2.1 Design of deployable unit

A single-closed-loop rigid origami mechanism model is proposed, which distributes along an equilateral triangle. The model consists of three congruent regular hexagon panels and six congruent rectangular panels. And it assumes that the thickness of the panel is zero and the stiffness is infinite. The centroid of hexagon panels S _U (U = 1,2,3) coincides with the three vertices of the equilateral triangle, and the three edges of the triangle can be regarded as three kinematic chains L _U connected by two rectangular panels F _U1 and F _U2 through the revolute joints R _UV (V = 1,2,3) with two adjacent regular hexagonal panels. The connection relationship of each component is shown in Fig. 1.

Figure 1.

Single-closed-loop origami mechanism.

Based on the axis-shift approach, it is possible to convert the origami mechanism shown in Fig. 1 into a thick panel model, as seen in Fig. 2a. The DOF of the deployable unit is analysed by using the screw theory [Reference Huang, Zhao and Zhao43]. To begin, the upper surface of panel S₁ is designated as the OXY plane, taking the intersection of R₁₁ and R₃₃ as the origin O, the X-axis coincides with the axis of the revolute joint R₃₃, the Z-axis is perpendicular to the OXY plane upwards, and that of the Y-axis is determined by the right-hand rule.

Figure 2.

Single-closed-loop deployable mechanism with thick panels.

The projection of the mechanism in the OXY plane is shown in Fig. 2b. According to the particular geometry and connection relationship of the components in the mechanism, the sufficient and necessary condition for the mechanism to form a closed loop is

(1) \begin{equation}\sum\limits_{U = 1}^3 {\angle {{\rm{A}}_U}{{\rm{B}}_U}{{\rm{C}}_U} = {{360}^ \circ }} \end{equation}

$\angle {{\rm{A}}_U}{{\rm{B}}_U}{{\rm{C}}_U}$ is derived from the projection of the panels S _i on the OXY plane, according to the basic properties of the projection, the value of ∠A _U B _U C _U satisfy

(2) \begin{equation}\angle {{\rm{A}}_U}{{\rm{B}}_U}{{\rm{C}}_U} \le 120^\circ \end{equation}

Since the upper surface of the panel S₁ is coplanar with the OXY plane, it can be obtained that

(3) \begin{equation}\angle {{\rm{A}}_1}{{\rm{B}}_1}{{\rm{C}}_1} = 120^\circ \end{equation}

If the panel S₂ is not parallel to the OXY plane, it can be obtained that

(4) \begin{equation}\angle {{\rm{A}}_2}{{\rm{B}}_2}{{\rm{C}}_2} \lt 120^\circ \end{equation}

From Equations (1), (3) and (4), it can be obtained that

(5) \begin{equation}\angle {{\rm{A}}_3}{{\rm{B}}_3}{{\rm{C}}_3} \gt 120^\circ \end{equation}

Equation (5) contradicts with Equation (2), thus proving that the S₂ is always parallel to the S₁ during the movement of the mechanism. Similarly, it can be proved that the S₃ is also always parallel to the panel S₁.

Alternatively, the deployable unit can be regarded as a spatial parallel mechanism having two limbs, with S₁ being a fixed platform and S₂ being a moving platform. The limb 1 can be regarded as the kinematic chain L₁, which contains three revolute joints R₁₁, R₁₂ and R₁₃. And the limb 2 contains the kinematic chain L₂ and L₃ connected in series, comprising six revolute joints of R₃₃, R₃₂, R₃₁, R₂₃, R₂₂ and R₂₁.

The screw system of limb 1 is

(6) \begin{align}\left\{ \begin{array}{c@{\quad}c@{\quad}c@{\quad}c@{\quad}c@{\quad}c@{\quad}c}{{\boldsymbol{$}}_{11}^{}} = ( 1 & {\sqrt 3 } & 0 & ; & 0 & 0 & 0 )\\[4pt]{{\boldsymbol{$}}_{12}^{}} = ( 1 & {\sqrt 3 } & 0 & ;& { - \sqrt 3 l{\rm{s}}{\gamma _{11}}}& {l{\rm{s}}{\gamma _{11}}}& {l{\mathop{\rm c}\nolimits} {\gamma _{11}}} )\\[4pt]{{\boldsymbol{$}}_{13}^{}} = ( 1 & {\sqrt 3 } & 0 & ; & { - \sqrt 3 l\left( {{\rm{s}}{\gamma _{11}} - {\mathop{\rm s}\nolimits} {\gamma _{12}}} \right)}& {l\!\left( {{\rm{s}}{\gamma _{11}} - {\mathop{\rm s}\nolimits} {\gamma _{12}}} \right)}& {2l\!\left( {{\mathop{\rm c}\nolimits} {\gamma _{11}} + {\mathop{\rm c}\nolimits} {\gamma _{12}}} \right)} )\end{array} \right.\end{align}

The screw system of limb 2 is

(7) \begin{equation}\left\{ \begin{array}{lc@{\quad}c@{\quad}c@{\quad}c@{\quad}c@{\quad}c@{\quad}c}{{\boldsymbol{$}}_{21}^{} } =& ( 1 & { - \sqrt 3 } & 0 & ; & { - 2\sqrt 3 l{\mathop{\rm s}\nolimits} {\gamma _{11}}} & {2l{\mathop{\rm s}\nolimits} {\gamma _{11}}} & { - l\!\left( {{\mathop{\rm c}\nolimits} {\gamma _{11}} + {\mathop{\rm c}\nolimits} {\gamma _{12}}} \right)} )\\[4pt]{{\boldsymbol{$}}_{22}^{} } =& ( 1 & { - \sqrt 3 } & 0 & ; & {\sqrt 3 l{A_\gamma }} & {l{A_\gamma }} & {l{B_\gamma }} )\\[4pt]{{\boldsymbol{$}}_{23}^{} } =& ( 1 & { - \sqrt 3 } & 0 & ; & {\sqrt 3 l\!\left( {{\mathop{\rm s}\nolimits} {\gamma _{32}} - {\mathop{\rm s}\nolimits} {\gamma _{31}}} \right)} & {l\!\left( {{\mathop{\rm s}\nolimits} {\gamma _{32}} - {\mathop{\rm s}\nolimits} {\gamma _{31}}} \right)} & {2l\!\left( {{\mathop{\rm c}\nolimits} {\gamma _{31}} + {\mathop{\rm c}\nolimits} {\gamma _{32}}} \right)} )\\[4pt]{{\boldsymbol{$}}_{31}^{} } =& ( { - 1} & 0 & 0 & ; & 0 & {l\!\left( {{\mathop{\rm s}\nolimits} {\gamma _{32}} - {\mathop{\rm s}\nolimits} {\gamma _{31}}} \right)} & { - l\!\left( {{\mathop{\rm c}\nolimits} {\gamma _{31}} + {\mathop{\rm c}\nolimits} {\gamma _{32}}} \right)} )\\[4pt]{{\boldsymbol{$}}_{32}^{} } =& ( { - 1} & 0 & 0 & ; & 0 & { - l{\mathop{\rm s}\nolimits} {\gamma _{32}}} & { - l{\mathop{\rm c}\nolimits} {\gamma _{32}}} )\\[4pt]{{\boldsymbol{$}}_{33}^{} } = & ( { - 1} & 0 & 0 & ; & 0 & 0 & 0 )\end{array} \right.\end{equation}

where

\begin{equation*}l = \sqrt {{d^2} + {c^2}} \end{equation*}

\begin{equation*}{\gamma _{i1}} = {\tau _{i1}} - \arctan \frac{d}{c}\end{equation*}

\begin{equation*}{\gamma _{i2}} = {\tau _{i2}} - \arctan \frac{d}{c}\end{equation*}

\begin{equation*}{A_\gamma } = {\rm{s}}{\gamma _{22}} - {\rm{s}}{\gamma _{31}} + {\rm{s}}{\gamma _{32}}\end{equation*}

\begin{equation*}{B_\gamma } = {\mathop{\rm c}\nolimits} {\gamma _{31}} + {\mathop{\rm c}\nolimits} {\gamma _{32}} - 2{\mathop{\rm c}\nolimits} {\gamma _{22}}\end{equation*}

τ _U1 and τ _U2 represent the acute angles between the panels F _U1, F _U2 and the OXY plane respectively; sγ and cγ represent sinγ and cosγ; c and d are the length and thickness of the rectangular panels.

