What type of bike has the most similar feel to a stationary bike?

Question: At the moment I cycle half an hour every evening, on a crappy indoor stationary bike.
BUT I'm pretty good, get some good speeds and distances etc.
I want to start cycling outside again, I want a bike similar to a stationary bike, one that rides similarly.
Maybe very few gears, or just one?
What would you guys recommend??
Comment: BTW, given how crappy even the best stationary bike is, I'm having trouble understanding why you'd want to duplicate the "feel" of one.
Answer: This really depends on the type of stationary bike you are talking about. There's the older style as well as bikes that resemble road bikes (at least in terms of body position), as well as some recumbent bikes. There are bicycles that match pretty well to all three of those stationary bikes.  However, it's worth mentioning that just because you are comfortable on a stationary bike, does not mean you will be comfortable on a bike with the same posture.  For instance the first bike I linked to is very much upright.  A bike like that would be fine for a quick trip to the corner store, but you would probably tire of it quickly if you were trying to get some serious exercise.
Comment: Ah, that's very helpful. I have a stationary bike very similar to the first one you linked to. So try to get serious exercise on it would probably be quite uncomfortable.
 I think I need to experiment. Thanks Kibbee!!
Comment: Yeah, the old upright style of bike would most closely resemble a cruiser such as this (http://www.trekbikes.com/us/en/bikes/town/cruiser/mens_cruiser/) which as the picture shows, are fine for cruising on the beach, or a ride through town, but for fitness they probably aren't the optimal choice.
Answer: A very significant factor with 99% of stationary bikes is that they have no momentum -- when you stop pedaling the bike stops, and it takes extra effort to resume pedaling, even if you only stopped for an instant.  You simply can't reproduce this behavior on a "real world" bike.
Beyond that, your basic stationary bikes provide no variability in resistance, whereas "in the real world" even riding on the flat there is variability as the road goes up and down slightly, the wind shifts, etc.  Reproducing the lack of variability requires a very smooth, flat road on a windless day.  (More sophisticated stationary bikes do provide variability, though it's rarely very realistic.)
And, as Kibbee stated, stationary bikes vary in posture, from very upright to quite recumbent.  The most upright stationary bikes probably resemble a track bike, while others more of a "cruiser".


Nikon D200 Auto Bracketing guidance

Question: Is there a good Nikon resource for explaining how to use after setting up the bracketing?  Do I set the shooting to single or continuous? If it's on continuous would it stop after the bracketing is over or does it continue on
Answer: The user-manual :)
Seriously, read the few paragraphs in the manual that tell you how to activate that feature first.
IIRC you can assign one of the buttons in the front to activate bracketing and speed up access.
Then, you basically specify the number of frames (3-9) and step size (1/3, 1/2, 1) you want. There generally two common reason for using bracketing:

To make sure you get the perfect exposure: Set the set size small, so that you do not miss it because it falls between steps. The number of frames should be set according to how confident you are about the meter.
For HDR / Exposure Blending / Simulating a Graduated ND-filter: You want as large steps as possible to cover the broadest range. The number of frames depends on how contrasty the scene is. The more contrasty, the more frames are needed.

EDIT - In response to question edits :)

Continuous drive mode lets you shoot the whole bracket continuously and stops after that. If you are talking portraits, action or anything that is fleeting, that is probably the best mode to use.
Single advance mode lets you shoot one image at a time. If you are taking long exposures where the timing is critical relative to moving objects, people, passing cars, etc, then you should use this one. This applies for HDR too, where you need to make sure no objects are moving in your frame. For long exposures, you can let vibrations settle between each shot to improve sharpness.
Answer: Setting up bracketing on the D200 is easy. Press and hold the BKT button. Choose the exposure step (0.3, 0.7, 1 stop) with the front dial. Choose the "pattern" (below, above, or around 0) using the rear dial. The top LCD will show a graphic representing the pattern and step.
Reset the bracketing with the rear dial.
Answer: The Manual or Thom Hogan's guides are probably the sources I would suggest.
Your other questions:
press and hold the BKT button (upper back left of the camera)
Turn the rear control dial to select number of shots (2 with one below, 2 with one above, or 3, 5, 7, or 9 evenly spaced) and the front control dial to select the distance each shot should be from what the camera metered for the scene (1/3 to 1/1 EV).
I would reccomend continuos, and yes, the camera will stop in continuous mode after the set is complete. To start another set release and then re-hold the button.
Answer: I think one of the keys here is that you have to hold the button down for the range of shots to complete. If you press and release, you get one at a time. That, it seems is mentioned nowhere in the literature. If it is, then I missed it.

Nikon D200 Auto Bracketing guidance


The NUCLEO F411RE Works with float point?

Question: I need a MCU to works with Wifi and Floating point for PID application. In the later I acquire a Wifi expansion board, but I wants know if the STM32 NUCLEO FR11RE works with a Floating Point. I don't find this information in web.
http://www.st.com/en/evaluation-tools/nucleo-f411re.html
Comment: I don't know what the PID applies to, but in most cases you are receiving data from an ADC and sending data to a DAC. Neither of these use floating point. In between, there is no need for floating point. If you organize your algorithms well, you can avoid its use entirely. And should. Fixed point often isn't correct, either. Instead, you examine your requirements and arrange the operations appropriately to maintain monotonic behavior and retain needed precision throughout. FP is useful when the dynamic range is insane (like galaxies and stars and atoms) or unknown. Or your skills are lacking.
Comment: Yes it does, it does doulbles and floats depending on your complier
Comment: Rather than add wifi to an MCU, consider if you can run your whole algorithm in something like an ESP8266.  But if you really want a Cortex M4 + WiFi, many of the WICED platforms are that.
Answer: I looked through the awful PDF's to see if there was some floating point arithmetic instructions, but then I realized that it doesn't really matter.
Because... is speed that important to you? There are libraries that implement the floating point data type and handles the floating point arithmetic through software. It's being done on every Arduino board. On microcontrollers that doesn't even have floating point arithmetic hardware.
But... if speed is important to you, then you wouldn't use floating points, you would use fixed point arithmetic.

This is what you would call an XY-problem.
Comment: Ridiculous answer. Bunch of your opinions. Not the answer, just the comment. What have you realised is an absolute rubbish. What is the point to take the part in the discussion on the topic  you have no knowledge? FYI it was enough to open STM website. No pdf needed, no instruction searching nessesary. All F3 & F3 have the float FPU + vector DSP. All F7 have the double precision FPU
Comment: @PeterJ - beware of attacking answers that may well provide more *practical* advice than your own.
Comment: @Chris Stratton This is not the answer this is the comment. About something else. Comparing fast FPU on fast 32bit uC with slow software implementation on the prehistoric 8 bit micro controller.
Comment: @PeterJ - no,  you have no basis to know that the speed matters *to that degree*.  This answer proposes two solutions, both of which may well be more suitable *to the actual need*.  For example, I've seen quadcopters fly with Cortex M0 running *multiple* soft float PIDs controlling tiny airframes that respond effectively instantly.  If I were writing it I'd probably use the integer option, but soft float worked fine.
Comment: @Chris Stratton and........ I saw in the computer museum the Apollo mission computer and it has computing power probably lower than 8051. Nowadays we have a huge choice of the uC and can choose the correct one. Stm32f303 with FPU cost less than $3 now. There is no point to limit ourselves.
Answer: In the NUCLEO-F411RE datasheet table 2 states that this particular model has STM32F411RET6. Datasheet of STM32F411RET6 clearly says:

The STM32F411XC/XE devices are based on the high-performance ARM® Cortex® -M4 32-
  bit RISC core operating at a frequency of up to 100 MHz. The Cortex®-M4 core features a
  Floating point unit (FPU) single precision which supports all ARM single-precision dataprocessing
  instructions and data types.

It is a common Cortex-M4F. It supports only single precision floats (ie. float in C, not double). Some tips.
Answer: All STM32F3, F4 & F7 micros have FPU. F3 & F4 single precision, F7 double precision. So you can choose which one you need. It is a bit tricky to configure compiler & linker to force use FPU instructions instead of math functions. Always check what has been generated by the compiler :)
Comment: While factually correct, it's not presently clear if this capability will be of any actual use to the poster.
Comment: @Chris Stratton - OP asked about about floating point, he knows the best what he needs.  So I struggle to understand your comment
Comment: No, Peter, posters who write questions worded this way generally don't know what they need - the question is from an inexperienced person who got an idea in their head - it's almost a given that someone who actually knew what they need wouldn't have had to ask this question at all.  It's not established either that floating point is the only solution, or that hard float is necessary.  One could even argue that it's not clear that a single precision float would be enough, though likely it would be.
Comment: (speed)On the architecture chosen by the OP it actually does not matter if he uses integer or float arithmetics. The speed will be the same. I am not going to judge the OPs knowledge, if he wants FP that is his choice.


How does Toxic set work in Raid : Shadow Legend?

Question: I started playing this game two days ago, so I'm pretty new to it.
I'm not that lost as I'm already playing Summoner wars for more than a year and a lots of mechanics are the same.
I've found some artifacts that don't have equivalent effect in SW, like toxic set, freeze set, sleep set and others sets like this, applying debuf when attack enemy.
I have a question about these sets :
Does multi-hit attacks give multiple chances to apply the debuf?
As an example, does a champion that attack twice with a toxic set has a chance to apply two DoT?
Answer: I need to clarify there is a difference between multi-hit attack and an attack that hits multiple foes (AOE).
A multi hit attack (one that say hits a single opponent 3x) has a single chance to trigger an artifact set debuff.
In your example of Banshee, her attack that hits multiple foes one time (AOE) That means each foe has a chance to receive the debuff.
a check is made on each opponent with % chance to occur, increased by your accuracy, decreased by your opponent's resistance.
Comment: Thank you for addressing both cases, it makes this answer very valuable.
Answer: Artifact Sets activate once per Skill use, so multi-hit attacks count as a single hit when applying Set debuffs
Comment: Does every target have same result ? Example : I got Banshee, which have two aoe, so only one hit but on every opponents. If one trigger the artifact effect, does all others also suffer the effect ?


Count below a comment is misleading

Question: When you are writing a comment, type something with 14 characters. It will tell you that you need one more at least. Leave a space. It will look like you can post the comment. Try posting it, it says you can't. But according to the counter just below the text-area, you should be fine.
Comment: because they are trimming the text?
Comment: @DarthVader Still..
Comment: Still what? if they dont trim the text, u can put 15 empty space.
Comment: Still, it is a bug.
Comment: no it s not. why would it be bug?
Comment: Ok whatever, if you say so..
Comment: put 15 space and try to submit.
Answer: The message showing the number of characters that are still necessary to reach the minimum comment length, or the number of characters that you can still type is shown on browser side, while the validation of the comment is done server-side.
The error message could be changed as the following, but then it should list all the characters that are removed.

