A closed-form formula for the moment arm matrix of a general musculoskeletal model with considering joint constraint and motion rhythm

Abstract

This paper introduces a new closed-form formulation for the moment arm matrix corresponding to the musculotendon unit in a general computational model of the musculoskeletal system. The novel approach uses matrix calculus to define a “Generalized Musculotendon Line of Action Vector (GMLAV)” in integration with the virtual work principle, hence leveraging both conventional geometry and tendon excursion methods. In contrast to previous methods, the concepts of motion rhythm and joint constraints have been incorporated without restrictions, where each joint variable is defined as an arbitrary smooth function of the generalized coordinate vector. The validity of our formulation was established by performing the simulations on two well-known musculoskeletal models from literature and comparing the outcomes with those obtained using OpenSim. The results presented in this paper show a high degree of fidelity between our novel simulation model and OpenSim (or SIMM).

Appendices

Appendix A

In this study, by employing the derivative of a matrix with respect to vectors, we managed to derive a closed-form formula for the moment arm matrix of a general musculoskeletal system in the presence of joint constraints and motion rhythm. However, as emphasized in the previous sections, the definition of matrix derivative used is different from that used by other studies such as Brewer [43] and Vetter [44]. In this appendix, we first introduce our definition for the derivative of a matrix with respect to a vector; and then, we have expatiated on this definition by stating and proving its properties.

Although the matrix calculus we use is different from the aforementioned studies, we have borrowed the definition of an elementary matrix from Brewer’s work [43]: $\mathcal{E}_{ik}^{m \times n}$ is a matrix which belongs to $\mathbb{R}^{m \times n}$ and is zero everywhere except at its $i$–$k$ element that is “1”. Using this matrix an arbitrary matrix like $\mathbf{A} \in \mathbb{R}^{m \times n}$, can be written as

$$ \mathbf{A} = \sum_{i = 1}^{m} \sum _{k = 1}^{n} A_{ik} \mathcal{E}_{ik}^{m \times n} $$

(A.1)

where $A_{ik}$ is an element of $\mathbf{A}$ located at its $i$th row and $k$th column.

We define the derivative of matrix $\mathbf{A}$ with respect to an arbitrary vector $\mathbf{q} \in \mathbb{R}^{r}$ as follows:

$$ \frac{\partial \mathbf{A}}{\partial \mathbf{q}} = \sum_{i = 1}^{m} \sum _{k = 1}^{n} \mathcal{E}_{ik}^{m \times n} \otimes \frac{\partial A_{ik}}{\partial \mathbf{q}}. $$

(A.2)

The following includes the various properties of the defined matrix calculus and associated proofs:

Property 1

$$\frac{\partial ( \mathbf{AB} )}{\partial \mathbf{q}} = \frac{\partial \mathbf{A}}{\partial \mathbf{q}}\mathbf{B} + ( \mathbf{A} \otimes \mathbf{I}_{r} )\frac{\partial \mathbf{B}}{\partial \mathbf{q}},\quad \mbox{where}\ \mathbf{B} \in \mathbb{R}^{m \times t}. $$

Proof

According to the definition of matrix multiplication, one can deduce

$$\frac{\partial ( \mathbf{AB} )}{\partial \mathbf{q}} = \sum_{i = 1}^{m} \sum _{k = 1}^{t} \mathcal{E}_{ik}^{m \times t} \otimes \sum_{j = 1}^{n} \biggl( \frac{\partial A_{ij}}{\partial \mathbf{q}}B_{jk} + A_{ij}\frac{\partial B_{jk}}{\partial \mathbf{q}} \biggr); $$

hence,

$$ \frac{\partial ( \mathbf{AB} )}{\partial \mathbf{q}} = \sum_{i = 1}^{m} \sum _{k = 1}^{t} \sum _{j = 1}^{n} \bigl( B_{jk} \mathcal{E}_{ik}^{m \times t} \bigr) \otimes \frac{\partial A_{ij}}{\partial \mathbf{q}} + \sum_{i = 1}^{m} \sum _{k = 1}^{t} \sum_{j = 1}^{n} \bigl( A_{ij}\mathcal{E}_{ik}^{m \times t} \bigr) \otimes \frac{\partial B_{jk}}{\partial \mathbf{q}}. $$

