Grand Staff using the Bass Clef and Treble Alto-Tenor Clef

Question: Does anybody have any additional information regarding the origins and usage of the Treble Alto-Tenor Clef symbol?  I found it in the late Gardner Read's book on Music Notation, and beyond that - NOTHING.  Does anybody know anything more about this symbol and where it may have come from?
Comment: Have you had a good look at http://en.wikipedia.org/wiki/Clef? It gives a pretty thorough explanation of all the various clefs and some information about their origins.
Comment: This C clef is not in the right position. It should be between the two staff, at the level of the middle C.
Comment: This is not a C clef. The clef on the top staff is a transposing G2 clef. Normally the transposition is indicated with a tiny 8 at the bottom of the clef, but in older notation, the bracket shown here indicates the location of middle C.
Answer: My best guess as to the intention of this notation would be that the entire grand staff is being transposed down an octave.
As I'm sure you know, the treble clef on its own places C5 on the third space. The C clef used for alto/tenor clef is meant to be centered upon C4, or middle C. So, the fact that a C clef is centered upon the third space of the upper staff would indicate that the entire staff has been shifted down by an octave.
Initially, I was considering the C clef to only be modifying the treble staff, but if that was the case, then the note on the ledger line in the middle of the grand staff would be ambiguous. Is it a C3 or a C4? And also in hindsight, I can't really think of a more efficient way to shift the entire grand staff in any direction than to simply designate which C is middle C with a C clef. You could use 8va notations but they would get tedious, and while you could try putting a little 8 on the bottom of each clef in the grand staff, I don't think that's any more reasonable than what is displayed here.
Answer: You have not exactly explained where you found this example: in Gardner Read's book (page 55) there is an example of the upper clef (G clef with funny bracket), which he explains as one form of the clef for a tenor part, meaning G clef 8vb (and not the same as the "tenor clef", which is a C clef). I can't see this particular combination of two staves.
On the face of it, this should mean a part for tenor voice and a part for bass (voice), which leaves no anomaly at all. I think it is unlikely that this indicates an entire grand staff transposed down an octave, because there is no indication that the F clef is 8vb, and no bracket suggesting that this is a grand staff.
(Edit) I have just seen that there is a note floating in the middle. Hmm. Somehow I find it difficult to believe that this is a real example from a score, rather than some sort of attempt to illustrate different clefs.
Comment: Brian, get a copy of Gardner Read's book on Music Notation.  Go to the chapter on Stringed Instruments, and also you will see it in the index.
Comment: https://www.facebook.com/ContraBasso.TrebleAltoTenorClef?ref=aymt_homepage_panel
Comment: Here is the FB site showing the Grand Staff using the Bass Clef stave in conjunction with the Alto-Tenor clef stave.  https://www.facebook.com/ContraBasso.TrebleAltoTenorClef?ref=aymt_homepage_panel
Comment: There is no floating C anywhere.  I am just showing you where ALL the Cs are located.  The Low B, slightly off to the left of the Low C is showing you an Open Low B string on a six string bass with a 24 fret double octave neck.  The whole point is to show you all the expansive range of the Cs involved.  On a six string bass with a double octave neck, there are five Cs developing four full octaves of C, three below middle C and on above.
Answer: Very interesting diagram. I hadn't ever seen the use of the "C-clef" to mark the pitch C in a space, rather than on a line. However, as @Old John points to in his comment, the Wikipedia page about clefs deals with just such a clef. At the top of the page this clef is first hinted at:

Only one clef that references a note in a space rather than on a line has ever been used.

And then later, this clef is revealed as being the C clef (most commonly used for the tenor and alto clefs, but also soprano, mezzo-soprano and baritone). However, in this case it indicates middle-C as being in the third space (from the bottom) of the stave. This, of course, means that all notes are identical in pitch to those when using the treble (G) clef, except that they sound an octave lower. It is not surprising that this clef is not commonly used to denote treble clef notes sounding an octave lower, as there are other, well established alternatives. Therefore, these two octave-transposing treble clefs, and the following "C" clef have the same function:

The first of these is commonly used, for instance for a tenor voice part or for an octave transposing instrument such as guitar (although a normal treble clef is also used, as the transposition is implicit). The second clef is less common. And, as I say, I had not seen the third clef before.
The Wikipedia page points out another good reason not to use this clef:

the C clef on the third space, [is] easily confused with the tenor clef

However, this doesn't explain a couple of aspects of the diagram you show.

It isn't clear if it is necessary to place a standard treble clef before this "C" clef. I suspect not, and that the treble clef is written there to show that they denote equivalent pitches, although at different octaves.
It isn't clear at which pitch the C between the staves should be: as @Brian Chandler points out, there is no indication that the lower bass (F) clef is an 8vb (octave lower) clef. For this reason, the C between the staves is in fact two different notes at the same time! (It is middle-C if relating it to the bass clef; it is an octave below middle-C if relating it to the clef on the upper stave.)

Although I'm sure the book you found this example in is a great resource, this example is certainly ambiguous, for the reasons stated above. (Having read the rest of @Brian Chandler's answer, I wonder if the musical example you show is in fact different to that found in the book?)
Answer: It looks like the strange Clefs you have provided are the result of a few musical situations which up until relatively recently, were somewhat confusing to formally notate, thus unusual clef suggestions in the late 19th through the early to mid 20th centuries.
Essentially, there are a few categories of instruments which transpose, usually down by an octave or two, but do not wish to think of themselves as transposing instruments. Two prominent examples would be the Tenor voice, and the Bass clarinet.
For the tenor voice, one would be tempted to simply write a treble clef, which is often good enough for the performer. Since the Tenor voice is higher than the bass, it would make sense to read the treble clef. However, for the accompanist, and the composer, a confusing situation is created. You often have a treble clef which reads an octave down, joined to a bass clef which does not. This makes the figures quite difficult to sight read on piano correctly, often causing confusing when the accompanist inevitably reads the bass part an octave lower than intended.
So, a few clefs were used over time to try and make it clear the first time, that the accompanist should only lower the treble clef. None of them really stuck, except perhaps the treble clef with the 8 below it - that has lived on quite well in choral music at least. Of course you still have the problem with middle C - which might be doubled, or an octave below what it should be depending on which way the note heads point.
For instruments like the bass clarinet, a composer attempting to sight read the score has to simply remember the transposition. It's not that bad.
Beyond this, I don't know where the C clef has been used this way. I wouldn't be surprised if there is a medieval example somewhere, but using the C clef in the way you demonstrate in your diagram just isn't used today.
Answer: There is nothing ambiguous about the different clef symbol denoting the 3rd space in the upper stave as being the location of middle C.  In fact it is quite clear.  Because people began writing bass music using only the single F Bass clef symbol, and not a Double F Bass clef symbol, this transposed the written bass music up and octave, which in turn moved the middle C up an octave to be located in the 3rd space of the upper stave.  The symbol denotes the space and not a line as being the location of middle C.  The famous movable C clef symbol only denotes a line as being the location of middle C, and not any of the spaces between the lines.  Originally the strange symbol was used in conjunction with the Treble clef symbol, but it was immediately demanded the Treble clef symbol be removed, with only the strange symbol being used alone for denoting middle C, in developing the Contra Basso/Alto-Tenor grand staff.  I found the symbol in the late Gardner Read's book on Music Notation, and when I showed it to a music teacher of mine, who holds a Masters in Music, he immediately demanded the Treble clef symbol be removed for the final adaption and adoption of this grand staff to be used by modern bass players.  I agreed and it is now presented to you all.
Answer: The C clefs originated from the clefs used to notate ancient Gregorian chants.  The C - shaped thing at the beginning of each barline showed where C was in the staff, and is also why the weird things we have today are called "C" clefs.  The clef show here would have been the predecessor to the C Baritone Clef (which is almost never used nowadays).  The "C" clef was put on other lines depending on the range of the chant, much like we do with alto and tenor clef today.
Answer:
This is the proposed Contra Basso/Alto-Tenor Grand Staff to be used by the modern bass player - World Wide.  Originally, the upper stave featured the use of the Treble clef symbol in conjunction with the Alto-Tenor clef symbol, but it was recommended the Treble clef symbol be removed in favor of just using only the Alto-Tenor clef symbol alone in the upper stave.  When using music in the treble range, all one has to do is add in the Alto-Tenor clef symbol and from then on the music goes from being in the Treble clef to the Alto-Tenor clef.
The Treble Alto-Tenor clef symbol was found in the section on stringed instruments in the now late Gardner Read's book on Music Notation.  Where he found it is a bit of a mystery, but it is valid, was used at one time, and just before becoming totally extinct has been brought back for a revival in usage.
As side from a website I set up on Facebook, and this posting here in Music Exchange, you won't find any listings or postings on the Alto-Tenor symbol anywhere on the internet.  There are a zillion postings showing the well known and famous Movable C clef symbol, but there is literally nothing on the Alto-Tenor clef symbol.  The intent is to bring this clef symbol to your consciousness towards reviving it for use by modern bass players - before it becomes totally and completely extinct.
Just as nobody can trade mark the alphabet or numerals, or just about all other symbols we use, for example the F Bass clef, G Treble clef symbols, this symbol is at least three hundred years old, and therefore cannot be trade marked and owned by - anybody.  The effort here is simply to adapt it and the adopt it for use by modern bass players.  Nobody can charge money for it's usage, but we can all use it in our music - for free, just as we use the alphabet to write the words we are reading.
The Alto-Tenor clef symbol denotes the third space (and not any of the lines) as being the location of Middle C.  While the famous Movable C indicates a line, the Alto-Tenor clef symbol denotes the space.
Should a new name for this symbol be found since it might be possible to move it also just as the movable C is moved?
Possibly.
For now, the symbol is simply known as the Alto-Tenor clef symbol.
Comment: So, it's simply a clef which writes the notes one octave higher than they're actually played. Rather like guitar music, with a small '8' on the treble clef. Instead of using leger lines beneath, it continues with the F clef.


Growing Money Plant purely in Vermiculite

Question: Is it possible to grow money plant only in Vermiculite? The plant grows in soil and water, so can it grow only in vermiculite, as vermiculite retains water really well?
Answer: Not without providing additional nutrients. Vermiculite is inert and contains negligible amounts of nutrients, and your plants would quickly show signs of nutrient deficiencies if provided with water alone. However, as a substrate in a hydroponics system where a nutrient solution is used, it can perform well although other substrates are more commonly used for economic reasons.
Comment: It might appear healthy at present but if you are giving it pure water and nothing else it will in time show symptoms of deficiencies. Some plants can last a very long time in just water - very often they won't grow much or they will "remodel" themselves perhaps getting thinner as they get taller. Categorically, all plants contain more elements than just Hydrogen, Carbon and Oxygen (which they can get from the atmosphere and water). Generally, the other elements are absorbed by the roots from soil.
Comment: But wouldn't the plant get nutrients from the water containing vermiculite? Just like how it grows in water alone?
Comment: Pure water does not contain nutrients. A plant grown in pure water will eventually get deficiencies and die. Some plants can tolerate this for a surprisingly long time but ultimately plants require elements other than Hydrogen (from water), Carbon (from Carbon dioxide) and Oxygen (from both).
Comment: But my money plant is growing in water only. I am not giving anything else, except changing water every week.