Comments must be at least 15 characters in length. Trailing spaces are removed.

For example, 15 spaces followed by a back-tick still cause the error to appear.
Probably the message should be similar to the following one, if it needs to really be changed.

Comments must be at least 15 characters in length. Some characters are not considered when calculating the comment lenght.
Comment: Surely javascript has a :strip" function that can be applied before the "count" function - so that it will accurately determine the *real* allowed character length?
Comment: It is more complicated than removing some characters from the end of the comment. Plus, parsing the comment is done server-side.

Count below a comment is misleading


Alternative to "immediately" given its positive and negative connotations

Question: What's a better way of saying "I'm willing to sign a confidentiality agreement immediately" that conveys that my willingness to sign an agreement immediately is for their benefit but doesn't sound desperate, urgent, or like I'm demanding that they send me the agreement immediately?
Comment: It sounds a bit urgent. It would be appropriate if the person was asking for a quick or immediate response, otherwise, you should say more appropriately "as requested" or "when needed."
Comment: Why not just end the sentence with *agreement*? Do you have any particular reason for having to specify a time frame?
Comment: Jason - I'm trying to convey that I'm willing to sign an agreement as soon as I receive it rather than dragging my feet because I'm not serious or not making it a priority, as has happened to them in the past and wasted their time.
Answer: Try at your convenience:

I can sign a confidentiality agreement at your convenience.

Here's one definition:

When one has the time to do something. No rush, you can fill out that paperwork at your convenience.

This phrase works because it puts considerations of timing to the other party. It focuses on their benefit rather than your own, while conveying that you'll make yourself available when it's convenient for them. It's polite and doesn't presume when is best for them.
This phrase is modifiable with earliest to signal that you'd be available as soon as they could get to it: at your earliest convenience. That said, readers will read this as more urgent, so I'd only use the additional adjective if you want to convey polite urgency.


Light effect on window

Question: I am modelling a building and I am having a hard time getting light to shine through windows.
This is the building - pay attention to the windows on the ground floor, specifically the ones on the right and on the sidewall:

There's a light fixture in the room, which is just a cylinder with emission + blackbody for colour temperature. Now the problem: only one window is bright, the other one is dark but you can see a bit of light cast on the ground outside the dark window.
The node setup as is simple as it gets:

The funny thing is that when I look in 'rendered view mode', both windows are bright however there's no shadow:

I've done some research on how to get light to shine through glass - there's tons of content (including on here) but 2 things:

None of them quite worked - they either fixed something but broke something else or just did not work at all
Most of them seem to use a combination of Glass BSDF + other nodes, which I think it's a legacy way of doing it. I read somewhere that glass in Blender 2.9x is fully supported by the Principled BSDF.

The closest I got to make it work properly was this:

The shadows on the ground look perfect but the window is not bright. Confusingly the dark windows on the side are reflecting the balconies - I don't get why. There's no other source of light except for the light fixtures inside the building.
The node setup was:

How can I get fix my windows so that light + shadows work properly?
BTW I am using Cycles.
Answer: I figured out what the issue was - the 'normals' were wrong.
I turned on 'face orientation' and noticed the outer faces of the window were red. I selected them, went into Edit Mode -> Mesh -> Normals -> Recalculate Inside.
Everything looks good now.

Light effect on window


Why are distant planets illuminated like stars, but when approached closely (by a space telescope for example) it's not illuminated?

Question: when I look up into the night sky I see planets, not twinkling like the other stars, but still illuminated. However, when the Hubble or James Webb telescopes approach planets we see the local features instead of an illuminated sphere in space. why is that?
Comment: HST is in low-Earth orbit, JWST is at L2. Neither, "approach planets" (which planets?). Please clarify your question.
Comment: [Related, possible duplicate](https://physics.stackexchange.com/q/128705/44126).
Comment: thank you but all posts linked above don't answer my question. i'm more questioning the optics behind it, not the photography. for example, even from the ISS they see a non-illuminated earth. im sure a distant earth would appear star-like though
Comment: I suspect you're using the word "illuminated" in some way differently from its usual definition ("lit up, exposed to light, not dark"), which is causing others not to understand what you're asking. Could you please try to clarify the distinction you're trying to make in other words, preferably in as plain and simple terms as you can? (FWIW, I do have a guess as to what you may mean, and if so, the answer comes down to the fact that _any_ small or far-away non-dark object on a dark background looks like a "point of light". But I'd prefer if you clarified your question in your own words first.)
Comment: Ilmari Karonen I suppose what I meant is that we see white light from a distance, meanwhile seeing distinct coloration with a larger view of the planet. the others did a great job at answering my question
Answer: The nearest star (Proxima Centaury) is hundreds of thousands of times farther away from us than the nearest planet (Mars). So even though stars are much larger than planets (the sun is roughly hundred times larger than Earth and many stars are much larger than the sun), they still appear much smaller than the largest visible planet (Jupiter). Our current telescopes have high enough resolution to resolve the planets in the solar system, but are still falling a little short of resolving other stars.
The angular size of an object in space is usually measured in seconds of arc, which is a 3600th part of one degree (and there are 360 degrees in a full circle). An arc second is a small quantity. The angular diameter of the moon is a little over 1800 arc seconds. The angular diameter of Jupiter is between 30 and 50 seconds of arc, depending on where Earth and Jupiter are in their orbits. Saturn appears with an angular diameter of between 14-20 seconds of arc.
The visually "largest" star, R Doradus, has an angular diameter of less than 0.06 seconds of arc.
The resolution of our best optical imaging telescopes, like the Hubble space telescope and the James Webb space telescope is on the order of 0.03-0.1 seconds of arc, so it's barely enough to resolve R Doradus as more than one pixel. There are interferometric telescopes that can measure the angular diameter of stars and reveal some surface details, but we are just beginning to build optical telescopes that will actually show the largest stars as more than just single dots. We also have radio telescopes that can make "images" of stars (and even black holes) with much higher resolution, but they don't show the visual appearance but merely the radio frequency emissions of these objects.
The twinkling of stars is directly related to their small angular size: you are seeing the motion of warm and cold air in the atmosphere, which distorts the light path. It is not an optical feature of the stars themselves. That planets "twinkle less" is due to their larger angular diameter: their light "averages" over more of the atmosphere before it reaches our eye, which reduces the variations in the changes of light intensity.
Disclaimer: all data taken from Wikipedia, so I apologize for any discrepancy with more credible scientific sources.
Comment: Just to add: sometimes, it may seem like the stars are twinkling less. This is due to good atmospheric conditions. When the air is very still and stable, and you are at higher altitude, Earth's atmosphere introduces less of the distortion that @FlatterMann mentioned above. This is one reason why major scientific telescopes tend to be built in these places.
Answer: For objects with significant angular size, optical systems (eyes, classical or digital cameras) adapt the image exposure to get the image brightness of the distant object within their working range, not being too bright nor too dark, losing detains. In such cases, the objects do not seem to be illuminated much.
For objects with negligible angular size on dark background, the same optical systems adapt to the background with near point-like objects as very illuminated. Note that their sensibility is set to be much higher.
In contrary to the former case, the optical systems cannot adapt properly in latter case, measuring proper exposure of the objects. And if they could, such objects would be much harder to notice in the image.
Answer: The planets reflect the radiation from the sun. At the night sky of the earth the planets look like point sources, all the reflected radiation seems to be coming from one point. If one approaches the planet  details of the reflection are seen, depending on the terrain.
We see details of the moon because it is close to earth.
The pictures taken on the moon itself do not show a shining landscape.

Why are distant planets illuminated like stars, but when approached closely (by a space telescope for example) it’s not illuminated?


What information should I provide in my answers?

Question: Is there any information that I have to necessarily provide when I answer to a question?
What information do you suggest to provide?
Answer: When the answer describes a hook that must be implemented to accomplish a task, it would help the OP if it is also provided a link to the documentation page. For Drupal core hooks, the link should take to the Drupal API Reference site; alternatively, DrupalContrib contains the documentation for functions implemented by third-party modules.
Similar link should be provided when referring a Drupal core function, or a function implemented by a module.
It is not information that must be provided, but it surely helps who asked the question (and any user who read the answers to that question) if the links to the documentation pages are provided.
Answer: When the question is about code, give example code, not just a description of what the code should do.
When possible, give links to the documentation for functions and methods used in the shown code; eventual links to blogs or documentation pages that explain the topic or a wider topic are welcome too, especially if the user needs to know something more than what asked in their questions.


Vagrant Docker refuses connection

Question: Stack
I'm running the following:

Metal

Ubuntu 18.04

vagrant

dockerprovider

gitlab dk

Expectation
I'm trying to curl localhost:3000 to reach gitlab.

When I do that from the vagrant box, it works:
vagrant@549f682a30a4:~/gdk$ curl localhost:3000
You are being redirected.vagrant@549f682a30a4:~/gdk$

I also see the request in the gdk tail

When I do that from the ubuntu, it doesnt work:
gdkuser@mymachine:~/gitlab-development-kit$ curl localhost:3000 -v --trace -
Warning: --trace overrides an earlier trace/verbose option
== Info: Rebuilt URL to: localhost:3000/
== Info:   Trying 127.0.0.1...
== Info: TCP_NODELAY set
== Info: Connected to localhost (127.0.0.1) port 3000 (#0)
=> Send header, 78 bytes (0x4e)
0000: 47 45 54 20 2f 20 48 54 54 50 2f 31 2e 31 0d 0a GET / HTTP/1.1..
0010: 48 6f 73 74 3a 20 6c 6f 63 61 6c 68 6f 73 74 3a Host: localhost:
0020: 33 30 30 30 0d 0a 55 73 65 72 2d 41 67 65 6e 74 3000..User-Agent
0030: 3a 20 63 75 72 6c 2f 37 2e 35 38 2e 30 0d 0a 41 : curl/7.58.0..A
0040: 63 63 65 70 74 3a 20 2a 2f 2a 0d 0a 0d 0a       ccept: */*....
== Info: Recv failure: Connection reset by peer
== Info: stopped the pause stream!
== Info: Closing connection 0
curl: (56) Recv failure: Connection reset by peer

I don't see anything in gdk tail

Analysis / Investigation
As I see the request received in gdk tail when it comes from inside vagrant and not, when it's from the host, I assume that it's stuck somewhere on the way. I'm not sure which route the request has to travel. I tried the following things to find out more:

Port Forwarding

Vagrantfile Network part at line 112
config.vm.network "forwarded_port", guest: 3000, host: 3000, auto_correct: true

Host (Ubuntu): netstat -tulpn | grep 3000
tcp6       0      0 :::3000                 :::*                    LISTEN      6508/docker-proxy

Vagrant: netstat -tulpn | grep 3000
tcp        0      0 127.0.0.1:3000          0.0.0.0:*               LISTEN      12387/gitlab-workho
Again, as internal calls are, but external calls are not in gdk tail I assume the request from host never arrives in vagrant.

docker
0.0.0.0:3000->3000/tcp, thus the ports should be forwarded and requests should be seen in docker
gdkuser@mymachine:~/gitlab-development-kit$ docker ps
CONTAINER ID        IMAGE               COMMAND                  CREATED             STATUS              PORTS                                            NAMES
549f682a30a4        cb5d370c7aaf        "/bin/sh -c 'supervi…"   About an hour ago   Up About an hour    0.0.0.0:3000->3000/tcp, 127.0.0.1:2222->22/tcp   gitlab-development-kit_default_1592131487

When attaching to the docker and executing curl on either Vagrant or Ubuntu, I don't get any info in the running Info List in the docker container.

