Description of the Patent Grant awaiting Weather Secure Agri Habitat Design by Dr.Mohideen Ibramsha

Weather-Secure Habitat Dr.Mohideen Ibramsha

1 of 27

The following specification particularly describes the invention and the manner in which

it is to be performed.

Weather_Secure Habitat 2 of 27

Introduction

Due to Global Warming the future weather is not expected to be similar to the past.

Earthquakes, hurricanes, tornadoes, and floods could occur anywhere on earth. We detail

secure habitats for the hostile weather.

Tornado has no effect on buildings with circular exteriors. A hurricane finds it extremely

difficult to uproot a hemisphere or a hemi ellipsoid in which the radius of an upper floor

is less than that of a lower floor. For such buildings the bending force due to the

hurricane decreases at higher levels while the weight of the building increases at the

lower levels. Further as shown in Figure 1 the slope of the building causes the force from

the hurricane winds push the building down increasing the stability of the building.

Floods do not affect a building with entry above ground when the flood height is less than

the entry. A building with triangles formed by pillars and beams is more resistant to

earthquake than without any triangle.

Our buildings have above the floor entry; have circular exteriors; have triangles in the

structural frame; and are hemisphere or hemi ellipsoid shaped with the radius of an upper

floor being less than that of a lower floor.

We demonstrate the feasibility of secure habitats from the smallest size plot of 40’ x 30’

to a plot of 400’ x 400’. Secure habitats could be on a plot of any size, larger than 400’ x

400’ also.

Housing plots are usually in grounds, 60’ x 40’. Some developers offer half a ground plot

also at 40’ x 30’. Until the hemispherical habitats become the norm, we would build

habitats on available plots. Usually the authorities require 10’ clearance abutting the road,

and 5’ clearance on the other three sides. We expect plots for an N feet radius hemisphere

to be (2N + 10)’ x (2n + 15)’ in future. For the present we consider 40’ x 30’; 60’ x 40’;

80’ x 60’; and 120’ x 120’ plots corresponding to half; one; two; and six grounds

respectively.

In India new material is laid on the old road resulting in the approach road going above

the ground floor of houses, even though we normally keep the ground floor about 1’

above the level of the approach road. We design for protection from 8’ high floods. Such

protection would remain valid only if the road repair is carried out by scraping the worn

out road and relaying the new road on the exposed soil.

Assuming that the approach road shall remain as it is at the time of building the habitat,

we design the habitats such that there is no entry point for water at a lesser height. One

way is to have the lower floor of the habitat surrounded by a sloping garden to the height

of 8’. Another way is to build the habitat sitting on pillars that are at least 8’ tall.

Weather-Secure Habitat 3 of 27

Most building authorities permit an FSI of 1.5. There are plans for raising the FSI to 2.

Where the plinth of the hemisphere is small, so that adding a ground level structure to be

surrounded by a garden as tall as the room still keeps the FSI low, a buried lower floor is

adopted.

We adopt this strategy for 40’ x 30’; 60’ x 40’; 80’ x 60’ plots. We consider plots other

than these with less than 6 grounds size at 120’ x 120’ to be rare and hence are not

discussed.

Before giving the detailed description, we consider the prior art and establish that our

invention is different from all the inventions so far.


Prior Art

Hemispherical domes are constructed for a number of years. Some of the famous ones

are considered now.

West Baden Springs Hotel opened for business in 15 September 19021. It has a 200 ft

skylight dome. This dome is supported by 12 trusses2. The distance between the trusses

works out to slightly more than 26’. Possibly the first monolithic dome was built In

Gatlinburg, TN, USA. The domes were built pouring concrete in forms specially

fabricated3. The very first dome built in Gatlinburg is not in use due to a moisture

problem. All domes have problem with refreshing the indoor air and thus well designed

windows need to be part of the domes. A 2,400 sq ft dome home at Cape Romano, FL,

USA lost one window in a storm4. Modern monolithic domes are constructed using an air

form. The air form is inflated; foam is sprayed inside; rebar attached to hooks placed in

foam; shotcrete is sprayed; the air blower is stopped after the concrete sets. The US

authorities have decided as follows:5

The dome, when finished, is earthquake, tornado and hurricane resistant (the US Federal Emergency Management Agency rates them as "near-absolute protection" from F5 tornadoes and Category 5 Hurricanes)

However monolithic domes do suffer from disadvantages. These are considered below.

http://forum.monolithic.org/viewtopic.php?f=2&t=961

A number of problems like cracks in the shell; difficulty in placing windows and doors

because the shotcrete was not uniform and had voids; and echo problems are discussed.

In a two dome garage dropping of a tool at the far end of one dome sounded as though it

was on their leg in the far end of the other dome.

http://en.wikipedia.org/wiki/Talk%3AMonolithic_dome

The above web page comments on the need to spray the shotcrete continuously. This

requirement of continuous shooting of shotcrete makes monolithic suitable for small

domes only; domes of radius 100’ or more could not be monolithic.

http://en.wikipedia.org/wiki/Monolithic_dome

This ‘Article’ part describes problems with the transition from rectangular prisms to

hemispheres like inefficient use of space as the curves of the dome and the straight edges

of furniture and equipments do not match.

1 http://en.wikipedia.org/wiki/West_Baden_Springs_Hotel

2 http://www.insideindianabusiness.com/newsitem.asp?ID=38338 We counted the trusses from the

photograph of the dome. 3 http://www.coastalbreezenews.com/2012/09/07/cape-romano-uncovered/

4 http://www.messynessychic.com/2013/06/12/the-mysterious-dome-homes-marching-into-the-sea-

before-after/domeinside/#main 5 http://en.wikipedia.org/wiki/Monolithic_dome

http://en.wikipedia.org/wiki/Earthquake

http://en.wikipedia.org/wiki/Tornado

http://en.wikipedia.org/wiki/Hurricane

http://en.wikipedia.org/wiki/Federal_Emergency_Management_Agency

http://forum.monolithic.org/viewtopic.php?f=2&t=961

http://en.wikipedia.org/wiki/Talk%3AMonolithic_dome


http://en.wikipedia.org/wiki/West_Baden_Springs_Hotel

http://www.insideindianabusiness.com/newsitem.asp?ID=38338

http://www.coastalbreezenews.com/2012/09/07/cape-romano-uncovered/

http://www.messynessychic.com/2013/06/12/the-mysterious-dome-homes-marching-into-the-sea-before-after/domeinside/#main

http://www.messynessychic.com/2013/06/12/the-mysterious-dome-homes-marching-into-the-sea-before-after/domeinside/#main



Possibly the very first patent for a dome was issued on August 26, 1930 to Mr. Conrad

Pantke. A number of patents are issued over the years for constructing the hemispherical

domes. None of these patents describe a dome that has a multi-floor structure inside the

dome except one. That patent describes a dome with two floors without any pillars. Our

domes do have pillars. Further those domes are not repairable, whereas our domes are.

Still we provide the following table indicating the difference between our design and that

of the listed patents.

The following Table considers the patents on domes.