Are experimental aircraft exempted from the 2 year transponder and avionics checks?

Question: Are experimental aircraft exempted from the 2 year verification test for avionics, transponder and altitude encoder? Or must they comply?
Comment: Which legislation?
Comment: What's BL?  Are you asking about the regulations under FAR 91.411 and 91.413?
Answer: When you operate in airspace that is under FAA regulations, FAR 91 applies (unless you are flying military)
FAR 91.413 a) states:

(a) No persons may use an ATC transponder that is specified in
  91.215(a), 121.345(c), or §135.143(c) of this chapter unless, within the preceding 24 calendar months, the ATC transponder has been tested
  and inspected and found to comply with appendix F of part 43 of this
  chapter;

So when you haven't performed the two year transponder check, you are not allowed to use the transponder. The way to fly then is to fly VFR in airspace where no transponder is required AND not to use the transponder at all.
This is not specific to experimental aircraft, there is no exemption.
There is good reason not to have an exemption for experimental aircraft; a malfunctioning transponder can cause many problems to the ATC radar systems.
Comment: @JScarry According to §91.215(c) the transponder must be turned on if it is maintained in accordance with §91.413. The case that is being discussed here is when the transponder is not maintained in accordance with §91.413.
Comment: If you have a transponder installed, it must be turned on (except in Class G airspace) §91.215 (c) Transponder-on operation. While in the airspace as specified in paragraph (b) of this section or in all controlled airspace, each person operating an aircraft equipped with an operable ATC transponder maintained in accordance with §91.413 of this part shall operate the transponder, including Mode C equipment if installed, and shall reply on the appropriate code or as assigned by ATC.
Comment: You stated that if you fly in airspace where it is not required then you can turn it off. That is not true. If you have a transponder, it must be maintained and turned on. You can't fly in controlled airspace with it turned off or not maintained. An interesting loophole is that if it was inspected and subsequently fails, you can turn it off and fly in controlled airspace.


Troubleshooting filament slip on MakerBot 5

Question: I run a high school 3D printer lab and we have several 5th generation MakerBot printers. On one of them I have considerable trouble with "thin" prints and filament slip warnings.
So far I've tried changing extruders and using different filament rolls with no luck. But, if I move the job and the extruder to a different printer it works.
I'd appreciate suggestions for how to sort this out. I would have expected the slipping problems to follow the extruder.
Comment: Do you have the Smart Extruder+ installed or still the original Smart Extruder? Also, what are your target parameters such as Layer Height, Infill %, Shell, etc.?
Comment: Not at school today, so some of this if from memory. All of the failures were with the original Smart Extruder. Parameters that I remember are: layer height = 0.2mm, infill = 10%, shells = 2, rafts = on.
Answer: Oh interesting. By slips, I take it you mean that the raw filament slips, not the print slips.
This will happen for a few reasons. First the tooth gear that grabs the plastic is either:

Worn out
Out of place
Not the correct distance from the guide wheel.

This is all part of the mechanism that the Smart Extruder attaches TO. Not the Smart Extruder itself. You might be able to fix this yourself, worst case you will need a replacement assembly from MakerBot. I would look into online auction sites for the part
Another option is to try thicker filament. Which you might be able to custom order. So instead of 1.5, maybe get 1.8. I am not sure where you can buy off sizes.
From there this machine might just be getting jammed. It happens to some machines. This again points to the base of the X axis assembly.
Last which I would say is not likely as you have tried multiple extruders, you might have the nozzle becoming clogged. I often pop open my extruders voids warranty and clean them out. Also micro hand drills are a good option here.
Comment: "This is all part of the mechanism that the Smart Extruder attaches TO" I don't think this is true. I have a Replicator+ extruder which based on YT videos looks very similar. The tooth gear is embedded inside the extruder and mechanically driven by the printer.
Answer: I posted an edit to the top answer in hopes that it will get changed for anyone looking at this post in the future but in case it gets rejected here is the actual issue and fix.
If you just want the solution skip down to the bolded text.
I came across this post while browsing for resources to document my own experiences with "filament slip" in Makerbots and every suggested solution was wrong and potentially malicious to the machine. I'm an expert in the field and have taken apart and fixed dozens of machines.
First a bit about the anatomy of the extruder so you can understand better. The way that the toolhead detects slips is with an encoder wheel placed above the actual drive mechanism, not attached to it or the idler whatsoever. It's also not even tensioned against the filament at all. The filament is "held" in place with an injection molded piece of plastic that attaches to the rest of the extruder body just by the pin connectors on the internal PCB and a small positioning groove on both sides of the case.
The encoder wheel and its sensor are very unlikely to wear but what happens to every older head is that either the casing or the injection molded part that holds the filament wears and the encoder wheel no longer presses firmly against the filament which causes the filament to move without turning the encoder. The encoder then not moving while the drive gear is, reads a filament slip and sends an error message even though nothing is mechanically or functionally wrong. The wear is also not clearly visible even on close inspection. This problem seems to arise in every Smart Extruder+ once it reaches 150-200 printing hours.
Actual Fix
As a temporary fix you can jam a piece of foam inside the right side of the extruder to push the poorly secured injection molded part against the encoder wheel. This will make the filament slip error much less frequent but won't completely solve the issue. I'm working on a 3D printed insert that should hold the part more securely, at least until the printed part wears after an additional 100 or so printing hours. I'm also working on a 3D printed replacement for the extruder body and some of the injection molded interior parts. It aims to be a drop in replacement which should greatly improve both reliability and functionality. If I can remember once I'm finished I'll come back and post a link here.
Comment: Hi Adam and welcome to SE.3D Printing! I would like to kindly ask you, please do not add *additional* information to someone else's answer. If you have something to add, please post your own answer (referring back to the answer that you wish to ameliorate, if need be, using the URL of the answer). Your suggested edit, while it may have been correct, was rather confusing, and caused some flags to be raised. Thank you for posting your *own* answer here... :-) If you wish to edit someone else's answer to correct typos, grammar, fix units, links, or incorrect formatting, please feel free to do so.
Comment: This was the cause in my case. I have my filament roll on the side and at just the right angle, the filament would bent slightly away from the wheel, losing contact. A bit of paper jammed behind the injection molded part solved the issue.
Answer: Okay, the original Smart Extruder has many issues. I've clocked about 60 hours on a new Replicator+ w/ Smart Extruder+ and haven't encountered any of the issues I heard about with the original extruder. So, I might begin by re-recommending what MakerBot Industries has recommended to their customers which is upgrade your Smart Extruder to the new Smart Extruder+.
However, the extruder may not be completely at fault. A picture says a thousand words, so if you have one, please add one. Based on your description of the part's layers "thinning out", double check your extrusion settings in MakerBot Desktop.
Another potential issue is too much tension on the filament as it is being fed into the extruder. Here are some things that can lead to too much tension:

Filament wound around spool/spool holder
Filament kinked
Too much friction between the spool and spool holder
Too much friction between the filament and the filament guide (the plastic tube)

Finally, and my reason for bringing up the Smart Extruder, is that your extruder may be clogged or "gumming up". In the past, if there was too much tension on the filament and the drive gear was not adjusted correctly, the drive gear would grind the filament. This would lead to small deposits of filament gumming up the drive gear, eventually leading to the filament slipping. Once the filament began slipping, it would slowly no longer extrude and heat up further up the extruder assembly. This inevitably leads to the extruder becoming clogged. In short:

{Too much filament tension} + {Too tight drive gear} = Gumming
{Too much gumming} + Time = Slipping/"Thinning"
{Too much slipping} + time + heat = Clogging
Answer: You could also try printing at a higher temperature.  Slips can happen when the filament is not melting and thus not pushing through the extruder consistently.
E.g. I had a spool of PLA filament that would slip constantly at 190°C.  I tried 220°C and then it printed smoothly after that.
Answer: I had the same issue, but my solution was a little different. Following along with what the others posted about the wheel, I had a jam inside the extruder around that wheel. I missed it the first time I tore it open so no matter how many times I tried the screwdriver trick it didn't work for me.
I figured I'd post so if anyone is coming along trying to find the solution, they can see that it might be a jam on the filament wheel and not just to check the nozzle area.
Answer: We had a similar issue with our Makerbot printer here in the CARIS lab.
We would have slip, after slip and tried oiling, hand feeding, and roughing up the filament none of it worked. But we found an answer!!!
With the smart extruder, it's as simple as inserting a flathead screwdriver between the filament wheel and the housing, and prying the wheel towards the pin-side of the print head. Solved all our problems! Seriously, not a slip since and perfect prints.
I've a short solution video for YouTube, Fixing filament slip on your 3D printer.
Check out the makerbot trouble shooting page for reference: MakerBot Support > Troubleshooting > MakerBot Replicator > Error Codes
ERROR 81
Answer: So I had this issue for months, was about to either give up and call my printer a paper weight, but I figured it out. And it doesn't cost anything.
I literally reprinted the same hose adapter 6 times (every time the filament slipped about 20 times and it was unusable). I changed 4 settings and since then I've printed the same print twice in a row without a single filament slip.
In "Device Settings" in Makerbot application:
Extruder Temp: 220 °C
Travel Speed (I think this is what is actually causing the slip): 115 mm/s
Then in the "Infill" section:
Infill Density: 30 %
Lastly in "Model Properties":
Shell Starting Point: 220 °C
Please let me know if this works for you as well.
(I should point out I just have the regular smart extruder, and I was planning on dumping the $350 for the top of the line model)
Comment: I use the "MakerBot Desktop" application, as opposed to the new application, because it won't work, so let me know if I need to modify based on application


7D Mark II or 80D

Question: I am an enthusiasts photographer specializing in family portraits, events, and now wanting to jump into live sports shots. I'm completely stuck between the 7D Mark II and the 80D. Those are the two within my budget that I am looking at. Knowing what I want to do. I really love the fps and autofocus on the 7D2 but I also love the higher MP and features on the 80D. What are the differences between these two cameras that would apply to the way I plan to use my new camera?
Comment: Which features of the 80D are important to you? 4 MP is probably neither here nor there.
Comment: Flip screen  and wifi
Comment: The WE-1 WiFi card for the 7D2 is $40 and fits into the 7D2's SD card slot (which is the secondary slot next to the primary CF card slot).
Answer: There are a lot of similarities between the 7D Mark II and the 80D. Here's a rundown of the most significant differences:

