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Armed against falls: the contribution of arm movements to balance recovery after tripping

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Abstract

Arm movements after perturbations like tripping over an obstacle have been suggested to be aspecific startle responses, serve a protective function or contribute to balance recovery. This study aimed at determining if and how arm movements play a functional role in balance recovery after a perturbation. We tripped young subjects using an obstacle that suddenly appeared from the floor at exactly mid-swing. We measured arm muscle EMG, quantified body rotations after tripping, and established the effects of arm movements by calculating how the body would have rotated without arms. Strong asymmetric shoulder muscle responses were observed within 100 ms after trip initiation. Significantly faster and larger responses were found in the contralateral arm abductors on the non-tripped (right) side. Mean amplitudes were larger in the ipsilateral retroflexors and contralateral anteflexors. The resulting asymmetric arm movements had a small effect on body rotation in the sagittal and frontal planes, but substantially affected the body orientation in the transverse plane. With the enlargement of the ongoing arm swing, the arms contributed to balance recovery by postponing the transfer of arm angular momentum to the trunk. This resulted in an axial rotation of the lower segments of the body towards the non-tripped side, which increases the length of the recovery step in the sagittal plane, and therefore facilitates braking the impending fall.

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Acknowledgments

The authors would like to thank Petra van der Burg and Maarten Bobbert for their contribution to this study and the Netherlands Organisation for Scientific Research (NWO) for their financial support (grant # 916.76.077).

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Correspondence to Mirjam Pijnappels.

Appendix

Appendix

Equations of motion for armless body

First, the body was separated in two segment sets, arms (two arms) and trunk + legs (trunk plus two legs). For these two segment sets, the equations of motion, with moment equations around the center of mass, are (see Fig. 8):

$$ {\mathbf{F}}_{{{\text{sh}},{\text{ua}}}} = \, m_{\text{arms}} \cdot \left( {{\mathbf{a}}_{\text{arms}} - {\mathbf{g}}} \right) $$
(5)
$$ {\dot{\mathbf{L}}}_{\text{arms}} = {\text{d}}(I\omega ){\text{d}}t = {\mathbf{M}}_{{{\text{sh}},{\text{ua}}}} + \left( {{\mathbf{r}}_{\text{sh}} - {\mathbf{r}}_{\text{arms}} } \right) \times {\mathbf{F}}_{{{\text{sh}},{\text{ua}}}} $$
(6)
$$ {\mathbf{F}}_{\text{grf}} = \, m_{\text{trunklegs}} \cdot \left( {{\mathbf{a}}_{\text{trunklegs}} - {\mathbf{g}}} \right) - {\mathbf{F}}_{{{\text{sh}},{\text{tr}}}} $$
(7)
$$ {\dot{\mathbf{L}}}_{\text{trunklegs}} = {\mathbf{M}}_{{{\text{sh}},{\text{tr}}}} + \left( {{\mathbf{r}}_{\text{sh}} - {\mathbf{r}}_{\text{trunklegs}} } \right) \times {\mathbf{F}}_{{{\text{sh}},{\text{tr}}}} + \left( {{\mathbf{r}}_{\text{COP}} - {\mathbf{r}}_{\text{trunklegs}} } \right) \times {\mathbf{F}}_{\text{grf}} + {\mathbf{M}}_{\text{grf}} $$
(8)

where g = gravity vector, m = mass, a = acceleration vector, r = position vector, \( {\dot{\mathbf{L}}} \) = rate of change of the angular momentum, x = vector product, F sh and M sh = reaction force and moment at the shoulder, with tr = on the trunk and ua = on the upper arm. Furthermore, F grf = ground reaction force, M grf = ground reaction moment (only non-zero around the vertical axis) and COP = center of pressure. Inserting Eq. 7 into 8 yields:

$$ \begin{gathered} {\dot{\mathbf{L}}}_{\text{trunklegs}} = {\mathbf{M}}_{{{\text{sh}},{\text{tr}}}} + \left( {{\mathbf{r}}_{\text{sh}} - {\mathbf{r}}_{\text{trunklegs}} } \right) \times {\mathbf{F}}_{{{\text{sh}},{\text{tr}}}} \hfill \\ + \left( {{\mathbf{r}}_{\text{COP}} - {\mathbf{r}}_{\text{trunklegs}} } \right) \times m_{\text{trunklegs}} \cdot \left( {{\mathbf{a}}_{\text{trunklegs}} - {\mathbf{g}}} \right) - \left( {{\mathbf{r}}_{\text{COP}} - {\mathbf{r}}_{\text{trunklegs}} } \right) \times {\mathbf{F}}_{{{\text{sh}},{\text{tr}}}} + {\mathbf{M}}_{\text{grf}} \hfill \\ \end{gathered} $$
(9)
Fig. 8
figure 8

