Jailbreak iPad 3 ("the new iPad") iOS 9.1

Question: Is it possible to jailbreak iOS 9.1 on an iPad 3, WiFi model? If not, I also have the possibility to update to iOS 9.3.5.
Answer: A beta of the Home Depot iOS 9.1-9.3.4 32-bit Mixtape Player supports your device on iOS 9.1.
There is no jailbreak for iOS 9.3.5, so take precautions with your jailbreak, especially since Prometheus does not support 32-bit devices.


Threads or asynch?

Question: How do you make your application multithreaded ?
Do you use asynch functions ?
or do you spawn a new thread ?
I think that asynch functions are already spawning a thread so if your job is doing just some file reading, being lazy and just spawning your job on a thread would just "waste" ressources...
So is there some kind of design when using thread or asynch functions ?
Answer: If you are talking about .Net, then don't forget the ThreadPool. The thread pool is also what asynch functions often use. Spawning to much threads can actually hurt your performance. A thread pool is designed to spawn just enough threads to do the work the fastest. So do use a thread pool instead of spwaning your own threads, unless the thread pool doesn't meet your needs.
PS: And keep an eye out on the Parallel Extensions from Microsoft
Answer: Spawning threads is only going to waste resources if you start spawning tons of them, one or two extra threads isn't going to effect the platforms proformance, infact System currently has over 70 threads for me, and msn is using 32 (I really have no idea how a messenger can use that many threads, exspecialy when its minimised and not really doing anything...)
Useualy a good time to spawn a thread is when something will take a long time, but you need to keep doing something else.
eg say a calculation will take 30 seconds. The best thing to do is spawn a new thread for the calculation, so that you can continue to update the screen, and handle any user input because users will hate it if your app freezes untill its finished doing the calculation.
On the other hand, creating threads to do something that can be done almost instantly is nearly pointless, since the overhead of creating (or even just passing work to an existing thread using a thread pool) will be higher than just doing the job in the first place.
Sometimes you can break your app into a couple of seprate parts which run in their own threads. For example in games the updates/physics etc may be one thread, while grahpics are another, sound/music is a third, and networking is another. The problem here is you really have to think about how these parts will interact or else you may have worse proformance, bugs that happen seemingly "randomly", or it may even deadlock.
Answer: I'll second Fire Lancer's answer - creating your own threads is an excellent way to process big tasks or to handle a task that would otherwise be "blocking" to the rest of synchronous app, but you have to have a clear understanding of the problem that you must solve and develope in a way that clearly defines the task of a thread, and limits the scope of what it does.
For an example I recently worked on - a Java console app runs periodically to capture data by essentially screen-scraping urls, parsing the document with DOM, extracting data and storing it in a database.
As a single threaded application, it, as you would expect, took an age, averaging around 1 url a second for a 50kb page. Not too bad, but when you scale out to needing to processes thousands of urls in a batch, it's no good.
Profiling the app showed that most of the time the active thread was idle - it was waiting for I/O operations - opening of a socket to the remote URL, opening a connection to the database etc.  It's this sort of situation that can easily be improved with multithreading.  Rewriting to be multi-threaded and with just 5 threads instead of one, even on a single core cpu, gave an increase in throughput of over 20 times.
In this example, each "worker" thread was explicitly limited to what it did - open the remote a remote url, parse the data, store it in the db.  All the "high level" processing - generating the list of urls to parse, working out which next, handling errors, all remained with the control of the main thread.
Answer: The use of threads makes you think more about the way your application needs threading and can in the long run make it easier to improve / control your performance.
Async methods are faster to use but they are a bit magic - a lot of things happen to make them possible - so it's probable that at some point you will need something that they can't give you. Then you can try and roll some custom threading code.
It all depends on your needs.
Answer: The answer is "it depends".
It depends on what you're trying to achieve. I'm going to assume that you're aiming for more performance.
The simplest solution is to find another way to improve your performance. Run a profiler. Look for hot spots. Reduce unnecessary IO.
The next solution is to break your program into multiple processes, each of which can run in their own address space. This is easiest because there is no chance of the individual processes messing each other up.
The next solution is to use threads. At this point you're opening a major can of worms, so start small, and only multi-thread the critical path of the code.
The next solution is to use asynch IO. Generally only recommended for people writing some of very heavily loaded server, and even then I would rather re-use one of the existing frameworks that abstract away the details e.g. the C++ framework ICE, or an EJB server under java.
Note that each of these solutions has multiple sub-solutions - there are different breeds of threads and different kinds of asynch IO, each with slightly different performance characteristics, but again, it's generally best to let the framework handle it for you.

Threads or asynch?


Can I connect my digital ammeter in parallel to a battery to check the maximum current It can supply?

Question: Can I connect my digital ammeter in parallel to a battery to check the maximum current It can supply? Will it burn my ammeter or its fine?
I have tried it and nothing happened so is the ammeter being damaged slowly from inside?
Comment: What do you mean? Can you draw a circuit? The only circuit that I can imagine that just contains a battery & a meter has them in series, in a loop.
Comment: Yes I mean that.
Answer: It'll most probably burn your ammeter (or at least blow the fuse, if the ammeter has one) unless your battery is tiny. And it might damage the battery as well.
E.g. car batteries even burn screwdrivers when used as short circuit, so your delicate digital ammeter definitely won't survive.
But in theory, your approach is correct, you "just" have to make sure the battery's expected short-circuit current is below the rated maximum of the ammeter. If e.g. the ammeter has a 2amps range, then even any decent AAA battery will exceed that.
Answer: The already given answer is sufficient. However i just wanted to point out that in order to measure maximum current possible you needn't do this. Connect it with any resistance whose value is known precisely well and then measure the current. You'll be able to find out the internal resistance of the battery. The voltage of the battery divided by the internal resistance would simply give the maximum possible current.
Comment: Your answer suggests that the classic model of a battery as an ideal voltage source in series with a resistor would be valid for whatever kind of battery the OP is asking about and, that it would be valid in the extreme case where the battery is "short circuited." I know that that model is a good approximation in many cases, but do you actually _know_ that it works well for any battery type, and even in the case of a short circuit? Or, do you merely assume that what works in most cases must work for all?


Exception 401 during export to Microsoft Exchange 2007

Question: I have a vb.net application that communicates with Exchange 2000 & 2003 using WebDAV, but it does not work with Exchange 2007 SP2
Test Environment and configuration

Windows 2003, AD and domain controller, Exchange 2007 SP2
With Public Folders
OWA can be accessed with a browser
https://myserver.com/public/Contacts can be access through browser
The credential used by the vb.net application is NetworkCredential
Using administrator credentials

When I run my code it throws exception 401 unathorized. Using EWS, it works. At this stage, I don't want to convert the code to EWS. Has anyone got a solution?
Answer: are you using the correct URL for SoapUI?
Point it to https://server/ews/exchange.asmx. Replace server with the dns name of your Exchange server.


How can you validate what you've learned as a UX designer and enhance your skills?

Question: For the past four years, I have been practicing interaction design, but I have recently encountered a bottleneck in my ability to improve. Currently, when I encounter a problem, I instinctively generate a solution, yet I am unsure if this is the correct approach. This uncertainty troubles me, and I am seeking guidance on how to enhance my problem-solving skills more efficiently.
Answer: Here are a few things I've done that have been incredibly helpful to me:

Connect with industry experts or experienced individuals on LinkedIn. Seek validation and feedback from them regarding what you've learned.

Research to determine whether what you've learned is already in use by someone. If it is, that indicates you're on the right path.

If you have clients or access to users, test your solution with them. If most of them find your solution effective, you're heading in the right direction. Remember that there isn't one fixed solution for every problem.

Validate your ideas with people within your immediate reach, such as family members, friends, and colleagues at work.

Consider sitting in a coffee shop and approaching people to spare a few minutes to discuss your solution and gather feedback.

Post about the problem and the solution you've devised on LinkedIn. Explain your approach and ask for feedback from your network.

Additionally, you can share your problem and solution on websites like https://ux.stackexchange.com/ to gather insights from a broader audience.
There are numerous ways to validate your solution or what you've learned.


turning shower off before turning water faucet off

Question: I had always been told to push shower knob down, directing the water back to the faucet, before pushing the water off. If you just push the water knob off, letting the shower knob fall on its own, is that hard on your shower equipment?
Answer: Welcome to DIY StackExchange!
I never heard of that. I always turn off the shower and let the spigot diverter fall when it's ready, if it ever does. I don't see how it would cause any extra wear.
Comment: If anything is going to cause wear, it would be pushing the diverter while there is pressure, as it means there is force of water pushing one surface against another. If the "let it fall" method is used, the force is reduced. If there is wear, it's microscopic, though!
Answer: As mentioned by @fred_do_u, wear is probably created a lot more when there is pressure, so when you pull it up or down while water is flowing. It'd still take a long time before it goes bad and need replacement.
I think the main reason for doing so is if you have knob that's rather sticky. I've experience those... where you start the water and you get all wet with cold water. If your knob comes down automatically, then that's definitely not an issue.


What is "Applied Harmonic Analysis"?

Question: Does it mean the application of the theorems of Harmonic Analysis to real-world problems, say, engineering or data science problems, and that Applied Harmonic Analysis is not about theory or proofs? And since it's "applied", does that mean there's no measure theory involved, except for the very minimal: monotone convergence and dominated convergence theorems?
Comment: Ddi you check [this](https://www.math.tu-berlin.de/fileadmin/i26_fg-kutyniok/Kutyniok/Papers/WiAHA.pdf)?
Comment: Possibly related: http://math.stackexchange.com/questions/1724290, http://math.stackexchange.com/questions/1002
Comment: See [signal processing](https://en.wikipedia.org/wiki/Signal_processing). Fourier analysis and hence distribution theory is used a lot, even if there is no need to be as rigorous as in pure maths.
Comment: live link for user99914's comment: DOI: [10.1515/dmvm-2008-0034](https://doi.org/10.1515/dmvm-2008-0034)
Answer: Just because a subject has an applied moniker attached to it does not mean that the subject does not require proofs and theorems. Applied mathematicians are mathematicians foremost, and as such they work to provide theoretical guarantees concerning their objects of study.
Applied Harmonic Analysis follows the tradition of classical Harmonic Analysis in that aims to represent acquired data by some decomposition of basis functions. Classical Harmonic Analysis deconstructs a signal into sinusoids of varying frequencies, whereas modern applied Harmonic analysis aims to represent signals in a variety of different bases. These bases are usually attached to some Hilbert space framework.
These bases include wavelets, orthogonal polynomials, reproducing kernels, and others. To provide a rigorous analysis of these representations, often convergence results require that the estimated functions reside in some Hilbert space (most frequently) or Banach space (such as $C[0,1]$). For example, Representer Theorems lean on projections available in a Hilbert space to yield guarantees as to the structure of the solution to certain regularized optimization problems.

