CN104376136A - Aircraft casing interlayer structure and implementation method thereof - Google Patents

Abstract

The invention discloses an aircraft casing interlayer structure and an implementation method thereof. The aircraft casing interlayer structure can be applied to casings of aircrafts such as passenger planes, unmanned aerial vehicles, rockets and guided missiles as well as casings of submersibles such as submarines.

Description

Aircraft casing sandwich construction and its implementation

Technical field

What the present invention relates to is a kind of method in aircraft manufacturing and computer image processing technology field, specifically a kind of be convenient to realize in computer program aircraft casing sandwich construction and its implementation.

Background technology

The three-dimensional paper folding structure of many research displays has good specific strength, energy absorption characteristics and sound damping.Based on these characteristics of three-dimensional paper folding structure, they have been applied in air-drop technology fast, as the buffering energy-absorbing structure delivered bottom thing.Meanwhile, these structures can as the interlayer of the crustless sandwich structure of a kind of excellent aircraft or underwater vehicle, thus replace at present with honeycomb sandwich construction more widely.The housing section of aircraft or underwater vehicle is rounded (such as rocket, guided missile) or the closed loop (such as passenger plane, submarine) that is made up of several sections of circular arcs generally.When paper folding structure is applied to these housings, an important technical matters is when the physical dimension of given housing section, designs the paper folding structure that cross section therewith matches.

Summary of the invention

The present invention is directed to prior art above shortcomings, propose a kind of aircraft casing sandwich construction and its implementation, this crustless sandwich structure can be used for aircraft, such as the housing of passenger plane, unmanned plane, rocket, guided missile etc., also can be used for underwater vehicle, the housing of such as submarine.

The present invention is achieved by the following technical solutions:

The present invention relates to a kind of implementation method of aircraft sandwich structures, this sandwich construction is made up of inner casing, shell and the multiple interlayers with paper folding structure be clipped between inside and outside shell.

Described interlayer realizes in the following manner:

Step one: according to the internal diameter R of sandwich construction _in, external diameter R _out, the thickness t of inner casing _in, shell thickness t _exand segment length l _segcalculate the external diameter r of required interlayer ₁, internal diameter r ₂and length l:r ₁=R _out-t _ex; r ₂=R _in+ t _in; L=l _seg;

Step 2: m the input point determining the x-z plane of three-dimensional cartesian coordinate system and the n+2 of a y-z plane input point

Step 3: m × n the summit V obtaining paper folding structure according to input point _{i, j}coordinate, $V_{i,j}=\left[\begin{array}{c}{x}_{i,j}\\ y_{i,j}\\ z_{i,j}\end{array}\right]=V_j^y+$ $\left[A_j\right]V_i^x,i=\mathrm{1,2},...,m;j=\mathrm{1,2},...,n,$ Wherein: $\left[A_j\right]=\left[\begin{array}{ccc}1& 0& 0\\ 0& 0& {(-1)}^j\frac{\cos {\mathrm{\θ}}_{j-1}+\cos {\mathrm{\θ}}_j}{\sin ({\mathrm{\θ}}_{j-1}-{\mathrm{\θ}}_j)}\\ 0& 0& {(-1)}^j\frac{\sin {\mathrm{\θ}}_{j-1}+\sin {\mathrm{\θ}}_j}{\sin ({\mathrm{\θ}}_{j-1}-{\mathrm{\θ}}_j)}\end{array}\right],$ $\sin {\mathrm{\θ}}_j=$ $\frac{i_z\·(V_{j+1}^y-V_j^y)}{\left|\right|V_{j+1}^y-V_j^y\left|\right|},$ i _y=[0 1 0] ^tfor the vector of unit length of y coordinate axis, i _z=[0 0 1] ^tfor the vector of unit length of z coordinate axle, || ■ || represent that subtend measures mould.

Step 4: definition { V _{i, j}v _{i+1, j}or { V _{i, j}v _{i, j+1}be a pair adjacent vertex.Coupled together by all adjacent vertex straight lines, namely the connecting line segment between these adjacent vertexs constitutes the scrimp of paper folding structure, and adopts area of computer aided to realize interlayer manufacture further.

Described adjacent vertex refers to: with V _{i, j}for summit, then its consecutive point are { V _{i, j}v _{i+1, j}or { V _{i, j}v _{i, j+1}.

Described computer-aided manufacturing comprises: manufacture the mould corresponding with paper folding structure, utilize mould to carry out shaping to flat sheet; Three-dimensional printing technology is utilized to print; Utilize and manufacture the similar way of comb core, namely first utilizing the material that can conveniently fold to carry out craft folding, and then immersing in glue and the structure folded is shaped and reinforces.

Accompanying drawing explanation

Fig. 1 is crustless sandwich structural representation.

Fig. 2 is adjacent vertex schematic diagram.

Fig. 3 is embodiment 1 paper folding structure three-dimensional schematic diagram.

Fig. 4 is the summit V of embodiment 1 _{i, j}subpoint on the y-z plane and radius are the circle of 19 and 21.

Fig. 5 is the schematic three dimensional views of embodiment 2 paper folding structure.

The schematic three dimensional views of the paper folding structure that Fig. 6 obtains for embodiment 3.

Fig. 7 is the partial enlarged drawing of Fig. 6 structure.

Fig. 8 is embodiment 3 paper folding structure partial zoomed-in view.

Fig. 9 is the schematic three dimensional views of embodiment 4 paper folding structure.

The 3-D view of the paper folding structure that Figure 10 obtains for embodiment 5.

Figure 11 is embodiment 5 paper folding structure three-dimensional view.

Figure 12 is the summit V of embodiment 5 _{i, j}subpoint on the y-z plane and radius are the round schematic diagram of 19 and 21.

Figure 13 is the schematic three dimensional views of embodiment 6 paper folding structure.

The 3-D view of the paper folding structure that Figure 14 obtains for embodiment 7.

Figure 15 is the partial enlarged view of structure shown in Figure 14.

Figure 16 is the summit V of embodiment 7 _{i, j}subpoint on the y-z plane and radius are the round schematic diagram of 19 and 21.

The 3-D view of the paper folding structure that Figure 17 obtains for embodiment 8.

Figure 18 is the partial enlarged view of structure shown in Figure 17.

Figure 19 is the summit V of embodiment 8 _{i, j}subpoint on the y-z plane and radius are the round schematic diagram of 19 and 21.

The 3-D view of the paper folding structure that Figure 20 obtains for embodiment 8.

Figure 21 is the partial enlarged view of structure shown in Figure 20.

Figure 22 is the summit V of embodiment 8 _{i, j}subpoint on the y-z plane and radius are the schematic diagram of the circle of 19 and 21.

The 3-D view of the paper folding structure that Figure 23 obtains for embodiment 9.

Figure 24 is the partial enlarged view of structure shown in Figure 22.

Figure 25 is the schematic three dimensional views of embodiment 12 paper folding structure.

Figure 26 is the 3-D view of obtained paper folding structure.

Figure 27 is the partial enlarged view of structure shown in Figure 26.

Figure 28 is all summit V _{i, j}subpoint on the y-z plane and radius are the schematic diagram of the circle of 19 and 21.

Figure 29 is the schematic three dimensional views of embodiment 14 paper folding structure.

Figure 30 is the 3-D view of obtained paper folding structure.

Figure 31 is the partial enlarged view of structure shown in Figure 30.

Figure 32 is all summit V _{i, j}subpoint on the y-z plane and radius are the schematic diagram of the circle of 19 and 21.

Figure 33 is the schematic three dimensional views of embodiment 16 paper folding structure.

Figure 34 is the 3-D view of obtained paper folding structure.

Figure 35 is the partial enlarged view of structure shown in Figure 34.

Figure 36 is the summit V of embodiment 17 _{i, j}subpoint on the y-z plane and radius are the schematic diagram of the circle of 19 and 21.

Figure 37 is the schematic three dimensional views of embodiment 17 paper folding structure.

Figure 38 is embodiment 19 housing section schematic diagram.

Figure 39 is the schematic three dimensional views of embodiment 19 paper folding structure.

Figure 40 is the sectional view of embodiment 19 paper folding structure.

Figure 41 is for comprising middle case face schematic diagram.

Figure 42 is not for comprise middle case face schematic diagram.

Embodiment

Elaborate to embodiments of the invention below, the present embodiment is implemented under premised on technical solution of the present invention, give detailed embodiment and concrete operating process, but protection scope of the present invention is not limited to following embodiment.

Embodiment 1

Make the internal diameter R of sandwich construction _in=18, external diameter R _out=22, the thickness t of inner casing _in=1, the thickness t of shell _ex=1, segment length l _seg=11.By r ₁=R _out-t _ex, r ₂=R _in+ t _in, l=l _segobtain the external diameter r of interlayer ₁=21, internal diameter r ₂=19, length l=11.

Get m=12, h=1, a=1, by ${\mathrm{\Σ}}_{k=1}^m(\frac{1-{(-1)}^k}{2}a+\frac{1+{(-1)}^k}{2}b)=l+a$ Known b=1.By $V_i^x={\left[\begin{array}{ccc}{\mathrm{\Σ}}_{k=1}^i(\frac{1-{(-1)}^k}{2}a+\frac{1+{(-1)}^k}{2}b)& 0& -h+{\mathrm{\Σ}}_{k=1}^i{(-1)}^{\frac{i-1}{2}}(1-{(-1)}^i)h\end{array}\right]}^T,i=1,...,m$ Obtain the input point of 12 x-z planes:

$V_1^x={\left[\begin{array}{ccc}1& 0& 1\end{array}\right]}^T;$ $V_2^x={\left[\begin{array}{ccc}2& 0& 1\end{array}\right]}^T;$ $V_3^x={\left[\begin{array}{ccc}3& 0& -1\end{array}\right]}^T;$ $V_4^x={\left[\begin{array}{ccc}4& 0& -1\end{array}\right]}^T;$ $V_5^x=$ ${\left[\begin{array}{ccc}5& 0& 1\end{array}\right]}^T;$ $V_6^x={\left[\begin{array}{ccc}6& 0& 1\end{array}\right]}^T;$ $V_7^x={\left[\begin{array}{ccc}7& 0& -1\end{array}\right]}^T;$ $V_8^x={\left[\begin{array}{ccc}8& 0& -1\end{array}\right]}^T;$ $V_9^x=$ ${\left[\begin{array}{ccc}9& 0& 1\end{array}\right]}^T;$ $V_{10}^x={\left[\begin{array}{ccc}10& 0& 1\end{array}\right]}^T;$ $V_{11}^x={\left[\begin{array}{ccc}11& 0& -1\end{array}\right]}^T;$ $V_{12}^x={\left[\begin{array}{ccc}12& 0& -1\end{array}\right]}^T;$ Get N=30, by obtain β=π/30.

According to $r_2^2={(r-\mathrm{\δ})}^2+h^2(1+{\left(\frac{(r+\mathrm{\δ})\sin \mathrm{\β}}{(r+\mathrm{\δ})\cos \mathrm{\β}-(r-\mathrm{\δ})}\right)}^2),$

$r_1^2={(r+\mathrm{\δ})}^2+h^2(1+{\left(\frac{(r-\mathrm{\δ})\sin \mathrm{\β}}{(r+\mathrm{\δ})-(r-\mathrm{\δ})\cos \mathrm{\β}}\right)}^2)$ Calculate r=19.9470 and δ=1.0084.

Get n=2N+1=61, by $V_j^y=[r+{(-1)}^j\mathrm{\δ}]\left[\begin{array}{c}0\\ \sin \left(\mathrm{j\β}\right)\\ \cos \left(\mathrm{j\β}\right)\end{array}\right],j=\mathrm{0,1},...,n+1$ Obtain the input point of 63 y-z planes: $V_j^y=[19.9470+{(-1)}^{j}1.0084]\left[\begin{array}{c}0\\ \sin \left(\frac{\mathrm{j\π}}{30}\right)\\ \cos \left(\frac{\mathrm{j\π}}{30}\right)\end{array}\right],j=\mathrm{0,1}...,62;$ By $V_{i,j}=\left[\begin{array}{c}{x}_{i,j}\\ y_{i,j}\\ z_{i,j}\end{array}\right]=V_j^y+\left[A_j\right]V_i^x,i=$ $\mathrm{1,2},...,m;j=\mathrm{1,2},...,n$ Calculate 12 × 61 summit V _{i, j}coordinate.

Finally define scrimp according to step 7.Fig. 3 is the 3-D view of obtained paper folding structure, and it joins end to end, and forms 1 closed loop configuration.Fig. 4 is all summit V _{i, j}subpoint on the y-z plane and radius are the circle of 19 and 21, and these two circles are all dropped on circumferentially in visible all summits.This paper folding structure shown in key diagram 3 reach radial dimension designing requirement.The axial length of structure shown in Fig. 3 equals 11, reaches axial dimension designing requirement.

Embodiment 2

Change the parameter n in embodiment 1 and keep other parameter constants, the paper folding structure of non-closed loop can be obtained.Such as when n gets 21, paper folding Structure composing 1/3 annulus obtained, as shown in Figure 5.

Step 1: according to the internal diameter R of sandwich construction _in, external diameter R _out, the thickness t of inner casing _in, shell thickness t _exand segment length l _seg, utilize formula (13-15) to calculate the external diameter r of required interlayer ₁and internal diameter r ₂and length l.

r ₁＝R _out-t _ex(13)；r ₂＝R _in+t _in(14)；l＝l _seg(15)；

Step 2: m the input point being determined the x-z plane of three-dimensional cartesian coordinate system by formula (16a), wherein, m is natural number.

$V_i^x={\left[\begin{array}{ccc}{\mathrm{\Σ}}_{k=1}^i(\frac{1-{(-1)}^k}{2}a+\frac{1+{(-1)}^k}{2}b)& 0& -h+{\mathrm{\Σ}}_{k=1}^i{(-1)}^{\frac{i-1}{2}}(1-{(-1)}^i)h\end{array}\right]}^T,i=$ $1,...,m---\left(16a\right);$ Wherein, parameter a and b meets: ${\mathrm{\Σ}}_{k=1}^m(\frac{1-{(-1)}^k}{2}a+\frac{1+{(-1)}^k}{2}b)=l+a---\left(16b\right);$

Step 3: selected Parameter N, wherein N be more than or equal to 3 natural number.Parameter beta is calculated according to formula (17) _set; ${\mathrm{\β}}_{\mathrm{set}}=\frac{\mathrm{\π}}{N}---\left(17\right);$

Step 4: selected parameter beta, β ₁and β ₂, make it meet β+β ₁+ β ₂=β _set, and meet β > β ₁and β > β ₂.

Step 5: calculate parameter r and δ according to formula (18-19).

$r_2^2={(r-\mathrm{\δ})}^2+h^2(1+{\left(\frac{(r+\mathrm{\δ})\sin \mathrm{\β}}{(r+\mathrm{\δ})\cos \mathrm{\β}-(r-\mathrm{\δ})}\right)}^2)---\left(18\right);$

$r_1^2={(r+\mathrm{\δ})}^2+h^2(1+{\left(\frac{(r-\mathrm{\δ})\sin \mathrm{\β}}{(r+\mathrm{\δ})-(r-\mathrm{\δ})\cos \mathrm{\β}}\right)}^2)---\left(19\right);$

Step 6: calculate parameter δ according to formula (20-21) ₁and δ ₂.

$\frac{(r+\mathrm{\δ})-(r+{\mathrm{\δ}}_1)\cos {\mathrm{\β}}_1}{\sqrt{{(r+\mathrm{\δ})}^2+{(r+{\mathrm{\δ}}_1)}^2-2(r+\mathrm{\δ})(r+{\mathrm{\δ}}_1)\cos {\mathrm{\β}}_1}}=\frac{(r+\mathrm{\δ})-(r-\mathrm{\δ})\cos \mathrm{\β}}{\sqrt{{(r+\mathrm{\δ})}^2+{(r-\mathrm{\δ})}^2-2(r+\mathrm{\δ})(r-\mathrm{\δ})\cos \mathrm{\β}}}---\left(20\right);$

$\frac{(r-\mathrm{\δ})-(r-{\mathrm{\δ}}_2)\cos {\mathrm{\β}}_2}{\sqrt{{(r-\mathrm{\δ})}^2+{(r-{\mathrm{\δ}}_2)}^2-2(r-\mathrm{\δ})(r-{\mathrm{\δ}}_2)\cos {\mathrm{\β}}_2}}=\frac{(r-\mathrm{\δ})-(r+\mathrm{\δ})\cos \mathrm{\β}}{\sqrt{{(r-\mathrm{\δ})}^2+{(r+\mathrm{\δ})}^2-2(r-\mathrm{\δ})(r+\mathrm{\δ})\cos \mathrm{\β}}}---\left(21\right);$

Step 7: n+2 the input point being determined the y-z plane of three-dimensional cartesian coordinate system by formula (22-23), wherein n is less than or equal to 6N+1.

$V_0^y={\left[\begin{array}{ccc}0& 0& r+{\mathrm{\δ}}_1\end{array}\right]}^T---\left(22\right);$ $V_j^y=u_j\left[\begin{array}{c}0\\ \sin \left({\mathrm{\Σ}}_{k=1}^j{\mathrm{\ω}}_k\right)\\ \cos \left({\mathrm{\Σ}}_{k=1}^j{\mathrm{\ω}}_k\right)\end{array}\right],j=1,...,n+1---\left(23\right);$ Wherein, u _jand ω _kjth and a kth element of two endless 1 dimensional vector U and Ω respectively.The definition of endless 1 dimensional vector U and Ω is provided by formula (24) and (25).

