Dome Cookbook of Geodesic Geometry
By David Kruschke — Second Edition, 1975
Overview
A practical, hand-lettered workbook that teaches the geometry needed to design and build geodesic domes. It walks from plane trigonometry through spherical trigonometry to the chord factors, dihedral angles, and great-circle relationships that turn a subdivided icosahedron into a buildable structure — every step illustrated with the author’s own diagrams and worked examples, and framed as a “cookbook” the reader follows with pencil, paper, and trig tables.
About this edition
The booklet is entirely hand-drawn — hand-lettered text, hand-drawn geometric diagrams, and hand-written formulas and tables. The page images below are the authoritative artifact; the text under each image is a best-effort transcription (produced with a vision model) provided for search and accessibility. Where a page is primarily a diagram or table, the transcription describes it and points to the image.
Pages
Page 1
Figure: Geometric diagram of geodesic dome geometry with labels for plane trigonometry, chord factors, spherical trigonometry, and dihedral angles.
PLANE TRIG.
CHORD FACTORS
SPHERICAL TRIG.
DIHEDRAL. ANGLES
DIAGRAMS & HELPFUL HINTS…
DOME COOKBOOK OF GEODESIC GEOMETRY
BY DAVID KRUSCHKE
SECOND EDITION
$3.00
Page 2
DOMe COOKBOOK OF GEODESIC GEOMETRY
BY DAVID KRUSCHKE
TO MY CHILDREN, LEE & JOHN
SPECIAL THANKS TO HANK PHILLIPS & WINDWORKS FOR CORRECTIONS & SUGGESTIONS…
PREFACE
THE PURPOSE OF THIS BOOK IS TO SHOW THE ACTUAL DERIVATION OF THE CHORD FACTORS AND DIHEDRAL ANGLES WITHOUT THE USE OF JARGON, CO-ORDINATES, AND STRANGE NAMES. ALSO, THERE HAS BEEN AN ATTEMPT TO USE AS FEW FORMULAS AND TRIG. FUNCTIONS AS POSSIBLE (FIVE SPHERICAL TRIG. FORMULAS AND THE SINE & COSINE FUNCTIONS ARE USED FOR THE CHORD FACTORS - THE LAW OF COSINES, SINE & COSINE FUNCTIONS, AND THE PYTHAGOREAN THEOREM FOR THE DIHEDRAL ANGLES).
UNLIKE SOME OTHER DOME BOOKS, THE CHORD FACTOR RESULTS HERE ARE IN CLOSE AGREEMENT WITH BUCKY FULLER’S (A FACT THAT IS IMPORTANT WHEN WORKING WITH THREE FREQUENCY DOMES - SEE PAGE 8).
CONTENTS
- HOW TO PROCEED … P1
- TRIG. REVIEW … P3
- FREQUENCY … P5
- SPHERICAL TRIG. … P10
- CHORD FACTORS … P.12
- COMPARISON WITH FULLER…P.29
- DIHEDRAL ANGLES … P.31
- REFERENCES … P.45
PUBLISHED BY
DAVID KRUSCHKE
RT 2, BOX 34 A
WILD ROSE, WI.
54984
COPYRIGHT © 1972, 1975
BY DAVID KRUSCHKE
ALL RIGHTS RESERVED
Page 3
HOW TO PROCEED:
THE READER SHOULD BE EQUIPPED WITH PENCILS, PAPER AND TRIG. TABLES. SOME OF THE MORE ELABORATE DIAGRAMS SHOULD BE SKETCHED OR TRACED ON A SEPARATE SHEET OF PAPER. THIS WILL ALLOW THE READER TO AVOID HAVING TO FLIP PAGES AROUND SEARCHING FOR A NEEDED OR RELEVANT DIAGRAM.
THE READER IS ALSO URGED TO MAKE UP AND SOLVE ADDITIONAL PROBLEMS (I.E. ANGLES, DISTANCES, & AREAS ASSOCIATED WITH TRIANGLES, GREAT CIRCLE DISTANCES ON A GLOBE, ETC.). EXPERIMENTING WITH SPECIFIC PROBLEMS WILL MAKE THE READER FEEL MORE COMFORTABLE WITH GENERAL FORMULAS. OFTEN, ONE CHECKS COMPLETED WORK BY EITHER DRAWING TRIANGLES TO SCALE ON PAPER USING A RULER AND A PROTRACTOR OR BY CONSTRUCTING THREE DIMENSIONAL MODELS TO SCALE USING PAPER, CARDBOARD, “GEO-D-STIX”, ETC.
Page 4
CAUTION:
FOR MOST PEOPLE (INCLUDING THIS
WRITER), LEARNING GEOMETRY IS
NOT A PASSIVE EXPERIENCE REQUIRING
ONLY CASUAL OBSERVATIONS. AS
A TEACHER, THIS WRITER HAS SEEN
MANY BRIGHT STUDENTS FAIL TO SOLVE
OR UNDERSTAND PROBLEMS BECAUSE
THEY WOULD NOT WRITE DOWN OR
SKETCH WHAT THEY WERE THINKING.
IT SEEMS THAT PEOPLE FEEL UNEASY
OR NERVOUS ABOUT WRITING DOWN
SOMETHING THAT MIGHT BE IN ERROR.
THIS RELUCTANCE TO WRITE OR
SKETCH SHOULD BE RESISTED AS
MUCH AS POSSIBLE. THE PERSON
WHO CAN WRITE AND SKETCH FREELY
AND EASILY WILL FIND IT MUCH
EASIER TO ANALYZE, DESIGN, OR
CREATE.
PAGE 2
Page 5
WIDELY USED PLANE TRIG
-
NOTATION: \overline{A} = 90^\circ - A, \overline{A} is called “COMPLEMENT OF A” or “A COMPLEMENT.” “A” of course, is an angle. Also, “ANGLE A” can be denoted by: \angle A.
-
FORMULAS:
GIVEN: ANGLE \overline{A}
$\begin
\sin A = \cos \overline \
\sin \overline = \cos A \
\tan A = \cot \overline \
\tan \overline = \cot A
\end$
EXAMPLE: \sin 30^\circ = \cos 60^\circ = 0.5
- FORMULAS:
GIVEN: TRIANGLE ABC, ANGLE C (\angle C) = 90^\circ
a = c \sin A OR \sin A = \frac{a}{c}
b = c \sin B OR \sin B = \frac{b}{c}
a = c \cos B OR \cos B = \frac{a}{c}
b = c \cos A OR \cos A = \frac{b}{c}
a = b \tan A OR \tan A = \frac{a}{b}
b = a \tan B OR \tan B = \frac{b}{a}
\tan A = \frac{\sin A}{\cos A}
\cot A = \frac{\cos A}{\sin A} = \frac{1}{\tan A}
\tan A = \frac{1}{\cot A}
EXAMPLE: GIVEN EQUILATERAL TRIANGLE WITH 10 FT. PER EDGE, FIND THE ALTITUDE.
Figure: Diagram of an equilateral triangle with 10 ft per edge, split into two right triangles to find the altitude.
ALT. \Rightarrow ALT. = 10 \sin 60^\circ
BUT TRIG. TABLES SHOW \sin 60^\circ = 0.86603 5 PLACES
\therefore ALT. = 10 \times 0.86603
OR ALTITUDE = 8.6603 FT.
Page 6
WIDELY USED PLANE TRIG. CON’T
- FORMULA: CALLED “THE LAW OF COSINES”.
GIVEN TRIANGLE ABC
0^\circ \leq \angle C \leq 180^\circ
c^2 = a^2 + b^2 - 2ab \cos C
NOTE: IF C IS GREATER THAN 90^\circ
THEN \cos C = -\cos(180^\circ - C)
AND -2ab \cos C = +2ab \cos(180^\circ - C)
OR c^2 = a^2 + b^2 + 2ab \cos(180^\circ - C)
ALSO IF C = 90^\circ, THEN \cos C = 0
AND THEN c^2 = a^2 + b^2
PYTHAGOREAN THEOREM
EXAMPLE 1: GIVEN TRIANGLE ABC WITH
a = 10 \text{ FT}, b = 10 \text{ FT}, & C = 60^\circ
FIND SIDE C.
CONSULTING TRIG. TABLES, WE FIND THAT
\cos 60^\circ = 0.5. THEN c^2 = 10^2 + 10^2 - 2(10)(10)(0.5)
\therefore c^2 = 10^2 + 10^2 - 10^2 = 10^2
c^2 = 10^2
\therefore c = 10 \text{ FT}
EXAMPLE 2: GIVEN TRIANGLE ABC WITH
a = 10 \text{ FT}, b = 10 \text{ FT}, & C = 120^\circ
FIND SIDE C.