What is "Applied Harmonic Analysis"?


How do you set up live preview for headless Craft CMS installs connected to Gatsby?

Question: How do you set up live preview for headless Craft CMS installs connected to Gatsby?
Answer: When you enter live preview in the cp, it's an iframe to the URL you've set up for that section, including a token=xxxxxxxx, you need to pass this along to your GraphQL (assuming that's what you use), to let it know it's not supposed to fetch data from the database.
Andrew Welch has a bit more info about it here: https://nystudio107.com/blog/headless-preview-in-craft-cms
As well as another question here (based on vue, so you probably need to change som stuff for Gatsby): Live preview using Vue.js
Comment: How should I configure the frontend application url on craft? I'm trying to do this, but I didn't find a way so far. The frontend application is already looking for a token parameter to send it back on graphql requests. Could you help me on this?


Discrepancy between calculus methodologies - Is it significant?

Question: Two of the ways of doing calculus with algebra are non-standard analysis NSA and smooth infinitesimal analysis SIA. NSA has a technique called 'taking the standard part' which neglects incremental (or infinitesimal) terms at the end of derivations, whereas SIA has the nilsquare rule which neglects higher power incremental terms during derivations. For example if we differentiate $y = x^2$ from first principles we have:
$$(x + h)^2 = x^2 + hx^{2'}$$
$$x^2 + 2hx + h^2 = x^2 + hx^{2'}$$
$$x^{2'} = 2x + h$$
$$x^{2'} = 2x$$
in NSA, whilst in SIA we have:
$$(x + h)^2 = x^2 + hx^{2'}$$
$$x^2 + 2hx + h^2 = x^2 + hx^{2'}$$
$$2hx = hx^{2'}$$
$$x^{2'} = 2x$$
It's tempting to think the two methods are equivalent. However, consider the formula for secant length:
$$s^2 = h^2 + (hy')^2$$
$$s = h \sqrt{1 + y'^2}$$
Taking the standard part of $y'$ while evaluating $s$ would result in the first RHS term being set to zero, rendering the equation useless, whereas the nilsquare rule doesn't have this effect. What does this tell us, if anything, about the validity and/or usefulness of NSA and SIA relative to eachother?
Answer: Taking the standard part is not really a technique for neglecting infinitesimal terms at the end of a derivation. Instead, non-standard analysis has a bona fide standard part function $st: \mathbb{R}^* \rightarrow \mathbb{R}$, a map that assigns an (extended) real number to each hyper-real number.
With this in mind, your first derivation, where you claim to differentiate the squaring function using NSA, is invalid. In particular, the claimed equality $(x + h)^2 = x^2 + h(x^2)'$ does not hold at all; even if it did, you are not allowed just neglect infinitesimals like you do when you pass from $(x^2)' = 2x + h$ to $(x^2)' = 2x$. None of this works.
What you should say instead is that the squaring function has derivative $2x$ because $$ st\left( \frac{(x+h)^2 - x^2}{h} \right) = 2x $$ for each real number $x \in \mathbb{R}$ and infinitesimal hyperreal $h \in \mathbb{R}^*$.
The same thing goes for your third derivation: you cannot just take the standard part in the middle of an equality and expect the resulting equation to hold. As an aside: in its present form, that third derivation does not hold up in Synthetic Differential Geometry (Smooth Infinitesimal Analysis) either - it amounts to inferring $s = h\sqrt{1 + (y')^2}$ from $s^2 = 0$.
This should answer your question: these arguments don't tell us anything about the validity and/or usefulness of NSA and SDG relative to each other, simply because they are not correct.
That said, the methods of NSA and SDG are far from equivalent: NSA strictly includes usual (standard) analysis, while SDG is specifically a way of doing differential calculus in a smooth topos - which excludes many mathematical artifacts that analysts are concerned with, but greatly simplifies a lot of differential-geometric concepts or computations. Then there are models of SDG in which both nilsquare SDG-style and proper NSA-style infinitesimals coexist. I'll stop here: a detailed look at these would be best suited as an answer to another (softer) question.
Comment: The conclusion of your third argument is used in SDG when deriving arc length (with $y'$ denoting the derivative of the function $y$ - what else would it denote?!) but that inference is not valid. You can find a valid derivation somewhere in Bell's A Primer of Infinitesimal Analysis.

As for NSA: it's the topic of my PhD thesis, so I'm fairly unlikely to be wrong about it, especially about such simple facts. None of those equations hold.
Comment: Firstly, if your take on NSA were correct then it's useless as a guide to calculus, which I don't believe even though on balance I prefer SIA (or SDG). Secondly, the equation for the secant $s$ is true by definition if $y'$ is the finite difference quotient and also for $s = 0$. The question is - is it true if $h$ is infinitesimal? Well SDG has nilpotent infinitesimals, so of course $s^2 = 0$; this observation doesn't tell us anything about $s$ itself.
Comment: It's on page 45 in this form: $s'(x) = (1 + f'(x)^2)^{1/2}$ ("Cancelling [h] yields the familiar equation"). So the answer is Yes - it does apply to infinitesimals, just by changing $s$ to $ds$ and dividing by $dh$. Also, if NSA can't even provide simple rules for differentiation then it's useless - why would you advocate for that?


I'm building a Deck. Is it cheaper to buy wood from a mill rather than a box store like Home Depot?

Question: I'm building a Deck. Is it cheaper to buy wood from a mill rather than a box store like Home Depot? Are there any benefits to choosing one over the other? Should I expect to see better quality from a mill vs a box store.
Comment: You found a mill that pressure treats their own wood?
Answer: Most mills do not produce and sell treated wood rated for contact with the ground, so any wood in your deck that will be in contact with the ground should be treated, and gotten wherever it is to be got.
Buying wood from a mill is often cheaper, though it is not always. A smaller operation with a smaller economic influence (like most mills open to the public) will feel fluctuations in the supplies and prices of the commodities they consume in the production of lumber (fuel, electricity, trees, etc.) sharply, and so the price of their own offered commodity (lumber...) will vary sharply according to these fluctuations. A larger corporation like The Home Depot, Lowes, or Menards, likely has multiple sources for the commodities they need, they have bigger and deeper lines of credit, and a huge buffer of stock not on the shelves, insulating them from small localized changes in economies, so their prices will be more stable over the long term.
So the advantages to buying lumber from a mill that you can really count on are customization of lumber dimensions, greater choice of species of wood, and personalized service. The disadvantages vary by mill and your own preferences: is the wood air dried, kiln dried, or not dried at all and sold green? Is the wood graded by the sawyer? Is the wood sold rough cut or has it been planed?
The advantage to buying lumber from a box store is that although they might sell an average generic lumber, this average generic lumber is always there, every day, at a reasonably average price, and you don't have to call ahead, place an order, or make an appointment, you can write a check or use a credit card, and they are often open earlier in the morning and later into the evening. The disadvantage is that the experience of buying lumber from these places is often just as bland and lifeless as the lumber they sell.
Either way you go you might save money, and you might spend more. If saving money is the crux of the issue, compare quotes and go with the cheapest. If saving money is not the crux of the issue, call your local sawmill and make friends with them. You won't regret it. Your deck wood will look and feel unique because it will be, it will have a good story to go along with it right off the bat, and, if you ever get really ambitious you'll have a place to bring logs from trees you've chopped down for sawing into lumber that you can call your own. ;-)
Comment: There is a lot to be said for personal service, it might turn out that paying more was actually a bargain. The workers at the "big box" have a tendency to run the other way if it looks like you might ask them a question.
Comment: If you find a mill that has/can get the lumber you're after, you may consider splitting your order, purchasing what you can from your local mill, and getting the pressure-treated posts from the big-box store.
Comment: Wow, beautiful answer: provides nice pros for both sides, very balanced, yet opinionated in just the right way. I'd upvote this twice if I could.
Comment: Cool, thanks for the thoughtful answer. I was just googling around and there were a lot of people saying you loose your shirt buying from a store like Home Depot so if there was a chance at saving a few dollars I figured I would reach out and se if anyone had experience buying elsewhere. I think there is something to say about buying wood hassle free from stores like Home Depot or Lowes even if it is a few dollars more


Weird Arduino math result

Question: I have a simple code, part of which contains the calculation 800/1000, which seems to return a value of zero for some reason (even though I store it in a float variable). When I replace the calculation with the expected result (0.8) the code works as expected, I just wonder why the calculation doesn't produce the same result.
Comment: The expression is first evaluated as an int, hence `0`, before converting it to float. If you want the expression computed as a float, then at least one argument must be a flost, e.g. `800.0 / 1000`.
Answer: A whole numbers like 800 or 1000 are integers, while a number like 800.4 is considered fractional and is treated as a float. Two integers divided gives an whole integer result, in your case 0. If one number in the division is an integer, and one is a float, the result is a float.
Answer: Both 800.0/1000 or ((float)800)/1000 will work.
The former works only with hardcoded constants, while the latter works also with variables:
<code>int a = 800;
int b = 1000;
float c = ((float)a) / b;
</code>

Weird Arduino math result


How can I fix the reflections on a glass window in Eevee?

