I'm an amateur radio operator, and part of this involves making antennas. One of the simplest is a dipole, which requires two pieces of wire. The total length of these is given by the formula:

Total length (in metres) = $143$ / Frequency (in MHz)

You then divide this number by two (and usually add a bit for safety as it's easier to cut wire off the end than add more) and cut your wires.

$143$ is a "standard constant". In the real world, this number varies due to a variety of factors and has to be measured by experimentation. Essentially by raising and lowering the antenna, cutting a bit off each time. It's really quite time-consuming.

Here's the problem I'd like to solve:

My measuring instruments are limited to a ruler and a metre stick. So I can measure long distances (metres) with poor accuracy, and short distances (up to $60$ cm) with good accuracy. In other words, if I've cut a wire to $14.2$m, it's probably $15$ or $14.7$ or something (but both wires will be equal because I've used one to cut the other).

Conversely, if I've cut $15$cm off the end of a wire, I know that I've cut $15$cm off, and that it's accurate, because it's a single measurement.

Here's the problem I'd like to solve:

  1. Let's say I have a magical instrument which tells me the frequency of my antenna (this is a real thing: it's called an antenna analyser). This tells my that my antenna works at $F1$. I don't know what the length truly is, so $L1$ is unknown.
  2. Now I cut a known length of wire off the end of each leg of my antenna (it has to be symmetrical). Let's say $15$cm. Now I have a new, higher, resonant frequency measurement, $F2$.
  3. Knowing $F1$, $F2$ and the change in length, is it possible to calculate the value of the "constant" $143$, to get a better idea what my next change in length should be?

In actuality, the length starts out as (L + n), then changes to (L + n - c); where L = the 'wanted' length, n is an unknown offset, and c is the amount we cut off (which we know because we measured it accurately).

My goal here is to go from time-consuming "successive approximation" to a method which allows the antenna to be raised and lowered an absolute minimum number of times.

Is this possible?

I've had a go with dividing the deltas between the lengths and frequencies, but the numbers I get don't make sense to me.

  • $\begingroup$ Check out ham.stackexchange.com - perhaps it is the better site for your question? $\endgroup$
    – Moritz
    Commented Oct 9, 2016 at 13:36
  • $\begingroup$ Thanks, Moritz - I will. I asked here because I was interested in it as a pure math problem, and wondered if there was some known method (which I don't know about) which could do what I want. $\endgroup$
    – philpem
    Commented Oct 9, 2016 at 14:37

2 Answers 2


Since this is trial-and-error approach, I suggest a binary search to minimize the number of trials.

Build each wire to a length which is more than adequate for the desired frequency. Fold-back one end of each wire to a 1/2 of its length which won't be long enough. Measure frequency. To change the length, fold or unfold 1/2 of the previous fold-length. Measure again. Do this until you hit the sweet spot and then cut the wire.


So you are given $$ L = k/F\quad \Leftrightarrow \quad F = k/L\quad \Leftrightarrow \quad F\,L = k $$ When you make a length cut, you will have a change in frequency such that the above still holds $$ \left( {F + \Delta F} \right)\,\left( {L + \Delta L} \right) = k\quad \Leftrightarrow \quad k = F\,L\left( {1 + \frac{{\Delta F}} {F}} \right)\,\left( {1 + \frac{{\Delta L}} {L}} \right) $$ Now, you can measure $F$ , $\Delta F$, $\Delta L$, but you are not sure about $L$ of which you just now it is some 100 times larger than the $\Delta L$ you are considering. So in the above equation you have two unknowns, $k$ and $\Delta L$, and you need to make at least two cuts to determine them (supposing $k$ is constant). Then $$ \left\{ \begin{gathered} k = F\,L\left( {1 + \frac{{\Delta F_{\,1} }} {F}} \right)\,\left( {1 + \frac{{\Delta L_{\,1} }} {L}} \right) \hfill \\ k = F\,L\left( {1 + \frac{{\Delta F_{\,2} }} {F}} \right)\,\left( {1 + \frac{{\Delta L_{\,2} }} {L}} \right) \hfill \\ \end{gathered} \right. $$ that is $$ \left\{ \begin{gathered} \left( {1 + \frac{{\Delta F_{\,2} }} {F}} \right)\,\left( {1 + \frac{{\Delta L_{\,2} }} {L}} \right) = \left( {1 + \frac{{\Delta F_{\,1} }} {F}} \right)\,\left( {1 + \frac{{\Delta L_{\,1} }} {L}} \right) \hfill \\ 2k = F\,L\left( {\left( {1 + \frac{{\Delta F_{\,2} }} {F}} \right)\,\left( {1 + \frac{{\Delta L_{\,2} }} {L}} \right) + \left( {1 + \frac{{\Delta F_{\,1} }} {F}} \right)\,\left( {1 + \frac{{\Delta L_{\,1} }} {L}} \right)} \right) \hfill \\ \end{gathered} \right. $$ Where, from the first you can compute $L$, insert into the second and compute $k$.
In the second, I expressed $k$ as the sum of the two terms, just to ... make an average.

Of course, you can take advantage of the fact that $\Delta F /F$ and $\Delta L/L$ are quite smaller than $1$, and simplify the expressions accordingly, if you do not require high precision in the computations.
Also, if you can afford to make more than two cuts, you can use the additional terms to smooth out the measurements errors, or to express $k$ as $k_0+k_1 \Delta F /F$, etc.


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