User talk:Mglg

-

Manual of style
You mentioned date links in the Manual of style. My position is that in most cases, the date links are unnecessary and only put there by people that misunderstand date preferences. Feel free to remove unnecessary date links.

You may wish to make use of a 'Dates' tab in edit mode that will help with unlinking unnecessary date links. Simply copy the entire contents of User:Bobblewik/monobook.js to your own monobook. Then follow the instructions in your monobook to clear the cache (i.e. press Ctrl-Shift-R in Firefox, or Ctrl-F5 in IE) before it will work. It also provides a 'Units' tab. Regards. bobblewik 14:26, 10 June 2006 (UTC)

Down syndrome
Hi, Mglg. You are correct, and I owe you an apology. Because of the way you juggled the text and the paragraphs &mdash; and because the vandalism along with everyone putting their hand into the pot has been out of control &mdash; I misread, was too quick on the trigger, and thought you were deleting a chunk of text, and replacing it with unreferenced text. (I'm not happy to see that the text wasn't referenced to begin with, but I didn't review that article when it was at FAC, as I was traveling.) Now that I've reviewed it more closely, your edit makes complete sense, except I'm wondering why you dropped the text that says (bolding mine), "Pregnant women can be screened for various complications in their pregnancy, or due to risk factors such as advanced maternal age" ? I think everything else stayed the same, just got moved around. Let me know, so I can re-instate your edit (so you won't run into 3RR). Or, if I don't hear from you tonight, maybe you can reinstate it before I get to it. Once again, I'm very sorry for misreading. Regards, Sandy (Talk) 01:39, 9 December 2006 (UTC)
 * Thanks for the explanation - since the original FA author isn't around to explain what was wanted in that portion, I'll reinstate your edit exactly as you had it. Thanks for understanding !  (Better check to be sure I get your edit reinstated correctly.)  Sandy (Talk) 02:32, 9 December 2006 (UTC)
 * OK, I think I got it - let me know if I goofed anything. Regards, Sandy (Talk) 02:39, 9 December 2006 (UTC)
 * Good - sorry for the all the extra typing you had to do! Sandy (Talk) 02:52, 9 December 2006 (UTC)

Ref Desk guidelines
Care to continue to help us develop the Ref Desk guidelines ? If so, they are here: Wikipedia_talk:Reference_desk/guidelines. StuRat 23:20, 30 April 2007 (UTC)

Dimension

 * I thank you for trying to explain, but your explaination is confusing.199.126.28.71 23:56, 6 May 2007 (UTC)
 * Sorry, 199.126, for being confusing. I didn't really intend for my comment about theories of 10 or 11 dimensions to be understandable by itself, I just thought you would be interested to know that such ideas exist. The main part of my answer was that there is nothing special about three dimensions of space. That is just how the world appears to be. Somebody may at some future time invent a theory that gives a reason why space has three observable dimensions and not seven, but right now there is no such theory. --mglg(talk) 00:12, 7 May 2007 (UTC)
 * But I don't understand, isn't matter 3 dimensional?199.126.28.71 01:07, 7 May 2007 (UTC)
 * "Matter" doesn't necessarily have any well-defined dimensionality. Space does. Think of matter as a collection of point-like elementary particles. Points have no dimension. They are just points in space. Objects (collections of such point particles) have the dimensionality of the space they are spread out in. --mglg(talk) 01:17, 7 May 2007 (UTC)
 * But I don't understand, matter has volume?!?!!199.126.28.71 01:25, 7 May 2007 (UTC)
 * Sorry, I assumed you were familiar with the idea that matter consists of atoms, and that an atom is made up of one or more electrons and a nucleus, and that the nucleus is made up of smaller elementary particles. The electrons and other elementary particles themselves don't necessarily have any volume. Think of the particles as a collection of sand grains in your hand (and think of each grain as a point without size). If you throw all the sand grains up into the air, they form a 3-dimensional sand cloud, but once they fall down on your floor they form a 2-dimensional pattern of sand grains on the 2-dimensional surface of the floor. It is the space, and the arrangement of the particles in the space, that has a dimensionality. --mglg(talk) 01:40, 7 May 2007 (UTC)
 * Well, I understand atomic theory, but I thought that matter had mass and volume?! I mean I know about protons, and electrons and neutrons, they have rest mass, where as photons don't, but don't protons, elcetrons and neutrons, \atomic particles\ have volume?199.126.28.71 02:02, 7 May 2007 (UTC)
 * Protons and neutrons themselves consist of smaller particles called quarks. Electrons and quarks are not thought to have any volume. They are considered to be point-like particles. --mglg(talk) 02:08, 7 May 2007 (UTC)
 * Thank you, I understand now. I like how you also mentioned '...10 or 11 dimensions...'.  Please tell me more about them, which is it 10 or 11, which theory is this?  Why 10 or 11? —The preceding unsigned comment was added by 199.126.28.71 (talk) 02:19, 7 May 2007 (UTC).
 * See string theory and why 10 dimensions. Don't expect to understand it (I don't, and I studied this stuff for a while in graduate school). The basic idea is that normal particle theory has some serious problems, and in string theories these problems happen to cancel out if the dimensionality of space has a certain value. Which value that is depends on the particular kind of string theory (there are several). --mglg(talk) 02:36, 7 May 2007 (UTC)
 * Which is the most correct /or right string¤ theory?199.126.28.71 05:02, 7 May 2007 (UTC)
 * If you can convince the world that any one of them is correct, you are in for a Nobel Prize. --mglg(talk) 19:34, 7 May 2007 (UTC)

