Structure Description

The kinematically redundant parallel mechanism studied in this paper is shown in Figure 1. This parallel mechanism consists of a fixed base, an equilateral triangular moving platform, three kinematic limbs with the same structure and a lifting platform. Qu et al. [17] proved that the moving platform has three independent movements, and the mobility of the whole mechanism is equal to 4. The 4 prismatic joints connected to the base are selected as the actuating joints, respectively L1, L2, L3, and L4.

Figure 1
figure 1

One side of the limbs of the mechanism is connected to the base by a prismatic joint, and three prismatic joints are all actuating joints. The other side is connected to the moving platform through a spherical joint. The rotation joints of each limb are passive and are parallel to each other. Ai, Ci, Di and Ei (i = 1, 2, 3) are the center points of the rotating joints in each limb; Bi is the center point of the spherical joint in each limb; P4 is the center point of the lifting platform; the link BiCi connects to the link DiEi and passes through the rotating joint Di, and Di is a point of the link BiCi. Links AiCi, DiEi, and BiCi are denoted as link li1, link li2, and link li3 (i = 1, 2, 3), respectively. The coordinate systems Oxyz and Puvw are establish. The z-axis is perpendicular to the base; the x-axis is towards the actuating joint P1; the w-axis is perpendicular to the moving platform, and the u-axis is towards the joint B1.

Inverse Kinematics Solution

The detailed process to solve the inverse kinematics solution of the mechanism is shown in Figure 2.

Figure 2
figure 2

The inverse kinematics solution process of the mechanism

As shown in Figure 3, it is the geometric structure of the mechanism. The base coordinate system {O} is Oxyz with the y-axis pointing out of the paper and x-axis aligned with the axis of prismatic joint Pi. The moving platform coordinate system {P} is Puvw with the w-axis perpendicular to the u-axis and the v-axis.

Figure 3
figure 3

The geometric structure of the mechanism

The position of the center point P of the moving platform can be defined as shown in Eq. (1).

$${}^{O}{varvec{P}} = left[ {begin{array}{*{20}c} {x_{p} } & {y_{p} } & {z_{p} } \ end{array} } right]^{{text{T}}} ,$$

(1)

Besides, the rotation matrix ORP is used to define the position of the moving platform relative to the base coordinate system {O}, using the expression of pitch-roll-yaw, as shown in Eqs. (2) and (3).

$${}^{O}{varvec{R}}_{P} = left[ {begin{array}{*{20}c} {u_{x} } & {v_{x} } & {w_{x} } \ {u_{y} } & {v_{y} } & {w_{y} } \ {u_{z} } & {v_{z} } & {w_{z} } \ end{array} } right]{, }$$

(2)

$$begin{gathered} {}^{O}{varvec{R}}_{P} { = }{varvec{R}}_{z} left( gamma right){varvec{R}}_{y} left( beta right){varvec{R}}_{x} left( alpha right) = left[ {begin{array}{*{20}c} {cbeta cgamma } & {cgamma salpha sbeta – calpha sgamma } & {calpha cgamma sbeta + salpha sgamma } \ {cbeta sgamma } & {salpha sbeta sgamma + calpha cgamma } & {calpha sbeta sgamma – cgamma salpha } \ { – sbeta } & {cbeta salpha } & {calpha cbeta } \ end{array} } right], hfill \ hfill \ end{gathered}$$

(3)

where, α, β and γ are respectively expressed as Euler angles of the moving platform about the x, y and z axes. Therefore, according to Eq. (4), the midpoint’s representation on the moving platform in {O} can be obtained.

$${}^{O}P = {}^{O}{varvec{R}}_{P} {}^{P}varvec{P} + {varvec{OP}}user2{.}$$

(4)

The positions of the three spherical joints in the moving platform coordinate system {B} are expressed in Eq. (5).

$$left{ {begin{array}{*{20}c} {{}^{P}varvec{B}_{1} = left[ {begin{array}{*{20}c} r & 0 & 0 \ end{array} } right]^{{text{T}}} ,} \ {{}^{P}varvec{B}_{2} = left[ {begin{array}{*{20}c} { – r/2} & {sqrt 3 r/2} & 0 \ end{array} } right]^{{text{T}}} ,} \ {{}^{P}varvec{B}_{3} = left[ {begin{array}{*{20}c} { – r/2} & { – sqrt 3 r/2} & 0 \ end{array} } right]^{{text{T}}} .} \ end{array} } right.$$

