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Article first time online July 2009.
Reworked October 2019.

The Cheops Pyramid: E - Ancient Hydraulics.

 

Unit of pressure:

Unit of pressure is the Pascal è1 Pa = 1 N/m² (Pa = Pascal, N = Newton, m = meter).
Pressure is usually displayed in bar or atmosphere (industrial applications).
1 bar = 1 Atm. = 9.8 N/cm² (cm² = area in square centimeters),
is usually rounded to 10 N/cm²


Unit of power:

Unity of power is the Newton.
In the past, the unit kgf (kilogram force) was used, also written as kg’.
1 kgf = 9.8 N sometimes rounded to 10 N.
1 kgf is the force that exerts a weight with a mass of 1 kg.

Although the kgf unit is outdated, it'll still be used here to keep things simple.


Formula: F = p x A

F = Force (Force) expressed in kgf or kg’
p = The applied pressure (here bar, atmosphere or kg/cm² is used).
A = Area (area expressed in cm²)

So, a pressure of 1 kg/cm² on a surface of 1 cm² gives a force of 1 kgf.

 

1 pound = 0,453 kilo (kg) è 1 kg = 2.2 pound

1 foot = 30.48 centimetres (cm) è 1 meter (m) = 3.28 feet.

1 inch = 2.54 centimetres (cm) è 1 cm = 0.3937 inch.

1 cubit = 52.36 cm = 20,62 inch [Petrie]

 

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E01 

The assumption is that the pyramid was built using hydraulic lifting equipment. It's about single hydraulic cylinders here whose piston was pushed up by water pressure.
 

Example 1: On a cylinder with a diameter of 10 cm (area 78.5 cm²) a water pressure of 4 atm. is applied (this is approx. 4 kg/cm² and about the pressure on our domestic tap water). The upward force exerted by this hydraulic cylinder is F = 4 kg/cm² x 78.5 cm² = 314 kgf. Theoretically, this cylinder can therefore lift up a weight of 314 kg.

Example 2: If, in the construction of the pyramid, a lifting device was used with 2 short hydraulic cylinders (2 short wooden posts according to Herodotos) with both an inside  diameter of 20 cm (section 314 cm²) and under a pressure of 10 bar (10 kg/cm²) then the device could theoretically lift a stone of 6.28 tons. F = 2 x (314 cm² x 10 kg/cm²) = 6280 kgf’ or 6.28 tons’

 

The pyramid filled with water.

How the pyramid could have been filled with water and where that water came from is a concern for later, it’s currently assumed here this indeed has been the case. The monolith once stood in the top of the great gallery and is in fact the piston of a hydraulic cylinder. This piston glided down and stands nowadays at the very bottom of that cylinder.

 

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Calculations: The piston (monolith) and the hydraulic cylinder.

If one creates passages and (burial) chambers in a pyramid, it will certainly be taken into account that the sarcophagus with some carriers must be able to pass through those corridors. Apart from the entrance itself, such a corridor must have a sufficient vertical height so that one can stand upright, i.e. about 4 to 5 cubit. Moreover, climbing a fairly steep corridor with a slope of 26° would certainly not be a superfluous luxury. Moreover, one would expect that the height of those corridors would be a whole number of cubits or palms!

Let’s calculate the monolith (piston) and the cylinder in which it stands.

 

E03
The shafts in Cheops’ pyramid, section 2 by 2 cubit.

 

This is certainly not the case in the pyramid of Cheops. It becomes already clear from the dimensions of the shafts alone that the height of it was of secondary importance. Indeed, these shafts are big enough to crawl through, but no more than that.


The vertical height of those shafts is 2,225 cubit, this is not even a whole number of palms (1 palm = 1/7 cubit). On the other hand, the perpendicular section of those shafts is exactly 2 by 2 cubit, which is indeed an exact number of cubit. This must demonstrate that the section of that channel was more important than the vertical height. That is why comparable dimensions can be expected for the great gallery.

 

E04
The great gallery, the vertical height and the perpendicular section.

 

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This is indeed also the case for the great gallery, the vertical height of it is approx. 16.23 cubit, the bottom layer has a height of around 4.35 cubit and the seven indenting layers have an average height of about 1.63 cubit (this according to Petrie, as an average of the top five layers). These aren’t really values that can be converted to an exact number of palms. Here too, in the great gallery, it wasn’t the vertical height that was important but instead the perpendicular section of it.

 

E05
Dimensions of the great gallery in cubit (seven indenting layers).