The 80D has slightly higher resolution, but it's not that significant. 6000 pixels wide vs. 5472 pixels is less than a 10% difference.
The 7D2 has two DiG!C 6 processors, the 80D has one. This enables the 7D2 to use iTR that incorporates info from the RGB+IR light meter with predictive Servo AF tracking. This is a distinct advantage when shooting sports and action.
7D2 has a 150,000 pixel RGB+IR light meter divided into 252 zones vs. 7,560 pixel RGB+IR meter with 63 zones for the 80D. The 7D2 will probably "guess" better when metering in very difficult lighting conditions, but both are very, very good in this respect.
80D has both ambience and white priority AWB, the 7D has only white priority WB. This only has any real effect if saving straight to jpeg since the WB can be totally changed in post with no loss of image quality with raw files.
7D2 has a slightly larger viewfinder. Both cameras show 100% of the frame, but the 7D2 is magnified at 1.00x vs. 0.95x for the 80D.
The 7D2 has user changeable focusing screens. Canon currently offers the Super Precision Matte screen in addition to the standard focusing screen. Third party makers may also offer other focusing screens for the 7D2 since it is user removable. The 80D's standard focusing screen is not user removable. If you plan on using manual focusing via the viewfinder, this gives the 7D2 an edge.
80D has a Vari Angle touch screen, the 7D2 has a fixed screen. Both are the same size and resolution. If you like using a movable screen the 80D is the clear winner here.
The 80D has a slightly stronger built-in flash (GN12 vs. GN11), the 7D's built in flash can do Multi flash (several distinct pulses during a single shutter opening). Neither is very powerful compared to external flash options.
80D has several 'Scene' exposure modes (Portrait, Landscape, Close-up, Sports, Night Portrait, Handheld Night Scene, HDR Backlight Control, Food, Kids, Candlelight) and creative filters (Grainy B/W, Soft focus, Fish-eye effect, Art
bold effect, Water painting effect, Toy camera effect,Miniature effect) but only one programmable 'Custom' mode. The 7D2 lacks the creative filters (which can be done better in post-processing with either camera) and 'Scene' modes but allows the user to save three 'Custom' modes.
The 7D2 includes in-camera HDR with 5 presets.
The 7D2 can shoot at 10fps for 31 raw image files or until the memory card is full when shooting jpegs with the fastest UDMA-7 CF cards. The 80D can shoot at 7fps for 25 raw or 110 jpeg images with the fastest UHS-1 SD cards. There's not much that is more frustrating when shooting sports/action than seeing a "money" shot in the viewfinder while waiting for the buffer to clear. The 7D2 has an edge here.
The 80D has user selectable adjustments for AF tracking in Live View Movie mode. The 7D2 has more user selectable options in PDAF mode shooting still images. In terms of AF, the 80D is a slightly better video camera, the 7D2 is a better sports/action camera.
The 7D2 has built-in GPS but requires the $40-50 WE-1 card in the SD card slot for WiFi. The 80D has built-in WiFi and NFC but lacks built-in GPS.
The 7D2 has a USB 3 interface, the 80D is USB 2.
The 7D2 has one CF and one SD card slot that can be used to either mirror all images, store all images in different forms on each card (i.e. raw on the CF card and jpeg on the SD card), or store different images on each card (i.e. fill one, then fill the other). The 80D has a single SD slot. For redundancy and data security the 7D2 is the clear leader.
The 80D is rated at about 40% more still images from the same LP-E6N battery as the 7D Mark II. Extra batteries are pretty cheap, especially high quality third party ones that can outperform the OEM batteries, but the 80D is the clear leader in this category.
The 7D2 has a magnesium alloy body, the 80D has a polycarbonate resin/glass fiber body. The 7D2 is built a bit tougher and can handle worse environmental conditions than the 80D.
The shutter durability rating for the 7D2 is 200,000 actuations vs. the 80D's 100,000 actuation rating. The 7D2's shutter should, on average, last twice as long.
The 7D2 is about 6.5 ounces heavier and slightly larger than the 80D.

In the end it basically comes down to which you would enjoy the benefit of more: More options when shooting video, a Vari angle touch screen, and 'Scene' modes when you're not sure how to set exposure yourself for a specific scenario from the 80D versus more shooting speed, better Servo AF tracking, and more user configurable custom modes with the 7D Mark II.
The 7D Mark II can take a little more punishment and should last a little longer under the same usage patterns than the 80D.
The 80D has a sensor that is two years newer technology and does slightly better in terms of dynamic range and signal-to-noise ratio at low ISO settings. At higher ISOs the 7D Mark II takes the lead in SNR and DR. This means the 80D might be a slightly better portrait camera in a controlled environment with studio lighting, especially when using low key styles of lighting (Both would be bested in this respect by a FF camera, such as the 6D Mark II). The 7D Mark II is better at shooting events and sports in existing low light.
Most of the 'creative filters' of the 80D can be done better with raw image files in post-processing. That same capability would also naturally apply to raw files from the 7D Mark II.
I would rarely if ever use the 'Scene' modes if any of my cameras (I primarily use a 5D Mark III and a 7D Mark II as well as a 5D Mark II) even had them. On the other hand, I often set up my cameras in advance with 2-3 'Custom' modes that allow me to switch between them almost instantly instead of having to change a large number of settings individually. This is especially helpful if you are shooting in an environment where you need to switch back and forth between two different approaches. For example, an outdoor event where some of the action is away from the sun and some is towards the sun, or a night athletic event where the field is much better lit than the stands but you need both action shots and crowd reaction shots (the best of which tend to happen immediately after the best action).
Answer: Both will give you equally good portraits, assuming you are using the same lens.
For action, the 7D Mark II is a much more sporty camera. Its 10 FPS drive will give you significantly more chances to capture action at its peak than the 80D still very good 7 FPS drive. What will also make a huge difference in favor of the 7D Mark II is its processing speed which lets you shoot up to 1090 JPEG at full-resolution in a single burst, compared to 110 for the 80D.
The final advantage of the 7D Mark II which is highly relevant is its much more sophisticated 65-point AF system, compared to 45-point for the 80D. As a bonus you get a weatherproof body and built-in GPS. They do not help with sports photos but are nice to have.
The 80D does have 4 MP more resolution which lets you print a little larger but the difference is not very big. More importantly, if focus or the moment is missed, it does not matter how many more MPs you have. The 80D has a good deal longer battery-life but buying an extra batter is sufficient to close the gap.
There are plenty of other minor differences between these cameras as listed in the other answer, but really they barely matter when shooting sports. What makes a difference is speed and focus, possible metering (also in favor of the 7D II). For sports photography you can forget about Flash, HDR, Scene modes, etc.
Comment: A camera does not become "obsolete" because a new model is released. The 7D Mark II will still be as good a camera as it is today when/if a potential 7D Mark III is released - and perhaps even better because it will be available at a lower price.
Comment: One can always wait if you do not need to but a replacement is not always better in every way, sometimes sacrifies get made along the way. Might get higher resolution at the expense of noise or dynamic-range, this happens often. Reality is that, if the camera fits your needs, a new camera being launched does not change that.
Comment: Also. With the 7D being out a couple years, would you wait for a newer model of the same? I don't want to buy it and it become obsolete right aftereards. Thanks
Comment: Although it could be sooner, all indications are that the 7D Mark III is still at least a year away from release. 4-5 years is the normal cycle with Canon's upper tier bodies. The 7D2 was introduced in late 2014. With Canon's announcement earlier this week of three new TS-E lenses and a new EF 85mm f/1.4 L IS (but no mention of a 7D2 replacement) that prediction is probably even stronger now than it was a week ago. With no Photokina in 2017, the late August/early Sept. product announcement is usually Canon's last of the year for their pro lines.
Comment: The release dates of the 60D (2010), 70D (2013), and 80D (2016) would indicate an expected 90D around 2019 which would only be a year or less later than a 7D3.

7D Mark II or 80D


How do I differentiate between decade and century in phrases like "the late 1600s"?

Question: If I wanted to say something occurred occurred between 1627 and 1629, I would use "the late 1620s". However, if I wanted to say something occurred between 1607 and 1609, I couldn't use "the late 1600s" because that would imply it occurred between 1670 and 1699.
Comment: You can say "the early 1600s"
Comment: There didn't used to be a problem until people started using "the 1600s" to mean "the 17th century". Then "the 1600s" meant 1600-1609. The shift in usage began because a lot of people were confused and thought that the 17th century started in 1700 (yes I know it ended then, not in 1699). This change in usage isn't a bad idea but does cause a problem when talking about the first 10 (or, to some people 9) years of a century.
Answer: There is no broadly well-understood way of expressing this unambiguously in the form you want. Some people use "aughts" to refer to the first decade of a century, as in "late sixteen-aughts", but I suspect that would be harder to parse for most people than just rephrasing as "occurred between 1605 and 1610". This is just a gap in the language.
Answer: Using "1600s" for the decade would stop the average reader in their tracks; I would interpret it as referring to the century. To clarify, a writer could take a few approaches:
1. Using parentheses
Using parentheses or other means, one might specify without being too explicit.

The "aughts" is one way of referring to the first decade of a century, in American English, such as 2000s (decade).
— Wikipedia (emphasis mine)

You could apply this as follows.

One of the first historic moments of the 1800s (the decade) was President John Adams moving into an unfinished White House.

2. Using a relatively new word
In appropriate context, you could use one of the newer words for "first decade of a century":

The aughts
The noughties
The oughties, etc.

For example, one may write,

The 1800s in U.S. Politics

President John Adams moved into the White House in the aughts.

These have an informal ring to them. We have only relatively recently started using these⁵, so they are discontinuous in regional popularity.
3. Being explicit
In my view, this extract states the best practice in this case, which I, as a reader, find concise enough.

When writing about the first two decades of a century, it's probably best to be a little wordy for the sake of clarity. For example:

History seemed to repeat itself in the decade 2000–2009.