U=[P _{6 × 1}p _{6 × 1}...] _{∞ × 1}(24); Ω=[Q _{6 × 1}q _{6 × 1}...] _{∞ × 1}(25); Namely 1 dimensional vector U is made up of numerous 6 × 1 vectorial P, and 1 dimensional vector Ω is made up of numerous 6 × 1 vectorial Q.6 × 1 vectorial P and Q are provided by formula (26) and (27) respectively.

P＝[r+δ r-δ r-δ ₂r-δ r+δ r+δ ₁] (26)；Q＝[β ₁β β ₂β ₂β β ₁] (27)；

Step 8: calculate m × n summit V by formula (28) _{i, j}coordinate.

$V_{i,j}=\left[\begin{array}{c}{x}_{i,j}\\ y_{i,j}\\ z_{i,j}\end{array}\right]=V_j^y+\left[A_j\right]V_i^x,i=\mathrm{1,2},...,m;j=\mathrm{1,2},...,n---\left(28\right);$ Wherein, 3 × 3 matrix [A _j] provided by formula (29).

$\left[A_j\right]=\left[\begin{array}{ccc}1& 0& 0\\ 0& 0& {(-1)}^j\frac{\cos {\mathrm{\θ}}_{j-1}+\cos {\mathrm{\θ}}_j}{\sin ({\mathrm{\θ}}_{j-1}-{\mathrm{\θ}}_j)}\\ 0& 0& {(-1)}^j\frac{\sin {\mathrm{\θ}}_{j-1}+\sin {\mathrm{\θ}}_j}{\sin ({\mathrm{\θ}}_{j-1}-{\mathrm{\θ}}_j)}\end{array}\right]---\left(29\right);$ Wherein; ${\sin \mathrm{\θ}}_j=\frac{i_z\·(V_{j+1}^y-V_j^y)}{\left|\right|V_{j+1}^y-V_j^y\left|\right|}---\left(30\right);$ ${\cos \mathrm{\θ}}_j=$ $\frac{i_y\·(V_{j+1}^y-V_j^y)}{\left|\right|V_{j+1}^y-V_j^y\left|\right|}---\left(31\right);$ Wherein, i _y=[0 1 0] ^tfor the vector of unit length of y coordinate axis, i _z=[0 0 1] ^tfor the vector of unit length of z coordinate axle, || ■ || represent that subtend measures mould.

M × n obtained above summit V _{i, j}namely the summit of paper folding structure is constituted.Step 9: definition { V _{i, j}v _{i+1, j}or { V _{i, j}v _{i, j+1}it is 1 pair of adjacent vertex.All adjacent vertex straight lines are coupled together, as shown in Figure 2.Namely connecting line segment between these adjacent vertexs constitutes the scrimp of paper folding structure.Can prove, equaled the internal diameter of shell structure housing to the external diameter that 9 design the paper folding structure obtained by step 1, the internal diameter of paper folding structure equals the external diameter of inner casing in shell structure, and the length of paper folding structure equals the length of shell structure.Therefore, this paper folding structure and inside and outside shell have geometry compatibility.Above-mentioned shell-paper folding structure-inner casing combination just constitutes the 2nd kind of aircraft casing sandwich construction of the present invention.

Embodiment 3

Make the internal diameter R of aircraft sandwich structures _in=18, external diameter R _out=22, the thickness t of inner casing _in=1, the thickness t of shell _ex=1, segment length l _seg=11.The external diameter r of interlayer is obtained by formula (13-15) ₁=21, internal diameter r ₂=19, length l=11.

Get m=12, h=1, a=1, by formula (4b) known b=1.

The input point of 12 x-z planes is obtained by formula (4a): $V_1^x={\left[\begin{array}{ccc}1& 0& 1\end{array}\right]}^T;$ $V_2^x={\left[\begin{array}{ccc}2& 0& 1\end{array}\right]}^T;$ $V_3^x=$ ${\left[\begin{array}{ccc}3& 0& -1\end{array}\right]}^T;$ $V_4^x={\left[\begin{array}{ccc}4& 0& -1\end{array}\right]}^T;$ $V_5^x={\left[\begin{array}{ccc}5& 0& 1\end{array}\right]}^T;$ $V_6^x={\left[\begin{array}{ccc}6& 0& 1\end{array}\right]}^T;$ $V_7^x=$ ${\left[\begin{array}{ccc}7& 0& -1\end{array}\right]}^T;$ $V_8^x={\left[\begin{array}{ccc}8& 0& -1\end{array}\right]}^T;$ $V_9^x={\left[\begin{array}{ccc}9& 0& 1\end{array}\right]}^T;$ $V_{10}^x={\left[\begin{array}{ccc}10& 0& 1\end{array}\right]}^T;$ $V_{11}^x=$ ${\left[\begin{array}{ccc}11& 0& -1\end{array}\right]}^T;$ $V_{12}^x={\left[\begin{array}{ccc}12& 0& -1\end{array}\right]}^T;$ Get N=30, obtain β by formula (17) _set=π/30.

Get β ₁=β ₂=π/300, then β=β _set-β ₁-β ₂=8 π/300.

R=19.9572 and δ=1.0051 are calculated according to formula (18-19).

δ is calculated according to formula (20-21) ₁=0.7227 and δ ₂=0.7824.

Get n=6N+1=181, obtained the input point of 183 y-z planes by formula (22-23):

$V_0^y={\left[\begin{array}{ccc}0& 0& 20.6798\end{array}\right]}^T;$ $V_j^y=u_j\left[\begin{array}{c}0\\ \sin \left({\mathrm{\Σ}}_{k=1}^j{\mathrm{\ω}}_k\right)\\ \cos \left({\mathrm{\Σ}}_{k=1}^j{\mathrm{\ω}}_k\right)\end{array}\right],j=1,...,182;$ Wherein, u _jand ω _kjth and a kth element of two endless 1 dimensional vector U and Ω respectively.Endless 1 dimensional vector U and Ω is; U=[P _{6 × 1}p _{6 × 1}...] _{∞ × 1}; Ω=[Q _{6 × 1}q _{6 × 1}...] _{∞ × 1}; Wherein;

P=[20.9623 18.9520 19.1748 18.9520 20.9623 20.6798]; Q=[π/300 8 π/300, π/300 π/300 8 π/300, π/300]; 12 × 181 summit V are calculated by formula (28) _{i, j}coordinate.

Finally define scrimp according to step 7.

Fig. 6 is the schematic three dimensional views of obtained paper folding structure.Fig. 7 is the partial enlarged drawing of Fig. 6 structure.Can see, compared to embodiment 1, the present embodiment can be regarded as and every bar scrimp " v " type groove in embodiment 1 is replaced.Fig. 8 is all summit V _{i, j}subpoint on the y-z plane and radius are the circle of 19 and 21, these two circles are all dropped on circumferentially in visible most of summit, all the other summits are then dropped on radius and are slightly less than the circle of 21 circumferentially slightly larger than 19 or radius, after this part vertex correspondence in the bottom of " v " type groove.This paper folding structure shown in key diagram 6 reach radial dimension designing requirement.In addition, the axial length of the structure shown in Fig. 6 equals 11, also reaches axial dimension designing requirement.

Embodiment 4

Change the parameter n in embodiment 3 and keep other parameter constants, the paper folding structure of non-closed loop can be obtained.Such as when n gets 61, paper folding Structure composing 1/3 annulus obtained, as shown in Figure 9.

Step 1: according to the internal diameter R of sandwich construction _in, external diameter R _out, the thickness t of inner casing _in, shell thickness t _exand segment length l _seg, utilize formula (32-34) to calculate the external diameter r of required interlayer ₁and internal diameter r ₂and length l.

r ₁＝R _out-t _ex(32)；r ₂＝R _in+t _in(33)；l＝l _seg(34)；

Step 2: m the input point being determined the x-z plane of three-dimensional cartesian coordinate system by formula (35a), wherein, m is natural number.

$V_i^x={\left[\begin{array}{ccc}{\mathrm{\Σ}}_{k=1}^i(\frac{1-{(-1)}^k}{2}a+\frac{1+{(-1)}^k}{2}b)& 0& -h+{\mathrm{\Σ}}_{k=1}^i{(-1)}^{\frac{i-1}{2}}(1-{(-1)}^i)h\end{array}\right]}^T,i=$ $1,...,m---\left(35a\right);$ Wherein, parameter a and b meets: ${\mathrm{\Σ}}_{k=1}^m(\frac{1-{(-1)}^k}{2}a+\frac{1+{(-1)}^k}{2}b)=l+a---\left(35b\right);$

Step 3: selected Parameter N, wherein N be more than or equal to 3 natural number.Parameter beta is calculated according to formula (36) _set; ${\mathrm{\β}}_{\mathrm{set}}=\frac{2\mathrm{\π}}{N}---\left(36\right);$

Step 4: selected parameter beta and β ₁, make it meet β+2 β ₁=β _set.Step 5: calculate parameter r and δ according to formula (37-39).

$r_2^2={(r-\mathrm{\δ})}^2+h^2(1+{\left(\frac{(r+\mathrm{\δ})\sin {\mathrm{\β}}_1}{(r+\mathrm{\δ})\cos {\mathrm{\β}}_1-(r-\mathrm{\δ})}\right)}^2)---\left(37\right);$ $r_1^2={((r+\mathrm{\δ})\cos \frac{\mathrm{\β}}{2}+h)}^2+$ ${((r+\mathrm{\δ})\sin \frac{\mathrm{\β}}{2}-\frac{2\mathrm{hs}}{\sqrt{1-s^2}})}^2---\left(38\right);$

$s\sin ({\mathrm{\β}}_1+\frac{\mathrm{\β}}{2})-\sqrt{1-s^2}\cos ({\mathrm{\β}}_1+\frac{\mathrm{\β}}{2})=\frac{(r-\mathrm{\δ})-(r+\mathrm{\δ})\cos {\mathrm{\β}}_1}{\sqrt{{(r+\mathrm{\δ})}^2+{(r-\mathrm{\δ})}^2-2(r+\mathrm{\δ})(r-\mathrm{\δ})\cos {\mathrm{\β}}_1}}---\left(39\right);$

Step 6: n+2 the input point being determined the y-z plane of three-dimensional cartesian coordinate system by formula (40-41), wherein n is less than or equal to 3N+1.

$V_0^y={\left[\begin{array}{ccc}0& 0& r-\mathrm{\δ}\end{array}\right]}^T---\left(40\right);$ $V_j^y=u_j\left[\begin{array}{c}0\\ \sin \left({\mathrm{\Σ}}_{k=1}^j{\mathrm{\ω}}_k\right)\\ \cos \left({\mathrm{\Σ}}_{k=1}^j{\mathrm{\ω}}_k\right)\end{array}\right],j=1,...,n+1---\left(41\right);$ Wherein, u _jand ω _kjth and a kth element of two endless 1 dimensional vector U and Ω respectively.The definition of endless 1 dimensional vector U and Ω is provided by formula (42) and (43).

U=[P _{3 × 1}p _{3 × 1}...] _{∞ × 1}(42); Ω=[Q _{3 × 1}q _{3 × 1}...] _{∞ × 1}(43); Namely 1 dimensional vector U is made up of numerous 3 × 1 vectorial P, and 1 dimensional vector Ω is made up of numerous 3 × 1 vectorial Q.3 × 1 vectorial P and Q are provided by formula (44) and (45) respectively.

P＝[r+δ r+δ r-δ] (44)；Q＝[β ₁β β ₁] (45)；

Step 7: calculate m × n summit V by formula (46) _{i, j}coordinate.

$V_{i,j}=\left[\begin{array}{c}{x}_{i,j}\\ y_{i,j}\\ z_{i,j}\end{array}\right]=V_j^y+\left[A_j\right]V_i^x,i=\mathrm{1,2},...,m;j=\mathrm{1,2},...,n---\left(46\right);$ Wherein, 3 × 3 matrix [A _j] provided by formula (47).

$\left[A_j\right]=\left[\begin{array}{ccc}1& 0& 0\\ 0& 0& {(-1)}^j\frac{\cos {\mathrm{\θ}}_{j-1}+\cos {\mathrm{\θ}}_j}{\sin ({\mathrm{\θ}}_{j-1}-{\mathrm{\θ}}_j)}\\ 0& 0& {(-1)}^j\frac{\sin {\mathrm{\θ}}_{j-1}+\sin {\mathrm{\θ}}_j}{\sin ({\mathrm{\θ}}_{j-1}-{\mathrm{\θ}}_j)}\end{array}\right]---\left(47\right);$ Wherein; ${\sin \mathrm{\θ}}_j=\frac{i_z\·(V_{j+1}^y-V_j^y)}{\left|\right|V_{j+1}^y-V_j^y\left|\right|}---\left(48\right);$ ${\cos \mathrm{\θ}}_j=$ $\frac{i_y\·(V_{j+1}^y-V_j^y)}{\left|\right|V_{j+1}^y-V_j^y\left|\right|}---\left(49\right);$ Wherein, i _y=[0 1 0] ^tfor the vector of unit length of y coordinate axis, i _z=[0 0 1] ^tfor the vector of unit length of z coordinate axle, || ■ || represent that subtend measures mould.

M × n obtained above summit V _{i, j}namely the summit of paper folding structure is constituted.

Step 8: definition { V _{i, j}v _{i+1, j}or { V _{i, d}v _{i, j+1}it is 1 pair of adjacent vertex.All adjacent vertex straight lines are coupled together, as shown in Figure 2.Namely connecting line segment between these adjacent vertexs constitutes the scrimp of paper folding structure.Can prove, equaled the internal diameter of shell structure housing to the external diameter that 8 design the paper folding structure obtained by step 1, the internal diameter of paper folding structure equals the external diameter of inner casing in shell structure, and the length of paper folding structure equals the length of shell structure.Therefore, this paper folding structure and inside and outside shell have geometry compatibility.Above-mentioned shell-paper folding structure-inner casing combination just constitutes the 3rd kind of aircraft casing sandwich construction of the present invention.

Embodiment 5

Make the internal diameter R of aircraft sandwich structures _in=18, external diameter R _out=22, the thickness t of inner casing _in=1, the thickness t of shell _ex=1, segment length l _seg=11.The external diameter r of interlayer is obtained by formula (32-34) ₁=21, internal diameter r ₂=19, length l=11.

Get m=12, h=0.5, a=1, by formula (35b) known b=1.

The input point of 12 x-z planes is obtained by formula (35a): $V_1^x={\left[0.5\right]}^T;$ $V_2^x={\left[\begin{array}{ccc}2& 0& 0.5\end{array}\right]}^T;$ $V_3^x=$ ${\left[\begin{array}{ccc}3& 0& -0.5\end{array}\right]}^T;$ $V_4^x={\left[\begin{array}{ccc}4& 0& -0.5\end{array}\right]}^T;$ $V_5^x={\left[\begin{array}{ccc}5& 0& 0.5\end{array}\right]}^T;$ $V_6^x={\left[\begin{array}{ccc}6& 0& 0.5\end{array}\right]}^T;$ $V_7^x=$ ${\left[\begin{array}{ccc}7& 0& -0.5\end{array}\right]}^T;$ $V_8^x={\left[\begin{array}{ccc}8& 0& -0.5\end{array}\right]}^T;$ $V_9^x={\left[\begin{array}{ccc}9& 0& 0.5\end{array}\right]}^T;$ $V_{10}^x={\left[\begin{array}{ccc}10& 0& 0.5\end{array}\right]}^T;$ $V_{11}^x=$ ${\left[\begin{array}{ccc}11& 0& -0.5\end{array}\right]}^T;$ $V_{12}^x={\left[\begin{array}{ccc}12& 0& -0.5\end{array}\right]}^T;$ Get N=30, obtain β by formula (36) _set=π/15.

Get β ₁=π/75, then β=β _set-2 β ₁=3 π/75.

R=19.7573 and δ=0.7660 is calculated according to formula (37-39).

Get n=3N+1=91, obtained the input point of 93 y-z planes by formula (40-41):

$V_0^y={\left[\begin{array}{ccc}0& 0& 18.9913\end{array}\right]}^T;$ $V_j^y=u_j\left[\begin{array}{c}0\\ \sin \left({\mathrm{\Σ}}_{k=1}^j{\mathrm{\ω}}_k\right)\\ \cos \left({\mathrm{\Σ}}_{k=1}^j{\mathrm{\ω}}_k\right)\end{array}\right],j=1,...,93;$ Wherein, u _jand ω _kjth and a kth element of two endless 1 dimensional vector U and Ω respectively.Endless 1 dimensional vector U and Ω is; U=[P _{3 × 1}p _{3 × 1}...] _{∞ × 1}; Ω=[Q _{3 × 1}q _{3 × 1}...] _{∞ × 1}; Wherein;

P=[20.5232 20.5232 18.9913]; Q=[π/75 3 π/75, π/75]; 12 × 91 summit V are calculated by formula (46) _{i, j}coordinate.

Finally define scrimp according to step 7.

Figure 10 is the 3-D view of obtained paper folding structure.Figure 11 is the partial enlarged view of structure shown in Figure 10.Figure 12 is all summit V _{i, j}subpoint on the y-z plane and radius are the circle of 19 and 21, and these two circles are all dropped on circumferentially in visible most of summit, and all the other summits are then dropped between above-mentioned two circumference.This illustrate the paper folding structure shown in Figure 10 reach radial dimension designing requirement.In addition, the axial length of the structure shown in Figure 10 equals 11, also reaches axial dimension designing requirement.

