Not to start an arguement just some thin s Ive learned over time.

[Needle Bender]

All the watt meters are reading is the same RF signal from the transmitter going through the entire system. I believe this is where the misconception from the original poster is coming into play. From reading Grim Reaper's synopsis, it appears there is a misconception that because each meter is reading the same power at the base of both antennas, the power output is doubling (left antenna 4 watts, right antenna 4 watts = double the power).

The meters are reading the voltage going up to the antennas and returning through the entire system. Just because someone is using two antennas, you are still dealing with the same 4 watts from the transmitter. Because the system is phased, the 4 watts of power is alternating between the two antennas, you just can't see it with a simple RF meter, because it is happening too fast at a cycle rate of 27 mhz.

As Doc pointed out, the dual radiating elements help direct the RF and receive in the desired direction which is "gain." There is no free lunch or magic power secrets when it comes to antennas or gain. Gain is not an increase in power, you are stuck with whatever power is coming from the transmitter, minus the losses. Gain is the ability to concentrate the transmit power and the area of receive, in the direction you want.

73

Hang on a minute, how does a parallel voltage divider circuit (two 100 ohm loads) act as a series circuit at 27mhz

 

[Marconi]

Hang on a minute, how does a parallel voltage divider circuit (two 100 ohm loads) act as a series circuit at 27mhz

I don't understand circuits very well, but where do you come up with two 100 ohm loads?

I thought this topic was about co-phasing two mobile antennas?

 

[W5LZ]

The antenna input impedance plus the impedance of the matching section for that antenna are approximately 100 ohms impedance. Parallel two of those antennas and the resulting impedance is approximately the desired 50 ohms impedance. That '100 ohms' and the resulting '50 ohms' are -only- approximations. There can be quite a variation in those impedances.
- 'Doc

 

[Needle Bender]

Fine, 98 ohms, 106.371 ohms, I don't really care if it's exact, I just don't understand how 4 watts can be divided by 2 but remain 4 watts into both parallel loads, now equaling 8 watts?

If that's the case, then let's set up 128 pairs and keep piggy-back-splitting them by using 75ohm coax phasing harnesses and then the original 4 watts will now provide over 1 kilowatt?

 

[office888]

I'd rather have a 1/4 wave vertical at 10,000 ft than a 10 element yagi at 10 ft.

 

[BOOTY MONSTER]

if you're using coax for the feed line on both the 1/4 and the yagi id rather have the beam at 10 ft. .
80 db of loss from 2 miles of RG8 LOL :laugh:

 

[Needle Bender]

Having just checked the price of tea in China, I've decided I'd rather have a bottle in front of me than a frontal lobotomy.
I hear the skiing is excellent this time of year.
Can you even imagine a supercharged V-8 mid-engine?
How 'bout them Knicks!
Never run with scissors.

 

[Marconi]

Below I hooked two 1/2 wave 50 ohm jumpers to two 50 ohm dummy loads. I connected the jumpers to a T-connector and hooked my Autek VA1 to the third port on the T and I saw 28 ohms.

I hooked one jumper direct to the VA1 and got 49 ohms, the other image is also 49 ohms with the VA1 direct into the dummy load.

I don't have a 75 ohm co-phase harness, but if I did and hooked it up in this setup then I would expect the TX'r end of the T-connector to show close to 50 ohms. Am I wrong?

The following is not a transformational harness so we should not see any transformation by the harness, but we do see an approximate halving of the impedance at the transmitter end of the harness.

50 ohm co-phase setup into two 50 ohm dummy loads.


VA1 into T-connector at the end of the harness.


VA1 direct into a 50 ohm non-reactive dummy load.


VA1 direct into one of the jumpers and a dummy load.


Sorry for the meter response no showing up well, this meter is old and worn somewhat.

If I'm not wrong, then where are ya'll getting 100 ohms? Is the discussion about a mobile antenna or some other antenna like a quad or maybe a 5/8 wave with an end impedance somewhere around a 100 ohms with a lot of reactance to boot?

NB, you're getting all incredulous again, like I called your momma a bad name. I asked a civil question, and it was not a trick question about the accuracy of the impedance.

BTW, I agree with you not seeing how 4 watts....however you said it.

 

[Needle Bender]

Below I hooked two 1/2 wave 50 ohm jump

NB, you're getting all incredulous again, like I called your momma a bad name. I asked a civil question, and it was not a trick question about the accuracy of the impedance.

Hi Marconi,

You turkey!

Sometimes you take things personally that are never meant that way, not by me anyway. You're the most enthusiastic user on this forum and the only one who addressed my question on topic.

...but stop calling my momma bad names! :tongue:

OK, the 100 ohms is what you should get at the other end of a 75ohm electrical 1/4 wave matching transformer coming from a 50ohm load. (Don't even begin to try to make me think you don't already know this)

Two of them, each from a 50ohm antenna into a T connector should show right about 50 ohms at the single input, like your test, except double the impedance back to 50ohms.

5 out of 4 Doctors recommend wine over Chicken McNuggets.