SINCE 120^\circ IS GREATER THAN 90^\circ, \cos 120^\circ = -\cos(180^\circ - 120^\circ)
\therefore \cos 120^\circ = -\cos 60^\circ = -0.5
THEN c^2 = 10^2 + 10^2 - 2(10)(10)(-0.5)
c^2 = 10^2 + 10^2 + 10^2 = 3 \times 10^2
c = \sqrt{3} \times 10 = 10\sqrt{3} = 10(1.732)
c = 17.32 \text{ FT}.
Page 7
FREQUENCY…
“FREQUENCY” IS ANOTHER TERM
THAT IS USED TO DESCRIBE A KEY
PROPERTY OF A DOME. FOR INSTANCE,
A DOME WITH A FIXED RADIUS, SAY
10 FT, COULD CONSIST OF A SMALL
NUMBER OF LONG STRUTS (LOW
FREQUENCY) OR A LARGE NUMBER
OF SHORT STRUTS (HIGH FREQUENCY).
WHEN SELECTING A PARTICULAR FREQUENCY
FOR A DOME OF A GIVEN DIAMETER,
ONE TRIES TO COMPROMISE IN ORDER
TO AVOID LARGE, BULKY, EXPENSIVE
STRUTS & PANELS (LOW FREQUENCY) AS
WELL AS NUMEROUS, FLIMSY SMALL STRUTS
& PANELS THAT ARE MADE WITH MANY
DIFFERENT DIMENSIONS(HIGH FREQUENCY).
MOST DOMES HAVE STRUTS THAT
ARE LESS THAN 8 OR 9 FEET. THIS
BOOKLET WILL COVER 1,2,3, & 4
FREQUENCY DOMES. A MORE
PRECISE DEFINITION OF “FREQUENCY”
FOLLOWS…
PAGE 5
Page 8
“FREQUENCY” (MORE ACCURATELY DEFINED)
TO BEGIN WITH, A ONE FREQUENCY “DOME” IS SIMPLY AN ICOSAHEDRON, A 20 FACED GEOMETRIC SOLID EACH FACE IS AN EQUILATERAL TRIANGLE. WITH AN ICOSAHEDRON, ALL ANGLES (BETWEEN FACES AND ON FACES) AND ALL EDGES OR “STRUTS” ARE EQUAL.
Figure: A solid icosahedron and a hollow icosahedron.
THE SOLID ICOSAHEDRON CAN BE CUT APART AND FLATTENED OUT THUSLY:
Figure: A flattened icosahedron net.
BY PROJECTING NATIONAL BOUNDARIES AS FOUND ON A GLOBE ONTO AN ICOSAHEDRON AND THEN FLATTENING OUT THE ICOSAHEDRON, ONE HAS A MAP THAT SHOWS COUNTRIES AT THE CORRECT SIZE IN RELATION TO EACH OTHER DEVELOPED BY R. BUCKMINSTER FULLER.
Page 9
“FREQUENCY” - CONTINUED …
A TWO FREQUENCY DOME IS A STRUCTURE THAT IS MADE BY TAKING EACH ICOSAHEDRON STRUT AND DIVIDING IT INTO TWO PARTS AND THEN INSERTING ANOTHER TRIANGLE WITHIN THE BASIC ICOSAHEDRON TRIANGLE.
Figure: Diagram showing progression from a single triangle to a two-frequency structure.
THIS MUST BE DONE IN SUCH A MANNER SO THAT EACH HUB THAT CONNECTS STRUTS IS TOUCHING THE SHELL OF A SPHERE THAT PASSES THROUGH ALL OF THE VERTICES OF THE ORIGINAL ICOSAHEDRON.
Figure: Illustration of a two-frequency dome.
2 FREQUENCY DOME
A THREE FREQUENCY DOME IS A STRUCTURE FORMED BY DIVIDING THE ICOSAHEDRON STRUTS INTO THREE PARTS. THE ICOSAHEDRON TRIANGLE CHANGES AS FOLLOWS:
Page 10
“FREQUENCY” - CONTINUED
1 FREQ. -> 3 FREQ.
EACH VERTEX OR “HUB” IS STILL IN THE SPHERE “GENERATED” BY THE BASIC ICOSAHEDRON.
Figure: Two dome illustrations labeled “3/8” cutoff and “5/8” cutoff.
3 FREQUENCY DOMES
ALL OF THE DOMEBOOKS THAT THIS WRITER HAS SEEN HAVE 3 FREQUENCY CHORD FACTORS THAT DO NOT ALLOW A PERFECT CUTOFF PLANE AT GROUND LEVEL (ALL GROUND LEVEL VERTICES EXACTLY IN THE SAME PLANE). IN FACT, BOTH DOMEBOOK 1 AND DOMEBOOK 2 HAVE STATEMENTS ABOUT 3 FREQUENCY DOMES THAT COULD BE CONFUSING AND MISLEADING.* NEVERTHELESS, THE CHORD FACTORS IN THIS BOOK YIELD TWO PERFECT CUT OFF PLANES FOR THE 3 FREQUENCY DOME AND THREE PERFECT CUTOFF PLANES FOR THE 4 FREQ. DOME (BUCKY FULLER ACCOMPLISHED THIS YEARS AGO).
- DOMEBOOK 1, PAGE 19, PACIFIC DOMES; AND DOMEBOOK 2, PAGES 24 & 76, PACIFIC DOMES.
PAGE 8
Page 11
“FREQUENCY” - CONTINUED
IN ANY CASE, THE READER CAN CHECK THIS OUT BY CONSTRUCTING A MODEL OF A THREE FREQUENCY DOME USING THE CHORD FACTORS HEREIN.
A FOUR FREQUENCY DOME IS A STRUCTURE FORMED BY DIVIDING THE ICOSAHEDRON STRUTS INTO FOUR PARTS. THE ICOSAHEDRON TRIANGLE CHANGES AS FOLLOWS:
Figure: Diagram showing the progression of an icosahedron triangle from a simple triangle to a more complex structure.
AS WITH THE OTHER DOMES MENTIONED, EACH VERTEX OR “HUB” IS IN THE SPHERE (SHELL) “GENERATED” BY THE BASIC ICOSAHEDRON.
Figure: Diagrams of three 4-frequency domes with different cutoffs: “3/8” cutoff, “hemisphere”, and “5/8” cutoff.
4 FREQUENCY DOMES
PAGE 9
Page 12
Spherical Trig.
TO DERIVE THE CHORD FACTORS FOR THE ONE (ICOSAHEDRON), TWO, THREE, AND FOUR FREQUENCY DOMES, WE NEED A FEW FORMULAS THAT ARE CONCERNED WITH RIGHT SPHERICAL TRIANGLES.
GIVEN: RIGHT SPHERICAL TRIANGLE ABC
NOTE: 1) ARCS A-bar B, B-bar C, & A-bar C ARE GREAT CIRCLE ARCS ON A SPHERE WITH RADIUS R AND CENTER O.
-
ANGLES A, B, & C ARE “SURFACE” ANGLES DETERMINED BY THE INTERSECTIONS OF THE THREE GREAT CIRCLE ARCS.
-
ANGLE C DETERMINED BY THE INTERSECTION OF ARCS A-bar C & B-bar C IS A RIGHT ANGLE.
-
a, b, & c ARE ANGLES, WHERE EACH ANGLE IS DETERMINED BY AN ARC AND TWO RADII. ANGLES a, b, & c ARE OFTEN CALLED “CENTRAL” ANGLES AND ARE THE ANGLES USED TO COMPUTE THE CHORD FACTORS OF A DOME.
Figure: A diagram of a right spherical triangle ABC with arcs and angles labeled.
Page 13
SPHERICAL TRIG. - CONTINUED …
THE CHORD FACTOR DERIVATIONS IN THIS BOOKLET USE ONLY FIVE SPHERICAL TRIG. FORMULAS:
-
\sin a = \sin c \sin A OR \sin A = \sin a \div \sin c \sin c = \sin a \div \sin A
-
\sin b = \sin c \sin B OR \sin B = \sin b \div \sin c \sin c = \sin b \div \sin B
-
\cos c = \cos a \cos b OR \cos a = \cos c \div \cos b \cos b = \cos c \div \cos a
-
\cos A = \cos a \sin B OR \cos a = \cos A \div \sin B \sin B = \cos A \div \cos a
-
\cos B = \cos b \sin A OR \cos b = \cos B \div \sin A \sin A = \cos B \div \cos b
Figure: A spherical triangle with sides labeled a, b, c and angles A, B, C, where C = 90°.
PAGE 11
Page 14
CHORD FACTORS
A CHORD FACTOR IS THE LENGTH OF A STRUT ON A DOME HAVING A ONE FOOT RADIUS. FOR EXAMPLE, A “TWO FREQUENCY” DOME
Figure: A hand-drawn illustration of a geodesic dome.
WITH A ONE FOOT RADIUS HAS TWO DIFFERENT STRUT LENGTHS: 0.547 FT. & 0.618 FT. THAT MEANS THAT THE CHORD FACTORS OF THIS DOME ARE .547 & .618. HOWEVER, IF WE WISH TO KNOW THE PROPER STRUT LENGTHS OF THIS TYPE OF DOME HAVING A 10 FOOT RADIUS INSTEAD OF A ONE FOOT RADIUS, WE WOULD JUST MULTIPLY EACH OF OUR CHORD FACTORS BY OUR DESIRED RADIUS (10 FT.). WE NOW HAVE STRUT LENGTHS OF 5.47 FT. & 6.18 FT. RESPECTIVELY. THUS: STRUT LENGTH = RADIUS x CHORD FACTOR FOR ALL DOMES DESCRIBED BY CHORD FACTORS.
PAGE 12
Page 15
CHORD FACTORS & CENTRAL ANGLES
AS MENTIONED PREVIOUSLY, “CENTRAL ANGLES” ARE THE ANGLES THAT ARE USED TO COMPUTE THE CHORD FACTORS OF A DOME. FOR EXAMPLE, CONSIDER AN ICOSAHEDRON WITH ONE FOOT RADIUS. IT SO HAPPENS THAT THE CENTRAL ANGLE OF AN ICOSAHEDRON LINE (SPHERICAL) EQUALS APPROXIMATELY 63.4^\circ OR 63^\circ 24'. USING THIS INFORMATION, LET US FIND THE CHORD FACTOR.
Figure: Diagrams illustrating the icosahedron line, chord, and central angle.
FROM PLANE TRIG. \frac{C}{2} = R \sin 31.7^\circ
\sin 31.7^\circ = \sin 31^\circ 42' = 0.525, R = 1 (OR 1 FT)
\therefore \frac{C}{2} = (1) (0.525) = 0.525
\therefore C = 1.05 (OR 1.05 FT)
\therefore CHORD FACTOR = 2 \sin \left(\frac{1}{2} \text{CENTRAL ANGLE}\right)
AND CHORD = 2R \sin \left(\frac{1}{2} \text{CENTRAL ANGLE}\right), R = \text{RADIUS}
Page 16
TRIG. TABLES - NOTE:
THE READER WILL NOTE THAT ALL ANGLES ARE IN DECIMAL FORM INSTEAD OF THE MORE CONVENTIONAL DEGREES AND MINUTES (I.E. 30.50^\circ INSTEAD OF 30^\circ 30'). ANY READER THAT WOULD PREFER TO USE DECIMAL ANGLES CAN PURCHASE THE FOLLOWING BOOK FOR A MERE 50¢: (IF ITS STILL IN PRINT)
TABLE OF SINES AND COSINES
TO FIFTEEN DECIMAL PLACES AT HUNDREOTHS OF A DEGREE.
NATIONAL BUREAU OF STANDARDS APPLIED MATHEMATICS SERIES 5 ISSUED MAY 2, 1949.
FOR SALE BY THE SUPERINTENDENT OF DOCUMENTS U.S. GOVERNMENT PRINTING OFFICE WASHINGTON, D.C. 20402
Page 17
CHORD FACTORS - 1 FREQUENCY DOME
(ICOSAHEDRON)
SOLID
HOLLOW
SINCE EVERY CHORD IS EQUAL
ON AN ICOSAHEDRON, ONLY ONE
CHORD FACTOR WILL BE FOUND.
ALSO, OTHER ANGLES DERIVED HERE
WILL BE USED IN THE TWO, THREE,
AND FOUR FREQUENCY DERIVATIONS.
TO BEGIN, THE ICOSAHEDRON
CONSISTS OF 20 EQUILATERAL
TRIANGLES. THEREFORE EACH PLANE
TRIANGLE HAS A 60^\circ ANGLE IN
EACH VERTEX. HOWEVER, THE
DERIVATIONS HERE REQUIRE THE
USE OF “SURFACE ANGLES.” Figure: Diagram illustrating surface angles.
SURFACE ANGLES ARE ANGLES
DETERMINED BY THE TANGENTS TO THE
INTERSECTING GREAT CIRCLE ARCS
THE POINT OF TANGENCY IS
THE SAME POINT THAT IS DETERMINED
BY THE INTERSECTING GREAT
CIRCLE ARCS]. THEREFORE, EACH
PLANE TRIANGLE HAS A CORRESPONDING
SPHERICAL TRIANGLE Figure: Diagram of a spherical triangle. AND
EACH PLANAR ANGLE HAS A
CORRESPONDING SPHERICAL TRIANGLE.
Page 18
ICOSAHEDRON - CONTINUED …
IF THE PLANAR ANGLES ARE ALL EQUAL TO 60^\circ, WHAT ARE SPHERICAL ANGLES EQUAL TO? TO FIND THIS OUT CONSIDER THE TOP VIEW OF A SPHERICAL ICOSAHEDRON.
Figure: Diagram showing great circle arcs and a “small” circle on a sphere.
BY STUDYING THE ABOVE DIAGRAM ONE CAN SEE THAT THE “SURFACE” ANGLES OF THE SPHERICAL ICOSAHEDRON EACH EQUAL TO 72^\circ. THIS 72^\circ ANGLE IS IMPORTANT.
Figure: Diagrams showing the progression from a spherical triangle to a right triangle.
RADIUS = 1 FT.
\therefore CHORD FACTOR EQUALS THE CHORD.
\cos A = \frac{c}{\sin B} \cos a
\cos 36^\circ = \sin 72^\circ \cos a
\cos a = \cos 36^\circ \div \sin 72^\circ
\cos a = (0.8090169949) \div (0.9510565158)
\cos a = 0.8506508093
a = 31.7174742623^\circ = \frac{c}{2}
c = 63.4349485246^\circ = 2a
CHORD FACTOR = CHORD = 2 \cdot R \sin \frac{a}{2} = 2 \cdot 1 \cdot \sin 31.7174742623^\circ
\sin 31.7174742623 = 0.5257311093
\therefore CHORD FACTOR = 1.0514622186
Page 19
2 FREQUENCY DOME
START WITH SPHERICAL TRIANGLE
OF AN ICOSAHEDRON AND CUT IT
UP AS FOLLOWS:
Figure: Diagrams showing the division of a spherical triangle.
SO THAT \overline{A_1G} = \overline{A_1C_2} = \overline{A_2C_2} = \overline{A_2C_3} = \overline{A_3C_1} = \overline{A_3C_3}
\therefore \overline{G_1D} = \overline{C_2D}
\cos A_1 = \sin A_3 \cos \overline{A_3C_3}
\cos 36^\circ = \sin 72^\circ \cos \overline{A_3C_3}
\cos \overline{A_3C_3} = \cos 36^\circ \div \sin 72^\circ
\cos \overline{A_3C_3} = (0.8090169949) \div (0.9510565158)
\cos \overline{A_3C_3} = 0.8506508093
\overline{A_3C_3} = 31.7174742623^\circ *
\frac{\overline{A_3C_3}}{2} = 15.8587371312^\circ
2 \sin \frac{\overline{A_3C_3}}{2} = 0.5465330548
\therefore CHORD FACTOR AC = CF_{Ac} = 0.5465330548
\sin \overline{C_1D} = \sin 36^\circ \sin \overline{A_3C_3}
\sin \overline{C_1D} = (0.5877852518)(0.5257311093)
\sin \overline{C_1D} = 0.3090169924
2 \sin \frac{\overline{C_1C_3}}{2} = 2 \sin \frac{3\overline{C_1D}}{2} = 0.6180339848
\therefore CHORD FACTOR CC = CF_{Cc} = 0.6180339848
\overline{C_1D} = 18.0000008671^\circ
\overline{C_2C_1} = 36.0000017143^\circ *
Page 20
3 FREQUENCY DOME
START WITH THE BASIC SPHERICAL ICOSA- TRIANGLE:
Figure: Diagrams showing the transformation of a spherical triangle.
NOTE: G IS THE “CENTER” (CENTROID) OF SPHERICAL TRIANGLE A_{1}A_{2}A_{3} AND F IS THE MIDPOINT OF ARC \overline{A_{2}A_{3}}.
