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Half-Wave Antenna Questions...

 
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Half-Wave Antenna Questions...
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Fission7x
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Post Half-Wave Antenna Questions... Reply with quote
I found these formulas somewhere on the net:

492/fo MHz = The length (feet) of a half wavelength dipole in free space.

The specifications call for multiplying the above result by a correction factor read from a graph which is based on the ratio of the length of the half-wavelength to the diameter of the radiating element.
All the correction factors appear to be less than 1.0 which would shorten the overall length of the radiating elements.
The result would further be shortened by the planned space between the 2 radiating elements (.25 inches) and then divided by 2 to the the length of each element.

Does this sound correct?

Does it matter if the radiating elements are solid copper or hollow pipes?
Thu Jul 08, 2004 2:20 am View user's profile Send private message
Storminorm
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Post Double edged Sword question Reply with quote
r u posting a lecture or are you really sincere in acquireing knowledge?

Well allow me to include one important factor that seems to be a variable..It is called 'SKIN' effect of Radio Frequency Currents flowing thru a Conductor of various Diameters versus the Frequency of the waves
Thu Jul 08, 2004 11:27 pm
Fission7x
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Post Re: Double edged Sword question Reply with quote
Storminorm wrote:
r u posting a lecture or are you really sincere in acquireing knowledge?

Well allow me to include one important factor that seems to be a variable..It is called 'SKIN' effect of Radio Frequency Currents flowing thru a Conductor of various Diameters versus the Frequency of the waves


Rest assured that I'm in it for the knowledge. I don't have the time nor the inclination to lecture the dwellers of this board on things I know very little about.
This "skin" effect sounds interesting. Where can I go to read about it?
Fri Jul 09, 2004 3:13 am View user's profile Send private message
Fission7x
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Post Re: Double edged Sword question Reply with quote
Fission7x wrote:
This "skin" effect sounds interesting. Where can I go to read about it?


I found a couple of articles on the net about it.
Seems like a hollow radiaiting element is actually better than solid one! Thanks for help!
Fri Jul 09, 2004 10:18 pm View user's profile Send private message
Storminorm_COMPULSIVEPOST
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Post THE K- FACTOR Reply with quote


Determining Your Proper Length for the Open Half-Wave Dipole Antenna



This webpage uses 2 formulas designed for half wavelength antennas (which includes the open half-wave dipole) resonating above 30 MHz.

If you take your time, follow the steps, and go back through certain steps that you do not understand...I am sure you will make out fine. So, without further ado, let's begin...

FORMULA I

492/MHz = The length of a half wavelength (feet) in free space




FORMULA II

K = The length of a half wavelength (in feet) in free space to the diameter of the radiating element(s) (conductor) used for the antenna.




Formula I states how long the half wavelength is to be, for your particular transmitting (resonating) frequency. Formula II accounts for the diameter of the conducting rod, used in making the antenna. With both Formulas taken into account, the final lengths will be a very close approximation of having the antenna resonate at the designated frequency.

Iusing the two formulas will give one example on how to derive the final answer, in stated aboveshould have no trouble in . Once you have understand the example, you finding the lengths of each of your rods for the dipole antenna.

The following example will will determine your rod length in decimal form. Once that is achieved, we will then change the decimal form into a fractional form. This way, it will then be easier to use a typical measuring tape; because typical measuring tapes are sectioned off in fractions of an inch.

Let's say that you wanted to transmit on 87.7 MHz, using a rod that has a diameter of 1/2 (.5") inch.


Step 1 CLICK ON THE LINK BELOW FOR MORE INFO

http://braincambre500.freeservers.com/Open%20Half%20Wave%20Dipole%20Calculation.htm
Sat Jul 10, 2004 5:01 am
Black_Hawk
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Post SKIN EFFECT "TEORY" Reply with quote
f we feed exactly equal and opposite current into the coax, currents have no choice but to flow INSIDE the cable shield! The outermost conductor can be treated as a single wire, since everything else is "hidden inside". There will be no "extra" current to flow over the outside of the shield. Because of this rule, we do not need to ground the shield to prevent radiation. The trick here is we need to have equal and opposite currents. When the shield has potential difference along the length we can greatly reduce outside currents by making the outside of the coax have a high impedance. (BALUN CONCEPT) We can do this by selecting a proper cable length, by adding sleeves of soft-iron magnetic materials or winding the cable in a coil with or without a core to form a choke. Skin depth prevents the inside of the cable from "seeing" what is done on the outside.

http://www.w8ji.com/coaxial%20currents/coaxial_line_and_shielded_wires.htm
Sat Jul 10, 2004 5:30 am View user's profile Send private message
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