Driver to drive?

[Spehro Pefhany]

On Sun, 05 Jul 2015 21:10:04 -0700, the renowned Jeff Liebermann
<jeffl@cruzio.com> wrote:

So, if 100% of the 1000 watts of power was dissipated by the driven
elements of the antenna, it would melt in 2 seconds. However, since
it's unlikely all 1000 watts actually made it to the antenna, and that
it's likely that the antenna actually radiated much of the power, the
actual meltdown times will be much longer. Still, it looks quite
possible to melt the elements.

Hey, some devices don't just melt, they **vaporize** their antennas.

--sp


--
Best regards,
Spehro Pefhany
Amazon link for AoE 3rd Edition: http://tinyurl.com/ntrpwu8
Microchip link for 2015 Masters in Phoenix: http://tinyurl.com/l7g2k48

 

[M Philbrook]

In article <apvlpadpg2tu6t0270i5moipkametuqj9j@4ax.com>,
speffSNIP@interlogDOTyou.knowwhat says...

On Sun, 05 Jul 2015 21:10:04 -0700, the renowned Jeff Liebermann
jeffl@cruzio.com> wrote:



So, if 100% of the 1000 watts of power was dissipated by the driven
elements of the antenna, it would melt in 2 seconds. However, since
it's unlikely all 1000 watts actually made it to the antenna, and that
it's likely that the antenna actually radiated much of the power, the
actual meltdown times will be much longer. Still, it looks quite
possible to melt the elements.

Hey, some devices don't just melt, they **vaporize** their antennas.

--sp

Sure, talk to any CB'er/Chicken_Choker and they'll tell you that!

Jamie

 

[Tim Wescott]

On Sun, 05 Jul 2015 21:10:04 -0700, Jeff Liebermann wrote:

On Sun, 05 Jul 2015 21:37:58 -0500, John S <Sophi.2@invalid.org> wrote:

On 7/5/2015 11:00 AM, Clifford Heath wrote:
On 06/07/15 07:59, makolber@yahoo.com wrote:
For sake of discussion, assume the tx line is 50 ohms.
If the tx output is an ideal 50 ohm match, any reflection from the
ant will be absorbed by the tx.
If the tx however, instead present a conjugate match to whatever z
the line present it,
Any reflection from the ant will be completely re- reflected by the
tx back to the ant.
The conjugate matched case will provide more radiated power.

At the cost, as Tim said, of possibly excessive voltages or currents.

A friend melted the aluminium tube driven elements on his six-metre
Yagi antenna with a 1000W HF amp. The amp, cable and connectors were
fine... the solid tube *elements* melted from excessive current.

How much current (power, temperature) does it take to melt a (solid?)
aluminium (sic) tube (rod?) driven element? An inquiring mind wants to
know.

Ok, let's see if my thermodynamics is still intact. Corrections welcome
as my math usually sucks and I always make at least one big mistake.


I'll assume a 3 element 6 meter beam. If it had more elements, this
would be a good time to disclose the details.
http://www.cushcraftamateur.com/Product.php?productid=A50-3S
3.2 kg for the entire antenna. I would guess(tm) that the boom is about
1.0 Kg, leaving 2.2 Kg of aluminum for the elements.

6061-T6 melts at about 580C = 853K I'll assume an ambient temp of 27C =
300K

Heat absorbed before melting:
Specific heat capacity of Al:
C = 0.897 J/(gm*K)
Heat_1 = 0.897 * m * change_in_temp Heat_1 = 0.897 * 2,200gm * 553K =
1100*10^3 KiloJoules

Heat of fusion for Al:
Lf = 0.395 kJ/g
Heat_2 = m * Lf = 2,200gm * 0.395 = 869 KiloJoules

Total heat required = 1100*10^3 + 869*10^3 = 1969 kJ I'll roundoff to
2000 KiloJoules = 2 kilowatt-seconds

So, if 100% of the 1000 watts of power was dissipated by the driven
elements of the antenna, it would melt in 2 seconds. However, since
it's unlikely all 1000 watts actually made it to the antenna, and that
it's likely that the antenna actually radiated much of the power, the
actual meltdown times will be much longer. Still, it looks quite
possible to melt the elements.

It's harder than you're making out, at least if the heating is at just
one spot.

