How to get Sitecore 10 certification material?

Question: I am totally new to Sitecore and would like to get study material for the certification.
Could someone assist me with this?
Answer: You can go with the Sitecore eLearning website for Sitecore 10 Certification material and details. There you will find the Sitecore 10 .NET Developer Certification Exam.
Prerequisites
This exam is intended for developers. We recommend you meet the following prerequisites before taking the Sitecore 10 .NET Developers Certification Exam:

At least one year of prior experience developing on the Sitecore XP solution.
Complete Learning@Sitecore's Developer's Fundamentals 9.3 Collection.
Complete Learning@Sitecore's Developer's Fundamentals 10 Collection.
Complete Learning@Sitecore's Sitecore Experience Solution Developer 9.1 Instructor-Led Training

To pass the Sitecore 10 .NET Developer Certification Exam, you will also need to be proficient in the exam competencies. We recommend you prepare by participating in Learning@Sitecore training opportunities, enrolling in, and completing our free study guide, reviewing Sitecore's developer documentation, and speaking with others within the Sitecore development community at large.
Exam competencies
To be certified in Sitecore 10 .NET Developer you are expected to have fundamental knowledge, skills, and abilities in seven distinct areas, known as competencies:

Competency 1: Sitecore structure & platform (12%)
Competency 2: Security & user management (16%)
Competency 3: Item management (20%)
Competency 4: Layout & placeholders (10%)
Competency 5: Components, controls, & renderings (26%)
Competency 6: Sitecore Content Serialization (10%)
Competency 7: Containers (6%)

You can also refer below links and youtube video for more insight:

Sitecore 10 Certification Tips and Tricks
Sitecore 10 .NET Developer Certification
Sitecore 10 Certification Crash Course

I would personally suggest that you complete the course Developer's Fundamentals 10 Collection from the Sitecore e-learning Portal. It will make you a better understanding of competencies points 5, 6, and 7 mentioned above.
Comment: It is more similar to Sitecore 9 in terms of questions. Only a few things are extra on containers & serializations.
Answer: Apart from the helpful points mentioned above, there is a wonderful crash course on Youtube from Sitecore MVP Navaneethakrishnan (NAVAN) Sundarrajan
The video link: Sitecore 10 Certification Crash Course by Navan


Install meter stops when installing apps in Ubuntu Software Center

Question: As the title says, install meter stops when trying to install software. I'm using Ubuntu 14.04. I get into the software Center, I find the app I want, click to install, enter the admin password, it's starts installing and then the process meter stops at like 10% or something. Therefore I can't install any new apps using this Ubuntu software Center. I've had to do everything using .deb manual installs and using sudo commands. I really want to fix this Ubuntu software Center problem. Anyone had this problem before and know how to fix?
Answer: Sometimes, there is a dialog box that opens behind the software center window that prevents installation from proceeding. When a selection is made, installation proceeds.
This isn't always the case, I've always been frustrated with how slow software center is.

Next, the synaptic package manager isn't quite as nice but it gets the job done and it's really fast. Run the following commands to install:
sudo apt-get update
sudo apt-get install synaptic

Now, search your apps for synaptic package manager.
Search for and select packages. Then, click apply and always review the list before accepting the changes.
help.ubuntu.com/community/SynapticHowto

Also, it isn't typically good practice to install packages using deb files unless the package is not available through apt-get.
You can search for packages using apt-get as well. Here's an example:
apt-cache search synaptic

To narrow the results, you can use grep to list results containing a keyword like in this example:
apt-cache search synaptic | grep "package manager"

Finally, install a package using the following command (this example installs synaptic):
sudo apt-get install synaptic

To update your list of packages:
sudo apt-get update

and to update all your packages:
sudo apt-get upgrade


How is /dev/zero created and how can I make variants like /dev/one?

Question: I read the mknod man page, which is (as far as I can tell) what you would use to make a character device like /dev/zero, but I don't see how you would get it to yield an infinite stream of zero bits (or another pattern). What is the procedure for creating such character devices?
Comment: Not necessarily, it depends on precisely what you want to do. Most likely, you would at least need to implement a module.
Comment: They're implemented in the kernel. Look at `zero_lseek`, `read_zero`, `write_zero`, `mmap_zero` in `drivers/char/mem.c`.
Comment: Oh…so I would have to modify the kernel?
Comment: What about `/dev/one`?
Answer: mknod creates the device node, but the VFS detects accesses to the device node and reroutes them to the appropriate driver within the kernel for handling. All device nodes, from /dev/null to /dev/sdX to /dev/ttyXX to /dev/videoX are handled this way.
Answer: All mknod does is associate a device file with a device driver. There are device drivers that implement interaction with actual devices, and there are device drivers that just react to read-write requests in useful ways.  If you want to you can sit down and write a driver that returns the lyrics of the Star Spangeled Banner. But it's a matter of coding, not finding the right arguments for mknod.
Answer: If you just want to re-create the /dev/zero abilities but with a character other than zero/null, you can use tr to change all the zero's ("\000" in octal) into something else.
So to spit out an endless stream of "a"s for example, you could do:
cat /dev/zero | tr "\000" "\141" | head -c 20
aaaaaaaaaaaaaaaaaaaa

Or skip the cat and output an M's worth (1024*1024) with
head -c 1M /dev/zero |tr "\000" "\141"

"\141" being the "a" character.
See this site http://www.asciitable.com/ for a quick short page of ASCII - Octal codes. It's actually this image here:

(I know, old "solved" question, but I found it while searching for an endless stream of different characters, so this "solves" the "how to make /dev/one or /dev/[different character]" problem.)
Comment: I dunno…based on unscientific I-just-tried-it-right-now, `head | tr >file` has varying degrees of overhead. Depending on the system, it takes either one, one and a half, or two times as long as just `head >file`.
Comment: So it does, tr's a workhorse I guess. Trying `head -c 3G ... > /dev/null` took under a second, but adding the pipe to `tr` took 21 seconds. Still about 150mb/sec so probably faster than most hard drives could keep up to. But it works, and there is no built-in /dev/one or /dev/a, etc
Comment: Hence my asking after one!


Can people live on Saturn's asteroid ring?

