explain xkcd - User contributions [en]

2034: Equations

2018-08-17T10:08:56Z

172.68.132.47: typos: adding "space" to alt text, fixing a rho in an equation.

{{comic
| number = 2034
| date = August 17, 2018
| title = Equations
| image = equations.png
| titletext = All electromagnetic equations: The same as all fluid dynamics equations, but with the 8 and 23 replaced with the permittivity and permeability of free space, respectively.
}}

==Explanation==
{{incomplete|Created by an EQUATION - Please change this comment when editing this page. Do NOT delete this tag too soon.}}
This comic gives a set of equations supposedly from different areas of mathematics and physics. To anyone not familiar with the field in question they look pretty similar to what you might find in research papers or on the relevent Wikipedia pages. To someone who knows even a little about the topic, they are clearly very wrong and only seem even worse the more you look at them.
{| class="wikitable"
!style="width:20%"|Equation
!style="width:20%"|Field
!style="width:60%"|Explanation
|-
|<math>E = K_0t + \frac{1}{2}\rho vt^2</math>
|All kinematics equations
|This equation literally states: "Energy equals a constant <math>K_0</math> multiplied by time plus half of density multiplied by speed multiplied by time squared". The first term here is hard to interpret - it could be correct if <math>K_0</math> is a constant power applied to the system, this symbol would more normally be used to denote an initial energy and so multiplying by <math>t</math> would be wrong. The second term looks similar to the traditional kinetic energy formula <math>\frac{1}{2}mv^2</math> but with a density instead of the mass. This is then wrong without some accompanying volume term (on either side of the equation).
|-
|<math>K_n = \sum_{i=0}^{\infty}\sum_{\pi=0}^{\infty}(n-\pi)(i-e^{\pi-\infty})</math>
|All number theory equations
|Taken literally the equation says: "The nth K-number is equal to for all i in 0 to infinity, for all pi in 0 to infinity; subtract pi from n and multiply it with i minus e (to the power of pi minus infinity)". A twofold misconception can be seen here. The first is the reassignment of pi as a variable instead of the constant (3.14). This might be a jab at how in number theory letters and numbers are used interchangeably, but where some letters are all of a sudden fixed constants. The second misconception is the use of infinity in the latter part of the formula. Naively this would signify that (with the reassigned pi values) the part in the power would range from minus infinity to zero. However infinity is not a number and cannot be used as one without using a limit construct.
|-
|<math>\frac{\partial}{\partial t}\nabla\cdot \rho = \frac{8}{23}

\int\!\!\!\!\!\!\!\!\!\;\;\bigcirc\!\!\!\!\!\!\!\!\!\;\;\int
\rho\,ds\,dt\cdot \rho\frac{\partial}{\partial\nabla}
</math>
|All fluid dynamic equations
|-
|<math>|\psi_{x,y}\rangle = A(\psi) A(|x\rangle \otimes |y\rangle)</math>
|All quantum mechanic equations
|
|-
|<math>\mathrm{CH}_4 + \mathrm{OH} + \mathrm{HEAT} \rightarrow \mathrm{H}_2\mathrm{O} + \mathrm{CH}_2 + \mathrm{H}_2 \mathrm{EAT}</math>
|All chemistry equations
| A modification of the combustion of methane. The correct form is often taught and a good example problem but obviously there are more chemistry problems.<math>\mathrm{HEAT}</math> is normally shorthand for {{w|activation energy}}, but in Randall's version it's jokingly used as a chemical ingredient and becomes <math>\mathrm{H}_2\mathrm{EAT}</math>, taking the hydrogen atom freed by the combustion equation shown. To deliver the punchline while maintaining proper stoichiometry, <math>\mathrm{OH}</math> (which should be <math>\mathrm{OH}^-</math>, since the oxygen keeps a free electron when it combines with a single hydrogen) is shown instead of <math>\mathrm{O}_2</math>. The proper methane combustion equation would be: <math>\mathrm{CH}_4 + 2 \mathrm{O}_2 \rightarrow 2 \mathrm{H}_2\mathrm{O} + \mathrm{CO}_2</math>
|-
|<math>\mathrm{SU}(2)\mathrm{U}(1) \times \mathrm{SU}(\mathrm{U}(2))</math>
|All quantum gravity equations
|This is more similar to experessions which appear in {{w|Grand_Unified_Theory|Grand Unified Theory}} (GUT) than general quantum gravity. Unlike some of the other equations, this one has no interpretation which could make it mathematically correct. This is similar to the notations used to describe the symmetry group of a particular phenomena in terms of mathematical {{w|Lie_Group|Lie Groups}}. A real example would be the Standard Model of particle physics which has symmetry according to <math>\rm{SU(3)\times SU(2) \times U(1)}</math>. Here, <math>\rm{SU}</math> and <math>\rm{U}</math> denote the special unitary and unitary groups respectively with the numbers indicating the dimension of the group. Loosely, the three terms correspond to the symmetries of the strong force, weak force and electromagnetism although the exact correspondence is muddied by symmetry breaking and the Higgs mechanism.

Of course, an expression missing an "=" sign, is difficult to interpret as an "equation", because equations normally express an "equality" of some kind. Nobody knows whether Randal refers to a horse here (equidae)

Randall's version clearly involves some similar groups although without the <math>\times</math> symbol it is hard to work out what might be happening. A term like <math>\rm{SU(U(2))}</math> has no current interpretation in mathematics, if anyone thinks otherwise and possibly has a solution to the quantum gravity problem they should probably get in touch with someone about that.
|-
|<math>S_g = \frac{-1}{2\bar{\varepsilon}}i\eth \hat{\big(} \zeta_0 \dotplus p_\epsilon \rho_v^{abc}\cdot \eta_0 \hat{\big)} f_a^0 a\lambda(\xi) \psi(0_a)</math> 
|All gauge theory equations
||This equation looks broadly similar to the sorts of things which appear in gauge theory such as the equations which define {{w|Yang–Mills_theory#Quantization|Yang-Mills Theory}}. By the time physics has got this far in, people have normally run out of regular symbols making a lot of the equations look very daunting. The actual equations in this field rarely go far beyond the greek alphabet though and no-one has yet to try putting hats on brackets. The appearence of many sub- and superscripts is normal (this links to the group theory origins of these equations) and for the layperson it can be impossible to determine which additions are labels on the symbols and which are indices for an {{w|Einstein_notation|Einstein Sum}}.