VirtualBox

I've read about troubles with virtualbox 5.2.xx, as I'm running virtualbox 6.1.10, I assume that doesn't apply here. Nevertheless I don't know how to verify, that virtualbox actually forwards the port.
I'm also not quite sure, if VirtualBox forwards to Docker, or vice versa.

Do you have insights how to proceed figuring out, what's wrong?
Thanks a lot!
Answer: Following this issue https://gitlab.com/gitlab-org/gitlab-development-kit/-/issues/545
Run this on the container:
sudo apt install net-tools redir
export EXTIF=`ifconfig eth0 | grep "inet " | awk -F'[: ]+' '{ print $3 }'`
redir $EXTIF:3000 127.0.0.1:3000

Vagrant Docker refuses connection


What distance apart should metal stud securing brackets be placed?

Question: I have hired a contractor to line the walls of an old apartment with metal stud and drywall as the walls are very uneven. The drywall is up but the walls have quite a lot of give in them, if I push against them with my hand the wall wobbles a little, is this acceptable?  I would expect there to be no give at all.
I suspect that this is because the contractor has not secured the stud to the wall with enough mounting brackets, either to save time or money. Is there a standard spacing for securing stud to the wall, my walls are approx 3 metres high and the contractor has used 1 bracket per stud, in some cases there seems to be no bracket, the stud is just screwed at the top and bottom.
Thanks a lot, I can't afford to have this fixed afterwards so need to make sure this guy does it right before it is plastered.
Comment: I would ask for a picture of the installed studs but if they are already covered with the drywall there should not be much to see.  I would think one issue is the uneven walls make it hard to secure each and every stud tight to the wall as this would possibly draw the new stud wall in to unevenness.
Comment: Metal studs do feel "wobbly" compared to wood when free standing. Once the sheetrock is screwed in place it will feel normal.
Answer: Typically the studs would be fastened to the wall with clips or L metal.
Spacing would be 24 inches vertical & 48 inches Horizontally.
Metal studs have a punch out for bridging typical 4' 6" vertical for cold rolled channel to be in stalled some bridging require clips screw attached and others just snap in place depending on the manufacture.
You also need to check on the Seismic rating as you stated plaster was being installed.
Answer: Normally finished walls with metal studs feel just as solid as wood framed.  They are supposedly stronger than wood framed, not sure I buy that, but I am confident they are strong enough.
As for the requirement of wall brackets - without knowing more about how the walls are supported you couldn't say home many are required.  After all it's very possible to construct metal framed partition walls that are attached only to each other and the floor that are perfectly sturdy and rigid.


Assigning attribute based on selection points on a graph?

Question: I have a bunch of points on a map on a river that branches. I want an easy way to select one branch from the other and was hoping I could just somehow "grab" those points on the map and add a column to the attribute table with those points that would delineate the two branches.
Comment: Have you tried `'Select by location'`? For instance, there is a related thread [How to build a query/expression to select specific points within a buffer in QGIS](https://gis.stackexchange.com/questions/179661/how-to-build-a-query-expression-to-select-specific-points-within-a-buffer-in-q). Or maybe this article can help you as well [Performing Spatial Queries](https://www.qgistutorials.com/en/docs/performing_spatial_queries.html)
Answer: Frist of all make an attribute to every single branch; every single branch of the river must be a single polygon's layer feature.
by your question I think you need to do the following steps:
Step 1
Generate a buffer poligon of every branch
Qgis~Menù -> Vector -> Geoprocessing Tool -> Buffer. Set "distance" value in a way that you are sure to cover the points around every branch.
Step 2
Use the new buffered polygon layer in a spatial join of the points to add specific attribute
Qgis~Menù -> Vector -> Data Management Tools -> Join attribute by location
Set "Input Layer = yourPointsLayer " and set "Join layer = buffered-branch-layer ", use these (check on) geometric predicate: "intersect, touch, are within", select the field to add (for example the field generated on step 1).
NOTE:
There are two important settings that are critical in the process

The buffer size (step 1) must be not too small or too big
The "join type" setting in spatial join (step 2)


Home Made Sauce comprising of other sauces

Question: I have made a sauce using my own chillies and spices sugar etc but there is 3 types of sauce by different companies in the recipe of the sauce. Can I patent my sauce or am I not allowed to use someone else's sauce as part of the make up of my own?
Answer: Yes, you could, in principle, patent your sauce.
A food product (such as a sauce) can be patented. In this case, food is not really any different from any other product. The food product simply needs to meet the requirements of novelty and non-obviousness. It is irrelevant whether the ingredients in the sauce are themselves patented.
Novelty is pretty easy to show for a food product—there just has to be a difference from what's been done before. However, non-obviousness is tough: how do you show that your new sauce is not an obvious combination of existing sauces? It is rare that there is a surprising and unintuitive effect from a combination of ingredients that would support an argument of non-obviousness. For example, combining hot sauce and mayonnaise obviously leads to delicious spicy mayonnaise, but not to anything patentable. This is why patents for food products which are just combinations of known ingredients are relatively rare.
However, it is still quite possible to patent a food product, as long as it is something more than an obvious combination of ingredients.
For example, it is now well known that adding a lot of sugar to a product tends to inhibit bacterial growth. However, assuming this was not well known, and that you had just discovered this (and thus had just invented jam), this would be novel and non-obvious, and therefore patentable. This goes much further than a simple recipe where you are just combining sugar (for sweetness) and fruit (for fruit flavour), which results in an obvious combination (sweetened fruit-flavoured product). There is instead a surprising result in the end (sweetened fruit-flavoured product which doesn't grow mould).
Answer: Recipes (for sauces, etc.) cannot be patented under US law.
Comment: This is true, but could you add some sources?
Comment: Recipes  cannot be patented but its something related to obviousness as mentioned by Maca, A new Sauce likely to have been created in a laboratory than on a kitchen counter can be patentable. For further reference, (https://www.uspto.gov/custom-page/inventors-eye-advice-1) which mentions about can Recipes Be Patented.
Comment: I like to add an example, chocolate cannot be patented but air infused chocolate can be patented
Answer: You may be interested in learning more about trade secrets from the USPTO website.


Why didn't Rutherford measure Bragg reflexion off the gold foil?

Question: As far as I know Rutherford measured only the
$$I\propto\frac{1}{\sin(\theta/2)^4}$$
intensity dependence, which we would expect for isolated gold atoms. Why didn't he measure Bragg scattering at the gold crystal?
Comment: Because the alpha particles were not diffracting off the gold.
Comment: Did they have too high energy and only interacted with the core or why is that? @JonCuster
Comment: Indeed, they were just scattering from nuclei. You can calculate the de Broglie wavelength and see just how small it is.
Comment: @JonCuster Of course.. Should have thought about that from the beginning! Thank you.
Answer: Rutherford used $^{222}$Ra as an alpha particle source.  The resulting $\sim 5.6$ MeV alpha particles have a de Broglie wavelength on the scale of femtometers. On the other hand, the Bragg condition for constructive interference is $n\lambda = d\sin(\theta)$, with $d\sim 1$ nm the spacing of the atomic planes. Since $\lambda/d \sim 10^{-6}$, the scattering maxima are far too close together to resolve.
As a general rule, the energy/momentum of a particular probe (in this case, the alpha particle) defines a length scale which corresponds to the size of features you can expect to observe.  If you want to resolve the atomic structure of a solid, you need to use a probe which is sensitive to nanometer-scale phenomena.  A $5.6$ MeV alpha is, in a sense, too small; from its point of view, the atoms which comprise a gold crystal may as well be isolated.
Comment: Now it makes perfect sense. Thank you. When reading your answer and Jon's comment I remembered that I thought about the very same thing a year ago and got the answer right back then, but totally forgot about it.


What exactly are dx and dy in differential equations?

