EP0362452A2 - Articulated structures - Google Patents

Abstract

The subject of the invention is articulated structures which can be covered with a stretched skin. Their definition is based on the preoccupation of improving their functioning under various stresses. In particular, with regard to their suspension function, they abandon the prismatic shapes which have edges parallel to the force field in favour of pyramids on the peripheral envelope. Likewise, for a radial force field with rotational guidance, the deformation will be axial, this being obtained by localising the elastic elements in the form of a helix. A line of structures is characterised by a ring of polygons having p sides distributed about a central polygon having p-1 sides, particularly being a pentagon. In their deformable version, these structures consisting of modules are used for the production of various rotors, even of variable diameter, and mechanical transmission cables. A line of structures with equal bars is characterised in that it contains a concentric assembly of frames, the height of which doubles from one to the other. A series of plates which are composed of an assembly of tetrahedra and pyramids made isostatic by the elimination of particular bars is characterised by the presence of a chamfer in corners of the serrated perimeter, of a hole and/or of an octahedron carefully placed. A method of using continuous material consists in changing the state of the ambient material by means of a heat pump located in a central region. A specific structure consists of two flanges, each formed from a pentagonal ring, the angular offset of which can be used for many purposes.

Description

The object of the invention is a range of isostatic spatial structures whose applications are essentially suspension systems and deformable rotating structures.

Two elements are considered rigid in this structure, the cockpit or the hanging machine, the base or the block - possibly the wheels - resting on the ground.

These two elements are connected by bars articulated in all directions by means of ball joints. Until now, the suspension was made by the direct interposition between the heavy element to be suspended and the element in contact with the ground of an elastic element working in compression, helical spring, or in bending, blades. If the interposition was not direct, it was done by means of lever, however, the elastic element was opposed most of the time to a vertical deformation.

There are exceptions: the CITROEN 2CV. This exception also relates to the non-independence of the wheels since there is a suspension pot for the wheels located on the same side.

However, the general rule in automotive matters as elsewhere is that each wheel, each support has an elastic element, spring or silent-block or specific support piece which is requested during the local expession of a movement of the suspended element. in relation to its support.

The proposed suspension allows a distribution of the movements of a support on the other supports. There is only one elastic element which is stressed whatever the support concerned by a displacement. Its deformation causes the displacement of the center of mass of the structure. We must be able to count on the same distribution with regard to efforts and therefore have equal reactions under each support. This equality is verified when there is a symmetry of revolution. We will propose a solution to achieve equality under each support when there is no symmetry of revolution

The obvious method for calculating this isostatic suspension is the reciprocity theorem.

The suspension is therefore a set of isostatic structures built around an elastic central element with adequate damping articulated on the suspended assembly and on its support so that, essentially, the joints are located in two parallel planes perpendicular to the field. of preponderant force.

These structures are from a geometric point of view three in number depending on whether the joints are arranged in each plane in a triangle, quadrilateral, or pentagon.

It should be noted that for the structure to be elastic and therefore to act as a suspension, it suffices that a single bar is elastic, this can a priori be located anywhere. This is not the choice we have made since we have located in the center the elastic bars which form a closed contour and therefore can be replaced by a solid of continuous material. However on the axis of the force field one can, in a variant have, in order, the suspended element, its support, the suspension, or even the reverse order.

Furthermore, it can be noted that to control the deformation it is necessary not only to locate the elements having an elastic function but also to ensure the rigidity of the others.

Before tackling the mechanical questions and then the choice of materials and the possible manufacturing processes, it is first necessary to define the structures geometrically. Structures whose joints located in parallel planes perpendicular to the force field are arranged in a triangle. As can be seen in FIG. 1a, the vertices of the two triangles are connected both by bars perpendicular to their plane thus forming a prism and by bars forming a pyramid on each face. The vertices of the pyramids are connected by three bars which compete at the center of gravity of the prism. Useless. (fig. 1/38 sheet). In this figure as in Figure 1b, it is an equilateral triangle, the center of gravity of the prism is also the center of gravity of the assembly. It should be said more precisely, the center of mass.

In Figure 1b, we see that the pyramids are possibly very flat. The elasticity of the structure can reside only in the drawbars or compressible springs GI, GJ, GK. There may indeed be either a force field, gravity tending to bring A B C closer to its support A'B'C ', or efforts tending to separate the two triangles. (Figure 1 b, sheet 1/38).

In Figure 2a, there is no longer a bar directly connecting the two triangles. The vertices of the pyramids are connected by bars forming a triangle. (fig. 2a, Fe. 2/38). As for the previous structure, we can check the isostaticity by calculating 3 number of joints - N number of bars.

Figure: 3x10-24 = 6 figure 2: 3x9-21 = 6

In Figure 2, the triangle 1 J K has become a circle. For a deformation tending to crush the structure, there is bending of the arcs.

In a variant, it can become a disc or better, an ellipsoid of revolution admitting the perpendicular cular to the plane of the figure as an axis, that is to say a form of onion. Figure 2b. from sheet 2/38.

We can already leave the geometric domain to say that this elastic core can be made of synthetic material or rubber. It can be made up of several skins.

For the damping of the structure it is necessary that the vibrations of the heart dampen. For this, use is made in a variant of the flow of a fluid, in particular through the network of channels internal to the material of the elastic core which has a sponge structure.

In Figure 3a, the vertices of the pyramids are outside the prism. Figure 3b shows that the bars can be spindles. Here the spindles can possibly be limited to two arcs whose section is vertical and evolves according to a Gauss curve in this direction when one traverses the arc. As the height of the section decreases from the middle of the arc towards the ends, its width increases in the same mode. Figures 3a and b of sheet 3/38.

Figures 4 differ from Figures 2 in that the vertices of the pyramids are outside.

de tour au niveau de la répartition des barres, sans tenir compte de leurs orientations pour les moitiés de côté du triangle équilatéral central, lesquelles n'ont aucune incidence sur la répartition des masses.In Figures 3 and 4, the IJK pyramid vertices are located on the circumscribed cylinder or prism. We can notice that in figure 4 there is a periodicity of

of rotation at the level of the distribution of the bars, without taking account of their orientations for the side halves of the central equilateral triangle, which have no effect on the distribution of the masses.

The arcs IJ, JK, KI do not pass through the vertices A, B, C they are located in a different plane, in principle parallel to A B C. (Figure 4 of sheet 4/38).

The more the periodicity in the mass distribution is a small fraction of a turn, the more the structure will be adapted thanks to its roundness to applications in which it is in rotation These applications will be developed later

Regarding the suspension, we can notice that it is all the more flexible as the pyramids are high.

However for the variants with bars perpendicular to the planes of the triangles therefore parallel to the main force field, these bars risk instantly opposing any movement without bringing into play the other bars in particular those having to have an elastic function which are not necessarily the same. These bars should therefore be inclined preferably by 45 relative to the vertical vertical plane passing through the middle of each of them. The pyramids are then a non-planar deformed base. In order to avoid the rotation of a triangle in relation to the other, in practice, the element suspended in relation to its support, between the fact that either one or the other can be in a variant mentioned integral with the tops of the pyramids, it is possible to double the suspension as shown in Figure 5 which shows chevrons: A A'A ", B B'B", C C'C ". (Figure 5 of sheet 5 / 38).

The second type of structure consists of a square on the sides of which articulate the sides of four adjacent pentagons so that only one bar leaves from each vertex of the square. The structure thus described, Figure 6, (sheet 6/38) has 16 bars and 12 joints so is not isostatic: 12x3-16 = 20 instead of 6.

There are lots of ways to make this structure isostatic depending on the desired applications.

In the case of a building frame, one does not want an elastic suspension but the possibility of a non-destructive deformation of this frame for the case where there is deformation of the ground. For earthquakes, the speed of the deformations will generate large inertia forces in the displacement of the bars. The structure must have elastic zones with adequate damping.

In this framework application, the ground can be considered as a set of bars; since there are four articulation supports the ground is equivalent to 6 bars number of sides of a tetrahedron. It will be preferable that these supports are not in the same plane.

Isostaticity 12x3-16-6 = 14

You have to find to place N'barres and n'articulations such that 3n'-N '= -8 to find 3n-N = 6

The four pentagons have their five vertices, each meeting in four pyramids. These can be inbound or outbound. Figure 7. (sheet 7/38).

Only three pentagons have a pyramid but also the square, which has a diagonal dividing the pyramid into two tetrahedra. Figure 8. (sheet 7/38). n = 2 N = 14 3x2-14 = -8

Two opposite pentagons have their five vertices united in two pyramids having their common vertex. A pyramid is built on the square. Figure 9. (sheet 8/38).

The pentagonal pyramids are connected by a bar, so their vertices are each connected by a bar at the top of the square pyramid. Hence: n '= 3 N' = 17 3x3-17 = -8 Figure 10. (sheet 8/38).