The left-hand side <math>S_g</math> is the symbol for some {{w|Action_(physics)|action}}, in Yang-Mills theory this is actually used for a so-called "ghost action". On the right-hand side we have a large number of terms, most of which are hard to interpret without knowing Randall's thought processes (this is why real research papers should all label their equations thoroughly). The <math>\frac{1}{2\bar{\varepsilon}}</math> looks like a constant of proportionality which often appears in gauge theories. The factor of <math>i = \sqrt{-1}</math> is not unusual as many of these equations use complex numbers. The <math>\eth</math> symbol looks similar to a <math>\partial</math> partial derivative symbol especially as the {{w|Dirac_equation#Covariant_form_and_relativistic_invariance|Dirac Equation}} uses a slashed version as a convenient shorthand.

The rest of the equation cannot be mathematically correct as the choice of indices used does not match that on the left-hand side (which has none). In particle physics subscripts (or superscripts) of greek letters (usually <math>\mu</math> or <math>\nu</math>) indicate terms which transform nicely under Lorentz transformations (special relativity). Roman indices from the beginning of the alphabet relate to various gauge transformation propetries, the triple index seen on <math>p^{abc}_v</math> would likely come from some <math>\rm{SU(3)}</math> transformation (related to the strong nuclear force). Since <math>S_g</math> has none of these (and is thus a scalar which remains constant under these operations), we would need the right-hand side to behave in the same way. Most of the indices which appear are unpaired and so will not result in a scalar making the equation very wrong. For those not familiar with this type of equation, it is a similar mistake messing up units and setting a distance equal to a mass.
|-
|<math>H(t) + \Omega + G \cdot \Lambda \, \dots \begin{cases} \dots > 0 & \text{(HUBBLE MODEL)} \\ \dots = 0 & \text{(FLAT SPHERE MODEL)} \\ \dots < 0 & \text{(BRIGHT DARK MATTER MODEL)} \end{cases}
</math>
|All cosmology equations
|This is a parody of equations defining the {{w|Hubble's_law#Derivation_of_the_Hubble_parameter|Hubble Parameter}} <math>H(t)</math> although it looks like Randall has become bored and not bothered to finish his equation. Such equations usually have several <math>\Omega</math> terms representing the contributions of different substances to the energy-density of the Universe (matter, radiation, dark energy etc.). In this context <math>G</math> could be Newton's constant and <math>\Lambda</math> is something dark energy related although seeing them appear multiplied and on the same footing as <math>H</math> is unusual (the dot is entirely unnecessary). Choosing to make <math>H</math> a function of time <math>t</math> and not of redshit <math>z</math> is also unusual.

The second section looks like the inequalities used to show how what shape the Universe, based on the value of the curvature parameter <math>\Omega_k</math>. A value of 0 indicates a flat Universe (this more or less what we observe) whilst a positive /negative value indicate a open /closed curved Universe. Randall's choice of labels further makes fun of the field as both a flat sphere and bright dark matter are oxymoronic terms which would involve some rather strange model universes.
|-
|<math>\hat H - u_{0} = 0</math>
|All truly deep physics equations
|<math>\hat H</math> is the hamiltonian operator, which when applied to a system returns the total energy. In this context U would usually be the potential energy. However there is also a subscript 0 and a diacritic making indicating some other variable. Much of physics is based on Lagrangian and Hamiltonian mechanics. The Lagrangian is defined as <math>\hat L = \hat K - \hat U </math> with K being the kinetic energy and U the potential. Hamiltonian mechanics uses the equation <math>\hat H = \hat K + \hat U </math>. The Hamiltonian must be conserved so taking the time derivative and setting it equal to zero is a powerful tool. The principle of least action says allows most modern physics to be derived by setting the time derivative of the Lagrangian to zero.
|-

|<math>\frac{\partial}{\partial t}\nabla\cdot p = \frac{\epsilon_0}{\mu_0}
\int\!\!\!\!\!\!\!\!\!\;\;\bigcirc\!\!\!\!\!\!\!\!\!\;\;\int
\rho\,ds\,dt\cdot \rho\frac{\partial}{\partial\nabla}
</math>
|All electromagnetic equations
|This equation has superficial resemblance to portions of [//en.wikipedia.org/wiki/Maxwell%27s_equations Maxwell's Equations], but just miscellaneous bits, some from the integral forms and some from the differential forms.
|}

==Transcript==
{{incomplete transcript|Do NOT delete this tag too soon.}}
[TODO: Avoid using math markup here because the images of these equations isn't helpful in a transcript. Sigh.] 
[Nine equations are listed and labeled as followed:]
 
E = K0t + 1/2 pvt2 
ALL KINEMATICS EQUATIONS 
 
<math>K_n = \sum_{i=0}^{\infty}\sum_{\pi=0}^{\infty}(n-\pi)(i-e^{\pi-\infty})</math> 
ALL NUMBER THEORY EQUATIONS 
 
∂/∂t ∇ ⋅ p = 8/23 (∯ ρ ds dt ⋅ ρ ∂/∂∇) 
ALL FLUID DYNAMIC EQUATIONS 
 
|ψx,y〉 = A(ψ) A(|x〉⊗ |y〉) 
ALL QUANTUM MECHANIC EQUATIONS 
 
CH4 + OH + HEAT → H2O + CH2 + H2EAT 
ALL CHEMISTRY EQUATIONS 
 
SU(2)U(1) × SU(U(2)) 
ALL QUANTUM GRAVITY EQUATIONS 
 
<math>S_g = \frac{-1}{2\bar{\varepsilon}}i\eth \hat{\big(} \zeta_0 \dotplus p_\epsilon \rho_v^{abc}\cdot \eta_0 \hat{\big)} f_a^0 a\lambda(\zeta) \psi(0_a)</math> 
ALL GAUGE THEORY EQUATIONS 
 
H(t) + Ω + G⋅λ ... > 0 (HUBBLE MODEL) ... = 0 (FLAT SPHERE MODEL) ... < 0 (BRIGHT DARK MATTER MODEL) 
ALL COSMOLOGY EQUATIONS 
 
Ĥ - u̧0 = 0 
ALL TRULY DEEP PHYSICS EQUATIONS
 

{{comic discussion}}
[[Category:Math]]

Talk:2030: Voting Software

2018-08-09T07:03:45Z

172.68.132.47:

