Does the autonomous-activator work on the grindstone?

Question: I'm currently not in the position to test this but i was wondering if this is possible? I know for a fact that the creator of the mod in which the grindstone is (applied-energistics) in didn't want to make the player able to automate the grindstone but since the autonomous-activator simulates the players clicks I thought it might be a viable option.
Answer: Yes and no. In 1.6 and previous versions, this was possible. However, from Minecraft 1.7 onwards, this will not work and the crank will drop instead.(Source: here) This is because the Autonomous Activator is a Fake Player and the code checks if it is a fake player. However, there are other options. They are:

Macerator from Industrial Craft
Pulverizer from Thermal Expansion
Answer: Yes. The autonomous activator acts as a fake player, simulating clicks as configured. It works on virtually anything that is activated by right clicking repeatedly, including the Applied Energistics grindstone. Using an Engineering Turtle from ComputerCraft also works. There's nothing in Applied Energistics that prevents it.
Keep in mind, the grindstone is not a fast way to double ores. It costs no power, sure, but it's really slow in comparison to, say, the pulverizer. So early-game, yes, you can definitely automate it, however, I wouldn't want to be in late-game with that kind of a setup.
Comment: Apparently, this is not *intended* to work. This is definitely removed in AE2, 1.7, and I will test that this works in AE1 in 1.6
Comment: This answer is *probably* invalid. I'm going to test it again.
Comment: @Unionhawk I think this is now disabled
Comment: I know its slow but i just want to use it to automate my wheat grinding.
Comment: @Tim I do believe you are correct. I'll work on an answer as to why (it's either because it's disabled or because right clicking continuously with no input breaks the handle)


Strong law of large numbers for continuous time martingales?

Question: Is there a theorem/reference that states if $M(t)$ is a martingale, then under certain mild conditions, $M(t)/t\to 0$ a.s. as $t\to\infty$?
Answer: There is Theorem 9 p.142 in Liptser and Shiryaev's "Theory of Martingales" of which a particular case is the following :
Let $M = (M_t)_{t \geq 0}$ be a local martingale and define the process $B = (B_t)_{t \geq 0}$ by :
$$B_t = \sum_{0 < s \leq t}\frac{(\Delta M_s / (1+s))^2}{1 + |\Delta M_s / (1 + s)|},$$
where $\Delta M_t$ is the jump of $M$ at time $t \geq 0$. Let $\tilde{B} = (\tilde{B}_t)_{t \geq 0}$ be the compensator of $B$ and denote by $\langle M^c \rangle = (\langle M^c \rangle_t)_{t \geq 0}$ the quadratic variation of the continuous martingale part of $M$. Then, if
$$\int_0^{\infty}(1+s)^{-2}d\langle M^c \rangle_s + \tilde{B}_{\infty} < +\infty \hspace{2mm} ({\bf P}-a.s.),$$
we have $\frac{M_t}{t} \to 0$ $({\bf P}-a.s.)$, as $t \to \infty$.


How does a median have a value that is a decimal which isn't exactly half of an integer if the data should consist of only integer values?

Question: I real an article which said the average man accumulated 6.1 sexual partners while the average woman accumulates 3.6. If the statistic talked about the average, surely the numbers would be equal-so it must talk about the median. But how is the median 6.1 or 3.6?
http://www.medicaldaily.com/pulse/whats-normal-breast-size-and-penis-size-find-out-if-youre-sexually-average-317324
Comment: People lie in surveys
Comment: Do the stats count gays?
Comment: They seem to be talking about the age range 25-44. What about cases where exactly one of the people involved is outside of that range?
Answer: For most of the reasons mentioned in the Comments, I do not
find it remarkable that means would be unequal (except
@GerryMyerson 's Comment happens to be explicitly foreclosed by the
description in your link). In every survey I have ever seen
on the issue of numbers of sexual partners, men report
more of them women--even in what appear to be populations
for which results should logically be exactly equal.
(In this particular case, I hope I can be excused for
harboring the suspicion that this survey was conducted
more with an eye towards ad sales than careful survey design.)
However, see the answer to recent question 1327801 on this site
to see how median values might not be integers or half
integers, when approximated from grouped data or a
histogram.
Note: On a closely related topic, most software packages agree on how to
find the median from original data, but they have very
different ways of finding the lower and upper quartiles.
These differences become negligible with increasing
sample size, but they can be confusing to students trying
to reconcile hand computation, answer book values, and values
from software. [In R statistical software, typing <code>? quantile</code>
gets you a help page that lists nine different formulas
in common use to find quantiles (hence quartiles). The median
is the 50th quantile, but the methods mainly agree for medians.]


How to change Hive Server Timezone

Question: I am using Apache Hive on RedHat. Hive version is 1.2.1. The current Timezone in Hive is EST and I want to convert it to GMT.
PS: I want to change the Hive Server Timezone and I am not asking about using a particular function in Hive that can convert EST timezone to GMT timezone.
Answer: Hive picks timezone of your JVM, so try changing JVM timezone. Setting "TZ" environment variable should help.
Comment: I'll check this in my production environment and let you know the updates.

How to change Hive Server Timezone


Default capacity of StringBuilder

Question: What is the default capacity of a StringBuilder?
And when should (or shouldn't) the default be used?
Answer: The default capacity of StringBuilder is 16 characters (I used .NET Reflector to find out).
Answer: Default is 16, which seems to be the default capacity of any type of array or list in the .NET framework.
The less number of reallocations you need on your StringBuilder, the better it is.
Meanwhile, it is unnecessary to allocate much more than is needed, too.
I usually instantiate the StringBuilder with some type of rough estimate as to the final size of the StringBuilder.
For instance, this could be based on some iteration count that you will use later on to build the string, times the size needed for each item in this iteration.
// where 96 is a rough estimate of the size needed for each item
StringBuilder sb = new StringBuilder ( count * 96 );
for ( int i = 0; i < count; i++ )
{
...
}

When the size of a StringBuilder is too small for writing the next string,
the internal char array of StringBuilder is reallocated to twice its current size.
Answer: This question came up today as a duplicate of another, but I notice one part wasn't answered. The default (assuming that means "when not created with a string that is large enough to require ) is 16 as people said, but I don't see anything here on when you should change it.
You change it when you can do so as a possible optimisation. Indeed, the choice of 16 is the opposite to an optimisation. Optimisation is picking values and approaches so as to particularly well suit a particular case or subset of possible cases, (not "making things faster" generally, though that's how we often use the word). Here the designer of the class had to deal with generalisation - picking values and approaches so as to give reasonably good performance across a broad range of cases.
The smaller they went, the less use of memory.
The larger they went, the less reallocation to deal with larger strings.
There's a few reasons why binary-round (whole powers of two) are likely to give better performance than other numbers in certain cases, so they went for one of those, but that aside the choice between 4 or 16 or 1024 was a matter of balancing different likely values.
Someone using StringBuilder rather than designing it, may have a better idea of what size they are likely to need.
If they are going to Append 5 1-digit numbers along with strings that total to 43 characters in length, then the total length of the StringBuilder is going to be 48 characters no matter what, so they should use a capacity of 48 as 48 is always the most efficient size for a string of 48 length.
If they are doing something where there could be any length between about 23 and 34 chars, they should use 34.
If they are doing something where there will likely never be more than 60 chars, but every now and then there could be, they should use 64 (don't reallocate for most parts, and gain the power-of-two benefit mentioned above for the few cases where you do).
If it's impossible to come to a conclusion on this, or at least hard to do so and not a performance hot-spot, then you should just use the default.
Answer: The Venerable J. Skeet has provided a good analysis of precisely this problem:
https://jonskeet.uk/csharp/stringbuilder.html
Comment: Good article, but doesn't answer either question.
Comment: That's a good reminder that I should edit the article to include the word "capacity" somewhere :)
Comment: This answer does not include any detail, just a link.
Comment: @JonSkeet - still missing the word capacity. Interesting article though.
Answer: Default capacity of a String-Builder is 16 characters ,
And Max capacity of String-Builder is 2147483647 characters.
So no need to worry for storing long response !!!
Comment: Can you add a link to the documentation where this is confirmed?
Comment: I have tried in my project. If u want to try out then u can practice it in dotnetfiddle.net.
Comment: [String Builder Class(System.Text) | Microsoft Docs](https://learn.microsoft.com/en-us/dotnet/api/system.text.stringbuilder?view=net-6.0#instantiating-a-stringbuilder-object) for reference
Answer: We can find the capacity using the capacity property of the StringBuilder Class:
StringBuilder builder = new StringBuilder();
var capacity = builder.Capacity;
var maxCapacity = builder.MaxCapacity;

Here capacity defines the default capacity of StringBuilder.
StringBuilder is Max Capacity is same as the maximum value of Int32.
Answer: [edit: at the time, the question asked about StringList]
Do you mean StringCollection? This uses an empty ArrayList initially, so the answer is 0. And you have no option to change it. When you first add an item, the capacity jumps to 4, then uses a doubling strategy when it fills.
If you mean List<string>, then it is similar (an empty T[], not an ArrayList), but you can initialize a known size if you need (i.e. you know how much data you expect). Again, the first time you Add to a List<T> the size jumps to 4, then doubles every time it fills up.
Comment: I think he did mean StringBuilder, nothing else.
Answer: read hear about Stringbuilder capacity , there is sample app to prove it too.
Comment: That linked article refers to maximum capacity. Not the default capacity (16).

Default capacity of StringBuilder


AIDE and OSSEC conflicts?

Question: Being fairly new to both AIDE and OSSEC I've been trying to find out if there are any potential conflicts in having them both installed on one host (CentOS 7.5). It seems like they could work as a multi-layered approach, but I haven't been able to find much about running the in tandem. And of course would there be a difference in running them OSSEC locally vs. an agent on the machine with AIDE? Thanks.
Answer: I have used AIDE in past and it does only file integrity monitoring. It creates a database when you first install the same, probably mapping hashes along with the respective file names. This comes handy when you want to see any Malware/Program changes which files.
Said that, OSSEC is more superset. Read "AIDE only does file integrity checks. It does not check for rootkits or parse logfiles for suspicious activity, like some other HIDS (such as OSSEC) do." from this site
I don't see any issue with both of them running on same system, but I don't see value in having both unless you find anything which proves OSSEC prone to hash-collision.
To answer the second part of your question, decision of deploying the OSSEC in Manager/Agent mode or standalone mode depends on the requirement. If there are hundreds of hosts, you want centralized management where all the database is stored on server side.
Comment: Thanks for the input on the first part of the question. I should clarify on the second part, I was curious more about running both in host mode. I know you can use just the agent and have the server instance running elsewhere managing a large implementation. But would running host mode (just the local instance) work differently? The first part of your answer would seem to suggest not though. Thanks and I appreciate the thoughts.
Comment: @jfran3 Yes, there is no difference in terms of functionality, it's just different deployment design.
Comment: Just to follow up... I understand the deployment design differences at the system level, but I think what I'm trying to get at (and I apologize if I wasn't clear) is whether at the host level the full local install of OSSEC would be more problematic with AIDE than a simple agent, while running in tandem. Thanks again for all your help.
Comment: not to mention AIDE is light in memory because it is a file monitoring system.

