This is Belden 9223. It is a 50-ohm coaxial cable, velocity factor 0.56


This is Belden 8524. It is a 22 AWG hookup wire. No velocity factor in the datasheet.

hookup wire

Why does the coax datasheet have a velocity factor, while the hookup wire datasheet does not?


3 Answers 3


A coaxial cable is a transmission line by itself, and therefore the velocity factor is known. It depends on the dielectric properties, geometry, and conductivity.

A single wire is not a transmission line. You need a return path for the current, and that will most probably be a second wire somewhere, or a ground plane.

So the geometry and dielectrics around the two conductors is unknown, and thus the velocity factor is unknown.

  • $\begingroup$ Well, that is not exactly true... Have a look at the Goubau line (en.wikipedia.org/wiki/Goubau_line). That is a single-conductor transmission line, which was quite popular at the beginning of the era of microwave experiments. The signal travels along the wire. The signal is 'launched' onto the write with a kind of corner reflector, and 'captured' at the other side. It was popular because it was simple and cheap to implement. It looses attractiveness when it rains or snows... $\endgroup$
    – jcoppens
    Dec 10, 2014 at 3:09
  • $\begingroup$ @jcoppens A Goubau line isn't just any old spool of hookup wire. It's a wire with a low-loss dielectric coating manufactured to controlled specifications. Put another way, a spool of hookup wire is a Goubau line about to the same extent that NM wire is twin-lead. $\endgroup$ Dec 11, 2014 at 11:34
  • $\begingroup$ Hey, nobody says it is practical. But it is a single-wire transmission line, isn't it? The dielectric doesn't make up for the second conductor does it? So, forgetting about it isn't good science. $\endgroup$
    – jcoppens
    Dec 11, 2014 at 17:31

Velocity factor is a property of electromagnetic wave propagation, not wire. A transmission line (like coax) is a conduit for an electromagnetic wave in itself, so the velocity factor can be defined. In fact, the velocity factor can be derived from the lumped model of a transmission line:


simulate this circuit – Schematic created using CircuitLab

Following the usual simplifying assumptions of a lossless transmission line, R and G become insignificant. If $L$ and $C$ are inductance (henry) and capacitance (farad) per meter, then the velocity of propagation in a transmission line is:

$$ v = \frac{1}{\sqrt{LC}} \tag{1} $$

The velocity factor is just this relative to the velocity of propagation in a vacuum: $v/c$.

A very similar equation is the characteristic impedance of the transmission line:

$$ Z_0 = \sqrt{\frac{L}{C}} \tag{2} $$

Thus, if we know any two of:

  • characteristic impedance,
  • capacitance per unit length, or
  • inductance per unit length

then we can calculate the velocity factor. Let's try it for Belden 9223, which specifies in the datasheet:

$$ Z_0 = 50\:\Omega \\ C = 37\:\mathrm{pF}/\mathrm{ft} = 1.21 \cdot 10^{-10}\:\mathrm{F/m} $$

So by equation (2):

$$ \begin{align} 50\Omega &= \sqrt{\frac{L}{ 1.21 × 10^{-10} }} \\ 50^2 &= \frac{L}{ 1.21 × 10^{-10} } \\ L &= 3.03 \cdot 10^{-7}\:\mathrm{H}/\mathrm{ft} \end{align} $$

And then by equation (1):

$$ \begin{align} v &= \frac{1}{\sqrt{( 3.03 \cdot 10^{-7} )( 1.21 × 10^{-10} )}} \\ v &= 165289256 \:\mathrm{m/s} \end{align} $$

Thus the velocity factor is:

$$ v/c = 165289256 / 299792458 = 0.55 $$

The datasheet says 0.56. I attribute the discrepancy to rounding error.

So what about a single wire? What values do we use for L and C?

That depends on the geometry of the wire. In the case of coax, the wave is propagating within the dielectric between the center conductor and the shield. This dielectric has a known geometry and composition, so the manufacturer can specify a velocity factor.