If the heating is in just one spot then whether it melts is a function of
power and time, because as the spot is receiving heat, the aluminum
surrounding the spot is conducting it away. Aluminum is pretty
conductive, but if the power is high enough then I would expect that
you'd only actually melt a section of rod that's roughly a diameter or
two.

For an antenna in still air I would expect that whether it melts or not
is ultimately more a function of power than of time, because the tips of
the antenna probably aren't ever going to get all that warm; you'd need
to do your calculations assuming that the antenna is always managing to
sink heat into the surrounding air.

As far as actually knowing how much power that would take -- I have no
clue, but having it happen at a current node in an antenna being excited
by a 1000W transmitter certainly seems plausible.

--

Tim Wescott
Wescott Design Services
http://www.wescottdesign.com

 

[Jasen Betts]

On 2015-07-06, Tim Wescott <seemywebsite@myfooter.really> wrote:

On Sun, 05 Jul 2015 21:10:04 -0700, Jeff Liebermann wrote:

On Sun, 05 Jul 2015 21:37:58 -0500, John S <Sophi.2@invalid.org> wrote:

On 7/5/2015 11:00 AM, Clifford Heath wrote:

At the cost, as Tim said, of possibly excessive voltages or currents.

A friend melted the aluminium tube driven elements on his six-metre
Yagi antenna with a 1000W HF amp. The amp, cable and connectors were
fine... the solid tube *elements* melted from excessive current.

How much current (power, temperature) does it take to melt a (solid?)
aluminium (sic) tube (rod?) driven element? An inquiring mind wants to
know.

Ok, let's see if my thermodynamics is still intact. Corrections welcome
as my math usually sucks and I always make at least one big mistake.

It's harder than you're making out, at least if the heating is at just
one spot.

If the heating is in just one spot then whether it melts is a function of
power and time, because as the spot is receiving heat, the aluminum
surrounding the spot is conducting it away. Aluminum is pretty
conductive, but if the power is high enough then I would expect that
you'd only actually melt a section of rod that's roughly a diameter or
two.

For an antenna in still air I would expect that whether it melts or not
is ultimately more a function of power than of time, because the tips of
the antenna probably aren't ever going to get all that warm; you'd need
to do your calculations assuming that the antenna is always managing to
sink heat into the surrounding air.

As far as actually knowing how much power that would take -- I have no
clue, but having it happen at a current node in an antenna being excited
by a 1000W transmitter certainly seems plausible.

For the same rigidity solid rod is worse than hollow tube,
as by skin effect the slimmer rod has a worse resistance and so
produces more heat, and having less surface will lose less of it.







--
umop apisdn

 

[John S]

On 7/6/2015 2:33 PM, rickman wrote:

On 7/6/2015 12:10 AM, Jeff Liebermann wrote:
On Sun, 05 Jul 2015 21:37:58 -0500, John S <Sophi.2@invalid.org
wrote:

On 7/5/2015 11:00 AM, Clifford Heath wrote:
On 06/07/15 07:59, makolber@yahoo.com wrote:
For sake of discussion, assume the tx line is 50 ohms.
If the tx output is an ideal 50 ohm match, any reflection from the ant
will be absorbed by the tx.
If the tx however, instead present a conjugate match to whatever z the
line present it,
Any reflection from the ant will be completely re- reflected by the tx
back to the ant.
The conjugate matched case will provide more radiated power.

At the cost, as Tim said, of possibly excessive voltages or currents.

A friend melted the aluminium tube driven elements on his six-metre
Yagi
antenna with a 1000W HF amp. The amp, cable and connectors were fine...
the solid tube *elements* melted from excessive current.

How much current (power, temperature) does it take to melt a (solid?)
aluminium (sic) tube (rod?) driven element? An inquiring mind wants
to know.

Ok, let's see if my thermodynamics is still intact. Corrections
welcome as my math usually sucks and I always make at least one big
mistake.


I'll assume a 3 element 6 meter beam. If it had more elements, this
would be a good time to disclose the details.
http://www.cushcraftamateur.com/Product.php?productid=A50-3S
3.2 kg for the entire antenna. I would guess(tm) that the boom is
about 1.0 Kg, leaving 2.2 Kg of aluminum for the elements.

