Gravity-Derived Structure For Optimal Response To Gravitational Forces

Abstract

A structure, such as a tower/building/column/beam/bridge/machine, is composed of elements that are arranged in a manner to optimally respond to any destructive force applied upon it. Using physical models that expose the effects of gravitational forces shows that the density of the structural elements of a structure should gradually increase from top to bottom in order for the structure to optimally respond to these and other forces. For optimal response to vibrational forces traveling through a structure/tower/bridge/machine, the structure should be divided in segments that satisfy the Formulas 1, 2, 3, Page 4 of this application. When a structure is divided into segments such that one, or two, or any ratio between these is an Irrational number, the vibration cannot travel through the structure because there isn't a wave length that fits through these different segments.

Description

CROSS-REFERENCE TO RELATED APPLICATIONS

This application claims the benefit of provisional patent application Ser. No. 62/391,203, field 2016 Apr. 23 by the present inventor.

FEDERALLY SPONSORED RESEARCH

None.

SEQUENCE LISTING

None.

BACKGROUND

This relates to any structure independent from the materials (wood, steal, concrete, etc.) that are used to construct the structure. The idea is to optimize the distribution of the structural elements of a structure with the goal of using the minimal amount of materials for a maximal response to all the forces that are applied to the structure.

I. Gravitational forces. Every structure on Earth is continually under the effects of gravitational forces. FIG. 1 is a diagram that illustrates the effects of gravitational forces on a structure. The building is horizontally divided into 20 equal sections. On each section there is a gravitational force “X”. On the top section the gravitational force is X, on the second section (from top) the gravitational force is 2λ, on the third section the gravitational force is 3× and so forth until reaching the bottom of the structure a gravitational force of 20λ. When moving from top to bottom, the weight of each section gradually increases because it holds the weight of all the sections above it. The structure of a building should respond to these additive forces; this can be achieved by increasing the strength of the lower sections. Increasing the strength can be achieved in two different ways. One way is to increase the density of the elements on the bottom, and a second way is to increase the width (bulk) of the elements.

II. Seismic forces and vibration. FIG. 2 is a diagram that explains the impact of seismic forces upon a structure/building. Seismic forces can be horizontal vibrations or vertical vibrations; these are depicted by the arrows. When these vibrations travel from the bottom to the top of the structure the amplitude of the vibrations increases. The vibrations are then reflected from the top to the bottom with a larger Moment=Force×Distance (M=F×D) at the bottom; (The Moment of a force is a measure of its tendency to cause a body to rotate about a specific point or axis) this is represented with the arrows. The structure of a building should respond these seismic forces and this can be achieved:

1) By increasing the strength of the bottom sections by increasing the density and/or width/bulk of the structure at the bottom.

2) By allocating the structural elements in the manner that reduces the travel path of the vibration and isolating the vibration forces in small segments in the structure. On other words: the structure should be divided in different segments in the manner that when a vibrational force is applied in one segment of the structure, these vibration force stays there until vanishes, and is not transmitted to the next segment of the structure. This effect can be achieved in three different ways claimed in this patent application:

a) The length of the segments. If a structure is divided in n segments (a, b, c, d etc.) Then one or more or all ratios between the lengths of these segments should equal an Irrational Number.

Example: the length of segment “a” divided by the length of segment “c” is an irrational number, the length of segment “b” divided by the length of segment “a” is an irrational number and so on. (Refer to FIG. 41)

a/c=I₁, b/a=I₂, a/d=I₃ etc. where I₁, I₂, I₃ are Irrational Numbers.  (Formula 1)

b) The composition of the structural segments. In any segment should exist at least one oblique element which defines an angle with the horizontal line. One or more or all ratios between the values of these angles should equal an Irrational Number.

Example: The angel m, n, o, p, q, r, etc. should satisfy the formula: value of “m” divided by value of “n” equals an Irrational number and so on. (Refer to FIG. 41)

m/n=I₄, n/o=I₅, o/p=I₆ etc. where I₄, I₅, I₆, is an Irrational number.  (Formula 2)

c) The length of the oblique elements in the same segment or in different segments should be in the manner that one or more or all the ratios between these lengths equals an irrational number and so on. (Refer to FIG. 41)

Example: The length 1, 2, 3, 4, etc. should satisfy the formula: length“1” divided by length “2” is an irrational number.