I think this comic is referencing [https://twitter.com/GossiTheDog/status/1026603800365330432 this twitter thread] and the controversy behind it.
[[Special:Contributions/172.69.190.4|172.69.190.4]] 17:59, 8 August 2018 (UTC)
: The [https://arstechnica.com/tech-policy/2018/08/experts-criticize-west-virginias-plan-for-smartphone-voting/ Experts criticize West Virginia’s plan for smartphone voting] article on ArsTechnica has more information (as much as possible when the company in question does not provide any details (note that it is about overseas voting). --[[User:JakubNarebski|JakubNarebski]] ([[User talk:JakubNarebski|talk]]) 19:44, 8 August 2018 (UTC)

Is he saying it's weird that we're so sophisticated in other areas of computer science but so far behind in voting technology, or is he making fun of the idea that electronic voting is somehow inherently unsafe?--[[Special:Contributions/108.162.216.106|108.162.216.106]] 18:10, 8 August 2018 (UTC)
: No i think he is saying computer science is a mess and we should not trust it with voting(he is not making fun of the idea of it being unsafe, he is pressing on the point of it being unsafe[saying that all experts agree on that])18:18, 8 August 2018 (UTC)

: I think he's commenting on how in most fields, the experts are very sure that they do their job well, and all the angles have been tried and tested, but in computer science the experts are more certain than anyone that there is ''absolutely no way'' for a person to actually build a complex software system with no flaws or vulnerabilities, even if they controlled every aspect of the system. in practice of course they control very little of the system and understand even less of it. [[Special:Contributions/172.68.34.88|172.68.34.88]] 18:22, 8 August 2018 (UTC)

: He's saying that software development is a terribly buggy process, most likely because the majority of software out there can have bugs without very dire real-world consequences (unlike aircraft or elevators).

::Not to mention the fact that there are incredibly smart people with great interest in undoing the work that software developers do, whereas that isn't at all the case with airplanes or elevators. [[Special:Contributions/108.162.219.214|108.162.219.214]] 18:29, 8 August 2018 (UTC)

:: Plus there's the general issue that the public as a whole takes the view that "Computers are majykal" (misspelling deliberate) and therefore somehow automatically safe & infallible, despite experts trying very hard to disillusion people about...pretty much all of that. Compare that to the common assumptions about aircraft and elevators--people need the safety verified, instead of assuming it like they do with computers. [[User:Werhdnt|Werhdnt]] ([[User talk:Werhdnt|talk]]) 19:08, 8 August 2018 (UTC)

:::There's a logical fallacy here. To compare airplaneS and elevatorS to a voting system program is comparing plural to singular. There would be significant opportunity to break/modify a single instance of those objects, although without the relative anonymity of electronic access involved. Once a computer system is infiltrated, the break-in can be replicated to all instances of that program relatively instantaneously, assuming communication pathways are available.[[Special:Contributions/162.158.75.130|162.158.75.130]] 19:12, 8 August 2018 (UTC)

::::No logical fallacy; there have been ''multiple'' attempts to get people to accept a voting system program, and the 'done by a computer=infallible' problem is '''''not''''' unique to voting programs. Mr. Babbage was being confused by people who were thinking it was possible to get the correct answers from a computer despite putting the wrong data in back in the 1860s (at least!), and the computer at the time was not much more than a fancy calculator. [[User:Werhdnt|Werhdnt]] ([[User talk:Werhdnt|talk]]) 20:23, 8 August 2018 (UTC)

A blockchain node doesn't technically need to be connected to the internet in order to function. It needs to have some method for receiving messages from other nodes on the blockchain network, and most blockchain nodes do indeed get these messages via the internet, but some bitcoin nodes (for example) get updates about new blocks and new transactions from the Blockstream satellite. An internet connection is therefore not intrinsically necessary for a blockchain to work, it's just the most convenient way to do it.

Do you think that this comic had anything to do with the debacle in Johnson County, KS last night? [[Special:Contributions/162.158.62.231|162.158.62.231]] 19:30, 8 August 2018 (UTC)

The comic ignores the fact that modern airplanes are heavily utilizing software of all kinds. A software failure in an aircraft could easily be fatal (and have been so various times in history already, while the consequences of a voting software working incorrect are ''relatively'' harmless), and still airplanes remain safe, as the comic recognizes. --[[User:YMS|YMS]] ([[User talk:YMS|talk]]) 21:05, 8 August 2018 (UTC)

:Airplanes are not connected to internet and reasonably well protected from people putting their USB devices in their control system. Also, they are NOT build by lowest bid contractor. There ARE people now capable of building offline voting machine which would be reasonable secure. They are working in banks and stock exchanges and at those companies providing switches for internet backbone, are extremely well paid and wouldn't ever promise they will get the machine finished in single year. Noone asks THEM to make the voting machines. Voting over internet? With consumer-grade devices? Impossible. (I'm also working in IT, although not on mentioned high-security systems.) -- [[User:Hkmaly|Hkmaly]] ([[User talk:Hkmaly|talk]]) 22:24, 8 August 2018 (UTC)

::Note that she is talking to aircraft designers, not to software engineers working on fly-by-wire systems (back when I took software engineering you got an answer similar to the one about voting machines when discussing fly-by-wire). I took this more as the aircraft designers glossing over the problems caused by software engineering. A voting system which uses paper ballots, with perhaps computer systems used for some stages of counting would be a reasonable analogy to the redundant systems used in aircraft. [[Special:Contributions/162.158.106.228|162.158.106.228]] 23:08, 8 August 2018 (UTC)

Seems to me that the last panel references the E.T for Atari Desert Burial (https://en.wikipedia.org/wiki/Atari_video_game_burial), perhaps to draw some analogy as to the potential quality or likelihood of success of a Block-chain solution as compared to the ill-fated video game. Anyone think that's worth explaining? [[User:Da_NKP|Da_NKP]] 10:15, 8 August 2018 (UTC) Da_NKP

What motive is there to "mine DemocracyCoin"? Who evaluates this blockchain? [[Special:Contributions/162.158.150.100|162.158.150.100]] 22:27, 8 August 2018 (UTC)
:That's simple, ideally it would be a private blockchain, and the evaluators would just be every voting computer in existence (They'd all be active for a similar fairly short time period). Presumably the evaluations would be ongoing during the voting process, then could be stopped once voting was complete. The last few votes of the night may not wind up being evaluated. [[Special:Contributions/162.158.74.225|162.158.74.225]]

