How to calculate increase of home wireless router range?

Never ever could bullshit its way out of a wet paper bag.

Never ever could bullshit its way out of a wet paper bag.

While your original was technically correct, its completely irrelevant to what was actually being discussed, WHETHER THE RECEIVE ANTENNA EVER RADIATES ENOUGH TO BE RELEVANT TO THE EIRP RESTRICTION.

Yes, BUT NOT IN THE EIRP RESTRICTION SENSE.

And that is what was being discussed when John made such a spectacular fool of himself mindlessly rabbiting on about what is no news to anyone with a clue about receiving antennas.

Duh. So John was mindlessly rabbiting on about a complete irrelevancy WHEN THE EIRP RESTRICTION WAS BEING DISCUSSED.

I've not been following this thread, but I can tell you straight off, I am significantly more inclined to believe the guys who are not shouting and hurling abuse.

Make of that what you will, but my suggestion is to counter with rational argument backed up by references and facts, rather than insults and obscenities.

Mark McIntyre wrote

You have always been, and always will be, completely and utterly irrelevant.

What you may or may not be inclined to believe in spades.

Anyone with a clue considers the facts, not the style stuff, f****it.

I flush it where it belongs.

Dont need 'references' on that basic fact that even when the receiving antenna does reradiate about half of what it receives, THAT IS COMPLETELY IRRELEVANT TO THE LEGISLATED ERIP LEVEL.

I suggest you take your stupid suggestion and shove it up your arse, where it belongs.

it receives. An

like the current in a

of power radiated

do. If you'd like a

text, often discussed

easier to make a

Maybe to some extent, it's a matter of nit-picking over what "received power" means. You can think of an antenna as having an effective aperture size over which it captures all the energy crossing that cross-sectional area. To me it's logical to think of "received power" as the power that actually gets scooped up and delivered.

Effective aperture increases with antenna gain. Obviously something like a dipole has a relatively small effective aperture. But the effective aperture of a high-gain horn antenna for instance, will approach its actual physical cross-sectional area.

For instance, look at Figure 13 in this pdf.

Don

Wrong.

On Sun, 9 Jul 2006 09:30:46 +1000, "Rod Speed" wrote, in part:

That assemblage does not appear to be a sentence.

But it is distracting when you choose to code your message content in the style of an illiterate A-hole. I also think you should have said "content" rather than "facts." Most of what you have been posting here recently seems to be opinion rather than fact.

Oh, I've just recognised your name. Conversation over, I don't waste my time talking to the sort of fool who thinks that because they're hiding on usenet they can emit language which would get them a severe slapping in real life.

Nobody cares what you think. Not even you.

*plonk*

Best get your seems machinery seen to then.

Its a fact that even when a receiving antenna does radiate back half of what it recieves, THAT IS NOT RELEVANT TO WHAT WAS BEING DISCUSSED, THE LEGISLATED EIRP LEVEL ALLOWED.

Not a shred of opinion involved what so ever.

Some terminal f****it claiming to be Mark McIntyre wrote just the puerile shit thats all it can ever manage.

Look at this part of the thread...

[John said:] If the receiver matching is for optimal noise figure, there may be some reflection and reradiation, but there's nothing pinning it to be half the received power. [Roy said:] John is correct. [Rod said:] Nope. [Roy said:] A receiving antenna, when matched, reradiates half the power it receives. [Rod said:] Yes but that ISNT ANY REAL POWER in the EIRP restriction sense.

Rex wrote

Not about that particular para of John's. I was saying that John was not correct on the original point about whatever the receiving anntenna radiates BEING RELEVANT TO THE LEGISLATED EIRP LEVEL.

No opinion there, just fact.

It wasnt relevant to what was actually being discussed, WHETHER WHATEVER THE RECEIVING ANTENNA RADIATES HAS ANYTHING TO DO WITH THE LEGISLATED EIRP WAS ACTUALLY BEING DISCUSSED.

It wasnt an opinion, it was a statement of fact that that comment John made WAS NOT RELEVANT TO THE LEGISLATED EIRP LEVEL THAT WAS BEING DISCUSSED.

The legislated EIRP level WAS WHAT WAS BEING DISCUSSED WHEN JOHN MADE SUCH A SPECTACULAR FOOL OF HIMSELF RABBITING ON ABOUT WHAT THE RECIEVING ANTENNA RADIATES.

Pity it was a comment made WHEN THE LEGISLATED EIRP LEVEL WAS BEING DISCUSSED WITH REFERENCE TO THE RECEIVING ANTENNA.

Wrong again. It is a FACT that John's comment had no relevance what so ever to what was being discussed, whether the receiving antenna has any relevance what so ever to the legislated EIRP level. It doesnt.

Nothing to do with my mind, it was what was being discussed.

Yeah, you just made a VERY spectacular fool of yourself, yet again.

[misc.consumers.frugal-living dropped from distribution list.]

Roy Lewallen hath wroth:

I'm not so sure. I couldn't find any specific references to this effect in several books I skimmed. Same with internet searches. If true, then the concept of converting solar power in an orbital satellite, converting it to microwaves, beaming it down to an antenna array in the middle of the desert, and converting it back to electricity, isn't going to work if the array re-radiates half the power. That's going to ruin quite a few nifty science fiction stories and innovative business plans.

I also note that the common microwave path analysis calculations don't take re-radiation into account. For example, if I start with an EIRP of perhaps XX dBm from a transmit antenna, -YY dB of path loss, and ZZ dB receive antenna gain, the power delivered to the receiver (ignoring coax losses) is calculated at (XX - YY + ZZ) dBm without any mention of the -3dB that would need to be subtracted if half the receive power is re-radiated from the rx antenna. It would seem that the common formula and web forms for link calculations are -3dB off.

