In article ,
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), 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 ! )
.


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.

--
Telamon
Ventura, California

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 ! )
.


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?

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 ! )
.


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