Algebra and Geometry of Phi
+ Phi in the Great Pyramid
by Tom Gilmore
All graphics by Tom Gilmore
Part II: Geometry of Phi
Part III: The Transcendent Equations
Part IV: Phi in Earth Measures
Part I: Algebra of Phi
The Fibonacci Series
Starting from 1 (and assuming a prior zero), iteratively generating the sum of the result with the previous number converges on a ratio of "phi" between successive numbers (each number in the series is equal to the sum of the previous two numbers). This iteration of sums is known as the Fibonacci Series.
Each new number has a ratio to the previous number that gets ever closer to the transcendent value of phi, alternating between being less than (<) and greater than (>) phi (to seven decimals 1.6180339). The ratios are shown below in parentheses, along with the greater or less symbol. As the fractional tail progresses into established values, the digits are shown in green. Underlined digits indicate infinitely repeating decimal strings.
1 (1 <) 1 (2 >) 2 (1.5 <) 3 (1.6 >) 5 (1.6 <) 8 (1.625 >) 13 (1.615384 <) 21 (1.619047 >) 34 (1.6176471 <) 55 (1.618 >) 89 (1.6179775 <) 144 (1.6185 >) 233 (1.6180258 <) 377 (1.6180371 >) 610 (1.6180328 <) 987, (1.6180345 >) 1597, (1.6180338 <) 2584, (1.6180340 >) 4181, (1.61803396 <) 6765, (1.618033998 >) 10946 (1.618033985 <) 17711 (1.61803399017 >) 28657 (1.618033988 <) 46368 (1.61803398895 >) 75025 (1.61803398867 <) 121393 (1.61803398878 >) 196418 .....
The violet ratio of 89/55 is utilized in the Great Pyramid (as developed in Part II - The Geometry of Phi).
Each ratio, when inverted, is equal to the decimal portion of the following ratio, for example, for the ratios 8/5, and 13/8,
8/5 inverted is:
5/8 = 0.625
13/8 = 1.625
Notice that (5/8 + 1 = 13/8) and since 8/5 is phi, 5/8 is 1/phi, this means
that (1/phi + 1 = phi).
This is the same as saying that the decimal tail of phi is equal to 1/phi.
Another way to express this relationship is that inverting a phi fraction and adding 1 results in the following phi fraction.
Iteration of Adding 1 to the Inverted Fraction
The general form of iteration takes a function (f) and applies it on a number (n), and repeatedly applies the function on the result.
The iteration f(n) = 1/n + 1 eventually converges on the value phi for any positive integer (n). For the purpose of this treatise, we use (n)=1.
Note that this produces the identical ratio series (bold fractions) resulting from the Fibonacci numbers.
This iterative inversion, followed by adding 1, is the same functionality as Euclid's “continued fraction of 1”.
The Continued Fraction of 1
In fact there is only one number and it is the number one (refer to the article on Number).
Euclid is credited with developing a "continued fraction expansion" utilizing the number 1 that is “equal” to Phi.
1 + 1 / (1 + 1 / (1 + 1 / (1 + ... …))) = Phi.
(This continued fraction is simply a transposition of (f(n) = 1/n + 1) to (f(n) = 1 + 1/n), for n=1).
In reducing this equation from a finite portion of the continued fraction (as detailed below), the final fractions of the 2-step reduction (shown in bold) are the Fibonacci ratios. In the 2-step reduction, ratios are inverted by the fractional pattern (1/n), and in the next step are added to 1, resulting in the next Fibonacci ratio in the series. In order to exactly follow the Fibonacci ratios, although the first reduction results in the number 2, it is replaced by (2 = 2/1).
1 + 1 / (1 + 1 / (1 + 1 / (1 + 1 / (1 + 1 / (1 + 1 / (1 + 1 / (1 + 1 / (1 +
1 + 1 / (1 + 1 / (1 + 1 / (1 + 1 / (1 + 1 / (1 + 1 / (1 + 1 / (1 + 1 / (2/1)))))))) =
1 + 1 / (1 + 1 / (1 + 1 / (1 + 1 / (1 + 1 / (1 + 1 / (1 + 1 / (1 + 1/2))))))) =
1 + 1 / (1 + 1 / (1 + 1 / (1 + 1 / (1 + 1 / (1 + 1 / (1 + 1 / (3/2))))))) =
1 + 1 / (1 + 1 / (1 + 1 / (1 + 1 / (1 + 1 / (1 + 1 / (1 + 2/3)))))) =
1 + 1 / (1 + 1 / (1 + 1 / (1 + 1 / (1 + 1 / (1 + 1 / (5/3)))))) =
1 + 1 / (1 + 1 / (1 + 1 / (1 + 1 / (1 + 1 / (1 + 3/5))))) =
1 + 1 / (1 + 1 / (1 + 1 / (1 + 1 / (1 + 1 / (8/5))))) =
1 + 1 / (1 + 1 / (1 + 1 / (1 + 1 / (1 + 5/8)))) =
1 + 1 / (1 + 1 / (1 + 1 / (1 + 1 / (13/8)))) =
1 + 1 / (1 + 1 / (1 + 1 / (1 + 8 /13))) =
1 + 1 / (1 + 1 / (1 + 1 / (21/13))) =
1 + 1 / (1 + 1 / (1 + 13 /21))
1 + 1 / (1 + 1 / (34/21)) =
1 + 1 / (1 + 21/34) =
1 + 1 / (55/34) =
1 + 34/55 =
89/55 = 1.618 = Phi
The ratio 89/55 is the Egyptian Royal Phi ratio used in constructing the Great Pyramid of Giza, as demonstrated below.