(5)

According to Eq. (5), the transformation formula of the three spherical joints can be obtained, as shown in Eq. (6).

$${}^{O}varvec{B}_{i} = {}^{O}R_{P} {}^{P}varvec{B}_{i} + {varvec{OP}} .$$

(6)

Through calculation, the basic coordinate vector representation of Bi (i = 1,2,3) can be obtained, as shown in Eqs. (79).

$${}^{O}{varvec{B}}_{1} = left[ {begin{array}{*{20}c} {x_{p} + ru_{x} } \ {y_{p} + ru_{y} } \ {z_{p} + ru_{z} } \ end{array} } right] = left[ {begin{array}{*{20}c} {x_{p} + r{text{cos}}beta {text{cos}}gamma } \ {y_{p} + r{text{cos}}beta {text{sin}}gamma } \ {z_{p} – r{text{sin}}beta } \ end{array} } right],$$

(7)

$${}^{O}{varvec{B}}_{2} { = }left[ {begin{array}{*{20}c} {x_{p} – frac{1}{2}rleft( {cos beta cos gamma + sqrt 3 ( – cosgamma sin alpha sin beta + cos alpha sin gamma )} right)} \ {y_{p} – frac{1}{2}rcos beta sin gamma + frac{sqrt 3 }{2}r(cosalpha cos gamma + sin alpha sin beta sin gamma )} \ {z_{p} + frac{1}{2}r(sqrt 3 cosbeta sinalpha + sin beta )} \ end{array} } right],$$

(8)

$${}^{O}{varvec{B}}_{3} = left[ {begin{array}{*{20}c} {x_{p} + frac{1}{2}rleft( { – cos beta cos gamma + sqrt 3 ( – cosgamma sin alpha sin beta + cos alpha sin gamma )} right)} \ {y_{p} – frac{1}{2}rleft( {sqrt 3 cos alpha cos gamma + (cosbeta + sqrt 3 sin alpha sin beta sin gamma )} right)} \ {z_{p} + frac{1}{2}r( – sqrt 3 cosbeta sinalpha + sin beta )} \ end{array} } right].$$

(9)

The base coordinate vector representation of the point Ai (i = 1,2,3) is shown in Eq. (10).

$$left{ {begin{array}{*{20}c} {{}^{O}{varvec{A}}_{1} = left[ {begin{array}{*{20}c} {L_{1} } & 0 & {d_{1} } \ end{array} } right]^{{text{T}}} ,} \ {{}^{O}{varvec{A}}_{2} = left[ {begin{array}{*{20}c} { – 1/2L_{2} } & {sqrt 3 {/2}L_{2} } & {d_{1} } \ end{array} } right]^{{text{T}}} ,} \ {{}^{O}{varvec{A}}_{3} = left[ {begin{array}{*{20}c} { – 1/2L_{3} } & { – sqrt 3 {/2}L_{3} } & {d_{1} } \ end{array} } right]^{{text{T}}} .} \ end{array} } right.$$

(10)

According to the position coordinates of Ai and Bi, the distance between the joint Ai and the joint Bi can be calculated, as shown in Eq. (11).

$${varvec{A}}_{i} {varvec{B}}_{i} = {varvec{B}}_{i} – {varvec{A}}_{i} .$$

(11)

The vector si is set in three directions on the base, as shown in Eq. (12).

$$left{ {begin{array}{*{20}c} {{}^{O}{varvec{s}}_{1} = left[ {begin{array}{*{20}c} 0 & 1 & 0 \ end{array} } right]^{{text{T}}} ,} \ {{}^{O}{varvec{s}}_{2} = left[ {begin{array}{*{20}c} {sqrt 3 } & 1 & 0 \ end{array} } right]^{{text{T}}} ,} \ {{}^{O}{varvec{s}}_{3} = left[ {begin{array}{*{20}c} { – sqrt 3 } & 1 & 0 \ end{array} } right]^{{text{T}}} .} \ end{array} } right.$$

(12)

Because si is perpendicular to AiBi, the dot product of the two vectors is equal to 0, and the expressions of xp, yp, and γ can be obtained, as shown in Eqs. (13)–(15).

$$x_{p} = – frac{{rleft( {cos^{2} alpha – cos^{2} beta + sin^{2} alpha sin^{2} beta } right)}}{{2sqrt {left( {1 + cos alpha cos beta } right)^{2} } }},$$

(13)

$$y_{p} = – frac{rcos beta sin alpha sin beta }{{sqrt {left( {1 + cos alpha cos beta } right)^{2} } }},$$

(14)

$$gamma = arctan left[ {frac{cos alpha + cos beta }{{sqrt {left( {1 + cos alpha cos beta } right)^{2} } }},frac{sin alpha sin beta }{{sqrt {left( {1 + cos alpha cos beta } right)^{2} } }}} right].$$

(15)

According to the above method, the coordinate representation of the remaining joints on the limb can be obtained sequentially.