AIDE and OSSEC conflicts?


Can I rewrite from subdomain to folder avoiding redirection?

Question: I want my subdomain to point directly to a folder.
That is if an HTTP request is made to apple.example.com,
I want the response to be the contents of example.com/apple/index.html, but not a 301 error.
Is it possible?
Answer: Yes, this is exactly the purpose of tools like ISAPI_REWRITE. My hosting company (orcsweb) uses exactly this technique.
-OIsin
Answer: Yes, but you need to make sure that your server is able to support virtual domains.  I don't know enough about this, but the mod_rewrite is possible without redirects.
Answer: <code># Internally rewrite <subdomain>.example.com/<URLpath> to example.com/subs/<subdomain/<URLpath>
RewriteCond $1 !^subs/
RewriteCond %{HTTP_HOST} !^www\.example\.com
RewriteCond %{HTTP_HOST} ^([^.]+)\.example\.com
RewriteRule (.*) /subs/%1/$1 [L]
</code>
Comment: I have already found this solution on the net.
But it does not work for me.
The following error is displayed in the browser when I navigate to apple.example.com
----
Not Found
The requested URL / was not found on this server.
Answer: You can serve content from any folder or you can send a redirect to the client, but AFAIK you cannot do both at the same time - you cannot serve content and somehow change the url in browser at the same time.
If it would be possible, it would most definitely lead to many security exploits.


algebraic curves , divisor

Question: In the definition of Divisor of a curve (Arithmetic of Elliptic Curves, Silverman)  what is the definition of "formal sum"? Adding points naturally  like: $((2,3)+(1,1)=(3,4))$ or adding like adding group law in elliptic curve? Thanks
Answer: Neither. Don't write co-ordinates. Use symbols. So a formal sum of $P$ and  $Q$ is $P+Q$, nothing more  nothing less. In particular if $P=Q$ and if $P$ is a point of order 2, $P+Q$, is still $P+Q$ and not the identity element of the curve. No simplification. Everything is considered a new expression.  Another way of stating  is, it is  a free abelian group on the elements of the curve.
Comment: so if we have adding on Elliptic curves then Div(C) is equal to curve itself? because every sum is in the curve , for example over complex numbers?! can you say an example to understand?
Comment: This formal sum is a starting point. One looks at what are called principal divisors. That is take a rational function on the curve. It leads to a divisor. Zeros of functions specifies the points, and multiplicities specifies the coefficient attached. Same way poles lead to negative coefficients leading to a divisor with sum of coeff = zero. The quotinet of formal sum group modulo principal divisors subgroup (it is a subgroup) will be interesting.

algebraic curves , divisor


In which pratical cases are used parallel and thevenin terminations?

Question: I've found some theory about various type of terminations but I can't figure in which cases are used parallel(pullup with half driving voltage and pulldowns) and thevenin termination. Since I'm used to simply boards with microcontrollers, where I can encounter this two type of termination and in which cases? Are something used in "extreme" cases like very long cables, critical or high power EMI applications?
This is an example of a Thevenin termination:
Comment: Are you asking for all signal types or for digital pulses?
Answer: Proper termination is needed at the beginning and end of a transmission line.
If the termination is not there or has the wrong impedance, the signal you're trying to transport will reflect and distort. This becomes an issue when the length of the transmission line is longer than the wavelength of the signal.
Termination is not always of the Thevenin variant. It depends what the next circuit, which is receiving the signal, needs. If that is a simple inverter then a DC voltage of 0 V (ground level) will not make it work. Inverters need around half the supply voltage to act as an amplifier. This is where the Thevenin termination comes in. It terminates the transmission line but at the same time provides the proper DC biasing voltage for the next stage.
Answer: Since you are talking about microcontrollers, I'm only addressing logic pulses here.  For these kinds of pulses you can singly terminate.  Let's assume the transmission line is 50 Ohms.  If a 50 Ohm resistor to ground was placed at the receiving end the line would be properly terminated.  The problem is when the send end was high, the current needed to maintain this level would be Voh/50, which is hard even on buffer chips. By switching to at 100 Ohm up and 100 Ohm down Thevenin terminator, the high state the current is reduced to (Voh-Vth)/50.  It is much easier to find drivers to do this.  Another way to terminate is to put the 50 Ohm resistor in series at the transmitting end.  Series termination is the lowest demand on the driver.  The way it works is on the edge of the pulse the signal is cut in half at the send end.  When it arrives at the receiving unterminated end, it will double providing the original pulse.  In this case the current seen by the driver is just the receiving gate Iil or Iih.  In series termination there will be a step in the pulse for the round trip delay but this distortion doesn't matter.
Answer: Open collectors are current sources with a pull-up are like voltage sources with a series R.
But more often drivers are lower than Zo so series R's need to be considered if not built in. You examine this by the incremental V/I specs like Vol/Iol and (Vcc-Von)/Ioh vs temp and Vcc.
When threshold is critical for noise , terminate with an active DC bias to threshold or use two R's to make the equivalent bias  to improve margin to binary errors from noise so that the margin is equal for each polarity. Matching the impedances from source to line to load help reduce ringing caused by mismatch.
To determine when to apply these rules you consider if the propagation delay is greater than the rise time. Then you must apply.


Colorbox doesn't seem to work with inline div

Question: I've a question about Colorbox.
I'm trying to load an inline div in a colorbox. I've literally copy-pasted the code from the example page that comes with it and it simply won't work - I have no idea why; the page loads correctly, the colorbox opens, but there's no content inside it. I'm using MVC with a header, index and footer views, with the script on the header view and the inline on the index view (the link which calls the colorbox is on the header view). Any idea why it could not be working?
(Also, I've tried some other code that makes the colorbox open on page load and that works as expected.)
Comment: how is this question related to MVC design pattern in any shape or form ?
Comment: Can you give an example fiddle or something?
Comment: Nevermind, I'm a moron. Got it, feel stupid right now. :)
Answer: Your html should be like this. Note: div with display none first
and then your content div

test content


Probability of selecting K elements of different probability from a set of N elemets

Question: I have a set of N elements. I want to select a few elements from the set. Now each element i of the set has a probability P(i) of getting selected. Then how can I figure out the probability of selecting at least K elements from the set?
Comment: Try this at http://math.stackexchange.com/
Comment: This an imprecise question. Are you selecting with exclusions, without ? It is best to give an example.
Answer: To get it exact, you need inclusion/exclusion:
ie for each subset of k elements (i_1...i_k), add P(i_1)*P(i_2)...P(i_k)
Then for each subset of k+1 elements subtract:
P(i_1)*P(i_2)...P(i_k+1)
Then for each subset of k+2 elements add:
P(i_1)*P(i_2)...P(i_k+2)
and so on so forth alternately adding and subtracting until you reach n or the terms become negligible.
Alternatively, if k is smaller than N/2 you can go the other way:
Start with 1
For all subsets of n-k+1 elements subtract:
(1-P(i_1))*(1-P(i_2))*(1-P(i_3))...(1-P(i_n-k+1))
then for all subsets of n-k+2 elements add:
(1-P(i_1))*(1-P(i_2))*(1-P(i_3))...(1-P(i_n-k+2))
then for all subsets of n-k+3 elements subtract:
(1-P(i_1))*(1-P(i_2))*(1-P(i_3))...(1-P(i_n-k+3))
and so on so forth alternately adding and subtracting until you reach n or the terms become negligible.


What is the reason for these 10-second gaps?

Question: I've been watching Supergirl for a while now, and something got me wondering. Why are there 10-second gaps between most scenes (not all)? I encountered the same thing years ago with No Ordinary Family, which was also produced by Greg Berlanti. At first I thought it was a Berlanti thing, but there are so many TV shows by him. So, what is the reason behind this?
Comment: Do you mean [What is the term called for the sudden blackout before the suspense scene?](http://movies.stackexchange.com/questions/14482/what-is-the-term-called-for-the-sudden-blackout-before-the-suspense-scene), then it will be a duplicate.
Comment: Maybe if you point out a few scenes, it could be easier to answer, but do not forget that TV shows are cut with ads. So after the ads end, it may be interesting to "restaure" the ambiance.
Comment: How are you watching the show? Could these blackouts be where commercial breaks would otherwise be? Is it a cut to ads?
Comment: Hrm, never thought of that. I'll try to find a website I can watch it in full and confirm. Then again, it never happens in Arrow or The Flash
Answer: I know what you're talking about here. I'm going to assume that you, like I, sometimes watch an MP4 file (or similar) that you have come by through a non-traditional DVR method. And that it's on these files where you're observing the long black gaps.
The long black gaps appear to occur at the time that the commercial breaks would have been inserted when the show was originally aired. I think the next question is, if the process or agent involved in creating the MP4 file had the intelligence to remove the commercials, then why didn't it also remove the weirdly long black silence?
Comment: If the process or agent involved in creating the MP4 file had the intelligence to remove the commercials, then why didn't it also remove the weirdly long black silence?
Comment: I'm afraid I don't know the answer to that. It's something I've wondered myself throughout the whole season.
Comment: Still, I keep wondering why the same thing doesn't happen on Arrow, The Flash or Legends of Tomorrow. I've only seen it on No Ordinary Family and Supergirl.
Comment: Because the cutting for commercials is done by a different company. Super girl aired on a different channel than the other three.