6061-T6 melts at about 580C = 853K
I'll assume an ambient temp of 27C = 300K

Heat absorbed before melting:
Specific heat capacity of Al:
C = 0.897 J/(gm*K)
Heat_1 = 0.897 * m * change_in_temp
Heat_1 = 0.897 * 2,200gm * 553K = 1100*10^3 KiloJoules

Please check your math. I think you made an error of 10^3.


Heat of fusion for Al:
Lf = 0.395 kJ/g
Heat_2 = m * Lf = 2,200gm * 0.395 = 869 KiloJoules

Total heat required = 1100*10^3 + 869*10^3 = 1969 kJ
I'll roundoff to 2000 KiloJoules = 2 kilowatt-seconds

But I guess you corrected it here.

What's going on here? Since one joule is one watt*second, is not a
kilojoule equal to 1000 watt*seconds? If so, then 2000 kilojoules is 2
million watt*seconds. Yes? No? What?

 

[John S]

On 7/5/2015 11:00 AM, Clifford Heath wrote:

On 06/07/15 07:59, makolber@yahoo.com wrote:
For sake of discussion, assume the tx line is 50 ohms.
If the tx output is an ideal 50 ohm match, any reflection from the ant
will be absorbed by the tx.
If the tx however, instead present a conjugate match to whatever z the
line present it,
Any reflection from the ant will be completely re- reflected by the tx
back to the ant.
The conjugate matched case will provide more radiated power.

At the cost, as Tim said, of possibly excessive voltages or currents.

A friend melted the aluminium tube driven elements on his six-metre Yagi
antenna with a 1000W HF amp. The amp, cable and connectors were fine...
the solid tube *elements* melted from excessive current.

A properly designed Yagi usually presents a 50 ohm match. To start with,
then, the maximum current into the feed point will not exceed 4.5A at
1kW. However, the gamma match is matching some resistance lower than
this at the center of the driven element. I will assume, for a 3-element
array that it is something like half that or maybe 25 ohms. So, the
current might be 9A. That doesn't seem to be very much for an aluminum
antenna element.

If the antenna is beginning to have some higher resistance, I think the
current will reduce because there are difficulties in driving it.

 

[rickman]

On 7/7/2015 5:55 AM, John S wrote:

On 7/6/2015 2:33 PM, rickman wrote:
On 7/6/2015 12:10 AM, Jeff Liebermann wrote:
On Sun, 05 Jul 2015 21:37:58 -0500, John S <Sophi.2@invalid.org
wrote:

On 7/5/2015 11:00 AM, Clifford Heath wrote:
On 06/07/15 07:59, makolber@yahoo.com wrote:
For sake of discussion, assume the tx line is 50 ohms.
If the tx output is an ideal 50 ohm match, any reflection from the
ant
will be absorbed by the tx.
If the tx however, instead present a conjugate match to whatever z
the
line present it,
Any reflection from the ant will be completely re- reflected by
the tx
back to the ant.
The conjugate matched case will provide more radiated power.

At the cost, as Tim said, of possibly excessive voltages or currents.

A friend melted the aluminium tube driven elements on his six-metre
Yagi
antenna with a 1000W HF amp. The amp, cable and connectors were
fine...
the solid tube *elements* melted from excessive current.

How much current (power, temperature) does it take to melt a (solid?)
aluminium (sic) tube (rod?) driven element? An inquiring mind wants
to know.

Ok, let's see if my thermodynamics is still intact. Corrections
welcome as my math usually sucks and I always make at least one big
mistake.


I'll assume a 3 element 6 meter beam. If it had more elements, this
would be a good time to disclose the details.
http://www.cushcraftamateur.com/Product.php?productid=A50-3S
3.2 kg for the entire antenna. I would guess(tm) that the boom is
about 1.0 Kg, leaving 2.2 Kg of aluminum for the elements.

6061-T6 melts at about 580C = 853K
I'll assume an ambient temp of 27C = 300K

Heat absorbed before melting:
Specific heat capacity of Al:
C = 0.897 J/(gm*K)
Heat_1 = 0.897 * m * change_in_temp
Heat_1 = 0.897 * 2,200gm * 553K = 1100*10^3 KiloJoules

Please check your math. I think you made an error of 10^3.