Question: Earth is overpopulated and needs somewhere to create a settlement. If they decided to settle on Saturn's asteroid ring and If the people who are making the settlement have the technology we currently have now and had a trillion dollar budget to build a settlement, could it actually work properly and will it be self-sustained? It is a city kind of settlement so no mining would be done for earth's economical purposes but only to help the settlement build so it can't buy. It has to be self-sustaining so it can't buy or bring anything else from earth. Pretty much no contact to earth after the settlement would be built.
Comment: And even if you go to space, Mars makes a better settlement and the Moon is much cheaper to get to. The problem with the rings is that they're not a solid object, they're a cloud of micrometeorites. Very dangerous to hulls and habitats. I can just about imagine someone building a space hotel "near" the rings for aesthetics, but not within the rings. Even the moons of Saturn and Jupiter would be easier.
Comment: I just don't see any reason the rings of Saturn would be selected over the moon, Mars, Ceres, the atmosphere of Venus, or even Earth orbit.
Comment: If you have a trillion dollars to spend, you will get an awful lot more settlement for your money if you build it somewhere on the Earth that's currently uninhabitable -- deserts, Antarctica, oceans, etc. The same goes for any kind of space-based settlement. It doesn't matter how crowded Earth is, it will _always_ be cheaper and easier to make more living space on Earth than in space.
Comment: Your question answers itself. If you bring "all the required materials" to a certain place then you can build precisely the kind of installation which will fit your needs and requirements to perfection. What you fail to do is give us any requirements, or a background for your story. In other words, yes, it's totally possible to build a self sustaining settlement anywhere, even in the heart of the Sun, as long as you bring "everything you need" with you. Since you're not telling us what tech is available, what the economic implications might be, we can't answer.
Comment: We can barely manage to build [closed ecological systems](https://en.wikipedia.org/wiki/Closed_ecological_system) (self-contained biospheres that can function without contact with the Earthen environment for significant time) *on Earth* with current technology, given easy access to all of the resources of Earth! I can't imagine how building something in -- *not "on", the pieces too small for that* -- what (as already pointed out) **basically amounts to a cloud of bits of ice and rock** would be significantly easier.
Comment: @CrazySlayaNinjaBear There's no such thing as "everywhere is inhabited". You can always build up or down.
Comment: Being self-sustaining in space requires many things, but the one thing with the strictest requirement that may be most difficult to find and/or synthensize in the required amounts is *water*. Somehow water must be available (out there we would look for ice) and/or hydrogen and oxygen must be available in some way to create water. Perhaps methane for hydrogen and silica for oxygen would suffice, and then sufficient power would be needed (solar?) and space to build the whole system that would create and/or mine water. That's your primary stumbling block - water.
Comment: I'd go next door, to Jupiter's moon Europa, which is apparently covered with an ocean. It's *cold frozen Hell* so you'd need to bring a heater, but you have water, and the system is much more stable - Jupiter's moons don't collide very often. Saturn's ring particles, however... besides, a civilization on Europa would experience an Earth-like climate in a few billion years, when the Sun becomes a red giant and swallows the Earth. Sustainability FTW!
Comment: @AndreiROM il edit it
Comment: @MikeScott what if everywhere on earth is inhabited
Comment: Two days ago explorer Juno reached Jupiter after a voyage of 5 years. Saturn is even further from Earth. How long would you like to travel in a space ship?
Comment: @Roland Juno used a [very indirect trajectory](https://en.wikipedia.org/wiki/Juno_(spacecraft)#Flight_trajectory) to get to Jupiter, including one Earth fly-by two years into the mission. While delta-v has a high cost in terms of fuel and payload capacity, a manned mission would almost certainly pick a very different flight profile that aims for a short travel time as opposed to minimizing delta-v budget requirements (which I *suspect* was a major motivator for picking Juno's trajectory).
Comment: @MichaelKjörling Thanks for the link, it has great trajectory animations. I don't know much about "best trajectory", and guess that StarTrek people would rather go the fast way in a straight line from A to B. I have never seen an episode where they ran out of gas and had to fill up. Also always plenty payload capacity. Also nobody ever got bored inside.
Answer: With today's technology the answer is most likely No".
There's two important factors at play here:
Technology
Going into space is a very specialized niche which very few companies deal with. There's not a lot of manufacturers for the sort of tech you're going to need to build a large habitat. It's debatable if we even posses the technology to build a large-ish habitat (we already have the ISS, but that is not adequate for long term habitation, and is protected by Earth's magnetosphere, which won't apply out in the depths of the solar system).
But even if we did have the theoretical knowledge to build such a construct, we lack the industry, and skilled labor to do so.
Politics and Economy
The second major aspect of this undertaking is politics and economy.
A trillion dollars sounds like a lot of money, but when it comes to setting up several new industries, and developing extremely advanced and expensive materials en masse, it just isn't enough. Furthermore, some of those materials can't be found in the quantities you're going to need - other industries need them, and they're eventually going to block you from buying out all the available stocks.
Even trickier are the politics of it all. If the Earth is already over-populated, and taking the current geo-political, and global economy into consideration, you are most likely looking at a planet in turmoil. There's already many conflicts being fought, and we're looking at a possible world war within our lifetimes. When you announce a major undertaking to leave the planet you may spark accusations of trying to build weapons in space, racism based on your colonist selection process, etc. etc. etc.
More than likely one govenrment or another would simply shut you down because they wouldn't want a massive space station to be in anyone else's control other than themselves.
Hope
However, assuming that the situation on Earth is very similar to what it is today, you wouldn't have to go to Saturn. Mars has a big "For Rent" sign on its front lawn, as it were, and no one has made it over there yet.
And while it's debatable whether we have the tech to allow colonists to survive the decade long journey to Saturn, we're pretty sure that we could get someone to Mars in one relatively healthy piece.
Add to that that other companies are already designing missions to Mars, and so some industry is already geared in that direction, and you may just have a working concept.
Comment: -1 The entire Apollo program (adjusted for inflation) cost 100 billion dollars. That's going from literally no space program to the moon. OP is saying that he has a **TRILLION** to spend. A budget 10 times as large. Yeah, it is enough to build industries from scratch. You have 10 times the budget as the *last* time you built those industries from scratch. Except this time, you aren't starting from nothing. A ton of the R&D has already been done for you. As Heinlein said "If you can get your ship into orbit, you're halfway to anywhere in the solar system"
Comment: @Shane - Ok, I missed your "adjusted for inflation" note. Other than that is there any particular reason for the sarcasm and tone?
Comment: We have the technology.  What we don't have is a purpose or the money.  And for going into space, the best bet is to go to the asteroids, not another planet that costs much $ to get off of.
Comment: @Shane - well, first of all, I'd like to see what 100 billion is adjusted for inflation. Second, once people know you have money and are purchasing specialized equipment they're going to start charging you through the nose for it. Quite simply, there's not that many suppliers for this high quality stuff, and only so many customers. They can charge whatever they want, and prices will go up fast once they realize how ambitious and rich you are. On top of that, think of the sheer magnitude of the project compared to the Apollo missions.
Comment: @Shane - that was also only a minor aspect of my overall answer, and hardly worth a downvote, but have it your way.
Comment: Moreover, and this is far more important than the rest: You mainly need steel, concrete and rocket fuel. None of which are supply constrained to such a degree that building a city will put an appreciable dent in supplies. If you think there is someother actually rare element, put that in the answer.
Comment: You said there isn't the tech available, that a trillion dollars isn't enough money, some nefarious government will stop you and that mars makes more sense. Mars *does* make more sense. But tech, government and funds are ***easily*** solved by money, money, money.
Comment: If it were the USA doing it, would they still be "shut down" by other governments?
Answer: On the more practical side of 'if it is possible?'...politics and sustainability aside.
The ring is not a solid structure and there would be no atmosphere on any of the ring debris that makes up the ring, ranging from dust particles up to fractured small moons. As far as I recall, if you were to compress the entire ring system you would only have an atmosphere a micron thick?! So pretty much you will be building a giant spaceship-city like stargate atlantis. No walking on the ring itself...unless you hopping from debris to debris in a spacesuit.
You would have to provide some sort of CONSTANT shielding/protection from the ring debris, from the larger city killers down to the tiny specks of dust. Look what happened to the ISS window when it hit a ?paintfleck? (or more accurately, the paintfleck hit the ISS)!
You would be safer on the smaller shepherd moons (but even these will have debris raining down on them occasionally) or better yet one of the larger moons. If you really insistent on being 'in' the rings, you could try for a settlement on the leeward side of the shepherd moons away from any bombardment. Then you get a view of the rings side-on.
Edit: that last suggestion about the leeward side of the shepherd moons would only work if your moon was tidally locked to Saturn. If it isn't tidally locked you could end up building your settlement in a sheltered area only for the moon to rotate your settlement into the 'highway to hell!'.
Comment: @Shane Every mass exerts a gravitational influence. It may be small, but it will be non-zero. Newton's famous apple does exert a gravitational influence on the planet Earth that is equal to that of the Earth on the apple -- though, of course, since the Earth is so enormously more massive than the apple, the *effect* of the Earth's gravity on the apple is far greater than the effect of the apple's gravity on Earth.
Comment: This is the right answer. There isn't enough 'ring' in the rings to support a colony. There's simply nowhere decent to plop the colony down on. The biggest problem will be the gravity. The elements of the ring don't have enough mass to have gravity. That means you have to create a spinning ring, or other artificial gravity source. At that point, you more or less have a self contained spaceship. What's the point of landing?
Comment: @MichaelKjörling When the next sentence is talking about spinning rings for artificial gravity, most reasonable people realize that we have a character limit and they can figure out the context. Since you couldn't, Let me rephrase. he elements of the ring don't have enough mass to have gravity suitable for a colony of humans. Having to have spelled that out for you.... *le sigh*
Answer: There is one very clear factor which speaks in favour of building in Saturn's rings; all the materials are already processed in bite sized pieces and it does not take a very great deal of energy to get around.
Going from planet to planet, or even from moon to moon in Saturn's system requires using a fair amount of energy to change orbits between the various starting and stopping points. In interplanetary space this can be measured in kilometres per second, in the rings, since everything is quite close and in similar orbits, you might only need to metres per second deltaV to move around.

So the ices and dust needed to feed your civilization are there for the taking with little more than a net and a very low powered rocket (or a tether if you like to save on reaction mass).
For the more advanced civilizations, the rings make a great place to set up a massive tether to reach down into the atmosphere of Saturn and draw 3He up as nuclear fusion fuel. Since the amount of sunlight is very feeble in Saturn's orbit, a solar powered civilization would have to build platoons of mirrors in orbits free of the rings and beam sunlight or microwave energy to the rings, since collisions between ring debris and fragile mirrors would be catastrophic. (The colonies will be well armoured anyway in order to withstand the intense radiation environment).
There will be a distinct lack of metals and many of the elements needed for advanced civilization, but the availability of cheap ice and volatiles would give Saturnian ring dwellers an advantage when trading with other space "nations".
Answer: Sure - why not? Here is some information about the rings from NASA:

The ring particle sizes range from tiny, dust-sized icy grains to a
  few particles as large as mountains. Two tiny moons orbit in gaps
  (Encke and Keeler gaps) in the rings and keep the gaps open. Other
  particles (10s to 100s of meters) are too tiny to see, but create
  propeller-shaped objects in the rings that let us know they are there.
  The rings are believed to be pieces of comets, asteroids or shattered
  moons that broke up before they reached the planet. Each ring orbits
  at a different speed around the planet.
The main rings are, working outward from the planet, known as C, B and
  A. The Cassini Division is the largest gap in the rings and separates
  Rings B and A. In addition a number of fainter rings have been
  discovered more recently. The D Ring is exceedingly faint and closest
  to the planet. The F Ring is a narrow feature just outside the A Ring.
  Beyond that are two far fainter rings named G and E. The rings show a
  tremendous amount of structure on all scales; some of this structure
  is related to gravitational perturbations by Saturn's many moons, but
  much of it remains unexplained.

http://solarsystem.nasa.gov/planets/saturn/rings
The fun the about science fiction is you have some measure of freedom to invent new technologies or scientific discoveries. They can mine the ice and other chemicals, establish a base on the larger rocks, or have a station orbiting in the Encke and Keppler gaps (maybe there are two groups and they are so named?)
Comment: Oh no I edited the question while you were writing the answer
Comment: But it shouldn't matter too much
Answer:
a trillion dollar budget to build a settlement

Well this alone would be your shortfall.

$100 billion for six astronauts?
A manned Mars mission would be incredibly expensive. NASA estimates peg the overall expenditures at about $100 billion over 30 or 40 years, Sherwood said, but those numbers may be too low.
The International Space Station (ISS), after all, was initially anticipated to cost $10 billion over 10 years. But it ended up costing 10 times that, and took nearly three decades to assemble.
  http://www.space.com/16918-nasa-mars-human-spaceflight-goals.html

