CMC algorithm

[Gary Lu]

Dear all, I am now studying the CMC algorithm. I read the paper https://www.sciencedirect.com/science/a ... via%3Dihub. And I am a little bit confused about the equation 1 about the muscle activation(a) and excitation(u), since we need to know the muscle activation level to compute the requirement muscle force to track the desired trajectory through the musculotendom contraction dynamics herehttps://simtk-confluence.stanford.edu:8 ... scle+Model ,how can we get the desired activation level. If we have to integrate the a_dot over a time step to get the activation(a), how can we get access to the excitation (u) that needed in the equation. Is the excitation signal(u) comes from other experiments data?

[Nicos Haralabidis]

Hey Gary,

I will try and answer your last point!

Is the excitation signal(u) comes from other experiments data?

The excitation signal is the control and it dictates the activation level of a muscle (which then determines how much force a muscle can produce; in combination with where the muscle is operating on its force-length and force-velocity curves). For example, in a forward dynamics simulation you provide the controls (excitation signal) into the muscle-activation dynamics equation, and by integrating this equation you generate the muscle's activation level across time. The excitation values are usually treated as design variables within an optimisation in order to satisfy some criterion (e.g., maximise jump height, minimise tracking error). Alternatively, you could use the EMG signal of a particular muscle as the excitation signal - have a search for EMG-driven modelling methods.

I hope that helps!

Nicos

[Gary Lu]

Hi, Nicos, thanks for your thoroughly explanation.
I think I understand now.

[Kaiwen Yang]

Hi Gary,

Here is my understanding of the answer of your first question. First, you solve activation using optimization solver at every instance of time. Second, you track activation using excitation. Since excitation-activation dynamics is just a first order system with very small time constant, tracking activation is not hard.

Kaiwen

[Gary Lu]

thanks Kaiwen. I realize that i was thinking it too complicated that I couldn't see through the algorithm clearly before. However I understand now, thanks again.