Free body diagram of the two segment sets arms and trunk + legs. Note that the arms segments set represents both arms. Forces and moments are represented in black arrows; linear and angular acceleration terms in grey arrows. Abbreviations are explained in the text of the Appendix

From the instant of trip initiation onward, we simulated arm removal so that F sh,tr and M sh,tr are zero and Eq. 9 simplifies to:

$$ {\dot{\mathbf{L}}}_{{{\text{trunklegs}}\_{\text{armless}}}} = \left( {{\mathbf{r}}_{\text{COP}} - {\mathbf{r}}_{\text{trunklegs}} } \right) \times m_{\text{trunklegs}} \cdot \left( {{\mathbf{a}}_{\text{trunklegs}} - {\mathbf{g}}} \right) + {\mathbf{M}}_{\text{grf}} $$
(10)

Theoretically, this equation can be used to achieve our objective, i.e., to calculate the angular displacement of the trunk + legs without arms between trip initiation and recovery foot landing. However, to calculate angular displacement, this equation would require double numerical integration, which strongly amplifies errors, such as the error in r COP during contact of the tripped foot with the obstacle. We therefore introduced an alternative solution. As M sh,tr = −M sh,ua, and F sh,tr = −F sh,ua, Eq. 6 can be used to rewrite Eq. 9 to:

$$ \begin{gathered} {\dot{\mathbf{L}}}_{\text{trunklegs}} = - \left( {{\dot{\mathbf{L}}}_{\text{arms}} - \left( {{\mathbf{r}}_{\text{sh}} - {\mathbf{r}}_{\text{arms}} } \right) \times \left( { - {\mathbf{F}}_{{{\text{sh}},{\text{tr}}}} } \right)} \right) + \left( {{\mathbf{r}}_{\text{sh}} - {\mathbf{r}}_{\text{trunklegs}} } \right) \times {\mathbf{F}}_{{{\text{sh}},{\text{tr}}}} \hfill \\ + \left( {{\mathbf{r}}_{\text{COP}} - {\mathbf{r}}_{\text{trunklegs}} } \right) \times m_{\text{trunklegs}} \cdot \left( {{\mathbf{a}}_{\text{trunklegs}} - {\mathbf{g}}} \right) - \left( {{\mathbf{r}}_{\text{COP}} - {\mathbf{r}}_{\text{trunklegs}} } \right) \times {\mathbf{F}}_{{{\text{sh}},{\text{tr}}}} + {\mathbf{M}}_{\text{grf}} \hfill \\ \end{gathered} $$
(11)

which can be simplified to:

$$ \begin{gathered} {\dot{\mathbf{L}}}_{\text{trunklegs}} = - {\dot{\mathbf{L}}}_{\text{arms}} + \left( {{\mathbf{r}}_{\text{arms}} - {\mathbf{r}}_{\text{COP}} } \right) \times {\mathbf{F}}_{{{\text{sh}},{\text{tr}}}} \hfill \\ + \left( {{\mathbf{r}}_{\text{COP}} - {\mathbf{r}}_{\text{trunklegs}} } \right) \times m_{\text{trunklegs}} \cdot \left( {{\mathbf{a}}_{\text{trunklegs}} - {\mathbf{g}}} \right) + {\mathbf{M}}_{\text{grf}} \hfill \\ \end{gathered} $$
(12)

Now Eq. 10 can be used to replace the right terms in Eq. 12 by \( {\dot{\mathbf{L}}}_{{{\text{trunklegs}}\_{\text{armless}}}} \):

$$ {\dot{\mathbf{L}}}_{\text{trunklegs}} = - {\dot{\mathbf{L}}}_{\text{arms}} + \left( {{\mathbf{r}}_{\text{arms}} - {\mathbf{r}}_{\text{cop}} } \right) \times {\mathbf{F}}_{{{\text{sh}},{\text{tr}}}} + {\dot{\mathbf{L}}}_{{{\text{trunklegs}}\_{\text{armless}}}} $$
(13)

which can be rearranged using Eq. 5 and F sh,tr = −F sh,ua:

$$ {\dot{\mathbf{L}}}_{{{\text{trunklegs}}\_{\text{armless}}}} = {\dot{\mathbf{L}}}_{\text{arms}} + {\dot{\mathbf{L}}}_{\text{trunklegs}} + \left( {{\mathbf{r}}_{\text{arms}} - {\mathbf{r}}_{\text{COP}} } \right) \times m_{\text{arms}} \cdot \left( {{\mathbf{a}}_{\text{arms}} - {\mathbf{g}}} \right) $$
(14)