 

The great gallery, with a height of about 8.5 meters and a floor under26 °, measuring completely correctly is an almost impossible task. On paper, or rather on papyrus, it may have been the intention to build the large gallery with dimensions as shown in the drawing above. There are small deviations in the actual dimensions that are impossible to avoid with such a construction. The above drawing is then only a theoretical approach taking into account the known data, it may have been the intention to build the great gallery with dimensions as shown on the above drawing.

 

E06

On the above drawing the monolith (piston) is still in the great gallery.
The monolith has six indenting layers, the great gallery itself has 7.
As a result, the monolith doesn’t reach the toothed ceiling.
The hydraulic cylinder at the bottom, still empty here, has 6 indenting layers!

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Dimensions of the monolith (piston) in cubit, 6 indented layers.

 

Not to forget that the monolith (piston) as well as the hydraulic cylinder count one indenting layer less than the great gallery itself. Both the piston and the hydraulic cylinder are 75 cubit long and 13 cubit high.

 

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On the above drawing both figures are identical,
they are both 75 cubit long and 13 cubit high.
Therefore it is much easier to make the calculations on the rectangle.

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Calculations.

 

E09

A: The perpendicular section of the large gallery, seven indenting layers.

B: The perpendicular section of the monolith (piston), six indenting layers.

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The above drawing shows the perpendicular section of the great gallery (A). As already mentioned, there are seven indenting layers. The floor has a width of 2 cubit and both ledges have a width of 1 cubit. Above the ledges the width is therefore 4 cubit and this remains so up to a height of again 4 cubit, this measured from the floor between the balustrades. Above it are seven indenting layers. Each of these layers jump inwards on each side of the great gallery over a distance of 1 palm (or 1/7 cubit = 7.48 cm). So, at the top of the ceiling there is still a width of 2 cubit left. The monolith (B) has 6 indenting layers, therefore one layer less than the great gallery.


It is more convenient to express the width in palms, for the great gallery it starts on the top with 2 cubit what then becomes 14/7. For each layer below it 2/7 are added, this is resp. 14/7, 16/7, 18/7, 20/7, 22/7, 24/7, 26/7 and 28/7 or 4 cubit. For the monolith (piston) it starts on the top with 16/7 cubit (16 palm). So, now it must be possible to calculate the area of perpendicular section of the piston.

 

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The area of ​​the perpendicular section of the piston.

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To calculate the perpendicular cross-section it's divided into three parts. Since the descending shaft in the monolith was closed with three granite plugs, this section must also be included. For part A it's an advantage to calculate in palms (1 palm = 1/7 tb), so all those pieces can be placed next to each other and their length added.


Length of part A = 16/7 + 18/7 + 20/7 + 22/7 + 24/7 + 26/7 = 126/7 or exactly 18 cubit.


The thickness of the layers is 1.5 cubit, so we can calculate the area.

Surface area A = 18 cubit x 1.5 cubit = 27 cubit²

Surface area B = 4 cubit x 3 cubit = 12 cubit²

Surface area C = 2 cubit x 1 cubit = 2 cubit²

The total area of ​​the perpendicular cross-section = part (A + B + C) = (27 + 12 + 2) cubit² = 41 cubit² and the length of that piston as well as the cylinder is 75 cubit.

Capacity cylinder = surface cross section x length = 41 cubit² x 75 cubit = 3075 cubit³

The displaced volume of hydraulic fluid through the piston is also 3075 cubit³

 

E11
Without the granite plug, the cross-sectional area is 37 cubit²

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We can also calculate the content of the monolith, because of the ascending shaft in the monolith (piston), the area of the cross-section is reduced to 37 cubit².

The volume of the piston = surface cross-section x length = 37 cubit² x 75 cubit = 2775 cubit³. We may also add the volume of the three granite plugs, let us take the total volume of the piston to 2,800 cubit³.


The weight of the monolith (piston).

The specific weight of granite is between 2,600 and 2,800 kg/m³ and for limestone this is between 2,600 and 2,900 kg/m³. We should not actually distinguish between the two types of stone. With a minimum specific weight of 2,600 kg/m³ we calculate how much 1 cubit³ weigh: 1 cubit³ = 0.1435 m³ (7 cubit³= 1 m³), the weight of 1 cubit³ granite or limestone is therefore: 2,600 kg/m³ x 0.1435 m³ = 373 kg , let's take an average of 375 kg per cubit³.


The piston has a volume of 2,800 cubit³, the weight is then: Content x specific weight = 2,800 cubit³ x 375 kg/cubit³ = 1,050,000 kg = 1,050 tons. It’s possible there was water in the recess of the monolith (ascending shaft) and even a large amount of water behind the monolith itself. This must have increased the total weight, but because we have no certainty about that, we do not include this weight.


