And after the last spattering of materials science related posts, I promised some more classical mechanics. So here we go!

I won't say which one I am. |

**Update, 4 March 2014:**fixed a typo in the force resolution. It should have been 6 N, not 0.6 N.

**I revisited the initial image analysis and found the approximate speed at time of impact to be about 2 m/s (~6.6 ft/s or ~4.5 mph), though this suffers from the same limitations as the other conclusions from the same analysis.**

**Update, 14 Feb 2014:****Update, 9 Feb 2014:**I went back and repeated the thrust force measurement with my Lichtenauer (same weapon as the test shown), and my Tinker-Hanwei Arming Sword and Longsword. The results for the Lichtenauer were consistent with those shown here. Woot for repeatability. The data from the Tinker-Hanwei swords exhibited similar characteristics as well, but at lower load ranges as expected based on the buckling load work I discuss here. Yay for predictive science! I am still working through the data, but felt this was worth an update.

When a fencer performs a thrust with a flexible trainer, against a rigid target or another fencer, at some point the blade begins to flex. The amount of force necessary and the extent of the bend depend on aspects such as:

- The incoming angle of the thrust relative to its target (the incident angle)
- The bending stiffness of the trainer
- The energy and force of the thrust
- The reaction of the fencer (collapsing arms or keeping them stiff, etc.)
- The reaction of the target (rigid, flexible, etc.)

- The target is rigid and does not move, but has sufficient friction to keep the point from sliding during impact. Rotation of the point is still possible.
- The incident angle of the thrust is perpendicular to the target (i.e., a perfectly horizontal thrust delivered to a vertical wall).
- For the moment, we'll ignore all aspects of the fencer's body save for the fact that there is some applied load on the sword and that the sword is constrained by the fencer's grip.
- The applied force on the sword grip is constant before and after impact.
- The sword can be modeled by an effective constant cross-section and Young's modulus beam as considered in the effective bending stiffness post.

Schematic of thrust simplification. |

*perfectly elastic collision*because no kinetic energy is lost due to deformation, etc. As a result, the sword would bounce off the target in much the same way as a ball being bounced off the ground.

The result of the collision would change, of course, if it was possible for the rigid sword to penetrate or deform the wall, as it would no longer be an elastic collision. Similarly if we allow the sword to deform but keep the wall rigid, we are no longer dealing with an elastic collision because a portion of the initial kinetic energy would go into deforming the sword. Since this last case is the far more useful one for our purpose, let's take a deeper look at it.

### Elastically Deformable Sword

At the moment of impact, the reaction force \(F_{reaction}\) acts on the x-axis of the sword causing an axial compression wave to travel down the length of the sword. This effect is much like if we were to push a spring into a wall, compressing it. If the impact happens at a speed approaching the speed of sound, \(c\), in the material (for steel alloys, \(c\approx6000\) m/s), localized deformation and some other interesting effects can occur.Luckily for us, our sword probably isn't moving anywhere near that fast (even if we were able to accelerate the sword to 100 mph during our thrust that's only 44 m/s), so the axial deformation and state of stress along the length of the blade can be assumed to be uniform.

If our sword is constrained to only deform along the x-axis, then we'd just be looking at the axial compression of a bar. Let's further assume we're not pressing hard enough to cause plastic deformation; that would require an applied stress of over 200 MPa for most steel alloys (some require over 1000 MPa - steel alloys have a large spread of yield strengths), for example. Given these assumptions, how much deformation and energy loss to that deformation should we expect?