I trust your judgement in such matters and you have far more expience than me, but something seems wrong or I'm missing something. Can you point me to any books or refernences? I just skimmed Chapter 2 (Fundamentals of Antennas) in "Antenna Engineering Handbook" by Jasik (1961) and found no obvious mention of this effect.

Antennas For All Applications by John D. Kraus and Ronald J. Marhefka Third Edition Page 746, Paragraph 21-15

"Prec=(Rr/(Ra+Rr))Pa

where

Rr=receiver impedance, ohms Ra=antenna radiation resistance, ohms

For a perfect match, Rr=Ra, so that

Prec=(Rr/(Rr+Rr))Pa=0.5Pa (W)

and the receiver gets 1/2 the power collected by the antenna. The other half is reradiated."

Also see "TV and Other Receiving Antennas" by Arnold B. Bailey published in

1950 by Rider Publishing. Beginning on page 235 near the bottom:

"In radio receiving antennas the predominant resistance is, strangely enough, largely due to the fact that no electrons can move on the antenna surface *without also sending radio energy back out into space*. So here we have the paradox of a receiving antenna, having the prime function of collecting or extracting energy from space, but unable to do so *without itself returning radio energy of like kind* into space. The amount which it returns is one-half of the total that it extracts under properly matched conditions. In a good installation, with the antenna properly connected to its receiver load, the receiving antenna will be able to *deliver to its load one-half of the energy* it extracts from the oncoming radio wave but, by necessity, *must return the other half to free space*. A receiving antenna, then, is itself a *new source* of radiation. This is not so surprising, since *any* reflecting surface, as we have seen, establishes a new source of radiation"

(Note: The emphasis in the book was italics)

Cheers, John

Antennas For All Applications by John D. Kraus and Ronald J. Marhefka Third Edition Page 746, Paragraph 21-15

"Prec=(Rr/(Ra+Rr))Pa

where

Rr=receiver impedance, ohms Ra=antenna radiation resistance, ohms

For a perfect match, Rr=Ra, so that

Prec=(Rr/(Rr+Rr))Pa=0.5Pa (W)

and the receiver gets 1/2 the power collected by the antenna. The other half is reradiated."

Also see "TV and Other Receiving Antennas" by Arnold B. Bailey published in

1950 by Rider Publishing. Beginning on page 235 near the bottom:

"In radio receiving antennas the predominant resistance is, strangely enough, largely due to the fact that no electrons can move on the antenna surface *without also sending radio energy back out into space*. So here we have the paradox of a receiving antenna, having the prime function of collecting or extracting energy from space, but unable to do so *without itself returning radio energy of like kind* into space. The amount which it returns is one-half of the total that it extracts under properly matched conditions. In a good installation, with the antenna properly connected to its receiver load, the receiving antenna will be able to *deliver to its load one-half of the energy* it extracts from the oncoming radio wave but, by necessity, *must return the other half to free space*. A receiving antenna, then, is itself a *new source* of radiation. This is not so surprising, since *any* reflecting surface, as we have seen, establishes a new source of radiation"

(Note: The emphasis in the book was italics)

Cheers, John

Scroggie's "Foundations of Wireless", the book I cut my teeth on (and my father before me...) mentions this in the chapter on Radiation and Aerials where he describes the reratiation as a fact of great importance in recieving aerial design.

I think however Rod's final remark is perhaps the most telling - if an antenna really did absorb all the energy landing on it, there would be highly curious side effects.

John - KD5YI hath wroth:

OK, that make sense. Thanks much.

So, why is NASA and other disreputable organizations still pushing beaming solar power down to earth via microwaves? |

| | | | | mention in any of these that half the delivered power gets re-radiated. If the receive array were anywhere as directional as the xmit array, then it would cook the satellite. The re-radiated power has to go somewhere. Oh, maybe because it's a government project the number don't need to be correct?

I'll spare you the usual anal joke about rectenna.

Also, why is there no -3dB added loss on the receive end of wireless path calculation? From what I can deduce from the path calcs, the receive antenna delivers all the power to the coax cable and then to the receiver. Where's the half power (-3dB) loss?

I'm still (half way) mystified.

Argh. That was published 2 years after I was born.

Resistance? Unless he's thinking of ohmic resistance, most antennas don't have any resistance. (Well, a rhombic has a 300-400 ohm load, but that's not what we're discussing). If the signal gets re-radiated, without any loss, there can't be any dissipative elements in the system. What resistance?

I could add more references to the ones already mentioned, but you should be able to find it in most antenna texts. Look in the index under aperture and scattering cross section. When dealing with path loss calculations, the effective aperture is used, and this has the reradiation already accounted for. In fact, the reradiated power has its own descriptive unit, the scattering aperture. A good and brief description of these can be found in Kraus' _Antennas_, p. 29ff, and many other texts.

You're right that the antennas used to receive beamed power will catch only half of it at best. But many, many business plans have been developed and billions in stock sold for schemes which are much less plausible. For starters, how about the current idea of hydrogen "fuel", "made from water"? (For those not acquainted with the harsh reality of thermodynamics, it takes more energy to extract hydrogen from water than you'll get back when you burn it. Charlatans notwithstanding, there's just flat no way around this little fact.) Then there's SDI. . .

Roy Lewallen, W7EL

On Sun, 09 Jul 2006 22:15:00 -0700, Roy Lewallen wrote in :

The point of hydrogen is a means of energy storage and transmission, not power generation. Typically the hydrogen will be released from water by solar or nuclear power. See: