CN1779144A

CN1779144A - Flip-out open expandable structure and manufacturing method thereof

Info

Publication number: CN1779144A
Application number: CN 200510060849
Authority: CN
Inventors: 罗尧治; 刘晶晶
Original assignee: Zhejiang University ZJU
Current assignee: Zhejiang University ZJU
Priority date: 2005-09-22
Filing date: 2005-09-22
Publication date: 2006-05-31
Anticipated expiration: 2025-09-22
Also published as: CN100346043C

Abstract

This invention is an outward retroflexion deployable structure and its process method. The structure comprises a movable connecting rod device, which is composed by several tetrahedrons that are connected by rotating joint to form a three- dimensional ring structure. At moving, the structure could deploy into an outspread structure. The process method is as followed: (1) decide the structure form and the basic geometry size and design the basic geometry parameters; (2) calculate out the tetrahedrons size according to the geometry size; (3) link the tetrahedrons through connecting joint at proper positions and design the wheel's position; (4) design the slide-rail's form and size; (5) link the structure and the slide-rail together. This structure's top part could be totally closed at closing and easy for control.

Description

Flip-out open expandable structure and manufacturing method thereof

技术领域technical field

本发明涉及一种向外翻转开启式可展结构及其制作方法。其包含一可运动的连杆机构，该机构由多个刚性构件组成，这些刚性构件通过旋转连接件连接，形成三维环形机构，通过机构的运动以成为开启式结构。The invention relates to an outwardly turning and opening expandable structure and a manufacturing method thereof. It includes a movable link mechanism, which is composed of a plurality of rigid components connected by rotating connectors to form a three-dimensional ring mechanism, which becomes an open structure through the movement of the mechanism.

背景技术Background technique

可展结构是近几年来一种新型工程结构，其几何形状能根据实际情况改变，依靠机构的大位移和运动来实现折叠和展开。机构组成的结构具有内在自由度，其折叠和展开过程是刚体运动，不会在其部件内部产生应变，有利于作为重复开启的结构。世界上已有的可开启屋盖结构的类型按屋盖开启运动方式可以分为水平运动开启、刚体转动开启、上下运动开启和径向开启。本发明提供了一种向外翻转开启式可展结构，这种可展结构以连杆机构^[1-2]和四面体旋转环^[3-4]的理论为基础。Deployable structure is a new type of engineering structure in recent years, its geometric shape can be changed according to the actual situation, relying on the large displacement and movement of the mechanism to realize folding and unfolding. The structure composed of the mechanism has an inherent degree of freedom, and its folding and unfolding process is a rigid body motion, which will not generate strain inside its components, which is conducive to being a structure that can be opened repeatedly. The existing types of openable roof structures in the world can be divided into horizontal movement opening, rigid body rotation opening, up and down movement opening and radial opening according to the roof opening movement mode. The present invention provides an outward-turning and openable expandable structure, which is based on the theory of a link mechanism ^[1-2] and a tetrahedral rotating ring ^[3-4] .

[1]Bricard.R.(1897).Mémoire sur la théorie de l’octacdre articulé.Journal demathématiques pures et appliquées，Liouville 3，113-148.[1] Bricard.R.(1897). Mémoire sur la théorie de l'octacdre articulé. Journal demathématiques pures et appliquées, Liouville 3, 113-148.

[2]Bricard.R.(1927).Lecons de cinématique.Tome II cinématique Appliquée，Gauthier-Villars，Paris，7-12.[2]Bricard.R.(1927).Lecons de cinématique.Tome II cinématique Appliquée, Gauthier-Villars, Paris, 7-12.

[3]Schattschneider，D.and Walker，W.(1977).M.C.Escher Kalerdocycles，Ballantin Books，New York.[3] Schattschneider, D. and Walker, W. (1977). M.C. Escher Kalerdocycles, Ballantin Books, New York.

[4]Schatz，P.(1998).Rhythmusforschung und Technik，2.Auflage.Verlag FreiesGeistesleben，Stuttgart.[4] Schatz, P. (1998). Rhythmusforschung und Technik, 2. Auflage. Verlag Freies Geistesleben, Stuttgart.

发明内容Contents of the invention

本发明的目的在于提供一种向外翻转开启式可展结构及其制作方法。The object of the present invention is to provide an expandable structure that can be turned outwards and opened and a manufacturing method thereof.

向外翻转开启式可展结构包括可运动的连杆机构，该连杆机构由N(N≥6，且N为偶数)个四面体刚性构件组成，这些四面体刚性构件通过屋面轴和支座轴的旋转连接件连接，形成三维环形机构，支座轴的底端为支座，当支座沿滑轨同时由内向外运动时，四面体绕其旋转连接件转动，可使结构的相邻四面体的内侧面相互接触，结构上部闭合；当支座沿滑轨同时由外向内运动时，四面体绕其旋转连接件转动，结构上部向外翻转开启。The outward-turning open-type expandable structure includes a movable link mechanism, which is composed of N (N≥6, and N is an even number) tetrahedral rigid members, and these tetrahedral rigid members pass through the roof shaft and the support The rotating connectors of the shafts are connected to form a three-dimensional ring mechanism. The bottom end of the bearing shaft is the support. When the bearing moves from the inside to the outside along the slide rail at the same time, the tetrahedron rotates around its rotating connector, which can make the adjacent structure The inner surfaces of the tetrahedron are in contact with each other, and the upper part of the structure is closed; when the support moves from the outside to the inside along the slide rail, the tetrahedron rotates around its rotating connector, and the upper part of the structure turns outwards to open.

N个四面体刚性构件组成的向外翻转开启式可展结构，其几何形式可以采用多种。结构闭合时，顶面在水平方向的投影可以为正N边形，每个四面体的四个面中作为结构顶面的三角形是相同的等腰三角形，闭合时每个三角形的顶点重合于一点；结构闭合时，顶面在水平方向的投影也可以为三角形、正方形、矩形或多边形。The outward-turning and open-type expandable structure composed of N tetrahedral rigid components can adopt various geometric forms. When the structure is closed, the projection of the top surface in the horizontal direction can be a regular N-gon, and the triangles used as the top surface of the structure among the four faces of each tetrahedron are the same isosceles triangles, and the vertices of each triangle coincide at one point when the structure is closed. ; When the structure is closed, the projection of the top surface in the horizontal direction can also be triangular, square, rectangular or polygonal.

N个四面体刚性构件组成的向外翻转开启式可展结构，以结构闭合时顶面在水平方向的投影为正N边形为例，制作方法的步骤如下：The outward-turning open-type expandable structure composed of N tetrahedral rigid components, taking the projection of the top surface in the horizontal direction as a regular N-gon when the structure is closed as an example, the steps of the manufacturing method are as follows:

(1)设计N个四面体刚性构件组成的可展结构的基本几何参数如下：结构闭合时屋面在水平方向投影的正N边形边长a，a＞0；结构闭合时上部矢高c，c＞0，设相对矢高ξ＝c/a；结构闭合时支座轴与水平方向的夹角φ， $0 < φ < π - \arccos (1 / \sqrt{1 + ξ^{2}});$ 支座轴的长度d，0＜d＜a/cosφ，设支座轴相对长度η＝d/a；结构完全展开时屋面轴与水平方向夹角 $γ^{'} (\arccos (1 / \sqrt{1 + ξ^{2}}) < γ^{'} < π / 2);$ (1) The basic geometric parameters of the design of a deployable structure composed of N tetrahedral rigid components are as follows: when the structure is closed, the side length of the regular N-gon projected by the roof in the horizontal direction is a, a >0; when the structure is closed, the upper sag height c, c >0, set the relative sagittal height ξ=c/a; the angle φ between the support axis and the horizontal direction when the structure is closed, $0 < φ < π - \arccos (1 / \sqrt{1 + ξ^{2}});$ The length d of the support shaft, 0<d<a/cosφ, set the relative length of the support shaft η=d/a; the angle between the roof axis and the horizontal direction when the structure is fully unfolded $γ^{'} (\arccos (1 / \sqrt{1 + ξ^{2}}) < γ^{'} < π / 2);$

