Stupid question of the day....

  • Thread starter AllTel - Jim Hubbard
  • Start date
John Fields wrote:
Violent little bitch, eh?
Click to expand...
Everyone is at some time.

Why don't you leave a snippet of the post you're replying to intact
so that those of us who aren't mind-readers don't have to go
searching to find out whom you're threatening?
Click to expand...
It was a vocatively-ambiguous imperative, duh.

BTW, don't be surprised if the law comes knocking at your door...
Threats of physical violence are looked on very seriously these
days.
Click to expand...
The law must be rubbed out then.

-Aut
 
"TokaMundo" <TokaMundo@weedizgood.org> wrote in message
news:f9e3f1hfp39l9jf7d0mffj2s0ha6a45psh@4ax.com...
On Sun, 31 Jul 2005 10:32:46 -0500, "DBLEXPOSURE"
celstuff@hotmail.com> Gave us:

It is all speculation of course. I have never seen an electron, Have
you?

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"
Click to expand...
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.



---

The visible effects are enough

---



I suppose that depends on how curious your are.
 
On Wed, 3 Aug 2005 22:37:11 -0500, "DBLEXPOSURE"
<celstuff@hotmail.com> wrote:


Thanks Don!

I have been watching this thread for days. I like your answer. It's
allways give and take, no hard and fast rules that fit every situation.
And even better yet, You made your point without slamming anybody..
Click to expand...
---
Geez, you don't know when to quit, do you, Mr Passive-Aggressive?

--
John Fields
Professional Circuit Designer
 
"Don Kelly" <dhky@shaw.ca> wrote in message
news:dZfIe.110141$s54.2240@pd7tw2no...
However, anyone wanting to do
the math from scratch better be familiar with Bessel functions. Are you?
Click to expand...
Or they need to be in Alameda. After all, Keptin, dis is vere dey keep
da nuclear Bessels, no?

:)

Bob M.
 
"John Fields" <jfields@austininstruments.com> wrote in message
news:1bm4f19qpofmgv534g4tg25lu5oqc53mnh@4ax.com...
On Wed, 3 Aug 2005 22:37:11 -0500, "DBLEXPOSURE"
celstuff@hotmail.com> wrote:


Thanks Don!

I have been watching this thread for days. I like your answer. It's
allways give and take, no hard and fast rules that fit every situation.
And even better yet, You made your point without slamming anybody..

---
Geez, you don't know when to quit, do you, Mr Passive-Aggressive?

--
John Fields
Professional Circuit Designer
Click to expand...

----

Passive-Aggressive

Hmmmm......
 
"TokaMundo" <TokaMundo@weedizgood.org> wrote in message
news:ag83f199paf7od11jt0progsts34cc7nuj@4ax.com...

If it were modeled after the earth, with the heat source in the
center, I would agree. I feel, however, that it is more closely
modeled after Io, which is heated by the magnetic forces of Jupiter,
and more closely approximates an evenly heated body.
Click to expand...
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:

http://www.planetaryexploration.net/jupiter/io/tidal_heating.html

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:

http://www.lpi.usra.edu/meetings/lpsc2004/pdf/2048.pdf

http://www.lpi.usra.edu/meetings/lpsc2003/pdf/2030.pdf

http://astron.berkeley.edu/~fmarchis/Information/CV/Archive/preprintIcarus.pdf


Bob M.
 
"TokaMundo" <TokaMundo@weedizgood.org> wrote in message
news:f9e3f1hfp39l9jf7d0mffj2s0ha6a45psh@4ax.com...
On Sun, 31 Jul 2005 10:32:46 -0500, "DBLEXPOSURE"
celstuff@hotmail.com> Gave us:

It is all speculation of course. I have never seen an electron, Have
you?

You have never seen lightning?
Click to expand...
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.


It is an entire stream of them, IN
MOTION.
Click to expand...
OK - according to you, how fast are the particles in question -
the "stream of them" - moving through the air?


Bob M.
 
