Talk:Radio propagation

message: Radio propagation is *influenced* by the conductive properties of the earth and ionosphere but does not rely on them. Radio waves propagate through a vacuum without movement of charge carriers.--Wtshymanski 20:37, 2 December 2005 (UTC)
replies: text: No, you are mistaken. Radio propagation does not occur by electrical conduction. In particular, when radio waves or light or any other EM radiation propagate in a vacuum there are no charge carriers moving, as Wtshymanski said. It's a little more complicated when a wave propagates through a medium, since then electrons in the medium do move in response to the wave, and this affects the propagation of the wave. The electrons do not, however, propagate along with the wave—they move perpendicularly to the direction of wave propagation. Fundamentally, propagation of EM waves is not an electrical conduction phenomenon. --Srleffler 22:49, 7 December 2005 (UTC)
replies: text: Precisely.  You can't see down a copper wire, for example. --Wtshymanski 03:44, 8 December 2005 (UTC)
thread_title: Conduction?

message: I have to disagree that doubling the distance from the transmitter means the signal strength is reduced to (exactly) one quarter - the same cannot inherently be true in radio signal propagation - being that electromagnetic transmission is subject to the same velocity constraints as the speed of light (providing no distortion) - radio signal strength (VSWR) and distance (distortion) go hand in hand with power signal integrity - therefore transmission is reduced to "NEARLY" one quarter.  --Lperez2029 23:06, 25 November 2006 (UTC)
replies: text: It sounds like most of that whole paragraph was lifted from a textbook of some description, see my edit below. MonstaPro 03:08, 23 April 2007 (UTC)
thread_title: VSWR

message: It would be good to have some information about radio propagation through other media than Earth's atmosphere, in particular submarine radio propagation. 131.111.8.103 15:05, 11 March 2007 (UTC)
thread_title:  Underwater propagation 

message: Thought I'd rearrange it so the mathematical formula is easier to read. MonstaPro 03:08, 23 April 2007 (UTC)
replies: text: Also added link to Google Maps as paradigm of increased stability in satellite links (yes, this occasionally conjecturous!). MonstaPro 03:15, 23 April 2007 (UTC)
replies: text:  What does Google Maps have to do with satellite link stability? --Drizzd 09:04, 14 June 2007 (UTC)
thread_title:  Inverse-square law, format of 