An estimated trip to mars is estimated to cost 100 billion dollars. However, the ISS was estimated at 10 billion and cost ten times that. So if this project followed suit you would have 100 billion x 10 = $1 Trillion.
...Well, looks like you're out of money and your 6 people (a few shy of a city) are 2 planets short and have no money to build a sustainable environment.
Comment: That price includes all the supporting infrastructure. Buying and shipping 40 years worth of food. Paying the 100s or 1000s of people working on the groundside for 40 years. OP is talking about single or few launches. The entire apollo program cost $20b ($100b in today's money)
Comment: And have no money to ... Spending money on Mars? Currency? LOL
Answer: I would have to say no, mostly due to the fact that with today's technology, we wouldn't be able to get anyone out there to live on it let alone build a habitat. Also, I don't think a sustainable settlement of any real significant size would be able to sustain itself out there without food supplies from earth, assuming they could terraform an asteroid to grow things I still don't think it would be enough. On top of that I don't think that whatever small settlement they could make out there would have a very big impact at all on the Earth's overpopulation problem.
Answer: With enough money, many MANY things can be made possible while still being completely impractical.
Economics
You provided a budget of \$1 Trillion US.  According to this stack exchange answer, the cost per KG to launch something into geo-synchronous orbit is approximately $56,000 with an Atlas V rocket.  That means that, within your budget, you could put about 1.5 billion KG (or 153,467 tons) of free material into geostationary orbit if you blew it all on launches.  That is just getting all of your materials/modules/fuel/food/oxygen/water into a place where it could be assembled together before then heading out towards the outer planets.  It doesn't account for their individual costs.
I don't have a calculation handy for figuring out how much fuel you'd need, but it would be significant (especially if you plan on landing your craft instead of crashing it into a planetary body at tens of thousands of miles per hour. Oh, and you are also going to have to carry some type of fuel to support life-support operations on your ship unless you are just planning on transporting corpses.  Fuel costs need to be subtracted from your available budget.
Maybe your ship consists of habitat modules that can be disassembled when you get to your destination and used as permanent living space for the settlers.  For that to be the case, they would have to be durable enough to survive the journey, land, be disassembled, and have a reasonable expectation of lasting long enough under daily living conditions to support your goal.  This probably means that they are both HEAVY and EXPENSIVE, so subtract that from your budget.
You are also go to have to purchase all of the supplies/food/oxygen/water/recreational equipment/exercise equipment/medical supplies (anyone priced an MRI machine lately?)/medicine/tools/manufacturing equipment (because you can't just run down to the parts store)/redundancies/computer equipment/educational tools/nuclear fuel for a permanent power source/etc. so subtract all those from your budget.
If, after purchasing all of your supplies you still have enough left to fly it into space and take some people along, then your journey has become economically possible.
Logistics
Assuming that, after all the math is done, we still have enough carrying capacity to put a few hundred people (still not "city-sized") on the ship to go along with the cargo, there are several logistical challenges to consider.
First, we know how to put things in space, how to fly something to Saturn's neck of the solar system, how to land "Earth things" on "space things," and so on.  What we haven't done is do all those things at once.  Yep, we've put people on the moon, and we've put landers on other planets, and we've done long space journeys with unmanned craft, and we're starting to understand the impact of long-term space flight on human health.  Now suddenly all of those things need to be put together for an extremely complex operation involving lots of people for an extremely long mission with no hope of rescue if something goes wrong.  Since we've never done it all at the same time, we don't know exactly what challenges we'll face.
Take for example the variety of passengers.  Astronauts are carefully vetted and extensively trained.  It might not be practical to do that with all of your residents, so, while best efforts might be made, you are taking risks just by putting your crew and passengers on the ship.  What if there are high instances of severe depression due to leaving an entire planet behind forever?  What if people reproduce onboard?  What if someone commits a criminal act?  What about terrorism? What if we find that space causes strange health issues in people that have a rare genetic defect?  What if the module holding the vitamins takes a meteor to the hull and everyone gets scurvy?  All of these contingencies have to be planned for with procedures in place to handle them.  Forget just one critical thing that pops up (like a flu-bug that mutates into something really nasty due to interaction with cosmic radiation), and everyone is dead.
Then there is the question of sheer time and distance.  It took Voyager 2 about 4 years to reach Saturn.  Assuming we could travel at that speed, there are a lot of things that could go really wrong in that amount of time while out in space.  It becomes a logistical problem because who wants to spend a trillion dollars on a mission, throw their hands up in the air, and hope it succeeds?  Plans need to be in place for rescue, repair, and resupply missions in the event that a problem occurs.  The further away and the longer it takes to get there, the more complicated and expensive (or impossible) these missions become.
Sustainability
According to NASA:

Saturn, the "Ringed Planet," is so far away from the Sun that it receives only about 1/80th the amount of sunlight that we receive here on Earth.

So, there is your figure for growing crops or producing solar power.  You are probably going to have to bring nuclear power sources to run lights, grow food, create heat, power factories, and process waste back into usable compounds.  Nuclear power, while lasting a long time, is still going to need expensive replenishment and maintenance.  If enough fuel isn't brought along, can it be readily obtained from the environment?  Will it weigh too much to take the quantities needed? Uranium and Plutonium are pretty DENSE stuff if I remember right.
How about basic things needed to maintain the habitat and systems?  To be completely self-sufficient, all of those will need to be manufactured onsite.  Otherwise you will need to plan supply missions with necessary components, and those missions will take 4 years to get there.  Forget to bring along a spare "main air scrubbing unit?"  Hope you can hold your breath a long, long time.
Conclusion
While (perhaps) technically possible, it wouldn't be at all feasible simply due to the extreme distance and the availability of far, far better choices.  The moon is a few days away and plenty big enough for small settlements.  You could have access to solar power 24/7, and the chances of something bad happening on the way are minuscule in comparison.  If something bad did happen, Earth is right around the corner to lend a hand.
If the moon wasn't available (maybe due to international politics), then Mars (or maybe one if its moons) would be a far better choice.  Or maybe Venus would be better?  The abundance of heat energy and an atmosphere from which to extract some raw material has to be advantageous.
Comment: According to spaceX, it costs $8000 per KG to go to Mars. http://www.spacex.com/about/capabilities
Comment: @Shane That's rather unspecific, since it doesn't specify *which* Mars. (No joke! Payload to orbit is likely very different from payload to surface, even more so since on Mars you don't get much benefit of aerobraking but still have to take aerodynamics into account for a lander and want to avoid significant lithobraking. Worst of both worlds.) Compare [What is the marginal cost of *landing* on the Moon?](http://space.stackexchange.com/q/4720/415) which ultimately concluded that Astrobotic (which aims for the Moon) can deliver 1 kg to lunar orbit for \$198k or 1 kg to lunar surface for \$1.2M.
Comment: @MichaelKjörling I just looked over the data. Maybe those figure are out of date and there's a new pdf, because they quote 1.2M$ per kg for both lunar orbit and landing. Ignoring that, even if it was 200K. per kg to lunar orbit and SpaceX's price to Martian orbit is 8k. per kg. Martian orbit is harder and SpaceX does it FAR cheaper than astrobotic does. They also include ground support services for you mission. Astrobotic is charging for data comms between earth and luna. They are charging you for the power your payload uses. Etc. Long story short, Those prices are vastly over inflated.
Answer: I'd generally tend towards a natural disaster or world war instead of overpopulation as the reasoning for off-world settlements - for a few reasons.
I agree with the sentiment expressed in earlier comments that given essentially unlimited resources, it's always going to be more economically feasible to build on earth than offworld. Unless earth is suffering from some global disaster.
Additionally, there's significant evidence to suggest that there's no such thing as global over-population. As currently accessible resources get used up and therefore get more expensive, there is a greater economic incentive to find new deposits of the same resources or find alternative methods of achieving the same goal.
For example, it took rising petroleum prices in the tried-and-true middle eastern markets for comparatively expensive oil shale and oil sand extraction in Canada to become economically viable, and for comparatively expensive hybrid and electric car development to become viable.
The same logic applies to livable space.
One option - a trope similar to the one used in Sunshine or Interstellar, where the world is slowly dying, and it makes economic sense to fly as many people and resources offworld as possible. And if Saturn's rings are still a bridge too far (considering places like Mars or Europa might be more livable sooner, you could say that those places are dying too.)
Another option - human civilization is vast and wealthy, and lots of people are eager to be frontiersman on the edge of civilization and deal all the inherent challenges.
Comment: That's an.... *interesting* interpretation of events. Our oil reserves are in such dire straights that mining tar sands (places where you barely get more energy than it takes to get the stuff) makes economic sense. Therefore we'll never run out!
Comment: Interpretation isn't the right word - and I'm not suggesting that a particular physical resource can somehow be in infinite abundance (at least for practical intents and purposes).

I'm saying simply that necessity is the mother of invention. As as a particular resource gets scarce we are incentivized to find novel alternatives.

(made multiple edits because hitting enter apparently submits the comment instead of creating a line break - whoops)
Comment: You take the fact that "as a particular resource gets scarce we are incentivized to find novel alternatives" and *interpret* it to mean that there is " significant evidence to suggest that there's no such thing as global over-population" The second statement is not supported by the first. Probability is not determined by desirability. Necessity is the mother of invention. Practicality is its father :)
Comment: Apologies - I didn't mean to imply that the 2nd statement was supported by the 1st. This isn't just my opinion (and it wasn't my opinion until very recently) - there is great amount of research and scholarship that has gone into debunking overpopulation. I can point to some authors if necessary; also the Population Research Institute has some information available on their site. Of note: for over 200 years dozens of thinkers have predicted starvation on a global scale during their lifetimes and none of those predictions have come to pass.
Comment: For an interesting related anecdote - look up the Simon / Ehrlich Wager.
Answer: I'm gonna say no, due to Kessler Syndrome. The relevant part of the syndrome states that once you have a sufficient number of debris you will be hit. Given that Saturn's ring is a massive amount of debris, staying there for a long time will guarantee your eventual obliteration. At which point we will have more debris than before, which is the famous part of Kessler syndrome.


Why is it bad if number of dimensions / factors > sample size?

