magnetic field

[N:dlzc D:aol T:com (dlzc)]

Dear Robert Clark:

"Robert Clark" <rgregoryclark@yahoo.com> wrote in message
news:1116781654.244742.187430@g47g2000cwa.googlegroups.com...
N:dlzc D:aol T:com (dlzc) wrote:

Dear Robert Clark:

"Robert Clark" <rgregoryclark@yahoo.com> wrote in message
news:1116780175.365218.161060@g44g2000cwa.googlegroups.com...
...
The point of this thread is the savings in power
you would get by using a lifter thruster method.
Look at the table near the bottom on this page:

Lifter Theory.
http://jnaudin.free.fr/html/lf­theory.htm

The last line in this table labled
Thrust(g)/Power(W) ratio gives the weight that
could be lifted for given power with the air
density available at ground level. It is given as
0.509, or about 2 to 1 for power in watts
required to lift a weight in grams.

Lift that is not countered in any sense by "V^2"
of drag through the atmosphere. An
atmosphere that becomes increasingly
conductive as it is heated, further reducing your
thrust.

This is a waste of time and effort. Rocket
engines are more efficient than 50% at turning
power into velocity, or even in sustaining
position.

I agree the calculation does not include the
effect of drag. It would probably be analogous
to the drag encountered by air-breathing
methods of space access, hypersonic craft for
instance.

OK.

Rockets are efficient but you have the problem
of the huge amount of fuel mass they have to
carry. At the very least lifters could provide a
low cost lower stage that could lift the craft to
high altitude

High altitude? No, since their thrust capacity is dependent on
gas density.

and to high velocity

High velocity? No, since their ability to thrust is dependent on
the gas being non-conductive. Heating destroys that.
Additionally, their thrusting capacity is dependent on the gas
pressure being uniform over the charged surface, which is
definately not true for something moving at more than a few tens
of miles per hour.

before a final rocket stage carried the craft to
orbit. This would result in signficant levels of
fuel savings.

Only in your dreams.

David A. Smith

 

[Ken Taylor]

"siliconmike" <siliconmike@yahoo.com> wrote in message
news:1116767278.076075.220420@g14g2000cwa.googlegroups.com...

John Miller wrote:
siliconmike wrote:
Right, thats understood. Now since the TV has 2000W exceptionally
good
audio amplifier, I would like to connect audio out from my PC to
audio
in of TV.

Has the brand of the TV been mentioned?

Samsung. (I'm in India)

Which model Samsung TV has 2000W output??

Ken

 

[N:dlzc D:aol T:com (dlzc)]

Dear bz:

"bz" <bz+sp@ch100-5.chem.lsu.edu> wrote in message
news:Xns965E853EE9B39WQAHBGMXSZHVspammote@130.39.198.139...

"Robert Clark" <rgregoryclark@yahoo.com> wrote in
news:1116780175.365218.161060@g44g2000cwa.googlegroups.com:

The next step is to test that this thrust ratio will
hold at the kilowatt range. Many people already
own electrical generators that can put out a few
kilowatts of power. They are used for example
for generators for RV's, stand-by generators,
picnic trips, etc. This page shows such
generators can be had for a few hundred dollars:

Electric Generator Store - Portable Generator,
Diesel Generators ...
http://www.electricgeneratorstore.com/

For example using the 1 to 1 thruster ratio, the
3250 watt generator advertised for $500 could
lift 3.25 kilos, about 7 pounds.

You need 30,000 VDC. That generator puts out
110 VAC.

That is the least of his problems. AC to high-VDC conversion is
not trivial, but it can be done. Getting ~3000 watts is slightly
more difficult. Getting flexible, light, cabling that can
deliver it at great length? Priceless...

His problems start with his belief that "lift" equates to
"overcoming air friction due to velocity through the very medium
being used to produce lift".

And the USAF charged the skin of an aircraft for the purpose of
being stealthy. No improvement in flight characterisitics was
noted. So I suspect that if we aren't talking MHD, then we are
talking about repelling the lifting body with the Earth as one
"capacitive plate". Anything else is window dressing, mere
slight-of-hand to distract the unwary.

David A. Smith

 

[Jim Logajan]

"tadchem" <thomas.davidson@dla.mil> wrote:

Charles Jean wrote:
"There are known knowns. These are things that we know we know.
There are known unknowns. That is to say, there are some
things that we know we don't know. But there are also unknown
unknowns. These are things we don't know we don't know."
-Secretary of Defense Donald Rumsfeld

If Rummy really said this, then he is much more widely read than I
would have believed. This is a variation of a quotation from Lady
Burton, attributed as an 'Arabian Proverb':
"Men are four:
He who knows not and knows not he knows not, he is a fool--shun him;
He who knows not and knows he knows not, he is simple--teach him;
He who knows and knows not he knows, he is asleep--wake him;
He who knows and knows he knows, he is wise--follow him!"