message: (I first asked this over at Watt, but it was suggested that this might be a better venue P=)
Ignoring gains, losses and free-space loss for the moment, field strength is typically defined as 
\scriptstyle{V/m=\frac{V}{4\pi\,m}=\sqrt{\frac{4\pi 30W}{4\pi\,m^2}}=\frac{\sqrt{30W}}{m}\,\!} or \scriptstyle{W/m^2=\frac{W}{4\pi\,m^2}.\,\!}  I realize field strength defines the surface area of a "power sphere" (inversely, \scriptstyle{W=\frac{W}{m^2}4\pi\,m^2\,\!}), but when attempting to define the strength of a signal at a receiver's input point (i.e., antenna connection point), isn't the situation like a ball resting on a flat surface?:  The ball's contact to the surface is only at a single point (or very small surface area around the point), not the whole surface area of the ball, thus the more relevant measurement is the "strength/intensity" of the radius <touching the flat surface (i.e., the particular "flux line").  Wouldn't the strength of the flux line be \scriptstyle{\frac{W}{m}\,\!} and be considered the "signal strength" (signal strength seems to confirm that, but other articles and outside sources seem to use signal and field strength interchangeably)?
::
Let W_l=W/m=\frac{W}{m};\qquad\;W_a=W/m^2=\frac{W}{4\pi\,m^2}.\,\!
Let's say you have a 100kW station that is 1km away, giving the following results:
::
100,000W @ 1000m	100Wl	.007958Wa
If you have ten other stations with different ERPs and distances, five equaling Wl and five Wa,
:::
               	Line 	Area
10,000W @  100m	100Wl	.07958Wa
1000W @   10m  	100Wl	.7958Wa
100W @    1m   	100Wl	7.958Wa
10W @   .1m    	100Wl	79.58Wa
1W @  .01m     	100Wl	795.8Wa
:::
               	Line  	Area
1000W @  100m  	10Wl  	.007958Wa
10W @   10m    	1Wl   	.007958Wa
.1W @    1m    	.1Wl  	.007958Wa
.001W @   .1m  	.01Wl 	.007958Wa
.00001W @  .01m	.001Wl	.007958Wa
and you measured the strength of the signal right at the antenna connection point (but not through any antenna, just "barefoot") of a receiver, which set of five stations would equal the same strength of the 100kW station, the 100Wl or .007958Wa set?  If it is Wl, would it be the same for any antenna attached or, as antenna length/size increases, is that when area (i.e., Wa) comes into play?  Now let's say there is an eleventh, distant station, say 500km away (well beyond its prescribed field strength), coming in via, say, either ducting or E-skip, wouldn't that be the same 100Wl or .007958Wa?
Finally, incorporating free-space loss (where λ is wavelength), W/m^2=\frac{W}{4\pi\,m^2}.\,\! becomes W/m^2=\frac{W}{4\pi\,m^2}\cdot\frac{\lambda^2}{4\pi}=W\left(\frac{\lambda}{4\pi\,m}\right)^2.\,\!.
Does that mean W/m\,\! becomes \frac{W\lambda}{\sqrt{4\pi}\,m}\,\! or just \frac{W\lambda}{m}\,\!? ~Kaimbridge~16:41, 24 April 2007 (UTC)
replies: text: The strength of the "flux" or field strength, is actually measured in volts per meter, which as you indicate drops off inversely proportionally to the distance.  So an antenna receving a signal will have a voltage produced that drops off proportionally to the distance from the transmitter.  However the current available from the antenna is proportional to the voltage.  The power available is proportional to the voltage multiplied by the current, so that the power from the receiving antennna still falls off inversely propoprtional to the square of the distance from the emiitter. GB
Right, but that's ignoring the actual question:  Yes, field strength is usually measured in volts, but it is a constant that \scriptstyle{W=\frac{V^2}{\Omega}\,\!}.  I guess the core question is why is it \scriptstyle{\frac{(V/m)^2}{\Omega}\,\!} and not the appearingly more obvious \scriptstyle{\frac{V^2}{\Omega\,m}\,\!}, as distance has nothing to do with the actual conversion of V to W?!?  Relatedly, I find it misleading to define \scriptstyle{\operatorname{dB}V/m=20\log_{10}(V/m)\,\!}, as \scriptstyle{20\log_{10}(V/m)=10\log_{10}(V^2/m^2)\,\!}, thus \scriptstyle{\operatorname{dB}V/m\,\!} should actually be \scriptstyle{\operatorname{dB}(V/m)^2\,\!}!
But back to the original question, which way is more fundamentally "right" depends on which of the above two sets of five signals (100Wl or .007958Wa) is true——since only one of them can be! P=) ~Kaimbridge~19:29, 7 May 2007 (UTC)
thread_title:  Signal strength(?) vs. field strength 

message: There are two inter-related parameters: Field Strength, which has the dimensions Volts per metre, and Power Flux Density, which has the dimensions Watts per square metre. They are related by the formula:
      E^2
PFD = --- 
      Zfs
Where:
*PFD = Power Flux Density in Watts per square metre
*E = Field Strength in Volts per metre
*Zfs = Characteristic Impedance of Free Space, a constant with a value of 377 ohms.
The terms Field Strength and Power Flux Density have these dimensions and are precise engineering terms. The term Signal Strength is more vague. It could mean the same thing as field strength (i.e. relating to an electromagnetic wave ), or it could refer to a signal voltage in a circuit. --Harumphy 13:08, 15 June 2007 (UTC)
thread_title:  Field Strength, Power Flux Density and Signal Strength

message: Radio waves has an section on propagation that needs this material.  There may be propagation of other than radio wave material that could be retained here?  Ideas?  
John 19:15, 15 August 2007 (UTC)
thread_title:  Proposal to Merge this with "'''Radio Waves'''" 

message: Removed the term 'base' and replaced with 'station' base sound's like 'CB slang' station is the proper term. Also removed reference to 'power input' satellite communications do not require allot of 'power' just allot of 'antenna' . --DP67 talk/contribs 14:32, 28 October 2007 (UTC)
replies: text: fix header. --DP67 talk/contribs 14:46, 28 October 2007 (UTC)
thread_title: Metor scatter

message: This is bullcrap. We're in a solar minimum now. CQWW, the largest ham radio contest of the year, was the weekend just gone, and there was PLENTY of actvity on 21 and 28 MHz bands. True, it's mostly single hop, but to imply the bands are unusable is patently false. When the results and reviews of the contest are published this should be eminently confirmed.
Replace with "During solar minima, propagation of higher frequencies is generally worse." 128.232.250.254 12:24, 30 October 2007 (UTC)
thread_title: "During solar cycle minimas, frequencies only up to about 15 MHz are usable"