From the Equation (5), it can be seen that the motion screw is linear independent, and the rank of the screw system of limb 1 is 3. According to Equation (7), $\boldsymbol{$}$ ₂₁ can be expressed as

(8) \begin{equation}{\boldsymbol{$}}_{21}^{} = {C_\gamma }\!\left( {{\mathop{\rm s}\nolimits} {\gamma _{31}}{\boldsymbol{$}}_{12}^{} - \left( {{\mathop{\rm s}\nolimits} {\gamma _{31}} - {\mathop{\rm s}\nolimits} {\gamma _{32}}} \right) {\boldsymbol{$}}_{32}^{} - {\mathop{\rm s}\nolimits} {\gamma _{32}}{\boldsymbol{$}}_{31}^{}} \right) + \left( {{D_\gamma } - 1} \right){\boldsymbol{$}}_{23}^{} - {D_\gamma }{\boldsymbol{$}}_{22}^{}\end{equation}

where

\begin{align*}{C_\gamma } & = \frac{{ {\mathop{\rm c}\nolimits} {\gamma _{11}} {\mathop{\rm s}\nolimits} {\gamma _{22}} - 4 {\mathop{\rm c}\nolimits} {\gamma _{22}} {\mathop{\rm s}\nolimits} {\gamma _{11}} + {\mathop{\rm c}\nolimits} {\gamma _{12}} {\mathop{\rm s}\nolimits} {\gamma _{22}} - 2 {\mathop{\rm c}\nolimits} {\gamma _{22}} {\mathop{\rm s}\nolimits} {\gamma _{31}} + {\mathop{\rm c}\nolimits} {\gamma _{31}}{\mathop{\rm s}\nolimits} {\gamma _{22}} + 2 {\mathop{\rm c}\nolimits} {\gamma _{22}}{\mathop{\rm s}\nolimits} {\gamma _{32}} + {\mathop{\rm c}\nolimits} {\gamma _{32}}{\mathop{\rm s}\nolimits} {\gamma _{22}}}}{{2 {\mathop{\rm s}\nolimits} \left( {{\gamma _{31}} + {\gamma _{32}}} \right) {\mathop{\rm s}\nolimits} {\gamma _{22}}}}\\[5pt]{D_\gamma } & = \frac{{2 {\mathop{\rm s}\nolimits} {\gamma _{11}} + {\mathop{\rm s}\nolimits} {\gamma _{31}} - {\mathop{\rm s}\nolimits} {\gamma _{32}}}}{{ {\mathop{\rm s}\nolimits} {\gamma _{22}}}}\end{align*}

Therefore, there are five linearity-independent screws in limb 2, which has six screws. According to ${\boldsymbol{$}} \circ {{\boldsymbol{$}}^r} = 0$ [Reference Huang, Zhao and Zhao43], it can be solved that the motion screw systems of two limbs have the same reciprocal screw:

\begin{equation*}{\boldsymbol{$} ^r} = \ \left( {0\ 0\ 0\ ;\ 0\ 0\ 1} \right)\end{equation*}

Hence, it is observed that the common constraint λ = 1. Furthermore, due to the presence of six motion screws in the motion screw system of limb 2, with a rank of 5, it can be deduced that there exists a passive DOF. In accordance with the modified G-K formula, the DOF of the mechanism can be determined:

\begin{align*}M & = \left( {6 - \lambda } \right)\left( {q - \varepsilon - 1} \right) + \sum\limits_{t = 1}^g {{T_e} - \xi } \\[4pt] & = \left( {6 - 1} \right) \times \left( {9 - 9 - 1} \right) + 9 - 1\\[4pt] & = 3\end{align*}

where λ represents the number of common constraints, q represents the number of parts of the mechanism containing the frame, $\varepsilon$ represents the number of kinematical pairs, T _e is the DOF of the eth motion pair, ξ represents the passive DOF.

Figure 3.

The construction of tetrahedral deployable unit (TDU).

In the process of modular expansion, the DOF of the deployable unit is usually required to be as small as possible. A method to reduce the DOF from three to one is proposed as shown in Fig. 3a. Revolute joints ^TR₁₁, ^TR₂₁ and ^TR₃₁ are used to connect rod G₁, G₂ and G₃ on panels S₁, S₂ and S₃, respectively. The other end of rod G₁, G₂ and G₃ is connected to the panel E through revolute joints ^TR₁₂, ^TR₂₂ and ^TR₃₂. The angular between the axes of the revolute joints ^TR₁₁, ^TR₂₁ and ^TR₃₁, as well as the angular between the axes of the revolute joints ^TR₁₂, ^TR₂₂ and ^TR₃₂, are 60°. The revolute joints on the same rod are parallel to each other. Figure 3 illustrates the equivalence of panel F to rod F, panel S to floral disc S, and panel E to floral disc E, leading to the creation of an equivalent rod model. In order to minimise the space occupied after the deployable unit is folded, the rod F is designed to be folded upward to obtain the mechanism shown in Fig. 3c. The S₁, S₂ and S₃ are always coplanar during the movement of the mechanism, and the △S₁S₂S₃ composed of its corresponding centroid is always an equilateral triangle, as shown in Fig. 4 [Reference Xu and Guan44]. In the Fig. 3c, c and g represent the length of F and G, c _p and g _p represent the length between the centroid and the revolute pair of F and G, respectively. In the Fig. 3d, r _s represents the distance between the centroid of the floral discs and the axis of the revolute joint, $\delta$ represents the acute angle between the F and the S, $\beta$ represents the acute angle between the G and the S.

2.2 Modular networking of deployable unit

A deployable mechanism with a large scale can be constructed by interconnecting a sequence of deployable units in accordance with specific theory of mechanism design. According to the motion characteristics of the TDU, a grid-based networking construction method is proposed as follows: firstly, as shown in Fig. 5, the deployable unit is represented by an equilateral triangle, where three vertices S₁, S₂ and S₃ of the triangle represent the centroid of the panels, respectively. Secondly, by sharing triangular vertices and adding RRR-constrained kinematic chains, a large-scale deployable mechanism can be formed so that the motion between modules can be transmitted without changing the DOF. Lastly, as shown in Fig. 6, according to the principle of planar mosaic, the plane can be divided into several graphic regions with the same area by using regular triangles, in which each regular triangle is a deployable unit. In the Fig. 6, ‘○’ represents the floral discs, and ‘—’ represents the generalised kinematic chain between the floral discs, which is similar to the topological graph of the mechanism.

Figure 4.

Deployable sequence of TDU.

Figure 5.

The sequence of rotational expansion.

Figure 6.

Schematic diagram of radial expansion.

