HAARP What Bandwidth? (was Recording of HAARP and Moon Echo)
 

On Jan 22, 9:06*pm, wrote:
In article 24335e74-5502-4e0c-b1a5-
,
says...





On Jan 22, 6:00*pm, Billy Burpelson wrote:
RHF wrote:
On Jan 22, 4:56 am, Billy Burpelson wrote:
wrote:
I think it is more like a long train (2 seconds long) on a short U
shaped track (1.25 seconds on each leg). *The engine coming back will
pass the caboose still on its way to the end of the U. *
An interesting and thoughtful response. However, it generates a question:

In your train analogy above, what is to keep the leading part of the
echo from being QRMed by the trailing part of the transmitted signal?

BP,

"*" In the 5 Second Time Cycle for the Two Signals; this is
the Two Periods of 'Silence' between the Earth Pulse and
the Lunar Reflection.

EP .= = = = * _ _ _ _ * 5s
LR ._ _ _ _ * -- -- -- -- * 5s

"." Start of the Time Sync for 5 Second Signal Cycle
"=" Earth Pluse Time
"_" Non-Signal Time
"--" Lunar Reflection Time

A 2 Second Earth Pulse with a 2.5 Second Lunar Reflection
Delay creates a One-Half (1/2) Second Period of Silence
between the two Signals.

~ RHF
*.

- Roy, I think you might still might be missing the point here.
- Yes, the ROUND-TRIP delay is 2.5 seconds;
- one-way is ~1.25 seconds.
-
- So, if you send a 2 second long pulse -to- the moon, it will
- arrive in 1.25 seconds and thus the reflected signal will start
- back at time 1.25 seconds; but there is still the remaining
- .75 second of the original terrestrial pulse still winging its way
- to the moon.

This is the way I see it also. *Well, technically, that .75 seconds of
pulse is not winging its way to the moon because it has not been
generated yet. *When the very first part of the pulse just touches the
moon, it will be at 1.25 seconds as you say, but only 1.25 seconds of
pulse will have been generated on earth and be winging its way to the
moon. *It will take an additional .75 seconds for the rest of the pulse
to be generated (transmitted) and be winging its way to the moon. When
the earth pulse ends there will still be 1.25 seconds of signal winging
its way to the moon, the last 1.25 seconds, but the first .75 seconds of
pulse will indeed be on its way back.

I know. Picky, picky.

- Thus my original question: Will the beginning of the echo
- (at time 1.25 sec) QRM the last .75 second of the original
- transmitted pulse still on the way?
-
- Inquiring minds want to know...

I have cut and pasted my response to the same question earlier
below:

The train analogy is good in so far as it shows the timing, but I have
to admit it is poor in that I used a solid object, the train, to
represent a wave, and their properties are very different. For example,
if two trains hit head on, you are going to have a mess. That is not the
case with waves. If you throw two rocks at the same time in a pond of
still water so that they land some distance apart, the waves from each
impact point move out in concentric rings. *When the rings from one
impact point spread out enough to meet the spreading rings of the
second, there is however no "wreck". *The rings of waves of one appear
to pass through the rings of the other with no harm done to either wave. *
It is the energy that is moving across the water, not the water.

Here is a good URL for seeing a wave reflecting.

http://www.bbc.co.uk/schools/gcsebit...ater_wavesrev3.
shtml

(http://tinyurl.com/2ykkdr)

In our case the pulse is much longer so the interaction is longer, and
it also is not physical water, but the wave theory is the same.

- I should also point out that although the returning part of the 2
second
wave will not interfere with that part of the wave still on its way,

- if you could set up your receiver
- where both parts exist
- at the same time
- (i.e. near the moon),

VERY SPECIAL CONDITIONS :

That would require that the First Antenna 'Existed' in Reality
and a Second Antenna 'Existing' in a 2.5 Second Time Shift.

That would be some set of diversity antennas. ~ RHF

- I think one might QRM the other as you would be
- trying to listen to both parts of the wave at the same.

Yes in Theory -if- Both Signals were Time Shifted
so that they did indeed 'exist' at the same time.

-*That is different than two waves just passing each other.

At the Mid-Point between the Earth-&-Moon With Two Antennas :
One for the Earth Pulse {In-Time} and a Second for the Lunar
Reflection {Shifted in Time to 'Be' In-Time} : Both Signals would
'be' "Passing Each Other" from Opposite Directions and thus
could QEM Each Other.

IIRC -IF- You were on the Surface of the Moon
at the Center of the Reflected Signal :

You would Hear the Earth Pulse very well.

You Would Not Hear Any Lunar Reflection At All [.]

