## 2 Gear and Motor Selection
### Calculations for Selecting an Appropriate Motor  
The motor selection process was as follows. The pressure required to achieve a therapeutic effect by compressing the body is 40 kPa [15; 16]. Assuming the target area of  
\begin{table}
\begin{tabular}{c c c} \hline \hline  & **Gear 1: Driver Gear** & **Gear 2: Driven Gear** \\ \hline Module & 1 & 1 \\ Press angle [\({}^{\circ}\)] & 20 & 20 \\ Helix angle [\({}^{\circ}\)] & 15 & 15 \\ Number of teeth & 19 & 76 \\ Face width [mm] & 8 & 8 \\ Center distance [mm] & 49.175 & \\ \hline \hline \end{tabular}
\end{table}
Table 1: Specifications of the gears.  
\begin{table}
\begin{tabular}{c c} \hline \hline  & **Motor Specifications** \\ \hline Power supply capacity [kVA] & 0.9 \\ Rated output [W] & 400 \\ Rated torque [N-m] & 1.3 \\ Rated rotational velocity [RPM] & 3000 \\ Maximum rotational velocity [RPM] & 5000 \\ \hline \hline \end{tabular}
\end{table}
Table 2: Specifications of the motor.  
Figure 2: Medical electric leg compression machine.  
the body to be a cylinder of 0.22 m in diameter and 0.15 m in height, the area to be pressed and required force (FA) can be calculated using Equation (1).  
\[\text{Required pressure [Pa]}\times\text{Contact area [m${}^{2}$]}=4146.9\text{ N} \tag{1}\]  
The torque of the selected motor is 1.3 N/m, and the gear ratio is 4. In this configuration, the torque of the gear can be estimated using Equation (2).  
\[\text{Motor torque [N/m]}\times\text{Gear ratio = 5.2 N/m} \tag{2}\]  
The radius of the motor axis is 0.05 m, and consequently, the force applied to the area (FB) is determined using Equation (3).  
\[\text{Gear torque [N/m]}\div\text{Motor shaft radius [m]}=104\text{ N} \tag{3}\]  
The diameter of the cuffs is designed to be 0.05 m, and the travel distance per one turn of the large gear is calculated using Equation (4).  
\[\text{Diameter of cuff drum [m]}\times\text{pi = 0.157 m} \tag{4}\]  
The rotational velocity of the driver gear directly connected to the motor is 3000 RPM (rated RPM of the motor), and then, the gear ratio sets the rotational velocity of the driven gear to 750 RPM. Based on this consideration, the stroke per second is determined by Equation (5).  
The rotational velocity of driven motor [RPM] \(\times\) Distance when cuff rotates one turn [m] \(\div\) 60 = 1.964 m/s  
The gears operate in a 1:2 ratio to inflate and deflate the cuffs per second. That is, it takes 0.33 s to deflate the cuffs. The stroke required to achieve the therapeutic pressure of 40 kPa is 0.06 cm. Based on this, the stroke per second is determined by Equation (6).  
\[\text{Required stroke [m]}\div\text{Operating time for shrinkage [s]}=0.18\text{ m/s} \tag{6}\]  
The product of the RPM and torque of the selected motor needs to be larger than that of the required RPM and torque to meet the specification established for this study. First, the product of the RPM and torque of the selected motor are determined using Equation (7).  
\[\text{Rotation velocity of Motor }\times\text{Motor torque = 3900 N/m\cdot\text{RPM}} \tag{7}\]  
The product of the RPM and torque required to obtain the therapeutic effect by compressing the body is determined using Equation (8).  
\[\text{Required stroke per sec }\times\text{Rotational velocity of driven motor }\div\text{ Stroke per sec }\times\text{Motor torque }\div(\text{F}_{\text{B}}/\text{F}_{\text{a}})\\ =3564\text{ N/m\cdot\text{RPM}} \tag{8}\]  
The results of Equations (7) and (8) indicate that the selected motor satisfied the specification, as the product of the RPM and torque of the selected motor exceeds that of the required RPM and torque required to obtain the therapeutic effect by compressing the body.