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Parkes large hyperbolic radio antenna

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alright1234

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I did a little calculation here. Do the values used seem responsible?


The power of the radio signal produce by the lander is estimated at 50 Watts. A radio signal's intensity is dependent on the inverse of the second order of the distance I = U2 = {[A cos(kr)]/r]}2 where A represents the power of the radio signal. After a radio signal propagates the distance of 238,000 miles (3.8 x 108 m) to the earth, a 50 W radio signal would diminish to I = A2 [cos2(kr)]/r2 = (50 W)2 (0.5) / (3.8 x 108 m)2 = 8.65 x 10-15 W2/m2. The Parkes large hyperbolic radio antenna located in Sydney Australia is used to communicate with the Apollo 11 mission. The sensitivity of the Parkes large hyperbolic radio antenna is extrapolated using the power of a 300 km height satellite that emits a 20 W radio signal that forms a detectable intensity at the surface of the earth I = A2 [cos2(kr)]/r2 = (20 W)2 (.5) / (3 x 105m)2 = 2.22 X 10-9 W2/m2. Adding two orders of magnitude to the satellite's power at the surface of the earth, the sensitivity of the Parkes radio antenna is estimated at 10-11 W2/m2. There is a four order of magnitude between the sensitivity of the Parkes and the 10-15 W2 /m2 s-band radio signal that originates from the Apollo 11 mission which suggest that NASA Apollo 11 lunar mission was hoax. Also, a radio signal cannot penetrate the Van Allen radiation belt that surrounds the earth. It is questionable how NASA communicated with the Apollo missions, Voyagers, and Mars probes using radio waves. It is argue that a satellite that is orbiting the earth at a height of 30,000 km above the earth that is passed the Van Allen belt justifies the functionality of the Apollo 11 communication system but the described satellite height is 10 % of the distance to the moon which is a doubtful magnitude for the height for orbiting communication satellite since there is no reason that a communication satellite would orbit at that height that would require more power to produce a usable signal at the earth. The space stations are orbiting the earth at the height of 130-250 miles above the earth. I predict that the maximum orbital height of a satellite is 500 miles.
 
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Nigel Goodwin

Super Moderator
Most Helpful Member
I did a little calculation here. Do the values used seem responsible?


The power of the radio signal produce by the lander is estimated at 50 Watts. A radio signal's intensity is dependent on the inverse of the second order of the distance I = U2 = {[A cos(kr)]/r]}2 where A represents the power of the radio signal. After a radio signal propagates the distance of 238,000 miles (3.8 x 108 m) to the earth, a 50 W radio signal would diminish to I = A2 [cos2(kr)]/r2 = (50 W)2 (0.5) / (3.8 x 108 m)2 = 8.65 x 10-15 W2/m2. The Parkes large hyperbolic radio antenna located in Sydney Australia is used to communicate with the Apollo 11 mission. The sensitivity of the Parkes large hyperbolic radio antenna is extrapolated using the power of a 300 km height satellite that emits a 20 W radio signal that forms a detectable intensity at the surface of the earth I = A2 [cos2(kr)]/r2 = (20 W)2 (.5) / (3 x 105m)2 = 2.22 X 10-9 W2/m2. Adding two orders of magnitude to the satellite's power at the surface of the earth, the sensitivity of the Parkes radio antenna is estimated at 10-11 W2/m2. There is a four order of magnitude between the sensitivity of the Parkes and the 10-15 W2 /m2 s-band radio signal that originates from the Apollo 11 mission which suggest that NASA Apollo 11 lunar mission was hoax. Also, a radio signal cannot penetrate the Van Allen radiation belt that surrounds the earth. It is questionable how NASA communicated with the Apollo missions, Voyagers, and Mars probes using radio waves. It is argue that a satellite that is orbiting the earth at a height of 30,000 km above the earth that is passed the Van Allen belt justifies the functionality of the Apollo 11 communication system but the described satellite height is 10 % of the distance to the moon which is a doubtful magnitude for the height for orbiting communication satellite since there is no reason that a communication satellite would orbit at that height that would require more power to produce a usable signal at the earth. The space stations are orbiting the earth at the height of 130-250 miles above the earth. I predict that the maximum orbital height of a satellite is 500 miles.
Needless to say, everything you posted is complete and utter rubbish, perhaps you might like to go on a conspiracy site instead?.

