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February 25

How are contactless credit cards protected against electronic pick-pocketing?

Couldn't a tech thief adapt a credit card processing machine to collect payments in crowded places?--Senteni (talk) 00:29, 25 February 2015 (UTC)[reply]

Yes, they certainly could. So a second level of authentication, like a PIN, is needed. StuRat (talk) 06:01, 25 February 2015 (UTC)[reply]

But the PIN is only occasionally required, so presumably a lot of small amounts could be collected before a PIN was needed. Presumably the issuers of the cards have some security measures to detect repeated use in unlikely places since they are liable for any money falsely debited, but their ultimate protection is simply to reclaim the money from the authenticated bank account to which money is credited. The fraud wouldn't last long and the perpetrators would be easy to trace through the identification required when they opened the account. Dbfirs 08:38, 25 February 2015 (UTC)[reply]

(ec)::The whole point of a contactless transaction is that you don't need a PIN. However, there is a limit of a small number of transactions of less than £20 each day and I suspect the banks have to pay for any that you can show were fraudulent. But there was a possibility of fraud on a larger scale - at least with some of them - although that may have been fixed by now see [1]. Richerman (talk) 08:46, 25 February 2015 (UTC)[reply]

Not having to slide credit cards through that annoying reader is another advantage. I often have it turned the wrong way, or not quite far enough in, or slide it too slowly or too quickly. StuRat (talk) 14:52, 25 February 2015 (UTC)[reply]

The limit will be raised to £30 this year. They are playing it safe. 217.158.236.14 (talk) 09:49, 25 February 2015 (UTC)[reply]

I also wonder about accidentally charging the wrong card. Consider people packed tightly in line, passing a street vendor and buying hot dogs as they pass. Your credit card may not be any closer to the reader than the next guy in line's credit card. StuRat (talk) 14:55, 25 February 2015 (UTC)[reply]

It's contactless in the sense that you don't need to swipe it. You still have to take it out of your pocket and wave it within an inch or so of the reader. The security comes from the fact that the RFID doesn't actually give you the credit card number (which the magnetic strip does, making it susceptible to skimming). So simply using an RFID scanner shouldn't give a thief anything useful. They would need an actual credit card payment terminal (or at least the software) which would presumably mean having to register with the credit card company (they have to know where to deposit the money), which should make spotting fraud rather easy. Mr.Z-man 15:28, 25 February 2015 (UTC)[reply]

I can think of a couple ways around that:

1) Hack the software.

2) Start some tiny business, say a pretzel stand (if they would even bother to check), and use that to get the scanner software, which you use to charge all the cards you've skimmed. Then withdraw the money as cash, leave town, and repeat the same thing in the next town. (If they have an alert out to not let you get another scanner, just use a bit of identity fraud to change your name each time.) StuRat (talk) 20:06, 25 February 2015 (UTC)[reply]

Hacking the software won't help. To be able to clone a card and use it, you need the account number, security code, expiration date, and possibly the cardholder's name and billing address. None of that is transmitted through the RFID chip on modern cards. The point of the software is to communicate with the processing network so that the credit card company can process the transaction. Only a portion of that transaction is done client-side on the terminal.

If you're going to go to the extent of #2, why even bother skimming RFID cards when you can just inflate charges on every transaction? It would probably be cheaper to just buy a gun and rob people directly. It doesn't protect against that either and the chances of getting caught are pretty similar. The money isn't deposited instantaneously, there can be a delay of up to a few days. Mr.Z-man 01:02, 26 February 2015 (UTC)[reply]

1) You're saying that the info is secure because it's stored on a remote server, but such remote servers have been repeatedly hacked to reveal supposedly secret info before, whether it's account numbers, social security numbers, phone texts, etc.

2) When you think of one individual doing it all, you may be right, but criminal gangs already have people doing all these various parts. StuRat (talk) 22:25, 27 February 2015 (UTC)[reply]

Are you sure about that? When I tested last year, I was still easily able to get I think the credit card number and possibly expiry date with this Android app [2]. The app details suggests new cards may not send the card holder name and transaction history feature is removed in new cards, but doesn't mention anything about removing the credit card number, nor for that matter expiry date. (These older refs confirm the app isn't lying about what has been transmitted [3] [4].) The card itself was relatively new (like 2014), but I am in NZ so may be a bit behind.

However as I understand it per [5] [6] [7] [8], the payment tokenisation spec was only published last year, it was rolled out in the US due to Apple Pay, but isn't in Europe or most of the world yet. As I understand it, it is the tokenisation process which will replace the transmission of the credit card numbers, even then it's as much to do with protecting terminals and for mobile payments, as protecting the contactless cards themselves from being "skimmed". (As per our contactless smart card article, the use did make contactless smart cards which were basically similar to magnetic stripe devices, but most of the world, and what I'm referring to are cards which are EMV.)