Question:
This is a render of an indoor environment. There is a glass window in the room that provides a view of the street outside. The scene is lit with an outdoor HDRI environment texture. The problem is that the outdoor HDRI is reflected on inside of the glass. What is the best way to remove or fix these reflections?
Comment: Moonboots's A is not solution for you. In reality gass viewed from int doesn't reflects ext, but int with light sources bright enough and ext dark enough to be seen thanks to a contrast between int light ray hitting the glass and background. If you are aiming to render with Eevee Ethan-Hall seems to me as answer for you.
Comment: @vklidu Thanks. This question has has been incorrectly marked as a duplicate and closed with a link to: [How to use multiple HDRIs](https://blender.stackexchange.com/questions/49337/how-to-use-multiple-hdris). Which is not what this question is asking, and is not the best way to address this problem.

This question is more similar to [Configuring eevee glass reflections](https://blender.stackexchange.com/questions/165029/configuring-eevee-glass-reflections).
Comment: Note that while similar, this question is still different form the post "[Configuring eevee glass reflections](https://blender.stackexchange.com/questions/165029/configuring-eevee-glass-reflections)" because this new question is asking about a planer glass which needs to be configured with different settings.
Answer: The solution is to set a (non-zero) refraction depth for the glass material and add a reflection plane just in front of the window pointing inward toward the room. The reflection plane will override the reflection of the background environment texture locally. Using this method, other objects in the scene will still be able to reflect the environment.
Notes: My glass material has screen space refraction enabled with a refraction depth set at 0.01 m. In the render properties I have also lowered the edge fading amount to 0.03 and enabled overscan at 10%.
(You may need to make further adjustments to these render property values if the refraction is still fading out toward the edge of the camera frame.)
Answer: To avoid the HDRI reflection you can create this node setup for the world: Mix the HDRI Background node with whatever Background you want into a Mix Shader, and use the Input > Light Path > Is Glossy Ray as the factor. The HDRI will still be the light source but the second Background will be the one that will be taken account for reflections:
Comment: you can put whatever you want in the other input of the Mix Shader, in my screenshot it's a grey color background but it can be another HDRI, or your current HDRI but with some tweakings, etc
Comment: Thanks, I tried this way, it looks so good. but can make this reflection translucent. it looks plastic right now.


Are passengers traveling together allowed to share the metromoney card in Tbilisi?

Question: In many cities you can share your metrocard (multiple persons validate the same card when getting on the metro or bus). Is this allowed for the metromoney card card?
Answer: No, you're not allowed to do this, given that the card always requires a minute between validations, and there are guards at the barriers to see to it that there is no abuse.
However, if you save the original receipt, you can get the 2 lari back when leaving Tbilisi (if less than 1 month after buying the card), by presenting your card, receipt and passport at a ticket counter
Comment: I don't think that is true, I am Georgian and from the day they introduced Metro Cards, you could use as many times as you want in a row. Now, I don't live in Georgia for 4 years anymore, but I was there in summer 2016 and it was still the same, remember very well I used for me and my friend, two validations in two seconds. Also, never heard any rule that says you are now allowed to do that.
Comment: @Leonardo I saw a foreign couple get shouted at at Marjanishvili station for trying to share a card, and they were forced to get another one. This was in November 2016. Also, once when I validated, I had to run back to fetch something quickly, and I couldn't enter the barrier for another 45 seconds
Answer: Seeing that the card itself costs a mere 2 lari - which is less than $1 - I don't see why you would need to share one. Simply get two (or more) cards for each person and top them up individually. A single ride is around $0.20 so it won't break your budget.
Comment: The problem with multi persons is not the 2 Lari. When you are in a trip with several persons, it takes more time to Top Up each time several cards, and most chances you will stay with more unused balance.
Comment: These cards are fun. For example, in Moscow you flat out may share a card, in St. Petersburg flat out can't, and in Kharkov you can't use it on the same turnstile but may use on the adjacent. In all cases the price is (was?) flat so neither ends up in cheating.


Altium Designer Rule: Between Silkscreen and Via

Question: I have not find a solution about between top overlay or bottom overlay and vias. Can you help me about this . I need identify a rule.I don't want the silkscreen to touch a via
Answer: For this, if you do not wish to tent your vias, it is safe to assume that there will be a solder mask layer present on the vias and any other pad for that matter which I also assume you will not want silkscreen to cover.
If this is the case, the answer is pretty simple, simply set a rule in the "SilkToSolderMaskClearance" option. See image below.

This is found under the manufacturing section of the rule list. Then you can select to create a new rule, and you can set a custom query under the "Where The First Object Matches" to "IsVia" - you can leave this as "All" if you want the rule to apply to all exposed copper regions, and then "Where The Second Object Matches" -> Layer -> Top Overlay.
You can then select whether the clearance is from the exposed copper, or the Solder Mask opening and set the distance that you require. If it is for a straight overlap, set this to 0 and then it will only detect if it overlapping, not if it is close to.

This will save you having to tent the vias if it is not something you want to do - I think some fab houses can get a bit funny about doing it as well. Or at least, in my experience.
Comment: This would be a good work-around -- using the soldermask aperture as your object rather than the via itself. Good thought
Comment: Unfortunately some board houses have relatively large minimum holes sizes for low volume prototyping tiers. I am stuck with 13.5mil vias which do not always tent well and the silk gets messy.
Answer: The following worked for me on Altium 17:

It did not work when the minimum clearance was 0 mil...likely a bug.
Answer: If you don't want the silkscreen to touch a via, you can use the following:
Where first object matches:
OnLayer('TOP OVERLAY')
Where second object matches:
IsVIa
Then set your clearance rule.
EDIT:
I just tried this at work and to my surprise it didn't work for me either. I also tried queries such as IsDesignator, and even played around with the Silk To Soldermask rules and had no success. This seems like it may be an Altium bug. I suggest you call their customer support and see if they can give you an answer that works. Otherwise I recommend doing what I suggested in the comments, and tent your vias with soldermask. Then it won't matter if the soldermask passes over the vias or not, they should still be visible. Most manufacturers can tent vias with up to a 12 mil (0.3mm) hole diameter.
Comment: The other thing is if you tent your vias, it won't matter if there's silkscreen over the via or not (assuming a reasonably sized via). I always tent my vias and never worry about silkscreen over them
Comment: Thank you Where will I write in this rule? Design Rules -> Manufacturing -> SilktoSilk Clearance ?
Comment: No, this will be a standard clearance rule: Design Rules -> Electrical -> Clearance
Comment: http://picresize.com/popup.html?images/rsz_20170330_091528.jpg  it is not working with this settings
Comment: That link isn't working for me. Could you try https://postimage.io/ ?


How can I check whether a point is inside of a plain equilateral shape? (given the center point of the shape and the length of the sides)