Heat of fusion for Al:
Lf = 0.395 kJ/g
Heat_2 = m * Lf = 2,200gm * 0.395 = 869 KiloJoules

Total heat required = 1100*10^3 + 869*10^3 = 1969 kJ
I'll roundoff to 2000 KiloJoules = 2 kilowatt-seconds

But I guess you corrected it here.

What's going on here? Since one joule is one watt*second, is not a
kilojoule equal to 1000 watt*seconds? If so, then 2000 kilojoules is 2
million watt*seconds. Yes? No? What?

From above...

Heat_1 = 0.897 * 2,200gm * 553K = 1100*10^3 KiloJoules

Is this right???

--

Rick

 

[John S]

On 7/7/2015 8:40 AM, rickman wrote:

On 7/7/2015 5:55 AM, John S wrote:
On 7/6/2015 2:33 PM, rickman wrote:
On 7/6/2015 12:10 AM, Jeff Liebermann wrote:
On Sun, 05 Jul 2015 21:37:58 -0500, John S <Sophi.2@invalid.org
wrote:

On 7/5/2015 11:00 AM, Clifford Heath wrote:
On 06/07/15 07:59, makolber@yahoo.com wrote:
For sake of discussion, assume the tx line is 50 ohms.
If the tx output is an ideal 50 ohm match, any reflection from the
ant
will be absorbed by the tx.
If the tx however, instead present a conjugate match to whatever z
the
line present it,
Any reflection from the ant will be completely re- reflected by
the tx
back to the ant.
The conjugate matched case will provide more radiated power.

At the cost, as Tim said, of possibly excessive voltages or currents.

A friend melted the aluminium tube driven elements on his six-metre
Yagi
antenna with a 1000W HF amp. The amp, cable and connectors were
fine...
the solid tube *elements* melted from excessive current.

How much current (power, temperature) does it take to melt a (solid?)
aluminium (sic) tube (rod?) driven element? An inquiring mind wants
to know.

Ok, let's see if my thermodynamics is still intact. Corrections
welcome as my math usually sucks and I always make at least one big
mistake.


I'll assume a 3 element 6 meter beam. If it had more elements, this
would be a good time to disclose the details.
http://www.cushcraftamateur.com/Product.php?productid=A50-3S
3.2 kg for the entire antenna. I would guess(tm) that the boom is
about 1.0 Kg, leaving 2.2 Kg of aluminum for the elements.

6061-T6 melts at about 580C = 853K
I'll assume an ambient temp of 27C = 300K

Heat absorbed before melting:
Specific heat capacity of Al:
C = 0.897 J/(gm*K)
Heat_1 = 0.897 * m * change_in_temp
Heat_1 = 0.897 * 2,200gm * 553K = 1100*10^3 KiloJoules

Please check your math. I think you made an error of 10^3.


Heat of fusion for Al:
Lf = 0.395 kJ/g
Heat_2 = m * Lf = 2,200gm * 0.395 = 869 KiloJoules

Total heat required = 1100*10^3 + 869*10^3 = 1969 kJ
I'll roundoff to 2000 KiloJoules = 2 kilowatt-seconds

But I guess you corrected it here.

What's going on here? Since one joule is one watt*second, is not a
kilojoule equal to 1000 watt*seconds? If so, then 2000 kilojoules is 2
million watt*seconds. Yes? No? What?

From above...

Heat_1 = 0.897 * 2,200gm * 553K = 1100*10^3 KiloJoules

Is this right???

Why are you asking ME? You should be asking the author of the assertions.

 

[Jeff Liebermann]

On Mon, 06 Jul 2015 09:29:27 -0500, John S <Sophi.2@invalid.org>
wrote:

On 7/6/2015 8:58 AM, John S wrote:
On 7/5/2015 11:10 PM, Jeff Liebermann wrote:

Total heat required = 1100*10^3 + 869*10^3 = 1969 kJ
I'll roundoff to 2000 KiloJoules = 2 kilowatt-seconds

I'm a bit confused. I thought one joule was one watt*second. If so,
wouldn't 2000 kilojoules be 2000 kilowatt*seconds or 2 million
watt*seconds?

You're not confused. I screwed up. It's 2 megawatt-seconds to melt
the aluminum elements. With 1,000 watts input, it will take
2*10^6 watt-seconds / 3600 hrs/sec / 1000 watts = 56 min
to melt the elements.

