Can't Enable Apple Watch Auto-Unlock

Question: With macOS Sierra installed on a Mac from at least 2013 and an Apple Watch (the very first one), users should be able to unlock their Mac with their Apple Watch. I have an Early 2013 MacBook Pro and an Apple Watch, both of which have the most updated version of the OS installed. I followed these instructions to get it to work (checked 2 factor authentication is on, I have a password on my Watch), but I do not see the option to use my Watch to unlock under System Preferences. Any ides on why it does not work? Could it have something to do with me having 2 Apple IDs, one for iTunes and one for iCloud? Could it be my Mac is somehow not ready for this feature (although it is from 2013)?
Comment: It could be a bug that will be resolved when the OS is officially released tomorrow.
Comment: You says you have an Early-2013 MBP, on the page that @Harv linked in his answer, https://support.apple.com/en-ca/HT206995, it says it needs an Mid-2013 or later MBP, maybe that is the case?
Comment: That article you linked was from a beta version so maybe they changed something for the release?  Take a look at [this recent article](http://www.macrumors.com/how-to/macos-sierra-apple-watch-unlock/) and see if following those steps help.
Comment: Yes, you need to be signed in to the same AppleID.
Comment: @TomShen Yes I think that is the problem.
Answer: Are you seeing the <code>"Allow your Apple Watch to unlock your Mac"</code> checkbox?
You may need to enable Continuity.  Make sure both devices are on the same WiFi network, and Bluetooth is enabled.  The article also had an error, it says

"[...] make sure that Disable Automatic Login is unchecked. If the
  option is grayed out, you might need to click the padlock icon at the
  bottom of the window to make the necessary changes."

I think they meant that the box is checked.
EDIT: Here's the Apple article you want to reference: https://support.apple.com/en-ca/HT206995
Comment: I do not see that checkbox. And I know why... The article you linked says I need a mid-2013 Mac. I have an early-2013 one... That is sad.

Can’t Enable Apple Watch Auto-Unlock


ASA5505 switchport command

Question: I've been learning how to use ASA5505 but I faced this thing: When I try to assign Ethernet0/1 with vlan 1, which is the inside network, I see no change in the answer of sh int ip bri or sh run. Why this could be happening? I do not see anything wrong with my firewall.
Comment: Did any answer help you? If so, you should accept the answer so that the question doesn't keep popping up forever, looking for an answer. Alternatively, you could provide and accept your own answer.
Answer: VLAN 1 (switchport access vlan 1) is the default. As such, it will not normally show in the config.
show switch vlan is the command to see what vlans are defined, and what ports are assigned to them.
The interface list will never show any IP configuration for the ethernet ports because they are part of a switch. (and you cannot no switchport them into routed interfaces.) All layer-3 configuration is via VLAN interfaces.
Comment: The default I got is VLAN2 is assigned to e0/0. However I want to assign VLAN1 to e0/1 but when I used the command switchport access vlan 2 on e0/1 nothing happens. Neither I see an error coming up nor the assignment is done.
Comment: Yes, the default system configuration (booted without any config) is E0/0 is vlan 2 (outside). The "ios" default for an interface is vlan 1, which is why `sh run` shows vlan 2 on e0/0 and nothing on the other interfaces. If you enter the correct CLI configuration commands, it will switch vlans: conf tint e0/1swi a v 2

ASA5505 switchport command


power analysis of a proposed study

Question: I am planning a study and for that I need to use power analysis to estimated the sample size. Now, I do not have any estimates for the desired effect and standard deviation required to do the power calculation. I also cannot do a pilot study.
Given these constraints, how do I estimate the sample size for my study?
Answer: The short answer is that you can't. Like other answers mentioned you need to assume a fixed effect size, fixed significance level and a desired statistical power in order to estimate the needed sample size.
Now if you don't know the mean difference and standard deviation - you can guess them based on some information.
And if you cannot reasonably do that then the next best thing is to calculate required sample sizes for a whole range of possible mean differences and standard deviations. This would allow you to at least imply on a range of possibilities. And you would be able to answer questions such as "what kind of effects can I reasonably (i.e. with power = 80%) expect to detect with a sample size of n".
Answer: While Bernhard is right that having some more info or a pilot would be ideal, a pragmatic approach in a situation where neither is available would be

Fix your question
Fix your statistical analysis
Decide on the minimal effect size that is of SCIENTIFIC significance, i.e. that you want to be able to pick up if it is there, and use this as effect size in the power analysis
Get an estimate about variability from the literature / colleagues (very unlikely that there is no precedence for a measurement method)
Comment: In some cases it may be possible to reduce steps 3 and 4 to a single number (e.g. "population signal to noise ratio", [Cohen's $d$](https://en.wikipedia.org/wiki/Effect_size#Standardized_and_unstandardized_effect_sizes)).
Comment: Right, assuming there are studies that have looked at similar effects.
Answer: I don't know, how hungry I will be, nor, how many people will join the meal. How much food should I buy?
You cannot do any reasonable computation without suitable data. You could probably try to guess reasonably. Or you need to look for similar investigations that have been done by others or you have yourself guided by what $n$ is reasonably achievable. Or maybe a descriptive analysis is in order?
Also look into How to choose the sample size for a pilot study for making a power analysis?
Comment: I tried to calculate sample size for a paired t-test study and I get the following results. I used the means and std's from an actual and similar study. The sample size I get is 16. I do not understand these results. Can someone please explain?

Input:  Tail(s)                        = Two
   Effect size dz                 = 0.7596495
   α err prob                     = 0.05
   Power (1-β err prob)           = 0.80
   Total sample size              = 16
Comment: Two-tailed tests are the standard. If you don't have very good reasons, you will want to do a two tailed test. $\alpha$ = .05 means that you consider your result significant, if p is below .05, the power is usually set to 80 or 90% giving the chance to find an effect as big as given with this $n$ is 80% or 90%. If you can gather more than 16, increasing the power over 80% isn't strictly bad. The effect size to be detected has been computed from the difference of the means and the standard deviations. Go with something slightly over  $n = 16$, maybe twenty, if you can afford that.

power analysis of a proposed study


Bus schedules in Israel

Question: I'm looking for paper or pdf schedules of isreali bus routes (that i can use with out internet coverage)
i found for the golan routes http://www.golanbus.co.il/Files/art_files/lozhdash72017.pdf
but im having trouble finding for the other bus companies
i do read hebrew so that helps
Comment: It seems like all companies have schedules on their sites, but you must choose the line number to get the info. I don't see a way to get all lines in one file.
Answer: No such PDF schedules currently exist, at least none that are up to date. You are therefore recommended to either write down the times in advance or buy a local SIM card to allow yourself to be permanently connected. I have traveled in the Golan Heights this year and connectivity was generally great.

Bus schedules in Israel


Can the value of the NFTs be added to the stake size of the pool?

Question: Many small stake pools struggle with increasing their stake. The stake could be increased by adding the value of all the NFTS held in wallets by both delegators and operators. How do we determine value? Once we have smart contracts, value will be the last sold price of the NFT, recorded in the Blockchain.
Answer: If the value of NFTs (their purchase price or last traded value) were to be added to staking then both the ADA used to purchase them and the NFT would counted.  This would make it possible to double your staking rewards by selling an NFT to yourself or a collaborator.
It would also affect the Cardano monetary policy.
Stake pool operation is a serious undertaking and gathering enough stake to establish a viable pool is part of that challenge.
Comment: There is also the appreciation of value of existing NFTs based on new traded prices... basically impossible to know what "total staked value" would be at any given time when counting NFTs. Many opportunities for adversarial behavior. "Affect the monetary policy" is an understatement, it would need a complete rewrite!
Comment: I completely agree with the remarks. Just an idea: is there a possibility the NFT to be 'locked' somehow (under another smart contract) within the stake pool and it's last sold value (written in the smart contract) to be added to the stake? By 'locked' I mean not to be traded in any way, or if traded its recorded value to be deducted from the stake.
Answer: Only ADA coins are participating in staking and securing the network. NFTs will not be part of stake for stake pools.


Preheating Diesel before injection?

Question: Significant amount of heat is lost from an engine by various means, so to recoup the energy, is it possible to increase efficiency via heating of diesel fuel prior to injection using waste engine heat in passenger cars similiar to marine diesels? Are there any recent journals or studies in this matter?
Answer: Consider how much energy is needed to heat the fuel that is injected each time compared to the energy that can be recovered from the turbo-charger for example. One thing to note is that many of the diesel pumps (style DPA / inline high pressure) will compensate for changes in the density of the fuel according to the temperature, the modern common rail injection systems will compensate for this temperature as well.
I did heat the diesel fuel on one of my cars using the radiator from the automatic version of the car and running the fuel through the oil cooler part (blocking the airflow through that bit), result : helped the running of the engine during the cold months but no appreciable change in fuel consumption was noted.
Edit: noticed that you tagged marine-engineering, so large marine diesels use a very different fuel and that is heated before injection so that the viscosity is reduced as the fuel very viscous when cold.
Answer: Everything that gets the mixture in the cylinder hotter makes for a more efficient combustion. That is because the power of the engine depends purely on the pressure exerted on the piston, and the effective pressure is caused by temperature rise, and nothing else. So yes, injecting hot diesel would make for a more efficient combustion, in theory. But it would be insignificant. You can heat it up to just below the boiling point, but you'd still need the energy to evaporate the diesel, which is far more than needed to heat it up. Installations for this purpose will cost more than the profit they'd give you.
There is an alternate solution though; heating up the air. Once the air is trapped inside e.g. a cylinder, you could heat it up, and it would gain pressure. That pressure will add up to any pressure caused by combustion. You would get a more efficient combustion. The problem is that you're dealing with a continuous air flow, which is hardly trappable. If you heat it up without trapping it, it will just expand and flow out of the cylinder.
You'd need a pressure tank in which you can trap bulk charges of air, you can then heat up that trapped air, and feed it to the engine once it's hot.
But also those installations would only perhaps be cost effective in large static installations. In cars, downsizing and turbochargers are the way to go for realistic and cost-effective gains in efficiency. That's why you see this more and more today.
For marine applications, heavy oils are preheated to reduce their viscosity, which only then enables them to be used for injection. The heaviest oils are so thick you can nearly walk on them when they're cold.
Comment: So, Removing intercooler?
Comment: No. The turbo doesn't lock up the air, so it will expand, the turbo will not be able to push as much air in the cylinder, and you'll end up with less mass air. It's a compromise to cool air to get more mass air in the cylinder, but the air itself being cooled only removes energy from it. If you could make the turbo fill a barrel with air, shut the barrel, heat it(pressure rises), and then feed it to the engine, then you'd be able to get more efficiency.
Answer: Heating the diesel will have a stronger effect on how it burns in the cylinder than how much energy it adds to the system by recovering waste heat. Diesels depend upon compression to heat and ignite the fuel adding heated fuel would cause it to vaporize sooner and predetonation would be possible.
Comment: Predetonation is not an issue in the dieselprocess. It burns as soon as it is injected. The problem would be more in the pressure created in the fuel lines, and in vapour locks in the fuel system.
Comment: So the fuel is injected at the desired time for detonation? I thought it was injected and detonated when pressure increased to a critical point. Your explanation makes more sense for timing ignition at or after the piston reaches top dead center. Thanks.
Comment: Air without fuel is compressed and is heated up by this, diesel is injected near top dead centre at the end of the compression stroke. There is no need for ignition. The diesel combusts as it is injected, because the air is hot enough to immediately react with the diesel. The diesel doesn't detonate or explode, but combusts gradually. The way you describe, where fuel is injected beforehand, is also possible, but currently still in development. It's called Homogeneous Compressed Charge Ignition or simply HCCI. It's not really in use yet since it's hard to control well enough to profit from.
Comment: Cool, do you have a link to am article on that where I could read more?
Comment: Google is your best friend, if you search for HCCI lots of hits show up ranging from WikiPedia to AutoWeek. Watch YouTube videos to get a better understanding of the working.
Comment: actually predetonation is more of an issue in gasoline engines because of engine knock and that is why the octane number is usually high to increase its activation energy.
whereas in the diesel engine, the cetane number describes how well the fuel oil burns in the engine. so in this case, preheating is carried out to improve combustion qualities in the diesel engine.