Find Max Date from a Distinct Row Then Add Value to a Table Column

Question: Quick disclaimer - I am familiar with Access, but a novice at MySql within Access.
I have a table with activity data that I pulled into a query to filter down to only necessary data.  The query table has DateTime, Call, and Designator columns.  The Call and Designators rows repeat numerous times. The Call is a numeric value which identifies a unique activity and there may be many Designators assigned to each Call. Within this query, I need to find the Max and Min DateTime values of distinct rows (based on the Call and Designator rows) and then add a "Start" value to a new column for the Min DateTime and "End" for the Max DateTime.
Once the Start and End times have been identified for each unique row (Call and Designator) I need to calculate the total time each Designator spent on each call.  I assume that I may have to create another query to include DateTime, Designator, Call, Start Time, End Time, Total Time on a single row to generate reports. But, it would be much easier if it can all be done with one script on the same query.
I've been searching for the past several days and haven't made any progress so I don't have any existing work to show.

Thank you in advance!!
Comment: Provide example data as table-formatted text which can be copypasted to the table.
Comment: Does output `Date/Time` and `Start Time` have the same value always?
Answer: Test
SELECT MIN([Date/Time]) [Date/Time],
       Call,
       Designator,
       MIN([Date/Time]) AS [Start Time],
       MAX([Date/Time]) AS [End Time],
       DateDiff("h",[Start Time],[End Time]) & ":" &
       DateDiff("n",[Start Time],[End Time]) Mod 60 & ":" &
       DateDiff("s",[Start Time],[End Time]) Mod 60 AS [Total Time]
FROM [table with activity data]
GROUP BY Call,
         Designator;

If you need in leading zeros in [Total Time] then add according FORMAT().


Is anything actually 1 meter long (or 1kg of weight)?

Question: I believe that no real objects are actually (exactly) 1 meter long, since for something to be 1.00000000... meters long, we would have to have the ability to measure with infinite precision. Obviously, this can be extended to any units of measurement. Am I wrong?
Comment: I cleaned up your question a bit. I removed the second bullet point because vague, open-ended questions like that aren't the kind of thing we do here.
Comment: It is impossible to know anything with infinite precision. Therefore to argue either way is just nonsense.
Comment: "for something to be 1.00000000... meters long, we would have to have the ability to measure with infinite precision": our inability to measure the length with infinite precision does not prevent the length from _being_ exactly one meter.
Answer: You're not wrong. However, there used to be an object exactly $1$ meter long until 1960, because a meter was defined to be the length of a certain platinum-iridium rod at certain conditions. Since then, the meter is defined in terms of interferometry, and now it is specifically the distance traversed by light in vacuum within a certain period of time.
Similarly, the kilogram has a prototype whose mass is $1$ kg by definition. There are some proposals to replace that definition, but it hasn't been done yet.
Comment: Yes, you do run into it. The only plausible way to get an experimental result of *exactly* 1 unit of length, duration, or mass is to define the unit in terms of whatever the outcome of the experiment happens to be. The second is defined as some number of transitions between two electron energy levels of cesium isotope... which at first glance looks like it could be exact, but uncertainty is introduced by noise: you don't just get the two levels you want in your measurement.
Comment: Stan, thanks for the answer. With the current definition of length in terms of time, don't we run into the same problem? In other words, is any event actually (*exactly*) one second/minute/hour long ??
Comment: @The Chaz, The meter is now defined in terms of the distance light goes in a certain fraction of a second. As measurements of the speed of light improve we change the definition of the metre since it is much more convenient to have a fixed second and a fixed speed of light. So the metre isn't really fixed anymore
Comment: Perhaps it's time for an update to this answer :)
Comment: Aparently, the standard kilogram is loosing weight in the order of micrograms. One more reason to redefine it.
Answer: 1 kg is defined as the mass of a particular reference mass in France.  So there is at least one thing that weighs exactly 1 kg.
Furthermore, it may in general be possible to construct an arbitrary number of objects which weigh 1kg exactly. Assuming all Si atoms have the same mass, one could simply define the kilogram to be the mass of a certain number of Si atoms.
Comment: In the real world, there is *exactly* one thing that weighs exactly one kilogram; the masses copies have drifted apart since they were manufactured.
Answer: In Nature the only exact realization of a number is thru $\mathbb{N}$ i.e. counting . I suspect that they are called 'Naturals' because of that. To all the other numbers we only get aproximations. Even when defined to be the exact length/weight of a certain physical object the measure of it will be within a certain range of values, $\pm$Angstrom.
Another example: there is not a single realization in the universe that correspond to the number $\pi$. Numbers are menthal constructs.
How do we measure ? Obtaining a ratio between the amount that is to be measured and the amount of a selected standard. If both amounts change equally, the measure (ratio) keep constant, i.e. the constancy of a measure is not a guarantee that the properties of 'objects' do not change.
Atomic measures are number counts:

We will now see that the measures of bodies' properties using atomic
  units are independent of the base quantities and dependent on the
  number of particles or atoms.
An atomic unit of mass is the mass of a certain number of baryons; the
  measure of the mass of a body using atomic units is therefore a number
  proportional to the number of baryons of the body (this is not an
  exact statement but it serves the needs of this work). If the mass of
  baryons changes, so will the mass unit and the mass of the body; the
  measure holds invariant because the number of baryons did not change.
  Therefore, a measure of the mass of a body using atomic units is
  basically a baryon count, holding invariant as long as the number of
  baryons does not change, independently of the eventual change of
  baryons' mass. The same kind of reasoning applies to charge measures.
  In what concerns length measures, the length unit is such that the
  measures of length of isolated bodies hold invariant; this is not the
  way length unit is formally defined, but this is a condition it has to
  obey to be acceptable, in order to fit Einstein's measuring rod or
  reference-body, translated in the time invariance of Bohr radius. So,
  we can say that the atomic length unit is a fixed multiple of the Bohr
  radius; if the latter varies, so will bodies' length and the unit of
  length, holding invariant the measures of bodies' length. Therefore,
  length measures are a way of counting atoms, the measures of the
  length of bodies holding invariant as long as the number of atoms does
  so, for bodies and measuring devices subject to the same conditions.
The above reasoning shows that the measures of mass, charge and length
  of bodies are independent of the mass and charge of elementary
  particles and of atoms' radii, tracing only the number of particles or
  atoms.

To any measure (ratio) that is not obtained thru direct counting an error margin is present (explicit or implicit) .
Answer: No, but for other reasons than you might suspect. For two endpoints of an object to be precisely one meter apart, one would have to be able to consistently measure that distance. But in reality, the outcome of any individual measurement is constrained by the Heisenberg inequality, and therefore you cannot consistently measure such a length. If you'd know the precise position of one endpoint, you'd have zero knowledge about it's speed.
Now there are also practical limitations to our ability to measure lengths, and they kick in much earlier (typically ~10-9 or worse; in comparison Planck's constant is ~10-34)
Answer: The definition of units in the SI are determined by the Bureau International des Poids et Mesures (BIPM). The official statement of the definition of the kilogram is

Le kilogramme est l'unité de masse ; il est égal à la masse du prototype
  international du kilogramme.

The English translation is unofficial, but it states

The kilogram is the unit of mass; it is equal to the mass of the international prototype of the kilogram.

Thus, as was stated earlier, there is an object exactly one kilogram in mass.  You can find the BIPM brochure that contains this definition here.


Do chainsaw kills count as non-melee damage for Peace Medic?

Question: The chainsaw benefits from the melee damage character skill, but doesn't benefit from the power fist. Can a chainsaw medic get the Peace Medic achievement (Heal 300 points of damage in a single mission without dealing any non-melee damage to the Swarm), provided they only use the chainsaw and melee attacks to kill aliens?
Comment: Probably not. Why is it so important?
Answer: I know this is an old question, but just in case someone else was wondering the same thing-
You HAVE to equip both the medical gun and healing beacons to be able to heal 300 points of damage, since just one of the above doesn't have sufficient healing power to unlock this achievement. So in effect, this rules out equipping the chainsaw. You can, however, have power-fist in your accessory slot to use in case of emergencies. Also note that teammates standing in the AoE of the Healing Beacon when already at full health doesn't count as healing.
Personally, I'd recommend getting Faith over Bastille for this achievement because she heals faster and runs faster. So you could get near your teammates who can cover for you and not get in the way of friendly fire.
Comment: You could get a second healing beacon/medigun or a chainsaw from a teammate.


Fix faulty 6x6 post install

Question: I mistakenly installed a post protector sheath on bottom of my 6x6 post without drilling the drainage holes. I'm not sure why manufacturer doesn't include the drainage holes to begin with. I buried the post and backfilled with concrete. Top of post protector sticking up from ground about 12 inches. 100% unwilling to dig up the post so what is next best option? Seal the border/gap between the post protector and the wood with exterior grade sealant? Build a little angled cap to send water away while leaving room for it to breathe?
Comment: A picture would be helpful.
Comment: Please provide information or a product link for a "post protector sheath". I don't know what that is.
Comment: Homedepot links cannot be viewed by anyone outside the US. [Never understood why they feel the need to put geofencing round a DIY store, but they do.]
Comment: Drainage holes?   Wouldn't they let moisture in?
Comment: https://www.homedepot.com/p/Post-Protector-6-in-x-6-in-x-42-in-In-Ground-Post-Decay-Protection-6642/203165686
Comment: Please put that information _in your question post_. It doesn't belong down here.
Comment: One reason to use a VPN.
Answer: Here's the claim by the manufacturer, which is debatable:

Moisture is not the cause of decay but is one of (3) conditions needed for decay to occur. The conditions: (1) Moisture, (2) Oxygen & (3) Suitable temperature. These conditions enable the soil-dwelling or atmospheric decay fungi, as well as termites & other subterranean insects, to feed on the wood fiber. So the Decay/Degradation of ground-contact wood is not CAUSED by moisture but by the action of wood destroying organisms feeding on the wood. The 'chemical preservatives' protection strategy is to 'poison' the wood fiber, rendering it an unsuitable food source for the wood destroying organisms. The downside to that approach is that all chemical preservatives eventually lose resistance to decay and insect infestation. Post Protector becomes the perfect complement to the pressure treated (or cedar) post by simply providing a physical "barrier" isolating the post so the wood-destroying organisms have NO access to the post.

The assertion is that by eliminating contact between soil and wood, decay is prevented. Since posts set in the ground are always wet anyway (except in the desert), the claimed functionality is not affected by lack of drainage. In fact, drainage holes would seem contrary to this product's stated mission.
Since you've opted to use the product, I would do nothing.
Answer: From the installation guide for the linked product:

Optional Drainage Holes- On building will have posts subjected to the
elements,  (pavilion, open sided building, porch post, etc.) Drill out
drainage hole dimples.  These dimples are centered on all (4) sides of
the base "horizontal" uplift rib. They  allow drainage for rainwater
running down the post and through the Post Protector.  We suggest 4" –
6" of drainage stone for initial backfill material.

