for $0$ coming at those places and ${1,2,3...9}$ coming at those places.
For $0$ we see it repeats $9×8×7$ times (but it doesnt affect the sum)
For digits ${1,2,...9}$ they repeat $8×8×7$ times ,now how does that come since you dont have $0$ at thousanda place so only $8$ possibilities and for other place $8$ possibilities again (since you can put $0$) now and $7$ possibilities for the last place .
$9×8×7(0)(111)+8×8×7(1+2+3+4+5+6+7+8+9)(111)$$
And the final answer-
$\big(1000(1+2+3+4+5+6+7+8+9)×{9\choose3}×3!\big)+
\big(8×8×7(1+2+3+4+5+6+7+8+9)(111)\big)$
$=1000×9×8×7×45+8×8×7×45×111=24917760$
Or According to your question
$A$ is $8×8×7$ and $B$ is $9×8×7$
Thank you
Comment: I know that and besides I used the digits 0,1,2,3 to find a formula for the real question which is the sum of all 4 digits with distinct digits.
Answer: You've done well so far.  You're getting stuck where you assign a digit to one of the places other than the $1000$'s place.  Note that if you assign $0$, it contributes nothing to the sum, so you can ignore that possibility.
Once you've assigned a non-zero digit, you have to assign $3$ more, but you can't assign $0$ to the $1000$'s place.  So, calculate the number of ways to assign the three digits, without regard to the restriction, and then subtract the number of ways to assign $0$ to the $1000$'s place.
EDIT
There are $9$ ways to assign a non-zero digit to a place other than the $1000$'s place.  That leaves you with $9$ digits to assign to $3$ places, which can be done in $9\cdot8\cdot7$ ways.  We have to subtract the $8\cdot7$ ways where we assigned $0$ to the $1000$'s place, giving $9\cdot8\cdot7-8\cdot7=8\cdot8\cdot7=448$ ways.
Answer: Take the $k$th column place.  For every digit $a$ in the $k$th columns there should be  $9*8*7$ digits in the remaining three columns.  $a$ appears in the $k$th column $9*8*7$ times.  So if you add up all the terms in the $k$ths column you should get $9*8*7(0+1+2+..... + 9) = 45*9*8*7$ and so the $k$th column provides $45*9*8*7*10^{k-1}$ to the sum.  So the total sum should be $45*9*8*7(10^3 + 10^2 + 10 + 1) = 45*9*8*7*1111$.
Except that adds up "four digit" numbers that have $0$ in the thousands column which should not be included.
So.... we subtract the sum of all strings that start with $0$.
Now by the same reasoning, if $a$ is in the $k$th column there are $8*7$  options for the remaining two columns (they can't be zero) so the sum of all those strings is $8*7*(1+2+3+ ..... + 9)*(10^2 + 10 + 1)= 45*8*7*111$.
So our sum is $  45*9*8*7*1111- 45*8*7*111 = 45*8*7(9999-111)=45*8*7*9888$
Comment: that is true and the answer also matches my book's.thank you!


Do Black Holes emit Hawking radiation while engulfing normal mass?

Question:
As I read on various sites ,all of them say that Black Holes emit Hawking radiation as one of the instantly created virtual particles near Event Horizon escapes from it.One goes inside it and other particle escapes and this particle(which escapes) constitutes Hawking radiation. So while engulfing Normal BODIES,does it emit something,so that the information preserves somehow?
Comment: No offense to Hawking, Susskind, John Rennie and Wikipedia, but not even a cup of coffee preserves information about the sugar falling into it, so take the "information paradox" with a strong cup of tea, you will need it to find anything of physical relevance in there... to this day I have not (and neither have any of the aforementioned).
Comment: @CuriousOne: are you claiming that quantum mechanics is non-unitary when you dissolve sugar in coffee? Because that's the technical definition of  "information loss". If you think the dissolution of sugar is non-unitary and you can prove it experimentally, go ahead and do it and get your Nobel Prize.
Comment: @PeterShor: I am not claiming anything of the sort. What I am saying is that a black hole is the worst system to test whether it is or not. The entire discussion is completely unphysical. Indeed, one can't even decide that question using a cup of coffee and sugar cubes, neither theoretically nor experimentally, so it's really just a matter of having the boring phrase "cup of coffee" replaced with "black hole" while ending up with the same non-physics. If somebody comes up with a precision tabletop unitarity experiment, then I am all ear.
Comment: @CuriousOne: just because it's impossible to ***test*** whether dissolving sugar in a cup of coffee is non-unitary, it doesn't mean that you can claim that the information is not preserved, as you did. The laws of quantum mechanics, which preserve information, have passed every experimental test performed so far.
Comment: @PeterShor: See the definition of science for the practical distinction between a hypothesis and "it's not even wrong". This is the latter.
Comment: @CuriousOne: it's not worth my time arguing with somebody who obviously does not understand science. No chemist would believe that quantum mechanics is violated when dissolving sugar in a cup of coffee.
Comment: @PeterShor: I didn't coin the phrase. Pauli did. Please take it up with him.
Answer: Have a read through the answers to An explanation of Hawking Radiation to get a better idea of what Hawking radiation is. The popular explanation of virtual particles falling into the black hole is just a metaphor and should not be taken literally. The proper explanation for Hawking radiation is I'm afraid far more complicated than that.
A black hole does not necessarily emit anything when an object falls into it. If the falling object has a non-zero angular momentum there will be gravitational waves emitted, but these are unrelated to the information paradox.
At the moment there is no widely accepted explanation for the information paradox. There are many suggestions including a recent one from Hawking himself, but none of them are sufficiently convincing to have become widely accepted.
Comment: Aren't there ***always*** gravitational waves emitted when something falls into a black hole? (Except when what is falling in is a perfectly symmetrical spherical shell of matter.) I thought that if the mass falling in is not symmetrical, the black hole has to redistribute the mass that falls in, and that this process creates gravitational waves. Certainly, black holes colliding head-on produce gravitational waves. [See this Caltech web page](http://www.tapir.caltech.edu/~sperhake/Research/BH_Simulations/head-on.html)
Comment: Thanks for explaining now I have 2 questions:   1) ingoing bodies have nothing to do with Hawking radiation? It is a spontaneous process?    2) I learnt in my school that red shift is just the loss of travelling wave's energy,so the wave expands but  this wave has lesser energy ,so the wave with Infinite red shift will become just NULL kind of thing,so it wont escape,but with finite Red Shift one can escape,but still it must have enough energy to escape Event Horizon? Is this correct?
Comment: @Śȟůbȟäm Ŗäĵpůț: There's no theoretical minimum to the energy a photon can have.
Comment: But at zero energy ,The photon simply doesn't exist so infinite red shift means zero energy,So the photon will be gone.