Due to △S₁S₂S₃ is a regular triangle, the lengths of its sides, namely S₁S₂, S₁S₃ and S₂S₃, will remain equal during the motion of the mechanism. Consequently, the RRR-constraint kinematic chain can be considered equivalent to a prismatic joint, as shown in Fig. 7b. The same equivalent unit is constructed between two units sharing the same vertex and two kinematic chains, so that the motion can be transmitted and the DOF of the expanded mechanism is maintained at 1. More shapes can be generated via the expansion method described above, as shown in Fig. 8.

Figure 7.

Equivalent expansion mechanism.

Figure 8.

The supporting mechanism.

The equipment is arranged at the centre O of the supporting mechanism shown in the Fig. 9, and the shading membrane is fixed on the supporting mechanism. During the launch phase of the spacecraft, the supporting mechanism is fully folded, and the shading membrane is folded according to specific pre-compression creases [Reference Kamaliya, Shukla, Upadhyay and Mallikarachchi45–Reference Arya and Pellegrino47]. During the work phase, the supporting mechanism deploys and lifts the shading membrane.

Figure 9.

The rendering of sunshield.

3.1 Centroid motion analysis

The supporting mechanism can be divided into six parts with a regular triangle profile, as shown in Fig. 6b. The initial part is represented by Fig. 10, whereas the six parts can be derived by rotating (K–1)π/3(K = 1∼6) rad in relation to the initial part. In the odd part (K = 1, 3 and 5) and the even part (K = 2, 4 and 6), the arrangement of E and G are different, as shown in Fig. 11.

Figure 10.

Schematic diagram of initial part of the n-ring supporting mechanism.

Figure 11.

The layout diagram of floral discs E and rod G in odd and even parts.

The position of any node (triangle vertex or the centroid of floral discs S) can be represented by vector S _i,j (i $\in$ [0, n], j $\in$ [1, n+1]), that is

(9) \begin{equation}{{\boldsymbol{S}}_{i,j}} = {\left[ {{X_{{{\rm{s}}_{i,j}}}},{Y_{{{\rm{s}}_{i,j}}}},{Z_{{{\rm{s}}_{i,j}}}}} \right]^{\rm{T}}}\end{equation}

The centroid of the S_0,1 coincides with the origin. The internal centroid of floral discs of the initial part can be expressed by the nodes on the two edges starting from the origin as follows:

(10) \begin{equation}{{\boldsymbol{S}}_{i,j}} = {{\boldsymbol{S}}_{i - j + 1,1}} + {{\boldsymbol{S}}_{j - 1,j}}\end{equation}

where

\begin{equation*}{{\boldsymbol{S}}_{i - j + 1,1}} = {\left[ {\sqrt 3 \left( {c\cos \delta + {r_S}} \right)\left( {i - j + 1} \right),\left( {c\cos \delta + {r_S}} \right)\left( {i - j + 1} \right),0} \right]^{\rm{T}}}\end{equation*}

\begin{equation*}{{\boldsymbol{S}}_{j - 1,j}} = {\left[ {0,2\!\left( {j - 1} \right)\left( {{r_S} + c\cos \delta } \right),0} \right]^{\rm{T}}}\end{equation*}

According to Fig. 10, each part of the mechanism contains two kinds of triangles with different layout arrangements as shown in Fig. 11. Let △S _i+1,j S _i+1,j+1S _i,j and △S _i+1,j S _i,j S _i+1,j+1 be named D _{i+1, j} and ^pD _{i, j} , respectively. Each side of D _{i+1, j} contains two rods F, whose position vector are

(11) \begin{equation}\left\{ \begin{array}{l}{}^{11}{{\boldsymbol{F}}_{i + 1,j}} = {{\boldsymbol{S}}_{i,j}} + {\left[ {\dfrac{{\sqrt 3 }}{2}\left( {{r_s} + {\xi _c}c\cos \delta } \right),\dfrac{1}{2}\left( {{r_s} + {\xi _c}c\cos \delta } \right), - {\xi _c}c\sin \delta } \right]^{\rm{T}}}\\[15pt]{}^{12}{{\boldsymbol{F}}_{i + 1,j}} = {{\boldsymbol{S}}_{i,j}} + {\left[ {\dfrac{{\sqrt 3 }}{2}\left( {{r_s} + \left( {1 + {\xi _c}} \right)c\cos \delta } \right),\dfrac{1}{2}\left( {{r_s} + \left( {1 + {\xi _c}} \right)c\cos \delta } \right), - {\xi _c}c\sin \delta } \right]^{\rm{T}}}\\[15pt]{}^{21}{{\boldsymbol{F}}_{i + 1,j}} = {{\boldsymbol{S}}_{i + 1,j}} + {\left[ { - \dfrac{{\sqrt 3 }}{2}\left( {{r_s} + {\xi _c}c\cos \delta } \right),\dfrac{1}{2}\left( {{r_s} + {\xi _c}c\cos \delta } \right), - {\xi _c}c\sin \delta } \right]^{\rm{T}}}\\[15pt]{}^{22}{{\boldsymbol{F}}_{i + 1,j}} = {{\boldsymbol{S}}_{i + 1,j}} + {\left[ { - \dfrac{{\sqrt 3 }}{2}\left( {{r_s} + \left( {1 + {\xi _c}} \right)c\cos \delta } \right),\dfrac{1}{2}\left( {{r_s} + \left( {1 + {\xi _c}} \right)c\cos \delta } \right), - {\xi _c}c\sin \delta } \right]^{\rm{T}}}\\[15pt]{}^{31}{{\boldsymbol{F}}_{i + 1,j}} = {{\boldsymbol{S}}_{i + 1,j + 1}} + {\left[ {0, - \left( {{r_s} + {\xi _c}c\cos \delta } \right), - {\xi _c}c\sin \delta } \right]^{\rm{T}}}\\[8pt]{}^{32}{{\boldsymbol{F}}_{i + 1,j}} = {{\boldsymbol{S}}_{i + 1,j + 1}} + {\left[ {0, - \left( {{r_s} + \left( {1 + {\xi _c}} \right)c\cos \delta } \right), - {\xi _c}c\sin \delta } \right]^{\rm{T}}}\end{array} \right.\end{equation}

where

\begin{equation*}{\xi _c} = \frac{{{c_{\rm{p}}}}}{c}\end{equation*}

Therefore, the centroid positions of all F in D _{i+1, j} of Part K can be expressed as

(12) \begin{equation}{}_{\rm{F}}^K{{\boldsymbol{I}}_{i + 1,j}} = \left[ {\begin{array}{*{20}{c}}{{{\boldsymbol{R}}_K}{}^{11}{{\boldsymbol{F}}_{i + 1,j}}}\quad {{{\boldsymbol{R}}_K}{}^{12}{{\boldsymbol{F}}_{i + 1,j}}}\quad {{{\boldsymbol{R}}_K}{}^{21}{{\boldsymbol{F}}_{i + 1,j}}} \quad {{{\boldsymbol{R}}_K}{}^{22}{{\boldsymbol{F}}_{i + 1,j}}}\quad {{{\boldsymbol{R}}_K}{}^{31}{{\boldsymbol{F}}_{i + 1,j}}}\quad {{{\boldsymbol{R}}_K}{}^{32}{{\boldsymbol{F}}_{i + 1,j}}}\end{array}} \right]\end{equation}

where

\begin{equation*}{{\boldsymbol{R}}_K} = \left[ \begin{array}{c@{\quad}c@{\quad}c}{\cos \!\left( {\dfrac{{\left( {K - 1} \right){\rm{\pi }}}}{3}} \right)} & -\sin \!\left( {\dfrac{{\left( {K - 1} \right){\rm{\pi }}}}{3}} \right) & 0\\[9pt] \sin \!\left( {\dfrac{{\left( {K - 1} \right){\rm{\pi }}}}{3}} \right) & \cos \!\left( {\dfrac{{\left( {K - 1} \right){\rm{\pi }}}}{3}} \right) & 0\\[9pt] 0 & 0 & 1\end{array} \right]\end{equation*}

Then the centroid positions of F in Part K can be expressed as

(13) \begin{equation}{}_{\rm{F}}^K{\boldsymbol{D}} = \left[ {{}_{\rm{F}}^K{{\boldsymbol{I}}_{1,1}}} \quad {{}_{\rm{F}}^K{{\boldsymbol{I}}_{2,1}}} \quad {{}_{\rm{F}}^K{{\boldsymbol{I}}_{2,2}}} \quad \cdots \quad {{}_{\rm{F}}^K{{\boldsymbol{I}}_{n,n}}} \right]\end{equation}

In addition, it can be seen that the arrangement of the odd part and the even part of G and E is different. Therefore, it is necessary to calculate the centroid position vector of G and E in the supporting mechanism in different part.