A double pyramid is built on the square. The top of one of them is connected to the four supports or to the four free pentagon vertices. Two bars join two vertices, free or intermediate, of pentagons between them, either in the plane of these or diagonally. Figure 11. The four edges starting from the top of one of the pyramids can be replaced by four bars, one in the plane of each pentagon. (Figure 11 of sheet 9/38). n '= 3 N' = 17

Three pentagonal pyramids plus two bars joining one of the vertices to the other two. Four pyramids with four bars, three adjacent, eight bars, one independent, four bars.

Two bars from two vertices of the square and meeting at a joint from which three bars leave, joining each of the vertices of the pyramids and the new joint.

Three pyramids adjacent to four bars, eight bars, the fourth to five bars its top is connected to that of the adjacent pyramids pae a bar.

This bar would be the bar 2 'in Figure 12 where 1 3' would be deleted, NB, ND. (Figure 12 of sheet 9/38).

Four pyramids, two to two adjacent each having four bars. 12 bars. Two bars end the tetrahedron formed by the vertices of the two sets of pyramids with two points common to these two unrelated sets. Two of these points can be diagonally opposite vertices of the square. The bars are then SS 'and AC. Figure 13. (from sheet 10/38). n '= 1

A central articulation from which a bar leaves to each each articulation minus one equals 11 bars.

A pyramid with a square base, four bars encircling the pentagons.

The relationships between the lengths of bars can a priori be arbitrary. So we can have a very large square and very flat irregular pentagons. The pyramids themselves can be very crushed. Structure built around a square, where a quadrilateral symmetrical about the plane passing through the top of the pentagons formerly the supports.

Let us suppose the structure having two opposite pyramids having a common vertex surmounted by the pyramid with square base already mentioned 3x2-14 = -8. Pyramids not shown.

It does not seem necessary to describe these isostatic structures without the contribution of the soil, a category consisting of already defined structures in which the soil between the four supports is replaced by the six bars of a tetrahedron.

In an important variant, four bars will surround the four supports, the floor acting as the two missing edges. Useless figures.

In another variant, the floor will act as four bars and only two bars will be installed. It seems that these intermediate solutions are quite judicious.

To make the structures constructed with a square isostatic without the ground, we can place six bars in each of the variants presented in different places, taking care not to have an internal hyperstaticity for a part of the structure whereas overall the structure would not be stable.

For example in the structure considered to define the symmetrical structure, it would suffice to triangulate, by putting two bars in each, three pentagons.

A structure of this type, that is to say built with a square and four pentagons is to be distinguished from the others. This is the structure in which the sides of the pentagon opposite the side common with the square are in line with one another. The structure then has the form of a frame, FIG. 16a, and can thus be better suited to implementation when it is a building or civil engineering works since 12 bars out of 16 are horizontal. (Figure 16a of sheet 11/38).

When it is a question before deformation of regular pentagons the structure is presented in the form of two squares of sides a and 2a double of the preceding one, having a joint in the middle, connected by four bars edges of a trunk of pyramid. The slope of these edges in the diagonal vertical plane is 1. They are therefore inclined at 45 in this plane. In the median plane the slope is the angle of 35 17.

We can have in a variant rectangles instead of a square; it is enough to make sure that the sides of the pentagons opposite to the side of the rectangle are in a constant ratio for all the pentagons.

Figures 16b and 16a show an interesting frame arrangement. The figure can be doubled by a symmetry with respect to the horizontal plane. (Figure 16b of sheet 12/38). In these figures are not shown the bars which allow the structure to be isostatic.

The bars connecting the four joints of a circle to four of the eight joints of the other circle are in a first variant unchanged.

To make the structure isostatic, two bars must be added in each quarter of the cone which must follow the conical surface.

Another solution is to place a joint in the center of the large circle and connect all but one.

• La longueur de ces barres résultant du même calcul est égale à 1,1107a. Cependant, en ce qui concerne les barres de stabilisation à placer,selon le mode décrit plus haut, seulement quatre barres auront cette longueur, les autres ayant leur longueur multipliée par √2. Il n'est pas possible de respecter notre préoccupation de longueurs égales. Les barres reliant les carrés devenus des cercles, doivent devenir des arcs de cercle de longueur 1, 107£. Les articulations appartenant aux deux cercles et situées dans un même plan vertical appartiennent à un cercle de rayon égal au rayon du plus grand cercle.
• La figure 16e présente une structure construite à partir de deux rectangles de largeur a et de longueur 2a situés dans deux plans parallèles et formant une croix en vue de dessus. Ces deux rectangles font apparaître entre eux les mêmes pentagones applatis que les carrés dans la figure 16a. (figure 16e de la feuille 13/38).
• La figure 16f met en évidence que l'enveloppe de la structure présentée ci-dessus s'obtient en prenant une tranche de tétraèdre régulier parallèlement à une arête laquelle correspond à la tranche centrale d'épaisseur un tiers de l'apothème. (figure 16f de la feuille 15/38).

For reasons of symmetry, two bars are removed in the extension of one another and parallel to a diameter of the small circle passing through two articulations which is then immediately materialized by a bar. These provisions are valid for the second variant where the bars are replaced by arcs of circles, quarter circles whose centers are located vertical to the joints of the small circle and whose radius is equal to a√2:

• The length of these bars resulting from the same calculation is equal to 1.1107a. However, as regards the stabilization bars to be placed, according to the mode described above, only four bars will have this length, the others having their length multiplied by √2. It is not possible to respect our concern of equal lengths. The bars connecting the squares which have become circles must become arcs of length 1, 107 £. The joints belonging to the two circles and located in the same vertical plane belong to a circle of radius equal to the radius of the largest circle.
• Figure 16e shows a structure constructed from two rectangles of width a and length 2a located in two parallel planes and forming a cross in top view. These two rectangles show between them the same flattened pentagons as the squares in figure 16a. (Figure 16e of sheet 13/38).
• FIG. 16f shows that the envelope of the structure presented above is obtained by taking a slice of regular tetrahedron parallel to an edge which corresponds to the central slice of thickness one third of the apothem. (Figure 16f of sheet 15/38).

Figures 16h and 16i show the same structure, which can have the thickness of a floor or that of a floor, made isostatic by stabilization bars. In Figure 16i, all the bars have the same length. In FIG. 16i, we sought before to have planes perpendicular to those of the two rectangles: AIB, BJC, C'D'B, DLA. (Figures 16h and i of sheet 14/38)

Figures 16j and 16k show unstable structures with only one bar missing, which will take its place when the module structures are assembled like ours in Figure 17, in order to make a plate that can find its place in any construction . It should be noted that the elementary structure can in itself find an application: it is the construction of a highway bridge. (Figures 16j and k sheet 16/38). (Figure 17 of sheet 17/38).

Before presenting this articulated plate, in order to encompass as many stabilization solutions of the elementary articulated structure, it should be said that one can adapt those presented in Figures 9 to 13, the square base pyramid being able to be built by joining, at the same point which will be the top of the pseudo-pyramid, the ends of two of the following segments: AB, CD, A'D ', B'C'.

Figure 17 therefore shows the assembly of four elementary structures or modules. All the designated points represent articulations of which are attached all the bars leading to it.

We note that there exists between these four modules a double pyramid with vertex S: S A "B'C D" above, S E "'F" G'H below. For this good view of the figure, you should know that the rectangles ABCD, A'B'C'D 'are above and that the rectangles EFGH are below.

To stabilize the structures 16j and 16k, it suffices to connect the vertex S to four homologous vertices which will make one more bar per module and restore the equality 3n-N = 6.

We will therefore have a square mesh. To be able to prestress the structure, it suffices to place the bars A "H, B'E ', C F, D" G' and to tension them. If these bars are shortened, then the thickness of the plate will thereby increase its surface.

It is now necessary to define the edges. The modules being arranged such that their edges are at 45 relative to those of the surface to be covered, the corner modules are such as that shown in FIG. 16g. The plate perpendicular to that to which the modules 16j belongs is sawtooth, its section is A 'D' DABB '. The left part of AD is outside the plate parallel to the plane of the sheet. Two elements identical to the 16g module which can have a greater length according to AD are assembled by their articulations a priori perpendicular to the plane of the sheet. These elements can be located both below and above this plane. We can notice that in B ^I , the wall will have an almost zero thickness but will be articulated. Error: remark canceled, there is continuity with the adjacent module. In a variant, there is continuity of the plate beyond the perpendicular plate. This is so as not to cause bending in this plate which is assumed to be vertical. There will therefore be balancing on the support and equal spans on each side. This implies having, beyond the support, for reasons of compatibility of the modules built on rectangles a 1.5a 1.5a 2a. This does not seem the best solution and it is preferable that the continuous plate is associated with posts ending in a pyramid such as SE "'F"B'H. These posts can be composed of pyramidal elements and paving stones such as A "B'C D" G 'H to avoid their buckling.