Question: In the notation for integrals, I was taught that we use a "dx" at the end as a delimiter and to tell us what variable we are integrating with respect to, and that it is not a quantity being multiplied by the integrand. But in differential equations, we multiply the denominator of dy/dx and receive dx multiplied by some function on the right hand side (what does that even mean?) and then we integrate that expression, which seems weird to me if dx isn't a factor of the integrand. What's going on here?
Comment: In undergraduate differential equations courses, calculations that involve manipulating $dx$ and $dy$ as independent quantities can always be rephrased easily to avoid doing so. (That said, "infinitesimal intuition" should not be thrown out the window. For intuition, we can think of $dx$ and $dy$ as tiny but finite quantities and do calculations with them, obtaining approximate equalities which "in the limit" become exact equalities. Infinitesimal intuition is very clear and powerful, so it's not hard to see why physicists and other people -- including mathematicians -- might use it often.)
Comment: This question might be helpful: https://math.stackexchange.com/questions/47092/physicists-not-mathematicians-can-multiply-both-sides-with-dx-why
Comment: And [this](https://math.stackexchange.com/questions/23902/what-is-the-practical-difference-between-a-differential-and-a-derivative) and [this](https://math.stackexchange.com/questions/21199/is-frac-textrmdy-textrmdx-not-a-ratio)
Comment: @teddy The question mentions differential forms - is that what dx and dy are "formally" in the context of my question too?
Comment: @hawexp I'm not sure since I don't have the mathematical background/formalism in that field, but that answer also avoided using differential forms, which I thought might be helpful.
Answer: If I understand correctly, you mean if we have something like:
$$\frac{dy}{dx}=f(y)g(x)$$
then we get:
$$\int\frac{1}{f(y)}\frac{dy}{dx}dx=\int g(x)dx$$
writing it like this shows that we integrate wrt the same variable on both sides but it can be simplified to:
$$\int\frac{dy}{f(y)}=\int g(x)dx$$
similarly if we have an expression of the form:
$$f(x,y)dx+g(x,y)dy=0$$
we can view this as:
$$\lim_{(\delta x,\delta y)\to(0,0)}f(x,y)\delta x+g(x,y)\delta y\to0$$
whilst this is not exactly true since the variables are dependent I think it helps to understand what they represent


Training Zombies in Black Ops 2

Question: I have the original zombies pack (green run) on my PS3 and I want to know how I can train Zombies properly to reach higher rounds. The lava in particular seems to be significant obstacle in training. In Tranzit, I end up dying in the bus around round 25, and the only place where I'm able to move around enough to get them in a bunch is the town area. Is there any other thing in Tranzit/single maps which I can exploit to my advantage? Any strategies are also welcome.
Answer: Town is the easiest best place to train zombies.  You are in an open area with multiple routes.
You can train a group in the area where the bus parks by running around the perimeter.  This is the area between the bank and the bar.  The bar also contains a pool table which you can use to turn and shoot at your train.  You can exit up the stairs, then fall down into the area below (therefore allowing you to repeat the building of a train).
You should be able to rinse and repeat with some practise.  You are next to the bank so have the option to drop off points for future games.  Jug is nearby too.  If you get into a jam you can run down into the mist or back into the main area around the corner.
Alternatively, you can avoid training zombies completely and just hold out under the bank.  You have access to a pack-a-punch machine under the bank.  Put 1-2 people at the top of the stairs facing the corridor that leads to the main bank area (don't go beyond the hole in the wall to the right).  Have the rest of your group in the tunnels below making sure nothing gets out of the windows.  The 1 or 2 guarding the vault entrance may rotate with people below to spread out the points being earned.
Update
Having watched several videos, the diner seems like another good place.  There is an open area where the bus exits the area, between the mist and the slightly open garage door.  You have plenty of room here to run circles.  If you are working in a full group then have people split up (maybe 2 outside bank, 2 outside diner).  It gets too overcrowded on later rounds for everyone in the same area.
As you have mentioned ammo being a problem, I would suggest you want to have 2 pack-a-punched weapons (something like starting pistol and ray gun for example) without relying on guns from the wall.  If the bank vault is open, power is up and the box is in the town you can rotate guns to increase your ammo (replace empty weapons, then pack a punch the new gun).  Other than that it is just a case of making the most of your weapons (explosive rounds into large groups of zombies, headshots etc) and picking up max ammo when it is flashing (emptying your gun whilst you wait, chucking grenades/emps too).  Having the knife or knuckles can help but loses its effectiveness during later rounds.  Its feasible to use knuckles only for the first 10 rounds at least, without ever needing to fire a bullet.  The only other tip I can give you is to control the rounds (leave a couple of zombies at the end of each wave) so that you can get perks/new weapons without a horde following you.
Comment: But what about ammo? On higher rounds, non-pack-a-punched guns are a licence to dying. How does one juggle around?
Answer: With my experience on town, where jug is there is a good point weapon on the wall and I have seen many people train  around there after buying it and you get bigger groups while training so points is not a problem and if you open the door to where the box can be, training is simple and on solo, you cant go to very high rounds without quick revive due to it reviving you if you get down on solo.Some You-tubers train before they get their pack-a-punched weapon.

Training Zombies in Black Ops 2


What AGC should I use in an Icom IC-705 when doing FT8 or other digital modes with WSJT-X?

Question: The User Guide for WSJT-X says  It is usually best to turn AGC off or reduce the RF gain control to minimize AGC action, but the Icom IC-705 seems not to have an option to turn AGC off, only to set it to fast, medium and slow. Should it be set to slow?
Answer: Actually you can get AGC off on the 705. The process is a bit roundabout, but it's the same as the other Icom touchscreen radios.

Set the mode you're interested in (i.e. USB-D for FT8) because AGC settings are remembered per mode.
Open the FUNCTION quick menu.
Press and hold AGC for one second to open the AGC settings menu.
Press whichever one of the AGC settings you want to be replaced by OFF.
Turn the main dial all the way left — the setting below the lowest AGC time constant should be OFF.
Press EXIT.
Now you can select AGC OFF at any time.

FT8 isn't really that sensitive to AGC, though; you'll do very little harm by just leaving the AGC at MED.
Comment: Thank you so much. I completely missed that, and I played with it and read the manual. I was expecting off to be at the other end of the spectrum. Do you want double the karma? https://ham.stackexchange.com/questions/22017/how-do-you-turn-off-agc-in-an-icom-ic-705


Install Rational Team Concert & Rational Insight

Question: I need to install both RTC and Rational Insight in my local machine but I am confuse that should I install Data base first for it. Even i don't know which database is right with Insight & RTC.
I don't know how to set up Dataware House?
Answer: The choice of database is important when combining RTC and Insight. With RTC by itself, especially with a small team, you can use Derby (included) for both the repository database (basically, the operational database for the product) and for the data warehouse (a separate database used for reporting). However, if you want to use Rational Insight, or RRDI (both of which include Cognos-based reporting), your must have a data warehouse on a database other than Derby. Typically, Oracle, SQL Server and DB2 are your choices, but please consult the platform requirements list for your version of Rational Insight to ensure you're using a supported configuration.
Regarding setting up the data warehouse, if you're using RTC with Insight, often you will share a warehouse between the two applications (that is, RTC can use the warehouse for its internal BIRT-based reporting, and Insight can use the warehouse for its Cognos-based reporting). In that case, you would set up the data warehouse as part of your RTC setup (the JTS setup wizard configures the data warehouse), and then point Insight at that warehouse when you set Insight up.
I hope this helps.
James
Technical Lead, IBM Rational Reporting


How can I put a string containing html text to be displayed in a javascript function?

Question: How can I put a string containing html text to be displayed in a JavaScript function?
String test = "DisplayNext(\"String containing html tags\")";

String NextLink = "
 Next";

JavaScript function:
function DisplayNext(Next){
    alert(Next);
}

I want to pass html tags to a JavaScript function and display it using an alert box..How do I do that ??
Answer: You'll need to encode your html tags.
> becomes >
< becomes <
" becomes "
You'll also need to encode your JavaScript.
' becomes \'
Then use single quotes to quote your JavaScript (so the double quotes don't close the onclick attribute prematurely):
String test = "DisplayNext('" + encodedHTML + "')";
String nextLink = "
 Next";

Answer: Javascript alert boxes aren't HTML viewers.  You can try replacing your alert calls with this:
http://plugins.jquery.com/plugin-tags/messagebox


Was Classical Latin spoken fast like in the Romance languages or slow like in English?

Question: Was Classical Latin spoken fast like in the Romance languages or slow like in English? In all the Nuntii Latini episodes, Classical Latin is spoken fast like in the Romance languages, https://areena.yle.fi/audio/1-1931339, Thomas Bervoets speaks Classical Latin both fast like in the Romance languages, https://www.stilus.nl/ce-geluid/47.mp3 and slow like in English! https://www.stilus.nl/ce-geluid/VergAen%20III,506-569%20Provehimur%20pelago.mp3
Comment: My results on this topic will hopefully be published soon, and then I'll have citable evidence for a proper answer; in the meantime, I can say it seems to have been spoken significantly slower than modern Romance languages, closer to English.
Comment: @TKR: one way of measuring speech rate is number of syllables/unit of time, and languages do have measurable differences on this metric (which is likely related a lot to the types of syllables that a language uses). Some relevant Language Log posts: "[Speech rate and per-syllable information across languages](https://languagelog.ldc.upenn.edu/nll/?p=22)", 2008, Mark Liberman, "[Speed vs. efficiency in speech production and reception](https://languagelog.ldc.upenn.edu/nll/?p=44386)", 2019, Victor Mair
Comment: @AnaMaria I sent my final revisions in to the editors yesterday, so hopefully soon!
Comment: @Jacqueline Indeed it has! I'd forgotten about this; let me add it as a new answer.
Comment: The assumptions in this question seem unfounded. Speaking rate in all languages varies between speakers and situations. It's not at all clear to me either that one can meaningfully talk about the speaking rate of a language, or that the Romance languages are generally spoken faster than English in any sense.
Comment: @Asteroides Thanks, I hadn't seen that research. But syllables/time doesn't seem like a good metric here because as you say, syllable types vary across languages. (And from the LL posts it looks like those papers end up arguing for a similar overall "rate".) Given the amount and more importantly the multiple dimensions of variation within any given language it seems like it would be hard even to define an "average". FWIW subjectively I don't share the OP's impression that Romance languages are generally spoken faster than English in any sense, so I'd like to see some basis for that assumption.
Comment: @Draconis When do you think your results on this topic will be published?
Comment: @Draconis Has it been published yet? It would be a great piece to read indeed.
Answer: Stelzer 2022 used information theory to estimate the speech rate of Classical Latin. Basically, there seems to be an optimal rate for information conveyed per second across all the world's (spoken) languages; if a language conveys more information per syllable, it's spoken with fewer syllables per second, and vice versa. Since we can measure the amount of information conveyed per syllable in Classical Latin, we can "reverse-engineer" a speech rate from that.
The result was an estimated 6.29 syllables per second, significantly slower than the modern Romance languages. (Bars show one standard deviation around the mean.)

However, it's fairly average for the languages of the world, and in fact very close to modern English.

Disclaimer: I am the author of this paper.
Comment: @AnaMaria Good question! It _probably_ didn't—generally poetry and speeches come a bit slower, to make sure the audience understands them, while reading a letter happens a bit faster, since the words are already prepared for you. Unfortunately I'm not familiar with any studies on this but it seems like a fertile area for further research!
Comment: Did the speech rate stay the same when reciting poetry, making speeches, and reading letters?
Comment: I remember now that there was a rhythmic figure in Classical Latin speeches called Clausula, used to add finality to the end of a sentence or phrase, which also existed in letters! The fact that there was rhythm implies that Clausula was fast spoken, and Wikipedia says that there is no doubt that the skilful use of clausulae was one of the techniques which an orator used to excite an audience. Cicero writes of one occasion when the use of a certain clausula (a dichoreus or double trochee – u – x) by the orator Carbo the Younger was so effective that the audience all gave a shout!
Comment: https://en.wikipedia.org/wiki/Clausula_(rhetoric)
Answer: I'm sure that the actual rate of speech would vary based on region, the age of the speaker, fluency, etc. just as in any language today.
But it's clear that elision of final vowels was common (as it is in many modern Romance languages) which may have made it sound fast to an English speaker, as that's one of the features of Romance languages that can make them sound "fast".


