Question

Motion capture data were collected in the Biomechanics Lab on the lower limb of a person during the swing phase of walking. At a certain point during the swing phase, the XY coordinates of the hip, knee, and ankle joints were: hip (0.52, 0.81); knee (0.43, 0.42); ankle (0.12, 0.16). Given these coordinate data, how many degrees of flexion were there at the knee joint at that point in the stride cycle?

Answer 1

given

xy coordinates of hip = (0.52, 0.81)
xy coordinates of knee = (0.43, 0.42)
xy coordinates of ankle = (0.12, 0.16)

hence
consider vector hip - knee AB
AB = (0.43-0.52)i + (0.42 - 0.81)j = -0.09i -0.39 j
consider vector knee ankle - BC
BC = (0.12 - 0.43)i + (0.16 - 0.42)j = -0.31i -0.26 j

now angle between these two vectors = phi

AB.BC = |AB||BC|cos(phi)
cos(phi) = (0.09*0.31 + 0.39(0.26))/sqrt(0.09^2 + 0.39^2)sqrt(0.31^2 + 0.26^2)
phi = 37.01849 deg of flexion