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BOW TIE LOOP: First, divide the frequency you wish to operate by 945 (945/FqMHz). Divide the result by 2, this is A,B,C. Now measure the distance between your supports 1 and 2 and divide by 2. Note that A is only relevant to B and C. Now determine "A" based upon the total of B and C. You should allow 2-3 FEET on each side for hanging.
In the example:
Attach an UPPER insulator at 47 FEET from one end (B). Now attach an insulator at 47 FEET from the opposing end (C). A = the result between insulator 1 and 2
(122.7-94). Connect the 4:1 BALUN at RED and a GIMMICK
at WHITE. Raise the top ends first and then adjust bottom.
If required use a CENTER POLE for additional support.
 
EXAMPLE: (945/3.850)/2 = 122.7
DISTANCE BETWEEN POLES = 100
B = 47 X 2 or total 94 FEETC = 47 X 2 or total 94 FEET
A = 122.7 - 94 or 28.7 FEET
VERTICAL component = 10 METER
K3HKR ON-LINE - SCIENTIFIC CALCULATOR and RADIO FORMULAS
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Instructions

Enter an expression into the tan bar and press enter to calculate the results.
This calculator remembers up to twenty past calculations in history. To save the history between visits you must have cookies enabled.
All results are calculated using a Javascript function. Syntax for expressions is the same as that for Javascript()eval.
This calculator can handle input numbers in several different bases:

^ is a bitwise xor operation. To raise a number to a power use pow() function.
The Display Result Selector will toggle your result automatically.
SCIENTIFIC CALCULATOR GNU LICENSE - DESIGN by: STEPHEN OSTERMILLER


FORMULAS:
 
(LC)
Calculating INDUCTANCE or CAPACITANCE for a known value...
LC= 25330/F2
First we SQUARE the Frequency desired (14.2 X 14.2) = 201.64
Now we DIVIDE 25330 by 201.64 = 125.6
To find "C" - If known INDUCTANCE is 5 uH, we divide 125.6 by 5 = 25.12 or 25 pF
To find "L" - If known CAPACITANCE is 25 pF we divide 125.6 by 25 = 5.0 uH
(In other words if I have a 25 pF CAPACITOR and I want an LC network to resonate on 14.2, I need a 5.0 uH INDUCTOR)
NOTE: The "Q" of an LC network is not calculated here, so all other "Q" factors must be used for consideration.


(DIPOLE)
Calculating a DIPOLE for HF is one of the first FORMULAS you should have learned.
DIPOLE = 468/FMHz
What frequency are we building the DIPOLE for ?? 3820 KHz or 3.82
Divide 468 by 3.82 = 122.6
Your half wave DIPOLE cut for 3820 KHz will be total 122.6 feet long, or 61 FT 3 IN on either side of center.
NOTE: As a flat top the impedance will be 75 OHM, and can be fed with 50 OHM or 75 OHM un-balanced feed line
As a INVERTED VEE, the impedance is lowered as the angle at the center is reduced. An INVERTED VEE usually
incorporates a 30 degree angle down from center and yields a 52 OHM impedance.
The FOLDED DIPOLE uses the same calculation, however the spread must be considered when calculating
antenna IMPEDANCE. The IMPEDANCE is 300 OHM or higher based upon the spread across the element.


(LOOP)
Calculating a DIPOLE for HF is one of the first FORMULAS you should have learned.
LOOP = 1005/FMHz
What frequency are we building the LOOP for ?? 3820 KHz or 3.82
Divide 1005 by 3.82 = 263.0
Your full wave LOOP cut for 3820 KHz will be total 263 feet long, or 65 FT 7 IN on any one of it's 4 sides.
Your full wave DELTA LOOP for 3820 KHz will be 87 FT 6IN on any one of it's 3 sides.
NOTE: As a flat top the impedance will be 250 OHM, and can be fed with 50 OHM un-balanced feed line
into a 4:1 BALUN. You can also feed the antenna with balanced feed-line and use a good tuner.
The BOW TIE loop calculation above was created by K3HKR taking into considerations, the final design.