### Samuel Haughton and the Twelve Faithless Hangmaids

In his study of humane methods of hanging, Samuel Haughton (1866) considered the earliest recorded account of execution by hanging (see Haughton’s Drop on this site). In the twenty-second book of the Odyssey, Homer described how the twelve faithless handmaids of Penelope “lay by night enfolded in the arms of the suitors” who were vying for Penelope’s hand in marriage. Her son Telemachus, with the help of his comrades, hanged all twelve handmaids on a single rope. The precise method of hanging is obscure, and Haughton considered a possible way the execution might have been accomplished. He imagined that Telemachus fastened one end of a rope to the top of a pillar and tied twelve slip-knots, which were placed around the necks of the women. The free end of the rope was then passed over another pillar, and the hangmen “all pulled together, in sailor fashion”, hoisting the women into the air. This formed what Haughton called a funicular polygon.

Mathematical Analysis

Simple Case: Four Handmaids

Haughton carried out a mechanical analysis of the funicular polygon, which we recount here. For simplicity, we first analyse the case of four handmaids. We assume all are of equal weight ${W = mg}$, there is symmetry about the central vertical line and the middle segment is horizontal. Let ${\theta_1}$ and ${\theta_2}$ be the angles from horizontal at the first two nodes and ${T_1}$, ${T_2}$ and ${T_3}$ the tension forces in the first three segments (numbered from the right; see Figure below). There is a balance of horizontal and vertical components of force at each of the polygonal nodes. This gives us four equations $\displaystyle \begin{array}{rcl} T_1 \sin\theta_1 - T_2 \sin\theta_2 = W & \qquad & T_1 \cos\theta_1 - T_2 \cos\theta_2 = 0 \\ T_2 \sin\theta_2 \phantom{ - T_3 \sin\theta_3} = W & \qquad & T_2 \cos\theta_2 - T_3\phantom{ \cos\theta_3} = 0 \end{array}$ The suspension of four masses, forming a funicular polygon.

These are four equations for five unknowns ${\{ \theta_1,\theta_2, T_1,T_2,T_3\}}$. There is also a geometric constraint: if ${d}$ is the distance between the points of suspension and all segments of the rope are of equal length ${\ell}$, then $\displaystyle ( 2\cos\theta_1 + 2\cos\theta_2 + 1 )\ell = d \,. \ \ \ \ \ (1)$

It is straightforward to combine equations (1)–(2) and solve for ${\{ \theta_1,\theta_2, T_1,T_2\}}$ in terms of ${T_3}$. We find that $\displaystyle \begin{array}{rcl} T_1^2 = 4W^2 + T_3^2 \,, &\qquad& \tan\theta_1 = 2W / T_3 \\ T_2^2 = W^2 + T_3^2 \,, &\qquad& \tan\theta_2 = W / T_3 \,. \end{array}$

We note that the first of these equations implies ${T_1 \ge 2W}$. If required, (3) can now then be used to get ${T_3}$ in term of known constants.

Homer’s Case: Twelve Handmaids

We now consider the case of twelve handmaids as described in the Odyssey and analysed by Haughton. The balance of forces yields twelve equations: $\displaystyle \begin{array}{rcl} T_1 \sin\theta_1 - T_2 \sin\theta_2 = W & \qquad & T_1 \cos\theta_1 - T_2 \cos\theta_2 = 0 \\ T_2 \sin\theta_2 - T_3 \sin\theta_3 = W & \qquad & T_2 \cos\theta_2 - T_3 \cos\theta_3 = 0 \\ T_3 \sin\theta_3 - T_4 \sin\theta_4 = W & \qquad & T_3 \cos\theta_3 - T_4 \cos\theta_4 = 0 \\ T_4 \sin\theta_4 - T_5 \sin\theta_5 = W & \qquad & T_4 \cos\theta_4 - T_5 \cos\theta_5 = 0 \\ T_5 \sin\theta_5 - T_6 \sin\theta_6 = W & \qquad & T_5 \cos\theta_5 - T_6 \cos\theta_6 = 0 \\ T_6 \sin\theta_6 \phantom{\ \ - T_7 \sin\theta_7} = W & \qquad & T_6 \cos\theta_6 - T_7\phantom{\,\cos\theta_7} = 0 \end{array}$

In addition to these, there is the geometric constraint: $\displaystyle [2(\cos\theta_1+\cos\theta_2+\cos\theta_3+\cos\theta_4+\cos\theta_5+\cos\theta_6) +1]\ell = d \,.$

This gives us thirteen equations for thirteen unknowns, six angles and seven tensions. However, all we really need to know is the value of the tension ${T_1}$: this is the force that must be exerted to suspend all twelve masses. The funicular polygon arrangement for hanging the twelve handmaids. Telemachus and his comrades all pull together, like sailors weighing an anchor.

We can easily rearrange the twelve equations, working from the bottom up: $\displaystyle \begin{array}{rcl} T_6 \sin\theta_6 = \ W & \qquad & T_6 \cos\theta_6 = T_7 \\ T_5 \sin\theta_5 = 2W & \qquad & T_5 \cos\theta_5 = T _7 \\ T_4 \sin\theta_4 = 3W & \qquad & T_4 \cos\theta_4 = T_7 \\ T_3 \sin\theta_3 = 4W & \qquad & T_3 \cos\theta_3 = T_7 \\ T_2 \sin\theta_2 = 5W & \qquad & T_2 \cos\theta_2 = T_7 \\ T_1 \sin\theta_1 = 6W & \qquad & T_1 \cos\theta_1 = T_7 \end{array}$

From these it is easy to find the angles and tensions in terms of ${T_7}$, the tension of the central, horizontal segment. In particular, $\displaystyle T_1^2 = 36 W^2 + T_7^2 \quad\mbox{so that}\quad T_1 \ge 6W \,.$

For small values of the ${\theta}$-angles, ${T_1}$ will be much greater than ${6W}$. Haughton concluded that the force required would be so great that the proposed method would be quite impractical: it would not have been possible for the hangmen to lift all the women into the air together. He wrote “We are therefore forced to the conclusion that the hanging of Penelope’s handmaids in a funicular polygon was mechanically impossible”.

Haughton surmised that an alternative method must have been employed. Perhaps a hawser was passed in a circle around the vault under the dome, and the twelve women were hanged from separate nooses tied to it. They would have to have been lifted up one by one to swing clear of the ground.

Author’s Comment: The above analysis follows Haughton’s arguments as described in the reference below. However, it seems that the solution applies only to the static case, where the unfortunate “hangmaids” are already suspended, and that a greater force is required during the dynamic process to put them in place.

Sources ${\bullet}$ Haughton, Samuel (1866): On hanging, considered from a mechanical and physiological point of view. The London, Edinburgh, and Dublin Philosophical Magazine and Journal of Science, 32 (213), 23-34. DOI: 10.1080/14786446608644122.