Question: I've got a programming task at hand, and I should give back a true or false, based on if a point (x,y) is inside of the equilateral shape (circle, triangle, square, hexagon). I've finished with the circle, but I cannot figure out how to calculate the other 3 properly. In the case of circle we've got the center point and the radius (instead of length of the side), but with the other 3 we get the center point and the length of the sides (which is all the same of course given they're equilateral). Lastly, we know that every shape has a side that is parallel to the X axis of the coordinate-system (except the cirlce of course).
Any ideas?
Comment: And for the square we get the following boolean: ((x is in between $c_x-\frac{l}{2}$ and $c_x+\frac{l}{2}$) and (y is in between $c_y-\frac{l}{2}$ and $c_y+\frac{l}{2}$)). Where $(c_x,c_y)$ is the center of the square and $l$ is the given side length
Comment: Thank you guys, but I don't need one single solution for all of them. I can use different methods for these, but as an update on the matter, I may have found solutions for all of them except the hexagon. Do you have any idea for that one?
Comment: What have you tried? I wonder if you can do something with the signed areas of the triangles formed by the point and pairs of consecutive vertices of the polygon
Comment: I have got a feeling that something can be done with the incircle and circumcircle of the given regular polygon. Just a suggestion…
Comment: By equilateral shape, do you actually mean regular polygon? And you know not just the centre and the length of the side but also how many sides it has? If so, you can rotate the point around the centre by any multiple of 360/n degrees (n is the number of sides) and this does not change whether it is insode or outside the polygon. After the right rotation, you can compare the y coordinate of the point with the y coordinate of the horizontal side to see if the point lies inside or outside.
Comment: For the polygons, the more general [Point in polygon](https://en.wikipedia.org/wiki/Point_in_polygon) test may be relevant.
Answer: Edit: Corrected condition to include checking $n^{\text{th}}$ side and angles
There's probably a simpler method for just the few simple shapes mentioned, but I decided to write a method for a general $n$-sided regular polygon. Let $C \equiv (x_c, y_c)$ be the centre point, $L$ the length of the sides, and $n$ the number of sides. In summary, a point $(x,y)$ lies within the shape if,
$$ f_k(x,y) f_k(x_c,y_c) \geq 0 \quad \forall \; k = 1,2, \dots, n$$
where,
$$
f_k(x,y) \equiv y - y_k - (x - x_k) \frac{y_{k+1} - y_k}{x_{k+1} - x_k}, \quad (x_{n+1}, y_{n+1}) \equiv (x_{1}, y_{1}) \\
\vec{V}_k = \vec{V}_{k-1} + \begin{bmatrix} \cos\left(\frac{2\pi}{n}\right) & - \sin\left(\frac{2\pi}{n}\right) \\ \sin\left(\frac{2\pi}{n}\right) & \cos\left(\frac{2\pi}{n}\right) \end{bmatrix} (\vec{V}_{k-1} - \vec{V}_{k-2}), \quad \vec{V}_k \equiv \begin{bmatrix} x_k \\ y_k \end{bmatrix} \\
V_1 \equiv \begin{bmatrix} x_c - \frac{L}{2} \\ y_c - \frac{L}{2} \cot \left(\frac{\pi}{n}\right) \end{bmatrix}, \quad V_2 \equiv \begin{bmatrix} x_c + \frac{L}{2} \\ y_c - \frac{L}{2} \cot \left(\frac{\pi}{n}\right) \end{bmatrix}
$$
Derivation
Consider one side, and draw lines from it's vertices to the centre point to form an isosceles triangle. Since congruent triangles can be formed in the same way with the other sides, the angle of the triangles at the centre are all equal and add up to $2 \pi$. So each angle is $\frac{2\pi}{n}$.
Now draw a line from the centre and perpendicular to the side parallel to the x-axis. Because of the isosceles triangle, this bisects the side (the two smaller and right-angled triangles are congruent) into $\frac{L}{2}$ and $\frac{L}{2}$. These $\frac{L}{2}$ segments are the opposite side for the two right-angled triangles to the centre angle which is also bisected into $\frac{\pi}{n}$ and $\frac{\pi}{n}$. So the length of the perpendicular is $\frac{L}{2} \cot \left(\frac{\pi}{n}\right)$
Because the side is parallel to the x-axis, the perpendicular is in the y-direction. That means the midpoint $M$ of this side is shifted from the centre in the y-direction only by the length of the perpendicular. Assuming the side is below the centre point,
$$M \equiv \left(x_c, y_c - \frac{L}{2} \cot \left(\frac{\pi}{n}\right)\right)$$
For even $n$, there will be a parallel side above and below. For odd $n$ (like triangle), if the side is actually above, then just change the $-$ to $+$. Now from the midpoint, we can get the two vertices of this side as,
$$V_1 \equiv \left(x_c - \frac{L}{2}, y_c - \frac{L}{2} \cot \left(\frac{\pi}{n}\right)\right), \quad V_2 \equiv \left(x_c + \frac{L}{2}, y_c - \frac{L}{2} \cot \left(\frac{\pi}{n}\right)\right)$$
Now the interior angle of an $n$-regular polygon is $\frac{(n - 2)\pi}{n}$, and the exterior angle is $\frac{2\pi}{n}$. Each side is a continuation of the previous side, but rotated by the exterior angle. So in vector notation and using the rotation matrix, one can calculate each vertex using the previous two vertices,
$$\vec{V}_k = \vec{V}_{k-1} + \begin{bmatrix} \cos\left(\frac{2\pi}{n}\right) & - \sin\left(\frac{2\pi}{n}\right) \\ \sin\left(\frac{2\pi}{n}\right) & \cos\left(\frac{2\pi}{n}\right) \end{bmatrix} (\vec{V}_{k-1} - \vec{V}_{k-2})$$
Now with the coordinates of all the vertices, we can construct equations of lines for each side. Since the $n^{\text{th}}$ side is formed by $V_n$ and $V_1$, define $V_{n+1} \equiv V_1$ for algorithmic simplicity.
$$\overline{V_k V_{k+1}} := f_k(x,y) = 0, \quad f_k(x,y) \equiv y - y_k - (x - x_k) \frac{y_{k+1} - y_k}{x_{k+1} - x_k}$$
So we know if a point lies on the line, it must make the above equation 0. But what if we plug in a point that doesn't lie on the line, that lies within one of the two half-spaces that the line divides space into? Then we get a non-zero value. Specifically, we get positive values for points on one half-space of the line and negative values for points on the other half-space of the line. Now we could figure out which half-space gives which sign, but that's not important.
Next, let's define what we mean by 'a point lying inside the shape'. Well, one way is if, for each side, the given point lies on the same half-space as the centre point. Meaning that when we plug in the point into all the side equations, it should have the same sign as when we plug in the centre point. So putting the condition together mathematically,
$$f_k(x,y) f_k(x_c,y_c) \geq 0 \quad \forall \; k = 1,2, \dots, n $$
If you want strictly inside (excluding points on the side), then it becomes,
$$f_k(x,y) f_k(x_c,y_c) > 0 \quad \forall \; k = 1,2, \dots, n $$
As an aside, this idea of lying on the same half-space of every side can be applied to any convex polygon. And if you didn't have the centre point, you could just use another vertex for checking the half-space.
Comment: The last paragraph suggests a way to describe your solution that might be easier to understand, perhaps to program. Find the $n$ equations of the lines containing the sides of the figure. Arrange the equations so that the center point is always in the positive half space. Then check that the given point is always in the positive half space.
Comment: Thank you but it's way more complicated for my brain though it might be good.
Answer: After some time, I came up with this simpler method (compared to my previous one) by changing the perspective of when a point is inside, by looking at maximal distance from centre. Let $C \equiv (x_c, y_c)$ be the centre point, $L$ the length of the sides, and $n$ the number of sides. In summary, a point $(x,y)$ lies within the shape if,
$$\sqrt{(x-x_c)^2 + (y-y_c)^2} \leq R_{max}(\theta)$$
where
$$
R_{max}(\theta) = \frac{L}{2} \cot \left(\frac{\pi}{n}\right) \sec \left( \min\left(\theta^*, \frac{2\pi}{n} - \theta^*\right)\right), \qquad \theta^* \equiv \theta \mod \frac{2\pi}{n} \\
\theta \equiv \tan^{-1} \left(\frac{x - x_c}{y-y_c}\right) + \pi (x < x_c \lor (x = x_c \land y > y_c))
$$
And if we let $L \to 0$ while $n \to \infty$ (in a certain manner), we'll get the natural check for a circle.
Derivation
A point can be no further than the circumradius from the circumcentre, as by definition the entire shape is contained within the circumcircle. But this is not a sufficient condition, as depending on the angle, the maximal distance a point can be away may be less.
For regular polygons, the circumcentre and incentre (for incircle) are the same, and is the centre referred to in the question. And so the maximal distance a point can be away varies between the in-radius and the circum-radius. The in-radius is the length of the perpendicular from the centre to any side bisecting it, and the circumradius is the length of the line from centre to any vertex, as shown below where $L$ is the length of the $n$ equal sides. See my previous answer for context on reaching these equations.
$$R_I = \frac{L}{2} \cot \left(\frac{\pi}{n}\right), \quad R_C = \frac{L}{2} \csc \left(\frac{\pi}{n}\right)$$
Now consider an arbitrary point $P(x,y)$, the line joining $P$ to the centre $C(x_c,y_c)$, and the angle $\theta$ this line makes with the y-axis, calculated as,
$$\theta \equiv \tan^{-1} \left(\frac{x - x_c}{y-y_c}\right) + \pi (x < x_c \lor (x = x_c \land y > y_c))$$
The logical expression is included to add $\pi$ if the point is on the left-side or exactly above of the centre, so that exactly below corresponds to $\theta = 0$. This is required as for odd $n$, there is only one side parallel to the x-axis, and so we are using it as reference (assuming it lies below; the expression can be adjusted if it lies above instead).
When $\theta = 0$, $P$ must lie vertically below the centre. By geometry, the furthest point on the shape it can be is the midpoint of the side parallel to the x-axis, which is at in-radius distance $R_I$ from the centre. If we increase $\theta$, then this maximal distance will increase upto the circumradius $R_C$ at $\theta = \frac{\pi}{n}$. Then it decreases back to $R_I$ at $\theta = \frac{2\pi}{n}$, and this cycle repeats. So using geometry and the modulus operation, we can calculate the maximal allowed distance inside the shape as,
$$R_{max}(\theta) =
\begin{cases}
   R_I \sec \theta^*, \; &0 \leq \theta^* \leq \frac{\pi}{n} \\
   R_I \sec \left(\frac{2\pi}{n} - \theta^*\right), \; &\frac{\pi}{n} \leq \theta^* \leq \frac{2\pi}{n} \\
\end{cases}, \qquad \theta^* \equiv \theta \mod \frac{2\pi}{n}$$
Since the value of each case's expression is greater than the other in the other's range, we can combine the two cases using a min function,
$$R_{max}(\theta) = R_I \sec \left( \min\left(\theta^*, \frac{2\pi}{n} - \theta^*\right)\right) $$
And so a point lies within the shape if its distance from the centre is less than the maximal distance for its angle,
$$||\overline{CP}|| \leq R_{max}(\theta)$$
Comment: Thank you, what is R though? I mentioned above that I solved like almost all of them except the hexagon, and I don't need necessarily one single solution for all of them at once, it's just as good if I solve them with different methods.
Comment: Each $R$ is different. $R_C$ is the circumradius, $R_I$ the inradius, $R_{max}(\theta)$ a function that tells the furthest distance you can go from the centre along a particular direction (indicated by $\theta$) while remaining within the shape. Are any of these $R$'s unclear?
Comment: It's clear now thank you!