Wouldn't it be possible to run said blockchain on one's personal computer, instead of running on a voting machine? and you could compile open source software yourself to perform the voting. That sounds like a solid enough way to keep security fine to me, but if I'm missing something, please tell me. [[User:Gjgfuj|TheSandromatic]] ([[User talk:Gjgfuj|talk]]) 03:25, 9 August 2018 (UTC)
:The bigger challenge in a voting system isn't making sure someone doesn't modify the record, it's making sure that each person only votes once and only for themselves -- think about past internet voting campaigns: Justin Bieber wasn't sent to North Korea by *changed* votes, but rather by flooding the system with *bogus* votes. [[Special:Contributions/172.68.132.47|172.68.132.47]] 06:26, 9 August 2018 (UTC)
::To be a bit clearer, bitcoin (for example), doesn't and can't enforce that wallets correspond one-to-one with people -- multiple people can share a wallet (if they all know the private key), and one person can have multiple wallets. If you want to guarantee one-to-one correspondence, you have to validate identities and issue unique, signed keys at some prior point. Leaving aside whether or not it's possible to do this part securely and without error (and how big of a target the root signing key would be), you then have millions of people doing their own key management, just like you do with bitcoin. When bitcoin wallets are stolen en masse by key compromises (which does happen), only the wallet owners (who were ostensibly using poor security practices which allowed the compromise) suffer, so the harm is limited. If voting system keys were stolen en masse, but the votes still counted, society as a whole would likely suffer. [[Special:Contributions/172.68.132.47|172.68.132.47]] 07:03, 9 August 2018 (UTC)

Talk:2030: Voting Software

2018-08-09T06:26:38Z

172.68.132.47:

803: Airfoil

2018-08-07T18:04:39Z

172.68.132.47: /* Explanation */

{{comic
| number = 803
| date = October 8, 2010
| title = Airfoil
| image = airfoil.png
| titletext = This is a fun explanation to prepare your kids for; it's common and totally wrong. Good lines include "why does the air have to travel on both sides at the same time?" and "I saw the Wright brothers plane and those wings were curved the same on the top and bottom!"
}}

==Explanation==
In the first panel a cross sectional drawing of a plane wing with the air moving around the wing showing a common teaching that an {{w|airfoil}} works because the air on top of the wing must travel faster to "keep up" with the air flowing across the bottom of the wing. The theory goes that, because the air on top of the wing is traveling faster, it must, as a result of {{w|Bernoulli's Principle}}, create an area of lower pressure above the wing; this causes {{w|lift (force)|lift}} (that is, the wing rises) because the higher pressure below the wing (symbolized by thick "up" arrow) pushes it up more than the low pressure above the wing. This is what the teacher [[Miss Lenhart]] is teaching as is revealed in the next panel.

As it turns out, this is, to put it mildly, a vast oversimplification of how lift is truly created. Because then a student asks a particularly insightful question: Why, if the theory is true, can planes fly upside down? (If the simple airfoil theory is all that permits planes to stay up in the air, then flying upside down should reverse the pressures — pushing the plane down and causing it to crash.) Miss Lenhart thinks about it and clearly has no answer.

The final set of panels posit three potential responses from Miss Lenhart, upon realizing her theory has been disproved:

In the '''right''' one, Miss Lenhart realizes that perhaps the model she's been using to explain how an airfoil works is wrong (or, at a minimum, too simple). She is curious about it and suggests that this is an area for further exploration, and encourages additional study — in effect, rewarding the student for their insight. It seems that Miss Lenhart has taken the right course as it is shown later in [[843: Misconceptions]] that she wished her students to generally avoid any {{w|List of common misconceptions|common misconceptions}}. The title text also mentions that this is a common misconception and it is actually the first mentioned on {{w|List_of_common_misconceptions#Physics|list of common physics misconceptions}} on Wikipedia.

In the '''wrong''' panel, Miss Lenhart, out of apparent embarrassment, avoids the question entirely, saying simply that it's complicated (and implying that such questions are outside the student's understanding). This way to continue a discussion where you wish to be right was much later used in [[1731: Wrong]].

In the '''very wrong''' panel, not only does Miss Lenhart avoid answering the question, she attempts to distract them (or even punish them for asking such an insightful question - note that in this panel, Miss Lenhart has clenched her fists, suggesting anger) by telling the kids that {{w|Santa Claus}} isn't real but in fact that he is really their parents — something that would obviously distress children if they still believe in Santa Claus (in addition to distracting them from the question they've asked) and constitute harsh punishment for pointing out the teacher's ignorance. Of course most children old enough to be taught about the airflow around plane wings should be too old to believe in Santa. However, if she just wished to tell them a bit about planes she may have drawn this drawing even in very early grades making the Santa trick effective.

The title text suggests additional reasons for re-thinking the common theory as to how airfoils create lift. It points out that (1) it is absurd to believe the air has to get across the airfoil's two sides in the same amount of time, and (2) the {{w|Wright brothers}} plane's wings were curved the same amount on both sides of the airfoil (which is not actually true; the Wright Flyer's wings were concave, like an arch), meaning that the distance that the air needs to travel to get across the wing is not the dispositive factor in creating lift.

The strip is correct in noting that lift is a far more complicated process than the simple theory posited by Miss Lenhart. While the role of Bernoulli's Principle (that is, the difference in pressures) cannot be entirely discounted, the theory here is vastly too simple. As an initial matter, as suggested by the title text, there is no reason that the air on top of the wing should be compelled to "keep up" with the air on the bottom of the wing. Indeed, as demonstrated by the illustration below, in the time that the air below the wing travels across, the air on top of the wing has not only traveled the length of the entire top of the wing (a distance that may be farther than the distance under the wing, due to its shape), but often additional distance.

[[File:Karman trefftz.gif]]

Lift may be more usefully described as resulting from the deflection of air, although this explanation still does not explain how symmetrical wings will work (at least, absent effects caused by a change in the "angle of attack") nor how a plane may fly upside down. The Wikipedia article on {{w|lift (force)|lift}} provides a more detailed explanation. It in fact gives an explanation as to these two issues. It explains that with zero angle of attack, a symmetrical wing will not generate lift (though it is possible that other factors may generate other slight upward force, such as updrafts, the shape of the plane, and the angle of the engine relative to the wings. It also explains that an asymmetrical (or "cambered") wing may adjust angle of attack to compensate and still generate lift.