#### Quasi-static Axial Deformation of Constrained Sword

First, let's assume that the sword is bring pressed into the rigid wall and neglect any impact and dynamic aspects. In that case, \(F_{reaction}\) is equal to \(F_{thrust}\) and the axial deformation of the sword, \(\Delta\), is:\[\Delta = \int_{0}^{l} \frac{xF_{thrust}}{A(x)E(x)} dx \]

Where \(l\) is the length of the sword, and \(A(x)\) and \(E(x)\) are the cross-sectional area and Young's modulus at a distance \(x\) along the blade. Using the assumption that an equivalent uniform cross-section can be constructed and that the Young's modulus is constant along the blade, then the above simplifies to:

\[\Delta =\frac{F_{thrust}\;l}{\bar{A}\bar{E}} \]

Where \(\bar{A}\) is the equivalent uniform cross-section and \(\bar{E}\) the constant Young's modulus. Note that this is only a deformation along the length of the blade, no flexing of the blade occurs. However, the blade doesn't simply shrink in length: there is also a corresponding transverse deformation due to the Poisson Effect, which we briefly discussed before. In order to determine the size of the transverse deformation, we need to look at the Poisson's Ratio, \(\nu\).:

\[\nu =-\frac{\epsilon_{trans}}{\epsilon_{axial}}\]

Where \(\epsilon_{axial}\) is the axial strain, the axial deformation per original unit length, and \(\epsilon_{trans}\) is the analogous strain in the transverse direction (e.g. change in radius divided by initial radius for a circular section). The negative sign is because an axial compression results in a transverse expansion (the usual sign convention has compression as a negative strain, while tension is positive). Like the Young's modulus, the Poisson's ratio is generally available in tables of mechanical properties for common materials. For steels undergoing elastic deformation, the Poisson's ratio is around 0.3. Re-arranging the last equation to get the relationship for the transverse deformation:

\[\epsilon_{trans} = -\nu \epsilon_{axial} \approx -0.3 \epsilon_{axial} \;\;\mathrm{(For\:steels)}\]

But how much energy is required to cause the axial deformation? In an inelastic impact, this would be the energy that gets 'lost': in our case, it can also be thought of as energy that doesn't get transmitted to the target. In classical mechanics, the energy required to cause a given deformation is called the strain energy (think: energy stored by compressing a spring). For our model sword, the equation for this quantity is as follows:

\[U_{static} = \frac{F_{thrust}{^2}\;l}{2\bar{A}\bar{E}} \]

Where \(U\) is the strain energy. Physically, this is the applied force on the blade times the distance it operates over, in other words: the work done on the blade by the applied force. Since this is a static system where only the sword is deformable, all of the energy put into the system goes into deforming the blade.

Before moving on, let's put some numbers into these equations, to get an idea of how things would look in this scenario.

#### Tossin' in Some Numbers!

Let's assume we have a sword with a uniform rectangular profile with \(l_{y}\)= 20 mm and \(l_z\)= 4 mm, and overall length \(l\)=0.95 m, made of AISI 6150 steel (as-produced). Further, let's assume the sword is pressed into our rigid wall with a compressive force of 1000 N. The cross-sectional area would be \(8\mathrm{x}10^{-5}\;\mathrm{m}^2\), the Young's Modulus is about 200 GPa and the Poisson's ratio is about 0.28. Plugging these values into the above equations, we arrive at:\[\Delta_{axial} = \frac{F_{thrust}\;l}{AE} = \frac{(-1000~\textrm{N}) \; (0.95~\mathrm{m})}{(8\mathrm{x}10^{-5}\mathrm{m^2}) \; (2\mathrm{x}10^{11}\mathrm{N/m^2})} \approx -5.9\mathrm{x}10^{-5}\;\mathrm{m}\]

\[U_{static} = \frac{F_{thrust}{^2}\;l}{2AE} = \frac{(-1000~\textrm{N})^2 \; (0.95~\mathrm{m})}{2\;(8\mathrm{x}10^{-5}\mathrm{m^2}) \; (2\mathrm{x}10^{11}\mathrm{N/m^2})} \approx 0.3\;\mathrm{J}\]

\[\epsilon_{axial} = \frac{\Delta_{axial}}{l} = \frac{-5.9\mathrm{x}10^{-5}\;\mathrm{m} }{0.95~\mathrm{m}} \approx -6.2\mathrm{x}10^{-5}\]