Relationship Problems

Question: In my plot, there are these 2 characters who were originally close to each other. However, they had an argument and broke up. In the end of the book, though, they get back together again. What I need help is for smooth transitions in their relationship. Every time I try to write it out, the flow of the story gets broken and confusing.
Comment: Is this for a short story, novel, or screenplay?  Each of those may have discrete answers.  Also, the age & sexes of the characters may matter.  What else is happening in the plot?  Is this romance a big part of it, or are these coworkers who are trying to keep a bagel stand afloat despite an alien attack?
Comment: The romance is a supporting line, and the characters are both in their teens
Comment: Edit the main question to include that information!
Comment: What @April said.  Also could you please add tags to indicate the type of work?  [tag:young-adult] if the work is *for* teens.  Then [tag:novel] or whatever type it is. Thanks.
Answer: On-again/off-again relationships don't usually have smooth transitions.  Especially not for outside observers (and often not for the participants).  Add in that this is a teen relationship, and forget about smooth.
Since the relationship is not central to the plot, your narrative can just describe status changes matter of fact and not linger on them.  I mean, show the emotions and the problems they have working together while broken up and anything else that you wish to show.  But don't explain too much.
Readers understand that relationships don't always have soft and gentle changes.  More like a stick thrown into the wheels.  People argue, they're upset for a while, they may or may not be able to be around each other, and then, sometimes all of a sudden, everything's cool again.
Some couples talk it out.  Some kiss it out.  Others slowly hang out more and resume doing relationship things, so there's no clear point where the status changed back.  Others just say "we cool?" and then they are.
Breakups are even more volatile.  They can happen in under a minute.  Or the anger and resentment can build up over time and then the break up happens after a long talk, or with yet another quick argument.  Or people can drift apart.
Just give enough tidbits for your reader to figure it out and don't dwell on it.
If the work were about the relationship that would be different.  But in this case, the relationship is background information and describing it has been slowing down your narrative.
Answer: If they are teens, I'd suggest you make the breakup a clear fault in one of the characters; preferably the more main character. By "clear fault" I mean one that most readers will agree with; a wrong-headed assumption of youth about "proper roles" or something like that; and something that with time the flawed character can overcome.
Hence making the flawed character the more main character; so you can show that personal growth. You don't need to show as much for the character that was in the right.
Then, follow the "try, try and try again" formula. As an example, I create Brittney as the main (flawed) character, and Alice (in the right); they are in love. They get into an argument over some principle Alice holds dear (religion, politics, reproductive rights, any hot-button issue you feel you can write about), and Brittney gets angry, Alice gets angry back, and they break up.
Alice (in the right) tries to get back together with Brittney because she misses her, but Brittney gloats that Alice came back to her and threw her principles aside. Alice doesn't want to be in that relationship, so End of Try 1.
Then Brittney, eventually sad at the loss, realizes she was wrong to gloat, and tries to get back together with Alice, but Britt won't admit she was wrong on the original point. Alice feels like this is all going to just repeat, so ... End of Try 2.
After enough time, Alice decides she doesn't care. She still loves Brittney, and she thinks she will just put her principles aside to be with the girl she loves, and will avoid the disagreement that broke them up in the first place. She comes to beg Brittney if they can reach détente for the sake of love, only to find out Brittney has changed and admits she was wrong, that Alice was always in her head, and she had seen first hand how wrong she was, but thought it was too late -- but if it isn't too late-- They kiss and make up.
You can, of course, make up your own reasons to break up, fail to get back together once, then twice, and then succeed on the third try. This is just the basic arc you can use to keep the relationship alive in the reader's mind, over the course of a long breakup. And in this time, the girls can have other romantic relationships that, in the reader's mind, threaten the Alice-Brittney prospects. Or make Alice or Brittney jealous and feel like they've lost their true love to somebody else. So the reader is relieved when these flings come to an end; it revives the Alice-Brittney true love possibilities.
This also doesn't have to be the main plot of your story; it can be sub-plot.
Answer: I suggest breaking the story into smaller scenes which you approach individually one at a time. Make each scene a connection between the characters that grows slowly over time, each scene bringing them that little bit closer.
Perhaps, at first, they are invited to the same party but refuse to speak to each other. Then something happens, an external event, that forces them to say just a few words.
Later, when they see each other again, the ice is broken. They are able to say a little more to each other, perhaps another event forces them to talk about what happened at the party.
On and on you go, each encounter, each external force, motivating them to come a little closer. That way, you keep the flow and it's not jarring, but believable.

One of my favourite examples of two characters coming together in
  spite of the odds against them is Eleanor Oliphant is Completely
  Fine. It's a fantastic example of character development and
  motivation.

Good luck!
Comment: @Jaystar, characters are #1.
Comment: Wouldn't this result in many events happening and drift from the plot? I mean, the characters are like 2nd priority, aren't they?
Comment: The secret is to be clever with your scenes: make them do more than one task at a same time. There should be no need to sacrifice plot to develop your characters and vice versa. Without knowing your story, I can't give more specific advice, above are just examples. If the argument is silly and meaningless, you could just have them apologise and move on...
Comment: ... But, I suspect, since you've asked the question, it's fairly crucial to your story. In which case, keep your plot moving forward but bring the characters together within it, addressing both at the same time. IMHO, characters are never 2nd priority to anything.
Comment: @Jaystar, in this group, we typically advise users to wait 24 hours before selecting an answer - often there are elements of many that are useful.  It's not like some Stacks where if I ask "what was this button" and someone accurately answers it in 10 minutes, there's no point in anyone else answering.

Relationship Problems


Is there a recommended method of maintaining multiple Photos libraries on same hard drive

Question: My main Master Photos Library is over 400GB, and getting too big to tote around on trips.  I keep it on removable HD.  So I started creating new library for each trip.  Problem now is managing them all, and creating a master.  I use Power Photos (from Fat Cat) to move photos between libraries, but they don't support Live Photos.  A given trip might be as many as 4000 photos.  Some from SLR some from iPhone (Live Photo) and I combine them in the library.  Any suggestions on managing these and combining the libraries?
Comment: Photos is happy to have multiple folders. You'll want a third party tool to merge / move files back and forth. What does managing look like to you in specific terms?
Comment: Already use 3rd party Power Photos to move photos back and forth.  Doesn't move Live Photos, so end up with just jpeg.  Looking for something that can do both.  'Managing' my photos would enable me to create separate library for a trip, do edits, then easily merge into the master library when I get home.  The suggestion below to use Lightroom won't help either; no support for Live Photos.  Apple created a new format standard but didn't really provide for the ability to manipulate or combine very easily.  Have to go back and forth to the iPhone to move the Live Photos.
Answer: Honestly Photos.app is not really designed as a pro or semi-pro photo managing application. It certainly seems to handle larger libraries better than iPhoto ever did but if you are taking that many pictures and want to manage them on your Mac you might want to consider an app designed specifically for the workflow and archiving of thousands of pictures.
Adobe has an app for that (Lightroom, I think it is called), and like many Adobe apps is the subject of much praise and anger online. Me, I just use photos.app because my needs are met with photos.app, but it sounds like you might be better served with something more capable than a consumer level photo managing app. IMHO


Color Scheme on Meta

Question: The second question should have a different color since I favorited 'tags'. While the RGB values I picked from the picture try to convince me of a difference between the background of the second question I can honestly not see it (I am a bit red-green color blind). Can we get something that looks actually a bit different for the 5% with a color deficiency?
Comment: Here's a feature I didn't know about. Looking at the code, the background of the tagged question is a light gray #fafafb. The background of other list items is transparent -- i.e., you can see the page's background pattern. Because the background pattern is also a light gray, they are virtually indistinguishable.
Comment: You might be interested in this: http://meta.stackexchange.com/questions/146005/can-the-meta-sites-have-non-color-based-visual-cues-to-help-distinguish-them-fro
Answer: The background color of page elements marked tagged-interesting should be changed from the color #fafafb. That color is basically indistinguishable from the default background. It should be darkened. The color #eff0f1, which is already in the scheme for code blocks and blockquotes, would do fine.
I have flagged this as a bug. The site design clearly intends questions marked tagged-interesting to have a contrasting color. This is a problem in implementation, i.e., a bug. Probably easily squashed.
Answer: I can't tell the difference in the original screenshot and I have regular color vision. However, things have changed since then and I can now see the highlight for followed tags. Looking at color blindness simulators, I can even see the difference in monochrome, though it's subtle. There's also now an eyeball on any tags that you're following for an additional visual cue.
Regular:

Greyscale:

Color Scheme on Meta


"How much are you offering "? "What is your offer"?

Question: If I am selling a house, Could you pls tell me that who could say these qestions? Seller or buyer or both(as in anyone)
Comment: Sellers offer their house for sale and set an *asking price*. Interested buyers say they have a different number in mind, usually lower. So, sellers ask them either "How much are you offering?" or "What is your offer?" Same question as for that rug in the bazaar.
Comment: But 'special offers' are normally on goods in shops, not on individual deals like house sales.
Comment: offers are always from sellers, not buyers. It's quite simple really.
Comment: @Yosef Baskin. "Do not you think, buyer can also say, 'What is your offer on TV'? ?" ( As buyer is looking for any discount or buy one get one free offer)
Answer: What is your offer?
This sentence can be said to anyone who is offering a deal or contract. In terms of buying/selling a house, this would almost always be the buyer, not the seller. (So the seller would say this to the buyer.) That's because culturally, in most English speaking countries, the buyer is the one who makes the offer on a house.
In other sale situations, this could be said by either the buyer or the seller -  to whoever is proposing a price or deal.
How much are you offering?
This is similar to the other sentence but different because the word "much" means amount, so the answer has to be an amount of money or something else that can have an amount.
In a house-buying situation, this would be almost always said by the seller to the buyer for the same reasons as above.
In other situations, it is less likely that a buyer would say this to a seller because the context would not usually apply.
Other Situations
In situations other than selling/buying a house, these sentences might be said when something is being sold or bought or when a different deal or trade is being offered. Here are some examples of that:
I am in a bazaar or market. I walk up to a shop selling shirts. The conversation might go like this:

Shopkeeper: This shirt would look great on you.
Me: My closet is full! I don't need more shirts. But how much are you asking for it?
Shopkeeper: It's $40, but it would match your eyes.
Me: I do kind of like it, but that's far too much.
Shopkeeper: Well then, how much are you offering?
Me: I could give you $25.

I'm walking home and stop at a corner market to pick up food for dinner. When I walk in, the shopkeeper greets me and wants to tell me about a special deal. The conversation might go like this:

Shopkeeper: Hello, good afternoon! We have a special deal on melons today. Are you interested?
Me: I do love melon! What is the offer?
Shopkeeper: 2 melons for $5.

I am looking for a new job and recently completed interviews for a job that I'm interested in. I get a phone call from the hiring manager.

Manager: Good morning, I'm calling to tell you that we'd love to offer you the job!
Me: That's great news! Thank you very much. Could you tell me more about your offer?
Manager: Based on your experience, we are able to offer you $70,000 / year and a full benefits package.
Comment: @BilalZafar Absolutely! I added a few more examples to the end. Could you take a look and see if that answered your question?
Comment: @ Friendly Racoon thank you for an indepth analysis with examples. Yes your examples made me understand. I appreciate.
Comment: Could you pls also explain your this statement? In other sale situations, this could be said by either the buyer or the seller - to whoever is proposing a price or deal.


Inexpensive way to determine mass of tiny seeds?

Question: My niece is doing a science project and she wants to weigh various seeds, determine their volume and density. She tried using my kitchen scale, but it is not sensitive enough.
I've suggested weighing 100 seeds then dividing to get an average weight, but she (she's 11) wants to weigh each one. I don't want to get in the way too much. I don't mind buying a scale, or other simple equipment as long as it isn't expensive.  Is there any way to do this inexpensively?
Answer: A pharmacy is a public business, and they should have a scale sensitive enough to weigh seeds.  They will have to let you in back to use it of course.  If you show up with her and she has the seeds and she asks the manager nicely, they might let you back and help you use it.  I would show up rather than call; if you show up you will see the manager / pharmacist but if you call you will not talk to the manager.
If this works she needs to be sure to credit the pharmacy in her project!
I recommend she weigh several seeds of each type - they might vary.
Your 100 seed idea is a better one, I think.  I understand your sentiment.

