Stupid question of the day....

On Mon, 01 Aug 2005 08:28:25 -0500, John Fields Gave us:

THIS GUY IS THE TROLL, FOLKS!!!

Reply to
TokaMundo
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On Sun, 31 Jul 2005 10:32:46 -0500, "DBLEXPOSURE" Gave us:

You have never seen lightning? It is an entire stream of them, IN MOTION. The visible effects are enough. Ever seen "Ball Lightning"? I suppose ionized air is just the visible result of the electron passing, and it moves so fast as not to be visible with the human eye, so what we see is air turned plasma.

Ever had a jolt fire into you? The pain at the entry site tells one actual movement occurs. The jolt through the body and out the exit point also concur.

See the photo at alt.binaries.pictures.misc titled "strike"

Reply to
TokaMundo

there

Reply to
Autymn D. C.

Yes, I have seen lightning, not ball lightning but I have heard of it. "Air turned plasma", something like that. Oh yes, I have felt the effects of the electron.

Reply to
DBLEXPOSURE

Reply to
John Fields

Or they need to be in Alameda. After all, Keptin, dis is vere dey keep da nuclear Bessels, no?

:-)

Bob M.

Reply to
Bob Myers

Reply to
DBLEXPOSURE

Is there no end of subjects in which you are willing to demonstrate your ignorance? Io is NOT primarily heated by "the magnetic forces of Jupiter," but rather by its gravity - or, more specifically, the tidal forces resulting from the pull of Jupiter and the other major moons. Heating due to internal currents generated by the moon's passage through the Jovian magnetic field occurs, but is small in comparison to the tidal heating. As the specifics of the tidal heating depends on the particular relationship between Io, its parent (Jupiter) and the other satellites at the time, it is by no means even.

For details, see:

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As a result, Io does NOT have a particularly uniform thermal profile, either on its surface or vs. distance from the moon's core; see in particular:

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Bob M.

Reply to
Bob Myers

Hardly the same as seeing an electron itself, any more than observing the glow of the phosphor screen of a CRT is direct observation of an electron. Both, in fact, are examples of materials emitting light because of the absorption of energy from SOME source - "electron" is just a convenient name for the model we use, a particle which transports that energy.

The visible appearance of the lightning bolt is also NOT a case of directly observing the "flow" of electric charge or current.

OK - according to you, how fast are the particles in question - the "stream of them" - moving through the air?

Bob M.

Reply to
Bob Myers

No, you still don't see it. The heat in the center is conducted outwards through an imaginary cylindrical surface separating the innermost core from the next outer layer. And the heat from the *that* layer, plus the heat from the inner core is transmitted through the next concentric imaginary cylindrical surface separating that layer from the area.

The surface area of each imaginary cylindrical surface increases proportionally with the radius out from the center, but the heat that must be conducted through each surface increases proportional to the radius squared. So the temperature gradient across each imaginary cylindrical surface gets stronger and stronger. Thus the *temperature* across the cross-section is parabolic, even though the heat generation is flat/level.

This has *nothing* to do with the heat transfer from the outer most surface to it's surroundings. Heat removal from the outer surface by convection, conduction, or radiation will *not* change the shape of the interior temperature gradient.

daestrom

Reply to
daestrom

That's very, very nearly true. But for any reasonably insulated wire, the thermal conductivity of the insulation (including air) will be minute compared to that of copper, so internal temp gradients will be very low. To force a decent grad, you'd need to run a lot of current through a wire directly in contact with water or something.

John

Reply to
John Larkin

OK, i'ts kinda counter-intuitive to us seat-of-the-pants techies, because we were raised with very small wire. 17 mm is a pretty hefty chunk o' wire! ;-)

It seems I've learned something here. With great big huge fat wires, skin effect in significant even at 60 Hz. :-)

Thanks! Rich

Reply to
Rich Grise

Once again, STFW to the rescue:

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In aluminum at 60 Hz, the skin depth is about 2 cm. And there's the formula right there.

Cheers! Rich

Reply to
Rich Grise

On 4 Aug 2005 05:36:24 -0700, "Autymn D. C." Gave us:

Over there?

Reply to
TokaMundo

On Thu, 4 Aug 2005 09:50:04 -0500, "DBLEXPOSURE" Gave us:

10kV AC at 1mA 60Hz. Kerosene ignition unit made into a Jacob's Ladder.

I got caught on one of the rungs. :-]

Reply to
TokaMundo

On Thu, 04 Aug 2005 13:09:19 -0500, John Fields Gave us:

Your lame ass also accuses people of "self aggrandizement". I have yet to see one post from you where you don't do the same. Funny, since you're no more than a fat tub of lard.

Reply to
TokaMundo

On Thu, 04 Aug 2005 19:05:30 GMT, "Bob Myers" Gave us:

That would be "wessels". It's a wessel function.

Reply to
TokaMundo

On Thu, 04 Aug 2005 19:24:11 GMT, "Bob Myers" Gave us:

I meant gravitational, not magnetic.

Your lame insult at the onset of your post is just that... lame.

I have the laser disc. It is nearly twenty years old. Likely knew about it before you did.

Reply to
TokaMundo

On Thu, 04 Aug 2005 20:59:08 GMT, "daestrom" Gave us:

No. You don't see "IT".

No. THAT is for cylinders where the heat source is at the center. This heat source is throughout the medium.

That makes the rest of your supposition incorrect so... SNIP.

Reply to
TokaMundo

Reply to
John Fields

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