He who knows what just ain't so, he works in government--to hell with him!

 

[George Dishman]

"Robert Clark" <rgregoryclark@yahoo.com> wrote in message
news:1116780175.365218.161060@g44g2000cwa.googlegroups.com...
<snipped due to indentation fault in OE>

The point of this thread is the savings in power you would
get by using a lifter thruster method.
Look at the table near the bottom on this page:

Lifter Theory.
http://jnaudin.free.fr/html/lf­theory.htm

The last line in this table labled Thrust(g)/Power(W) ratio gives the
weight that could be lifted for given power with the air density
available at ground level. It is given as 0.509, or about 2 to 1 for
power in watts required to lift a weight in grams.
This is at ground level.

This is also at zero speed. Energy is force times
distance and power is force times speed so 0.5g
rising vertically at 202 m/s would double the power
needed. You seem to be neglecting that, but at
orbital speeds it is going to be far greater than
the figures you are quoting.

This is the same problem you had with the thrust
equations, you are forgetting the basic conservation
laws for momentum and energy.

On a practical front, it is trivial to reach high
altitude, just launch from a balloon, they need no
power at all. The hard part is reaching orbital
velocity once you get up there, and lifters aren't
going to work well in a near vacuum.

the only benefit they give you is the reaction mass
of the surrounding air at low altitude but this
advantage will soon be outweighed by the inefficiency
of burning fuel to run a generator which in turn
powers the lifter as the air gets thinner.

George

 

[bz]

"N:dlzc D:aol T:com \(dlzc\)" <N: dlzc1 D:cox T:net@nospam.com> wrote in
news:Sy6ke.904$rr.807@fed1read01:

You need 30,000 VDC. That generator puts out
110 VAC.

That is the least of his problems. AC to high-VDC conversion is
not trivial, but it can be done. Getting ~3000 watts is slightly
more difficult. Getting flexible, light, cabling that can
deliver it at great length? Priceless...

His problems start with his belief that "lift" equates to
"overcoming air friction due to velocity through the very medium
being used to produce lift".

And the USAF charged the skin of an aircraft for the purpose of
being stealthy. No improvement in flight characterisitics was
noted. So I suspect that if we aren't talking MHD, then we are
talking about repelling the lifting body with the Earth as one
"capacitive plate". Anything else is window dressing, mere
slight-of-hand to distract the unwary.

The effect is apparently due to 'ion wind' induced in the air by 30kv on a
set of emission points spaced some distance above a grid of wires.

Electrons emitted by the points ionize the air. The negative ions are then
attracted to the positive grid, producing lift.

The article referenced earlier in the thread explains it better than I can.


--
bz

please pardon my infinite ignorance, the set-of-things-I-do-not-know is an
infinite set.

bz+sp@ch100-5.chem.lsu.edu remove ch100-5 to avoid spam trap

 

[Pat Flannery]

Robert Clark wrote:

Electric Generator Store - Portable Generator, Diesel Generators ...
http://www.electricgeneratorstore.com/

For example using the 1 to 1 thruster ratio, the 3250 watt generator
advertised for $500 could lift 3.25 kilos, about 7 pounds.

This might be a good point to mention Major de Seversky's Ion -
Propelled Aircraft:
http://www.geocities.com/CapeCanaveral/Hall/1805/ion.html
The first generation of the Lifters.

Pat

 

[Pat Flannery]

Robert Clark wrote:

I agree the calculation does not include the effect of drag. It would
probably be analogous to the drag encountered by air-breathing methods
of space access, hypersonic craft for instance.


I don't know if you'd get much drag at all, as long as the whole surface

of the Lifter that is encountering the air is accelerating it. Drag is
induced by air encountering and flowing around a fixed surface- the
lifter would suck the air toward it from above and use it s acceleration
downwards to draw itself upwards. The difference being similar to the
difference between the bow of a ship cutting through water and a paddle
wheel pushing it backwards. In this case the vast majority of the
vehicle could be the aerodynamic equivalent of a paddle wheel.
The use of ion propulsion in regards to drag reduction gets discussed in
this article by Bill Gunston in the section about the B-2's propulsive
technique, in the section called "Stealth' about 3/5's of the way down
the webpage: http://www.aeronautics.ru/nws001/ai014.htm

Pat

 

[Pat Flannery]

bz wrote:

You need 30,000 VDC. That generator puts out 110 VAC.

