Category: 12 Technical sheets

12-12 Pyramids’ angles

On By khufuliteIn 12- Angles in the pyramidsLeave a comment

Everybody would have understood by now that a pyramid is a matter of angles.

However, the angles of the pyramid have ordinary values compared to what we are used to. For example, the angle of the inner slopes of the pyramid of Cheops is very close to 26°56’, and the faces’ angle is close to 51°84’ with the horizontal, and the angle of the edges is close to 52° with the horizontal in the median plane.

This is due to the way ancient Egyptians measured angles indirectly using the cotangent of the angles, the  , or seked; they did not use degree. This information comes from the Middle Kingdom of Egypt, but the sine, the cosine or else the tangent could have been used from time to time.

More precisely, they used to express length in Palm and height in Cubits, so an SDK was 7 * cotangent of the same angle.

So the chosen angles had at least a value, whether it was the sine, the cosine or the tangent, as simply as possible using the measure scale based on the royal cubit for which the smallest subdivision is the finger, i.e. 1/28.

As for numerical writing, the Ancient Egyptians used fractional notation and not decimal notation.

We can then expect that one, or several, of these trigonometric values of an Egyptian angle will be expressed as a whole multiple of the fraction 1/28.

Many archaeologists have endeavored to record the pyramids’ angles in degrees, minutes and seconds. They better should have recorded the cotangent, the sine or cosine of these angles, which is a ration expressed by the Egyptian fraction for length based on the royal cubit; for example, a 45° angle would equal a seked of 7.

If they had proceeded that way, as they did for the lengths expressed by royal cubits, measurements would have been more meaningful.

The royal cubit (MH NSWT)

  • The royal cubit is divided into 28 segments of 1 digit.
  • The palm is 4/28 or 1/7; the double palm is the double.
  • The small span is 12/28 or 3/7, which equals 3 palms.
  • The djeser is 16/28 or 4/7, which equals 4 palms.
  • The remen is 20/28 or 5/7, which equals 5 palms.
  • The short cubit is 27/28 or 6/7, which equals 6 palms.

As a result, when an angle is found in a pyramid, its SKD should be checked to ensure that it can be expressed in fractions of the units of measurement.

Let’s take, for example, the angle of the face with the horizontal measured by Petrie which used different measurement methods that all gave a different result:

In conclusion, he suggested retaining the value 51°52’ with an error of more or less 2’, i.e., an interval between 50’ and 54’.

He deduced the height from this value, after measuring the base.

Petrie measurements for the base give 9068.8 inches, or 230.34 meters, which, with the addition of a royal cubit of 0.5235 m (1,717 ft.), would give 440 cubits exactly, so the height is 220 * tangent 51°52’, i.e., 280 or 24 cubits.

It seems the constructors made sure the base measurement was a whole number, so we could venture that they did the same for the height, i.e., 280 cubits. Consequently, the tangent or seked would be 22 * 7/28, or 11/2, which probably could be expressed by the value 5 + a half-cubit or big span.

To conclude, in our notation, the angle is 51°50’35’’ as the nearest approximation, which falls within the interval given by Petrie.

However, this value is theoretical because the faces’ median is “hollowed” by 2 cubits from the base, giving the real SKD value at the median level to be 7 * (220-2)/280, or 109/20, or 5 + 9/20, or else 5+1/4+1/5 to use the old Egyptian notation system, or 52°5’48’’ in our current notation system.

We can guess that any angle of the pyramid is at least a trigonometric function (7 x cotangent) expressed by a whole multiple of the fraction 1/28, in other words, a whole number of “digits”.

The angle of the slopes is 26°56’ with the horizontal, i.e., a SKD of 14.

Throughout the construction time, there were two days per year when the shadows on the North, East and West faces, as well as the shadows on the north edges, were invisible, as if they “swallowed their shadows”. First, it permitted to position the pyramids in South North and East West with great accuracy using a model, and also to check the correct pyramid alignment thanks to the faces and edges’ shadows, so the edges would meet at the summit, invisible until the last day.

The sun was thus used as a giant instrument of measure to the pyramid’s scale!

Furthermore, the 51.8° angle gives a ratio of 22/7 between the base perimeter and the height, a value near π to 1 per thousand, and gives a ratio between the apothem of one face and the half-base close to the golden ratio with higher accuracy, which is no coincidence!

12-11 Understanding the pendulum operation

On By khufuliteIn 11-PendulumLeave a comment

Nothing is simpler than a pendulum: a bob suspended from a rope,

but the back-and-forth oscillations of the pendulum bob can be very helpful.

As it swings, the pendulum follows a reciprocating sine wave movement, which can be calculated approximately with the following formula: , where L is the length between the axis and the center of gravity of the bob.

Thus, if L = 1 m (3.2 ft.), the period will be 2 seconds.

If the oscillation amplitude increases, the period then slightly varies as follows:, where A is expressed in radians.

Thus, if L = 1m and the oscillation amplitude makes a 45° angle compared to the vertical position, the period is 2.08 seconds and 2.14 seconds for an amplitude of 60°, which is a slight difference.

The benefit of the pendulum is that it acts as a weight amplifier.

In the resting position, the tension on the rope supporting the bob equals the force exerted by the pendulum bob. For a mass M of 1 ton, the tension is T = 9.82 kN.

But the mass swinging movement creates additional tension on the rope, has a sine wave shape, and is a function of the angle at a given point in time and of the angle of the maximum amplitude A.

Thus, with an oscillation amplitude of 90°, the tension force in the rope is multiplied by 3 when the pendulum bob swings low.

On the other hand, when it swings high, the tension force is null.

This tension force exerts on the rope and triggers a reaction in the axis supporting the rope that equals the tension force.

This reaction can be decomposed into two projections: a horizontal projection and a vertical one.

Applied to the pyramid, the horizontal component is the most useful since it can be used as thrust.

This component also exhibits a sinusoidal pattern illustrated below for a motion amplitude of 45°, and another one of 60°.

It can be noticed that for an amplitude of 60°, the maximum horizontal projection is close to 85 % of the force produced by the bob’s mass and is 60 %for an amplitude of 45°.

This graph only shows one period of the motion; for the following one, the same pattern is reproduced but in reverse.

This means that the pendulum’s horizontal force component is zero on average.

If we want to take advantage of the force produced during a period to create motion, the motion of the pendulum for the next period must be blocked naturally; if not, the pendulum has a reciprocating motion, which may be interesting for some uses, like moving a saw, for instance.

If we aim to use a pendulum to shift a stone over a path, the stone must be blocked to avoid moving back; the stone is jerk-moved: it moves forward during one period, stops during the other period, and then moves forward again, and so on.

It is crucial to understand that the horizontal force exerted by the pendulum axis begins at the core of matter.

IT NEEDS NO ANCHOR POINT TO EXERT ITSELF.

For example, if a load had to be pushed down on an icy slope, there would be no need for studded tires or caterpillar tracks to exert forward motion, but the backward movement would have to be blocked.

Energy management:

As long as the pendulum axis is fixed, the pendulum motion consumes little energy—only the friction of the air and the friction in the rope. The force exerted moves along the axis; it works by consuming the energy stored in the mass of the pendulum bob. If let alone, the amplitude of the oscillations will quickly decrease, and motion will stop.

So, the energy produced by the pendulum bob has to be given back to it by giving potential energy to the system. For instance, workers could climb on the mass when in a high position and leave when in a low position. At these two points, vertical speed equals zero, so all the potential energy of the workers is transferred to the pendulum.

But another solution can be more efficient: workers stay on the pendulum bob and ask them to do an up-and-down movement in rhythm with the bob motion with their legs. Anyone who has ever been on a swing understands that!

As a matter of fact, this is the generic, basic solution to put to use that I suggest in my study.

Now the following question remains: how can a single worker doing swings produce a maximum power of 200 W worthy of a trained professional athlete?

The answer is simple, but it depends on the worker’s morphology.

For example, let’s take an 80 kg worker; both their femur and tibia are 0.4 m (1.31 ft) long. Between the sitting position on their heels and the standing position, their center of gravity shifted. To simplify, their body can be divided into three different masses: the feet and tibiae, representing, say, 10 %; the thighs, 30 %; and what is left of the body, 60 %.

Between the sitting position and the standing one, the center of gravity of the tibiae did not move, while that of the things did shift by 40 cm (15.74 in.) and that of the rest of the body by 80 cm (31.49 in.). Everything happens as if 75 % of their weight shifted by 0.8 m (2.62 ft.), so they produce a workforce of 470 joules at each elevation. A period of 470/200 = 2.35s is needed to produce a 200-w power, so the length of the rope had to measure 1.4 m (4.59 ft.) long, which is a rather short length for a big, strong man. And if the worker were ballasted with 100kg (220.46 lbs.), they would produce 590 kJ at each elevation, the period would be 2.95 s, and the length of the rope would be 2.17 m (7.11 ft.).

