Input Impedance of off center fed dipoles

[Jackson Richter]

My name is Jackson Richter. I am retired from Rockwell-Collins and worked in the engineering lab for 31 years. Since my retirement my interested have peaked in radio communications. Especially in antenna design. Thanks to anyone here that would give me a little more insight with my questions. The input impedance of a half wave dipole fed at the center in free space is 73 ohms with some reactance. I have been looking through a lot of data for a formula showing the input Impedance at various distances from the feed point. One in particular states from (Wikipedia)
Radiation resistance = 73.1/ sin squared(kx)
If the dipole is not driven at the center then the feed point will be higher. If the feed point is distance x from one end of the half wave (lambda/2) dipole, the radiation resistance relative to the feed point will be given by the above equation.
My question is if the feed point is at the center which is the 90 degree point, since a have wave dipole would be 180 degrees. then the sin of 90 =1 and K would also have to be 1 for this formula to work.
What is this value for K?

thanks JR

To be more exact the value for sin squared would have to be 1, for the 73 ohms to be valid in free space.

 

[tech99]

Hello JR. Not sure what factor K is - it looks as if it should be 1 - but the formula looks approximately correct, assuming a sinusoidal current distribution. If feeding near to the end, the current distribution is not sinusoidal, as the impedance does not rise to infinity.
Also be aware that it is best to have the antenna a little shorter so it is resonant. This will improve the equality of the balance. It is also likely that a current balun or RF isolating transformer will be needed at the feedpoint to avoid common mode current on the feeder. The feeder may also require a second current balun some distance from the antenna. This is because with off-centre feeding, the feeder is not located at a neutral plane and will strongly pick up RF.

 

[Jackson Richter]

Thanks so much for the info. Yes I know a current balan should be used at the feed point here. If a 4:1 balan is used at the feed point of approx. 33% of the over all length of 134ft. With a 4:1 balan then the impedances seen at different frequencies are around 120 ohms to 220 ohms. And this would equate to desirable impedance of around 50 ohms at the source input to the balan. And yes sir you are right this could cause some common mode currents on the 50 ohm source feed line. The current balan used if done professionally would offer some common mode rejection, haven't tested that out yet. With the formulas I suspect derived from Maxwell's theorems would allow to evaluate the feed point impedance at any position along the have wave antenna. That's where I am stuck, my friend the impedance equation should work for every degree on the half wave dipole. I am currently using this setup now very successfully and have worked many foreign station in my log, even with the band conditions being at its worst. So again thanks for your help.

JR

 

[tech99]

Be sure that the balun you are using is a current, not voltage, type.
It is also my thinking that at these frequencies, matching at the antenna is not the top priority because feeder loses with a Hi Z feeder are small. Matching can take place at the transmitter. The antenna needs to be chosen primarily to have the optimum radiation pattern in vertical and horizontal planes, and maximum gain in the desired direction(s).

 

[Jackson Richter]

Thanks again for your input. Yes I am using a current balun ( sorry for the poor spelling in previous posts, hard to see screen) not a voltage type. The antenna is shaped like an inverted V with the apex about 44ft high. The great feature of this antenna, it is multiband and the Q is surprisingly low, making it broadband as well. Further investigating is needed for the input impedance at various distances from center of half wave dipole. Thank you again for your input

JR Richter

 

[Jackson Richter]

Sorry forgot to ask, What is the major difference between a current and voltage balun? A side from the fact that the current type will support higher currents and voltage type have higher insulating properties. I found that OCF antenna impedance is also dependent on height. Unlike the dipole that will target a 73 ohm value the higher it is. I believe it follows the dipole impedance vs height charts some what, only if the height is below half wave length. The optimal height according to others is 15 meters, which also sets the feed point impedance as well. Also the length found to optimal is 134 ft, targeting a frequency around 3.6 Mhz. The second harmonic falls on 7.2 and then 14.4 etc. Still wondering why this length works as well as it does. I am trying to equate everything here that I using as of today. I hoped to make sense of all of it and put it down on paper for proof. Thanks again for any help here.

JR Richter

 

[Baluncore]

For any length dipole I expect the radiation resistance will be a function of the distance in quarter wavelengths from the nearest maximum current node of the standing wave. The k in your equation will depend on the units used for x. At first sight I think k will only be 1, or π/2 when x is measured in quarter wavelengths. This might be a transmission line problem for solution on a Smith Chart.