Preheating Diesel before injection?


A picture taken appears in the camera but does not appear when it is copied to the computer

Question: I have a Canon EOS 60D and I tried the copyright function and, after that, I copied the pictures to my laptop but they did not appear. So I put my memory card back into my camera and I then tried viewing the pics and realized that I could view them from the camera but not from the computer... please help.
Comment: Can you clarify what you mean by "tried this copyright function"? How did you copy your pictures to the laptop? Did you do something different from what you normally do (and did that work?) What OS are you running?
Comment: Did you read the manual of canon ?
Comment: Further to @mattdm's questions, are you shooting in RAW or JPEG?
Comment: @Aristos I love how that sounds all regal and religious. But, yes: RTM.
Comment: Possibly related: http://photo.stackexchange.com/questions/11127/why-cant-i-read-a-2gb-cf-card-from-a-mac-or-pc-after-having-formatted-it-on-a-30
Answer: What did you see on your SD card? Did you change anything else? Maybe you updated from JPEG mode to RAW and have no utility installed yet on your pc for viewing RAW files? If so try picasa or any other RAW viewer. Your canon should come with a cd with software. A viewer for instance with support for RAW images.
Answer: From the information provided, it sounds like you might be shooting in RAW mode, and haven't yet installed the RAW codec from Canon yet (caveat: this will only resolve the issue if you have a 32-bit operating system; if you have a 64-bit operating system, you would need to buy a third party codec like Ardfry's which has a 15 day trial)


Pion decay in particle physics

Question: I'm taking a particle physics course and we're using Perkins Introduction to High Energy Physics as the text. I am looking at problem 1.7. It asks whether $$\pi^0\rightarrow e^- + e^+$$ is allowed or forbidden by the Standard Model based off of conservation laws. I feel it doesn't violate any of them. Energy and momentum can clearly be conserved, charge is conserved, lepton number is conserved, etc. However, I know Dalitz decay is a very similar process, but that the pion decays into $$\pi^0 \rightarrow e^- + e^+ + \gamma$$ so feel as though I must have overlooked something. I would be greatly appreciative if someone could point me in the right direction.
Comment: Consider parity and angular momentum, too.
Comment: Thank you! I'll look at it a bit more and let you know if I come up with anything.
Answer: Since $\pi^0$ is a pseudoscalar particle, we have $$\langle 0|J^\mu_{em}|\pi^0 \rangle =0,$$ and the pion cannot decay into two leptons with a simple photon exchange. In the Standard Model, the leading-order contributions for this process are a box diagram and a $Z^0$ exchange, as you can see in fig. 1 of arXiv:0806.4782 (replacing a $c$ quark by a light quark). Therefore, this process is allowed in the SM, but highly suppressed.
On the other hand, if you have a photon in the final state, you can have an eletromagnetic decay at tree-level, as shown in the third diagram of the same figure.
The difference between these two processes can easily be seen from the experimental measurements, because the branching ratio is of order $10^{-8}$ for the first decay, while it is of order $10^{-2}$ for the Dalitz decay (c.f. PDG).

Pion decay in particle physics


How to cut an edge at specific point?

Question: I have an object representing a room. I want to cut a door on it. I planed to use knife tool, because I do not know any better for this - any other (or "proper") idea would be good too BTW. I want to move X units from the corner on the bottom edge and make a cut from there (green dot) to the top. How can I do this?
Comment: I would simply use loop cut and position it about at the right place, then select the cut. After that, go to the right-hand-side panel in the 3D view. (Aka. Properties, hit N to open.) Scroll around (it's somewhere up, I don't exactly remember where). You should find three numeric inputs for the x, y and z of the selection. There you can enter the exact value you want. Pro tip: For any numeric input in blender, you can use basic mathemathical operators in them.
Answer: Use the knife tool or loop cuts to build your mesh, and then here is a trick to move a vertex or edge to a precise position relative to the corner:

select the vertex or edge you want to move
move it all the way to the corner, validate
move it back (for example with GX), enter the distance on the keyboard.
Answer: There is no direct way, sadly, afaik. One different way is to cut anywhere on the edge, and then move the vertices where you need precisely (in your image, along the x axis). Or you could also just create a vertex in the middle of the edge (by subdividing) then move it on x axis and then cut with the knife tool using snapping, to that precisely positioned vertex.
Note, you can also position a vertex on the edge by using "vertex slide" SHIFTV, and see its precise position on the edge in the viewport activating, in the properies panel (press N) mesh display > edge info > length

How to cut an edge at specific point?


PNG-8 upload to S3 hangs

Question: We're having trouble uploading PNG-8s. PNG-24s and JPGs work fine, but after selecting a PNG-8 to upload, the progress bar appears and fills to max in a couple of seconds, but then hangs indefinitely. Sometimes a message appears saying "Upload failed for 'filename.jpg'", sometimes it just hangs with no error message. If we then refresh the page, the image is in the assets list but no thumbnail has been created for it. The PHP error log gets this:
<code>[03-May-2017 14:03:48 UTC] PHP Fatal error: Maximum execution time of 30 seconds exceeded in /home/sitename/webapps/sitename/craft/app/vendor/pixelandtonic/imagine/lib/Imagine/Imagick/Image.php on line 313
</code>
We could of course make them all PNG-24s but the file size is often way higher.
We're on Craft 2.6.2958, and the assets are on Amazon S3. Any advice on how to diagnose this would be much appreciated!
Answer: To fix that particular error, you'd need to bump your php.ini file's max_execution_time setting to something higher than 30 seconds to allow more time for the request to complete.