\[\epsilon_{y} = \epsilon_{x} = -\nu\epsilon_{axial} = -0.28 (-6.2\mathrm{x}10^{-5}) \approx 1.7\mathrm{x}10^{-5}\]

\[\Delta_{y} = \epsilon_{y} l_y = 1.7\mathrm{x}10^{-5} (0.02\;\mathrm{m}) \approx 3.4\mathrm{x}10^{-7}\;\mathrm{m}\]

\[\Delta_{z} = \epsilon_{z} l_z = 1.7\mathrm{x}10^{-5} (0.004\;\mathrm{m}) \approx 6.8\mathrm{x}10^{-8}\;\mathrm{m}\]

For our sword, pressed into the rigid wall with a stout force (1000 N is approximately 225 lbs-force), we should expect a very small axial compression: 0.059 mm (an object this size would just be visible to the human eye), and a correspondingly small amount of energy being stored in the deformed sword. Now, experience tells us that if we lean onto our swords with all our body weight, they deform much more than what we calculated above. But more importantly, they flex significantly in the transverse direction. So how does this occur?

### Column Buckling and You!

Anyone who's performed a thrust at a target with a flexible sword trainer knows that they tend to flex perpendicular to the plane of the flat of the blade, even if you thrust straight at a target. In principle this phenomenon is no different than standing a plastic ruler on end, pressing down on it and seeing it bow under the load, possibly even breaking when loaded too much. This bowing of a long thin member due to an axially applied compressive load is referred to as simple buckling or just buckling (there are several types of buckling). This is a phenomenon of great importance to a number of engineering applications, and has been studied for over two centuries.Consider a sword: a long, thin object whose cross-sectional dimension in at least one direction is much smaller than its length and is placed under a compressive axial load (during a thrust, at least). In engineering terms, this type of object is referred to as a column (in the architectural sense). When a perfectly straight column is gradually loaded with increasing force, there comes a point when axial deformation no longer occurs and instead the column buckles and may even collapse under the load.

The point where the behavior changes is referred to as a

*bifurcation point*, and at this point a slight disturbance of the system will result in in the lateral deflection of the column instead of continued axial deformation. Because of this change in behavior, buckling is referred to as an

*instability*meaning that there exist multiple deformation states for a given set of conditions. For our purpose, a good question to ask is: for the static axial loading of a flexible trainer (such as our example in the previous section), what is the load at which the sword is on the verge of buckling?

First, we need to consider what sort of boundary conditions we may have for our sword model. If we assume that the point is free to rotate at the contact point, we have the following basic possibilities, which I'll talk about in turn:

#### Pinned Ends

Let's consider the case in which the point (\(x=l\)) of the sword does not penetrate the wall, there is sufficient friction at the surface that it cannot move parallel to the wall (\(z(l) = 0\)), but the point can freely rotate in the x-z plane. Similarly, let's assume that the hilt cannot move in the plane parallel to the wall so that \(z(0) = 0\), but like the point it is free to rotate. This type of boundary condition is called a*pinned support*('A' in the above figure), because it reacts as though you had drilled a hole through the member and pinned it in place with a round pin.

The differential equation that relates the lateral deformation (\(z(x)\)) of our model sword to the internal bending moment and applied force \(P\) is as follows:

\[EI \frac{d^2z}{dx^2} = -Pz\]

Re-arranging, we obtain:

\[\frac{d^2z}{dx^2} + \frac{P}{EI}z = 0 \]

Which has the following general solution:

\[z(x) = C_{1}\;\sin\left(\sqrt{\frac{P}{EI}}\;x\right) + C_{2}\;\cos\left(\sqrt{\frac{P}{EI}}\;x\right)\]

Since our boundary conditions have \(z(0)=0\), \(C_2 = 0\), leaving us with the following to satisfy:

\[z(l) = C_{1}\;\sin\left(\sqrt{\frac{P}{EI}}\;l\right) = 0\]