Length1/length2=I₇, length2/length3=I₈ etc. where I₇, I₈ is an Irrational number.  (Formula 3)

III. Wind and other lateral forces. FIG. 3 is a diagram depicting the overturn forces of the wind on a structure/building. Wind forces cause an overturn moment upon the building. Moment=force×distance, or M=(F)(D). These forces increase from top to bottom. The structure should respond to these forces in two ways, by increasing the density or by increasing the width of the lower structural elements.

SUMMARY

The above ideas suggest that a structure/building should be 1) stronger on the bottom 2) composed by different non identical segments, in order to withstand the forces applied on it. One way to achieve this is to gradually increase the density of the structural elements from top to bottom. How gradual should the increase in density be? The optimal way to solve this problem is by using gravity to allocate the structural elements.

EXPERIMENT

FIG. 4 to FIG. 21 show how gravity can be used to allocate the structural element of a structure/building. On FIG. 4, FIG. 5, and FIG. 6 a strip of cardboard is bent in a zig-zag manner. When the top segment of the strip is pulled up, the effect of gravity upon the strip can be visualized. The segments of the zig-zag are denser at the bottom. FIG. 7, FIG. 8, FIG. 9, and FIG. 10 shows different models that are suspended over a gridded background; again, the effect of gravity is shown. The models are made from different materials, and with a different number of segments. FIG. 11, FIG. 12, FIG. 13, and FIG. 14 shows the zigzag strip of cardboard suspended through two metal rods, again the effect of the gravity upon the structure is visually depicted. FIG. 15, FIG. 16, FIG. 17, FIG. 18, FIG. 19, FIG. 20, and FIG. 21 show a wider strip that is bent in a zigzag manner (the center of each segment is subtracted) suspended through four metal rods.

Conclusion

These experiments demonstrate 1) how gradual the density of the structural elements should be increased from top to bottom and 2) how to divide a structure in n different non identical segments that can be calculated to satisfy the above formulas 1, 2, 3. (the length “a” divided by the length “b” equals an irrational number, the value of angle “m” divided by the value of angle “n” equals an irrational number, and length “1” divided by length “2” equals an irrational number. (FIG. 41.) For optimal response of the forces that are applied upon a structure/building.

Process

The FIG. 22 shows a model, which contains 8 equal segments. FIG. 23 is a diagram of the above model with measured distances between the zig-zag points; it is clear that these distances are increasing from the bottom up. On the left side the distances between points are 20, 80, 124, 147, units, and on the right side they are 49, 110, 135, 163, units, (I used the grid from FIG. 7) to generate these distances. FIG. 24 shows the distances a, b, c, d, e, f, g, h, which are the distances from one bent to the next one. Using these distances I produced the graphic displayed at the FIG. 25. The vertical lines a, b, c, d, e, f, g, and h, are equally arranged on a horizontal line AB, and connected at the top with the spline DC. Experimenting with physical materials will always have some limitations or errors which lead to the production of a non-idealistic (non-perfect) experiment. As a result the graphic line DC on FIG. 25 is not smooth. To correct these limitations I smoothed out spline DC and got the new graphic on FIG. 26. This graphic represent an idealistic experiment without the limitations of physical materials. The result from the new graphic are the segments a′, b′, c′, d′, e′, f′, g′, h′. Using these segments I constructed the diagram on FIG. 27, which is the new (idealistic) zig-zag line. FIG. 28 shows both zigzag lines. It is clear that the difference is small and the new zigzag is smoother. FIG. 29 shows this zigzag line bolder which represents how the structure element should be allocated at a vertical building/structure. FIG. 30, FIG. 31, and FIG. 32 show how this structure is used to brace a tower/building with variations on building height. On FIG. 30 the bracing starts with a horizontal element at the bottom and finishes with a horizontal element at the top. On FIG. 31 the building is a segment shorter. On FIG. 32 the top is oblique.

FIG. 33, FIG. 34, and FIG. 35 show the process of using two symmetrical zigzag lines to produce a new structure which has two zigzag elements. FIG. 35 shows the elevation of a building using the double structure, the other three elevations are similar to the elevation on FIG. 35.