This example teaches us two things: it is easy to adapt the length of the rope and the weight of the ballast in accordance with the morphology to benefit the required force from it. Several workers could stand on the same pendulum, but they had to have compatible morphologies.

The swinging platforms would be 2 m (6.56 ft.) high and have a 4 m (13.12 ft.) long base to prevent them from tipping over under the torque of the horizontal component, and the width depends on the number of workers placed on the pendulum. These frames could range between 1.5 m (4.92 ft.) wide for two workers, 2.3 m (7.54 ft.) wide for four, 3 m (9.84 ft) wide for six, and 4 m (13.12 ft.) wide for eight.

We can ask ourselves: Why use a pendulum?

The answer is simple and obvious: the pendulum exerts a force proportional to its mass. To produce great force, a heavy pendulum bob, a strong rope to support it, and a sturdy frame are needed.

So considerable force, both horizontal and vertical, could be produced and concentrated.

It can be used, for example, in a device whose “course pusher” function requires little power and little movement, but that could produce high and calibrated force.

Let’s take, for the sake of the example, a 6-t cladding stone ready to be embedded in mortar; it has a force of resistance between the order of 20 and 30 kN to shift on its base. A pendulum made of copper, between 3 and 4 t, operated by a single worker could easily do it, but more importantly, with ACCURACY.

However, its primary application could have been in limestone quarries, where it would simultaneously load and advance the blade digging the furrows.
With eight operators driving the pendulum, this “excavator” had a power output on the order of 2 kW.

12-01 BASIC PHYSICS NOTIONS REVIEW

On By khufuliteIn 01- Elementary physics reviewLeave a comment

All the world’s energy is in matter.
Gravity, the mother of all energy, never stops, nowhere,

originating from matter’s core, it exerts itself on any matter particle anytime, anywhere in the Universe.


Two bodies are attracted to each other proportionally by the product of their weights, which is, in return, proportional to the square of the distance separating them.

So is how gravity acts.


These two bodies both have available gravitational potential energy and attract each other.
They get closer as the speed increases; a fraction of the potential energy is transformed into kinetic energy proportional to their weight and to the square of their relative speed.
They collide; their kinetic energy cancels out as well as their potential energy, both turning into heat, a degraded form of energy.

Thus aggregated, they form a new mass, slowly cooling down, while somewhere else another mass exerts its force of attraction.


The potential energy of matter is the highest form of energy.


Numerous bodies agglomerate with each other, forming huge masses, the protostars, around which protoplanets revolve.


Heat increases so that a thermonuclear reaction occurs, releasing a new form of energy called the sun. This energy is transmitted to planets through radiation.


Vegetation can grow thanks to this radiation through photosynthesis, and provide wildlife.
Over the millennia, decaying plant matter turned into coal while decaying wildlife turned into oil, both fossil fuels that store the sun’s energy—sons of gravity.


Human beings can exert a physical effort of hundreds of watts through muscular cell metabolism, a process during which glucose in the blood “burns” and is limited by the oxygen transported by the red cells of the blood.


A heat engine, which is a hundred times more powerful, burns oil or coal with oxygen.


Both consume solar energy, the daughter of gravity; one is renewable within human timescales, the other is not.

 

At the time of the pyramids, the only available energy was muscular energy; it was weak but renewable endlessly over the millennia.
Builders had therefore to take the best advantages from it, reducing energetic waste as much as possible, unlike we do today.


To do so, they elaborated work methods based on energy conservation.

The ancient Egyptians left the pyramids as a testimony to their knowledge.
What heritage will we leave with the fossil energies we excessively consume today?

Potential energy:
Let’s take a reader sitting on their chair. They just acquire a potential energy, PE, expressed by the product of the height, h, its mass, m, and the gravitational acceleration constant, g.


PE= mhg


PE is expressed in joules, m in kg, h in m, and g = 9.81 meters/s2.

Laws of inertia:

The reader is bold enough to let themselves fall to the ground. During the short period of time they are falling, they are gaining kinetic energy that is the product of their mass, m, and the square of the fall’s speed, v, all divided by two.


KE = (1/2)mv2


KE is expressed in joules, m in kg and v in meters/second.


Brave, but not reckless, they’d like to know the speed of their fall before touching the ground.


They apply the simplified first thermodynamic law stated by Sadi Carnot, which states:


Energy can neither be created nor destroyed.


So, as they reach the ground, their kinetic energy equals the potential energy they had on the chair.


mgh=(½)mv2


Since m is present on both sides of the equation, m can be canceled from it, so we have:


V=√2.9,81.h
or V = 4,43.√h, where h is in meters and v in m/s.

Or V = 16.√h, h in meters and v in kmph.


Curious, the reader would like to know how long their fall will take:
For linear motion at constant acceleration, velocity is the product of acceleration over time:


v=gt


If v is replaced by g.t in the equality, we obtain:

t=√(2•h/g) or t=0.45.√h


The reader carefully softens their fall, using their muscles as they touch the ground.
By doing so, and due to the law of energy conservation, the kinetic energy is transformed into heat within their body. Anyone who did stepping knows it well!

To put it simply, a reader weighing 70 kg sitting on a 0.6 m highchair acquired a potential energy of 70 x 9.81 x 0.6 = 412 joules, with a time fall before touching ground of 3.43 m/s or 12.35 km/h; the fall only lasted 0.35 s.


If they repeat this little game every second, they will produce a power of 0.421 kw, tiring themselves rapidly!
On the other hand, if they do it every 4 seconds, they will produce 0.1 kw of power and could do that for a whole working day with some practice.


The Archimedes’ principle, which everyone knows, is missing for the physics review to be completed, but this principle was not called or expressed as we know in the pharaohs’ time, under the Fourth Dynasty:


Any object immersed in a liquid is buoyed up by a vertical force equal to the weight of the fluid displaced by the object.


This principle can be translated into dynamics as follows:
A float whose draught is h in meters starts to oscillate if moved upward with a time period t=2.√h, where t is in seconds.


I just explained all the basic notions of physical sciences put into practice by the pyramid’s builders; nowadays, every student who finishes their secondary education is supposed to know that much.

12-19 Perfect geometry thanks to the light and shadow effect

On By khufuliteIn 18- A perfect geometry thanks to playing ligth and shadowLeave a comment

For 4 millennia, the Great Pyramid of Cheops has been an architectural feat, with the hollowing of its faces in their middle. It is a silent message that has not been deciphered yet; this article will aim to do it.

The Great Step Pyramid of Djoser, at Saqqara, was already an achievement to build as it was the first gigantic pyramid ever built but being a step pyramid made it easy to properly erect the summit; from one step to the next, builders could rectify any drift in the orientation of the lower step without that being noticed once the construction finished.

From Meidum, builders took up a much more difficult challenge; not only were the pyramids bigger (100 meters, or 328.08 feet, high against 60 meters, or 196.85 feet, at Saqqara), but also the sharp edges had to be perfect straight lines and faultlessly oriented from the start to ensure that the four sides will all precisely meet at the summit, only visible at the very end of the construction. Any default in the construction would be as plain as the nose on one’s face, and moreover, it could not be rectified. But it is clear that all these great pyramids have perfect geometry, such as an almost perfect North-South orientation, which is proof that this issue has been brilliantly resolved.

This could not have been possible without precise measurement and survey tools compatible with the gigantic size of these pyramids to guarantee throughout the construction the perfect orientation of the pyramid and its faces toward the virtual summit and to rectify any occasional drifts in time.

This article will intend to explain the method of measurement that would have enabled builders to take on the challenge with the help of Ra, the guardian God of Ancient Egypt.

Orientation from circumpolar stars?

Every author has noticed the perfect North-South orientation of the six Great Pyramids, with smooth faces. Some authors have thought that the builders used the North Star at that millennia, Thuban, whose brightness is fragile though, for the orientation (2.5 times less bright than our current North Star), either by using it directly or by using its vertical alignment with two other nearby brighter stars, such as Mizar and Pherkad.

However, this hypothesis has several major weaknesses which, in my view, are disqualifying in view of the extreme precision obtained.

At the alleged time of construction, no stars marked the true North; the closest was Thuban, which circled a 9′ arc radius. The northern passage of these stars was therefore a transition that required two things: a prior measurement to determine the moment of northern passage and an association with a reference point in the sky, such as the vertical alignment of Mizar and Pherkad with Thuban, which deviated a little from true north. Therefore, the referential was a moving point that could only be used for a short period of time.

Because of the axial precession, the azimuth of these stars has changed over time, not to mention that the Egyptian chronology of the ancient empire is poorly established and differs by several centuries depending on the author. It is therefore risky to associate the orientation of a pyramid with a star.