Embodiment 6

Change the parameter n in embodiment 5 and keep other parameter constants, the paper folding structure of non-closed loop can be obtained.Such as when n gets 31, paper folding Structure composing 1/3 annulus obtained, as shown in figure 13.

Step 1: according to the internal diameter R of sandwich construction _in, external diameter R _out, the thickness t of inner casing _in, shell thickness t _exand segment length l _seg, utilize formula (50-52) to calculate the external diameter r of required interlayer ₁and internal diameter r ₂and length l.

r ₁＝R _out-t _ex(50)；r ₂＝R _in+t _in(51)；l＝l _seg(52)；

Step 2: m the input point being determined the x-z plane of three-dimensional cartesian coordinate system by formula (53a), wherein, m is natural number.

$V_i^x={\left[\begin{array}{ccc}{\mathrm{\Σ}}_{k=1}^i(\frac{1-{(-1)}^k}{2}\mathrm{a\α}+\frac{{1+(-1)}^k}{2}b)& 0& -h+{\mathrm{\Σ}}_{k=1}^i{(-1)}^{\frac{i-1}{2}}(1-{(-1)}^i)h\end{array}\right]}^T,$ I=1 ..., m (53a); Wherein, parameter a and b meets: ${\mathrm{\Σ}}_{k=1}^m(\frac{1-{(-1)}^k}{2}a+\frac{1+{(-1)}^k}{2}b)=l+a---\left(53b\right);$

Step 3: selected Parameter N, wherein N be more than or equal to 3 natural number.Parameter beta is calculated according to formula (54) _set; ${\mathrm{\β}}_{\mathrm{set}}=\frac{2\mathrm{\π}}{N}---\left(54\right);$

Step 4: selected parameter beta and β ₁, make it meet β+2 β ₁=β _set.

Step 5: calculate parameter r and δ according to formula (55-57).

$r_1^2={(r+\mathrm{\δ})}^2+h^2(1+{\left(\frac{(r-\mathrm{\δ})\sin {\mathrm{\β}}_1}{(r+\mathrm{\δ})-(r-\mathrm{\δ})\cos {\mathrm{\β}}_1}\right)}^2)---\left(55\right);$ $r_2^2={((r-\mathrm{\δ})\cos \frac{\mathrm{\β}}{2}-h)}^2+$ ${((r-\mathrm{\δ})\sin \frac{\mathrm{\β}}{2}-\frac{2\mathrm{hs}}{\sqrt{1-s^2}})}^2---\left(56\right);$ $-s\sin \frac{\mathrm{\β}}{2}-\sqrt{1-s^2}\cos \frac{\mathrm{\β}}{2}=\frac{(r-\mathrm{\δ})-(r+\mathrm{\δ})\cos {\mathrm{\β}}_1}{\sqrt{{(r+\mathrm{\δ})}^2+{(r-\mathrm{\δ})}^2-2(r+\mathrm{\δ})(r-\mathrm{\δ})\cos {\mathrm{\β}}_1}}---\left(57\right);$

Step 6: n+2 the input point being determined the y-z plane of three-dimensional cartesian coordinate system by formula (58-59), wherein n is less than or equal to 3N+1.

$V_0^y={\left[\begin{array}{ccc}0& 0& r-\mathrm{\δ}\end{array}\right]}^T---\left(58\right);$ $V_j^y=u_j\left[\begin{array}{c}0\\ \sin \left({\mathrm{\Σ}}_{k=1}^j{\mathrm{\ω}}_k\right)\\ \cos \left({\mathrm{\Σ}}_{k=1}^j{\mathrm{\ω}}_k\right)\end{array}\right],$ J=1 ..., n+1 (59); Wherein, u _kand ω _kjth and a kth element of two endless 1 dimensional vector U and Ω respectively.The definition of endless 1 dimensional vector U and Ω is provided by formula (60) and (61).

U=[P _{3 × 1}p _{3 × 1}...] _{∞ × 1}(60); Ω=[Q _{3 × 1}q _{3 × 1}...] _{∞ × 1}(61); Namely 1 dimensional vector U is made up of numerous 3 × 1 vectorial P, and 1 dimensional vector Ω is made up of numerous 3 × 1 vectorial Q.3 × 1 vectorial P and Q are provided by formula (62) and (63) respectively.

P＝[r-δ r-δ r+δ](62)；Q＝[β ₁β β ₁](63)；

Step 7: calculate m × n summit V by formula (64) _{i, j}coordinate.

$V_{i,j}=\left[\begin{array}{c}{x}_{i,j}\\ y_{i,j}\\ z_{i,j}\end{array}\right]=V_j^y+\left[A_j\right]V_i^x,$ I=1,2 ..., m; J=1,2 ..., n (64); Wherein, 3 × 3 matrix [A _j] provided by formula (65).

$\left[A_j\right]=\left[\begin{array}{ccc}1& 0& 0\\ 0& 0& {(-1)}^j\frac{\cos {\mathrm{\θ}}_{j-1}+\cos {\mathrm{\θ}}_j}{\sin ({\mathrm{\θ}}_{j-1}-{\mathrm{\θ}}_j)}\\ 0& 0& {(-1)}^j\frac{\sin {\mathrm{\θ}}_{j-1}+\sin {\mathrm{\θ}}_j}{\sin ({\mathrm{\θ}}_{j-1}-{\mathrm{\θ}}_j)}\end{array}\right]---\left(65\right)\text{;}$ Wherein; $\sin {\mathrm{\θ}}_j=\frac{i_z\·(V_{j+1}^y-V_j^y)}{\left|\right|V_{j+1}^y+V_j^y\left|\right|}---\left(66\right);$ $\cos {\mathrm{\θ}}_j=$ $\frac{i_y\·(V_{j+1}^y-V_j^y)}{\left|\right|V_{j+1}^y-V_j^y\left|\right|}---\left(67\right);$ Wherein, i _y=[0 1 0] ^tfor the vector of unit length of y coordinate axis, i _z=[0 0 1] ^tfor the vector of unit length of z coordinate axle, || ■ || represent that subtend measures mould.

M × n obtained above summit V _{i, j}namely the summit of paper folding structure is constituted.

Step 8: definition { V _{i, j}v _{i+1, j}or { V _{i, j}v _{i, j+1}it is 1 pair of adjacent vertex.All adjacent vertex straight lines are coupled together, as shown in Figure 2.Namely connecting line segment between these adjacent vertexs constitutes the scrimp of paper folding structure.Can prove, equaled the internal diameter of shell structure housing to the external diameter that 8 design the paper folding structure obtained by step 1, the internal diameter of paper folding structure equals the external diameter of inner casing in shell structure, and the length of paper folding structure equals the length of shell structure.Therefore, this paper folding structure and inside and outside shell have geometry compatibility.Above-mentioned shell-paper folding structure-inner casing combination just constitutes the 4th kind of aircraft casing sandwich construction of the present invention.

Embodiment 7

Make the internal diameter R of fuselage sandwich construction _in=18, external diameter R _out=22, the thickness t of inner casing _in=1, the thickness t of shell _ex=1, segment length l _seg=11.The external diameter r of interlayer is obtained by formula (50-52) ₁=21, internal diameter r ₂=19, length l=11.

Get m=12, h=0.5, a=1, by formula (53b) known b=1.

The input point of 12 x-z planes is obtained by formula (53a): $V_1^x={\left[0.5\right]}^T;$ $V_2^x={\left[\begin{array}{ccc}2& 0& 0.5\end{array}\right]}^T;$ $V_3^x=$ ${\left[\begin{array}{ccc}3& 0& -0.5\end{array}\right]}^T;$ $V_4^x={\left[\begin{array}{ccc}4& 0& -0.5\end{array}\right]}^T;$ $V_5^x={\left[\begin{array}{ccc}5& 0& 0.5\end{array}\right]}^T;$ $V_6^x={\left[\begin{array}{ccc}6& 0& 0.5\end{array}\right]}^T;$ $V_7^x=$ ${\left[\begin{array}{ccc}7& 0& -0.5\end{array}\right]}^T;$ $V_8^x={\left[\begin{array}{ccc}8& 0& -0.5\end{array}\right]}^T;$ $V_9^x={\left[\begin{array}{ccc}9& 0& 0.5\end{array}\right]}^T;$ $V_{10}^x={\left[\begin{array}{ccc}10& 0& 0.5\end{array}\right]}^T;$ $V_{11}^x=$ ${\left[\begin{array}{ccc}11& 0& -0.5\end{array}\right]}^T;$ $V_{12}^x={\left[\begin{array}{ccc}12& 0& -0.5\end{array}\right]}^T;$ Get N=30, obtain β by formula (54) _set=π/15.

Get β ₁=π/75, then β=β _set-2 β ₁=3 π/75.

R=20.2617 and δ=0.7305 is calculated according to formula (55-57).

Get n=3N+1=91, obtained the input point of 93 y-z planes by formula (58-59):

$V_0^y={\left[\begin{array}{ccc}0& 0& 20.9922\end{array}\right]}^T;$ $V_j^y=u_j\left[\begin{array}{c}0\\ \sin \left({\mathrm{\Σ}}_{k=1}^j{\mathrm{\ω}}_k\right)\\ \cos \left({\mathrm{\Σ}}_{k=1}^j{\mathrm{\ω}}_k\right)\end{array}\right],$ J=1 ..., 93; Wherein, u _jand ω _kjth and a kth element of two endless 1 dimensional vector U and Ω respectively.Endless 1 dimensional vector U and Ω is; U=[P _{3 × 1}p _{3 × 1}...] _{∞ × 1}; Ω=[Q _{3 × 1}q _{3 × 1}...] _{∞ × 1}; Wherein;

P=[19.5312 19.5312 20.9922]; Q=[π/75 3 π/75, π/75]; 12 × 91 summit V are calculated by formula (64) _{i, j}coordinate.

Finally define scrimp according to step 7.

Figure 14 is the 3-D view of obtained paper folding structure.Figure 15 is the partial enlarged view of structure shown in Figure 14.Figure 16 is all summit V _{i, j}subpoint on the y-z plane and radius are the circle of 19 and 21, and these two circles are all dropped on circumferentially in visible most of summit, and all the other summits are then dropped between above-mentioned two circumference.This illustrate the paper folding structure shown in Figure 14 reach radial dimension designing requirement.In addition, the axial length of the structure shown in Figure 14 equals 11, also reaches axial dimension designing requirement.

Embodiment 8

Change the parameter n in embodiment 7 and keep other parameter constants, the paper folding structure of non-closed loop can be obtained.Such as when n gets 31, paper folding Structure composing 1/3 annulus obtained, as shown in figure 17.

Step 1: according to the internal diameter R of sandwich construction _in, external diameter R _out, the thickness t of inner casing _in, shell thickness t _exand segment length l _seg, utilize formula (68-70) to calculate the external diameter r of required interlayer ₁and internal diameter r ₂and length l.

r ₁＝R _out-t _ex(68)；r ₂＝R _in+t _in(69)；l＝l _seg(70)；

Step 2: m the input point being determined the x-z plane of three-dimensional cartesian coordinate system by formula (71a), wherein, m is natural number.

$V_i^x={\left[\begin{array}{ccc}{\mathrm{\Σ}}_{k=1}^i(\frac{1-{(-1)}^k}{2}a+\frac{{1+(-1)}^k}{2}b)& 0& -h+{\mathrm{\Σ}}_{k=1}^i{(-1)}^{\frac{i-1}{2}}(1-{(-1)}^i)h\end{array}\right]}^T,$ I=1 ..., m (71a); Wherein, parameter a and b meets: ${\mathrm{\Σ}}_{k=1}^m(\frac{1-{(-1)}^k}{2}a+\frac{1+{(-1)}^k}{2}b)=l+a---\left(71b\right);$

Step 3: selected Parameter N, wherein N be more than or equal to 3 natural number.Parameter beta is calculated according to formula (72) _set; ${\mathrm{\β}}_{\mathrm{set}}=\frac{2\mathrm{\π}}{N}---\left(72\right);$

Step 4: selected parameter beta, β ₁and β ₂, make it meet 2 β+β ₁+ β ₂=β _set.

Step 5: calculate parameter r and δ according to formula (73-76).

$r_1^2={((r+\mathrm{\δ})\cos \frac{{\mathrm{\β}}_1}{2}+h)}^2+{((r+\mathrm{\δ})\sin \frac{{\mathrm{\β}}_1}{2}-\frac{2\mathrm{hs}}{\sqrt{1-s^2}})}^2---\left(73\right);$ $s\sin (\mathrm{\β}+\frac{{\mathrm{\β}}_1}{2})-\sqrt{1-s^2}\cos (\mathrm{\β}+\frac{{\mathrm{\β}}_1}{2})=\frac{(r-\mathrm{\δ})-(r+\mathrm{\δ})\cos \mathrm{\β}}{\sqrt{{(r+\mathrm{\δ})}^2+{(r-\mathrm{\δ})}^2-2(r+\mathrm{\δ})(r-\mathrm{\δ})\cos \mathrm{\β}}}---\left(74\right);$ $r_2^2=\left(\right(r-$ $\mathrm{\δ})\cos \frac{{\mathrm{\β}}_2}{2}-h{)}^2+{((r-\mathrm{\δ})\sin \frac{{\mathrm{\β}}_2}{2}-\frac{2\mathrm{ht}}{\sqrt{1-t^2}})}^2---\left(75\right);$ $-t\sin \frac{{\mathrm{\β}}_2}{2}-\sqrt{1-t^2}\cos \frac{{\mathrm{\β}}_2}{2}=\frac{(r-\mathrm{\δ})-(r+\mathrm{\δ})\cos \mathrm{\β}}{\sqrt{{(r+\mathrm{\δ})}^2+{(r-\mathrm{\δ})}^2-2(r+\mathrm{\δ})(r-\mathrm{\δ})\cos \mathrm{\β}}}---\left(76\right);$

Step 6: n+2 the input point being determined the y-z plane of three-dimensional cartesian coordinate system by formula (77-78), wherein n is less than or equal to 4N+1.

$V_0^y={\left[\begin{array}{ccc}0& 0& r-\mathrm{\δ}\end{array}\right]}^T---\left(77\right);$ $V_j^y=u_j\left[\begin{array}{c}0\\ \sin \left({\mathrm{\Σ}}_{k=1}^j{\mathrm{\ω}}_k\right)\\ \cos \left({\mathrm{\Σ}}_{k=1}^j{\mathrm{\ω}}_k\right)\end{array}\right],$ J=1 ..., n+1 (78); Wherein, u _jand ω _kjth and a kth element of two endless 1 dimensional vector U and Ω respectively.The definition of endless 1 dimensional vector U and Ω is provided by formula (79) and (80).

U=[P _{4 × 1}p _{4 × 1}...] _{∞ × 1}(79); Ω=[Q _{4 × 1}q _{4 × 1}...] _{∞ × 1}(80); Namely 1 dimensional vector U is made up of numerous 4 × 1 vectorial P, and 1 dimensional vector Ω is made up of numerous 4 × 1 vectorial Q.4 × 1 vectorial P and Q are provided by formula (81) and (82) respectively.

P＝[r+δ r+δ r-δ r-δ](81)；Q＝[β β ₁β β ₂](82)；

Step 7: calculate m × n summit V by formula (83) _{i, j}coordinate.

$V_{i,j}=\left[\begin{array}{c}{x}_{i,j}\\ y_{i,j}\\ z_{i,j}\end{array}\right]=V_j^y+\left[A_j\right]V_i^x,$ I=1,2 ..., m; J=1,2 ..., n (83); Wherein, 3 × 3 matrix [A _j] provided by formula (84).

$\left[A_j\right]=\left[\begin{array}{ccc}1& 0& 0\\ 0& 0& {(-1)}^j\frac{\cos {\mathrm{\θ}}_{j-1}+\cos {\mathrm{\θ}}_j}{\sin ({\mathrm{\θ}}_{j-1}-{\mathrm{\θ}}_j)}\\ 0& 0& {(-1)}^{\text{j}}\frac{\sin {\mathrm{\θ}}_{j-1}+\sin {\mathrm{\θ}}_j}{\sin ({\mathrm{\θ}}_{j-1}-{\mathrm{\θ}}_j)}\end{array}\right]---\left(84\right);$ Wherein; $\sin {\mathrm{\θ}}_j=\frac{i_z\·(V_{j+1}^y-V_j^y)}{\left|\right|V_{j+1}^y+V_j^y\left|\right|}---\left(85\right);$ $\cos {\mathrm{\θ}}_j=$ $\frac{i_y\·(V_{j+1}^y-V_j^y)}{\left|\right|V_{j+1}^y-V_j^y\left|\right|}---\left(86\right);$ Wherein, i _y=[0 1 0] ^tfor the vector of unit length of y coordinate axis, i _z=[0 0 1] ^tfor the vector of unit length of z coordinate axle, || ■ || represent that subtend measures mould.

M × n obtained above summit V _{i, j}namely the summit of paper folding structure is constituted.

Step 8: definition { V _{i, j}v _{i+1, j}or { V _{i, j}v _{i, j+1}it is 1 pair of adjacent vertex.All adjacent vertex straight lines are coupled together, as shown in Figure 2.Namely connecting line segment between these adjacent vertexs constitutes the scrimp of paper folding structure.

Can prove, equaled the internal diameter of shell structure housing to the external diameter that 8 design the paper folding structure obtained by step 1, the internal diameter of paper folding structure equals the external diameter of inner casing in shell structure, and the length of paper folding structure equals the length of shell structure.Therefore, this paper folding structure and inside and outside shell have geometry compatibility.Above-mentioned shell-paper folding structure-inner casing combination just constitutes the 5th kind of aircraft casing sandwich construction of the present invention.

