BONK! - Analysis of Blunt Impact Forces

This article is the first in a very long time (apparently like 2 years), owing mainly to life happening.

Not too long ago, a friend approached me with a question: “How much of a force multiplier is a sword?”, in the context of getting hit by blunt swords during training.

It’d been a while since I did much mechanics, so I set down to answering that question.

24 Sept 2017: Edit to clarify Kinetic Energy/Momentum relationship
11 Nov 2017: Finally fixed the inaccurate statement regarding inelastic collisions, and re-explained a bit. Hopefully this is a bit more accurate.

What Does Force Multiplication Mean to You?

The first question I had to get answered was “what do you mean by force multiplication?”

For the purposes of this analysis, we’ll look at a really simple working definition of the ratio between two different force measures, and call it the Force Multiplication Factor, or FMF:
\[\mathrm{FMF} = \frac{F_{Cut}}{F_{Thrust}}\]

We’ll get more into the details of the above ratio as we go, but this definition is important: right now, we’re just talking about forces. This means that blade geometry is largely irrelevant, and we’re not looking at pressure distributions (which are typically more important to injury correlations, or material failures). But, this definition will let us quickly and easily explore some ideas around how much harder we can hit with a sword than without one.

Laying Out the Problem

In general, there are two particular actions we’re concerned about: thrusts and cuts. Slicing isn’t really an impact problem, but definitely can lead to nasty lacerations due to sliding contact.

Without getting into the, at times controversial, topic of cutting mechanics from a martial arts perspective, we can still talk about some basic ideas of sword cuts and thrusts. Thrusts, from an engineering mechanics perspective, are fairly straightforward - they are largely a translational motion (when something moves entirely without rotation) phenomenon, along some given line of action.

Cuts, on the other hand, tend to be a bit more complex: as they include both rotational and translational motion in three-dimensional space. This is called Generalized Three Dimensional Motion.

So if we were to imagine a hypothetical, but typical, scenario for a thrust we’d end up with something where the point of contact was the point of the sword and that point of contact moved from its initial position to the target through some more or less straight line path. 

Simple schematic for a thrust.

Meanwhile, for a cut, the point of contact is typically a bit further back from the point and that point of contact reaches the target through a path that has a translational and rotational component.

Simple schematic for a cut. Note that the motion is both translational and rotational in nature.

For these attacks, the sword is powered by a force applied at the grip, and at the moment of impact some interesting stuff happens. First, the sword applies a force to the target, and by Newton’s Third Law, the target applies a force on the sword. The force applied by the sword to the target leads to displacement and deformation of the target (if we want to be really fancy, since this all happens due to energy transfer, which means that noise and heat are produced as well). The force applied by the target to the sword acts to decelerate the sword. And the important thing is that this is all happening in a dynamic manner - things are accelerating and decelerating.

Now, it’s possible to perform the calculations we’d like to using the fairly complicated arrangements of actual trajectories using dynamic mechanical simulations… but ain’t no one got time for that.

So, how can we approach this problem?

Simplifying the problem - turning swords into balls.

Well, let’s take a look at some aspects of this problem.

First, we have two things we want to consider:

1. A translational trajectory that results in an impact (thrust)
2. A combined rotational and translational trajectory that results in an impact (cut)

If we assume that the sword and the target are rigid bodies, things get a lot simpler: we won’t need to look at the deformation part of the problem. Similarly, we can assume the energy transfer is perfect with no noise or heat produced. Another simplification we can make is that the target is an immovable object, which lets us basically remove its characteristics from at least our initial calculations.

We can further assume since we care about total forces, not force distributions or stress, that we can consider it an ideal point-contact problem (where the sword and target contact at a point, rather than some finite space like a particular edge geometry).

We can also simplify the problem down into one plane (say the vertical plane), since the mechanics of a vertical cut (from a physics perspective) are the same as one made on an angled line… and it makes the math easier.