No Patent No P date Title Comment

1 1773851 26/08/30 Space Covering Structure Discussed

Quoting from the patent we have: [subsequent quotes without this explanation]

It will thus be seen that, when these various joints and elements are assembled,

they provide a safe structure, which even tough certain members may be

eliminated or destroyed due to fire, or other causes, will stand up safely, in as

much as the stresses will arch around the space previously occupied by such

eliminated members.

In contrast, we are concerned about the fidelity of the shape of the dome: a hemisphere or

a hemi ellipsoid.

2 2682235 29/06/54 Building Construction Discussed

In contrast to equilateral triangles used in the geodesic design, we use non-equilateral

triangles in our invention.

3 2905113 22/09/59 Self-strutted Geodesic

Plydome

Discussed

If desired, the openings between the sheets can be closed up, this being merely a

function of the selected frequency of the grid in relation to sheet size.

During the 50s openings in domes could be tolerated. With hostile weather we cannot

have any opening in the domes.

4 2914074 24/11/59 Geodesic Tent Discussed

This patent produces conical shapes hanging all over the geodesic dome and such cones

would be blown away by hurricanes and tornadoes.

5 2918992 29/12/59 Building Structure Discussed

the triangular faces of the pieces define regular pentagons and hexagons, with

each pentagon surrounded by hexagons.

We have no pentagons or hexagons in our structure.

6 2929473 22/03/60 Structural Framework Structural member

7 2934075 26/04/60 Inflatable Structure Inflatable

8 2962034 29/11/60 Shelter and method of

making same

Regular polygon

9 2976968 28/03/61 Wall Construction Equilateral triangle

10 2978074 04/04/61 Spherical Building

Structure with Curved

Beams

Isosceles triangles



11 2978704 04/04/61 Radome Structural

Devices

Discussed

Parallelism is substantially non-existent in the present embodiment, and instead,

the term triangular incrementation is used to describe the orientation which an

antenna device energized by a source of electromagnetic wave energy included

within the radome device sees insofar as transmissibility is concerned.

In our design the ribs that are more or less parallel and the rings that are parallel are

fabricated. Thus electromagnetic energy is possibly blocked from entering the dome.

12 2979064 11/04/1961 Inflatable Building

Construction

Inflatable

13 3004302 17/10/61 Building Construction Strips

14 3006670 31/10/61 Frame for supporting

Domed Structures

Spherical balls

required

15 3018858 30/01/62 Shelter Frame Strut with 2 sections

16 3026651 27/03/62 Building Construction Tetrahedron

17 3042052 03/07/62 Portable Tepee Cone

18 3043054 10/07/62 Spherical Self-supporting

Enclosures

Discussed

In constructing the modules, a flat body sheet of honeycomb material or foamed

plastic may be laid between two skin sheets of fiber resin laminate in wet

condition and the laminate or sandwich so formed may be confined between

spherically curved molding plates where it is held while the materials are bonded

together in any well known manner as by polymerizing agents in the composition

or by heat applied thereto or developed therein. After the spherical laminate sheet

is formed it is cut to the shape and size of the desired module …

The above method of preparing spherically curved sheets requires the spherically curved

molding plates. Without the molding plates the spherically curved sheets of any shape

could not be made. Isosceles triangles forming regular pentagons and irregular hexagons

are used by Mr. Clarence to make the sphere. If we do not use the idea of another

inventor, we are forced to make the first set of spherically curved molding plates using

regular pentagons and isosceles triangles. Once the molding plates are made arbitrary

shapes could be used to make future spherical objects. We agree that once the original

spheres are made using regular pentagons and isosceles triangles, the sphere could be cut

to get molding plates of any shape. Hence the invention requires regular pentagons and

isosceles triangles. Our design has no pentagon at all.

19 3049201 14/08/62 Base Details No overlap



20 3058550 16/10/62 Structural Unit Discussed

The invention of the geodesic dome by Fuller uses planar sheets to form the triangles,

pentagons, and hexagons. The surface of the dome is not smooth. This patent uses sheets

pressed to approximate the curvature of the dome. In Fuller’s design equilateral and

isosceles triangles are used. Any two triangles in Fuller’s design with a common edge

form a diamond. This patent uses sheets that are diamonds. In the illustration given in

Figure 3A a single diamond is divided into 16 quadrilaterals or triangles and thus reduces

the area of flatness. In general a diamond could be divided into m x n areas where m. n

are greater than or equal to 1. As the numbers m and n increase the flat area decreases

and the diamond sheet approaches the desired curvature. Still it does not become the

desired curvature. Monolithic domes and our invention give the desired curvature to the

elements that make the dome.

21 3061977 06/11/62 Spherically Domed

Structures

Isosceles triangles

22 3063519 13/11/62 Building Structure Like 3058550

23 3063521 13/11/62 Tensile-Integrity

Structures

Discussed

… the length of all booms will be such as to subtend an angle of 25°14’30” of the

sphere. … all utilize one common length of strut …

Since there is just one length of strut, the triangles, pentagons, and hexagons of the dome

are regular. Our design has none of these regular elements.

24 3083793 02/04/63 Membrane Sustained Roof

Structure

Two surfaces

25 3094708 25/06/63 Indoor-Outdoor

Swimming Pool and

Enclosure Therefor

Diamonds of equal

length sides

changing widths

26 3114176 17/12/63 Wood Building

Construction

Plane triangles

27 3129531 21/04/64 Reinforced Building

Structure

Central Post

28 3185164 25/05/65 Three Dimensional

Reticular Structure

Discussed

The element comprises of a core 2 provided with three arms 4, 5. And 6. In

practice the number of arms may vary, depending on the desired geometrical form

of the contour of the structure. … the arms projecting from the core 2 must be

equal …

We have no such requirement.

29 3192669 06/07/65 Skylight Construction Related to

supporting the dome

and not the design

of the dome.



30 3203144 31/08/65 Laminar Geodesic Dome Spherical equilateral

and isosceles

triangles

31 3220152 30/11/65 Truss Structure Tetrahedron

32 3221464 07/12/65 Tetrahelical Structure Tetrahedron

33 3292316 20/12/66 Self-supporting roof Rhombus, isosceles

triangle

34 3304669 21/02/87 Containers and

Space_Enclosing

Structures

Discussed

Robert C. Geschwender uses tetrahedrons, isosceles triangles, and kites in his

constructions. Our design has none of these structures as the distance between a central

rib and its left neighbor rib need not be same as the distance with the right neighbor rib.

Similarly the distance between a central ring and its bottom neighbor ring need not be

same as its distance with its upper neighbor ring.

35 3314199 18/04/67 Prefabricated Sectional

Building Construction

Discussed

Three butterfly units are used by C. A. Wulf Jr. et al to form a tetrahedron with an

equilateral triangle as base. They form a base rhombus using two equilateral triangles. By

placing a number of rhombuses side by side they create a passage. By keeping the valley

forming edge of the butterfly large enough, the passage is wide to park an automobile.

Thus they create drive-in spaces. There are no tetrahedrons in our design.