Your later reply above is partially correct. I think one thing which definitely isn't transmitted is the CVV. Or rather a CVV is transmitted, but it's unique to the transaction. And as my post mentioned, possibly the name isn't in some cards either. The CVV and general communication is the other important security as you sort of mentioned too. The problem with skimming on magnetic cards is that you not only can get all the data, you can easily generate a card (clone) which to someone else's system seems the same as a genuine card. So if you're using signatures like in the US people can use this to make rogue transactions in stores etc.

This has two problems from tbe banks POV, one is that antifraud measures are generally reduced for such transactions since they are considered lower risk, the other is that in many cases the bank is the one responsible for fraud in such card present transactions. (If you use PIN, you at least have to work out how to steal the PIN as well, which as shown in places like NZ, is possible but does take more work.)

I'm not disagreeing skimming all the details including CVV is a concern. E.g it can be used for card not present transactions. Hence why stuff such as the Target breach gets attention and why PCI-DSS etc are generally mandatory. But I think also something difficult to protect against, particularly since the details are generally published on the card. (Numberic billing address details are not, but my understand is they're not checked in many places besides the US.) With modern high speed, high definition cameras and good image recognition technologies, it's probably getting to the stage where a strategicly placed camera could capture both sides and the software could identify where the cards are exposed. (Perhaps OCRing them would be a step too far, but if you can reduce the amount of video that has to be scanned, you could easily hire someone in a developing country to steal the numbers for you.) Notably while most would advise you to protect PIN entry, it's rare that you're told to hide the card details, and for some people it may be difficult anyway. Of course even then this is still relatively expensive per card compared to stealing 40 million (or whatever) by breaching a Target's payment systems, but I'm not sure whether hiring someone to go around with a RFIC reader to steal details is much cheaper. (Contactless and modern contact systems often mean the card doesn't have to leave you hand, which is an advantage since you may hide the details even unintenionally and at least the staff can't just put your card in front of a camera.) Stuff like 3-D Secure and other antifraud measures by the bank, and stores (including using third party payment gateways like PayPal with their own security measures) who will take most of the cost for the fraud are likely to be an important part of stopping fraud for for card-not-present for quite a while I suspect.

Nil Einne (talk) 14:29, 26 February 2015 (UTC)[reply]

Have you read Contactless smart card, contactless payment and chip and pin? There are sections about security, as well as privacy concerns. You can make an aluminum foil wallet if you want to feel extra protected, or check out some other info at RFID skimming. SemanticMantis (talk) 17:20, 25 February 2015 (UTC)[reply]

Can one get his entire/whole vas deferens removed?

As in, is it practically possible (for a doctor) to do this?

After all, vasectomies can fail, and thus, I am curious about whether or not it is practically possible to do this. Futurist110 (talk) 04:33, 25 February 2015 (UTC)[reply]

No matter how much of it they removed, they would still have to cauterize at each end, and if the cauterization somehow fails at both ends, theoretically some of the wiggly little guys might be able to get through. ←Baseball Bugs ^{What's up, Doc?} carrots→ 14:17, 25 February 2015 (UTC)[reply]

Yes, which, in turn, is why I asked about removing the whole vas deferens, rather than about only removing a part of it. Futurist110 (talk) 21:38, 25 February 2015 (UTC)[reply]

And I'm saying no matter how much of it you remove, you still have seal it, and that's where the risk for error comes in. And how would you define precisely where it starts and stops (or vice versa)? Are you talking about removing it all the way into the abdomen? I suppose that would do it. But the work could run into money. ←Baseball Bugs ^{What's up, Doc?} carrots→ 21:41, 25 February 2015 (UTC)[reply]

If removing it all the way up to the abdomen (I am sorry, but I don't know where exactly the vas deferens starts and stops) is what it takes to do this, then Yes, I am talking about doing that. Futurist110 (talk) 21:54, 25 February 2015 (UTC)[reply]

I'm at pretty much the limit of my anecdotal info. Aside from reading the vas deferens article, you should probably contact your urologist and pose this question to him, to get a thorough answer. ←Baseball Bugs ^{What's up, Doc?} carrots→ 22:53, 25 February 2015 (UTC)[reply]

OK; also, though, out of curiosity--how much do you think that surgically removing the *entire* vas deferens will cost? Any thoughts on this, Baseball_Bugs? Futurist110 (talk) 07:51, 13 November 2015 (UTC)[reply]