The above assumptions still leave us in a situation where the sword has a complex mass distribution, and gravity is still a thing. These are a bit inconvenient, but we can do something about them. First, let’s talk about gravity: it’s effect would be as an additional component of the force the sword applies to the target, and its influence would depend on angle relative to the direction of the gravitational field. So let’s toss it aside for now, since it would vary more as a function of direction of action than because of the presence of a sword or not.

Mass is a bit trickier, but since we’ve removed gravity it gets a lot easier to deal with, since we won’t need to worry as much about that component of the problem.

One final consideration to help simplify the problem to get to an idea of what’s happening is to assume that as a result of the collision, the sword is stopped. This isn’t that absurd of an idea (rarely do swords bounce back after a strike), but again it will let us greatly simplify our considerations. In order to do this, however, we will also have to assume that the sword had some initial force applied to get it moving, but that same force doesn’t continue to act after the actual moment of contact.

So we end up with two systems that can be modeled as follows, where we’re looking at the moment just before and just after the collision occurs:

Moving from an FBD for a thrust to a lumped mass approximation of the same contact event.

Moving from an FBD for the sword strike to a lumped mass approximation of the same contact event.

Which are some pretty fun little diagrams. Let’s now talk about how to analyze these systems

Kinetic Energy, Momentum, and Impulse, Oh My!

You may be familiar with a Newton’s Cradle, the cool little hanging ball thing you probably played with at some point in your life or at least saw on a desk somewhere:

A Newton's Cradle. Image courtesy of Wikimedia Commons.

In broad strokes, that toy works by having the balls you lift strike the others and transferring energy to them. Because the balls are all essentially the same, and are fairly rigid at the levels of force involved, meaning that the energy you put into the balls gets transferred through the chain, moving the same number of balls the same distance on the other side, this is called Conservation of Kinetic Energy - because once you put some energy into the system (lifting a set of balls), it stays in the system through the collisions. Another thing that happens is the balls you move hit at the same speed as the set that moves away after the collision - this is an example of the Conservation of Momentum. To get a bit fancier, momentum (\(m*v\)) is the first derivative of kinetic energy (\(1/2 m*v^2\)), with respect to velocity. In other words: momentum is the change in an object's kinetic energy due to a change in velocity.

This type of impact behavior is called an elastic collision, and is governed by both the conservation of momentum and kinetic energy. However, in our simplifications above, we made it so that the target will not move after the collision. Therefore, the total kinetic energy in the system before and after the collision is different because the target doesn’t move and the sword doesn’t bounce back. Also, as a result of our assumptions we’ve also ended up with a system in which momentum is not conserved. So it doesn't sound like an example of inelastic collision mechanics either… Oh no, Sparky what have you done!?

Well, what I did was sort of gloss over some nuances that are worth discussing. What we have in our model is a scenario in which a sword collides with an initially stationary target, and we said it stops, but let's step back from that assumption for a moment. Let's just say the two objects stick together after the collision and move with speed \(v_f\). So, let's begin with the idea that this is an elastic collision, because we have no reason to believe otherwise. In an elastic collision, momentum is conserved, so we end up with a momentum balance before and after that looks like the following:
\[m_s v_s  = (m_s + m_t) v^f \]
Which we can solve for the final speed with some rearrangement:
\[v^f = \frac{m_s}{(m_s + m_t)} v_s\]

And again, for an elastic collision, there's a conservation of kinetic energy. So, the kinetic energy balance works out to be:
\[ \frac{1}{2} m_s v_s^2  = \frac{1}{2}(m_s + m_t) (v^f)^2 \]

Let's plug in our solution for the final speed above, and do some rearranging:
\[ \frac{1}{2} m_s v_s^2  = \frac{1}{2}(m_s + m_t) (\frac{m_s}{(m_s + m_t)} v_s)^2 \]
\[ \frac{1}{2} m_s v_s^2  = \frac{1}{2}\frac{m_s^2}{(m_s + m_t)} v_s^2 \]

... but Sparky! Yep, that's right: that doesn't actually work unless \(m_t = 0\), though it would also work if \(v_s = 0\), but that's not terribly interesting.  Since \(m_t > 0\),  it means that kinetic energy can't be conserved in this collision. Thus, it's an inelastic collision!