— Daily Writing Tips (emphasis mine)

Some might even prefer an even longer version:

Documented here is the first decade of the 1800's with duels, battles, explorations, and births in the U.S. and abroad.
— ThoughtCo (emphasis mine)


Activity and radioactive decay

Question: So,
I am pretty new to that concept, which i am learning about in the context of physical chemistry. I am a mathematician, so a physicists approach can be confusing at times. I have two questions :

In my reference book, it is said that activity $A$ is the number of decays per second  $$ A = -\frac{dN}{dt} \tag{1}$$ where $N$ is the number of particles. It is also said that $$ A = kN. \tag{2}$$
Yet it is said that $(2)$ is always true, whereas $(1)$ isn't.
But, $(1)$ is the mathematical definition of number of decays per second. So isn't it always true, if you look at the activity of one particular particle and not for example a system ?
 And now, this is what confuses me the most. For some exercise, I am supposed to look at a Room, where some decay is happening (222Rn) and you want it's activity to remain constant, by adding some amount of the decaying particle (Rn) in the Room at the end of each day. So, mathematically speaking, the decay is exponential, so although it is true that asymptotically, it can be said to be constant, every time you add some amount, you ruin that behaviour.
So is it meant that the average over the whole day is supposed to be constant ?
Comment: physicist consider a single radon atom (or nucleus). It has a fixed rate of decay, which is an infinitesimal probability per infinitesimal time period that never changes nor is influenced by anything else, and then construct an ensemble of many of them which can then be treated statistically.
Comment: @JEB is your name a reference to one of the Kerman's ? On a more serious note. I get that $N(t) = N(t_0)exp(-k(t-t_0))$ for that exact reason. And also that $N(t_0)$ is what the value number of particles you know are there at $t_0$ . But how on Earth are you supposed, by adding once a day some Radon quantity, get a constant rate ? If you add what you have lost during your day, then on the span of the whole day, i believe the average activity would indeed be constant.
Answer: You're getting too bogged down in precise definition. This is physics, not math.
What you're looking at is a probability distribution. The chance that any one atom decays fits an exponential distribution. When you start with N items, the sum is not memoryless and so not exponential, though when N is large and the time is short enough that there won't be many decays, it's an excellent approximation.
Still, at any instant, kN is the rate where Nis the number of survivors at that instant.
For your second question, yes, they're talking about restoring the previous state.
The assumption was that the decay rate k would be constant over time because they couldn't find anuthing that affected it. Not temperature, ot electric or magnetic fields. Now I think some kinds of radiation are supposed to be affected by the density of neutrinos. The assumption that it was constant came because we couldn't find anything we could change in labs that would change it. Maybe more things will turn up that affect the rate.

Activity and radioactive decay


Are $x \cdot 0 = 0$, $x \cdot 1 = x$, and $-(-x) = x$ axioms?

Question: Context: Rings.

Are $x \cdot 0 = 0$ and $x \cdot 1 = x$ and $-(-x) = x$ axioms?

Arguably three questions in one, but since they all are properties of the multiplication, I'll try my luck...
Comment: your second one isn't usually true...
Comment: Axioms for what?
Answer: I will assume this is in the context of rings (e.g., real numbers, integers, etc).  In this case, the axiom defining $0$ is that $x + 0 = x$ for all $x$.  $x*0 = 0$ is a result of this since we have $x*0 = x*(0+0) = x*0 + x*0$ which implies $x*0 = 0$ (canceling one of the $x*0$'s).
I am guessing that for the second one you mean $x*1 = x$.  This is a definition (axiom).
The third one is a consequence of the definition of $-x$ being the element such that $x + (-x) = 0$.  For then we have $(-x) + x$ is also zero so that $x$ is the negative of $-x$.
Comment: @guillermooo - Better to edit your question to state the assumption explicitly. P.S. That's you from Sublime right? :)
Comment: in other words, only the second one is an axiom.
Comment: Your assumption is correct!
Comment: It's me, yeah! :)
Answer: The question is more profound than is initially seems, and is really about algebraic structures. The first question you have to ask yourself is where you're working:
In general, addition and multiplication are defined on a structure, which in this case  is a set (basically a collection of "things") with two operators we call addition (marked $+$) and multiplication (marked $\cdot$ or $\times$ or $\ast$ or whatever). If this structure holds some properties, which are sometimes called axioms, then it is called a unit ring. The properties are:

The set is closed under the operator $+$. That is, if $a$ and $b$ are in $R$, then $a+b$ is also in $R$.
The set has a member which we mark as $0$. It has the properties that for every $a$ in $R$, $a+0 = 0$ and $0+a = 0$.
The operation $+$ is commutative: $a+b = b+a$.
The operation $+$ is associative: $(a+b)+c = a+(b+c)$.
Every member has an additive inverse: for every $a$ in $R$ there is some $b$ in $R$ such that $a+b = 0$ (we mark $b$ as $-a$).
The set is closed under the operator $*$. That is, if $a$ and $b$ are in $R$, then $a*b$ is also in $R$.
The set has a member which we mark as $1$. It has the properties that for every $a$ in $R$, $a*1 = a$ and $1*a = a$.
The operation $*$ is associative: $(a*b) * c = a * (b*c)$.
Multiplication is distributive over addition: $a * (b+c) = a*b + a*c$ and $(a+b) * c = a*b + a*c$.

While this is a long list, and introduces the operator $+$ which is not even explicitly mentioned in the question, these properties are quite natural. For example, the integers $\{ \ldots, -2, -1, 0, 1, 2, \ldots\}$ we all know and love indeed form a ring. The real numbers also form a ring  (in fact they form a field, which means they hold even more properties).
In regard to your question, the identity $x * 1 = x$ (I assume that's what you meant) is in fact an axiom - it is axiom 7. However, the other two identities are results of the other axioms.
First identity: We use axioms 2 and 9 to get  $0 * x = (0+0) * x = 0*x + 0*x$  and then by adding $-(0*x)$ (the additive inverse of $0*x$, from axiom 5) to both sides,  $0 = 0*x$.
Second identity: As stated in axiom 5, $-(-x)$ is just a notation used which means "the additive inverse of $-x$". To show that $-(-x) = x$ we need to show that $x$ is in fact the additive inverse of $-x$, or in other words that $x + -x = 0$ and $-x + x = 0$. But that's just what axiom 5 says, so we're done.
Last point: You might be wondering why did we have to go and introduce addition to answer a question about multiplication? Well, it so happens that without addition the other two identities are simply not true. For example, if we look at the positive integers $\{1, 2, 3, \ldots\}$ with only multiplication, then there is no $0$ there! Simply put, this is because the positive integers do not form a ring.
Answer: Let $A$ be a ring (which I will assume commutative).
Only the second sentence is an axiom: multiplication has a (provably unique) neutral element, id est, $a \in A$ such that $ax = xa = x$ for every $x \in A$; it's called one and its symbol is $1$.
The others are consequences of axioms:

The 3rd sentence comes from the existence of a symmetric to every element in the ring: for every $x \in A$, there is a (unique) element $y \in A$ such that $x + y = y + x = 0$. It is denoted $–x$. From this equality, also comes that the symmetrical of $y$ can only be $x$.
Analogously to the multiplication axiom above, addition has a neutral element, a (unique) number $b$ such that $x + b = b + x = x$ for each $x \in A$; it's called zero and its symbol is $0$. The first sentence is deduced by using the inverse additive (for some $a$), distributivity and multiplicative associativity axioms: $x · 0 = x · (a + (–a)) = xa + x(–a) = xa + (–xa) = 0$.
Answer: x*1 =x is axiom.
The rest is not.
Let me pencil push the rest :D. It's been a while.
<code>X*0
=x*0+0 ' adding 0 does no harm
=x*0+(x*0+(-(x*0))) ' 0 is the sum of x* 0 and it's additive inverse
=(x*0+x*0)+(-(x*0)) ' associativity.
=x*(0+0)+(-(x*0)) 'distributive
=x*(0)+(-(x*0)) '0+0 is 0
=x*0+(-(x*0)) '(0) is just 0
=0 ' the sum of x*0 and it's own additive inverse is 0
</code>
Notice that I add +(-(x*0)) to mean add by the additive inverse of (x*0).
From now one I would write +(-x) by simply -. So we define a subtraction as adding by the additive inverse.  (is this the standard definition of substraction?)
Now, let's do -(-x)
<code>-(-x)
=-(-x)+0 'adding 0 does no harm
=-(-x)+(-x+x) '0 is (-x+x) by definition of -x.
</code>
I wonder if the axiom only specify that only (x+ -x) is zero, then things can get more tricky. We would need the commutative axiom as reinforcement
<code>=(-(-x)+(-x))+x ' by associative.
=0+x ' the additive inverse of (-x) added by (-x) is 0
=x 'by definition of 0
</code>
I sort of like this kind of derivation because it follows the pattern of
A=...=...=...=...=...C which is better than to go half way and then use logic to imply stuffs.
Answer: Answering what you asked, the question is poorly formed because you didn't specify the theory you are talking about. To do that, well, you must define symbols, deduction rules... and axioms.
Answering what you probably meant to ask, no. Those are properties that $-$ and $\times$ have in the $\mathbb Z$ set, not axioms. They can be deduced from the operations themselves.


How to hang painting frames?

Question: I know there are many styles of painting frames out there, is there a "best" one? And what are the various usage techniques of them?
Comment: There is no best type of way to hang a painting frame. I've closed this because it is very subjective.
Comment: This question is not about best type of a frame - but how to hang a frame!!! Usually, frames are used for paintings. You know, like art pieced with paint on them...
Answer: Thats a great question, few days ago I asked something similar but got put "on hold".
For the question I had, there was a "vintage" old frame that did not have a back cover and I was interested to know how to use it:

The answer for my question at least was that there is a ridge in the inside frame where the glass cover can sit, and the art piece will be placed behind it. In turn, behind the art piece, a carboard or something solid will be placed to hold it in place. Now, because it is an older frame, the only way to secure the glass, art, and backing all together will be to drive small nails into the frame, bend them, and use them as stoppers.
Hope it helps.

How to hang painting frames?


Does anyone know how to install drivers for Canon MG3200 from web?

Question: Any help greatly appreciated, fairly new to both Ubuntu & linux
Answer: You will need to download the drivers from the canon website, since Ubuntu does not provide these drivers. at least for me the drivers Ubuntu provided for canon were not doing their job correctly.
What you need to do:
Browse to the canon website and look for support & downloads for your printer
Select Linux as your OS and search for the correct drivers. You will need the debian package of the IJ printer driver and the Scangear driver, which is needed for printing and scanning respectively.
If you downloaded these, you will need to run them via the terminal, which should happen automatically if you double click on it.
If you did that for both drivers, printing will function correctly. If you want to scan something, you will need to type "scangearmp" in the terminal, which causes the scanning software to run.
Furthermore: Some websites of canon don't have the drivers for Linux. You will need to look on other websites of canon (of other countries) to find them.
Feel free to ask for clarification if it was not clear enough!


How to create a new branch from a tag?