THINK ABOUT IT { Here on Earth } : Do You Hear ? "The Reflection"
From Your Receiving Antenna 'On' the Signal Being Received ?

it is the nature {physics} of things at one instance in time ~ RHF


IMHO - To the Radio receiving the Two Separate Signals
the Answer would be a : "NO" ~ RHF

The receiving Radio is Earth 'based' -and- The Two Signals
come by 'different' Paths to it :

1 - First comes the 2-Second Earth Pulse*
* The Earth Pulse has a 'Short Path' via Inner Atmosphere SkyWave.
- - - followed by a 1/2 Second of Silence.

I would guess the path delay for the terrestrial signal is no more than
.04 seconds here in the US, probably less, so transmitted time can be
considered received time.



2 - Second comes the 2-Second Lunar Reflection**
**The Lunar Reflection has a 'Long Path via the Trans-Atmosphere
Earth-to-Moon-to-Earth (EME) One-Big-Bounce.
- - - followed by a 1/2 Second of Silence.

NOTE - The Two Signal 'arrive' at the Antenna of the receiving
Radio at "Separate" Times in a recuring 5-Second Cycle.

Exactly so. At least I have one person that agrees with me ;-).



*.
Per the X-Files : The Answer {Truth} Is Out There ! :o)
*.- Hide quoted text -

- Show quoted text -- Hide quoted text -

- Show quoted text -

HAARP What Bandwidth? (was Recording of HAARP and Moon Echo)
 

In article telamon_spamshield-4766A5.22462822012008
@newsclstr03.news.prodigy.net,
lid says...
SNIP

- Roy, I think you might still might be missing the point here.
- Yes, the ROUND-TRIP delay is 2.5 seconds;
- one-way is ~1.25 seconds.
-
- So, if you send a 2 second long pulse -to- the moon, it will
- arrive in 1.25 seconds and thus the reflected signal will start
- back at time 1.25 seconds; but there is still the remaining
- .75 second of the original terrestrial pulse still winging its way
- to the moon.

This is the way I see it also. Well, technically, that .75 seconds of
pulse is not winging its way to the moon because it has not been
generated yet. When the very first part of the pulse just touches the
moon, it will be at 1.25 seconds as you say, but only 1.25 seconds of
pulse will have been generated on earth and be winging its way to the
moon. It will take an additional .75 seconds for the rest of the pulse
to be generated (transmitted) and be winging its way to the moon. When
the earth pulse ends there will still be 1.25 seconds of signal winging
its way to the moon, the last 1.25 seconds, but the first .75 seconds of
pulse will indeed be on its way back.

I know. Picky, picky.


- Thus my original question: Will the beginning of the echo
- (at time 1.25 sec) QRM the last .75 second of the original
- transmitted pulse still on the way?
-

- Inquiring minds want to know...

I have cut and pasted my response to the same question earlier
below:

The train analogy is good in so far as it shows the timing, but I have
to admit it is poor in that I used a solid object, the train, to
represent a wave, and their properties are very different. For example,
if two trains hit head on, you are going to have a mess. That is not the
case with waves. If you throw two rocks at the same time in a pond of
still water so that they land some distance apart, the waves from each
impact point move out in concentric rings. When the rings from one
impact point spread out enough to meet the spreading rings of the
second, there is however no "wreck". The rings of waves of one appear
to pass through the rings of the other with no harm done to either wave.
It is the energy that is moving across the water, not the water.

Here is a good URL for seeing a wave reflecting.

http://www.bbc.co.uk/schools/gcsebit...ater_wavesrev3.
shtml

(http://tinyurl.com/2ykkdr)

In our case the pulse is much longer so the interaction is longer, and
it also is not physical water, but the wave theory is the same.

I should also point out that although the returning part of the 2 second
wave will not interfere with that part of the wave still on its way, if
you could set up your receiver where both parts exist at the same time
(i.e. near the moon), I think one might QRM the other as you would be
trying to listen to both parts of the wave at the same. That is
different than two waves just passing each other.



IMHO - To the Radio receiving the Two Separate Signals
the Answer would be a : "NO" ~ RHF

The receiving Radio is Earth 'based' -and- The Two Signals
come by 'different' Paths to it :



1 - First comes the 2-Second Earth Pulse*
* The Earth Pulse has a 'Short Path' via Inner Atmosphere SkyWave.
- - - followed by a 1/2 Second of Silence.

I would guess the path delay for the terrestrial signal is no more than
.04 seconds here in the US, probably less, so transmitted time can be
considered received time.


2 - Second comes the 2-Second Lunar Reflection**
**The Lunar Reflection has a 'Long Path via the Trans-Atmosphere
Earth-to-Moon-to-Earth (EME) One-Big-Bounce.
- - - followed by a 1/2 Second of Silence.

NOTE - The Two Signal 'arrive' at the Antenna of the receiving
Radio at "Separate" Times in a recuring 5-Second Cycle.