Thuban shines dully, making it difficult to spot, so using it for angular measurements (sights) required dark nights. In addition, taking sight requires that it be done at a height of 30°, but there would remain the issue of adapting them on Earth, on a horizontal surface, and over a distance of approximately 200 m (656.16 ft.) on a perfectly dark night.

Assuming that the North-South axis could have been given from the stars, what about the East-West axis?

Orientation from the Sun

On the other hand, using the Sun was quite the opposite, and I will demonstrate how it could have intelligently been used, not only to orient the pyramid from the start, but also to monitor the alignment of the edges and the faces throughout the construction.

The Sun could be used to determine both axes, North-South and East-West, with great accuracy, provided the right method was used.

However, it should be well known that these axes remain nonetheless virtual because they are defined by the daily position of the Sun at its zenith for the South, but what about the East?

To determine the East direction, it should be understood that the Sun’s azimuth as its rising on an equinox day, does not precisely indicate the East as, every four years, the calendar takes an additional day. As a result, from one year to the next, the East shifts by 1/4 of a day (and so by a degree), and the accumulation of this drift over the years could reach one degree, but the accuracy of the pyramid’s orientation is within 3 minutes of arc. So another method had to be found.

The other reference point that could have been used to orientate the pyramids is the Sun’s azimuth at its zenith, but unfortunately, at this point, the sun’s trajectory is horizontal, making it impossible to determine the precise position by direct observation of its height. An indirect method was therefore required, unless a precise “mechanical” clock was available to determine solar noon, of which we have no proof in those days.

The ancient Egyptians therefore found a way of accurately determining solar noon, as evidenced by the Great Pyramid with its eight faces, and I’m about to show how. But in the meantime, I’d like to take stock of the ancient Egyptians’ degree of competence in measuring time.

Measuring time:

To determine the succession of days, the Egyptians of that period had no less than 3 calendars at their disposal: the civil or “vague” calendar, the ” scholar” or Sothiac calendar and the lunar calendar.

These calendars divided time into years, seasons, months, decades and days with a relentless precision that I won’t dwell on, but thanks to which it was impossible for them to lose a day even over millennia.

On the other hand, we know very little about the measurement of time during the day, other than that it was divided into 12 hours of daylight and 12 hours of night, but what about the measurement of the hour?

We did find the Clepsydra of Karnak, dating from Amenhotep III, whose regularly spaced subdivisions divided the night between 12 and 14 hours, depending on the month of the year, which suggests that they considered the hour to be of a fixed length.

Two or three rather “rustic” sundials have been found to measure the time during the day, so we can assume that the pyramid builders – and this was the least of their duties to Ra – knew how to design a sundial.

But these objects are far too imprecise to meet the specific needs of the pyramid builders. On the other hand, we can assume that they would have been able to design a specialized, ultra-precise sundial to meet their construction needs.

Turn the interaction of a scale model of the pyramid and the Sun into a compass and a clock:

For this purpose, I’m going to speculate that they could have built a reduced-scale wooden model of the pyramid, perhaps 1/28, to use as a sort of compass + sun clock combination.

By placing it on a perfectly level and horizontal terrace, and orienting NS as best they could to begin with, they would have had plenty of time, well in advance of starting work, to observe over the course of a year or even several, the play of light and shadow that the Sun made on the faces of this mini-pyramid.

From this observation, they were able to derive particular events, enabling them first to adjust the orientation of the model with near-absolute precision, and then to use it as a clock that gave, at certain times and on certain days of the year only, but unequivocally and with great precision, a visual signal giving the starting signal to take action, but not only for the orientation of the pyramid to be built, which only happens once, but also enabling them to check regularly that the pyramid was indeed rising straight towards its summit, which would remain invisible until the last day.

Analyzing the Sun’s light on the faces:

Every day of the year, the east, south and west faces go from shadow to sunlight within a variable time period. The North face, on the other hand, only emerges from shadow for a certain period of the year.

The phenomena described below have been repeated endlessly for millennia, and are visible to all. Thousands of visitors have stood before this pyramid, some of them seeing it every day, but apart from a vague mention here and there of the phenomenon often referred to as “lightning”, no one has yet described precisely what happens or what use the builders might have made of it.

We’re going to analyze the evolution of the shadow of the 4 NE, NW, SE and SW edges on the eight half-sides, observing the effect of the sun’s rays rotating around each edge.

In its daily course, the sun passes from West to East, taking an azimuth measured from North, i.e., 90° for East to 270° for West, passing through 180° for South. Its height on the horizon varies from zero at sunrise and sunset to a variable maximum every day as it passes over the zenith, with a maximum at the summer solstice and a minimum at the winter solstice, between around 83° and 36° at that time.

To illuminate a face, a ray of sunlight must shine on it with what I’ll call a positive incidence, whereas it is negative when the face is in the shadow of its edges. There is therefore, except at sunrise and sunset, a moment when the sun’s light shines on the half-faces, and it’s this moment that I’m going to study in detail in what follows.

The common case is therefore the moment when, for a given azimuth, the sun’s height equals and then exceeds the slope of the half-face, i.e., its illumination, and then falls below it, i.e., its transition to shadow. To describe this interaction properly, I’m going to start on any day of the year when all the pyramid’s faces will successively go from shadow to light, and vice versa.

https://videopress.com/v/CNt6s9Cj?preloadContent=metadata

Then, for the East, West and North faces, I’ll describe a special day that occurs only twice a year, giving the exact position of the Sun, due East, due West or due South…

On a graph, I’ll plot the slope of a straight line drawn on the half-face from a point on the edge of this half-face and/or the opposite half-face, having the same azimuth as that of the sun in its course, so that I can compare this slope with the height of the Sun.

The East face is passing into shadow:

As soon as the sun rises, it illuminates this face, and we’re going to examine how it passes into shadow.

On the graph below, we can see how the sun’s height varies with the hours for 4 particular days: the summer and winter solstices, any day close to the spring equinox, and finally the day when the sun passes due west with a height very close to 52.1°, which is the slope of the median line of the pyramid’s faces.

From noon to a given hour, a ray of sunlight passing over the ES edge meets the slope of this half-face, which increases as the sun moves westwards while its height decreases.

When the slope of the half-face is steeper than the height of the sun, the shadow of the edge covers it.

Let’s take the sun’s path on 05-04 (yellow curve): the East face is still entirely exposed to the Sun; at around 3:15 pm, the Sun’s height becomes equal to, then less than, the slope of the ES half-face, so the shadow of the ES ridge immediately covers the entire half-face.

However, at this moment, the slope of the other half-side EN is still slightly lower than the height of the sun, so the shadow of the ES edge sweeps across this half-side on its way from the EN edge to the median, and once this is reached, the East face is completely in shadow.

Because of the abrupt transition from light to shadow on the ES half-face, this phenomenon has been referred to as “lightning”, although the time taken to sweep the opposite half-face is of the order of a minute, slower when the Sun is low (around 2 min), and faster around the summer solstice (around 30 sec).

During the time it is swept by the shadow of the edge, the light is grazing, highlighting any defects in flatness and slope.

There are only two days in the year, around 15 days before and after the summer solstice, when at precisely 6 p.m. the Sun is due east at a height of just over 52.1° and the shadow of the ES edge, after having covered the ES half-face, no longer meets the EN half-face. A moment later, the shadow of the EN edge suddenly covers this half-face.

https://videopress.com/v/jxEsUcKD?preloadContent=metadata

This sudden event signals the right moment to exploit the Sun’s shadow, which is then exactly due West, to trace a perfectly aligned East-West axis.

Before that day, the shadow of the EN edge will first cover the EN half-side, then sweep across the ES half-side before covering it completely.

https://videopress.com/v/ptC5vmpT?preloadContent=metadata

Ultimately, the sun’s passage due west is an event “phoned in” in advance as the days go by, with an increasingly rapid sweep of the EN half-face, followed by a simultaneous “extinction” of both half-faces.

However, the sun’s ” due west ” was in relation to the scale model of the pyramid, which had been oriented “as best as possible ” using the north face a few months earlier. To be sure that this corresponded to a ” true ” due west of the sun, we had to observe the exactly symmetrical phenomenon on the west face 12 hours earlier, which translated into illumination rather than the face passing into shadow; if this were not the case, the scale model’s orientation would have to be rectified.

This symmetry could be precisely assessed on the days preceding the ” due west “, because then the sweep times of the face opposite the edge had to be exactly identical on the east and west faces, which was easy to measure over a duration of the order of a minute.

Illumination of the West face:

The West face illuminates symmetrically to the East face and goes out as the sun sets, whereas it used to be illuminated in the morning.

Before noon, as the Sun moves westward, its azimuth and height increase, while the slope of the west face decreases for the same azimuth.