Question: I'd like to create a new master branch from an existing tag. Say I have a tag v1.0. How to create a new branch from this tag?
Comment: possible duplicate of [Checkout GIT tag](http://stackoverflow.com/questions/5582208/checkout-git-tag)
Answer: Wow, that was easier than I thought:
git checkout -b newbranch v1.0

If running the above command you get an error

warning: refname 'v1.0' is ambiguous.
fatal: ambiguous object name: 'v1.0'

this is because you also have a branch with name v1.0. In this case specify tags/ prefix:
git checkout -b newbranch tags/v1.0

Comment: Correct.  Note you also could have just set the master branch back to the point of the tag with `git reset --hard v1.0`
Comment: The suggestion from @wadesworld could work, but if anyone reads this and isn't 100% sure what it means to reset the master branch, *don't do that*.
Comment: If this doesn't work because of " is not a valid commit" or a similar error (often when working on a shared repository), refer to https://stackoverflow.com/questions/35979642/what-is-git-tag-how-to-create-tags-how-to-checkout-git-remote-tags#35979751
Comment: This is a better answer that I found here - https://stackoverflow.com/a/35979751/3145960
Comment: `git switch -C` should be the more "modern" way; see the [below answer](https://stackoverflow.com/a/73153620/3429115)
Comment: If you get an error  " is not a valid commit" or a similar error", please make sure you have executed "git fetch --tags" first.
Answer: If you simply want to create a new branch without immediately changing to it, you could do the following:
git branch newbranch v1.0

Comment: I think this should be the accepted answer as it does exactly what is required. The accepted answer does something else not stated in the question.
Answer: I used the following steps to create a new hot fix branch from a Tag.
Syntax
git checkout -b <New Branch Name> <TAG Name>

Steps to do it.

git checkout -b NewBranchName v1.0
Make changes to pom / release versions
Stage changes
git commit -m "Update pom versions for Hotfix branch"
Finally push your newly created branch to remote repository.

git push -u origin NewBranchName

I hope this would help.
Answer: I have resolve the problem as below
1. Get the tag from your branch
2. Write below command
Example: git branch <Hotfix branch> <TAG>
    git branch hotfix_4.4.3 v4.4.3
    git checkout hotfix_4.4.3

or you can do with other command
git checkout -b <Hotfix branch> <TAG>
-b stands for creating new branch to local

once you ready with your hotfix branch, It's time to move that branch to github, you can do so by writing below command
git push --set-upstream origin hotfix_4.4.3

Comment: If you last release was `v4.4.3`, shouldn't your new branch be called hotfix_4.4.4
Answer: An exemple of the only solution that works for me in the simple usecase where I am on a fork and I want to checkout a new branch from a tag that is on the main project repository ( here upstream )
git fetch upstream --tags

Give me
From https://github.com/keycloak/keycloak
   90b29b0e31..0ba9055d28  stage      -> upstream/stage
 * [new tag]    11.0.0     -> 11.0.0

Then I can create a new branch from this tag and checkout on it
git checkout -b tags/<name> <newbranch>
git checkout tags/11.0.0 -b v11.0.0

Comment: I think you have this backwards in your first example: `git checkout -b  tags/`
Answer: At the time of writing this answer, the most up-to-date idiomatic git use is this command:
git switch -C branch tag

Answer: The situation becomes a little bit problematic if we want to create a branch from a tag with the same name.
In this, and in similar scenarios, the important thing is to know: branches and tags are actually single-line text files in .git/refs directory, and we can reference them explicitly using their pathes below .git. Branches are called here "heads", to make our life more simple.
Thus, refs/heads/master is the real, explicit name of the master branch. And refs/tags/cica is the exact name of the tag named cica.
The correct command to create a branch named cica from the tag named cica is:
git branch cica refs/tags/cica

Answer:
Open branches/tags dropdown
Switch to 'Tags' tab
Select the required tag

Redirected to the tag repository—

Open branches/tags dropdown
Switch to 'Branches' tab
Start typing to create new branch with name
Comment: This question is about Git, not GitHub.
Answer: My branch list (only master now)

My tag list (have three tags)

Switch to new branch feature/codec from opus_codec tag
git checkout -b feature/codec opus_codec

Comment: Please don't use screenshots if also text copy-paste would be enough.
Comment: Downvoted because this answer doesn't add anything to the accepted answer from 2012.
Answer: i was doing this like the other answers state with git checkout <tag> -b <new-branch-name> (and not the order of the tag vs branch being entered doesnt matter), even after doing a git fetch and was still getting a
fatal: 'my-tag' is not a commit and a branch 'temp-release-xxx' cannot be created from it

when i ran the more explicit git fetch --tags, then i was able to run the above command successfully
i thought git fetch would fetch all tags too, but sometimes you gotta nudge git i guess

How to create a new branch from a tag?


Automated daily summary of top stories from Google Reader?

Question: I love Google Reader, but am tiring of constant, all day long updates. Does anyone know of a way to get an automated list of the top stories from all my feeds, by number of links or some other statistic, so at the end of each day I can see a news paper like report on the days most important stories to appear in my feeds?
Answer: If you change "View Settings" (or "Folder Settings" for a particular folder) to "Sort by magic", GReader will pull to the top the items it thinks you'll be most interested in, based on your past activity. (What you've read, clicked on, etc.)


iOS 13 VoIP push notification - didReceiveIncomingPushWith not getting called when the application is terminated

Question: I am implementing a VoIP calling functionality in an iOS application. In the following method, I am immediately reporting an incoming call with CallKit:
func pushRegistry(_ registry: PKPushRegistry, didReceiveIncomingPushWith payload: PKPushPayload, for type: PKPushType, completion: @escaping () -> Void)
When the application is in the foreground or background, this method gets triggered and the call gets reported perfectly. However, when the application is terminated, this method does not get called at all.
If I run the application from XCode with the "Launch: Wait for the execurable to be launched" option added in the scheme, I can see, that something gets run when the VoIP push arrives, but not that method. Instead, I get a couple of seconds later the following message in the console: Message from debugger: Terminated due to signal 9
Any help/idea/input is very much appreciated.
Answer: Additional information: this is what I have for "Background Modes".
Answer: You need to check Background processing in the background modes


When was the shoulder rest invented?

Question: When was the shoulder rest for violins invented, and when did it come into widespread use?
Comment: I found a date only for the chinrest, invented by Louis Spohr around 1820.
Comment: As far as I know actual shoulder rests as such appeared in the first half of the 20th century - but haven't found any particular date. That may also be because it is not all that easy to state at which point exactly some kind of cushion you use instead becomes an actual chinrest. The question of widespread use is similar - for that I'd actually suggest looking at the history of some companies producing shoulder rests or so.
Answer: According to this article, the shoulder rest was a mid 20th Century invention. However, the author also seems to imply that the 'unofficial version' of the shoulder rest was in use even in the time of the invention of the chin rest - the "shoulder pad."

Should we decide rather to take Prinner's and Leopold Mozart's advice and use the chin-on position, there is one very important consideration, which is quite simply not mentioned in any of the 17th or 18th century sources; - namely, how to get the chin close enough to the violin to support it, without the use of some kind of shoulder pad (first mentioned by Baillot in 1834), a chin-rest (invented around 1820 by Spohr), or even more extreme, a shoulder rest (a mid 20th century invention).

This professor seems to think it was invented in the 1920's-30's, more specifically.
Ultimately, the fact that there seems to be no specific date to 'pin-down' the invention of a thing that has become such a staple of use and debate in the music world seems to give credence to the idea that the old shoulder pad had already been in use to achieve the same effect long before the 'shoulder rest' was an actual thing.
Comment: Mulling over this after the fact, I speculate that the invention became a necessity as the use of shoulder-pads waned in fashion...
Answer: I didn't write this, I'm just copy pasting. Taken from here:
https://maestronet.com/forum/index.php?/topic/321907-shoulder-rest-history/
Here's the post:
The first actual (American) patent I can find for something that is actually a specific shoulder rest (actually a shoulder rest/chin rest combo) and not a pad, is dated 2/9/1909, issued to G. Becker as patent number 908541. You can find it at the US Patent search site. In the text it is referred to as an improvement to such rests already available! From that we can deduce that such rests, at least like this one, were already available on the market, if not in common use (but one assumes that they would not be on the market if they didn't sell, and therefore must have been used by some players). Unfortunately, Becker does not reference the patent or patents or device that he is improving on, and considering the date, it's possible that some similar devices actually had no patents (though not likely).
An interesting side note is that, if one studies patents, it will be seen that often the actual device eventually sold does not always resemble the drawings patented, although they must include the basic features to protect the patent..
This assumes a lot I admit, but to me it indicates that specific shoulder rests were in use at least on or before that date.
Another interesting patent issued to a Colbertson on 4/29/1930 as patent number 1756676, seems to be designed to give a padded edge to the bottom plate for players that did not normally use a rest, but didn't like the feel of the plate edge on the collarbone. It is similar to something I tried to come up with a few years ago but failed. There are many other patents listed as one approaches the present.
Again, I am assuming a lot here, but this info seems to indicate that specific use shoulder rests (not pads) were in use at least by the turn of the last century, and are not a recent development, although considering the lifespan of the violin, I suppose it could be considered recent.

When was the shoulder rest invented?


This is Quite Strange!