Exactly so. At least I have one person that agrees with me ;-).

.
Per the X-Files : The Answer {Truth} Is Out There ! :o)
.


Situation #1
Lets say the direct earth signal "1" takes 0.0 seconds to reach you and
the moon bounce signal "2" takes 2.5 seconds. One frame sequence would
be:
Each number represents .1 second
12345678901234567890123456789012345678901234567890
11111111111111111111000002222222222222222222200000
The cycle starts and you get the 2 seconds direct signal, then 0.5
seconds noise, the 2 seconds moon reflection, then 0.5 second noise,
then the cycle repeats. The leading edge of the direct to reflected is
2.5 seconds.

Situation #2
Lets say the direct earth signal "1" takes 0.1 seconds to reach you and
the moon bounce signal "2" takes 2.5 seconds. One frame sequence would
be:
Each number represents .1 second
12345678901234567890123456789012345678901234567890
01111111111111111111100002222222222222222222200000
The cycle starts and you get 0.1 second noise, then the 2 second direct
signal, then 0.4 noise, the 2 second moon reflection, then .5 second
noise, then the cycle repeats. The leading edge of the direct to
reflected is 2.4 seconds.

Do you see how this works? The echo does not overlap the direct signal.
You could think of situation #1 being close to the HAARP station and #2
that you have a magic trigger with no time delay and you are a long
distance away from HAARP. #2 is just an example as you can't get far
enough away from HAARP for the 0.1 second direct time of flight.

If you were 1860 miles away time of flight would be 0.01 seconds for
example.

You don't live next to HAARP and you don't have the magic trigger so the
error you would measure would likely be 0.01 seconds. The error would
only be in one direction causing the moon measurement to be closer. You
could correct this error by adding the time of direct flight from you to
HAARP to the moon reflected signal in #2 situation. In #2 you saw the
moon bounce as 2.4 seconds + 0.1 seconds direct brings you back to 2.5
seconds in the #1 situation.



I agree with this analysis completely. Was there something posted
previously that you did not agree with that prompted this?

HAARP What Bandwidth? (was Recording of HAARP and Moon Echo)
 

In article ,
wrote:

In article telamon_spamshield-4766A5.22462822012008
@newsclstr03.news.prodigy.net,
lid says...
SNIP

- Roy, I think you might still might be missing the point here.
- Yes, the ROUND-TRIP delay is 2.5 seconds;
- one-way is ~1.25 seconds.
-
- So, if you send a 2 second long pulse -to- the moon, it will
- arrive in 1.25 seconds and thus the reflected signal will start
- back at time 1.25 seconds; but there is still the remaining
- .75 second of the original terrestrial pulse still winging its way
- to the moon.

This is the way I see it also. Well, technically, that .75 seconds of
pulse is not winging its way to the moon because it has not been
generated yet. When the very first part of the pulse just touches the
moon, it will be at 1.25 seconds as you say, but only 1.25 seconds of
pulse will have been generated on earth and be winging its way to the
moon. It will take an additional .75 seconds for the rest of the pulse
to be generated (transmitted) and be winging its way to the moon. When
the earth pulse ends there will still be 1.25 seconds of signal winging
its way to the moon, the last 1.25 seconds, but the first .75 seconds of
pulse will indeed be on its way back.

I know. Picky, picky.


- Thus my original question: Will the beginning of the echo
- (at time 1.25 sec) QRM the last .75 second of the original
- transmitted pulse still on the way?
-

- Inquiring minds want to know...

I have cut and pasted my response to the same question earlier
below:

The train analogy is good in so far as it shows the timing, but I have
to admit it is poor in that I used a solid object, the train, to
represent a wave, and their properties are very different. For example,
if two trains hit head on, you are going to have a mess. That is not the
case with waves. If you throw two rocks at the same time in a pond of
still water so that they land some distance apart, the waves from each
impact point move out in concentric rings. When the rings from one
impact point spread out enough to meet the spreading rings of the
second, there is however no "wreck". The rings of waves of one appear
to pass through the rings of the other with no harm done to either wave.
It is the energy that is moving across the water, not the water.

Here is a good URL for seeing a wave reflecting.

http://www.bbc.co.uk/schools/gcsebit...ater_wavesrev3.
shtml

(http://tinyurl.com/2ykkdr)

In our case the pulse is much longer so the interaction is longer, and
it also is not physical water, but the wave theory is the same.

I should also point out that although the returning part of the 2 second
wave will not interfere with that part of the wave still on its way, if
you could set up your receiver where both parts exist at the same time
(i.e. near the moon), I think one might QRM the other as you would be
trying to listen to both parts of the wave at the same. That is
different than two waves just passing each other.