36 3323820 06/06/67 Space Frame Structures Unrelated

37 3333375 01/08/67 Frame for supporting a

dome shaped building

Regular hexagon

38 3341989 19/09/67 Construction of

stereometric domes

Regular polygons

39 3380203 30/04/68 Modular Free-span

Curvilinear Structures

Strut connector

40 3407558 29/10/68 Self-supporting structural

unit having a series of

repetitious geometrical

modules

Module has legs

41 3766959 23/10/73 Pivoted Panel Connection

and Hinge

Unrelated

42 3863659 04/02/75 Shelter Structure Pulley, roller, spring

43 3889433 17/06/75 Structural Frame No dome; cylinder

and semi-cylinder



44 3955329 11/05/76 Hollow Structure interlocking flat

shapes

The following four patents are referred by 3955329

45 604277 17/05/1898 Knockdown House No rings

46 791149 30/05/1905 Building Blocks for

Cylindrical Structures

No rings for the

hemisphere

47 3359694 26/12/67 Domical Building

Structures

Discussed

Walter R. Hein designs a dome given the number of vertices on the dome. Quoting, we

have:

… once the pattern of vertices is established so that the total number of vertices in

the imaginary spheroid is known, the polyhedral angle for each vertex is

calculated by the present invention. Essential to this calculation is an

understanding that in accordance with the present invention, the sum of all such

polyhedral angles of the spheroid is 720° less than the product of 360° times the

total number of vertices on the spheroid, i.e., the polyhedral angles are less than

360° each by an average amount equal to 720° divided by the total number of

vertices.

If we change the number of vertices even by 1 by deleting a vertex or adding a vertex we

need to redesign the complete dome. With 8 vertices with one vertex at the top, one

vertex at the bottom, and 3 vertices each in two planes in between, we get a cube as a

sphere. Our domes are smooth.

48 3485000 23/12/69 Dome Structure and

method for building the

same

Discussed

I claim: 1. A dome structure comprising:

A plurality of superposed edge-overlapping concentric ring members of gradually

smaller diameter, …

In our design we have rings. However their diameters do not gradually decrease. The above

description of gradually decreasing diameter members would form a spiral when connected as

described in the patent.

49 4027449 07/06/77 System for Constructing

Spatial Structures

Spherical nucleus

between struts

50 4233794 18/11/80 Adjustable Connecting

Means for Ceiling

Constructions and the like

Connector; not

related to dome

shape



51 4357118 02/11/82 Connecting Assembly for

Geodesic Dome

Framework Construction

Connector not dome

52 4491437 01/01/85 Connector for Geodesic

Dome

Two concentric

geodesic domes one

inside another

53 4566818 28/01/86 Ledger Hanger for

Geodesic Domes

Discussed

Second floor inside the dome without pillar from ground transferring the load to the

dome. Our dome need not carry even its own load.

54 4608789 02/09/86 Star Dome Structure Neither hemisphere

nor hemi-ellipsoid

55 4679361 14/07/87 Polyhedral Structures that

approximate a Sphere

Neither sphere or

ellipsoid

56 4796394 10/01/89 Geodesic Structure Growing dome

57 4923544 08/05/90 Method of manufacturing

tetrahexogonal truss

structure

Parallelogram

58 4945603 07/08/90 Concentric Dome Energy

Generating Building

Enclosure

Two concentric

domes

59 5040966 20/08/91 Die for making a

Tetrahexagonal Truss

Structure

Parallelogram

60 5070673 10/12/91 Tetrahexagonal Truss

Structure

Cavity,

Parallelogram

61 5230196 27/07/93 Polyhedron Building

System

Cable

62 5305564 26/04/94 Hemispherical Dome

Building Structure

[lapsed]6

Cells are same size;

frustopyramidal

shape

63 5782035 21/07/98 Multi-purpose automatic

filling and leveling fluid

basin with water transfer

Unrelated

64 5918415 06/07/99 Multi-purpose self-

watering system

Unrelated

65 7464900 16/12/08 Folding Retractable

protective dome for space

vehicle equipment

Collapsible ribs

66 7624705 01/12/09 Small animal habitat Unrelated

6 http://www.google.com/patents/US5305564

http://www.google.com/patents/US5305564



67 7735265 15/06/10 Foam rigidized inflatable

structural assemblies

Inflatable

68 8397678 19/03/13 Small animal habitat Unrelated

69 8782965 22/07/14 Universal Hub and Strut

system for a Geodesic

enclosure

Universal links to

make domes; no

dome design detail

We consider the invention in detail now.


Invention in detail

We consider 40’ x 30’ first. As 5’ clearance is required from 3 sides, the largest circle has

a radius of 10’. We provide a 3’ x 3’ foyer at the 8’ above ground landing for the 10’

radius habitat. The lower floor has a full circle of 10’ radius with a uniform height of 7’8”

which gives the height of the lower floor to be 8’ including the 4” thickness of the

ceiling/floor concrete slab. Where should we place the kitchen, dining, and living space?

The 10’ radius hemisphere has reduced space where a 6’ high individual could stand

straight. This space has a radius of 8’ only.

The occupants of the habitat could tolerate slight inconvenience by sleeping under the

hemisphere. We decide to provide kitchen, dining, and living spaces at the lower level.

The layout of the First Floor is given in Figure 2. [1 is the master bed of size 6’ x 5’; 2

child bed 6’ x 3’; 3 Wash Basin; 4 bath tub with shower; 5 western commode; 6 flush

tank; 7 entry to the apartment; 8 black circle with 0’ clearance; 9 blue circle with 2’

clearance; 10 magenta circle with 4’ clearance; and 11 green circle with 6’ clearance.]

Figure 3 [1 common wash basin, shared by the family and the visitors; 2 stove and oven;

3 spice rack; 4 refrigerator and freezer; 5 washer dryer; 6 six seated dining table; 7 three

seated sofa; 8 four seated sofa; 9 television and other entertainment systems; 10 powder

room under the stairs; and 11 open due to stairs] depicts the arrangement in the lower

floor. In the Figure the space allotted for wash basin; stove and oven; refrigerator and

freezer; and washer dryer are large enough to accommodate the current non-circular

appliances. Having established the feasibility of a couple with two children living inside a

10’ radius hemisphere we consider the plot use alone in what follows.

The area of the floor of a 10’ radius hemisphere is 314 sq ft. The area of a 40’ x 30’ plot

is 1,200 sq ft. The FSI [Floor Space Index] of a home with the lower and upper floors is

(314 + 314)/1,200 = 0.52 only. Accordingly we build the lower floor on ground without

digging below the ground level. To protect from floods up to a height of 8’ we maintain a

garden around the lower level of the house. Instead of providing a door to enter the living

area at the ground level, we enter at the upper level and have provided the internal

staircase. The garden and the steps outside the house to reach the upper floor are depicted

in Figure 4. The elevation is along the line cutting through the steps. The plan shows two

car parks marked by number 1.

Positioning of the 15’; and 25’ hemispheres are displayed in Figures 5 and 6 respectively.

The area under the 15’ hemisphere is 706 sq ft in each floor giving a total of 1412 sq ft.

