Antenna Software?
- Thread starter Murphy625
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What are you looking for the software to do.
Measurements / calculations?
Building ideas?
Dipole antennas are the easiest antennas to build, by far.
There are thousands of web pages with instructions and
calculators. Just google dipole antennas for whichever band
you are interested in
Here is a good calculator along with some other good info
http://www.westmountainradio.com/antenna_calculator.php
Measurements / calculations?
Building ideas?
Dipole antennas are the easiest antennas to build, by far.
There are thousands of web pages with instructions and
calculators. Just google dipole antennas for whichever band
you are interested in
Here is a good calculator along with some other good info
http://www.westmountainradio.com/antenna_calculator.php
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iMONITOR
Silent Key
Got it.. I liked the WestMountainRadio link.. It has a chart for all the bands.
One question though, when you loop the antenna wire around an eye-hook, do you count the extra length of wire being used to wrap around the hook? Or do you just measure to the very end?
Is the calculated cut length even that important that one needs to worry about it?
When building these antennas, I noticed that the overall length (both legs) seems to be important, or at least it seems to be.
With that in mind, does this mean that the connection points where wires meet in the center tee need to as close as they can be?
How critical is this stuff? I'm usually the type to pay attention to detail and sometimes end up going to far with it.
One question though, when you loop the antenna wire around an eye-hook, do you count the extra length of wire being used to wrap around the hook? Or do you just measure to the very end?
Is the calculated cut length even that important that one needs to worry about it?
When building these antennas, I noticed that the overall length (both legs) seems to be important, or at least it seems to be.
With that in mind, does this mean that the connection points where wires meet in the center tee need to as close as they can be?
How critical is this stuff? I'm usually the type to pay attention to detail and sometimes end up going to far with it.
KC4RAF
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There's another site, and it also is free:
Martin E. Meserve - K7MEM
He covers antenna theory for those who want an in-depth knowledge.
Martin E. Meserve - K7MEM
He covers antenna theory for those who want an in-depth knowledge.
Will your antenna be for transmitting & receiving or for receiving only?
For transmit, the length is pretty important. Twisting the wire where it fastens to the insulators can have an affect. But generally when you're installing a dipole for transmitting, you'll almost always need to adjust it by changing the length.
In practice, for transmitting antennas, you cut the wire longer than what is specified in the formula for a 1/2 wave at your desired frequency. Then you cut each end back slightly checking the SWR until you attain an acceptable ratio. Doing it this way is easier than cutting the antenna too short initially and then having to add wire.
For receiving, length isn't near as important. You can be a up to a few feet or so off the calculated length with no discernible difference in performance.
And where the coax meets the antenna... Practice is to keep the distance between the wires at at 1 to 3 inches. That's generally the length of the typical antenna insulator, or the connect points of a balun. (Google for dipole insulators and you'll see a large variety of commercial and home made ones).
For transmit, the length is pretty important. Twisting the wire where it fastens to the insulators can have an affect. But generally when you're installing a dipole for transmitting, you'll almost always need to adjust it by changing the length.
In practice, for transmitting antennas, you cut the wire longer than what is specified in the formula for a 1/2 wave at your desired frequency. Then you cut each end back slightly checking the SWR until you attain an acceptable ratio. Doing it this way is easier than cutting the antenna too short initially and then having to add wire.
For receiving, length isn't near as important. You can be a up to a few feet or so off the calculated length with no discernible difference in performance.
And where the coax meets the antenna... Practice is to keep the distance between the wires at at 1 to 3 inches. That's generally the length of the typical antenna insulator, or the connect points of a balun. (Google for dipole insulators and you'll see a large variety of commercial and home made ones).
Thank you! That gives me the information I was looking for.
For right now, until I learn more, I'm just receiving, but eventually I'm going to want to go through the steps to be able to transmit. So with that in mind, I'd rather build the antenna right the first time.
Correct me if I'm mistaken, but I'm assuming that the longer the wavelength, the bigger the "adjustment cuts" and the less perfection is needed in the antenna's length yes? And if I have that correct, what kind of numbers are we looking at?
If I cut an antenna for 80 meters, do I make each side a foot longer and take off four inches at a time? Or do I make it four inches longer and take off an inch at a time?
And what about at 20 meters?
I was watching a DIY build of a 3/4 copper pipe j-pole for 2 meters on youtube and the guy used hose clamps to temporarily connect the center conductor and shield so he could adjust their contact point before soldering. His SWR adjustments were on the order of 1/8 of an inch. I can't imagine doing that with a wire that's 40 some feet long.
Thanks for your help!.
All the dipole theory you read deals with dipoles in free space. Practically speaking there will always be other factors that come into play that will skew the results. (Height above ground, nearby metal objects etc).
Longer wavelengths require more cutting.. For example go to this site: KWARC Dipole Calculator
If you plug in the value 3.750 in the 'desired frequency' box and calculate, you get 124.8 feet.
If you plug in the value 3.760 you get 124.47 feet. 124.8 - 124.47 = .33 feet or 3.96 inches
Now lets go to 10 meters.. -- Entering 28.400 Mhz results in 16.48 feet. Entering 28.410 nets 16.47 feet. A difference of .01 feet or .12 inches..
So.. On 80 meters a 3.96 inches results in a 10 Khz change in the resonant frequency. On 10 meters 0.12 inches results in a 10 khz change... The lower the frequency, the more you need to prune (or add) in order to change the resonant point.
But again, this is in free space with no external influences. In reality there will be some variance. The 3.96 inches on 80 meters or the .36 inches on 10 meters for every 10 Khz is just a number to start with.
Longer wavelengths require more cutting.. For example go to this site: KWARC Dipole Calculator
If you plug in the value 3.750 in the 'desired frequency' box and calculate, you get 124.8 feet.