(A.3)

Since (see Table 1 of Brewer [43])

$$B_{jk}\mathcal{E}_{ik}^{m \times t} = \mathcal{E}_{ij}^{m \times n} \mathbf{B} \mathcal{E}_{kk}^{t \times t} $$

and

$$A_{ij}\mathcal{E}_{ik}^{m \times t} = \mathcal{E}_{ii}^{m \times m} \mathbf{A} \mathcal{E}_{jk}^{n \times t}, $$

Eq. (A.3) transforms as follows:

$$ \begin{aligned}[b] \frac{\partial ( \mathbf{AB} )}{\partial \mathbf{q}} &= \sum_{i = 1}^{m} \sum_{k = 1}^{t} \sum _{j = 1}^{n} \bigl( \mathcal{E}_{ij}^{m \times n} \mathbf{B} \mathcal{E}_{kk}^{t \times t} \bigr) \otimes \frac{\partial A_{ij}}{\partial \mathbf{q}} \\ &\quad + \sum_{i = 1}^{m} \sum _{k = 1}^{t} \sum_{j = 1}^{n} \bigl( \mathcal{E}_{ii}^{m \times m} \mathbf{A} \mathcal{E}_{jk}^{n \times t} \bigr) \otimes \frac{\partial B_{jk}}{\partial \mathbf{q}} \end{aligned}. $$

(A.4)

Applying the mixed-product property of the Kronecker multiplication on Eq. (A.4) yields

$$\begin{aligned} \frac{\partial ( \mathbf{AB} )}{\partial \mathbf{q}} &= \sum_{i = 1}^{m} \sum _{j = 1}^{n} \biggl( \mathcal{E}_{ij}^{m \times n} \otimes \frac{\partial A_{ij}}{\partial \mathbf{q}} \biggr) \Biggl( \mathbf{B}\sum _{k = 1}^{t} \mathcal{E}_{kk}^{t \times t} \otimes 1 \Biggr) \\ &\quad + \Biggl( \Biggl( \sum_{i = 1}^{m} \mathcal{E}_{ii}^{m \times m}\mathbf{A} \Biggr) \otimes \mathbf{I}_{r} \Biggr)\sum_{k = 1}^{t} \sum_{j = 1}^{n} \biggl( \mathcal{E}_{jk}^{n \times t} \otimes \frac{\partial B_{jk}}{\partial \mathbf{q}} \biggr). \end{aligned}$$

Since $\mathbf{I}_{m} = \sum_{i = 1}^{m} \mathcal{E}_{ii}^{m \times m}$ and $\mathbf{I}_{t} = \sum_{k = 1}^{t} \mathcal{E}_{kk}^{t \times t}$, the first property has been proven. □

Property 2

$$\begin{aligned} \frac{\partial ( a\mathbf{A} )}{\partial \mathbf{q}} = \mathbf{A} \otimes \frac{\partial a}{\partial \mathbf{q}} + a\frac{\partial \mathbf{A}}{\partial \mathbf{q}}\quad \mbox{where $a$ is a scalar real-valued function}. \end{aligned}$$

Proof

According to (A.2), we have

$$\begin{aligned} \frac{\partial ( a\mathbf{A} )}{\partial \mathbf{q}} &= \sum_{i = 1}^{m} \sum _{k = 1}^{n} \mathcal{E}_{ik}^{m \times n} \otimes \frac{\partial ( aA_{ik} )}{\partial \mathbf{q}} \\ &= \sum_{i = 1}^{m} \sum _{k = 1}^{n} \mathcal{E}_{ik}^{m \times n} \otimes \biggl( \frac{\partial a}{\partial \mathbf{q}}A_{ik} + a\frac{\partial A_{ik}}{\partial \mathbf{q}} \biggr); \end{aligned}$$

therefore,

$$\frac{\partial ( a\mathbf{A} )}{\partial \mathbf{q}} = \sum_{i = 1}^{m} \sum _{k = 1}^{n} \bigl( \mathcal{E}_{ik}^{m \times n}A_{ik} \bigr) \otimes \frac{\partial a}{\partial \mathbf{q}} + a\sum_{i = 1}^{m} \mathcal{E}_{ik}^{m \times n} \otimes \frac{\partial A_{ik}}{\partial \mathbf{q}}. $$