FIG. 36 to, FIG. 37, FIG. 38, FIG. 39, FIG. 40, FIG. 41, FIG. 42, FIG. 43, FIG. 44, FIG. 45, and FIG. 46 is the same process repeated with a 9 segments zigzag strip of cardboard. The results are similar with the first process. FIG. 44 shows the double zigzag structure and FIG. 46 shows quadruple zigzag structure. The graphic at the FIG. 45 shows how we can generate the extra 2 zigzag lines. At the graphic on the FIG. 45 we add extra segments in the middle of the space between the existing ones. With these segment we construct one other zigzag line and generate the symmetrical line of it. Overlying the four zig-zag lines we get the structure FIG. 46. All other elevations can be similar to the elevation at the FIG. 46. This process can be repeated multiply times as e result to produce structures with multiply zigzag lines.

Adjustments.

1) In FIG. 41 we can make small adjustments in the length of the segments a, b, c . . . in order to achieve the result: one or more or all the ratios of any segment a, b, c . . . with any other segment a, b, c . . . equals an irrational number. 2) In FIG. 41 we can make small adjustments in the value of the angles m, n, o, p, q, r . . . in order to achieve the result: one or more or all the ratios between the value of the angles m, n, p, o, q, r . . . equals an irrational number. 3) In the FIG. 41 we can make small adjustments in the length of the segments 1, 2, 3, 4, 5 . . . in order to achieve the result: one or more or all the ratios between the length of the segments 1, 2, 3, 4, 5, . . . equals an irrational number. (Formulas 1, 2, and 3. Page 4)

FIG. 47, FIG. 48, FIG. 49, showing the diagrams of a single bracing element, double bracing element, and quadruple bracing element. FIG. 50 shows a computer generated rendering of a quadruple structure. FIG. 51, FIG. 52, FIG. 53, and FIG. 54 shows computer generated renderings of buildings that are produced by connecting the zig-zag structure from elevation to elevation in three different ways.

FIG. 55, FIG. 56, FIG. 57, FIG. 58, FIG. 59, FIG. 60, FIG. 61, FIG. 62, and FIG. 63 shows the process repeated again with a different strip of cart board. The result again is similar to the above processes. FIG. 64, FIG. 65, FIG. 66 showing the elevations of a building that has the front and the back elevations identical and the left and right are produced by adding horizontal beams which connect the bending point of the front zigzag with the bending point of the back zigzag. FIG. 67, FIG. 68, FIG. 69 showing the single, double, and quadruple zigzag structure applied on a tower. FIG. 70, FIG. 71 showing the computer generated rendering of a building with to elevators single zigzag structure and two others horizontal beams connecting the zigzag structures, the top finishes at the last segment of the zigzag.

The process is repeated again with a new 13-segments model FIG. 72, FIG. 73, FIG. 74, FIG. 75, FIG. 76, FIG. 77, FIG. 78, FIG. 79 FIG. 80, FIG. 81, FIG. 82, and FIG. 83. FIG. 80 shows the diagram of the density of a shading device on the façade of a building where the density is reversed; top denser bottom less dense. FIG. 83 shows the elevation of a building with a quadruple zigzag structure combine with diamond shaped windows and with a secondary zigzag structure. FIG. 84, FIG. 85, FIG. 86, FIG. 87, and FIG. 88 show different renderings.

FIG. 89, FIG. 90 shows a physical model which is constructed using 18 plastic (elastic) frames connected diagonally at the corner, and the connection is repeated using the other two corners. FIG. 89 the model is suspended on a gridded background, and at the FIG. 90 the model is suspended through the metal rods. FIG. 91, FIG. 92, FIG. 93, FIG. 95, and FIG. 96 are the diagrams generated based on the models in the FIG. 89, and FIG. 90. I repeated the process again as a result generating the building depicted in the FIG. 94, FIG. 97, and FIG. 98. FIG. 99 is a rendering of the quadruple zigzag model generated by the above diagrams.

Next model FIG. 100, and FIG. 102 is constructed using a wider cardboard strip which contains 20 segments (the central part of each segment is removed in the manner each segment forms a frame). This model is suspended through the four metal roads. Based on this model are generated the diagrams in the FIG. 101, and FIG. 103. These diagrams are used to generate the new structure/building shown in the FIG. 104, FIG. 105, FIG. 106, and FIG. 107 showing the four elevations of the new structure/building. FIG. 108 shows the rendering of the zig-zag structure. FIG. 109 shows the combination of the zig-zag structure with other structural elements in this case horizontal and vertical elements. FIG. 110 is a rendering of the Building/structure.

FIG. 111, FIG. 112, FIG. 113, and FIG. 114 shows the combination of the different elements that form the façade of the building.

FIG. 115 to FIG. 138 showing different structures with different segments vertically, from 2 to multiply, and horizontally different number of zigzags from one, two, three, to multiply.