In these conditions, so that the elements at the edge of the plate do not work differently from the others, it is advisable to separate the squares of which a post is the center. For the edge elements, there is instead of connecting the square mesh lattice to the top of the flat pyramid whose base is the trapezoid. An isostatic solution does not require the trellis but only the vertical bars used for prestressing when it is present. In this case, on the edges we will only place the bars A "G and B'F".

In the design based on squares supported in the center on posts, they are consoles so we can be led to vary the thickness of the post plate going outwards, however, this will change the ratio between the bars until then equal. In fact, this choice of equal bars supposes equal elongations for equal normal forces and identical elastic moduli therefore a constant geometry if there are the same forces at the joints.

However, these real or prestressed forces are exerted perpendicular to the plate and not tangentially - it will be necessary to be wary of these forces - therefore, the ratio between the sides of the rectangle is not decisive. The height of the module will then be such that the element is a trunk of the tetrahedron. The angles will also result. If you want the plate to thin, it is necessary that the planes cutting the trunk of tetrahedron are not parallel and that in two directions. Anyway, there will be a problem with the diagonals. Two solutions: dissociate, make a joint or cut the module in the tetrahedron using a plane and a very flat cone having its point on the other side than the plane relative to its base. The edges of the plate will however remain straight and the latter square although one can envisage a variant leading to a round plate from crown elements as a module assembling in the same way as previously supported by a central cone. We can consider with more studies designing hulls.

The implementation of the modules must respect their isostatic design even if the joints take various forms and in particular for so-called fixed works, they are only areas of material foreshadowing with regard to the others of the joints.

For mechanical works, it is advisable to make the best articulations, that is to say conventional ball joints, however the number of bars coming from them poses a problem. For this, it is planned to split the male sphere of the ball to make it a yoke and thus be able to accommodate another ball. The problem is then twofold: that of allowing sufficient amplitudes for the desired operation, that of having ball joints of acceptable size with respect to their size or their fragility. It is conceivable that these joints are quite large relative to the body of the bar. which will decrease towards the middle to resemble a form of bone.

For the articulations of modules intended for the realization of plates, one can conceive a module whose two rectangles are solid plates which must work like the eight diagonal and median bars of figure 16j, therefore in which of the plates will have to have a central articulation . To do this, choose a rectangle with a ratio of 1 L = 1.618.

The unstable figure 16k becomes hyperstatic when we join the points M and M therefore, it is this prestressing solution which seems preferable to us. It will necessarily require a surplus of material thickness in M and M '.

In general, the design of the module is focused on joints with a lot of material and a canvas stretched between them whose thickness decreases with distance. In the trapezoids, there will in principle be two holes or a light in the shape of a bean seed leaving material in its hollow for the joint.

We must correct an error in our first way of applying the prestressing: the points A "and C are not joints of the same E and G 'at least for the bar on which they are located.

If we consider them as such, it is necessary to add bars, three per articulation which amounts, as there is one bar per module, 16k for example, to triangulate the prism with square base A "B'C D" G'H EF "therefore to make a solid block crossed by four bars passing through its center of mass and not preventing its corner joints from functioning, that is to say having in fact a soft heart. Its production may require heat treatment. Its implementation including this possible treatment can be done after assembling the modules.In another variant, four quarter blocks are manufactured at the same time as the module

From this technological point of view, these are tests which will make it possible to find the compromise, even the ideal, between the concept of mechanical equilibrium as we have defined it, and the manufacturing possibilities that may require new processes.

As it stands, it appears that, for a stretched canvas design, two processes seem to be essential: stamping and blowing. For this reason, the material to be used will preferably be plastic, possibly hot, and, in addition to thermosetting or thermoplastic plastics, we can try the glasses. In the perspective of the deformation of the surface of a rectangle following the application of a normal central effect, it is also necessary to consider the rectangles of ratio L 1 = 1.817 whose ellipse of inertia is inscribed in the rectangle itself as well as the ratio L 1 = 2.29 whose ellipse of inertia is inscribed to the rectangle. One can adapt one of the structures described in 16 and arranged in 17 for the fixing of facade elements.

The articulated structure consisting of a pentagon and six adjacent hexagons is shown in Figures 18a and 18b. In the figure the structure is flattened, the hexagons are no longer regular hexagons.

Isostaticity: additional bars will be necessary

barres

n = 20 joints N = 25 bars 10 joints are on the ground. This is equivalent to 6 = 3x10-Ns Ns = 24 Five bars exist at ground level therefore:

bars

It takes n and N such that 3n'-N '= -10

One solution is to put five bars to build a pyramid on the pentagon, five bars forming a belt in the support plane and three bars to trianaulate a hexaaone.

Another solution consists of a pentagonal pyramid, preferably re-entrant, from the top of which 8 bars come from joining the vertices of hexagons which do not belong to the pentagon and which can form, for example, two hexagonal pyramids distinct from each other.

n '= N' = 13 Figure 18c. (from sheet 20/38).

In a variant, the base of the pentagonal pyramid is triangulated, two bars, five bars joining its vertex to five intermediate vertices of pentagons, one of them has a diagonal.

Yet another solution consists of two separate hexagonal pyramids and the central pentagonal pyramid, the base of which is trian ^q ulate.

We can notice that taking into account the inclination of the planes of hexagons, the reactions to the joints will be great also one cannot perhaps count the ground for 19 bars. To make less use of it, it is advisable to put the radial bars articulated in the center, ten bars. In which case the ground would only intervene for: 19-10 + 3x1 = 12 bars. If the role of the soil is not taken into account at all, N 'bars and n joints must be placed such that:

For this, it is necessary to connect a central articulation to the 20 vertices of the figure, place ten bars perpendicular to the sides of the pentagon and two bars to triangulate it Figure 19 (sheet 20/38).

In this variant, the ten bars perpendicular to the sides of the pentagon can be replaced by two crowns of five bars connecting, one, the intermediate vertices, the other, those located at the supports.

The articulated structures pose the problem of making all-round joints. These will be studied later.

At least for building construction, it is necessary to plug the polygons formed by bars.

A variant envisages panels inside which the material would be easily deformable, the edges, themselves, having undergone a treatment, thermal, for example, having the resistant function and being subjected, since articulated, only to normal forces . The angles can also be subject to a special treatment, plasticization of the material, to assemble so that they can orient themselves without resistance other than limited and constant, so as not to compromise the isostaticity.

s'il est isocèle en A avec AB-AC=1.The presence of mass inside the polygons will change the distribution of masses and the balance of the structure. Indeed, unless there is a center of symmetry, the center of mass of an empty polygon formed only of an articulated contour, is not the same as that of a solid polygon. For a triangle for example, G is at hh from the vertex for an empty triangle only if it is equilateral if not

if it is isosceles in A with AB-AC = 1.

Although we have presented a process allowing the structure to be adjusted so as to have equal actions on each support, it cannot correct everything and, in this case, it cannot bring the center of mass of the structure to the vertical from the center of mass of the polygon formed by the supports if this structure is not already relatively balanced.

It has been shown from FIG. 5 that the center of mass of the triangle AIB J C K is the center of mass of the triangle A B C, the triangles A I B, B J C, C K A being equilateral triangles which have the center of mass G1, G2, G3. In addition, the relations between the sides and the angles of a triangle make it possible to establish G1 G2 = G1 G3 = G2 G3 therefore that G1 G2 G3 is equilateral.

For the structure, these are pyramids and not triangles but these must have their center of mass in G1 G2 G3.

After calculations, we find that if these pyramids are square-based, they must have for structure 3, with any central triangle, oblique edge lengths equal to: 1 = ₁ , ₂₅₅ a for structure 4 to: 1 = 1.1a.

It can be noted that an isostatic structure therefore having no elastic energy to release during a possible deformation, having two polygons in parallel planes, sees these remain hemothetic to themselves when it undergoes a perpendicular load at the planes of the polygons and distributed between their joints.

The center of mass of the structure remains on the same straight line with efforts, the center of gravity descends on this straight line corresponding to a decrease in potential energy while the bars forming the polygons elastically deform. This deformation being proportional to their length with the same elongation percent A% for all the bars, we can deduce that if they have the same Young's modulus E, the forces in them are proportional to their lengths. Thus, they are equal in a rectangle a, 2a presenting, as it should be an articulation in the middle, or more simply, in a square an equilateral triangle, nn regular pentagon. It is remarkable to note that in the structure corresponding to the model 16, by blocking the articulations in the middle of the length of a rectangle and by creating two other articulations for example in the width, one multiplies the effort in a bar by two while that we divide it in the other by two.

It should be noted that in the same operation we still had to move the application of the forces from the old to the new applications.

This remark may give rise to applications aimed at releasing tensions in the material by removing a part of it to create a joint. The basic structure built concentrically around an equilateral triangle, a square or a pentagon, in two parallel planes connected by bars forming lateral pyramids, finds applications in its deformable version.