Finally, to answer the question in the second panel in a general sense: most planes ''can't'' fly upside down for an extended period of time. While many aerobatic aircraft can sustain inverted flight with negative g forces, some others can achieve an inverted attitude only momentarily, and are experiencing positive g forces. Usually the reason for this is not the wings, which function perfectly fine upside down (albeit sometimes at lower efficiency), but the engines, which may not get fuel or oil under such conditions. It has to also be noted that if angle of attack were ignored, movable control surfaces would be useless. Almost any airplane can do a {{w|barrel roll}} or {{w|Aileron roll}}, given sufficient altitude (a {{w|Boeing 707#Model 367-80 origins|Boeing 707 prototype}} once did this, and so did the Concorde in a demonstration).

==Transcript==
:[In a frame-less picture to the left of the first panel there's a picture of a cross section of an airfoil (a plane wing), with a small black arrow pointing down on the wing from above and similar but larger arrow pointing up on the wing from below. Two lines beginning close to each other at the right respectively moves over and under the wing ending in arrow heads to the left. Just before and after the wing four small lines crossing the long arrows indicate approximately where the path of the lines stop being parallel. Above the drawing there is a caption. Below, in a speech bubble with an arrow pointing towards the next panel to the right, is the text that the teacher Miss Lenhart has just used to describe the drawing.]
:Handling a student who challenges your expertise with an insightful question:
:Miss Lenhart: So, kids, the air above the wing travels a longer distance, so it has to go faster to keep up. Faster air exerts less pressure, so the wing is lifted upward.

:[Miss Lenhart is shown standing while a student asks a question from off-panel.]
:Student (off-panel): But then why can planes fly upside down?

:[Miss Lenhart is pondering the question. Beat panel. Three long and curved arrows point out from the right frame of this panel, leading to each of the next three panels which are arranged vertically above each other, making the comic much deeper in this column than in the first two.]

:[In the top panel Miss Lenhart turns away from the students taking a hand to her chin. Overlaid on the top of the panel there is a small frame with a caption:]
:Right:
:Miss Lenhart: Wow, good question! Maybe this picture is simplified - or wrong! We should learn more.

:[In the middle panel Miss Lenhart stands as before. Overlaid on the top of the panel there is a small frame with a caption:]
:Wrong:
:Miss Lenhart: It's... complicated.
:Miss Lenhart: And we need to move on.

:[In the bottom panel Miss Lenhart visibly ball her hands in to fists and leans a little forward looking more down. Overlaid on the top of the panel there is a small frame with a caption:]
:Very wrong:
:Miss Lenhart: Santa Claus is your parents.

{{comic discussion}}

[[Category:Comics featuring Miss Lenhart]]
[[Category:Physics]]
[[Category:Christmas]]

Talk:2027: Lightning Distance

2018-08-05T07:28:55Z

172.68.132.47:

Calculations I used:

<math>t_1=\frac{s}{v_1}</math>

<math>t_2=\frac{s}{v_2}</math>

Substract:

<math>t_1-t_2=\Delta t=\frac{s}{v_1}-\frac{s}{v_2}=\frac{sv_2-sv_1}{v_1v_2}=s\frac{v_2-v_1}{v_1v_2}=s\frac{\Delta v}{v_1v_2}</math>

Therefore

<math>s=\Delta t\frac{v_1v_2}{\Delta v}</math>

I evaluated <math>\frac{v_1v_2}{\Delta v}</math> and it came to be 13.6 billion. Can someone verify it's correct? [[Special:Contributions/172.68.51.112|172.68.51.112]] 13:08, 1 August 2018 (UTC)

:The comic begins with the question "how many miles away", so converting to kilometers isn't the right calculation.[[Special:Contributions/172.69.71.24|172.69.71.24]] 17:06, 1 August 2018 (UTC)

I used refractive index for visible light of 1.000277 (air at STP as opposed to 0C 1atm) and arrived at around 7.9 billion instead. Refractive index of 1.000337 is then required for the radio waves for the comic to be correct. [[Special:Contributions/172.68.11.221|172.68.11.221]] 13:46, 1 August 2018 (UTC)

:Do you mean 7.9 billion to convert to miles or to kilometers? Because my 13.6 bilion is to kilometers.

::I'm sure the actual comic is referring to miles and 5 billion was picked to match with the "divide by five" rule for miles. [[Special:Contributions/172.69.70.131|172.69.70.131]] 13:59, 1 August 2018 (UTC)

:::I did mean kilometers. If we use miles, 1.000314 fits almost precisely! (5.04 billion) [[Special:Contributions/172.68.11.17|172.68.11.17]] 14:42, 1 August 2018 (UTC)

If you can count several seconds, as is suggested in the comic, the flash is still billions of miles away, the widest possible distance between Earth and Neptune is about 5 billion km. Sebastian --[[Special:Contributions/172.68.110.40|172.68.110.40]] 14:51, 1 August 2018 (UTC)

:Not to mention that there's not a lot of air within a few billion miles of earth, so the dispersion will be much lower for all but the last 100-ish miles, AFAIK.[[Special:Contributions/172.68.54.142|172.68.54.142]] 20:12, 1 August 2018 (UTC)

::Also, while Jupiter has {{w|Great Red Spot|VERY gigantic storms}}, they are still too small to see the lightning from them from Earth. -- [[User:Hkmaly|Hkmaly]] ([[User talk:Hkmaly|talk]]) 23:17, 1 August 2018 (UTC)

Do you really need to know the spectrum of the flash? If we assume that a flash contains UV and X-ray radiation and that the visible light is generated at the same time as the UV or X-ray radiation then you only need to know the refractive index of light/UV/X-ray in air under the same temperature conditions and not the exact spectrum. [[User:Condor70|Condor70]] ([[User talk:Condor70|talk]])

:I initially made the mistake of thinking this referred to time difference between visible and UV/X-ray, but it specifically says "brightness." If you want to compare the brightness at a distance to the brightness at the source you'll need to know the brightness at the source, i.e. the spectrum of the flash itself. With this technique you don't need to know the dispersion "only" the relative attenuation, but I suspect that would be a more error-prone measurement.[[Special:Contributions/172.68.54.142|172.68.54.142]] 18:54, 1 August 2018 (UTC)

I understand the joke Randall was going for, but have a problem with the wording. "Count the number of seconds" won't work for fractions of anything. "Measure" would work, but spoils the gag a bit. Counting numbers are integers; counting the seconds between the visible and radio frequency flashes will give you zero. [[Special:Contributions/172.69.71.24|172.69.71.24]] 17:00, 1 August 2018 (UTC)

:You're certainly correct, but the joke works (for me at least) by its comparison to the standard rule of counting seconds, and humans are not generally precise enough to resolve better than one second. By keeping Megan's wording as close to the customary rule as possible I think it optimizes the humor. That "Billion" at the end is the whole joke for me, the replacement of "sound" with "radio wave" can be glossed-over on first reading, until you get to the unexpected extra 9 orders of magnitude in the conversion.[[Special:Contributions/172.68.54.142|172.68.54.142]] 18:54, 1 August 2018 (UTC)
::Actually they would've been correct if she'd said "count the number of seconds" but she said "'''count the seconds'''".It's part of the joke, because it's correct, just completely impractical, because you'd be "counting" something like 10^(-10) seconds--[[Special:Contributions/172.68.132.47|172.68.132.47]] 07:23, 5 August 2018 (UTC)

::Just realized I also glossed-over the replacement of "divide" with "multiply." The brain is a funny thing.[[Special:Contributions/172.68.54.142|172.68.54.142]] 20:07, 1 August 2018 (UTC)

Do these account for the air pressure variability common in most thunderstorms?