I have no doubt that a dipole in free space has a neat equation for off-centre complex feed impedance, but that equation you are looking for is hypothetical once you are in the real world. There was a time when I was running portable 3.5MHz links from tall forests. I found the inverted V to be an excellent antenna. I modeled it numerically before building it, then found I could not improve the design by changing wirelength in the field. I always centre-fed the dipole because that was where the hoist up the tree and the feedline were best positioned to avoid tangles with other trees. The two arms of the dipole sloped down at an angle designed to optimise a one skip reflection to base, it was changed for different ranges. Tuning was dependent on ground conditions. I tuned the reactance by adjusting the height of the ends above the ground rather than trimming the wire length or transmatch, that also reduced feed-line losses.

Even with a centre fed inverted V, the length of the wire, the feed height and the end-ground effects all conspire to make it non-theoretical. As the V becomes steeper the ground reflection becomes more important in determining radiation pattern because low angle end-fire rather than dipole field begins to become significant.

 

[tech99]

Sorry forgot to ask, What is the major difference between a current and voltage balun?

The voltage balun may be thought of as an RF transformer having a secondary which has a grounded centre tap and which is connected to the antenna. It will produce equal voltages on the two legs of the load, but the currents will differ if the legs have different impedances. It can also allow common mode currents (i.e. push-push currents) to pass unimpeded.
The current balun can be thought of as an RF transformer with an untapped floating secondary. It will force the current in the two legs to be equal no matter what their impedances are and will not allow common mode currents to pass.

 

[Jackson Richter]

Yes your right, using the antenna like this in the real world compared to free space is difficult to pin down. I do believe though with that being said, a design parameter can be used to put this antenna system up with fairly good results. Most of the people using this technique are also having very good luck with higher efficiencies with their antenna systems. Again this is with a 4:1 balun 90ft on one leg and 44ft on the other. The balun is raised to 35 to 45 ft with the wires spread 180 degrees apart from each other and the ends of the wires are at least 15ft off the ground. Feeding this antenna with 50 ohm coax (half wave lengths) with a velocity factor (.666) very good results have been achieved. on 80 meters the lowest VSWR is around 3.7 Mhz. Across the ban the highest VSWR is 2 from one end of the band to the other. On 40 meters pretty much the same if now lower. etc.

I am trying to put all of this in a logical order with starting from the basic Lambda= speed of light(C)/ F (Mhz). to the wire size vs length for the k factor in this wavelength formula. Along with that why does this 134 length make sense or is there another length (on paper and real world) that would be better. The position of the feed point vs impedance along with height above ground would make a very good antenna system and also multiband as well.

So to Clarify:
1. Calculated half wave length used for OCF antenna? What is the optimal length for 80,40,20,10 and 6 meters
half wave (meters) =.5* k * 300/f(Mhz), 12awg wire the k factor is approx. equal to length (meters)/dia wire (meters)
41.7/.00202574 = 20,268
l/d ratio equates to .975 from charts
half wave (meters)= .5 *.975* 300/3.5
half wave (meters)= 41.7
half wave (feet) =136.8



k (harmonic) 1 at 80 meters x ( degrees where feed point is located) example below
2 at 40 meters x = 90 degrees if put in middle of dipole
3 at 20 meters x = 60 degrees when feed point is at 1/3 length of dipole (44ft point)
4 at 10 meters

2. Radiation resistance at feed point = 73/ sin squared ( kx) free space 3. Antenna height vs impedance : It is very apparent that this ocf antenna doesn't follow the standard dipole
input impedance vs height relationship. The standard dipole will target to 73 ohms
above 1 wave length. Where as the ocf impedance will continue to increase as the
height is increased. I assume that's why the recommended height is 15 meters,
to make the feed point 100 to 200 ohms. Again with the 4:1 balun this would bring
the impedance down to or close to 50 ohms, ideal for coax usage.

This antenna again is a multiband antenna, which makes it very nice to use. Its know also for being very flat across the bands on the different frequencies.
If anyone can find holes in this please advise. I know there are a lot of unknowns here especially feed line, soil data, and proximity etc. Also known is this antenna is very sensitive to metal objects around or near it. What I am trying to do is to approximate these solutions so as to make sense of this antenna
Thanks to all who might add to this. Also will respect any criticism.

