Dipole Calculation

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This webpage uses 2 formulas designed for half wavelength antennas (which includes the open half-wave dipole) resonating above 30 MHz.

If you take your time, follow the steps, and go back through certain steps that you do not understand...I am sure you will make out fine. So, without further ado, let's begin...

FORMULA I

492/MHz = The length of a half wavelength (feet) in free space

FORMULA II

K = The length of a half wavelength (in feet) in free space to the diameter of the radiating element(s) (conductor) used for the antenna.

Formula I states how long the half wavelength is to be, for your particular transmitting (resonating) frequency. Formula II accounts for the diameter of the conducting rod, used in making the antenna. With both Formulas taken into account, the final lengths will be a very close approximation of having the antenna resonate at the designated frequency.

Iusing the two formulas will give one example on how to derive the final answer, in stated aboveshould have no trouble in . Once you have understand the example, you finding the lengths of each of your rods for the dipole antenna.

The following example will will determine your rod length in decimal form. Once that is achieved, we will then change the decimal form into a fractional form. This way, it will then be easier to use a typical measuring tape; because typical measuring tapes are sectioned off in fractions of an inch.

Let's say that you wanted to transmit on 87.7 MHz, using a rod that has a diameter of 1/2 (.5") inch.

Let's use the Formula I to find the length of a half wave th at is oscillating at 87.7 MHz. Since we know that f (frequency) is 87.7 MHz, we can then plug that into Formula I.

The next thing to do is to convert 5.61 feet into inches. This can be done by changing 5.61 feet into an improper fraction...

And then multiplying the above number by 12, since there are twelve inches in 1 foot.

So our answer is now 67.32 inches...which is the length of a half wave resonating at 87.7 MHz in free space.

But this answer does not account for the diameter of the conductor used in the dipole antenna. That is why Formula II is needed; in order to get a more precise rendering of the length of the conductor.

We come now to using Formula II. Since we know that our answer up to now, is 67.32 inches for the conductor (the two rods) on the dipole...we will know acquire a more precise answer, by introducing the diameter of the rods used for the antenna. The thickness of the conductor plays a vital role in the overall determination of the required length for the conductor of the antenna. The larger the diameter of the conductor, the less overall length is needed for the conductor.

In the chart below, The Multiplying Factor K, is the result of its' intersection with a ratio. This ratio is...the length of a half wave resonating at your desired transmitting frequency to the diameter of the inductor (rods) used for the dipole antenna.

Since we now know both values in the above paragraph, we can now plug them into Formula II.

Look at the figures on the bottom portion of the chart. These figures are...the 'Ratio of Half Wavelength to Conductor Diameter'. Approximate where 134.64 would be in the figures, then traced an imaginary line straight up...until the line intersects with the K line.

At that point when your imaginary line has touched the K line, draw another imaginary line directly to the left...and all the way until it reaches the figures on the left side of the graph (Multiplying Factor, K). We see that 134.54 will interect the K line in between .96 and .97. So our Multiplying Factor K would be approximately .965

So now we take our Formula I answer, which was 67.32 inches, and multiplying it times K.

So our final answer is 64.96 inches...which is the length of a half wave resonating at 87.7 MHz in free space, by using a conductor whose diameter is 1/2 inch.

Nexr, we have to allow for a 1/2 inch spacing between the two conductors, when the time comes to actually make the dipole antenna. (Refer to the dipole webpage for a detailed explanation on construction of the antenna. ) Hence, a total of .5 inches must be taken away from our answer.

Also, since our dipole antenna consist of 2 conductors (rods), each conductor will be 1/2 of 64.46 inches. So we must divide 2 into 64.96...

This final step converts our answer, which is 32.23 inches in decimal form, to a fractional form. This is necessary because typical measuring tapes are sectioned off in fractions of an inch. We will be rounding off our 'fractional form' answer...to the nearest 16th of an inch. Let's begin...

We are only concerned (referring to 32.23) with the figures to the right of the decimal point...so we will drop off the 32 (and use it later)...and be concerned just with...

Let's convert .23 into a fractional form. So .23 (twenty three hundredths) is the same as...

We will now multiply 23/100 times 16, because we want our answer to come out in 16ths of an inch.

Then, go ahead and just add our denominator to the 4, which would look like this...

There you have it my friend. Now you can plug in your own frequency and diameter of your conductor...to come up with the precise length of each rod.

Should you need more help with the math, do not hesitate to send me a letter. I will gladly help you through this...