IMHO - To the Radio receiving the Two Separate Signals
the Answer would be a : "NO" ~ RHF

The receiving Radio is Earth 'based' -and- The Two Signals
come by 'different' Paths to it :



1 - First comes the 2-Second Earth Pulse*
* The Earth Pulse has a 'Short Path' via Inner Atmosphere SkyWave.
- - - followed by a 1/2 Second of Silence.

I would guess the path delay for the terrestrial signal is no more than
.04 seconds here in the US, probably less, so transmitted time can be
considered received time.


2 - Second comes the 2-Second Lunar Reflection**
**The Lunar Reflection has a 'Long Path via the Trans-Atmosphere
Earth-to-Moon-to-Earth (EME) One-Big-Bounce.
- - - followed by a 1/2 Second of Silence.

NOTE - The Two Signal 'arrive' at the Antenna of the receiving
Radio at "Separate" Times in a recuring 5-Second Cycle.

Exactly so. At least I have one person that agrees with me ;-).

.
Per the X-Files : The Answer {Truth} Is Out There ! :o)
.


Situation #1
Lets say the direct earth signal "1" takes 0.0 seconds to reach you and
the moon bounce signal "2" takes 2.5 seconds. One frame sequence would
be:
Each number represents .1 second
12345678901234567890123456789012345678901234567890
11111111111111111111000002222222222222222222200000
The cycle starts and you get the 2 seconds direct signal, then 0.5
seconds noise, the 2 seconds moon reflection, then 0.5 second noise,
then the cycle repeats. The leading edge of the direct to reflected is
2.5 seconds.

Situation #2
Lets say the direct earth signal "1" takes 0.1 seconds to reach you and
the moon bounce signal "2" takes 2.5 seconds. One frame sequence would
be:
Each number represents .1 second
12345678901234567890123456789012345678901234567890
01111111111111111111100002222222222222222222200000
The cycle starts and you get 0.1 second noise, then the 2 second direct
signal, then 0.4 noise, the 2 second moon reflection, then .5 second
noise, then the cycle repeats. The leading edge of the direct to
reflected is 2.4 seconds.

Do you see how this works? The echo does not overlap the direct signal.
You could think of situation #1 being close to the HAARP station and #2
that you have a magic trigger with no time delay and you are a long
distance away from HAARP. #2 is just an example as you can't get far
enough away from HAARP for the 0.1 second direct time of flight.

If you were 1860 miles away time of flight would be 0.01 seconds for
example.

You don't live next to HAARP and you don't have the magic trigger so the
error you would measure would likely be 0.01 seconds. The error would
only be in one direction causing the moon measurement to be closer. You
could correct this error by adding the time of direct flight from you to
HAARP to the moon reflected signal in #2 situation. In #2 you saw the
moon bounce as 2.4 seconds + 0.1 seconds direct brings you back to 2.5
seconds in the #1 situation.



I agree with this analysis completely. Was there something posted
previously that you did not agree with that prompted this?

What other people posted indicated they did not understand what was
happening so I posted the examples.

--
Telamon
Ventura, California

HAARP What Bandwidth? (was Recording of HAARP and Moon Echo)
 

wrote:
In article ,
says...
wrote:

I think it is more like a long train (2 seconds long) on a short
U shaped track (1.25 seconds on each leg). The engine coming
back will pass the caboose still on its way to the end of the U.

An interesting and thoughtful response. However, it generates a
question:

In your train analogy above, what is to keep the leading part of
the echo from being QRMed by the trailing part of the transmitted
signal?



The analogy is good in so far as it shows the timing, but I have to
admit it is poor in that I used a solid object, the train, to
represent a wave and their properties are very different. For
example, if two trains hit head on, you are going to have a mess.
That is not the case with waves. If you throw two rocks at the same
time in a pond of still water so that they land some distance apart,
the waves from each impact point move out in concentric rings. When
the rings from one impact point spread out enough to meet the
spreading rings of the second, there is however no "wreck". The
rings of waves of one appear to pass through the rings of the other
with no harm done to either wave.

I don't think that is true. I believe the waves would add, subtract or
be somewhere in between depending on their phase relationship.

It is the energy that is moving across the water, not the water.

Here is a good URL for seeing a wave reflecting.

http://www.bbc.co.uk/schools/gcsebit...ater_wavesrev3.
shtml

(http://tinyurl.com/2ykkdr)

In our case the pulse is much longer so the interaction is longer,
and it also is not physical water, but the wave theory is the same.

What you say is interesting, but...wave theory notwithstanding, what we
have are TWO RF signals on -essentially- the same frequency (ignoring
Doppler, libration, etc), the incident and the reflected. And I think we
both agree that the signals "overlap" for .75 seconds (as so aptly
stated in your 'train' analogy). Just like the old USSR jamming the VOA
-- two signals on the same frequency. Why wouldn't they interfere for
the .75 seconds in question?