It should be noted right away that the symmetry of revolution is here a requirement, in any case when one wants to move in translation the central polygons perpendicular to their plane, because otherwise there would not be, as we saw more above, that a position of the pyramids for which their center of mass would be arranged in a centered equilateral triangle.

It is easy to see that FIGS. 2a, 16a, 14, 18 and others, may be suitable for producing a hydraulic cylinder because they are deformable by a fluid. The problem is then to find the structure which will have a degree of instability while retaining a center of symmetry. Figure 20 ^q satisfies this exi ence. by doubling it.

Therefore, a degree of instability allowing the deformation perpendicular to its desired plane.

From a technological point of view, an enclosure must be produced which transmits the forces it receives well to the joints.

In a variant, it can be made of ribbed rubber. The valve must be placed in the center. Such a jack can serve as a bridge support piece.

The deformable structures studied have other applications in which the planes of the central polygons do not remain parallel. It suffices, for example in FIG. 1, to delete the bars AA 'BB CC' and to move G to obtain an orientation of one triangle with respect to the other. G can for example make a rotational movement of axis z'Goz perpendicular to the initial plane of the triangles. We can thus obtain a perhaps silent vibrator - we can undoubtedly thus obtain movements in which the mobile points move on a surface with double curvature or not geometrically identified. For example, the shape of a windshield for a wiper, a car fender for a paint sprayer.

It is possible by superimposing the structures of type 1 or 6 or 18, possibly taking symmetrical structures for the two latter so as to maintain a mean section for the arm thus formed, constant, therefore producing articulated arms whose movements would be controlled by the rotation eccentric or cams specific to each elementary structure.

In a variant, the rotation of these cams can be driven by a pneumatic system, a compressed air duct actuating a turbine secured to the cam. This duct is formed of several sections fitted into each other by conical fitting for example: when there is tightness, the compressed air goes to the end of the pipe and turns the second cam, when the fitting is disconnected , it lets pass the air which will activate the turbine located immediately after it. By an axial control comprising a cable with an extendable lead at its end, a decreasing progressiveness in the effort necessary to separate the shanks, one can rotate the elementary articulation structure which one has chosen. It is enough to position the lead in the pipe and to draw it from the sufficient distance. Figure 21 of sheet 32/38.

Depending on the starting point of the end of the arm and the desired end point, a computer will schedule the rotations to be made, will order the sequence allowing the least movement. Each elementary structure can give even more possibilities of movement. It is enough to connect the three vertices of the three pyramids for a triangular structure to the three vertices of three pyramids belonging to an identical structure inside the first. One can even imagine, if necessary, a series of cells or structures comprising a large number of elements thus connected.

Without requiring as many elements, it seems that a structure formed of two elements can replace the universal joint.

As far as rotating structures are concerned, it can be said that isostaticity allows natural balancing and prevents the formation of unbalances. Indeed, a rotating part is never perfectly symmetrical with respect to its center of rotation, taking into account the tolerances of execution of the differences in structures existing even to a lesser degree in the material, which lead to different elasticities therefore to deformations.

It is probably for this reason that aircraft propellers do not have two blades because it would seem that the propeller which is a twisted ellipse and does not have a branch like the propellers with four blades for example, can distribute naturally its material on this convex course which is at the same tension in all points.

A helicopter rotor comprising three blades, no doubt for reasons of lift possibly incorporating questions of turbulence caused by the first blade and which should disappear before the passage of the next, most certainly presents problems of balancing.

It can be presented as the structure admitting a central square and four pentagons, which will have significant lengths radially. This leads to sharp blades. If reasons of turbulence are not opposed, the rotor with five hexagonal blades seems to present a more interesting surface. A less superficial study then tests are necessary.

For an isostatic rotor, it will be necessary to be sure of its geometrical position also, is it preferable to design a rotor with variable geometry.

In the same type of application, we can cite the wheel rim, the flywheel, rotors of all kinds and multi-blade propellers.

The deformable structure in rotation has other applications when it does not admit a center of symmetry. Indeed, in Figure 1 rotating along an axis perpendicular to the triangles if one of the vertices of the pyramid is further from the axis of rotation, vertex having a mass, there will be deformation of the structure. If the vertex in question is in the plane parallel to the triangle containing the other vertices but offset angularly with respect to the perpendicular passing through the center of its base and if, in fact, all the vertices have the same angular phase shift relative to this straight line , the phase shift in question will vary when the structure becomes flattened following an increase in the speed of rotation. It is this variation of phase shift which will be used to obtain a variable ignition advance which system seems a priori better than vacuum systems.

Among the other applications of these simple rotating deformable structures, are those which allow an axial displacement following the radial displacement of the vertices of pyramids due to centrifugal force. Here we have the principle of a centrifugal clutch.

Concentric articulated structures

The invention presented below synthesizes, at least a synthesis, necessarily not exhaustive, of concentric structures with equal bars as well as their applications and their production methods.

Since this is a final synthesis, the invention will give an overview of its field of application ranging beyond the applications that it can define precisely, limiting itself for others to locate the object. From a geometric point of view there are two types of structures, those using square-based pyramids, those using pentagonal-based pyramids, those using pentagonal frames.

In addition, for an application which perhaps - this is not yet acquired - may escape this exception, the bars are not strictly equal going outward, at least in a variant, when the cohesion of together. These are tire carcasses and a line of structures among those using pentagons.

From an embodiment point of view, the bars are therefore articulated and equal; they may therefore not exist materially but only by the distance between the centers of two articulation balls. The cohesion of the structure necessarily calls for specific methods of implementing the invention.

It is the deposit of material by condensation or solidification or transformation associating the two preceding ones to obtain a solid binder possibly occupying all the space between balls or being limited to an envelope covering the bars like continuous sleeves, consisting for example of resin . This deposit is made concentrically thanks to a heat pump located in the center and a heat transfer fluid or better. This will be presented later.

Among the applications, and this ties in with the processes for producing these concentric structures with equal bars, there are structures responding to this geometric order at the very level of their atoms. Thus, we already know polymerization, cyclic molecules, we do not yet know the chains of molecules forming rings, no advantage of the chains of cyclic molecules forming independent circles, or even concentrically associated. These are therefore chemical bodies and their process of production.

Let us define the geometry of concentric structures integrating square base pyramids.

• - structures ayant seulement des discontinuités dans la triangulation;des barres en moins
• - structures comportant des trous carrés traversant la structure de part en part, ou structures formées de modules cadres.

These fall into two categories:

• - structures with only discontinuities in triangulation; fewer bars
• - structures with square holes passing right through the structure, or structures formed from framework modules.

For the first, the keystone is formed by a tetrahedron framed by two adjacent pyramids, that is to say coinciding by one face with it and two other inverted pyramids, adjacent in the perpendicular direction and coinciding each by one of the two faces remained free on the tetrahedron.

We therefore obtain a cross which is stable when we bring together the four vertices of the rectangle that form a branch to the four vertices of the rectangle perpendicular to the first and located in another plane, corresponding to the other branch. Thus has figure 1 of sheet 24/38, representing the basic isostatic structure generated concentrically by the tetrahedron. It is presented in top view as an octagon with unequal sides, the smallest sides being the oblique bars of connection between the rectangles.

To obtain the associated isostatic square structure like the first, comprising 30 bars and 12 joints, it suffices to place its four vertices by connecting each of them by three bars, at the top of a pyramid and two vertices at its base. The structure shown in Figure 2 of sheet 24/38 is obviously not prismatic.

The following structure, to which the desired concentric development leads, is obtained as follows: the three bars are placed, two of which form the base of a pyramid having a common angle with the previous square. This operation is carried out eight times to obtain central symmetry, the structure is then always isostatic, it becomes hyperstatic when four bars are applied, making it possible to obtain an octagonal outline from above. If you want isostaticity and in order to keep the central symmetry, you can remove one of the following two bars as desired: the parallel slash, in top view, to the slash belonging to the outer contour for example AF or BG, the perpendicular and mediating horizontal bar in top view of a bar forming the end of the central cross example Il or JJ '. The following extension leads to an octagonal contour, always in projection, with alternation on a horizontal unit side and on a side equal to 3 /, / 2, since it is formed by three oblique bars. The bar to be removed is always designated, AB and its counterparts. The square associated with this figure 6 of the sheet 26/38, is obtained without problem: it is U V W X and its side is equal to 5√2. Figure 7 of sheet 27/38.

The following structure is octagonal of course, and in projection, its sides are alternately equal to 2 or 3 /, / 2. In the same way as previously, the installation of the central bar p8 between two basic angles of newly formed pyramids, requires to remove the bar which is parallel to it in projection.