Embodiment 9

Make the internal diameter R of aircraft sandwich structures _in=18, external diameter R _out=22, the thickness t of inner casing _in=1, the thickness t of shell _ex=1, segment length l _seg=11.The external diameter r of interlayer is obtained by formula (68-70) ₁=21, internal diameter r ₂=19, length l=11.

Get m=12, h=0.2, a=1, by formula (71b) known b=1.

The input point of 12 x-z planes is obtained by formula (71a):

$V_1^x={\left[\begin{array}{ccc}1& 0& 0.2\end{array}\right]}^T;$ $V_2^x={\left[\begin{array}{ccc}2& 0& 0.2\end{array}\right]}^T;$ $V_3^x={\left[\begin{array}{ccc}3& 0& -0.2\end{array}\right]}^T;$ $V_4^x={\left[\begin{array}{ccc}4& 0& -0.2\end{array}\right]}^T;$ $V_5^x=$ ${\left[\begin{array}{ccc}5& 0& 0.2\end{array}\right]}^T;$ $V_6^x={\left[\begin{array}{ccc}6& 0& 0.2\end{array}\right]}^T;$ $V_7^x={\left[\begin{array}{ccc}7& 0& -0.2\end{array}\right]}^T;$ $V_8^x={\left[\begin{array}{ccc}8& 0& -0.2\end{array}\right]}^T;$ $V_9^x=$ ${\left[\begin{array}{ccc}9& 0& 0.2\end{array}\right]}^T;$ $V_{10}^x={\left[\begin{array}{ccc}10& 0& 0.2\end{array}\right]}^T;$ ${V_{11}^x=\left[\begin{array}{ccc}11& 0& -0.2\end{array}\right]}^T;$ $V_{12}^x={\left[\begin{array}{ccc}12& 0& -0.2\end{array}\right]}^T;$ Get N=30, obtain β by formula (72) _set=π/15.

Get β ₁=β ₂=π/45, then β=(β _set-β ₁-β ₁)/2=π/90.

R=20.0053 and δ=0.7994 is calculated according to formula (73-76).

Get n=4N+1=121, obtained the input point of 123 y-z planes by formula (77-78):

$V_0^y={\left[\begin{array}{ccc}0& 0& 19.2059\end{array}\right]}^T;$ $V_j^y=u_j\left[\begin{array}{c}0\\ \sin \left({\mathrm{\Σ}}_{k=1}^j{\mathrm{\ω}}_k\right)\\ \cos \left({\mathrm{\Σ}}_{k=1}^j{\mathrm{\ω}}_k\right)\end{array}\right],$ J=1 ..., 122; Wherein, u _jand ω _kjth and a kth element of two endless 1 dimensional vector U and Ω respectively.Endless 1 dimensional vector U and Ω is; U=[P _{4 × 1}p _{4 × 1}...] _{∞ × 1}; Ω=[Q _{4 × 1}q _{4 × 1}...] _{∞ × 1}; Wherein; P=[20.8048 20.8048 19.2059 19.2059]; Q=[π/45, π/90, π/45, π/90]; 12 × 121 summit V are calculated by formula (83) _{i, j}coordinate.

Finally define scrimp according to step 8.

Figure 18 is the 3-D view of obtained paper folding structure.Figure 19 is the partial enlarged view of structure shown in Figure 18.Figure 20 is all summit V _{i, j}subpoint on the y-z plane and radius are the circle of 19 and 21, and these two circles are all dropped on circumferentially in visible most of summit, and all the other summits are then dropped between above-mentioned two circumference.This illustrate the paper folding structure shown in Figure 18 reach radial dimension designing requirement.In addition, the axial length of the structure shown in Figure 18 equals 11, also reaches axial dimension designing requirement.

Embodiment 10

Change the parameter n in embodiment 9 and keep other parameter constants, the paper folding structure of non-closed loop can be obtained.Such as when n gets 41, paper folding Structure composing 1/3 annulus obtained, as shown in figure 21.

Step 1: according to the internal diameter R of sandwich construction _in, external diameter R _out, the thickness t of inner casing _in, shell thickness t _exand segment length l _seg, utilize formula (87-89) to calculate the external diameter r of required interlayer ₁and internal diameter r ₂and length l.

r ₁＝R _out-t _ex(87)；r ₂＝R _in+t _in(88)；l＝l _seg(89)；

Step 2: m the input point being determined the x-z plane of three-dimensional cartesian coordinate system by formula (90a), wherein, m is natural number.

$V_i^x={\left[\begin{array}{ccc}{\mathrm{\Σ}}_{k=1}^i(\frac{1-{(-1)}^k}{2}a+\frac{{1+(-1)}^k}{2}b)& 0& -h+{\mathrm{\Σ}}_{k=1}^i{(-1)}^{\frac{i-1}{2}}(1-{(-1)}^i)h\end{array}\right]}^T,$ I=1 ..., m (90a); Wherein, parameter a and b meets: ${\mathrm{\Σ}}_{k=1}^m(\frac{1-{(-1)}^k}{2}a+\frac{1+{(-1)}^k}{2}b)=l+a---\left(90b\right);$

Step 3: selected Parameter N, wherein N be more than or equal to 3 natural number.Parameter beta is calculated according to formula (91) _set; ${\mathrm{\β}}_{\mathrm{set}}=\frac{2\mathrm{\π}}{N}---\left(91\right);$

Step 4: selected parameter beta, β ₁and β ₂, make it meet β+2 β ₁+ 2 β ₂=β _set, and meet β ₁> β ₂.Step 5: calculate parameter r and δ according to formula (92-94).

$r_2^2={(r-\mathrm{\δ})}^2+h^2(1+{\left(\frac{(r+\mathrm{\δ})\sin {\mathrm{\β}}_1}{(r+\mathrm{\δ})\cos {\mathrm{\β}}_1-(r-\mathrm{\δ})}\right)}^2)---\left(92\right);$ $r_1^2={((r+\mathrm{\δ})\cos \frac{\mathrm{\β}}{2}+h)}^2+$ ${((r+\mathrm{\δ})\sin \frac{\mathrm{\β}}{2}-\frac{2\mathrm{hs}}{\sqrt{1-s^2}})}^2---\left(93\right);$ $s\sin ({\mathrm{\β}}_1+\frac{\mathrm{\β}}{2})-\sqrt{1-s^2}\cos ({\mathrm{\β}}_1+\frac{\mathrm{\β}}{2})=\frac{(r-\mathrm{\δ})-(r+\mathrm{\δ})\cos {\mathrm{\β}}_1}{\sqrt{{(r+\mathrm{\δ})}^2+{(r-\mathrm{\δ})}^2-2(r+\mathrm{\δ})(r-\mathrm{\δ})\cos {\mathrm{\β}}_1}}---\left(94\right);$

Step 6: calculate parameter δ according to formula (95) ₂.

$\frac{(r-\mathrm{\δ})-(r-{\mathrm{\δ}}_2)\cos {\mathrm{\β}}_2}{\sqrt{{(r-\mathrm{\δ})}^2+{(r-{\mathrm{\δ}}_2)}^2-2(r-\mathrm{\δ})(r-{\mathrm{\δ}}_2)\cos {\mathrm{\β}}_2}}=\frac{(r-\mathrm{\δ})-(r+\mathrm{\δ})\cos {\mathrm{\β}}_1}{\sqrt{{(r-\mathrm{\δ})}^2+{(r+\mathrm{\δ})}^2-(r-\mathrm{\δ})(r+\mathrm{\δ})\cos {\mathrm{\β}}_1}}---\left(95\right);$

Step 7: n+2 the input point being determined the y-z plane of three-dimensional cartesian coordinate system by formula (96-97), wherein n is less than or equal to 5N+1.

$V_0^y={\left[\begin{array}{ccc}0& 0& r-{\mathrm{\δ}}_2\end{array}\right]}^T---\left(96\right);$ $V_j^y=u_j\left[\begin{array}{c}0\\ \sin \left({\mathrm{\Σ}}_{k=1}^j{\mathrm{\ω}}_k\right)\\ \cos \left({\mathrm{\Σ}}_{k=1}^j{\mathrm{\ω}}_k\right)\end{array}\right],$ J=1 ..., n+1 (97); Wherein, u _jand ω _kjth and a kth element of two endless 1 dimensional vector U and Ω respectively.The definition of endless 1 dimensional vector U and Ω is provided by formula (98) and (99).

U=[P _{5 × 1}p _{5 × 1}...] _{∞ × 1}(98); Ω=[Q _{5 × 1}q _{5 × 1}...] _{∞ × 1}(99); Namely 1 dimensional vector U is made up of numerous 5 × 1 vectorial P, and 1 dimensional vector Ω is made up of numerous 5 × 1 vectorial Q.5 × 1 vectorial P and Q are provided by formula (100) and (101) respectively.

P＝[r-δ r+δ r+δ r-δ r-δ ₂](100)；Q＝[β ₂β ₁β β ₁β ₂](101)；

Step 8: calculate m × n summit V by formula (102) _{i, j}coordinate.

$V_{i,j}=\left[\begin{array}{c}{x}_{i,j}\\ y_{i,j}\\ z_{i,j}\end{array}\right]=V_j^y+\left[A_j\right]V_i^x,$ I=1,2 ..., m; J=1,2 ..., n (102); Wherein, 3 × 3 matrix [A _j] provided by formula (103).

$\left[A_j\right]=\left[\begin{array}{ccc}1& 0& 0\\ 0& 0& {(-1)}^j\frac{\cos {\mathrm{\θ}}_{j-1}+\cos {\mathrm{\θ}}_j}{\sin ({\mathrm{\θ}}_{j-1}-{\mathrm{\θ}}_j)}\\ 0& 0& {(-1)}^{\text{j}}\frac{\sin {\mathrm{\θ}}_{j-1}+\sin {\mathrm{\θ}}_j}{\sin ({\mathrm{\θ}}_{j-1}-{\mathrm{\θ}}_j)}\end{array}\right]---\left(103\right);$ Wherein; $\sin {\mathrm{\θ}}_j=\frac{i_z\·(V_{j+1}^y-V_j^y)}{\left|\right|V_{j+1}^y+V_j^y\left|\right|}---\left(104\right);$ $\cos {\mathrm{\θ}}_j=$ $\frac{i_y\·(V_{j+1}^y-V_j^y)}{\left|\right|V_{j+1}^y-V_j^y\left|\right|}---\left(105\right);$ Wherein, i _y=[0 1 0] ^tfor the vector of unit length of y coordinate axis, i _z=[0 0 1] ^tfor the vector of unit length of z coordinate axle, || ■ || represent that subtend measures mould.

M × n obtained above summit V _{i, j}namely the summit of paper folding structure is constituted.

Step 9: definition { V _{i, j}v _{i+1, j}or { V _{i, j}v _{i, j+1}it is 1 pair of adjacent vertex.All adjacent vertex straight lines are coupled together, as shown in Figure 2.Namely connecting line segment between these adjacent vertexs constitutes the scrimp of paper folding structure.Can prove, equaled the internal diameter of shell structure housing to the external diameter that 9 design the paper folding structure obtained by step 1, the internal diameter of paper folding structure equals the external diameter of inner casing in shell structure, and the length of paper folding structure equals the length of shell structure.Therefore, this paper folding structure and inside and outside shell have geometry compatibility.Above-mentioned shell-paper folding structure-inner casing combination just constitutes the 6th kind of aircraft casing sandwich construction of the present invention.

Embodiment 11

Make the internal diameter R of aircraft sandwich structures _in=18, external diameter R _out=22, the thickness t of inner casing _in=1, the thickness t of shell _ex=1, segment length l _seg=11.The external diameter r of interlayer is obtained by formula (87-89) ₁=21, internal diameter r ₂=19, length l=11.

Get m=12, h=0.5, a=1, by formula (90b) known b=1.

The input point of 12 x-z planes is obtained by formula (90a): $V_1^x={\left[0.5\right]}^T;$ $V_2^x={\left[\begin{array}{ccc}2& 0& 0.5\end{array}\right]}^T;$ $V_3^x=$ ${\left[\begin{array}{ccc}3& 0& -0.5\end{array}\right]}^T;$ $V_4^x={\left[\begin{array}{ccc}4& 0& -0.5\end{array}\right]}^T;$ $V_5^x={\left[\begin{array}{ccc}5& 0& 0.5\end{array}\right]}^T;$ $V_6^x={\left[\begin{array}{ccc}6& 0& 0.5\end{array}\right]}^T;$ $V_7^x=$ ${\left[\begin{array}{ccc}7& 0& -0.5\end{array}\right]}^T;$ $V_8^x={\left[\begin{array}{ccc}8& 0& -0.5\end{array}\right]}^T;$ $V_9^x={\left[\begin{array}{ccc}9& 0& 0.5\end{array}\right]}^T;$ $V_{10}^x={\left[\begin{array}{ccc}10& 0& 0.5\end{array}\right]}^T;$ $V_{11}^x=$ ${\left[\begin{array}{ccc}11& 0& -0.5\end{array}\right]}^T;$ $V_{12}^x={\left[\begin{array}{ccc}12& 0& -0.5\end{array}\right]}^T;$ Get N=30, obtain β by formula (91) _set=π/15.

Get β ₂=π/300, β ₁=π/75, then β=β _set-2 β ₁-2 β ₂=π/30.

R=19.7551 and δ=0.7638 is calculated according to formula (92-94).

Parameter δ is calculated according to formula (95) ₂=0.4068.

Get n=5N+1=151, obtained the input point of 153 y-z planes by formula (96-97):

$V_0^y={\left[\begin{array}{ccc}0& 0& 19.3483\end{array}\right]}^T;$ $V_j^y=u_j\left[\begin{array}{c}0\\ \sin \left({\mathrm{\Σ}}_{k=1}^j{\mathrm{\ω}}_k\right)\\ \cos \left({\mathrm{\Σ}}_{k=1}^j{\mathrm{\ω}}_k\right)\end{array}\right],$ J=1 ..., 152; Wherein, u _jand ω _kjth and a kth element of two endless 1 dimensional vector U and Ω respectively.Endless 1 dimensional vector U and Ω is; U=[P _{5 × 1}p _{5 × 1}...] _{∞ × 1}; Ω=[Q _{5 × 1}q _{5 × 1}...] _{∞ × 1}; Wherein; P=[18.9913 20.5189 20.5189 18.9913 19.3483]; Q=[π/300, π/75, π/30, π/75, π/300]; 12 × 151 summit V are calculated by formula (102) _{i, j}coordinate.

Finally define scrimp according to step 9.

Figure 22 is the 3-D view of obtained paper folding structure.Figure 23 is the partial enlarged view of structure shown in Figure 22.Can see, compared to embodiment 5, the present embodiment can be regarded as and inner side scrimp " v " type groove in embodiment 5 is replaced.Figure 24 is all summit V _{i, j}subpoint on the y-z plane and radius are the circle of 19 and 21, and these two circles are all dropped on circumferentially in visible most of summit, and all the other summits are then dropped between above-mentioned two circumference.This illustrate the paper folding structure shown in Figure 22 reach radial dimension designing requirement.In addition, the axial length of the structure shown in Figure 22 equals 11, also reaches axial dimension designing requirement.

Embodiment 12

Change the parameter n in embodiment 11 and keep other parameter constants, the paper folding structure of non-closed loop can be obtained.Such as when n gets 51, paper folding Structure composing 1/3 annulus obtained, as shown in figure 25.

Step 1: according to the internal diameter R of sandwich construction _in, external diameter R _out, the thickness t of inner casing _in, shell thickness t _exand segment length l _seg, utilize formula (106-108) to calculate the external diameter r of required interlayer ₁and internal diameter r ₂and length l.

r ₁＝R _out-t _ex(106)；r ₂＝R _in+t _in(107)；l＝l _seg(108)；

Step 2: m the input point being determined the x-z plane of three-dimensional cartesian coordinate system by formula (109a), wherein, m is natural number.

$V_i^x={\left[\begin{array}{ccc}{\mathrm{\Σ}}_{k=1}^i(\frac{1-{(-1)}^k}{2}a+\frac{{1+(-1)}^k}{2}b)& 0& -h+{\mathrm{\Σ}}_{k=1}^i{(-1)}^{\frac{i-1}{2}}(1-{(-1)}^i)h\end{array}\right]}^T,$ I=1 ..., m (109a); Wherein, parameter a and b meets: ${\mathrm{\Σ}}_{k=1}^m(\frac{1-{(-1)}^k}{2}a+\frac{1+{(-1)}^k}{2}b)=l+a---\left(109b\right);$

Step 3: selected Parameter N, wherein N be more than or equal to 3 natural number.Parameter beta is calculated according to formula (110) _set; ${\mathrm{\β}}_{\mathrm{set}}=\frac{2\mathrm{\π}}{N}---\left(110\right);$

Step 4: selected parameter beta, β ₁and β ₂, make it meet β+2 β ₁+ 2 β ₂=β _set, and meet β ₁> β ₂.

Step 5: calculate parameter r and δ according to formula (111-113).