I got walloped by 20,000 VAC (or DC...I'm not sure which, except you
could do one hell of a Jacob's ladder set-up with it.) out of a furnace
ignition coil that ran on 120 VAC.
Didn't most early lifters run on flyback transformers off of computer
monitor screens?
The important thing is how much total electrical energy is being used-
not its voltage, but its wattage.

Pat

 

[Pat Flannery]

N:dlzc D:aol T:com (dlzc) wrote:

And the USAF charged the skin of an aircraft for the purpose of
being stealthy. No improvement in flight characterisitics was
noted.

Of course if they were doing it for stealth (are you referring to the

A-11 test under project Kempster?)
the results wouldn't be relevant to propulsion, as the system wouldn't
be designed to propel the aircraft, merely to cloak it in an ion cloud
as Kempster attempted:
"Such concerns led the Agency to an entirely different approach to
antiradar efforts in Project KEMPSTER. This project attempted to develop
electron guns that could be mounted on the OXCART to generate an ion
cloud in front of the plane that would reduce it's radar cross section.
Although this project proved unsuccessful, the CIA also developed a
number of more conventional ECM devices for use in the OXCART. OSA
History, chap. 20, pp. 149-151 [13 spaces] Notes on the OXCART project
by [14 spaces], OSA records, [13 spaces]"
http://www.blackbirds.net/sr71/oxcart/successortou2.html

Pat

 

[Pat Flannery]

George Dishman wrote:

the only benefit they give you is the reaction mass
of the surrounding air at low altitude but this
advantage will soon be outweighed by the inefficiency
of burning fuel to run a generator which in turn
powers the lifter as the air gets thinner.



If one were going to build something along these lines, this would be a

real good argument for the use of some sort of beamed power solution
that leaves the generator on the ground- say a high-powered microwave beam.

Pat

 

[bz]

Pat Flannery <flanner@daktel.com> wrote in news:1192h0gfm5ffrf3
@corp.supernews.com:

bz wrote:

You need 30,000 VDC. That generator puts out 110 VAC.



I got walloped by 20,000 VAC

Ouch. Well, high voltage AC is slightly less dangerous than 110 VAC because
it is more likely to throw you clear of the circuit.

110 AC is especially dangerous because you tend to 'freeze on'.

(or DC...I'm not sure which, except you
could do one hell of a Jacob's ladder set-up with it.) out of a furnace
ignition coil that ran on 120 VAC.

Probabaly AC. Ignition coils are transformers and those produce AC. I doubt
that it was then rectified and filtered to produce DC.

Didn't most early lifters run on flyback transformers off of computer
monitor screens?

I don't know the details of how they produce the DC. Just saw that the
lifters needed 30,000 VDC.

The important thing is how much total electrical energy is being used-
not its voltage, but its wattage.

You don't generate streams of electrons and ion wind with low voltages.

Lifting will take power and high voltage.





--
bz

please pardon my infinite ignorance, the set-of-things-I-do-not-know is an
infinite set.

bz+sp@ch100-5.chem.lsu.edu remove ch100-5 to avoid spam trap

 

[N:dlzc D:aol T:com (dlzc)]

Dear Robert Clark:

"Robert Clark" <rgregoryclark@yahoo.com> wrote in message
news:1116812266.009025.315290@z14g2000cwz.googlegroups.com...
N:dlzc D:aol T:com (dlzc) wrote:

...
The "endoatmospheric ion-propulsion" engine
is clearly the same thing as the lifter drive.

Hardly. They don't need wires to the craft.

David A. Smith

 

[N:dlzc D:aol T:com (dlzc)]

Dear Robert Clark:

"Robert Clark" <rgregoryclark@yahoo.com> wrote in message
news:1116812934.280578.83020@f14g2000cwb.googlegroups.com...
George Dishman wrote:
....

"Robert Clark" <rgregoryclark@yahoo.com> wrote in message
news:1116780175.365218.161060@g44g2000cwa.googlegroups.com...
snipped due to indentation fault in OE

The point of this thread is the savings in power
you would get by using a lifter thruster method.
Look at the table near the bottom on this page:

Lifter Theory.
http://jnaudin.free.fr/html/lf­theory.htm

The last line in this table labled Thrust(g)/Power(W)
ratio gives the weight that could be lifted for given
power with the air density available at ground level.
It is given as 0.509, or about 2 to 1 for
power in watts required to lift a weight in grams.
This is at ground level.

