# Development and Biomechanical Evaluation of a Back Exoskeleton to Support Lifting Tasks
## 2 Exoskeleton Functionality and Design
### Concept and Functionality  
The developed exoskeleton is designed to support lifting movements from the ground to the hip height, such as squat lifting. By providing mechanical support to the lumbar region, the exoskeleton aims to reduce the risk of lower back pain and injuries associated with lifting, bending, and standing for prolonged periods. Considering the primary involvement of the lower back muscles in lifting tasks, the system follows a biomimetic design using a cable-driven mechanism that runs parallel to the involved muscle chain during upper body extension. This biologically inspired approach allows for the use of smaller actuators and a more compact, form-fitting design compared to systems that use direct-mounted actuators positioned parallel to the hip joint, which requires high torque. To replicate the function of the back muscles and provide effective support, the system generates tensile forces that, together with a lever arm, produce torque around the hip joint. However, these tensile forces also introduce compression forces that must be redirected away from sensitive body areas, such as the spine, to avoid discomfort or unintended strain. To address this, the exoskeleton incorporates materials that are rigid in compression to effectively transmit support forces while remaining flexible along the vertical axis to allow for natural upper body flexion. To estimate the required dynamics of the actuation units, the tension force and shortening velocity of the steel cable were approximated based on insights from the literature. To calculate the hip joint torque, the relative weights of the upper body segments and their respective distances from the lumbosacral joint were obtained from [PERSON] and Pheasant [20]. Figure 1 illustrates an example of the absolute values for a 62 kg person. The right side depicts the main concept of the exoskeleton. The force required to be applied to the steel cables is determined using the following formula, which is derived from the torque equation at the hip joint considering the weight distribution of the upper body, arms, and head:  
\[F_{Act}=\frac{(F_{B}\ \times r_{2})+(F_{A}\ \times r_{3})+(F_{H}\ \times r_{4})}{r_{1}} \times sin(x)\times 30\% \tag{1}\]where \(F_{B,A,H}\) represent the weights of the body (B), arms (A), and head (H), and \(r_{2,3,4}\) denote the corresponding lever arms. The parameter \(r_{1}\) refers to the lever arm from the hip joint to the cable with approximately \(r_{1}=0.15\) m, considering a pulley system, while \(sin(x)\) accounts for the upper body bending angle. Since the goal is to support 30% of the body weight against gravity, the required actuation force \(F_{Act}\) is reduced by 70%. Consequently, at a bending angle of \(x=90^{\circ}\), the minimum force required to provide support during lifting is \(F_{Act}=301\) N. Considering a more moderate bending angle of \(x=45^{\circ}\) during a semi-squat lift, the required force is \(F_{Act}=213\) N.  
In addition to the required tension, the overall dynamics of the actuation unit depend on the shortening velocity of the steel cable. This velocity determines how quickly the system can adapt to the movement and provide effective support. In the context of nursing tasks, [PERSON] et al. [21] provide insights into the kinematics of everyday caregiving activities. Based on their findings, the mean sagittal angular velocity of the upper body is reported as \(v=36.40\ \pm 16.46\ \mathrm{deg/s}\). To approximate the required linear shortening velocity of the steel cable, the following equation is used, which is based on the calculation of the chord length:  
\[SV=\frac{\Delta l}{\Delta t}=2r1*sin\left(\frac{x_{v}}{2}\right)\ \ \ (for\ \Delta t=1\mathrm{s}) \tag{2}\]  
where \(\Delta l\) is the shortening of the steel cable, \(\Delta t\) is the time interval, and \(x_{v}\) represents the angular displacement during one second, derived from the mean angular velocity \(v\). This results in a required minimum shortening velocity of \(SV=0.13\ \frac{\mathrm{m}}{8}\).