 

[Marconi]

if you're using coax for the feed line on both the 1/4 and the yagi id rather have the beam at 10 ft. .
80 db of loss from 2 miles of RG8 LOL :laugh:

Boy it don't take much to get you boys all bent out of shape! ^can not stand it^

Booty is right, if old 888 could get it up at 10000' feet his signal from his CB would probably peter out with about 200' - 300' feet up the coax. Plus if he was that high above everyone around him for about 200> miles, and could make a signal, nobody within that range would even know he was out there.

 

[bob85]

eddie,
if you used odd multiple electrical 1/4waves of 75ohm coax terminated with 50ohm pure resistance input impedance seen looking into each line becomes

Zin = Zo^2
---------
Zl

the characteristic impedance of the line (75ohm) squared, divided by (50ohm X=0) load

if the loads are not 50ohm X=0 the phase relationship between current and voltage along the 75ohm lines shifts and all bets are off.

 

[Marconi]

Hi Marconi,

You turkey!

Sometimes you take things personally that are never meant that way, not by me anyway. You're the most enthusiastic user on this forum and the only one who addressed my question on topic.

...but stop calling my momma bad names! :tongue:

OK, the 100 ohms is what you should get at the other end of a 75ohm electrical 1/4 wave matching transformer coming from a 50ohm load. (Don't even begin to try to make me think you don't already know this)

Two of them, each from a 50ohm antenna into a T connector should show right about 50 ohms at the single input, like your test, except double the impedance back to 50ohms.

5 out of 4 Doctors recommend wine over Chicken McNuggets.

I'll have to dig around in the shop and see if I can find one of my old harnesses, and see how it acts hooked up to my dummy loads.

You may be right, but I have never seen a 50 ohm complex load from a 1/4 wave radiator attached to a mobile. Sometimes it's even difficult for me to get such a load on a real 1/4 wave base radiator, with slanted down radials unless I have real good control over the slant of the radials below 50* degrees.

Check out this 1/4 wave with 50* degree radials. This is the Z I usually see and worse when reading the match at the feed point on a mobile. When I check through the feed line, this match does often get better due to transformation. This is why I think guys typically tell us they have a perfect match in their mobiles with a flat SWR that is low and hardly moves the needle. IMO many mobiles can't come close to this angle and provide the necessary ground plane. We found that installing these co-phase setup closer to the ground, on the bumper of pickup trucks seem to improve performance just using old Mother Earth on the roadway. This was the reason for my work with the 102" whip over the years, long before I could make a model.

View attachment 6665

 

[Marconi]

eddie,
if you used odd multiple electrical 1/4waves of 75ohm coax terminated with 50ohm pure resistance input impedance seen looking into each line becomes

Zin = Zo^2
---------
Zl

the characteristic impedance of the line (75ohm) squared, divided by (50ohm X=0) load

if the loads are not 50ohm X=0 the phase relationship between current and voltage along the 75ohm lines shifts and all bets are off.

Good morning Bob. I use this test formula that I've used for years, and it is base on the feed point impedance for a single installed radiator. Of course both antennas should be the same of very similar for this to prove out and work right the way I understand it.

I don't know if it's good or not, and I don't know the source. Could be my old antenna Mentor buddy gave it to me when we were building co-phase harnesses years ago.

Z = SQR (load impedance ohm x transformer line impedance)

Example of 37 ohm 102" antennas using RG6, 75 ohm, or RG71U, 93 ohm.

37x75=2775 SQR = 52.7 ohms results at the TX end of harness. So, if using a 50 ohm feed line in system, then further transformation can occur, and the following is supposed to calculate the new impedance value using the results from the first calculation.
52.7x50=2635 SQR = 51.3 ohms.

Then we trimmed the pig tail ends of our harness to the lowest SWR and I figured, if everything was good, maybe the Z value for the TX end of the match should be close to 51.3 ohms.

I never questioned the idea, and I really don't understand all the nuances of this business. So, if you find this in error or totally off base...then let me know.

Do you see anything wrong with the images I posted above? Is that pretty much what I should be seeing with such a setup?

 

[bob85]

im not familiar with that formula eddie,

the only thing i see wrong with your pics of the 50ohm lines are the rubber and glass where your lcd display connects needs cleaning with alchohol to get the missing segments working.

 

[Marconi]

im not familiar with that formula eddie,

the only thing i see wrong with your pics of the 50ohm lines are the rubber and glass where your lcd display connects needs cleaning with alchohol to get the missing segments working.

Well I see my parallel 50 ohm harness shows about half of the two parallel 50 ohm loads in my example above, so I guess 'Doc and NB were correct. I stand corrected, a transformer should show 100 ohms if the load is 50 ohms, and when run in parallel the net load results will be 50 ohms.

Sounds reasonable to me, but I always imagined this thing backward I suppose.

What is a great tip Bob, thanks. No wonder Autek is so willing to fix it for a nominal charge, over $50 dollars plus shipping both ways. I'll try to clean mine.

Are you telling me that this part of the meter is rubber, like latex?

Thanks for the tip and the advice.