<Volis> Planck_, I guess i am screwed by that grid resistor problem.
<Planck_> The infinite one?
<Volis> For equivalent resistance across the square in the grid.
<Planck_> Ah, the finite one then?
<Volis> Planck_, I took a 3x3 grid and looked for equivalent resistance across the square in the middle of it.
<Planck_> Oh, yes that gets messier.
<Volis> and with the way i'm trying(superposing two configurations), i get same answer for 3x3 and infinite lattice.
<Volis> Can i be wrong?
<Planck_> I don't actually recall. I'll see if I can dig up my notes
<Volis> For grid with resistors of resiatnce R, the equivalent resistance came out to be R/8.
<Planck_> Though if the superposition method you're using is the one I'm thinking, it requires an infinite grid to work (and isn't mathemtically solid)
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<Volis> Planck_, ok, so i can't distribute current as I/4 in all four connections from A? http://www.twiddla.com/546890
<Volis> Unless its infinite grid.
<Planck_> You can't use symmetry to conclude that they're equal in each direction
<Planck_> In an infinite grid you can (assuming they're well-defined at all, which is the mathematical issue there)
<Volis> So what happens to be the case is solving it for a finite grid is trickier than the infinite grid.
<Planck_> Yes
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<Planck_> At least, in the sense that there isn't a dirty trick that happens to give the right answer.
<Volis> How would be the current distribution here now, http://www.twiddla.com/546890 ?
<dli> Volis, all segment with resistance R, want to know resistance between A B?
<Planck_> One thing you can determine pretty quickly is that the diagonal corners have no current flow by symmetry
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<Volis> dli, Exactly.
<Volis> Planck_, Right.
<Volis> Oh, i think i can give it a try right now. Kirchoff's rule would work well.
<dli> Volis, 2*/(1/1+1/1+1/(5/2))=5/6 R
<Planck_> There's more symmetries to use too
<dli> Volis, no, need, just equivalent circult
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<Volis> dli, Where did that 5/2 came from?
<Volis> And how did you are they in parallel?
<dli> Volis, say, from B to bottom, then, to left, then, both (left and up), you get 2+(1/2)
<dli> Volis, if you apply A at potential U, and B and potential -U, then, all the diagonal points are at 0 potential, so, you treat all diagonal as grounded
<Volis> dli, Right
* Disconnected (Connection reset by peer).
[IDENTD] Servicing request from 208.71.169.36 [ouseasel,50199,6667]
* Volis already in use. Retrying with Volis...
* Volis already in use. Retrying with sollysid...
>nickserv< identify ****
* sollysid sets mode +i sollysid
-NickServ- sollysid is not a registered nickname.
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<Planck_> Once you apply symmetries you get the diagram in the lower-right
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<Planck_> Which you can collapse pretty quickly
>nickserv< ghost volis access
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-NickServ- volis has been ghosted.
>nickserv< identify ****
-NickServ- You are now identified for Hugo_Drax.
* You are now known as Volis
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<Volis> oh, ok
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<Volis> Planck_, Can you redraw it a bit a bit up, i can't see the whole diagram, not even in fullscreen.
<Planck_> I'm not very good with this site :)
<Planck_> I'm surprising there isn't a scroll function
<Volis> Same
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<dli> Volis, each red segment means equivalent or R
<dli> Volis, each red segment means equivalent of R
<dli> Volis, 3 in parallel, so, 1/(1/5 + 1/1 +1/5)=5/7 R
<Volis> dli, Oh so by symmetry we concluded current won't cross the junction P and Q?
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<Volis> And the red segement is the equivalent of the green segment i drew?
<dli> Volis, P and Q are at the same potential, if you connect P and Q, the circult is equivalent
<dli> Volis, so, yes, since 1= 1/(1/2+1/2)
<Volis> ah, I got it dli
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<Volis> Thank you so much dli and Planck_
<Volis> That was really helpful.
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<Volis> :)
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<kokorra> ummm
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<Windshield> The level will be very low so your eyes will be safe
<kokorra> i cant do the setup anymore
<Windshield> oh?