Likewise the installation of barsap and -y5 requires it to remove SS'and TT ', or UI And XL. Figure 8 of sheet 28/38. We can generalize by saying that the placement of bars leading to the octagon requires deletions, among which solutions is always the possibility of removing the bar parallel to that

• 4 (1hz+10b) octogone 1 hz: horizontal
• 4 (3ob) carré ob: oblique
• 4 (2hz + 1 ob) octogone 2
• 4 (hz + 3ob) octogone 1
• 4 (5ob) carré
• 4 (2hz + 3ob) octogone 2
• Il vient les structures non construites suivantes:
• 4 (hz + 5ob) octogone 1
• 4 (7ob) carré
• 4 (2hz + 5ob) octogone 2
• 4 (hz + nob) octogone 1
• 4 (n + 2)ob carré
• 4 (2hz + nob) octogone 2

belonging to the external contour, transforming during its installation an isostatic structure into a hyperstatic structure. The sequence of concentric structures described above is as follows, noting that there is an angular periodicity of a quarter turn for the projected polygon studied in this sequence:

• 4 (1hz + 10b) octagon 1hz: horizontal
• 4 (3ob) square ob: oblique
• 4 (2hz + 1 ob) octagon 2
• 4 (hz + 3ob) octagon 1
• 4 (5ob) square
• 4 (2hz + 3ob) octagon 2
• It comes the following unconstructed structures:
• 4 (hz + 5ob) octagon 1
• 4 (7ob) square
• 4 (2hz + 5ob) octagon 2
• 4 (hz + nob) octagon 1
• 4 (n + 2) ob square
• 4 (2hz + nob) octagon 2

The structure shown in Figure 9 of sheet 29/38 keeps the interior structure intact, such as in Figure 7 of sheet 27/38 the installation of three bars leading to a joint leaves the isostaticity intact. The outer contour is serrated. All the vertices in projection drawing the serrated polygon are not in the same plane. The vertices circled in the drawing are in a plane parallel to that of the sheet but sideways

This structure in a bearing plate application, requires an arrangement of the supports, possibly the geometry of the enclosure that it covers.

The concentric structures with equal bars based on pyramids and having holes are square and constructed with frames of the type shown in the figure. These frames combine by triangulation, squares of unit side with squares of double sides. This is the frame of Figure 16h of sheet 14/38.

barres

articulations

pour le critère d'isostaticité Ci au lieu de 6.1 frame = 8 bars (on the lathe) + 4 + 8 (triangulation) + 4 (central square) = 24 bars and 12 joints. It is not stable, it must be doubled by its symmetrical with respect to its plane.

bars

joints

for the isostaticity criterion Ci instead of 6.

We obtain isistaticity by putting an octahedron admitting the small square as a plane of symmetry. Figure 10 of sheet 30/38.

barres

articulations

manquent 10 stabilisationsWe can have a different double frame for which it is the small square that is common.

bars

joints

missing 10 stabilizations

Solution: the central octahedron and two times four median segments from its vertices. Figures 11a and 11b b of sheet 30/38.

Although this was not the object of the structures which were to follow, let us develop a single framework into a larger framework structure before coming to the association of executives. One can place around the previous double frame two larger frames having the largest square in common, ie 12 additional joints and 12 bars.

Continuing and adding two side frames 4 on either side of the plane passing through the square on side 3:

It is the same problem as before, just extend the four medians on each side. Figure 12a and 12b of sheet 30/38.

The development of double frames is obviously not a problem.

• D'abord: 2 x 2 cadres Figure 13 de la feuille 31/38.
• Barres 24 x 4 - 8 = 88
• Arti. 4 x 4 + 4 x 4 + 5 37
• Ci = 23
• Solution: une pyramide centrale 1
• 8 barres de liaison entre carré 8
• 4 octaèdres sur ces petits carrés 8
• Ci = 23 - 17 = 6
• Remarque: il n'y a plus de trous . Figure 14 de la feuille 31/38.
• Ensuite: 3 x 3 cadres
• Barres 24 x 9 - (2 + 2) x 6 = 192
  
• La solution préférable est la suivante: figure 15 de la feuille 31/38.
• Ci = -30 - 24 barres de liaison entre petits carrés
• - 4x4 arêtes de pyramide ayant un point commun avec le carré central: gain 4
• - l'octaèdre central: gain 2

Let us return to the structures obtained by the association of simple frames.

• First: 2 x 2 frames Figure 13 of sheet 31/38.
• Bars 24 x 4 - 8 = 88
• Arti. 4 x 4 + 4 x 4 + 5 37
• Ci = 23
• Solution: a central pyramid 1
• 8 connecting bars between square 8
• 4 octahedra on these small squares 8
• Ci = 23 - 17 = 6
• Note: there are no more holes. Figure 14 of sheet 31/38.
• Then: 3 x 3 frames
• Bars 24 x 9 - (2 + 2) x 6 = 192
  
• The preferable solution is as follows: Figure 15 of sheet 31/38.
• Ci = -30 - 24 connecting bars between small squares
• - 4x4 pyramid edges having a common point with the central square: gain 4
• - the central octahedron: gain 2

The first two dashes can be described in another way: four pyramids surrounding the central octahedron, a belt of eight bars.

We can consider other variants, for example by replacing bars with others which will be the edges of pyramids or octahedra blocking the holes in the frames. For example, we can remove the eight connecting bars connecting the central frame to those surrounding it by eight pyramids or rather four octahedra centered on the squares concerned by the removed links. Figure 16 of sheet 31/38. It is also possible, in another variant, to remove the four cross-pieces forming the four edges of pyramids having a common point with the central octahedron to place them on the holes of the corner frames. Figure 17 of sheet 31/38.

However, in all these manipulations, care must be taken not to create regional imbalances or hyperstatities.

Thus, it seems little advisable to put octahedra on all the frames, unless perhaps to have removed the double bars connecting their square bases between them, except those connecting them to the central octahedron.

Then: 4 x 4 frames

Bars 24 x 16 - (3 + 3) x 8 = 336 bars

THE best solution is as follows: 24 connecting bars between small peripheral tiles, 16 connecting bars perpendicular to the sides of the square formed by the structure, 5 cross-pieces forming the edges of a pyramid. This solution is shown in Figure 18 of sheet 32/38. It is not obvious that the above solution is the best because, apart from the central cross, it too favors the frame added on the 2x2 to make the 4x4 in bar density at the expense of the latter. However, it is also not obvious that a solution keeping 2x2 in the state it was defined above is a good solution. This solution would connect the latter to the outer frame of the 4x4 which would have 24 connecting bars by four crosspieces located at each corner.

Figure 19 of sheet 32/38 summarizes the two trends previously mentioned by their extreme solution. We kept the bars of 2x2 bridges, we deleted the octahedrons, resulting in a loss of Ci of 8 compensated by 8 connecting bars with the 4x4 located on the medians. The previous solution may present an apparently improved variant: it involves replacing an external connecting bar of the 2x2 with a cross, four edges of a pyramid resting on the ends of the removed bar as well as on two points forming a square with these. This solution with nine braces is shown in Figure 20 of sheet 32/38.

Another solution with nine braces is shown in Figure 21 of sheet 32/38. She is characterized by the fact that it comprises a simple chaining of 12 bars.

Each quarter has a variant. The upper half has a central pyramid having no base square, therefore being a simple cross, but connecting bars along the medians between the 2x2 and the 4x4. The lower half does not have these bars but a square base of the central pyramid and connecting bars outside the 2x2, variant of the left quarter, or inside the 4x4, variant of the right.

Then: 5x5 frames

• Partons de la solution du 3x3 ACio = 30
• Trouver ΔCi_o = 32
• Solution simple: double chaînage 8x 4 Satisfaisante?:

Bars 24x25 - (4 + 4) x 10 = 520

• Let's start from the solution of 3x3 ACio = 30
• Find ΔCi _o = 32
• Simple solution: double chaining 8x 4 Satisfactory ?:

Move the four 3x3 braces to the corners Figure 22 of sheet 33/38.

We put the corner pyramids in the corners of the 5x5, as for figure 17 (sheet 31/38) of the 3x3.

Another solution trying to respond to the trend "the inside holds the outside", is shown in Figure 24 of sheet 34/38 and has a variant with simple chaining.

Before moving on to the 6X6 frames structure, we can notice that if n is the number of frames on one side, the sequence Ci "evolves as follows:

So we can easily find the isostaticity criterion for a 20x20 structure.

A few bars to place judiciously ... Then: 6x6 frames

Solution: 2x16 connections between 4x4 and 6x6.

Figure 25 of sheet 35/38. 9 braces

40 double chains

Back on the 4x4: it is advisable to consider the first two variants drawn by moving the crosspieces to put them on the small squares located in the angles to thus have four pyramids. Another solution for the 6x6 provides figure 26 of the sheet 35/38, a simple chaining 20 bars

25 braces 25 stabilizations (3x1-4 = -1)

4 pyramids in the corners 4 stabilizations of the successive interior chaining of 4,4,12,12 bars

total 81

The outermost 12-bar chaining is either outside the 4x4 or inside the 6x6 by not joining the corner frames.