$r_1^2={(r+\mathrm{\δ})}^2+h^2(1+{\left(\frac{(r-\mathrm{\δ})\sin {\mathrm{\β}}_1}{(r+\mathrm{\δ})-(r-\mathrm{\δ})\cos {\mathrm{\β}}_1}\right)}^2)---\left(111\right);$ $r_2^2={((r-\mathrm{\δ})\cos \frac{\mathrm{\β}}{2}-h)}^2+$ ${((r-\mathrm{\δ})\sin \frac{\mathrm{\β}}{2}-\frac{2\mathrm{hs}}{\sqrt{1-s^2}})}^2---\left(112\right);$ $-s\sin \frac{\mathrm{\β}}{2}-\sqrt{1-s^2}\cos \frac{\mathrm{\β}}{2}=\frac{(r-\mathrm{\δ})-(r+\mathrm{\δ})\cos {\mathrm{\β}}_1}{\sqrt{{(r+\mathrm{\δ})}^2+{(r-\mathrm{\δ})}^2-2(r+\mathrm{\δ})(r-\mathrm{\δ})\cos {\mathrm{\β}}_1}}---\left(113\right);$

Step 6: calculate parameter δ according to formula (114) ₁.

$\frac{(r-\mathrm{\δ})-(r-{\mathrm{\δ}}_2)\cos {\mathrm{\β}}_2}{\sqrt{{(r-\mathrm{\δ})}^2+{(r-{\mathrm{\δ}}_2)}^2-2(r-\mathrm{\δ})(r-{\mathrm{\δ}}_2)\cos {\mathrm{\β}}_2}}=\frac{(r-\mathrm{\δ})-(r+\mathrm{\δ})\cos {\mathrm{\β}}_1}{\sqrt{{(r-\mathrm{\δ})}^2+{(r+\mathrm{\δ})}^2-(r-\mathrm{\δ})(r+\mathrm{\δ})\cos {\mathrm{\β}}_1}}---\left(114\right);$

Step 7: n+2 the input point being determined the y-z plane of three-dimensional cartesian coordinate system by formula (115-116), wherein n is less than or equal to 5N+1.

$V_0^y={\left[\begin{array}{ccc}0& 0& r+{\mathrm{\δ}}_1\end{array}\right]}^T---\left(15\right);$ $V_j^y=u_j\left[\begin{array}{c}0\\ \sin \left({\mathrm{\Σ}}_{k=1}^j{\mathrm{\ω}}_k\right)\\ \cos \left({\mathrm{\Σ}}_{k=1}^j{\mathrm{\ω}}_k\right)\end{array}\right],$ J=1 ..., n+1 (116); Wherein, u _jand ω _kjth and a kth element of two endless 1 dimensional vector U and Ω respectively.The definition of endless 1 dimensional vector U and Ω is provided by formula (117) and (118).

U=[P _{5 × 1}p _{5 × 1}...] _{∞ × 1}(117); Ω=[Q _{5 × 1}q _{5 × 1}...] _{∞ × 1}(118); Namely 1 dimensional vector U is made up of numerous 5 × 1 vectorial P, and 1 dimensional vector Ω is made up of numerous 5 × 1 vectorial Q.5 × 1 vectorial P and Q are provided by formula (119) and (120) respectively.

P＝[r+δ r-δ r-δ r+δ r+δ ₁](119)；Q＝[β ₂β ₁β β ₁β ₂](120)；

Step 8: calculate m × n summit V by formula (121) _{i, j}coordinate.

$V_{i,j}=\left[\begin{array}{c}{x}_{i,j}\\ y_{i,j}\\ z_{i,j}\end{array}\right]=V_j^y+\left[A_j\right]V_i^x,$ I=1,2 ..., m; J=1,2 ..., n (121); Wherein, 3 × 3 matrix [A _j] provided by formula (122).

$\left[A_j\right]=\left[\begin{array}{ccc}1& 0& 0\\ 0& 0& {(-1)}^j\frac{\cos {\mathrm{\θ}}_{j-1}+\cos {\mathrm{\θ}}_j}{\sin ({\mathrm{\θ}}_{j-1}-{\mathrm{\θ}}_j)}\\ 0& 0& {(-1)}^j\frac{\sin {\mathrm{\θ}}_{j-1}+\sin {\mathrm{\θ}}_j}{\sin ({\mathrm{\θ}}_{j-1}-{\mathrm{\θ}}_j)}\end{array}\right]\left(122\right);$ Wherein; $\sin {\mathrm{\θ}}_j=\frac{i_{z^{\·}}(V_{j+1}^y-V_j^y)}{\left|\right|V_{j+1}^y-V_j^y\left|\right|}\left(123\right);$ wherein, i _y=[0 1 0] ^tfor the vector of unit length of y coordinate axis, i _z=[0 0 1] ^tfor the vector of unit length of z coordinate axle, || ■ || represent that subtend measures mould.

M × n obtained above summit V _{i, j}namely the summit of paper folding structure is constituted.

Step 9: definition (V _{i, j}y _{i+1, j}) or (V _{i, j}v _{i, j+1}) be 1 pair of adjacent vertex.All adjacent vertex straight lines are coupled together, as shown in Figure 2.Namely connecting line segment between these adjacent vertexs constitutes the scrimp of paper folding structure.

Can prove, equaled the internal diameter of shell structure housing to the external diameter that 9 design the paper folding structure obtained by step 1, the internal diameter of paper folding structure equals the external diameter of inner casing in shell structure, and the length of paper folding structure equals the length of shell structure.Therefore, this paper folding structure and inside and outside shell have geometry compatibility.Above-mentioned shell one paper folding structure one inner casing combination just constitutes the 7th kind of aircraft casing sandwich construction of the present invention.

Embodiment 13

Make the internal diameter R of aircraft sandwich structures _in=18, external diameter R _out=22, the thickness t of inner casing _in=1, the thickness t of shell _ex=1, segment length l _seg=11.The external diameter r of interlayer is obtained by formula (106-108) ₁=21, internal diameter r ₂=19, length l=11.

Get m=12, h=0.5, a=1, by formula (109b) known b=1.

The input point of 12 x-z planes is obtained by formula (109a): $V_2^x={\left[\begin{array}{ccc}2& 0& 0.5\end{array}\right]}^T;$ $V_3^x=$ ${\left[\begin{array}{ccc}3& 0& -0.5\end{array}\right]}^T;$ $V_4^x={\left[\begin{array}{ccc}4& 0& -0.5\end{array}\right]}^T;$ $V_5^x={\left[\begin{array}{ccc}5& 0& 0.5\end{array}\right]}^T;$ $V_6^x={\left[\begin{array}{ccc}6& 0& 0.5\end{array}\right]}^T;$ $V_7^x=$ ${\left[\begin{array}{ccc}7& 0& -0.5\end{array}\right]}^T;$ $V_8^x={\left[\begin{array}{ccc}8& 0& -0.5\end{array}\right]}^T;$ $V_9^x={\left[\begin{array}{ccc}9& 0& 0.5\end{array}\right]}^T;$ $V_{10}^x={\left[\begin{array}{ccc}10& 0& 0.5\end{array}\right]}^T;$ $V_{11}^x=$ ${\left[\begin{array}{ccc}11& 0& -0.5\end{array}\right]}^T;$ $V_{12}^x={\left[\begin{array}{ccc}12& 0& -0.5\end{array}\right]}^T;$ Get N=30, obtain β by formula (110) _set=π/15.

Get β ₂=π/300, β ₁=π/75, then β=β _set-2 β ₁-2 β ₂=π/30.

R=20.2580 and δ=0.7343 is calculated according to formula (111-113).

Parameter δ is calculated according to formula (114) ₁=0.3434.

Get n=5N+1=151, obtained the input point of 153 y-z planes by formula (115-116):

$V_0^y={\left[\begin{array}{ccc}0& 0& 20.6013\end{array}\right]}^T;$ $V_j^y=u_j\left[\begin{array}{c}0\\ \sin \left({\mathrm{\Σ}}_{k=1}^j{\mathrm{\ω}}_k\right)\\ \cos \left({\mathrm{\Σ}}_{k=1}^j{\mathrm{\ω}}_k\right)\end{array}\right],j=1,...,152;$ Wherein, u _jand ω _kjth and a kth element of two endless 1 dimensional vector U and Ω respectively.Endless 1 dimensional vector U and Ω is; U=[P _{5 × 1}p _{5 × 1}...] _{∞ × 1}; Ω=［ Q _{5 × 1}q _{5 × 1}...] _{∞ × 1}; Wherein; P=[20.9922 19.5237 19.5237 20.9922 20.6013]; Q=[π/300, π/75, π/30, π/75, π/300]; 7 × 151 summit V are calculated by formula (121) _{i, j}coordinate.

Finally define scrimp according to step 9.

Figure 26 is the 3-D view of obtained paper folding structure.Figure 27 is the partial enlarged view of structure shown in Figure 26.Can see, compared to embodiment 7, the present embodiment can be regarded as and outside scrimp " v " type groove in embodiment 7 is replaced.Figure 28 is all summit V _{i, j}subpoint on the y-z plane and radius are the circle of 19 and 21, and these two circles are all dropped on circumferentially in visible most of summit, and all the other summits are then dropped between above-mentioned two circumference.This illustrate the paper folding structure shown in Figure 26 reach radial dimension designing requirement.In addition, the axial length of the structure shown in Figure 26 equals 11, also reaches axial dimension designing requirement.

Embodiment 14

Change the parameter n in embodiment 13 and keep other parameter constants, the paper folding structure of non-closed loop can be obtained.Such as when n gets 51, paper folding Structure composing 1/3 annulus obtained, as shown in figure 29.

Step 1: according to the internal diameter R of sandwich construction _in, external diameter R _out, the thickness t of inner casing _in, shell thickness t _exand segment length l _seg, utilize formula (125-127) to calculate the external diameter r of required interlayer ₁and internal diameter r ₂and length l.

r ₁＝R _out-t _ex(125)；r ₂＝R _in+t _in(126)；l＝l _seg(127)；

Step 2: m the input point being determined the x-z plane of three-dimensional cartesian coordinate system by formula (128a), wherein, m is natural number.

$V_i^x={\left[\begin{array}{ccc}{\mathrm{\Σ}}_{k=1}^i(\frac{1-{(-1)}^k}{2}a+\frac{1+{(-1)}^k}{2}b)& 0& -h+{\mathrm{\Σ}}_{k=1}^i{(-1)}^{\frac{i-1}{2}}(1-{(-1)}^{i}h)\end{array}\right]}^T,$ I=1 ..., m (128a); Wherein, parameter a and b meets: ${\mathrm{\Σ}}_{k=1}^m(\frac{1-{(-1)}^k}{2}a+\frac{1+{(-1)}^k}{2}b)=l+a\left(128b\right);$

Step 3: selected Parameter N, wherein N be more than or equal to 3 natural number.Parameter beta is calculated according to formula (129) _set; ${\mathrm{\β}}_{\mathrm{set}}=\frac{2\mathrm{\π}}{N}\left(129\right);$

Step 4: selected parameter beta, β ₁and β ₂, make it meet 2 β+2 β ₁+ 2 β ₂=β _set.

Step 5: calculate parameter r and δ according to formula (130-131).

$r_2^2={(r-\mathrm{\δ})}^2+h^2(1+{\left(\frac{r\sin {\mathrm{\β}}_2}{r\cos {\mathrm{\β}}_2-(r-\mathrm{\δ})}\right)}^2)\left(130\right);$

$r_1^2={(r+\mathrm{\δ})}^2+h^2(1+{\left(\frac{r\sin {\mathrm{\β}}_1}{(r+\mathrm{\δ})-r\cos {\mathrm{\β}}_1}\right)}^2)\left(131\right);$

Step 6: n+2 the input point being determined the y-z plane of three-dimensional cartesian coordinate system by formula (132-133), wherein n is less than or equal to 6N+1.

$V_0^y={\left[\begin{array}{ccc}0& 0& r-\mathrm{\δ}\end{array}\right]}^T\left(132\right);$ $V_j^y=u_j\left[\begin{array}{c}0\\ \sin \left({\mathrm{\Σ}}_{k=1}^j{\mathrm{\ω}}_k\right)\\ \cos \left({\mathrm{\Σ}}_{k=1}^j{\mathrm{\ω}}_k\right)\end{array}\right],j=1,...,n+1\left(133\right);$ Wherein, u _jand ω _kjth and a kth element of two endless 1 dimensional vector U and Ω respectively.The definition of endless 1 dimensional vector U and Ω is provided by formula (134) and (135).

U=[P _{6 × 1}p _{6 × 1}...] _{∞ × 1}(134); Ω=[Q _{6 × 1}q _{6 × 1}...] _{∞ × 1}(135); Namely 1 dimensional vector U is made up of numerous 6 × 1 vectorial P, and 1 dimensional vector Ω is made up of numerous 6 × 1 vectorial Q.6 × 1 vectorial P and Q are provided by formula (136) and (137) respectively.

P＝[r r r+δ r r r-δ](136)；Q＝[β ₂ ββ ₁β ₁β β ₂](137)；

Step 7: calculate m × n summit V by formula (138) _{i, j}coordinate.

$V_{i,j}=\left[\begin{array}{c}{x}_{i,j}\\ y_{i,j}\\ x_{i,j}\end{array}\right]=V_j^y+\left[A_j\right]V_i^x,i=\mathrm{1,2},...,m;j=\mathrm{1,2},...n\left(138\right);$ Wherein, 3 × 3 matrix [A _j] provided by formula (139).

$\left[A_j\right]=\left[\begin{array}{ccc}1& 0& 0\\ 0& 0& {(-1)}^j\frac{\cos {\mathrm{\θ}}_{j-1}+\cos {\mathrm{\θ}}_j}{\sin ({\mathrm{\θ}}_{j-1}-{\mathrm{\θ}}_j)}\\ 0& 0& {(-1)}^j\frac{\sin {\mathrm{\θ}}_{j-1}+\sin {\mathrm{\θ}}_j}{\sin ({\mathrm{\θ}}_{j-1}-{\mathrm{\θ}}_j)}\end{array}\right]\left(139\right);$ Wherein; $\sin {\mathrm{\θ}}_j=\frac{i_{z^{\·}}(V_{j+1}^y-V_j^y)}{\left|\right|V_{j+1}^y-V_j^y\left|\right|}\left(140\right);$ wherein, i _y=[0 1 0] ^tfor the vector of unit length of y coordinate axis, i _z=[0 0 1] ^tfor the vector of unit length of z coordinate axle, || ■ || represent that subtend measures mould.

M × n obtained above summit V _{i, j}namely the summit of paper folding structure is constituted.

Step 8: definition { V _{i, j}v _{i+1, j}or { Vi _dv _{i, j+1}it is 1 pair of adjacent vertex.All adjacent vertex straight lines are coupled together, as shown in Figure 2.Namely connecting line segment between these adjacent vertexs constitutes the scrimp of paper folding structure.

Can prove, equaled the internal diameter of shell structure housing to the external diameter that 8 design the paper folding structure obtained by step 1, the internal diameter of paper folding structure equals the external diameter of inner casing in shell structure, and the length of paper folding structure equals the length of shell structure.Therefore, this paper folding structure and inside and outside shell have geometry compatibility.Above-mentioned shell-paper folding structure-inner casing combination just constitutes the 8th kind of aircraft casing sandwich construction of the present invention.

Embodiment 15

Make the internal diameter R of aircraft sandwich structures _in=18, external diameter R _out=22, the thickness t of inner casing _in=1, the thickness t of shell _ex=1, segment length l _seg=11.The external diameter r of interlayer is obtained by formula (125-127) ₁=21, internal diameter r ₂=19, length l=11.

Get m=12, h=0.5, a=1, by formula (128b) known b=1.

The input point of 12 x-z planes is obtained by formula (128a): $V_2^x={\left[\begin{array}{ccc}2& 0& 0.5\end{array}\right]}^T;$ $V_3^x=$ ${\left[\begin{array}{ccc}3& 0& -0.5\end{array}\right]}^T;$ $V_4^x={\left[\begin{array}{ccc}4& 0& -0.5\end{array}\right]}^T;$ $V_5^x={\left[\begin{array}{ccc}5& 0& 0.5\end{array}\right]}^T;$ $V_6^x={\left[\begin{array}{ccc}6& 0& 0.5\end{array}\right]}^T;$ $V_7^x=$ ${\left[\begin{array}{ccc}7& 0& -0.5\end{array}\right]}^T;$ $V_8^x={\left[\begin{array}{ccc}8& 0& -0.5\end{array}\right]}^T;$ $V_9^x={\left[\begin{array}{ccc}9& 0& 0.5\end{array}\right]}^T;$ $V_{10}^x={\left[\begin{array}{ccc}10& 0& 0.5\end{array}\right]}^T;$ $V_{11}^x=$ ${\left[\begin{array}{ccc}11& 0& -0.5\end{array}\right]}^T;$ $V_{12}^x={\left[\begin{array}{ccc}12& 0& -0.5\end{array}\right]}^T;$ Get N=30, obtain β by formula (129) _set=π/15.

Get β ₂=β ₁=π/90, then β=(β _set-2 β ₁-2 β ₂)/2=π/90.

R=19.9907 and δ=1.0005 are calculated according to formula (130-131).

Get n=6N+1=181, obtained the input point of 183 y-z planes by formula (132-133):

$V_0^y={\left[\begin{array}{ccc}0& 0& 18.9901\end{array}\right]}^T;$ $V_j^y=u_j\left[\begin{array}{c}0\\ \sin \left({\mathrm{\Σ}}_{k=1}^j{\mathrm{\ω}}_k\right)\\ \cos \left({\mathrm{\Σ}}_{k=1}^j{\mathrm{\ω}}_k\right)\end{array}\right],j=1,...,182;$ Wherein, u _jand ω _kjth and a kth element of two endless 1 dimensional vector U and Ω respectively.Endless 1 dimensional vector U and Ω is; U=［ P _{6 × 1}p _{6 × 1}...] _{∞ × 1}; Ω=[Q _{6 × 1}q _{6 × 1}...] _{∞ × 1}; Wherein;

P=[19.9907 19.9907 20.9912 19.9907 18.9901]; Q=[π/90, π/90, π/90, π/90, π/90, π/90]; 12 × 181 summit V are calculated by formula (138) _{i, j}coordinate.