<Windshield> why
<kokorra> well i could if i needed to redo it
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<Windshield> I reckon you might see a faint red spot at about 20 degrees, but nevermind
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<Volis> Windshield, I solved that circuit. Surprisingly a finite circuit is more complex to solve.
<Windshield> I was thinking about it today
<Windshield> Don't tell me yet
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<Volis> Windshield, As a practice try a finite 3x3 lattice of sqaures, and find the equivalent resistance when connection is made from the opposite corners of the square in centre.
<Windshield> ok
<Volis> I did that today with much help from the good folks here. Really nice configuration.
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<Quack> help me lord
<Quack> oh lord of Antennae
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<benjamin_frankli> hey Quack, i'm here!
<Quack> oh hey whatsup
<Quack> how do i find the B field and E field intensity of a traveling wave at some distance from a monopole antenna
<Volis> I think you're wandering here for much time with this question.
<Quack> well, i did some simulations
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<Quack> but i don't understand derivation still
<Quack> just stay out of it Volis
<Quack> unless ur a genius
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<Volis> Quack, weren't you the one you wrote Quack something in that resistor grid we were discussing. hehe
<Quack> uhh ? no?
<Quack> someone else man....
<Quack> when is my White Knight gonna show up
<kokorra> i need help with my science fair project
<Quack> kokorra i'll only help u with your science fair project if you let me pick a different project
<kokorra> um
<Quack> j/k man
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<Quack> well whadduya need
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<Quack> i hope it's not really complicated
<Quack> HS students these days tend to do PhD level work -_-
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<artypig78> hey guys!
<artypig78> anyone up for a pretty cool question i need help with?
<artypig78> A lift ascends with an upward acceleration of 1.25ms^-2. At the instant its upward speed is 2.50ms^-1. A loose bolt falls from the ceiling of the lift, 2.75m from the floor.
<artypig78> Calculate the time of flight of the bolt from the ceiling to floor
<artypig78> can anyone point me in the right direction for this?
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-ChanServ- [##physics] Don't ask to ask... ask!
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<Bilz> anyone else here worked with htmlagilitypack
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<Bilz> seems so so buggy
<Bilz> oops
<Bilz> wrong channel
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<Volis> dli, Have a few minutes for help?
<dli> Volis, if it's not too complicated
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<Volis> dli, Same question, i thought about it for quite a while.
<Volis> Just one glitch, how we did it, The diagram here http://www.twiddla.com/545996
<Volis> green lines are 1 ohm resistors, red lines are 2 ohm resistors.
<Volis> How did we concluded that current won't actually 'cross' the place where i have drawn the pink line?
<Windshield> Volis, that last point is significant
<Windshield> I got that far
<Windshield> kind of
<Volis> I don't get it after this point.
<Windshield> You don't get what?
<Volis> Now the circuit on right, thats what you calculated the resistance of dli .
<Volis> I don't see how these two configuration of resistors are the same.
<Windshield> I think that pink line however, lies on the boundary if infinity.
<Windshield> *of*
<Volis> Windshield, The circuit i've drawn is equivalent to "a finite 3x3 lattice of sqaures, when connection is made from the opposite corners of the square in centre."
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<Volis> Ignore the pink line.
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<Volis> Windshield, Its not a infinite grid, do you still think it lies on the boundary?
<Windshield> Yeah
<dli> Volis, 1/(1/1+2/(4+1/(2+1/2))=11/16 ?
<Windshield> It is always possible to connect any point to any other point, no matter where you are.
<stiltwater> If I can jump in - on the right hand circuit, we can see that on the four middle points (between two green resistors), the voltage needs to be V/2, so since they're all at the same voltage, it doesn't matter if they're connected, so the circuits are the same
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<Volis> dli, Where does that 4 came in?
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<Volis> stiltwater, Can you mark the middle points by some color?
<stiltwater> Volis, I've tried to do so.
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<Volis> stiltwater, I see that V1 = V1 and so is the case with V2, how is V1 = V2?
<Volis> "the voltage needs to be V/2" How?
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<stiltwater> Volis, It's a symmetry property. Consider the voltage drop V0-V2. Now if you turn the circuit around, it looks exactly the same, but the voltage drop needs to be V2-Vf, because that's what the voltage drop of the right hand of the circuit (now on the left side after the flip) is. So V0-V2=V2-Vf, so V2 is the mean of the two voltages
<stiltwater> The key is that flipping the grid results in the same grid, so the left-hand drop and the right-hand drop need to be the same.