Thanks for the correction and for ruining my day.

Also, the flaw in the ointment is that this method of estimating implies
that the aluminum will melt with one watt applied for 2 million seconds.
I don't think so.

If the system was adiabatic (thermally insulated) that's exactly what
will happen. Of course, nobody wraps their 6 meter antenna in
fiberglass insulation, so that's an unlikely situation. Still, if you
insulate the antenna and somehow block all forms of heat radiation and
conduction, it should theoretically melt the elements with 1 watt in
555 hrs. Well, somewhat longer because it would also need to melt the
additional mass of the boom and matching section.

When I was about 16, I decided to see if I really could burn my finger
with only 1 watt. I put a resistor inside a vacuum insulated glass
Thermos bottle, applied power, and watched the temperature rise on an
oven thermometer shoved through the stopper. After about 10 hrs, the
experiment was deemed over when the plastic stopper melted. Learn by
Destroying(tm).



--
Jeff Liebermann jeffl@cruzio.com
150 Felker St #D http://www.LearnByDestroying.com
Santa Cruz CA 95060 http://802.11junk.com
Skype: JeffLiebermann AE6KS 831-336-2558

 

[Jeff Liebermann]

On Mon, 06 Jul 2015 17:57:49 -0500, Tim Wescott
<seemywebsite@myfooter.really> wrote:

How much current (power, temperature) does it take to melt a (solid?)
aluminium (sic) tube (rod?) driven element? An inquiring mind wants to
know.

It's harder than you're making out, at least if the heating is at just
one spot.

If the heating is in just one spot then whether it melts is a function of
power and time, because as the spot is receiving heat, the aluminum
surrounding the spot is conducting it away. Aluminum is pretty
conductive, but if the power is high enough then I would expect that
you'd only actually melt a section of rod that's roughly a diameter or
two.

Agreed. I calculated how many Joules would be required to melt the
antenna elements into a puddle of molten metal. Just having the
elements sag because the aluminum softened at the boom will take much
less energy. I'll do the math if necessary, but with my track record
of arithmetic errors, it will probably be wrong.

For an antenna in still air I would expect that whether it melts or not
is ultimately more a function of power than of time, because the tips of
the antenna probably aren't ever going to get all that warm; you'd need
to do your calculations assuming that the antenna is always managing to
sink heat into the surrounding air.

I could probably do that if I assume the sagging elements were of
uniform cross section over their length, still air, and thermally
insulated from the boom. It would then be a simple thermal resistance
calculation as in a common heat sink.

As far as actually knowing how much power that would take -- I have no
clue, but having it happen at a current node in an antenna being excited
by a 1000W transmitter certainly seems plausible.

If there were hot spots in the elements at the boom, it's unlikely
that they would have occurred simultaneously and equally for all the
elements. Most likely is the driven element sagging, with the other
elements remaining intact.

The only time I've seen aluminum melt in an antenna is when there's
arcing.

Drivel:
<http://www.microwaves101.com/encyclopedias/microwave-mortuary>
Also try links to previous years at bottom of page.

--
Jeff Liebermann jeffl@cruzio.com
150 Felker St #D http://www.LearnByDestroying.com
Santa Cruz CA 95060 http://802.11junk.com
Skype: JeffLiebermann AE6KS 831-336-2558

 

On Tue, 07 Jul 2015 05:39:45 -0500, John S <Sophi.2@invalid.org>
wrote:

On 7/5/2015 11:00 AM, Clifford Heath wrote:
On 06/07/15 07:59, makolber@yahoo.com wrote:
For sake of discussion, assume the tx line is 50 ohms.
If the tx output is an ideal 50 ohm match, any reflection from the ant
will be absorbed by the tx.
If the tx however, instead present a conjugate match to whatever z the
line present it,
Any reflection from the ant will be completely re- reflected by the tx
back to the ant.
The conjugate matched case will provide more radiated power.

At the cost, as Tim said, of possibly excessive voltages or currents.

A friend melted the aluminium tube driven elements on his six-metre Yagi
antenna with a 1000W HF amp. The amp, cable and connectors were fine...
the solid tube *elements* melted from excessive current.

