Below are well known equations from Wikipedia for the resistance and reactance in the impedance of a dipole antenna.

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where $a$ is the radius of the conductors, $k$ is the wave number 2πf/c, $η_0$ denotes the impedance of free space = 377Ω, and $γ_e$ is Euler's constant = 0.57721566.

Wikipedia doesn't specify where the feed point is, but since the feed point impedance is a function of the position of the feed point along the antenna, and there are no variable terms in the equations for feed point position, I assume the equations are for a feed point in the center.

Notice how there are no terms anywhere in the equations for anything related to the self inductance of the elements, or related to the fact that radio waves travel slower in metal than in air.

Every one knows that a center fed dipole exactly 1/2 λ in length has about + 45 Ω of reactance at the feed point impedance so to make it resonant it has to be shortened about 5 %.

I proved this by writing a program using Microsoft Visual Studio .NET which allows me to input different values for frequency, length and radius and which then calculates R and X using these equations. The program showed that a dipole which is exactly 1/2 λ in length does in fact have + 45 Ω of reactance in the feed point impedance. I also added the reactance caused by self inductance and subtracted that from the rest of the reactance to see if they are the same, which they are not except for one value of radius.

It's undeniable that the presence of self inductance in the elements of a dipole will affect the reactance in the feed point impedance, and similarly the velocity factor of the metal will affect the resonant length and so also affect the feed point reactance, so it seems strange that the above equation for reactance doesn't appear to include terms for these factors.

In order to be resonant, a dipole must have to be shortened also to allow for self inductance of the elements and for the fact that the radio waves travel slower in the elements than in air, right?

So considering the above equations, does the self inductance and velocity factor of the elements of a dipole change its resonant frequency?

Addition. -----------------------------------------------------------

I suppose what i'm trying to point out is this.

Either one of the two following statements must be true.

  1. The above equations are correct and take into account the self inductance and velocity factor of the dipole elements, and in this case the extra + 45 Ω of reactance present in the impedance for a dipole of exactly 1/2 λ in length can be attributed at least in part to those characteristics.

  2. The above equations assume an ideal dipole with no self inductance and a velocity factor of 1, and so the reactance caused by these characteristics for a real dipole is present in addition to the + 45 Ω and the cause for this rogue 45 Ω to me remains a mystery.

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    $\begingroup$ I couldn't resist inserting some MathJax, I hope you don't mind my edits! $\endgroup$
    – rclocher3
    Aug 18, 2021 at 16:09
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    $\begingroup$ Hi Rclocher3 i don't mind, the changes improve the question ! $\endgroup$
    – Andrew
    Aug 18, 2021 at 22:05

2 Answers 2


Notice how there are no terms anywhere in the equations for anything related to the self inductance of the elements or related to the fact that radio waves travel slower in metal than in air.

Indeed, but they do contain a term for the length. These equations describe the impedance of a center-fed dipole made of an ideal conductor of negligible diameter in free space. How can you construct any two dipoles meeting these criteria that differ in any way except by length?

A circle's diameter impacts it's area, and yet there is no diameter term in $A = \pi r^2$. Likewise, the self inductance and velocity factor of a dipole are relevant to the impedance; that does not mean those terms need to appear in the equation.

The above equations assume an ideal dipole with no self inductance

An "ideal dipole with no self inductance" is probably not a thing. The particular definition of "ideal" varies of course, but typically it describes some simplified model that can be realistically approached, though not quite reached.

For example, an "ideal resistor" is one that obeys the equation $V=IR$. Real resistors don't quite follow this equation: they also have some inductance, and their resistance varies with temperature, and so on. But these effects are small, and can be made smaller though appropriate design, and are negligible in a great many applications.

Likewise, an "ideal dipole" can be considered to be a 2-dimensional line with no thickness, because real dipoles can be made of thin wire where the thickness is negligible. An "ideal dipole" can be considered to be in free space, because dipoles can be mounted high enough above the ground and the feedline routed such that these things have negligible consequence to the model.