The above can be satisfied if \(C_1 = 0\) leaving us with the solution of \(z(x) = 0\), meaning there is no lateral deformation (no buckling). This is called the

*trivial solution*, because it isn't terribly enlightening. However, another solution exists as well:

\[\sin\left(\sqrt{\frac{P}{EI}}\;l\right) = 0\]

Which is possible if the following is true:

\[\sqrt{\frac{P}{EI}}= \frac{n\pi}{l}\]

Where \(n=1,2,…\), since \(n=0\) would also result in the trivial solution. Manipulating this to determine the buckling load, \(P\), we have:

\[P= \frac{n^2\pi^2 EI}{l^2}\]

The minimum value of \(P\) occurs when \(n=1\) (also called the

*first buckling mode*), giving what is referred to as the critical buckling load, \(P_{cr}\), for buckling of a simple column with pinned ends:

\[P_{cr}^{p-p}= \frac{\pi^2 EI}{l^2}\]

\(P_{cr}\) is the maximum load that can be carried before reaching the bifurcation point, and therefore the onset of buckling. Going back to our sword example, \(P_{cr}\) is the load at which we would expect to see it just begin to flex. Note that determining this load from the above equation only requires knowledge of the effective bending stiffness \(EI\) and the length of the blade (both of which we obtained earlier for a variety of trainers).

#### Pinned/Guided Ends

Now imagine the sword point is still allowed to rotate but not translate (pinned support) and the base of the blade is not allowed to rotate, but is allowed to translate in the z-direction (*a guided*support), as in 'B' in the schematic above. This would be the case if we pushed our point against the wall, the hilt moved to the side as the blade buckled but the point stayed in place. To make the math a bit simpler, I'll choose the point to be \(x=0\), while the base of the blade is \(x=l\). The boundary conditions for this scenario are then \(z(0)= 0\), \(z(l) = z_{max}\), and \(dz/dx|_{x=l} = 0\). The governing differential equation is:

\[EI \frac{d^2z}{dx^2} = -Pz\]

The general solution to this differential equation is as follows:

\[z(x) = C_{1}\;\sin\left(\sqrt{\frac{P}{EI}}\;x\right) + C_{2}\;\cos\left(\sqrt{\frac{P}{EI}}\;x\right)\]

Because \(z(0) = 0\), \(C_2 = 0\), and the boundary conditions at the hilt (\(x=l\)) result in the following:

\[z(x) = z_{max}\sin\left(\sqrt{\frac{P}{EI}}\;x\right)\]

where the following holds for \(n = 1,2,3…\):

\[\sqrt{\frac{P}{EI}}= \frac{(2n-1)\pi}{2l}\]

Again manipulating terms to get the critical load, we arrive at:

\[P_{cr}^{p-g}= \frac{\pi^2 EI}{4l^2}\]

We can see an interesting thing here: the equation for the critical load for pinned/guided ends takes the same form as the one for the pinned ends, but with an effective length of \(2l\). In other words, our pinned/guided column is mathematically one half of a longer pinned/pinned column. It also means that a pinned/guided column can only support a quarter of the load of a pinned/pinned column of the same length before buckling.

#### Fixed/Pinned Ends

The final case I'll present is the case in which the base of the column is fixed while the top is pinned, labeled 'C' in the figure above. This would be the case if the sword's point rotated but did not translate and there was no translation or rotation of the hilt. Without going through the equations again, the critical buckling load for such a column is as follows:\[P_{cr}^{c-p}= \frac{2\pi^2 EI}{l^2}\]

In other words, this column is capable of supporting

*twice*the load of a pinned/pinned column of the same length before buckling.

