Is it possible to have frozen clouds floating in the sky?

Question: Not really a spoiler, but in the movie Interstellar there was a planet that had frozen clouds. Is this actually possible? I know ice is less dense than water and so can float, but I'm having a hard time maintaining suspension of disbelief in this case as you would need a solid that was less dense than a gas AND was solid at the same temperature as the gas.
Comment: What you're looking for are [hailstorm clouds](https://en.wikipedia.org/wiki/Hail), I think.
Answer: Yes, there are high-altitude clouds made of ice crystals, for example cirrus clouds. You do not need to fulfill any condition of having a lower density than surrounding air, much like regular clouds (composed of water droplets) do not require the density of water to be lower than that of air. The trick is that the individual droplets or crystals are so small that they do not fall at a significant speed (which can be 100 mph updrafts in the center of thunderstorm clouds, supporting huge hail-like balls of ice).
Comment: Given the explicit reference, the question seems to be about ice clouds solid enough to put a dent in a spacecraft that flies into them.
Comment: Yeah, in the movie noticeable chunks of ice shattered off of the cloud. I imagine a cirrus cloud would be a dusty puff of ice crystals (much like kicking snow that is too powdery to stick) when the star ship flew through it.
Comment: I think [hailstorm clouds](https://en.wikipedia.org/wiki/Hail) are a better example of icy clouds with very solid projectiles inside.
Answer: I think the answer is, in principle, yes, although in practice no.
First of all what you are not talking about, I think, is clouds of ice crystals: these are, as other answers have pointed out, common, but they are not what was depicted in the movie because they are not in any sense solid, they are just aerosols where the particles are ice.  I'm also going to claim that the cloud doesn't need to be water-based: it just needs to be some kind of solid.
What you are after, in fact, is a cloud which is an aerogel.  Aerogels are essentially made from gels by removing the liquid component, leaving a very low-density matrix.  They are really odd materials.  As far as I know you can't make aerogels from ice (the liquid component would have to be ethanol or something), although you might be able to: that's why this answer only works if the clouds don't need to be water-based.
Well, aerogels are denser than air, if only just, so they don't look like candidates.  But I can get around this using a couple of tricks.  First of all construct the thing so it has huge drag (I will leave the details to my aeronautical engineering minions -- my role here is merely to be the mad scientist who bosses people around): if it falls it will now fall only very slowly.  Secondly heat it: it will now trap a lot of hot, low-density air.  The resulting thing is essentially a solid hot-air balloon.  It will rise, and then when it cools it will fall extremely slowly.  It may in fact be kept aloft by thermal air currents or by heating from the Sun (it may be necessary to paint it black, sadly).  It may need to be hollow, to exploit surface-to-volume gains as it becomes larger (a completely solid cloud of dimension $r$ has mass which goes like $r^3$, a hollow one has mass which goes a bit more than $r^2$ in the usual way).  If it is large enough it will be able to support the evil genius lair which I will construct on top of it, of course.
I think, therefore, that this is possible in theory.  In practice there are some problems: would it be strong enough to maintain integrity?  would it be possible to construct a cubic mile of aerogel?  could it arise naturally?  The answers to all these questions are probably no.  But I hope they are yes, because I want one.
Comment: "The answers to all these questions are probably no. But I hope they are yes, because I want one."  Wait... you want one Rabid Matt Damon's Planet?!? You ARE a mad scientist!
Answer: As noted above, yes, ice crystals are present in high-altitude clouds such as cirrus and cirrocumulus, but importantly, these clouds are by no means solid, like the ones in the movie. Real ice clouds are incredibly diffuse and composed of tiny free-floating particles - therefore they would present no more of a physical obstacle than a regular cloud.


What ways can I get Tower Bux?

Question: In what ways can I get Tower Bux in Tiny Towers (besides buying it with RLM)?
Answer: You can get Bux by:

Locating a person when someone asks you to find them: 1 Bux (this is easiest when you have very few floors and not worth the time when you have very very many floors)
Fully Stocking a floor (most of the time you just get coins): 1 Bux
Tips from bitizens using the elevator (usually coins, again): 1 Bux
Employing a bitizen in his or her dream job (always, the first time per bitizen only though): 2 Bux
Completing missions: Rate and task depends on the mission
Birthdays: Your bitizens have birthdays, and on their birthday you get a Bux: 1 Bux

The easiest way to get lots of bux, if you have a large number of residential/commercial floors, is to evict bitizens that don't have their dream job and pay Bux to get new bitizens to replace them. If you have the majority of different floor types, more often than not these bitizens will be able to give you 2 bux, netting 1 bux after losing one to instantly move in the new bitizen. Even if you have all skill level 9 employees in their dream job, you can still swap out one of your level 9 employees for the new one and then immediately put the first one back again, evicting the inferior employee, but still getting your 2 bux for giving them their dream job. In this way you can slowly but steadily accumulate bux, as long as you have empty apartment slots to move bitizens into. This is why the wiki recommends keeping as many residential floors as possible with only four residents, as any bitizen who arrives at a floor with vacancies will move in and save you the one bux you would otherwise pay to move someone in.
Comment: @Retrosaur oooohh That's not just for Christmas, they always do that...I just totally forgot about it.
Comment: Don't forget about birthdays. Apparently the wiki says on the bit's birthday you get one towerbux
Comment: Also (i) the free bux (or coins) gifted daily by NimbleBit over the recent holidays (just by opening the app daily) and (ii) I remember long ago that NB gifted bux when it was one of the developers birthdays!
Comment: Apparently, you can also get Bux from doing minor jobs for the Christmas event (I think()
Comment: @Retrosaur oh huh I haven't played since they introduced that
Comment: Ahh. Well, they just ask you to find a "person", and once you tap on that person, you get 1 bux.
Answer: Well, If you have a Sushi Bar, You can do a glitch with a free job space,
Tap the job place and quickly tap it again with no names, It will say (null) has his/her dream job, If you give someone/null their dream job, you get 2 tower bux.
It doesn't take a spot, so you can repeat over and over and over again.
Answer: You can get them by getting a sushi bar and then hire 2 people then stock the best one with 2 people then delete one finally u have to click on hire really fast then straight after click on the Newley stocked 2nd item

What ways can I get Tower Bux?


How do I show how many products are remaining from the bought package

Question: I am new to this. What I want to do is create a package that gives the customer liberty to buy 10 of any of my product at anytime. Now when a user buys this package I want my site to show him that there are 10 products he can buy and when he buys any of the product the number of the products remaining will be deducted by the amount of the product he buys. for E.g If he has 10 products remaining and he buys two then I want to show him that there are still 8 products remaining. Now how do I distinguish normal customer from the one that has already bought the package and how do I not show the payment page to the one that has already bought the package. Here I want to sell both the packages and singles products to customer. I have a little idea about wordpress so is there any plugin for this and if not then where do I code this. I am using twentyseventeen theme. Sorry for the inconvenience if this is not the right question for this site.
Answer: May this code will helpful for you. Ihave use a code to get the current logged in users completed (or you may use your status) order data. Then get their product count, now we can obtain the product which will purchased before. Just gone through this code.
$customer_orders = get_posts( array(
        'numberposts' => - 1,
        'meta_key'    => '_customer_user',
        'meta_value'  => get_current_user_id(),
        'post_type'   => 'shop_order', // WC orders post type
        'post_status' => 'wc-completed' // Only orders with status "completed"
        //use status such as "wc-processing"/"wc-on-hold"etc.
    ) );
    $previous_count = 0;

    // Going through each current customer orders
    foreach ( $customer_orders as $customer_order ) {
        $order    = wc_get_order( $customer_order );
        $items    = $order->get_items();

        // Going through each current customer products bought in the order
        foreach ( $items as $item ) {
            $id = $item['product_id'];

            // If product not in array, add it
            if ( ! array_key_exists( $item['product_id'], $products ) ) {
                $products[ $id ] = array(
                    'name' => $item['name'],
                    'count' => 0,
                );
            }
            // Increment Product count from cart quantity
            //$products[ $id ]['count'] += $item->get_quantity();// get the total product
            $previous_count += $item->get_quantity();
        }
    }
    $cart_count = WC()->cart->get_cart_contents_count();
    $remaining_product_count = $previous_count+$cart_count;
    if($remaining_product_count < 10){
        $remaining_product_count = 10-$remaining_product_count;
        echo $remaining_product_count;
    }

Comment: Thanks for the coding but since I am new I have no idea where to write this code?Will a custom plugin be a good idea?
Comment: please refer this plugin  https://wordpress.org/plugins/woo-advanced-discounts/ - @Bhaumik Bhatt


Multiple Dropdowns Lists and Removing Value Of Observable Array On Select

Question: I have a situation where I have multiple dropdowns on my page sharing the same observable array.  Lets say 5 dropdowns lists, bound to an observable array of [1,2,3,4,5] with a optionsCaption of - Game - .  After they select a value, I would like the other dropdown lists to remove that item from their list.  So if dropdown 1 selected 2, all the other dropdowns would show 1,3,4,5, and if they select - Game - again, all dropdwons go back to their original state.
Answer: They'll have to have separate data sources since they have different data. Two ways I can think of to do this:
Have one array of all choices, and computed arrays for each select - each containing the contents of the choices array minus the selected option from the other selects.
You could also have each select's options be an observable array, and then on select you call remove on every other option with the selected item, and push the old value to every other select box. You would make this happen with subscribe.


Does Subsonic 3 ActiveRecord Handle Many To Many Relationships?

Question: Does the ActiveRecord template files create classes that handle many to many relationships?
Answer: You will get an IQueryable foreign key reference object when the class is generated.
Comment: I assume in that instance, lets say ClassA has an IQueryable to ClassB that ClassB has a foreign key IQueryable to ClassC? That way you can return properties of ClassC when querying stuff in ClassA
Comment: That is correct.  I'd recommend downloading the latest bits and playing around.  It really only takes about 5 minutes to get up and running.  The docs are located here http://subsonicproject.com/docs/  and here is a great 5 minute demo screencast on getting started with subsonic3 http://subsonicproject.com/docs/The_5_Minute_Demo
Comment: Is there a naming convention for Many to Many tables?
Comment: No it sniffs out the foreign key references
Comment: I think this might be a new question but I'm trying to create a foreign key to the ASP.Net Membership Roles table but not sure how to query it as you don't have access to the GUID in the Membership Roles class
Comment: Yeah probably should be a new question.  For asp.net membership stuff probably better to just use builtin IsUserInRole stuff from System.Web.Security.Roles but Subsonic can handle this if you got the FKs setup.


Why is the carry of any n base system equal to the radix in arithmetic?