As the drainage holes are listed "optional" I would recommend doing nothing unless you want to dig it up and do it properly (you indicated you do not).
The danger is that water running down the face of the wood post and inside the Post Protector has nowhere to go, this is in contrast to direct-buried posts and concrete-enclosed posts where water can drain away. This may cause problems in the future.
The manufacturer includes a phone number to call for special custom installation guidance, you're likely to get a better answer from them than from DIY.se users who have never used this product: 877-966-8768
Comment: Nice job of reading the instructions for the OP. :/
Answer: I spoke with the manufacturer and they indicated there is no issue they also recommended applying some permanent waterproof sealant where the top of the post protector meets the post

Fix faulty 6x6 post install


Could μονογενής mean simply "unique"?

Question: In this article Hebrew scholar Michael Heiser makes the claim that the term μονογενής (monogenēs) in John 3:16 means "unique" rather than "only-begotten." However, he does not provide citations, referring to "Scholars of Greek."
Comment: See also: [What does μονογενὴς Θεὸς means in John 1:18?](http://hermeneutics.stackexchange.com/q/9140/3555) and [What are the translations of the Greek word "monogenes" in Ancient Versions? Is it "only" or "only-begotten"?](http://hermeneutics.stackexchange.com/q/12836/3555)
Comment: @H3br3wHamm3r81 Nice! I'd be happy to see you pull that over here (assuming it's yours)....appears to be fully relevant to answering the question.
Comment: "The Greek word μονογενής is typically (although, not always) translated in the King James Version of the Bible by the English phrase "only-begotten."
Answer: Heiser's statement (from the document you link) is:

The Greek word μονογενής (monogenēs) doesn't actually mean "only begotten." It presents a problem neither with respect to Jesus having a beginning, nor with respect to divine "sons of God" who are called gods (אֱלֹהִים , elohim) in the Old Testament. The confusion extends from a misunderstanding of the root of the Greek word. For many years, μονογενής (monogenēs) was thought to have derived from two Greek terms, μόνος (monos, "only") and γεννάω (gennaō, "to beget, bear"). Scholars of Greek eventually discovered, though, that the second part of the word μονογενής (monogenēs) does not come from the Greek verb γεννάω (gennaō), but rather the noun γένος (genos, "class, kind"). The term literally means "one of a kind" or "unique" with no connotation to time, origin or solitary existence. The validity of this understanding is shown by the New Testament itself. In Hebrews 11:17, Isaac is called Abraham's μονογενής (monogenēs)—but Isaac was not the only son Abraham fathered, since he fathered Ishmael prior to Isaac. The term must mean that Isaac was Abraham's unique son—the son of the covenant promises and the line through which the messiah would come. Just as Yahweh is an אֱלֹהִים (elohim), and no other אֱלֹהִים (elohim) are Yahweh, so Jesus is the unique son, and no other sons of God are like Him.

To summarize his argument:

(a) Certain Greek scholars
'discovered" (sic.) that the word μονογενής was derived the noun γένος, and not γεννάω
(b) γένος means "class" or "kind"
(c) Isaac is called Abraham's μονογενής in Hebrews 11:17, but Isaac was not Abraham's only son.

Therefore

(d) μονογενής means one of a kind or class (i.e. "unique") and nothing else

First, I would say that the argument is invalid, since the conclusion could still be false even if all the premises are true.  Bauer's A Greek-English Lexicon of the New Testament and Other Early Christian Literature, for example, lists several different meanings of μονογενής in both secular and sacred Greek sources (i.e. Greek New Testament plus Septuagint) where on some occasions the word means "unique", as Heiser suggests (e.g. Cornutus - a Roman who sometimes wrote in Greek), but on many others clearly means "only child" (i.e. "only begotten").  The word may, in fact, have the etymology that certain Greek scholars "discovered", but that does not prevent the word from having other multiple meanings that may not be entirely consistent with its supposed roots.

Heiser's point about Isaac is valid and I think what he claims is correct:  that the term here meant that Isaac was unique with respect to being the son through whom the Messiah would come, and not that Isaac was an only child of Abraham.  The Greek texts here read:

... τὸν μονογενῆ ... πρὸς ὃν ἐλαλήθη ὅτι ἐν Ἰσαὰκ κληθήσεταί σοι σπέρμα

Heiser has a point in that many English translations render τὸν μονογενῆ here as only begotten or only son or only begotten son, which many readers would take at face value to mean "only child". But that is not what the Greek says:  τὸν μονογενῆ is qualified by the phrase πρὸς ὃν ἐλαλήθη ... ("of whom was said").  The NIV goes so far as to say his one and only son, which is clearly false.  A couple of versions avoid the phrase only begotten or only son.  The Eastern Orthodox Bible: New Testament, for example, translates v.17-18:

By faith, Abraham, when he was tested, offered up Isaac [in sacrifice]. Yes, he who had joyfully received the promises was now offering up his uniquely loved son, the very one of whom it was said, "In Isaac will your seed be called."

The late Eastern Orthodox Bishop Dmitri Royster explains Hebrews 11:17-18:

Isaac here is called the "only-begotten" son of the one "that had received the promises."  Actually, he had another son, Ishmael, born of Sara's maid Hagar (Genesis 16:15), but it was to be the son of Sara, Abraham's wife, that would be counted as the only-begotten, or "beloved son" (ton huion sou ton agapēton Genesis 22:2 LXX). God Himself confirmed Sara's insistence that Ishmael would not be the one through whom Abraham's seed would be called (Genesis 21:10-12).*

* The Epistle to the Hebrews: A Commentary (St. Vladimir's Seminary Press, 2003), p.187
Answer: even if you do not believe the second stem of μονογενής is not from γεννάω (gennaō), but rather the noun γένος Kittle describes this use IV 737 as the only descendent from a man or a woman.  This is in fact the only way the first three uses can be interpreted in the New Testament.  Luke 7:12 Luke 8:42 and Luke 9:38.  The two uses in Psalms if literalized to one of kind make the Psalm ridiculous.  Deliver me from the power of the dog, my only one of kind life.  No Deliver me from the power of the dog my only begotten life. My only descended life.  Isaac is the only begotten from Sarah and Abraham. that combination.
Comment: Welcome to BH. Please see the Tour and the Help (below, bottom left) as to the purpose and the functioning of the site. You may like to look at my question on the other site as to the meaning of μονογενησ [Stack Exchange - Christianity](https://christianity.stackexchange.com/questions/71904/a-question-regarding-arguments-against-the-nicene-creed) in which I refer to the valuable work of Denny Burk _et al_ in the book _Retrieving Eternal Generation_. (Up-voted +1.)
Answer: Short Answer: The word is best translated "one-of-a-kind" or simply "unique". ("Only" would also work, though it could be misunderstood more easily.) The old translation "only-begotten" was based on an honest mistake in parsing the Greek word.

Background on "only-begotten"
The Greek word in question is μονογενη. It is pretty clearly a compound word formed from combining "μονο" with "γενη". Since μονο means "one", or "only", and γενη looks like the verb "γενναω", which means "to beget", the translation "only-begotten" seemed to make sense, and was the preferred interpretation for a long time.
Background on "one-of-a-kind"
However, as our understanding of the mechanics of Koine Greek have improved, it was eventually recognized that γενη cannot be a derivation of γενναω, and is instead from the word γενος (genos, meaning "kind"). If John had intended to say "only-begotten", he would have said "μονογεννητος" (single + begotten), not "μονογενης" (single + kind). (cf. VGNT).
Γενος has a semantic range including "people", "race", "descendant", or in its most basic sense, simply "kind". The term obviously isn't claiming that He is a "people", "race", or "descendant", and so the most basic sense of the word ("kind") makes the most sense here.
Thus we have the corrected interpretation of "one-of-a-kind", as supported by such instances as Psalm 21:21 (LXX) and Psalm 24:16 (LXX) where it clearly cannot be understood as "only-begotten", even if the grammatical difficulties with that interpretation were overcome.

Deliver my soul from the sword,
  My only life from the power of the dog. -Psalm 21:21 (LXX) / 22:20 (NASB)
Turn to me and be gracious to me,
  For I am lonely and afflicted. -Psalm 24:16 (LXX) / 25:16 (NASB)
Comment: Has our understanding of the mechanics of  Koine Greek improved over that of the folks who lived with it, red it, spoke it, and translated it up until 600 AD?  Wouldn't the early church have better known how the word was derived?
Comment: *The term obviously isn't claiming that He is a "people", "race", or "descendant"* - Why isn't the term not referring to his identity as a descendant from God. Is that not what son of God implies?

Could μονογενής mean simply "unique"?


how to stream Webrtc's camera output to the Wikitude sdk on native android

Question: In my webrtc app i am sending the captured VideoStream to a remote browser. My goal is now that i want to augment some AR Stuff with wikitude (i am using Wikitude 7.2.1 for compatibility reasons) on that same videostream on the android part and show it on the display (it currently only shows the camera output). I know that i can't just open another Camera Instance since Webrtc already uses one so i need to pass the output from the webrtc VideoCapturer (or alternatively the contents of some surface that output gets rendered too?) to the Wikitude SDK, however, i have difficulties doing so.
I have seen that the Input Plugin wikitude needs for an external Camera Input uses an ImageReader with an ImageAvaibleListener like this:
mWikitudeCamera2.start(new ImageReader.OnImageAvailableListener() { ... }

and on the WikitudeCamera2 side:
 mManager.openCamera(getCamera(), cameraStateCallback, null);
 mImageReader = ImageReader.newInstance(mFrameWidth, mFrameHeight, ImageFormat.YUV_420_888, 2);
 mImageReader.setOnImageAvailableListener(onImageAvailableListener, null);

that Image Readers surface then gets attached to the actual camera output:
CaptureRequest.Builder builder = mCameraDevice.createCaptureRequest(CameraDevice.TEMPLATE_RECORD);
builder.addTarget(mImageReader.getSurface());

However since i need to use the provided CameraCapturer from Webrtc i am not able to do that.
Is there a way to get the camera stream from maybe the Webrtc's SurfaceViewRender, or render it to another (maybe fake?) SurfaceView and attach that to an Image reader. Or some other way to pass the output that gets sended to the client to wikitude without the use of an Image Reader?
Or generally some other way i could get this to work that i havent thought off. I would really appreciate some help since im stuck on that part as of right now.
Answer: You're on the right path but unfortunately, since the Wikitude's SDK is consuming the camera API, it is not possible (yet, SDK 8.2) to access the same stream via the WebView / WebRTC.
You have to stick to the ImageReader approach for now and find a way to process the frames in a native environment.
Best regards