#### Critical Buckling Loads for Select Trainers

Because of the work done for a previous post, I can readily put some numbers into the above buckling loads. By doing this, we can take a look at the buckling loads for the different end conditions. The calculated values for each weapon, based on the pinned/guided, pinned/pinned, and clamped/pinned end conditions in the table below:Critical buckling loads for a variety of HEMA trainers. Yep, I got lazy and did the table in LaTeX. |

To provide a bit of context, I also calculated the critical load range for FIE homologated blades, based on the allowed stiffness range:

Critical buckling loads for FIE weapons, based on allowed stiffness range. |

So how can we apply the values in the above tables to a flexible trainer? First, recall that the critical buckling loads were calculated based on a straightforward static scenario, as was the effective bending stiffness. Like the effective bending stiffness, the static buckling load provides us a simple 'ballpark' measure for understanding how much force is required before the blade begins to flex. We can take the clamped/pinned case to be an upper bound on the onset of blade flex and think of the pinned/guided case as a lower bound. With this range, we can assume that force required to bend a trainer during a straight thrust lies somewhere in that range. While it is possible for us to propose scenarios in which the load to cause blade flex is less than these values (such as a high incident angle thrust), or in some short-duration dynamic cases higher, the static buckling load values provide us an estimate with which we can perform some simple equipment analysis. Further, we've been able to easily examine the role of the hilt conditions and the measured stiffness of the trainer.

There is a drawback to the analysis we just performed, however: while we can estimate the buckling load, we cannot calculate the exact deflection or the strain energy of the blade as it buckles, nor the reaction forces at the ends. One method to obtain that information is a so-called 'numerical approach', in which we use an iterative scheme to determine the displacement and forces at discrete points along the blade. This sort of approach can also be used to explore more realistic boundary conditions (flexible target and end constraints, for example). Another method we could use is experimental measurement: a rigorous experimental setup for buckling studies is somewhat complex, involving a number of force transducers, a load cell for applying a known load and displacement sensors or detailed image analysis of high-speed imagery. A basic image-processing method, however, can still be instructive and can provide some limited validation of our analytical work. So let's do that very thing.

### Basic Image Analysis of a Thrust

The most basic form of image analysis of a dynamic event is to look at individual frames of a video of the event. From a sequence of images during key phases of the event, we can extract some basic idea of what occurred. This approach is not without caveats, however: the frame rate and resolution of the capture as well as actual method of capture (e.g. global or rolling shutter, digital or film, etc.) can greatly influence the conclusions that can be drawn. Further, any changes in the distance and angle between the object being captured and the lens of the camera can make frame-to-frame comparisons difficult. But if the distance effects are minimized, and the framerate and resolution are at least moderately high, a fairly useful qualitative description of the event can be gained.Schematic of the über high-tech image capture setup. |

I carried out 3 different sets of thrusts and static pushes of the sword against the target. For each set, I started the camera, took up my starting position and performed the actions. Once the actions were completed I stopped the video. The two most clear cases I chose to discuss here, as they were representative of the thrusts performed and had the best image quality. One case is a 'natural' thrust, in which I use a centerline guard with the pommel low (i.e. Posta Breve) and extend into an extended guard (i.e. Posta Longa) with a single step of the lead foot. This position results in a downward angle from the point to the cross. In the second case, I begin from a higher position, such that the extension has a minimal downward angle (a straight thrust like we considered in the earlier sections). The key frames of these two thrusts can be seen in the images below:

The agreement between the overlaid analytical displacements is quite good, once we take the various assumptions into account. We solved the buckling equations based on an equivalent uniform cross section with a constant bending stiffness, but we know the profile of the Albion Lichtenauer changes with position and that the bending stiffness will vary as well (due to changes in the cross-sectional moment, the elastic modulus probably doesn't vary much). There are some additional measurement errors due to the angle of the blade in the natural thrust, as the hilt is lower than the point. Overall, the main difference between the overlaid curves and the blade is that the deflection of the blade is weighted more toward the regions with a narrower blade profile while the overlaid curves are symmetric about the maximum deflection. Secondly, we assumed much simpler end conditions than we knew would exist in reality: the simple conditions ignore aspects like the compliance of the hand. Finally, our calculations were based on static assumptions while the reality appears to be a progressive buckling event. The second frame above for both Thrust 1 and 2 appears similar to the fixed/pinned form shown earlier. This effect may be an artifact of the image capture method as it progresses to something more like the final form by the third frame. A set of accelerometers and a higher frame rate video camera would be needed to clarify this part of the process. The quality of agreement gives us fairly good confidence that the pinned/pinned buckling model is reasonable as a first approximation for this weapon.