PNG-8 upload to S3 hangs


Gray values on QP card 201

Question: I have a colour standard card QP 201 which I am hoping to use to calibrate gray values in some scientific photos. For this purpose I will need to know the percentage gray values of the gray squares (e.g. 18% gray, 95% gray), but the card does not come with that information. The closest that I can find is this chart - how do these RGB values correspond to percentages? Is it simply a linear transformation, so that 255 equals 100%, hence the corner squares (RGB 179, 180, 178) would be 70% gray?
Answer: Photography is approaching 200 years of age, first permanent image Joseph Niepce 1826. The digital era starts in 1975 when Steven Sasson at Kodak invented the first digital camera. In the early days, exactly how to expose and develop was hit and miss. It was Hurter & Driffield who conceived the science of Sensitometry. This is the science of how film and photo paper react to exposure and chemical developing. Thus applying numerical values to shades of gray or hues is rooted in this science. Their math is the basis of exposure calibration and film speed.
It's a complicated subject because engineering of that era relies on logarithmic notation. This math notation, still used extensively, was the mainstay before engineers had electric calculators (computers). The slide rule was used every day for multiplication and division. Before the slide rule, books filled with tables were used. The basis of the tables and the slide rule is logarithmic notation. This math method substitutes addition for multiplication and subtraction for division. Its use permitted ships, buildings, airplanes, etc. to be designed with pen and paper. Photography also appreciated the use of logarithmic notation.
The increment of exposure is the f-stop. We open up or close down the circular opening of the aperture to control the light energy that plays on film or digital sensor. The f-stop adjustment changes the light energy of exposure in 2x increments, a doubling or halving of the exposing light energy. In modern cameras we often work in 1/2 or 1/3 f-stop increments.
Sensitometry (stemming from Hurter & Driffield) is the science of exposing film (now digital) using precise exposure changes, usually 1 stop or ½ f-stop. Typically a black & white film is exposed to make a gray scale --- clear film to max black.  Each step is measured by passing a beam of light of known strength through the developed film. The amount of light blocked due to the density of each step is recorded. The results, called film density, is graphed on ruled paper. The result is a plot of the film's response to exposure and development.
If the math used to record the density of each step is ordinary -- 10 base numbers, the graph is too long to be practical (several yards long). This is because each f-stop is a 2x change in exposure, so after fourteen steps, the scale is 8,192 the first step. To make the graph practical, some say elegant, logarithmic notation is used.
Sorry no time or space for a logarithmic lesson, but here is a taste of the math of photography:
T = Transmission the amount of light that gets through the film  ÷ Amount of light that hit the film. A transmission of 18% means 18% of the light that hits gets through.
Percent means per 100 so 18% in decimal form is 0.18.
O = Opacity the amount of light that hit ÷ Amount of light that gets through.  O = 1/T we don't use the word percentage thus 1 ÷ 0.18 = 5.56.
D = Density this value is the opacity in logarithmic form log(5.57) = 0.75
The key tone of photo film and paper is a gray patch with a density of 0.75 (log base 10). This is middle gray of the scale of film and photo print paper (battleship gray).
A patch of film with a density of 0.75 is the center value of the tonal range of a pictorial black & white negative (patch # 7 of photo below) A print made from this negative, properly exposed on photo paper and properly developed yields a gray patch, we say zone V, the pivotal point, and both properly exposed film and a print of that film have the same density. The film's density is 0.75 transmission density, the print's density is 0.75 reflected density. This is the value of the 18% gray card used to calibrate exposure meters.
Comment: An H &D graph of pictorial negative film reveals that film density resembles the left half of a bell curve. The increasing density (blackening) is depicted as an upward clime. A normal contract film upsweeps at an angle of about 38 degrees. The tan of that angle = 0.8. This is what we commonly call gamma. Prints from this negative are represented as the right half of a bell curve. the typical gamma of this upward sweep is about  2. The two gamma columns, left one is has elevated contrast the second lest contrast. Likely the table is to aid in the control of two specific Kodak photo papers.
Comment: Thank you very much for such a detailed answer, it really helps with understanding what is the meaning of 18% gray and is far more thorough than what I was expecting! The chart that you've shown gives some idea of how the reflectance and density relate to RGB values, but why are there two different columns - what do the two gamma values represent?
Answer: Your example square is actually about 53%.
Upon reference image download exiftool shows that the ICC profile is ProPhoto RGB:
Kali:~ kelsolaar$ exiftool ***/rgbvalues.jpg
ExifTool Version Number         : 10.80
File Name                       : rgbvalues.jpg
Directory                       : ***
File Size                       : 316 kB
File Modification Date/Time     : 2019:09:08 14:06:30+12:00
File Access Date/Time           : 2019:09:08 14:18:20+12:00
File Inode Change Date/Time     : 2019:09:08 14:09:20+12:00
File Permissions                : rw-r--r--
File Type                       : JPEG
File Type Extension             : jpg
MIME Type                       : image/jpeg
JFIF Version                    : 1.02
Exif Byte Order                 : Big-endian (Motorola, MM)
Orientation                     : Horizontal (normal)
X Resolution                    : 72
Y Resolution                    : 72
Resolution Unit                 : inches
Software                        : Adobe Photoshop CS2 Macintosh
Modify Date                     : 2007:09:12 11:12:29
Color Space                     : Uncalibrated
Exif Image Width                : 1742
Exif Image Height               : 566
Compression                     : JPEG (old-style)
Thumbnail Offset                : 332
Thumbnail Length                : 4057
Current IPTC Digest             : 460cf28926b856dab09c01a1b0a79077
Application Record Version      : 2
IPTC Digest                     : 460cf28926b856dab09c01a1b0a79077
Displayed Units X               : inches
Displayed Units Y               : inches
Print Style                     : Centered
Print Position                  : 0 0
Print Scale                     : 1
Global Angle                    : 120
Global Altitude                 : 30
Copyright Flag                  : False
URL List                        :
Slices Group Name               : ProPhoto
Num Slices                      : 1
Pixel Aspect Ratio              : 1
Photoshop Thumbnail             : (Binary data 4057 bytes, use -b option to extract)
Has Real Merged Data            : Yes
Writer Name                     : Adobe Photoshop
Reader Name                     : Adobe Photoshop CS2
Photoshop Quality               : 12
Photoshop Format                : Progressive
Progressive Scans               : 3 Scans
XMP Toolkit                     : 3.1.1-112
Format                          : image/jpeg
Creator Tool                    : Adobe Photoshop CS2 Macintosh
Create Date                     : 2007:09:12 11:12:29+02:00
Metadata Date                   : 2007:09:12 11:12:29+02:00
Document ID                     : uuid:1EB2585362B311DC9D2ADDF10FA72CD1
Instance ID                     : uuid:1EB2585462B311DC9D2ADDF10FA72CD1
Derived From Instance ID        : uuid:1EB2585062B311DC9D2ADDF10FA72CD1
Derived From Document ID        : uuid:4162C64161E811DC9D2ADDF10FA72CD1
Native Digest                   : 256,257,258,259,262,274,277,284,530,531,282,283,296,301,318,319,529,532,306,270,271,272,305,315,33432;80CC530BEFCFB6E95CD01DEF050420F7
Color Mode                      : RGB
ICC Profile Name                : ProPhoto RGB
History                         :
Profile CMM Type                : Unknown (KCMS)
Profile Version                 : 2.1.0
Profile Class                   : Display Device Profile
Color Space Data                : RGB
Profile Connection Space        : XYZ
Profile Date Time               : 1998:12:01 18:58:21
Profile File Signature          : acsp
Primary Platform                : Microsoft Corporation
CMM Flags                       : Not Embedded, Independent
Device Manufacturer             : Kodak
Device Model                    : ROMM
Device Attributes               : Reflective, Glossy, Positive, Color
Rendering Intent                : ICC-Absolute Colorimetric
Connection Space Illuminant     : 0.9642 1 0.82487
Profile Creator                 : Kodak
Profile ID                      : 0
Profile Copyright               : Copyright (c) Eastman Kodak Company, 1999, all rights reserved.
Profile Description             : ProPhoto RGB
Media White Point               : 0.9642 1 0.82489
Red Tone Reproduction Curve     : (Binary data 14 bytes, use -b option to extract)
Green Tone Reproduction Curve   : (Binary data 14 bytes, use -b option to extract)
Blue Tone Reproduction Curve    : (Binary data 14 bytes, use -b option to extract)
Red Matrix Column               : 0.79767 0.28804 0
Green Matrix Column             : 0.13519 0.71188 0
Blue Matrix Column              : 0.03134 9e-05 0.82491
Device Mfg Desc                 : KODAK
Device Model Desc               : Reference Output Medium Metric(ROMM)
Make And Model                  : (Binary data 40 bytes, use -b option to extract)
DCT Encode Version              : 100
APP14 Flags 0                   : [14]
APP14 Flags 1                   : (none)
Color Transform                 : YCbCr
Image Width                     : 1742
Image Height                    : 566
Encoding Process                : Progressive DCT, Huffman coding
Bits Per Sample                 : 8
Color Components                : 3
Y Cb Cr Sub Sampling            : YCbCr4:4:4 (1 1)
Image Size                      : 1742x566
Megapixels                      : 0.986
Thumbnail Image                 : (Binary data 4057 bytes, use -b option to extract)

With that knowledge and Colour, it is possible to decode the values as given in your image and convert them to Luminance values which for practical purposes would be equivalent to reflectance values for the various gray swatches:
Kali:~ kelsolaar$ python
Python 3.6.4 |Anaconda, Inc.| (default, Jan 16 2018, 12:04:33)
[GCC 4.2.1 Compatible Clang 4.0.1 (tags/RELEASE_401/final)] on darwin
Type "help", "copyright", "credits" or "license" for more information.
>>> import colour
>>> RGB = colour.models.eotf_ROMMRGB([179, 180, 178], in_int=True)
>>> colour.models.RGB_luminance(RGB, colour.models.ROMM_RGB_COLOURSPACE.primaries, colour.models.ROMM_RGB_COLOURSPACE.whitepoint) * 100
53.26811741384234

I made a Google Colab notebook so that you don't need to install Colour on your machine: https://colab.research.google.com/drive/1OuaAnxqct3E4OAbddcA74vcNF4uwQyx4

Gray values on QP card 201


Modern Warfare 2: Pistol run distance

Question: Does holding a pistol make you run further (or faster) than your primary weapon?
Answer: I wasn't able to find an 'official answer' but there were a number of sources that said only the primary affects your speed.
From this site

On movement speed (player speed is
  determined by the weapon in his
  primary slot): SMGs and Sniper Rifles
  have a Speed of 100% Assault Rifles
  have a Speed of 95% Machineguns have a
  Speed of 87.5% Riot Shield has a Speed
  of 80% A player can pick up another
  weapon and keep the movement speed of
  the weapon he spawned with.

From game FAQs

Only primaries affect running speed so
  pistols don't count.  Also, it goes
  like this: SMG's/Snipers > AR's >
  LMG's > Riot shield

On a semi related note: Lightweight will multiply the player's movement speed by 1.1x ( 8% to 10% increase depending on the weapon).
Comment: Thats interesting, so spawning with an SMG or Sniper its possible to pick up a Riot Shield or LMG and run full speed?
Comment: Going by the information I found that would be possible.
Comment: This is really useful information. Great data visual presentation of this complex information.
Comment: @Aardvark No. The speed you move at is the slowest speed among the weapons you are carrying. If you pick up a riot shield, you move at riot shield speed, regardless of what other weapon you have or what you spawned with. Also, I could've sworn the lightweight speed was x 1.15, but maybe not.

Modern Warfare 2: Pistol run distance


How to pronounce words starting with a "v" fast?

Question: I just learned that recently from this video, how do words like "one" are pronounced differently from "van", the latter need to using the bottom lip to touch the upper teeth to produce the /v/ sound. Then I realise that it is much easier to pronounce a single word, compare to pronounce the word in a sentence, especially if I want to say something fast, for example "in the van were the foremost chiefs", to me it feels like I don't have enough time to touch my teeth with my lip. So my question is, how do native English speakers or linguists pronounce the /v/ sound fast in a sentence? Is there any skips during the pronunciation when speaking fast? Or I just need more practice?
Answer: Just like many other languages, (I assume) your native language doesn't contrast [v] with [w]. There are a number of languages where [v] and [w] are treated as the allophones (different realisations) of the same phoneme i.e. it doesn't matter whether you use [v] or [w], the meaning doesn't change. For example, most Indian and Pakistani English speakers use [w] for both /w/ and /v/, so they might pronounce vet and wet the same.
In English, however, those two sounds—[v] and [w]—are completely different sounds and distinguish the meaning of the word (/vet/ and /wet/ are totally different words and are distinguished by a single sound). Native speakers pronounce them easily and there's no pause while switching from /v/ to another sound (or vice versa). So if your native language doesn't have a [v] sounds, the only thing you need is practice.
[v] is made by touching your lower lip with upper teeth and by creating friction. [f] is made the same way; the only difference is voicing i.e. there's no vibration in the throat while producing [f].
Answer: When learning a new sound, in your case /v/ you begin by practising individual words. Make a list of words containing /v/. Split your list into 3 parts as follows because you will need to practise /v/ in all positions of the word:
Words beginning with v:
vet
van
vest
vase
voice
Words with /v/ in the middle position:
over
never
having
reveal
severe
Words ending with a /v/:
have
love
give
leave
move
Practise saying your list of words aloud everyday. Gradually your lip will become more comfortable slightly pressing against your teeth.
You can then start to practise /v/ words in phrases. Here are a few to get started:
"Vegetarians and vegans have lots of veggies in their diet".
"My valuables are kept in private vault, where thieves will never find them".
"She lives in an expensive villa in the vicinity of Venice".


