### Toolbox Talk: Dipoles

I gave a number of Toolbox Talks over Field Day weekend. One was done with Ron (WQ1Z) on dipole basics. I did a little write-up to go with the talk that discussed the "magic" formulas cited by all the radio books. The point of the talk was: use the formulas, but understand they are

There are standard formulas for computing the length of a dipole or dipole arm. A typical one is length (in feet) = 234 / frequency (in megahertz). Why 234? Let’s see where they got that number:

Light speed (c) is 3 x 10

also

Let’s test what we have so far with a 20m signal. We know the frequency (14 MHz). We know the speed of light in feet per microsecond (980). We can compute the wavelength (in feet) by dividing 980 by 14 to get 70 feet. A full wave of a 14 MHz signal is about 70 feet long. That seems about right!

We don’t want to cut a full wavelength of wire, though. We want to cut only 1/4 of that. So, we divide 980 by 4 to get how far light goes in a quarter microsecond, 980/4 = 245 feet. So, 245/frequency gives us the size of a quarter wave of light at that frequency in feet. That is pretty close to the “magic” 234/f. Why the difference?

All of the above was for light traveling in a vacuum. But, our signal is going in a wire, where the shockwave of the electrons travels a little slower than light. We call this reduction in speed the

Let’s assume that the speed needs to be reduced to only 95% of c to account for the velocity factor of the wire we are using. That would change our formula to be (245*0.95)/frequency which is 233/frequency = length of wire for a quarter wave. That is suspiciously close to our 234/frequency cited by all the books you’ve read!

Of course, we had a horse-sized assumption in here: the velocity factor of the wire. This is why this formula

__guidelines__and not absolute recipies. Enjoy.There are standard formulas for computing the length of a dipole or dipole arm. A typical one is length (in feet) = 234 / frequency (in megahertz). Why 234? Let’s see where they got that number:

Light speed (c) is 3 x 10

^{8}meters per second. Let’s convert that into feet per second. Multiplying by 3 x 10

^{8}meters per second by 3.28 (feet per meter) to get 9.8 x 10

^{8}feet per second. Since we know we are interested in frequencies as megahertz, it would be convenient to get rid of all these zeros and talk about feet per microsecond (and we’ll toss the MHz part of our frequency later to make up for it). So, light travels about 980 feet per microsecond. A wavelength is related to the speed of light and frequency with the following familiar formula:

**nu = c / lambda**meaning

**frequency = speed of light / wavelength**

also

**wavelength = speed of light / frequency**

Let’s test what we have so far with a 20m signal. We know the frequency (14 MHz). We know the speed of light in feet per microsecond (980). We can compute the wavelength (in feet) by dividing 980 by 14 to get 70 feet. A full wave of a 14 MHz signal is about 70 feet long. That seems about right!

We don’t want to cut a full wavelength of wire, though. We want to cut only 1/4 of that. So, we divide 980 by 4 to get how far light goes in a quarter microsecond, 980/4 = 245 feet. So, 245/frequency gives us the size of a quarter wave of light at that frequency in feet. That is pretty close to the “magic” 234/f. Why the difference?

All of the above was for light traveling in a vacuum. But, our signal is going in a wire, where the shockwave of the electrons travels a little slower than light. We call this reduction in speed the

*velocity factor of the wire*. This value can vary widely. Ladder line has a velocity factor of 95% of c. RG-58 has a velocity factor of only 66% of c!

Let’s assume that the speed needs to be reduced to only 95% of c to account for the velocity factor of the wire we are using. That would change our formula to be (245*0.95)/frequency which is 233/frequency = length of wire for a quarter wave. That is suspiciously close to our 234/frequency cited by all the books you’ve read!

Of course, we had a horse-sized assumption in here: the velocity factor of the wire. This is why this formula

*234/f*is just a guideline. If the velocity factor is lower, you’ll need to make the wire shorter (as the signal doesn’t travel so far). The formula

*234/f*actually provides a worst-case length and a dipole arm will often be shorter than the size you compute. Then again, it is easier to make a wire shorter than longer! The 234/f is no “magic formula” and “234” is no “magic number”. They are guidelines. Use them as such.

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