Part II: Geometry of Phi
The Phi Right-Triangle
Phi in the Great Pyramid
Two exact Royal measures are given by the Nile Scholars for the Great Pyramid dimensions.
Pyramid = 440 Royal Cubits
Apothem of Pyramid = 356 Royal Cubits
The "apothem" of a pyramid is the line from the mid-base to the
The measures of the pyramid in Royal Cubits are diagrammed below left.
Applying the Pythagorean formula (Hypotenuse2 = Base2 + Height 2) to the right-triangle of height, apothem, and half-base, (height2 = 3562 - 2202), so height = ~279.8857 Royal Cubits. This is approximately 0.0004% short of the Royal approximate measure of 280 Royal Cubits.
The relative Royal dimensions of the right triangle consisting of the
apothem, height, and half-base, are the values Phi, root-of-Phi, and 1
(illustrated above right).
The relative lengths are found by dividing all 3 sides by 220.
Half Base = 220/220 =1
Apothem =356/220 = 89/55 = 1.6181818 = Phi.
Height = 280/220 = 1.272727 = square-root of Phi (when Phi =1.6181818)
The measures of the apothem and half-base are 4 times the Phi ratio of 89/55 (89x4=356 and 55x4=220).
The slope of the faces of the Great Pyramid is 51 degrees 52 minutes, the same as the angle between the right triangle sides of 1 and Phi
Approximate Squaring of the Circle
It is mathematically impossible to exactly square the circle because the number Pi is transcendent.
The ratio of base to height in the Great Pyramid squares the
(Squaring the circle is essential to projecting global maps onto flat surfaces).
The circumference of the square base approximately equals the circumference
of the circle with a radius of the height (see diagram above).
The height of the pyramid is ~279.8857 Royal Cubits, and for this proof is rounded to 280.
The circumference of a circle is (Pi x 2r).
Taking the height of the pyramid (280) as the radius(r) of a circle, the circumference is (Pi x 560).
Using Pi = 22/7,
(Pi x 560) = (22/7 x 560) = (12320 / 7) = 1760
The circumference of the base is 440 x 4 = 1760.
Phi as a Rectangle
Geometrically, the number Phi is defined as a ratio represented by a point on a line positioned such that the two segments have the property that the short segment has the same proportion to the long segment as the long segment has to the entire line.
The formula is Phi = AC/BC = BC/AB ~ 1.618
Da Vinci named the rectangle with sides the proportion of 1 to 1.618 (AB to
BC) the Golden Section.
The phi-rectangle is shown below.
Phi in the Pentagram
Connecting the 5 points that are seperated by 72 degrees with all possible lines, creates a pentagram (5-pointed star) within a pentagon, as shown below.
At the center of the pentagram is an inverted pentagon, and connecting the vertices of that pentagon produces an inverted pentagram. As the symbol of magic, the outer pentagram represents “white magic”, meaning majic used to help others (outer), and the inner pentagram represents “black magic”, meaning magic used for selfish purposes (inner).
The pentagram illustrates two equations
involving Phi. The phi ratios are
integral to the proportions of the line segments.
In the diagram above, two equations are shown on the upper right, and are illustrated (color coded) by lined-up spanned lengths between intersections of a pentagram line, such that the sums match the equations.
Powers of Phi
The two equations above are specific examples
of the general equation that
Phi n = Phi n-1 + Phi n-2, or that each power of Phi is the sum of the previous 2 powers of Phi.
Phi 1 = Phi 0 + Phi -1 means that Phi = 1 + 1/Phi, the first equation in the graphic above.
Phi 2 = Phi 1 + Phi 0 means that Phi 2 = Phi + 1, the second equation in the graphic above.
The following proof of the general equation
uses (for an example) the series numbers and ratios of
(5, 8/5, 8, 13/8, 13, 21/13, 21).