Now the angular momentum L trunklegs_armless at the time range from trip to recovery foot landing can be calculated by integrating Eq. 14:

$$ {\mathbf{L}}_{{{\text{trunklegs\_armless}}}} = {\mathbf{L}}_{\text{arms}} + {\mathbf{L}}_{\text{trunklegs}} + \int\limits_{{t = {\text{trip}}}}^{{t = {\text{lift - off}}}} {\left( {\left( {{\mathbf{r}}_{\text{arms}} - {\mathbf{r}}_{\text{COP}}} \right) \times m_{\text{arms}} \cdot ({\mathbf{a}}_{\text{arms}} - {\mathbf{g}})} \right)} {\text{d}}t $$
(15)

Note that the integral term on the right is only non-zero between trip initiation and liftoff of the non-tripped leg. This is not the case for and L arms and L trunklegs, which are non-zero at the initiation of the trip so that:

$$ \begin{aligned} {\mathbf{L}}_{\text{arms}} & = {\mathbf{L}}_{\text{arms,trip}} + \int\limits_{{t = {\text{trip}}}}^{{t = {\text{lift - off}}}} {\left( {{\mathbf{L}}_{\text{arms}} } \right)} {\text{d}}t \\ {\mathbf{L}}_{\text{trunklegs}} &= {\mathbf{L}}_{\text{trunklegs,trip}} + \int\limits_{{t = {\text{trip}}}}^{{t = {\text{lift - off}}}} {\left( {{\mathbf{L}}_{\text{trunklegs}} } \right)} {\text{d}}t \\ \end{aligned} $$
(16)

Importantly, L arms and L trunklegs can be calculated directly from the kinematics using Eq. 1 rather than by using Eq. 16. Therefore, only the rightmost term in Eq. 15 requires integration, so that application of Eq. 15 is more robust than application of Eq. 10, in that the effect of the r COP error during contact with the obstacle, as outlined before, is smaller than in Eq. 10.

As can be seen from Eq. 16, Eq. 15 takes into account the angular momentum of the arms and trunk + legs at the instant of trip as well. Effectively, application of Eq. 15 therefore means that, at the instant of tripping, L arms is transferred to the trunk + legs, prior to ‘cutting away’ the arms. We will further denote this as the ‘transfer&cut’ condition. This transfer of L arms can have substantial effects. In normal gait, the arm swing causes substantial angular momentum in the arms, the direction of which is reversed at each step by exchange of angular momentum with the rest of the body (Bruijn et al. 2008). At mid-swing, i.e., at the instant our subjects were tripped, the angular momentum of the arms reaches a maximum.

To establish the effect of the transfer of the angular momentum of the arms in the ‘transfer&cut’ condition, we performed an alternative calculation. In this condition, we ignored the angular momentum of the arms at the instant of tripping:

$$ {\mathbf{L}}_{{{\text{trunklegs\_armless}}}} = {\mathbf{L}}_{\text{arms}} - {\mathbf{L}}_{\text{arms,trip}} + {\mathbf{L}}_{\text{trunklegs}} + \int\limits_{{t = {\text{trip}}}}^{{t = {\text{lift - off}}}} {\left( {\left( {{\mathbf{r}}_{\text{arms}} - {\mathbf{r}}_{\text{COP}} } \right) \times m_{\text{arms}} \cdot ({\mathbf{a}}_{\text{arms}} - {\mathbf{g}})} \right)} {\text{d}}t $$
(17)

This calculation, to be further denoted as ‘cut’ condition, effectively simulates that the arms would be cut off at the instant of tripping, but would keep on rotating, i.e., would keep their own angular momentum rather than transferring it to the trunk + legs segment.

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Pijnappels, M., Kingma, I., Wezenberg, D. et al. Armed against falls: the contribution of arm movements to balance recovery after tripping. Exp Brain Res 201, 689–699 (2010). https://doi.org/10.1007/s00221-009-2088-7

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