In the odd part ( K = 1, 3 and 5):

G and E in the odd part are only arranged in the deployable unit D _{i+1, j} , so the centroid position vectors of G are

(14) \begin{equation}\left\{ \begin{array}{l}{}_{}^1{{\boldsymbol{G}}_{i + 1,j}} = {{\boldsymbol{S}}_{i,j}} + {\left[ {\dfrac{1}{2}\left( {{r_s} + {\xi _g}g\cos \beta } \right),\dfrac{{\sqrt 3 }}{2}\left( {{r_s} + {\xi _g}g\cos \beta } \right), - {\xi _g}g\sin \beta } \right]^{\rm{T}}}\\[16pt]{}_{}^2{{\boldsymbol{G}}_{i + 1,j}} = {{\boldsymbol{S}}_{i + 1,j}} + {\left[ { - \left( {{r_s} + {\xi _g}g\cos \beta } \right),0, - {\xi _g}g\sin \beta } \right]^{\rm{T}}}\\[9pt]{}_{}^3{{\boldsymbol{G}}_{i + 1,j}} = {{\boldsymbol{S}}_{i + 1,j + 1}} + {\left[ {\dfrac{1}{2}\left( {{r_s} + {\xi _g}g\cos \beta } \right), - \dfrac{{\sqrt 3 }}{2}\left( {{r_s} + {\xi _g}g\cos \beta } \right), - {\xi _g}g\sin \beta } \right]^{\rm{T}}}\end{array} \right.\end{equation}

where

\begin{equation*}{\xi _g} = \frac{{{g_p}}}{g}\end{equation*}

The centroid positions of all G in D _i+1,j of Part K can be expressed as

(15) \begin{equation}{}_{\rm{G}}^K{{\boldsymbol{I}}_{i + 1,j}} = \left[ {{{\boldsymbol{R}}_K}{}_{}^1{{\boldsymbol{G}}_{i + 1,j}}} \quad {{{\boldsymbol{R}}_K}{}_{}^2{{\boldsymbol{G}}_{i + 1,j}}} \quad {{{\boldsymbol{R}}_K}{}_{}^3{{\boldsymbol{G}}_{i + 1,j}}} \right]\end{equation}

G of the odd part do not exist in the deployable unit ^pD _i,j , which are expressed as

(16) \begin{equation}{}_{{\rm{Gp}}}^K{{\boldsymbol{I}}_{i,j}} = \left[ {{{\boldsymbol{R}}_K}\textbf{0}}\quad {{{\boldsymbol{R}}_K}\textbf{0}} \quad {{{\boldsymbol{R}}_K}\textbf{0}} \right]\end{equation}

where 0 represents a null matrix with 3 rows and 1 column.

The centroid positions of E can be expressed as

(17) \begin{equation}{}^{}{{\boldsymbol{E}}_{i + 1,j}} = {{\boldsymbol{S}}_{i + 1,j}} + {\left[ { - \left( {{r_s} + {r_E} + g\cos \beta } \right),0, - g\sin \beta } \right]^{\rm{T}}}\end{equation}

Therefore, the centroids position of E in D _i+1,j of Part K can be expressed as

(18) \begin{equation}{}_{\rm{E}}^K{{\boldsymbol{I}}_{i + 1,j}} = \left[ {{{\boldsymbol{R}}_K}{{\boldsymbol{E}}_{i + 1,j}}} \right]\end{equation}

E of the odd part do not exist in the deployable unit ^pD _{i, j} , which are expressed as

(19) \begin{equation}{}_{{\rm{Ep}}}^K{{\boldsymbol{I}}_{i,j}} = \left[ {{{\boldsymbol{R}}_K}\textbf{0}} \right]\end{equation}

In the even part ( K = 2, 4 and 6):

G and E of the even part are only arranged in the deployable unit ^pD _{i, j} , so the centroid positions vector of G are

(20) \begin{equation}\left\{ \begin{array}{l}{}_{\rm{p}}^1{{\boldsymbol{G}}_{i,j}} = {{\boldsymbol{S}}_{i,j}} + {\left[ { - \dfrac{1}{2}\left( {{r_s} + {\xi _g}g\cos \beta } \right),\dfrac{{\sqrt 3 }}{2}\left( {{r_s} + {\xi _g}g\cos \beta } \right), - {\xi _g}g\sin \beta } \right]^{\rm{T}}}\\[15pt]{}_{\rm{p}}^2{{\boldsymbol{G}}_{i,j}} = {{\boldsymbol{S}}_{i + 1,j + 1}} + {\left[ { - \dfrac{1}{2}\left( {{r_s} + {\xi _g}g\cos \beta } \right), - \dfrac{{\sqrt 3 }}{2}\left( {{r_s} + {\xi _g}g\cos \beta } \right), - {\xi _g}g\sin \beta } \right]^{\rm{T}}}\\[15pt]{}_{\rm{p}}^3{{\boldsymbol{G}}_{i,j}} = {{\boldsymbol{S}}_{i + 1,j}} + {\left[ {\left( {{r_s} + {\xi _g}g\cos \beta } \right),0, - {\xi _g}g\sin \beta } \right]^{\rm{T}}}\end{array} \right.\end{equation}

Therefore, the centroid positions of G in the even part in ^p D _{i, j} of Part K can be expressed as

(21) \begin{equation}{}_{{\rm{Gp}}}^K{{\boldsymbol{I}}_{i,j}} = \left[ {{{\boldsymbol{R}}_K}\,{}_{\rm{p}}^1{{\boldsymbol{G}}_{i,j}}} \quad{{{\boldsymbol{R}}_K}\,{}_{\rm{p}}^2{{\boldsymbol{G}}_{i,j}}} \quad {{{\boldsymbol{R}}_K}\,{}_{\rm{p}}^3{{\boldsymbol{G}}_{i,j}}} \right]\end{equation}

G of the even part do not exist in the deployable unit D _i+1,j , which are expressed as

(22) \begin{equation}{}_{\rm{G}}^K{{\boldsymbol{I}}_{i + 1,j}} = \left[ {{{\boldsymbol{R}}_K}\textbf{0}} \quad {{{\boldsymbol{R}}_K}\textbf{0}} \quad {{{\boldsymbol{R}}_K}\textbf{0}} \right]\end{equation}

The centroid positions of E can be expressed as

(23) \begin{equation}{}^{\rm{p}}{{\boldsymbol{E}}_{i,j}} = {{\boldsymbol{S}}_{i + 1,j}} + {\left[ {\left( {{r_s} + {r_E} + g\cos \beta } \right),0, - c\sin \beta } \right]^{\rm{T}}}\end{equation}