You did not ask about volume but this could be done with a graduated cylinder.  Put some water in it up to, say, 5 ml.  Then drop in a seed.  How much farther does the water level rise?  That is the volume of the seed.


Access to XMLHttpRequest has been bloked by CORS policy

Question: Access to XMLHttpRequest at 'http://localhost:8080/ws' from origin 'http://localhost:3001' has been blocked by CORS policy: Response to preflight request doesn't pass access control check: No 'Access-Control-Allow-Origin' header is present on the requested resource.
Comment: are you using express js?
Comment: I got this error last week. I solved the problem by adding the following phrase to the package.json.  "proxy": "YourAPIUrl". Please try it and return back.
Comment: Does this answer your question? [XMLHttpRequest cannot load XXX No 'Access-Control-Allow-Origin' header](https://stackoverflow.com/questions/35553500/xmlhttprequest-cannot-load-xxx-no-access-control-allow-origin-header)
Answer: When invoking an XMLHttpRequest, the browser makes a preflight request and checks for an Access-Control-Allow-Origin header to determine whether the request should be allowed. You should edit your server code to send that header with a value that allows the domain of your client (or just * to allow CORS requests from any origin). For example, if using a Node server with Express, you could do res.set('Access-Control-Allow-Origin', '*')


prove that this expression equals the area of a specific triangle

Question: Prove that the area of a triangle with one of its angles $\frac{\pi}4$ rad, and the side opposite to this angle is $2$cm equals $\sin(2\theta)-\cos(2\theta)+1$, where $\theta$ is the angle adjacent to the $2$cm side.
and here is a picture to make things cleaner:
Comment: Use the sine rule to find another side.
Answer: Let $a = 2, A = \frac {\pi} 4, B = \theta$
Then,
$$
\frac{b}{\sin \theta} = \frac{2}{\frac{\sqrt{2}}2}
\implies b = 2\sqrt 2 \sin \theta
$$
Then,
$$
\sin C = \sin(A+B) = \frac {\sqrt{2}}2 (\sin \theta + \cos \theta)
$$
Then,
$$
S = \frac 12 ab\sin C =\frac 12 \times 2\times 2\sqrt 2\sin \theta\times \frac {\sqrt{2}}2 (\sin \theta + \cos \theta)\\
= 2\sin \theta (\sin \theta + \cos \theta)=2\sin^2\theta+2\sin \theta \cos \theta = 1-\cos 2\theta + \sin 2\theta
$$
Comment: I really appreciate for your answer!
Comment: @AmrWaleed You didn't mark the points with letters so I marked them. A is the known angle, a is the known side and B is $\theta$, mark these on the picture you provided and  you are set.
Comment: I asked about the step of law of sines, but i just recognized what you did there, thanks anyway


Lowering pitch of a piezoelectric buzzer

Question: I have tried to drastically lower the pitch of a piezoelectric buzzer, but to no avail. All the piezo buzzers I have emit a very high pitched whistle-like tone. I've tried changing all the parameters of my pulse generator, i.e., frequency, duty width, spacing, amplitude, etc. but the pitch of the tone remains the same, very high! Is it even possible to make a base tone from a piezo, or is the high pitch an intrinsic characteristic of all piezo units?
Comment: https://electronics.stackexchange.com/ would be a good place to ask.  Some buzzers are fixed pitch (you just apply power) and can't be changed.  Some should be controllable.  Recommend you ask there and give the specific information about the unit you have.
Answer: a plain, unadorned piezo disc transducer with no other circuitry affixed to it can be driven across a broad range of frequencies, which will make it vibrate. However, for it to vibrate strongly enough to easily make a loud noise, it needs to be driven at a frequency that coincides with its fundamental frequency of mechanical vibration.
Most piezo "beeper modules" contain a simple one-transistor oscillator that applies a sine or square wave to the disc at its fundamental resonant frequency. If you apply power to one of these, all you get out of it is that characteristic BEEEEEEEEEEEEP no matter what you do to it.
Answer: You can epoxy a weight to the center of the buzzer disk to lower the resonant frequency. A steel nut could be used.


E-bass tuned correctly but E string too loose

Question: I'm self-taught and new to bass, and I tuned my bass today (E-A-D-G, E being the lowest), but the E String seems a bit loose (It hits the neck sometimes, and I can bend it much further than the other strings). Is that normal? Have I done something wrong?
Comment: A clear photo of the fingerboard near the nut would be helpful to see if the strings are at least in the right positions and approximately the right gauges. A short audio or video of you playing the open strings would also be helpful. It could be as simple as a truss rod adjustment but more info is needed.
Answer: You probably did nothing wrong, and normality is more a matter of taste.
If you feel that this string is too loose for you to be happy with maybe you should consider getting an higher gauge string. It will require more tension in order to achieve the same pitch.
Answer: Has it been like that all the time you've had it? How did you tune it - with a tuner, by ear, against another instrument? What bass is it, and what gauge strings are on now? All points which are needed to give any helpful answers.
Going on the sparse information rendered, all I can suggest is you tuned the bottom string an octave too low. It wouldn't be the first time something like this has happened - the tuner says 'E', but that E could be an octave too high or an octave too low.


Query Bulk Data Load Jobs once completed

Question: Can someone tell me how to query the data which is displayed in " Monitor Bulk Data Load Jobs" Section "Completed last 7 days"
Answer: It seems the functionality is not exposed by Salesforce at all.
Those objects are not avialable in APEX. As I checked all Apex supported objects(with globalDescribe)  and none of them have "750" as object prefix.
Also as far as I know the bulk API endpoints also do not support listing of the jobs. You only can access the jobs by providing JobId.
Bulk API Walkthrough
Answer: I don't have any orgs with bulk jobs to look at off hand to figure out exactly which fields are the ones displayed, but this should help you get started:
SELECT CompletedDate, Id, JobItemsProcessed, Status FROM AsyncApexJob
WHERE CompletedDate = LAST_N_DAYS:7

Comment: AsyncApexJobs doe not include bulk data load jobs
Answer: Send an HTTP callout to the Bulk API v2.0, REST and JSON enabled.  Not that bad.
final String VERSION_BULKJOB_WAS_MADE_WITH = 'v53.0';
final Id ASYNCAPIJOB_ID = '750000123456789';
HttpRequest req = new HttpRequest();
String endPointUrl = URL.getOrgDomainUrl().toExternalForm() + '/services/data/' + VERSION_BULKJOB_WAS_MADE_WITH + '/jobs/ingest/' + ASYNCAPIJOB_ID;
req.setEndpoint(endPointUrl);
req.setMethod('GET');
req.setHeader('Authorization', 'Bearer ' + UserInfo.getSessionId());
Http http = new Http();
HTTPResponse response = http.send(req);
String respBody = response.getBody();
System.debug('respBody: ' + respBody);

Sample response:
{
    "id": "750000123456789",
    "operation": "insert",
    "object": "MyObject__c",
    "createdById": "005000000012345AAA",
    "createdDate": "2022-02-19T03:00:36.000+0000",
    "systemModstamp": "2022-02-19T03:01:46.000+0000",
    "state": "JobComplete",
    "concurrencyMode": "Parallel",
    "contentType": "CSV",
    "apiVersion": 53,
    "jobType": "V2Ingest",
    "lineEnding": "LF",
    "columnDelimiter": "PIPE",
    "numberRecordsProcessed": 10293,
    "numberRecordsFailed": 0,
    "retries": 0,
    "totalProcessingTime": 66100,
    "apiActiveProcessingTime": 59388,
    "apexProcessingTime": 0
}

Hope that helps!


Are attachments a built-in feature in SharePoint 2013?

Question: I've seen some very mixed results while searching on the internet of whether attaching documents to a blog post is a built-in feature for SharePoint 2013 and I'm not sure what to believe.  Most people say it's not built-in but that might be for previous versions of SharePoint.  Somewhere I read there should be an option in List Settings -> Advanced Settings to enable attachments on the Posts list but I do not see this and can't find any sort of built-in feature.
Answer: Attachments are technically available on any SharePoint list.  Depending on the definition of the list, they can sometimes be obscured.
With the blog posts list, they are enabled (just not visible from the settings).  If you browse to the list item view of the Posts list and select the checkmark for a post, the Attach File option is available on the ribbon under Items.
Comment: Thanks, this does work but my users don't have access to that.  Is there any way to attach a document on creation of the post?
Comment: So, there isn't really a simple way to do this.  And the attachments wouldn't display when viewing the post anyway.  I would suggest going a different route:  with the cursor in the post body while creating it, select Upload File from the Insert ribbon tab.  This will let you upload a file to a document library and insert a link into the post to it.
Comment: And you could have a library named "Post Attachments" available to upload the documents to.
Answer: I agree, on 2013 i am not seeing the option through the UI.  If you go through the API, you might can change the EnableAttachments property to true and see what happens.  Most likely, there is a good reason why they have it turned off.
But with the new 2013 features of being able to easily insert images and upload files into the html of the body, i don't really see why you need attachments.
Answer: I agree with both John and Steve. Attachments can be done this way BUT there is a caveat to this approach. If you decide to remove the file link from the post after uploading the file, the file will not be automatically deleted from the library where it was uploaded. You will have make sure you manually delete that specific file otherwise it will stay there for the rest of its life as an "orphaned" item!


How to erase all data on android tablet?

Question: My cousin locked my tablet trying to figure out the passcode and locked me out! Im trying to erase all data. please help.
Comment: If you want to re-gain access to your tablet, please see the [locked-out tag-wiki](http://android.stackexchange.com/tags/locked-out/info). If you really just want to reset the device to factory settings, you might need to boot into recovery-mode and perform a factory-reset. Usually that's done by powering up the device while holding the vol+ or vol- keys; the combination often is device specific.
Answer: There are a couple of things you could do.