(2)确定组成结构的N个四面体刚性构件的尺寸，其中表示一个四面体A₁ABG六条边长度的公式如下：(2) Determine the size of the N tetrahedral rigid members that make up the structure, where the formula for expressing the length of the six sides of a tetrahedron A ₁ ABG is as follows:

BG＝ηa； $A_{1} A = A_{1} B = a \sqrt{1 + ξ^{2}}; AB = a \sqrt{2 (1 - \cos \frac{2 π}{N})};$ BG = ηa; $A_{1} A = A_{1} B = a \sqrt{1 + ξ^{2}}; AB = a \sqrt{2 (1 - \cos \frac{2 π}{N})};$

${A A}_{11} G G = = a a \sqrt{11 + + {ξ ξ}^{22} + + {η η}^{22} - - 22 η η ((cos cos φ φ - - ξ ξ sin sin φ φ))};;$

$AG AG = = a a \sqrt{22 ((11 - - η η cos cos φ φ)) ((11 - - cos cos \frac{22 π π}{N N})) + + {η η}^{22}}$

根据以上公式也可确定其余(N-1)个四面体的几何形状和尺寸；The geometry and size of the remaining (N-1) tetrahedrons can also be determined according to the above formula;

(3)确定四面体刚性构件之间的连接形式，相邻四面体刚性构件通过可替换的旋转连接件分别在屋面轴和支座轴的位置连接，形成三维环形机构，支座轴的底端为支座，在支座的位置安装滑轮，滑轮可滑动方向沿正N边形外接圆的径向方向；(3) Determine the connection form between tetrahedral rigid components. Adjacent tetrahedral rigid components are connected at the positions of the roof axis and the support axis through replaceable rotating connectors to form a three-dimensional ring mechanism. The bottom end of the support axis As a support, a pulley is installed at the position of the support, and the sliding direction of the pulley is along the radial direction of the circumscribed circle of the regular N-gon;

(4)设计滑轨的形式，(N/2)条滑轨在同一平面，相互夹角为(4π/N)且延长线可交于一点，这点作为滑轨中心点。滑轨最外侧点的位置相当于结构闭合时支座的位置，距离中心点距离公式(4) Design the form of the slide rail, (N/2) slide rails are on the same plane, the mutual angle is (4π/N) and the extension lines can intersect at one point, which is used as the center point of the slide rail. The position of the outermost point of the slide rail is equivalent to the position of the support when the structure is closed, and the distance formula from the center point

J₁＝a-dcosφJ ₁ ＝a-dcosφ

滑轨最内侧点的位置相当于结构完全展开时支座的位置，距离中心的距离公式The position of the innermost point of the slide rail is equivalent to the position of the support when the structure is fully expanded, and the distance formula from the center is

${J J}_{22} = = \frac{22 ((sin sin {τ τ}^{' '} cos cos {γ γ}^{' '} + + cos cos {τ τ}^{' '} sin sin {γ γ}^{' '} cos cos ((22 π π / / N N))))}{22 sin sin α α sin sin ((22 π π / / N N))} l l - - ((d d - - h h)) cos cos {τ τ}^{' '};;$

其中，in,

$α α = = arccos arccos \frac{22 ξ ξ sin sin φ φ - - 22 cos cos φ φ cos cos ((22 π π / / N N))}{22 \sqrt{11 + + {ξ ξ}^{22}} cos cos ((22 π π / / N N))}$

$l l = = \frac{a a \cdot &Center Dot; ((sin sin φ φ + + ξ ξ cos cos φ φ)) sin sin ((22 π π / / N N))}{\sqrt{11 + + {ξ ξ}^{22}} sin sin α α}$

$h h = = \frac{\sqrt{\frac{{a a}^{22} {sin sin}^{22} ((22 π π / / N N)) + + {22 c c}^{22} [[11 - - cos cos ((22 π π / / N N))]]}{(({a a}^{22} + + {c c}^{22}))} \cdot &Center Dot; {a a}^{22} - - {l l}^{22}}}{sin sin α α}$

${τ τ}^{' '} = = arccos arccos \frac{- - cos cos {γ γ}^{' '} cos cos α α cos cos ((22 π π / / N N)) + + sin sin {γ γ}^{' '} \sqrt{{sin sin}^{22} α α - - {cos cos}^{22} {γ γ}^{' '} {sin sin}^{22} ((22 π π / / N N))}}{11 - - {cos cos}^{22} {γ γ}^{' '} {sin sin}^{22} ((22 π π / / N N))};;$

(5)把滑轮和滑轨对应安装，保证结构中心轴通过滑轨的中心点，滑轮可以在滑轨的轨道中滑动。(5) Install the pulley and the slide rail correspondingly to ensure that the central axis of the structure passes through the center point of the slide rail, and the pulley can slide in the track of the slide rail.

本发明的优点在于直接利用三维环形机构构造可展结构，结构形式美观，且闭合时结构屋顶中央完全封闭，保证支座运动对称性，实现其向所需要的结构形态展开。采用六个四面体构件组成可展结构时，机构本身是单自由度过约束三维机构，单自由度使结构运动非常易于控制，避免了展开所需的复杂操作。此外，机构的过约束特性提供了相对较高的刚性和结构强度，这些优点使得所述可展结构尤为有用。The advantage of the present invention is that the expandable structure is directly constructed by the three-dimensional ring mechanism, and the structural form is beautiful, and the center of the roof of the structure is completely closed when it is closed, ensuring the symmetry of the movement of the support, and realizing its deployment to the required structural form. When six tetrahedral components are used to form a deployable structure, the mechanism itself is a single-degree-of-freedom three-dimensional mechanism, which makes the structure movement very easy to control and avoids the complicated operations required for deployment. In addition, the overconstrained nature of the mechanism provides relatively high rigidity and structural strength, which make the deployable structures particularly useful.

附图说明Description of drawings

下面将参照附图对本发明进一步说明。在附图中The present invention will be further described below with reference to the accompanying drawings. in the attached picture

图1(a)～(c)是向外翻转开启式可展结构闭合状态结构示意图；Figure 1 (a) to (c) are schematic diagrams of the closed state of the outward-turning open-type expandable structure;

图2(a)～(c)是向外翻转开启式可展结构展开过程中结构示意图；Figure 2(a)-(c) are structural schematic diagrams during the unfolding process of the outward-turning open-type expandable structure;

图3是N＝6时可展结构用垂直相邻轴且与相邻轴相交的杆件替换四面体的连杆机构的几何关系示意图；Fig. 3 is a schematic diagram of the geometric relationship of the link mechanism in which the expandable structure replaces the tetrahedron with a bar perpendicular to the adjacent axis and intersecting with the adjacent axis when N=6;

图4(a)和(b)是连接两个四面体构件的可替换的连接件形式简图；Fig. 4 (a) and (b) are the alternative connector form sketches that connect two tetrahedral members;

图5(a)～(c)是向外翻转开启式可展结构闭合状态、展开过程中和完全展开状态结构模型的平面图、立面图和轴测图；Figure 5(a)-(c) are the plan view, elevation view and axonometric view of the structure model of the outwardly flip open expandable structure in the closed state, in the process of unfolding and in the fully unfolded state;

图6是向外翻转开启式可展结构闭合状态四面体A₁ ABG计算示意图；Fig. 6 is a schematic diagram of the calculation of the tetrahedron A ₁ ABG in the closed state of the open expandable structure turned outward;

图7是向外翻转开启式可展结构展开过程中四面体A₁ ABG计算示意图；Fig. 7 is a schematic diagram of calculation of tetrahedron A ₁ ABG during the unfolding process of the outwardly flipped open deployable structure;

图8(a)、(b)、(c)、(d)和(e)是分别是向外翻转开启式可展结构闭合时顶面在水平方向投影为正六边形、三角形、正八边形、正方形和矩形的平面示意图；Figure 8 (a), (b), (c), (d) and (e) are the top surface projected in the horizontal direction as a regular hexagon, triangle and regular octagon when the open expandable structure is turned outwards and closed. , planar schematic diagrams of squares and rectangles;

图9(a)、(b)、(c)和(d)是向外翻转开启式可展结构闭合和展开状态网格结构图。Figure 9(a), (b), (c) and (d) are grid structure diagrams of the closed and unfolded states of the expandable structure turned outwards.

