calbiterol

Sources for flame being a plasma:

From http://www.madsci.org/posts/archives/apr99/923460606.Ch.r.html

A flame is a plasma that contains,

among other things, a lot of hydroxyl (OH) free radicals.

From http://www2.abc.net.au/science/k2/stn-archive1/posts/topic18198.shtm

Fire in itself is an oxidisation reaction. This is the bit you can put out. The visible flame is a plasma (gas of ions and electrons), which is the bit that releases the light. Heat is not movement of molecules, but radiated energy. The heat is also radiated from the plasma. The plasma is also very hot (this is the movement of the particles, the energy coming from the reaction) and this energy can be trasfered to other objects which come into the plasma, and start up a 'fire' reaction in that object if it gets hot enough.

From http://www.osaka-gu.ac.jp/php/nakagawa/TRIZ/eTRIZ/electures/eSalamatovTextbook001122/eErrataQ&AChap8.htm

This constitutes the answer to the problem, because flame is a plasma, that is ionized gas, it serves as a perfect conductor of electricity.

You have to have heat, fuel, and oxygen (in any form, oxidizers included) to have a fire - I'm sure most if not all of you know that - and flames aren't the actual reaction, the chemical change in your fuel is (basic combustion). Transferring flames has nothing to do with blocking ignition, because flames are plasma. With Trans's saucepan thingy, if you did that and put a sparker (like from a butane lighter, without the butane) in it, there will be no fire, because there's no fuel, and as such, no flame. Note that in this case, the spark is supplying the heat/energy, not the flame below.

I think Callipigous got what the original question was asking (I don't know how he did it, but he did), and I think yourdadonapogostick/akcapr got the explanation for it correct. It has nothing to do with the ignition of the fire, but rather that plasma cannot permeate through a glass lattice.

I could be wrong, but I think that's how it was explained to me.

A much better possibility would be modifying a sword to have a "powersword" (I 'borrowed' that term from games workshop) that you can turn on/off but that permanantly has a blade. But it still wouldn't be anything like a lightsaber.

And don't even ask about it deflecting bullets.

That's the internal IP one, right?

If you talk crap to the indians 101 times in 4 days, you'll get an answer.

Think about it. No tool is needed.

Finally just got it by guessing. Oh well, if it works, it works. Don't have much of an idea for level nine, though.

[EDIT: Got through to level ten. Level nine, I just kind of... solved? I started just trying things in connection with the wording they give you, and... Well, surprisingly, something I tried existed... And had the name of the page... Was I supposed to use [hide]spiders/bots/crawlers?[/hide] If yeah, could someone explain the concept to me, cause I'd still like to learn! Thanks...]

Yep, but now I'm having problems with novice level 8. I thought I saw it a while ago, but I can't remember what it was.

Well, I wasn't going to even try if there wasn't a possibility of it being feasible, and if it is, I'll prolly track down a new xbox just for the reason. And I'm dropping the CD unit and not considering a screen unit - kind of like an uberpowerful PalmPilot.

Kind of, I was wondering about the feasibility of turning it into a handheld (with my own cooling mechanism that's totally overkill and shouldn't take up much space) and how much space the actual architecture takes up.

That's the idea, I don't want to void the warranties... Do those expire after a year?

Hmm. The architechture one on Novice (level 4) is... Annoying, and I'm stuck on it.

I just started using Firefox and it just keeps getting better and better. How could I ever have gotten by with plain, annoying old IE?!

That's actually quite useful.

Any ideas on number ten? I know exactly what I need to do, but not how to do it.

[hide]You have to intercept a cookie before they delete it, but I don't know how. Or use a packet sniffer - which I also don't know how to use.[/hide]

Or you could do it with telnet and get the same usefull info without downloading anything.

But how can that be? The function [math](z^2-1)/(z-1)[/math] and the function [math]z+1[/math] have two different domains. Of course, the two functions are[/b'] equal for z not equal to 1. But we were asked about when z=1.

To be honest, I guess I was so used to simplifying it that I never though of the implications given this particular case. I seem to remember my math teacher saying something about this a while ago, but seeing as I'm out of school, there's no way to ask him, except perhaps emailing him.

Check these out:

 

Indeterminate

Undefined

Can't say the definition of indeterminate is particularly easy to comprehend with the wording they use' date=' but I think I have the idea, and thanks for the links.