Hin and her prepositional contractions

Question:
What are all the hin and her contractions people use? z.B.
raus, rauf, usw. Is it as simple as every her-preposition, or
is it more of a subset? What about hin? For instance, I've seen
nüber but can't find much documentation on it online. I realize
this may be regional.
In which situations / settings are they used? Only spoken /
informal?
Even in such settings, are there places that you would still use
the full word (z.B. heraus) rather than the contraction form? Or
is it pretty much if you're going to use the contractions you use
them every time?
Comment: Related: http://german.stackexchange.com/questions/7794/how-to-use-the-hin-and-her-prefixes
Answer: How to use the "hin" and "her" prefixes? provides pretty good information, but I'll try to provide you with a better "feel" for it nevertheless.
The German terms you mention are mostly used in similar way as you would use them in English...

"hin und her" = "back and forth"
"Es geht hin und her in der Diskussion." = "The discussion goes back and forth."

The word "raus" could be used with "rein" the same way you would use the English:

"Die rennen rein und raus…" = "They're running in and out…"

Same for

"Er läuft rauf und runter…" = "He's walking up and down…"

And yes, there are situations where you wouldn't use contradictions (just like you wouldn't use them in English). An example:

"Finde es raus!" = "Find out!"

Now, here are some examples where "heraus" is frequently used, but please note that these lines are only used colloquially:

heraus mit euch!
heraus aus dem Bett
heraus mit dem Geld!
er ist aus dem Knast heraus
aus dem Trubel der Stadt heraus sein
aus diesem Alter bin ich langsam heraus
aus einer schwierigen Situation heraus sein
endlich ist sie mit ihrem Anliegen heraus
die ganze Geschichte, Wahrheit ist heraus

Last but not least:
You're absolutely correct in thinking that most of the time, you'll end up using the contraction form instead of the full words… as the above examples show. Sure, you could use "Finde es herraus!" instead of "Finde es raus!"; but in the end, the later is more natural and used more often. Using the prepositions is more of a remnant of the older German language use.
Answer: ad 1)
Only "her" can be contracted to "r" if it is a prefix:

herauf - rauf
  herunter - runter
  heraus - raus
  herein - rein
  herüber - rüber

This is not allowed:
Komm doch her zu mir - wrong: Komm doch r zu mir.

Contactions following the same pattern with "hin" are used less frequently. I also think the words "nauf", "nunter", "naus" and "nüber" are only used in southern parts of Germany. They are very rarely used in Austria and I guess that this is true for northern parts of Germany too.
I never heard "nein" as contraction of "hinein".
ad 2)
This contractions are not part of Standard German. They are used in informal speech and they are part of many german dialects.
ad 3)
You can use the standard forms in informal settings to emphasize the word.

Hin and her prepositional contractions


Could clouds on aircraft wings produce lightning?

Question: As air is accelerated over the wing, pressure drops and so does temperature, which is reduced to dew point and finally a cloud forms, which can be seen on humid days. My question is, can this cloud formed over the wing of an airplane produce lightning activity just like some other kinds of "normal clouds"?
Comment: For a cloud to produce lightning, it needs to be charged, and as clouds don't form over aircraft wings, it can't attract lightning activity, or be charged.
Comment: Normal clouds do not produce lightning. Only cumulonimbus clouds.
Comment: @J.Hougaard Surely that can't be right?  I've frequently seen lightning produced simply through the effect of heavy rain fall from clouds that would be better classified as nimbostratus.
Comment: @Michael That is one of the very first things you well learn in any meteorology course. What you have probably seen is a cumulonimbus embedded in other stratiform clouds.
Answer: Lightning occurs when an electrical charge is built up within a cloud, due to static electricity generated by supercooled water droplets colliding with ice crystals near the freezing level. When a large enough charge is built up, a large discharge will occur and can be seen as lightning (from Wikipedia).
So you need two things for lightning to occur- a charge buildup in a place and a charge gradient (difference in charge, like between two clouds or cloud to surface) for lightning to occur.
Based on this, there is little chance of any charge buildup and discharge in the condensation over wings- it is too short lived for any charge buildup within itself and the charges over the aircraft wings are  dissipated by the static wicks in the trailing edges- so no chance of 'cloud' to wing surface either.
Comment: Don't helicopters build up enough of a static charge to shock people though?  And that's an airfoil...  To be fair, that's not technically lightning (an arc that goes between a cloud and either the ground or another cloud), and as I recall it's just the blade passing through the air, with or without moisture.  So I guess that's not what we are looking for, eh?
Comment: @JayCarr, the static wicks don't eliminate the resistence, only reduce it, and they can only equalise the charge towards the ambient potential. Therefore as the aircraft—fixed wing, rotorcraft or aerostat (it _was_ static discharge that set the Hindenburg on fire)—flies near or through a storm, it will collect charge and some will still be there upon landing. Normally it will discharge through the landing gear; only when attaching or detaching under-slung cargo to a helicopter (and when tieing an aerostat) can there be a problem. Either way it requires already charged air though.
Comment: @JanHudec Ah, and now I understand that one scene from "The Hunt for Red October" when they were trying to collect Jack Ryan from a helicopter to a submarine....  Thanks!
Comment: @JayCarr you are correct. All aircraft can build up a static charge if not designed correctly. As the answer says, fixed wings have wicks to dissipate this charge. I would assume and think it highly likely rotary aircraft have similar mechanisms.
Answer: Friction creates static electricity
On the principle you're right: Electrons are separated from air or water vapor molecules due to friction occurring in the boundary layer and other friction areas. Instantaneously there are two types of static electricity created: Negative where some ripped off electrons are located and positive where air molecules with a missing electron are located (these molecules are named "ions" to indicate their electronic unbalance).

Ions are naturally neutralized
However opposite charges are attracted to each other, the ions recombine soon with their separated electrons behind the aircraft, the number of isolated ions or electrons –the "electric potential"– at any time doesn't build up so that a difference of potential with the ground, another mass, or between ions and electrons is sufficient to create any significant electrostatic discharge.
Note the recombination is accelerated by the water in the air, as water –with impurities– is a better conductor of electricity than dry air. Condensation doesn't help for static charge creation on aircraft.
Huge potential differences are required for lightning
The difference of potential is measured in volts*. It requires 3,000 V to initiate an electrostatic breakdown for each millimeter of dry air –the voltage used in an automotive sparking plug is about 10 times this value to ensure a good spark– or 3 GV per km.

Source
If the electrons were trapped on the aircraft surface, the potential of the aircraft would reach the value required to create lightning with the ground :-)
So, why lightning happens with clouds but not with aircraft?
Friction surface from wings is ridiculously small compared to friction surface in the huge cumulonimbus cloud. Static electricity created is not of the same order of magnitude. That's a first reason, but there is also a significant difference in how charges behave.
In natural lightning the electrons and the positive ions are created by the violent wind inside specific clouds. However, contrary to what happens with an aircraft, and for reasons not yet really understood, charged particles are then physically separated: Positive ions move at the top of the cloud while negative charges move at the center where they combine temporary with neutral molecules to form negative ions.
A smaller quantity of positive ions also moves at the bottom. This positive cloud bottom also moves negative charges of the ground at its nadir, which in turn reduces the thickness of the air layer to move through, and at the same time increases the electrical field gradient.

Source
As friction continues in the cloud, the electrical potential increases until the difference of potential between the bottom of the cloud and the ground reaches the voltage threshold required for breakdown. The huge avalanche breakdown heat creates plasma used by particles in excess to join the attractive ground. This phenomena is complex and still not entirely explained either.
*: The volt is the electrical potential required to create one joule of work with a charge of one coulomb. See this nice page for more details.
Answer: What you see above the wings is condensation, but not technically a cloud.
Clouds are suspended in the atmosphere due to prolonged saturation. Not a localized momentary pressure drop.
Those condensations dissipate as quickly as they form, there isn't enough time for them to be charged.
Comment: The question probably can be edited for 'fog' instead of 'cloud'.
Answer:
can this cloud formed over the wing of an airplane produce lightning activity just like some other kinds of "normal clouds"?

No. Lightning requires large, long-lived clouds in which warm, wet air can rise, condense and fall again. That kind of structure can't exist in a cloud so small and temporary as a contrail.
Wikipedia's page on thunderstorms contains lots of information about what's needed.
Comment: Thanks for mentioning cloud size. This is the key enabler for lightning.
Answer: TL-DR: lightning is made possible not by the availability of free electrons (a very common occurrence, in Nature and in the atmosphere in particular), but by the existence of some non-electric force which separates positive and negative ions (despite their electrostatic attraction), over scales large enough  to give rise to a large, sudden increase in the number of free, energetic electrons, once a threshold value of the electric field is reached. No such forces exist over the surface of a metal.
All of the previous answers are correct, but they touch in passing (and not all of them, at this) on the key prerequisite for the formation of lightning: the maintenance of large-scale charge separation, i.e. on the scale of kilometers.
Ionizing gases is reasonably easy, and all sorts of forces are able the sever the links between the outermost (and least bound) electrons from their (aptly called) donor atoms. But this occurs rather uniformly across an average gas like the Earth's atmosphere, which means that, in any given microscopic volume, there are as many positive as negative ions. The ensuing electric field is nearly vanishing, and achieving the large electric field necessary for a lightning discharge (roughly 3GV per km, as pointed out by mins) is impossible.
For instance, on an airplane wing, which is an excellent conductor, electric forces will tend to keep the positive and the negative (=electrons) ions very close to each other, and no large-scale electric field will ever be produced.
The key to explain lightning is that there are other forces which tend to pull electrons and positive ions apart, over very large distances. This has been stated already by mins (although I believe less clearly), and by David Rickerby. The fact that the nature of these forces is not clearly understood does not detract from the fact that this is what makes lightning  possible.
It is easy to understand why such large electric fields are necessary: an electron hitting a neutral atom can ionize it, thus freeing one more electron which will go on to produce twice more free electrons as our initial situation where we only has an electron. But in ionizing a neutral atom, the original electron will lose energy, and might be unable to ionize an another atom when the next collision occurs, and/or it might produce a new electron which also does not have enough kinetic energy to ionize the next neutral it hits: thus no cascade, thus no increase in the number of electrons, thus no lightning.
Unlesss... the electric field is so strong that it replenishes the energy of the original electron (and of course also of its newly-formed brethren) of enough energy between collisions to ionize the next neutral as well. In other words, the electrons re-supply themselves of energy in between collisions, courtesy of the electric field, so as to start the process again. This explains the 3GV per km: the energy necessary to ionize a loosely bound electron is roughly 3eV, and the path between collisions in the atmosphere is roughly 0.0001cm. This implies that a single electron will acquire 3GeV over a km (thanks to the electric field), which corresponds to a potential drop of 3GV over 1 km, exactly mins' figure.
The above also explains the sudden discharge: when then electric field exceeds the above threshold value, the number of electrons doubles every collision: the typical distance in air between collisions is 0.0001 cm, so that, over the course of 1km, a good value for the cloud thickness, one single electron would produce a cascade of 2^10^9 electrons. This number is unrealistically large (the total number of free electrons produced is limited the finite amount of energy available, roughly that of a capacitor of size 1km and electric field 3GV per km), but it gives a good idea of how a lightning is suddenly released. This is in fact a cascade, a sudden vacuum breakdown. For lower values of the elctric field, the newly formed electrons will not have sufficient energy to ionize the next neutral they hit; for larger values they will, thus causing a sudden increase in their numbers.
Answer: Airplane produce lightning ? YES. The thunderstorm electric field is multiplied by the airplane and may trigger lightning.
https://www.scientificamerican.com/article/what-happens-when-lightni/
Cloud formed over the wing of an airplane produce lightning activity just like some other kinds of "normal cloud"? NO. The charge it may hypotethically produce is order of magnitudine smaller if even created.