Read first, do no harm. A physician is not going to do an unnecessary and damaging procedure for no reason, unless he's Doctor Robert. Given cutting the duct and cauterizing it works, there's no reason to remove the entire duct, and possibly do seven damage as a result. See also, Michael Jackson for the result of such unnecessary surgeries. μηδείς (talk) 22:54, 26 February 2015 (UTC)[reply]

The "first, do no harm" principle might have already been thrown out of the window by some/many doctors who are willing to turn women into Barbie dolls, et cetera. After all, if one follows the "first, do no harm" principle, then one shouldn't perform any elective cosmetic surgeries which carry any risk at all, correct? Also, I would argue that this surgery is necessary due to the fact that regular vasectomies, including with cauterization, can fail, and if a vasectomy fails, then this can certainly result in extremely unpleasant consequences for the male who gets it done. Plus, this is not to mention the possibility of individuals trying to perform surgery themselves and/or trying to get surgery performed in "back-alleys" if they are unable to get these surgeries performed in a safe, medical setting. Futurist110 (talk) 23:52, 26 February 2015 (UTC)[reply]

Which reminds me that there's a vas deferens between men and women. -- Jack of Oz ^{[pleasantries]} 23:02, 26 February 2015 (UTC) [reply]

Yes, well you've gotten my point about barbie dolls which I illustrated using Michael Jackson. But you are talking about removing the entirety of two foot-long ductsthat are intimately involved with the testicles, bladder, and prostate. I'll leave it up to you to contact a reputable surgeon and ask his advice on removing the entire duct, and the risks involved. μηδείς (talk) 00:17, 27 February 2015 (UTC)[reply]

If you are worried about a vasectomy failing, then also use a condom and spermicide. Insist that your partner takes oral contraceptives plus uses a diaphragm. Have intercourse only in the days immediately following her menstrual period. And practice coitus interruptus. That should do the trick. It is very rare that a guy who wears a belt and suspenders experiences the embarassment of his pants falling down. Cullen³²⁸ Let's discuss it 07:06, 1 March 2015 (UTC)[reply]

In my honest opinion, though, pulling out defeats the purpose of pulling in in the first place. :( Plus, even 2+ forms of birth control can simultaneously fail. :( Futurist110 (talk) 21:33, 13 November 2015 (UTC)[reply]

Are there quantum effects having unexpected "memory"?

For instance: can a piece of information, which apparently got lost because of strong entropy, be recovered / restored - in quantum systems?

In macro-systems, I can hypothetically think about a "magician" who takes a random card out of a deck of cards, then puts it back in the deck, shuffles, and then takes out a random card which turns out to be the first card. Are there similar phenonema in quantum systems? HOOTmag (talk) 10:55, 25 February 2015 (UTC)[reply]

Not a quantum system, but in a macro-system, a better example than a magician is two cylinders with a thin layer of glycerin between them. Place a drop on ink in the glycerin, rotate one of the cylinders, and the ink disappears. Rotate the cylinders back into place and the ink drop reappears. This happens because the ink didn't evenly distribute itself in the glycerin, but rather was spread out over multiple layers of glycerin, due to laminar flow. StuRat (talk) 15:40, 25 February 2015 (UTC)[reply]

Quantum entanglement may be relevant. StuRat (talk) 15:44, 25 February 2015 (UTC)[reply]

My main purpose, is to inquire whether any quantum effect can overcome the increase in entropy, so I suspect your ink example is irrelevant - because it involves no real increase of entropy. As for quantun entanglement: yes, I'd thought about that, but I couldn't think about hypothetical experiments involving quantum entanglement that are relevant to "preserving memory" (by overcoming the increase of entropy). HOOTmag (talk) 16:51, 25 February 2015 (UTC)[reply]

either you should be more specific about what you mean when you say "piece of information " or I would say any actual memory that a person has in their brain would qualify. At the very least any recorded information that is stored on memory devices would qualify. Microprocessors are very small. And probably employ some quantum effects to record memory. 66.87.82.51 (talk) 17:50, 25 February 2015 (UTC)[reply]