Boot into recovery mode, and perform a factory reset that way. Google something like "recovery mode " and the volume buttons act as the arrow keys and the power/home button as the back/select button.
Log into your Google account on your computer (if you aren't already logged in) ensuring that it is the same account as the one on your tablet. And go to https://www.google.com/android/devicemanager
From there, if your device is supported, you should be able to wipe your device, or even better, change the device's password so that you can easily unlock it again.


iPad game transfer

Question: I have an iPad and my son plays his games on it.
We are getting him his own iPad.
How can I transfer his games over to his iPad with out giving him the same Apple ID as mine?
Answer: You can create an Apple ID for him when setting up his iPad.
For him to be able to use the same apps that you have paid for, and to -mostly- avoid buying things twice, you can set his Apple ID as part of your family on Family Sharing.
The progress of each game will be more difficult or impossible to transfer, depending on how and where it is saved.
The information on the feasibility of that transfer it's probably on each developer support site.
You can also try your luck asking for each separately here, but for some reason most games questions are shunned.

iPad game transfer


Trying to identify what I saw in the sky

Question: This occurred at around 12:45 AM on Saturday August 12th somewhere in the center of Nantucket Island Massachusetts. We were trying to spot the Perseids meteor shower when I noticed a strange green thing in the corner of my eye. I could only se it in the corner of my eye for some reason because anytime I tried looking at it directly it disappeared. I took out my binoculars to try and see it and all I saw was six greenish lights looking like they were entering the atmosphere. Anyone know what this may have been? Or should my entire family see the eye doctor?
Comment: I don't want to say this is a real phenomenon because I'm not 100% sure, but I've read that near the horizon, if the conditions of the sky are just right, bright objects can appear green, if the red and blue lights are sufficiently scattered.  This is the cause of the often reported but rarely documented "green flash".    I gather it's possible, but rare.  It requires sufficient water vapor in the air between you and the object.   No reason it couldn't happen with stars, in fact, the smaller size might make it easier, but would need to be very close to the horizon.
Comment: Just to say it, the "can't see when looking right at it" is pretty normal for very dim lights. Your peripheral vision is much more rich in rod-cells than your cone-focused central vision. Rods are highly sensitive to luminance in exchange for not picking up on color, so your central vision needs more light but gives you color discrimination, while your peripheral vision is better for seeing very dim things. Astronomy students are often advised to look slightly to the side of whatever they're trying to see.
Answer: Sorry, but my answer may be a little speculative
I used Stellarium to find out where and what exactly it was, 12:45 AM on Saturday August 12th in the center of Nantucket Island Massachusetts. There were 6 things in the constellation Perseus, which were Starlink satellites e.x Starlink-2447 and few others.
And I think they were glowing bright green, because of some Satellite flare that may have occurred due to the angle of reflection from neighboring sources of light like the meteor shower itself.
It also may be nearby star cluster like Pleides (NE), if you were not looking in the right direction, which is also known as the seven sisters
Thanks, Hope it helps you!
Precaution: The retina was not adapted for the light, and light adaptation is essential for stargazing or else glares may occur like in the question
Answer: Were you looking NE? The Pleiades were just above the horizon at 12:45am.
If not that, then unlikely to be any astronomical phenomenon.
Comment: The Pleiades solution is simple to prove or disprove as they aren't going anywhere anytime soon.  OP could try to replicate the event and let us know if the six lights were still there.  My guess is they will be.


Lutris doesn't see Vulkan

Question: I've just installed Lutris through flathub and I'd like to run a DXVK game but Lutris can't see Vulkan for some reason. Is there a solution to that?
Comment: Hello @Qushy. Please provide more information (with examples if possible) about problems you're experiencing and what have you tried so far to solve them.
Answer: I'm also having this issue, did you get anywhere with this?
Vulkan drivers are installed and working on the Lutris version installed via apt but when I use the Flatpak version I get the following error:

This is also true of wine, I got a similar message when installing a new game.
Comment: I actually think I was missing a runtime. If you go to the thread at the Lutris Forums I've actually posted the command to install the runtime there by it may be outdated by now.

Lutris doesn't see Vulkan


Is the function in the $L^2$-sobolev space $H^{\frac{1}{2}} \ [- \pi, \pi]$ of order 1?

Question: My professor and as a consequence the rest of my class are saying that the specified function is not in $L^2$-sobolev space $H^{\frac{1}{2}} \ [- \pi, \pi]$ of order 1.
TLDR: Just skip ahead to $(\star \star \star)$ this initial part is an explanation of my professor's notation as well as proving some trivial conditions.
$L^2$-sobolev space $H^{\frac{1}{2}} \ [- \pi, \pi]$

In order to be in $L^2$-sobolev space $H^{\frac{1}{2}} \ [- \pi, \pi]$ of order 1 a function $f(x)$ must satisfy three conditions:

The function must be continuous on the interval $[-\pi,\pi]$
The function must satisfy $f(-\pi) = f(\pi)$
The function's derivative, $f'(x)$, must exist and be continuous on the interval $[-\pi,\pi]$ except at a finite number of points.
The function must satisfy: $\int_{-\pi}^{\pi}(f'(x))^2dx$ < $\infty$, ie the integral of the derivative of the function squared over the interval must be finite.

If all these conditions are satisfied then the function is in $L^2$-sobolev space $H^{\frac{1}{2}} \ [- \pi, \pi]$ of order 1, and therefore the function's Fourier series converges uniformly and absolutely on the interval.

Now the function I am dealing with is the following:
$$ f(x) = \begin{cases} \frac{\sin x}{x} \ \ \text{for} \ \ x \neq 0 \\  \ \ \ 1 \ \ \ \  \text{for} \ \ x =0 \end{cases}$$
Now (1) is satisfied since the function is continuous everywhere except perhaps at $x = 0$ however taking the limit at 0 and then applying L'Hopital shows that the function approaches $1$ as $x \rightarrow 0$. By definition of continuity and the piece-wise condition that $f(x) = 1 \ \ \text{for} \ \ x =0 $, that is:
$$ \lim_{x \rightarrow 0} f(x) = f(0)$$
Therefore the function is continuous everywhere on the interval.
For (2) the condition is easily shown it is trivial.
For(3) the functions derivative exists it is:
$$f'(x) = \frac{x \cos x - \sin x}{x^2}$$
and it is continuous everywhere including $x = 0$ where the function is defined as:
$$ f'(x) = \begin{cases} \frac{x \cos x - \sin x}{x^2} \ \ \text{for} \ \ x \neq 0 \\  \ \ \ \ \ \ \ \  \ \ \ \ \ \ 0 \ \ \ \  \text{for} \ \ x =0 \end{cases}$$
Using L'Hopital it is shown just like it was done before that the derivative is continuous at $x = 0$ and therefore the derivative exists and is continuous on the interval.
($\star \star \star$)
Finally (4) (this is where the disagreement comes in) My professor says that the integral condition diverges to infinity his argument being that the integral can be written as:
$$\int_{-\pi }^{\pi } \left(\frac{x \cos x-\sin x}{x^2}\right)^2 dx = \int_{-\pi }^{\pi } \frac{x^2 \cos^2(x)}{x^4}dx + \int_{-\pi }^{\pi } \frac{\sin^2(x)}{x^4}dx - 2 \int_{-\pi }^{\pi } \frac{ x \cos(x) \sin x}{x^4}dx  $$
His argument is that if you take the limit of each integral as $x$ goes to $0$ each integral goes to infinity therefore the integral diverges.
I disagree, my argument being that let there be a function $g(x)$, where $g(x) = (f'(x))^2$ $$\rightarrow \ \ \ \ \ \ \ \ \ \ \ \ \  g(x) = \begin{cases} \left(\frac{x \cos x - \sin x}{x^2}\right)^2 \ \ \text{for} \ x \neq0 \\  \ \ \ \ \ \ \  0 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \  \text{for} \ \ x = 0
 \end{cases} $$
This function is clearly what is inside the integral, and it is continuous on the interval except perhaps at $0$, using 3 applications of L'Hopital we show that the function $g(x)$ is continuous everywhere on the interval including at $x = 0$ (this being done in similar fashion as before). Then from a previous question I posted and confirmed on SE I determined via the extreme value theorem that this function has a maximum on the interval, (doesn't go to $\infty$), is continuous (due to piece wise condition), and the integral is evaluated over finite bounds. Therefore the definite integral must be finite, and thus the 4th condition is satisfied and therefore the function is in  $L^2$-sobolev space $H^{\frac{1}{2}} \ [- \pi, \pi]$ of order 1. Who is right? I believe my arguments are carefully constructed and follow logically, and I disagree with my professor's method since $\infty + \infty - \infty$ does not necessarily imply the function goes to infinity it says nothing of the 'finiteness' or lack thereof of the integral and also this argument does not take into account the piecewise nature of the function. Please an expert on this help me out and is the integral finite in this case, Thank you!

Also just for some graphical grounding in my argument (although not proof) you can also look at the graph of $g(x)$ and see that considering that if the function is continuous at $x = 0 $ which it was shown (due to piece wise) the integral of $g(x)$ must be finite.
$g(x)$">
Answer: The problem with the integral is that the formula
$$\int_a^b\,(f+g)\,dx=\int_a^b\,f\,dx+\int_a^b\,g\,dx$$
is only valid under the assumption that either $\int_a^b\,f\,dx$ or $\int_a^b\,g\,dx$ is finite. The finiteness assumption can only be dropped if both $f$ and $g$ are non-negative or nonpositive. This is not the case in the way that your professor is dividing the integral.
An equivalente way of viewing it, it is the same as arguing that $\int_0^1\,x\,dx$ is not finite because it equals $\int_0^1\,(x-1/x)\,dx+\int_0^1\,1/x\,dx$ where each summand diverges to infinity.
Comment: Thanks for the great insight, I will definitely bring this up. However is my argument showing that the integral is finite correct, as done above?
Comment: Your argument is correct, continuous functions on a compact interval have always a finite integral.


Positive harmonic function on $\mathbb{R}^n$ is a constant?