具体实施方式Detailed ways

本发明提供了一种可向外翻转开启式可展结构，以连杆机构和四面体旋转环的理论为基础。连杆是一种特殊形式的机构，它由一些相互连接的刚性构件组成，两个构件之间的连接体称为铰。铰有很多类型，本发明采用的是旋转铰，这种铰只有一个转动自由度；刚性构件可以是直杆，曲杆甚至是体。The invention provides an expandable structure that can be turned outwards and opened, based on the theory of a link mechanism and a tetrahedral rotating ring. A linkage is a special form of mechanism that consists of some interconnected rigid members, and the connection between two members is called a hinge. There are many types of hinges, and the present invention uses a rotary hinge, which has only one rotational degree of freedom; the rigid member can be a straight rod, a curved rod or even a body.

四面体旋转环可以看作四面体作为刚性构件的连杆机构。4N(N≥6，且N为偶数)个相同等边或等腰三角形组成N个相同四面体，每个四面体的两个对边分别与其他四面体的边相连形成环形，四面体可绕相邻公共边转动，形成三维机构，这个环形机构可以向内或向外连续转动。注意到，N＝6时四面体环的运动具有单自由度，如采用正四面体形成环，机构运动到某一状态时，两个面相互接触使运动停止。The tetrahedral rotating ring can be regarded as a linkage mechanism with tetrahedrons as rigid members. 4N (N ≥ 6, and N is an even number) identical equilateral or isosceles triangles form N identical tetrahedrons, and the two opposite sides of each tetrahedron are respectively connected with the sides of other tetrahedrons to form a ring, and the tetrahedron can be wound around Adjacent common sides rotate to form a three-dimensional mechanism, and this ring mechanism can continuously rotate inwards or outwards. Note that when N=6, the motion of the tetrahedral ring has a single degree of freedom. If a regular tetrahedron is used to form a ring, when the mechanism moves to a certain state, the two surfaces contact each other to stop the motion.

四面体旋转环中相邻两公共边相互垂直，运动到任何状态都有(N/2)个对称平面。如保证机构对称平面不变，改变四面体的形状，使相邻公共边不一定垂直，且不平行也不相交，四面体环仍然可以运动。它的特征在于，包括N个四面体构件并且借助于具有旋转轴线的铰沿特定方向可运动，且运动时保证机构在任何运动状态都有(N/2)个对称平面，在某些几何条件的限制下做单自由度运动。设一直线垂直于相邻两轴且与相邻两轴相交，把相邻两交点连线作为一个刚性杆，铰的轴线方向不变，则相邻铰的轴线距离即杆长a_i(i+1)、相邻铰的轴线夹角α_i(i+1)和铰i处相邻两杆端点沿轴线的距离R_i满足以下几何条件(图3表示N＝6时的连杆机构几何关系示意图)Two adjacent common sides in the tetrahedral rotating ring are perpendicular to each other, and there are (N/2) symmetry planes in any state. If the symmetry plane of the mechanism is kept unchanged and the shape of the tetrahedron is changed so that the adjacent common sides are not necessarily vertical, nor parallel nor intersecting, the tetrahedral ring can still move. It is characterized in that it includes N tetrahedral components and can move in a specific direction by means of a hinge with a rotation axis, and when moving, it is guaranteed that the mechanism has (N/2) symmetry planes in any state of motion. Under certain geometric conditions single-degree-of-freedom motion. Assuming a straight line is perpendicular to the two adjacent axes and intersects with the two adjacent axes, the line connecting two adjacent intersection points is regarded as a rigid rod, and the axis direction of the hinge remains unchanged, then the axial distance between adjacent hinges is the rod length a _{i(i +1)} , the axis angle α _i(i+1) of adjacent hinges and the distance R _i between the ends of two adjacent rods at hinge i along the axis satisfy the following geometric conditions (Figure 3 shows the geometry of the linkage mechanism when N=6 relationship diagram)

a₁₂＝a₂₃＝……＝a_N1＝la ₁₂ ＝a ₂₃ ＝...＝a _N1 ＝l

α₁₂＝α₃₄＝……＝α_(N-1)N＝α，α₂₃＝α₄₅＝……＝α_N1＝2π-α (1)α ₁₂ =α ₃₄ =...=α _(N-1)N =α, α ₂₃ =α ₄₅ =...=α _N1 =2π-α (1)

R₁＝R₂＝……＝R_N＝0R ₁ ＝R ₂ ＝...＝R _N ＝0

四面体旋转环吸引着许多人进行了研究和分析，但是到目前为止，对四面体旋转环的研究主要集中在四面体是规则四面体(作为公共边的四面体的两个对边是垂直的)且环形机构可以连续转动的情况，主要用于玩具设计等方面，很少有人把其用于结构。The tetrahedral rotating ring has attracted many people to study and analyze, but so far, the research on the tetrahedral rotating ring has mainly focused on the fact that the tetrahedron is a regular tetrahedron (the two opposite sides of the tetrahedron as the common side are perpendicular ) and the ring mechanism can rotate continuously, it is mainly used in toy design, etc., and few people use it for structure.

本发明结合四面体旋转环和连杆机构的概念，设计了一种向外翻转开启式可展结构，并提出其制作方法。步骤如下：The present invention combines the concepts of tetrahedral rotating ring and link mechanism, designs an outwardly turning and opening expandable structure, and proposes its manufacturing method. Proceed as follows:

一、本发明向外翻转开启式可展结构的几何形状和开启方式描述。1. Description of the geometric shape and opening method of the outwardly flip-open expandable structure of the present invention.

可展结构包括一个连杆机构，该机构由多个刚性四面体构件(1)组成，这些四面体构件(1)通过旋转连接件(例如，分别在图4(a)和图4(b)中的可替换连接件2)连接。The deployable structure consists of a linkage mechanism consisting of multiple rigid tetrahedral members (1) connected by rotating joints (e.g., in Fig. 4(a) and Fig. 4(b) 2) to connect the replaceable connectors in.

向外翻转开启式可展结构是由N(N≥6，且N为偶数)个四面体组成的环形机构时，图5是N＝6时结构模型。N个四面体间隔相同，相邻四面体构件(1)的边通过旋转连接件(2)连接形成环形(连接方式如图4)。相邻公共边不一定垂直，且不平行也不相交。结构闭合时相邻四面体的内侧面(6)接触，结构可以在自重和外力作用下保持稳定。(N/2)个公共边作为屋面轴(3)，另外(N/2)个公共边作为支座轴(4)，支座轴(4)向内倾斜一定角度，(N/2)个支座(7)分别位于支座轴(4)的底端。屋面轴(3)和支座轴(4)交替位于N个相互夹角为(2π/N)的竖直平面内，结构闭合和展开时屋面轴(3)和支座轴(4)在这一平面内运动。When the expandable structure flipped outward is a ring mechanism composed of N (N≥6, and N is an even number) tetrahedrons, Fig. 5 is a structure model when N=6. N tetrahedrons are at the same interval, and the sides of adjacent tetrahedron components (1) are connected to form a ring through a rotating connector (2) (the connection method is shown in Figure 4). Adjacent common sides are not necessarily perpendicular, nor parallel nor intersecting. When the structure is closed, the inner surfaces (6) of adjacent tetrahedrons are in contact, and the structure can remain stable under the action of its own weight and external force. The (N/2) common sides are used as the roof axis (3), and the other (N/2) common sides are used as the support axis (4), and the support axis (4) is inclined inward at a certain angle, (N/2) The bearings (7) are respectively located at the bottom of the bearing shaft (4). The roof axis (3) and the support axis (4) are alternately located in N vertical planes with a mutual angle of (2π/N). When the structure is closed and unfolded, the roof axis (3) and the support axis (4) are movement in a plane.