A properly designed Yagi usually presents a 50 ohm match. To start with,
then, the maximum current into the feed point will not exceed 4.5A at
1kW. However, the gamma match is matching some resistance lower than
this at the center of the driven element. I will assume, for a 3-element
array that it is something like half that or maybe 25 ohms. So, the
current might be 9A. That doesn't seem to be very much for an aluminum
antenna element.

An alternative way of looking at the problem is to use a slightly
below 1/2 lambda tubing as a dipole, fed by a 4:1 balun from 50 ohms
unbal feed. With Y-feed (200 ohm balanced) you move the connection
point on the dipole sufficiently far from the boom to find the 200 ohm
points. Clearly, closer to the boom, ther are points with impedance
level significantly below 200 ohms and possibly even below 10 ohms.

Anyway, I haven't seen the OP, but a 15 m vertical is 1/4 lamda at 5
MHz. However, if the frequency is well below this, the real issue is
the effectiveness of the ground plane. At LF and VLF such small
antenna might have an efficiency between 1 % and 0.1 %.

 

[rickman]

On 7/7/2015 12:38 PM, John S wrote:

On 7/7/2015 8:40 AM, rickman wrote:
On 7/7/2015 5:55 AM, John S wrote:
On 7/6/2015 2:33 PM, rickman wrote:
On 7/6/2015 12:10 AM, Jeff Liebermann wrote:
On Sun, 05 Jul 2015 21:37:58 -0500, John S <Sophi.2@invalid.org
wrote:

On 7/5/2015 11:00 AM, Clifford Heath wrote:
On 06/07/15 07:59, makolber@yahoo.com wrote:
For sake of discussion, assume the tx line is 50 ohms.
If the tx output is an ideal 50 ohm match, any reflection from the
ant
will be absorbed by the tx.
If the tx however, instead present a conjugate match to whatever z
the
line present it,
Any reflection from the ant will be completely re- reflected by
the tx
back to the ant.
The conjugate matched case will provide more radiated power.

At the cost, as Tim said, of possibly excessive voltages or
currents.

A friend melted the aluminium tube driven elements on his six-metre
Yagi
antenna with a 1000W HF amp. The amp, cable and connectors were
fine...
the solid tube *elements* melted from excessive current.

How much current (power, temperature) does it take to melt a (solid?)
aluminium (sic) tube (rod?) driven element? An inquiring mind wants
to know.

Ok, let's see if my thermodynamics is still intact. Corrections
welcome as my math usually sucks and I always make at least one big
mistake.


I'll assume a 3 element 6 meter beam. If it had more elements, this
would be a good time to disclose the details.
http://www.cushcraftamateur.com/Product.php?productid=A50-3S
3.2 kg for the entire antenna. I would guess(tm) that the boom is
about 1.0 Kg, leaving 2.2 Kg of aluminum for the elements.

6061-T6 melts at about 580C = 853K
I'll assume an ambient temp of 27C = 300K

Heat absorbed before melting:
Specific heat capacity of Al:
C = 0.897 J/(gm*K)
Heat_1 = 0.897 * m * change_in_temp
Heat_1 = 0.897 * 2,200gm * 553K = 1100*10^3 KiloJoules

Please check your math. I think you made an error of 10^3.


Heat of fusion for Al:
Lf = 0.395 kJ/g
Heat_2 = m * Lf = 2,200gm * 0.395 = 869 KiloJoules

Total heat required = 1100*10^3 + 869*10^3 = 1969 kJ
I'll roundoff to 2000 KiloJoules = 2 kilowatt-seconds

But I guess you corrected it here.

What's going on here? Since one joule is one watt*second, is not a
kilojoule equal to 1000 watt*seconds? If so, then 2000 kilojoules is 2
million watt*seconds. Yes? No? What?

From above...

Heat_1 = 0.897 * 2,200gm * 553K = 1100*10^3 KiloJoules

Is this right???


Why are you asking ME? You should be asking the author of the assertions.

You replied the first time I asked about it. Why did you respond then?

--

Rick

 

On Saturday, September 26, 2009 at 6:17:33 AM UTC+3, smtbsupport wrote:

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smtbsupport@gmail.com

 

The Thorium genie is now well and truly out of the bottle.

And what does China do with the nuclear waste? Perhaps we can send ours
to let them dispose of?

Simple, they don't have to deal with it.

