Specifications

149
F0 = 100.0;
};
AnyMuscleModel ReacModel = {
F0 = 10000.0;
};
AnyGeneralMuscle WallReaction = {
ForceDirection = -1;
AnyKinMeasureOrg Org = {
AnyKinMeasure &wall = ..HandPos;
MeasureOrganizer = {0};
};
AnyMuscleModel &Model = .ReacModel;
};
There are two things to notice here
1. The muscle model for the reaction, ReacModel, is much stronger than the joint muscles. This is
because the wall is presumed to be very strong.
2. The ForceDirection property equals -1. This means that the force is working in the opposite direction
of the Kinematic measure, i.e. in the negative global x direction, just like a contact force with the
wall would do.
Running the InverseDynamicAnalysis again and plotting the two joint torques provides the following graph
(notice they can be plotted simultaneously with the specification line
Main.ArmStudy.Output.Model.*Torque.Fm):
The red curve is the shoulder joint torque, and the green curve is the elbow torque. Notice that the envelope
of these two curves is in fact identical to the MaxMuscleActivity curve we plotted above for the case of no
support. You would think that the support would be beneficial in the final stages of the movement where the
arm could rest a bit against the wall. Actually, it is beneficial for the elbow, but the reaction force also
increases the torque about the shoulder, and since the shoulder (red curve) has the higher load of the two,
this limits the benefit of the support. Let us see what happens if we turn the reaction force the other way
like if the hand could pull against the far side of the wall:
AnyGeneralMuscle WallReaction = {
ForceDirection = 1;
AnyKinMeasureOrg Org = {
AnyKinMeasure &wall = ..HandPos;
MeasureOrganizer = {0};
};
AnyMuscleModel &Model = .ReacModel;