Let me give you one more analogy:

Assume the moon is -totally- absorbent (no RF reflections).

Next assume an earth station sends a 2 second long 7.0000 MHz CW signal
towards the moon.

Finally, assume there is a radio operator on the moon. The -instant- he
hears the earth signal, he turns on -his- 7.0000 MHz CW transmitter.

Again, because of the 2 second earth signal and only a 1.25 second
transit time, the two signals will overlap by 0.75 seconds.

How can they NOT interfere with each other for the .75 second overlap?
This is -exactly- like SWBC jamming (only unintentional) and all the
'wave theory' in the world doesn't mean jamming doesn't work. Sorry, but
I believe two signals on the same frequency during the same period of
time will interfere with each other.

HAARP What Bandwidth? (was Recording of HAARP and Moon Echo)
 

Telamon wrote:

Do you understand the direct and moon reflected signals do not exist at
the same time anywhere on the earth?

Do -you- understand what I wrote? Do you grasp the concept? Please pay
attention.

I -never- said the direct and reflected signals would exist at the
-same- time 'anywhere on earth'.

What I -did- speculate on is that a *portion* of the echo (the echo's
leading edge) will be QRM'd *right as it leaves the moon* (the first
0.75 seconds of the echo) by the trailing edge of the incident wave (its
last 0.75 seconds). Therefore, either a shortened echo (due to full
cancellation of the 'overlap', which is unlikely) or an echo with a
distorted or weakened leading edge (more likely) will ultimately reach
the earth, depending on how much out-of-phase cancellation at the moon
end of the circuit occured.

HAARP What Bandwidth? (was Recording of HAARP and Moon Echo)
 

In article ,
Billy Burpelson wrote:

Telamon wrote:

Do you understand the direct and moon reflected signals do not exist at
the same time anywhere on the earth?

Do -you- understand what I wrote? Do you grasp the concept? Please pay
attention.

Yes I understand what you wrote loser and I'm doing my best to explain
that your thinking is incorrect. I'm the second person tell you this.

I -never- said the direct and reflected signals would exist at the
-same- time 'anywhere on earth'.

I didn't say you did. I was trying to give you the bigger picture.

What I -did- speculate on is that a *portion* of the echo (the echo's
leading edge) will be QRM'd *right as it leaves the moon* (the first
0.75 seconds of the echo) by the trailing edge of the incident wave (its
last 0.75 seconds). Therefore, either a shortened echo (due to full
cancellation of the 'overlap', which is unlikely) or an echo with a
distorted or weakened leading edge (more likely) will ultimately reach
the earth, depending on how much out-of-phase cancellation at the moon
end of the circuit occured.

OK you misunderstood what I wrote, which is these EM waves can mix as
you think in a antenna in the part of space where the wave going to the
moon exists in the same space time as the reflected signal. Without the
antenna these waves do not interact. With the antenna they do interact.
Is that redundant enough for you?

Logically, since no antennas exist in this area of space time your idle
speculation is moot. Now you have had two people explain this to you so
if this does not sink into your very dense skull give up.

Let me explain something else to you O'clueless wonder and that's if you
don't get it your problem not mine.

--
Telamon
Ventura, California

HAARP What Bandwidth? (was Recording of HAARP and Moon Echo)
 

On Jan 24, 7:08*pm, Billy Burpelson wrote:
Telamon wrote:
Do you understand the direct and moon reflected signals do not exist at
the same time anywhere on the earth?

Do -you- understand what I wrote? Do you grasp the concept? Please pay
attention.

I -never- said the direct and reflected signals would exist at the
-same- time 'anywhere on earth'.
-
- What I -did- speculate on is that a *portion* of the echo
- (the echo's leading edge) will be QRM'd *right as it leaves
- the moon* (the first 0.75 seconds of the echo) by the
- trailing edge of the incident wave (its last 0.75 seconds).
- Therefore, either a shortened echo (due to full cancellation
- of the 'overlap', which is unlikely) or an echo with a
- distorted or weakened leading edge (more likely) will
- ultimately reach the earth, depending on how much
- out-of-phase cancellation at the moon end of the
- circuit occured.
-