For the area of 2,400 sq ft, the FSI is 0.58. In contrast to a 15’ radius hemisphere, we

have two floors in addition to the floor at ground for a 25’ hemisphere. The magenta

circle marked 1 represents the area shielded by the garden and the First Floor through

which entry is made. These two floors have a plinth of 1,963 sq ft. We could have a floor

at a height of 8’ and 16’ above the entry level. These have floors with radii of 23.68’ and

19.21’ with plinths of 1,762 sq ft and 1,159 sq ft respectively. The red circle is the floor


with 23.68’ radius marked 2 in the figure. The black circle marked 3 represents the

19.21’ radius floor. The total plinth for the 4 floors is 1,963 x 2 + 1,762 + 1,159 sq ft

adding to 6,847 sq ft. The plot area being 60’ x 80’ = 4,800 sq ft, the FSI is 1.42. We are

justified in having a floor on the ground shielded by the garden.

In a plot of 80’ x 120’ we could have just one hemisphere of 35’ radius. In addition to the

floor guarded by the garden, we have floors with 35’; 34.07’; 31.12’ and 25.47’ radii. The

plinths are 3,848; 3,848; 3,647; 3,044; and 2,038 sq ft giving a total of 16,425 sq ft of

plinth. The FSI is 16,425 / (80 x 120) = 1.71. The 35’ radius floor is shown in magenta;

34.07’ radius floor in red; 31.12’ radius floor in blue and the 25.47’ radius floor in black

in Figure 7. In addition we have six car parks. We add more hemispheres instead of car

parks as shown in Figure 8. In addition to the 35’ radius indicated by 1, 2, 3, and 4, we

added two floors of 15’ radius marked 5 and two floors of radius 10’ indicated by 6. The

total plinth now is 16,425 + 1408 + 628 = 18,461 sq ft. The FSI increases to 1.92.

A plot of 120’ x 120’ gives an FSI of 2.50 with 8’ high floors and a FSI of 2.07 with 10’

high floors. We recommend the 10’ high floors permitting large distances between

pillars. With a height of 8’ the beams cannot be deeper than 12” for a clearance of 7’ for

the occupants. However the beams could be 36” deep offering the same 7’ clearance for

the occupants of 10’ high apartments.

A 50’ radius hemisphere in a 120’ x 120’ plot has 7,853 sq ft at a height of 10’; 7,539 sq

ft at 20’ height; 6,597 sq ft at 30’; 5,026 sq ft at 40’; and 2,827 sq ft at 50’ height for a

total plinth of 29,842 sq ft. For protection from floods we leave the ground completely

for car park. All apartments are above 10’ from ground. The layout of the 120’ x 120’

plot is shown in Figure 9.

A family might build a super luxury home at Rs. 10,000 per sq ft and build it for Rs.

29,84,20,000 or Rupees 30 Crores. We consider building 21 family apartments and 37

single person apartments with quality at Rs. 5,000 per sq ft. A young couple would have

a 300 sq ft family apartment at an affordable cost of Rupees 15 lakhs while a single

person has a 100 sq ft apartment costing just Rupees 5 lakhs excluding the cost of

common spaces.

A sample layout of the family apartment is given in Figure 10. We use a bay of 15’ width

and allocate 20.59’ for family apartment as that happens to be the smallest family

apartment in the hemisphere with family apartments. The family apartment [ 1 3’ x 6’

door; 2 3’ x 4’ window starting 4’ above the floor; 3 3’ x 4’ window starting 2’ above the

floor; 4 4-seat dining table; 5 3-seat sofa; 6 2-seat sofa; 7 TV; 8 refrigerator; 9 double

bed; 10 4’ x 4’ window starting 2’ above the floor; 11 safe; 12 bath; 13 western

commode; 14 Wash basin; 15 Microwave oven; 16 cooking space with gas/ electric

stove; and 17 sink] is shown.


Figure 11 gives the layout of a person apartment with 108 sq ft plinth, which is the

smallest person apartment in the hemisphere [ 1 2’ x 6’ door; 2 1’ x 3’ window from 3’6”

above the floor; 3 two-seat dining table; 4 TV above the dining table; 5 3’ x 6’ single bed;

6 refrigerator; 7 wash basin; 8 shower; and 9 Western Commode ].

We find that a total of 24 cars could be parked as depicted in Figure 12. Out of these 10

cars do not have free access. These restricted access parks cannot be sold alone. For

example, park 23 could be owned with park 12 only. Any restricted car park shown in red

in Figure 12 could be owned along with parks that lead to access only. Unsold restricted

access car parks could become flower beds or children playground if properly barricaded.

There are 6 dotted circles with radii 50.00’; 48.98’; 45.82’; 40.00’; 30.00’; and 18.90’in

Figure 13. The 50.00’ radius magenta circle is the floor at a height of 10’ above the car

park. The 48.98’ radius violet circle is the ceiling of the First Floor and floor of the

Second Floor. The ceiling of the Second Floor and the floor of the Third Floor is the blue

circle with radius 45.82’. It is enough to place the apartments within the ceiling circle to

maintain vertical walls. Still no apartment is outside the 45.00’ radius full red circle. This

is because there are no pillars outside the 45.00’ radius circle and we desire to avoid

loading the cantilever supported parts of the hemisphere. The 40.00’ radius orange circle

is the ceiling of the Third Floor and floor of the Fourth Floor. The green circle with

30.00’ radius is the terrace and the ceiling of the Fourth Floor.

The First Floor and the Second Floor have all the 17 apartments numbered 3 to 19. The

Third Floor has 13 apartments numbered 3 to 15. The apartments numbered 3 to 9, a total

of 7 apartments only are found in the Fourth Floor. The plinths of these apartments are

listed in the following table.


No Apartment No First Second Third Fourth

1 3 175 sq ft 175 sq ft 175 sq ft 175 sq ft







8 10 179 sq ft 179 sq ft 179 sq ft

9 11 179 sq ft 179 sq ft 179 sq ft

10 12 179 sq ft 179 sq ft 179 sq ft

11 13 179 sq ft 179 sq ft 179 sq ft

12 14 172 sq ft 172 sq ft 172 sq ft

13 15 129 sq ft 129 sq ft 129 sq ft

14 16 108 sq ft 108 sq ft

15 17 108 sq ft 108 sq ft

16 18 172 sq ft 172 sq ft

17 19 129 sq ft 129 sq ft

Floor total 2,572 sq ft 2,572 sq ft 2,055 sq ft 1,038 sq ft

Hemisphere total 8,237 sq ft

It could be seen that all the apartments are less than 200 sq ft and hence are for an

individual only. We could combine more than one person apartment and use it as family

apartment provided the area of an apartment is at least 200 sq ft. We expect that a just

married couple would have no problem in living in a 200 sq ft apartment. Couples with a

few years behind them might need larger apartments.

To demonstrate the feasibility of different floor layouts in different levels of the

hemisphere, we show different layouts with family apartments. It must be recorded that

some apartments could not be combined and thus would be retained as person apartments

only.

In Figure 13 we see that apartments 4 and 5 share a wall. By removing this wall, we get a

larger apartment. Likewise we get family apartments by combining apartment 6 with 7;

and apartment 8 with 9. The result is shown in Figure 14. To demonstrate that the

apartments in different floors could have different plinth distributions, we combine

different person apartments in Third Floor. The Third Floor layout is found in Figure 15.