N个四面体刚性构件组成的向外翻转开启式可展结构，其几何形式可以采用多种。结构闭合时，顶面在水平方向的投影可以为正N边形(如图1(a)中ABCDEF)，每个四面体的四个面中作为结构顶面(5)的三角形是相同的等腰三角形，闭合时每个三角形的顶点重合于一点(8)；结构闭合时，顶面在水平方向的投影也可以为三角形、正方形、矩形或多边形(如图8)。The outward-turning and open-type expandable structure composed of N tetrahedral rigid components can adopt various geometric forms. When the structure is closed, the projection of the top surface in the horizontal direction can be a regular N-gon (ABCDEF in Figure 1(a)), and the triangles that serve as the top surface (5) of the structure among the four faces of each tetrahedron are the same, etc. When the waist triangle is closed, the vertices of each triangle coincide at one point (8); when the structure is closed, the projection of the top surface in the horizontal direction can also be triangle, square, rectangle or polygon (as shown in Figure 8).

当支座同时沿滑轨(9)由内向外运动时，四面体(1)绕其旋转连接件(2)转动，可使结构的相邻四面体的内侧面(6)相互接触，结构上部闭合；当支座同时沿滑轨(9)由外向内运动时，四面体(1)绕其旋转连接件(2)转动，结构上部向外翻转开启。When the support moves from inside to outside along the slide rail (9) at the same time, the tetrahedron (1) rotates around its rotating connector (2), which can make the inner surfaces (6) of adjacent tetrahedrons of the structure contact each other, and the upper part of the structure Closed; when the support moves from outside to inside along the slide rail (9) at the same time, the tetrahedron (1) rotates around its rotating connector (2), and the upper part of the structure turns outwards to open.

二、本发明向外翻转开启式可展结构中涉及到的几何参数。2. The geometric parameters involved in the outwardly flip-open expandable structure of the present invention.

设定结构几何参数如下：Set the structural geometry parameters as follows:

(1)a表示结构闭合时屋面在水平方向投影的正N边形边长(如图1(a)，a＞0)；(1) a represents the side length of the regular N-gon projected on the roof in the horizontal direction when the structure is closed (as shown in Figure 1(a), a > 0);

(2)c表示结构闭合时上部矢高(如图1(b)，OO’＝c，c＞0)，设相对矢高ξ＝c/a；(2) c represents the upper sagittal height when the structure is closed (as shown in Figure 1(b), OO’=c, c>0), and the relative sagittal height ξ=c/a;

(3)φ表示结构闭合时支座轴与水平方向的夹角 $(0 < φ < π - \arccos (1 / \sqrt{1 + ξ^{2}})$ 如图6)；(3) φ represents the angle between the support axis and the horizontal direction when the structure is closed $(0 < φ < π - \arccos (1 / \sqrt{1 + ξ^{2}})$ As shown in Figure 6);

(4)d表示支座轴长度(如图1(c)，BG＝DM＝FN＝d，0＜d＜a/cosφ)，设支座轴相对长度η＝d/a；(4) d represents the length of the support shaft (as shown in Figure 1(c), BG=DM=FN=d, 0<d<a/cosφ), and the relative length of the support shaft is set to η=d/a;

(5)γ′表示结构完全展开时屋面轴与水平方向夹角 $(\arccos (1 / \sqrt{1 + ξ^{2}}) < γ^{'} < π / 2)$ (5) γ′ represents the angle between the roof axis and the horizontal direction when the structure is fully expanded $(\arccos (1 / \sqrt{1 + ξ^{2}}) < γ^{'} < π / 2)$

(6)γ表示结构运动时支座轴与水平方向夹角(如图7)；(6) γ represents the angle between the support axis and the horizontal direction when the structure moves (as shown in Figure 7);

(7)τ表示结构运动时屋面轴与水平方向夹角(如图7)；(7) τ represents the angle between the roof axis and the horizontal direction when the structure moves (as shown in Figure 7);

(8)α表示相邻铰的轴线之间的夹角(如公式(1))；(8) α represents the angle between the axes of adjacent hinges (such as formula (1));

(9)l表示相邻铰的轴线之间的距离(如图公式(1))；(9) l represents the distance between the axes of adjacent hinges (as shown in formula (1));

(10)e、f和h表示垂直于结构相邻两铰轴线且与两轴相交的杆的端点沿轴线方向到四面体公共边端点的距离(如图6和图7，A₁A₀＝e，AA₀＝f，BB₀＝h，GB₀＝d-h)。(10) e, f and h represent the distance from the end point of the bar perpendicular to the axis of two adjacent hinges of the structure and intersecting the two axes to the end point of the common side of the tetrahedron (as shown in Figure 6 and Figure 7, A ₁ A ₀ = e, AA ₀ =f, BB ₀ =h, GB ₀ =dh).

(1)～(5)是结构的基本几何参数，决定了结构的几何形状和尺寸，结构的其他几何参数都可以由基本几何参数得到。(6)和(7)表示结构展开过程的参数，这两个变量随结构运动而发生变化。(8)～(10)是结构的间接几何参数，表示四面体的轴之间的关系以及四面体与垂直于两轴且与两轴相交的杆之间的关系。(1)～(5) are the basic geometric parameters of the structure, which determine the geometric shape and size of the structure. Other geometric parameters of the structure can be obtained from the basic geometric parameters. (6) and (7) represent the parameters of the structure unfolding process, and these two variables change with the structure movement. (8) to (10) are indirect geometric parameters of the structure, representing the relationship between the axes of the tetrahedron and the relationship between the tetrahedron and the rod perpendicular to and intersecting the two axes.

三、本发明向外翻转开启式可展结构中涉及的公式。3. The formulas involved in the outward-turning open-type expandable structure of the present invention.

本发明可以采用多种几何形式，以N个四面体组成的可展结构，且闭合时结构顶面在水平方向投影为正N边形的情况为例。The present invention can adopt a variety of geometric forms, taking the case of an expandable structure composed of N tetrahedrons, and the top surface of the structure projected as a regular N-gon in the horizontal direction when closed, as an example.