• Dans cette variante, il n'y a pas de base de pyramide centrale ou de pyramide dans les coins.
• On peut extraire de cette variante équilibrée une structure 4x4 cadres.
• Chaînage simple de 4, de 4, de 12 et 12 barres
• 9 croisillons 9 stabilisations
• 4 pyramides de coin 4 stabilisations
• total 45

Another variant has a distribution close to the previous one, however the last single chaining inside the 6x6 is complete. It has 20 bars.

• In this variant, there is no base of central pyramid or pyramid in the corners.
• We can extract from this balanced variant a 4x4 frame structure.
• Single chaining of 4, 4, 12 and 12 bars
• 9 braces 9 stabilizations
• 4 corner pyramids 4 stabilizations
• total 45

Then: 7x7 frames

One solution is to link the 5x5 as it is already defined to the frame, therefore comprising 62 stabilizations, 5x2x4 links Move the crosspieces to the corners.

This solution which is perhaps not optimal, works to pass from a square of odd side to the square of odd side immediately higher, therefore for all odd.

Then: 8X8 frames

We have already seen that for even numbers, there is no solution, if only only satisfactory from the point of view of concentricity, if the lower structure is kept intact. Ci for 6x6 = 81 ACi ⁼ 50, not divisible by 4. We can at best add 12 link bars on each side and two crossbars at diagonal angles.

One solution: we keep 16 connections 8 crosses plus the central: Ci = 106 therefore 100 stabilizations to find

We will preferably put 16 additional braces instead of 24, the four sides of the central pyramid, as well as four braces in the corners forming pyramids. Figure 27 of sheet 36/38. In this figure, we can replace the 48 connecting bars with a simple chain of 20 bars around the 6x6 and 4 times 7 cross-pieces on one side.

To make an attempt at synthesis, it seems that the structures comprising an odd number of frames, are built around a center which holds the unit, it is the octahedron.

As for the even structures, it is more the rigid frame which holds the interior, at the limit, like a canvas. Before testing, no dimension is excluded, provided that it is a multiple of the side of a basic frame. This insofar as the criterion of global isostaticity Ci = 3n-N = 6, is satisfied. However, it is not excluded that there are local instabilities or unavoidable hyperstabilities for certain dimensions which have the consequence of banishing them.

Thus, a series of dimensions would be born, on condition of knowing the dimension of the elementary mesh when it is a matter continuous responding to an arrangement in crystals or molecules.

entre le côté et l'épaisseur, voire

si elle est formée de l'empilement de structures telles que celles étudiées, lequel empilement devrait obligatoirement suivre certaines règles pour satisfaire l'isostaticité.
- une plaque doit toujours être formée de carrés. Il faut donc que le plus grand diviseur commun à sa longueur et à sa largeur soit un multiple, et même un multiple validé si certains apparaîssaient ne pas convenir, de la dimension du cadre élémentaire.Already, for the plates, we can draw two lessons:
- there is always a connection

between side and thickness, even

if it is formed by the stack of structures such as those studied, which stack should necessarily follow certain rules to satisfy isostaticity.
- a plate must always be formed of squares. It is therefore necessary that the largest divider common to its length and its width is a multiple, and even a multiple validated if some appeared not to be appropriate, of the dimension of the elementary frame.

The structures presented, unstable, can present interesting deformations, in particular when I miss only one, two or four bars. They are therefore not to be excluded. Ci = 10.

The structures can be frames formed by several elementary frames 8 or 16 with a side of 5 modules and an interior vacuum of 3 modules.

It is, of course, necessary to study isostaticity.

The structures that will be presented are obtained by stacking regular concentric polygons, the number of sides doubling each time, while, at the same time, their length is reduced by half.

We can make a first remark: their perimeter is constant. This will allow us to calculate π, after making a second: this mathematical result is more a discovery, if it is, than an invention.

However, it cannot be ignored.

It is of course a question of being able to express their rays, that is to say that of the inscribed and circumscribed circles. Consider a square with side 2 and perimeter 8:

Or an octagon on side 1, therefore with the same perimeter 8

We can deduce:

Note:

We check just as well:

It suffices to express Rc _n as a function of Ri _n . By calling on Pythagoras

thus, we obtain the following sequence:

So, starting from a square with side 2 P = 8
Ro = 1

• - le théorème de Pythagore est juste
• - la convention a° = 1 est justifiée
• On peut faire le programme suivant: (sur CASIO 7000 G)
  
  

In these calculations, the validity of the assumptions must be taken into account:

• - Pythagoras' theorem is correct
• - the convention a ° = 1 is justified
• You can do the following program: (on CASIO 7000 G)
  
  

For 15 iterations, i.e. 2 ¹⁶ side

π = 3.141592654 (1 ^st iteration 2 sides)

We can do the same calculation leading to the same result from the pentagon

Better make a program

The same readable ten decimal values are obtained with the same number of iterations.

We can make identical programs from the polygon whose number of sides is a prime number.

Inversion: forget that there was a program 0.

The other decimals of π can be calculated as follows:

First line of the photo taken by the magazine N ° 2 "Hypothesis" of Texas-Instruments at the Palais de la Découverte.

We note that if we take a range of nine digits, the product of these digits is always the number whose sum of digits is always reduced to 9.

Thus, it is said, a priori that the number 7 placed under "CAUCHY" is wrong: The previous product is good

The one including the 7 is bad

You have to put a 9 instead

• - celle de l'isostaticité, il faut rajouter des barres
• - celle de la génératrice, quelle forme a-t'elle?

Let us return to the structures constructed by a stack of polygons used to calculate ₇ r. They are linked together by a triangulation. Several questions arise:

• - that of isostaticity, it is necessary to add bars
• - that of the generator, what shape is it?

If we consider a symmetrical column structure with a circular section in the middle, it is likely that the generator has the shape of a convex or concave bell curve according to the direction of the observation. As for stabilization, it is of course necessary to add bars while being inspired by the patent of 02/12. Pyramids on the end squares and between, arrangements to study.

The object of these new constructions, therefore eligible for invention, relates to structures based on a pentagon, pentagonal pyramid or pentagonal frame made up of two pentagons, one on side 1, the other on side 2 and connected by a triangulation.

It is a crown of pentagons in any number a priori although the numbers 5 and 8 seem closer to offer results than the others.

A crown of 5 pentagons can be the central crown of a structure uniting 5 dodecahedra. Two identical structures joined and phase shifted by a tenth of a turn, can provide the framework of a tire, carcass completed by pyramids on the pentagons sufficiently rigid to not require internal pressure.

It is impossible, as we believed too quickly without checking this evidence, to put four regular pentagonal pyramids in coincidence with the four faces of a tetrahedron. Furthermore, the distance between the vertices of pyramids of two pentagons adjacent to a crown of 8 pe pentagons, is not equal to a side or 2 or a number that can give some hope of development of the structure. It is equal, it seems to

Can we wedge pentagons?

This study having been made before that of construction using a frame, we deposit the construction based on crowns of pentagonal frames.

We also envisaged the construction by planar crown of 10 pentagons, the second crown, difficult to define is in the best of cases formed of pentagons of double side, connected by points to the first, the top of a small pentagon of the first crown coinciding with the middle of a large pentagon of the second crown.

• les barres sont creuses et contiennent un gaz sous pression toute cette tuyautérie aboutie au centre où se trouve une vanne.

For the implementation of these various structures, the following method can be used:

• the bars are hollow and contain a pressurized gas all this piping terminated in the center where there is a valve.

= 0,98%. Ceci est donc acceptable.When this is opened, there is expansion of the compressed gas in the bars comprising "gelule" shaped enclosures and condensation of the material which it is desired to deposit on the framework. It is necessary to find the body for which the frost has an interesting mechanical resistance and although the sleeves of ice have proven themselves (in 82 on the electric wires). Let's go back to the geometry and answer the question asked at the beginning of this page. Yes, this is possible and we will make sure to place an inverted pyramid with a pentagonal base, two vertices of the base of which would be the vertices of the pyramids of the crown of 8 pentagons and the apex of which would be the outer end of the segment belonging to two adjacent pyramids of the crown. The new inverted pyramid has the space between two vertices of the basic pentagon separated by another 1.586606634 instead of 1.6180559 for the crown pyramids. If we share the difference, we

= 0.98%. This is therefore acceptable.

In addition, the plane of this new intercalated inverted pyramid is 1.26356127 ⁰ with the horizontal plane. This is very interesting as to form a circular plate to be developed concentrically.

Where we stopped is the edge of these inverted pyramids: between them there remain spaces equal to 1.2621 - but it is possible that the distance between a peripheral apex of the base of the inverted pyramids and a peripheral apex of the base pyramids of the first crown is distant from 1 in which case a third pyramid can be built immediately if not we can consider putting pyramids edge to edge with the inverted pyramids even if leaving diamond-shaped holes.

Another possibility of independent construction consists in putting five pyramids with square base in crown, the point in the center. There is no big angle deviation. The base should be rather a little trapezoidal than square. We have an angle at the top of 70.528779 ° instead of 72 °.

