Figure 1 shows the free body diagram of a segment. Identified in the figure are the external forces such as the joint reaction forces (J's), muscle forces (M's), weight (W), and the additional external forces from the environment. Points C, P, and D in the figure are the center of mass, the proximal end, and the distal end of the segment, respectively.
Force vector M is the muscle force which is the sum of all the forces produced by the muscles. Force vector J stands for the joint reaction force which is exerted by the bones forming joints with the current segment. In reality, M can be viewed as the sum of all the eccentric forces (that do not pass through the joint center) while J is the sum of all the concentric forces. m and j are the relative position vectors of the application points of M and J, respectively, to the segment center of mass (CM; point C). M, J, m, and j are identified for both the proximal and the distal ends.
Force vector FE in the figure is the additional external force acting on the segment from the environment. One may have several FE's or none depending on the situation. Similarly, vector TE stands for the external torque exerted on the segment by the environment. A good example of this is the free vertical torque (Tz) from the force plate. See the Force Plate Issues section for the concept of the free vertical torque. This torque is required in some cases because FE alone can not sufficiently reflect the interaction between the body and the environment. A circling arrow instead of a straight arrow is used here as torque vector for better visual effect. Vector e shows the relative position of the application point of FE to the CM.
Application of the Newton's equations of motion to the mechanical system shown in Figure 1 gives
where p = the momentum of the segment, and H = the angular momentum of the segment.  and  basically says that the sum of all the external forces acting on the system is the same to the rate of change in the momentum of the system and the sum of all the external torques acting on the system is equal to the rate of change in the angular momentum of the system, respectively. Rates of change in the linear and angular momentums can be computed from the motion analysis data. Note here that the weight of the segment does not produce any torque about its CM since it is a concentric force.
From  and :
Since  and  together contain 3 groups of unknowns, M, J, and m, they provide an underdetermined system: number of the unknowns exceed that of the equations. In order to solve this system, one should either reduce the number of unknowns or add an equation to the system. Let's take the former strategy. Let
In other words, F is the sum of both the muscle force and the joint reaction force. Substitution of  into  and  yields
Using , one can reduce  to
 and  now have two groups of unknowns, F and T. Assuming that FD and TD are somehow known, one can solve the system for FP and TP.
Even though F and T in  and  were invented for computational convenience, these quantities may have some practical meaning. Firstly, T is a sort of torque since it is in the form of (position vector) x (force). Since vector m - j is the relative position of the application point of M to the joint center, as shown on Figure 2, T becomes the torque produced by the muscle force about the joint center (joint torque). This quantity is meaningful since it gives the amount of torque produced by the muscle about the joint. On the other hand, F (net joint force) is simply the sum of J and M and it does not provide any specific information in terms of muscle activation or the stress on the bones involved in the articulation. Therefore, one must not give any major meaning to the net joint force (F) while interpreting data.
Figure 3 shows two equivalent systems with the new system represented by the joint torque and the net joint force. M, J and m are simply replaced by F and T. Again, circling arrows instead of straight arrows are used here as the torque vectors for better visual effect.
Now, let's get back to  and . In order to solve this system for FP and TP, one must know FD and TD. There are two possible cases:
The foot segment is a good example of the first case. There is no segment linked to the foot distally as shown in Figure 4.  and  therefore reduce to
where FAK/FT = the net joint force acting on the foot at the ankle, TAK/FT = the joint torque produced by the muscles attached to the foot around the ankle, and jAK/FT = the relative position vector of the ankle to the foot CM. GRF stands for the ground reaction force.
Now, let's move to the shank (Figure 5). In the shank, one can not ignore F and T at the distal end and this falls into the second case listed above. Assuming all muscles included in this model are uni-articular, one can come up with
Both  and  are obtained from the Newton's 3rd Law of Motion: Law of Reaction. "For every action there is an equal and opposite reaction."
As shown in Figure 6, muscle forces MAK/SH and MAK/FT have an action-reaction relationship since all muscles are uni-articular. They have the same moment arm about the ankle joint, thus torques TAK/SH and TAK/FT also have an action-reaction relationship.
Based on  and , one can reduce  and  to
where FKN/SH = the net joint force acting on the shank at the knee, TKN/SH = the joint torque produced by the muscles attached to the shank around the knee, jKN/SH = the relative position vector of the knee to the shank CM, and jAK/SH = the relative position vector of the ankle to the shank CM. Here, of course, it was assumed that there is no additional external force acting on the shank from the environment.
Similarly, the net joint force and the joint torque of the hip-thigh can be written as
Similar approach can be used in computing the joint torques of the arm joints, starting from the hand. For a complex whole body motion one must compute the joint torques using the sequence shown in Figure 7.
Rate of Change in Linear and Angular Momentum
The rates of change in momentum and angular momentum used in the equations can be computed based on the kinematic data obtained from motion analysis. The rate of change in linear momentum in  is
where m = the mass of the segment, a = the acceleration of the segment CM. m can be obtained through the BSP estimation while a can be computed from the position of the segment CM.
The rate of change in angular momentum of the segment is the net torque acting on the segment:
where = the angular velocity of the segment, ()rot = a vector observed in the rotating reference frame (segmental reference frame). See the Rotating Reference Frame section for the details. The inertia tensor of the segment (I), the angular velocity (), and the angular momentum of the segment (H) are given as
x, y & z in the equations stand for the three axes of the segment reference frame in rotation. Inertia tensor I can be obtained from the BSP data while can be found through motion analysis. See the Angular Velocity vs. Orientation Angles page for the details of how to compute using the orientation angles.
As shown in , one can compute the net torque of the segment from the inertia tensor and the angular velocity. One must not forget to transform the net torque vector to the global frame after the computation if all other vectors are described in the global frame. Another simple way to compute the net torque is to obtain it directly from the global angular momentum data through time-derivation.
© Young-Hoo Kwon, 1998-