Created custom keyboard, installed it, but can't start using it

Question: I'm currently on Windows while my MacBook gets repaired. I used Microsoft Keyboard Layout Creator to tweak Arabic (101) so that layout matched the layout of Arabic letters on my Macbook.
I installed the keyboard, Arabic (101) - Custom, as per the instructions here, but I'm now stuck. When I try and add the keyboard under Preferred Languages in Settings, I can choose one of several Arabic language by country, and when I click on options, I have the ability to 'Add a keyboard', which is where my custom keyboard turns up. If I choose to add Arabic (101) - Custom as a keyboard, and then switch to Arabic, the keyboard is still layed out as per the original Arabic (101). If I remove the original and add my Custom keyboard, the whole language disappears from my Preferred Languages.
Can someone help with what I'm doing wrong?
Answer: Apparently there were updates pending on my Windows machine, and the custom keyboard now seems to work after a restart. When I removed Arabic (101) and added the Custom, the keyboard layout was applied successfully without the whole language being removed.


In Ethereum: Who does keep track of users Ether balance and update them?

Question: It's not clear to me how the user's account is updated, and who does it. For instance, who update users account balance?
Answer: The miners do.  It's how the whole system works, whoever gets to add the next block gets to say what changed regarding account balances.
Answer: Everyone running a node keeps track of and updates every account's balance. However, updates to balances do not occur unless they have been confirmed by being included in a block. Blocks are created through the process of mining; mining is performed by miners.
This is why one can consider Ethereum to be trustless: everyone can see for themselves that miners are doing their job (or not) and no node needs to  worry that other nodes are making up chain state since every node is performing (more or less) the same calculations and should always arrive at a consistent conclusion about the state of the blockchain. However, nodes do need to rely on other nodes not to be hiding newer information from them.


How accurate is the saying that fat loss occurs first in the place it was put on last, and vice versa?

Question: I've heard that you lose fat in the areas it was last gained. Similarly, the last place you'll lose it is the area you first put on the weight. What is the reason behind this mechanism?
Answer: It is both very accurate, and very inaccurate.
When you gain fat, you gain fat in all the fat repositories. On the same coin, when you lose fat, you lose it in all the same places at once.
Important: You can't pick and choose where to gain/lose fat.
The reason why it seems like you gain it in certain areas more than others (like belly and butt), is because these places have the largest fat storage capacity. But equally, this is where you lose more as well, when you lose fat.


How $X$ noise propagates through controlled-$S$ gate

Question: I am struggling to find how an $X$ noise propagates through a controlled-$S$ gate.
Here`s the circuit.
Answer: If I understand you correctly, you would like to find how an $X$ error before a $cS$ (controlled-S, where $S=diag(1,i)$) gate behave after a $cS$ gate.
What you should solve is simply the equation (I assume an $X$ error on the control as in your circuit):
$$cS X_1 = E cS \Leftrightarrow E=cS X_1 cS^{\dagger}$$
Solving it on mathematica gives the following two-qubit matrix for $E$ (in the two-qubit computational basis):

In practice if you decompose $E$ on the Pauli-matrices you will have a non-trivial expression (linear combination of tensor products of $I$, $X$, $Y$ and $Z$ operators) as $cS$ is non-Clifford. Finally, given the shape of the $E$ matrix above, I think in that case the error propagates on the two registers after the gate.
Comment: Thanks!! I sent you a request.
Comment: May you share the code on wolframalpha.com?
Comment: I am not sure of how I could give it to you through wolfram alpha website. However, you can contact me on linkedin and I can provide you the code.
Answer: In Quirk, the trick to finding the equivalent operation after an intermediate operation is to check that their phase kickback exactly cancels even when operating on entangled qubits:

Note that the bottom qubit ends up OFF. This is a proof that X before CS is the inverse of $S \cdot X \cdot CZ$ after CS.


What happens when a MOSFETs gate is left open?