But I talked about our sword-target interaction earlier in terms of a rigid, unmoving target. That rigid unmoving target is essentially an infinite mass, which is why it doesn't move. For example, imagine a large steel pole driven deep into the ground, making it so that the strike is trying to move the earth. Good luck with that. So what does this mean to the above inelastic collision, since the end momentum would be zero and the initial momentum is non-zero?

Well, let's look at that equation for our final velocity, based on the momentum balance, in the limit that the target mass, \(m_t\) goes to infinity, assuming the magnitude of \(m_s v_s\) is much smaller than \(m_t\):
\[ v^f = \lim_{m_t \rightarrow \infty} \frac{m_s}{(m_s + m_t)} v_s = \frac{m_s}{\infty} v_s = 0\]

So, we actually *do* conserve momentum, somewhat paradoxically, in our model sword-target interaction where we assumed the final speed was zero, but only if we assume that our target is infinitely massive. In reality, there'd be target motion - but this assumption will make our lives a bit easier.

Since we’re in the land of Newtonian Mechanics,  the Impulse-Momentum Theorem (though we could arrive at the same conclusions with the work-energy principle) says that the change in momentum of a system is a force which acts on our sword over a period of time, called the impulse (\(J\)). Mathematically, that means that our collision looks something like:
\[ J = \int^{t_c}_c{F_c} \mathrm{d}t = \Delta p = m_s v_s\]

Where \(m_s\) and \(v_s\) are the (effective) mass and speed of the sword just before impact, and \(F_c\) is the force acting on the sword during the impact and \(t_c\) is the collision time. If we assume that some average force acts over the collision time we can simplify this to:
\[ m_s v_s - \bar{F_c} \delta t_c = 0\]

or

\[ m_s v_s = \bar{F_c} \delta t_c\]

Which helps us start getting closer to the average force we are interested in, so we can compare it to something else. Before we go on, however, I want to point out a little trick I used here: I’ve made a lumped mass approximation, which assumes that it’s possible to calculate an effective point mass that will mimic the behavior of the actual full object. In our case, since we’re looking at rigid body contact in a very simple way - this isn’t too crazy of an assumption. And we’ll see shortly that the assumption will basically cancel out because we’re going to calculate a ratio of quantities.

For a thrust, with our linear motion assumption, it’s pretty easy to write down the ‘thrust’ or purely translational contact equations:
\[ m_s v_s = \bar{F_c}’ \delta t_c\]

Where \(\bar{F_c}’\) is the average reaction force resulting from the purely linear contact of an object with the same effective mass as our sword, moving at speed \(v_s\). This could be a thrust, or it could be a punch with a gauntlet, giant ball, whatever.

For a cut, we have a combined translational and rotational motion that we need to consider. We can decompose the speed of our contact point into two parts: the translational speed of the pivot point and the rotational speed about that pivot point. If our contact point c is a distance l from the pivot, we end up with an equation for the contact point speed that looks like this:
\[ v_s =  v_p + \omega_s * l\]

Where I’ve denoted the speed of the pivot as \(v_p\), and the rotational speed of the sword just prior to contact is \(\omega_s\). That means we can now write the equation of motion for the cut case as follows:
\[ m_s v_s = m_s (v_p + \omega_s * l) = \bar{F_c} \delta t_c\]

Now, we wanted to find a force multiplication factor, which we defined earlier to be the ratio of the force at impact from a cut to that due to a thrust:
\[\mathrm{FMF} = \frac{F_{Cut}}{F_{Thrust}}\]
\[\mathrm{FMF} = \frac{m_s (v_p + \omega_s * l)}{m_s v_s} = \frac{\bar{F_c}\delta t_c}{\bar{F_c}’ \delta t_c}\]

If we assume that the speed of the contact point in the thrust/linear motion case is the same as the pivot point, we can simplify this:
\[\frac{m_s (v_p + \omega_s * l)}{m_s v_s} = \frac{m_s (v_p + \omega_s * l)}{m_s v_p}= \frac{\bar{F_c}\delta t_c}{\bar{F_c}’ \delta t_c}\]

With some re-arranging we arrive at:
\[\frac{v_p + \omega_s * l}{v_p}= \frac{\bar{F_c}}{\bar{F_c}’}\]

Or

\[\mathrm{FMF}  = 1 + \frac{\omega_s * l}{v_p}\]

So what does this equation mean?