The inverse solution of the mechanism is that the parameters α, β and zp of the moving platform are known, and the actuating lengths L1, L2 and L3 of the input prismatic joints are calculated. According to the vector closed-loop principle, a vector closed-loop equation is established in a limb of this mechanism, as shown in Figure 4. The four vector combinations OA1, A1C1, C1B1, B1O are selected as the closed-loop.

Figure 4
figure 4

According to Figure 4, the closed-loop equation 1 of limb 1 can be obtained, as shown in Eq. (16).

$${varvec{OP}}_{1} + {varvec{P}}_{1} {varvec{A}}_{1} + {varvec{A}}_{1} {varvec{C}}_{1} + {varvec{C}}_{1} {varvec{B}}_{1} = {varvec{OB}}_{1} .$$

(16)

Then Eq. (16) is expressed with the structural parameters of the mechanism, as shown in Eq. (17).

$$left[ {begin{array}{*{20}c} {x_{a1} } \ 0 \ {z_{a1} } \ end{array} } right] + left[ {begin{array}{*{20}c} {a_{1} cos theta_{11} } \ 0 \ {a_{1} sin theta_{11} } \ end{array} } right] + left[ {begin{array}{*{20}c} { – left( {a_{2} + a_{3} } right)cos theta_{12} } \ 0 \ {left( {a_{2} + a_{3} } right)sin theta_{12} } \ end{array} } right] = left[ {begin{array}{*{20}c} {x_{b1} } \ 0 \ {z_{b1} } \ end{array} } right].$$

(17)

Because the mechanism is a parallel mechanism with kinematic redundancy, its redundant structure is a lifting platform. The two closed-loop equations of the limb are solved simultaneously, as shown in Figures 5 and 6.

Figure 5
figure 5
Figure 6
figure 6

According to Figure 5, the closed-loop equation 2 of limb 1 can be obtained as Eq. (18).

$${varvec{OP}}_{1} + {varvec{P}}_{1} {varvec{A}}_{1} + {varvec{A}}_{1} {varvec{C}}_{1} + {varvec{C}}_{1} {varvec{D}}_{1} + {varvec{D}}_{1} {varvec{E}}_{1} = {varvec{OE}}_{1} .$$

(18)

Then Eq. (18) is expressed with the structural parameters of the mechanism, as shown in Eq. (19).

$$left[ {begin{array}{*{20}c} {x_{a1} } \ 0 \ {z_{a1} } \ end{array} } right] + left[ {begin{array}{*{20}c} {a_{1} cos theta_{11} } \ 0 \ {a_{1} sin theta_{11} } \ end{array} } right] + left[ {begin{array}{*{20}c} { – a_{2} cos theta_{12} } \ 0 \ {a_{2} sin theta_{12} } \ end{array} } right] + left[ {begin{array}{*{20}c} { – a_{4} cos theta_{13} } \ 0 \ { – a_{4} sin theta_{13} } \ end{array} } right] = left[ {begin{array}{*{20}c} {x_{e1} } \ 0 \ {z_{e1} } \ end{array} } right].$$

(19)

As shown in Figure 6, the closed-loop equation 3 of limb 1 can be obtained, as shown in Eq. (20).

$${varvec{E}}_{1} {varvec{D}}_{1} + {varvec{D}}_{1} {varvec{B}}_{1} = {varvec{E}}_{1} {varvec{B}}_{1} .$$

(20)

Then Eq. (20) is expressed with the structural parameters of the mechanism, as shown in Eq. (21).

$$left[ {begin{array}{*{20}c} {a_{4} cos theta_{13} } \ 0 \ {a_{4} sin theta_{13} } \ end{array} } right] + left[ {begin{array}{*{20}c} { – a_{3} cos theta_{12} } \ 0 \ {a_{3} sin theta_{12} } \ end{array} } right] = left[ {begin{array}{*{20}c} {x_{b1} – x_{e1} } \ 0 \ {z_{b1} – z_{e1} } \ end{array} } right].$$