Question: I was working with an interface with a MOSFET and I ran into a case where high impedence might be an input to the MOSFET's gate, essentially leaving the gate open. Would the MOSFET exhibit the same electrical & physical behavior as if the gate was connected to GND?
Comment: You can wave your hand over it and the voltage from your hand through the air can turn it off and on...
Answer: Not even close. The complete opposite, in fact.
A high-impedance floating N-channel gate can pick up charge from a variety of sources, (via the parasitic gate-to-drain capacitance, via handling - touching the gate with your finger, etc.) and spontaneously turn itself on. Any gate that could have a high-impedance control (a digital I/O from a micro, for instance) really needs a gate-to-source resistor to bleed off these spurious charges and keep the device state defined (=off) when the control is high-Z.
Answer: No, it will not behave the same.  It depends on the local environment and local leakage currents, stray electric fields etc.  In short it's not very predictable at what state the gate will be and then subsequently what state the channel will be underneath.  The bigger MOSFET's having larger gate capacitances will respond more slowly.
Put in a high value pulldown/up resistor to deal with this High "Z" state.


Occupy, enslave, or exterminate?

Question: What are the positive and negative consequences of each of these choices? The difference between enslavement and occupation is particularly puzzling to me: Enslavement makes a new town much easier to control and distributes the population among older, more developed centers. At a glance, there doesn't seem to exist a good reason to ever choose occupation.
Comment: That's the thing though, occupation preserves the old population; enslavement halves it, distributing the other half to your other settlements. New conquests are always unstable, and the troops you garrison scale in efficiency with the settlement size, hence enslaving allows you to leave half the garrison you'd normally need. The slaves themselves go to regions already secure, and I don't think they contribute to unrest by themselves; there is however some negative squalor thing going on, which isn't explained by the tooltips.
Comment: EXTERMINATE THE INFIDELS
Comment: Once slaves are added to your population, they are counted as full citizens (there is no difference between them and the previous inhabitants). They bring the full unrest/squalor that happens with increasing population.
Comment: @ChargingPun slaves only "count" as half a man. As Scipii I take Syracuse, it has 6000 pop, i enslave, it goes down to 3000, and my capital gets +1500 (my other city Messana had no general in it so it got nothing). If both my cities have generals (or none) they get +750 each.
Comment: In real life, I'd much rather my city be occupied than being made a slave to do dangerous manual labour for little to no food, so maybe (I haven't played the game so this is just a guess) Occupation gives a better morale/less chance of an uprising?
Comment: All of them. In that order..
Answer: Exterminate

Maximum Loot
Reduce population by 3/4 (to a minimum of 400)

PRO:

highest money option
easiest way to control large enemy settlements

CON:

loss of population
could take a while for the settlement to recover/become useful

Enslave

Regular Loot
Reduce the population by 1/2 (to a minimum of 400)
The population that this settlement is reduced by gets immediately distributed to all Governed settlements.
This settlement now has Slaves as a resource (for 20 turns), which boosts economy and population

PRO:

Distribute Population to specific areas
Create Slaves Resource to further boost populations
easier to control large populations
no loss of population

CON:

need to have govornors in place before taking the settlement to ensure that the extra population goes where you want it to

Occupy

Regular Loot
All population stays in new settlement

PRO:

no loss of population
new settlement stays at highest population

CON:

hardest to control

Summary
Exterminate when you need cash now or need to control a large foreign settlement.
Enslave when you want to increase the population of your core cities (both by slaves now and slaves for the next 20 turns).
Occupy when you have taken a town or other small settlement (especially one that will grow slowly), and you don't want to hobble its development.
Comment: You know, the trade resource thing actually makes enslavement seem better than occupation at both maintaining order AND boosting economy. You get the same income (no actual, net loss of citizens), and on top of that you also get a trade resource you wouldn't benefit from otherwise, not to mention the retainers your generals obtain (might want to add them to your answer). I used to think that a squalor penalty exists just because you enslave, but apparently it has nothing to do with that. Unless the conquest has some crazy good buildings I lack and won't get soon, I see no reason to occupy.
Comment: Enslaved pop actually counts as half only. For example if you Enslave a city with 6000, it gets reduced to 3000 and your other cities gain 1500 pop (because slaves are half a man apparently)
Comment: Personally I prefer enslaving a settlement to occupying, except when the population is around 1200-1999. If the town would be a candidate for growth soon, I take my benefit from having another Large Town at my disposal, instead of a less-useful Town. Most of the benefit comes by having access to the 2nd tier of economy buildings: port, market, mines +1, paved roads.
Comment: I always used exterminate on settlements that had a revolt and involved me getting them. Works great for quelling that city for ALONG time.
Answer: You exterminate if the town's infrastructure is of no use to you and you just want to loot and pillage, this one goes without saying.
You occupy if the city possesses buildings and resources that you actually want to use and get the most out of in the long run. If you capture a huge city with amazing infrastructure and buildings, you'll want to have as many people living there as possible to generate max tax revenue, etc.
You enslave if you would rather just distribute the population amongst your other settlements.
So the amount of unrest that results is inversely proportional to how much you want to get out of the city in the long run. To completely assimilate it with minimal damage costs you the biggest up front investment of garrisoned troops, whereas totally destroying it for some quick cash now will pretty much eliminate any resistance but also make it very hard to actually take advantage of the city's pre-existing buildings in the future.
Answer: Exterminating is most useful in the larger cities. Obviously, you're not going to gain any immediate income for exterminating a tiny town. But if the city either has a huge population, or a ridiculously high growth rate, you're better off exterminating the populace than dealing with rebellion and riots. Unless you plan to use the new city to replenish thousands of soldiers in your armies, of course.
One of my biggest problems in-game is controlling the growth of population. Particularly when you get farther from your capital, you already need to sacrifice construction of more useful buildings to build things to increase happiness. It's virtually impossible to control a far-flung empire with complete culture penalties and massive populations that are so unhappy that there's almost no way to not have them rioting and revolting.