Finally define scrimp according to step 8.

Figure 30 is the 3-D view of obtained paper folding structure.Figure 31 is the partial enlarged view of structure shown in Figure 30.Figure 32 is all summit V _{i, j}subpoint on the y-z plane and radius are the circle of 19 and 21, visible all summits all drop on these two circles circumferentially or drop between above-mentioned two circumference.This illustrate the paper folding structure shown in Figure 30 reach radial dimension designing requirement.In addition, the axial length of the structure shown in Figure 30 equals 11, also reaches axial dimension designing requirement.

Embodiment 16

Change the parameter n in embodiment 15 and keep other parameter constants, the paper folding structure of non-closed loop can be obtained.Such as when n gets 61, paper folding Structure composing 1/3 annulus obtained, as shown in figure 33.

Step 1: according to the internal diameter R of sandwich construction _in, external diameter R _out, the thickness t of inner casing _in, shell thickness t _exand segment length l _seg, utilize formula (142-144) to calculate the external diameter r of required interlayer ₁and internal diameter r ₂and length l.

r ₁＝R _out-t _ex(142)；r ₂＝R _in+t _in(143)；l＝l _seg(144)；

Step 2: m the input point being determined the x-z plane of three-dimensional cartesian coordinate system by formula (145a), wherein, m is natural number.

$V_i^x={\left[\begin{array}{ccc}{\mathrm{\Σ}}_{k=1}^i(\frac{1-{(-1)}^k}{2}a+\frac{1+{(-1)}^k}{2}b)& 0& -h+{\mathrm{\Σ}}_{k=1}^i{(i-1)}^{\frac{i-1}{2}}(1-{(-1)}^i)h\end{array}\right]}^T,$ I=1 ..., m (145a); Wherein, parameter a and b meets: ${\mathrm{\Σ}}_{k=1}^m(\frac{1-{(-1)}^k}{2}a+\frac{1+{(-1)}^k}{2}b)=l+a\left(145b\right);$

Step 3: selected Parameter N, wherein N be more than or equal to 3 natural number.Parameter beta is calculated according to formula (146) _set; ${\mathrm{\β}}_{\mathrm{set}}=\frac{2\mathrm{\π}}{N}\left(146\right);$

Step 4: selected parameter beta, β ₁, β ₂, β ₃and β ₄, make it meet 2 β+2 β ₁+ 2 β ₂+ 2 β ₃+ 2 β ₄=β _set, and meet β ₁> β ₃and β ₂> β ₄.Step 5: calculate parameter r and δ according to formula (147-148).

$r_2^2={(r-\mathrm{\δ})}^2+h^2(1+{\left(\frac{r\sin {\mathrm{\β}}_2}{r\cos {\mathrm{\β}}_2-(r-\mathrm{\δ})}\right)}^2)\left(147\right);$

$r_1^2={(r+\mathrm{\δ})}^2+h^2(1+{\left(\frac{r\sin {\mathrm{\β}}_1}{(r+\mathrm{\δ})-r\cos {\mathrm{\β}}_1}\right)}^2)\left(148\right);$

Step 6: calculate parameter δ according to formula (149-150) ₁and δ ₂.

$\frac{(r+\mathrm{\δ})-(r+{\mathrm{\δ}}_1)\cos {\mathrm{\β}}_3}{\sqrt{{(r+\mathrm{\δ})}^2+{(r+{\mathrm{\δ}}_1)}^2-2(r+\mathrm{\δ})(r+{\mathrm{\δ}}_1)\cos {\mathrm{\β}}_3}}=\frac{(r+\mathrm{\δ})-r\cos {\mathrm{\β}}_1}{\sqrt{{(r+\mathrm{\δ})}^2+r^2-2(r+\mathrm{\δ})r\cos {\mathrm{\β}}_1}}\left(149\right);$

$\frac{(r-\mathrm{\δ})-(r-{\mathrm{\δ}}_2)\cos {\mathrm{\β}}_4}{\sqrt{{(r-\mathrm{\δ})}^2+{(r-{\mathrm{\δ}}_2)}^2-2(r-\mathrm{\δ})(r-{\mathrm{\δ}}_2)\cos {\mathrm{\β}}_4}}=\frac{(r-\mathrm{\δ})-r\cos {\mathrm{\β}}_2}{\sqrt{{(r-\mathrm{\δ})}^2+r^2-2(r-\mathrm{\δ})r\cos {\mathrm{\β}}_2}}\left(150\right);$

Step 7: n+2 the input point being determined the y-z plane of three-dimensional cartesian coordinate system by formula (151-152), wherein n is less than or equal to 10N+1.

$V_0^y={\left[\begin{array}{ccc}0& 0& r-{\mathrm{\δ}}_2\end{array}\right]}^T\left(151\right);$ $V_j^y=u_j\left[\begin{array}{c}0\\ \sin \left({\mathrm{\Σ}}_{k=1}^j{\mathrm{\ω}}_k\right)\\ \cos \left({\mathrm{\Σ}}_{k=1}^j{\mathrm{\ω}}_k\right)\end{array}\right],j=1,...,n+1\left(152\right);$ Wherein, u _jand ω _kjth and a kth element of two endless 1 dimensional vector U and Ω respectively.The definition of endless 1 dimensional vector U and Ω is provided by formula (153) and (154).

U=[P _{10 × 1}p _{10 × 1}...] _{∞ × 1}(153); Ω=[Q _{10 × 1}q _{10 × 1}...] _{∞ × 1}(154); Namely 1 dimensional vector U is made up of numerous 10 × 1 vectorial P, and 1 dimensional vector Ω is made up of numerous 10 × 1 vectorial Q.10 × 1 vectorial P and Q are provided by formula (155) and (156) respectively.

P＝[r-δ r r r+δ r+δ ₁r+δ r r r-δ r-δ ₂](155)；

Q＝[β ₄β ₂β β ₁β ₃β ₃β ₁β β ₂β ₄](156)；

Step 8: calculate m × n summit V by formula (157) _{i, j}coordinate.

$V_{i,j}=\left[\begin{array}{c}{x}_{i,j}\\ y_{i,j}\\ z_{i,j}\end{array}\right]=V_j^y+\left[A_j\right]V_i^x,i=\mathrm{1,2},...,m;j=\mathrm{1,2},...,n\left(157\right);$ Wherein, 3 × 3 matrix [A _j] provided by formula (158).

$\left[A_j\right]=\left[\begin{array}{ccc}1& 0& 0\\ 0& 0& {(-1)}^j\frac{\cos {\mathrm{\θ}}_{j-1}+\cos {\mathrm{\θ}}_j}{\sin ({\mathrm{\θ}}_{j-1}-{\mathrm{\θ}}_j)}\\ 0& 0& {(-1)}^j\frac{\sin {\mathrm{\θ}}_{j-1}+\sin {\mathrm{\θ}}_j}{\sin ({\mathrm{\θ}}_{j-1}-{\mathrm{\θ}}_j)}\end{array}\right]\left(158\right);$ Wherein; $\sin {\mathrm{\θ}}_j=\frac{i_{z^{\·}}(V_{j+1}^y-V_j^y)}{\left|\right|V_{j+1}^y-V_j^y\left|\right|}\left(159\right);$ Cos wherein, i _y=[0 1 0] ^tfor the vector of unit length of y coordinate axis, i _z=[0 0 1] ^tfor the vector of unit length of z coordinate axle, || ■ || represent that subtend measures mould.

M × n obtained above summit V _{i, j}namely the summit of paper folding structure is constituted.

Step 9: definition { V _uv _{i+1, j}or { V _{i, j}v _{i, j+1}it is 1 pair of adjacent vertex.All adjacent vertex straight lines are coupled together, as shown in Figure 2.Namely connecting line segment between these adjacent vertexs constitutes the scrimp of paper folding structure.Can prove, equaled the internal diameter of shell structure housing to the external diameter that 9 design the paper folding structure obtained by step 1, the internal diameter of paper folding structure equals the external diameter of inner casing in shell structure, and the length of paper folding structure equals the length of shell structure.Therefore, this paper folding structure and inside and outside shell have geometry compatibility.Above-mentioned shell-paper folding structure-inner casing combination just constitutes the 9th kind of aircraft casing sandwich construction of the present invention.

Embodiment 17

Make the internal diameter R of aircraft sandwich structures _in=18, external diameter R _out=22, the thickness t of inner casing _in=1, the thickness t of shell _ex=1, segment length l _sex=11.The external diameter r of interlayer is obtained by formula (142-144) ₁=21, internal diameter r ₂=19, length l=11.

Get m=12, h=0.5, a=1, by formula (145b) known b=1.

The input point of 12 x-z planes is obtained by formula (145a): $v_1^x={\left[0.5\right]}^T;$ $V_2^x={\left[\begin{array}{ccc}2& 0& 0.5\end{array}\right]}^T;$ $V_3^x=$ ${\left[\begin{array}{ccc}3& 0& -0.5\end{array}\right]}^T;$ $V_4^x={\left[\begin{array}{ccc}4& 0& -0.5\end{array}\right]}^T;$ $V_5^x={\left[\begin{array}{ccc}5& 0& 0.5\end{array}\right]}^T;$ $V_6^x={\left[\begin{array}{ccc}6& 0& 0.5\end{array}\right]}^T;$ $V_7^x=$ ${\left[\begin{array}{ccc}7& 0& -0.5\end{array}\right]}^T;$ $V_8^x={\left[\begin{array}{ccc}8& 0& -0.5\end{array}\right]}^T;$ $V_9^x={\left[\begin{array}{ccc}9& 0& 0.5\end{array}\right]}^T;$ $V_{10}^x={\left[\begin{array}{ccc}10& 0& 0.5\end{array}\right]}^T;$ $V_{11}^x=$ ${\left[\begin{array}{ccc}11& 0& -0.5\end{array}\right]}^T;$ $V_{12}^x={\left[\begin{array}{ccc}12& 0& -0.5\end{array}\right]}^T;$ Get N=30, obtain β by formula (146) _set=π/15.

Get β ₃=β ₄=π/420, β ₁=β ₂=π/105, then β=(β _set-2 β ₁-2 β ₂-2 β ₃-2 β ₄)/2=π/105.

R=19.9915 and δ=1.0005 are calculated according to formula (147-148).

Parameter δ is calculated according to formula (149-150) ₁=0.7393 and δ ₂=0.7615.

Get n=10N+1=301, obtained the input point of 303 y-z planes by formula (151-152):

$V_0^y={\left[\begin{array}{ccc}0& 0& 19.2300\end{array}\right]}^T;$ $V_j^y=u_j\left[\begin{array}{c}0\\ \sin \left({\mathrm{\Σ}}_{k=1}^j{\mathrm{\ω}}_k\right)\\ \cos \left({\mathrm{\Σ}}_{k=1}^j{\mathrm{\ω}}_k\right)\end{array}\right],j=1,...,302;$ Wherein, u _jand ω _kjth and a kth element of two endless 1 dimensional vector U and Ω respectively.Endless 1 dimensional vector U and Ω is;

U=[P _{10 × 1}p _{10 × 1}...] _{∞ × 1}; Ω=[Q _{10 × 1}q _{10 × 1}...] _{∞ × 1}; Wherein,

P＝

[18.9910 19.9915 19.9915 20.9920 20.7308 20.9920 19.9915 19.9915 18.9910 19.2300]；

Q＝

π/420, π/105, π/105, π/105, π/420, π/420, π/105, π/105, π/105, π/420]; 12 × 301 summit V are calculated by formula (157) _{i, j}coordinate.

Finally define scrimp according to step 9.

Figure 34 is the 3-D view of obtained paper folding structure.Figure 35 is the partial enlarged view of structure shown in Figure 34.Can see, compared to embodiment 15, the present embodiment can be regarded as and medial and lateral scrimp " v " type groove in embodiment 15 is replaced.Figure 36 is all summit V _{i, j}subpoint on the y-z plane and radius are the circle of 19 and 21, visible all summits all drop on these two circles circumferentially or drop between above-mentioned two circumference.This illustrate the paper folding structure shown in Figure 34 reach radial dimension designing requirement.In addition, the axial length of the structure shown in Figure 34 equals 11, also reaches axial dimension designing requirement.

Embodiment 18

Change the parameter n in embodiment 17 and keep other parameter constants, the paper folding structure of non-closed loop can be obtained.Such as when n gets 101, paper folding Structure composing 1/3 annulus obtained, as shown in figure 37.

Above-described embodiment gives the aircraft casing sandwich construction and its implementation that are applicable to have round section.Said method can be applied in the housing of noncircular cross section easily.

When being applied to the aircraft casing with noncircular cross section, first this housing section being approximated to and being made up of some sections of circular arc housings.For every 1 section of arc section, the external diameter r of required interlayer can be determined ₁with internal diameter r ₂, the method design of recycling described in above-described embodiment goes out to be applicable to the interlayer based on paper folding structure of this arc section.

Embodiment 19

Consider aircraft casing cross section as shown in figure 38, this housing is made up of 4 parts, is respectively: top is with O ₁for the arc section in the center of circle, outer, the internal diameter of its interlayer are respectively r ' ₁with r ' ₂; Left side is with O ₂for the arc section in the center of circle, outer, the internal diameter of its interlayer are respectively r " ₁with r " ₂; Bottom is with O ₃for the arc section in the center of circle, outer, the internal diameter of its interlayer are respectively r " ' ₁with r " ' ₂; Right side is with O ₄for the arc section in the center of circle, outer, the internal diameter of its interlayer are respectively r " " ₁with r " " ₂.

Make r ' ₁=21, r ' ₂=19, r " ₁=51, r " ₂=49, r " ' ₁=29.786, r " ' ₂=27.786, r " " ₁=51, r " " ₂=49.Shell length l=11.Utilize the method for embodiment 1, design the interlayer in top, right side, bottom, left side respectively.Specific as follows:

I () top interlayer: get m=12, h=1, a=1, by formula (4b) known b=1.The input point of 12 x-z planes is obtained by formula (4a): $V_1^x={\left[\begin{array}{ccc}1& 0& 1\end{array}\right]}^T;$ $V_2^x={\left[\begin{array}{ccc}2& 0& 1\end{array}\right]}^T;$ $V_3^x={\left[\begin{array}{ccc}3& 0& -1\end{array}\right]}^T;$ $V_4^x=$ ${\left[\begin{array}{ccc}4& 0& -1\end{array}\right]}^T;$ $V_5^x={\left[\begin{array}{ccc}5& 0& 1\end{array}\right]}^T;$ $V_6^x={\left[\begin{array}{ccc}6& 0& 1\end{array}\right]}^T;$ $V_7^x={\left[\begin{array}{ccc}7& 0& -1\end{array}\right]}^T;$ $V_8^x=$ ${\left[\begin{array}{ccc}8& 0& -1\end{array}\right]}^T;$ $V_9^x={\left[\begin{array}{ccc}9& 0& 1\end{array}\right]}^T;$ $V_{10}^x={\left[\begin{array}{ccc}10& 0& 1\end{array}\right]}^T;$ $V_{11}^x={\left[\begin{array}{ccc}11& 0& -1\end{array}\right]}^T;$ $V_{12}^x=$ ${\left[\begin{array}{ccc}12& 0& -1\end{array}\right]}^T;$ Get N=40, obtain β=π/40 by formula (5).

R=19.9593 and δ=1.0045 are calculated according to formula (6-7).

Get n=21, obtained the input point of 23 y-z planes by formula (8):

$V_j^y=[19.9593+{(-1)}^{j}1.0045]\left[\begin{array}{c}0\\ \sin \left(\frac{\mathrm{j\π}}{40}\right)\\ \cos \left(\frac{\mathrm{j\π}}{40}\right)\end{array}\right],j=\mathrm{0,1},...,22;$ 12 × 21 summit V are calculated by formula (9) _{i, j}coordinate.

(ii) left and right side interlayer: because left and right side interlayer has identical physical dimension, therefore only need the interlayer of design 1 side.

Get m=12, h=1, a=1, by formula (4b) known b=1.The input point of 12 x-z planes is obtained by formula (4a):

$V_1^x={\left[\begin{array}{ccc}1& 0& 1\end{array}\right]}^T;$ $V_2^x={\left[\begin{array}{ccc}2& 0& 1\end{array}\right]}^T;$ $V_3^x={\left[\begin{array}{ccc}3& 0& -1\end{array}\right]}^T;$ $V_4^x={\left[\begin{array}{ccc}4& 0& -1\end{array}\right]}^T;$ $V_5^x=$ ${\left[\begin{array}{ccc}5& 0& 1\end{array}\right]}^T;$ $V_6^x={\left[\begin{array}{ccc}6& 0& 1\end{array}\right]}^T;$ $V_7^x={\left[\begin{array}{ccc}7& 0& -1\end{array}\right]}^T;$ $V_8^x={\left[\begin{array}{ccc}8& 0& -1\end{array}\right]}^T;$ $V_9^x=$ ${\left[\begin{array}{ccc}9& 0& 1\end{array}\right]}^T;$ $V_{10}^x={\left[\begin{array}{ccc}10& 0& 1\end{array}\right]}^T;$ $V_{11}^x={\left[\begin{array}{ccc}11& 0& -1\end{array}\right]}^T;$ $V_{12}^x={\left[\begin{array}{ccc}12& 0& -1\end{array}\right]}^T;$ Get N=80, obtain β=π/80 by formula (5).