Question: In binary the carry/borrow is 2, in hex the carry is 16. What's the reason for this?
Answer: Background
Writing a number in base $r$ means to write it in the form
$$a_0+a_1r+a_2r^2+\ldots+a_nr^n$$
where $0\leq a_i< r$, for all $i=0,1,\ldots,n$. The positional notation that we use consists in just writing the digits $a_i$ from right to left
$$(a_na_{n-1}\ldots a_2a_1a_0)_r$$
Some people write parentheses and maybe the radix as a subscript. Some people write a bar over the digits, and in the decimal positional system, which is the most popular, we just write the digits.
You can think of the digits as buckets that can only be filled up to the value $r-1$.
The carry actually depends on the position of the digit, only that when you compute you imagine that it is the degree zero position, the units. Actually, when you are inside algorithms like adding or converting numbers from one radix to another in each iteration you have already divided enough by $r$ such that the digit that you are treating has become the degree zero position or you are treating it as such.
The carry at the $k$-th position
For example, suppose that you are looking at position $k$ and the digits at that position and the one that follows are
$$a_kr^k+a_{k+1}r^{k+1}$$
You can divide by $r^k$ to get $a_k+a_{k+1}r$, which implicitly is what happened if we just write the two digits $(a_ka_{k+1})_r$.
Now, the maximum carry will happen when $a_k$ is as full as it can be and we add to it the maximum digit that we can add. This is $a_k=r-1$ and we add $r-1$. Then we get
$$\begin{align}
(a_k+a_{k+1}r)+(r-1)&=(r-1)+a_{k+1}r+(r-1)\\
&=(r-2)+\mathbf{r}+a_{k+1}r\\
&=(r-2)+(a_{k+1}+1)r
\end{align}$$
As you can see, it is $\mathbf{r}$ the amount that we have to move out of the $k$-th position to re-write the number as a sum of powers of $r$ in which the coefficients are all between $0$ and $r-1$. Well, it is really $r\cdot r^k$, because we divided by $r^k$. In terms of the contribution to the next ($k+1$-th) digit the carry contributes $1$, when carry over happens. Otherwise it contributes $0$ to the next digit.
Potentially, $a_{k+1}+1$ is larger than $r-1$, so $(a_{k+1}+1)r$ will have to spit out an $r^2$ (actually $r^2\cdot r^k$), but well that's the carry generated by the $(k+1)$-th position.


Mongo DB Site to Site Migration

Question: I am somewhat new to MongoDB but I do have some experience with MySQL and PostgreSQL. Currently, I need to migrate Mongo DB from one site to another. The sites should be identical (development and production) but the production ran a bit forward in terms of content so we want to make sure they are mirroring each other. How do I do it? The mongo is run on the CentOS Linux system. Will this be as easy as doing a dump and restore? Or is there more involved?
Comment: Are we talking of the data? Quite some legislations prohibit  the use your customers data for other things than processing their orders / service requests / etc. If this isn't a concern, use mongodump on a secondary, since this prevents almost any impact on your production environment.
Answer: I think just -- exporting db and taking out the json file and import in another system.
Or You can directly copy the db from one location to another----
http://docs.mongodb.org/manual/reference/method/db.copyDatabase/

Mongo DB Site to Site Migration


Issue with If and VLOOKUP functions in two sheets of data

Question: I'm struggling to write down the correct function to perform a calculation in excel.
In my first sheet, I have a list of towns in Albania, with a corresponding letter "U" or "R" if they are located either in an urban or rural environment. In sheet 2, I have the corresponding literacy rate of either rural or urban environments for a set of countries, including Albania.
Therefore, in Sheet 1, under "Literacy Rate %" I'd like to attribute for each Albanian town an urban or rural literacy rate using the values of sheet 2. I was combining if function and vlookup but I'm failing somewhere... My data is illustrated in this picture :
Comment: what have you tried so far? And why use a `VLOOKUP` if you ony use the data from Albania and no other country?
Comment: @cybernetic.nomad I've found out how to do it. I did: =IF("townA="R";VLOOKUP(ISOCODE;Sheet2"Literacy rate for rural";)VLOOKUP(ISOCODE;"literacy rate for urban"))
Comment: And it works alright. I used VLOOKUP as I am focusing on other countries as well, Albania was just to illustrate the example. Thanks for your input
Answer: <code>=IFERROR(VLOOKUP($A3,'Sheet2'!$A$2:$D$7,IF(I3="R",4,IF(I3="U",3,0)),0),"MISSING INFO")
</code>
Place the above formula in J3 of sheet1 and copy down as required.
The above formula can be broken in to 3 separate formulas.  Let start with VLOOKUP.  Its going to look up you code from cell A3 in the table on sheet2 and its looking for an exact match based on the last zero in the formula  In order to determine which column to return , either 3 or 4, an IF function is used to determine what is I3.  A single IF could have being used if it is assumed that when I3 does not equal R the only other possibility is U.  In the equation above, I used a nested IF function to cover a third possibility of a blank where the U or R was not filled in.  In hindsight it also covers the possibility of data entry error where a letter other than R or U was entered.  In the third case 0 is returned in order to produce an error.
The whole thing is placed in an IFERROR function.  When the VLOOKUP produces an error it could be because the U or R was not properly set, or that the country code was not found in the table.   Either way you a missing proper information.  I choose that the error message of MISSING INFO be displayed in the cell.
Edit the cell ranges, formulas, and message to suit your needs.
Comment: Thanks for your input, I've found my way around another formula, explained above. It works just fine.


Why is Brans-Dicke Theory considered as a failed attempt to incorporate Mach's principle in a relativistic theory of gravity?

Question: In Generalized Brans-Dicke theory: A dynamical systems analysis by Nandan Roy and Narayan Banerjee, Brans-Dicke theory is described as a failed attempt to incorporate Mach's principle in a relativistic theory of gravity. How did it fail?
Answer: Roy and Banerjee seem to be trying to be intentionally provocative with the first sentence of their paper:

Brans-Dicke (BD) theory[1], a failed attempt to incorporate Mach's principle in a relativistic theory of gravity,[...]

The following text doesn't clarify what they mean. There are several ways that this could be interpreted:

When Brans and Dicke originally proposed the theory, they said that there was no reason for the dimensionless constant $\omega$ to have any particular value, so the only reasonable thing to assume was that $\omega\sim1$. On the other hand, $\omega\rightarrow\infty$ reproduces GR. Since solar system tests force $\omega$ to be large, BD gravity could be said to be a failure on its own terms -- the terms stated by Brans and Dicke.
We could read this sentence as a claim that BD gravity doesn't fully incorporate Mach's principle. To me, this would seem off base, because Mach's principle was never a well-defined thing in Mach's formulation, and in fact BD gravity gave it a more clear-cut embodiment in which it could actually be tested.
They claim that there are problems with the $\omega\rightarrow\infty$ limit, so maybe this is why they call it a failure. They reference their own 1997 paper, which I couldn't find, and also this one: https://arxiv.org/abs/gr-qc/9902083
Answer: Brans-Dickie theory, can not be yet declared failed. Here is a very readable article on it. The evidence of variable G is low (it doesn't couple with matter).

Why is Brans-Dicke Theory considered as a failed attempt to incorporate Mach’s principle in a relativistic theory of gravity?


I have a 9v battery that is only needed to output power during night time/when the sun is down, how can i do this?

Question: My main question is, am I able to use a solar panel to only allow power flow when it is dark outside?/When the Solar panel isn't receiving any light? What would the circuit for that look like? In the same way a garden light works?
Comment: This question is not answerable.  You need to use the "edit" button to clarify what exactly you are trying to accomplish, or the question will soon be closed.
Answer: This is a simple PNP circuit with a light dependent resistor. When the sun goes down the resistance of the LDR increases and the voltage drop across the LDR  forward biases the PNP transistor and illuminating the LED or whatever you like within the operating range of the transistor.
This should be a good start

Here is another circuit that I found at
http://www.talkingelectronics.com/projects/SolarLight/SolarLight.html
Comment: @TonyEErocketscientist for the fun of learning and enjoyment of building something with your own hands. Or you could just buy it
Comment: Yah buy it and make it bigger and better  by adding LEDs and bigger battery.
Comment: @TonyEErocketscientist sorry to bother, but could you throw a link my way of something that is like what is mentioned above?
Comment: Where do you live near? How big is Solar Panel [Watts, Voc]
Comment: would I also be able to add a 3v rechargeable battery that is connected to the solar panel?
Comment: Why make when U can buy these for a few bucks


div position changes when zoom in /out

Question: I had spent many hours in figuring out the solution to this problem and looking badly for help. Any pointers would be of great help.
I have an image positioned at the top right corner of the page. On click of this i have to show a drop down below the image.
Below is the code.

          " +

The div Count is positioned next to the image and div Table is positioned below the image.
It works fine in 100% zoom. But when zoom in/out, the div moves away from the image. Also the content in the zoom overflows and the div size gets changed.
Please suggest what can be done.. Thanks..
Answer: Hi you can use this style with image:
 "

This must solve your issue.
Comment: Hi, Tried the solution but no luck. The div still moves away from the image when zoom in /out

div position changes when zoom in /out


What are some actually evil gods around the world?

Question: I mean gods that are truly evil, not ones who are reflecting the social norms of the era they were worshiped in. (Zeus for example, he wasn't evil, his actions were seen as acceptable and normal at the time.)
Comment: Apart from the obvious answer that *all* gods ard evil because they promote hatred towards non-believers, this is a list question.
Answer: In Zoroastrianism, the Angra Mainyu could be seen as an evil god.

Angra Mainyu...is the Avestan-language name of Zoroastrianism's hypostasis of the "destructive/evil spirit" and the main adversary in Zoroastrianism either of the Spenta Mainyu, the "holy/creative spirits/mentality", or directly of Ahura Mazda, the highest deity of Zoroastrianism. The Middle Persian equivalent is Ahriman.

I don't love Wikipedia's description, but it gets across the heart of the matter.
Also note that daevas are a whole host of malevolent gods. Interestingly, the evil daevas of Zoroastrianism battle Ahura, meanwhile the good devi in Hinduism battle the "less good" asuras. My contention, which I sadly never followed up on when I could have, is that this reflects actual hostility between India and Persia 3000+ years ago.
However, the devi in Hinduism aren't really evil, especially in early Hinduism. Some texts do treat them as totally corrupt though.
Comment: You'll have to define "god" and define what separates "gods" from "demons". At some point you'll run up against arbitrary distinctions.
Comment: @Orionixe *Daevas* definitely fall into the former category. First, etymologically, the word comes from the PIE root that means "god". Wikipedia notes that the Gathas literally calls them "gods to be rejected." Zoroastrianism is often understood as a dualistic religion in which two equal sets of gods fight each other, although getting a proper chronology for this can be tricky.
Comment: I thought the daeva are a race of shadow demons in Zoroastrianism, not gods. I could be mistaken though
Comment: I would define god as divine higher beings that have power over natural and human concepts.