Therefore the centroid position in the even part of the E in ^p D _{i, j} of Part K can be expressed as

(24) \begin{equation}{}_{{\rm{Ep}}}^K{{\boldsymbol{I}}_{i,j}} = \left[ {{{\boldsymbol{R}}_K}{}^{\rm{p}}{{\boldsymbol{E}}_{i,j}}} \right]\end{equation}

E of the even part do not exist in the deployable unit D _{i+1, j} , which are expressed as

(25) \begin{equation}{}_{\rm{E}}^K{{\boldsymbol{I}}_{i + 1,j}} = \left[ {{{\boldsymbol{R}}_K}\textbf{0}} \right]\end{equation}

Then the centroid positions of G and E in Part K can be expressed as

(26) \begin{equation}{}_{\rm{G}}^K{\boldsymbol{D}} = \left[ {{}_{\rm{G}}^K{{\boldsymbol{I}}_{1,1}}} \quad {{}_{\rm{G}}^K{{\boldsymbol{I}}_{2,1}}} \quad {{}_{\rm{G}}^K{{\boldsymbol{I}}_{2,2}}}\quad \cdots \quad {{}_{\rm{G}}^K{{\boldsymbol{I}}_{n,n}}} \quad {{}_{{\rm{Gp}}}^K{{\boldsymbol{I}}_{1,1}}} \quad {{}_{{\rm{Gp}}}^K{{\boldsymbol{I}}_{2,1}}} \quad {{}_{{\rm{Gp}}}^K{{\boldsymbol{I}}_{2,2}}} \quad \cdots \quad {{}_{{\rm{Gp}}}^K{{\boldsymbol{I}}_{n - 1,n - 1}}} \right]\end{equation}

(27) \begin{equation}{}_{\rm{E}}^K{\boldsymbol{D}} = \left[ {{}_{\rm{E}}^K{{\boldsymbol{I}}_{1,1}}} \quad {{}_{\rm{E}}^K{{\boldsymbol{I}}_{2,1}}} \quad {{}_{\rm{E}}^K{{\boldsymbol{I}}_{2,2}}} \quad \cdots \quad {{}_{\rm{E}}^K{{\boldsymbol{I}}_{n,n}}} \quad {{}_{{\rm{Ep}}}^K{{\boldsymbol{I}}_{1,1}}} \quad {{}_{{\rm{Ep}}}^K{{\boldsymbol{I}}_{2,1}}} \quad {{}_{{\rm{Ep}}}^K{{\boldsymbol{I}}_{2,2}}} \quad \cdots \quad {{}_{{\rm{Ep}}}^K{{\boldsymbol{I}}_{n - 1,n - 1}}} \right]\end{equation}

Then the centroid positions of S can be expressed as

(28) \begin{equation}{}_{\rm{S}}^K{\boldsymbol{D}} = \left[{{{\boldsymbol{R}}_K}{}^K{{\boldsymbol{S}}_{0,1}}} \quad {{{\boldsymbol{R}}_K}{}^K{{\boldsymbol{S}}_{1,1}}} \quad {{{\boldsymbol{R}}_K}{}^K{{\boldsymbol{S}}_{1,2}}} \quad {{{\boldsymbol{R}}_K}{}^K{{\boldsymbol{S}}_{2,1}}} \quad {{{\boldsymbol{R}}_K}{}^K{{\boldsymbol{S}}_{2,2}}} \quad {{{\boldsymbol{R}}_K}{}^K{{\boldsymbol{S}}_{2,3}}} \quad \cdots \quad {{{\boldsymbol{R}}_K}{}^K{{\boldsymbol{S}}_{n,n + 1}}}\right]\end{equation}

In summary, the centroid positions of each component in the Part K can be expressed as

(29) \begin{equation}{{\boldsymbol{D}}_K} = \left[ {{}_{\rm{F}}^K{\boldsymbol{D}}} \quad {{}_{\rm{G}}^K{\boldsymbol{D}}} \quad {{}_{\rm{E}}^K{\boldsymbol{D}}} \quad {{}_{\rm{S}}^K{\boldsymbol{D}}} \right]\end{equation}

In the n-ring supporting mechanism, the centroid positions of F, G and E can be expressed by a matrix with 18 rows and ((15n ²+13n–2)/2) columns:

(30) \begin{equation}{\boldsymbol{M}} = {\left[ {{{\boldsymbol{D}}_1}} \quad {{{\boldsymbol{D}}_2}} \quad \cdots \quad {{{\boldsymbol{D}}_6}} \right]^{\rm{T}}}\end{equation}

The velocity and acceleration can be expressed as

(31) \begin{equation}\dot{\boldsymbol{M}} = {\left[ {{{\dot{\boldsymbol{D}}}_1}} \quad {{{\dot{\boldsymbol{D}}}_2}} \quad \cdots \quad {{{\dot{\boldsymbol{D}}}_6}}\right]^{\rm{T}}}\end{equation}

and

(32) \begin{equation}\ddot{\boldsymbol{M}} = {\left[ {\ddot{\boldsymbol{D}}}_1 \quad {\ddot{\boldsymbol{D}}}_2 \quad \cdots \quad {\ddot{\boldsymbol{D}}}_6 \right]^{\rm{T}}}\end{equation}

Figure 12.

Centroid motion law of F_3,2 in D_2,2 of Part 3.

Taking the solid supporting mechanism with n = 2 as an example, the active component is set to ^1,1F_1,1, and the angular velocity is 18°/s. The virtual prototype is established in Solidworks, and the motion simulation is compared with the calculation results of Matlab. The motion curves of the centroid of F_3,2 in D_2,2 of Part 3 are obtained, as shown in Fig. 12. By analysing the motion curves, the theoretical values of displacement, velocity and acceleration are completely consistent with the simulation values, which verifies the correctness of the kinematic model of the mechanism established in this paper, and the velocity and acceleration values are not mutated. The motion process of the mechanism is stable and has no large impact.

3.2 Analysis of folding ratio

For the deployable mechanism, the folding ratio is one of the important indexes to evaluate the performance of the mechanism. The mechanism with higher folding ratio reach a larger workspace after fully deploying in a limited initial space. Some deployable mechanisms evaluate the size of the workspace with the expanded envelope area, such as the deployable supporting truss mechanism. The folding ratio can be expressed as

(33) \begin{equation}{\eta _{\rm{L}}} = \frac{{{S_{\rm{D}}}}}{{{S_{\rm{F}}}}}\end{equation}

where S _D and S _F represent the maximum envelope area and the minimum envelope area of the supporting mechanism on the OXY plane in the deployable state and the folded state, respectively.