I think explanation and transcript are pretty complete now. [[Special:Contributions/172.68.51.112|172.68.51.112]] 20:58, 1 August 2018 (UTC)

:There is the additional problem that a flash is no instantaneous, but progresses at a fraction of the speed of light. Who says that radio waves and light at different wavelenghts or xrays have their maximum at the same moment? ;-) --[[Special:Contributions/162.158.91.59|162.158.91.59]] 08:05, 2 August 2018 (UTC)

:: I added a few words about the problem that a flash is not instantaneous and removed the 'incomplete' tag. Hope that's OK. [[User:Chrisahn|Chrisahn]] ([[User talk:Chrisahn|talk]]) 19:41, 2 August 2018 (UTC)

Here's a variation of the calculation above that simplifies numeric evaluation:

The {{w|refractive index}} is defined as <math>n=\frac{c}{v}</math>, so <math>v=\frac{c}{n}</math> and thus <math>t=\frac{s}{v}=\frac{s\,n}{c}</math>.

<math>t_1=\frac{s\,n_1}{c}</math>

<math>t_2=\frac{s\,n_2}{c}</math>

Subtract:

<math>t_1-t_2=\Delta t=\frac{s\,n_1}{c}-\frac{s\,n_2}{c}=s\frac{n_1-n_2}{c}=s\frac{\Delta n}{c}</math>

Therefore

<math>s=\Delta t\frac{c}{\Delta n}</math>

and the factor we want to calculate is

<math>\frac{c}{\Delta n}</math>

With the numbers given in the sources in the main text:

<math>n_1=1.000315</math>

<math>n_2=1.000277</math>

<math>\Delta n=n_1-n_2=0.000038</math>

For kilometers: <math>\frac{c}{\Delta n}\approx\frac{300,000\,km/s}{0.000038}\approx7.9\cdot10^9\,km/s</math>

For miles: <math>\frac{c}{\Delta n}\approx\frac{186,000\,mi/s}{0.000038}\approx4.9\cdot10^9\,mi/s</math>

[[User:Chrisahn|Chrisahn]] ([[User talk:Chrisahn|talk]]) 18:28, 2 August 2018 (UTC)

:Nice one. I didn't think to use the refractive indicies directly. [[Special:Contributions/172.68.51.118|172.68.51.118]] 22:22, 2 August 2018 (UTC)

== Assumptions on the medium properties sound? ==

Refractive index of *dry* air might be pretty close to 1 for both light and RF EM waves, but:

Let's assume that the air is humid, if not even full of water drops. After all, lightning.

Let's further assume that an air/water mixture or solution has electromagnetic properties between these two materials.

In water, refractive index for light is about <math>n_{\text{water, optical}}=1.33 n_{\text{air, optical}}</math>, (as easily demonstrated by the optical refractive effects); for RF, we typically use values of <math>\frac{n_{\text{water, RF}}^2}{\mu_r}=\epsilon\approx 80</math>. So, <math>n_{\text{water, RF}}\approx \sqrt{80}n_{\text{air, RF}}</math>.

Let's assume a 10⁻³ "EM-effective" water content in the comic air.

That would lead to <math>\frac{v_{\text{opt.}}}{v_{\text{RF}}} = \frac{\frac34}{\sqrt{80}^{-1}}= \frac34\sqrt{80}=6.7</math>.

:While the humidity (amount of water vapor) is certainly higher during the rain, I don't think that would count as a proper "water-air mixture". Wikipedia says that "Violent rain" is above 5 cm/h. If you divide it by 3600 (to get cm/s), and then imagine stretching that all the way to the cloud, you'll find out there's not that much water at given moment in the air. [[Special:Contributions/172.68.51.112|172.68.51.112]] 19:12, 1 August 2018 (UTC)

::Great point. To finish the calculation let's use a typical terminal velocity for a large raindrop (it's a big storm, I'm sure) of 9m/s. 0.05 m/hr / 3600 s/hr / 9 m/s = 0.00015% water by volume. Sure seems like more than that when I have to drive through it! Then it seems more like [http://what-if.xkcd.com/12/].[[Special:Contributions/172.68.54.142|172.68.54.142]] 20:32, 1 August 2018 (UTC)

<code>we can't detect radiation outside the visible spectrum without very specialized instruments</code> Something that I think was overlooked in the explanation is that while humans can't *directly* sense radio waves, there are devices called "radios" which at one point in time were fairly commonly owned by humans, whose whole purpose is to detect encoded radio waves and convert them into sounds which humans can sense. I.e. you hear static during an electrical storm. So you could listen for the static and compare that to the flash... if you were fast enough. [[Special:Contributions/172.68.54.64|172.68.54.64]] 14:22, 3 August 2018 (UTC) (newbie)

:But radios ARE specialized equipment. [[Special:Contributions/172.68.51.52|172.68.51.52]] 10:35, 4 August 2018 (UTC)

== Whoops! ==

I always thought it was 1 second per mile. I didn't know about the 'divide by 5" part. [[User:These Are Not The Comments You Are Looking For|These Are Not The Comments You Are Looking For]] ([[User talk:These Are Not The Comments You Are Looking For|talk]]) 01:23, 5 August 2018 (UTC)

Talk:2027: Lightning Distance

2018-08-05T07:23:21Z

172.68.132.47:

Calculations I used:

<math>t_1=\frac{s}{v_1}</math>

<math>t_2=\frac{s}{v_2}</math>

Substract:

<math>t_1-t_2=\Delta t=\frac{s}{v_1}-\frac{s}{v_2}=\frac{sv_2-sv_1}{v_1v_2}=s\frac{v_2-v_1}{v_1v_2}=s\frac{\Delta v}{v_1v_2}</math>