Demons (as not in the Abrahamic sense) I would define as evil spirits that are powerless compared to gods and often act as their servants/enforcers or spends time preying humans on earth


Where can I access and manage downloaded themes in Google Chrome

Question: I downloaded and installed several themes on Google Chrome. Now how can I see the list of themes that I just downloaded? And how can I switch to one of them without downloading it again?
Answer: Here are the steps to see installed/downloaded extensions/themes:

Open Google Chrome browser and login into it or alternatively login into Google Account by using which you have downloaded the extensions/themes.
Open My Extensions & Apps.
Under "Installed" tab, you will find the list of active extensions/themes. The currently installed theme shows up here. Under "Library" tab you will find the list of downloaded but not in use extensions/themes. The previously installed themes shows up here.
Comment: This is the actual solution, guys. Old themes will show up under Library. Thanks!
Answer: AFAIK there is no way to manage themes like you would apps that you have downloaded. Once you click on "Choose Theme" it will download the crx and then overwrite whatever theme you had previously installed.
By going to Settings you will see a section called Appearance, however the only revert option is to revert to default.
One way of keeping, managing, and sharing the themes you choose is to go tot he themes page and share the theme you are installing with either your Gmail or Google+. This will at least provide a way to track which themes you like, and it will keep all of the links to the themes in 1 place. Here are the steps to share with a blank circle to keep it private.

Go to Settings > Appearance
Select "Get Themes"
Find one you want and hover over the card. Next to the "Choose Theme"  button, select the "Share" button
Share the theme with Google+
Choose to share the theme with a Circle that only you are a member of, that way all the posts will be visible only by you

Now when you visit Google+ (or Gmail if you choose to share via email) you will have a list of all the themes you've tried or want to try in the future.
Comment: Clever workaround!
Comment: There is no Google+ anymore
Answer:
Go to the Chrome Web Store
Click the settings icon in the top right:

Select Your Apps
Comment: not shown anymore i guess(?),now you can see only extensions and apps.
Answer: There is no way to browse your installed themes as you can only install one theme at a time, the closest you can get currently is the following:

Chrome web store: https://chrome.google.com/webstore/user/library
Press the cog on the left of your email on the top right
This reveals installed extensions and themes (but only the currently installed theme)
Press the "Library" tab to see extensions and themes you've used in the past
This list contains your previously used themes, use F3 and type "Theme" and F3 again to step through your themes :-)

Hope this helps until Google adds a native feature!
Answer: You can create a "theme manager" of a sort. Find the theme you want in the store. Click it one time. The description window will open. Bookmark this link. In the bookmark tab create a new folder and label it themes. Now when you want to go back to a previous theme all you need to do is go to your bookmarks. You are all finished. Simply bookmark each theme into this folder as you find them. Oh, its easier to access them if you enable the bookmarks bar. I sincerely hope this helps.
Answer: Sign in to  your Google account.
Either go to https://chrome.google.com/webstore/user/purchases#themes > Themes directly or through chrome://settings  > Get themes button.
Your history of the installed themes are in plain sight.
The screenshot of the extensions list, with the URL visible and the themes higlighted
Comment: Can you be a little more specific with your detail, consider adding some reference and proof supporting what you state, and confirming this answer is not already answered in one of the existing answers on the post if applicable.
Comment: Thanks this was exactly what I needed!
Comment: Not true! This just show the current installed theme
Comment: currently on that link - https://chrome.google.com/webstore/user/purchases#themes it shows only installed apps and extensions, there not installed\downloaded themes...
Answer: To se all downloaded Themes and Extensions
*(you must be logged in to your google account), simply go to web store, on top right corner there will be your gmail address and next to it is setting button click on it and then select My Extensions and Apps.
Chrome Web Store > Settings > My Extensions and Themes
is you login in to google recently and trying this you might find your themes in My Extensions and Themes, but you need to download them if you haven't downloaded it on the computer.
Answer: Create a new folder in bookmarks and name it Themes.  Save the bookmark of each theme you want to use, or to keep available.
When you want to change themes:

Go to settings.
Click appearance.
Click reset to default.
Go to the bookmark for the theme you wish to use and click "install".  All user restrictions will have been removed when you reset the last theme to default.

It is quick and works for me.   ☺
Answer: Sorry for posting on such an old thread.
There is a trick that may work to see what Themes you have installed on your Chrome.
Go to the Search box on the left column, and type keywords that you use when you look for a Theme.
Then select the Themes button.
The difference is that now, in this listing your already installed themes have a green flag saying 'ADDED', and a 'rate it' green button on the right.
Hope this helps!
Answer: Themes aren't saved to the browser.
When one is overwritten it's gone and must be redownloaded at the Chome Web Store if desired again.
However, if you input a valid keyword into the Chrome Web Store search bar at the top left of the screen (ex. Disney or Cats) and have 'themes' ticked, those you've downloaded before will appear with a green 'added' ribbon over the thumbnail and a 'rate' button in place of the 'add to Chrome' button.
Answer:
Now how can I see the list of themes that I just downloaded?

Themes do show up in chrome://system > Extensions row > click Expand button. However, they are jumbled with regular extensions and may not be easy to distinguish by name.
Assuming you're using Windows, here's a cmd you can copy and paste that should give you a  list:
for /f "usebackq tokens=*" %a in (`where /F /R "%LOCALAPPDATA%\Google\Chrome\User Data" manifest.json`) do @findstr /I /M /C:"theme" %a >nul 2>&1 && @findstr /C:"\"name\"" %a

For me, I get this output:
"name": "Morpheon Dark",

I've assumed manifests containing the word theme means it is a theme, but this might raise some false positive if normal extensions also contain theme.
Answer: I feel your frustrations, and I know I had seen away to mix and match themes with base colors, etc..  Then I open a new tab, in the lower right corner there is a button that says "customize chrome".  That opens a panel on the right and in there you can switch themes, change the colors and still keep backgrounds, etc..
The button:

The panel:
Answer: If u have crx file then right click on it choose open with chrome and then choose to save at the below section of chrome. Then a pop up appears. Select add. The theme will be applied
Answer: Search for and install from the Chrome Webstore (Extension Manager/ Switcher) that allows viewing of Installed Extensions and browse through your extensions and hopefully find one that links to your Installed Themes.
Comment: Does not work. Themes are not listed under Extensions.


Potential inside a uniformly charged spherical shell

Question: If the magnitude of the electric field inside a uniformly charged spherical shell is zero then is how potential a non-zero constant equal to the potential of shell itself? How does a non-zero potential exist given that there is no need to do work in moving a charge in forceless field?
Comment: The fact that the field is zero indicates that the potential is constant. So I'm a bit unclear what you are asking.
Comment: I have a small confusion that whether electric field is zero exactly at centre or within shell everywhere.
Comment: The electric field is zero throughout the interior of the shell (in other words, there is no force field). The potential at a point in space is a property of that location. The amount of work that has to be done to move a charge $q$ from A to B is equal to $W = q\Delta V$. Since the potential in the interior of the spherical shell does not change (because the field is zero, $E = -\frac{dV}{dx}$), the difference in potential between any two points in the interior is zero; this in other words means that no work is done in moving a charge inside the spherical shell.
Comment: Your conception of work seems to be wrong. Work done is **change in potential energy** (in this case; it can also be kinetic energy or other things). The fact that potential is same everywhere inside a charged shell means that you won't have to do work to move charges from one place to another, **only because you won't encounter a *change* in potential energy (V×q)**. This implies that you don't need to work unless you want to change the potential of something. **No change of potential does not mean no potential**, particularly for this case. I hope this solves your problem.
Answer: We can first determine the electric field within the shell using Gauss' law, one of Maxwell's equations. Consider a thin shell of radius $R$ which has total surface charge $Q$. For a spherical Gaussian surface $\Sigma$ within the shell, radius $r$, Gauss' law indicates that
$$ \oint_\Sigma \mathbf{E} \cdot d\mathbf{a} = \frac{Q_{\rm enc}}{\epsilon_0} = 0,$$
since we know that $Q_{\rm enc}$, the charged enclosed by our Gaussian surface, is zero. It follows that if $Q_{\rm enc}$, it must be that $\mathbf{E} = \mathbf{0}.$
Since $\mathbf{E}=\mathbf{0}$, this implies that $V = \rm constant$ because of the relationship $\mathbf{E} = -\nabla V$. That is, the (vector) derivative of a constant is zero. This means that the interior is equipotential everywhere, and it takes no work to move a charge anywhere within the shell.
This is much like how it takes no work (against the gravitational field) to move an object horizontally, since there is no change in $mgh$. Every horizontal position along a certain altitude is at a gravitational equipotential.
Answer: The potential is defined relative to the infinity - not relative to the center of the shell.
Since there is no field inside the shell, the potential at any point inside the shell is equal to the potential on the surface of the shell, $V=\frac Q {4\pi\epsilon_0}$.
This is illustrated for a positively charged sphere on the diagram below copied from this Hyperphysics page.

The potential in the infinity is defined as zero and it increases as we move toward a positively charged sphere as a positive work would have to be done moving a positive charge against the electric field produced by the sphere.
Inside the sphere, the field is zero, therefore, no work needs to be done to move the charge inside the sphere and, therefore, the potential there does not change.
Comment: Is this field is microscopic or macroscopic?
Comment: If in a microscopic field the Electric field vary from point to point inside shell?
Comment: @SRIVISHNUBHARAT I am not sure what you mean by microscopic field. In any case though, there is no field inside the shell.
Comment: This "field" does not have a real existence, in the sense, you can't "see" it (not yet, as of 2020). It's a theoretical understanding; a framework rather, that serves very helpful in studying how charges *physically* (not sure if that's the term) effect each other - i.e. understanding the extent of "push" and "pull" effects produced by/on various charges.


Algorithm to compute a Voronoi diagram on a sphere?

