Mechanics of Breaking

Non-Idealized Holding Conditions

The above analysis describes ideal holding conditions. In reality, even the best holders will let the board move backward slightly. In addition, 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. Here the 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. 

E=1/2mav2 – (mav )L*pi/2