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Author Topic: 4-Wire Ladder Line (Not Parrallel 2-Wire)  (Read 18545 times)
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aa5wg
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« on: October 27, 2014, 02:39:41 AM »

Hi to all,

Does anyone know of a 4-wire ladder line calculator?

Or, does anyone have a 4-wire ladder line design chart
they could upload as a picture.

Thank you.
Chuck
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« Reply #1 on: October 27, 2014, 01:19:47 PM »

Do you mean a box configuration, not four abreast?
I guess you could model it on Eznec.
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aa5wg
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« Reply #2 on: October 27, 2014, 01:40:31 PM »

Rick,

I mean square or box.
Please attachment.

Chuck

* 4 Wire Ladder Line Insulator.PDF (63.25 KB - downloaded 439 times.)
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AB2EZ
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« Reply #3 on: October 27, 2014, 02:27:34 PM »

Note: this post has been deleted because it was made obsolete by later posts, below.


Stu
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« Reply #4 on: October 27, 2014, 03:22:04 PM »

http://www.rfcafe.com/references/electrical/transmission-lines.htm

Formula on RFCafe from the Handbook for (RF) Engineers

I recall a discussion of four wire lines a number of years ago, relating to feeding antennas on top of ridges/mountains, i.e. long distance.

I and thought there was also an ARRL article, but can't find anything in QST/QEX archives.  May be in one of the "Wire Antenna Compendiums".   Unfortunately, I'm traveling and not at my library.  bill
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AB2EZ
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« Reply #5 on: October 27, 2014, 04:01:37 PM »

The information on the RF Cafe page, provided by Bill, kb4qaa, indicates the following:

If Z(ladder line) is the characteristic impedance of a ladder line, with a given wire size, and a given wire spacing,

Then the characteristic impedance of a square array with the same wire size, and where each side of the square array has the same spacing as that of the ladder line, is:

Z(square)= [0.5 x Z(ladder line)] - [138 Ohms x 0.15]

where 0.15 is approximately the logarithm (base 10) of the square root of 2

For example, if Z(ladder line)= 300 ohms, then the characteristic impedance of the square array is: Z(square) = 150 ohms - 20.8 ohms = 129.2 ohms

Note also:

The impedance of the 4-wire transmission line is lower by more than a factor of 2 vs the 2-wire transmission line. Therefore the total current in each direction must be higher by more than a factor of the square root of 2 = 1.414, if the power propagating through the transmission line remains the same. Power = I x I x Z.

This implies that the current in each wire of the 4-wire transmission line is larger than 0.707 x the current in each wire of the 2 wire transmission line. But, the skin effect resistance of each wire is the same. Therefore, the power dissipated, per unit length, in each wire in the 4 wire transmission line is larger than 1/2 the power dissipated in each wire of the 2-wire transmission line.

This means that the loss (dB/km) of the 4-wire tranission line is (perhaps non intuitively) somewhat higher than the loss in the 2-wire transmission line... even though it employs twice as much copper per unit length!

A fairer comparison might be to compare the loss of a 4-wire balanced transmission line to the loss of ladder line of the same characteristic impedance (having the same wire size, but closer wire spacing). In that case, the loss of the 4-wire balanced transmission line would be half of the loss of the ladder line.

Stu
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« Reply #6 on: October 27, 2014, 05:07:33 PM »

This makes the third post in a week that didn't make it to the BBS.

The line your speaking of is referred to as 'quadro line'.   It's covered in the 1st or 3rd ARRL Antenna Compendium editions.

--Shane
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« Reply #7 on: October 27, 2014, 05:41:08 PM »

Chuck...
Aside from intellectual curiosity, why do you need to know?  What do you plan on doing with your 4-wire line?
In the high power short wave broadcast world, four wire line is typically set up as the left two wires versus the right two wires.  The left two are connected together periodically with a vertical bar that grabs both, and the right two are done similarly.  It's basically a big honking two-wire line, and easier to hang on supports.  If you need to fine tune the line, usually to optimize the higher frequency end,  you can add extra vertical (capacitor) plates at locations determined with your network analyzer.  There are hundreds of feet of this at any given SWBC site.  An interesting fact, should you ever be a contestant on Jeopardy, is that while most of the western bloc broadcasters use 300 ohm line, the BBC sites  use 328 ohm lines.