Example of an “ordinary” sunny day 20 days before the equinox (blue curve):

At around 8:30 a.m., when the sun’s height exceeds the slope of the ON half-side, the shadow of the OS edge that covered it moves towards the median, gradually illuminating the ON half-side with a grazing light over a period of around 2 minutes. At the end of this movement, the sun’s height reaches the slope of the OS half-side, which is then suddenly illuminated in its entirety.

https://videopress.com/v/9XBailkj?preloadContent=metadata

Example when Sun is “due East”:

On the same day that the East face marks the Sun as “due West” by this face passing suddenly into the shadow, the symmetrical phenomenon occurs with the Sun being “due east”. When its height reaches 52.1° at around 6 a.m, the ON half-side suddenly lights up, followed a moment later by the OS half-side.

https://videopress.com/v/O8bGQTQS?preloadContent=metadata

As with the East face, this “full East” position of the sun is in relation to the model of the pyramid, so we trace the shadow of the same plumb line, which may be at an angle to the first trace. By taking the bisector of this angle, we obtain the true East-West axis, which will be used, if necessary, to rectify the orientation of the scale model before its next passage in about a month’s time.

So, on the same day, the builders had two very precise visual cues to take advantage of the full west and full east orientation of the sun to first orient the model exactly and directly on the east-west axis, i.e., precisely but indirectly north-south.

South Face:

From the winter solstice to the equinoxes, this face remains illuminated from morning to evening; from the equinoxes to the summer solstice, we have to wait until after sunrise that the sun reaches a height equal to the slope of the south face, which happens for a variable azimuth depending on the day, which is represented on the diagram below for the summer solstice and an ordinary day limiting ourselves to the SE half-face.

When, on a given day, the height of the sun reaches the slope of the south face, the shadow of the SE edge of the SW half-face gradually moves towards the median of the face in about one minute, at which point the SE half-face is suddenly and totally illuminated.

12 hours later, the opposite phenomenon occurs, with the SW half-face suddenly and totally shaded by the SW edge, which then moves onto the SE half-face to cover it in about one minute.

Thus, for the East, South and West faces, every day between the equinoxes and the summer solstice, from sunrise to sunset, we can observe this phenomenon of illumination and shading of the faces, which, thanks to the hollowing of the faces, lasts for some time, during which the EN, SE, SW, WN half-faces remain in low-angled light long enough to inspect the flatness and slope.

The North face “engulfs” its shadow:

Starting from the winter solstice, when the north face is in shadow all day, there is one day in the year when the sun is at its zenith, with a height slightly exceeding the slope of the north face at its median 52. 1° begins to illuminate this face in low-angled light. At the time of construction, this day was situated around two decans before the vernal equinox and had its symmetrical counterpart two decans after the autumnal equinox, opening up a very short window of time during which the effects of shadow could be used to determine the sun’s passage to zenith with great precision.

To illustrate how this works, I’ve chosen the appropriate date in the year 2479 BC, which, according to some authors, could have been the starting year for the construction of the Great Pyramid.

In fact, the exact year doesn’t matter; changing the year only changes the day of the event. For example, this phenomenon will be observable on February 28, 2021, in the Julian calendar, i.e., March 13 in our calendar. However, the deterioration of the north face may slightly reduce the precision of the phenomenon.

The times given below are solar times.

To analyze the Sun’s effect on this north face, we need to consider the two half-faces, NE and NW, and analyze the Sun’s effect on each half-face.

On this particular day, at around 10 a.m., the Sun’s rays, rotating around the NW and NE edges, reach a height that a moment later equals the slope of the NW half-face, whereas for this azimuth, the NE half-face is steeper, and the NW face is suddenly illuminated (point A). Some time later, at around 10:35 a.m., the sun’s height exceeds the slope of the NE half-face, which in turn is suddenly illuminated.

As the Sun continues its course, reaching its zenith, its height falls below the slope of the NW half-face (point B), which suddenly falls into the shadow of the NW edge. Some time later, at around 2.05 p.m., the NE half-face falls into the shadow of the NE edge.

https://videopress.com/v/B4fbk3JY?preloadContent=metadata

As the graph shows, the duration of this phenomenon depends on the height difference between the Sun at its zenith and the 51.2° slope of the median. Whereas for the East and West faces, the phenomenon lasts only a short time.

From one day to the next, the Sun’s height at zenith varies by 0.37° of arc, which represents 45 minutes of sun movement. Therefore, for the sake of precision, it was crucial to be able to vary the model’s slope slightly by tilting it NS on its base, so that for the daytime sun, the slope of the rectified north face would be only slightly less than the Sun’s height.

We can imagine that there must have been a certain amount of trial and error involved in making this adjustment, and that this work on the scale model must have taken place well before construction began.

Nevertheless, although adjusted, the time between illumination and extinction of the north face could not have been less than a few minutes.

To obtain the precise zenith passage, it was therefore necessary to mark the shadow of a plumb line on the platform at point B and to mark the new shadow at point C, probably using a clepsydra to count the time between these two passages. Ultimately, the median of the two plots and the half-time of the count gave the precise NS axis, as well as the moment when the Sun’s south-facing shadow could be exploited.

After this measurement, if necessary, the orientation of the pyramid scale model could be rectified thanks to the NS axis traced on the platform.

One minute of arc equals to a sun passage of 4 seconds, so time counting over a few minutes had to be accurate to the nearest second – a far cry from the precision of Swiss chronometers – and it’s likely that a clepsydra would have done the trick for this short duration.

We now fully understand the mute “message” of the hollowed median, which is to create a predictable visual effect on certain days – a kind of lightning, instantaneous and significant, observable by all—which marks unequivocally and precisely the passage of the sun due south as well as due east and west.

It’s possible that this 8-face scale model was developed for the pyramids preceding Cheops, but that the builders at the time didn’t see fit to leave any trace of it on the actual pyramid, so as not to complicate the already immense task of building a gigantic, perfect pyramid.

Precision of the method:

It took at least one year with 5 days of measurement, and perhaps one or two more inclining the model slightly, to obtain a “perfect” orientation of the pyramid model.

If, at the end of this year, the result obtained on the last day was not satisfactory, the builders have to wait for the following year to start the operation again to reach the perfection that would enable the exact orientation tops to be given for the base of the pyramid when the time came.

However, the sun is not a punctual source of light but a disc whose apparent size is half a degree of arc, or a passage time of 120 sec.

In the above method, it was not the center of the sun, but the middle of the NW or NE quarter of the periphery of the solar disk that first lit up or extinguished a face. Since both shadows were symmetrical, the bisector indicated the exact NS orientation of the sun’s axis.

Depending on the method used by the builders to align the base of the pyramid with the Sun, for the sake of precision, they had to make a time correction to take account of any offset caused by the method used. They had to be fast: every 4 seconds, the sun drifted by one minute of arc.

This may explain why the alignment “error” between the 6 great smooth-sided pyramids ranged from 3 to 20 arc minutes, which could have been the consequence of varying speeds in exploiting the sun’s shadow.

Orienting the base of the future pyramid:

The SE angle could have served as the starting point, as the Sun’s shadow from there pointed towards the NE angle at noon and SW at 8 o’clock.

While we’ll probably never know the exact method they used to make such an alignment, I’ll simply suggest one that combines simplicity with accuracy and is fully compatible with the technology of the time.

They could have used the shadow cast by a plumb bob suspended from a tripod a few meters high, leaving a shadow on the ground whose exact direction is NS at noon, and EO at 8 o’clock.

But since this shadow is only a few meters long, whereas the distance to be covered was 440 cubits, they could have aligned several of these tripods along one side, connected by a cord, giving a perfectly straight shadow on the ground, which would have covered the shadows cast by the plumb bobs connecting them. Such a cast shadow is centered on the center of the solar disc.

Workers could have prepositioned this device to a satisfactory approximation using an “ordinary” estimate of solar noon and 8 o’clock, then used the five days of year zero from the start of construction to rectify the alignment with the ultra-precise signal given by the model. If the result hadn’t been satisfactory, they would have had to wait for the following year to start again.

Surveying the faces and edges during the construction:

While these faces were being illuminated with low-angled light, it was possible for the builders to ensure that the slope of the faces under construction was identical to that of the scale model based on its lighting. On the pyramid under construction, on the same day at the same time, the faces had to be lit or darkened at the same time as on the scale model.

However, even though the grazing light highlights any defects in flatness, this phenomenon had a limit because the shadow cast by an edge could only partially sweep a half-side when that edge was also partially completed. For the rest of the half-face, builders had to make do with the short time provided by the sun’s grazing light.

The builders only had short periods of time in which to carry out these checks, which could nevertheless be carried out over many days of the year.

Pointed at the top of the model, the builders could have left a plumb line, which, by projecting its shadow, gave the exact azimuth of the sun, thus enabling them to check the direction of the edges on the pyramid under construction, again using a plumb line, particularly for azimuths 135° and 225°, where this shadow fell on the NW and NE edges.