Question: My last posting relating to the finite/infinite content of the universe has been deleted without informing me.At least I can't find it.Its not there on the user page also.My current posting incidentally is "On Parallel Transport".
Comment: Some of your questions leave something to be desired, but you sometimes hit on interesting points. Just try to be clearer and terser, it takes a lot of effort to follow your arguments sometimes, and the payoff is often not great enough to do so. It was probably closed and deleted by a mod. What did it ask? Can you paste the content in the question here so we know if it was justified?
Comment: The justification issue could have been reasonably handled at the time of closure by informing me. Keeping the original poster uninformed is quite unethical.I suddenly noticed that my posting was not there.
Comment: The question was definitely not deleted by a moderator; that would have been recorded in its history. It had to be some sort of automated action, as dmckee said in his answer. If you would like to argue that you should be emailed when a deletion of this sort happens, you could certainly bring it up on [metaSO].
Comment: Anamitra - 'Unethical'? Really? - there is no ethical reason to inform the poster. Sure, it would be handy, but ethics don't come into it.
Comment: I have an email subscription by which they could have informed me of the closure in case it was closed in a legitimate manner. I have received emails in relation to comments regarding the same posting earlier.But nothing in relation to the closure!
Comment: I'm not sure the argument that you were "uninformed" holds much weight. If the question was negatively voted, then that's a pretty clear indication that it needs improvement. Not everyone wants to receive notifications of SE activity through email, although, as was mentioned, this is certainly something that can be discussed on Meta Stack Overflow.
Comment: Voters don't have to give explanations for a negative vote----it is a secret ballot.Justification is not mandatory. Deletion by software in this case is based on such negative voting which does not require any reasoning. And you call it ethics!
Answer: It appears that the question may have been the victim of an automated "bad question" reaper.
As I recall the qualification for automatic removal are

one month old
negative vote total on the question and
no answers

If it is not that I'm stumped as it does not appear to have been done by anyone (including Community who get the "credit" for closure due to spam flags).
Comment: @AnamitraPalit This is another feature of the network that exists to help manage the multi-thousand question per day load on Stack Overflow. To be sure, it is not really necessary on Physics.SE, but on the other hand your posting was negative voted, and not interesting enough o have attracted any answer in month. Were we supposed to assume that on day 35 you were going to come back and try to fix it? In any case, the *time* of deletion was noted (that's how I was able to learn that was 30 days on and guess what happened), and Kevin has already said that the breadcrumbs will be improved.
Comment: This is correct, although the next build will start recording and displaying that Community did these deletions explicitly.
Comment: If the software deletes an entire posting with comments without informing the concerned poster[who has an email subscription] I have to say that the software was "POORLY" developed.Proper considerations were missing while creating the software.The "software" did not notify the moderators of the deletion nor did it register the date of deletion.
Comment: @ dmckee --- ex-moderator kitten: Your comment does not qualify for the fact that a thread should be deleted without informing the original poster[one who started the discussion], In any case prior to a deletion the poster should be given the opportunity to reason in favor of the discussion started by him.Negative or positive voters do not have to provide reasons  for their action. Considering a thread to be uninteresting  in the face of a deletion stands to a suspicion without a debate on it . Subsequent  deletion due to machine constraints as suggested is quite regrettable!
Comment: (in continuation)Considering a thread to be uninteresting  in the face of a deletion stands to a suspicion  . Subsequent  deletion due to  machine constraints(without informing the poster ) as suggested is quite regrettable!

This is Quite Strange!


iMac and MacBook Air shared monitor

Question: I have used a thunderbolt connection and pressed Command F2 to allow my track pad and keyboard on my MBA to run both screens. However, I want to use the wireless track pad and keyboard that comes with the iMac instead of the laptop track/keyboard. In other words, I would like the iMac to be the primary computer and use the laptop as a secondary screen (with all control remaining with the iMac). Is this possible?
Answer: Simple answer, no it's not possible. The Target Display Mode is a function specific to the iMac and this functionality isn't incorporated into MacBooks.
You could of course pair the wireless keyboard and trackpad to your MacBook and use TDM on the iMac, but if you're trying to leverage the iMac's CPU power you're out of luck I'm afraid. You'd be better off buying a small cheap monitor if you need a secondary display for use with the iMac.

iMac and MacBook Air shared monitor


Headphones, MacBook Pro, Parallels and no sound on the headphones

Question: I have Parallels installed on my MacBook Pro.  I have a headset plugged in.  There is no option to choose to play sound on the headset.  My options are Internal Speakers and Internal Sound or Parallels Access Sound for both input and output.  What do I need to change in order for the headphones to be usable?
Answer: Go to System Preferences > Sound Click on headphones or internal speakers


Testing for stationarity in biosignal data stream

Question: I am looking for the state-of-the-art in stationarity testing. I have a number of streams of data acquired from EMG recordings and I would like to know which one is the most "stable" or stationary among the streams. I've read a number of articles and presentations, for example here and here. Since such signals are contaminated by a high amount of noise and may include various artifacts, such as body motion artifacts, I am not sure which methods are valid and robust. I would like to know if anyone has hands-on experience with such statistical measures and can point me in the best direction. Thank you!
Answer: The probability density function (pdf) of an electromyography (EMG) signal provides useful information for choosing an appropriate feature extraction technique. The pdf is influenced by many factors, including the level of contraction force, muscle type, and noise. In this paper, we investigated the pdfs of noisy EMG signals artificially contaminated with five different noise types: 1) Electrocardiography (ECG) interference; 2) many spurious background spikes; 3) white Gaussian noise; 4) motion artifact; and 5) power line interference at various levels of signal-to-noise ratio (SNR)
detail explanation on probability density function for Stationary


Singular covariance matrix, understanding the beginning of a proof

Question: Let $\left\{X_t,t\in T\right\}$ be a stationary process such that $\text{Var}(X_t)<\infty$ for each $t\in T$. The autocovariance function $\gamma_X(\cdot)=\gamma(\cdot)$ of $\left\{X_t\right\}$ is defined to be
$$
\gamma(h)=\text{Cov}(X_{h+t},X_t)~\forall h,t\in\mathbb{Z}.
$$
Moreover, assume that $EX_t=0$ for each $t\in T$.
There is the following statement:

If $\gamma(0)>0$ and $\gamma(h)\to 0$ as $h\to\infty$, then the covariance matrix $\Gamma_n=[\gamma(i-j)]_{i,j=1,\ldots,n}$ of the column vector $(X_1,\ldots,X_n)'$ is non-singular for every $n$.

The proof of this starts as follows:
Suppose that $\Gamma_n$ is singular for some $n$. Then since $EX_t=0$ there exists an integer $r\geq 1$ and real constants $a_1,\ldots,a_r$ such that $\Gamma_r$ is non-singular and
$$
X_{r+1}=\sum_{j=1}^r a_j X_j.~~~~~(*)
$$
I do not completely see this argument. In particular, there are two inaccuricies to my opinion.
Here's how I do understand it:
Suppose $\Gamma_n$ is singular. This means, its determinant is zero. So there is at least one column (one row) that is a linear combinations of the other columns (rows). If we delete this column (row), we get $\Gamma_{n-1}$, if its determinant is again zero, we repeat this. Eventually, we get $\Gamma_r$, for some $r\geq 1$, such that all columns (rows) are independent, meaning that $\Gamma_r$ has positive determinant, i.e. is non-singular.
So far so good.
It remains to argue, how to get $(*)$.
In the proof, there is said to look at the following statement that one probably should use for my problem:

If $X=(X_1,\ldots,X_n)'$ is a random vector with covariance matrix $\Sigma$, then $\Sigma$ is singular if and only if there exists a non-zero vector $b=(b_1,\ldots,b_n)'\in\mathbb{R}^n$ such that $\text{Var}(b'X)=0$.

So, I apply this as follows:
As done above, let $\Gamma_r$ be non-singular. Since $r$ was the smallest $r\geq 1$ such that all columns are linear independent, the covariance matrix $\Gamma_{r+1}$ of $X=(X_1,X_2,\ldots,X_{r+1})'$ is singular. By the cited statement, there is some $b=(b_1,b_2,\ldots,b_{r+1})'\in\mathbb{R}^{r+1}$ such that
$$
\text{Var}(b'X)=0.
$$
But this means that
$$
b'X=\sum_{i=1}^{r+1}b_iX_i=E(b'X)=0\text{ almost surely },
$$
implying that
$$
b_{r+1}X_{r+1}=-\sum_{i=1}^r b_iX_i.
$$
Imho, we can suppose that $b_{r+1}\neq 0$ since if $b_{r+1}=0$, we have that $0=\text{Var}(b_1X_1+b_2X_2+\ldots + b_rX_r+b_{r+1}X_{r+1})=\text{Var}(b_1X_1+b_2X_2+\ldots + b_rX_r)$, meaning by the cited statement that $\Gamma_r$ is singular. But $\Gamma_r$ is non-singular. Hence
$$
X_{r+1}=\sum_{i=1}^r a_iX_i,~~~a_i:=-\frac{b_i}{b_{r+1}}
$$
or, more precisely, this identity holds almost surely. I don't know why the "almost surely" is omitted in the proof.
Is this okay?
Answer: No, it's straightforward from the first statement.
$ \Gamma_n $ is singular $\Rightarrow \exists b_n s.t. \Gamma_nb_n=0 \Rightarrow b_n'\Gamma_nb_n = 0 \iff Var(b_n'X)=0 $ where $X = (X_1,X_2,...,X_n)'$
Since $E(X)=0$, $Var(b_n'X) = E((b_n'X)'(b_n'X))=0 \Rightarrow b_n'X = 0 $ a.s.
And the result follows.


What is this on two boards in crawl space?

Question:
These two boards are in a dry crawl space. Whatever is on them is not present on any other boards. Is this mold? Or something else?
Comment: It's a cardboard box. It appears to contain puzzles and games.
Comment: It is probably those two boards were used somewhere else before being used there.  Too neat for mold/fungus/water stains.
Comment: @crip659 I agree it's likely from those boards being used elsewhere before being installed in their current location.  But it could still be mold from a previous location.  That said, and it's real hard to tell from the photo, I think it may be concrete residue.
Answer: I've seen similar patterns where apparently some of the boards were first used in a concrete "form", then the form was disassembled (after the concrete was poured and hardened) and the lumber reused.
Comment: TL;DR: It's nothing to worry about.
Comment: The person who built my garage back in the 50's used a reeeeaallllly long 2x4 (I've never seen one this long anywhere else) as the screed for the concrete slab, then used it to brace the wonky truss system they built.
Answer: I'd guess you have two boards that are pressure treated (or possibly paint-on treated) with a copper-containing mixture sharing the diagonal sheathing job with a bunch of un-treated boards. Hard to think that water or mold would be so contained to just two boards, not running onto their neighbors, and that they would happen to appear cuprous green.
Comment: I agree that there is almost no chance that those 2 slats would look like that from mold that occurred *after* installation.  It *is* possible that the boards have mold from a *previous* installation, but I can't tell from the photo.  The 2x4 looks a bit suspect, but maybe that's concrete residue?  Really hard to tell from the single photo, which is blurry for the 2x4.  If the 2x4 isn't holding any weight, I would maybe consider replacing it because it would be so simple and inexpensive to replace.  But I have a hunch that's concrete residue on it.
Answer: It looks a little too blue-green for this, but boards left in the sun will eventually turn grey due to UV weathering.
Answer: It looks like you can visually inspect the cut end of the 2x4.  If so, does it show the same discoloration?  If there is no discoloration on the cut end, then, as many others have said, it is probably residual concrete or mold from a previous installation.


How do I select Edit Mode after I imported an SVG into Blender 2.79b?