Coins problem at RD/S
I added my own take on this problem (once I finally cracked it), and get results that are tantalizingly similar to but different from yours. Care to take a look? --Tardis 23:21, 31 May 2007 (UTC)


 * Looks like we are in complete agreement about the physics. One of us just made some algebra error. I'll admit that I didn't do the algebra by hand but had Mathematica solve the equations, and trusted the result. I did convince myself, when the OP challenged it, that my answer is consistent with conservation of energy. But that could of course be wrong too. Have you plugged the two answers back into the original equations to check? --mglg(talk) 23:49, 31 May 2007 (UTC)


 * Tardis, I tried to reproduce your calculation using Mathematica (I stopped thinking for myself long ago), and I get $$\vec a=\begin{bmatrix}\frac7{15}\\\frac2{15}\end{bmatrix}\frac F{m}$$, in agreement with my earlier results. Your value of $$\vec a$$ corresponds to row 1 instead of column 1 out of my result for the matrix $$\left(M^{-1}+M^T\right)^{-1}$$. --mglg(talk) 01:34, 1 June 2007 (UTC)


 * I feel silly now. I used Mathematica too, but destroyed the file so I can't see what I did.  I know that on paper I had (incorrectly) transposed M at one point, but I was certain I had repaired that before posting; evidently only partially!  Anyway, my answer gives the slightly wrong (0.9911,0.0361)F/m as the total acceleration, and what is now our answer gives (F/m,0), so it seems you were right all along.  Anyway, I largely thought it would be neat to approach this from "first principles", as it were, without the "what-if" of considering $$t\not=0$$, and the resulting matrix equation is just pretty.  I hope you enjoyed fixing it!  --Tardis 22:22, 1 June 2007 (UTC)
 * Also silly: my first commit to RD/S didn't even have F in the final answers, and even when I thought to fix it I forgot m. They just seem so trivial, I can't even remember to include them when I want to!  --Tardis 22:25, 1 June 2007 (UTC)
 * Yes, I enjoyed the clean aesthetics of your approach. It is of course much more general than mine and can be used to simulate (or conceivably solve for?) the motion for $$t\not=0$$, until contact is broken, whereas I only considered t=0. --mglg(talk) 22:35, 1 June 2007 (UTC)
 * I actually did simulate it (with a simple reduction of order solver: forward Euler for velocity and midpoint for position), and found that the 1-3 connection is broken at about $$t=0.8\sqrt{\frac{mr}F}$$ (where the dependence on the parameters I have guessed entirely from dimensional analysis). Obviously one could detect this (the coins actually become separated even under the wrong equations, so you don't even have to go looking for negative forces) and continue to simulate until 2 slides off, but even making a pretty animation of the three-coin-chain problem doesn't make it look any more interesting than you'd think. --Tardis 20:08, 6 June 2007 (UTC)


 * Cool. Good to hear that there even worse compulsives than me out there... ;-). Seriously, I suspect the OP Aepryus would be quite interested in seeing your full solution, since that is the sort of thing he is trying to do. I would be curious to see your movie myself (if it is in a format I can read, like quicktime or mathematica). --mglg(talk) 20:23, 6 June 2007 (UTC)
 * Since you obviously can run this, I'll just provide you my code. It needs a trivial   function, which I haven't included because my version is relatively complicated.