As a conclusion, thanks to the model and the Sun, the pyramid was perfectly oriented from the very start, with its faces and edges pointing exactly towards the summit, which would remain virtual until the very last day!

12-17 Why a Royal Cubit = PI/6 meter

On By khufuliteIn 17- Why the royal cubic = PI/6 meterLeave a comment

The fact that a royal cubit equals exactly PI/6 = 0.5236 m (1,717 ft) is not a mere coincidence.

We can nevertheless have the following reasoning:

The book «The Measure of the World » by Denis Guedj narrates two astronomers’ adventures, Pierre Méchain and Jean-Baptiste Delambre, at the beginning of the 1789 French Revolution. It tells us that at that time, for the sake of universalism, this revolution wanted to provide humanity with a unique, indisputable unit of measure of distance, as it is based on a steady physical quantity that is accessible and that can be divided: the length of the terrestrial meridian.

It is accepted, but not necessarily well known, that the ideology of the French Revolution was mainly influenced by freemasonry, which claimed its heritage from Ancient Egypt.

We could think then that they followed an approach to establishing the meter that could have been the same followed by the ancient Egyptians to establish the royal cubit, a measure of unit used for distance.

A civilization that lasted 3 millennia knew how to cultivate constant features, and what is more constant, although cyclic, than the Sky and the Earth?

However, I strongly doubt that the ancient Egyptians, and later the French, had the right instruments to undertake the direct surveying of the earth’s circumference!

They could simply choose two locations points on the same meridian, far enough apart from each other to give two latitudes sufficiently different for their deviation to be known with a good degree of accuracy.

The latitude of a location can be calculated from the Sun’s height at its zenith on a certain special day, such as a solstice.

Nobody questions their ability to measure the distance between two locations using a temporary unit of length.

They could have carried out one or several coordinated measurement surveys using their calendar to establish the latitude of the points of location and their terrestrial distance to determine the relationship between a distance along a meridian and an angle in the sky.

With this relationship established, all there was left to do was measure an angle in the sky to calculate the distance to Earth.

But then, the question was to identify a constant in the changing sky. The height of the Sun at its zenith during solstices could have been an adequate candidate; however, due to the axial precession, the earth’s axis shifts and leads to a variation that could be observed and measured by a thousand-year-old civilization.

On the other hand, the difference in height between two solstices could have been a better candidate, as this value equals exactly twice the obliquity of the ecliptic, a much more constant value as it only varies from 24.5° to 22. 4° over a cycle of 41,000 years. At the time, this value was 23.98° (it is 23.44° today).

Thus, the arc of the circle of reference could have been 2∗23.98 = 47.96°

A reverse calculation shows that to make the sun’s height vary by 47.96°, a distance of 10 183 000 royal cubits had to be crossed along the meridian; that can be rounded to 10 million if we admit a double imprecision in the angle’s and length’s measurements of 2 %, which might be within the confidence interval of the time.

All that was left was to derive a definitive benchmark from this temporary unit of measure through a rule of three to obtain the royal cubit considered a gift for humanity from Ra, God of the Sun, and from Geb, God of the Earth, which makes it sacred and eternal.

Thus, after millennia, the rule used to derive the royal cubit from the movements of the Sun related to the meridian might be almost the same rule for the meter four millennia later!

The relationship PI/6 between the meter and the royal cubit would derive from the relation of the angle of reference 47.96°, measured in the ancient Egyptian Empire by the Sun worshipers as opposed to 90°, an abstract and universal value chosen by mathematicians during the French Revolution.

Within the freemasonry symbol, there is a funny detail: the compass, opened to a 45°angle, is opposed to the 90° triangle!

12 – 15 The Egyptian chronology, at the risk of its calendar

On By khufuliteIn 15- Egyptian calendarLeave a comment

Egyptian chronology is confusing as it is divided into eras matching the reigns of the pharaohs. Any attempt to make sense of it 4,000 years later would require knowledge of all the kings and the length of their reigns, but this is still a matter of debate. To this day, there are not one but several chronologies, all of which are based on the Julian calendar rather than the Egyptian calendar.

For a long time, I have asked myself: how could these “priests,” known to be accurate and meticulous and aware that a year has 365 days and a half, promote a 365-days calendar without a leap year, which is thus a quarter of a day shorter compared to the sky.

After months of thinking, I found the answer: dating from the heliacal rising of the star Sirius.

This star is the brightest star in the night sky and can be seen during a certain period of the year on the horizon at dawn. It marks the beginning of a new year and happens every 365.25 days.

The height difference in the sky between Sirius and the Sun is called “arcus visionis“. Despite sunlight, this difference permits to see, for a short period of time, the heliacal rising of the star Sirius on the horizon before the Sun reaches it. This value ranges between 8 and 9 degrees of arc.

The Egyptian civil calendar, sometimes referred to as “the vague year“, has 365 days and takes an additional day every 4 years on the heliacal rising of Sirius.

Therefore, after 1460 years (the Sothic cycle), the day of the heliacal rising of Sirius will coincide again with the beginning of the 365-days year. This means that each passing day of the heliacal rising of this star marks an accumulation of years since I Akhet 1, the first day of the first year of the Egyptian calendar when implemented.

Observing the heliacal rising of Sirius permits the calculation of the year, but with an imprecision of more or less 3 years.

Thus, from the start, with this calendar it was possible to write the entire future chronology according to the date Egyptians observed the heliacal rising of Sirius.

There are three seasons called Akhet, Peret and Shemu; each lasts 4 months and has 30 days, and there are 5 additional days called epagomenal complete the 365-day year.

The date I Akhet 1 corresponds to the year 2920, the starting point of the third Sothic period of 1460 years.

For example, on the day II Shemu 1 a heliacal rising of Sirius occurred, which corresponds to the year 1080 or the year 2540 (Egyptian calendar) but is imprecise by more or less 3 years.

However, we lack a piece of information: the starting date of the Egyptian calendar according to the Julian calendar.

Nevertheless, both calendars can be synchronized from the date of Sirius’ heliacal rising, knowing the date in both the Julian and Egyptian calendars.

In the year 139 of the Julian calendar, the Grammarian Censorinus observed a heliacal rising on I Akhet 1 of the Egyptian calendar, which was still in use in the Empire at the time.

At that time, Censorinus would have seen it at Alexandria, the seat of the Roman and Egyptian power, which falls on the 19th of July, with an arcus visionis of 9° and six days after the new Moon.

If Censorinus’ observation had been accurate, the Egyptian calendar could have started -2920 years earlier, corresponding to two Sothic periods.

The Egyptian chronology can then be expressed in Julian years.

The Egyptian calendar would have started in -2781 of the Julian calendar, the date on which occurred a heliacal rising of Sirius in the geographical zone of Alexandria; due to the axial precession, the observation happened on the 17th of July, not on the 19th of July, the new moon day.

And what tops it all is that on the 17 of July -2781 BC was also the summer solstice.

Therefore, I Akhet 1 corresponds to the 19 of July 139 BC, day Censorinus made his observation.

Another example: a heliacal rising of Sirius was reported on II Shemu 1, which corresponds to the year -1701 BC or -241 BC, with an imprecision of more or less 3 years.

If the few Sirius’s heliacal rising, reported throughout the Egyptian history, had been noticed in another location, the date of these observations would have had to be corrected as it depends on the location.

The Ancient Egyptians were “precisions fanatics”, it is unconceivable they elaborated a dating system imprecise by more or less 3 years.

To resolve this uncertainty, they also had a lunar calendar in addition to the Sothic calendar. The lunar cycle, also called synodic, lasts 29.53058885 days between two lunations, so the 12-month lunar year is 354.3670662 days and steadily shifts by 10.6329338 days every year compared to the civil calendar; in other words, every year it shifts by 0.36006508 days according to the lunar cycle, or close by one moon quarter and a half every heliacal rising.

This means that the successive heliacal risings featured a different quarter of the moon each time. After 14 years, five complete lunar cycles had been observed on the first day of the year, and the Moon returned approximately, with a slight shift of 1.6 tenth quarter, to the same position it had on the first day of implementation of the calendar.

This slight shift meant that after 25 years, nine complete cycles had been observed, along with a totally invisible shift of 4/1000 quarters.

The Ancient Egyptian civilization did not last long enough to observe and measure this slight shift; therefore, the 25-years cycle could be used without worrying about dating errors.

During the first 25 years of use of the calendar, priests had the time to conceive a matching table between Sirius’s heliacal rising and the Moon phase on this day; this table replicated itself every 25 years.

Thanks to both the Moon and the star Sirius, the civil calendar was a past and future chronology, which gave on each heliacal rising of Sirius the precise year since the beginning of the Egyptian calendar relative to the date of the observation according to the “civil,” or “vague,” calendar and the Moon phase.

