There are two major models that specify the point at which a sword should ideally hit its target. One model focusses on the sword’s vibration, particularly the nodes of the fundamental flexural vibration; the node which is closer to the sword’s point is often called *centre of percussion*. [1, 2] The other model, which is the subject of this work, considers translation and rotation as the components of a rigid body’s motion [3] upon impact. Several comprehensive articles on the rigid-body dynamics of swords exist, such as by Turner [4], Denny [5] and Le Chevalier [6]; those articles are highly recommended for a further reading on the consequences and possible applications of a sword’s analysis on the basis of rigid-body dynamics. The physical principles have been known since the 17th century. [7] Rigid-body dynamics were already applied in some historical sources on fencing [8, 9] to describe the point where a cutting weapon should ideally hit its target.

This article is meant as a brief introduction of the subject for those who are not yet aware of the physics of fencing. Additionally, this article emphasises the importance of the moment of inertia for the behaviour of a sword, and encourages sword researchers and makers to determine and include this fundamental measure from which other parameters can be calculated.

## Motion of a Sword

At any given time, a sword’s motion can be described as rotation about an axis, translation along a trajectory and as vibration. As we focus on sword hits with the edge, i. e. sword use with no significant bending of the blade, we neglect vibration and consider the sword an ideally rigid body. Every body has a resistance against a change of its translation, i. e. mass, and against a change of its rotation, i. e. moment of inertia. Each mass point \(m_i\) with the distance \(r_i\) from the axis has a moment of inertia \(I_i = m_i r_i^2\). The total moment of inertia \(I\) is the sum of the moments of inertia of its parts:

\begin{equation}

I = \sum_i m_i r_i^2

\end{equation}

The moment of inertia not only depends on the mass distribution of the body but also on the body’s position relative to the axis about which it rotates. The axis about which the moment of inertia is smallest passes through the centre of mass and is called the third principal axis. The perpendicular axes which intersect the third principal axis in the centre of mass are the first and the second principal axis. The first principal axis is the principal axis with the largest moment of inertia. It is also the first principal axis which is most relevant for the sword’s cutting motion. While the sword does not necessarily rotate about the first principal axis during a strike, it is commonly an axis which is parallel to the first principal axis. For an axis parallel to an axis through the centre of mass \(M\) for which the moment of inertia \(I_M\) is known, the moment of inertia \(I_x\) can be calculated according to the parallel-axis theorem with the distance \(r_x\) between the axis of rotation and the parallel axis through the centre of mass:

\begin{equation}\label{eq02}

I_x = I_M + m r_x^2

\end{equation}

Note that the moment of inertia \(I_x\) is larger when the axis of rotation is situated further away from the centre of mass \(M\). As the moment of inertia describes the resistance to a change of rotation, this means that weapons like axes or maces with more distance between the axis of rotation and the centre of mass are in general more difficult for the fencer to manoeuvre but also more difficult for the opponent’s armour to stop. The larger distance between axis of rotation and centre of mass and the thus larger moment of inertia is what makes the difference between a *Mordhau* and an *Oberhau*.

### The Sword as a Physical Pendulum

A point mass which can rotate about an axis in a homogeneous gravitational field is called a mathematical pendulum. In contrast, an extended rigid object of arbitrary shape which can rotate about an axis in a homogeneous gravitational field is called a physical pendulum. The angular eigenfrequency \(\omega_0\) of a mathematical pendulum is defined by its length \(l\) and the gravitational acceleration \(g\). The length of a mathematical pendulum which oscillates at the same angular eigenfrequency \(\omega_0\) as a physical pendulum is called *radius of oscillation* or *equivalent pendulum length* \(l_r\). The point which is situated \(l_r\) below the pivot \(P\) when the pendulum is at rest is called the *centre of oscillation*. In a mathematical pendulum, the centre of mass and the centre of oscillation are the same point, while in a physical pendulum, these points are generally in different positions.