Question: I've been told (read) this many times, but I never understood why it's bad for the number of dimensions in your data, or the number of explanatory variables in  your model to be higher than your number of samples.  Why is this the case?
Answer: First of all, this is not a "hard" limit in the way that a model with sample size $n =$ number of variates $p$ is bad and adding one sample or case (model with $n = p + 1$) would turn this into a good model. The rule rather focuses on the idea that the model quality depends on sample size compared to the number of variates, as opposed e.g. to the quality of validation results which often depends on the absolute number of test cases.
You can think of your data as spanning a $p$-dimensional space: each variate opens a new direction. The model describes the points (samples) in this space; often it is some kind of (hyper)surface.
In that description, you can think of model quality as the answer to how certain you can be about the hypersurface: that depends on the sample density. The tighter are the data points which you know, the more certain can you be about the surface.
In a very general setting, this leads to the conclusion that with growing number of variates you may need exponential growth in the number of samples in order to have the same certainty in fitting your model. However, if you impose restrictions on your model (i.e. you reduce the degrees of freedom) such as allowing planes only (linear model), you can reduce the needed growth in sample size (e.g. to linear for linear models).
However, for sample sizes $n \leq p$ even a linear model is not even unique any more: there are infinitely many planes that perfectly fit the given points (including their noise) in $p$ dimensions.
Adding more samples will first make the model mathematically unique and then overdetermined (so that degrees of freedom are left after fitting the model that can be used to estimate and reduce uncertainty). Of course the step from underdetermined to determined (at $n = p + 1$) is a huge improvement. But the improvement to $n = p + 2$ which for the first time gives a (still very rough) idea about the uncertainty is almost as big IMHO. Hence, the model will gradually improve by each new case.
The Elements of Statistical Learning give a nice discussion in chapter 2.
Comment: Well, with $n = p$, the surface is not yet fixed. So it is *obvious* that the sample size is too small. With $n = p + 1$, the plane is fixed (unique, determined), and it goes through *all* points. But there is still no indication about the uncertainty related to noise
(uncertainty) of the data points. IMHO this is almost as bad as not being able to fix the plane. Each new data point now helps to estimate this uncertainty (and refine the estimated plane). Overdetermined: there are more data points than are needed to mathematically define the plane (the ones used for estimating the uncertainty)
Comment: I'll start reading the chapter you mentioned, but this  is illuminating.  Just thinking about the most simple case helps, e.g. fitting a linear model with two independent variables beginning with 1 sample point, and then 2, then 3... etc.  However I don't follow what you mean by the improvement to $n = p + 2$ is almost as big as the step from $n <= p$ to $n = p + 1$.  Also what do you mean by "overdetermined?"
Answer: To comment on @cbeleites' answer, there are exceptions to the rule (sort of). Specifically, often your predictors are correlated, so they can be replaced by only one variable; or otherwise many of the predictors can be removed. Thus, even though you cannot create, say, a straightforward linear regression model with $n \leq p$, you can work with data sets that have many more variables than samples. This is a case for dimension reduction strategies, variable selection and other related tools.
Comment: Yes, that is true. Dimension reduction often helps very much (though it is no magic bullet).
Comment: Yeah, I totally agree. I'm a biologist. My latest data set has 16 samples and over 400,000 variables.


Can someone explain how to find and use kext files?

Question: The post I found on this forum: "Apple Cinema Display Brightness no longer works in Mavericks or Yosemite" explains how to fix it using kext files but I can't figure out how to get the kext file to use with the KextBeast program. I bought a used Apple LED Cinema Display and need to be able to adjust brightness to calibrate it.
Comment: to find all kext files type "mdfind .kext" in Terminal. Beyond that you are entering dangerous ground, but you know that! do you ?
Answer: I found out that by also plugging in the usb cord from my monitor to the computer, the brightness controls on my keyboard now work! No need for kext files after all.


GRASS GIS change raster output size?

Question: I have a large inut raster geotiff map of size 12000x9000.
I am running r.watershed to output a stream raster map. However it is much lower resolution, only about 960x711.
How do I make GRASS GIS / r.watershed command output a matching size raster map of same resolution of my input?
Comment: You need to use `g.region` here (or the corresponding entry in the map legend context menu). For the concept, see https://grasswiki.osgeo.org/wiki/Computational_region
Comment: @markusN Please consider adding your comment as an answer.
Comment: Which parameter is it - rows/cols or res or nsres/ewres?
Comment: if your are concerned about 2D grid resolution and using square cells, go to "Resolution" tab > "2D grid resolution" box put 0.5 for a 50cmx50cm resolution. In commandline, this is 'res=value' parameter
Answer: In general, GRASS GIS is ready for massive data analysis.
In your case, you need to use g.region, by using the corresponding entry in the map legend context menu (right mouse button on legend entry) or on command line (generic example):
<code>g.region raster=your_raster_map -p
</code>
It will then use bounding box and raster resolution to set the computational region.
For the underlying concept, see https://grasswiki.osgeo.org/wiki/Computational_region .

GRASS GIS change raster output size?


How do i get the ocean modifier to animate?

Question: I have been going through many tutorials and even asked in BlenderArtists but no help. When I get to the point to insert keyframes a keyframe drop down menu appears instead. It has been very frustrating to figure it out
Comment: Possible dupe? https://blender.stackexchange.com/questions/164/can-modifiers-such-as-subdivision-and-array-be-animated
Answer: Insert keyframes for the "Time". Set the frame to 1, move your mouse on "Time" and press I

Change the frame, add some time and press I again.
Comment: For OP, also worth noting you can animate other properties, play around with them.
Answer: Here is a short video tutorial on how to animate using expressions in Blender 2.80: https://youtu.be/2oUK0n5prRY
Right click on Time
Select Add Driver
Make sure you remove the Input Variable
Then type in this expression and you are good to go:
frame * 0.1 / 10

Answer: I found adding a driver and changing the expression in graph editor (not sure why no one thought to allow expression editing directly in the field?!), with a reference to the current frame got it moving when playing fine.
(bpy.data.scenes[0].frame_current / 10)

The warning TIP is not important here and I just delete the variable below.
You'll have to open a node panel with the 'n' key in the graph etc. in the end it's easier just to keyframe the value, if we can't put expressions directly into the value via backticks ` like Houdini etc.
You should be able to do something directly in the field like @curframe or call a global function frame()?... alas it's free software.
Comment: Interesting. Not the easy way but no doubt it's powerful. You should explain with more details how to connect this to the ocean modifier, what can be done and how, inside blender of course.
Comment: too long? see below... or above? it's hoping around.
Answer: Sure if you right click on any field there is "Add Driver" CTRL-D Choose Single Manually Later.
In a Graph Editor you change to view Drivers from F-Curve drop down.
Select and use 'n' to view Drivers TAB on right panel. The Scripted Expression Type has the Expr: field below that so you can then put in the Expression.
A better option is to use the C.scene where C is bpy.context but it's not set there, so (bpy.context.scene.frame_current / 10) would drive the ocean.
You can delete the var and don't worry about the warning like I said before.
Seems it also says it's unsafe for RenderFarm so the previous scenes[0] might be better it's all new to me so I find it lacks professionalism and clarity.
Again one should be able to just put that expression into any field directly via back-ticks or what-have-you and skip all this super confusing poorly designed hidden driver linking to graph editor expression crap.
Better yet have the ocean include a checkbox (or time link button) by default to drive it with an appropriate setting or adjustable speed slider by default [expected behavior].


Web Site Design Based on User Catagory

Question: I am starting a building design consultancy (think architect, but not physical building design, building automation, AV instead, eco buildings, etc...).
I want a website to push prospective clients towards to get inquiries. I have defined 4 target customer types, each with a tailored marketing & sales strategy. I plan to test which customer type is best for my company.
The specific sales content differs depending on client type, so I would like the website to ask each user who they are first, then present them a dedicated page for their customer type.
First draft wire frame is below, explaining (if I was to pursue this idea) how the site could look/work. It also accounts for people who do not want to define their "customer type" and want to remain "anonymous".
Is this a good idea, or will people get turned off if they arrive on a site and get hassled to say what type of customer they are?
I am no UX designer, so any info, especially research that is A/B tested, would be awesome.
Answer: I strongly caution against this plan. While role-based navigation might be appropriate in certain, very specific scenarios, it usually isn't. It has been demonstrated that role-based information architecture increases cognitive load and user anxiety (which is the opposite of a good user experience).
In most cases, you want to do research to develop clearly-labelled, mutually-exclusive cardinal categories, then base your navigation on what the user wants to buy from you, not who the user is.
Some of the problems of role-based navigation include (from the Nielsen Norman group):

Users don't know which group to choose.
Users question whether the category will have information about that
group or for that group.
Forcing people to self-identify creates an additional step and takes
people out of their task mindset.
Users feel anxious that the information they see might be incomplete
or incorrect.
Websites with audience-based navigation often have overlapping
content, which creates a greater workload for users (and content
maintainers).

https://www.nngroup.com/articles/audience-based-navigation/
Answer: Asking what kind of customer someone is, is very important to you, but in UX design you need to think about what is important to your users. Focus on your offerings, your users have an idea of what they're looking for, they're just trying to figure out if you're offering it.
The "Who are you?" model is off-putting to users because it creates pressure and uncertainty of which category they fit into. I think you already know that, since you included a "?" for those who don't know. So go with your gut, this isn't a good idea! Think about whether it is really critical for you to create different experiences for different people. Are they really that different to warrant a different site experience? How else might you be able to collect information on what people visiting your site care about?
Rather than a potentially confrontational, "Who are you?" you can organize your site by what you offer. I see that below with your "Projects" section. That's the right idea and this is where you can create segmentation based on what your customers care about.
A helpful trick is to use real world scenarios, because they are very powerful. Imagine walking into a store, but before you went in, someone was standing there with a clipboard asking you "So are you looking for something specific, a whole new outfit, browsing, or going to the sale rack?" Yikes! You'd run in the other direction. A better experience would be to let a user walk in freely, let them notice the layout of the store and observe that they went to the sales rack. You can make helpful suggestions once you see where they're looking, "You know anything you buy that's on sale is another 10% off today." Relaxed and helpful when needed, that's how you want your site to feel when you present your services. So go with categorizing your offerings in ways that might reflect what kind of customer they might be and meet them where they are once they get there.
On this note, you can try to include action items in contextual ways to drive users toward contacting you with inquiries. You might think about including a "Like what you see? Send us an inquiry!" link at the bottom of each project page. :) Drive, drive, drive. In context, in context, in context.
Finally, once they get to the inquiry form, you could do something very direct to get the data you seek. Include a drop down menu item asking the customer who they are. :) You never know what kind of customers you might be attracting so let them tell you who they are, but do this in the form-like setting of the inquiry itself, where they user will expect it, not in the design of the site.
Answer: I agree with Paula.
You can also use usual categories that users would freely compare to find out what matches their needs (without stressing them).
However, considering your business model, if you offer very different products matching different client groups, you may consider crafting different websites. Each website would have its specificity for each group, and perhaps a specific graphic identity as well.

Web Site Design Based on User Catagory


Why application notes advise series resistance to be added on clock lines for high-speed interfaces, like SDIO, and not on data lines?