FIG. 139, FIG. 140, FIG. 141, FIG. 142, FIG. 143, FIG. 144, FIG. 145, and FIG. 146 showing that different zig-zag elements can be combined 2 or more to produce new structure/building.

FIG. 147, FIG. 148, FIG. 149, FIG. 150, and FIG. 151 showing that the structure can be used in a horizontal manner, as a cantilever beam or regular beam or a bridge.

FIG. 152, FIG. 153 showing diagrams how the vibrating waives travel in the structure. From physics we know that when the frequency of the vibrating waives that are applied upon a building/structure equals the internal frequency of the structure itself the vibration is maximized (this is called resonance). In this instance the building can collapse. The benefit of the structure described in this application is that because of the structure has different in length and form segments there is difficult for an exterior vibrations that is applied upon the structure/building to reach the resonance. On the other hand the resonance can easier reached in a building/structure with similar segments FIG. 152.

It is very difficult for the structure in the FIG. 153 to reach the resonance, because there is no wave length to fit in two segments a and b, that satisfy a/b=I, where I is an irrational number (Formulas 1, 2, 3 Page 4).

Claims

1. Composition of a structure with at least one zigzag element the form of this (these) element is generated using the experiments explained above (using gravity to allocate the structural elements FIG. 4, FIG. 5, FIG. 6, to FIG. 21). For different appearances of these structures see FIG. 30, FIG. 31, FIG. 32, FIG. 35, FIG. 42, FIG. 44, FIG. 46, FIG. 50. FIG. 51, FIG. 52, FIG. 53, FIG. 54, FIG. 64, FIG. 65, FIG. 66, FIG. 67, FIG. 68, FIG. 69, FIG. 70, FIG. 71, FIG. 79, FIG. 80, FIG. 81, FIG. 82, FIG. 83, FIG. 84, FIG. 85, FIG. 86, FIG. 87, FIG. 88, FIG. 94, FIG. 97, FIG. 98, FIG. 99, FIG. 104, FIG. 105, FIG. 106, FIG. 107, FIG. 109, FIG. 110. These are some examples how the structure may look. Using the experiments explain above (FIG. 4-FIG. 21) we can generate unlimited appearances.

2. The structure of the claim one should be divided in segments that one or more or all ratios between the lengths of these segments should equal an Irrational Number. FIG. 41. Length of segment “a” divided by of length of segment “b” equals an Irrational number a/b=I1 where I1 is an Irrational Number. (In mathematics golden section of a segment is one example of the statement above.) (Formula 1 Page 4 of this application).

3. The structure of the claim one can be divided in segments that contain one or more oblique elements which defines angles with the horizon in the manner that one or more or all ratios between the values of these angles should equal an Irrational Number. The statement above can apply to different angles in the same segment or angels in different segments. FIG. 41 the value of angle “m” divided by the value of angle “n” equals an irrational number.

m/n=irrational number. (Formula 2 page 4 of this application)

4. In the structure of the claim one, the length of the oblique elements in the same segment or in different segments should be in the manner that one or more or all the ratios between these lengths equals an irrational number. FIG. 41

Length of segment “1” divided by length of segment “2” equals an irrational number. (Formula 3 Page 4 of this application)

5. There is no limitation in the materials that the “Gravity derivate structure” can be constructed. Some instances are wood, steal, other metals, concrete, plastic, engineered materials, or combination of any materials, etc.

6. There is no limitation in size and how the “Gravity derivate structure” can be used. The structure can be vertical, horizontal or oblique. For instance this structure can be used as a column, as a beam, as a cantilever beam, as an oblique beam, as a tower or any other shape building, as a bridge, FIG. 147-FIG. 151 or any other way.

7. There is no limitation on the number of the segments, in a zigzag FIG. 115 to FIG. 138.

8. There is no limitation on the number of the zigzags. There is no limitations how segments can overlap each other. There is no limitations on the combination of different zigzags. FIG. 139-FIG. 146.

9. There is no limitation on the combination of the zigzags structure with other structural or not structural elements of a structure/building/bridge.

10. There is no limitation on the profile of the structural elements, these can be square, rectangular, diamond, triangle, pentagons, hexagons, etc. regular or irregular.

11. There is no limitation how the structural elements are connected together.

12. The formulas expressed on the claims two, three, and four can be used in any other object or machine with the goal to isolate, reduce, and eliminate vibrations in this objects or machines.