In order to obtain this relation, we have used the mixed-product property of the Kronecker multiplication. □

Property 3

$$\begin{aligned} &\mbox{(a)}\quad d\mathbf{A} = \bigl( \mathbf{I}_{m} \otimes d \mathbf{q}^{T} \bigr)\frac{\partial \mathbf{A}}{\partial \mathbf{q}}; \\ &\mbox{(b)}\quad d\mathbf{A} = \biggl( \frac{\partial \mathbf{A}^{T}}{\partial \mathbf{q}} \biggr)^{T} ( \mathbf{I}_{n} \otimes d\mathbf{q} ). \end{aligned}$$

Proof

(a) Using Eq. (A.1), we have

$$ d\mathbf{A} = \sum_{i = 1}^{m} \sum _{k = 1}^{n} \mathcal{E}_{ik}^{m \times n} dA_{ik}; $$

(A.5)

moreover,

$$dA_{ik} = \sum_{j = 1}^{r} \frac{\partial A_{ik}}{\partial q_{j}} dq_{j} = d\mathbf{q}^{T}\frac{\partial A_{ik}}{\partial \mathbf{q}}. $$

Substituting this relation into Eq. (A.5) results in

$$d\mathbf{A} = \sum_{i = 1}^{m} \sum _{k = 1}^{n} \mathcal{E}_{ik}^{m \times n} \biggl( d\mathbf{q}^{T}\frac{\partial A_{ik}}{\partial \mathbf{q}} \biggr). $$

Since $d\mathbf{q}^{T}\frac{\partial A_{ik}}{\partial \mathbf{q}}$ is a scalar, one may rewrite the aforementioned equation into what follows:

$$d\mathbf{A} = \sum_{i = 1}^{m} \sum _{k = 1}^{n} \mathcal{E}_{ik}^{m \times n} \otimes \biggl( d\mathbf{q}^{T}\frac{\partial A_{ik}}{\partial \mathbf{q}} \biggr). $$

Utilizing the mixed-product property, this relation is rearranged into the following structure:

$$d\mathbf{A} = \sum_{i = 1}^{m} \sum _{k = 1}^{n} \bigl( \mathbf{I}_{m} \otimes d\mathbf{q}^{T} \bigr) \biggl( \mathcal{E}_{ik}^{m \times n} \otimes \frac{\partial A_{ik}}{\partial \mathbf{q}} \biggr); $$

therefore,

$$d\mathbf{A} = \bigl( \mathbf{I}_{m} \otimes d\mathbf{q}^{T} \bigr)\frac{\partial \mathbf{A}}{\partial \mathbf{q}}. $$

(b) The main process of proving this claim is similar to that of part (a). However, for computing $dA_{ik}$ we write

$$dA_{ik} = \biggl( \frac{\partial A_{ik}}{\partial \mathbf{q}} \biggr)^{T}d \mathbf{q}; $$

therefore,

$$d\mathbf{A} = \sum_{i = 1}^{m} \sum _{k = 1}^{n} \mathcal{E}_{ik}^{m \times n} \otimes \biggl( \biggl( \frac{\partial A_{ik}}{\partial \mathbf{q}} \biggr)^{T}d\mathbf{q} \biggr). $$

Employing the same plan as we did in part (a), we obtain the following result:

$$d\mathbf{A} = \sum_{i = 1}^{m} \sum _{k = 1}^{n} \biggl( \mathcal{E}_{ik}^{m \times n} \otimes \biggl( \frac{\partial A_{ik}}{\partial \mathbf{q}} \biggr)^{T} \biggr) ( \mathbf{I}_{n} \otimes d\mathbf{q} ). $$