Set Nautilus as default file manager in Xubuntu

Question: How do I set Nautilus to be the default file manager in Xubuntu? Should I expect any trouble with this choice compared to normal gnome nautilus setup?
Answer: Install the package called exo-utils. Then press Alt+F2 and run the command exo-preferred-applications. In Preferred Applications, click on the Utilities tab, and under File Manager, select Nautilus. (You must install Nautilus first, if you haven't already.)
Comment: The only problem with this is that thunar still loads at startup, I'll make a startup script to correct for that, since thunar shuts down when nautilus starts.
Comment: This way Thunar is also the browser that's opened when a new external drive/memory card is inserted -- I don't know how to stop that happening, but actually removing the thunar package doesn't seem to be an (easy) option as that results in removing the whole xbuntu-desktop package...
Comment: @andybuckley Sorry I didn't reply before. I'm not sure why Xubuntu still opens when new media are mounted; you might want to ask a separate question about that. Regarding removing Thunar: if you're not going to use it, and removing it wouldn't cause any *other* packages (besides `xubuntu-desktop`) to be removed, then you can probably go ahead and remove it without problems. `xubuntu-desktop` is a [metapackage](https://help.ubuntu.com/community/MetaPackages) that doesn't provide any software *directly*. Just reinstall `xubuntu-desktop` before upgrading to the next [X]Ubuntu release.
Answer: I see these are all for older versions, but here is the answer for 12.04+
In menu/settings/preferred applictions (In 12.04) or, I think, in menu/system settings, preferred applications in 12.10+ there is a tab "utilities."
Change the preferred file manager to any in the list you have installed, or type one in (this works for say, Marlin.)
If you get an error and are using a non-native file manager such as Dolphin (KDE) or Nautilus (Gnome) make sure you have "start gnome services" and/or "start KDE services" in the session and startup part of system settings manager, it's in the advanced tab on the far right. Be aware this will start ALL Gnome and/or KDE services so marked for startup.
Comment: In 21.10, open Settings Manager and click System->Default Applications. Select Utilities->File Manager.
Answer: You can also find a good how-to on the wiki page (changing to thunar for example):

https://help.ubuntu.com/community/DefaultFileManager

Where you can change the /usr/share/applications/defaults.list or better ~/.local/share/applications/defaults.list. You can even modify your .desktop files if you want.
Answer: I've used Nautilus safely in Xubuntu under the 11.04 release. Bare in mind however that by default, launching Nautilus may also load several Gnome desktop features.
At the time I used gconf-editor to modify the variable /apps/nautilus/preferences/show_desktop to "false" and /desktop/gnome/background/draw_background to "false". This stopped the desktop showing.
I haven't played with Xubuntu 11.10 yet so I cannot confirm what the experience is like there. I imagine that given the Gnome stack has been bumped to 3.x there will be differences.
I hope this at least has given you an idea of what you might encounter and how to deal with it.


What does the Deduction skill do?

Question: It states in the skill tree "One more move to collect a candy on puzzles". The way i understand this seems to make this skill a PitA. It would take 1 more moves to get a candy?
Comment: Is there a limit to how many moves you can take?  If so, this probably increases that limit by 1, rather than increasing the difficulty.
Comment: If that is so, not yet there but soon will, it's pretty useless for such a high skill. Cause most cases give you between 30 and 40 moves to complete... +1 would be pretty useless, thanks for trying at least!
Answer: I got the skill after much progress in the game and i now understand what it does.
It allows you to see in advance what is the secret candy's requirement. So having it level 1 is really helpful if the requirement is based on a number of evidence. When it is based off clues, it was already pretty simple.
Getting more than a level is IMO useless unless you have finished the game and have nothing left to spend on as 2 moves is more than enough to get practically any evidence...

What does the Deduction skill do?


What is the albedo of a habitable planet?

Question: What is the albedo range for an Earth-like planet? I need to know because it's important for determining the inner and outer edges of the habitable zone. I'm thinking in terms of cloud/water-ratio (I'm assuming that water has a low average albedo of 0.10 and clouds have a high average albedo of 0.70, correct me if these estimates are way off).
Comment: check out Enceladus, Saturn's moon suspected to contain the building blocks for life but it is also one of the highest bond albedo of 0.81 or 81% of all sunlight are reflected back into space for comparison Earth is 0.29
Answer: Simple answer: Somewhere around 0.3 is good.
Longer answer: It really depends on greenhouse effect.
The following is an equation for surface temperature from https://web.archive.org/web/20210605120431/https://scied.ucar.edu/earth-system/planetary-energy-balance-temperature-calculate

Working out the equation with earth's values, we get a global temp of -19 Celsius. This is obviously not earth's temperature, and the equation is wrong because it doesn't account for greenhouse effect at all. Plugging the equation into a graphing software like Desmos, it can be seen that even at 0 albedo, the temp is only 5 degrees (although at ~99% albedo it reaches -200 degrees.) As such, with enough fidgeting with greenhouse gas levels, you can basically make the albedo anything you like, and I doubt any combination of water and cloud/ice would produce an albedo extreme enough that it couldn't be accounted for.
To cool the planet (for albedos <0.3), you could replace some of the ~80% nitrogen (a greenhouse gas) in the atmosphere with something like argon, which allows humans to breathe in the same way nitrogen does (although some nitrogen is necessary for DNA to form), and to warm the planet (for albedos >0.3), you could either increase carbon dioxide or water vapour (water vapour being a potent greenhouse gas).
Basically, don't worry about albedo too much unless your atmospheric composition is already set in stone.
Hope that helps!
Answer: Though I don't know much about this subject but you made me curious. So after a little research I found out the albedo range for an earth-like planet is around 0.07 and 0.8. This is based on the minimum and maximum albedo surface for planets covered by water. The 0.07 represents the ocean and the 0.8 is sea ice + snow cover. Hope this helps a little! ^^
Website:
Astronomy and Astrophysics
Photochemical and Climate Models
Comment: This is potentially a great answer. To improve it you could link to the articles you're referring to.
Comment: Good point I'll do that ^^


How to overcome fear to fail squat?

Question: If one is it to fail the low bar squat, the way to fail it is given as you remove the hand grip and walk forward. , I have a feeling that I'll have an injury because I can't see the weight I'm dropping. Could some advice be given on how to overcome this?
Comment: I'm guessing using safety rails is not an option?
Comment: Sort of is , but may not always be available as the single squat rack can get busy sometimes in my gym @DeeV
Answer: Practice bailing out of the squat with lower weights that you can handle safely and build up to your working weight.
Set up that squat rack with safety rails when you can and practice bailing out with the empty barbel a few times, once you're confident add some weight. This way you'll be prepared when it happens.
Or just don't go to failure when you can't do it safely. Squat with lighter weights for more reps or stop with more reps in reserve and do more sets for the same volume. I wouldn't go doing singles and doubles for PR if I can't safely fail.

How to overcome fear to fail squat?


Can this phenomenon be explained in ground frame?

Question: Suppose there is a disc which is rotating about its centre with constant angular velocity, the surface of disc is rough and there is a small block at some distance away from centre (at rest initially), now if one observes what happens in the foregoing motion to block in rotating frame it can be explained using centrifugal and coriolis force but in ground frame how to explain what force is causing the block to move in a straight path leaving the disc finally?
Answer: TL;DR When both the box and the disc are at rest, there are two (significant) forces acting on the box: (i) weight from the gravitational force, and (ii) normal force from the contact with the disc surface. If disc starts rotating, there are two possible scenarios: (i) if there is no friction force, the box remains at the same position as seen by ground observer; (ii) if there is friction force, the box rotates together with the disc up until the disc (angular) velocity is large enough to overcome static friction force between the box and the disc surface, after which the box leaves the disc.
Below I discuss these two scenarios, both seen by a ground observer. I would suggest always to analyze free-body diagrams (forces that act on the body) from an inertial reference frame. Otherwise, you will just get confused by non-existent forces such as a centrifugal force.

No friction between box and disc surface
The only two forces acting on the box are the gravitational force (the weight) exerted by the Earth and normal force exerted by the disc surface. Since the box does not move in vertical direction, these two forces are equal in magnitude and opposite in direction. Even if the disc rotates, the box remains at rest, because there are still only two forces exerted on the box.

There is friction between box and disc surface
Imagine both box and disc rotate at some angular velocity $\omega$. There are three forces exerted on the object: (i-ii) weight and normal force as already discussed, and (iii) static friction force exerted by the disc surface in direction parallel to the surface. Since the box rotates about an axis that goes through disc center, the net force on the box is (always) directed towards the center of rotation (study uniform circular motion to understand why):
$$\vec{F}_\text{net} = m \frac{v^2}{R} \hat{r} = m \omega^2 R \hat{r}$$
where $v$ is linear velocity, $\omega$ is angular velocity, $R$ is distance of the box from the center of rotation, and $\hat{r}$ is a unit vector that points towards the center of rotation. This net force equals vector sum of all forces that are exerted on the object:
$$\vec{F}_\text{net} = \vec{w} + \vec{n} + \vec{f}_s$$
where $\vec{w}$ and $\vec{n}$ are weight and normal force which cancel as already discussed, and $\vec{f}_s$ is static friction force. From this it is obvious that the static friction provides a radial force component which makes the box rotate together with the disc. In other words, the static friction opposes relative motion between the box and the disc surface.
However, the static friction force has a maximum value it can take which depends on the coefficient of static friction $\mu_s$. If the velocity ($v$ or $\omega$) is too large such that the net force is larger than the maximum static friction force, the box starts slipping on the disc surface and eventually falls off.
Comment: Sir consider the situation of friction being there then is it possible to explain the effect in ground frame ?
Comment: Understood Sir thanks
Answer: the surface of disc is smooth and there is a small block at some distance away from centre (at rest initially) means that in the ground frame the box does not move.
In the rotating frame which travels with the disc the box will be seen to undergo a circular motion with acceleration $r\omega^2$ towards the centre of rotation of the disc.
In the accelerated (relative to ground) rotating frame one would include a fictitious centripetal force of magnitude $mr\omega^2$ towards the centre of rotation if one wanted the circular motion of the box to be consistent with Newton's laws of motion.
Answer: what force is causing the block to move in a straight path
It's not a force, it's the absence of force. By Newton's first law, an object in motion will naturally continue to move in a straight line. The only reason the block moves in a circle in the first place is because friction provides sufficient centripetal force to keep the block moving in a circle. If the disc spins fast enough, friction cannot provide sufficient force to keep the block moving in a circle, at which point it slips. It is the sudden absence of the frictional force that results in the block moving in a straight line.