#### Radius of Oscillation

To calculate the radius of oscillation and thus the position of the centre of oscillation, we start with the conservation of energy. In the frictionless case, the sum of rotational energy \(E_\text{rot}\) and potential energy \(E_\text{pot}\) remains constant:

\begin{equation}\label{eq03}

E_\text{total} = E_\text{rot} + E_\text{pot}

\end{equation}

The rotational energy is \(E_\text{rot} = I_P \dot{\alpha}^2/2\) with the moment of inertia \(I_P\) about the pivot \(P\) and the angular velocity \(\dot{\alpha} = d\alpha/dt\). The potential energy of a mass \(m\) in the gravitational field with a gravitational acceleration \(g\) and an angular displacement \(\alpha\) is \(E_\text{pot} = m g r_P(1-\cos{\alpha})\), which is approximated as \(E_\text{pot} \approx m g r_P\alpha^2/2\) for small angular displacements \(\alpha \ll \pi/2\). With Eq. (3), this yields:

\begin{equation}

E_\text{total} = \frac{I_P}{2}\dot{\alpha}^2 + \frac{m g r_P}{2}\alpha^2

\end{equation}

Deriving with respect to time \(t\) results in the simple oscillation equation:

\begin{equation}

0 = I_P \ddot{\alpha} + m g r_P\alpha

\end{equation}

Thus, the angular displacement is \(\alpha = \alpha_0 \sin(\omega_0 t + \varphi)\) with the amplitude \(\alpha_0\), the phase angle \(\varphi\) and the angular eigenfrequency

\begin{equation}\label{eq06}

\omega_0 = \sqrt{\frac{m g r_P}{I_P}}

\end{equation}

A mathematical pendulum with the same eigenfrequency \(\omega_0\) has the length \(l = g/\omega_0^2\). Thus, using Eq. (6), the radius of oscillation \(l_r\) of a physical pendulum, which is the distance between the pivot \(P\) and the centre of oscillation \(O\), is

\begin{equation}\label{eq07}

l_r = \frac{I_P}{m r_P}

\end{equation}

It can be proven [3, 10] that the relation of the pivot \(P\) and the centre of oscillation \(O\) is symmetric, i. e. if \(O\) is the centre of oscillation for the pivot \(P\), then \(P\) is the centre of oscillation for an oscillation about the pivot \(O\). Thus, the oscillations about \(P\) and \(O\) have the same radius of oscillation \(l_r\) and the same angular eigenfrequency \(\omega_0\).

A mathematical pendulum with the radius of oscillation \(l_r\) of a physical pendulum has the same eigenfrequency \(\omega_0\) and – as the eigenfrequency of a mathematical pendulum is independent of its mass – can have either the same mass or the same moment of inertia as a physical pendulum, but usually not both of them. The motion of a physical pendulum is defined by three parameters, e. g. mass \(m\), distance \(r_P\) between the pivot and the centre of mass, and the distance \(l_r\) between the pivot and the centre of oscillation. In a mathematical pendulum, \(r_P\), \(l_r\) and the radius of gyration \(R_g\) degenerate into the pendulum length \(l\). As a mathematical pendulum is defined by two parameters (length and mass), it cannot model every aspect of a physical pendulum defined by three parameters. Therefore, Turner [4] and Le Chevalier [6] suggest a two-mass model to describe the dynamic aspects of a sword.

#### Side Note: Radius of Gyration

The radius of oscillation \(l_r\) is not to be confused with the *radius of gyration* \(R_g\). The radius of gyration \(R_g\) is defined as the distance from the pivot \(P\), where all the mass \(m\) of a physical pendulum would be situated to result in a body with the same mass \(m\) and the same moment of inertia \(I_P\) as the original physical pendulum, that is

\begin{equation}

R_g = \sqrt{\frac{I_P}{m}}

\end{equation}

and thus \(R_g^2 = r_P l_r\). The radius of gyration \(R_g\) characterises the *mass distribution* about a given pivot \(P\). Substituting \(I_P\) with \(m R_g^2\) in Eq. (6), we find

\begin{equation}

\omega_0^2 = \frac{m g r_P}{I_P} = \frac{m g r_P}{m R_g^2} = \frac{g r_P}{R_g^2}

\end{equation}

Thus, the eigenfrequency of a physical pendulum depends not on the mass itself, but on the mass distribution characterised by the radius of gyration \(R_g\). Two homogeneous physical pendulums of different density but identical geometric dimensions and equivalent axes have the same eigenfrequency.

### The Sword and the Impact

A force \(F\) that is exerted on a rigid body in a point different from the centre of mass \(M\) can be substituted with a force \(F’\) of the same value in the centre of mass \(M\) and a moment of force \(\tau’\) about the centre of mass \(M\). If a sword is considered a physical pendulum with the pivot \(P\), the centre of oscillation \(O\), the radius of oscillation \(l_r\) and mass \(m\), then the acceleration \(a_{F’}\) through the force \(F’\) and the acceleration \(a_{\tau’}\) through the moment of force \(\tau’\) cancel out each other in the pivot \(P\), when the force \(F\) is exerted in the centre of oscillation \(O\).