I predict that the maximum orbital height of a satellite is 500 miles.
To deal just with your last ludicrous statement, I suggest you try looking up geo-stationary orbit, or the 'Clarke Belt', which is FAR above your suggested maximum.
 

rjenkinsgb

Active Member
I did a little calculation here. Do the values used seem responsible?
Sorry, total nonsense.

Some examples;
The satellites used for satellite television, in geostationary orbit at about 22,500 miles above the earth's surface and using quite small antennas with transmit powers in the tens of watts up to 100W or so, can be picked up with dish antennas of about 18" to 2ft diameter.

Each doubling of antenna diameter (with the same frequency in use) give 4x signal gain so approx. double the working range for the same power levels.
With dish antennas, gain is related to frequency; the higher the frequency the smaller the wavelength so the more wavelengths a given size of dish relates to and the higher the gain.


Re. space communications in general, one of the more specialist modes of Amateur radio (ham) operation is "Moonbounce" [formally "EME"] - using VHF or UHF equipment and directional aerials with full steering capability to bounce signals off the moon, so being able to communicate with people in different parts of the world where direct VHF or UHF communications would be impossible due to the signal otherwise having to pass through the earth.
https://www.contesting.com/articles/294

Space communications is actually a lot less problematic in some ways than terrestrial, as there are less problems with multi-path signals and terrain etc.
Some of the Skylab mission astronauts have been radio amateurs and I've talked to one of them some years ago, using a handheld with 5W transmit power. It's the perfect "clear path" so losses are quite low.


Edit.
Just to clarify something you mention.
The various layers of the ionosphere react very differently to different frequencies of electromagnetic radiation.
Mid-Low frequencies such as medium wave radio & "short wave" are variously absorbed or reflected depending on angle of incidence etc. and these effects also vary between day and night.

Higher frequencies are affected less or not at all - VHF through microwave frequencies, and obviously visible light, can pass through.
Also VLF is little affected (but not much use for space comms).

More info:
https://www.ibm.com/support/libraryserver/handheld/Connected/BOOKS/QBKA2J00/1.3.4?DT=19940909135003
https://radiojove.gsfc.nasa.gov/education/educ/radio/tran-rec/exerc/iono.htm
 
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JimB

Super Moderator
Most Helpful Member
In addition to what has already been said:

The "Dish Antenna" will be in the form of a parabola, not an hyperbola. There is a big difference.

To put some real world engineering numbers to the signal strength to a signal transmitted from the moon.

For the gain of a parabolic antenna:
\[ Gain\;=\;10\times Log_{10}.k.\left(\frac{\pi.D}\lambda\right)^2 \]
Where
Gain in dB
k is antenna efficiency
D is diameter in Metres
lambda is the wavelength in Metres

Using k =50%
D = 64 metres
lambda = 0.125 metres
Gives a gain of 61dB

There is a well known formula for radio path loss in free space:
\[ Loss\;=\;32.45\;+\;20.Log_{10}F\;+\;20.Log_{10}D\; \]
Where
Loss in dB
F is frequency in Mhz
D is distance in kilometres

Using F = 2400MHz
D = 400000 kilometers
Gives a path loss of 212dB

To find the signal strength of the signal received on earth from the moon:
Transmitter power = 50 watts or +47dBm
Transmitter antenna gain = 10dB (best guess)
Receiver antenna gain = 61dB
Path loss = 212dB

Add it all up +47 +10 +61 -212 = -94dBm
Which is a nice strong signal.

JimB
 

alright1234

New Member
Dear beloved Friends and colleagues


A cell phone has the power of about 1 W and a range of 1 mile. How can a 50 W radio wave from the moon that is 238,000 miles from the earth form a detectable signal? It just does not make sense to me. Thank you.


You dear friend

alright1234

Love, peace and the Grateful Dead

"may the force be with you"
 

JimB

Super Moderator
Most Helpful Member
A cell phone has the power of about 1 W and a range of 1 mile. How can a 50 W radio wave from the moon that is 238,000 miles from the earth form a detectable signal?
The antenna in a cell phone is not very efficient, usually just a slab of ceramic which has been metalised.
The cellphone antenna is designed to be compact, it is not designed to be very efficient, id does not need to be.
The cell base station antenna has no where near as much gain as the large parabolic antenna, it does not need it.
A cellular phone system is designed to have many closely spaced (a few miles) base stations, it does not need high sensitivity.

