How to Design a Shortened Vertical Antenna

Practical Guide for the Design of Shortened Verticals

Keith Greiner

2025

Although every HF amateur radio operator would like to have a full-length antenna, there are frequencies and locations where a full-length HF antenna just won’t work.   Those who live in an urban environment frequently comment that backyards or apartment buildings, or HOA restrictions present a considerable limitation.  Wouldn’t it be fun to see a full-sized 160-meter antenna anywhere you want it?  Maybe on your car?  Fun fantasy, but of course it won’t work. there is simply not enough room and if you put a 160 meter quarter wavelength of 40-meters  (aka 131 feet)  antenna onto your car, and drive around town, there’s a good chance it’s not going to fit under something.  The minimum height of an interstate overpass is 16 feet in most areas.  As a result, radio operators often need to turn to shortened designs that will fit into a room, on a deck, under a porch, or on the back end of a vehicle.  This commentary focuses on some practical ways to build a shorter vertical antenna. 

Shortened vertical antennas have been a necessity for amateur radio operators for years since the beginnings of the service/hobby in the first decade of the 20th century.  I’ve seen projects that specify exact amounts for lengths of wire and loading coils. I once had a vertical that came with the parts all figured out and all I had to do was assemble, but without any idea of the how and why of those designs.  All we did (and still do) was plug-in and use.  Where’s the learning in that?   Amateur radio is about learning RF, antennas, amplifiers, oscillators and such, so it’s worthwhile to take a few minutes to better understand a shortened antenna, and maybe even make your own.  

The basic idea of a shortened antenna, is to figure out how long you need it to be, calculate the inductance and inductive reactance of the section you need to remove to make it the length you need, and then figure out what inductance coil value will replace the section that you are removing.   The example below is based on a 20-meter vertical that is ¼ wavelength.  I used a quarter wavelength because that is the size of a vertical that results in a 50-ohm impedance at the bottom.  A half-wave dipole is a quarter on each size and is fed with a matching impedance of about 75 ohms in the middle. 

Now back to the vertical.  If you remove 10 feet from the bottom of the vertical, and substitute a coil, the inductance of that coil will have a value of let’s say,  Xa.  If you remove 10 feet from the top of the vertical, and substitute a coil, the inductance of that coil will have a value of Xb.  Xa and Xb will not be the same.  The value of a coil placed at or near the top is not the same as the value of a coil placed at or near the bottom.  In short, the same coil won’t work in both places.    So, when you calculate the required coil size it is important to remember that the location of removal is an important part of the calculation.  You’ll also see that I recommend that you remove the section that is near the top, but about a foot from the very top. You remove from the top to keep the greatest possible amount of magnetic aperture.  You don’t go all the way to the top because that would require a loading coil of infinite impedance.     

Some authors have put the calculations into a single formula.  I approach it as a series of mathematical unit conversions, calculations steps and tasks.  That helps me understand the process and helps me arrange task so the calculations can be programmed into a computer using your preferred language.  My preferred language is C++.

Let’s work with a 20-meter vertical.  For the 20-meter CW band, of 14,050,000 Hz, a full wavelength “l” = the speed of light / frequency.  Using the typical notation,  l = c/f.  That is 299,792,458 meters/second divided by 14,050,000 Hz/second = 21.34 meters, and that is 70.4 feet.   Now divide that by 4 because a quarter wavelength vertical, having a 50-ohm resistive impedance at the base, would be  l/4 = 70.4/4 = 17.5 feet.  For me, that’s too tall to fit inside a room with an 8-foot ceiling.   So, let’s remove 10 feet.  

To accomplish that shortening, we need to replace 10 feet with a coil.  What should the size of the coil?  The rest of this document will show how to find the inductance of the coil, and show how many turns would be needed on a 2.125”  diameter by 4” long piece of schedule 40 PVC. 

The basic idea is that we calculate the inductive reactance (XL1) from the top of the antenna to the bottom of the section to be removed.   Then we calculate the inductive reactance from the top of the antenna to the top of the section to be removed (XL2).  The inductive reactance of the section to be removed is:  XL = XL1 – XL2.   The value of XL is then used to calculate the size of the coil.  

The following image shows a drawing of a l/4 vertical showing that X = XL1 -  XL2. 

The calculations required to find the inductance may be accomplished on a calculator, spreadsheet, or computer program.  If you want a good calculator, on the date of this writing there is one at: https://www.homepages.ed.ac.uk/jwp/radio/software/   by Jack Ponton.  Even if you use the calculator, it is worthwhile to study the following discussion and learn how the values are obtained so that you have a better understanding of the theory and the context of the calculations.  That helps when you find that your application of the calculation doesn’t seem to fit the circumstances.

The following image shows a drawing of a vertical antenna with the coil, where the inductive reactance of the coil is X = X1 – X2.   These drawings are not to scale.

For our calculations, there are several equations.  The Ponton website shows a comprehensive formula that describes the calculations.  That is helpful.  As I look at it I need to create a series of steps that can be implemented in a spreadsheet or a C++ program. 

Below are the steps I used.

First is the calculation to find the characteristic impedance of a wire in space.

Equation 1.

Where:

Zo = Characteristic impedance of the wire segment of interest.

Ɛ = is relative permittivity, and I used a value of 1 for air.

h = height above ground (1/2 the height of the vertical antenna).

d = diameter of the wire.

h and d must be in the same units of feet or meters.