According to the Equation (43), the folding ratio of the n-ring supporting mechanism can be expressed as

(34) \begin{equation}\eta = \frac{{6\sqrt 3 {{\left( {\dfrac{{\sqrt 3 }}{3}a + \left( {{r_{\rm{S}}} + c} \right)n} \right)}^2}}}{{6\sqrt 3 {{\left( {\dfrac{{\sqrt 3 }}{3}a + \left( {{r_{\rm{S}}} + d} \right)n} \right)}^2}}}\end{equation}

and further obtained

(35) \begin{equation}\eta = {\left( {\frac{{\sqrt 3 a + 3n\!\left( {{r_{\rm{S}}} + c} \right)}}{{\sqrt 3 a + 3n\!\left( {{r_{\rm{S}}} + d} \right)}}} \right)^2}\end{equation}

According to the Equation (35), the folding ratio grows with the increase of c and the decrease of d. Derivation of a on Equation (35) yields

(36) \begin{equation}\frac{{{\rm{d}}\eta }}{{{\rm{d}}a}}{\rm{ = }} - \frac{{6 n ( {c - d})( {3 a + c n + n {r_{\rm{s}}}} )}}{{\left( {3 a + d n + n {r_{\rm{s}}}} \right)}^3}\end{equation}

Equation (34) shows that when the thickness d of the rod is greater than the length c of the rod, the folding ratio of the mechanism is less than 1. Therefore, the design should ensure that c > d, which can make

\begin{equation*}\frac{{{\rm{d}}\eta }}{{{\rm{d}}a}} \lt 0\end{equation*}

Therefore, the folding ratio decreases with the increase of d. Derivation of r _s and n, respectively, on Equation (35) yields

(37) \begin{equation}\frac{{{\rm{d}}\eta }}{{{\rm{d}}{r_{\rm{s}}}}} = - \frac{{2 {n^2} ( {c - d}) \left( {3 a + c n + n {r_{\rm{s}}}} \right)}}{{{{\left( {3 a + d n + n {r_{\rm{s}}}} \right)}^3}}} \lt 0\end{equation}

and

(38) \begin{equation}\frac{{{\rm{d}}\eta }}{{{\rm{d}}n}} = \frac{{6 a ( {c - d}) \left( {3 a + c n + n {r_{\rm{s}}}} \right)}}{{{{\left( {3 a + d n + n {r_{\rm{s}}}} \right)}^3}}} \gt 0\end{equation}

Equations (37) and (38) show that the folding ratio grows with the decrease of r _s and the increase of n. The definition domains of each parameter are defined as: n $\in$ [1,3,6], c $\in$ [0.1,0.35], r _s $\in$ [0.01,0.035], a $\in$ [0.04,0.06] and d $\in$ [0.005,0.015]. According to Equation (35) and the parameter of components, the diagram which can express the relationship among $\eta$ , n, c, r _s , a and d can be obtained by Matlab. As shown in Fig. 13, the folding ratio is positively correlated to c and n, and negatively correlated to d, a and r _s, which verify the correctness of above analysis.

Figure 13.

The diagram of folding ratio.

Figure 14.

Connection relationship.

Figure 15.

The parameters of components.

3.3 Prototype verification

A hollow two-ring physical prototype of supporting mechanism with folding ratio of 45.8 was established based on 3D-print technology. The rod is assembled from joints and prefabricated stainless steel rods. The joints and floral discs are both 3D printed from poly lactic acid (PLA), with a printing accuracy of 0.02 mm. The diameter of the stainless steel rod is 2 mm. The construction process of physical prototype is as follows:

1. STEP 1: The joints and the stainless steel rods are assembled to construct the rods G and the rods F, as shown in Fig. 14. The parameters of the components are shown in Fig. 15.

2. STEP 2: The basic unit is assembled according to the connection relationship of each component in Section 2.1 as shown in Fig. 16(c), and the specific connection method is shown in the Fig. 16(a), (b), (d) and (e). The deployable process of the basic deployable unit is shown in Fig. 17.

3. STEP 3: The basic deployable units are expanded according to the method in Section 2.2, as shown in Fig. 18. By the same way, a hollow two-ring physical prototype of supporting mechanism is obtained, and the deployable process is shown in Fig. 19.

The prototype moves smoothly and reliably during the deployable process, which verified the correctness of the proposed deployable unit expansion method for constructing supporting mechanisms in this paper, providing a reference for the subsequent research work of the supporting mechanism of the sunshield.

Figure 16.

Assembly relationship of basic deployable unit.

Figure 17.

Deployable process of basic deployable unit.

Figure 18.

The basic deployable units expansion.

Figure 19.

Deployable process of physical prototype.

4.1 Dynamic modeling and simulation verification of supporting mechanism

On account of supporting mechanism has one DOF, according to the mechanisms and machine theory, it can be known that only one drive device can be added to make the determined motion. However, in order to improve the deployment reliability of the mechanism, multiple torsion springs are generally added, and the elastic potential energy stored in the folded torsional springs is used to drive the mechanism to deploy. The mechanism has only one set of independent generalised coordinates, the input angle $\delta$ is selected as the generalised coordinate, and the dynamic equation can be established according to Lagrange equation as follows:

(39) \begin{equation}\frac{{\rm{d}}}{{{\rm{d}}t}}\frac{{\partial {E_v}}}{{\partial \dot \delta }} - \frac{{\partial {E_v}}}{{\partial \delta }} + \frac{{\partial {E_p}}}{{\partial \delta }} = Q\end{equation}

where E _v , E _p and Q are the kinetic energy, potential energy, and generalised force of the supporting mechanism, respectively.

If a floral discs S is fixed of supporting mechanism, such as S_0,1 shown in Fig. 10, the mechanism in the process of deployment, all the moving components can be divided into three categories according to the characteristics of motion: translational motion (S and E), fixed-axis rotation (F and G, also called inner rods, are connected to S_0,1.) and compound motion (except for the F and G which are connected to the S_0,1, also called outer rods). In order to facilitate the calculation of the total kinetic energy between various components, the compound motion of the rod is divided into movement and rotation.

The total kinetic energy produced by the translation of F can be expressed as

(40) \begin{equation}{}^{\rm{F}}{E_{v1}} = \frac{1}{2}{m_{\rm{F}}}\sum\limits_{{A_{\rm{F}}}{\rm{ = }}1}^3 {\sum\limits_{{B_{\rm{F}}} = 1}^{3\!\left( {1 + n} \right)n} {{}^{\rm{F}}w{}_{{A_{\rm{F}}},{B_{\rm{F}}}}} } \end{equation}

where ^F w _AF,BF $\in$ W _F

(41) \begin{equation} \boldsymbol{W}_{\textrm{F}} = \sum^{6}_{k=1} \left({}^{K}_{\textrm{F}}\boldsymbol{B}\bullet \boldsymbol{H}_{\textrm{F}}\right)\end{equation}

(42) \begin{equation}{}^{K}_{\textrm{F}}\boldsymbol{B} = {}^{K}_{\textrm{F}}\boldsymbol{D}\bullet {}^{K}_{\textrm{F}}\boldsymbol{D}\end{equation}

H _F represents a matrix with the same dimension as the matrix ^F _K D and its internal elements correspond to the internal elements of ^K _F D one by one. In the matrix H _F, the element ‘1’ represents the component in the mechanism, and the element ‘0’ represents the component that needs to be deleted.

The total kinetic energy produced by the rotation of F can be expressed as

(43) \begin{equation}{}^{\rm{F}}{E_{v2}} = \frac{1}{2}\left( {{N_{{\rm{F1}}}}{J_{{\rm{F}}1}} + {N_{{\rm{F2}}}}{J_{{\rm{F}}2}}} \right){\dot \delta ^2}\end{equation}

where J _F1 and J _F2 represents the rotational inertia of F belonging to the inner rods and the rotational inertia of F belonging to the outer rods in the global coordinate system, N _F1 is the quantity of F belonging to the inner rods, N _F2 is the quantity of F belonging to the outer rods.