JR Richter

 

[Jackson Richter]

The voltage balun may be thought of as an RF transformer having a secondary which has a grounded centre tap and which is connected to the antenna. It will produce equal voltages on the two legs of the load, but the currents will differ if the legs have different impedances. It can also allow common mode currents (i.e. push-push currents) to pass unimpeded.
The current balun can be thought of as an RF transformer with an untapped floating secondary. It will force the current in the two legs to be equal no matter what their impedances are and will not allow common mode currents to pass.

If that's the case, if a current balun was used in a dipole configuration with unequal lengths of wire on each side, would it still have the common mode rejection, assuming fed with unbalanced feed line.

Thanks for your response.

JR Richter

 

[Jackson Richter]

For any length dipole I expect the radiation resistance will be a function of the distance in quarter wavelengths from the nearest maximum current node of the standing wave. The k in your equation will depend on the units used for x. At first sight I think k will only be 1, or π/2 when x is measured in quarter wavelengths. This might be a transmission line problem for solution on a Smith Chart.

I have no doubt that a dipole in free space has a neat equation for off-centre complex feed impedance, but that equation you are looking for is hypothetical once you are in the real world. There was a time when I was running portable 3.5MHz links from tall forests. I found the inverted V to be an excellent antenna. I modeled it numerically before building it, then found I could not improve the design by changing wirelength in the field. I always centre-fed the dipole because that was where the hoist up the tree and the feedline were best positioned to avoid tangles with other trees. The two arms of the dipole sloped down at an angle designed to optimise a one skip reflection to base, it was changed for different ranges. Tuning was dependent on ground conditions. I tuned the reactance by adjusting the height of the ends above the ground rather than trimming the wire length or transmatch, that also reduced feed-line losses.

Even with a centre fed inverted V, the length of the wire, the feed height and the end-ground effects all conspire to make it non-theoretical. As the V becomes steeper the ground reflection becomes more important in determining radiation pattern because low angle end-fire rather than dipole field begins to become significant.

When using the value for K you said that K could be 1 or pi/2. I assume pi is in radian measure. Later posts I tried k as a variable for bands such as 80m =1 and 40m=2, 20m=3. Just guessing here? probably way off the mark lol. So thank you again will try it out too. When drooping the ends of the antenna I assume your increasing the capacitance reactance. This antenna is super sensitive to objects around it, but works very well when put up properly. thanks again for your input
JR Richter

 

[Baluncore]

Zo = 72 ohms
k = 2π / λ
x measured from nearest end of half wave dipole.
θ = k * x
R = Zo / ( Sin θ * Sin θ )

  Xposn  Rohms
  0.010 18261.8
  0.020  4583.5
  0.030  2050.6
  0.040  1164.2
  0.050   754.0
  0.060   531.3
  0.070   397.2
  0.080   310.2
  0.090   250.8
  0.100   208.4
  0.110   177.2
  0.120   153.6
  0.130   135.5
  0.140   121.3
  0.150   110.0
  0.160   101.0
  0.170    93.8
  0.180    87.9
  0.190    83.3
  0.200    79.6
  0.210    76.7
  0.220    74.6
  0.230    73.1
  0.240    72.3
  0.250    72.0

 

[tech99]

If that's the case, if a current balun was used in a dipole configuration with unequal lengths of wire on each side, would it still have the common mode rejection, assuming fed with unbalanced feed line.

Thanks for your response.

JR Richter

Yes

 

[Jackson Richter]

Zo = 72 ohms
k = 2π / λ
x measured from nearest end of half wave dipole.
θ = k * x
R = Zo / ( Sin θ * Sin θ )

  Xposn  Rohms
  0.010 18261.8
  0.020  4583.5
  0.030  2050.6
  0.040  1164.2
  0.050   754.0
  0.060   531.3
  0.070   397.2
  0.080   310.2
  0.090   250.8
  0.100   208.4
  0.110   177.2
  0.120   153.6
  0.130   135.5
  0.140   121.3
  0.150   110.0
  0.160   101.0
  0.170    93.8
  0.180    87.9
  0.190    83.3
  0.200    79.6
  0.210    76.7
  0.220    74.6
  0.230    73.1
  0.240    72.3
  0.250    72.0

Thanks for your input here. Not sure what the x value is. Is it the % center, or x/lambda, degrees from lambda/2. Please clarify if you would, thanks

JR Richter

 

[Baluncore]

Thanks for your input here. Not sure what the x value is. Is it the % center, or x/lambda, degrees from lambda/2. Please clarify if you would, thanks

Both λ and x are in metres. x is measured from the end of the half wave dipole.