What units is strong nuclear charge measured in?

Question: Do particles have a strong nuclear charge in the same way as they have electric charge?  If so what unit would be used to measure this?  Would it be measured in Coulomb for instance?
Comment: Does this answer your question? [Base quantities and charges](https://physics.stackexchange.com/q/64568/)
Answer: The electromagnetic force couples to electric charge, the gravitational force couples to mass charge and the strong nuclear force couples to color charge. Photons are the exchange bosons in the electromagnetic force, gravitons - if they exist- are the exchange bosons in the gravitational force.
The strong nuclear force is mediated by the exchange bosons called gluons.

Do particles have a strong nuclear charge in the same way as they have electric charge?

No, not really, since (as below), all particles (that exist as natural particle states) that take part in the strong nuclear force, will always have a zero net color charge, but can have net electric charge.

Would it be measured in Coulomb for instance?

Color charge is not really analogous to electric charge in this sense. Electric charge has one value (which is some number multiplied by the electron charge).
But color charge can be one of six possible values (red, blue, green, and antired, antiblue, antigreen). These are quantum numbers we assign to the six different flavor of  quarks which combine to form hadrons. While particles like $\pi^-$, K$^-$, protons, $\Delta^-$, $\Delta^{++}$ etc., can have a net electric charge, all particles containing quarks must always have zero, or neutral, net color charge due to quark confinement
For more about this, see quantum chromodynamics
which describes the behavior of the strong nuclear force and strong interactions.


Orderable columns

Question: I am using a apex:pageBlockTable with apex:columns on VF page. Is there any tag I can ad to have sortable columns. Note: I would like the column sorting to be in - memory rather than hitting the server again.
Comment: I think you can do it only with building some custom sort function. you can use JS or JQuery to do it if you dont want to do the server call. There are many open source examples on web for this. One of them is http://blogforce9.blogspot.in/2012/08/sort-able-pageblock-table-component-for_23.html
Answer: While I understand the desire to stick with pure VF, for column sorting and other advanced datagrids features you have to branch out to a Javascript framework.
I've personally used the following datagrid plugins, even some with in-line editing.
http://www.datatables.net/ - Requires jQuery, and can ajax load the data. While you can go to great lengths to rewrite parts of the plugin to utilize the ajax proxy I find it much simpler to build out a simple visualforce page returning json data for the grid. The downside to this one is the rather arcane syntax used to configure the datatable. Definitely a swiss-army-tool-gone-wild.
http://demos.kendoui.com/web/grid/editing-custom.html - Kendo isn't free but it's amongst the best data grids out there. easy to configure with a fallback option of always contacting support.
http://docs.sencha.com/ext-js/4-2/extjs-build/examples/build/KitchenSink/ext-theme-neptune/#array-grid - Ext.js from Scencha is also not free, and it looks a little 200&late to me, but it is a well done data grid.
Lastly, and honestly this is my personal recommendation though the learning curve is quite steep, http://angular-ui.github.com/ng-grid/ This one requires not only jQuery, but also Angular.js Angular is a relative sleeper-cell amongst the JS frameworks and works radically differently than anything else I've ever seen. That said, it's by far the most powerful, most concise, and most semantic of all these data grids. In the Angular world (ng for short) you develop "directives" which act as either Elements, Attributes, Classes or Comments to standard HTML markup. In other words you can develop a directive called dataGrid, specify that it's an Element directive and then utilize the brand new-materialized-from-your-js-code <data-grid> </data-grid> tags in your visual force pages. Neat.
Answer: Unfortunately as of today there isn't any native support for sorting with the apex:pageBlockTable, you've got to roll your own.  This can usually be done pretty easily with an javascript ui framework (i.e. jQuery UI or YUI).

Orderable columns


How to create a productrepository using object manager in Magento 2?

Question: I want to save/update the product in phtml file and I need to create productRepository using object manager.
I don't want any controller to be called. Is there any possibility to save product in phtml file without calling controller or helper class?
Answer: No. Noooo. Not the right approach.
You haven't given much information about exactly what problem you are trying to solve, but you should never be loading or saving any model directly in a template.
The correct process depends on the circumstances, but you should be running your product logic in a context where you can take advantage of Dependency Injection (DI)--a block, a controller, an observer, but preferably a model. In that class, you would use DI to get an instance of \Magento\Catalog\Api\ProductRepositoryInterface (ProductRepository), and then you could use that to create/load your product and then to save it.
Using ObjectManager to directly create and save a Product object, in a template, is breaking just about every rule and code standard in the book. I cannot overstate this: What you are asking to do is very bad practice. Don't do it.

Don't use ObjectManager. The only place it should be used is in a Factory class.
Don't use business logic in a template. Templates are for displaying content only; they should never contain any code that changes the state of the system. Some logic can be contained in a template's Block class, but that block also should not change any system state. Changes (like creating a product) belong in a Model.
Use dependency injection and inject the objects you need in your class constructors. It makes it clear what your dependencies are, and allows you/others to change them as well via DI.
ORM objects (anything stored/loaded in the database, like Products) are non-injectable. You need to inject its Factory (literally append Factory to the class name you're getting), and use that to create your object instance.
Use the service layer. That means use DI to get interfaces corresponding to objects, not the objects themselves.

Ignoring these standards can give you code that works (like the other answer), but it's code that will probably break in future versions, it's code that won't be findable (it's not where standards say it should be), it's code that isn't maintainable (does things wrong), and you're making a mess of your site in the process.
Taken all together: Here is how to inject, create, and save a product correctly in the context of dependency injection and M2 code standards.
<?php
/**
 * MyClass Class
 */
class MyClass
{
    /**
     * @var \Magento\Catalog\Api\Data\ProductInterfaceFactory
     */
    private $productFactory;
    /**
     * @var \Magento\Catalog\Api\ProductRepositoryInterface
     */
    private $productRepository;

    /**
     * MyClass constructor
     *
     * @param \Magento\Catalog\Api\Data\ProductInterfaceFactory $productFactory
     * @param \Magento\Catalog\Api\ProductRepositoryInterface $productRepository
     */
    public function __construct(
        \Magento\Catalog\Api\Data\ProductInterfaceFactory $productFactory,
        \Magento\Catalog\Api\ProductRepositoryInterface $productRepository
    ) {
        $this->productFactory = $productFactory;
        $this->productRepository = $productRepository;
    }

    /**
     * Create my new simple product.
     *
     * @return void
     */
    public function createProduct()
    {
        $product = $this->productFactory->create();
        $product->setSku('MyNewProduct');
        // Set other product data here
        $this->productRepository->save($product);
    }
}

There are a lot of complex concepts at play here, but it's worth the time to learn and understand them and do things right. You'll be able to work with M2 much more effectively when you do.
Comment: Thank you sir and sorry for not using the Magento 2 coding standards
Comment: Don't sir me. :) I'm not trying to discourage or yell at you, hope it didn't come across that way--I just care a lot about things being done right (because I spend a lot of time fixing sites and extensions where they weren't). I hope the info I gave was enough to get you headed in the right direction. If you don't already use it, check out PhpStorm, and definitely make use of the official docs and other resources out there.
Comment: for testing this can be useful to use object manager to quickly see, for example, an output  product detail etc.


Products and coproducts in the simplex category

Question: I wonder if the category $\Delta$ of finite totally ordered sets and monotone functions has binary products and coproducts. In particular, is the ordinal sum a categorical coproduct and/or is there any comparable "ordinal product"?
Comment: Ah, but now I think my $h$ is not necessarily monotone :(
Comment: If $[n]=\{1,2,..,n\}$, and $f:[n]\rightarrow [k]$ and $g:[m]\rightarrow [k]$ are monotone functions, can you always find a $h:[n+m]=[n]+[m]\rightarrow [k]$ which is a monotone function on and which, when restricted to the first $n$ elements is $f$ and last $m$ elements is equal to $g$?
Comment: I believe we can. Define $h(x) = f(x)$ if $x \le n$ and $h(x) = g(x - n)$ otherwise.
Comment: Products have a similar problem - specifically, there has to be a category isomorphism between $[n]\times[m]$ and $[m]\times[n]$ which commutes with projections to $[n]$ and $[m]$.
Comment: @saintali: Take Thomas' example with $n=m=k$. Can there be a monotone function from $n+n$ into $n$?
Answer: Rewritten completely, same conclusions.
Let $\mathbf{TOrd}$ be the category of totally ordered sets with monotone maps (maps such that $a\leq b\Rightarrow f(a)\leq f(b)$); let $\mathbf{TOrd}_s$ b the category of totally ordered sets with strictly monotone maps (maps such that $a\lt b \leftrightarrow f(a)\lt f(b)$); and let $\mathbf{FTord}$ and $\mathbf{FTord}_s$ be the full subcategories of finite totally ordered sets. (The diference between $\mathbf{TOrd}$ and $\mathbf{TOrd}_s$ is that in the former we can have $a\lt b$ and $f(a)=f(b)$, whereas in the latter we cannot.
Theorem 1. Let $\mathcal{C}$ be any of $\mathbf{TOrd}$, $\mathbf{TOrd}_s$, $\mathbf{FTOrd}$, or $\mathbf{TOrd}_s$. If $A$ and $B$ each have at least two elements, then there is no product of $A$ and $B$ in $\mathcal{C}$.
Proof. Assume that $A$ and $B$ are totally ordered sets, and each has at least two elements. Let $a_1,a_2\in A$, $a_1\lt a_2$; and let $b_1,b_2\in B$, with $b_1\lt b_2$. Consider the maps from the singleton totally ordered set $S=\{\bullet\}$ to $A$ and $B$ given by $f_A\colon S\to A$, $f_B\colon S\to A$, $f_A(\bullet)=a_2$, $f_B(\bullet)=b_1$; and $g_A\colon S\to A$, $g_B\colon S\to B$ given by $g_A(\bullet)=a_1$, $g_B(\bullet)=b_2$.
If there is a totally ordered set $X$ with monotone maps $p_A\colon X\to A$ and $p_B\colon X\to B$ such that both $f_A,f_B$ and $g_A,g_B$ factors through $(X,p_A,p_B)$ via maps $F,G\colon S\to X$, then note that since $p_A(F(\bullet))=a_2\gt a_1= p_A(G(\bullet))$, we must have $F(\bullet)\gt G(\bullet)$. But then we must have $b_1 = p_B(F(\bullet)) \gt p_B(G(\bullet))=b_2$, contradicting the choice of $b_1$ and $b_2$. Thus, no such $X$ exists, and therefore there is no product of $A$ and $B$ in $\mathcal{C}$. $\Box$
Theorem 2. Let $\mathcal{C}$ be any of $\mathbf{TOrd}$, $\mathbf{TOrd}_s$, $\mathbf{FTOrd}$, or $\mathbf{TOrd}_s$. If $A$ and $B$ each have at least one element each, then there is no coproduct of $A$ and $B$ in $\mathcal{C}$.
Proof. Let $X = \{1\}\times A\cup \{2\}\times B\cup \{3\}\times A$; order $X$ lexicographically; that is, $(i,x)\leq (j,y)$ if and only if $i\lt j$, or $i=j$ and $x\leq y$. Let $f_1,f_3\colon A\to X$ be given by $f_i(a) = (i,a)$, and let $g\colon B\to X$ be given by $g(b)=(2,b)$.
If there is an object $C$ in $\mathcal{C}$ and maps $\iota_A\colon A\to C$ and $\iota_B\colon B\to C$ such that both $f_1,g$ and $f_3,g$ factor through $(C,\iota_A,\iota_B)$, then there exist maps $F_1\colon C\to X$ and $F_3\colon C\to X$ such that $f_1=F_1\circ\iota_A$, $f_3=F_3\circ\iota_A$, and $g=F_1\circ\iota_B = F_3\circ\iota_B$. Let $a\in A$ and $b\in B$. Since $f_1(a)\lt g(b)$, then $\iota_A(a)\lt iota_B(b)$; but since $g(b)\lt f_3(a)$, we must have $\iota_B(b)\lt \iota_A(a)$. This is impossible, so no such object $C$ exists. That is, $A$ and $B$ do not have a coproduct in $\mathcal{C}$. $\Box$
For the remaining cases, we have:
Theorem 3. Let $\mathcal{C}$ be $\mathbf{TOrd}$ or $\mathbf{FTOrd}$.