This is also at zero speed. Energy is force times
distance and power is force times speed so 0.5g
rising vertically at 202 m/s would double the power
needed. You seem to be neglecting that, but at
orbital speeds it is going to be far greater than
the figures you are quoting.

This is the same problem you had with the thrust
equations, you are forgetting the basic conservation
laws for momentum and energy.

On a practical front, it is trivial to reach high
altitude, just launch from a balloon, they need no
power at all. The hard part is reaching orbital
velocity once you get up there, and lifters aren't
going to work well in a near vacuum.

the only benefit they give you is the reaction mass
of the surrounding air at low altitude but this
advantage will soon be outweighed by the inefficiency
of burning fuel to run a generator which in turn
powers the lifter as the air gets thinner.

If you look at the pages describing the drive, the
electrical power is a means to create airflow.
The thrust is due to this airflow. I'm calculating
the thrust as you do with the rocket equation.
What would be dependent on the speed is the
drag. That would be a rather complicated
dependence on the shape of the vehicle.

Especially since the shape of the vehicle would maximize the
surface for thrust, and incidentally pull a partial vaccuum to
counteract your thrust.

As a first guess you could give the craft the
aerodynmic shape of a supersonic or
hypersonic vehicle and use the same type of
intakes on those vehicles.

If you are doing the thrusting in the engine cowl, this is MHD
propulsion. It is not a "lifter".

David A. Smith

 

[Pat Flannery]

bz wrote:

I got walloped by 20,000 VAC



Ouch. Well, high voltage AC is slightly less dangerous than 110 VAC because
it is more likely to throw you clear of the circuit.

110 AC is especially dangerous because you tend to 'freeze on'.

Oh, I froze on real good- I froze on so good that I went jumping
violently down the boulevard with the electrified rod clamped tightly in
my hand and sparks coming out of my feet on each impact with the ground-
as the connection to the rod stuck in next to the anthill was completed.
I finally got far enough away that I pulled the wire off the coil, but
my whole right arm's muscles were contracted, and it was sore for a few
hours.
That was the last time I used a bread bag over an oven mit as a means of
high voltage protection.

(or DC...I'm not sure which, except you
could do one hell of a Jacob's ladder set-up with it.) out of a furnace
ignition coil that ran on 120 VAC.



Probabaly AC. Ignition coils are transformers and those produce AC. I doubt
that it was then rectified and filtered to produce DC.

I suspect it was AC also; It was basically just a big transformer, but
boy could it generate a hot arc..it had a superheated purple/yellow
plasma flame coming off the top of it.

You don't generate streams of electrons and ion wind with low voltages.

Lifting will take power and high voltage.

You know those Ionic Breeze air cleaners? If we were to take around ten
thousand of those and weld them to a spaceship, and we were in a gaseous
nebula... did I ever tell you about my plan for a rocket engine that
generates almost no thrust, but nevertheless creates an incredible
amount of noise? ;-)

Pat

 

In article <qLOdnazj6NSs4Q3fRVn-vQ@rcn.net>, jmfbahciv@aol.com writes:

In article <jeSje.115$25.24310@news.uchicago.edu>,
mmeron@cars3.uchicago.edu wrote:
In article <GeidncrxZrNTuBLfRVn-3w@rcn.net>, jmfbahciv@aol.com writes:
In article <a9Oie.79$25.17760@news.uchicago.edu>,
mmeron@cars3.uchicago.edu wrote:
In article <1116446501.892789.156400@f14g2000cwb.googlegroups.com>,
"tadchem" <thomas.davidson@dla.mil> writes:

Charles Jean wrote:

snip

"There are known knowns. These are things that we know we know.
There are known unknowns. That is to say, there are some
things that we know we don't know. But there are also unknown
unknowns. These are things we don't know we don't know."
-Secretary of Defense Donald Rumsfeld

If Rummy really said this,

He did.

I dropped my jaw to hear this from a dais located in Washington D. C.
He also did other things that were fraught with intelligence.

Yep.

then he is much more widely read than I
would have believed. This is a variation of a quotation from Lady
Burton, attributed as an 'Arabian Proverb':
"Men are four:
He who knows not and knows not he knows not, he is a fool--shun him;
He who knows not and knows he knows not, he is simple--teach him;
He who knows and knows not he knows, he is asleep--wake him;
He who knows and knows he knows, he is wise--follow him!"

Yep. Closer to home, note that the first category above is endowed
with what we (meaning sci.physics regulars) refer to as "second order
ignorance", which is the characteristic of many of our cranks.