Therefore

<math>s=\Delta t\frac{v_1v_2}{\Delta v}</math>

I evaluated <math>\frac{v_1v_2}{\Delta v}</math> and it came to be 13.6 billion. Can someone verify it's correct? [[Special:Contributions/172.68.51.112|172.68.51.112]] 13:08, 1 August 2018 (UTC)

:The comic begins with the question "how many miles away", so converting to kilometers isn't the right calculation.[[Special:Contributions/172.69.71.24|172.69.71.24]] 17:06, 1 August 2018 (UTC)

I used refractive index for visible light of 1.000277 (air at STP as opposed to 0C 1atm) and arrived at around 7.9 billion instead. Refractive index of 1.000337 is then required for the radio waves for the comic to be correct. [[Special:Contributions/172.68.11.221|172.68.11.221]] 13:46, 1 August 2018 (UTC)

:Do you mean 7.9 billion to convert to miles or to kilometers? Because my 13.6 bilion is to kilometers.

::I'm sure the actual comic is referring to miles and 5 billion was picked to match with the "divide by five" rule for miles. [[Special:Contributions/172.69.70.131|172.69.70.131]] 13:59, 1 August 2018 (UTC)

:::I did mean kilometers. If we use miles, 1.000314 fits almost precisely! (5.04 billion) [[Special:Contributions/172.68.11.17|172.68.11.17]] 14:42, 1 August 2018 (UTC)

If you can count several seconds, as is suggested in the comic, the flash is still billions of miles away, the widest possible distance between Earth and Neptune is about 5 billion km. Sebastian --[[Special:Contributions/172.68.110.40|172.68.110.40]] 14:51, 1 August 2018 (UTC)

:Not to mention that there's not a lot of air within a few billion miles of earth, so the dispersion will be much lower for all but the last 100-ish miles, AFAIK.[[Special:Contributions/172.68.54.142|172.68.54.142]] 20:12, 1 August 2018 (UTC)

::Also, while Jupiter has {{w|Great Red Spot|VERY gigantic storms}}, they are still too small to see the lightning from them from Earth. -- [[User:Hkmaly|Hkmaly]] ([[User talk:Hkmaly|talk]]) 23:17, 1 August 2018 (UTC)

Do you really need to know the spectrum of the flash? If we assume that a flash contains UV and X-ray radiation and that the visible light is generated at the same time as the UV or X-ray radiation then you only need to know the refractive index of light/UV/X-ray in air under the same temperature conditions and not the exact spectrum. [[User:Condor70|Condor70]] ([[User talk:Condor70|talk]])

:I initially made the mistake of thinking this referred to time difference between visible and UV/X-ray, but it specifically says "brightness." If you want to compare the brightness at a distance to the brightness at the source you'll need to know the brightness at the source, i.e. the spectrum of the flash itself. With this technique you don't need to know the dispersion "only" the relative attenuation, but I suspect that would be a more error-prone measurement.[[Special:Contributions/172.68.54.142|172.68.54.142]] 18:54, 1 August 2018 (UTC)

I understand the joke Randall was going for, but have a problem with the wording. "Count the number of seconds" won't work for fractions of anything. "Measure" would work, but spoils the gag a bit. Counting numbers are integers; counting the seconds between the visible and radio frequency flashes will give you zero. [[Special:Contributions/172.69.71.24|172.69.71.24]] 17:00, 1 August 2018 (UTC)

:You're certainly correct, but the joke works (for me at least) by its comparison to the standard rule of counting seconds, and humans are not generally precise enough to resolve better than one second. By keeping Megan's wording as close to the customary rule as possible I think it optimizes the humor. That "Billion" at the end is the whole joke for me, the replacement of "sound" with "radio wave" can be glossed-over on first reading, until you get to the unexpected extra 9 orders of magnitude in the conversion.[[Special:Contributions/172.68.54.142|172.68.54.142]] 18:54, 1 August 2018 (UTC)
::Actually they would've been correct if she'd said "count the number of seconds" but she said '''count the seconds'''".It's part of the joke, because it's correct, just completely impractical, because you'd be "counting" something like 10^(-11) seconds--[[Special:Contributions/172.68.132.47|172.68.132.47]] 07:23, 5 August 2018 (UTC)

::Just realized I also glossed-over the replacement of "divide" with "multiply." The brain is a funny thing.[[Special:Contributions/172.68.54.142|172.68.54.142]] 20:07, 1 August 2018 (UTC)

Do these account for the air pressure variability common in most thunderstorms?

I think explanation and transcript are pretty complete now. [[Special:Contributions/172.68.51.112|172.68.51.112]] 20:58, 1 August 2018 (UTC)

:There is the additional problem that a flash is no instantaneous, but progresses at a fraction of the speed of light. Who says that radio waves and light at different wavelenghts or xrays have their maximum at the same moment? ;-) --[[Special:Contributions/162.158.91.59|162.158.91.59]] 08:05, 2 August 2018 (UTC)

:: I added a few words about the problem that a flash is not instantaneous and removed the 'incomplete' tag. Hope that's OK. [[User:Chrisahn|Chrisahn]] ([[User talk:Chrisahn|talk]]) 19:41, 2 August 2018 (UTC)

Here's a variation of the calculation above that simplifies numeric evaluation:

The {{w|refractive index}} is defined as <math>n=\frac{c}{v}</math>, so <math>v=\frac{c}{n}</math> and thus <math>t=\frac{s}{v}=\frac{s\,n}{c}</math>.

<math>t_1=\frac{s\,n_1}{c}</math>

<math>t_2=\frac{s\,n_2}{c}</math>

Subtract:

<math>t_1-t_2=\Delta t=\frac{s\,n_1}{c}-\frac{s\,n_2}{c}=s\frac{n_1-n_2}{c}=s\frac{\Delta n}{c}</math>

Therefore

<math>s=\Delta t\frac{c}{\Delta n}</math>

and the factor we want to calculate is

<math>\frac{c}{\Delta n}</math>

With the numbers given in the sources in the main text:

<math>n_1=1.000315</math>

<math>n_2=1.000277</math>

<math>\Delta n=n_1-n_2=0.000038</math>

For kilometers: <math>\frac{c}{\Delta n}\approx\frac{300,000\,km/s}{0.000038}\approx7.9\cdot10^9\,km/s</math>

For miles: <math>\frac{c}{\Delta n}\approx\frac{186,000\,mi/s}{0.000038}\approx4.9\cdot10^9\,mi/s</math>

[[User:Chrisahn|Chrisahn]] ([[User talk:Chrisahn|talk]]) 18:28, 2 August 2018 (UTC)

:Nice one. I didn't think to use the refractive indicies directly. [[Special:Contributions/172.68.51.118|172.68.51.118]] 22:22, 2 August 2018 (UTC)

== Assumptions on the medium properties sound? ==

Refractive index of *dry* air might be pretty close to 1 for both light and RF EM waves, but:

Let's assume that the air is humid, if not even full of water drops. After all, lightning.