Isostaticity of a simple crown of 5 dodecahedra.

• Barres 90
• Articulations 32
• Couronne 450 barres (90x5)
• - 5x5 5 faces communes 425
• Ci = -20
• On peut oter 5 pyramides des faces communes:
• ΔCi₁ =5x2= + 10 ou laisser deux pyramides à quatre arêtes.
• On peut enlever une barre sur 3 pyramides de chaque dodécaèdre.
• ΔOC₁ = +15
• Ci =-20+25=+5

A dodecahedron whose all faces are triangulated by a pyramid.

• 90 bars
• Joints 32
• Crown 450 bars (90x5)
• - 5x5 5 common faces 425
• Ci = -20
• We can remove 5 pyramids from the common faces:
• ΔCi ₁ = 5x2 = + 10 or leave two pyramids with four edges.
• We can remove a bar on 3 pyramids of each dodecahedron.
• ΔOC ₁ = +15
• Ci = -20 + 25 = + 5

A bar remains to be removed so that the structure is isostatic. It can be removed in a pyramid with 5 edge bars, or even in the central pentagon.

Double crown

• bars 2x400 Crowns such that C _i = 5
• joints 130
• By joining two crowns, we put 10 faces in coincidence: 2 bars and 15 joints
• bars = 800-20 = 780
• arti. = 2x130-15 = 245
• C ² i = -45
• By removing the pyramids on the ten sides in coincidence, we gain Ci = + 20 - C ₂ i = -25

• Ainsi C"_i =-29
• On doit enlever 35 barres pour obtenir l'isostaticité:
• 5 formant un des pentagones cité ci-dessus.

Note: the double crown encloses a dodecahedron. We can regain 2 or 4 in Ci by putting pyramids with 4 or 5 bars leaning on the two pentagons of opposite face of this central dodecahedron, at least on the two pentagons located in the center of the double crown.

• So C " _i = -29
• 35 bars must be removed to obtain isostaticity:
• 5 forming one of the pentagons cited above.

There remains a bar to remove on three of the faces of each dodecahedron. Thus, this will have 6 faces comprising pyramids with 4 edges, 6 faces comprising pyramids with 5 edges.

We can for example leave 5 edges on the two central faces of each dodecahedron, that is to say those closest to the center, as well as on the common faces with other dodecahedra of the double crown, 2 on the same crown and two inter-crowns, which is good 6.

• d = D-2h (hauteur de pyramide)
• distance entre faces opposées
• D = 2,21911308 h = 0,5257311119
• d=1,167650856
• Cette distance est égale à environ s de la longueur d'une barre.

We can also do the opposite: 4 edges on these last faces and 5 on the others. In addition, since the structure is isostatic, we can remove the five bars from the remaining central pentagon and put five radial bars joining the top of a pyramid on the central face to the top of the pyramid on the opposite face. The distance between these peaks

• d = D-2h (pyramid height)
• distance between opposite faces
• D = 2.21911308 h = 0.5257311119
• d = 1.167650856
• This distance is equal to approximately s of the length of a bar.

It is true that the dodecahedra can and even must widen in the structure version with modules coinciding without a slot. However, this distance seems a bit large only tests can confirm this solution.

The solution where the dodecahedrons of the same crown are not joined is valid although it requires leaving more bars. In fact, it is the only truly isostatic structure since we calculated on the figure: IJ = 1.456202674 which is the distance between two vertices of dodecahedra, which is less than the side of the decagon on which their attachment to the pentagon , more exactly, at the central hemi-dodecahedron, places them.

The provision providing tirans between the vertices of the pyramid is also envisaged for concentric structures formed of frames. In general, depending on the structure, you can prestress at the corners or at the center. In this category of structures, it is envisaged to have two thicknesses of the same structure coincidence by the small or by the large squares. We can then remove bars in the frames and obtain diamonds.

To return to structures composed of dodecahedral modules, there are, of course, one of the faces of which has no pyramid but a hole.

These structures can be much more important than the double crowns since one can imagine a partition of the space in dodecahedra.

Axially, there is no problem in stacking crowns, so a pipe is easy to build.

Peripherally and in all directions, this is possible thanks to the remark already made. The center of a double crown is occupied by a dodecahedron with the necessary deformations, lengthening or shortening, which can be estimated at: 1/1 = (1.46619-1.45620): 1.461: 2 divide by two to share the distortion.

1/1 = 0.34% which can easily be obtained for most materials.

It is certain that there is an infinite number of ways and technical solutions to accommodate this figure. Relationship to% carbon in steel? Therefore, any dodecahedron can be considered as the center of a double crown, and therefore all the material space can be partitioned as well.

To return to the structures consisting of a stack of polygons whose number of sides doubles, while their length is divided by two, it must be seen that we can go from a segment of length 2 to a perimeter circle 2 to find at the other end, a side segment 2 perpendicular to the first. Thus one can associate several link structures of this type and obtain rotations of a half turn of a turn and more. We made a pipe that could serve as a conductor for a lot of fluid, energy in general. A wave tube in which light can propagate by simple reflection. Optical filter. Concerning the dodecahedral structure, it remains a centered structure. There is a progression of bar lengths from the center, there is no non-localized partition of space.

Concentric articulated structures with equal bars

The invention groups together all concentric structures with equal bars constructed using the following module: This module is composed of a pentagon base of a pyramid with five edges, connected to a square by eight bars, two of which are parallel, the other six from the other three vertices form a triangulation.

The planes of the square and the pentagon are parallel and distant by 0.85065 ... x C.

We can consider this module as an icosahedron from which we would have removed 9 bars before deformation. The module is therefore composed of 21 bars if the square is not closed. It has twelve faces in equilateral triangle and two square faces. It is formed by four modules as described in claim 1 of the patent of 29/01/88, this before deformation.

These modules can be assembled into crowns in two ways: either the square of the inner polygon of the crown is the side common to the two squares AB, or it is an outer side of one of the squares DC or FE. The module admits two planes of symmetry, therefore, an axis of symmetry. In each case, the common sides for associating two modules in a crown are the sides coming from the ends of the side considered for the distinction of the two cases, which make between them an angle of 36 °.

The crown can include 10 modules, or even 8 or even 9 or even 7 or 6. In some cases, we have isosceles trapezoids instead of squares. In the case of 10 modules, the plane of symmetry, mediating plane on the common side of the squares, partitions the space into 10 dihedrons of 36 °. The plane of the crown is, in one of the ways distinguished at the start, the other plane of symmetry of the modules, in the other way, its plane is the plane of a pentagon.

• Modules:
• barres: 22
• arti. : 10
• Ci = 30-22 = 8 instable
• Couronne de 10:
• barres: 220-10= 210
• arti. : 100-20 = 80
• Ci = 240-210 = 30

The first type of crown is not stable as it is:

• Modules:
• bars: 22
• arti. : 10
• Ci = 30-22 = 8 unstable
• Crown of 10:
• bars: 220-10 = 210
• arti. : 100-20 = 80
• Ci = 240-210 = 30

We can thus stabilize 20 square base pyramids plus two double pyramids (octahedron) on two diametrically opposite modules.

If two crowns Ci = 10, four diametrically opposite pyramids as before; if three crowns stacked Ci = -10. Remove 10 middle bars from the modules on the diametrically opposite crowns of these same crowns (same diameter or different diameter).

An interesting variant by these developments is the construction by double modules. These are made up of two simple modules with a square in common. There are two ways of joining the squares, one giving rise to a symmetry, the most studied, a second allowing assemblies whose general shape is a wavy bar.

• Ci = -(22x2-4) + 3(10x2-4) = 8
• Il y a deux façons d'assembler les modules: soit le Vé formé par deux carrés à l'intérieur de la couronne, soit celui-ci à l'extérieur.
• Isostaticité Ci = 3x120-380 = -20
• On peut enlever les barres du carré appartenant au plan de symétrie d'un module Ci = -10. Dix arêtes de pyramides pentagonales sur chaque flasque Ci = + 10 sauf sur deux modules doubles diamétralement opposés.
• Ci= +16Cif= +6

There are also other variants in which the planes of symmetry of the two associated modules are perpendicular. For the first double modules studied:

• Ci = - (22x2-4) + 3 (10x2-4) = 8
• There are two ways to assemble the modules: either the Vee formed by two squares inside the crown, or this one outside.
• Isostaticity Ci = 3x120-380 = -20
• We can remove the bars from the square belonging to the plane of symmetry of a module Ci = -10. Ten edges of pentagonal pyramids on each flange Ci = + 10 except on two diametrically opposite double modules.
• Ci = + 16Cif = +6

Note: the set of two consecutive modules is isostatic. Ci = 84-78 = 6. When we add a third module by making it coincide at four points, we obtain a hyperstatic structure. Too much less articulation

One finds indeed Ci = 4. For each assembly of additional modules one has ACi = -2.

