Punch physics
Intro
Some of the physics behind the punch.Energy in a punch
To quantify the potential of a punch 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 Emax = 1/(2Y) Alsb2, which works 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 focused punch.
Let’s consider two people: the puncher and the opponent. If the collision between the puncher's fist and the opponent's internal organs is completely inelastic, conservation of momentum tells us that M1V1 = (M1 + M2)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' = M1V1/(M1 + M2). 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 M1V12 - 1/2 (M1 + M2)V'2.
Plugging in the above expression for V' yields DE = 1/2 M1V12 - 1/2 M12V12/(M1 + M2)DE = 1/2 M1V12(1 - M1/(M1 + M2))DE = 1/2 M1V12(M2/(M1 + M2)DE = 1/2 M1M2V12/(M1 + M2).
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 the same mass, we have M1 ~ 0.1M2, and the above expression reduces to DE = 1/22 M2V12.
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 Joules.
At 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 and are more susceptible to breaking.
Let's consider the amount of force needed to break a horizontal board supported at both ends struck in the center by a force F. The downward force will be shared by the two supports so each will supply an upward force of F/2.
Imagine the board is deflected downwards by the force., This means the top surface of the board will be in compression and the bottom surface will be in tension. This force will produce a torque about an axis through the middle of the board. To understand this, consider the forces on the left half of the board that are due to the right half. The top of the board is in compression, so the right half is pushing to the left. The bottom of the board is in tension, so here the right half is pulling to the right. This means the forces will produce a counter-clockwise torque about the center.
The board will break if this torque is greater than tmax = 1/6 Wh2sb, where W is the width of the board, h is the thickness, and sb is the modulus of rupture of wood. Notice that the upward force from the left support will tend to produce a torque in the opposite direction from the torque due to the stresses. The magnitude of this torque is simply r*F (since r and F are perpendicular in this case), or F/2 * L/2. If this torque is less than tmax noted above, then the board will not break. Thus, the minimum force needed to break the board can be calculated by setting these two torques equal to each other FL/4 = 1/6 Wh2sbF = 2/3 sb Wh2/L.
Notice that the force is proportional to h2; boards get much harder to break as they get thicker. This is why breakers will break stacks of several thin boards rather than a single piece of wood with the same total thickness. If we put in some typical dimensions for a board, such as L x W x h = 30 x 20 x 2 cm, and look up the rupture modulus, we get a force of around 711 N, or 160 lbs. This force sounds feasible for a punch to generate, but let’s check to be sure.
If we assume, as above, that we have a 7-kg arm traveling at 7 m/s then we have a total momentum of P ~ 49 kg m/s. We assume that the fist and arm come to rest during the blow, so this is the total DP in the collision. Since the force is DP/Dt, we need to know how quickly the fist stops.
In a graph of the downward hammer fist strike, you would notice that the velocity starts out fairly constant (and negative) and then quickly changes. The maximum acceleration upon contact with the target is upwards of 350 g!
By looking at the width of the peak in the acceleration curve, we may get an idea of the interaction time. It looks to be a little over 5 ms. Since this situation is a bit different from ours, we will be conservative and assume an interaction time of 10 ms. A change in momentum of 49 kg m/s in a time of 10 ms would require a force of 4900 N, which is more than sufficient to break the board described above.
Won't this force break the bones in the hand?
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 punchers break bones, or boards, without damaging their hands?There is a bit of a safety margin since bone is inherently stronger than wood (or even concrete). A torque analysis like the one above for a small, short cylinder of bone (similar to the bones in the hand) indicates that it should take a force of about 1500 N to break a hand bone. But the bones in the hand are protected because they are not rigidly supported like the board discussed above.
The soft connective tissue (muscles, tendons, etc.) in the hand may absorb much of the energy of the strike if the hand is held in the proper position. Also, recall that bone is much stronger in compression than in tension or torsion, so, if the hand is held in a position such that the bones are exposed to compression rather than tension, the bones will be further protected. Therefore, the board should break before the hand does.
However, the ~4900 N available in the punch is much greater than the 1500 N required to break a hand bone. Therefore, even with the safety margins, there is still a risk. Most of the safety margin depends on how the punch is executed, such as holding the fist in a proper shape, snapping it out with great speed, the striking area of the fist, etc. If the punch is executed properly, the board will break; otherwise, the hand will break.
Energy is lost in the joints
In punching, some of the energy generated by the torso is lost during its transfer through joints, primarily the shoulder, elbow, and wrist joints. Bones function as perfect transmitters of force since they have more strength longitudinally than in cross-section. Force moves readily down a bone until the bone ends, then it must transfer across a joint before traveling down the next bone. Misaligned, over-tensed, or under-tensed joints cause some of the force to be dissipated into the body. For example, people who wing their punches, tend to lose power at the shoulder and elbow. More power is transferred through joints when the bones they connect are aligned in a straight line.Sources
- Steiger. T. (1999). Physics 208: Physics of Sports. [Online]. Available: http://neutrino.phys.washington.edu/~wilkes/post/temp/phys208/
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