Can I create a stockpile for rotting items only?

Question: I want to make a stockpile for only rotting items. Is it possible to do this? I already have a refuse pile, but the only options I see is Rotten Raw Hide. What if I have rotten meat or rotten prepared meals? How do I make my refuse pile take only the rotting items
Answer: What you are looking for is known as a refuse pile. Any rotting material, including rotten food will be moved by your dwarves to this pile. If you go into the stockpile settings you can exclude other refuse items such as bones or rotten hide. Be aware that items such as clothing or armor placed on this pile will wear(out) at an accelerated rate.
A refuse pile is not the same thing as a garbage dump. A garbage dump is a activity zone rather than a stockpile and will accept only items with the "dumped" flag on them. Any object can be designated as garbage, while your dwarves will only put specific types of items like bone or rotting items on a refuse pile.
Comment: Sorry, I did not make myself clear enough. I already have a refuse pile, I want to make it rotten items only (ie no shells or non-rotten corpses)
Answer: In the refuse pile, keep all (or most) of the "Items" turned on, and turn off the bones, hair, rodents remains, shells, and what-not. Regular items (from the 'refuse' stockpile menu) will only be taken to the refuse pile if they are rotten or otherwise extremely damaged. (Rotten food is an "Item" in this instance.)


Using the verb oversleep

Question: Can we use the verb 'oversleep' with a direct object?
For example can we say "a person has overslept something" (such as work, school or plane)?
If no, what should we say instead of 'I have overslept my plane'?
Answer: The verb oversleep is normally intransitive (it doesn't take an object), but the full OED does have this for their second definition...

oversleep To sleep beyond (a particular time); to miss (a train, etc.) by sleeping too long; to sleep through (something). Also figurative.

So in principle you can say I overslept my plane. But I think that's a rather unusual usage - if you're going to oversleep anything, it's almost always the alarm. For OP's context I might say I overslept [the alarm] and missed my plane.
Answer: Yes, you can use oversleep with a direct object. Here is one that gives an example with "appointment."

v.tr.  {transitive verb}
  To sleep beyond the time for: I overslept my appointment.

American Heritage
Note, however, that a plane is not something that has a scheduled time. A flight has a scheduled time, so in my experience it would be more natural to say:

I overslept my flight

for instance

I overslept my flight and had to take another one.

Real life examples:

My mind is racing and I am trying to figure out how the hell I am going to tell my mother that I over-slept the flight. (Source)

Another one I found on Google is

a more accurate use of your hypothetical: in your example, if i oversleep the flight (as opposed to it being canceled), AA could possibly charge me hundreds and hundreds of dollars (a lot more than your $150)

I found no examples of oversleep with plane. I am simply saying that in my experience, it seems more natural to use flight and not plane. And I can only go by my experience.
The OED's example of oversleep a train may be natural to some, because trains themselves are often referred to by their departure times:

I overslept the 9:55 train

I would never say

I overslept the 9:55 plane

but I would say

I overslept the 9:55 flight.

Flights, appointments, alarm clocks (and, for some, trains) all have a particular time connected with them and the main definition, in the OED and elsewhere is

To sleep beyond (a particular time)
Comment: I think you're overthinking this one when you say *a plane is not something that has a scheduled time*, and that the usage would therefore only be "valid" if you say you overslept your *flight*. In which context it's worth noting that OED's definition specifically mentions oversleeping *a **train,** etc.* as an example.
Comment: I did not say it was invalid to say *overslept my plane*, I said it was more natural to say *overslept my flight*.

Using the verb oversleep


Will using portable heaters in the attic during heavy snow help melt it?

Question: If I use garage-type heaters in the attic during the winter, will this help
keep snow and ice off the roof?
Comment: ssaltman's answer is correct. Your attic is **SUPPOSED to be cold to keep the snow from melting and creating ice damns at the eve's**. If you have ice at the eve's then you need more insulation in your attic and possibly better venting to keep it cold enough in there to prevent the roof from melting the snow and running down to the eve's where it will freeze and create damage.
Answer: Very unlikely.
what would happen is that the snow or ice on the roof would melt somewhat, flow down to the edge of the roof where the attic ends, and refreeze, forming large ice dams. The ice and snow would continue to melt, backing up liquid water against the roof and evenutually finding its way under the shingles and through the sheathing into your home where it would drip down the ceilings and walls onto your dog, your kids and your computer. You would then curse and try to find a solution, but by then its too late. Your kids would hate you, your computer would be ruined.
Answer: Should you do this, when you replace your roof, put the rubber sheet material under the shingles. It is specifically designed to help prevent the ice dam issue.
The snow is a great insulator, so it would do its best to hold the heat inside of your attic space. Just mocking your attempts to melt it.
And the electric bills from the heaters would likely give you a heart attack, so then there would be medical bills.
I understand exactly what you are trying to do, but sadly physics is against you on this one.
It just kills me when there is a bright sunny day, and my solar panels are buried under 5 inches of snow.


Von Mangoldt function estimate

Question: In many treatments of Vinogradov's three prime theorem, one considers the summation $S(\alpha) = \sum_{k \leq N}\Lambda(k)e^{2\pi i\alpha k}$ in place of $T(\alpha) = \sum_{p \leq N}e^{2\pi i\alpha p}$ (where this sum runs over primes). I assume these two expressions are roughly the same, but I can't seem to be able to estimate $|S(\alpha) - T(\alpha)|$. Can anyone give an estimate?
Comment: Do you mean $T(\alpha) = \sum_{p \leq N} \log p e^{2 \pi i \alpha p}$?
Comment: I do not actually, I do know that $|\sum_{p \leq N}\log p e^{2\pi i\alpha p} - S(\alpha)| \ll N^{1/2}$.
Comment: I see, I am then just curious about why $S$ and $T$ should be roughly the same.
Answer: The reason we consider $\Lambda(n)$ rather that just the sum over primes is because this new sum is slightly nicer to work with.  The idea here is that we may freely weight by a smooth function without affecting the end result significantly.
We should expect that $|S(\alpha,N)-\log N T(\alpha,N)|$ to be small, and one can try to manipulate it with partial summation.  However that misses the point, and is in some sense the wrong "time" to compare the two sums.
In Vinogradov's theorem, we are looking at the primes which satisfy $k_1+k_2+k_3=N$, and we look at the slightly weighted version $$\sum_{k_1+k_2+k_3=N}\Lambda(k_1)\Lambda(k_2)\Lambda(k_3).$$  Let $\chi_{\mathcal{P}}$ be the characteristic function for the primes.  Then lets it is not hard to see how the above relates to $$\sum_{k_1+k_2+k_3=N}\chi_{\mathcal{P}}(k_1)\chi_{\mathcal{P}}(k_2)\chi_{\mathcal{P}}(k_3),$$ at least in terms of the order of magnitude. This connection is easier to prove things about, rather than comparing the exponential sums.
Comment: Can you say a word on why $S(\alpha, N) \sim \log N T(\alpha, N)$, and what is the general statement behind?
Comment: Dear Eric, Should the last word of line two be "without"?  Regards,

Von Mangoldt function estimate


What is a re-manufactured CV axle?

Question: Today I replaced the passenger side CV axle of my 1994 Camry LE 4 Cyl. The package of the CV axle was saying 'remanufactured'. Should I assume that it was not new?
Answer: They change the boot and make sure joints are within tolerances and maybe remove some rust and maybe spray or rust treat them.
They are not new but recycled and or refurbished to a "new" like state.
Answer: I don't believe remanufactured has a legal definition in most states. It is important that you know the company that rebuilds the part. If possible check their web site or literature for specifics. Most of the large reputable rebuilders will state what they replace, what they inspect and what they reuse. You want to understand what they did so you can compare different vendors. A vendor may change all the wearable parts or just the worn parts. You want to be sure you are getting the best value for the money.
Comment: Here's a company that specializes is remanufacturing OEM parts http://cvjaxles.com/, you might reach out to them over the phone or email them for some questions.

What is a re-manufactured CV axle?


Are there any differences between 〜けらず and 〜ざりけり? Are they interchangeable?

Question: Was ~ざりけり used in place of 〜けらず, for example in 知りけらず, to indicate the past tense in the negative form? Did 知らざりけり exist?
Answer: 〜ざりけり certainly existed and was very widely used. This definition includes a couple of examples:
https://kobun.weblio.jp/content/ざりけり
I believe that 〜けらず was actually far less common than 〜ざりけり, at least from the Heian period on, and the entry for けり in 大辞林 seems to confirm this – in fact, it suggests that 〜けらず disappeared entirely. (The main definition indicates that けり has no 未然形, so there would be nothing to attach 〜ず to; a note at the end of the definition clarifies that 未然形「けら」は、「けらず」「けらく」の形で上代にのみ用いられる and cites an example from 万葉集.) Also note that 〜ざりけり can and often does have an exclamatory nuance. I'm not sure 〜けらず, on its own, conveys quite the same feeling.