Question: Many application notes (like this one) advise series resistance to be added on the clock line, close to the source. I do understand that this resistance is added there to match the source impedance to the characteristic impedance of the line (usually 50 Ohm). However, what bugs me is that when it comes to data lines (which can switch just as fast as the clock line for DDR cases) no one advises to place series resistance. I would expect engineers to put series impedance on the drivers of data lines to provide matched source impedance just as well. Any idea why is this the case?
Comment: The biggest problem with short-term reflections is that they can make transitions non-monotonic, with a "hook" in the waveform that crosses the logic threshold more than once. This is very bad in a clock signal (logic can get randomly double-clocked), but has almost no effect on data signals.
Comment: To reduce the slew rate at clock signal, ringing and all. Used in low speed designs.
Answer: The receiving circuit only cares about voltage levels on data lines during a time interval centered on the clock edge, with setup/hold times on both sides.

So... even if the data lines are unterminated and suffer from poor signal integrity, if the clock is slow enough versus trace length to allow the levels to settle before the clock edge minus Tsetup, the receiver will sample correct data.
However, if you want higher throughput from your trace or twisted pair, then the clock has to be faster. Then, you don't want to wait for the mess on the data lines to settle, and receiver Tsetup+Thold becomes a significant portion of the clock period. In this case, you absolutely need to terminate the data lines. For an example, check some USB eye diagrams. Termination resistors can be internal to the chip (for example DDR RAM) so you won't always see them on the schematic.
Resistors on data lines can also reduce EMI: bandwidth is inversely proportional to rise time, thus slower edges emit less HF noise. Less ringing also means lower EMI amplitude. Ferrite beads also work.
Sometimes the chip will have internal terminations, or offer several drive strength or slew rate settings for the IO's. Some other chips like 74HC or AHC have rather weak drivers, this is done on purpose both to limit slew rate and source-terminate the trace with the output FET's resistance. Other chips like 74AC and some micros have fast and strong output drivers, which make a lot more noise. So maybe you'll get a clean signal without resistors, or maybe not, it really depends on what's driving the trace.
Comment: Yes, that's true (also slowing down the edges means less EMI since bandwidth is inversely proportional to rise time).
Comment: That was a very useful answer, thank you! Based on your answer, I am assuming, that adding series resistance on the data lines can be beneficial for EM radiation even for lower speed interfaces, as there will be less ringing (functionally it wouldn't matter). Am I correct?
Answer:
However, what bugs me is that when it comes to data lines (which can
switch just as fast as the clock line for DDR cases) no one advices to
place series resistance.

If the data gets clocked into a chip (such as a DDR SRAM as per your question) then it's important that the clock line does have a proper source-end series-termination resistor to ensure that the signal integrity at the DDR CLK pin is good.
It's less important for data because, if the data has a little wobble on it before it is clocked in then, that wobbly event has settled before the clock transfers it (it's all in the timing sequence of data being made available prior to clock activating).
Having said that there are a couple of things to consider: -

DDR Data lines are normally bidirectional so, if you are going to series terminate them, which end do you choose?
It's likely you'll choose both ends (for bidirectional data) and add series resistors to achieve a speedier and high-integrity data throughput.
The worse-case scenario is when the host is clocking data from the distant DDR memory. Due to the extra delay (the returning data is delayed relative to the CLK signal originating at the host). Hence, if you were going to use data-line termination resistors at only one end, you'd probably use them at the DDR chips.
Answer: You put series termination only close to source. SD specification says:

CLK: Host to card clock signal
DAT0 - DAT3: 4 Bidirectional data signals.

Clock source is always host. Data source can be SD card or Host, depending whether is it read or write operation.
Because Data is bidirectional, series termination can only make things worse.
Comment: You say this: `Because Data is bidirectional, series termination can only make things worse.` - a series termination **does not** make things worse when placed at both ends of a bidirectional data line. Transmission line theory properly indicates this fact.
Comment: @Andyaka you mean, if a series termination is added close to the receiver, it would not have much influence. Signal will still see infinite impedance (termination resistance+infinite receiver impedance)??
Comment: Exactly and yes.
Comment: @Andyaka you are right. I removed the text. Thank You
Answer: If you consider DDR2-DDR5 SRAM, the standard uses on die termination (ODT) on the data lines. The termination is typically configurable, but it is there, just internal to the chips. Since it's bidirectional, the ODT gets turned on and off whether you're doing a read or a write. I've only done designs with DDR3 and DDR4, but in this case I don't think I've ever had series resistors on the clock lines. They are differential, sourced by the controller, and are AC terminated to VTT from what I remember.
For other types of signals, I don't know if this is common, but for any new prototype I design, I almost always add series jumpers to every data line. This allows me to replace them with snubbers during testing if I see reflections causing any issues. The only exception to this is if I have very high speed signals or bidirectional data, in which case I will use Hyperlynx or another SI tool to optimize the layout prior to fab, which may or may not include series snubbers.


Proving the countability of algebraic numbers

Question: I am trying to prove that algebraic numbers are countably infinite, and I have a hint to use: after fixing the degree of the polynomial, consider summing the absolute values of its integer coefficients, and setting the sum less than or equal to $m$, for each $m \in \mathbb{N}$. I am also allowed to use the fact that every polynomial has a finite number of roots. I am not sure where to begin... Perhaps after fixing the degree of the polynomials, I could try enumerating them, and after proving countability of the set that contains the roots of all polynomials of a certain degree, I could state that the union of infinitely many ($\bigcup_{n=1}^{\infty}$, $n \in \mathbb{N}$) countably infinite sets is countable, and thus the algebraic numbers are countably infinite, too, but I am not sure if this is a good approach/how to enumerate the polynomials. Thanks!
Comment: +1 for showing your thoughts about the problem - this is unfortunately quite rare.
Comment: An alternative approach to showing that polynomials with integer coefficients are countable: consider the bijection $\phi: \mathbb{Z}[x] \rightarrow \mathbb{N}$ that sends the polynomial $a_0 + a_1x + a_2x^2 + \ldots + a_nx^n$ to the natural number $2^{b_0}3^{b_1}5^{b_2}\cdots p_n^{b_n}$ (notation: $p_i$ is the $i$-th prime number and $b_i$ is the image of $a_i$ under any bijection from the integers to the natural numbers).
Answer: The hint pretty much tells you what to do.
First, consider the polynomials of degree $1$ in which the sum of the absolute value of the coefficients is $1$; there's only finitely many of them, each with finitely many roots. (Think of this collection as corresponding to the pair $(1,1)$, the first $1$ giving the degree, the second $1$ giving the sum of the absolute value of the coefficients).
Then, consider the polynomials of degree $2$ whose sum of the absolute value of the coefficients is $1$; again, only finitely many, each with finitely many roots; make this set correspond to $(2,1)$.
Then, the polynomials of degree $1$ whose sum of the absolute value of the coefficients is $2$; only finitely many, each with finitely many roots. The set corresponds to the pair $(1,2)$.
Then consider the pair $(3,1)$; then the pair $(2,2)$; then the pair $(1,3)$. Then move on to the pair $(4,1)$, followed by $(3,2)$, followed by $(2,3)$, followed by $(1,4)$. Etc.
This will ensure that you cover each and every polynomial of positive degree with integer coefficients after finitely many steps; many algebraic numbers will occur more than once, but that's not a problem to showing they are at most countable (that they are infinite should be trivial).
Comment: Because there are only so many ways to add up to 1, to 2, to 3, etc using positive numbers.  I've got it now....
Comment: You wrote:  First, consider the polynomials of degree 1 in which the sum of the absolute value of the coefficients is 1; there's only finitely many of them, each with finitely many roots.  My question is why are there only finitely many of them?
Answer: That's exactly the right approach! The hint provided to you is about how might you go about enumerating the polynomials over $\mathbb{Z}$. Can you show there are finitely many polynomials over $\mathbb{Z}$ of degree $n$ and with sum-of-absolute-values-of-coefficients less than $m$, for any $m$? How many polynomials of degree $n$ are there over $\mathbb{Z}$, then? How many polynomials over $\mathbb{Z}$ are there, then? How many roots of polynomials over $\mathbb{Z}$ (i.e. algebraic numbers) are there, then?
Answer: HINT $\ $ Enumerate the polynomials by "height" := max(degree, max |coefficient|)
Comment: Maybe I am missing something but this problem seems really easy.  For each positive integer $n$ , there are countably many polynomials of degree $n$ with integer coefficients.  Each such polynomial has finitely many roots.  There are countably many integers.  Is that good enough?
Comment: My comment above uses that fact that countable unions of countable sets are countable.  I think this requires some form of the Axiom of Choice.  Are the proofs above intended to avoid AC?


My push-pull converter uses a flyback transformer with an air gap. How does changing air gap size affect the secondary output voltage?

Question: My push-pull converter circuit uses a flyback transformer with an air gap.  If that air gap is changed (increased), the primary inductance is decreased.  How should I expect the secondary voltage and the output waveform to change after increasing the flyback air gap?
Answer: There are other variables, you have to specify what you are keeping constant to get a valid comparison.
Is anything else other than the total output capacitance limiting the output voltage? If not, then more energy stored in the flyback means a higher voltage developed into that capacitance.
Does an increased airgap mean more or less energy stored in the flyback? It depends what else stays constant.
If the primary voltage and pulse width stays constant, then the lower inductance will mean a faster rate of rise of current. Fortunately, this increased current will not saturate the inductor, as the input volt.seconds are the same. The current is higher by the factor that the inductance is lower, but as stored energy goes as \$0.5I^2L\$, the energy is increased by this factor. The output voltage will be higher.
If the primary is switched on until the current gets to a threshold, or like in a car coil the current is limited by the coil resistance and supply voltage, the current stays the same. With reduced inductance, the stored energy is lower, and the output voltage will be lower.
The increased airgap will also lower the secondary inductance, and may well reduce the primary - secondary coupling. This means that the secondary waveform will tend to have shorter time constants.
Comment: Everything else remains constant in my push-pull converter, input waveform, and output load... So only thing changing is the gap width in my "flyback" transformer.   My simulations show that increased gap does increase the output voltage, but I didn't understand well why this was the case.  Your explanation makes sense, thanks.
Answer:
My push-pull converter circuit uses a flyback transformer

If your converter is a push pull type then, although the transformer you used may have been called a flyback transformer when you acquired it, the new application (as part of a push pull circuit) puts it into operating as a conventional transformer.
If your converter primary voltage(s) remain as they were after the gap is increased, then the output voltage will probably not be altered but, bear in mind that enlarging the gap can mean a larger (usually unwanted) primary side current and, depending on other primary protection circuits, the secondary voltage may fall.
If you are running it as a true flyback circuit then Neil's answer is valid.
Comment: Yes, my converter is a transistorized push-pull and the transformer is operating conventionally (not in flyback mode).
Comment: Maybe someone can explain the down vote?
Answer: All the energy is stored in the gaps between the magnetic core particles or gap in this case. As the gap size increases, flux will start to 'spray' into the area of the windings in proximity to the gap. This will generate high levels of eddy current losses in the wire. To avoid this problem you can wind on a few turns of insulating tape around the middle of the coil former to exclude wire from the region.  The main advantage is it reduces the sensitivity to thermal runaway by moving the losses from the core towards the glass gap when increased.
There is an optimal gap to minimize the sum of  all tradeoffs for increased winding loss, reduced core loss , increased gap coupling leakage  and Eddy current losses.