1.结构中四面体构件几何尺寸的确定1. Determination of geometric dimensions of tetrahedral members in the structure

结构N个四面体相同或对称，只需计算四面体A₁ABG的几何尺寸，如图6。以O’为原点，O’B为X轴，O’Q(O’Q垂直平面OO’B)为Y轴，O’O为Z轴建立直角坐标系O’XYZ。结构闭合时，各点坐标分别为A₁(0，0，c)，B(a，0，0)，A(acos(2π/N)，asin(2π/N)，0)，G(a-dcosφ，0，-dsinφ)。四面体A₁ABG的六条边的长度为The N tetrahedra of the structure are the same or symmetrical, and only need to calculate the geometric dimensions of the tetrahedron A ₁ ABG , as shown in Figure 6. Take O' as the origin, O'B as the X axis, O'Q (the vertical plane OO'B of O'Q) as the Y axis, and O'O as the Z axis to establish a rectangular coordinate system O'XYZ. When the structure is closed, the coordinates of each point are A ₁ (0, 0, c), B(a, 0, 0), A(acos(2π/N), asin(2π/N), 0), G(a -dcosφ, 0, -dsinφ). The lengths of the six sides of the tetrahedron A ₁ ABG are

BG＝ηa； $A_{1} A = A_{1} B = a \sqrt{1 + ξ^{2}}; AB = a \sqrt{2 (1 - \cos (2 π / N))};$ BG = ηa; $A_{1} A = A_{1} B = a \sqrt{1 + ξ^{2}}; AB = a \sqrt{2 (1 - \cos (2 π / N))};$

${A A}_{11} G G = = a a \sqrt{11 + + {ξ ξ}^{22} + + {η η}^{22} - - 22 η η ((cos cos φ φ - - ξ ξ sin sin φ φ))};; - - - - - - ((22))$

$AG AG = = a a \sqrt{22 ((11 - - η η cos cos φ φ)) ((11 - - cos cos ((22 π π / / N N)))) + + {η η}^{22}}$

2.确定结构间接几何参数α、l、e、f和h的值2. Determine the values of the structural indirect geometric parameters α, l, e, f and h

如图6，设直线A₀B₀与四面体两轴A₁A和BG垂直，且分别交于A₀和B₀，相邻轴的轴线夹角为α，距离为l，则A₀B₀＝l；杆延伸为四面体的延伸长度A₁A₀＝e，AA₀＝f，BB₀＝h。过A作AR∥BG且交平面O’XY于R，则∠OAR＝π-α，R点坐标为(0，asin(2π/N)，-atanφcos(2π/N))，满足几何关系As shown in Figure 6, let the straight line A ₀ B ₀ be perpendicular to the two axes A ₁ A and BG of the tetrahedron, and intersect A ₀ and B ₀ respectively, the angle between the adjacent axes is α, and the distance is l, then A ₀ B ₀ =l; the rod is extended into a tetrahedron with extension lengths A ₁ A ₀ =e, AA ₀ =f, BB ₀ =h. Make AR∥BG through A and intersect the plane O'XY at R, then ∠OAR=π-α, the coordinate of point R is (0, asin(2π/N), -atanφcos(2π/N)), satisfying the geometric relationship

${A A}_{11} A A = = a a \sqrt{11 + + {ξ ξ}^{22}}$

${A A}_{11} R R = = a a \sqrt{{sin sin}^{22} ((22 π π / / N N)) + + {tan the tan}^{22} φ φ {cos cos}^{22} ((22 π π / / N N)) + + 22 ξ ξ tan the tan φ φ cos cos ((22 π π / / N N)) + + {ξ ξ}^{22}} - - - - - - ((33))$

$AR AR = = a a \sqrt{11 + + {tan the tan}^{22} φ φ} \cdot \cdot cos cos ((22 π π / / N N))$

且 $\cos α = - \frac{A_{1} A^{2} + A R^{2} - A_{1} R^{2}}{2 \cdot A_{1} A \cdot AR} - - - (4)$ and $\cos α = - \frac{A_{1} A^{2} + A R^{2} - A_{1} R^{2}}{2 \cdot A_{1} A \cdot AR} - - - (4)$

把公式(3)代入公式(4)Substitute formula (3) into formula (4)

$α α = = arccos arccos \frac{22 ξ ξ sin sin φ φ - - 22 cos cos φ φ cos cos ((22 π π / / Ν Ν))}{22 \sqrt{11 + + {ξ ξ}^{22}} cos cos ((22 π π / / Ν Ν))} - - - - - - ((55))$

轴A₁A和BG的夹角α，距离l，则四面体A₁ABG的体积The angle α between axis A ₁ A and BG, distance l, then the volume of tetrahedron A ₁ ABG

$V V = = \frac{11}{66} {A A}_{11} A A \cdot \cdot BG BG \cdot \cdot l l \cdot &Center Dot; sin sin α α - - - - - - ((66))$

过A作直线AH⊥平面OBG交O’B于H，四面体A₁ABG的体积Make a straight line AH⊥ plane OBG intersect O'B at H through A, the volume of tetrahedron A ₁ ABG

$V V = = \frac{11}{33} AH AH \cdot &Center Dot; \frac{11}{22} BG BG \cdot \cdot {A A}_{11} B B \cdot &Center Dot; sin sin &angle; &angle; OBG OBG - - - - - - ((77))$

公式(6)和(7)相等，且有Formulas (6) and (7) are equal, and we have

$\sin &angle; OBG = \frac{ξ \cos φ + \sin φ}{\sqrt{1 + ξ^{2}}},$ AH＝asin(2π/N)，A₁A＝A₁B $\sin &angle; OBG = \frac{ξ \cos φ + \sin φ}{\sqrt{1 + ξ^{2}}},$ AH＝asin(2π/N), A ₁ A＝A ₁ B

得have to

$l l = = \frac{a a \cdot &Center Dot; ((sin sin φ φ + + ξ ξ cos cos φ φ)) sin sin ((22 π π / / N N))}{\sqrt{11 + + {ξ ξ}^{22}} sin sin α α} - - - - - - ((88))$

过A₀作AP∥BB₀且AP＝BB₀＝h，则BP＝A₀B₀＝l，∠A₁A₀P＝α，满足Make AP∥BB ₀ through A ₀ and AP=BB ₀ ＝h, then BP＝A ₀ B ₀ ＝l, ∠A ₁ A ₀ P＝α, satisfy

${A A}_{11} B B = = \sqrt{{l l}^{22} + + {e e}^{22} + + {h h}^{22} - - 22 eh eh cos cos α α} = = \sqrt{{a a}^{22} + + {c c}^{22}}$

$AB AB = = \sqrt{{l l}^{22} + + {f f}^{22} + + {h h}^{22} + + 22 fh fh cos cos α α} = = a a \sqrt{22 ((11 - - cos cos ((22 π π / / N N))))} - - - - - - ((99))$

${A A}_{11} A A = = e e + + f f = = \sqrt{{a a}^{22} + + {c c}^{22}}$

解得Solutions have to

$e e = = \frac{{a a}^{22} cos cos ((22 π π / / N N)) + + {c c}^{22}}{\sqrt{{a a}^{22} + + {c c}^{22}}} + + h h cos cos α α - - - - - - ((1010))$

$f f = = \frac{{a a}^{22} [[11 - - cos cos ((22 π π / / N N))]]}{\sqrt{{a a}^{22} + + {c c}^{22}}} - - h h cos cos α α$

3、确定结构展开过程中屋面轴与水平方向夹角γ和支座轴与水平方向夹角τ之间的关系3. Determine the relationship between the angle γ between the roof axis and the horizontal direction and the angle τ between the axis of the support and the horizontal direction during the unfolding of the structure

结构展开过程中示意图如图2，屋面轴A₁A、C₁C和E₁E的延长线交于O₁，支座轴BG、DM和FN的延长线交于O₂，平面A₀C₀E₀交轴O₁O₂于O₀₁，平面B₀D₀F₀交轴O₁O₂于O₀₂。结构N个四面体相同或对称，取四面体A₁ABG计算结构展开时的几何关系，如图7。The schematic diagram of the structure unfolding process is shown in Figure 2, the extension lines of the roof axes A ₁ A, C ₁ C and E ₁ E intersect at O ₁ , the extension lines of the support axes BG, DM and FN intersect at O ₂ , and the plane A ₀ C ₀ E ₀ intersects O ₁ O ₂ at O ₀₁ , plane B ₀ D ₀ F ₀ intersects O ₁ O ₂ at O ₀₂ . The N tetrahedrons of the structure are the same or symmetrical, and the geometric relationship when the structure is expanded is calculated using the tetrahedron A ₁ ABG , as shown in Figure 7.