Therefore, the total kinetic energy generated by F is

(44) \begin{equation}{}^{\rm{F}}{E_v} = {}^{\rm{F}}{E_{v1}} + {}^{\rm{F}}{E_{v2}}\end{equation}

The total kinetic energy produced by the translation of G can be expressed as

(45) \begin{equation}{}^{\rm{G}}{E_v} = \frac{1}{2}{m_{\rm{G}}}\sum\limits_{{A_{\rm{G}}}{\rm{ = }}1}^3 {\sum\limits_{{B_{\rm{G}}} = 1}^{\frac{{3\!\left( {1 + n} \right)n}}{2}} {{}^{\rm{G}}w{}_{{A_{\rm{G}}},{B_{\rm{G}}}}} } \end{equation}

where ^G w _AG,BG $\in$ W _G

(46) \begin{equation} \boldsymbol{W}_{\textrm{G}} = \sum^{6}_{k=1} \left({}^{K}_{\textrm{G}}\boldsymbol{B}\bullet \boldsymbol{H}_{\textrm{G}}\right)\end{equation}

(47) \begin{equation}{}^{K}_{\textrm{G}}\boldsymbol{B} = {}^{K}_{\textrm{G}}\boldsymbol{D}\bullet {}^{K}_{\textrm{G}}\boldsymbol{D}\end{equation}

The total kinetic energy produced by the rotation of G can be expressed as

(48) \begin{equation}{}^{\rm{G}}{E_{v2}} = \frac{1}{2}\left( {{N_{{\rm{G1}}}}{J_{{\rm{G}}1}} + {N_{{\rm{G2}}}}{J_{{\rm{G}}2}}} \right){\dot \beta ^2}\end{equation}

where J _G1 and J _G2 represents the rotational inertia of G belonging to the inner rods and the rotational inertia of G belonging to the outer rods in the global coordinate system, N _G1 is the quantity of G belonging to the inner rods, N _G2 is the quantity of G belonging to the outer rods.

Therefore, the total kinetic energy generated by G is

(49) \begin{equation}{}^{\rm{G}}{E_v} = {}^{\rm{G}}{E_{v1}} + {}^{\rm{G}}{E_{v2}}\end{equation}

The total kinetic energy of S is expressed as

(50) \begin{equation}{}^{\rm{S}}{E_v} = \frac{1}{2}{m_{\rm{S}}}\sum\limits_{{A_{\rm{S}}}{\rm{ = }}1}^3 {\sum\limits_{{B_{\rm{S}}} = 1}^{\frac{{\left( {1 + n} \right)\left( {n + 2} \right)}}{2}} {{}^{\rm{S}}w{}_{{A_{\rm{S}}},{B_{\rm{S}}}}} } \end{equation}

where ^S w _{AS, BS} $\in$ W _S

(51) \begin{equation} \boldsymbol{W}_{\textrm{S}} = \sum^{6}_{k=1} \left({}^{K}_{\textrm{S}}\boldsymbol{B}\bullet \boldsymbol{H}_{\textrm{S}}\right)\end{equation}

(52) \begin{equation}{}^{K}_{\textrm{S}}\boldsymbol{B} = {}^{K}_{\textrm{S}}\boldsymbol{D}\bullet {}^{K}_{\textrm{S}}\boldsymbol{D}\end{equation}

The total kinetic energy of E is expressed as

(53) \begin{equation}{}^{\rm{E}}{E_v} = \frac{1}{2}{m_{\rm{E}}}\sum\limits_{{A_{\rm{E}}}{\rm{ = }}1}^3 {\sum\limits_{{B_{\rm{E}}} = 1}^{\frac{{{n^2} + n}}{2}} {{}^{\rm{E}}w{}_{{A_{\rm{E}}},{B_{\rm{E}}}}} } \end{equation}

where ^E w _{AE, BE} $\in$ W _E

(54) \begin{equation} \boldsymbol{W}_{\textrm{E}} = \sum^{6}_{k=1} \left({}^{K}_{\textrm{E}}\boldsymbol{B}\bullet \boldsymbol{H}_{\textrm{E}}\right)\end{equation}

(55) \begin{equation}{}^{K}_{\textrm{E}}\boldsymbol{B} = {}^{K}_{\textrm{E}}\boldsymbol{D}\bullet {}^{K}_{\textrm{E}}\boldsymbol{D}\end{equation}

In summary, the total kinetic energy of the n-ring supporting mechanism is

(56) \begin{equation}{E_v} = {}^{\rm{F}}{E_v} + {}^{\rm{G}}{E_v} + {}^{\rm{S}}{E_v} + {}^{\rm{E}}{E_v}\end{equation}

Assuming that the torsion springs are added to the revolute joints between the floral discs S and the rod F of the supporting mechanism, and considering that the application environment of the supporting mechanism has no gravity effect, only the elastic potential energy of the mechanism is considered here. Consequently, the total potential energy of the supporting mechanism is expressed as

(57) \begin{equation}{E_{\rm{p}}} = \frac{1}{2}\sum\limits_{u = 1}^\varsigma {{k_u}{{\left( {{\mu _{u0}} - {\delta _0} + \delta } \right)}^2}} \end{equation}

where k _u and μ _u0 represent the stiffness and initial compression of the u-th torsion spring, $\delta_0$ represents the initial angle, ζ represents the quantity of torsion springs installed.

The following is an example of a double-ring supporting mechanism (n = 2). Since F of the six parts of the supporting mechanism is rotated from the initial part, when calculating the total kinetic energy generated by the movement of the F, there will be ¹¹F _i+1,1, ¹²F _i+1,1 repeated calculations at the junction of the two adjacent parts. The elements in H _F are

(58) \begin{equation}{}^{\rm{F}}{h_{P,1 + Q}} = {}^{\rm{F}}{h_{P,2 + Q}} = {}^{\rm{F}}{h_{P,6}} = 0\end{equation}

where ${}^{\rm{F}}h \in {{\boldsymbol{H}}_{\rm{F}}}$ , $Q = \sum\limits_{q = 0}^i {6q} $ , $i \in \left[ {0,n} \right]$ , $P \in \left[ {1,3} \right]$ , and the others are 1. When calculating the kinetic energy generated by the movement of G, the G connected to the S_0,1 is not calculated. Therefore, the elements in H _G are

(59) \begin{equation}{}^{\rm{G}}{h_{P,1}} = 0\end{equation}

where ${}^{\rm{G}}{h_{P,1}} \in {{\boldsymbol{H}}_{\rm{G}}}$ , $P \in \left[ {1,3} \right]$ , and the others are 1. When calculating the kinetic energy generated by the movement of S, there will be S _i,1 repeated calculations at the junction of the two adjacent parts. The elements in H _S are

(60) \begin{equation}{}^{\rm{S}}{h_{P,M}} = 0\end{equation}

where ${}^{\rm{S}}{h_{P,M}} \in {{\boldsymbol{H}}_{\rm{S}}}$ , $P \in \left[ {1,3} \right]$ , $M = \sum\limits_{q{\rm{ = 0}}}^{{i_{\rm{s}}}} q $ , ${i_{\rm{s}}} \in \left[ {1,n} \right]$ , and the others are 1, all elements in matrix H _E are 1.

The quantities of F and G with compound motions are

(61) \begin{equation}{N_{{\rm{F2}}}} = 18{n^2} + 6n - 6\end{equation}

(62) \begin{equation}{N_{{\rm{G2}}}} = 9{n^2} - 3\end{equation}

N _F1 and N _G1 are 6 and 3, respectively.

The Equations (58), (59), and (60) are substituted into Equations (41), (46) and (51), respectively; Equation (61) and N _F1, Equation (62) and N _G1 are substituted into Equations (43) and (48), respectively, H _E is substituted into Equation (54), so that the total kinetic energy of each component can be calculated. Finally, Equations (56) and (57) are substituted into Equation (39). The change value of $\delta$ with time under the driving action of ζ torsion springs can be obtained by solving the differential equation. A set of structural parameters and dynamic parameters of the supporting mechanism are given, as shown in Table 1. The simulation model is established, and the torsion springs are installed at the revolute joints of the mechanism at six positions, as shown in Fig. 20, and the simulation is performed for 5.75 s. The theoretical and simulation law of motion about $\delta$ can be obtained by Matlab and Solidworks, respectively, as shown in Fig. 21.

