User talk:KlappCK

Scratchpad

---+Problem Statement: The purpose of this page is to discuss a hypothetical scenario where one wishes to find the average spontaneous and stimulated monochromatic emission, ${\displaystyle \Phi }$ at any point on the curved surface of a unit-half-spherical object (a sphere cut in half) illuminated by a monochromatic columnated light source (such as a laser or some other emissive object infinitely far away from the unit-half-sphere) when the total surface emission function is ${\displaystyle \phi \left(c,v,{\hat {s}},{\hat {n}}(x,y)\right)\equiv c+v{\hat {s}}\cdot {\hat {n}}(x,y)}$, where ${\displaystyle c}$ is the (constant) spontaneous surface emission, ${\displaystyle v}$ is the (relative) stimulated surface emission, ${\displaystyle {\hat {s}}\equiv s_{x}{\hat {x}}+s_{y}{\hat {y}}+{\sqrt {1-s_{x}^{2}+s_{y}^{2}}}{\hat {z}}}$ is the unit vector pointing in the direction of the monochromatic columnated light source (assumed to be on the curved side of the surface), and ${\displaystyle {\hat {n}}(x,y)\equiv x{\hat {x}}+y{\hat {y}}+{\sqrt {1-x^{2}-y^{2}}}{\hat {z}}}$ is the parameterized surface normal on the curved side of the unit-half-sphere, so that ${\displaystyle \Phi \equiv {\overline {\phi \left(c,v,{\hat {s}},{\hat {n}}(x,y)\right)}}\equiv {\tfrac {1}{2\pi }}\int \limits _{x=-1}^{x=1}\int \limits _{y=-{\sqrt {1-x^{2}}}}^{y={\sqrt {1-x^{2}}}}{\tfrac {c+v{\hat {s}}\cdot {\hat {n}}(x,y)}{\sqrt {1-x^{2}-y^{2}}}}{\text{d}}x{\text{d}}y=c+{\tfrac {vs_{z}}{2}}}$ for ${\displaystyle s_{z}\geq 0}$ . Derivation:

${\displaystyle \Phi \equiv {\tfrac {1}{2\pi }}\int \limits _{x=-1}^{x=1}\int \limits _{y=-{\sqrt {1-x^{2}}}}^{y={\sqrt {1-x^{2}}}}{\tfrac {c+v{\hat {s}}\cdot {\hat {n}}(x,y)}{\sqrt {1-x^{2}-y^{2}}}}{\text{d}}x{\text{d}}y={\tfrac {1}{2\pi }}\int \limits _{x=-1}^{x=1}\int \limits _{y=-{\sqrt {1-x^{2}}}}^{y={\sqrt {1-x^{2}}}}{\tfrac {c+v\left(s_{x}x+s_{y}y+{\sqrt {1-s_{x}^{2}+s_{y}^{2}}}{\sqrt {1-x^{2}-y^{2}}}\right)}{\sqrt {1-x^{2}-y^{2}}}}{\text{d}}x{\text{d}}y=c+{\tfrac {v}{2\pi }}\int \limits _{x=-1}^{x=1}\int \limits _{y=-{\sqrt {1-x^{2}}}}^{y={\sqrt {1-x^{2}}}}{\tfrac {s_{x}x+s_{y}y+{\sqrt {1-s_{x}^{2}+s_{y}^{2}}}{\sqrt {1-x^{2}-y^{2}}}}{\sqrt {1-x^{2}-y^{2}}}}{\text{d}}x{\text{d}}y=c+{\tfrac {v{\sqrt {1-s_{x}^{2}+s_{y}^{2}}}}{2}}+{\tfrac {v}{2\pi }}\int \limits _{x=-1}^{x=1}\int \limits _{y=-{\sqrt {1-x^{2}}}}^{y={\sqrt {1-x^{2}}}}{\tfrac {s_{x}x+s_{y}y}{\sqrt {1-x^{2}-y^{2}}}}{\text{d}}x{\text{d}}y=c+{\tfrac {v{\sqrt {1-s_{x}^{2}+s_{y}^{2}}}}{2}}=c+{\tfrac {vs_{z}}{2}}}$

The Cement Garden

No problem - and many thanks for expanding it. Lugnuts (talk) 14:05, 1 September 2011 (UTC)[reply]

Hehe, glad you liked! Pretty much everything on here isn't completed, so there's no need for the tag. Lugnuts (talk) 17:04, 20 September 2011 (UTC)[reply]

RE: Joe Swanson Debate

Haven't absorbed the debate enough to have an opinion. I'm an inclusionist, and I believe the Family Guy article and all its related articles have been taken over by deletionists. However, I can't jump in because it would simply by a biased opinion. I've pretty much given up on these articles and now just spend my Family Guy time at the Family Guy wiki. Thanks for the consideration though. Geeky Randy (talk) 05:43, 2 September 2011 (UTC)[reply]

But I don't have an opinion. It wouldn't be fair of me to give my input... just yet. One part of me thinks that since it has been established, maybe it could be put in the article; on the other hand, it's already established that he's a paraplegic so going into that kind of detail is redundant since his incontinence is kind of a given. I haven't decided which side I lean towards, and given that Family Guy Wiki is a much better database for the show than wikipedia is, quite frankly I don't care much about the discussion. Lastly, you should be careful not to WP:CANVAS, it could lead to a lot of trouble and will certainly be your downfall in your current debate. Geeky Randy (talk) 16:45, 2 September 2011 (UTC)[reply]