(21)

Eqs. (16)–(21) are expanded to get six equations, as shown in Eq. (22).

$$left{ {begin{array}{*{20}c} {x_{a1} + a_{1} cos theta_{11} – left( {a_{2} + a_{3} } right)cos theta_{12} = x_{b1} , } \ {z_{a1} + a_{1} sin theta_{11} + left( {a_{2} + a_{3} } right)sin theta_{12} = z_{b1} , } \ {x_{a1} + a_{1} cos theta_{11} – a_{2} cos theta_{12} – a_{4} cos theta_{13} = x_{e1} , } \ {z_{a1} + a_{1} sin theta_{11} + a_{2} sin theta_{12} – a_{4} sin theta_{13} = z_{e1} , } \ {a_{4} cos theta_{13} – a_{3} cos theta_{12} = x_{b1} – x{}_{e1}, } \ {a_{4} sin theta_{13} + a_{3} sin theta_{12} = z_{b1} – z_{e1} . } \ end{array} } right.$$

(22)

According to the coordinate expressions of B1 and E1, D1 is taken as the center of the circle, B1D1, and E1D1 as the radius. The equations of the two circles are listed respectively, and the coordinates of D1 can be solved, as shown in the Eqs. (23) and (24).

$$left( {x_{d1} – x_{b1} } right)^{2} + left( {z_{d1} – z_{b1} } right)^{2} = a_{3}^{2} ,$$

(23)

$$left( {x_{d1} – x_{e1} } right)^{2} + left( {z_{d1} – z_{e1} } right)^{2} = a_{4}^{2} .$$

(24)

According to the coordinates of B1 and D1 and the tangent formula of the triangle, the size of θ13 can be solved, as shown in Eq. (25).

$$theta_{13} = arctan frac{{z_{d1} – z_{e1} }}{{x_{d1} – x_{e1} }}.$$

(25)

According to the closed-loop equation (22), the size of θ12 can be solved, as shown in Eq. (26).

$$theta_{12} = arccos frac{{ – x_{b1} + x_{e1} + a_{4} cos theta_{13} }}{{a_{3} }}.$$

(26)

Simultaneously eliminating θ12 in Eqs. (22) and (27) can be obtained.

$$begin{gathered} a_{1}^{2} + left( {x_{a1} – x_{b1} } right)^{2} + 2a_{1} left( {x_{a1} – x_{b1} } right)cos theta_{11} + left( {z_{a1} – z_{b1} } right)^{2} { + }2a_{1} left( {z_{a1} – z_{b1} } right)sin theta_{11} = left( {a_{2} + a_{3} } right)^{2} . hfill \ hfill \ end{gathered}$$

(27)

According to Eq. (27), the expression of θ11 can be obtained, as shown in Eq. (28).

$$begin{gathered} theta_{11} = arctan hfill \ left( {frac{{ – a_{1} left( {x_{a1} – x_{b1} } right)left( {a_{1}^{2} – left( {a_{2} + a_{3} } right)^{2} + left( {x_{a1} – x_{b1} } right)^{2} + left( {z_{a1} – z_{b1} } right)^{2} } right)}}{{a_{1}^{2} left( {left( {x_{a1} – x_{b1} } right)^{2} + left( {z_{a1} – z_{b1} } right)^{2} } right)}}} right. + hfill \ left. begin{gathered} frac{{sqrt { – a_{1}^{2} left( {left( { – a_{1} + a_{2} + a_{3} } right)^{2} – left( {x_{a1} – x_{b1} } right)^{2} – left( {z_{a1} – z_{b1} } right)^{2} } right)} }}{{a_{1}^{2} left( {left( {x_{a1} – x_{b1} } right)^{2} + left( {z_{a1} – z_{b1} } right)^{2} } right)}} hfill \ frac{{sqrt {left( {left( {a_{1} + a_{2} + a_{3} } right)^{2} – left( {x_{a1} – x_{b1} } right)^{2} – left( {z_{a1} – z_{b1} } right)^{2} } right)left( {z_{a1} – z_{b1} } right)^{2} } }}{{a_{1}^{2} left( {left( {x_{a1} – x_{b1} } right)^{2} + left( {z_{a1} – z_{b1} } right)^{2} } right)}} hfill \ end{gathered} right). hfill \ end{gathered}$$

(28)

