- Energy Structures, Inc.
- Earthworks.
- Garden Dome Homes.
- Monolithic Dome.
- Geodesic DomesHomes.
- Natural Spaces Domes.
- Oregon Domes Inc.
- Manufacturers, Vendors, and Consultants.
- Concrete Domes.
- New Age Construction.
- American Ingenuity Domes.
- Timberline Geodesics.
- Graphcomp.
- Geometrica.
- Dome Connections.
- Fuller Dome FAQ
- Reality Sculptors.
- 2V Simple Dome.
- Sci-toys Dome.
- Conduit Dome.
- Dome Math.
- Fuller Dome Math.
- How to Make a Geodesic Sphere Out of Wire.
- Dome Home Messages.
- Onlinemetals Perforated panel.
- Mcnichols.
- Alibaba 43c/m2 Galvanized Perforated Panel
- Alibaba Galvanized Perforated Panel $1 to $6/pcs.
- Alibaba Galvanized Perforated Panel: 1.0 min 20 pcs Model Number CE4-020.
- PVC Perforated Plastic Sheet.
- Clear Polycarbonate Lexan Sheet.
- Aerogel.
- Aerogel_Panels.
- Alibaba Iron aluminum alloy zinc channel.
- Lexan glue.

Materials:

My first geodesic dome:

**How to build a simple Geodesic Dome with two pentagon centers:**

Prepare to cut 20 plastic hexagons and 12 plastic pentagons at the same side C. Attach 5 pairs of hexagons to each center pentagon. One pentagon at each of two center poles of the spherical or geodesic dome.

Each flat side C of 5 pairs of hexagons, and the flat side C of center hexagon are in parallel. Temporary attach these 2 sides with the masking tape on the outside.

Each top (North) two pairs of hexagons having two bottom hexagons which are contacting one top hexagon of each bottom (South) pairs of hexagons at 3 sides C. These 3 outer hexagons form a pentagon with the same side C.

Similarly 2 top hexagons of 2 bottom pairs of hexagons are contacting 1 bottom hexagon of each top pairs of hexagons, and these 3 hexagons also form a pentagon with the same side C. Repeat until you finish 5 pairs of hexagon for each center pentagon.

You can apply hot glue inside the matching sides, then remove the masking tape, and seal them with metal tape on the outside and inside. Remember to leave 1 or 2 pentagonal holes to be used as the flapped door and for easy sealing. If you want to use this geodesic dome as a hot air balloon, leave one bottom center hexagon unseal.

For a perfect sphere, you need to mold each plastic polygon as a contact lense with radius R=15C/(2PI)=2.39C, according to an angle of pentagon is (360/5) =72 degrees, which will have a 3C lenght. For a metal dome you need to cast it, or heat then uniformly hammering each metal hexagon to a fraction of the spherical shape.

The radius rp and altitude Hp of a pentagon are shorter than the radius rh and altitude Hh of a hexagon with the same side C. We skip the strut or radius of pentagon for using the perforated panel. However, we need it to calculate the radius R of a geodesic dome:

Find the radius of a pentagon rp:

C/2=rp*sin(72/2)

**rp=C/1.17557**

C/2=rh*sin(60/2)

rp=rh*sin(30)/sin(36)

**rp=0.85rh**

Altitude of pentagon Hp:

Hp=rp*sin(54)

Hp=(0.85rh)*sin(54)

Hp=0.688*rh

Similarly altitude of hexagon Hh:

Hh=rh*sin(60)

Hh=0.866*rh

Solve for Hp:

**Hp=0.79Hh**

To define the exact radius R of the dome, use a wire to wrape around the dome, and measure the largest perimeter P:

P=42"=2*PI*R

R=42/(2*PI)=6.68"

For each side C equal 3" of the geodesic dome in photo above, we have formula for R of sphere:

R=6.6845/3=2.228C

Compare to the above estimated result from visual check for R=2.39C. We have 2.39-2.228=0.162"=5/32" error.

Find the perimeter of the geodesic dome:

From the altitude Hh of hexagon having radius rh above Hh=0.866*rh, and the altitude Hp of pentagon having radius rp was Hp=0.688*rh :

Hp=rp*sin((180-(360/5))/2)=rp*sin(54)=0.81rp.

P=4Hp + 8Hh+4rp+2C.

P=4*0.79Hh + 8Hh + 4*0.85rh + 2C

P=4*0.79*0.866C +8*0.866C + 4*0.85C+ 2C

P=15.06C

P=2*PI*R

Radius of geodesic dome:

R=15.06C/(2*PI)

**R=2.397C**

The base line BD in FIG5, which connect the 2nd and 3rd (vertices) angles of hexagon or pentagon. This base line and the top 1st (vertix) angle make a triangle.

