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For consistency and easier expansion of the equations, let

,    [1]

where CP stands for the center of pressure. See Center of Pressure for the details. In other words, e_CP/FT is the vector drawn from the foot center of mass to the center of pressure.

From [10] of Joint Torque:

.    [2]

Substitute [2] into [11]:

,    [3]

where j_FT/AK = the relative position of the ankle to the foot CM, and e_CP/AK = the relative position of the center of pressure to the ankle:

    [4]

Substitute [2] & [3] into [14] & [15] of Joint Torque:

    [5]

,    [6]

where

    [7]

Substitute [3] to [6] into [16] and [17] of Joint Torque:

    [8]

    [9]

Generalize [2], [3], [4], [5], [8], & [9]:

    [10]

    [11]

where j = joint of interest, and i's = segments distal to the joint of interest. The inertial force of the segment relative to its weight is given as

    [12]

The net joint force is basically the sum of the relative inertial forces of the distal segments minus the ground reaction force, as shown in [10]. The joint torque shown in [11] has 4 terms: sum of the inertial torques, sum of the torques caused by the relative inertial forces, the torque due to the ground reaction force, and the external torque. The first term is due to the angular motion of the segments while the second term is due to the linear motion of the segment CM.

An easy way to compute the net joint force and the joint torque is to compute them as if there is no ground reaction force acting and later subtract the terms due to the ground reaction force and/or the external torque for each joint.

© Young-Hoo Kwon, 1998-