Editing Driving and Car Scenes

Question: I'm editing a video just to do for a fun challenge, its of a rally car doing various manoeuvres and drifting, just using library material, no recordings I've made.  Was curious to know of any strategies or approaches people have when editing sound for a scene like that, or a chase scene.   One thing i'm struggling with is to make the sounds of the different cars in the library sound like they are only one car...
Answer: In that situation, where you're having trouble getting sounds to match, layering is going to be your friend. If you can find some elements to lay in underneath that will augment the actual sound, you'll at least have some commonalities between the multiple sounds that will help dupe the audience.
Also, camera/perspective changes are where you'll want to put in the sounds with the most radical tonal changes, if they're all you have to work with. If you have those layered elements in there still, you will still have that connection to the previous sounds.  We are much more forgiving of sonic changes that occur with visual cuts than we are of ones that occur independent of the picture.
Answer: I always work thinking in terms of stems, and have sets of source tracks named for each set of stems. So for editing a car chase, I would have a set of (10-32) tracks and a stem dedicated to each of:

Vehicle movement (tyre movement, screeches, suspension etc)
Exterior vehicle moves
Onboard vehicle
Interior vehicle

This helps clarify cutting sound as per the picture cuts. It depends on the context and picture cuts of the film as to how strongly you want to make the sequence 'smooth' or 'cutty' - it can sometimes be more dramatic to accentuate picture cuts/perspective changes...
Once you have the sequence working it can be valuable to solo a stem/set of tracks eg vehicle movement, and focus soley on it to make it more exciting/dynamic/interesting. Tyres do not only make sound when they screech - watch a great car chase film like Ronin and listen to the different tyre sounds for different surfaces... Think about G Forces on the car interior as its moving fast, when it skids around a corner the suspension may be under a lot of load as well.
As for making different car motor sounds work for a single car, that is not a process I would ever pursue. Good vehicle editing is like good ADR editing: you start with a good performance! For a film, access to vehicles is something to insist on.
Comment: Totally agree with you. I usually spend more time on the tires than the engine sounds as they sell movement more than anything.
Comment: "making different car motor sounds work for a single car, that is not a process I would ever pursue" +1 for that - each car has a voice - and you need a shed load of good source material from that one car to make it work. coincidentally http://www.pole.se/ is a good place for engine source material :)

Editing Driving and Car Scenes


Normalization of the action in Special Relativity

Question: The action for a massive point particle in Special Relativity is given as
$$A =-mc^2\int d\tau,$$
Where $\tau$ represents the proper time, and $m$ represents the (rest) mass. From what I could understand, the Action must not change with respect to the reference frame, and hence it can be written as $$something\int d\tau$$ but why must the something need to be proportional to mass?
Comment: It comes from dimensional analysis - The action has to be dimensionless and $\tau$ has dimensions of length so you need to multiply with something that has dimensions of mass.
Answer: I prefer to think like this. Inertial observer ($X$) is at rest in his/hers reference frame. The world-line of $X$ is the longest possible route between any two events. This follows since, in its frame, $X$ is moving fully along the temporal axis, therefore $ds=cd\tau$ (displacement in time; $c$ is the speed of light), and $d\mathbf{r}=\mathbf{0}$, and one should bear in mind that, in general, the 4d distance between two events, within the causality cone, is $ds=\sqrt{c^2 d\tau^2 - dr^2}$.
We can therefore describe the path taken by the observer $\bar{x}^\mu=\bar{x}^\mu\left(\tau\right)$ as the (unique) path that is longest possible path between events $A$ and $B$ (that lie on the world-line). Since length of a curve (in 4d sense) is an invariant quantity, the universal description of the path of the observer is, a world-line that maximizes/minimizes:
$S\left[\bar{x}\right]\propto\int^B_A cd\tau$
Where $cd\tau$ gives a length of a small segment of the world-line. We can then fix the sign of the scaling constant in order to demand, that this quantity, which we shall call `action', needs to be minimized to arrive at the correct world-line:
$S\left[\bar{x}\right]=-\alpha\int^B_A cd\tau,\quad \alpha>0$
The above prescription must be valid for all observers, including those that observe $X$ as moving at small velocity $\mathbf{v}\: \left(\left|\mathbf{v}\right|\ll c\right)$. Call any such observer as being in the lab-rame. In lab-frame we can describe $X$ with classical action, thus, in the limit $\left|\mathbf{v}\right|/c\to 0$:
$S=-\alpha\int^B_A cd\tau \to \int^B_A \frac{m\left|\mathbf{v}\right|^2}{2} dt$
Next, the world-line of $X$ in the lab-frame is, by construction $\bar{x}^\mu=\left(ct,\,\bar{\mathbf{r}}\left(t\right)\right)^\mu,\:d\bar{\mathbf{r}}/dt=\mathbf{v}$. The 4d distance between any two (close) events on $\bar{x}$ is:
$cd\tau=\sqrt{c^2-\left|\mathbf{v}\right|^2}dt$
Thus, in the limit $\left|\mathbf{v}\right|/c\to 0$
$-\alpha\int^B_A \sqrt{c^2-\left|\mathbf{v}\right|^2}dt \to \int^B_A \frac{m\left|\mathbf{v}\right|^2}{2} dt$
This fixes $\alpha$, though, of course, the specific value of $\alpha$ is more of a convention.
Comment: Since the answer proved to be popular, I have expanded it, whilst (hopefully) keeping the original flavour
Answer: For a single particle, it does not matter what prefactor you use, the equations of motion and everything else stays the same. The factors only start to matter when you couple different systems to each other. For example, consider a charged particle in an electromagnetic field described by a vector potential $A_\mu$. The right action describing its movement is
$$ S = S_{\text{EM}}+\int d\tau \left[ -m - q A_\mu \frac{dX^\mu}{d\tau} \right] , $$ where $S_\text{EM}$ is the action for the electromagnetic field.
This only works out to give the correct dynamics if the proportionality factor $m$ here corresponds to the mass of the particle.
Similarly, when one tries to extract meaningful quantities like the energy-momentum tensor from the action, only one choice of the factor will give the correct result.
I personally like to think of it as the particle having an always present static contribution to its potential energy, coming from its rest-mass energy (hence the negative sign), but don't take that too seriously.
Answer: The equation you mention is the action of a single point particle.
$$S =-mc^2\int d\tau$$
The unit of action is energy multiplied by time, in the present case the rest energy which corresponds to the mass of the particle multiplied by the proper time of the particle.
This equation refers to the action from the point of view of the reference frame of the point particle along its own worldline (which is parameterized by its own proper time), that means that the point particle is its own observer. From its own point of view, its velocity and its displacement in space is always zero, and also its momentum and its kinetic energy is zero. You can say that the equation describes the aging of the particle, its mass (= its rest energy) is transported through time.
Comment: You define action in terms of energy (I think). Does it not create a circular argument? IMHO, energy is the quantity conserved as a result of time-invariance of the laws of motion, thus you would need Lagrangian to get the expression for energy, and for that you need action.
Answer: Since action has the units of angular momentum, the proportionality constant needs the units of energy by dimensional analysis. It must also be Lorentz-invariant, so is $mc^2$ times some real number. This number's modulus can be varied without changing the resulting equations of motion, but @Cryo's answer shows a modulus of $1$ recovers the usual $\int(T-V)dt$ outcome of Newton's physics. The sign, on the other hand, is fixed by the requirement that the action be minimized.


What are some examples of indirect mechanisms that are not auctions?

Question: I'd like to know if there are any examples of indirect mechanisms that are not auctions. Auctions seem like a very controlled environment. I'd like to see some examples of indirect mechanisms "out in the wild".
Answer: Voting mechanisms would seem to be a good example.  So are matching mechanisms such as those used to assign students to public schools.
Answer: Here is an example of a mechanism "out in the wild". I take "the wild" literary, but I won't be specific because I am not a biologist and I am making this up from the top of my head.
Suppose two predators spotted some easy to get food source, and their value for the prey is their private information, i.e., a random variable $v_i$. It would be efficient to auction off the food with an appropriately designed auction. Instead the predators could fight and the one who exerts more effort in the fight wins the food as a price.
Each player $i$ maximizes his payoff
$$G(e_i)v_i - e_i - \delta(e_j),$$
where $e_i$ is the effort of player $i$, and this player wins the fight with probability $G(e_i)$, the probability that he put in more effort than player $j$. The (by assumption) linear cost of effort has to be incurred independent of winning or losing the fight. On top, some damage $\delta(e_j)$ from the opponent is incurred.
To solve for the optimal effort as a function of the value $v_i$, you can approach the problem like an all-pay auction.
An all-pay auction can also be interpreted as other forms of contests. It can also be used to model political lobbying activity. You can use insights from auction theory to solve for the optimal strategy in a plethora of allocation mechanisms. Think about queeuing or wars of attrition and so on.


1&1 hosting + azure cname + google apps mx

Question: I have a problem while hosting my website on azure :
I need to have an url mydomain.com, when deploying in azure it gives the url mydomain.cloudapp.net, so I do a cname record from mydomain.com to map with mydomain.cloudapp.net.
But I also have set up mx records for mydomain.com to google mail servers in order to have emails like email@example.com, and the cname record drops these mx records... Then I try to do the following :
 1) keep mx records for mydomain.com and give up with the cname record here
 2) create a subdomain www.mydomain.com and do a cname to map this subdomain to mydomain.cloudapp.net
 3) do a redirect from mydomain.com to www.mydomain.com
then I could keep my pretty mails and have a pretty url for my azure website.
BUT it seems to not work with 1&1 hosting, I can't map www subdomain to a cname.
Does anyone know if there is a workaround, or if I'm doing wrong ?
Answer: The issue is that you cannot have a CNAME record for a domain that has other record types.  It's for this reason that you should never have a CNAME record for your root domain (mydomain.com).
It seems like you have the right answers:

Keep the MX record for mydomain.com pointed at Google.
CNAME alias your www record to mydomain.cloudapp.net.
Create a redirect from mydomain.com to www.mydomain.com (via <code>.htaccess</code> rewrite-rule, or a basic HTML redirect page hosted on mydomain.com).
Comment: I've done this but it seems to not work all the time : it was working yesterday evening, but this morning when I go to www.mydomain.com or mydomain.com, the browser shows a dns error... but mydomain.cloudapp.net is reached, so it seems to be a pb with the dns mapping. may be sometimes one record overwrite the second one ? it is very weird
Comment: more precisely, I have a 105 error sometimes
Comment: That sounds like you either have a problem with your local DNS servers, or the DNS server that is authoritative for your domain.


