*3.3. Statics Analysis of the Underactuated Mechanism*

As shown in Figure 9, the unilateral clamping mechanism drives rod *BG* to move through spiral rotation and drives the four-bar mechanism to complete the whole moving process. The specific meanings of mathematical symbols in Figure 9 can be referred to in Table 1. The red characters *F*1, *F*2, and *F*<sup>3</sup> in Figure 9 respectively represent the pressure on the surface of the flower ball, the force on the *DK* rod, and the cutting force of the stem.

**Figure 9.** Statics analysis diagram of underactuated mechanism.

In Figure 9, rod *GE* is pushed down by the left thrust *FT0* to rotate the driving rod *GB*, making the four-bar mechanism rotate around base point *A.* The total thrust *FT* and the force of rod *BG* are shown in Equations (4) and (5):

$$F\_T = 2F\_{T\_{0'}} \tag{4}$$

$$F\_0 = F\_{T\_0} \cos \alpha\_\prime \tag{5}$$

The force *Fb* of rod *BC* is the component force *F*0*<sup>y</sup>* of rod *BG* as shown in Equation (6):

$$\begin{cases} \begin{array}{l} F\_{0y} = F\_{0} \sin\left(\frac{\pi}{2} - \alpha\right) \\\ F\_{b} = F\_{0y} \sin q\_{1} \end{array} \end{cases} \tag{6}$$

The underactuated mechanism approaches the flower ball under the force *Fcy* perpendicular to the rod *BC* as shown in Equation (7):

$$\begin{cases} \quad F\_c = F\_{bx} \\ \quad F\_{bx} = F\_b \sin \varphi\_1 \\ \quad F\_{cy} = F\_c \sin \varphi\_1 \end{cases} \tag{7}$$

Therefore, the torque *T*<sup>0</sup> applied by linkage *AB* around point *A* is shown in Equation (8):

$$rT\_0 = aF\_{T\_0} \cos \alpha \sin(\frac{\pi}{2} - \pi) \cos(\alpha + \theta\_1) = aF\_{0\chi} \cos(\alpha + \theta\_1),\tag{8}$$

When picking is completed, rod *AD* and cutting blade *KP* are constrained by *F*<sup>1</sup> and *F*3. The mechanical model of the unilateral underdrive is [19,20]:

$$T^T \omega = \mathbf{F}^T \mathbf{v},\tag{9}$$

Each vector in Equation (9) can be obtained according to Figure 9.

$$\begin{cases} \boldsymbol{T}^{\mathrm{T}} = \begin{bmatrix} T\_0 & T\_1 \end{bmatrix}^{\mathrm{T}} \\ \boldsymbol{\omega}^{\mathrm{T}} = \begin{bmatrix} \dot{\boldsymbol{\theta}}\_1 & \dot{\boldsymbol{\theta}}\_3 \end{bmatrix}^{\mathrm{T}} \\ \boldsymbol{F}^{\mathrm{T}} = \begin{bmatrix} F\_1 & F\_2 \end{bmatrix}^{\mathrm{T}'} \\ \boldsymbol{\upsilon}^{\mathrm{T}} = \begin{bmatrix} \upsilon\_1 & \upsilon\_2 \end{bmatrix}^{\mathrm{T}} \end{cases} \tag{10}$$

In Equation (10), the torsional spring torque is *<sup>T</sup>*<sup>1</sup> = −(*kθ*<sup>3</sup> + *<sup>T</sup>*<sup>1</sup> <sup>0</sup> ), *k* is the stiffness coefficient of the torsional spring (*k* = 52.85 N·mm/◦), and *T*<sup>0</sup> <sup>1</sup> is the initial torque of the torsional spring (*T*<sup>0</sup> <sup>1</sup> = 317.1 N·mm).