Question: I'm looking for a simple (if exists) algorithm to find the Voronoi diagram for a set of points on the surface of a sphere. Source code would be great. I'm a Delphi man (yes, I know...), but I eat C-code too.
Answer: Update in July 2016:
Thanks to a number of volunteers (especially Nikolai Nowaczyk and I), there is now far more robust / correct code for handling Voronoi diagrams on the surface of a sphere in Python. This is officially available as scipy.spatial.SphericalVoronoi from version 0.18 of scipy onwards. There's a working example of usage and plotting in the official docs.
The algorithm follows quadratic time complexity. While loglinear is the theoretical optimum for Voronoi diagrams on the surfaces of spheres, this is currently the best we've been able to implement. If you'd like to find out more and help with the development effort there are some open issues related to improving the way Python handles spherical Voronoi diagrams and the related data structures:

Effort to improve the plotting of spherical polygons in matplotlib
Effort to improve the handling of spherical polygon surface area calculations in scipy

For further background on the theory / development / challenges related to this Python code and related computational geometry efforts you can also check out some talks from Nikolai and I:

Nikolai PyData London 2016 talk
Tyler PyData London 2015 talk
Tyler PyCon 2016 Computational Geometry tutorial

Original Answer:
I've actually recently written some open source Python code for Voronoi diagrams on the surface of a sphere: https://github.com/tylerjereddy/py_sphere_Voronoi
The usage, algorithm, and limitations are documented on readthedocs (http://py-sphere-voronoi.readthedocs.org/en/latest/voronoi_utility.html). There are some detailed examples there but I'll place one or two below as well. The module also handles the calculation of the Voronoi region surface areas, albeit with some numerical weaknesses in the current development version.
I haven't seen many well-documented open source implementations for spherical Voronoi diagrams, but there has been a bit of buzz about the JavaScript implementation on Jason Davies' website (http://www.jasondavies.com/maps/voronoi/). I don't think his code is open though. I also saw a blog post about using Python to deal with part of the problem (http://jellymatter.com/2014/01/29/voronoi-tessellation-on-the-surface-of-a-sphere-python-code/). Many of the primary literature sources cited in the above posts seemed very challenging to implement (I tried some of them) but maybe some people will find my implementation useful or even suggest ways to improve it.
Examples:
1) Produce a Voronoi diagram for a pseudo-random set of points on the unit sphere:
import matplotlib
import matplotlib.pyplot as plt
import matplotlib.colors as colors
from mpl_toolkits.mplot3d import Axes3D
from mpl_toolkits.mplot3d.art3d import Poly3DCollection
import numpy as np
import scipy as sp
import voronoi_utility
#pin down the pseudo random number generator (prng) object to avoid certain pathological generator sets
prng = np.random.RandomState(117) #otherwise, would need to filter the random data to ensure Voronoi diagram is possible
#produce 1000 random points on the unit sphere using the above seed
random_coordinate_array = voronoi_utility.generate_random_array_spherical_generators(1000,1.0,prng)
#produce the Voronoi diagram data
voronoi_instance = voronoi_utility.Voronoi_Sphere_Surface(random_coordinate_array,1.0)
dictionary_voronoi_polygon_vertices = voronoi_instance.voronoi_region_vertices_spherical_surface()
#plot the Voronoi diagram
fig = plt.figure()
fig.set_size_inches(2,2)
ax = fig.add_subplot(111, projection='3d')
for generator_index, voronoi_region in dictionary_voronoi_polygon_vertices.iteritems():
   random_color = colors.rgb2hex(sp.rand(3))
   #fill in the Voronoi region (polygon) that contains the generator:
   polygon = Poly3DCollection([voronoi_region],alpha=1.0)
   polygon.set_color(random_color)
   ax.add_collection3d(polygon)
ax.set_xlim(-1,1);ax.set_ylim(-1,1);ax.set_zlim(-1,1);
ax.set_xticks([-1,1]);ax.set_yticks([-1,1]);ax.set_zticks([-1,1]);
plt.tick_params(axis='both', which='major', labelsize=6)

2) Calculate the surface areas of the Voronoi region polygons and verify that the reconstituted surface area is sensible:
import math
dictionary_voronoi_polygon_surface_areas = voronoi_instance.voronoi_region_surface_areas_spherical_surface()
theoretical_surface_area_unit_sphere = 4 * math.pi
reconstituted_surface_area_Voronoi_regions = sum(dictionary_voronoi_polygon_surface_areas.itervalues())
percent_area_recovery = round((reconstituted_surface_area_Voronoi_regions / theoretical_surface_area_unit_sphere) * 100., 5)
print percent_area_recovery
97.87551 #that seems reasonable for now

Comment: It seems to me that your  `Voronoi_Sphere_Surface` accepts points expressed in a three dimensional Cartesian coordinate system, am I right? Maybe accepting points expressed in spherical coordinate (just longitude and latitude) could make your code easier to use in case of points that are geographical locations.
Comment: I suppose I could add a flag to allow spherical coordinates. To be fair, my voronoi_utility module does already provide the function convert_spherical_array_to_cartesian_array(), which could be used easily enough prior to running the Voronoi code.
Comment: @RutgerHofste True, and even if we assume a perfect sphere it is a challenging problem; I know there are some efforts in the GIS world to handle the ellipsoid in various ways but my focus was just to see if we can get things to work on a sphere well initially.
Comment: In most projections the earth is not a sphere but an ellipsoid. Any thoughts on that?
Comment: Is there already a C++ port for this?
Comment: @Noob Not sure, but certainly the algorithm in SciPy has been improving over the years. It is certainly not pure Python--many sections are abstracted to Cython for compilation, or offloaded to linear algebra libraries, etc.
Answer: Here's a paper on spherical Voronoi diagrams.
Or if you grok Fortran (bleah!) there's this site.
Original link (dead): https://people.sc.fsu.edu/~jburkardt/f_src/sxyz_voronoi/sxyz_voronoi.html
Comment: For me, there's a download link for the paper here: http://citeseerx.ist.psu.edu/viewdoc/summary?doi=10.1.1.19.6737 - it is not that helpful though...
Comment: There is a great paper from 2011 describing a plane sweep algorithm (based on Fortune's algorithm for the plane): http://www.e-lc.org/tmp/Xiaoyu__Zheng_2011_12_05_14_35_11.pdf
Comment: The Fortran page doesn't load, nor does it on WayBack Machine.  Does anyone know where I can find this content?
Comment: @ThomasHodges: I followed a similar path and it appears that the author updated his website, which broke the original link. Looking on his updated site, it appears that this link: https://people.sc.fsu.edu/~jburkardt/f_src/sphere_voronoi/sphere_voronoi.html is the new one. I've updated the link in the post. I hope this helps (although it's over a year late)!
Answer: Notice that Delaunay triangulation on a sphere is just the convex hull.
Thus you can compute the 3D convex hull (e.g. using CGAL)
and take the dual.
Comment: concerning special sphere package, it is not yet ready. Hopefully in CGAL 4.3
Answer: In short, try cssgrid from NCAR Graphics. I wrote a longer answer for a similar question at codereview.stackexchange.com.
Answer: There is a paper from INRIA about the Delaunay Triangulation (DT) of points lying on a sphere: CAROLI, Manuel, et al. Robust and Efficient Delaunay triangulations of points on or close to a sphere. 2009. where they talk about an implementation in CGAL.
The paper refers to various available implementation of DT algorithms.
Quoting from the paper:

An easy and standard answer consists in computing the 3D convex hull
  of the points, which is notoriously equivalent.

for computing the convex hull the paper suggests:

Hull, a program for convex hulls.
Qhull.
Three-dimensional convex hulls. in FORTRAN.Three-dimensional convex hulls.
STRIPACK in FORTRAN.

The DT C++ class of CGAL has the method dual to get the Voronoi Diagram.
According to this post by Monique Teillaud (one of the author of the above mentioned paper) it seems to me that in November 2012 the implementation was not still ready.
Answer: It's been a while since the question has been answered, but I've found two papers that implement Fortune's algorithm (efficiency O(N lg N), memory O(N)) over the surface of the sphere.  Maybe a future viewer will find this information useful.

"Sweeping the Sphere" by Dinis and Mamede, published in the 2010 International Symposium on Voronoi Diagrams in Science and Engineering.  Can be purchased at http://dx.doi.org/10.1109/ISVD.2010.32
"A Plane Sweep Algorithm for the Voronoi Tessellation of the Sphere" by Zheng et al.  I'm not sure it was published because of the first one, but it is dated 13 December 2011.  It is available for free at http://www.e-lc.org/tmp/Xiaoyu__Zheng_2011_12_05_14_35_11.pdf

I'm working through them myself at the moment, so I can't explain it well.  The basic idea is that Fortune's algorithm works on the surface of the sphere so long as you calculate the points' bounding parabolas correctly.  Because the surface of the sphere wraps, you can also use a circular list to contain the beach line and not worry about handling cells at the edge of rectangular space.  With that, you can sweep from the north pole of the sphere to the south and back up again, skipping to sites that introduce new points to the beach line (adding a parabola to the beach line) or the introduction of cell vertices (removing a parabola from the beach line).
Both papers expect a high level of comfort with linear algebra to understand the concepts, and they both keep losing me at the point they start explaining the algorithm itself.  Neither provide source code, unfortunately.
Answer: I think the Voronoi plane for each point can be constructed using non-Euclidian geometry. What was normally a line on a 2d plane, is now a 'great circle' on the sphere (see Wikipedia:elliptic geometry). It is easy to find which points are on the wrong side of any great circle between two points, by simply rotating the sphere such that the dividing great circle is the equator, and then it's all points on the other hemisphere than the point you're constructing the Voronoi plane for.
This is not the entire answer, but this is where I'd start..
Answer: There's a nice Voronoi diagram example program here (including source code for Delphi 5/6).
I think "points on the surface of a sphere" means that you first have to remap them to 2D-coordinates, create the Voronoi diagram and then remap them to sphere surface coordinates. Are the two formulas from Wikipedia UV mapping article working here?
Also notice that the Voronoi diagram will have the wrong topology (it is inside a rectangle and does not "wrap around"), here it could help to copy all the points from (0,0)-(x, y) to the neighbour regions above (0, -y * 2)-(x, 0), below (0, y)-(x, y * 2), left (-x, 0)-(0, y) and right (x, 0)-(x*2, y). I hope you know what I mean, feel free to ask :)
Comment: I thought about remapping to a 2D plane but that doesn't work; AFAIK there's no sphere-to-plane mapping which preserves distance. Simple example: distances between any pair of vertices a regular tetrahedron are the same, but you can't draw this in a 2D plane.
Thanks for your reply anyway.
Comment: @stevenh: Technically you don't have to preserve distances, you just have preserve the ordering of all distances.  Of course, I don't think even this is possible.
Comment: @stevenvh: The Delaunay triangulation seems to work as is after mapping the sphere to a plane (with eg polar projection: great circles go to circles, and being inside circumscribed circle will work equally for spherical and planar triangles). Then you go back to the sphere and dualize the graph. This would probably work.
Answer: CGAL is working on the "spherical kernel" package, which would allow to compute exactly these kind of things. Unfortunately, is not released yet, but maybe it will be in their next release, since they already mentioned it in a google tech talk in march
Comment: Has it accomplished by now?
Answer: If your points are within one hemisphere, you could do a gnomonic projection from spherical to planar coordinates, and then triangulate, since great-circles become straight lines of shortest distance.
Answer: Quoting from this reference: http://www.qhull.org/html/qdelaun.htm

To compute the Delaunay triangulation of points on a sphere, compute their convex hull. If the sphere is the unit sphere at the origin, the facet normals are the Voronoi vertices of the input.
Comment: Generally, _the convex hull is different_ problem than the Delaunay triangulation. But QHull's algorithm for the convex hull delivers also the Delaunay triangulation which in is dual to the Voronoi diagram.