According to Eq. (22), only the actuating parameter L1 is unknown in the equation. By solving the equation, the expression equation of L1 can be obtained, as shown in Eq. (29).

$$L_{1} = x_{e1} – a_{1} cos theta_{11} – a_{2} cos theta_{12} – a_{4} cos theta_{13} .$$

(29)

In the same way, the expressions for actuating lengths L2 and L3 can be obtained by the closed-loop equations of limb 2 and limb 3, as shown in Eqs. (30) and (31).

$$L_{2} = – 2x_{e2} – a_{1} cos theta_{21} + a_{2} cos theta_{22} + a_{4} cos theta_{23} ,$$

(30)

$$L_{3} = – 2x_{e3} – a_{1} cos theta_{31} + a_{2} cos theta_{32} + a_{4} cos theta_{33} .$$

(31)

Singularity Analysis

The positive Jacobian matrix Jq is the relationship between the twist χ of the motion platform expressed by the six-dimensional vector and the rate of changing Li of the actuating length, as shown in Eq. (32).

$${varvec{L}}_{i} { = }{varvec{J}}_{q} {varvec{chi}}user2{.}$$

(32)

The Jacobian matrix of the parallel mechanism with kinematic redundancy is the relationship between the angular velocity ω of the moving platform and the vector L1, L2, and L3 of the actuating joint change rate, as shown in Eq. (33).

$${varvec{L}} = {varvec{J}} cdot {varvec{omega}}user2{.}$$

(33)

The relationship among the inverse Jacobian matrix Jx, the three-dimensional twist vector χ of the moving platform, and the angular velocity ω is shown in Eq. (34).

$${varvec{chi}}{ = }left[ {begin{array}{*{20}c} {{}^{A}{varvec{v}}_{p} } \ {varvec{omega}} \ end{array} } right]{ = }{varvec{J}}_{x} {varvec{omega}}user2{.}$$

(34)

where χ represents the moving platform’s twist; Avp is the speed of the moving platform, and ω is the angular velocity of the moving platform.

The closed-loop 2 of the limb1, 2, 3 is selected. According to the result of the inverse solution, the constraint equations of limbs 1, 2, and 3 are shown in Eqs. (35)–(37).

$$begin{gathered} left( {x_{a1} – x_{e1} – a_{2} times cos theta_{12} – a_{4} cos theta_{13} } right)^{2} + left( {z_{a1} – z_{e1} + a_{2} sin theta_{12} – a_{4} sin theta_{13} } right)^{2} = a_{1}^{2} , hfill \ hfill \ end{gathered}$$

(35)

$$begin{gathered} left( {2x_{a2} – 2x_{e2} – a_{2} cos theta_{22} + a_{4} cos theta_{23} } right)^{2} + left( {z_{a2} – z_{e2} + a_{2} sin theta_{22} – a_{4} sin theta_{23} } right)^{2} = a_{1}^{2} , hfill \ hfill \ end{gathered}$$

(36)

$$begin{gathered} left( {2x_{a3} – 2x_{e3} + a_{2} cos theta_{32} + a_{4} cos theta_{33} } right)^{2} + left( {z_{a3} – z_{e3} + a_{2} sin theta_{32} – a_{4} sin theta_{33} } right)^{2} = a_{1}^{2} . hfill \ hfill \ end{gathered}$$

(37)

The partial derivative of the time t is calculated in the actuating parameters L1, L2, L3, and the moving platform motion parameters α, β, zp. After simplification, the Eqs. (38)–(40) can be obtained.

$$e_{1} = e_{11} times dot{alpha } + e_{12} times dot{beta } + e_{13} times dot{z}_{p} + e_{14} times dot{L}_{1} ,$$

(38)

$$e_{2} = e_{21} times dot{alpha } + e_{22} times dot{beta } + e_{23} times dot{z}_{p} + e_{24} times dot{L}_{2} ,$$

(39)

$$e_{3} = e_{31} times dot{alpha } + e_{32} times dot{beta } + e_{33} times dot{z}_{p} + e_{34} times dot{L}_{3} .$$

(40)

In Eqs. (38)–(40), eij (i = 1,2,3 and j = 1,2,3,4) respectively represent the result of the combination of the same type and the simplified expressions of the result in Eqs. (35)–(37). Because these expressions are too long, they are not shown in the paper.