Table 1.

Parameters of the components

Figure 20.

Position of the torsion springs.

Figure 21.

The relationship among $\delta$ , $\dot{\delta}$ and t.

It can be seen from Fig. 21 that the theoretical and simulation value of $\delta$ under the action of torsion spring drive are completely consistent, which verifies the correctness of the dynamic model of the mechanism established in this paper. The angular velocity increases slowly in the first 5 s and suddenly increases from 5 s.

When calculating the total kinetic energy of the supporting mechanism with ribs, the kinetic energy of some components can be subtracted from the solid supporting mechanism, which corresponds to the change of the matrix H _F, H _G, H _S and H _E, the quantity of F and G in the equation. For example, the mechanism shown in Fig. 22 can be formed by removing the components in the wireframe from the mechanism. The parameters of each component are shown in Table 1, and the masses of the solid supporting mechanism and the supporting mechanism with ribs are 102.704 kg and 76.043 kg, respectively. The mass of supporting mechanism with ribs relative to solid supporting mechanism is reduced by 26%. Some parameters can be obtained: n = 5, N _F2 = 366, N _G2 = 141, N _F1 = 6 and N _G1 = 3. All the parameters of components are the same as in Table 1, expect for the torsion springs. The torsion springs are installed at the revolute joints connecting S_4,5 and ³¹F_4,4 shown in Fig. 22, and the parameters of the torsion springs are shown in Table 2.

Table 2.

The parameters of torsion springs

Figure 22.

Construction of the 5-ring supporting mechanism with six ribs.

The elements in matrix H _F are

\begin{equation*}{}^{\rm{F}}{h_{P,{Q_{\rm{F}}}}} = 0\end{equation*}

where P $\in$ [1,3], Q _F $\in$ ([1,2] ∪ [6,…,10] ∪ [19,…,28] ∪ [37,38] ∪ [41,…,44] ∪ [47,…,50] ∪ [61,62), and the others are 1.

The elements in matrix H _G are

\begin{equation*}{}^{\rm{G}}{h_{P,{Q_{\rm{G}}}}} = 0\end{equation*}

where P $\in$ [1,3], Q _G $\in$ ([1] ∪ [4–6] ∪ [10,…,15] ∪ [19,27] ∪ [49,51] ∪ [55,60), and the others are 1.

The elements in matrix H _E are

\begin{equation*}{}^{\rm{E}}{h_{P,{Q_{\rm{E}}}}} = 0\end{equation*}

where P $\in$ [1,3], Q _E $\in$ ([2] ∪ [4,5] ∪ [7–9] ∪ [17] ∪ [19,20), and the others are 1.

The elements in matrix H _S are

\begin{equation*}{}^{\rm{G}}{h_{P,{Q_{\rm{S}}}}} = 0\end{equation*}

where P $\in$ [1,3], Q _S $\in$ ([1,2] ∪ [4] ∪ [7,8] ∪ [11,12]), and the others are 1.

Lastly, under the same driving conditions, the motion law diagrams of two kinds of mechanisms are drawn, as shown in Fig. 23. The supporting mechanism with ribs and the solid supporting mechanism took 4.21 s and 4.90 s from the folded state to the deployable state. Therefore, the supporting mechanism with ribs is easier to be driven than solid supporting mechanism.

Figure 23.

The relationship among $\delta$ , $\dot{\delta}$ and t.

4.2. The influence of torsion springs on the deployment of the supporting mechanism

Since the supporting mechanism of the sunshield is driven by the torsion springs, it is necessary to study the influence of parameters on the deployment of the mechanism. The basic parameters of the torsion springs in the mechanism include the quantity, position, stiffness and initial compression.

In order to analyse the influence of the combination of torsion springs with different quantities and different stiffnesses on the deployment motion of the mechanism, the total stiffness of the torsion springs in the simulation model is unchanged, and different quantities (1, 3, 6, 12 and 18) and different stiffnesses of torsion springs are selected to drive mechanism. The stiffnesses and position of the torsion springs are shown in Table 3, and the curves of $\delta$ is shown in Fig. 24. It shows that when the initial compression and total stiffness of torsion springs are the same, the combination of torsion springs with different quantities and different stiffnesses have the same driving effect on the deployment of the mechanism.

Table 3.

Quantities, stiffnesses and position of torsion springs

Let the quantity, initial compression and the position of torsion springs in the mechanism remain unchanged, the total stiffnesses of torsion springs are 0.006 N·m/°, 0.012 N·m/° and 0.018 N·m/°, to obtain the effect of the total stiffnesses of torsion springs on the deployment of mechanism as shown in Fig. 25. The figure shows that the greater the total stiffness of the torsion spring contained in the mechanism, the shorter the time taken for the mechanism to deploy in place, and the faster the mechanism deploys. Therefore, by reasonably selecting the total stiffness of the torsion spring, it is possible to control the time required for the mechanism to deploy in place, that is, to control the deploying speed of the mechanism, and thus reduce the impact of the mechanism when deploying in place.

In order to analyse the effect of the initial compression of torsion springs on the deployment of the mechanism, the quantity of torsion spring, the total stiffness and the position are kept constant, and the torsion springs with the initial compression of 90°, 100°, and 110° are selected to drive the mechanism to deploy, respectively. The torsion spring parameters are shown in Gr. 1, Gr. 2 and Gr. 3 in Table 4, and the relationship between $\delta$ and t is shown in Fig. 26. It can be seen that as the initial compression increases, the time for the mechanism to deploy in place is shorter.

Table 4.

Quantities of torsion springs, initial compression, stiffness and position

Figure 24.

The influence of the number and position of torsion springs on the deployment of the mechanism.

Figure 25.

The influence of the total stiffness on the deployment of the mechanism.

Figure 26.

The influence of the initial compression on the deployment of the mechanism.

To analyse the effect of torsion spring stiffness on the deployment of the mechanism when the initial compression of the torsion spring is less than $\delta_0$ (to ensure that the folding ratio of the mechanism is as large as possible, it should be $\delta_0$ = 90°), six torsion springs with different initial compression are added to the simulation model of the mechanism, and the deployment of the mechanism is analysed. The parameters of the torsion springs are shown in Gr. 1, Gr. 4 and Gr. 5 in Table 4, and the relationship between $\delta$ and t is shown in Fig. 27. It can be seen that when there are a number of torsion springs with initial compression less than $\delta_0$ in the mechanism, the deployment will be hindered. When the stiffness k ₄∼k ₆ of the torsion spring with initial compression of 30° is increased to 0.01 N·m/°, the parameters of the torsion springs as shown in Gr. 5, the mechanism deploys at first and then closes up, but $\delta$ does not reach 0°, as shown in Fig. 27a.That indicates when there are some torsion springs with initial compression less than $\delta_0$ in the mechanism, as the stiffness of these torsion springs increase or the initial compression decreases, the impediment to the deployment of the mechanism is strengthened resulting in the mechanism not being able to complete deploy. During the deployable process of the mechanism, $\delta$ decreases from 90° to 0°. The initial compression of some torsion springs (μ _u0) is less than 90°, when the change in $\delta$ is greater than μ _u0, the torsion springs will be compressed again, doing negative work in the mechanical system, which will hinder the deployment of the mechanism and cause it to fail to deploy.

Figure 27.

The influence of the initial compression less than $\delta_0$ on the deployment of the mechanism.