Next is the inductive reactance a segment of the wire in space, based on the characteristic impedance.  

Equation 2.

Where:

B = is the angle to the point of interest measuring from the highest end of the vertical where the top is zero degrees and the bottom is 90 degrees.

o = Characteristic impedance of the segment of wire per the equation.  

cot = the cotangent of the angle, which is cosine(B)/sine(B). Remember that B may need to be converted from degrees to radians.

-j = Square root of -1.  Just ignore it.

The reactance calculations require the cotangent of an angle because the reactance has different values along the length of an antenna.  Reactance is smallest at the bottom feed point and greatest at the top.     

Remember that some calculators, Excel, and computer programs require that the angle used in the sine and cosine functions use the angle in radians.    Degree angles can be converted to radians by multiplying the degrees by 𝜋/180. 

We will also be using the cotangent trigonometric function.  The cotangent is the inverse of the tangent.  Recall that the tangent is the opposite side of a triangle over (divided by) the adjacent side.  The cotangent is the adjacent over the opposite.  It is also, the cosine of an angle divided by the sine of an angle.  The cotangent is designated by the string, “cot”.

There are four ways to indicate the height of a quarter wavelength antenna.  

1.     Meters (for the example that is 21.34 meters).

2.     Feet (for the example that is 17.5 feet)

3.     Degrees of a circle ranging from 0 to 90.  If we start at the top, that would be zero at the top, and 90 at the bottom, and if we start measuring from the bottom that would be 0 at the bottom and 90 at the top. 

4.     Radians of ¼ of a circle ranging from 0 to 1.570963.  If we start at the top, that would be 0 at the top and 1.570963 at the bottom. 

We know that the intensity of RF radiation is directly related to the amount of current in an antenna.  With a vertical antenna, the current is maximum at the bottom feed point, and is minimum at the top.  Conversely, the voltage is minimum at the bottom and is maximum at the top.  Because the maximum current is at the bottom, any shortening of the antenna will need to preserve as much of the length at the bottom as possible.  As the same time, we know that we’ll need to leave a short piece at the top. If we put an inductor at the exact top it will need to have infinite impedance.  I put a coil in position and found that the math is correct.  Let’s put about two feet at the top, above the coil.  That means, we can take out about 10 feet beginning two feet from the top of a 17.5-foot antenna, and extending downward until there are 5.5 feet remaining at the bottom.  These 5.5 feet (1.676 meters, ) will form the aperture of our shortened quarter wave antenna.   This will result in 51.5% of the area under the cosine current curve, which is 51.5% of the total magnetic field radiated by the quarter wave antenna.  If the exposed radiator is the same 5.5 feet positioned at the top of the antenna, that is only 16.6% of the total magnetic field. See the Ampere-Maxwell equation for an explanation of the concept.  See the method of moments and Note 1.

Let’s take a moment to better understand why we want the inductor near the top.  When a voltage is connected to an inductor the voltage increases and a few micro-moments later the current follows.  When the voltage is alternating from positive to negative, as in an RF signal, the voltage leads the current by 90 degrees.   When the alternating voltage is connected to a wire that has a length equal to the wavelength of the alternating voltage, the distributions of voltage and current are sinusoidal along the wire and are 90 degrees apart.  At the end point, the voltage is maximum and the current is minimum. At points that are 90 degrees (1/4 wavelength) in from each end, the current is maximum and the voltage is minimum. 

Half wave dipole antennas take advantage of this relationship by connecting the antenna to a feed line at the 90-degree point where the voltage is the minimum and the current is the maximum.  Quarter wavelength vertical antennas take advantage of this by feeding an antenna at the bottom where the voltage is minimum and the current is maximum. 

The amount and location of maximum current is important for a radio antenna because it is the current — the moving electrons — that create the magnetism of the radio wave. 

If we divide the antenna into infinitely small portions and add the cosine values at each of the points, we have an indicator of the total signal being radiated from the antenna.  If we want to shorten a quarter wavelength antenna, where the quarter wavelength is 17 feet, we can remove a segment from near the top or we can remove a segment from the bottom.  Let’s shorten our antenna to ten feet using two scenarios.  We will analyze the two scenarios by summing the cosine values that are remaining after the removal and then by dividing that by the sum of all cosine values in the 90-degree sine curve segment. 

1.     We leave two feet at the top, and then remove the next 10 feet, and then leave five feet at the bottom.  In this scenario, the sum of cosine values is 51.6% of the sum of all cosine values over the 90-degree range.  

2.     We leave 10 feet at the top and remove 7 feet from the bottom, the sum of cosine values is 16.6% of the sum of all cosine values over the 90-degree range.  

This analysis suggests that if we want to have the maximum possible signal, we need the inductance near, but not exactly at, the top.  Scenario 1 is better. Figure 4 summarizes the discussion. In Scenario 1, the sections that radiate are marked in blue.  There are two sections one at the top 2’ and the other 5’ at the bottom.  In Scenario 2, the one section that radiates is marked in grey, including the 2’ at the top that is also blue. Considering the sum of the cosine values, that is the selected areas under the curve (blue or grey).  It is possible to visualize that the total area of the two blue sections is considerably larger than the grey.

Also see the report on magnetic loop antennas at:  https://sites.google.com/view/magneticloopantenna/home

(c) Keith Greiner  2025