Moreover, since $( \mathcal{E}_{ik}^{m \times n} )^{T} = \mathcal{E}_{ki}^{n \times m}$, one may reformulate this equation into the following one:

$$d\mathbf{A} = \sum_{i = 1}^{m} \sum _{k = 1}^{n} \biggl( \mathcal{E}_{ki}^{n \times m} \otimes \frac{\partial A_{ik}}{\partial \mathbf{q}} \biggr)^{T} ( \mathbf{I}_{n} \otimes d\mathbf{q} ); $$

hence,

$$d\mathbf{A} = \biggl( \frac{\partial \mathbf{A}^{T}}{\partial \mathbf{q}} \biggr)^{T} ( \mathbf{I}_{n} \otimes d\mathbf{q} ). $$

Using this property, one can easily extract similar formulae for $\frac{d\mathbf{A}}{dt}$ and $\delta \mathbf{A}$. □

Appendix B

The purpose of this section is to substantiate the following equality:

$$ \bigl( \bar{\mathbf{0}}^{ T} \otimes \delta \mathbf{q}^{T} \bigr)\mathbf{D}_{i} = \delta \mathbf{q}^{T} \bigl( \bar{ \mathbf{0}}^{ T} \otimes \mathbf{I}_{n_{q}} \bigr) \mathbf{D}_{i} $$

(B.1)

where $\mathbf{D}_{i} \in \mathbb{R}^{4n_{q} \times 4}$ which in Sect. 2.3.1 has been defined as

$$\mathbf{D}_{i} = \biggl( \frac{\partial \mathbf{B}_{i}}{\partial \mathbf{q}} \biggr){}_{i}^{0} \bar{\mathbf{A}}_{4}. $$

Starting from the left-hand side of Eq. (B.1), we have

$$\bigl( \bar{\mathbf{0}}^{ T} \otimes \delta \mathbf{q}^{T} \bigr)\mathbf{D}_{i} = \bigl( \bar{\mathbf{0}}^{ T} \otimes \delta \mathbf{q}^{T} \bigr)\mathbf{I}_{4n_{q}} \mathbf{D}_{i}. $$

Since $\mathbf{I}_{4n_{q}} = \mathbf{I}_{n_{q}} \otimes \mathbf{I}_{4}$, one may rewrite the right-hand side of this equation into the following form:

$$({\overline{\mathbf{0}}}^T\otimes \delta {\mathbf{q}}^T){\mathbf{D}}_i=({\overline{\mathbf{0}}}^T\otimes \delta {\mathbf{q}}^T)(\{\left[\begin{array}{cc}{\mathbf{I}}_3& {\mathbf{0}}_{3\times 1}\\ {\mathbf{0}}_{3\times 1}^T& 0\end{array}\right]+\left[\begin{array}{cc}{\mathbf{0}}_3& {\mathbf{0}}_{3\times 1}\\ {\mathbf{0}}_{3\times 1}^T& 1\end{array}\right]\}\otimes {\mathbf{I}}_{n_q}){\mathbf{D}}_i.$$

Expanding the right-hand side of this equation results in

$$\begin{array}{rl}({\overline{\mathbf{0}}}^T\otimes \delta {\mathbf{q}}^T){\mathbf{D}}_i& =({\overline{\mathbf{0}}}^T\otimes \delta {\mathbf{q}}^T)(\left[\begin{array}{cc}{\mathbf{I}}_3& {\mathbf{0}}_{3\times 1}\\ {\mathbf{0}}_{3\times 1}^T& 0\end{array}\right]\otimes {\mathbf{I}}_{n_q}){\mathbf{D}}_i\\ & +({\overline{\mathbf{0}}}^T\otimes \delta {\mathbf{q}}^T)(\left[\begin{array}{cc}{\mathbf{0}}_3& {\mathbf{0}}_{3\times 1}\\ {\mathbf{0}}_{3\times 1}^T& 1\end{array}\right]\otimes {\mathbf{I}}_{n_q}){\mathbf{D}}_i.\end{array}$$