Occupy, enslave, or exterminate?


database knowledge in ruby on rails

Question: I have learned the language itself as well as html/css/javascript. But with no database knowledge and (a bit)network knowledge. When I read the Pragmatic book on ROR, I found it confusing at the very beginning where they creates a project and setting up databases and models. Should I learn some database(and network knowledge)knowledge first in order to fully understanding the code and to fully grasp rails? If so, points out some book suited for these subject. Thanks in advance.
sorry for my bad english..
Answer: The short answer is YES. Even though ActiveRecord abstracts away much of the dirty work with the database, it is still important to understand what is going on. It will be useful, sometimes vital, when debugging, deploying, and/or maintaining.
Comment: thanks Ben, but should I learn some specific database like MySQL or rather learn the database theory in general? Thanks
Answer: Unless you're doing network specific stuff, all you need to know concerning networking is a basic understanding of how the HTTP protocol works.
Models have nothing to do with databases per se, they're just regular classes in your language, that happen to map to a database table in Object Relational Mapping (ORM) type frameworks such as Rails.
You can view them as the gateway to your database.
The database you'll be using is most likely a relational database. You don't need to know much about the theory, but you can get a very basic overview on wikipedia.
Probably the thing you'd need the most is a basic understanding of SQL, although the point of an ORM is to abstract that SQL away. You'll see glimpse of it when you look at the log file where you see the actual database queries. And when you get to more complicated, specific stuff, you might need some SQL.
In any case, you'll need general knowledge about databases and SQL in pretty much any type of development.
Answer: You almost certainly will need some general understanding of SQL to work with RoR (and as others have said, probably for programming itself).
I'm a fan of the book Simple SQL. It covers SQL generically, as much as it can. That is, it goes over the basics of SQL itself rather than the details of one or the other SQL implementation. Also, the majority of the examples are web related which should fit you well.
Answer: I'd agree that a basic (at least) understanding of SQL is essential when doing any sort of database work. I found the "Sams Teach Yourself SQL in 24 Hours" book helpful when i first started dipping into database work. It covers pretty much everything you need to get you started.
My advice would be to pick a DB (MySQL, Sqlite, MsSQL etc.), learn some of the basics for that particular DB (i found it useful to know how to create new users and set permissions), learn some generic SQL and really get to know what ActiveRecord can and can't do.
Hope this helps a little.
Answer: What is really important is to understand the concept of MVC (model view controller)
I recommend you reading this getting started guide
RailsGuides in general are really helpful and more condensed than the book. You might prefer it

(source: rubyonrails.org)

database knowledge in ruby on rails


Rectangle and circle shapes alignment/sizing

Question: I am having a hard to figure out what are the right proportions between a rectangular and a circular shapes and how to horizontally align them. These two shapes are two icons but they always look wrong to me. I wonder if there is a clear rule to avoid perception influence and align and size them right.
Comment: I always align different shapes to the exact center. Only way to do it if it's the same size but a different shape imo.
Comment: As it stands there seems to be a mix of (1) a specific question about horizontal alignment of those two shapes (2) a general question about sizing and alignment of two classes of shapes. I think for this to be answerable within the SE schema, you need to focus down to a single question.
Answer: I think you are asking about the graphical implementation of an overshoot, commonly used in typography. In short, overshoot is added to letters like A and O (pointy or round—like circle in your example)  to visually make them look the same height as f.ex. H and X ("flat", rectangle-like).
The sources of the Wikipedia article suggest overshoot of 1–3 % or 5 % for O. Keep in mind these are suggestions for typefaces and differ from font to font. You're better of trying different ratios and choosing one that is the most æsthetically pleasing in your opinion. Of course you could use the given percents as a starting point.
In practice:

Align your rectangle and circle by their centers
Choose a ratio—say, 5 %
Resize the circle by its center using double the value you've chosen (ie. 10 % in this example) as in typography the 5 % is added only vertically with the baseline as pivot point.
Comment: Yes, I am looking for something like this but I don't get why you want me to resize the circle back to its size (step 4)
Comment: @piede true, me neither. Must've been a brainfart of some sort.


Difference between machine image and instance in virtualization?

Question: These two terms have been used frequently in virtualization. Is there a subtle difference? Or are they totally different things in the context of virtualization.
Answer: Image is static data containing the software (the OS and applications together with their configuration and data files etc) that the virtual machine will run once started. It is usually stored on disk (though you can store it anywhere you like).
Instance is a running virtual machine. It has been started from an image and it is capable of  running an OS and processes, performing computations, I/O etc. Unlike static image, instance is a dynamic thing which does some work and with which you can interact.
Answer: Typically, "instance" refers to a specific, well, instance of a server. It is one virtual machine. The "image" is the filesystem that is used to initialize, or create that instance.


What did H.L. Mencken mean when he wrote this?

Question: Most people are unable to write because they are unable to think, and they are unable to think because they congenitally lack the equipment to do so, just as they congenitally lack the equipment to fly over the moon.
Comment: It means most people are too stupid to write.  And "congenital" means they were born that way.
Comment: A cursory search doesn't turn up the quote in Mencken's writing. Most likely it's from this: "I doubt that the art of thinking can be taught at all.... It is not acquired, but congenital. Some persons are born with it. Their ideas flow in straight channels; they are capable of lucid reasoning; ... when they write anything it is clear and persuasive. ... The rest of God's children are just as incapable of logical thought as they are incapable of jumping over the moon. Trying to teach them to think is as vain an enterprise as trying to teach a streptococcus the principles of Americanism.
Answer: This hinges on congenital:

having a particular trait from birth or by firmly established habit

So, the phrase simply means that people can't write because they can't think, and they can't think because they lack the equipment (brain) from birth/habit, just as they lack the equipment (rockets/etc.) to fly over the moon. It's an insult: they clearly lack the equipment necessary to fly over the moon, which tells you how little brain they have, and thus they have zero ability to think, which is why they can't write.
Somewhat brutal, in my opinion, but that's the intention!
Answer: Prof Yaffle has the literal meaning of the quote, but I believe Mencken meant

"You cannot teach a person to write".