For simplicity, we retain the same layout for the Second Floor and the First Floor, shown

in Figure 16. The apartments and their plinths of the hemisphere with family and person

apartments are listed in the table below.


No First Floor Second Floor Third Floor Fourth Floor

No Plinth sq ft No Plinth sq ft No Plinth sq ft No Plinth sq ft

1 3 355 3 355 3 355 3 175

2 4 108 4 108 4 355 4 348

3 5 355 5 355 5 172 5 258

4 6 172 6 172 6 129 6 258

5 7 129 7 129 7 308

6 8 308 8 308 8 129

7 9 108 9 108 9 308

8 10 129 10 129 10 172

9 11 308 11 308 11 129

10 12 172 12 172

11 13 129 13 129

12 14 172 14 172

13 15 129 15 129

Floor total 2,574 2,574 2,057 1,039

Hemisphere total 8,244

The difference of 7 sq ft between the totals [8,237 as single person apartments and 8,244

as mixed apartments] is due to ignoring the fractions of the plinths of the apartments.

We consider the person apartments only for the location of the pillars, beams, pillar-

beams now. Pillars from ground are permitted only when the pillar does not affect the

movement of cars on ground. However the lift and the stairs are essential and we have

bifurcated one park way and located them. The pillars that support the lift are at

(48.5’,60’) where the X coordinate is 48.5’ and the Y coordinate is 60’. The X coordinate

is zero at the left and increases in the right direction. The Y coordinate is zero at the

bottom and increases upwards. The other three lift pillars are: (56.5’, 60’); (48.5’, 68’);

and (56.5’, 68’). The stairs are supported by the lift pillars and the four pillars: (45’, 60’);

(45’, 71.5’); (60’, 60’); and (60’, 71.5’). These pillars are not shown in Figure 17 in

which we show another 40 potential pillars. The pillars that have the same X coordinate

have their number displayed for the bottom most and the top most pillars. The X and Y

coordinates along with their heights of the pillars are listed in the table below. The pillars

reach the floor of the circle enclosing the pillar. The table lists a total of 66 pillars

including the pillars that were added as the design continued till end.

No X coordinate Y coordinate Height

1 20.00’ 39.38’ 30.00’

2 15.40’ 54.00 30.00’

3 15.00’ 60.00’ 30.00’

4 20.00’ 80.61’ 30.00’

5 39.00’ 20.20’ 30.00’

6 39.00’ 45.39’ 50.00’

7 39.00’ 54.00’ 50.00’



8 39.00’ 60.00’ 50.00’

9 39.00’ 71.50’ 50.00’

10 39.00’ 99.80’ 30.00’

11 60.00’ 15.00’ 30.00’

12 60.00’ 17.57’ 30.00’

13 60.00’ 24.80’ 40.00’

14 60.00’ 36.78 50.00’

15 60.00’ 45.39’ 50.00’

16 60.00’ 54.00’ 50.00’

17 60.00’ 83.21’ 50.00’

18 60.00’ 95.19’ 40.00’

19 60.00’ 102.42’ 30.00’

20 60.00’ 105.00’ 30.00’

21 79.00’ 19.20’ 30.00’

22 79.00’ 24.89’ 40.00’

23 79.00’ 36.78’ 50.00’

24 79.00’ 45.39’ 50.00’

25 79.00’ 54.00’ 50.00’

26 79.00’ 60.00’ 50.00’

27 79.00’ 71.50’ 50.00’

28 79.00’ 83.21 50.00’

29 79.00’ 95.19 40.00’

30 79.00’ 100.79 30.00’

31 100.00’ 39.38’ 30.00’

32 100.00’ 45.39’ 30.00’

33 100.00’ 54.00’ 40.00’

34 100.00’ 60.00’ 40.00’

35 100.00’ 71.50’ 30.00’

36 100.00’ 80.61’ 30.00’

37 104.00’ 50.56’ 30.00’

38 104.59 54.00’ 30.00’

39 105.00’ 60.00’ 30.00’

40 104.00’ 89.43’ 30.00’

The following pillars are added for earthquake protection.

41 39.00’ 24.80’ 30.00’

42 39.00’ 36.78’ 40.00’

43 39.00’ 83.21’ 40.00’

44 39.00’ 95.19’ 30.00’

45 20.00’ 45.39’ 30.00’

46 20.00’ 71.50’ 30.00’

The following virtual pillars were added for Fourth Floor

47 26.25’ 81.47’ 30’ 40’



48 39.00’ 89.20’ 30’ 40’

49 60.00’ 100.00’ 30’ 40’

50 79.00’ 95.19’ 30’ 40’

51 95.40’ 78.61’ 30’ 40’

52 100.00’ 60.00’ 30’ 40’

53 96.67’ 44.02’ 30’ 40’

54 79.00’ 24.80’ 30’ 40’

55 60.00’ 20.00’ 30’ 40’

56 39.00’ 30.79’ 30’ 40’

57 22.76’ 45.39’ 30’ 40’

58 20.00’ 60.00’ 30’ 40’

To support the Terrace we add the following virtual pillars

59 30.00’ 60.00’ 40’ 50’

60 39.00’ 81.42’ 40’ 50’

61 60.00’ 90.00’ 40’ 50’

62 90.00’ 60.00’ 40’ 50’

63 89.39’ 54.00’ 40’ 50’

64 60.00’ 30.00’ 40’ 50’

65 39.00’ 38.57’ 40’ 50’

66 30.60’ 54.00’ 40’ 50’

To facilitate the drive way in the car park, no pillars were allowed outside the circle of

45’ radius. However the First Floor has a 50’ radius floor. We support the additional 5’

by cantilevers. There are main beams at a distance of 19’ and 21’. Normally the support

beams are drawn perpendicular to the main beams. The cantilevers are drawn radial to

minimize the length of the cantilever. The beams supporting the First Floor with a radius

of 50.00’ are shown in Figure 18. The green beams join the adjacent pillars on the 45’

radius circle. They are sixteen in number. The longest of these beams connects pillars 30

and 36 and has a length of 29.12’. A beam of length 30’ is expected to have a depth of

30”. Since we have 10’ between floors, we decide to have uniform depth of all beams to

be 36” allowing the longest beam to be 36’ in length. We need to add additional pillars to

reduce the beam length if necessary.

Normally we run beams from pillar to pillar. In Figure 18 these beams are shown in

green. We have major beams – 2 in the X direction and 5 in the Y direction – that also

run pillar to pillar. However these are shown in red to emphasize that they are major

beams. Every apartment has a non-load bearing brown beam. These brown beams are

beam to beam connections and support the weight of the partition wall only. These

partition walls could be as light as two plywood sheets with fluffy cotton in between for

sound insulation. It is found that the interconnections between the pillars and beams form

a number of rectangles. If a tornado creates a twist on the structure, the indeterminate

rectangles become parallelograms and the floor does not remain circular. For the floor to

maintain its shape we introduce a number of beams shown blue in Figure 19. We have


introduced 37 blue beams just to retain the shape of a floor in case an earthquake occurs.