NOW FROM THE 2 FREQUENCY DERIVATION WE KNOW THAT:
\overline{A_{3}F} = 31.7174742623^\circ
\overline{A_{1}A_{2}} = 63.4349485246^\circ
AND BY INSPECTION: \angle A_{3}A_{1}F = \angle GA_{3}F = 36.00...^\circ
NOW \sin \overline{A_{1}F} = \sin 72^\circ \sin \overline{A_{1}A_{2}}
\sin \overline{A_{1}F} = (0.8090169943)(0.8944271878)
\sin \overline{A_{1}F} = 0.8506508048
\overline{A_{1}F} = 58.2825267415^\circ
NEXT, SOLVE TRIANGLE GA_{3}F.
Figure: Diagram of triangle GA_{3}F.
\cos \angle A_{3}GF = \sin \angle GA_{3}F \cos \overline{A_{3}F}
\cos \angle A_{3}GF = (0.5877852518)(0.8506508093)
\cos \angle A_{3}GF = 0.500000001
\angle A_{3}GF = 60.0000005958^\circ \approx 60^\circ
(NOT SURPRISING)
Page 21
3 FREQUENCY DOME - CONTINUED …
SOLUTION OF TRIANGLE G A₃ F - CONTINUED
\sin \overline{A_{3}F} = \sin \overline{A_{3}G} \sin \angle A_{3}GF
\sin \overline{A_{3}G} = \sin \overline{A_{3}F} \div \sin \angle A_{3}GF
\sin \overline{A_{3}G} = (0.5257311093) \div (0.8660254089)
\sin \overline{A_{3}G} = 0.6070619917
\overline{A_{3}G} = 37.3773671524^\circ
\sin \overline{GF} = \sin \angle GA_{3}F \sin \overline{A_{3}G}
\sin \overline{GF} = (0.877852518)(0.6070619917)
\sin \overline{GF} = 0.5368220856
\overline{GF} = 20.9051575539^\circ
Figure: A diagram showing the geodesic dome structure with labeled points and arcs.
NOW \overline{GF} = \overline{FG'}
\overline{A_{1}G'} = \overline{A_{1}F} + \overline{GF}
\overline{A_{1}F} = 58.2825267415^\circ
\overline{FG'} = 20.9051575539^\circ
\overline{A_{1}G'} = 79.1876842954^\circ
NOW ARC \overline{A_{1}G'} IS THE ARC FROM “VERTICAL” (POINT A₁) TO THE “\frac{3}{8}” CUTOFF PLANE. SO FOR A “PERFECT” CUTOFF PLANE (NO EXTRA “BASE” PANELS), LET POINT B₅ (& PT. B₂’) BE IN THE CUTOFF PLANE BY SETTING ARC \overline{A_{1}B_{5}} = \overline{A_{1}G'}.
Page 22
3 FREQUENCY DOME - CONTINUED
CONSIDER TRIANGLE F’ A₁ B₅
WHERE F’ IS LIKE F AND IS
LOCATED ON THE GREAT CIRCLE
EXTENSION OF ARC \overline{A_{1}B_{6}B_{5}A_{3}}.
\therefore \angle A_{1}F'B_{5} = 90^\circ & \Delta A_{1}F'B_{5}
IS A RIGHT SPHERICAL TRIANGLE.
THAT MEANS THAT:
\overline{F'A_{1}} = 31.7174742623^\circ
\overline{F'A_{3}} = 58.2825267415^\circ
\overline{A_{1}A_{3}} = 63.4349485246^\circ
\overline{A_{1}B_{5}} = 79.1876842954^\circ
\angle F'A_{1}A_{3} = 72.00...^\circ
$$
\cos \overline} = \cos \overlineB_} \div \cos \overline}
$$
$$
\cos \overline} = (0.1875924541) \div (0.8506508093)
$$
$$
\cos \overline} = 0.2205281556
$$
$$
\overline} = 77.2599437366^\circ
$$
$$
\overline} = 58.2825267415^\circ
$$
$$
\overlineB_} = 18.9774169951^\circ
$$
BUT \overline{A_{3}B_{5}} = \overline{A_{3}B_{5}} = \overline{A_{1}B_{6}} = \overline{A_{1}B_{1}} = \overline{A_{2}B_{2}} = \overline{A_{2}B_{3}} = \overline{A_{3}B_{4}}
\therefore REFER TO \overline{A_{3}B_{5}} AS \overline{AB}.
\therefore \overline{AB} = 9.4887084975^\circ
AND 2 \sin \frac{\overline{AB}}{2} = 0.3297064628
\therefore CF_{AB} = 0.3297064628
Page 23
3 FREQUENCY DOME - CONTINUED
NOW \overline{B_5B_6} = \overline{A_1A_3} - 2 \overline{AB}
\overline{B_5B_6} = 25.4801145344^\circ*
BUT \overline{B_5B_6} = \overline{B_1B_2} = \overline{B_3B_4}
\therefore REFER TO \overline{B_5B_6} AS \overline{BB_A}
WHERE \overline{BB_A} IS AN ARC ALONG
AN ICOSALINE (ONE OF THIRTY LINES
FORMING A SPHERICAL ICOSAHEDRON).
\frac{\overline{BB_A}}{2} = 12.7400572672^\circ
2 \sin \frac{\overline{BB_A}}{2} = .4410563540
\therefore CF \overline{BB_A} = .4410563540
FROM DWG. A (DRAWING A) WE HAVE:
\overline{A_1B_6} = \overline{AB} = 18.9774169951^\circ
\sin \overline{B_6H} = \sin 36^\circ \sin \overline{A_1B_6}
\sin \overline{B_6H} = (.5877852518)(.3251954548)
\sin \overline{B_6H} = .1911450927
\overline{B_6H} = 11.0196182050^\circ
\overline{B_6B_1} = 22.0392364100^\circ*
BUT \overline{B_6B_1} = \overline{B_2B_3} = \overline{B_4B_5}
\therefore REFER TO \overline{B_6B_1} AS \overline{BB_c}
WHERE \overline{BB_c} IS AN ARC THAT “CUTS”
OR SUBDIVIDES THE ICOSALINES.
2 \sin \frac{\overline{BB_c}}{2} = .3822901844
\therefore CF \overline{BB_c} = .3822901844
PAGE 21
Page 24
3 FREQUENCY DOME CONTINUED …
REFERRING BACK TO A F’A₁B’₅ (DWG. A),
WE HAVE: \sin \angle F'A₁B'₅ = \sin F'B'₅ \div \sin A₁B'₅
\sin \angle F'A₁B'₅ = (.9753806079) \div (.9822469503)
\sin \angle F'A₁B'₅ = .993009558
\angle F'A₁B'₅ = 83.2213490485^\circ
- \angle F'A₁A₃ = -72.00...^\circ
\angle A₃A₁B'₅ = 11.2213490485^\circ
\angle A₃A₁G' = 36.00...^\circ
- \angle A₃A₁B'₅ = -11.2213490485^\circ
\angle B'₅A₁G' = 24.7786509515^\circ
LET POINT P BE LOCATED ON GREAT
CIRCLE LINE \overline{B'₅G'}, SUCH THAT \angle B'₅A₁P = \angle PA₁G' = 12.3893254757^\circ NOTICE THAT
POINT P IS NOT IN THE CUT OFF PLANE].
Figure: A diagram showing points A₁, B’₅, P, and G’ forming angles.
BUT \angle A₁G' = 79.1876842954^\circ
\therefore \sin \overline{PG'} = \sin A₁G' \sin \angle CPA₁G'
\sin \overline{PG'} = (.9822469503)(.2145533638)
\sin \overline{PG'} = .2107443872
\therefore \overline{PG'} = 12.1659797541^\circ
AND \overline{B'₅G'} = 24.3319595081^\circ *
BUT \overline{B'₅G'} = \overline{B₁G} = \overline{B₂G} = \overline{B₃G} = \overline{B₄G} = \overline{B₅G} = \overline{BG}
\therefore REFER TO \overline{B'₅G'} AS \overline{BG}.
2 \sin \frac{\overline{BG}}{2} = .4214887744
\therefore CF_{BG} = .4214887744
PAGE 22
Page 25
ADDITIONAL 3 FREQUENCY DOME DATA
Spherical (Max.) Radius = R = 1.00...