Can a stellar mass black hole swallow Earth?

Question: In the Astronomy magazine I read that the stellar mass black holes are only the size of Rhode Island. Can they swallow Earth?
Comment: Relevant: [How do black holes grow?](http://hubblesite.org/explore_astronomy/black_holes/encyc_mod3_q9.html). Mass of the black hole is more important than its size. So, a very crude answer would be a "yes", but it'll depend on how far apart the black hole and earth are from each other.
Comment: If the Earth is one AU away.
Comment: To quote a few lines from the linked article: _For example, imagine replacing the Sun by a black hole of the same mass. Permanent darkness would fall on Earth, but the planets would continue to revolve around the black hole at the same distance and speed as they do now. None of the planets would be sucked into the black hole. Our Earth would be in danger only if it came within some 10 miles of the black hole, much less than the actual distance of Earth from the Sun (a comforting 93 million miles)._
Comment: At 1 AU, Earth would orbit a stellar mass black hole, not be swallowed by it.  It's not even inside the Roche limit where the planet would begin to break apart.
Comment: What if the Earth was right next to it?
Answer: There's a fairly common misconception in this question that black holes "suck everything in".   They do - at distances close to their event horizon but as noted, the event horizon's diameter is about the distance across rhode island (different masses have different sizes), but many stellar mass black holes would fit inside the border of rhode island.  Virtually all of them would fit inside pennsylvania for example.   They're tiny.    I'd rate my chances of survival as much higher in a space-ship 1 million miles from a stellar mass black hole than in a space-ship 1 million miles from our Sun.
At normal planetary distances, a planet would just orbit a black hole and likely maintain a stable orbit for billions of years.   Gravitationally, they're no different than a star at those distances.  The orbital mechanics is based on mass and a stellar mass black hole has stellar mass.   What makes it different is it's shrunk inside it's schwarzchild radius, so compared to stars, it's incredibly small and very dense.   The "weird stuff" only happens at very close distances - basically at distances you can't reach around a star because you'd be inside the star.
So . . . what would happen if you put the Earth next to a black hole.
I'm not sure why you'd want to do that, but the Earth would begin to break apart form tidal forces before it got that close.  The gravity would be stronger on the near side than the far side and the Earth would stretch.  This is called spagettification.
If you theoretically made a black hole just appear near the Earth then it would begin to eat the Earth.   Think of a black hole as a drain, a drain some 20-30 miles across, and think of the Earth as a liquid trying to get down that drain.   It's a lot of material squeezing into a comparatively tiny space and in the process the Earth material would get very hot, and likely form into an accretion disk trying to get inside the black hole and the formation of the accretion disk would eject two cosmic jets along the axes.   Black holes are sloppy eaters.
I couldn't tell you the precise amount of Earth that would be absorbed if they were placed right next to each other, but that's also an impossible scenario.   Everything in space is moving and gravity guarantees that if the Earth did meet or closely orbit a stellar mass black hole, that it would be moving enormously fast.
Earth could in theory, but extremely unlikely, either orbit a black hole or collide with one, or pass very close to one, close enough to break apart.  Each scenario would have a different outcome with a different percentage of Earth falling into the black hole.
I don't feel like going through each scenario individually, but the short answer is that the Earth would get quickly ripped apart and some of it, but not all, would fall into the black hole.  Some would be ejected as high speed charged particles.   The matter spiralling in would create a very strong magnetic field and because protons and electrons are charged, much of the matter wouldn't entirely follow the gravitational forces.  Some would get shot out as jets of charged particles.
That's a rough answer anyway.


n-cube connectivity problem

Question: Given $n\geq 2$, let us consider the n-cube $H_n=(V,E)$, i.e. vertex set $V$ is $\{0,1\}^n$.
Here, the edges of $H_n$ are directed, oriented by set inclusion, i.e., $(x,y)\in E$ iff $x\subseteq y$ and $\text{dist}_{Hamm}(x,y)=1$).
Let $0^n := (0, 0, \ldots, 0)$ and $1^n := (1, 1, \ldots, 1)$.
Let $S\subseteq V\setminus \{0^n, 1^n\}$ such that $|S|=n$.
Assume that in $H_n\setminus S$ there is no directed path from $0^n$ to $1^n$.
Then, is it true that $H_n\setminus S$, now viewed as undirected graph, is disconnected (i.e. "no directed path" implies "no undirected path" in this setting) ?
Comment: What is the orientation of the edges, even for $H_3$? The $n-1$ dimensional faces of $H_n$ inherit an orientation from $H_n$, but already the $n-2$ dimensional edges inherit opposite orientations from the two adjacent $n-1$ dimensional faces. Also, do you want to assume $0^n,1^n\notin S$?
Comment: the orientation is by set inclusion: (x,y)\in E iff x\subset y and hamming_dist(x,y)=1. And yes $0^n,1^n$ are not in $S$. Thank you.
Answer: Every vertex removed from the $k$th layer prohibits $k!(n-k)!\leq (n-1)!$ paths. Thus, if $n$ removed verices prohibit all $n!$ paths, then each of them prohibits exactly $(n-1)!$ paths, and the sets of prohibited paths are disjoint. This means that either all the vertices are on the first layer, or all the vertices are on the $(n-1)$th layer. In both cases their removal disconnects the unoriented graph as well.
Comment: I think so, thanks for confirming Ilya.

n-cube connectivity problem


How to index related (or linked) items in a page using Coveo and display them in the search results?

Question: How do I index related or linked items in a certain page and then use UnderscoreJS and OOTB Coveo features to display those items underneath the parent page in the results?
Comment: What about writing your own Computed Field?
Comment: Harsh, Gatogordo, Justin, Alan and Dmytro Can you please explain how this is too broad? I guess JF got it right and provided a great solution to the issue.
Answer: The result your are trying to achieve is called folding. It needs a few fields with special values to link parent and child results together. It also needs UI components to alter the search query and display the results. All of this is supported by the Coveo index, Search API and JavaScript Search Framework.
First, have a look at the required fields. If your related items are also Sitecore items, you will need to code your own Sitecore Computed Index Fields to fill those field values with the required information. If your related items are in an external system, it will be harder for you to fill the fields but doable. You will need to ensure your Coveo for Sitecore edition supports the type of external source you need. Enterprise editions include the Connector Framework (on-premises) or Push API (Cloud) for you to develop connectivity if no out of the box Coveo connector exists for your repository.
Then, you will need to add a Folding component in your search interface to alter the search query. And finally, a ResultFolding component in your result templates to display the pages and their media library items together.
While not Coveo for Sitecore specific, this blog post explains the process from A to Z.
Comment: Thank you very much JF. We are in the process of trying this solution and will let you know once we achieve our goal.
Comment: Hi JF, just a couple clarifying questions ...

1) When we index a Parent Item (page) I'm assuming we are to put the page's ID in the filterField, and the parentField fields and a list of the Child Item IDs (the related document IDs) in the childField field.

2) Do we also need to put some values in these fields on the Child Items (the related documents)?  If so, what? Is the parentField on the Child Items supposed to be set to the ID of the Parent Page? What if the Child Item is related to multiple parents?
Comment: 1) I think you are right. But I think ChildField is not needed. 2) Yes you have to fill in those fields on the child items. FilterField will contain the page ID. ChildField will contain the document ID. ParentField is not needed I think. If the child item is related to multiple parents, folding cannot be used as the child item can only be returned once in the search results (no matter if the list is flat or folded).
Comment: Thanks JF. In the case where we can have more than one parent item for a document how would you suggest we display them in the search result for the parent documents?   Would we, perhaps, stick some Jason data in a computed field on the parent and parse it in the results?
Comment: A child search result cannot have multiple parents in the Coveo search results. If you have to live with this constraint, I would recommend to: 1) Avoid folding and treat each parent and related document independently. 2) On the PDF/Doc documents, index all their parent Sitecore item IDs in a computed index field. 3) At query time, process the value of this field to compute the URLs of the items like the `ResolveResultClickableUriProcessor` in the `coveoProcessParsedRestResponse` pipeline. Store this value in a new field. 4) Display links to parents in a PDF/Doc result template.
Answer: Try creating a computed field for PDF and Word documents as shown here https://developers.coveo.com/display/public/SitecoreV4/Creating+a+Computed+Field+for+a+Referenced+Item
Then you can access the field property in the results template and will also be able to format/massage the clickuri for these documents.
Comment: Storing the content of the PDF/Word documents in a field is not a good practice and a very complex one. You shouldn't have to understand those file formats from withing Sitecore code to extract their text.
Comment: I dont think that is what I was hinting at. I meant to store the path of the document.
Comment: Ok, in this case it is simpler. But the content of the DOC/PDF documents aren't searchable which limit the possibility they would be found by the visitor. Also, the visitor will have no clue if the words he searched for are contained in the documents.