As a result, the Ancient Egyptians could base their chronology on regnal years, naming years according to the pharaoh’s accession to the throne for dating events -, without risking losing track of the time, of the vagaries of power, or of periods called “intermediates”.

We can now verify if both calendars synchronize with each other with other observation of a heliacal rising.

The accepted consensus reports the observation of Sirius’s heliacal rising during the ninth year of Amenhotep I, on III Shemu 9, on a full moon.

What would be the matching date according to the Julian calendar?

Since the location of the observation is not given, I assume it took place at Thebes, seat of power at that time.

At Thebes, the heliacal rising happens six days earlier than at Alexandria, I must then correct the date to III Shemu 15 for Alexandria.

As III Shemu 15 corresponds to the 315th day of the year, 1260 years (315 x 4) have passed since the beginning of the Egyptian calendar.

On the Julian calendar, it corresponds to the year -1521 BC (-2781 + 1260), within three years.

At that time, the full moon and Sirius’s heliacal rising occurred at the same time at Thebes on July 12th, -1523 BC.

In conclusion, the observation of Sirius heliacal rising during the reign of Amenhotep I at Thebes happened on July 12th, -1523 BC, in Julian years, placing the beginning of his reign in -1532 BC.

This observation disputes the “official” chronology, which dates it back from -1514 to -1493 BC.

If the heliacal rising had been observed at Heliopolis, and not at Thebes, the same calculations give the date July 16th, -1549 BC dating further back to Amenhotep’s reign, and if it had been at Aswan, it could not have occurred during a full moon.

Another known survey of a heliacal rising happened during the fourth year of reign of Seti I, on I Akhet 1, but there is no mention of the moon phase. If it had been observed at Thebes, it means that it happened on I Akhet 7 at Alexandria, i.e., 1486 years passed (1460 + 28) since the start of the calendar, falling on the 16th of July -1295 BC, within three years. This signifies that Seti I’s reign started in -1299 BC (within three years), whereas it is accepted that it is dated between -1294 and -1283 BC. This observation shows a minimum discrepancy of two years with the “official” chronology.

Another observation, inscribed on an ivory object dating from the Djer’s reign, also reports a heliacal rising on I Akhet 1 and does not mention the phase of the moon, nor the geographical location. If it had occurred at Alexandria, this event would have happened on 17th of July -2781 BC (within three years), date for the beginning of the Egyptian calendar.

During the reign of Mentuhotep II, another heliacal rising was reported on II Peret 21, but there was still no mention of the phase of the Moon or the location, which, if it were Alexandria, would correspond in Julian years to some day in August -2097 BC (within three years). If it had occurred at Thebes, it would have added seven days to this date, so adding 28 years gives the year -2069 (within three years), but the reign of Mentuhotep II is commonly dated between -2045 and -1994 BC. Therefore, there is an incompatibility between this observation date and the “official” chronology.

Let’s verify the observation made under the reign of Thutmose III, on III Shemu 28 year 25, a full moon day.

If the heliacal rising had been seen at Thebes, seven days had to be added, which gives IV Shemu 5 which corresponds to the 335th day of the year and to 1340 years since the beginning of the Egyptian calendar, placing this event in -1441 (within three years).

The heliacal rising having been observed at Thebes means that it occurred on July 12th, -1444 BC, and the full moon had two days. The reign would have then started 25 years earlier, in -1469 BC, while the “official” chronology places it in -1472 BC, i.e., a three-year discrepancy.

Now let’s verify the observation made in Augustus’s fifth year of reign, on III Shemu 25, probably at Alexandria at the time. III Shemu 25 corresponds to the 325th day of the calendar, meaning 1300 years have passed since its beginning. Here, a Sothic period (1460 years) needs to be added, so a total of 2760 years, giving -2781 + 2760 = -21 (with an imprecision of +/- 3), placing the beginning of his reign in -26 BC (within three years). Therefore, there is little discrepancy with the official dates of -27 BC to 14 AD.

Finally, in Ptolemy III’s ninth year of reign, a heliacal rising of Sirius took place on II Shemu 1, without mentioning the phase of the moon or the location, but it could have been at Alexandria. This date is the 271st day of the year, meaning 1084 years have passed, plus the addition of a Sothic cycle gives a total of 2541 years since the beginning of the Egyptian calendar. According to this observation, Ptolemy’s reign started in -245 (within three years), which coincides with the official chronology placing it between -246 and -221 BC.

Except for the reigns of Mentuhotep II and Amenhotep I, the Egyptian calendar is close to the “official“ chronology.

The Egyptian calendar being a trustworthy and accurate instrument, it needs to be taken into account to adjust the Egyptian chronologies.

12-04 Emerging floats

On By khufuliteIn 04- Oscillating floatsLeave a comment

The emerging float was the final evolution of the submersible float, invented by Imhotep for the pyramid of Saqqara after more than a century of R&D and after the construction of four pyramids.

It should be remembered that floats were implemented to lift stone blocks too large to be hauled up by a rope along the face of the pyramid; this process used to place millions of filling and casing stone blocks. These floats were expected to deliver 50 tons per hour, while emerging floats were expected to lift a 72-ton beam to a height of 60 meters.

Unfortunately, this device is difficult to imagine as it is very long and narrow. In the Cheops’ Pyramid, the float of the first floor measures 37 m high and is 4 m² in section; it slides inside a vertical well shaft of same dimensions. This well shaft goes from 3 m above the base to a depth of 34 m under the base, where a large volume of water, 126 m², is stored in the subterranean chamber to prevent the displacement of the float to modify the water level of the well shaft.

It is extended by a 3x4m cage, which starts from the loading station (at the 3 m level) to a height of 60 m from the base.

The submersible float used in the Saqqara pyramid was made to have a large section and a low draught. To properly operate, it had a low freeboard (the height to sink the deck). To increase its tonnage capacity, its section had to be enlarged as well as its size (weakening it) to make room for a large, empty, hollowed, and airtight volume.

The emerging float, on the other hand, had a full section of smaller size and a high freeboard. Its wetted length could be lengthened to increase its load capacity, and as a result, its span was increased as well. This type of float presented an advantage regarding the lifting height. From its static equilibrium point, workers could make it oscillate by several meters, increasing its span.

In Cheops’s Pyramid, the float of the first floor measured 37 m (121.39 ft.) long, had a 4M² cross-section and weighted 32 t; unloaded its draught was 32/4 = 8 m (26.24 ft.), its “deck” would therefore raise by 37-8 = 29 m (95.14 ft.) in relation to the waterline, in other words to the water level in the well shaft.

Loaded with a 16-ton stone block, it sank an additional 4 meters. At its new static equilibrium, the deck was only 25 m above the water level but could be swung by +/- (16/4) = 4 m; at its highest oscillation point, it could have a span of 29 m (95.14 ft.).

Loaded with a 32-ton stone block, it sank additional 8 meters (26.24 ft). At this new static equilibrium point, its deck was at 21 m (68.89 ft.), but could be swung by +/- 8 m and still had the same span as when unloaded.

As a result, the emerging float had a maximal span regardless of its load within certain construction limit.

But this performance advantage had a cost: as it sank into water, the float moved such a significant volume of water that it would raise the water level, overflowing the well shaft. To resolve this issue, a large surface of open water had to be provided in the well shaft’s water circuit to absorb the volume of water displaced by the float, so that as it sank, it would only change the water level slightly.

The harrows chamber, remains of the well shaft of the third floor, is 1.5 m2 in cross-section. The upper chamber that supplied it has a surface area of 50 m2; therefore, the builders chose a relation of 50/1.5 = 33. When the float sinks by 33 meters (108.26 ft.), the water level inside the well shaft rises by 1 meter. To not overflow it, builders had to accept losing 1 meter of span.

By complying with this relation, the lower chamber with an open water surface of 80 m2 (the total sum of the chamber’s, the horizontal gallery’s, and the niche’s surfaces) could supply a well shaft of 2.4 m2 in cross-section, and the subterranean chamber of 126 m2 could supply a well shaft of 4 m2.

Builders had to be able to lower the first floor float’s platform to the level of the loading station, meaning sinking it by 29 m (95.14 ft.), static equilibrium point height, either by decreasing the water level by 29 m by emptying the subterranean chamber, or by loading the platform with (29 x 4) 116 ton of copper bars with a volume of 13 m3. In both cases, the same volume of water was displaced, either by means of a chain of buckets in the descending gallery to empty the subterranean chamber, or by means of a course winch loading the 116 tons of bars onto the platform.

Everything we see inside the pyramid tells us that the water levels were keep stable with accuracy, at least when it came to the ordinary procedure. We can conclude that weighting down the float to lower it to its loading station was the solution retained by builders.

This option functions like a shaduf: displacing relatively low masses, but multiple times to arrive at a significant mass in the end.