Quantum mechanics, in the basic sense that matter and energy levels are discrete, is an essential foundation to statistical mechanics and the modern understanding of entropy. In that regard, quantum mechanics is at the heart of entropy and what it actually means for a system to become more or less ordered over time. In the spirit of "If you think you understand quantum mechanics, you don't understand quantum mechanics" (widely attributed to Richard Feynman), I am not sure if there are examples of quantum mechanical systems that would satisfy you. However, I will note that even in the classical formulation, entropy always increases is only a statistical maxim and not an absolute law. Firstly, one can always locally decrease entropy through an expenditure of energy and a willingness to increase entropy elsewhere. Most complex living systems rely on the energy they bring in to maintain their complexity against entropy. Secondly, even in the absence of external energy, spontaneous declines in entropy do and will occur; however, the larger the decrease the more improbable the event. Atoms allowed to wander randomly will sometimes, by mere chance, return to their point of origin. The larger the number of atoms involved the less likely they all return at once, but there is nothing in principle that prevents it. In this sense, the increase of entropy is really a statement that the most probable spontaneous outcome is one that tends to maximize entropy. There is nothing about quantum mechanics that occurs to me that would tend to disagree with that or necessarily lead to magic tricks where one can intentionally reduce global entropy. Dragons flight (talk) 19:30, 25 February 2015 (UTC)[reply]

I'm not sure, but I think Ergodic theory and Poincaré_recurrence_theorem are relevant. If you have a box that is half full of air and half full of vacuum at time zero, eventually all the gas will happen to end up where it started, leaving vacuum in the other half (at least in terms of the mathematical model). SemanticMantis (talk) 22:14, 25 February 2015 (UTC)[reply]

I'm thinking of quantum entanglement and quantum teleportation. (These effects are less impressive than they sound, but basically they amount to information seeming to disappear yet being interrogable somewhere else) Wnt (talk) 19:34, 26 February 2015 (UTC)[reply]

@HOOTmag: the more I think about it, the more I think your question is precisely addressed by the Poincaré_recurrence_theorem#Quantum_mechanical_version - information can be lost due to entropy and other fluctuations, but still sort of spontaneously re-emerge. SemanticMantis (talk) 19:35, 26 February 2015 (UTC)[reply]

question (optics)

is F=R/2 is applicable in lenses — Preceding unsigned comment added by 117.198.29.91 (talk) 14:25, 25 February 2015 (UTC)[reply]

I added to the title to make it actually useful. StuRat (talk) 14:46, 25 February 2015 (UTC) [reply]

Assuming he's referring to the focal length of a spherically curved mirror, Wikipedia's article titled Focal length contains all the information one needs to calculate the focal length of various lenses and mirrors. --Jayron 32 15:31, 25 February 2015 (UTC)[reply]

(edit conflict) Looks like a variation on the Lensmaker's equation, for certain combinations of lense geometry (only one "R", so both faces have same curvature or else one face is planar) and certain types of materials (some unstated "n" optical density). DMacks (talk) 15:34, 25 February 2015 (UTC)[reply]

animal

How many classes there are in Kingdom Animalia?--217.200.200.222 (talk) 19:29, 25 February 2015 (UTC)[reply]

List of animal classes says roughly 108 classes in 34 phyla, though with the caveat that different authorities give slightly different classification schemes and hence the exact number of classes is somewhat ambiguous. Dragons flight (talk) 19:33, 25 February 2015 (UTC)[reply]

Physics: Gravitational Fields

Calculate the total energy of a communication satellite of mass 100kg in a stable circular orbit with a time period of 25 hr.

i have managed all the questions in the book but the question above is giving me problem, i keep getting the wrong answer and there are no solutions in the book.

I know that total energy = 1/2mv^2 + (-GMm)/r^2 Period(T)= (24 * 60^2) i have used this to find the radius = 1.03

if i could get some guidance or a clue on how to approach this it will help a lot. -Thanks86.183.65.110 (talk) 23:17, 25 February 2015 (UTC)[reply]

A few suggestions:

Keep better track of units. You found a radius of 1.03 whats? Is that a number that makes sense?
You have a 25-hour period in the problem statement, and a 24-hour period in your calculation. One of them must be incorrect.
You appear to be mixing up the force of gravitational attraction (with r^2 in the denominator) with gravitational potential energy (with r^1 in the denominator).

If you check the units in your calculation, you will see that you're adding together two things with different dimensions -- one is an energy and one is a force. Make a habit of carrying units through the entire calculation so that you can easily check your work and spot this type of mistake. --Amble (talk) 23:26, 25 February 2015 (UTC)[reply]

What Amble said above is very good advice. You are right that the first step is finding the radius of the orbit. If you haven't already, maybe it's a good idea checking out Kepler's laws of planetary motion for this. - Lindert (talk) 23:43, 25 February 2015 (UTC)[reply]

If you use Google Calculator, WolframAlpha, or the units program, they will do the dimensional analysis for you, and show if you've used the wrong power somewhere. LongHairedFop (talk) 13:39, 26 February 2015 (UTC)[reply]