BP,

What you are saying is the 'theoretical' Signal Time-Line
on the Moon 'might-be' :
[ Roughly 0.1 Second Time Increments ]
EEEEEEEEEE 1 { 'E' 100% Earth Pulse }
EEEEEEEEEE 2
EEEEEEEEEE 3
EEEEEEEEEE 4
EEEEEEEEEE 5
EEEEEEEEEE 6
EEEEEEEEEE 7
EEEEEEEEEE 8
EEEEEEEEEE 9
EEEEEEEEEE 10
EEEEEEEEEE 11
EEEEEEEEEE 12
eeeee l l l l l 13 { 'e l' Mixed Signal Conditions}
eeeee l l l l l 14
eeeee l l l l l 15
eeeee l l l l l 16
eeeee l l l l l 17
eeeee l l l l l 18
eeeee l l l l l 19
eeeee l l l l l 20
LLLLLLLLLL 21 { 'L' 100% Lunar Reflection }
LLLLLLLLLL 22
LLLLLLLLLL 23
LLLLLLLLLL 24
LLLLLLLLLL 25
LLLLLLLLLL 26
LLLLLLLLLL 27
LLLLLLLLLL 28
LLLLLLLLLL 29
LLLLLLLLLL 20
LLLLLLLLLL 21
LLLLLLLLLL 22

IWFT - The 'potential' Exists for these "Mixed Signal Conditions"
to be Observed in the Signal Received by the Earth Radio.

OK - It is 'possible' ~ RHF

(OT) : HAARPing With Telamon - Do We Have Enough Bandwidth ? - The Never Ending Lunar echo, Echo. ECHO !
 

In article
,
RHF wrote:

On Jan 25, 4:55*pm, Telamon
wrote:
In article
,





*RHF wrote:
On Jan 24, 7:08*pm, Billy Burpelson wrote:
Telamon wrote:
Do you understand the direct and moon reflected signals do not exist
at
the same time anywhere on the earth?

Do -you- understand what I wrote? Do you grasp the concept? Please pay
attention.

I -never- said the direct and reflected signals would exist at the
-same- time 'anywhere on earth'.
-
- What I -did- speculate on is that a *portion* of the echo
- (the echo's leading edge) will be QRM'd *right as it leaves
- the moon* (the first 0.75 seconds of the echo) by the
- trailing edge of the incident wave (its last 0.75 seconds).
- Therefore, either a shortened echo (due to full cancellation
- of the 'overlap', which is unlikely) or an echo with a
- distorted or weakened leading edge (more likely) will
- ultimately reach the earth, depending on how much
- out-of-phase cancellation at the moon end of the
- circuit occured.
-

BP,

What you are saying is the 'theoretical' Signal Time-Line
on the Moon 'might-be' :
[ Roughly 0.1 Second Time Increments ]
EEEEEEEEEE 1 *{ 'E' 100% Earth Pulse }
EEEEEEEEEE 2
EEEEEEEEEE 3
EEEEEEEEEE 4
EEEEEEEEEE 5
EEEEEEEEEE 6
EEEEEEEEEE 7
EEEEEEEEEE 8
EEEEEEEEEE 9
EEEEEEEEEE 10
EEEEEEEEEE 11
EEEEEEEEEE 12
eeeee l l l l l 13 *{ 'e l' Mixed Signal Conditions}
eeeee l l l l l 14
eeeee l l l l l 15
eeeee l l l l l 16
eeeee l l l l l 17
eeeee l l l l l 18
eeeee l l l l l 19
eeeee l l l l l 20
LLLLLLLLLL 21 { 'L' 100% Lunar Reflection }
LLLLLLLLLL 22
LLLLLLLLLL 23
LLLLLLLLLL 24
LLLLLLLLLL 25
LLLLLLLLLL 26
LLLLLLLLLL 27
LLLLLLLLLL 28
LLLLLLLLLL 29
LLLLLLLLLL 20
LLLLLLLLLL 21
LLLLLLLLLL 22

IWFT - The 'potential' Exists for these "Mixed Signal Conditions"
to be Observed in the Signal Received by the Earth Radio.

OK - It is 'possible' ~ RHF
*.

(OT) : HAARPing With Telamon : Do We Have Enough Bandwidth ?
The Never Ending Lunar echo, Echo. ECHO !

- For God's sake it's not possible.

OK - We have your word for that.

-ps- For God's Sake : Chill-Out Dude !
This was a Hypothetical Discussion based on the
'potential' for 'possibilities' given the condition of
'what if' where the 'if' beyond both the Scientific and
practical limits of the Experiment.
Remember - There is an 'if' in the word : "Scient-if-ic"

- Didn't you read my examples?

Yes But - Maybe I did not understand them
in the same the way that you did.

- Can't you think at all?

thinking, Thinking. THINKING ! - OK - My Trible Redundant
System Check Shows That : I Can Think !

- EM waves don't directly interact with each other anyway.

Hey Score One For You ! {In FreeSpace}

However - I Did Write :

IWFT - The 'potential' Exists for these "Mixed Signal Conditions"
to be Observed in the Signal Received by the Earth Radio.

Once Two Signals are inside a Radio/Recever "It Ain't Free Space"

- Think what would the result would be
if this luny-tunes idea were true.

A Universal Mixed Product RF/EM Environment.
http://en.wikipedia.org/wiki/Universe

- How about the people that designed the experiment?