"Ideal" does not mean "unrealistic". For example, you will not find "an ideal ingot of lead with no mass" because this is just a pointless thing to model. All lead has mass. One can not make "more ideal lead" with less mass. Modeling lead with no mass is an utterly pointless thing to do.

Likewise you will not find any model describing "an ideal dipole with no self-inductance". All circuits of nonzero size have some inductance. You can't reduce this inductance any more than you can reduce the mass of lead. You can make the dipole shorter which reduces its inductance, but then there is a term for length in the equation to account for exactly that.

  • $\begingroup$ Hi Phil that's a very good point. But, for a circle, the diameter is directly related to r so A = π x (d/2)², there are no other variables which determine A. Does this mean that all the factors which determine self inductance and velocity factor are included in these equations ? I had another look, the equation for velocity factor depends on inductance, and the formula for inductance depends on the permeability of free space ... does permeability of free space depend on the impedance of free space ? ... $\endgroup$
    – Andrew
    Aug 18, 2021 at 22:39
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    $\begingroup$ @Andrew the impedance of free space, the permeability of free space, the permittivity of free space, and the speed of light are all interrelated by Maxwell's equations; any two are enough to give the rest. $\endgroup$ Aug 19, 2021 at 1:32
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    $\begingroup$ @Andrew Yes, all those factors are included. $\endgroup$ Aug 19, 2021 at 3:33
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    $\begingroup$ @Andrew Can I ask why you think they are not? $\endgroup$ Aug 19, 2021 at 14:46
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    $\begingroup$ @Andrew the self inductance, velocity factor, and impedance are all closely related. I don't know that we can separate reactance into a part caused by self inductance and velocity factor, and another part caused by something else. But, you certainly can say the reactance when the wire is a physical half-wave is due to the velocity factor being less than 1. When you shorten the antenna to make the reactance zero, you're finding the point where the dipole is electrically a half-wave. The ratio between the physical half-wavelength and the electrical half-wavelength is the velocity factor. $\endgroup$ Aug 20, 2021 at 17:29

Note the velocity factor of the metal also doesn't appear in generic equations for inductors. The inductance is more about the permeability of the free space surrounding the conductive elements, not how the current travels inside the conductive elements.

A dipole with capacitive hats can be made shorter than one without. The shortening of a resonant dipole from the free space half wave length could be less about inductance and more about the tips of the dipole having some capacitance to free space (and/or the rest of the antenna), and thus acting like capacitive hats, thus lowering the resonant frequency.

  • $\begingroup$ I'm thinking that velocity factor and self inductance as two separate characteristics, not that one causes the other, it makes sense that the impedance of free space is in the equations and this could account for the effect velocity factor has on impedance. But surely if inductance makes the antenna too long then capacitance would have the opposite effect ? Also the idea that free space is the other plate of a capacitor doesn't make sense to me and if the total reactance is 45 ohms it seems unlikely that the tiny amount of capacitance the antenna has to itself would cause that ? $\endgroup$
    – Andrew
    Aug 27, 2021 at 6:57
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    $\begingroup$ Consider capacitance to free space to be the capacitance to every other conductive object in the entire visible universe, a trillions of suns and free electrons. Increase the value of a capacitor in a tuned circuit, and it usually lowers the frequency of resonance, same as making an antenna longer. $\endgroup$
    – hotpaw2
    Aug 27, 2021 at 15:32
  • $\begingroup$ Could it be that capacitive hats make the antenna shorter because the antenna has larger diameter at the ends, also resulting in lower Q and therefor larger bandwidth, and not because there is capacitance to anything ? $\endgroup$
    – Andrew
    Aug 28, 2021 at 23:27
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    $\begingroup$ It's not either/or. The EM field equations indicate multiple effects to multiple parameters. $\endgroup$
    – hotpaw2
    Aug 29, 2021 at 4:00

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