Radius of “small” circle (base) = R'
Height of “\frac{3}{8}” dome = H_1
Height of “\frac{5}{8}” dome = H_2
\phi = 79.1876842954^\circ
R' = .9822469503
\Delta H = .1875924541
H_1 = R - \Delta H = .8124075459
H_2 = R + \Delta H = 1.1875924541
Figure: A diagram showing the spherical geometry with angles and radii.
Figure: A floor plan showing the layout with points labeled B and G, and a radius R' indicated.
Page 26
THE FOUR FREQUENCY DOME
Figure: A diagram of a geodesic dome with various labeled points and planes.
AGAIN, WE START WITH ONE OF THE SPHERICAL TRIANGLES OF THE SPHERICAL ICOSAHEDRON.
\cos \overline{A_{3}C_{2}} = \cos 36^\circ \div \sin 72^\circ
\cos \overline{A_{3}C_{2}} = (0.8090169949) \div (0.9510565158)
\cos \overline{A_{3}C_{2}} = 0.8506508093
\overline{A_{3}C_{2}} = 31.7174742623^\circ
\overline{A_{1}A_{3}} = 2 \overline{A_{3}C_{2}} = 63.4349485246^\circ
\sin \overline{A_{1}C_{2}} = \sin \overline{A_{1}A_{3}} \sin 72^\circ
\sin \overline{A_{1}C_{2}} = (0.8944271878) (0.9510565158)
\sin \overline{A_{1}C_{2}} = 0.8506508048
\overline{A_{1}C_{2}} = 58.2825267415^\circ
SO FAR, THE PROCEDURE HERE IS THE SAME AS THE 3 FREQUENCY DOME.
NOTE: \overline{A_{1}C_{3}} = \overline{C_{3}A_{3}} = \overline{A_{3}C_{2}} = \overline{C_{2}A_{2}} = \overline{A_{2}C_{1}} = \overline{A_{1}C_{1}}
Page 27
4 FREQUENCY DOME - CONTINUED
\sin \overline{C_3D_1} = \sin 36^\circ \sin \overline{A_1C_3}
\sin \overline{C_3D_1} = (0.5877852518)(0.5257311093)
\sin \overline{C_3D_1} = 0.3090169924
\overline{C_3D_1} = 18.0000008571^\circ *
BUT \overline{C_3D_1} = \overline{D_1C_1} = \overline{C_1D_2} = \overline{D_2C_2} = \overline{C_2D_3} = \overline{D_3C_3}
\therefore REFER TO \overline{C_3D_1} AS \overline{CD}
2 \sin \frac{\overline{CD}}{2} = 0.3128689446
\therefore CF_{\overline{CD}} = 0.3128689446
\overline{CC} = 2 \overline{CD} = 36.0000017143^\circ
\sin \angle C_3C_2D_1 = \sin \overline{C_3D_1} \div \sin \overline{C_2C_3}
\sin \angle C_3C_2D_1 = (0.3090169924) \div (0.5877852760)
\sin \angle C_3C_2D_1 = 0.5257310875
\angle C_3C_2D_1 = 31.7174726672^\circ
\sin \overline{D_3E} = \sin \overline{D_3C_2} \sin \angle D_3C_2E
\sin \overline{D_3E} = (0.3090169924)(0.5257310875)
\sin \overline{D_3E} = 0.1624598394
\overline{D_3E} = 9.3497038433^\circ
\overline{D_3D_2} = \overline{D_3E} \overline{D_2} = 2 \overline{D_3E} = 18.6994076866^\circ *
BUT \overline{D_3D_2} = \overline{D_2D_1} = \overline{D_1D_3}
\therefore REFER TO \overline{D_3D_2} AS \overline{DD}.
2 \sin \frac{\overline{DD}}{2} = 0.3249196788
\therefore CF_{\overline{DD}} = 0.3249196788
\cos \overline{C_2E} = \cos \overline{D_3C_2} \div \cos \overline{D_3E}
\cos \overline{C_2E} = (0.9510565118) \div (0.9867151546)
\cos \overline{C_2E} = 0.9638612597
\overline{C_2E} = 15.4504390478^\circ
PAGE 25
Page 28
4 FREQUENCY DOME - CONTINUED …
DWG C. (FROM DWG. B)
FROM PREVIOUS PAGE \overline{C_{2}E} = 15.4504390478^\circ
\overline{A_{1}C_{2}} = \overline{A_{1}C_{2}} = 58.2825267415^\circ
\overline{A_{1}E} = \overline{A_{1}EC_{2}} = 73.7329657893^\circ
\cos \overline{A_{1}D_{3}} = \cos \overline{A_{1}E} \cos \overline{D_{3}E}
\cos \overline{A_{1}D_{3}} = (.2801144289)(.9667151546)
\cos \overline{A_{1}D_{3}} = .2763931520
\overline{A_{1}D_{3}} = 73.9549467026^\circ
\sin \angle D_{3}A_{1}E = \sin \overline{D_{3}E} \div \sin \overline{D_{3}A_{1}}
\sin \angle D_{3}A_{1}E = (.1624598394) \div (.9610446568)
\sin \angle D_{3}A_{1}E = .1690450472
\angle D_{3}A_{1}E = 9.7323021117^\circ
NOW ARC \overline{A_{1}D_{3}} IS EQUAL TO THE ARC THAT GOES FROM “VERTICAL” (POINT A_{1}) TO THE \frac{3}{8} CUTOFF PLANE. SO FOR A "PERFECT CUTOFF PLANE (NO EXTRA BASE PANELS), LET B_{5} (& PT. B_{2}) BE IN THE SAME PLANE AS D_{3} & D_{2} BY SETTING ARC \overline{A_{1}B_{5}} = \overline{A_{1}D_{3}}. IT MIGHT HELP TO MOMENTARILY THINK OF POINT A_{1} AS THE TOP OF THE DOME AND ARC \overline{B_{5}D_{3}D_{2}B_{2}} AS PART OF THE \frac{3}{8} CUTOFF PLANE. IN ANY CASE,
\overline{A_{1}D_{3}} = \overline{A_{1}B_{5}} = 73.9549467026^\circ
\overline{A_{1}A_{3}} = 63.4349485246^\circ
\overline{FA_{3}} = 58.2825267415^\circ
\overline{A_{1}F} = 31.7174742623^\circ
Page 29
4 FREQUENCY DOME - CONTINUED…
USING DWG. B & DWG. C, CONSIDER
RIGHT TRIANGLE B_5 F A'_1 :
\cos \overline{B_5 F} = \cos \overline{B_5 A'_1} \div \cos \overline{F A'_1}
\cos \overline{B_5 F} = (2763931520) \div (8506508093)
\cos \overline{B_5 F} = 0.3249196367
\overline{B_5 F} = 71.0392954727^\circ
BUT - \overline{F A_3} = -58.2825267415^\circ
\therefore \overline{A_3 B_5} = 12.7567687312^\circ *
BUT \overline{A_3 B_5} = \overline{A_1 B_6} = \overline{A_1 B_1} = \overline{A_2 B_2} = \overline{A_2 B_3} = \overline{A_3 B_4}
\therefore REFER TO \overline{A_3 B_5} AS \overline{A B}.
2 \sin \frac{\overline{A B}}{2} = 0.2221880226
\therefore CF_{A B} = 0.2221880226
NOW \overline{A_3 C_3} = 31.71474742623^\circ
AND - \overline{A_3 B_5} = -12.7567687312^\circ
\overline{B_5 C_3} = 18.9607055311^\circ *
BUT \overline{B_5 C_3} = \overline{B_6 C_3} = \overline{B_1 C_1} = \overline{B_2 C_1} = \overline{B_3 C_2} = \overline{B_4 C_2}
\therefore REFER TO \overline{B_5 C_3} AS \overline{B C}
2 \sin \frac{\overline{B C}}{2} = 0.3294187798
\therefore CF_{B C} = 0.3294187798
ALSO \sin \angle F A'_1 B_5 = \sin \overline{B_5 F} \div \sin \overline{B_5 A'_1}
\sin \angle F A'_1 B_5 = (0.9457416380) \div (0.9610446568)
\sin \angle F A'_1 B_5 = 0.9840766838
\angle F A'_1 B_5 = 79.7615938264^\circ
Page 30
4 FREQUENCY DOME - CONTINUED
NOW \angle FA'_1B_5 = 79.7615938264^\circ \angle A_3A'_1E = 36.00^\circ
AND \angle FA'_1A_3 = -72.00^\circ \therefore \angle A_3A'_1B_5 + \angle D_3A'_1E = 17.4938959381^\circ
\therefore \angle A_3A'_1B_5 = 7.7615938264^\circ \angle B_5A'_1D_3 = 18.5061040619^\circ
\angle D_3A'_1E = 9.7323021117^\circ \angle B_5A'_1D_3 = 9.2530520309^\circ
\angle A_3A'_1B_5 + \angle D_3A'_1E = 17.4938959381^\circ
LET POINT P BE LOCATED ON GREAT
CIRCLE LINE \overline{B_5D_3} SUCH THAT \angle B_5A'_1P = \angle PA'_1D_3 = \frac{\angle B_5A'_1D_3}{2} = 9.2530520309^\circ.