Destroying the TSA's equipment

Question: So let's say that you were detained by the TSA on your way out of the US. They demand your phone password and you give it too them. Now when they attempt to plug your phone, you phone breaks their computer (either by accident or on purpose). Would the TSA be able of charging you for anything and would they be able to inflict any repercussions on you (except for detaining you for longer)?
Comment: Setting aside the legality of the TSA demanding your phone password (which is debatable)...Was the damage by accident, or on purpose? That matters.
Comment: This is almost completely in theory, but I heard someone did this to the TSA so I was just wondering about it. Thanks
Comment: Was it the TSA or customs?  Customs officers are far more likely to demand access to electronic devices than are TSA officers.
Answer: Intentional sabotage of a TSA computer system is almost certainly a serious crime and would also almost surely give rise to civil liability, although you might avoid both if you warned the TSA that the phone was set up to intentionally break their system, in which case it might be confiscated as contraband.
If the product had a "feature" unknown to you and that you had no reasonable reason to know of that caused the harm, you would ultimately have no criminal or civil liability, although the manufacturer might be strictly liable to the TSA under a product liability theory, and you would probably be detained as a witness to figure out what happened.
If the product had a "feature" that broke the TSA computer that could be de-activated and that you meant to de-active but carelessly failed to, you would have negligence liability to the government and might or might not have criminal liability (I'm not enough of an expert in the relevant statutes to know). You might be liable for a strict liability Federal Communications Commission offense for having a device that is in violation of their regulations.
Comment: @Harper What is possible is a job for other SE forums. Here in Law.SE we assume that the question's technical assumptions are true whether or not this is the case, for the sake of argument, unless the possibility to totally absurd to a non-expert. And, the kind of stuff proposed in the OP certainly happens in fiction all the time.
Comment: Just to mention...no standard phone has a "fry a computer when plugged in" feature. :)
Comment: And what would a surge do? Burn out one USB port, maybe all of them, maybe the PCB the USB port is on.  They have spares.   A $20 USB hub would firewall them from even that.
Comment: @cHao I would imagine that the notion is that there is malware on the phone which the user may or may not be aware of, and may or may not be in control of. I doubt that a hardware based electrical surge, for example, is what the OP is really asking about. I also added scare quotes to the word "feature" in my answer to clarify what I mean.
Comment: A hardware based electrical surge, though, is one of the very few situations that don't require malice or gross negligence on the OP's part. Malware doesn't generally have that kind of control over a USB port, even on a rooted phone.
Answer: There's an old military saying:
<code> Once is happenstance.
 Twice is coincidence.
 Three times is enemy action.
</code>
To sue you for money damages and win, they'd need to show (more likely than not) that the damage was your mistake. To prosecute you criminally under system tampering, or destruction of property, they would have to prove  (beyond a reasonable doubt) that you meant to damage their system, and they can't even really start seizing computers and subpoenaing your online activities until they show reasonable suspicion to a DA, judge and grand jury. Here is where probability matters: it is borderline absurd that this would even be possible, so it having happened before is very important to penetrate that disbelief.
On the other hand, they can sue you for anything, on the pretense of recovering the cost of the equipment. That might actually be the better tactic here because they can subpoena anything without court oversight - the judge only looks if you ask him to (quash), but you'd have to know what you were doing, and move pretty fast.
Once the civil discovery exposes a reasonable suspicion of criminality, they cross-table their work to the D.A. and it becomes a criminal investigation. However in civil law, they are supposed to "CC you" on everything, and the moment you see them going after the incriminating stuff, you could be in the wind.  That doesn't matter if you have nowhere to go.
One place I differ with my colleague is you having some duty to actively do something to protect their system, e.g. Turning off a feature known to damage computers.  If you had asked to use TSA's system for your own purposes, sure.  But TSA didn't ask. They seized, so they handle it at their own risk.  Heck, if you said "hey, you need to turn off X so it doesn't damage your computer, here, let me see it for a minute", they would assume you were trying to hit the "wipe" button.

Destroying the TSA's equipment


Bottom Bracket Height

Question: I am just wondering if BB height from the ground is the same for all bikes or for all mountain bikes, all road bikes, etc? Or does every brand have its own design for that dimension?
Answer: No, it is not. Different bikes in the same class (mountain, cross, road, etc.) may even have different bottom bracket heights even among comparably sized bikes. It is a per manufacturer design decision where to place the bottom bracket.
Mountain bikes need higher bottom brackets to clear obstacles vs road bikes or cross bikes, for example.
Answer: Most fixed gear frames and track frames have higher bottom brackets to mitigate the possibility of pedal strike on the down stroke since you cannot avoid having your turning-side foot at the six o'clock position while making a turn. Some frame manufacturers make a point to design their frames with a higher bottom bracket for this particular reason. In addition to what others have answered, mountain bikes have higher bottom brackets too (as per the frame manufacturers/engineers' discretion) to clear obstacles.
Answer: I would think that the height of the bottom bracket is completely up to the frame designer.  However, there must be practical constraints: most cranks are about the same length for adults, 172.5 mm for road bikes or 175 mm for mountain bikes, although other sizes are available (130 mm–220 mm according to this source).
The higher the bottom bracket the taller the bike, other things being equal, so I'd think that raising the bottom bracket reduces the stability of the bike and increases the aerodynamic drag.  As @Batman says, mountain bikes need higher clearance than road bikes to avoid obstacles.  If the bottom bracket is too low, then riders risk striking the pedals if they pedal through a moderate turn.  Because designers nearly universally want to maximize stability, minimize drag, and avoid pedal strikes through moderate turns, and because the vast majority of cranks are nearly the same length, I'd think that the heights of bottom brackets of popular bikes for average-sized adults are all fairly similar, for the same class of bike (road, mountain, etc.).  (Of course I'd think that many bikes for children and small people that come with smaller-than-average cranks would have lower-than-average bottom brackets, and that many bikes designed for tall people would have higher-than-average bottom brackets.)

Bottom Bracket Height


Can sweetened condensed milk be substituted for whole milk and sugar?

Question: I am making a fudge recipe that calls for 1cup whole milk and 5 cups sugar. It also uses 2 sticks of butter and 25 marshmallows. I'm wondering if I can omit the milk and cut back on the sugar by using sweetened condensed milk.
Comment: Condensed milk has a lot of water removed. How are you planning to incorporate that back in?
Comment: Hello JennyJ71, and welcome! Our site suggests related questions in the sidebar when you ask a new one, and the top question is quite relevant in this case: http://cooking.stackexchange.com/questions/18443. Basically, sugar is the main ingredient in fudge, so while the substitution itself might be possible (the answers should clear up that), I doubt that you can reduce the sugar.
Answer: There are recipes that specifically use condensed and or evaporated milk as a shortcut to making fudge. It's not just a drop in replacement, it's a whole change in method. Just do a search for "easy fudge recipe" on Google to get examples of these fudge recipes that use condensed milk.
To add my opinion however, the old fashioned cooking and churning method produces the best fudge, but the condensed milk method although isn't as good at least produces a more consistent result. I do the old fashioned method. It's worth the extra effort and the extra learning curve (to recognize when the fudge is about to setup).
Comment: Just so you know, the it's not a reduction process. It a candy making process. The sugar in condensed milk is already candied somewhat, where as traditional fudge you need to candy it yourself.
Comment: Thanks everyone! I typically use the fast and easy method found on the sweetened condensed milk label. But I found a Jack Daniels fudge recipe on Facebook that I'm about to try my hand at. Seems like the substitution process just adds work instead of reducing it, so I'm going to follow the recipe and keep my fingers crossed! I appreciate all the comments! They certainly helped.


How to get out of boot loop on Mi 4i?

Question: I got stuck in a boot loop. Can access Mi Recovery. Can't access fastboot for more than 10 seconds. I had accidentally wiped everything from my phone, without any backup. Also, I had USB Debugging off in my device. Now, how do I flash a new ROM and reboot into my system?
Comment: Can you access mi recovery without connecting usb cable or charger?
Comment: No. In fact, the recovery crashes too, if I disconnect the charger.
Comment: So, I assume you can access recovery only by connecting a charger and no other functionality is accessible. If you disconnect the charger, it power off. Am I right?
Comment: Yes, you are absolutely right. Sorry to reply so late. Now, what should I do?
Comment: check the answer. It is based on experience. after my experience, I adviced this to many people and almost 90 % will work
Comment: Thank you so much. I was thinking the same too. Will try that ASAP.
Comment: Yes. Thanks a million again Sagar. I'm in that 90%.
Comment: Consider marking it answer for future visitors.
Answer: From Experience : The problem is with your battery. The battery can't produce stable power.
When you go to recovery mode, the power consumption is very low. So, it will stay for some time.
When you turn on the phone, the brand logo is shown, may or may not the Booting is done. Suddenly the phone turns off. Then again turns on and repeat this automatically about 15-20 times.
I have the same issue and after checking for a weak , I found that the battery can't produce the power in a stable manner.
How to check.
Since your phone has an inbuilt battery, I am not suggesting you to remove cover, which may remove the sensor, then taking out the battery and doing so.
So, if the phone is under warranty, go to a service centre. Else, go to a local mobile mechanic and ask him to test the phone by putting an alternate battery.
80% it will work.
Try it and if you are in the 20% part, let me know.

How to get out of boot loop on Mi 4i?


Roshi is over 300 years old because he used Dragon Balls, but how is Master Shen still alive?