Meaning of Simplify's tolerance parameter

Question: The standard parameters of Douglas-Peucker's simplify algorithm are geometry and tolerance (e.g ST_Simplify in PostGIS). What's the meaning of the tolerance parameter? I know that the bigger the value, the coarser the geometry will be. But does the number has any unit or it is just arbitrary?
Answer: The tolerance is a distance.  Roughly, any "wiggles" in a curve that vary from a straight line by less than this amount will be straightened out.  The algorithm finds the most extreme wiggles that exceed the tolerance, pins down the points where they deviate the most from a straight path, and then recursively applies itself to the arcs between the pinned-down wiggles.
The tolerance must be expressed in the same units used by the software to execute the algorithm.  (This will depend on whether it uses the coordinates as stored or as projected "on the fly" for display or analysis.)  An illustrated description appears in the Wikipedia article on the Douglas-Peucker algorithm.
Comment: Usually, this tolerance parameter value should be equal to the target resolution.
Comment: @julien That makes sense.  I always like to understand the reasoning behind such rules of thumb, because in my experience many such conclusions don't hold up when more closely examined.  (And it's delightful to be surprised like that: you always learn something.)  Could you therefore share your thoughts about why the tolerance ought to equal the "target resolution" (and what exactly that resolution is)?
Comment: Great explanation, +1.
Comment: How to **check sensibility**? Testing `ST_Simplify(geom,0.00001)=geom` is false.. But even `ST_AsGeoJSON(geom,3)=ST_AsGeoJSON(ST_Simplify(geom,0.0000001),3)`  is false!!
Answer: Didn't see it in the link you posted but found this:

The units of tolerance are the same as the projection of the input geometry.

http://revenant.ca/www/postgis/workshop/advanced.html#processing-functions
Comment: I see. But what does it exactly means when say the geometry is WGS84 (lat/lon) and set the tolerance to 1.0? Error within 1 degree? Still a bit confused.
Comment: I wouldn't say "error within 1 degree" ... take a look at the wikipedia page for the algorithm, there's a nice graphic and good explanation there:  http://en.wikipedia.org/wiki/Ramer%E2%80%93Douglas%E2%80%93Peucker_algorithm
Comment: How to **check sensibility**? Testing `ST_Simplify(geom,0.00001)=geom` is false.. But even `ST_AsGeoJSON(geom,3)=ST_AsGeoJSON(ST_Simplify(geom,0.0000001),3)`  is false!!
Answer:
All points in the simplified object will be within the tolerance
distance of the original geometry.

from https://shapely.readthedocs.io/en/stable/manual.html


is coin-tossing an instance of secure computation (MPC, 2PC, etc.)?

Question: Coin-tossing protocol is defined as a protocol between at least 2 non-trusting parties that allows them to obtain an unbias random bit in such a way that no involved party can influence the output. Hence, does it pertains to the notion of secure computation?
Answer: Yes it does.
Here is a brief summary from https://eprint.iacr.org/2009/214.pdf:
When a majority of the parties are honest, efficient and completely fair coin-flipping protocols are known as a special case of secure multiparty computation with an honest majority (assuming a broadcast channel) as in

M. Ben-Or, S. Goldwasser, and A. Wigderson Completeness theorems for non-cryptographic fault-tolerant distributed computation. STOC 1988.

When an honest majority is not available, and in particular when there
are only two parties, the situation is more complex. Blum's two-party coin-flipping protocol from 1982

M. Blum. Coin flipping by telephone - A protocol for solving impossible problems. 25th IEEE Computer Society International Conference, 1982.

guarantees that the output of the honest party is unbiased only if the malicious party does not abort prematurely (note that the malicious party can decide to abort after learning the result of the coin
flip).
This satisfies a rather weak notion of fairness in which once the malicious party is labeled as a "cheater" the honest party is allowed to halt without outputting any value. Blum's protocol relies on any one-way function and Impagliazzo and Luby (see linked paper) later on showed that one-way functions are in fact essential even for such a seemingly weak notion.
Comment: looks sciency, but its actually an album cover of the pop band Joy Division


Doesn't look like an automorphism to me...

Question: I am self-studying algebra using Dummit and Foote.  One early exercise asks me to prove that for each fixed nonzero $k \in \mathbb{Q}$, the map $\varphi:\mathbb{Q} \to \mathbb{Q}$ defined as $\varphi(q) = kq$ is an automorphism.  It's certainly a bijection, but I am having difficulty persuading myself it is a homomorphism.  For let $p,q \in \mathbb{Q}$.  Then
\begin{align}
\varphi(pq) & = kpq
\end{align}
while
\begin{align}
\varphi(p)\varphi(q) & = k^2pq,
\end{align}
and unless $k = 1$, these are not equal.  Evidently I'm missing something obvious, but I can't figure out what.
Comment: The group you're asked to do prove should be $(\mathbb{Q},+)$ and not $(\mathbb{Q}\setminus\{0\},\times)$?
Comment: Well, that's clearly the obvious thing I was missing.  Thanks, all.
Comment: You are right, this is not a $\mathbb{Q}$-automorphism if we're talking about field automorphisms. Maybe they were talking about group automorphisms if you see  $\mathbb{Q}$ as a group under addition.
Comment: As a group, under addition, this is an automorphism, since scalar multiplication distributes over addition, etc. However, you're correct in that it is **not** a homomorphism when lookin gat $\mathbb{Q}\setminus\{0\}$ under multiplication, since for any $k\ne 1,$ we have $k1\ne 1,$ and so it does not preserve the identity of $\mathbb{Q}\setminus\{0\}.$
Answer: I presume the question implicitly means for the semigroup operation to be addition, not multiplication.


How do I curve a textbox on the lower part of a circle without the text having to be upside down, in Illustrator CC?

Question: I used the "type on a path" to make a curved textbox on the top of a circle but when I want to have the text curved on the lower part of the circle it has to be upside down.. Any help is appreciated.
Comment: You can type on the other side of the path (inside the circle). If I remember correctly you can just drag the curser to the other side of the path.
Answer: Use the 'type on path' tool to type out your words on the circle, then while you have that circular type selected with the selection tool, double-click the 'type on path' tool to pull up the 'type on path' options dialog box. There you will see a dropdown menu and a checkbox. Check the box to invert the text (something like that), then play around with the 'ascender/baseline/etc.' options in the dropdown menu to get the text on the outside/inside/etc. of your circle.


What is the difference between "Factory New" and "Minimal Wear" items?

Question: I was wondering what the difference (in looks) was between 'factory new' and 'minimal wear' in Counter Strike: GO.
Answer: A random exterior quality is chosen when a weapon skin is dropped, uncrated or received from a Trade Up Contract. The exterior quality simulates randomized wear and tear on the skin, and is determined when the skin drops. A weapon's exterior quality will not degrade over time and can never be changed.
The exterior quality hierarchy is the following, from best to worst:

Factory New
Minimal Wear
Field-Tested
Well-Worn
Battle-Scarred

Source: http://counterstrike.wikia.com/wiki/Skins
Overall you can say that Factory New looks best and the skin quality will get worse from quality to quality.
Answer: Factory new is a fully clean gun while the minimal wear has a few scratches but not really anything significant. If the prices vary too much, I would recommend getting the minimal wear as it is cheaper and still looks as good.
Comment: Thats not 100% correct, a factory new weapon can still have a few scratches and sometimes they can look worse than minimal wear


Can we add user stories in the middle of an Iteration, and what is the likely impact of doing so?

Question: Development of planned user stories is complete. Testing is pending as a part of our Definition of Done (DoD). Can the Product Owner still accept new user stories for the Iteration? What are the scenarios to consider when adding or removing user stories to the current Iteration, and how should we account for this when measuring velocity?
Comment: we are using scaled agile framework with tightly coupled Definition of done where our user stories are blocked for testing and await approval from PO, past iteration closure. I understand that user story is a Boolean whether it is accepted or not but what to do in these scenarios with the story point deviation ( all the User stories are above described pattern in each iteration) need a suggestion to handle the story points
Comment: I've added the SAFe tag to your question, but it's obvious from your comment that your company has implemented SAFe improperly or incompletely. You need to address issues like this with your Release Train Engineer right away, assuming you have one. If not, I hope your parachute is gold-colored.
Answer: TL;DR
You didn't tag your question with any specific framework, but "Product Owner" is generally related to the Scrum framework, so that's how I will answer your question. There are some minor edge cases, but in general the answer to your question is that "new work" is only selected during Sprint Planning, so regardless of when the Product Owner adds items to the Product Backlog there should generally be no impact to the current Sprint.
Accountability for Product and Sprint Backlogs
The Product Backlog
The Product Owner can accept Product Backlog Items for the Product Backlog at any time. While the whole Scrum Team generally participates in Backlog Refinement activities, when and how to accept items for that particular artifact is the Product Owner's responsibility. Remember, the Product Owner is a single-person role accountable for the contents and ordering of the Product Backlog. While that person can delegate or collaborate with others the Product Owner alone is ultimately accountable for how the Product Backlog is managed.
The Sprint Backlog
The Sprint Backlog is owned by the Developers. It represents the plan the Developers have created to meet the current Sprint Goal. Since the Sprint Goal must not change during an iteration unless the goal becomes obsolete and an early termination is declared by the Product Owner, new items from the Product Backlog should be considered as potential work for future Sprints, not candidates for adding post facto to the current Sprint.
Because the Sprint Backlog belongs to the Developers, the Product Owner cannot put new work into the Sprint Backlog, nor should any Sprint Backlog items be added by anyone but the Developers as they discover or decompose work needed to meet the Sprint Goal the Scrum Team defined during Sprint Planning.
Pragmatic Summary
In other words, it's up to the Product Owner to decide how and when user stories (or Product Backlog items in any format) are added to the Product Backlog, but the Product Owner may not add additional work items to the Sprint Backlog, change the scope of Product Backlog items already accepted into the Sprint, or modify the Sprint Goal in any way without collaborating closely with the Developers. The Sprint Backlog represents a plan with a singular goal, and Sprint Planning is the appropriate event for creating new plans or new goals.
The Product Backlog is a queue for future work, and that queue is selected from by the Developers during Sprint Planning in order to meet a singular Sprint Goal that contributes to the current Product Goal. As a general rule, this should happen only during Sprint Planning. Unless the Product Owner wishes the Scrum Team to abandon work in progress and restart the planning process from scratch then they should not move the goalposts while work is already underway. This is an important facet of Scrum precisely because it makes the cost of unplanned change visible to the Scrum Team and to the rest of the organization.
Don't set your process up for failure by treating ongoing changes to the Product Backlog as cascading changes to the Sprint Backlog. I will again stress that there are edge cases, but for the most part if you have to ask the question then the answer should be "new work can only be selected during Sprint Planning." Anything else is likely to lead to scope creep, lack of focus on a cohesive Sprint Goal, the inability of the Developers to commit to meeting the Sprint Goal, or frequent failures in delivering planned Increments on a reliable cadence.
Answer: Sprint starts with a fixed set of stories. You should not add new stories to the sprint in the middle of it. If the work is completed earlier than expected, then it means that the estimations are not correctly provided or there's something else affecting the velocity. This must be investigated at the retrospective and addressed for future sprint estimations.
Answer:
Can the Product Owner still accept new user stories for the Iteration?