When closing a crown ΔCi = -10, if it is a simple crown.

• Ci = 3x170-500 =10

Double module with connection by the tip of the pyramids based on the squares of the single modules that are parallel to each other. Crown of 10 double modules:

• Ci = 3x170-500 = 10

So, four pyramids to place, two on each of the two diametrically opposite modules, either octahedra or based on free squares.

In a variant, the two flanges are connected by a cube. Ci = 3x160-460 = 20. You need a pyramid on certain sides of the cube. Preferably, on the faces parallel to the plane of the crown, 5 on each flange for example. It is necessary to add four diametrically opposite.

• Pour 10 modules: Ci =100.

One can, in a variant, have stacks of cubes between flanges. Ex: 2cubes Ci = 72-52 = 20 for a module.

• For 10 modules: Ci = 100.

Eight pyramids on the sides of the lateral surface plus one in the median plane ΔCi = 90, plus four diametrically opposite.

• ( détail figure 2 bis de la page 37/38).
• = 79,18
• 2θ=158,36
• angle entre les plans de deux pentagones consécutifs d'une même couronne:_j8= 152,53
• Ci = 3(8x10-6) - (8x22-8) = 24
• Huit modules doubles: Ci =-16.
• Ce cas correspond au cas Vé entre carrés à l'extérieur de la couronne. On voit que 20>,e

Crown of eight simple modules:

• (detail figure 2 bis on page 37/38).
• = 79.18
• 2θ = 158.36
• angle between the planes of two consecutive pentagons of the same crown: _j 8 = 152.53
• Ci = 3 (8x10-6) - (8x22-8) = 24
• Eight double modules: Ci = -16.
• This case corresponds to the case Vé between squares outside the crown. We see that 20>, e

So, the modules are stuck or better, you have to have a bar on the common side of the squares which is on fire. It is it which will constrain the structure or which will play a role of spring.

If we maintain these 8 bars which in a variant can be removed, we must still remove 14 bars to have Ci = 6-8 = -2. We can remove 8 edges of radial pentagonal pyramids except 2.

Another variant consists in connecting the flanges not with cubes, but with rhombohedra. We can thus obtain a phase shift between the flanges and bars, similar to the teeth of a helical gear.

We thus have a rotor which can have all the mechanical applications that we know about them: turbine, pneumatic motor, hydraulic, electric, transmission.

The rhombohedrons have diamond faces, composed of two equilateral triangles, it is easy to block or unblock them by removing the diagonal.

In a deformable variant, it can be seen that the phase shift of the flanges is linked to the volume of the square prisms which evolve from the square to any rhombohedra. This fact can be used to make pumps, press control systems and other applications.

Still in a controlled deformable variant, there is the possibility of varying the angle between the flanges, and thus of transmitting rotations between non-collinear axes. Of course, it is not a single rhombohedron as a connection, but a serrated sequence to obtain angles between interesting axes. As a static application, the crowns envisaged are designed to act as a tire carcass. They suppress inflation.

Figure 4 (sheet 38/38) shows a ball itself a sphere between two pairs of cones, each pair is obtained by rolling a sheet metal, each cone in a different direction from the other. The cones are spread to accommodate the ball and therefore pinch it.

Claims (19)

1. Articulated structure introduced into many construction works having a static or dynamic function which can be assembled, combined, covered with soft skin in these various applications, characterized in that it is based on a polygon central in general regular, triangle, square, pentagon and polygons adjacent to it and adjacent to each other, all being articulated by ball joints, the latter polygons being in number equal to the number of sides of the central polygon having themselves a side more and each being the base of a pyramid hinged at its top. 2. Concentric articulated structures with equal bars or equidistant nodes characterized in that they comprise pyramids whose base is any regular polygon, to which one or more bars are missing so as to obtain the isostaticity of the structure or its placement judiciously planned constraint. 3. Concentric articulated structures with equal bars or equidistant nodes according to claim 1, the thickness of the plates is equal to the length of the bars divided by J2 characterized in that its center is occupied by a tetrahedron surrounded by four square pyramids oriented towards up or down depending on the cardinal direction in which they are in relation to the central tetrahedron. 4. Concentric articulated structures with equal bars according to claim 2 characterized in that they are obtained from the central cross by constructing tetrahedra on the available triangles which can appear spontaneously or after having completed the bases of pyramids located in the plane of structure and connected at their top. 5. concentric articulated structure according to claims 2 and 3 characterized in that their isostaticity is guaranteed by a serrated polygonal contour occurring in the two parallel planes containing the plate structure as stair treads, the two paths being phase shifted by half walking and connected by bars thus forming tetrahedrons this pattern repeating every quarter turn where there are peripheral or internal interruptions as well as bars can be removed inside the structure without modifying the concentricity and in order to obtain isostaticity. 6. concentric articulated structures with equal bars according to claim 1 characterized in that they comprise an even or odd number equal to N ² ₀ u 1 or 8 identical frame modules formed by a square on the side unit connected to a square on the side double, parallel, by a triangulation the small square being able to be the base of a pyramid or an octahedron, the large square being able to include its medians, the large squares are juxtaposed and therefore have common sides, the small squares are possibly connected by bars extending their sides. 7. concentric articulated structure with equal bars according to claim 5 comprising an odd number of framework modules characterized in that only that of the center which comprises an octahedron and those of the corners which comprise a pyramid are modules whose small square is not a hole. 8. concentric articulated structures with equal bars according to claim 5 comprising an even number of frame modules, n = 2p per side, characterized in that they comprise a central pyramid, 2 "' ^1- 1 cross-pieces that is in fact say together of the four edges of a pyramid, a double chainage connecting the small squares of the peripheral modules, parallel to the outer contour. 9. A method of dimensioning structures formed from those described according to claim 5 characterized in that it fixes the ratio between the side and the thickness of a plate at 2n 2 -ence that it excludes certain values of n for which isostaticity could not be obtained by considering all the distributions corresponding to the criterion of global isostaticity Ci = -in that the PGDC of length and width of a rectangular plate is a multiple of 2x, / 2 x thickness of the plate. 10. Concentric articulated structures with equal bars characterized in that they are composed of a stack of regular polygons whose number of sides increases by doubling while their dimension is divided by two, connected by a triangulation whose bars are equal to smaller side, the first polygon being a segment 1 in the extreme case the volume between this one and the square of side half is formed of two tetrahedrons and a pyramid, the median or extreme polygon being a circle of radius l / π, the interior volume being partitioned by the edges of a pyramid and real or fictitious tetrahedra. 11. Structures with specific links which can in particular serve as a conductor characterized in that they are composed of modules described according to claim 9 connected by identical polygons the periodicity being equal to the length of one or more links if these are incomplete and have no plane of symmetry perpendicular to their axis. 12. Concentric articulated structures with equal bars according to claim 9, characterized in that they do not comprise a triangulation equivalent to a double helix having a right-hand thread a left-hand thread but a simple one thus showing diamonds. 13. Concentric articulated structures with equal bars according to claim 1, characterized in that they are composed of a crown of pentagons located in the center of the structure and the stability of which is ensured by pyramids with four or five bars on the pentagons and in the center by spokes, if necessary, or again by a triangulation reducing by half the number of sides and rays and that this structure extends by various concentric rings. 14. Concentric articulated structures with equal bars according to claim 12 used in particular as a tire carcass characterized in that they are composed of a double crown of five dodecahedra including pentagons carrying the pyramids useful for isostaticity or prestressing desired, the two crowns being in coincidence by ten pentagonal faces of their side. 15. Structures used in particular as a tire carcass according to claim 13, characterized in that the pentagons of the tread exactly opposite to those of the pentagons bordering the central pentagon, that for each crown are slightly larger so that the five dodecahedra therefore slightly conical coincide by five radial faces instead of having a radial slot between two dodecahedra. 16. Concentric articulated structures with equal bars according to claim 12, characterized in that the crown comprises 8 pentagons comprising the appropriate pyramids, two faces of successive pyramids belonging to another pyramid, the point of which is the outer end of the side common to two pentagons, the structure being extended concentrically in a slightly conical manner still using pyramids. 17. Thermal process for implementing and stressing the material applicable for concentric structures with hollow bars, characterized in that it causes the deposition of material originally in the gaseous or liquid phase on the hollow bars during the expansion of a gas which is discharged through the center of the structure. 18. Module made up of equal bars hinged together being worth its geometrical characteristics and the numerous construction possibilities it offers, characterized in that it is composed of a pentagon carrying a pyramid with five edges connected by eight bars to the three sides of a square two of which are parallel, the other six forming a triangulation, thus thus comprising 21 bars or 22 if we close the square and have 12 faces in equilateral triangle and two square faces. 19. Concentric crown formed of equal bars characterized in that it is formed of several modules, simple like that described in claim 1 or double, 8 or 10 in particular, depending, for double modules, that their concave part is inside or outside the crown.

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