Question: After I imported the SVG file, it automatically entered Object Mode. When I select the Object Mode drop-down, there are no options other than Object Mode. I am specifically trying to correct the faces/spaces reversal problem, but can't get out of Object Mode to Edit Mode. Tab doesn't switch between the modes.
Comment: You can only edit one object at the time. Selecting all will not work you need a single active object. With an active object you can enter edit mode.
Comment: How are you selecting it, is there an active object? Try clicking on one of the imported objects first
Comment: @DuarteFarrajotaRamos I chose Select > Select All By Layer and the object is selected. This is on a Mac, if that makes any difference.
Comment: @DuarteFarrajotaRamos I now have Edit Mode in the Object dropdown. I have no idea what made the difference. I did duplicate the object.
Comment: Selecting with patterns via Shift+G or selecting all doesn't make any object as active which is needed in order to start editing it
Answer: After importing the SVG you need to right click it in 3D view or left click it in Outliner. Then you will be able to Tab into Edit Mode without issues.
Comment: Clicking makes it active. Select all, even with only one object present in the scene, does not? OK.


Are Certain Magical Item Not Disenchantable Skyrim?

Question: I have a magic staff that has an enchant that summons a familiar for 60 seconds wherever the caster is pointing but when I go to the arcane enchanter table, it does not show for for disenchanting.  Are there certain things you can disenchant or do I need to be certain level in enchanting to disenchant these items?
Answer: You may neither enchant nor disenchant staves.
Additionally, you may not disenchant unless the item has an enchamtment you don't already know.
Additionally, there are some quest rewards you may not disenchant (for example: Ghost Sword).
And finally, there are some items (such as the Amulet of [Insert Aedra here]) that you simply can't disenchant.
Comment: additionally there are apparel items such as some Dark Brotherhood items that cannot be disenchanted. I havent confirmed this totally but it seems to me that if there isnt an 'of [effect]' then it can't be disenchanted.


How do you get the monster masher achievement in the ipad version?

Question: I am close to getting all the achievements in the ipad version but I am having trouble getting the monster masher achievement. No matter what I try, I just squash 1 or 2 zombies. Has anyone got this achievement and can tell me how?
Answer: Once you completed the game you can go another run, but then with all plants unlocked from the start.
At an early level take pumpkin, Tall nut and the regular wall nut. Fill the top lane with those and play the other 4 lanes regularly.
This will pile up the zombies behind your top lane wall and then you can take them out with a single squash. You can do something similar with a cherry bomb for an other achievement.
Comment: would go for pumpkin imitator seeing you can never have enough of those.
Comment: Should I bring Imitator for any of those plants?
Comment: What if my other lanes need a defensive plant?


Is BeginInvoke/EndInvoke good practice for invoking on main thread?

Question: Is it good practice to invoke delegate for MainForm thread - this way?:
Txt.MainForm.EndInvoke(
Txt.MainForm.BeginInvoke(
    new MethodInvoker(delegate()
       { // code here }
)));

Answer: No - because if you're calling EndInvoke, that will block until the delegate has completed. If you want that behaviour, just use Invoke instead.
To put it another way: if you're trying to do something other than blocking until your (presumably UI-modifying) delegate has executed in the UI thread, you should explain what that something is. If there isn't anything else, then Invoke will give you simpler code.
Answer: It doesn't make a lot of sense as the code fires up an asynchronous call and then immediately waits for the call to finish. I.e. you end up waiting on the calling thread.
Answer: Not considering the thing that other mentioned (I believe this EndInvoke - BeginInvoke chain is just an example usage of delegate): Using delegates is 100% OK. If this is the only usage of the delegate body, there's no need to define it as a named method. It is cleaner in the code and there's no need to jump through the file. Consider using newer syntax for delegates:
new MethodInvoker(() => { // code here })


Using DataLoader (Command line) on Mac

Question: Is it possible that Data Loader command line can work in os x?  After all it is "just java."  I'm sure some of the JDBC stuff might not work but is it possible that simple CSV extracts could work? I'm hoping that with the right shell scripts it can be done but if people know otherwise I won't waste my time. Thanks.
Comment: See also the answer for this question: https://salesforce.stackexchange.com/questions/30484/command-line-data-loading-on-mac-os-x
Answer: I used it a while back ... not sure if it still works.  Here are some tools to use the Mac DataLoader from the command line.  https://github.com/sthiyaga/dataloader
Answer: Command Line for Dataloader is not supported by Mac, only Windows.
Comment: Thanks. I get that it is not supported, I was asking if it is known to work at all.
Answer: Yes, it does work.
I created tool a while back to help with it:
https://github.com/theswamis/dataloadercliq
Let me know if you run into any trouble.
Comment: It would be helpful to include the overview text in the answer as well as the link in case the link breaks.

Using DataLoader (Command line) on Mac


Metric in dilatation transformation of massless scalar field

Question: The lagrangian density of the massless real scalar field is
\begin{align}
L = \frac{1}{2}\eta^{\mu\nu}\partial_\mu\Phi\partial_\nu\Phi = \frac{1}{2}\partial_\mu\Phi\partial^\mu\Phi.
\end{align}
I want to check if the action is invariant under dilatation transformation
\begin{align}
x'^\mu=e^\alpha x^\mu \\
\Phi'(x')=e^{-\alpha}\Phi(x).
\end{align}
Since $d^4x'=e^{4\alpha}d^4x$ the lagrangian density must thus transform as $L' = e^{-4\alpha}L$. My question here is, how the metric $\eta^{\mu\nu}$ transforms.
If I have a look at the line element
\begin{align}
ds^2=\eta_{\mu\nu}dx^\mu dx^\nu = e^{-2\alpha}\eta_{\mu\nu}dx'^\mu dx'^\nu .
\end{align}
I find, that the transformed metric is $g_{\mu\nu}=e^{-2\alpha}\eta_{\mu\nu}$. When I use this to transform the Lagrangian I get
\begin{align}
L' = \frac{1}{2}g^{\mu\nu}\partial'_\mu\Phi'(x')\partial'_\nu\Phi'(x') = e^{-6\alpha} L.
\end{align}
So this differs from the correct result by a factor of $e^{-2\alpha}$. I think my mistake here is, that I also transform the metric, but I dont get why the metric should stay the same since it obviously transforms under this coordinate transformation.
Comment: What about the transformation of $\sqrt{|g|}$?
Answer: We are in flat spacetime, so the action is
$$
S=\int d^4x \, \frac{1}{2}(\partial \phi)^2
$$
The Lagrangian transforms as
$$
\mathcal{L}'=\frac{1}{2}(\partial \phi')^2=\frac{1}{2}\partial'_{\mu}\Phi'\partial'^{\mu}\Phi'=e^{-4\alpha}\mathcal{L}
$$
which cancels with
$d^4x'=e^{4\alpha}d^4x$.
Note: this is not a Lorentz transformation, so $ds^2$ does not remain invariant. It changes as
$$ ds'^2=\eta_{\mu\nu}dx'^{\mu}dx'^{\nu}=e^{2\alpha}ds^2 $$
Comment: Why do you integrate over $(\partial_\mu\Phi)^2$ and not over $\partial_\mu\Phi\partial^\mu\Phi$? These two differ by a sign on the 0th component
Comment: When I wrote $(\partial \Phi)^2$ I meant $\partial_{\mu}\Phi\partial^{\mu}\Phi$, it is just another way to write the same thing.


What is an unprivileged LXC container?

Question: What does it mean if a Linux container (LXC container) is called "unprivileged"?
Answer: Unprivileged LXC containers are the ones making use of user namespaces (userns). I.e. of a kernel feature that allows to map a range of UIDs on the host into a namespace inside of which a user with UID 0 can exist again.
Contrary to my initial perception of unprivileged LXC containers for a while, this does not mean that the container has to be owned by an unprivileged host user. That is only one possibility.
Relevant is:

that a range of subordinate UIDs and GIDs is defined for the host user (usermod [-v|-w|--add-sub-uids|--add-sub-gids])
... and that this range is mapped in the container configuration (lxc.id_map = ...)

So even root can own unprivileged containers, since the effective UIDs of container processes on the host will end up inside the range defined by the mapping.
However, for root you have to define the subordinate IDs first. Unlike users created via adduser, root will not have a range of subordinate IDs defined by default.
Also keep in mind that the full range you give is at your disposal, so you could have 3 containers with the following configuration lines (only UID mapping shown):

lxc.id_map = u 0 100000 100000
lxc.id_map = u 0 200000 100000
lxc.id_map = u 0 300000 100000

NB: as per a comment recent versions call this lxc.idmap!
assuming that root owns the subordinate UIDs between 100000 and 400000. All documentation I found suggests to use 65536 subordinate IDs per container, some use 100000 to make it more human-readbable, though.
In other words: You don't have to assign the same range to each container.
With over 4 billion (~ 2^32) possible subordinate IDs that means you can be generous when dealing the subordinate ranges to your host users.
Unprivileged container owned and run by root
To rub that in again. An unprivileged LXC guest does not require to be run by an unprivileged user on the host.
Configuring your container with a subordinate UID/GID mapping like this:
lxc.id_map = u 0 100000 100000
lxc.id_map = g 0 100000 100000

where the user root on the host owns that given subordinate ID range, will allow you to confine guests even better.
However, there is one important additional advantage in such a scenario (and yes, I have verified that it works): you can auto-start your container at system startup.
Usually when scouring the web for information about LXC you will be told that it is not possible to autostart an unprivileged LXC guest. However, that is only true by default for those containers which are not in the system-wide storage for containers (usually something like /var/lib/lxc). If they are (which usually means they were created by root and are started by root), it's a whole different story.
lxc.start.auto = 1

will do the job quite nicely, once you put it into your container config.
Getting permissions and configuration right
I struggled with this myself a bit, so I'm adding a section here.
In addition to the configuration snippet included via lxc.include which usually goes by the name /usr/share/lxc/config/$distro.common.conf (where $distro is the name of a distro), you should check if there is also a /usr/share/lxc/config/$distro.userns.conf on your system and include that as well. E.g.:
lxc.include = /usr/share/lxc/config/ubuntu.common.conf
lxc.include = /usr/share/lxc/config/ubuntu.userns.conf

Furthermore add the subordinate ID mappings:
lxc.id_map = u 0 100000 65535
lxc.id_map = g 0 100000 65535

which means that the host UID 100000 is root inside the user namespace of the LXC guest.
Now make sure that the permissions are correct. If the name of your guest would be stored in the environment variable $lxcguest you'd run the following:
# Directory for the container
chown root:root $(lxc-config lxc.lxcpath)/$lxcguest
chmod ug=rwX,o=rX $(lxc-config lxc.lxcpath)/$lxcguest
# Container config
chown root:root $(lxc-config lxc.lxcpath)/$lxcguest/config
chmod u=rw,go=r $(lxc-config lxc.lxcpath)/$lxcguest/config
# Container rootfs
chown 100000:100000 $(lxc-config lxc.lxcpath)/$lxcguest/rootfs
chmod u=rwX,go=rX $(lxc-config lxc.lxcpath)/$lxcguest/rootfs