Data science and MapReduce programming model of Hadoop

Question: What are the different classes of data science problems that can be solved using mapreduce programming model?
Answer: Let's first split it into parts.
Data Science is about making knowledge from raw data. It uses machine learning, statistics and other fields to simplify (or even automate) decision making. Data science techniques may work with any data size, but more data means better predictions and thus more precise decisions.
Hadoop is a common name for a set of tools intended to work with large amounts of data. Two most important components in Hadoop are HDFS and MapReduce.
HDFS, or Hadoop Distributed File System, is a special distributed storage capable of holding really large data amounts. Large files on HDFS are split into blocks, and for each block HDFS API exposes its location.
MapReduce is framework for running computations on nodes with data. MapReduce heavily uses data locality exposed by HDFS: when possible, data is not transferred between nodes, but instead code is copied to the nodes with data.
So basically any problem (including data science tasks) that doesn't break data locality principle may be efficiently implemented using MapReduce (and a number of other problems may be solved not that efficiently, but still simply enough).

Let's take some examples. Very often analyst only needs some simple statistics over his tabular data. In this case Hive, which is basically SQL engine over MapReduce, works pretty well (there are also Impala, Shark and others, but they don't use Hadoop's MapReduce, so more on them later).
In other cases analyst (or developer) may want to work with previously unstructured data. Pure MapReduce is pretty good for transforming and standardizing data.
Some people are used to exploratory statistics and visualization using tools like R. It's possible to apply this approach to big data amounts using RHadoop package.
And when it comes to MapReduce-based machine learning Apache Mahout is the first to mention.

There's, however, one type of algorithms that work pretty slowly on Hadoop even in presence of data locality, namely, iterative algorithms. Iterative algorithms tend to have multiple Map and Reduce stages. Hadoop's MR framework reads and writes data to disk on each stage (and sometimes in between), which makes iterative (as well as any multi-stage) tasks terribly slow.
Fortunately, there are alternative frameworks that can both - use data locality and keep data in memory between stages. Probably, the most notable of them is Apache Spark. Spark is complete replacement for Hadoop's MapReduce that uses its own runtime and exposes pretty rich API for manipulating your distributed dataset. Spark has several sub-projects, closely related to data science:

Shark and Spark SQL  provide alternative SQL-like interfaces to data stored on HDFS
Spark Streaming makes it easy to work with continuous data streams (e.g. Twitter feed)
MLlib implements a number of machine learning algorithms with a pretty simple and flexible API
GraphX enables large-scale graph processing

So there's pretty large set of data science problems that you can solve with Hadoop and related projects.
Comment: I'm not so sure about what you say on MapReduce, especially `So basically any problem that doesn't break data locality principle may be efficiently implemented using MapReduce`. From my understanding, you can only solve problems which can be expressed with the MapReduce pattern.
Comment: @fxm: MapReduce framework may be used for pretty much different tasks. For example, Oozie - workflow scheduler for different Hadoop components - has so-called Java action, that simply creates one mapper and runs custom Java code in it. With this approach you can essentially run _any_ code. This, however, won't give you any MR advantages compared to simple Java app. The only way to gain these advantages is to run computations in parallel (using Map) on *nodes with data*, that is, locally. To summarize: you can run _any_ code with MR, but to obtain performance, you need to hold  data locality.
Comment: I agree, but what I meant is a developer should not worry about managing data locality (which is managed by hadoop), but rather about expressing the wanted algorithms with the MapReduce pattern.
Comment: @fxm: developer ensures locality by using mapper. Roughly speaking, "map(f, data)" means "transfer f() to data nodes and run it locally".  If developer doesn't consider locality (e.g. puts all computations to reducer), he loses all advantages of MR. So expressing algo as (efficient) MR job implies utilization of data locality anyway. That said, I would still suggest using more flexible tools than pure MR such as Spark.
Answer: Data Science has many different sub-areas as described in my post). Nearly for each area, scientists and developer has significant contributions. To learn more about what can be done, please look at following websites:

Data Mining Algorithms & Machine Learning -> Apache Mahout
Statistics -> RHadoop
Data Warehousing & Database Querying -> SQL-MapReduce
Social Network Analysis -> Article
Bio-informatics -> Article - 1 , Article - 2

Also, there are some work on MapReduce + Excel + Cloud combination but I have not found the link.

What are the different classes of Data Science problems ...

Each "classes" is not purely homogeneous problem domain, i.e. some problem cannot be solved via map and reduce approach due to its communication cost, or algorithm behavior. What I mean by behavior is that some problem wants to have control on all data sets instead of chunks. Thus, I refuse to list type of problem "classes".
Do not forget that knowing what MapReduce can do is not enough for Data Science. You should also aware of What MapReduce can't do, too.
Answer: There is a paper you should look into:
MapReduce: Distributed Computing for Machine Learning
They distinguish 3 classes of machine-learning problems that are reasonable to address with MapReduce:

Single pass algorithms
Iterative algorithms
Query based algorithms

They also give examples for each class.
Answer: map/reduce is most appropriate for parallelizable offline computations. To be more precise, it works best when the result can be found from the result of some function of a partition of the input. Averaging is a trivial example; you can do this with map/reduce by summing each partition, returning the sum and the number of elements in the partition, then computing the overall mean using these intermediate results. It is less appropriate when the intermediate steps depend on the state of the other partitions.


ArcGis Online address related pop-up

Question: I am a new user of ArcGIS online. I need to build a map that allows to see the name of the inspector in charge of the area.
So far I succeeded in creation a map that shows the boundaries of the neighborhoods and a pop-up (very basic) that shows the information required.
I would like to know if there is a way to show the same pop-up (for the corresponding area) when I use the address locator.
Example: If I type the address of my house I would like to see (right away) a pop-up showing the name of the inspector that I should contact.

Answer: This would require a custom solution built with the ArcGIS JS API if you wanted the reverse geocode result and the feature query result in the same popup.
Answer: It's not possible to achieve it in ArcGIS Online, but there is a workaround.

Do a search with the address locator
The locator will zoom to the area
Click on the feature layer in the area and the custom popup will appear

In this case there is no need for any custom made webapps or such, but just the basic is enough.

ArcGis Online address related pop-up


What is a tidy algorithm to find overlapping intervals?

Question: I'm sure this must have been asked before, but I'm not finding it: I'm only finding related, but harder questions.
I've got four points, representing two lines like this:
       A         C      B   D
|------*---|-----+----|-*---+---|----------|
0          10         20        30         40

So in the example, AB = {7, 21} and CD = {16,26}. (The lines could be in any relation to each other, and any size.) I want to find out whether or not they overlap, and by how much if so. (In the example, the answer would be 5.) My current solution involves a bunch of complicated if/then steps, and I can't help but think there's a nice arithmetical solution. Is there?
(P.S. Really, I'm doing bounding-box intersection, but if I can get it in one dimension, the other will be the same, obviously.)
Answer: Try this:
intersects = (max(a,b) > min(c,d)) && (min(a,b) < max(c,d))
overlap = min(max(a,b), max(c,d)) - max(min(c,d), min(a,b))

If you can assume a <= b and c <= d:
intersects = (b > c) && (a < d)
overlap = min(b, d) - max(c, a)

You can also calculate intersects as follows:
intersects = (overlap > 0)

Comment: I want the amount of overlap, not just whether or not they do. Thanks, though.
Comment: @sprugman: it would be pretty easy to extrapolate Mark's code to compute the amount of intersection.
Comment: Which he did for me, Matt. Thanks, Mark! :)
Comment: won't `max(a,b)` always be `b`, and at least as I've set up the problem? and `min(c,d)` is always `c`? etc.?
Comment: @sprugman: If you assume that `a <= b` and `c <= d` then yes. I didn't see anywhere in your question where you stated that it was safe to make this assumption, so I didn't make it. I've updated my answer.
Comment: @Mark, good point. In which case we get `overlap = min(b, d) - max(c, a)` which is the kind if tidiness I was talking about! :^,
Answer: A line segment is the intersection of two opposing rays (two half-infinite lines in opposite directions). You have two line segments to intersect -- the result is the intersection of all 4 rays. So you can factor your code as three successive ray-intersections: the leftward of the left-facing rays intersected with the rightward of the right-facing rays.
(This is another way of stating the now-accepted answer.)


Is Vader's Darth title known to ordinary Imperials and general public?

Question:     Vader is a member of Sith Order, therefore he has a title of Darth - Darth Vader. But most Imperials address him as Lord Vader. Obi-Wan calls him Darth, but Obi-Wan knows a lot more than others. Was title of Darth known to ordinary Imperials and general public ? If it was, someone more educated could surely connect the dots between this Darth, his Force powers and look, with ancient Sith lords.
Answer: Yes, the title of Darth was generally known throughout the Empire, and its meaning was at least somewhat known, but whether Vader was himself a Sith was a matter or rumour rather than fact. The new canon novel Tarkin gives us some insight into the general perception of Darth Vader:

...that had given rise to so many rumours about Vader: that he was a counterpart to the Confederacy's General Grievous the Emperor had been holding in reserve; that he was an augmented human or near-human who had been trained or had trained himself in the ancient dark arts of the Sith; that he was nothing more than a monster fashioned in some clandestine laboratory.