I don't know the context of this quote, but I strongly suspect it was not so general as it sounds, and carried subtext something like this:

"Go away and stop asking me how to become a writer. If you can't
  think, you can't write, and the fact you are asking me how to write
  strongly suggests you cannot think."
Comment: Yeah, it's likely that Mencken was besieged by people wanting advice for their writing.


URL Structure Without WWW

Question: I've recently generated a sitemap.xml for my website to help with its visibility on search engines. With this in mind, I have a specific concern about the structure of the URLs listed within the sitemap. If I list URLs using the format https://example.com (without www), will this particular format have any implications on my website's traffic, its ability to be crawled efficiently by search engines, and its subsequent indexing? I'm keen to understand if the structure of the URLs in the sitemap can impact these aspects and if there are any best practices I should be following.
Answer: Whether or not you include the www in the URLs is not important. It won't affect how well or easily Google indexes your site. What does matter is consistency.
Pick one URL format and stick to it. All internal links, canonical URLs, and sitemaps should use that URL format.
Best practice is to redirect traffic from one format to the other, so all https://www.example.com URLs should redirect to https://example.com equivalents, for example (or vice versa, depending on which you prefer).
The reason for this is that if both variants of the URLs are indexed, Google will see them as distinct URLs with duplicate content. This isn't necessarily problematic (if you have proper canonical URLs defined), but it's not ideal. Stick to the one URL structure improves your chances of good SEO performance.
Comment: Just want to add, URLs in the sitemap should match how they exist. If you use www. the links in the sitemap should match and not need a redirect. If you use a crawler to make the sitemap it should match already. Same with WordPress plugins.

URL Structure Without WWW


How do you know when your chain breaks while fishing for shinies?

Question: Is there a way of telling? Maybe a difference in music?
Comment: Unfortunately no.
Answer: You have no definite way of telling. This is kinda why shiny hunting is so hard.
This said shiny fishing is the easiest way of getting a shiny because all you have to do is fish over and over again. As long as you get a Pokemon on each rod swing and you don't lose it, your chain still goes on. Get a Pokémon with the "Suction Cups" or "Sicky Hold" ability in the first slot, fish in a "corner" (where you block the only square to get out of there) and you should have a pretty good chance to satisfy this condition.
The best way you have to know is to keep fishing until you do get a shiny.


Returning a JTextField to editing after a warning dialog

Question: In my application I have a JTextField that will only allow the user to fill it with input formatted in a certain way.  I have a method that checks if input into the field is valid and if not it displays a dialog.  After this occurs I would like the focus to return to the text field so that the user can correct there mistake and type in the appropriately formatted input.  I tried to do this by having the JTextField call requestFocus() on itself but this seems to only half work.  The JTextField displays the flashing cursor but I cannot actually type in any text until I click on the text field again.  Is there another method I need to call?  I could not find anything else in the documentation.
Comment: In the API it says "use of this method is discouraged because its behavior is platform dependent. Instead we recommend the use of `requestFocusInWindow()`". Did you tried this one ?
Answer: It's not exactly the solution you asked for, but you might want to look at javax.swing.InputVerifier. This object allows you to verify the input to a field and prevent taking focus away from it if the format isn't right. I don't think it would allow you to put up a dialog explaining the error (though it might - try it) but it would certainly allow you to put a message in a 'message area'.
What you do is set an InputVerifier on the JComponent, have it test the contents for validity; if it isn't right then the JComponent won't release focus (and can write an error message).


Why is iridium so important?

Question: Why is iridium so important to everybody that I see on YouTube. Is it something magical?
Answer: Iridium Ore is used to make the Quantum set of Armor, aswell as the Iridium Drill(the upgraded version of the Diamond Drill). First you will need to find one in a chest, and use it to be scanned so that your Mass Fabricator trough UU-Matter can produce more of them.
In FTB Infinity Evolved Expert Mode the Iridium have more uses though.
Answer: You use iridium ore to make Quantum Armour, the most powerful armour in IC2. It can be found in dungeons or produced with a Mass Fabricator. Quantum armour is extremely powerful, and makes you virtually invincible (along with other effects such as super fast sprinting and breathing underwater) while it is charged, so iridium ore is very desirable.

Why is iridium so important?


Reference needed - Who first introduced linear models, fixed and random effects models?

Question: I am writing an essay which briefly discusses linear models as well as models with fixed and random effects. I am googling since ages to find a reference which says who first introduced linear models and especially later mixed effects models. I found on wikipedia that in 1918 Ronald Fisher introduced random effects models to study the correlations of trait values between relatives.[1] Also, as far as I can see  Laird, Nan M.; Ware, James H. (1982). "Random-Effects Models for Longitudinal Data". was the first one who applied those models to longitudinal data. My data is also of type longitudinal and I therefore can use that reference later on. However, it would be great to found evidence of who first introduced linear models as wells the fixed effets models.
Answer: There is a "Brief History of LMMs" in "Linear mixed Models - A Practical Guide Using Statistical Software" (West, Welch, Galecki).
They say: "First known formulation of a one-way random-effects model" was in 1861 by Airy. Sadly there is no more precise information about the corresponding paper/book/whatever.
HTH
Answer: According to my understanding: least squares was proposed by Legendre in 1806. The linear model (simple linear regression) was introduced by Galton in 1895. Multiple linear regression was formulated formally by Pearson in 1938.
See 1st and 8th chapters of Stigler (1986) and Stanton (2001) for references. BTW, if you're interested in history of statistics, Stigler is a beautiful book.