Yes they can, but it is not an ideal situation.
The issues with adding stories to an in-progress sprint include:

It may invalidate or disrupt the planning the team has done for the sprint
It may push the work in the sprint beyond the capacity of the team to deliver
Context switching to accomodate the new work may reduce the time available to deliver value
It may impact on other planned work which in turn may result in stakeholders being disappointed

I would say there are three common scenarios when work is added to an in-progress sprint:

The story is small and has little impact on the sprint
The story is medium sized and the team has to remove other planned work to accomodate it
The story is large and the team is forced to replan or restart the sprint

how should we account for this when measuring velocity?

Velocity is a measure of the work that the team gets done in a sprint. If adding new stories during a sprint causes disruption then it is likely the team will get less done. This will then result in a reduction in velocity.
It may be tempting to ignore the reduction in velocity and treat is as unusual circumstances. However, if work is regularly being introduced to in-progress sprints I would suggest you accept the impact on the team's velocity and reflect this in your team's reporting. You might say something like:

The team's velocity has dropped from 25 points to 22 points as a result of work being brought in to in-progress sprints and the disruption this has caused.


Android Development

Question: Is it possible for a program to mimic the Android Homepage?  Can widgets, contacts, and canvases all interact at the appropriate FPS?  Is it possible for an app to make calls and IMs?  Or are the ISOs used to make the original too different from Java to mimic correctly?
Comment: Too many questions in one question. What research have you done? Try to focus your question and you'll get better results.
Answer: To a large extent yes . For a start have a look at the home app sample provided by google . http://developer.android.com/resources/samples/Home/index.html .
Once you get your home app working perfectly you can add modules which practically do everything , but a better design strategy is to split the modules which do different work , so you calling module can be a different app.
I dont think there should be a fps issue if you have good code :) .

Android Development


Is the density of the unnormalised posterior distribution the same as the density for normalised posterior?

Question: The posterior distribution is proportional to the likelihood times the prior distribution $p(\theta|D) \propto p(D|\theta)p(\theta)$. Computing $p(D|\theta)p(\theta)$ would give the un-normalised posterior distribution and the normalising constant gives rise to a valid probability distribution of 1. However, the normalising constant is just a scaling factor for the height of the distribution and it does not in any way affect the density of the distribution because the height of the posterior are all scaled proportionately. What is the purpose of techniques like MCMC sampling to estimate the posterior distribution ?
Answer:
Computing p(D|θ)p(θ) would give the un-normalised posterior

That sounds as if MCMC was an alternative to compute the posterior distribution. However, MCMC starts to shine whenever p(D|θ)p(θ) is to complicated and cannot be computed (with reasonable effort).
Bayes books often start with simple cases where there is a conjugate prior and the result can be computed in closed form. In many real life cases such a conjugate prior does not exist or is for other reasons not feasible. Then Monte Carlos simulation can still produce usable results. It is not primarily a technique to determine scaling factors.
Comment: does the density of the posterior remain the same for the un-normalised posterior, because it is just a scaling factor applied. I have read that $Z$ does not need to be computed if we want a MAP estimate of $\theta$, because the $\theta_{MAP}$ does not shift when scaled. However, wouldn't all $\theta$ values have the same density as the un normalised posterior since the height of the distribution is just scaled by some constant factor ?
Comment: I am not quite shure how much of that I understand. For the point estimate no scaling is necessary. Maybe this explains the difference well https://towardsdatascience.com/mle-map-and-bayesian-inference-3407b2d6d4d9
Comment: In the sense that the normalising constant only scales $P(\theta|D)$ by a factor of $Z$. Since all values are scaled, the density for the unnormalised posterior is the same as the normalised posterior ?
Answer: MCMC methods approximate the posterior by drawing samples from it. Using these samples, one can

form an empirical approximation to the true posterior
calculate certain statistics like posterior mean/variance/moments
numerically calculate hard integrals

and so on. So, finding the posterior might be seen a byproduct of what you really want to achieve via MCMC.
Analytically, if you have the unnormalized version of a density, i.e. $p(x)=cq(x)$ where $c$ is unknown, integral calculations (moments) will include $c$ as well:
$$\mathbb E[X]=\int xp(x)dx=c\int x q(x)dx$$
Even if you can easily evaluate $\int x q(x)dx$, you still need to find $\int q(x)dx$.


Where to Learn Cycles Material Nodes from basics

Question: Is there any video tutorial series learning cycles material nodes. Where can i learn material nodes for cycles right from the basics. What are different nodes and in what order i can connect. What effect i get to what node i connect. etc. All the video on Youtube i have seen are base on topic. For example how to give Steel effect or plastic,etc. I did not find a complete series on cycles node material. Any help or suggestion where can i learn. Thanks
Comment: check out Grant Aabbitt's youtube channel. Node school tutorial series, where he goes through explaining how to construct an entire node tree from scratch using principle BSDF. You can also check out blender insight youtube channel for more in-depth videos on cycle's nodes.
Answer: Each node is very basic and very descriptive. For example, A Mix node mixes 2 input based on a factor value, a SplitXYZ node splits the input into XYZ. A simple online search should help if you still need clarification.
I think what you need to learn is the high level concept of shading and texturing. Look up "PBR material" basics, That should get you started learning about how the shaders behave, how inputs such as albedo, normal, roughness affects the surface, and how to create steel, plastic, etc.
I also suggest you focus on learning about the principled shader(as oppose to individual shaders such as glossy or diffuse), as it's the most artist friendly, complete, and industry-accepted pbr shading model.
Comment: Thanks but i want to learn each nodes right from the basic how can i connect the nodes in order and what effect i get if i add some nodes. Some time i watch video they just show like professional add this node then color ramp etc. But im confuse when color ram or light weight etc node used and each node have small options. I have search but did not get help or video tutorial how to start from basic.
Answer: I suggest you to take a look at blender Manual Docs. There you would description of each and every node and you can learn more things there as well.


Can someone explain how to program i2C communications without using the library?

Question: I am starting to get into i2C, SPI, and other forms of communication. I have read about them and have a fuzzy idea of how they work electrically, but can someone explain how I would turn the electronic portion into the coding/controlling portion? I really want to get a great understanding of it.
Comment: If you have Arduino Uno, you can try my free I2C terminal. I2C communication is quite complex. Check out https://iobtoolkit.com/s/i2c/
and https://www.youtube.com/watch?v=4x3NNWLgtTU .
The idea is that you hook up I2C device and you can chat with the device using a console, exploring the protocol and device functions, without writing a single line of C.
Comment: `I really want to get a great understanding of it.` - reading the existing library would help (`Wire.cpp` and `twi.c` amongst other files).
Answer: The AVR microcontrollers that Arduino uses have built-in hardware support for the said protocols.
That means that you don't have to write program to control the transmission/reception process but you only need to configure the appropriate registers and the hardware will initiate transfer and inform you when it's done.  I suggest you carefully read the datasheet of the microcontroller you are using with your Arduino and you should find all the information there.  It even has code examples, so you can actually copy-paste functions and use them in your program, here is one of the examples:
<code>void SPI_MasterInit(void)
{
    /* Set MOSI and SCK output, all others input */
    DDR_SPI = (1<<DD_MOSI)|(1<<DD_SCK);
    /* Enable SPI, Master, set clock rate fck/16 */
    SPCR = (1<<SPE)|(1<<MSTR)|(1<<SPR0);
}

void SPI_MasterTransmit(char cData)
{
    /* Start transmission */
    SPDR = cData;
    /* Wait for transmission complete */
    while(!(SPSR & (1<<SPIF)));
}
</code>
The keywords with the capital letters are usually register and bit names: you can look them up in the datasheet.
Comment: Would that work for I2C?
Answer:
Can someone explain how to program i2C communications without using the library?

The AVR has hardware support for this but you can implement it with the pin functions. I2C is a lot about timing the signals and detecting the reply signal from the slave.
The hardware support for I2C requires understanding of a complex state-machine. If you have implemented the I2C protocol it becomes easy to understand the hardware support and how this help remove a lot of work from the processor core. The state machine, control and status registers and interrupts, are described in the AVR product specification. There are also lots of excellent descriptions on the web. Do not miss the AVR application notes for the I2C. Most of the code in the Arduino core originated from them.
Cheers!
Comment: So its based all on timing, and hardware application? Sounds like it has to almost be like clockwork
Comment: I2C hardware is present in almost any modern microcontroller. It's all about writing a driver for it. Hardware interface will take care of low-level details, such as clock generation and timing.
Answer: A good way to learn how to code for these interfaces is to look at the source code for the libraries themselves. In the PC world, these will be in the directory
Program files (x86)\Arduino\hardware\arduino\avr\libraries


Straight, dashed horizontal and vertical lines in GIMP

Question: I need to create a horizontal and vertical dashed line. I used the paths tool and then I edit the line to do it dashed. The problem is that I set the two ends of the line manually and I cannot be sure if the line is perfectly horizontal/vertical. Is there a way to set the coordinate of the two points defining the line?
Comment: Have you tired configuring the grid, enable the grid, then enable snap to grid?  It works with the pen tool.
Answer: You can set guides, that will help you put the two extremities on the same horizontal/vertical line.
Another solution is to do a rectangle selection where the edges of the rectangle are the lines you want to create. Then either:

Edit>Stroke selection (with a dashed line)
Remove the unwanted parts (eraser)

Or

Select>To path (makes a path with 4 points, where everything is perfectly horizontal and vertical)
Delete the sides and anchor point you don't need (Ctrl-Shift-click on them in the path editor)
Select>None (otherwise only half of the line will be drawn)
Edit>Stroke path (with a dashed line)
Answer: You can use Ctrl after you press Shift to use Angle Lock (which works at 15 degree increments). It works very nicely for vertical and horizontal lines.


