On 6/30/2010 10:35 AM, Joe wrote:
I agree,
It's not the meteor or even the ionized portion near the head that you bounce signals off of, it's the inonized trail that it leaves behind. and that can be many many miles long.
Joe - It would seem that the plasma cloud in front of the meteor would be moving at the same velocity as the meteor itself and would cause the fast Doppler shift we see prior to the trail formation.
Tony -K2MO
Now about reflecting off of odd shaped things. A buddy an myself used to in the late 70's have fun on 2 meters just playing around. we were too far apart for direct communication, I was in the northwest suburbs of Chicago, he was down in central indiana.
We would point our 4 ele beams at eachother and run like a meteor scatter transmission sequesnce. and would make a qso. it was flaky and wide variations in signal strength. But what did we use?
It was all the Jets coming in to land at Ohare! between us was the main route for all the jets to com in on. The route killed TV all the time by making signals in phase and then out then in then out, so we thought if it's strong enough reflection to cancel out a direct megawayy signal, hmmm. so we tried and it worked and worked faily well. each "Opening" lasted between 10 and 15 seconds. and would go most times from s zero to peg the meter.
was fun and thats an odd shaped reflector. that for me was ohhh 60 miles away, and for him close to 100 miles.
Joe WB9SBD
The Original Rolling Ball Clock Idle Tyme Idle-Tyme.com http://www.idle-tyme.com
On 6/30/2010 3:31 AM, Tony wrote:
Dominico,
Your path loss calculations were very interesting. At first glance, I assumed it would be possible to hear 28MHz echoes off of the 10M sphere since smaller targets such as the ionized head of a meteor can reflect such signals with enough signal power to be heard.
But, I think this is one of those apples to oranges comparisons since: a. the ionization around the front of the meteor head is responsible for signal reflection, not the rock itself. And b. the diameter of the ionized atmosphere in front of the meteor may be a lot larger than one would imagine.
The 10M sphere brought another question to mind -- Mike Wantanabe, JH1KRC, managed to hear his own EME echoes on the 21MHz using a KW and a 6 element Yagi. Details and recordings are on his website (see below). I was wondering how the path loss calculations compare with his EME results.
http://eme.dokidoki.ne.jp/sound/jh1krc/index.html
Thanks Dominico...
Tony -K2MO
On 6/29/2010 9:52 PM, i8cvs wrote:
----- Original Message ----- From: "Robert Bruninga"bruninga@usna.edu To:amsat-bb@amsat.org Sent: Monday, June 28, 2010 7:44 PM Subject: [amsat-bb] HF Satellite Relay
Heard today of a Passive HF relay satellite being proposed. Wondered if Hams could relay off of it.
It's a 10m diameter sphere. I assumed a 10m signal and 1000 Watts And antenna gains at both ends of 10 dB. Unless I made a dumb error, it looks impossible? I get a received signal of -170 dBm Compared to a good HF receiver of -122 dBm So its 48 dB down in the noise. Going to narrow band, could improve things, but the Doppler of +/- 600 Hz would make that difficult.
Anyway, if someone else wants to double check the link budget using the radar range equation, go for it.
The beauty of this system is that it is perfectly spherical, so the reflection coefficient would be constant within 1 dB. That is the advantage over trying to use the ISS or other large rocket body... They vary by 20 dB making communication by reflection impossible.
Oh, and it would be in space for 30 years or more. So with something that reliable, it would be worth developing an amateur capability to use it. It is not designed for comms, but as a calibration sphere for over the horizon radars that have LOTS more power and LOTS more gain than we do.
Bob, Wb4APR
Hi Bob, WB4APR
I have assumed that the altitude of the Passive HF relay satellite over the earth is 1500 km and as we know the diameter of the sphere is 10 meters. Also I assumed that the reflectivity coefficient of the sphere is 50%
The 28 MHz Round Trip Isotropic Attenuation using the concept of Radar Equation is as follows:
Pt x Gt x Ar x SigmaPr = ------------------------------ (4 x 3.14 x R^2)^2
where :
Pr = received power
Pt = transmitted power = 1watt
Gt = gain of a 28 MHz isotropic antenna = 1 in power ratio
Ar = Aperture of the isotropic antenna at 28 MHz in square meters.
R = Radius of a sphere wich distance from the earth is 1500^3 i.e the distance from the Passive HF relay satellite and the earth expressed in meters.
Sigma = Surface of the target in square meters i.e. of the Passive HF relay satellite as seen as a radar target disc multiplied by the reflectivity coefficient of 50%
Computing:
/ 2 2 /\ 10.71 Ar = ---------- = ----------- = 9.13 square meters 4 x 3,14 4 x 3,14Sigma = 5^2 x 3.14 x 0.5 = 39.2 square meters
1 x 1 x 9.13 x 39.2Pr = --------------------------------------- = 4.47 ^ -25 watt (4 x 3.14 x 1500000^2) ^2
1Round trip attenuation = 10 log --------------- = 243.5 dB 4.47^ -25
Link budged calculation:
Assuming that we are using a good HF receiver with a NF= 8 dB equivalent to 1539 kelvin we must consider in addition that the receiver sensitivity is limited by the external available noise power.For quiet,rural locations the galactic noise is the limiting factor and at 28 MHz the noise temperature is around 29.000 kelvin so that reducing the Noise Figure belove 8 dB at 28 MHz do not improve too much the S/N ratio.
With the above data the noise floor of this receiver for SSB into a bandwidth of 2500 Hz can be calculated as follows:
Noise Floor = KTB = 1.38 x 10^-23 ( 1539 + 29.000 ) x 2500 = - 151dBW or - 121 dBm
TX power 1000 watt.............................+30 dBW TX Antenna gain....................................+10 dBi ----------- Transmitted EIRP .................................+40 dBW Round trip attenuation 1500 km..........- 243.5 dB ----------- Received power Pr on isotropic antenna on the earth ..............................-203.5 dBW RX antenna gain....................................+ 10 dB ----------- Available power at RX input............... - 193.5 dBW RX noise floor...................................... - 151 dBW ----------- Signal received with a S/N ratio.......... - 42.5 dB
So according with Bob calculations the signal is 42.5 dB under the noise and so it is not detectable.
Best 73" de
i8CVS Domenico
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Sent via AMSAT-BB@amsat.org. Opinions expressed are those of the author. Not an AMSAT-NA member? Join now to support the amateur satellite program! Subscription settings: http://amsat.org/mailman/listinfo/amsat-bb
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