No external force was exerted on the piston, the force acting on it was caused solely by the weight of the monolith itself, a weight of 1,050 tons.

 

E12
The weight of the monolith dissolved in 2 vectoral forces.

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Because the monolith rests on an inclined surface that makes an angle of 26°, the applied force is dissolved according to two vectors. From the drawing above it becomes clear the weight of 1,050 tons could only exert a force on the water of 460 tons’ or thus 460,000 kg’.

 

F = 1,050 tons’ x sin26° = 1,050 tons’ x 0.4438 = 460.29 tons’

The cross-section of the monolith has a surface of 41 cubit², a pressure of 11.225 tons per cubit² (or 11,225 kg per cubit²) was applied. Most of the force pressed on the sloping surface over which the monolith slid down and was therefore lost. The loss was very large precisely because the angle of the slope is relatively small. However, that slope could not be taken too steeply because the speed of the monolith would have become much too great.

 

The main data of the hydraulic piston summarized:

 

E13

Displacement = 75 cubit. Cross-section of the piston = 41 cubit²
Displaced hydraulic fluid = 75 cubit x 41 cubit² = 3075 cubit³

Weight of the hydraulic piston = 1050 tons.
Force F1 on a slope of 26° = 1050 ton' x sin 26° = 460.29 ton'

Pressure on the water = 11.227 ton per cubit²
(this is approx. 4 kg/cm² or 4 atm.)
= roughly the same pressure as on our tap water.

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Djed's in pairs.

Inside the pyramid it was precisely the reverse effect, no water pressure was applied to the piston from outside, it was  just the weight of the monolith (piston) itself which put pressure on the water in the cylinder. Because of its own weight, the piston pushed away all the water that stood in the hydraulic cylinder below, this was a volume water of 3075 cubit³.

Where did all that water go? Was it sprayed out through the shaft at the entrance of the pyramid? Indeed, this would have been a totally useless project! So, this isn’t the right answer!

 

E14

Djed pillars are often depicted in pairs, every now and then there are reliefs to see with another symbol besides the two Djeds.


For Egyptian hieroglyphs see Wikipedia:
https://nl.wikipedia.org/wiki/Egyptische_hi%C3%ABrogliefen

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According to the current general view, that symbol stood for 'bolt (latch) or folded clothing' and became later the hieroglyph for the letter S. No one will deny that this is the letter S, but that the original meaning of this hieroglyph would be a bolt (latch) might be incorrect. What significance could two Djeds now have in common with a bolt (drop shutter)?

 

E15

Wouldn't it be much more logical to assume that this is a siphon? Water that is transferred from one djed to the other, from one cylinder to the other. That symbol, a siphon together with two djeds, could be the symbolic representation of a hydraulic press.

This principle is often applied in our current society, such as in a hydraulic jack, a fork lift, the brakes in a car, etc.

 

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F1 = force placed on the small piston with section S1,
movement of the piston over a distance h1.
F2 = force developed by the large piston with a section S2,
movement of the piston over a distance h2.
F2 = F1 x (S2/S1)
A hydraulic press is a simple tool to increase a force.
The multiplication factor is
S/S1
The displacement of the large piston is much less than that of the small one.
h2 = h1 x (S1/S2)

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The great gallery together with the hydraulic cylinder below it have a total length of 89 + 75 = 164 cubit. If this were the great cylinder it would have had the smallest displacement, this means that somewhere else in the pyramid there would be a smaller cylinder which is much longer. This is not possible because it would no longer fit within the pyramid. However huge the monolith may be, this is the piston of the small cylinder of this hydraulic press. There must therefore be a larger cylinder somewhere in the pyramid, with a piston that has a much larger cross-section and whose displacement has been much smaller.

The only possible candidate for the great cylinder is the king’s chamber!

 

E17

 

The king's chamber! This is completely absurd! And yet, after a lot of calculations, this appears to be possible.

But, if the king’s chamber is the large cylinder then all the hydraulic fluid (water), that the monolith has displaced from the small cylinder, was pushed under the king's chamber.

Big problem! Until today, not one single shaft has been found that reaches just to below the king's chamber. Although such a shaft has not yet been discovered, it really must exist. The most logical is a channel (X) that starts from the subterranean chamber and ends just below the king's chamber. If no such channel exists the entire internal construction wouldn’t have the least meaning, then the interior construction of shafts and chambers would have been totally useless.

 

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