I am trying to figure out the force delivered behind a right rear hand punch as you twist your hips and shoulders into the punch. I have looked at several sources to find most of the information but I am stuck on one part and would appreciate any help. I found an article that helped me determine the approximate moment of inertia (I) for that of a thin wide rectangle (kind of shaped like a glass) with the top being the torso shoulders and the narrower bottom being the torso waist: I = 1/12 M (d1^2 + d2^2) d1 = the distance from center of torso to the edge of the shoulder d2 = the distance from the center of the torso to the outer edge of the back Assume: M = 35 kg, d1 = 0.22 m, d2= 0.07 m Therefore: I = 0.1554 kg*m^2 Find total energy (Et) by adding the potential (Ep), kinetic (Ek) and rotational (Er) components: Et = Ep + Ek + Er = m*g*h + 0.5*m*v^2 + 0.5*I*w^2 m = M = mass, assume arm is 8 kg g = gravitational acceleration = 9.8 m/sec^2 h = height, assume slightly downward punch of 0.16 meters v = velocity of the punch, assume 7.3 m/sec w = angular velocity The article I found then has this equation: Et = (8 kg)(9.8 m/s^2)(0.16 m) + 0.5(8 kg)(7.3 m/s)^2 + 0.5(0.1554 kg m^2)(1.9^2/0.22^2) = 207.03 Joules I can see where all parts come from but the 1.9 valve. The 1.9 is in the term for angular velocity, which is: w = velocity/radius Can anyone tell me where 1.9 comes from, or is information missing? Thanks in advance for any help!