The acceleration \(a_{F’}\) through the force \(F’\) is \(a_{F’} = F’/m\) in \(P\) just as in \(M\). The angular acceleration \(\ddot{\alpha}\) about the centre of mass \(M\) through the moment of force \(\tau’ = F r_O\) is \(\ddot{\alpha} = \tau’/I_M = F r_O/I_M\). The linear acceleration in the pivot \(P\) through \(\ddot{\alpha}\) is \(a_{\tau’} = \ddot{\alpha} r_P = F r_P r_O/I_M\). The moment of inertia \(I_M\) about the centre of mass \(M\) is, according to the parallel-axis theorem (2), expressed as \(I_M = I_P – m r_P^2\). Thus,

\begin{equation}

a_{\tau’} = \frac{F r_P r_O}{I_P – m r_P^2}

\end{equation}

According to Eq. (7), we can substitute \(I_P\) with \(m r_P l_r\) and get the result

\begin{equation}

a_{\tau’} = \frac{F r_P (l_r – r_P)}{m r_P l_r – m r_P^2} = \frac{F}{m} = a_{F’}

\end{equation}

This means that a sword which hits in the centre of oscillation \(O\) will not impart any acceleration (but a moment of force) in the corresponding pivot \(P\) upon impact.

## Measurement

As shown above, a rigid body’s response to external forces can be calculated if the mass \(m\), the centre of mass \(M\) and the moment of inertia \(I\) are known. Without the moment of inertia, a body’s response to non-central forces cannot be calculated. From Eqs. (2) and (7), we know that the moment of inertia can be deduced from the radius of oscillation and vice versa.

Turner [4] and Le Chevalier [6] suggest the so-called “waggle test” as a method to determine the radius of oscillation \(l_r\). For this procedure, the sword is loosely held at a pivot point and carefully shaken in the directions of the edges. The point of the sword that remains without acceleration is the corresponding centre of oscillation. The advantage of this experiment is that it does not require additional equipment. However, despite the authors’ thorough explanations on how to properly conduct the measurement, it should be kept in mind that the accuracy and precision of the corresponding points’ positions depend on the experimenter’s ability to manually shake a sword with no radial component while spotting the point that moves least on a shaking sword.

The moment of inertia is relevant not only for the dynamics of swords, but also for the automotive industry, the assessment of golf clubs and various other applications. While the moment of inertia can be easily calculated from exact density values using CAD, the precision of the input values may be insufficient to produce an accurate value for the moment of inertia. Thus, it is convenient to measure the moment of inertia. Usually, the object is therefore oscillated using a trifilar torsion pendulum or a spiral spring pendulum. From the period of oscillation, the object’s moment of inertia can be calculated. [10]

## Conclusion

Mass and the position of the centre of mass together with the geometric dimensions are not sufficient to define a rigid body’s dynamic properties. Researchers and makers who want to characterise blunt or cutting weapons should also determine the weapon’s moment of inertia for at least the first principal axis or a parallel axis in order to provide meaningful numbers that define the weapon’s motion. The moment of inertia can be measured using a rotating pendulum. [10]

Alternatively, the moment of inertia can be deduced from the radius of oscillation (and vice versa), which can be measured with the “waggle test” with no additional equipment. However, the accuracy of this method heavily relies on the experimenter’s skill and should thus not be preferred.

In the context of sword characterisation, the vibrational node near the point as well as the centre of oscillation are mistakably both called *centre of percussion*. To avoid confusion, it is suggested to use non-ambiguous terms such as *vibrational node* or *centre of oscillation* with respect to a given pivot, respectively.