"As our own article on paraplegia indicates, paraplegia does not necessarily imply incontinence..." This is your overriding factor in the debate. I never even bothered to consider this. Give me a day or so and I'll provide my input. Well done. Geeky Randy (talk) 04:14, 3 September 2011 (UTC)[reply]

September 2011

Please do not add or change content without verifying it by citing reliable sources, as you did to Nicole Polizzi. Please review the guidelines at Wikipedia:Citing sources and take this opportunity to add references to the article. Thank you. SummerPhD (talk) 15:12, 14 September 2011 (UTC)[reply]

I have responed to your comment on your talk page.KlappCK (talk) 16:02, 14 September 2011 (UTC)[reply]

October 2011

Welcome to Wikipedia. Please be aware of Wikipedia's policy that biographical information about living persons must not include unsupported or inaccurate statements. Whenever you add possibly controversial statements about a living person to an article or any other Wikipedia page, as you did to Helena Bonham Carter with this edit, you must include proper sources. If you don't know how to cite a source, you may want to read Wikipedia:Referencing for beginners for guidelines. Thank you. DVdm (talk) 13:32, 6 October 2011 (UTC)[reply]

See the talk page. This appears well sourced to me, given the nature of the information. FOX news doesn't typically run articles about urinary incontinence, mostly (in my opinion) because it is a taboo subject and not exactly a ratings booster. Consider the prominence of the cited sources given the subject matter and then consider the rulesKlappCK (talk) 13:37, 6 October 2011 (UTC)[reply]

Replied on talk page. - DVdm (talk) 13:48, 6 October 2011 (UTC)[reply]

Struck comment. Do feel free to remove section. Owe you an apology. - DVdm (talk) 14:03, 6 October 2011 (UTC)[reply]

M203 buckshot round

Regarding M203 Buckshot round...

You must be a mathematician…and as a mathematician you like precision. Unfortunately, a Buckshot round is anything but.

1) The M203 has a large cone of fire, that expands very quickly. While it would be nice to hit the target with every pellet. The Army knows that most of the pellets will miss the target anyway. They just want the soldier to hit the target with some of the pellets.

2) The sight radius of the M16 is about 5 inches above the center line of the M203’s bore. This causes a significant parallax problem.

3) When using the M16 sights to fire an M203 Buckshot round, the shooter has is a natural tendency to look down over the sights. This causes the M203’s bore to point up. At 10 feet this is not a problem. But, at 100 feet it becomes a major issue. As a result, the shooter can actually fire over the targets head.

4) When shooting at more distance targets, if the shooter aims low or at the foot of the target, it tends to correct these issues.

5) If any pellets hit the ground before striking the target, they may bounce into the target anyway. While any pellets that fly over the targets head have no possibility (or second chance) to hit the target. — Preceding unsigned comment added by 71.22.156.40 (talk) 01:28, 19 October 2011 (UTC)[reply]

This explanation is elucidating. Perhaps it should be included somewhere on the main page of M203 Grenade Launcher.KlappCK (talk) 13:51, 19 October 2011 (UTC)[reply]

We have a bad apple

User:Mike Searson...has a habit of violating civility rules, insulting other users, using profanity and basically treating Wikipedia as his own personal web-site. If you read his contribution page you will see what I mean. Unfortunately, I do not have the enough Wiki knowledge to challenge him in any meaningful way. I believe that you do. At the very least you know how to send this to the next level. Please, look in to this and tell me what you think. Thank You. — Preceding unsigned comment added by 71.22.156.40 (talk) 02:27, 13 November 2011 (UTC)[reply]

Thank you for the info. — Preceding unsigned comment added by 71.22.156.40 (talk) 16:26, 14 November 2011 (UTC)[reply]

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Chain rule in absolute value

Thanks for the explanation. I don't completely agree, but I don't feel strongly about it.... --Macrakis (talk) 16:44, 10 February 2012 (UTC)[reply]

Thanks for your note. --Macrakis (talk) 15:02, 24 February 2012 (UTC)[reply]

re "Absolute Value"

Hi KlappCK. I've replied to your message on my talk page here. Paul August ☎ 20:26, 28 March 2012 (UTC)[reply]

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Whisks

I was also wondering about the whisks. –Roscelese (talk ⋅ contribs) 00:12, 12 October 2012 (UTC)[reply]

Comparison of the AK-47 and M16

I request that you review the happenings on the Comparison of the AK-47 and M16. There is a user who is suppressing manufacture technical specifications simply because he disagrees with them. He seems unwilling or unable to produce hard evidence that they are wrong. There is also more info on the “talk page”. I’m afraid that I have done all that I can do without being block. Thank you for your attention — Preceding unsigned comment added by 71.22.156.40 (talk) 15:33, 20 November 2012 (UTC)[reply]

How does someone file a complaint against an editor who claims he does not have to prove anything...see talk page Comparison of the AK-47 and M16.--71.22.156.40 (talk) 21:56, 22 November 2012 (UTC)[reply]

I am sorry to keep coming to you but G PViB has now posted the following comment "Do I need to call into question your mental adequacy to be a Wikipedia editor?" I believe qualifies as a civility violation. And, still refuses to provide any reference to support his positions.--71.22.156.40 (talk) 21:08, 26 November 2012 (UTC)[reply]

Cauchy_formula_for_repeated_integration needs your help

There is a broken equation, just when it was getting to the interesting bit.

Thanks Thepigdog (talk) 08:15, 10 February 2014 (UTC)[reply]

Works for me.KlappCK (talk) 02:59, 25 February 2014 (UTC)[reply]

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