Question: Is it true that a positive harmonic function on $\mathbb{R}^n$ must be a constant? How might we show this? The mean value property seems not to be the way...for that we would need boundedness.
Comment: The whole proof from Nelson's classic paper is given here: http://en.wikipedia.org/wiki/Harmonic_function#Liouville.27s_theorem
Comment: If $f$ is positive, then $f = |f|$. Try computing the Laplacian of $|f|$.
Answer: Since it is not clear whether the Wikipedia proof uses boundedness or not, please allow me to give a detailed proof that only uses nonnegativity.
Let $u$ be a nonnegative harmonic function in $\mathbb{R}^n$, and let $x,y\in\mathbb{R}^n$. Denote by $B_R(y)$ the ball of radius $R>0$ centred at $y$, and similarly by $B_r(x)$ the ball of radius $r>0$ centred at $x$. We consider such balls with $R>0$ large and $r=R-|x-y|$, that is, $B_r(x)\subset B_R(y)$ and the boundary of $B_r(x)$ is tangent to the boundary of $B_R(y)$.
Then we have
$$
u(x) = \frac1{|B_r|}\int_{B_r(x)}u \leq \frac1{|B_r|}\int_{B_R(y)}u = \frac{|B_R|}{|B_r|}u(y),
$$
where we have used the mean value property in the first and last equalities, and the nonnegativity of $u$ in the second inequality. Now we send $R\to\infty$ and get
$$
u(x) \leq u(y).
$$
As $x$ and $y$ were arbitrary, we conclude that $u$ is constant.
Note that the Harnack inequality is lurking behind this proof. Also, it was crucial to consider balls with different radii in order to make use of nonnegativity. The Wikipedia proof uses balls of equal radii.
Comment: @user135520 Note that $r$ is tied to $R$, so when $R\to\infty$, so does $r$.
Comment: @Chubwagon: The middle inequality is due to monotonicity of the integral, as the integrand is nonnegative.
Comment: Sorry, I'm confused, so we have that $|B_{R}| \geq |B_{r}|$, but how do we get $u(y) \geq u(x)$ from the inequality $u(x) \leq \frac{|B_{R}|}{|B_{r}|}u(y)$?
Comment: Oh, I see, maybe one can think of $|B_{R}|$ and $|B_{r}|$ being the same size as $R \to \infty$.
Comment: In the right side of the inequality, why do we have $\frac{1}{|B_r|}$, as opposed to $\frac{1}{|B_R|}$. Also, how do we know that the first inequality holds? Are we using that $r<R$?
Comment: Oh, I see, we just used the equality and multiplied by $\frac{|B_R|}{|B_r|}$, though I still don't see why the middle inequality is true.
Answer: Taking $n=2$ just to simplify:
Note that, given $y\in\mathbb{R}^2$, let $r=||y||$. Then, $\forall n\in\mathbb{N}$:
$$B(y,n)\subset B(0,2n)$$
Thus
$$\int_{B(y,n)}u(z)dz\leq\int_{B(0,n+r)}u(z)dz$$
$$u(y) = \frac{1}{\pi n^2}\int_{B(y,n)}u(z)dz\leq\frac{(n+r)^2}{n^2}\frac{1}{\pi(n+r)^2}\int_{B(0,n+r)}u(z)dz=\left(\frac{n+r}{n}\right)^2u(0)$$
That is
$$u(y)\leq\left(\frac{n+r}{n}\right)^2u(0),\qquad\forall n \in\mathbb{N}$$
Takin $lim_n\rightarrow \infty$, we have that $u(y)\leq u(0)$. But since $y\in\mathbb{R}^2$ was arbitrary, we conclude that $u(z)\leq u(0),\forall z\in\mathbb{R}^2$.
But then, given $R>0$, we have that $\max_{z\in\overline{B(0,R)}}u(z) \leq u(0)$, thus, $\max_{z\in\overline{B(0,R)}}u(z) = u(0)$. Since $0$ is an interior point, by the strong maximum principle, $u$ is constant equal to $u(0)$ in the ball. Since $R$ was arbitrary, it follows that $u(z) = u(0),\forall z\in\mathbb{R}^2$.
Answer: One can use Louisville's Theorem for harmonic functions to prove this. Note that if $u$ is a harmonic function then so is $e^{-u}$ and if $u$ is bounded below, then $e^{-u}$ is bounded and thus constant. This then forces $u$ to be constant. This shows the slightly stronger fact that if $u$ is a harmonic function bounded below, then $u$ is constant.
Comment: Why is $e^-u$ harmonic? If I take the Laplacian of $e^{-u}$ I get $- e^{-u} \triangle u + e^{-u} |\nabla u|^2 $.
Comment: You're right. I don't know what I was thinking. See https://math.stackexchange.com/questions/2441878/is-euv-a-harmonic-function for a proof that the only time $e^f$ is harmonic for $f$ harmonic is when $f$ is constant.


Which is less DB intensive in the loop: custom taxonomies or meta fields?

Question: If I'm checking for a certain variable for each post in the loop, is it less DB intensive to specify that variable using custom taxonomies or meta fields?
Comment: Are you developing the code for future use or on top of existing data? This is kind of question that is better answered with profiling in specific scenario than theory.
Comment: Are you just displaying the value or are you filtering out the items in the loop? They are both about the same for a relatively small set of posts (less than 500).
Comment: Filtering items in the loop. What about for many thousands of posts?
Answer: I had the same question here and it came down to this...
It depends! (yep that's it). This is what Otto said...

Your question cannot be answered because you don't give a proper
  scenario or framework by which to answer it.
Searching by meta value only is inefficient because it is not an
  indexed value. Searching by key and value is not as bad, because the
  key is indexed. Still bad, but not as bad for small sets. For it to be
  noticably bad, you would need to have multiple tens of thousands of
  posts and millions of meta. A large news site, for example.
If you're talking about a rather small blog, then do what you like.
  Everything depends on scale.
No quantitative data exists because the answer is self-evident to
  anybody who understands databases and basic SQL queries. There is no
  need to measure that which is obvious.

If you're talking about many 1000s of posts then taxonomies might be better.
Answer: On the Database the Taxonomies should be less intensive for sure, meta table gets crowded fast!
I mean, if you are dealing with an operation that doesn't change that much all the time it's ways better to try to cache stuff in anyway possible.
Comment: What is your basis for stating meta fills up faster than taxonomy _in general_? It might be your practical experience, but I would be hesitant to say this is reliable generic scenario.


how to ask manager to evaluate me as product manager

Question: I am decently good developer. I have been toying with the idea of becoming a technical program manager/developer for a long time. what has been holding me back is the fear whether i am suited for that role and whether i would succeed. I was wondering about asking my manager's feedback about whether he sees me as a good program/product manager.
Answer: The best way to handle this is directly:

Hey boss, Can I have a moment of your time? I wanted your input on how you feel I would be as a product manager. I think this is where I want to take my career but I'm not sure what my strengths and weaknesses in this area would be.

Do it when you will have time to talk. Maybe arrange a follow up.


What, technically, is the rule on whether to pronounce a "d" in a german word as a 't' or a 'd'?

Question: I know the "d" in German tends to be voiced when its at the start of a words and voiceless when its at the end, but I've never really known what the rule technically is.
Comment: Can you give some examples?
Comment: See [What are the pronunciation rules for the consonant d?](https://german.stackexchange.com/questions/37395)
Answer: German d does not tend to be voiced at the start of words. If anything, then it tends to be voiceless everywhere. In the entire Southern part of the German speaking area, it is always voiceless. In the Northern part, it is also voiceless unless entirely surrouned by voiced sounds. In that sense, voice is not a distinctive feature of German obstruents, but only an accidental feature that may be present under certain conditions in certain regions.
What you are probably asking about is the phenomen called Auslautverhärtung. In English, this is usually translated as final devoicing, which is somewhat of a misnomer when speaking about German, since voicing is not a distinctive feature of obstruents to start with. Literally, it means 'final hardening', which more accurately characterizes the phenomenon. In German varieties that have Auslautverhärtung, final "soft" obstruents like d (traditionally called lenis sounds) are pronounced exactly as if they were "hard" obstruents like t (traditionlly called fortis sounds):

The word Rad is pronounced exactly like the word Rat.
The word Bund is pronounced exactly like the word bunt.

Auslautverhärtung is typical of Northern German varieties. There is a strong prescriptivist tradition that elevates the Northern German pronunciation to the one and only true pronunciation of standard German. According to this prescriptivist point of view, Auslautverhärung is a characteristic of German proper. In a more neutral point of view, all we can say is that some pronunciations of standard German have Auslautverhärtung and others don't, just like some pronunciations of English are rhotic and others are not – the times when only the Queen's non-rhotic Received Pronunciation counted as English proper have long passed.
Comment: You should emphasize **voicing is not a distinctive feature of obstruents**. The English *voiced-unvoiced* concept simply does not match German, it's *fortis-lenis*, a more general concept.
Comment: @Janka: Thanks for the suggestions. I have now emphasized it, and I have also mentioned the terms *fortis* and *lenis.*
Comment: This is probably all correct. I do not think that it answers the question for the OP, though.
Comment: There may actually be a caveat about that "exactly" though: even though there is a perceptual merger for a great many speakers between "Rad" and "Rat", there are supposedly certain phonetic measurements that indicate that it is nonetheless an incomplete rather than a categorical merger: http://www.bodowinter.com/papers/winter_rottger_IN_grazer.pdf
Comment: Rather, I should have said "...that some people interpret as indicating..."
Answer: Well, let's try and observe our pronunciation.

Dunkelkammer vs. Tunnelbohrung
Dachstuhl vs. Tannenbaum
Drückeberger vs. Trottel

We see (or hear): the D- are more "voiced" than the T-, but this seems not to be the most important difference. Rather, it is the strength of the obstruction to the stream of air which then is released. The T is just more powerful in all aspects. I would say, this fits well into the concept of "lenis" and "fortis" mentioned in Janka's earlier comment.

Petersilienbund vs. kunterbunt
Tellerrand vs. rumgerannt
Weiderind vs. spinnt

I would say, these -d and -t are rather similar (at last in what I would call standard German), they are voiceless, but importantly all of them, both d and t, are aspirated, i.e. an audible stream of air passes through the mouth after it forced by a contraction of the lung (or other respiratory organs).