BoundingBox[g_Graphics]:= AbsoluteOptions[g,PlotRange]\[LeftDoubleBracket]1,2\[RightDoubleBracket] BoundingBox[l:{__Graphics}]:= With[{b=BoundingBox/@l}, Transpose[{Min/@ Transpose[b\[LeftDoubleBracket]All,All,1\[RightDoubleBracket]], Max/@Transpose[b\[LeftDoubleBracket]All,All,2\[RightDoubleBracket]]}]] With[{F={1,0},m=1,r=1/2,dt=1/100000,T=6/10,f=800}, Module[{x1={0,0},x2={2r,0},x3=2r{Cos[-\[Pi]/3],Sin[-\[Pi]/3]},v1={0,0}, v2={0,0},v3={0,0},c=0,hist={}}, Do[With[{\[Alpha]=UnitVector[x2-x1],\[Beta]=UnitVector[x3-x1]}, With[{M={\[Alpha],\[Beta]}}, With[{a=LinearSolve[Inverse[M]+Transpose[M],F]/m//N}, With[{a1=LinearSolve[M,a], a2=a\[LeftDoubleBracket]1\[RightDoubleBracket]\[Alpha], a3=a\[LeftDoubleBracket]2\[RightDoubleBracket]\[Beta]},{x1,x2, x3}+={v1,v2,v3}dt/2;{v1,v2,v3}+={a1,a2,a3}dt;{x1,x2, x3}+={v1,v2,v3}dt/2; If[Mod[c++,f]\[Equal]0,AppendTo[hist,{x1,x2,x3}]]]]]],{t,0,T, dt}];hist]] With[{r=1/2}, With[{g=Graphics[Circle[#,r]&/@#]&/@%}, Show[#,PlotRange\[Rule]BoundingBox[g],AspectRatio\[Rule]Automatic, ImageSize\[Rule]1000]&/@g]];
 * Note that this uses  to render the graphics; the   function is present to make a smooth movie.  Enjoy!  --Tardis 21:49, 6 June 2007 (UTC)


 * Thanks. I'm only sort of following the code - you use Mathematica at a higher level than I do – but it runs fine. I'm being really dense right now, though: Why do the coins separate in the simulation, given that you are not explicitly turning the forces off when they go negative? Isn't your (our) assumption of maintained contact still active? Shouldn't the coins thus stick together like magnets in the simulation, with coin 3 (and later coin 2) maybe oscillating back and forth like a wagging dog-tail still in contact behind coin 1? --mglg(talk) 22:48, 6 June 2007 (UTC)
 * I wondered about that myself for a bit, and I haven't a rigorous answer. But I think it's actually pretty simple: in equation 3 I force the accelerations to be orthogonal to the separation, not the velocities.  The real question is how this can allow the coins to separate but not to overlap, and this is where I have no rigor.  The acceleration constraint prevents the force from causing radial velocities, and there are none to start with.  However, the coins' inertias are not so constrained (since they are not accelerations!) and so the coins can drift apart when the tangential velocity swings the coins around so that pre-existing velocities become radial.  (This can't cause overlap because the swinging is in the wrong direction then.)  However, the constraint is still non-physical, of course; coin 1 can't accelerate away from coin 3, so coin 3 will be dragged behind it (although presumably at ever-increasing distance; run the code farther than I did and see!).  It is certainly an interesting model that somehow prohibits overlap without strength and yet allows such (partial) disassociation!  --Tardis 23:41, 6 June 2007 (UTC)


 * OK, but doesn't that mean that setting $$\ddot\vec x_2-\ddot\vec x_1\perp\hat\alpha, \ddot\vec x_3-\ddot\vec x_1\perp\hat\beta$$ is the wrong constraint in the first place? Consider the vector $$\vec c=\vec x_3-\vec x_1$$ and its norm $$g=||\vec c||=\vec c\cdot\vec c$$; the norm's time derivative is $$\dot g=2\vec c\cdot\dot \vec c$$, and its second time derivative is $$\ddot g=2\dot\vec c\cdot\dot \vec c + 2\vec c\cdot\ddot \vec c$$. Your perpendicularity constraint forces the second term $$\vec c\cdot\ddot \vec c$$ of $$\ddot g$$ to be zero, but what I think you are trying to constrain is the sum of both terms, to maintain contact by forcing $$g\,\!$$ to stay constant through requiring $$\ddot g$$ and therefore $$\dot g$$ to vanish. --mglg(talk) 00:32, 7 June 2007 (UTC)


 * For what it's worth, That was my biggest question. I couldn't figure out where that constraint came from.  I have been sick and traveling the last week so haven't been able to analyze this stuff, but I'll work on it today.  I'm guessing conservation of energy must come into play somehow in the solution.  Also, I don't have access to Mathematica these days if there does happen to a simple-timeless way to convert the animation that would be appreciated. Aepryus 13:10, 7 June 2007 (UTC)