Maybe 'if' you would first read and attempt to understand
-vice- Being So Judgemental : Remember I wrote in this
very thread :

Newsgroups: rec.radio.shortwave
From: RHF
Date: Tue, 22 Jan 2008 08:03:28 -0800 (PST)
Local: Tues, Jan 22 2008 8:03 am
Subject: HAARP What Bandwidth?
(was Recording of HAARP and Moon Echo)
http://groups.google.com/group/rec.r...46e18812e433e1
.
I 'Trust' to the Fact that the HAARP Scientists
are Equal to Rocket Scientists and that :
They Do Know What They Are Doing. ~ RHF
http://en.wikipedia.org/wiki/Rocket_science

- Do they have a functioning brain?

IMHO - Clearly They Do.

- I would say yes!

On That We Can Agree.

- Do you?

On That We Clearly Disagree.

- I would say no and that also goes for Billy boy.

You Are Entitle To You Opinion.
EVEN WHEN YOU ARE WRONG [.]

To bad for you that you don't understand. By all means don't take my
word for it as you can go read up on the subject.

- How embarrassing.

Yes - Indeed "How Embarrassing" For You. ~ RHF


You don't get it even after I posted about it?

You can go educate yourself but that takes work doesn't it? So you won't
do the work you will instead continue to remain ignorant either for the
reason of ability or laziness but that's your problem not mine. So do
continue to make foolish posts. They are very entertaining.

--
Telamon
Ventura, California

HAARP What Bandwidth? (was Recording of HAARP and Moon Echo)
 

In article ,
says...

SNIP
In your train analogy above, what is to keep the leading part of
the echo from being QRMed by the trailing part of the transmitted
signal?



The analogy is good in so far as it shows the timing, but I have to
admit it is poor in that I used a solid object, the train, to
represent a wave and their properties are very different. For
example, if two trains hit head on, you are going to have a mess.
That is not the case with waves. If you throw two rocks at the same
time in a pond of still water so that they land some distance apart,
the waves from each impact point move out in concentric rings. When
the rings from one impact point spread out enough to meet the
spreading rings of the second, there is however no "wreck". The
rings of waves of one appear to pass through the rings of the other
with no harm done to either wave.

I don't think that is true. I believe the waves would add, subtract or
be somewhere in between depending on their phase relationship.

What you say is true, but what I said is also true. The waves are not
"wrecked" and after the waves pass each other, they are are the same as
before they overlapped.

I found a reference on the web

(http://www.smgaels.org/physics/amsco.../chapter04.pdf)

that says it much better than I could. On page 109 the author states:


(Beginning of Quote)
"Two or more waves may pass through a medium at the same time. When this
occurs two rules apply. First, the total displacement experienced at any
point where waves meet is equal to the sum of the displacements of the
individual waves at that point. This is known as the principle of
superposition. Second, waves pass through each other, with each wave
unaffected by the passage of the others. After meeting, the individual
waves continue traveling in their original directions and with the same
characteristics as before."
(End of Quote)


SNIP

What you say is interesting, but...wave theory notwithstanding, what we
have are TWO RF signals on -essentially- the same frequency (ignoring
Doppler, libration, etc), the incident and the reflected. And I think we
both agree that the signals "overlap" for .75 seconds (as so aptly
stated in your 'train' analogy).

Actually they overlap for very close to 2 seconds. Thinking back to the
train analogy, the instant the pulse hits the moon (cow catcher on the
engine) and starts back the other way as an echo, we have an overlap.
That means 99.99% of the pulse (train), or almost 2 seconds worth, still
has to hit the surface and reflect before there is no longer an overlap.

Just like the old USSR jamming the VOA
-- two signals on the same frequency. Why wouldn't they interfere for
the .75 seconds in question?

I think they do to the extent of the "superposition" of the waves, and
if you were in a space capsule with a receiver 1000 miles from the
surface of the moon, I think you might indeed hear that interference.


Let me give you one more analogy:

Assume the moon is -totally- absorbent (no RF reflections).

Next assume an earth station sends a 2 second long 7.0000 MHz CW signal
towards the moon.

Finally, assume there is a radio operator on the moon. The -instant- he
hears the earth signal, he turns on -his- 7.0000 MHz CW transmitter.

Again, because of the 2 second earth signal and only a 1.25 second
transit time, the two signals will overlap by 0.75 seconds.

How can they NOT interfere with each other for the .75 second overlap?
This is -exactly- like SWBC jamming (only unintentional) and all the
'wave theory' in the world doesn't mean jamming doesn't work. Sorry, but
I believe two signals on the same frequency during the same period of
time will interfere with each other.


