Back when I was a newly winged nugget, I departed the advanced flight training squadron for the fleet replenishment squadron to begin learning to ply my trade as a B/N. Being the newbie in the squadron, one of my direct duties was operating the radios and sending/receiving all sorts of message traffic. I did not have to worry about radio or antenna maintenance. I had the AT, ET and Air-frames shops to keep everything up and running. Antenna theory, what’s that?

Fast forward two or three lifetimes and while I have a solid handle on radio operation and communications protocols, planning radio nets and antenna theory not so much. In an effort to get up speed, I attended the BrushBeater Basic RTO course. Our host, NC Scout did and does a phenomenal job of presenting the course material. The course answers numerous questions and fills in many of the knowledge gaps one may have on the topic of communications. However, in many ways, the material leads to more questions. Scout teaches that if you remember the constant 936, you can build most any antenna needed to meet your needs at any particular time. As a good instructor does, Scout references the material he teaches in class in his blog posts and articles found on and American Scout makes reference to the 936 constant in his articles on the Jungle Antenna and The Dipole Atenna. But, why 936?

While reading and studying the Special Forces Radio Antenna manual (by the way, Scout has some for sale) I found the answer to my question. The length of any ½ wavelength antenna, be it a Marconi or a Hertz antenna, can be found using the following formula:

Antenna length (in meters) = (N-.05)150 megameters/frequency (megahertz)

Converting that formula into feet provides the following:

Antenna length (in feet) = (N-.05)492/frequency (megahertz)

where N = the number of desired 1/2 wavelengths and .05 is a correction factor.

For those who’ve taken the Basic RTO course, this formula should look somewhat familiar. However, it still does not answer the question of where 936 comes from. If we were looking at an antenna in a vacuum, a correction factor would not be needed. However, several factors in our environment such as wire diameter, wire material, proximity to the ground and other influences force us to use a correction factor. As per the Special Forces Radio Antenna Manual, “.05 is a good approximation for the correction factor.” By the way, it is the correction factor which forces us to trim our antenna wire to the proper length.

If we solve the numerator of our formula for a single ½ wavelength we obtain the following (1-.05)492/frequency (megahertz) yields 468/frequency (megahertz) which is the length in feet of a single ½ wavelength antenna. Scout teaches us that 936 is the constant used for a one wavelength antenna (1 wavelength = 2 halves) or (468×2)/frequency (megahertz) or 936/frequency (megahertz), the exact formula taught in the Basic RTO course!

Knowing this information gives rise to the following table:

Often Quoted Antenna Constants




Length, in feet, of a half wavelength in free space at 1.0 MHz


Length, in feet, of a half wavelength dipole in typical wire at 1.0 MHz


Length, in feet, of a half wavelength inverted vee in typical wire at 1.0 MHz


Length, in feet, of an extended double Zepp (EDZ) in typical wire at 1.0 MHz


Length, in feet, of a quarter wavelength vertical in typical wire at 1.0 MHz

Given the above discussion, you now have the basic theory behind how antenna lengths are calculated. Now, come to class and get some hands on training. The way things are going, you’ll need these skills sooner than later.