"TokaMundo" <TokaMundo@weedizgood.org> wrote in message
news:ag83f199paf7od11jt0progsts34cc7nuj@4ax.com...
On Wed, 03 Aug 2005 15:54:26 -0500, John Fields
jfields@austininstruments.com> Gave us:

---
Sorry, Charlie, there'll still be a temperature gradient across the
diameter of the wire. There has to be, since the surface of the
wire will be radiating heat and being cooled by convection.
---

If it were modeled after the earth, with the heat source in the
center, I would agree. I feel, however, that it is more closely
modeled after Io, which is heated by the magnetic forces of Jupiter,
and more closely approximates an evenly heated body.

In the wire, since the heat is generated throughout the medium via
current flow, even from low currents on up to my cherry red scenario
would show the wire at the same temp from center to outer surface.
Click to expand...
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
 
On Thu, 04 Aug 2005 20:59:08 GMT, "daestrom"
<daestrom@NO_SPAM_HEREtwcny.rr.com> wrote:

"TokaMundo" <TokaMundo@weedizgood.org> wrote in message
news:ag83f199paf7od11jt0progsts34cc7nuj@4ax.com...
On Wed, 03 Aug 2005 15:54:26 -0500, John Fields
jfields@austininstruments.com> Gave us:

---
Sorry, Charlie, there'll still be a temperature gradient across the
diameter of the wire. There has to be, since the surface of the
wire will be radiating heat and being cooled by convection.
---

If it were modeled after the earth, with the heat source in the
center, I would agree. I feel, however, that it is more closely
modeled after Io, which is heated by the magnetic forces of Jupiter,
and more closely approximates an evenly heated body.

In the wire, since the heat is generated throughout the medium via
current flow, even from low currents on up to my cherry red scenario
would show the wire at the same temp from center to outer surface.


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.
Click to expand...
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
 
On Tue, 02 Aug 2005 12:27:47 -0700, John Larkin wrote:
On Tue, 02 Aug 2005 18:28:32 GMT, TokaMundo <TokaMundo@weedizgood.org
On Tue, 02 Aug 2005 10:54:26 -0700, John Larkin
Before any difference could even be noted, the wired diameter would
have to be over 16 mm.

Not so. At 0.85 cm depth, current density is down to 1/e (ie, only
0.37 of) the surface density. That's pretty significant.

0.85 cm is pretty thick. 8.5 mm in fact. Double that to get 17mm.

Unless the wire is larger than 17mm at 60Hz, the entire wire will
carry current. VERY simple math.

Current begins to fall off monotonically from the very surface for any
wire size at any AC frequency. There's no hard "skin boundary", and
the 1/e density is just a handy if arbitrary measurement point.

I don't see why this needs arguing over. In a given situation, you
just calculate the effects and decide how they affect things.
Sometimes a 200% increase in resistance doesn't matter, and sometimes
a 1% increase does. But skin effect does often matter in real
situations at 60 Hz, and shouldn't be always/automatically discounted.
Click to expand...
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
 
On Wed, 03 Aug 2005 18:31:32 +0200, Alexander wrote:

"TokaMundo" <TokaMundo@weedizgood.org> schreef in bericht
news:ri70f1hhphqbr56v86alsaafq1feldqi8r@4ax.com...
On Wed, 03 Aug 2005 01:24:11 GMT, Repeating Rifle
salmonegg@sbcglobal.net> Gave us:

Some people say that there is no such thing as a stupid question.
Obviously
there seems to be no shortage of stupid answers.

Bill

Yours certainly contributed absofuckinglutely nothing, and would
certainly fall into the "stupid answer" category.

You remind me of a Firesign Theatre quote:

"Who wona second world war.. you so smart?"

Perhaps, if you are so informed, you should try giving an answer
that actually has facts in it that are in sync with the topic of the
thread, not merely its title.

Someone once tolf me something, I believe it wass: "Don't feed the trolls",
it seems to me that they are having a ball with food in abundance.
Click to expand...
It's a game to them. I don't know if it has a name, but I'd call it
something like troll-baiting.