If you plug in the value 3.760 you get 124.47 feet. 124.8 - 124.47 = .33 feet or 3.96 inches
Now lets go to 10 meters.. -- Entering 28.400 Mhz results in 16.48 feet. Entering 28.410 nets 16.47 feet. A difference of .01 feet or .12 inches..
So.. On 80 meters a 3.96 inches results in a 10 Khz change in the resonant frequency. On 10 meters 0.12 inches results in a 10 khz change... The lower the frequency, the more you need to prune (or add) in order to change the resonant point.
But again, this is in free space with no external influences. In reality there will be some variance. The 3.96 inches on 80 meters or the .36 inches on 10 meters for every 10 Khz is just a number to start with.
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ko6jw_2
Member
I suggest buying an ARRL Antenna book. You can learn about practical antenna designs and antenna theory. You are asking good questions, but there is much more to know. Antennas are the one area of radio where DIY is still viable as opposed to building radios etc. For a simple dipole you don't need software, just a formula. However, no antenna will be right on just from a formula. They always need to be trimmed and, because they are influenced by their surroundings, height above ground and ground conductivity to name a few factors, you just have to experiment.
Good luck and have fun.
Good luck and have fun.
When the antenna is in place the variables are now constant. Only the length and frequency are now variable. Adjustments are merely by ratio to hit the desired resonant frequency. You only need to take down the antenna one time.
Continuing on jwt873’s example: the target frequency is 3.750 MHz but the antenna at 124.8 feet is a bit long at 3.760 MHz. It needs to be shortened by the ratio 3.750/3.760=0.99734.
Therefore 124.8 X 0.99734=124.468 feet long.
Continuing on jwt873’s example: the target frequency is 3.750 MHz but the antenna at 124.8 feet is a bit long at 3.760 MHz. It needs to be shortened by the ratio 3.750/3.760=0.99734.
Therefore 124.8 X 0.99734=124.468 feet long.
majoco
Stirrer
The length of the wire is not that critical - it won't just be able to receive one single frequency. If it were me, I'd do some more research into an "Off-centre Fed Dipole" and cover a multitude of frequency bands. Used to be called a "Carolina Windom" when it was fed with open wire.
For the most part,a dipole can be 'cut' using the formula 468/(frequency in MHz)=length in feet. The length is overall length (end to end). As mentioned, in practice, most people start with the antenna about 10% longer than the formula, then start trimming for lowest SWR. If using the antenna for transmitting,you get a 2:1 SWR 'bandwidth of about 2-4%. For receiving, just use the formula for the lowest frequency you plan to use.
There are software packages ( I use EZNEC) for designing antennas, but they would be overkill for a dipole.
To answer your question, the formula is 'end to end', if you 'wrap' the wire back on itself it 'acts' like part of the conductor it is around.
Have fun tinkering-Warning- antenna building can become a hobby in itself and antenna analyzers and software can cost as much as radios-I know.
73
There are software packages ( I use EZNEC) for designing antennas, but they would be overkill for a dipole.
To answer your question, the formula is 'end to end', if you 'wrap' the wire back on itself it 'acts' like part of the conductor it is around.
Have fun tinkering-Warning- antenna building can become a hobby in itself and antenna analyzers and software can cost as much as radios-I know.
73
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Probably the one question that I haven't gotten a consistent
answer on yet. I've been wanting to know this for about 4 or 5
years now and I think I'm seeing a 60% 40% split on the answer.
Some people say the point where the wire loops back around
(the U turn) is considered the end. Others have said that the "end"
is the point where the wire begins wrapping around itself after making
the U turn. And some say it is the total length of the wire.
For a simple wire made dipole it shouldn't be that confusing. But it is!
I always use the end of the antenna (meaning total length or end of loop). For HF frequencies, the detuning caused by nearby objects and/ground will throw things off more than weather the wire is wraped back 2-3 inches. Remember with antennas we are talking about the electrical 'field' around the wire, as well as the current in the wire. Because the antenna can be affected by anything in the near field(out to about a wavelength), antenna building is not all science- there comes a point where you make assumptions to simplify things and then 'trim' for best results.
I'm in need of clarification on something concerning the standard calculation formula results.
the formula is: 468\ Frequency in MHZ.
for instance, to use 18.050 MHZ the formula is 468\18.050 MHZ = 25.9279
Is that to be understood as 25 feet .9279 "inches"?
or do you now have to convert the .9279 to inches Like .9279 X 12 = 11.134" (or a little over 11 1/8" )?
the formula is: 468\ Frequency in MHZ.
for instance, to use 18.050 MHZ the formula is 468\18.050 MHZ = 25.9279
Is that to be understood as 25 feet .9279 "inches"?
or do you now have to convert the .9279 to inches Like .9279 X 12 = 11.134" (or a little over 11 1/8" )?
You would convert the fractional part to inches by multipling by 12.
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Sent from my SM-T580 using Tapatalk
Sooooo in other words I am Correct?
(if what you mean by the "Fractional Part" is the .134 after 11")
(if what you mean by the "Fractional Part" is the .134 after 11")
If the math says the antenna needs to be 48.5 feet long then .5 times 12 = 6 inches so it needs to be 48 ft 6 inches.
Sounds like you're right.
ko6jw_2
Member
If you're just going to receive, don't sweat the fine tuning. It makes a difference it you're transmitting, but for reception it won't make any difference. There is no sweet spot where a few inches will make a dramatic difference in reception. Get an antenna analyzer if you want to obsess over fractions of inches.
It worked for me!
I can now hear more distant transmissions, and so that's "empirical evidence" (at least for me anyway!)
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