Half base line of hexagon BHh=C*sin 60°

Base line hexagon diagonal BH=2*C*sin 60°

**BH=1.732*C**

Base line pentagon diagonal BP=2*C*sin 54°

**BP=1.618*C**

Circumference of circle P1 circumscribes abound 5 top hexagon at the bottom sides:

P1=(C+BP)*5=(C+1.618*C)*5=13.09*C

Minimum diameter D1 of above circle is not total arc lengths:

**D1=13.09*C/π**

Circumference of circle P2 circumscribes about 5 bottom hexagons at the bottom sides:

P2 =(C+2C)*5 = 15*C

Minimum diameter D2 of above circle:

**D2=15*C/π**

Do the math, we found that the trigonometric calculations gave an extra 0.007" from the ancient errors on the proportional angles, and PI number is not a constant number.

All connections of the geodesic dome having configuration or sequence from center North or South pole: Pentagon center-6-6-5-1.

Estimate diameter D of the geodesic dome:

P=42"=2*PI*R

Diameter=42/PI

**D=13.37"**

Add the isolated foam boxes with the sealed polycarbonate lexan hexagons and pentagons, or using the aerogel polygons between the perforated galvanized metal polygons for visibility, safety, and lower wind pressure due to Bernouilli and Venturi principles, as we have seen all the airplane noses are round.

The firefighters have had hardtimes to break the car windows which are made of the composite glass, the floating dome will have the same quality, if the heaxagons are also made of colorful composite glass in the triple-hexagonal shape above to prevent fire.

We were ready for the Great Flood Noah from the galactic alignment day 21/12/2012. We can also use this simple dome in hemisphere for the tea room or the social gathering on weekend. If you have this dome in your back yard, you will not worry about hurricanes or flood like Katrina.

1/2 inch EMT conduit connector for the conduit dome. The 3 hole equilateral triangle plate at the bottom, or 3 120 degree angle with 3 hose clamps at 3 sides for fast installation.

If you have 10 hexagons and 5 pentagons as the panels cut out, you can install from the top center pentagon and finish at the base on the ground.

However, you can start from the base upward to the top pentagon center, provided you have one equilateral triangle side C which attaches to each bottom side of the bottom hexagon, to form 5 uniform pentagons to avoid misalignment problems. Two equilateral triangles from two consecutive bottom hexagons form one half hexagon at the base.

You can use a wire or tube in 2C lenght to connect two bottom hexagons which made one half hexagon.

To use the geodesic dome as a basket boat for fishing or for life boat, we may need at leat 5 galvanizd steel EMT tubes, each tube has 3/4 inch diameter, 10 ft long, and fasteners, a total of 50 short tubes at 12" for each side of hexagon, for a 57" diameter dome, 2.375 ft radius.

We have volume of sphere Vs=4*π*(R^3)/3

Volume of hemisphere V=2*π*(R^3)/3 = 2*3.14*(2.375^3)/3

V=28 cubic ft or 0.794 m^3.

The bouyant force Fb of the dome submerging in water having volume V, density of water d, and gravity g:

In English units:

Fb = 28.057 cu ft *(1.936 slug/cu ft)*(32.2 ft/s^2)

Fb = 1749.0 slug ft/s^2 or lbf at 70° C.

For MKS units volume V in m, density d in kg/m^3, and gravity g in m/s^2:

Fb = V*d*g = 0.794*1000*9.81 = 7789 N.

Convert 7789 N for Fb=1751 lbf.

This value 1751 and the calculated value above 1749 is having error +2 by the round up.

With the closed ends tubing and 5 EMT tubes with 10 ft long each tubes for under 5 lb weight, we have about 50 lbf buoyant force, so if this lifeboat is having a pecking hole by collision, you can use the metal tape to repair.

This lifeboat will handle stronger waves, and the inside galvanized steel tube struts and spaces between the outside and inside surfaces of the dome are filled up with foam, so water can not occupy inside the tubes and the polygon spaces .

If they used the basket boats (photo link below) or above domes and stacked them like the hats, they may have more rooms to save lives. Since Titanic had only 20 lifeboats to save 1178 persons from a total of 3327 peoples:

3327 /(1178/20)=56.48 lifeboats.

The lifeboat alone will not sink to the bottom as Titanic.

Formula to convert cubic inch to buoyant force FB in lbs. For number N of cylinders submerging in water having radius R, lenght L in inch:

lbs=N*pi*R*R*L/27.67

For N=6, R=3", L=60".

We have the buoyant force FB=367.86 lbs.

For N=7, R=3", L=96:".

We have FB=686 lbs, with a Trihulls Heptatubes Lifeboat, we can have 2,060 lbs: