Stupid question of the day....

there

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.

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

:-)

Bob M.

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:

Bob M.

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.

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

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

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

Once again, STFW to the rescue:

Cheers! Rich

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

Over there?

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

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

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.

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

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

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.

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.