R=49.9803 and δ=1.0011 are calculated according to formula (6-7).

Get n=21, obtained the input point of 23 y-z planes by formula (8):

$V_j^y=[49.9803+{(-1)}^{j}1.0011]\left[\begin{array}{c}0\\ \sin \left(\frac{\mathrm{j\π}}{80}\right)\\ \cos \left(\frac{\mathrm{j\π}}{80}\right)\end{array}\right],j=\mathrm{0,1},...,22;$ 12 × 21 summit V are calculated by formula (9) _{i, j}coordinate.

(iii) bottom interlayer: get m=12, h=1, a=1, by formula (4b) known b=1.The input point of 12 x-z planes is obtained by formula (4a): $V_1^x={\left[\begin{array}{ccc}1& 0& 1\end{array}\right]}^T;$ $V_2^x={\left[\begin{array}{ccc}2& 0& 1\end{array}\right]}^T;$ $V_3^x={\left[\begin{array}{ccc}3& 0& -1\end{array}\right]}^T;$ $V_4^x=$ ${\left[\begin{array}{ccc}4& 0& -1\end{array}\right]}^T;$ $V_5^x={\left[\begin{array}{ccc}5& 0& 1\end{array}\right]}^T;$ $V_6^x={\left[\begin{array}{ccc}6& 0& 1\end{array}\right]}^T;$ $V_7^x={\left[\begin{array}{ccc}7& 0& -1\end{array}\right]}^T;$ $V_8^x=$ ${\left[\begin{array}{ccc}8& 0& -1\end{array}\right]}^T;$ $V_9^x={\left[\begin{array}{ccc}9& 0& 1\end{array}\right]}^T;$ $V_{10}^x={\left[\begin{array}{ccc}10& 0& 1\end{array}\right]}^T;$ $V_{11}^x={\left[\begin{array}{ccc}11& 0& -1\end{array}\right]}^T;$ $V_{12}^x=$ ${\left[\begin{array}{ccc}12& 0& -1\end{array}\right]}^T;$ Get N=40, obtain β=π/40 by formula (5).

R=28.7462 and δ=1.0049 are calculated according to formula (6-7).

Get n=41, obtained the input point of 43 y-z planes by formula (8):

$V_j^y=[28.7462+{(-1)}^{j}1.0049]\left[\begin{array}{c}0\\ \sin \left(\frac{\mathrm{j\π}}{80}\right)\\ \cos \left(\frac{\mathrm{j\π}}{80}\right)\end{array}\right],j=\mathrm{0,1},...,42;$ 12 × 41 summit V are calculated by formula (9) _{i, j}coordinate.

Joined end to end in order by four part interlayers obtained above, namely obtain paper folding structure as shown in figures 39 and 40, it has the physical dimension identical with the interlayer of housing shown in Figure 38.

Above-mentioned aircraft casing is only made up of a layer interlayer.But, utilize the method that above-described embodiment provides, can be easy to design the aircraft casing with two-layer or two-layer above interlayer.Can be separated by middle case face between interlayer, as shown in figure 41, middle case face can not had to separate, as shown in figure 42 yet.

For each layer interlayer, determine the external diameter r of required interlayer ₁with internal diameter r ₂, the either method of recycling described in above-described embodiment is designed accordingly based on the interlayer of paper folding structure.

There is the crustless sandwich structure of two-layer or two-layer above interlayer, can be separated by middle case face between interlayer; There is the crustless sandwich structure of two-layer or two-layer above interlayer, separate without middle case face between interlayer; The interlayer based on paper folding structure that above-mentioned design obtains can be made by various suitable material, includes but not limited to: metal, synthetic material, carbon fibre material, paper.

Manufacture method includes but not limited to: design the mould corresponding with paper folding structure, utilize mould to carry out shaping to flat sheet; Three-dimensional printing technology is utilized to print; Utilize and manufacture the similar way of comb core, namely first utilizing the material (such as paper) that can conveniently fold to carry out craft folding, and then immersing in glue and the structure folded is shaped and reinforces.

After making interlayer, interlayer is placed among housing, utilizes glue or welding to be connected with shell face, aircraft casing sandwich construction of the present invention can be produced.

Claims (7)

1. an implementation method for aircraft sandwich structures, this structure is made up of inner casing, shell and the multiple interlayers with paper folding structure be clipped between inside and outside shell, and described interlayer realizes in the following manner:

Step one: according to the internal diameter R of sandwich construction _in, external diameter R _out, the thickness t of inner casing _in, shell thickness t _exand segment length l _segcalculate the external diameter r of required interlayer ₁, internal diameter r ₂and length l:r ₁=R _out-t _ex; r ₂=R _in+ t _in; L=l _seg;

Step 2: m the input point determining the x-z plane of three-dimensional cartesian coordinate system and the n+2 of a y-z plane input point wherein: $V_i^x={\left[\begin{array}{ccc}{\mathrm{\Σ}}_{k=1}^i(\frac{1-{(-1)}^k}{2}a+\frac{1+{(-1)}^k}{2}b)& 0& -h+{\mathrm{\Σ}}_{k=1}^i{(-1)}^{\frac{i-1}{2}}\end{array}(1-{(-1)}^i)h\right]}^T,$ I=1 ..., m, wherein: ${\mathrm{\Σ}}_{k=1}^m(\frac{1-{(-1)}^k}{2}a+\frac{1+{(-1)}^k}{2}b)=l+a,$ M is the natural number much larger than m, 2mT=l;

the following either type of concrete employing realizes:

2.1) $V_j^y=[r+{(-1)}^j\mathrm{\δ}]\left[\begin{array}{c}0\\ \sin \left(\mathrm{j\β}\right)\\ \cos \left(\mathrm{j\β}\right)\end{array}\right],$ J=0,1 ..., n+1, wherein: n be more than or equal to 3 natural number;

2.2) $V_0^y={\left[\begin{array}{ccc}0& 0& r+{\mathrm{\δ}}_1\end{array}\right]}^T,$ $V_j^y=u_j\left[\begin{array}{c}0\\ \sin \left({\mathrm{\Σ}}_{k=1}^j{\mathrm{\ω}}_k\right)\\ \cos \left({\mathrm{\Σ}}_{k=1}^j{\mathrm{\ω}}_k\right)\end{array}\right],$ J=1 ..., n+1, wherein: n is less than or equal to 6N+1, N be more than or equal to 3 natural number, $r_2^2={(r-\mathrm{\δ})}^2+h^2(1+{\left(\frac{(r+\mathrm{\δ})\sin \mathrm{\β}}{(r+\mathrm{\δ})\cos \mathrm{\β}-(r-\mathrm{\δ})}\right)}^2),$ $r_1^2={(r+\mathrm{\δ})}^2+$ $h^2(1+{\left(\frac{(r-\mathrm{\δ})\sin \mathrm{\β}}{(r+\mathrm{\δ})-(r-\mathrm{\δ})\cos \mathrm{\β}}\right)}^2);$ $\frac{(r+\mathrm{\δ})-(r+{\mathrm{\δ}}_1)\cos {\mathrm{\β}}_1}{\sqrt{{(r+\mathrm{\δ})}^2+{(r+{\mathrm{\δ}}_1)}^2-2(r+\mathrm{\δ})(r+{\mathrm{\δ}}_1)\cos {\mathrm{\β}}_1}}=\frac{(r+\mathrm{\δ})-(r-\mathrm{\δ})\cos \mathrm{\β}}{\sqrt{{(r+\mathrm{\δ})}^2+{(r-\mathrm{\δ})}^2-2(r+\mathrm{\δ})(r-\mathrm{\δ})\cos \mathrm{\β}}},$ $\frac{(r-\mathrm{\δ})-(r-{\mathrm{\δ}}_2)\cos {\mathrm{\β}}_2}{\sqrt{{(r-\mathrm{\δ})}^2+{(r-{\mathrm{\δ}}_2)}^2-2(r-\mathrm{\δ})(r-{\mathrm{\δ}}_2)\cos {\mathrm{\β}}_2}}=\frac{(r-\mathrm{\δ})-(r+\mathrm{\δ})\cos \mathrm{\β}}{\sqrt{{(r-\mathrm{\δ})}^2+{(r+\mathrm{\δ})}^2-2(r-\mathrm{\δ})(r+\mathrm{\δ})\cos \mathrm{\β}}},$ ${\mathrm{\β}}_{\mathrm{set}}=\frac{\mathrm{\π}}{N},$ β+β ₁+ β ₂=β _set, and meet β > β ₁and β > β ₂; u _jand ω _kjth and a kth element of two endless one-dimensional vector U and Ω respectively; U=[P _{6 × 1}p _{6 × 1}...] _{∞ × 1}, Ω=[Q _{6 × 1}q _{6 × 1}...] _{∞ × 1}; 6 × 1 vectorial P and Q are P=[r+ δ r-δ r-δ ₂r-δ r+ δ r+ δ ₁], Q=[β ₁β β ₂β ₂β β ₁];

2.3) $V_0^y={\left[\begin{array}{ccc}0& 0& r-\mathrm{\δ}\end{array}\right]}^T,$ ${V_j^y=u}_j\left[\begin{array}{c}0\\ \sin \left({\mathrm{\Σ}}_{k=1}^j{\mathrm{\ω}}_k\right)\\ \cos \left({\mathrm{\Σ}}_{k=1}^j{\mathrm{\ω}}_k\right)\end{array}\right],$ J=1 ..., Ω+1, wherein: n is less than or equal to 3N+1, N be more than or equal to 3 natural number, β+2 β ₁=β _set;

$r_2^2={(r-\mathrm{\δ})}^2+h^2(1+{\left(\frac{(r+\mathrm{\δ})\sin {\mathrm{\β}}_1}{(r+\mathrm{\δ})\cos {\mathrm{\β}}_1-(r-\mathrm{\δ})}\right)}^2),$ $r_1^2={\left(\left(r+\mathrm{\δ}\right)\cos \frac{\mathrm{\β}}{2}+h\right)}^2+((r+\mathrm{\δ})\sin \frac{\mathrm{\β}}{2}-$ ${\frac{2\mathrm{hs}}{\sqrt{1-s^2}})}^2,$ $s\sin ({\mathrm{\β}}_1+\frac{\mathrm{\β}}{2})-\sqrt{1-s^2}\cos ({\mathrm{\β}}_1+\frac{\mathrm{\β}}{2})=\frac{(r-\mathrm{\δ})-(r+\mathrm{\δ})\cos {\mathrm{\β}}_1}{\sqrt{{(r+\mathrm{\δ})}^2+{(r-\mathrm{\δ})}^2-2(r+\mathrm{\δ})(r-\mathrm{\δ})\cos {\mathrm{\β}}_1}};$ U _jand ω _kjth and a kth element of two endless one-dimensional vector U and Ω respectively, U=[P _{3 × 1}p _{3 × 1}...] _{∞ × 1}, Ω=[Q _{3 × 1}q _{3 × 1}...] _{∞ × 1}, 3 × 1 vectorial P and Q are respectively P=[r+ δ r+ δ r-δ], Q=[β ₁β β ₁];

2.4) $V_0^y={\left[\begin{array}{ccc}0& 0& r-\mathrm{\δ}\end{array}\right]}^T,$ $V_j^y=u_j\left[\begin{array}{c}0\\ \sin \left({\mathrm{\Σ}}_{k=1}^j{\mathrm{\ω}}_k\right)\\ \cos \left({\mathrm{\Σ}}_{k=1}^j{\mathrm{\ω}}_k\right)\end{array}\right],$ J=1 ..., n+1, wherein: n is less than or equal to 3N+1, N be more than or equal to 3 natural number, β+2 β ₁=β _set; $r_1^2={(r+\mathrm{\δ})}^2+h^2(1+$ ${\left(\frac{(r-\mathrm{\δ})\sin {\mathrm{\β}}_1}{(r+\mathrm{\δ})-(r-\mathrm{\δ})\cos {\mathrm{\β}}_1}\right)}^2),$ $r_2^2={((r-\mathrm{\δ})\cos \frac{\mathrm{\β}}{2}-h)}^2+{((r-\mathrm{\δ})\sin \frac{\mathrm{\β}}{2}-\frac{2\mathrm{hs}}{\sqrt{1-s^2}})}^2,$ $-s\sin \frac{\mathrm{\β}}{2}-\sqrt{1-s^2}\cos \frac{\mathrm{\β}}{2}=\frac{(r-\mathrm{\δ})-(r+\mathrm{\δ})\cos {\mathrm{\β}}_1}{\sqrt{{(r+\mathrm{\δ})}^2+{(r-\mathrm{\δ})}^2-2(r+\mathrm{\δ})(r-\mathrm{\δ})\cos {\mathrm{\β}}_1}},$ U _jand ω _kjth and a kth element of two endless one-dimensional vector U and Ω respectively, U=[P _{3 × 1}p _{3 × 1}...] _{∞ × 1}, Ω=[Q _{3 × 1}q _{3 × 1}...] _{∞ × 1}, 3 × 1 vectorial P and Q are respectively P=[r-δ r-δ r+ δ], Q=[β ₁β β ₁];

2.5) $V_0^y={\left[\begin{array}{ccc}0& 0& r+\mathrm{\δ}\end{array}\right]}^T,$ $V_j^y=u_j\left[\begin{array}{c}0\\ \sin \left({\mathrm{\Σ}}_{k=1}^j{\mathrm{\ω}}_k\right)\\ \cos \left({\mathrm{\Σ}}_{k=1}^j{\mathrm{\ω}}_k\right)\end{array}\right],$ J=1 ..., n+1, wherein: n is less than or equal to 4N+1, N be more than or equal to 3 natural number, 2 β+β ₁+ β ₂=β _set; $r_1^2={((r+\mathrm{\δ})\cos \frac{{\mathrm{\β}}_1}{2}+h)}^2+$ ${((r+\mathrm{\δ})\sin \frac{{\mathrm{\β}}_1}{2}-\frac{2\mathrm{hs}}{\sqrt{1-s^2}})}^2,$ $s\sin (\mathrm{\β}+\frac{{\mathrm{\β}}_1}{2})-\sqrt{1-s^2}\cos (\mathrm{\β}+\frac{{\mathrm{\β}}_1}{2})=\frac{(r-\mathrm{\δ})-(r+\mathrm{\δ})\cos \mathrm{\β}}{\sqrt{{(r+\mathrm{\δ})}^2+{(r-\mathrm{\δ})}^2-2(r+\mathrm{\δ})(r-\mathrm{\δ})\cos \mathrm{\β}}},$ $r_2^2={((r-\mathrm{\δ})\cos \frac{{\mathrm{\β}}_2}{2}-h)}^2+{((r-\mathrm{\δ})\sin \frac{{\mathrm{\β}}_2}{2}-\frac{2\mathrm{ht}}{\sqrt{1-t^2}})}^2,$ $-t\sin \frac{{\mathrm{\β}}_2}{2}-\sqrt{1-t^2}\cos \frac{{\mathrm{\β}}_2}{2}=\frac{(r-\mathrm{\δ})-(r+\mathrm{\δ})\cos \mathrm{\β}}{\sqrt{{(r+\mathrm{\δ})}^2+{(r-\mathrm{\δ})}^2-2(r+\mathrm{\δ})(r-\mathrm{\δ})\cos \mathrm{\β}}},$ U _jand ω _kjth and a kth element of two endless one-dimensional vector U and Ω respectively, U=[P _{4 × 1}p _{4 × 1}...] _{∞ × 1}, Ω=[Q _{4 × 1}q _{4 × 1}...] _{∞ × 1}, 4 × 1 vectorial P and Q are respectively P=[r+ δ r+ δ r-δ r-δ], Q=[β β ₁β β ₂];

2.6) $V_0^y={\left[\begin{array}{ccc}0& 0& r-{\mathrm{\δ}}_2\end{array}\right]}^T,$ $V_j^y=u_j\left[\begin{array}{c}0\\ \sin \left({\mathrm{\Σ}}_{k=1}^j{\mathrm{\ω}}_k\right)\\ \cos \left({\mathrm{\Σ}}_{k=1}^j{\mathrm{\ω}}_k\right)\end{array}\right],$ J=1 ..., n+1, wherein: n is less than or equal to 5N+1, N be more than or equal to 3 natural number, β+2 β ₁+ 2 β ₂=β _set, and meet β ₁> β ₂; $r_2^2={(r-\mathrm{\δ})}^2+h^2(1+{\left(\frac{(r+\mathrm{\δ})\sin {\mathrm{\β}}_1}{(r+\mathrm{\δ})\cos {\mathrm{\β}}_1-(r-\mathrm{\δ})}\right)}^2),$ $r_1^2={\left(\left(r+\mathrm{\δ}\right)\cos \frac{\mathrm{\β}}{2}+h\right)}^2+((r+\mathrm{\δ})\sin \frac{\mathrm{\β}}{2}-$ ${\frac{2\mathrm{hs}}{\sqrt{1-s^2}})}^2,$ $s\sin ({\mathrm{\β}}_1+\frac{\mathrm{\β}}{2})-\sqrt{1-s^2}\cos ({\mathrm{\β}}_1+\frac{\mathrm{\β}}{2})=\frac{(r-\mathrm{\δ})-(r+\mathrm{\δ})\cos {\mathrm{\β}}_1}{\sqrt{{(r+\mathrm{\δ})}^2+{(r-\mathrm{\δ})}^2-2(r+\mathrm{\δ})(r-\mathrm{\δ})\cos {\mathrm{\β}}_1}},$ $\frac{(r-\mathrm{\δ})-(r-{\mathrm{\δ}}_2)\cos {\mathrm{\β}}_2}{\sqrt{{(r-\mathrm{\δ})}^2+{(r-{\mathrm{\δ}}_2)}^2-2(r-\mathrm{\δ})(r-{\mathrm{\δ}}_2)\cos {\mathrm{\β}}_2}}=\frac{(r-\mathrm{\δ})-(r+\mathrm{\δ})\cos {\mathrm{\β}}_1}{\sqrt{{(r-\mathrm{\δ})}^2+{(r+\mathrm{\δ})}^2-2(r-\mathrm{\δ})(r+\mathrm{\δ})\cos {\mathrm{\β}}_1}};$ U _jand ω _kjth and a kth element of two endless one-dimensional vector U and Ω respectively, U=[P _{5 × 1}p _{5 × 1}...] _{∞ × 1}; Ω=[Q _{5 × 1}q _{5 × 1}...] _{∞ × 1}, 5 × 1 vectorial P and Q are respectively P=[r-δ r+ δ r+ δ r-δ r-δ ₂], Q=[β ₂β ₁β β ₁β ₂];

