Page 1 - Page 2 - Page 3

Non-Idealized Holding Conditions

The above analysis describes ideal holding conditions. In reality, even the best holders will let the board move backward slightly. Also, speed breaks are performed with a board held only at one end. How does this affect the required breaking energy?

One-Side Hold

In a one-side hold, the board is modeled as a cantilever beam since the board holder's affect on the break is negligible. The martial artist must deliver enough energy to cause any deflections of the board in addition to the energy required to break the board. This will be equal to the displacement of the beam multiplied by the force applied to cause that displacement. With a point load P applied in the middle of the board, the maximum deflection is given by:

D=5PL3/(24EI)

Assuming the deflection is small and linear, multiplying this by the force that caused this displacement will give a reasonable approximation to the energy ‘wasted’ . Since E= Force times Distance, now the energy delivered to a board during a particular strike is:

E=1/2mav2 - 5P2L3/(24EI)

But how do we find P? For this analysis we consider the impulse applied to the board. Since its original momentum is zero, we have the momentum of the strike (mav) equal to the force of the strike multiplied by the time of application. Herethe time of application is taken to be 0.25 seconds.  This leads to the final equation for a one sided hold:

E=1/2mav2 – 5(mav )L3/(6EI)

For the given parameters over a normal range of strikes, less than one joule is lost to move the board, which moves only 7 mm. However, a perfectly fixed cantilever beam is a unrealistic assumption. A loosely held board sweeps out a quarter circle with a radius of its length, so the average board particle moves pi*L/8 meters. A better equation is graphed in Figure 2.

E=1/2mav2 – (mav )L*pi/2

Two-Side Hold

In a two-side hold, in which the board holders ability to hold the board stationary may affect the break, can be modeled in several different ways. The amount that they allow the board to move backward could be expressed as a fixed distance, or a fixed or percentage amount of energy or force absorbed.  A percentage of the force absorbed is the best option, because the holders are applying a force. This will let the distance that the board moves be a function of the strike.  With F as a fraction of the force absorbed by the holders, the board now absorbs the net force Fnet=(1-F) (mav )/0.25 s, which gives an acceleration to the board of Fnet/mb. Since the board has this acceleration for the same 0.25 seconds and W=Fnet*1/2ab*t2, the energy imparted to the board during a strike is:

E=1/2mav2 – ((1-F) (mav ))2 /2 mb

The results are plotted in Figure 3 (F=0.85) and Figure 4 (F=0.80). In general, for F values greater than 0.9 the difference is small. For 0.78<F<0.90, the difference becomes significant and only a very powerful strike can fracture the board. Below this range the model breaks down as the board undergoes an acceleration that causes it to move away from the strike faster than the strike itself, so the board cannot be broken regardless of the strike. It is interesting to note that this model can also handle F values slightly greater than one, such as when the holders push the board into the strike, such as when an adult pushes the board against a child’s strike to make it easier for them to break.

Figure 1. The energy delivered to a board is a function of striking speed and apparent mass. The horizontal plane represents the energy needed to break a standard adult board.

Page 1 - Page 2 - Page 3