"Search Setting" missing from 'Search And MetaData' section on configuration page

Question: I have a problem. I have a project that uses Domain Access.
I recently installed the "Search API" module and attempted to set up an indexing configuration.  That all went well, but when I tried to index the items I tried going to Drupal 'configuration' page, and under "Search And MetaData" section, there should be a link called "Search Settings", but there is not.
In a different project that did not use Domain Access I set this up and it works fine. Here is an example of the project that has search api set up properly and "Search Settings" shows up under "Search and Metadata"

I can click on "Search Settings" and it takes me to the "Search Settings" page where I can index my nodes, as well as a few other configuration options.
For my new project that uses Domain Access the "Search Settings" section is missing from "Search and Metadata", behold:

I initially thought it might be a permissions problem, and went to permissions section and the both have 'administrator' set under permissions for "administer search api".
Does anyone have any idea why my new project that uses Domain Access might be missing this page?  I tried going to the "Search Settings" page by manually typing the url into my browser, but it said "Access Denied". (I also have Access Denied backtrace set up and no reports were being generated after it gave me the access denied.
The only reason I mention that my new site uses Domain Access is because that is the only major difference between the 2 projects that I can see.
I would really love some help with this, and would consider it my Xmas present.  :D
Answer: I found the answer out.
I thought that the "Search Setting" configuration section was created for the "Search API" module, but it is actually for the core "Search" module.
Re-enabling the core "Search" module caused the "Search Setting" configuration section to reappear.


Accessing arcpy functions from IronPython

Question: I made a few python scripts to automatically optimize and publish my data..
Now i'm thinking about writing a .NET program to manage and run these scripts: the obvious way is invoking them in a different process from the .NET code but this seems downright ugly
I was thinking about using IronPython and migrate my scripts into .NET.
Does anyone know if it's possible to invoke arcpy functions from IronPython?
Thanks!
Answer: At the Esri Discussion Forums I found a 2011 response from @JasonScheirer which says:

Arcpy is built on arcgisscripting, which is in turn a CPython
  extension. You will not be able to use arcpy in IronPython.
Answer: An update to this question, this can now be accomplished. You can follow this guide to complete your task in IronPython. Using ArcPy with Visual Studio 2010 and 2012


asp.net Dropdownlist conditional postback

Question: I have a dropdownlist and have an event selectedIndexChanged which postsback, i want to be able to show a message to the user whenever he changes the value in dropdownlist, based on the input from the message i will decide if i have to postback or not.
The message shown would be are you sure? if he selects yes i would continue with postback, if he says no, i would cancel the postback and assign the previous value as selected.
I have searched alot but cant figure out a solution to this, i think if there is a javascipt function which determines if a postback is required or not that could help i guess
Thanks
Answer: <code>// get a reference to the DropDownList
var selectlistId = '<%= ddlYourList.ClientID %>',
    selectlist = document.getElementById(selectlistId);

// attach to the onchange event
selectlist.onchange = function() {

  // decide whether to execute the __doPostBack event, which submits the
  // form back to the server
  if(confirm("Are you sure you want to do this?")){
     __doPostBack(selectlistId, '');
  }
};
</code>
Answer: You can stop can cancel postback of dropdownlist very simply.Just add this javascript on page load event.
<code>protected void Page_Load(object sender, EventArgs e)
{
    DropDownList1.Attributes.Add("OnChange", "if (!confirm('Change this?')){return};");
}
</code>
Answer: In order to accomplish this task you will need to use a CustomValidator with custom client side Javascript to control the post back.
You can read this article on 4Guys discussing the different validators with a client side validator JavaScript sample to get an idea.
But the core solution would be using a custom validator to control the post back only when the form is valid.


Why it's best to buy fixed-rate annuities when long-term interest rates are significantly higher than short-term interest rates?

Question: Why it's best to buy fixed-rate annuities when long-term interest rates are significantly higher than short-term interest rates?
Comment: Because when predicting long-term costs of servicing the annuity, long-term rates are more directly applicable?
Answer: I would not use the term "best" as it implies that you should time the market, but if short-term rates are higher than long-term rates, it's usually a temporary condition, and the market will eventually adjust by either either raising longer term rates or lowering shorter term rates.
In either case, it would be better to buy instruments that take more advantage of the higher short-term rates than annuities (annuities are fixed at an average of all projected rates over the life of the annuity) and wait to buy annuities until the market is back in its "normal" state.


How to revive an over-discharged DJI Mavic battery?