Are we justified in increasing the cost? It is estimated that an earthquake could reduce the

GDP of a nation.7 At least for the future constructions we should introduce additional

beams to convert rectangles to triangles so that the GDP of the nation does not suffer due

to an earthquake. We strongly feel that the additional cost due to the beams for

earthquake protection is highly justified.

Under an earthquake a floor of a building could slip with respect to the floor above or

below that floor. Even though the floor retains its shape during an earthquake, such

slippage between floors results in destruction of the building. We add inclined pillar-

beams between the adjacent floors to avoid such slippage. Figure 20 shows the 4 pillar-

beams between the First Floor and the Second Floor along a main beam in the X

direction. The pillar-beams connect the bottom of pillar 2 with the top of pillar 7. This

pillar beam crosses a service lane near the floor. We expect the service personnel to jump

over the exposed pillar-beam. The rest of the pillar-beam is inside the partition wall of

Apartment 15 adjoining the corridor. The next pillar-beam starts at the top of pillar 7 and

connects the bottom of pillar 16. The exposed part of this pillar-beam in the corridor is

above 7’ and thus causes no difficulty to the occupants. Likewise the other 2 pillar-beams

do not cause any difficulty. These 4 pillar-beams divide the vertical rectangle into 8

triangles and these triangles resist any slippage between the floors of First Floor and

Second Floor in the X direction. Similar pillar-beams are placed between pillars (41, 13,

22); (42, 14, 23); (45, 6, 15, 24, 32); (46, 9, pillar at the corner of the stairs, 27, 35); (43,

17, 28); and (44, 18, 29).

To avoid slippage along the Y direction, we place pillar-beams using a different strategy

to ensure proper placing of doors and windows of the apartments using the corridor. We

show the pillar-beams along the corridor connecting pillars 21 to 30. On this corridor are

apartments 11, 10, 6, and 5. The pillar-beams are shown in Figure 21. There are no

apartments between pillars (21, 22) and (29, 30). For the other 6 apartments the entry is

in the triangles shown in red. The windows are in the triangles shown in blue and green.

The doors and windows are sized such that they do not touch the pillar-beams. Like these

4 apartments, apartments 14 and 15 have three triangles at the entry. The other

apartments have the main beam running at the back of the apartment. The windows of the

kitchen and bathroom of the remaining apartments are adjusted after deciding the pillar-

beams.

The First and Second Floors have all apartments because the ceiling of First Floor has

radius of 48.98’ greater than 45’. Likewise the ceiling of Second Floor having a radius of

45.82’ is greater than 45’. Thus both the First Floor and the Second Floor have all 17

apartments numbered 3 to 19 When it comes to the Third Floor, the ceiling of Third Floor

has radius of 40’ only. Because this is less than 45’, the Third Floor has 13 apartments

7 http://www.stuff.co.nz/business/rebuilding-christchurch/4984173/Quake-rebuild-will-eat-into-GDP

Reconstruction after the Canterbury earthquakes is likely to eat up about 7.5 per cent of 2011 GDP, according to the International Monetary Fund.

http://www.stuff.co.nz/business/rebuilding-christchurch/4984173/Quake-rebuild-will-eat-into-GDP


only numbered 3 to 15. All the pillars are within the 45’ radius and the Third Floor is

fully supported.

When we consider the Fourth Floor with a radius of 40’, the pillars 1, 2, 3, 4, 45, 46, 5,

41, 44, 10, 11, 12, 19, 20, 21, 30, 31, 32, 33, 35, 36, 37, 38, 39, and 40 do not support the

Fourth Floor. The result is the Fourth Floor needs a cantilever of 20’ if we require all

pillars to start from ground. A cantilever of 20’ is not acceptable.

A beam could support distributed and concentrated loads. As a beam could support

concentrated load also, we have the freedom to start a pillar above the ground level sitting

on a beam. We call such pillars ‘virtual pillars.’ The virtual pillars that support the Fourth

Floor start from the beams of the Third Floor. These virtual pillars are shown in Figure

22. These virtual pillars are numbered 47 to 58. The virtual pillars 48 and 56 do not lie on

the 40’ radius circle. As the apartments of the Third Floor are within the 40’ radius circle,

we need to check whether virtual pillars 48 and 56 obstruct any apartment. From Figure

13 we find that there are no apartments in the places occupied by the virtual pillars 48

and 56. The beams supporting the Fourth Floor are shown in Figure 23. Any beam from a

virtual pillar is shown in brown,

Similar to the Fourth Floor, the Terrace also needs virtual pillars. The virtual pillars

numbered 59 to 66 along with the pillars from the ground inside the 30’ radius circle

support the Terrace. The virtual pillars and the beams that support the Terrace are given

in Figure 24.

Now we consider the rich who normally live in ‘Farm Houses.’ The attraction for living

in such houses is the greenery that surrounds the house on 4 sides. If the occupants

maintain a terrace garden the house is surrounded by greens on 5 sides. In a multistory

hemispherical complex we provide greens on all six sides of the house. We use 40’ wide

bay for the apartments. One set of pillars support an apartment and the entry corridor of

6’. Another set of pillars support the 4’ wide service lane and an apartment of 40’. Thus

the main beams are at a distance of 46’ and 44’ alternately.

We design the support beams to be 54” deep permitting the support beams to have a

maximum length of 54’. To provide a clearance of 7’ 6” for the occupants, we keep the

floor at heights of 12’ each. The design of the pillars and support beams is similar to that

of the 50’ radius hemisphere just described. We do not consider the design of pillars and

beams for the sake of brevity.

We consider the location of the apartments only now. To avoid the possibility of the

plinth of the apartments decreasing as we move up, we start the design with the top floor.

We have selected a hemisphere of 200’ radius. The terrace is the 15th

floor at a height of

180’ above the bottom of the hemisphere. The hemisphere starts at a height of 12’ leaving

a gap of 12’ for the car park. As the main beams have a minimum gap of 44’ we do not

design the car park as it would be very easy to design it for those who design car parks.


There are no apartments in a bay adjoining a bay with apartments to ensure that the

apartment has greens in front and back. We leave a gap of 12’ between apartments in the

same bay so that the apartments have greens on all the 4 sides in the same floor. The bay

above and the bay below a bay with apartment are left for greens to satisfy the

requirement that all 6 sides of an apartment has greens.

As the ground is used for car park, the bottom most of the floors, called First Floor at 12’

above ground is fully used for greens. The apartments in the top floor, Fourteenth Floor

are given in Figure 25. As a luxury apartment we desired to have at least 3,200 sq ft of

plinth. At the Fourteenth Floor there are only two apartments of plinth 1,600 sq ft each in

the displayed quadrants. Only one of the two could be selected. We arbitrarily chose the

one in the first quadrant in preference to the one in the fourth quadrant. We show the

apartments in the first and fourth quadrant only. There are apartments in the second and

third quadrants that are symmetric to those shown in the figures. As 1,600 sq ft is small,

we hope to link the apartment in the Twelfth Floor with the one identified in the

Fourteenth Floor. Hence the 1,600 sq ft apartment in the Fourteenth Floor is called ‘A1

top.’