Figure: A diagram showing points B5, P, and D3 on a great circle line.
BUT \overline{A'_1D_3} = 73.9549467026^\circ
\therefore \sin \overline{PD_3} = \sin \angle PA'_1D_3 \sin \overline{A'_1D_3}
\sin \overline{PD_3} = (0.1607951406)(0.9610446568)
\sin \overline{PD_3} = 0.1545313107
\overline{PD_3} = 8.8896164913^\circ *
\therefore \overline{B_5D_3} = 2\overline{PD_3} = 17.7792329826^\circ *
BUT \overline{B_5D_3} = \overline{B_4D_3} = \overline{B_6D_1} = \overline{B_1D_1} = \overline{B_2D_2} = \overline{B_3D_3}
\therefore REFER TO \overline{B_5D_3} AS \overline{BD}
2\sin \frac{\overline{BD}}{2} = 0.3090626214
\therefore CF_{BD} = 0.3090626214
CONSIDER
Figure: A diagram showing points A1, Bc, and H.
\sin \overline{B_6H} = \sin 36^\circ \sin \overline{A_1B_6}
\sin \overline{B_6H} = (0.5877852518)(0.2208126570)
\sin \overline{B_6H} = 0.1297904231
\overline{B_6H} = 7.4574840633^\circ
\overline{B_6B_1} = 2\overline{B_6H} = 14.8149681266^\circ *
BUT \overline{B_6B_1} = \overline{B_2B_3} = \overline{B_4B_5} \therefore REFER TO \overline{B_6B_1}
AS \overline{BB}. 2\sin \frac{\overline{BB}}{2} = 0.2585808462
\therefore CF_{BB} = 0.2585808462
PAGE 28
Page 31
A FREQUENCY DOME
SUMMARY AND COMPARISON
| CHORD FACTOR |
SUMMARY |
COMPARISON WITH R. BUCKMINSTER FULLER’S BOSTON BLUE PRINT DATA ARCHIVE, ITEM “M”. |
| CB |
.3294187798 |
.3294 |
| DD |
.3249196788 |
.3249 |
| CD |
.3128689446 |
.3129 |
| BD |
.3090626214 |
.3091 |
| BB |
.2595808462 |
.2596 |
| AB |
.2221880226 |
.2215 |
Figure: A triangular diagram labeled “ICOSA-TRIANGLE” with points A, B, C, and P, showing a subdivided icosahedral triangle.
ICOSA - TRIANGLE
PAGE 29
Page 32
ADDITIONAL 4 FREQUENCY DOME DATA
Spherical (max.) radius = R = 1.00...
Radius of “small” circle (base) = R'
Height of hemisphere dome = R = 1.00...
Height of “\frac{3}{8}” dome = H_1
Height of “\frac{5}{8}” dome = H_2
\phi = 73.9549467026^\circ
R' = .961044659
\Delta H = .276393134
H_1 = R - \Delta H = .72360687
H_2 = R + \Delta H = 1.27639313
Figure: A diagram showing the spherical geometry with angles \Delta H and \phi and radii R and R'.
Floor Plan (see DWG. B)
Figure: A floor plan diagram with points labeled B and D and a central radius R'.
Page 33
DIHEDRAL ANGLES FOR THE FACES
OF 1,2,3, & 4 FREQUENCY DOMES
CONSIDER THE ONE FREQUENCY
DOME (ICOSAHEDRON):
Figure: A simple diagram of an icosahedron.
HERE, EACH FACE MAKES APPROXIMATELY
A 138^\circ ANGLE WITH ANY OTHER FACE.
Figure: An oblique view of a triangular face with points labeled A, B, C, and D.
Figure: An end view showing a 138^\circ angle between two faces labeled A, B, C, and D.
Figure: An end view with a plane passing through line AB and center O of icosahedron. Note: OA = OB = OC = OD = 1
\therefore IN FIGURE 3, THE PLANE
DETERMINED BY POINTS O,A,B,&C
CUTS THE 138^\circ ANGLE IN HALF.
FOR THIS, HOWEVER, IS NOT TRUE ANY PAIR OF
TRIANGLES ON A SPHERE THAT HAPPEN TO
SHARE AN EDGE. AS A RESULT, WE
MUST CONCERN OURSELVES WITH THE
ANGLES A FACE MAKES WITH PLANES
THAT PASS THROUGH EACH EDGE AND
THE CENTER OF THE DOME. THIS
MEANS THAT WE ARE CONCERNED
WITH PYRAMIDS. FOR EXAMPLE, THE
ICOSAHEDRON.
Figure: A pyramid with points O, A, B, C, and P, where OA = OB = OC = 1 & AB = BC = CA = 1.05 & \angle OPC = 69^\circ
THIS IS AN EXAMPLE OF ONE OF THE DIHEDRAL
ANGLES THAT WE ARE LOOKING FOR.
Page 34
DIHEDRAL ANGLES : GENERAL CASE
ROUGHLY SPEAKING, WE UNFOLD THE PYRAMIDS FORMED BY THE DOME CENTER AND FACES AND THEN AFTER NOTING CERTAIN DIMENSIONS, FOLD THE PYRAMID BACK UP AGAIN AND THEN CALCULATING THE DIHEDRAL ANGLES BY “LOOKING DOWN” EACH EDGE OF EACH TRIANGLE. AN EXACT UNDERSTANDING WILL REQUIRE CAREFUL STUDY OF THE FOLLOWING DIAGRAMS AND FORMULAS (AND A MODEL ??):
Figure: Diagrams of a pyramid with labeled points and angles (Fig. 4, Fig. 5, Fig. 6)
$$
\begin
X_1 &= \overline \
X_2 &= \overline \
X_3 &= \overline
\end
$$
XY = \text{DISTANCE BETWEEN PT. X \& PT. Y.}
$$
\begin
\overline &= X_2 \cos \angle A_3 \
\overline &= X_1 \cos \angle A_3 \
\overline &= X_1 \cos \angle A_2
\end
$$
LET c_N = \overline{O'_N A_N}
BUT \overline{A_N O_N} = 1, N = 1, 2, \text{OR } 3
AND \Delta_N = \overline{D_N P_N}
ALSO \overline{O_N O'_N} = \overline{P_N D_N}
\therefore c_N^2 = 1 - \Delta_N^2
\therefore \cos \theta_N = \frac{a_N^2 + h_N^2 - c_N^2}{2 a_N h_N}
OR \cos \theta_N = \frac{a_N^2 + h_N^2 + \Delta_N^2 - 1}{2 a_N h_N}
Page 35
UNIQUE PANELS AND THEIR ASSOCIATED
CHORD FACTORS FOR THE 1,2,3, & 4 FREQ. DOMES.
ICOSA
\overline{AA} = 1.0514622186
2 FREQ
\overline{AC} = .5465330548
\overline{CC} = .6180339848
3 FREQ.
\overline{AB} = .3297064628
\overline{BB_C} = .3822901844
\overline{BG} = .4214887744
\overline{BB_A} = .4410563540
4 FREQ.
\overline{AB} = .2221880226
\overline{BB} = .2595808462
\overline{BD} = .3090626214
\overline{CD} = .3128689446
\overline{DD} = .3249196788
\overline{BC} = .3294187798
Figure: Diagrams of unique panels for 1, 2, 3, and 4 frequency domes with labeled points and chord factors.
Page 36
ICOSA
\overline{A_2A_3} = X_1 = 1.0514622186
\overline{A_1A_3} = X_2 = \text{""}
\overline{A_1A_2} = X_3 = \text{""}
\frac{X_1}{2} = 0.5257311093
\frac{X_2}{2} = \text{""}
\frac{X_3}{2} = \text{""}
\cos \angle 1 = 0.5000000000
\cos \angle 2 = \text{""}
\cos \angle 3 = \text{""}
\angle 1 = 60.00^\circ
\angle 2 = \text{""}
\angle 3 = \text{""}
\overline{P_1A_3} = 0.5257311093
\overline{P_2A_3} = \text{""}
\overline{P_3A_2} = \text{""}
Figure: A diagram of a triangular section of an icosahedron with labeled points A_1, A_2, A_3, P_1, P_2, P_3, and heights h_1, h_2, h_3.
h_1 = 0.9105929924
h_2 = \text{""}
h_3 = \text{""}
a_1 = 0.8506508101
a_2 = \text{""}
a_3 = \text{""}
\Delta_1 = 0.000...