If $A$ is a singleton, and $B$ is arbitrary, then $(B,p_A,\mathrm{id}_B)$ is a product of $A$ and $B$ in $\mathcal{C}$, where $p_A\colon B\to A$ is the unique map from $B$ to $A$.
If $A$ is empty, and $B$ is arbitrary, then $(A,\mathrm{id}_A,n)$ is a product of $A$ and $B$ in $\mathcal{C}$, where $n$ is the empty map with domain $A$ and codomain $b$.

Proof.

Let $X$ be any object of $\mathcal{C}$, and let $f\colon X\to A$ and $g\colon X\to B$ be maps. Then the maps factor uniquely through $B$ via $f$, so $(B,p_A,\mathrm{iid}_B)$ is a product of $A$ and $B$.
Let $X$ be an object of $\mathcal{C}$ and $f\colon X\to A$ and $g\colon X\to B$ be maps; since $A$ is empty, then $X$ must be empty, so the unique map $X\to A$ shows that $(A,\mathrm{id}_A,n)$ has the desired universal property. $\Box$

Theorem 4. Let $\mathcal{C}$ be $\mathbf{TOrd}_s$ or $\mathbf{FTOrd}_s$.

If $A$ is a singleton and $B$ is not empty and not a singleton, then there is no product of $A$ and $B$ in $\mathcal{C}$.
If $A$ and $B$ are both singletons, then $(A,\mathrm{id}_A,p)$ is a product of $A$ and $B$, where $p$ is the unique map $p\colon A\to B$.
If $A$ is empty, and $B$ is arbitrary, then $(A,\mathrm{id}_A,n)$ is a product of $A$ and $B$ in $\mathcal{C}$, where $n\colon A\to B$ is the empty map.

Proof.

Since maps in $\mathcal{C}$ must be one-to-one, a product of $A$ and $B$ must be either empty or a singleton (otherwise, there are no maps into $A$). But then $p_B(C)\neq B$; letting $S=\{\bullet\}$ be a singleton, let $b\in B-p_B(C)$; then the pair $(f,g)$, where $f\colon S\to A$ is the unique map between singletons and $g\colon S\to B$ has $g(\bullet)=b$ cannot factor through $(C,p_A,p_B)$. So $A$ and $B$ do not have a product in $\mathcal{C}$.
The only objects that can map to $A$ and $B$ are the emptyset and singletons, and the maps clearly factor through $(A,\mathrm{id}_A,p)$.
The only object that can map to $A$ is the empty set. $\Box$

Theorem 5. Let $\mathcal{C}$ be $\mathbf{TOrd}$, $\mathbf{FTOrd}$, $\mathbf{TOrd}_s$, or $\mathbf{FTOrd}_s$. If $A$ is empty and $B$ is arbitrary, then $(B,n,\mathrm{id}_B)$ is a coproduct of $A$ and $B$, where $n\colon A\to B$ is the empty map.
Proof. Let $X$ be an object of $\mathcal{C}$, $f\colon A\to X$, and $g\colon B\to X$. Then $f$ must be the empty map, so $f$ and $g$ factor uniquely through $(B,n,\mathrm{id}_B)$. Thus, this is a coproduct for $A$ and $B$. $\Box$


Can windows viruses affect my encrypted linux partition on the same disk?

Question: Can Windows viruses affect my encrypted Linux partition on the same disk? I just want to use Windows but I am concerned.
Answer: Only indirectly: Windows also has BIOS/UEFI malware. That could affect your ability to boot your system including non-Windows systems like Ubuntu. A full disk wipe is also possible.
Only viruses targeted at Linux would affect Ubuntu. There are currently zero viruses active that are targetting Linux the way Windows can be infested. We desktop users are still a lot more better protected out of the box than Windors users.
A fully updated current Ubuntu could have an unknown bug in a software someone could abuse to get a virus onto your system but there is currently not a method where a malware infects millions of systems by releasing that virus. Your system would be targeted specifically to attack you.
You can install a virus yourself by downloading it as part of another software but that would be from an unsecure source and we just do not do that. All of our ways of installing software are through trusted sources.
If possible I would suggest to use two separate systems if you want to use Windows and Ubuntu. And you can then also use the Ubuntu system to check all incoming traffic to the Windows system.
When you ...

use a decent admin password and do not input it when you do not expect it
use backups for your essential data and store them with a timestamp on another machine
keep your system updated
do not install software you do not need
use it as a desktop (so no server tools like an ftp server, fileserver, website on yours system). If you use those you are in need of some extra tools.
preferably stay away from Windows; but you can use clamav for instance to scan Windows

the chance of having a virus are close to zero. But these are basically normal things for an Ubuntu user and not specifically to prevent a virus.
Comment: Well, a Windows virus which wipes the disk device will certainly directly affect your Linux partition, encrypted or not.  Good backups are especially recommended these days with a Cyberwar ongoing.
Comment: if the partition is encrypted no; at most someone can copy the whole encrypted partition. Then they would still need your key. Brute forcing a good key would take a lot of years. If the partition can be active when windows is active yes.
Comment: "all known viruses need to be installed by a user" - That's just not true... There are plenty of RCE vulnerabilities and therefore malware that can exploit them. Sure, they're patched quickly, but not everyone is up-to-date. The thrust of your answer is still correct though.
Comment: can it read my data on the linux partition?
Comment: Brute forcing a bad key could take seconds, however.
Comment: Ill rewrite it. I forgot to reread what I wrote and I made a few errors in my sentences  :-)
Answer: Yes: malware on Windows could modify the bootloader of you Linux system. This modified bootloader could then for example save the encryption key to your disk once you provide the password to unlock your disk or it could further infect the then unlocked disk. There are no known virusses doing this, but it's entirely possible this exists and just has not been detected yet.
A way to defend against malware modifying the bootloader is Secure Boot. Unfortunately, a lot of Linux install tutorials make you turn Secure Boot off.
Comment: 'Unfortunately a lot of Linux install tutorials make you turn Secure Boot off' Mostly because it provides little benefit for a typical user on a single boot Linux setup and is known to cause issues on a number of systems.
Comment: Malware *could* modify the bootloader, but unless you're being targeted by someone like the NSA, you're not going to encounter such malware.  Almost all malware targets the common case, which is a pure-Windows system.
Comment: If you're turning off secure boot then you won't be able to boot modern Windows anyway. I personally haven't had a problem with secure boot on either of my Ubuntu systems, other than you cannot hibernate.
Comment: @Rodney, I am running Windows 11 with BIOS legacy mode enabled and secure boot disabled. Rufus disables the UEFI, TPM, and secure boot check in the installer.
Comment: @XP1 ah OK, but I think if you want to dual boot along side the factory install of Windows you won't be able to disable secure boot?
Answer: Short answer: yes.
Longer answer: yes, but unlikely (too rare target to bother) and if they do, they will more likely destroy/damage your Linux partition than access data there (it's possible, but way too tricky for most malware).
About why and how is said enough in other answers.
Answer: Yes, absolutely. Any program running on your user account has the ability to do anything that you can do - that includes modifying disks, as long as you have access to them in Explorer. Raw disk access requires administrator permissions, but there are still a lot of evil things that you can do as a standard user account to corrupt a Linux installation, like formatting the partition.
Note that Windows is a fundamentally insecure OS, and if you are concerned about it being able to tamper with your Linux installation, you should be running it in a sufficiently sandboxed virtual machine.
If your computer has a TPM (most recent machines do), it might be possible for you to enable Secure Boot with disk encryption, which would prevent a bare-metal Windows from reading your Linux data (but not corrupting it). It would also prevent it from installing a rootkit in kernel-space (...by bricking your Linux install if it tries), but it wouldn't prevent reflashing/infection of the BIOS like some of the other answers here describe.
Note that UEFI is not separate from the BIOS as some answerers claim. UEFI is simply an interface that some modern BIOSes implement as an alternative (or the only) way of booting the system. Therefore any "UEFI malware" is actually just BIOS malware.
Comment: In what ways is Windows "fundamentally insecure"?
Comment: @PCLuddite There are many things that can be corrected manually, like user accounts being administrator by default, and Windows Update destroying your system constantly, but there are also some things that can't, like the general architecture on top of NT being kind of shoddy. The vast majority of the operating system itself has also proven to be quite insecure in practice. NSA backdoors/exploits for example. Linux and macOS are not subject to the same kinds of vulnerabilities that are found in Windows, mostly because the architecture is not only different, but generally more secure.
Comment: Note that my comment about Windows being fundamentally insecure was mostly just in passing. There are reasons why you might want or need to have Windows installed, but I prefer to foster a healthy understanding of the fact that Windows is not optimal for any problem. Most of my interactions with it tend to be trying to get it to stay out of the way
Answer: Malware, yes.  Viruses yes.  If it had access to the disk it can wipe it, or just tracts of it, even.
Bear in mind a virus must live long enough for the host to reproduce and spread it.
So can badly written software though.
In the mid 1980's I used Norton Utilities to wipe the free space on my C: drive. I had a D: drive on the the same physical disk. It erased the D: drive entirely as free space. Recovering from 360K floppies was not fun, and ultimately not successful. Lesson learned.


choice of parity bits for linear block codes

Question: We know that in linear block codes the codeword for a given a message is generated by multiplying the message with the generator matrix G . The generator matrix consists of a sub-matrix which is the transpose of the submatrix of H matrix which contains the parity bits. and another sub-matrix which is an identity matrix.I am curious to know how is the set of row vectors for the H matrix chosen . For example in the case of a (6,3) linear block code the H matrix contains a sub-matrix whose row vectors are made up of the parity bits . However only 3 possible combinations of 1's and 0's for three bit numbers are allowed to form the row vector of this sub-matrix . How do we decide which three bit numbers are to e chosen to form the row vector ?
Answer: I am a bit rusty on error correction, but I think the answer is that parity bits are chosen to maximize the hamming distance between the codewords. Otherwise, the parity bits do not provide any coding gain


Can I boost/improve reception on my Droid?