But what about the forces during a thrust?

### Preliminary Force Measurements

To get some more insight into what occurs during a thrust, I set up a contact pad wired with a set of load cells and, well, hit it with a sword. The goal was to get a ballpark idea of thrust forces (to validate the above work), and to test the equipment, which will see a few different uses. I'll write a post specifically about the equipment (a follow on to an older post), but the short summary is as follows:- 4 load cells, approximately 2kN maximum force with roughly 6 N resolution. The load cells only measure the force normal to the contact plate.
- Data acquisition system with approximately 15 ms time resolution.
- Camera setup similar to that used in the simple image analysis described above.
- Contact pad was a 3/4" thick pine board with 16 ga steel epoxied to either side to provide a rigid reinforcement. To provide friction, sole leather was then taped onto the front face, to prevent the point from easily glancing.

Behold my works, ye mighty and well-funded, and despair. |

Data collected during preliminary run. Colored blocks indicate thrust type: orange are straight thrusts, purple are natural thrusts, green are static pushes and blue are mixed. |

Data from first cycle of thrusts, used as a representative set from the preliminary measurements. |

Now that I've run through some of the reasons this data is far from the last word on the topic of thrust forces, let's use these measurements to take another look at the analytically determined buckling loads. From the analytical work, we have 3 critical buckling loads for the Albion Lichtenauer:

- \(P_{cr}^{C-P} = 1023~\mathrm{N}\), for the Clamped/Pinned constraints
- \(P_{cr}^{P-P} = 511.4~\mathrm{N}\), for the Pinned/Pinned constraints
- \(P_{cr}^{G-P} = 127.9~\mathrm{N}\), for the Guided/Pinned constraints

I found peaks that are above both the pinned/pinned and guided/pinned critical loads, though the majority are between these two loads. This is likely due to the following things:

- The real end conditions are likely somewhere between a pinned/pinned and guided/pinned support
- The critical load is the onset of transverse deformation, but it is possible that higher loads can be momentarily supported during dynamic loading
- The variations in motion during the thrust may lead to transverse loading on the blade, lowering the axial load required to buckle the blade.

To get an idea of how human force generation may vary, a paper which discusses measurements taken of boxers provides some useful context. The paper is Head Impact Biomechanics in Sport by D.C. Viano, and in one section they discuss their measurements of boxers striking a specific type of test dummy. They found a range of peak forces of 1990-4741 N and an average force range of 588-1164 N, with part of the wide variation able to be correlated to weight class. I would not expect to find any more consistency in sword thrusts, though I am not sure if we'd see the same correlation to weight. It is certainly something that would be interesting to study.

#### Thoughts on future work

Since this topic could provide a great deal of useful information to the community, like the effective bending stiffness, I have thought about ways to extend the work I laid out here. The main constraints are: money, time and subjects for the thrust experiment. From a safety equipment perspective, getting a hold of the same sort of anthropomorphic dummy that they use in crash testing would be great, but prohibitively expensive. Same goes for the sort of equipment necessary to get very fine time and force resolution.My current thoughts, however, are:

- Same equipment, lots more thrusts. Use the same process with another weapon and subject(s). Look for trends.
- New DAQ, same load cells: may improve time resolution.
- Set up way to better time-align DAQ and video capture.
- Reinforce the support structure for the contact pad (so as not to wreck my wall), and make the height adjustable.

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