2.7) $V_0^y={\left[\begin{array}{ccc}0& 0& r+{\mathrm{\δ}}_1\end{array}\right]}^T,$ $V_j^y=u_j\left[\begin{array}{c}0\\ \sin \left({\mathrm{\Σ}}_{k=1}^j{\mathrm{\ω}}_k\right)\\ \cos \left({\mathrm{\Σ}}_{k=1}^j{\mathrm{\ω}}_k\right)\end{array}\right],$ J=1 ..., n+1, wherein: n is less than or equal to 5N+1, N be more than or equal to 3 natural number, β+2 β ₁+ 2 β ₂=β _setand meet β ₁> β ₂; $r_1^2={(r+\mathrm{\δ})}^2+h^2(1+{\left(\frac{(r-\mathrm{\δ})\sin {\mathrm{\β}}_1}{(r+\mathrm{\δ})-(r-\mathrm{\δ})\cos {\mathrm{\β}}_1}\right)}^2),$ $r_2^2={((r-\mathrm{\δ})\cos \frac{\mathrm{\β}}{2}-h)}^2+((r-\mathrm{\δ})\sin \frac{\mathrm{\β}}{1}-$ ${\frac{2\mathrm{hs}}{\sqrt{1-s^2}})}^2,$ $-s\sin \frac{\mathrm{\β}}{2}-\sqrt{1-s^2}\cos \frac{\mathrm{\β}}{2}=\frac{(r-\mathrm{\δ})-(r+\mathrm{\δ})\cos {\mathrm{\β}}_1}{\sqrt{{(r+\mathrm{\δ})}^2+{(r-\mathrm{\δ})}^2-2(r+\mathrm{\δ})(r-\mathrm{\δ})\cos {\mathrm{\β}}_1}},$ $\frac{(r+\mathrm{\δ})-(r+{\mathrm{\δ}}_1)\cos {\mathrm{\β}}_2}{\sqrt{{(r+\mathrm{\δ})}^2+{(r+{\mathrm{\δ}}_1)}^2-2(r+\mathrm{\δ})(r+{\mathrm{\δ}}_1)\cos {\mathrm{\β}}_2}}=\frac{(r+\mathrm{\δ})-(r-\mathrm{\δ})\cos {\mathrm{\β}}_1}{\sqrt{{(r+\mathrm{\δ})}^2+{(r-\mathrm{\δ})}^2-2(r+\mathrm{\δ})(r-\mathrm{\δ})\cos {\mathrm{\β}}_1}};$ U _jand ω _kjth and a kth element of two endless one-dimensional vector U and Ω respectively, U=[P _{5 × 1}p _{5 × 1}...] _{∞ × 1}; Ω=[Q _{5 × 1}q _{5 × 1}...] _{∞ × 1}, 5 × 1 vectorial P and Q are respectively P=[r+ δ r-δ r-δ r+ δ r+ δ ₁]; Q=[β ₂β ₁β β ₁β ₂];

2.8) $V_0^y={\left[\begin{array}{ccc}0& 0& r-\mathrm{\δ}\end{array}\right]}^T,$ $V_j^y=u_j\left[\begin{array}{c}0\\ \sin \left({\mathrm{\Σ}}_{k=1}^j{\mathrm{\ω}}_k\right)\\ \cos \left({\mathrm{\Σ}}_{k=1}^j{\mathrm{\ω}}_k\right)\end{array}\right],$ J=1 ..., n+1, wherein: n is less than or equal to 6N+1, N be more than or equal to 3 natural number, 2 β+2 β ₁+ 2 β ₂=β _set; $r_2^2={(r-\mathrm{\δ})}^2+h^2(1+$ ${\left(\frac{r\sin {\mathrm{\β}}_2}{r\cos {\mathrm{\β}}_2-(r-\mathrm{\δ})}\right)}^2),$ $r_1^2={(r+\mathrm{\δ})}^2+h^2(1+{\left(\frac{r\sin {\mathrm{\β}}_1}{(r+\mathrm{\δ})-r\cos {\mathrm{\β}}_1}\right)}^2);$ U _jand ω _kjth and a kth element of two endless one-dimensional vector U and Ω respectively, U=[P _{6 × 1}p _{6 × 1}...] _{∞ × 1}; Ω=[Q _{6 × 1}q _{6 × 1}...] _{∞ × 1}, 6 × 1 vectorial P and Q are respectively P=[r r r+ δ r r r-δ]; Q=[β ₂β β ₁β ₁β β ₂];

2.9) $V_0^y={\left[\begin{array}{ccc}0& 0& r-{\mathrm{\δ}}_2\end{array}\right]}^T,$ $V_j^y=u_j\left[\begin{array}{c}0\\ \sin \left({\mathrm{\Σ}}_{k=1}^j{\mathrm{\ω}}_k\right)\\ \cos \left({\mathrm{\Σ}}_{k=1}^j{\mathrm{\ω}}_k\right)\end{array}\right],$ J=1 ..., n+1, wherein: n is less than or equal to 10N+1, N be more than or equal to 3 natural number, 2 β+2 β ₁+ 2 β ₂+ 2 β ₃+ 2 β ₄=β _set, and meet β ₁> β ₃and β ₂> β ₄; $r_2^2={(r-\mathrm{\δ})}^2+h^2(1+{\left(\frac{r\sin {\mathrm{\β}}_2}{r\cos {\mathrm{\β}}_2-(r-\mathrm{\δ})}\right)}^2),$ $r_1^2={(r+\mathrm{\δ})}^2+h^2(1+$ ${\left(\frac{r\sin {\mathrm{\β}}_1}{(r+\mathrm{\δ})-r\cos {\mathrm{\β}}_1}\right)}^2),$ $\frac{(r+\mathrm{\δ})-(r+{\mathrm{\δ}}_1)\cos {\mathrm{\β}}_3}{\sqrt{{(r+\mathrm{\δ})}^2+{(r+{\mathrm{\δ}}_1)}^2-2(r+\mathrm{\δ})(r+{\mathrm{\δ}}_1)\cos {\mathrm{\β}}_3}}=\frac{(r+\mathrm{\δ})-r\cos {\mathrm{\β}}_1}{\sqrt{{(r+\mathrm{\δ})}^2+r^2-2(r+\mathrm{\δ})r\cos {\mathrm{\β}}_1}},$ $\frac{(r-\mathrm{\δ})-(r-{\mathrm{\δ}}_2)\cos {\mathrm{\β}}_4}{\sqrt{{(r-\mathrm{\δ})}^2+{(r-{\mathrm{\δ}}_2)}^2-2(r-\mathrm{\δ})(r-{\mathrm{\δ}}_2)\cos {\mathrm{\β}}_4}}=\frac{(r-\mathrm{\δ})-r\cos {\mathrm{\β}}_2}{\sqrt{{(r-\mathrm{\δ})}^2+r^2-2(r-\mathrm{\δ})r\cos {\mathrm{\β}}_2}};$ U _jand ω _kjth and a kth element of two endless one-dimensional vector U and Ω respectively, U=[P _{10 × 1}p _{10 × 1}...] _{∞ × 1}; Ω=[Q _{10 × 1}q _{10 × 1}...] _{∞ × 1}, 10 × 1 vectorial P and Q are respectively P=[r-δ r r r+ δ r+ δ ₁r+ δ r r r-δ r-δ ₂], Q=[β ₄β ₂β β ₁β ₃β ₃β ₁β β ₂β ₄];

Step 3: m × n the summit V obtaining paper folding structure according to input point _{i, j}coordinate, $V_{i,j}=\left[\begin{array}{c}{x}_{i,j}\\ y_{i,j}\\ z_{i,j}\end{array}\right]=V_j^y+$ ${[A}_j]V_i^x,$ I=1,2 ..., m; J=1,2 ..., n, wherein: $\left[A_j\right]=\left[\begin{array}{ccc}1& 0& 0\\ 0& 0& {(-1)}^j\frac{{\cos \mathrm{\θ}}_{j-1}+{\cos \mathrm{\θ}}_j}{\sin ({\mathrm{\θ}}_{j-1}-{\mathrm{\θ}}_j)}\\ 0& 0& {(-1)}^j\frac{{\sin \mathrm{\θ}}_{j-1}+{\sin \mathrm{\θ}}_j}{\sin ({\mathrm{\θ}}_{j-1}-{\mathrm{\θ}}_j)}\end{array}\right],$ i _y=[0 1 0] ^tfor the vector of unit length of y coordinate axis, i _z=[0 0 1] ^tfor the vector of unit length of z coordinate axle, ‖ ■ ‖ represents that subtend measures mould;

Step 4: definition { V _{i, j}v _{i+1, j}or { V _{i, j}v _{i, j+1}be a pair adjacent vertex; Coupled together by all adjacent vertex straight lines, namely the connecting line segment between these adjacent vertexs constitutes the scrimp of paper folding structure, and adopts area of computer aided to realize interlayer manufacture further.

2. method according to claim 1, is characterized in that, described adjacent vertex refers to: with V _{i, j}for summit, then its consecutive point are { V _{i, j}v _{i+1, j}or { V _{i, j}v _{i, j+1}.

3. method according to claim 1, is characterized in that, described housing is made up of four parts, is respectively: top is with O ₁for the arc section in the center of circle, outer, the internal diameter of its interlayer are respectively r ' ₁with r ' ₂; Left side is with O ₂for the arc section in the center of circle, outer, the internal diameter of its interlayer are respectively r " ₁with r " ₂; Bottom is with O ₃for the arc section in the center of circle, outer, the internal diameter of its interlayer are respectively r " ' ₁with r " ' ₂; Right side is with O ₄for the arc section in the center of circle, outer, the internal diameter of its interlayer are respectively r " " ₁with r " " ₂;

Make r ' ₁=21, r ' ₂=19, r " ₁=51, r " ₂=49, r " ' ₁=29.786, r " ' ₂=27.786, r " " ₁=51, r " " ₂=49; Shell length l=11, adopts step 2.1) mode design the interlayer in top, right side, bottom, left side respectively; Specific as follows:

(i) top interlayer: get m=12, h=1, a=1, known b=1; Obtain the input point of 12 x-z planes: $V_1^x={\left[\begin{array}{ccc}1& 0& 1\end{array}\right]}^T,$ $V_2^x={\left[\begin{array}{ccc}2& 0& 1\end{array}\right]}^T,$ $V_3^x={\left[\begin{array}{ccc}3& 0& -1\end{array}\right]}^T,$ $V_4^x={\left[\begin{array}{ccc}4& 0& -1\end{array}\right]}^T,$ $V_5^x={\left[\begin{array}{ccc}5& 0& 1\end{array}\right]}^T,$ $V_6^x={\left[\begin{array}{ccc}6& 0& 1\end{array}\right]}^T,$ $V_7^x={\left[\begin{array}{ccc}7& 0& -1\end{array}\right]}^T,$ $V_8^x={\left[\begin{array}{ccc}8& 0& -1\end{array}\right]}^T,$ $V_9^x={\left[\begin{array}{ccc}9& 0& 1\end{array}\right]}^T,$ $V_{10}^x=$ ${\left[\begin{array}{ccc}10& 0& 1\end{array}\right]}^T,$ $V_{11}^x={\left[\begin{array}{ccc}11& 0& -1\end{array}\right]}^{\text{T}},$ $V_{12}^x={\left[\begin{array}{ccc}12& 0& -1\end{array}\right]}^T;$ ‘

Get N=40, obtain β=π/40; Calculate r=19.9593 and δ=1.0045; Get n=21, obtain the input point of 23 y-z planes: $V_j^y=[19.9593+{(-1)}^{j}1.0045]\left[\begin{array}{c}0\\ \sin \left(\frac{\mathrm{j\π}}{40}\right)\\ \cos \left(\frac{\mathrm{j\π}}{40}\right)\end{array}\right],$ J=0,1 ..., 22, calculate 12 × 21 summit V _{i, j}coordinate;

(ii) left and right side interlayer: because left and right side interlayer has identical physical dimension, therefore only needs the interlayer designing side; Get m=12, h=1, a=1, known b=1, obtain the input point of 12 x-z planes: $V_1^x={\left[\begin{array}{ccc}1& 0& 1\end{array}\right]}^T,$ $V_2^x={\left[\begin{array}{ccc}2& 0& 1\end{array}\right]}^T,$ $V_3^x={\left[\begin{array}{ccc}3& 0& -1\end{array}\right]}^T,$ $V_4^x={\left[\begin{array}{ccc}4& 0& -1\end{array}\right]}^T,$ $V_5^x={\left[\begin{array}{ccc}5& 0& 1\end{array}\right]}^T,$ $V_6^x=$ ${\left[\begin{array}{ccc}6& 0& 1\end{array}\right]}^T,$ $V_7^x={\left[\begin{array}{ccc}7& 0& -1\end{array}\right]}^T,$ $V_8^x={\left[\begin{array}{ccc}8& 0& -1\end{array}\right]}^T,$ $V_9^x={\left[\begin{array}{ccc}9& 0& 1\end{array}\right]}^T,$ $V_{10}^x=$ ${\left[\begin{array}{ccc}10& 0& 1\end{array}\right]}^T,$ $V_{11}^x={\left[\begin{array}{ccc}11& 0& -1\end{array}\right]}^T,$ $V_{12}^x={\left[\begin{array}{ccc}12& 0& -1\end{array}\right]}^T;$ Get N=80, obtain β=π/80; Calculate r=49.9803 and δ=1.0011; Get n=21, obtain the input point of 23 y-z planes: $V_j^y=[49.9803+$ ${(-1)}^{j}1.0011]\left[\begin{array}{c}0\\ \sin \left(\frac{\mathrm{j\π}}{80}\right)\\ \cos \left(\frac{\mathrm{j\π}}{80}\right)\end{array}\right],$ J=0,1 ..., 22, calculate 12 × 21 summit V _{i, j}coordinate;

(iii) bottom interlayer: get m=12, h=1, a=1, known b=1, obtain the input point of 12 x-z planes: $V_1^x={\left[\begin{array}{ccc}1& 0& 1\end{array}\right]}^T,$ $V_2^x={\left[\begin{array}{ccc}2& 0& 1\end{array}\right]}^T,$ $V_3^x={\left[\begin{array}{ccc}3& 0& -1\end{array}\right]}^T,$ $V_4^x={\left[\begin{array}{ccc}4& 0& -1\end{array}\right]}^T,$ $V_5^x={\left[\begin{array}{ccc}5& 0& 1\end{array}\right]}^T,$ $V_6^x={\left[\begin{array}{ccc}6& 0& 1\end{array}\right]}^T,$ $V_7^x={\left[\begin{array}{ccc}7& 0& -1\end{array}\right]}^T,$ $V_8^x={\left[\begin{array}{ccc}8& 0& -1\end{array}\right]}^T,$ $V_9^x={\left[\begin{array}{ccc}9& 0& 1\end{array}\right]}^T,$ $V_{10}^x=$ ${\left[\begin{array}{ccc}10& 0& 1\end{array}\right]}^T,$ $V_{11}^x={\left[\begin{array}{ccc}11& 0& -1\end{array}\right]}^T,$ $V_{12}^x={\left[\begin{array}{ccc}12& 0& -1\end{array}\right]}^T;$ Get N=40, obtain β=π/40; Calculate r=28.7462 and δ=1.0049; Get n=41, obtain the input point of 43 y-z planes: $V_j^y=[28.7462+$ ${(-1)}^{j}1.0049]\left[\begin{array}{c}0\\ \sin \left(\frac{\mathrm{j\π}}{80}\right)\\ \cos \left(\frac{\mathrm{j\π}}{80}\right)\end{array}\right],$ J=0,1 ..., 42, calculate 12 × 41 summit V _{i, j}coordinate; Four part interlayers obtained above are joined end to end in order, namely obtains described housing.

4. method according to claim 1, is characterized in that, described computer-aided manufacturing comprises: manufacture the mould corresponding with paper folding structure, utilize mould to carry out shaping to flat sheet; Three-dimensional printing technology is utilized to print; Utilize and manufacture the similar way of comb core, namely first utilizing the material that can conveniently fold to carry out craft folding, and then immersing in glue and the structure folded is shaped and reinforces.

5. an aircraft casing sandwich construction, is characterized in that, comprises method according to above-mentioned arbitrary claim and prepares.

6. sandwich construction according to claim 5, is characterized in that, is separated between described interlayer by middle case face.

7. sandwich construction according to claim 5, is characterized in that, described interlayer adopts metal, synthetic material, carbon fibre material or paper to make.