According to the relation between the contact force and input torque, the contact force vector of the underdriven mechanism can be obtained using the virtual work principle as follows:

$$F = T\!\!\!f\_{\mathbf{v}}^{-1}\!\!\!f\_{\mathbf{w}}^{-T}\!\!\!f\_{\mathbf{w}} \,\tag{11}$$

Clamping pressure *F*<sup>1</sup> and cutting force *F*<sup>3</sup> can be obtained from Equation (11).

$$\begin{cases} \quad F\_1 = -\frac{(1 - Al\_1 \cos \theta\_3) \, T\_0 + (d \cos \theta\_3 + l\_2) \, T\_1}{I\_1 l\_2} \\\quad F\_2 = \frac{AT\_0 (d \cos \theta\_3 - l\_2) + (l\_1 \cos \theta\_3 + l\_2) \, T\_1}{I\_1 l\_2} \\\quad F\_3 = \frac{AT\_0 (d \cos \theta\_3 - l\_2) + (l\_1 \cos \theta\_3 + l\_2) \, T\_1}{I\_1 l\_2 \cos \theta} \end{cases} \tag{12}$$

According to Equation (12), the clamping pressure *F*<sup>1</sup> and cutting force *F*<sup>3</sup> are related to *T*<sup>0</sup> and *T*1, whereas *θ*1, *θ*2, *θ*3, and other angles affect the motion position of the mechanism, as shown in Figure 8. The related parameters are listed in Table 1.

### *3.4. Parameter Determination of Each Rod of the Underactuated Mechanism*

In this paper, a stepper motor, model 57HD5401-110, with a torque of 1.2 N·m, was selected as the driving actuator of the manipulator, together with the 4 mm lead of a TBI high-precision linear screw. The thrust was calculated by Equation (13):

$$F\_T = \frac{2\pi n T\_{stepper}}{S},\tag{13}$$

The transmission efficiency *n* of the lead screw is generally 85–90%. Therefore, the thrust of the nut on the lead screw driven by the motor is 1601.4–1695.6 N.

When the motor is not driven, *ϕ*<sup>2</sup> in the unilateral underactuated mechanism is 120◦ and *θ* = 26◦ as shown in Figure 9. Therefore, according to the requirements of the design scheme, a torsional spring was placed in the distal knuckle *D* to restrict the relative rotation of the two joints. According to the picking requirements and previous design experience [31], rods *AB*, *BG*, and *LCF* were set as 50, 50, and 65 mm, respectively. The length of the bottom plate was set to 130 mm to avoid affecting the enveloping effect of the manipulator with a short-frame bottom plate. The length and width of the cutting blade were designed to be 80 and 50 mm, respectively, to ensure that the manipulator can cut at one time. The objective function was designed to optimize the length of rod *CF* and determine its parameters, as shown in Equation (14).

$$\begin{cases} \quad F\_2 l\_2 > T\_1\\ \quad F\_2 = F\_3 \cos \theta \end{cases} \tag{14}$$

When the unilateral underactuated mechanism contacts broccoli, rod *AD* remains stationary, and the driving force overcomes the binding force of the torsional spring and continues to drive the swing rod *DK* to move. At this time, *ϕ*<sup>2</sup> is 102◦, and the change of ∠*CDA* at joint *D* is 18◦. Therefore, the torsional spring torque *T*<sup>1</sup> is –1268.4 N·mm and the constraint condition is set as 30 N ≤ *F*<sup>3</sup> ≤ 35 N. The following results were obtained according to the constraints of *F*3.

$$\begin{cases} F\_3 = 35 \text{ N}, l\_2 > 38.81 \text{ mm} \\ F\_3 = 30 \text{ N}, l\_2 > 45.28 \text{ mm}' \end{cases}$$

Therefore, *l*<sup>2</sup> is rounded to 50 mm, and *LDK* is 50 mm.