changing subsite template type

Question: I have a subsite in SharePoint that was originally created as a Publishing Enterprise Wiki. The problem with this is that all of the new pages added to the site are wiki pages, which in not ideal. As far as I know, the site templates are all basically the same thing but with different features enabled. What features do I need to modify to change the subsite from a wiki to a standard team site (so that added pages are just normal pages and not wiki pages).
Answer: I understand that you want to migrate enterprise wiki site to a standard team site. As far as I know , there are no PowerShell script which can achieve this. When use Import-SPWeb command, The site or subsite that you are importing must have a template that matches the template of the site specified by Identity.
You can get the difference between both from below given URL.
Enterprise Wiki vs Team Site in Sharepoint 2010
BTW you can refer below mentioned link & check the answers on the post. You might get some help from there.
Migrate a SharePoint Team Wiki to an Enterprise Wiki
Comment: thank you for your answer. now that I have additional info, I realize that what I want isn't necessarily to switch the site types, but to have newly created pages be standard wiki pages rather than enterprise wiki pages. For some reason, the subsite has both a site pages folder and a pages folder (while all other subsites only have a site pages folder). When I create a new page the subsite uses the pages list(?) which has a default type as an enterprise wiki page. Do you know how I can change the settings to default to the site pages list(?) to add standard wiki pages?
Comment: You can check here for some problem...https://social.technet.microsoft.com/Forums/office/en-US/febb84b1-5598-4a09-b18d-ba95c1761f77/deactivating-the-sharepoint-server-publishing-feature-at-my-enterprise-wiki-site-collection-will?forum=sharepointgeneral
Comment: Deactivate the Publishing feature at my your wiki subsite. So you will get rid of the Pages library from site pages. In your case, If you try to create a new page from the page available under the pages library than it will create the page in pages library only.
Comment: If you try to create a page from wiki page (Site Pages) than it will create a wiki page for you within site pages.
Comment: so deactivating the Publishing feature did not do what I wanted it to do. I still have both a Pages and Site Pages folder, but now the "Add a Page" option is gone from the Site menu. I could already add standard wiki pages in the Site Pages list before (and still can), but what I wanted was for the Add a Page option to generate pages in the Site Pages list, which defaults to standard wiki pages that allow me to restructure the page layout. Now I don't have the option to quickly add a page without navigating to the Site Pages list

changing subsite template type


Locally trivial deformations of surfaces with quotient singularities

Question: Let us consider the surface $\mathbb{A}^{2}/\mu_{6}$ where the action is given by
$$
\begin{array}{ccc}
\mu_{6}\times\mathbb{A}^{2} & \longrightarrow & \mathbb{A}^{2}\\
(\epsilon,x_{1},x_{2}) & \longmapsto & (\epsilon^{2}x_{1},\epsilon^{4}x_{2})
\end{array}
$$
The invariant polynomials with respect to this action are $x_{1}^{3},x_{2}^{3},x_{1}x_{2}$. Therefore we can interpret the surface $\mathbb{A}^{2}/\mu_{6}$ as
$$S = \{f(x,y,z) = z^{3}-xy = 0\}\subset\mathbb{A}^{3}.$$
We have
$$Ext^{1}(\Omega_{S},\mathcal{O}_{S})\cong k[x,y,z]/(f,\frac{\partial f}{\partial x},\frac{\partial f}{\partial y},\frac{\partial f}{\partial z}) = k[x,y,z]/(z^{3}-xy,-y,-x,3z^{2})\cong k[z]/(z^{2}).$$
Therefore $Ext^{1}(\Omega_{S},\mathcal{O}_{S})\neq 0$ and $S$ admits non-trivial infinitesimal first order deformations. Is any deformation in $Ext^{1}(\Omega_{S},\mathcal{O}_{S})$ locally trivial ?
Comment: If you consider the local to global spectral sequence for $Ext$, it follows that the locally trivial deformations are classified by an $H^1$, so these will always vanish on affine schemes.  (This is with the natural meaning of locally trivial -- below someone has suggested that perhaps you mean equisingular.)
Answer: This is a $A_2$ surface singularity, in fact it is isomorphic to the quotient $\mathbb{A}^{2}/\mu_{3}$ where the action is given by
$$
\begin{array}{ccc}
\mu_{3}\times\mathbb{A}^{2} & \longrightarrow & \mathbb{A}^{2}\\
(\epsilon,(x_{1},x_{2})) & \longmapsto & (\epsilon x_{1},\epsilon^{2}x_{2}).
\end{array}
$$ The explicit expression of its versal deformation is given by $$a+xy-3bz+z^3=0,$$
where $(a, \, b) \in \mathbb{C}^2$.
Taking derivatives, one easily checks that if $a= \pm 2 b \sqrt{b}$ and $(a, b) \neq (0,0)$, the corresponding germ has a ordinary double point at $P_{\pm}=(0, \, 0, \, \pm \sqrt{b})$ and no further singularities, whereas if $a^2-4b^3 \neq 0$ the germ is smooth.
So we have a $A_2$-singularity if and only if $(a, \,b)=(0, \, 0)$, which corresponds to the trivial deformation. In other words there are no equisingular deformations apart from the trivial one, hence no locally trivial deformations.
Comment: It seems that your argument implies that there are not deformations at all i.e. that $S$ is rigid. Is this true?
Comment: Not at all: look at the versal deformation given by Francesco, it depends on two parameters. And the general deformation is smooth.
Comment: Let me interpret the answer in another way. Consider the exact sequence $0\mapsto H^{1}(S,T_{S})\rightarrow Ext^{1}(\Omega_{S},\mathcal{O}_{S})\rightarrow H^{0}(S,\mathcal{E}xt^{1}(\Omega_{S},\mathcal{O}_{S}))$. Then $H^{1}(S,T_{S}) = 0$ and $H^{0}(S,\mathcal{E}xt^{1}(\Omega_{S},\mathcal{O}_{S}))\cong Ext^{1}(\Omega_{S},\mathcal{O}_{S})\cong k[z]/(z^{2})$. Is this correct?
Comment: Could you give me an example of a locally trivial deformation of a surface singularity? Thank you.
Comment: By "locally trivial" you mean equisingular? In this case, some cones do the job, since they have non-trivial, equisingular deformations. See
http://link.springer.com/article/10.1007%2FBF01303329


sshd not writing any kind of log Debian 6

Question: I just installed Debian 6 on my server and noticed that sshd isn't writing any log to /var/log. I understand this is a huge security issue as I won't be able to know who logged in to my server. Is there anyway to fix this?
Answer: By default, sshd's SyslogFacility is set to AUTH.
If you haven't changed this in /etc/ssh/sshd_config, you should be getting sshd log information in /var/log/auth.log. You'll also see messages here from pam and sudo, so all access information is logged in the same place.
Comment: Odd. All auth messages go there by default. You could check `/etc/rsyslog.conf` to confirm that it hasn't been directed elsewhere. Are your other logs being updated correctly? Is `rsyslogd` running?
Comment: Looks good in sshd_config:
# Logging
SyslogFacility AUTH
LogLevel INFO

/var/log/auth.log doesn't exist.
Comment: Thank you so much! Yep you're right, for some reason rsyslogd isn't even installed. I installed rsyslogd and the ssh access log is in auth.log like you said. Is there a way to install all of essential packages? I had to install nano and tcpdump as it isn't included in my version of Debian. I think the version of Debian I have is the bare minimum.
Comment: I haven't gone through the installation recently, but I'd expect any default configuration to include rsyslogd. Unless it's **really** bare-bones, I'd just install things as you need them. Pretty soon you'll find it's all there.
Comment: I am tailing every file in /var/log and all I am getting is:
Feb 15 07:21:40 zmijevik sshd[8764]: Connection closed by 188.8.131.52 [preauth], why is it not logging failed connections?


Prerequisite could not be found for bootstrapping (All listed)

Question: Like with Wallys Post I cannot seem to fix the problem with broken links in my prerequisites list.  I am deploying a vb.net desktop application using clickonce but the required prerequisites are showing an exclamation warning icon.  I have checked the location in the registry and ensured the packages are in the correct folder (C:\Program Files\Microsoft SDKs\Windows\v7.0A\Bootstrapper\Packages) but each time I close and re-open the studio, I do not see the added packages and there are only .Net v2.0 to 3.5 listed. (with warnings)
I experienced this in the past when changing development machine but once I added the packages to the correct folder they appeared in the prerequisites list.
This problem occured after installing vs2010 SP1.  I removed the SP and re-installed the studio, everything else seems ok.
Has anybody else encountered this problem and can you help?
Thanks
Answer: I eventually fixed it.
Uninstalled everything related to .Net and vs2008/vs2010.
Cleaned the program files directory to remove all related to visual studio
Logged in as Administrator
Ran full install.
All prerequisites now visible for deployment.


Calculating values in field to round up to 1 if value is greater than 0 using ArcGIS Field Calculator?

Question: I am prepping vector data for raster conversion and analysis. In order to prep the data I need to have a field that represents data/no data (1/0). I built a model that outputs the necessary features but I need to calculate the output field (based on Shape_Area) so that if a row value is greater than 0 (represented as a decimal number), it needs to be rounded to "1". Field type is double.
Comment: Are you looking for math.ceil() https://www.geeksforgeeks.org/python-math-ceil-function/ in a python field calculation? int() performs a floor() by truncating every decimal place. To use in field calculator you will need to use the advanced dialog and start with import math so that the function loads the correct lib to use this function.
Answer: Using Python in the field calculator, you could try the following expression:
1 if !myField! > 0.0 else 0
Or if you want it based on the polygon area, try:
1 if !shape.area! > 0.0 else 0
Since booleans are automatically cast to integers (1 or 0), you can do the same thing more simply as:
!myField! > 0
or
!shape.area! > 0
If there is a risk of empty (NULL) geometries when using area in a polygon feature class, then you can cater for this with:
1 if !shape! and !shape.area! > 0.0 else 0
Comment: This assumes there are no empty geometries; often a safe assumption, but an assumption nonetheless.
Comment: Empty geometries manifest as `None`, so there would be a `NoneType object does not have attribute area` exception.
Comment: !field! >0 will suffice.
Comment: Looks right, though a proper polygon ought to have non-zero area.
Comment: Thank you. I ran with the above expression in my model using the field calculator. It is converting values > 0.0 to 1 BUT it is also converting values = 0 to 1 as well. Need to keep the 0 values as 0. --> Field Name = WeightValue; Expression = 1 if !Shape_Area! > 0.0 else 0
Comment: Should be `!shape.area!`, NOT `!Shape_Area!`.  'Shape_Area' is a virtual field and should not be used directly in field calculator.  It will always resolve to True in this context (unless there is an empty geometry).