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<Volis> stiltwater, Are we using the fact that there are two axis of mirror symmetry(the lines i've drawn).
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<stiltwater> Volis, only the vertical axis is relevant. For example, if you delete the uppermost green path, it's still true that the drop from V0 to V2 must be the same as the drop from V2 to Vf.
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<stiltwater> For example, consider a circuit A(1ohm)B(2ohm)C. Now flip the circuit: A'(2ohm)B'(1ohm)C'. The drop AB must be the same as B'C'.
<stiltwater> Here, it so happens that the two circuits are the same, so saying the drop over AB is the same as the drop B'C' is the same as saying the drop over AB is the same as the drop over BC.
<Volis> stiltwater, How about the same configuration i've drawn at upper left corner?
<stiltwater> Volis, I'm not sure what I'm looking at (the yellow dots), or which question you're asking about it. Clarify please?
<Volis> Shouldn't i conclude that those two yellow points must also have same potential?
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<stiltwater> Volis, there's something a little weird about that - if you were actually building this circuit with wires, where would those points be? It seems like they're connected by a wire, so they definitely must be at the same potential.
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<stiltwater> Volis, it looks like the lower-right corner, yes? (Black lines wires, and resistors green)
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<Windshield> What you're looking for is not the boundary where current *no longer* flows, but the boundary beyond which *resistance is no higher*
<stiltwater> So now in the upper left, the top purple equals the top blue, and the bottom purple equals the bottom blue - you can see this by flipping the circuit over the vertical axis
<Volis> stiltwater, I'm actually tried quite hard to learn but still it was what i was saying, but those dots have same potential(current won't flow)?
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<Volis> stiltwater, I agree "the top purple equals the top blue, and the bottom purple equals the bottom blue" but the trouble is i think the those two purple dots too are at same potential.
<Volis> Wait, i think i'm getting this.
<stiltwater> They can't be the same, because the top one needs to be between V0 and V1, and the other between V2 and Vf
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<Volis> stiltwater, Okay so what i can conclude is that the configuration in upper left isn't the same as that in center but the configuration in lower right is same as that in the center circuit.
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<Volis> am i right? stiltwater
<stiltwater> Volis, that's right
<Volis> ok
<stiltwater> Now, be careful with 'is the same' though - if you add or remove bits, the two circuits that currently behave the same (as in, have the same currents and voltages everywhere) may begin behaving differently.
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<Volis> What are bits?
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<Macarisil> very small pieces of things.
<Volis> oh, bits and pieces.
<stiltwater> Volis, I mean, if you add or remove resistors or wires
<stiltwater> Yes, sorry.
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<Volis> stiltwater, So should we only consider the axis of symmetry which is in the direction of current flow?
<stiltwater> Volis, it depends what you want, of course. For example, by noting the other axis of symmetry, you get V1=V1 and V2=V2 that you had before.
<stiltwater> The axis of symmetry that flips input and output voltages lets you detect when some particular point is at the mean of those voltages, on the other hand.
<Volis> stiltwater, Yes, there were two axis of symmetry in the case, but we only considered the one which is along the line of current flow, the axis along line joining V0 and Vf.
<stiltwater> no, we considered flipping along the vertical axis
<stiltwater> This made the V2-Vf voltage drop line up with the V0-V2 voltage drop, so we got them as equal.
<Volis> Oh, ok.
<Volis> I just want to be as clear as possible, as the red lines are 2 ohm resistors. Had the configuration been like in the upper left corner, the equivalent resistance would be 3 ohm right?
<Windshield> current will increase from A to B forever
<stiltwater> In the upper left, there are a total of eight 1-ohm resistors?
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<Volis> stiltwater, As green path won't be conducting current.
<Volis> Ignore the dots for a moment.
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<stiltwater> Volis, current will flow from V0 to V1, for example, through the green path
<FreeNslaved> http://www.youtube.com/watch?v=qiQj0o8BOYw
<stiltwater> (and also through the red path)
<Volis> oh
<Volis> Thanks stiltwater