Page 2 of 4

To quantify the potential of a taekwondo strike to do damage, we need to evaluate how much deformation energy is delivered by the blow. The amount of energy that a leg bone may absorb before breaking is represented by:

E_{max} = 1/(2Y) Als_{b}^{2}

which worked out to about 350 joules. This result is proportional to the cross-sectional area of the bone. If we consider a smaller bone, like an arm bone or a rib, then a proportionately smaller amount of energy will be required to break the bone. An arm bone has a diameter of about half that of a leg bone, so the energy it may absorb will be a quarter as much as that of a leg bone, or about 88 joules. Keep this in mind as we calculate the amount of energy delivered in a taekwondo punch.

Lets consider two people: the puncher and the opponent. Assuming that the collision between the puncher's fist and the opponent's internal organs is completely inelastic, conservation of momentum tells us that:

M_{1}V_{1} = (M_{1} + M_{2})V'

Where M1 is the mass of the puncher's arm and V1 is the speed of the fist at the instant of contact; M2 is the mass of the opponent, and V' is the speed of the puncher's fist and the opponent's internal organs just after the collision. This implies that:

V' = M_{1}V_{1}/(M_{1} + M_{2})

The total energy available to transfer to the opponent will be the difference in the total kinetic energy before and after the collision. This difference is:

DE = 1/2 M_{1}V_{1}^{2} - 1/2 (M_{1} + M_{2})V'^{2}

Plugging in the above expression for V' yields:

DE = 1/2 M_{1}V_{1}^{2} - 1/2 M_{1}^{2}V_{1}^{2}/(M_{1} + M_{2})DE = 1/2 M_{1}V_{1}^{2}(1 - M_{1}/(M_{1} + M_{2}))DE = 1/2 M_{1}V_{1}^{2}(M_{2}/(M_{1} + M_{2})DE = 1/2 M_{1}M_{2}V_{1}^{2}/(M_{1} + M_{2})

In general, the mass of a person's arm is about 10% of the total mass of his or her body. So if we assume that the puncher and the opponent have about the same mass we have M1 ~ 0.1M2, and the above expression reduces to:DE = 1/22 M_{2}V_{1}^{2}If we assume a standard 70-kilogram person, and that the punch makes contact at its maximum velocity (~7 meters per second for a black belt) then we have:DE = 156 JAt first, this looks more than sufficient to break an arm bone. But the real-life situation is more complicated; in most cases an arm is more likely to just move aside when hit, rather than deform and break. Ribs, however, may move very little. It is advantageous to break the bones of your opponent, but you would prefer to avoid breaking the bones in your own hand—these bones are even smaller (and thus more vulnerable) than arm or rib bones. So how do taekwondo punchers break bones, or boards, without damaging their hands?