During lecture, we learned about the mathematical process of inverse dynamics and how to determine joint moments, joint

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answerhappygod
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During lecture, we learned about the mathematical process of inverse dynamics and how to determine joint moments, joint

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During Lecture We Learned About The Mathematical Process Of Inverse Dynamics And How To Determine Joint Moments Joint 1
During Lecture We Learned About The Mathematical Process Of Inverse Dynamics And How To Determine Joint Moments Joint 1 (45.81 KiB) Viewed 44 times
During Lecture We Learned About The Mathematical Process Of Inverse Dynamics And How To Determine Joint Moments Joint 2
During Lecture We Learned About The Mathematical Process Of Inverse Dynamics And How To Determine Joint Moments Joint 2 (35.98 KiB) Viewed 44 times
During lecture, we learned about the mathematical process of inverse dynamics and how to determine joint moments, joint reaction forces, and total torques occurring for a body segment. Using the information on the next page, you will go through the process of mathematically solving for several variables at three different body segments. On the next page, you are provided with segment coordinates (proximal and distal ends and center of gravity), linear accelerations, ground reaction force, angular accelerations, segment lengths, segment percentage of body mass, and radius of gyration percentage. This is all the information you need to subsequently solve for the following pieces of information. Fill out the following information and along with this sheet, provide a separate sheet of the math work you used to solve these problems. Foot Proximal horizontal force Proximal vertical force Proximal joint moment Shank Proximal horizontal force Proximal vertical force Proximal joint moment Thigh Proximal horizontal force Proximal vertical force Proximal joint moment
Hip(0.646, 0.918) Subject Mass = 80 kg Thigh Length = 0.404 m Shank Length -0.074 m/s2 = 0.438 m -7.939 m/s2 Foot Length = 0.227 m (0.716, 0.758) -15.07 r/s2 Seg Mass as % Body Mass Thigh = 10% Leg = 4.65% Foot = 1.45% Knee (0.807,0.548) Radius of Gyration as % Seg Length 0.102 m/s2 Thigh = 32.3% Leg = 30.2% -9.263 m/s2 Foot = 47.5% (0.822, 0.359) 10.9 r/52 5.544 m/s2 36.862[r/s2 (0.841, 0.111) -5.749 m/s2 (0.926, 0.068) (1.059,0.048) -202.19N (0.82,0.0) 829.94N Key Equation: ΣΤ=lα Dm + Pm + (FDx * dDy) + (FDy * dDx) + (FPx* dPy) + (FPy* dPx) = la
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