Norm  W1ITT
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« Reply #8 on: October 27, 2014, 05:57:02 PM »

I'd always thought of 4-wire line as a way to power a pair of crossed dipoles, possibly steering the pattern of same. Beyond my current practice but the line is real interesting.
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aa5wg
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« Reply #9 on: October 27, 2014, 07:05:49 PM »

Stue, kb4qaa, Shane, Norm and opcom and all,

Thank you for your inputs.  Stue, thank you for the math explanation.  I can now calculate the desired
impedance.  

The RF Cafe formula and pictorial showed joining of two wires by crossing them rather than two left wires tied and two right wires tied.

(1) Can I still use the above formula Stue provided if I connect the two lefts wires together and the two right wires together (at the antenna tuner) rather than tie the opposite cross wires together?

The Radio handbook, William Orr, 16th edition, page 435, left column explained the advantages of using 300 ohm open feeders over 600 ohm.  Advantage - lowering the average impedance needed to be matched from 70 - 5000 ohms with 600 ohm ladder line to 75 - 1200 ohms with 300 ohms open wire feeders.  "With this much lowered impedance variation it is usually possible to use series tuning on all bands..."  If not I could shunt some reactance to achieve series tune/current feed.  

To achieve 300 ohms with #10 wire requires .625 inch (5/8 inch) spacing for parallel line.  This wont work because of snow, ice, rain, etc.  This spacing is to close for bad weather.  With 4-wire (quadra wire) feeders I can achieve 300 ohms with much wider spacing to accommodate the snow, ice, rain, etc.

Chuck
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aa5wg
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« Reply #10 on: October 27, 2014, 09:16:25 PM »

Stu,

In the formula: Z(ladder line) = 300 ohms, then the characteristic impedance of the square array is:
Z(square) = 150 ohms - 20.8 ohms = 129.2 ohms

Is 20.8 ohms used by default for all calculation of different values of Z(ladder line)?

i.e. Z(ladder line) = 450 ohms, then the characteristic impedance of the square array is:
Z(square) = 225 ohms - 20.8 ohms = 204.2 ohms.

Is this correct?

Chuck
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AB2EZ
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« Reply #11 on: October 27, 2014, 09:28:28 PM »

Chuck

Yes, that is what the formulas on the RF Cafe web page imply.

Stu

P.S. I don't know what the formula for the 2 wire x 2 wire parallel configuration is, but I expect it to be different from the 4 wire balanced configuration.


Stu,

In the formula: Z(ladder line) = 300 ohms, then the characteristic impedance of the square array is:
Z(square) = 150 ohms - 20.8 ohms = 129.2 ohms

Is 20.8 ohms used by default for all calculation of different values of Z(ladder line)?

i.e. Z(ladder line) = 450 ohms, then the characteristic impedance of the square array is:
Z(square) = 225 ohms - 20.8 ohms = 204.2 ohms.

Is this correct?

Chuck
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« Reply #12 on: October 28, 2014, 08:29:36 AM »

Chuck

Thinking about the 2-wire x 2-wire parallel configuration vs the 4-wire balanced configuration... after a good night's sleep:

Going back to basic principles, and mentally transposing the positions of two adjacent wires of the balanced 4-wire configuration... to obtain the 2-wire x 2-wire parallel configuration... I now believe that the impedance formula for the 2-wire x 2-wire parallel configuration is probably:

Z(parallel) = [0.5 x Z(ladder line)] + 20.8 ohms

Recall that the impedance formula for the 4-wire balanced configuration is (ref: RF Cafe)

Z(square) = [0.5 x Z(ladder line)] - 20.8 ohms

Stu
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WA1ICN
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« Reply #13 on: October 28, 2014, 01:04:12 PM »

The attached graph for characteristic impedance of 4 wire transmission line appeared in the 1949 ARRL Antenna Book. There was no corresponding formula given.

Spacing is measured diagonally, so multiply by 1.414, the square root of 2, to convert from spacing measured on a side.

A 6 inch spacing measured on a side would be approximately 8.5 inches when measured on the diagonal.

A 6 inch, side measured, 4 wire open wire line with 10 AWG diameter wire, would be approximately 255 ohms, according to the graph.