Anyway, Rummy did say this and many seemingly intelligent people
jumped on this as a "dumb statement", not realizing that it was their
own stupidity they were thus proclaiming.

Yes, these types fixed his behaviour. Now anything that is decided
or accomplished using brains, as they should be used, will not
be documented.

Indeed. Would be dangerous to act otherwise.

This irritates the hell out of me. How am I supposed to learn
stuff if those who have the knowledge can't make it available?
Politics have already sewn distribution of certain doors shut.
When will the intelligentsia finish with mundane things like
science and personal finances?

Well, our Western intelligentsia turns out to be a net liability,
rather than an asset. A cancer on society, pretty much.

Imposing this kind of behaviour is extremely dangerous because
one of the things that these idiots jumped on was a memo
where he asked that people speculate about scenarios that
could happen and backup plans if they did occur.

Yes, and as you say, things like this are jumped on. The only useful
aspect of this is that it enables me to mark anyboody who expresses
outrage, *be it a world class scientist or a Nobel prize winner*, as
*moron* to be ignored.

....

I got that book. I've read the first two chapters. I spent my
dreamtime last night editing my mental data base. I'm so tired.

Well, let me know what you think, when you finish.

I've already used it. I have a criticism but
I wish to wait until I finish because there had to be
a purpose the author organized it the way he did.

So far, most of what he has written, I've known instinctively
but could never describe in English ASCII. I'm blessing that
man's mother because he's done this work for me.

Well, just to avoid confusion, I'm assuming that this is still John

Ruscio's book you're talking about. If so, then yes, his mother is a
very nice woman, you would like her. And as for his wife, she's one
of my favorites.

Mati Meron | "When you argue with a fool,
meron@cars.uchicago.edu | chances are he is doing just the same"

 

[Vidar Lřkken]

Lorraine wrote:

Hi,

Supposed you have a dc source of 1 ampere and you short
it... from I=V/R, R=0, what then is the current produced??

Also suppose the voltage is 10 volts, and I want the
maximum current, from I=V/R. What is the least ohms
value can I use, 10 ohms? 1 ohm? Suppose the maximum
current of the source is 1 ampere. What value of current
would I get when I use a 1 ohm resistor in the 10
volt circuit? The I=v/r gives out I=10/1ohm= 10 Ampere
which doesn't make sense. What is the rule of units
used? Thanks.

What you probably want, is maximum effect over the connected load. Then,
you have to know the internal resistance in the source. Most power will
go into the load, if the internal and load resistance is the same.
However, you'll then make as much effect in the source as the load...

--
MVH,
Vidar

www.bitsex.net

 

[Vidar Lřkken]

Lorraine wrote:

Suppose you have a 10 meter long wire, does the current alternative
(rise or fall) at the same time in the entire stretch of the 10 meter
long wire, or does a portion rise up while a portion fall down?

With the speed of light, yes it will be different. If you have long
enough wire, you could measure it...get 150,000KM of wire, then it
should be inverted from end to end


--
MVH,
Vidar

www.bitsex.net

 

[Mark VB]

Lorraine wrote:

Hi,

Supposed you have a dc source of 1 ampere and you short
it... from I=V/R, R=0, what then is the current produced??

R will never become 0, but for instance 1*10^-15. With 10 volts,

I=10/(1*10^-15)=1*10^16. However, if your dc source is current-limited
at 1 amp., drawn current will be 1 amp while the voltage is lowered to
1*10^-15, so V/R will become 1

Also suppose the voltage is 10 volts, and I want the
maximum current, from I=V/R. What is the least ohms
value can I use, 10 ohms? 1 ohm? Suppose the maximum
current of the source is 1 ampere. What value of current
would I get when I use a 1 ohm resistor in the 10
volt circuit? The I=v/r gives out I=10/1ohm= 10 Ampere
which doesn't make sense. What is the rule of units
used? Thanks.

I=V/R <=> R=V/I <=> R=10/1 <=> R=10 Ohms

Lorraine

P.S. May I know what are other electronic newsgroups
with lots of participants, tnx.

 

[Severi Salminen]

Vidar Lřkken wrote:

Suppose you have a 10 meter long wire, does the current alternative
(rise or fall) at the same time in the entire stretch of the 10 meter
long wire, or does a portion rise up while a portion fall down?


With the speed of light, yes it will be different. If you have long
enough wire, you could measure it...get 150,000KM of wire, then it
should be inverted from end to end

At 50Hz the wavelength is 300000km/50 = 6000km. So half of it is only
3000km of wire (Or maybe I did a mistake?)