Question: I left my drone on until the battery shutdown.  The Mavic battery bottom meter light flashes 16 times when checking charge.  When i place the battery on DJI chargers it will not charge and also flashes 16 times.  Keep in mind i did not store it this way and the battery depleted this low in the past 24 hours.  Is there a way to force the battery to charge up from a depleted shutdown state?
Answer: Short answer. Buy a new battery.
I heavily researched this and although there may be another solution I will share my experience.
The Mavic batteries have complex circuitry in them to maintain stability and longevity.  This is partly why the Mavic charger will not revive the battery.
Is there a way to manually charge it?  Well the answer is yes and no.   The cells inside the battery are probably okay if the battery did not deplete this far while in storage.  The problem is the circuitry will not allow a manual charge above failsafe levels restoring to a functioning DJI battery.   There are some videos and reports of people cracking open their batteries and doing various physical things to the circuitry to fix this issue.  I do not recommend this.  I found that opening the Mavic battery was not easy and left the battery case in disrepair due to it being glued or affixed with single use fasteners.  If there is a tool that can open the battery case gracefully i don't believe it is something in a standard tool box.  On top of that the Mavic circuitry is precise to avoid radio interference and modding that could increase radio interference.
Once the battery was open, poorly, I was able to charge directly to the main leads.  The battery stack read about 10.5 to 10.6 volts on 3 cells.  Very low.  My hobby charger pulled it up to 11.1 volts static.  I was still unable to charge using the Mavic charger.  This led me to believe the circuitry remembers the charge state of the battery and will not continue if that battery has inconsistent untracked voltage changes.
I saw another video that shows desoldering the positive terminal while the balance lines are disconnected.   Then soldering it back.  I tried this and it did seem to refresh the charge level readout, but i still could not charge using the Mavic charger.  With my battery split open and looking like it would never be allowed in checked or carry on baggage i gave up and decommissioned the battery using a method that worked for me.  Once the battery became inert it could be disposed of properly.
I suggest contacting DJI before opening the case. Maybe they have a trade in policy or something when the Mavic depletes the battery too low?
Best bet is to get a new battery.
Comment: _"decommissioned the battery by placing it in a bucket of salted water"_ Please don't recommend this method as it brings a lot of risks while not even being a good method to start with.
Comment: Why is that?  Explain the risks.  If the battery is discharged to sub usable voltages, "over-discharged", what are the risks to depleting the battery to zero volts in a controlled reaction?   Please explain the risks since we are all about learning here. I want to know the details.   Thanks.
Comment: @cliffclof Dumping an over-discharged LiPo battery into a bucked of salt water is far from a _controlled_ reaction. The general idea is old and predates back to the first months of consumer available LiPo batteries. A lot of (post-)accidents happened using this seemingly sound method and Fred Marks (FMA) himself, from whom this idea originated and was made popular in those early days, no longer recommends this. At least wear safety glasses. Or even better, bring them to an expert for disposal. Almost any electronics shop will also take LiPo batteries.
Comment: Right on and thanks for sharing. I've been doing it for a while and you led me to look into other ways. I read a lightbulb method that sounds pretty good. Given my motivation to bring a single battery to a disposal facility I mostly want a way to set it and forget it. Until i get enough to send to the recycler/destroyer. You make some excellent points.
Comment: Important to note that a fully depleted battery or one in a similar state should not be considered flight safe as it will have sag issues, voltage will drop very fast and unexpected, some risk of fire, risk of component damage, and the drone will not perform the same with a compromised battery. There is nothing you can do to the chemistry (that is reasonable) to change this fact.  Generally we keep old batteries for desk use, better ones to scout with (short flight), and recycle the chemistry.
Answer: I just had this experience with my first generation Mavic Pro.  I have very low charge counts on the batteries.  I flew it last a week ago, and when I went to charge one of the batteries, I saw the dreaded rapidly flashing lights.  I wasn't sure what to do, so I plugged in my other batteries that had not been charged or flown recently.  The others charged normally, and as the problem battery was awaiting it's turn to charge, the multi charger indicator light die turn green.  I was thinking this may be some sort of trickle charge.  I cannot confirm that, however the battery that would not accept a charge did accept a charge after the first battery was done.  I do not know if I was just lucky.
Answer: For a normal, "raw", unprotected LiPo you just hook it up to a charger and select the NiCad or NiMh setting (yes, for a LiPo), and set it to 1/20C charge rate (Capacity / 20). Then, you charge it until it's > 3.0V/cell. Then, you switch to 1/10 C until past 3.5V/cell or so, then maybe do 0.5C until full. I wrote extensively about that on my personal website here in 2014: www.ElectricRCAircraftGuy.com: Restoring/Recharging Over-discharged LiPo (Lithium Polymer) Batteries!.
For a Mavic battery, however, it has protection circuitry and may stop that. You might try my approach above anyway, or you might try bypassing the circuitry to hook up to the raw battery cells to do my approach, but be careful and study my article first.
Answer: DJI Mavic batteries include a firmware self-destruct switch; if the battery becomes "unsafe" (according to the software - e.g. - charge drops too low) or if it is disconnected from the circuit - this trips the fuse and from then-on it will never work again (even if the batteries are OK or replaced with good ones).
DJI say this is for "safety", but it also effectively blocks repairs, forcing you to buy new DJI batteries of course.
There are some hacks around where you can pull dead batteries apart and re-flash the fuse manually, but they're hard to find and very difficult to perform.


How can I get my Canon RT speedlite to go off when using a Godox transmitter?