\Delta_2 = \text{""}
\Delta_3 = \text{""}
\cos \theta_1 = 0.3568220869
\cos \theta_2 = \text{""}
\cos \theta_3 = \text{""}
\theta_1 = 69.0948437982^\circ
\theta_2 = \text{""}
\theta_3 = \text{""}
Page 37
2 FREQ.
\overline{A_2A_3} = X_1 = .5465330548
\overline{A_1A_3} = X_2 = \text{same}
\overline{A_1A_2} = X_3 = .6180339848
\frac{X_1}{2} = .2732665274
\frac{X_2}{2} = \text{same}
\frac{X_3}{2} = .3090169924
\cos \angle 1 = .5654131796
\cos \angle 2 = \text{same}
\cos \angle 3 = .3606158719
\angle 1 = 55.5690107900^\circ
\angle 2 = \text{same}
\angle 3 = 68.8619781722^\circ
\overline{P_1A_3} = .1970884940
\overline{P_2A_3} = \text{same}
\overline{P_3A_2} = .3090169924
h_1 = .5097592622
h_2 = \text{same}
h_3 = .4507847362
a_1 = .9619383582
a_2 = \text{same}
a_3 = .9510565169
\Delta_1 = .0761780334
\Delta_2 = \text{same}
\Delta_3 = .000 \ldots
\cos \theta_1 = .1947387387
\cos \theta_2 = \text{same}
\cos \theta_3 = .1256238733
\theta_1 = 78.7705404132^\circ
\theta_2 = \text{same}
\theta_3 = 82.7832160269^\circ
Figure: A diagram showing a triangular structure with labeled points A_1, A_2, A_3, and P_1, P_2, P_3, along with various angles and heights.
Page 38
2 FREQ
\overline{A_2A_3} = X_1 = .6180339848
\overline{A_1A_3} = X_2 = \text{''}
\overline{A_1A_2} = X_3 = \text{''}
\frac{X_1}{2} = .3090169924
\frac{X_2}{2} = \text{''}
\frac{X_3}{2} = \text{''}
\cos \angle 1 = .5000000000
\cos \angle 2 = \text{''}
\cos \angle 3 = \text{''}
\angle 1 = 60.0000000000^\circ
\angle 2 = \text{''}
\angle 3 = \text{''}
\overline{P_1A_3} = .3090169924
\overline{P_2A_3} = \text{''}
\overline{P_3A_2} = \text{''}
Figure: A diagram showing a triangular structure with points A_1, A_2, A_3, and P_1, P_2, P_3.
h_1 = .5352331312
h_2 = \text{''}
h_3 = \text{''}
a_1 = .9510565169
a_2 = \text{''}
a_3 = \text{''}
\Delta 1 = .000...
\Delta 2 = \text{''}
\Delta 3 = \text{''}
\cos \theta_1 = .1875924724
\cos \theta_2 = \text{''}
\cos \theta_3 = \text{''}
\theta_1 = 79.1876841121^\circ
\theta_2 = \text{''}
\theta_3 = \text{''}
PAGE 36
Page 39
3 FREQ.
\overline{A_2A_3} = X_1 = .3297064628
\overline{A_1A_3} = X_2 = .3297064628
\overline{A_1A_2} = X_3 = .3822901844
\frac{X_1}{2} = .1648532314
\frac{X_2}{2} = .1648532314
\frac{X_3}{2} = .1911450922
\cos \angle 1 = .5797432371
\cos \angle 2 = .5797432371
\cos \angle 3 = .3277955570
\angle 1 = 54.56751365900^\circ
\angle 2 = 54.56751365900^\circ
\angle 3 = 70.8649716474^\circ
\overline{P_1A_3} = .1080763136
\overline{P_2A_3} = .1080763136
\overline{P_3A_2} = .1911450922
Figure: A diagram showing a triangular structure with points A_1, A_2, A_3, and P_1, P_2, P_3.
h_1 = .3114897463
h_2 = .3114897463
h_3 = .2686445707
a_1 = .9863181089
a_2 = .9863181089
a_3 = .9815617931
\Delta_1 = .0567769178
\Delta_2 = .0567769178
\Delta_3 = .000...
\cos \theta_1 = .1189229130
\cos \theta_2 = .1189229130
\cos \theta_3 = .0675666318
\theta_1 = 83.1700558513^\circ
\theta_2 = 83.1700558513^\circ
\theta_3 = 86.1257657222^\circ
PAGE 37
Page 40
3 FREQ.
\overline{A_2 A_3} = X_1 = .4214887744
\overline{A_1 A_3} = X_2 = \text{""}
\overline{A_1 A_2} = X_3 = .3822901844
\frac{X_1}{2} = .2107443872
\frac{X_2}{2} = \text{""}
\frac{X_3}{2} = .1911450922
\cos \angle 1 = .4534998407
\cos \angle 2 = \text{""}
\cos \angle 3 = .5886757882
\angle 1 = 63.0315470000^\circ
\angle 2 = \text{""}
\angle 3 = 53.93690470216^\circ
\overline{P_1 A_3} = .2481202364
\overline{P_2 A_3} = \text{""}
\overline{P_3 A_2} = .1911450922
Figure: A diagram showing a triangular structure with points A_1, A_2, A_3, and P_1, P_2, P_3.
h_1 = .3407185571
h_2 = \text{""}
h_3 = .3756545497
a_1 = .9775412028
a_2 = \text{""}
a_3 = .9815617931
\Delta_1 = .0373758492
\Delta_2 = \text{""}
\Delta_3 = .000...
\cos \theta_1 = .1096971744
\cos \theta_2 = \text{""}
\cos \theta_3 = .1418116478
\theta_1 = 83.7021403811^\circ
\theta_2 = \text{""}
\theta_3 = 81.8473082873^\circ
Page 41
3 FREQ.
\overline{A_2 A_3} = X_1 = .4214887744
\overline{A_1 A_3} = X_2 = \text{""}
\overline{A_1 A_2} = X_3 = .4410563540
Figure: A triangular diagram with points A_1, A_2, A_3, and P_1, P_2, P_3.
\frac{X_1}{2} = .2107443872
\frac{X_2}{2} = \text{""}
\frac{X_3}{2} = .2205281770
\cos \angle 1 = .5237124563
\cos \angle 2 = \text{""}
\cos \angle 3 = .4524974506
\angle 1 = 58.4520155000^\circ
\angle 2 = \text{""}
\angle 3 = 63.0959685143^\circ
\overline{P_1 A_3} = .1907225958
\overline{P_2 A_3} = \text{""}
\overline{P_3 A_2} = .2205281770
h_1 = .3758692304
h_2 = \text{""}
h_3 = .3591936944
a_1 = .9775412028
a_2 = \text{""}
a_3 = .9753806042
\Delta_1 = .0200217914
\Delta_2 = \text{""}
\Delta_3 = .000...
\cos \theta_1 = .1323598684
\cos \theta_2 = \text{""}
\cos \theta_3 = .1147242811
\theta_1 = 82.3940192833^\circ
\theta_2 = \text{""}
\theta_3 = 83.4122778320^\circ
PAGE 39
Page 42
4 FREQ.
\overline{A_2A_3} = X_1 = .2221880226
\overline{A_1A_3} = X_2 = \text{''}
\overline{A_1A_2} = X_3 = .2595808462
\frac{X_1}{2} = .1110940113
\frac{X_2}{2} = \text{''}
\frac{X_3}{2} = .1297904231
\cos \angle 1 = .5841468028
\cos \angle 2 = \text{''}
\cos \angle 3 = .3175450230
\angle 1 = 54.2572610000^\circ
\angle 2 = \text{''}
\angle 3 = 71.4854770386^\circ
\overline{P_1A_3} = .0705547007
\overline{P_2A_3} = \text{''}
\overline{P_3A_2} = .1297904231
Figure: A diagram of a triangular section with points A_1, A_2, A_3, and P_1, P_2, P_3.
h_1 = .2106882806
h_2 = \text{''}
h_3 = .1803384692
a_1 = .9938099016
a_2 = \text{''}
a_3 = .9915414495
\Delta_1 = .0405393106
\Delta_2 = \text{''}
\Delta_3 = .000...