Let's further assume that an air/water mixture or solution has electromagnetic properties between these two materials.

In water, refractive index for light is about <math>n_{\text{water, optical}}=1.33 n_{\text{air, optical}}</math>, (as easily demonstrated by the optical refractive effects); for RF, we typically use values of <math>\frac{n_{\text{water, RF}}^2}{\mu_r}=\epsilon\approx 80</math>. So, <math>n_{\text{water, RF}}\approx \sqrt{80}n_{\text{air, RF}}</math>.

Let's assume a 10⁻³ "EM-effective" water content in the comic air.

That would lead to <math>\frac{v_{\text{opt.}}}{v_{\text{RF}}} = \frac{\frac34}{\sqrt{80}^{-1}}= \frac34\sqrt{80}=6.7</math>.

:While the humidity (amount of water vapor) is certainly higher during the rain, I don't think that would count as a proper "water-air mixture". Wikipedia says that "Violent rain" is above 5 cm/h. If you divide it by 3600 (to get cm/s), and then imagine stretching that all the way to the cloud, you'll find out there's not that much water at given moment in the air. [[Special:Contributions/172.68.51.112|172.68.51.112]] 19:12, 1 August 2018 (UTC)

::Great point. To finish the calculation let's use a typical terminal velocity for a large raindrop (it's a big storm, I'm sure) of 9m/s. 0.05 m/hr / 3600 s/hr / 9 m/s = 0.00015% water by volume. Sure seems like more than that when I have to drive through it! Then it seems more like [http://what-if.xkcd.com/12/].[[Special:Contributions/172.68.54.142|172.68.54.142]] 20:32, 1 August 2018 (UTC)

<code>we can't detect radiation outside the visible spectrum without very specialized instruments</code> Something that I think was overlooked in the explanation is that while humans can't *directly* sense radio waves, there are devices called "radios" which at one point in time were fairly commonly owned by humans, whose whole purpose is to detect encoded radio waves and convert them into sounds which humans can sense. I.e. you hear static during an electrical storm. So you could listen for the static and compare that to the flash... if you were fast enough. [[Special:Contributions/172.68.54.64|172.68.54.64]] 14:22, 3 August 2018 (UTC) (newbie)

:But radios ARE specialized equipment. [[Special:Contributions/172.68.51.52|172.68.51.52]] 10:35, 4 August 2018 (UTC)

== Whoops! ==

I always thought it was 1 second per mile. I didn't know about the 'divide by 5" part. [[User:These Are Not The Comments You Are Looking For|These Are Not The Comments You Are Looking For]] ([[User talk:These Are Not The Comments You Are Looking For|talk]]) 01:23, 5 August 2018 (UTC)

801: Golden Hammer

2018-07-26T23:20:44Z

172.68.132.47: Forgot my end brackets like a moron!

{{comic
| number = 801
| date = October 4, 2010
| title = Golden Hammer
| image = golden hammer.png
| titletext = Took me five tries to find the right one, but I managed to salvage our night out--if not the boat--in the end.
}}

==Explanation==
{{incomplete|In English please! This first paragraph is gobbledegook to the non-Java savvy.}}
{{w|Java}} is a programming language touted for its Portability™, which sometimes leads to it being used in systems where it really just shouldn't be used. [[Cueball]] laments that the hardware he's tinkering with, despite being used for a single purpose, has its firmware written in Java; since the microprocessor is unknown, it's quite possible the {{w|Java Virtual Machine}} (JVM) had to be ported over to the processor before the hardware designers could write firmware for it. Presumably, they considered this worthwhile to be able to write the control code in a language they're comfortable with, even though it probably would have been much simpler to just write the control code in whatever language they used to port the JVM in the first place.

[[Black Hat]] explains that this is really an example of an age-old adage: "When all you have is a hammer, everything looks like a nail", also referred to as the "{{w|law of the instrument}}" or, as in the title, the "golden hammer". The hardware developers probably only knew Java, and when they thought about how to write firmware for their new device, "Java" was the only solution that occurred to them.

Of course, instead of a hammer and a nail, Black Hat's analogy is about using bolt-cutters and vodka to get through the lock on {{w|Wolf Blitzer}}'s boathouse. Not-so-coincidentally, Black Hat is holding a pair of bolt-cutters and a bottle of vodka. The implication is that Black Hat ''did'', in fact, break into Wolf Blitzer's boathouse the previous night, which is why he has just now entered the door at the start of the strip. The changes he makes to the adage implies that he believes vodka and boltcutters are designed specifically to be used on Wolf Blitzer's boathouse, an interpretation that fits Black Hat's warped and anarchic disposition. As he is carrying both of these items, it also implies that he has just used those instruments for exactly that purpose. Cueball however, being extremely jaded by the (mis)use of Java, can only bring himself to tell that he's glad that Black Hat had a nice night.

The title text implies that Black Hat had to break into a number of boathouses before he found Wolf's, and that his boat did not survive the evening. The use of the phrase 'our night' allows us to infer that Black Hat may have been with [[Danish]], his partner in crime.

==Transcript==
:[Black Hat is going through a door, a bottle in his hand. A voice speaks to him from off panel.]
:Cueball: Seriously? This thing runs ''Java?'' It's single-purpose hardware!

:[Cueball is sitting at a computer, holding some device which is wired to a box, and pointing at the screen.]
:Cueball: I bet they actually hired someone to spend six months porting this JVM so they could write their 20 lines of code in a familiar setting.

:[Black Hat has a pair of bolt cutters in the hand that had been obscured in the first panel.]
:Black Hat: Well, you know what they say— When all you have is a pair of bolt cutters and a bottle of vodka, everything looks like the lock on the door of Wolf Blitzer's boathouse.
:Cueball: I'm glad ''you ''had a nice night.

{{comic discussion}}
[[Category:Comics featuring Cueball]]
[[Category:Comics featuring Black Hat]]
[[Category:Programming]]
[[Category:Language]]