Convert array element to float

Question: I have a one dimensional array called monteCarloPerf which can look something like:
monteCarloPerf  [[113.4848779294831], [169.65800173373898], [211.35999049731927], [169.65800173373901], [229.66974328119005]]

I am retrieving a single element from the array using:
finalValue = monteCarloPerf[arrayValue]

where arrayValue is an integer.
Say arrayValue = 0, at the moment I am getting returned : [113.4848779294831].  Is there a way to just return the float without the brackets please? So I would be returned just 113.4848779294831.
Many thanks
Comment: moteCarloPerf[arrayValue][0]?
Answer: Your object monteCarloPerf is a one dimensional array containing elements of one dimensional arrays, or a list of lists. In order to access the value of the first element of the object you have to change your access to that element to the following:
finalValue = monteCarloPerf[arrayValue][0]

Answer: In fact, that is a 'TWO dimensional' array.
To get the float value you can do the following:
finalValue = monteCarloPerf[arrayValue][0]

Or you can transform the two dimensional array to a one dimensional array:
one_dim = [item[0] for item in monteCarloPerf]

I hope this helps.
Answer: monteCarloPerf is a list of list. When you are using monteCarloPerf[index] it is returning list at index position. Based on the symmetry in your list, in each sub-list, item at [0] position is the actual value you are trying to fetch.
Use this to fetch the value
finalValue = monteCarloPerf[arrayValue][0]

Answer: Here's a weird way to do this without using list.__getitem__() method:
float(''.join(i for i in str(monteCarloPerf[arrayValue])))

Convert array element to float


Sort a multidimensional list by a variable number of keys

Question: I've read this post and is hasn't ended up working for me.
Edit: the functionality I'm describing is just like the sorting function in Excel... if that makes it any clearer
Here's my situation, I have a tab-delimited text document. There are about 125,000 lines and 6 columns per line (columns are separated by a tab character). I've split the document into a two-dimension list.
I am trying to write a generic function to sort two-dimensional lists. Basically I would like to have a function where I can pass the big list, and the key of one or more columns I would like to sort the big list by. Obviously, I would like the first key passed to be the primary sorting point, then the second key, etc.
Still confuzzled?
Here's an example of what I would like to do.
Joel    18  Orange  1
Anna    17  Blue    2
Ryan    18  Green   3
Luke    16  Blue    1
Katy    13  Pink    5
Tyler   22  Blue    6
Bob     22  Blue    10
Garrett 24  Red 7
Ryan    18  Green   8
Leland  18  Yellow  9

Say I passed this list to my magical function, like so:
sortByColumn(bigList, 0)

Anna    17  Blue    2
Bob     22  Blue    10
Garrett 24  Red 7
Joel    18  Orange  1
Katy    13  Pink    5
Leland  18  Yellow  9
Luke    16  Blue    1
Ryan    18  Green   3
Ryan    18  Green   8
Tyler   22  Blue    6

and...
sortByColumn(bigList, 2, 3)

Luke    16  Blue    1
Anna    17  Blue    2
Tyler   22  Blue    6
Bob     22  Blue    10
Ryan    18  Green   3
Ryan    18  Green   8
Joel    18  Orange  1
Katy    13  Pink    5
Garrett 24  Red 7
Leland  18  Yellow  9

Any clues?
Answer: import operator:
def sortByColumn(bigList, *args)
    bigList.sort(key=operator.itemgetter(*args)) # sorts the list in place

Comment: That's fabulous.  I had never heard of itemgetter (or attrgetter, which I also now see).
Answer: This will sort by columns 2 and 3:
a.sort(key=operator.itemgetter(2,3))

Answer: The key idea here (pun intended) is to use a key function that returns a tuple.
Below, the key function is lambda x: (x[idx] for idx in args)
x is set to equal an element of aList -- that is, a row of data. It returns a tuple of values, not just one value. The sort() method sorts according to the first element of the list, then breaks ties with the second, and so on. See http://wiki.python.org/moin/HowTo/Sorting#Sortingbykeys
#!/usr/bin/env python
import csv
def sortByColumn(aList,*args):
    aList.sort(key=lambda x: (x[idx] for idx in args))
    return aList

filename='file.txt'
def convert_ints(astr):
    try:
        return int(astr)
    except ValueError:
        return astr
biglist=[[convert_ints(elt) for elt in line]
         for line in csv.reader(open(filename,'r'),delimiter='\t')]

for row in sortByColumn(biglist,0):
    print row

for row in sortByColumn(biglist,2,3):
    print row

Comment: The numbers have to be converted to ints
Answer: Make sure you have converted the numbers to ints, otherwise they will sort alphabetically rather than numerically
# Sort the list in place
def sortByColumn(A,*args):
    import operator
    A.sort(key=operator.itemgetter(*args))
    return A

or
# Leave the original list alone and return a new sorted one
def sortByColumn(A,*args):
    import opertator
    return sorted(A,key=operator.itemgetter(*args))


How can I fill the empty gap between two parts of my photoshop document

Question: I have a website page in photoshop. In the middle I deleted an element and I want the empty space between the top and bottom of the document to be closed. So the top and bottom part merge into one document again. How can this be done?
Comment: This is a very basic Photoshop operation and there are several tools and methods you could use in this situation... Have you tried anything at all?? I'm voting to close this question as tech support. It shows no effort or research at all.
Answer: Make a fill layer which is white and then add it to the background
If you need to you can merge the layers to flatten the artwork
Answer: Create a New layer below the layer of the image.
Keep the Foreground color as white. Then select the new layer and press Alt+Delete.
This will fill the up the space with white color.


Can I independently verify the integrity of the network with only block headers?

Question: Section 7 of the official Bitcoin paper "Reclaiming Disk Space" explains that only block headers are needed to verify transactions. Essentially by making sure that the transaction has been buried behind enough blocks. Why is the block chain so big then? It seems like we could all just store headers and reclaim a lot of disk space. Can I independently verify the integrity of the network with only block headers?
Answer: Reclaiming disk space describes what is nowadays called pruning. After downloading the entire block chain, you can safely throw most of it away because old spent transaction outputs are never needed again. This is not yet fully supported in Bitcoin Core, but it probably will be possible in the next major release.
From the paper (my emphasis):

Once the latest transaction in a coin is buried under enough blocks, the spent transactions before it can be discarded to save disk space

You're talking about simplified payment verification (SPV), the next section in the paper. With SPV, you use just headers and Merkle trees to verify that a transaction has a certain number of confirmations. A transaction with a high number of confirmations is probably safe, but you can't be 100% sure without downloading and verifying the entire block chain. SPV clients trust that the majority of miners are honest, which full nodes do not need to do in most cases. Today, SPV is used by many lightweight clients such as MultiBit and Bitcoin Wallet for Android.
Comment: Wouldn't I be able to mine only with the headers? When I get new transactions I could validate transactions by checking the inputs with the merkle root of the blocks. I guess you can't calculate if the amounts add up with only the merkle root?
Comment: @LucaMatteis It's possible to mine without being a full node (you don't need to include any transactions), but you'll end up on the wrong side of forks in the chain caused by invalid blocks because you won't know which fork is valid. Just a single invalid block could cost you a lot of money. No one "mines" like that because it's very bad for the network, the profits are likely to be much lower on average, and running a full node is fairly cheap.
Answer: In order to validate a new block with a new set of transactions, your node will also need to ensure that the source account for each new transaction actually has a balance to spend. For this, you will need the entire blockchain.
Comment: Thanks Greg that's what I figured. I was missing/forgot the part that transaction depend entirely on the balance of other past transactions. Nonetheless, using the Merkle Root in the headers you can still verify if a specific transaction exists in a block. So you could validate the set of new transactions this way no?
Comment: @LucaMatteis How could you verify that there didn't exist a transaction that spent those funds unless you either check every transaction or maintain a table of all unspent outputs?
Comment: @DavidSchwartz so with the Merkle Root you can only verify that a transaction is part of that block but you can't verify that enough funds are available, right?
Comment: @LucaMatteis Assuming you mean this, you are correct: "You can verify that the past transactions that a new transaction depends on were part of a block, but you can't verify that the funds the new transaction uses weren't consumed by some other past transaction."


Can I use downloaded content with another account?

Question: I had an account on PSN and downloaded lots of stuff on that account. But when the PSN hack happened, I forgot my password. I got my account back but still forgot my password. I got a new account that I use more often and I want all my downloads like avatars and games on that account.
Answer: Well, since you have your original back, you can download all the stuff back onto your system, and you should be able to use it with all accounts on that system.
Comment: That's correct, the only exception are avatars, that can be used only by the downloading user and not other users on the system
Answer: Also, if your original account had any Playstation Plus content, you must have at least one active Playstation Plus account on the PS3, but the Plus content can still be used on non-plus accounts.
(I was going to add this as a comment, but don't have enough Rep)
Answer: As far as I know, you can still play the games you download as long as you can access one of the following:
 - your psn account
 - your console account
If for some reason you can't use any of them, then no. I think the same works for DLC's and PSN games, though I'm not sure about the avatars.


Size, depth, and time of circuits

Question: Can polynomial depth circuits (with, let's say AND, OR, and NOT gates) be simulated in polynomial time?
Answer: No.  Every function can be implemented with a polynomial-depth circuit, but not all function families can be implemented with a polynomial-time algorithm.
In particular, if we're allowed unbounded fan-in/fan-out, you can implement every function in depth 2 using a CNF or DNF formula for the function (but the size of the circuit is exponential).  If we're only allowed constant fan-in/fan-out, you can simulate that with linear depth (use a $n$-layer tree to make exponentially many copies of each input; then 1 layer to evaluate the CNF clauses; then a $n$-layer AND tree).

Size, depth, and time of circuits


Adobe Lightroom Slow after Installing Develop Presets

Question: After installing all VSCO Film Presets on 2 different Macs on Mac OS Mavericks(using Lr 5.4), I've noticed constant hangs and some slowdowns when using it.
Prior to installation, software was pretty quick. Does anybody have the same issue and knows the solution? I didn't find anything useful searching on Google.
Answer: You say you installed on 2 different Macs. Do they both have the slowness problem?
Two possible causes:

The installation didn't go well and some files were incorrectly installed. When LR tries to load them, internal failures happen, and the whole process becomes slower. Can you open/use every single preset?
You installed too many presets. The more presets you have the slower to open the Develop module.

My advice: Uninstall/Delete the new presets; Install again; Delete the ones you don't like; Optimize Catalog.
Good luck.
Comment: Well, I guess there are just too many presets installed. When disabled, Lr is blazing fast again :/
What I'm doing is: When scrolling through my library, I let them disabled, and when editing, enable them again. It works.
Thanks!


I'm unable to remember my old phone's passcode. I've tried just about everything I know of.

Question: I have a Samsung Galaxy Note 4 and it is a typing/word password-my carrier is AT&T. This phone isn't currently activated (as in I'm currently using a different phone, an iPhone to be specific, so my account hasn't been active in a while). I don't think I had the debugging enabled. I also attempted to try "Find my Mobile" and "Android Device Manager", but these require having already downloaded and activated these or require you to be able to have access to your settings which i am unable to do. Is there a way to do a backup with it like this? or is there a way besides  paying apps or factory rest to unlock it? I have very important old scripts, photos/videos, and I don't remember if i did a backup so I would truly prefer not to have to do a factory reset. Please help!
Comment: Android Device Manager requires no special setup, but you do have to be logged into the same account you have registered on the device. If it has a working network connection (like WiFi), you should be able to set a passcode and access the device. Does it even show up in ADM?
Answer: On TWRP Mount the system,use file manager,navigate to data/system remove all files ending with .key
Done.You got your phone unlocked.
If you like than you can try xda developers site


Immediate braking mechanism?