Phi 1 = 8/5
Phi 2 = Phi x Phi = (8/5 x 13/8) = 13/5, and
Phi 3 = Phi x Phi x Phi = (8/5 x 13/8 x 21/13) = (13/5 x 21/13) = 21/5.
The product of any number of consecutive phi ratios always reduces to a ratio with the denominator of the first ratio's denominator and the numerator of the last ratio's numerator.
Logical proof of the exponential equation: Since all consecutive powers of Phi have the same denominator, the general exponential equation (Phi n = Phi n-1 + Phi n-2), reduces to the Fibonacci series numbers (see example below). In other words, the powers of the ratio Phi reduce to the Fibonacci series numbers, which by definition sum the previous 2 numbers.
Example: Using Phi n = Phi
n-1 + Phi n-2, where Phi n =21/5, Phi
n-1 =13/5, and Phi n-2 =8/5,
(21/5 = 13/5 + 8/5), or (21 = 13 + 8).
Phi by Compass and Ruler
derivation of Phi is shown below.
It is half the length of a unit square plus the diagonal of the half-square rectangle.
The diagonal is the hypotenuse of a right
triangle with sides of 0.5 and 1.
Thus, the diagonal is sqrt(1x1+.5x.5) = sqrt(1.25) = sqrt(5/4).
So (Phi = .5 + sqrt (5/4)) or
Phi = (1+sqrt5)/2
Part III: The Transcendent Equations
The 3 transcendent numbers of Phi, Pi, and sqrt5 have a common transcendent root. This is established by the equations between them.
Phi = (1+sqrt5)/2
Pi = Phi 2 x 6/5
The (Approximate) Transcendent Equation Proof
Pi = Phi 2 x 6/5
Proof using Pi ~ 22/7 and the Fibonacci series numbers (21, 34, 55).
Since each number in the series increases by Phi (because Phi is the ratio
between the numbers), 2 numbers in the series increase by Phi2.
This means that, using the series numbers (21, 34, 55), Phi = 34/21, and Phi2 = 55/21.
Substituting Phi2 = 55/21 in the formula (X = Phi 2
X = (55/21 x 6/5) = 330/105 = 22/7 ~ Pi.
Using higher numbers in the series increases the accuracy of both Phi and Pi in the equation, but because Pi and Phi are transcendent there is no exact proof.
The 6/5 in the equation is also the
disparate measure of approximation of Pi~22/7 using Phi~89/55, because
Pi=(Phi x Phi x 6/5) =
(89/55 x 89/55 x 6/5) =
(7921/3025 x 6/5) = 47526/15125,
and 22/7 x 2160 = 47520/15120
(leaving a residual 00006/00005 (the 6/5 in the formula).
Solving for Sqrt5 in terms of Phi
Demonstrating that sqrt5 = Phi + 1/Phi
Phi = (1 + sqrt5) /2
2Phi = 1+ sqrt5
sqrt5 = 2Phi –1
sqrt5 = Phi+(Phi –1)
Phi = 1 + 1/Phi
(Phi – 1) = 1/Phi), and substituting 1/Phi for (Phi – 1) above,
sqrt5 = Phi + 1/Phi
Solving for Pi in terms of Sqrt5
Pi = Phi2 x 6/5, and Phi2 = Phi +1, so
Pi = (Phi + 1) x 6/5 and Phi = (1+sqrt5)/2 , so
Pi = ((1+sqrt5)/2) +1) x 6/5
= (1 + sqrt5 +2)/2) x 6/5
= (1.5 + (sqrt5)/2) x 6/5
= 9/5 + (sqrt5 x 6/10), so
Pi = 9/5 + .6 sqrt5
Confirmation usimg the
sqrt5 ~2.236, Pi ~3.1416)
Pi = 9/5 + .6 x 2.236 = 1.8 + 1.3416 = 3.1416
Part IV: Phi in Earth Measures
The table below records the various measures of distance by fractions of the circumference of the Earth at the equator. The fractions are cumulative, so for example the English foot is based on 24 hourly divisions of one full rotation, with 1000 miles per hour by definition, and 5280 feet per mile, for a circumference total of 126,720,000 feet. The geographic measure uses 360 degrees and the time measures use 24 hours. The advantage of using the time measure of the cubit is that it is commensurate with clocks, and thus convenient in calculating longitudinal position on the Earth’s surface using the positions of the stars in the night sky as compared to the clock time.
Earth and Time Measures (Base-12)
Geographic Royal Roman English
Foot Cubit Cubit Foot Statute Foot
360 degrees 24
hours 24 24
60 minutes 60 minutes 60 1000 miles 1000 miles
60 seconds 60 seconds 60 chains (60x100) chains (33x160)
100 feet 1000 cubits 880 6000 Roman feet 5280 Statute feet
129,600,000 86,400,000 76,032,000 144,000,000 126,720,000
Measures of the Base of the Great Pyramid
The base measures exactly 750
A Cubit is 1.5 Geographic Feet (129,600,000 / 86,400,000 = 1.5)
Thus, the base measures 500 Cubits (750/1.5)
1,000 cubits equals 1 second of time, so 500 cubits equals ½ second.