* 4wireChart1949AntBook.jpg (71.82 KB, 516x900 - viewed 792 times.)
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aa5wg
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« Reply #14 on: October 30, 2014, 10:48:21 PM »

Gentleman,

I have a couple of questions.  But, I want to do
some reading. When I complete the reading
I will give a shout.  Again, thank you to all.

Chuck
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aa5wg
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« Reply #15 on: November 15, 2014, 12:23:18 PM »

Hi to all,

Which graph is correct?

(1) The antenna manual, 1948, page 118

or

(2) ARRL Antenna Book, 1949, pape 82

73,
Chuck


* 4-Wire, 1948 ant. manual pg. 118.JPG (479.71 KB, 2550x3299 - viewed 748 times.)

* ARRL 1949 Ant. Book pg. 82.JPG (552.76 KB, 2550x3299 - viewed 675 times.)
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« Reply #16 on: November 16, 2014, 12:17:12 PM »

In response to Your question, both are correct. Please note that in the first, the dimension is based on a square arrangement depicting the wire-to-wire spacing while the second is based on the diameter of the enscribed circle, wherein, for comparison purposes, the square wire-to-wire spacing arrangement is .707 of the diameter.

Anthony
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« Reply #17 on: November 16, 2014, 08:06:59 PM »

Anthony and all,

Thanks for your input!  Yes, I agree that .707 times 6 inches gives us 4.242 inches for the square "D",
Distance, needed for the "antenna manual" chart.

(1) When I graph 4.242 inches for a number 12 wire the result on the left side of graph shows about 260 ohms.

For the ARRL "Diameter" graph:

(2) I plugged in 6 inches for the diameter for a number 12 wire and the results are about 250  ohms.

Can some one check my above calculations from these two different graphs?

I have attached the formula to calculate a two wire parallel open wire feeder.  This formula came from
"Radio antenna Engineering" book, 1952, page 399, McGraw-Hill Book Company, Inc.

p = .0404 radius of #12 wire (the formula needs radius and not diameter of wire).
a = 4.242
b = 4.242

a and b are the two sides of the square making up the 4-wire open wire feeder.  

I am using the bottom formula on page 299, see attachment.

When plugging above data in to the formula I get 299.7 ohms (300 ohms)

Graph wise 250 ohms is close enough to 260 ohms.  But, these two numbers are way off from calculated 300 ohms.

Am I doing the above comparison/calculation correctly?

Thank you.
Chuck



* Two Wire Ladder Line Formula.JPG (722.19 KB, 2550x3299 - viewed 752 times.)
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« Reply #18 on: November 16, 2014, 09:48:53 PM »

Chuck.

Be careful regarding the directions of the currents in the four wires:

In the earlier two of your references, the currents in diagonally opposite wires are in the same direction.

In the last reference, the currents in diagonally opposite wires are in opposite directions.

See my previous post. The difference is -20.8 ohms vs. + 20.8 ohms

Stu
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« Reply #19 on: November 16, 2014, 10:13:06 PM »

Stu,

Take a look at this page 400 attachment.

Text line 4 states "four-wire side-connected" of this type when the diagonal spacing ...

Is this what you are referring to? 

Do you think this means two wires on each of their respected sides are connect vs. the two wires opposite each other are connected?

Chuck


* 4-wire line page 400.JPG (574.86 KB, 2550x3299 - viewed 716 times.)
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« Reply #20 on: November 16, 2014, 10:27:57 PM »

Stu.
I double checked page 399 attachment and line 7 states "Each diagonal pair is in the neutral plane of the other with no intercoupling.  This means the same layout for the original diagonally 4 wire line.

What do you think?

Chuck
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« Reply #21 on: November 16, 2014, 10:43:26 PM »

Chuck

What that means is that diagonally opposite wires have currents flowing in opposite directions.

What is meant by the two diagonal pairs being in each other's neutral plane is that the configuration can be thought of as two 2-wire balanced lines that are perpendicular to each other, and centered on the same axis.

Stu

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« Reply #22 on: November 16, 2014, 11:24:01 PM »

Stu,

Thank you.

Chuck
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« Reply #23 on: November 17, 2014, 11:05:51 PM »

Chuck, For your Information.  Tony


* Terman1.jpg (3086.08 KB, 3342x4818 - viewed 822 times.)
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« Reply #24 on: November 18, 2014, 09:46:29 AM »

Tony,

Thank you for the feed line chart.
I will put it to use.

Chuck
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