## Notation

symbol | explanation |
---|---|

\(a\) | acceleration |

\(E_\text{pot}\) | potential energy |

\(E_\text{rot}\) | rotational energy |

\(F, F’\) | force |

\(g\) | gravitational acceleration |

\(I_x\) | moment of inertia with respect to axis \(x\) |

\(l\) | length of a mathematical pendulum |

\(l_r\) | radius of oscillation; effective length [4]; dynamic length [6] |

\(m\) | mass |

\(M\) | centre of mass; centre of gravity; point of balance |

\(O\) | centre of oscillation; centre of percussion |

\(P\) | pivot point of a pendulum |

\(R_g\) | radius of gyration |

\(r_x\) | distance between \(x\) and the centre of mass |

\(t\) | time |

\(\alpha\) | angular displacement |

\(\dot{\alpha}\) | angular velocity |

\(\ddot{\alpha}\) | angular acceleration |

\(\alpha_0\) | amplitude; maximum angular displacement |

\(\pi\) | ratio of a circle’s circumference to its diameter; 3.14159… |

\(\tau, \tau’\) | moment of force; torque |

\(\varphi\) | phase angle |

\(\omega_0\) | angular eigenfrequency |

## References

- [1] T. Laible. Das Schwert – Mythos und Wirklichkeit. 2nd ed. Wieland Verlag, 2008.
- [2] H. Schmidt. Vibration of the blade and how to use it. Hroarr. 2014. URL: http://hroarr.com/vibration-of-the-blade-and-how-to-use-it/.
- [3] H. Vogel. Gerthsen Physik. 20th ed. Springer, 1999.
- [4] G. L. Turner. Dynamics of Hand-Held Impact Weapons. Association for Renaissance Martial Arts. 2002. URL: http://armor.typepad.com/bastardsword/sword_dynamics.pdf.
- [5] M. Denny. “Swordplay: an exercise in rotational dynamics”. In: European Journal of Physics 27.4 (July 2006), pp. 943–950. DOI: 10.1088/0143-0807/27/4/025.
- [6] V. Le Chevalier. A dynamic method for weighing swords. Ensis Sub Caelo. 2010. URL: http://www.subcaelo.net/ensis/weighing/weighing.pdf.
- [7] C. Huygens. Horologium oscillatorium sive de motu pendulorum ad horologia aptato demonstrationes geometricæ. F. Muguet, 1673.
- [8] A. Fehn. Die Fechtkunst mit Stoß- und Hiebwaffen. Carl Rümpler, 1851.
- [9] R. F. Burton. The Book of the Sword. Chatto & Windus, 1884.
- [10] T. Dorfmüller, W. T. Hering and K. Stierstadt. Bergmann/Schaefer Lehrbuch der Experimentalphysik. 11th ed. Vol. 1. Walter de Gruyter, 1998.

Nice article!

It’s true that the waggle test can be hard to get right. It also has limited precision, that can be checked by doing several on the same sword from different points. I estimate that I am able to get to within 5% on the value of the moment of inertia, which is big for scientific purposes but seems enough for our needs, given that we only measure one value and not the whole tensor.

The other methods that you point out are more accurate (even too accurate for the analysis of hand-made objects I think) and I have done some measurements with a pendulum test. But they are tricky to set up and the result is harder to compute, so for the general sword lover they are not appropriate yet. They also do not relate to our perception in the same direct way. I’ll have to detail how I did this somewhere!

Thanks for the citation by the way 🙂

Vincent Le Chevalier

Thank you, Vincent!

The main issue that I have with the waggle test ist that it is so difficult to judge the accuracy of the measurement from the measurement alone. It’s somewhat similar to weighing something by holding it in one’s hand: you can get reliable results from this quick measurement if it is carried out by a skilled experimenter, but at some point you have to evaluate the results by comparing at least some of them with those of scales. While you can check the precision by repeating the waggle test with different pivots, if you add a (reproducible) radial component for instance, you will have reproducibly too large or too small radii of oscillation. Also, hand-and-eye measurements tend to yield the results one expects to see, so the accuracy might be affected by the looks of the sword.

Of course, the waggle test has its applications, just as weighing a sword with bare hands has. But for the guys who visit museums or private collections in order to thoroughly examine original swords, I advocate to rely on a torsion pendulum or a spiral spring pendulum instead. While it is more effort to set up, it is very easy to use once it is done. Also, in my opinion, the calculations are not very difficult; it’s nothing that couldn’t be done with a very simple spreadsheet application. Considering the general sword lover, I guess that any decent swordsman who can find a sword’s centre of mass could learn how to use a readily working pendulum setup within ten minutes, while, on the other hand, I wouldn’t trust myself to achieve your precision of 5 % with the waggle test despite having spent at least an hour of waggling swords and several hours of thinking about it. 😉

I agree with you that, in comparison with length, moment of inertia doesn’t relate overly well with our perception – yet. However, a car’s power and torque also don’t correlate well with human perception, but people get used to the numbers and learn to interpret them by driving cars and correlating them to the values they are given. If, say, Albion or some other makers measured and published their swords’ moments of inertia or if the sword enthusiasts who take data sheets of originals included the moment of inertia, the fencers would eventually learn to understand the values because they would at occasions handle the swords and relate the handling characteristics to the numbers. And even if not – you can still easily calculate the radius of oscillation from the moment of inertia.

Reading and citing your articles is a pleasure!

