People learning Morse code often use what is called “Farnsworth” timing to make recognising the sound patterns easier. Rather than slowing down the whole sound (which runs the risk of encouraging the counting of the dits and dahs), it is thought better to keep the character sounds at a moderately fast speed so that each character is heard as a whole sound and just increase the gaps between the characters and words to give more recognition time. Farnsworth timing was named after Donald R. "Russ" Farnsworth (W6TTB).
The page describing basic Morse code timing shows the example of the standard word “PARIS ”. Here's a version of the chart again showing the timing of the word “PARIS ” with a character speed of 20 WPM:
The “Farnsworth” toggle button shows how the timing changes when a 10 wpm Farnsworth speed is applied: changing the total length of “PARIS ” (including the final inter-word space) from 3000 ms to 6000 ms while leaving the timing of the individual characters in tact.
The speed of the characters is determined by the normal WPM speed ($s_{wpm}$) but we need to define the Farnsworth speed ($s_{farn}$) as well and use this to determine the actual number of words per minute. That is:
where one Farnsworth-unit is longer than the basic unit. In this way the inter-character space and word space are kept in the normal proportion to each other, but longer than they would be with standard spacing. How can the inter-character space and the word space be calculated though?
To keep the characters at the $s_{wpm}$ speed, of the 50 dit units in “PARIS ” the part that we need to stretch out is the 4 inter-character spaces (12 units) and the 1 word space (7 units) and the part that needs to stay at the same speed is 31 standard $t_{dit}$ units within the characters (the dits, dahs and intra-character spaces).
The length of a dit is therefore the same as before:
$$ t_{dit} = {60 \over 50 s_{wpm}} \label{tdit} $$For the length of the word “PARIS ” we use the slower Farnsworth words per minute speed, or $s_{farn}$. One word (“PARIS ”) should take this many seconds:
$$ \text{time to transmit `PARIS '} = {60 \over s_{farn}} \text{ seconds} \label{paris} $$The time for the 31 standard units in “PARIS ” should be this many seconds:
$$ \text{time for the 31 standard units in `PARIS '} = 31 \times t_{dit} \text{ seconds} \label{paris31} $$Subtracting \eqref{paris31} from \eqref{paris} leaves the amount of seconds for the inter-character spaces and the word space:
$$ \text{time for 19 Farnsworth units} = {60 \over s_{farn}} - 31 t_{dit} \text{ seconds} \nonumber $$There are 19 Farnsworth units, so each one takes this many seconds:
$$ \text{time for 1 Farnsworth unit} = t_{fdit} = {(60 / s_{farn}) - 31 t_{dit} \over 19} \text{ seconds} \label{tfdit} $$Using \eqref{tfdit} we can then look at the ratio of $t_{fdit} / t_{dit}$
\begin{align} {t_{fdit} \over t_{dit}} &= {(60 / s_{farn}) - 31 t_{dit} \over 19 t_{dit}} \nonumber \\ &= {60 / s_{farn} \over 19 t_{dit}} - {31 t_{dit} \over 19 t_{dit}} \nonumber \\ &= {60 \over 19 s_{farn} t_{dit}} - {31 \over 19} \label{ratio} \end{align}Substituting \eqref{tdit} into \eqref{ratio}:
\begin{align} {t_{fdit} \over t_{dit}} &= {60 \times 50 s_{wpm} \over 19 s_{farn} \times 60} - {31 \over 19} \nonumber \\ &= {50 \over 19}\cdot{s_{wpm} \over s_{farn}} - {31 \over 19} \nonumber \\ &= {50s_{wpm} - 31s_{farn} \over 19s_{farn}} \label{ratio2} \end{align}It’s easy to check that if $s_{farn} = s_{wpm}$ then, using \eqref{ratio2} we see $t_{fdit} / t_{dit} = 1$ which makes sense.
Alternatively, we can substitute \eqref{tdit} into \eqref{tfdit} to get an expression for $t_{fdit}$ in terms of the two speeds:
\begin{align} t_{fdit} &= { {60 \over s_{farn}} - 31 t_{dit} \over 19} \nonumber \\ &= { {60 \over s_{farn}} - {31 \times {60 \over 50 s_{wpm}}} \over 19} \nonumber \\ &= {60 \times 50 s_{wpm} - 31 \times 60 s_{farn} \over 19 \times 50 s_{wpm} s_{farn}} \nonumber \\ &= {300 s_{wpm} - 186 s_{farn} \over 95 s_{wpm} s_{farn}} \text{ seconds}\label{tfdit2} \end{align}We can check that if $s_{farn} = s_{wpm}$ then \eqref{tfdit2} works out the same as \eqref{tdit}.
Although this derivation is a little different, you will find it comes out the same as the equations described by Jon Booom (KE3Z) in an ARRL article which Ronald L. pointed me to and which is about as definitive as I can find. ARRL is the (USA) national association for Amateur Radio.
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