Tarkin's private thoughts later make it clear that he, at least, knew that Vader was indeed a Sith:

Disdain crept into his thoughts. That Vader was a Sith shouldn't factor into the question of authority

Of course, Tarkin was an exceptional individual of great intelligence and education, but it does show that the knowledge of the Darth title was not unusual in the galaxy.
While it is no longer canon, it's worth noting that the title of Darth has quite a rich history in Legends. In that continuity, the Sith were not always as secretive as they have been for the last thousand years; the history of the Republic was littered with various Darths who invaded or attacked it across the millennia. It's also important to note that in Legends, the title was sometimes taken by people with no real connection to the Sith at all (Darth Revan and Darth Malak, who were renegade Jedi rather than true Sith).
All of this gives us a pretty clear idea - Vader was a mysterious figure in the Empire, and while the significance of the Darth title was known, the significance of Vader using it was uncertain.
Comment: I doubt that meaning of word Darth was unknown to professional historians thought the Galaxy, or even to any more educated person. After all, Sith supposed to be much more open organization before Darth Bane's reform.
Comment: Indeed. They had even an Empire of their own, in Legends.
Answer: Yes, people called him Darth Vader.
Princess Leia calls him Darth Vader when she meets him for the very first time in A New Hope.
Comment: Of course, when that scene was filmed, Darth was his first name.
Comment: Yes, but were citizens of the Empire generally aware of the meaning of the title?
Comment: @Essen That's the same as asking if the public knew "Darth" was a Sith title. Would the general public know about the Sith, given how secretive the Sith were?
Comment: Leia was part of Imperial Senate, but she was not in Palpatine's inner circle. I guess if she knew, general public must have known. Of course, at that moment Darth was still personal name, out of universe.
Comment: @RichS That's exactly what I was after, and I assume the original question, too. If people knew Darth was a Sith title, and they knew what Sith actually were, then they had clear understanding what they were dealing with Vader. Like **Werrf** says in his answer, at least in Legends, the (Old) Republic had a long and bloody history with the Sith, but were the citizens aware of these over a thousand years old facts.


Rotary Sensor Potentiometer SP280X series

Question: I have a novatechnik rotary sensor potentiometer SP2801 100 002 001.Th datasheet shows that it has a maximum electrical angle of 100 degree.But I saw that we can move the pot in any direction any number of times.Then what does this maximum angle mean and how to locate it if it is there(the datasheet mentions the 0 degree and 100 degree angle).
Answer: The electrical angle is independent of the pot's physical ability to turn. In other words, there are no physical stops.
Within the 100 degree operational range, it behaves like a regular pot - with the wiper traversing from the resistor terminal at one end of the "electrical" range to the terminal at the other end (100 degrees apart). When you go outside this range, the wiper would be open.
These pots are often used on the output shafts of gear-motors for position sensing, and a physical stop would break the pot if the motor ran outside of the 100 degree range.


I2C - bit banging query

Question: When bitbanging I2C, rather than switching the LOW/HIGH states of the pin, switch between LOW/HiZ by simply altering the DDR/DIR registers.Why is it like this? Why can't we switch LOW/HIGH states of data pin? My controller is communicating with only one slave.
Comment: Have you studied how I2C works? It should be pretty obvious why you can't just drive the lines high or low.
Comment: Actually Iam trying to implement I2C in my controller which doest have I2C hardware driver. I am aware of I2C protocol (start,write, read, stop). According to my understanding, I need to configure 'DDR pins' of GPIO (SCL And SDA) as output, and to set/clear the SCL and SDA i need to change the 'data pins' accordingly. I am very much new to this module. Please guide me if I am wrong.
Comment: What would happen if you were driving high when the peripheral needed to drive low?  Review the operation of I2C and you will see how that, and your original question, are solved.
Comment: SDA **must** be high-Z when high as either the master or slave may need to set it low. If the master is pulling high (not high-Z) and the slave needs to send a zero and pulls down the line, you will end up with the two chips fighting. For SCL, even if you only have one master, a slave may cause the master to pause by pulling the line low (clock streaching). Please read up on I2C: it appears you have a lot to learn.
Comment: This was already answered in the comment at your earlier question: http://electronics.stackexchange.com/questions/103440/bit-banging-i2c-question
Answer: Your code change the DDR/DIR bit of that pin because in I2C you want either an high impedance pin (pin set as input pin -> send a "1") or a super low impedance pin (output pin set to 0 -> send a "0"). By design I2C forbids to pull up the bus directly from the pin as it will cause very BAD things (one device setting an high, another device setting a low and... bang!!! a great short-circuit!). Pull up is done by the external 4.7K pull-ups. Any pin used for I2C should be open-drain.
Comment: Thanks... this solves my query.One last question.Lets say, I have initially configured a port(GPIO) pin as input, pin(HighZ), Now I want to write '0' to that port. So
changing the pin data to '0'and then changing its DDR port to output is correct or,
changing DDR port to output and then writing '0' is correct. I know what I am asking something silly, but I want to improve my basics regarding port configurations.
Comment: you got two registers: DDR and PORT register. When you set the DDR bit to "output" the pin will act depending on the content of the corresponding bit in the PORT register. So yes, you can set the PORT register to zero at init time and leave it as it is and then toggle the DDR bit as you want to be.

I2C - bit banging query


About numeral systems

Question: I have a question
about conversion in number systems . For converting binary to decimal ,why do we need to multiply each of the coefficients by powers of 2 ? And while converting decimal (such as100.12 ) back to binary ,why do we need to multiply the numbers after decimal point by 2 ? And this happens in almost all the conversions . Can somebody please explain me why ?
For example  11 in Binary would be equal to  1+2=3(in decimal)
But why do we need to raise them to the powers of two ?
Answer: For your last Q, a sequence $d_n...d_0$ of $1$'s and $0$'s, in base two, represents the number $2^nd_n+...+2^0d_0.$
Answer: That follows immediately from the definition of a "base n number system".  The numeral "abcd" in base n means $a\cdot n^3+ b\cdot n^2+ c\cdot n^1+ d\cdot n^0$.  (Of course, each of a, b, c, d must be less than n and $n^1= n$ and $n^0= 1$.)
  In our "usual" base 10 number system "abcd" would be $a\cdot 10^3+ b\cdot 10^2+ c\cdot 10^1+ d\cdot 10^0= a(1000)+ b(100)+ c(10)+ c(1)$.  For example, $3215= 3(1000)+ 2(100)+ 1(10)+ 5$.
In base 2, we can use only "0" and "1" as "digits as they are the only positive integers less than 2.  $11001= 1(2^4)+ 1(2^3)+ 0(2^2)+ 0(2)+ 1(1) which would be 1(16)+ 1(8)+ 0(4)+ 0(2)+ 1= 25 in base 10.
"Base 16" is also commonly used in computer science because it is a power of 2 (and computer registers are typically 4, 9, or 16 bites long) and results in smaller numerals than base 2.  Of course, we also need symbols for the integers 10, 11, 12, 13, 14, and 15 which are also less than 16. The standard is A= 10, B= 11, C= 12, D= 13, E= 14, and F= 15.  The numeral A5B3 is the same as  $A(16^3)+ 5(16^2)+ B(16)+ 3(1)= 10(4096)+ 5(225)+ 12(16)+ 3= 42280 in base 10.
Answer: The whole thing about having different number systems is having different number of available symbols to express the quantity.
The concept of 11 things is not the same in base 3, as it is in base 10, or in base 2.
For your question, I am doing a simple computation below. In base 2, you have only two digits (symbols) available, $\big\{0,1\big\}$, while in decimal system you have 10 digits (symbols) available $\big\{0,1,\cdots,9\big\}$.
Now we write the following correspondence with the decimal numbers on the left side, and the binary numbers on the right.
0$\to$0
1$\to$1
What do we do now? We have more symbols left in the decimal system, but no more symbols in the binary system. The solution is, do exactly as what we do when the symbols run out in decimal system.
When we reach 9, we write the first symbol 0 again, but since we already have a number 0, we elevate it by writing the next bigger digit beside it.
The consequence?
We get 10. Now we proceed in the same way as before.
So we get.
2$\to$10
3$\to$11
4$\to$100
5$\to$101
.
.
.
So in base b, we are using b digits, and suppose we have an n-digit number in base b, $a_na_{n-1}a_{n-2}\cdots a_1$. Now let us try to find out how we arrived at this number.
(Please note that all the following calculations have been done in the familiar decimal system that we use.)
Note that we have the symbol $a_n$ in the leftmost place. From the generic rule outlined above, we can see that $a_n$ comes to play only if we have already used up all the symbols occurring before it.
So, for the $n^{th}$ place we have used up all the symbols in $\big\{0,1,\cdots a_n-1\big\}$, which contains $a_n$ symbols. So, we have used up $a_n$ symbols for the $n^{th}$ place.
Now, we move on to the $n-1^{st}$ place, and we check that, in a similar way, we have used up all of the $a_{n-1}$ symbols in $\big\{0,1,\cdots a_{n-1}-1\big\}$. But for each of the symbols $a_i$ in the $n^{th}$ place, we have actually used up all the $b$ symbols in the $n-1^{st}$ place, before we could come to $a_{n-1}$ in the $n-1^{st}$ place for $a_n$ in the $n^{th}$ place.
Thus to reach $a_n$ in the $n^{th}$ place and $a_{n-1}$ in the $n-1^{st}$ place, we have crossed $b\times a_n + a_{n-1}$ numbers before it.
By a similar argument, we reach $a_n$ in the $n^{th}$ place, $a_{n-1}$ in the $n-1^{st}$ place and $a_{n-2}$ in the $n-2^{nd}$ place after crossing $b^2\times a_n + b\times a_{n-1} + a_{n-2}$ numbers before it.
We proceed in the same way, till we have crossed $b^{n-1}\times a_n+ b^{n-2}\times a_{n-1} + \cdots + a_1$.
Thus, to reach $a_na_{n-1}a_{n-2}\cdots a_1$ in base b, we need to cross $b^{n-1}\times a_n+ b^{n-2}\times a_{n-1} + \cdots + a_1$ numbers (calculation in decimal numbers). So it is equivalent to saying that $0+b^{n-1}\times a_n+ b^{n-2}\times a_{n-1} + \cdots + a_1$ in decimal is same as $a_na_{n-1}a_{n-2}\cdots a_1$ in base b.
I hope this answers your question, because b (base) is 2 in the binary system since we use two symbols only, 0 and 1.