Question: Roshi and Master Shen (Crane) were rivals while growing up. But why did Crane not die of old age?
Answer: A major thing to keep in mind is that Roshi is an unreliable narrator (and his turtle isn't terribly smart by human standards). He routinely lies and twists half-truths to help people get through something tough, motivate his students to work harder, or get something he wants that he thinks people otherwise might not give him. It's a generally benign sort of thing, serving his mostly comedic perversions or serving some sort of "greater good".
As Gary mentions, in the original series there is mention of an Elixir of immortality/Fountain of Youth, and Roshi at one point tries to console Tien by pointing out that he (Roshi) is immortal, so there's nothing to really worry about.  This was another lie or half-truth, though, as Tien was soon knocked out and Roshi was killed by the strain of the technique he used to try to save the day. So, as indicated, he twisted the truth, or outright lied, to help someone out. Most people at this point presumed that in fact the elixir/fountain only stopped him from dying of old age (and possibly other natural causes), but suicide/exhaustion/getting eaten/laser beams to the face/etc. would still be lethal in the usual fashion.
In Super, Roshi sends Goku and Krillin out to fetch some paradise herb in exchange for teaching them a super secret technique.  The technique was a half-truth, though. It was meant to motivate them to do the quest, and thereby help Goku increase his control of his energy and for Krillin to overcome his (very justified) fears. So they were taught something useful for combat, but not really a "technique" (and certainly not one of Roshi's martial arts techniques). He had also stated that eating the herb would increase his lifespan by another thousand years.  Whether that was another lie, or a bit of dramatic license, or not...we don't know.  But it stands as the current best explanation for Roshi's long life.  Presumably, Master Shen and his brother also partook of this herb.
Alternative options do exist, however.  Demon King Piccolo used the dragon balls to acquire Eternal Youth, which was inherited by Piccolo. This prevented Piccolo from aging once he hit his physical prime (a mere 4 years later), but neither of them had true immortality and could be killed by means other than aging. Conceivably Roshi, Shen, and others may also have made such a wish at some point.
A standard martial arts trope is the long-lived master, where their supreme skill and spiritual development gives them an unnaturally long lifespan; this may even be a specific "technique/skill" of theirs. Such a technique may have been found by Roshi/Shen/General Tao/Baba and used amongst them, only they found it later in life and it does not revert your body to an age before you acquired it. Part of Roshi's character is how he at once serves this role of a transcendent master of martial arts while also being a perverted, lazy, goofball.
And there's the fact that if he is in fact several hundred years old, then he may have gone through a lot of stuff, and the reason he has lived so long is buried somewhere in all of those adventures, possibly stretched across several of them in a complex fashion, and he may not even remember them all. After all, he once had true immortality by owning a Phoenix (which ironically died, and the immortality was lost). Which is to say there's plenty of story telling room to "retcon" the reason for his long life and insert some new justification(s) later.  All of which serves the handy purpose of making others distrusting of his claims, and difficult for others to achieve what he has (which could be dangerous to the world, as well as those attempting the feat).
Comment: I think this is a good answer but one point I disagree with. I don't think Master Roshi used the dragonballs ever before as he had no idea the dragonballs could be used for wishes at the start of dragonball when he meets Goku
Comment: @EricF That could just fall under the unreliable narrator thing.  He could be playing dumb. Or have forgotten things that have happened over such a long life. Or he really might not have ever seen or heard of them before that moment. The answer is part conjectural as Roshi's character and backstory are left a bit murky by his tendency to fabricate things, change up stories, and on occasion act like an addled old man.
Comment: Fair enough. I agree and wish we could see more of his back story. The most we got was the Fan Fan episode really
Answer: This is a mistake which was overlooked and never addressed in the series. It was initially assumed that they drank the Elixer of immortality from the Fountain of Youth. However, Roshi later claims this doesn't really exist. So yes! There is no logical reason explaining why the Crane Hermit is as old as he is.
Answer: It is important to remember that Immortality doesnt always= unkillable it often means you wont die from age sometimes refered to as agelessness. In roshis case this is the most likely senario from one of the above mentioned possible sources of said immortality


Scraping multiple cryptocurrencies

Question: I would like to scrap data from Bitcoin and mix of alternative cryptocurrencies which are fundamentally like Bitcoin. The data I am looking for are blocks and transactions.
I found a pretty awesome (very old and not maintained) project called MultiCoin: https://bitcointalk.org/index.php?topic=24209.0
I have noticed cool projects such as DNS Seeding which do things like connect to peer nodes simply to request more peer nodes (without transmitting data), but do not seem to transmit transactions or blocks or any other data.
Are there any more modern implementations of this? As the alternate cryptocurrency space grows I would like to get data from a lot of coins without having to consume more resources than strictly necessary.
Comment: The obvious way to collect blocks and transactions is to just run a client for each coin. If that is not feasible, can you explain why? Also, I don't really see what DNS seeding has to do with this.
Answer: The BlockCypher web services supports 3 different cryptocurrencies right now, more may be added in the future. All coins have the exact same services exposed so it's easy to develop for all of them at once.

Scraping multiple cryptocurrencies


Can I use two crates to crate train my new puppy?

Question: We are getting ready to crate train our new puppy when she comes home.  We would like to have a crate on the first floor of our home for daytime use and another crate in our bedroom for night time.  Will this confuse her?  Once completely housebroken, she can sleep in a dog bed in our bedroom, but while she is little, we would rather have her in a crate at night.
Answer: You should simply use 2 different commands, one for the daytime crate and another one for sleep time.
In general, you should choose your commands according to the result you expect, not according to the things that might (or might not) yield the result. Some exaples are:

You want your dog to go into the daytime crate, tell her "into the crate".
You want her to go into the nighttime crate or her dog bed, tell her "sleep time".
You want her to stay put and stop following you around, tell her "stay" instead of "sit"
You want her to stop doing whatever she is doing, tell her "no" instead of "come here"

That way you avoid her confusing what you mean and ultimately stop responding when she doesn't seem to get it right.
Answer: I am presuming that you are going to bed at the same time as your puppy. The easiest way is to close the door. Continue to put the puppy in the proper pen with the door closed until it knows where it should be when you are in the room or not.My presumption hangs facts that you are in the bedroom when you want the puppy in the night pen and you are not in it when you want it in the day pen.Close the door with the puppy on the proper side of the door and continue to put it in the proper pen until it learns which pen to go in. How long it will take depends on the learning curve of the breed of dog.


How do I get my name removed from a co-signed car loan?

Question: I am the primary signer for a car loan consigned and being paid now by my son.  Is there any way I can get my name removed from this loan?
Comment: This is getting close to a legal question and the answer will depend on the exact wording of the loan agreement. Even if your son is paying off the loan because you cannot afford to do so, you are still legally responsible for payment of the loan and you still presumably have title to the car. So, the only way to get your name off the loan would seem to require the cooperation of your son in taking over the loan **and** the title of the car (effectively buying the car from you and taking out a new loan in his name only for the balance still due on your loan.)
Comment: Does your son have enough income for the bank to approve the loan based on his income, debts, and credit score alone?
Comment: Pay the loan off.
Answer: I believe that your son will need to get a new loan for the car in his name only and use the proceeds of that loan to pay off the one you co-signed on.  The only way that will happen is if he can find a lender willing to loan him the money based on his credit only.
From the current lender's perspective, if your son isn't a good credit risk, then why would they let someone out of the loan who might be able to pay if your son defaults?  If he is a good credit risk, then they, or someone else, should be willing to lend to him without you as co-signer.
Also, as Dilip Sarwate mentioned, you might have to do something with the title, depending on whose name it is in.
Answer: The wording of this question is very confusing because "primary signer" would,
in ordinary parlance, mean the person borrowing the money and the co-signer
(not consigner) would mean the one who is guaranteeing the repayment of the loan:
if the borrower does not pay, the co-signer is liable for making the payments.

Whose name is on the title of the car?
Who borrowed the money to buy the car? Is the loan in your name and your
son co-signed the loan to induce the bank to loan you money
to purchase the car, or is it the other way around, that your son borrowed
the money and you co-signed the loan in order to induce the bank to loan
your son the money?
If the car title and the loan are in your name, are you defaulting
on the loan and so your son is making the loan payments that should have
come from you? Or is it that your son borrowed the money to buy the car,
his name is on the title, he is making the payments, and you are no
longer interested in backing him up in case he defaults and the bank
comes after you for the money?
Comment: Primary, co-signer.  It really doesn't matter.  If this thing does not get paid both credit scores will get hammered.


What is the difference between moisture regain and water holding capacity?

Question: Water retention is how much water the fibre can retain. Water holding is function of hydrophilicity and construction of the fibre (fabric). What is the difference between those two? Because to retain is also to hold, in my knowledge.
Answer: According to the FabricLink Textile Dictionary:

Moisture Regain - The amount of water a completely dry fiber will absorb from the air at a standard condition of 70 degrees F and a relative humidity of 65%. Expressed as a % of the dry fiber weight.

Water Holding Capacity is the amount of water a completely dry fiber will retain after soaking in liquid water and allowed to "drip dry".


Is a two-opt move guaranteed to produce a non-worse tour?

Question: I have a little doubt with respect to the two-opt improvement heuristic for TSP. I understand how it works, from the Wikipedia article on 2-opt and "Heuristics for the Traveling Salesman Problem". However, from my understanding and subsequent experiments, it appears this heuristic does not give any guarantee to at least not lead to a worse tour. If this is so, I don't understand how a two-opt move is any better than a random swap of cities. I would really appreciate if this can be clarified for me. Thanks!
Comment: 2-opt is indeed just a random change and the new tour may be worse, but it can be incorporated in e.g. simulated annealing. If the new tour is shorter, the chance is bigger that simulated annealing will accept the 2-opt change, otherwise it might undo it.
Comment: Unfortunately, I am unable to access any of your links.
Comment: @AlbertHendriks Thanks for the answer. That's quite interesting, though! One wonders then, why it's so much touted? And, sorry to bother you, but could you expatiate on why SA would have a greater chance of accepting a better 2-opt move than a better random swap?
Comment: @AlbertHendriks PS: Sorry about the lateness in my reply! :)
Comment: @ayePete SA just checks if the tour is shorter after applying the random 2-opt change. If so, it accepts the change. If not, it randomly undoes or accepts the change. It's basically just an "if"-statement like that. 2-opt is powerful with simulated annealing, since you can find good solutions this way.
Answer: 2-Opt is a move that doesn't guarantee to give a better tour.
We use such moves ($k$-Opt, swap, insertion,..) in local searches to look for a better tour in the neighbor of the input (which is a tour).
Local search consists of browsing all (feasible) tours that can be produced with only one move from the input.
If the input tour is not improved, we call it a local optimum which may be the optimal solution.