 * OK, I have modified Tardis's code to constrain the actual coin-coin distance, and to let go of the coins when they loose contact:

asolve[r_, vr22_, vr32_, alpha_, beta_, fom_] = Flatten[{a2, a3} /. Solve[{2 a2 + vr22/(2r) + a3 (beta.alpha) == fom.alpha, 2 a3 + vr32/(2r) + a2 (beta.alpha) == fom.beta}, {a2,a3}]] a2solve[r_, vr22_, alpha_, fom_] = (1/2)(fom.alpha - vr22/(2 r)) seq = With[{F = {1, 0}, m = 1, r = 1/2, dt = 1/5000, T = 5, f = 200}, Module[{x1 = {0, 0}, x2 = {2r, 0}, x3 = 2r{Cos[-\[Pi]/3], Sin[-\[Pi]/3]}, v1 = {0, 0},v2 = {0, 0}, v3 = {0, 0}, c = 0, hist = {}, contact = {True, True}, a, a1}, Do[With[{\[Alpha] = UnitVector[x2 - x1] // N,            \[Beta] = UnitVector[x3 - x1] // N}, With[{M = {\[Alpha], \[Beta]}}, With[{vr22 = (v2 - v1).(v2 - v1), vr32 = (v3 - v1).(v3 - v1)}, a = If[contact\[LeftDoubleBracket]2\[RightDoubleBracket], asolve[r, vr22, vr32, \[Alpha], \[Beta], F/m] // N,                   If[contact\[LeftDoubleBracket]1\[RightDoubleBracket], {a2solve[r, vr22, \[Alpha], F/m], 0}, {0, 0}]];           With[{ a2 = Max[a\[LeftDoubleBracket]1\[RightDoubleBracket],0]\[Alpha], a3 = Max[a\[LeftDoubleBracket]2\[RightDoubleBracket],0]\[Beta]}, a1 = (F/m) - a2 - a3; {x1, x2, x3} += {v1, v2, v3}dt/2; {v1, v2, v3} += {a1, a2, a3}dt; {x1, x2, x3} += {v1, v2, v3}dt/2; contact\[LeftDoubleBracket]1\[RightDoubleBracket] = Evaluate[a\[LeftDoubleBracket]1\[RightDoubleBracket] > 0]; contact\[LeftDoubleBracket]2\[RightDoubleBracket] = Evaluate[a\[LeftDoubleBracket]2\[RightDoubleBracket] > 0]; If[Mod[++c, f] == 0, AppendTo[hist, {x1, x2, x3}]]]]]], {t, 0,T, dt}]; hist]]; With[{r = 1/2}, With[{g = Graphics[Circle[#, r] & /@ #] & /@ seq}, Show[#, PlotRange -> BoundingBox[g], AspectRatio -> Automatic, ImageSize -> 500] & /@ g]];
 * I have a Quicktime animation of this, but I don't know how to convert it to any format (like animated gif) that Wikipedia will allow me to upload. Maybe Tardis can make something? --mglg(talk) 19:08, 7 June 2007 (UTC)
 * I just uploaded two quicktime animations to Yousendit: this depicts the real situation, this one shows (for your amusement) what happens if the coins remain stuck together even when the contact force goes negative. In the latter movie, coins 2 and 3 are allowed to pass through one another. --mglg(talk) 17:00, 8 June 2007 (UTC)


 * Thanks for your help on this problem. I have tried to clarify my thinking on the conservation of energy.  I have been out of the physics thing for a long time so, I may be way off on all this, but any further clarification would be greatly appreciated. Aepryus 02:55, 5 June 2007 (UTC)


 * I added another post to the archived RD page. --mglg(talk) 16:43, 5 June 2007 (UTC)
 * Thanks a lot for your help and patience in this. Aepryus 18:40, 5 June 2007 (UTC)

Ref desk answers
Just to say I appreciate your responses on the math and science desks. (Recently, the penny and probability ones.) iames 14:04, 25 June 2007 (UTC)


 * Thanks iames! --mglg(talk) 17:57, 25 June 2007 (UTC)

Signing Comments
Hi mglg! I noticed my signature on a comment I didn't make (twice!) in the last few days. It seems that one of these occurences is from your edit. I certainly don't mind you referencing my comments, but please use caution and avoid adding my signature to comments I didn't make! Thanks, Nimur 06:59, 13 July 2007 (UTC)

Radiative shield for cryogen dewar
(from Science ref-desk…) As a bonus, all the crinkly foil makes great confetti when the dewar explodes. I can try to find the video clip if you want. DMacks 22:50, 13 July 2007 (UTC)

Barnstar

 * Thanks FusionKnight! --mglg(talk) 02:08, 1 February 2008 (UTC)

How to put up a straight pole by pushing it at an angle
Thanks for your comment at Wikipedia talk:How to put up a straight pole by pushing it at an angle. Thought-provoking. I've replied there. --Dweller (talk) 18:05, 26 February 2008 (UTC)

Mugabe
Thanks for your patience in helping my ignorance, and for the context. It's really Mugabe who's ancient at 84-not out. : ) Julia Rossi (talk) 23:46, 7 April 2008 (UTC)

ArbCom elections are now open!
MediaWiki message delivery (talk) 13:58, 23 November 2015 (UTC)