I just usually ignore those symbiotic threads. ;-)

Cheers!
Rich
 
On Tue, 02 Aug 2005 18:40:17 +0000, Bob Myers wrote:
"TokaMundo" <TokaMundo@weedizgood.org> wrote in message
news:i2dve15rqf7183m844bjidvu0v98rn9cjf@4ax.com...
On Tue, 02 Aug 2005 10:54:26 -0700, John Larkin
jjlarkin@highNOTlandTHIStechnologyPART.com> Gave us:

Before any difference could even be noted, the wired diameter would
have to be over 16 mm.

Not so. At 0.85 cm depth, current density is down to 1/e (ie, only
0.37 of) the surface density. That's pretty significant.


0.85cm is 8.5 mm. That means that the wire has to be bigger than
that number as a radius before the current flow anywhere else besides
the entire wire.

Wrong again. You seem to think that the current is uniform
down to the "skin depth," and THEN it somehow starts to
fall off. As John already pointed out, with seemingly unwarranted
patience, that ain't so.

Once again: "do the math." And this time, go beyond just using
the skin-depth calculator on your favorite web site, and actually
figure out what the EFFECTS would be (in terms of resistive
loss, heating, whatever) of the skin depth at 60 Hz in a conductor
otherwise seemingly-properly-sized for the 800A service that
John mentioned as an example.

You might be surprised by the result.

Once again, STFW to the rescue:
Click to expand...
http://www.physics.ubc.ca/~mattison/Courses/Phys454/lecture17.pdf

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

Cheers!
Rich
 
On 4 Aug 2005 05:36:24 -0700, "Autymn D. C." <lysdexia@sbcglobal.net>
Gave us:

John Fields wrote:
And where are you, my dear?

there
Click to expand...
Over there?
 
On Thu, 4 Aug 2005 09:50:04 -0500, "DBLEXPOSURE"
<celstuff@hotmail.com> Gave us:

"TokaMundo" <TokaMundo@weedizgood.org> wrote in message
news:f9e3f1hfp39l9jf7d0mffj2s0ha6a45psh@4ax.com...
On Sun, 31 Jul 2005 10:32:46 -0500, "DBLEXPOSURE"
celstuff@hotmail.com> Gave us:

It is all speculation of course. I have never seen an electron, Have
you?

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"

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.


The visible effects are enough


I suppose that depends on how curious your are.
Click to expand...
10kV AC at 1mA 60Hz. Kerosene ignition unit made into a Jacob's
Ladder.

I got caught on one of the rungs. :-]
 
On Thu, 04 Aug 2005 13:09:19 -0500, John Fields
<jfields@austininstruments.com> Gave us:

On Wed, 3 Aug 2005 22:37:11 -0500, "DBLEXPOSURE"
celstuff@hotmail.com> wrote:


Thanks Don!

I have been watching this thread for days. I like your answer. It's
allways give and take, no hard and fast rules that fit every situation.
And even better yet, You made your point without slamming anybody..

---
Geez, you don't know when to quit, do you, Mr Passive-Aggressive?

--
John Fields
Professional Self Aggrandizer
Click to expand...
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"
<nospamplease@address.invalid> Gave us:

"Don Kelly" <dhky@shaw.ca> wrote in message
news:dZfIe.110141$s54.2240@pd7tw2no...
However, anyone wanting to do
the math from scratch better be familiar with Bessel functions. Are you?

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

:)

Bob M.


That would be "wessels". It's a wessel function.
Click to expand...
 
On Thu, 04 Aug 2005 19:24:11 GMT, "Bob Myers"
<nospamplease@address.invalid> Gave us:

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
Click to expand...
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"
<daestrom@NO_SPAM_HEREtwcny.rr.com> Gave us:

No, you still don't see it.
Click to expand...
No. You 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.
Click to expand...
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.
 
On Fri, 05 Aug 2005 04:36:15 GMT, TokaMundo
<TokaMundo@weedizgood.org> wrote:


John Fields
Professional Self Aggrandizer

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.
Click to expand...
---
LOL, if I were to write the single line: "Tokamundo is a good guy."
and post it, you'd critcise it in your boring, predictable way.

Probably by calling me a liar.

--
John Fields
Professional Circuit Designer
 
You must log in or register to reply here.

Welcome to EDABoard.com

Sponsor

Back
Top