I agree. I thought your opening question, "what is to keep the leading
part of the echo from being QRMed by the trailing part of the
transmitted signal?", was asking why the overlapping signals
(transmitted and echo) didn't QRM the echo at the receivers on earth.
That is why I gave the example of the stones to show that did not
happen. I did not realize you were indicating a hypothetical receiver
where the signals are overlapping. My apologies.

(OT) : HAARPing With Telamon - Do We Have Enough Bandwidth ? - TheNever Ending Lunar echo, Echo. ECHO !
 

On Jan 25, 5:19*pm, Telamon
wrote:
In article
,





*RHF wrote:
On Jan 24, 7:08*pm, Billy Burpelson wrote:
Telamon wrote:
Do you understand the direct and moon reflected signals do not exist at
the same time anywhere on the earth?

Do -you- understand what I wrote? Do you grasp the concept? Please pay
attention.

I -never- said the direct and reflected signals would exist at the
-same- time 'anywhere on earth'.
-
- What I -did- speculate on is that a *portion* of the echo
- (the echo's leading edge) will be QRM'd *right as it leaves
- the moon* (the first 0.75 seconds of the echo) by the
- trailing edge of the incident wave (its last 0.75 seconds).
- Therefore, either a shortened echo (due to full cancellation
- of the 'overlap', which is unlikely) or an echo with a
- distorted or weakened leading edge (more likely) will
- ultimately reach the earth, depending on how much
- out-of-phase cancellation at the moon end of the
- circuit occured.
-

BP,

What you are saying is the 'theoretical' Signal Time-Line
on the Moon 'might-be' :
[ Roughly 0.1 Second Time Increments ]
EEEEEEEEEE 1 *{ 'E' 100% Earth Pulse }
EEEEEEEEEE 2
EEEEEEEEEE 3
EEEEEEEEEE 4
EEEEEEEEEE 5
EEEEEEEEEE 6
EEEEEEEEEE 7
EEEEEEEEEE 8
EEEEEEEEEE 9
EEEEEEEEEE 10
EEEEEEEEEE 11
EEEEEEEEEE 12
eeeee l l l l l 13 *{ 'e l' Mixed Signal Conditions}
eeeee l l l l l 14
eeeee l l l l l 15
eeeee l l l l l 16
eeeee l l l l l 17
eeeee l l l l l 18
eeeee l l l l l 19
eeeee l l l l l 20
LLLLLLLLLL 21 { 'L' 100% Lunar Reflection }
LLLLLLLLLL 22
LLLLLLLLLL 23
LLLLLLLLLL 24
LLLLLLLLLL 25
LLLLLLLLLL 26
LLLLLLLLLL 27
LLLLLLLLLL 28
LLLLLLLLLL 29
LLLLLLLLLL 20
LLLLLLLLLL 21
LLLLLLLLLL 22

IWFT - The 'potential' Exists for these "Mixed Signal Conditions"
to be Observed in the Signal Received by the Earth Radio.

OK - It is 'possible' ~ RHF
*.
-
- Here get yourself one of these and don't take it off for any reason.
-
- http://www.dailymail.co.uk/pages/live/articles/health/
healthmain.html?in
- _article_id=510172&in_page_id=1774
-
- --
- Telamon
- Ventura, California
-

Gee - Telamon - A Personage of Your Great Knowledge :
Don't You Know How To Provide and Un-Truncated Link/URL ?

-=WRT=- Your Link from the Daily Mail [UK]
The Helmet That Could Turn Back The Symptoms of Alzheimer's
{ Electro Gizmo Beanie - http://tinyurl.com/3xem75 }
http://www.dailymail.co.uk/pages/liv...page_id=17 74
-by- David Derbyshire - The Daily Mail - 25th January 2008

Telamon - I take it that this Recommendation is base on your
own Personal Experience : Oops I've Got News For You . . .
It Ain't Working ! ~ RHF

OBTW - I prefer the Inverted "CK" Antenna : After all it is a
passive device and natually : No Batteries Are Required.
http://groups.google.com/group/rec.r...260fb0b38afd68

The Inverted "CK" Antenna (Patent Pending) is the Ultimate
Solution for All Your Cranial Reception and Cerebral
Interference Needs. CAUTION: When Properly Installed.
http://groups.google.com/group/rec.r...1dbdfa5566478c

YES - The Inverted "CK" Antenna - For the ever Increasing
Number of Senior-Moments that are getting ever and forever
longer and longer . . .

WOW - When The Tin Foil Hat Just Can't Get The Crazy Job Done.
It's time to put the Inverted "CK" Antenna back-on !
Remember the Solid Copper Kettle Pot Keeps the Brain from Boiling.

telamon -wrt- alzheimer's disease - I honestly hope that no
member of you family is so afflected. - it ain't no joke [.] ~ RHF
http://en.wikipedia.org/wiki/Alzheimer's_disease