Figures 26 and 27 display the apartments in the Thirteenth and Twelfth Floors. On the

Twelfth Floor we find ‘A1 bottom’ in the first quadrant with a plinth of 3,200 sq ft.

Combining ‘A1 top’ and ‘A1 bottom’ the plinth of apartment A1 is 4,800 sq ft making

apartment A1 possibly a super luxury. Apartment A3 is a trapezoid and not a rectangle.

As a rectangle it would have just 2,800 sq ft area. We decided to relax the rule of having

a rectangle, hoping that such occasions would be rare. Indeed apartment A3 is the only

one which is not a rectangle. It has a plinth of 3,000 sq ft.

Figures 28 to Figure 36 display the apartments in Eleventh Floor to Third Floor.

Apartment 32 in the Third Floor has a plinth of 2,395 sq ft only. Still it is not called ‘A32

top’ as the First Floor is reserved for Flower Bed only. Likewise the Second Floor

displayed in Figure 37 also has no top part of any apartment. Apartment 40 is displayed

in Figure 37. The total number of apartments in the first and fourth quadrants is 40 giving

the total number of apartments inside the 200’ radius hemisphere to be 80. The plinths of

these apartments are listed in the following table.

No Apartment No Symmetric Apartment No Plinth sq ft Bottom plinth sq ft

1 1 41 1,600 3,200 (4,800)

2 2 42 3,200

3 3 43 3,000

4 4 44 3,200

5 5 45 3,200

6 6 46 1,600 3,200 (4,800)

7 7 47 3,200

8 8 48 3,200

9 9 49 3,200

Weather-Secure Habitat 22 0f 27

No Apartment No Symmetric Apartment No Plinth sq ft Bottom plinth sq ft

10 10 50 3,200

11 11 51 1,600 2,400 (4,000)

12 12 52 1,600 3,200 (4,800)

13 13 53 3,200

14 14 54 1,600 2,400 (4,000)

15 15 55 3,200

16 16 56 3,200

17 17 57 1,600 1,600 (3,200)

18 18 58 3,200

19 19 59 3,200

20 20 60 3,200

21 21 61 3,200

22 22 62 1,600 2,400 (4,000)

23 23 63 3,200

24 24 64 3,200

25 25 65 2,400 2,400 (4,800)

26 26 66 3,200

27 27 67 3,200

28 28 68 3,200

29 29 69 2,400 2,400 (4,800)

30 30 70 3,200

31 31 71 1,600 2,400 (4,000)

32 32 72 2,400

33 33 73 3,200

34 34 74 2,400

35 35 75 3,200

36 36 76 3,200

37 37 77 3,200

38 38 78 3,200

39 39 79 3,200

40 40 80 2,400

1,11,000 25,600

Quadrants 1 and 4 1,36,600 sq ft

All 4 quadrants 2,73,200 sq ft

We have used 200’ radius on the ground. The area of circle with 200’ radius is 1,25,663

sq ft. If the rules permit building on a circular plot, then FSI = 2,73,200 / 1,25,663 = 2.17.

Given that the FSI is 2 in most places, we need to add 10,937 sq ft of plot to the circular

plot. If we take a square plot of 400’ side, the FSI is 2,73,200 / 1,60,000 = 1.70. We have

used pillars at a distance of 46’ or 44’ using beams of depth 54” or 4’6”. If we have used

pillars at 21’ or 19’ distance, we use beams of depth 2’ only. Then every floor is just 10’

high. We would have 18 floors instead of 14. With the Ground Floor dedicated for car


park and the First Floor for flower beds, we have apartments in 12 floors now. With 18

floors, we would have apartments in 16 floors. Assuming a linear relation between the

number of floors and the area of the apartments, the FSI of 1.70 becomes 1.70 x 16 / 12 =

2.26. We conclude that a hemisphere of radius 200’ offers luxury apartments comparable

to the rectangular prisms. The major difference is the rectangular prism is not as strong as

the hemisphere with respect to earthquakes, floods, hurricanes, and tornadoes.

The apartments described from the smallest with a 10’ radius hemisphere to the 200’

radius hemisphere need to be provided with the hemispherical dome. We detail the steps

involved in designing the dome.

The smallest dome has a radius of 10’ and a surface area of 628.31 sq ft. Figure 38 shows

the plan of a cloth that provides for half the hemisphere. We take another cloth that is a

mirror reflection of the wing shaped cloth. By first stitching the two vertical lines

together we get a butterfly shape. In the butterfly stitch the points with the same color.

We get a closed shape with the bottom open. Arrange the bottom on a 10’ radius circle

and inflate. The cloth would assume a hemispherical shape. The area of the single wing

shown is 312.61 sq ft close to the actual 314.15 sq ft. The difference is because of the

straight line connections between the points. We have added the two cones and their

cloths nearby. These cones are given to show that the procedure used to get the

hemisphere’s cover works for other shapes as well as long as the shapes have a circular

cross section on every horizontal plane. Monolithic Domes are made by preparing air

forms like the one shown in Figure 38 and blowing concrete slurry on the cloth from

inside. Such a monolithic dome does suffer some drawbacks.

The air pressure under the cloth must be just right; over pressure causes the shape to

approach an ellipsoid and under pressure leads sagging at the top. To avoid such

problems we build a cage using metallic strips. For ease of manual assembly we use

strips of about 2’ length. The rib of a 10’ radius hemisphere is 15.70’ in length on the

surface of the hemisphere.

The elevation and plan view of the cage is given in Figure 39. Just to demonstrate that the

ribs and the rings could be at arbitrary gaps, we use 2’2” and 1’10” alternately for the

location of the rings. The ribs are at 2’2” and 1’10” at the bottom and they shrink

proportionately as they approach the top.

In the plan view of Figure 39 a scaffold is shown in blue. A worker standing on this

scaffold cannot reach any of the points shown on the dotted orange line or above. To

reach them we need to provide 6 ‘maintenance windows,’ three of which displayed.

Another three in the shadow and are not shown. The plan and elevation are shown in the

same color. These windows are removable so that a worker standing inside the

hemisphere could reach each of the 8 cells touching the maintenance window. Any

damage caused by debris in a hurricane or tornado could be repaired through the

maintenance window. It is possible that a window is designed for some other purpose by


the architect. In that case we do not provide any maintenance window for the cells that

are close to the useful window. The useful window must be removable to perform the

maintenance work after a natural event. Figure 39 shows three maintenance windows in

blue, green, and red. Their plans and elevations have the same color. Through the

windows we reach down to the dotted red line. We could reach the brown line above also

through the windows. The space above the brown line could be reached from the top. The

top is covered by a two piece lid which could be removed if desired. Thus the whole

hemisphere is reachable for repair and maintenance.

We concentrate on making the surface of the hemisphere a true hemisphere. We fix a

cloth of appropriate shape against the desired space; blow air from inside so that the cloth

assumes the spherical gradient all over; and spray cement solution or some other binder

to freeze the cloth as true spherical surface. Once the cloth dries and becomes rigid, we

apply plaster inside to the desired thickness. We recommend leaving the metal strips

inside the plastered wall to form a Faraday Cage offering protection from an EMP

[Electro-Magnetic Pulse] attack as well. If desired the strips could be designed to have

protrusions to pull them out after the plaster has set. Before considering the metallic

strips, we detail the process of preparing the cloth.