Maybe I should have included these obvious initial lines.
MHz = 10⁶
c = 299792458.
freq = 300. * MHz
λ = c / freq
Then generate the table by looping For x = λ/100 to λ/4 step λ/100

Since the example table was computed for 300MHz, where λ = 1 m, the example x was also in wavelengths.
If you change freq to 1 MHz, where λ = 300 metres, then R will still range from 18261 ohms to 72 ohms, but x will then range from x=2.998m to 74.948m.

 

[tech99]

Both λ and x are in metres. x is measured from the end of the half wave dipole.

Maybe I should have included these obvious initial lines.
MHz = 10⁶
c = 299792458.
freq = 300. * MHz
λ = c / freq
Then generate the table by looping For x = λ/100 to λ/4 step λ/100

Since the example table was computed for 300MHz, where λ = 1 m, the example x was also in wavelengths.
If you change freq to 1 MHz, where λ = 300 metres, then R will still range from 18261 ohms to 72 ohms, but x will then range from x=2.998m to 74.948m.

Please remember that the current distribution is only approximately sinusoidal. If you feed the wire at the end, the impedance depends on the conductor diameter, but for a wire it may be 5000 ohms or so and not infinity. Also, resonance occurs when the length is a little less than half a wavelength.

 

[Baluncore]

If you feed the wire at the end, the impedance depends on the conductor diameter, but for a wire it may be 5000 ohms or so and not infinity.

It is quite farcical to consider driving a dipole with a current source at the end of the dipole.

Ideally, the wire diameter of an inverted V antenna should be tapered in proportion to the distance from the ground. Where the wires approach the balun they should also taper towards the balun terminals. For narrow band HF that requires a cage dipole, but no one seems to acknowledge the advantage of that impedance matching these days.

 

[Jackson Richter]

Thank you for your input again. I know from the past that when a wire is end fed its called an end fed zep. A matching system must be used to match the output impedance from the transmitter and feed line to the input impedance of the wire antenna. Typically these antennas are multiple wave lengths long and have directivity when mounted the right distance off the ground. As far as the OCF antenna, I have read so many descriptions of the this antenna and no one is getting the same results, different equations, different lengths, different heights and different feed points across the 1/2 wave dipole. The height plays a very important factor for determining input impedance of this antenna. unlike the standard dipole that targets 73 ohms above 1 wave length. I am trying to validate data and put the results in sensible order and achieve the optimal feed point position to cover the maximum band usage along with proper height to achieve ground wave and single hop communications. Maybe a simple solution can't cover all the real world installations. I am trying to cover the basics so others can understand and add to the knowledge base of this particular antenna. I not saying others are wrong, but what I am saying, I just want to validate their findings with my data and experimentation. So thank you for your input.

JR Richter

 

[Jackson Richter]

It is quite farcical to consider driving a dipole with a current source at the end of the dipole.

Ideally, the wire diameter of an inverted V antenna should be tapered in proportion to the distance from the ground. Where the wires approach the balun they should also taper towards the balun terminals. For narrow band HF that requires a cage dipole, but no one seems to acknowledge the advantage of that impedance matching these days.

I thought that the tapering would cause higher Q and lower bandwidth, making it a much narrower operating window for single frequency.

 

[Baluncore]

I thought that the tapering would cause higher Q and lower bandwidth, making it a much narrower operating window for single frequency.

Once you tune a 3.5 MHz dipole with a Q of 100, you still have a bandwidth of 35 kHz.

 

[Jackson Richter]

It is quite farcical to consider driving a dipole with a current source at the end of the dipole.

Ideally, the wire diameter of an inverted V antenna should be tapered in proportion to the distance from the ground. Where the wires approach the balun they should also taper towards the balun terminals. For narrow band HF that requires a cage dipole, but no one seems to acknowledge the advantage of that impedance matching these days.

It is quite farcical to consider driving a dipole with a current source at the end of the dipole.

Ideally, the wire diameter of an inverted V antenna should be tapered in proportion to the distance from the ground. Where the wires approach the balun they should also taper towards the balun terminals. For narrow band HF that requires a cage dipole, but no one seems to acknowledge the advantage of that impedance matching these days.

Yes of course, no one would use a current balun at the end of the line.

Supposedly the optimal length of the antenna to be 134 ft and fed at the 1/3 the distance from the end. Using a 4:1 balun at this point. A construction practice is to connect a 90ft wire to one side of this balun and 44ft to the other side. When raised to a 15 metre (aprox.) height in an inverted V configuration. This antenna is used at (3.5 to 4 Mhz), ( 7 to 7.3 Mhz), (14 to 14.35Mhz), (28 to 28.7Mhz) and (50 to 54Mhz) with great success.

I currently have this antenna now and it works surprisingly well, I am not sure if this is the optimal configuration. The questions I have are:
What is the optimal length? What is the optimal height? What is the optimal feed point location? What is the optimal wire size? Currently I am using #12 awg silver stranded wire with Teflon coating for the wire lengths. The antenna appears to have a little lower noise floor while also reducing the IR losses, making the antenna a little more efficient.

 

[Jackson Richter]

Both λ and x are in metres. x is measured from the end of the half wave dipole.

Maybe I should have included these obvious initial lines.
MHz = 10⁶
c = 299792458.
freq = 300. * MHz
λ = c / freq
Then generate the table by looping For x = λ/100 to λ/4 step λ/100

Since the example table was computed for 300MHz, where λ = 1 m, the example x was also in wavelengths.
If you change freq to 1 MHz, where λ = 300 metres, then R will still range from 18261 ohms to 72 ohms, but x will then range from x=2.998m to 74.948m.

When using C / F(Mhz) to find length, this is for free space and doesn't take in account vf or Length vs diameter of conductor used. Shouldn't it be: k * speed of light/ F (Mhz). or appox or .975 * C/ F(Mhz). lambda (metres) = 292.3 / F(Mhz)?

 

[Baluncore]

Shouldn't it be: k * speed of light/ F (Mhz). or appox or .975 * C/ F(Mhz). lambda (metres) = 292.3 / F(Mhz)?

The VF of bare copper wire is close to 99.7%, I once measured it because my computer model did not fit reality. But it depends slightly on the moisture content of the brown copper chemical surface which might be an acetate or an oxide. If you have insulation on the wire, then things will be slightly slower.

What is the optimal length? What is the optimal height? What is the optimal feed point location? What is the optimal wire size?

Optimal for what? The OSF inverted V is a messy antenna. The radiation patterns of the two legs make the pattern quite complex on the higher frequencies. Any big antenna works OK if you have propagation and a tuner.
Why do you use feet for the only thing that is related to the wavelength in metres.
90ft + 44ft = 134ft = 40.843 m = λ/2, λ = 81.686 so Freq = 3.67 MHz.

The impedance of the feedpoint will be different for every band since the distance from the nearest current maximum will be wavelength dependent. Is the 1/3 distance from the end measured along the wire or one of the legs? Is the feed at the top of the V or half way up the long leg? Maybe a picture with dimensions would help.

 

[Jackson Richter]

Yes, your right this is a messy antenna. There are a lot of variables. Different frequencies, different heights, different lengths. Also. I should maintain standard measuring lengths, thank you. Again the length used for this antenna is 40.843 m. The feed point is at 13.411m. The half wave length at 3.6 Mhz is 40.843 m. The second harmonic falls in the 40 m band, 4th harmonic falls on the 20 m band etc. And yes their impedances are different. But If a correct length is used a common impedance around 95 to 250 can be obtained. So far, 1/3 the length of the half wave seems to work very nicely. G80DE has done a lot of work on this but I am not sure if these is best lengths. With the 4:1 current balun, fed at this x length a multiband antenna can used effectivity. The point is to determine what the best length is supposed to be. Also, what is the optimal height to make this work and why. Maybe your right, maybe there is no pin point accuracy to all of this, but I have to try. Thanks again for your time.