把直角坐标系原点平移到O₀₁点，建立直角坐标系O₀₁X₀₁Y₀₁Z₀₁。过A₀作A₀S∥BB₀交平面O₀₁Y₀₁Z₀₁于S，设A₀O₁＝e’，各点坐标为A₀(e′cosγcos(2π/N)，e′cosγsin(2π/N)，0)，O₁(0，0，e′sinγ)，S(0，e′cosγsin(2π/N)，-e′cosγcos(2π/N)tanτ)，满足几何关系Translate the origin of the rectangular coordinate system to point O ₀₁ to establish the rectangular coordinate system O ₀₁ X ₀₁ Y ₀₁ Z ₀₁ . Make A ₀ S∥BB ₀ through A ₀ and intersect the plane O ₀₁ Y ₀₁ Z ₀₁ in S, let A ₀ O ₁ = e', the coordinates of each point are A ₀ (e′cosγcos(2π/N), e′cosγsin( 2π/N), 0), O ₁ (0, 0, e′sinγ), S(0, e′cosγsin(2π/N), -e′cosγcos(2π/N)tanτ), satisfying the geometric relationship

O₁A₀＝e′O ₁ A ₀ =e'

${A A}_{00} S S = = {e e}^{' '} cos cos γ γ \sqrt{11 + + {tan the tan}^{22} τ τ} \cdot \cdot cos cos ((22 π π / / N N)) - - - - - - ((1111))$

${O o}_{11} S S = = {e e}^{' '} \sqrt{11 + + {cos cos}^{22} γ γ (({tan the tan}^{22} τ τ - - 11)) {cos cos}^{22} ((22 π π / / N N)) + + 22 cos cos γ γ sin sin γ γ tan the tan τ τ cos cos ((22 π π / / N N))}$

且 $\cos α$ $= - \frac{O_{1} {A_{0}}^{2} + A_{0} S^{2} - O_{1} S^{2}}{2 A_{0} S \cdot O_{1} A_{0}} - - -$ $(12)$ and $\cos α$ $= - \frac{o_{1} {A_{0}}^{2} + A_{0} S^{2} - o_{1} S^{2}}{2 A_{0} S &Center Dot; o_{1} A_{0}} - - -$ $(12)$

把公式(11)代入公式(12)，得到结构展开过程中的几何参数关系Substituting formula (11) into formula (12), the relationship of geometric parameters in the process of structure expansion can be obtained

sinγsinτ-cosγcosτcos(2π/N)＝cosα (13)sinγsinτ-cosγcosτcos(2π/N)＝cosα (13)

结构闭合时， $τ_{0} = φ, γ_{0} = \arccos (1 / \sqrt{1 + ξ^{2}}) - - - (14)$ When the structure is closed, $τ_{0} = φ, γ_{0} = \arccos (1 / \sqrt{1 + ξ^{2}}) - - - (14)$

结构完全展开时，γ＝γ′，则When the structure is fully unfolded, γ=γ′, then

${τ τ}^{' '} = = arccos arccos \frac{- - cos cos {γ γ}^{' '} cos cos α α cos cos ((22 π π / / N N)) + + sin sin {γ γ}^{' '} \sqrt{{sin sin}^{22} α α - - {cos cos}^{22} {γ γ}^{' '} {sin sin}^{22} ((22 π π / / N N))}}{11 - - {cos cos}^{22} {γ γ}^{' '} {sin sin}^{22} ((22 π π / / N N))} - - - - - - ((1515))$

公式(13)描述了结构展开过程中变量τ和γ的关系。结构从闭合状态展开时，屋面轴与水平面夹角γ从γ₀到γ′逐渐增大，而支座轴与水平面夹角τ从τ₀到τ′逐渐减小。Equation (13) describes the relationship between the variables τ and γ during the structure unfolding process. When the structure unfolds from the closed state, the angle γ between the roof axis and the horizontal plane increases gradually from γ ₀ to γ′, while the angle τ between the support axis and the horizontal plane decreases gradually from τ ₀ to τ′.

4、确定滑轨的尺寸和位置4. Determine the size and position of the slide rail

设结构展开到某一状态时，正(N/2)边形A₀C₀E₀……边长为a₀，正(N/2)边形B₀D₀F₀……边长为b₀，平面A₀C₀E₀和B₀D₀F₀的距离为c₀，即O₀₁O₀₂＝c₀。When the structure is expanded to a certain state, the regular (N/2) polygon A ₀ C ₀ E ₀ ... has a side length of a ₀ , and the regular (N/2) polygon B ₀ D ₀ F ₀ ... has a side length of b ₀ , the distance between the planes A ₀ C ₀ E ₀ and B ₀ D ₀ F ₀ is c ₀ , that is, O ₀₁ O ₀₂ =c ₀ .

在直角坐标系O₀₁X₀₁Y₀₁Z₀₁中，各点坐标为 $A_{0} (\frac{a_{0} \cos (2 π / N)}{2 \sin (2 π / N)}, \frac{a_{0}}{2}, 0),$ In the Cartesian coordinate system O ₀₁ X ₀₁ Y ₀₁ Z ₀₁ , the coordinates of each point are $A_{0} (\frac{a_{0} \cos (2 π / N)}{2 \sin (2 π / N)}, \frac{a_{0}}{2}, 0),$

${B B}_{00} ((\frac{{b b}_{00}}{22 sin sin ((22 π π / / N N))},, 00,, - - {c c}_{00})),, {C C}_{00} ((\frac{{a a}_{00} cos cos ((22 π π / / N N))}{22 sin sin ((22 π π / / N N))},, - - \frac{{a a}_{00}}{22},, 00)),,$

$D_{0} (\frac{b_{0} \cos (4 π / N)}{2 \sin (2 π / N)}, - \frac{b_{0} \sin (4 π / N)}{2 \sin (2 π / N)}, - c_{0}), \dots \dots,$ 满足几何关系 ${D.}_{0} (\frac{b_{0} \cos (4 π / N)}{2 \sin (2 π / N)}, - \frac{b_{0} \sin (4 π / N)}{2 \sin (2 π / N)}, - c_{0}), \dots \dots,$ satisfies the geometric relationship

$tan the tan τ τ = = \frac{{b b}_{00} - - {a a}_{00} cos cos ((22 π π / / N N))}{{22 c c}_{00} sin sin ((22 π π / / N N))}$

$tan the tan γ γ = = \frac{{a a}_{00} - - {b b}_{00} cos cos ((22 π π / / N N))}{22 {c c}_{00} sin sin ((22 π π / / N N))} - - - - - - ((1616))$

$l l = = \sqrt{\frac{{a a}_{00}^{22} - - 22 {a a}_{00} {b b}_{00} cos cos ((22 π π / / N N)) + + {b b}_{00}^{22}}{{44 sin sin}^{22} ((22 π π / / N N))} + + {c c}_{00}^{22}}$

公式(16)解得Formula (16) can be solved

${a a}_{00} = = \frac{22 ((tan the tan γ γ + + tan the tan τ τ cos cos ((22 π π / / N N))))}{(({tan the tan}^{22} τ τ + + 22 tan the tan τ τ tan the tan γ γ cos cos ((22 π π / / N N)) + + {tan the tan}^{22} γ γ)) + + {sin sin}^{22} ((22 π π / / N N))} l l$

${b b}_{00} = = \frac{22 ((tan the tan γ γ cos cos ((22 π π / / N N)) + + tan the tan τ τ))}{\sqrt{(({tan the tan}^{22} τ τ + + 22 tan the tan τ τ tan the tan γ γ cos cos ((22 π π / / N N)) + + {tan the tan}^{22} γ γ)) + + {sin sin}^{22} ((22 π π / / N N))}} l l - - - - - - ((1717))$