I fully agree that getting as many measurements of the moment of inertia (or other related quantity) as possible is necessary, but given the current lack of data I don’t think we can afford to be difficult on how people measure 🙂 It’s true that I’ve been doing that for so long that I may not realize how difficult the waggle test is now… It is more reliable than estimating weights by hand, at least for me, as it is not much affected by my level of fatigue or even excitement when touching a new sword for the first time. It is in between just weighing something by hand and putting it on a scale.

I understand that setting up a spiral spring pendulum device isn’t something that the average fencer does for the couple of wasters he owns. But, on the other hand, there are groups who make considerable efforts to examine swords and provide a variety of data. Assume you wanted to do some meta-analysis with these data, like determining the correlation of radius of oscillation and Oakeshott type, how would you know if the guy who executed the waggle test really knew what to do?

Considerung the people who visit collections and assess swords on an almost regular basis, I don’t think that the lack of data on the dynamics (moment of inertia, radius of oscillation, radius of gyration or equivalent data) is due to the difficulty of the measurement, but because they are not yet aware of the possibility to determine a quantity that actually describes how an object reacts to external forces. This is why I wanted to try to have this article on HROARR where it is found not just by the few who already know about the physics of swords.

By the way, just out of curiosity, have you ever checked your accuracy with the waggle test by measuring dynamic lengths of simple objects like uniform thin staves that have moments of inertia that you can actually calculate?

I’ve done more 🙂

I have of course checked on simple staves as soon as I became aware of these notions. It works decently well, but then staves are not balanced like swords and there are distinc challenges with swords.

I have measured the full profile and distal thicknesses of my ATrim type XI sword, then computed the moment of inertia from these, and got figures comparables with my waggle test results.

I have also measured a rapier with a pendulum test, which I know is accurate or at least a lot more consistent than the waggle test. Again, the result that I found wasn’t too far from the waggle test. The difference is that there is a lot more of confidence.

The key to the waggle test, in order to get trustable results, is to do it from several points, because the results should be correlated, and how far they are tells you how imperfect your test is. It’s true that you could have bad luck as you say, and have a systematic bias, but my experience so far is that you get uncertainty but not a big bias.

So, would you recommend training the waggle test with a simple object like a staff where you can easily verify your results with a pocket calculator? Or is it too different from testing swords, so that accurate results with a staff won’t help you with testing a sword, anyway?

I see that you put a lot of brains, dedication and time into the accuracy of the waggle test and its verification, so I would certainly trust your measurements – within the error margins. 😉 I still think, however, that you are an exception and that systematic measurements of sword properties require a setup that does not depend on the individual experimenter as much as the waggle test does.

You wrote earlier that we cannot afford to be difficult and meanwhile I agree with you. But I don’t think that this necessarily precludes a pendulum device. There are occasions for the waggle test and there are occasions for the pendulum. I hope I will find the time to write a follow-up article with step-by-step instructions on how to set up and use a pendulum device and how to relate the results with the moment of inertia and the radius of oscillation/dynamic length.

Oh no testing simple known objects is certainly useful at the start. If the results are not accurate on staves, then on swords they certainly won’t be much better. But there are more problems with swords. Mainly because they have hilts which can get in the way and because their mass is more concentrated, making them relatively more sensitive to experimental errors such as torque in the fingers.

Writing about pendulum tests is also in my projects, sadly I have not found the time yet. I must admit that I’m not necessarily aware of the state of the art with them: although my pendulum device works for my needs it’s a bit artisanal, surely others have come up with different solutions… I’d be interested to read your take on it!

So far, I’ve used a trifilar pendulum, which is quite easy to set up, but a little tricky to use because the axis is not fixed. So I would like to compare the results to those from a spiral spring pendulum, similar to the ones that are used in some physics lab classes at universities. The spiral spring pendulum is more effort to set up, though (I haven’t started yet), but it is more accurate and easier to use because of the fixed axis and the considerably larger Hooke range, so it could almost immediately be operated by anyone with a sword and a stopwatch.

In our club, we did a small waggle test experiment: Eight of our experienced fencers tried to determine the centre of oscillation of a simple stick, holding it at a given point. Despite thorough instructions on how to do the waggle test according to Le Chevalier and the aid ouf a slidable hair tie to mark the right spot on the staff, they all measured a too short distance. Unfortunately, I only told them to not believe what the others measure, but I did not think of separating them for the waggle test to make it a blind trial. So, I don’t say it’s impossible to find the centre of oscillation using the waggle test, but if you haven’t done so yet, really check blindly and double-blindly to make sure that you are actually able to reliably find an unknown centre of oscillation and not just reproduce what you believe is the centre of oscillation.