message: Correct me if I'm behind the latest scientific research, although I don't think I am, but doesn't the section on Sproadic E contain many speculative and unproven hypotheses about the cause of Es?  Gerry Lynch (talk) 18:00, 17 July 2008 (UTC)
replies: text: This is a well established phenomena with many papers on the topic showing in an internet search.  The gravity waves are air moving up and down not gravitational waves. 
*http://www.agu.org/pubs/crossref/2004/2004JA010508.shtml
*http://www.amfmdx.net/propagation/Es.html Review
*http://www.nature.com/nature/journal/v435/n7043/full/nature03638.html
03:36, 18 July 2008 (UTC)
replies: text:  If someone doesn't beat me to it, I'll put it back (with a mention of air movements, not gravitational waves).  I must say they are stunningly badly named. What's waving in a "gravity wave" ? --Wtshymanski (talk) 12:10, 18 July 2008 (UTC)
thread_title: Sporadic E

message: I thought I understood a little about radio propagation but I don't understand the diagram at all. Does this explain *anything* to the reader without experience in the field? --Wtshymanski (talk) 14:45, 15 February 2009 (UTC)
thread_title: Does anyone understand the diagram?

message: I removed the reference to google maps. The original sentence read "(see Google Maps for a "real-world" application)". However, this is not an example of satellite communication. It is an example of satellite imaging. Further, information on satellite communication is more accurately included in the satellite article, and not in here. Once the difference is explained and the corresponding article has been linked that should be sufficient. So. I took it out. Wilson Harron (talk) 03:36, 11 April 2009 (UTC)
thread_title: Google maps

message: The reason I came here is that someone had an open req for a wiki page on the F1 and F2 layers etc. It occured to me to see what existed. Certainly some plasma stuff and diagrams of atmosphere must exist somewhere, something like an electromagnetic cross section of the atmosphere. 
While I'm here, I think their are plenty of good sites to ref here but haven't looked in a while. Obviously ARRL plays to my soapbox but I would expect them to have
good refs here. Just skimming these, I guess qsl.net and hfradio.org do have a good selection of stuff, 
http://images.google.com/images?hl=en&safe=off&um=1&q=radio+propagation&sa=N&start=40&ndsp=20
Allocation is not equivalent to prop but factors influence it. Cool charts like this,
http://www.ntia.doc.gov/osmhome/allochrt.pdf
Nerdseeksblonde (talk) 00:31, 6 September 2009 (UTC)
thread_title:  any links to ionosphere or plasma stuff etc?  

message: I'd like to add some information to wikipedia about the two ray path loss model. It's hinted at here when ground reflections and the inverse fourth law are mentioned. Presumably this wants to go in its own article, and just be linked from this page?
Thelastbert (talk) 11:18, 21 December 2010 (UTC)
replies: text: Well, here's my take on it.  If you're planning on a couple of paragraphs, with maybe a diagram and a few references, as a sort of overview explaining what the model is and where you'd go to find out more about it, I'd like to see it in here.  I think then you won't have to give so much context to explain what you're talking about (what's a radio wave? what's an ionosphere? etc.). If you're planning a longish article, with maybe a brief introductory section that provides the background for a general reader and allows the keener to wade deeply into the volume integrals and relativistic effects, then it should be mentioned here and linked when it's ready. I would err on the side of a short, accurate, comprehensive introductory section here - Einstein said we should be able to explain physics to a barmaid - rather than getting into retyping someone's graduate thesis project. I, of course, am a dangerous and non-consensus-holding minority.  I love reading well-written material in Wikipedia on things that I know only a little or nothing about. I'm looking forward to anything on the two-ray path loss model. A good explanation of how these things work would help the reader who wants to know why, even though there's a cell tower every three blocks, his phone that worked fine *over there* quits working *over here*. --Wtshymanski (talk) 14:27, 21 December 2010 (UTC)
thread_title:  Two Ray Path Loss Model 

message: T.F. Giella is a well known name in propagatio studies, but the block of text attributed to him needs a formal citation to a reliable source, per WP:V. Also this block of text, after resolving the reference issue, is more suited for the main article Sporadic E propagation. If no reference can be found I shall delete it fro here and copy it to the discussion page of the main article. SV1XV (talk) 06:41, 13 September 2012 (UTC)
replies: text: This text was from the old site of T.F. Giella. It seems that it is no longer available at the original site, but it was copied by the author to the Propagation mailing list and is archived at contesting.com as http://lists.contesting.com/archives//html/Propagation/2005-04/msg00075.html  Therefore I have to delete it as copyvio and replace it with a link to the contesting.com archive. SV1XV (talk) 07:43, 13 September 2012 (UTC)
thread_title:  T.F. Giella on Es propagation