Instead of selecting a cell different from the maintenance windows, we describe the

process for the green maintenance window. In Figure 38 we have developed a cloth to

form a 10’ radius hemisphere. For a cone we form a circular segment with radius equal to

the line joining the apex with a point at the bottom. We extend this process to any

structure which has the same distance to all points at a horizontal plane. The intersection

of the structure with every horizontal plane is a circle, with all circles having the same

center in plan. Instead of the straight line connecting the apex with a point on any

horizontal circular intersection, we find the length of the string along the curvature of the

object between the apex and the selected point. In Figure 40 the apex is 1 and the points

of interest are those inside the green polygon in the plan along the dotted red line. The

string length is given by the arc 12 shown in the elevation. The length of the circular

arc between 34 in the plan should be the length of the arc with a center and radius

equal to the string length. To maintain symmetry, we have half of 34 on the left and the

other half on the right. We divide the length along the rib into 10 equal parts, giving 11

horizontal circles. The bottom most circle is the blue circle while the top most circle is

the green one. We list the string length, the arc length, and the subtending angle on each

side corresponding to the half arc length for the 11 rings below.

No Ring arc string Angle

1 Blue 2.786826837388657’ 7.707963267948968’ 10.357691809463495˚

2 Blue + 0.1 2.730981625703501’ 7.515135224880602’ 10.410572557799377˚

3 Blue + 0.2 2.672911662015854’ 7.318242587158437’ 10.463342491101492˚

4 Blue + 0.3 2.612468595145274’ 7.116935359881187’ 10.516002818858988˚

5 Blue + 0.4 2.549483646342933’ 6.910814989896926’ 10.568554728105412˚

6 Blue + 0.5

[Green - 0.5]

2.483763446866371’ 6.699424304112744’ 10.620999384002745˚


No Ring arc string Angle

7 Green – 0.4 2.415084709006854’ 6.482234606217968’ 10.673337930406004˚

8 Green – 0.3 2.343187303234138’ 6.258628885370287 10.725671490409121˚

9 Green – 0.2 2.267765115497280’ 6.027879602208813’ 10.777701166873005˚

10 Green – 0.1 2.188453745173699’ 5.789118746146116’ 10.829728042936114˚

11 Green 2.104813593915855’ 5.541296601282292’ 10.881653182509153˚

The cloth to cover the green window is shown in red in Figure 41. The interface between

the blue and the red region looks straight. Is that a straight line? There is a very slight

curvature giving a distance of 0.00719902394601’ between the straight line joining the

ends and the middle of the curve. This gap is 0.0863” and is negligible in engineering

terms. In any case the hemispherical shape is produced by the metallic strips that define

the ribs and rings to be described. In practice the edge would be cut as a straight line and

an error so small would not affect the resulting hemispherical surface. The resulting

hemispherical surface would be as true as possible.

The blue and the magenta strips are pasted over the left rib and bottom ring of a selected

quadrilateral. When the next quadrilateral of the ring in the right side gets the blue strip

pasted, the green strip is pasted on the cloth of the next quadrilateral. Thus we paste the

blue, green, and magenta strips of all the quadrilaterals of a ring. The orange strips are

clipped to the metallic strips of the upper ring. Now we blow air on the cloth forcing the

cloth to assume the true hemispherical curvature. While the airflow is maintained, we

spray the cement solution or some other drying solution on the cloth. We apply plaster to

more than half the depth of the metallic strips. If we desire to reuse the metallic strips, we

paste the full hemisphere to half depth of the strips, allow the hemisphere to dry and

harden, and pull the metallic strips out. The plaster remains in place because we have

allowed holes on the outer edges of the metallic strips permitting the plaster to be

connected across the strip.

The first rib strip connected to the foundation strip is shown in Figure 42. The three 1”

diameter holes at the bottom are used to align the first rib strip with the vertical

foundation strip. There is a 0.1” cut half way at the top of rib in the horizontal direction to

accommodate a 0.2” thick ring part. The three 1” diameter holes at the top of the rib strip

are used to align the next rib strip to achieve true 10’ radius hemisphere rib structure. The

lowest two ribs are shown in Figure 43. All other rib and ring elements are formed

likewise. The smallest hemisphere with 10’ radius could be formed with bamboo strips as

well.

Erect a temporary pole in the center with a circle of radius 1.5253’ of cane. Take 14’2”

long thin bamboo strips and attach one end with the circle at the top and the other end at

the corresponding point at the edge of the 10’ radius floor slab. The strip would assume

the shape of a circular curve of 10’ radius with the centre at the centre of the floor. Place

as many strips as desired with spacing of 6” or about at the edge of the floor. Use ropes or

bamboo strips to form rings tying the bamboo strips to form rings. Once the cage is

formed using the bamboo ribs and bamboo or rope rings, the circle at the top would be


held in position by the hemisphere. Remove the temporary pole and cover the top hole

with a circular lid woven with bamboo strips. Cut windows and doors as required. Cover

the hemisphere with waterproof cloth all over except the door and windows. Use curtains

for the door and the windows.

For hemispheres with two or more floors use the metallic or bamboo strips of appropriate

length to form that section of the cage between the floor and ceiling. Plaster or cover with

waterproof cloth as desired.

In the description we have the skin enclosing the building inside. It is very easy to

visualize a stricture with floors enclosing the described circular floors and providing the

skin as described on the circles in the floors and ceilings with extended floors jutting

outside the skin. Such trivial extensions do not exempt the buildings from the protection

offered to this patent.


I Claim:

1. We claim a hemispherical or hemi ellipsoidal habitat of at least two levels with or

without plants inside meant for human occupancy.

2. The hemispherical or hemi ellipsoidal skin of the habitat of claim 1 could touch or

not touch the internal building, but carries no load of the internal building.

3. The skin of claim 2 could be with holes allowing natural ventilation or without

holes allowing ventilation through windows only.

4. The radius of the hemisphere or of the hemi ellipsoid of claim 1 could be from 10’

upwards.

Ibramsha Mohideen


Abstract

Due to Global Warming the future weather is not expected to be similar to the past.

Earthquakes, hurricanes, tornadoes, and floods could occur anywhere on earth. We detail

secure habitats for the hostile weather.

Tornado has no effect on buildings with circular exteriors. A hurricane finds it extremely

difficult to uproot a hemisphere or a hemi ellipsoid in which the radius of an upper floor

is less than that of a lower floor. Floods do not affect a building with entry above ground

when the flood height is less than the entry. A building with triangles formed by pillars

and beams is more resistant to earthquake than without any triangle.

Our buildings have above the floor entry; have circular exteriors; have triangles in the

structural frame; and are hemisphere or hemi ellipsoid shaped with the radius of an upper

floor being less than that of a lower floor.

Description of the Patent Grant awaiting Weather Secure Agri Habitat Design by Dr.Mohideen Ibramsha

Engineering

Transcript of Description of the Patent Grant awaiting Weather Secure Agri Habitat Design by Dr.Mohideen Ibramsha