As i understand it, they don't have much actual experiences with nuclear reactors, only academic experiences. But that doesn't stop them from offering assistances/advices for building nuclear reactors. So, it's builder beware.

 

[rickman]

On 8/6/2015 3:38 AM, Mark Wills wrote:

On Wednesday, 5 August 2015 17:25:38 UTC+1, rickman wrote:
On 8/5/2015 3:53 AM, Mark Wills wrote:
On Wednesday, 5 August 2015 08:23:16 UTC+1, polymorph self wrote:
On Friday, July 31, 2015 at 3:42:00 PM UTC-4, rickman wrote:
On 7/31/2015 2:13 PM, Jason Damisch wrote:
On Thursday, July 30, 2015 at 4:38:11 PM UTC-7, polymorph self wrote:

check youtube for thorium atomic power, its the answer to all energy needs,

https://www.youtube.com/watch?v=xIDytUCRtTA

About 6:40
"they had to have something that was so slick,
that was so safe, that was so simple"

I'm 14 minutes into the video and I still haven't seen an explanation of
it. My concerns are not just for the inherent safety of operating a
reactor without creating a disaster. The real problem with nuclear
power is the generation of waste products. Interesting also is that the
video is about a molten salt reactor, but not a thorium salt reactor...

"As only two liquid-core fluoride salt reactors have been built (the
ORNL ARE and MSRE) and neither have used thorium, it is hard to validate
the exact benefits.[4]As only two liquid-core fluoride salt reactors
have been built (the ORNL ARE and MSRE) and neither have used thorium,
it is hard to validate the exact benefits.[4]"

https://en.wikipedia.org/wiki/Aircraft_Nuclear_Propulsion#Direct_Air_Cycle

--

Rick
https://www.youtube.com/watch?v=qLk46BZfEMs watch and learn

....<snip>...

Yep. Molten salt reators are inherently "safe" requiring no human
intervention to prevent a meltdown. I spent a while about a year
ago watching numerous videos and presentations about MSR and I must
admit that it did convince me to the point that I thought someone
should just go ahead and build one.

And that's exactly what China is doing right now.

US helping China to build MSR:
http://fortune.com/2015/02/02/doe-china-molten-salt-nuclear-reactor/

Chinese going for broke on thorium nuclear power:
http://blogs.telegraph.co.uk/finance/ambroseevans-pritchard/100026863/china-going-for-broke-on-thorium-nuclear-power-and-good-luck-to-them/

Lots of coverage of this on the internet. Just do a google.

The Thorium genie is now well and truly out of the bottle.

And what does China do with the nuclear waste? Perhaps we can send ours
to let them dispose of?

--

Rick

You can't "dispose" of radio-active waste. You can only move
it to somewhere where it (hopefully) won't do any ^M^M^M
much harm.

What I haven't garnered from the various videos on MLSR (perhaps
I missed it) is what the expected half-life of the waste material
is.

I learned a lot from the video and would be happy discussing it, but
this is not the right group. I am crossposting to s.e.d and once
replies show up there will only reply to that group. This thread is not
any more on topic there, but in that group anything goes. lol

--

Rick

 

[rickman]

On 8/7/2015 1:12 PM, rickman wrote:

On 8/6/2015 3:38 AM, Mark Wills wrote:
On Wednesday, 5 August 2015 17:25:38 UTC+1, rickman wrote:
On 8/5/2015 3:53 AM, Mark Wills wrote:
On Wednesday, 5 August 2015 08:23:16 UTC+1, polymorph self wrote:
On Friday, July 31, 2015 at 3:42:00 PM UTC-4, rickman wrote:
On 7/31/2015 2:13 PM, Jason Damisch wrote:
On Thursday, July 30, 2015 at 4:38:11 PM UTC-7, polymorph self
wrote:

check youtube for thorium atomic power, its the answer to all
energy needs,

https://www.youtube.com/watch?v=xIDytUCRtTA

About 6:40
"they had to have something that was so slick,
that was so safe, that was so simple"

I'm 14 minutes into the video and I still haven't seen an
explanation of
it. My concerns are not just for the inherent safety of operating a
reactor without creating a disaster. The real problem with nuclear
power is the generation of waste products. Interesting also is
that the
video is about a molten salt reactor, but not a thorium salt
reactor...