Now we prove that the first part on the right-hand side of this relation vanishes. To this end, we use the mixed-product property of the Kronecker multiplication:

$$\begin{array}{rl}({\overline{\mathbf{0}}}^T\otimes \delta {\mathbf{q}}^T)(\left[\begin{array}{cc}{\mathbf{I}}_3& {\mathbf{0}}_{3\times 1}\\ {\mathbf{0}}_{3\times 1}^T& 0\end{array}\right]\otimes {\mathbf{I}}_{n_q}){\mathbf{D}}_i& =({\overline{\mathbf{0}}}^T\left[\begin{array}{cc}{\mathbf{I}}_3& {\mathbf{0}}_{3\times 1}\\ {\mathbf{0}}_{3\times 1}^T& 0\end{array}\right]\otimes \delta {\mathbf{q}}^T{\mathbf{I}}_{n_q}){\mathbf{D}}_i\\ & =({\mathbf{0}}_{1\times 4}\otimes \delta {\mathbf{q}}^T){\mathbf{D}}_i\\ & ={\mathbf{0}}_{1\times 4};\end{array}$$

hence,

$$({\overline{\mathbf{0}}}^T\otimes \delta {\mathbf{q}}^T){\mathbf{D}}_i=({\overline{\mathbf{0}}}^T\otimes \delta {\mathbf{q}}^T)(\left[\begin{array}{cc}{\mathbf{0}}_3& {\mathbf{0}}_{3\times 1}\\ {\mathbf{0}}_{3\times 1}^T& 1\end{array}\right]\otimes {\mathbf{I}}_{n_q}){\mathbf{D}}_i.$$

Since

$$\left[\begin{array}{cc}{\mathbf{0}}_3& {\mathbf{0}}_{3\times 1}\\ {\mathbf{0}}_{3\times 1}^T& 1\end{array}\right]={\overline{\mathbf{0}}}^T\otimes \overline{\mathbf{0}},$$

we can rewrite this equation into the following one:

$$\bigl( \bar{\mathbf{0}}^{ T} \otimes \delta \mathbf{q}^{T} \bigr)\mathbf{D}_{i} = \bigl( \bar{\mathbf{0}}^{ T} \otimes \delta \mathbf{q}^{T} \bigr) \bigl( \bigl( \bar{\mathbf{0}}^{ T} \otimes \bar{\mathbf{0}} \bigr) \otimes \mathbf{I}_{n_{q}} \bigr) \mathbf{D}_{i}. $$

Employing the distributive property of the Kronecker multiplication, we have

$$\bigl( \bar{\mathbf{0}}^{ T} \otimes \delta \mathbf{q}^{T} \bigr)\mathbf{D}_{i} = \bigl( \bar{\mathbf{0}}^{ T} \otimes \delta \mathbf{q}^{T} \bigr) ( \bar{\mathbf{0}} \otimes \mathbf{C} ) \mathbf{D}_{i} $$

where $\mathbf{C} = \bar{\mathbf{0}}^{ T} \otimes \mathbf{I}_{n_{q}}$.

Since $\mathbf{D}_{i} = 1 \otimes \mathbf{D}_{i}$,

$$\bigl( \bar{\mathbf{0}}^{ T} \otimes \delta \mathbf{q}^{T} \bigr)\mathbf{D}_{i} = \bigl( \bar{\mathbf{0}}^{ T} \otimes \delta \mathbf{q}^{T} \bigr) ( \bar{\mathbf{0}} \otimes \mathbf{CD}_{i} ). $$

Taking the mixed-product property of the Kronecker multiplication into consideration results in

$$\bigl( \bar{\mathbf{0}}^{ T} \otimes \delta \mathbf{q}^{T} \bigr)\mathbf{D}_{i} = \bar{\mathbf{0}}^{ T}\bar{\mathbf{0}} \otimes \delta \mathbf{q}^{T}\mathbf{CD}_{i}. $$

Since $\bar{\mathbf{0}}^{ T}\bar{\mathbf{0}} = 1$, we come into the following conclusion:

$$\bigl( \bar{\mathbf{0}}^{ T} \otimes \delta \mathbf{q}^{T} \bigr)\mathbf{D}_{i} = \delta \mathbf{q}^{T} \bigl( \bar{ \mathbf{0}}^{ T} \otimes \mathbf{I}_{n_{q}} \bigr) \mathbf{D}_{i}. $$