Question: Not sure if this is the right place to ask, but why don't automobiles have some sort of immediate braking mechanism?  Even one that plows into the ground to stop the car faster than if you just slammed on the brakes.
If we were worried about abuse, it could be one time use before some costly replacement, so it would only be used in emergencies.
Comment: "It's not the speed that kills you, it's the sudden stop"
Comment: Are you familiar with Newton's Laws of Motion?
Comment: You can definitely decrease braking time, even given laws of physics.
Comment: If the safety of your occupants is your primary concern, read this; https://en.wikipedia.org/wiki/G-force#Horizontal_axis_g-force paying particular attention of the "Short duration shock, impact, and jerk" section.  Remember your occupants aren't all test pilots.  Some may be children or the elderly.
Answer: Simple answer - because that would be a very bad thing!
As Chenmunka says, look at Newton's Laws of Motion. Or, if you don't believe them, look at videos of crash-testing - you'll see what happens to the occupants of a car in a very rapid deceleration - it isn't nice... The car stops, quite quickly, but you don't, until you hit a solid thing (like the dashboard or steering wheel) - this is why we have airbags. Then your internal organs keep going until they smash into the inside of your ribcage, which also is a bad thing!
Comment: The damage caused by a land anchor being deployed at motorway speed would be catastrophic.  The forces would likely snap the chassis of the vehicle, not to mention break the necks of the occupants.  Now consider the vehicle traveling behind which had been maintaining a safe braking distance at speed but is now faced with the remains of the vehicle in the form of a stationary object.  Stopping becomes tricky as the land anchor has ruined the road surface too.  It's a recipe for disaster.
Comment: @Zabari, in rehab the damages we see are not just because of the forward movement of the passenger/airbag impact, its the jerking motion that results from it when the occupant moves back into (and past) the original position. If you are familiar with what a whiplash injury is you can visualize the movement. A sudden immediate stop at high speed, save for the car damage, would be of the same effect of the car hitting a wall at the same speed. There are limitations in what the system can do and a car needs to be safe for the occupants, the driver needs to use it safely for everybody else.
Comment: Assuming airbags were immediately deployed as well, the largest danger would be to internal organs?
Comment: Airbag deployment is still risky, certainly not something you want to happen if you can avoid it....
Comment: I guess the question is, which is worse, the damage caused to the people in the car, or to the people the car would hit?
Answer: Lets ignore the aforementioned pitfalls of an emergency quick-stop system, and instead try to design one. (For realistic answer, see option 4)
Front Mounted

Option 1: Pneumatic Piston

Hollywood rollover/flip stunts are usually performed, in the absence of a ramp (though a deploy-able ramp that launches you over the obstacle isn't a bad idea..), by installing a large diameter pneumatic cylinder that fires a heavy steel piston (3-4" Ø) at high speed directly into the ground, which serves to vault the vehicle into the air.
Flipping the vehicle onto its roof in order to stop, while checking the 'one-time use' and 'expensive to repair' boxes, will probably not provide enough friction to stop the car faster than the tires would have, not to mention the air time involved, and reliability issues of takeoff/landing (see aviation.SE).
For that reason, we would want to install such a devise so as to fire forward into the ground at a shallow angle. Too steep an angle and the rod could penetrate the road surface, leading to what track & field experts refer to as the 'pole vaulting' effect.
Increasing the surface area of the piston in order to spread the force along the road might solve that problem. It may also be a good idea to make the end of the piston deformable, to gradually slow the vehicle before the piston really digs in. This system would also require a high strength yet deformable cage so as to avoid damaging the chassis of the vehicle when it literally stops you on a dime (USD).

Option 2: Rocket Thrusters

An aviation style engine mounted on the front of the vehicle could effectively increase the wind resistance by providing reverse thrust to the vehicle. The only reverse-propulsion method that could react quick enough for an emergency stop would be a rocket thruster. These thrusters would be lighter than option 1, and would require less body modification to mount securely. Plus it'd look like you have some James Bond type s#it in your grill. Dope.
Shortfalls of this system include the risk of flame grillin' whatever obstacle you didn't want to hit. The systems endless applications would best be utilized by food trucks, as it would allow them to turn what would have been roadkill into a locally sourced roadside barbecue dining experience!
Top Mounted

Option 3: The Claw

If you've ever had to manually thatch your lawn or rake wet leafs, you know exactly how difficult it is to pull a toothed comb over unforgiving ground; that is the basic concept behind The Claw™.
The Claw™ is a spring loaded, articulating arm bolted to the (reinforced) roof of your vehicle. In an emergency stop situation, The Claw™ will deploy toward the rear of the vehicle (thus avoiding the 'pole vaulting' problems encountered in the prototype stages) with enough force to penetrate the tarmac. Upon penetration, the tips of the teeth will further dig themselves into the asphalt from the force of the spring loading, as they drag your car to a stop.
While not hugely popular among road repair crews, state budget officials, or any poor sap who happens to travel that stretch of road in the next 10-12 weeks (thanks, government inertia!), you can rest assured that your family will be able to return home safely to recover from their severe neck trauma, together.
Rear Mounted

Option 4: Hydraulic Spoiler

Relatively modern "hyper cars" employ the use of hydraulically actuated rear spoilers. Take, for example, the Bugatti Veyron 16.4. In the event of a hard stop, the Bugatti's spoiler will pivot up to a 55° angle (wiki source). Through aerodynamic loading, this action provides down force, allowing greater braking traction for the rear tires, while also creating massive aerodynamic drag such that at speeds over 120 mph, the spoiler alone provides as much braking force as a compact car (roughly .70 G). To maintain such performance, the Veyron has a separate radiator to dissipate the heat created due to the immense pressure on the hydraulic system.
While average cars on public roads will (well... should) never (uhh.... rarely) see speeds that would benefit hugely from this type of spoiler, the technology has begun to trickle down into mere super car, and even sports car territory, and there is no question it has a positive effect on braking distances.

Option 5: Parachute

Land speed record and drag racing vehicles have for years used parachutes to slow down after high speed passes. The basic concept, as seen in option 4, is to increase aerodynamic drag, but while the Bugatti is putting along at a mere 260 mph, top fuel drag cars are passing the 1/4 mile marker at 330 mph, and land speed cars are cruising through Nevada's Black Rock Desert at 760 mph. Saucy stuff.
While a small, short throw parachute would be terribly effective in preventing that fender bender on your commute home from work, the lack of sight lines for the car behind you might leave you worse off than had you accepted your fate. You might not have hit that car had you put your damn phone down.

Immediate braking mechanism?


disable upload in rtorrent

Question: Is it somehow possible to disable upload in rtorrent? One would think that upload_rate = 0 might do the trick, but based on my tests and commented rtorrent.rc file in rtorrent source tarball, 0 is for unlimited speed. Or is BitTorrent protocol built in a way that at least some upload bandwidth is always required?
Answer: Another alternative, you can setup upload connection to 0 or put 1kbps speed so it will use the minimum rate or even you can recompile the program by adding the file provided here.
I do not want to be rude but did you consider what happen if everybody start doing same thing as you. The torrent work's with peer-to-peer connection. Therefore, by using a little upload from everyone, it help's other to download the file. Other's are depending on your upload to download the file as you are depending on others upload.
As explain the link above if you are worried about privacy, please note that even if you are not uploading, others can still tack you.

Remember: people will STILL see you are connected to the tracker and if a client is using Superseeding, you will NOT GET ANY OTHER CHUNK of the file until you upload what you just downloaded to someone else! (which won't occur since you just disabled sharing).

Consider yourself warned.

disable upload in rtorrent


Maximising volume increases brightness

Question: Why does maximising the volume on a Fender Mustang Amp increase the perceived brightness?
Answer: One reason may be the loudness curve that represents how sensitive we are to sounds at different volumes. Compared to midrangey sounds, we are more much more sensitive to bass and slightly more sensitive to treble sounds at higher volumes.
Another reason would be that amplifiers tend to distort more at higher volumes. This is true even of amplifiers that are set to 'clean', i.e., to not deliberately distort. These distortions will create more higher harmonics and a brighter sound. If you are using distortion deliberately the same principle may apply.
On the guitar (rather than the amp), maximizing volume can give you a brighter sound due to capacitance effects. This seems much less likely in an amplifier though.
Of course if you're seeing a dramatic increase in brightness with no obvious increase in distortion, there's a possibility of a bad design or a faulty amp.
Comment: Thank You for a very informative answer.  I was under the misconception that increasing the volume did just and only that.
Comment: @DonaldFisher That's usually the engineer's intention, but it's hard to achieve perfectly. The distortion effects we love today are a result of interesting failures to do so!

Maximising volume increases brightness


How do these in-house letterboxes work?

Question: I just moved to Prague and my new house has the letter boxes in the hallway, meaning that in order to throw letters in you have to enter through the main door which is usually locked. I did receive a letter before, so the post officer seems to be able to get in somehow. How does this work here? Does the post get keys to all the houses or is there a person in charge of letting the post officer in every day (there is no reception)? Will UPS and similar services be able to throw in we-could-not-deliver-your-package-notes?
Answer: This will probably differ from city to city or post office organisation.
In the past (in Germany), when there was only one postal service, they did recieve keys for each house on their route. When the letters were picked up at the local post office, they would take the (often gigantic) key ring with them and return it the post office later for storage.
When alternative postal services started, they generally didn't recieve keys and they had to ring someone to get in.
Today, after many of the local post offices have been closed, the letters are often stored in local storages on the street. I have noticed that they often ring to get in, so I assume they no longer have the key rings since the local storages are not very secure.

Update:
I just met our deliverer and asked about this. She confirmed that they (German Post), but also other postal services, receives the keys for most houses. Some houses, however, are not cooperative and they must ring to get in. If this is not possible, they report this. (I did not ask where the keys are stored)
Comment: So what do they do if no one is there to let them in? Would I have to wait for my letter until the next day? What if no one is home the entire week because they all have to work?
Comment: @Squirrel With the 24 parties in my house, I have never noticed that ever happend. Some modern Apartment houses have outside slits, that you retrieve from the inside. The post office would probably inform the house owner that it is reaccuring problem for which must find a solution for.
Comment: @Squirrel In the US, letter carriers have a key that opens letter boxes (that is, the boxes on the street where one deposits letters one wants to send), and apartment buildings have a switch that accepts that key to operate the door release mechanism.  They also use this key to open the mailboxes to deposit the mail they're delivering.  I don't know how it works in Prague.

How do these in-house letterboxes work?