${c c}_{00} = = \frac{sin sin ((22 π π / / N N))}{\sqrt{(({tan the tan}^{22} τ τ + + 22 tan the tan τ τ tan the tan γ γ cos cos ((22 π π / / N N)) + + {tan the tan}^{22} γ γ)) + + {sin sin}^{22} ((22 π π / / N N))}} l l$

把公式(13)代入公式(17)得Substitute formula (13) into formula (17) to get

${a a}_{00} = = \frac{22 ((sin sin τ τ cos cos γ γ cos cos ((22 π π / / N N)) + + cos cos τ τ sin sin γ γ))}{sin sin α α} l l$

${b b}_{00} = = \frac{22 ((sin sin τ τ cos cos γ γ + + cos cos τ τ sin sin γ γ cos cos ((22 π π / / N N))))}{sin sin α α} l l - - - - - - ((1818))$

${c c}_{00} = = \frac{cos cos τ τ cos cos γ γ sin sin ((22 π π / / N N))}{sin sin α α} l l$

N个四面体刚性构件组成的可展结构，(N/2)条滑轨(9)在同一平面，相互夹角为(4π/N)且延长线可交于一点，这点作为滑轨中心点。滑轨(9)最外侧点(图1(a)和(b)中G、M和N)的位置相当于结构闭合时支座(7)的位置，距离中心点距离公式A deployable structure composed of N tetrahedral rigid members, (N/2) sliding rails (9) are on the same plane, the mutual angle is (4π/N) and the extension lines can intersect at one point, which is used as the center of the sliding rail point. The position of the outermost point of the slide rail (9) (G, M and N in Figure 1(a) and (b)) is equivalent to the position of the support (7) when the structure is closed, and the distance from the center point is given by the formula

J₁＝a-dcosφ (19)J ₁ ＝a-dcosφ (19)

滑轨(9)最内侧的点(图1(a)、(b)和图2(a)、(b)中S_G、S_M和S_N)的位置相当于结构完全展开时支座(7)的位置，距离中心的距离公式The position of the innermost point of the slide rail (9) (S _G , S _M and S _N in Fig. 1(a), (b) and Fig. 2(a), (b)) is equivalent to the position of the support when the structure is fully expanded ( 7) The position, the distance formula from the center

${J J}_{22} = = \frac{{b b}_{00}}{22 sin sin ((22 π π / / N N))} - - ((d d - - h h)) cos cos {τ τ}^{' '} - - - - - - ((2020))$

把公式(18)代入公式(20)得Substitute formula (18) into formula (20) to get

${J J}_{22} = = \frac{22 ((sin sin {τ τ}^{' '} cos cos {γ γ}^{' '} + + cos cos {τ τ}^{' '} sin sin {γ γ}^{' '} cos cos ((22 π π / / N N))))}{22 sin sin α α sin sin ((22 π π / / N N))} l l - - ((d d - - h h)) cos cos {τ τ}^{' '} - - - - - - ((21 twenty one))$

四、本发明向外翻转开启式可展结构的制作方法4. The manufacturing method of the outwardly flipped open expandable structure of the present invention

1、根据前面叙述，本发明可展结构的基本形状已经确定。选取合适的几何参数：结构闭合时顶面在水平方向的正N边形投影边长a，结构闭合时上部矢高c，结构闭合时支座轴(4)与水平方向夹角φ，支座轴(4)的长度d和结构完全展开时屋面轴(3)与水平方向的夹角γ′。1. According to the foregoing description, the basic shape of the deployable structure of the present invention has been determined. Select appropriate geometric parameters: when the structure is closed, the side length a of the regular N-gon projection of the top surface in the horizontal direction is a, when the structure is closed, the upper saggy height c, when the structure is closed, the angle between the support axis (4) and the horizontal direction φ, the support axis The length d of (4) and the angle γ' between the roof axis (3) and the horizontal direction when the structure is fully expanded.

2、根据公式(2)，确定组成结构的N个四面体刚性构件的几何尺寸，根据尺寸设计N个四面体构件。2. According to the formula (2), determine the geometric dimensions of the N tetrahedral rigid components that make up the structure, and design the N tetrahedral components according to the dimensions.

3、正确排列N个四面体构件(1)之间的环形位置关系，确定四面体刚性构件之间的连接形式，把相邻四面体构件(1)通过可替换的旋转连接件(2)分别在屋面轴(3)和支座轴(4)的位置连接，支座轴的底端为支座(4)，在支座的位置安装滑轮，滑轮可滑动方向沿正N边形外接圆的径向方向。注意确认四面体构件(1)的屋面轴(3)和支座轴(4)的正确位置。3. Correctly arrange the annular positional relationship between the N tetrahedral components (1), determine the connection form between the tetrahedral rigid components, and separate the adjacent tetrahedral components (1) through the replaceable rotary connectors (2) Connect at the position of the roof shaft (3) and the support shaft (4), the bottom end of the support shaft is the support (4), and a pulley is installed at the position of the support, and the sliding direction of the pulley is along the circumcircle of the regular N-gon radial direction. Pay attention to confirm the correct position of the roof axis (3) and support axis (4) of the tetrahedron member (1).

4、根据公式(19)和(21)，确定滑轨(9)最外侧的点(图1(a)和(b)中G、M和N)和滑轨(9)最内侧的点(图1(a)、(b)和图2(a)、(b)中S_G、S_M和S_N)距中心的距离，按照图1和图2中滑轨(9)基本示意图设计滑轨，(N/2)条滑轨在同一平面，相互夹角为(4π/N)且延长线可交于一点，这点作为滑轨中心点。4. According to the formulas (19) and (21), determine the outermost point of the slide rail (9) (G, M and N in Fig. 1 (a) and (b)) and the innermost point of the slide rail (9) ( The distance between S _G , S _M and S _N in Fig. 1(a), (b) and Fig. 2(a), (b) and the center, according to the basic schematic diagram of slide rail (9) in Fig. 1 and Fig. 2, design the slide Rail, (N/2) slide rails are on the same plane, the mutual angle is (4π/N) and the extension lines can intersect at one point, which is used as the center point of the slide rail.

5、把滑轮和滑轨对应安装，保证结构中心轴通过滑轨的中心点，滑轮可以在滑轨的轨道中滑动。5. Install the pulley and the slide rail correspondingly to ensure that the central axis of the structure passes through the center point of the slide rail, and the pulley can slide in the track of the slide rail.

五、在前面所述中，可展结构的N个四面体构件当作N个实体来考虑。实际上，本发明中的N个四面体还可以是杆件构成的网格组成的四面体(如图9)，也可以与膜材结合，这样可大幅度降低结构的密度和重量，对工程应用非常有利。5. In the foregoing, the N tetrahedral components of the deployable structure are considered as N entities. In fact, the N tetrahedrons in the present invention can also be tetrahedrons composed of grids formed by rods (as shown in Figure 9), and can also be combined with membrane materials, which can greatly reduce the density and weight of the structure. Application is very beneficial.

六、本发明的优点在于直接利用三维环形机构构造可展结构，结构形式美观，且闭合时结构屋顶中央完全封闭，保证支座运动对称性，实现其向所需要的结构形态展开。采用六个四面体构件组成可展结构时，机构本身是单自由度过约束三维机构，单自由度使结构运动非常易于控制，避免了展开所需的复杂操作。此外，机构的过约束特性提供了相对较高的刚性和结构强度，这些优点使得所述可展结构尤为有用。6. The advantage of the present invention is that the expandable structure is directly constructed by the three-dimensional ring mechanism, and the structural form is beautiful, and the center of the roof of the structure is completely closed when it is closed, ensuring the symmetry of the support movement and realizing its deployment to the required structural form. When six tetrahedral components are used to form a deployable structure, the mechanism itself is a single-degree-of-freedom three-dimensional mechanism, which makes the structure movement very easy to control and avoids the complicated operations required for deployment. In addition, the overconstrained nature of the mechanism provides relatively high rigidity and structural strength, which make the deployable structures particularly useful.