The platform was left at the loading station, at the 3 m level, loaded with 116 tons of copper bars, and placed in a way to leave a space oriented east-west, large enough to accommodate the biggest stone blocks mounted on their roller skids.

Let’s examine two examples to illustrate the operation:

The 22-ton entrance lintel, at 15 m high, and the 32-ton chevron of the lower chamber, at 25 m high.

Lintel, L = 0.86 m; W = 2.82 m; H = 3.7m; Wt = 24 t; the float traveled 15–3 = 12 m.

The platform carrying the lintel and the ballast measured 3 x 4 m. In order to accommodate the stone block, the load had to be evenly spread over the platform for its stability as it was lifted by the float of 4 m2 in section. For instance, the lintel rested on one of its faces of 0.86 x 2.82 m (2.82 x 9.25 ft.) and was mounted on its four roller skids in the center of the platform, oriented east-west. On each side of the lintel, there were two spaces of (3- 0.86)/2 = 1 x 4 m available to accommodate 13 m3 of ballast, i.e., two stacks of 1.6 m high. The ballast could be done with 116 bars of 1 ton (1 x 0.33 x 0.33 m or 3.2 x 1.08 x 1.08 ft.), easy to maneuver with the course winch.

The lintel was loaded with its skids onto the platform at the loading station. Once this was done, the ballast bars were removed one by one from the course, employing a winch to raise the platform, pushed by the float, by 12 m (39.37 ft.).

When the 24th bar was removed, the platform didn’t move yet; it would start raising by 1/4 = 0.25 m (0.82 ft.) once the 25th bar was removed. It would keep rising by 0.25 m with each bar removed.

The platform would reach the 15-meter (49.21 ft.) level after the removal of 48 bars; these were then stored on the course, on the edge of the cage, sticking out a little bit into the cage.

That way, when the lintel would be pushed out of the platform before being placed in its final position, the platform would stay stuck. To go back to the loading station, it had to be loaded again.

The dimensions of the vault chevrons in the lower chamber are only partially known, but we can expect them to be cut to the same dimensions as the entrance chevrons: thickness = 0.8 m (2.62ft.), width = 2.4 m (7.87 ft.), length = 7 m (22.96 ft.), and weight = 36 tons, to raise to a height of 25 m (82.02 ft.).

To statically lift the chevron to this height, as was the case for the entrance lintel, the platform would have to be completely unloaded of its ballast; the float would have weighted 36 + 32 = 68 tons and would have sunk by 68/4 = 17 m (55.77 ft.). It would therefore have been raised by 37 – 17 = 20 m (121.39 – 55.77 ft = 65.62 ft.) above the water level (at 3 m, or 9.84 ft.). This represents a total height of 20 + 3 = 23 m (65.61 + 9.84 = 75.45 ft.), whereas the targeted height was 25 m (82.02 ft.).

This is where the benefits of this architecture come in: from the new static equilibrium of 23 m (75.45 ft.), a combination of ballast and worker movements could be used to set the float in oscillation, eventually reaching the level of 25 meter (82.02 ft.), at which point it would remain locked on check-valves.

Following which, the chevron could be unloaded from the platform, but some ballast bars were left astride on the cage to prevent the platform from going up under the float’s thrust during the unloading.

Through this example, we understand that the solution offered by the emerging float, implemented in Cheops’s Pyramid, allowed, theoretically through oscillations, a span equals to the float’s length; here, 37 m + 3 m = 40 m (or 121.39 + 9.84 =131.23 ft.), but let’s say a span of 37 m (121.39 ft.) to maintain a safety margin to prevent the float from dislodging itself on reaching the high point!

This 37 m level would have been reached with a static equilibrium point at 20 m (65.61 ft.); the float oscillated with a maximal amplitude of 17 m (55.77 ft.).

When it came to lifting the 72-ton beam to the 50-meter (164.04 ft.) level to top the upper chamber, the float would weight 32 + 72 = 104 tons, with a draught of 104/4 = 26 m (85.3 ft.), and the platform’s static equilibrium point would be at 37-26 = 11 meters from the water level at 11 + 3 = 14 meters (45.931 ft.); when oscillating, the float would reach the maximal height of 14 + 11 = 26 m (85.3 ft.). 24 meters (78.74 ft.) would still miss reaching the targeted height.

We will examine in details in the megaliths’ procedure how, using check-valves and extensions, the builders could have hauled this monster to a height of 50 m (164.04 ft.) in successive stages.

To deliver these performances, the float obviously had to be correctly guided within the well shaft and the cage.

12 – 03 Second-generation submersible float

On By khufuliteIn 03- Second generation of submersible floatsLeave a comment

 The submersible float invented by the builders of the Saqqara pyramid has proven its efficiency but has some limitations.

In particular, it required a ballast under the float to guarantee the stability of the movable structure (the float and its load). This disposition, inherited from naval architecture, functions perfectly but weighs down the movable structure, which weights around thirty times more than the load to raise, slowing down the float’s movement as it would take around 2 minutes between two ascents.

This slow execution led to the installation of 11 floats in parallel inside 11 well shafts, still visible today, inside the first pyramid.

It came to them as the idea to get rid of the keel and its ballast, entrusting the vertical stability function to a guiding float system using guides in the shaft and the cage that extends it, thus gaining, all other things being equal, in both span length and load capacity.

However, with the next pyramid, Meidum, there is an architectural change because it contains only three well shafts that can still be seen but are doctored, all supplied by the same water circuit that includes a “common” slope of 26°, a horizontal corridor, a chamber, and two antechambers. This disposition is found again inside the two following pyramids: the Red and the Rhomboidal (the latter being more complex than the previous two pyramids).

Inside the Pyramid of Meidum, these well shafts have, certainly, been partially plugged and doctored into chambers or antechambers.

One well shaft has been doctored into a funerary chamber and has 15 m2 of cross-section.

Two more well shafts resemble two small parallelepiped antechambers, having a cross-section of 5.5 m2, but were well identified as well shafts by G. Dormion and J.Y. Verd’hurt, who found the corbeled vault that tops them.

These pieces of equipment were mainly used to haul the same heavy stone block, from the base to the funerary chamber. We can be surprised by the difference in size of the cross-section of the well shafts, from 5.5 m2 to 15 m2 for the “vault”. The two well shafts/”antechambers” being identical, we can venture they might have functioned in parallel, leaving a difference of 11 to 15 m2. The float of the “vault” well shaft should have been loaded at the zero level to bring the stone blocks to the course’s level to a maximal height of 18 m (59.05 ft).

This float follows a guiding system in its shaft and cage, which causes friction, but this friction is under reduced load and therefore consumes little energy. It will, however, require the shaft and cage to be built in fine masonry, with great care taken to ensure regularity of dimensions, parallelism, and a surface finish of quality, a quality abundantly demonstrated by the builders of the pyramids.

How high should the float be? What will its load capacity be?

The float is made of a finely laced rod of 18 m (59.05 ft.) long (span of 18 m or 59.05 ft.) assembled on a waterproof hull of 5 m (16.4 ft.) long, hollowed at the bottom (air pocket). The whole measures 23 m (75.45 ft.) long; therefore, it requires a 23 m (75.45 ft.) deep well shaft to hold it completely, so loads can be placed on its plateau at the zero level.

This air pocket compresses itself as the float sinks, and its decreasing volume compensates for the increasing buoyancy caused by the immersion of the rod, perfectly.

The structure rod/float weighs 11 tons and displaces a volume of water of 18 m3. When it is completely immersed, the water level inside the well shaft is then at zero level. When it floats unloaded in the well shaft of 11 m2 in cross-section, it moves a volume of water of 11 m3 as it sinks by 1 m (3.28 ft.). The water level inside the well shaft is then at -0.6 m (-1.96 ft.) level. When the hull is sunk by 1 m (3.28 ft.) in relation to this water level, the load-bearing plateau, but unloaded, will be at a height of 18 + 4 – 0.64 = 21.36 m (59.05 + 13.12 – 1.96 = approximately 70.21 ft).

To make the float sink to the bottom of the well shaft, it needs to be sunk by 4 meters (13.12 ft), meaning it has to be weighed down by 44 tons. The float will then be completely immersed and will sink to the bottom.

If the builders had wanted to increase the load capacity of the float, its hull would have to be lengthened as well as the depth of the well shaft; by lengthening it by 1 m (3.28 ft.), the float could have lifted an additional weight of 11 tons.

The platform might have been ballasted with copper bars and placed on it in a way to leave enough space in the middle for placing the stone block.

44 tons of copper bars represent a volume inferior to 5 m3. Stacked 1 meter high, these bars would have only occupied 5 m2 out of the 11 m2 of the platform, leaving a space of 6 m2 for the stone block.

Weighting 44 tons, the stone block could have had a height of 2.7 m (8.85 ft.) for a base of 6 m2.