Extracting the common parameter items in the above equations, the positive Jacobian matrix Jq and the inverse Jacobian matrix Jx can be obtained as Eqs. (41) and (42).

$${varvec{J}}_{q} = left[ {begin{array}{*{20}c} {e_{14} } & 0 & 0 \ 0 & {e_{24} } & 0 \ 0 & 0 & {e_{34} } \ end{array} } right],$$

(41)

$${varvec{J}}_{x} = left[ {begin{array}{*{20}c} {e_{11} } & {e_{12} } & {e_{13} } \ {e_{21} } & {e_{22} } & {e_{23} } \ {e_{31} } & {e_{32} } & {e_{33} } \ end{array} } right].$$

(42)

The singularity of the mechanism is solved. The structural parameters of the mechanism are set, d1 = 30 mm, a1 = 114 mm, a2 = 92 mm, a3 = 115 mm, a4 = 132 mm, d4 = 62.68 mm, r = 78 mm. The range of zp is set as [150, 210] mm, and the range of α and β is set as [−2, 2] rad.

Let Det(Jx) = 0, the actuating parameter L4 is taken as the structure parameter of the mechanism. When L4  =  80 mm, the height of the lifting platform is located between the base and the moving platform, so that the mechanism as a whole maintains a normal working state. When the actuating joint changes, the height of the lifting platform will not have a significant impact on the position and posture of the mechanism. And the singularity diagram of the mechanism under L4  =  80 mm is measured, as shown in Figure 7.

Figure 7
figure 7

Singular diagram of mechanism when L4 = 80 mm

Set zp = 150/170/190/ 210 mm, respectively, and the positive motion singular images of the mechanism under different values of zp are measured, as shown in Figure 8.

Figure 8
figure 8

The singularity diagram of the mechanism when zp takes different values

By analyzing the Jacobian matrix and singular image of the mechanism, the singular position of the mechanism can be obtained [31,32,33]. When θi1 = π/2 and θi2 = 0°of a limb, the mechanism will have a singular position 1, as shown in Figure 20a. When θi1 = θi2 = π/2 of the three limbs, the mechanism will have a singular position 2, as shown in Figure 24a. When θi1 = 0° of a limb, the mechanism produces a singular position 3, as shown in Figure 26a. The redundant components of the mechanism called lifting platform, its position will affect the singularity of the mechanism, which will be analyzed in the content.

Workspace Analysis

The structural parameters of the mechanism are set, d1 = 30 mm, a1 = 114 mm, a2 = 92 mm, a3 = 115 mm, a4 = 132 mm, d4 = 62.68 mm, r = 78 mm.

The workspace of the mechanism is analyzed by adopting the workspace search method. The length of the actuating joint L4 is set as a fixed value, and the point set is obtained where the workspace satisfies the inverse solution of the mechanism. Finally, we can get the workspace diagram of the mechanism.

Analyzing limb 1, according to the Eqs. (25)–(28) of θ11, θ12, θ13 obtained by the inverse solution, substituting the above three equations into Eq. (29), we can get L1’s expression only have the moving platform kinematic parameters include α, β, and zp. Similarly, we can get L1, and L2’s expression only has the moving platform kinematic parameters including α, β, and zp.

The value range of L1, L2, and L3 is set as [50, 200] mm; the range of zp is set as [50, 350] mm, and the value range of α and β is set to [−π/2, π/2]. L4 is used as a structural parameter, and the workspace of the mechanism can be calculated when L4 = 90 mm.

The surface image [34] of the workspace of the analyzed mechanism can be obtained, as shown in Figure 9, by using the method of fitting a surface with a set of points and fitting the workspace point set of the mechanism.

Figure 9
figure 9

Three-dimensional curve diagram of the workspace

Based on the obtained workspace in Figure 9, the workspace diagram of the mechanism with zp = 230/250/270 /290 mm can be sketched, as shown in Figure 10. The moving platform parameters α and β are variables, and zp is a fixed value.

Figure 10
figure 10

Workspace of mechanism with different values of zp

By analyzing the workspace of the mechanism, we can get the changes of the moving platform kinematics parameters α and β when the moving platform parameter zp is a fixed value, as shown in Table 1.

Table 1 Changes of kinematic parameters α and β of the moving platform

Table 1 shows that when the moving platform is in working condition, the maximum angle α and β of its rotation around the x and y axes can reach 66°. The moving platform of this mechanism has the characteristics of a large rotation angle compared with the general parallel mechanism, which can meet the needs of specific working conditions.

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