七、本发明可应用于各种尺寸的可开启屋盖结构，这中可开启屋盖结构的开启方式是向外翻转开启式。7. The present invention can be applied to openable roof structures of various sizes, wherein the opening method of the openable roof structure is an outward-turning opening type.

Claims

1. An outwards-overturning opening type deployable structure is characterized by comprising a movable link mechanism, wherein the link mechanism consists of N tetrahedral rigid components (1), the tetrahedral rigid components are connected through a roof shaft (3) and a rotary connecting piece (2) of a support shaft (4) to form a three-dimensional annular mechanism, a support (7) is arranged at the bottom end of the support shaft (4), when the support moves from inside to outside along a sliding rail (9), the tetrahedrons (1) rotate around the rotary connecting piece (2) of the tetrahedrons, so that the inner side surfaces (6) of adjacent tetrahedrons of the structure are in contact with each other, and the upper part of the structure is closed; when the support moves from outside to inside along the sliding rail (9), the tetrahedron (1) rotates around the rotating connecting piece (2), and the upper part of the structure is turned outwards and opened.

2. A flip-out open deployable structure according to claim 1, wherein the N tetrahedral rigid members are an even number of tetrahedral rigid members equal to or greater than six.

3. A structure deployable in an outwardly turned open configuration according to claim 1 or 2, wherein when the structure of N tetrahedron-shaped rigid members is closed, the projection of the top surface (5) in the horizontal direction is a regular N-sided polygon, the triangles of the four faces of each tetrahedron as the top surfaces (5) of the structure are identical isosceles triangles, and the vertices (8) of the N triangles of the top surface coincide at a point when the structure is closed.

4. A structure deployable in an outwardly turned open position according to claim 1 or 2, wherein the horizontal projection of the top surface (5) when the structure of N tetrahedral rigid members is closed is triangular, square, rectangular or polygonal.

5. A method of making an outwardly turned open deployable structure according to claim 1, further comprising the steps of:

(1) the basic geometrical parameters for designing a developable structure consisting of N tetrahedral rigid members are as follows: when the structure is closed, the side length a of the regular N-shaped polygon projected in the horizontal direction of the roof is larger than 0; when the structure is closed, the upper rise c is larger than 0, and the relative rise xi is c/a; when the structure is closed, the included angle phi between the support shaft (4) and the horizontal direction,

<math> <mrow> <mn>0</mn> <mo><</mo> <mi>φ</mi> <mo><</mo> <mi>π</mi> <mo>-</mo> <mi>arccos</mi> <mrow> <mo>(</mo> <mn>1</mn> <mo>/</mo> <msqrt> <mn>1</mn> <mo>+</mo> <msup> <mi>ξ</mi> <mn>2</mn> </msup> </msqrt> <mo>)</mo> </mrow> <mo>;</mo> </mrow> </math>

the length d of the support shaft (4) is more than 0 and less than a/cos phi, and the relative length eta of the support shaft is d/a; when the structure is completely unfolded, the roof axis (3) forms an included angle with the horizontal direction

<math> <mrow> <msup> <mi>γ</mi> <mo>′</mo> </msup> <mrow> <mo>(</mo> <mi>arccos</mi> <mrow> <mo>(</mo> <mn>1</mn> <mo>/</mo> <msqrt> <mn>1</mn> <mo>+</mo> <msup> <mi>ξ</mi> <mn>2</mn> </msup> </msqrt> <mo>)</mo> </mrow> <mo><</mo> <msup> <mi>γ</mi> <mo>′</mo> </msup> <mo><</mo> <mi>π</mi> <mo>/</mo> <mn>2</mn> <mo>)</mo> </mrow> <mo>;</mo> </mrow> </math>

(2) Determining the dimensions of N tetrahedral rigid members constituting the structure, wherein one tetrahedron A is represented₁The formula for the length of the six sides of ABG is as follows:

the geometry and dimensions of the remaining (N-1) tetrahedra can also be determined according to the above formula;

(3) determining a connection form between tetrahedral rigid members, wherein adjacent tetrahedral rigid members (1) are respectively connected at positions of a roof shaft (3) and a support shaft (4) through replaceable rotary connectors (2) to form a three-dimensional annular mechanism, the bottom end of the support shaft (4) is provided with a support (7), a pulley is arranged at the position of the support (7), and the sliding direction of the pulley is along the radial direction of a circumscribed circle of a regular N-edge;

(4) the form and the size of the slide rail are designed, and (N/2) slide rails (9) are in the same plane, the mutual included angle is (4 pi/N), and the extension lines can intersect at one point which is used as the center point of the slide rail. The position of the outermost point of the slide rail (9) is equivalent to the position of the support (7) when the structure is closed, and the distance formula from the central point

J₁＝a-dcosφ

The position of the innermost point of the slide rail (9) is equivalent to the position of the support (7) when the structure is completely unfolded, and the distance formula from the center

Wherein,

<math> <mrow> <mi>α</mi> <mo>=</mo> <mi>arccos</mi> <mfrac> <mrow> <mn>2</mn> <mi>ξ</mi> <mi>sin</mi> <mi>φ</mi> <mo>-</mo> <mn>2</mn> <mi>cos</mi> <mi></mi> <mi>φ</mi> <mi>cos</mi> <mrow> <mo>(</mo> <mn>2</mn> <mi>π</mi> <mo>/</mo> <mi>N</mi> <mo>)</mo> </mrow> </mrow> <mrow> <mn>2</mn> <msqrt> <mn>1</mn> <mo>+</mo> <msup> <mi>ξ</mi> <mn>2</mn> </msup> </msqrt> <mi>cos</mi> <mrow> <mo>(</mo> <mn>2</mn> <mi>π</mi> <mo>/</mo> <mi>N</mi> <mo>)</mo> </mrow> </mrow> </mfrac> </mrow> </math>

<math> <mrow> <msup> <mi>τ</mi> <mo>′</mo> </msup> <mo>=</mo> <mi>arccos</mi> <mfrac> <mrow> <mo>-</mo> <mi>cos</mi> <msup> <mi>γ</mi> <mo>′</mo> </msup> <mi>cos</mi> <mi></mi> <mi>α</mi> <mi>cos</mi> <mrow> <mo>(</mo> <mn>2</mn> <mi>π</mi> <mo>/</mo> <mi>N</mi> <mo>)</mo> </mrow> <mo>+</mo> <mi>sin</mi> <msup> <mi>γ</mi> <mo>′</mo> </msup> <msqrt> <msup> <mi>sin</mi> <mn>2</mn> </msup> <mi>α</mi> <mo>-</mo> <msup> <mi>cos</mi> <mn>2</mn> </msup> <msup> <mi>γ</mi> <mo>′</mo> </msup> <mi>si</mi> <msup> <mi>n</mi> <mn>2</mn> </msup> <mrow> <mo>(</mo> <mn>2</mn> <mi>π</mi> <mo>/</mo> <mi>N</mi> <mo>)</mo> </mrow> </msqrt> </mrow> <mrow> <mn>1</mn> <mo>-</mo> <msup> <mi>cos</mi> <mn>2</mn> </msup> <msup> <mi>γ</mi> <mo>′</mo> </msup> <mrow> <mo>(</mo> <mn>2</mn> <mi>π</mi> <mo>/</mo> <mi>N</mi> <mo>)</mo> </mrow> </mrow> </mfrac> <mo>;</mo> </mrow> </math>

(5) the pulley and the slide rail are correspondingly installed, so that the central shaft of the structure is ensured to pass through the central point of the slide rail, and the pulley can slide in the track of the slide rail.