It just does not make sense to me
Probably because you do not know enough about radio stuff yet.

JimB
 

MikeMl

Well-Known Member
Most Helpful Member
...Probably because you do not know enough about radio stuff yet...
That never seems to slow down the science deniers.
 

Ylli

Active Member
100 miles straight up, certainly. Parallel to the ground..... The level of terrestrial 'noise' might be higher than the level of the desired signal.
 

alright1234

New Member
Straight up is difficult to determine since we have to rely on the integrity of NASA which I am contesting. What about a 3 meter radio telescope on a mountain would that detect a 1 W cell phone signal 100 miles away? Do you believe in climate change?
 

Ylli

Active Member
Depends on the gain of the said antenna - would need to do some path loss calculations. What has climate change to do with it?
 

MikeMl

Well-Known Member
Most Helpful Member
Can a large radio telescope detect a 1 W cell phone signal 100 miles away?
Easily. The transmitter on Voyager I is only 22W, and NASA can receive its signal from 22 Billion km out in space, beyond the edge of our planetary system.

I have personally stood outside, and talked to a ham astronaut aboard the ISS during a near overhead pass using a 1.5W 146MHz walky-talky with a crappy 6" rubber-duck antenna on it. The ham radio in the ISS is a nothing-special mobile two-way radio with a simple monopole (almost no gain) antenna.

I routinely talk to hams from the ham rig mounted in my airplane over distances of ~400mi. Again, the ham on the ground might be using a 25W radio with a simple 19" whip antenna. The antenna mounted on the belly of the airplane is a 19" taxi cab antenna...

My best 2meter band airplane-to-airplane contact is >1400mi. He was at 47,000ft in a Lear Jet; I was at ~12,000ft agl in my Cessna.
 
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Nigel Goodwin

Super Moderator
Most Helpful Member
I suggest you go away and find a forum for conspiracy nutters?, you're just making an utter fool of yourself here with your ludicrous ideas.

As someone with no understanding of science of electronics, you probably don't even know that radio hams have been bouncing radio signals off the moon (using it as a passive reflector) for many decades, in order to get around the curvature of the earth.

I can only mention to you that your time here is VERY limited, you've nothign to contribute and nothing useful to say.
 

Ylli

Active Member
Straight up is difficult to determine since we have to rely on the integrity of NASA which I am contesting.
So am I trying to discuss factual data with someone who doesn't believe we landed on the moon and that the Earth is flat?
 

rjenkinsgb

Active Member
Can a large radio telescope detect a 1 W cell phone signal 100 miles away?
As I've already stated, I've worked one of the astronauts while they were in earth orbit, using just a handheld "walkie talkie" stye ham radio - one of these, to be exact, with the little helical antenna as in that photo.
https://img.yumpu.com/45875827/1/500x640/icom-ic-32-user-manual-radiomanualeu.jpg

(I was thinking it was Skylab, but was a slightly later series of space shuttle flights in the early 80s).
https://en.wikipedia.org/wiki/Shuttle_Amateur_Radio_Experiment

Very little RF power is needed for a near-vertical path of a couple of hundred miles and you don't need any fancy antennas etc.


A cell phone has the power of about 1 W and a range of 1 mile.
Again, nonsense.

The first 4G cell tower in this area was 20 miles from where I live and that was giving a reasonable signal, totally functional with a less-than-1W phone.

The GSM system itself uses time division multiplex and for the phone transmitters to "slot in" without interference, it's continuously calculating distance from the cell tower and telling the phone how much to adjust the transmit timing so all the phone transmitters on that channel synchronise.

The time offset range allows for a phone to be up to 35 miles from a cell, if I remember right.
Most coverage limitations are due to terrain & obstructions, not RF power or sensitivity or the system itself.


Re. NASA - Remember that the Apollo lunar missions took place during the cold war and the USSR would have like nothing more that to show any part of the missions being faked; they were not far behind the USA planning lunar missions.

The USSR put the first astronaut in space and the had full facilities to monitor and track the lunar transfers etc.
Plus many other countries watching things extremely closely - and later unmanned probes sent by various countries which have independently taken photos of the Apollo landing sites...