About numeral systems


Why does it seem like there are many restrictions regarding asking questions?

Question: It feels like the only acceptable questions I can write are those that fit to a narrow ideal outlined by the website administrators.
Why does it seem like there are many restrictions regarding asking questions?
Comment: I think opening the discussion of your question by asserting an answer without qualification might discourage people from participating.  Perhaps you could consider rephrasing the question; what about "what is the purpose of a Philosophy stack exchange site, given the implicit background knowledge, beliefs and standards of participants that may not be accessible to an inquisitive audience asking questions?"
Comment: It can often be frustrating to fit your question to another's frame of reference, but it is often necessary in our world.  It is worth putting in the effort to articulate your questions well, something which I am still working on.
Comment: @felino Have you tried any of the *other* Q&A sites used for questions based in philosophy? e.g., https://answers.yahoo.com/dir/index?sid=396545231 http://www.quora.com/Philosophy http://www.answerbag.com/category/philosophy_535

Philsophy.SE is far more clearly geared towards people with an interest in *academic* philosophy, as its studied, which is a much more narrow field of interest than you may have been expecting.

Also FWIW, rhetorical questions aren't a very good fit for this site in general.
Comment: @user4894 I don't think that's quite right... or rather if you ask the big philosophical questions, I agree there's no clear answers. But if you ask for say an interpretation of a passage or an explanation of a quote, many (not quite all) of those sorts of questions are answerable. I take those sorts of question to be useful. I think the bigger issue is that the word philosophy refers both to a professional discipline and to a popular concept of talking about non-practical things...
Comment: I just want to say that I don't understand this place either. Philosophy asks questions that only lead to other questions. StackExchange is about questions that have answers. The two are wholly incompatible; hence the contradictions and frustrations of this forum. The OP's phrasing may not be artful; but I understand where he's coming from.
Answer: This question is better addressed in meta as its not really a question about Philosophy but about the 'friendliness' of the site.
There are biases on this site; as many other sites will have; for example, I find myself frustrated by the inordinate attention that questions around mathematics enjoy, whereas there are many more important topics to discuss or subjects to raise; also its important to articulate your questions well - so that its easily understood by a wide and diverse audience; or if it is directed towards a specialist audience to use specialist jargon correctly.
I recall the Physicist Feynman when he was studying Physics in advance of his tutors developing a specialised language of his own - it took him a while to understand that standardised technical language is there for a reason - to facilitate communication in what is a demanding discipline; the same goes for philosophy; it too does have a technical vocabulary - and one should at least attempt to learn something of this language when one is asking questions in this subject.
But given the low visibility of philosophy in the Anglophone world (as opposed to the Francophone one) and due to the prominence of science as opposed to the humanities, particularly now which is an exceedingly technocratic age and on the web which by its very medium (McLuhans 'the medium is the message' comes to mind) tends to foreground the sciences and the physical & mathematical sciences in particular; and the well-publicised fight between them (the Science wars, which actually goes much further back if one is to judge by Snows two cultures and some of Simone Weils writings) this technical language isn't used much here (Philosophy according to Hawking is dead, for example).
Still it is possible (despite the obscurities of the Continental tradition) to ask questions in a clear way - one of the advantages of the anglo-analytical tradition (one might even call it a Socratic virtue) - without burdening oneself with a technical tradition.
If the question is unclear, then one is passing the burden onto someone else to make some sense of the question as well as answering it; this is different to a wide-ranging question which is difficult to answer as one doesn't know quite where to start.
Both of these difficulties can be gotten over, if one is prepared to ask and accept  help in clarifying the question and/or focusing it.
Comment: Could you provide references to the claim that the Francophone world has a high visibility of philosophy compared to the Anglophone world?
Answer: Take a look at some of the questions on meta which address the issues you raise. Note that a new user is not necessarily an amateur philosopher. We have new users who are already familiar with philosophy who fit right in rather quickly. The issue here primarily is with people who are new to philosophy and unaware as to how to phrase their question or narrow it sufficiently for this Q&A site.
Helping amateur philosophers (who often have difficulty with the format):
Are amateur users that have little knowledge about philosophy not "ideal" for philosophy.SE?
New to Philosophy and learning how to ask good questions
Summer of Love (of Wisdom): How to drive friendliness?
Veteran users having problems question quality (particularly from amateur philosophers):
Subjective answers and questions stimulating subjectiveness (subjectivity and lack of focus)
Do questions have to be answerable? (unreasonably answerable)
Should we restrict this site to academic questions only...? (a preference for 'academic' questions)
Options for other websites which may allow the ambiguous/open-ended/unfocused content Philosophy.SE does not:
What other sites are available for debating issues?


VLAN links for IP Cameras

Question: I currently have fiber vlan trunks between all of our buildings. We are adding another ip camera server and am looking to segregate traffic. I am thinking of dedicating a fiber link and putting it on the specific ip camera vlan so traffic for the cameras in those buildings would only use that fiber and not the fiber trunk back to the main office. I have attached a a drawing of the current layout plus additional links to be added. What is the best way to do this.
Comment: What make and model of switches are you using?  How they handle Spanning Tree will greatly affect your design choices.
Comment: Core is a stack of 3 - Dell PowerConnect 7048P's. Each building has Dell PowerConnect 5548P's.
Comment: Why do you want to put camera traffic on separate fiber? Is it performance?  Bandwidth?  How many cameras do you have?  I would be reluctant to do that unless there was a compelling reason.
Comment: There are over 60 cameras connected to the Packaging, Shipping, and Fiberglass switches. For bandwidth it would seem better to have there traffic go over dedicated links...
Comment: Did any answer help you? if so, you should accept the answer so that the question doesn't keep popping up forever, looking for an answer. Alternatively, you could provide and accept your own answer.
Answer: The short answer is, your proposed additional fiber in the diagram will not gain you any real advantage for VLAN 3.
All traffic from Packaging or Fiberglass, destined to the server in Shipping on VLAN 3, will still traverse the Core.  (Presuming that you have already properly tuned your Spanning Tree settings to make the Core switch stack the root bridge.)
This is due to the Dell 5500 series switches only supporting regular 802.1D Spanning Tree and 802.1w Rapid Spanning Tree.  What this means, is that in your entire switched domain in the diagram, there is only one Spanning Tree for all VLANs.

If you are set on adding additional links, and dealing with VLAN 3 in this way, you have several alternatives (viable or not, depending on your circumstances):

Install switches capable of running MST (Multiple Spanning Tree), which allows you to create VLAN to Spanning Tree mappings.  (Your core switches (Dell 7048P) DO support MST, but the 5548P switches do not.) You could assign all other VLANS to MST Instance 1, and then assign VLAN 3 to MST Instance 3 (instance numberings are just for example).  Then you would tune the settings so that the root bridge for MST Instance 1 is your core stack, and root bridge for MST Instance 3 is the Shipping switch.
Install another vendor's switches which are capable of creating a Spanning Tree instance for each VLAN.  It is called different things by different vendors, PVST+ in Cisco, VLAN Spanning Tree in Juniper, etc.  Similar to MST, tune the Per VLAN Spanning Tree instance for VLAN 3 to make the Shipping switch the root bridge for that instance.

With all of that said, there are still one main concern in my mind.
Why locate the server in the building with the Shipping switch?  You would be better served by placing it in the main location with your Core switches.
If it MUST be attached to the Shipping switch, and you can't make changes to the switches to get a more robust Spanning Tree configuration, you still have a few options.
You could increase your bandwidth to each switch in order to handle the hair-pining of VLAN 3 traffic through the core.  All of your Dell switches support LACP link aggregation, this would allow you to gain bandwidth easily between the switches to support the IP-Cameras.
Comment: The Dell 5500 series switches do suppost MSTP. http://www.dell.com/us/business/p/powerconnect-5500-series/pd So it may be an option to keep those in place. The reason this server is in the shipping building is because I was trying to segregate some of the traffic. I also have some free fiber pairs between these buildings. There is another camera server in the main office where the core switch is that 80 more cameras connect too.
Comment: @ravelsol if I were you, I would verify that the 5500's have MST capability in their configuration (it might be software version dependent).  The source of my assertion that they were not MST capable, was the actual spec sheet pdf (dated from January 2014) for the 5500 switches.  I have longs since stopped trusting most vendor websites, as they tend to bend the truth more than the actual spec sheets.  As for segregating the traffic, I'm still not understanding why you're attempting to do that.  Are there throughput issues you're running into or anything of that nature?

VLAN links for IP Cameras


Command line glob to view multiple images using ImageMagick's display

Question: How can I open multiple images files using ImageMagick's display command? If I enter say display *.png in a directory containing multiple PNG files, only one is displayed. I think I'd prefer them to appear one at a time, but all at once could be useful too.
Answer: If you don't specify a delay, imagemagick will load the first image and wait for your input, that is
SPACE - load next image
BACKSPACE - load previous image
If you want to display the images in sequence without interaction, use a delay:
display -delay 50 *.png

To display all of them you could try with
display 'vid:*.png'

Answer: If you would like to see all your images at once, you can use montage to group them into a single, temporary image in ImageMagick's MIFF format (which is lossless regardless of whether image has alpha channels or 8/16/32 bit integer/float samples) and tell display to show that:
display <(montage *.png MIFF:-)

The nice thing about this is that montage will lay out your images sensibly according to how many you have:
If you have 3 images:

If you have 4 images:

If you have 8 images:

If you have 9 images:

If you prefer to layout your images yourself, rather than let montage do them automagically, you can specify how many columns (i.e. how many images across the screen) you'd like and it will make as many rows as necessary. So, say you wanted 3 images in each row:
display <(montage *.png -tile 3x MIFF:-)