\cos \theta_1 = .0804528797
\cos \theta_2 = \text{''}
\cos \theta_3 = .0438346301
\theta_1 = 85.3854028807^\circ
\theta_2 = \text{''}
\theta_3 = 87.4876557449^\circ
Page 43
4 FREQ.
\overline{A_2A_3} = X_1 = .3090626214
\overline{A_1A_3} = X_2 = \text{""}
\overline{A_1A_2} = X_3 = .2595808462
Figure: A triangular diagram with points A_1, A_2, A_3, and P_1, P_2, P_3.
\frac{X_1}{2} = .1545313107
\frac{X_2}{2} = \text{""}
\frac{X_3}{2} = .1297904231
\cos \angle 1 = .4199486254
\cos \angle 2 = \text{""}
\cos \angle 3 = .6472863031
\angle 1 = 65.1686560000^\circ
\angle 2 = \text{""}
\angle 3 = 49.66268726480^\circ
\overline{P_1A_3} = .2000520016
\overline{P_2A_3} = \text{""}
\overline{P_3A_2} = .1297904231
h_1 = .2355820463
h_2 = \text{""}
h_3 = .2804891264
a_1 = .9879878916
a_2 = \text{""}
a_3 = .9915414495
\Delta_1 = .0455206909
\Delta_2 = \text{""}
\Delta_3 = .000 \ldots
\cos \theta_1 = .0723754839
\cos \theta_2 = \text{""}
\cos \theta_3 = .1111558911
\theta_1 = 85.8495613864^\circ
\theta_2 = \text{""}
\theta_3 = 83.6180483304^\circ
PAGE 41
Page 44
4 FREQ.
\overline{A_2A_3} = X_1 = .3249196788
\overline{A_1A_3} = X_2 = "
\overline{A_1A_2} = X_3 = "
Figure: A triangular diagram with points A_1, A_2, A_3, and P_1, P_2, P_3.
\frac{X_1}{2} = .1624598394
\frac{X_2}{2} = "
\frac{X_3}{2} = "
\cos \angle 1 = .5000000000
\cos \angle 2 = "
\cos \angle 3 = "
\angle 1 = 60.0000000000^\circ
\angle 2 = "
\angle 3 = "
\overline{P_1A_3} = .1624598394
\overline{P_2A_3} = "
\overline{P_3A_2} = "
h_1 = .2813886961
h_2 = "
h_3 = "
a_1 = .9867151567
a_2 = "
a_3 = "
\Delta 1 = .000\ldots
\Delta 2 = "
\Delta 3 = "
\cos \theta_1 = .0950590761
\cos \theta_2 = "
\cos \theta_3 = "
\theta_1 = 84.5452808475^\circ
\theta_2 = "
\theta_3 = "
PAGE 42
Page 45
4 FREQ
\overline{A_{2}A_{3}} = X_{1} = .3128689446
\overline{A_{1}A_{3}} = X_{2} = \text{''}
\overline{A_{1}A_{2}} = X_{3} = .3249196788
Figure: A diagram of a triangular section of a geodesic dome with labeled points A_{1}, A_{2}, A_{3}, P_{1}, P_{2}, P_{3}, and angles \alpha_{1}, \alpha_{2}, \alpha_{3}.
\frac{X_{1}}{2} = .1564344723
\frac{X_{2}}{2} = \text{''}
\frac{X_{3}}{2} = .1624598394
\cos \angle 1 = .5192584374
\cos \angle 2 = \text{''}
\cos \angle 3 = .4607413488
\angle 1 = 58.7174780000^\circ
\angle 2 = \text{''}
\angle 3 = 62.5650438344^\circ
\overline{P_{1}A_{3}} = .1441516595
\overline{P_{2}A_{3}} = \text{''}
\overline{P_{3}A_{3}} = .1624598394
h_{1} = .2776819683
h_{2} = \text{''}
h_{3} = .2673832027
a_{1} = .9876883394
a_{2} = \text{''}
a_{3} = .9867151567
\Delta_{1} = .0122828128
\Delta_{2} = \text{''}
\Delta_{3} = .000...
\cos \theta_{1} = .0962330898
\cos \theta_{2} = \text{''}
\cos \theta_{3} = .0854724562
\theta_{1} = 84.4777040985^\circ
\theta_{2} = \text{''}
\theta_{3} = 85.0968081725^\circ
Page 46
4 FREQ.
\overline{A_2A_3} = X_1 = .3090626214
\overline{A_1A_3} = X_2 = .3128689446
\overline{A_1A_2} = X_3 = .3294187798
\frac{X_1}{2} = .1545313107
\frac{X_2}{2} = .15643444723
\frac{X_3}{2} = .1647093899
\cos \angle 1 = .5379329604
\cos \angle 2 = .5213062778
\cos \angle 3 = .4389523013
\angle 1 = 57.4569630156^\circ
\angle 2 = 58.5800869096^\circ
\angle 3 = 63.9629500748^\circ
\overline{P_1A_3} = .1373345432
\overline{P_2A_3} = .1356637489
\overline{P_3A_3} = .1611162847
h_1 = .2811159897
h_2 = .2776959689
A_1h_3 = .2637446619
a_1 = .9879878916
a_2 = .9876883394
a_3 = .9863181089
\Delta_1 = .0171967675
\Delta_2 = .0207707234
\Delta_3 = .0035931052
\cos \theta_1 = .0998094673
\cos \theta_2 = .0967538582
\cos \theta_3 = .0814911169
\theta_1 = 84.2718026907^\circ
\theta_2 = 84.4477269475^\circ
\theta_3 = 85.3257193814^\circ
Figure: A diagram showing a triangular section of a geodesic dome with labeled points A_1, A_2, A_3, and P_1, P_2, P_3, along with various angles and distances.
PAGE 44
Page 47
NOTE
SINCE THIS IS A “COOK BOOK” AND
NOT A MATH BOOK, PROOFS OF FORMULAS
ALONG WITH EXTRA FORMULAS & DIAGRAMS
HAVE NOT BEEN INCLUDED HERE. ANYONE
DESIRING MORE THAN A COOK BOOK
APPROACH SHOULD SEEK OUT A “PLANE
& SPHERICAL TRIG.” BOOK.
THE RESOURCE BOOK USED BY
THIS WRITER WAS:
PLANE AND SPHERICAL
TRIGONOMETRY
BY
KELLS, KERN, & BLAND.
SECOND EDITION
McGRAW-HILL BOOK CO., INC.
1940
THIS BOOK APPEARED RATHER
COMPLETE AND HAS EXCELLENT DIAGRAMS.
IT ALSO CONTAINS APPLIED TECHNIQUES
OF GLOBAL NAVIGATION & ASTRONOMY.
Page 48
DOME COOKBOOK OF GEODESIC GEOMETRY
REVIEWS:
“SHOWS ACTUAL DERIVATION OF CHORD FACTORS AND ANGLES”… LLOYD KAHN, WHOLE EARTH EPILOG.
“HERE IS DAVID’S 3 - FREQUENCY TRUNCATABLE DOME, WHICH SITS FLAT ON THE GROUND, UNLIKE THE DOMEBOOK 2 - 3 FREQ. DOMES”. LLOYD KAHN, SHELTER.
“FOR FULLER FREAKS OF A MATHEMATICAL BENT. AUTHOR GIVES CORRECT CONSTRUCTION OF CHORD FACTORS FOR A 3 FREQUENCY DOME ALL OF WHOSE GROUND LEVEL VERTICES ARE COPLANAR…” “THIS CONSTRUCTION PUTS TO SHAME THE AUTHORS OF VARIOUS BOOKS ON DOMES (EXCEPT FULLER), ESPECIALLY DOME BOOK TWO, WHICH CLAIMED SUCH A DOME WAS IMPOSSIBLE.” PJC, THE AMERICAN MATHEMATICAL MONTHLY (AUG-SEPT, 1974).
APPLICATIONS:
CHORD FACTOR DATA, DIHEDRAL ANGLE DATA, UNDERSTANDING DOME DERIVATIONS, INTRODUCTION TO SPHERICAL MATH, AND THREE DIMENSIONAL MODEL MAKING.
AUTHOR:
A DESIGNER / TEACHER CURRENTLY WORKING ON SOLAR ENERGY TECHNIQUES FOR HOUSING.
Source: David Kruschke, Dome Cookbook of Geodesic Geometry, 2nd edition, 1975. Page images rendered with pdftoppm; page text transcribed with a vision model (qwen2.5-vl). Diagrams and tables are reproduced as page images.
File: DomeCookbookofGeodesicGeometry-Kruschke2ndEdition1975.pdf