That would be quite a neat trick, considering that division has nothing to do with taking a limit. The former is an issue of algebra, the latter an issue of analysis.

I guess I meant that AFAIK the concept of division by zero being undefined has something to do with the limit being infinity. I.e., [math]\lim_{x\rightarrow 0}=\frac{k}{x}[/math] when k is any real number, equals infinity, but this is strictly the limit. I'll be the first to admit I don't know the relationship between the function's limit and its interpretation as "undefined." I wasn't referring to division as a whole, only to the divide by zero case.

I think that the function is "undefined"' date=' not "indeterminate" at z=1, but I would defer to a mathematician. The reason I say this is that in my (limited) experience with pure mathematics, I have not seen the term "indeterminate forms" used except to refer to the result of a limit.[/quote']

After finding out the definition of indeterminate (because of your links, I might add, and by the way, thanks) I would have to agree.

Well, I can't see any other way around it. I think I saw something on their forum about doing it "manually," but I don't think there's any difference. From the same point, in theory, someone could have done the whole thing in Telnet to begin with. Then it wouldn't require them to change anything, now would it?

Anytime.

I guess I was just under the impression that you could simplify something and have it still be equal, even under the condition stated here.

By the way, what's the difference between interdeterminate and undefined?

And technically, I could bring in limits when talking about dividing by zero. Would it be correct to say that the limit of the above function at f(1) is 2, while the value of the function at f(1) is undefined/interdeterminate?

Forgive me if I'm misunderstanding you, but then how do explain the process behind finding the derivative of sine?

Cancelling it out. Hence the "When we do limits, we're allowed to simplify - if you couldn't simplify, then you couldn't find the derivative of sin or cos, etc, and you can - so, we'll go about it this way." part. We're allowed to do this in this case, just like you are when finding the derivative of sin or cos.

Hitting enter twice submits it.

This one won't work, but run it anyway. Scan the message given for the reason behind the error. What does it say?

[hide]It should say something about a malformed header. The header is the thing you're faking, not just the referer. What do you think that means?[/hide]

Johnny5, there's another way of saying what you just said. Tom, I'm assuming you've been through limits before. Bear with me, I'm new to using [math]LaTeX[/math].

So, first off, through the definition of a limit, the limit of a whole number is the whole number.

Now...

[math]lim_{z\to 1}f(z) = lim_{z\to 1} \frac{z^2 - 1}{z - 1}[/math]

When we do limits, we're allowed to simplify - if you couldn't simplify, then you couldn't find the derivative of sin or cos, etc, and you can - so, we'll go about it this way.

[math]lim_{z\to 1}f(z) = lim_{z\to 1}\frac{(z - 1)(z + 1)}{z - 1}[/math]

Just as before, and so on:

[math]lim_{z\to 1}f(z) = lim_{z\to 1} \frac{(1)(z + 1)}{1}[/math]

Now, by definition of limits, you can break this apart into the following:

[math]lim_{z\to 1}f(z) =\frac{lim_{z\to 1}1 * lim_{z\to 1}(z + 1)}{lim_{z\to 1}1}[/math]

Which becomes

[math]lim_{z\to 1}f(z) = lim_{z\to 1}(z + 1)[/math]

...

[math]lim_{z\to 1}f(z) = (lim_{z\to 1}z) + 1[/math]

... and by definition of limits, we can just plug in 1 for z...

[math]lim_{z\to 1}f(z) = 1 + 1=2[/math]

AFAIK, you can do this.

It's a wonderful resource.

I don't think there's anything wrong with your syntax... You could try variatons of http://www.dievo.org or http://www.dievo.org (that's what worked for me) or dievo.org - just make sure your "host" field matches later on (you'll get there eventually, that's actually a hint if you think about it).

I use the telnet program CRT. I too have XP, so it should work with your computer as well, but I don't have a link. It's free, though - google for CRT telnet download or some such thing and it should appear. Are you logged in as admin? That might also affect the telnet'ing. You're going about this the right way, just play with it until it works.

I tried this twice and had it. If you think about it, 1 and 8 only have to worry about being next to 1 nuber each, so put them in the middle. All the rest, just put somewhere, going down starting with 7 and up starting with 2 (at the same time), making sure none touch. Or at least, that's how I did it.

This gave me

    [u]
| |[/u][u]4|6| |[/u][u]
|7|1|8|2|
| |3|5| |[/u]