In article , wrote:
In article ffd0d3c9-18d9-4b7b-95dc-

I think that echo you hear is the one that results from the terrestrial
pulse you just heard, and not 1 or 2 back. My guess is that the leading
edge of the reflected pulse arrives approx. 2.5 seconds after it leaves
and the same for the trailing edge of the pulse. That means if you look
at entire 5 second time line, at 0 seconds the earth pulse starts and at
2 seconds it ends. At 2.5 seconds the beginning part of the echo pulse
arrives and at 4.5 seconds (2.5 seconds after the terrestrial pulse
ended) the end of the echo pulse arrives. There is always 2.5 seconds
between any reference point on the timeline of the sent pulse to the
corresponding point on the timeline of the return pulse. One complete
cycle of send/receive should be approximately 4.5 seconds. I think that
is why they selected 5 seconds between sending 2 second pulses. At
least that is the way I view it. I am open to other views.


On the first transmission, (Jan 19, 05:00 UTC), I tried using an
oscilloscope to see if I could see the echo. It was a 4 second cycle
with 2 seconds on and 2 off. (Which made syncing the scope (with a 5
second sweep) a real pain. The next pulse was going when the first one
ended, triggering it again).

Anyway, with the narrowest (2 kHz) bandwidth on my old R-1000, it
just didn't cut it. When the outgoing signal was S9+10 (in Seattle)
I thought I might have heard an echo or two, but just couldn't be
sure.

What bandwidth were you successful listeners using?

Mark Zenier
Googleproofaddress(account:mzenier provider:eskimo domain:com)

On Jan 21, 9:51 am, Arne wrote:
On Sun, 20 Jan 2008 19:35:37 GMT, (Mark Zenier)
wrote:

On the first transmission, (Jan 19, 05:00 UTC), I tried using an
oscilloscope to see if I could see the echo. It was a 4 second cycle
with 2 seconds on and 2 off. (Which made syncing the scope (with a 5
second sweep) a real pain. The next pulse was going when the first one
ended, triggering it again).
Anyway, with the narrowest (2 kHz) bandwidth on my old R-1000, it
just didn't cut it. When the outgoing signal was S9+10 (in Seattle)
I thought I might have heard an echo or two, but just couldn't be
sure.
What bandwidth were you successful listeners using?

2 kHz in USB mode on a NRD-535D tuned to 6792.0 kHz and 7407.0 kHz.
The cycle (per the HAARP website) was 5 seconds. 2 seconds on,
followed by 3 seconds off.

Arne (AzUSA)

It seems those that heard the reflected signal were in the south.

I used a crystal CW filter. 250hz BW. My recollection is that is at
-6dB, not -3dB.

On Jan 21, 9:51*am, Arne wrote:
On Sun, 20 Jan 2008 19:35:37 GMT, (Mark Zenier)
wrote:

On the first transmission, (Jan 19, 05:00 UTC), I tried using an
oscilloscope to see if I could see the echo. *It was a 4 second cycle
with 2 seconds on and 2 off. *(Which made syncing the scope (with a 5
second sweep) a real pain. *The next pulse was going when the first one
ended, triggering it again).
Anyway, with the narrowest (2 kHz) bandwidth on my old R-1000, it
just didn't cut it. *When the outgoing signal was S9+10 (in Seattle)
I thought I might have heard an echo or two, but just couldn't be
sure.
What bandwidth were you successful listeners using?

2 kHz in USB mode on a NRD-535D tuned to 6792.0 kHz and 7407.0 kHz.
The cycle (per the HAARP website) was 5 seconds. * *2 seconds on,
followed by 3 seconds off.

Arne (AzUSA)

Note - There is a 2 1/2 Second Round-Trip Lunar Propagation Delay.
http://www.setileague.org/eme/emepix3.htm

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

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

~ RHF

In article 76070442-c8f1-4043-8437-d9e3cbaad885
@y5g2000hsf.googlegroups.com, says...

Note - There is a 2 1/2 Second Round-Trip Lunar Propagation Delay.
http://www.setileague.org/eme/emepix3.htm

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

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

~ RHF
.

I don't think that is quite correct. If the pulse were only a
millisecond long, 0 to .001 on the time line, you would receive the echo
at between 2.500 and 2.501 seconds on the time line. If you think of a
two second pulse as only a series of millisecond pulses, the first pulse
(or the beginning of a two second pulse) would be hitting the moon after
1.25 seconds and starting on its way back before the last millisecond
pulse (or the end of the 2 second pulse) even left.

I think you are treating the pulse as a single discrete object that
actually doesn't leave until the pulse is complete, like bouncing a ball
off the wall. I don't think that is the case.

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. The engine will
reach the end of the return leg 2.5 seconds after it left. The caboose
will also reach the end of the return leg 2.5 seconds after it left, but
it will be 4.5 seconds after the engine left making the total time for
the train (2 second pulse) to make the trip (one cycle) 4.5 seconds.

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?

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

On Jan 22, 7:32*am, 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

Look at it as a Repetitive Linear Event :
[ Repetitive Signal Interval Timing ]

.= = = = * -- -- -- -- *.= = = = *-- -- -- -- *.= = = = * -- -- -- --
* ~

"." Start of the Time Sync for 5 Second Signal Cycle {Instant}

"=" Earth Pluse Time
{Four 1/2 Second Marks = 2 Seconds}

"*" "Silence" in between Non-Signal Time
{One 1/2 Second Mark = 1/2 a Second}

"--" Lunar Reflection Time
{Four 1/2 Second Marks = 2 Seconds}

"*" "Silence" in between Non-Signal Time
{One 1/2 Second Mark = 1/2 a Second}

This Ends the 5 Second Time Cycle
and the process repeats itself.

Do-the-Math : = 2 + 1/2 + 2 + 1/2 = 5

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

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. 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...

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.
-
- 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...

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.

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.

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