Additionally, the objective function was designed according to the design scheme and picking requirements, as shown in Equation (15), to optimize and determine the length of rod *AD*.

$$T\_0 \ge 2F\_{\rm cy}l\_1 + F\_2l\_2 + T\_{1\prime} \tag{15}$$

Substituting the cutting force into the above equation and setting the constraint condition to 30 N ≤ *F*<sup>3</sup> ≤ 35 N obtained the following results:

$$\begin{cases} F\_3 = 30 \text{ N}, l\_1 \le 46.44 \text{ mm} \\ F\_3 = 35 \text{ N}, l\_1 \le 44.36 \text{ mm}^{-1} \end{cases}$$

The height of the plants left after cutting was ensured to be 12–15 cm, and the structure was kept compact by setting *l*<sup>1</sup> to 45 mm and rods *BC* and *AD* to 90 mm.

Substituting *l*<sup>1</sup> = 45 mm, *l*<sup>2</sup> = 50 mm, *T*<sup>0</sup> = 5357.13 N·mm, and *T*<sup>1</sup> = –1268.4 N·mm into Equation (12) yielded the following results:

$$\begin{cases} F\_1 = 23.08 \text{ N} \\ F\_2 = 32.57 \text{ N} \\ F\_3 = 34.89 \text{ N} \end{cases}$$

In this result, the flower ball surface pressure *F*<sup>1</sup> is 23.08 N, whereas the pressure range in the previous paper is 25–30 N. Therefore, this result is less than the pressure range and meets the requirement of low-loss picking. In addition, the cutting force *F*<sup>3</sup> in this result is 34.89 N, which conforms to the cutting force range of 30–35 N in the stem-cutting test and meets the requirements for picking.

Therefore, the parameters of each rod are shown in Table 2.

**Table 2.** Parameters of the connecting rods of the underactuated mechanism (unit: mm).


*3.5. Determination of Underactuated Mechanism Thrust*

*F*<sup>1</sup> and *F*<sup>2</sup> in Equation (11) can be obtained by Equation (16):

$$T\_0 = F\_1 l\_1 + (d \cos \theta\_{\, 3} + l\_2) F\_{2\prime} \tag{16}$$

The following can be obtained from Equation (8):

$$F\_{T\_0} = \frac{T\_0}{a \cos a \sin \left(\frac{\pi}{2} - a\right)},\tag{17}$$

Substituting *l*<sup>1</sup> = 45 mm, *l*<sup>2</sup> = 50 mm, *d* = 90 mm, *θ* = 21◦, *θ*<sup>3</sup> = 18◦, *α* = 55◦, 25 N ≤ *F*<sup>1</sup> ≤ 30 N, and 25 N ≤ *F*<sup>3</sup> ≤ 30 N into Equations (16) and (17) resulted in:

> ⎧ ⎪⎪⎨ ⎪⎪⎩ *F*<sup>1</sup> = 25 N, *F*<sup>2</sup> = 28.01 N, *FT*<sup>0</sup> = 299.28 N *F*<sup>1</sup> = 25 N, *F*<sup>2</sup> = 32.68 N, *FT*<sup>0</sup> = 337.77 N *F*<sup>1</sup> = 30 N, *F*<sup>2</sup> = 28.01 N, *FT*<sup>0</sup> = 312.95 N *F*<sup>1</sup> = 30 N, *F*<sup>2</sup> = 32.68 N, *FT*<sup>0</sup> = 351.44 N .

The manipulator has a symmetrical structure of the manipulator; therefore, thrust *FT* was calculated as shown in Equation (18):

$$F\_T = 2F\_{T\_0 \prime} \tag{18}$$

The required thrust *FT* range of the manipulator is 598.66–702.88 N. According to the result, the thrust of the motor drive nut is greater than that required by the underactuated mechanism:

$$F\_{\mathbf{r}} \rhd F\_T = 2F\_{T\_0}.\tag{19}$$

Therefore, the thrust of the stepper motor meets the requirements of picking and the requirements of the scheme design of the underactuated mechanism.