Question: With a rooted phone and an app you can overclock or underclock your processor.  Also, off Android, with some hacks you can boost power of a wifi router and wifi antenna.  It seems to me that if you can do those things you could boost the reception of your cell antenna some how.  Am I right?  If so, how?
Comment: Put your droid in a can? ;) http://en.wikipedia.org/wiki/Cantenna
Answer: Not really to do with actually altering your reception per-se, but you can update your PRL (preferred roaming list) of your phone for newer cell towers which might give you better reception.  On Verizon, dial <code>*228</code> and follow the prompts.
Answer: As a telecom engineer student, I know that one of the main limitation for power is the EIRP (http://en.wikipedia.org/wiki/Equivalent_isotropically_radiated_power).
Overpassing the EIRP is one of the must known way to increase wifi router power (by adding a better antenna, which leads to an increase in gain, and in power RX/TX). This power is perhaps fixed by the software in Android handsets, and you could use this parameter to leverage the emission power.
But, as far as I know it's illegal (at least in France).
Answer: updating your radio may help (if there is update for your phone).
after some search its seems there is.
Comment: BE VERY CAREFUL: you can brick your phone flashing your radio if do not know what you are doing
Comment: ...and I don't know what I'm doing, but luckily I knew that before bricking anything.


Crashed upgrade form 13.10 to 14.04

Question: Today, I turned on my computer and upgrade Ubuntu to 14.04, since it is so long, so I went to have dinner. When I'm back, the screen freeze(I tried to move my mouse, but the mouse icon moved a few minutes later). I left it for an hour and finally force reboot the computer.
I searched in forums and found a way to continue the upgrade(enter recovery mode shell and dpkg --configure -a however, when I reboot the computer again, the screen resolution seemed to be increased(the images looked to be more meticulous) but after I login, the mouse responded slower than that before update. The system also seemed to be laggy. After few minute my cpu reached 70°C!!!(That's my bios alarm) ,also, there is a small icon on the upper bar showing something like "...packet damaged...type apt-get in terminal" some thing like this, after I did sudo aptitude build-dep ,showing a lot on unsolved dependence.
I don't know what to do to or if there is "reupgrade command" Please help!
Edit:After successfully solve dependence, I found out the following problems:
1. No hardware acceleration(poor graphics with high cup usage, I can only set the resolution lower to prevent cup burn), however I tried to install driver again, not helping.
2. Play.ape file in cmus will crackling, noise will come out when playing .m4a (however, they can be played in vlc, but my bios scream for high temperature again after I play a few songs...)
I think I left some important packages during this damaged upgrade.
Answer: I don't know you're level of experience, or knowledge of Linux, so if it seems I'm being simple, don't take it personally.
In my opinion, the best way to keep a certain distro (that is not a rolling release) is to make sure the / (root directory) and /home (home directory), this way when a new version comes out, I can install the new version on the same drives/partitions, formatting the root partition, and NOT formatting the home partition. This way you can install it, and you still have all you files and settings, you just need to install your apps again, which is the only real downside.
Now if you used the Ubuntu update tool, when you left, and let it go by itself, there may have been issues which you were not there to see, and correct. You should always stay when doing something as important as  upgrading your entire OS.
Comment: Thank you very much for your answer, however, I was inspired by your response, I cam wait for next release of ubuntu and so I can upgrade again to repair it.
Answer: I've had a similar issue when going from 13.04 to 13.10.
To fix unmet depedency hell, boot into a terminal and type
sudo apt-get -f install
The f parameter here means it will try to fix broken dependencies. It is likely this will end with something like and x not upgraded. Now you must run:
sudo apt-get dist-upgrade
This will install new packages for packages that changed their dependencies between 13.10 and 14.04.
While installing, it is unfortunately important to keep an eye on it! There will be quite many packages who will ask some configuration options during installation (just use the default if you're unsure what to do). Installation will hang until you give an answer.
After everything is finished, run
sudo apt-get autoclean and sudo apt-get autoremove to remove now-redundant files and packages.
Hope this helps :-).

Crashed upgrade form 13.10 to 14.04


Why are pre-main sequence stars brighter than they will be on the main sequence?

Question:
Why is a pre-main sequence star brighter than it will be when it reaches the main sequence?
Comment: That depends on the mass of the system. Large systems contract adiabatically-ish, meaning the temperature and brightness rises with contraction. lower mass systems contract isothermally-ish, meaning the temperature is largely unaffected by the contraction.  See these figures. https://upload.wikimedia.org/wikipedia/commons/thumb/3/3a/PMS_evolution_tracks.svg/220px-PMS_evolution_tracks.svg.png
Comment: What is the typical temperature of a gas cloud?
Comment: @Thriveth What is the thing that gives energy to the molecules of a gas could and allows them to aquire kinetic energy thus increasing it's temperature?
Comment: The contraction converts the gravitational potential energy of the system into thermal energy. This is in fact true through the entire lifetime of a star; Hydrogen fusion only contributes a part of the energy. I do not remember the percentages, though.
Comment: What research have you done before asking this question?
Comment: @sammygerbil   What inspired my question was seeing the image I posted. I asked to some of my professors but couldn't get a compete answer about it. Personally, I didn't look on the internet for an explanation (except for physics StackExchange )
Answer: It is not generally true that a Pre-Main Sequence (PMS) star is brighter than the corresponding Zero-Age Main Sequence (ZAMS) star - whether this is the case depends on the mass. The main source of energy in this phase is the gravitational potential energy of the gas cloud being converted into random kinetic, i.e. thermal, energy. The main question is how fast this energy is transported out of the protostar, compared to how fast the star is contracting and the temperature rising.
The figure in the question shows, correctly, an evolutionary track in the H-R diagram of a PMS object. However, this track is only valid for a certain mass.

The figure above, from the Wikipedia Hayashi Track entry, shows representative PMS evolutionary tracks for different masses. The blue lines are tracks through the H-R diagram for PMS objects of different masses; they start on the upper diagonal, called the birth line (the time when the surrounding clouds get cleared away and the system becomes visible), and end on the lower black diagonal, the Zero Age Main Sequence (the time when Hydrogen fusion sets in). The blue numbers below the ZAMS show the final mass of the star in Solar masses. The red lines show isochrones; where the different evolutionary tracks crosses the same isochrone, the corresponding objects have the same age. You can see that a $6.0 M_{\odot}$ star reaches the ZAMS after only a hundred thousand years, a time when a $0.1 M_{\odot}$ star has not left the birthline yet. It takes 10 million years for a $2.0 M_{\odot}$ star and 100 million years for stars of $M \lesssim 1.5 M_{\odot}$ to reach the ZAMS.
The almost-vertical parts of the tracks are called Hayashi tracks. In this phase, the contraction happens more or less isothermally. The objects on these stages are convective, such that the heat generated by contraction is transported from the core to the outer layers efficiently enough that the object stays roughly at the same temperature. Therefore, its surface brightness also stays unchanged, and the luminosity simply scales with the surface area.
For higher mass systems, however, the contraction happens too fast, the temperature in the core becomes so hot that a radiative zone develops, and convection becomes inefficient and surplus energy is trapped in the system. As it contracts, the temperature therefore rises steeply, and the surface brightness rises with it. This is partly or completely cancelled by the shrinking surface area, meaning that the total luminosity stays roughly unchanged or growing slightly, but the temperature rises steeply. This is seen as a more or less horizontal track going right-to-left in the diagram; these are called Heyney Tracks
As the figure shows, intermediate mass systems are relatively cool in the beginning, allowing for convection, so they start on the Hayashi Track, but as the temperature gets larger, convection is broken, and it turns onto a Heyney track as the heat cannot escape efficiently anymore.

EDIT: i found this figure from this publication which shows some more detail.