And the base measures 1/2 second of time (by Earth’s rotation at the equator).
A Cubit is 0.88 of a Royal Cubit (76,032,000 / 86,400,000
500x.88=440 Royal Cubits (The measure given by the Nile Scholars)
A Roman Foot is 0.9 Geographic Feet (129,600,000 /
144,000,000 = 0.9)
750/.9=833.33 Roman Feet (The measure given by the Roman historian Pliny)
English Statute Foot and Freemasonry
The English could have retained the time-based mile-measure and their foot be exactly equal to the Geographic Foot by changing the Roman Mile of 6000 feet to being 5400 feet (24x1000x5400=129,600,000), but the Masons sought to incorporate the ratio of Phi into the time measure of feet, and the predominant Masonic presence in British government allowed them to influence the British Parliament in 1593 to declare there to be 5,280 feet to a mile by statute. The Masons chose this number because it preserved the Phi-Measure of the Royal Cubit, and also closely approximated the Geographic Foot.
5,280 feet to a mile accomplishes two major objectives:
Since (126,720/129,600 =
a Statute foot is 0.97… of a geographic foot.
Thus the Statute foot correlates closely with the Geographic foot.
Since (76,032/126,720 = 0.6),
a Statute foot is exactly 0.6 of a Royal cubit.
Thus, in the Statute foot, the measure of Phi is preserved.
The Remen (a Phi-Measure)
Phi in the Great Pyramid
The Remen is defined as the diagonal of a square Royal Cubit. Mathematically a Royal Cubit is 1.7045 Geographic feet, so using the diagonal of a square Royal Cubit is approximately (sqrt (2 x 1.70452)) ~ (2.4106) feet in length. However since Phi measures are by definition inexact, and a standard Remen rod would have been used, some measures of the interior of the Great Pyramid support the notion that the standard measure of the Remen was 2.3885 feet (and 2 Remen was 4.7777 feet). The following are examples of the 2-Remen measure of 4.7777 feet.
There are 3 stone plugs 2 Remen long lodged at the base of the ascending passage.
The 3 plugs lodged in the passage have been estimated to be just under 4.8 feet in length, which is 2 Remen (4.777 feet).
The Grand Gallery floor is of equal width as the ascending corridor, but there are raised shoulders on each side that are slotted at 27 intervals. These slots extend horizontally across the shelves and vertically up the walls. It is currently thought by many that these slots were intended for wooden beams suspending the granite plugs for release into the ascending passage, completely filling it. There are 27 paired slots, while the ascending passage is 129 feet, and (27x4.777=129).
All the passages, shafts, and chambers are estimated to be offset 23.9 feet east of the pyramid centerline.
This would be a 10 Remen offset, since 5x4.777 = 23.888 feet.
One explanation of the offset is that it was intentional and intended to help obscure the location of the sealed entrance. An alternate scenario is that initially there was a smaller pyramid (shown in green above) where the Great Pyramid now stands. The Queen's Chamber would have been centered, both north/south and east/west in this initial pyramid. When the truncated pyramid (at the 50th layer) was built over the initial one, it was too close to the eastern end of the plateau, and this did not allow the larger pyramid to be exactly centered east to west over it.
It is likely that the Descending passage was cut at ground level from outside the Queen’s Chamber pyramid (portion of passage in green above), and when the truncated pyramid was built over it, the in-line extension of the descending passage caused the entrance to be elevated. The descending angle of 260 31' 23" points exactly at the intersection of the axis of Earth's rotation with the celestial globe (for the 30th parallel where the pyramid is located). The passage was most certainly used to align the interior Grand Gallery (an astronomical observatory) exactly north/south, using the then North Star shining down the passage, and reflected up the equally inclined ascending passage at the junction.
The Purpose of the “Queen’s Chamber pyramid”
Early explorers reported a half inch of a salty encrustation on the walls of the Queen's Chamber, and that this chamber originally had a putrid smell. The Queen’s Chamber shafts are about 8 x 5 inches. The shafts end 5 inches behind the walls, with a narrow slot connecting to the interior. These shafts might have stored chemicals to be combined in the chamber, producing Hydrogen gas. This would also account for the niche in the Queen’s Chamber that appears to have housed a cooling tower.
After the King’s Chamber was added to the pyramid, the combustion (Queen’s) chamber may have been used to generate Hydrogen to be piped to the King’s chamber, where it was excited to emit microwaves. The microwave wavelength from hydrogen emission is 8.309 inches and the south shaft is slightly less than 8.4 inches wide (running straight to the pyramid surface).