Question: First, I am using my Canon Camera 6Dmark2.  I have a two flashes: one is the Godox V860IIC and a Canon Flash 600 EX-RT.  I also have a wireless trigger Godox XProC.  I can easily trigger the Godox Flash with my Godox trigger using channel 1 group A for the flash and having that set on the trigger.  Now, the problemo is how to get the Canon flash to trigger as well!  I can't seem to understand the manual for the Godox X2Tc and I thought I had to have this in order to get the canon flash to go off.  Does anyone know how to do this??  I have been searching high and low on Google and Youtube and can't figure it out.  Normally, I am pretty swift with technical material, this one eludes me.  Thank you!
Answer: Your XPro-C and X2T-C Godox triggers speak a different "language" than the radio receiver in the Canon 600EX-RT flash. You can't control the 600EX-RT directly via radio with a Godox trigger.
To control the 600EX-RT with your Godox transmitters you need a compatible Godox receiver attached to the hot foot of the Canon flash. The only such receiver available is the X1R-C.
The X2T-C has a bluetooth receiver, but that is only for using select smartphones to control the X2T-C via the Godox app. It's not capable of acting as a receiver for other Godox triggers.
Be sure the channel numbers on the XPro-C and the V860II-C (or any X1R-C receiver you are using attached to another flash) match. All wireless ID numbers should also match.¹
¹ You may need to turn wireless ID to "off" on all devices to get things to work with the V860II-C, which does not appear to support wireless ID.
Answer: The 600EX-RT requires a Godox X1R-C receiver on its foot to work in the Godox X radio system.  The X2T you purchased is a dedicated transmitter unit, like the XPro. The hotshoe on top is if you plan to use a non-Godox speedlight on-camera. stack a different system transmitter, or use an AD200 extension head on top of the X2T.
You could, of course, also switch out the Godox gear for RT-compatible gear instead, if you wanted to use the built-in Canon RT transceiver in the 600EX-RT. Aside from Canon's ST-E3-RT/ST-E3-RT v2, there are a number of 3rd-party transmitters, like the Jinbei TR-Q7/Westcott FJ-X2m, Yongnuo YN-E3-RT, and Phottix Laso that "speak" RT. As for RT-compatible 3rd-party speedlights, Jinbei makes the HD2-Pro/Westcott FJ80 and Caler 600EX-RT, while Yongnuo has the YN-600EX-RT and YN-600EX-RT II models. Keep in mind, however, that Yongnuo and Jinbei's RT systems do not work with each other, even though they both work with Canon's gear.
If you go with Jinbei/Westcott, there's the advantage that the speedlight and transmitter have a "Universal TTL" foot that can be used on any of the supported brands (Canon, Nikon, Sony, Fuji) for the Jinbei RT/Westcott FJ system. But the larger strobe choices are only the one speedlight and 200 Ws and 400 Ws li-ion powered TTL/HSS strobes and that's it (unless you're willing to go for older Jinbei RT/Adorama Orlit products).
If you go with Yongnuo, you are limiting yourself to Canon and speedlights only for future expansion.


Can I put multiple resistors in series to get higher resistance?

Question: If I put 3 1MΩ resistors in series, is that the equivalent of one 3MΩ resistor?
Comment: @vicatu: It may be simple to YOU, but everyone has to start somewhere - and the best start anyone interested in a subject can make is to ask questions.
Comment: this question seems awfully simple... do you have some context that makes this answer less obvious?
Comment: @Linker3000 nope, the best start anyone interested in a subject can make is to *read books and/or articles* first, then *search for relevant data there*, then *think about what they read*, and **then and only then** ask questions. You learn by cognitive introspection, by searching for answers in existing sources first - not by mindlessly asking every question you can imagine to everybody.
Comment: You should read an introduction to electronics.
Answer: The short answer: Yes. It's the same.

The current through resistors in
  series stays the same, but the voltage
  across each resistor can be different.
  The sum of the potential differences
  (voltage) is equal to the total
  voltage. To find their total
  resistance:

By Wikipedia

But this the ideal value. In the real word, resistors have tolerances and you have take in count. E.g., tolerance of 10%. The final value can vary between 2.7MΩ to 3.3MΩ.
Comment: yeah, the tolerance part makes sense because it is multiplied as well. thanks for the answer!
Answer: If you place resistors in series the total resistance is the sum of the resistances of all resistors in the chain. There are two main reasons why you would do this.

to obtain a value you don't have in you box. For instance a 30k resistor is not an E12 series value, but you can obtain it by putting two 15k resistors in series
to allow a higher voltage. 0603 resistors have a working voltage of 50V, So if you want to use them in a 70V circuit you'll have to place at least two of them in series. Care should be taken if you pick unequal values. Divide the voltage by the total resistance to get the current flowing through the resistors, and multiply this current by each of the resistances in turn to get the voltage over this resistor. In the 0603 example none of the voltages should exceed 50V.
Comment: 3. More power dissipation before the resistors burn.
Answer: Yes of course, and to add some marginally more interesting content to my answer, you can also put resistors in parallel to get a lower resistance, though the effective resistance is not quite as straightforward as the simple sum for series resistance. Parallel resistance is calculated as (R1 * R2) / (R1 + R2), which equals R/2 for R = R1 = R2...
More formally, for n resistors in series: R_effective = SUM(i=1, i=n, R_i)
...and for n resistors in parallel: R_effective = 1 / SUM(i=1, i=n, 1/R_i)
Answer:
If I put 3 1MΩ resistors in series, is that the equivalent of one 3MΩ resistor?

I don't know exactly whether your question contains a trap.
If you put at most 2 resistors in super cold environment and the remaining one(s) in hotter place, the combined resistance will probably no longer equal to 3 mega ohms.
Comment: A question on E&R wouldn't be a trap. If it was, it would be edited by another member to remove the confusion. Questions have to be as clear as possible.
Comment: +1 for noticing that different operating conditions can affect the values of Rs and make the "equal" R values be unequal in reality... it's a common caveat in power circuit's design, where linear parts laying close to Joule heat sources can (and usually will) exhibit nonlinear behaviour, making it necessary to compensate in some way (e.g. by pairing NTC/PTC parts or by introducing a thermal feedback loop) for thermal value drift.