"As only two liquid-core fluoride salt reactors have been built (the
ORNL ARE and MSRE) and neither have used thorium, it is hard to
validate
the exact benefits.[4]As only two liquid-core fluoride salt reactors
have been built (the ORNL ARE and MSRE) and neither have used
thorium,
it is hard to validate the exact benefits.[4]"

https://en.wikipedia.org/wiki/Aircraft_Nuclear_Propulsion#Direct_Air_Cycle


--

Rick
https://www.youtube.com/watch?v=qLk46BZfEMs watch and learn

...<snip>...

Yep. Molten salt reators are inherently "safe" requiring no human
intervention to prevent a meltdown. I spent a while about a year
ago watching numerous videos and presentations about MSR and I must
admit that it did convince me to the point that I thought someone
should just go ahead and build one.

And that's exactly what China is doing right now.

US helping China to build MSR:
http://fortune.com/2015/02/02/doe-china-molten-salt-nuclear-reactor/

Chinese going for broke on thorium nuclear power:
http://blogs.telegraph.co.uk/finance/ambroseevans-pritchard/100026863/china-going-for-broke-on-thorium-nuclear-power-and-good-luck-to-them/


Lots of coverage of this on the internet. Just do a google.

The Thorium genie is now well and truly out of the bottle.

And what does China do with the nuclear waste? Perhaps we can send ours
to let them dispose of?

--

Rick

You can't "dispose" of radio-active waste. You can only move
it to somewhere where it (hopefully) won't do any ^M^M^M
much harm.

What I haven't garnered from the various videos on MLSR (perhaps
I missed it) is what the expected half-life of the waste material
is.

I learned a lot from the video and would be happy discussing it, but
this is not the right group. I am crossposting to s.e.d and once
replies show up there will only reply to that group. This thread is not
any more on topic there, but in that group anything goes. lol

Might as well change the subject...

--

Rick

 

[Skybuck Flying]

Ok,

Let's get serious about this new vuln.

A very serious question:

How easy or hard would it be for virus/malware scanners to detect this ?

I will add my own favorite link/document about this exploit to this posting
to inform some people:

https://www.blackhat.com/docs/us-15/materials/us-15-Domas-The-Memory-Sinkhole-Unleashing-An-x86-Design-Flaw-Allowing-Universal-Privilege-Escalation-wp.pdf

Bye,
Skybuck.

 

[Skybuck Flying]

Why is there "wp" in the filename at the end ? I don't get that...

WP ? Work in progress ?

WP ? Word Perfect ? LOL

WP ? WordPress ?

Hmmm...

Bye,
Skybuck.

Ok,

Let's get serious about this new vuln.

A very serious question:

How easy or hard would it be for virus/malware scanners to detect this ?

I will add my own favorite link/document about this exploit to this posting
to inform some people:

https://www.blackhat.com/docs/us-15/materials/us-15-Domas-The-Memory-Sinkhole-Unleashing-An-x86-Design-Flaw-Allowing-Universal-Privilege-Escalation-wp.pdf

Bye,
Skybuck.

 

[M Philbrook]

In article <ad348$55c53a62$5419aafe$41548@news.ziggo.nl>, skybuck2000
@hotmail.com says...

Why is there "wp" in the filename at the end ? I don't get that...

WP ? Work in progress ?

WP ? Word Perfect ? LOL

WP ? WordPress ?

Hmmm...

Bye,
Skybuck.

You forgot one, one that you should know about;
Worst Programmer

 

[ceg]

On Sun, 16 Aug 2015 05:16:39 -0700, trader_4 wrote:

Click on your link
and there is a listing for "distracted driving":

You have to realize what you just intimated.

Bear in mind, it's the PARADOX that we're trying to resolve.

If distracted driving statistics were reliable (they're not), then the
paradox is EVEN WORSE!

Remember, the accidents don't seem to exist in the reliable statistics.
The accidents only exist in the highly unreliable statistics, and they
don't show up in the reliable ones - so - you and I both know what that
means.

Even so, if, as you and I assume, cellphone use causes accidents, then we
should be able to *see* those accidents in the aggregate statistics.

But we don't.

The fact that it's virtually impossible to determine whether a cellphone
was the primary (or even secondary) cause of an accident isn't really
part of the equation - because the accident count is going down (not up).

Hence the paradox.
Where are the accidents?