To hoist the stone block onto the platform, a winch, placed on the course, had to be used to haul up the copper bars one by one; they could weight around 1 ton apiece.

The float gradually went up by 0.1 m (0.328 ft.) with the bar removed, leaving enough time for the workers to put it onto the course. There will arrive a moment when the whole ballast/float will weigh less than 44 tons, and the float will go up towards its destination at the speed of the ascent of the last bar removed.

To gradually reach the 18-meter (59.05-foot) span, it would simply have been necessary to add a sufficient amount of water to the shaft with each new course.

As the pyramid was thus completed, the builders only had to fill in cages, with a cross-section of the order of 4 to 5 M², to close off the chamber ceilings, conceal the shafts and cages, and reveal only the “funerary” chambers and antechambers.

A large 28 M2 chamber, like the antechambers of the Red Pyramid, could have contained several floats and several cages.

As with the first generation, the float was made up of 4 parts: a load-bearing plate, lifted by a long rod made of the thinnest possible mesh (to limit its weight), a watertight float body above an air bell pocket, which compensates for the additional volume of waterline caused by the rod sinking into the water by a reduction in the volume of the air bell pocket caused by an increase in pressure as the float sinks.

As a result, the volume of buoyancy is constant throughout the float’s travel (as is the Archimedes’ thrust), and consequently the weight of the movable equipment (float, ballast and payload) remains constant throughout the construction of the pyramid, only the distribution between ballast, stone and operators varies.

The ballast on the platform is adjustable “day by day”, while the ballast on the float is adjustable only in exceptional cases, such as the large stones in the mortuary complex.

Of course, the level of elevation within the maximum span of the shaft is adjusted as the foundations rise by adding water to the shaft.

Float technology:

The float was obviously made of wood, but how watertight has the flotation part been over the years, and even more, how airtight was the bell? The solar boat discovered at the foot of the Cheops pyramid has a hull made of wooden planks assembled with ropes!

Everything is left to the imagination, as no archaeological remains have been found.

Egypt at the time lacked many species of wood, but its inhabitants knew how to trade with the countries of the Mediterranean and those beyond the southern frontier, the land of Punt.

They could have achieved buoyancy simply by using low-density wood, such as poplar or even better balsa, without having to build a watertight hollow volume, and for the air volume part, using naturally airtight reservoirs, such as pig bladders or other animals, which they would have inflated and placed in the float structure. The rod’s latticework could have been made from hardwood rods or, who knows, bamboo?

For handling reasons, these floats were probably made up of several sections assembled on site.

They only had to withstand a vertical compressive force, and being guided by cages and shaft walls, they ran no risk of buckling.

12 – 05 Impermeability between the float and the well shaft.

On By khufuliteIn 05- Sealing float against wellLeave a comment

This only concerns the float of the first floor, that of the subterranean chamber.

When the water level in the water circuit is higher than the well shaft, where the float is, can hold, an impermeability between the float and the well shaft’s walls needs to be installed. If not, the water in the circuit will flow out of the circuit and into the access gallery for the stones.

Guaranteeing the float’s watertightness is as good as turning the float into a piston, as only its top part can be in contact with water while the rest is exposed to air and no longer needs to be watertight.

Yet, the well shaft is 36 meters long and might be fine masonry, as we can notice it on the casing of the pyramid or inside the Upper Chamber of Cheops, but can it really provide a contact surface flat enough between the moving float and the wall for it to move without water infiltrating it? I highly doubt that.

The only solution that comes to mind, and that complies with the five criteria of the pyramid, which are audacity, simplicity, efficiency, reliability and low cost, is to close the cylinder with a membrane, like a kind of sock, that is fixed at the bottom of the well shaft in a way to ensure impermeability, large enough to contain the float, and made to keep, at the same time, pressing against the wall of the well shaft and against the float as it goes up and down.

Waterproof sock

A membrane made of a fabric, that resembles that of skins, is not completely watertight, but it only lets slip through a minimal amount of water, which is sufficient to guarantee water tightness.

By leaving a slack of a few centimeters between the float and the well shaft on each side, the membrane will, under the effect of pressure, press against the float, wrapping it, and against the well shaft’s wall. As a result, the membrane suffers no stress on these contact surfaces.

Membrane wrapping the float

The cross-section of the float, and therefore of the well shaft, had to be circular to avoid the factory turning back on an angle, which would have led to premature wear.

The only area of the membrane subjected to pressure is the part located in the float/well slack, which can be, for example, 5 cm (1.96 in.) wide, i.e., a surface area of 5 cm2 per cm of fabric perimeter around the float.

With the water level at its highest point at +21 m (68.89 ft.) and the half-height of the well at -18 m, the maximum pressure on the membrane is equivalent to a water column 39 m (127.95 ft.) high, resulting in a pressure of 39 N/cm2, which produces a tensile force on the fabric at the point where it comes into contact with the open air of around 200 N per cm length. This force is divided between the two sheets: the one pressed against the cylinder and the one pressed against the float, i.e., 100 N per cm of length, which the membrane fabric is designed to withstand. The fabric is not subjected to great stress and can resist for a long time.

In addition, the fabric is wetted and thus lubricated as it deforms during the slow movement of the float, which increases its longevity.

Over time, however, it will have to be replaced regularly as a wearing part.

That way, the water tightness between the float and well shaft is guaranteed without friction; all it takes is for the float to be guided inside the cage and its walls to be well flat, easy to obtain by masonry, thus avoiding any rough surface.

In this configuration, the oscillations of the float consume very little energy.

It should be remembered, however, that the float on the subterranean chamber would have moved back and forth some 4,000 times a day, and 20 million times over the duration of the project; no membrane can last that long.

This solution was reserved to lift megaliths with the subterranean chamber elevator to increase the elevator’s efficiency, i.e., when it comes to the construction of 13-meter (42.65 ft.) high courses; otherwise, all other well shafts functioned without installing impermeability between the float and the well shaft.

12-17 Pi and Phi

On By khufuliteIn 12 Technical sheets, 17- Why the royal cubic = PI/6 meterLeave a comment

Many significant authors have associated the Great Pyramid with the number PI and the golden ratio, or PHI.

PI:

For Cheops’s Pyramid, the relation of the 1/2 base perimeter to the height (440×2)/280 can be reduced to 22/7, equaling 3.142857, which is very close to the Pi number, at more or less 4/10 000. This is why numerous authors have claimed that the ancient Egyptians knew Pi.

It makes no sense regarding the way ancient Egyptians calculated. They did not use decimal notation. Non-natural numbers were always represented in the form of an addition of natural numbers and the reciprocal of the fraction of a number.

So, 3.14159 and other decimal numerals did not exist for them.

If they wanted to write the value 22/7, they would have written 3 + a palm; a palm is 1/7 of a cubit.

The Rhind Mathematical Papyrus gives two significant pieces of information:

Ancient Egyptians did not use the number π; for example, in the papyrus, we find the cunning way to calculate the surface area of the circle: “To calculate the surface area of a 9-diameter circle, take that of a square of 8”. This formula gives 0.6% more than the actual surface, but they had no way of knowing.

For other diameters, they used the rule of three.

If we decompose “the Egyptian π” of this formula, it results in the fraction 256/81, or 3 + 1/7 + 1/81 + 1/567, but they had no reason to count like that!

In the Great Pyramid, height could have been the diameter of a circle whose perimeter equals that of the half-base. In Royal cubit, the half base is 440 + 440 = 880, and the height is 280, so (1/2 base) / height = 880 / 280, or 22/7, a value close to π more or less 4/10 000. We could word differently Rhind’s formula about the perimeter of a circle as follows:

“If you want the perimeter of a 14-circle, take that of a square of 11”

Archaeologists generally evaluate a royal cubit to equal 0.5236 meters. If we divide by 6 the value of “the Egyptian π”, it results in 22/7. We find the fraction 11/21, or 0.5238, i.e., exactly the value of a royal cubit.

This relation is unsettling, and I suggest an explanation for it.

This observation weirdly and unexpectedly links the metric system of the French Revolution with the measurement system of the Old Empire of Egypt.

If we represent the face of a clock with its 12 hours having a radius of 1m, the arc line from one hour to the next would have the same length as a royal cubit.

Another amusing relation can be made: if we measured the third of a quarter of a meridian of the earth, we would find its radius in meters!

PHI:

Contrary to PI, the golden ratio is defined by a relation between two lengths:

Hence, two segments AB and AD with AD > AB,

So φ equals (AB + AD) / AD, which equals AD/AB.

PHI

In decimal notation, its value is close to 1.618034, or (1 + √5)/2, which made no sense to the ancient Egyptians. But it was easy for them to draw a rectangle respecting the golden ratio without making calculations, like in the example above.

By the way, we will notice that the ABC triangle has the same proportions as the gallery slope in both pyramids.