 

[Baluncore]

Is the 1/3 distance from the end measured along the wire or one of the legs? Is the feed at the top of the V or half way up the long leg? Maybe a picture with dimensions would help.

1/3 of the way along what? Measuered from which end? Where?

With the 4:1 current balun, fed at this x length a multiband antenna can used effectivity.

Because your feed-point has highly variable impedance over frequency, you will need an antenna matching unit at the balun to prevent circulating reactive current on the lossy coaxial feedline. Would you not be better using an open wire parallel transmission line which will have a better matched impedance with much lower losses than coax. You can then ignore the matching unit at the antenna by using a matching unit in the shack where it will be more accessible.

 

[Baluncore]

You will notice that the cheaper is the coax used for the feedline, the lower is the SWR. That is because the reflected wave is being absorbed twice by the cheap coax. When you use a balanced ladder-line you will begin to recognise the real SWR and so can see the mismatch of the antenna to the feedline. Eliminating the coax and balun is probably the single greatest improvement you could make to the performance and to your understanding of the antenna.

 

[Jackson Richter]

1/3 of the way along what? Measuered from which end? Where?

Because your feed-point has highly variable impedance over frequency, you will need an antenna matching unit at the balun to prevent circulating reactive current on the lossy coaxial feedline. Would you not be better using an open wire parallel transmission line which will have a better matched impedance with much lower losses than coax. You can then ignore the matching unit at the antenna by using a matching unit in the shack where it will be more accessible.

Open wire feed line is very low loss for sure. Its more difficult to work with and matching is definitely required at the transmitter site like you mentioned. When very long runs of feed line are needed, the two options are open wire feed or hard line. My particular situation, I running about 1/2 wave length feed line on 75 metres. You asked "1/3 length to what?" The feed point is located 1/3 length of 40.623metres which is 13.5 metres. The other leg is 27.123 metres. The over wave length is a 1/2 wave length at 3.6 Mhz which is 40.623 metres. Making a feed point impedance of 97.69 ohms, but when raised to a higher level the impedance also increases. Determining what over all height and feed point positions are my goal if at all possible. The rule in the past has been the 15 metre level. The question comes in about common mode currents? A well constructed 4:1 balun (unbalanced to balanced) high power, should reduce the common mode currents. Balun Design makes such a balun its 4114 ocf, rated at 5kw. Thanks again for your time and information.

 

[Baluncore]

The more I consider your offset fed dipole, with a bend close to 90° at the feed point, the more unpredictable it becomes at the shorter wavelengths and the less like a dipole it appears to be at the longer wavelengths.

It might be better to analyse it as the sum of two random wires driven against two counterpoises.

 

[Jackson Richter]

The more I consider your offset fed dipole, with a bend close to 90° at the feed point, the more unpredictable it becomes at the shorter wavelengths and the less like a dipole it appears to be at the longer wavelengths.

It might be better to analyse it as the sum of two random wires driven against two counterpoises.

Well I guess that's the beauty of it. Yes it is a complicated antenna, but yet simplistic to build and works very well. I know you say compared to what? If I wanted to build a single antenna for one frequency then I wouldn't worry about all of this and design it by the book. If I had a huge area to put it on, or a tall tower or tall trees to hang it. We all don't have this option, but this is one antenna that has it all. It appears to be very efficient, even with the balun. It also is very broad band, with small deviation in VSWR across the bands. Its also a multiband antenna that works at the fundamental, 2nd, 4th, 8th harmonic of the original low band frequency, (if properly picked). I personally have heard many amateur stations change from being a weak station to strong station with this antenna. Yes sir you are right, analysis of this antenna would be my next step. I think there is still more to learn about this OCF antenna and possibly make it better. Thank you for your input.

 

[Jackson Richter]

I was looking at a earlier study and I am frankly stumped. In the equation where we were looking at a certain feed point location on a half wave dipole and moving this location away from center, and plotting the different impedance values for the x distance from end of the antenna, works well for a single frequency. I am have trouble doubling the frequency and plotting the same impedance point. For instance; We know the impedance of a half wave dipole fed in the middle is about 73 ohms in free space. If we tried to use this same antenna, same feed point location, but double the frequency, what happens to the impedance? According to the prior study this impedance will be 4000 ohms. I agree that it will be hi, but not sure how to enter that in the equation below. It obvious that I don't understand the equation that you nicely showed in your explanation for impedance vs feed point location. I know this antenna only works on 2nd, 4th, 8th harmonics. Where does that fit into the equation if you change frequency and keep the same original length? Is that just a multiplier of the angle?