What is normal flow coming from icemaker tubing?

Question: Could someone give me a description of what normal flow of water should be coming from water supply through icemaker tubing. It is 1/4 od plastic tubing that is traveling approximately 10 feet from water supply. What I mean is should water be coming out in a steady stream, fast drip, slow drip, etc?
Comment: It should be a solid Pee many fridg's use saddle valves to make the connection, this cheap method of connection has been the cause of almost all the low flow , no flow and leaks that I have found over a few decades.
Comment: I would say mine is a slow-medium drip. It would take a while to fill a cup.
Answer: It should come in a steady stream. Some refrigerator user manuals state "filling a cup of water in 10 seconds".
Typically it is recommended (or even stated) to use copper tubing and not plastic. I do not know if this has any real impact on flow, especially with just 10' away.
Comment: Plastic lines are fine, pex is plastic many of the supplies I have installed have a plastic liner in stainless mesh , I also use soft copper tubing but sometimes the plastic lines are more feasible , most the problems 95%+ has been with the saddle valves used to tap the lines only a few % have been the solenoid on the fridge.
Comment: If you have a saddle valve supplying the water, try turning the saddle valve past the closed point. When you close the valve, it has a sharp end that punctures the copper. You may need to turn the saddle valve a little further closed to pierce a larger hole in the copper it is mounted on.
Comment: There was a kink in part of line that was under house. I cut it out of line and redid fitting and have good flow now.


Please identify this ceiling texture

Question:
Please identify this ceiling texture
Thank you!
Comment: Which stipple brush will match my pattern?
Comment: Is this in a manufactured home?
Answer: OK, i can identify it.
It is That ugly texture for the 70's that the neighbor lady talked my mom into letter her put on my bedroom ceiling.
She also did one on the living room ceiling that was like a swirl, she would dip a brush in mud, stick it on the ceiling (poke the ceiling)  and turn it before pulling straight away from the ceiling.
I think she had stock in the mud company because it uses a lot of mud.
I also had red white and blue shag carpet in 1976. And you wonder where i get my warped sense of humor.

Which stipple brush will match my pattern?

One possibility is they used a straight brush, loaded it with mud and poked the ceiling, pulled straight back and then turned the brush and poke and turned the brush and poked and tur- - - ked.
Answer: That is a stippled ceiling.
Supporting link
Comment: Could be a round supple brush or sponge matching textures is one of the hardest things to do in my opinion.
Answer: It reminds me of a plastic bag texture - where the plaster is spread on smooth-ish, then a sponge or cloth is put inside a clean plastic bag, lightly touched to the wet plaster, and then pulled straight down.
I'd suggest doing a practice run on a scrap board, both to confirm the right technique and to get your hand in.
Or don't - stippling is no longer trendy.  For a modern look, scrape it all off with a hoe, sand flat and prime/paint.  This is a super messy task.

Please identify this ceiling texture


Can a quotient field ever be finitely generated as an algebra?

Question: If A is a commutative integral domain that's not a field, and let $K$ be the quotient field of A. We know that $K$ is not finitely generated as an A-module. But can $K$ ever be finitely generated as an A-algebra?
Answer: Sure. Let $A$ be the integers localized at $(2)$; that is,
$$A = \left\{ \frac{a}{b}\in\mathbb{Q}\;\Bigm|\; a,b\in\mathbb{Z}, b\gt 0, \gcd(a,b)=\gcd(2,b)=1\right\}.$$
The field of quotients of $A$ is $\mathbb{Q}$, and is equal to $A[\frac{1}{2}]$, so it is generated as an $A$-algebra by $1$ and $\frac{1}{2}$.
More generally, any UFD $R$ with only finitely many pairwise non-associated irreducibles will have a field of fractions that is finitely generated as an $R$-algebra: just take $m_1,\ldots,m_k$ to be a maximal list of pairwise non-associates irreducibles in $R$, and the field of fractions will be equal to the subalgebra $R[\frac{1}{m_1},\ldots,\frac{1}{m_k}]$.
Added. In fact:
Theorem. Let $R$ be a UFD. The following are equivalent:

The field of fractions $K$ of $R$ is finitely generated as an $R$-algebra.
$R$ has only finitely many pairwise non-associate irreducible elements.

Proof. If $R$ has only finitely many non-associate irreducible elements (possibly $0$), $m_1,\ldots,m_k$, then the subalgebra $R[\frac{1}{m_1},\ldots,\frac{1}{m_k}]$ in $K$ equals all of $K$: every element of $K$ can be written as $\frac{a}{b}$ with $a,b\in R$, $b\neq 0$, and $b$ can be factored into irreducibles $b=um_1^{\alpha_1}\cdots m_k^{\alpha^k}$, where $\alpha_1,\ldots,\alpha_k$ are nonnegative integers and $u$ is a unit of $R$. Then
$$\frac{a}{b} = au^{-1}\left(\frac{1}{m_1}\right)^{\alpha_1}\cdots\left(\frac{1}{m_k}\right)^{\alpha_k}\in R\left[\frac{1}{m_1},\ldots,\frac{1}{m_k}\right].$$
Conversely, suppose that $K$ is finitely generated as an $R$-algebra. We may assume that the set that generated $K$ is made up entirely of fractions of the form $\frac{1}{b}$, because any element $\frac{a}{b}$ can be replaced with $\frac{1}{b}$ and we still get the same $R$-subalgebra. Moreover, we can assume that $b$ is irreducible, because if $b=m_1^{\alpha_1}\cdots m_r^{\alpha_r}$, then we can replace $\frac{1}{b}$ with $\frac{1}{m_1},\ldots,\frac{1}{m_r}$. Thus, we may assume that $K$ is generated as an $R$-algebra by the multiplicative inverses of a finite set of pairwise non-associated irreducible elements of $R$, $m_1,\ldots,m_k$. Now let $m\in R$ be any irreducible. We can express $\frac{1}{m}$ as a sum of multiples of powers of the $m_i^{-1}$, so we have
$$\frac{1}{m} = \frac{a_1}{m_1^{\alpha_1}} + \cdots + \frac{a_k}{m_k^{\alpha_k}} = \frac{b_1a_1+\cdots+b_ka_k}{m_1^{\alpha_1}\cdots m_k^{\alpha_k}}$$
where
$$b_j = \frac{m_1^{\alpha_1}\cdots m_k^{\alpha_k}}{m_i^{\alpha_i}}.$$
Then we must have that $(b_1a_1+\cdots+b_km_k)m = m_1^{\alpha_1}\cdots m_k^{\alpha_k}$, so $m$ divides $m_1^{\alpha_1}\cdots m_k^{\alpha_k}$, and hence $m$ is an associate of one of $m_1,\ldots,m_k$. Thus, $R$ has only finitely many pairwise non-associate irreducibles, as claimed. $\Box$
Comment: @Pete: Take a finite set $\frac{a_1}{b_1},\ldots,\frac{a_r}{b_r}$ of elements of $K$ that generate $K$ as an $A$-algebra. If these elements generate, then so do $\frac{1}{b_1},\ldots,\frac{1}{b_r}$, because we may obtain the original list from these by suitable multiplication by elements of $A$. If these elements generate, then we can replace, say, $\frac{1}{b_1}$ by $\frac{1}{m_1},\ldots,\frac{1}{m_r}$ as in the argument, because from these elements we can obtain $\frac{1}{b_1}$ by suitable multiplication of powers of elements of the new set. Am I making a mistake somewhere?
Comment: @Pete L. Clark: I'm unclear on something about your comment: what does "every overring is a localization" mean? $\mathbb{Z}$ is Dedekind with torsion Picard group, so this would imply that the overring $\mathbb{Z}[x]$ is "a localization"... not of $\mathbb{Z}$, surely, so of what? Or do you mean "every overring inside the field of quotients"? In any case, I'm not dealing with *every* overring, just the full ring of quotients, which can always be obtained by localizing at the nonzero elements, $S^{-1}D$ with $S=D-\{0\}$.
Comment: sorry, the term "overring" has a technical meaning in this context which is not what one would otherwise guess: an overring of a domain is ring intermediate between that domain and its field of fractions.  But about your argument --- oops!  You're right.  I somehow forgot that we were looking at the full fraction field.  I apologize for the (i.e., my) confusion.
Comment: @Pete: Phew! Thanks for the clarification.
Answer: In fact, the class of such $A$ is not only nonempty, but important enough to have a standard name: Goldman domains.  See page 117 of Pete L. Clark's notes on commutative algebra here.
Comment: Funny, I was going to say something very similar.  Oh, well: +1.


plane bounding box intersection

Question: I am developing a visualization where I have already calculated the best fit plane from points and have a bounding box (from the voxel structure). The plane is defined by its normal and the centroid and the bounding box by center and the size. I have to specify the end points of the plane. Is there any easy way to find out the bounds of the best-fit plane?
Comment: So you're after the corners of the rectangle resulting from intersecting the plane $(c, \vec{n})$ with the bounding box $(p_1,$ $p_2,$ $p_3,$ $p_4,$ $p_5,$ $p_6,$ $p_7,$ $p_8),$ right? Now let $\{ p_1, p_2, p_3, p_4 \}$ be the upper corner points of the bounding box. I'm a bit confused, but here's an idea not necessarily the easiest: project $\{ p_1, p_2, p_3, p_4 \}$ onto the plane. Doesn't this give you the corners of the intersection rectangle? (edit: Ops, I've just noticed that your (axis-aligned) bounding box is *not* defined in terms of its corner points.)
Comment: Thanks for your response. Yeah the bounding box is defined as (center, size). But I can think about it.
Answer: The intersection of the plane and the bounding box is going to be a polygon that has between three and six sides (I'm ignoring some special cases that are probably not interesting).
It's not clear what you mean by the "bounds" of the plane.
One interpretation would be that the "bounds" of the plane are just the edges of the polygon mentioned above. You can find these edges by intersecting the plane with each of the faces of the bounding box.
If you want a simpler bounding shape, you can construct this from the aforementioned polygon.

plane bounding box intersection