In the case of a wire, the wave will be propagating between the wire, and something else. Maybe the ground. Maybe another wire. Maybe the same wire some distance away, as in the case of a dipole. Are you stretching the wire in a straight line, or winding it into a coil? The capacitance will be depend on the permittivity of the space containing the electric field. Is it air? A tree? Because there are so many variables, a wire datasheet can not possibly specify a velocity factor. And while we might measure one, we must be careful to specify what propagation mode we are talking about, and the conditions under which it was measured.

  • $\begingroup$ Velocity factor is indeed a property of a conductor, and can sometimes be the property of electromagnetic wave propagation. For an electromagnetic wave, the velocity will typically be nearly the speed of light in a vacuum, or a "velocity factor" of almost 1.0, while for a given conductor it will typically be much less. $\endgroup$
    – Noji
    Feb 11, 2022 at 15:02

Have a look here: for inductance http://chemandy.com/calculators/round-wire-inductance-calculator.htm

$L=0.002\times l[ln{\frac{4.0\times l}{d}}-1.0+\frac{d}{2.0\times l}+\frac{\mu_r\times T(x)}{4.0}]$ (in $\mu H$)

This formula is an approximation, and $T(x)$ can be had in several degrees of precision, one value for $T(x)$ is shown in the referenced article.

For capacitance we have a similarly complicated formula (http://en.wikipedia.org/wiki/Capacitance):

$C=\frac{2 \pi \epsilon l}{\Lambda}[1+\frac{1}{\Lambda}(1-ln(2))+\frac{1}{\Lambda^2}(1+(1-ln(2))^2-\frac{\pi^2}{12})+O\frac{1}{\Lambda^3}]$

Unluckily, most of my basic electronics textbooks were lost in a flood, many years ago, and I've spent most of my career in the digital world. Never felt the necessity to buy the books again - I spent more than my share on computer books. I might be able to access the mentioned papers at the university, but it'll take a few days - we're in the exams.

As in the inductance case, I suspect 'O' is again some magic constant...

$\Lambda$ is just the ln(l/a) where l is the length, and a the radius.

Though self-inductance of a wire is relatively understood, capacitance of a 'lone' conductor is more mysterious. It helps if you think about what capacitance represents from another point of view: even if alone in in the universe, adding another electron to the conductor still takes some work, because you have to re-distribute the ones already on the conductor. After all, having the same sign, they don't like each other.

  • $\begingroup$ These formulas assume what...a wire in a straight line? And what do we use for $\epsilon_r$ and $\mu_r$? $\endgroup$ Dec 10, 2014 at 12:53
  • $\begingroup$ Also, each of these equations have a non-linear dependence on length. So if velocity factor were to be included on a wire datashet, for what length would it be specified? $\endgroup$ Dec 10, 2014 at 13:02
  • $\begingroup$ I don't think there will be a manufacturer (or a customer for that matter) who is interested in the datasheet with specs for a component in purely theoretical conditions. And yes, from the few graphs I did find on the internet, velocity (factor) is not constant. And, for lack of contradictions, I suspect $\epsilon_r$ and $\mu_r$ to be the 'normal' expected values. You could have a wire in an infinite space filled with, say, water. a wire in a straight line? as mentioned in the docs I referenced: "Thin straight wire, finite length" $\endgroup$
    – jcoppens
    Dec 10, 2014 at 16:17
  • $\begingroup$ OK, but that doesn't answer the question. You are basically saying "given a complete description of all the electrical properties of all matter in all space, the propagation of a wave can be calculated". The question is "what is the velocity factor of a wire?" not "what is the velocity of wave propagation in a completely homogeneous space with known permittivity and permeability with a straight wire of known diameter and length in it?" $\endgroup$ Dec 10, 2014 at 16:26
  • $\begingroup$ If you apply a voltage to a wire, in reference to another wire already connected to the opposite polarity of the same voltage source, the voltage value (minus some ohmic loss) will appear at the end of the wire opposite the voltage source, after a finite period of time. Does this finite time, compared with the speed of light, not represent the velocity factor of the wire? $\endgroup$
    – Noji
    Feb 7, 2022 at 5:13

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