Rr = 73 / Sin squared ( 2 * pi * X/ lambda)

where x = location ratio to lambda (if at center of half wave antenna) = .25
2 * pi = 360 degrees (if at center .25 * 360) = 90 degrees
sin of 90 = 1 and that squared is 1 in the denominator, which just leaves the numerator of 73.
Thanks for any help here

 

[Baluncore]

It is not a dipole, it is not a multi-band antenna. It is an amateurs random wire against a counterpoise.
It cannot be modeled generally as it is highly dependent on ground and nearby structures.
It will work sometimes if you have a good antenna tuner.

 

[davenn]

It is not a dipole, it is not a multi-band antenna. It is an amateurs random wire against a counterpoise.

the best statement so far made in this thread

Well I guess that's the beauty of it. Yes it is a complicated antenna, but yet simplistic to build and works very well. I know you say compared to what? If I wanted to build a single antenna for one frequency then I wouldn't worry about all of this and design it by the book. If I had a huge area to put it on, or a tall tower or tall trees to hang it.

Hi Jackson

Have you considered a probably much better alternative ... a trapped dipole ?
They are great for multiband use. And you can please yourself if you use it as a flat ( horizontal )
dipole or as an inverted V. The inverted-V just gives a better non-directive radiation pattern than a horiz. dipole that you cannot rotate

EDIT: or if you don't like trap building ... go for a G5RV broad band antennaDave

 

[Jackson Richter]

Thank you Dave for your input. I have considered many antennas for my projects. I do believe strongly that the antenna is the most important part of the communications process. With in the last year I have been using this OCF antenna. There are so many positive aspects about it. 1st, its very broad band. 2nd, its multiband. 3rd, it has a very flat VSWR across these bands. 4th, it actually blows the g5rv and others like it, out of the water. It appears to be a very efficient antenna. I talk to many others that have this same antenna, who sound strong and do much better now since they have changed to this antenna. So what I am trying to do is to quantify it and put the data in a form that makes it more predictable. I guess you would have to see for yourself, being from Missouri, lol. All kidding aside, There are a few quirks about this antenna that have to pinned down and analyzed. So with that being said, that's what I am trying to do. There are 4 things that make this antenna work. 1st, A very good current balun, designed for high power and built unbalanced to balanced configuration. (to help with common mode currents) and impedance matching. 2nd, This antenna is put in an inverted V configuration. 3rd, The height is also very important, because of the near field radiation. This antenna if mounted 30 to 40 ft high, the OCF input impedance falls within limits of the 4:1 balun and nicely transfers that impedance down to a proper value that output transmitter and coax can drive. 4th, This antenna must be clear of any large metal objects within the near field proximity. That's pretty much it. Sorry, I hope this makes sense to you. Seeing is believing and I should have tackled this antenna years ago. I talk every day on this antenna and have worked Europe on 20 meters many times even with the terrible band conditions. I also talk to California every night on 80meters with this antenna. Its like having a super nice car, you want to keep it fine tuned. I am also trying to fine tune the OCF as well.

Jackson WB0USA

 

[tech99]

If you drive it at one end and place a terminating resistor at the other, it forms a half rhombic, or Bruce, antenna, and will have gain and F/B ratio for frequencies of about 14MHz upwards. It will also have a fairly uniform driving point impedance.

 

[Baluncore]

I do believe strongly that the antenna is the most important part of the communications process.

You have a lot to learn. Without propagation, communication is impossible. With propagation, just about any antenna will work when tuned.

There are so many positive aspects about it. 1st, its very broad band. 2nd, its multiband. 3rd, it has a very flat VSWR across these bands.

If that flat VSWR is close to 1.0 then it indicates that you do not have a resonant antenna attached, you simply have high losses in your feedline, tuner and/or environment.

A dummy load also has a flat VSWR. It radiates in the IR band as heat. A low SWR may indicate that little energy is returning to the SWR meter because it is lost as heat before it gets back.