Here, the isochrones are gone, but the diagram is separated into regions corresponding to four different scenarios: Fully convective, fully radiative, Radiative core with convective outer layers, and convective core with radiative outer layers. All stars with $M \leq 3.5 M_{\odot}$ go through some phase with a convective core. All stars of $M \geq 1.5 M_{\odot}$ go through several phases, from possibly being fully convective, over having a radiative core and convective outer layers, over being fully radiative, to developing a convective core and radiative outer layers, before settling on the ZAMS.
Answer: There are two questions to answer: Why is a pre-main sequence star so bright, and why does it get dimmer?
The first question is easy. A pre-main sequence star directly follows a protostar, and protostars can get quite large from accretion. From the virial theorem and Kelvin-Helmholtz contraction, we know that the thermal energy that can be released from a contracting mass is
$$U_{\text{thermal}}=\frac{3}{10}\frac{GM^2}{R}$$
If we assume, prior to pre-main sequence contraction, that none of this energy has been radiated away, we can calculate the maximum radius of the body. This turns out to be quite substantial (see e.g. Eqn 5.10 here).
We can also approximate the pre-main sequence star as a blackbody, which has a luminosity of
$$L=4\pi\sigma R^2T_e^4$$
There is indeed a temperature-radius relationship, but the dependence is very small. Thus,
$$\frac{dL}{dt}\simeq4\pi\sigma T^4\left(2R\frac{dR}{dt}\right)$$
and, given that $\frac{dR}{dt}<0$, the luminosity of the pre-main sequence star decreases until the core becomes strongly radiative, at which point the star heats up and moves left on the H-R diagram until it reaches the main sequence.
Comment: @Flaffo The virial theorem *is* applied to stars of all kinds. What is important is that the star is in "quasi-equilibrium" - i.e. it is not collapsing on a *dynamical* timescale.
Comment: @Flaffo What Rob Jeffries said regarding the virial theorem. Additionally, I stated that there is a very weak temperature=radius relationship, meaning that while $\frac{dT}{dR}\neq0$, it can be approximated as $0$.
Comment: I don't think you could use the virial theorem 2K=U, because this should only apply to a system of molecules orbiting each other, which would not collapse. Also in the Stefan-Boltzmann equation the temperature is not constant...
Answer: The virial theorem can be applied to stars in quasi-equilibrium and tells us that the internal kinetic energy and gravitational potential energy of a star are directly proportional to each other. Crudely speaking, this means the pressure (which is proportional to kinetic energy density) times the volume in the star is proportional to gravitational potential energy:
$$PR^3 \propto M^2/R.$$
 For an ideal gas, pressure is proportional to density and temperature, so
$$ \frac{M}{R^3}T  R^3 \propto \frac{M^2}{R}$$
$$ T \propto \frac{M}{R}$$
Thus for a star of a given mass, the central temperature increases as the radius shrinks. Why does the radius decrease? Because the star is losing energy from its surface and this is supplied by decreasing its gravitational potential energy by becoming smaller (see here for more detail).
In a low mass star the shrinkage continues at roughly constant surface temperature because the temperature gradient is limited by convective instability. This means the luminosity shrinks as $R^2$ (along the Hayashi track). Eventually the core temperature becomes high enough to start hydrogen fusion. This increases smoothly until it exactly replaces the luminosity supplied by gravitational potential energy, thus halting the collapse at (roughly) a minimum in radius and luminosity.
In larger stars (more than about 0.7 solar masses) things are more complicated because the core is less dense, opacities are lower and at some core temperature (prior to H fusion), convective instability ends in favour of radiative energy transport. This lowers the temperature gradient, allowing the PMS star to be larger for a similar core temperature. This keeps the luminosity roughly constant for a short time whilst the radius more slowly contracts and the surface temperature increases (along the Henyey track) until core hydrogen burning stabilises the contraction as before.


Is there a community for showing stuff you've made?

Question: Is there a Stack Exchange network site for showing stuff you've made?
Comment: This question really doesn't below on SE Meta.
Comment: The whole SE ecosystem is a Q&A site. How exactly could "showing stuff you've made" ever become a question? (I suppose it could become an answer, as long as someone asked "How to make X" and you were willing to provide a detailed description of how to do so and "showing the stuff" as a proof.)
Comment: @BentheCoder I disagree. I think if this question hasn't been asked before, then it should be asked at least once so we can give it an answer (or multiple answers) once and for all.
Comment: @user Ben meant the question OP trying to ask, i.e. question about showing stuff. I think.
Answer: In general, no, we don't really do that kind of thing. That doesn't really sound like it fits within Stack Exchange's general framework for Q&A. There are exceptions, which I'll get to later.
For something to be on-topic for Stack Exchange, there needs to be a question that solicits (generally) objective, fact-based answers. For exceptions, see the "good subjective" section of /help/dont-ask. For most sites here, that means we wouldn't accept the question even if it was requesting critique on your work. (though there are narrow exceptions, such as the Code Review site). There are sites that don't accept critique requests even if they could be made into good subjective questions or objective questions (Ex. the Writing site, the English site, the English Language Learners site).
We also generally recommend that questions be based on a problem that you want solved (see /help/dont-ask: "You should only ask practical, answerable questions based on actual problems that you face.")
If you just want to show something you've made, what's the question? What's the problem to be solved?
If what you've made could solve a problem for someone else, or you imagine someone else might genuinely be curious on their own volition about something like what you've made, you can post it in the form of a question asking for something like what you've made (see /help/how-to-ask) and then answer it with what you've made (see /help/how-to-answer) (making sure that the question is actually on-topic for the network site you post it on in the first place (see the site's individual /help/on-topic page)), but we also discourage doing too much self promotion here (see /help/promotion).
For one example of a /help/on-topic page you might want to look at to see what our standards are like, see the one for the Arts & Crafts site and the one for the Game Dev site.
If you just want to share something you've made without solving a problem for someone else, or answering a question that you had or think someone else might have, this isn't really the place for that. The internet is a big place though. You can probably easily find another site outside of Stack Exchange to do that :)
However

If you like making puzzles, see the Puzzling site. See also their Sandbox area.

The Math site also allows math puzzle questions (see their [puzzle] tag, and this meta Q&A).

If you do end up posting on MSE, please make it clear in your question that you are "puzzling" the community and that you will be answer the question yourself if no one in the community posts your desired solution.

If you like making code golf challenges, see the Code Golf site. See also their Sandbox area.

The Game Dev site does (yearly?) showcase threads on their meta site, which you can find with this search query.

If you like writing userscripts / browser extensions for augmented usage of the Stack Exchange sites, see the Stack Apps site. See also their guide on sharing your content.

Quoting from F1Krazy's comment:

As a mod on Writing.SE, I'd like to add that we do have a chatroom for getting critiques or beta readers (albeit a very inactive one), as well as monthly writing challenges. So you can't show off your writing on the main site, but you can show it off elsewhere on Writing.

I'm guessing the room being referred to is this one: https://chat.stackexchange.com/rooms/115415/beta-reviewers-reviewing (of course, access to chat is a privilege, so you need to earn it by contributing good content to the network).
Comment: As a mod on Writing.SE, I'd like to add that we *do* have a chatroom for getting critiques or beta readers (albeit a very inactive one), as well as monthly writing challenges. So you can't show off your writing on the main site, but you *can* show it off elsewhere on Writing.
Answer: If "stuff" include applications written with the Stack Exchange API, or scripts that work on Stack Exchange sites, then one could post on the Stack Apps site. It would be wise to read pages in the Help Center before posting there - including How to list your application/library/wrapper/script here.


Making a tunnel follower robot with 2 IR side sensors and 1 ultrasonic sensor

Question: We have a tunnel and the distance from the walls to the robot is about 30cm when the robot is centered, the tunnel curves to the right. The robot has an infrared sensor on its left and one on its right and has an ultrasonic sensor at its front (so basically its eyes). How can I make the robot turn appropriately while its moving so that it's centered in the tunnel. I just need an idea for this algorithm because I can't find any. All what the infrared sensors can do is measure the distance to the walls.
Comment: Walk slowly down a hallway at your house, or at your school.  ... Think about what the robot does. ... Measure distance to each wall. ... What do you do with your motors if the distance is too small? ... Write it down in a flowchart form.
Answer: One possible method is to find the width of your robot, subtract that from 30 cm, then half the remaining distance and if one of the IR sensors reads less than that number, turn in the other direction.
Answer: I agree with @NikolaPi in principle.
But the controlling/driving mechanism is important here. If you are using PID controlling, then steering the robot can be very smooth. All you have to do is to tune the variables so that when robot starts getting closer to lets say, left wall, left motor's pwm increases w.r.t. the right one and vice versa. This way the robot will remain in an imaginary boundary/lane that is slightly wider than the robot but significantly smaller than the Tunnel's width. But the tuning will take a little effort to find the sweet spot where the correction neither occurs too often nor too late (Hint: You robot should move almost straight without compromising on speed. If its is moving in the form of the letter "S", your limits are too generous).
On the other hand, if you are using the simpler On/Off Controlling technique (also known as band bang approach), you take a limit of IR sensor reading from one wall and program according to that. If it crosses the limit, stop the other sides motor for some time so that the robot comes back to its center position. But beware that in this approach, with the battery voltage drop, the response from the motors changes. What I mean is that if you tune the robot to switch off motor for 100 ms in case of limit detection and it comes back to the good position, when the battery would get drained a little, the robot would need 130 or 150 ms of switching off to make the same correction. Of course it won't be able to as you have set it to 100 ms so there will be an error in the robots's position every time an adjustment is made. Eventually, these errors will pile up and the robot will collide with a wall.
Answer: If I understand correctly, all your sensors look ahead, but you want safety to the sides.
Maybe it would be better to turn the IR sensors 90deg so they check the distance to the side walls, while you keep the ultrasound sensor to look ahead.
Using the data from the three sensors, you should be able to create a limited map of the robot inside the tunnel, which will be the basis for controlling the movement.
Answer: Ahmed talked about PID (proportional integral differential, which is very complicated) and bang-bang control (which is too simple and often over-corrects). I would recommend proportional control and an example of this is on:
https://www.instructables.com/How-to-Make-a-Robot-Car-Drive-Straight-and-Turn-Ex/
The website gives an example of a car trying to drive straight using a gyroscope on a flat surface, which is similar to your tunnel situation.
Proportional control is much simpler than PID, since you only need to change tune one variable, instead of three in PID. If you want a bit more complexity, you can try proportional integral control, which uses two variables.
Additionally, I am skeptical about the ultrasonic sensor, since the tunnel might form an echo chamber, creating a lot of noise interference.
Answer: You can use a simple constant step controller that does not need tuning like a PID controller.  A step controller will simply start adding or subtracting a constant control value based on the input signal.
In this case the control signal involves steering.  If you are too close to the left wall, then start turning right; if you are too close to the right wall then start turning left.  What 'turning' means depends on your robot.  If you are using a differential drive configuration then turning means slowing one wheel down and speeding the other up; the turn will be in the direction of the slower wheel.  If you are using an car-like Ackermann steering, then just change the angle of the steering servo.
The input signal is the difference in distances to each wall (there would be a separate controller for the forward distance to stop the robot).  We use the difference because we just want to know which wall we are closer to (and so should move away from).  This handles the fact that the robot will often not be driving parallel to the walls, so the distance to the walls as measured by the sensor may be greater than the actual distance because the sensors are at an angle to the walls.
So if we use (left - right) distance, then a negative value indicates we are closer to the left wall, a positive value indicates we are closer to the right wall.  We can use some zero-tolerance to treat very small differences as zero.
If we are within the zero-tolerance, do nothing.  If we are closer to the left wall, then add a small constant (the step) to steering. If we are closer to the right wall then subtract that small constant from steering.  What that step value is and it's sign and how it's actually applied (how it's actuated) will depend on your robot.