This should allow you to run the container after your first attempt may have given some permission-related errors.
Comment: Good answer - but `lxc` is not a necessity for this kind of thing. You can create a namespace container of any kind using the `util-linux` tool `unshare`. You can enter said container using the `util-linux` tool `nsenter`. The latter tool also allows you to add running processes to an already created container from without it. The namespace support is implemented in-kernel.
Comment: @mikeserv: you mean you don't need LXC to make use of [tag:userns]? I knew that. I also know Docker now has its own library making use of these facilities. But how would you containerize a whole system without the help of facilities offered by LXC? And why would you do it? I mean to contain a single application and combined with `chroot` this can help, but LXC combines various of the name spaces (UTS, mount etc ...) to containerize the whole system.
Comment: Well... As i said, `unshare` already does this admirably for any/all of the various namespaces - and will even get you a separate, private `/proc` mount with a single cli-switch. If your *single application* is `init` and your `chroot` is `initramfs` then you get a whole container in seconds flat.
Comment: Note that in recent LXC the key is called `lxc.idmap`
Answer: To followup on 0xC0000022L, whose solution worked fine for me, I wrote a increase-uid-gid.pl perl script to automatize the necessary ownership changes required so files within the LXC containers are properly mapped.
Without it, with this proposed setup, a file within LXC container rootfs belonging to 0/root on the main host will, within the LXC container itself, be mapped to 65534/nobody.
To be mapped to 0/root within the LXC container, they must belong 100000 on the host.
This is described here https://yeupou.wordpress.com/2017/06/23/setting-up-lxc-containers-with-mapped-giduid/ and the script can be directly obtained on gitlab https://gitlab.com/yeupou/stalag13/blob/master/usr/local/bin/increase-uid-gid.pl

What is an unprivileged LXC container?


$O(n \log n)$ simple polygon triangulation via divide and conquer

Question: I am looking for the simplest possible $O(n \log n)$ algorithm for triangulating a simple polygon.  It seems like there should be a simple divide and conquer variant that would fit the bill, ideally one avoiding auxiliary data structures such as trapezoidal decompositions.  Indeed, all that would be required is an $O(n)$ algorithm for finding a roughly balanced diagonal.  However, I am unable to find any algorithms that are as simple as I (perhaps naively) imagine should exist; nearly all the research is focused on $o(n \log n)$ algorithms, nearly all of which operate via trapezoidal decomposition.
Comment: A polygon can be triangulated in *linear* time, though the algorithm (due to Chazelle) is not particularly simple: http://www.cs.princeton.edu/~chazelle/pubs/polygon-triang.pdf.
Comment: Yes, linear time is $o(n \log n)$.
Comment: Triangulation by decomposing into monotone pieces avoids the trapezoidal decomposition.
Answer: Garey, Johnson, Preparata and Tarjan came up with a simple $O(n\log n)$ algorithm back in 1978. It is described in many lecture notes, for example these lecture notes of Piotr Indyk.
Comment: Thanks, I've seen that one, though it's essentially doing trapezoidal decomposition.  It's also not really divide and conquer.  It's possible I should get over my allergy to non-rotationally invariant algorithms.
Answer: I was looking for such a diagonal as well. As I could not find a simple way to obtain it we applied a linear time cutting of the polygon (Sutherland–Hodgman approach). Given a simple polygon $P$ we take a vertical line $\ell$ and cut through $P$. This can be done if $O(n)$ time by simply iterating over the boundary list of $P$ and verify if $\ell$ is crossed. This produces a set of sub-polygons. Clearly the number of vertices of the polygons to the left of $\ell$ is equal to the number of vertices on the right side of $\ell$, up to a constant at least. Then we triangulated each sub-polygon. Since there are vertices in the sub-polygons due to the cutting we had to also add a repair step to get rid of these additional vertices. For more details look directly at our work.


eXist Native XML database

Question: I have a simple Wordpress site with phpMyAdmin panel for creating a database. The database which I have to use is so called native XML db eXist-DB and I don't know how to connect it to my website or how to import it. Is there any software or may be plugin to do this?
Comment: Nikki, I'm curious as to whether you got the answer to this in such a fashion that you were able to connect existDB to WordPress. Thanks
Answer: You cannot use Phpmyadmin to connect to eXist. It's a completely different Technology than mySQL and mariaDB.  Don't use eXist if you are beginner. You'd need to learn an absurdly complex technology stack that almost nobody else uses. (Read "Recent News" section how one Java/XML Expert wrestled with eXist in 2010. I doubt it has become much easier).
If you must:
Read the docs on http://exist-db.org/. Download eXist. Ask a Java Programmer how to  setup the tooling infrastructure (JRE/JDK, Maven, Tomcat/Jetty.... whatever else) and connect to the database.
If your website is simple, maybe your legacy data stored in eXist is simple too. Just let an expert extract the data in a one-off task, and store the data as CSV file, perhaps. Then forget about eXist.
Comment: Thank you very much for your answer. The main problem is that I am a humanitarian and I have to operate with xml xsl and xquery and for this purposes Exist-Db is the one that I have to use. I will try to follow your tips. Thank you once again

eXist Native XML database


Is a sequel allowed to start before the end of the first book?

Question: I'm writing 2 stories simultaneously. The first one spans 17 years. The second picks up where the first left off... but only about 25% of the way in.
The sequel seems to work best if I introduce the new characters about 5 years before the end of book 1, but most of the characters don't fit into the first story. (One of them does appear, but only for a single chapter.)
Am I allowed to have a sequel that starts 5 years before the first book ends, or will I need to do something else with that 25%?
Comment: Of course you can do that, there's no set rules. In the sequel, are you going to cover the same events from a different point of view, or the plot is mostly independent at the beginning?
Comment: @Alexander One of the new PoV characters starts out retelling a side event. The others start out being mostly independent.
Answer: As has been pointed out, there are no "rules" stopping you from having a sequel starting within the time frame of another book in the series. "The Horse and His Boy" takes place wholly within the last few chapters of "The Lion, The Witch, and The Wardrobe", for example.
Thinking like a reader, if I loved a book and its characters and I read another book where I can see different characters tangentially intersecting with events I'm already aware of - I would LOVE it. Not only will I be reading a fresh new story with a great new character (or two) and different stakes but I will get a chance to briefly revisit ones I loved without the story directly continuing the last one.
Far from saying this is ill-advised, I would go so far as to say this is a brilliant idea. I wish more authors did things like this.
Advice for writing the sequel
Disclaimer: Your mileage may vary, this is just my opinion.
I would strongly recommend setting out early to anchor your readers with a sense of where and when the new story takes place within the wider fictional universe. That's not to say that an "ah-ha" moment where they realise halfway through couldn't work, just that it carries more risk. It really is up to you if you think you have the skills to pull a late reveal off or not.
Let us get attached to the new characters long before we meet anyone central to the previous story. The reason being is that readers may expect the established character to come in and take over. Meeting established support characters early is unlikely to carry that same impression.
Let us see how the events in the previous story ripple out into this story. I would caution against solving any problems for your characters with those ripples but making additional problems will probably make for a gripping read.
My final advice: in all things writing, if you wonder "should I do this?" or "am I even allowed?" do it anyway. Try stuff. Break rules. If it works, great; if it sucks, tuck it away and pretend it never happened (or figure out what's wrong and fix it). Whatever you write is not set in stone - so experiments are always entirely recommended.
Comment: In the Marvel Cinimatic Universe, The Hulk, Iron Man II, Thor, and Captain America: The First Avenger's modern setting scenes all occur within the same week (News Footage from The Hulk's ramapge at a college is playing in the background of Iron Man II indicating it's on going situation, and Coulson is assigned to the landing site of Mjnoir at the end of Iron Man II.  A deleted scene from The Hulk is the incident that allows for the discovery of Captain America's frozen body and his revival.  There is also a chronological order of the films that does not fit the release order.
Answer: A "sequel" that takes place during the events of its predecessor is called a midquel (or more precisely, an "intraquel"). In your case, only the first 25% of your second book is intraquel, but the point is: not only is this allowed, but it's so common there's even a term for it.
To give you another example, that's probably more relevant to your own story: Shrek Forever After begins during the events of Shrek 1, with Fiona's parents being informed that she's been rescued from the tower just before they were due to sign a deal with Rumpelstiltskin. This is what sparks Rumpelstiltskin's vendetta with Shrek, and sets up the main storyline.
So if your second book needs to be part-intraquel, go ahead and make it part-intraquel.
Comment: other examples are several of the "Ender's Game" series books occur during the events of the original book


"Majority" is defined as the larger part of something. Is the LARGEST part of something better, instead?

Question: Cambridge Dictionary defines "majority" as the larger part of something.
I wonder why they use "larger" instead of "largest".
Could you explain with examples why "larger" fits better than "largest" for the definition of "majority"?
Comment: If you are talking about voting, you could also investigate the word "plurality", which is related to "majority"
Answer: They use "larger" because they are in the comparative.  It involves two parts.  The larger is the majority, the smaller is the minority.
If it is the largest, it may not be the majority if there are three or more parts.  If something is 40%, while two other things are 30% apiece, it's the largest, but it does not constitute a majority, merely a plurality.
Answer: You can use either larger or largest. Logically, if you partition a group between all that are A, and all that are not A, you have two groups, and you can use the simple comparative larger to identify one of the parts.
You can also logically use the term largest.
There is some discussion of this here:
stackexchange ELU *

Use of the superlative when only two items are present.

Most dictionaries indicate superlative for majority. (The term most is superlative.)
Comment: A majority is always *more than half*, regardless of how the minority *(less than half)* might be subdivided.
Comment: We say 'the larger' of two sections, 'the largest' of more than two. If the majority of a string of beads are red, there are more red ones than any other colour, regardless of how many different colours there are.
Comment: @JiHyunLee Yes, the word *majority* implies more than half, and implies a group partitioned in two. You can partition the group into as many subgroups as you like, but if A is more than half, that in itself creates a partition into two: A, and not A.
Comment: Is "Majority" always used to indicate the larger part in a two partitioned group? I wonder whether it can be used in a group divided into more than two, and it also means the larger part in the case.


DevExpress MVC control bind to upshot datasource

Question: I would like to know if its possible to use the DevExpress MVC controls with Upshot and knockout
Answer: I don't believe this is easy to do. DevExpress developed their DXtreme product to interface with the server using Knockout and Upshot. The product is good, but is missing the GridView control, so we aren't able to use it.
If you have a simple form, you could get this to work by using the SetValue() and GetValue() properties on all of the controls. So, instead of using the Knockout data-bind behavior, you could use the normal DevExpress binding, then when you want to save with Ajax you can create a class that takes the values from the forms and sends them back to the server. But that really defeats the purpose of using Knockout for the UI.