It means that if we strike with a sword cut, the additional speed at the contact point due to the rotational motion of the sword results in an increase in impact force relative to an attack which does not have the rotational component but is traveling at the same speed as the pivot point of the cut and has the same effective mass. This also states that if there is no rotational component (\(\omega_s = 0\)), there is no force multiplication - which makes sense since that means we have purely translational motion and the top and bottom of the equations are the same. Not surprisingly, this also says that the further away from the pivot the point of contact is, the greater the difference in force because of the speed (which ties into some ideas around particular cutting mechanics which seek to increase this distance by pivoting outside the length of the sword). The converse of that is that by decreasing the speed at the moment of contact, you can reduce the force assuming nothing else changes… more on that later.

There’s some gotchas to this though:

1. Force isn’t the only thing that matters: force distribution matters when we want to talk of the response of materials, or even our bodies.
2. This is assuming there’s no additional forces acting during the contact, like gravity - which can increase the impact force or decrease them depending on the specific scenario.
3. Our ability to move a sword in a linear action like a thrust isn’t exactly the same as cutting - so saying the speeds are the same is a bit of a simplification that could change quite a bit in reality.
4. This ignored energy being absorbed by the sword at the point of contact (essentially assuming the point of contact was the center of percussion).

For the sake of argument, let’s actually use some numbers from my previous posts and try to get some sort of numbers we can more easily think about.

Putting The Results in Perspective

First, let’s actually fill in some numbers so that we can get some appreciation for how much force multiplication we might expect a cut to deliver.

Let’s assume our pivot point is just below the cross, and the contact point is 1 meter away from that point, so l = 1 m. Let’s assume that the pivot point is traveling at about the same speed as some of the thrusts I did in my thrust measurement post, so \(v_p) = 2 m/s, and let’s assume that you step 1 m at that speed (so 0.5 s time), and the sword traces an arc of 90 degrees in that time, resulting in a rotational speed of the sword of about 3 rad/s. So that means that our force multiplication factor is (1 + (3/2)), or about 2.5 - so my hypothetical cut based on pretty reasonable values is going to hit with 2.5 times more force than the hypothetical thrust - which is a pretty significant increase. If my hypothetical thrust hits with 400 N of force (approximately what I measured in an earlier post), then the hypothetical cut will hit with a force of 1000 N. To put 1000 N into perspective, if that force is applied to an edge 2 mm wide by 5 cm long, the resulting average pressure is 10 MPa: about the ultimate compressive strength of ice at about -15 C, and about 1/4 of the maximum recommended axial loading for neck of a 50th percentile male for crash safety from the NHTSA (see one of my earlier posts). So, in short: you can hit really hard with a cut.

Now, let’s go back for a moment and look at the governing equation and talk about some things:
\[ m_s v_s = m_s (v_p + \omega_s * l) = \bar{F_c} \delta t_c\]

We’ve already talked about how the speed of the point of contact not surprisingly affects the impact force. But let’s look at that contact time, \(\delta t_c\). if we increase that time, even if everything else on the left side of the equation stays the same, the contact force goes down. This is the principle by which things like helmet padding, car air bags and even moving with a blow received function: by increasing the contact time you can decrease the average force imparted during the contact. Since this is something the target can affect, that’s why it gets used so much in impact protection methods.

However, i think the left hand side of the equation is important to point out too - as that’s where good control can lead to safer practice. Slowing down a blow just before making contact can significantly reduce the forces.