Edit - correction, the timing range of the original GSM spec allowed a range of 35 Km not 35 miles.
One of the early revision to the specifications added an "extended range" bit however, which relaxes the timeslot requirement and allows ranges of up to 120Km between the phone and a cell.
http://www.telecomhall.com/parameter-timing-advance-ta.aspx
 
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alright1234

New Member
I've re-written the subject to: (and did not add the cell phone). If a satellite can orbit at 30,000 km above the surface why do not the the space stations and there is also the problem with the Van Allen belt. How do radio waves propagate through the Van Allen belt and what is the physical maximum range of a radio wave? According to NASA it is infinite yet cell phone require a tower. Why can not a cell phone work from china to the USA or from Paris to NY without a tower? There has to be a maximum range of the radio signal. Using a cell phone, at 1 W and 100 miles then a 50 W signal from the moon would have a maximum range of 5,000 miles yet the moon is 383,000 miles from the earth which represents three order of magnitude difference and it is 4.8 billion miles to Pluto. So how did the Japn communicate with that probe on the asteroid. To communicate with the moon you would have to send out communication satellites along the path to the moon but I may be wrong since I never meet Neil but I did use the bathroom at the NASA space center (last stall).


The power of the radio signal produce by the Apollo 11 lander is estimated at 50 Watts. A radio signal's intensity is dependent on the inverse of the second order of the distance I = U2 = {[A cos(kr)]/r]}2 where A represents the power of the radio signal. After a radio signal propagates the distance of 238,000 miles (3.8 x 108 m) to the earth, a 50 W radio signal's intensity would diminish to I = A2 [cos2(kr)]/r2 = (50 W)2 (0.5) / (3.8 x 108 m)2 = 8.65 x 10-15 W2/m2. The Parkes large hyperbolic radio antenna located in Sydney Australia was used to communicate with the Apollo 11 mission. The sensitivity of the Parkes radio antenna is extrapolated using the radio signal intensity formed by a 300 km height communication satellite that emits a 20 W radio signal and produces an radio signal intensity at the surface of the earth of I = A2 [cos2(kr)]/r2 = (20 W)2 (.5) / (3 x 105m)2 = 2.22 X 10-9 W2/m2; adding two orders of magnitude to the satellite radio signal's intensity that forms at the surface of the earth, the sensitivity of the Parkes radio antenna is estimated at 10-11 W2/m2. There is a four order of magnitude difference between the extrapolated sensitivity of the Parkes radio telescope and the 10-15 W2 /m2 s-band radio signal that originates from the Apollo 11 mission which suggest that NASA Apollo 11 lunar mission was hoax. Also, a radio signal cannot penetrate the Van Allen radiation belt that surrounds the earth. It is questionable how NASA communicated with the Apollo missions, Voyagers, and Mars probes using radio waves. It is argue that a satellite that is orbiting the earth at a height of 30,000 km above the earth that is passed the Van Allen belt justifies the functionality of the Apollo 11 communication system but the described satellite height is 10 % of the distance to the moon which is a doubtful magnitude for the height for orbiting communication satellite.
 
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rjenkinsgb

Active Member
Why can not a cell phone work from china to the USA or from Paris to NY without a tower?
Line of sight!
VHF / UHF / Microwave signals need to have a pretty much direct path between the two stations.

Very similar to visible light.

If anyone said "Why can't you see a torch/flashlight in New York from China", you would think them stupid...


Space does not have obstructions or limits due to curvature of the Earth.

As I've already pointed out, the various atmospheric stratospheric layers etc. have different properties dependant on frequency. Some frequencies are strongly affected, others pass with minimal or no effect.

Light is EM radiation and it's pretty obvious that light can travel through space.
Your arguments are nonsensical.
 

alright1234

New Member
There has to be some sort of limit to the range of a cell phone. A cell phone cannot communicate with the Apollo 11 lander at the moon and there is also the problem of the Voyager since how can NASA communicate with the Voyager at 1 billion miles. Something just seems to be wrong and satellite phones work but how can it communicate with a satellite that is 30 k from the earth and all physics books are concealing the intensity of a radio wave that should be known fact. I should not be even having this discussion. I should be able to just look it up in a physics book but it is not there.
 
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